Strong Convergence Theorems for Infinitely Nonexpansive Mappings in Hilbert Space

Abstract

We introduce a modified Ishikawa iterative process for approximating a fixed point of two infinitely nonexpansive self-mappings by using the hybrid method in a Hilbert space and prove that the modified Ishikawa iterative sequence converges strongly to a common fixed point of two infinitely nonexpansive self-mappings.

1. Introduction

Let be a nonempty closed convex subset of a Hilbert space , a self-mapping of . Recall that is said to be nonexpansive if for all .

Construction of fixed points of nonexpansive mappings via Mann's iteration [1] has extensively been investigated in literature (see, e.g., [2–5] and reference therein). But the convergence about Mann's iteration and Ishikawa's iteration is in general not strong (see the counterexample in [6]). In order to get strong convergence, one must modify them. In 2003, Nakajo and Takahashi [7] proposed such a modification for a nonexpansive mapping .

Consider the algorithm,

(1.1)

where denotes the metric projection from onto a closed convex subset of . They prove the sequence generated by that algorithm (1.1) converges strongly to a fixed point of provided that the control sequence is chosen so that .

Let be a sequence of nonexpansive self-mappings of , a sequence of nonnegative numbers in . For each , defined a mapping of into itself as follows:

(1.2)

Such a mapping is called the -mapping generated by and ; see [8].

In this paper, motivated by [9], for any given ( is a fixed number), we will propose the following iterative progress for two infinitely nonexpansive mappings and in a Hilbert space :

(1.3)

and prove, converges strongly to a fixed point of and .

We will use the notation:

for weak convergence and for strong convergence.

denotes the weak -limit set of .

2. Preliminaries

In this paper, we need some facts and tools which are listed as lemmas below.

Lemma 2.1 (see [10]).

Let be a Hilbert space, a nonempty closed convex subset of , and a nonexpansive mapping with Fix. If is a sequence in weakly converging to and if converges strongly to , then .

Lemma 2.2 (see [11]).

Let be a nonempty bounded closed convex subset of a Hilbert space . Given also a real number and . Then the set is closed and convex.

Let be a sequence of nonexpansive self-mappings on , where is a nonempty closed convex subset of a strictly convex Banach space . Given a sequence in , one defines a sequence of self-mappings on by (1.2). Then one has the following results.

Lemma 2.3 (see [8]).

Let be a nonempty closed convex subset of a strictly convex Banach space , a sequence of nonexpansive self-mappings on such that and let be a sequence in for some . Then, for every and the limit exists.

Remark 2.4.

It can be known from Lemma 2.3 that if is a nonempty bounded subset of , then for there exists such that for all .

Remark 2.5.

Using Lemma 2.3, we can define a mapping as follows:

(2.1)

for all Such a is called the -mapping generated by and Since is nonexpansive mapping, is also nonexpansive. Indeed, observe that for each ,

(2.2)

If is a bounded sequence in , then we put . Hence, it is clear from Remark 2.4 that for there exists such that for all This implies that

(2.3)

Lemma 2.6 (see [8]).

Let be a nonempty closed convex subset of a strictly convex Banach space . Let be a sequence of nonexpansive self-mappings on such that and let be a sequence in for some . Then, .

3. Strong Convergence Theorem

Theorem 3.1.

Let be a closed convex subset of a Hilbert space and let and be defined as (1.2). Assume that for all and for some and for all and . If then generated by (1.3) converges strongly to .

Proof.

Firstly, we observe that is convex by Lemma 2.2. Next, we show that for all .

Indeed, for all ,

(3.1)

Therefore,

(3.2)

That is for all . Next we show that for all .

We prove this by induction. For , we have Assume that for all since is the projection of onto so

(3.3)

As by the induction assumption, the last inequality holds, in particular, for all . This together with definition of implies that . Hence for all .

Notice that the definition of implies . This together with the fact further implies for all

The fact asserts that implies

(3.4)

We now claim that and . Indeed,

(3.5)

since , we have

(3.6)

Thus

(3.7)

We now show . Let be any subsequence of . Since is a bounded subset of , there exists a subsequence of such that

(3.8)

Since

(3.9)

it follows that . By (3.1), we have

(3.10)

Hence

(3.11)

Thus,

(3.12)

Using (3.1) again, we obtain that

(3.13)

This imply that . For the arbitrariness of , we have and

(3.14)

Thus, by (3.4), (3.7) and (3.14), we have

(3.15)

Since and we have

(3.16)

Thus, using (3.16), Lemma 2.1, and the boundedness of , we get that . Since and , we have where . By the weak lower semicontinuity of the norm, we have for all . However, since , we must have for all . Hence and

(3.17)

That is, converges to .

This completes the proof.