1. Introduction

Iteration methods for nonexpansive mappings have recently been applied to solve convex minimization problems; see, for example, [1–4] and the references therein. A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space : where is the fixed points set of a nonexpansive mapping on , is a given point in , and is strongly positive operator, that is, there exists a constant with the property

Recall that is nonexpansive if for all . Throughout the rest of this paper, we denote by the fixed points set of and assume that is nonempty. It is well known that is closed convex (cf. [5]). In [4] (see also [2]), it is proved that the sequence defined by the iteration method below, with the initial guess chosen arbitrarily, converges strongly to the unique solution of minimization problem (1.1) provided the sequence satisfies certain conditions.

On the other hand, Moudafi [6] introduced the viscosity approximation method for nonexpansive mappings (see [7] for further developments in both Hilbert and Banach spaces). Let be a contraction on . Starting with an arbitrary initial guess , define a sequence recursively by where is a sequence in . It is proved [6, 7] that under certain appropriate conditions imposed on , the sequence generated by (1.4) converges strongly to the unique solution in of the variational inequality

Recently (2006), Marino and Xu [2] combine the iteration method (1.3) with the viscosity approximation method (1.4) and consider the following general iteration method: they have proved that if the sequence of parameters satisfies appropriate conditions, then the sequence generated by (1.6) converges strongly to the unique solution of the variational inequality which is the optimality condition for the minimization problem where is a potential function for (i.e., for ).

The purpose of this paper is to present a general viscosity iteration process which is defined by and to study the convergence of , where is a nonexpansive mapping and is a strongly positive linear operator, if , satisfy appropriate conditions, then iteration sequence converges strongly to the unique solution of variational inequality (1.7). Meanwhile, an approximate iteration algorithm is presented which is used to calculate the fixed point of nonexpansive mapping and solution of variational inequality; the convergence rate estimate is also given. The results presented in this paper extend, generalize and improve the results of Xu [7], Marino and Xu [2], and some others.

2. Preliminaries

This section collects some lemmas which will be used in the proofs for the main results in the next section. Some of them are known; others are not hard to derive.

Lemma 2.1 (see [3]). 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 2.2 (see [5]). Let be a Hilbert space, a closed convex subset of , and a nonexpansive mapping with nonempty fixed points set . If is a sequence in weakly converging to and if converges strongly to , then .

The following lemma is not hard to prove.

Lemma 2.3. Let be a Hilbert space, a closed convex subset of , a contraction with coefficient , and a strongly positive linear bounded operator with coefficient . Then, for , That is, is strongly monotone with coefficient .

Recall the metric (nearest point) projection from a real Hilbert space to a closed convex subset of is defined as follows: given , is the only point in with the property is characterized as follows.

Lemma 2.4. Let be a closed convex subset of a real Hilbert space . Given that and . Then if and only if there holds the inequality

Lemma 2.5. Assume that is a strongly positive linear-bounded operator on a Hilbert space with coefficient and . Then .

Proof. Recall that a standard result in functional analysis is that if is linear bounded self-adjoint operator on , then Now for with , we see that (i.e., is positive). It follows that

The following lemma is also not hard to prove by induction.

Lemma 2.6. Assume that is a sequence of nonnegative real numbers such that where is a nonnegative constant and are sequences in such that(i); (ii). Then is bounded.

Notation. We use for strong convergence and  for weak convergence.

3. A general Iteration Algorithm with Bounded Linear Operator

Let be a real Hilbert space, be a bounded linear operator on , and be a nonexpansive mapping on . Assume that the fixed point set of is nonempty. Since is closed convex, the nearest point projection from onto is well defined.

Throughout the rest of this paper, we always assume that is strongly positive, that is, there exists a constant such that (Note: is throughout reserved to be the constant such that (3.1) holds.)

Recall also that a contraction on is a self-mapping of such that where is a constant which is called contractive coefficient of .

For given contraction with contractive coefficient , and such that and , consider a mapping on defined by Assume that it is not hard to see that is a contraction for sufficiently small , indeed, by Lemma 2.5 we have Hence, has a unique fixed point, denoted by , which uniquely solves the fixed point equation: Note that indeed depends on as well, but we will suppress this dependence of on for simplicity of notation throughout the rest of this paper. We will also always use to mean a number in .

The next proposition summarizes the basic properties of .

Proposition 3.1. Let be defined via (3.6).(i) is bounded for .(ii).(iii) defines a continuous surface for into .

Proof. Observe, for , that by Lemma 2.5.
To show (i) pick . We then have It follows that Hence is bounded.
(ii) Since the boundedness of implies that of and , and observe that we have
To prove (iii) take and calculate which implies that as . Note that it is obvious that This completes the proof of Proposition 3.1.

Our first main result below shows that converges strongly as to a fixed point of which solves some variational inequality.

Theorem 3.2. One has that converges strongly as to a fixed point of which solves the variational inequality: Equivalently, One has , where is the nearest point projection from onto .

Proof. We first shows the uniqueness of a solution of the variational inequality (3.15), which is indeed a consequence of the strong monotonicity of . Suppose and both are solutions to (3.15), then Adding up (3.16) gets The strong monotonicity of implies that and the uniqueness is proved. Below we use to denote the unique solution of (3.15).
To prove that converges strongly to , we write, for a given , to derive that It follows that which leads to Observe that condition (3.4) implies as . Since is bounded as , then there exists real sequences in such that and converges weakly to a point . Using Proposition 3.1 and Lemma 2.2, we see that , therefore by (3.21), we see . We next prove that solves the variational inequality (3.15). Since we derive that so that It follows that, for , since is monotone (i.e., for ). This is due to the nonexpansivity of . Now replacing in (3.26) with and letting , we, noticing that for , obtain That is, is a solution of (3.15), hence by uniqueness. In a summary, we have shown that each cluster point of equals . Therefore, as .
The variational inequality (3.15) can be rewritten as This, by Lemma 2.4, is equivalent to the fixed point equation This complete the proof.

Taking in Theorem 3.2, we get

Corollary 3.3 (see [7]). One has that converges strongly as to a fixed point of which solves the variational inequality: Equivalently, One has , where is the nearest point projection from onto .

Next we study a general iteration method as follows. The initial guess is selected in arbitrarily, and the th iterate is recursively defined by where are sequences satisfying following conditions: ; ;  either or ; .Below is the second main result of this paper.

Theorem 3.4. Let be general by Algorithm (3.31) with the sequences of parameters satisfying conditions –. Then converges strongly to that is obtained in Theorem 3.2.

Proof. Since by condition , we may assume, without loss of generality, that for all .
We now observe that is bounded. Indeed, pick any to obtain where is a constant. By Lemma 2.6 we see that is bounded.
As a result, noticing and , we obtain But the key is to prove that To see this, we calculate Since and condition holds, an application of Lemma 2.1 to (3.36) implies (3.35) which combined with (3.34), in turns, implies Next we show that where is obtained in Theorem 3.2.
To see this, we take a subsequence of such that We may also assume that . Note that in virtue of Lemma 2.2 and (3.38). It follows from the variational inequality (3.15) that So (3.39) holds, thanks to (3.38).
Finally, we prove . To this end, we calculate where That is, Since is bounded, by the conditions of Theorem 3.4, we get and , this together with (3.39) implies that Now applying Lemma 2.1 to (3.44) concludes that . This complete the proof of Theorem 3.4.