Monotone Relations in Hadamard Spaces

Monotone Relations in Hadamard Spaces Ali Moslemipoura, Mehdi Roohib arXiv:1906.00396v2 [math.FA] 26 Nov 2019 a Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran b Department of Mathematics, Faculty of Sciences, Golestan University, Gorgan, Iran Abstract. In this paper, the notion of W-property for subsets of X × X♦ is introduced and investigated, where X is an Hadamard space and X♦ is its linear dual space. It is shown that an Hadamard space X is flat if and only if X × X♦ has W-property. Moreover, the notion of monotone relation from an Hadamard space to its linear dual space is introduced. A characterization result for monotone relations with W-property (and hence in flat Hadamard spaces) is given. Finally, a type of Debrunner-Flor Lemma concerning extension of monotone relations in Hadamard spaces is proved. 1. Introduction and Preliminaries Let (X, d) be a metric space. We say that a mapping c : [0, 1] → X is a geodesic path from x ∈ X to y ∈ X if c(0) = x, c(1) = y and d(c(t), c(s)) = |t − s|d(x, y), for each t, s ∈ [0, 1]. The image of c is said to be a geodesic segment joining x and y. A metric space (X, d) is called a geodesic space if there is a geodesic path between every two points of X. Also, a geodesic space X is called uniquely geodesic space if for each x, y ∈ X there exists a unique geodesic path from x to y. From now on, in a uniquely geodesic space, we denote the set c([0, 1]) by [x, y] and for each z ∈ [x, y], we write z = (1 − t)x ⊕ ty, where t ∈ [0, 1]. In this case, we say that z is a convex combination of x and y. Hence, [x, y] = {(1 − t)x ⊕ ty : t ∈ [0, 1]}. More details can be found in [3, 5]. Definition 1.1. [9, Definition 2.2] Let (X,Pd) be a geodesic space, v1 , v2 , v3 , . . . , vn be n points in X and {λ1 , λ2 , λ3 , . . . , λn } ⊆ (0, 1) be such that ni=1 λi = 1. We define convex combination of {v1 , v2 , v3 , . . . , vn } inductively as following:  λ1 λ2 λn−1 v1 ⊕ v2 ⊕ · · · ⊕ vn ⊕ λn vn . 1 − λn 1 − λn 1 − λn   P n Note that for every x ∈ X, we have d x, ⊕i=1 λi vi ≤ ni=1 λi d(x, vi). ⊕ni=1 λi vi := (1 − λn )  (1) According to [3, Definition 1.2.1], a geodesic space (X, d) is a CAT(0) space, if the following condition, so-called CN-inequality, holds: d(z, (1 − λ)x ⊕ λy)2 ≤ (1 − λ)d(z, x)2 + λd(z, y)2 − λ(1 − λ)d(x, y)2 for all x, y, z ∈ X, λ ∈ [0, 1]. (2) One can show that (for instance see [3, Theorem 1.3.3]) CAT(0) spaces are uniquely geodesic spaces. An Hadamard space is a complete CAT(0) space. Let X be an Hadamard space. For each x, y ∈ X, the ordered pair (x, y) is called a bound vector and is → Indeed, X2 = {− → : x, y ∈ X}. For each x ∈ X, we apply 0 := → − denoted by − xy. xy xx as zero bound vector at x and x 2010 Mathematics Subject Classification. Primary 47H05; Secondary 47H04 Keywords. (Monotonicity, geodesic space, flat Hadamard spaces, monotone relations, Lipschitz semi-norm.) Email addresses: [email protected] (Ali Moslemipour), [email protected] (Mehdi Roohi) → as the bound vector − → → and → − −− xy yx. The bound vectors − xy uz are called admissible if y = u. Therefore the sum − → → − →+→ − − of two admissible bound vectors xy and yz is defined by − xy yz = → xz. Ahmadi Kakavandi and Amini in [2] have introduced the dual space of an Hadamard space, by using the concept of quasilinearization of abstract metric spaces presented by Berg and Nikolaev in [4]. The quasilinearization map is defined as following: h·, ·i : X2 × X2 → R o → − → − 1n hab, cdi := d(a, d)2 + d(b, c)2 − d(a, c)2 − d(b, d)2 ; a, b, c, d ∈ X. 2 (3) → (z) = 1 (d(x, z)2 − d(y, z)2 ); for each z ∈ X. We will see → : X → R by ϕ− Let x, y ∈ X, we define the mapping ϕ−xy xy 2 → possess attractive properties that simplify some calculations. We observe that (3) can be rewritten that ϕ−xy as following: → − → − − (b) − ϕ→ − (a) = ϕ→ − (d) − ϕ→ − (c). hab, cdi = ϕ→ cd cd ab ab The metric space (X, d) satisfies the Cauchy-Schwarz inequality if → − → − hab, cdi ≤ d(a, b)d(c, d) for all a, b, c, d ∈ X. This inequality characterizes CAT(0) spaces. Indeed, it follows from [4, Corollary 3] that a geodesic space (X, d) is a CAT(0) space if and only if it satisfies in the Cauchy-Schwarz inequality. For an Hadamard space (X, d), consider the mapping Ψ : R × X2 → C(X, R) → − − (t, a, b) 7→ Ψ(t, a, b)x = thab, → axi; a, b, x ∈ X, t ∈ R, where C(X, R) denotes the space of all continuous real-valued functions on R × X2 . It follows from CauchySchwarz inequality that Ψ(t, a, b) is a Lipschitz function with Lipschitz semi-norm L(Ψ(t, a, b)) = |t|d(a, b), for all a, b ∈ X, and all t ∈ R, (4) where the Lipschitz semi-norm for any function ϕ : (X, d) → R is defined by L(ϕ) = sup  ϕ(x) − ϕ(y) d(x, y)  : x, y ∈ X, x , y . A pseudometric D on R × X2 induced by the Lipschitz semi-norm (4), is defined by D((t, a, b), (s, c, d)) = L(Ψ(t, a, b) − Ψ(s, c, d)); a, b, c, d ∈ X, t, s ∈ R. For an Hadamard space (X, d), the pseudometric space (R × X2 , D) can be considered as a subspace of the pseudometric space of all real-valued Lipschitz functions Lip(X, R). Note that, in view of [2, Lemma 2.1], → − → → − → D((t, a, b), (s, c, d)) = 0 if and only if thab, − xyi = shcd, − xyi for all x, y ∈ X. Thus, D induces an equivalence relation on R × X2 , where the equivalence class of (t, a, b) ∈ R × X2 is → − → − [tab] = {scd : s ∈ R, c, d ∈ X, D((t, a, b), (s, c, d)) = 0}. → − The dual space of an Hadamard space (X, d), denoted by X∗ , is the set of all equivalence classes [tab] where → − → − (t, a, b) ∈ R × X2 , with the metric D([tab], [scd]) := D((t, a, b), (s, c, d)). Clearly, the definition of equivalence → − − − aa], where a ∈ X and t ∈ R classes implies that [→ aa] = [bb] for all a, b ∈ X. The zero element of X∗ is 0 := [t→ are arbitrary. It is easy to see that the evaluation h0, ·i vanishes for any bound vectors in X2 . Note that in 2 general X∗ acts on X2 by − − − → = th→ → where x∗ = [t→ → ∈ X2 . hx∗ , − xyi ab, xyi, ab] ∈ X∗ and − xy The following notation will be used throughout this paper. n DX i=1 n X →E := → α ∈ R, x∗ ∈ X∗ , n ∈ N, x, y ∈ X. αi x∗i , − xy αi hx∗i , − xyi, i i i=1 For an Hadamard space (X, d), Chaipunya and Kumam in [7], defined the linear dual space of X by X♦ = n X i=1  αi x∗i : αi ∈ R, x∗i ∈ X∗ , n ∈ N . Therefore, X♦ = span X∗ . It is easy to see that X♦ is a normed space with the norm kx♦ k♦ = L(x♦ ) for all x♦ ∈ X♦ . Indeed: Lemma 1.2. [14, Proposition 3.5] Let X be an Hadamard space with linear dual space X♦ . Then kx k♦ := sup ♦ ( → − → − hx♦ , abi − hx♦ , cdi d(a, b) + d(c, d) ) : a, b, c, d ∈ X, (a, c) , (b, d) , → − is a norm on X♦ . In particular, k[tab]k♦ = |t|d(a, b). 2. Flat Hadamard Spaces and W-property Let M be a relation from X to X♦ ; i.e., M ⊆ X × X♦ . The domain and range of M are defined, respectively, by n o Dom(M) := x ∈ X : ∃ x♦ ∈ X♦ such that (x, x♦) ∈ M , and n o Range(M) := x♦ ∈ X♦ : ∃ x ∈ X such that (x, x♦ ) ∈ M . Definition 2.1. Let X be an Hadamard space with linear dual space X♦ . We say that M ⊆ X × X♦ satisfies W-property if there exists p ∈ X such that the following holds: D −−−−−−−−−−−−−−−→E −→i + λhx♦ , − −→i, ∀λ ∈ [0, 1], ∀x♦ ∈ Range(M), ∀x , x ∈ Dom(M). x♦ , p((1 − λ)x1 ⊕ λx2 ) ≤ (1 − λ)hx♦ , − px px 1 2 1 2 Proposition 2.2. Let X be an Hadamard space with linear dual space X♦ and let M ⊆ X × X♦ . Then the following statements are equivalent: (i) M ⊆ X × X♦ satisfies the W-property for some p ∈ X. (ii) M ⊆ X × X♦ satisfies the W-property for any q ∈ X. (iii) For any q ∈ X, n D −−−−−−−−→E X →i, for all x♦ ∈ Range(M), {x }n ⊆ Dom(M), {λ }n ⊆ [0, 1]. (W (q)) x♦ , q(⊕ni=1 λi xi ) ≤ λi hx♦ , − qx i i i=1 i i=1 n i=1 3 (iv) For some p ∈ X, (Wn (p)) holds. Proof. (i) ⇒ (ii): Let q ∈ X be any arbitrary element of X, λ ∈ [0, 1], x♦ ∈ Range(M), and x1 , x2 ∈ Dom(M). Then D −−−−−−−−−−−−−−−→E D → −−−−−−−−−−−−−−−→E x♦ , q((1 − λ)x1 ⊕ λx2 ) = x♦ , − qp + p((1 − λ)x1 ⊕ λx2 ) D −−−−−−−−−−−−−−−→E − = hx♦ , → qpi + x♦ , p((1 − λ)x1 ⊕ λx2 ) − −→i) + λ(hx♦ , → − −→i) ≤ (1 − λ)(hx♦ , → qpi + hx♦ , − px qpi + hx♦ , − px 1 2 − −→i + λhx♦ , → − −→i = (1 − λ)hx♦ , → qp + − px qp + − px 1 2 − → − → ♦ − ♦ − = (1 − λ)hx , qx i + λhx , qx i, 1 2 as required. (ii) ⇒ (iii): We proceed by induction on n. By Definition 2.1 the claim is true for n = 2. Now assume that (Wn−1 (q)) is true. In view of equation (1), −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− →E D −−−−−−−−→E D −  −−−−−− λ1 λ2 λn−1 x♦ , q(⊕ni=1 λi xi ) = x♦ , q((1 − λn ) x1 ⊕ x2 ⊕ · · · ⊕ xn−1 ⊕ λn xn ) 1 − λn 1 − λn 1 − λn −−−−−−→E D −−−−−λ−1−−−−−−−−−−λ−2−−−−−−−−−−−−−−λ−n−1 −→i ♦ qx x1 ⊕ x2 ⊕ · · · ⊕ xn−1 + λn hx♦ , − ≤ (1 − λn ) x , q n 1 − λn 1 − λn 1 − λn n X λi →i + λ hx♦ , − −→i ≤ (1 − λn ) hx♦ , − qx qx i n n 1 − λn i=1 = n−1 X →i + λ hx♦ , − −→i λi hx♦ , − qx qx i n n = n X →i. λi hx♦ , − qx i i=1 i=1 (iii) ⇒ (iv): Clear. (iv) ⇒ (i): Take n = 2 in (Wn (p)). We are done. Remark 2.3. It should be noticed that Proposition 2.2 implies that W-property is independent of the choice of the element p ∈ X. Definition 2.4. [11, Definition 3.1] An Hadamard space (X, d) is said to be flat if equality holds in the CN-inequality, i.e., for each x, y ∈ X and λ ∈ [0, 1], the following holds: d(z, (1 − λ)x ⊕ λy)2 = (1 − λ)d(z, x)2 + λd(z, y)2 − λ(1 − λ)d(x, y)2, for all z ∈ X. Proposition 2.5. Let X be an Hadamard space. The following statements are equivalent: (i) X is a flat Hadamard space. − − −−−−−−−−−−−−−−→ → →→ (ii) hx((1 − λ)x ⊕ λy), abi = λh− xy, abi, for all a, b, x, y ∈ X and all λ ∈ [0, 1]. (iii) X × X♦ has W-property. 4 (iv) Any subset of X × X♦ has W-property. − (v) For each p, z ∈ X, the mapping ϕ→ pz is convex. − (vi) For each p, z ∈ X, the mapping ϕ→ pz is affine, in the sense that: − − − ϕ→ pz (y), ∀ x, y ∈ X, ∀ λ ∈ [0, 1]. pz (x) + λϕ→ pz ((1 − λ)x ⊕ λy) = (1 − λ)ϕ→ Proof. (i) ⇔ (ii): [11, Theorem 3.2]. (ii)⇒ (iii): Let x, y ∈ X, λ ∈ [0, 1] and (x, x♦ ) ∈ X × X♦ . Then x♦ = we get: Pn i=1 −−→ αi [ti ai bi ] ∈ X♦ , and hence by using (ii) n D−−→ − −−−−−−−−−−−−−−→E D −−−−−−−−−−−−−−→E X px + x (1 − λ)x ⊕ λy x♦ , p (1 − λ)x ⊕ λy = αi ti ai bi , → i=1 = n X i=1 D−−→ − E D−−→ −−−−−−−−−−−−−−→E px + ai bi , x (1 − λ)x ⊕ λy αi ti ai bi , → = n X i=1 D−−→ E D−−→ E − → px + λ ai bi , − xy αi ti ai bi , → = n X i=1 D−−→ − E D−−→ → → E px + λ ai bi , − py − − px αi ti ai bi , → = n X  D−−→ − E D−−→ →E px + λ ai bi , − py αi ti (1 − λ) ai bi , → i=1 n X D−−→ E D−−→ E → − py px + λ αi ti ai bi , − αi ti ai bi , → = (1 − λ) n X = (1 − λ) n DX i=1 i=1 i=1 n DX −−→ − E −−→ →E px + λ αi [ti ai bi ], → py αi [ti ai bi ], − − → = (1 − λ)hx♦ , → pxi + λhx♦ , − pyi. i=1 Therefore X × X♦ has W-property. (iii) ⇔ (iv): Straightforward. (iv) ⇒ (v): Let x, y ∈ X and λ ∈ [0, 1], then − − − − − − − (1 − λ)ϕ→ pz (x) + λϕ→ pz (y) − ϕ→ pz ((1 − λ)x ⊕ λy) = λ(ϕ→ pz (y) − ϕ→ pz (x)) + ϕ→ pz (x) − ϕ→ pz ((1 − λ)x ⊕ λy) −−−−−−−−−−−−−−→ → − − → → − = λhpz, xyi + hpz, ((1 − λ)x ⊕ λy)xi −−−−−−−−−−−−−−→ − →−→ − − − = λh→ pz, − py pxi + h→ pz, → px − p((1 − λ)x ⊕ λy)i −−−−−−−−−−−−−−→ − → + (1 − λ)h→ − − − = λh→ pz, − pyi pz, → pxi − h→ pz, p((1 − λ)x ⊕ λy)i ≥ 0. − Therefore, ϕ→ pz is convex. (v) ⇒ (vi): It is easy. 5 (vi) ⇒ (iii): Let x, y, p ∈ X, λ ∈ [0, 1] and x♦ = D Pn i=1 ♦ αi [ti − p−→ i zi ] ∈ X be given. Then n −−−−−−−−−−−−−−→E −−−−−−−−−−−−−−→E D X αi x∗i , p (1 − λ)x ⊕ λy x , p (1 − λ)x ⊕ λy = ♦ i=1 = n X i=1 D −→ −−−−−−−−−−−−−−→E pi zi , p (1 − λ)x ⊕ λy αi ti − = n X i=1   ((1 − λ)x ⊕ λy) − ϕ−p−→ (p) αi ti ϕ−p−→ i zi i zi = n X i=1   (x) + λϕ−p−→ αi ti (1 − λ)ϕ−p−→ (y) − ϕ−p−→ (p) i zi i zi i zi = n X i=1   αi ti (1 − λ)(ϕ−p−→ (x) − ϕ−p−→ (p)) + λ(ϕ−p−→ (y) − ϕ−p−→ (p)) i zi i zi i zi i zi = n X  D −→ → E D −→ − →E p i zi , − px + λ − pi zi , py αi ti (1 − λ) − i=1 n X D −→ − D−−→ → E →E − pi zi , py αi ti − αi ti pi zi , px + λ = (1 − λ) n X = (1 − λ) n DX i=1 i=1 i=1 n DX − →E → −E αi [ti − p−→ αi [ti − p−→ i zi ], py i zi ], px + λ − → = (1 − λ)hx , → pxi + λhx♦ , − pyi; i=1 ♦ i.e., X × X♦ has W-property. (iii)⇒ (ii): For a, b, x, y ∈ X and λ ∈ [0, 1], we have: →  → → − → → − −−−−−−−−−−−−−−→ − → → − −−−−−−−−−−−−−−→ − λhab, − xyi − hab, x((1 − λ)x ⊕ λy)i = λ hab, − py − − pxi − hab, p((1 − λ)x ⊕ λy) − → pxi →  − − → − → − → − − − − − − − − − − − − − − − − → → − hab, → − = λ hab, pyi pxi − hab, p((1 − λ)x ⊕ λy)i + hab, → pxi → − → → − → − − − − − − − − − − − − − − − → → − hab, p((1 − λ)x ⊕ λy)i = (1 − λ)hab, − pxi + λhab, − pyi −−−−−−−−−−−−−→ − → − hx♦ , − = (1 − λ)hx♦ , → pxi + λhx♦ , − pyii p((1 − λ)x ⊕ λy), → − where x♦ = [ab] ∈ X♦ . Since X × X♦ has W-property, one can deduce that: → − −−−−−−−−−−−−−−→ → − → xyi ≥ hab, x((1 − λ)x ⊕ λy)i. λhab, − (5) Hence, by interchanging the role of a and b in (5), we obtain: → − −−−−−−−−−−−−−−→ → − → hab, x((1 − λ)x ⊕ λy)i ≥ λhab, − xyi. (6) Finally, (5) and (6) yield: → → − −−−−−−−−−−−−−−−→ − → hab, p((1 − λ)x ⊕ λxy)i = λ hab, − xyi . We are done. 6 The next example shows that there exists a relation M ⊆ X × X♦ in the non-flat Hadamard spaces which doesn’t have the W-property. Example 2.6. Consider the following equivalence relation on N × [0, 1]: (n, t) ∼ (m, s) ⇔ t = s = 0 or (n, t) = (m, s). Set X := N×[0,1] ∼ and let d : X × X → R be defined by    |t − s| d([(n, t)], [(m, s)]) =   t + s n = m, n , m. The geodesic joining x = [(n, t)] to y = [(m, s)] is defined as follows:  t   [(n, (1 − λ)t − λs)] 0 ≤ λ ≤ t+s , (1 − λ)x ⊕ λy :=   [(m, (λ − 1)t + λs)] t ≤ λ ≤ 1, t+s whenever x , y and vacuously (1 − λ)x ⊕ λx := x. It is known that (see [1, Example 4.7]) (X, d) is an R-tree space. It follows from [3, Example 1.2.10], that any R-tree space is an Hadamard space. Let x = [(2, 12 )], y = [(1, 21 )], 3 a = [(3, 13 )], b = [(2, 12 )] and λ = 51 . Then 45 x ⊕ 15 y = [(2, 10 )] and −−−−−−−−−→  −E − 4 1 → −1 −1 1 D− →→ ab . x( x ⊕ y), ab = , = xy, 5 5 6 10 5 Now, Proposition 2.5(ii) implies that (X, d) is not a flat Hadamard space. For each n ∈ N, set xn := [(n, 21 )] and yn := [(n, n1 )]. Now, we define n o −−−→ M := (xn , [− y−n+1 yn ]) : n ∈ N ⊆ X × X♦ . D → −E Take p = [(1, 1)] ∈ X, [− y−5−→ y4 ] ∈ Range(M) and λ = 13 . Clearly, x̃ := (1 − λ)x1 ⊕ λx3 = [(1, 61 )] and [− y−5−→ y4 ], px̃ = while, 1 24 , 2 −−−→ −−→ 1 −−→ −−→ 1 h[ y5 y4 ], px1i + h[− y5 y4 ], px3i = . 3 3 40 Therefore, M doesn’t have the W-property. 3. Monotone Relations Ahmadi Kakavandi and Amini [2] introduced the notion of monotone operators in Hadamard spaces. In [10], Khatibzadeh and Ranjbar, investigated some properties of monotone operators and their resolvents and also proximal point algorithm in Hadamard spaces. Chaipunya and Kumam [7] studied the general proximal point method for finding a zero point of a maximal monotone set-valued vector field defined on Hadamard spaces. They proved the relation between the maximality and Minty’s surjectivity condition. Zamani Eskandani and Raeisi [14], by using products of finitely many resolvents of monotone operators, proposed an iterative algorithm for finding a common zero of a finite family of monotone operators and a common fixed point of an infinitely countable family of non-expansive mappings in Hadamard spaces. In this section, we will characterize the notation of monotone relations in Hadamard spaces based on characterization of monotone sets in Banach spaces [8, 12, 13]. Definition 3.1. Let X be an Hadamard space with linear dual space X♦ . The set M ⊆ X × X♦ is called → monotone if hx♦ − y♦ , − yxi ≥ 0, for all (x, x♦),(y, y♦ ) in M. 7 Example 3.2. Let xn , yn and M be the same as in Example 2.6. Let (u, u♦ ), (v, v♦) ∈ M. There exists m, n ∈ N such −−−→ −−−y→]. Then that u = xn , u♦ := [− y−n+1 yn ], v = xm and v♦ := [− y−m+1 m −−−−−−→ii − h−−−−1−−− ih−−−−1−→iE Dh− h−−−−−−−−−1−−−−ih 1 − → → → ♦ − ♦ − ) (n, ) , (m, ) (n, ) hu − v , vui = hu , vui − hv , vui = (n + 1, n+1 n 2 2 − − − − − − − − − − − − −−−ih −−−−−−−→ii − −−−−−−− −−−−−→iE Dhh h ih 1 1 1 1 − (m + 1, ) (m, ) , (m, ) (n, ) m+1 m 2 2   0, n = m,      1 1 1   n = m + 1,  m+1 + n + m , =  1  + n1 + m1 , n = m − 1,  n+1     1 + 1 , n < {m − 1, m, m + 1}. ♦ ♦ n m → ≥ 0 which shows that, M is a monotone relation. Therefore, hu♦ − v♦ , − vui In the sequel, we need the following notations. Let X be an Hadamard space and Y ⊆ X. Put   X η(x) = 1 ςY := η : Y → [0, +∞[ supp η is finite and x∈Y where supp η = {y ∈ Y : η(y) , 0}. Clearly, for each ∅ , A ⊂ Y, ςA = {η ∈ ςY : supp η ⊆ A}. It is obvious that ςA is a convex subset of RY . Moreover, if ∅ , A ⊆ B, then ςA ⊆ ςB . Suppose u ∈ Y be fixed. Define δu ∈ ςY by ( 1 x = u, δu (x) = 0 x , u. Let M ⊆ X × X♦ and η ∈ ςA . Then suppη = {λ1 , . . . , λn } where λi = η(xi , x♦i ), for each 1 ≤ i ≤ n. Let p ∈ X be fixed. Define α : ςX×X♦ → X (resp. β : ςX×X♦ → X♦ and θp : ςX×X♦ → R) by α(η) = n M i=1 n n X X  →i. λi xi , resp. β(η) = λi x♦i and θp (η) = λi hx♦i , − px i i=1 i=1 Proposition 3.3. Let X be an Hadamard space, M ⊆ X × X♦ and p ∈ X. Set n −−−−→ o Θp,M := η ∈ ςM : θp (η) ≥ hβ(η), pα(η)i . (7) Then Θp,M = Θq,M for any q ∈ X. Proof. It is enough to show that Θp,M ⊆ Θq,M . Let η ∈ Θp,M be such that suppη = {λ1 , . . . , λn } where λi = η(xi , x♦i ), for each 1 ≤ i ≤ n. Then θq (η) = n X →i = λi hx♦i , − qx i i=1 n X =h n X − λi hx♦i , → qpi + i=1 n X →i λi hx♦i , − px i i=1 − − λi x♦i , → qpi + θp (η) = hβ(η), → qpi + θp (η) i=1 − −−−→ − ≥ hβ(η), → qpi + hβ(η), pα(η)i − −−− → = hβ(η), qα(η)i. Therefore, η ∈ Θq,M , i.e., Θp,M ⊆ Θq,M . 8 According to Proposition 3.3, for each M ⊆ X × X♦ , the set Θp,M is independent of the choice of the element p ∈ X and hence we denote the set Θp,M by ΘM . Theorem 3.4. Let X be an Hadamard space and M ⊆ X × X♦ satisfies the W-property. Then M is a monotone set if and only if ΘM = ςM . Proof. Let M be a monotone set. In view of (7), it is enough to show that ςM ⊆ ΘM . Let η ∈ ςM be such that suppη = {λ1 , . . . , λn } where λi = η(xi , x♦i ), for each 1 ≤ i ≤ n. By using Proposition 2.2, we obtain: n n X X −− −−−−→E −−− −−−−→ n →i − D θp (η) − hβ(η), pα(η)i = λi hx♦i , − px λ j x♦j , p ⊕i=1 λi xi i i=1 = n X j=1 →i − λi hx♦i , − px i i=1 n X ≥ n X →i − λi hx♦i , − px i n X n X = n X n X →i − λi λ j hx♦i , − px i = n X n X →i λi λ j hx♦i − x♦j , − px i = n X n X →i px λi λ j hx♦j − x♦i , − j j=1 −−−−→E D −− −−− n λ j x♦j , p ⊕i=1 λi xi →i λi λ j hx♦j , − px i j=1 i=1 i=1 j=1 i=1 n X n X →i λi λ j hx♦j , − px i j=1 i=1 j=1 i=1 j=1 i=1 n = n 1 XX →−− →i λi λ j hx♦i − x♦j , − px px i j 2 i=1 j=1 = n n 1 XX 2 λi λ j hx♦i − x♦j , − x−→ j xi i ≥ 0. i=1 j=1 Then ςM ⊆ ΘM and hence ςM = ΘM . For the converse, let (x, x♦ ), (y, y♦) ∈ M and set η := 12 δ(x,x♦ ) + 21 δ(y,y♦ ) ∈ ςM . By using W-property, we get: 1 ♦ 1 → − → hx − y♦ , − yxi = hx♦ − y♦ , → px − − pyi 4 4 1 − → = (hx♦ − y♦ , → pxi − hx♦ − y♦ , − pyi) 4 1 1 1 − → − 1 hy♦ , → − → = hx♦ , → pxi + hy♦ , − pyi pxi − hx♦ , − pyi 4 4 4 4 1 1 1 1 − → − 1 hx♦ , → − → − 1 hy♦ , → − → = hx♦ , → pxi + hy♦ , − pyi pxi − hx♦ , − pyi pxi − hy♦ , − pyi 2 2 4 4 4 4 −−−−−−−−−→E 1 D →E D 1 ♦ 1 ♦ 1 1 1 − py − x + y , p( x ⊕ y) pxi + y♦ , − ≥ hx♦ , → 2 2 2 2 2 2 −−−−−−−−−→E −−−−−−−−−→ D 1 1 − → − 1 x♦ , p( 1 x ⊕ 1 y) − 1 D y♦ , p( 1 x ⊕ 1 y)E pxi + hy♦ , − pyi = hx♦ , → 2 2 2 2 2 2 2 2 −−−−→ = θp (η) − hβ(η), pα(η)i ≥ 0. Therefore, M is monotone. 9 Corollary 3.5. Let X be a flat Hadamard space and M ⊆ X × X♦ . Then M is a monotone set if and only if ΘM = ςM . Proof. Since X is flat, Proposition 2.5 implies that M ⊆ X × X♦ satisfies the W-property. Then the conclusion follows immediately from Theorem 3.4. A fundamental result concerning monotone operators is the extension theorem of Debrunner-Flor (for a proof see [6, Theorem 4.3.1] or [15, Proposition 2.17]). In the sequel, we prove a type of this result for monotone relations from an Hadamard space to its linear dual space. First, we recall some notions and results. Definition 3.6. [2, Definition 2.4] Let {xn } be a sequence in an Hadamard space X. The sequence {xn } is said w −→, − → to be weakly convergent to x ∈ X, denoted by xn −→ x, if limn→∞ h− xx n xyi = 0, for all y ∈ X. One can easily see that convergence in the metric implies weak convergence. Lemma 3.7. [14, Proposition 3.6] Let {xn } be a bounded sequence in an Hadamard space (X, d) with linear dual k·k♦ − x−→zi → hx♦ , → xzi, space X♦ and let {x♦ } be a sequence in X♦ . If {x } is weakly convergent to x and x♦ −−→ x♦ , then hx♦ , − n n n n n for all z ∈ X. Theorem 3.8. Let X be an Hadamard space and M ⊆ X × X♦ be a monotone relation satisfies the W-property. Let C ⊆ X♦ be a compact and convex set, and ϕ : C → X be a continuous function. Then there exists z♦ ∈ C such that {(ϕ(z♦), z♦ )} ∪ M is monotone. Proof. Let x ∈ X, u♦ , v♦ ∈ X♦ be arbitrary and fixed element. Consider the function τ : C → R defined by −−−−−→ τ(x♦ ) = hx♦ − v♦ , xϕ(u♦ )i, x♦ ∈ C. k·k♦ Let {x♦n } ⊆ C be such that x♦n −−→ x♦ , for some x♦ ∈ C. By Lemma 3.7, −−−−−→ −−−−−→ hx♦n − v♦ , xϕ(u♦ )i → hx♦ − v♦ , xϕ(u♦ )i. Thus τ(x♦n ) → τ(x♦ ). Hence τ is continuous. For every (y, y♦ ) ∈ M, set −−−−−→ U(y, y♦) := {u♦ ∈ C : hu♦ − y♦ , yϕ(u♦ )i < 0}. Continuity of τ implies that U(y, y♦) is an open subset of C. Suppose that the conclusion fails. Then for each ♦ u♦ ∈ C there exists (y, y♦) ∈ M such that u♦ ∈ U(y, y♦). This means that the family Sn of open ♦sets {U(y, y )}(y,y♦ )∈M is an open cover of C. Using the compactness of C, we obtain that C = i=1 U(yi , yi ). In addition, [15, Page 756] implies that there exists a partition of unity associated with this finite subcover. Hence, there are continuous functions ψi : X♦ → R (1 ≤ i ≤ n) satisfying P (i) ni=1 ψi (x♦ ) = 1, for all x♦ ∈ C. (ii) ψi (x♦ ) ≥ 0, for all x♦ ∈ C and all i ∈ {1, . . . , n}. (iii) {x♦ ∈ C : ψi (x♦ ) > 0} ⊆ Ui := U(yi , y♦i ) for all i ∈ {1, . . . , n}. Set K := co({y♦1 , . . . , y♦n }) ⊆ C and define ι :K → K n X u♦ 7→ ψi (u♦ )y♦i . i=1 10 Let {u♦m } ⊆ K be such that u♦m → u♦ , ι(u♦m ) − ι(u♦ ) n X = ♦ i=1 n X = ≤ ≤ i=1 n X i=1 n X ψi (u♦m )y♦i − n X ψi (u♦ )y♦i i=1 (ψi (u♦m ) − ψi (u♦ ))y♦i ♦ (ψi (u♦m ) − ψi (u♦ ))y♦i ♦ ♦ ψi (u♦m ) − ψi (u♦ ) ky♦i k♦ . i=1 By continuity of ψi (1 ≤ i ≤ n), letting m → +∞, then ψi (u♦m ) → ψi (u♦ ) and this implies that ι(u♦m ) → ι(u♦ ) and so ι is continuous. One can identify K with a finite-dimensional convex and compact set. By using Brouwer fixed point theorem [15, Proposition 2.6], there exists w♦ ∈ K such that ι(w♦ ) = w♦ . Moreover, by using Proposition 2.2 we get: D −−−−−−−−−−−−−−−−→E 0 = ι(w♦ ) − w♦ , ϕ(w♦ )(⊕ j ψ j (w♦ )y j ) DX −−−−−−−−−−−−−−−−→E = ψi (w♦ )(y♦i − w♦ ), ϕ(w♦ )(⊕ j ψ j (w♦ )y j ) i = DX i −−−−−−−−−−−→ E D X −−−−−→E ψi (w♦ )(y♦i − w♦ ), p(⊕ jψ j (w♦ )y j ) − ψi (w♦ )(y♦i − w♦ ), pϕ(w♦ ) i −−−−−→E ψi (w♦ )(y♦i − w♦ ), pϕ(w♦) ψ j (w♦ ) j DX − →E − D py ψi (w♦ )(y♦i − w♦ ), − j = X ψ j (w♦ ) DX X DX −−−−−→E − →E − ψi (w♦ )(y♦i − w♦ ), pϕ(w♦) ψ j (w♦ ) py ψi (w♦ )(y♦i − w♦ ), − j = X ψ j (w ) j DX −−−−→E − →−− py pϕ(w♦ ) ψi (w )(yi − w ), − j = X ψ j (w♦ ) DX −−−−−−→E ψi (w♦ )(y♦i − w♦ ), ϕ(w♦ )y j = X = XX = X ≤ X i i j ♦ i i j ♦ ψ j (w ) j j i, j X i i X i i j ♦ ♦ (p ∈ X) ♦ D −−−−−−→E ψi (w♦ ) y♦i − w♦ , ϕ(w♦ )y j D −−−−−−→E ψ j (w♦ )ψi (w♦ ) y♦i − w♦ , ϕ(w♦ )y j D −−−−−−→E ψi (w♦ )ψ j (w♦ ) y♦i − w♦ , ϕ(w♦ )y j . (8) −−−−−−→ Set ai j = hy♦i − w♦ , ϕ(w♦ )y j i. It follows from monotonicity of M that −−−−−−→ −−−−−−→ aii + a j j − ai j − a ji = hy♦i − w♦ , ϕ(w♦ )yi i + hy♦j − w♦ , ϕ(w♦ )y j i −−−−−−→ −−−−−−→ − hy♦i − w♦ , ϕ(w♦ )y j i − hy♦j − w♦ , ϕ(w♦ )yi i −−−−−−→ −−−−−−→ = hy♦i − y♦j , ϕ(w♦ )yi − ϕ(w♦ )y j i = hy♦ − y♦ , − y−→ y i ≥ 0; i j j i 11 i.e., aii + a j j ≥ ai j + a ji . (9) Applying (8) and (9), we obtain: 0≤ n X ψi (w♦ )ψ j (w♦ )ai j n X ψi (w♦ )ψ j (w♦ )ai j + i, j = i< j = i=1 n X ψi (w♦ )2 aii + n X ψi (w♦ )2 aii + i=1 ≤ n X ψi (w♦ )2 aii + n X ψi (w♦ )ψ j (w♦ )ai j i> j n X ψi (w♦ )ψ j (w♦ )(ai j + a ji ) (10) n X ψi (w♦ )ψ j (w♦ )(aii + a j j ). (11) i< j i=1 i< j o n Set I(w♦ ) := i ∈ {1, . . . , n} : w♦ ∈ Ui . Applying property (iii) of the partition of unity in (11) we get: 0≤ X i∈I(w♦ ) ψi (w♦ )2 aii + X ψi (w♦ )ψ j (w♦ )(aii + a j j ). (12) i< j i, j∈I(w♦ ) By using property (iii) of the partition of unity and the definition of Ui , one deduce that all terms in the right-hand side of (12) are nonpositive. 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