Monotonicity of sets in Hadamard spaces from polarity point of view
aa r X i v : . [ m a t h . F A ] S e p Monotonicity of sets in Hadamard spaces from polarity point of view
Ali Moslemipour a , Mehdi Roohi b a Department of Mathematics, Islamic Azad University, Science and Research Branch, Tehran, Iran b Department of Mathematics, Faculty of Sciences, Golestan University, Gorgan, Iran
Abstract.
This paper is devoted to introduce and investigate the notion of monotone sets in Hadamardspaces. First, flat Hadamard spaces are introduced and investigated. It is shown that an Hadamard space X is flat if and only if X × X ♦ has F l -property, where X ♦ is the linear dual of X . Moreover, monotone andmaximal monotone sets are introduced and also monotonicity from polarity point of view is considered.Some characterizations of (maximal) monotone sets, specially based on polarity, is given. Finally, it isproved that any maximal monotone set is sequentially bw ×k · k ♦ -closed in X × X ♦ .
1. Preliminaries
In this section, we collect some fundamental definitions and general notations of Hadamard spaces thatwill be used throughout of this paper. For background materials on Hadamard spaces, we refer to thestandard texts and literatures such as [2, 3, 5, 6].Given a metric space ( X , d ), a geodesic path from x ∈ X to y ∈ X is a map c : [0 , → X such that c (0) = x , c (1) = y and d ( c ( t ) , c ( s )) = | t − s | d ( x , y ), for each t , s ∈ [0 , c is called a geodesic segment joining x and y . We say that metric space ( X , d ) is a geodesic space if for each x , y ∈ X , there exists a geodesic path c : [0 , → X from x to y . Also, a geodesic space ( X , d ) is called uniquely geodesic space if for every pair ofpoints x , y ∈ X there exists a unique geodesic from x to y .A geodesic space ( X , d ) is called a CAT(0) space if we have: d ( z , c ( t )) ≤ (1 − t ) d ( z , x ) + td ( z , y ) − t (1 − t ) d ( x , y ) , (1)for each geodesic path c : [0 , → X from x to y , each z ∈ X and each t ∈ [0 , CN-inequality . Moreover, the notion of CAT(0) spaces is characterized by CN-inequality. It is known [3,Theorem 1.3.3] that CAT(0) spaces are uniquely geodesic spaces. Furthermore, the image of unique geodesicpath c from initial point x to terminal point y is denoted by [ x , y ], i.e., c ([0 , = [ x , y ]. For each z ∈ [ x , y ], wewrite z = (1 − t ) x ⊕ ty and say that z is a convex combination of x and y . A complete CAT(0) space is called an Hadamard space . Basic examples of Hadamard spaces are: Hilbert spaces, Hadamard manifolds, EuclideanBuildings, R -trees, (see [6, Chapter II.1, 1.15] for many other examples).In 2008, Berg and Nikolaev [5] introduced the concept of quasilinearization in abstract metric spaces.Ahmadi Kakavandi and Amini [2] defined the dual space for an Hadamard space ( X , d ) by using the conceptof quasilinearization of it. More precisely, let X be an Hadamard space. For each x , y ∈ X , the ordered pair( x , y ) ∈ X is denoted by −→ xy and will be called a bound vector in which x and y are called tail and head of −→ xy ,respectively. For each x ∈ X , the zero bound vector at x ∈ X will be written as x : = −→ xx . We identify two boundvectors −−→ xy and −→ yx . The bound vectors −→ xy and −→ uz are called admissible if y = u . The operation of addition oftwo admissible bound vectors −→ xy and −→ yz is defined by −→ xy + −→ yz = −→ xz . The quasilinearization map is defined by h· , ·i : X × X → R , (2) Mathematics Subject Classification . Primary 47H05; Secondary 47H04; 49J40.
Keywords . (Monotone sets, maximal monotone sets, polarity, flat Hadamard space, R -tree) Email addresses: [email protected] (Ali Moslemipour), [email protected] (Mehdi Roohi) −→ ab , −→ cd i : = n d ( a , d ) + d ( b , c ) − d ( a , c ) − d ( b , d ) o , a , b , c , d , ∈ X . For each a , b , c , d , u ∈ X we have:(i) h−→ ab , −→ cd i = h−→ cd , −→ ab i ,(ii) h−→ ab , −→ cd i = −h−→ ba , −→ cd i ,(iii) h−→ ab , −→ cd i = h−→ au , −→ cd i + h−→ ub , −→ cd i .An approach to characterization of the CAT(0) spaces have been proved in [5, Corollary 3] based on Cauchy-Schwarz inequality . More precisely, a geodesic space ( X , d ) is a CAT(0) space if and only if for each a , b , c , d ∈ X we have: h−→ ab , −→ cd i ≤ d ( a , b ) d ( c , d ) . (3)Consider the map Θ : R × X → C ( X , R )( t , a , b ) Θ ( t , a , b ) x = t h−→ ab , −→ ax i , a , b , x ∈ X , t ∈ R , where C ( X , R ) is the space of all continuous real-valued functions on R × X . It follows from Cauchy-Schwarzinequality (3) that Θ ( t , a , b ) is a Lipschitz function with Lipschitz semi-norm L ( Θ ( t , a , b )) = | t | d ( a , b ) , a , b ∈ X , t ∈ R . (4)Recall that the Lipschitz semi-norm on C ( X , R ) is defined by L : C ( X , R ) → R ϕ sup n ϕ ( x ) − ϕ ( y ) d ( x , y ) : x , y ∈ X , x , y o . The Lipschitz semi-norm (4) induces a pseudometric D on R × X , which is defined by D (( t , a , b ) , ( s , c , d )) = L ( Θ ( t , a , b )) − L ( Θ ( s , c , d )) , a , b , c , d ∈ X , t , s ∈ R . Now, the pseudometric space ( R × X , D ) can be considered as a subspace of the pseudometric space of allreal-valued Lipschitz functions (Lip( X , R ) , L ). It is known that (see [2, Lemma 2.1]), D (( t , a , b ) , ( s , c , d )) = t h−→ ab , −→ xy i = s h−→ cd , −→ xy i for all x , y ∈ X . It is easily seen that D induces an equivalence relation on R × X . Indeed, the equivalence class of( t , a , b ) ∈ R × X is given by [ t −→ ab ] = n s −→ cd : D (( t , a , b ) , ( s , c , d )) = o . The set of all equivalence classes equipped with the metric D , defined by D ([ t −→ ab ] , [ s −→ cd ]) : = D (( t , a , b ) , ( s , c , d )) , is called the dual space of the Hadamard space X , and is denoted by X ∗ . By using the definition of equivalenceclasses we get [ −→ aa ] = [ −→ bb ] for each a , b ∈ X . In general, X ∗ acts on X by h x ∗ , −→ xy i = t h−→ ab , −→ xy i , where x ∗ = [ t −→ ab ] ∈ X ∗ and −→ xy ∈ X . * n X i = α i x ∗ i , −→ xy + : = n X i = α i D x ∗ i , −→ xy E , α i ∈ R , x ∗ i ∈ X ∗ , n ∈ N . In [8], Chaipunya and Kumam introduced the concept of linear dual space of an Hadamard space ( X , d ), asfollows: X ♦ = ( n X i = α i x ∗ i : α i ∈ R , x ∗ i ∈ X ∗ , n ∈ N ) . Indeed, X ♦ = span X ∗ . The zero element of X ♦ is denoted by X ♦ : = [ t −→ aa ], where a ∈ X and t ∈ R . One can seethat the evaluation h X ♦ , ·i vanishes on X . It is worth mentioned that X ♦ is a normed space with the norm k x ♦ k ♦ = L ( x ♦ ), for all x ♦ ∈ X ♦ . Indeed: Proposition 1.1. [15, Proposition 3.5]
Let X be an Hadamard space with linear dual space X ♦ and let x ♦ ∈ X ♦ bearbitrary. Then k x ♦ k ♦ : = sup ( (cid:12)(cid:12)(cid:12) h x ♦ , −→ ab i−h x ♦ , −→ cd i (cid:12)(cid:12)(cid:12) d ( a , b ) + d ( c , d ) : a , b , c , d ∈ X , ( a , c ) , ( b , d ) ) , is a norm on X ♦ . In particular, k [ t −→ ab ] k ♦ = | t | d ( x , y ) . Definition 1.2. [2, Definition 2.4] Let { x n } be a sequence in an Hadamard space X . The sequence { x n } is saidto be weakly convergent to x ∈ X , denoted by x n w −→ x , if lim n →∞ h−−→ xx n , −→ xy i =
0, for all y ∈ X .One can easily see that convergence in the metric implies weak convergence. Proposition 1.3. [15, Proposition 3.6]
Let { x n } be a bounded sequence in an Hadamard space ( X , d ) with linear dualspace X ♦ and let { x ♦ n } be a sequence in X ♦ . If { x n } is weakly convergent to x and x ♦ n k·k ♦ −−→ x ♦ , then h x ♦ n , −−→ x n z i → h x ♦ , −→ xz i ,for all z ∈ X. Definition 1.4.
Let X be an Hadamard space with linear dual space X ♦ .(i) A sequence { x n } ⊆ X is bw-convergent to x ∈ X , if { x n } is bounded and x n w −→ x . In this case, we write x n bw −→ x .(ii) A sequence { ( x n , x ♦ n ) } ⊆ X × X ♦ is called bw × k · k ♦ -convergent to ( x , x ♦ ) ∈ X × X ♦ , if x n bw −→ x and x ♦ n k·k ♦ −−→ x ♦ .In this case, we write ( x n , x ♦ n ) bw ×k·k ♦ −−−−−→ ( x , x ♦ ).(iii) M ⊆ X × X ♦ is sequentially bw ×k·k ♦ -closed if the limit of every bw ×k·k ♦ -convergent sequence { ( x n , x ♦ n ) } ⊆ M is in M .(iv) The mapping ϕ : X × X ♦ → ] − ∞ , ∞ ] is said to be sequentially bw × k · k -continuous at ( x , x ♦ ) ∈ X × X ♦ iffor every { ( x n , x ♦ n ) } ⊆ X × X ♦ , with ( x n , x ♦ n ) bw ×k·k ♦ −−−−−→ ( x , x ♦ ) we have ϕ ( x n , x ♦ n ) → ϕ ( x , x ♦ ). Moreover, ϕ is sequentially bw × k · k -continuous if it is sequentially bw × k · k -continuous at each point of X × X ♦ . Definition 1.5.
For an arbitrary point p ∈ X , we define the p-coupling function of the dual pair ( X , X ♦ ) by π p : X × X ♦ → R ; ( x , x ♦ )
7→ h x ♦ , −→ px i . This function is useful in the formulation of some basic results of the monotone sets in Hadamard spaces(see Proposition 3.19).
Proposition 1.6.
Let X be an Hadamard space with linear dual space X ♦ and p ∈ X. Then the following hold:(i) π p is sequentially bw × k · k ♦ -continuous and hence continuous. ii) π p is linear with respect to it’s second variable.Proof. (i): Let { ( x n , x ♦ n ) } ⊆ X × X ♦ be such that ( x n , x ♦ n ) bw ×k·k ♦ −−−−−→ ( x , x ♦ ), where ( x , x ♦ ) ∈ X × X ♦ . It follows fromProposition 1.3 that h x ♦ n , −−→ px n i → h x ♦ , −→ px i , which implies that π p is sequentially bw × k · k ♦ -continuous. Moreover, since convergence in the metricimplies weak convergence, we conclude that π p is continuous.(ii): Let x ♦ , y ♦ ∈ X ♦ , x ∈ X and α, β ∈ R . Then π p ( x , α x ♦ + β y ♦ ) = h α x ♦ + β y ♦ , −→ px i = α h x ♦ , −→ px i + β h y ♦ , −→ px i = απ p ( x , x ♦ ) + βπ p ( x , y ♦ ) . The proof is completed.
Definition 1.7. [3, Section 2.2] Let ( X , d ) be an Hadamard space and f : X → ] − ∞ , ∞ ] be a function. Then(i) The domain of f is defined by dom( f ) = { x ∈ X : f ( x ) < ∞} . Moreover, f is called proper if dom( f ) , ∅ .(ii) f is lower semi-continuous (briefly l.s.c. ) if for each α ∈ R , the set { x ∈ X : f ( x ) ≤ α } is closed.(iii) f is convex if f ((1 − λ ) x ⊕ λ y ) ≤ (1 − λ ) f ( x ) + λ f ( y ) , for each x , y ∈ X and λ ∈ [0 , . (iv) f is strongly convex with parameter κ > f ((1 − λ ) x ⊕ λ y ) ≤ (1 − λ ) f ( x ) + λ f ( y ) − κλ (1 − λ ) d ( x , y ) , for each x , y ∈ X and λ ∈ [0 , . (v) When f is proper, an element x ∈ X is said to be a minimizer of f , if f ( x ) = inf z ∈ X f ( z ).The set of all proper, l.s.c. and convex extended real-valued functions on X is denoted by Γ ( X ). Definition 1.8.
Let ( X , d ) be an Hadamard space with linear dual space X ♦ and f : X → ] − ∞ , ∞ ] be afunction. Define the mapping I f : X × X ♦ × X ♦ → [ −∞ , + ∞ ] by I f ( x , x ♦ , y ♦ ) = inf y ∈ X n f ( y ) + π y ( x , x ♦ + y ♦ ) o . Remark 1.9.
For each x ∈ X and each x ♦ , y ♦ , u ♦ , v ♦ ∈ X ♦ , we have:(i) I f ( x , x ♦ , y ♦ ) = I f ( x , y ♦ , x ♦ ) .(ii) I f ( x , x ♦ , y ♦ ) = I f ( x , u ♦ , v ♦ ) , provided that x ♦ + y ♦ = u ♦ + v ♦ .(iii) I f ( x , x ♦ , y ♦ ) = I f ( x , X ♦ , x ♦ + y ♦ ) . Definition 1.10.
Let f : X → ] − ∞ , ∞ ] be a function where X is an Hadamard space with linear dual space X ♦ . For any y ♦ ∈ X ♦ , set M fy ♦ : = n ( x , x ♦ ) ∈ X × X ♦ : I f ( x , x ♦ , y ♦ ) ≥ f ( x ) o . We use the notation M f : = M f X ♦ . Lemma 1.11.
Let X be an Hadamard space with linear dual space X ♦ , p ∈ X and y ♦ ∈ X ♦ . ThenM fy ♦ = n ( x , x ♦ − y ♦ ) ∈ X × X ♦ : ( x ♦ , y ♦ ) ∈ M f o = M ˜ f , where, ˜ f ( · ) : = f ( · ) − π p ( · , y ♦ ) . roof. By using Remark 1.9(i)&(iii), we obtain:( x , x ♦ − y ♦ ) ∈ M fy ♦ ⇔ I f ( x , x ♦ − y ♦ , y ♦ ) ≥ f ( x ) ⇔ I f ( x , x ♦ , X ♦ ) ≥ f ( x ) ⇔ ( x , x ♦ ) ∈ M f . On the other hand, for each z ∈ X and ( x , x ♦ ) ∈ X × X ♦ ,˜ f ( z ) + π z ( x , x ♦ ) − ˜ f ( x ) = f ( z ) − h y ♦ , −→ pz i + π z ( x , x ♦ ) − f ( x ) + h y ♦ , −→ px i = f ( z ) + h y ♦ , −→ zx i + π z ( x , x ♦ ) − f ( x ) = f ( z ) + π z ( x , x ♦ + y ♦ ) − f ( x ) . Now, by taking the infimum over z ∈ X , we obtain I ˜ f ( x , x ♦ − y ♦ , y ♦ ) − ˜ f ( x ) = I f ( x , x ♦ , y ♦ ) − f ( x ). This impliesthat M fy ♦ = M ˜ f .
2. Flat Hadamard spaces
Let X be an Hadamard space with linear dual space X ♦ and M ⊆ X × X ♦ . The domain and range of M aredefined, respectively, by Dom( M ) : = { x ∈ X : ∃ x ♦ ∈ X ♦ s.t. ( x , x ♦ ) ∈ M } , and Range( M ) : = { x ♦ ∈ X ♦ : ∃ x ∈ X s.t. ( x , x ♦ ) ∈ M } . Definition 2.1.
Let X be an Hadamard space with linear dual space X ♦ and p ∈ X be fixed. We say that M ⊆ X × X ♦ satisfies F l -property if for each λ ∈ [0 , , x ♦ ∈ Range( M ) and each x , y ∈ Dom( M ), D x ♦ , −−−−−−−−−−−−−−→ p ((1 − λ ) x ⊕ λ y ) E ≤ (1 − λ ) h x ♦ , −→ px i + λ h x ♦ , −→ py i . (5) Remark 2.2.
Let M ⊆ X × X ♦ satisfies in F l -property for some p ∈ X. Let q ∈ X be arbitrary. Then for each λ ∈ [0 , , x ♦ ∈ Range( M ) and each x , y ∈ Dom( M ) , D x ♦ , −−−−−−−−−−−−−−→ q ((1 − λ ) x ⊕ λ y ) E = D x ♦ , −→ qp + −−−−−−−−−−−−−−→ p ((1 − λ ) x ⊕ λ y ) E = h x ♦ , −→ qp i + D x ♦ , −−−−−−−−−−−−−−→ p ((1 − λ ) x ⊕ λ y ) E ≤ (1 − λ )( h x ♦ , −→ qp i + h x ♦ , −→ px i ) + λ ( h x ♦ , −→ qp i + h x ♦ , −→ py i ) = (1 − λ ) h x ♦ , −→ qp + −→ px i + λ h x ♦ , −→ qp + −→ py i = (1 − λ ) h x ♦ , −→ qx i + λ h x ♦ , −→ qy i . Thus F l -property is independent of the choice of the point p. Lemma 2.3. [14, Theorem 3.2]
Let X be an Hadamard space. Then X is flat if and only if for each x , y ∈ X and λ ∈ [0 , , h−−−−−−−−−−−−−−→ x ((1 − λ ) x ⊕ λ y ) , −→ ab i = λ h−→ xy , −→ ab i , a , b ∈ X . (6) Proposition 2.4.
An Hadamard space X is flat if and only if X × X ♦ has F l -property. Proof.
Assume that X is flat. Let x , y , p ∈ X , λ ∈ [0 ,
1] and ( x , x ♦ ) ∈ X × X ♦ . Then x ♦ = P ni = α i [ t i −−→ a i b i ] ∈ X ♦ ,5here n ∈ N , 1 ≤ i ≤ n , α i , t i ∈ R and a i , b i ∈ X . Hence, by using Lemma 2.3 we get: D x ♦ , −−−−−−−−−−−−−−→ p (cid:16) (1 − λ ) x ⊕ λ y (cid:17)E = D x ♦ , −→ px E + D x ♦ , −−−−−−−−−−−−−−→ x (cid:16) (1 − λ ) x ⊕ λ y (cid:17)E = D x ♦ , −→ px E + n X i = α i t i D −−→ a i b i , −−−−−−−−−−−−−−→ x (cid:16) (1 − λ ) x ⊕ λ y (cid:17)E = D x ♦ , −→ px E + λ n X i = α i t i D −−→ a i b i , −→ xy E = D x ♦ , −→ px E + λ h x ♦ , −→ xy i = D x ♦ , −→ px E + λ (cid:16) h x ♦ , −→ py i − h x ♦ , −→ px i (cid:17) = (1 − λ ) h x ♦ , −→ px i + λ h x ♦ , −→ py i . Therefore X × X ♦ has F l -property. For the converse, it is su ffi cient to consider p : = x and x ♦ : = ± [ −→ ab ] in(5).The next example shows that there exists a relation M ⊆ X × X ♦ , in a non-flat Hadamard space X , whichdoesn’t have the F l -property. Moreover, it is easy to check that in any Hadamard space ( X , d ) with lineardual space X ♦ , for each ( x , x ♦ ) ∈ X × X ♦ , the singleton set { ( x , x ♦ ) } has F l -property. Example 2.5.
Consider the following equivalence relation on N × [0 , n , t ) ∼ ( m , s ) : ⇔ t = s = n , t ) = ( m , s ) . Set X : = N × [0 , ∼ and let d : X × X → R be defined by d ([( n , t )] , [( m , s )]) = | t − s | n = m , t + s n , m . The geodesic joining x = [( n , t )] to y = [( m , s )] is defined as follows:(1 − λ ) x ⊕ λ y : = [( n , (1 − λ ) t − λ s )] 0 ≤ λ ≤ tt + s , [( m , ( λ − t + λ s )] tt + s ≤ λ ≤ , whenever, x , y and vacuously (1 − λ ) x ⊕ λ x : = x . It is known that [1, Example 4.7] ( X , d ) is an R -tree space.It follows from [3, Example 1.2.10], that R -tree spaces are Hadamard space. Let x = [(2 , )], y = [(1 , )], a = [(3 , )] and b = [(2 , )]. Then(1 − λ )[(2 , )] ⊕ λ [(1 , )] = [(2 , − λ )] 0 ≤ λ ≤ , [(1 , λ − )] ≤ λ ≤ . For each λ ∈ (0 , ] we obtain, D −−−−−−−−−−−−−−→ x ((1 − λ ) x ⊕ λ y ) , −→ ab E = (cid:28) −−−−−−−−−−−−−−−−→ [(2 ,
12 )][(2 , − λ )] , −−−−−−−−−−−−→ [(3 ,
13 )][(2 ,
12 )] (cid:29) = − λ, while λ h−→ xy , −→ ab i = − λ . Now, Lemma 2.3 implies that ( X , d ) is not a flat Hadamard space. For each n ∈ N ,set x n : = [( n , )] and y n : = [( n , n )]. Now, we define M : = n ( x n , [ −−−−−→ y n + y n ]) : n ∈ N o ⊆ X × X ♦ . Take λ = , p = [(1 , ∈ X and [ −−−→ y y ] ∈ Range( M ). Clearly, (1 − λ ) x ⊕ λ x = [(1 , )] and D [ −−−→ y y ] , −−−−−−−−−−−−−−−→ p ((1 − λ ) x ⊕ λ x ) E = , h [ −−−→ y y ] , −−→ px i + h [ −−−→ y y ] , −−→ px i = . Therefore, M doesn’t have the F l -property. Lemma 2.6. [3, Proposition 2.2.17]
Let ( X , d ) be an Hadamard space and f : X → ] − ∞ , ∞ ] be l.s.c. and stronglyconvex with κ > . Then there exists a unique minimizer x ∈ X of f and each minimizing sequence converges to x.Moreover, f ( x ) + κ d ( x , y ) ≤ f ( y ) , for each y ∈ X . Lemma 2.7.
Let X be an Hadamard space with linear dual space X ♦ , p ∈ X and y ♦ ∈ X ♦ . Then X × { y ♦ } has F l -property if and only if π p ( · , y ♦ ) is convex.Proof. Let a , b ∈ X and λ ∈ [0 , π p ((1 − λ ) a ⊕ λ b , y ♦ ) = h y ♦ , −−−−−−−−−−−−−→ p ((1 − λ ) a ⊕ λ b ) i , (7)and (1 − λ ) π p ( a , y ♦ ) + λπ p ( b , y ♦ ) = (1 − λ ) h y ♦ , −→ pa i + λ h y ♦ , −→ pb i . (8)Now, inequalities (7), (8) and (5) imply that X × { y ♦ } has F l -property if and only if π p ( · , y ♦ ) is convex. Proposition 2.8.
Let ( X , d ) be an Hadamard space with linear dual space X ♦ and f ∈ Γ ( X ) . Let p , y ∈ X be fixed andarbitrary and let y ♦ ∈ X ♦ be such that X × { y ♦ } has F l -property. Then(i) The mapping : X → ] − ∞ , ∞ ] defined by ( x ) = f ( x ) + π p ( x , y ♦ ) is proper, l.s.c. and convex.(ii) Define the mapping h : X → ] − ∞ , ∞ ] by h ( x ) = ( x ) + d ( x , y ) . Then h is proper, l.s.c. and strongly convexwith the parameter κ = . Moreover, h has a unique minimizer x ∈ X such that h ( z ) ≥ h ( x ) + d ( x , z ) , foreach z ∈ X.Proof . (i): It is clear that is proper. Lower semi-continuity of follows from lower semicontinuity of f andProposition 1.6(i). It follows from Lemma 2.7 and convexity of f that is convex.(ii): Similarly, h is proper and lower semi-continuous. Let a , b ∈ X and λ ∈ [0 , and CN-inequality (1), we get: h ((1 − λ ) a ⊕ λ b ) = ((1 − λ ) a ⊕ λ b ) + d ((1 − λ ) a ⊕ λ b , y ) ≤ (1 − λ ) ( a ) + λ ( b ) + (cid:16) (1 − λ ) d ( a , y ) + λ d ( b , y ) − λ (1 − λ ) d ( a , b ) (cid:17) = (1 − λ )( ( a ) + d ( a , y ) ) + λ ( ( b ) + d ( b , y ) ) − λ (1 − λ ) d ( a , b ) = (1 − λ ) h ( a ) + λ h ( b ) − λ (1 − λ ) d ( a , b ) . Thus h is strongly convex with the parameter κ = . In view of Lemma 2.6, there exists a uniqueminimizer x ∈ X such that h ( z ) ≥ h ( x ) + d ( x , z ) , for each z ∈ X . This completes the proof. Corollary 2.9.
Let ( X , d ) be a flat Hadamard space and f : X → ] − ∞ , ∞ ] be proper, l.s.c. and convex. Let p , y ∈ Xbe fixed and arbitrary and y ♦ ∈ X ♦ . Then(i) The mapping , defined in Proposition 2.8(i), is proper, l.s.c. and convex.(ii) Define the mapping h as in Proposition 2.8(ii). Then h is proper, l.s.c. and strongly convex with the parameter κ = . Moreover, h has a unique minimizer x ∈ X such that h ( z ) ≥ h ( x ) + d ( x , z ) , for each z ∈ X.Proof.
Since X is flat, Proposition 2.4 implies that X × X ♦ has F l -property and hence for each y ♦ ∈ X ♦ , X × { y ♦ } as a subset of X × X ♦ has this property too. Now, Proposition 2.8 completes the proof.7 . Monotonicity from polarity point of view The concept of monotone operators in Hadamard spaces is introduced in [2]. Moreover, some propertiesof monotone operators, their resolvents and proximal point algorithm are discussed in [8, 10, 15]. Thenotions of monotone sets and maximal monotone sets in Hadamard spaces are introduced in [13]. In thissection, fundamental properties of (maximal) monotone sets in Hadamard spaces from polarity point ofview are considered. Also, some important results of [12] are proved in Hadamard spaces.
Definition 3.1.
Let X be an Hadamard space with linear dual space X ♦ . We say that ( x , x ♦ ) ∈ X × X ♦ and( y , y ♦ ) ∈ X × X ♦ are monotonically related , if h x ♦ − y ♦ , −→ yx i ≥ x , x ♦ ) µ ( y , y ♦ ). It is easy to seethat µ is a reflexive and symmetric relation on X × X ♦ . This motivates that monotonically relatedness canbe define for a subset M ⊆ X × X ♦ . An element ( x , x ♦ ) is monotonically related to M if ( x , x ♦ ) µ ( y , y ♦ ), for each( y , y ♦ ) ∈ M and this will be denoted by ( x , x ♦ ) µ M . Moreover, M ⊆ X × X ♦ is said to be a monotone set if every( x , x ♦ ) , ( y , y ♦ ) ∈ M are monotonically related. The monotone polar of M is M µ : = { ( x , x ♦ ) ∈ X × X ♦ : ( x , x ♦ ) µ M } . In addition, we often use the notations M µµ : = ( M µ ) µ and M µµµ : = ( M µµ ) µ . Example 3.2.
Let ( X , d ) be an Hadamard space with linear dual space X ♦ and f : X → ] − ∞ , ∞ ] be an arbitraryfunction. Let ( u , u ♦ ) ∈ M f and ( v , v ♦ ) ∈ M f . According to definition of M f , we get I f ( u , u ♦ , X ♦ ) ≥ f ( u ) and I f ( v , v ♦ , X ♦ ) ≥ f ( v ) . Hence, for each y ∈ X, f ( y ) + π y ( u , u ♦ ) ≥ f ( u ) , (9) and f ( y ) + π y ( v , v ♦ ) ≥ f ( v ) . (10) Now, put y : = v and y : = u in (9) and (10) , respectively, to obtain: π v ( u , u ♦ ) ≥ f ( u ) − f ( v ) , (11) and π u ( v , v ♦ ) ≥ f ( v ) − f ( u ) . (12) Adding inequalities (11) and (12) and since π u ( v , v ♦ ) = − π v ( u , v ♦ ) , we get: π v ( u , u ♦ ) − π v ( u , v ♦ ) ≥ , this means that h u ♦ − v ♦ , −→ vu i ≥ , that is M f is monotone. Finally, Lemma 1.11 implies that for each y ♦ ∈ X ♦ , M fy ♦ isa monotone set. Example 3.3.
Let X be the same as in Example 2.5. For each n ∈ N , set M : = n(cid:16) x n , h −−−−−→ x n + y n i(cid:17) : n ∈ N o , wherex n = [( n , and y n = [( n , . Then M is monotone. Indeed, for each n , m ∈ N , Dh −−−−−→ x n + y n i − h −−−−−→ x m + y m i , −−−→ x m x n E = , n ∈ { m − , m + } , , o . w . Definition 3.4.
Let X be an Hadamard space. A monotone set M ⊆ X × X ♦ is called maximal if there is nomonotone set L ⊆ X × X ♦ that properly contains M . Example 3.5. (i) Let X and M be the same as in Example 3.3. It is shown that M is monotone. We claim that itis not maximal monotone. To see this, let (cid:16) x = [( n , , x ♦ = −−−−−−−−−−−−−−−→ [( n + , n , (cid:17) ∈ M be arbitrary, z = [(1 , nd z ♦ = h −−−−−−−−−−−→ [(1 , )][(1 , i . Therefore, ( z , z ♦ ) < M, but h z ♦ − x ♦ , −→ xz i = D −−−−−−−−−−−→ [(1 , )][(1 , , −−−−−−−−−−−→ [( n , , E − D −−−−−−−−−−−−−−−→ [( n + , n , , −−−−−−−−−−−→ [( n , , E = , n = , , n , . This implies that M is not maximal.(ii) Let ( X , d ) be a flat Hadamard space with linear dual space X ♦ , y ♦ ∈ X ♦ and and f ∈ Γ ( X ) . It follows fromExample 3.2 that M f is monotone. Let ( y , y ♦ ) ∈ X × X ♦ \ M f . By using Lemma 1.11, we get: ( y , X ♦ ) < M , (13) where ( z ) = f ( z ) + π p ( z , − y ♦ ) , z ∈ X. Now, consider the mapping h ( z ) = ( z ) + π z ( x , [ −→ xy ]) , z ∈ X.It follows from Corollay 2.9(ii) that h has unique minimizer x ∈ X. Hence, for each z ∈ X we obtainh ( z ) + π z ( x , X ♦ ) = h ( z ) ≥ h ( x ) , therefore, I h ( x , [ −→ xy ] , [ −→ yx ]) ≥ h ( x ) . Consequently, by using Lemma 1.11, weconclude that ( x , [ −→ xy ]) ∈ M h [ −→ yx ] = M ˜ h = M , where ˜ h ( · ) = h ( · ) + π x ( · , [ −→ xy ]) . Thus ( x , [ −→ xy ]) ∈ M . (14) It follows from (13) that x , y. On the other hand, we derive from Lemma 1.11 and (14) that ( x , [ −→ xy ] + y ♦ ) ∈ M f .Moreover, h [ −→ xy ] + y ♦ − y ♦ , −→ yx i = h−→ xy , −→ yx i = − d ( x , y ) < , which implies that ( y , y ♦ ) is not monotonically related to M f . Hence, M f ⊆ X × X ♦ is a maximal monotone set.Finally, it follows from Lemma 1.11 and Corollary 2.9(i) that M fy ♦ is a maximal monotone set. The following well-known fact states that any monotone set in Hadamard spaces can be extended to amaximal monotone set. The proof is similar to that proof of [4, Theorem 20.21]. For the sake of completeness,we add a proof.
Proposition 3.6.
Let M ⊆ X × X ♦ be a monotone set. Then there exists a maximal monotone extension (which is notnecessarily unique) of M, i.e., a maximal monotone set e M ⊆ X × X ♦ such that M ⊆ e M.Proof.
There are two cases:
Case I: M , ∅ ; In this case, consider the following set M : = { L ⊆ X × X ♦ : L is a monotone set and M ⊆ L } . It is clear that ( M , (cid:22) ) is a partially ordered set, where for every L , L ∈ M , L (cid:22) L : ⇔ L ⊆ L . Let A be a chain in M . One can see that ∪ A ∈ A A is an upper bound of A . Now, by using the Zorn’s lemma,there exists a maximal element e M ∈ M . Case II: M = ∅ ; In this case, let p ∈ X be fixed. By Case I, { ( p , X ♦ ) } has a maximal monotone extension, say e M . Obviously, e M is a maximal monotone extension of M .Then in any cases M ⊆ X × X ♦ has a maximal monotone extension. Remark 3.7.
Let X be an Hadamard space and M ⊆ X × X ♦ be a monotone set. In view of Proposition 3.6, thereexists e M ⊆ X × X ♦ as a maximal monotone extension of M. Set e M ( M ) : = n e M ⊆ X × X ♦ : e M is a maximal monotone extension of M o . he set of all maximal monotone subsets of X × X ♦ is denoted by MS ( X ) ; i.e., MS ( X ) = e M ( X × X ♦ ) . Proposition 3.8.
Let M ⊆ X × X ♦ be a monotone set. Then ( x , x ♦ ) ∈ M µ if and only if M ∪ { ( x , x ♦ ) } is monotone.Proof. It is straightforward.
Definition 3.9. [11, Definition 2.1] Let µ be a relation from A to B . Define two functions σ : P ( A ) → P ( B )and τ : P ( B ) → P ( A ) as follows: σ ( U ) = { b ∈ B : u µ b , ∀ u ∈ U } ,τ ( V ) = { a ∈ A : a µ v , ∀ v ∈ V } . Then ( σ, P ( A ) , µ, P ( B ) , τ ) or simply ( σ, µ, τ ) is called a polarity . Lemma 3.10. [11, Proposition 2.4]
Let ( σ, P ( A ) , µ, P ( B ) , τ ) be a polarity. Then(i) P ( A ) and P ( B ) are partially ordered sets, ordered by set inclusion, and σ and τ are order reversing functions.(ii) στ and τσ are order increasing; i.e.,U ⊆ τσ ( U ) and V ⊆ στ ( V ) , ∀ U ⊆ A , ∀ V ⊆ B . (iii) τ is a quasi-inverse for σ and σ is a quasi-inverse for τ ; i.e., στσ = σ and τστ = τ . Definition 3.11. [7, Definition 5.1] A function σ : P ( A ) → P ( A ) is a closure operator on A , if σ has thefollowing properties:(i) σ ( U ) ⊆ σ ( V ), ∀ U , V ⊆ A with U ⊆ V .(ii) σ ( σ ( U )) = σ ( U ), ∀ U ⊆ A .(iii) U ⊆ σ ( U ), ∀ U ⊆ A . Proposition 3.12.
Let X be an Hadamard space and M ⊆ X × X ♦ . Then(i) M ⊆ M µµ .(ii) M µµµ = M µ .(iii) M ⊆ M ⇒ M µ ⊆ M µ , ∀ M , M ⊆ X × X ♦ .Proof. Consider the mapping ζ : P ( X × X ♦ ) → P ( X × X ♦ ) M M µ . It follows from Definition 3.9 that ( ζ, µ, ζ ), is a polarity. Consequently, the items (i), (ii) and (iii) follow fromLemma 3.10.
Proposition 3.13.
Let ( X , d ) be an Hadamard space and { M i } i ∈ I ⊆ P ( X × X ♦ ) be a family of monotone sets. Then(i) ( S i ∈ I M i ) µ = T i ∈ I M µ i . (ii) ∅ µ = X × X ♦ .(iii) ( X × X ♦ ) µ = ∅ , provided that card( X ) > .Proof. (i) By using Definition 3.1, for each ( x , x ♦ ),( x , x ♦ ) ∈ (cid:16) [ i ∈ I M i (cid:17) µ ⇔ ( x , x ♦ ) µ (cid:16) [ i ∈ I M i (cid:17) ⇔ ( x , x ♦ ) µ M i , ∀ i ∈ I ⇔ ( x , x ♦ ) ∈ M µ i , ∀ i ∈ I ⇔ ( x , x ♦ ) ∈ \ i ∈ I M µ i . ∅ µ ⊆ X × X ♦ . Conversely, assume to the contrary that there exists ( x , x ♦ ) ∈ X × X ♦ \ ∅ µ . Hence,there is ( y , y ♦ ) ∈ ∅ such that ( x , x ♦ ) and ( y , y ♦ ) are not monotonically related, yields a contradiction.(iii) It is enough to prove that ( X × X ♦ ) µ ⊆ ∅ . Let ( x , x ♦ ) ∈ ( X × X ♦ ) µ . It follows from Proposition 3.8 that( X × X ♦ ) ∪ { ( x , x ♦ ) } = X × X ♦ is a monotone set. On the other hand, for every a , b ∈ X with a , b , put a ♦ = [ −→ ab ] and b ♦ = [ −→ bb ]. Hence, h a ♦ − b ♦ , −→ ba i = h−→ ab , −→ ba i = − d ( a , b ) < , a contradiction. Definition 3.14.
For each M ⊆ X × X ♦ , the closure operator induced by the polarity µ is defined by themapping M M µµ . Moreover, we say that M µµ is the µ -closure of M and M is µ -closed if M µµ = M . Remark 3.15.
It follows from Proposition 3.12(ii) that the family of all µ -closed sets is the family of polars { A µ : A ⊆ X × X ♦ } . One can see that M µµ is the smallest µ -closed set containing M . Proposition 3.16.
Let X be an Hadamard space and M ⊆ X × X ♦ be monotone. Then the following statements areequivalent:(i) M is monotone.(ii) M ⊆ M µ .(iii) M µµ ⊆ M µ .(iv) M µµ is monotone.In addition, M ∈ MS ( X ) if and only if M = M µ . Moreover, every element of MS ( X ) is µ -closed.Proo f . (i) ⇔ (ii): Clearly, M is monotone if and only if every two members of M are monotonically related,or equivalently M ⊆ M µ . Hence (i) and (ii) are equivalent.(ii) ⇒ (iii): An immediate consequence of Proposition 3.12(iii).(iii) ⇒ (iv): By assumption and Proposition 3.12(iii), we obtain: M µµ = ( M µ ) µ ⊆ ( M µµ ) µ . Now, equivalence of (i) and (ii), implies that M µµ is monotone.(iv) ⇒ (ii): Let M µµ be a monotone set. By using Proposition 3.12(i), applying (i) ⇒ (ii) to M µµ and Proposition3.12(ii) we get: M ⊆ M µµ ⊆ M µµµ = M µ . Therefore, M is monotone by (ii) ⇒ (i).We know that M is monotone if and only if M ⊆ M µ . Moreover, maximality of M is equivalent to M µ ⊆ M . Therefore, M ∈ MS ( X ) if and only if M = M µ . Finally, by using this fact, for each M ∈ MS ( X )we have: M = M µ = ( M µ ) µ = M µµ . Now, it follows from Definition 3.14 that M is a µ -closed set. Proposition 3.17.
Let X be an Hadamard space and M ⊆ X × X ♦ be a monotone set. Then the following hold:(i) M µ = S e M ∈ M ( M ) e M.(ii) M µµ = T e M ∈ M ( M ) e M. roof . (i): Let ( x , x ♦ ) ∈ M µ . It follows from Proposition 3.8 that M ∪ { ( x , x ♦ ) } is monotone. Now, by usingProposition 3.6, there exists maximal monotone extension e M for M ∪ { ( x , x ♦ ) } . Hence, e M ∈ M ( M )and ( x , x ♦ ) ∈ e M . Therefore, M µ ⊆ S e M ∈ M ( M ) e M . Conversely, let ( x , x ♦ ) ∈ S e M ∈ M ( M ) e M . Then thereexists e M ∈ M ( M ) such that ( x , x ♦ ) ∈ e M . By using Proposition 3.16 and Proposition 3.12(iii), we get( x , x ♦ ) ∈ e M = (cid:16) e M (cid:17) µ ⊆ M µ . Consequently, S e M ∈ M ( M ) e M ⊆ M µ .(ii): By using (i), Proposition 3.13(i) and Proposition 3.6, we obtain: M µµ = ( M µ ) µ = (cid:18) [ e M ∈ M ( M ) e M (cid:19) µ = \ e M ∈ M ( M ) (cid:16) e M (cid:17) µ = \ e M ∈ M ( M ) e M . We are done.
Lemma 3.18.
Let X be an Hadamard space and M ⊆ X × X ♦ . Then M is monotone if and only if there exists e M ∈ MS ( X ) such that e M ⊆ M µ .Proof. Let M be a monotone set. It follows from Proposition 3.17(i) that M µ contains a maximal monotoneset. Conversely, let there exists e M ∈ MS ( X ) such that e M ⊆ M µ . Now, by using Proposition 3.12(i)&(iii) andProposition 3.16, we conclude that: M ⊆ ( M µ ) µ ⊆ (cid:16) e M (cid:17) µ = e M ⊆ M µ . The claim therefore follows from Proposition 3.16((i) ⇔ (ii)). Proposition 3.19.
Let X be an Hadamard space and M ⊆ X × X ♦ be a maximal monotone set. Then(i) For each u ∈ X, M ♦ u : = { u ♦ ∈ X ♦ : ( u , u ♦ ) ∈ M } is a closed and convex subset of X ♦ .(ii) M is sequentially bw × k · k ♦ -closed in X × X ♦ , (and hence d × k · k ♦ -closed).(iii) If Dom( M ) ⊆ X is bounded, then M is sequentially weakly ×k · k ♦ -closed in X × X ♦ .Proof. (i): Let u ♦ ∈ M ♦ u . Then there exists { u ♦ n } ⊆ M ♦ u such that u ♦ n → u ♦ . For each ( x , x ♦ ) ∈ M , by monotonicityof M , we have h u ♦ n − x ♦ , −→ xu i ≥
0. Accordance to Proposition 1.6(i), by taking limit as n → ∞ , we get h u ♦ − x ♦ , −→ xu i ≥
0. Therefore ( u , u ♦ ) ∈ M µ . Now, maximality of M implies that ( u , u ♦ ) ∈ M or u ♦ ∈ M ♦ u .Consequently, M ♦ u is closed. For proving convexity of M ♦ u , let u ♦ , v ♦ ∈ M ♦ u and λ ∈ [0 ,
1] be arbitraryand fixed. Then for each ( x , x ♦ ) ∈ M , by monotonicity of M , we have: h (1 − λ ) u ♦ + λ v ♦ − x ♦ , −→ xu i = (1 − λ ) h u ♦ − x ♦ , −→ xu i + λ h v ♦ − x ♦ , −→ xu i ≥ . Consequently, ( u , λ u ♦ + (1 − λ ) v ♦ ) ∈ M µ . Again, maximality of M implies that ( u , λ u ♦ + (1 − λ ) v ♦ ) ∈ M ;i.e., λ u ♦ + (1 − λ ) v ♦ ∈ M ♦ u and so M ♦ u is convex.(ii) Let { ( x n , x ♦ n ) } ⊆ M be a sequence such that ( x n , x ♦ n ) bw ×k·k ♦ −−−−−→ ( x , x ♦ ). For each ( y , y ♦ ) ∈ M , h x ♦ n − y ♦ , −−→ yx n i ≥ . It follows from Proposition 1.3 that, h x ♦ − y ♦ , −→ yx i = lim n → + ∞ h x ♦ n − y ♦ , −−→ yx n i ≥ . This implies that ( x , x ♦ ) ∈ M µ = M . Thus M is sequentially bw ×k · k ♦ -closed in X × X ♦ .(iii): It is an immediate consequence of (ii). 12 eferences [1] B.Ahmadi Kakavandi, Weak topologies in complete
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