Uniformly closed replaced AKTT or ^∗AKTT condition to get strong convergence theorems for a countable family of relatively quasi-nonexpansive mappings and systems of equilibrium problems

Abstract

The purpose of this paper is to construct a new iterative scheme and to get a strong convergence theorem for a countable family of relatively quasi-nonexpansive mappings and a system of equilibrium problems in a uniformly convex and uniformly smooth real Banach space using the properties of generalized f-projection operator. The notion of uniformly closed mappings is presented and an example will be given which is a countable family of uniformly closed relatively quasi-nonexpansive mappings but not a countable family of relatively nonexpansive mappings. Another example shall be given which is uniformly closed but does not satisfy condition AKTT and ^∗AKTT. Our results can be applied to solve a convex minimization problem. In addition, this paper clarifies an ambiguity in a useful lemma. The results of this paper modify and improve many other important recent results.

MSC:47H05, 47H09, 47H10.

1 Introduction and preliminaries

Let E be a real Banach space and C be a nonempty closed convex subset of E. A mapping $T:C\to C$ is called nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

Let E be a real Banach space and C be a nonempty closed convex subset of E. A point $p\in C$ is said to be an asymptotic fixed point of T if there exists a sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}\subset C$ such that $x_n\rightharpoonup p$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. The set of asymptotic fixed point is denoted by $\hat{F}(T)$. We say that a mapping T is relatively nonexpansive (see [1–4]) if the following conditions are satisfied:

• (I) $F(T)\ne \mathrm{\varnothing }$;

• (II) $\varphi (p,Tx)\le \varphi (p,x)$, $\mathrm{\forall }x\in C$, $p\in F(T)$;

• (III) $F(T)=\hat{F}(T)$.

If T satisfies (I) and (II), then T is said to be relatively quasi-nonexpansive. It is easy to see that the class of relatively quasi-nonexpansive mappings contains the class of relatively nonexpansive mappings.

Let E be a real Banach space. The modulus of smoothness of E is the function ${\rho }_E:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ defined by

$${\rho }_E(\tau )=sup\{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )-1:\parallel x\parallel \le 1,\parallel y\parallel \le \tau \}.$$

E is uniformly smooth if and only if

$$\underset{\tau \to 0}{lim}\frac{{\rho }_E\tau }{\tau }=0.$$

Let $dimE\ge 2$. The modulus of convexity of E is the function ${\delta }_E(ϵ):=inf\{1-\parallel \frac{x+y}{2}\parallel :\parallel x\parallel =\parallel y\parallel =1;ϵ=\parallel x-y\parallel \}$. E is uniformly convex if for any $ϵ\in (0,2]$, there exists $\delta =\delta (ϵ)>0$ such that if $x,y\in E$ with $\parallel x\parallel \le 1$, $\parallel y\parallel \le 1$ and $\parallel x-y\parallel \ge ϵ$, then $\parallel \frac{1}{2}(x+y)\parallel \le 1-\delta$. Equivalently, E is uniformly convex if and only if ${\delta }_E(ϵ)>0$ for all $ϵ\in (0,2]$. A normed space E is called strictly convex if for all $x,y\in E$, $x\ne y$, $\parallel x\parallel =\parallel y\parallel =1$, we have $\parallel \lambda x+(1-\lambda )y\parallel <1$, $\mathrm{\forall }\lambda \in (0,1)$.

Let $E^{\ast }$ be the dual space of E. We denote by J the normalized duality mapping from E to $2^{E^{\ast }}$ defined by

$$J(x)=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^2={\parallel f\parallel }^2\}.$$

The following properties of J are well known (see [5–7] for more details):

• (1) If E is uniformly smooth, then J is norm-to-norm uniformly continuous on each bounded subset of E.

• (2) If E is reflexive, then J is a mapping from E onto $E^{\ast }$.

• (3) If E is smooth, then J is single valued.

Throughout this paper, we denote by ϕ the functional on $E\times E$ defined by

$$\varphi (x,y)={\parallel x\parallel }^2-2〈x,J(y)〉+{\parallel y\parallel }^2,\mathrm{\forall }x,y\in E.$$

(1.1)

Let E be a smooth, strictly convex, and reflexive real Banach space and let C be a nonempty closed convex subset of E. Following Alber [8], the generalized projection ${\mathrm{\Pi }}_C$ from E onto C is defined by

$${\mathrm{\Pi }}_C(x)=\underset{y\in C}{arg\hspace{0.17em}min}\varphi (y,x),\mathrm{\forall }x\in E.$$

The existence and uniqueness of ${\mathrm{\Pi }}_C$ follows from the property of the functional $\varphi (x,y)$ and strict monotonicity of the mapping J. It is obvious that

$${(\parallel x\parallel -\parallel y\parallel )}^2\le \varphi (x,y)\le {(\parallel x\parallel +\parallel y\parallel )}^2,\mathrm{\forall }x,y\in E.$$

(1.2)

Next, we recall the notion of generalized f-projection operator and its properties. Let $G:C\times E^{\ast }\to R\cup \{+\mathrm{\infty }\}$ be a functional defined as follows:

$$G(\xi ,\phi )={\parallel \xi \parallel }^2-2〈\xi ,\phi 〉+{\parallel \phi \parallel }^2+2\rho f(\xi ),$$

(1.3)

where $\xi \in C$, $\phi \in E^{\ast }$, ρ is a positive number and $f:C\to R\cup \{+\mathrm{\infty }\}$ is proper, convex, and lower semi-continuous. From the definitions of G and f, it is easy to see the following properties:

• (i) $G(\xi ,\phi )$ is convex and continuous with respect to φ when ξ is fixed;

• (ii) $G(\xi ,\phi )$ is convex and lower semi-continuous with respect to ξ when φ is fixed.

Definition 1.1 [9]

Let E be a real Banach space with its dual $E^{\ast }$. Let C be a nonempty, closed, and convex subset of E. We say that ${\mathrm{\Pi }}_C^f:E^{\ast }\to 2^C$ is a generalized f-projection operator if

$${\mathrm{\Pi }}_C^f\phi =\{u\in C:G(u,\phi )=\underset{\xi \in C}{inf}G(\xi ,\phi )\},\mathrm{\forall }\phi \in E^{\ast }.$$

For the generalized f-projection operator, Wu and Huang [9] proved in the following theorem some basic properties.

Lemma 1.2 [9]

Let E be a real reflexive Banach space with its dual $E^{\ast }$. Let C be a nonempty, closed, and convex subset of E. Then the following statements hold:

• (i) ${\mathrm{\Pi }}_C^f$ is a nonempty closed convex subset of C for all $\phi \in E^{\ast }$.

• (ii) If E is smooth, then for all $\phi \in E^{\ast }$, $x\in {\mathrm{\Pi }}_C^f\phi$ if and only if

  $$〈x-y,\phi -Jx〉+\rho f(y)-\rho f(x)\ge 0,\mathrm{\forall }y\in C.$$

• (iii) If E is strictly convex and $f:C\to R\cup \{+\mathrm{\infty }\}$ is positive homogeneous (i.e., $f(tx)=tf(x)$ for all $t>0$ such that $tx\in C$ where $x\in C$), then ${\mathrm{\Pi }}_C^f$ is a single-valued mapping.

Fan et al. [10] showed that the condition f is positive homogeneous which appeared in Lemma 1.2 can be removed.

Lemma 1.3 [10]

Let E be a real reflexive Banach space with its dual $E^{\ast }$ and C a nonempty, closed, and convex subset of E. Then if E is strictly convex, then ${\mathrm{\Pi }}_C^f$ is a single-valued mapping.

Recall that J is a single-valued mapping when E is a smooth Banach space. There exists a unique element $\phi \in E^{\ast }$ such that $\phi =Jx$ for each $x\in E$. This substitution in (1.3) gives

$$G(\xi ,Jx)={\parallel \xi \parallel }^2-2〈\xi ,Jx〉+{\parallel x\parallel }^2+2\rho f(\xi ).$$

(1.4)

Now, we consider the second generalized f-projection operator in a Banach space.

Definition 1.4 [11]

Let E be a real Banach space and C a nonempty, closed, and convex subset of E. We say that ${\mathrm{\Pi }}_C^f:E\to 2^C$ is a generalized f-projection operator if

$${\mathrm{\Pi }}_C^{f}x=\{u\in C:G(u,Jx)=\underset{\xi \in C}{inf}G(\xi ,Jx)\},\mathrm{\forall }x\in E.$$

Obviously, the definition of relatively quasi-nonexpansive mapping T is equivalent to

• (1) $F(T)\ne \mathrm{\varnothing }$;

• (2) $G(p,JTx)\le G(p,Jx)$, $\mathrm{\forall }x\in C$, $p\in F(T)$.

Lemma 1.5 [12]

Let E be a Banach space and $f:E\to R\cup \{+\mathrm{\infty }\}$ be a lower semi-continuous convex functional. Then there exist $x\in E^{\ast }$ and $\alpha \in R$ such that

$$f(x)\ge 〈x,x^{\ast }〉+\alpha ,\mathrm{\forall }x\in E.$$

We know that the following lemmas hold for operator ${\mathrm{\Pi }}_C^f$.

Lemma 1.6 [13]

Let C be a nonempty, closed, and convex subset of a smooth and reflexive Banach space E. Then the following statements hold:

• (i) ${\mathrm{\Pi }}_C^f$ is a nonempty, closed, and convex subset of C for all $x\in E$;

• (ii) for all $x\in E$, $\hat{x}\in {\mathrm{\Pi }}_C^{f}x$ if and only if

  $$〈\hat{x}-y,Jx-J\hat{x}〉+\rho f(y)-\rho f(x)\ge 0,\mathrm{\forall }y\in C;$$

• (iii) if E is strictly convex, then ${\mathrm{\Pi }}_C^{f}x$ is a single-valued mapping.

Lemma 1.7 [13]

Let C be a nonempty, closed, and convex subset of a smooth and reflexive Banach space E. Let $x\in E$ and $\hat{x}\in {\mathrm{\Pi }}_C^{f}x$. Then

$$\varphi (y,\hat{x})+G(\hat{x},Jx)\le G(y,Jx),\mathrm{\forall }y\in C.$$

The fixed points set $F(T)$ of a relatively quasi-nonexpansive mapping is closed convex as given in the following lemma.

Lemma 1.8 [14, 15]

Let C be a nonempty closed convex subset of a smooth, uniformly convex Banach space E. Let T be a closed relatively quasi-nonexpansive mapping of C into itself. Then $F(T)$ is closed and convex.

Also, this following lemma will be used in the sequel.

Lemma 1.9 [16]

Let C be a nonempty closed convex subset of a smooth, uniformly convex Banach space E. Let ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ be sequences in E such that either ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ or ${\{y_n\}}_{n=0}^{\mathrm{\infty }}$ is bounded. If ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,y_n)=0$, then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$.

Lemma 1.10 [17]

Let $p>1$ and $r>0$ be two fixed real numbers. Then a Banach space X is uniformly convex if and only if there is a continuous, strictly increasing and convex function $g:R^{+}\to R^{+}$, $g(0)=0$, such that

$${\parallel \lambda x+(1-\lambda )y\parallel }^p\le \lambda {\parallel x\parallel }^p+(1-\lambda ){\parallel y\parallel }^p-W_p(\lambda )g(\parallel x-y\parallel )$$

for all $x,y\in B_r$ and $0\le \lambda \le 1$, where $W_p(\lambda )=\lambda {(1-\lambda )}^p+{\lambda }^p(1-\lambda )$.

Remark We can see from the Lemma 1.10 that the function g has no relation with the selection of x, y and λ. However, the key point above, in the process of generalization and application about this lemma, has been ambiguous gradually. For instance, the following lemma states that the function g has something to do with λ, which always leads to failure in the proof.

Lemma (stated in [[11], Lemma 2.10])

Let E be a uniformly convex real Banach space. For arbitrary $r>0$, let $B_r(0):=\{x\in E:\parallel x\parallel \le r\}$ and $\lambda \in [0,1]$. Then there exists a continuous strictly increasing convex function

$$g:[0,2r]\to R,g(0)=0$$

such that for every $x,y\in B_r(0)$, the following inequality holds:

$${\parallel \lambda x+(1-\lambda )y\parallel }^2\le \lambda {\parallel x\parallel }^2+(1-\lambda ){\parallel y\parallel }^2-\lambda (1-\lambda )g(\parallel x-y\parallel ).$$

Let F be a bifunction of $C\times C$ into R. The equilibrium problem is to find $x^{\ast }\in C$ such that $F(x^{\ast },y)\ge 0$, for all $y\in C$. We shall denote the solutions set of the equilibrium problem by $EP(F)$. Numerous problems in physics, optimization, and economics reduce to find a solution of equilibrium problem. The equilibrium problems include fixed point problems, optimization problems, and variational inequality problems as special cases.

For solving the equilibrium problem for a bifunction $F:C\times C\to R$, let us assume that F satisfies the following conditions:

(A1) $F(x,x)=0$ for all $x\in C$;

(A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

(A3) for each $x,y\in C$, ${lim}_{t\to 0}F(tz+(1-t)x,y)\le F(x,y)$;

(A4) for each $x\in C$, $y\mapsto F(x,y)$ is convex and lower semi-continuous.

Lemma 1.11 [18]

Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space E and let F be a bifunction of $C\times C$ into R satisfying (A1)-(A4). Let $r>0$ and $x\in E$. Then there exists $z\in C$ such that

$$F(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0,\mathrm{\forall }y\in K.$$

Lemma 1.12 [19]

Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space E. Assume that $F:C\times C\to R$ satisfies (A1)-(A4). For $r>0$ and $x\in E$, define a mapping $T_r^F:E\to C$ as follows:

$$T_r^F(x)=\{z\in C:F(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0,\mathrm{\forall }y\in C\}$$

for all $z\in E$. Then the following hold:

• (1) $T_r^F$ is single valued;

• (2) $T_r^F$ is a firmly nonexpansive-type mapping, i.e., for any $x,y\in E$,

  $$〈T_r^{F}x-T_r^{F}y,J{T}_r^{F}x-J{T}_r^{F}y〉\le 〈T_r^{F}x-T_r^{F}y,Jx-Jy〉;$$

• (3) $F(T_r^F)=EP(F)$;

• (4) $EP(F)$ is closed and convex.

Lemma 1.13 [19]

Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space E. Assume that $F:C\times C\to R$ satisfies (A1)-(A4) and let $r>0$. Then for each $x\in E$ and $q\in F(T_r^F)$,

$$\varphi (q,T_r^{F}x)+\varphi (T_r^{F}x,x)\le \varphi (q,x).$$

Let $\{T_n\}$ be a sequence of mappings from C into E, where C is a nonempty closed convex subset of a real Banach space E. For a subset B of C, we say that

• (i) $(\{T_n\},B)$ satisfies condition AKTT (see [15]) if

  $$\sum _{n=1}^{\mathrm{\infty }}sup\{\parallel T_{n+1}x-T_{n}x\parallel :x\in B\}<\mathrm{\infty };$$

• (ii) $(\{T_n\},B)$ satisfies condition ^∗AKTT (see [15]) if

  $$\sum _{n=1}^{\mathrm{\infty }}sup\{\parallel J{T}_{n+1}x-J{T}_{n}x\parallel :x\in B\}<\mathrm{\infty }.$$

Recently, Shehu [11] proved strong convergence theorems for approximation of common element of set of common fixed points of countably infinite family of relatively quasi-nonexpansive mappings and set of common solutions to a system of equilibrium problems in a uniformly convex and uniformly smooth real Banach space using the properties of generalized f-projection operator. The author obtained the following theorem.

Theorem 1.14 [11]

Let E be a uniformly convex real Banach space which is also uniformly smooth. Let C be a nonempty closed convex subset of E. For each $k=1,2,\dots ,m$, let $F_k$ be a bifunction from $C\times C$ satisfying (A1)-(A4) and let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be an infinite family of relatively quasi-nonexpansive mappings of C into itself such that $F:=({\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n))\cap ({\bigcap }_{k=1}^{m}EP(F_k))\ne \mathrm{\varnothing }$. Let $f:E\to R$ be a convex and lower semi-continuous mapping with $C\subset int(D(f))$ and suppose ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is iteratively generated by $x_0\in C$, $C_1=C$, $x_1={\mathrm{\Pi }}_{C_1}^f{x}_0$,

$$\{\begin{array}{l}{y}_n=J^{-1}({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n),\\ u_n=T_{r_{m,n}}^{F_m}{T}_{r_{m-1,n}}^{F_{m-1}}\cdots T_{r_{2,n}}^{F_2}{T}_{r_{1,n}}^{F_1}{y}_n,\\ C_{n+1}=\{w\in C_n:G(w,J{u}_n)\le G(w,J{x}_n)\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}^f{x}_0,n\ge 1,\end{array}$$

(1.5)

where J is the duality mapping on E. Suppose ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence in $(0,1)$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ ${\{r_{k,n}\}}_{n=1}^{\mathrm{\infty }}\subset (0,\mathrm{\infty })$ ($k=1,2,\dots ,m$) satisfying ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{k,n}>0$ ($k=1,2,\dots ,m$). Suppose that for each bounded subset B of C, the ordered pair $(\{T_n\},B)$ satisfies either condition AKTT or condition ^∗ AKTT. Let T be the mapping from C into E defined by $Tx={lim}_{n\to \mathrm{\infty }}{T}_{n}x$ for all $x\in C$ and suppose that T is closed and $F(T)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$. Then ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to ${\mathrm{\Pi }}_F^f{x}_0$.

In this paper we will construct a new iterative scheme and will get strong convergence theorem for a countable family of relatively quasi-nonexpansive mappings and a system of equilibrium problems in a uniformly convex and uniformly smooth real Banach space using the properties of generalized f-projection operator. The notion of uniformly closed mappings is presented and an example will be given which is a countable family of uniformly closed relatively quasi-nonexpansive mappings but not a countable family of relatively nonexpansive mappings. Another example shall be given which is uniformly closed but not satisfy condition AKTT and ^∗AKTT.

2 Main results

Now, we shall first introduce the notion of uniformly closed mappings and give an example which is a countable family of uniformly closed relatively quasi-nonexpansive mappings but not a countable family of relatively nonexpansive mappings in the sense of G. Another example shall be given which is uniformly closed but not satisfy condition AKTT and ^∗AKTT.

Definition 2.1 Let E be a Banach space, C be a nonempty closed convex subset of E. Let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}:C\to E$ be a sequence of mappings of C into E such that ${\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$ is nonempty. ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ is said to be uniformly closed, if $p\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$, whenever $\{x_n\}\subset C$ converges strongly to p and $\parallel x_n-T_n{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Example 1 Let $E=l^2$, where

$$\begin{array}{c}{l}^2=\{\xi =({\xi }_1,{\xi }_2,{\xi }_3,\dots ,{\xi }_n,\dots ):\sum _{n=1}^{\mathrm{\infty }}{|{\xi }_n|}^2<\mathrm{\infty }\},\hfill \\ \parallel \xi \parallel ={\left(\sum _{n=1}^{\mathrm{\infty }}{|{\xi }_n|}^2\right)}^{\frac{1}{2}},\mathrm{\forall }\xi \in l^2,\hfill \\ 〈\xi ,\eta 〉=\sum _{n=1}^{\mathrm{\infty }}{\xi }_n{\eta }_n,\mathrm{\forall }\xi =({\xi }_1,{\xi }_2,{\xi }_3,\dots ,{\xi }_n,\dots ),\eta =({\eta }_1,{\eta }_2,{\eta }_3,\dots ,{\eta }_n,\dots )\in l^2.\hfill \end{array}$$

It is well known that $l^2$ is a Hilbert space, so that ${(l^2)}^{\ast }=l^2$. Let $\{x_n\}\subset E$ be a sequence defined by

$$\begin{array}{c}{x}_0=(1,0,0,0,\dots ),\hfill \\ x_1=(1,1,0,0,\dots ),\hfill \\ x_2=(1,0,1,0,0,\dots ),\hfill \\ x_3=(1,0,0,1,0,0,\dots ),\hfill \\ \cdots \hfill \\ x_n=({\xi }_{n,1},{\xi }_{n,2},{\xi }_{n,3},\dots ,{\xi }_{n,k},\dots )\hfill \\ \cdots ,\hfill \end{array}$$

where

$${\xi }_{n,k}=\{\begin{array}{cc}1,\hfill & \text{if }k=1,n+1,\hfill \\ 0,\hfill & \text{if }k\ne 1,k\ne n+1,\hfill \end{array}$$

for all $n\ge 1$.

Define a countable family of mappings $T_n:E\to E$ as follows:

$$T_n(x)=\{\begin{array}{cc}\frac{n}{n+1}{x}_n,\hfill & \text{if }x=x_n,\hfill \\ -x,\hfill & \text{if }x\ne x_n,\hfill \end{array}$$

for all $n\ge 0$.

Conclusion 2.2 ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ has a unique fixed point 0, that is, $F(T_n)=\{0\}\ne \mathrm{\varnothing }$, $\mathrm{\forall }n\ge 0$.

Proof The conclusion is obvious. □

Let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a countable family of quasi-relatively quasi-nonexpansive mappings, if

$$\bigcap _{n=0}^{\mathrm{\infty }}F(T_n)=\hat{F}\left({\{T_n\}}_{n=0}^{\mathrm{\infty }}\right),$$

the ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ is said to be a countable family of relatively nonexpansive mappings in the sense of functional G, where

$$\hat{F}\left({\{T_n\}}_{n=0}^{\mathrm{\infty }}\right)=\{p\in C:\mathrm{\exists }{x}_n\rightharpoonup p,\parallel x_n-T_n{x}_n\parallel \to 0,x_n\in C\}$$

is said to be the asymptotic fixed point set of ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$.

Conclusion 2.3 ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ is a countable family of relatively quasi-nonexpansive mappings but not a countable family of relatively nonexpansive mappings in the sense of functional G.

Proof By Conclusion 2.2, we only need to show that $G(0,J{T}_{n}x)\le G(0,Jx)$, $\mathrm{\forall }x\in E$. Note that $E=l^2$ is a Hilbert space, for any $n\ge 0$ we can derive

$$\begin{array}{c}G(0,J{T}_{n}x)\le G(0,Jx)\mathrm{\forall }x\in E\hfill \\ \iff \varphi (0,T_{n}x)\le \varphi (0,x)\hfill \\ \iff {\parallel 0-T_{n}x\parallel }^2\le {\parallel 0-x\parallel }^2\hfill \\ \iff {\parallel T_{n}x\parallel }^2\le {\parallel x\parallel }^2.\hfill \end{array}$$

It is obvious that $\{x_n\}$ converges weakly to $x_0=(1,0,0,\dots )$, and

$$\parallel x_n-T_n{x}_n\parallel =\parallel \frac{n}{n+1}{x}_n-x_n\parallel =\frac{1}{n+1}\parallel x_n\parallel \to 0,$$

as $n\to \mathrm{\infty }$, so $x_0$ is an asymptotic fixed point of ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$. Joining with Conclusion 2.2, we can obtain ${\bigcap }_{n=0}^{\mathrm{\infty }}F(T_n)\ne \hat{F}({\{T_n\}}_{n=0}^{\mathrm{\infty }})$.

Thus, ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ is a countable family of relatively quasi-nonexpansive mappings but not a countable family of relatively nonexpansive mappings in the sense of G. □

Conclusion 2.4 ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ is a countable family of uniformly closed relatively quasi-nonexpansive mappings in the sense of functional G.

Proof In fact, for any strong convergent sequence $\{z_n\}\subset E$ such that $z_n\to z_0$ and $\parallel z_n-T_n{z}_n\parallel \to 0$ as $n\to \mathrm{\infty }$, there exists a sufficiently large nature number N, such that $z_n\ne x_m$ for any $n,m>N$ (since $x_n$ is not a Cauchy sequence it cannot converge to any element in E). Then $T_n{z}_n=-z_n$ for $n>N$, it follows from $\parallel z_n-T_n{z}_n\parallel \to 0$ that $2z_n\to 0$ and hence $z_n\to z_0=0$.

Therefore, ${\{T_n\}}_{n=0}^{\mathrm{\infty }}$ is a countable family of uniformly closed relatively quasi-nonexpansive mappings but not a countable family of relatively nonexpansive mappings in the sense of functional G. □

Now, we give an example which is a countable family of uniformly closed quasi-nonexpansive mappings but not satisfied condition AKTT and ^∗AKTT.

Example 2 Let $X={\mathrm{\Re }}^2$. For any complex number $x=r{e}^{i\theta }\in X$, define a countable family of quasi-nonexpansive mappings as follows:

$$T_n:r{e}^{i\theta }\to r{e}^{i(\theta +n\frac{\pi }{2})},n=1,2,3,\dots .$$

Proof It is easy to see that ${\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)=\{0\}$. We first prove that $\{T_n\}$ is uniformly closed. In fact, for any strong convergent sequence $\{x_n\}\subset X$ such that $x_n\to x_0$ and $\parallel x_n-T_n{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$, there must be $x_0=0\in {\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$. Otherwise, if $x_n\to x_0\ne 0$, and

$$\parallel x_{4n+1}-T_{4n+1}{x}_{4n+1}\parallel \to 0,$$

since $T_1$ is continuous, we have

$$\begin{array}{r}\parallel x_{4n+1}-T_{4n+1}{x}_{4n+1}\parallel \\ =\parallel x_{4n+1}-T_1{x}_{4n+1}\parallel \to \parallel x_0-T_1{x}_0\parallel \ne 0.\end{array}$$

This is a contradiction. Therefore, $\{T_n\}$ is uniformly closed.

Besides, take any $x=r{e}^{i\theta }\ne 0$. For any n by the definition of $T_n$, we have

$$\parallel T_{n}x-T_{n+1}x\parallel =\parallel r{e}^{\frac{\pi i}{2}}\parallel =r>0$$

and

$$\parallel J{T}_{n}x-J{T}_{n+1}x\parallel =\parallel r{e}^{\frac{\pi i}{2}}\parallel =r>0.$$

That is to say, $\{T_n\}$ does not satisfied condition AKTT and ^∗AKTT. □

Now we are in a position to present our main theorems.

Theorem 2.5 Let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a countable family of uniformly closed relatively quasi-nonexpansive mappings of C into itself and other conditions are the same as Theorem 1.14 except for condition AKTT, ^∗ AKTT and condition ‘Let T be the mapping from C into E defined by $Tx={lim}_{n\to \mathrm{\infty }}{T}_{n}x$ for all $x\in C$ and suppose that T is closed and $F(T)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$ ’. Then the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ generated by (1.5) converges strongly to ${\mathrm{\Pi }}_F^f{x}_0$.

Proof We first show that $C_n$, $\mathrm{\forall }n\ge 1$, is closed and convex. It is obvious that $C_1=C$ is closed and convex. Suppose that $C_n$ is closed convex for some $n>1$. From the definition of $C_{n+1}$, we have $z\in C_{n+1}$ implies $G(z,J{u}_n)\le G(z,J{x}_n)$. This is equivalent to

$$2(〈z,J{x}_n〉-〈z,J{u}_n〉)\le {\parallel x_n\parallel }^2-{\parallel u_n\parallel }^2.$$

This implies that $C_{n+1}$ is closed convex for the same $n>1$. Hence, $C_n$ is closed and convex for all $n\ge 1$. This shows that ${\mathrm{\Pi }}_{C_{n+1}}^f{x}_0$ is well defined for all $n\ge 0$.

By taking ${\theta }_n^k=T_{r_{k,n}}^{F_k}{T}_{r_{k-1,n}}^{F_{k-1}}\cdots T_{r_{2,n}}^{F_2}{T}_{r_{1,n}}^{F_1}$, $k=1,2,\dots ,m$ and ${\theta }_n^0=I$ for all $n\ge 1$, we obtain $u_n={\theta }_n^m{y}_n$.

We next show that $F\subset C_n$, $\mathrm{\forall }n\ge 1$. From Lemma 1.12, one sees that $T_{r_{k,n}}^{F_k}$, $k=1,2,\dots ,m$, is relatively nonexpansive mapping. For $n=1$, we have $F\subset C=C_1$. Now, assume that $F\subset C_n$ for some $n\ge 2$. Then for each $x^{\ast }\in F$, we obtain

$$\begin{array}{rl}G(x^{\ast },J{u}_n)=& G(x^{\ast },J{\theta }_n^m{y}_n)\le G(x^{\ast },J{y}_n)\\ =& G(x^{\ast },({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n))\\ =& {\parallel x^{\ast }\parallel }^2-2{\alpha }_n〈x^{\ast },J{x}_n〉-2(1-{\alpha }_n)〈x^{\ast },J{T}_n{x}_n〉\\ +{\parallel {\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n\parallel }^2+2\rho f\left(x^{\ast }\right)\\ \le & {\parallel x^{\ast }\parallel }^2-2{\alpha }_n〈x^{\ast },J{x}_n〉-2(1-{\alpha }_n)〈x^{\ast },J{T}_n{x}_n〉\\ +{\alpha }_n{\parallel J{x}_n\parallel }^2+(1-{\alpha }_n){\parallel J{T}_n{x}_n\parallel }^2+2\rho f\left(x^{\ast }\right)\\ =& {\alpha }_{n}G(x^{\ast },J{x}_n)+(1-{\alpha }_n)G(x^{\ast },J{T}_n{x}_n)\le G(x^{\ast },J{x}_n).\end{array}$$

(2.1)

So, $x^{\ast }\in C_n$. This implies that $F\subset C_n$, $\mathrm{\forall }n\ge 1$ and the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ generated by (1.5) is well defined.

We now show that ${lim}_{n\to \mathrm{\infty }}G(x_n,J{x}_0)$ exists. Since $f:E\to R$ is a convex and lower semi-continuous, applying Lemma 1.5, we see that there exist $u^{\ast }\in E^{\ast }$ and $\alpha \in R$ such that

$$f(y)\ge 〈y,u^{\ast }〉+\alpha ,\mathrm{\forall }y\in E.$$

It follows that

$$\begin{array}{rl}G(x_n,J{x}_0)& ={\parallel x_n\parallel }^2-2〈x_n,J{x}_0〉+{\parallel x_0\parallel }^2+2\rho f(x_n)\\ \ge {\parallel x_n\parallel }^2-2〈x_n,J{x}_0〉+{\parallel x_0\parallel }^2+2\rho 〈x_n,u^{\ast }〉+2\rho \alpha \\ ={\parallel x_n\parallel }^2-2〈x_n,J{x}_0-\rho u^{\ast }〉+{\parallel x_0\parallel }^2+2\rho \alpha \\ \ge {\parallel x_n\parallel }^2-2\parallel x_n\parallel \parallel J{x}_0-\rho u^{\ast }\parallel +{\parallel x_0\parallel }^2+2\rho \alpha \\ ={(\parallel x_n\parallel -\parallel J{x}_0-\rho u^{\ast }\parallel )}^2+{\parallel x_0\parallel }^2-{\parallel J{x}_0-\rho u^{\ast }\parallel }^2+2\rho \alpha .\end{array}$$

(2.2)

Since $x_n={\mathrm{\Pi }}_{C_n}^f{x}_0$, it follows from (2.2) that

$$G(x^{\ast },J{x}_0)\ge G(x_n,J{x}_0)\ge {(\parallel x_n\parallel -\parallel J{x}_0-\rho u^{\ast }\parallel )}^2+{\parallel x_0\parallel }^2-{\parallel J{x}_0-\rho u^{\ast }\parallel }^2+2\rho \alpha$$

for each $x^{\ast }\in F(T)$. This implies that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded and so is ${\{G(x_n,J{x}_0)\}}_{n=0}^{\mathrm{\infty }}$. By the construction of $C_n$, we have $C_m\subset C_n$ and $x_m={\mathrm{\Pi }}_{C_m}^f{x}_0\in C_n$ for any positive integer $m\ge n$. It then follows from Lemma 1.7 that

$$\varphi (x_m,x_n)+G(x_n,J{x}_0)\le G(x_m,J{x}_0).$$

(2.3)

It is obvious that

$$\varphi (x_m,x_n)\ge {(\parallel x_m\parallel -\parallel x_n\parallel )}^2\ge 0.$$

In particular,

$$\varphi (x_{n+1},x_n)+G(x_n,J{x}_0)\le G(x_{n+1},J{x}_0)$$

and

$$\varphi (x_{n+1},x_n)\ge {(\parallel x_{n+1}\parallel -\parallel x_n\parallel )}^2\ge 0,$$

and so ${\{G(x_n,J{x}_0)\}}_{n=0}^{\mathrm{\infty }}$ is nondecreasing. It follows that the limit of ${\{G(x_n,J{x}_0)\}}_{n=0}^{\mathrm{\infty }}$ exists.

By the fact that $C_m\subset C_n$ and $x_m={\mathrm{\Pi }}_{C_m}^f{x}_0\in C_n$ for any positive integer $m\ge n$, we obtain

$$\varphi (x_m,u_n)\le \varphi (x_m,x_n).$$

Now, (2.3) implies that

$$\varphi (x_m,u_n)\le \varphi (x_m,x_n)\le G(x_m,J{x}_0)-G(x_n,J{x}_0).$$

(2.4)

Taking the limit as $m,n\to \mathrm{\infty }$ in (2.4), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_m,x_n)=0.$$

It then follows from Lemma 1.9 that $\parallel x_m-x_n\parallel \to 0$ as $m,n\to \mathrm{\infty }$. Hence, ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is a Cauchy sequence. Since E is a Banach space and C is closed and convex, there exists $p\in C$ such that $x_n\to p$ as $n\to \mathrm{\infty }$.

Now since $\varphi (x_m,x_n)\to 0$ as $m,n\to \mathrm{\infty }$ we have in particular that $\varphi (x_{n+1},x_n)\to 0$ as $n\to \mathrm{\infty }$ and this further implies that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$. Since $x_{n+1}={\mathrm{\Pi }}_{C_{n=1}}^f{x}_0\in C_{n+1}$ we have

$$\varphi (x_{n+1},u_n)\le \varphi (x_{n+1},x_n),\mathrm{\forall }n\ge 0.$$

Then we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},u_n)=0.$$

Since E is uniformly convex and smooth, we have from Lemma 1.9

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0=\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-u_n\parallel .$$

So,

$$\parallel x_n-u_n\parallel \le \parallel x_{n+1}-x_n\parallel +\parallel x_{n+1}-u_n\parallel .$$

Hence,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_n\parallel =0.$$

(2.5)

Since J is uniformly norm-to-norm continuous on bounded sets and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u_n\parallel =0$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_n-J{u}_n\parallel =0.$$

(2.6)

Let $r={sup}_{n\ge 1}\{\parallel x_n\parallel ,\parallel T_n{x}_n\parallel \}$. Since E is uniformly smooth, we know that $E^{\ast }$ is uniformly convex. Then from Lemma 1.10, we have

$$\begin{array}{rl}G(x^{\ast },J{u}_n)=& G(x^{\ast },J{\theta }_n^m{y}_n)\le G(x^{\ast },J{y}_n)\\ =& G(x^{\ast },({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n))\\ =& {\parallel x^{\ast }\parallel }^2-2{\alpha }_n〈x^{\ast },J{x}_n〉-2(1-{\alpha }_n)〈x^{\ast },J{T}_n{x}_n〉\\ +{\parallel {\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n\parallel }^2+2\rho f\left(x^{\ast }\right)\\ \le & {\parallel x^{\ast }\parallel }^2-2{\alpha }_n〈x^{\ast },J{x}_n〉-2(1-{\alpha }_n)〈x^{\ast },J{T}_n{x}_n〉\\ +{\alpha }_n{\parallel J{x}_n\parallel }^2+(1-{\alpha }_n){\parallel J{T}_n{x}_n\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n)g(\parallel J{x}_n-J{T}_n{x}_n\parallel )+2\rho f\left(x^{\ast }\right)\\ =& {\alpha }_{n}G(x^{\ast },J{x}_n)+(1-{\alpha }_n)G(x^{\ast },J{T}_n{x}_n)\\ -{\alpha }_n(1-{\alpha }_n)g(\parallel J{x}_n-J{T}_n{x}_n\parallel )\\ \le & G(x^{\ast },J{x}_n)-{\alpha }_n(1-{\alpha }_n)g(\parallel J{x}_n-J{T}_n{x}_n\parallel ).\end{array}$$

It then follows that

$${\alpha }_n(1-{\alpha }_n)g(\parallel J{x}_n-J{T}_n{x}_n\parallel )\le G(x^{\ast },J{x}_n)-G(x^{\ast },J{u}_n).$$

But

$$\begin{array}{rl}G(x^{\ast },J{x}_n)-G(x^{\ast },J{u}_n)& ={\parallel x_n\parallel }^2-{\parallel u_n\parallel }^2-2〈x^{\ast },J{x}_n-J{u}_n〉\\ \le {\parallel x_n\parallel }^2-{\parallel u_n\parallel }^2+2\left|〈x^{\ast },J{x}_n-J{u}_n〉\right|\\ \le |\parallel x_n\parallel -\parallel u_n\parallel |(\parallel x_n\parallel +\parallel u_n\parallel )+2\parallel x^{\ast }\parallel \parallel J{x}_n-J{u}_n\parallel \\ \le \parallel x_n-u_n\parallel (\parallel x_n\parallel +\parallel u_n\parallel )+2\parallel x^{\ast }\parallel \parallel J{x}_n-J{u}_n\parallel .\end{array}$$

From (2.5) and (2.6), we obtain

$$G(x^{\ast },J{x}_n)-G(x^{\ast },J{u}_n)\to 0,n\to \mathrm{\infty }.$$

Using the condition ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, we have

$$\underset{n\to \mathrm{\infty }}{lim}g(\parallel J{x}_n-J{T}_n{x}_n\parallel )=0.$$

By the properties of g, we have ${lim}_{n\to \mathrm{\infty }}\parallel J{x}_n-J{T}_n{x}_n\parallel =0$. Since $J^{-1}$ is also uniformly norm-to-norm continuous on bounded sets, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_n{x}_n\parallel =0.$$

Since ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ are uniformly closed, and ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a Cauchy sequence. Then $p\in F(T)={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)$.

Next, we show that $p\in {\bigcap }_{k=1}^{m}EP(F_k)$. From (2.1), we obtain

$$\begin{array}{rl}\varphi (x^{\ast },u_n)& =\varphi (x^{\ast },{\theta }_n^m{y}_n)=\varphi (x^{\ast },T_{r_{m,n}}^{F_m}{\theta }_n^{m-1}{y}_n)\\ \le \varphi (x^{\ast },{\theta }_n^{m-1}{y}_n)\le \varphi (x^{\ast },x_n).\end{array}$$

(2.7)

Since $x^{\ast }\in EP(F_m)=F(T_{r_{m,n}}^{F_m})$ for all $n\ge 1$, it follows from (2.7) and Lemma 1.13 that

$$\begin{array}{rl}\varphi (u_n,{\theta }_n^{m-1}{y}_n)& =\varphi (T_{r_{m,n}}^{F_m}{\theta }_n^{m-1}{y}_n,{\theta }_n^{m-1}{y}_n)\\ \le \varphi (x^{\ast },{\theta }_n^{m-1}{y}_n)-\varphi (x^{\ast },u_n)\le \varphi (x^{\ast },x_n)-\varphi (x^{\ast },u_n).\end{array}$$

From (2.5) and (2.6), we obtain ${lim}_{n\to \mathrm{\infty }}\varphi ({\theta }_n^m{y}_n,{\theta }_n^{m-1}{y}_n)={lim}_{n\to \mathrm{\infty }}\varphi (u_n,{\theta }_n^{m-1}{y}_n)=0$. From Lemma 1.9, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel {\theta }_n^m{y}_n-{\theta }_n^{m-1}{y}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-{\theta }_n^{m-1}{y}_n\parallel =0.$$

(2.8)

Hence, we have from (2.8) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{\theta }_n^m{y}_n-J{\theta }_n^{m-1}{y}_n\parallel =0.$$

(2.9)

Again, since $x^{\ast }\in EP(F_{m-1})=F(T_{r_{m-1,n}}^{F_{m-1}})$ for all $n\ge 1$, it follows from (2.7) and Lemma 1.13 that

$$\begin{array}{rl}\varphi ({\theta }_n^{m-1}{y}_n,{\theta }_n^{m-2}{y}_n)& =\varphi (T_{r_{m-1,n}}^{F_{m-1}}{\theta }_n^{m-2}{y}_n,{\theta }_n^{m-2}{y}_n)\\ \le \varphi (x^{\ast },{\theta }_n^{m-2}{y}_n)-\varphi (x^{\ast },{\theta }_n^{m-1}{y}_n)\le \varphi (x^{\ast },x_n)-\varphi (x^{\ast },u_n).\end{array}$$

Again, from (2.5) and (2.6), we obtain ${lim}_{n\to \mathrm{\infty }}\varphi ({\theta }_n^{m-1}{y}_n,{\theta }_n^{m-2}{y}_n)=0$. From Lemma 1.9, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel {\theta }_n^{m-1}{y}_n-{\theta }_n^{m-2}{y}_n\parallel =0$$

(2.10)

and hence,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{\theta }_n^{m-1}{y}_n-J{\theta }_n^{m-2}{y}_n\parallel =0.$$

(2.11)

In a similar way, we can verify that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel {\theta }_n^{m-2}{y}_n-{\theta }_n^{m-3}{y}_n\parallel =\cdots =\underset{n\to \mathrm{\infty }}{lim}\parallel {\theta }_n^1{y}_n-y_n\parallel =0.$$

(2.12)

From (2.8), (2.10), and (2.12), we can conclude that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel {\theta }_n^k{y}_n-{\theta }_n^{k-1}{y}_n\parallel =0,k=1,2,\dots ,m.$$

(2.13)

Since $x_n\to p$, $n\to \mathrm{\infty }$, we obtain from (2.5) that $u_n\to p$, $n\to \mathrm{\infty }$. Again, from (2.8), (2.10), (2.12), and $u_n\to p$, $n\to \mathrm{\infty }$, we have that ${\theta }_n^k{y}_n\to p$, $n\to \mathrm{\infty }$ for each $k=1,2,\dots ,m$. Also, using (2.13), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{\theta }_n^k{y}_n-J{\theta }_n^{k-1}{y}_n\parallel =0,k=1,2,\dots ,m.$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{k,n}>0$, $k=1,2,\dots ,m$,

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel J{\theta }_n^k{y}_n-J{\theta }_n^{k-1}{y}_n\parallel }{r_{k,n}}=0.$$

(2.14)

By Lemma 1.12, we have for each $k=1,2,\dots ,m$

$$F_k({\theta }_n^k{y}_n,y)+\frac{1}{r_{k,n}}〈y-{\theta }_n^k{y}_n,J{\theta }_n^k{y}_n-J{\theta }_n^{k-1}{y}_n〉\ge 0,\mathrm{\forall }y\in C.$$

Furthermore, using (A2) we obtain

$$\frac{1}{r_{k,n}}〈y-{\theta }_n^k{y}_n,J{\theta }_n^k{y}_n-J{\theta }_n^{k-1}{y}_n〉\ge F_k(y,{\theta }_n^k{y}_n).$$

(2.15)

By (A4), (2.14), and ${\theta }_n^k{y}_n\to p$, we have for each $k=1,2,\dots ,m$

$$F_k(y,p)\le 0,y\in C.$$

For fixed $y\in C$, let $z_t=ty+(1-t)p$ for all $t\in (0,1]$. This implies that $z_t\in C$. This yields $F_k(z_t,p)\le 0$. It follows from (A1) and (A4) that

$$0=F_k(z_t,z_t)\le t{F}_k(z_t,y)+(1-t)F_k(z_t,p)\le t{F}_k(z_t,y)$$

and hence

$$0\le F_k(z_t,y).$$

From condition (A3), we obtain

$$F_k(p,y)\ge 0,y\in C.$$

This implies that $p\in EP(F_k)$, $k=1,2,\dots ,m$. Thus, $p\in {\bigcap }_{k=1}^{m}EP(F_k)$. Hence, we have $p\in F={\bigcap }_{k=1}^{m}EP(F_k)\cap ({\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n))$.

Finally, we show that $p={\mathrm{\Pi }}_F^f{x}_0$. Since $F={\bigcap }_{k=1}^{m}EP(F_k)\cap ({\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n))$ is a closed and convex set, from Lemma 1.6, we know that ${\mathrm{\Pi }}_F^f{x}_0$ is single valued and denote $w={\mathrm{\Pi }}_F^f{x}_0$. Since $x_n={\mathrm{\Pi }}_{c_n}^f{x}_0$ and $w\in F\subset C_n$, we have

$$G(x_n,J{x}_0)\le G(w,J{x}_0),\mathrm{\forall }n\ge 0.$$

We know that $G(\xi ,J\phi )$ is convex and lower semi-continuous with respect to ξ when φ is fixed. This implies that

$$G(p,J{x}_0)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}G(x_n,J{x}_0)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}G(x_n,J{x}_0)\le G(w,J{x}_0).$$

From the definition of ${\mathrm{\Pi }}_F^f{x}_0$ and $p\in F$, we see that $p=w$. This completes the proof. □

Corollary 2.6 Let E be a uniformly convex and uniformly smooth real Banach space, and let C be a nonempty closed convex subset of E. For each $k=1,2,\dots ,m$, let $F_k$ be a bifunction from $C\times C$ satisfying (A1)-(A4) and let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a countable family of uniformly closed relatively quasi-nonexpansive mappings of C into itself such that $F:=({\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n))\cap ({\bigcap }_{k=1}^{m}EP(F_k))\ne \mathrm{\varnothing }$. Suppose ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is iteratively generated by $x_0\in C$, $C_1=C$, $x_1={\mathrm{\Pi }}_{C_1}^f{x}_0$,

$$\{\begin{array}{l}{y}_n=J^{-1}({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n),\\ u_n=T_{r_{m,n}}^{F_m}{T}_{r_{m-1,n}}^{F_{m-1}}\cdots T_{r_{2,n}}^{F_2}{T}_{r_{1,n}}^{F_1}{y}_n,\\ C_{n+1}=\{w\in C_n:\varphi (w,u_n)\le \varphi (w,x_n)\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 1,\end{array}$$

where J is the duality mapping on E. Suppose ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence in $(0,1)$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, and ${\{r_{k,n}\}}_{n=1}^{\mathrm{\infty }}\subset (0,\mathrm{\infty })$ ($k=1,2,\dots ,m$) satisfying ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{k,n}>0$ ($k=1,2,\dots ,m$). Then ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to ${\mathrm{\Pi }}_F{x}_0$.

Proof Take $f(x)=0$ for all $x\in E$ in Theorem 2.5, then $G(\xi ,Jx)=\varphi (\xi ,x)$ and ${\mathrm{\Pi }}_C^f{x}_0={\mathrm{\Pi }}_C{x}_0$. Then Corollary 2.6 holds. □

Take $F_k\equiv 0$ ($k=1,2,\dots ,m$), it is obvious that the following holds.

Corollary 2.7 Let E be a uniformly convex and uniformly smooth real Banach space, and let C be a nonempty closed convex subset of E. Let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a countable family of uniformly closed relatively quasi-nonexpansive mappings of C into itself such that $F=({\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n))\ne \mathrm{\varnothing }$. Let $f:E\to R$ be a convex and lower semi-continuous mapping with $C\subset int(D(f))$ and suppose ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is iteratively generated by $x_0\in C$, $C_1=C$, $x_1={\mathrm{\Pi }}_{C_1}^f{x}_0$,

$$\{\begin{array}{l}{y}_n=J^{-1}({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n),\\ C_{n+1}=\{w\in C_n:G(w,J{y}_n)\le G(w,J{x}_n)\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}^f{x}_0,n\ge 1,\end{array}$$

where J is the duality mapping on E. Suppose ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence in $(0,1)$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$, and ${\{r_{k,n}\}}_{n=1}^{\mathrm{\infty }}\subset (0,\mathrm{\infty })$ ($k=1,2,\dots ,m$) satisfying ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{k,n}>0$ ($k=1,2,\dots ,m$). Then ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to ${\mathrm{\Pi }}_F{x}_0$.

3 Applications

Let $\phi :C\to R$ be a real-valued function. The convex minimization problem is to find $x^{\ast }\in C$ such that

$$\phi \left(x^{\ast }\right)\le \phi (y),$$

(3.1)

$\mathrm{\forall }y\in C$. The set of solutions of (3.1) is denoted by $CMP(\phi )$. For each $r>0$ and $x\in E$, define the mapping

$$T_r^{\phi }(x)=\{z\in C:\phi (y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge \phi (z),\mathrm{\forall }y\in C\}.$$

Theorem 3.1 Let E be a uniformly convex and uniformly smooth real Banach space, and let C be a nonempty closed convex subset of E. For each $k=1,2,\dots ,m$, let ${\phi }_k$ be a bifunction from $C\times C$ satisfying (A1)-(A4) and let ${\{T_n\}}_{n=1}^{\mathrm{\infty }}$ be a countable family of uniformly closed relatively quasi-nonexpansive mappings of C into itself such that $F:=({\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n))\cap ({\bigcap }_{k=1}^{m}CMP({\phi }_k))\ne \mathrm{\varnothing }$. Let $f:E\to R$ be a convex and lower semi-continuous mapping with $C\subset int(D(f))$ and suppose ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is iteratively generated by $x_0\in C$, $C_1=C$, $x_1={\mathrm{\Pi }}_{C_1}^f{x}_0$,

$$\{\begin{array}{l}{y}_n=J^{-1}({\alpha }_{n}J{x}_n+(1-{\alpha }_n)J{T}_n{x}_n),\\ u_n=T_{r_{m,n}}^{{\phi }_m}{T}_{r_{m-1,n}}^{{\phi }_{m-1}}\cdots T_{r_{2,n}}^{{\phi }_2}{T}_{r_{1,n}}^{{\phi }_1}{y}_n,\\ C_{n+1}=\{w\in C_n:G(w,J{u}_n)\le G(w,J{x}_n)\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}^f{x}_0,n\ge 1,\end{array}$$

where J is the duality mapping on E. Suppose ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence in $(0,1)$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${\{r_{k,n}\}}_{n=1}^{\mathrm{\infty }}\subset (0,\mathrm{\infty })$ ($k=1,2,\dots ,m$) satisfying ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{k,n}>0$ ($k=1,2,\dots ,m$). Then ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ converges strongly to ${\mathrm{\Pi }}_F^f{x}_0$.

Proof Define $F_k(x,y)={\phi }_k(y)-{\phi }_k(x)$, $x,y\in C$ and $k=1,2,\dots ,m$. Then $F(T_{r_k}^{F_k})=EP(F_k)=CMP({\phi }_k)=F(T_{r_k}^{{\phi }_k})$ for each $k=1,2,\dots ,m$, and therefore ${\{F_k\}}_{k=1}^m$ satisfies conditions (A1) and (A2). Furthermore, one can easily show that ${\{F_k\}}_{k=1}^m$ satisfies (A3) and (A4). Therefore, from Theorem 2.5, we obtain Theorem 3.1. □