A strong convergence theorem for Bregman quasi-noexpansive mappings with applications

Abstract

In this paper we propose a new iterative algorithm for finding a common fixed point of an infinitely countable family of Bregman quasi-nonexpansive mappings in reflexive Banach spaces. We give some applications of our algorithm to solving equilibrium problems, convex feasibility problems, variational inequalities of monotone operators and finding zeros of maximal monotone operators.

1 Introduction and preliminaries

Throughout this paper, let C be a nonempty closed and convex subset of a reflexive real Banach space X. we shall denote the dual space of X by $X^{*}$. The norm and the duality pairing between X and $X^{*}$ are denoted by $\Vert \cdot \Vert$ and $\langle \cdot ,\cdot \rangle$, respectively. Let $f:X\rightarrow (-\infty ,\infty ]$ be a proper convex and lower semicontinuous function. We denote by dom(f), the domain of f, that is, the set $dom(f) =\{x\in X:f(x)<\infty \}.$ The subdifferential of f at $x\in int\, dom(f)$ is the convex set defined by:

$$\begin{aligned} \partial f(x)=\left\{ \xi \in X^{*}: f(x) +\langle y-x,\xi \rangle \le f(y), \forall y\in X\right\} , \end{aligned}$$

where intA is the interior of the set A, and the Fenchel conjugate of f is the convex function $f^{*}:X^{*}\rightarrow (-\infty ,\infty ]$ defined by:

$$\begin{aligned} f^{*}(\xi )=\sup \left\{ \langle x,\xi \rangle - f(x): x\in X\right\} . \end{aligned}$$

It is not difficult to check that $f^{*}$ is proper convex and lower semicontinuous function. The function f is said to be cofinite if dom $(f^{*})=X^{*}.$

We denote by $f^{\circ }(x,y)$ the right-hand derivative of a convex mapping $f:X\rightarrow (-\infty ,+\infty ]$ at $x\in int\, dom(f)$ in the direction y, that is,

$$\begin{aligned} f^{\circ }(x,y):= \displaystyle \lim _{t\downarrow 0}\frac{f(x+ty)- f(x)}{t}. \end{aligned}$$

(1.1)

If the limit as $t\rightarrow 0$ in (1.1) exists for each y, then the function f is said to be G$\hat{a}$teaux differentiable at x. In this case, the gradient of f at x is the linear function $\nabla f(x),$ which is defined by $\langle y,\nabla f(x)\rangle := f^{\circ }(x,y)$ for all $y\in X$. The function f is said to be G$\hat{a}$teaux differentiable if it is G$\hat{a}$teaux differentiable at each $x \in int\, dom(f).$ When the limit as $t\rightarrow 0$ in (1.1) is attained uniformly for any $y \in X$ with $\Vert y\Vert =1$, we say that f is Fr$\acute{e}$chet differentiable at x. Finally, f is said to be uniformly Fr$\acute{e}$chet differentiable on a subset E of X if the limit is attained uniformly for $x \in E$ and $\Vert y\Vert = 1$.

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

• (L1) $int \, dom(f)\ne \emptyset$ and $\partial f$ is single-valued on its domain.

• (L2) $int \, dom(f^{*})\ne \emptyset$ and $\partial f^{*}$ is single-valued on its domain.

Since X is reflexive, we know that $(\partial f)^{-1}=\partial f^{*}$ (see [7]). This fact, when combined with conditions (L1) and (L2), implies the following equalities:

$$\begin{aligned} \nabla f= & {} (\nabla f^{*})^{-1},\\ ran (\nabla f)= & {} dom (\nabla f^{*})= int\, dom(f^{*}),\\ ran (\nabla f^{*})= & {} dom (\nabla f)= int\, dom (f). \end{aligned}$$

Conditions (L1) and (L2), in conjunction with [5, Theorem 5.4], imply that the functions f and $f^{*}$ are strictly convex on the interior of their respective domains. Also, we know that f is Legendre if and only if $f^{*}$ is Legendre. Several interesting examples of Legendre functions are presented in [3, 5]. Among them are the functions $\frac{1}{p}\Vert \cdot \Vert ^{p}$ with $p\in (1,\infty )$, where the Banach space X is smooth and strictly convex. Information regarding fully Legendre functions can be found in the paper by Reem and Reich [29].

In 1967, Bregman [8] introduced the concept of Bregman distance, and he has discovered an elegant and effective technique for the use of the Bregman distance in the process of designing and analyzing feasibility and optimization algorithms.

Let $f : X \rightarrow (-\infty ,+\infty ]$ be a convex and G$\hat{a}$teaux differentiable function. The Bregman distance with respect to f, or simply, Bregman distance is the bifunction $D_{f}: dom(f) \times int\, dom(f) \rightarrow [0,+\infty ]$ defined by

$$\begin{aligned} D_{f} (y, x) := f (y) - f (x) -\langle y-x,\nabla f(x)\rangle . \end{aligned}$$

It should be noted that $D_{f}$ is not a distance in the usual sense of the term. Clearly, $D_{f}(x, x) = 0$, but $D_{f}(y, x) = 0$ may not imply x = y. In our case when f is Legendre this indeed holds (see [5, Theorem 7.3(vi)]). In general, $D_{f}$ is not symmetric and does not satisfy the triangle inequality. However, $D_{f}$ satisfies the three point identity

$$\begin{aligned} D_{f} (x, y)+D_{f} (y, z)-D_{f} (x, z) = \langle x-y,\nabla f(z)-\nabla f(y)\rangle , \end{aligned}$$

for any $x\in dom(f)$ and $y,z\in int \, dom(f).$ During the last 30 years, Bregman distances have been studied by many researchers (see [4, 5, 9, 10, 17, 28]). Over the last 10 years the usage of this concept has been extended to many fields like clustering, image reconstruction, information theory and machine learning.

The modulus of total convexity at x is the bifunction $\upsilon _{f}:int \, dom(f) \times [0,+\infty )\rightarrow [0,\infty ]$ defined by:

$$\begin{aligned} \upsilon _{f}(x,t):= \inf \left\{ D_{f}(y,x):y \in dom (f), \Vert y-x\Vert =t\right\} . \end{aligned}$$

The function f is called totally convex at $x\in int \, dom(f)$ if $\upsilon _{f}(x,t)$ is positive for any $t > 0$. This notion was first introduced by Butnariu and Iusem in [10]. Let E be a nonempty subset of X. The modulus of total convexity of f on E defined by:

$$\begin{aligned} \upsilon _{f}(E,t):=\inf \left\{ \upsilon _{f}(x,t): x\in E\cap int\, dom (f)\right\} . \end{aligned}$$

The function f is called totally convex on bounded subsets if $\upsilon _{f}(E,t)$ is positive for any nonempty and bounded subset E and for any $t > 0.$ (see also [12]).

We will use the following propositions.

Proposition 1.1

[10] The function $f: X\rightarrow (-\infty ,+\infty ]$ is totally convex on bounded subsets of X if and only if for any two sequences $\{x_{n}\}$ and $\{y_{n}\}$ in $intdom f$ and domf,  respectively, such that the first one is bounded,

$$\begin{aligned} \lim _{n\rightarrow \infty } D_{f} (y_{n} , x_{n})=0\Longrightarrow \lim _{n\rightarrow \infty }\Vert y_{n} - x_{n}\Vert =0. \end{aligned}$$

Proposition 1.2

[33] Let $f : X \rightarrow {\mathbb {R}}$ be a G$\hat{a}$teaux differentiable and totally convex function. If $x_{0}\in X$ and the sequence $\{D_{f} (x_{n} , x_{0})\}$ is bounded, then the sequence $\{x_{n}\}$ is also bounded.

Proposition 1.3

[36] Let $f : X \rightarrow {\mathbb {R}}$ be a Legendre function such that $\nabla f^{*}$ is bounded on bounded subsets of $int \, dom(f)$. If for $x_{0} \in X$, $\{D_{f}(x_{0},x_{n}) \}$ is bounded, then the sequence $\{x_{n}\}$ is bounded too.

Let $C \subset int dom(f)$, the Bregman projection [8] with respect to f of $x \in int\, dom(f)$ onto C is defined as the necessarily unique vector $\overleftarrow{Proj}_{C}^{f}(x)\in C$, which satisfies

$$\begin{aligned} D_{f}\left( \overleftarrow{Proj}_{C}^{f}(x),x\right) = \inf \left\{ D_{f}(y,x): y \in C\right\} . \end{aligned}$$

Similarly to the metric projection in Hilbert spaces, the Bregman projection with respect to totally convex and G$\hat{a}$teaux differentiable functions has a variational characterization [13, Corollary 4.4].

Proposition 1.4

Let f be a G$\hat{a}$teaux differentiable and totally convex function on int  dom(f). Let $x \in$$int \, dom(f)$ and $C\subset int \, dom(f)$. If $\hat{x} \in C,$ then the following conditions are equivalent:

• (i) The vector $\hat{x} \in C$ is the Bregman projection of x onto C with respect to f.

• (ii) The vector $\hat{x} \in C$ is the unique solution of the variational inequality

  $$\begin{aligned} \langle z-y,\nabla f(x) - \nabla f(z) \rangle \ge 0,\quad \forall y\in C. \end{aligned}$$

• (iii) The vector $\hat{x}$ is the unique solution of the inequality

  $$\begin{aligned} D_{f} (y, z) + D_{f} (z, x) \le D_{f} (y, x),\quad \forall y\in C. \end{aligned}$$

Proposition 1.5

[33] Let $f: X\rightarrow {\mathbb {R}}$ be a G$\hat{a}$teaux differentiable and totally convex function and $x_{0}\in X$. Suppose that the sequence $\{x_{n}\}$ is bounded and any weak subsequential limit of $\{x_{n}\}$ belongs to C. If $D_{f}(x_{n},x_{0})\le D_{f}(\overleftarrow{Proj}^{f}_{C}(x_{0}),x_{0})$ for any $n \in {\mathbb {N}}$, then $\{x_{n}\}$ converges strongly to ${Proj}^{f}_{C}(x_{0}).$

Definition 1.6

An operator $T : C \rightarrow int \, dom(f)$ is said to be:

• (i) Bregman firmly nonexpansive (BFNE) if

  $$\begin{aligned} \left\langle Tx - Ty,\nabla f(Tx)- \nabla f(Ty)\right\rangle \le \left\langle Tx - Ty,\nabla f(x)- \nabla f(y)\right\rangle , \end{aligned}$$

  for any $x,y \in C,$ or equivalently,

  $$\begin{aligned} D_{f}(Tx,Ty)+D_{f}(Ty,Tx)+D_{f}(Tx,x)+D_{f}(Ty,y)\le D_{f}(Tx,y)+ D_{f}(Ty,x). \end{aligned}$$

• (ii) Bregman quasi-firmly nonexpansive (BQFNE) if

  $$\begin{aligned} \left\langle Tx - p,\nabla f(x)- \nabla f(Tx)\right\rangle \ge 0,\quad \forall x\in C,\quad p\in F(T), \end{aligned}$$

  or equivalently,

  $$\begin{aligned} D_{f}(p,Tx)+ D_{f}(Tx,x)\le D_{f}(p,x), \end{aligned}$$

  where, F(T) is the set of all fixed points of T.

• (iii) Bregman quasi-nonexpansive (BQNE) if

  $$\begin{aligned} D_{f}(p,Tx)\le D_{f}(p,x),\quad \forall x\in C,p\in F(T). \end{aligned}$$

Definition 1.7

[25] An operator $T :C \rightarrow int \, dom(f)$ is said to be Bregman strongly nonexpansive (BSNE) if

$$\begin{aligned} D_{f} (p, Tx) \le D_{f} (p, x), \end{aligned}$$

for all $p \in F(T)$ and $x \in C,$ and if whenever $\{x_{n}\} \subset C$ is bounded, $p\in F(T)$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\big (D_{f} (p, x_{n})- D_{f} (p, Tx_{n})\big )=0, \end{aligned}$$

it follows that

$$\begin{aligned} \lim _{n\rightarrow \infty }D_{f} (Tx_{n}, x_{n})=0. \end{aligned}$$

Definition 1.8

[30] (Asymptotic fixed point). A point $u\in C$ is said to be an asymptotic fixed point of $T :C\rightarrow C$ if there exists a sequence $\{x_{n}\}$ in C such that $x_{n}\rightharpoonup u$ and $\Vert x_{n}- Tx_{n}\Vert \rightarrow 0$. We denote the set of asymptotic fixed points of T by ${\widehat{F}}(T)$.

Proposition 1.9

[34] Let $f: X\rightarrow (-\infty ,+\infty ]$ be a Legendre function, $C\subset int \, dom(f)$ and $T:C\rightarrow C$ be a BQNE operator. Then F(T) is closed and convex.

Let $f:X\rightarrow {\mathbb {R}}$ be a convex and G$\hat{a}$teaux differentiable function. From [1, 14], we make use of the function $V_{f}: X\times X^{*}\rightarrow [0,+\infty ]$ associated with f which is defined by:

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

Then we have

$$\begin{aligned} V_{f}(x,x^{*})= D_{f}(x,\nabla f^{*}(x^{*})), \end{aligned}$$

for all $x\in X$ and $x^{*}\in X^{*}.$ Moreover, by the subdifferential inequality, we have

$$\begin{aligned} V_{f}(x,x^{*}) + \langle \nabla f^{*}(x^{*})-x,y^{*}\rangle \le V_{f}(x,x^{*}+y^{*}), \end{aligned}$$

for all $x\in X$ and $x^{*},y^{*}\in X^{*}$ [24]. In addition, if $f:X\rightarrow (-\infty ,+\infty ]$ is a proper lower semicontinuous function, then $f^{*}:X^{*}\rightarrow (-\infty ,+\infty ]$ is a proper weak$^{*}$ lower semicontinuous and convex function [27]. Hence $V_{f}$ is convex in the second variable. Thus, for all $z\in X$,

$$\begin{aligned} D_{f}\left( z,\nabla f^{*}\left( \sum _{i=1}^{N} t_{i}\nabla f(x_{i})\right) \right) \le \sum _{i=1}^{N} t_{i}D_{f}(z,x_{i}), \end{aligned}$$

where $\{x_{i}\}_{i=1}^{N}\subset X$ and $\{t_{i}\}_{i=1}^{N}\subset (0,1)$ with $\sum _{i=1}^{N} t_{i}=1.$

Let B and S be the closed unit ball and the unit sphere of a Banach space X, respectively. Let $B_r =\{z\in X:\Vert z\Vert \le r \}$ for all $r> 0$. Then a function $f:X\rightarrow {\mathbb {R}}$ is said to be uniformly convex on bounded sets [37] if $\rho _{r}(t)>0$ for all $r,t>0$, where $\rho _{r}:[0,\infty )\rightarrow [0,\infty ]$ is defined by:

$$\begin{aligned} \rho _{r}(t)= \inf _{x,y\in B_r,\Vert x-y\Vert =t, \alpha \in (0,1)}\frac{\alpha f(x)+ (1-\alpha )f(y)- f(\alpha x+ (1-\alpha )y)}{\alpha (1-\alpha )}, \end{aligned}$$

for all $t\ge 0.$ The function $\rho _{r}$ is called the gauge of uniform convexity of f. It is known that $\rho _{r}$ is a nondecreasing function. If f is uniformly convex, then the following proposition is known.

Proposition 1.10

[26] Let X be a Banach space, $r>0$ be a constant and $f:X\rightarrow {\mathbb {R}}$ be a uniformly convex function on bounded subsets of X. Then

$$\begin{aligned} f(\sum _{k=0}^{n}\alpha _{k}x_{k})\le \sum _{k=0}^{n}\alpha _{k}f(x_{k})- \alpha _{i}\alpha _{j}\rho _{r}(\Vert x_{i}-x_{j}\Vert ), \end{aligned}$$

for all $i,j\in \{0,1,2,\ldots ,n\}$, $x_{k}\in rB$, $\alpha _{k}\in (0,1)$ and $k=0,1,2,\ldots ,n$ with $\sum _{k=0}^{n}\alpha _{k}=1,$ where $\rho _{r}$ is the gauge of uniform convexity of f.

The function f is also said to be uniformly smooth on bounded subsets (see [37]) if $\lim _{t\downarrow 0}\frac{\sigma _{r}(t)}{t}=0$ for all $r>0,$ where $\sigma _{r}:[0,\infty ) \rightarrow [0,\infty ]$ is defined by:

$$\begin{aligned} \sigma _{r}(t) = \sup _{x\in rB,y\in S, \alpha \in (0,1)} \frac{\alpha f(x+(1-\alpha )ty)+ (1-\alpha )f(x-\alpha ty)- f(x)}{\alpha (1-\alpha )}, \end{aligned}$$

for all $t\ge 0.$ A function f is said to be super coercive if $\lim _{\Vert x\Vert \rightarrow \infty }\frac{f(x)}{\Vert x\Vert }= +\infty .$

We will use the following theorems:

Theorem 1.11

[37] Let $f:X\rightarrow {\mathbb {R}}$ be a continuous convex and super coercive function. Then the following statements are equivalent:

• (i) f is bounded on bounded sets and uniformly smooth on bounded sets;

• (ii) f is Fr$\acute{e}$chet differentiable and $\nabla f$ is uniformly norm-to-norm continuous on bounded sets;

• (iii) dom$(f^{*})= X^{*},$$f^{*}$ is super coercive and uniformly convex on bounded sets.

Theorem 1.12

[37] Let $f: X\rightarrow {\mathbb {R}}$ be a continuous convex function which is bounded on bounded sets. Then the following statements are equivalent:

• (i) f is super coercive and uniformly convex on bounded sets;

• (ii) dom($f^{*}) = X^{*},$$f^{*}$ is bounded on bounded sets and uniformly smooth on bounded sets;

• (iii) dom($f^{*}) = X^{*},$$f^{*}$ is Fr$\acute{e}$chet differentiable and $\nabla f^{*}$ is uniformly norm-to-norm continuous on bounded sets.

Theorem 1.13

[11] Let $f:X\rightarrow (-\infty ,+\infty ]$ be a Legendre function. The function f is totally convex on bounded subsets if and only if f is uniformly convex on bounded subsets.

The duality mapping $J :X \rightarrow 2^{X^{*}}$ of X is defined by:

$$\begin{aligned} J(x) =\left\{ f\in X^{*}: <f,x>=\parallel x\parallel ^{2}=\parallel f\parallel ^{2}\right\} , \end{aligned}$$

for every $x \in X$. It is known that if X is strictly convex and reflexive, then the duality mapping J of X is one-to-one and onto, and $J ^{-1}:X^{*} \rightarrow X$ is the duality mapping of $X^{*}$.

The function $\phi : X ^{2}\rightarrow {\mathbb {R}}$ is defined by:

$$\begin{aligned} \phi (y, x) = \parallel y\parallel ^{2}-2<y,Jx>+\parallel x\parallel ^{2}, \end{aligned}$$

for all $x ,y \in X$. In [15], the authors introduced the following algorithm:

$$\begin{aligned} \left\{ \begin{array}{lr} x_{0}\in C_{0}\, is \, chosen \, arbitrarily ,\\ z_{n}= J^{-1} (\beta _{n}J(x_{n}) + (1- \beta _{n})J(Tx_{n})),\\ y_{n} = J^{-1}(\alpha _{n}\nabla f(x_{0}) + (1- \alpha _{n})J(z_{n})),\\ u_{n}\in T_{r_{n}}y_{n}=\left\{ z\in C : f(z,y)+\frac{1}{r_{n}}\langle y-z,J(z)-J(x_{n})\rangle \ge 0\, \forall y\in C\right\} ,\\ C _{n} = \left\{ z\in C_{n-1}: \langle z- x_{n},J(x_{0})- J(x_{n}) \rangle \le 0\right\} ,\\ Q_{n} = \left\{ z\in C: \phi (z,u_{n} )\le \alpha _{n} \phi (z,x_{0} )+ (1-\alpha _{n})\phi (z,x_{n} )\right\} ,\\ x_{n+1}= \Pi _{C_{n}\cap Q_{n}}(x_{0}), \end{array} \right. \end{aligned}$$

where $\Pi _{C}(x) = argmin_{y\in C}\phi (y,x)$ and $T:C\rightarrow C$ is a firmly nonexpansive-type mapping. This means that

$$\begin{aligned} \phi (Tx,Ty)+\phi (Ty,Tx)+\phi (Tx,x)+\phi (Ty,y)\le \phi (Tx,y)+ \phi (Ty,x), \end{aligned}$$

for all $x,y\in C$. They proved that $\{x_{n}\}$ converges strongly to $\Pi _{F(T)\cap EP(f)}x_{0}$.

In this paper, we introduce the following new algorithm:

$$\begin{aligned} \left\{ \begin{array}{lr} x_{0}\in C_{0} = C \,is \, chosen\, arbitrarily ,\\ z_{n}^{i} = \nabla f^{*} (\beta _{n}\nabla f(x_{n}) + (1- \beta _{n})\nabla f(T_{i}x_{n})),\\ y_{n}^{i} = \nabla f^{*} (\alpha _{n}\nabla f(x_{0}) + (1- \alpha _{n})\nabla f(z_{n}^{i})),\\ C _{n} = \left\{ z\in C_{n-1}: \langle z- x_{n},\nabla f(x_{0})- \nabla f(x_{n}) \rangle \le 0\right\} ,\\ Q_{n}^{i} = \left\{ z\in C: D_{f}(z,y_{n}^{i} )\le \alpha _{n} D_{f}(z,x_{0} )+ (1-\alpha _{n})D_{f}(z,x_{n} )\right\} ,\\ Q_{n}= \bigcap _{i=1}^{\infty } Q_{n}^{i},\\ x_{n+1}= \overleftarrow{Proj}_{C_{n}\cap Q_{n}}^{f}(x_{0}), \end{array} \right. \end{aligned}$$

(1.2)

and establish a strong convergence theorem. Finally, we mention some applications of our algorithm to solving equilibrium problems, convex feasibility problems, variational inequalities of monotone operators and finding zeros of maximal monotone operators.

2 Main results

Now we are in a position to introduce and prove the strong convergence theorem for the new iterative algorithm (1.2) as our main result. A common fixed point of an infinitely countable family of Bregman quasi-nonexpansive mappings is approximated by Theorem 2.1 in reflexive Banach spaces. It is worth noting that the class of Bregman quasi-nonexpansive mappings contains important operators such as the resolvents of bifunctions and the resolvents of maximal monotone operators.

Theorem 2.1

Let $\{T_{i}\}_{i=1}^{\infty }$ be an infinitely countable family of Bregman quasi nonexpansive mappings from C into itself with $F(T_{i}) = {\widehat{F}}(T_{i}).$ Let $f : X \rightarrow {\mathbb {R}}$ be a super coercive Legendre function which is bounded, uniformly Fr$\acute{e}$chet differentiable and totally convex on bounded subsets of X. Assume that$\Omega := \bigcap _{i=1}^{\infty }F(T_{i})\ne \emptyset ,$$\{\alpha _{n}\},\{\beta _{n}\}$$\subset (0,1)$ such that $\lim _{n\rightarrow \infty } \alpha _{n}=0$ and $\liminf _{n\rightarrow \infty } \beta _{n}(1-\beta _{n})> 0.$ Then the sequence $\{x_{n}\}$ generated by (1.2) converges strongly to $\overleftarrow{Proj}_{\Omega }^{f}(x_{0})$.

Proof

Since $T_{i}$ is BQNE, it follows from Proposition 1.9 that $F(T_{i})$ is closed and convex for all $i\ge 1$ and hence $\Omega$ is closed and convex. Therefore $\overleftarrow{Proj}_{\Omega }^{f}$ is well defined. Let $n\in \mathbb {Z_{+}}:={\mathbb {N}}\cup$$\{0\}$. It is not difficult to check that the sets $Q_{n}^{i}$ and $C_{n}$ are closed and convex for all $i\ge 1$. Hence $C_{n}\cap Q_{n}$ is closed and convex.

Next, let us show that $\Omega \subset C_{n}\cap Q_{n}$ for all $n\in \mathbb {Z_{+}}$. Let $u\in \Omega$. Since

$$\begin{aligned} D_{f}(u,z_{n}^{i})= & {} D_{f} \left( u,\nabla f^{*} \left( \beta _{n}\nabla f(x_{n}) + (1- \beta _{n})\nabla f(T_{i}x_{n})\right) \right) \\\le & {} \beta _{n} D_{f}(u,x_{n}) + (1-\beta _{n}) D_{f}(u,T_{i}x_{n})\\\le & {} D_{f}(u,x_{n}), \end{aligned}$$

and

$$\begin{aligned} D_{f}(u,y_{n}^{i})= & {} D_{f}\left( u,\nabla f^{*} \left( \alpha _{n}\nabla f(x_{0}) + (1- \alpha _{n})\nabla f(z_{n}^{i})\right) \right) ,\\\le & {} \alpha _{n}D_{f}(u,x_{0}) + (1-\alpha _{n}) D_{f}(u,z_{n}^{i})\\\le & {} \alpha _{n}D_{f}(u,x_{0}) + (1-\alpha _{n})D_{f}(u,x_{n}). \end{aligned}$$

Thus $u\in Q_{n}^{i}$ for all $i\ge 1$. Therefore $u\in Q_{n}$ for all $n\in \mathbb {Z_{+}}$. Now we show by induction that $\Omega \subset C_{n}$ for all n in $\mathbb {Z_{+}}$. From $C_{0} = C$, we get $\Omega \subset C_{0}$. Suppose that $\Omega \subset C_{k}$ for some $k \in \mathbb {Z_{+}}$. From $x_{k+1}= \overleftarrow{Proj}_{C_{k}\cap Q_{k }}^{f}(x_{0})$ and Proposition 1.4, we get

$$\begin{aligned} \langle u-x_{k+1},\nabla f(x_{0})- \nabla f(x_{k+1}) \rangle \le 0,\forall u\in \Omega \subset C_{k}\cap Q_{k}, \end{aligned}$$

which implies that $\Omega \subset C_{k+1}$. As a consequence, we have $\Omega \subset C_{n}$ for all n in $\mathbb {Z_{+}}$. Therefore $\Omega \subset C_{n} \cap Q_{n}$ and so $C_{n} \cap Q_{n}$ is nonempty, closed and convex. This means that $\{x_{n} \}$ is well defined. It follows from the definition of $C_{n}$ and Proposition 1.4 that $x_{n}= \overleftarrow{Proj}_{C_{n}}^{f}(x_{0}).$ Again by using Proposition 1.4, we have

$$\begin{aligned} D_{f}(x_{n}, x_{0} )\le D_{f}(u,x_{0})- D_{f}(u,x_{n})\le D_{f}(u,x_{0}), \end{aligned}$$

(2.1)

for each $u\in \Omega$. Hence $\{D_{f}(x_{n}, x_{0})\}$ is bounded. Therefore by Proposition 1.2, $\{x_{n} \}$ is bounded. By Theorem 1.12, $f^{*}$ is bounded on bounded subset of $X^{*}$ and hence $\nabla f^{*}$ is bounded on bounded subsets of $X^{*}$ [10]. On the other hand $D_{f}(u,T_{i}x_{n})\le D_{f}(u,x_{n})$ for each $u\in \Omega$. So by Proposition 1.3, $\{T_{i}(x_{n})\}_{n=1}^{\infty }$ is bounded for all $i\ge 1$ and $\{z_{n}^{i}\}_{n=1}^{\infty }$, $\{y_{n}^{i}\}_{n=1}^{\infty }$ are also bounded for all $i\ge 1$. Since $x_{n+1}= \overleftarrow{Proj}_{C_{n}\cap Q_{n}}^{f}(x_{0}),$ we have

$$\begin{aligned} D_{f}(x_{n+1},y_{n}^{i} )\le \alpha _{n} D_{f}(x_{n+1},x_{0} )+ (1-\alpha _{n})D_{f}(x_{n+1},x_{n} ). \end{aligned}$$

(2.2)

It follows from $x_{n}= \overleftarrow{Proj}_{C_{n}}^{f}(x_{0})$ and $x_{n+1}\in C_{n}$ that

$$\begin{aligned} D_{f}(x_{n},x_{0} )\le D_{f}(x_{n+1},x_{0} ), \end{aligned}$$

(2.3)

for each $n\in \mathbb {Z_{+}}$ and $i\in {\mathbb {N}}.$ Thus, $\{ D_{f}(x_{n},x_{0} )\}$ is nondecreasing. It follows from the boundedness of $\{ D_{f}(x_{n},x_{0} )\}$ that $\lim _{n\rightarrow \infty } D_{f}(x_{n},x_{0} )$ exists. By using Proposition 1.4, we get

$$\begin{aligned} D_{f}(x_{n+1},x_{n} )= D_{f}\left( x_{n+1},\overleftarrow{Proj}_{C_{n}}^{f}(x_{0})\right) \le D_{f}(x_{n+1},x_{0} )- D_{f}(x_{n},x_{0} ). \end{aligned}$$

So

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

(2.4)

By Proposition 1.1, we have

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

(2.5)

Using $\lim _{n\rightarrow \infty }\alpha _{n}=0$, (2.2), (2.4) and boundedness of $\{ D_{f}(x_{n},x_{0} )\}$, we get

$$\begin{aligned} \lim _{n\rightarrow \infty } D_{f}(x_{n+1},y_{n}^{i} ) =0, \quad (\forall i\in {\mathbb {N}}), \end{aligned}$$

and hence

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert x_{n+1}-y_{n}^{i} \Vert =0, \quad (\forall i\in {\mathbb {N}}). \end{aligned}$$

(2.6)

From (2.5) and (2.6), we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert x_{n}-y_{n}^{i} \Vert =0,\quad (\forall i\in {\mathbb {N}}). \end{aligned}$$

(2.7)

Since $\nabla f$ is uniformly continuous on bounded subsets [31, Proposition 2], we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert \nabla f(x_{n})-\nabla f(y_{n}^{i}) \Vert =0, \quad (\forall i\in {\mathbb {N}}). \end{aligned}$$

(2.8)

Let $r_{i}= \sup \{\Vert \nabla f(x_{n})\Vert ,\Vert \nabla f(T_{i}x_{n})\Vert :n\in \mathbb {Z_{+}} \}.$ Since the sequences $\{x_{n}\}$ and $\{T_{i}x_{n}\}$ are bounded and $\nabla f$ is bounded on bounded subsets of X [10], we have $r_{i}<\infty$ for all $i\in {\mathbb {N}}.$ In view of Theorem 1.11, we have $dom(f^{*})=X^{*}$, $f^{*}$ is super coercive and uniformly convex on bounded sets. Then by using Proposition 1.10, for $u\in \Omega$ we have

$$\begin{aligned} D_{f}(u,z_{n}^{i})= & {} V_{f}\left( u,\beta _{n}\nabla f(x_{n}) + (1- \beta _{n})\nabla f(T_{i}x_{n})\right) \\\le & {} f(u) - \beta _{n}\langle u,\nabla f(x_{n})\rangle - (1-\beta _{n})\langle u,\nabla f(T_{i}x_{n})\rangle + \beta _{n} f^{*}(\nabla f(x_{n})) \\&+ (1-\beta _{n}) f^{*}(\nabla f(T_{i}x_{n}))- \beta _{n} (1- \beta _{n}) \rho ^{*}_{r_{i}}\left( \Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert \right) \\= & {} \beta _{n} V_{f}(u,\nabla f(x_{n}))+(1-\beta _{n})V_{f}(u,\nabla f(T_{i}x_{n}))\\&-\beta _{n} (1- \beta _{n}) \rho ^{*}_{r_{i}}\left( \Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert \right) \\= & {} \beta _{n} D_{f}(u,x_{n})+(1-\beta _{n})D_{f}(u,T_{i}x_{n})\\&-\beta _{n} (1- \beta _{n}) \rho ^{*}_{r_{i}}\left( \Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert \right) \\\le & {} D_{f}(u,x_{n})-\beta _{n} (1- \beta _{n}) \rho ^{*}_{r_{i}}\left( \Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert \right) , \end{aligned}$$

and so

$$\begin{aligned} D_{f}(u,y_{n}^{i})\le & {} \alpha _{n}D_{f}(u,x_{0}) + (1-\alpha _{n}) D_{f}(u,z_{n}^{i})\\\le & {} \alpha _{n}D_{f}(u,x_{0})+ D_{f}(u,x_{n})-\beta _{n}(1-\beta _{n})\rho ^{*}_{r_{i}}\left( \Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert \right) . \end{aligned}$$

Since $\liminf _{n\rightarrow \infty }\beta _{n}(1-\beta _{n})>0,$ there exists $a>0$ such that

$$\begin{aligned} a\rho ^{*}_{r_{i}}\left( \Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert \right)\le & {} \beta _{n}(1-\beta _{n}) \rho ^{*}_{r_{i}}\left( \Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert \right) \\\le & {} \alpha _{n} D_{f}(u,x_{0}) +D_{f}(u,x_{n})- D_{f}(u,y_{n}^{i})\\= & {} \alpha _{n} D_{f}(u,x_{0}) + f(u) - f(x_{n})- \left\langle u-x_{n},\nabla f(x_{n})\right\rangle \\&- f(u) + f(y_{n}^{i})+ \left\langle u-y_{n}^{i},\nabla f(y_{n}^{i})\right\rangle \\&+\left\langle u-x_{n},\nabla f(y_{n}^{i})\right\rangle -\left\langle u-x_{n},\nabla f(y_{n}^{i})\right\rangle \\= & {} \alpha _{n} D_{f}(u,x_{0}) + f(y_{n}^{i}) - f(x_{n}) \\&+ \left\langle u-x_{n}, \nabla f(y_{n}^{i})-\nabla f(x_{n})\rangle +\langle x_{n} - y_{n}^{i}, \nabla f(y_{n}^{i}) \right\rangle . \end{aligned}$$

Therefore, by boundness of $\nabla f$, uniform continuity of f [2, Theorem 1.8], uniform continuity of $\nabla f$ on bounded subsets, (2.7) and (2.8), we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \rho ^{*}_{r_{i}}\big (\Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert \big )=0, \quad (\forall i\in {\mathbb {N}}). \end{aligned}$$

Now we show that

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert \nabla f(x_{n})-\nabla f(T_{i}x_{n})\Vert =0, \quad (\forall i\in {\mathbb {N}}). \end{aligned}$$

In fact, if not, there exist $\epsilon _{0}>0$ and a subsequence $\{n_{j}\}$ of $\{n\}$ such that

$$\begin{aligned} \Vert \nabla f(x_{n_{j}})-\nabla f(T_{i}x_{n_{j}})\Vert \ge \epsilon _{0}, \end{aligned}$$

for all $j\in {\mathbb {N}}.$ Since $\rho ^{*}_{r_{i}}$ is nondecreasing, we have

$$\begin{aligned} \rho ^{*}_{r_{i}}\left( \Vert \nabla f(x_{n_{j}})-\nabla f(T_{i}x_{n_{j}})\Vert \right) \ge \rho ^{*}_{r_{i}}(\epsilon _{0}) \end{aligned}$$

for all $j\in {\mathbb {N}}.$ Letting $j\rightarrow \infty ,$ we have $0\ge \rho ^{*}_{r_{i}}(\epsilon _{0})$ for all $i\in {\mathbb {N}}.$ This contradicts the uniform convexity of $f^{*}$ on bounded subsets of $X^{*}$.

Since $\nabla f^{*}$ is uniformly continuous on bounded subset of $X^{*}$, we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert x_{n}- T_{i}x_{n}\Vert = 0, \quad (\forall i\in {\mathbb {N}}). \end{aligned}$$

(2.9)

Suppose that $\{x_{n_{k}}\}$ be a weakly convergent subsequence of $\{x_{n}\}$ and denote its weak limit by ${\bar{x}}.$ From (2.9), ${\bar{x}}\in {\widehat{F}}(T_{i})= F(T_{i})$ for all $i\in {\mathbb {N}}$ and hence ${\bar{x}}\in \Omega .$

Let $\hat{u}= \overleftarrow{Proj}_{\Omega }^{f}(x_{0}).$ Since $x_{n+1} = \overleftarrow{Proj}_{C_{n}\cap Q_{n}}^{f}(x_{0})$ and $\Omega \subset C_{n}\cap Q_{n},$ we have $D_{f}(x_{n+1},x_{0})\le D_{f}(\hat{u},x_{0})$. Therefore Proposition 1.5 implies that $\{x_{n}\}$ converges strongly to $\hat{u}.$ This completes the proof. $\square$

Let X be a uniformly smooth and uniformly convex Banach space and $f=\frac{1}{2}\Vert \cdot \Vert ^{2}.$ So $\nabla f=J$ and hence

$$\begin{aligned} D_{\frac{1}{2}\Vert \cdot \Vert ^{2}}(x,y)= & {} \dfrac{1}{2}\left( \Vert x\Vert ^{2} -\Vert y\Vert ^{2} - 2\langle x-y, Jy\rangle \right) \\= & {} \dfrac{1}{2}\left( \Vert x\Vert ^{2} +\Vert y\Vert ^{2} - 2\langle x, Jy\rangle \right) \\= & {} \dfrac{1}{2}\phi (x,y), \end{aligned}$$

therefore $\overleftarrow{Proj}_{C}^{\frac{1}{2}\Vert \cdot \Vert ^{2}}=\Pi _{C}$. Thus we get the following corollary.

Corollary 2.2

Let X be a uniformly smooth and uniformly convex Banach space, $\{T_{i}\}_{i=1}^{\infty }$ be an infinitely countable family of relatively nonexpansive mappings from C into itself, that is, $F(T_i)={\widehat{F}}(T_i)\ne \emptyset$ and $\phi (u,T_{i}x)\le \phi (u,x),$ for all $u\in F(T_i), x\in C.$ Assume that$\Omega := \bigcap _{i=1}^{\infty }F(T_{i})\ne \emptyset$ and $\{\alpha _{n}\},\{\beta _{n}\}$$\subset (0,1)$ such that $\lim _{n\rightarrow \infty } \alpha _{n}=0$ and $\liminf _{n\rightarrow \infty } \beta _{n}(1-\beta _{n})>0$. Then the sequence $\{x_{n}\}$ generated by:

$$\begin{aligned} \left\{ \begin{array}{lr} x_{0}\in C_{0} = C \, is \, chosen \, arbitrarily, \\ z_{n}^{i} = J^{-1} (\beta _{n}J(x_{n}) + (1- \beta _{n})J(T_{i}x_{n})),\\ y_{n}^{i} = J^{-1} (\alpha _{n}J(x_{0}) + (1- \alpha _{n})J(z_{n}^{i})),\\ C _{n} = \left\{ z\in C_{n-1}: \langle z- x_{n},J(x_{0})- J (x_{n}) \rangle \le 0\right\} ,\\ Q_{n}^{i} = \left\{ z\in C: \phi (z,y_{n}^{i} )\le \alpha _{n} \phi (z,x_{0} )+ (1-\alpha _{n})\phi (z,x_{n} )\right\} ,\\ Q_{n}= \bigcap _{i=1}^{\infty } Q_{n}^{i},\\ x_{n+1}= \Pi _{C_{n}\cap Q_{n}}(x_{0}), \end{array} \right. \end{aligned}$$

converges strongly to $\Pi _{\Omega }(x_{0}).$

Remark 2.3

In particular, if X is a Hilbert space, then $\nabla f= I$ and hence

$$\begin{aligned} D_{f}(x,y)=\dfrac{1}{2}\Vert x-y\Vert ^{2}\quad and\quad \overleftarrow{Proj_{C}^{f}}= P_{C}, \end{aligned}$$

where $P_{C}$ is the metric projection. Therefore algorithm (1.2) takes the following form:

$$\begin{aligned} \left\{ \begin{array}{lr} x_{0}\in C_{0} = C \, is \, chosen \, arbitrarily, \\ z_{n}^{i} = \beta _{n}x_{n} + (1- \beta _{n})T_{i}x_{n},\\ y_{n}^{i} = \alpha _{n}x_{0} + (1- \alpha _{n})z_{n}^{i},\\ C _{n} = \left\{ z\in C_{n-1}: \langle z- x_{n},x_{0}- x_{n} \rangle \le 0\right\} ,\\ Q_{n}^{i} = \left\{ z\in C: \Vert z-y_{n}^{i}\Vert ^{2} \le \alpha _{n} \Vert z-x_{0} \Vert ^{2}+ (1-\alpha _{n})\Vert z-x_{n}\Vert ^{2} \right\} ,\\ Q_{n}= \bigcap _{i=1}^{\infty } Q_{n}^{i},\\ x_{n+1}= P_{C_{n}\cap Q_{n}}(x_{0}), \end{array} \right. \end{aligned}$$

3 Applications

Equilibrium problem: Let $g:C\times C\rightarrow {\mathbb {R}}$ be a bifunction. An equilibrium problem is to find $x \in C$ such that for all $y \in C$,

$$\begin{aligned} g(x, y) \ge 0. \end{aligned}$$

(3.1)

Denote the set of solutions of problem (3.1) by EP(g). Equilibrium problems and their generalizations have been important tools for solving problems arising in the fields of linear or nonlinear programming, variational inequalities, complementary problems, optimization problems, fixed point problems, and have been widely applied to physics, structural analysis, management sciences and economics, etc. (see, for example, [6, 16, 19,20,21,22,23, 35]).

For solving equilibrium problem (3.1), let us assume that a bifunction g satisfies the following conditions:

• (C1) $g(x,x)=0,$ for all $x\in C,$

• (C2) g is monotone, i.e., $g(x,y) + g(y,x)\le 0,$ for all $x,y \in C,$

• (C3) for each $x,y,z \in C,$$\limsup _{t\downarrow 0}g(tz+(1-t)x,y)\le g(x,y),$

• (C4) for each $x\in C,$$g(x,\cdot )$ is convex and lower semicontinuous.

The resolvent operator of a bifunction $g : C \times C\rightarrow {\mathbb {R}}$ is $Res_{g}^{f}:X\rightarrow 2^{C}$ defined by

$$\begin{aligned} Res_{g}^{f}(x)= \left\{ z\in C: g(z,y) + \langle y-z,\nabla f(z) - \nabla f(x)\rangle \ge 0, \forall y\in C\right\} . \end{aligned}$$

(3.2)

Theorem 3.1

[32] Let $f : X \rightarrow (-\infty ,+\infty ]$ be a super coercive Legendre function. If the bifunction $g : C\times C \rightarrow {\mathbb {R}}$ satisfies conditions $(C1)-(C4)$, then

• (i) dom$(Res_{g}^{f}) = X$,

• (ii) $Res_{g}^{f}$ is single-valued,

• (iii) $Res_{g}^{f}$ is a BFNE operator,

• (iv) $F(Res_{g}^{f}) = EP(g)$,

• (v) EP(g) is a closed and convex subset of C.

If the Legendre function f is uniformly Fr$\acute{e}$chet differentiable and bounded on bounded subsets of X,  then $F(Res_{g}^{f})={\widehat{F}}(Res_{g}^{f})$ (see [34]). Hence by using Theorems 2.1 and 3.1, we get the following theorem.

Theorem 3.2

Let $g_{i} : C \times C \rightarrow {\mathbb {R}}$$(i= 1,2 , \ldots )$ be an infinitely countable family of bifunctions satisfying conditions $(C1)-(C4)$. Let $f : X \rightarrow {\mathbb {R}}$ be a super coercive Legendre function which is bounded, uniformly Fr$\acute{e}$chet differentiable and totally convex on bounded subsets of X. Assume that$\Omega := \bigcap _{i=1}^{\infty } EP(g_{i})\ne \emptyset ,$ and $\{\alpha _{n}\},\{\beta _{n}\}$$\subset (0,1)$ such that $\lim _{n\rightarrow \infty } \alpha _{n}=0$ and $\liminf _{n\rightarrow \infty } \beta _{n}(1-\beta _{n})> 0.$ Then the sequence $\{x_{n}\}$ generated by:

$$\begin{aligned} \left\{ \begin{array}{lr} x_{0}\in C_{0} = C \,is\, chosen\, arbitrarily ,\\ z_{n}^{i} = \nabla f^{*} (\beta _{n}\nabla f(x_{n}) + (1- \beta _{n})\nabla f(Res_{g_{i}}^{f}x_{n})),\\ y_{n}^{i} = \nabla f^{*} (\alpha _{n}\nabla f(x_{0}) + (1- \alpha _{n})\nabla f(z_{n}^{i})),\\ C _{n} = \left\{ z\in C_{n-1}: \langle z- x_{n},\nabla f(x_{0})- \nabla f(x_{n}) \rangle \le 0\right\} ,\\ Q_{n}^{i} = \left\{ z\in C: D_{f}(z,y_{n}^{i} )\le \alpha _{n} D_{f}(z,x_{0} )+ (1-\alpha _{n})D_{f}(z,x_{n} )\right\} ,\\ Q_{n}= \bigcap _{i=1}^{\infty } Q_{n}^{i},\\ x_{n+1}= \overleftarrow{Proj}_{C_{n}\cap Q_{n}}^{f}(x_{0}), \end{array} \right. \end{aligned}$$

converges strongly to $\overleftarrow{Proj}_{\Omega }^{f}(x_{0})$.

Convex feasibility problem: Let $\{C_{i} \}$ be a family of nonempty closed and convex subsets of X such that $\cap _{i=1}^{\infty }C_{i}\ne \emptyset$. The convex feasibility problem (CFP) is to find $x\in \cap _{i=1}^{\infty }C_{i}.$ Obviously,

$$\begin{aligned} F(\overleftarrow{Proj}_{C_{i}}^{f})= C_{i},\quad \forall i\ge 1. \end{aligned}$$

If the Legendre function is uniformly Fr$\acute{e}$chet differentiable and bounded on bounded subsets of X,  then the Bregman projection $\overleftarrow{Proj}_{C_{i}}^{f}$ is BFNE, hence BQNE and $F(\overleftarrow{Proj}_{C_{i}}^{f})={\widehat{F}}(\overleftarrow{Proj}_{C_{i}}^{f})$ (see [34]). Thus, if we take $T_{i}=\overleftarrow{Proj}_{C_{i}}^{f}$ in Theorem 2.1, we get a strong convergence theorem for approximating a solution of convex feasibility problems.

Theorem 3.3

Let $\{C_{i} \}$ be a family of nonempty closed and convex subsets of C such that $\Omega :=\bigcap _{i=1}^{\infty }C_{i}\ne \emptyset .$ Let $f : X \rightarrow {\mathbb {R}}$ be a super coercive Legendre function which is bounded, uniformly Fr$\acute{e}$chet differentiable and totally convex on bounded subsets of X. Assume that $\{\alpha _{n}\},\{\beta _{n}\}$$\subset [0,1]$ such that $\lim _{n\rightarrow \infty } \alpha _{n}=0$ and $\liminf _{n\rightarrow \infty } \beta _{n}(1-\beta _{n})>0$. Then the sequence $\{x_{n}\}$ generated by:

$$\begin{aligned} \left\{ \begin{array}{lr} x_{0}\in C_{0} = C \, is \, chosen \, be \, arbitrarily, \\ z_{n}^{i} = \nabla f^{*} (\beta _{n}\nabla f(x_{n}) + (1- \beta _{n})\nabla f( \overleftarrow{Proj}_{C_{i}}^{f}(x_{n})),\\ y_{n}^{i} = \nabla f^{*} (\alpha _{n}\nabla f(x_{0}) + (1- \alpha _{n})\nabla f(z_{n}^{i})),\\ C _{n} = \left\{ z\in C_{n-1}: \langle z- x_{n},\nabla f(x_{0})- \nabla f(x_{n}) \rangle \le 0\right\} ,\\ Q_{n}^{i} = \left\{ z\in C: D_{f}(z,y_{n}^{i} )\le \alpha _{n} D_{f}(z,x_{0} )+ (1-\alpha _{n})D_{f}(z,x_{n} )\right\} ,\\ Q_{n}= \bigcap _{i=1}^{\infty } Q_{n}^{i},\\ x_{n+1}= \overleftarrow{Proj}_{C_{n}\cap Q_{n}}^{f}(x_{0}) \end{array} \right. \end{aligned}$$

converges strongly to $\overleftarrow{Proj}_{\Omega }^{f}(x_{0})$.

Variational inequality: Let C be a nonempty, closed and convex subset of a Banach space X and $A : X \rightarrow 2^{X^{*}}$ be a multi-valued mapping. Then the corresponding variational inequality is defined as follows: find ${\bar{x}} \in C$ such that exists $\xi \in A({\bar{x}})$ with

$$\begin{aligned} \langle y-{\bar{x}},\xi \rangle \ge 0,\quad \forall y\in C. \end{aligned}$$

(3.3)

The mapping $A: X\rightarrow X^{*}$ is called hemicontinuous if for any $x\in dom(A)$, $x+t_{n}y\in dom(A),$ for $y\in X$ and $\lim _{n\rightarrow \infty }t_{n}= 0^{+}$ implies that $A(x+t_{n}y)$ converges weakly to Ax.

Remark 3.4

Let $A_{j} : X\rightarrow X^{*}$ be monotone mappings such that $C= dom(A_{j})$$(j= 1,2,\cdots )$. Assume that $A_{j}$ is bounded on bounded subsets and hemicontinuous on C. Then $g_{j} (x,y) := \langle y-x,A_{j}x\rangle$ satisfies conditions $(C1)-(C4)$ [18]. Thus if we take $g_{j} (x,y) = \langle y-x,A_{j}x\rangle$ in Theorem 2.1, we get a strong convergence theorem for approximating a common solution of an infinite system of variational inequalities.

Zeros of maximal monotone operators: Let $A:X\rightarrow 2^{X^*}$ be a maximal monotone mapping. The problem of finding an element $x\in X$ such that $0^{*}\in Ax$ is very important in optimization theory and related field. The resolvent of A, denoted by $Res_{A}^{f}: X\rightarrow 2^{X^{*}},$ is defined as follows:

$$\begin{aligned} Res_{A}^{f}(x)=(\nabla f+ A)^{-1}\circ \nabla f(x). \end{aligned}$$

This resolvent is a single-valued BFNE operator [4]. In addition, if the Legendre function f is uniformly Fr$\acute{e}$chet differentiable and bounded on bounded subsets of X, then $F(Res_{A}^{f})= {\widehat{F}}(Res_{A}^{f})$ [34]. It is well known that the fixed point of $Res_{A}^{f}$ is equal to the zeros of the operator A, that is, $F(Res_{A}^{f})= A^{-1}(0^{*}).$ If we take $T_{i}= Res_{A_{i}}^{f}$ in Theorem 2.1, then we get a strong convergence theorem for finding a common zero of an infinite family of maximal monotone operators.