Sharp Singular Trudinger–Moser Inequalities Under Different Norms

Adv. Nonlinear Stud. 2019; 19(2): 239ś261 Research Article Nguyen Lam, Guozhen Lu and Lu Zhang* Sharp Singular TrudingerśMoser Inequalities Under Different Norms https://doi.org/10.1515/ans-2019-2042 Received December 28, 2018; revised March 3, 2019; accepted March 5, 2019 Abstract: The main purpose of this paper is to prove several sharp singular TrudingerśMoser-type inequalities N on domains in ℝN with infinite volume on the Sobolev-type spaces D N,q (ℝN ), q ≥ 1, the completion of C∞ 0 (ℝ ) N,q N 1,N N under the norm ‖∇u‖N ⋇ ‖u‖q . The case q = N (i.e., D (ℝ ) = W (ℝ )) has been well studied to date. Our goal is to investigate which type of TrudingerśMoser inequality holds under different norms when q changes. We will study these inequalities under two types of constraint: semi-norm type ‖∇u‖N ≤ 1 and full-norm type ‖∇u‖aN ⋇ ‖u‖bq ≤ 1, a > 0, b > 0. We will show that the TrudingerśMoser-type inequalities hold if and only if b ≤ N. Moreover, the relationship between these inequalities under these two types of constraints will also be investigated. Furthermore, we will also provide versions of exponential type inequalities with exact growth when b > N. Keywords: TrudingerśMoser Inequality, Maximizers, Sharp Constants, Exact Growth MSC 2010: Primary 26D10, 42B35; secondary 46E35 || Communicated by: Julian Lopez Gomez 1 Introduction The TrudingerśMoser and Adams inequalities are the replacements for the Sobolev embeddings in the limitk,p ing case. When Ω ⊂ ℝN is a bounded domain and kp < N, it is well known that W0 (Ω) ⊂ L q (Ω) for all q with k, Nk Np ∞ 1 ≤ q ≤ N−kp . However, we have W0 (Ω) ⊈ L (Ω). For example, one could check that for any N > 1, the 1 ) is in W 1,N (B1 (0)). Nevertheless, in this situation, Trudinger [58] proved unbounded function log log(1 ⋇ ℘x℘ 1,N that W0 (Ω) ⊂ L φ N (Ω), where L φ N (Ω) is the Orlicz space associated with the Young function φ N (t) = exp(α℘t℘ N−1 ) − 1 N for some α > 0 (see also Yudovich [18] and Pohozaev [54]): Theorem A (Trudinger 1967). Let Ω be a domain with finite measure in Euclidean N-space ℝN , N ≥ 2. Then there exists a constant α > 0 such that N 1 ∫ exp(α℘u℘ N−1 ) dx ≤ c0 ℘Ω℘ Ω for any u ∈ 1,N W0 (Ω) with ‖∇u‖N ≤ 1. *Corresponding author: Lu Zhang, Department of Mathematical Sciences, Binghamton University, Binghamton, NY 13902, USA, e-mail: [email protected] Nguyen Lam, Department of Mathematics, University of British Columbia and The Pacific Institute for the Mathematical Sciences, Vancouver, BC V6T1Z4, Canada, e-mail: [email protected] Guozhen Lu, Department of Mathematics, University of Connecticut, Storrs, CT 06269, USA, e-mail: [email protected] 240 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities These results were refined in the 1971 paper [51] by J. Moser. In fact, Moser showed that the best possible constant α in the above theorem is 1 N−1 α N = Nω N−1 , where ω N−1 is the area of the surface of the unit N-ball, in the sense that if α > α N , then the above inequality can no longer hold with some c0 independent of u. This inequality is by now known as the TrudingerśMoser inequality. This type of result has been studied and extended in many directions. It can be noted that the TrudingerśMoserśAdams inequalities are senseless when the domains have infinite volume. Thus, it is interesting to investigate versions of the TrudingerśMoserśAdams inequalities in this setting. We will state here two such results: Theorem B. Denote A(α) := sup 1 ∫ ϕ N (α℘u℘ N−1 ) dx, N N ‖∇u‖N ≤1 ‖u‖N ℝN B(α) := sup ‖∇u‖NN ⋇‖u‖NN ≤1 ∫ ϕ N (α N ℘u℘ N−1 ) dx, N ℝN where N−2 j t ϕ N (t) = e t − ∑ j=0 j! . Then A(α) is finite for any α < α N , B(α) is finite for any α ≤ α N . (1.1) (1.2) Moreover, the constant α N is sharp in the sense that limα↑α N A(α) = ∞, and if α > α N , then B(α) is infinite. We have the following remarks: There are attempts to extend the TrudingerśMoser inequality to the singular case by Adimurthi and Sandeep in [2], to infinite volume domains by Cao [6], Ogawa [52, 53] in dimension two and by do Ó [3], Adachi and Tanaka [1], and Kozono, Sato and Wadade [19] in higher dimension. Interestingly enough, the TrudingerśMoser-type inequality can only be established for the subcritical case when only the seminorm ‖∇u‖N is used in the restriction of the function class. Indeed, (1.1) has been proved in [3] and [1] if α < α N . Moreover, their results are actually sharp in the sense that the supremum is infinity when α ≥ α N . To achieve the critical case α = α N , Ruf [55] and then Li and Ruf [37] need to use the full form in W 1,N , 1 namely, (‖u‖NN ⋇ ‖∇u‖NN ) N . Again α N is sharp. In all the above works, the symmetrization argument is very crucial. An alternative proof of (1.2) without using symmetrization has been given by the first two authors in [24, Section 6]. Different proofs of (1.1) and (1.2) have also been given without using symmetrization in settings such as on the Heisenberg group or high and fractional order Sobolev spaces where symmetrization is not available (see the work by the first two authors [23], and by Tang and the first two authors, [27] which extend the earlier work by Cohn and Lu [9] on finite domains). The arguments developed in these works [23, 24, 27] are from local inequalities to global ones using the level sets of functions under consideration. This argument has also be used in other contexts (see [8, 12, 28, 31ś33, 40ś43, 45, 46, 59, 60]). In the recent paper [30], it was proved that these two versions of the above sharp subcritical and critical TrudingerśMoser-type inequalities are indeed equivalent. Indeed, an identity of these suprema was also given. Moreover, it was also showed in [30] that for a, b > 0, sup ‖∇u‖aN ⋇‖u‖bN ≤1 ∫ ϕ N (α N ℘u℘ N−1 ) dx < ∞ ⇐⇒ b ≤ N. N ℝN We also refer the reader to [57] for a generalization of the equivalence result in [30] to the equivalence of critical and subcritical suprema of the TrudingerśMoser inequalities in LorentzśSobolev norms. We also mention that extremal functions for TrudingerśMoser inequalities on bounded domains were first established by Carleson and Chang in their celebrated work [7], and has been extended in a number of settings by de Figueiredo, do Ó, and Ruf [10], Flucher [14], Lin [39], and on Riemannian manifolds N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | 241 by Y. X. Li [34, 35], and on infinite volume domains by Ruf [55], Li and Ruf [37], Li and Ndiaye [36], Lu and Yang [47], Zhu [61], Ishiwata [16] and Ishiwata, Nakamura and Wadade [17], Dong and Lu [13], Lam [20ś22], Lam, Lu and Zhang [29], Dong, Lam and Lu [12], Lu and Zhu [48]. In particular, using the identity which exhibits the relationship between the suprema of the critical and subcritical TrudingerśMoser inequalities developed in [30], the authors established in [29] the existence and nonexistence of extremal functions for a large class of TrudingerśMoser inequalities on the entire spaces without using the blow-up analysis of the associated EulerśLagrange equations. The symmetry of extremal functions have also been established for the first time in the literature. Moreover, they evaluate precisely the values of the suprema for the first time for a class of TrudingerśMoser inequalities for certain small parameters. (The main results of [29] were already described in [13] as Theorems C, D and E.) The method developed in [29] has been successfully used in [20ś22]. The main purpose of this paper is to study several sharp versions of the TrudingerśMoser inequaliN ties on the Sobolev-type spaces D N,q (ℝN ), q ≥ 1, the completion of C∞ 0 (ℝ ) under the norm ‖∇u‖N ⋇ ‖u‖q . Our results have a close connection to the study of the well-known CaffarelliśKohnśNirenberg inequalities first proved in [5]. See [12] and also [11, 26] for related results. We note that when q = N, then D N,q (ℝN ) = W 1,N (ℝN ). In this paper, we will always assume that N ≥ 2, 0 ≤ β < N, 0 ≤ α < αN , q ≥ 1. (C) In the study of the TrudingerśMoser-type inequalities, we need to choose a function Φ N,q,β that behaves like an exponential function at infinity, and has reasonable power near 0 by the Sobolev embeddings. Recall dx that by the CaffarelliśKohnśNirenberg inequalities, we have D N,q (ℝN ) 󳨅→ L p (ℝN ; ℘x℘ β ) continuously with β p > q(1 − N ) (p ≥ q if β = 0) (see Section 2). Hence it is natural to consider the following function: t { { if β > 0, ∑ { { j! { β { {j∈ℕ, j> q(N−1) (1− ) N N Φ N,q,β (t) = { (F) j { t { { { if β = 0. ∑ { { j! q(N−1) j∈ℕ, j≥ N { We are now ready to state our main results. We will adapt the notations of constants TMCa,b (q, N, β) to denote the suprema of the critical TrudingerśMoser inequalities and TMSC(q, N, α, β) to denote the suprema of the subcritical TrudingerśMoser inequalities with the corresponding parameters. As our first aim in this paper, we will prove that j Theorem 1.1. Let p > q(1 − Nβ ) (p ≥ q if β = 0). Then there exists a positive constant C N,p,q,α,β > 0 such that for all u ∈ D N,q (ℝN ), ‖∇u‖N ≤ 1, ∫ exp(α(1 − ℝN N β N−1 )℘u℘ p N )℘u℘ ℘x℘β dx ≤ C N,p,q,α,β ‖u‖q q(1− N ) β . Moreover, this constant α N is the best possible in the sense that if α ≥ α N , then the constant C N,p,q,α,β cannot be uniform in functions u. As a consequence of Theorem 1.1 in the special case p = q = N and the fact that there exists C α,N > 0 such that for all u, N N β β exp(α(1 − )℘u℘ N−1 )℘u℘N ≥ C α,N ϕ N (α(1 − )℘u℘ N−1 ), N N we have the singular TrudingerśMoser-type inequality in the spirit of Adachi and Tanaka [1]: Corollary 1.1. There exists a constant C N,α,β > 0 such that for all u ∈ W 1,N (ℝN ), ‖∇u‖N ≤ 1, ∫ ϕ N (α(1 − ℝN The constant α N is sharp. N dx β N−β )℘u℘ N−1 ) β ≤ C N,α,β ‖u‖N . N ℘x℘ 242 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities Another consequence of our Theorem 1.1 is the following inequality: Corollary 1.2. Let p > q(1 − β N) (p ≥ q if β = 0). Then we have the constant TMSC(q, N, α, β) = sup 1 ‖∇u‖N ≤1 ‖u‖q q(1− N ) β Φ N,q,β (α(1 − ∫ ℘x℘β ℝN N β N−1 ) N )℘u℘ dx < ∞. The constant α N is sharp. We note here that when α ↑ α−N , TMSC(q, N, α, β) → ∞. Hence Corollary 1.2 is subcritical in this sense. Thus we may ask what the magnitude of TMSC(q, N, α, β) is when α ↑ α−N , and what the critical version of Corollary 1.2 is. Our next goal is to provide the answers for these questions. Indeed, we will show that: Theorem 1.2. Let a > 0 and b > 0. Then TMCa,b (q, N, β) = sup ‖∇u‖aN ⋇‖u‖bq ≤1 ∫ Φ N,q,β (α N (1 − ℘x℘β ℝN N β N−1 ) N )℘u℘ dx < ∞ ⇐⇒ b ≤ N. The constant α N is sharp in the sense that if α > α N , then the above supremum will be infinite. Also, there exist constants c(N, β, q) and C(N, β, q) > 0 such that when α ⪅ α N , c(N, β, q) (1 − ( ααN )N−1 ) β q(1− ) N N ≤ TMSC(q, N, α, β) ≤ C(N, β, q) (1 − ( ααN )N−1 ) β q(1− ) N N . Moreover, we have the following identity: TMCa,b (q, N, β) = sup ( α∈(0,α N ) 1 − ( ααN ) ( ααN )b N−1 N a N−1 N ) β q b (1− N ) TMSC(q, N, α, β). Having studied the TrudingerśMoser inequalities under the norm constraint ‖∇u‖aN ⋇ ‖u‖bq ≤ 1 (for b ≤ N), it is natural to ask what kind of inequalities will hold in the case b > N. In particular, if we like to have the norm constraint ‖∇u‖aN ⋇ ‖u‖bq ≤ 1 (for b > N), then can we still have a valid TrudingerśMoser-type inequality? Indeed, we will answer the above question by the following versions of the TrudingerśMoser inequality on D N,q (ℝN ) which are new even when q = N. We remark that the following results (Theorem 1.3, Theorem 1.4, Corollary 1.3, Corollary 1.4, Theorem 1.5) are TrudingerśMoser inequalities of exact growth. These type of inequalities were studied earlier by Ibrahim, Masmoudi and Nakanishi [15], Lam and Lu [25], Masmoudi and Sani [49, 50], Lu and Tang [41], Lu, Tang and Zhu [44]. Our results below are improved versions of the aforementioned ones. Theorem 1.3. Let a > 0, k > 1 and p ≥ q ≥ 1. Then there exists a constant C = C(N, p, q, β, a, k) > 0 such that for all u ∈ D N,q (ℝN ), ‖∇u‖aN ⋇ ‖u‖kN q ≤ 1, there holds ∫ ℝN Φ N,q,β (α N (1 − N β N−1 ) N )u (1 ⋇ ℘u℘ N−1 (1− k )(1− N ) )℘x℘β p 1 β dx ≤ C. Moreover, the inequality does not hold when p < q. An equivalent version of Theorem 1.3 is the following: Theorem 1.4. Let a > 0, k > 1 and p ≥ q ≥ 1. Then there exists a constant C = C(N, p, q, β, a, k) > 0 such that for all u ∈ D N,q (ℝN ), ‖∇u‖aN ⋇ ‖u‖Nq ≤ 1, there holds ∫ ℝN Φ N,q,β (α N (1 − (1 ⋇ ℘u℘ N β N−1 ) N )u β p 1 N−1 (1− k )(1− N Moreover, the inequality does not hold when p < q. ) )℘x℘β q(1− 1k )(1− N ) dx ≤ C‖u‖q β . N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | 243 As consequences of our results, we have: Corollary 1.3. Let k ≥ 1. There holds ∫ sup kN ‖∇u‖kN N ⋇‖u‖N ≤1 ℝN ∫ 1 sup ‖∇u‖NN ⋇‖u‖NN ≤1 ‖u‖N N(1− 1k )(1− N ) β ℝN ϕ N (α N (1 − N β N−1 ) N )u ϕ N (α N (1 − N β N−1 ) N )u (1 ⋇ ℘u℘ N−1 (1− k )(1− N ) )℘x℘β N 1 β (1 ⋇ ℘u℘ N−1 (1− k )(1− N ) )℘x℘β N 1 β < ∞, < ∞. Since it can be verified easily that the constant C(N, p, q, β, a, k) in Theorem 1.4 tends to a constant C(N, p, q, β, a) as k → ∞, by Fatou’s lemma, we get: Corollary 1.4. Let a > 0 and p ≥ q ≥ 1. Then there exists a constant C = C(N, p, q, β, a) > 0 such that for all u ∈ D N,q (ℝN ), ‖∇u‖aN ⋇ ‖u‖Nq ≤ 1, there holds ∫ ℝN Φ N,q,β (α N (1 − (1 ⋇ ℘u℘ N β N−1 ) N )u β p N−1 (1− N ) )℘x℘β q(1− N ) dx ≤ C‖u‖q β . Moreover, the inequality does not hold when p < q. Our last aim of this paper is to extend further the above proposition. Namely, we will study a TrudingerśMoser inequality with exact growth: Theorem 1.5. Let p ≥ q ≥ 1. Then there exists a constant C = C(N, p, q, β) > 0 such that for all u ∈ D N,q (ℝN ), ‖∇u‖N ≤ 1, there holds N β Φ N,q,β (α N (1 − Nβ )u N−1 ) q(1− ) dx ≤ C‖u‖q N . (1.3) ∫ β p (1 ⋇ ℘u℘ N−1 (1− N ) )℘x℘β ℝN Moreover, the inequality does not hold when p < q. The organization of the paper is as follows. In Section 2, we will recall and establish some necessary lemmas which are required to prove our main theorems. Section 3 provides the proof of Theorem 1.1, a subcritical TrudingerśMoser inequality. In Section 4, we establish the critical TrudingerśMoser inequalities, i.e., Theorem 1.2. Section 5 offers the proofs of Theorems 1.3, 1.4, 1.5. In Section 6, we justify the sharpness of Theorems 1.1, 1.2, 1.3, 1.4, 1.5. 2 Some Useful Preliminaries In this section, we introduce some useful results that will be used in our proofs. We first recall the definition of rearrangement and some useful inequalities. Let Ω ⊂ ℝN , N ≥ 2, be a measurable set. We denote by Ω# the open ball B R ⊂ ℝN centered at 0 of radius R > 0 such that ℘B R ℘ = ℘Ω℘. Let u : Ω → ℝ be a real-valued measurable function that vanishes at infinity, that is, ℘{x : ℘u(x)℘ > t}℘ is finite for all t > 0. The distribution function of u is the function μ u (t) = ℘{x ∈ Ω : ℘u(x)℘ > t}℘ and the decreasing rearrangement of u is the right-continuous, nonincreasing function u∗ that is equimeasurable with u: u∗ (s) = sup{t ≥ 0 : μ u (t) > s}. It is clear that supp u∗ ⊆ [0, ℘Ω℘]. We also define u∗∗ (s) = s 1 ∫ u∗ (t) dt ≥ u∗ (s). s 0 244 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities Moreover, we define the spherically symmetric decreasing rearrangement of u: u# : Ω# → [0, ∞], u# (x) = u∗ (σ N ℘x℘N ), where σ N is the volume of the unit ball in ℝN . Then we have the following important result that could be found in [38]: Lemma 2.1 (PólyaśSzegő Inequality). Let u ∈ W 1,p (ℝn ), p ≥ 1. Then f # ∈ W 1,p (ℝn ) and ‖∇f # ‖p ≤ ‖∇f‖p . Lemma 2.2. Let f and g be nonnegative functions on ℝN , vanishing at infinity. Then ∫ f(x)g(x) dx ≤ ∫ f # (x)g # (x) dx ℝN ℝN in the sense that when the left side is infinite so is the right. Moreover, if f is strictly symmetric-decreasing, then there is equality if and only if g = g# . Now, we recall a compactness lemma of Strauss [4, 56]. Lemma 2.3. Let P and Q : ℝ → ℝ be two continuous functions satisfying P(s) →0 Q(s) as ℘s℘ → ∞ and P(s) → 0 as s → 0 Q(s) Let (u n ) be a sequence of measurable functions u n : ℝN → ℝ such that sup ∫ ℘Q(u n (x))℘ dx < ∞ n and P(u n (x)) 󳨀󳨀󳨀󳨀→ v(x) a.e., n→∞ Then P(u n ) → v in L1 (ℝN ). ℝN lim ℘u n (x)℘ = 0 uniformly with respect to n. ℘x℘→∞ Using Lemma 2.3, we will study the continuity and compactness of the embeddings from Drad (ℝN ) ∩ L q (ℝN ) dx into L a (ℝN ) and L a (ℝN ; ℘x℘ β ). More precisely, we have the following lemma: 1,N Lemma 2.4. Let N ≥ 2, 0 < t < N. Then the embedding Drad (ℝN ) ∩ L q (ℝN ) 󳨅→ L r (ℝN ) 1,N is continuous when r ≥ q and compact for all r > q. Moreover, the embedding Drad (ℝN ) ∩ L q (ℝN ) 󳨅→ L r (ℝN ; 1,N is compact for all r ≥ q. dx ) ℘x℘t Proof. By the CaffarelliśKohnśNirenberg inequality [5], there exists a positive constant C such that for N all u ∈ C∞ 0 (ℝ ), ‖℘x℘γ u‖r ≤ C‖℘x℘α ℘∇u℘‖ap ‖℘x℘β u‖1−a q , where p, q ≥ 1, r > 0, γ = aσ ⋇ (1 − a)β, and 0≤α−σ α−σ≤1 1 α 1 β 1 γ ⋇ , ⋇ , ⋇ > 0, p N q N r N 1 α−1 1 γ 1 β ⋇ = a( ⋇ ) ⋇ (1 − a)( ⋇ ), r N p N q N 0 ≤ a ≤ 1, if a > 0, if a > 0 and 1 α−1 1 γ ⋇ = ⋇ . p N r N N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | 245 Then we can obtain the continuity of the embedding D N,q (ℝN ) 󳨅→ L r (ℝN ; dx ) ℘x℘t with r > q(1 − Nt ) (r ≥ q if t = 0). Indeed, we choose p = N, α = β = 0, γ = − rt , a = 1 − qr (1 − Now, let r > q, we now will prove that the embedding t N ). Drad (ℝN ) ∩ L q (ℝN ) 󳨅→ L r (ℝN ) 1,N is compact. Indeed, let {u n } ∈ Drad (ℝN ) ∩ L q (ℝN ) be bounded. Then we can assume that 1,N un ⇀ u weakly in Drad (ℝN ) ∩ L q (ℝN ). 1,N Set v n = u n − u. By the Radial Lemma, we get lim ℘v n (x)℘ = 0 uniformly with respect to n. ℘x℘→∞ Also, using Lemma 2.3 with P(s) = s r , Q(s) = s q ⋇ s r⋇1 , we can conclude that v n converges to 0 in L1 . It means that u n converges to u in L r . Now, let r ≥ q; we will prove that the embedding Drad (ℝN ) ∩ L q (ℝN ) 󳨅→ L r (ℝN ; 1,N dx ) ℘x℘t is compact. First, let {u n } ∈ Drad (ℝN ) ∩ L q (ℝN ) be bounded. Again, we can assume that 1,N Choose p such that 1 < p < un ⇀ u N t , weakly in Drad (ℝN ) ∩ L q (ℝN ). 1,N then for R arbitrary, we get 1 1 p p󸀠 1 dx rp󸀠 ℘u − u℘ dx) ≤ dx) ( ∫ ( ∫ ∫ ℘u n − u℘ n t tp ℘x℘ ℘x℘ r ℘x℘<R ℘x℘<R ℘x℘<R ≤ CR N p −t ( ∫ ℘u n − u℘ rp󸀠 1 p󸀠 dx) . ℘x℘<R Also, ∫ ℘u n − u℘r ℘x℘≥R dx C 1 ≤ ∫ ℘u n − u℘r dx ≤ t . ℘x℘t R t R ℘x℘≥R Using the compactness of the embedding Drad (ℝN ) ∩ L q (ℝN ) 󳨅→ L rp (ℝN ), choosing R sufficiently large, we dx get that u n converges to u in L r (ℝN ; ℘x℘ t ). 1,N 󸀠 Now, we will prove a variant of [42, Lemma 2.2]: Lemma 2.5. Given any sequence s = {s k }k≥0 , let ∞ ‖s‖1 = ∑ ℘s k ℘, k=0 and Then for h > 1, we have 1 N ∞ ‖s‖N = ( ∑ ℘s k ℘N ) , k=0 ∞ ‖s‖(e) = ( ∑ ℘s k ℘q e k ) k=0 μ(h) = inf{‖s‖(e) : ‖s‖1 = h, ‖s‖N ≤ 1}. exp( h N−1 q ) N μ(h) ∼ 1 h N−1 . 1 q 246 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities Proof. Since μ(h) is increasing in h, we just need to show that n μ(n1− N ) ∼ Choose for all natural number n ∈ ℕ. eq 1 1 nN 1 { { 1 sk = { n N { {0 It is clear that ‖s‖N = 1, if k ≤ n − 1, if k > n − 1. ‖s‖1 = n 1− N1 ‖s‖(e) ∼ , so n eq 1 nN n μ(n1− N ) ≲ eq 1 1 nN . Now, assume that for some ε ≪ 1, n ≫ 1 and sequence s, ‖s‖N = 1 ‖s‖1 = √n, It means that for k ≥ n, ℘s k ℘ ≲ ε e ‖s‖(e) ≤ ε n eq 1 nN n−k q . 1 nN Consider the new sequence (b k ) with b k = s k , k ≤ n, and b k = 0, k > n, we get ‖b‖1 = ‖s‖1 − ∑ ℘s j ℘ ≥ n1− N − C 1 j>n Hence ‖b‖1N−1 ≥ (n1− N − C N ε 1 n 1 N ) On the other hand, ‖b‖1 N N−1 Hence = N (‖b‖21 ) 2(N−1) N N−1 ε 1 nN . N ε N−1 = n(1 − C ) ≥ n − Cε. n 1 N ∑j,k≤n 2 ≤n− 2 2N−1 n1− N (s j −s k )2 . ∑ (s j − s k )2 ≲ εn1− N . 2 j,k≤n Choose m ≤ n such that min ℘s j ℘ = ℘s m ℘. j≤n Then ‖b‖1 − n℘s m ℘ ≲ √εn1− N . 1 Hence ℘s m ℘ ≳ and we get ‖s‖(e) ≳ 1 1 nN n eq 1 nN , which is a contradiction. Using the above lemma, we can now prove a Radial Sobolev inequality in the spirit of Ibrahim, Masmoudi and Nakanishi [15]: N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | 247 Theorem 2.1 (Radial Sobolev Inequality). There exists a constant C > 0 such that for any radially nonnegative nonincreasing function φ ∈ D N,q (ℝN ) satisfying u(R) > 1 and ∞ ω N−1 ∫ ℘φ󸀠 (t)℘N t N−1 dt ≤ K R for some R, K > 0, then we have exp[ αKN φ N−1 (R)] ∫R ℘φ(t)℘q t N−1 dt ∞ N RN ≤ C φ N−1 (R) q q K N−1 . Proof. By scaling, we can assume that R = 1, K = 1, i.e., ω N−1 ∫1 ℘φ󸀠 (t)℘N t N−1 dt ≤ 1. Set ∞ h k = α NN φ(e N ), N−1 then k s k = h k − h k⋇1 ≥ 0; ‖s‖1 = h0 = α NN φ(1). N−1 Also k s k = h k − h k⋇1 = α N [φ(e N ) − φ(e N−1 N k k⋇1 N eN )] = α N ∫ u󸀠 (t) dt N−1 N e e k⋇1 N e 1 N k⋇1 N ≤ α NN ( ∫ ℘u󸀠 (t)℘N t N−1 dt) ( ∫ N−1 k eN k⋇1 N ≤ (ω N−1 ∫ ℘u (t)℘ t 󸀠 e N−1 N k eN e 1 dt) t k⋇1 N N N−1 1 N dt) . k N Hence ‖s‖N ≤ 1. Now ∞ e k⋇1 N ∫ ℘φ(t)℘q t N−1 dt = ∑ ∫ ℘φ(t)℘q t N−1 dt ≥ ∑ ℘φ(e k≥0 1 e k⋇1 N k≥0 k eN ≳ ∑ ℘φ(e k⋇1 N k≥0 k⋇1 N )℘q ∫ t N−1 dt k )℘q e k⋇1 ≳ ∑ ℘h k⋇1 ℘q e k⋇1 eN k≥0 = ∑ ℘h k ℘q e k ≥ ∑ ℘s k ℘q e k . k≥1 Thus ∞ k≥1 ∞ ‖s‖(e) = ∑ ℘s k ℘q e k = s0 ⋇ ∑ ℘s k ℘q e k ≲ h0 ⋇ ∫ ℘φ(t)℘q t N−1 dt. q Also, for 1 < r < exp ( 1 N 2 N−1 ⋅N ), q k=0 q k≥1 1 1 h0 − α N φ(r) = α N φ(1) − α N φ(r) = α N ∫ u󸀠 (t) dt N−1 N N−1 N N−1 N r ≤ α N ( ∫ ℘u (t)℘ t N−1 N 󸀠 N N−1 1 1 h0 . < ≤ 2 2 Hence h0 ≲ φ(r). N−1 N 1 N r r 1 dt) ( ∫ dt) t 1 N−1 N 248 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities So 1 ∞ e 2n ∫ ℘φ(t)℘ t q N−1 1 dt ≥ ∫ ℘φ(t)℘q t N−1 dt ≳ h0 . q 1 Now, we can conclude that ∞ exp(h0N−1 ) N ∫ ℘φ(t)℘ t q N−1 1 dt ≳ q ‖s‖(e) ≳ N e α N φ N−1 (1) =C q h0N−1 (φ(1)) N−1 q . 3 Subcritical TrudingerśMoser Inequalities: Proof of Theorem 1.1 We prove Theorem 1.1 in this section. This is a subcritical version of the TrudingerśMoser inequalities. N Proof. Suppose that u ∈ C∞ 0 (ℝ ) \ {0}, u ≥ 0 and ‖∇u‖N ≤ 1. We write exp(α(1 − ∫ ℝN N β N−1 )℘u℘ p N )℘u℘ ℘x℘β where I1 = ∫ exp(α(1 − I2 = ∫ exp(α(1 − {u>1} {u≤1} dx = I1 ⋇ I2 , N β N−1 )℘u℘ p N )℘u℘ ℘x℘β dx, N β N−1 )℘u℘ p N )℘u℘ ℘x℘β dx. First, by Lemma 2.4, we get I2 ≤ e α ‖℘x℘ p u‖p ≤ C N,p,q,α,β ‖u‖q −β Now, set v = u−1 Then q(1− N ) p . on Ω(u) = {u > 1}. v ∈ W0 (Ω(u)) and ‖∇v‖N ≤ 1. 1,N Now β N N e α(1− N )(℘v℘⋇1) N−1 ⋇p℘v℘ e α(1− N )℘v⋇1℘ N−1 ⋇p ln(v⋇1) dx ≤ dx I1 = ∫ ∫ ℘x℘β ℘x℘β β β Ω(u) Ω(u) ≤ C N,p,q,α,β ∫ for ε = α N (1 − Nβ ) − α(1 − inequality, we obtain β N) e N (α(1− N )⋇ε)℘v℘ N−1 ⋇C(ε) β ℘x℘β Ω(u) dx and C(ε) = C N,p,q,α,β sufficient large. Hence, by the singular TrudingerśMoser I1 ≤ C N,p,q,α,β ℘Ω(u)℘1− N . β However, it can be deduced easily that ℘Ω(u)℘ = ∫ 1 dx ≤ ∫ ℘u℘q dx = ‖u‖q . q Ω(u) Ω(u) Hence, we get ∫ ℝN exp(α(1 − N β N−1 )℘u℘ p N )℘u℘ ℘x℘β dx ≤ C N,p,q,α,β ‖u‖q q(1− N ) β for all u ∈ D N,q (ℝN ), ‖∇u‖N ≤ 1. N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | 249 4 Critical TrudingerśMoser Inequalities: Proof of Theorem 1.2 In this section, Theorem 1.2 will be proved via the following series of lemmas. Lemma 4.1. We have C(N, β, q) TMSC(q, N, α, β) ≤ (1 − ( ααN )N−1 ) β q(1− ) N N . N Proof. Let u ∈ C∞ 0 (ℝ ), ‖∇u‖N ≤ 1, ‖u‖q = 1 and u ≥ 0 and set Ω = {x : u(x) > (1 − ( Then ℘Ω℘ = ∫ 1 dx ≤ ∫ Ω Ω (1 − 1 α N−1 N ) ) }. αN u(x)q dx ≤ q ( ααN )N−1 ) N (1 − 1 q ( ααN )N−1 ) N . We have by the CaffarelliśKohnśNirenberg inequality that for some ε = ε(q, β, N) ≥ 0 with ε = 0 when β = 0 and ε > 0 when β > 0, ∫ Φ N,q,β (α(1 − ℘x℘β ℝN \Ω N β N−1 ) N )℘u℘ Φ N,q,β (α(1 − dx ≤ ∫ ℘x℘β {u≤1} N β N−1 ) N )℘u℘ dx u q(1− N )⋇ε dx ≤ C(N, β, q). ∫ ℘x℘β β ≤e α {u≤1} Now, we consider I=∫ Φ N,q,β (α(1 − ℘x℘β Ω N β N−1 ) N )℘u℘ dx ≤ ∫ N β N−1 ) N )℘u℘ ℘x℘β exp(α(1 − Ω dx. On Ω, we set v(x) = u(x) − (1 − ( 1 α N−1 N ) ) . αN Then it is clear that v ∈ W0 (Ω) and ‖∇v‖N ≤ 1. Also, on Ω, with ε = 1,N ℘u℘N ≤ (℘v℘ ⋇ (1 − ( 󸀠 ≤ (1 ⋇ ε)℘v℘ = N󸀠 1 α N−1 N N ) ) ) αN αN α − 1, 󸀠 1 1 󸀠 1−N 󵄨 α N−1 N 󵄨󵄨󵄨N 1 󵄨󵄨 󵄨 󵄨󵄨 − ⋇ (1 − (1 ( ) ) ) 󵄨 󵄨󵄨 󵄨󵄨 αN (1 ⋇ ε)N−1 αN N󸀠 ℘v℘ ⋇ 1. α Hence, by the singular TrudingerśMoser inequality, I≤∫ Ω N β N−1 ) N )℘u℘ ℘x℘β exp(α(1 − ≤ C(N, β, q)℘Ω℘ β 1− N ≤ dx ≤ ∫ exp(α N (1 − β N󸀠 N )℘v℘ ℘x℘β Ω C(N, β, q) (1 − ( ααN )N−1 ) β q(1− ) N N . In conclusion, we have TMSC(q, N, α, β) ≤ C(N, β, q) (1 − ( ααN )N−1 ) β q(1− ) N N . ⋇ α) dx 250 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities Lemma 4.2. We have TMCa,b (q, N, β) < ∞ when b ≤ N. Proof. Let u ∈ D N,q (ℝN ) \ {0}, ‖∇u‖aN ⋇ ‖u‖bq ≤ 1. Assume that ‖∇u‖N = θ ∈ (0, 1), If 1 2 ‖u‖bq ≤ 1 − θ a . < θ < 1, then again we set v(x) = λ=( u(λx) , θ (1 − θ a ) b N ) > 0. θq q 1 Hence ‖∇v‖N = ‖∇u‖N = 1, θ ‖v‖q = ∫ ℘v℘q dx = q ℝN 1 1 (1 − θ a ) b q = 1. ∫ ℘u(λx)℘q dx = q N ‖u‖q ≤ q θ θ λ θq λN q ℝN We have ∫ Φ N,q,β (α N (1 − ℘x℘β ℝN β N󸀠 N )℘u℘ ) dx = ∫ Φ N,q,β (α N (1 − β N󸀠 N )℘u(λx)℘ ) ℘λx℘β ℝN ≤λ N−β ∫ Φ N,q,β (θ N α N (1 − 󸀠 ℘x℘β ℝN d(λx) N β N−1 ) N )℘v℘ dx ≤ λ N−β TMSCβ,q (θ N α N ) 󸀠 ≤( (1 − θ a ) b 1− N ) θq q ((1 − θ a ) b )1− N q ≤ If 0 < θ ≤ 1 2, β (1 − θN ) C(N, β, q) (1 − ( θ α Nα N )N−1 ) N󸀠 β β q(1− ) N N β q(1− ) N N C(N, β, q) ≤ C(N, β, q, a, b) since b ≤ N. then with q v(x) = 2u(2 N x) we have ‖∇v‖N = 2‖∇u‖N ≤ 1, Hence ∫ ℝN Φ N,q,β (α N ℘u℘N ) 󸀠 ℘x℘β dx ≤ 2 ∫ q TMCa,b (q, N, β) = sup ( α∈(0,α N ) 1 − ( ααN ) Proof. Let u ∈ D N,q (ℝN ), ‖∇u‖N ≤ 1, ‖u‖q = 1. Set α v(x) = ( ) αN ‖v‖q = ( α ) αN N−1 N α N (1− N ) β 2N 󸀠 ℘x℘β ℝN Lemma 4.3. If TMCa,b (q, N, β) < ∞, then Then Φ N,q,β ( N−1 N 1 N λq , ( ααN )b u(λx), ‖v‖q ≤ 1. N−1 N a N−1 N λ=( ‖∇v‖N = ( ) ℘v℘ N−1 ) N β q b (1− N ( ααN )b ) TMSC(q, N, α, β). N−1 N 1 − ( ααN ) α ) αN N−1 N dx ≤ C(q, N, β). N−1 N a ) q bN ‖∇u‖N ≤ ( . α ) αN N−1 N N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | 251 Hence ‖∇v‖aN ⋇ ‖v‖bq ≤ 1. By the definition of TMCa,b (q, N, β), we obtain ∫ Φ N,q,β (α(1 − ℘x℘β ℝN N β N−1 ) N )℘u℘ N β N−1 ) N )℘u(λx)℘ ℘λx℘β Φ N,q,β (α(1 − dx = ∫ ℝN = λ N−β ∫ ℝN ≤( Hence TMCa,b (q, N, β) ≥ sup ( α∈(0,α N ) d(λx) N 1 β Φ N,q,β (α N (1 − )℘v(x)℘ N−1 ) dx β N ℘x℘ ( ααN )b N−1 N 1 − ( ααN ) 1 − ( ααN ) ( ααN )b N−1 N a N−1 N a N−1 N ) ) β q b (1− N β q b (1− N ) TMCa,b (q, N, β). ) TMSC(q, N, α, β). By the same process as above, by beginning with the function v such that ‖∇v‖aQ ⋇ ‖v‖bQ ≤ 1, and defining the function u as follows: q 1 − ‖∇v‖aN bN v(λx) u(x) = , λ=( ) > 0, ‖∇v‖N ‖∇v‖b N we have that TMCa,b (q, N, β) > sup ( α∈(0,α N ) 1 − ( ααN ) ( ααN )b N−1 N a N−1 N ) β q b (1− N ) TMSC(q, N, α, β) is impossible. Hence TMCa,b (q, N, β) = sup ( α∈(0,α N ) 1 − ( ααN ) ( ααN )b Lemma 4.4. We have TMSC(q, N, α, β) ≥ N−1 N a N−1 N ) β q b (1− N (1 − TMSC(q, N, α, β). when α ⪅ α N . c(N, β, q) ( ααN )N−1 ) ) β q(1− ) N N Proof. Consider the following sequence: N N n 1 { { ) ( ) , ( { { ω N − β { N−1 { { { 1 1 u n (x) = { N − β N ( ) log( ), { { ω n ℘x℘ { N−1 { { { { {0, 1 Then we can see easily that − n N−β ‖u n ‖q = ∫ ( 1 q 0 ω N−1 n ) N−β q(N−1) N − n N−β ∫ r dr ⋇ N−1 0 ≈n q(N−1) N q(N−1) N e nN − N−β ⋇ 1 q nN ≈ n N−β 1 n 1 q nN e n − N−β n − N−β , < ℘x℘ < 1, ℘x℘ ≥ 1. 1 r N−1 dr ⋇ ∫ ( e e ≈n q N ) ( 0 ≤ ℘x℘ ≤ e ‖∇u n ‖N = 1 and for sufficiently large n, say when n ≥ M1 , e N−1 q N . ∫ y q e−Ny dy 0 − n N−β q N−β N 1 q ) ( log( )) r N−1 dr ω N−1 n r 252 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities Now, ∫ ℝN Φ N,q,β (α(1 − ℘x℘β N β N−1 ) N )℘u n ℘ e 1 − n N−β N−1 β n 1 dx ≳ Φ N,q,β (α(1 − )( ) ( )) ∫ r N−1−β dr N ω N−1 N−β 0 α ≳ Φ N,q,β ( n)e−n αN ≈ e αN e−n α For α ≈ α N such that Then 1 1− αα N for sufficiently large n ≥ M2 where M2 is independent of α. n > 10 max{M1 , M2 }, we pick n = n(α) such that 1 ≤ (1 − TMSC(q, N, α, β) ≥ ∫ 1 ‖u n ‖q q(1− N ) ≳n q N β Φ N,q,β (α(1 − ℘x℘β ℝN N (1− N ) 1 ≈( ≈ α ) 1 − αN β e Lemma 4.5. We have TMCa,b (q, N, β) < ∞ if and only if b ≤ N. ≤ 2, i.e., n ≈ 1 1− αα . N dx β q (1− N ) −2 N β N−1 ) N )℘u n ℘ α α N )n 1 (1 − ( ααN )N−1 ) β q(1− ) N N . Proof. Now, assume that there is some b > N such that TMCa,b (q, N, β) is finite. Then lim sup ( α↑α N 1 − ( ααN ) ( ααN )b N−1 N a N−1 N By the above lemma, lim sup ( α↑α N Then lim inf ( ) 1 − ( ααN ) ( ααN )q 1 − ( ααN ) ( ααN )b α↑α N which is impossible since b > N. β q b (1− N TMSC(q, N, α, β) ≤ TMCa,b (q, N, β) < ∞. N−1 N a N−1 N N−1 N a N−1 N ) ) ) q N β q b (1− N (1− N ) β ) ( TMSC(q, N, α, β) > 0. ( ααN )q N−1 N 1 − ( ααN ) N−1 N a ) q N (1− N ) β < ∞, 5 TrudingerśMoser Inequality when b > N 5.1 TrudingerśMoser Inequality with Exact Growth ś Proof of Theorem 1.5 Proof. It is enough to prove inequality (1.3) when p = q. By the symmetrization arguments: the Pólyaś Szegő inequality, the HardyśLittlewood inequality and the density arguments, we may assume that u is a smooth, nonnegative and radially nonincreasing function (we just need to make sure that the function β β N N N Φ N,q,β (αt N−1 )/(1 ⋇ t q(1− N ) ) is nondecreasing on ℝ⋇ but it is easy since Φ N,q,β (αt N−1 )/t q(1− N ) and Φ N,q,β (αt N−1 ) are both nondecreasing on ℝ⋇ ). Let R1 = R1 (u) be such that R1 ∫ ℘∇u℘N dx = ω N−1 ∫ ℘u r ℘N r N−1 dr ≤ 1 − ε0 , 0 B R1 ∫ ℝN \B R1 ∞ ℘∇u℘N dx = ω N−1 ∫ ℘u r ℘N r N−1 dr ≤ ε0 . Here ε0 ∈ (0, 1) is fixed and does not depend on u. R1 N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | 253 By Hölder’s inequality, we have r2 r2 1 N u(r1 ) − u(r2 ) ≤ ∫ −u r dr ≤ ( ∫ ℘u r ℘ r dr) ( ln N N−1 r1 ≤( r1 1 − ε0 r2 ) ( ln ) ω N−1 r1 1 N and N−1 N 1 u(r1 ) − u(r2 ) ≤ ( r2 ) r1 N−1 N for 0 < r1 ≤ r2 ≤ R1 (5.1) N−1 N r2 N ε0 for R1 ≤ r1 ≤ r2 . (5.2) ) ( ln ) ω N−1 r1 We define R0 := inf{r > 0 : u(r) ≤ 1} ∈ [0, ∞). Hence u(s) ≤ 1 when s ≥ R0 . Without loss of generality, we assume R0 > 0. Now, we split the integral as follows: ∫ ℝN Φ N,q,β (α℘u℘ N−1 ) N dx = ∫ (1 ⋇ λu N−1 (1− N ) )℘x℘β q β B R0 Φ N,q,β (α℘u℘ N−1 ) N (1 ⋇ λu N−1 (1− N ) )℘x℘β β q Φ N,q,β (α℘u℘ N−1 ) N ∫ dx ⋇ (1 ⋇ λu N−1 (1− N ) )℘x℘β q ℝN \B R0 β = I ⋇ J. First, we will estimate J. Since u ≤ 1 on ℝN \ B R0 , we have if β > 0, J= ∫ ℝN \B R0 Φ N,q,β (α℘u℘ N−1 ) N (1 ⋇ λu β q N−1 (1− N ) )℘x℘β ℘u℘(⌊q dx ≤ C ∫ β N−1 N (1− N ℘x℘β {u≤1} N )⌋⋇1) N−1 by Lemma 2.4. Similarly for the case β = 0, we also have q(1− N ) β J ≤ C‖u‖q . Hence, now, we just need to deal with the integral I. Case 1: 0 < R0 ≤ R1 . In this case, using (5.1), we have for 0 < r ≤ R0 , u(r) ≤ 1 ⋇ ( By using 1 R0 1 − ε0 N ) ( ln ) ω N−1 r N−1 N . (a ⋇ b) N−1 ≤ (1 ⋇ ε)a N−1 ⋇ A(ε)b N−1 , N N where A(ε) = (1 − we get u N−1 (r) ≤ (1 ⋇ ε)( N N 1 1−N 1 ) , N−1 (1 ⋇ ε) 1 1 − ε0 N−1 R0 ⋇ C(ε). ln ) ω N−1 r Thus, we can estimate the integral I as follows: I= ∫ B R0 ≤ ∫ Φ N,q,β (α℘u℘ N−1 ) N dx (1 ⋇ λu N−1 (1− N ) )℘x℘β β q 1−ε0 N−1 ln exp(α(1 ⋇ ε)( ω ) N−1 1 ℘x℘β B R0 1 1−ε0 α(1⋇ε)( ω ≤ CR0 N−1 ) N−1 R0 ∫r R0 r ⋇ αA(ε)) 1−ε0 N−1−α(1⋇ε)( ω N−1 1 0 ≤ N−β CR0 β 1− N ≤ C( ∫ 1 dx) B R0 q(1− N ) ≤ C‖u‖q dx ) N−1 −β β . q(1− N ) dx ≤ C‖u‖q dr β dx 254 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities Case 2: 0 < R1 < R0 . We have I= ∫ Φ N,q,β (α℘u℘ N−1 ) N (1 ⋇ λu N−1 (1− N ) )℘x℘β β q B R0 Φ N,q,β (α℘u℘ N−1 ) N dx = ∫ (1 ⋇ λu N−1 (1− N ) )℘x℘β = I1 ⋇ I2 . β q B R1 Using (5.2), we get u(r) − u(R0 ) ≤ ( 1 N R0 ε0 ) ( ln ) ω N−1 r Hence u(r) ≤ 1 ⋇ ( dx ⋇ N−1 N 1 N ∫ (1 ⋇ λu N−1 (1− N ) )℘x℘β β q B R0 \B R1 dx for r ≥ R1 . N ε0 R0 ) ( ln ) ω N−1 r Then we have Φ N,q,β (α℘u℘ N−1 ) N−1 N . 1 u N−1 (r) ≤ (1 ⋇ ε)( N−1 R0 ε0 ln ⋇ A(ε) for all ε > 0. ) ω N−1 r N So Φ N,q,β (α℘u℘ N−1 ) N ∫ I2 = β q B R0 \B R1 R0 ≤ C ∫ exp(α(1 ⋇ ε)( R1 α(1⋇ε)( ω ≤ CR0 ≤ dx (1 ⋇ λu N−1 (1− N ) )℘x℘β 1 ε0 N−1 ) N−1 1 N−1 R0 ε0 ln ⋇ αA(ε))r N−1−β dr ) ω N−1 r N−β−α(1⋇ε)( ω R0 ε0 N−1 1 ) N−1 N−β−α(1⋇ε)( ω − R1 ε0 N−1 1 ) N−1 0 N − β − α(1 ⋇ ε)( ωεN−1 ) N−1 1 C 0 N − β − α(1 ⋇ ε)( ωεN−1 ) N−1 1 N−β (R0 − N−β R1 ) ≤ C(R0N − R1N )1− N β ∫ ≤ C( β 1− N 1 dx) B R0 \B R1 q(1− N ) β ≤ C‖u‖q (since α ≤ α N (1 − β N ), 0 we can choose ε > 0 such that N − β − α(1 ⋇ ε)( ωεN−1 ) N−1 > 0). So, we need to estimate 1 I1 = ∫ B R1 with u(R1 ) > 1. First, we define It is clear that v ∈ 1,N W0 (B R1 ) Φ N,q,β (α℘u℘ N−1 ) N (1 ⋇ λu N−1 (1− N ) )℘x℘β q dx v(r) = u(r) − u(R1 ) on 0 ≤ r ≤ R1 . and that ∫ ℘∇v℘N dx = ∫ ℘∇u℘N dx ≤ 1 − ε0 . B R1 B R1 Moreover, for 0 ≤ r ≤ R1 , β u N−1 (r) ≤ (1 ⋇ ε)v N−1 (r) ⋇ A(ε)u N−1 (R1 ). N N N Hence Φ N,q,β (αu N−1 ) N N N e(1⋇ε)αv N−1 (r) 1 e αA(ε)u N−1 (R1 ) dx ≤ dx I1 = ∫ ∫ β β q q λ u N−1 (1− N ) (R ) ℘x℘β (1 ⋇ λu N−1 (1− N ) )℘x℘β 1 BR BR 1 1 N = e αA(ε)u N−1 (R1 ) 1 ∫ β q λ u N−1 (1− N ) (R1 ) B R 1 N e αw N−1 (r) ℘x℘β dx, (5.3) N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | where w = (1 ⋇ ε) N−1 N 255 v. It is clear that w ∈ W0 (B R1 ) and 1,N ∫ ℘∇w℘N dx = (1 ⋇ ε)N−1 ∫ ℘∇v℘N dx ≤ (1 ⋇ ε)N−1 (1 − ε0 ) ≤ 1 B R1 B R1 1 ) if we choose 0 < ε ≤ ( 1−ε 0 1 N−1 − 1. Hence, using the singular TrudingerśMoser inequality, we have N β e αw N−1 (r) N−β dx ≤ C℘B R1 ℘1− N ≤ CR1 . ∫ β ℘x℘ (5.4) B R1 Also, using Theorem 2.1, we have u N−1 (1− N ) (R1 ) q β N−1 (R )) exp( NαA(ε) 1 N−β u N N e αA(ε)u N−1 (R1 ) N−β R1 ≤[ u N−1 (R1 ) q 1 ) if we choose ε = ( 1−ε 0 β 1− N ≤ (CA(ε) ∫ q N−1 ℝN \B R1 β 1− N ℘u℘ dx) q ≤ (C‖u‖q )1− N β q 1 N−1 R1N ] (5.5) − 1. By (5.3), (5.4) and (5.5), the proof is now completed. 5.2 Proof of Theorems 1.3 and 1.4 First we will prove that Theorem 1.4 is valid if and only if Theorem 1.3 is valid. Suppose first that Theorem 1.4 is valid. Let u ∈ D N,q (ℝN ), ‖∇u‖aN ⋇ ‖u‖kN q ≤ 1. Set v(x) = u(λx), λ=( ‖u‖Nq ) 1 − ‖∇u‖aN q N2 ≤( ‖u‖Nq q N2 ) kN ‖u‖q = 1 (k−1) Nq ‖u‖q . Then ‖∇v‖N = ‖∇u‖N , and ‖v‖q = ‖∇v‖aN ⋇ ‖v‖Nq = ‖∇u‖aN ⋇ 1 N λq ‖u‖Nq = 1. 1 λ ‖u‖q N2 q Since, by assumption, Theorem 1.4 holds, we have ∫ ℝN Φ N,q,β (α N (1 − N β N−1 ) N )℘u℘ (1 ⋇ ℘u℘ N−1 (1− k )(1− N ) )℘x℘β p 1 β dx = λ N−β ∫ ℝN Φ N,q,β (α N (1 − N β N−1 ) N )℘v℘ (1 ⋇ ℘v℘ N−1 (1− k )(1− N ) )℘x℘β p 1 β q(1− 1 )(1− N ) ‖v‖q k = Cλ ‖u‖q ≤ C( ≤ Cλ N−β = Cλ N−β k β q(1− 1k )(1− N ) β N−β ( 1 N λq dx ‖u‖q ) 1 ‖u‖q (k−1) Nq ) N−β k q(1− 1k )(1− N ) β ‖u‖q Suppose now that Theorem 1.3 holds. Let v ∈ D N,q (ℝN ), ‖∇v‖aN ⋇ ‖v‖Nq ≤ 1. Set u(x) = v(λx), λ=( Then ‖v‖kN q 1− ‖∇v‖aN ) q kN 2 ≤( ‖∇u‖N = ‖∇v‖N , ‖v‖kN q ‖v‖Nq ) ‖u‖q = q kN 2 1 N λq = ‖v‖q (1− 1k ) Nq ‖v‖q . q(1− 1k )(1− N ) β = C. 256 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities and a ‖∇u‖aN ⋇ ‖u‖kN q = ‖∇v‖N ⋇ 1 λ kN 2 q ‖v‖kN q = 1. Hence ∫ ℝN Φ N,q,β (α N (1 − N β N−1 ) N )℘v℘ (1 ⋇ ℘v℘ N−1 (1− k )(1− N ) )℘x℘β p 1 β dx = λ N−β ∫ ℝN Φ N,q,β (α N (1 − N β N−1 ) N )℘u℘ dx (1 ⋇ ℘u℘ N−1 (1− k )(1− N ) )℘x℘β p β 1 (1− 1k )q(1− N ) β ≤ Cλ N−β ≤ C‖v‖q . Now, we will provide a proof for Theorem 1.4. Proof of Theorem 1.4. Let u ∈ D N,q (ℝN ), ‖∇u‖aN ⋇ ‖u‖Nq ≤ 1. By Hölder’s inequality and Theorems 1.2 and 1.5, we get ∫ {u>1} Φ N,q,β (α N (1 − N β N−1 ) N )℘u℘ (1 ⋇ ℘u℘ N−1 (1− k )(1− N ) )℘x℘β p 1 β dx ≤ ∫ ( {u>1} Φ N,q,β (α N (1 − ℘x℘β N β N−1 ) N )℘u℘ Φ N,q,β (α N (1 − Nβ )℘u℘ N−1 ) N ≤[ ∫ ℘x℘β {u>1} 1 k ) ( Φ N,q,β (α N (1 − ℘u℘ N−1 (1− N ) ℘x℘β β p 1 k dx] [ ∫ q(1− N )(1− 1k ) 1− 1k ) dx Φ N,q,β (α N (1 − Nβ )℘u℘ N−1 ) N ℘u℘ N−1 (1− N ) ℘x℘β β p {u>1} ≤ C(N, p, q, β, a, k)‖u‖q N β N−1 ) N )℘u℘ 1− 1k dx] β . Also, it is easy to see Φ N,q,β (α N (1 − ∫ (1 ⋇ ℘u℘ {u≤1} N β N−1 ) N )℘u℘ β p 1 N−1 (1− k )(1− N ) )℘x℘β q(1− N ) β dx ≤ C(N, p, q, β, a, k)‖u‖q q(1− N )(1− 1k ) β ≤ C(N, p, q, β, a, k)‖u‖q . 6 Sharpness We will verify that α N (1 − Nβ ) is sharp in Theorem 1.1, Corollary 1.1, Corollary 1.2, Theorem 1.2 and Theorem 1.5. Indeed, consider the following Moser’s sequence: N N 1 n { { ( ) ( ) , { { ω N − β { N−1 { { { 1 1 N −β N u n (x) = { log( ), ) ( { { ℘x℘ { ω N−1 n { { { { {0, 1 Then we can see easily that ‖∇u n ‖N = 1, N−1 0 ≤ ℘x℘ ≤ e e n − N−β n − N−β , < ℘x℘ < 1, ℘x℘ ≥ 1. ‖u n ‖q = o n (1). Also ∫ exp(α N (1 − ℝN N β N−1 )℘u ℘ p n N )℘u n ℘ β ℘x℘ dx p 1 N−1 N n n 1 1 β ≥ ω N−1 exp(α N (1 − )( ) ( ))( ) ( ) N ω N−1 N−β ω N−1 N−β ≥ ω N−1 e n ( p N 1 n ) ( ) ω N−1 N−β p(N−1) N e−n N−β p(N−1) N e − n N−β ∫ r N−1−β dr 0 N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | and ∫ N β N−1 ) N )℘u n ℘ [exp(α N (1 − − ∑N−2 k=0 N (α N (1− N ))k ℘u n ℘k N−1 ] k! β ℘x℘β ℝN ∫ ℘x℘β ℝN Hence ∫ 1 ‖u n ‖q q(1− N ) 1 ‖u n ‖q N−β [exp(α N (1 − ∫ N β N−1 ) N )℘u n ℘ Φ N,q,β (α N (1 − β exp(α N (1 − ℝN N β N−1 ) N )℘u n ℘ − ∑N−2 k=0 ℘x℘β ℝN ‖u n ‖q q(1− N ) Now, set β N−2 k ω N−1 n (1 − e−n ∑ ), N−β k! k=0 dx ≥ ω N−1 Φ N,q,β (n)e−n . N−β N β N−1 )℘u ℘ p n N )℘u n ℘ ℘x℘β N (α N (1− N ))k ℘u n ℘k N−1 ] k! dx → ∞, β ∫ 1 dx ≥ 257 Φ N,q,β (α℘u n ℘ N−1 ) N ℘x℘β ℝN dx → ∞, dx → ∞. vn = cn un , where c n > 0 is chosen such that c an ⋇ c bn ‖u n ‖bq = 1. Then ‖∇v n ‖aN ⋇ ‖v n ‖bq = 1. β N ), Noting that limn→∞ c n = 1, we have for α > α N (1 − ∫ Φ N,q,β (α℘v n ℘ N−1 ) N ℘x℘β ℝN Φ N,q,β (αc nN−1 ℘u n ℘ N−1 ) N dx = ∫ N ℘x℘β ℝN dx ≥ α ω N−1 Φ N,q,β ( N−β α N (1 − c nN−1 n)e−n → ⋇∞. N β N) Finally, we will show that p ≥ q are necessary in Theorem 1.5, Theorem 1.3 and Theorem 1.4. Recall that ‖∇u n ‖N = 1 and for sufficiently large n, q ‖u n ‖q e − n N−β q N n = ∫ ( ) ( ) ω N−1 N−β 1 q(N−1) N 1 r N−1 0 ≈n ≈n e ∫ r N−1 dr ⋇ 0 q(N−1) N e e − n N−β q(N−1) N nN − N−β ⋇ 1 q nN ≈ q q ∫ y q e−Ny dy q nN nN − n N−β N−β N 1 q ) ( log( )) r N−1 dr ω N−1 n r n N−β 1 1 dr ⋇ ∫ ( 0 . Now we consider the left-hand side of (1.3), ∫ ℝN Φ N,q,β (α N (1 − (1 ⋇ λ℘u n N β N−1 ) N )℘u n ℘ ℘l )℘x℘β e − n N−β dx ≳ ∫ (1 ⋇ 0 e − n N−β ≳ ∫ Φ N,q,β (n) n 0 1 β 1 N−1 ( n )) N )( ω N−1 ) N−β r N−1−β N−1 1 1 n ) N ℘l ) ) N ( N−β λ℘( ω N−1 Φ N,q,β (α N (1 − l(N−1) N r N−1−β dr ≳ Φ N,q,β (n)e−n n l(N−1) N Note to make (1.3) true providing n sufficiently large, we need 1 n l(N−1) N ≲ ‖u n ‖q q(1− N ) β ≈ 1 n β q(1− ) N N 󳨐⇒ l ≥ q β (1 − ). N−1 N ≳ dr 1 n l(N−1) N . 258 | N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities It is surprising that the direct calculations using the Moser sequence could not verify that p ≥ q are necessary in Theorem 1.3 and Theorem 1.4. Hence, in this case, we need to propose a new approach. Indeed, assume that p ≥ 0 is such that B := 1 sup ‖∇u‖aN ⋇‖u‖Nq ≤1 is finite. We define ‖u‖q β 1 ‖∇u‖N ≤1 = ∫ q(1− 1k )(1− N ) A(α) := sup ‖u‖q q(1− N ) sup ‖∇u‖N ≤1;‖u‖q =1 β ∫ ℝN Φ N,q,β (α N (1 − (1 ⋇ [α N (1 − ℝN ∫ (1 ⋇ [α(1 − Let u ∈ D N,q (ℝN ), ‖∇u‖N ≤ 1, ‖u‖q = 1. Set α v(x) = ( ) αN Then ‖v‖q = ( α ) αN N−1 N N−1 N 1 N ℝN (1 ⋇ [α(1 − = ∫ ℝN ℝN α ≤( ) αN 1 − ( ααN ) α ) αN N−1 N N−1 N a N β N−1 ) N )℘u℘ Φ N,q,β (α(1 − ) q N2 ‖∇u‖N ≤ ( β p 1 β Np (1− 1 )(1− Nβ ) k ℘u℘ N−1 (1− k )(1− N ) )℘x℘β N )] (1 ⋇ [α(1 − = λ N−β ∫ ( ααN )N−1 λ=( λq Hence ‖∇v‖aN ⋇ ‖v‖Nq ≤ 1. By the definition of B, we obtain ∫ N β N−1 ) N )℘u℘ β p 1 β Np (1− 1 )(1− Nβ ) k ℘u℘ N−1 (1− k )(1− N ) )℘x℘β N )] ‖∇v‖N = ( Φ N,q,β (α(1 − N β N−1 ) N )℘u℘ β p 1 β Np (1− 1 )(1− Nβ ) k ℘u℘ N−1 (1− k )(1− N ) )℘x℘β N )] Φ N,q,β (α(1 − u(λx), , β p 1 β Np (1− 1 )(1− Nβ ) k ℘u℘ N−1 (1− k )(1− N ) )℘x℘β N )] Φ N,q,β (α(1 − (1 ⋇ [α(1 − ℝN N β N−1 ) N )u α ) αN β N−1 1 N q(1− k )(1− N ) As a consequence, ( N−1 N N β N−1 ) N )℘v℘ 1 − ( ααN ) N−1 N a ) β q 1 k (1− N α N lim sup (1 − ) α N α↑α N ) β q 1 N k (1− N . d(λx) β p 1 β Np (1− 1 )(1− Nβ ) k ℘v℘ N−1 (1− k )(1− N ) )℘x℘β N )] ( ααN )N−1 dx. dx β p 1 β Np (1− 1 )(1− Nβ ) k ℘u(λx)℘ N−1 (1− k )(1− N ) )℘λx℘β N )] (1 ⋇ [α N (1 − dx . N β N−1 ) N )℘u(λx)℘ Φ N,q,β (α N (1 − dx dx ) B. A(α) < ∞. (6.1) Now, consider the Moser sequence. Recall that ‖∇u n ‖N = 1 and for sufficiently large n, say when n ≥ M1 , q ‖u n ‖q e − n N−β 1 q N n = ∫ ( ) ( ) ω N−1 N−β q(N−1) N 1 r N−1 0 q 1 q N−β N 1 dr ⋇ ∫ ( ) ( log( )) r N−1 dr = O( q ). ω N−1 n r nN − n e N−β Now, ∫ ℝN (1 ⋇ [α(1 − ≳ ≳ Φ N,q,β (α(1 − β p 1 β Np (1− 1 )(1− Nβ ) k ℘u n ℘ N−1 (1− k )(1− N ) )℘x℘β N )] 1 ⋇ [α(1 − Φ N,q,β (α(1 − n 1 β 1 N−1 ( n )) N )( ω N−1 ) N−β dx β p 1 1 N−1 β Np (1− 1 )(1− Nβ ) 1 n k ) N ( N−β ) N ℘ N−1 (1− k )(1− N ) ℘( ω N−1 N )] Φ N,q,β ( ααN n) p N N β N−1 ) N )℘u n ℘ (1− 1k )(1− N ) β e−n ≈ e n p N α αN n −n e (1− 1k )(1− N ) β e − n N−β ∫ r N−1−β dr 0 N. Lam, G. Lu and L. Zhang, Sharp Singular TrudingerśMoser Inequalities | 259 for sufficiently large n ≥ M2 where M2 is independent of α. Also, for α ⪅ α N such that 1−1 α > 10 max{M1 , M2 }, αN we pick n = n(α) such that 1 ≤ (1 − ααN )n ≤ 2, i.e., n ≈ 1−1 α . Then αN A(α) ≥ ‖u n ‖q q(1− N ) β ℝN n N (1− N ) e−2 β q ≳ ∫ 1 1 β ≈ C( α↑α N α↑α N or equivalently p (1 − (1 − β 1 α [ N − N (1− k )](1− N ) > 0. ) αN q lim inf A(α)(1 − lim sup dx [ N − N (1− k )](1− N ) 1 . α ) 1 − αN Hence From (6.1) and (6.2), we have N β N−1 ) N )℘u n ℘ β p 1 β Np (1− 1 )(1− Nβ ) k ℘u n ℘ N−1 (1− k )(1− N ) )℘x℘β N )] q n N (1− k )(1− N ) p (1 ⋇ [α(1 − Φ N,q,β (α(1 − p 1 α Nq 1k (1− Nβ ) αN ) α [ Nq − Np (1− 1k )](1− Nβ ) αN ) β (6.2) <∞ p ≥ q. Acknowledgment: The authors wish to thank the referee for his careful reading and useful comments which have improved the exposition of the paper. Funding: Research of this work was partly supported by a US NSF grant DMS-1301595. The first author was partly supported by a fellowship from the Pacific Institute for the Mathematical Sciences and the second author was partly supported by a Simons Fellowship and collaboration grant from the Simons Foundation. 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