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 inversestrongly 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.