An inertial s-iteration process for a common fixed point of a family of quasi-Bregman nonexpansive mappings

Abstract

In this paper, an inertial S-iteration iterative process for approximating a common fixed point of a finite family of quasi-Bregman nonexpansive mappings is introduced and studied in a reflexive Banach space. A strong convergence theorem is proved. Some applications of the theorem are presented. The results presented here improve, extend, and generalize some recent results in the literature.

1 Introduction

Let E be a real reflexive Banach space with dual space $E^{*}$. Throughout this paper we shall assume that $f:E \to (- \infty, + \infty )$ is a proper, lower semicontinuous, and convex function. We denote by $\operatorname{dom} f:=\{x \in E: f(x)< + \infty \}$, the domain of f. Let $x \in \operatorname{int} \operatorname{dom} f$, then the subdifferential of f at x is the convex function defined by

$$\begin{aligned} \partial {f}(x)= \bigl\{ x^{*} \in E^{*}: f(x) + \bigl\langle x^{*},y-x\bigr\rangle \leq f(y), \forall y \in E\bigr\} . \end{aligned}$$

The Fenchel conjugate of f is the function $f^{*}:E^{*} \to (- \infty, + \infty ]$ defined by

$$\begin{aligned} f^{*}\bigl(x^{*}\bigr)=\sup \bigl\{ \bigl\langle x^{*},x \bigr\rangle - f(x):x \in E \bigr\} . \end{aligned}$$

It is known that the Young–Fenchel inequality,

$$\begin{aligned} \bigl\langle x^{*},x \bigr\rangle \leq f(x)+f\bigl(x^{*} \bigr),\quad \forall x \in E, x^{*} \in E^{*}, \end{aligned}$$

holds. A function f is coercive [12] if the sublevel set of f is bounded; equivalently,

$$\begin{aligned} \lim_{\|x\| \rightarrow \infty } f(x)=+ \infty. \end{aligned}$$

A function f is said to be strongly coercive if

$$\begin{aligned} \lim_{ \Vert x \Vert \rightarrow \infty } \frac{f(x)}{ \Vert x \Vert }=+ \infty. \end{aligned}$$

For any $x \in \operatorname{int} \operatorname{dom} f$ and $y \in E$, the derivative of f at x in the direction of y is defined by

$$\begin{aligned} {f}^{0}(x,y)= \lim_{t \rightarrow 0} \frac{f(x+ty) - f(x)}{t}. \end{aligned}$$

(1.1)

The function f is said to be Gâteaux differentiable at x if the limit (1.1) exists for any y. In this case, the gradient of f at x is the function $\nabla f(x):E \to (- \infty, + \infty ]$ defined by $\langle \nabla f(x),y \rangle =f^{0}(x,y)$ for any y ∈E. The function f is said to be Gâteaux differentiable if it is Gâteaux differentiable at every point $x\in \operatorname{int} \operatorname{dom} f$. Furthermore, f is said to be Fréchet differentiable at x if this limit (1.1) is attained uniformly in y, $\|y\|=1$; f is said to be uniformly Fréchet differentiable on a subset C of E if the limit (1.1) is attained uniformly for $x \in C$ and $\|y\|=1$. It is well known that if f is Gâteaux differentiable (respectively Fréchet differentiable) on $\operatorname{int} \operatorname{dom} f$, then f is continuous and its Gâteaux derivative ∇f is norm-to-weak^∗ continuous (respectively continuous) on $\operatorname{int} \operatorname{dom} f$, see, for example, [2, 3, 6]. Let $f:E \to (- \infty, + \infty )$ be a convex and Gâteaux differentiable function. The Bregman distance with respect to f, $D_{f}:\operatorname{dom} f \times \operatorname{int} \operatorname{dom} f \to [0, + \infty )$ is defined as

$$D_{f}(x,y)= f(x)-f(y)- \bigl\langle \nabla f(y),x-y \bigr\rangle . $$

Let C be a nonempty closed and convex subset of E. Let $T: C \rightarrow E$ be a mapping, then

• A point $v \in C$ is said to be an asymptotic fixed point of T if for any sequence $\{x_{n}\} \subset C$ which converges weakly to v, $\lim_{n \to \infty }\|x_{n}-Tx_{n}\|=0$. The set of asymptotic fixed points of T is denoted by $\hat{F}(T)$;

• T is said to be Bregman relatively nonexpansive if $F(T) \neq \emptyset,F(T)=\hat{F}(T)$, and $D_{f}(x,Ty) \leq D_{f}(x,y)$ for any $x \in C, y \in F(T)$;

• T is said to be quasi-Bregman nonexpansive if $F(T) \neq \emptyset$ and $D_{f}(x,Ty) \leq D_{f}(x,y)$ for any $x\in C, y \in F(T)$;

• $(I -T)$ is demiclosed at $y\in E$ if having a sequence $\{v_{n}\}$ in C converging weakly to u and $\{ v_{n} - Tv_{n}\}$ converging strongly to y implies that $(I -T)u =y$ where I is the identity mapping. From this we get that $(I-T)$ is demiclosed at zero if whenever a sequence $\{v_{n}\}$ in C converges weakly to u and $\{ v_{n} - Tv_{n}\}$ converges strongly to 0 then $u \in F(T)$.

Agarwal et al. [1] introduced and studied a two-step iterative process called the S-iteration process. They proved a convergence theorem for fixed points of nearly asymptotically nonexpansive mappings. Since then various modifications of the S-iteration scheme and also multistep schemes were studied by many authors for solutions of some nonlinear problems, see, for example, [10, 11, 15] and the references therein.

Suparatulatorn et al. [24] introduced and studied an iteration method called modified S-iteration process which is defined by

$$\begin{aligned} \textstyle\begin{cases} x_{0} \in C; \\ y_{n}=(1-\beta _{n})x_{n} + \beta _{n}S_{1}x_{n}; \\ x_{n+1}=(1-\alpha _{n})S_{1}x_{n} + \alpha _{n}S_{2}y_{n}, \end{cases}\displaystyle \end{aligned}$$

where C is a nonempty closed convex subset of a real Banach space, $S_{1},S_{2}$ are G-nonexpansive mappings, and $\{\alpha _{n}\}, \{\beta _{n}\}$ $\subset (0,1)$. They proved that the sequence generated by the iterative algorithm converges weakly to a common fixed point of two G-nonexpansive mappings in a uniformly convex Banach space.

Recently, Phon-on et al. [17] studied the following inertial modified S-iteration process by combining the inertial extrapolation and modified S-iteration process to speed up the convergence of the modified S-iteration process:

$$\begin{aligned} \textstyle\begin{cases} w_{n}=x_{n} + \gamma _{n}(x_{n}-x_{n-1}); \\ y_{n}=(1-\beta _{n})w_{n} + \beta _{n}S_{1}w_{n}; \\ x_{n+1}=(1-\alpha _{n})S_{1}w_{n} + \alpha _{n}S_{2}y_{n}, \end{cases}\displaystyle \end{aligned}$$

$n\geq 1$, where $S_{1},S_{2}$ are nonexpansive mappings, $\{S_{i}w_{n}-w_{n}\}$ bounded for $i=1,2$, $\{S_{i}w_{n}-y\}$ is bounded for $i=1,2$, and for any $y\in F(S_{1}) \cap F(S_{2})$, $\sum_{n=1}^{\infty } \gamma _{n} < \infty$, $\{\gamma _{n}\}\subset [0,\gamma ]$, $0\leq \gamma <1$, $\{\alpha _{n}\},\{\beta _{n}\} \subset [\delta,1-\delta ]$ for some $\delta \in (0,0.5)$.

They proved, under some assumptions, that the sequence generated by the algorithm converges weakly to a common fixed point of two nonexpansive mappings in a uniformly convex Banach space. Several inertial algorithms were studied by numerous authors to speed up the convergence processes of iterative schemes, see, for example, [13, 18–20] and the references contained therein.

Motivated by the results of Phon-on et al. [17] and Suparatulatorn et al. [24], we raised the following interesting questions:

• 1. Can one iteratively approximate solutions of inertial modified S-iteration process in real Banach spaces more general than uniformly convex spaces?

• 2. Can the result also be proved for a common fixed point of a finite family of quasi-Bregman nonexpansive mappings?

• 3. Can a strong convergence theorem be proved without assuming that the operator is semicompact?

In this paper, we answer the questions in the affirmative. We introduce and study the following algorithm:

$$\begin{aligned} \textstyle\begin{cases} x_{0},x_{1}\in C, \quad C=C_{1}; \\ w_{n}= x_{n}+\gamma _{n} (x_{n}- x_{n-1}); \\ y_{1n}=\nabla f^{*}(\beta _{n} \nabla fw_{n} + (1- \beta _{n}) \nabla f{S_{1}}w_{n}); \\ y_{in}= \nabla f^{*}(\beta _{n} \nabla f{S_{i-1}}w_{n}+ (1- \beta _{n}) \nabla f{S_{i}}y_{(i-1)n}),\quad 2\le i\le m; \\ C_{in}=\{ v \in C_{n}: D_{f}(v,y_{in})\leq D_{f}(v,w_{n})\}; \\ C_{n+1}=\bigcap_{i=1}^{{m}}C_{in}; \\ x_{n+1}={\Pi _{C_{n+1}}}^{f}x_{0}, \end{cases}\displaystyle \end{aligned}$$

(1.2)

where C is a nonempty, closed, and convex subset of a reflexive Banach space E, for some natural number $m\ge 2$, $\{S_{i}\}_{i=1}^{m}$ is a finite family of quasi-Bregman nonexpansive self-mappings of C, $\{\gamma _{n}\}, \{\beta _{n}\}\subset (a,b)$ are sequences such that $0< a< b<1$. Then we prove that the sequence generated by the algorithm (1.2) converges to a common fixed point of a finite family of quasi-Bregman nonexpansive mappings. Furthermore, we apply our theorem to solution of some equilibrium problem and zeros of some maximal monotone operators.

2 Preliminaries

Let $f:E \to (- \infty, + \infty )$ be a convex and Gâteaux differentiable function. The modulus of total convexity of f at $x \in \operatorname{int} \operatorname{dom} f$ is the function $v_{f}(x,\cdot ):[0,+\infty ) \to [0,+\infty )$ defined by

$$\begin{aligned} v_{f}(x,t):= \inf \bigl\{ D_{f}(y,x): y \in \operatorname{dom} f, \Vert y-x \Vert =t\bigr\} . \end{aligned}$$

The function f is called totally convex at x if $v_{f}(x,t)>0$ whenever $t>0$. The function f is called totally convex if it is totally convex at every point $x \in \operatorname{int} \operatorname{dom} f$ and is said to be totally convex on bounded subsets if $v_{f}(B,t)>0$ 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 $v_{f}:\operatorname{int} \operatorname{dom} f \times [0,+\infty ) \to [0,+ \infty )$ defined by

$$\begin{aligned} v_{f}(B,t):= \inf \bigl\{ v_{f}(x,t): x \in B \cap \operatorname{dom} f\bigr\} . \end{aligned}$$

The function f is said to be Legendre if it satisfies the following conditions:

• (1) $\operatorname{int} \operatorname{dom} f \neq \emptyset$ and the subdifferential ∂f is single-valued on its domain;

• (2) $\operatorname{int} \operatorname{dom} f^{*} \neq \emptyset$ and $\partial {f^{*}}$ is single-valued on its domain.

If E is a reflexive Banach space, we have the following:

• (i) f is Legendre if and only if $f^{*}$ is Legendre (see [4, Corollary 55]).

• (ii) If f is Legendre, then ∇f is a bijection satisfying $\nabla f=(\nabla f^{*})^{-1}$, $\operatorname{ran} \nabla f=\operatorname{dom} \nabla f^{*}= \operatorname{int} \operatorname{dom} f^{*}$ and $\operatorname{ran} \nabla f^{*}=\operatorname{dom} f=\operatorname{int} \operatorname{dom} f$(see [4, Theorem 5.10]).

If the Banach space E is smooth and strictly convex, the function $\frac{1}{p}\|\cdot \|^{p}$ with $p \in (1, \infty )$ is Legendre.

The Bregman projection [7] with respect to f of $x \in \operatorname{int} \operatorname{dom} f$ onto a nonempty closed convex subset $C \subset \operatorname{int} \operatorname{dom} f$ is defined as the unique vector ${\Pi _{C}}^{f}x \in C$, which satisfies

$$\begin{aligned} D_{f}\bigl({\Pi _{C}}^{f}x,x\bigr)= \inf \bigl\{ D_{f}(y,x), y \in C\bigr\} . \end{aligned}$$

Lemma 2.1

([8])

Let C be a nonempty closed and convex subset of a reflexive Banach space E. Let $f:E \to \mathbb{R}$ be a Gâteaux differentiable and totally convex function and let $x \in E$. Then

• (1) $z={\Pi _{C}}^{f}x$ if and only if $\langle \nabla fx- \nabla fz, y-z \rangle \leq 0, \forall y \in C$;

• (2) $D_{f}(y, {\Pi _{C}}^{f}x)+ D_{f}({\Pi _{C}}^{f}x,x) \leq D_{f}(y,x), \forall x \in E, y \in C$.

Lemma 2.2

([8, 14])

Let E be a reflexive Banach space. Let $f:E \to \mathbb{R}$ be a strongly coercive Bregman function and let V be the function defined by

$$\begin{aligned} V_{f}\bigl(x,x^{*}\bigr)=f(x)- \bigl\langle x,x^{*} \bigr\rangle +f^{*}\bigl(x^{*}\bigr),\quad x \in E, x^{*} \in E^{*}. \end{aligned}$$

Then the following hold:

• (1) $D_{f}(x, \nabla f^{*}(x^{*}))= V(x,x^{*}), \forall x\in E, x^{*} \in E^{*}$;

• (2) $V_{f}(x,x^{*})+ \langle \nabla f^{*}(x^{*})-x, y^{*} \rangle \leq V_{f}(x,x^{*}+y^{*})$.

Lemma 2.3

([22])

If $f:E \to \mathbb{R}$ is uniformly Fréchet differentiable and bounded on bounded subsets of E, then ∇f is uniformly continuous on bounded subsets of E from strong topology of E to the strong topology of $E^{*}$.

Theorem 2.4

([25])

Let E be a reflexive Banach space and let $f:E \to \mathbb{R}$ be a convex function which is bounded on bounded subsets of E. Then the following are equivalent:

• (1) f is strongly coercive and uniformly convex on bounded subsets of E.

• (2) $\operatorname{dom} f^{*}=E^{*}$, $f^{*}$ is bounded and uniformly smooth on bounded subsets of $E^{*}$.

• (3) $\operatorname{dom} f^{*}=E^{*}$, $f^{*}$ is Fréchet differentiable and $\nabla f^{*}$ is norm-to-norm uniformly continuous on bounded subsets of $E^{*}$.

Theorem 2.5

([25])

Let E be a reflexive Banach space and let $f:E \to \mathbb{R}$ be a continuous convex function which is strongly coercive. Then the following are equivalent:

• (1) f is bounded and uniformly smooth on bounded subsets of E.

• (2) $f^{*}$ is Fréchet differentiable and $f^{*}$ is norm-to-norm uniformly continuous on bounded subsets of $E^{*}$.

• (3) $\operatorname{dom} f^{*}=E^{*}$, $f^{*}$ is strongly coercive and uniformly convex on bounded subsets of $E^{*}$.

Lemma 2.6

Let E be a reflexive Banach space, let $r>0$ be a constant, let $\rho _{r}$ be the gauge of uniform convexity of f, and let $f:E \to \mathbb{R}$ be a convex function which is bounded and uniformly convex on bounded subsets of E. Then, for any $x \in E, y^{*}, z^{*} \in B_{r}$ and $\alpha \in (0,1)$,

$$\begin{aligned} V_{f}\bigl(x, \alpha y^{*} + (1- \alpha ) z^{*} \bigr)\leq \alpha V_{f}\bigl(x,y^{*}\bigr) + (1- \alpha )V_{f}\bigl(x, z^{*}\bigr)- \alpha (1- \alpha ) {\rho _{r}}^{*}\bigl( \bigl\Vert y^{*}-z^{*} \bigr\Vert \bigr). \end{aligned}$$

Lemma 2.7

([16])

Let E be a Banach space and $f:E \to \mathbb{R}$ be a Gâteaux differentiable function which is uniformly convex on bounded subsets of E. Let $\{x_{n}\}$ and $\{y_{n}\}$ be bounded sequences in E. Then

$$\begin{aligned} \lim_{n\rightarrow \infty }D_{f}(x_{n},y_{n})=0 \quad\textit{if and only if}\quad \lim_{n \rightarrow \infty } \Vert x_{n}-y_{n} \Vert =0. \end{aligned}$$

Lemma 2.8

([21])

Let $f:E \to \mathbb{R}$ be a Gâteaux differentiable and totally convex function. If $x_{0} \in E$ and the sequence $\{D_{f}(x_{n},x_{0})\}$ is bounded, the sequence $\{x_{n}\}$ is bounded, too.

The function f is called sequentially consistent if for any two sequences $\{u_{n}\}$ and $\{v_{n}\}$ in E such that the first one is bounded:

$$\begin{aligned} \lim_{n\rightarrow \infty }D_{f}(u_{n},v_{n})=0\quad \text{implies}\quad \lim_{n \rightarrow \infty } \Vert u_{n}-v_{n} \Vert =0. \end{aligned}$$

Lemma 2.9

([9])

The function f is totally convex on bounded subsets if and only if the function f is sequentially consistent.

3 Main results

Theorem 3.1

Let C be a nonempty, closed, and convex subset of a reflexive Banach space E, and let $f:E\to \mathbb{R}$ be a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let $\{S_{i}\}_{i=1}^{m}$ be a finite family of quasi-Bregman nonexpansive self mappings of C such that $S_{i}$ is $L_{i}$-Lipschitz and $(I-S_{i})$ is demiclosed at 0 for each $i\in \{1,2, \dots,m\}$. Assume $\Gamma = \bigcap_{i=1}^{m} F(S_{i})\neq \emptyset$. Let a sequence $\{x_{n}\}$ be generated by (1.2), then the sequence $\{x_{n}\}$ converges to ${\Pi _{\Gamma }}^{f}x_{0}$.

Proof

We divide the proof into six steps.

Step 1. We show that $C_{n}$ is closed and convex for any $n \geq 1$.

Since $C=C_{1}$, $C_{1}$ is closed and convex.

Assume $C_{n}$ is closed and convex for some $n\geq 1$. Since for any $y\in C_{n}$, $i=1$,

$$\begin{aligned} &D_{f}(y,y_{1n}) \leq D_{f}(y,w_{n}) \\ & \quad\Leftrightarrow\quad f(w_{n})-f(y_{1n})+\bigl\langle \nabla f(w_{n}),y-w_{n} \bigr\rangle -\bigl\langle \nabla f(y_{1n}),y-y_{1n}\bigr\rangle \leq 0 \\ &\quad\Leftrightarrow\quad f(w_{n})-f(y_{1n})+\bigl\langle \nabla f(y_{1n}),y_{1n} \bigr\rangle -\bigl\langle \nabla f(w_{n}),w_{n}\bigr\rangle \leq \bigl\langle \nabla f(y_{1n})- \nabla f(w_{n}),y \bigr\rangle \end{aligned}$$

and, for $2\leq i \leq m$,

$$\begin{aligned} &D_{f}(y,y_{in})\leq D_{f}(y,w_{n}) \\ &\quad\Leftrightarrow \quad f(w_{n})-f(y_{in})+\bigl\langle \nabla f(w_{n}),y-w_{n} \bigr\rangle -\bigl\langle \nabla f(y_{in}),y-y_{in}\bigr\rangle \leq 0 \\ &\quad \Leftrightarrow\quad f(w_{n})-f(y_{in})+\bigl\langle \nabla f(y_{in}),y_{in} \bigr\rangle -\bigl\langle \nabla f(w_{n}),w_{n}\bigr\rangle \leq \bigl\langle \nabla f(y_{in})- \nabla f(w_{n}),y \bigr\rangle , \end{aligned}$$

we have that $C_{n+1}$ is closed and convex. Therefore, $C_{n}$ is closed and convex for any $n \geq 1$.

Step 2. We show that $\Gamma \subset C_{n}$ for any $n \geq 1$.

For $n=1$, $\Gamma \subset C=C_{1}$.

Now assume $\Gamma \subset C_{n}$ for some $n\geq 1$. Let $u\in \Gamma$, then by Lemma 2.6, we have for $i=1$,

$$\begin{aligned} D_{f}(u,y_{1n}) ={}& D_{f} \bigl(u,\nabla f^{*}\bigl(\beta _{n} \nabla f(w_{n})+ (1- \beta _{n}) \nabla f\bigl(S_{1} (w_{n})\bigr)\bigr) \bigr) \\ ={}&V_{f} \bigl(u, \beta _{n} \nabla f(w_{n})+ (1- \beta _{n}) \nabla f\bigl(S_{1} (w_{n})\bigr) \bigr) \\ ={}&f(u) - \bigl\langle u,\beta _{n} \nabla f(w_{n})+ (1- \beta _{n}) \nabla f\bigl(S_{1} (w_{n})\bigr) \bigr\rangle \\ &{} +f^{*} \bigl(\beta _{n} \nabla f(w_{n})+ (1- \beta _{n}) \nabla f\bigl(S_{1} (w_{n})\bigr) \bigr) \\ ={}& \beta _{n} f(u)+ (1- \beta _{n}) f(u) - \beta _{n} \bigl\langle u, \nabla f(w_{n}) \bigr\rangle - (1- \beta _{n}) \bigl\langle u,\nabla f\bigl(S_{1} (w_{n})\bigr) \bigr\rangle \\ &{} +f^{*} \bigl(\beta _{n} \nabla f(w_{n})+ (1- \beta _{n}) \nabla f\bigl(S_{1} (w_{n})\bigr) \bigr) \\ \leq{}& \beta _{n} f(u)+ (1- \beta _{n}) f(u) - \beta _{n} \bigl\langle u, \nabla f(w_{n}) \bigr\rangle - (1- \beta _{n}) \bigl\langle u,\nabla f\bigl(S_{1} (w_{n})\bigr) \bigr\rangle \\ &{} + \beta _{n} f^{*} \bigl( \nabla f(w_{n}) \bigr)+ (1- \beta _{n}) f^{*} \bigl(\nabla f \bigl(S_{1} (w_{n})\bigr) \bigr) \\ ={}& \beta _{n} \bigl[f(u) - \bigl\langle u, \nabla f(w_{n}) \bigr\rangle + f^{*} \bigl( \nabla f(w_{n}) \bigr) \bigr] \\ &{} +(1- \beta _{n}) \bigl[f(u) - \bigl\langle u,\nabla f \bigl(S_{1} (w_{n})\bigr) \bigr\rangle + f^{*} \bigl(\nabla f\bigl(S_{1} (w_{n})\bigr) \bigr) \bigr] \\ = {}&\beta _{n} D_{f}(u,w_{n}) +(1- \beta _{n})D_{f}\bigl(u,S_{1}(w_{n})\bigr) \\ \leq{}& \beta _{n} D_{f}(u,w_{n}) +(1- \beta _{n})D_{f}(u,w_{n}) \\ ={}& D_{f}(u,w_{n}) \end{aligned}$$

(3.1)

Now for $2\le i\le m$, we have

$$\begin{aligned} & D_{f}(u,y_{in}) \\ &\quad= D_{f} \bigl(u,\nabla f^{*} \bigl(\beta _{n} \nabla f(S_{i-1}w_{n})+ (1- \beta _{n}) \nabla f(S_{i} y_{(i-1)n}) \bigr) \bigr) \\ &\quad=V_{f} \bigl(u, \beta _{n} \nabla f(S_{i-1}w_{n})+ (1- \beta _{n}) \nabla f(S_{i} y_{(i-1)n}) \bigr) \\ &\quad=f(u) - \bigl\langle u,\alpha _{n} \nabla f(S_{i-1}w_{n})+ (1- \beta _{n}) \nabla f(S_{2} y_{(i-1)n}) \bigr\rangle \\ &\qquad{} +f^{*} \bigl(\beta _{n} \nabla f(S_{i-1}w_{n})+ (1- \beta _{n}) \nabla f(S_{i} y_{(i-1)n}) \bigr) \\ &\quad= \beta _{n} f(u)+ (1- \beta _{n}) f(u) - \beta _{n} \bigl\langle u, \nabla f(S_{i-1}w_{n}) \bigr\rangle \\ &\qquad{} - (1- \beta _{n}) \bigl\langle u,\nabla f(S_{i}y_{(i-1)n}) \bigr\rangle \\ & \qquad{}+f^{*} \bigl(\beta _{n} \nabla f(S_{i-1}w_{n})+ (1- \beta _{n}) \nabla f(S_{i} y_{(i-1)n}) \bigr) \\ &\quad\leq \beta _{n} f(u)+ (1- \beta _{n}) f(u) - \beta _{n} \bigl\langle u, \nabla f(S_{i-1}w_{n}) \bigr\rangle \\ &\qquad{} - (1- \beta _{n}) \bigl\langle u,\nabla f(S_{i} y_{(i-1)n}) \bigr\rangle \\ &\qquad{} + \beta _{n} f^{*} \bigl( \nabla f(S_{i-1}w_{n}) \bigr)+ (1- \beta _{n}) f^{*} \bigl(\nabla f(S_{i} y_{(i-1)n}) \bigr) \\ &\quad= \beta _{n} \bigl[f(u) - \bigl\langle u, \nabla f(S_{i-1}w_{n}) \bigr\rangle + f^{*} \bigl( \nabla f(S_{i-1}w_{n}) \bigr) \bigr] \\ & \qquad{}+(1- \beta _{n}) \bigl[f(u) - \bigl\langle u,\nabla f(S_{i} y_{(i-1)n}) \bigr\rangle + f^{*} \bigl(\nabla f(S_{i} y_{(i-1)n}) \bigr) \bigr] \\ &\quad= \beta _{n} D_{f}(u,S_{i-1}w_{n}) +(1- \beta _{n})D_{f}(u,S_{i}y_{(i-1)n}) \\ &\quad\leq \beta _{n} D_{f}(u,w_{n}) +(1- \beta _{n})D_{f}(u,y_{(i-1)n}) \\ &\quad\leq \beta _{n} D_{f}(u,w_{n}) +(1- \beta _{n}) \bigl[ \beta _{n} D_{f}(u,w_{n}) + (1-\beta _{n})D_{f}(u,y_{(i-2)n}) \bigr] \\ &\quad= \bigl(\beta _{n} + \beta _{n} (1-\beta _{n}) \bigr) D_{f}(u,w_{n}) + (1-\beta _{n})^{2} D_{f}(u,y_{(i-2)n}) \\ &\quad\leq \beta _{n} \bigl( 1+ (1- \beta _{n}) \bigr) D_{f}(u,w_{n}) \\ &\qquad{} +(1- \beta _{n})^{2} \bigl[ \beta _{n} D_{f}(u,w_{n}) + (1- \beta _{n})D_{f}(u,y_{(i-3)n}) \bigr] \\ &\quad= \beta _{n} \bigl(1+ (1-\beta _{n}) + (1-\beta _{n})^{2} \bigr) D_{f}(u,w_{n}) + (1- \beta _{n})^{3} D_{f}(u,y_{(i-3)n}) \\ &\quad\leq \\ &\quad \vdots \\ &\quad\leq \beta _{n} \bigl(1+ (1-\beta _{n}) + (1-\beta _{n})^{2}+ \cdots +(1-\beta _{n})^{i-1} \bigr) D_{f}(u,w_{n}) \\ & \qquad{}+ (1-\beta _{n})^{i} D_{f}(u,w_{n}) \\ &\quad=\beta _{n} \biggl[ \frac{1-(1-\beta _{n})^{i}}{1-(1-\beta _{n})} \biggr]D_{f}(u,w_{n}) + (1-\beta _{n})^{i} D_{f}(u,w_{n}) \\ &\quad= D_{f}(u,w_{n}). \end{aligned}$$

(3.2)

Hence $\Gamma \subset C_{n}$ for any $n \geq 1$.

Step 3. We shall show that $\{x_{n}\}$ is a Cauchy sequence.

Since $\Gamma \subset C_{n+1} \subset C_{n}$ and $x_{n}={\Pi _{C_{n}}}^{f} x_{0} \subset C_{n}$, by Lemma 2.1, we have that $D_{f}(x_{n}, x_{0}) \leq D_{f} (x_{n+1},x_{0})$ and also $D_{f}(x_{n},x_{0}) \leq D_{f} (u,x_{0}), u \in \Gamma$. Hence $D_{f}(x_{n}, x_{0})$ is nondecreasing and bounded. So, $\lim_{n\to \infty } D_{f} (x_{n}, x_{0})$ exists. Furthermore, by Lemma 2.8, $\{x_{n}\}$ is bounded. Also, since $x_{n}={\Pi _{C_{n}}}^{f} x_{0}$, it follows from Lemma 2.1 that $D_{f}(x_{k}, x_{n}) = D_{f} (x_{k},{\Pi _{C_{n} }}^{f}x_{0}) \leq D_{f} (x_{k},x_{0})-D_{f} (x_{n},x_{0}) \rightarrow 0$ as $n,k \rightarrow \infty$. Since f is totally convex on bounded subsets of E, f is sequentially consistent. Therefore $\|x_{n}-x_{k}\| \rightarrow 0$ as $n,k \rightarrow \infty$. Hence, $\{x_{n}\}$ is a Cauchy sequence.

Step 4. We show that

$$\begin{aligned} \lim_{n\rightarrow \infty } \Vert x_{n}-w_{n} \Vert &=\lim_{n \rightarrow \infty } \Vert x_{n}-y_{in} \Vert \\ &= \lim_{n\rightarrow \infty } \Vert y_{(i+1)n}-y_{in} \Vert \\ &=\lim_{n\rightarrow \infty } \bigl\Vert (I-S_{1})w_{n} \bigr\Vert \\ &=\lim_{n\rightarrow \infty } \bigl\Vert (I-S_{i})y_{(i-1)n} \bigr\Vert =0, \end{aligned}$$

for each $i\in \{1,2,\dots,m\}$.

Since $x_{n+1} \in C_{n+1} \subset C_{n}$, by Lemma 2.1, we have $D_{f} (x_{n+1},x_{n}) \leq D_{f} (x_{n+1},x_{0})-D_{f} (x_{n},x_{0})$. Taking the limit as $n \rightarrow \infty$, we have $\underset{n\to \infty }{\lim } D_{f} (x_{n+1},x_{n}) =0$.

Since f is totally convex on bounded subsets of E, f is sequentially consistent. Therefore

$$\begin{aligned} \Vert x_{n+1}-x_{n} \Vert \rightarrow 0\quad \text{as } n \rightarrow \infty. \end{aligned}$$

(3.3)

From (1.2) we get

$$\begin{aligned} \Vert x_{n} - w_{n} \Vert = \bigl\Vert \gamma _{n}(x_{n} - x_{n-1}) \bigr\Vert \leq \Vert x_{n}-x_{n-1} \Vert , \end{aligned}$$

which implies

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert x_{n}-w_{n} \Vert = 0. \end{aligned}$$

(3.4)

Since $\{x_{n}\}$ is bounded, (3.4) implies that $\{w_{n}\}$ is also bounded and

$$\begin{aligned} \Vert x_{n+1} -w_{n} \Vert \leq \Vert x_{n+1}-x_{n} \Vert + \Vert x_{n} - w_{n} \Vert . \end{aligned}$$

Thus, we get

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert x_{n+1}-w_{n} \Vert = 0. \end{aligned}$$

By Lemma 2.7,

$$\begin{aligned} \lim_{n \rightarrow \infty } D_{f}(x_{n+1},w_{n})= 0. \end{aligned}$$

Since $x_{n+1} \in C_{n}$, for $1 \leq i \leq m$, from (1.2) we have $D_{f}(x_{n+1},y_{in}) \leq D_{f}(x_{n+1},w_{n})$. Hence $\lim_{n \to \infty } D_{f}(x_{n+1},y_{in})= 0$, $\forall i \in \{1,2,3,\dots,m \}$. Since f is totally convex on bounded subsets of E, f is sequentially consistent. Therefore

$$\begin{aligned} \Vert x_{n+1}-y_{in} \Vert \rightarrow 0 \quad\text{as } n \rightarrow \infty, \forall i \in \{1,2,3,\dots,m\}. \end{aligned}$$

(3.5)

Observe that $\| x_{n} -y_{in}\| \leq \|x_{n}-x_{n+1}\|+ \|x_{n+1} - y_{in}\|, \forall i \in \{ 1,2,3,\dots,m\}$, which implies

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert x_{n}-y_{in} \Vert = 0,\quad \forall i \in \{1,2,3,\dots,m\}. \end{aligned}$$

(3.6)

Also, $\| y_{in} -w_{n}\| \leq \|y_{in}-x_{n}\|+ \|x_{n} - w_{n}\|$. Thus,

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert y_{in}-w_{n} \Vert = 0, \quad \forall i \in \{1,2,3,\dots,m\}. \end{aligned}$$

(3.7)

Since ∇f is norm-to-norm uniformly continuous on bounded subsets of E, we have

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert \nabla fy_{in}- \nabla fw_{n} \Vert = 0, \quad\forall i \in \{1,2,3,\dots,m\}. \end{aligned}$$

(3.8)

Since $\{w_{n}\}$ is bounded, (3.7) implies that $\{y_{in}\}$ is also bounded.

Thus, for $1\leq i \leq m-1$, we have $\| y_{(i+1)n} -y_{in}\| \leq \|y_{(i+1)n}-x_{n+1}\|+ \|x_{n+1} - y_{in} \|$, so that

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert y_{(i+1)n}-y_{in} \Vert = 0. \end{aligned}$$

(3.9)

Since ∇f is norm-to-norm uniformly continuous on bounded subsets of E, we have

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert \nabla fy_{(i+1)n}- \nabla fy_{in} \Vert = 0,\quad \forall i \in \{1,2,3,\dots,m-1\}. \end{aligned}$$

(3.10)

From (1.2)

$$\begin{aligned} \Vert \nabla fy_{1n}- \nabla fw_{n} \Vert =(1- \beta _{n}) \Vert \nabla f{S_{1}}w_{n}- \nabla fw_{n} \Vert . \end{aligned}$$

From (3.7), we have

$$\begin{aligned} 0=\lim_{n \rightarrow \infty } \Vert \nabla fy_{1n}- \nabla fw_{n} \Vert = \lim_{n \rightarrow \infty } (1- \beta _{n}) \Vert \nabla f{S_{1}}w_{n}- \nabla fw_{n} \Vert . \end{aligned}$$

Hence

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert \nabla f{S_{1}}w_{n}- \nabla fw_{n} \Vert = 0. \end{aligned}$$

(3.11)

This implies that as $\nabla f^{*}$ is norm-to-norm uniformly continuous on bounded subsets of $E^{*}$,

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert w_{n}- {S_{1}}w_{n} \Vert = 0. \end{aligned}$$

(3.12)

Now

$$\begin{aligned} \Vert y_{1n}-S_{1}w_{n} \Vert \leq \Vert y_{1n}-w_{n} \Vert + \Vert w_{n}-S_{1}w_{n} \Vert , \end{aligned}$$

which implies

$$\begin{aligned} \lim_{n\rightarrow \infty } \Vert y_{1n} -S_{1}w_{n} \Vert =0. \end{aligned}$$

Thus

$$\begin{aligned} \Vert y_{2n}-S_{1}w_{n} \Vert \leq \Vert y_{2n}-y_{1n} \Vert + \Vert y_{1n}-S_{1}w_{n} \Vert \end{aligned}$$

gives

$$\begin{aligned} \lim_{n\rightarrow \infty } \Vert y_{2n} -S_{1}w_{n} \Vert =0. \end{aligned}$$

Since ∇f is norm-to-norm uniformly continuous on bounded subsets of E, we have

$$\begin{aligned} \lim_{n\rightarrow \infty } \Vert \nabla f y_{2n} -\nabla fS_{1}w_{n} \Vert =0. \end{aligned}$$

Again, from (1.2), we have

$$\begin{aligned} \Vert \nabla fy_{2n}- \nabla f{S_{1}} w_{n} \Vert =(1-\beta _{n}) \Vert \nabla f{S_{2}}y_{1n}- \nabla f{S_{1}}w_{n} \Vert . \end{aligned}$$

Therefore,

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert \nabla fS_{2}y_{1n}- \nabla f{S_{1}}w_{n} \Vert = 0. \end{aligned}$$

Since $\nabla f^{*}$ is norm-to-norm uniformly continuous on bounded subsets of $E^{*}$, we have

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert S_{2}y_{1n}- S_{1} w_{n} \Vert = 0. \end{aligned}$$

Thus

$$\begin{aligned} \Vert y_{1n}- {S_{2}}y_{1n} \Vert \leq \Vert y_{1n}-w_{n} \Vert + \Vert w_{n}-{S_{1}}w_{n} \Vert + \Vert {S_{1}}w_{n}- {S_{2}}y_{1n} \Vert \end{aligned}$$

gives

$$\begin{aligned} \lim_{n \rightarrow \infty } \bigl\Vert (I-{S_{2}})y_{1n} \bigr\Vert = 0. \end{aligned}$$

(3.13)

Now

$$\begin{aligned} \Vert y_{3n}- S_{2}w_{n} \Vert &\leq \Vert y_{3n}- y_{2n} \Vert + \Vert y_{2n}-y_{1n} \Vert + \Vert y_{1n}-S_{2}y_{1n} \Vert + \Vert S_{2}y_{1n}- S_{2}w_{n} \Vert \\ &\leq \Vert y_{3n}- y_{2n} \Vert + \Vert y_{2n}-y_{1n} \Vert + \Vert y_{1n}-S_{2}y_{1n} \Vert + L_{2} \Vert y_{1n}- w_{n} \Vert . \end{aligned}$$

This implies $\lim_{n\to \infty }\|y_{3n}-S_{2}w_{n}\|=0$.

From this and the fact that ∇f is norm-to-norm uniformly continuous on bounded subsets of E, we have

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert \nabla fy_{3n}-\nabla fS_{2}w_{n} \Vert = 0. \end{aligned}$$

Similarly, from (1.2) we have

$$\begin{aligned} \Vert \nabla fy_{3n}- \nabla f{S_{2}} w_{n} \Vert =(1-\beta _{n}) \Vert \nabla f{S_{3}}y_{2n}- \nabla f{S_{2}}w_{n} \Vert . \end{aligned}$$

Therefore,

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert \nabla fS_{3}y_{2n}- \nabla f{S_{2}}w_{n} \Vert = 0. \end{aligned}$$

Since $\nabla f^{*}$ is norm-to-norm uniformly continuous on bounded subsets of $E^{*}$, we have

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert S_{3}y_{2n}- S_{2} w_{n} \Vert = 0. \end{aligned}$$

From the following inequality:

$$\begin{aligned} \Vert y_{2n}- {S_{3}}y_{2n} \Vert &\leq \Vert y_{2n}-y_{1n} \Vert + \Vert y_{1n}-{S_{2}}y_{1n} \Vert + \Vert {S_{2}}y_{1n}- {S_{2}}w_{n} \Vert + \Vert S_{2}w_{n}- S_{3}y_{2n} \Vert \\ &\leq \Vert y_{2n}-y_{1n} \Vert + \Vert y_{1n}-{S_{2}}y_{1n} \Vert + L_{2} \Vert y_{1n}- w_{n} \Vert + \Vert S_{2}w_{n}- S_{3}y_{2n} \Vert , \end{aligned}$$

we get

$$\begin{aligned} \lim_{n \rightarrow \infty } \bigl\Vert (I-{S_{3}})y_{2n} \bigr\Vert = 0. \end{aligned}$$

(3.14)

Also,

$$\begin{aligned} \Vert y_{4n}- S_{3}w_{n} \Vert &\leq \Vert y_{4n}- y_{3n} \Vert + \Vert y_{3n}-y_{2n} \Vert + \Vert y_{2n}-S_{3}y_{2n} \Vert + \Vert S_{3}y_{2n}- S_{3}w_{n} \Vert \\ &\leq \Vert y_{4n}- y_{3n} \Vert + \Vert y_{3n}-y_{2n} \Vert + \Vert y_{2n}-S_{3}y_{2n} \Vert + L_{3} \Vert y_{2n}- w_{n} \Vert , \end{aligned}$$

implies $\lim_{n\to \infty }\|y_{4n}-S_{3}w_{n}\|=0$.

Since ∇f is norm-to-norm uniformly continuous on bounded subsets of E, we have $\lim_{n\to \infty }\|\nabla f y_{4n}-\nabla fS_{3}w_{n}\|=0$.

From (1.2) we have

$$\begin{aligned} \Vert \nabla fy_{4n}- \nabla f{S_{3}} w_{n} \Vert =(1-\beta _{n}) \Vert \nabla f{S_{4}}y_{3n}- \nabla f{S_{3}}w_{n} \Vert . \end{aligned}$$

Therefore,

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert \nabla fS_{4}y_{3n}- \nabla f{S_{3}}w_{n} \Vert = 0. \end{aligned}$$

Since $\nabla f^{*}$ is norm-to-norm uniformly continuous on bounded subsets of $E^{*}$, we have

$$\begin{aligned} \lim_{n \rightarrow \infty } \Vert S_{4}y_{3n}- S_{3} w_{n} \Vert = 0. \end{aligned}$$

From the inequality

$$\begin{aligned} \Vert y_{3n}- {S_{4}}y_{3n} \Vert &\leq \Vert y_{3n}-y_{2n} \Vert + \Vert y_{2n}-{S_{3}}y_{2n} \Vert + \Vert {S_{3}}y_{2n}- {S_{3}}w_{n} \Vert + \Vert S_{3}w_{n}- S_{4}y_{3n} \Vert \\ &\leq \Vert y_{3n}-y_{2n} \Vert + \Vert y_{2n}-{S_{3}}y_{2n} \Vert + L_{3} \Vert y_{2n}- w_{n} \Vert + \Vert S_{3}w_{n}- S_{4}y_{3n} \Vert , \end{aligned}$$

we get

$$\begin{aligned} \lim_{n \rightarrow \infty } \bigl\Vert (I-{S_{4}})y_{3n} \bigr\Vert = 0. \end{aligned}$$

(3.15)

Continuing in this fashion, we get

$$\begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert (I-S_{1})w_{n} \bigr\Vert &=\lim_{n \rightarrow \infty } \bigl\Vert (I-S_{2})y_{1n} \bigr\Vert \\ &=\lim_{n\rightarrow \infty } \bigl\Vert (I-S_{3})y_{2n} \bigr\Vert \\ &=\lim_{n\rightarrow \infty } \bigl\Vert (I-S_{4})y_{3n} \bigr\Vert \\ & \vdots \\ & =\lim_{n\rightarrow \infty } \bigl\Vert (I-S_{m})y_{(m-1)n} \bigr\Vert =0. \end{aligned}$$

Step 5. We show that $\{x_{n}\}$ converges to an element of Γ.

Since $\{x_{n}\}$ is a Cauchy sequence, we assume that $x_{n} \rightarrow x^{*}$ as $n\to \infty$. From the fact that

$$\begin{aligned} \lim_{n\rightarrow \infty } \Vert x_{n}-w_{n} \Vert = \lim_{n\rightarrow \infty } \Vert x_{n}-y_{in} \Vert =0,\quad \forall i \in \{1,2,3, \ldots ,m\}, \end{aligned}$$

we have that

$$\begin{aligned} w_{n}\rightarrow x^{*}, \qquad y_{in} \rightarrow x^{*} \quad\text{as } n \to \infty, \forall i \in \{1,2,3, \ldots ,m \}. \end{aligned}$$

Since $I-S_{i}$, $i \in \{1,2,3, \ldots ,m\}$ are demiclosed at 0 and

$$\begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert (I-S_{1})w_{n} \bigr\Vert = \lim_{n\rightarrow \infty } \bigl\Vert (I-S_{i})y_{(i-1)n} \bigr\Vert =0 \quad\text{for } 2\leq i \leq m, \end{aligned}$$

we have $x^{*} \in \bigcap_{i=1}^{m} F(S_{i})$. Therefore, $x^{*} \in \Gamma$.

Step 6. We show that $x^{*} ={\Pi _{\Gamma }}^{f}x_{0}$.

Let $y={\Pi _{\Gamma }}^{f}x_{0}$. Since $x^{*} \in \Gamma$, we have that

$$\begin{aligned} \begin{aligned} D_{f}(y,x_{0}) & \leq D_{f}\bigl(x^{*},x_{0}\bigr). \end{aligned} \end{aligned}$$

(3.16)

Since $y \in \Gamma \subset C_{n}$ and $x_{n}= {\Pi _{C_{n}}}^{f}x_{0}$, we have

$$\begin{aligned} D_{f}(x_{n},x_{0}) \leq D_{f}(y,x_{0}) \end{aligned}$$

and, taking into account that $x_{n} \rightarrow x^{*}$, obtain

$$\begin{aligned} D_{f}\bigl(x^{*},x_{0}\bigr) \leq D_{f}(y,x_{0}). \end{aligned}$$

(3.17)

Combining (3.16) and (3.17) yields

$$\begin{aligned} D_{f}(y,x_{0})= D_{f}\bigl(x^{*},x_{0} \bigr). \end{aligned}$$

Hence, $x^{*}=y= {\Pi _{\Gamma }}^{f}x_{0}$. □

Corollary 3.2

Let C be a nonempty, closed, and convex subset of a reflexive Banach space E, and let $f:E\to \mathbb{R}$ a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let $\{S_{i}\}_{i=1}^{m}$ be a finite family of Bregman relatively nonexpansive mappings such that $S_{i}$, $i=1,2,3,\dots,m$ are $L_{i}$-Lipschitz and $(I-S_{i}), i=1,2, \ldots ,m$ are demiclosed at 0. Assume $\Gamma = \bigcap_{i=1}^{m} F(S_{i})\neq \emptyset$. Let a sequence $\{x_{n}\}$ be generated by

$$\begin{aligned} \textstyle\begin{cases} x_{0},x_{1}\in C,\quad C=C_{1}; \\ w_{n}= x_{n}+\gamma _{n} (x_{n}- x_{n-1}); \\ y_{1n}=\nabla f^{*}(\beta _{n} \nabla fw_{n} + (1- \beta _{n}) \nabla f{S_{1}}w_{n}); \\ y_{in}= \nabla f^{*}(\beta _{n} \nabla f{S_{i-1}}w_{n}+ (1- \beta _{n}) \nabla f{S_{i}}y_{(i-1)n}); \\ C_{in}=\{ v \in C_{n}: D_{f}(v,y_{in})\leq D_{f}(v,w_{n}) \}; \\ C_{n+1}=\bigcap_{i=1}^{{m}}C_{in}; \\ x_{n+1}={\Pi _{C_{n+1}}}^{f}x_{0}; \end{cases}\displaystyle \end{aligned}$$

(3.18)

where $\{\gamma _{n}\}$ and $\{\beta _{n}\}\subset (a,b)$, $0< a< b<1$, are sequences. Then the sequence $\{x_{n}\}$ converges to a point $z\in \Gamma$, where $z={\Pi _{\Gamma }}^{f}x_{0}$.

Corollary 3.3

Let E be a uniformly convex real Banach space. Let $\{S_{i}\}_{i=1}^{m}$ be a finite family of nonexpansive mappings. Assume $\Gamma = \bigcap_{i=1}^{m} F(S_{i})\} \neq \emptyset$. Let a sequence $\{x_{n}\}$ be generated by

$$\begin{aligned} \textstyle\begin{cases} x_{0},x_{1}\in C,\quad C=C_{1}; \\ w_{n}= x_{n}+\gamma _{n} (x_{n}- x_{n-1}); \\ y_{1n}=(\beta _{n} w_{n} + (1- \beta _{n}){S_{1}}w_{n}); \\ y_{in}= (\beta _{n} {S_{i-1}}w_{n}+ (1- \beta _{n}){S_{i}}y_{(i-1)n}); \\ C_{in}=\{ v \in C_{n}: \Vert y_{in}-v \Vert \leq \Vert w_{n}-v \Vert \}; \\ C_{n+1}=\bigcap_{i=1}^{{m}}C_{in}; \\ x_{n+1}=P_{C_{n+1}}x_{0}, \end{cases}\displaystyle \end{aligned}$$

(3.19)

where $\{\gamma _{n}\}$ and $\{\beta _{n}\}$ are sequences in $(0,1)$. Then the sequence $\{x_{n}\}$ converges to a point $z\in \Gamma$, where $z=P_{\Gamma }x_{0}$.

4 Applications

4.1 Application to the equilibrium problem

Let C be a nonempty closed convex subset of a real Banach space E, and let $F:C \times C \to \mathbb{R}$ be a bifunction.

The equilibrium problem with respect to F and C is to find $z\in C$ such that

$$\begin{aligned} F(z,y) \geq 0,\quad \forall y \in C. \end{aligned}$$

The set of solutions of the equilibrium problem above is denoted by $EP(F)$. For solving the equilibrium problem, we assume that F satisfies the following conditions:

• (A1) $F(x,x)=0$ for all $x \in C$;

• (A2) F is monotone, i.e., $F(x,y)+F(y,x) \leq 0$, $\forall x,y \in C$;

• (A3) for each $x,y,z$ ∈C, $\lim_{t\downarrow 0} F(tz + (1-t)x,y) \leq F(x,y)$;

• (A4) for each $x \in C$, $y\mapsto F(x,y)$ is convex and lower semicontinuous.

The resolvent of a bifunction F is the operator ${\operatorname{Res}_{f}}^{F}:E \to 2^{C}$ defined by

$$\begin{aligned} {\operatorname{Res}_{f}}^{F} x=\bigl\{ z \in C: F(z,y)+ \bigl\langle \nabla f(z)- \nabla f(x), y-z\bigr\rangle \geq 0, \forall y \in C \bigr\} . \end{aligned}$$

Lemma 4.1

([23])

Let E be a reflexive Banach space, and C be a nonempty closed convex subset of E. Let $f:E \to (- \infty, + \infty )$ be a Legendre function. If the bifunction $F:C \times C \to \mathbb{R}$ satisfies conditions (A1)–(A4), then the following holds:

• (1) ${\operatorname{Res}_{f}}^{F}$ is single-valued;

• (2) ${\operatorname{Res}_{f}}^{F}$ is Bregman firmly nonexpansive;

• (3) $\operatorname{Fix}(\operatorname{Res}^{F})=EP(F)$;

• (4) $EP(F)$ is a closed and convex subset of C;

• (5) For all $x \in E$ and for all $q \in \operatorname{Fix}(\operatorname{Res}^{F})$,

  $$\begin{aligned} D_{f}\bigl(q, {\operatorname{Res}_{f}}^{F}x\bigr)+ D_{f}\bigl({\operatorname{Res}_{f}}^{F}x,x\bigr) \leq D_{f}(q,x). \end{aligned}$$

Theorem 4.2

Let C and Q be nonempty, closed, and convex subsets of a reflexive Banach space E, and Let $f:E\to \mathbb{R}$ be a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let $F_{i}:C \times C \to \mathbb{R}$, $i=1,2,3,\dots,m$ be bifunctions satisfying conditions $(A1)$–$(A4)$ such that ${\operatorname{Res}_{f}}^{F_{i}}$ are $L_{i}$-Lipschitz for $1\leq i \leq m$. Assume $\Gamma =\bigcap_{i=1}^{m}EP(F_{i}) \neq \emptyset$. Let a sequence $\{x_{n}\}$ be generated by

$$\begin{aligned} \textstyle\begin{cases} x_{0},x_{1}\in C,\quad C=C_{1}; \\ w_{n}= x_{n}+\gamma _{n} (x_{n}- x_{n-1}); \\ y_{1n}=\nabla f^{*}(\beta _{n} \nabla fw_{n} + (1- \beta _{n}) \nabla f{\operatorname{Res}_{f}}^{F_{1}}w_{n}); \\ y_{in}= \nabla f^{*}(\beta _{n} \nabla f{\operatorname{Res}_{f}}^{F_{i-1}}w_{n}+ (1- \beta _{n})\nabla f{\operatorname{Res}_{f}}^{F_{i}}y_{(i-1)n}); \\ C_{in}=\{ v \in C_{n}: D_{f}(v,y_{in})\leq D_{f}(v,w_{n}) \}; \\ C_{n+1}=\bigcap_{i=1}^{{m}}C_{in}; \\ x_{n+1}={\Pi _{C_{n+1}}}^{f}x_{0}, \end{cases}\displaystyle \end{aligned}$$

(4.1)

where $\{\gamma _{n}\},\{\beta _{n}\}\subset (a,b)$, $0< a< b<1$, are sequences and ${\operatorname{Res}_{f}}^{F_{i}}$ are the resolvents of $F_{i}$, $i\in \{1,2,\dots,m\}$. Then the sequence $\{x_{n}\}$ converges to $z={P_{\Gamma }}^{f}x_{0}$.

Proof

Putting $S_{i}={\operatorname{Res}_{f}}^{F_{i}}$ in Theorem 3.1, we get the desired result. □

4.2 Application to the maximal monotone operator

A set-valued mapping $B\subset E\times E^{*}$ with domain $D(B)=\{x \in E: Bx \neq \emptyset \}$ and range $R(B)=\cup \{Bx:x \in D(B)\}$ is said to be monotone if $\langle x-y,x^{*}-y^{*}\rangle \geq 0$ whenever $(x,x^{*}),(y,y^{*}) \in B$, see, for example, [2]. A monotone mapping $B \subset E \times E^{*}$ is said to be maximal monotone if its graph $G(B)=\{ (x,y): y \in Bx\}$ is not properly contained in the graph of any other monotone mapping. We know that if B is maximal monotone, then the zero of B, $B^{-1}(0)=\{x \in E: 0 \in Bx\}$ is closed and convex. Define the resolvent of B, ${\operatorname{Res}_{B}}^{f}:E\to 2^{E}$ by

$$\begin{aligned} {\operatorname{Res}_{B}}^{f}x=(\nabla f +B)^{-1}\circ \nabla fx. \end{aligned}$$

We know the following (see [5]):

• (1) ${\operatorname{Res}_{B}}^{f}$ is single valued;

• (2) $\operatorname{Fix}({\operatorname{Res}_{B}}^{f})=B^{-1}0$.

Lemma 4.3

([21])

Let $B:E \to {2}^{E*}$ be a maximal monotone mapping such that $B^{-1}(0) \neq \emptyset$. Then for all $x \in E$ and $q \in B^{-1}(0)$, we have

$$\begin{aligned} D_{f}\bigl(q, {\operatorname{Res}_{B}}^{f}x\bigr)+ D_{f}\bigl(\operatorname{Res}^{f}x,x\bigr) \leq D_{f}(q,x). \end{aligned}$$

Theorem 4.4

Let C be a nonempty, closed, and convex subset of a reflexive Banach space E, and let $f:E\to \mathbb{R}$ be a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let $B_{i}:E\to 2^{E*}$ $i=1,2,3,\dots,m$ be maximal monotone operators such that ${\operatorname{Res}_{B_{i}}}^{f}$ are $L_{i}$-Lipschitz for $1\leq i \leq m$. Assume $\Gamma = \bigcap_{i=1}^{m} {B_{i}}^{-1}(0)\neq \emptyset$. Let a sequence $\{x_{n}\}$ be generated by

$$\begin{aligned} \textstyle\begin{cases} x_{0},x_{1}\in C,\quad C=C_{1}; \\ w_{n}= x_{n}+\gamma _{n} (x_{n}- x_{n-1}); \\ y_{1n}=\nabla f^{*}(\beta _{n} \nabla fw_{n} + (1- \beta _{n}) \nabla f{\operatorname{Res}_{B_{1}}}^{f}w_{n}); \\ y_{in}= \nabla f^{*}(\beta _{n} \nabla f{\operatorname{Res}_{B_{i-1}}}^{f}w_{n}+ (1- \alpha _{n})\nabla f{\operatorname{Res}_{B_{i}}}^{f}y_{(i-1)n}), \quad 2 \le i\le m; \\ C_{in}=\{ v \in C_{n}: D_{f}(v,y_{in})\leq D_{f}(v,w_{n}) \}; \\ C_{n+1}=\bigcap_{i=1}^{{m}}C_{in}; \\ x_{n+1}={\Pi _{C_{n+1}}}^{f}x_{0}, \end{cases}\displaystyle \end{aligned}$$

(4.2)

where $\{\gamma _{n}\}, \{\beta _{n}\}\subset (a,b)$, $0< a< b<1$, are sequences and ${\operatorname{Res}_{B_{i}}}^{f}$ are the resolvents of $B_{i}$. Then the sequence $\{x_{n}\}$ converges to a point $z\in \Gamma$, where $z={P_{\Gamma }}^{f}x_{0}$.

Proof

Putting $S_{i}={\operatorname{Res}_{B_{i}}}^{f}$ in Theorem 3.1, we get the desired result. □