In this section, we will prove our main result. First, we give some assumptions on the operators and the parameters. Subsequently, we introduce our iterative algorithm for finding a common element of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion. Finally, we will show that the proposed algorithm has strong convergence.
Let
be a nonempty closed convex subset of a real Hilbert space
. Let
be a lower semicontinuous and convex function and let
be a bifunction satisfying conditions (H1)–(H5). Let
be a strongly positive bounded linear operator with coefficient
and let
be a maximal monotone mapping. Let the mappings
be
-inverse strongly monotone and
-inverse strongly monotone, respectively. Let
and
be two constants such that
and
.
Now we introduce the following iteration algorithm.
Algorithm 3.1.
For given
arbitrarily, compute the sequences
and
as follows:
where
is a real sequence in
.
Now we study the strong convergence of the algorithm (3.1).
Theorem 3.2.
Suppose that
. Assume the following conditions are satisfied:
(i)
;
(ii)
;
(iii)
.
Then the sequence
generated by (3.1) converges strongly to
which solves the following variational inequality:
Proof.
Take
. It is clear that
We divide our proofs into the following five steps:
(1)the sequences
and
are bounded;
(2)
;
(3)
;
(4)
;
(5)
.
Proof.
Since
is
-inverse strongly monotone and
is
-inverse strongly monotone, we have
It is clear that if
and
, then
and
are all nonexpansive. Set
. It follows that
By Lemma 2.1, we have
for all
. Then, we have
Hence, we have
Since
is linear bounded self-adjoint operator on
, then
Observe that
that is to say
is positive. It follows that
From (3.1), we deduce that
Therefore,
is bounded. Hence,
,
, and
are all bounded.
Proof.
From (3.1), we have
Note that
Substituting (3.14) into (3.13), we get
Notice that
This together with the last inequality and Lemma 2.3 implies that
Proof.
From (3.5) and (3.7), we get
By (3.1), we obtain
where
is some constant satisfying
. From (3.17) and (3.18), we have
Thus,
which imply that
Proof.
Since
is firmly nonexpansive, we have
Hence, we have
Since
is 1-inverse strongly monotone, we have
which implies that
Thus, by (3.23) and (3.25), we obtain
Substitute (3.26) into (3.18) to get
Then we derive
So, we have
Proof.
We note that
is a contraction. As a matter of fact,
for all
. Hence
has a unique fixed point, say
. That is,
. This implies that
for all
. Next, we prove that
First, we note that there exists 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
.
We next show that
. By
, we know that
It follows from (H2) that
Hence,
For
and
, let
. From (3.35) we have
Since
, we have
. Further, from the inverse strongly monotonicity of
, we have
. So, from (H4) and the weakly lower semicontinuity of
,
and
weakly, we have
From (H1), (H4), and (3.37), we also have
and hence
Letting
, we have, for each
,
This implies that
.
Next, we show that
. In fact, since
is
-inverse strongly monotone,
is Lipschitz continuous monotone mapping. It follows from Lemma 2.2 that
is maximal monotone. Let
, that is,
. Again since
, we have
, that is,
. By virtue of the maximal monotonicity of
, we have
and so
It follows from
,
and
that
It follows from the maximal monotonicity of
that
, that is,
. Therefore,
. It follows that
Proof.
First, we note that
; then for all
, we have
.
From (3.1), we have
that is,
where
and
. It is easy to see that
and
. Hence, by Lemma 2.3, we conclude that the sequence
converges strongly to
. This completes the proof.