Theorem 2.1.

Let be a nonempty closed convex subset of a Hilbert space and a uniformly continuous asymptotically -strict pseudocontractive mapping in the intermediate sense with sequence such that . Let be a sequence in generated by the following Ishikawa iterative process:

where and are sequences in . Assume that the following restrictions are satisfied:

(i) and ,

(ii) for some and .

Then the sequence given by (2.1) converges weakly to an element of .

Proof.

Let . From (1.13) and Lemma 1.4, we see that

Without loss of generality, we may assume that for all . Since

it follows from Lemma 1.6 that

By (2.2) and (2.4), we obtain that

where . It follows from (2.5) and that

From the condition (ii) and , we see that there exists such that

By (2.6), we have

In view of Lemma 1.1 and the condition (i), we obtain that exists. For any , it is easy to see from (2.6) and (2.7) that

which implies that

Note that

From (2.10), we have

Since is uniformly continuous, we obtain from (2.10), (2.12) and Lemma 1.7 that

By the boundedness of , there exist a subsequence of such that . Observe that is uniformly continuous and as , for any we have as . From Lemma 1.8, we see that .

To complete the proof, it suffices to show that consists of exactly one point, namely, . Suppose there exists another subsequence of such that converges weakly to some and . As in the case of , we can also see that . It follows that and exist. Since satisfies the Opial condition, we have

which is a contradiction. We see and hence is a singleton. Thus, converges weakly to by Lemma 1.2.

Corollary 2.2.

Let be a nonempty closed convex subset of a Hilbert space and a uniformly continuous asymptotically -strict pseudocontractive mapping with sequence such that . Let be a sequence in generated by the following Ishikawa iterative process:

where and are sequences in . Assume that the following restrictions are satisfied:

(i),

(ii) for some and .

Then the sequence given by (2.15) converges weakly to an element of .

Next, we modify Ishikawa iterative process to get a strong convergence theorem.

Theorem 2.3.

Let be a nonempty closed convex subset of a Hilbert space and a uniformly continuous asymptotically -strict pseudocontractive mapping in the intermediate sense with sequence such that and bounded. Let and are sequences in . Let be a sequence in generated by the modified Ishikawa iterative process:

where , , and for each . Assume that the control sequences and are chosen such that for some and . Then the sequence given by (2.16) converges strongly to an element of .

Proof.

We break the proof into six steps.

Step 1 ( is closed and convex for each ).

It is obvious that is closed and convex and is closed for each . Note that the defining inequality in is equivalent to the inequality

it is easy to see that is convex for each . Hence, is closed and convex for each .

Step 2 ( for each ).

Let . Following (2.6), (2.7) and the algorithm (2.16), we have

where , , and for each . Hence for each .

Next, we show that for each . We prove this by induction. For , we have . Assume that for some . Since is the projection of onto , we have

By the induction consumption, we know that . In particular, for any we have

This implies that . That is, . By the principle of mathematical induction, we get and hence for all . This means that the iteration algorithm (2.16) is well defined.

Step 3 ( exists and is bounded).

In view of (2.16), we see that and . It follows that

for each . We, therefore, obtain that the sequence is nondecreasing. Noticing that and , we have

This shows that the sequence is bounded. Therefore, the limit of exists and is bounded.

Step 4 ().

Observe that and which imply

Using Lemma 1.4, we obtain

Hence, we obtain that as .

Step 5 ( as ).

In view of , we have

On the other hand, we see that

Combing (2.25) and (2.26) and noting , we obtain that

From the assumption and (2.7), we see that there exists such that

For any , it follows from the definition of and (2.27) that

Noting that as and Step 4, we obtain that

It follows from Step 4, (2.30) and Lemma 1.7 that as .

Step 6 ( as , where ).

Since is reflexive and is bounded, we get that is nonempty. First, we show that is a singleton. Assume that is subsequence of such that . Observe that is uniformly continuous and as , for any we have as . From Lemma 1.8, we see that .

Since , we obtain that

for each . Observe that as . By the weak lower semicontinuity of norm, we have

This implies that

Hence by the uniqueness of the nearest point projection of onto . Since is an arbitrary weakly convergent subsequence, it follows that and hence . It is easy to see as (2.34) that . Since has the Kadec-Klee property, we obtain that , that is, as . This completes the proof.