In this section, we prove the strong convergence theorem for infinitely many nonexpansive mappings in a real Hilbert space.

Theorem 3.1.

Let be a nonempty closed convex subset of a real Hilbert space , let be a bifunction from to satisfying (A1)–(A4), let be an infinitely many nonexpansive of into itself, and let be an -inverse-strongly monotone mapping of into such that . Let be a contraction of into itself with and let be a strongly positive linear bounded operator on with coefficient and . Let , , and be sequences generated by (1.25), where is the sequence generated by (1.23), , are three sequences in and is a real sequence in satisfying the following conditions:

(i)

(ii) and

(iii)

(iv) and

(v) for some and .

Then, and converge strongly to a point which is the unique solution of the variational inequality

Equivalently, one has

Proof.

Note that from the condition (i), we may assume, without loss of generality, that for all . From Lemma 2.2, we know that if , then . We will assume that . First, we show that is nonexpansive. Indeed, from the -inverse-strongly monotone mapping definition on and condition (v), we have

which implies that the mapping is nonexpansive. On the other hand, since is a strongly positive bounded linear operator on H, we have

Observe that

and this show that is positive. It follows that

Let , where . Note that is a contraction of into itself with . Then, we have

Since , it follows that is a contraction of into itself. Therefore by the Banach Contraction Mapping Principle, which implies that there exists a unique element such that .

We will divide the proof into five steps.

Step 1.

We claim that is bounded. Indeed, pick any . From the definition of , we note that . If follows that

Since is nonexpansive and from (1.6), we have

Put . Since , we have . Substituting and in (1.5), we can write

Using the fact that is -inverse-strongly monotone mapping, and is a solution of the variational inequality problem , we also have

It follows from (3.9) and (3.10) that

Substituting by and in (1.4), we obtain

It follows that

Substituting (3.13) into (3.11), we have

Setting , we can calculate

By induction,

Hence, is bounded, so are , , , , , and .

Step 2.

We claim that .

Observing that and we get

Putting in (3.17) and in (3.18), we have

So, from (A2) we have

and hence

Without loss of generality, let us assume that there exists a real number such that for all Then, we have

and hence

where . Note that

Setting

we have . It follows that

It follows from (3.24) and (3.26) that

Since and are nonexpansive, we have

where is a constant such that for all

Combining (3.27) and (3.28), we have

which implies that (noting that (i), (ii), (iii), (iv), (v), and

Hence, by Lemma 2.8, we obtain

It follows that

Applying (3.32) and (ii), (iv), and (v) to (3.23) and (3.24), we obtain that

Since , we have

that is

By (i), (iii), and (3.32) it follows that

Step 3.

We claim that the following statements hold:

(i)

(ii)

For any and (3.14), we have

Observe that

where

It follows from condition (i) that

Substituting (3.37) into (3.38), and using (v), we have

It follows that

Since and from (3.32), we obtain

Note that

Since , we have

As is -Lipschitz continuous, we obtain

then, we get

from

Applying (3.43), (3.45), and (3.47), we have

For any , note that is firmly nonexpansive (Lemma 2.4), then we have

and hence

which together with (3.38) gives

So

Using , as , (3.32), and (3.49), we obtain

Since , we obtain

Observe that

Applying (3.36), (3.49), and (3.54) to the last inequality, we obtain

Let be the mapping defined by (2.6). Since is bounded, applying Lemma 2.7 and (3.57), we have

Step 4.

We claim that where is the unique solution of the variational inequality

Since is a unique solution of the variational inequality (3.1), to show this inequality, we choose a subsequence of such that

Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume that From we obtain . Next, We show that , where . First, we show that . Since , we have

If follows from (A2) that

and hence

Since and it follows by (A4) that for all For with and let Since and we have and hence So, from (A1) and (A4) we have

and hence . From (A3), we have for all and hence

Next, we show that By Lemma 2.6, we have . Assume Since and it follows by the Opial's condition that

which derives a contradiction. Thus, we have . By the same argument as that in the proof of [35, Theorem 2.1, Pages 10–11], we can show that Hence . Since , it follows that

It follows from the last inequality, (3.36), and (3.54) that

Step 5.

Finally, weshow that and convergestrongly to . Indeed, from (1.25) , we have

Since , and are bounded, we can take a constant such that

for all . It then follows that

where

Using (i), (3.65), and (3.66), we get . Applying Lemma 2.9 to (3.69), we conclude that in norm. Finally, noticing

we also conclude that in norm. This completes the proof.

Corollary 3.2 ([28, Theorem 3.1]).

Let be nonempty closed convex subset of a real Hilbert space , let be a bifunction from to satisfying (A1)–(A4), and let be an infinitely many nonexpansive of into itself such that . Let be a contraction of into itself with and let be a strongly positive linear bounded operator on with coefficient and . Let and are the sequences generated by

where is the sequence generated by (1.23), is a sequences in and is a real sequence in satisfying the following conditions:

(i)

(ii) and

Then, and converge strongly to a point which is the unique solution of the variational inequality

Equivalently, one has

Proof.

Put , and in Theorem 3.1., then . The conclusion of Corollary 3.2 can obtain the desired result easily.

Corollary 3.3.

Let be nonempty closed convex subset of a real Hilbert space , let be a bifunction from to satisfying (A1)–(A4) and let be an -inverse-strongly monotone mapping of into such that . Let be a contraction of into itself with and let be a strongly positive linear bounded operator on with coefficient and . Let , , and be sequences generated by

where , and are three sequences in and is a real sequence in satisfying the following conditions:

(i) and

(ii) and

(iii)

(iv) and

(v) for some and . Then, and converge strongly to a point which is the unique solution of the variational inequality

Equivalently, one has

Proof.

Put for all and for all . Then for all . The conclusion follows from Theorem 3.1.

Corollary 3.4.

Let be nonempty closed convex subset of a real Hilbert space , let be an infinitely many nonexpansive of into itself, and let be a -inverse-strongly monotone mapping of into such that . Let be a contraction of into itself with and let be a strongly positive linear bounded operator on with coefficient and . Let , , and be sequences generated by

where is the sequences generated by (1.23), and , , are three sequences in satisfying the following conditions:

(i) and

(ii) and

(iii)

(iv) for some and .

Then, converges strongly to a point which is the unique solution of the variational inequality

Equivalently, one has

Proof.

Put for all and for all in Theorem 3.1. Then, we have . So, by Theorem 3.1, we can conclude the desired conclusion easily.

If and in Theorem 3.1, then we can obtain the following result immediately.

Corollary 3.5.

Let be nonempty closed convex subset of a real Hilbert space , let be a bifunction from to satisfying (A1)–(A4), let be an infinitely many nonexpansive of into itself, and let be an -inverse-strongly monotone mapping of into such that . Let be a contraction of into itself with . Let , , and be sequences generated by

where is the sequences generated by (1.23), , , are three sequences in and is a real sequence in satisfying the following conditions:

(i)

(ii) and

(iii) and ;

(iv)

(v) and

(vi) for some and .

Then, and converge strongly to a point which is the unique solution of the variational inequality

Equivalently, one has

Corollary 3.6.

Let be nonempty closed convex subset of a real Hilbert space , let be a bifunction from to satisfying (A1)–(A4) and let be an infinite family of nonexpansive of into itself such that . Let be a contraction of into itself with . Let and be sequences generated by

where , , and are three sequences in , and is a real sequence in satisfying the following conditions:

(i)

(ii) and

(iii)

(iv) and

Then, and converge strongly to a point which is the unique solution of the variational inequality

Equivalently, one has

Proof.

Put and in Corollary 3.5. then . The conclusion of Corollary 3.6 can obtain the desired result easily.