Abstract

Two hybrid algorithms for the variational inequalities over the common fixed points set of nonexpansive semigroups are presented. Strong convergence results of these two hybrid algorithms have been obtained in Hilbert spaces. The results improve and extend some corresponding results in the literature.

1. Introduction

Let 𝐻 be a real Hilbert space and 𝐶 a nonempty closed convex subset of 𝐻. Recall that a mapping 𝑇𝐶𝐶 is called nonexpansive if 𝑇𝑥𝑇𝑦𝑥𝑦(1.1) for every 𝑥,𝑦𝐶. A family 𝑆={𝑇(𝜏)0<𝜏<} of mappings from 𝐶 into itself is called a nonexpansive semigroup on 𝐶 if it satisfies the following conditions:(i)𝑇(0)𝑥=𝑥 for all 𝑥𝐶,(ii)𝑇(𝑠+𝑡)=𝑇(𝑠)𝑇(𝑡) for all 𝑠,𝑡0,(iii)𝑇(𝑠)𝑥𝑇(𝑠)𝑦𝑥𝑦 for all 𝑥,𝑦𝐶,𝑠0,(iv)for all 𝑥𝐶,𝑠𝑇(𝑠)𝑥 is continuous.We denote by Fix(𝑆) the set of all common fixed points of 𝑆, that is, Fix(𝑆)=0𝜏<Fix(𝑇(𝜏)). It is known that Fix(𝑆) is closed and convex.

Approximation of fixed points of nonexpansive mappings has been considered extensively by many authors, see, for instance, [118]. Nonlinear ergodic theorem for nonexpansive semigroups have been researched by some authors, see, for example, [1923]. Our main purpose in the present paper is devoted to finding the common fixed points of nonexpansive semigroups.

Let 𝐹𝐶𝐶 be a nonlinear operator. The variational inequality problem is formulated as finding a point 𝑥𝐶 such that VI(𝐹,𝐶)𝐹𝑥,𝜐𝑥0,𝜐𝐶.(1.2) Now it is well known that VI problem is an interesting problem and it covers as diverse disciplines as partial differential equations, optimal control, optimization, mathematical programming, mechanics, and finance. Several numerical methods including the projection and its variant forms have been developed for solving the variational inequalities and related problems, see [2441].

It is clear that the VI(𝐹,𝐶) is equivalent to the fixed point equation 𝑥=𝑃𝐶𝑥𝑥𝜇𝐹,(1.3) where 𝑃𝐶 is the projection of 𝐻 onto the closed convex set 𝐶 and 𝜇>0 is an arbitrarily fixed constant. So, fixed point methods can be implemented to find a solution of the VI(𝐹,𝐶) provided 𝐹 satisfies some conditions and 𝜇>0 is chosen appropriately. The fixed point formulation (1.3) involves the projection 𝑃𝐶, which may not be easy to compute, due to the complexity of the convex set 𝐶. In order to reduce the complexity probably caused by the projection 𝑃𝐶, Yamada [24] (see also [42]) recently introduced a hybrid steepest-descent method for solving the VI(𝐹,𝐶).

Assume that 𝐹 is an 𝜂-strongly monotone and 𝜅-Lipschitzian mapping with 𝜅>0,𝜂>0 on 𝐶. An equally important problem is how to find an approximate solution of the VI(𝐹,𝐶) if any. A great deal of effort has been done in this problem; see [43, 44].

Take a fixed number 𝜇 such that 0<𝜇<2𝜂/𝜅2. Assume that a sequence {𝜆𝑛} of real numbers in (0,1) satisfies the following conditions:(C1)lim𝑛𝜆𝑛=0,(C2)𝑛=0𝜆𝑛=,(C3)lim𝑛(𝜆𝑛𝜆𝑛+1)/𝜆2𝑛=0.

Starting with an arbitrary initial guess 𝑥0𝐻, one can generate a sequence {𝑥𝑛} by the following algorithm: 𝑥𝑛+1=𝑇𝑥𝑛𝜆𝑛+1𝜇𝐹𝑇𝑥𝑛,𝑛0.(1.4) Yamada [24] proved that the sequence {𝑥𝑛} generated by (1.4) converges strongly to the unique solution of the VI(𝐹,𝐶). Xu and Kim [30] proved the strong convergence of {𝑥𝑛} to the unique solution of the VI(𝐹,𝐶) if {𝜆𝑛} satisfies conditions (C1), (C2), and (C4): lim𝑛𝜆𝑛/𝜆𝑛+1=1, or equivalently, lim𝑛(𝜆𝑛𝜆𝑛+1)/𝜆𝑛+1=1. Recently, Yao et al. [25] presented the following hybrid algorithm: 𝑦𝑛=𝑥𝑛𝜆𝑛𝐹𝑥𝑛,𝑥𝑛+1=1𝛼𝑛𝑦𝑛+𝛼𝑛𝑊𝑛𝑦𝑛,𝑛0,(1.5) where 𝐹 is a 𝜅-Lipschitzian and 𝜂-strongly monotone operator on 𝐻 and 𝑊𝑛 is a 𝑊-mapping. It is shown that the sequences {𝑥𝑛} and {𝑦𝑛} defined by (1.5) converge strongly to 𝑥𝑛=1𝐹(𝑇𝑛), which solves the following variational inequality: 𝐹𝑥,𝑥𝑥0,𝑥𝑛=1𝐹𝑇𝑛.(1.6) Very recently, Wang [26] proved that the sequence {𝑦𝑛} generated by the iterative algorithm (1.5) converges to a common fixed point of an infinite family of nonexpansive mappings under some weaker assumptions.

Motivated and inspired by the above works, in this paper, we introduce two hybrid algorithms for finding a common fixed point of a nonexpansive semigroup {𝑇(𝜏)}𝜏0 in Hilbert spaces. We prove that the presented algorithms converge strongly to a common fixed point 𝑥 of {𝑇(𝜏)}𝜏0. Such common fixed point 𝑥 is the unique solution of some variational inequality in Hilbert spaces.

2. Preliminaries

In this section, we will collect some basic concepts and several lemmas that will be used in the next section.

Suppose that 𝐻 is a real Hilbert space with inner product , and norm . For the sequence {𝑥𝑛} in 𝐻, we write 𝑥𝑛𝑥 to indicate that the sequence {𝑥𝑛} converges weakly to 𝑥. 𝑥𝑛𝑥 means that {𝑥𝑛} converges strongly to 𝑥. We denote by 𝜔𝑤(𝑥𝑛) the weak 𝜔-limit set of {𝑥𝑛}, that is 𝜔𝑤𝑥𝑛=𝑥𝐻𝑥𝑛𝑖𝑥𝑥forsomesubsequence𝑛𝑖𝑥of𝑛.(2.1) Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. A mapping 𝐹𝐶𝐶 is called 𝜅-Lipschitzian if there exists a positive constant 𝜅 such that 𝐹𝑥𝐹𝑦𝜅𝑥𝑦,𝑥,𝑦𝐶.(2.2)𝐹 is said to be 𝜂-strongly monotone if there exists a positive constant 𝜂 such that 𝐹𝑥𝐹𝑦,𝑥𝑦𝜂𝑥𝑦2,𝑥,𝑦𝐶.(2.3) The following equalities are well known: 𝑥𝑦2=𝑥22𝑥,𝑦+𝑦2,𝜆𝑥+(1𝜆)𝑦2=𝜆𝑥2+(1𝜆)𝑦2𝜆(1𝜆)𝑥𝑦2(2.4) for all 𝑥,𝑦𝐻 and 𝜆[0,1] (see [45]).

In the sequel, we will make use of the following well-known lemmas.

Lemma 2.1 (see [46]). Let 𝐶 be a nonempty bounded closed convex subset of 𝐻 and let 𝑆={𝑇(𝑠)0𝑠<} be a nonexpansive semigroup on 𝐶. Then, for any 0, lim𝑡sup𝑥𝐶1𝑡𝑡01𝑇(𝑠)𝑥𝑑𝑠𝑇()𝑡𝑡0𝑇(𝑠)𝑥𝑑𝑠=0.(2.5)

Lemma 2.2 (see [47]). Assume that 𝑇𝐻𝐻 is a nonexpansive mapping. If 𝑇 has a fixed point, then 𝐼𝑇 is demiclosed. That is, whenever {𝑥𝑛} is a sequence in 𝐻 weakly converging to some 𝑥𝐻 and the sequence {(𝐼𝑇)𝑥𝑛} strongly converges to some 𝑦, it follows that (𝐼𝑇)𝑥=𝑦. Here, 𝐼 is the identity operator of 𝐻.

Lemma 2.3 (see [27]). Let {𝛾𝑛} be a real sequence satisfying 0<liminf𝑛𝛾𝑛limsup𝑛𝛾𝑛<1. Assume that {𝑥𝑛} and {𝑧𝑛} are bounded sequences in Banach space 𝐸, which satisfy the following condition: 𝑥𝑛+1=(1𝛾𝑛)𝑥𝑛+𝛾𝑛𝑧𝑛. If limsup𝑛(𝑧𝑛+1𝑧𝑛)𝑥𝑛+1𝑥𝑛)0, then lim𝑛𝑥𝑛𝑧𝑛=0.

Lemma 2.4 (see [48]). Let 𝐹 be a 𝜅-Lipschitzian and 𝜂-strongly monotone operator on a Hilbert space 𝐻 with 0<𝜂𝜅 and 0<𝑡<𝜂/𝜅2. Then, 𝑆=(𝐼𝑡𝐹)𝐻𝐻 is a contraction with contraction coefficient 𝜏𝑡=1𝑡(2𝜂𝑡𝜅2).

Lemma 2.5 (see [49]). Let {𝑎𝑛} be a sequence of nonnegative real numbers satisfying 𝑎𝑛+11𝜆𝑛𝑎𝑛+𝜆𝑛𝛿𝑛+𝛾𝑛,𝑛0,(2.6) where {𝜆𝑛} and {𝛾𝑛} satisfy the following conditions:(i)𝜆𝑛[0,1] and 𝑛=0𝜆𝑛=,(ii)limsup𝑛𝛿𝑛0or𝑛=0𝜆𝑛𝛿𝑛<, (iii)𝛾𝑛0(𝑛0),𝑛=0𝛾𝑛<.
Then, lim𝑛𝑎𝑛=0.

3. Main Results

In this section we will show our main results.

Theorem 3.1. Let 𝐻 be a real Hilbert space. Let 𝑆={𝑇(𝜏)0𝜏<}𝐻𝐻 be a nonexpansive semigroup such that Fix(𝑆). Let 𝐹 be a 𝜅-Lipschitzian and 𝜂-strongly monotone operator on 𝐻 with 0<𝜂𝜅. Let {𝛾𝑡}0<𝑡<1 be a continuous net of positive real numbers such that lim𝑡0+𝛾𝑡=+. Putting 𝜏𝑡=1𝑡(2𝜂𝑡𝜅2), for each 𝑡(0,𝜂/𝜅2), let the net {𝑥𝑡} be defined by the following implicit scheme: 𝑥𝑡=1𝛾𝑡𝛾𝑡0𝑇(𝜏)(𝐼𝑡𝐹)𝑥𝑡𝑑𝜏.(3.1) Then, as 𝑡0+, the net {𝑥𝑡} converges strongly to a fixed point 𝑥 of 𝑆, which is the unique solution of the following variational inequality: 𝐹𝑥,𝑥𝑢0,𝑢Fix(𝑆).(3.2)

Proof. First, we note that the net {𝑥𝑡} defined by (3.1) is well defined. We define a mapping 𝑃𝑡1𝑥=𝛾𝑡𝛾𝑡0[]𝜂𝑇(𝜏)(𝐼𝑡𝐹)𝑥𝑑𝜏,𝑡0,𝜅2,𝑥𝐻.(3.3) It follows that 𝑃𝑡𝑥𝑃𝑡𝑦1𝛾𝑡𝛾𝑡0[][]𝑇(𝜏)(𝐼𝑡𝐹)𝑥𝑇(𝜏)(𝐼𝑡𝐹)𝑦𝑑𝜏(𝐼𝑡𝐹)𝑥(𝐼𝑡𝐹)𝑦.(3.4) Obviously, 𝑃𝑡 is a contraction. Indeed, from Lemma 2.4, we have 𝑃𝑡𝑥𝑃𝑡𝑦(𝐼𝑡𝐹)𝑥(𝐼𝑡𝐹)𝑦𝑥𝑦,(3.5) for all 𝑥,𝑦𝐶. So it has a unique fixed point. Therefore, the net {𝑥𝑡} defined by (3.1) is well defined.
We prove that {𝑥𝑡} is bounded. Taking 𝑢Fix(𝑆) and using Lemma 2.4, we have 𝑥𝑡=1𝑢𝛾𝑡𝛾𝑡0𝑇(𝜏)(𝐼𝑡𝐹)𝑥𝑡=1𝑑𝜏𝑢𝛾𝑡𝛾𝑡0𝑇(𝜏)(𝐼𝑡𝐹)𝑥𝑡1𝑑𝜏𝛾𝑡𝛾𝑡01𝑇(𝜏)𝑢𝑑𝜏𝛾𝑡𝛾𝑡0(𝑇(𝜏)𝐼𝑡𝐹)𝑥𝑡𝑑𝜏𝑇(𝜏)𝑢𝑑𝜏(𝐼𝑡𝐹)𝑥𝑡(𝑢𝐼𝑡𝐹)𝑥𝑡(𝐼𝑡𝐹)𝑢𝑡𝐹𝑢𝜏𝑡𝑥𝑡𝑢+𝑡𝐹𝑢.(3.6) It follows that 𝑥𝑡𝑡𝑢1𝜏𝑡𝐹𝑢.(3.7) Observe that lim𝑡0+𝑡1𝜏𝑡=1𝜂.(3.8) Thus, (3.7) and (3.8) imply that the net {𝑥𝑡} is bounded for small enough 𝑡. Without loss of generality, we may assume that the net {𝑥𝑡} is bounded for all 𝑡(0,𝜂/𝜅2). Consequently, we deduce that {𝐹𝑥𝑡} is also bounded.
On the other hand, from (3.1), we have 𝑥𝑡𝑇(𝜏)𝑥𝑡𝑇(𝜏)𝑥𝑡1𝑇(𝜏)𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡+1𝑑𝜏𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡𝑑𝜏𝑥𝑡+1𝑇(𝜏)𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏2𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡𝑑𝜏𝑥𝑡+1𝑇(𝜏)𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏=2𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏𝛾𝑡𝛾𝑡0[]𝑥𝑇(𝜏)(𝐼𝑡𝐹)𝑡+1𝑑𝜏𝑇(𝜏)𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏2𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡𝑇(𝜏)(𝐼𝑡𝐹)𝑥𝑡+𝑇1𝑑𝜏(𝜏)𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡𝑑𝜏2𝑡𝐹𝑥𝑡+1𝑇(𝜏)𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡1𝑑𝜏𝛾𝑡𝛾𝑡0𝑇(𝜏)𝑥𝑡.𝑑𝜏(3.9) This together with Lemma 2.1 implies that lim𝑡0+𝑥𝑡𝑇(𝜏)𝑥𝑡=0.(3.10) Let {𝑡𝑛}(0,1) be a sequence such that 𝑡𝑛0 as 𝑛. Put 𝑥𝑛=𝑥𝑡𝑛. Since {𝑥𝑛} is bounded, without loss of generality, we may assume that {𝑥𝑛} converges weakly to a point ̃𝑥𝐶. Noticing (3.10), we can use Lemma 2.2 to get ̃𝑥Fix(𝑆).
Again, from (3.1), we have 𝑥𝑡𝑢2=1𝛾𝑡𝛾𝑡0𝑇(𝜏)(𝐼𝑡𝐹)𝑥𝑡𝑑𝜏𝑢21𝛾𝑡𝛾𝑡0𝑇(𝜏)(𝐼𝑡𝐹)𝑥𝑡𝑇(𝜏)𝑢𝑑𝜏2(𝐼𝑡𝐹)𝑥𝑡(𝐼𝑡𝐹)𝑢𝑡𝐹𝑢2𝜏2𝑡𝑥𝑡𝑢2+𝑡2𝐹𝑢2+2𝑡(𝐼𝑡𝐹)𝑢(𝐼𝑡𝐹)𝑥𝑡,𝐹𝑢𝜏𝑡𝑥𝑡𝑢2+𝑡2𝐹𝑢2+2𝑡𝑢𝑥𝑡,𝐹𝑢+2𝑡2𝐹𝑥𝑡𝐹𝑢,𝐹𝑢𝜏𝑡𝑥𝑡𝑢2+𝑡2𝐹𝑢2+2𝑡𝑢𝑥𝑡,𝐹𝑢+2𝜅𝑡2𝑥𝑡𝑢𝐹𝑢.(3.11) Therefore, 𝑥𝑡𝑢2𝑡21𝜏𝑡𝐹𝑢2+2𝑡1𝜏𝑡𝑢𝑥𝑡,𝐹𝑢+2𝑡2𝜅1𝜏𝑡𝑥𝑡𝑢𝐹𝑢.(3.12) It follows that 𝑥𝑛̃𝑥2𝑡2𝑛1𝜏𝑡𝑛𝐹̃𝑥2+2𝑡𝑛1𝜏𝑡𝑛̃𝑥𝑥𝑛,𝐹̃𝑥+2𝑡2𝑛𝜅1𝜏𝑡𝑛𝑥𝑛̃𝑥𝐹̃𝑥.(3.13) Thus, 𝑥𝑛̃𝑥 implies that 𝑥𝑛̃𝑥.
Again, from (3.12), we obtain 𝑥𝑛𝑢2𝑡2𝑛1𝜏𝑡𝑛𝐹𝑢2+2𝑡𝑛1𝜏𝑡𝑛𝑢𝑥𝑛,𝐹𝑢+2𝑡2𝑛𝜅1𝜏𝑡𝑛𝑥𝑛𝑢𝐹𝑢.(3.14) It is clear that lim𝑛(𝑡2𝑛/(1𝜏𝑡𝑛))=0, lim𝑛(2𝑡𝑛/(1𝜏𝑡𝑛))=2/𝜂, and lim𝑛(2𝑡2𝑛𝜅/(1𝜏𝑡𝑛))=0. We deduce immediately from (3.14) that 𝐹𝑢,̃𝑥𝑢0,(3.15) which is equivalent to its dual variational inequality 𝐹̃𝑥,̃𝑥𝑢0.(3.16) That is, ̃𝑥Fix(𝑆) is a solution of the variational inequality (3.2).
Suppose that 𝑥Fix(𝑆) and ̃𝑥Fix(𝑆) both are solutions to the variational inequality (3.2); then 𝐹𝑥,𝑥̃𝑥0,𝐹̃𝑥,̃𝑥𝑥0.(3.17) Adding up (3.17) and the last inequality yields 𝐹𝑥𝐹̃𝑥,𝑥̃𝑥0.(3.18)
The strong monotonicity of 𝐹 implies that 𝑥=̃𝑥 and the uniqueness is proved. Later, we use 𝑥Fix(𝑆) to denote the unique solution of (3.2).
Therefore, ̃𝑥=𝑥 by uniqueness. In a nutshell, we have shown that each cluster point of {𝑥𝑡}(𝑡0) equals 𝑥. Hence 𝑥𝑡𝑥 as 𝑡0. This completes the proof.

Next we introduce an explicit algorithm for finding a solution of the variational inequality (3.2).

Algorithm 3.2. For given 𝑥0𝐶 arbitrarily, define a sequence {𝑥𝑛} iteratively by 𝑦𝑛=𝑥𝑛𝜆𝑛𝐹𝑥𝑛,𝑥𝑛+1=1𝛼𝑛𝑦𝑛+𝛼𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑑𝜏,𝑛0,(3.19) where {𝜆𝑛} and {𝑡𝑛} are sequences in (0,) and {𝛼𝑛} is a sequence in [0,1].

Theorem 3.3 3.3. Let 𝐻 be a real Hilbert space. Let 𝐹 be a 𝜅-Lipschitzian and 𝜂-strongly monotone operator on 𝐻 with 0<𝜂𝜅. Let 𝑆={𝑇(𝜏)0𝜏<}𝐻𝐻 be a nonexpansive semigroup with Fix(𝑆). Assume that(i)limsup𝑛𝜆𝑛<𝜂/𝜅2 and 𝑛=1𝜆𝑛=,(ii)lim𝑛𝑡𝑛= and lim𝑛(𝑡𝑛+1/𝑡𝑛)=1,(iii)0<𝛾liminf𝑛𝛼𝑛limsup𝑛𝛼𝑛<1, for some 𝛾(0,1).
Then, the sequences {𝑥𝑛} and {𝑦𝑛} generated by (3.19) converge strongly to 𝑥Fix(𝑆) if and only if 𝜆𝑛𝐹(𝑥𝑛)0, where 𝑥 solves the variational inequality (3.2).

Proof. The necessity is obvious. We only need to prove the sufficiency. Suppose that 𝜆𝑛𝐹(𝑥𝑛)0. First, we show that 𝑥𝑛 is bounded. In fact, letting 𝑢Fix(𝑆), we have 𝑥𝑛+1=𝑢1𝛼𝑛𝑦𝑛+𝛼𝑛𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛=𝑑𝜏𝑢1𝛼𝑛𝑦𝑛𝑢+𝛼𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑑𝜏𝑢1𝛼𝑛𝑦𝑛𝑢+𝛼𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑑𝜏𝑢1𝛼𝑛𝑦𝑛𝑢+𝛼𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑇(𝜏)𝑢𝑑𝜏1𝛼𝑛𝑦𝑛𝑢+𝛼𝑛𝑦𝑛=𝑦𝑢𝑛.𝑢(3.20) From condition (i), without loss of generality, we can assume that 𝜆𝑛𝑎<𝜂/𝜅2 for all 𝑛. By (3.19) and Lemma 2.4, we have 𝑦𝑛=𝑥𝑢𝑛𝜆𝑛𝐹𝑥𝑛=𝑢𝐼𝜆𝑛𝐹𝑥𝑛𝐼𝜆𝑛𝐹𝑢𝜆𝑛𝐹𝑢𝜏𝜆𝑛𝑥𝑛𝑢+𝜆𝑛𝐹𝑢,(3.21) where 𝜏𝜆𝑛=1𝜆𝑛(2𝜂𝜆𝑛𝜅2)(0,1).
Then, from (3.20) and (3.21), we obtain 𝑥𝑛+1𝑢𝜏𝜆𝑛𝑥𝑛𝑢+𝜆𝑛=𝐹𝑢11𝜏𝜆𝑛𝑥𝑛+𝑢1𝜏𝜆𝑛𝜆𝑛1𝜏𝜆𝑛𝑥𝐹𝑢max𝑛,𝜆𝑢𝑛𝐹𝑢1𝜏𝜆𝑛.(3.22) Since lim𝑛(𝜆𝑛/(1𝜏𝜆𝑛))=1/𝜂, we have by induction 𝑥𝑛+1𝑥𝑢max0𝑢,𝑀1𝐹𝑢,(3.23) where 𝑀1=sup𝑛{𝜆𝑛/(1𝜏𝜆𝑛)}<. Hence, {𝑥𝑛} is bounded. We also obtain that {𝑦𝑛},{𝑇(𝜏)𝑦𝑛}, and {𝐹𝑥𝑛} are all bounded.
Define 𝑥𝑛+1=(1𝛼𝑛)𝑥𝑛+𝛼𝑛𝑧𝑛 for all 𝑛. Observe that 𝑧𝑛+1𝑧𝑛=𝑥𝑛+21𝛼𝑛+1𝑥𝑛+1𝛼𝑛+1𝑥𝑛+11𝛼𝑛𝑥𝑛𝛼𝑛=1𝛼𝑛+1𝑦𝑛+1+𝛼𝑛+11/𝑡𝑛+1𝑡𝑛+10𝑇(𝜏)𝑦𝑛+1𝑑𝜏1𝛼𝑛+1𝑥𝑛+1𝛼𝑛+11𝛼𝑛𝑦𝑛+𝛼𝑛1/𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑑𝜏1𝛼𝑛𝑥𝑛𝛼𝑛=𝛼𝑛+11/𝑡𝑛+1𝑡𝑛+10𝑇(𝜏)𝑦𝑛+1𝑑𝜏1𝛼𝑛+1𝜆𝑛+1𝐹𝑥𝑛+1𝛼𝑛+1𝛼𝑛1/𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑑𝜏1𝛼𝑛𝜆𝑛𝐹𝑥𝑛𝛼𝑛1𝛼𝑛+1𝛼𝑛+1𝜆𝑛+1𝐹𝑥𝑛+1+1𝛼𝑛𝛼𝑛𝜆𝑛𝐹𝑥𝑛+1𝑡𝑛+1𝑡𝑛+10𝑇(𝜏)𝑦𝑛+11𝑑𝜏𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛.𝑑𝜏(3.24) Next, we estimate (1/𝑡𝑛+1)𝑡𝑛+10𝑇(𝜏)𝑦𝑛+1𝑑𝜏(1/𝑡𝑛)𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑑𝜏. As a matter of fact, we have 1𝑡𝑛+1𝑡𝑛+10𝑇(𝜏)𝑦𝑛+11𝑑𝜏𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛1𝑑𝜏𝑡𝑛+1𝑡𝑛+10𝑇(𝜏)𝑦𝑛+11𝑑𝜏𝑡𝑛+1𝑡𝑛+10𝑇(𝜏)𝑦𝑛+1𝑑𝜏𝑡𝑛+1𝑡𝑛+10𝑇(𝜏)𝑦𝑛1𝑑𝜏𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑦𝑑𝜏𝑛+1𝑦𝑛+1𝑡𝑛+1𝑡𝑛+10𝑇(𝜏)𝑦𝑛1𝑑𝜏𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑦𝑑𝜏𝑛+1𝑦𝑛+||||1𝑡𝑛+11𝑡𝑛||||𝑡𝑛0𝑇(𝜏)𝑦𝑛+1𝑑𝜏𝑡𝑛+1𝑡𝑛+1𝑡𝑛𝑇(𝜏)𝑦𝑛𝑥𝑑𝜏𝑛+1𝜆𝑛+1𝐹𝑥𝑛+1𝑥𝑛+𝜆𝑛𝐹𝑥𝑛+||||𝑡𝑛𝑡𝑛+1||||𝑀12𝑥𝑛+1𝑥𝑛+𝜆𝑛+1𝐹𝑥𝑛+1+𝜆𝑛𝐹𝑥𝑛+𝑀2||||𝑡𝑛𝑡𝑛+1||||,1(3.25) where 𝑀2=sup𝑛{2𝑇(𝜏)𝑦𝑛}<. From (3.24) and (3.25), we have 𝑧𝑛+1𝑧𝑛1𝛾𝛾𝜆𝑛+1𝐹𝑥𝑛+1+1𝛾𝛾𝜆𝑛𝐹𝑥𝑛+𝑥𝑛+1𝑥𝑛+𝜆𝑛+1𝐹𝑥𝑛+1+𝜆𝑛𝐹𝑥𝑛+𝑀2||||𝑡𝑛𝑡𝑛+1||||11𝛾𝜆𝑛+1𝐹𝑥𝑛+1+1𝛾𝜆𝑛𝐹𝑥𝑛+𝑥𝑛+1𝑥𝑛+𝑀2||||𝑡𝑛𝑡𝑛+1||||.1(3.26) Namely, 𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛1𝛾𝜆𝑛+1𝐹𝑥𝑛+1+1𝛾𝜆𝑛𝐹𝑥𝑛+𝑀2||||𝑡𝑛𝑡𝑛+1||||.1(3.27) Since 𝜆𝑛𝐹(𝑥𝑛)0 and (𝑡𝑛/𝑡𝑛+1)10, we get limsup𝑛𝑧𝑛+1𝑧𝑛𝑥𝑛+1𝑥𝑛0.(3.28) Consequently, by Lemma 2.3, we deduce lim𝑛𝑧𝑛𝑥𝑛=0. Therefore, lim𝑛𝑥𝑛+1𝑥𝑛=lim𝑛𝛼𝑛𝑧𝑛𝑥𝑛=0.(3.29)
Next, we claim that lim𝑛𝑥𝑛𝑇(𝜏)𝑥𝑛=0. Observe that 𝑇(𝜏)𝑥𝑛𝑥𝑛𝑇(𝜏)𝑥𝑛1𝑇(𝜏)𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛+1𝑑𝜏𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛𝑑𝜏𝑥𝑛+1𝑇(𝜏)𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛1𝑑𝜏𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛1𝑑𝜏2𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛𝑑𝜏𝑥𝑛+1𝑇(𝜏)𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛1𝑑𝜏𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛.𝑑𝜏(3.30) Note that 𝑥𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛𝑥𝑑𝜏𝑛𝑥𝑛+1+𝑥𝑛+11𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛𝑥𝑑𝜏𝑛𝑥𝑛+1+1𝛼𝑛𝑦𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛𝑑𝜏+𝛼𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛1𝑑𝜏𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑥𝑑𝜏𝑛𝑥𝑛+1+1𝛼𝑛𝑦𝑛𝑥𝑛+1𝛼𝑛𝑥𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛𝑑𝜏+𝛼𝑛𝑦𝑛𝑥𝑛𝑥𝑛𝑥𝑛+1+𝑦𝑛𝑥𝑛+1𝛼𝑛𝑥𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛.𝑑𝜏(3.31) It follows that 𝑥𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑥𝑛1𝑑𝜏𝛼𝑛𝑥𝑛𝑥𝑛+1+𝑦𝑛𝑥𝑛=1𝛼𝑛𝑥𝑛𝑥𝑛+1+𝜆𝑛𝐹𝑥𝑛0.(3.32) By Lemma 2.1, (3.30), and (3.32), we derive lim𝑛𝑇(𝜏)𝑥𝑛𝑥𝑛=0.(3.33)
Next, we show that limsup𝑛𝐹𝑥,𝑥𝑥𝑛0, where 𝑥=lim𝑛𝑥𝑡𝑛 and 𝑥𝑡𝑛 is defined by 𝑥𝑡𝑛=(1/𝑡𝑛)𝑡𝑛0𝑇(𝜏)[(𝐼𝑡𝑛𝐹)𝑥𝑡𝑛]𝑑𝜏. Since {𝑥𝑛} is bounded, there exists a subsequence {𝑥𝑛𝑘} of {𝑥𝑛} that converges weakly to 𝜔. It is clear that 𝑇(𝜏)𝑥𝑛𝑘𝜔. From Lemma 2.2, we have 𝜔Fix(𝑆). Hence, by Theorem 3.1, we have limsup𝑛𝐹𝑥,𝑥𝑥𝑛=lim𝑘𝐹𝑥,𝑥𝑥𝑛𝑘=𝐹𝑥,𝑥𝜔0.(3.34)
Finally, we prove that {𝑥𝑛} converges strongly to 𝑥Fix(𝑆). From (3.19), we have 𝑥𝑛+1𝑥21𝛼𝑛𝑦𝑛𝑥2+𝛼𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑑𝜏𝑥21𝛼𝑛𝑦𝑛𝑥2+𝛼𝑛1𝑡𝑛𝑡𝑛0𝑇(𝜏)𝑦𝑛𝑇(𝜏)𝑥2𝑦𝑑𝜏𝑛𝑥2=𝑥𝑛𝜆𝑛𝐹𝑥𝑛𝑥2=𝐼𝜆𝑛𝐹𝑥𝑛𝐼𝜆𝑛𝐹𝑥𝜆𝑛𝐹𝑥2𝜏2𝜆𝑛𝑥𝑛𝑥2+𝜆2𝑛𝐹𝑥2+2𝜆𝑛𝐼𝜆𝑛𝐹𝑥𝐼𝜆𝑛𝐹𝑥𝑛𝑥,𝐹𝜏𝜆𝑛𝑥𝑛𝑥2+𝜆2𝑛𝐹𝑥2+2𝜆𝑛𝑥𝑥𝑛,𝐹𝑥+2𝜆𝑛𝜆𝑛𝐹𝑥𝑛,𝐹𝑥2𝜆2𝑛𝐹𝑥211𝜏𝜆𝑛𝑥𝑛𝑥2+2𝜆𝑛𝑥𝑥𝑛,𝐹𝑥+2𝜆𝑛𝜆𝑛𝐹𝑥𝑛𝐹𝑥𝜆2𝑛𝐹𝑥211𝜏𝜆𝑛𝑥𝑛𝑥2+1𝜏𝜆𝑛2𝜆𝑛1𝜏𝜆𝑛𝑥𝑥𝑛,𝐹𝑥+2𝜆𝑛𝐹𝑥1𝜏𝜆𝑛𝜆𝑛𝐹𝑥𝑛=1𝛿𝑛𝑥𝑛𝑥2+𝛿𝑛𝜎𝑛,(3.35) where 𝛿𝑛=1𝜏𝜆𝑛 and 𝜎𝑛=(2𝜆𝑛/1𝜏𝜆𝑛)𝑥𝑥𝑛,𝐹𝑥+(2𝜆𝑛𝐹𝑥/(1𝜏𝜆𝑛))𝜆𝑛𝐹𝑥𝑛. Obviously, we can see that 𝑛=1𝛿𝑛= and limsup𝑛𝜎𝑛0. Hence, all conditions of Lemma 2.5 are satisfied. Therefore, we immediately deduce that the sequence {𝑥𝑛} converges strongly to 𝑥Fix(𝑆).
Observe that 𝑦𝑛𝑥𝑦𝑛𝑥𝑛+𝑥𝑛𝑥𝜆𝑛𝐹𝑥𝑛+𝑥𝑛𝑥0(𝑛).(3.36)
Consequently, it is clear that {𝑦𝑛} converges strongly to 𝑥Fix(𝑆). From 𝑥=lim𝑡0𝑥𝑡 and Theorem 3.1, we get that 𝑥 is the unique solution of the variational inequality 𝐹𝑥,𝑥𝑢0,𝑢Fix(𝑆).(3.37) This completes the proof.

Acknowledgments

Y. Yao was supported in part by NSFC 11071279 and NSFC 71161001-G0105. Y.-C. Liou was partially supported by NSC 100-2221-E-230-012. R. Chen was supported in part by NSFC 11071279.