#### Abstract

We propose synchronal algorithm and cyclic algorithm based on the general iterative method for solving a hierarchical fixed point problem. Under suitable parameters, the iterative sequence converges strongly to a common fixed point of nonexpansive mappings and also the unique solution of a variational inequality. The results presented in this paper improve and extend the corresponding results reported recently by some authors. Furthermore, a numerical example is given to demonstrate the effectiveness of our iterative schemes.

#### 1. Introduction

Let be a real Hilbert space with an inner product and its induced norm . Let be a nonempty, closed, and convex subset of .

Let be a nonlinear mapping; we denote the set of fixed points of by (i.e., ). A mapping is called -Lipschitzian continuous if there exists a constant such that In particular, is said to be a nonexpansive mapping if . A mapping is called -strongly monotone on , if there exists a constant such that

A variational inequality (short for VI) is formulated as finding a point such that

If is a monotone operator, then VI (3) is known as a monotone variational inequality. If the set is replaced by the set of of fixed points of a mapping , then the VI (3) is called a hierarchical variational inequality problem.

Many practical problems in applied sciences such as signal processing [1], beamforming [2], and power control [3] are formulated as the monotone variational inequality with a fixed point constrained. In recent years, several authors paid attention toward this kind of problem. Some methods have been proposed to solve the hierarchical fixed point problems and variational inequalities; see for instance [4–10] and the references therein.

In 2010, Tian [11] proposed a general iterative method and revealed the inner contact of the Yamada’s algorithm [12] and viscosity iterative algorithm; then he obtained the following result in a real Hilbert space.

Theorem 1. *Let be generated by algorithm with the sequence of parameters satisfying conditions (C1)–(C3):*(C1)*, *(C2)*, *(C3)* either or .**Then converges strongly to a fixed point of which solves the variational inequality
*

Recently, Yao et al. [10] investigated an iterative method for solving a hierarchical fixed point problem by where , are nonexpansive mapping with and is a contraction; the sequence converges strongly to the unique solution of the variational inequality

Very recently, on this basis, Wang and Xu [8] introduced a new modified iterative method for solving a hierarchical fixed point problem. To be more precise, they proposed the following algorithm: where are nonexpansive mappings with , is a Lipschitzian mapping, and is a Lipschitzian and strongly monotone operator. They proved the sequence generated by the above algorithm converges strongly to the unique solution of the variational inequality

On the other hand, Tian and Di [13] established a synchronal algorithm and a cyclic algorithm for fixed point problems and variational inequalities. In 2012, Ceng et al. [4] proposed an iterative method to solve a special form of VI (3), where the constraint set is the set of common fixed points of nonexpansive mappings .

Motivated and inspired by the above works, in this paper, we combine the hybrid steepest descent algorithm and hierarchical variational inequalities to propose a synchronal algorithm and a cyclic algorithm involving finite family of nonexpansive mappings. Under certain assumptions, we will prove that the sequences converge strongly. Further an example will be given to demonstrate the effectiveness of our iterative schemes.

#### 2. Preliminaries

Recall that given a nonempty, closed and convex subset of a real Hilbert space , for any , there exists a unique nearest point in , denoted by , such that for all . Such a is called the metric (or the nearest point) projection of onto . As we all know, if and only if there holds the relation

In the sequel, we will make use of the following lemmas in a real Hilbert space .

Lemma 2. *Let be a real Hilbert space; the following inequalities hold:*(i)*, , *(ii)*, , .*

Lemma 3 (see [13]). *Let be a -Lipschitzian and -strongly monotone operator on a Hilbert space with , , , and . Then is a contraction with contractive coefficient and . *

Lemma 4 (see [5]). *Let be an -Lipschitz mapping with coefficient and a -Lipschitzian continuous operator and -strongly monotone operator with , . Then for ,
**
That is, is strongly monotone with coefficient .*

Lemma 5 (see [9]). *Assume that is a sequence of nonnegative real numbers such that
**
where is a sequence in and is a sequence such that *(i)*,*(ii)* or .**Then, .*

Lemma 6 (see [14]). *Let be a real Hilbert, and let be all nonexpansive mappings with . Let , where such that . Then is a nonexpansive mapping with . *

Lemma 7 (see [13]). *Let be a Hilbert space, and let be a nonempty closed convex subset of and a nonexpansive mapping with . If is a sequence in weakly converging to and if converges strongly to , then .*

We adopt the following notations:(1) stands for the weak convergence of to ,(2) stands for the strong convergence of to .

#### 3. Synchronal Algorithm

Throughout the rest of this paper, we always assume that is an -Lipschitzian mapping with coefficient and is a -Lipschitzian continuous operator and -strongly monotone with , . Let be an integer. Let, for each , be a nonexpansive mapping and also nonexpansive. Assume that and .

Define a mapping . Since is nonexpansive, it is easy to get that is also nonexpansive. Consider the following mapping on defined by where , with , and . By Lemmas 2 and 3, we obtain

Since , it follows that is a contraction. Therefore, by the Banach contraction principle, has a unique fixed point such that

For simplicity, we will write for provided that no confusion occurs. Next we prove that the sequence converges strongly to a point which solves the variational inequality By the property of the projection, we can get equivalently.

Theorem 8. *Let be a nonempty, closed, and convex subset of a real Hilbert space and an -Lipschitzian mapping with . Let be an integer. Let, for each , be a nonexpansive mapping and let be also nonexpansive. Assume the set . Let be a -Lipschitzian continuous operator and -strongly monotone with , , and . Given , let be the sequence generated by the following algorithm:
**
If and satisfy the following properties: *(i)*, and ,*(ii)*, , *(iii)* and . ** Then, converges strongly to , which solves the variational inequality (16). *

*Proof. *The proof is divided into several steps.*Step 1*. Show first that is bounded.

Take any ; we have

Further we get

By induction, we obtain , . Hence, is bounded, so is . It follows from the Lipschitz continuity of and that , , and are also bounded. From the nonexpansivity of and , it follows that , , and are also bounded.*Step 2*. Show that

By (17), we have

Observe that

Together with (21) and (22), we get
where .

By Lemma 5, we obtain
*Step 3*. Show that

Observe that
From condition (i) and (ii), we obtain
*Step 4*. Show that
where is a unique solution of the variational inequality (16).

Indeed, take a subsequence of such that

Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume and

By Lemma 7, we have . From Lemma 6, we get

Since , it follows that
*Step 5*. Show that

Denote , then . From (17), we have

This implies that
where , . Put , . It is easy to see that . Hence by Lemma 5, the sequence converges strongly to .

*Remark 9. *Let in Theorem 8; we can get Theorem 3.1 of [8].

*Remark 10. *Let , , , and , be a contraction in Theorem 8; it is easy to get the theorem of [10].

#### 4. Cyclic Algorithm

In this section, we consider the cyclic algorithm of nonexpansive mappings . Similarly, we can get that the mapping on defined by is a contraction, where with taking values in .

Theorem 11. *Let be a nonempty, closed, and convex subset of a real Hilbert space , and let be an -Lipschitzian mapping with . Let be an integer. Let, for each , be a nonexpansive mapping and let be also nonexpansive. Assume the set . Let be a -Lipschitzian continuous operator and -strongly monotone with , , , and . Given , let be the sequence generated by the following algorithm:
**
If and satisfy the following properties:*(i)*, and , *(ii)*, ,*(iii)* and .** Assume in addition that
**
Then, converges strongly to , which solves the variational inequality (16). *

*Proof. *The proof is divided into several steps.*Step 1*. Show first that is bounded.

The proof of Step 1 is similar to that of Theorem 8.*Step 2*. Show that

By (37), we have

Observe that

Together with (40) and (41), we have
where .

By Lemma 5, we get
*Step 3.* Show that

Observe that

From conditions (i) and (ii) of Theorem 11, we obtain

Recursively,

Since every is nonexpansive, it is easy to get

Similarly, we obtain

Thus we get

Since
we obtain (44).*Step 4*. Show that
where is a unique solution of the variational inequality (16).

Indeed, take a subsequence of such that

Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume and

Notice that, for each , is some permutation of the mappings . Since are finite, all the finite permutations are ; there must be some permutation that appears infinite times. Without loss of generality, we can assume this permutation is . We obtain

Obviously, is nonexpansive. By Lemma 7, we have . Further by the assumption in Theorem 11, we get

Since , it follows that
*Step 5*. Show that

Denote ; then . From (37), we have

This implies that
where , . Put , . It is easy to see that . Hence, by Lemma 6, the sequence converges strongly to .

#### 5. Numerical Result

In this section, we consider the following simple example to demonstrate the effectiveness, realization, and convergence of the algorithms in Theorems 8 and 11.

*Example 12. *Let , . Define , , . Take with Lipschitz constant and strongly monotone constant , , , with Lipschitz coefficient . Give the parameters ; for every ; fix and . Then by Theorems 8 and 11, respectively, the sequence is generated by
As , we have .

Let , ; then we have , if is odd and if is even. Put ; then (61) is equivalent to Using the same method to treat (62), we can get similar equation as the above formula.

Now we turn to numerical simulation using the algorithms (17) and (37), respectively. Take the initial guess ; using software Matlab R2012, we obtain the numerical experiment results in Tables 1 and 2.

From the computer programming point of view, the algorithms are easier to implement in this paper.

#### Acknowledgments

The authors would like to thank the referee for valuable suggestions to improve the manuscript and the Fundamental Research Funds for the Central Universities (Grant 3122013k004).