In this section, we prove a strong convergence theorem for finding a common element of the set of solutions for a generalized equilibrium problem and the set for two maximal monotone operators and .

Theorem 3.1.

Suppose that Assumption 2.1 is satisfied. Let be chosen arbitrarily. Consider the sequence

where

is defined by (2.14), satisfy

and satisfies . Then, the sequence converges strongly to , where is the generalized projection of onto .

Proof.

For the sake of simplicity, we define

so that

We divide the proof into several steps.

Step 1.

We claim that is closed and convex for each .

Indeed, it is obvious that is closed and is closed and convex for each . Let us show that is convex. For and , put . It is sufficient to show that . We first write for each . Next, we prove that

is equivalent to

Indeed, from (2.1) we deduce that there hold the following:

which combined with (3.6) yield that (3.6) is equivalent to (3.7). Thus we have

This implies that . Therefore, is closed and convex.

Step 2.

We claim that for each and that is well defined.

Indeed, take arbitrarily. Note that is equivalent to

Then from Lemma 2.12, we obtain

Moreover, we have

and hence by Proposition 2.11,

So for all . Now, let us show that

We prove this by induction. For , we have . Assume that . Since is the projection of onto , by Lemma 2.3 we have

As by the induction assumption, the last inequality holds, in particular, for all . This, together with the definition of implies that . Hence (3.14) holds for all . So, for all . This implies that the sequence is well defined.

Step 3.

We claim that is bounded and that as .

Indeed, it follows from the definition of that . Since and , so for all , that is, is nondecreasing. It follows from and Lemma 2.4 that

for each for each . Therefore, is bounded, which implies that the limit of exists. Since

so is bounded. From Lemma 2.4, we have

for each . This implies that

Step 4.

We claim that , , and .

Indeed, from , we have

Therefore, from and , it follows that .

Since and is uniformly convex and smooth, we have from Lemma 2.2 that

and, therefore, . Since is uniformly norm-to-norm continuous on bounded subsets of and , then .

Let us set . Then, according to Lemma 2.5 and Proposition 2.11, we know that is a nonempty closed convex subset of such that . Fix arbitrarily. As in the proof of Step 2, we can show that ,

Hence it follows from the boundedness of that , and are also bounded. Let . Since is a uniformly smooth Banach space, we know that is a uniformly convex Banach space. Therefore, by Lemma 2.6 there exists a continuous, strictly increasing, and convex function , with , such that

for and . So, we have that

and hence

for all . Consequently, we have

Since and is uniformly norm-to-norm continuous on bounded subsets of , we obtain . From and , we have

Therefore, from the properties of , we get

recalling that is uniformly norm-to-norm continuous on bounded subsets of . Next let us show that

Observe first that

Since , and is bounded, so it follows that . Also, observe that

Since , and the sequences are bounded, so it follows that . Meantime, observe that

and hence

Since and , it follows from the boundedness of that . Thus, in terms of Lemma 2.2, we have that and so . Furthermore, it follows from (3.25) that

and hence

Since is uniformly norm-to-norm continuous on bounded subsets of , it follows from that . Thus from , , and the boundedness of both and , we deduce that . Utilizing the properties of , we have that . Since is uniformly norm-to-norm continuous on bounded subsets of , it follows that .

Step 5.

We claim that , where

Indeed, since is bounded and is reflexive, we know that . Take arbitrarily. Then there exists a subsequence of such that . Hence it follows from , , and that and converge weakly to the same point . On the other hand, from (3.28a), (3.28b) and , we obtain that

If and , then it follows from (2.17) and the monotonicity of the operators that for all

Letting , we have that and . Then the maximality of the operators implies that and .

Next, let us show that . Since

from and Proposition 2.11 it follows that

Also, since

so we get

So, from (3.39), , and , we have .

Since is uniformly convex and smooth, we conclude from Lemma 2.2 that

From , , and (3.42), we have and .

Since is uniformly norm-to-norm continuous on bounded subsets of , from (3.42) we derive

From , it follows that

By the definition of , we have

where

Replacing by , we have from (A2) that

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting in the last inequality, from (3.44) and (A4), we have

For , with , and , let . Since and , then and hence . So, from (A1) we have

Dividing by , we have

Letting , from (A3) it follows that

So, . Therefore, we obtain that by the arbitrariness of .

Step 6.

We claim that converges strongly to .

Indeed, from and , it follows that

Since the norm is weakly lower semicontinuous, then

From the definition of , we have . Hence , and

which implies that . Since has the Kadec-Klee property, then . Therefore, converges strongly to .

Remark 3.2.

In Theorem 3.1, let , , and . Then, for all and , we have that

Moreover, there hold the following

and hence

In this case, Theorem 3.1 reduces to [17, Theorem ].