In this section, we show a strong convergence of an iterative algorithm based on both viscosity approximation method and extragradient method which solves the problem of finding a common element of the set of solutions of a mixed equilibrium problem, the set of fixed points of a finite family of nonexpansive mappings, and the set of solutions of the variational inequality for a monotone, Lipschitz continuous mapping in a Hilbert space.
Theorem 3.1.
Let
be a nonempty closed convex subset of a real Hilbert space
. Let
be a bifunction from
to
satisfying (A1)–(A5) and let
be a proper lower semicontinuous and convex function. Let
be a monotone and
-Lipschitz continuous mapping of
into
. Let
be a finite family of nonexpansive mappings of
into
such that
. Let
be sequences in
with
. Let
be the
-mapping generated by
and
. Assume that either (B1) or (B2) holds. Let
be a contraction of
into itself and let
,
, and
be sequences generated by
for every
where
,
,
,
,
, and
are sequences of numbers satisfying the conditions:
(C1)
and
;
(C2)
;
(C3)
;
(C4)
and
;
(C5)
for all
.
Then,
,
, and
converge strongly to
.
Proof.
We show that
is a contraction of
into itself. In fact, there exists
such that
for all
. So, we have
for all
Since
is complete, there exists a unique element
such that
.
Put
for every
Let
and let
be a sequence of mappings defined as in Lemma 2.1. Then
. From
, we have
From (2.3), the monotonicity of
, and
, we have
Further, Since
and
is
-Lipschitz continuous, we have
So, it follows from (C3) that the following inequality holds for
, where
is a positive integer:
Put
It is obvious that
Suppose
By Lemma 2.5, we know that
is nonexpansive and
. From (3.3), (3.6) and
, we have
and
for every
. Therefore,
is bounded. From (3.3) and (3.6), we also obtain that
and
are bounded.
From
and the monotonicity and the Lipschitz continuity of
, we have
Hence, we obtain that
is bounded. It follows from the Lipschitz continuity of
that
,
, and
are bounded. Since
and
are nonexpansive, we know that
and
are also bounded. From the definition of
, we get
where
is an approximate constant such that
On the other hand, from
and
, we have
Putting
in (3.11) and
in (3.12), we have
So, from the monotonicity of
, we get
and hence
Without loss of generality, let us assume that there exists a real number
such that
for all
Then,
and hence
where 
It follows from (3.9) and (3.7) that
Define a sequence
such that
Then, we have
Next we estimate
. It follows from the definition of
that
where
is an approximate constant such that
Since
for all
and
, we have
It follows that
Substituting (3.24) into (3.21) yields that
Hence, we have
From (3.20), (3.26), and (3.18), we have
It follows from (C1)–(C5) that
Hence by Lemma 2.4, we have
. Consequently
Since
, we have
and thus
It follows from (C1) and (C2) that
.
Since
for
, it follows from (3.3) and (3.6) that
from which it follows that
It follows from (C1)–(C3) and
that
.
By the same argument as in (3.6), we also have
Combining the above inequality and (3.32), we have
and thus
which implies that
.
From
we also have
. As
is
-Lipschitz continuous, we have
.
For
, we have, from Lemma 2.1,
Hence,
By(3.3), (3.6), (3.32), and (3.38), we have
Hence,
It follows from (C1), (C2), and
that
.
Since
It follows that
Next we show that
where
. To show this inequality, we can choose 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
From
, we obtain that
. From
, we also obtain that
. From
, we also obtain that
. Since
and
is closed and convex, we obtain
.
In order to show that
, we first show
By
we know that
It follows from (A2) that
Hence,
It follows from (A4), (A5), and the weakly lower semicontinuity of
,
and
that
For
with
and
, let
Since
and
, we obtain
and hence
. So by (A4) and the convexity of
, we have
Dividing by
, we get
Letting
, it follows from (A3) and the weakly lower semicontinuity of
that
for all
and hence
.
Now we show that
Put
where
is the normal cone to
at
. We have already mentioned that in this case the mapping
is maximal monotone, and
if and only if
. Let
. Then
and hence
. So, we have
for all
. On the other hand, from
and
we have
and hence
Therefore, we have
Hence we obtain
as
. Since
is maximal monotone, we have
and hence
.
We next show that
To see this, we observe that we may assume (by passing to a further subsequence if necessary)
for
Let
be the
-mapping generated by
and
. By Lemma 2.5, we know that
is nonexpansive and
. it follows from Lemma 2.6 that
Assume
Since
and
, it follows from the Opial condition, (3.42), and (3.56) that
which is a contradiction. Hence, we have
. This implies
Therefore, we have
Finally, we show that
, where
.
From Lemma 2.3, we have
and thus
It follows from Lemma 2.2, (3.58), and (3.60) that
. From
and
, we have
and
. The proof is now complete.