In this section, we prove strong convergence theorem which is our main result.
Theorem 3.1.
Let be a nonempty, closed and convex subset of a 2uniformly convex and uniformly smooth Banach space , let be an inversestrongly monotone mapping of into with for all and . Let and be two finite families of closed relatively weak quasinonexpansive mappings from into itself with , where . Assume that and are uniformly continuous for all . Let be a sequence generated by the following algolithm:
where , and is the normalized duality mapping on . Assume that , and are the sequences in satisfying the restrictions:
(C1);
(C2) for some with , where is the 2uniformly convexity constant of ;
(C3) and if one of the following conditions is satisfied
(a) and and
(b) and .
Then converges strongly to , where is the generalized projection from onto .
Proof.
By the same method as in the proof of Cai and Hu [22], we can show that is closed and convex. Next, we show for all . In fact, is obvious. Suppose for some . Then, for all , we know from Lemma 2.5 that
Since and is inversestrongly monotone, we have
Therefore, from Lemma 2.1 and the assumption that for all and , we obtain that
Substituting (3.3) and (3.4) into (3.2) and using the condition that , we get
Using (3.5) and the convexity of , for each , we obtain
It follows from (3.6) that
So, . Then by induction, for all and hence the sequence generated by (3.1) is well defined. Next, we show that is a convergent sequence in . From , we have
It follows from for all that
From Lemma 2.4, we have
for each and for all . Therefore, the sequence is bounded. Furthermore, since = and = , we have
This implies that is nondecreasing and hence exists. Similarly, by Lemma 2.4, we have, for any positive integer , that
The existence of implies that as . From Lemma 2.2, we have
Hence, is a Cauchy sequence. Therefore, there exists a point such that as
Now, we will show that .

(I)
We first show that . Indeed, taking in (3.12), we have
It follows from Lemma 2.2 that
This implies that
The property of the function implies that
Since , we obtain
It follows from the condition (3.14) and (3.17) that
From Lemma 2.2, we have
Combining (3.15) and (3.20), we have
Since is uniformly normtonorm continuous on any bounded sets, we have
On the other hand, noticing
Since is uniformly normtonorm continuous on any bounded sets, we have
Using (3.15), (3.20), and (3.24) that
Taking the constant , we have, from Lemma 2.6, that there exists a continuous strictly increasing convex function satisfying the inequality (2.7) and .
Case 1.
Assume that (a) holds. Applying (2.7) and (3.5), we can calculate
This implies that
We observe that
It follows from (3.15), (3.22), (3.23) and (3.25) that
From and (3.27), we get
By the property of function , we obtain that
Since is uniformly normtonorm continuous on any bounded sets, we have
From (3.15) and (3.32), we have
Noticing that
for all . By the uniformly continuity of , (3.16) and (3.33), we obtain
Thus
From the closeness of , we get . Therefore . In the same manner, we can apply the condition to conclude that
Again, by (C2) and (3.26), we have
It follows from (3.29) and that
Since , we have
From Lemmas 2.4, 2.5, and (3.4), we have
It follows from (3.40) that
Lemma 2.2 implies that
Since is uniformly normtonorm continuous on any bounded sets, we have
Combining (3.37) and (3.43), we also obtain
Moreover
By (3.43), (3.15), we have
This implies that
Noticing that
for all . Since is uniformly continuous, we can show that . From the closeness of , we get . Therefore . Hence .
Case 2.
Assume that (b) holds. Using the inequalities (2.7) and (3.5), we obtain
This implies that
It follows from (3.21), (3.24) and the condition that
By the property of function , we obtain that
Since is uniformly normtonorm continuous on any bounded sets, we have
On the other hand, we can calculate
Observe that
It follows from (3.53) and (3.54) that
Applying and (3.57) and the fact that is bounded to (3.55), we obtain
From Lemma 2.2, one obtains
We observe that
This together with (3.25) and (3.59), we obtain
Noticing that
for all . By the uniformly continuity of , (3.16) and (3.61), we obtain
Thus
From the closeness of , we get . Therefore . By the same proof as in Case 1, we obtain that
Hence as for each and
Combining (3.54), (3.61), and (3.65), we also have
Moreover
By (3.43), (3.15), we have
This implies that
Noticing that
for all . Since is uniformly continuous, we can show that . From the closeness of , we get . Therefore . Hence .

(II)
We next show that .
Let be an operator defined by:
By Lemma 2.7, is maximal monotone and . Let , since , we have . From , we get
Since is inversestrong monotone, we have
On other hand, from and Lemma 2.3, we have , and hence
Because is constricted, it holds from (3.74) and (3.75) that
for all . By taking the limit as in (3.76) and from (3.43) and (3.44), we have as . By the maximality of we obtain and hence . Hence we conclude that
Finally, we show that . Indeed, taking the limit as in (3.9), we obtain
and hence by Lemma 2.3. This complete the proof.
Remark 3.2.
Theorem 3.1 improves and extends main results of Iiduka and Takahashi [15], Xu and Ori [19], Qin et al. [21], and Cai and Hu [22] because it can be applied to solving the problem of finding the common element of the set of common fixed points of two families of relatively weak quasinonexpansive mappings and the set of solutions of the variational inequality for an inversestrongly monotone operator.
Strong convergence theorem for approximating a common fixed point of two finite families of closed relatively weak quasinonexpansive mappings in Banach spaces may not require that is 2uniformly convex. In fact, we have the following theorem.
Corollary 3.3.
Let be a nonempty, closed, and convex subset of a uniformly convex and uniformly smooth Banach space . Let and be two finite families of closed relatively weak quasinonexpansive mappings from into itself with , where . Assume that and are uniformly continuous for all . Let be a sequence generated by the following algolithm:
where , and is the normalized duality mapping on . Assume that , and are the sequences in satisfying the following restrictions:
(C1) ;
(C2) and if one of the following conditions is satisfied
(a) and and
(b) and .
Then converges strongly to , where is the generalized projection from onto .
Proof.
Put in Theorem 3.1. Then, we get that . Thus, the method of the proof of Theorem 3.1 gives the required assertion without the requirement that is 2uniformly convex.
Remark 3.4.
Corollary 3.3 improves Theorem 3.1 of Cai and Hu [22] from a finite family of of relatively weak quasinonexpansive mappings to two finite families of relatively weak quasinonexpansive mappings.
If , a Hilbert space, then is uniformly convex (we can choose ) and uniformly smooth real Banach space and closed relatively weak quasinonexpansive map reduces to closed weak quasinonexpansive map. Furthermore, , identity operator on and , projection mapping from into . Thus, the following corollaries hold.
Corollary 3.5.
Let be a nonempty, closed and convex subset of a Hilbert space . Let and be two finite families of closed weak quasinonexpansive mappings from into itself with , where with for all and . Assume that and are uniformly continuous for all . Let be a sequence generated by the following algorithm:
where , and is the normalized duality mapping on . Assume that , , , , and are the sequences in satisfying the restrictions:
(C1) ;
(C2) for some with , where is the 2uniformly convexity constant of ;
(C3) and if one of the following conditions is satisfied
(a) and and
(b) and .
Then converges strongly to , where is the metric projection from onto .
Let be a nonempty closed convex cone in , and let be an operator from into . We define its polar in to be the set
Then an element in is called a solution of the complementarity problem if
The set of all solutions of the complementarity problem is denoted by . Several problem arising in different fields, such as mathematical programming, game theory, mechanics, and geometry, are to find solutions of the complementarity problems.
Theorem 3.6.
Let be a nonempty, closed and convex subset of a 2uniformly convex and uniformly smooth Banach space , let be an inversestrongly monotone mapping of into with for all and . Let and be two finite families of closed relatively weak quasinonexpansive mappings from into itself with , where . Assume that and are uniformly continuous for all . Let be a sequence generated by the following algorithm:
where = = , and is the normalized duality mapping on . Assume that , and are the sequences in satisfying the restrictions:
(C1);
(C2) for some with , where is the 2uniformly convexity constant of ;
(C3) and if one of the following conditions is satisfied
(a) and and
(b) and .
Then converges strongly to , where is the generalized projection from onto .
Proof.
From [25, Lemma ], we have . From Theorem 3.1, we can obtain the desired conclusion easily.