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
-inverse-strongly monotone mapping, and
is a multivalued maximal monotone mapping;
is a nonexpansive semigroup;
is a finite family of bifunctions satisfying conditions (H1)–(H4), 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 semi-group
,
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 self-adjoint 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 (H2) 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
-inverse-strongly 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.