An implicit method for finding a common fixed point of a representation of nonexpansive mappings in Banach spaces

Abstract

We introduce an implicit method for finding an element of the set of common fixed points of a representation of nonexpansive mappings. Then we prove the strong convergence of the proposed implicit scheme to the common fixed point of a representation of nonexpansive mappings.

MSC:90C33, 47H10.

1 Introduction

Let C be a nonempty closed and convex subset of a Banach space E and $E^{\ast }$ be the dual space of E. Let $〈\cdot ,\cdot 〉$ denote the pairing between E and $E^{\ast }$. The normalized duality mapping $J:E\to E^{\ast }$ is defined by

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

for all $x\in E$. In the sequel, we use j to denote the single-valued normalized duality mapping. Let $U=\{x\in E:\parallel x\parallel =1\}$. E is said to be smooth or to have a Gâteaux differentiable norm if the limit

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists for each $x,y\in U$. E is said to have a uniformly Gâteaux differentiable norm if for each $y\in U$, the limit is attained uniformly for all $x\in U$. E is said to be uniformly smooth or is said to have a uniformly Féchet differentiable norm if the limit is attained uniformly for $x,y\in U$. It is known that if the norm of E is uniformly Gâteaux differentiable, then the duality mapping J is single-valued and uniformly norm to weak^∗ continuous on each bounded subset of E. A Banach space E is smooth if the duality mapping J of E is single-valued. We know that if E is smooth, then J is norm to weak-star continuous; for more details, see [1].

Let C be a nonempty closed and convex subset of a Banach space E. A mapping T of C into itself is called nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in C$, and a mapping f is an α-contraction on E if $\parallel f(x)-f(y)\parallel \le \alpha \parallel x-y\parallel$, $x,y\in E$ such that $0\le \alpha <1$.

In this paper, motivated by Lashkarizadeh Bami and Soori [2] and Hussain and Takahashi [3], we introduce the following general implicit algorithm for finding a common element of the set of fixed points of a representation $\mathcal{S}=\{T_t:t\in S\}$ of a semigroup S as nonexpansive mappings from C into itself, with respect to a left regular sequence of means defined on an appropriate subspace of bounded real-valued functions of the semigroup. On the other hand, our goal is to prove that there exists a sunny nonexpansive retraction P of C onto $Fix(\mathcal{S})$ and $x\in C$ such that the following sequence $\{z_n\}$ converges strongly to Px:

$$z_n={ϵ}_{n}f(z_n)+(1-{ϵ}_n)T_{{\mu }_n}{z}_n(n\in \mathbb{N}).$$

2 Preliminaries

Let S be a semigroup. We denote by $B(S)$ the Banach space of all bounded real-valued functions defined on S with supremum norm. For each $s\in S$ and $f\in B(S)$, we define $l_s$ and $r_s$ in $B(S)$ by

$$(l_{s}f)(t)=f(st),(r_{s}f)(t)=f(ts)(t\in S).$$

Let X be a subspace of $B(S)$ containing 1, and let $X^{\ast }$ be its topological dual. An element μ of $X^{\ast }$ is said to be a mean on X if $\parallel \mu \parallel =\mu (1)=1$. We often write ${\mu }_t(f(t))$ instead of $\mu (f)$ for $\mu \in X^{\ast }$ and $f\in X$. Let X be left invariant (resp. right invariant), i.e., $l_s(X)\subset X$ (resp. $r_s(X)\subset X$) for each $s\in S$. A mean μ on X is said to be left invariant (resp. right invariant) if $\mu (l_{s}f)=\mu (f)$ (resp. $\mu (r_{s}f)=\mu (f)$) for each $s\in S$ and $f\in X$. X is said to be left (resp. right) amenable if X has a left (resp. right) invariant mean. X is amenable if X is both left and right amenable. As is well known, $B(S)$ is amenable when S is a commutative semigroup (see p.29 of [1]). A net $\{{\mu }_{\alpha }\}$ of means on X is said to be left regular if

$$\underset{\alpha }{lim}\parallel l_s^{\ast }{\mu }_{\alpha }-{\mu }_{\alpha }\parallel =0$$

for each $s\in S$, where $l_s^{\ast }$ is the adjoint operator of $l_s$.

Let f be a function of the semigroup S into a reflexive Banach space E such that the weak closure of $\{f(t):t\in S\}$ is weakly compact, and let X be a subspace of $B(S)$ containing all the functions $t\to 〈f(t),x^{\ast }〉$ with $x^{\ast }\in E^{\ast }$. We know from [4] that for any $\mu \in X^{\ast }$, there exists a unique element $f_{\mu }$ in E such that $〈f_{\mu },x^{\ast }〉={\mu }_t〈f(t),x^{\ast }〉$ for all $x^{\ast }\in E^{\ast }$. We denote such $f_{\mu }$ by $\int f(t)\mathrm{d}\mu (t)$. Moreover, if μ is a mean on X, then from [5], $\int f(t)\mathrm{d}\mu (t)\in \overline{co}\{f(t):t\in S\}$.

Let C be a nonempty closed and convex subset of E. Then a family $\mathcal{S}=\{T_s:s\in S\}$ of mappings from C into itself is said to be a representation of S as a nonexpansive mapping on C into itself if $\mathcal{S}$ satisfies the following:

• (1) $T_{st}x=T_s{T}_{t}x$ for all $s,t\in S$ and $x\in C$;

• (2) for every $s\in S$, the mapping $T_s:C\to C$ is nonexpansive.

We denote by $Fix(\mathcal{S})$ the set of common fixed points of $\mathcal{S}$, that is, $Fix(\mathcal{S})={\bigcap }_{s\in S}\{x\in C:T_{s}x=x\}$.

Theorem 2.1 [6]

Let S be a semigroup, let C be a closed, convex subset of a reflexive Banach space E, $\mathcal{S}=\{T_s:s\in S\}$ be a representation of S as a nonexpansive mapping from C into itself such that weak closure of $\{T_{t}x:t\in S\}$ is weakly compact for each $x\in C$, and let X be a subspace of $B(S)$ such that $1\in X$ and the mapping $t\to 〈T(t)x,x^{\ast }〉$ be an element of X for each $x\in C$ and $x^{\ast }\in E$, and μ be a mean on X. If we write $T_{\mu }x$ instead of $\int T_{t}x\mathrm{d}\mu (t)$, then the following hold.

• (i) $T_{\mu }$ is a nonexpansive mapping from C into C.

• (ii) $T_{\mu }x=x$ for each $x\in Fix(\mathcal{S})$.

• (iii) $T_{\mu }x\in \overline{co}\{T_{t}x:t\in S\}$ for each $x\in C$.

• (iv) If X is $r_s$-invariant for each $s\in S$ and μ is right invariant, then $T_{\mu }{T}_t=T_{\mu }$ for each $t\in S$.

Remark From Theorem 4.1.6 in [1], every uniformly convex Banach space is strictly convex and reflexive.

Let D be a subset of B, where B is a subset of a Banach space E, and let P be a retraction of B onto D, that is, $Px=x$ for each $x\in D$. Then P is said to be sunny if for each $x\in B$ and $t\ge 0$ with $Px+t(x-Px)\in B$, $P(Px+t(x-Px))=Px$. A subset D of B is said to be a sunny nonexpansive retract of B if there exists a sunny nonexpansive retraction P of B onto D. We know that if E is smooth and P is a retraction of B onto D, then P is sunny and nonexpansive if and only if for each $x\in B$ and $z\in D$, $〈x-Px,J(z-Px)〉\le 0$. For more details, see [1].

Lemma 2.2 [7]

Let S be a semigroup, and let C be a compact convex subset of a real strictly convex and smooth Banach space E. Suppose that $\mathcal{S}=\{T_s:s\in S\}$ is a representation of S as a nonexpansive mapping from C into itself. Let X be a left invariant subspace of $B(S)$ such that $1\in X$, and the function $t\mapsto 〈T_{t}x,x^{\ast }〉$ is an element of X for each $x\in C$ and $x^{\ast }\in E^{\ast }$. If μ is a left invariant mean on X, then $Fix(T_{\mu })=T_{\mu }C=Fix(\mathcal{S})$ and there exists a unique sunny nonexpansive retraction from C onto $Fix(\mathcal{S})$.

Throughout the rest of this paper, the open ball of radius r centered at 0 is denoted by $B_r$. Let C be a nonempty closed convex subset of a Banach space E. For $ϵ>0$ and a mapping $T:C\to C$, we let $F_{ϵ}(T)$ be the set of ϵ-approximate fixed points of T, i.e., $F_{ϵ}(T)=\{x\in C:\parallel x-Tx\parallel \le ϵ\}$.

3 Main result

In this section, we deal with a strong convergence approximation scheme for finding a common element of the set of common fixed points of a representation of nonexpansive mappings.

Theorem 3.1 Let S be a semigroup. Let C be a nonempty compact convex subset of a real strictly convex and reflexive and smooth Banach space E. Suppose that $\mathcal{S}=\{T_s:s\in S\}$ is a representation of S as a nonexpansive mapping from C into itself such that $Fix(\mathcal{S})\ne \mathrm{\varnothing }$. Let X be a left invariant subspace of $B(S)$ such that $1\in X$, and the function $t\mapsto 〈T_{t}x,x^{\ast }〉$ is an element of X for each $x\in C$ and $x^{\ast }\in E^{\ast }$. Let $\{{\mu }_n\}$ be a left regular sequence of means on X. Suppose that f is an α-contraction on C. Let ${ϵ}_n$ be a sequence in $(0,1)$ such that ${lim}_n{ϵ}_n=0$. Then there exists a unique sunny nonexpansive retraction P of C onto $Fix(\mathcal{S})$ and $x\in C$ such that the following sequence $\{z_n\}$ generated by

$$z_n={ϵ}_{n}f(z_n)+(1-{ϵ}_n)T_{{\mu }_n}{z}_n(n\in \mathbb{N})$$

(1)

strongly converges to Px.

Proof By Proposition 1.7.3 and Theorem 1.9.21 in [8], any compact subset C of a reflexive Banach space E is weakly compact, and from Proposition 1.9.18 in [8], any closed convex subset of a weakly compact subset C of a Banach space E is itself weakly compact, and by Proposition 1.9.13 in [8], any convex subset C of a normed space E is weakly closed if and only if C is closed. Therefore, weak closure of $\{T_{t}x:t\in S\}$ is weakly compact for each $x\in C$.

We divide the proof into five steps.

Step 1. The existence of $z_n$ which satisfies (1).

This follows immediately from the fact that for every $n\in \mathbb{N}$, the mapping $N_n$ given by

$$N_{n}x:={ϵ}_{n}f(x)+(1-{ϵ}_n)T_{{\mu }_n}x(x\in C)$$

is a contraction. To see this, put ${\beta }_n=(1+{ϵ}_n(\alpha -1))$, then $0\le {\beta }_n<1$ ($n\in \mathbb{N}$). Then we have

$$\begin{array}{rl}\parallel N_{n}x-N_{n}y\parallel & \le {ϵ}_n\parallel f(x)-f(y)\parallel +(1-{ϵ}_n)\parallel T_{{\mu }_n}x-T_{{\mu }_n}y\parallel \\ \le {ϵ}_n\alpha \parallel x-y\parallel +(1-{ϵ}_n)\parallel x-y\parallel \\ =(1+{ϵ}_n(\alpha -1))\parallel x-y\parallel ={\beta }_n\parallel x-y\parallel .\end{array}$$

Therefore, by the Banach contraction principle [1], there exists a unique point $z_n\in C$ such that $N_n{z}_n=z_n$.

Step 2. ${lim}_{n\to \mathrm{\infty }}\parallel z_n-T_t{z}_n\parallel =0$ for all $t\in S$.

Consider $t\in S$ and let $ϵ>0$. By Lemma 1 in [9], there exists $\delta >0$ such that $\overline{co}{F}_{\delta }(T_t)+2B_{\delta }\subseteq F_{ϵ}(T_t)$. By Corollary 2.8 in [10], there also exists a natural number N such that

$$\parallel \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y-T_t\left(\frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y\right)\parallel \le \delta$$

(2)

for all $s\in S$ and $y\in C$. Let $p\in Fix(\mathcal{S})$ and $M_0$ be a positive number such that ${sup}_{y\in C}\parallel y\parallel \le M_0$. Let $t\in S$, since $\{{\mu }_n\}$ is strongly left regular, there exists $N_0\in \mathbb{N}$ such that $\parallel {\mu }_n-l_{t^i}^{\ast }{\mu }_n\parallel \le \frac{\delta }{(3M_0)}$ for $n\ge N_0$ and $i=1,2,\dots ,N$. Then we have

$$\begin{array}{r}\underset{y\in C}{sup}\parallel T_{{\mu }_n}y-\int \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y\mathrm{d}{\mu }_n(s)\parallel \\ =\underset{y\in C}{sup}\underset{\parallel x^{\ast }\parallel =1}{sup}|〈T_{{\mu }_n}y,x^{\ast }〉-〈\int \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y\mathrm{d}{\mu }_n(s),x^{\ast }〉|\\ =\underset{y\in C}{sup}\underset{\parallel x^{\ast }\parallel =1}{sup}|\frac{1}{N+1}\sum _{i=0}^N{({\mu }_n)}_s〈T_{s}y,x^{\ast }〉-\frac{1}{N+1}\sum _{i=0}^N{({\mu }_n)}_s〈T_{t^{i}s}y,x^{\ast }〉|\\ \le \frac{1}{N+1}\sum _{i=0}^N\underset{y\in C}{sup}\underset{\parallel x^{\ast }\parallel =1}{sup}|{({\mu }_n)}_s〈T_{s}y,x^{\ast }〉-{\left(l_{t^i}^{\ast }{\mu }_n\right)}_s〈T_{s}y,x^{\ast }〉|\\ \le \underset{i=1,2,\dots ,N}{max}\parallel {\mu }_n-l_{t^i}^{\ast }{\mu }_n\parallel (M_0+2\parallel p\parallel )\\ \le \underset{i=1,2,\dots ,N}{max}\parallel {\mu }_n-l_{t^i}^{\ast }{\mu }_n\parallel (3M_0)\\ \le \delta (n\ge N_0).\end{array}$$

(3)

By Theorem 2.1 we have

$$\int \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y\mathrm{d}{\mu }_n(s)\in \overline{co}\{\frac{1}{N+1}\sum _{i=0}^N{T}_{t^i}(T_{s}y):s\in S\}.$$

(4)

It follows from (2)-(4) that

$$\begin{array}{rl}{T}_{{\mu }_n}y& \in \overline{co}\{\frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y:s\in S\}+B_{\delta }\\ \subset \overline{co}{F}_{\delta }(T_t)+2B_{\delta }\subset F_{ϵ}(T_t)\end{array}$$

for all $y\in C$ and $n\ge N_0$. Therefore, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{sup}_{y\in C}\parallel T_t(T_{{\mu }_n}y)-T_{{\mu }_n}y\parallel \le ϵ$. Since $ϵ>0$ is arbitrary, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{y\in C}{sup}\parallel T_t(T_{{\mu }_n}y)-T_{{\mu }_n}y\parallel =0.$$

(5)

Let $t\in S$ and $ϵ>0$, then there exists $\delta >0$ which satisfies (2). Take $L_0=(1+\alpha )2M_0+\parallel f(p)-p\parallel$. Now, from the condition ${lim}_n{ϵ}_n=0$ and from (5), there exists a natural number $N_1$ such that $T_{{\mu }_n}y\in F_{\delta }(T_t)$ for all $y\in C$ and ${ϵ}_n<\frac{\delta }{2L_0}$ for all $n\ge N_1$. Since $p\in Fix(\mathcal{S})$, we have

$$\begin{array}{rcl}{ϵ}_n\parallel f(z_n)-T_{{\mu }_n}{z}_n\parallel & \le & {ϵ}_n(\parallel f(z_n)-f(p)\parallel +\parallel f(p)-p\parallel +\parallel T_{{\mu }_n}p-T_{{\mu }_n}{z}_n\parallel )\\ \le & {ϵ}_n(\alpha \parallel z_n-p\parallel +\parallel f(p)-p\parallel +\parallel A\parallel \parallel z_n-p\parallel )\\ \le & {ϵ}_n(\alpha \parallel z_n-p\parallel +\parallel f(p)-p\parallel +\parallel z_n-p\parallel )\\ \le & {ϵ}_n((1+\alpha )\parallel z_n-p\parallel +\parallel f(p)-p\parallel )\\ \le & {ϵ}_n((1+\alpha )2M_0+\parallel f(p)-p\parallel )\\ =& {ϵ}_n{L}_0\le \frac{\delta }{2}\end{array}$$

for all $n\ge N_1$. Observe that

$$\begin{array}{rl}{z}_n& ={ϵ}_{n}f(z_n)+(1-{ϵ}_n)T_{{\mu }_n}{z}_n\\ =T_{{\mu }_n}{z}_n+{ϵ}_n(f(z_n)-T_{{\mu }_n}{z}_n)\\ \in F_{\delta }(T_t)+B_{\frac{\delta }{2}}\\ \subseteq F_{\delta }(T_t)+2B_{\delta }\\ \subseteq F_{ϵ}(T_t)\end{array}$$

for all $n\ge N_1$. This shows that

$$\parallel z_n-T_t{z}_n\parallel \le ϵ(n\ge N_1).$$

Since $ϵ>0$ is arbitrary, we get ${lim}_{n\to \mathrm{\infty }}\parallel z_n-T_t{z}_n\parallel =0$.

Step 3. $\mathfrak{S}\{z_n\}\subset Fix(\mathcal{S})$, where $\mathfrak{S}\{z_n\}$ denotes the set of strongly limit points of $\{z_n\}$.

Let $z\in \mathfrak{S}\{z_n\}$, and let $\{z_{n_j}\}$ be a subsequence of $\{z_n\}$ such that $z_{n_j}\to z$,

$$\begin{array}{rl}\parallel T_{t}z-z\parallel & \le \parallel T_{t}z-T_t{z}_{n_j}\parallel +\parallel T_t{z}_{n_j}-z_{n_j}\parallel +\parallel z_{n_j}-z\parallel \\ \le 2\parallel z_{n_j}-z\parallel +\parallel T_t{z}_{n_j}-z_{n_j}\parallel ,\end{array}$$

then by Step 2,

$$\parallel T_{t}z-z\parallel \le 2\underset{j}{lim}\parallel z_{n_j}-z\parallel +\underset{j}{lim}\parallel T_t{z}_{n_j}-z_{n_j}\parallel =0,$$

therefore $z\in Fix(\mathcal{S})$.

Step 4. There exists a unique sunny nonexpansive retraction P of C onto $Fix(\mathcal{S})$ and $x\in C$ such that

$$\mathrm{\Gamma }:=\underset{n}{lim\hspace{0.17em}sup}〈x-Px,J(z_n-Px)〉\le 0.$$

(6)

By Lemma 2.2 there exists a unique sunny nonexpansive retraction P of C onto $Fix(\mathcal{S})$. The Banach contraction mapping principle guarantees that fP has a unique fixed point $x\in C$. We show that

$$\mathrm{\Gamma }:=\underset{n}{lim\hspace{0.17em}sup}〈x-Px,J(z_n-Px)〉\le 0.$$

Note that from the definition of Γ and the fact that C is a compact subset of E, we can select a subsequence $\{z_{n_j}\}$ of $\{z_n\}$ with the following properties:

• (i) ${lim}_j〈x-Px,J(z_{n_j}-Px)〉=\mathrm{\Gamma }$;

• (ii) $\{z_{n_j}\}$ converges strongly to a point z.

By Step 3, we have $z\in Fix(\mathcal{S})$. Since E is smooth, we have

$$\mathrm{\Gamma }=\underset{j}{lim}〈x-Px,J(z_{n_j}-Px)〉=〈x-Px,J(z-Px)〉\le 0.$$

Since $fPx=x$, we have $(f-I)Px=x-Px$. From Theorem 4.2.1(v) in [1], for $x,y\in E$ and $f\in J(y)$, ${\parallel x\parallel }^2-{\parallel y\parallel }^2\ge 2(x-y,f)$. Therefore, for each $n\in \mathbb{N}$, we have

$$\begin{array}{c}{ϵ}_n(\alpha -1){\parallel z_n-Px\parallel }^2\hfill \\ \ge {[{ϵ}_n\alpha \parallel z_n-Px\parallel +(1-{ϵ}_n)\parallel z_n-Px\parallel ]}^2-{\parallel z_n-Px\parallel }^2\hfill \\ \ge {[{ϵ}_n\parallel f(z_n)-f(Px)\parallel +(1-{ϵ}_n)\parallel T_{{\mu }_n}{z}_n-Px\parallel ]}^2-{\parallel z_n-Px\parallel }^2\hfill \\ \ge 2〈{ϵ}_n(f(z_n)-f(Px))+(1-{ϵ}_n)(T_{{\mu }_n}{z}_n-Px)-(z_n-Px),J(z_n-Px)〉\hfill \\ =-2{ϵ}_n〈(f-I)Px,J(z_n-Px)〉\hfill \\ =-2{ϵ}_n〈x-Px,J(z_n-Px)〉.\hfill \end{array}$$

Hence

$${\parallel z_n-Px\parallel }^2\le \frac{2}{1-\alpha }〈x-Px,J(z_n-Px)〉.$$

(7)

Step 5. $\{z_n\}$ strongly converges to Px.

Indeed, from (6), (7) and $Px\in Fix(\mathcal{S})$, we conclude

$$\underset{n}{lim\hspace{0.17em}sup}{\parallel z_n-Px\parallel }^2\le \frac{2}{1-\alpha }\underset{n}{lim\hspace{0.17em}sup}〈x-Px,J(z_n-Px)〉\le 0.$$

That is, $z_n\to Px$. □

Remark 3.2 It would be an interesting problem to prove Theorem 3.1 for continuous representations instead of nonexpansive.