Two New Iterative Methods for a Countable Family of Nonexpansive Mappings in Hilbert Spaces

Abstract

We consider two new iterative methods for a countable family of nonexpansive mappings in Hilbert spaces. We proved that the proposed algorithms strongly converge to a common fixed point of a countable family of nonexpansive mappings which solves the corresponding variational inequality. Our results improve and extend the corresponding ones announced by many others.

1. Introduction

Let be a real Hilbert space and let be a nonempty closed convex subset of . Recall that a mapping is said to be nonexpansive if . We use to denote the set of fixed points of . A mapping is called -Lipschitzian if there exists a positive constant such that

(1.1)

is said to be -strongly monotone if there exists a positive constant such that

(1.2)

Let be a strongly positive bounded linear operator on , that is, there exists a constant such that

(1.3)

A typical problem is that of minimizing a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space :

(1.4)

where is a given point in .

Remark 1.1.

From the definition of , we note that a strongly positive bounded linear operator is a -Lipschitzian and -strongly monotone operator.

Construction of fixed points of nonlinear mappings is an important and active research area. In particular, iterative algorithms for finding fixed points of nonexpansive mappings have received vast investigation (cf. [1, 2]) since these algorithms find applications in variety of applied areas of inverse problem, partial differential equations, image recovery, and signal processing; see [3–8]. One classical way to find the fixed point of a nonexpansive mapping is to use a contraction to approximate it. More precisely, take and define a contraction by , where is a fixed point. Banach's Contraction Mapping Principle guarantees that has a unique fixed point in , that is,

(1.5)

The strong convergence of the path has been studied by Browder [9] and Halpern [10] in a Hilbert space.

Recently, Yao et al. [11] considered the following algorithms:

(1.6)

and for arbitrarily,

(1.7)

They proved that if and satisfying appropriate conditions, then the defined by (1.6) and defined by (1.7) converge strongly to a fixed point of .

On the other hand, Yamada [12] introduced the following hybrid iterative method for solving the variational inequality:

(1.8)

where is a -Lipschitzian and -strongly monotone operator with , , . Then he proved that generated by (1.8) converges strongly to the unique solution of variational inequality , .

In this paper, motivated and inspired by the above results, we introduce two new algorithms (3.3) and (3.13) for a countable family of nonexpansive mappings in Hilbert spaces. We prove that the proposed algorithms strongly converge to which solves the variational inequality: , .

2. Preliminaries

Let be a real Hilbert space with inner product and norm . For the sequence in , we write to indicate that the sequence converges weakly to . implies that converges strongly to . For every point , there exists a unique nearest point in , denoted by such that

(2.1)

The mapping is called the metric projection of onto . It is well know that is a nonexpansive mapping. In a real Hilbert space , we have

(2.2)

In order to prove our main results, we need the following lemmas.

Lemma 2.1 (see [13]).

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

Lemma 2.2 (see [14]).

Let and be bounded sequences in Banach space and a sequence in which satisfies the following condition:

(2.3)

Suppose that , and . Then .

Lemma 2.3 (see [15, 16]).

Let be a sequence of nonnegative real numbers satisfying

(2.4)

where , , and satisfy the following conditions: (i) and , (ii) or , (iii) . Then .

Lemma 2.4 (see [17, Lemma ]).

Let C be a nonempty closed convex subset of a Banach space E. Suppose that

(2.5)

Then, for each , converges strongly to some point of C. Moreover, let T be a mapping of C into itself defined by , for all . Then .

Lemma 2.5.

Let be a -Lipschitzian and -strongly monotone operator on a Hilbert space with and . Then is a contraction with contraction coefficient .

Proof.

From (1.1), (1.2), and (2.2), we have

(2.6)

for all . From and , we have

(2.7)

where . Hence is a contraction with contraction coefficient .

3. Main Results

Let be a -Lipschitzian and -strongly monotone operator on with and a nonexpansive mapping. Let and ; consider a mapping on defined by

(3.1)

It is easy to see that is a contraction. Indeed, from Lemma 2.5, we have

(3.2)

for all . Hence it has a unique fixed point, denoted , which uniquely solves the fixed point equation

(3.3)

Theorem 3.1.

Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive mapping such that . Let be a -Lipschitzian and -strongly monotone operator on with . For each , let the net be generated by (3.3). Then, as , the net converges strongly to a fixed point of which solves the variational inequality:

(3.4)

Proof.

We first show the uniqueness of a solution of the variational inequality (3.4), which is indeed a consequence of the strong monotonicity of . Suppose and both are solutions to (3.4); then

(3.5)

Adding up (3.5) gets

(3.6)

The strong monotonicity of implies that and the uniqueness is proved. Below we use to denote the unique solution of (3.4).

Next, we prove that is bounded. Take ; from (3.3) and using Lemma 2.5, we have

(3.7)

that is,

(3.8)

Observe that

(3.9)

From , we may assume, without loss of generality, that . Thus, we have that is continuous, for all . Therefore, we obtain

(3.10)

From (3.8) and (3.10), we have that is bounded and so is .

On the other hand, from (3.3), we obtain

(3.11)

To prove that . For a given , by (2.2) and using Lemma 2.5, we have

(3.12)

Therefore,

(3.13)

From , we have and . Observe that, if , we have .

Since is bounded, we see that if is a sequence in such that and , then by (3.13), we see . Moreover, by (3.11) and using Lemma 2.1, we have . We next prove that solves the variational inequality (3.4). From (3.3) and , we have

(3.14)

that is,

(3.15)

Now replacing in (3.15) with and letting , we have

(3.16)

That is is a solution of (3.4); hence by uniqueness. In a summary, we have shown that each cluster point of (as ) equals . Therefore, as .

Setting in Theorem 3.1, we can obtain the following result.

Corollary 3.2.

Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive mapping such that . Let be a strongly positive bounded linear operator with coefficient . For each , let the net be generated by . Then, as , the net converges strongly to a fixed point of which solves the variational inequality:

(3.17)

Setting , the identity mapping, in Theorem 3.1, we can obtain the following result.

Corollary 3.3.

Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive mapping such that . For each , let the net be generated by (1.6). Then, as , the net converges strongly to a fixed point of which solves the variational inequality:

(3.18)

Remark 3.4.

The Corollary 3.3 complements the results of Theorem in Yao et al. [11], that is, is the solution of the variational inequality: .

Theorem 3.5.

Let be a nonempty closed convex subset of a real Hilbert space . Let be a sequence of nonexpansive mappings of into itself such that . Let be a -Lipschitzian and -strongly monotone operator on with . Let and be two real sequences in and satisfy the conditions:

(A1) and ;

(A2) .

Suppose that for any bounded subset of . Let be a mapping of into itself defined by for all and suppose that . For given arbitrarily, let the sequence be generated by

(3.19)

Then the sequence strongly converges to a which solves the variational inequality:

(3.20)

Proof.

We proceed with the following steps.

Step 1.

We claim that is bounded. From , we may assume, without loss of generality, that for all . In fact, let , from (3.19) and using Lemma 2.5, we have

(3.21)

where . Then from (3.19) and (3.21), we obtain

(3.22)

By induction, we have

(3.23)

where . Therefore, is bounded. We also obtain that , , and are bounded. Without loss of generality, we may assume that , , , and , where is a bounded set of .

Step 2.

We claim that . To this end, define a sequence by . It follows that

(3.24)

Thus, we have

(3.25)

From and (3.25), we have

(3.26)

By (3.26), (A2), and using Lemma 2.2, we have . Therefore,

(3.27)

Step 3.

We claim that . Observe that

(3.28)

that is,

(3.29)

Step 4.

We claim that . Observe that

(3.30)

Hence, from Step 3 and using Lemma 2.4, we have

(3.31)

Step 5.

We claim that , where and is defined by (3.3). Since is bounded, there exists a subsequence of which converges weakly to . From Step 4, we obtain . From Lemma 2.1, we have . Hence, by Theorem 3.1, we have

(3.32)

Step 6.

We claim that converges strongly to . From (3.19), we have

(3.33)

where , , and . It is easy to see that , , and . Hence, by Lemma 2.3, the sequence converges strongly to . From and Theorem 3.1, we have that is the unique solution of the variational inequality:

Remark 3.6.

From Remark of Peng and Yao [18], we obtain that is a sequence of nonexpansive mappings satisfying condition for any bounded subset B of H. Moreover, let be the mapping; we know that for all and that . If we replace by in the recursion formula (3.19), we can obtain the corresponding results of the so-called mapping.

Setting and in Theorem 3.5, we can obtain the following result.

Corollary 3.7.

Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive mapping such that . Let be a strongly positive bounded linear operator with coefficient . Let and be two real sequences in and satisfy the conditions (A1) and (A2). For given arbitrarily, let the sequence be generated by

(3.34)

Then the sequence strongly converges to a fixed point of which solves the variational inequality:

(3.35)

Setting and in Theorem 3.5, we can obtain the following result.

Corollary 3.8.

Let be a nonempty closed convex subset of a real Hilbert space . Let be a nonexpansive mapping such that . Let and be two real sequences in and satisfy the conditions (A1) and (A2). For given arbitrarily, let the sequence be generated by (1.7). Then the sequence strongly converges to a fixed point of which solves the variational inequality:

(3.36)

Remark 3.9.

The Corollary 3.8 complements the results of Theorem in Yao et al. [11], that is, is the solution of the variational inequality: .