# On Mixed Equilibrium Problems in Hadamard Spaces

Authors: Chinedu Izuchukwu; Kazeem Olalekan Aremu; Olawale Kazeem Oyewole; Oluwatosin Temitope Mewomo; Safeer Hussain Khan
Journal: Journal of Mathematics
 (2019)

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## Abstract

The main purpose of this paper is to study mixed equilibrium problems in Hadamard spaces. First, we establish the existence of solution of the mixed equilibrium problem and the unique existence of the resolvent operator for the problem. We then prove a strong convergence of the resolvent and aΔ-convergence of the proximal point algorithm to a solution of the mixed equilibrium problem under some suitable conditions. Furthermore, we study the asymptotic behavior of the sequence generated by a Halpern-type PPA. Finally, we give a numerical example in a nonlinear space setting to illustrate the applicability of our results. Our results extend and unify some related results in the literature.


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## Body

## 1. Introduction

LetC be a nonempty set and Ψ be any real-valued function defined on C. The minimization problem (MP) is defined as(1)findx∗∈Csuch thatΨx∗≤Ψy,∀y∈C.In this case,x∗ is called a minimizer of Ψ and argminy∈CΨy denotes the set of minimizers of Ψ. MPs are very useful in optimization theory and convex and nonlinear analysis. One of the most popular and effective methods for solving MPs is the proximal point algorithm (PPA) which was introduced in Hilbert space by Martinet [1] in 1970 and was further extensively studied in the same space by Rockafellar [2] in 1976. The PPA and its generalizations have also been studied extensively for solving MP (1) and related optimization problems in Banach spaces and Hadamard manifolds (see [3–7] and the references therein), as well as in Hadamard and p-uniformly convex metric spaces (see [8–13] and the references therein).An important generalization of Problem (1) is the following equilibrium problem (EP), defined as(2)findx∗∈Csuch thatFx∗,y≥0,∀y∈C.The pointx∗ for which (2) is satisfied is called an equilibrium point of F. The solution set of problem (2) is denoted by EPC,F. The EP is one of the most important problems in optimization theory that has received a lot of attention in recent time since it includes many other optimization and mathematical problems as special cases, namely, MPs, variational inequality problems, complementarity problems, fixed point problems, and convex feasibility problems, among others (see, for example, [5, 14–18]). Thus, EPs are of central importance in optimization theory as well as in nonlinear and convex analysis. As a result of this, numerous authors have studied EPs in Hilbert, Banach, and topological vector spaces (see [19, 20] and the references therein), as well as in Hadamard manifolds (see [3, 21]).Very recently, Kumam and Chaipunya [5] extended these studies to Hadamard spaces. First, they established the existence of an equilibrium point of a bifunction satisfying some convexity, continuity, and coercivity assumptions, and they also established some fundamental properties of the resolvent of the bifunction. Furthermore, they proved that the PPA Δ-converges to an equilibrium point of a monotone bifunction in a Hadamard space. More precisely, they proved the following theorem.Theorem 1.
LetC be a nonempty closed and convex subset of an Hadamard space X and F:C×C⟶ℝ be monotone and Δ-upper semicontinuous in the first variable such that DJλF⊃C for all λ>0 (where DJλF means the domain of JλF). Suppose that EPC,F≠∅ and for an initial guess x0∈C, the sequence xn⊂C is generated by(3)xn≔JλnFxn−1,n∈ℕ,where λn is a sequence of positive real numbers bounded away from 0. Then, xnΔ-converges to an element of EPC,F.
Other authors have also studied EPs in Hadamard spaces (see, for example, [14, 15]).
In the linear settings (for example, in Hilbert spaces), EPs have been generalized into what is called the mixed equilibrium problem (MEP), defined as(4)findx∗∈Csuch thatFx∗,y+Ψy−Ψx∗≥0,∀y∈C.
The MEP is an important class of optimization problems since it contains many other optimization problems as special cases. For instance, ifF≡0 in (3), then the MEP (4) reduces to MP (1). Also, if Ψ≡0 in (3), then the MEP (4) reduces to the EP (2). The existence of solutions of the MEP (4) was established in Hilbert spaces by Peng and Yao [22] (see also [23]). More so, different iterative algorithms have been developed by numerous authors for approximating solutions of MEP (4) in real Hilbert spaces (see, for example, [22–24] and the references therein).
Since MEPs contain both MPs and EPs as special cases in Hilbert spaces, it is important to extend their study to Hadamard spaces, so as to unify other optimization problems (in particular, MPs and EPs) in Hadamard spaces. Moreover, Hadamard spaces are more suitable frameworks for the study of optimization problems and other related mathematical problems since many recent results in these spaces have already found applications in diverse fields than they do in Hilbert spaces. For instance, the minimizers of the energy functional (which is an example of a convex and lower semicontinuous functional in a Hadamard space), called harmonic maps, are very useful in geometry and analysis (see [9]). Also, the gradient flow theorem in Hadamard spaces was employed to investigate the asymptotic behavior of the Calabi flow in Kahler geometry (see [25]). Furthermore, the study of the PPA for optimization problems has successfully been applied in Hadamard spaces, for computing medians and means, which are very important in computational phylogenetics, diffusion tensor imaging, consensus algorithms, and modeling of airway systems in human lungs and blood vessels (see [26, 27], for details). It is also worthy to note that many nonconvex problems in the linear settings can be viewed as convex problems in Hadamard spaces (see Section 4 of this paper).
Therefore, it is our interest in this paper to extend the study of the MEP (4) to Hadamard spaces. First, we establish the existence of solution of the MEP (4) and the unique existence of the resolvent operator associated withF and Ψ. We then prove a strong convergence of the resolvent and a Δ-convergence of the PPA to a solution of MEP (4) under some suitable conditions on F and Ψ. Furthermore, we study the asymptotic behavior of the sequence generated by the Halpern-type PPA. Finally, we give a numerical example in a nonlinear space setting to illustrate the applicability of our results. Our results extend and unify the results of Kumam and Chaipunya [5] and Peng and Yao [22].
The rest of this paper is organized as follows: In Section2, we recall the geometry of geodesic spaces and some useful definitions and lemmas. In Section 3, we establish the existence of solution for MEP (4) and the unique existence of the resolvent operator associated with F and Ψ. Some fundamental properties of the resolvent operator are also established in this section. In Section 4, we prove a strong convergence of the resolvent and a Δ-convergence of the PPA to a solution of MEP (4) under some suitable conditions on F and Ψ. In Section 5, we study the asymptotic behavior of the sequence generated by the Halpern-type PPA. In Section 6, we generate some numerical results in nonlinear setting for the PPA and the Halpern-type PPA, to show the applicability of our results.


## 2. Preliminaries

### 2.1. Geometry of Geodesic Spaces

Definition 1.
LetX,d be a metric space, x,y∈X and I=0,dx,y be an interval. A curve c (or simply a geodesic path) joining x to y is an isometry c:I⟶X such that c0=x, cdx,y=y, and dct,ct′=t−t′ for all t,t′∈I. The image of a geodesic path is called a geodesic segment, which is denoted by x,y whenever it is unique.Definition 2 (see [28]).
A metric spaceX,d is called a geodesic space if every two points of X are joined by a geodesic path, and X is said to be uniquely geodesic if every two points of X are joined by exactly one geodesic path. A subset C of X is said to be convex if C includes every geodesic segments joining two of its points. Let x,y∈X and t∈0,1, and we write tx⊕1−ty for the unique point z in the geodesic segment joining from x to y such that(5)dx,z=1−tdx,yanddz,y=tdx,y.
A geodesic triangleΔx1,x2,x3 in a geodesic metric space X,d consists of three vertices (points in X) with unparameterized geodesic segment between each pair of vertices. For any geodesic triangle, there is comparison (Alexandrov) triangle Δ¯⊂ℝ2 such that dxi,xj=dℝ2x¯i,x¯j for i,j∈1,2,3. Let Δ be a geodesic triangle in X and Δ¯ be a comparison triangle for Δ¯, then Δ is said to satisfy the CAT(0) inequality if for all points x,y∈Δ and x¯,y¯∈Δ¯:(6)dx,y≤dℝ2x¯,y¯.
Letx,y,andz be points in X and y0 be the midpoint of the segment y,z; then, the CAT(0) inequality implies(7)d2x,y0≤12d2x,y+12d2x,z−14dy,z.
Inequality (7) is known as the CN inequality of Bruhat and Titis [29].Definition 3.
A geodesic spaceX is said to be a CAT(0) space if all geodesic triangles satisfy the CAT(0) inequality. Equivalently, X is called a CAT(0) space if and only if it satisfies the CN inequality.
CAT(0) spaces are examples of uniquely geodesic spaces, and complete CAT(0) spaces are called Hadamard spaces.Definition 4.
LetC be a nonempty closed and convex subset of a CAT(0) space X. The metric projection is a mapping PC:X⟶C which assigns to each x∈X, the unique point PCx in C such that(8)dx,PCx=infdx,y:y∈C.Definition 5 (see [30]).
LetX be a CAT(0) space. Denote the pair a,b∈X×X by ab→ and call it a vector. Then, a mapping .,.:X×X×X×X⟶ℝ defined by(9)ab→,cd→=12d2a,d+d2b,c−d2a,c−d2b,d,∀a,b,c,d∈X,is called a quasilinearization mapping.
It is easy to check thatab→,ab→=d2a,b, ba→,cd→=−ab→,cd→, ab→,cd→=ae→,cd→+eb→,cd→, and ab→,cd→=cd→,ab→ for all a,b,c,d,e∈X. A geodesic space X is said to satisfy the Cauchy–Swartz inequality if ab→,cd→≤da,bdc,d∀a,b,c,d∈X. It has been established in [30] that a geodesically connected metric space is a CAT(0) space if and only if it satisfies the Cauchy–Schwartz inequality. Examples of CAT(0) spaces include Euclidean spaces ℝn, Hilbert spaces, simply connected Riemannian manifolds of nonpositive sectional curvature [31], ℝ-trees, and Hilbert ball [32], among others.
We end this section with the following important lemmas which characterize CAT(0) spaces.Lemma 1.
LetX be a CAT(0) space, x,y,z∈X, and t,s∈0,1. Then,(i)
dtx⊕1−ty,z≤tdx,z+1−tdy,z (see [28])(ii)
d2tx⊕1−ty,z≤td2x,z+1−td2y,z−t1−td2x,y (see [28])


### 2.2. Notion ofΔ-Convergence

Definition 6.
Letxn be a bounded sequence in a geodesic metric space X. Then, the asymptotic center Axn of xn is defined by(10)Axn=v¯∈X:limsupn⟶∞dv¯,xn=infv∈Xlimsupn⟶∞dv,xn.
A sequencexn in X is said to be Δ-convergent to a point v¯∈X if Axnk=v¯ for every subsequence xnk of xn. In this case, we write Δ‐limn⟶∞xn=v¯ (see [33]). The concept of Δ-convergence in metric spaces was first introduced and studied by Lim [34]. Kirk and Panyanak [35] later introduced and studied this concept in CAT(0) spaces and proved that it is very similar to the weak convergence in Banach space setting.
We now end this section with the following important lemmas which are concerned withΔ-convergence.Lemma 2 (see [28, 36]).
LetX be an Hadamard space. Then,(i)
Every bounded sequence inX has a Δ-convergent subsequence(ii)
Every bounded sequence inX has a unique asymptotic centerLemma 3 ([37], Opial’s Lemma).
LetX be an Hadamard space and xn be a sequence in X. If there exists a nonempty subset F in which(i)
limn⟶∞dxn,z exists for every z∈F(ii)
ifxnk is a subsequence of xn which is Δ-convergent to x, then x∈F
Then, there is ap∈F such that xn is Δ-convergent to p.Lemma 4 ([14], Proposition 4.3).
Suppose thatxn is Δ-convergent to q and there exists y∈X such that limsupdxn,y≤dq,y, then xn converges strongly to q.


## 2.1. Geometry of Geodesic Spaces

Definition 1.
LetX,d be a metric space, x,y∈X and I=0,dx,y be an interval. A curve c (or simply a geodesic path) joining x to y is an isometry c:I⟶X such that c0=x, cdx,y=y, and dct,ct′=t−t′ for all t,t′∈I. The image of a geodesic path is called a geodesic segment, which is denoted by x,y whenever it is unique.Definition 2 (see [28]).
A metric spaceX,d is called a geodesic space if every two points of X are joined by a geodesic path, and X is said to be uniquely geodesic if every two points of X are joined by exactly one geodesic path. A subset C of X is said to be convex if C includes every geodesic segments joining two of its points. Let x,y∈X and t∈0,1, and we write tx⊕1−ty for the unique point z in the geodesic segment joining from x to y such that(5)dx,z=1−tdx,yanddz,y=tdx,y.
A geodesic triangleΔx1,x2,x3 in a geodesic metric space X,d consists of three vertices (points in X) with unparameterized geodesic segment between each pair of vertices. For any geodesic triangle, there is comparison (Alexandrov) triangle Δ¯⊂ℝ2 such that dxi,xj=dℝ2x¯i,x¯j for i,j∈1,2,3. Let Δ be a geodesic triangle in X and Δ¯ be a comparison triangle for Δ¯, then Δ is said to satisfy the CAT(0) inequality if for all points x,y∈Δ and x¯,y¯∈Δ¯:(6)dx,y≤dℝ2x¯,y¯.
Letx,y,andz be points in X and y0 be the midpoint of the segment y,z; then, the CAT(0) inequality implies(7)d2x,y0≤12d2x,y+12d2x,z−14dy,z.
Inequality (7) is known as the CN inequality of Bruhat and Titis [29].Definition 3.
A geodesic spaceX is said to be a CAT(0) space if all geodesic triangles satisfy the CAT(0) inequality. Equivalently, X is called a CAT(0) space if and only if it satisfies the CN inequality.
CAT(0) spaces are examples of uniquely geodesic spaces, and complete CAT(0) spaces are called Hadamard spaces.Definition 4.
LetC be a nonempty closed and convex subset of a CAT(0) space X. The metric projection is a mapping PC:X⟶C which assigns to each x∈X, the unique point PCx in C such that(8)dx,PCx=infdx,y:y∈C.Definition 5 (see [30]).
LetX be a CAT(0) space. Denote the pair a,b∈X×X by ab→ and call it a vector. Then, a mapping .,.:X×X×X×X⟶ℝ defined by(9)ab→,cd→=12d2a,d+d2b,c−d2a,c−d2b,d,∀a,b,c,d∈X,is called a quasilinearization mapping.
It is easy to check thatab→,ab→=d2a,b, ba→,cd→=−ab→,cd→, ab→,cd→=ae→,cd→+eb→,cd→, and ab→,cd→=cd→,ab→ for all a,b,c,d,e∈X. A geodesic space X is said to satisfy the Cauchy–Swartz inequality if ab→,cd→≤da,bdc,d∀a,b,c,d∈X. It has been established in [30] that a geodesically connected metric space is a CAT(0) space if and only if it satisfies the Cauchy–Schwartz inequality. Examples of CAT(0) spaces include Euclidean spaces ℝn, Hilbert spaces, simply connected Riemannian manifolds of nonpositive sectional curvature [31], ℝ-trees, and Hilbert ball [32], among others.
We end this section with the following important lemmas which characterize CAT(0) spaces.Lemma 1.
LetX be a CAT(0) space, x,y,z∈X, and t,s∈0,1. Then,(i)
dtx⊕1−ty,z≤tdx,z+1−tdy,z (see [28])(ii)
d2tx⊕1−ty,z≤td2x,z+1−td2y,z−t1−td2x,y (see [28])


## 2.2. Notion ofΔ-Convergence

Definition 6.
Letxn be a bounded sequence in a geodesic metric space X. Then, the asymptotic center Axn of xn is defined by(10)Axn=v¯∈X:limsupn⟶∞dv¯,xn=infv∈Xlimsupn⟶∞dv,xn.
A sequencexn in X is said to be Δ-convergent to a point v¯∈X if Axnk=v¯ for every subsequence xnk of xn. In this case, we write Δ‐limn⟶∞xn=v¯ (see [33]). The concept of Δ-convergence in metric spaces was first introduced and studied by Lim [34]. Kirk and Panyanak [35] later introduced and studied this concept in CAT(0) spaces and proved that it is very similar to the weak convergence in Banach space setting.
We now end this section with the following important lemmas which are concerned withΔ-convergence.Lemma 2 (see [28, 36]).
LetX be an Hadamard space. Then,(i)
Every bounded sequence inX has a Δ-convergent subsequence(ii)
Every bounded sequence inX has a unique asymptotic centerLemma 3 ([37], Opial’s Lemma).
LetX be an Hadamard space and xn be a sequence in X. If there exists a nonempty subset F in which(i)
limn⟶∞dxn,z exists for every z∈F(ii)
ifxnk is a subsequence of xn which is Δ-convergent to x, then x∈F
Then, there is ap∈F such that xn is Δ-convergent to p.Lemma 4 ([14], Proposition 4.3).
Suppose thatxn is Δ-convergent to q and there exists y∈X such that limsupdxn,y≤dq,y, then xn converges strongly to q.


## 3. Existence and Uniqueness of Solution

In this section, we establish the existence of solution for MEP (4). We also establish the unique existence of the resolvent operator associated with the bifunctionF and the convex functional Ψ. In addition, we study some fundamental properties of this resolvent operator. We begin with the following known results.Definition 7.
LetX be a CAT(0) space. A function Ψ:DΨ⊆X⟶ℝ (where DΨ means the domain of Ψ) is said to be convex, if(11)Ψtx⊕1−ty≤tΨx+1−tΨy,∀x,y∈X,t∈0,1.
Ψ is lower semicontinuous (or upper semicontinuous) at a point x∈DΨ, if(12)Ψx≤liminfn⟶∞ΨxnorΨx≥limsupn⟶∞Ψxn,for each sequence xn in DΨ such that limn⟶∞xn=x. We say that Ψ is lower semicontinuous (or upper semicontinuous) on DΨ, if it is lower semicontinuous (or upper semicontinuous) at any point in DΨ.Lemma 5 (See [9]).
LetX be a Hadamard space and Ψ:C⟶ℝbe a convex and lower semicontinuous function. Then, Ψis Δ-lower semicontinuous.
For a nonempty subsetC of X, we denote by convC, the convex hull of C. That is, the smallest convex subset of X containing C. Recall that the convex hull of a finite set is the set of all convex combinations of its points.Theorem 2 (the KKM principle) (see [5], Theorem 3.3; see also [14], Lemma 1.8).
LetC be a nonempty, closed, and convex subset of an Hadamard space X and G:C⟶2C be a set-valued mapping with closed values. Suppose that for any finite subset x1,x2,…,xn of C,(13)convx1,x2,…,xm⊂∪i=1mGxi.
Then, the familyGxx∈C has the finite intersection property. Moreover, if Gx0 is compact for some x0∈C, then ∩x∈CGx≠∅.


### 3.1. Existence of Solution for Mixed Equilibrium Problem

Theorem 3.
LetC be a nonempty closed and convex subset of an Hadamard space X. Let Ψ:C⟶ℝbe a real-valued function and F:C×C⟶ℝ be a bifunction such that the following assumptions hold:(A1)
Fx,x=0,∀x∈C(A2)
For everyx∈C, the set y∈C:Fx,y+Ψy−Ψx<0is convex(A3)
There exists a compact subsetD⊂C containing a point y0∈D such that Fx,y0+Ψy0−Ψx<0whenever x∈C/D
Then, the MEP (4) has a solution.Proof.
For eachy∈C, define the set-valued mapping G:C⟶2C by(14)Gy≔x∈C:Fx,y+Ψy−Ψx≥0.
By (A1), we obtain that, for eachy∈C,Gy≠∅ since y∈Gy. Also, we obtain from (A2) that Gy is a closed subset of C for all y∈C.
We claim thatG satisfies the inclusion (13). Suppose for contradiction that this is not true, then there exist a finite subset y1,y2,…,ymofC and αi≥0,∀i=1,2,…,mwith∑i=1mαi=1 such that y∗=∑i=1mαiy1∉Gyi for each i=1,2,…,m. That is, there exists y∗∈convy1,y2,…,ym such that y∗∉Gyi, for each 1,2,…,m. By (14), we obtain for each i=1,2,…,m that(15)Fy∗,yi+Ψyi−Ψy∗<0.
Thus, for eachi=1,2,…,m,yi∈y∈C:Fy∗,y+Ψy−Ψy∗<0, which is convex by (A2). Since convy1,y2,…,ym is the smallest convex set containing y1,y2,…,ym, we have that convy1,y2,…,ym⊂y∈C:Fy∗,y+Ψy−Ψy∗<0, which implies that y∗∈y∈C:Fy∗,y+Ψy−Ψy∗<0. That is, 0=Fy∗,y∗+Ψy∗−Ψy∗<0, which is a contradiction. Therefore, G satisfies the inclusion (13).
Now, observe that (A3) implies that there exists a compact subsetD of C containing y0∈D such that for any x∈C/D, we have(16)Fx,y0+Ψy0−Ψx<0,which further implies that(17)Gy0=x∈C:Fx,y0+Ψy0−Ψx≥0⊂D.
Thus,Gy0 is compact. It then follows from Theorem 2 that ∩y∈CGy≠∅. This implies that there exists x∗∈C such that(18)Fx∗,y+Ψy−Ψx∗≥0,∀y∈C.
That is, MEP (4) has a solution.


### 3.2. Existence and Uniqueness of Resolvent Operator

Definition 8.
LetX be an Hadamard space and C be a nonempty subset of X. Let F:C×C⟶ℝ be a bifunction, Ψ:C⟶ℝ be a real-valued function, x¯∈X, and λ>0; then, we define the perturbation F˜x¯:C×C⟶ℝ of F and Ψ, by(19)F˜x¯x,y≔Fx,y+Ψy−Ψx+1λxy→,x¯x→,∀x,y∈C.
In the next theorem, we shall prove the existence and uniqueness of solution of the following auxiliary problem: findx∗∈C such that(20)F˜x¯x∗,y≥0,∀y∈C,where F˜x¯ is as defined in (19). The proof for existence is similar to the proof of Theorem 3. But for completeness, we shall give the proof here.Theorem 4.
LetC be a nonempty closed and convex subset of an Hadamard space X. Let Ψ:C⟶ℝbe a convex function and F:C×C⟶ℝ be a bifunction such that the following assumptions hold:(A1)
Fx,x=0,∀x∈C(A2)
F is monotone, i.e., Fx,y+Fy,x≤0,∀x,y,∈C(A3)
Fx,.:C⟶ℝ is convex ∀x∈C(A4)
For eachx¯∈X and λ>0, there exists a compact subset Dx¯⊂C containing a point yx¯∈Dx¯ such that Fx,yx¯+Ψyx¯−Ψx+1/λxyx¯→,x¯x→<0whenever x∈C/Dx¯.
Then, (20) has a unique solution.Proof.
Letx¯ be a point in X. For each y∈C, define the set-valued mapping G:C⟶2C by(21)Gy=x∈C:Fx,y+Ψy−Ψx+1λxy→,x¯x→≥0.
Then, it is easy to see thatGy is a nonempty closed subset of C. As in the proof of Theorem 3, we claim that G satisfies the inclusion (13). Suppose for contradiction that this is not true, then there exists y∗=∑i=1mαiyi∈convy1,y2,…,ym such that(22)Fy∗,yi+Ψyi−Ψy∗+1λy∗yi→,x¯y∗→<0,i=1,2,…,m.
By (A3) and the convexity ofΨ, we obtain that(23)0=Fy∗,y∗+Ψy∗−Ψy∗+1λy∗y∗→,x¯y∗→≤∑i=1mαiFy∗,yi+Ψyi−Ψy∗+1λ∑i=1mαiy∗yi→,x¯y∗→<0,which is a contradiction. Therefore, G satisfies the inclusion (13). By (A4), we obtain that Gyx¯⊂Dx¯. Thus, Gyx¯ is compact and by Theorem 2, we get that ∩y∈CGy≠∅. Therefore, (20) has a solution.
Next, we show that this solution is unique. Suppose thatx and x∗ solve (20). Then,(24)0≤F˜x¯x,x∗=Fx,x∗+Ψx∗−Ψx+1λx¯x→,xx∗→,0≤F˜x¯x∗,x=Fx∗,x+Ψx−Ψx∗+1λx¯x∗→,x∗x→.
Adding both inequalities and noting thatF is monotone, we obtain that(25)0≤−1λxx¯→,xx∗→+x¯x∗→,xx∗→=−1λdx,x∗2,which implies that x=x∗.Definition 9.
LetX be an Hadamard space and C be a nonempty closed and convex subset of X. Let F:C×C⟶ℝ be a bifunction and Ψ:C⟶ℝ be a convex function. Assume that (20) has a unique solution for each λ>0 and x∈X. This unique solution is denoted by JλFΨx, and it is called the resolvent operator associated with F and Ψ of order λ>0 and at x∈X. In other words, the resolvent operator associated with F and Ψ is the set-valued mapping JλFΨ:X⟶2C defined by(26)JλFΨx≔EPC,F˜x=z∈C:Fz,y+Ψy−Ψz+1λzy→,xz→≥0,∀y∈C,for allxinX.
Under the assumptions of Theorem 4, we have the unique existence ofJλFΨx. Therefore, JλFΨ is well defined.


### 3.3. Fundamental Properties of the Resolvent Operator

In the following theorem, we shall study some fundamental properties of the resolvent operator. First, we recall the following definitions which will be needed for our study.Definition 10.
LetX be a CAT(0) space. A point x∈X is called a fixed point of a nonlinear mapping T:X⟶X, if Tx=x. We denote the set of fixed points of T by FixT. The mapping T is said to be(i)
Firmly nonexpansive, if(27)d2Tx,Ty≤TxTy→,xy→,∀x,y∈X.(ii)
Nonexpansive, if(28)dTx,Ty≤dx,y,∀x,y∈X.Theorem 5.
LetC be a nonempty closed and convex subset of an Hadamard space X. Let Ψ:C⟶ℝbe a convex function and F:C×C⟶ℝ be a bifunction satisfying assumptions (A1)–(A4) of Theorem 4. For λ>0, we have that JλFΨis single valued. Moreover, if C⊂DJλFΨ,then(i)
JλFΨis firmly nonexpansive restricted to C(ii)
ForFJλFΨ≠∅,we have(29)d2JλFΨx,x≤d2x,v−d2JλFΨx,v,∀x∈C,∀v∈fixJλFΨ,(iii)
For0<λ≤μ, we have dJμFΨx,JλFΨx≤1−λ/μdx,JμFΨx,which implies that dx,JλFΨx≤2dx,JμFΨx,∀x∈C(iv)
FixJλFΨ=MEPC,F,ΨProof.
For eachx∈DJλFΨ and λ>0, let z1,z2∈JλFΨx. Then from (26), we have(30)Fz1,z2+Ψz2−Ψz1+1λz1z2→,xz1→≥0,Fz2,z1+Ψz1−Ψz2+1λz2z1→,xz2→≥0.
Adding both inequalities and using assumption (A2), we obtain that(31)z2z1→,z1z2→≥0,which implies that d2z1,z2≤0. This further implies that z1=z2. Therefore, JλFΨ is single valued.(i)
Letx,y∈C, then(32)FJλFΨx,JλFΨy+ΨJλFΨy−ΨJλFΨx+1λJλFΨxJλFΨy→,xJλFΨx→≥0,and(33)FJλFΨy,JλFΨx+ΨJλFΨx−ΨJλFΨy+1λJλFΨyJλFΨx→,yJλFΨy→≥0.Adding (32) and (33), and noting that F is monotone, we obtain(34)1λxJλFΨx→,JλFΨxJλFΨy→+yJλFΨy→,JλFΨyJλFΨx→≥0,which implies that(35)xy→,JλFΨxJλFΨy→≥JλFΨxJλFΨy→,JλFΨxJλFΨy→.That is,(36)xy→,JλFΨxJλFΨy→≥d2JλFΨx,JλFΨy.(ii)
It follows from (36) and the definition of quasilinearization that(37)d2x,JλFΨx≤d2x,v−d2v,JλFΨx,∀x∈C,v∈fixJλFΨ.(iii)
Letx∈C and 0<λ≤μ, then we have that(38)FJλFΨx,JμFΨx+ΨJμFΨx−ΨJλFΨx+1λxJλFΨx→,JλFΨxJμFΨx→≥0,and(39)FJμFΨx,JλFΨx+ΨJλFΨx−ΨJμFΨx+1μxJμFΨx→,JμFΨxJλFΨx→≥0.Adding (38) and (39), and using the monotonicity of F, we obtain that(40)JλFΨxx→,JμFΨxJλFΨx→≥λμJμFΨxx→,JμFΨxJλFΨx→.By quasilinearization, we obtain that(41)λμ+1d2JμFΨx,JλFΨx≤1−λμd2x,JμFΨx+λμ−1d2x,JλFΨx.Sinceλ/μ≤1, we obtain that(42)λμ+1d2JμFΨx,JλFΨx≤1−λμd2x,JμFΨx,which implies that(43)dJμFΨx,JλFΨx≤1−λμdx,JμFΨx.Moreover, we obtain by triangle inequality and (43) that(44)dx,JλFΨx≤2dx,JμFΨx.(iv)
Observe that(45)x∈fixJλFΨ⟺Fx,y+Ψy−Ψx+1λxx→,xy→≥0,∀y∈C⟺Fx,y+Ψy−Ψx≥0,∀y∈C⟺x∈MEPC,F,Ψ.Remark 1.
It follows from Cauchy–Schwartz inequality that firmly nonexpansive mappings are nonexpansive, and it is well known that the set of fixed points of nonexpansive mappings is closed and convex. Thus, by (i) and (iv) of Theorem 5, we have thatMEPC,F,Ψ is closed and convex.


## 3.1. Existence of Solution for Mixed Equilibrium Problem

Theorem 3.
LetC be a nonempty closed and convex subset of an Hadamard space X. Let Ψ:C⟶ℝbe a real-valued function and F:C×C⟶ℝ be a bifunction such that the following assumptions hold:(A1)
Fx,x=0,∀x∈C(A2)
For everyx∈C, the set y∈C:Fx,y+Ψy−Ψx<0is convex(A3)
There exists a compact subsetD⊂C containing a point y0∈D such that Fx,y0+Ψy0−Ψx<0whenever x∈C/D
Then, the MEP (4) has a solution.Proof.
For eachy∈C, define the set-valued mapping G:C⟶2C by(14)Gy≔x∈C:Fx,y+Ψy−Ψx≥0.
By (A1), we obtain that, for eachy∈C,Gy≠∅ since y∈Gy. Also, we obtain from (A2) that Gy is a closed subset of C for all y∈C.
We claim thatG satisfies the inclusion (13). Suppose for contradiction that this is not true, then there exist a finite subset y1,y2,…,ymofC and αi≥0,∀i=1,2,…,mwith∑i=1mαi=1 such that y∗=∑i=1mαiy1∉Gyi for each i=1,2,…,m. That is, there exists y∗∈convy1,y2,…,ym such that y∗∉Gyi, for each 1,2,…,m. By (14), we obtain for each i=1,2,…,m that(15)Fy∗,yi+Ψyi−Ψy∗<0.
Thus, for eachi=1,2,…,m,yi∈y∈C:Fy∗,y+Ψy−Ψy∗<0, which is convex by (A2). Since convy1,y2,…,ym is the smallest convex set containing y1,y2,…,ym, we have that convy1,y2,…,ym⊂y∈C:Fy∗,y+Ψy−Ψy∗<0, which implies that y∗∈y∈C:Fy∗,y+Ψy−Ψy∗<0. That is, 0=Fy∗,y∗+Ψy∗−Ψy∗<0, which is a contradiction. Therefore, G satisfies the inclusion (13).
Now, observe that (A3) implies that there exists a compact subsetD of C containing y0∈D such that for any x∈C/D, we have(16)Fx,y0+Ψy0−Ψx<0,which further implies that(17)Gy0=x∈C:Fx,y0+Ψy0−Ψx≥0⊂D.
Thus,Gy0 is compact. It then follows from Theorem 2 that ∩y∈CGy≠∅. This implies that there exists x∗∈C such that(18)Fx∗,y+Ψy−Ψx∗≥0,∀y∈C.
That is, MEP (4) has a solution.


## 3.2. Existence and Uniqueness of Resolvent Operator

Definition 8.
LetX be an Hadamard space and C be a nonempty subset of X. Let F:C×C⟶ℝ be a bifunction, Ψ:C⟶ℝ be a real-valued function, x¯∈X, and λ>0; then, we define the perturbation F˜x¯:C×C⟶ℝ of F and Ψ, by(19)F˜x¯x,y≔Fx,y+Ψy−Ψx+1λxy→,x¯x→,∀x,y∈C.
In the next theorem, we shall prove the existence and uniqueness of solution of the following auxiliary problem: findx∗∈C such that(20)F˜x¯x∗,y≥0,∀y∈C,where F˜x¯ is as defined in (19). The proof for existence is similar to the proof of Theorem 3. But for completeness, we shall give the proof here.Theorem 4.
LetC be a nonempty closed and convex subset of an Hadamard space X. Let Ψ:C⟶ℝbe a convex function and F:C×C⟶ℝ be a bifunction such that the following assumptions hold:(A1)
Fx,x=0,∀x∈C(A2)
F is monotone, i.e., Fx,y+Fy,x≤0,∀x,y,∈C(A3)
Fx,.:C⟶ℝ is convex ∀x∈C(A4)
For eachx¯∈X and λ>0, there exists a compact subset Dx¯⊂C containing a point yx¯∈Dx¯ such that Fx,yx¯+Ψyx¯−Ψx+1/λxyx¯→,x¯x→<0whenever x∈C/Dx¯.
Then, (20) has a unique solution.Proof.
Letx¯ be a point in X. For each y∈C, define the set-valued mapping G:C⟶2C by(21)Gy=x∈C:Fx,y+Ψy−Ψx+1λxy→,x¯x→≥0.
Then, it is easy to see thatGy is a nonempty closed subset of C. As in the proof of Theorem 3, we claim that G satisfies the inclusion (13). Suppose for contradiction that this is not true, then there exists y∗=∑i=1mαiyi∈convy1,y2,…,ym such that(22)Fy∗,yi+Ψyi−Ψy∗+1λy∗yi→,x¯y∗→<0,i=1,2,…,m.
By (A3) and the convexity ofΨ, we obtain that(23)0=Fy∗,y∗+Ψy∗−Ψy∗+1λy∗y∗→,x¯y∗→≤∑i=1mαiFy∗,yi+Ψyi−Ψy∗+1λ∑i=1mαiy∗yi→,x¯y∗→<0,which is a contradiction. Therefore, G satisfies the inclusion (13). By (A4), we obtain that Gyx¯⊂Dx¯. Thus, Gyx¯ is compact and by Theorem 2, we get that ∩y∈CGy≠∅. Therefore, (20) has a solution.
Next, we show that this solution is unique. Suppose thatx and x∗ solve (20). Then,(24)0≤F˜x¯x,x∗=Fx,x∗+Ψx∗−Ψx+1λx¯x→,xx∗→,0≤F˜x¯x∗,x=Fx∗,x+Ψx−Ψx∗+1λx¯x∗→,x∗x→.
Adding both inequalities and noting thatF is monotone, we obtain that(25)0≤−1λxx¯→,xx∗→+x¯x∗→,xx∗→=−1λdx,x∗2,which implies that x=x∗.Definition 9.
LetX be an Hadamard space and C be a nonempty closed and convex subset of X. Let F:C×C⟶ℝ be a bifunction and Ψ:C⟶ℝ be a convex function. Assume that (20) has a unique solution for each λ>0 and x∈X. This unique solution is denoted by JλFΨx, and it is called the resolvent operator associated with F and Ψ of order λ>0 and at x∈X. In other words, the resolvent operator associated with F and Ψ is the set-valued mapping JλFΨ:X⟶2C defined by(26)JλFΨx≔EPC,F˜x=z∈C:Fz,y+Ψy−Ψz+1λzy→,xz→≥0,∀y∈C,for allxinX.
Under the assumptions of Theorem 4, we have the unique existence ofJλFΨx. Therefore, JλFΨ is well defined.


## 3.3. Fundamental Properties of the Resolvent Operator

In the following theorem, we shall study some fundamental properties of the resolvent operator. First, we recall the following definitions which will be needed for our study.Definition 10.
LetX be a CAT(0) space. A point x∈X is called a fixed point of a nonlinear mapping T:X⟶X, if Tx=x. We denote the set of fixed points of T by FixT. The mapping T is said to be(i)
Firmly nonexpansive, if(27)d2Tx,Ty≤TxTy→,xy→,∀x,y∈X.(ii)
Nonexpansive, if(28)dTx,Ty≤dx,y,∀x,y∈X.Theorem 5.
LetC be a nonempty closed and convex subset of an Hadamard space X. Let Ψ:C⟶ℝbe a convex function and F:C×C⟶ℝ be a bifunction satisfying assumptions (A1)–(A4) of Theorem 4. For λ>0, we have that JλFΨis single valued. Moreover, if C⊂DJλFΨ,then(i)
JλFΨis firmly nonexpansive restricted to C(ii)
ForFJλFΨ≠∅,we have(29)d2JλFΨx,x≤d2x,v−d2JλFΨx,v,∀x∈C,∀v∈fixJλFΨ,(iii)
For0<λ≤μ, we have dJμFΨx,JλFΨx≤1−λ/μdx,JμFΨx,which implies that dx,JλFΨx≤2dx,JμFΨx,∀x∈C(iv)
FixJλFΨ=MEPC,F,ΨProof.
For eachx∈DJλFΨ and λ>0, let z1,z2∈JλFΨx. Then from (26), we have(30)Fz1,z2+Ψz2−Ψz1+1λz1z2→,xz1→≥0,Fz2,z1+Ψz1−Ψz2+1λz2z1→,xz2→≥0.
Adding both inequalities and using assumption (A2), we obtain that(31)z2z1→,z1z2→≥0,which implies that d2z1,z2≤0. This further implies that z1=z2. Therefore, JλFΨ is single valued.(i)
Letx,y∈C, then(32)FJλFΨx,JλFΨy+ΨJλFΨy−ΨJλFΨx+1λJλFΨxJλFΨy→,xJλFΨx→≥0,and(33)FJλFΨy,JλFΨx+ΨJλFΨx−ΨJλFΨy+1λJλFΨyJλFΨx→,yJλFΨy→≥0.Adding (32) and (33), and noting that F is monotone, we obtain(34)1λxJλFΨx→,JλFΨxJλFΨy→+yJλFΨy→,JλFΨyJλFΨx→≥0,which implies that(35)xy→,JλFΨxJλFΨy→≥JλFΨxJλFΨy→,JλFΨxJλFΨy→.That is,(36)xy→,JλFΨxJλFΨy→≥d2JλFΨx,JλFΨy.(ii)
It follows from (36) and the definition of quasilinearization that(37)d2x,JλFΨx≤d2x,v−d2v,JλFΨx,∀x∈C,v∈fixJλFΨ.(iii)
Letx∈C and 0<λ≤μ, then we have that(38)FJλFΨx,JμFΨx+ΨJμFΨx−ΨJλFΨx+1λxJλFΨx→,JλFΨxJμFΨx→≥0,and(39)FJμFΨx,JλFΨx+ΨJλFΨx−ΨJμFΨx+1μxJμFΨx→,JμFΨxJλFΨx→≥0.Adding (38) and (39), and using the monotonicity of F, we obtain that(40)JλFΨxx→,JμFΨxJλFΨx→≥λμJμFΨxx→,JμFΨxJλFΨx→.By quasilinearization, we obtain that(41)λμ+1d2JμFΨx,JλFΨx≤1−λμd2x,JμFΨx+λμ−1d2x,JλFΨx.Sinceλ/μ≤1, we obtain that(42)λμ+1d2JμFΨx,JλFΨx≤1−λμd2x,JμFΨx,which implies that(43)dJμFΨx,JλFΨx≤1−λμdx,JμFΨx.Moreover, we obtain by triangle inequality and (43) that(44)dx,JλFΨx≤2dx,JμFΨx.(iv)
Observe that(45)x∈fixJλFΨ⟺Fx,y+Ψy−Ψx+1λxx→,xy→≥0,∀y∈C⟺Fx,y+Ψy−Ψx≥0,∀y∈C⟺x∈MEPC,F,Ψ.Remark 1.
It follows from Cauchy–Schwartz inequality that firmly nonexpansive mappings are nonexpansive, and it is well known that the set of fixed points of nonexpansive mappings is closed and convex. Thus, by (i) and (iv) of Theorem 5, we have thatMEPC,F,Ψ is closed and convex.


## 4. Convergence Results

For the rest of this paper, we shall assume thatC is a nonempty closed and convex subset of an Hadamard space X and that DJλFΨ⊃C.


### 4.1. Convergence of Resolvent

In the following theorem, we shall prove thatJλFΨx converges strongly to a solution of MEP (4) as λ⟶0.Theorem 6.
LetΨ:C⟶ℝbe a convex and lower semicontinuous function and F:C×C⟶ℝ be Δ-upper semicontinuous in the first argument which satisfies assumptions (A1)–(A4) of Theorem 4. If MEPC,F,Ψ≠∅, then JλnFΨxconverges strongly to q∈MEPC,F,Ψ,which is the nearest point of MEPC,F,Ψto x as λ⟶0.Proof.
Letv∈MEPC,F,Ψ, since JλFΨ is nonexpansive (by Remark 1), we obtain that JλFΨx is bounded. Let λn be a sequence that converges to 0 as n⟶∞. Then, JλnFΨx is bounded. Thus, by Lemma 2(i), there exists a subsequence JλnkFΨx of JλnFΨx that Δ-converges to q∈C.
Now, observe that, by the definition ofJλFΨ, the Δ-upper semicontinuity of F, lower semicontinuous of Ψ, and Lemma 5, we obtain that(46)Fq,y+Ψy−Ψq≥0.
Therefore,q∈MEPC,F,Ψ. Hence, we obtain from Theorem 5(ii) that(47)d2JλnkFΨx,x≤d2x,v,∀v∈MEPC,F,Ψ.
Sinced2.,x is Δ-lower semicontinuous, we obtain that(48)d2q,x≤liminfk⟶∞d2JλnkFΨx,x≤d2x,v,∀v∈MEPC,F,Ψ,which implies that(49)dq,x≤dx,v,∀v∈MEPC,F,Ψ.
Thus,q=PΓx, where PΓ is the metric projection of X onto Γ, and Γ=MEPC,F,Ψ. Therefore, by taking λnk=λ, we have that JλFΨxΔ-converges to q=PΓx as λ⟶0.
Now, observe also that Theorem 5(ii) implies that(50)dJλFΨx,x≤dq,x.
It then follows from Lemma 4 thatJλFΨx converges strongly to q=PΓx as λ⟶0.
By settingΨ≡0 in Theorem 6, we obtain the following result which is similar to ([14], Theorem 4.4).Corollary 1.
LetF:C×C⟶ℝ be Δ-upper semicontinuous in the first argument which satisfies assumptions (A1)–(A4) of Theorem 4. If MEPC,F≠∅, then JλFxconverges strongly to q∈MEPC,F, which is the nearest point of MEPC,F to x as λ⟶0.


### 4.2. Proximal Point Algorithm

In this section, we study theΔ-convergence of the sequence generated by the following PPA for approximating solutions of MEP (4): For an initial starting point x1 in C, define the sequence xn in C by(51)xn+1=JλnFΨxn,n≥1,where λn is a sequence in 0,∞, F:C×C⟶ℝ is a bifunction, and Ψ:C⟶ℝ is a convex function.Recall that the PPA does not converge strongly in general without additional assumptions even for the case whereF≡0. See for example, the question of interest raised by Rockafella as to whether the PPA can be improved from weak convergence (an analogue of Δ-convergence) to strong convergence in Hilbert space settings. Several counterexamples have been constructed to resolve this question in the negative (see [38, 39]). Therefore, only weak convergence of the PPA is expected without additional assumptions. For this reason, we propose the following Δ-convergence theorem for the PPA (51).Theorem 7.
LetΨ:C⟶ℝ be a convex and lower semicontinuous function and F:C×C⟶ℝ be Δ-upper semicontinuous in the first argument which satisfies assumptions (A1)–(A4) of Theorem 4. Let λn be a sequence in 0,∞ such that 0<λ≤λn,∀n≥1. Suppose that MEPC,F,Ψ≠∅, then, the sequence given by (51) Δ-converges to an element of MEPC,F,Ψ.Proof.
Letv∈MEPC,F,Ψ. Then, by Remark 1 and Theorem 5(iv), we obtain that(52)dv,xn+1=dv,JλnFΨxn≤dv,xn,which implies that limn⟶∞dxn,v exists for all v∈MEPC,F,Ψ. Hence xn is bounded. It then follows from Theorem 5(ii) that(53)d2xn+1,xn≤d2xn,v−d2xn+1,v⟶0,asn⟶∞.
That is,(54)limn⟶∞dxn+1,xn=0.
Sincexn is bounded, then there exists a subsequence xnk of xn that Δ-converges to a point, say q∈C. From (51) and (26), we obtain that(55)Fxnk+1,y+Ψy−Ψxnk+1≥−1λnkxnkxnk+1→,xnk+1y→≥−1λnkdxnk+1,xnkdxnk+1,y.
Since0<λ≤λnk,xn is bounded, F is Δ-upper semicontinuous in the first argument and Ψ is lower semicontinuous, we obtained from (54) and (55) that(56)Fq,y+Ψy−Ψq≥limsupk⟶∞Fxnk+1,y+Ψy−liminfk⟶∞Ψxnk+1≥−Mλlimsupk⟶∞dxnk+1,xnk=0,for some M>0 and for all y∈C. This implies that q∈MEPC,F,Ψ.
It then follows from Lemma 3 thatxnΔ-converges to an element of MEPC,F,Ψ.
By settingΨ≡0 in Theorem 7, we obtain the following result which coincides with ([5], Theorem 7.3).Corollary 2.
LetF:C×C⟶ℝ be Δ-upper semicontinuous in the first argument which satisfies assumptions (A1)–(A4) of Theorem 4 and λn be a sequence in 0,∞ such that 0<λ≤λn∀n≥1. Suppose that EPC,F≠∅; then, the sequence given for x1∈C by(57)xn+1=JλnFxn,n≥1.
Δ-converges to an element of EPC,F.
By settingF≡0 in Theorem 7, we obtain the following corollary which is similar to ([9], Theorem 1.4).Corollary 3.
LetΨ:C⟶ℝ be a convex and lower semicontinuous function and λn be a sequence in 0,∞ such that 0<λ≤λn,∀n≥1. Suppose that argminy∈CΨy≠∅; then, the sequence given for x1∈C by(58)xn+1=JλnΨxn,n≥1.
Δ-converges to an element of argminy∈CΨy.


## 4.1. Convergence of Resolvent

In the following theorem, we shall prove thatJλFΨx converges strongly to a solution of MEP (4) as λ⟶0.Theorem 6.
LetΨ:C⟶ℝbe a convex and lower semicontinuous function and F:C×C⟶ℝ be Δ-upper semicontinuous in the first argument which satisfies assumptions (A1)–(A4) of Theorem 4. If MEPC,F,Ψ≠∅, then JλnFΨxconverges strongly to q∈MEPC,F,Ψ,which is the nearest point of MEPC,F,Ψto x as λ⟶0.Proof.
Letv∈MEPC,F,Ψ, since JλFΨ is nonexpansive (by Remark 1), we obtain that JλFΨx is bounded. Let λn be a sequence that converges to 0 as n⟶∞. Then, JλnFΨx is bounded. Thus, by Lemma 2(i), there exists a subsequence JλnkFΨx of JλnFΨx that Δ-converges to q∈C.
Now, observe that, by the definition ofJλFΨ, the Δ-upper semicontinuity of F, lower semicontinuous of Ψ, and Lemma 5, we obtain that(46)Fq,y+Ψy−Ψq≥0.
Therefore,q∈MEPC,F,Ψ. Hence, we obtain from Theorem 5(ii) that(47)d2JλnkFΨx,x≤d2x,v,∀v∈MEPC,F,Ψ.
Sinced2.,x is Δ-lower semicontinuous, we obtain that(48)d2q,x≤liminfk⟶∞d2JλnkFΨx,x≤d2x,v,∀v∈MEPC,F,Ψ,which implies that(49)dq,x≤dx,v,∀v∈MEPC,F,Ψ.
Thus,q=PΓx, where PΓ is the metric projection of X onto Γ, and Γ=MEPC,F,Ψ. Therefore, by taking λnk=λ, we have that JλFΨxΔ-converges to q=PΓx as λ⟶0.
Now, observe also that Theorem 5(ii) implies that(50)dJλFΨx,x≤dq,x.
It then follows from Lemma 4 thatJλFΨx converges strongly to q=PΓx as λ⟶0.
By settingΨ≡0 in Theorem 6, we obtain the following result which is similar to ([14], Theorem 4.4).Corollary 1.
LetF:C×C⟶ℝ be Δ-upper semicontinuous in the first argument which satisfies assumptions (A1)–(A4) of Theorem 4. If MEPC,F≠∅, then JλFxconverges strongly to q∈MEPC,F, which is the nearest point of MEPC,F to x as λ⟶0.


## 4.2. Proximal Point Algorithm

In this section, we study theΔ-convergence of the sequence generated by the following PPA for approximating solutions of MEP (4): For an initial starting point x1 in C, define the sequence xn in C by(51)xn+1=JλnFΨxn,n≥1,where λn is a sequence in 0,∞, F:C×C⟶ℝ is a bifunction, and Ψ:C⟶ℝ is a convex function.Recall that the PPA does not converge strongly in general without additional assumptions even for the case whereF≡0. See for example, the question of interest raised by Rockafella as to whether the PPA can be improved from weak convergence (an analogue of Δ-convergence) to strong convergence in Hilbert space settings. Several counterexamples have been constructed to resolve this question in the negative (see [38, 39]). Therefore, only weak convergence of the PPA is expected without additional assumptions. For this reason, we propose the following Δ-convergence theorem for the PPA (51).Theorem 7.
LetΨ:C⟶ℝ be a convex and lower semicontinuous function and F:C×C⟶ℝ be Δ-upper semicontinuous in the first argument which satisfies assumptions (A1)–(A4) of Theorem 4. Let λn be a sequence in 0,∞ such that 0<λ≤λn,∀n≥1. Suppose that MEPC,F,Ψ≠∅, then, the sequence given by (51) Δ-converges to an element of MEPC,F,Ψ.Proof.
Letv∈MEPC,F,Ψ. Then, by Remark 1 and Theorem 5(iv), we obtain that(52)dv,xn+1=dv,JλnFΨxn≤dv,xn,which implies that limn⟶∞dxn,v exists for all v∈MEPC,F,Ψ. Hence xn is bounded. It then follows from Theorem 5(ii) that(53)d2xn+1,xn≤d2xn,v−d2xn+1,v⟶0,asn⟶∞.
That is,(54)limn⟶∞dxn+1,xn=0.
Sincexn is bounded, then there exists a subsequence xnk of xn that Δ-converges to a point, say q∈C. From (51) and (26), we obtain that(55)Fxnk+1,y+Ψy−Ψxnk+1≥−1λnkxnkxnk+1→,xnk+1y→≥−1λnkdxnk+1,xnkdxnk+1,y.
Since0<λ≤λnk,xn is bounded, F is Δ-upper semicontinuous in the first argument and Ψ is lower semicontinuous, we obtained from (54) and (55) that(56)Fq,y+Ψy−Ψq≥limsupk⟶∞Fxnk+1,y+Ψy−liminfk⟶∞Ψxnk+1≥−Mλlimsupk⟶∞dxnk+1,xnk=0,for some M>0 and for all y∈C. This implies that q∈MEPC,F,Ψ.
It then follows from Lemma 3 thatxnΔ-converges to an element of MEPC,F,Ψ.
By settingΨ≡0 in Theorem 7, we obtain the following result which coincides with ([5], Theorem 7.3).Corollary 2.
LetF:C×C⟶ℝ be Δ-upper semicontinuous in the first argument which satisfies assumptions (A1)–(A4) of Theorem 4 and λn be a sequence in 0,∞ such that 0<λ≤λn∀n≥1. Suppose that EPC,F≠∅; then, the sequence given for x1∈C by(57)xn+1=JλnFxn,n≥1.
Δ-converges to an element of EPC,F.
By settingF≡0 in Theorem 7, we obtain the following corollary which is similar to ([9], Theorem 1.4).Corollary 3.
LetΨ:C⟶ℝ be a convex and lower semicontinuous function and λn be a sequence in 0,∞ such that 0<λ≤λn,∀n≥1. Suppose that argminy∈CΨy≠∅; then, the sequence given for x1∈C by(58)xn+1=JλnΨxn,n≥1.
Δ-converges to an element of argminy∈CΨy.


## 5. Asymptotic Behavior of Halpern’s Algorithm

To obtain strong convergence result, we modify the PPA into the following Halpern-type PPA and study the asymptotic behavior of the sequence generated by it: Forx1,u∈C, define the sequence xn⊂C by(59)xn+1=αnu⊕1−αnJλnFΨxn,where αn is a sequence in 0,1 and λn,F and Ψ are as defined in (51).We begin by establishing the following lemmas which will be very useful to our study.Lemma 6.
LetΨ:C⟶ℝ be a convex and lower semicontinuous function and F:C×C⟶ℝ be a bifunction satisfying (A1)–(A4) of Theorem 4. If λ,μ>0 and x,y∈C, then the following inequalities hold:(60)d2JλFΨx,JμFΨy≤2λFJλFΨx,JμFΨy+2λΨJμFΨy−ΨJλFΨx+d2x,JμFΨy−d2x,JλFΨx,λ+μd2JλFΨx,JμFΨy+μd2JλFΨx,x+λd2JμFΨy,y≤λd2JλFΨx,y+μd2JλFΨy,x.Proof.
We first prove (60). Let λ,μ>0 and x,y∈C. Then, by (26), we obtain that(61)FJλFΨx,z+Ψz−ΨJλFΨx+1λxJλFΨx→,JλFΨxz→≥0,∀z∈C,which implies that(62)2λΨJλFΨx≤2λFJλFΨx,z+2λΨz+2xJλFΨx→,JλFΨxz→=2λFJλFΨx,z+2λΨz+d2x,z−d2x,JλFΨ−d2JλFΨx,z≤2λFJλFΨx,z+2λΨz+d2x,z−d2x,JλFΨx.
Now, setz=tJμFΨy⊕1−tJλFΨx for all t∈0,1 in (5). Since Ψ is convex and F satisfies conditions (A1) and (A3) of Theorem 4, we obtain that(63)2λΨJλFΨx+d2x,JλFΨx≤2λtFJλFΨx,JμFΨy+1−tFJλFΨx,JλFΨx+2λtΨJμFΨy+1−tΨJλFΨx+td2x,JμFΨy+1−td2x,JλFΨx−t1−td2JμFΨy,JλFΨx=2λtFJλFΨx,JμFΨy+2λtΨJμFΨy+1−tΨJλFΨx+td2x,JμFΨy+1−td2x,JλFΨx−t1−td2JμFΨy,JλFΨx,which implies that(64)2λΨJλFΨx+d2x,JλFΨx≤2λFJλFΨx,JμFΨy+2λΨJμFΨy+d2x,JμFΨy−1−td2JμFΨy,JλFΨx.
Ast⟶0 in (64), we obtain (60).
Next, we prove (60). From (60), we obtain that(65)μd2JλFΨx,JμFΨy≤2λμFJλFΨx,JμFΨy+ΨJμFΨy−ΨJλFΨx+μd2x,JμFΨy−μd2x,JλFΨx.
Similarly, we have(66)λd2JμFΨy,JλFΨx≤2μλFJμFΨy,JλFΨx+ΨJλFΨx−ΨJμFΨy+λd2y,JλFΨx−λd2y,JμFΨy.
Adding both inequalities and noting thatF is monotone, we get(67)λ+μd2JλFΨx,JμFΨy+μd2x,JλFΨx+λd2y,JμFΨy≤μd2x,JμFΨy+λd2y,JλFΨx.Lemma 7.
LetΨ:C⟶ℝ be a convex and lower semicontinuous function and F:C×C⟶ℝ be a bifunction satisfying (A1)–(A4) of Theorem 4. Let λn be a sequence in 0,∞ and v¯ be an element of C. Suppose that limn⟶∞λn=∞ and AJλnΨxn=v¯ for some bounded sequence xn in X, then v¯∈MEPC,F,Ψ.
Proof. From (60), we obtain that(68)λn+1d2JλnFΨxn,JFΨv¯+d2JλnFΨxn,xn+λnd2JFΨv¯,v¯≤d2JFΨv¯,xn+λnd2JλnFΨxn,v¯,which implies that(69)d2JλnFΨxn,JFΨv¯≤1λnd2JFΨv¯,xn+d2JλnFΨxn,v¯2.
Sincelimn⟶∞λn=∞,xn is bounded and AJλnΨxn=v¯, we obtain that(70)limsupn⟶∞dJλnFΨxn,JFΨv¯≤limsupn⟶∞dJλnFΨxn,v¯=infy∈Xlimsupn⟶∞dJλnFΨxn,y,which by Lemma 2(ii) and Theorem 5(iv) implies that v¯∈fixJFΨ=MEPC,F,Ψ.Lemma 8.
(Xu, [40]). Let an be a sequence of nonnegative real numbers satisfying the following relation:(71)an+1≤1−αnan+αnσn+γn,n≥0,where iαn⊂0,1,∑αn=∞;iilimsupσn≤0;iiiγn≥0;n≥0,∑γn<∞. Then, an⟶0 as n⟶∞.Theorem 8.
LetΨ:C⟶ℝ be a convex and lower semicontinuous function and F:C×C⟶ℝ be a bifunction satisfying (A1–A4) of Theorem 4. Let xn be a sequence defined by (59), where αn is a sequence in 0,1 and λn is a sequence in 0,∞ such that limn⟶∞λn=∞. Then, we have the following:(i)
The sequenceJλnFΨxn is bounded if and only ifMEPC,F,Ψ≠∅(ii)
Iflimn⟶∞αn=0,∑n=1∞αn=∞ andΓ≔MEPC,F,Ψ≠∅, then xn and JλnFΨxnconverge to v¯=PΓu,wherePΓis the metric projection of X onto ΓProof.
(i) Suppose thatJλnfxn is bounded. Then by Lemma 2(ii), there exists v¯∈X such that AJλnfxn=v¯. From (59) and Lemma 1(i), we obtain that(72)dxn+1,v¯≤αndu,v¯+1−αndJλnFΨxn,v¯,which implies that xn is bounded. Also, since limn⟶∞λn=∞ and AJλnFΨxn=v¯, we obtain by Lemma 7 thatMEPC,F,Ψ≠∅.
Conversely, letMEPC,F,Ψ≠∅. Then, we may assume that v¯∈MEPC,F,Ψ≠∅. Thus, by (59) and Lemma 1, we obtain that(73)dxn+1,v¯≤αndu,v¯+1−αndJλnFΨxn,v¯≤αndu,v¯+1−αndxn,v¯≤maxdu,v¯,dxn,v¯,which implies by induction that(74)dxn,v¯≤maxdu,v¯,dx1,v¯,∀n≥1.
Therefore,xn is bounded. Consequently, JλnFΨxn is also bounded.
(ii) SinceΓ≔MEPC,F,Ψ≠∅, we obtain from (74) that xn and JλnFΨxn are bounded. Furthermore, we obtain from Lemma 1(ii) that(75)d2xn+1,v¯≤αnd2u,v¯+1−αnd2JλnFΨxn,v¯−αn1−αnd2u,JλnFΨxn≤αnd2u,v¯+1−αnd2xn,v¯−αn1−αnd2u,JλnFΨxn=1−αnd2xn,v¯+αnδn,∀n≥1,where δn=d2u,v¯+αn−1d2u,JλnFΨxn. Now, set vn=JλnFΨxn,∀n≥1. Then, by the boundedness of vn and Lemma 2(i), we obtain that there exists a subsequence vnk of vn that Δ-converges to some v^∈C. Thus, by Lemma 2(ii), we obtain that Avnk=v^. Moreover, limk⟶∞λnk=∞ and xnk is bounded. Hence, by Lemma 7, we obtain that v^∈MEPC,F,Ψ.
Next, we show thatxn converges to v^. By the Δ-lower semicontinuity of d2u,., we obtain that(76)d2u,v^≤liminfk⟶∞d2u,vnk=limk⟶∞d2u,vnk=liminfn⟶∞d2u,vn.
Sinceδn=d2u,v¯+αn−1d2u,vn, limn⟶∞αn=0, v¯=PΓu, and v^∈Γ, we obtain from the definition of PΓ and (76) that(77)limsupn⟶∞δn≤d2u,v¯−liminfn⟶∞d2u,vn≤d2u,v^−liminfn⟶∞d2u,vn≤0.
Thus, applying Lemma 8 to (75) gives that xn converges to v¯=PΓu. It then follows that JλnFΨxn is convergent to v¯=PΓu.
By settingΨ≡0 in Theorem 8, we obtain the following new result for equilibrium problem in an Hadamard space.Corollary 4.
LetF:C×C⟶ℝ be a bifunction satisfying (A1–A3) of Theorem 4 and xn be a sequence defined for u,x1∈C, by(78)xn+1=αnu⊕1−αnJλnFxn,where αn is a sequence in 0,1 and λn is a sequence in 0,∞ such that limn⟶∞λn=∞. Then, we have the following:(i)
The sequenceJλnFxn is bounded if and only if EPC,F≠∅(ii)
Iflimn⟶∞αn=0,∑n=1∞αn=∞ and Γ≔EPC,F≠∅, then xn and JλnFxn converge to v¯=PΓu,where PΓ is the metric projection of X onto Γ
By settingF≡0 in Theorem 8, we obtain the following result which coincides with ([41], Theorem 5.1).Corollary 5.
LetΨ:C⟶C be a proper convex and lower semicontinuous function and xn be a sequence defined for u,x1∈C, by(79)xn+1=αnu⊕1−αnJλnΨxn,where αn is a sequence in 0,1 and λn is a sequence in 0,∞ such that limn⟶∞λn=∞. Then, we have the following:(i)
The sequenceJλnΨxn is bounded if and only if argminy∈CΨy≠∅(ii)
Iflimn⟶∞αn=0,∑n=1∞αn=∞ and Γ≔argminy∈CΨy≠∅,then xn and JλnΨxn converge to v¯=PΓu, where PΓis the metric projection of X onto Γ


## 6. Numerical Results

In this section, we generate some numerical results in nonlinear setting for Algorithms (58) and (79).LetX=ℝ2 be endowed with a metric dX:ℝ2×ℝ2⟶0,∞ defined by(80)dXx,y=x1−y12+x12−x2−y12+y22,∀x,y∈ℝ2.Then,ℝ2,dX is an Hadamard space (see ([42], Example 5.2)) with the geodesic joining x to y given by(81)1−tx⊕ty=1−tx1+ty1,1−tx1+ty12−1−tx12−x2−ty12−y2.Now, defineΨ:ℝ2⟶ℝ by(82)Ψx1,x2=100x2−2−x1−222+x1−32.Then, it follows from ([42], Example 5.2) that Ψ is a proper convex and lower semicontinuous function in ℝ2,dX but not convex in the classical sense (Figure1).Figure 1
Errors vs iteration numbersn: Case 1 (a); Case 2 (b); Case 3 (c); Case 4 (d).
(a)
(b)
(c)
(d)Now, takeαn=1/n+1 and λn=n+1 for all n≥1, then all the conditions of Corollaries 4.5 and 5.6 are satisfied. Hence, by considering the following initial vectors, we obtain the numerical results for Algorithms (58) and (79) as shown by the graphs as follows:Case 1:x1=0.5,−0.25T and u=0.5,3TCase 2:x1=−1.5,−3T and u=0.5,3TCase 3:x1=0.5,3T and u=−1.5,−3TCase 4:x1=0.5,3T and u=0.5,−0.25T



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*Source: 3210649-2019-10-13.xml*