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Kyungpook Mathematical Journal 2021; 61(3): 523-558

Published online September 30, 2021

Convergence Theorem for Finding Common Fixed Points of N-generalized Bregman Nonspreading Mapping and Solutions of Equilibrium Problems in Banach Spaces

Lateef Olakunle Jolaoso, Oluwatosin Temitope Mewomo*

School of Mathematics, Statistics and Computer Science, University of KwaZulu-Natal, Durban, South Africa
e-mail : jollatanu@yahoo.co.uk

School of Mathematics, Statistics and Computer Science University of Kwazulu-Natal Durban, South Africa
e-mail : mewomoo@ukzn.ac.za

Received: May 4, 2019; Revised: July 28, 2020; Accepted: August 4, 2020

In this paper, we study some fixed point properties of n-generalized Bregman nonspreading mappings in re exive Banach space. We introduce a hybrid iterative scheme for finding a common solution for a countable family of equilibrium problems and fixed point problems in re exive Banach space. Further, we give some applications and numerical example to show the importance and demonstrate the performance of our algorithm. The results in this paper extend and generalize many related results in the literature.

Keywords: nonspreading mapping, Bregman distance, equilibrium problem, fixed point prolem, reflexive Banach space.

Let E be a real Banach space, and C be a nonempty, closed and convex subset of E. Let $g:C×C→ℝ$ be a bifunction, the Equilibrium Problem with respect to g denoted by EP(g) is define as finding a point $z∈C$ such that

$(z,y)≥0, ∀y∈C.$

The EP(g) was shown by Blum and Oettli [7] to cover several other optimization problems such as monotone inclusion problems, saddle point problems, minimization problems, variational inequality problems and Nash equilibria in non-cooperative games. In addition, there are many other important problems, for example, the complementarity problem and fixed point problems, which can be written in the form of EP(g) (1.1). Thus, the EP(g) is a unifying model for several problems arising in physics, engineering, science, optimization, economics etc.

In the last two decades, the existence of solutions of the EP(g) have been mentioned in many papers, see for instance [7, 11, 13, 26, 36, 39, 40 ], and several iterative methods have been proposed for solving EP(g) and related optimization problems, see for instance [1, 2, 4, 14, 15, 17, 18, 19, 28, 29, 32, 30, 31, 38, 41, 42] and reference therein. In solving the EP(g) (1.1) it is necessary to assume that the bifunction g satisfies the following assumptions:

• (A1) g(x,x) =0 for all x ∈ C;

• (A2) g is monotone, that is g(x,y)+g(y,x) ≤ 0 for all x,y ∈ C;

• (A3) For all x,y, z∈ C

$limsupt↓0+g(tz+(1−t)x,y)≤g(x,y);$

• (A4) For all x ∈ C, g(x,⋅) is convex and lower semicontinuous.

Definition 1.1. Let $f:E→(−∞,+∞]$ be a convex and Gâteaux differentiable function. The function defined by

$Df(y,x)=f(y)−f(x)−⟨∇f(x),y−x⟩$

is called a Bregman distance with respect to f.

From the definition, we know that the following properties are satisfied (see [6]):

(i) The three points identity, for any and

$Df(x,y)+Df(y,z)−Df(x,z)=⟨∇f(z)−∇f(y),x−y⟩;$

(ii) Four point identity, for any and

$Df(x,y)−Df(x,z)−Df(w,y)+Df(w,z)=⟨∇f(z)−∇f(y),x−w⟩.$

Definition 1.2. Let C be a nonempty closed convex subset of $int(domf)$ and $T:C→C$ be a mapping. A point x ∈ C is called a fixed point of T if Tx =x. We denote the set of all fixed points of T by F(T). The mapping $T:C→C$ is called

(a) Bregman nonexpansive [33] if

$Df(Tx,Ty)≤Df(x,y) ∀x,y∈C;$

Bregman nonspreading [23] if

$Df(Tx,Ty)+Df(Ty,Tx)≤Df(Tx,y)+Df(Ty,x), ∀x,y∈C,$

(c) $(α,β,γ,δ)$-generalized Bregman nonspreading [3, 16] if there exist $α,β,γ,δ∈ℝ$ such that

$αDf(Tx,Ty)+(1−α)Df(x,Ty)+γ{Df(Ty,Tx)−Df(Ty,x)} ≤βDf(Tx,y)+(1−β)Df(x,y)+δ{Df(y,Tx)−Df(y,x)}, ∀x,y∈C.$

for all $x,y∈C$.

Next, we introduce a n-generalized Bregman nonspreading mapping in Banach spaces.

Definition 1.3. Let $f:E→ℝ∪{+∞}$ be a convex and Gâteaux differentiable function and C be a nonempty closed convex subset of $int(domf)$. A mapping $T:C→C$ is called a n-generalized Bregman nonspreading mapping if there exist $αi,βi,γi,δi∈ℝ$ ($i=1,2,…,n$) such that

$∑k=1nαkDf(Tn+1−kx,Ty)+(1−∑k=1nαk)Df(x,Ty)+∑k=1nγkDf(Ty,Tn+1−kx)−Df(Ty,x)≤∑k=1nβkDf(Tn+1−kx,y)+(1−∑k=1nβk)Df(x,y)+∑k=1nδkDf(y,Tn+1−kx)−Df(y,x),$

for all $x,y∈C$.

Remark 1.4. From Definition 1.3,

(a) when n=2, (1.4) becomes

$α1Df(T2x,Ty)+α2Df(Tx,Ty)+(1−α1−α2)Df(x,Ty)+γ1(Df(Ty,T2x)−Df(Ty,x))+γ2(Df(Ty,Tx)−Df(Ty,x))≤β1Df(T2x,y)+β2Df(Tx,y)+(1−β1−β2)Df(x,y)+δ1(Df(y,T2x)−Df(y,x))+δ2(Df(y,Tx)−Df(y,x)),$

which is called 2-generalized Bregman nonspreading in the sense of [44], where $f(x)=12||x||2.$

(b) When n=1, then (1.4) becomes

$α1Df(Tx,Ty)+(1−α1)Df(x,Ty)+γ1(Df(Ty,Tx)−Df(Ty,x)) ≤β1Df(Tx,y)+(1−β1)Df(x,y)+δ1(Df(y,Tx)−Df(y,x)),$

which is the generalized Bregman nonspreading mapping in the sense of [3, 16]. Note that, the 2-generalized Bregman nonspreading mapping reduces to the generalized Bregman nonspreading mapping if $α1=β1=γ1=δ1=0.$

(c) The class of generalized Bregman nonspreading mapping reduces to Bregman nonspreading [23] if $α1=β1=γ1=1$ and $δ1=0.$

(d) The class of generalized Bregman nonspreading mapping reduces to Bregman nonexpansive [35] if $α1=1$ and $β1=γ1=δ1=0.$

We now present an example of Bregman nonspreading mapping which is not nonspreading in the usual Hilbert space setting.

Example 1.5. Let $E=ℝ$ with the usual metric. Let $f:E→ℝ$ be defined by $f(x)=x10$ for all $x∈ℝ$ and $T:[0,0.85]→[0,0.85]$ be defined by $Tx=x2.$ We first show that T is not nonspreading, i.e.,

does not hold. Taking x = 0.5 and y = 0.85, then

$||Tx−Ty||2=(x2−y2)2=[(0.5)2−(0.85)2]2=0.22325625,$

while

$||x−y||2+2⟨x−Tx,y−Ty⟩=(x−y)2+2(x−x2)(y−y2) =(0.5−0.85)2+2(0.5−0.52)(0.85−0.852) =0.18625.$

Hence, T is not nonspreading. Put

By simple calculations, we obtain

$Df(Tx,Ty)=x20+9y20−10x2y18,Df(Ty,Tx)=y20+9x20−10x18y2,Df(Tx,y)=x20+9y10−10y2x9,Df(Ty,x)=y20+9x10−10x9y2.$

Then

$h(x,y)=9y10(y10−1)+9x10(x10−1)−10x2y9(y9−1)−10x9y2(x9−1) ≤0,$

for all $x,y∈[0,0.85].$ Thus T is Bregman nonspreading.

We further give an example of 2-generalized Bregman nonspreading mapping which is not necessarily 1-generalized Bregman nonspreading.

Example 1.6. Let $E=ℝ$ and $f(x)=x22$ then the associated Bregman distance is given by

$Df(x,y)=f(x)−f(y)−⟨x−y,∇f(y)⟩ =12x2−12y2−(x−y)(y) 12(x−y)2, ∀x,y∈ℝ.$

Define $T:[0,2]→[0,2]$ by

$Tx=0, if x∈[0,2),1, if x=2.$

It is easy to see that $F(T)={0}.$ Let

$h(x,y)=α1Df(T2x,Ty)+α2Df(Tx,Ty)+(1−α1−α2)Df(x,Ty) +γ1(Df(Ty,T2x)−Df(Ty,x))+γ2(Df(Ty,Tx)−Df(Ty,x)) −β1Df(T2x,y)−β2Df(Tx,y)−(1−β1−β2)Df(x,y) −δ1(Df(y,T2x)−Df(y,x))−δ2(Df(y,Tx)−Df(y,x)),$

for all $x,y∈[0,2].$ We consider the following possible cases.

Case I: Suppose $x=y=2,$ then $Tx=Ty=1$ and $T2x=0.$ Thus

$Df(Tx,Ty)=Df(Ty,Tx)=Df(x,y)=Df(y,x)=0,Df(x,Ty)=Df(Ty,x)=Df(Tx,y)=Df(y,Tx)=12,Df(T2x,Ty)=Df(Ty,T2x)=12,Df(T2x,y)=Df(y,T2x)=2.$

Hence

Case II: Suppose x = 2 and $y∈[0,2)$, then Tx = 1 and Ty = T2x = 0. Thus

$Df(Tx,Ty)=Df(Ty,Tx)=12,Df(x,Ty)=Df(Ty,x)=2,Df(Tx,y)=Df(y,Tx)=12(y−1)2,Df(x,y)=Df(y,x)=12(y−2)2,Df(T2x,y)=Df(y,T2x)=y22,Df(T2x,Ty)=Df(Ty,T2x)=0.$

Hence

$h(x,y)=−12(y2−4y)−2(α1+γ1)−32(α2+γ2) −2(y−2)(β2+δ1)−12(2y−3)(β2+δ2).$

Case III: Suppose $x,y∈[0,2)$ then $Tx=Ty=T2x=0.$ Thus

$Df(Tx,Ty)=Df(Ty,Tx)=Df(T2x,Ty)=Df(Ty,T2x)=0,Df(x,y)=Df(y,x)=12(x−y)2,Df(x,Ty)=Df(Ty,x)=x22,Df(Tx,y)=Df(y,Tx)=Df(T2x,y)=Df(y,T2x)=y22.$

Hence

$h(x,y)=(1−α1−α2)x22−x22(γ1+γ2)−y22(β1+β2) −12(1−β1−β2)(x−y)2−δ1xy−x22−δ2xy−y22.$

Choosing suitable choices of $α1,α2,β1,β2,γ1,γ,δ1,δ2∈ℝ,$ for instance, $α1=α2=β1=β2=γ1=γ2=1$ and $δ1=δ2=−1,$ we see that $h(x,y)≤0$ for all the cases. Hence, T is 2-generalized Bregman nonspreading. However, in this case, T is not 1-generalized Bregman nonspreading (since $α1≠0,β1≠0,γ1≠0,δ1≠0$).

In 2010, by making use of the Bregman projection, Reich and Sabach [33] studied some approximation methods for finding common zeros of maximal monotone operators in reflexive Banach spaces. They also studied some approximation techniques for finding common solutions of finitely many Bregman nonexpansive operators, see [35]. In the same sense, Kassay et al. [20] studied the approximation of solutions of system of variational inequalities in reflexive Banach spaces. It is worth noting that extension of many theory from Hilbert space to general Banach space suffer some difficulties because many of the useful techniques employed in Hilbert space (for instance the inner product and the nonexpansiveness of resolvent operators) are no longer valid in Banach spaces setting.

Motivated by the works given in [21, 35, 46], we prove some properties of the n-generalized Bregman nonspreading mappings in reflexive Banach space. Further, we introduce a hybrid method for finding a common solution of countable family of equilibrium problem and finite family of fixed points of n-generalized Bregman nonspreading mapping in reflexive Banach space. We also discuss some applications and numerical example to demonstrate the applicability of our iterative algorithm and result. The method and results present in this paper generalized and unify many previously known related results, see for instance [21, 22, 35, 45, 46].

In this section, we recall some definitions and preliminary results which will be used in the sequel. We denote the strong convergence (resp. weak convergence) of a sequence ${xn}⊂E$ to a point x ∈ E by $xn→x$ (resp. $xn⇀x$).

Let E be a real reflexive Banach space with the dual space E* and C a nonempty closed convex subset of E. Throughout this paper, we shall assume that the mapping $f:E→ℝ∪{+∞}$ is proper, convex and lower semi-continuous and also denote the domain of f by domf, where dom $f={x∈E:f(x)<∞}$. Let x∈ int(domf), the subdifferential of f at x is the convex set defined by

and the Frénchet conjugate of f is the function $f*:E*→(−∞,+∞]$ defined by

$f*(y*)=sup{⟨y*,x⟩−f(x):x∈E}.$

Let x ∈ int(domf), for any y ∈ E, the directional derivative of f at x is defined by

$fo(x,y):=limh→0f(x+hy)−f(x)h.$

If the limit in (2.1) exists as $h→0$ for each y, then the function f is said to be G$a^$teaux differentiable at x. In this case, the gradient of f at x is the linear function $∇f(x)$, which is defined by $⟨∇f(x),y⟩:=fo(x,y)$ for all y ∈ E. The function f is said to be G$a^$teaux differentiable if it is G$a^$teaux differentiable at each x ∈ int(dom f). When the limit as $h→0$ in (2.1) is attained uniformly for any y ∈ E with ||y||=1, we say that f is Fréchet differentiable at x. It is well known that f is G$a^$teaux (resp. Fréchet) differentiable at x ∈ int(dom f) if and only if the gradient ∇ f is norm-to-weak* (resp. norm-to-norm) continuous at x (see [6]).

Let E be a reflexive Banach space. The function f is called Legendre if and only if it satisfies the following two conditions:

(L1) f is Gâteaux differentiable, int(dom $f)≠∅$ and dom $∇f$ = int(dom f),

(L2) f* is Gâteaux differentiable, int(dom $f*)≠∅$ and dom $∇f*$ = int(dom f*).

Since E is reflexive, we know that $(∇f)−1=∇f*$, this together with conditions (L1) and (L2) implies that

$ran∇f=dom∇f*=int(domf*),$

and

$ran∇f*=dom∇f=int(domf).$

The notion of Legendre function in infinite dimensional spaces was first introduced by Bauschke, Borwein and Combettes in [6]. By their definition, the conditions (L1) and (L2) also yield that f and f* are Gâteaux differentiable and strictly convex in the interior of their respective domains. It follows that f is Legendre if and only if f* is Legendre (see [6], Corollary 5.5, p. 634).

One important and interesting example of Legendre function is $1p||⋅||p$ $(1 when E is a smooth and strictly convex Banach space. In this case, the gradient $∇f$ of f coincide with the generalized duality mapping of E. More examples of Legendre functions can be found in [5, 6]. In the rest of this paper, we always assume that $f:E→ℝ∪{+∞}$ is a Legendre function.

Definition 2.1. Let $f:E→(−∞,+∞]$ be a convex and Gâteaux differentiable function. The Bregman projection of $x∈int(domf)$ onto the nonempty, closed and convex subset $C⊂domf$ is the necessarily unique vector Proj$Cf(x)∈C$ satisfying

Remark 2.2.

1. If E is a Hilbert space and $f(x)=12||x||2$, then the Bregman projection $ProjCf(x)$ is reduced to the metric projection of x onto C.

2. If E is smooth and strictly convex and $f(x)=1p||x||p$ $(1, then the Bregman projection $ProjCf(x)$ reduces to the generalized projection $ΠC(x)$, which is defined by

$Dp(ΠC(x),x):=inf{Dp(z,x):z∈C}.$

It is known from [10] that z = Proj$Cf(x)$ if and only if

We also have

Similar to the metric projection in Hilbert space, the Bregman projection also has a variational characterization which is given below.

Lemma 2.3. [33] (Characterization of Bregman Projection)) Let f be totally convex on int(domf). Let C be a nonempty, closed and convex subset of int(domf) and x ∈ int(domf), if ω ∈ C, then the following conditions are equivalent:

(i)the vector ω is the Bregman projection of x onto C, with respect to f,

(ii) the vector ω is the unique solution of the variational inequality

$⟨∇f(x)−∇f(z),z−y⟩≥0 ∀y∈C,$

(iii) the vector ω is the unique solution of the inequality

$Df(y,z)+Df(z,x)≤Df(y,x) ∀y∈C.$

Definition 2.4. Let $f:E→(−∞,+∞]$ be a convex and G$a^$teaux differentiable function. The function f is called:

(i) totally convex at x if its modulus of totally convexity at x ∈ int(domf), that is, the bifunction vf: int(dom$f)×[0,+∞)→[0,+∞)$ defined by

$vf(x,t):=inf{Df(y,x):y∈domf,||y−x||=t}$

is positive for any $t>0,$

(ii) totally convex if it is totally convex at every point x ∈ int(dom f),

(iii) totally convex on bounded subset B of E, if vf(B,t) is positive for any nonempty bounded subset B, where the function vf: int(dom $f)×[0,+∞)→[0,+∞]$ is defined by

$vf(B,t):=inf{vf(x,t):x∈B∩int(domf)}, t>0.$

(iv) cofinite if dom$f*=E*$,

(v) coercive if $lim||x||→+∞(f(x)||x||)=+∞,$

(vi) sequentially consistent if for any two sequences ${xn}$ and ${yn}$ in E such that ${xn}$ is bounded,

$limn→∞Df(yn,xn)=0⇒limn→∞||yn−xn||=0.$

For further details and examples on totally convex functions see [8, 9, 10].

Lemma 2.5. ([9]) The function $f:E→ℝ$ is totally convex on bounded subsets if and only if it is sequentially consistent.

Lemma 2.6. ([34]) Let $f:E→ℝ$ be a G$a^$teaux differentiable and totally convex function. If $x0∈E$ and the sequence ${Df(x0,xn)}$ is bounded, then the sequence ${xn}$ is also bounded.

Lemma 2.7. ([10]) Let $f:E→(−∞,+∞]$ be a convex function whose domain contains at-least two points. Then the following statements holds:

(i) f is sequentially consistent if and only if it is totally convex on bounded subsets.

(ii) If f is lower semicontinuous, then f is sequential consistent if and only if it is uniformly convex on bounded subsets.

(iii) If f is uniformly strictly convex on bounded subsets, then it is sequentially consistent and the converse implication holds when f is lower semicontinuous, Fréchet differentiable on its domain, and the Fréchet derivative $∇f$ is uniformly continuous on bounded subsets.

Lemma 2.8. ([33]) If $f:E→ℝ$ 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*.

Let $f:E→ℝ$ be a convex Legendre and Gâteaux differentiable function. The function $Vf:E×E*→[0,∞)$ associated with f defined by

$Vf(x,x*)=f(x)−⟨x*,x⟩+f*(x*), ∀x∈E,x*∈E*.$

Then, Vf is non-negative and $Vf(x,x*)=Df(x,∇f*(x*))$ for all $x∈E$ and $x*∈E*$. More so, by the subdifferential inequality,

$Vf(x,x*)+⟨y*,∇f*(x*)−x⟩≤Vf(x,x*+y*)$

for all $x∈E$ and $x*,y*∈E*$ (see [24]). In addition, if $f:E→(−∞,+∞]$ is a proper lower semicontinuous function, then $f*:E*→(−∞,+∞]$ is a proper weak* lower semicontinuous and convex function. Hence, Vf is convex in the second variable. Thus, for all z ∈ E

$Df(z,∇f*(∑ i=1Nti∇f(xi)))≤∑ i=1NtiDf(z,xi),$

where ${xi}⊂E$ and ${ti}⊂(0,1)$ with $∑ i=1Nti=1$.

Let E be a Banach space and let $Br:={z∈E:||z||≤r}$ for all r>0. Then, a function $f:E→ℝ$ is said to be uniformly convex on bounded subsets of E if $ρr(t)>0$ for all t≥ 0, where $ρr:[0,+∞)→[0,∞]$is defined by

$ρr(t)=infx,y∈Br,||x−y||=t,α∈(0,1)αf(x)+(1−α)f(y)−f(αx+(1−α)y)α(1−α).$

The function ρr is called the gauge of uniform convexity of f. More so, the function $f:E→(−∞,+∞]$ is called totally coercive if

$lim||x||→+∞(f(x)||x||)=+∞.$

Lemma 2.9. ([27]) Let r > 0 be a constant and let $f:E→ℝ$ be a continuous uniformly convex function on bounded subsets of E. Then

$f(∑ k=0∞αkxk)≤∑ k=0∞αkf(xk)−αiαjρr*(||xi−xj||),$

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.

Let $l∞$ be the Banach lattice of bounded real sequences with the supremum norm. It is well known that there exists a bounded linear functional µ on $l∞$ such that the following three conditions hold:

(i) if ${tn}$ in $l∞$ and $tn≥0$ for every $n∈ℕ$, then $μ({tn})≥0$,

(ii) if $tn=1$ for every $n∈ℕ$, then $μ({tn})=1$,

(iii) $μ({tn+1})=μ({tn})$ for all ${tn}$ in $l∞$.

Here, ${tn+1}$ denotes the sequence $(t2,t3,…,tn,tn+1,…)$ in $l∞$. Such a functional µ is called a Banach limit and the value of µ at ${tn}$ in $l∞$ is denoted by $μntn$. Therefore, condition (3) means $μntn=μntn+1$. If µ satisfies conditions (1) and (2), we call µ a mean on $l∞$ (see, for example, [43] for more details).

Lemma 2.10. ([12]) Let C be a nonempty, closed and convex subset of a real reflexive Banach space E. Let $f:E→ℝ$ be strictly convex, continuous, strongly coercive, Gâteaux differentiable, locally bounded and local uniformly convex on E. Let $T:C→C$ be a mapping and ${xn}$ be a bounded sequence of C and µ be a mean on $l∞$. Supposet that

$μnDf(xn,Ty)≤μnDf(xn,y) ∀y∈C.$

Then, T has a fixed point in C.

Let T be a mapping from C into itself. A point x ∈ C is said to be an asymptotic fixed point of T if there exists a sequence ${xn}$ in C which converges weakly to p and $limn→∞||xn−Txn||=0.$ We denote the set of all asymptotic fixed points of T by $F^(T).$

Recall that a mapping $T:C→C$ is said to be Bregman quasi-nonexpansive [27] if $F(T)≠∅$ and

$Df(p,Tx)≤Df(p,x) ∀x∈C,p∈F(T).$

A mapping $T:C→C$ is to be Bregman relatively nonexpansive [27] if the following conditions are satisfied:

(i) F(T) is nonempty;

(ii) $Df(p,Tv)≤Df(p,v),$ $∀p∈F(T)$, $v∈C;$

(iii) $F^(T)=F(T).$

Lemma 2.11. ([37]) Let C be a nonempty, closed and convex subset of a real reflexive Banach space E and let $f:E→ℝ$ be a strictly convex and Gâteaux differentiable function. Let $g:C×C→ℝ$ be a bifunction satisfying conditions (A1)-(A4). For all $λ>0$ be any given number and x ∈ E, there exists z ∈ C such that

Define the resolvent mapping $Tr:E→2C$ as follows

then, $Resλ,gf$ has the following properties:

(i) $Resλ,gf$ is single-valued;

(ii) $Resλ,gf$ is a firmly nonexpansive mapping, that is;

$⟨Resλ,gfz−Resλ,gfy,∇f(Resλ,gfz)−∇f(Resλ,gfy)⟩≤⟨Resλ,gfz−Resλ,gfy,∇f(z)−∇f(y)⟩$

$∀z,y∈E$;

(iii) $F(Resλ,gf)=EP(g)$;

(iv) EP(g) is closed and convex.

It is easy to see that the resolvent operator satisfies the following inequality: for all r > 0, u ∈ EP(g) and x ∈ E, then

$Df(x,Resλ,gfx)+Df(Resλ,gfx,u)≤Df(x,u).$

In this section, we present the existence and some properties of fixed points of n-generalized Bregman nonspreading mapping in a reflexive Banach space. This result extend the corresponding results of [45] and [25] to reflexive Banach space.

Proposition 3.1. Let E be a real reflexive Banach space and $f:E→ℝ$ be a strictly convex and Gâteaux differentiable function. Let $C⊂int(domf)$ be a nonempty, closed and convex set and $T:C→C$ be a n-generalized Bregman nonspreading mapping. Then, the following are equivalent

(i) F(T) is nonempty;

(ii) ${Tmz}$ is bounded for some z ∈ C and $m∈ℕ$.

proof First we show that (i) implies (ii). Suppose $F(T)≠∅$, then ${Tmz}={z}$ for $z∈F(T)$. So ${Tmz}$ is bounded. Next, we show that (ii) implies (i). Let ${Tmz}$ be bounded for some z ∈ C. Since T is n-Bregman generalized nonspreading, then there exist $αi,βi,γi,δi∈ℝ$ for $i=1,2,…,n$,

such that

$∑k=1nαkDf(Tn+1−kx,Ty)+(1−∑k=1nαk)Df(x,Ty)+∑k=1nγk{Df(Ty,Tn+1−kx)−Df(Ty,x)}≤∑k=1nβkDf(Tn+1−kx,y)+(1−∑k=1nβk)Df(x,y)+∑k=1nδk{Df(y,Tn+1−kx)−Df(y,x)},$

for all x,y ∈ C. Replacing x by Tm-1z in (3.1), we have that for any y,z ∈ C,

$∑k=1nαkDf(Tn+1−kTm−1z,Ty)+(1−∑k=1nαk)Df(Tm−1z,Ty)+∑k=1nγk{Df(Ty,Tn+1−kTm−1z)−Df(Ty,Tm−1z)}≤∑k=1nβkDf(Tn+1−kTm−1z,y)+(1−∑k=1nβk)Df(Tm−1z,y)+∑k=1nδk{Df(y,Tn+1−kTm−1z)−Df(y,Tm−1z)}.$

Since ${Tmz}$ is bounded, we can apply Banach limit µ to both sides of (3.2), then we have

$μm(∑k=1nαkDf(Tm+n−kz,Ty)+(1−∑k=1nαk)Df(Tm−1z,Ty)+∑k=1nγk{Df(Ty,Tm+n−kz)−Df(Ty,Tm−1z)})≤μm(∑k=1nβkDf(Tm+n−kz,y)+(1−∑k=1nβk)Df(Tm−1z,y)+∑k=1nδk{Df(y,Tm+n−kz)−Df(y,Tm−1z)}).$

Thus, we obtain

$∑k=1nαkμmDf(Tm+n−kz,Ty)+(1−∑k=1nαk)μmDf(Tm−1z,Ty)+∑k=1nγk{μmDf(Ty,Tm+n−kz)−μmDf(Ty,Tm−1z)}≤∑k=1nβkμmDf(Tm+n−kz,y)+(1−∑k=1nβk)μmDf(Tm−1z,y)+∑k=1nδk{μmDf(y,Tm+n−kz)−μmDf(y,Tm−1z)}.$

Then

$∑k=1nαkμmDf(Tmz,Ty)+(1−∑k=1nαk)μmDf(Tmz,Ty)+∑k=1nγk{μmDf(Ty,Tmz)−μmDf(Ty,Tmz)}≤∑k=1nβkμmDf(Tmz,y)+(1−∑k=1nβk)μmDf(Tmz,y)+∑k=1nδk{μmDf(y,Tmz)−μmDf(y,Tmz)}.$

Hence

Therefore by Lemma 2.10, T has a fixed point in C. This completes the proof.

The following results follow as direct consequences of Theorem 3.1.

Corollary 3.2. Let C be a nonempty, closed and convex subset of a smooth, strictly convex Banach space E, let p be a real number such that $1 and let f be a function defined by $f(x)=1p||x||p$ and $T:C→C$ be a n-generalized Bregman nonspreading mapping. Then, the following assertions are equivalent:

(i) F(T) is nonempty;

(ii) ${Tmz}$ is bounded for some z ∈ C.

Corollary 3.3. Let C be a nonempty bounded closed convex subset of a real reflexive Banach space E and $f:E→ℝ$ be a strictly convex and Gâteaux differentiable function. Let $T:C→C$ be a n-generalized Bregman nonspreading mapping. Then, T has a fixed point.

Remark 3.4. Corollary 3.2 is a generalization of the corresponding result in Theorem 3.2 of [45], where the equivalence between the two assertions was shown for p=2.

We now show another important property of the fixed points of n-generalized Bregman nonspreading mapping.

Proposition 3.5. Let C be a nonempty, closed and convex subset of a real reflexive Banach space E and $f:E→ℝ$ be a strictly convex and Gâteaux differentiable function. Let $T:C→C$ be a n-generalized Bregman nonspreading mapping such that $F(T)≠∅$. Then F(T) is closed and convex.

Proof. Let $u∈F(T)$, then putting $u=x∈F(T)$ in (1.4), we have

$∑k=1nαkDf(u,Ty)+(1−∑k=1nαk)Df(u,Ty)+∑k=1nγk{Df(Ty,u)−Df(Ty,u)}≤∑k=1nβkDf(u,y)+(1−∑k=1nβk)Df(u,y)+∑k=1nδk{Df(y,u)−Df(y,u)},$

which implies that

$Df(u,Ty)≤Df(u,y), ∀u∈F(T),y∈C.$

This means that T is quasi-Bregman nonexpansive. Now let ${xn}⊂F(T)$ such that $xn→p$. Then

Hence, $p∈F(T)$. Therefore F(T) is closed.

Next, we show that F(T) is convex. For any $x,y∈F(T)$ and $λ∈(0,1)$, let $z=λx+(1−λ)y$. Then

$Df(z,Tz)=f(z)−f(Tz)−⟨∇f(Tz),z−Tz⟩ =f(z)−f(Tz)−⟨∇f(Tz),λx+(1−λ)y−Tz⟩ =f(z)+λDf(x,Tz)+(1−λ)Df(y,Tz)−λf(x)−(1−λ)f(y) ≤f(z)+λDf(x,z)+(1−λ)Df(y,z)−λf(x)−(1−λ)f(y) =f(z)−f(z)−⟨∇f(z),λx+(1−λ)y−z⟩ =f(z)−f(z)−⟨∇f(z),z−z⟩ =0.$

Hence, z = Tz. Therefore, F(T) is convex.

Using Corollary 3.3 and Proposition 3.5, we prove the following common fixed point theorem for a commutative family of n-generalized Bregman nonspreading mapping in a reflexive Banach space.

Theorem 3.6. Let $f:E→ℝ$ be a strictly convex and Gâteaux differentiable function, C be a nonempty bounded closed convex subset of a real reflexive Banach space E and let ${Tα}α∈I$ be a commutative family of n-generalized Bregman nonspreading mappings from C into itself. Then ${Tα}α∈I$ has a common fixed point.

Proof. By Theorem 3.5, we know that F(Tα) is a closed convex subset of C. Since E is reflexive and C is a bounded closed and convex subset, C is weakly compact. To show that $∩α∈IF(Tα)$ is nonempty, it is sufficient to show that ${F(Tα)}α∈I$ has a nonempty finite intersection property.

Now, let ${T1,T2,…,TN}$ be a commutative finite family of n-generalized Bregman nonspreading mapping from C into itself. We prove by induction that ${T1,T2,…,TN}$ has a common fixed point. To do this, we start by showing the case for N=2. By Corollary 3.3 and Theorem 3.5, F(T1) is nonempty, bounded, closed and convex. Let $u∈F(T1)$, since $T1T2=T2T1$, then we have $T1T2u=T2T1u=T2u$. This implies that $T2u∈F(T1)$. Hence, F(T1) is T2-invariant. Thus, the restriction of T2 to F(T1) is a n-generalized Bregman nonspreading self mapping. By Corollary (3.3), T2 has a fixed point in F(T1), that is, we have $z∈F(T1)$ such that $T2z=z$. Hence, $z∈F(T1)∩F(T2)$.

Suppose that for some N ≥ 2, $Γ=∩k=1NF(Tk)$ is nonempty. Then Γ is a nonempty, bounded, closed and convex subset of C and the restriction of $TN+1$ to Γ is a n-generalized Bregman nonspreading self mapping. By Corollary 3.3, $TN+1$ has a fixed point in Γ. This implies that $Γ∩F(TN+1)$ is nonempty. Hence, $∩k=1N+1F(Tk)$ is nonempty. This completes the proof.

The following result will be used in the sequel.

Proposition 3.7. Let E be a real reflexive Banach space and let C be a nonempty, closed and convex subset of E. Let $f:E→ℝ$ be a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let $T:C→C$ be a n-generalized Bregman nonspreading mapping. Then, for any $x,y∈C$, $αi,βi,γi,δi∈ℝ$, for $i=1,2,…,n$, we have

$0≤∑k=1n(βk−αk)(Df(Tn+1−kx,Ty)−Df(x,Ty))+Df(Ty,y)+⟨∇f(Ty)−∇f(y),∑k=1nβk(Tn+1−kx−x)+x−Ty⟩+∑k=1nδk{Df(y,Tn+1−kx)−Df(y,x)}−∑k=1nγk{Df(Ty,Tn+1−kx)−Df(Ty,x)}.$

Proof. From the definition of n-generalized Bregman nonspreading mapping, we have

$∑k=1nαkDf(Tn+1−kx,Ty)+(1−∑k=1nαk)Df(x,Ty)+∑k=1nγk{Df(Ty,Tn+1−kx)−Df(Ty,x)}≤∑k=1nβkDf(Tn+1−kx,y)+(1−∑k=1nβk)Df(x,y)+∑k=1nδk{Df(y,Tn+1−kx)−Df(y,x)},$

for all $x,y∈C$. This implies that

$0≤∑k=1nβkDf(Tn+1−kx,y)+(1−∑k=1nβk)Df(x,y)+∑k=1nδk{Df(y,Tn+1−kx)−Df(y,x)}−∑k=1nαkDf(Tn+1−kx,Ty)−(1−∑k=1nαk)Df(x,Ty)−∑k=1nγk{Df(Ty,Tn+1−kx)−Df(Ty,x)}.$

Hence, from the three points identity (1.2), we have

$0≤∑k=1nβk(Df(Tn+1−kx,Ty)+Df(Ty,y)+⟨∇f(Ty)−∇f(y),Tn+1−kx−Ty⟩)+(1−∑k=1nβk)(Df(x,Ty)+Df(Ty,y)+⟨∇f(Ty)−∇f(y),x−Ty⟩)−∑k=1nαkDf(Tn+1−kx,Ty)−(1−∑k=1nαk)Df(x,Ty)−∑k=1nγk{Df(Ty,Tn+1−kx)−Df(Ty,x)}+∑k=1nδk{Df(y,Tn+1−kx)−Df(y,x)}.$

Therefore

$0≤∑k=1n(βk−αk)(Df(Tn+1−kx,Ty)−Df(x,Ty))+Df(Ty,y)+⟨∇f(Ty)−∇f(y),∑k=1nβk(Tn+1−kx−x)+x−Ty⟩+∑k=1nδk{Df(y,Tn+1−kx)−Df(y,x)}−∑k=1nγk{Df(Ty,Tn+1−kx)−Df(Ty,x)}.$

The following result is another important property which characterized the n-generalized Bregman nonspreading mapping.

Proposition 3.8. Let $T:C→C$ be a n-generalized Bregman nonspreading mapping. Suppose $F(T)≠∅,$ then T is Bregman relatively nonexpansive.

Proof. It is clear that

$Df(p,Tx)≤Df(p,x) ∀p∈F(T),x∈C.$

We show that $F^(T)=F(T).$ It is easy to see that $F(T)⊂F^(T).$ Now let $p∈F^(T),$ that is, there exist a sequence ${xn}⊂C$ such that $xn⇀p$ and $||xn−Txn||→0.$ Since f is uniformly Frćhet differentiable on bounded subsets of E, then ∇f is uniformly continuous and thus

$limn→∞||f(xn)−f(Txn)||=limn→∞||∇f(xn)−∇f(Txn)||=0.$

Putting x = xn and y = q in Proposition 3.7, we have

$0≤∑k=1n(βk−αk)(Df(Tn+1−kxn,Tq)−Df(xn,Tq))+Df(Tq,q)+⟨∇f(Tq)−∇f(q),∑k=1nβk(Tn+1−kxn−xn)+xn−Tq⟩+∑k=1nδk{Df(q,Tn+1−kxn)−Df(q,xn)}−∑k=1nγk{Df(Tq,Tn+1−kxn)−Df(Tq,xn)}.$

Observe that

$Df(Tn+1−kxn,Tq)−Df(xn,Tq)=f(Tn+1−kxn)−f(Tq) −⟨∇f(Tq),Tn+1−kxn−Tq⟩ −f(xn)+f(Tq)+⟨∇f(Tq),xn−Tq⟩ =f(Tn+1−kxn)−f(xn) +⟨∇f(Tq),xn−Tq⟩ −⟨∇f(Tq),Tn+1−kxn−Tq⟩ =f(Tn+1−kxn)−f(xn) +⟨∇f(Tq),xn−Tn+1−kxn⟩.$

Similarly

$Df(q,Tn+1−kxn)−Df(q,xn)=f(xn)−f(Tn+1−kxn)+⟨∇f(xn),Tn+1−kxn−xn⟩ +⟨∇f(xn)−∇f(Tn+1−kxn),q−xn⟩,$

and

$Df(Tq,Tn+1−kxn)−Df(Tq,xn)=f(xn)−f(Tn+1−kxn) +⟨∇f(xn),Tn+1−kxn−xn⟩ +⟨∇f(xn)−∇f(Tn+1−kxn),Tq−xn⟩.$

Substituting (3.10), (3.11) and (3.12) into (3.9), we have

$0≤∑k=1n(βk−αk)(f(Tn+1−kxn)−f(xn)+⟨∇f(Tq),xn−Tn+1−kxn⟩)+Df(Tq,q)+⟨∇f(Tq)−∇f(q),∑k=1nβk(Tn+1−kxn−xn)+xn−Tq⟩+∑k=1nδk{f(xn)−f(Tn+1−kxn)+⟨∇f(xn),Tn+1−kxn−xn⟩+⟨∇f(xn)−∇f(Tn+1−kxn),q−xn⟩}−∑k=1nγk{f(xn)−f(Tn+1−kxn)+⟨∇f(xn),Tn+1−kxn−xn⟩+⟨∇f(xn)−∇f(Tn+1−kxn),Tq−xn⟩}.$

Taking limit as $n→∞$ in (3.13) and using (3.8), we have

Using the four points identity (1.3), we have

$0≤Df(Tq,q)+Df(Tq,Tq)−Df(Tq,q)−Df(q,Tq)+Df(q,q)=−Df(q,Tq).$

Thus $Df(q,Tq)≤0$ and then $Df(q,Tq)=0$. Since f is strictly convex, we have q = Tq. Hence, $q∈F(T)$. Therefore $F^(T)⊂F(T)$. This thus implies that $F^(T)=F(T).$

In this section, we introduce a hybrid algorithm for finding common solutions of countable family of equilibrium problem and finite fixed points of n-generalized Bregman nonspreading mapping in reflexive Banach space.

Let ${αn,i:n,i∈ℕ,1≤i≤ℕ}$ be sequences of real numbers such that ${αn,i}⊂(0,1).$ We define the following $Wn:C→C$ mapping generated by $Ti, i=1,2,…,N$ and ${αn,i},$ where $Ti:C→C$ is a finite family of n-generalized Bregman nonspreading mappings.

$Sn,0x=x,Sn,1x=∇f*[αn,1∇f(T1x)+(1−αn,1)∇f(x)]Sn,2x=∇f*[αn,2∇f(T2Sn,1x)+(1−αn,2)∇f(Sn,1x)]Sn,3x=∇f*[αn,3∇f(T3Sn,2x)+(1−αn,3)∇f(Sn,2x)] ⋮Sn,N−1x=∇f*[αn,N−1∇f(TN−1Sn,N−2x)+(1−αn,N−1)∇f(Sn,N−2x)]Wn=Sn,N=∇f*[αn,N∇f(TNSn,N−1x)+(1−αn,N)∇f(Sn,N−1x)].$

Using the above definition, we have the following lemma.

Proposition 4.1. Let C be a nonempty, closed and convex subset of a real reflexive Banach space E and let $f:E→ℝ$ be a coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let ${Ti}i=1N$ be a finite famiy of n-generalized Bregman nonspreading mapping of C into itself such that $∩ i=1NF(Ti)≠∅.$ Let ${αn,i}$ be real sequence in (0,1) such that $liminfn→∞αn,i>0,$ $∀i∈{1,2,…,N}.$ Let Wn be a Bregman W-mapping generated by $T1,T2,…,TN$ in (4.1). Then

(i) $∩i=1NF(Ti)=F(Wn),$

(ii) Wn is Bregman quasi-nonexpansive,

(iii) If in addition, Ti is Bregman relatively nonexpansive mapping, for each i, then Wn is Bregman relatively nonexpansive.

Proof. Let $x∈∩i=1NF(Ti).$ Then $Tix=x,$ $i=1,2,…,N.$ From (4.1), we have that $Sn,1x=x,$ $Sn,2x=x,…,$ $Sn,Nx=x.$ Thus $∩i=1NF(Ti)⊂F(Wn).$ Conversely, let $y∈F(Wn)$ and $x∈∩i=1NF(Ti).$ Then

$Df(x,y)=Df(x,Wny) =Df(x,∇f*(αn,N∇f(TNSn,N−1y)+(1−αn,N)∇f(Sn,N−1y))) =f(x)−⟨x,αn,N∇f(TNSn,N−1y)⟩+(1−αn,N)∇f(Sn,N−1y)⟩ +f*(αn,N∇f(TNSn,N−1y)+(1−αn,N)∇f(Sn,N−1y)) ≤αn,N(f(x)−⟨x,∇f(TNSn,N−1y)+f*(∇f(TNSn,N−1y))) +(1−αn,N)(f(x)−⟨x,∇f(Sn,N−1y)⟩+f*(∇f(TNSn,N−1y))) −αn,N(1−αn,N)ρr*(||∇f(TNSn,N−1y)−∇f(Sn,N−1y)||) =αn,NDf(x,TNSn,N−1y)+(1−αn,N)Df(x,Sn,N−1y) −αn,N(1−αn,N)ρr*(||∇f(TNSn,N−1y)−∇f(Sn,N−1y)) ≤Df(x,Sn,N−1y) −αn,N(1−αn,N)ρr*(||∇f(TNSn,N−1y)−∇f(Sn,N−1y)||) ⋮ ≤Df(x,y)−αn,1(1−αn,1)ρr*(||∇f(T1y)−∇f(y)||) −αn,2(1−αn,2)ρr*(||∇f(T2Sn,1y)−∇f(Sn,1y)||) −…−αn,N(1−αn,N)ρr*(||∇f(TNSn,N−1y)−∇f(Sn,N−1y)||).$

This implies that

$αn,1(1−αn,1)ρr*(||∇f(T1y)−∇f(y)||) =αn,2(1−αn,2)ρr*(||∇f(T2Sn,1y)−∇f(Sn,1y)||) =…=αn,N(1−αn,N)ρr*(||∇f(TNSn,N−1y)−∇f(Sn,N−1y)||)=0.$

Then by the property of $ρr*$ from Lemma 2.9 and the norm-to-norm continuity of ∇ f*, we have

$T1y=y,T2Sn,1y=Sn,1y, ⋮TNSn,N−1=Sn,N−1y.$

It follows that

$Df(y,Sn,1y)=Df(y,∇f*(αn,1∇f(T1y)+(1−αn,1)∇f(y))) ≤αn,1Df(y,T1y)+(1−αn,1)Df(y,y)=0.$

Therefore $y∈F(Sn,1)$ and consequently, $y∈F(T1).$ Following similar argument, we have that $y∈F(Ti)$ for $i=1,2,…,N$ and hence $y∈∩ i=1NF(Ti).$

(ii) Let $y∈F(Wn).$ Then

$Df(y,Wnx)=Df(y,∇f*(αn,N∇f(TNSn,N−1x)+(1−αn,N)∇f(Sn,N−1x))) ≤αn,NDf(y,TNSn,N−1x)+(1−αn,N)Df(y,Sn,N−1x) ≤αn,NDf(y,Sn,N−1x)+(1−αn,N)Df(y,Sn,N−1x) =Df(y,Sn,N−1x)=Df(y,∇f*(αn,N−1∇f(TN−1Sn,N−2x) +(1−αn,N−1)∇f(Sn,N−2x))) ≤αn,N−1Df(y,TN−1Sn,N−2x)+(1−αn,N−1)Df(y,Sn,N−2x) ≤Df(y,Sn,N−2x) ⋮ ≤Df(y,x).$

(iii) Let ${xn}⊂C$ such that $xn⇀x¯$ and $||Wnxn−xn||→0$ as $n→∞.$ From (1.2), we have

$Df(x¯,Wnxn)≤Df(x¯,xn)−αn,1(1−αn,1)ρr*(||∇f(T1xn)−∇f(xn)||) −αn,2(1−αn,2)ρr*(||∇f(T2Sn,1xn)−∇f(Sn,1xn)||) −…−αn,N(1−αn,N)ρr*(||∇f(TNSn,N−1xn)−∇f(Sn,N−1xn)||).$

Using three points identity (1.2), we obtain

$Df(x¯,xn)−Df(x¯,Wnxn)=⟨x¯−xn,∇f(Wnxn)−∇f(xn)⟩ −Df(xn,Wnxn).$

Since $xn⇀x¯$ and $limn→∞||xn−Wnxn||=0,$ we obtain

Therefore from (4.3), we have

$αn,1(1−αn,1)ρr*(||∇f(T1xn)−∇f(xn)||)+αn,2(1−αn,2)ρr*(||∇f(T2Sn,1xn) −∇f(Sn,1xn)||)+… +αn,N(1−αn,N)ρr*(||∇f(TNSn,N−1xn)−∇f(Sn,N−1xn)||) ≤Df(x¯,xn)−Df(x¯,xn).$

Taking limit as $n→∞$, using (4.5) and property of $ρr*$, yields

$limn→∞||∇f(T1xn)−∇f(xn)||=limn→∞||∇f(T2Sn,1xn)−∇f(Sn,1xn)||= …=limn→∞||∇f(TNSn,N−1xn)−∇f(Sn,N−1xn)||=0.$

By the norm-to-norm uniform continuity of ∇f on bounded subset of E*, it follows that

$limn→∞||T1xn−xn||=limn→∞||T2Sn,1xn−Sn,1xn||=⋯ =limn→∞||TNSn,N−1xn−Sn,N−1xn||=0.$

We next prove that $Sn,ixn−xn→0$ for each $i=1,2,…,N−1.$ From (4.6), we get

$Dp(xn,Sn,1xn)=Df(xn,∇f*[αn,1∇f(T1xn)+(1−αn,1)∇f(xn)]) ≤αn,1Df(xn,T1xn)+(1−αn,1)Df(xn,xn).$

Taking limit as $n→∞$ and using (4.6), we have

hence

$limn→∞||Sn,1xn−xn||=0.$

Thus

Similarly, we have

$Df(xn,Sn,2xn)=Df(xn,∇f*[αn,2∇f(T2Sn,1xn)+(1−αn,2)∇f(Sn,1xn)]) ≤αn,2Df(xn,T2Sn,1xn)+(1−αn,2)Df(xn,Sn,1xn)$

Taking limit as $n→∞,$ we have

and hence

Following similar approach as above, we have

Therefore

This together with the Bregman relative nonexpansiveness of each Ti for $i=1,2,…,N,$ implies that $x¯∈F(Sn,i)$ for $i=1,2,…,N.$ Hence $x¯∈F(Wn).$ This therefore implies that Wn is Bregman relatively nonexpansive.

We are now in position to introduce our iterative algorithm.

Theorem 4.2. Let C be a nonempty, closed and convex subset of a real reflexive Banach space E and $f:E→ℝ$ be a coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. For $i=1,2,…,N,$ let ${αn,i}⊂(0,1),$ $Ti:C→C$ be finite family of n-generalized Bregman nonspreading mappings and $Wn:C→C$ be a Bregman W-mapping generated by ${αn,i}$ and $T1,T2,…,TN$ in (4.1). Let $gj:C×C→ℝ$ be bifunctions satisfying assumptions (A1)-(A4) and suppose $Γ:=∩ i=1NF(Ti)∩∩ j=1∞EP(gi)≠∅.$ Define the sequence ${xn}$ by the following process

$x0=x∈C,C0=Q0=C,zn=∇f*[βn,0∇f(xn)+∑ j=1∞βn,j∇f(Resλn,gjfxn)],yn=∇f*[δn∇f(xn)+(1−δn)∇f(Wnzn)],Cn={z∈C:Df(z,yn)≤Df(z,xn)},Qn={z∈C:⟨∇f(x)−∇f(xn),xn−z⟩≥0},xn+1=ProjCn∩Qnfx,$

for all n ≥ 0, where ${λn}⊂(0,∞),$ ${βn,j}$ and ${δn}$ are sequences in [0,1) satisfying the following control conditions:

(i) $∑ j=0∞βn,j=1,$

(ii) There exists $k∈ℕ$ such that $liminfn→∞βn,jβn,k>0,$ $∀j∈ℕ∪{0};$

(iii) $0≤δn<1,$ $∀n∈ℕ$ and $liminfn→∞δn<1;$

(iv) $liminfn→∞λn>0.$

Then, the sequence ${xn}$ converges strongly to Proj$Γfx$ as $n→∞.$

Proof. We divide the proof into several steps.

Step 1: We show that $Γ⊂Cn∩Qn$ and $xn+1$ is well defined.

It is clear that Cn and Qn are closed and convex. Then $Cn∩Qn$ is closed and convex for $n≥0.$ Obviously, $Γ⊂C0∩Q0.$ Suppose $Γ⊂Cm∩Qm$ for some $m∈ℕ.$

Let $p∈Γ,$ then

$Df(p,ym)=Df(p,∇f*[δm∇f(xm)+(1−δm)∇f(Wmzm)]) =Vf(p,δm∇f(xm)+(1−δm)∇f(Wmzm)) =f(p)−⟨p,δm∇f(xm)+(1−δm)∇f(Wmzm)⟩ +f*(δm∇f(xm)+(1−δm)∇f(Wmzm)) ≤δm[f(p)−⟨p,∇f(xm)⟩+f*(xm)] +(1−δm)[f(p)−⟨p,∇f(Wmzm)⟩+f*(Wmzm)] −δm(1−δm)ρr*(||xm−Wmzm||) ≤δmDf(p,xm)+(1−δm)Df(p,zm)−δm(1−δm)ρr*(||xm−Wnzm||) =δnDf(p,xm)+(1−δm)Df(p,∇f*[βm,0∇f(xm) +∑ j=1∞βm,j∇f(ResEP(g)fxm)]) −δm(1−δm)ρr*(||xm−Wmzm||).$

Hence

$Df(p,ym)≤δmDf(p,xm)+(1−δm)[βm,0Df(p,xm) +∑ j=1∞βm,jDf(p,ResEP(g)fxm) −βm,0∑ j=1∞βm,jρr*(||xm−ResEP(g)fxm||)] −δm(1−δm)ρr*(||xm−Wmzm||) ≤δmDf(p,xm)+(1−δm)[βm,0Df(p,xm)+∑ j=1∞βm,jDf(p,xm)] −(1−δm)βm,0∑ j=1∞βm,jρr*(||xm−ResEP(g)fxm||) −δn(1−δm)ρr*(||xm−Wmzm||) =Df(p,xm)−(1−δm)βm,0∑ j=1∞βm,jρr*(||xm−ResEP(g)fxm||) −δn(1−δn)ρr*(||xm−Wmzm||) ≤Df(p,xm).$

Hence $p∈Cm$, which implies that $Γ∈Cm.$ Since $xm+1=ProjCm∩Qmfx,$ then $⟨∇f(x)−∇f(xm+1),z−xm+1⟩≤0$ In particular, $⟨∇f(x)−∇f(xm+1),p−xm+1⟩≤0$ $∀p∈Γ.$ Thus $p∈Qm+1.$ This proves that $Γ⊂Cm+1∩Qm+1.$ Therefore $Γ⊂Cn∩Qn$ Consequently, since $Cn∩Qn$ is closed and convex, then $xn+1=ProfCn∩Qnfx$ is well-defined.

Step 2: We prove that ${xn},{yn},{zn},{Resλn,gjfxn}$ and ${Wnzn}$ are bounded.

Since $Γ⊂Cn∩Qn$ for every $n≥0$ and $xn+1=ProjCn∩Qnfx,$ then

So ${Df(p,xn)}$ is bounded and hence there exists a constant $M>0$ such that

In view of Lemma 2.6, we conclude that the sequence {xn} is bounded. Similarly, the sequences ${yn},{zn},{Resλn,gjfxn}$ and ${Wnzn}$ are bounded.

Step 3: Next, we show that $limn→∞||xn+1−xn||=0,$ $limn→∞||Resλn,gjfxn−xn||=0$ and $limn→∞||Wnzn−zn||=0.$

Since $xn+1∈Cn∩Qn⊂Qn$ and $xn=ProjQnf(x)$, we have

$Df(xn+1,ProjQnf(x))+Df(ProjQnf(x1),x)≤Df(xn+1,x).$

Thus

$Df(xn+1,xn)+Df(xn,x)≤Df(xn+1,x).$

Therefore the sequence ${Df(xn,x)}$ is non-decreasing and thus $limn→∞Df(xn,x)$ exists. Hence, it follows that $limn→∞Df(xn+1,xn)=0,$ and by Lemma 2.5, we have

$limn→∞||xn+1−xn||=0.$

Also, since $xn+1∈Cn,$ we have

$Df(xn+1,yn)≤Df(xn+1,xn).$

This yields that $limn→∞Df(xn+1,yn)=0$ and thus

$limn→∞||xn+1−yn||=0.$

Therefore from (4.11) and (4.12), we get

$limn→∞||yn−xn||=0.$

By the uniform continuity of f and ∇f on bounded subsets of E and E* respectively, we have

$limn→∞||f(yn)−f(xn)||=0$

and

$limn→∞||∇f(yn)−∇f(xn)||*=0.$

Furthermore,

$Df(p,xn)−Df(p,yn)=f(p)−f(xn)−⟨p−xn,∇f(xn)⟩ −f(p)+f(yn)+⟨p−yn,∇f(yn)⟩ =f(yn)−f(xn)+⟨p−yn,∇f(yn)⟩−⟨p−xn,∇f(xn)⟩ =f(yn)−f(xn)+⟨xn−yn,∇f(yn)⟩ −⟨p−xn,∇f(yn)−∇f(xn)⟩.$

Therefore from (4.12) - (4.14), we get

$limn→∞[Df(p,xn)−Df(p,yn)]=0.$

Note that from (4.8), we have

$Df(p,yn)≤Df(p,xn)−(1−δn)βn,0∑ j=1∞βn,jρr*(||xn−Resλn,gjfxn||) −δn(1−δn)ρr*(||xn−Wnzn||).$

Using the property of $ρr*$ and conditions (ii) and (iii) together with (4.15), we have

$limn→∞||xn−Resλn,gjfxn||=0$

and

$limn→∞||xn−Wnzn||=0.$

By the uniform continuity of ∇f on bounded subsets of E*, we have

Hence from (4.7), we get

Furthermore, since f is Fréchet differentiable on bounded subset of E, then ∇f* is uniformly continuous on bounded subsets of E*. Thus

$limn→∞||zn−xn||=0.$

Therefore

$limn→∞||Wnzn−zn||=limn→∞[||Wnzn−xn||+||xn−zn||]=0.$

Since ${xn}$ is bounded, there exists a subsequence ${xnk}$ of ${xn}$ which converges weakly to $q∈E.$ Since $||Wnzn−zn||→0$ and $||zn−xn||→0$ as $n→∞,$ then from Lemma 2.11 we have that $q∈F(Wn).$ Hence $q∈∩ i=1NF(Ti).$

Also from Lemma 2.11, we have for each $j=1,2,…$

Hence

From the assumption (A2), we have

$1λnk||y−Resλnk,gjfxnk||||∇f(Resλnkgjfxnk)−∇f(xnk)|| ≥1λnk⟨y−Resλnk,gjfxnk,∇f(Resλnk,gjfxnk)−∇f(xnk)⟩ ≥−gj(Resλnkgjfxnk,y)≥gj(y,Resλnk,gjfxnk) ∀y∈C.$

Taking the limit as $k→∞$ in the above inequality, from (A4) and condition (iv), we have $xnk→q,||∇f(Resλ nk,gjfxnk)−∇f(xnk)||→0,$ we have that $gj(y,q)≤0$ for all $y∈C.$ For $0 and $y∈C,$ define $yt=ty+(1−t)q$. Noting that $yt∈C,$ which yields $gj(yt,q)≤0.$ It therefore follows from (A1) that

That is $gj(yt,y)≥0.$

Let $t↓0,$ from (A3), we obtain $gj(q,y)≥0$ for any $y∈C,$ $j=1,2,….$ This implies that $q∈∩ j=1∞EP(gj)$. Therefore $q∈Γ:=∩ i=1NF(Ti)∩∩ j=1∞EP(gj).$

Now since $xn+1=ProjCn∩Qnfx,$ we have

$⟨∇f(x)−∇f(xn+1),xn+1−z⟩≥0 ∀z∈Cn∩Qn.$

Since $Γ⊂Cn∩Qn,$ we have

$⟨∇f(x)−∇f(xn+1),xn+1−z⟩≥0 ∀z∈Γ.$

Taking the limit of the above inequality, we have

$⟨∇f(x)−∇f(q),q−z⟩≥0 ∀z∈Γ.$

Therefore $q=ProjΓfx.$ This completes the proof.

5. Application to Zeros of Maximal Monotone Operators

Sabach [37] showed that under some properties of the function f, the solution set of the equilibrium problem is equivalent to the set of zeros of a maximal monotone operator, that is the points x* ∈ dom A such that

$0*∈Ax*,$

where $A:E→2E*$ is a maximal monotone operator. We denotes the set of zeros of A by $A−1(0*).$ An operator $A:E→2E*$ is said to be monotone if for any we have

A monotone operator A is said to be maximal if the graph of A, $Gr(A):={(x,ξ):ξ∈Ax}$ is not contained in the graph of any other monotone operator. The problem of finding the zeros of monotone operators is very important due to its applications in differential equations, evolution equations, optimization and other related fields. Many algorithms have also been introduced to find its solutions in Hilbert and Banah spaces.

Let $g:C×C→ℝ$ be a bifunction and define the following operator $Ag:E→2E*$ in the following manner

The following result was proved for the mapping Ag in [37].

Proposition 5.1. (Sabach [37]) Let C be a nonempty, closed and convex subset of a reflexive Banach space E and let $f:E→ℝ$ be a coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Assume that the bifunction $g:C×C→ℝ$ satisfies conditions (A1)-(A4), then:

(i) $EP(g)=Ag−1(0*);$

(ii) Ag is maximal monotone operator;

(iii) $Resgf=ResAgf.$

Based on the above result, we propose the following which can be obtain from Theorem 4.2 for finding common fixed point of finite family of n-generalized Bregman nonspreading mapping and zeros of maximal monotone operators in reflexive Banach space.

Theorem 5.2. Let C be a nonempty, closed and convex subset of a real reflexive Banach space E and $f:E→ℝ$ be a coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. For $i=1,2,…,N,$ let ${αn,i}⊂(0,1),$ $Ti:C→C$ be finite family of n-generalized Bregman nonspreading mappings and $Wn:C→C$ be a Bregman W-mapping generated by ${αn,i}$ and $T1,T2,…,TN$ in (4.1). Let $gj:C×C→ℝ$ be bifunctions satisfying assumptions (A1)-(A4), $Agj:E→2E*$ be as defined in (5.3) for $j=1,2,…$ and suppose $Γ:=∩ i=1NF(Ti)∩∩ j=1∞Agj−1(0*)≠∅.$ Define the sequence ${xn}$ by the following process

$x0=x∈C,C0=Q0=C,zn=∇f*[βn,0∇f(xn)+∑ j=1∞βn,j∇f(ResAgjfxn)],yn=∇f*[δn∇f(xn)+(1−δn)∇f(Wnzn)],Cn={z∈C:Df(z,yn)≤Df(z,xn)},Qn={z∈C:⟨∇f(x)−∇f(xn),xn−z⟩≥0},xn+1=ProjCn∩Qnfx,$

for all n ≥ 0, where ${βn,j}$ and ${δn}$ are sequences in [0,1) satisfying the following control conditions:

(i) $∑ j=0∞βn,j=1,$

(ii) There exists $k∈ℕ$ such that $liminfn→∞βn,jβn,k>0,$ $∀j∈ℕ∪{0};$

(iii) $0≤δn<1,$ $∀n∈ℕ$ and $liminfn→∞δn<1.$

Then, the sequence ${xn}$ converges strongly to Proj$Γfx$ as $n→∞.$

We give a numerical example to demonstrate the performance of our algorithm (4.7).

Example 6.1. Let $E=ℝ$, $C=[−10,10]$ and let $f:ℝ→ℝ$ be defined by $f(x)=23x2.$ Let $g:C×C→ℝ$ be defined by $g(x,y)=x(y−x)$, $∀x,y∈C$ and $T:C→C$ be defined by $Tix=13ix,$ $i=1,2,…,N.$ It is easy to observe that f is coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subset of $ℝ$ and $∇f(x)=43x.$ Also since $f*(x*)=sup{⟨x*,x⟩−f(x):x∈ℝ},$ then $f*(z)=38z2$ and $∇f*(z)=34z.$ Further, Ti is 1-generalized Bregman nonspreading mapping and $Resλn,gjfz=z2−3λnj.$

Choose ${αn,i}=1(n+i)2,$ ${δn}=1(n+1)2,$ ${λn}=12$ and for each $n∈ℕ∪{0}$, and $j≥0$, let ${βn,j}$ be defined by

$βn,j=13j+1nn+1, n>j,1−nn+1∑ k=1 n13k n=j,0 n

Observe that g satisfy Assumption (A1)-(A4) and $Γ={0}≠∅.$ After simplification, the hybrid iterative scheme (4.7) reduces to the following: Given x0,

$zn=34βn,043(xn)+∑ j=1∞βn,j2xn3(2−3j);yn=3443(n+1)2(xn)+1−1(n+1)243(Wnzn);Cn=0,2(xn2+yn2)3;Qn=0,xn;xn+1=ProjCn∩Qnfx0,$

where $Wnzn$ is computed as follow:

$Sn,0zn=zn, Sn,1zn=zn3(n+1)2+1−1(n+1)2zn; Sn,2zn=zn6(n+2)2+1−1(n+2)2Sn,1zn; ⋮Wnzn=Sn,N=zn3N(n+N)2+1−1(n+N)2Sn,N−1zn.$

Finally, we select the following values

Case(i): N=10 and x0 = -1,

Case(ii): N=50 and x0 = 0.5,

Case(iii): N= 100 and x0 = 2.

Using Matlab 2016(b) and $ϵ=10−6$ as stopping criterion, we plot the graphs of error $||xn+1−xn||$ against number of iteration in each case. The computational results can be found in Figure 1.

Figure 1. Example 6.1, Top-Left: Case(i); Top-Right: Case(ii); Bottom: Case(iii).

The authors sincerely thank the reviewer for his careful reading, constructive comments and fruitful suggestions that substantially improved the manuscript. The first author acknowledges with thanks the bursary and financial support from Department of Science and Innovation and National Research Foundation, Republic of South Africa Center of Excellence in Mathematical and Statistical Sciences (DSI-NRF COE-MaSS) Doctoral Bursary. The second author is supported by the National Research Foundation (NRF) of South Africa Incentive Funding for Rated Researchers (Grant Number 119903). Opinions expressed and conclusions arrived are those of the authors and are not necessarily to be attributed to the NRF.

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