Evolution Equations Lecture Notes

Lecture Notes Evolution Equations Roland Schnaubelt These lecture notes are based on my course from winter semester 2018/19, though there are small corrections and improvements, as well as minor changes in the numbering. Typically, the proofs and calculations in the notes are a bit shorter than those given in class. The drawings and many additional oral remarks from the lectures are omitted here. On the other hand, the notes contain very few proofs (of peripheral statements) and a short chapter not presented during the course. Occasionally I use the notation and definitions of my lecture notes Analysis 1–4 and Functional Analysis without further notice. I want to thank Heiko Hoffmann for his support in the preparation of an earlier version of these notes. Karlsruhe, June 17, 2019 Roland Schnaubelt Contents Chapter 1. Strongly continuous semigroups and their generators 1.1. Basic concepts and properties 1.2. Characterization of generators 1.3. Dissipative operators 1.4. Examples with the Laplacian 1 1 15 22 35 Chapter 2. The evolution equation and regularity 2.1. Wellposedness and the inhomogeneous problem 2.2. Mild solution and extrapolation 2.3. Analytic semigroups and sectorial operators 46 46 51 55 Chapter 3. Perturbation and approximation 3.1. Perturbation of generators 3.2. The Trotter-Kato theorems 3.3. The Lax–Chernoff product formula 72 72 80 86 Chapter 4. Long-term behavior 4.1. Exponential stability and dichotomy 4.2. Spectral mapping theorems 90 90 99 Chapter 5. Stability of positive semigroups Bibliography 106 112 ii CHAPTER 1 Strongly continuous semigroups and their generators Throughout, X and Y are complex Banach spaces of positive dimension, where we mostly write k · k instead of k · kX etc. for their norms. The space of all bounded linear maps T : X → Y is denoted by B(X, Y ) and endowed with the operator norm kT kB(X,Y ) = kT k = supx6=0 kT xk/kxk. We abbreviate B(X) = B(X, X). Further, X ∗ is the dual space of X acting as hx, x∗ i, and I is the identity map on X. For ω ∈ R, we denote R≥0 = [0, ∞), R+ = (0, ∞), Cω = {λ ∈ C | Re λ > ω}, R≤0 = (−∞, 0], C+ = C0 , R− = (−∞, 0), C− = {λ ∈ C | Re λ < 0}. In this course we study linear evolution equations such as u′ (t) = Au(t), t ≥ 0, u(0) = u0 , (EE) on a state space X for given linear operators A and initial values u0 ∈ D(A). (For a moment we assume that A is closed and densely defined.) We are looking for the state u(t) ∈ X describing the system governed by A at time t ≥ 0. A reasonable description of the system requires a unique solution u of (EE) that continuously depends on u0 . In this case (EE) is called wellposed, cf. Definitions 1.10 and 2.1. We will show in Section 2.1 that wellposedness is equivalent to the fact that A generates a C0 -semigroup T (·) which yields the solutions via u(t) = T (t)u0 . In the next section we will define and investigate these concepts, before we characterize generators in Sections 1.2 and 1.3. In the final section the theory is then applied to the Laplacian. In three intermezzi we present basic notions and facts from the lecture notes [ST] on spectral theory. The omitted proofs are not needed later on. 1.1. Basic concepts and properties We introduce the fundamental notions of these lectures. Definition 1.1. A map T (·) : R≥0 → B(X) is called a strongly continuous operator semigroup or just C0 -semigroup if it satisfies (a) T (0) = I and T (t + s) = T (t)T (s) for all t, s ∈ R≥0 , (b) for each x ∈ X the orbit T (·)x : R≥0 → X; t 7→ T (t)x, is continuous. Here, (a) is the semigroup property and (b) the strong continuity of T (·). The generator A of T (·) is given by o n 1 (T (t)x − x) exists , D(A) = x ∈ X | the limit lim t t→0, t∈R≥0 \{0} Ax = lim t→0, t∈R≥0 \{0} 1 t (T (t)x − x) 1 for x ∈ D(A). 1.1. Basic concepts and properties 2 If one replaces throughout R≥0 by R, one obtains the concept of a C0 -group with generator A. Observe that the domain of the generator is defined in a ‘maximal’ way, in the sense that it contains all elements for which the orbit is differentiable at t = 0. In view of the introductory remarks, usually the generator is the given object and T (·) describes the unknown solution. We will first study basic properties of C0 -semigroups, starting with simple observations. Remark 1.2. a) Let A generate a C0 -semigroup or a C0 -group. Then its domain D(A) is a linear subspace and A is a linear map. b) Let (T (t))t∈R be a C0 -group with generator A. Then its restriction (T (t))t≥0 is a C0 -semigroup whose generator extends A. (Actually these two operators coincide by Theorem 1.30.) c) Let T (·) : R≥0 → B(X) be a semigroup. We then have T (t)T (s) = T (t + s) = T (s + t) = T (s)T (t), Pn
Y n T (t) = T (t)n T (nt) = T j=1 t = j=1 for all t, s ≥ 0 and n ∈ N. If T (·) is even a group, these properties hold for all s, t ∈ R and hence T (t)T (−t) = T (0) = I = T (−t)T (t). There thus exists the inverse T (t)−1 = T (−t) for every t ∈ R. ♦ We next construct a C0 -group with a bounded generator, which is actually differentiable in operator norm. Conversely, an exercise shows that a C0 -semigroup with T (t) → I in B(X) as t → 0+ must have a bounded generator. Example 1.3. Let A ∈ B(X). For t ∈ C with |t| ≤ b for some b > 0, the numbers tn n (b kAk)n A ≤ n! n! are summable in n ∈ N0 . As in Lemma 4.23 of [FA], the series ∞ n X t An , t ∈ C, T (t) = etA := n! n=0 thus converges in B(X) uniformly for |t| ≤ b. In the same way one sees that AetA N N N −1 n=0 n=1 k=0 X tk d X tn n X tn−1 A = An = A Ak dt n! (n − 1)! k! tends to in B(X) as N → ∞ locally uniformly in t ∈ C. As in Analysis 1 one then sees that the map C → B(X); t 7→ etA , is continuously differentiable with derivative AetA . Moreover, (etA )t∈C is a group (where one replaces R≥0 by C in Definition 1.1(a)). ♦ The case of a matrix A on X = Cm was treated in Section 4.4 of [A4]. For a semigroup a mild extra assumption implies its exponential boundedness. This assumption is valid if kT (t)k is uniformly bounded on an interval [0, b] with b > 0 or if T (·) is strongly continuous. (We need both cases below.) 1.1. Basic concepts and properties 3 Lemma 1.4. Let T (·) : R≥0 → B(X) satisfy condition (a) in Definition 1.1 as well as lim supt→0 kT (t)xk < ∞ for all x ∈ X. Then there are constants M ≥ 1 and ω ∈ R such that kT (t)k ≤ M eωt for all t ≥ 0. Proof. 1) We first claim that there are constants c, t0 > 0 with kT (t)k ≤ c for all t ∈ [0, t0 ]. To show this claim, we suppose that there is a null sequence (tn ) in R≥0 such that limn→∞ kT (tn )k = ∞. The principle of uniform boundedness (Corollary 4.5 in [FA]) then yields a vector x ∈ X with kT (tn )xk → ∞ as n → ∞. This fact contradicts the assumption, and so the claim is true. 2) Let t ≥ 0. Then there are numbers n ∈ N0 and τ ∈ [0, t0 ) such that t = nt0 + τ . Setting ω = lnt0c ∈ R and M = max{1, c}, we obtain kT (t)k = kT (τ )T (t0 )n k ≤ cn+1 = e(nt0 +t0 )ω = etω e t0 −τ t0 ln c ≤ M etω , where we used Remark 1.2.  The above considerations lead to the following concept, which is discussed below and will be explored more thoroughly in Section 4.1. Definition 1.5. Let T (·) be a C0 -semigroup with generator A. The quantity ω0 (T ) = ω0 (A) := inf{ω ∈ R | ∃ Mω ≥ 1 ∀ t ≥ 0 : kT (t)k ≤ Mω eωt } ∈ [−∞, ∞) is called its (exponential) growth bound. If supt≥0 kT (t)k < ∞, then T (·) is bounded. (Similarly one defines ω0 (f ) ∈ [−∞, +∞] for any map f : R≥0 → Y .) Remark 1.6. Let T (·) be a C0 -semigroup. a) Lemma 1.4 implies that ω0 (T ) < ∞. b) There are C0 -semigroups with ω0 (T ) = −∞, see Example 1.9. c) In general the infimum in Definition 1.5 is not a minimum.
For instance, 0 1 . We then have let X = C2 be endowed with the 1-norm | · | and A = 1 00
T (t) = etA = 10 1t and kT (t)k = 1 + t for t ≥ 0. As a result, Mε := supt≥0 e−εt kT (t)k = supt≥0 (1 + t)e−εt = ε−1 eε−1 tends to infinity as ε → 0+ . d) For X = Cm and A ∈ Cm×m , the semigroup etA is bounded if and only if s(A) ≤ 0 and all eigenvalues of A on iR are semi-simple, see Satz 4.19 in [A4]. This indicates that boundedness of C0 -semigroups is a more subtle property. ♦ The next auxiliary result will often be used to check strong continuity. We set ω+ = max{ω, 0} for ω ∈ R. Lemma 1.7. Let T (·) : R≥0 → B(X) be a map satisfying condition (a) in Definition 1.1. Then the following assertions are equivalent. a) T (·) is strongly continuous (and thus a C0 -semigroup). b) T (t)x → x in X as t → 0+ for all x ∈ X. c) There is a time t0 > 0 and a dense subspace D of X such that sup0≤t≤t0 kT (t)k < ∞ and T (t)x → x in X as t → 0+ for all x ∈ D. For groups one has analogous equivalences. 1.1. Basic concepts and properties 4 Proof. Assertion c) follows from a) because of Lemma 1.4, and b) from c) by Lemma 4.10 in [FA]. Let statement b) be true. Take x ∈ X and t > 0. For h > 0, the semigroup property and b) imply the limit kT (t + h)x − T (t)xk = kT (t)(T (h)x − x)k ≤ kT (t)k kT (h)x − xk −→ 0 as h → 0+ . Let h ∈ (−t, 0). Lemma 1.4 yields the bound kT (t + h)k ≤ M eω(t+h) ≤ M eω+ t for some constants M ≥ 1 and ω ∈ R. We then derive kT (t + h)x − T (t)xk ≤ kT (t + h)k kx − T (−h)xk ≤ M eω+ t kx − T (−h)xk → 0 as h → 0− , so that a) is true. The final assertion is shown similarly.  Remark 1.8. In the above lemma the implication ‘c) ⇒ a)’ can fail if one omits the boundedness assumption, cf. Exercise I.5.9(4) in [EN]. ♦ We now examine translation semigroups, which are easy to grasp and still illustrate many of the basic features of C0 -semigroups. Another important class of simple examples are multiplication semigoups as discussed in the exercises. We recall that supp f is the support of a function f : M → Y on a metric space M ; i.e., the closure in M of the set {s ∈ M | f (s) 6= 0}, Example 1.9. a) Let X = C0 (R) := {f ∈ C(R) | f (s) → 0 as |s| → ∞}, f ∈ X, and t, r, s ∈ R. We define the translations (T (t)f )(s) = f (s + t), which shift the graph of f to the left if t > 0. Clearly, T (0) = I and T (t) is a linear isometry on X so that kT (t)k = 1. We further obtain T (t)T (r) = T (t+r) noting


T (t)T (r)f (s) = T (r)f (s + t) = f (r + s + t) = T (t + r)f (s). We claim that Cc (R) = {f ∈ C(R) | supp(f ) is compact} is dense in C0 (R). Indeed, take f ∈ C0 (R) and cut-off functions ϕn ∈ Cc (R) satisfying ϕn = 1 on [−n, n] and 0 ≤ ϕn ≤ 1. Then the maps ϕn f belong to Cc (R) and kf − ϕn f k∞ ≤ sup |(1 − ϕn (s))f (s)| ≤ sup |f (s)| |s|≥n |s|≥n tends to 0 as n → ∞. Take f ∈ Cc (R) and a number a > 0 such that supp f ⊆ [−a, a]. If |s| > a + 1 and |t| ≤ 1, we have |s + t| > a and thus f (s + t) = 0; i.e., supp T (t)f is contained in [−a − 1, a + 1] for all t ∈ [−1, 1]. It follows kT (t)f − f k∞ ≤ sup |f (s + t) − f (s)| −→ 0 |s|≤a+1 as t → 0, since f is uniformly continuous on [−a − 1, a + 1]. Lemma 1.7 implies that T (·) is a C0 -group. Similarly, one shows that T (·) is an (isometric) C0 -group on X = Lp (R) with 1 ≤ p < ∞, see Example 4.12 in [FA]. 1.1. Basic concepts and properties 5 In contrast to these results, T (·) is not strongly continuous on X = L∞ (R). Indeed, consider f = ✶[0,1] and observe that ( ) 1, s + t ∈ [0, 1] T (t)f (s) = ✶[0,1] (s + t) = = ✶[−t,1−t] (s) 0, else for s, t ∈ R. Thus, kT (t)f − f k∞ = 1 for every t 6= 0. In addition, T (·) is not continuous as a B(X)-valued function for X = C0 (R) (and neither for X = Lp (R) by Example 4.12 in [FA]). In fact, pick
functions fn ∈ Cc (R) with 0 ≤ fn ≤ 1, fn (n) = 1, and supp fn ⊆ n −
n1 , n + n1 for n ∈ N. Then the suppurt of T ( n2 )fn is contained in n − n3 , n − n1 , implying kT ( n2 ) − Ik ≥ kT ( n2 )fn − fn k∞ = 1 for all n ∈ N. b) For an interval that is bounded from above, one has to prescribe the behavior of the left translation at the right boundary point. Here we simply prescribe the value 0. Let X = C0 ([0, 1)) := {f ∈ C([0, 1)) | lims→1 f (s) = 0} be endowed with the supremum norm, which is a Banach space by Example 1.14 in [FA]. Let t, r ≥ 0, f ∈ X, and s ∈ [0, 1). We define ( f (s + t), s + t < 1, (T (t)f )(s) := 0, s + t ≥ 1. Since f (s + t) → 0 as s → 1 − t if t < 1, the function T (t)f belongs to X. Clearly, T (t) is linear on X and kT (t)k ≤ 1. We stress that T (t) = 0 whenever t ≥ 1. (One says that T (·) is nilpotent.) As a consequence, ω0 (T ) = −∞ and T (·) cannot be a group by Remark 1.2. We next compute T (t)T (r)f (s) = = ( ( (T (r)f )(s + t), 0, f (s + t + r), 0, s < 1 − t, s ≥ 1 − t, s < 1 − t, else, s + t < 1 − r, = (T (t + r)f )(s). Hence, T (·) is a semigroup. As in part a) or in Example 1.19 of [FA], one sees that Cc ([0, 1)) := {f ∈ C([0, 1)) | ∃ bf ∈ (0, 1) : supp f ⊆ [0, bf ]} is a dense subspace of X. For f ∈ Cc ([0, 1)) and t ∈ (0, 1 − bf ) we compute ( f (s + t) − f (s), if s ∈ [0, 1 − t), T (t)f (s) − f (s) = 0, if s ∈ [1 − t, 1) ⊆ [bf , 1), and deduce limt→0 kT (t)f − f k∞ = 0 using the uniform continuity of f . Ac♦ cording to Lemma 1.7, T (·) is a C0 -semigroup on X. We now introduce one solution concept for the problem (EE). Other ones will be discussed in Section 2.2. 1.1. Basic concepts and properties 6 Definition 1.10. Let A be a linear operator on X with domain D(A) and let x ∈ D(A). A function u : R≥0 → X solves the homogeneous evolution equation (or Cauchy problem) u′ (t) = Au(t), t ≥ 0, u(0) = x, (1.1) if u belongs to C 1 (R≥0 , X) and satisfies u(t) ∈ D(A) and (1.1) for all t ≥ 0. The next result provides the fundamental regularity properties of C0 semigroups. Recall the domain D(A) was ‘maximally’ defined as the set of all initial values for which the orbit is differentiable at t = 0. We now use the semigroup law to transfer this property to later times. The crucial invariance of the domain under the semigroup then directly follows from its definition. Proposition 1.11. Let A generate the C0 -semigroup T (·) and x ∈ D(A). Then T (t)x belongs to D(A), T (·)x to C 1 (R≥0 , X), and we have d dt T (t)x = AT (t)(x) = T (t)Ax for all t ≥ 0. Moreover, the function u = T (·)x is the only solution of (1.1). Proof. 1) Let t > 0, h > 0 and x ∈ D(A). Remark 1.2 and the continuity of T (t) then imply the convergence

1 1 h T (h) − I T (t)x = T (t) h T (h)x − x −→ T (t)Ax as h → 0. By Definition 1.1 of the generator, the vector T (t)x belongs to D(A) and satisfies AT (t)x = T (t)Ax. Next, let 0 < h < t. We then compute

1 1 −h T (t − h)x − T (t)x = T (t − h) h T (h)x − x −→ T (t)Ax as h → 0, by means of Lemma 1.13 below (with S(τ, σ) = T (τ − σ)). Together we have shown that the orbit u = T (·)x has the derivative AT (·)x. Since T (·)Ax is continuous, u is contained in C 1 (R≥0 , X). Summing up, u solves (1.1). 2) Let v be another solution of (1.1). Take t > 0 and set w(s) = T (t − s)v(s) for s ∈ [0, t]. Let h ∈ [−s, t − s] \ {0}. We write 1 h (w(s+h)−w(s)) 1 = T (t−s−h) h1 (v(s+h)−v(s))− −h (T (t−s−h)−T (t−s))v(s). Using v ∈ C 1 , Lemma 1.13, v(s) ∈ D(A) and the first step, we infer that w is differentiable with derivative w′ (s) = T (t − s)v ′ (s) − T (t − s)Av(s) = 0, where the last equality follows from (1.1) for v. Hence, for each x∗ ∈ X ∗ the scalar function hw(·), x∗ i is differentiable with vanishing derivative and thus constant, which leads to the equality hT (t)x, x∗ i = hw(0), x∗ i = hw(t), x∗ i = hv(t), x∗ i for all t > 0. The Hahn-Banach theorem (Corollary 5.10 of [FA]) now yields T (·)x = v as asserted.  Remark 1.12. Let f ∈ C0 (R) \ C 1 (R). Then the orbit T (·)f = f (· + t) of the translation semigroup on C0 (R) is not differentiable (cf. Example 1.9). ♦ 1.1. Basic concepts and properties 7 The following simple lemma is used in the above proof and also later on.1 Lemma 1.13. Let D = {(τ, σ) | a ≤ σ ≤ τ ≤ b} for some a < b in R, S : D → B(X) be strongly continuous, and f be contained in C([a, b], X). Then the function g : D → X; g(τ, σ) = S(τ, σ)f (σ), is also continuous. Proof. Observe that sup(τ,σ)∈D kS(τ, σ)xk < ∞ for every x ∈ X by continuity. The uniform boundedness principle thus says that c := supD kS(τ, σ)k is finite. For (t, s), (τ, σ) ∈ D we then obtain kS(t, s)f (s) − S(τ, σ)f (σ)k ≤ k(S(t, s) − S(τ, σ))f (s)k + c kf (s) − f (σ)k. The right-hand side of this inequality tends to 0 as (τ, σ) → (t, s).  Remark 1.14. Let xn → x in X and Tn → T strongly. As in the proof of ♦ Lemma 1.13 one then shows that Tn xn → T x as n → ∞. Intermezzo 1: Closed operators, spectrum, and X-valued Riemann integrals. As noted above, generators of C0 -semigroups are unbounded unless the semigroup is continuous in B(X). However, we will see in Proposition 1.20 that they still respect limits to some extent. We introduce the relevant concepts here. Let D(A) ⊆ X be a linear subspace and A : D(A) → X be linear. We often endow D(A) with the graph norm kxkA := kxk + kAxk. We write [D(A)], X1A , or X1 for (D(A), k · kA ) and also kxk1 instead of kxkA . Observe that [D(A)] is a normed vector space and that A is an element of B([D(A)], X). Moreover, a function f ∈ C([a, b], X) belongs to C([a, b], [D(A)]) if and only if f takes values in D(A) and Af : [a, b] → X is continuous. The operator A is called closed if for every sequence (xn ) in D(A) possessing the limits lim xn = x and lim Axn = y in X n→∞ n→∞ we obtain x ∈ D(A) and Ax = y. For this definition and the following results we refer to Chapter 1 in [ST]. We first present prototypical examples. Example 1.15. a) Every operator A ∈ B(X) with D(A) = X is closed, since here Axn → Ax if xn → x in X as n → ∞. b) Let X = C([0, 1]) and Af = f ′ with D(A) = C 1 ([0, 1]). Let (fn ) be a sequence in D(A) such that (fn ) and (fn′ ) converge in X to f and g, respectively. By Analysis 1, the limit f then belongs to C 1 ([0, 1]) and satisfies f ′ = g; i.e., A is closed. Next, consider the map A0 f = f ′ with D(A0 ) = {f ∈ C 1 ([0, 1]) | f ′ (0) = 0}. Let (fn ) be a sequence in D(A) such that fn → f and fn′ → g in X as n → ∞. We again obtain f ∈ C 1 ([0, 1]) and f ′ = g. It further follows f ′ (0) = g(0) = limn→∞ fn′ (0) = 0, so that also A0 is closed. ♦ 1In the lectures a somewhat weaker version was presented which fitted to Proposition 1.11. The present variant is needed below. 1.1. Basic concepts and properties 8 Before we discuss basic properties of closed operators, we define the Riemann integral for X-valued functions. Let a < b be real numbers. A (tagged) partition Z of the interval [a, b] is a finite subset of [a, b] containing both a and b written in the form a = t0 < t1 < . . . < tm = b together with a finite sequence (τk )m k=1 satisfying tk−1 ≤ τk ≤ tk for all k ∈ {1, . . . , m}. We write δ(Z) = maxk∈{1,...,m} (tk − tk−1 ). For a function f ∈ C([a, b], X) and a partition Z we define the Riemann sum by S(f, Z) = m X k=1 f (τk )(tk − tk−1 ) ∈ X. As for real valued functions it can be shown that for any sequence (Zn ) of (tagged) partitions with limn→∞ δ(Zn ) = 0 the sequence (S(f, Zn ))n converges in X and that the limit J does not depend on the choice of such (Zn ). In this sense, we say that S(f, Z) converges in X to J as δ(Z) → 0. The Riemann integral is now defined as Z b f (t) dt = lim S(f, Z). a Ra δ(Z)→0 Rb We also set b f (t) dt = − a f (t) dt. As in the real-valued case, one shows the basic properties the integral (except for monotony), e.g., linearity, additivity and validity of the standard estimate. Moreover, the same definition and results work for piecewise continuous functions. The fundamental theorem of calculus and a result on dominated convergence are shown in the next remark. Remark 1.16. For a linear operator A in X the following assertions hold. a) The operator A is closed if and only if its graph Gr(A) = {(x, Ax) | x ∈ D(A)} is closed in X × X (endowed with the product metric) if and only if D(A) is a Banach space with respect to the graph norm k · kA . b) If A is closed with D(A) = X, then A is even continuous (closed graph theorem). c) Let A be injective. Set D(A−1 ) := R(A) = {Ax | x ∈ D(A)}. Then A is closed if and only if A−1 is closed. d) Let A be closed and f ∈ C([a, b], [D(A)]). We then have Z b Z b Z b Af (t) dt. f (t) dt = f (t) dt ∈ D(A) and A a a a An analogous result holds for piecewise continuous f and Af . e) Let fn , f ∈ C([a, b], X) for n ∈ N such that fn (s) → f (s) in X as n → ∞ for each s ∈ [a, b] and kfn (·)k ≤ ϕ for a map ϕ ∈ L1 (a, b) and all n ∈ N. Then there exists the limit Z b Z b lim fn (s) ds = f (s) ds. n→∞ a a The assumptions are satisfied if fn → f uniformly as n → ∞, of course. 1.1. Basic concepts and properties f) For f ∈ C([a, b], X), the function [a, b] → X; t 7→ Z 9 t f (s) ds, a is continuously differentiable with derivative Z d t f (s) ds = f (t) dt a (1.2) for each t ∈ [a, b]. For g ∈ C 1 ([a, b], X), we have Z b g ′ (s) ds = g(b) − g(a). (1.3) a g) Let J ⊆ R be an interval and a ∈ J. Take a sequence (fn ) in C 1 (J, X) and maps f, g ∈ C(J, X) such that fn → f and fn′ → g uniformly on J as n → ∞. We then obtain f ∈ C 1 (J, X) and f ′ = g. Proof. Parts lished in Theorem To prove d), let [a, b] the Riemann a) and c) are shown in Lemma 1.4 of [ST], and b) is estab1.5 of [ST]. f be as in the statement. Note that for each partition Z of sum S(f, Z) belongs to D(A). We further obtain Z b m X AS(f, Z) = Af (t) dt (Af )(τk )(tk − tk−1 ) = S(Af, Z) −→ a k=1 as ∆(Z) → 0, because Af is continuous. Assertion d) now follows from the closedness of A. Dominated convergence with majorant kf k∞ ✶ + ϕ yields claim e) since Z b Z b Z b kf (s) − fn (s)k ds. fn (s) ds ≤ f (s) ds − a a a For f), take t ∈ [a, b] and h 6= 0 such that t + h ∈ [a, b]. We can then estimate Z Z Z t  1 t+h 1  t+h (f (s) − f (t)) ds (1.4) f (s) ds − f (s) ds − f (t) = h a h t a ≤ sup kf (s) − f (t)k −→ 0 |s−t|≤h as h → 0. So we have shown (1.2). In the proof of Proposition 1.11 we have seen that a function in C 1 ([a, b]) is constant if its derivative vanishes. Equation (1.3) can thus be deduced from (1.2) as in Analysis 1. Let fn , f , and g be as in part g). Formula (1.3) yields the identity Z t fn′ (s) ds fn (t) = fn (a) + a for all t ∈ J. Letting n → 0, from e) we deduce Z t g(s) ds f (t) = f (a) + a for all t ∈ J. Due to (1.2), the map f belongs C 1 (J, X) and satisfies f ′ = g.  1.1. Basic concepts and properties 10 For a closed operator A we define the resolvent set ρ(A) = {λ ∈ C | λI − A : D(A) → X is bijective}. If λ ∈ ρ(A), we write R(λ, A) for (λI − A)−1 and call it resolvent. The spectrum of A is the set σ(A) = C \ ρ(A). The point spectrum σp (A) = {λ ∈ C | ∃ v ∈ D(A) \ {0} with Av = λv} is a subset of σ(A) which can be empty if dim X = ∞, see Example 1.25 in [ST]. We discuss basic properties of spectrum and resolvent which will be used throughout these lectures. Remark 1.17. a) Let A be closed and λ ∈ ρ(A). It is easy to check that also the operator λI − A is closed, and hence R(λ, A) is closed by Remark 1.16 c). Assertion d) of this remark then shows the boundedness of R(λ, A). b) Let A be a linear operator and λ ∈ C such that λI − A : D(A) → X is bijective with bounded inverse. Then (λI −A)−1 is closed, so that Remark 1.16 c) implies the closedness of A. In particular, λ belongs to ρ(A). c) We list several important statements of Theorem 1.13 of [ST]. The set ρ(A) is open and so σ(A) is closed. More precisely, for λ ∈ ρ(A) all µ with |µ − λ| < 1/kR(λ, A)k are also contained in ρ(A) and we have the power series R(µ, A) = ∞ X n=0 (λ − µ)n R(λ, A)n+1 . (1.5) This series converges absolutely in B(X, [D(A)]) and uniformly for µ with |µ − λ| ≤ δ/kR(λ, A)k and δ ∈ (0, 1), where one also obtains the inequality kR(µ, A)k ≤ kR(λ, A)k/(1 − δ). The resolvent has the derivatives
n n+1 d n (1.6) dλ R(λ, A) = (−1) n!R(λ, A) for all λ ∈ ρ(A) and n ∈ N0 . It further fulfills the resolvent equation R(µ, A) − R(λ, A) = (λ − µ)R(λ, A)R(µ, A) = (λ − µ)R(µ, A)R(λ, A). (1.7) d) Let T ∈ B(X). By Theorem 1.16 of [ST], the spectrum σ(T ) is even compact and always non-empty, and the spectral radius of T is given by 1 1 r(T ) := max{|λ| | λ ∈ σ(A)} = inf kT n k n = lim kT n k n . n∈N n→∞ e) Example 1.22 provides closed operators A with σ(A) = ∅ or σ(A) = C. ♦ This ends the intermezzo, and we come back to the investigation of C0 semigroups. We first note a simple rescaling lemma which is often used to simplify the reasoning. Lemma 1.18. Let T (·) be a C0 -semigroup with generator A, λ ∈ C, and a > 0. Set S(t) = eλt T (at) for t ≥ 0. Then S(·) is a C0 -semigroup and has the generator B = λI + aA with D(B) = D(A). 1.1. Basic concepts and properties 11 Proof. For t, s ≥ 0 we compute S(t + s) = eλt eλs T (at)T (as) = S(t)S(s). The strong continuity of S(·) and S(0) = I are clear. Let B be the generator of S(·). Because of 1 1 1 (S(t)x − x) = aeλt (T (at)x − x) + (eλt − 1)x, t at t x belongs to D(B) if and only if x ∈ D(A), and we then have Bx = aAx+λx.  Below we will derive key features of generators, which are consequences of the next fundamental lemma. Lemma 1.19. Let T (·) be a C0 -semigroup with generator A, λ ∈ C, t > 0, Rt and x ∈ X. Then the integral 0 e−λs T (s)x ds belongs to D(A) and satisfies Z t e−λs T (s)x ds. (1.8) e−λt T (t)x − x = (A − λI) 0 Furthermore, for x ∈ D(A) we have Z t −λt e−λs T (s)(A − λI)x ds. e T (t)x − x = (1.9) 0 Proof. We only consider λ = 0 since the general case then follows by means of Lemma 1.18. For h > 0 we compute Z Z t Z t  1 t 1 T (s + h)x ds − T (s)x ds (T (h) − I) T (s)x ds = h h 0 0 0 Z t Z t+h   1 = T (s)x ds T (r)x dr − h h 0 Z t+h Z h 1 1 T (s)x ds − T (s)x ds, (1.10) = h t h 0 where we substituted r = s + h. The last line tends to T (t)x − x as h → 0 due to the continuity of the orbits and (1.4). By the definition of the generator, Rt this means that 0 T (s)x ds is an element of D(A) and (1.8) holds. If x ∈ D(A), Proposition 1.11 shows that T (·)x belongs to C 1 ([a, b], X) with derivative d  dt T (·)x = T (·)Ax. Hence, formula (1.9) follows from (1.3). We can now show basic properties of generators. Recall that they commute with their semigroup by Proposition 1.11. Proposition 1.20. Let A generate a C0 -semigroup T (·). Then A is closed and densely defined. Moreover, T (·) is the only C0 -semigroup generated by A. If λ ∈ ρ(A), then we have R(λ, A)T (t) = T (t)R(λ, A) for all t ≥ 0. Proof. 1) To show closedness, we take a sequence (xn ) in D(A) with limit x in X such that (Axn ) converges to some y in X. Equation (1.9) yields Z 1 1 t T (s)Axn ds (T (t)xn − xn ) = t t 0 for all n ∈ N and t > 0. Letting n → ∞, we infer Z 1 1 t T (s)y ds (T (t)x − x) = t t 0 1.1. Basic concepts and properties 12 by means of Remark 1.16 e). Because of (1.4), the right-hand side tends to y as t → 0. This exactly means that x belongs D(A) and Ax = y; i.e., A is closed. 2) Let x ∈ X. For n ∈ N, we define the vector Z 1 n T (s)x ds xn = n 0 which belongs to D(A) by Lemma 1.19. Formula (1.4) shows that (xn ) tends to x, and hence the domain D(A) is dense in X. 3) Let A generate another C0 -semigroup S(·). The function S(·)x then solves (1.1) for each x ∈ D(A) by Proposition 1.11. The uniqueness statement in this result thus implies that T (t)x = S(t)x for all t ≥ 0 and x ∈ D(A). Since these operators are bounded, step 2) leads to T (·) = S(·) as desired. 4) Let λ ∈ ρ(A), t ≥ 0 and x ∈ X. Set y = R(λ, A)x ∈ D(A). Proposition 1.11 implies the identity T (t)(λy − Ay) = (λI − A)T (t)y. Applying R(λ, A), we conclude that R(λ, A)T (t)x = T (t)R(λ, A)x.  We next derive important information about spectrum and resolvent of generators. Actually we show a bit more than needed later on. Proposition 1.21. Let A generate the C0 -semigroup T (·) and λ ∈ C. Then the following assertions hold. a) If the improper integral Z ∞ Z t −λs e T (s)x ds := lim e−λs T (s)x ds R(λ)x := 0 t→∞ 0 exists in X for all x ∈ X, then λ ∈ ρ(A) and R(λ) = R(λ, A). b) The integral in a) exists even absolutely for all x ∈ X if Re λ > ω0 (T ). Hence, the spectral bound (of A) s(A) := sup{Re λ | λ ∈ σ(A)} (1.11) is less than or equal than ω0 (T ). c) Let M ≥ 1 and ω ∈ R with kT (t)k ≤ M eωt for all t ≥ 0. Take n ∈ N and λ ∈ Cω (i.e., Re λ > ω). We then have M . kR(λ, A)n k ≤ (Re λ − ω)n The integral in part a) is called the Laplace transform of T (·)x. It can be used for alternative approaches to the theory of C0 -semigroups (and their generalizations), cf. [ABHN]. In Section 4.1 we will study whether the equality s(A) = ω0 (T ) can be shown in b). This property would allow to control the growth (or decay) of the semigroup in terms of the given object A. We recall from Definition 1.5 and Lemma 1.4 that the exponent ω in part c) has to satisfy ω ≥ ω0 (T ) and that any number ω ∈ (ω0 (T ), ∞) fulfills the conditions in c). Proof of Proposition 1.21. a) Let h > 0 and x ∈ X. By Lemma 1.18, we have the C0 -semigroup Tλ (·) = (e−λs T (s))s≥0 with generator A − λI on the domain D(A). Equation (1.10) yields Z t 1 1 Tλ (s)x ds (Tλ (h) − I)R(λ)x = lim (Tλ (h) − I) t→∞ h h 0 1.1. Basic concepts and properties 13 Z Z 1 t+h 1 h Tλ (s)x ds − Tλ (s)x ds t→∞ h t h 0 Z 1 h Tλ (s)x ds, =− h 0 R∞ due to the convergence of 0 Tλ (s)x ds. The right-hand side tends to −x as h → 0 by (1.4), so that R(λ)x belongs to D(A − λI) = D(A) and satisfies (λI − A)R(λ)x = x. Let x ∈ D(A). Proposition 1.11 says that T (s)Ax = AT (s)x for s ≥ 0, and A is closed due to Proposition 1.20. Using also Remark 1.16 d), we deduce Z t Z t e−λs T (s)(λI − A)x ds = lim (λI − A) e−λs T (s)x ds R(λ)(λI − A)x = lim t→∞ t→∞ 0 0 Z t = (λI − A) lim e−λs T (s)x ds = (λI − A)R(λ)x. = lim t→∞ 0 Hence, part a) is shown. b) Let x ∈ X. Fix a number ω ∈ (ω0 (T ), Re λ). It follows ke−λs T (s)xk ≤ M e(ω−Re λ)s for some M ≥ 1 and all s ≥ 0. For 0 < a < b we can thus estimate Z a Z b Z b Z b Tλ (s)x ds ≤ Tλ (s)x ds− e(ω−Re λ)s ds → 0 kTλ (s)xk ds ≤ M kxk 0 0 Rt a a as a, b → ∞. Consequently, 0 Tλ (s)x ds converges (absolutely) in X as t → ∞ for all x ∈ X, and thus assertion b) follows from a). c) Let n ∈ N and x ∈ X. Arguing as in Analysis 2, one can differentiate Z t  d n−1 Z t e−λs T (s)x ds = (−1)n−1 sn−1 e−λs T (s)x ds. dλ 0 0 As in part b), the integrals converge as t → ∞ uniformly for Re λ ≥ ω + ε > ω > ω0 (T ) and any ε > 0. Hence, (1.6) and a variant of Remark 1.16 f) imply Z t (−1)n−1  d n−1 n lim Tλ (s)x ds R(λ, A) x = t→∞ 0 (n − 1)! dλ Z t Z ∞ 1 1 sn−1 Tλ (s)x ds = sn−1 e−λs T (s)x ds. = lim t→∞ (n − 1)! 0 (n − 1)! 0 Computing an elementary integral, one can now estimate Z ∞ M kxk M n kR(λ, A) xk ≤ sn−1 e(ω−Re λ)s ds = kxk (n − 1)! 0 (Re λ − ω)n for all Re λ > ω > ω0 (T ) since ε is arbitrary.  We calculate the generators of the translation semigroups from Example 1.9 and discuss their spectra. They turn our to be the first derivative endowed with appropriate domains. We also use the above necessary conditions to show that on certain domains the first derivative fails to be a generator. Example 1.22. a) Let T (t)f = f (· + t) be the translation group on X = C0 (R). We compute its generator A. Below we use that a function g ∈ X 1.1. Basic concepts and properties 14 is uniformly continuous since Cc (R) is dense in X and uniform continuity is preserved by uniform limits. For f ∈ D(A) and s ∈ R, there exist the pointwise limits 1 1 Af (s) = lim (T (t)f (s) − f (s)) = lim (f (s + t) − f (s)) = f ′ (s) t→0 t t→0 t so that f is differentiable with f ′ = Af ∈ C0 (R). We have shown the inclusion D(A) ⊆ C01 (R) := {f ∈ C 1 (R) | f, f ′ ∈ X}. Conversely, let f ∈ C01 (R). For s ∈ R, we compute 1 1 (T (t)f (s) − f (s)) − f ′ (s) = (f (s + t) − f (s)) − f ′ (s) t t Z 1 t ′ (f (s + τ ) − f ′ (s)) dτ = t 0 ≤ sup |f ′ (s + τ ) − f ′ (s)|. 0≤|τ |≤|t| The right-hand side tends to 0 as t → 0 uniformly in s ∈ R since f ′ ∈ C0 (R) is d uniformly continuous. As a result, f ∈ D(A) and so A = ds with D(A) = C01 (R). Proposition 1.21 yields the inquality s(A) ≤ ω0 (A) = 0 because kT (t)k = 1. In Theorem 1.30 we will see that −A generates the contractive C0 -semigroup (S(t))t≥0 = (T (−t))t≥0 . Hence, s(−A) ≤ 0. Observing that −λI − A = −(λI − (−A)), we conclude σ(−A) = −σ(A) as well as −R(λ, −A) = R(−λ, A). So we have proven the inclusion σ(A) ⊆ iR. To show the converse, let λ ∈ C+ , f ∈ X, and s ∈ R. Since all of the following limits exist with respect to the supremum norm in s, Proposition 1.21 yields Z b Z b   −λt e−λt (T (t)f )(s) dt e T (t)f dt (s) = lim (R(λ, A)f )(s) = lim b→∞ 0 Z b b→∞ 0 Z b+s e−λt f (t + s) dt = lim = lim b→∞ 0 b→∞ Z ∞ = eλ(s−τ ) f (τ ) dτ. eλ(s−τ ) f (τ ) dτ s s We pick functions ϕn ∈ Cc (R) with 0 ≤ ϕn ≤ 1 and ϕn = 1 on [0, n] for n ∈ N, and set α = Re λ > 0, β = Im λ, as well as fn (τ ) = eiβτ ϕn (τ ). Since kfn k∞ = 1, the above formula leads to the lower bound Z ∞ kR(λ, A)k ≥ kR(λ, A)fn k∞ ≥ |R(λ, A)fn (0)| = e−ατ e−iβτ fn (τ ) dτ 0 Z ∞ Z n 1 − e−αn = e−ατ ϕn (τ ) dτ ≥ e−ατ dτ = . α 0 0 Letting n → ∞, we arrive at kR(λ, A)k ≥ Re1 λ . Proposition 1.21 then yields the equality kR(λ, A)k = Re1 λ (take M = 1, ω = 0 and n = 1 there). If iβ belonged to ρ(A) for some β ∈ R, then we would infer 1 = kR(α + iβ, A)k → kR(iβ, A)k α as α → 0, which is impossible. We thus obtain σ(A) = iR. 1.2. Characterization of generators 15 b) We treat the nilpotent left translation semigroup on X = C0 ([0, 1)); i.e., ( f (s + t), s + t < 1, (T (t)f )(s) = 0, s + t ≥ 1, for f ∈ X, t ≥ 0 and s ∈ [0, 1). Let A be its generator. Take f ∈ D(A). + + As in part a), one shows that the right derivative dds f exists and dds f = Af . (Here we can only consider t → 0+ .) However, since f and Af are continuous, Corollary 2.1.2 of [Pa] says that f ∈ C 1 ([0, 1)), and so we have the inclusion D(A) ⊆ C01 ([0, 1)) := {f ∈ C 1 ([0, 1)) | f, f ′ ∈ X} as well as Af = f ′ . Let f ∈ C01 ([0, 1)) and note that its 0-extension f˜ to R≥0 belongs to C01 (R≥0 ). As in part a), it follows ( 1 (f (s + t) − f (s)), 0 ≤ s < 1 − t, 1 (T (t)f (s) − f (s)) = t 1 t 1 − t ≤ s < 1, − t f (s), 1 = (f˜(s + t) − f˜(s)) −→ f˜′ (s) = f ′ (s) t as t → 0 uniformly in s ∈ [0, 1), since f˜′ is uniformly continuous. Hence, D(A) = C01 ([0, 1)) and Af = f ′ . Because of ω0 (A) = −∞, Proposition 1.20 yields σ(A) = ∅ and ρ(A) = C. c) The operator Af = f ′ with D(A) = C 1 ([0, 1]) on X = C([0, 1]) has the spectrum σ(A) = C. In fact, for each λ ∈ C the function t 7→ eλ (t) := eλt belongs to D(A) with Aeλ = λeλ so that even λ ∈ σp (A). Hence, A is not a generator in view of Proposition 1.21. d d) Let X = C0 (R≤0 ) := {f ∈ C((−∞, 0]) | f (s) → 0 as s → −∞} and A = ds 1 1 ′ with D(A) = C0 (R≤0 ) := {f ∈ C (R≤0 ) | f, f ∈ X}. Then A is not a generator. Indeed, for all λ ∈ C+ we have eλ ∈ D(A) and Aeλ = λeλ so that λ ∈ σ(A), violating s(A) < ∞ in Proposition 1.21. d with D(A) = {f ∈ C 1 ([0, 1]) | f (1) = 0} e) On X = C([0, 1]) the map A = ds is not a generator as D(A) = {f ∈ X | f (1) = 0} = 6 X, cf. Proposition 1.20. ♦ We stress that in parts c) and d) one does not impose conditions at the upper bound of the spatial domain, in contrast to a) and b). This lack of boundary conditions leads to spectral properties of A ruling out that it is a generator. We will come back to this point in Example 1.37. 1.2. Characterization of generators Proposition 1.20 and 1.21 contain necessary conditions to be a generator. In this section we want to show their sufficiency. This is the content of Hille– Yosida Theorem 1.27 which is the core of the theory of C0 -semigroups. Our approach is based on the so called Yosida approximations which are defined by Aλ := λAR(λ, A) = λ2 R(λ, A) − λI ∈ B(X). for λ ∈ ρ(A). Here we note the basic identities AR(λ, A) = λR(λ, A) − I and AR(λ, A)x = R(λ, A)Ax (1.12) (1.13) 1.2. Characterization of generators 16 for x ∈ D(A). The next lemma is stated in somewhat greater generality than needed later on. In view of Proposition 1.20 and 1.21, for a generator A it says that Aλ approximates A strongly on D(A) as λ → ∞. Lemma 1.23. Let A be a closed operator satisfying (ω, ∞) ⊆ ρ(A) and kR(λ, A)k ≤ M λ for some M, ω ≥ 0 and all λ ≥ ω + 1. We then have λR(λ, A)x → x as λ → ∞ for all x ∈ D(A) and λAR(λ, A)y → Ay as λ → ∞ for all y ∈ D(A) with Ay ∈ D(A). Proof. Let x ∈ D(A). Equation (1.13) and the assumption yield that M kλR(λ, A)x − xk = kR(λ, A)Axk ≤ kAxk −→ 0 λ as λ → ∞. Since λR(λ, A) is uniformly bounded for λ ≥ ω+1, the first assertion follows. Taking x = Ay and using (1.13), one then deduces the second assertion from the first one.  For linear operators A, B on X we write A ⊆ B if Gr(A) ⊆ Gr(B); i.e., if D(A) ⊆ D(B) and Ax = Bx for all x ∈ D(A). In this case we call B an extension of A. Equality of A and B is then often shown by means of the next observation, requiring that D(A) is not ‘too small’ and D(B) is not ‘too large.’ Lemma 1.24. Let A and B be linear operators with A ⊆ B such that A is surjective and B is injective. We then have A = B. In particular, A and B are equal if they satisfy A ⊆ B and ρ(A) ∩ ρ(B) 6= ∅. Proof. We have to prove the inclusion D(B) ⊆ D(A). Let x ∈ D(B). By the assumptions, there is a vector y ∈ D(A) with Bx = Ay = By. Since B is injective, it follows x = y ∈ D(A). Let λ ∈ ρ(A) ∩ ρ(B). The first part then shows the equality λI − A = λI − B, and hence A = B.  We introduce a class of C0 -semigroups which is easier to handle in many respects, cf. Theorem 1.40. Definition 1.25. Let ω ∈ R. An ω-contraction semigroup is a C0 -semigroup T (·) satisfying kT (t)k ≤ eωt for all t ≥ 0. Such a semigroup is said to be quasicontractive. If ω = 0, we call T (·) a contraction semigroup. This concept depends on the choice of the norm on X as described below. Remark 1.26. a) Let A ∈ B(X). Estimating the power series in Example 1.3, we derive ketA k ≤ etkAk so that A generates a kAk-contractive semigroup. However, its growth bound ω0 (A) is possibly much smaller, as can be seen from Remark 1.6 d). b) There are unbounded generators A of a C0 -semigroup having norms kT (t)k ≥ M for all t > 0 and some M > 1. Hence, they cannot be ω-contractive for any ω ∈ R. As an example, let X = C0 (R) be endowed with the norm  kf k = max sups≥0 |f (s)|, M sups<0 |f (s)| for some M > 1, which is equivalent to the supremum norm. The translations T (t)f = f (·+t) thus yield a C0 -semigroup on (X, k·k). Take any t > 0. Choose 1.2. Characterization of generators 17 a function f ∈ C0 (R) such that kf k∞ = 1 and supp f ⊆ (0, t). We then obtain kf k = 1, supp T (t)f ⊆ (−t, 0), and so kT (t)k ≥ kT (t)f k = M sup |f (s + t)| = M. −t≤s≤0 Since kT (t)k ≤ M , we actually have kT (t)k = M for all t > 0. c) However, for each C0 -semigroup T (·) on a Banach space X one can find an equivalent norm on X for which T (·) becomes ω-contractive. Indeed, take numbers M ≥ 1 and ω ∈ R such that kT (t)k ≤ M eωt for all t ≥ 0. We set 9x9 = sup e−ωs kT (s)xk s≥0 for x ∈ X, which defines an equivalent norm since kxk ≤ 9x9 ≤ M kxk. We further obtain 9e−ωt T (t)x9 = sup e−ω(s+t) kT (s + t)xk ≤ 9x9 s≥0 so that T (·) is ω-contractive for this norm. However, this renorming can destroy additional properties as the Hilbert space structure, and in general one cannot do it for two C0 -semigroups at the same time. See Remark I.2.19 in [Go]. ♦ The following major theorem characterizes the generators of C0 -semigroups. It was shown in the contraction case independently by Hille and Yosida in 1948. Yosida’s proof extends very easily to the general case and is presented below. As we see in Theorem 2.2, the generator property of A is equivalent to ‘wellposedness’ of (1.1). In other words, the Hille–Yosida Theorem describes the class of operators for which (1.1) is solvable in a reasonable sense. It is thus the fundament of the theory of linear evolutions equations, which is actually concerned with many topics beyond wellposedness – below we treat regularity, perturbation, approximation, and long-time behavior, for instance. Theorem 1.27. Let M ≥ 1 and ω ∈ R. A linear operator A generates a C0 -semigroup on X satisfying kT (t)k ≤ M eωt for all t ≥ 0 if and only if A is closed, D(A) = X, (ω, ∞) ⊆ ρ(A), M . ∀ n ∈ N, λ > ω : kR(λ, A)n k ≤ (λ − ω)n (1.14) In this case one even has Cω = {λ ∈ C | Re λ > ω} ⊆ ρ(A) and ∀ n ∈ N, λ ∈ Cω : kR(λ, A)n k ≤ M . (Re λ − ω)n (1.15) The operator A generates an ω-contraction semigroup if and only if 1 . (1.16) A is closed, D(A) = X, R+ ⊆ ρ(A), ∀ λ > 0 : kR(λ, A)k ≤ (λ − ω) In this case (1.15) is true with M = 1. In applications it is of course much easier check the assumptions in the quasicontractive case. Based on the above result, Theorem 1.40 will provide another, even more convenient characterization of generators in this case. 1.2. Characterization of generators 18 Proof of Theorem 1.27. It is clear (1.16) implies (1.14) for M = 1. The other addendum and the necessity of (1.14) follow from Propositions 1.20 and 1.21. If (1.14) is true, then the shifted operator A − ωI satisfies (1.14) with ‘ω = 0.’ Below we show that A − ωI generates a bounded semigroup. Lemma 1.18 then implies the assertion. We establish the sufficiency of (1.14) in two steps. We first use the semigroups etAn generated by the (bounded) Yosida approximations An = n2 R(n, A) − nI for n ∈ N and prove that they converge to a C0 -semigroup T (·) as n → ∞. In a second step we show that it is generated by A. 1) Let (1.14) be true with ω = 0. Take n, m ∈ N and t ≥ 0. Employing Lemma 1.18, the powers series representation of etAn in Example 1.3 and (1.14), we estimate ∞ ∞ X X (nt)j (nt nkR(n, A)k)j 2 ≤ M e−tn ketAn k = ke−tn en R(n,A)t k ≤ e−tn j! j! j=0 j=0 = M. (1.17) We further have An Am = Am An and hence ∞ j ∞ j X X t j t j Am = A An = etAm An . An etAm = An j! j! m j=0 j=0 Take t0 > 0, y ∈ D(A), and t ∈ [0, t0 ]. Using (1.3), we then compute Z t Z t d (t−s)Am sAn tAm tAn y= e y−e e(t−s)Am esAn (An − Am )y ds. e y ds = e ds 0 0 Estimate (1.17) and Lemma 1.23 thus yield ketAn y − etAm yk ≤ t0 M 2 kAn y − Am yk −→ 0 (1.18) as n, m → ∞. Because of the density of D(A) and the bound (1.17), we can apply Lemma 4.10 of [FA]. Since t0 > 0 is arbitrary, it yields operators T (t) ∈ B(X) such that etAn x → T (t)x as n → ∞ and kT (t)k ≤ M for all t ≥ 0 and x ∈ X. Clearly, T (0) = I and T (t + s)x = lim e(t+s)An x = lim etAn esAn x = T (t)T (s)x n→∞ n→∞ for all t, s ≥ 0 (use Remark 1.14). Letting m → ∞ in (1.18), we further deduce ketAn y − T (t)yk ≤ t0 M 2 kAn y − Ayk for all t ∈ [0, t0 ]. This means that etAn y converges to T (t)y uniformly for t ∈ [0, t0 ], and hence T (·)y is continuous for all y ∈ D(A). Lemma 1.7 and the density of D(A) then imply that T (·) is a (bounded) C0 -semigroup. 2) Let B be the generator of T (·). Observe that (0, ∞) ⊆ ρ(A) ∩ ρ(B) due to Proposition 1.21 and the assumptions. In view of Lemma 1.24 it thus remains to show A ⊆ B. For t > 0 and y ∈ D(A), Lemma 1.19 and Remarks 1.14 and 1.16 e) yield Z Z 1 t 1 t sAn 1 tAn 1 e An y ds = T (s)Ay ds. (T (t)y − y) = lim (e y − y) = lim n→∞ t 0 n→∞ t t t 0 1.2. Characterization of generators 19 As t → 0, from (1.4) we conclude that y ∈ D(B) and By = Ay; i.e., A ⊆ B.  We illustrate the above theorem by some examples. Applications to partial differential operators will be discussed in Section 1.4. d with D(A) = C01 (R≤0 ), Example 1.28. a) Let X = C0 (R≤0 ) and A = − ds cf. Example 1.22. Then A generates the C0 semigroup given by T (t)f = f (· − t) for t ≥ 0 and f ∈ X. It has the spectrum σ(A) = C− .2 Proof. We first check in several steps the conditions (1.16). 1) Let f ∈ X and ε > 0. We extend f to a function f˜ ∈ C0 (R). As in Example 1.9 one finds a map g̃ ∈ Cc (R) with kf˜ − g̃k∞ ≤ ε. By the proof of Proposition 4.13 in [FA] there is function h̃ ∈ Cc∞ (R) with kg̃ − h̃k∞ ≤ ε. As a result the restriction h of h̃ to R≤0 belongs to D(A) and satisfies kf −hk∞ ≤ 2ε, so that A is densely defined.3 2) Let the sequence (un ) in D(A) tend in X to a function u, and (Aun ) to some f in X. The map u is thus differentiable with −u′ = f ∈ X. As a result u ∈ D(A) and Au = f ; i.e., A is closed. 3) Let f ∈ X and λ > 0. To show the bijectivity of λI − A, we note that a function u belongs to D(A) and satisfies λu − Au = f if and only if u′ = −λu + f, u ∈ C 1 (R− ), and u∈X (using that then u′ = −λu + f ∈ X). This condition is equivalent to Z s 1 −λ(s−t0 ) u ∈ C (R− ) ∩ X, ∀ t0 ≤ s ≤ 0 : u(s) = e u(t0 ) + e−λ(s−τ ) f (τ ) dτ. t0 Since u and f are bounded and λ > 0, here one can let t0 → −∞ and derive Z s e−λ(s−τ ) f (τ ) dτ =: R(λ)f (s) for all s ≤ 0, lim u(s) = 0. u(s) = −∞ s→−∞ Conversely, the same reasoning yields that if the function v := R(λ)f belongs to X, then it is an element of D(A) and satisfies λv − Av = f . We show R(λ)f ∈ X, where the continuity is clear. Let ε > 0. There is a number sε ≤ 0 such that |f (τ )| ≤ ε for all τ ≤ sε . For s ≤ sε we then estimate Z ∞ Z s ε −λ(s−τ ) |R(λ)f (s)| ≤ e |f (τ )| dτ ≤ ε e−λr dr = , λ −∞ 0 substituting r = s − τ . As a result, R(λ)f (s) → 0 as s → −∞ so that λ ∈ ρ(A) and R(λ) = R(λ, A). 4) Employing the above formula for the resolvent, we calculate Z ∞ Z s kf k∞ −λ(s−τ ) e−λr dr = e kf k∞ dτ = kf k∞ kR(λ, A)f k∞ ≤ sup λ s≤0 −∞ 0 for all f ∈ X and λ > 0. Theorem 1.27 now implies that A generates a contraction semigroup T (·). In particular, σ(A) is contained in C− . For λ ∈ C− , the function e−λ belongs to D(A) and satisfies Ae−λ = −e′−λ = λe−λ so that C− ⊆ σ(A). The closedness of σ(A) then implies the second assertion. 2The last assertion was not part of the lectures. 3This argument was given in the lectures at a different place. 1.2. Characterization of generators 20 5) To determine T (·), we take ϕ ∈ D(A). We set u(t, s) = (u(t))(s) = (T (t)ϕ)(s) and for t ≥ 0 and s ≤ 0. By Proposition 1.11, the function u belongs to C 1 (R+ , X) ∩ C(R+ , [D(A)]) and solves the problem ∂t u(t, s) = −∂s u(t, s), t ≥ 0, s ≤ 0, s ≤ 0. u(0, s) = ϕ(s), (Note that D(A) includes the ‘boundary condition’ u(t, s) → 0 as s → −∞.) It is straighforward to see that via v(t, s) = ϕ(s−t) one defines another solution in the same function spaces. The uniqueness statement in Proposition 1.11 then yields u = v and hence T (t)ϕ = ϕ(· − t) for all t ≥ 0. The last claim now follows from the density of D(A).  b) We provide an operator A which fulfills (1.14) for n = 1 and some M > 1, but which is not generator. So one cannot omit the powers n in (1.14). Let X = C0 (R) × C0 (R) with k(f, g)k = max{kf k∞ , kgk∞ }, m(s) = is, and        u m m mu + mv u = = A v 0 m mv v for (u, v) ∈ D(A) = {(u, v) ∈ X | (mu, mv) ∈ X}. Since Cc (R) × Cc (R) ⊆ D(A), the domain D(A) is dense in X. Take (un , vn ) in D(A) such that (un , vn ) → (u, v) and A(un , vn ) → (f, g) in X as n → ∞. By pointwise limits, we infer that mu + mv = f and mv = g ∈ C0 (R), so that also mu ∈ C0 (R). As a result, the vector (u, v) belongs to D(A) and A is closed. Let λ ∈ C+ . Since 1/(λ − m) and m/(λ − m) are bounded, the operator ! R(λ) = 1 λ−m 0 m (λ−m)2 1 λ−m maps X into D(A). We further compute   λ − m −m (λI − A)R(λ) = 0 λ−m 1 λ−m 0 m (λ−m)2 1 λ−m ! = I, and similarly R(λ)(λw − Aw) = w for w ∈ D(A). So we have shown that C+ ⊆ ρ(A) and R(λ) = R(λ, A). For λ > 0 and k(f, g)k ≤ 1 we next estimate   n o f g mg f R(λ, A) ≤ max , + g λ−m ∞ (λ − m)2 ∞ λ − m ∞  1 |s| |s|  1 + ≤ + sup 2 ≤ sup 2 |λ − is| |λ − is| λ λ + s2 s∈R s∈R 3/2 = . λ On the other hand, for a > 0 and n ∈ N we choose gn ∈ C0 (R) such that gn (n) = 1 and kgn k∞ = 1. It then follows   m 0 kR(a + in, A)k ≥ R(a + in, A) gn ≥ gn (a + in − m)2 ∞ 1.2. Characterization of generators ≥ 21 in n gn (n) = 2 . 2 (a + in − in) a The resolvent R(λ, A) is thus unbounded on every imaginary line Re λ = a, violating Proposition 1.21 c); i.e., A does not generate a C0 -semigroup. c There are operators satisfying even kR(λ, A)k ≤ Re(λ) for some c > 1 and all λ ∈ C+ which fail to be a generator (see Example 2 in § 12.4 of [HP]). ♦ We now turn our attention to the generation of groups. We will reduce this question to the semigroup case, using the following simple fact. Lemma 1.29. Let T (·) be a C0 -semigroup and t0 > 0 such that T (t0 ) is invertible. Then T (·) can be extended to a C0 -group (T (t))t∈R . Proof. Take constants M ≥ 1 and ω ∈ R such that kT (t)k ≤ M eωt for all t ≥ 0. Set c = kT (t0 )−1 k. Let 0 ≤ t ≤ t0 . We then compute T (t0 ) = T (t0 − t)T (t) = T (t)T (t0 − t), I = T (t0 )−1 T (t0 − t)T (t) = T (t)T (t0 − t)T (t0 )−1 . The operator T (t) thus has the inverse T (t0 )−1 T (t0 − t) with norm less than or equal to M1 := cM eω+ t0 . Next, let t = nt0 + τ for some n ∈ N and τ ∈ [0, t0 ). In this case T (t) = T (τ )T (t0 )n has the inverse T (t0 )−n T (τ )−1 . We now define T (t) := T (−t)−1 for t ≤ 0. This definition gives a group, since for t, s ≥ 0 we can calculate T (−t)T (−s) = T (t)−1 T (s)−1 = (T (s)T (t))−1 = T (s + t)−1 = T (−s − t), T (−t)T (s) = (T (s)T (t − s))−1 T (s) = T (t − s)−1 T (s)−1 T (s) = T (t − s)−1 = T (s − t) T (−t)T (s) = T (t) −1 for t ≥ s, T (t)T (s − t) = T (s − t) for s ≥ t, and similarly for T (s)T (−t). Let t ∈ [0, t0 ] and x ∈ X. We then obtain kT (−t)x − xk = kT (−t)(x − T (t)x)k ≤ M1 kx − T (t)xk → 0 as t → 0. So (T (t))t∈R is a C0 -group by Lemma 1.7.  The next theorem characterizes generators of C0 -groups in the same way as in the Hille–Yosida Theorem 1.27, but now requiring resolvent bounds also for negative λ. Moreover, A generates the C0 -group (T (t))t∈R if and only if A and −A generate the C0 -semigroups (T (t))t≥0 and (T (−t))t≥0 , respectively. Theorem 1.30. Let A be a linear operator, M ≥ 1, and ω ≥ 0. The following assertions are equivalent. a) A generates a C0 -group (T (t))t∈R with kT (t)k ≤ M eω|t| for all t ∈ R. b) A generates a C0 -semigroup (T+ (t))t≥0 , and −A with D(−A) := D(A) generates a C0 -semigroup (T− (t))t≥0 with kT± (t)k ≤ M eωt for all t ≥ 0. c) A is closed, D(A) = X, and for all λ ∈ R with |λ| > ω we have λ ∈ ρ(A) and k(|λ| − ω)n R(λ, A)n k ≤ M for all n ∈ N. If one (and thus all) of these conditions is (are) fulfilled, one has T+ (t) = T (t) and T− (t) = T (−t) for every t ≥ 0. Moreover, in part c) one can then replace ‘λ ∈ R’ by ‘λ ∈ C’ and ‘|λ|’ by ‘|Re λ|’. 1.3. Dissipative operators 22 Proof. 1) We first deduce statement b) from a). Assuming a), we set T+ (t) = T (t) and T− (t) = T (−t) for each t ≥ 0. Recall from Remark 1.2 that T (−t) = T (t)−1 . It is easy to check that one thus obtains two C0 -semigroups. We denote their generators by A± . d T (0)x = Ax implying A ⊆ A+ and A ⊆ −A− . For x ∈ D(A), there exists dt To show the inverse inclusion, let x ∈ D(A− ) and t > 0. We then compute 1 1 (T (−t)x − x) = (T− (t)x − x) → −A− x, −t −t 1 1 (T (t)x − x) = −T (t) (T− (t)x − x) → −A− x t t as t → 0, so that x ∈ D(A) and hence A = −A− . One proves A = A+ similarly. Therefore, claim b) and the first adddendum are true. 2) Let b) be valid. For λ > ω, assertion c) follows from Theorem 1.27. For λ < −ω, we use that σ(−A) = −σ(A) with R(−λ, −A) = −R(λ, A), cf. Example 1.22 a). Theorem 1.27 thus also yields the estimate in part c) for λ < −ω since here −λ = |λ|. The second addendum is shown in the same way. 3) We assume property c) and derive part a). Theorem 1.27 implies that A generates a C0 -semigroup (T+ (t))t≥0 and −A generates a C0 -semigroup (T− (t))t≥0 (arguing for −A as in the previous step). Let x ∈ D(A) = D(−A) and t ≥ s ≥ 0. Proposition 1.11 and its proof imply d ds T+ (s)T− (s)x = T+ (s)AT− (s)x + T+ (s)(−A)T− (s)x = 0, and thus T+ (t)T− (t)x = x. Analogously, one obtains T− (t)T+ (t)x = x. It follows that I = T+ (t)T− (t) = T− (t)T+ (t) since D(A) is dense. By Lemma 1.29, T+ (·) can thus be extended to a C0 -group. Let B be its generator. We have B ⊆ A by definition and s(B) < ∞ by step 1) and Proposition 1.21. Condition c)  and Lemma 1.24 then yield A = B and hence assertion a). 1.3. Dissipative operators Even in the contraction case, the Hille-Yosida Theorem 1.27 poses the difficult task to show a resolvent estimate for all λ > 0. In this section we prove the Lumer-Phillips Theorem 1.40 which reduces this task to checking the dissipativity and a certain range condition of A. The former property can often be verified by direct computations and for the latter there are powerful functional analytic tools to solve the occuring equations. Below these matters will be illustrated by the first derivative again, more involved applications will be treated in the following section. We start with an auxiliary concept. The duality set J(x) of a vector x ∈ X is defined by J(x) = {x∗ ∈ X ∗ | hx, x∗ i = kxk2 , kxk = kx∗ k}, where hx, x∗ i = x∗ (x) for all x ∈ X and x∗ ∈ X ∗ . The Hahn-Banach theorem ensures that J(x) 6= ∅, cf. Corollary 5.10 in [FA]. In standard function spaces one can compute elements in the duality set explicitely. Example 1.31. a) Let X be a Hilbert space with scalar product (·|·). By Riesz’ Theorem 3.10 in [FA], for each functional y ∗ ∈ X ∗ there is a unique 1.3. Dissipative operators 23 vector y ∈ X satisfying hx, y ∗ i = (x|y) for all x ∈ X, and one has kyk = ky ∗ k. As a result, y ∗ ∈ J(x) is equivalent to kxk = kyk and (x|y) = kxk2 , or to kxk = kyk and (x|y) = kxk kyk. These conditions are valid if and only if y = αx for some α ∈ C with |α| = 1 (due to the characterization of equality in the Cauchy-Schwarz inequality). Inserting this expression in (x|y) = kxk2 , we see that x = y. The converse implication is clear. Consequently, J(x) = {ϕx } for the functional given by ϕx (z) = (z|x). b) Let X = Lp (µ) for an exponent p ∈ [1, ∞) and a measure space (S, A, µ) ′ which has to be σ-finite if p = 1. We identify X ∗ with Lp (µ) via the usual p for p > 1 and 1′ = ∞, see Theorem 5.4 in [FA]. duality pairing, where p′ = p−1 Take f ∈ X \ {0}. We set g = kf k2−p f |f |p−2 p writing 0 0 := 0. For p = 1, we have kgk∞ = kf k1 . For p > 1, we compute Z  p−1 p p (p−1)· p−1 2−p ′ |f | kgkp = kf kp dµ = kf k2−p kf kp−1 = kf kp . p p S Since also hf, gi = kf kp2−p Z S f f |f |p−2 dµ = kf kp2−p kf kpp = kf k2p , we obtain g ∈ J(f ). It follows from an exercise that J(f ) = {g} if p ∈ (1, ∞). Note that g = f for p = 2 which corresponds to part a). c) Let ∅ = 6 U ⊆ Rm be open and E = C0 (U ) with C0 (U ) := {f ∈ C(U ) | f (x) → 0 as x → ∂U and as |x| → ∞ for unbounded U }, which is a Banach space for the supremum norm. For f ∈ E there is a point x0 ∈ U with |f (x0 )| = kf k∞ . Set ϕ(g) = f (x0 )g(x0 ) for g ∈ E; i.e., ϕ = f (x0 )δx0 . As in Example 2.8 of [FA] one checks that ϕ ∈ E ∗ , kϕk = |f (x0 )| = kf k∞ , and ϕ(f ) = |f (x0 )|2 = kf k2∞ . Hence, ϕ belongs to J(f ). The same construction works on E = C(K) for a compact metric space K. ♦ We now state the core concept of this section. Definition 1.32. A linear operator A is called dissipative if for each vector x ∈ D(A) there is a functional x∗ ∈ J(x) such that RehAx, x∗ i ≤ 0. The operator A is called accretive if −A is dissipative. The next fundamental characterization provides the link between dissipativity and the resolvent condition (1.16) in the Hille–Yosida theorem. We also show that a generator of a contraction semigroup is dissipative in a somewhat stronger sense, which will be used in Theorem 3.8. Proposition 1.33. A linear operator A is dissipative if and only if it satisfies kλx − Axk ≥ λkxk for all λ > 0 and x ∈ D(A). If A generates a contraction semigroup, then we have RehAx, x∗ i ≤ 0 for every x ∈ D(A) and all x∗ ∈ J(x). 1.3. Dissipative operators 24 Proof. 1) Let A generate the contraction semigroup T (·). Take x ∈ D(A) and x∗ ∈ J(x). Using x∗ ∈ J(x) and the contractivity, we estimate D1 E
1 RehAx, x∗ i = lim Re (T (t)x − x), x∗ = lim RehT (t)x, x∗ i − kxk2 t t→0+ t→0+ t 1 ≤ lim sup (kxk kx∗ k − kxk2 ) = 0. t→0+ t 2) Let A be dissipative. Take x ∈ D(A) and λ > 0. There thus exists a functional x∗ ∈ J(x) with RehAx, x∗ i ≤ 0. These facts imply the inequalities λkxk2 ≤ Re(λhx, x∗ i) − RehAx, x∗ i ≤ |hλx − Ax, x∗ i| ≤ kλx − Axk kx∗ k. Since kxk = kx∗ k, it follows λkxk ≤ kλx − Axk. 3) Conversely, let kλx − Axk ≥ λ kxk be true for all λ > 0 and x ∈ D(A). If x = 0 we can take x∗ = 0 in the definition of dissipativity. Otherwise, we replace x by kxk−1 x, and will thus assume that kxk = 1. Take yλ∗ ∈ J(λx − Ax). This functional is not zero since kyλ∗ k = kλx − Axk ≥ λkxk = λ > 0 by the assumptions. We now set x∗λ = kyλ∗ k−1 yλ∗ and note that kx∗λ k = 1. Using the assumptions again, we deduce 1 λ ≤ kλx − Axk = ∗ hλx − Ax, yλ∗ i = Rehλx − Ax, x∗λ i kyλ k ∗ = λ Rehx, xλ i − RehAx, x∗λ i ≤ min{λ − RehAx, x∗λ i, λ Rehx, x∗λ i + kAxk}. This inequality implies the core bounds 1 kAxk ≤ Rehx, x∗λ i. λ Let x̃∗λ be the restriction of x∗λ to the space E = lin{x, Ax} equipped with the norm of X. Because of kx̃∗λ k ≤ kx∗λ k = 1, the Bolzano–Weierstraß theorem yields a sequence (λj ) in R+ and a vector y ∗ ∈ E ∗ such that λj → ∞ and x̃∗λj → y ∗ as j → ∞. Applying these limits to the above estimates, we derive RehAx, x∗λ i ≤ 0 ky ∗ k ≤ 1, and RehAx, y ∗ i ≤ 0 1− and 1 ≤ Rehx, y ∗ i. The Hahn-Banach theorem allows us to extend y ∗ to a functional x∗ ∈ X ∗ with kx∗ k = ky ∗ k ≤ 1. It then satisfies RehAx, x∗ i ≤ 0 and 1 ≤ Rehx, x∗ i ≤ |hx, x∗ i| ≤ kx∗ k ≤ 1 as kxk = 1. So we have equalities in the above formula, which means that kx∗ k = 1 = kxk and hx, x∗ i = 1 = kxk2 ; i.e., x∗ ∈ J(x) and A is dissipative.  The dissipativity of differential operators A heavily depends on the boundary conditions, as we now discuss for first-order operators on an interval. Example 1.34. a) Let X = C0 (R), b, c ∈ Cb (R) be real-valued, and Au = bu′ + cu with D(A) = C01 (R). Take u ∈ D(A) and some s0 ∈ R with |u(s0 )| = kuk∞ . Then ϕ = u(s0 )δs0 belongs to J(u) by Example 1.31. We then compute r := RehAu − kc+ k∞ u, ϕi = b(s0 ) Re(u′ (s0 )u(s0 )) + (c(s0 ) − kc+ k∞ ) Re(u(s0 )u(s0 )) ≤ b(s0 ) Re(u′ (s0 )u(s0 )). 1.3. Dissipative operators 25 We set h(s) = Re(u(s0 )u(s)) for s ∈ R. Clearly, h ∈ C01 (R) is real-valued and |u(s0 )|2 = h(s0 ) ≤ khk∞ ≤ |u(s0 )| kuk∞ = |u(s0 )|2 so that h attains its maximum at s0 . Therefore, h′ (s0 ) = 0 and r ≤ 0. This means that A − kc+ k∞ I is dissipative. b) Let X = C([0, 1]), b, c ∈ X be real-valued, b(0) ≥ 0 for simplicity, and Aj = bu′ + cu with D(Aj ) = {u ∈ C 1 ([0, 1]) | u′ (j) = 0} for j ∈ {0, 1}. Then A1 − kc+ k∞ I is dissipative. If b(1) ≤ 0, also A0 − kc+ k∞ I is dissipative. On the other hand, if b(1) > 0 the operator A0 − ωI is not dissipative for any ω ∈ R. Proof. For u ∈ D(Aj ), we use the functional ϕ(v) = u(s0 )v(s0 ) on X, where |u(s0 )| = kuk∞ for some s0 ∈ [0, 1]. We also set h(s) = Re(u(s0 )u(s)) for s ∈ [0, 1]. As in a), one sees that ϕ belongs to J(u), h ∈ C 1 ([0, 1]) attains its maximum at s0 , and r := RehAj u − kc+ k∞ u, ϕi ≤ b(s0 ) Re(u′ (s0 )u(s0 )) = b(s0 )h′ (s0 ). If s0 ∈ (0, 1), this inequality again yields r ≤ 0. Similarly, for s0 = 0 we obtain h′ (0) = lim s→0+ 1 s (h(s) − h(0)) ≤ 0 since h(0) is a maximum of h. Using b(0) ≥ 0, we infer r ≤ 0. Finally, let s0 = 1. In this case the above argument yields h′ (1) ≥ 0. We first look at j = 0. For b(1) ≤ 0, we derive r ≤ b(1)h′ (1) ≤ 0 so that A0 − kc+ k∞ I is dissipative in this case. Next, let b(1) > 0. Fix ω ∈ R. Choose a real-valued function u ∈ D(A0 ) with maximum u(1) = 1 and u′ (1) > (ω − c(1))/b(1). Since then ϕ = δ1 , we obtain the inequality RehA0 u − ωu, ϕi = b(1)u′ (1) + c(1) − ω > 0. Hence, A0 − ωI is not dissipative. For j = 1 we have the boundary condition u′ (1) = 0 and thus h′ (1) = 0. It follows that r ≤ b(1)h′ (1) = 0 and so A1 − kc+ k∞ I is dissipative.  d 1 2 c) Let X = L (R) and A = ds with D(A) = Cc (R). For u ∈ D(A) we have u ∈ J(u) by Example 1.31. Integration by parts yields Z Z ′ u u ds + u′ u ds = 0; 2 RehAu, ui = hAu, ui + hAu, ui = R R i.e., A is dissipative (but surely not closed). The same applies to −A. d d) Let X = L2 (0, 1), Aj = ds , and D(Aj ) = {u ∈ C 1 ([0, 1]) | u(j) = 0} for j ∈ {0, 1}. For u ∈ D(Aj ) we take again u ∈ J(u) and obtain Z 1 Z 1 1 2 RehAu, ui = u′ u ds + u′ u ds = uu 0 = |u(1)|2 − |u(0)|2 . 0 0 It follows that A1 is dissipative. However, A0 − ωI is not dissipative for any ω ∈ R, since we can find a map u in D(A0 ) satisfying |u(1)|2 > 2ωkuk22 and so 1 ♦ RehA0 u − ωu, ui = |u(1)|2 − ωkuk22 > 0. 2 Examples c) and d) can be extended to Lp with p ∈ [1, ∞), cf. Example 1.49. Above we have encountered rather natural dissipative, but nonclosed operators. To treat such operators, we introduce a concept extending closedeness. 1.3. Dissipative operators 26 Intermezzo 2: Closable operators. Definition 1.35. A linear operator A is called closable if it possesses a closed extension B. Note that a closed operator is closable since A ⊆ A. We first characterize closability and construct the closure A of a closable operator A, which is the smallest closed extension of A. Lemma 1.36. For a linear operator A, the following statements are equivalent. a) The operator A is closable. b) Let (xn ) be a sequence in D(A) such that xn → 0 and Axn → y in X as n → ∞. Then y = 0. c) In the set D(A) = {x ∈ X| ∃ (xn ) in D(A), y ∈ X : xn → x, Axn → y, n → ∞} the vector y is uniquely determined by x. Letting A : D(A) → X; Ax = y, one thus defines a map. The operator A is linear, closed, and extends A. If one and hence all of the properties a)–c) are valid, then Gr(A) = Gr(A), D(A) is dense in [D(A)], and we have A ⊆ B for every closed operator B ⊇ A. Proof. Clearly, part c) implies a). Let a) be true and B be a closed extension of A. Take (xn ) as in statement b). Then the vectors Axn = Bxn tend to y = B0 = 0 since B is closed. We assume that property b) holds. Let (xn )n and (zn )n be sequences in D(A) with limit x in X such that (Axn )n converges to y and (Azn )n to w in X. Then (xn − zn ) is a null sequence in X with A(xn − zn ) = Axn − Azn → y − w as n → ∞. Part b) thus implies y = w, so that A is a mapping. One easily verifies that A is linear and that Gr(A) = Gr(A), which shows the first part of the addendum. Hence, A is closed due to Remark 1.16 and A extends A. Therefore assertion c) is shown. Let B be another closed extension of A. We then have Gr(A) ⊆ Gr(B) and so Gr(A) = Gr(A) ⊆ Gr(B) because of the closedness of B. In particular, B extends A. The density assertion is an immediate consequence of Gr(A) =  Gr(A) and the definition of the graph norm. As consequence of this lemma, a linear operator is closed if and only if it is its own closure. We illustrate the concepts of extensions and closure by the first derivative, where we again stress the role of the boundary conditions. Example 1.37. a) Let X = L1 (0, 1) and Af = f (0)✶ with D(A) = C([0, 1]). This operator is not closable. In fact, the functions fn ∈ D(A) given by fn (s) = max{1 − ns, 0} satisfy kfn k1 = 1/2n → 0 as n → ∞, but Afn = ✶ for all n ∈ N, contradicting Lemma 1.36 b). b) Let X = C([0, 1]) and A0 u = u′ with D(A0 ) = Cc1 (0, 1) := Cc1 ((0, 1)), as well as Au = u′ with D(A) = C01 (0, 1) := C01 ((0, 1)). As in Example 1.15 we see that A is closed. Hence, A0 is closable and A0 ⊆ A since A0 ⊆ A. To check equality, let f ∈ C01 (0, 1). Take ϕn ∈ Cc1 (0, 1) such that ϕ = 1 on [1/n, 1 − 1/n], 0 ≤ ϕn ≤ 1 and kϕ′n k∞ ≤ cn for some c > 0 and all n ∈ N with n ≥ 2. (For 1.3. Dissipative operators 27 instance, one can take   0,  
 1 2 2   8n s − 4n ,
2 3 ϕn (s) = 1 − 8n2 4n −s ,    1,   ϕ (1 − s), n 1 0 < s < 4n , 1 1 4n ≤ s ≤ 2n , 1 3 2n ≤ s ≤ 4n , 1 3 4n < s ≤ 2 , 1 2 < s < 1, where c = 4.) Then the function fn = ϕn f belongs to D(A0 ), and we have kfn − f k∞ = kϕn f ′ − f ′ k∞ ≤ sup 1 1 [0, n ]∪[1− n ,1] sup 1 1 ]∪[1− n ,1] [0, n |(ϕn (s) − 1)f (s)| ≤ sup 1 1 [0, n ]∪[1− n ,1] |f (s)| −→ 0, |(ϕn (s) − 1)f ′ (s)| −→ 0 as n → ∞ since f, f ′ ∈ C0 (0, 1). We further obtain kϕ′n f k∞ ≤ sup |ϕ′n (s)f (s)| + 1 s∈[0, n ] ≤ sup cn 1 s∈[0, n ] ≤ cn Z 1 n 0 Z s sup 1 s∈[1− n ,1] ′ f (τ ) dτ + |f ′ (τ )| dτ + cn sup 1 ,1] s∈[1− n 0 Z |ϕ′n (s)f (s)| 1 1 1− n cn Z 1 f ′ (τ ) dτ s |f ′ (τ )| dτ −→ 0 as n → ∞, because of (1.4) and f ′ ∈ C0 (0, 1). Hence, A0 (ϕn f ) = ϕ′n f + ϕn f ′ converges to Af = f ′ . This means that A ⊆ A0 and thus A0 = A. In particular A0 is not closed and thus fails to be a generator. We discuss further closed extensions of A0 given by Aj u = u′ for j ∈ {1, 2, 3}. 1) Let D(A1 ) = {u ∈ C 1 ([0, 1]) | u′ (1) = 0}. By an exercise, A1 generates a C0 -semigroup on X and σ(A1 ) = {0}. Observe that A1 is a strict extension of A. Lemma 1.24 thus implies that ρ(A) ∩ ρ(A1 ) = ∅ and hence C \ {0} ⊆ σ(A). (Actually, we have σ(A) = C since ✶ ∈ / AD(A).) As a result, A is not generator – it has too many boundary conditions, namely four instead of one as in D(A1 ). 2) Let D(A3 ) = C 1 ([0, 1]). Example 1.22 says that σ(A3 ) = C. So A3 is not a generator because it has not enough boundary conditions, namely none. We have A ( A1 ( A3 . 3) Let D(A2 ) = {u ∈ C 1 ([0, 1]) | u(0) = 0}. Also A2 is ‘sandwiched’ between A and A3 ; i.e., A ( A2 ( A3 , but A1 and A2 are not comparable. The operator A2 is not a generator as its domain is not dense, see Example 1.22. Summing up, the ‘minimal’ operator A and the ‘maximal’ operator A3 do not generate C0 -semigroups. Between them there are various, partly noncomd parable operators (so–called ‘realizations’ of ds ) which may or may not be generators. Their domains are often determined by boundary conditions. ♦ We come back to the investigation of semigroups. Below we use closures in a generation result, but at first we establish sufficient conditions for a subspace D to be dense in D(A) in the graph norm. Such a subspace is called core of a 1.3. Dissipative operators 28 closed operator A, since one can often extend properties from cores to the full domain. (Observe that A|D = A if and only if D is a core.) Proposition 1.38. Let A generate the C0 -semigroup T (·) on X. Let D be a linear subspace of D(A) which is dense in X and invariant under the semigroup; i.e., T (t)D ⊆ D for all t ≥ 0. Then D is dense in [D(A)]. Proof. Set C = sup0≤t≤1 kT (t)k. Let x ∈ D(A). The map T (·)x : R≥0 → [D(A)] is continuous by Proposition 1.11. Take ε > 0. There is a time τ = τ (ε, x) ∈ (0, 1] with kT (t)x − xkA ≤ ε for all t ∈ [0, τ ]. It follows Z Z 1 τ 1 τ T (t)x dt − x ≤ kT (t)x − xkA dt ≤ ε. τ 0 τ 0 A Using the density of D in X, we find a vector y ∈ D with  C + 1 −1 ε. kx − yk ≤ C + τ Let D̃ be the closure of D in [D(A)]. We want to replace y by a vector z in D̃ that is close to x for k · kA . To this aim, we set Z 1 τ z= T (t)y dt. τ 0 The integrand T (t)y takes values in D by assumption, and as above it is continuous in [D(A)]. In view of the definition of the integral, z thus belongs to D̃. The previous inequalities and Lemma 1.19 imply the bound Z τ Z 1 τ 1 T (t)(x − y) dt kx − zkA ≤ x − T (t)x dt + τ 0 τ A 0 Z τ 1 + A T (t)(x − y) dt τ 0 Z 1 C τ kx − yk dt + k(T (τ ) − I)(x − y)k ≤ε+ τ 0 τ  C + 1 kx − yk ≤ 2ε. ≤ε+ C + τ Finally, there is a vector w ∈ D with kz−wkA ≤ ε, and hence kx−wkA ≤ 3ε.  The next result shows further important properties of dissipative operators following from the characterization in Proposition 1.33. In particular, the Hille– Yosida estimate (1.16) is reduced to a range condition, and a densely defined, dissipative operator has a dissipative closure. Proposition 1.39. Let A be dissipative. The following assertions hold. a) Let λ > 0. Then the operator λI − A is injective and for y ∈ R(λI − A) := (λI − A)(D(A)) we have k(λI − A)−1 yk ≤ λ1 kyk. b) Let λ0 I −A be surjective for some λ0 > 0. Then A is closed, (0, ∞) ⊆ ρ(A), and kR(λ, A)k ≤ λ1 for all λ > 0. c) Let D(A) be dense in X. Then A is closable and A is also dissipative. 1.3. Dissipative operators 29 Proof. Assertion a) immediately follows from Proposition 1.33 where y = λx − Ax for some x ∈ D(A). Let the assumptions in b) hold. Part a) then implies that λ0 I − A has an inverse with norm less than or equal to λ10 . In particular, A is closed by Remark 1.17 b). Let λ ∈ (0, 2λ0 ). Since |λ − λ0 | < λ0 ≤ kR(λ0 , A)k−1 , Remark 1.17 shows that λ belongs to ρ(A). Step a) also yields the estimate kR(λ, A)k ≤ 1/λ. We can now iterate this argument, deriving assertion b). c) Assume that D(A) is dense in X. To check the closability of A, we choose a sequence (xn ) in D(A) with limit 0 in X such that (Axn ) converges in X to some y ∈ X. By density, there are vectors yk in D(A) tending to y in X as k → ∞. Take λ > 0 and n, k ∈ N. Proposition 1.33 implies the lower bound kλ2 xn − λAxn + λyk − Ayk k = k(λI − A)(λxn + yk )k ≥ λkλxn + yk k. Letting n → ∞, we deduce k−λy + λyk − Ayk k ≥ λkyk k, k−y + yk − λ−1 Ayk k ≥ kyk k. As λ → ∞, it follows that k−y + yk k ≥ kyk k. Taking the limit k → ∞, we conclude y = 0. Due to Lemma 1.36, the operator A is closable and for x ∈ D(A) there are vectors zn ∈ D(A) such that zn → x and Azn → Ax in X as n → ∞. Using Proposition 1.33, we now infer the estimate kλx − Axk = lim kλzn − Azn k ≥ lim λkzn k = λkxk, n→∞ n→∞ and thus the dissipativity of A.  The following theorem by Lumer and Phillips from 1961 is the most important tool to verify the generator property in concrete cases (besides Theorem 2.25 below). To show that an operator A (or its closure) generates a contraction semigroup, one only has to establish the density of D(A), the dissipativity of A, and that λ0 I − A is surjective (or has dense range) for some λ0 > 0. The first two properties can often be checked by direct computations using the given information on A. The final range conditions are usually harder to show. One has to solve the ‘stationary problem’ ∃ u ∈ D(A) : λ0 u − Au = f at least for f from a dense set of ‘good’ vectors. Fortunately, there are various tools to solve this problem which we partly discuss in the next section. Based on our preparations, the Lumer–Phillips theorem can easily be deduced from the contraction case of the Hille–Yosida Theorem 1.27. In Example 1.50 we will see that one cannot omit the range conditions in parts a) or b). Theorem 1.40. Let A be a linear and densely defined operator. The following assertions hold. a) Let A be dissipative and λ0 > 0 such that λ0 I − A has dense range. Then A generates a contraction semigroup. b) Let A be dissipative and λ0 > 0 such that λ0 I − A is surjective. Then A generates a contraction semigroup. 1.3. Dissipative operators 30 c) Let A generate a contraction semigroup. Then A is dissipative, C+ ⊆ ρ(A), and kR(λ, A)k ≤ 1/ Re(λ) for λ ∈ C+ . One can replace ‘contraction’ by ‘ω-contraction’ and A by A − ωI for ω ∈ R. Operators satisfying the assumptions in assertion b) are called maximally dissipative or m-dissipative. (Such maps cannot have non-trivial dissipative extensions because of Lemma 1.24 and Proposition 1.39.) If a closed operator A satisfies the hypotheses of part a), then A generates a contraction semigroup since A = A. This variant of the result is often very useful in applications. Concerning the addendum, one can easily check that the closure of A − ωI is equal to A − ωI. Proof of Theorem 1.40. Let the conditions in a) be true. Proposition 1.39 then tells us that A possesses a dissipative closure A. Let y ∈ X. By assumption, there are vectors xn ∈ D(A) such that the images yn = λ0 xn − Axn tend to y in X as n → ∞. The dissipativity of A yields the inequality 1 1 kxn − xm k ≤ k(λ0 − A)(xn − xm )k = kyn − ym k λ0 λ0 for all n, m ∈ N thanks to Proposition 1.33. This means that (xn ) has a limit x in X, and hence the vectors Axn = Axn = λ0 xn − yn tend to λ0 x − y as n → ∞. Since A is closed, x belongs to D(A) and satisfies Ax = λ0 x − y so that λ0 I − A is surjective. Proposition 1.39 and Theorem 1.27 now imply the assertion. By Proposition 1.39, A is closed if λ0 I −A is surjective, and then part a) shows that A = A generates a contraction semigroup. Assertion c) is a consequence of Propositions 1.33 and 1.21. The addendum follows by a rescaling argument based on Lemma 1.18.  We will reformulate the range condition in the Lumer–Phillips theorem using duality. To this aim, we recall the following concept from the lecture Spectral Theory. For a densely defined linear operator A, we define its adjoint A∗ by A∗ x ∗ = y ∗ ∗ ∗ for all x∗ ∈ D(A∗ ), where ∗ ∗ ∗ (1.19) ∗ ∗ D(A ) = {x ∈ X | ∃ y ∈ X ∀ x ∈ D(A) : hAx, x i = hx, y i}. This means that hAx, x∗ i = hx, A∗ x∗ i for all x ∈ D(A) and x∗ ∈ D(A∗ ). Recall from Remark 1.23 in [ST] that A∗ is a closed linear operator. The domain D(A∗ ) in (1.19) is defined in a ‘maximal way’ which is convenient for the theory, but for concrete operators it is often very difficult to calculate D(A∗ ) explicitly. The next result replaces the range condition by the injectivity of λ0 I − A∗ (or the dissipativity of A∗ ), cf. Theorem 1.24 in [ST]. In Example 1.50 we present a closed and densely defined dissipative operator having a non-dissipative adjoint. Corollary 1.41. Let A be dissipative and densely defined, and let λ0 I − A∗ be injective for some λ0 > 0. Then A generates a contraction semigroup. If A∗ is dissipative, then λI − A∗ is injective for all λ > 0. Proof. The addendum follows from Proposition 1.39. Let λ0 I − A∗ be injective. Take a functional x∗ ∈ X ∗ such that hλ0 x − Ax, x∗ i = 0 for all x ∈ D(A). From (1.19) we then deduce that x∗ belongs to D(A∗ ) and A∗ x∗ = λ0 x∗ . 1.3. Dissipative operators 31 Hence, x∗ = 0. The Hahn-Banach theorem now implies the density of R(λ0 I − A), see Corollary 5.13 in [FA]. Theorem 1.40 thus yields the assertion.  Examples 1.34 c) and d) indicate that integration by parts is a very convenient tool to check dissipativity for differential operators in an L2 -context. To tackle such problems, we briefly discuss concepts and basic facts which are treated Chapter 3 of [ST] in much greater detail. Intermezzo 3: Weak derivatives and Sobolev spaces. Let ∅ 6= U ⊆ Rm be open, k ∈ N, j ∈ {1, . . . , m}, and p ∈ [1, ∞]. A function u ∈ Lp (U ) has a weak derivative in Lp (U ) with respect to the jth coordinate if there is a map v ∈ Lp (U ) satisfying Z Z u∂j ϕ dx = − vϕ dx U U for all ϕ ∈ Cc∞ (U ). (Hence, by definition weak derivatives can be integrated by parts against ‘test functions’ ϕ ∈ Cc∞ (U ).) The function v is (up to a null function) uniquely determined by Lemma 3.4 in [ST]. We set ∂j u := v in this case and define the Sobolev space W 1,p (U ) = {u ∈ Lp (U ) | ∀ j ∈ {1, . . . , m} ∃ ∂j u ∈ Lp (U )}. It is a Banach space when endowed with the norm  1 P  p p kukpp + m k∂ uk , p j j=1 kuk1,p = max j∈{1,...,m} {kuk∞ , k∂j uk∞ }, p < ∞, p = ∞, see Proposition 3.7 of [ST]. (As usual we identify functions which are equal almost everywhere.) This norm is equivalent to the norm given by kukp + m X j=1 k∂j ukp due to Remark 3.2 in [ST]. Analogously one defines the Sobolov spaces W k,p (U ) αm for α ∈ Nm and |α| = and higher-order weak derivatives ∂ α = ∂1α1 . . . ∂m 0 0 α1 + · · · + αm ≤ k. We put u = ∂ u. One often writes H k instead of W k,2 which is a Hilbert space. We summarize some properties of Sobolev spaces and weak derivatives. Remark 1.42. a) Let u ∈ C k (U ) such that u and all its derivatives up to order k belong to Lp (U ). Then u belongs to W k,p (U ) and its classical and weak derivatives coincide. See Remark 3.2 in [ST]. α b) Let u, un , v ∈ Lp (U ) and α ∈ Nm 0 such that un → u and ∂ un → v in Lp (U ) as n → ∞. Then u possesses the weak derivative ∂ α u = v by Lemma 3.5 of [ST] and its proof. c) Let p < ∞. Theorem 3.15 of [ST] says that Cc∞ (Rm ) is dense in W k,p (Rm ) and that C ∞ (U ) ∩ W k,p (U ) is dense in W k,p (U ). d) Let −∞ ≤ a < b ≤ ∞ and u ∈ Lp (a, b). Then the function u belongs to W 1,p (a, b) := W 1,p ((a, b)) if and only if (a representative of) u is continuous 1.3. Dissipative operators and there is a map v ∈ Lp (a, b) satisfying Z t v(τ ) dτ u(t) = u(s) + s 32 for all t, s ∈ (a, b). (1.20) We then have u′ = ∂u := ∂1 u = v and u has a continuous extension to a (or b) if a > −∞ (or b < ∞). See Theoren 3.11 in [ST]. Actually, W 1,p (a, b) is continuously embedded into Cb (J) for J = (a, b).4 We show the last claim first for the case a = −∞. In (1.20) we take s ∈ [t − 2, t − 1]. Integrating over s and using Hölder’s inequality, we derive Z t−1 Z t−1 Z t |u(t)| ≤ |u(s)| ds + |u′ (τ )| dτ ds t−2 t−1 ≤ Z t−2 t−2 |u(s)|p ds 1 p + Z s t t−2 1 |u′ (τ )| dτ ≤ kukp + 2 p′ ku′ kp . which yields the claim. The case (a, ∞) is treated in the same way using s ∈ [t + 1, t + 2]. If (a, b) is bounded, we set c = (a + b)/2 and δ = (b − a)/2. Let t ∈ [c, b]. Taking the integral over s ∈ [t − δ, t] we derive Z t Z t Z t |u′ (τ )| dτ ds. |u(s)| ds + δ |u(t)| ≤ t−δ t−δ s We can now estimate as above. If t ∈ [a, c), we use s = t + δ. The claim follows. As an example for a weak derivative take a function u ∈ Cc (R) whose restrictions u+ and u− to R≥0 and R≤0 , respectively, are continuously differentiable. The map u then belongs to W 1,p (R) for all p ∈ [1, ∞] and its derivative is given by (u± )′ on R± . ′ e) Let u ∈ W 1,p (U ) and v ∈ W 1,p (U ) with p1 + p1′ = 1. Proposition 3.8 of [ST] yields that the product uv is an element of W 1,1 (U ) and satisfies the product rule ∂j (uv) = u∂j v + v∂j u. Analogous results hold for higher-order derivatives. f) Let U have a compact boundary ∂U of class C 1 . By the Trace Theorem 3.32 in [ST], the map W 1,p (U ) ∩ C(U ) → Lp (∂U, dσ); u 7→ u|∂U , has a continuous extension tr : W 1,p (U ) → Lp (∂U, dσ) called the trace operator. Its kernel is the closure W01,p (U ) of the test functions Cc∞ (U ) in W 1,p (U ). If tr u = 0, one says that u vanishes on ∂U ‘in the sense of trace.’ ′ Let f ∈ W 1,p (U )m and u ∈ W 1,p (U ). The Divergence Theorem 3.34 in [ST] then yields Z Z Z u div f dx = − f · ∇u dx + tr(u) ν · tr(f ) dσ. (1.21) U U ∂U Here ν is the unit outer normal and the dot denotes the scalar product in Rm . We usually omit the trace operator in the boundary integral. If U = Rm the formula is true without the boundary integral. ♦ Coming back to semigroups, we illustrate the above concepts by a simple d example concerning generation properties of ds in L2 (R). 4This fact was omitted in the lectures. 1.3. Dissipative operators 33 d with D(A) = Cc1 (R). The operaExample 1.43. Let X = L2 (R) and A = ds tors ±A are densely defined and dissipative by Example 1.34. Proposition 1.39 then yields their closability and the dissipativity of their closures, where −A has the closure −A. For each u ∈ D(A) there are functions un ∈ Cc1 (R) such that un → u and u′n = Aun → Au in L2 (R) as n → ∞. In view of Remark 1.42, the map u thus belongs to W 1,2 (R) and Au = u′ ; i.e., A = ∂ and D(A) = W 1,2 (R). ∗ To compute A , we take u, v ∈ W 1,2 (R). Formula (1.21) then yields Z Z ′ hAu, vi = u v ds = − uv ′ ds = hu, −∂vi, R R ∗ so that (−∂, W 1,2 (R)) is a restriction of A , see (1.19). Conversely, let v ∈ ∗ ∗ D(A ). The functions v and A v thus belong to L2 (R) and satisfy Z Z ∗ ∗ uA v ds = hu, A vi = hAu, vi = u′ v ds R R Cc∞ (R) for all u ∈ ⊆ D(A) ⊆ D(A), which means that v ∈ W 1,2 (R) and ∗ ∗ A v = −v ′ = −Av. As a result, A = −A. Corollary 1.41 now shows that ±A generate contraction semigroups. On the other hand, the translation group T (t)f = f (· + t) on X has a generator B, cf. Example 1.9. For f ∈ D(A) the functions w(t) = 1t (T (t)f − f ) converge uniformly to f ′ as t → 0+ . Moreover, the supports supp w(t) are contained in the bounded set supp f + [−1, 0] for all 0 ≤ t ≤ 1, so that w(t) tends to f ′ in X. This means A ⊆ B and so A ⊆ B. Lemma 1.24 now yields A = B and hence A generates T (·).5 ♦ We conclude this section with a discussion of isometric groups. Corollary 1.44. For a linear operator A, the following statements are equivalent. a) The operator A generates an isometric C0 -group T (·); i.e., kT (t)xk = kxk for all x ∈ D(A) and t ∈ R. b) The operator A is closed, densely defined, ±A are dissipative, and λ0 I ± A are surjective for some λ0 > 0. c) The operator A is closed, densely defined, R \ {0} belongs to ρ(A), and 1 kR(λ, A)k ≤ |λ| for all λ ∈ R \ {0}. In this case, one can replace in c) the set R \ {0} by C \ iR and |λ| by |Re λ|. Proof. The Lumer-Phillips Theorem 1.40 says that b) holds if and only if A and −A generate contraction semigroups. Theorem 1.30 thus implies the equivalence of assertions b) and c), the addendum, and that b) is true if and only if A generates a contractive C0 -group T (·). It remains to show that a contractive C0 -group T (·) is already isometric. Indeed, in this case we have kT (t)xk ≤ kxk = kT (−t)T (t)xk ≤ kT (−t)k kT (t)xk ≤ kT (t)xk for all x ∈ X and t ∈ R, so that T (t) is isometric. 5The semigroup generated by A was not computed in the lectures.  1.3. Dissipative operators 34 We want to show an important variant of the above corollary on Hilbert spaces which requires a few more concepts from [ST]. Let X be a Hilbert space. For a linear operator on X with dense domain we define the Hilbert space adjoint A′ of A as in (1.19) replacing the duality pairing hx, x∗ i by the inner product (x|y). A linear operator A on X is called symmetric if ∀ x, y ∈ D(A) : (Ax|y) = (x|Ay), which means that A ⊆ A′ if D(A) is dense. If A is densely defined, we say that it is selfadjoint if A = A′ ; i.e., if A is symmetric and D(A) = {y ∈ X | ∃ z ∈ X ∀ x ∈ D(A) : (Ax|y) = (x|z)} = {y ∈ X | (D(A), k · k) → C; x 7→ (Ax|y), is continuous}. (The last equality is a consequence Riesz’ representation Theorem 3.10 in [FA].) A densely defined, linear operator A is called skewadjoint if A = −A′ which is equivalent to the selfadjointness of iA. Finally, T ∈ B(X) is unitary if it has the inverse T −1 = T ′ . We recall a very useful criterion from Theorem 4.7 of [ST]. A symmetric, densely defined, closed operator A is selfadjoint if and only if its spectrum σ(A) belongs to R, which in turn follows from the condition ρ(A) ∩ R 6= ∅. As in Remark 1.23 in [ST] one can check that A′ is a closed linear map. Hence, every densely defined, symmetric operator is closable with A ⊆ A′ (cf. Lemma 1.36) and each selfadjoint operator is closed. Let A be symmetric and densely defined. Take u, v ∈ D(A). There are sequences (un ) and (vn ) in D(A) with limits u and v in X, respectively, such that Aun → Au and Avn → Av in X as n → ∞. We then compute (Au|v) = lim (Aun |vn ) = lim (un |Avn ) = (u|Av), n→∞ n→∞ so that also the closure A is symmetric. There are densely defined, symmetric, closed operators that are not selfadjoint. (By Example 4.8 of [ST] this is the case for A = i∂ with D(A) = {u ∈ W 1,2 (0, ∞) | u(0) = 0} on X = L2 (0, ∞). Here one has D(A′ ) = W 1,2 (0, ∞).) The next result due to Stone from 1930 belongs to the mathematical foundations of quantum mechanics. Theorem 1.45. Let X be a Hilbert space and A be a linear operator on X with a dense domain. Then A generates a C0 -group of unitary operators if and only if A is skewadjoint. Proof. 1) Let A′ = −A. Hence, A is closed. For x ∈ D(A), we have J(x) = {ϕx } with ϕx = (·|x) by Example 1.31. We thus obtain hAx, ϕx i = (Ax|x) = −(x|Ax) = −(Ax|x) and so RehAx, ϕx i = 0. Therefore A, A′ = −A, and (−A)′ = A are dissipative From Corollary 1.41 we then deduce that A and A′ generate contraction semigroups. Corollary 1.44 now shows that A generates a C0 -group T (·) of invertible isometries, implying that each T (t) is unitary by Proposition 5.52 in [FA]. 1.4. Examples with the Laplacian 35 2) Let A generate a unitary C0 -group T (·). We infer T (t)′ = T (t)−1 = T (−t) for all t ∈ R by Remark 1.2, and hence T (·)′ is a unitary C0 -group with the generator −A. For x, y ∈ D(A) we thus obtain

(Ax|y) = lim 1t (T (t)x − x) y = lim x 1t (T (t)′ y − y) = (x|−Ay). t→0 t→0 A′ . This means that −A ⊆ We further know from Theorem 1.30 that σ(A) and σ(−A) are contained in iR. Equation (4.2) in [ST] then yields σ(A′ ) = σ(A) ⊆  iR. The assertion −A = A′ now follows from Lemma 1.24. 1.4. Examples with the Laplacian In this section we discuss generation and related properties of the Laplacian 2 = div ∇ ∆ = ∂12 + · · · + ∂m in various settings. To apply the Lumer–Phillips Theorem 1.40, we have to check three conditions. The density of the domain often follows from standard results on function spaces. With the right tools one can usually verify dissipativity in a straightforward way (imposing appropriate boundary conditions). For the range condition one has to solve the ‘elliptic problem’ u − ∆u = f plus boundary conditions for given f . Using differing methods, this will be done first on Rm , then on intervals, and finally with Dirichlet boundary conditions on bounded domains. As we will see in the next chapter, these results will allow us to solve diffusion equations, actually with improved regularity. We will further use the Dirichlet–Laplacian in the wave equation, cf. Example 1.53. We strive for a self-contained presentation (employing the lectures Functional Analysis and Spectral Theory), but for certain additional facts we have to cite deeper results from the theory of partial differential equations. A) The Laplacian on Rm . Since the Laplacian has constant coefficients, on the full space Rm the Fourier transform is a very powerful tool to deal with it, for instance, to check the range condition. We first recall relevant results from Spectral Theory, taken from Sections 5.1 and 5.2 of [FA14]. For a function f ∈ L1 (Rm ) we define its Fourier transform Z 1 ˆ e−i ξ·x f (x) dx, ξ ∈ Rm , (Ff )(ξ) = f (ξ) := m (2π) 2 Rm P where we put ξ · x = m j=1 ξj xj . This formula clearly defines a function Ff : m m R → R which is bounded by (2π)−m/2 kf k1 . Actually, Ff belongs to C0 (Rm ) by Corollary 5.8 in [FA14]. For further investigations the Schwartz space k α Sm = {f ∈ C ∞ (Rm ) | ∀ k ∈ N0 , α ∈ Nm 0 : pk,α (f ) := sup |x|2 |∂ f (x)| < ∞}. x∈Rm turns out to be very useful. By Remark 5.6 of [FA14] the family of seminorms {pk,α | k ∈ N0 , α ∈ Nm 0 } yields a complete metric on Sm . The space Cc∞ (Rm ) and also the Gaussian 2 γ(x) = e−|x|2 are contained in Sm . Proposition 5.10 and Lemma 5.7 of [FA14] 1.4. Examples with the Laplacian 36 show that the restriction F : Sm → Sm is bijective and continuous with the continuous inverse given by Z 1 −1 F g(y) = (Fg)(−y) = ei y·ξ g(ξ) dξ, y ∈ Rm , m (2π) 2 Rm for g ∈ Sm . The crucial fact in our context is Plancherel’s theorem which says that one can extend F : Sm → Sm to a unitary map F2 : L2 (Rm ) → L2 (Rm ) satisfying F2 f = Ff for f ∈ L2 (Rm ) ∩ L1 (Rm ), see Theorem 5.11 in [FA14]. We stress that F2 f is not given by the above integral formula if f ∈ L2 (Rm ) is not integrable; but we still write F instead of F2 and fˆ instead of F2 f . We recall some of the facts proved in Theorem 5.11 and Proposition 5.10 of [FA14]. Let f ∈ L2 (Rm ), h ∈ L1 (Rm ), and ϕ, ψ ∈ Sm . First, we again have the inversion formula F −1 f (y) = Ff (−y) for y ∈ Rm . We define the convolution Z (h ∗ f )(x) = h(x − y)f (y) dy, x ∈ Rm . Rm The function h ∗ f belongs to L2 (Rm ) and satisfies m F(h ∗ f ) = (2π) 2 ĥfˆ, m F(F −1 (h)f ) = (2π)− 2 h ∗ fˆ. (1.22) Moreover, the convolution ϕ ∗ ψ is an element of Sm . To apply the Fourier transform to differential operators, one needs the following properties. Lemma 5.7 of [FA14] yields the differentiation formulas F(∂ α u) = i|α| ξ α Fu ∂ α Fu = (−i)|α| F(xα u) and (1.23) α α for u ∈ Sm and α ∈ Nm 0 , where we write ξ for the map ξ 7→ ξ and so on. Plancherel’s theorem and (1.23) imply the inequalities Z X X X 2 α 2 α 2 kukk,2 = kF∂ uk2 = kξ ûk2 = |ξ α |2 |û|22 dξ ( |α|≤k Rm |α|≤k |α|≤k ≤ c1 (kuk22 + k |ξ|k2 ûk22 ), ≥ c2 (kuk22 + k |ξ|k2 ûk22 ) for u ∈ Sm and constants cj > 0. Taking into account the density of Sm in W k,2 (Rm ), see Remark 1.42, one can then deduce the crucial description W k,2 (Rm ) = {u ∈ L2 (Rm ) | |ξ|k2 û ∈ L2 (Rm )}, ∼ kuk2 + k|ξ|k ûk2 kukk,2 = (1.24) 2 for k ∈ N0 and also that first part of (1.23) is true for u ∈ W |α|,2 (Rm ). Actually, the inclusion ‘⊇’ requires another argument, see Theorem 5.21 in [FA14]. To check the range condition for the Laplacian on Rm , we take f ∈ Sm and λ ∈ C \ R≤0 . We look for a function u ∈ Sm satisfying λu − ∆u = f . (Observe that ∆u belongs to Sm in u ∈ Sm .) Because of formula (1.23), it is equivalent to solve the problem fˆ = λû − m X k=1 i2 ξk2 û = (λ + |ξ|22 )û 1.4. Examples with the Laplacian 37 for u ∈ Sm . The unique solution of this equation is given by û = (λ + |ξ|22 )−1 fˆ, which is an element of Sm by Lemma 5.7 of [FA14]. We now set  fˆ  (1.25) u := R(λ)f = F −1 λ + |ξ|22 Since F is bijective on Sm , this function belongs to Sm ⊆ W 2,2 (Rm ). From (1.23) and the formula for F −1 we thus deduce   λ 2 ˆ − i2 |ξ| fˆ = f. f (1.26) λu − ∆u = F −1 λ + |ξ|22 λ + |ξ|22 (Here we need f ∈ Sm , unless we extend the second part (1.23) to a suitable larger class of functions.) Based on these observations we can now establish our first generation result for the Laplacian. Example 1.46. Let E = L2 (Rm ), A = ∆, and D(A) = W 2,2 (Rm ). The operator A generates a contraction semigroup on E and it is selfadjoint. Moreover, its graph norm is equivalent to that of W 2,2 (Rm ). Proof. The asserted norm equivalence follows from (1.24) and Plancherel’s theorem since F(∆u) = −|ξ|22 û by (1.23) for u ∈ W 2,2 (Rm ). The domain D(A) is dense in E since it contains Cc∞ (Rm ), see Proposition 4.13 of [FA]. Let f ∈ E and λ ∈ C \ R≤0 . To check the range condition, we estimate ( 1 Re λ ≥ 0, fˆ |λ| , ˆ ≤ c | f |, with c := λ λ 1 2 λ + |ξ|2 |Im λ| , Re λ < 0. Because of Plancherel’s theorem, the term in parentheses in (1.25) thus belongs to E, so that we can define u = R(λ)f ∈ E as in (1.25). Using Plancherel once more, we also obtain kuk2 = kûk2 ≤ cλ kf k2 ; i.e., kR(λ)kB(E) ≤ cλ . (1.27) We further compute |ξ|22 |û| = |ξ|22 |(λ + |ξ|22 )−1 | |fˆ| ≤ c′λ |fˆ| for some constants c′λ . Formula (1.24) thus implies that u belongs to W 2,2 (Rm ) with norm kuk2,2 ≤ c(kuk2 + k|ξ|22 ûk2 ) ≤ c̃λ kf k2 . As a result, R(λ) maps E continously into W 2,2 (Rm ). To use (1.26), we take functions fn ∈ Sm tending to f in E as n → ∞. The maps un := R(λ)fn ∈ Sn then converge to u in W 2,2 (Rm ) and satisfy λun −∆un = fn by (1.26). Letting n → ∞, we derive the equation λu−∆u = f so that λI − A is bijective with the bounded inverse R(λ). The spectrum σ(A) is thus contained in R≤0 ,6 and inequality (1.27) implies the Hille-Yosida estimate for λ > 0. As a result, E generates a contraction semigroup on A by Theorem 1.27. Let u, v ∈ W 2,2 (Rm ). Formulas (1.21) and ∆ = div ∇ yield Z Z Z (Au|v) = div(∇u)v dx = − ∇u · ∇v dx = u div(∇v) dx = (u|Av), Rm Rm Rm so that A is symmetric. Since σ(A) ⊆ R≤0 , the selfadjointness of A now follows from Theorem 4.7 of [ST].  6Actually we have the equality σ(A) = R ≤0 by Example 3.40 in [ST]. 1.4. Examples with the Laplacian 38 We stress that the above norm equivalence says that one can bound in L2 (Rm ) each derivative of u ∈ D(A) up to order 2 just by u and the sum ∆u of unmixed second derivatives. In particular, if m ≥ 2 the possible cancellations in ∆u do not play a role! On C0 (Rm ) the situation is quite different. Here we use of the version of the Lumer–Phillips theorem involving the closure. With the available tools we can compute its domain only for m = 1, see the comments below. Example 1.47. Let E = C0 (Rm ), D(A0 ) = {u ∈ C 2 (Rm ) | u, ∆u ∈ E}, and A0 = ∆. The operator A0 has a closure A that generates a contraction semigroup on E. If m = 1, we have Au = u′′ and D(A) = D(A0 ) = C02 (R) := {u ∈ C 2 (R) | u, u′ , u′′ ∈ E}. Proof. The domain of A0 is dense in E because of Cc∞ (Rm ) ⊆ D(A0 ), cf. the proof of Proposition 4.13 in [FA]. Let u ∈ D(A0 ). Example 1.31 says that the functional ϕ = u(x0 )δx0 belongs to J(u), where x0 ∈ Rm satisfies |u(x0 )| = kuk∞ . Setting h = Re(u(x0 )u) ∈ D(A0 ), we obtain RehA0 u, ϕi = Re(u(x0 )∆u(x0 )) = ∆h(x0 ). As in Example 1.34 we see that h(x0 ) is a maximum of h. By Analysis 2, the matrix D2 h(x0 ) is thus negative semidefinite and hence ∆h(x0 ) = tr(D2 h(x0 )) ≤ 0; i.e., A0 is dissipative. Equation (1.26) next shows that the range of I − A0 contains the dense subspace Sm . The first assertion now follows from the Lumer– Phillips Theorem 1.40. Let m = 1 and u ∈ D(A). Since A = A0 there are functions un ∈ D(A0 ) such that un → u and u′′n → Au in E as n → ∞. We further need to control the first derivative. To achieve this aim, we look at an interval J of length |J| > 0, a function v ∈ C 2 (J) with bounded v and v ′′ , δ ∈ (0, |J|), and points r, s ∈ J with δ < s − r < 2δ. Taylor’s theorem provides a number σ ∈ (r, s) such that v(s) = v(r) + v ′ (r)(s − r) + v ′′ (σ) v ′ (r) = The last equation yields (s − r)2 , 2 v(s) − v(r) s−r − v ′′ (σ) . s−r 2 |v ′ (r)| ≤ 2 δ kv ′ k∞ ≤ 2 δ max |v(τ )| + δ max |v ′′ (τ )|, τ ∈[r,r+δ] kvk∞ + δ kv ′′ k∞ . τ ∈[r,r+δ] (1.28) Inserting v = un , we infer that u′n ∈ E. With v = un − um , it also follows that u′n converges in E to a function f in E. As a result, u belongs to C 1 (R) with  u′ = f ∈ E. Using u′′n → Au, we then conclude u ∈ C02 (R) and Au = u′′ . For m ≥ 2 the domain D(A) is not C02 (Rm ). To make this fact plausible, we look at the function ( (x2 − y 2 ) ln(x2 + y 2 ), (x, y) 6= (0, 0), ũ(x, y) = 0, (x, y) = (0, 0). 1.4. Examples with the Laplacian 39 By a straightforward computation, the second derivative ∂xx ũ(x, y) = 2 ln(x2 + y 2 ) + 4x2 (6x2 − 2y 2 )(x2 + y 2 ) − 4x2 (x2 − y 2 ) + x2 + y 2 (x2 + y 2 )2 2 2 −y is unbounded on B(0, 1), but the functions ũ, ∇ũ, and ∆ũ(x, y) = 8 xx2 +y 2 are bounded on B(0, 2). We take a smooth map ϕ with supp ϕ ⊆ B(0, 2) which is equal to 1 on B(0, 1). Then the maps u = ϕũ and u − ∆u = ϕ(ũ − ∆ũ) − 2∇ϕ · ∇ũ − ũ∆ϕ are bounded and have compact support on Rm , but u does not belong to W 2,∞ (Rm ). One can construct an analogous example in C0 (Rm ) instead of L∞ (Rm ) using ln ln. With much more effort and deeper tools, Corollary 3.1.9 in [Lu] shows that the operator A1 = ∆ with domain D(A1 ) = {u ∈ C0 (Rm ) | ∀ p ∈ (1, ∞), r > 0 : u ∈ Wp2 (B(0, r)), ∆u ∈ C0 (Rm )} is closed in E and that ρ(A1 ) contains a halfline (ω, ∞). Since D(A0 ) ⊆ D(A1 ), we first obtain A = A0 ⊆ A1 , and then A = A1 by Lemma 1.24. B) The second derivative on an interval. In the one-dimensional case the equation λu − ∆u = f with boundary conditions becomes an ordinary boundary value problem, which we can solve explicitly and thus obtain a concrete (a bit lenghty) formula for the resolvent. We only look at Dirichlet conditions, others can be treated similarly. We start with the sup-norm case. Example 1.48. Let E = C0 (0, 1), D(A) = {u ∈ C 2 (0, 1) | u, u′′ ∈ E}, and Au = u′′ . The operator A generates a contraction semigroup on E, and its graph norm is equivalent to that of C 2 ([0, 1]). Proof. The equivalence of the norms can be deduced from (1.28). Let f ∈ E. Take ε > 0. As in Example 1.9 we find a map f˜ ∈ Cc (0, 1) with kf − f˜k∞ ≤ ε. Moreover, proceeding as in the proof of Proposition 4.13 in [FA] one constructs a function g ∈ Cc∞ (0, 1) ⊆ D(A) satisfying kf˜ − gk∞ ≤ ε. Hence, A is densely defined. The dissipativity of A is shown as in Example 1.47, where the argument x0 of the maximum of h belongs to (0, 1) since the cases x0 ∈ {0, 1} are excluded by the boundary conditions. Let f ∈ E.√We extend it by 0 to a function f in Cc (R) ⊆ L2 (R). Let λ > 0 and set µ = λ > 0. We then define 1 v = R(λ)f = F −1 ((µ2 +ξ 2 )−1 fˆ) = (2π)− 2 F −1 (µ2 +ξ 2 )−1 ∗f =: k∗f ∈ W 2,2 (R) as in (1.25) and Example 1.46, where we also use (1.22). Using the transformation η = ξ/µ and Example 3.15 of [A4], we compute Z Z Z 1 eisξ eisξ eiµsη 1 1 e−µ|s| k(s) = dξ = dξ = dη = 2π R µ2 + ξ 2 2πµ2 R 1 + ξ 2 µ−2 2πµ R 1 + η 2 2µ for s ∈ [0, 1]. We thus obtain Z 1 1 e−µ|s−τ | f (τ ) dτ, v(s) = 2µ 0 s ∈ [0, 1], 1.4. Examples with the Laplacian 40 recalling that supp f ⊆ [0, 1]. As in Example 1.49 it is easy to check that this function belongs to C 2 ([0, 1]) and solves the equation λv − v ′′ = f even for λ ∈ C \ R≤0 , but it does not satisfy the boundary conditions v(0) = 0 = v(1) except for special f . To fulfill them, we make the ansatz u(s) = a(f, µ)eµs + b(f, µ)e−µs + v(s) for s ∈ [0, 1] and unknown coefficients a(f, µ), b(f, µ) ∈ C. Observe that we still have u ∈ C 2 ([0, 1]) and λu−u′′ = f even for λ ∈ C\R≤0 . We now want to choose a(f, µ) and b(f, µ) such that u ∈ D(A) which means that u(0) = 0 = u(1). This condition is equivalent to the linear system Z −1 1 −µτ a(f, µ) + b(f, µ) = e f (τ ) dτ, 2µ 0 Z −1 1 µ(τ −1) e f (τ ) dτ, a(f, µ)eµ + b(f, µ)e−µ = 2µ 0 which has the unique solution   1 a(f, µ) = −µ b(f, µ) 2µ(e − eµ ) ! R1 e−µ 0 (eµτ − e−µτ )f (τ ) dτ R 1 µ −µτ . − e−µ eµτ )f (τ ) dτ 0 (e e As a result, λI − A is bijective even for λ ∈ C \ R≤0 . The Lumer–Phillips Theorem 1.40 now implies that A is closed and generates a contraction semigroup on E. We also obtain the formula Z 1 1 e−µ|s−τ | f (τ ) dτ (1.29) R(λ, A)f (s) = a(f, µ)eµs + b(f, µ)e−µs + 2µ 0 for s ∈ [0, 1], f ∈ C0 (0, 1), and λ ∈ C \ R≤0 .  We next show the analogous result for Lp (0, 1). Here we check dissipativity on Lp now also for p 6= 2. Example 1.49. Let E = Lp (0, 1), 1 ≤ p < ∞, Au = u′′ , and D(A) = {u ∈ W 2,p (0, 1) | u(0) = u(1) = 0} = W 2,p (0, 1) ∩ W01,p (0, 1). (Remark 1.42 yields W 1,p (0, 1) ֒→ C([0, 1]).) The operator A generates a contraction semigroup on E and its graph norm is equivalent to k · k2,p . Proof. The last assertion follows from Proposition 3.30 of [ST], cf. (1.28). The domain D(A) is dense due to Proposition 4.13 in [FA] since it contains Cc∞ (0, 1). One can extend the operator R(λ, A) from (1.29) to a map R(λ) on E = Lp (0, 1) for λ ∈ C \ R≤0 . We rewrite Z 1 Z s Z 1 e−µτ f (τ ) dτ eµτ f (τ ) dτ + eµs e−µ|s−τ | f (τ ) dτ = e−µs ṽ(s) := s 0 0 for f ∈ E and s ∈ [0, 1]. Using (1.20), we can now differentiate Z s Z 1 ′ −µs µτ µs ṽ (s) = −µe e f (τ ) dτ + f (s) + µe e−µτ f (τ ) dτ − f (s). 0 s 1.4. Examples with the Laplacian 41 Since the summands ±f (s) cancel, ṽ = 2µv belongs to C 1 ([0, 1]). Analogously one checks that v ′′ ∈ Lp (0, 1) satisfies λv − v ′′ = f . As in the previous example one then shows that u = R(λ)f is an element of D(A) and fulfills λu − u′′ = f . To apply the Lumer Phillips theorem it remains to check the dissipativity. To avoid certain technical problems we restrict ourselves to p ∈ [2, ∞), see Example 2.29 for the case p ∈ [1, 2). Let u ∈ D(A). We set w = |u|p−2 u which belongs to J(u) by Example 1.31. Note that v(0) = 0 = v(1) by the boundary conditions. Remark 1.42 yields the embedding W 2,p (0, 1) ֒→ C 1 ([0, 1]). Since p ≥ 2, we can now compute
p−2 2 p−2 −1 d 2 (uu) 2 u = |u|p−4 |u|2 u′ + p−2 (u′ u + uu′ )u w′ = ds 2 (|u| ) = |u|p−4 (|u|2 u′ + (p − 2) Re(uu′ )u.) Formula (1.21) and w(0) = 0 = w(1) now imply Z 1 Z 1 1 ′′ u′ w′ ds + u′ w 0 u w ds = − Re RehAu, wi = Re 0 0 Z 1
|u|p−4 |uu′ |2 + (p − 2)(Re(uu′ ))2 ds =− 0 Z 1
=− |u|p−4 (Im(uu′ ))2 + (p − 1)(Re(uu′ ))2 ds ≤ 0. 0 Theorem 1.40 now implies the assertion, and R(λ) is the resolvent of A.  We add an example where A is dissipative but not a generator and A∗ is not dissipative, cf. Corollary 1.41. This behavior occurs because we impose too many (four) boundary conditions instead of two (for two derivatives) as in the above examples. Example 1.50. Let E = L2 (0, 1), Au = u′′ , and D(A) = W02,2 (0, 1) = {u ∈ W 2,2 (0, 1) | u(0) = u′ (0) = u(1) = u′ (1) = 0}. Then A is closed, densely defined, dissipative, and symmetric, but not a generator, and A∗ is not dissipative. Proof. The density of D(A) follows again from Proposition 4.13 in [FA]. To check closedness, take maps un ∈ D(A) such that un → u and u′′n → v in E as n → ∞. Proposition 3.30 in [ST] then shows that also (u′n ) converges in E, cf. (1.28). From Remark 1.42 we now deduce that u belongs to W 2,2 (0, 1) and un → u in W 2,2 (0, 1). The boundary conditions for un transfer to u via the limits of (un ) and (u′n ) since W 1,2 (0, 1) ֒→ C([0, 1]) by Remark 1.42. Hence, u belongs to D(A) and A is closed. Let u ∈ D(A) and v ∈ W 2,2 (0, 1). Using integration by parts (1.21) and the boundary conditions of u, we compute Z 1 Z 1 Z 1 1 ′ ′ ′ 1 ′′ u v ds + u v 0 = uv ′′ ds + uv ′ 0 = (u|Av) u v ds = − (Au|v) = 0 0 0 for u, v ∈ D(A). Hence, A is symmetric (take v ∈ D(A)) and dissipative (take v = u). Moreover, the operator ∂ 2 with domain W 2,2 (0, 1) is a restriction of A∗ . Let v ∈ D(A∗ ). As in Example 1.43 one can see that A∗ v ∈ E is the second 1.4. Examples with the Laplacian 42 weak derivative of v ∈ E.7 Lemma 3.5 in [ST] yields smooth functions vn such that vn → v and vn′′ → v ′′ in L2 (a, b) for all 0 < a < b < 1. Then vn′ tends in the same sense to a function g ∈ E by Proposition 3.30 in [ST]. From Lemma 3.5 in [ST] we deduce that g is the weak derivative of v, and thus v belongs to W 2,2 (0, 1). It follows A∗ = ∂ 2√with D(A∗ ) = W 2,2 (0, 1). Since ∂ 2 eµs = λeµs for µ = λ and λ ∈ C \ R≤0 , the operator λI − A∗ is not injective. As a result, A∗ is not dissipative in view of Proposition 1.39 and the spectrum of A contains C \ R≤0 by Theorem 1.24 of [ST]. In particular, A is not a generator.  C) The Dirichlet–Laplacian and the wave equation. In many applications one looks at the Laplacian on a domain in R3 . In an L2 context we can show generation properties of this operator, though it is not possible to describe its domain precisely by our means. (This point is discussed below.) We restrict ourselves again to Dirichlet boundary conditions, others are treated in the exercises. The main tool is the Lax–Milgram lemma which is a core consequence of Riesz’ representation of Hilbert space duals.8 Theorem 1.51. Let Y be a Hilbert space and a : Y 2 → C be sesquilinear map which is bounded and strictly accretive; i.e., |a(x, y)| ≤ c kxk kyk and Re a(y, y) ≥ δ kyk2 for all x, y ∈ Y and some constants c, δ > 0. Then for each functional ψ ∈ Y ∗ there is a unique vector z ∈ Y satisfying a(y, z) = ψ(y) for all y ∈ Y . The map ψ 7→ z is antilinear and bounded. Proof. Let y ∈ Y . The map ϕy := a(·, y) belongs to Y ∗ with kϕy k ≤ c kyk. Riesz’ Theorem 3.10 in [FA] yields a unique element Sy of Y satisfying kSyk = kϕy k ≤ c kyk and (· |Sy) = ϕy . Moreover, S : Y → Y is linear. We next estimate δ kyk2 ≤ Re a(y, y) = Re(y|Sy) ≤ |(y|Sy)| ≤ c kyk kSyk, δ c kyk ≤ kSyk ≤ c kyk for every y ∈ Y . As a consequence, S is bounded, injective and has a closed range R(S) by Corollary 4.31 in [FA]. For a vector y ⊥ R(S) we also obtain 0 = (y|Sy) = Re(y|Sy) = Re a(y, y) ≥ δ kyk2 (1.30) so that y = 0. It follows that R(S) = R(S) = Y by Theorem 3.8 in [FA] and so S is invertible with kS −1 k ≤ δc . Let ψ ∈ Y ∗ . There is a unique vector v ∈ Y such that ψ = (·|v) thanks to Riesz’ theorem. The above construction implies the identity a(y, S −1 v) = (y|SS −1 v) = (y|v) = ψ(y) (1.31) for all y ∈ Y . We set z = S −1 v = S −1 T ψ, where T : Y ∗ → Y denotes the antilinear isomorphism from Riesz’ theorem.  7The following argument was omitted in the lectures. 8The next proof was not given in the lectures. 1.4. Examples with the Laplacian 43 We can now construct the Dirichlet–Laplacian ∆D and show its main properties. Here will need Poincaré’s inequality. For any bounded open nonempty set G ⊆ Rm and any p ∈ [1, ∞), there is a constant c = c(G, p) > 0 such that kf kp ≤ c k|∇f |kp (1.32) for all f ∈ W01,p (G), see Corollary D.19 in [ISem]. We set W01,2 (G)∗ =: W −1,2 (G). Since W01,2 (G) is densely embedded into L2 (G) via the inclusion I, Proposition 5.46 in [FA] shows that L2 (G) is densely embedded into W −1,2 (G) with the emdding I ∗ . Here we identify as usual L2 (G) with L2 (G)∗ by means of the Riesz’ isomorphism, but we do not identify W01,2 (G) with W −1,2 (G) (which would require a different Riesz’ isomorphism).9 Example 1.52. Let G ⊆ Rm be open and bounded with boundary ∂G of class C 1 , E = L2 (G), and A0 = ∆ with D(A0 ) = W 2,2 (G) ∩ W01,2 (G). Then A0 is densely defined, symmetric, and dissipative. The operator A0 has an extension ∆D which is selfadjoint, invertible and generates a −δ-contraction semigroup, where δ = 1/c(G, 2) > 0 is given by (1.32). Moreover, [D(∆D )] is densely embedded in W01,2 (G). The domain D(∆D ) contains all maps u ∈ W01,2 (G) for which there is a function f =: ∆D u in L2 (G) such that Z 1,2 ∇v · ∇u dx. ∀ v ∈ W0 (G) : (v|∆D u) = (v|∆D u)L2 = − G The operator ∆D has a bounded invertible extension ∆D : W01,2 (G) → W −1,2 (G) (the weak Dirichlet–Laplacian) which acts as Z 1,2 ∀ u, v ∈ W0 (G) : hv, ∆D uiW 1,2 (G) = − ∇v · ∇u dx. 0 G Proof. The density of D(A0 ) in E again is a consequence of Proposition 4.13 in [FA]. Let u, v ∈ D(A0 ). Using formula (1.21) and v, u ∈ W01,2 (G), we deduce Z Z Z (A0 u|v) = div(∇u)v dx = − ∇u · ∇v dx + (ν · ∇u)v dσ G ∂G ZG Z = u∆v dx − u(ν · ∇v) dσ = (u|A0 v). G Estimate (1.32) similarly yields (A0 u|u) = − ∂G Z G |∇u|2 dx ≤ −δ kuk22 ≤ 0 for δ = 1/c(G, 2) > 0. Hence, A0 is symmetric and dissipative. In order to construct the extension, we introduce the sesquilinear form Z a(u, v) = ∇u · ∇v dx G W01,2 (G). for u, v ∈ The form a is bounded with a constant c by Hölder’s inequality and strictly accretive by (1.32). Let f ∈ L2 (U ). The map g 7→ 9In the lectures some of the following assertions were not stated. 1.4. Examples with the Laplacian R 44 dx defines an element in L2 (G)∗ ֒→ W −1,2 (G). Theorem 1.51 now yields R a unique function uf in W01,2 (G) such that a(v, uf ) = G vf dx for all v ∈ W01,2 (G). We introduce G gf D(Ã) = {u ∈ W01,2 (G) | ∃ c > 0 ∀ v ∈ W01,2 (G) : |a(v, u)| ≤ c kvk2 }. Note that the function uf belongs to D(Ã) with c = kf k2 and that D(Ã) is the set of all u ∈ W01,2 (G) such that the map v 7→ a(v, u) can be extended to an element of L2 (G)∗ . Take u ∈ D(Ã). By Riesz’ Theorem 3.10 in [FA], there then exists a unique function g ∈ L2 (G) such that a(·, u) = (· |g) on W01,2 (G). We then define Ãu = g. Observe that Ãuf = f and so à is surjective. Let v ∈ W01,2 (G) and u ∈ D(A0 ). Using (1.21) and tr v = 0, we compute Z Z Z ∇v · ∇u dx = − a(v, u) = v∆u dx + v(ν · ∇u) dσ = (v| − ∆u); (1.33) G G ∂U i.e., à extends −A0 . Let u, v ∈ D(Ã). Our definitions imply (v|Ãu) = a(v, u) = a(u, v) = (u|Ãv) = (Ãv|u), (−Ãu|u) = −(u|Ãu) = −a(u, u) ≤ −δ kuk22 = −δ (u|u), so that à is symmetric and δI − à is dissipative. Moreover, δI − (δI − Ã) = à is surjective. Thanks to the Lumer-Phillips Theorem 1.40, the operator δI − à generates a contraction semigroup which means that −à generates a −δ–contraction semigroup by Lemma 1.18. In particular, à is invertible, and hence selfadjoint due to Theorem 4.7 in [ST]. We set ∆D = −Ã. To show the other claims, we take u ∈ D(∆D ). Our construction first yields kuk21,2 = kuk22 + a(u, u) = kuk22 − (u|∆D u) ≤ kuk22 + 12 kuk22 + 12 k∆D uk22 so that [D(∆D )] ֒→ W01,2 (G). Since Cc∞ (G) ⊆ D(A0 ) ⊆ D(∆D ), the set D(∆D ) is dense in W01,2 (G). We further compute k∆D ukW −1,2 (G) = ≤ sup v∈W01,2 (G),kvk1,2 ≤1 sup v∈W01,2 (G),kvk1,2 ≤1 |(v|∆D u)| = sup v∈W01,2 (G),kvk1,2 ≤1 |a(v, u)| c kvk1,2 kuk1,2 = c kuk1,2 . We can thus extend ∆D to a bounded map ∆D : W01,2 (G) → W −1,2 (G) given as in the statement, using the density of D(∆D ). Its range is dense because it contains L2 (G). Employing also (1.32), we finally infer k∆D ukW −1,2 (G) kuk1,2 ≥ |hu, ∆D uiW 1,2 (G) | = |a(u, u)| ≥ δ kuk21,2 0 W01,2 (G). u∈ This means that ∆D is injective and has closed range in W −1,2 (G)  by Remark 2.11 in [FA]; i.e., it is invertible. So far we only know that A0 ⊆ ∆D ; i.e., ∆D u = ∆u for u ∈ W 2,2 (G) ∩ W01,2 (G) ⊆ D(∆D ). The equality A0 = ∆D is not true in the above example, in general. If ∂G ∈ C 2 , however, Theorem 6.4 of [Ev] implies that A0 is surjective, and hence A0 = ∆D in this case by Lemma 1.24. We clearly have kuk2 + k∆uk2 ≤ c kuk2,2 for u ∈ D(A0 ). Since A0 is closed in this case, the 1.4. Examples with the Laplacian 45 space [D(A0 )] is complete. The graph norm of A0 = ∆D is thus equivalent to that of W 2,2 (G) by the open mapping theorem, see Corollary 4.29 in [FA]. The next operator will be used to solve the wave equation as explained in Example 2.4. Because of (1.32), we can endow W01,2 (G) with the equivalent scalar product Z (u|v)Y = ∇u · ∇v dx (1.34) G for u, v ∈ W01,2 (G). We write Y for W01,2 (G) with this scalar product. Example 1.53. Let G ⊆ Rm be open and bounded with boundary ∂G of class C 1 and ∆D be given on L2 (G) by Example 1.52. Set E = Y × L2 (G), D(A) = D(∆D ) × Y , and   0 I A= . ∆D 0 Then A is skewadjoint, and thus generates a unitary C0 -group on E due to Stone’s Theorem 1.45. Observe that D(A) and D(∆D ) × Y also have equivalent norms. Proof. Let (u1 , v1 ) and (u2 , v2 ) belong to D(A). We compute Z  
u2


u2

v1 u1 ∇v1 · ∇u2 + ∆D u1 v 2 dx A v 1 v 2 E = ∆ D u1 v 2 E = U Z  
v2

=− v1 ∆D u2 + ∇u1 · ∇v 2 dx = − uv11 ∆D u2 E U using the scalar product of Y and the definition of ∆D . We thus arrive at          u u1 u u2 A 1 −A 2 = v1 v1 v2 v2 E E Hence, −A ⊆ A′ and so iA ⊆ (iA)′ . We define   0 ∆−1 D R= : E → D(∆D ) × Y = D(A), I 0 where ∆−1 D exists thanks to Example 1.52. It is easy to see that AR = I and RAw = w for every w ∈ D(A). Hence, A is invertible so that 0 ∈ ρ(iA) and iA is selfadjoint by Theorem 4.7 in [ST]; i.e., A is skewadjoint.  CHAPTER 2 The evolution equation and regularity 2.1. Wellposedness and the inhomogeneous problem In this section we come back to the relationship between generation properties of A and the solvability of the corresponding differential equation. In a second part we treat inhomogeneous problems in which one adds a given input function to the evolution equation. Let A be a closed operator on X and x ∈ D(A). We study the Cauchy problem or evolution equation u′ (t) = Au(t), t ≥ 0, u(0) = x. (2.1) Recall from Definition 1.10 that a (classical) solution of (2.1) is a function u ∈ C 1 (R≥0 , X) taking values in D(A) and satisfying (2.1) for all t ≥ 0. Observe that then Au belongs to C(R≥0 , X) and thus u to C(R≥0 , [D(A)]). Let the states u(t) ∈ X describe a physical system whose properties are encoded in the operator A and its domain. We then want to predict the future behaviour of the system by means of (2.1). To this aim, we need solutions for ‘many’ initial values x. Moreover, the solutions have to be uniquely determined by x since otherwise we do not really predict the behavior. In addition, one will know the initial value only approximately, so that for a reasonable prediction the solutions should not vary too much under small changes of the data.1 In the next definition we make these requirements precise. Definition 2.1. Let A be closed. The Cauchy problem (2.1) is called wellposed if D(A) is dense in X, if for each x ∈ D(A) there is a unique solution u = u(· ; x) of (2.1), and if the solutions depend continuously on the data; i.e., ∀b > 0 : (D(A), k · kX ) → C([0, b], X); x 7→ u(· ; x), is continuous. (2.2) The next theorem says that for closed A the wellposedness of (2.1) and the generation property of A are equivalent. This fact justifies the definitions made at the beginning of Chapter 1. Theorem 2.2. Let A be a closed operator. It generates a C0 -semigroup T (·) if and only if (2.1) is wellposed. In this case, the function u = T (·)x solves (2.1) for each given x ∈ D(A). Proof. If A generates T (·), then T (·)x is the unique solution of (2.1) according to Proposition 1.11. The solution continuously depends on the initial data since kT (t)k is bounded for t ∈ [0, b] and any fixed b > 0 by Lemma 1.4. 1Actually, the same applies to the dependence on the operator A, but this will be discussed in Section 3.2. 46 2.1. Wellposedness and the inhomogeneous problem 47 Conversely, let (2.1) be wellposed. Let x ∈ D(A) and t ≥ 0. We define the operator T (t) by T (t)x = u(t; x) ∈ D(A) for x ∈ D(A) and t ≥ 0 using uniqueness. Clearly, T (0) = I and T (·)x : R≥0 → X is continuous. For x, y ∈ D(A) and α, β ∈ C, the function αu(· ; x) + βu(· ; y) solves the problem (2.1) with initial value αx + βy since A is linear. Hence, T (t) : D(A) → D(A) is linear for every t ≥ 0. Let t, s ≥ 0 and x ∈ D(A). Then u(s; x) belongs to D(A) so that v(t) := T (t)u(s; x) = T (t)T (s)x for t ≥ 0 is the unique solution of (2.1) with initial value u(s; x). On the other hand, u(t + s; x) = T (t + s)x for t ≥ 0 also solves this problem. Uniqueness then shows that T (t)T (s)x = T (t + s)x. For each b > 0 there is a constant c(b) > 0 such that kT (t)xk ≤ c(b) kxk for all x ∈ D(A) and t ∈ [0, b]. In fact, if this assertion was wrong, there would exist a time b > 0, a sequence (xn ) in D(A), and times tn ∈ [0, b] such that kxn k = 1 and kT (tn )xn k =: cn → ∞ as n → ∞. Set yn = c1n xn ∈ D(A) for every n ∈ N. The initial values yn then tend to 0 as n → ∞, but the norms ku(tn ; yn )k = c1n kT (tn )xn k = 1 do not converge to 0. This contradicts assumption (2.2), and thus T (·) is locally uniformly bounded. Lemma 2.13 of [FA] now allows us to extend each single map T (t) to a bounded linear operator on D(A) = X (also denoted by T (t)) having the same operator norm. Lemma 4.10 in [FA] yields the strong continuity of the family (T (t))t≥0 . By approximation, the semigroup law extends from D(A) to X so that T (·) is a C0 -semigroup. Let B be the generator of T (·). We have A ⊆ B since T (·) solves (2.1). Because D(A) is dense in X and T (t)D(A) ⊆ D(A) for all t ≥ 0, Proposition 1.38 shows that D(A) is dense in [D(B)]. So for each x ∈ D(B) there are vectors xn in D(A) such that xn → x and Axn = Bxn → Bx in X as n → ∞. The closedness of A now implies that x ∈ D(A) and A = B.  We discuss variants of the above result. Remark 2.3. a) One cannot drop condition (2.2) in Theorem 2.2: Let B be a closed, densely defined,
unbounded operator in a Banach space Y . Set X = Y × Y and A = 00 B0 with D(A) = Y × D(B). Observe that A is closed and D(A) is dense in X. For (x, y) ∈ D(A) one has the unique solution u(t) = (x + tBy, y) of (2.1) with u(0) = (x, y). But for t > 0 one cannot continuously extend the map T (t) : (x, y) 7→ u(t) to a map on X since T (t)(0, y) = (tBy, y). b) By Proposition II.6.6 in [EN], problem (2.1) has a unique solution for a closed operator A and each x ∈ D(A) if and only if the operator A1 on X1 = [D(A)] given by A1 x = Ax with D(A1 ) = {x ∈ X1 | Ax ∈ X1 } generates a C0 -semigroup on X1 . Moreover, if ρ(A) 6= ∅ and (2.1) has a unique solution for each x ∈ D(A), then A is a generator on X, see Theorem II.6.7 in [EN]. ♦ We now use Example 1.53 to solve the wave equation. Example 2.4. Let G ⊆ Rm be open and bounded with boundary ∂G of class C 1 . We study the wave equation ∂tt u(t, x) = ∆u(t, x), u(t, x) = 0, u(0, x) = u0 (x), t ≥ 0, x ∈ G, t ≥ 0, x ∈ ∂G, ∂t u(0, x) = u1 (x), (2.3) x ∈ G, 2.1. Wellposedness and the inhomogeneous problem 48 with Dirichlet boundary conditions and given functions (u0 , u1 ). Let ∆D on L2 (G) be given by Example 1.52. We take u0 ∈ D(∆D ) and u1 ∈ Y = W01,2 (G), where Y is endowed with the scalar product from (1.34). We interpret the partial differential equation (2.3) as the second order evolution equation u′′ (t) = ∆D u(t), t ≥ 0, u(0) = u0 , u′ (0) = u1 , (2.4) in L2 (G). Here we look for solutions u ∈ C 2 (R≥0 , L2 (G)) ∩ C 1 (R≥0 , Y ) ∩ C(R≥0 , [D(∆D )]). In particular, the boundary condition in (2.3) is understood in the sense of trace u(t) ∈ W01,2 (G) and the Laplacian in the form sense of Example 1.52. To obtain a first order evolution equation, we set E = Y ×L2 (G), D(A) = D(∆D ) × Y , and   0 I A= . ∆D 0 From Example 1.53 we know that A generates a unitary C0 -group T (·) on E. We claim that (2.4) has a solution u if and only if the problem (2.1) on E for the above A and w0 = (u0 , u1 ) ∈ D(A) has a solution w = (w1 , w2 ), which is then given by w = (u, u′ ). To show the claim, let w solve (2.1) for A. The function u := w1 then belongs to C 1 (R≥0 , Y ) ∩ C(R≥0 , [D(∆D )]) and w2 to C 1 (R≥0 , L2 (G)) ∩ C(R≥0 , Y ). Equation (2.1) for A also yields that u′ = w1′ = w2 so that u is also an element of C 2 (R≥0 , L2 (G)) and satisfies u′′ = w2′ = ∆D w1 = ∆D u as required. Conversely, if u solves (2.4), then we set w = (u, u′ ). This function is contained in C(R≥0 , [D(A)]) ∩ C 1 (R≥0 , E) and fulfills  ′  ′  u u = w′ = = Aw. u′′ ∆D u For each (u0 , u1 ) ∈ D(∆D ) × Y we thus have a unique solution u of (2.4). ♦ Inhomogeneous evolution equations. To the problem (1.1) we now add a given function f : J → X on a time interval J with inf J = 0. In view of applications to nonlinear problems, cf. [Lu], we allow here for very general J. In our linear problem, f can model a force in a wave equation or a source-sink term in a diffusion problem. We require that f ∈ C(J, X) satisfies Z δ kf (s)k ds < ∞ for some δ ∈ J \ {0}. (2.5) 0 Observe that in this case f satisfies (2.5) for all δ ∈ J \{0} and that (2.5) is true for closed J by continuity. Let A generate the C0 -semigroup T (·) and u0 ∈ X. We study the inhomogeneous evolution equation u′ (t) = Au(t) + f (t), t ∈ J, u(0) = u0 . (2.6) Our first solution concept is similar to the homogeneous case in Definition 1.10, where we set J ′ = J ∪ {0} and require continuity of u at t = 0 in view of the initial condition. Definition 2.5. A map u : J ′ → X is a (classical) solution of (2.6) on J if u belongs to C 1 (J, X) ∩ C(J ′ , X), u(t) ∈ D(A) for t ∈ J, and u satisfies (2.6). 2.1. Wellposedness and the inhomogeneous problem 49 Again a solution is contained in C(J, [D(A)]). We first show uniqueness of such solutions and that they are given by Duhamel’s formula (2.7). If 0 ∈ / J, in (2.7) one uses an X-valued improper Riemann integral which is defined as in Analysis 1 and which exists if the norm of the integrand is integrable. Proposition 2.6. Let A generate the C0 -semigroup T (·), u0 ∈ X, and f ∈ C(J, X) satisfy (2.5). If u solves (2.6) on J, then u is given by Z t T (t − s)f (s) ds, t ∈ J. (2.7) u(t) = T (t)u0 + 0 In particular, solutions of (2.6) are unique. Proof. Let t ∈ J \ {0} and set v(s) = T (t − s)u(s) for 0 ≤ s ≤ t, where u solves (2.6) on J. As in the proof of Proposition 1.11 and using (2.6), one shows that v is continuously differentiable with derivative v ′ (s) = T (t − s)u′ (s) − T (t − s)Au(s) = T (t − s)f (s) for all 0 < s ≤ t. Let ε ∈ (0, t). By integration we deduce Z t T (t − s)f (s) ds = v(t) − v(ε) = u(t) − T (t − ε)u(ε). ε Condition (2.5) and Lemma 1.4 yield the bound kT (t − s)f (s)k ≤ M eω+ t kf (s)k whose right hand side is integrable by (2.5). So we can let ε → 0 in the above integral. Lemma 1.13 and (2.6) further imply that T (t − ε)u(ε) → T (t)u0 .  Note that Duhamel’s formula (2.7) defines a function u for all x ∈ X and f ∈ C(J, X) with (2.5). One can thus ask whether u still solves the equation (2.6) for such data. In the present setting, this is not true in general as the next example shows, but we continue to discuss this point in the following section. d with D(A) = C01 (R), and ϕ ∈ X \ Example 2.7. Let X = C0 (R), A = ds C 1 (R). The operator A generates the C0 -group T (·) given by T (t)g = g(· + t). The function T (t)ϕ then does not belong to D(A) for each t ≥ 0 and for some s ∈ R the map t 7→ (T (t)ϕ)(s) = ϕ(s + t) is not differentiable. Define f ∈ C(R, X) by f (t) = T (t)ϕ and let x = 0. Formula (2.7) then yields Z t T (t − r)T (r)ϕ dr = tT (t)ϕ, t ≥ 0. u(t) = 0 So u does not solve (2.6) as u(t) ∈ / D(A) and u is not differentiable for t > 0. ♦ We now show criteria on f implying that Duhamel’s formula (2.7) provides a solution of (2.6). We start with the core step that says that time and ‘space’ regularity are equivalent. Lemma 2.8. Let A generate the C0 -semigroup T (·), u0 ∈ D(A), and f ∈ Rt C(J, X) satisfy (2.5). Define v(t) = 0 T (t − s)f (s) ds for t ∈ J and v(0) = 0 if 0 ∈ / J. Then the following assertions are equivalent. a) v ∈ C 1 (J, X). b) v(t) ∈ D(A) for all t ∈ J and Av ∈ C(J, X). In this case, (2.7) gives the unique solution of (2.6) on J. If (2.6) has a solution on J, then properties a) and b) are true. 2.1. Wellposedness and the inhomogeneous problem 50 Proof. By Proposition 1.11, the orbit T (·)u0 belongs to C 1 (R≥0 , X) ∩ d T (t)u0 = AT (t)u0 for all t ∈ J, since u0 ∈ C(R≥0 , [D(A)]) with derivative dt D(A). Let u solve (2.6). We then deduce v = u − T (·)x from Proposition 2.6, so that v satisfies properties a) and b). Proposition 2.6 yields uniqueness. Let a) or b) be valid. It remains to show that v solves (2.6) with u0 = 0, since then u defined by (2.7) is a solution of (2.6) for the given initial value Rt u0 . We first note that kv(t)k ≤ M eω+ t 0 kf (s)k ds tends to 0 as t → 0 since s 7→ kf (s)k is integrable near 0 by (2.5). It is then easy to check that v : J ′ → X is continuous, for instance using dominated convergence. We next fix t ∈ J and take h 6= 0 such that t + h ∈ J. We compute Z 1 1 1 t+h T (t + h − s)f (s) ds D1 (h) := (T (h) − I)v(t) = (v(t + h) − v(t)) − h h h t =: D2 (h) − I(h). Since f ∈ C(J, X), it follows Z  1 t+h  kI(h) − f (t)k = T (t + h − s)f (s) − f (t) ds h t ≤ max kT (t + h − s)f (s) − f (t)k −→ 0 |s−t|≤|h| as h → 0, thanks to Lemma 1.13. As a result, D1 (h) converges if and only if D2 (h) converges as h → 0. The convergence of D1 means that v(t) ∈ D(A) and D1 (h) → Av(t) as h → 0, and that of D2 is equivalent to the differentiability of v at t with D2 (h) → v ′ (t) as h → 0. We further obtain that Av(t) = v ′ (t)−f (t); i.e., v satisfies the differential equation in (2.6) for this t. For each t ∈ J the properties a) and b) imply the convergence of D2 and D1 , respectively, and hence the function v solves (2.6) with u0 = 0.  The next theorem is the fundamental existence result for the inhomogeneous evolution equation (2.6). For simplicity, we restrict ourselves to closed J. Theorem 2.9. Let A generate the C0 -semigroup T (·), u0 ∈ D(A), and J be closed. Assume either that f ∈ C 1 (J, X) or that f ∈ C(J, [D(A)]). Then the function u given by (2.7) is the unique solution of (2.6) on J. Proof. Since J is closed, f satisfies condition (2.5) and so uniqueness Rt follows from Proposition 2.6. Let f ∈ C 1 (J, X). Writing v(t) = 0 T (s)f (t − s) ds for t ∈ J, we see that v has the continuous derivative Z t T (s)f ′ (t − s) ds v ′ (t) = T (t)f (0) + 0 as in Analysis 2 or Remark 1.16 f). Hence, property a) in Lemma 2.8 is satisfied. Let f ∈ C(J, [D(A)]). Proposition 1.11 and Lemma 1.13 imply that the vector T (t−s)f (s) belongs to D(A) and the map (t, s) 7→ AT (t−s)f (s) = T (t−s)Af (s) is continuous in X for s ≤ t in J. Remark 1.16 d) now yields that v(t) belongs Rt to D(A) and Av(t) = 0 T (t − s)Af (s) ds. As in Analysis 2 one then checks that Av is an element of C(J, X), and so statement b) of Lemma 2.8 is fulfilled. The theorem is now follows from Lemma 2.8.  2.2. Mild solution and extrapolation 51 Variants for more regular solutions are discussed in the exercises. We apply the above result to the wave equation with a given force.2 Example 2.10. In the setting of Example 2.4, we consider the inhomogeneous wave equation ∂tt u(t, x) = ∆u(t, x) + g(t, x), u(t, x) = 0, u(0, x) = u0 (x), t ≥ 0, x ∈ ∂G, t ≥ 0, x ∈ G, (2.8) x ∈ G, ∂t u(0, x) = u1 (x), for given u0 ∈ D(∆D ), u1 ∈ Y = W01,2 (G) and g ∈ C(R≥0 , L2 (G)), where we set g(t, x) = (g(t))(x) for all t ≥ 0 and almost every x ∈ G. As in Example 2.4 we write these equations as u′′ (t) = ∆D u(t) + g(t), C 2 (R t ≥ 0, u(0) = u0 , , L2 (G)) u′ (0) = u1 , (2.9) C 1 (R and look for solutions u ∈ ∩ ≥0 ≥0 , Y ) ∩ C(R≥0 , [D(∆D )]). Again the second order problem is equivalent to the first order problem w′ (t) = A(t)w(t) + f (t), on E = Y × A= L2 (U )  0 ∆D (u, u′ ), t ≥ 0, with w =  I on D(A) = D(∆D ) × Y, 0 w0 = (u0 , u1 ), and   0 f= . g In view of Theorem 2.9 and Example 1.53, we obtain a unique solution u of (2.9) if either g belongs to C 1 (R≥0 , L2 (G)) (then f ∈ C 1 (R≥0 , E)) or to C(R≥0 , Y ) (then f ∈ C(R≥0 , [D(A)])). ♦ 2.2. Mild solution and extrapolation So far we have considered solutions of (2.1) or (2.6) taking values in D(A), which is surely a natural choice. However, in many situations one wants to admit solutions and initial values in X. For instance, in the wave equation from Examples 2.4 and 2.10 the squared norm of the state space E is (up to factors) equal to the physical energy, and it is often desirable only to require that the solutions have finite energy. We first introduce a concept that is motivated by Proposition 2.6 and which plays an important role for certain nonlinear evolution equations. Let J be an interval with inf J = 0 and J ′ = J ∪ {0}. Definition 2.11. Let A generate the C0 -semigroup T (·), u0 ∈ X, and f ∈ C(J, X) satisfy (2.5). The function u ∈ C(J ′ , X) given by Z t T (t − s)f (s) ds, t ∈ J ′, u(t) = T (t)u0 + 0 is called mild solution (on J ′ ) of (2.6). The continuity of the mild solution was shown in the proof of Lemma 2.8. This definition has the obvious draw-back that one does not directly see the connection to A and to (2.6). For f = 0, Lemma 1.19 suggests the following notion which involves A explicitly. 2The function g in (2.8) corresponds to a force if the mass density of the vibrating object is equal to 1. 2.2. Mild solution and extrapolation 52 Definition 2.12. Let A generate the C0 -semigroup T (·), u0 ∈ X, J be closed, and f ∈ C(J, X). A function u ∈ C(J, X) is called an integrated solution (on Rt J) of (2.6) if the integral 0 u(s) ds belongs to D(A) and satisfies Z t Z t u(t) = u0 + A u(s) ds + f (s) ds (2.10) 0 0 for all ∈ J. The questions arise whether integrated solutions are unique, how they relate to mild ones, R t and whether they solve a differential equation. At least, the function t 7→ 0 u(s) ds is differentiable though in X instead of [D(A)]. Moreover, for mild solutions it is not clear at all how to differentiate t 7→ T (t − s)f (s). The key idea to solve these problems is to enlarge the state space X suitably. Definition 2.13. Let A be closed operator with µ ∈ ρ(A). We define the extrapolated norm kxk−1 = kR(µ, A)xk for x ∈ X and the extrapolation space A as the completion of (X, k · k ). X−1 = X−1 −1 Here is k · k−1 a coarser norm on X than the original one (which is not complete if A is unbounded). We recall from Section 2.2 D) of [FA] that for a normed vector space Y there exists the completion Ỹ . It is a Banach space such that there is an isometry J : Y → Ỹ with dense range which is unique up to isometric isomorphisms. We thus identify X with a dense subspace of X−1 . The norm k · k−1 does not depend on the choice of µ ∈ ρ(A) (up to equivalence): Let λ ∈ ρ(A) \ {µ}. Using the resolvent equation (1.7), we compute kR(λ, A)xk ≤ kR(µ, A)xk + |µ − λ| kR(λ, A)R(µ, A)xk ≤ (1 + |µ − λ| kR(λ, A)k) kR(µ, A)xk, (2.11) and one can interchange λ and µ here. In Example 2.17 we compute X−1 in one case. But actually one can quite often use X−1 to ‘legalize illegal computations’ without knowing a precise description of it. The next result shows that we can extend the C0 -semigroup A keeping many of its properties. generated by A to X−1 Proposition 2.14. Let A generate the C0 -semigroup T (·) on X. For t ≥ 0, A , which form the operators T (t) have a bounded extension T−1 (t) to X−1 = X−1 a C0 -semigroup on X−1 . It is generated by the extension A−1 ∈ B(X, X−1 ) of A, where D(A−1 ) = X, and k · kX is equivalent to the graph norm of A−1 . Moreover, for all λ ∈ ρ(A) the resolvent R(λ, A) has an extension in B(X−1 , X) which is the resolvent of A−1 . The maps R := R(µ, A−1 ) : X−1 → X and R−1 = µI − A−1 : X → X−1 are isometric isomorphisms satisfying A = RA−1 R−1 on D(A), and thus σ(A) = σ(A−1 ). Analogous facts are true for R(λ, A) and T (t). Proof. 1) Let λ ∈ ρ(A) and x ∈ X. By estimate (2.11) we have kR(λ, A)xk ≤ cλ kxk−1 for a constant cλ . Because X is dense in X−1 , we can extend R(λ, A) to a map Rλ in B(X−1 , X) using Lemma 2.13 in [FA]. We note that Rµ is an isometry. For x ∈ D(A) we have kAxk−1 = k(A − µI + µI)R(µ, A)xkX ≤ (1 + |µ| kR(µ, A)k)kxk 2.2. Mild solution and extrapolation 53 so that A has an extension A−1 ∈ B(X, X−1 ). The identity IX = (λIX1 − A)R(λ, A) on X can thus be extended to IX−1 = (λIX − A−1 )Rλ on X−1 , and analogously one obtains IX = Rλ (λIX − A−1 ) on X. This means that λ ∈ ρ(A−1 ) and Rλ = R(λ, A−1 ). We next compute R(µ, A−1 )A−1 (µI − A)x = A−1 R(µ, A)(µI − A)x = Ax for x ∈ D(A), obtaining that A and A−1 are similar. It follows that σ(A) = σ(A−1 ) since R(λI − A−1 )R−1 = λI − A on D(A). Using X ֒→ X−1 , we show the asserted norm equivalence by kxkA−1 = kxk−1 + kA−1 xk−1 ≤ c kxkX + kA−1 k kxkX , kxkX = kRR−1 xkX = kµx − A−1 xk−1 ≤ max{|µ|, 1} kxkA−1 . 2) It is easy to see that A−1 = R−1 AR with D(A−1 ) = X generates the C0 semigroup on X−1 given by T−1 (t) := R−1 T (t)R for t ≥ 0, cf. Paragraph II.2.1 in [EN]. This semigroup extends T (·) since we have T−1 (t) = (µI − A)T (t)R(µ, A)x = T (t)x for x ∈ X. The other assertions are shown similarly.  Part 1) of the proof also works if one only assumes that A is closed with µ ∈ ρ(A). Using these concepts and results, we can now easily show that mild and integrated solutions coincide and that they are just the unique (classical) solutions in X−1 of the extrapolated problem u′ (t) = A−1 u(t) + f (t), t ∈ J, u(0) = u0 ∈ X. (2.12) Proposition 2.15. Let A generate the C0 -semigroup T (·) on X, u0 ∈ X, J be closed, and f ∈ C(J, X). Then the mild solution u ∈ C(J, X) given by (2.7) also belongs to C 1 (J, X−1 ) and u is the (classical) solution of (2.12) in X−1 . It is also the unique integrated solution of (2.6) in the sense of (2.10). Proof. The first assertion follows from Theorem 2.9 and Proposition 2.14 employing that X = D(A−1 ) and T−1 (t)|X = T (t). Let u ∈ C(J, X) be the (unique) solution of (2.12). Integrating this differential equation, we derive the identity Z t Z t f (s) ds A−1 u(s) ds + u(t) − u0 = 0 0 for t ∈ J. We can take A−1 ∈ B(X, X−1 ) out of the integral, resulting in Z t Z t Z t f (s) ds. u(s) ds − u(s) ds = u(t) − u0 − µ (A−1 − µI) 0 0 0 Since the right hand side belongs to X and R(µ, A−1 ) extends R(µ, A), the Rt integral 0 u(s) ds thus belongs to D(A) and u is an integrated solution of (2.1). Let u ∈ C(J, X) be an integrated solution of (2.12). As A−1 ∈ B(X, X−1 ), Rt we can differentiate t 7→ A 0 u(s) ds in X−1 with derivative A−1 u(t). Equation (2.10) then implies that u is contained in C 1 (J, X−1 ) and solves (2.12).  2.2. Mild solution and extrapolation 54 For any Banach space X we have the isometry JX : X → X ∗∗ ; JX (x) = hx, · iX×X ∗ , see Proposition 5.24 of [FA]. The space X is called reflexive if JX is surjective. By Example 5.27 in [FA], a Hilbert space X is reflexive with JX = ΦX ∗ ΦX for the Riesz isomorphisms. In the reflexive case one can describe the extrapolation ∗ instead of (X )∗ .3 by duality in a convenient way. We write X−1 −1 Proposition 2.16. Let A be closed with µ ∈ ρ(A). Then there is an isomor∗ satisfying (Ψx∗ )(x) = hx, x∗ i phism Ψ : [D(A∗ )] → X−1 X×X ∗ for x ∈ X ֒→ X−1 ∗ ∗ and x ∈ D(A ). Let also X be reflexive. Then there is an isomorphism Φ : X−1 → [D(A∗ )]∗ extending JX : X → X ∗∗ . Proof. Replacing A − µI by A we can restrict ourselves to the case µ = 0. Let x∗ ∈ D(A∗ ). For x−1 ∈ X−1 we set We first observe that ∗ ∗ (Ψx∗ )(x−1 ) = hA−1 −1 x−1 , A x iX×X ∗ . ∗ ∗ ∗ |(Ψx∗ )(x−1 )| ≤ kA−1 −1 x−1 kX kA x kX ∗ ≤ kx−1 kX−1 kx kA∗ , ∗ with norm less or equal kx∗ k ∗ and hence Ψ : so that Ψx∗ belongs to X−1 A ∗ is a linear contraction. Since A−1 extends A−1 on X, this map [D(A∗ )] → X−1 −1 acts as (Ψx∗ )(x) = hx, x∗ iX×X ∗ for x ∈ X. ∗ . Let x ∈ X. We then To show surjectivity, we take a functional ϕ ∈ X−1 estimate ∗ kA−1 xkX ∗ |ϕ(A−1 x)| ≤ kϕkX−1 −1 = kϕkX−1 kxkX , and hence ϕ ◦ A−1 is contained in X ∗ . There thus exists an element y ∗ of X ∗ ∗ . We set such that ϕ(A−1 x) = hx, y ∗ iX for all x ∈ X and ky ∗ kX ∗ ≤ kϕkX−1 ∗ ∗ −1 ∗ ∗ ∗ x = (A ) y ∈ D(A ) recalling that σ(A ) = σ(A) by Theorem 1.24 of [ST]. It follows A∗ x∗ = y ∗ and ∗ . kx∗ kA∗ = k(A∗ )−1 A∗ x∗ kX ∗ + kA∗ x∗ kX ∗ ≤ c ky ∗ kX ∗ ≤ c kϕkX−1 Moreover, the definitions of Ψ and y ∗ yield −1 ∗ ∗ (Ψx∗ )(x−1 ) = hA−1 −1 x−1 , A x iX = ϕ(A−1 A−1 x−1 ) = ϕ(x−1 ) for all x−1 ∈ X−1 ; i.e., ϕ = Ψx∗ . We have thus shown the surjectivity of Ψ. It is also injective by the above lower estimate. So Ψ is invertible by the open mapping theorem 4.28 in [FA]. Let X be reflexive so that also the isomorphic space X−1 is reflexive, see Corollary 5.51 in [FA]. We then define the isomorphism Φ = Ψ∗ JX−1 : X−1 → [D(A∗ )]∗ . For x ∈ X and x∗ ∈ D(A∗ ) we compute ∗ hx∗ , ΦxiD(A∗ ) = hΨx∗ , JX−1 xiX−1 = hx, Ψx∗ iX−1 = hx, x∗ iX = hx∗ , JX xiX ∗ , using the above properties. This shows the last assertion.  By extrapolation we now obtain solutions u of the wave equation (2.9) such that (u(t), u′ (t)) only take values in the space W01,2 (G) × L2 (G) of finite energy. 3This proposition was stated in a less precise way in the lectures. 2.3. Analytic semigroups and sectorial operators 55 Example 2.17. As in Examples 2.4 and 2.10 we study the wave equation (2.9) but now with data (u0 , u1 ) ∈ W01,2 (G) × L2 (G) and g ∈ C(J, L2 (G)). We thus look for solutions u in Z := C(J, W01,2 (G)) ∩ C 1 (J, L2 (G)) ∩ C 2 (J, W −1,2 (G)) ˜ D : W 1,2 (G) → W −1,2 (G) from Examusing the weak Dirichlet–Laplacian ∆ 0 ple 1.52, where we include a tilde for a moment. Again we look at the first order formulation of (2.9) on E = W01,2 (G)×L2 (U ) with w = (u, u′ ),     0 0 I and f= . A= on D(A) = D(∆D ) × W01,2 (G), ∆D 0 g Example 1.53 provides the inverse   0 ∆−1 −1 D A = : E → D(A). I 0 ˜ D : W 1,2 (G) → From Example 1.52 we obtain the invertible extension ∆ 0 ˜ −1 extends ∆−1 , take ϕ ∈ L2 (G) ֒→ W −1,2 (G). W −1,2 (G) of ∆D . To see that ∆ D D ˜ −1 ϕ ∈ Y and v = ∆−1 ϕ ∈ D(∆D ) both satisfy ∆ ˜ D ṽ = ϕ and The maps ṽ = ∆ D D ˜ D v = ∆D v = ϕ, so that ṽ = v as ∆ ˜ D is injective. We now drop the tilde. ∆ For (u, v) ∈ E we thus obtain k(u, v)kE−1 = k(∆−1 D v, u)kE = k(u, v)kF where we set F = L2 (G) × W −1,2 (G). This means that E−1 = F since we identify E with a dense subspace of E−1 . The extension of A to E is given by   0 I A−1 = : E → L2 (G) × W −1,2 (G). ∆D 0 It generates a C0 -semigroup on E−1 = L2 (G) × W −1,2 (G) by Proposition 2.14. We thus have a unique solution w of (2.12) in E−1 for our data. As in Examples 2.4 and 2.10, one now obtains a unique solution u ∈ Z of (2.9) given by w = (u, u′ ). ♦ 2.3. Analytic semigroups and sectorial operators So far we have treated C0 -semigroups and groups without requiring further properties of them. However, both from the view point of applications and from a more theoretical perspective, it is natural and rewarding to study classes of C0 -semigroups with specific properties. (In [EN] such questions are treated in detail.) For instance, compact semigroup or resolvent operators often occur in concrete problems, and they have special properties, of course. If the Banach space X carries an order structure (e.g., X = Lp (µ) or X = C0 (G)), then ‘positive’ semigroups preserving the order are important, and they are used to describe diffusion or transport phenomena. Compactness does not play a role below, but occasionally we will come back to positivity later in the course. Another possible property of C0 -semigroups T (·) is the improved regularity of the map R+ ∋ t 7→ T (t) beyond strong continuity.4 In this section we study the strongest case in this context, namely analyticity of the map R+ → B(X); t 7→ T (t). This class turns out to be of great importance in applications to diffusion 4Recall from the exercises that the generator A is bounded if T (t) → I in B(X) as t → 0. 2.3. Analytic semigroups and sectorial operators 56 problems, for instance. We first introduce and discuss a class of operators which is crucial to determine the generators of such ‘analytic semigroups.’ Let φ ∈ (0, π]. We write Σφ = {λ ∈ C \ {0} | | arg λ| < φ} for the open sector with (half) opening angle φ. Observe that Σπ/2 = C+ is the open right halfplane and Σπ = C \ R≤0 is the plane with cut R≤0 . Definition 2.18. A closed operator A is called sectorial of type (K, φ) if for some constants φ ∈ (0, π] and K > 0 the sector Σφ belongs to ρ(A) and the resolvent satisfies the inequality kR(λ, A)k ≤ K |λ| for all λ ∈ Σφ . (2.13) The supremum ϕ(A) = ϕ ∈ (0, π] of all such φ is called the angle of A.5 Often we will look at maps A such that the shifted operator A−ωI is sectorial for some ω ∈ R, which can be treated by rescaling arguments. Clearly, if A is sectorial with angle ϕ, then it has type (Kφ , φ) for all φ ∈ (0, ϕ). Typically, Kφ becomes unbounded as φ → ϕ as we will see below in several examples. We also note that several variants of the above concepts are used in literature; in particular many authors consider operators whose resolvent set contains a sector opening to the left. We first discuss several relatively simple examples which are typical nevertheless, starting with the arguably ‘nicest’ class of operators. Example 2.19. Let X be a Hilbert space and A be closed, densely defined, and selfadjoint on X satisfying σ(A) ⊆ R≤0 . Then A is sectorial of angle π. Proof. Let φ ∈ ( π2 , π) and λ ∈ Σφ ⊆ ρ(A). Since R(λ, A)′ = R(λ, A) by (4.2) in [ST], the operator R(λ, A) is normal. Propositions 4.3 and 1.20 of [ST] then yield ( 1 , Re λ ≥ 0, 1 1 ≤ = |λ|1 kR(λ, A)k = r(R(λ, A)) = d(λ, σ(A)) d(λ, R≤0 ) |Im λ| , Re λ < 0. If Re λ < 0, we can write λ = |λ| e±iθ for some θ ∈ ( π2 , φ). Elementary properties λ| of sine thus imply |Im |λ| = sin θ ≥ sin φ > 0, and hence kR(λ, A)k ≤ 1 sin φ |λ| =: Kφ |λ| for all λ ∈ Σφ .  2 d 2 Example 2.20. Let X = C([0, 1]), A = ds 2 , and D(A) = {u ∈ C ([0, 1]) | u(0) = u(1) = 0}. The closure of D(A) is X0 = C0 (0, 1). We set A0 u = u′′ for u ∈ D(A0 ) = {u ∈ D(A) | u′′ ∈ X0 }. The operators A in X and A0 in X0 are sectorial of angle π. Proof. The closure of D(A) can be determined as in Example 1.48. We only treat A, as A0 is handled similarly. Let f ∈ X and λ ∈ Σπ so that λ = µ2 5In the lectures we used slightly different concepts. 2.3. Analytic semigroups and sectorial operators 57 for some µ ∈ C+ . Formula (1.29) in Example 1.48 implies that λ ∈ ρ(A) and Z 1 1 µs −µs e−µ|s−τ | f (τ ) dτ R(λ, A)f (s) = a(f, µ)e + b(f, µ)e + 2µ 0 for s ∈ [0, 1] and the coefficients   1 a(f, µ) = −µ b(f, µ) 2µ(e − eµ ) ! R1 e−µ 0 (eµτ − e−µτ )f (τ ) dτ R 1 µ −µτ . − e−µ eµτ )f (τ ) dτ 0 (e e Fix φ ∈ ( π2 , π). Take λ = µ2 ∈ Σφ and hence µ ∈ Σφ/2 . Let θ = arg µ. It follows 0 ≤ |θ| < φ2 and Re µ = |µ| cos θ ≥ |µ| cos φ2 . So we can estimate Z s kf k∞ kR(λ, A)f k∞ ≤ |a(f, µ)| eRe µ + |b(f, µ)| + sup e− Re µ |τ | dτ 2 |µ| s∈[0,1] s−1 Z 1
kf k∞ eRe µτ + e− Re µτ dτ ≤ µ −µ 2|µ| |e − e | 0 Z 1
 kf k∞ eRe µ e− Re µτ + e− Re µ eRe µτ dτ + + |µ| Re µ 0 kf k∞ (eRe µ − 1 + 1 − e− Re µ ) = 2|µ| Re µ |eµ − e−µ |
kf k∞ + eRe µ (1 − e− Re µ ) + e− Re µ (eRe µ − 1) + |µ| Re µ 1   (eRe µ − e− Re µ ) + (eRe µ − e− Re µ ) cos(φ/2) kf k + 1 ≤ ∞ |µ|2 2(eRe µ − e− Re µ ) = 2 cos(φ/2) |λ| kf k∞ .  Example 2.21. Let X = C0 (R) and Au = u′ for D(A) = C01 (R). Then A is sectorial of angle π2 . (The analogous result for X = Lp (R) is shown in Example 5.10 of [ST].) Proof. By Example 1.22, we have σ(A) = iR and kR(λ, A)k = 1/ Re λ for λ ∈ C+ . Take φ ∈ (0, π/2). Let λ ∈ Σφ . We obtain Re λ ≥ |λ| cos φ and hence kR(λ, A)k ≤ 1 cos φ |λ| , which shows sectoriality of angle greater or equal π/2. Since iR ⊆ σ(A) the angle cannot be greater that π/2.  To study analytic semigroup we need a bit of complex analysis in Banach spaces, see also Section 5.1 of [ST]. Let Y be a Banach space, J ⊆ R be a closed interval, and γ : J → Y be piecewise C 1 . If J = [a, b] and γ(a) = γ(b), the curve γ is called closed. We set Γ = γ(J). For f ∈ C(J, Y ) we introduce the complex contour integral Z Z f dz = f (γ(s))γ ′ (s) ds. γ J 2.3. Analytic semigroups and sectorial operators 58 If J is not compact, here it is assumed that the right-hand side exists as an improper Riemann integral in Y . As in the proof of Proposition 1.21, one sees that this improper integral exists if the function kf ◦ γk |γ ′ | is integrable on J. Let U ⊆ C be open and starshaped, f be complex differentiable on U , Γ be closed, and z ∈ U \ Γ. We then have Cauchy’s theorem Z f (w) dw = 0, (2.14) Γ and Cauchy’s formula Z f (w) 1 dw, n(Γ, z)f (z) = 2πi Γ w − z where 1 n(Γ, z) = 2πi Z Γ dw . w−z (2.15) In fact, by Theorems 2.6 and 2.8 and formula (3.3) in [A4] these equations are f replaced by hf, x∗ i for each x∗ ∈ X ∗ . We hence obtain R true with ∗ = 0 for every x∗ ∈ X ∗ , implying (2.14) due to the Hahn-Banach Γ f dz, x theorem. Formula (2.15) is shown similiarly. If Y = C, identity (2.15) yields Z 1 za e = eλz (λ − a)−1 dλ for a ∈ C and z ∈ C. (2.16) 2πi ∂B(a,1) We want to imitate this formula for sectorial A. To this aim, we need a curve Γ encircling the (typically unbounded) spectrum of A counterclockwise. This curve has to be contained in Σφ for some φ < ϕ(A) in order to use the resolvent estimate (2.13). As a result, Γ has to be unbounded. In view of the occuring exponential function, the real part of λ ∈ Γ has to tend to −∞ to guarantee the convergence of the integral. We thus assume that A is sectorial with angle ϕ larger than π/2. For given numbers R > r > 0 and θ ∈ (π/2, ϕ) we define the paths Γ1 = Γ1 (r, θ) = {λ = γ1 (s) = −se−iθ | s ∈ (−∞, −r]}, Γ2 = Γ2 (r, θ) = {λ = γ2 (s) = reiα | α ∈ [−θ, θ]}, Γ3 = Γ3 (r, θ) = {λ = γ3 (s) = seiθ | s ∈ [r, ∞)}, Γ = Γ(r, θ) = Γ1 ∪ Γ2 ∪ Γ3 , (2.17) ΓR = Γ ∩ B(0, R). R We write Γ instead of γ since the maps γj are injective. We first show that the relevant integral exists in B(X). R Lemma 2.22. Let A be sectorial of type (K, φ) with φ > π2 , t > 0, θ0 ∈ ( π2 , φ), θ ∈ [θ0 , φ), r > 0, and Γ = Γ(r, θ) be defined by (2.17). Then the integral Z Z 1 1 etλ R(λ, A) dλ (2.18) etλ R(λ, A) dλ = lim etA = R→∞ 2πi ΓR 2πi Γ converges absolutely in B(X). The resulting operator etA ∈ B(X) does not depend on the choice of r > 0 and θ ∈ ( π2 , φ). Moreover, ketA k ≤ M for all t > 0 and a constant M = M (K, θ0 ) > 0. 2.3. Analytic semigroups and sectorial operators 59 K Proof. Since kR(λ, A)k ≤ |λ| on Γ by (2.13), we can estimate Z R Z exp(ts Re e−iθ ) −iθ tλ |e | ds ke R(λ, A)k dλ ≤ K |se−iθ | r ΓR Z θ exp(tr Re eiα ) |ireiα | dα +K |reiα | −θ Z R exp(ts Re eiθ ) iθ |e | ds +K |seiθ | r Z θ Z ∞ ts cos θ e ds + K etr cos α dα ≤ 2K s −θ r  Z ∞  e−σ dσ ≤K 2 (−t cos θ) + 2θetr −t cos θ rt |cos θ| σ  Z ∞  −σ e dσ + 2πetr =: Kc(r, t, θ0 ), ≤K 2 rt |cos θ0 | σ for all R > r and t > 0, where we substituted σ = −st cos θ and used that cos θ ≤ cos θ0 < 0. The limit in (2.18) thus exists absolutely in B(X) by the majorant criterium, and ketA k ≤ Kc(r, t, θ0 ). If we take r = 1/t, then c(t−1 , t, θ0 ) =: c(θ0 ) does not depend on t > 0. So it remains to check that the integral in (2.18) is independent of r > 0 and θ ∈ ( π2 , φ). To this aim, we define Γ′ = Γ(r′ , θ′ ) for some r′ > 0 and θ′ ∈ ( π2 , φ), where we may assume that θ′ ≥ θ. We further set Γ′R = Γ′ ∩ B(0, R) and choose + − R > max{r, r′ }. Let CR and CR be the circle arcs from the endpoints of ΓR ′ to that of ΓR in {λ ∈ C | Im λ > 0} and {λ ∈ C | Im λ < 0}, respectively. (If ± + − θ = θ′ , then CR contain just one point.) Then SR = ΓR ∪ CR ∪ (−Γ′R ) ∪ (−CR ) is a closed curve in the starshaped domain Σφ . So (2.14) shows that Z etλ R(λ, A) dλ = 0. SR We further estimate Z Z θ′ iα tλ etR Re e e R(λ, A) dλ ≤ + CR θ K |iReiα | dα ≤ K(θ′ − θ)etR cos θ → 0, |Reiα | − as R → ∞ since cos α ≤ cos θ < 0, and analogously for CR . So we conclude Z Z Z etλ R(λ, A) dλ etλ R(λ, A) dλ = lim etλ R(λ, A) dλ = lim R→∞ ΓR Γ = Z etλ R(λ, A) dλ. R→∞ Γ′ R  Γ′ We next establish some of the fundamental properties of the operators etA . We stress that we do not assume that A is densely defined here. Theorem 2.23. Let A be sectorial of angle ϕ > π2 . Define etA as in (2.18) for t > 0, and set e0A = I. Then the following assertions hold. a) etA esA = esA etA = e(t+s)A for all t, s ≥ 0. 2.3. Analytic semigroups and sectorial operators 60 b) The map t 7→ etA belongs to C 1 (R+ , B(X)). Moreover, etA X ⊆ D(A), d tA = AetA and kAetA k ≤ C/t for a constant C > 0 and all t > 0. We also dt e have AetA x = etA Ax for all x ∈ D(A) and t ≥ 0. c) Let x ∈ X. Then etA x converges as t → 0 in X if and only if x is contained in D(A). In this case, etA x tends to x as t → 0. d) Let D(A) be dense. Then (etA )≥0 is a C0 -semigroup generated by A. Proof. a) We proceed as in the holomorphic functional calculus in Theorem 5.1 of [ST]. Let t, s > 0. We use that etA does not depend on the choice of r and θ by Lemma 2.22. Take 0 < r < r′ and π2 < θ′ < θ < φ < ϕ. Set Γ = Γ(r, θ) and Γ′ = Γ(r′ , θ′ ) as in (2.17). Using the resolvent equation (1.7) and Fubini’s theorem, we compute Z Z 1 tλ tA sA e e e = esµ R(λ, A)R(µ, A) dµ dλ (2πi)2 Γ ′ Γ Z Z 1 1 esµ tλ dµ dλ e R(λ, A) = 2πi Γ 2πi Γ′ µ − λ Z Z etλ 1 1 sµ dλ dµ. + e R(µ, A) 2πi Γ′ 2πi Γ λ − µ (One shows Fubini in this context by inserting the parametrizations and applying a functional y ∗ ∈ B(X)∗ . The integrability in (λ, µ) is checked as in Lemma 2.22 or below.) ′ = {z = R′ eiα | α ∈ Fix λ ∈ Γ and take R′ > max{r, r′ , |λ|}. We set CR ′ ′ ′ ′ ′ ′ [θ , 2π − θ ]} and SR′ = ΓR′ ∪ CR′ . Cauchy’s formula (2.15) yields Z 1 esµ dµ = esλ 2πi S ′ ′ µ − λ R since ′ , λ) n(SR ′ Z Γ′R′ Z ′ CR ′ = 1. As in Lemma 2.22, we further compute Z esµ esµ dµ −→ dµ and µ−λ Γ′ µ − λ esµ 2πR′ es Re µ ′ ′ dµ ≤ 2πR′ sup ≤ esR cos θ ′ −→ 0 ′ |µ − λ| µ−λ R − |λ| µ∈CR as R′ → ∞, using that cos α ≤ cos θ′ < 0. It follows Z 1 esµ esλ = dµ. 2πi Γ′ µ − λ Next, fix µ ∈ Γ′ and take R > max{r, r′ , |µ|}. Set CR = {z = Reiα | α ∈ [θ, 2π − θ]} and SR = ΓR ∪ CR . We now have n(SR , µ) = 0 and derive as above Z eλt dλ = 0. Γ λ−µ The above equalities imply that Z 1 tA sA eλt esλ R(λ, A) dλ = e(t+s)A = e(s+t)A = esA etA . e e = 2πi Γ 2.3. Analytic semigroups and sectorial operators 61 b) Let x ∈ X, t > 0, ε > 0, and R > r. Since λ 7→ R(λ, A) is continuous in B(X, [D(A)]), also the integral Z TR (t) = eλt R(λ, A) dλ ΓR belongs to B(X, [D(A)]). Recall from (2.18) that TR (t) tends to 2πietA in B(X) as R → ∞. We further compute Z Z Z tλ λt ATR (t) = e AR(λ, A) dλ = e λR(λ, A) dλ − etλ dλ I. ΓR ΓR ΓR Take again CR = {µ = Reiα | α ∈ [θ, 2π − θ]}. Using Cauchy’s theorem (2.14), one shows as in step a) that Z Z tλ etλ dλ ≤ 2πR sup etR cos α ≤ 2πR eεR cos θ −→ 0 e dλ = − CR ΓR α∈[θ,2π−θ] as R → ∞, uniformly for t ≥ ε. As in the proof of Lemma 2.22 (with r = 1/t and kλR(λ, A)k ≤ K), we then estimate Z θ Z ∞ Z tλ ts cos θ recos α dα e ds + K kλe R(λ, A)k dλ ≤ 2K 1 t ΓR ≤ C′ 2K 2eKθ + =: . t |cos θ| t t −θ Hence, ATR (t) converges to the integral Z λeλt R(λ, A) dλ, Γ in B(X) as R → ∞. Since A is closed, it follows that etA X ⊆ D(A), Z 1 tA λeλt R(λ, A) dλ, Ae = 2πi Γ (2.19) ′ C for all t > 0. and kAetA k ≤ 2πt Observe that TR (·) belongs to C 1 (R≥0 , B(X, [D(A)])) with derivative Z d λeλt R(λ, A) dλ dt TR (t) = ΓR for t ≥ 0. In a similar way as above, one sees that Z Z ∞ tλ λe R(λ, A) dλ ≤ 2K ets cos θ ds ≤ Γ\ΓR R 2K eRε cos θ −→ 0 ε |cos θ| as R → ∞ for t ≥ ε. As a result, Z Z d λt eλt R(λ, A) dλ λe R(λ, A) dλ = dt ΓR ΓR converges in B(X) to AetA uniformly for t ≥ ε, see (2.19). We infer that d tA t 7→ etA ∈ B(X) is continuously differentiable for t > 0 with dt e = AetA . For x ∈ D(A), we further obtain Z 1 tA Ae x = lim eλt R(λ, A)Ax dλ = etA Ax. R→∞ 2πi ΓR 2.3. Analytic semigroups and sectorial operators 62 c) Let x ∈ D(A), R > r, and t > 0. As in part a), Cauchy’s formula (2.15) yields Z λt Z e 1 eλt 1 dλ = lim dλ = 1 R→∞ 2πi ΓR λ − 0 2πi Γ λ Observing that λR(λ, A)x − x = R(λ, A)Ax, we conclude Z Z λt  1 e 1 1 tA λt e x−x= R(λ, A)Ax dλ. e R(λ, A) − x dλ = 2πi Γ λ 2πi Γ λ Since the right integrand is bounded by c |λ|−2 kAxk on Γ for all t ∈ (0, 1], Lebesgue’s convergence theorem implies the existence of the limit Z 1 1 tA R(λ, A)Ax dλ =: z. lim e x − x = t→0 2πi Γ λ Let KR = {Reiα − θ ≤ α ≤ θ}. Cauchy’s theorem (2.14) shows that Z 1 R(λ, A)Ax dλ = 0. λ ΓR ∪(−KR ) Since also Z −KR 2πRK 1 R(λ, A)Ax dλ ≤ kAxk −→ 0 λ R2 as R → ∞, we arrive at z = 0. Because of the uniform boundedness of etA , it follows that etA x → x as t → 0 for all x ∈ D(A). Conversely, if etA x → y as t → 0, then y ∈ D(A) by part b). Moreover, R(1, A)etA x = etA R(1, A)x tends to R(1, A)x as t → 0, since R(1, A)x ∈ D(A). We thus obtain R(1, A)y = R(1, A)x, and so x = y ∈ D(A). d) Let D(A) be dense. The above results then imply that (etA )≥0 is a C0 semigroup. Let B be its generator. To check that A = B, take x ∈ D(A). For t > s > 0, part b) yields that Z t eτ A Ax dτ. etA x − esA x = s Since the semigroup is strongly continuous, we can let s → 0 resulting in Z
1 t τA 1 tA e Ax dτ. e x−x = t t 0 The right-hand side tends to Ax as t → 0 by strong continuity; i.e., A ⊆ B. As the spectra of A and B are contained in C− , Lemma 1.24 yields A = B.  We next establish a converse to the above theorem and study further regularity properties of etA , assuming that D(A) is dense for simplicity. There are variants of the following results without the density of the domain, see Section 2.1 of [Lu]. We first introduce a basic concept. Definition 2.24. An analytic C0 -semigroup of angle ψ ∈ (0, π/2] is a family of operators {T (z) | z ∈ Σψ ∪ {0}} such that (a) T (0) = I and T (w)T (z) = T (w + z) for all z, w ∈ Σψ ; (b) the map T : Σψ → B(X); z 7→ T (z), is differentiable; (c) T (z)x → x in X as z → 0 in Σψ′ for all x ∈ X and each ψ ′ ∈ (0, ψ). 2.3. Analytic semigroups and sectorial operators 63 The generator of T (·) is defined as the generator of the C0 -semigroup (T (t))t≥0 . If kT (z)k is bounded for all z ∈ Σψ′ and each ψ ′ ∈ (0, ψ), the analytic C0 semigroup is called bounded. We now establish the fundamental characterization theorem of bounded analytic C0 -semigroups which goes back to Hille in 1948. Basically it says that a densely defined operator A generates such a semigroup if and only if A is sectorial of angle gerater than π/2. Moreover, it gives two useful characterizations of sectoriality and describes the class of bounded analytic C0 -semigroups in a different, very convenient way. For n ∈ N with n ≥ 2 we inductively define the powers of linear operator D(An ) = {x ∈ D(An−1 ) | An−1 x ∈ D(A)} and An x = A(An−1 x). Theorem 2.25. Let A be a closed linear operator on X. Then the following assertions are equivalent. a) A is densely defined and sectorial of angle ϕ > π/2. b) A is densely defined, C+ ⊆ ρ(A), and there is a constant C > 0 such that kR(λ, A)k ≤ C/|λ| for all λ ∈ C+ . c) For some ϑ ∈ (0, π/2), the maps e±iϑ A generate bounded C0 -semigroups. d) A generates a bounded C0 -semigroup (T (t))t≥0 such that T (t)X ⊆ D(A) and kAT (t)k ≤ M1 /t for all t > 0 and a constant M1 > 0. e) A generates a bounded analytic C0 -semigroup with angle ψ ∈ (0, π/2]. If this is the case, T (t) is given by (2.18) and we have T (t)X ⊆ D(An ), dn n kAn T (t)k ≤ Mn t−n , T (·) ∈ C ∞ (R+ , B(X)), and dt n T (t) = A T (t) for all t > 0, n ∈ N, and some constants Mn > 0. Proof. We prove the chain of implications e) ⇒ c) ⇒ b) ⇒ a) ⇒ d) ⇒ e) going from analyticity to sectoriality and back via claim d) using Theorem 2.23. 1) Let statement e) be true. Take ϑ ∈ (0, ψ). The operators T (e±iϑ t) for t ≥ 0 then yield two bounded C0 -semigroups. As in Remark 1.18 one sees that they are generated by e±iϑ A, and so c) has been established. 2) We assume property c) and that the semigroups generated by e±iϑ A are bounded by M . Proposition 1.20 shows the density of D(A). Because of Proposition 1.20 in [ST] and Proposition 1.21, condition c) first yields that ρ(A) = e∓iϑ ρ(e±iϑ A) ⊇ e∓iϑ C+ and hence ρ(A) ⊇ Σ π2 +ϑ ⊇ C+ . To check the resolvent estimate in b), we write eiϑ = a + ib for a, b > 0 and take r > 0 and s ≥ 0. We set c = min{a, b} > 0. Employing again Proposition 1.20 in [ST], assumption c) and Proposition 1.21, we estimate M kR(r + is, A)k = ke−iϑ R(e−iϑ (r + is), e−iϑ A)k ≤ Re((a − ib)(r + is)) M M/c M/c = ≤ ≤ . ar + bs r+s |λ| The case s < 0 can similarly be treated using eiϑ A. Hence, b) is valid. 3) Under assumption b), Remark 1.17 first implies that iR \ {0} belongs to ρ(A). By continuity, we infer the bound kR(is, A)k ≤ C/|s| for s ∈ R \ {0}. 2.3. Analytic semigroups and sectorial operators 64 Take q ∈ (0, 1) and λ = r+is with s 6= 0 and |r| ≤ q |s|/C. Set θ = arctan(q/C). Remark 1.17 then also shows that λ ∈ ρ(A) and the inequality C/(1 − q) ≤ kR(λ, A)k ≤ |s| C (1−q) cos θ |λ| . Condition a) also yields the bound kR(λ, A)k ≤ c(α)/|λ| for λ ∈ Σα and α ∈ (0, π/2). These two resolvent estimates show the sectoriality of A with angle greater than π/2. 4) The implication ‘a) ⇒ d)’ was shown in Theorem 2.23 together with T (·) ∈ d T (t) = AT (t) for t > 0, where T (t) is given by (2.18). C 1 (R+ , B(X)) and dt 5) Let property d) be valid. Let t > 0 and n ∈ N. Since AT (t) = T (t − t/n)AT (t/n), we obtain that T (t)X ⊆ D(A2 ). Iteratively, it follows that T (t)X ⊆ D(An ) and An T (t) = (AT ( nt ))n . Condition d) then implies the bound kAn T (t)k ≤ (M1 n)n t−n . P nk nn Observe that en = ∞ k=0 k! ≥ n! . Let q ∈ (0, 1). We take z ∈ C+ with tan |arg z| = q |Im z| ≤ . Re z eM1 Set t = Re z. The power series T (z) = ∞ X (z − t)n n=0 n! An T (t) around t converges absolutely in B(X) and uniformly for the above z, since ∞ ∞ X 1 |z − t|n M1n nn X  qt n M1n en ≤ = . kT (z)k ≤ n n n! t eM1 t 1−q n=0 n=0 We have thus extended T (·) to a bounded differentiable map T : Σϑ → B(X) q and every q ∈ (0, 1). for ϑ = arctan eM 1 Let x ∈ X and x∗ ∈ X ∗ . For fixed t > 0, we note that the holomorphic functions hT (z)T (t)x, x∗ i and hT (z+t)x, x∗ i coincide for z ∈ R+ . Consequently, they are the same for all z ∈ Σϑ thanks to the Identity Theorem 2.21 of [A4]. The Hahn-Banach theorem now yields that T (z)T (t) = T (z + t) for all z ∈ Σϑ . In the same way one can replace here t > 0 by any w ∈ Σϑ . It remains to check the strong continuity as z → 0. Let z ∈ Σϑ , x ∈ X, and ε > 0. We fix h > 0 such that kT (h)x − xk < ε. Using the boundedness and the continuity of T (·) on Σϑ , we estimate kT (z)x − xk ≤ kT (z)k kx − T (h)xk + kT (z + h)x − T (h)xk + kT (h)x − xk ε ≤ + kT (z + h) − T (h)k kxk + ε, 1−q  1  ε. lim kT (z)x − xk ≤ 1 + z→0 1−q As a result, T (z)x → x as z → 0 in Σϑ and claim e) is proved. 6) The first three assertions in the addenddum were shown in steps 4) and 5). d In step 4) we have also seen that T (·) ∈ C 1 (R+ , B(X)) and dt T (t) = AT (t) for 2.3. Analytic semigroups and sectorial operators 65 t > 0. Writing An−1 T (t) = T (t − δ)An−1 T (δ) for some δ ∈ (0, t) and n ∈ N, an dn n induction yields that T (·) belongs to C n (R+ , B(X)) with dt  n T (t) = A T (t). We provide additional information concerning the above theorem. Remark 2.26. a) Let ω ∈ R and A be closed. By rescaling one sees that A generates an analytic C0 -semigroup (T (z))z∈Σψ ∪{0} for some ψ > 0 such that e−ωz T (z) is bounded on all smaller sectors if and only if A is densely defined and A − ωI is sectorial of angle greater than π/2, cf. Section 2.1 in [Lu]. b) Let A be sectorial of angle ϕ. In (2.18) one can then replace t > 0 by z ∈ Σϕ+π/2 obtaining an analytic semigroup, see Proposition 2.1.1 of [Lu] or Proposition II.4.3 in [EN]. This means that in Theorem 2.25 the angle ψ of the semigroup is at least ϕ − π/2. On the other hand, a variant of step 2) in the above proof shows that ϕ ≥ ψ + π/2. We thus obtain the equality ϕ = ψ + π/2 for the angles. In a similar way can check that ψ is the supremum of all ϑ for which statement c) of Theorem 2.25 is true. c) In view of property a) or d) in Theorem 2.25, the shift T (t)f = f (· + t) cannot be extended to an analytic semigroup on C0 (R), cf. Example 2.21. The same is true for every C0 -group T (t) with an unbounded generator since T (t) : X → X is then a bijection. ♦ In the next result we combine Theorem 2.25 c) with the Lumer–Phillips Theorem 1.40 to obtain a very convenient sufficient condidition for the generation of a bounded analytic C0 -semigroup. In this case it is actually contractive on a sector. We note that the corresponding angle can be smaller than the angle ψ of analyticity, and that there are analytic semigroups which are contractive only on R+ or not even there. Corollary 2.27. Let A be closed, densely defined and dissipative. Assume that there are numbers λ0 > 0 such that λ0 I − A is surjective and ϑ ∈ (0, π/2) such that also the operators e±iϑ A are dissipative. Then A generates a bounded analytic C0 -semigroup T (·) of angle ϕ ≥ ϑ with kT (z)k ≤ 1 for | arg(z)| ≤ ϑ. Proof. Theorem 1.40 implies that C+ ⊆ ρ(A). The operators I − e±iϑ A = − A) are thus surjective, and so e±iϑ A generate contraction semigroups again by Theorem 1.40. Hence, A generates a bounded analytic C0 semigroup of angle ϕ ≥ ϑ due to Theorem 2.25 and Remark 2.26. We show the contractivity of T (z) with |arg z| = α ∈ (0, ϑ). Take x ∈ D(A) and x∗ ∈ J(x). Set ζ = hAx, x∗ i. By assumption, the numbers e±iϑ ζ belong to C− so that ζ is an element of C \ Σϑ+π/2 . It follows that Re(e±iα ζ) ≤ 0, thus the contractivity of T (z) follows as above.  e±iϑ (e∓iϑ I We now show a somewhat improved version of Example 2.19 combined with Theorem 2.25. Corollary 2.28. Let X be a Hilbert space and A be densely defined and selfadjoint with (Ax|x) ≤ 0 for all x ∈ D(A). (In this case one writes A = A′ ≤ 0 and says that A is non-positive (definite)). Then σ(A) ⊆ R≤0 and A generates a contractive analytic C0 -semigroup of angle π2 . 2.3. Analytic semigroups and sectorial operators 66 Proof. Let x ∈ D(A) and λ > 0. Using the dissipativity of A, we compute λkxk2 ≤ Re(λx − Ax|x) ≤ kλx − Axk kxk, and infer the lower bound kλx − Axk ≥ λkxk. Since A is symmetric, this bound shows that λ ∈ ρ(A) due to Theorem 4.7 of [ST]. The spectrum of A = A′ is real by the same theorem, and hence σ(A) is contained in R≤0 . We already know that the operator A is dissipative. For ϑ ∈ (0, π/2) the number (e±iπ/2 Ax|x) belongs to C− as (Ax|x) is real. The operators e±iϑ A are thus dissipative. Taking the supremum over ϑ < π/2, the second assertion  follows from Corollary 2.27. We now discuss the prototypical example for analytic semigroups, the Dirichlet-Laplacian on Lp (G). Example 2.29. Let p ∈ (1, ∞) and A = ∆ for E = Lp (Rm ) and D(A) = W 2,p (Rm ) or for E = Lp (G) and D(A) = W 2,p (G) ∩ W01,2 (G), assuming that G ⊆ Rm is open and bounded with ∂G ∈ C 2 . Then A generates a bounded analytic C0 -semigroup on E which is contractive on Σκp for  |p − 2|   π i  |p − 2|  π = arccot √ ∈ 0, . κp = − arctan √ 2 2 2 p−1 2 p−1 Moreover, the graph norm of A and k · k2,p are equivalent. In particular, for p = 2 the semigroup has angle π/2. Here one can allow for open and bounded G ⊆ Rm with ∂G ∈ C 1 where one replaces A by ∆D from Example 1.52. Proof. The norm equivalence can be shown as after Example 1.52 once the generator property has been proven. For p = 2 the result follows from Corollary 2.28 since then A is selfadjoint and dissipative by Examples 1.46 and 1.52. For p 6= 2 we use Corollary 2.27, allowing for G = Rm . The domain D(A) is dense by Proposition 4.13 in [FA]. Theorems 9.9 and 9.15 in [GT] show that I − A is surjective.6 Below we check that the operators e±iϑ A are dissipative for ϑ ∈ [0, κp ].7 Let u ∈ D(A) \ {0}. First, take p ≥ 2. We define u∗ = |u|p−2 u. Recall that kukp2−p u∗ ∈ J(u) by Example 1.31. Assume for a moment that u ∈ C 1 (G) so p that u∗ ∈ C 1 (G). Using u∗ = (uu) 2 −1 u, we compute ∂k u∗ = |u|p−2 ∂k u + = |u| p−4 p p−2 −2 2 (u∂k u 2 (uu) + u∂k u)u
uu∂k u + (p − 2)u Re(u∂k u) ′ for k ∈ {1, . . . , m}. The functions on the right are bounded in Lp by c k∂k ukp kukp−2 due to Hölder’s inequality. We can approximate u ∈ W01,p (G) in p W01,p by functions u ∈ Cc∞ (G), see Remark 1.42. The function u∗ thus belongs ′ to W01,p (G) and the above formula is true for u ∈ D(A). It leads to
∂k u∂k u∗ = |u|p−4 |u∂k u|2 + (p − 2)(Re(u∂k u))2 + i(p − 2) Im(u∂k u) Re(u∂k u) 6These are a deeper results based on harmonic analysis for G = Rm and p 6= 2 (the so-called Calderón–Zygmund theory) and also on PDE methods for bounded G. 7In the lectures the following arguments were presented only for m = 1 and p ≥ 2. 2.3. Analytic semigroups and sectorial operators 67
= |u|p−4 (p−1)(Re(u∂k u))2 + (Im(u∂k u))2 + i(p − 2) Im(u∂k u) Re(u∂k u) . Formula (1.21) and u ∈ D(A) then yield Z ∗ h∆u, u i = − ∇u · ∇u∗ dx G Z =− |u|p−4 (p − 1)|Re(u∇u)|2 + |Im(u∇u)|2 ) G
+ i(p − 2)|u|p−4 Im(u∇u) Re(u∇u) dx, and A = ∆ is dissipative. The inequalities of Hölder and Young further imply Z p p ∗ | Imh∆u, u i| ≤ |p − 2| |u| 2 −2 |Re(u∇u)| |u| 2 −2 |Im(u∇u)| dx G Z hp i1 h 1 Z i1 2 2 2 2 p−4 p−4 √ ≤ |p − 2| p − 1 |u| |Re(u∇u)| dx |u| |Im(u∇u)| dx p−1 G G √ Z Z |p − 2| p − 1 |p − 2| |u|p−4 |Re(u∇u)|2 dx + √ ≤ |u|p−4 |Im(u∇u)|2 dx 2 2 p − 1 G G |p − 2| Reh∆u, u∗ i. =− √ 2 p−1 We set z = −h∆u, u∗ i ∈ C+ , where we may assume that z 6= 0. We have shown the inequality |Im z| |p − 2| π ≤ arctan √ |arg z| = arctan = − κp . Re z 2 2 p−1 For ϑ ∈ [0, κp ] it follows |arg(e±iϑ z)| ≤ π2 and the dissipativity of the operators e±iϑ A. Corollary 2.27 thus implies the assertion for p > 2. ∗ Next, let p ∈ (1, 2). To avoid psingularities at zeros of u, we replace u by √ p−2 ∗ uε = uε u with ε ≤ uε := ε + |u|2 for ε > 0. Since uε ≥ |u|, we have ′ |uε |p−2 ≤ |u|p−2 and thus |u∗ε | ≤ |u∗ | ∈ Lp (G). Again we first take u ∈ Cc1 (G), p so that up−2 = (ε + uu) 2 −1 belongs to Cc1 (G). As above, we calculate ε
∂k u∗ε = |uε |p−4 ε∂k u + uu∂k u + (p − 2) Re(u∂k u)u . ′ By approximation we can extend this formula in Lp to u ∈ D(A) similar as above, estimating the first summand by |uε |p−4 ≤ ε−1 |u|p−2 . For u ∈ D(A), we ′ thus obtain uε ∈ W01,p (G) and ∂k u∂k u∗ε = |uε |p−4 ε|∂k u|2 + (p − 1)(Re(u∂k u))2 + (Im(u∂k u))2
+ i(p − 2) Im(u∂k u) Re(u∂k u) . Formula (1.21) thus yields as above the inequalities Z
∗ − Reh∆u, uε i = |uε |p−4 ε |∇u|2 + (p − 1)|Re(u∇u)|2 + |Im(u∇u)|2 dx ZG
≥ |uε |p−4 (p − 1)|Re(u∇u)|2 + |Im(u∇u)|2 dx, G Z
|p − 2| ∗ | Imh∆u, uε i| ≤ √ |uε |p−4 (p − 1)|Re(u∇u)|2 + |Im(u∇u)|2 dx. 2 p−1 G 2.3. Analytic semigroups and sectorial operators 68 ′ Observe that |u∗ε | ≤ |u∗ | ∈ Lp (G) and that uε converges pointwise to u as ′ ε → 0. So u∗ε tends to u∗ in Lp (G) by dominated convergence. Using also Fatou’s lemma, we deduce that − Reh∆u, u∗ i = − lim Reh∆u, u∗ε i ε→0 Z   ≥ lim inf (p − 1)|Re(u∇u)|2 + |Im(u∇u)|2 |uε |p−4 dx ε→0 Rm Z   ≥ (p − 1)|Re(u∇u)|2 + |Im(u∇u)|2 |u|p−4 dx =: J. G Because of |uε |p−4 ≤ |u|p−4 , we also obtain |p − 2| |p − 2| J ≤− √ Reh∆u, u∗ i. | Imh∆u, u∗ i| = lim | Imh∆u, u∗ε i| ≤ √ ε→0 2 p−1 2 p−1 We can now proceed as for p ≥ 2 and conclude that e±iϑ ∆ are dissipative for 0 ≤ ϑ ≤ κp . The assertion for p < 2 then also follows from Corollary 2.27.  For more general generation result we refer to [Pa] and Chapter 3 of [Lu], which focusses on the sup-norm setting. The case p = 1 is treated in [Ta]. These works make heavy use of results from partial differential equations. In Example 1.52 we have studied the Dirichlet–Laplacian on L2 (G) in a self-contained way using functional analytic methods, though without computing the domain explicitly. This approach can be extended to more general operators and with more efforts to Lp (G), see [Ou]. Inhomogeneous evolution equations. If A generates an analytic semigroup, then the inhomogeneous problem (2.6) exhibits better regularity properties than in the general case. The mild solution is ‘almost’ differentiable in X for continuous inhomogeneities f , and one needs very little extra regularity of f to obtain the differentiability of the solution Let x ∈ X, b > 0, f ∈ C([0, b], X) and A−ωI be densely defined and sectorial of angle ϕ > π2 for some ω ∈ R. We study the inhomogeneous evolution equation u′ (t) = Au(t) + f (t), t ∈ (0, b] =: J, u(0) = x. It has the mild solution Z t u(t) = T (t)x + T (t − s)f (s) ds =: T (t)x + v(t), 0 t ∈ [0, b], (2.20) (2.21) where A generates the analytic C0 -semigroup T (·). By Definition 2.5, a solution of (2.20) on J is a map u ∈ C(J, X) ∩ C 1 (J, X) with u(t) ∈ D(A) for all t ∈ J which satisfies (2.20). We need the Hölder space C α ([a, b], X) with exponent α ∈ (0, 1). It contains all functions u ∈ C([a, b], X) fulfilling [u]α := ku(t) − u(s)k < ∞, (t − s)α a≤s<t≤b sup and it becomes a Banach space when endowed with the norm kukα := kuk∞ + [u]α . 2.3. Analytic semigroups and sectorial operators 69 For 0 < α < β < 1 we have the embeddings C 1 ([a, b], X) ֒→ C β ([a, b], X) ֒→ C α ([a, b], X) ֒→ C([a, b], X). (2.22) We now show the results indicated above. Theorem 2.30. Let x ∈ X, b > 0, f ∈ C([0, b], X), and A − ωI be densely defined sectorial of angle ϕ > π2 for some ω ∈ R. Then the mild solution u of (2.20) satisfies the following assertions. a) We have u ∈ C β ([ε, b], X) for all β ∈ (0, 1) and ε ∈ (0, b). If also x ∈ D(A), we can even take ε = 0 here. b) If f ∈ C α ([0, b], X) for some α ∈ (0, 1), then u solves (2.20) on (0, b]. If also x ∈ D(A), then u solves (2.20) on [0, b]. Remark 2.31. For α = 0, Theorem 2.30 b) is wrong due to Example 4.1.7 in [Lu]. One thus needs a bit of extra regularity of f . Much more detailed and deeper information on the regularity of u can be found in Chapter 4 of [Lu], where also ‘spatial regularity’ is studied (and not only time regularity as above), see also the exercises. ♦ Proof. Due to Theorem 2.25 and Remark 2.26, the function T (·)x solves (2.20) on R+ with f = 0 if x ∈ X and on R≥0 if x ∈ D(A). In particular, T (·)x belongs to the space C 1 ([ε, b], X) for all ε > 0 (and for ε = 0 if x ∈ D(A)). In view of (2.22), thus we only have to consider the function v from (2.21). To show assertion a), let 0 ≤ s < t ≤ b. Theorem 2.25 and Remark 2.26 yield constants cj = cj (b) with j ∈ {0, 1} such that kT (t)k ≤ c0 and ktAT (t)k ≤ c1 . We first note that kvk∞ ≤ c0 b kf k∞ . The increment of v is split into the terms Z s Z t (T (t − τ ) − T (s − τ ))f (τ ) dτ =: I1 + I2 . T (t − τ )f (τ ) dτ + v(t) − v(s) = s 0 It follows kI1 k ≤ c0 |t − s| kf k∞ ≤ c0 b1−β |t − s|β kf k∞ . For t > s > τ ≥ 0, we further compute Z t−s T (σ)AT (s − τ ) dσ, T (t − τ ) − T (s − τ ) = (T (t − s) − I)T (s − τ ) = 0 using that T (s − τ )X ⊆ D(A) by Theorem 2.25. This formula leads to the inequality c0 c1 |t − s| , kT (t − τ ) − T (s − τ )k ≤ |s − τ | which is not good enough since the denomimator is not integrable in τ < s. Since it also gives more than the needed factor |t − s|β , we only apply the above bound to a fraction of the integrand in I2 , obtaining Z s kI2 k ≤ kT (t − τ ) − T (s − τ )kβ kT (t − τ ) − T (s − τ )k1−β kf (τ )k dτ ≤ Z 0 s 0 cβ0 cβ1 |t − s|β 21−β c0 cβ1 b1−β 1−β (2c ) dτ kf k = kf k∞ |t − s|β . 0 ∞ 1−β (s − τ )β Hence, v belongs to C β ([ε, b], X) and there is a constant c = c(β, b, c0 , c1 ) such that kvkC β ≤ c kf k∞ . (Observe that c explodes as β → 0.) 2.3. Analytic semigroups and sectorial operators 70 We next treat part b). In view of Lemma 2.8 (with u0 = 0), we have to show that v ∈ C([0, b], [D(A)]). Let t ∈ [0, b]. Inserting the constant vector f (t) and substituting τ = t − s, we obtain Z t Z t T (τ )f (t) dτ =: v1 (t) + v2 (t). T (t − s)(f (s) − f (t)) ds + v(t) = 0 0 As in Lemma 2.8, one checks the continuity of the maps v1 , v2 : [0, b] → X. By Lemmas 1.19 and 1.13, the function v2 takes values in D(A) and Av2 = T (·)f (·) − f (·) is continous in X. For 0 < ε < ε0 ≤ t ≤ b, Theorem 2.25 further implies that the truncated integral Z t−ε Z t−ε v1,ε (t) := T (t − s)(f (s) − f (t)) ds = T (ε) T (t − ε − s)(f (s) − f (t)) ds 0 0 is an element of D(A) and that Av1,ε ∈ C([ε, b], X). Moreover, v1,ε (t) tends to v1 (t) as ε → 0, and from AT (ε) ∈ B(X) we infer Z t−ε Z t−ε Av1,ε (t) = AT (ε) T (t−ε−s)(f (s)−f (t)) ds = AT (t−s)(f (s)−f (t)) ds. 0 0 Next, let 0 < ε < η < ε0 ≤ t. As above it follows Z t−ε Av1,ε (t) − Av1,η (t) = A T (t − s)(f (s) − f (t)) ds = Z t−η t−ε t−η AT (t − s)(f (s) − f (t)) ds. From Theorem 2.25 we then deduce that Z t−ε (t − s)−1 (t − s)α [f ]α ds kAv1,ε (t) − Av1,η (t)k ≤ c1 t−η c1 c1 = [f ]α (t − s)α |t−η [f ]α (η α − εα ). t−ε = α α Hence, Av1,ε converges in C([ε0 , b], X) as ε → 0. Since A is closed, the vector v1 (t) is contained in D(A) and (Av1,ε )ε has the limit Av1 in C([ε0 , b], X) for all ε0 > 0, so that v1 ∈ C((0, b], [D(A)]). Finally, v1 (0) = 0 ∈ D(A) and Z t−ε c1 |t − s|−1 [f ]α (t − s)α ds kAv1 (t)k = lim kAv1,ε (t)k ≤ lim ε→0 ε→0 0 c1 ≤ [f ]α tα α tends to 0 as t → 0. We conclude that Av ∈ C([0, b], X) as required.  The following example is a straighforward consequence of our results. Example 2.32. Let G ⊆ Rm be bounded and open with a C 1 boundary, u0 ∈ L2 (G), and f ∈ C α ([0, b], L2 (G)) for some α ∈ (0, 1). Theorem 2.30 and Example 2.29 then yield a unique solution u in C 1 ((0, b], L2 (G)) ∩ C((0, b], [D(∆D )]) ∩ C([0, b], L2 (G)) of the inhomogeneous diffusion equation u′ (t) = ∆D u(t) + f (t), 0 < t ≤ b, u(0) = u0 , where ∆D is the Dirichlet–Laplacian from Example 1.52 with D(A) ֒→ (2.23) W01,2 (G). 2.3. Analytic semigroups and sectorial operators 71 Next, we also assume that ∂G ∈ C 2 so that D(A) = W 2,2 (G) ∩ W01,2 (G) as noted in this example. Set f (t, x) = (f (t))(x) for all 0 < t ≤ b and almost every x ∈ G. Then we can interpret (2.23) more concretely as the partial differential equation ∂t u(t, x) = ∆u(t, x) + f (t, x), u(t, x) = 0, u(0, x) = u0 (x), t > 0, x ∈ ∂G, x ∈ U. t > 0, x ∈ G, In general, here the first and third equality hold almost everywhere and the second one in the sense of trace. The solutions become more regular if we ♦ improve the regularity of u0 , f or ∂G, see Section 5 of [Lu]. CHAPTER 3 Perturbation and approximation So far we have only looked at one given generator A. In this chapter we add another operator to A or we approximate it. Both procedures are of great importance both from a theoretical perspective and for applications. 3.1. Perturbation of generators Let A generate a C0 -semigroup T (·) and B be linear. We study the question whether ‘A + B’ generates a C0 -semigroup S(·), and then also whether S(·) inherits properties of T (·). Positive results in this direction will allow us to transfer our knowledge about A to larger classes of operators. In this setting one faces two basic problems. First, how to define ‘A + B’ if D(A) ∩ D(B) is ‘small’ (e.g., equal to {0} as in Example III.5.10 in [EN])? In this section we only treat the basic case that D(A) ⊆ D(B). Unless something else is said, we then put D(A + B) = D(A). Second, if B with D(B) ⊇ D(A) is ‘too large’, it can happen that A + B fails to be a generator. For instance, let A be a generator whose spectrum is unbounded to the left (e.g., d/ds on C0 (R≤0 ) with D(A) = C01 (R≤0 ) or ∆ on L2 (Rm ) as in Example 1.28, resp. 1.46), and B = −(1 + δ)A for any δ > 0. The sum A + B = −δA thus has the spectral bound s(A + B) = ∞ and so A + B is not a generator by Proposition 1.21. Below we restrict ourselves to ‘small’ perturbations B employing the following concept. Definition 3.1. Let A and B be linear operators with D(A) ⊆ D(B). The map B is called A-bounded (or relatively bounded) if there are constants a, b ≥ 0 such that ∀ y ∈ D(A) : kByk ≤ a kAyk + b kyk. (3.1) In this case we set D(A + B) = D(A) (unless something else is specified). The A-bound of B is the infimum of the numbers a ≥ 0 for which (3.1) is valid with some b = b(a) ≥ 0. We note that B is A-bounded if and only B belongs to B([D(A)], X). We derive a quantitative version of this equivalence. First, let A be closed with λ ∈ ρ(A). Assume that D(A) ⊆ D(B), and set γ = kBR(λ, A)k < ∞. Let y ∈ D(A). We compute kByk = kBR(λ, A)(λy − Ay)k ≤ γ kAyk + γ|λ| kyk, (3.2) which is (3.1) with a = γ. Conversely, let B be A-bounded and x ∈ X. Using (3.1) and AR(λ, A) = λR(λ, A) − I, the vector BR(λ, A)x ∈ X can be estimated by kBR(λ, A)xk ≤ a kAR(λ, A)xk + b kR(λ, A)xk 72 3.1. Perturbation of generators 73
≤ a|λ| kR(λ, A)k + a + b kR(λ, A)k kxk; (3.3) i.e., BR(λ, A) ∈ B(X). The next result gives a widely used criterion for relative boundedness. Lemma 3.2. Let A and B be linear operators satisfying D(A) ⊆ D(B) and kByk ≤ c kAykα kyk1−α for all y ∈ D(A) and some constants c ≥ 0 and α ∈ (0, 1). The map B then has the A-bound 0. In the assumption one can replace kAyk by kykA . ′ Proof. Recall Young’s inequality ab ≤ ap /p+bp /p′ from Analysis 1, where p . We use it with p = 1/α and p′ = 1/(1 − α). a, b ≥ 0, p ∈ (1, ∞) and p′ = p−1 For y ∈ D(A) and ε > 0 it follows 1 1 1 c kByk ≤ ε kAykα kyk1−α ≤ αε α kAyk + c 1−α (1 − α)ε− 1−α kyk. ε 1 If one replaces kAyk by kykA , one only obtains an extra summand αε α kyk.  Our arguments below are based on the following perturbation result for the resolvent taken from in Theorem 1.27 in [ST], see also Proposition 4.24 in [FA]. Lemma 3.3. Let A be closed with λ ∈ ρ(A) and B be A-bounded with kBR(λ, A)k < 1. Then the sum A + B with D(A + B) = D(A) is closed, λ is contained in ρ(A + B), and the resolvent satisfies ∞ X R(λ, A + B) = R(λ, A) (BR(λ, A))n = R(λ, A)(I − BR(λ, A))−1 , n=0 kR(λ, A)k . kR(λ, A + B)k ≤ 1 − kBR(λ, A)k Moreover, the graph norms of A and A + B on D(A) are equivalent. Proof. We only have to show the last assertion. For y ∈ D(A) we estimate kykA = kyk + kAR(λ, A + B)(λI − A − B)yk = kyk + kAR(λ, A)(I − BR(λ, A))−1 (λy − (A + B)y)k 1 ≤ kyk + (|λ| kR(λ, A)k + 1) 1−q (|λ| kyk + k(A + B)yk) =: c kykA+B , where q = kBR(λ, A)k. The converse inequality is shown similarly, and also follows from the open mapping theorem, see Corollary 4.29 in [FA].  We start with the bounded perturbation theorem which is the prototype for the desired results. It also characterizes the perturbed semigroup S(·) in terms of an integral equation and describes it by a series expansion, both only involving T (·) and B. These formulas allow us to transfer certain properties from T (·) to S(·), cf. Example 3.6, the exercises or Section 3.1 in [EN]. Theorem 3.4. Let A generate a C0 -semigroup T (·) satisfying kT (t)k ≤ M eωt for all t ≥ 0 and constants M ≥ 1 and ω ∈ R. Let B ∈ B(X). Then the sum A + B with D(A + B) = D(A) generates the C0 -semigroup S(·) which fulfills kS(t)k ≤ M e(ω+M kBk)t , (3.4) 3.1. Perturbation of generators S(t)x = T (t)x + S(t)x = T (t)x + S(t) = ∞ X Z Z 74 t 0 T (t − s)BS(s)x ds, (3.5) S(t − s)BT (s)x ds, (3.6) t 0 Sn (t), S0 (t) := T (t), Sn+1 (t)x := n=0 Z t 0 T (t − s)BSn (s)x ds (3.7) for all t ≥ 0, n ∈ N0 , and t ≥ 0. The Dyson-Phillips series in (3.7) converges in B(X) uniformly on compact subsets of R≥0 . The operator family (S(t))t≥0 is the only strongly continuous family of operators solving (3.5). The graph norms of A and A + B on D(A) are equivalent. Proof. 1) Observe that A + B is densely defined. The operator A − ωI generates the C0 -semigroup T̃ (·) = (e−ωt T (t))t≥0 by Lemma 1.18. As in Remark 1.26 we define the norm 9x9 = sups≥0 ke−ωs T (s)xk on X satisfying kxk ≤ 9x9 ≤ M kxk for x ∈ X and for which T̃ (·) becomes contractive. (We also denote the induced operator norm by triple bars.) For x ∈ X, we estimate 9Bx9 ≤ M kBxk ≤ M kBk kxk ≤ M kBk 9x9 . Take λ > M kBk ≥ 9B9. The Hille-Yosida estimate (1.16) thus implies the inequality 9BR(λ, A − ωI)9 ≤ 9B9 /λ < 1. From Lemma 3.3 we then deduce that λ belongs to ρ(A + B − ωI), the bound 9BR(λ, A + B − ω)9 ≤ 1 λ−1 = , 1 − λ−1 9B9 λ − 9B9 and the equivalence of the graph norms. The Hille–Yosida Theorem 1.27 now shows that A + B − ωI generates a contraction semigroup S̃(·) on (X, 9 · 9) with 9S̃(t)9 ≤ e9B9 t ≤ eM kBk t for all t ≥ 0. Lemma 1.18 now tells us that A + B generates the semigroup given by S(t) = eωt S̃(t) fulfilling kS(t)xk ≤ 9S(t)x9 ≤ eωt eM kBk t 9 x9 ≤ M e(ω+M kBk)t kxk for all t ≥ 0 and x ∈ X, as asserted. 2) We next prove (3.5), (3.6), and uniqueness. For every x ∈ D(A), the function u = S(·)x solves the problem u′ (t) = (A + B)u(t) = Au(t) + f (t), t ≥ 0, u(0) = x, where f := BS(·)x : R≥0 → X is continuous. Proposition 2.6 then shows that u is given by Z t Z t T (t − s)BS(s)x ds T (t − s)f (s) ds = T (t)x + S(t)x = u(t) = T (t)x + 0 0 for t ≥ 0. We derive (3.5) for all x ∈ X by approximation since D(A) is dense in X and all operators (in particular B) are bounded uniformly in s ∈ [0, t]. 3.1. Perturbation of generators 75 Equation (3.6) is established in the same way, using that v = T (·)x solves v ′ (t) = (A + B)v(t) − Bv(t), t ≥ 0, v(0) = x ∈ D(A). Let U (·) be another strongly continuous solution of (3.5). For t ≥ 0, t0 > 0 and t ∈ [0, t0 ], we estimate Z t T (t − s)B(S(s)x − U (s)x) ds kS(t)x − U (t)xk = 0 Z t ω+ t0 kS(s)x − U (s)xk ds. kBk ≤ Me 0 Gronwall’s inequality from Satz 4.14 in Analysis 2 now yields that S(t)x − U (t)x = 0, and hence U (·) = S(·). 3) Let t ≥ 0 and all x ∈ X. Concerning (3.7), we note that S1 (·) is strongly continuous and satisfies Z t kS1 (t)xk ≤ M eω(t−s) kBk M eωs kxk ds = M 2 teωt kBk kxk. 0 By induction one further deduces the strong continuity Sn (·) and the inequality M n+1 kBkn n ωt t e n! for all n ∈ N. The series in (3.7) thus converges in B(X) uniformly on compact subsets of R≥0 to some operator R(t). Hence, R(·) is strongly continuous and fulfills the equations Z t ∞ ∞ Z t X X T (t − s)BSn (s)x ds = T (t − s)BR(s)x ds = Sj (t)x kSn (t)k ≤ 0 n=0 0 j=1 = R(t)x − T (t)x. The uniqueness of (3.5) says that R(t) = S(t).  Using also −A, we extend the above result to the group case. Corollary 3.5. Let A generate the C0 -group T (·) satisfying kT (t)k ≤ M eω |t| for all t ∈ R and constants M ≥ 1 and ω ≥ 0. Let B ∈ B(X). Then the sum A + B with D(A + B) = D(A) generates the C0 -group S(·) which fulfills kS(t)k ≤ M e(ω+M kBk)|t| and (3.5)–(3.7) for all t ∈ R. Proof. Theorem 1.30 says that the operators ±A generate C0 -semigroups with kT± (t)k ≤ M eωt for t ≥ 0. From Theorem 3.4 we then deduce that A + B and −(A + B) generate C0 -semigroups S± (·) with kS± (t)k ≤ M e(ω+M kBk)t . By Theorem 1.30, A + B thus generates a C0 -group S(·) with the asserted bound. Formulas (3.5)–(3.7) can be shown for t ∈ R as in the previous proof.  If a model involves the mass density of a substance, it is natural to require that a non-negative initial function leads to a non-negative solution. We will come back to this issue at the end the chapter. Here we first discuss whether such a behavior is inherited under perturbations. 3.1. Perturbation of generators 76 Example 3.6. Let E = C0 (U ) or E = Lp (µ) for an open set U ⊆ Rm , respectively for a measure space (S, A, µ) and 1 ≤ p < ∞. We set E+ = {f ∈ E | f ≥ 0}. Let T (·) be a C0 -semigroup on E with generator A such that T (t)f ≥ 0 for all f ∈ E+ and t ≥ 0. We call such operators or semigroups positive.1 We look at two classes of perturbations. a) Let B ∈ B(E) be also positive. Take f ∈ E+ . The function T (t−s)BT (s)f is then non-negative for each s ∈ [0, t]. Since E+ is closed in E, we infer that S1 (t)f ≥ 0 and, by induction, that all terms Sn (t)f in the Dyson-Phillips series (3.7) belong to E+ . So the semigroup S(·) generated by A + B is positive and satisfies S(t) ≥ T (t) = S0 (t); i.e., S(t)f ≥ T (t)f for all f ∈ E+ . b) Let Bf = bf for a real-valued function b ∈ Cb (U ) if E = C0 (U ), resp. b ∈ L∞ (µ) if E = Lp (µ). For all f ∈ E+ we then have (B + kb− k∞ I) ≥ b+ f ≥ 0 so that B0 := B + kb− k∞ I is positive. By part a), A + B0 generates a positive C0 -semigroup S̃(·) ≥ T (·) and so A + B = A + B0 − kb− k∞ I generates the positive C0 -semigroup S(·) given by S(t) = e−kb− k∞ t S̃(t) ≥ e−kb− k∞ t T (t). d with D(A) = C01 (R) As a simple example, we take U = S = R and A = ds 1,p p if E = C0 (U ), resp. D(A) = W (R) if E = L (µ). Because A generates the positive translation semigroup on E, the operator Cu = u′ + bu with D(C) = D(A) also generates a positive C0 -semigroup. ♦ We next use Corollary 3.5 to treat a damped or excited wave equation. Example 3.7. Let G ⊆ R3 be bounded and open with a C 1 -boundary and ∆D be the Dirichlet–Laplacian on L2 (G) given by Example 1.52. We set E = Y × L2 (G), where Y = W01,2 (G) is endowed with the norm kvkY = k |∇v|2 k2 from (1.34). As in Example 1.53 we define the operator   0 I A= with D(A) = D(∆D ) × Y ∆D 0 on E. It is skewadjoint and thus generates a unitary C0 -group T (·).
We further let b ∈ L∞ (G) and introduce the bounded operator B = 00 0b on X. Corollary 3.5 now yields that A + B with D(A + B) = D(A) generates a C0 -group S(·) on E which is bounded by ekbk∞ |t| . Let (u0 , u1 ) ∈ D(A). Following Example 2.4, we can show that (u, u′ ) = S(·)(u0 , u1 ) yields the unique solution u ∈ C 2 (R≥0 , L2 (G)) ∩ C 1 (R≥0 , Y ) ∩ C(R≥0 , [D(∆D )]) of the perturbed wave equation u′′ (t) = ∆D u(t) + bu′ (t), bu′ t ≥ 0, u(0) = u0 , u′ (0) = u1 . (3.8) 2 The term acts as a damping if b ≤ 0. As in Example 2.17 we also want to allow for data in E. To determine the A+B for A + B, we fix some λ > 3 kbk∞ . Lemma 3.3 extrapolation space E−1 then yields the bound kR(λ, A + B)wkE ≤ 32 kR(λ, A)wkE . On the other hand, from the Hille–Yosida estimate (1.16) we obtain kBR(λ, A + B)k ≤ kbk∞ /(λ − kbk∞ ) ≤ 12 . Writing R(λ, A) = R(λ, A + B − B), Lemma 3.3 also 1We note that these concepts do not fit to our usual notation such as R = (0, ∞) for + the set of positive real numbers. 2The following part of the example was only sketched in the lectures. 3.1. Perturbation of generators 77 leads to the ineqality kR(λ, A)wkE ≤ 2 kR(λ, A + B)wkE . These expressions thus define equivalent norms on E, which are also equivalent to w 7→ kA−1 wkE A+B by (2.11). From Example 2.17 we now infer that E−1 is isomorphic to 2 −1,2 A ∼ L (G)×W (G) = E−1 where the isomorphisms extend the identity on E. Let (u0 , u1 ) ∈ E. As in Example 2.4 and Example 2.17, we finally obtain a unique solution u of (3.8) in C 2 (R≥0 , W −1,2 (G)) ∩ C 1 (R≥0 , L2 (G)) ∩ C(R≥0 , Y ). Here ♦ we consider the operator ∆D as a map from Y = W01,2 (G) to W −1,2 (G). We now turn our attention to unbounded perturbations B of a generator A. As noted above, we should impose a smallness assumption on B. We restrict ourselves to two very useful theorems for contraction and analytic semigroups, employing the simpler characterizations of the generation properties available here.3 We start with the dissipative perturbation theorem. Theorem 3.8. Let A generate the contraction semigroup T (·) and B be dissipative. Assume that B is A-bounded with a constant a < 1 in (3.1). Then A + B with D(A + B) = D(A) generates a contraction semigroup S(·) which also satisfies formulas (3.5) and (3.6) for all x ∈ D(A). The graph norms of A and A + B on D(A) are equivalent. Proof. 1) Observe that A + B is densely defined and that we have RehAx, x∗ i ≤ 0 for all x ∈ D(A) and x∗ ∈ J(x) due to Proposition 1.33. Since B is dissipative, for each x ∈ D(A) there is a functional y ∗ ∈ J(x) such that RehBx, y ∗ i ≤ 0. Hence, RehAx + Bx, y ∗ i ≤ 0 and A + B is dissipative. By the assumption there are constants a ∈ [0, 1) and b ≥ 0 with kBxk ≤ a kAxk + b kxk for all x ∈ D(A). First, assume that a < 21 . Fix λ0 > (3.3) and the Hille-Yosida estimate (1.16) yield b 1−2a ≥ 0. Inequality kBR(λ0 , A)k ≤ aλ0 kR(λ0 , A)k + a + b kR(λ0 , A)k ≤ a + a + bλ−1 0 < 1. Lemma 3.3 now implies that A + B is closed, its graph norm is equivalent to k · kA , and λ0 ∈ ρ(A + B). The sum A + B thus generates a contraction semigroup by the Lumer–Phillips Theorem 1.40. 2a 2) Let a ∈ [ 21 , 1). We take k ∈ N with k > 1−a . Then k1 B is dissipative and 1 1 A-bounded with a constant a′ = ka < 1−a 2 ≤ 2 . Step 1) yields that A + k B generates a contraction semigroup. We inductively assume that Cj := A + kj B generates a contraction semigroup for some j ∈ {1, . . . , k − 1}. It follows aj kByk + b kyk, kByk ≤ a kAyk + b kyk ≤ a kCj yk + k  aj  (1 − a)kByk ≤ 1 − kByk ≤ a kCj yk + b kyk, k b a 1 k By ≤ k(1 − a) kCj yk + k(1 − a) kyk 3In Section III.3 of [EN] one can find results for general generators A based on the fixed point equation (3.5) for S(·). 3.1. Perturbation of generators 78 a for all y ∈ D(A). Since ã := k(1−a) < 12 , by step 1) the sum Cj + k1 B = Cj+1 is a generator of a contraction semigroup. By induction, A+Ck = A+B generates a contraction semigroup. The last assertion can be shown as in Theorem 3.4. But note that it is not clear that (3.5) and (3.6) hold for all x ∈ X by approximation since B may be unbounded.  If X is reflexive and one has a = 1 in the above theorem, the closure of A + B generates a contraction semigroup by Corollary III.2.9 in [EN]. We now use Theorem 3.8 to solve the Schrödinger equation for the Coulomb potential, see also Example 4.19 in [ST]. Example 3.9. Let E = L2 (R3 ) and A = i∆ with D(A) = W 2,2 (R3 ). Example 1.46 implies that A is skewadjoint, and so it generates a unitary C0 group T (·) by Stone’s Theorem 1.45. We further set Bv(x) = ib|x|−1 2 v(x) =: −iV (x)v(x) for some b ∈ R, where V (0) := 0. Sobolev’s Theorem 3.17 in [ST] yields the embedding W 2,2 (R3 ) ֒→ C0 (R3 ). Let ε > 0. For v ∈ W 2,2 (R3 ) we thus estimate Z Z |v(x)|2 |v(x)|2 2 2 2 dx + b dx kBvk2 = b 2 2 B(0,ε) |x| R3 \B(0,ε) |x| Z ε Z kvk2∞ 2 b2 2 ≤ 4πb r dr + 2 |v(x)|2 dx 2 r ε 3 0 R \B(0,ε) b2 kvk22 . ε2 Since the graph norm of A is equivalent to k · k2,2 by Example 1.46, we conclude that B has the A-bound 0. Theorem 3.8 thus says that A + B generates a contraction semigroup, and one sees in the same way that this is also true for −(A + B). In view of Corollary 1.44, these semigroups yield an isometric group S(·) which is unitary by Proposition 5.52 of [FA]. This semigroup then solves the Schrödinger equation ib u′ (t) = i∆u(t) + 2 u(t), t ∈ R, ( ⇐⇒ iu′ (t) = −(∆ − V )u(t), ) |x| u(0) = u0 . R For a suitable constant b > 0 and appropriate units, the integral G |u(t, x)|2 dx describe the probability that the electron in the hydrogen atom is contained in the (Borel) set G ⊆ R3 at time t. ♦ ≤ 4πb2 CSob ε kvk22,2 + We now come to the core sectorial perturbation theorem. We note that we keep the angle φ from Definition 2.18, but increase the shift ω. Theorem 3.10. Let A be densely defined and closed. Assume there are constants ω ≥ 0, K > 0 and φ ∈ (0, π] such that ω + Σφ ⊆ ρ(A) and ∀ λ ∈ Σφ : kR(λ + ω, A)k ≤ K . |λ|
 1 in (3.1). Then there is a number Let B be A-bounded with constant a ∈ 0, K+1 ω ≥ 0 such that A + B − ωI is sectorial of type (K ′ , φ) for some K ′ > K. In 3.1. Perturbation of generators 79 particular, if φ > π/2, the sum A + B generates an analytic C0 -semigroup., where [D(A + B)] = D(A) with equivalent norms. Proof. Let a ≥ 0 and b > 0 as in (3.1). Take q ∈ (a(K + 1), 1) and set K(aω+b) r := q−a(K+1) > 0. Let λ ∈ Σφ \B(0, r) and x ∈ X. Using (3.1), the assumption and |λ| ≥ r, we estimate kBR(λ, A − ωI)xk ≤ a kAR(λ + ω, A)xk + b kR(λ + ω, A)xk bK ≤ a k(λ + ω)R(λ + ω, A)xk + a kxk + kxk |λ|   K(|λ| + ω) bK ≤a + 1 kxk + kxk |λ| |λ| ≤ a(K + 1) kxk + (q − a(K + 1)) kxk = qkxk. Lemma 3.3 thus implies that λ ∈ ρ(A + B − ωI), k · kA+B ∼ = k · kA , and kR(λ, A + B − ωI)k ≤ kR(λ + ω, A)k K/(1 − q) ≤ 1−q |λ| for all λ ∈ Σφ \ B(0, r). Taking γ = r if φ ≤ π/2 and γ = r sin φ if φ > π/2, for instance, we obtain the inclusion γ + Σφ ⊆ Σφ \ B(0, r) and the inequality kR(µ, A + B − (ω + γ)I)k = kR(µ + γ, A + B − ωI)k ≤ K′ K/(1 − q) ≤ |µ + γ| |µ| K K if φ ≤ π/2 and K ′ = (1−q) for all µ ∈ Σφ , with K ′ = 1−q sin φ if φ > π/2. Here −1 we use that |1 + γµ | is larger than the distance between −1 and Σφ which is 1, resp. sin φ. Setting ω = γ + ω, we arrive at the first assertion. The second  one is a consequence of Theorem 2.25 and Remark 2.26. The following example contains several important techniques which often occur in applications to partial differential equations. It says that first-order perturbations B have the ∆D -bound 0 if the coefficients are not too bad. Example 3.11. Let G ⊆ Rm be bounded and open with a C 2 -boundary, p ∈ (1, ∞), E = Lp (G), A = ∆D with D(A) = W 2,p (G) ∩ W01,p (G). By Example 2.29, the operator A is sectorial with angle ϕ > π/2 and its graph norm is equivalent to k · k2,p . Theorem 3.25 of [ST] yields the Sobolev embedding W 2,p (G) ֒→ W 1,q1 (G) ∩ Lq2 (G) for any qk ∈ (p, ∞) if p = m/k and ( ∞, p> m k, qk := mp m m−kp , p < k , m m where k ∈ {1, 2}. (In the first case one has k − m p > 0, and k − p = − qk in the second.) Note that qk > p. We take a number θ ∈ (0, 1) close to 1 and introduce the exponents q̃k ∈ (p, qk ) and rk ∈ (p, ∞) by 1 1−θ θ = + q̃k p qk and 1 1 1 = − . rk p q̃k 3.2. The Trotter-Kato theorems 80 Let v ∈ W 2,p (G). For given coefficients b ∈ Lr1 (G)m and b0 ∈ Lr2 (G), the operator B is defined by Bv = b · ∇v + b0 v = b0 v + Using Hölder’s inequality twice, we first derive m X bj ∂ j v . j=1 kBvkp ≤ k |b|r1 kr1 k |∇v|q̃1 kq̃1 + kb0 kr2 kvkq̃2 1−θ θ kvkθq2 . ≤ kbkr1 kvk1−θ 1,p kvk1,q1 + kb0 kr2 kvkp Proposition 3.30 of [ST] yields constants c, ε0 > 0 such that c kvk1,p ≤ ε kvk2,p + kvkp ε for all ε ∈ (0, ε0 ]. Sobolev’s embedding, the equivalence of k · kA and k · k2,p , and the elementary Young’s inequality then imply   1−θ −1 1−θ θ θ θ 1−θ θ 1−θ kBvkp ≤ c(b) ε kvk2,p kvk2,p + ε kvkp ε kvk2,p + kvkp kvk2,p   −1 ≤ ĉ(b) ε1−θ kvkA + (1 − θ)ε 1−θ kvkp + θε kvkA + kvkp1−θ kvkθA for constants c(b), ĉ(b) > 0 depending kbkr1 and kb0 kr2 . In view of Lemma 3.2, the operator B : D(A) → Lp (G) has A-bound 0. Theorem 3.10 thus shows that A + B with domain D(A) generates an analytic semigroup on Lp (G). ♦ 3.2. The Trotter-Kato theorems In applications one often knows the parameters in a problem only approximately since the rely on measurements. As in the case of inital values one can then argue that the solution should depend continuously on the parameters. In other words, let An and A generate C0 –semigroups Tn (·) and T (·) for n ∈ N. Assume that ‘An → A’ as n → ∞ in some sense. Do we obtain ‘Tn (t) → T (t)’ ? This question also occurs if one wants to regularize a problem in order to ‘legalize’ certain calculations, and also in numerical analysis where the operators An are actually matrices on subspaces of finite dimensions mn with mn → ∞ (if dim X = ∞). In the easiest case one has D(An ) = D(A) and Bn := An − A is bounded. We assume that c := supn∈N kBn k < ∞ and let kT (t)k ≤ M eωt for all t ≥ 0 and some contstants M ≥ 0 and ω ∈ R. Duhamel’s formula (3.5) and estimate (3.4) yield Z t kTn (t)x − T (t)xk = T (t − s)Bn Tn (s)x ds 0 Z t 2 ≤ M kBn k eω(t−s) e(ω+cM )s kxk ds ≤ c(t0 )kBn k 0 for all x ∈ X with kxk ≤ 1, t ∈ [0, t0 ], t0 > 0, and a constant depending on t0 . This means that Tn (t) tends to T (t) in B(X) locally uniformly in t if An → A in B(X). We give a typical example for which the question cannot be settled just by a bounded perturbation. 3.2. The Trotter-Kato theorems 81 Example 3.12. Let G ⊆ Rm be open and bounded with a C 1 -boundary, E = L2 (G), ∆D is the Dirichlet Laplacian in E from Example 1.52, and n ∈ N0 . Recall that ∆D is invertible and generates a contraction semigroup. Let an ∈ L∞ (G) satisfy 1δ ≥ an (x) ≥ δ > 0 and an (x) → a0 (x) as n → ∞ for a.e. x ∈ G and a constant δ > 0. We define An = an ∆D on D(An ) = D(∆D ) and note that this domain is dense in E. To treat An , we use the weighted scalar products Z 1 (u|v)n = uv dx a G n for u, v ∈ E. The induced norm satisfies δ kvk2L2 ≤ kvk2n ≤ δ −1 kvk2L2 for v ∈ E. For u ∈ D(An ) we obtain Z an ∆D u u dx = (∆D u|u)L2 ≤ 0, (An u|u)n = G an so that An is dissipative for (·|·)n . To check the range condition, take f ∈ E. −1 The map fn = a−1 n f also belongs to E and we can set un = ∆D fn ∈ D(An ) which satisfies An vn = an fn = f . By the Lumer-Phillips Theorem 1.40, the operators An generate contraction semigroups Tn (·) for k · kn . The same arguments works for the operators e±iϑ An for all ϑ ∈ (0, π2 ) so that the semigroups are contractive analytic of angle π2 for k · kn by Corollary 2.27 and taking the supremum over ϑ. For the equivalent 2-norm, z ∈ C+ and v ∈ E, it follows kTn (z)vkL2 ≤ δ −1/2 kTn (z)vkn ≤ δ −1/2 kvkn ≤ δ −1 kvkL2 . Observe that An u tends to A0 u pointwise a.e. as n → ∞ and moreover |An u| ≤ δ −1 |∆u|. Dominated convergence then yields the limit An u → A0 u in E for each u ∈ D(∆D ). Does Tn (T ) tends to T0 (t) strongly? ♦ The next example indicates that one needs a uniform bound on the semigroups Tn (·) to obtain a general result. Example 3.13. Let X = ℓ2 , n ∈ N, A((xk )k ) = (ikxk )k with D(A) = {x ∈ 2 k )k ∈ ℓ } and An ((xk )k ) = (ikxk + δk,n kxk )k with D(An ) = D(A) for the Kronecker delta δk,n . As in the exercises, one sees that the multiplication operators A and An generate the C0 -semigroup on X given by T (t)x = (eikt xk )k and Tn (t)x = (eikt ekδk,n t xk )k , respectively. For x ∈ D(A) the distance kAn x − Axk2 = |nxn | = |(Ax)n | tends to 0 as n → ∞; i.e.; An converges on the common domain strongly to A. On the other hand, we have ℓ2 | (kx kTn (t)k ≥ kTn (t)en k2 = |eint ent | = ent −→ ∞ as n → ∞ for all t > 0. So Tn (t) cannot converge strongly, since strong convergence would imply uniform boundedness of {Tn (t) | n ∈ N}. ♦ The first Trotter–Kato theorem from 1958/59 shows that the convergence of resolvents and semigroups are equivalent and that it follows from the convergence of the generators, provided that the C0 -semigroups Tn (·) are exponentially bounded uniformly in n. 3.2. The Trotter-Kato theorems 82 Theorem 3.14. Let An and A generate C0 -semigroups Tn (·) and T (·), respectively, which satisfy kTn (t)k, kT (t)k ≤ M eωt for all t ≥ 0 and n ∈ N and some M ≥ 1 and ω ∈ R. Let D be a core of D(A). Then the implications a) ⇒ b) ⇔ c) ⇔ d) hold among the following claims, where we always let n → ∞. a) D ⊆ D(An ) for all n ∈ N and An y → Ay for all y ∈ D. b) For all y ∈ D and n ∈ N there are yn ∈ D(An ) with yn → y and An yn → Ay. c) For some λ ∈ Cω , we have R(λ, An )x → R(λ, A)x for all x ∈ X. d) For each t ≥ 0 we have Tn (t)x → T (t)x for all x ∈ X. If c) or d) are true, then c) is valid for all λ ∈ ω + C+ = Cω and the limit in d) is uniform on all compact subsets of R≥0 . Proof. The implication from a) to b) is trivial (take yn = y). Let statement b) be true. Take any λ ∈ Cω . Since λI − A : [D(A)] → X is an isomorphism, the set (λI − A)D is dense in X. The Hille–Yosida estimate (1.14) and the assumption yield the uniform bound kR(λ, An )k ≤ ReM λ−ω for all n ∈ N. By Lemma 4.10 of [FA] we thus have to show property c) only for all x = λy − Ay with y ∈ D. Let y ∈ D. Due to condition b), there are vectors yn ∈ D(An ) such that yn → y and An yn → Ay in X as n → ∞. These limits imply xn := λyn − An yn −→ x = λy − Ay as n → ∞. Estimating kR(λ, An )x − R(λ, A)xk ≤ kR(λ, An )(x − xn )k + kR(λ, An )xn − R(λ, A)xk M ≤ kx − xn k + kyn − yk −→ 0, n → ∞, Re λ − ω we conclude assertion c) for all λ ∈ Cω . Next, let property c) be valid for some λ ∈ Cω . Let y ∈ D. We set x = λy−Ay and yn = R(λ, An )x ∈ D(An ). It follows that yn → y and An yn = λR(λ, An )x − x −→ λR(λ, A)x − x = λy − x = Ay as n → ∞; i.e., claim b) holds. We assume condition d). Take x ∈ X and λ ∈ Cω . Proposition 1.21 yields Z ∞ kR(λ, A)x − R(λ, An )xk ≤ ke− Re λt (T (t)x − Tn (t)x)k dt. 0 2M kxke(ω−Re λ)t The integrand is bounded by and tends to 0 for each t ≥ 0 as n → ∞. Part c) now results from dominated convergence, for all λ ∈ Cω . Finally, let again c) be true for some λ ∈ Cω . Take x ∈ X, t0 > 0, t ∈ [0, t0 ], and ε > 0. Since D(A) is dense, there is a vector y ∈ D(A) with kx − yk ≤ ε. Set z = λy − Ay ∈ X. We then compute kTn (t)x − T (t)xk ≤ kTn (t)k kx − yk + kTn (t)y − T (t)yk + kT (t)k ky − xk ≤ 2M eω+ t0 ε + k(Tn (t) − T (t))R(λ, A)zk. Commuting resolvents and semigroups, the last term is split in three terms k(Tn (t) − T (t))R(λ, A)zk ≤ kTn (t)(R(λ, A)z − R(λ, An )z)k + kR(λ, An )(Tn (t)z − T (t)z)k 3.2. The Trotter-Kato theorems 83 + k(R(λ, An ) − R(λ, A))T (t)zk =: d1,n (t) + d2,n (t) + d3,n (t). Because of c), the summand d1,n (t) ≤ M eω+ t0 kR(λ, A)z − R(λ, An )zk tends 0 uniformly for t ∈ [0, t0 ] as n → ∞. Since the set {T (t)y | t ∈ [0, t0 ]} is compact, the same holds for d3,n by an exercise in Functional Analysis. It remains to show this convergence for d2,n . As above we find an element w ∈ X satisfying kz − R(λ, A)wk ≤ ε. Inserting v = R(λ, A)w, we compute d2,n (t) ≤ kR(λ, An )(Tn (t) − T (t))(z − R(λ, A)w)k + kR(λ, An )(Tn (t) − T (t))vk M ≤ 2M eω+ t0 ε + k(Tn (t)R(λ, An ) − R(λ, An )T (t))R(λ, A)wk. Re λ − ω We denote the last summand by dˆ2,n (t). To dominate also this term, we write Z t ˆ ∂s [Tn (t − s)R(λ, An )T (s)R(λ, A)w] ds d2,n (t) = − 0 Z t = Tn (t − s)An R(λ, An )T (s)R(λ, A)w 0 = ≤ Z t
− Tn (t − s)R(λ, An )T (s)AR(λ, A)w ds Tn (t 0 M e ω + t 0 t0 − s)[R(λ, An ) − R(λ, A)]T (s)w ds sup k[R(λ, An ) − R(λ, A)]T (s)wk. s∈[0,t0 ] The right-hand side converges to 0 uniformly for t ∈ [0, t0 ] as n → ∞, again due to c) and the compctness of {T (s)w | s ∈ [0, t0 ]}. Combining these estimates, we derive assertion d) with local uniform convergence.  Example 3.15. In the setting of Example 3.13, the above theorem implies that the semigroup generated by An = an ∆D converges strongly on L2 (G) to the C0 -semigroup generated by A = a∆D . Here we have D(∆D ) = D = D(A) = D(An ), ω = 0, and M = δ −1 . ♦ In Theorem 3.14 we have assumed that the limit operator A is a generator. We want to replace this assumption by a range condition as in the Lumer– Phillips theorem. In the main step of our argument we start with strongly converging resolvents and have to show that the limit operators form again the resolvent of a map (which then turns out to be a generator thanks to the Hille-Yosida theorem). In this step we employ the next concept. Definition 3.16. Let ∅ 6= Λ ⊆ C. A set {R(λ) | λ ∈ Λ} in B(X) is called pseudo-resolvent if it satisfies R(λ) − R(µ) = (µ − λ)R(λ)R(µ) for all λ, µ ∈ Λ. (3.9) We first show that pseudo-resolvents occur as strong limits of resolvents, which only have to converge for one point λ0 . Lemma 3.17. Let R(λ, An ) be resolvents satisfying kR(λ, An )k ≤ ReM λ−ω for all n ∈ N and λ ∈ Cω and some ω ∈ R and M > 0. If R(λ0 , An ) strongly tends 3.2. The Trotter-Kato theorems 84 to an operator R(λ0 ) in B(X) for some λ0 ∈ Cω , then all operators R(λ, An ) strongly converge to a pseudo-resolvent {R(λ) | λ ∈ Cω } for λ ∈ Cω as n → ∞. Proof. We show the strong convergence for all λ ∈ Cω below. Then the resolvent equation (1.7) for R(λ, An ) implies (3.9) in the strong limit. Let µ ∈ Cω . Remark 1.17 yields the expansion ∞ X R(λ, An ) = (µ − λ)k R(µ, An )k+1 k=0 Re µ−ω 2M for all λ ∈ Cω with |µ − λ| ≤ ≤ 12 k(µ, An )k−1 . If R(µ, An ) converges strongly as n → ∞, then also the partial sums of the above series have strong limits. The norms of the remainder terms ∞ X (µ − λ)k R(µ, An )k+1 P∞ k=N +1 are bounded by c k=N +1 2−k = c2−N with c = M/(Re µ − ω), which tends to 0 as N → ∞ independently of n. As a result, the operator R(λ, An ) converges 1 (Re µ − ω)). The radii of these balls are strongly as n → ∞ for λ ∈ B(µ, 2M greater than a number r(δ) > 0 for all µ ∈ Cω+δ and each δ > 0. Starting from λ0 , for each µ ∈ Cω+δ we can thus show the strong convergence of (R(µ, An ))n by a finite iteration. The result follows since δ > 0 is arbitrary.  We note that in Lemma 3.17 the limits R(λ) do not need to form a resolvent. For instance, the bounded generators An = −nI satisfy ketAn k = e−nt ≤ 1 for 1 all t ≥ 0 and n ∈ N, and their resolvent R(λ, An ) = λ+n I tends to 0 = R(λ) as n → ∞ for all λ ∈ C+ . Before we deal with this problem, we derive important properties of pseudo-resolvents. Lemma 3.18. For a pseudo-resolvent {R(λ) | λ ∈ Λ} and all λ, µ ∈ Λ, we have a) R(λ)R(µ) = R(µ)R(λ), b) N(R(λ)) = N(R(µ)), c) R(λ)X = R(µ)X. Proof. Interchanging λ and µ, equation (3.9) implies assertion a). These facts further yield the formulas R(λ) = R(µ)(I + (µ − λ)R(λ)) = (I + (µ − λ)R(λ))R(µ), which lead to the inclusions R(λ)X ⊆ R(µ)X and N(R(µ)) ⊆ N(R(λ)). The converse inclusions are shown analogously.  We now establish sufficient conditions for a pseudo-resolvent to be a resolvent. Lemma 3.19. Let {R(λ) | λ ∈ Λ} be a pseudo-resolvent. a) Let R(λ0 ) be injective with dense range for some λ0 ∈ Λ. Then there is a closed operator A with dense domain D(A) = R(λ0 )X such that Λ ⊆ ρ(A) and R(λ) = R(λ, A) for all λ ∈ Λ. b) Let R(µ) have dense range for some µ ∈ Λ and let there be λj ∈ Λ with |λj | → ∞ as j → ∞ such that kλj R(λj )k ≤ M for all j ∈ N and some constant M > 0. Then R(λ) is injective for all λ ∈ Λ (and thus a resolvent by part a)). 3.2. The Trotter-Kato theorems 85 Proof. a) The assumption allows us to define the closed operator A = λ0 I − R(λ0 )−1 with dense domain D(A) = R(λ0 )X. It satisfies the equations (λ0 I − A)R(λ0 ) = R(λ0 )−1 R(λ0 ) = I, R(λ0 )(λ0 y − Ay) = R(λ0 )R(λ0 )−1 y = y for all y ∈ D(A), so that λ0 ∈ ρ(A) and R(λ0 ) = R(λ0 , A). Lemma 3.18 shows that R(λ)X = D(A) for all λ ∈ Λ. Using this fact and (3.9), we further compute (λI − A)R(λ) = [(λ − λ0 )I + (λ0 I − A)]R(λ0 )[I − (λ − λ0 )R(λ)] = I + (λ − λ0 )(R(λ0 )[I − (λ − λ0 )R(λ)] − R(λ)) = I, and similarly R(λ)(λy − Ay) = y for y ∈ D(A). Assertion a) is thus proved. b) We have λj 6= µ for all sufficiently large j ∈ N. Equation (3.9) and the assumptions then yield the limit λj (R(λj ) − R(µ)) − R(µ) µ − λj λj µ R(λj ) − R(µ) = µ − λj µ − λj M + kµR(µ)k ≤ −→ 0 |µ − λj | k(λj R(λj ) − I)R(µ)k = as j → ∞. Since the set R(µ)X is dense and the operators λj R(λj ) are uniformly bounded, it follows that λj R(λj )x → x as j → ∞ for all x ∈ X. Now, let R(λ)x = 0 for some x ∈ X and λ ∈ Λ. From Lemma 3.18 deduce that 0 = λj R(λj )x → x as j → ∞ and hence x = 0.  With these preparations we can now show the second Trotter–Kato theorem, which adds a generation result to the first one. Theorem 3.20. Let An generate C0 -semigroups Tn (·) such that kTn (t)k ≤ M eωt for all t ≥ 0 and n ∈ N and some constants M ≥ 1 and ω ∈ R. We then obtain the implications a) ⇒ b) ⇔ c) among the following statements. a) There exists a densely defined operator A0 such that D(A0 ) ⊆ D(An ) for all n ∈ N and An y → A0 y as n → ∞ for all y ∈ D(A0 ), and the range (λ0 I − A0 )D(A0 ) is dense in X for some λ0 ∈ Cω . b) For some λ0 ∈ Cω the operators R(λ0 , An ) converge strongly to a map R ∈ B(X) with dense range. c) There is a C0 -semigroup T (·) with generator A such that Tn (t) converges strongly to T (t) for all t ≥ 0 as n → ∞. If property b) is true, then R = R(λ0 , A). If part a) holds, then A = A0 . The semigroups Tn (·) and T (·) satisfy the assertions of Theorem 3.14 if we assume conditions a), b) or c). Proof. The implication ‘c) ⇒ b)’ is a consequence of Theorem 3.14 with R = R(λ0 , A) since kT (t)k ≤ M eωt follows from the assumptions. 3.3. The Lax–Chernoff product formula 86 Let statement a) be true. Take any y ∈ D(A0 ) and set x = λ0 y − A0 y. Using the assumption and the Hille-Yosida estimate (1.14), we compute
kR(λ0 , An )x − yk = kR(λ0 , An ) (λ0 y − A0 y) − (λ0 I − An )y k M kA0 y − An yk −→ 0 ≤ Re λ0 − ω as n → ∞. Since the range (λ0 I −A0 )D(A0 ) is dense and R(λ0 , An ) is uniformly bounded, the resolvents R(λ0 , An ) thus converge strongly to a map R ∈ B(X). The range of R contains the dense set D(A0 ); so that claim b) is shown. Assume condition b). Due to Lemma 3.17, the operators R(λ, An ) converge strongly to a pseudo-resolvent {R(λ) | λ ∈ Cω } as n → ∞, where R(λ0 ) = R has dense range by b). Therefore also the terms (λ − ω)k R(λ, An )k tend to (λ − ω)k R(λ)k strongly for all k ∈ N and all λ ∈ Cω as n → ∞. By assumption, the resolvents satisfy the Hille–Yosida estimate (1.14) with uniform constants and hence the pseudo-resolvent inherits it. Lemma 3.19 now provides a closed operator A with dense domain R(λ0 )X such that R(λ) = R(λ, A). From the Hille–Yosida Theorem 1.27 we also infer that A generates a C0 -semigroup T (·). Theorem 3.14 now yields statement c) and the last addendum. Finally, we have to show that A0 has the closure A if property a) is true. Let y ∈ D(A0 ). Assertions a) and b) yield y = lim R(λ0 , An )(λ0 y − An y) = R(λ0 , A)(λ0 y − A0 y), n→∞ so that Ay = A0 y and A0 ⊆ A. Therefore, A0 possesses the closure A0 ⊆ A. On the other hand, the range (λ0 I − A0 )D(A0 ) is dense in X since it contains the set (λ0 I − A0 )D(A0 ). Let y ∈ D(A0 ). There exist vectors yk ∈ D(A0 ) such that yk → y and A0 yk → A0 y in X as k → ∞. Above we have seen that yk = R(λ0 , A)(λ0 yk − A0 yk ) which tends to y = R(λ0 , A)(λ0 y − A0 y). Hence, kyk is bounded by a constant times kλ0 y − A0 yk. Proposition 1.19 of [ST] then implies that the range (λ0 I − A0 )D(A0 ) is closed and so λ0 I − A0 is surjective. Because of λ0 ∈ ρ(A), Lemma 1.24 yields the quality A0 = A.  3.3. The Lax–Chernoff product formula Based on the Trotter–Kato theorems, we now discuss further approximation results for C0 -semigroups. We start with an auxiliary fact. Lemma 3.21. Let S ∈ B(X) satisfy kS n k ≤ M for all n ∈ N and some M > 0. We then obtain √ ken(S−I) − S n k ≤ M n kS − Ik for all n ∈ N. Proof. For n ∈ N, we compute ∞ ∞ ∞ X X nj j nj j X nj −n n S − e S = e−n (S − S n ), en(S−I) − S n = e−n j! j! j! j=0 kS m − S l k = m−1 X j=l j=0 j=0 S j (S − I) ≤ M (m − l)kS − Ik. 3.3. The Lax–Chernoff product formula 87 Computing an elementary series, we then estimate ∞ q q X nj nj ken(S−I) − S n k ≤ M e−n kS − Ik j! j! |n − j| j=0 ∞ X 1 nj  12  X nj 2 ≤ M e kS − Ik (n − j)2 j! j! j=0 j=0 √ n n√ ≤ M e−n kS − Ik e 2 n e 2 = M n kS − Ik. −n ∞  We next show the Lax–Chernoff product formula which is the core of this section. It was proved by Lax and Richtmyer in 1957 without its generation part, which was added by Chernoff in 1972 (who also discussed further variants of the result). The theorem says that consistency and stability imply convergence, which is a fundamental principle in numerical analysis. In this context one has to combine it with finite dimensional approximations, cf. Section 3.6 of [Pa]. In the exercises we treat convergence rates for vectors x in suitable subspaces. Theorem 3.22. Let V : R≥0 → B(X) be a function such that V (0) = I and kV (t)k k ≤ M ekωt for all t ≥ 0 and k ∈ N and some ω ∈ R and M ≥ 1. Assume that the limit A0 y := limt→0 1t (V (t)x − x) exists for all y in a dense subspace D(A0 ). Let the range (λI − A0 )D(A0 ) be dense in X for some λ ∈ Cω . Then A0 is closable and its closure A generates the C0 -semigroup T (·). The products V ( nt )n strongly converge to T (t) locally uniformly in t ≥ 0 as n → ∞. Proof. By rescaling, we may assume that ω = 0. For s > 0 we define the bounded operator As = 1s (V (s) − I) on X. The assumption says that As x → A0 x for all x ∈ D(A0 ) as s → 0 and that ke tAs k=e −t s ke t V s (s) k≤e −t s ∞ X −t t tk kV (s)k k ≤ e s e s M = M sk k! k=0 for all t ≥ 0. Theorem 3.20 thus shows that A0 has a closure which generates the C0 -semigroup T (·) and for any null sequence (sn ) the operators etAsn strongly tend to T (t) as n → ∞, uniformly for t ∈ [0, t0 ] and each t0 > 0. We claim that etAt/n converges strongly to T (t) locally uniformly in t as n → ∞. If the claim was wrong, there would exist a vector x ∈ X and times tn ∈ [0, t0 ] for some t0 > 0 such that inf ketn Atn /n x − T (tn )xk > 0. n∈N Since sn := tn /n → 0 as n → ∞, we obtain a contradiction. Let t0 > 0, t ∈ [0, t0 ], and y ∈ D(A0 ). Lemma 3.21 then yields √ ketAt/n x − V (t/n)n xk = ken(V (t/n)−I) x − V (t/n)n xk ≤ M n kV (t/n)x − xk tM t0 M = √ kAt/n xk ≤ √ sup kAs xk. n n 0≤s≤t0 The right hand side tends to 0 as n → ∞, and the assertion follows.  3.3. The Lax–Chernoff product formula 88 We add two special cases of the above general approximation result. (More examples are discussed in the exercises.) The first one is the Lie-Trotter product formula, shown by Trotter 1959 in a more direct way. It is of great importance in numerical analysis for problems where one can compute approximations of T (·) and S(·) in an efficient way, cf. the exercises. Note that the assumptions after (3.10) are satisfied if we know that (a closure of) C is a generator. Corollary 3.23. Assume that A and B generate C0 -semigroups T (·) and S(·), respectively, subject to the stability bound


n ≤ M eωt (3.10) T nt S nt for all n ∈ N and t ≥ 0 and some M ≥ 1 and ω ∈ R. Let D := D(A) ∩ D(B) and (λI −(A+B))D be dense in X for some λ ∈ Cω . Then the sum C := A+B on D(C) := D has a closure C which generates a C0 -semigroup U (·) given by


n U (t)x = lim T nt S nt x n→∞ uniformly on all compact subsets of R≥0 and for all x ∈ X. Proof. Define V (t) = T (t)S(t) for t ≥ 0. For x ∈ D, the vectors 1 t (V (t)x − x) = T (t) 1t (S(t)x − x) + 1t (T (t)x − x) converge to Bx+Ax as t → 0+ . The result now follows from Theorem 3.22.  The stability condition (3.10) holds if both semigroups are ω/2-contractive. In general, one cannot find an equivalent norm for which both semigroups become quasi-contractive, cf. Remark 1.26. In fact, there are generators A and B such that A + B exists and generates a C0 -semigroup, but (3.10) is violated, and thus the Lie-Trotter product formula fails, see [KW]. The Lie-Trotter formula can be used to give an alternative proof of the positivity assertion in Example 3.6. It also yields a rigorous mathematical interpretation for the ‘Feynman path integral formula’ in quantum mechanics for the Schrödinger group eit(∆−V ) , see Paragraph 8.13 in [Go]. By Proposition 1.21, the resolvent of the generator is the Laplace transform Z ∞ e−λt T (t)x dt = R(λ, A)x, Re λ > ω0 (A). (3.11) L(T (·)x)(λ) = 0 of the semigroup. In the next corollary we invert this transformation (for semigroup orbits) and thus approximate T (t) by powers of the resolvent. In numerics the resulting formula is called ‘implicit Euler scheme.’ By these formulas one can often transfer properties from the resolvent to the semigroup and back, see e.g. Corollary 3.25. This is an important fact since the resolvent is closely related to the generator, which is usually the given object in applications. We use this link in Example 3.26. Corollary 3.24. Let A generate the C0 -semigroup T (·). We then have

n
−n T (t)x = lim nt R nt , A x = lim I − nt A x n→∞ n→∞ uniformly on all compact subsets of R≥0 and for all x ∈ X. 3.3. The Lax–Chernoff product formula 89 Proof. Take M, ω > 0 with kT (t)k ≤ M eωt for all t ≥ 0. Fix δ ∈ 1 ). We then define (0, ω(ω+1)   t = 0, I, 1 1 V (t) = t R( t , A), 0 < t ≤ δ,   0, t > δ. The Hille-Yosida estimate (1.14) yields kV (t)n k = t−n kR( 1t , A)n k ≤ for 0 < t ≤ δ < 1 t (V 1 ω M M = ≤ M en(1+ω)t tn (t−1 − ω)n (1 − ωt)n by our choice of δ. From Lemma 1.23 we deduce the limit


(t)x − x) = 1t 1t R 1t , A x − x = 1t R 1t , A Ax −→ Ax as t → 0 for all x ∈ D(A). Theorem 3.22 implies the assertion.  We note that one can show the resolvent approximation directly without involving Chernoff’s product formula, see Theorem 1.8.3 in [Pa]. In the next result we use notions introduced in Example 3.6. Corollary 3.25. Let U ⊆ Rm be open and E = C0 (U ) or let (S, A, µ) be a measure space and E = Lp (µ) for some 1 ≤ p < ∞. We assume that A generates a C0 -semigroup T (·) on E. Then T (·) is positive for all t ≥ 0 if and only if R(λ, A) is positive for all λ ≥ ω and some ω > ω0 (A). Proof. Let the resolvent be positive and t > 0. For all f ∈ E+ and large n ∈ N, the functions ( nt R( nt , A))n f are non-negative and hence their limit T (t)f also belongs to E+ . (Here we use Corollary 3.24.) For λ > ω0 (A), the converse follows in a similiar way from formula (3.11).  Employing the above result and the ‘weak maximum principle’, we show that the Dirichlet Laplacian generates a positive semigroup. Example 3.26. Let G ⊆ Rm be open and bounded with a C 2 boundary, p E∞ = C0 (G), ET p = L (G) for 1 < p < ∞, and Ap = ∆ for p ∈ (1, ∞] with D(A∞ ) = {u ∈ p>1 W 2,p (G) | u, ∆u ∈ E∞ } on E∞ and D(Ap ) = W 2,p (G) ∩ W01,p (G) on Ep for 1 < p < ∞. These operators generate bounded analytic C0 -semigroup Tp (·) on Ep , see Corollary 3.1.21 in [Lu] and Example 2.29. Let λ > 0 and 0 ≤ f ∈ C0 (G). Note that C0 (G) ⊆ Ep for all 1 < p ≤ ∞. Set u = R(λ, A∞ )f ∈ D(A∞ ). Then u also belongs to D(Ap ) and λu − ∆u = f on G so that u = R(λ, Ap )f for all p ∈ (1, ∞] as λ ∈ ρ(Ap ). We show that u ≥ 0. Clearly, v = Im u is contained in D(A∞ ) and λv − ∆v = Im f = 0. It follows that v = 0 and so u is real-valued. Suppose there was a point x0 ∈ G such that u(x0 ) < 0. Since u = 0 on ∂G, the function u has a minimum u(x1 ) < 0 for some x1 ∈ G. Proposition 3.1.10 in [Lu] then yields ∆u(x1 ) ≥ 0, implying f (x1 ) = λu(x1 ) − ∆u(x1 ) < 0 which is impossible. Hence, u = R(λ, Ap )f is non-negative. Since C0 (G) is dense in Ep , we obtain the positivity of R(λ, Ap ), and thus Tp (t) is positive for all t ≥ 0 and p ∈ (1, ∞] by Corollary 3.25. ♦ CHAPTER 4 Long-term behavior In this chapter we study the long-term behavior of C0 -semigroups focusing on exponential stability and dichotomy. The main aim is to derive these properties from conditions on the spectrum and the resolvent of the (given) generator. 4.1. Exponential stability and dichotomy We first introduce the most basic property concerning the long-time behavior. Definition 4.1. A C0 -semigroup T (·) is called (uniformly) exponentially stable if there exist constants M, ε > 0 such that kT (t)k ≤ M e−εt for all t ≥ 0. The above concept can be reformulated as ω0 (T ) < 0 or equivalently as kT (t)xk ≤ M e−εt kxk for all x ∈ X and t ≥ 0. Let A generate the C0 -semigroup T (·) and ε > 0. Observe that we have kT (t)k ≤ e−εt for all t ≥ 0 if and only if A − εI is dissipative by the Lumer– Phillips Theorem 1.40. A simple, but typical example is the Dirichlet Laplacian ∆D on L2 (G) for a bounded domain, see Example 1.52. We first characterize exponential stability by properties of the semigroup itself. To this aim, we recall from Theorem 1.16 in [ST] that an operator T ∈ B(X) satisfies 1 1 r(T ) = max{|λ| | λ ∈ σ(T )} = lim kT n k n = inf kT n k n ≤ kT k. n→∞ n∈N (4.1) By the next result, a C0 -semigroup automatically decays exponentially if it tends to 0 in operator norm as t → 0. Proposition 4.2. Let T (·) be a C0 -semigroup with generator A. Then the following assertions are equivalent. a) T (·) is exponentially stable. b) kT (t0 )k < 1 for some t0 > 0. c) r(T (t1 )) < 1 for some t1 > 0. d) ω0 (A) < 0. If this is the case, then statement b) is valid for all sufficiently large t0 > 0, assertion c) is true for all t1 > 0, and s(A) < 0, cf. (1.11). We further have ets(A) ≤ etω0 (A) = r(T (t)) for all t ≥ 0 and (with ln 0 := −∞). 1 t→∞ t ω0 (A) = lim 1 t>0 t ln kT (t)k = inf 90 ln kT (t)k. 4.1. Exponential stability and dichotomy 91 Proof. Since ln kT (t + s)k ≤ ln kT (t)k + ln kT (s)k, the elementary Lemma IV.2.3 in [EN] shows that the limit limt→∞ 1t ln kT k exists and equals ω := inf t>0 1t ln kT k. This equality yields etω ≤ kT (t)k for all t ≥ 0 and thus ω ≤ ω0 (A). Take any ω1 > ω. By the description via the limit, there is a time t0 ≥ 0 such that kT (t)k ≤ eω1 t for all t ≥ t0 so that kT (t)k ≤ M eω1 t for all t ≥ 0 and the number M := max{eω1 t kT (t)k | 0 ≤ t ≤ t0 } ≥ 1. This means that ω1 ≥ ω0 (A) and so ω = ω0 (A). Using (4.1), we infer the identities     1 1 r(T (t)) = lim exp t ln kT (nt)k = exp t lim ln(kT (nt)k) = etω0 (A) n→∞ n→∞ nt nt for all t > 0. All other assertions about T (·) now follow. Proposition 1.21 says that s(A) ≤ ω0 (A), which yields the remaining inequality ets(A) ≤ etω0 (A) .  For bounded A, Example 5.4 of [ST] yields the equality s(A) = ω0 (A). The next example due to Arendt (1993) shows that s(A) < ω0 (A) is possible for unbounded generators. See also Examples IV.2.7 and IV.3.4 as well as Exercises IV.2.13 and IV.3.5 in [EN]. Example 4.3. Let X = Lp (1, ∞) ∩ Lq (1, ∞) for 1 < p ≤ q < ∞ which is a reflexive Banach space for the norm kf k = kf kp + kf kq . We look at the positive operators (T (t)f )(s) = f (set ) for t ≥ 0, f ∈ X and s > 1. Let also τ ≥ 0. Computing (T (t)T (τ )f )(s) = (T (τ )f )(set ) = f (set eτ ) = (T (t + τ )f )(s), we see that T (·) is a semigroup. Let r ∈ (1, ∞) and f ∈ Lr (1, ∞). We estimate Z ∞ Z ∞ t r r |f (τ )|r e−t dτ ≤ e−t kf kr , |f (se )| dr = kT (t)f kr = 1 et where we substituted τ = set . For f ∈ X it follows kT (t)f k = kT (t)f kp + kT (t)f kq ≤ e−t/p kf kp + e−t/q kf kq ≤ e−t/q kf k so that T (t) belongs to B(X) with growth bound ω0 (T ) ≤ −1/q. Let f ∈ Cc (1, ∞) and t ∈ (0, 1]. There is a number s0 > 1 such that f (set ) = 0 for all s ≥ s0 . By uniform continuity, the maps T (t)f tend to f uniformly as t → 0, and thus in X due to the bounded support. Lemma 1.7 now yields that T (·) is C0 -semigroup. Let A be its generator. Let r ∈ (1, ∞). Taking p = q = r, we also obtain a C0 -semigroup Tr (·) on Lr (1, ∞) with generator Ar . Let ft = ✶[et ,et +1] for t ≥ 0. Observe that kft kr = 1 and so kft k = 2. Since T (t)ft (s) = ✶[et ,et +1] (set ) = ✶[1,1+e−t ] (s) for s > 1, we have kT (t)ft kr = e−t/r . It follows that kT (t)ft k ≥ kT (t)ft kq = e−t/q = 12 e−t/q kft k and hence ω0 (T ) = ω0 (A) = −1/q. To determine s(A), we look at the functions gα (s) = s−α for s > 1 and α > 1/r. Then gα belongs to Lr (1, ∞) and
1 −αt 1 − 1) + α gα . t (T (t)gα − gα ) + αgα = t (e 4.1. Exponential stability and dichotomy 92 These maps clearly tend to 0 in Lr (1, ∞) as t → 0 so that gα belongs to D(Ar ) with Ar gα = −αgα . This means that −α ∈ σp (Ar ) and so s(Ar ) ≥ −1/r. As ω0 (Ar ) = −1/r, Proposition 1.21 shows that s(Ar ) = ω0 (Ar ) = −1/r. We now pass to X. Since X ֒→ Lp (1, ∞) and T (t) = Tp (t)|X , A is the ‘part of Ap in X’ (i.e., Af = Ap f and D(A) = {f ∈ D(Ap ) ∩ X | ApRf ∈ X}) by ∞ Proposition II.2.3 in [EN]. Proposition 1.21 yields R(0, Ap )f = 0 Tp (t)f dt. We first take f ∈ Cc (1, ∞) with f (s) = 0 for s ≥ s0 . Since Tp (t)f = 0 for all t > ln s0 , the integral converges not only in Lp (1, ∞) but in C0 (1, ∞). We infer Z ∞ Z ∞  Z ∞ dτ f (τ ) , (4.2) f (set ) dt = Tp (t)f dt (s) = R(0, Ap )f (s) = τ s 0 0 substituting τ = set . Hölder’s inequality now implies  s1−p′  1′  1′ Z ∞ s−1/p p p −p′ kf kp . = kf kp ′ = ′ τ dτ |R(0, Ap )f (s)| ≤ kf kp p −1 (p − 1)1/p′ s We then approximate a given function f ∈ Lp (1, ∞) by fn ∈ Cc (1, ∞) and derive (4.2) for such f and a.e. sR> 1. ∞ We finally take q > p. Then 1 s−q/p ds is finite, so that R(0, Ap ) continuously maps into X; i.e., [D(Ap )] ֒→ X ֒→ Lp (1, ∞). Proposition IV.2.17 of [EN] thus shows that σ(A) = σ(Ap ), and so s(A) = −1/p < −1/q = ω0 (A) in view of the above results. Rescaling with a number ω ∈ (1/q, 1/p), we hence obtain a generator A + ωI of an exponentially growing C0 -semigroup with the negative spectral bound ω − 1/p. ♦ As the best possible identity s(A) = ω0 (A) fails in general, one can try to show exponential stability under stronger assumptions. We will first establish it assuming an additional bound of the resolvent. In the next section we actually prove s(A) = ω0 (A) (and more) for a class of C0 -semigroups with better regularity properties including analytic ones. We will also comment on results about weaker convergence properties. In infinite dimensions it is often more approriate to complement spectral conditions by resolvent estimates. To establish a corresponding stability theorem, we need some properties of the Bochner integral and the Fourier transform, Let J ⊆ R be an interval. Simple functions f : J → X and their integral are defined as in the case X = R. A function f : J → X is called strongly measurable if there are simple functions fn : J → X converging to f pointwise almost everywhere. Observe that then the function t 7→ kf (t)k is measurable. By Theorem X.1.4 in [AE], the map f is strongly measurable if and only if f is Borel measurable and there is a null set N ⊆ J such that f (J \ N ) is separable. (The latter is true for separable X, of course.) We then define the space Lp (J, X) = {f : J → X | f is strongly measurable, kf (·)kX ∈ Lp (J)}, 1 Z p p kf (t)kX dt kf kp = k kf (·)kX kLp (J) = J for p ∈ [1, ∞) and analogously for p = ∞. Here we identify functions that coincide almost everywhere. Bochner’s Theorem X.3.14 in [AE] says that f belongs to L1 (J, X) if and only if there are simple functions converging to f pointwise 4.1. Exponential stability and dichotomy 93 a.e. such that the sequence (fn )n is a Cauchy R for k · k1 , see Theorem X.3.14 in [AE]. This fact implies that the integrals J fn (t) dt converge in X and that their limit isR independent of the choice of such a sequence (fn )n . This limit is denoted by J f (t) dt and called the (Bochner) integral of f . It can be shown that (Lp (J, X), k·kp ) is a Banach space and that the Bochner integral satisfies the analogues of Hölder’s inequality and the theorems of RieszFischer, Lebesgue and Fubini, see Chapter X of [AE]. We note that the dual ′ of Lp (J, X) for p ∈ [1, ∞) coincides with Lp (J, X ∗ ) only for certain classes of Banach spaces X, e.g., reflexive ones. Otherwise the dual is larger. (See Sections 8.18 and 8.20 of [Ed].) 1 Let A be closed and R f ∈ L (J, X) take values in D(A) a.e. and Af be integrable. The integral J f dt then belongs to D(A) and fulfills Z Z A f (t) dt = Af (t) dt, J J by Theorem C.4 of [EN]. For f ∈ L1 (R, X) we define the Fourier transform Z 1 fb(τ ) = Ff (τ ) = √ e−iτ t f (t) dt, τ ∈ R. 2π R As in the scalar case one shows that fb ∈ C0 (R, X) and the convolution and inversion theorems, see Theorem 1.8.1 of [ABHN]. Let X be Hilbert space. By Plancherel’s Theorem 1.8.2 of [ABHN], the Fourier transform extends from L1 (R, X) ∩ L2 (R, X) to a unitary operator F : L2 (R, X) → L2 (R, X) where L2 (R, X) is a Hilbert space with the inner product Z (f |g) = (f (t)|g(t))X dt, f, g ∈ L2 (R, X). R In the theorem below we also need the next auxiliary result by Datko (1970). Lemma 4.4. Let T (·) be a C0 -semigroup and 1 ≤ p < ∞. If T (·)x ∈ Lp (R≥0 , X) for all x ∈ X, then T (·) is exponentially stable. Proof. Define the bounded operator Φn : X → Lp (R≥0 , X); x 7→ ✶[0,n] T (·)x, for each n ∈ N. The assumption shows that supn∈N kΦn (x)k is finite for all x ∈ X, and hence C := supn∈N kΦn k < ∞ thanks to the principle of uniform Rt boundedness. As a result, 0 kT (s)xkp ds ≤ C p kxkp for all t ≥ 0 and x ∈ X. Fix constants M ≥ 1 and ω > 0 such that kT (t)k ≤ M eωt for all t ≥ 0. Let t ≥ 1 and x ∈ X. We calculate Z t 1 − e−pωt 1 − e−pω kT (t)xkp ≤ kT (t)xkp = e−pωs kT (s)T (t − s)xkp ds pω pω 0 Z t M p eωsp e−ωsp kT (t − s)xkp ds ≤ 0 4.1. Exponential stability and dichotomy =M p Z t 0 94 kT (τ )xkp dτ ≤ (CM )p kxkp , so that kT (t)k ≤ N for all t ≥ 0, where N := max{M eω , (pω)1/p CM (1 − e−pω )−1/p }. It follows Z t Z t p p p kT (s)xkp ds ≤ (CN )p kxkp , tkT (t)xk = kT (t − s)T (s)xk ds ≤ N 0 0 and hence kT (t)k ≤ CN . t1/p Proposition 4.2 now implies the assertion.  We fist give a heuristic argument for the following stability theorem. Let A generate the C0 -semigroup T (·) on a Hilbert space X. Assume that s(A) < 0. Pick a number ω > ω0 (A). We set ( e−ωt T (t), t ≥ 0, Tω (t) = 0, t < 0. Then there are constants M ≥ 1 and ε > 0 such that kTω (t)k ≤ M e−εt for all t ≥ 0. Take x ∈ X and τ ∈ R. The map Tω (·)x belongs to L1 (R, X) ∩ L2 (R, X) with 2-norm less or equal M (2ε)−1/2 kxk. Using Proposition 1.21, we compute Z ∞ 1 1 e−iτ t e−ωt T (t)x dt = √ R(ω + iτ, A)x. (4.3) F(Tω (·)x)(τ ) = √ 2π 0 2π Plancherel’s theorem then yields p √ kR(ω + i·, A)xkL2 (R,X) = 2π kTω (·)xkL2 (R,X) ≤ M π/ε kxk. (4.4) We want to transform this inequality to the imaginary axis. From the resolvent equation (1.7) we infer R(iτ, A)x = R(ω + iτ, A)x + ωR(iτ, A)R(ω + iτ, A)x. (4.5) Assuming the boundedness kR(i·, A)k on R, from the above results we deduce that R(i·, A)x is an element of L2 (R, X). It is now tempting to use Plancherel’s theorem once more and to conclude √ ∞ > kR(i·, A)kL2 (R,X) = kF(T0 (·)x)kL2 (R,X) = 2π kT (·)xkL2 (R+ ,X) . Dakto’s Lemma would then yield ω0 (A) < 0. However, we need the assertion ω0 (A) < 0) before to use (4.3) for ω = 0 and to apply F to T0 (·)x. These problems can actually be settled using a refined version of (4.3) and an approximation arguiment, see the proof of Theorem V.1.11 of [EN]. Below we instead use a shorter argument taken from Theorem 5.2.1 of [ABHN]. The resulting stability theorem of Gearhart is special case of Theorem 4.17. Theorem 4.5. Let X be a Hilbert space. A C0 -semigroup T (·) with generator A is exponentially stable if and only if s(A) ≤ 0 and C := supλ∈C+ kR(λ, A)k < ∞. If this is the case, then s(A) is negative. 4.1. Exponential stability and dichotomy 95 Proof. The necessity of the conditions and the addendum follow from Proposition 1.21. Let the conditions in display be true. We set ω+ = max{0, ω0 (A)}. Take ω > ω+ , α > 0, x ∈ X, and define Tα (·) as above. For τ ∈ R, we also abbreviate rα (τ ) = R(α + iτ, A)x. Fix ω > ω+ . There exist constants M ≥ 1 and ε > 0 such that kT (t)k ≤ M e(ω−ε)t for all t ≥ 0 and so Tω (·)x is an element L1 (R, X) ∩ L2 (R, X) with 2-norm less or equal M (2ε)−1/2 kxk. As in (4.3)–(4.5) we thus obtain p krω kL2 (R,X) ≤ M π/ε kxk, krα kL2 (R,X) ≤ krω kL2 (R,X) + |ω − α| krα kL∞ (R,X) krω kL2 (R,X) p √ ≤ M π/ε (1 + |ω − α| C) kxk =: 2π c(α) kxk, kTω (·)xkL2 (R,X) = √1 2π krω kL2 (R,X) ≤ c(ω) kxk. for all α > 0 and ω > ω+ . Fatou’s lemma thus yields Z ∞ 2 lim e−2ωt kT (t)xk2 dt ≤ lim kTω (·)xk2L2 (R,X) kTω+ (·)xkL2 (R,X) = 0 ω→ω+ ω→ω+ M 2 (1 + ωC)2 kxk2 . ω→ω+ 2ε Dakto’s Lemma 4.4 thus implies that (Tω+ (t))t≥0 is exponentially stable. This is impossible if ω+ = ω0 (A) so that ω0 (A) has to be negative.  ≤ lim c(ω)2 kxk2 = In a general Banach space X the boundedness of the resolvent R(·, A) on C+ only implies the existence of some constants M, ε > 0 such that we have kT (t)xk ≤ M e−εt kxkA (4.6) for all t ≥ 0 and x ∈ D(A) by a result due to Weis and Wrobel, see Proposition 5.1.6 and Theorem 5.1.7 in [ABHN]. We thus obtain exponential decay of classical solutions only. In Example 4.3, the resolvent of A + ωI is bounded on C+ by Theorem 5.3. There are generators A with s(A) < 0 such that (4.6) fails, see Remark 5.5. We add a typical example for Theorem 4.5, concerned with wave equations having a strictly positive damping.1 Example 4.6. We first recall the setting and the results of Example 3.7. Let G ⊆ R3 be bounded and open with a C 1 -boundary, ∆D be the Dirichlet– Laplacian on L2 (G), and b ∈ L∞ (G) satisfy b(x) ≥ β for almost every x ∈ G and some β > 0. We set E = Y × L2 (G), where Y = W01,2 (G) is endowed with the norm kvkY = k |∇v|2 k2 from (1.34), and define the operator   0 I A= with D(A) = D(∆D ) × Y ∆D −b on E. It generates a C0 -group T (·) solving the damped wave equation u′′ (t) = ∆D u(t) − bu′ (t), t ≥ 0, u(0) = u0 , u′ (0) = u1 . (4.7) More precisely, for (u0 , u1 ) ∈ E the orbit w(t) = T (t)(u0 , u1 ) has the form w = (u, u′ ) for the unique solution u of (4.7) in C 2 (R≥0 , W −1,2 (G)) ∩ 1In the lectures we presented a different version of the proof. 4.1. Exponential stability and dichotomy 96 C 1 (R≥0 , L2 (G)) ∩ C(R≥0 , Y ). Here we consider the operator ∆D also as a map from Y = W01,2 (G) to W −1,2 (G). We first check that A is dissipative. For b = 0 it is skewadjoint by Example 1.53. For w = (ϕ, ψ) ∈ D(A) we can thus comupte Z Re(Aw|w)E = − b |ψ|2 dx ≤ 0. G as desired. We assert that T (·) is exponentially stable, and thus the ‘energy’ kT (t)(u0 , u1 )k2E = k |∇u|2 k22 + k∂t u(t)k22 of the solution decays as ce−2εt k(u0 , u1 )k2E for some c, ε > 0. This claim is proved by means of Theorem 4.5. To this end, we first we first note that A is invertible with bounded inverse    −1  ∆D (bf + g) −1 f A = , (f, g) ∈ E. g f We next show that iR ⊆ ρ(A) and supiτ ∈R kR(iτ, A)k =: κ < ∞. (4.8) In  of Remark 1.17, by inequality (4.8) each number λ ∈ C with |Re λ| ∈  view 1 0, 2κ is an element of ρ(A) and the resolvent is bounded by kR(λ, A)k ≤ 2C. Due this bound and the Hille-Yosida estimate (1.14), the assumptions of Theorem 4.5 are fulfilled and the assertion follows. We establish (4.8). Since s(A) ≤ 0, any point iτ ∈ σ(A) would belong to ∂σ(A) so that Proposition 1.19 of [ST] (or (4.15) below) would yield m(τ ) := inf{kiτ w − AwkE | w ∈ D(A), kwkE = 1} = 0. Note that kR(iτ, A)k ≤ 1/m(τ ) if m(τ ) > 0. Therefore the lower bound inf τ ∈R m(τ ) =: m0 > 0 implies our claim (4.8) with κ = 1/m0 . Since 0 ∈ ρ(A) and ρ(A) is open, there is a number τ0 > 0 such that [−iτ0 , iτ0 ] ⊆ ρ(A). For τ ∈ [−τ0 , τ0 ] and w ∈ D(A) with kwkE = 1, we set iτ w − Aw = z and obtain the first bound kiτ w − AwkE = kzkE ≥ kR(iτ, A)k−1 kR(iτ, A)zkE = kR(iτ, A)k−1 ,  −1 > 0. inf m(τ ) ≥ max kR(iτ, A)k |τ |≤τ0 |τ |≤τ0 3εβ < τ0 . Suppose there are |τ | ≥ τ0 and w = (ϕ, ψ) ∈ Fix ε ∈ (0, β2 ) with 0 < β−2ε 2 D(A) such that kwkE = k|∇ϕ|k22 + kψk22 = 1 a nd kiτ w − AwkE ≤ ε. We infer      ϕ ϕ ε ≥ (iτ I − A) ψ ψ Z Z
= − ∆D ϕ + (iτ + b)ψ ψ) dx ∇(iτ ϕ − ψ) · ∇ϕ dx + G G Z Z Z 2 2 ∇ϕ · ∇ψ dx + ∇ϕ · ∇ψ dx + b |ψ|2 dx = iτ (k|∇ϕ|k2 + kψk2 ) − G G G Z Z b|ψ|2 dx , = i(τ + 2 Im ∇ϕ · ∇ψ dx) + G G 4.1. Exponential stability and dichotomy 97 using the definition of ∆D . The imaginary and real parts thus satisfy Z Z ε ≥ τ + 2 Im ∇ϕ · ∇ψ dx and ε ≥ b |ψ|2 dx ≥ β kψk22 , G G The second estimate yields k|∇ϕ|k22 =1− 1 − 2 k|∇ϕ|k22 ≤ kψk22 ≥ 1 − βε , and hence 2ε −1<0 β because of ε < β2 . We conclude that   Z 2ε 2 |τ | 1 − ≤ |τ | 1 − 2 k|∇ϕ|k2 = τ + 2 Im ∇ϕ · iτ ∇ϕ dx β G Z Z
∇ϕ · ∇ψ dx + 2 Im ≤ τ + 2 Im ∇ϕ · iτ ∇ϕ − ∇ψ dx G G ≤ ε + 2 k|∇ϕ|k2 k|∇(iτ ϕ − ψ)|k2 ≤ ε + 2 k(iτ I − A)wkE ≤ 3ε by the choice of w = (ϕ, ψ) and the definition of A. It follows |τ | ≤ This contradiction yields m(τ ) ≥ ε > 0 for all |τ | ≥ τ0 , as needed. 3εβ β−2ε < τ0 . ♦ We next introduce a more sophisticated concept for the long-time behavior. Definition 4.7. A C0 -semigroup T (·) has an exponential dichotomy if there are constants N, δ > 0 and a projection P = P 2 ∈ B(X) such that T (t)P = P T (t), T (t) : N (P ) → N (P ) has an inverse denoted by Tu (−t), and we have the estimates kT (t)P k ≤ N e−δt and kTu (−t)(I − P )k ≤ N e−δt for all t ≥ 0. Setting Q = I − P , we recall from Lema 2.16 in [FA] that N(P ) = QX and Q = Q2 . Observe that exponential dichotomy coincides with exponential stability if P = I. Moreover, exponential dichotomy means that T (t)Xj ⊆ Xj for all t ≥ 0 where j = {s, u}, Xs := P X and Xu := QX, that Ts (·) := T (·)|Xs is an exponentially stable C0 -semigroup on Xs and that T (·) induces a C0 -group Tu (·) on Xu which is exponentially stable in backwards time. (Use Lemma 1.29 for the group property.) We first characterize this notion in terms of the spectrum of T (t). Proposition 4.8. A C0 -semigroup T (·) has an exponential dichotomy if and only if S1 := {λ ∈ C | |λ| = 1} ⊆ ρ(T (t)) for some (and hence all) t > 0. Proof. Let T (·) have an exponential dichotomy. Take t > 0 and λ ∈ S1 . Then the series ∞ ∞ X X −1 −n −1 Rλ = λ λ T (nt)P − λ λn Tu (−nt)Q n=0 n=1 converges in B(X). We then compute ∞  ∞  −n  n   X X −1 −1 Q λ−1 Tu (t) λ T (t) P − (λI − T (t))Rλ = I − λ T (t) n=1 n=0 = ∞  X n=0 λ−1 T (t) n P− ∞  X k=1 λ−1 T (t) k P 4.1. Exponential stability and dichotomy − ∞  X λ −1 Tu (t) n=1 = P + Q = I. −n Q+ ∞  X λ k=0 −1 Tu (t) 98 −k Q Similarly one sees that Rλ (λI − T (t)) = I, and hence T is contained in ρ(T (t)) for all t > 0. Conversely, let T ⊆ ρ(T (t)) for some t > 0. We define the ‘spectral projection’ Z 1 R(λ, T (t)) dλ. P := 2πi T Theorem 5.5 in [ST] implies that P 2 = P ∈ B(X) commutes with T (t), σ(Ts (t)) = σ(T (t)) ∩ B(0, 1), and σ(Tu (t)) = σ(T (t)) \ B(0, 1) for all t > 0. Since r(Ts (t)) < 1, Proposition 4.2 yields the exponential stability of Ts (·) on P X. Moreover, Tu (t) is invertible and σ(Tu (t)−1 ) = σ(Tu (t))−1 ⊆ B(0, 1) by Proposition 1.20 in [ST]. As for Ts (·), we infer that (Tu (t)−1 )t≥0 is exponentially stable on QX. Consequently, T (·) has an exponential dichotomy.  In Corollary 4.16 and Theorem 4.17 we characterize exponential dichotomy in terms of A in certain situations. Here we give a typical implication of this property on the long-time behavior of inhomogeneous problems. Proposition 4.9. Let A generate the C0 -semigoup T (·) having an exponential dichotomy with projections P and Q = I − P . Assume that u0 ∈ X and f ∈ C(R≥0 , X) satisfy Z ∞ Qu0 = − Tu (−t)Qf (t) dt. 0 Then the mild solution u of the inhomogeneous problem (2.6) on R≥0 also belongs to C(R≥0 , X) and fulfills Z t Z ∞ u(t) = T (t)P u0 + T (t − s)P f (s) ds − Tu (t − s)Qf (s) ds, t ≥ 0. (4.9) 0 t Proof. Let t ≥ 0. We first note that the integrals in the displayed equations above and those below exist because of the exponenial dichotomy. Using Duhamel’s formula (2.7) and P + Q = I, we compute Z t Z ∞ u(t) = T (t)u0 + T (t − s)P f (s) ds + Tu (t − s)Qf (s) ds 0 0 Z ∞ Tu (t − s)Qf (s) ds − t Z t Z ∞ = T (t)u0 + T (t − s)P f (s) ds + T (t) Tu (−s)Qf (s) ds 0 0 Z ∞ Tu (t − s)Qf (s) ds, − t so that the assumption yields the second assertion. 4.2. Spectral mapping theorems 99 Let ε > 0. There is a time sε such that kf (s)k ≤ ε for all s ≥ s0 . Let t ≥ s0 . Formula (4.9) and the exponential dichotomy lead to the estimate Z t Z s0 −δt −δ(t−s) ku(t)k ≤ N e ku0 k + Ne kf k∞ ds + N e−δ(t−s) ε ds s0 0 Z ∞ −δ(s−t) Ne kf k∞ ds + t
≤ N e−δt ku0 k + δ −1 (eδs0 − 1)kf k∞ + 2N δ −1 ε, which easily implies that u(t) → 0 as t → ∞.  4.2. Spectral mapping theorems Let A generate the C0 -semigroup T (·). We say that T (·) or A satisfy the spectral mapping theorem if σ(T (t)) \ {0} = etσ(A) for all t ≥ 0, (4.10) where we put et∅ = ∅. Observe that we have to exclude 0 on the left-hand side since 0 does not belong to etσ(A) . Theorem 3.5 of [ST] shows even the identity σ(T (t)) = etσ(A) for A ∈ B(X). Assume for a moment that spectral mapping theorem is true. It then implies r(T (t)) = max{|etµ | | µ ∈ σ(A)} = max{et Re µ | µ ∈ σ(A)} = ets(A) (4.11) T (·) has exp. dichotomy ⇐⇒ S1 ⊆ σ(T (1)) ⇐⇒ iR ⊆ ρ(A). (4.12) for all t ≥ 0 and hence the equality s(A) = ω0 (A) by Proposition 4.2. Using also Proposition 4.8, we also deduce from (4.10) the equivalence Example 4.3 thus tells us that the spectral mapping theorem is not valid for all C0 -semigroups. We first explore which partial results are still true. For this purpose, we recall the following concepts and results from spectral theory for a closed operator A. We define by σp (A) = {λ ∈ C | λI − A is not injective}, σap (A) = {λ ∈ C | ∀ n ∈ N ∃ xn ∈ D(A) : kxn k = 1, λxn − Axn → 0 (n → ∞)}, σr (A) = {λ ∈ C | (λI − A)D(A) is not dense} the point spectrum, the approximate point spectrum and the residual spectrum of A, respectively. We call the elements of σap (A) approximate eigenvalues and the corresonding vectors approximate eigenvectors. Proposition 1.19 of [ST] shows that σap (A) = σp (A) ∪ {λ ∈ C | (λI − A)D(A) is not closed}, σ(A) = σap (A) ∪ σr (A), ∂σ(A) ⊆ σap (A). (4.13) (4.14) (4.15) Let A be also densely defined. Theorem 1.24 of [ST] then says that σr (A) = σp (A∗ ), σ(A) = σ(A∗ ), and R(λ, A)∗ = R(λ, A∗ ) (4.16) for λ ∈ ρ(A). The following spectral inclusion theorem provides the easy inclusion in (4.10) and in related formulas for the parts of the spectrum. 4.2. Spectral mapping theorems 100 Proposition 4.10. Let A generate the C0 -semigroup T (·). We then have etσ(A) ⊆ σ(T (t)) and etσj (A) ⊆ σj (T (t)) for all t ≥ 0 and j ∈ {p, ap, r}. (Approximate) Eigenvectors of A for the (approximate) eigenvalue λ are (approximate) eigenvectors of T (t) for the (approximate) eigenvalue etλ . Proof. Let λ ∈ C and t ≥ 0. Recall from Lemma 1.19 that Z t λt e x − T (t)x = (λI − A) eλ(t−s) T (s)x ds for x ∈ X, 0 Z t eλ(t−s) T (s)(λx − Ax) ds for x ∈ D(A). = 0 Hence, if λx = Ax for some x ∈ D(A) \ {0}, then eλt x = T (t)x and x is an eigenvector of T (t) for the eigenvalue eλt ∈ σp (T (t)). If (λI − A)D(A) is not dense or not equal to X, then R(eλt I − T (t)) has the same property. Finally, let xn be approximate eigenvectors of A for λ ∈ σap (A). It follows that keλt xn − T (t)xn k ≤ c kλxn − Axn k −→ 0 as n → ∞ so that xn are approximate eigenvectors for eλt ∈ σap (T (t)).  We have thus shown the inequality s(A) ≤ ω0 (A) from Proposition 4.2 again. We also obtain the analogous implication for exponential dichotomy. Corollary 4.11. Let A generate the C0 -semigroup T (·) having an exponential dichotomy. We then have iR ⊆ ρ(A) since S1 ⊆ ρ(T (1)) ⊆ C \ eσ(A) by Propositions 4.8 and 4.10. In the following example we use the spectral inclusion to compute the spectra of the translation semigroup on 1-periodic functions. Here the spectral mapping theorem fails for irrational t, but a variant with an additional closure holds. Example 4.12. Let X = {f ∈ C(R) | ∀ t ∈ R : f (t) = f (t + 1)} be endowed with the supremum norm and T (t)f = f (· + t) for t ∈ R and f ∈ X. It is easy to see that X is a Banach space and that T (·) is an isometric C0 -group on X (since each f ∈ X is uniformly continuous). As in Example 1.22 one can verify that the generator A of T (·) is given by Af = f ′ with D(A) = C 1 (R) ∩ X. Let Γk = {λ ∈ C | λk = 1} for k ∈ N. We claim that σ(A) = σp (A) = 2πiZ, ( S1 = exp(tσ(A)), σ(T (t)) = Γk = etσ(A) , t ∈ R≥0 \ Q, t = j/k, j, k ∈ N, without common divisors. Proof. Clearly, e2πin belongs to D(A) and Ae2πin = 2πine2πin for all n ∈ Z. Note that T (n) = I for all n ∈ N0 . Proposition 4.10 thus yields eσ(A) ⊆ σ(T (1)) = {1} so that σ(A) ⊆ 2πiZ. The first assertion is proved. Since T (t) is isometric and invertible, Proposition 4.2 implies that r(T (t)) = 1 = r(T (t)−1 ) = min{|λ| | λ ∈ σ(T (t))}, 4.2. Spectral mapping theorems 101 where we also use Proposition 1.20 of [ST]. This means that σ(T (t)) is included in S1 for t ≥ 0. If t ∈ R≥0 \ Q, it is known that etσ(A) = et2πiZ is dense in S1 . The second claim now follows from Proposition 4.10 and the closedness of the spectra because of S1 = etσ(A) ⊆ σ(T (t)) ⊆ S1 . Let t = j/k for some j, k ∈ N without common divisors. The spectral mapping theorem for bounded operators from Theorem 5.3 of [ST] then yields
k
σ(T (t))k = σ T kj = σ(T (j)) = {1}; i.e., σ(T (t)) ⊆ Γk . On the other hand, the set etσ(A) = exp(2πi kj Z) is equal to Γk and contained in σ(T (t)) by Proposition 4.10, establishing the last assertion.  In order to use spectral information on A to show exponential stability or dichotomy, we need the converse inclusions in Proposition 4.10. As we have seen they fail in general for the spectrum itself. We next show them for the point and residual spectrum, starting with the spectral mapping theorem for the point spectrum. Theorem 4.13. Let A generate the C0 -semigroup T (·). We then have σp (T (t)) \ {0} = etσp (A) for all t ≥ 0. Proof. We have to prove σp (T (t)) \ {0} ⊆ etσp (A) since the other inclusion was shown in Proposition 4.10. Let t > 0, λ ∈ C and x ∈ X \ {0} such that eλt x = T (t)x. Hence, the function u(s) = e−λs T (s)x has period t > 0. Suppose that all Fourier coefficients √ Z t 2πin 2π √ e− t s u(s) ds, n ∈ Z, t 0 would vanish. Therefore all Fourier coefficients of the scalar function ϕ(t) = hu(t), x∗ i are 0 for any x∗ ∈ X ∗ . Parseval’s formula (see Example 3.17 of [FA]) then yields ϕ = 0, and so u = 0 by the Hahn-Banach theorem. This is wrong and thus there exists an index m ∈ Z with Z t 2πims y := e− t e−λs T (s)x ds 6= 0. 0 Lemma 1.19 shows that y ∈ D(A) and

2πim A − λ + 2πim y = e−λt e− t t T (t)x − x = 0. t Therefore the number µ := λ+ 2πim belongs to σp (A) and so eλt = eµt to etσp (A) . t We have shown σp (T (t)) ⊆ etσp (A) , as needed.  Formula (4.16) now suggests to use duality and derive a spectral mapping theorem for the residual spectrum from Theorem 4.13. Unfortunately, in general T (·)∗ fails to be strongly continuous. (For instance, the adjoint T (·)∗ of the left translations T (·) on L1 (R) are the right translations on L∞ (R) which are not strongly continuous by Example 1.9.) To deal with this problem, we introduce a new concept. 4.2. Spectral mapping theorems 102 Let A generate the C0 -semigroup T (·) and set C = sup0≤t≤1 kT (t)k. We define the sun dual X ⊙ = {x∗ ∈ X ∗ | T (t)∗ x∗ → x∗ as t → 0}. We first check that X ⊙ is a closed subspace of X ∗ being invariant under T (·)∗ . Let x∗n ∈ X ⊙ with x∗n → x∗ in X ∗ as n → ∞. Take ε > 0. There is an index n ∈ N with kx∗n −x∗ k ≤ ε. We fix a time tε ∈ (0, 1] such that kT (t)∗ x∗n −x∗n k ≤ ε for all t ∈ [0, tε ]. Since kT (t)k = kT (t)∗ k by Proposition 4.52 of [FA], it follows kT (t)∗ x∗ − x∗ k ≤ kT (t)∗ k kx∗ − x∗n k + kT (t)∗ x∗n − x∗n k + kx∗ − x∗n k ≤ (2 + C)ε, so that x∗ ∈ X ⊙ and X ⊙ is closed. Clearly, T (·)∗ is a semigroup on X ∗ . Let t, τ ≥ 0 and x∗ ∈ X ⊙ . We then obtain the invariance of X ⊙ by computing T (t)∗ T (τ )∗ x∗ − T (τ )∗ x∗ = T (τ )∗ (T (t)∗ x∗ − x∗ ) −→ 0, T (t)⊙ t → 0, = T (t)|X ⊙ for t ≥ 0 thus form a C0 By Lemma 1.7, the operators semigroup on X ⊙ , endowed with k · kX ∗ . Its generator is denoted by A⊙ . We have to show that the point spectra of the duals and sun duals are the same. Let x∗ ∈ D(A⊙ ). Take x ∈ D(A). We derive hx, A⊙ x∗ i = lim x, 1t (T (t)∗ − I)x∗ = lim t→0 t→0 ∗ 1 t (T (t) − I)x, x∗ = hAx, x∗ i, A⊙ ⊆ A . (4.17) As restrictions, the operators A⊙ and T (t)⊙ satisfy the inclusions σp (A⊙ ) ⊆ σp (A∗ ) and σp (T (t)⊙ ) ⊆ σp (T (t)∗ ) for t ≥ 0. Let x∗ ∈ D(A∗ ) and t ∈ [0, 1]. Lemma 1.19 yields kT (t)∗ x∗ − x∗ k = sup x∈X,kxk≤1 |hx, T (t)∗ x∗ − x∗ i| = sup |hT (t)x − x, x∗ i| kxk≤1 E DZ t D Z t E ∗ ∗ ∗ T (s)x ds, A x = sup = sup A T (s)x ds, x kxk≤1 ∗ ∗ kxk≤1 0 ≤ C kA x k t. 0 This means that x∗ belongs to X ⊙ and hence D(A∗ ) ⊆ X ⊙ . (4.18) Let T (t)∗ x∗ = eλt x∗ for some x∗ ∈ X ∗ \ {0} and t ≥ 0. Take µ ∈ ρ(A∗ ) = ρ(A), cf. (4.16). Note that R(µ, A)∗ = R(µ, A∗ ) is injective and maps X ∗ into D(A∗ ) ⊆ X ⊙ and that it commutes with T (t)∗ . Hence, R(µ, A∗ )x∗ is an eigenvector for T (t)⊙ and the eigenvalue eλt . Let x∗ ∈ D(A∗ ) \ {0} with A∗ x∗ = λx∗ . As above, we obtain the limit D 1Z t E 1 ⊙ ∗ ∗ ∗ A sup T (s)x ds, x∗ − hx, λx∗ i (T (t) x − x ) − λx = t t 0 x∈X,kxk≤1 D 1Z t E = sup x, T (s)∗ A∗ x∗ ds − λx∗ t 0 kxk≤1 Z t 1 T (s)λx∗ ds − λx∗ −→ 0 ≤ t 0 4.2. Spectral mapping theorems 103 as t → 0, using A∗ x∗ = λx∗ and (4.18). We have thus shown σp (A⊙ ) = σp (A∗ ) and σp (T (t)⊙ ) = σp (T (t)∗ ) for all t ≥ 0. (4.19) These equalities also hold for the full spectra. For this and further information we refer to Proposition IV.2.18 and §II.2.6 of [EN]. We now easily obtain the spectral mapping theorem for the residual spectrum. Theorem 4.14. Let A generate the C0 -semigroup T (·). We then have σr (T (t)) \ {0} = etσr (A) for all t ≥ 0. Proof. Let t ≥ 0. Combining (4.16), (4.19) and Theorem 4.13, we obtain σr (T (t)) \ {0} = σp (T (t)∗ ) \ {0} = σp (T (t)⊙ ) \ {0} = etσp (A ⊙) = etσp (A ∗) = etσr (A) .  As a result, the spectral mapping theorem can only fail if we are not able to transport approximate eigenvectors from T (t) to A. This can be done if the semigroup has some additional regularity, as stated in the spectral mapping theorem for eventually norm continuous semigroups. Besides analytic C0 semigroups, this class includes various geenrators arising in mathematical biology, see e.g. Example 5.6 and the comments before Theorem 5.8. Theorem 4.15. Let A generate the C0 -semigroup T (·) and let the map (t0 , ∞) → B(X); t 7→ T (t), (4.20) be continuous (in operator norm) for some t0 ≥ 0. Then T (·) satisfies the spectral mapping theorem σ(T (t)) \ {0} = etσ(A) for all t ≥ 0. Assumption (4.20) is true if T (·) is analytic (then t0 = 0) or if T (t0 ) is compact for some t0 > 0. Proof. Let T (t0 ) be compact. Then the set T (t0 )B X (0, 1) is compact. By an exercise in Functional Analysis, the map [t0 , ∞) → B(X); t 7→ T (t)x = T (t − t0 )T (t0 )x, thus is uniformly continuous for x ∈ B X (0, 1) and so (4.20) is true. In view of Proposition 4.10, Theorem 4.14 and formula (4.14), it remains to show that σap (T (t)) \ {0} ⊆ etσap (A) for all t > 0. To this aim, let λ ∈ C, τ > 0 and xn ∈ X satisfy kxn k = 1 for all n ∈ N and λxn − T (τ )xn → 0 as n → ∞. We look for a number µ ∈ σap (A) with λ = eτ µ . Considering the C0 -semigroup (e−νs T (sτ ))s≥0 with λ = eν and its generator B = τ A − νI, see Lemma 1.18, we can assume that λ = 1 and τ = 1. Fix some k ∈ N with k > t0 . Let n ∈ N. By (4.20), the map [0, 1] → X; s 7→ T (s)T (k)xn , is continuous uniformly for n; i.e., equi-continuous. Moreover, kT (k)xn − xn k ≤ kT (k − 1)(T (1)xn − xn )k + · · · + kT (1)xn − xn k tends to 0 as n → ∞. This fact implies also that the functions [0, 1] → X; s 7→ T (s)(T (k)xn − xn ), are equi-continuous. Hence, the same is true for the maps [0, 1] → X; s 7→ T (s)xn . 4.2. Spectral mapping theorems 104 Choose x∗n ∈ X ∗ such that kx∗n k ≤ 1 and hxn , x∗n i ≥ 21 for all n ∈ N, using the Hahn-Banach theorem. Since the functions ϕn : [0, 1] → C; s 7→ hT (s)xn , x∗n i, are equi-continuous and uniformly bounded, the Arzelà-Ascoli theorem (see Theorem 1.47 in [FA]) says that a subsequence (ϕnj )j converges in C([0, 1]) to a function ϕ. Observe that kϕk∞ ≥ |ϕ(0)| = lim |ϕnj (0)| = lim |hxnj , x∗nj i| ≥ j→∞ j→∞ 1 2 showing that ϕ 6= 0. Example 3.17 of [FA] thus implies that ϕ has a nonzero Fourier coefficient; i.e., there exists an index mR ∈ Z such that for µ := 2πim R 1 −µs 1 we have 0 e ϕ(s) ds 6= 0. We now set zn = 0 e−µs T (s)xn ds. Lemma 1.19 leads to zn ∈ D(A) and (µI − A)zn = (I − e−µ T (1))xn = xn − T (1)xn −→ 0 as n → ∞. We further compute |hznj , x∗nj i| lim inf kznj k ≥ lim inf ≥ lim inf j→∞ j→∞ j→∞ Z 1 e−µs ϕ(s) ds > 0 = Z 1 0 e−µs hT (s)xnj , x∗nj i ds 0 so that µ ∈ σap (A), completing the proof.  The above theorem yields the desired characterizations (4.11) and (4.12). Corollary 4.16. Let A generate the C0 -semigroup T (·) satisfying (4.20). Then the following equivalences hold. a) The semigroup T (·) is exponentially stable if and only if s(A) < 0. b) The semigroup T (·) has an exponential dichotomy if and only if iR ⊆ ρ(A). We add three other important results on the long-time behavior of semigroups without proof, starting with Gearhart’s spectral mapping theorem. It was shown by Gearhart in 1978 for quasi-contraction semigroups and independently by Herbst (1983), Howland (1984), and Prüss (1984) for general C0 -semigroups. It says that spectral information on A combined with resolvent estimates yield the corresponding spectra for the semigroup, provided that X is Hilbert space. For a proof we refer to Theorem 2.5.4 in [vN]. Theorem 4.17. Let A generate the C0 -semigroup T (·) on a Hilbert space X. Let t > 0 and λ ∈ C. Then 2πik ∈ σ(A), sup kR(λk , A)k < ∞. eλt ∈ σ(T (t)) ⇐⇒ ∀ k ∈ Z : λk := λ + t k∈Z We add two results on weaker decay properties, assuming that the semigroup is bounded. As in (4.6), the first one deals with classical solutions; i.e., initial values in D(A). Since one looks at estimates of T (t) in B(X1 , X), one can obtain decay rates which are not exponential in contrast to convergence in B(X), see Proposition 4.2. To obtain polynomial decay, one can allow for a corresponding increase of the resolvent along iR. 4.2. Spectral mapping theorems 105 Theorem 4.18. Let A generate the bounded C0 -semigroup T (·) on a Hilbert space X and let α > 0. The follwing two assertions are equivalent. a) kT (t)xk ≤ N t−1/α kxkA for some N > 0 and all t ≥ 1 and x ∈ D(A). b) σ(A) ⊆ C− and kR(iτ, A)k ≤ C |τ |α for some C > 0 and all τ ∈ R with |τ | ≥ 1. Property b) and Remark 1.17 imply that | Im λ| ≥ c | Re λ|−1/α for all λ ∈ σ(A) with Re λ ≤ −δ for some c, δ > 0. The implication ‘b) ⇒ a)’ is due to Borichev and Tomilov (see [BT] from 2010), who also constructed an example saying that it fails in an L1 -space. The converse implication was shown by Batty and Duyckaerts in [BD] from 2008 even for general X and other rates. In this more general framework they also proved a variant of ‘b) ⇒ a)’ with logarithmic corrections. In the setting of the above theorem, by density one obtains strong stability of T (·); i.e., T (t)x tends to 0 as t → ∞ for all x ∈ X. But this fact is true in much greater generality, as established already in 1988 by Arendt and Batty as well as, with a different proof, by Lyubich and Vũ. Theorem 4.19. Let A generate the bounded C0 -semigroup T (·) on a Banach space X. Assume that σ(A) ∩ iR is countable and that σ(A∗ ) ∩ iR = ∅. (The latter is true if σ(A) ∩ iR = ∅.) Then T (·) strongly stable. The Lyubich-Vũ can be found in Theorem V.2.21 of [EN], see Lemma V.2.20 for the addendum. A variety of related results are discussed in [ABHN]. CHAPTER 5 Stability of positive semigroups Evolution1 equations often describe the behavior of positive quantities, such as the concentration of a species or the distribution of mass or temperature. It is then a crucial property of the system that non-negative initial functions lead to non-negative solutions. This property of positivity has to be verified in the applications, of course, and we will see below that it implies many additional useful features of the semigroup solving the equation. To deal with positivity, we consider as state spaces only the following classes of Banach spaces E consisting of scalar-valued functions. Standing hypothesis. In this chapter, E denotes a function space of the type Lp (µ), C0 (U ) or C(K), where p ∈ [1, ∞), (S, A, µ) is a σ–finite measure space, U is a locally compact metric space (e.g., an open subset of Rm ), or K is a compact metric space, respectively. We stress that we still take C as the scalar field in order to use spectral theory. Actually, we could work in the more general class of (complex) Banach lattices E, but for simplicity we restrict ourselves to the above indicated setting. It suffices for the typical applications; however for certain deeper investigations one actually needs the more abstract framework. We refer to the monograph [Na-Ed] for a thorough discussion of positive C0 –semigroups in Banach lattices. In the spaces E given by the standing hypothesis, we have the usual concept of non-negative functions f ≥ 0, of positive and negative parts f± and domination f ≤ g of real-valued functions, and of the absolute value |f |. We write E+ = {f ∈ E | f ≥ 0} for the cone of non-negative functions, which is closed in E. For all f, g ∈ E, it holds k |f | k = kf k, and 0 ≤ f ≤ g implies that kf k ≤ kgk. Recall from Example 3.6 that an operator T ∈ B(E) is called positive if T f ≥ 0 for every f ∈ E+ . One then writes T ≥ 0. A C0 –semigroup T (·) is positive if each operator T (t), t ≥ 0, is positive. We discuss a few basic properties of positive operators T, S ∈ B(E) which are used below without further notice. First, products of positive operators are positive. Next, for all f, g ∈ E with f ≥ g we have T (f − g) ≥ 0 ⇐⇒ T f ≥ T g. For real-valued f , also the image T f = T f+ − T f− has real values. Moreover, T f ≤ |T f | ≤ T f+ + T f− = T |f |. For complex-valued f , we take a point x in Ω ∈ {S, U, K}. Take a number α such that |α| = 1 and |T f (x)| = α T f (x), where we fix a representative of T f if E = Lp . It follows that |T f (x)| = α T f (x) = T (Re(αf ))(x) + i T (Im(αf ))(x) = T (Re(αf ))(x) 1This chapter was not part of the lectures. 106 107 ≤ T (| Re(αf )|)(x) ≤ T (|αf |)(x) = T (|f |)(x). Consequently, |T f | ≤ T |f | holds for all f ∈ E. We further write 0 ≤ T ≤ S if 0 ≤ T f ≤ Sf for all f ∈ E+ . Let 0 ≤ T ≤ S. Then |T f | ≤ T |f | ≤ S |f | is true for all f ∈ E, and hence kT k = sup kT f k = sup k |T f | k ≤ sup kS |f | k ≤ kSk. kf k≤1 kf k≤1 kf k≤1 We recall from Corollary 3.25 that the semigroup is positive if and only if there exists a number ω ≥ ω0 (A) such that R(λ, A) ≥ 0 for all λ > ω. In Example 3.26 we have seen that the Dirichlet Laplacian ∆D generates a positive C0 -semigroup on Lp (G) for p ∈ (1, ∞) and C0 (G), where G = Rm or G ⊆ Rm is bounded and open with ∂G ∈ C 2 . To discuss the Neumann Laplacian we need Hopf ’s lemma. For w ∈ C 2 (B) ∩ 1 C (B), it is a special case of the lemma in Section 6.4.2 in [Ev]. Our result can be shown in the same way using Proposition 3.1.10 of [Lu]. Lemma 5.1. Let B = B(y, ρ) ⊂ Rm be an open ball and w belong to Wp2 (B) for all p ∈ (1, ∞) and satisfy 0 ≤ ∆w ∈ C(B). Assume that there is an x0 ∈ ∂B such that w(x0 ) > w(x) for all x ∈ B. Then ∂ν w(x0 ) > 0 for the outer normal ν(x) = ρ−1 (x − y) of ∂B. Example 5.2. Let G ⊆ Rm be open and bounded with boundary of class or let G = Rm . Set E = Lp (G) for p ∈ (1, ∞). The Neumann Laplacian on E is given by ∆N u = ∆u on D(∆N ) = {u ∈ Wp2 (G) | ∂ν u = 0}. One sees as in Example 2.29 that the operator eiθ ∆N is dissipative on Lp (G), if √ 0 ≤ |θ| ≤ arccot( 2|p−2| ) ∈ (0, π/2]. Theorem 9.3.5 in [Kr] further implies that p−1 that I − ∆N is surjective. Consequently, ∆N generates a contractive analytic C0 –semigroup on E by Corollary 2.27. To show positivity, let λ > 0 and 0 ≤ f ∈ C0 (G). Set u = R(λ, ∆N )f . Corollary 3.1.24 in [Lu] implies that u belongs to D(∆N ) for all p ∈ (1, ∞) and ∆u to C(G). As in Example 3.26, we see that u takes real values. Suppose there was a point x0 ∈ G such that u(x0 ) < 0. The function u thus has a minimum u(x1 ) < 0 for some x1 ∈ G. We then have ∆u(x1 ) = λu(x1 ) − f (x1 ) < 0 and so ∆u(x) ≤ 0 for all x in a neighborhood of x1 in G. If x1 ∈ G, Proposition 3.1.10 in [Lu] then yields ∆u(x1 ) ≥ 0 which is impossible. So all such minimina occur on ∂G. Since ∂G is C 2 , we can find an open ball B ⊆ G with B ∩ ∂G = {x1 } on which −u satuisfies the assumptions of Lemma 5.1. Hence, ∂ν v(x1 ) < 0 contradicting u ∈ D(∆N ). We have shown that R(λ, ∆N )f ≥ 0 and by density the resolvent is positive. The positivity of the semigroup then follows from Corollary 3.25. ♦ C 2, The next result collects the basic features of the spectral theory of positive semigroups. For a gerenerator A we define two more quantities s0 (A) = inf{r > s(A) | supµ∈Cr kR(µ, A)k < ∞}, ω1 (A) = inf{ω ∈ R | ∃ Mω ≥ 1 ∀ t ≥ 0, x ∈ D(A) : kT (t)xk ≤ Mω eωt kxkA }. 108 Theorem 5.3. Let A generate the positive C0 –semigroup T (·) on E. Then the following assertions hold. a) Let Re λ > s(A) and f ∈ E. Then the improper Riemann integral Z ∞ e−λt T (t)f dt = R(λ, A)f (5.1) 0 exists. Moreover, kR(λ, A)k ≤ kR(Re λ, A)k. b) s(A) = s0 (A). c) If σ(A) 6= ∅, then s(A) ∈ σ(A). d) For λ ∈ ρ(A), the resolvent R(λ, A) is positive if and only if λ > s(A). e) s(A) = ω1 (A). In particular, if s(A) < 0, then there are N, δ > 0 such that kT (t)xk ≤ N e−δt kxkA for all x ∈ D(A) and t ≥ 0. Proof. a) For λ > ω0 (A), Corollary 3.25 yields that R(λ, A) ≥ 0. If µ ∈ (s(A), λ) with 0 < λ − µ < kR(λ, A)k−1 , the Neumann series gives R(µ, A) = ∞ X n=0 (λ − µ)n R(λ, A)n+1 ≥ 0. Since kR(r, A)k is bounded for r ≥ s(A) + ε and any fixed ε > 0, we deduce the positivity of R(µ, A) for all µ > s(A) (establishing one implication of assertion d)). Let µ > s(A), Re α > 0, f ∈ E and t ≥ 0. We set Z t e−µs T (s)f ds. V (t)f = 0 From Lemma 1.19 we deduce that 0 ≤ V (t)f = R(µ, A)f − R(µ, A)e−µt T (t)f ≤ R(µ, A)f for all f ∈ E+ . Hence, kV (t)k ≤ kR(µ, A)k for all t ≥ 0, and thus the function R+ ∋ t 7→ e−αt V (t)f is integrable. Integrating by parts, we deduce Z t Z t αe−αs V (s)f ds + e−αt V (t)f = e−αs e−µs T (s)f ds 0 0 for all f ∈ E. We can now let t → ∞, obtaining the integral in (5.1) with λ = µ + α on the right hand side. Proposition 1.21 then yields λ ∈ ρ(A) and (5.1). Since we can vary µ > s(A), these results also hold for all Re α ≥ 0. It further follows that Z ∞ Z ∞ −(µ+Re α)t e−µt T (t)|f | dt = R(µ, A)|f |. e |T (t)f | dt ≤ |R(µ + α, A)f | ≤ 0 0 This inequality implies that kR(µ + α, A)k ≤ kR(µ, A)k, and thus the second assertion in a) is true. b) It is clear that s(A) ≤ s0 (A). The converse inequality follows from a) and the fact that kR(r, A)k is bounded for r ≥ s(A) + ε and any fixed ε > 0. c) Assume that σ(A) 6= ∅. We can find λn ∈ ρ(A) tending to σ(A) with Re λn > s(A) > −∞. Assertion a) and Theorem 1.13 in [ST] imply that kR(Re λn , A)k ≥ kR(λn , A)k ≥ d(λn , σ(A))−1 −→ ∞ 109 as n → ∞. If s(A) ∈ ρ(A), then R(Re λn , A) would converge to R(s(A), A) leading to a contradiction. The spectral bound thus belongs to σ(A). d) Let R(λ, A) be positive for some λ ∈ ρ(A). Take 0 6= f ∈ E+ . The function 0 6= u := R(λ, A)f is also positive and Au = limt→0 1t (T (t)f − f ) is real valued. Hence, λu = f + Au is real, so that λ ∈ R. Let µ > max{λ, s(A)}. Part a) of the proof shows that R(µ, A) ≥ 0, and thus R(λ, A) = R(µ, A) + (µ − λ)R(µ, A)R(λ, A) ≥ R(µ, A) ≥ 0. Using s(A) ∈ σ(A) and Theorem 1.13 in [ST], we deduce that 1 1 ≤ ≤ kR(µ, A)k ≤ kR(λ, A)k. µ − s(A) d(µ, σ(A)) If λ ≤ s(A), the limit µ → s(A) would give a contradiction. Hence, d) holds. e) Let λ > s(A) and f ∈ D(A). Assertion a) then implies that Z t −λt e−λs T (s)(A − λI)f ds −→ f + R(λ, A)(A − λI)f = 0 e T (t)f = f + 0 as t → ∞. Hence, e−λt T (t) is bounded in B([D(A)], X) uniformly for t ≥ 0 by the principle of uniform boundedness. This fact implies that ω1 (A) ≤ s(A). Conversely, let Re λ > ω1 (A) and f ∈ D(A). Then the integral Z t e−λt T (t)f dt =: Rλ f 0 converges in E. As in the proof of Proposition 1.21, it follows that Rλ f ∈ D(A) and (λI − A)Rλ f = f . Moreover, Rλ (λI − A)f = f if f ∈ D(A2 ). We denote by A1 the restriction of A to X1 = [D(A)] with domain D(A1 ) = D(A2 ). We have shown that λ ∈ ρ(A1 ). Since A and A1 are similar via the ismorphism R(λ, A) : D(A) → D(A2 ), we arrive at λ ∈ ρ(A); i.e., s(A) = ω1 (A).  The next corollary immediately follows from part b) of the above theorem and Gearhart’s stability Theorem 4.5. Corollary 5.4. Every generator A of a positive semigroup on E = L2 (µ) satisfies s(A) = ω0 (A). Remark 5.5. The above corollary actually holds for all our spaces E, see Section 5.3 in [ABHN], but it fails already on Lp ∩ Lq by Example 4.3. For any generator A, one has s(A) ≤ ω1 (A) ≤ s0 (A) ≤ ω0 (A). (These inequalities follow from the proof of Theorem 5.3 e), Proposition 5.1.6 and Theorem 5.1.7 in [ABHN], and Proposition 1.21.) Hence, in Theorem 5.3 assertion e) follows directly from b) thanks to the (more difficult) general result in [ABHN], which is due to Weis and Wrobel. The positive semigroup in Example 4.3 satisfies s0 (A) < ω0 (A), see Example 5.1.11 in [ABHN]. Moreover, there are (non positive) semigroups on Banach spaces X such that s(A) < ω1 (A) < s0 (A), see ♦ Example 5.1.10 in [ABHN]. As an application we look at a cell division problem. Rb Example 5.6. Let a u(t, s) ds be the number of cells of a certain species at time t ≥ 0 of size s ∈ [a, b]. We make the following assumptions on this species. 110 • Each cell grows linearly with time at (normalized) velocity 1. • Cells of size s ≥ α > 0 divide with per capita rate b(s) ≥ 0 in two daughter cells of equal size, where b = 0 on [1, ∞) and on [α/2, α]. • Cells of size s die with per capita rate µ(s) ≥ 0. • The functions b 6= 0 and µ are continuous, and α > 1/2. • There are no cells at size α/2. It is just a normalization that the cells divide up size s = 1. The assumptions of linear growth and that α > 1/2 are made for simplicity, see [GN2] for the general case. The assumptions on b indicate that the interesting cell sizes belong to J = [α/2, 1] (for others one only has growth and death), so that we choose as state space E = L1 (J). Hence, the norm ku(t)k1 equals the number of (relevant) cells at time t, if u ≥ 0. It can be shown that under the above assumptions smooth cell size distributions u satisfy the equations ∂t u(t, s) = −∂s u(t, s) − µ(s)u(t, s) − b(s)u(t, s) + 4b(2s)u(t, 2s), u(t, α 2) = 0, t ≥ 0, u(0, s) = u0 (s), t ≥ 0, s ∈ J, (5.2) s ∈ J. Note that b(2s) = 0 for s ≥ 1/2. For such s we put v(2s) := 0 for any function v on J. We take 0 ≤ u0 ∈ D(A) := {v ∈ W 1,1 (J) | v(α/2) = 0} and define Av = −v ′ − µv − bv + Bv, Bv(s) = 4b(2s)v(2s), (5.3) for v ∈ D(A), respectively v ∈ E and s ∈ J. Observe that B is a bounded (and positive) operator on E because Z 1/2 |v(2s)| ds ≤ 2 kbk∞ kvk1 . kBvk1 ≤ 4 kbk∞ α/2 d with domain D(A) generates a positive C0 –semigroup on E (the Since − ds nilpotent translations), Example 3.6 shows that also A generates a positive C0 – semigroup T (·) on E. It is clear that thenon-negative map u(t, s) = (T (t)u0 )(s) with t ≥ 0 and s ∈ J belongs to C 1 (R+ , E) ∩ C(R+ , W 1,1 (J)) and satisfies the system (5.2), where the first line holds for a.e. s ∈ J. On the other hand, each solution u ∈ C 1 (R+ , E) ∩ C(R+ , W11 (J)) of (5.2) is given by T (·). ♦ In the above example the embedding D(A) ֒→ E is compact due to Theorem 3.28 in [ST]. Therefore the resolvent of A is compact and σ(A) consists of eigenvalues only, see Remark 2.14 and Theorem 2.16 of [ST]. We can even determine the eigenvalues by the zeros of a holomorphic function ξ. (The assumption α > 21 is only needed to obtain the simple formula of ξ below.) Lemma 5.7. Let A be given by (5.3). Then a number λ ∈ C belongs to σ(A) if and only if Z 1/2   Z 2σ (λ + µ(τ ) + b(τ )) dτ dσ. 4b(2σ) exp − 0 = ξ(λ) := −1 + α/2 σ Proof. As noted above, we have σ(A) = σp (A). Hence, λ ∈ C belongs to σ(A) if and only if there is a map 0 6= v ∈ D(A) with λv = v ′ . Equivalently, 111 0 6= v ∈ W 1,1 (J) satisfies v ′ (s) = −(λ + b(s) + µ(s))v(s), 1/2 ≤ s ≤ 1, ′ v (s) = −(λ + b(s) + µ(s))v(s) + 4b(2s)v(2s), v(α/2) = 0. α/2 ≤ s < 1/2, These equations are only fulfilled by the function given by  Z 1 1 (λ + b(σ) + µ(σ)) dσ , v(s) = c exp 2 ≤ s ≤ 1, s  Z 1 (λ + b(σ) + µ(σ)) dσ v(s) = c exp h · 1− s Z  Z 4b(2σ) exp − 2σ 1/2 s σ  i (λ + µ(τ ) + b(τ )) dτ dσ , α 2 ≤ s < 21 , for any constant c 6= 0. Clearly, this map v belongs to W 1,1 (J), and it satisfies v(α/2) = 0 if and only if ξ(λ) = 0.  Theorem 5.3 shows that ω1 (A) = s(A), and Remark 5.5 even yields ω0 (A) = s(A). In Proposition VI.1.4 of [EN] it is further shown that t 7→ T (t) is continuous in operator norm for t > 1 − α2 . (Here one uses the nilpotency of the semigroup generated by A0 := A − B and the Dyson–Phillips series (3.7) for A = A0 + B.) Therefore the spectral mapping theorem σ(T (t)) = etσ(A) \ {0} is true implying again ω0 (A) = s(A), see Theorem 4.10 and Corollary 4.16. Positivity even yields a very simple criterion for ω0 (A) = s(A) < 0. Theorem 5.8. The semigroup generated by A from (5.3) is exponentially stable on E if and only if Z 1/2  Z 2σ  ξ(0) = −1 + 4b(2σ) exp − (µ(τ ) + b(τ )) dτ dσ < 0. α/2 σ In particular, there are constants N, δ > 0 such that ku(t)k1 ≤ N e−δt ku0 k1 for all t ≥ 0 and all solutions u ∈ C 1 (R+ , E) ∩ C(R+ , W 1,1 (J)) of (5.2). Proof. In view of Lemma 5.7 and the discussion above the statement of the theorem, we have to show that all zeros of ξ have strictly negative real parts. To characterize this property, we use the positivity of the semigroup in a crucial way. Theorem 5.3 says that s(A) ∈ σ(A). 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