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.