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
].