In this section, we show a strong convergence of an iterative algorithm based on both viscosity approximation method and extragradient method which solves the problem of finding a common element of the set of solutions of a mixed equilibrium problem, the set of fixed points of a finite family of nonexpansive mappings, and the set of solutions of the variational inequality for a monotone, Lipschitz continuous mapping in a Hilbert space.

Theorem 3.1.

Let be a nonempty closed convex subset of a real Hilbert space . Let be a bifunction from to satisfying (A1)–(A5) and let be a proper lower semicontinuous and convex function. Let be a monotone and -Lipschitz continuous mapping of into . Let be a finite family of nonexpansive mappings of into such that . Let be sequences in with . Let be the -mapping generated by and . Assume that either (B1) or (B2) holds. Let be a contraction of into itself and let , , and be sequences generated by

for every where , , , , , and are sequences of numbers satisfying the conditions:

(C1) and ;

(C2);

(C3);

(C4) and ;

(C5) for all .

Then, , , and converge strongly to .

Proof.

We show that is a contraction of into itself. In fact, there exists such that for all . So, we have

for all Since is complete, there exists a unique element such that .

Put for every Let and let be a sequence of mappings defined as in Lemma 2.1. Then . From , we have

From (2.3), the monotonicity of , and , we have

Further, Since and is -Lipschitz continuous, we have

So, it follows from (C3) that the following inequality holds for , where is a positive integer:

Put It is obvious that Suppose By Lemma 2.5, we know that is nonexpansive and . From (3.3), (3.6) and , we have and

for every . Therefore, is bounded. From (3.3) and (3.6), we also obtain that and are bounded.

From and the monotonicity and the Lipschitz continuity of , we have

Hence, we obtain that is bounded. It follows from the Lipschitz continuity of that , , and are bounded. Since and are nonexpansive, we know that and are also bounded. From the definition of , we get

where is an approximate constant such that

On the other hand, from and , we have

Putting in (3.11) and in (3.12), we have

So, from the monotonicity of , we get

and hence

Without loss of generality, let us assume that there exists a real number such that for all Then,

and hence

where

It follows from (3.9) and (3.7) that

Define a sequence such that

Then, we have

Next we estimate . It follows from the definition of that

where is an approximate constant such that

Since for all and , we have

It follows that

Substituting (3.24) into (3.21) yields that

Hence, we have

From (3.20), (3.26), and (3.18), we have

It follows from (C1)–(C5) that

Hence by Lemma 2.4, we have . Consequently

Since , we have

and thus

It follows from (C1) and (C2) that .

Since for , it follows from (3.3) and (3.6) that

from which it follows that

It follows from (C1)–(C3) and that .

By the same argument as in (3.6), we also have

Combining the above inequality and (3.32), we have

and thus

which implies that .

From we also have . As is -Lipschitz continuous, we have .

For , we have, from Lemma 2.1,

Hence,

By(3.3), (3.6), (3.32), and (3.38), we have

Hence,

It follows from (C1), (C2), and that .

Since

It follows that

Next we show that

where . To show this inequality, we can choose a subsequence of such that

Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume that From , we obtain that . From , we also obtain that . From , we also obtain that . Since and is closed and convex, we obtain .

In order to show that , we first show By we know that

It follows from (A2) that

Hence,

It follows from (A4), (A5), and the weakly lower semicontinuity of , and that

For with and , let Since and , we obtain and hence . So by (A4) and the convexity of , we have

Dividing by , we get

Letting , it follows from (A3) and the weakly lower semicontinuity of that

for all and hence .

Now we show that Put

where is the normal cone to at . We have already mentioned that in this case the mapping is maximal monotone, and if and only if . Let . Then and hence . So, we have for all . On the other hand, from and we have

and hence

Therefore, we have

Hence we obtain as . Since is maximal monotone, we have and hence .

We next show that To see this, we observe that we may assume (by passing to a further subsequence if necessary) for Let be the -mapping generated by and . By Lemma 2.5, we know that is nonexpansive and . it follows from Lemma 2.6 that

Assume Since and , it follows from the Opial condition, (3.42), and (3.56) that

which is a contradiction. Hence, we have . This implies Therefore, we have

Finally, we show that , where .

From Lemma 2.3, we have

and thus

It follows from Lemma 2.2, (3.58), and (3.60) that . From and , we have and . The proof is now complete.