In this section, we prove a strong convergence theorem of a new iterative method (3.4) for an infinite family of nonexpansive mappings and relaxed
-cocoercive mappings in a real Hilbert space.
We first prove the following lemmas.
Lemma 3.1.
Let
be a real Hilbert space, let
be a nonempty closed convex subset of
, and let
be
-inverse-strongly monotone. It
, then
is a nonexpansive mapping in
.
Proof.
For all
and
, we have
So,
is a nonexpansive mapping of
into
.
Lemma 3.2.
Let
be a real Hilbert space, let
be a nonempty closed convex subset of
and let
be a relaxed
-cocoercive and
-Lipschitz continuous. If
,
, then
is a nonexpansive mapping in
.
Proof.
For any
and
,
.
Putting
, we obtain
that is,
. It follows that
for all
. Thus
.
So,
is a nonexpansive mapping of
into
.
Now, we prove the following main theorem.
Theorem 3.3.
Let
be a nonempty closed convex subset of a real Hilbert space
, and let
be a bifunction satisfying (A1)–(A4). Let
(1)
be an infinite family of nonexpansive mappings of
into
;
(2)
be an
-inverse strongly monotone mappings of
into
;
(3)
be relaxed
-cocoercive and
-Lipschitz continuous mappings of
into
.
Assume that
. Let
be a contraction mapping with
and let
be a strongly positive linear bounded operator on
with coefficient
and
. Let
,
,
and
be sequences generated by
where
is the sequence generated by (1.24) and
,
,
and
are sequences in
satisfy the following conditions:
and 
,
,
,
for some
with
,
,
for some
with
.
Then,
and
converge strongly to a point
, where
, which solves the variational inequality
which is the optimality condition fot the minimization problem
where
is a potential function for
(i.e.,
for
).
Proof.
Since
by the condition (C1) and
, we may assume, without loss of generality, that
. Since
is a strongly positive bounded linear operator on
, then
Observe that
that is to say
is positive. It follows that
We will divide the proof of Theorem 3.3 into six steps.
Step 1.
We prove that there exists
such that
.
Let
. Note that
is a contraction mapping of
into itself with coefficient
. Then, we have
Therefore,
is a contraction mapping of
into itself. Therefore by the Banach Contraction Mapping Principle guarantee that
has a unique fixed point, say
. That is,
.
Step 2.
We prove that
is bounded.
Since
we obtain
From Lemma 2.6, we have
for all
.
For any
, it follows from
that
So, we have
By Lemma 2.6 again, we have
for all
. If follows that
If we applied Lemma 3.2, we get
and
are nonexpansive. Since
and
is a nonexpansive, we have
, and we have
It follows that
which yields that
This in turn implies that
Therefore,
is bounded. We also obtain that
,
,
,
,
,
,
,
,
and
are all bounded.
Step 3.
We claim that
and
.
From Lemma 2.6, we have
and
. Let
, we get
,
, and so
Putting
in (3.20) and
in (3.21), 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, we have
and hence
where
.
Put
,
and
. Since
,
and
are nonexpansive, then we have the following some estimates:
Similarly, we can prove that
Since
and
are nonexpansive, we deduce that, for each
,
where
is a constant such that
for all 
Similarly, we can obtain that there exist nonnegative numbers
,
such that
and so are
Observing that
we obtain
which yields that
Substitution of (3.27) and (3.30) into (3.35) yields that
where
is an appropriate constant such that
.
Observing that
we obtain
which yields that
Substitution of (3.28) and (3.32) into (3.39) yields that
where
is an appropriate constant such that
.
Substituting (3.26) and (3.36) into (3.40), we obtain
where
is an appropriate constant such that
.
Substituting (3.41) into (3.29), we obtain
where
is an appropriate constant such that
.
Define
Observe that from the definition
, we obtain
It follows from (3.32), (3.42), and (3.44) that
where
is an appropriate constant such that
.
It follows from conditions (C1), (C2), (C3), (C4), (C5), and
for all 
Hence, by Lemma 2.11, we obtain
It follows that
Applying (3.48) and conditions in Theorem 3.3 to (3.26), (3.41), and (3.42), we obtain that
From (3.49), (C2), (C5), and
for all
, we also have
Since
, we have
that is,
By (C1), (C3), and (3.48) it follows that
Step 4.
We claim that the following statements hold:
(i)
;
(ii)
;
(iii)
.
Since
is relaxed
-cocoercive and
-Lipschitz continuous mappings, by the assumptions imposed on
for any
, we have
Similarly, we have
Observe that
where
It follows from condition (C1) that
Substituting (3.54) into (3.56), and using condition (C6), we have
It follows that
Since
as
and (3.48), we obtain
Note that
Using (3.56) again, we have
Substituting (3.62) into (3.64) and using condition (C2) and (C6), we have
It follows that
Since
as
and (3.48), we obtain
In a similar way, we can prove
By (2.3), we also have
which yields that
Substituting (3.70) into (3.56), we have
It follows that
Applying
,
and
as
to the last inequality, we have
On the other hand, we have
which yields that
Similarly, we can prove
Substituting (3.75) into (3.56), we have
which yields that
Applying (3.48) and (3.61) to the last inequality, we have
Using (3.64) again, we have
which implies that
From (3.48) and (3.67), we obtain
By using the same argument, we can prove that
Note that
Since
and
as
, respectively, we also have
On the other hand, we observe
Applying (3.73), (3.83), (3.84), and (3.86), we have
On the other hand, we have
Substituting (3.89) into (3.64) and using conditions (C2) and (C7), we have
This implies that
In view of the restrictions (C2) and (C7), we obtain that
Let
. Since
and
is firmly nonexpansive (Lemma 2.6), then we obtain
So, we obtain
Therefore, we have
It follows that
Using
,
as
, (3.48), (3.88), and (3.92), we obtain
Since
, we obtain
Note that
and thus from (3.88) and (3.97), we have
Observe that
Applying (3.53) and (3.100), we obtain
Let
be the mapping defined by (2.11). Since
is bounded, applying Lemma 2.10 and (3.102), we have
Step 5.
We claim that
where
is the unique solution of the variational inequality
for all 
Since
is a unique solution of the variational inequality (3.5), to show this inequality, we 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
. Next, We show that
, where
.
-
(a)
First, we prove
.
Since
, we know that
From (A2), we also have
Replacing
by
, we have
For any
with
and
let
Since
and
we have
So, from (3.107) we have
Since
is Lipschitz continuous, from (3.97), we have
as
.
Further, from the monotonicity of
, we get that
It follows from (A4) and (3.108) that
From (A1), (A4), and (3.110), we also have
and hence
Letting
in the above inequality, we have, for each
,
Thus 
-
(b)
Next, we show that 
By Lemma 2.9, we have
. Assume
Since
we know that
and
and it follows by the Opial's condition (Lemma 2.3) that
that is a contradiction. Thus, we have
.
-
(c)
Finally, Now we prove that
Define,
Since
is relaxed
-cocoercive and condition (C6), we have
which yields that
is monotone. Then,
is maximal monotone. Let
Since
and
we have
On the other hand, from
we have
and hence
Therefore, we have
which implies that
Since
is maximal monotone, we have
and hence
That is,
, where
. Since
, it follows that
On the other hand, we have
From (3.53) and (3.121), we obtain that
Step 6.
Finally, we show that
and
converge strongly to
.
Indeed, from (3.4) and Lemma 2.4, we obtain
Since
,
and
are bounded, we can take a constant
such that
for all
. It then follows that
where
Using (C1), (3.121), and (3.123), we get
,
and
. Applying Lemma 2.13 to (3.126), we conclude that
in norm. Finally, noticing
we also conclude that
in norm. This completes the proof.
Corollary 3.4.
Let
be a nonempty closed convex subset of a real Hilbert space
. Let
be a bifunction satisfying (A1)–(A4), let
be relaxed
-cocoercive and
-Lipschitz continuous mappings, and let
be an infinite family of nonexpansive mappings of
into itself such that
. Let
be a contraction mapping of
into itself with
. Let
,
,
and
be sequences generated by
where
is the sequence generated by (1.24) and
,
,
, and
are sequences in
and
is a real sequence in
satisfying the following conditions:
,
,
, 
for some
with
,
.
Then,
and
converge strongly to a point
, where
.
Proof.
Put
,
,
(:the zero mapping) and
in Theorem 3.3. Then
and for any
, we see that
Let
be a sequence satisfying the restriction:
, where
. Then we can obtain the desired conclusion easily from Theorem 3.3.
Corollary 3.5.
Let
be a nonempty closed convex subset of a real Hilbert space
. Let
be an infinite family of nonexpansive mappings of
into itself and let
be relaxed
-cocoercive and
-Lipschitz continuous mappings such that
. Let
be a contraction mapping with
and let
be a strongly positive linear bounded operator on
with coefficient
and
. Let
,
and
be sequences generated by
where
is the sequence generated by (1.24) and
,
,
and
are sequences in
satisfying the following conditions:
,
,
, 
for some
with
,
.
Then,
converges strongly to a point
, where
, which solves the variational inequality
which is the optimality condition fot the minimization problem
where
is a potential function for
(i.e.,
for
).
Proof.
Put
,
for all
and
for all
in Theorem 3.3. Then, we have
. So, by Theorem 3.3, we can conclude the desired conclusion easily.