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 .
Suppose that Assumption 2.1 is satisfied. Let be chosen arbitrarily. Consider the sequence
is defined by (2.14), satisfy
and satisfies . Then, the sequence converges strongly to , where is the generalized projection of onto .
For the sake of simplicity, we define
We divide the proof into several steps.
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.
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.
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
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
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
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
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 .
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
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
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 .
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 .
In Theorem 3.1, let , , and . Then, for all and , we have that
Moreover, there hold the following
In this case, Theorem 3.1 reduces to [17, Theorem ].