In order to prove our main results, we need the following lemmas.

Lemma 2.1 (see [13]).

Let be bounded sequences in a Banach space and be a sequence in which satisfies the following condition: . Suppose that for all and . Then, .

Lemma 2.2 (see Xu [14]).

Assume that is a sequence of nonnegative real numbers such that , where is a sequence in (0, 1) and is a sequence in such that

(i),

(ii) or .

Then .

Lemma 2.3 (see [15] demiclosedness principle).

Let be a nonempty closed convex subset of a reflexive Banach space which satisfies Opial's condition, and suppose is nonexpansive. Then the mapping is demiclosed at zero, that is, , implies .

Lemma 2.4 (see [16, Lemmas 3.1, 3.3]).

Let be real smooth and strictly convex Banach space, and be a nonempty closed convex subset of which is also a sunny nonexpansive retraction of . Assume that is a nonexpansive mapping and is a sunny nonexpansive retraction of onto , then .

Lemma 2.5 (see [17, Lemma 2.2]).

Let be a nonempty convex subset of a real -uniformly smooth Banach space and be a -strict pseudocontraction. For , we define . Then, as , , is nonexpansive such that .

Lemma 2.6 (see [12, Remark 2.6]).

When is non-self-mapping, the Lemma 2.5 also holds.

Lemma 2.7 (see [12, Lemma 2.8]).

Assume that is a strongly positive linear bounded operator on a smooth Banach space with coefficient and . Then,

Lemma 2.8 (see [18, Lemma 2.3]).

Let be an MKC on a convex subset of a Banach space . Then for each , there exists such that

Lemma 2.9.

Let be a closed convex subset of a reflexive Banach space which admits a weakly sequentially continuous duality mapping from to . Let be a nonexpansive mapping with and be a MKC, is strongly positive linear bounded operator with coefficient . Assume that . Then the sequence define by converges strongly as to a fixed point of which solves the variational inequality:

Proof.

The definition of is well definition. Indeed, from the definition of MKC, we can see MKC is also a nonexpansive mapping. Consider a mapping on defined by

It is easy to see that is a contraction. Indeed, by Lemma 2.8, we have

Hence, has a unique fixed point, denoted by , which uniquely solves the fixed point equation

We next show the uniqueness of a solution of the variational inequality (2.3). Suppose both and are solutions to (2.3), not lost generality, we may assume there is a number such that . Then by Lemma 2.8, there is a number such that . From (2.3), we know

Adding up (2.7) gets

Noticing that

Therefore and the uniqueness is proved. Below, we use to denote the unique solution of (2.3).

We observe that is bounded. Indeed, we may assume, with no loss of generality, , for all , fixed , for each .

Case 1 ().

In this case, we can see easily that is bounded.

Case 2 ().

In this case, by Lemmas 2.7 and 2.8, there is a number such that

therefore, . This implies the is bounded.

To prove that as .

Since is bounded and is reflexive, there exists a subsequence of such that . By . We have , as . Since satisfies Opial's condition, it follows from Lemma 2.3 that . We claim

By contradiction, there is a number and a subsequence of such that . From Lemma 2.8, there is a number such that , we write

to derive that

It follows that

Therefore,

Using that the duality map is single valued and weakly sequentially continuous from to , by (2.15), we get that . It is a contradiction. Hence, we have .

We next prove that solves the variational inequality (2.3). Since

we derive that

Notice

It follows that, for ,

Now replacing in (2.19) with and letting , noticing for , we obtain . That is, is a solution of (2.3); Hence by uniqueness. In a summary, we have shown that each cluster point of (at ) equals , therefore, as .

Lemma 2.10 . (see, e.g., Mitrinović [19, page 63]).

Let . Then the following inequality holds:

for arbitrary positive real numbers , .

Lemma 2.11.

Let be a -uniformly smooth Banach space which admits a weakly sequentially continuous duality mapping from to and be a nonempty convex subset of . Assume that is a countable family of -strict pseudocontraction for some and such that . Assume that is a positive sequence such that . Then is a -strict pseudocontraction with and .

Proof.

Let

and . Then, is a -strict pseudocontraction with . Indeed, we can firstly see the case of .

which shows that is a -strict pseudocontraction with . By the same way, our proof method easily carries over to the general finite case.

Next, we prove the infinite case. From the definition of -strict pseudocontraction, we know

Hence, we can get

Taking , from (2.24), we have

Consquently, for all , if , and , then strongly converges. Let

we have

Hence,

So, we get is -strict pseudocontraction.

Finally, we show . Suppose that , it is sufficient to show that . Indeed, for , we have

where . Hence, for each , this means that .