In order to prove the main result, we first give the following lemma.
Lemma 3.1 (see [5]).
is a solution of variational inclusion (1.2) if and only if , that is,
If , then is a closed convex subset in .
In the sequel, we assume that satisfy the following conditions:
is a real Hilbert space, is a nonempty closed convex subset;
is a strongly positive linear bounded operator with a coefficient is a contraction mapping with a contraction constant , , is an inversestrongly monotone mapping, and is a multivalued maximal monotone mapping;
is a nonexpansive semigroup;
is a finite family of bifunctions satisfying conditions (H_{1})–(H_{4}), and is a finite family of lower semicontinuous and convex functional;
is a finite family of Lipschitz continuous mappings with constant such that
is affine in the first variable,
for each fixed , is sequentially continuous from the weak topology to the weak topology;
is a finite family of strongly convex with constant , and its derivative is not only continuous from the weak topology to the strong topology but also Lipschitz continuous with constant .
In the sequel we always denote by the set of fixed points of the nonexpansive semigroup , the set of solutions to the variational inequality (1.2), and MEP() the set of solutions to the following auxiliary problem for a system of mixed equilibrium problems:
where
and , is the mapping defined by (2.8).
In the sequel we denote by for and .
Theorem 3.2.
Let be the same as above. Let be a finite family of positive numbers, , and . If and the following conditions are satisfied:
for each , there exists a bounded subset and such that for any
, , and , then
for each , there is a unique such that
the sequence converges strongly to some point , provided that is firmly nonexpansive;
is the unique solution of the following variational inequality
Proof.
We observe that from condition (ii), we can assume, without loss of generality, that .
Since is a linear bounded selfadjoint operator on , then
Since
this implies that is positive. Hence we have
For each given , let us define the mapping
Firstly we show that the mapping is a contraction. Indeed, for any , we have
This implies that is a contraction mapping. Let be the unique fixed point of . Thus,
is well defined.
Letting , , and , then
We divide the proof of Theorem 3.2 into 8 steps.
Step 1.
First prove that the sequences , and are bounded.

(a)
Pick , since and , we have
() Since and , we have , and so
Letting , , we have
Similarly, we have
Form (3.5), (3.9), (3.14), (3.15), (3.16), and (3.17) we have
So, . This implies that is a bounded sequence in . Therefore , and are all bounded.
Step 2.
Next we prove that
Since , then
Hence
From condition (ii), we have
Let , then is a nonempty bounded closed convex subset of and invariant. Since and is bounded, there exists such that ; it follows from Lemma 2.6 that
From (3.22) and (3.23), we have
Step 3.
Next we prove that
In fact, for any given and , since is firmly nonexpansive, we have
It follows that
From (3.5), we have
Since
and this together with (3.27) and (3.28), it yields
Simplifying it we have
Since and , by condition (ii), it yields .
Step 4.
Now we prove that for any given
In fact, it follows from (3.15) that
Substituting (3.33) into (3.28), we obtain
Simplifying it, we have
Since , and are bounded, these imply that .
Step 5.
Next we prove that
In fact, since
for the purpose, it is sufficient to prove
() First we prove that . In fact, since
we have
Substituting (3.40) into (3.28), it yields that
Simplifying it we have
Since , , , and are bounded, these imply that .
() Next we prove that
In fact, since , so . This together with (3.25) shows that .
Step 6.
Next we prove that there exists a subsequence of such that , and is the unique solution of the variational inequality (3.6).

(a)
We first prove that . In fact, since is bounded, there exists a subsequence of such that . From Lemma 2.7 and Step 2, we obtain .

(b)
Now we prove that .
Since and noting Step 3, without loss of generality, we may assume that . Hence for any and for any , we have
By the assumptions and by condition (H_{2}) we know that the function and the mapping both are convex and lower semicontinuous, hence they are weakly lower semicontinuous. These together with and , we have
That is,
for all and , hence .

(c)
Now we prove that .
In fact, since is inversestrongly monotone, it follows from Proposition 1.1 that is a Lipschitz continuous monotone mapping and (where is the domain of ). It follows from Lemma 2.4 that is maximal monotone. Let , that is, . Since and noting Step 3, without loss of generality, we may assume that ; in particular, we have . From , we can prove that . Again since , we have
By virtue of the maximal monotonicity of , we have
So,
Since , , and , we have
Since is maximal monotone, this implies that , that is, , and so .

(d)
Now we prove that is the unique solution of variational inequality (3.6).
We first prove that .
Since
It follows that
Therefore,
Now, replacing in (3.53) with and letting and , we have .
Next we prove that is the unique solution of the variational inequality (3.6).
Since
we have
Hence for any we have,
then
It is easily seen that is monotone. Thus from (3.57) we have that
Now, in (3.58) replacing by and letting and , from (3.36), we have
So, we have
It follows from [18, Theorem ] that the solution of the variational inequality (3.6) is unique, that is, is a unique solution of (3.6).
Step 7.
Next we prove that
() First, we prove that
Indeed, there exists a subsequence of such that
We may also assume that . This together with (3.22) and (3.36) shows that . Since , we have . Again by the same method as given in Step 6 we can prove that . So, we have
() Now we prove that
From and (a), we have
Step 8.
Finally we prove that
Indeed, from (3.5), (3.15), and (3.17), we have
This implies that
Combining (3.61) and (3.69), we obtain that .
This completes the proof of Theorem 3.2.
Corollary 3.3.
Let be the same as in Theorem 3.2. Let be a finite family of positive parameter, and . If and conditions (i) and (ii) in Theorem 3.2 are satisfied, then
for each there is a unique such that
the sequence converges strongly to some point , provided that is firmly nonexpansive;
is the unique solution of variational inequality (3.6).
Proof.
Taking in Theorem 3.2, where is the indicator function of , that is,
then the variational inclusion problem (1.2) is equivalent to variational inequality (1.4), that is, to find such that
Again, since , then . Therefore we have
The conclusion of Corollary 3.3 can be obtained from Theorem 3.2 immediately.