# Strong Convergence Theorem for Bregman Strongly Nonexpansive Mappings and Equilibrium Problems in Reflexive Banach Spaces

Authors: Jinhua Zhu; Shih-sen Chang; Min Liu
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

By using a new hybrid method, a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of Bregman strongly nonexpansive mappings in a reflexive Banach space is proved.


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## Body

## 1. Introduction

Throughout this paper, we denote byℝ and ℝ+ the set of all real numbers and all nonnegative real numbers, respectively. We also assume that E is a real reflexive Banach space, E* is the dual space of E, C is a nonempty closed convex subset of E, and 〈·,·〉 is the pairing between E and E*. Let Θ be a bifunction from C×C→ℝ. The equilibrium problem is to find
(1)x*∈CsuchthatΘ(x*,y)≥0,∀y∈C.
The set of such solutions x* is denoted by EP(Θ).Recall that a mappingT:C→C is said to be nonexpansive, if
(2)∥Tx-Ty∥≤∥x-y∥,∀x,y∈C.
We denote by F(T) the set of fixed points of T.Numerous problems in physics, optimization, and economics reduce to find a solution of the equilibrium problem. Some methods have been proposed to solve the equilibrium problem in a Hilbert spaces; see, for instance, Blum and Oettli [1], Combettes and Hirstoaga [2], and Moudafi [3]. Recently, Tada and Takahashi [4, 5] and S. Takahashi and W. Takahashi [6] obtained weak and strong convergence theorems for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space. In particular, Tada and Takahashi [4] established a strong convergence theorem for finding a common element of two sets by using the hybrid method introduced by Nakajo and Takahashi [7]. The authors also proved such a strong convergence theorem in a uniformly convex and uniformly smooth Banach space.In this paper, motivated by Takahashi et al. [8], we prove a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a Bregman strongly nonexpansive mapping in a real reflexive Banach space by using the shrinking projection method. Using this theorem, we obtain two new strong convergence results for finding a solution of an equilibrium problem and a fixed point of Bregman strongly nonexpansive mappings in a real reflexive Banach space.


## 2. Preliminaries and Lemmas

In the sequel, we begin by recalling some preliminaries and lemmas which will be used in the proof.LetE be a real reflexive Banach space with the norm ∥·∥ and E* the dual space of E. Throughout this paper, f:E→(-∞,+∞] is a proper, lower semicontinuous, and convex function. We denote by domf the domain of f, that is, the set {x∈E:f(x)<+∞}.Letx∈  intdomf. The subdifferential of f at x is the convex set defined by
(3)∂f(x)={x*∈E*:f(x)+〈x*,y-x〉≤f(y),∀y∈E},
where the Fenchel conjugate of f is the function f*:E*→(-∞,+∞] defined by
(4)f*(x*)=sup{〈x*,x〉-f(x):x∈E}.
We know that the Young-Fenchel inequality holds:
(5)〈x*,x〉≤f(x)+f*(x*),∀x∈E,x*∈E*.A functionf on E is coercive [9] if the sublevel set of f is bounded; equivalently,
(6)lim∥x∥→+∞f(x)=+∞.A functionf on E is said to be strongly coercive [10] if
(7)lim∥x∥→+∞f(x)∥x∥=+∞.For anyx∈intdomf and y∈E, the right-hand derivative of f at x in the direction y is defined by
(8)f∘(x,y)∶=limt→0+f(x+ty)-f(x)t.The functionf is said to be Gâteaux differentiable at x if limt→0+((f(x+ty)-f(x))/t) exists for any y. In this case, f∘(x,y) coincides with ∇f(x), the value of the gradient ∇f of f at x. The function f is said to be Gâteaux differentiable if it is Gâteaux differentiable for any x∈intdomf. The function f is said to be Fréchet differentiable at x if this limit is attained uniformly in ∥y∥=1. Finally, f is said to be uniformly Fréchet differentiable on a subset C of E if the limit is attained uniformly for x∈C and ∥y∥=1. It is known that if f is Gâteaux differentiable (resp., Fréchet differentiable) on intdomf, then f is continuous and its Gâteaux derivative ∇f is norm-to-weak* continuous (resp., continuous) on intdomf (see also [11, 12]). We will need the following result.Lemma 1 (see [13]).
Iff:E→ℝ is uniformly Fréchet differentiable and bounded on bounded subsets of E, then ∇f is uniformly continuous on bounded subsets of E from the strong topology of E to the strong topology of E*.Definition 2 (see [14]).
The functionf is said to be(i)
essentially smooth, if∂f is both locally bounded and single valued on its domain,(ii)
essentially strictly convex, if(∂f)-1 is locally bounded on its domain and f is strictly convex on every convex subset of dom∂f,(iii)
Legendre if it is both essentially smooth and essentially strictly convex.Remark 3.
LetE be a reflexive Banach space. Then we have the following.(i)
f is essentially smooth if and only if f* is essentially strictly convex (see [14, Theorem 5.4]).(ii)
(∂f)-1=∂f* (see [12]).(iii)
f is Legendre if and only if f* is Legendre (see [14, Corollary 5.5]).(iv)
Iff is Legendre, then ∇f is a bijection satisfying ∇f=(∇f*)-1, ran ∇f= dom∇f*=intdomf*, and ran ∇f*=dom∇f=intdomf (see [14, Theorem 5.10]).Examples of Legendre functions were given in [14, 15]. One important and interesting Legendre function is (1/p)∥·∥p(1<p<∞) when E is a smooth and strictly convex Banach space. In this case, the gradient ∇f of f is coincident with the generalized duality mapping of E; that is, ∇f=Jp(1<p<∞). In particular, ∇f=I the identity mapping in Hilbert spaces. In the rest of this paper, we always assume that f:E→(-∞,+∞] is Legendre.Letf:E→(-∞,+∞] be a convex and Gâteaux differentiable function. The function Df:domf×intdomf→[0,+∞) defined as
(9)Df(y,x)∶=f(y)-f(x)-〈∇f(x),y-x〉
is called the Bregman distance with respect to f [16].Recall that the Bregman projection [17] of x∈intdomf onto the nonempty closed and convex set C⊂domf is the necessarily unique vector PCf(x)∈C satisfying
(10)Df(PCf(x),x)=inf{Df(y,x):y∈C}.
Concerning the Bregman projection, the following are well known.Lemma 4 (see [18]).
LetC be a nonempty, closed, and convex subset of a reflexive Banach space E. Let f:E→ℝ be a Gâteaux differentiable and totally convex function and let x∈E. Then(a)
z=PCf(x) if and only if 〈∇f(x)-∇f(z),y-z〉≤0, for all y∈C.(b)
(11)Df(y,PCf(x))+Df(PCf(x),x)≤Df(y,x),∀x∈E,y∈C.Letf:E→(-∞,+∞] be a convex and Gâteaux differentiable function. The modulus of total convexity of f at x∈intdomf is the function νf(x,·):[0,+∞)→[0,+∞] defined by
(12)νf(x,t)∶=inf{Df(y,x):y∈domf,∥y-x∥=t}.The functionf is called totally convex at x if νf(x,t)>0 whenever t>0. The function f is called totally convex if it is totally convex at any point x∈intdomf and is said to be totally convex on bounded sets if νf(B,t)>0 for any nonempty bounded subset B of E and t>0, where the modulus of total convexity of the function f on the set B is the function νf:intdomf×[0,+∞)→[0,+∞] defined by
(13)νf(B,t)∶=inf{νf(x,t):x∈B∩domf}.The next lemma will be useful in the proof of our main results.Lemma 5 (see [19]).
Ifx∈domf, then the following statements are equivalent.(i)
The functionf is totally convex at x.(ii)
For any sequence{yn}⊂domf,
(14)limn→+∞Df(yn,x)=0⟹limn→+∞∥yn-x∥=0.Recall that the functionf is called sequentially consistent [18] if, for any two sequences {xn} and {yn} in E such that the first one is bounded,
(15)limn→+∞Df(yn,xn)=0⟹limn→+∞∥yn-xn∥=0.Lemma 6 (see [20]).
The functionf is totally convex on bounded sets if and only if the function f is sequentially consistent.Lemma 7 (see [21]).
Letf:E→ℝ be a Gâteaux differentiable and totally convex function. If x0∈E and the sequence {Df(xn,x0)} is bounded, then the sequence {xn} is bounded too.Lemma 8 (see [21]).
Letf:E→ℝ be a Gâteaux differentiable and totally convex function, x0∈E, and let C be a nonempty, closed, and convex subset of E. Suppose that the sequence {xn} is bounded and any weak subsequential limit of {xn} belongs to C. If Df(xn,x0)≤Df(PCfx0,x0) for any n∈ℕ, then {xn} converges strongly to PCfx0.LetC be a convex subset of intdomf and let T be a self-mapping of C. A point p∈C is called an asymptotic fixed point of T (see [22, 23]) if C contains a sequence {xn} which converges weakly to p such that limn→∞∥xn-Txn∥=0. We denote by F^(T) the set of asymptotic fixed points of T.Definition 9.
A mappingT with a nonempty asymptotic fixed point set F^(T) is said to be(i)
Bregman strongly nonexpansive (see [24, 25]) with respect to F^(T) if
(16)Df(p,Tx)≤Df(p,x),∀x∈C,p∈F^(T),
and if, whenever {xn}⊂C is bounded, p∈F^(T) and
(17)limn→∞(Df(p,xn)-Df(p,Txn))=0,
it follows that
(18)limn→∞Df(xn,Txn)=0.(ii)
Bregman firmly nonexpansive [26] if, for all x,y∈C,
(19)〈∇f(Tx)-∇f(Ty),Tx-Ty〉≤〈∇f(x)-∇f(y),Tx-Ty〉
or, equivalently,
(20)Df(Tx,Ty)+Df(Ty,Tx)+Df(Tx,x)+Df(Ty,y)≤Df(Tx,y)+Df(Ty,x).The existence and approximation of Bregman firmly nonexpansive mappings were studied in [26]. It is also known that if T is Bregman firmly nonexpansive and f is Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E, then F(T)=F^(T) and F(T) is closed and convex (see [26]). It also follows that every Bregman firmly nonexpansive mapping is Bregman strongly nonexpansive with respect to F(T)=F^(T).Lemma 10 (see [27]).
LetE be a real reflexive Banach space and f:E→(-∞,+∞] a proper lower semicontinuous function; then f*:E*→(-∞,+∞] is a proper weak* lower semicontinuous and convex function. Thus, for all z∈E, we have
(21)Df(z,∇f*(∑i=1N‍ti∇f(xi)))≤∑i=1N‍tiDf(z,xi).In order to solve the equilibrium problem, let us assume that a bifunctionΘ:C×C→ℝ satisfies the following conditions [28]:(A1)
Θ(x,x)=0, for all x∈C.(A2)
Θ is monotone; that is, Θ(x,y)+Θ(y,x)≤0, for all x,y∈C.(A3)
limsupt↓0Θ(x+t(z-x),y)≤Θ(x,y) for all x,z,y∈C.(A4)
The functiony↦Θ(x,y) is convex and lower semicontinuous.The resolvent of a bifunctionΘ [29] is the operator ResΘf:E→2C defined by
(22)ResΘf(x)={z∈C:Θ(z,y)+〈∇f(z)-∇f(x),y-z〉≥0,∀y∈C}.From Lemma 1 in [24], if f:E→(-∞,+∞] is a strongly coercive and Gâteaux differentiable function and Θ satisfies conditions (A1–A4), then dom (ResΘf)=E. We also know the following lemma which gives us some characterizations of the resolvent ResΘf.Lemma 11 (see [24]).
LetE be a real reflexive Banach space and C a nonempty closed convex subset of E. Let f:E→(-∞,+∞] be a Legendre function. If the bifunction Θ:C×C→ℝ satisfies the conditions (A1)–(A4), then the followings hold:(i)
ResΘf is single-valued;(ii)
ResΘf is a Bregman firmly nonexpansive operator;(iii)
F(ResΘf)=EP(Θ);(iv)
EP(Θ) is a closed and convex subset of C;(v)
for allx∈E and for all q∈F(ResΘf), we have
(23)Df(q,ResΘf(x))+Df(ResΘf(x),x)≤Df(q,x).


## 3. Strong Convergence Theorem

In this section, we proved a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and a fixed point of Bregman strongly nonexpansive mapping in a real reflexive Banach space by using the shrinking projection method.Theorem 12.
LetC be a nonempty, closed, and convex subset of a real reflexive Banach space E and f:E→ℝ a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of E. Let g be a bifunction from C×C to ℝ satisfying (A1)–(A4) and let T be a Bregman strongly nonexpansive mapping from C into itself such that F(T)=F^(T) and G=F(T)∩EP(g)≠∅. Let {xn} be a sequence generated by x0=x∈C,C0=C and
(24)yn=∇f*(αn∇f(xn)+(1-αn)∇f(Txn)),un∈Csuchthatg(un,y)+〈∇f(un)-∇f(yn),y-un〉≥0,∀y∈C,Cn+1={z∈Cn:Df(z,un)≤Df(z,xn)},xn+1=PCn+1fx
for every n∈ℕ∪{0}, where {αn}⊂[0,1] satisfies liminfn→∞(1-αn)>0. Then, {xn} converges strongly to PF(T)∩EP(g)fx, where PF(T)∩EP(g)f is the Bregman projection of E onto F(T)∩EP(g).Proof.
We divide the proof of Theorem12 into five steps.
(I) We first prove thatG and Cn both are closed and convex subset of C for all n≥0. In fact, it follows from Lemma 11 and by Reich and Sabach [26] that EP(g) and F(T) both are closed and convex. Therefore, G is a closed and convex subset in C. Furthermore, it is obvious that C0=C is closed and convex. Suppose that Cn is closed and convex for some n≥1. Since the inequality Df(z,un)≤Df(z,xn) is equivalent to
(25)〈∇f(xn),z-xn〉-〈∇f(un),z-un〉≤f(un)-f(xn).
Therefore, we have(26)Cn+1={z∈Cn:〈∇f(xn),z-xn〉-〈∇f(un),z-un〉≤f(un)-f(xn)}.
This implies thatCn+1 is closed and convex. The desired conclusions are proved. These in turn show that PF(T)∩EP(g)fx and PCnfx are well defined.
(II) we prove thatG:=F(T)∩EP(g)⊂Cn for all n≥0.
Indeed, it is obvious thatG=F(T)∩EP(g)⊂C0=C. Suppose that G⊂Cn for some n∈ℕ. Let u∈G⊂Cn; since un=Resgf(yn), by Lemma 11 and (21), we have
(27)Df(u,un)=Df(u,Resgfyn)≤Df(u,yn)=Df(u,∇f*(αn∇f(xn)+(1-αn)∇f(Txn)))≤αnDf(u,xn)+(1-αn)Df(u,Txn)≤αnDf(u,xn)+(1-αn)Df(u,xn)=Df(u,xn).
Hence, we have u∈Cn+1. This implies that
(28)F(T)∩EP(g)⊂Cn,∀n∈ℕ∪{0}.
So, {xn} is well defined.
(III) We prove that{xn} is a bounded sequence in C.
By the definition ofCn, we have xn=PCnfx for all n≥0. It follows from Lemma 4(b) that
(29)Df(xn,x)=Df(PCnfx,x)≤Df(u,x)-Df(u,PCnfx)≤Df(u,x),∀n≥0,u∈G.
This implies that {Df(xn,x)} is bounded. By Lemma 7, {xn} is bounded. Since f:E→ℝ is uniformly Fréchet differentiable and bounded on bounded subsets of E, by Lemma 1∇f is uniformly continuous and bounded on bounded subsets of E. This implies that {∇f(xn)} is bounded.
(IV) Now we proved thatlimn→∞∥xn-Txn∥=0.
Fromxn+1∈Cn+1⊂Cn and xn=PCnfx, we have
(30)Df(xn,x)≤Df(xn+1,x),∀n∈ℕ∪{0}.
Thus, {Df(xn,x)} is nondecreasing. So, the limit of {Df(xn,x)} exists. Since Df(xn+1,xn)=Df(xn+1,PCnfx)≤Df(xn+1,x)-Df(PCnfx,x)=Df(xn+1,x)-Df(xn,x) for all n≥0, we have limn→∞Df(xn+1,xn)=0. From xn+1=PCn+1fx∈Cn+1, we have
(31)Df(xn+1,un)≤Df(xn+1,xn),∀n∈ℕ∪{0}.
Therefore, we have
(32)limn→∞Df(xn+1,un)=0.
From Lemma 5, we have
(33)limn→∞∥xn+1-xn∥=limn→∞∥xn+1-un∥=0.
So, we have
(34)limn→∞∥xn-un∥=0.
This means that the sequence {un} is bounded. Since f is uniformly Fréchet differentiable, it follows from Lemma 1 that ∇f is uniformly continuous. Therefore, we have
(35)limn→∞∥∇f(xn)-∇f(un)∥=0.
Sincef is uniformly Fréchet differentiable on bounded subsets of E, then f is uniformly continuous on bounded subsets of E (see [30, Theorem 1.8]). It follows that
(36)limn→∞|f(xn)-f(un)|=0.
From the definition of the Bregman distance, we obtain that(37)Df(u,xn)-Df(u,un)=[f(u)-f(xn)-〈∇f(xn),u-xn〉]-[f(u)-f(un)-〈∇f(un),u-un〉]=(f(un)-f(xn))+〈∇f(un)-∇f(xn),u-un〉+〈∇f(xn),xn-un〉
for any u∈G.
It follows from (34)–(37) that
(38)limn→∞(Df(u,xn)-Df(u,un))=0.
On the other hand, from un=Resgfyn and Lemma 11(v), for any u∈G we have that
(39)Df(un,yn)=Df(Resgfyn,yn)≤Df(u,yn)-Df(u,Resgfyn)≤Df(u,xn)-Df(u,Resgfyn)=Df(u,xn)-Df(u,un).
So, we have from (38) that
(40)limn→∞Df(un,yn)=0.
From Lemma 5, we have
(41)limn→∞∥un-yn∥=0.
So, from (34) and (41),we have
(42)limn→∞∥xn-yn∥=0.
This means that the sequence {yn} is bounded. Since f is uniformly Fréchet differentiable, it follows from Lemma 1 that
(43)limn→∞∥∇f(xn)-∇f(yn)∥=0.
Since f is uniformly Fréchet differentiable on bounded subsets of E, then f is uniformly continuous on bounded subsets of E (see [30]). It follows that
(44)limn→∞|f(xn)-f(yn)|=0.
From the definition of the Bregman distance, we obtain that(45)Df(u,yn)-Df(u,xn)=[f(u)-f(yn)-〈∇f(yn),u-yn〉]-[f(u)-f(xn)-〈∇f(xn),u-xn〉]=(f(xn)-f(yn))-〈∇f(yn)-∇f(xn),u-yn〉+〈∇f(xn),yn-xn〉
for any u∈G.
It follows from (42) to (45) that
(46)limn→∞(Df(u,yn)-Df(u,xn))=0.
On the other hand, for anyu∈G we have
(47)Df(u,yn)-Df(u,xn)=Df(u,∇f*(αn∇f(xn)+(1-αn)∇f(Txn)))-Df(u,xn)≤αnDf(u,xn)+(1-αn)Df(u,Txn)-Df(u,xn)=(1-αn)(Df(u,Txn)-Df(u,xn)).
This together with (46), (16), and limn→∞αn<1 shows that
(48)limk→∞(Df(u,Txn)-Df(u,xn))=0.
Since T is Bregman strongly nonexpansive, it follows from (48) that
(49)limn→∞∥xn-Txn∥=0.
(V) Next, we prove that every weak subsequential limit of{xn} belongs to G=F(T)∩EP(g).
Since{xn} is bounded, there exists a subsequence {xnk} of {xn} such that xnk⇀x*. Since T is a Bregman strongly nonexpansive mapping with F(T)=F^(T), we have x*∈F(T).
Fromxnk⇀x* and (34), we have unk⇀x*.
Byun=Resgfyn, we have
(50)g(un,y)+〈∇f(un)-∇f(yn),y-un〉≥0,∀y∈C.
Replacing n by nk, we have from (A2) that
(51)〈∇f(unk)-∇f(ynk),y-unk〉≥-g(unk,y)≥g(y,unk),∀y∈C.
Since g(x,·) is convex and lower semicontinuous, it is also weakly lower semicontinuous. So, letting k→∞, we have from (35), (43), and (A4) that
(52)g(y,x*)≤0,∀y∈C.
For t∈(0,1] and y∈C, letting yt=ty+(1-t)x*, there are yt∈C and g(yt,x*)≤0. By condition (A1) and (A4), we have
(53)0=g(yt,yt)≤tg(yt,y)+(1-t)g(yt,x*)≤tg(yt,y).
Dividing both sides of the above equation by t, we have g(yt,y)≥0, for all y∈C. Letting t↓0, from condition (A3), we have
(54)g(x*,y)≥0,∀y∈C.
Therefore, x*∈EP(g).
(VI)  Now, we provexn→PF(T)∩EP(g)fx.
Letw=PF(T)∩EP(g)fx. From w∈F(T)∩EP(g)⊂Cn+1, we have Df(xn+1,x)≤Df(w,x). Therefore, Lemma 8 implies that {xn} converges strongly to w=PF(T)∩EP(g)fx, as claimed. This completes the proof of Theorem 12.Corollary 13.
LetC be a nonempty, closed, and convex subset of a real reflexive Banach space E and f:E→ℝ a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of E. Let g be a bifunction from C×C to ℝ satisfying (A1)–(A4). Let {xn} be a sequence generated by x0=x∈C,C0=C, and
(55)un∈Csuchthatg(un,y)+〈∇f(un)-∇f(xn),y-un〉≥0,∀y∈C,Cn+1={z∈Cn:Df(z,un)≤Df(z,xn)},xn+1=PCn+1fx
for every n∈ℕ∪{0}. Then, {xn} converges strongly to PEP(g)fx, where PEP(g)f is the Bregman projection of E onto EP(g).Proof.
PuttingT=I in Theorem 12, we obtain Corollary 13.Corollary 14.
LetC be a nonempty, closed, and convex subset of a real reflexive Banach space E and f:E→ℝ a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of E. let T be a Bregman strongly nonexpansive mapping from C into itself such that F(T)=F^(T) and G=F(T)∩EP(g)≠∅. Let {xn} be a sequence generated by x0=x∈C,C0=C, and
(56)un=PCf∇f*(αn∇f(xn)+(1-αn)∇f(Txn)),Cn+1={z∈Cn:Df(z,un)≤Df(z,xn)},xn+1=PCn+1fx
for every n∈ℕ∪{0}, where {αn}⊂[0,1] satisfies liminfn→∞(1-αn)>0. Then, {xn} converges strongly to PF(T)fx, where PF(T)f is the Bregman projection of E onto F(T).Proof.
Puttingg(x,y)=0 for all x,y∈C in Theorem 12, we obtain Corollary 14.



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