Throughout this section, unless otherwise stated, we assume that is a maximal monotone operator with domain , is a relatively nonexpansive mapping, is an *α*-inverse-strongly monotone mapping and is a bifunction satisfying (A1)–(A4), where is a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space . In this section, we study the following algorithm.

Algorithm 3.1.

where is a sequence in and , are sequences in .

First we investigate the condition under which the Algorithm 3.1 is well defined. Rockafellar [40] proved the following result.

Lemma 3.2 (Rockafellar [40]).

Let be a reflexive, strictly convex, and smooth Banach space and let be a multivalued operator, then there hold the following:

(i) is closed and convex if is maximal monotone such that ;

(ii) is maximal monotone if and only if is monotone with for all .

Utilizing this result, we can show the following lemma.

Lemma 3.3.

Let be a reflexive, strictly convex, and smooth Banach space. If , then the sequence generated by Algorithm 3.1 is well defined.

Proof.

For each , define two sets and as follows:

It is obvious that is closed and are closed convex sets for each . Let us show that is convex. For and , put . It is sufficient to show that . Indeed, observe that

is equivalent to

Note that there hold the following:

Thus we have

This implies that . Therefore, is convex and hence is closed and convex.

On the other hand, let be arbitrarily chosen, then and . From Algorithm 3.1, it follows that

So for all . Now, from Lemma 3.2 it follows that there exists such that and . Since is monotone, it follows that , which implies that and hence . Furthermore, it is clear that , then , and therefore is well defined. Suppose that and is well defined for some . Again by Lemma 3.2, we deduce that such that and , then from the monotonicity of we conclude that , which implies that and hence . It follows from Lemma 2.4 that

which implies that . Consequently, and so . Therefore is well defined, then, by induction, the sequence generated by Algorithm 3.1, is well defined for each integer .

Remark 3.4.

From the above proof, we obtain that

for each integer .

We are now in a position to prove the main theorem.

Theorem 3.5.

Let be a uniformly smooth and uniformly convex Banach space. Let be a sequence in and be sequences in such that

Let . If is uniformly continuous, then the sequence generated by Algorithm 3.1 converges strongly to .

Proof.

First of all, if follows from the definition of that . Since , we have

Thus is nondecreasing. Also from and Lemma 2.3, we have that

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

we conclude that is bounded. Thus, we have that exists. From Lemma 2.3, we derive the following:

for all . This implies that . So it follows from Lemma 2.1 that . Since , from the definition of , we also have

Observe that

At the same time,

Since and , it follows that

and hence that . Further, from , we have , which yields

Then it follows from that . Hence it follows from Lemma 2.1 that . Since from (3.15) we derive

we have

Thus, from , , and , we know that . Consequently from (3.16), , and it follows that

So it follows from (3.15), , and that . Utilizing Lemma 2.1 we deduce that

Furthermore, for arbitrarily fixed, it follows from Proposition 2.8 that

Since is uniformly norm-to-norm continuous on bounded subsets of , it follows from (3.23) that and , which hence yield . Utilizing Lemma 2.1, we get . Observe that

due to (3.23). Since is uniformly norm-to-norm continuous on bounded subsets of , we have that

On the other hand, we have

Noticing that

we have

From (3.26) 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 (3.27), (3.31) and that .

Now let us show that , where

Indeed, since is bounded and is reflexive, we know that . Take arbitrarily, then there exists a subsequence of such that . Hence . Let us show that . Since , we have that . Moreover, since is uniformly norm-to-norm continuous on bounded subsets of and , we obtain

It follows from and the monotonicity of that

for all and . This implies that

for all and . Thus from the maximality of , we infer that . Therefore, . Further, let us show that . Since and , from we obtain that and .

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

From , it follows that

By the definition of , we have

where

Replacing by , we have from (A2) that

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting in the last inequality, from (3.38) and (A4) we have

For , with , and , let . Since and , then and hence . So, from (A1) we have

Dividing by , we have

Letting , from (A3) it follows that

So, . Therefore, we obtain that by the arbitrariness of .

Next, let us show that and .

Indeed, put . From and , we have . Now from weakly lower semicontinuity of the norm, we derive for each

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 arbitrary weakly convergent subsequence of , we know that converges strongly to . This completes the proof.

Theorem 3.5 covers [25, Theorem 3.1] by taking and . Also Theorem 3.5 covers [24, Theorem 2.1] by taking , and .

Theorem 3.6.

Let be a nonempty closed convex subset of a uniformly smooth and uniformly convex Banach space . Let be a maximal monotone operator with domain , be a relatively nonexpansive mapping, be an *α*-inverse-strongly monotone mapping and be a bifunction satisfying (A1)–(A4). Assume that is a sequence in satisfying and that is a sequences in satisfying .

Define a sequence .

Algorithm 3.7.

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

Proof.

For each , define two sets and as follows:

It is obvious that is closed and are closed convex sets for each . Let us show that is convex and so is closed and convex. Similarly to the proof of Lemma 3.3, since

is equivalent to

we know that is convex and so is . Next, let us show that for each . Indeed, utilizing Proposition 2.8, we have, for each ,

So for all and . As in the proof of Lemma 3.3, we can obtain and hence . It follows from Lemma 2.4 that

which implies that . Consequently, and so for all . Therefore, the sequence generated by Algorithm 3.7 is well defined. As in the proof of Theorem 3.5, we can obtain . Since , from the definition of we also have

As in the proof of Theorem 3.5, we can deduce not only from that but also from , and that

Since , from the definition of , we also have

It follows from (3.55) and that

Utilizing Lemma 2.1 we have

Furthermore, for arbitrarily fixed, it follows from Proposition 2.8 that

Since is uniformly norm-to-norm continuous on bounded subsets of , it follows from (3.58) that and , which together with , yield . Utilizing Lemma 2.1, we get . Observe that

due to (3.58). Since is uniformly norm-to-norm continuous on bounded subsets of , we have

Note that

Therefore, from we get

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

It follows that

Since is uniformly continuous, it follows from (3.58) and (3.64) that .

Finally, we prove that . Indeed, for arbitrarily fixed, there exists a subsequence of such that , then . Now let us show that . Since , we have that . Moreover, since is uniformly norm-to-norm continuous on bounded subsets of , and , we obtain that . It follows from and the monotonicity of that for all and . This implies that for all and . Thus from the maximality of , we infer that . Further, let us show that . Since and , from we obtain that and .

Since is uniformly norm-to-norm continuous on bounded subsets of , from we derive . From it follows that

By the definition of , we have

where . Replacing by , we have from (A2) that

Since is convex and lower semicontinuous, it is also weakly lower semicontinuous. Letting in the last inequality, from (3.66) and (A4) we have , for all . For , with , and , let . Since and , then and hence . So, from (A1) we have

Dividing by , we have , for all . Letting , from (A3) it follows that , for all . So, . Therefore, we obtain that by the arbitrariness of .

Next, let us show that and .

Indeed, put . From and , we have . Now from weakly lower semicontinuity of the norm, we derive for each

It follows from the definition of that and hence . So we have . Utilizing the Kadec-Klee property of , we know that . Since is an arbitrary weakly convergent subsequence of , we know that . This completes the proof.

Theorem 3.6 covers [25, Theorem 3.2] by taking and . Also Theorem 3.6 covers [24, Theorem 2.2] by taking and .