Now we are in a position to prove the main theorems of this paper.

Theorem 3.1.

Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , let be a continuous mapping and, let be a relatively nonexpansive mapping such that . Assume that are sequences in and is a sequence in such that and . Define a sequence in by the following algorithm:

where is assumed to exist for each , If is uniformly continuous and , then converges strongly to , which is a solution of the (1.1).

Proof.

First of all, let us show that and are closed and convex for each . Indeed, from the definition of and , it is obvious that is closed and is closed and convex for each . We claim that is convex. For any and any , put . It is sufficient to show that . Note that the inequality

is equivalent to the one

Observe that there hold the following:

and . Thus, we have

This implies that . So, is convex. Next let us show that for all . Indeed, we have for all

So for all . Next let us show that

We prove this by induction. For , we have . Assume that . Since is the projection of onto , by Lemma 2.3, we have

As by the induction assumption, the last inequality holds, in particular, for all . This together with the definition of implies that . Hence (3.7) holds for all . This implies that is well defined.

On the other hand, it follows from the definition of that . Since , we have

Thus is nondecreasing. Also from and Lemma 2.4, it follows that

for each for each . Consequently, is bounded. Moreover, according to the inequality

we conclude that is bounded and so is . Indeed, since is relatively nonexpansive, we derive for each

and hence is bounded. Again from we know that is also bounded.

On account of the boundedness and nondecreasing property of we deduce that exists. From Lemma 2.4, we derive

for all . This implies that . So it follows from Lemma 2.2 that . Since , from the definition of , we also have

Observe that

On the other hand, since from (3.1) we have for each

utilizing Lemma 2.3 we obtain . Thus, in terms of Lemmas 2.4 and 2.6 we conclude that

Since and , we obtain . Thus by Lemma 2.2 we have . From it follows that is bounded. At the same time, observe that

and hence

From and the boundedness of and , we derive . Note that is uniformly continuous. Hence by virtue of . Since

it is known that is bounded. Consequently, from (3.15), and it follows that

Further, it follows from (3.14), and that

Utilizing Lemma 2.2, we obtain

Since is uniformly norm-to-norm continuous on bounded subsets of , we have

Furthermore, we have

It follows from and that

Noticing that

we have

From (3.24) and , we obtain

Since is also uniformly norm-to-norm continuous on bounded subsets of , we obtain

Observe that

Since is uniformly continuous, it follows from (3.26), (3.30) and that .

Finally, let us show that converges strongly to , which is a solution of the (1.1). Indeed, assume that is a subsequence of such that . Then . Next let us show that and convergence is strong. Put . From and , we have . Now from weakly lower semicontinuity of the norm, we derive

It follows from the definition of that and hence

So we have . Utilizing the Kadec-Klee property of , we conclude that converges strongly to . Since is an arbitrarily weakly convergent subsequence of , we know that converges strongly to . Now observe that from (3.1) we have for each

Since is uniformly norm-to-norm continuous on bounded subsets of , from we infer that . Noticing that and is a continuous mapping, we obtain that and . Therefore, from it follows that

that is,

Letting we conclude from (3.34) that

and hence

This shows that is a solution of the (1.1). This completes the proof.

Corollary 3.2 ([11, Theorem?2.1]).

Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , and let be a relatively nonexpansive mapping such that . Assume that and are sequences in such that and . Define a sequence in by the following algorithm:

where is the single-valued duality mapping on . If is uniformly continuous, then converges strongly to .

Proof.

In Theorem 3.1, we know from (3.1) and Lemma 2.3 that

is equivalent to . Now, put for all . Then we have

for all . Thus algorithm (3.1) reduces to algorithm (3.39). By Theorem 3.1 we obtain the desired result.

Theorem 3.3.

Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , let be a continuous mapping, and let be a relatively nonexpansive mapping such that . Assume that satisfies and satisfies . Define a sequence in by the following algorithm:

where is assumed to exist for each , If is uniformly continuous and , then converges strongly to , which is a solution of the (1.1).

Proof.

We only derive the difference. First, let us show that is closed and convex for each . From the definition of , it is obvious that is closed for each . We prove that is convex. Similarly to the proof of Theorem 3.1, since

is equivalent to

we know that is convex. Next, let us show that for each . Indeed, we have for each

So for all and . Similarly to the proof of Theorem 3.1, we also obtain for all . Consequently, for all . Therefore, the sequence generated by (3.42) is well defined. As in the proof of Theorem 3.1, we can obtain . Since , from the definition of , we also have

As in the proof of Theorem 3.1, we can deduce from and that and hence by Lemma 2.2. Further, it follows from and the boundedness of and that

Since , from the definition of , we also have

It follows from (3.47) and that

Utilizing Lemma 2.2, we have

Since is uniformly norm-to-norm continuous on bounded subsets of we have

Note that

Therefore, from we have

Since is also uniformly norm-to-norm continuous on bounded subsets of , we obtain

It follows that

Since is uniformly continuous, it follows from (3.50) and (3.54) that .

Finally, let us show that converges strongly to , which is a solution of the (1.1). Indeed, assume that is a subsequence of such that . Then . Next let us show that and convergence is strong. Put . From and , we have . Now from weakly lower semicontinuity of the norm, we derive

It follows from the definition of that and hence . So, we have . Utilizing the Kadec-Klee property of , we conclude that converges strongly to . Since is an arbitrarily weakly convergent subsequence of , we know that converges strongly to . Now observe that from (3.1), we have for each

Since is uniformly norm-to-norm continuous on bounded subsets of , from we infer that . Noticing that and is a continuous mapping, we obtain that and . Observe that

It follows from that

Letting we conclude from (3.34) that

This shows that is a solution of the (1.1). This completes the proof.

Corollary 3.4 ([11, Theorem?2.2]).

Let be a uniformly convex and uniformly smooth Banach space, let be a nonempty closed convex subset of , and let be a relatively nonexpansive mapping. Assume that is a sequence in such that . Define a sequence in by the following algorithm:

where is the single-valued duality mapping on . If is nonempty, then converges strongly to .

Proof.

In Theorem 3.3, we know from (3.42) and Lemma 2.3 that

is equivalent to . Now, put for all . Then we have

for all . Thus algorithm (3.42) reduces to algorithm (3.61). Thus under the lack of the uniform continuity of it follows from (3.55) that . By the careful analysis of the proof of Theorem 3.3, we can obtain the desired result.