# A Halpern-Type Iteration Method for Bregman Nonspreading Mapping and Monotone Operators in Reflexive Banach Spaces

Authors: F. U. Ogbuisi; L. O. Jolaoso; F. O. Isiogugu
Journal: Journal of Mathematics
 (2019)

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## Abstract

In this paper, we introduce an iterative method for approximating a common solution of monotone inclusion problem and fixed point of Bregman nonspreading mappings in a reflexive Banach space. Using the Bregman distance function, we study the composition of the resolvent of a maximal monotone operator and the antiresolvent of a Bregman inverse strongly monotone operator and introduce a Halpern-type iteration for approximating a common zero of a maximal monotone operator and a Bregman inverse strongly monotone operator which is also a fixed point of a Bregman nonspreading mapping. We further state and prove a strong convergence result using the iterative algorithm introduced. This result extends many works on finding a common solution of the monotone inclusion problem and fixed-point problem for nonlinear mappings in a real Hilbert space to a reflexive Banach space.


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## Body

## 1. Introduction

LetE be a real reflexive Banach space with a norm ⋅ and E∗ be the dual space of E. We denote the value of x∗∈E∗ at x∈E by x∗,x. A mapping A is called a monotone mapping if for any x,y∈domA, we have(1)μ∈Ax,ν∈Ay⟹μ−ν,x−y≥0.A monotone mappingA:E⟶2E∗ is said to be maximal monotone if its graph, GA≔x,u∈E×E∗:u∈Ax, is not properly contained in the graph of any other monotone operator. A basic problem that arises in several branches of applied mathematics [1–7] is to find x∈E such that(2)0∈Ax.One of the methods for solving this problem is the well-known proximal point algorithm (PPA) introduced by Martinet [8]. Let H be a Hilbert space and let I denote the identity operator on H. The PPA generates for any starting point x0=x∈H, a sequence xn in H by(3)xn+1=I+λnA−1xn,n=1,2,…,where A is a maximal monotone mapping and λn is a given sequence of positive real numbers. It has been observed that (3) is equivalent to(4)0∈Axn+1+1λnxn+1−xn,n=1,2,….This algorithm was further developed by Rockafellar [5], who proved that the sequence generated by (3) converges weakly to an element of A−10 when A−10 is nonempty and liminfn⟶∞λn>0. Furthermore, Rockafellar [5] asked if the sequence generated by (3) converges strongly in general. This question was answered in the negative by Güler [9] who presented an example of a subdifferential for which the sequence generated by (3) converges weakly but not strongly. Also, the works of Bruck and Reich [10] and Bauschke et al. [11] are very important in this direction. For more recent results on PPA, see [12–14].The problem of finding the zeros of the sum of two monotone mappingsA and B, is to find a point x∗∈E such that(5)0∈A+Bx∗,has recently received attention due to its significant importance in many physical problems. One classical method for solving problem (5) is the forward-backward splitting method [15], which is as follows: for x1∈E,(6)xn+1=I+rB−1xn−rAxn,n≥1,where r>0. This method combines the proximal point algorithm and the gradient projection algorithm. In [16], Lions and Mercier introduced the following splitting iterative methods in a real Hilbert space H:(7)xn+1=2JrA−I2JrB−Ixn,n≥1,xn+1=JrA2JrB−Ixn+I−JrBxn,n≥1,where JrT=I+rT−1. The first one is called Peaceman–Rachford algorithm and the second one is called Douglas–Rachford algorithm [15]. It was noted that both algorithms converge weakly in general [16, 17].Many authors have studied the approximation of zero of the sum of two monotone operators (in Hilbert space) and accretive operators (in Banach spaces), but the approximation of the sum of two monotone operators in more general Banach spaces other the Hilbert spaces has not enjoyed such popularity.Throughout this paper,f:E⟶−∞,+∞ is a proper lower semicontinuous and convex function, and the Fenchel conjugate of f is the function f∗:E∗⟶−∞,+∞ defined by(8)f∗x∗=supx∗,x−fx:x∈E.We denote by domf the domain of f, that is, the set x∈E:fx<+∞. For any x∈intdomf and y∈E, the right-hand derivative of f at x in the direction of t is defined by(9)fox,y≔limt⟶0+fx+ty−fxt.The functionf is said to be Gâteaux differentiable at x if the limit as t⟶0+ in (9) exists for any y. In this case, fox,y coincides with ∇fx, the value of the gradient ∇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 Fréchet differentiable at x if the limit is attained with y=1 and uniformly Fréchet differentiable on a subset C of E if the limit is attained uniformly for x∈C and y=1.The functionf is said to be Legendre if it satisfies the following two conditions:(L1) intdomf≠∅ and the subdifferential ∂f is single-valued in its domain(L2) tdomf∗≠∅ and ∂f∗ is single-valued on its domainThe class of Legendre functions in infinite dimensional Banach spaces was first introduced and studied by Bauschke et al. in [18]. Their definition is equivalent to conditions (L1) and (L2) because the space E is assumed to be reflexive (see [18], Theorems 5.4 and 5.6, p. 634). It is well known that in reflexive Banach spaces, ∇f=∇f∗−1 (see [19], p. 83). When this fact is combined with conditions (L1) and (L2), we obtain(10)ran∇f=dom∇f∗=intdomf∗,ran∇f∗=dom∇f=intdomf.It also follows thatf is Legendre if and only if f∗ is Legendre (see [18], Corollary 5.5, p. 634) and that the functions f and f∗ are Gâteaux differentiable and strictly convex in the interior of their respective domains.Several interesting examples of the Legendre functions are presented in [18, 20, 21]. A very important example of Legendre function is the function 1/s⋅s with s∈1,∞, where the Banach space E is smooth and strictly convex, and in particular, a Hilbert space. Throughout this article, we assume that the convex function f:E⟶−∞,+∞ is Legendre.Definition 1.
Letf:E⟶−∞,+∞ be a convex and Gâteaux differentiable function, the function Df: domf× intdomf⟶0,∞ which is defined by(11)Dfy,x≔fy−fx−∇fx,y−x,is called the Bregman distance [22–24].The Bregman distance does not satisfy the well-known metric properties, but it does have the following important property, which is called the three-point identity: for anyx∈domf and y,z∈intdomf,(12)Dfx,y+Dfy,z−Dfx,z=∇fz−∇fy,x−y.LetC be a nonempty subset of a Banach space E and T:C⟶C be a mapping, then a point x is called fixed point of T if Tx=x. The set of fixed point of T is denoted by FT. Also, a point x∗∈C is said to be an asymptotic fixed point of T if C contains a sequence xnn=1∞ which converges weakly to x∗ and limn⟶∞xn−Txn=0 [25]. The set of asymptotic fixed points of T is denoted by F^T.Definition 2 [26, 27].
LetC be a nonempty, closed, and convex subset of E. A mapping T:C⟶intdomf is called(i)
Bregman firmly nonexpansive (BFNE for short) if(13)∇fTx−∇fTy,Tx−Ty≤∇fx−∇fy,Tx−Ty,∀x,y∈C.(ii)
Bregman strongly nonexpansive (BSNE) with respect to a nonemptyF^T if(14)Dfp,Tx≤Dfp,x,for allp∈F^T and x∈C and if whenever xnn=1∞⊂C is bounded, p∈F^T and(15)limn⟶∞Dfp,xn−Dfp,Txn=0,it follows that(16)limn⟶∞DfTxn,xn=0.(iii)
Bregman quasi-nonexpansive ifFT≠∅ and(17)Dfp,Tx≤Dfp,x,∀x∈Candp∈FT.(iv)
Bregman skew quasi-nonexpansive ifFT≠∅ and(18)DfTx,p≤Dfx,p,∀x∈Candp∈FT.(v)
Bregman nonspreading if(19)DfTx,Ty+DfTy,Tx≤DfTx,y+DfTy,x,∀x,y∈C.
It is easy to see that every Bregman nonspreading mappingT with FT≠∅ is Bregman quasi-nonexpansive. Also Bregman nonspreading mappings include, in particular, the class of nonspreading functions studied by Takahashi et al. in [28, 29]. For more information on Bregman nonspreading mappings, see [30].
In a real Hilbert spaceH, the nonlinear mapping T:C⟶C is said to be(i)
Nonexpansive if(20)Tx−Ty≤x−y,∀x,y∈C.(ii)
Quasi-nonexpansive ifFT≠∅ and(21)Tx−p≤x−p,∀x∈Candp∈FT.(iii)
Nonspreading if(22)2Tx−Ty2≤Tx−y2+Ty−x2,∀x,y∈C.
Clearly, every nonspreading mappingT with FT≠∅ is also quasi-nonexpansive mapping. The class of nonspreading mappings is very important due to its relation with maximal monotone operators (see, e.g., [28]).
LetB:E⟶2E∗ be a maximal monotone operator. The resolvent of B, ResBf:E⟶2E, is defined by (see [26])(23)ResBf≔∇f+B−1∘∇f.
It is known thatResBf is a BFNE operator, single-valued, and FResBf=B−10∗ (see [26]). If f:E⟶ℝ is a Legendre function which is bounded, uniformly Fréchet differentiable on bounded subsets of E, then ResBf is BSNE and F^ResBf=FResBf (see [31]).
Assume that the Legendre functionf satisfies the following range condition:(24)ran∇f−A⊆ran∇f.
An operatorA:E⟶2E∗ is called Bregman inverse strongly monotone (BISM) if (domA) (domf), and for any x,y∈intdomf and each u∈Ax and v∈Ay, we have(25)u−v,∇f∗∇fx−u−∇f∗∇fy−v≥0.
The class of BISM mappings is a generalization of the class of firmly nonexpansive mappings in Hilbert spaces. Indeed, iff=1/2⋅2, then ∇f=∇f∗=I, where I is the identity operator and (25) becomes(26)u−v,x−u−y−v≥0,which means(27)u−v2≤x−y,u−v.
Observe that(28)domAf=domA∩intdomf,ranAf⊂intdomf.
In other words,T is a (single-valued) firmly nonexpansive operator.
For any operatorA:E⟶2E∗, the antiresolvent operator Af:E⟶2E of A is defined by(29)Af≔∇f∗∘∇f−A.
It is known that the operatorA is BISM if and only if the antiresolvent Af is a single-valued BFNE (see [32], Lemma 3.2(c) and (d), p. 2109) and FAf=A−10∗. For examples and further information on BISM, see [32].
Since the monotone inclusion problems have very close connections with both the fixed-point problems and the equilibrium problems, finding the common solutions of these problems has drawn many people’s attention and has become one of the hot topics in the related fields in the past few years [33, 34]. Furthermore, interest in finding the common solution of these problems has also grown because of the possible application of these problems to mathematical models whose constraints can be present as fixed points of mappings and/or monotone inclusion problems and/or equilibrium problems. Such a problem occurs, in particular, in the practical problems as signal processing, network resource allocation, and image recovery (see [35, 36]).
In this paper, we introduce an iterative method for approximating a common solution of monotone inclusion problem and fixed point of Bregman nonspreading mapping in a reflexive Banach space and prove a strong convergence of the sequence generated by our iterative algorithm. This result extends many works on finding common solution of monotone inclusion problem and fixed problem of nonlinear mapping in a real Hilbert space to a reflexive Banach space.


## 2. Preliminaries

The Bregman projection [22] of x∈intdomf onto the nonempty, closed, and convex subset C ⊂int (domf is defined as the necessarily unique vector ProjCfx∈C satisfying(30)DfProjCfx,x=infDfy,x:y∈C.It is known from [37] that z=ProjCfx if and only if(31)∇fx−∇fz,y−z≤0,for ally∈C.We also have(32)Dfy,ProjCfx+DfProjCfx,x≤Dfy,x,for allx∈E,y∈C.Note that ifE is a Hilbert space and fx=1/2x2, then the Bregman projection of x onto C, i.e., argminy−x:y∈C, is the metric projection PC.Lemma 1 [37].
Letf be totally convex on int (domf). Let C be a nonempty, closed, and convex subset of int (domf) and x∈intdomf; if z∈C, then the following conditions are equivalent:(i)
z=ProjCfx(ii)
∇fx−∇fz,z−y≥0forally∈C(iii)
Dfy,z+Dfz,x≤Dfy,xforally∈CLetf:E⟶ℝ∪+∞ be a convex and Gâteaux differentiable function. The function f is said to be totally convex at x∈intdomf if its modulus of totally convexity at x, that is, the function vf: intdomf×0,+∞ defined by(33)vfx,t≔infDfy,x:y∈domf,y−x=t,is positive for any t>0. The function f is said to be totally convex when it is totally convex at every point x∈intdomf. In addition, the function f is said to be totally convex on bounded set if vfB,t is positive for any nonempty bounded subset B, where the modulus of total convexity of the function f on the set B is the function vf: intdomf×0,+∞ defined by(34)vfB,t≔infvfx,t:x∈B∩domf.For further details and examples on totally convex functions, see [37–39].Letf:E⟶ℝ be a convex, Legendre, and Gâteaux differentiable function and let the function Vf:E×E∗⟶0,∞ associated with f (see [23, 40]) be defined by(35)Vfx,x∗=fx−x∗,x+f∗x∗,∀x∈E,x∗∈E∗.ThenVf is nonnegative and Vfx,x∗=Dfx,∇f∗x∗,∀x∈E,x∗∈E∗. Furthermore, by the subdifferential inequality, we have (see [41])(36)Vfx,x∗+y∗,∇f∗x∗−x≤Vfx,x∗+y∗,∀x∈E,x∗,y∗∈E∗.In addition, iff:E⟶−∞,+∞ is a proper lower semicontinuous function, then f∗:E∗⟶−∞,+∞ is a proper weak∗ lower semicontinuous and convex function (see [42]). Hence, Vf is convex in the second variable. Thus, for all z∈E,(37)Dfz,∇f∗∑i=1Nti∇fxi≤∑i=1NtiDfz,xi,where xii=1N⊂E and ti⊂0,1 with ∑i=1Nti=1.Lemma 2 (see [43]).
Letr>0 be a constant and let f:E⟶ℝ be a continuous uniformly convex function on bounded subsets of E. Then(38)f∑k=0∞αkxk≤∑k=0∞αkfxk−αiαjρrxi−xj,for all i,j∈ℕ∪0, xk∈Br, αk∈0,1, and k∈ℕ∪0 with ∑k=0∞αk=1, where ρr is the gauge of uniform convexity of f.Recall that a functionf is said to be sequentially consisted (see [37]) if for any two sequences xn and yn in E such that the first one is bounded,(39)limn⟶∞Dfyn,xn=0⟹limn⟶∞yn−xn=0.The following lemma follows from [44].Lemma 3.
If domf contains at least two points, then the function f is totally convex on bounded sets if and only if the function f is sequentially consistent.Lemma 4 (see [45]).
Letf:E⟶−∞,+∞ be a Legendre function and let A:E⟶2E∗ be a BISM operator such that A−10∗≠∅. Then the following statements hold:(i)
A−10∗=FAf(ii)
For anyw∈A−10∗ and x∈domAf, we have(40)Dfw,Afx+DfAfx,x≤Dfw,x.Remark 1.
If the Legendre functionf is uniformly Fréchet differentiable and bounded on bounded subsets of E, then the antiresolvent Af is a single-valued BSNE operator which satisfies FAf=F^Af (cf. [31]).Lemma 5 (see [46]).
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∗.Lemma 6 (see [44]).
Letf:E⟶ℝ be a Gâteaux differentiable and totally convex function. If x1∈E and the sequence Dfxn,x1 is bounded, then the sequence xn is also bounded.Lemma 7 (see [45]).
Assume thatf:E⟶ℝ is a Legendre function which is uniformly Fréchet differentiable and bounded on bounded subset of E. Let C be a nonempty, closed, and convex subset of E. Let Ti:1≤i≤N be BSNE operators which satisfy F^Ti=FTi for each 1≤i≤N and let T≔wnTN−1…T1. If(41)∩FTi:1≤i≤N,and FT are nonempty, then T is also BSNE with FT=F^T.Lemma 8 (Demiclosedness principle [30]).
LetC be a nonempty subset of a reflexive Banach space. Let g:E⟶ℝ be a strict convex, Gâteaux differentiable, and locally bounded function. Let T:C⟶E be a Bregman nonspreading mapping. If xn⇀p in C and limn⟶∞Txn−xn=0, then p∈FT.Lemma 9 (see [47]).
Assumean is a sequence of nonnegative real numbers satisfying(42)an+1≤1−tnan+tnδn∀n≥0,where tn is a sequence in 0,1 and δn is a sequence in ℝ such that(i)
∑n=o∞tn=∞(ii)
limsupn⟶∞δn≤0
Then,limn⟶∞an=0.Lemma 10 [48].
Letan be a sequence of real numbers such that there exists a nondecreasing subsequence ni of n, that is, ani≤ani+1 for all i∈ℕ. Then there exists a nondecreasing sequence mk⊂ℕ such that mk⟶∞, and the following properties are satisfied for all (sufficiently large number k∈ℕ): amk≤amk+1 and ak≤amk+1, mk=maxj≤k:aj≤aj+1.


## 3. Main Results

Theorem 1.
LetC be a nonempty, closed, and convex subset of a real reflexive Banach space E and f:E⟶ℝ a Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of E. Let A:E⟶2E∗ be a Bregman inverse strongly monotone operator, B:E⟶2E∗ be a maximal monotone operator, and T:C⟶C be a Bregman nonspreading mapping. Suppose Γ≔FResAf∘Af∩FT≠∅. Let γn⊂0,1 and αn,βn, and δn be sequences in 0,1 such that αn+βn+δn=1. Given u∈E and x1∈C arbitrarily, let xn and yn be sequences in E generated by(43)yn=∇f∗γn∇fxn+1−γn∇fTxn,xn+1=ProjCf∇f∗αn∇fu+βn∇fyn+δn∇fResBf∘Afyn,n≥1.Suppose the following conditions are satisfied:(i)
limn⟶∞αn=0 and ∑n=1∞αn=∞(ii)
1−αna<δn,αn≤b<1,a∈0,1/2(iii)
0≤c<liminfn⟶∞γn≤limsupn⟶∞γn<1Thenxn converges strongly to ProjΓfu, where ProjΓf is the Bregman projection of E onto Γ.Proof.
First we observe thatFResBf∘Af=A+B−10 and FResBf∘Af=FResBf∩FAf. Thus, since ResBf and Af are BSNE operators and FResBf∩FAf=A+B−10≠∅, it then follows from Lemma 7 that ResBf∘Af is BSNE and FResBf∘Af=F^ResBf∘Af.
We next show thatxn and yn are bounded.
Letp∈Γ, then from (43), we have(44)Dfp,yn=Dfp,∇f∗γn∇fxn+1−γn∇fTxn≤γnDfp,xn+1−γnDfp,Txn≤γnDfp,xn+1−γnDfp,xn=Dfp,xn.
Also(45)Dfp,xn+1≤Dfp,∇f∗αn∇fu+βn∇fyn+δn∇fResBf∘Afyn≤αnDfp,u+βnDfp,yn+δnDfp,ResBf∘Afyn≤αnDfp,u+βnDfp,yn+δnDfp,yn=αnDfp,u+1−αnDfp,yn=αnDfp,u+1−αnDfp,xn≤maxDfp,u,Dfp,xn⋮≤maxDfp,u,Dfp,x1.
HenceDfp,xn is bounded. Therefore, by Lemma 6, xn is also bounded, and consequently, yn is also bounded.
We now show thatxn converges strongly to x¯=ProjΓfu. To do this, we first show that if there exists a subsequence xni of xn such that xni⇀q∈C, then q∈Γ.
Lets=sup∇fxn,∇fTxn and ρs∗:E∗⟶ℝ be the gauge of uniform convexity of the conjugate function f∗. From Lemma 2 and (9), we have(46)Dfp,yn≤Dfp,∇f∗γn∇fxn+1−γn∇fTxn=Vfp,γn∇fxn+1−γn∇fTxn=fp−p,γn∇fxn+1−γn∇fTxn+f∗γn∇fxn+1−γn∇fTxn≤γnfp−γnp,∇fxn+γnf∗∇fxn+1−γnfp−1−γnp,∇fTxn+1−γnf∗∇fTxn−γn1−γnρs∗∇fxn−∇fTxn=γnDfp,xn+1−γnDfp,Txn−γn1−γnρs∗∇fxn−∇fTxn≤Dfp,xn−γn1−γnρs∗∇fxn−∇fTxn.
Thus, from (45), we have(47)Dfp,xn+1≤αnDfp,u+1−αnDfp,xn−γn1−γnρs∗∇fxn−∇fTxn.
We consider the following two cases for the rest of the proof.Case A.
SupposeDfp,xn is monotonically nonincreasing. Then, Dfp,xn converges and Dfp,xn−Dfp,xn+1⟶0 as n⟶∞. Thus, from (47), we have(48)1−αn1−γnγnρs∗∇fxn−∇fTxn≤αnDfp,u−Dfp,xn+Dfp,xn−Dfp,xn+1.
Sinceαn⟶0, n⟶∞, then we have(49)limn⟶∞γn1−γnρs∗∇fxn−∇fTxn=0,and hence, by condition (iii) and the property of ρs∗, we have(50)limn⟶∞∇fxn−∇fTxn=0.
Since∇f∗ is uniformly norm-to-norm continuous on bounded subset of E∗, we have(51)limn⟶∞xn−Txn=0.
Again(52)∇fxn−∇fyn=∇fxn−γn∇fxn+1−γn∇fTxn=1−γn∇fxn−∇fTxn⟶0,n⟶∞.
Since∇f∗ is uniformly norm-to-norm continuous on bounded subsets of E∗, we have that(53)limn⟶∞xn−yn=0.
Now, letwn=∇f∗βn/1−αn∇fyn+δn/1−αn∇fResBf∘Afyn, then(54)Dfp,wn=Dfp,∇f∗βn1−αn∇fyn+δn1−αn∇fResBf∘Afyn≤βn1−αnDfp,yn+δn1−αnDfp,ResBf∘Afyn≤βn+δn1−αnDfp,yn=Dfp,yn.
Therefore, we have(55)0≤Dfp,xn−Dfp,wn=Dfp,xn−Dfp,xn+1+Dfp,xn+1−Dfp,wn≤Dfp,xn−Dfp,xn+1+αnDfp,u+1−αnDfp,wn−Dfp,wn=Dfp,xn−Dfp,xn+1+αnDfp,u−Dfp,wn⟶0,asn⟶∞.
More so(56)Dfp,wn≤βn1−αnDfp,yn+δn1−αnDfp,ResBf∘Afyn=Dfp,yn−1−βn1−αnDfp,yn+δn1−αnDfp,ResBf∘Afyn≤Dfp,xn+δn1−αnDfp,ResBf∘Afyn−Dfp,yn.
Since1−αna<δn and αn≤b<1, we have(57)aDfp,yn−Dfp,ResBf∘Afyn<δn1−αnDfp,yn−Dfp,ResBf∘Afyn≤Dfp,xn−Dfp,wn⟶0,asn⟶∞.
Thus,(58)Dfp,yn−Dfp,ResBf∘Afyn⟶0,asn⟶∞.
Therefore, sinceResBf∘Af is BSNE, we have that limn⟶∞Dfyn,ResBf∘Afyn=0, which implies that(59)limn⟶∞yn−ResBf∘Afyn=0.
Settingun=∇f∗αn∇fu+βn∇fyn+δn∇fResBf∘Afyn, for each n≥1, we have(60)Dfyn,un=Dfyn,∇f∗αn∇fu+βn∇fyn+δn∇fResBf∘Afyn≤αnDfyn,u+βnDfyn,yn+δnDfyn,ResBf∘Afyn⟶0.
Thus,(61)limn⟶∞yn−un=0.
Therefore, from (47), we have(62)un−xn≤un−yn+yn−xn⟶0,n⟶∞.
Moreover, sincexn+1=ProjCfun, then(63)Dfp,xn+1+Dfxn+1,un≤Dfp,un,and therefore, we have that(64)Dfxn+1,un≤Dfp,un−Dfp,xn+1≤αnDfp,u+βnDfp,yn+δnDfp,ResBf∘Afyn−Dfp,xn+1=αnDfp,u+1−αnDfp,yn−Dfp,xn+1≤αnDfp,u−Dfp,xn+Dfp,xn−Dfp,xn+1⟶0,n⟶∞,which implies(65)xn+1−un⟶0,n⟶∞.
Hence,(66)xn+1−xn≤xn+1−un+un−xn⟶0,n⟶∞.
Sincexn is bounded, there exists a subsequence xni of xn such that xni converges weakly to q∈C as n⟶∞. Since limn⟶∞xni−Txni=0, it follows from Lemma 8 that q∈FT. Also, since xni−yni⟶0, it implies that yni also converges weakly to q∈E. Therefore, from (59), we have that q∈FResBf∘Af, and hence, q∈Γ=FT∩FResBf∘Af.
Next, we show thatxn converges strongly to x¯=ProjΓfu.
Now from (43), we have(67)Dfx¯,xn+1≤Dfx¯,∇f∗αn∇fu+βn∇fyn+δn∇fResBf∘Afyn=Vfx¯,αn∇fu+βn∇fyn+δn∇fResBf∘Afyn≤Vfx¯,αn∇fu+βn∇fyn+δn∇fResBf∘Afyn−αn∇fu−∇fx¯−−αn∇fu−∇fx¯,∇f∗αn∇fu+βn∇fyn+δn∇fResBf∘Afyn−x¯=Vfx¯,αn∇fx¯+βn∇fyn+δn∇fResBf∘Afyn+αn∇fu−∇fw,un−x¯=Dfx¯,∇f∗αn∇fx¯+βn∇fyn+δn∇fResBf∘Afyn+αn∇fu−∇fx¯,un−x¯=αnDfx¯,x¯+βnDfx¯,yn+δnDfx¯,ResBf∘Afyn+αn∇fu−∇fx¯,un−x¯≤βnDfx¯,yn+δnDfx¯,yn+αn∇fu−∇fx¯,un−x¯=1−αnDfx¯,yn+αn∇fu−∇fx¯,un−x¯≤1−αnDfx¯,xn+αn∇fu−∇fx¯,un−x¯.
Choose a subsequencexnj of xn such that(68)limsupn⟶∞∇fu−∇fx¯,xn−x¯=limj⟶∞∇fu−∇fx¯,xnj−x¯.
Sincexnj⇀q, it follows from Lemma 1(ii) that(69)limsupn⟶∞∇fu−∇fx¯,xn−x¯=limj⟶∞∇fu−∇fx¯,xnj−x¯=∇fu−∇fx¯,q−x¯≤0.
Sinceun−xn⟶0, n⟶∞, then,(70)limsupn⟶∞∇fu−∇fx¯,un−x¯≤0.
Hence, by Lemma 9 and (67), we conclude that Dfx¯,xn⟶0, n⟶∞. Therefore, xn converges strongly to x¯=ProjΓfu.Case B.
Suppose that there exists a subsequencenj of n such that(71)Dfxnj,w<Dfxnj+1,w,for all j∈ℕ. Then, by Lemma 10, there exists a nondecreasing sequence mk⊂ℕ with mk⟶∞ as n⟶∞ such that(72)Dfp,xmk≤Dfp,xmk+1,Dfp,xk≤Dfp,xmk+1,for all k∈ℕ. Following the same line of arguments as in Case I, we have that(73)limk⟶∞Txmk−xmk=0,limk⟶∞ResBfAfymk−ymk=0,limk⟶∞wmk−xmk=0,limsupk⟶∞∇fu−∇fp,wmk−p≤0.
From (67), we have(74)Dfp,xmk+1≤1−αmkDfp,xmk+αmk∇fu−∇fp,wmk−p.
SinceDfp,xmk≤Dfp,xmk+1, it follows from (74) that(75)αmkDfp,xmk≤Dfp,xmk−Dfp,xmk+1+αmk∇fu−∇fx∗,wmk−p≤αmk∇fu−∇fp,wmk−p.
Sinceαmk>0, we obtain(76)Dfp,xmk≤∇fu−∇fp,wmk−p.
Then from (73), it follows that Dfp,xmk⟶0 as k⟶∞. Combining Dfp,xmk⟶0 with (74), we obtain Dfp,xmk+1⟶0 as k⟶∞. Since Dfp,xk≤Dfp,xmk+1 for all k∈ℕ, we have xk⟶p as k⟶∞, which implies that xn⟶p as n⟶∞.
Therefore, from the above two cases, we conclude thatxn converges strongly to x¯=ProjΓfu.
This completes the proof.Corollary 1.
LetC be a nonempty, closed, and convex subset of a real reflexive Banach space E and f:E⟶ℝ a Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of E. Let A:E⟶2E∗ be a Bregman inverse strongly monotone operator, B:E⟶2E∗ be a maximal monotone operator, and T:C⟶C be a Bregman firmly nonexpansive mapping. Suppose Γ≔FResAf∘Af∩FT≠∅. Let γn⊂0,1 and αn,βn, and δn be sequences in 0,1 such that αn+βn+δn=1. Given u∈E and x1∈C arbitrarily, let xn and yn be sequences in E generated by(77)yn=∇f∗γn∇fxn+1−γn∇fTxn,xn+1=ProjCf∇f∗αn∇fu+βn∇fyn+δn∇fResBf∘Afyn,n≥1.Suppose the following conditions are satisfied:(i)
limn⟶∞αn=0 and ∑n=1∞αn=∞(ii)
1−αna<δn,αn≤b<1,a∈0,1/2(iii)
0≤c<liminfn⟶∞γn≤limsupn⟶∞γn<1Then,xn converges strongly to ProjΓfu, where ProjΓf is the Bregman projection of E onto Γ.Corollary 2.
LetC be a nonempty, closed, and convex subset of a real Hilbert space H. Let A:H⟶H be a single-valued 1-inverse strongly monotone operator, B:E⟶2E∗ be a maximal monotone operator, and T:C⟶C be a firmly nonexpansive mapping. Suppose Γ≔FI+B−1I−A∩FT≠∅. Let γn⊂0,1 and αn,βn, and δn be sequences in 0,1 such that αn+βn+δn=1. Given u∈E and x1∈C arbitrarily, let xn and yn be sequences in E generated by(78)yn=γnxn+1−γnTxn,xn+1=PCαnu+βnyn+δnI+B−1I−Ayn,n≥1.Suppose the following conditions are satisfied:(i)
limn⟶∞αn=0 and ∑n=1∞αn=∞(ii)
1−αna<δn,αn≤b<1,a∈0,1/2(iii)
0≤c<liminfn⟶∞γn≤limsupn⟶∞γn<1Then,xn converges strongly to PΓu, where PΓ is the metric projection of H onto Γ.


## 4. Application

In this section, we apply our result to obtain a common solution of variational inequality problem (VIP) and equilibrium problem (EP) in real reflexive Banach spaces.LetC be a nonempty, closed, and convex subset of a real reflexive Banach space E. Suppose g:C×C⟶ℝ is a bifunction that satisfies the following conditions:A1gx,x=0,∀x∈CA2gx,y+gy,x≤0,∀x,y∈CA3limsupt↓0gtz+1−tx,y≤gx,y,∀x,y,z∈CA4gx,. is convex and lower semicontinuous, for each x∈C.The equilibrium problem with respect tog is to find x¯∈C such that(79)gx¯,y≥0,∀y∈C.We denote the set of solutions of (79) by EP(g). The resolvent of a bifunction g:C×C⟶ℝ that satisfies A1−A4 (see [49]) is the operator Tgf:E⟶2C defined by(80)Tgfx≔z∈C:gz,y+∇fz−∇fx,y−z≥0,∀y∈C.Lemma 11 ([27], Lemma 1, 2).
Letf:E⟶−∞,∞ be a coercive Legendre function and let C be a nonempty, closed, and convex subset of E. Suppose the bifunction g:C×C⟶ℝ satisfies A1−A4, then(1)
domTgf=E.(2)
Tgf is single valued(3)
Tgf is Bregman firmly nonexpansive(4)
FTgf=EPg(5)
EPg is a closed and convex subset of C(6)
Dfu,Tgfx+DfTgfx,x≤Dfu,x, for all x∈E and for all u∈FTgfLetA:E⟶E∗ be a Bregman inverse strongly monotone mapping and let C be a nonempty, closed, and convex subset of domA. The variational inequality problem corresponding to A is to find x∈C, such that(81)Ax∗,y−x∗≥0,∀y∈C.The set of solutions of (81) is denoted by VIC,A.Lemma 12 (see [25, 46]).
LetA:E⟶E∗ be a Bregman inverse strongly monotone mapping and f:E⟶−∞,∞ be a Legendre and totally convex function that satisfies the range condition. If C is a nonempty, closed, and convex subset of domA∩intdomf, then(1)
PCf∘Af is Bregman relatively nonexpansive mapping(2)
FPCf∘Af=VIC,ANow letiC be the indicator function of a closed convex subset C of E, defined by(82)iCx=0,x∈C,+∞,otherwise.The subdifferential of the indicator function∂iCx¯=NCx¯, where C is a closed subset of a Banach space E and NC⊂E∗ is the normal cone defined by(83)NCx¯=v∈E∗:v,x−x¯≤0,for allx∈C,x¯∈C,∅,x∉C.The normal coneNC is maximal monotone and the resolvent of the normal cone corresponds to the Bregman projection (see [50], Example 4.4) that is ResNCf=ProjCf.Therefore, if we letB=NC and T=Tgf, then the iterative algorithm (77) becomes(84)yn=∇f∗γn∇fxn+1−γn∇fTgfxn,xn+1=ProjCf∇f∗αn∇fu+βn∇fyn+δn∇fProjCf∘Afyn,n≥1.Thus, from Corollary 1, we obtain a strong convergence result for approximating a pointx∈VIC,A∩EPg.



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*Source: 8059135-2019-09-02.xml*