On solving Lipschitz pseudocontractive operator equations

Abstract

We analyze the convergence of the Mann-type double sequence iteration process to the solution of a Lipschitz pseudocontractive operator equation on a bounded closed convex subset of arbitrary real Banach space into itself. Our results extend the result in (Moore in Comp. Math. Appl. 43: 1585-1589, 2002).

MSC:47H10, 54H25.

1 Introduction

Let E be a real Banach space and $E^{\ast }$ be the dual space of E. Let J be the normalized duality mapping from E to $2^{E^{\ast }}$ defined by

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

for all $x\in E$ where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. A single-valued duality map will be denoted by j.

An operator $T:E\to E$ is said to be

• pseudocontractive if there exists $j(x-y)\in J(x-y)$ such that

  $$〈Tx-Ty,j(x-y)〉\le {\parallel x-y\parallel }^2$$

for any $x,y\in E$;

• accretive if for any $x,y\in E$, there exists $j(x-y)\in J(x-y)$ satisfying

  $$〈Tx-Ty,j(x-y)〉\ge 0;$$

• strongly pseudocontractive if there exist $j(x-y)\in J(x-y)$ and a constant $\lambda \in (0,1)$ such that

  $$〈Tx-Ty,j(x-y)〉\le \lambda {\parallel x-y\parallel }^2$$

for any $x,y\in E$;

• strongly accretive if for any $x,y\in E$, there exist $j(x-y)\in J(x-y)$ and a constant $t\in (0,1)$ satisfying

  $$〈Tx-Ty,j(x-y)〉\ge t{\parallel x-y\parallel }^2$$

for all $x,y\in E$.

As a consequence of a result of Kato [1], the concept of pseudocontractive operators can equivalently be defined as follows:

T is strongly pseudocontractive if there exists $\lambda \in (0,1)$ such that the inequality

$$\parallel x-y\parallel \le \parallel x-y+r[(I-T-\lambda I)x-(I-T-\lambda I)y]\parallel$$

(1.1)

holds for all $x,y\in E$ and $r>0$. If $\lambda =0$ in the inequality (1.1), then T is pseudocontractive.

It is easy to see that T is pseudocontractive if and only if $I-T$ is accretive where I denotes the identity mapping on E.

Let C be a compact convex subset of a real Hilbert space and let $T:C\to C$ be a Lipschitz pseudocontraction. It remains as an open question whether the Mann iteration process always converges to a fixed point of T. In [2] it was proved that the Ishikawa iteration process converges strongly to a fixed point of T. In 2001, Mutangadura and Chidume [3] constructed the following example to demonstrate that the Mann iteration process is not guaranteed to converge to a fixed point of a Lipschitz pseudocontraction mapping a compact convex subset of a real Hilbert space H into itself.

Example [3]

Let $H={\mathrm{\Re }}^2$ with the usual Euclidean inner product, and for $x=(a,b)\in H$ define $x^{\perp }=(b,-a)$. Now, let $C=B_1(o)$; the closed unit ball in H and let $C_1=\{x\in H:\parallel x\parallel \le \frac{1}{2}\}$, $C_2=\{x\in H:\frac{1}{2}\le \parallel x\parallel \le 1\}$. Define the map $T:C\to C$ by

$$Tx=\{\begin{array}{cc}x+x^{\perp },\hfill & \text{if }x\in C_1;\hfill \\ \frac{x}{\parallel x\parallel }-x+x^{\perp },\hfill & \text{if }x\in C_2.\hfill \end{array}$$

Observe that T is pseudocontractive, Lipschitz continuous (with Lipschitz constant 5) and has the origin $(0,0)$ as its unique fixed point; C is compact and convex. However, for any $x\in C_1$, we have

$${\parallel (1-\lambda )x+\lambda Tx\parallel }^2=(1+{\lambda }^2){\parallel x\parallel }^2>{\parallel x\parallel }^2,\mathrm{\forall }\lambda \in (0,1),$$

while for any $x\in C_2$, we have

$${\parallel (1-\lambda )x+\lambda Tx\parallel }^2\ge \frac{1}{2}{\parallel x\parallel }^2,\mathrm{\forall }\lambda \in (0,1),$$

and therefore no Mann sequence can converge to $(0,0)$, the unique fixed point of T, unless the initial guess is the fixed point itself.

Moore [4] introduced the concept of a Mann-type double sequence iteration process and proved that it converges strongly to a fixed point of a continuous pseudocontraction which maps a bounded closed convex nonempty subset of a real Hilbert space into itself.

Definition 1.1 [4]

Let $\mathcal{N}$ denote the set of all nonnegative integers (the natural numbers) and let E be a normed linear space. By a double sequence in E is meant a function $f:\mathcal{N}\times \mathcal{N}\to E$ defined by $f(n,m)=x_{n,m}\in E$. A double sequence $\{x_{n,m}\}$ is said to converge strongly to $x^{\ast }$ if given any $ϵ>0$, there exist $N,M>0$ such that $\parallel x_{n,m}-x^{\ast }\parallel <ϵ$ for all $n\ge N$, $m\ge M$. If $\mathrm{\forall }n,r\ge N$, $\mathrm{\forall }m,t\ge M$, we have $\parallel x_{n,r}-x_{m,t}\parallel <ϵ$, then the double sequence is said to be Cauchy. Furthermore, if for each fixed n, $x_{n,m}\to x_n^{\ast }$ as $m\to \mathrm{\infty }$ and then $x_n^{\ast }\to x^{\ast }$ as $n\to \mathrm{\infty }$, then $x_{n,m}\to x^{\ast }$ as $n,m\to \mathrm{\infty }$.

Theorem 1.1 [4]

Let C be a bounded closed convex nonempty subset of a (real) Hilbert space H, and let $T:C\to C$ be a continuous pseudocontractive map. Let ${\{{\alpha }_n\}}_{n\ge 0},{\{a_k\}}_{k\ge 0}\subset (0,1)$ be real sequences satisfying the following conditions:

• (i) ${lim}_{k\to \mathrm{\infty }}{a}_k=1$,

• (ii) ${lim}_{k,r\to \mathrm{\infty }}(a_k-a_r)/(1-a_k)=0$, $\mathrm{\forall }0<r\le k$,

• (iii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$,

• (iv) ${\sum }_{n\ge 0}{\alpha }_n=\mathrm{\infty }$.

For an arbitrary but fixed $\omega \in C$, and for each $k\ge 0$, define $T_k:C\to C$ by $T_{k}x=(1-a_k)\omega +a_{k}Tx$, $\mathrm{\forall }x\in C$. Then the double sequence ${\{x_{k,n}\}}_{k\ge 0,n\ge 0}$ generated from an arbitrary $x_{0,0}\in C$ by

$$x_{k,n+1}=(1-{\alpha }_n)x_{k,n}+{\alpha }_n{T}_k{x}_{k,n},k,n\ge 0,$$

converges strongly to a fixed point $x_{\mathrm{\infty }}^{\ast }$ of T in C.

The following lemma will be useful in the sequel.

Lemma 1.2 [5]

Let $\{{\delta }_n\}$ and $\{{\sigma }_n\}$ be two sequences of nonnegative real numbers satisfying the inequality

$${\delta }_{n+1}\le \gamma {\delta }_n+{\sigma }_n,n\ge 0.$$

Here $\gamma \in [0,1)$. If ${lim}_{n\to \mathrm{\infty }}{\sigma }_n=0$, then ${lim}_{n\to \mathrm{\infty }}{\delta }_n=0$.

It is our purpose in this paper to extend Theorem 1.1 from Hilbert space to an arbitrary real Banach space with no further assumptions on the real sequences ${\{{\alpha }_n\}}_{n\ge 0}$, ${\{a_k\}}_{k\ge 0}$.

2 Main results

Theorem 2.1 Let C be a bounded closed convex subset of a Banach space E and $T:C\to C$ be a Lipschitz pseudocontraction with $F(T)\ne \mathrm{\varnothing }$. Let ${\{{\alpha }_n\}}_{n\ge 0},{\{a_k\}}_{k\ge 0}\subset (0,1)$ be real sequences satisfying the following conditions:

• (i) ${lim}_{k\to \mathrm{\infty }}{a}_k=1$,

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$.

For an arbitrary but fixed $\omega \in C$, and for each $k\ge 0$, define $T_k:C\to C$ by $T_{k}x=(1-a_k)\omega +a_{k}Tx$, $\mathrm{\forall }x\in C$. Then the double sequence ${\{x_{k,n}\}}_{k\ge 0,n\ge 0}$ generated from an arbitrary $x_{0,0}\in C$ by

$$x_{k,n+1}=(1-{\alpha }_n)x_{k,n}+{\alpha }_n{T}_k{x}_{k,n},k,n\ge 0$$

(2.1)

converges strongly to a fixed point $x^{\ast }$ of T in C.

Proof Since T is Lipschitzian, there exists $L>0$ such that

$$\parallel Tx-Ty\parallel \le L\parallel x-y\parallel \text{for all }x,y\in C.$$

Since T is pseudocontractive, for each $k\ge 0$, we have

$$〈T_{k}x-T_{k}y,j(x-y)〉=a_k〈Tx-Ty,j(x-y)〉\le a_k{\parallel x-y\parallel }^2.$$

Hence, $T_k$ is Lipschitz and strongly pseudocontractive. Also, C is invariant under $T_k$ for all $k\ge 0$, by convexity. Thus, for each $k\ge 0$, $T_k$ has a unique fixed point $x_k^{\ast }$, say, in C.

Now, we proceed in the following steps.

• (I) for each $k\ge 0$, $x_{k,n}\to x_k^{\ast }\in C$ as $n\to \mathrm{\infty }$.

• (II) $x_k^{\ast }\to x^{\ast }\in C$ as $k\to \mathrm{\infty }$.

• (III) $x^{\ast }\in F(T)$.

Proof of (I). In fact, it follows from (2.1) that

$$\begin{array}{rcl}{x}_{k,n}& =& x_{k,n+1}+{\alpha }_n{x}_{k,n}-{\alpha }_n{T}_k{x}_{k,n}\\ =& (1+{\alpha }_n)x_{k,n+1}+{\alpha }_n(I-T_k-\lambda I)x_{k,n+1}-(2-\lambda ){\alpha }_n{x}_{k,n+1}+{\alpha }_n{x}_{k,n}\\ +{\alpha }_n(T_k{x}_{k,n+1}-T_k{x}_{k,n})\\ =& (1+{\alpha }_n)x_{k,n+1}+{\alpha }_n(I-T_k-\lambda I)x_{k,n+1}-(2-\lambda ){\alpha }_n[(1-{\alpha }_n)x_{k,n}+{\alpha }_n{T}_k{x}_{k,n}]\\ +{\alpha }_n{x}_{k,n}+{\alpha }_n(T_k{x}_{k,n+1}-T_k{x}_{k,n})\\ =& (1+{\alpha }_n)x_{k,n+1}+{\alpha }_n(I-T_k-\lambda I)x_{k,n+1}-(1-\lambda ){\alpha }_n{x}_{k,n}\\ +(2-\lambda ){\alpha }_n^2(x_{k,n}-T_k{x}_{k,n})+{\alpha }_n(T_k{x}_{k,n+1}-T_k{x}_{k,n}).\end{array}$$

Thus, if $x_k^{\ast }$ is a fixed point of $T_k$, $k\ge 0$, then

$$\begin{array}{rcl}{x}_{k,n+1}-x_k^{\ast }& =& (1+{\alpha }_n)(x_{k,n+1}-x_k^{\ast })+{\alpha }_n(I-T_k-\lambda I)(x_{k,n+1}-x_k^{\ast })\\ -(1-\lambda ){\alpha }_n(x_{k,n}-x_k^{\ast })+(2-\lambda ){\alpha }_n^2(x_{k,n}-T_k{x}_{k,n})+{\alpha }_n(T_k{x}_{k,n+1}-T_k{x}_{k,n}).\end{array}$$

Using inequality (1.1), it follows that

$$\begin{array}{rcl}\parallel x_{k,n+1}-x_k^{\ast }\parallel & \ge & (1+{\alpha }_n)\parallel x_{k,n+1}-x_k^{\ast }\parallel -(1-\lambda ){\alpha }_n\parallel x_{k,n}-x_k^{\ast }\parallel \\ -(2-\lambda ){\alpha }_n^2\parallel x_{k,n}-T_k{x}_{k,n}\parallel -{\alpha }_n\parallel T_k{x}_{k,n+1}-T_k{x}_{k,n}\parallel .\end{array}$$

(2.2)

On the other hand, by (2.1) we obtain

$$\begin{array}{rcl}\parallel x_{k,n+1}-x_{k,n}\parallel & =& {\alpha }_n\parallel T_k{x}_{k,n}-x_{k,n}\parallel \\ \le & {\alpha }_n(\parallel T_k{x}_{k,n}-x_k^{\ast }\parallel +\parallel x_{k,n}-x_k^{\ast }\parallel )\\ =& {\alpha }_n(a_k\parallel T{x}_{k,n}-x_k^{\ast }\parallel +\parallel x_{k,n}-x_k^{\ast }\parallel )\\ \le & {\alpha }_n(a_{k}L\parallel x_{k,n}-x_k^{\ast }\parallel +\parallel x_{k,n}-x_k^{\ast }\parallel )\\ \le & {\alpha }_n(L+1)\parallel x_{k,n}-x_k^{\ast }\parallel .\end{array}$$

Therefore,

$$\begin{array}{rcl}\parallel T_k{x}_{k,n+1}-T_k{x}_{k,n}\parallel & =& a_k\parallel T{x}_{k,n+1}-T{x}_{k,n}\parallel \\ \le & a_{k}L\parallel x_{k,n+1}-x_{k,n}\parallel \\ \le & L\parallel x_{k,n+1}-x_{k,n}\parallel \\ \le & {\alpha }_{n}L(L+1)\parallel x_{k,n}-x_k^{\ast }\parallel .\end{array}$$

(2.3)

Substituting (2.3) into (2.2), we arrive at

$$\begin{array}{rcl}\parallel x_{k,n}-x_k^{\ast }\parallel & \ge & (1+{\alpha }_n)\parallel x_{k,n+1}-x_k^{\ast }\parallel -(1-\lambda ){\alpha }_n\parallel x_{k,n}-x_k^{\ast }\parallel \\ -(2-\lambda ){\alpha }_n^2\parallel x_{k,n}-T_k{x}_{k,n}\parallel -L(L+1){\alpha }_n^2\parallel x_{k,n}-x_k^{\ast }\parallel ,\end{array}$$

which implies that

$$\begin{array}{rcl}{\alpha }_n\parallel x_{k,n+1}-x_k^{\ast }\parallel & \le & (1-\lambda ){\alpha }_n\parallel x_{k,n}-x_k^{\ast }\parallel +{\alpha }_n^2[L(L+1)\parallel x_{k,n}-x_k^{\ast }\parallel \\ +(2-\lambda )\parallel x_{k,n}-T_k{x}_{k,n}\parallel ],\end{array}$$

and so

$$\begin{array}{rcl}\parallel x_{k,n+1}-x_k^{\ast }\parallel & \le & (1-\lambda )\parallel x_{k,n}-x_k^{\ast }\parallel +{\alpha }_n[L(L+1)\parallel x_{k,n}-x_k^{\ast }\parallel \\ +(2-\lambda )\parallel x_{k,n}-T_k{x}_{k,n}\parallel ].\end{array}$$

(2.4)

Since C is bounded, there exists $M>0$ such that

$$M=max\{L(L+1)\underset{n\ge 0}{sup}\parallel x_{k,n}-x_k^{\ast }\parallel ,(2-\lambda )\underset{n\ge 0}{sup}\parallel x_{k,n}-T_k{x}_{k,n}\parallel \}.$$

Hence, it follows from (2.4) that

$$\parallel x_{k,n+1}-x_k^{\ast }\parallel \le (1-\lambda )\parallel x_{k,n}-x_k^{\ast }\parallel +{\alpha }_{n}M.$$

Since $\lambda \in (0,1)$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, it follows from Lemma 1.2 that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{k,n}-x_k^{\ast }\parallel =0,$$

i.e., $x_{k,n}\to x_k^{\ast }$ as $n\to \mathrm{\infty }$.

Proof of (II). We prove that ${\{x_k^{\ast }\}}_{k=0}^{\mathrm{\infty }}={\{T_k{x}_k^{\ast }\}}_{k=0}^{\mathrm{\infty }}$ converges to some $x^{\ast }\in C$. For this purpose, we need only to prove that ${\{x_k^{\ast }\}}_0^{\mathrm{\infty }}$ is a Cauchy sequence.

In fact, we have

$$\begin{array}{rcl}{\parallel x_l^{\ast }-x_m^{\ast }\parallel }^2& =& 〈x_l^{\ast }-x_m^{\ast },j(x_l^{\ast }-x_m^{\ast })〉\\ =& 〈T_l{x}_l^{\ast }-T_m{x}_m^{\ast },j(x_l^{\ast }-x_m^{\ast })〉\\ =& 〈(1-a_l)\omega +a_{l}T{x}_l^{\ast }-(1-a_m)\omega -a_{m}T{x}_m^{\ast },j(x_l^{\ast }-x_m^{\ast })〉\\ =& (a_m-a_l)〈\omega ,j(x_l^{\ast }-x_m^{\ast })〉+a_l〈T{x}_l^{\ast }-T{x}_m^{\ast },j(x_l^{\ast }-x_m^{\ast })〉\\ +(a_l-a_m)〈T{x}_m^{\ast },j(x_l^{\ast }-x_m^{\ast })〉\\ \le & |a_l-a_m|(\parallel \omega \parallel \parallel x_l^{\ast }-x_m^{\ast }\parallel +\parallel T{x}_m^{\ast }\parallel \parallel x_l^{\ast }-x_m^{\ast }\parallel )\\ +a_l〈T{x}_l^{\ast }-T{x}_m^{\ast },j(x_l^{\ast }-x_m^{\ast })〉\\ \le & |a_l-a_m|(\parallel \omega \parallel +\parallel T{x}_m^{\ast }\parallel )\parallel x_l^{\ast }-x_m^{\ast }\parallel +a_l\lambda {\parallel x_l^{\ast }-x_m^{\ast }\parallel }^2\\ \le & |a_l-a_m|(\parallel \omega \parallel +\parallel T{x}_m^{\ast }\parallel )\parallel x_l^{\ast }-x_m^{\ast }\parallel +\lambda {\parallel x_l^{\ast }-x_m^{\ast }\parallel }^2,\end{array}$$

that is,

$$\parallel x_l^{\ast }-x_m^{\ast }\parallel \le [|a_l-a_m|(\parallel \omega \parallel +\parallel T{x}_m^{\ast }\parallel )+\lambda \parallel x_l^{\ast }-x_m^{\ast }\parallel ],$$

hence

$$\parallel x_l^{\ast }-x_m^{\ast }\parallel \le 2\frac{|a_l-a_m|}{1-\lambda }d,$$

where $d=diamC$. If follows from condition (i) that

$$\underset{l,m\to \mathrm{\infty }}{lim}\parallel x_l^{\ast }-x_m^{\ast }\parallel =0.$$

This completes step (II) of the proof.

Proof of (III). In order to accomplish step (III), we first have to prove that ${\{x_k^{\ast }\}}_{k=0}^{\mathrm{\infty }}$ is an approximate fixed point sequence for T. In fact, from $T_k{x}_k^{\ast }=(1-a_k)\omega +a_{k}T{x}_k^{\ast }$, we have

$$\begin{array}{rcl}\parallel x_k^{\ast }-T{x}_k^{\ast }\parallel & =& \parallel x_k^{\ast }-\frac{1}{a_k}{T}_k{x}_k^{\ast }+\frac{1-a_k}{a_k}\omega \parallel \\ =& \parallel x_k^{\ast }-\frac{1}{a_k}{x}_k^{\ast }+\frac{1-a_k}{a_k}\omega \parallel \\ =& \parallel \frac{1-a_k}{a_k}(\omega -x_k^{\ast })\parallel \\ \le & \frac{1-a_k}{a_k}(\parallel \omega \parallel +\parallel x_k^{\ast }\parallel )\\ \le & \frac{1-a_k}{a_k}\cdot 2d,\end{array}$$

where $d=diamC$. Hence ${lim}_{k\to \mathrm{\infty }}\parallel x_k^{\ast }-T{x}_k^{\ast }\parallel =0$. Since $x_k^{\ast }\to x^{\ast }$ as $k\to \mathrm{\infty }$, T is continuous and using continuity of the norm, we get ${lim}_{k\to \mathrm{\infty }}\parallel x^{\ast }-T{x}^{\ast }\parallel =0$, i.e., $x^{\ast }=T{x}^{\ast }$. This completes the proof. □

Corollary 2.2 Let C be a bounded closed convex subset of a Banach space E and $T:C\to C$ be a nonexpansive mapping with $F(T)\ne \mathrm{\varnothing }$. Let ${\{{\alpha }_n\}}_{n\ge 0},{\{a_k\}}_{k\ge 0}\subset (0,1)$ be real sequences satisfying conditions (i)-(ii) in Theorem  2.1. For an arbitrary but fixed $\omega \in C$, and for each $k\ge 0$, define $T_k:C\to C$ by $T_{k}x=(1-a_k)\omega +a_{k}Tx$, $\mathrm{\forall }x\in C$. Then the double sequence ${\{x_{k,n}\}}_{k\ge 0,n\ge 0}$ generated from an arbitrary $x_{0,0}\in C$ by

$$x_{k,n+1}=(1-{\alpha }_n)x_{k,n}+{\alpha }_n{T}_k{x}_{k,n},k,n\ge 0,$$

converges strongly to a fixed point of T in C.

Proof Obvious, observing the fact that every nonexpansive mapping is Lipschitz and pseudocontractive. □

The following corollary follows from Theorem 2.1 on setting $\omega =0\in C$.

Corollary 2.3 Let C, E, T, ${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{a_k\}}_{k=0}^{\mathrm{\infty }}$ be as in Theorem  2.1. For an arbitrary but fixed $\omega \in C$, and for each $k\ge 0$, define $T_k:C\to C$ by $T_{k}x=a_{k}Tx$ for all $x\in C$. Then the double sequence ${\{x_{k,n}\}}_{k\ge 0,n\ge 0}$ generated from an arbitrary $x_{0,0}\in C$ by

$$x_{k,n+1}=(1-{\alpha }_n)x_{k,n}+{\alpha }_n{T}_k{x}_{k,n},k,n\ge 0,$$

converges strongly to a fixed point of T in C.

Remark 2.1 Theorem 2.1 improves and extends Theorem 3.1 of Moore [3] in three respects:

• (1) It abolishes the condition that ${lim}_{r,k\to \mathrm{\infty }}\frac{a_k-a_r}{1-a_k}=0$.

• (2) It abolishes the condition that ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

• (3) The ambient space is no longer required to be a Hilbert space and is taken to be the more general Banach space instead.

Remark 2.2

• (1) Whereas the Ishikawa iteration process was proved to converge to a fixed point of a Lipschitz pseudocontractive mapping in compact convex subsets of a Hilbert space, we imposed no compactness conditions to obtain the strong convergence of the double sequence iteration process to a fixed point of a Lipschitz pseudocontraction.

• (2) Our results may easily be extended to the slightly more general classes of Lipschitz hemicontractive and Lipschitz quasi-nonexpansive mappings.

• (3) Prototypes of the sequences ${\{a_k\}}_{k=0}^{\mathrm{\infty }}$ and ${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}$ are

  $$a_k=\frac{k}{1+k}\text{and}{\alpha }_n=\frac{1}{{(n+1)}^2}.$$