Theorem 3.1.

Let be a nonempty closed convex subset of a Hilbert space and let be a sequence of -strict pseudocontractions mappings on into itself with . Assume that Let be a bifunction satisfying (A1), (A2), (A3), (A4), and . Let and be sequence generated by and

Assume that for all , where is a small enough constant, and is a sequence in with Let for any bounded subset of and let be a mapping of into itself defined by and suppose that Then the sequences and converge weakly to an element of .

Proof.

Pick . Then from the definition of in Lemma 2.4, we have , and therefore . It follows from (3.1) that

Since for all , we get that is, the sequence is decreasing. Hence exists. In particular, is bounded. Since is firmly nonexpensive, is also bounded. Also (3.2) implies that

Taking the limit as yields that

Since is bounded, it follows that

We apply Lemma 2.2 to get

Next, we claim that . Indeed, let be an arbitrary element of . Then as above

and hence

Therefore, from (3.2), we have

and hence

So, from the existence of , we have

Next, we claim that . since is bounded and is reflexive, is nonempty. Let be an arbitrary element. Then a subsequence of converges weakly to . Hence, from (3.11) we know that As , we obtain that . Let us show . Since , we have

By (A2), we have

and hence

From (A4), we have

Then, for and , from (A1), and (A4), we also have

Taking and using (A3), we get

and hence . Since is a strict pseudocontraction mapping, by Lemma 2.1() we know that the mapping is demiclosed at zero. Note that and . Thus, . Consequently, we deduce that . Since is an arbitrary element, we conclude that .

To see that and are actually weakly convergent, we take Since exist for every , by (2.2), we have

Hence and proof is completed.