Throughout the rest of this paper, we always assume that is a -lipschitzian continuous and -strongly monotone operator with and assume that . . Let be mappings defined as Lemma 2.1. Define a mapping by , for all , where , then, by Lemma 2.3, is nonexpansive. We consider the mapping on defined by

where . By Lemmas 2.1 and 2.3, we have

It is easy to see that is a contraction. Therefore, by the Banach contraction principle, has a unique fixed-point such that

For simplicity, we will write for provided no confusion occurs. Next, we prove that the sequence converges strongly to a which solves the variational inequality

Equivalently, .

Theorem 3.1.

Let be a nonempty closed convex subset of a real Hilbert space and a bifunction from into satisfying (A1), (A2), (A3), and (A4). Let be a -strictly pseudocontractive nonself mapping such that . Let be an -Lipschitzian continuous and - monotone operator on with and , . Let be asequence generated by

where , , and satisfy if and satisfy the following conditions:

(i), ,

(ii) and ,

then converges strongly to a point which solves the variational inequality (3.4).

Proof.

First, take . Since and , from Lemma 2.1, for any , we have

Then, since , we obtain that

Further, we have

It follows that .

Hence, is bounded, and we also obtain that and are bounded. Notice that

By Lemma 2.1, we have

It follows that

Thus, from Lemma 2.4, (3.7), and (3.11), we obtain that

It follows that

Since , therefore

From (3.9), we derive that

Define by , then is nonexpansive with by Lemma 2.3. We note that

So by (3.15) and , we obtain that

Since is bounded, so there exists a subsequence which converges weakly to . Next, we show that . Since is closed and convex, is weakly closed. So we have . Let us show that . Assume that , Since and , it follows from the Opial's condition that

This is a contradiction. So, we get and .

Next, we show that . Since , for any , we obtain

From (A2), we have

Replacing by , we have

Since and , it follows from (A4) that , for all . Let for all and , then we have and hence . Thus, from (A1) and (A4), we have

and hence . From (A3), we have for all and hence . Therefore, . On the other hand, we note that

Hence, we obtain

It follows that

This implies that

In particular,

Since , it follows from (3.27) that as . Next, we show that solves the variational inequality (3.4).

As a matter of fact, we have

and we have

Hence, for ,

Since is monotone (i.e., , for all . This is due to the nonexpansivity of ).

Now replacing in (3.30) with and letting , we obtain

That is, is a solution of (3.4). To show that the sequence converges strongly to , we assume that . Similiary to the proof above, we derive . Moreover, it follows from the inequality (3.31) that

Interchange and to obtain

Adding up (3.32) and (3.33) yields

Hence, , and therefore as ,

This is equivalent to the fixed-point equation

Theorem 3.2.

Let be a nonempty closed convex subset of a real Hilbert space and a bifunction from into satisfying (A1), (A2), (A3) and (A4). Let be a -strictly pseudocontractive nonself mapping such that . Let be an -Lipschitzian continuous and -strongly monotone operator on with . Suppose that , . Let and be sequences generated by and

where , if ,, and satisfy the following conditions:

(i), , , ,

(ii) and , ,

(iii), and ,

then and converge strongly to a point which solves the variational inequality(3.4).

Proof.

We first show that is bounded. Indeed, pick any to derive that

By induction, we have

and hence is bounded. From (3.6) and (3.7), we also derive that and are bounded. Next, we show that . We have

where

On the other hand, we have

From and , we note that

Putting in (3.43) and in (3.44), we have

So, from (A2), we have

and hence

Since , without loss of generality, let us assume that there exists a real number a such that for all . Thus, we have

where . Next, we estimate . Notice that

From (3.48), (3.49), and (3.42), we obtain that

where is an appropriate constant such that

From (3.41) and (3.50), we obtain

where . Hence, few by Lemma 2.5, we have

From (3.48) and (3.50), and , we have

Since

it follows that

From and (3.53), we have

For , we have

This implies that

Then, from (3.7) and (3.59), we derive that

Since , , we have

From (3.57) and (3.61), we obtain that

Define by , then is nonexpansive with by Lemma 2.3. Notice that

By (3.62) and , we obtain that

Next, we show that , where is a unique solution of the variational inequality (3.4). Indeed, take 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 (3.61) and (3.64), we obtain and . By the same argument as in the proof of Theorem 3.1, we have . Since , it follows that

From , we have

This implies that

where , , and .

It is easy to see that , , and by (3.66). Hence by Lemma 2.5, the sequence converges strongly to .