Algorithmic Approach to the Split Problems

Abstract

This paper deals with design algorithms for the split variational inequality and equilibrium problems. Strong convergence theorems are demonstrated.

1. Introduction

Let ℍ be a real Hilbert space. Let ℂ and ℚ be two nonempty closed convex subsets of ℍ. Consider the following problem.

Problem 1. Find a point 𝑢^§∈ℂ such that Ψ𝑢^§∈ℚ. (1)
This problem is called split feasibility problem when Ψ is a bounded linear operator. In this case, Problem 1 can be applied to many practical problems such as signal processing and image reconstruction. Specifically, we can find the prototype of Problem 1 in intensity-modulated radiation therapy; see, for example, [1–3]. Based on this relation, many mathematicians were devoted to study the split feasibility problem and develop its iterative algorithms. Related works can be found in [4–8] and the references therein.
Let 𝔸,Ψ∶ℂ→ℍ be two mappings. Consider the variational inequality of finding 𝑢^†∈ℂ,Ψ(𝑢^†)∈ℂ such that 𝔸𝑢^†,Ψ(𝑢)−Ψ𝑢^†≥0, (2)
for all Ψ(𝑢)∈ℂ. We use VI(𝔸,Ψ) to denote the set of solutions of (2). Variational inequality problems have important applications in many fields such as elasticity, optimization, economics, transportation, and structural analysis, and various numerical methods have been studied by many researchers; see, for instance, [9–17].
Let 𝜚∶ℂ×ℂ→ℝ be an equilibrium bifunction; that is, 𝜚(𝑢,𝑢)=0 for each 𝑢∈ℂ. Consider the equilibrium problem which is to find 𝑢^∗∈ℂ such that 𝜚𝑢^∗,𝑣≥0,∀𝑣∈ℂ. (3)
Denote the set of solutions of (3) by EP(𝜚,ℂ). The equilibrium problems include fixed point problems, optimization problems, and variational inequality problems as special cases. Some algorithms have been proposed to solve the equilibrium problems; see, for example, [18–22]. Thus it is an interesting topic associated with algorithmic approach to the variational inequality and equilibrium problems. In this paper, our main purpose is to study the following split problem involved in the variational inequality and equilibrium problems. Find a point 𝑥^♮ such that 𝑥^♮(),Ψ𝑥∈VI𝔸,Ψ_♮∈EP𝜚,ℂ. (4)
We are devoted to study (4) with operator Ψ being a nonlinear mapping. For this purpose, we develop an iterative algorithm for solving the split problem (4). We can compute 𝑥^♮ iteratively by using our algorithm. Convergence analysis is given under some mild assumptions.

2. Basic Concepts

Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. An operator 𝔹∶ℂ→ℍ is said to be (i)monotone ↠⟨𝑢−𝑣,𝔹𝑢−𝔹𝑣⟩≥0 for all 𝑢,𝑣∈ℂ;(ii)strongly monotone ↠⟨𝑢−𝑣,𝔹𝑢−𝔹𝑣⟩≥𝜁‖𝑢−𝑣‖² for some constant 𝜁>0 and for all 𝑢,𝑣∈ℂ;(iii)inverse-strongly monotone ↠⟨𝑢−𝑣,𝔹𝑢−𝔹𝑣⟩≥𝜍‖𝔹𝑢−𝔹𝑣‖² for some 𝜍>0 and for all 𝑢,𝑣∈ℂ; in this case, 𝔹 is called 𝜍-inverse strongly monotone;(iv)𝜍-inverse strongly 𝜃-monotone ↠⟨𝜃(𝑢)−𝜃(𝑣),𝔹𝑢−𝔹𝑣⟩≥𝜍‖𝔹𝑢−𝔹𝑣‖² for all 𝑢,𝑣∈ℂ and for some 𝜍>0, where 𝜃∶ℂ→ℂ is a mapping.

A mapping 𝜗∶ℂ→ℍ is said to be(i)nonexpansive ↠‖𝜗𝑢−𝜗𝑣‖≤‖𝑢−𝑣‖ for all 𝑢,𝑣∈ℂ;(ii)firmly nonexpansive ↠‖𝜗𝑢−𝜗𝑣‖²≤⟨𝑢−𝑣,𝜗𝑢−𝜗𝑣⟩ for all 𝑢,𝑣∈ℂ;(iii)𝐿-Lipschitz continuous ↠‖𝜗𝑢−𝜗𝑣‖≤𝐿‖𝑢−𝑣‖ for some constant 𝐿>0 and for all 𝑢,𝑣∈ℂ. In such a case, 𝜗 is said to be 𝐿-Lipschitz continuous.

In the sequel, we use Fix(𝜗) to denote the set of fixed points of 𝜗.

Let 𝔸∶ℍ→2^ℍ be a multivalued mapping. The effective domain of 𝔸 is denoted by dom(𝔸). 𝔸 is said to be (i)monotone ↠⟨𝑥−𝑦,𝑢−𝑣⟩≥0 for all 𝑥,𝑦∈dom(𝔸), 𝑢∈𝔸𝑥, and 𝑣∈𝔸𝑦;(ii)maximal monotone ↠𝔸 is monotone and its graph is not strictly contained in the graph of any other monotone operator on ℍ.

A function 𝑓∶ℍ→ℝ is said to be convex if for any 𝑢,𝑣∈ℍ and for any 𝜏∈[0,1], 𝑓(𝜏𝑢+(1−𝜏)𝑣)≤𝜏𝑓(𝑢)+(1−𝜏)𝑓(𝑣).

Let proj_ℂ∶ℂ→ℍ be the metric projection from ℍ onto ℂ. It is known that proj_ℂ satisfies the following inequality: 𝑥−proj_ℂ𝑥,𝑦−proj_ℂ𝑥≤0. (5) for all 𝑥∈ℍ and 𝑦∈ℂ. From this characteristic inequality, we can deduce that proj_ℂ is firmly nonexpansive.

3. Useful Lemmas

In this section, we present several lemmas which will be used in the next section.

Lemma 2 (see [19]). Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. Let 𝜚∶ℂ×ℂ→ℝ be a bifunction. Assume that 𝜚 satisfies the following conditions:  (𝔉1)𝜚(𝑢,𝑢)=0 for all 𝑢∈ℂ;  (𝔉2)𝜚 is monotone, that is, 𝜚(𝑢,𝑣)+𝜚(𝑣,𝑢)≤0 for all 𝑢,𝑣∈ℂ;  (𝔉3)for each 𝑢,𝑣,𝑤∈ℂ, lim_𝑡↓0𝜚(𝑡𝑤+(1−𝑡)𝑢,𝑣)≤𝜚(𝑢,𝑣);  (𝔉4)for each 𝑢∈ℂ, 𝑣↦𝜚(𝑢,𝑣) is convex and lower semicontinuous.
Let 𝜛>0 and 𝑢∈ℂ. Then there exists 𝑤∈ℂ such that 𝜚()+1𝑤,𝑣𝜛⟨𝑣−𝑤,𝑤−𝑢⟩≥0,∀𝑣∈ℂ. (6)
Set ϝ_𝜛(𝑢)={𝑤∈ℂ∶𝜚(𝑤,𝑣)+(1/𝜛)⟨𝑣−𝑤,𝑤−𝑢⟩≥0 for all 𝑣∈ℂ}. Then one have the following:(i)ϝ_𝜛 is single valued and ϝ_𝜛 is firmly nonexpansive,(ii)EP(𝜚,ℂ) is closed and convex and EP(𝜚,ℂ)=Fix(ϝ_𝜛).

Lemma 3 (see [23]). Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. For 𝑥∈ℍ, let the mapping ϝ_𝜛 be the same as in Lemma 2. Then for 𝜇,𝜈>0 and 𝑥∈ℍ, one has ϝ_𝜇(𝑥)−ϝ_𝜈(𝑥)²≤𝜇−𝜈𝜇ϝ_𝜇(𝑥)−ϝ_𝜈(𝑥),ϝ_𝜇(𝑥)−𝑥. (7)

Lemma 4 (see [24]). Let {𝑢_𝑛} and {𝑣_𝑛} be two bounded sequences in a Banach space 𝔼, and let {𝜅_𝑛} be a sequence in [0,1] satisfying 0<liminf_𝑛→∞𝜅_𝑛≤limsup_𝑛→∞𝜅_𝑛<1. Suppose 𝑢_𝑛+1=(1−𝜅_𝑛)𝑣_𝑛+𝜅_𝑛𝑢_𝑛 for all 𝑛≥0 and limsup_𝑛→∞(‖𝑣_𝑛+1−𝑣_𝑛‖−‖𝑢_𝑛+1−𝑢_𝑛‖)≤0. Then, lim_𝑛→∞‖𝑢_𝑛−𝑣_𝑛‖=0.

Lemma 5 (see [25]). Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. Let 𝕊∶ℂ→ℂ be a nonexpansive mapping with Fix(𝕊)≠∅. Then 𝕊 is demiclosed on ℂ.

Lemma 6 (see [26]). Let {𝑎_𝑛}⊂[0,∞) be a sequence. Assume that 𝑎_𝑛+1≤(1−𝛾_𝑛)𝑎_𝑛+𝛿_𝑛𝛾_𝑛, where {𝛾_𝑛} is a sequence in (0,1), and {𝛿_𝑛} is a sequence satisfying ∑^∞_𝑛=1𝛾_𝑛=∞ and limsup_𝑛→∞𝛿_𝑛≤0 (or ∑^∞_𝑛=1|𝛿_𝑛𝛾_𝑛|<∞). Then lim_𝑛→∞𝑎_𝑛=0.

4. Main Results

In this section, we firstly present our problem and algorithm constructed. Consequently, we give the convergence analysis of the presented algorithm.

Problem 7. Let ℂ be a nonempty closed convex subset of a real Hilbert space ℍ. Assume that (1)Ψ∶ℂ→ℂ is a weakly continuous and 𝜁-strongly monotone mapping such that 𝑅(Ψ)=ℂ; (2)𝔸∶ℂ→ℍ is an 𝜍-inverse strongly Ψ-monotone mapping; (3)𝜚∶ℂ×ℂ→ℝ is a bifunction satisfying conditions (𝔉1)--(𝔉4) in Lemma 2. Our objective is to find𝑥^♮∈VI(𝔸,Ψ)suchthatΨ𝑥^♮∈EP𝜚,ℂ. (8)
We use Υ to denote the set of solutions of (8). In the following, we assume that Υ is nonempty. For solving Problem 7, we introduce the following algorithm.

Algorithm 8.   
Step  0 (initialization). Let 𝑢₀∈ℂ. (9)
Step  1. For given {𝑢_𝑛}, let the sequence {𝑣_𝑛} be generated iteratively by 𝑣_𝑛=proj_ℂΨ𝑢_𝑛−𝜇_𝑛𝔸𝑢_𝑛,𝑛≥0, (10) where proj_ℂ is the metric projection and {𝜇_𝑛} is a real number sequence.
Step  2. For given {𝑣_𝑛}, find {𝑧_𝑛} such that 𝜚𝑧_𝑛+1,𝑦𝜛_𝑛𝑦−𝑧_𝑛,𝑧_𝑛−1−𝛼_𝑛𝑣_𝑛≥0,∀𝑦∈ℂ, (11)
where {𝜛_𝑛}⊂(0,∞) and {𝛼_𝑛}⊂[0,1] are two real number sequences.
Step  3. For the previous sequences {𝑢_𝑛} and {𝑧_𝑛}, let the (𝑛+1)th sequence {𝑢_𝑛+1} be generated by Ψ𝑢_𝑛+1=𝜅_𝑛Ψ𝑢_𝑛+1−𝜅_𝑛𝑧_𝑛,𝑛≥0, (12) where {𝜅_𝑛}⊂[0,1] is a real number sequence.

Theorem 9. Assume that the following conditions are satisfied:  (𝐶1)lim_𝑛→∞𝛼_𝑛=0 and ∑_𝑛𝛼_𝑛=∞;  (𝐶2)0<liminf_𝑛→∞𝜅_𝑛≤limsup_𝑛→∞𝜅_𝑛<1;  (𝐶3)𝜛_𝑛∈(𝜂₁,𝜂₂)⊂(0,∞), 𝜇_𝑛∈(𝜉₁,𝜉₂)⊂(0,2𝜍), and 𝜁∈(0,2𝜍);  (𝐶4)lim_𝑛→∞(𝜇_𝑛+1−𝜇_𝑛)=0 and lim_𝑛→∞(𝜛_𝑛+1−𝜛_𝑛)=0. Then the sequence {𝑢_𝑛} generated by Algorithm 8 converges strongly to 𝑥^∗∈Υ.

Proof. Let 𝑥̆∈Υ. Hence 𝑥̆∈VI(𝔸,Ψ) and Ψ(𝑥̆)∈EP(𝜚,ℂ), noting that 𝑥̆∈VI(𝔸,Ψ) implies Ψ(𝑥̆)=proj_ℂ(Ψ(𝑥̆)−𝜈𝔸𝑥̆) for all 𝜈>0. Hence Ψ(𝑥̆)=proj_ℂ(Ψ(𝑥̆)−𝜇_𝑛𝔸𝑥̆) for all 𝑛≥0. Thus, from (10), we have 𝑣_𝑛−Ψ(𝑥̆)²=proj_ℂΨ𝑢_𝑛−𝜇_𝑛𝔸𝑢_𝑛−proj_ℂΨ(𝑥̆)−𝜇_𝑛𝔸𝑥̆²≤Ψ𝑢_𝑛−𝜇_𝑛𝔸𝑢_𝑛−Ψ(𝑥̆)−𝜇_𝑛𝔸𝑥̆²=Ψ𝑢_𝑛−Ψ(𝑥̆)²−2𝜇_𝑛𝔸𝑢_𝑛−𝔸𝑥̆,Ψ𝑢_𝑛−Ψ(𝑥̆)+𝜇²_𝑛𝔸𝑢_𝑛−𝔸𝑥̆²≤Ψ𝑢_𝑛−Ψ(𝑥̆)²−2𝜇_𝑛𝜍𝔸𝑢_𝑛−𝔸𝑥̆²+𝜇²_𝑛𝔸𝑢_𝑛−𝔸𝑥̆²≤Ψ𝑢_𝑛−Ψ(𝑥̆)²+𝜇_𝑛𝜇_𝑛−2𝜍𝔸𝑢_𝑛−𝔸𝑥̆². (13) Condition (𝐶3) and (13) imply that 𝑣_𝑛−Ψ(𝑥̆)≤Ψ𝑢_𝑛−Ψ(𝑥̆). (14) From Lemma 2 and (11), we get 𝑧_𝑛=ϝ_𝜛𝑛(1−𝛼_𝑛)𝑣_𝑛 for all 𝑛≥0. Since Ψ(𝑥̆)∈EP(𝜚,ℂ), from Lemma 2 we deduce that Ψ(𝑥̆)=ϝ_𝜛𝑛Ψ(𝑥̆) for all 𝑛≥0. So, 𝑧_𝑛(𝑥̆=ϝ−Ψ)_𝜛𝑛1−𝛼_𝑛𝑣_𝑛−ϝ_𝜛𝑛Ψ(𝑥̆≤)1−𝛼_𝑛𝑣_𝑛(𝑥̆≤−Ψ)1−𝛼_𝑛𝑣_𝑛−Ψ(𝑥̆)+𝛼_𝑛‖Ψ(𝑥̆()≤)‖by141−𝛼_𝑛Ψ𝑢_𝑛−Ψ(𝑥̆)+𝛼_𝑛‖Ψ(𝑥̆)‖. (15) It follows that Ψ𝑢_𝑛+1−Ψ(𝑥̆)≤𝜅_𝑛Ψ𝑢_𝑛−Ψ(𝑥̆)+1−𝜅_𝑛𝑧_𝑛−Ψ(𝑥̆)≤𝜅_𝑛Ψ𝑢_𝑛(𝑥̆+−Ψ)1−𝜅_𝑛1−𝛼_𝑛Ψ𝑢_𝑛(𝑥̆+−Ψ)1−𝜅_𝑛𝛼_𝑛‖Ψ(𝑥̆=)‖1−1−𝜅_𝑛𝛼_𝑛Ψ𝑢_𝑛(𝑥̆+−Ψ)1−𝜅_𝑛𝛼_𝑛‖Ψ(𝑥̆)‖. (16) By induction Ψ𝑢_𝑛−Ψ(𝑥̆)≤maxΨ𝑢₀−Ψ(𝑥̆),‖Ψ(𝑥̆)‖. (17) Hence, {Ψ(𝑢_𝑛)} is bounded. Since Ψ is 𝜁-strongly monotone, we can get 𝜁‖𝑢_𝑛−𝑥̆‖≤‖Ψ(𝑢_𝑛)−Ψ(𝑥̆)‖. So, ‖𝑢_𝑛−𝑥̆‖≤(1/𝜁)‖Ψ(𝑢_𝑛)−Ψ(𝑥̆)‖≤(1/𝜁)max{‖Ψ(𝑢₀)−Ψ(𝑥̆)‖,‖Ψ(𝑥̆)‖}. This implies that {𝑢_𝑛} is bounded. Next, we show ‖𝑢_𝑛+1−𝑢_𝑛‖→0. From 𝑧_𝑛=ϝ_𝜛𝑛(1−𝛼_𝑛)𝑣_𝑛, we have 𝑧_𝑛+1−𝑧_𝑛=ϝ_𝜛𝑛+11−𝛼_𝑛+1𝑣_𝑛+1−ϝ_𝜛𝑛1−𝛼_𝑛𝑣_𝑛≤ϝ_𝜛𝑛+11−𝛼_𝑛+1𝑣_𝑛+1−ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛+ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−ϝ_𝜛𝑛1−𝛼_𝑛𝑣_𝑛≤1−𝛼_𝑛+1𝑣_𝑛+1−1−𝛼_𝑛𝑣_𝑛+ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−ϝ_𝜛𝑛1−𝛼_𝑛𝑣_𝑛. (18) Using Lemma 3, we obtain ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−ϝ_𝜛𝑛1−𝛼_𝑛𝑣_𝑛²≤𝜛_𝑛+1−𝜛_𝑛𝜛_𝑛+1×ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−ϝ_𝜛𝑛1−𝛼_𝑛𝑣_𝑛,ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−1−𝛼_𝑛𝑣_𝑛≤𝜛_𝑛+1−𝜛_𝑛𝜛_𝑛+1ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−ϝ_𝜛𝑛1−𝛼_𝑛𝑣_𝑛×ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−1−𝛼_𝑛𝑣_𝑛. (19) Then ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−ϝ_𝜛𝑛1−𝛼_𝑛𝑣_𝑛≤𝜛_𝑛+1−𝜛_𝑛𝜛_𝑛+1ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−1−𝛼_𝑛𝑣_𝑛. (20) By condition (𝐶3), we have 𝜛_𝑛>𝜂₁>0. So, 𝑧_𝑛+1−𝑧_𝑛≤1−𝛼_𝑛+1𝑣_𝑛+1−1−𝛼_𝑛𝑣_𝑛+𝜛_𝑛+1−𝜛_𝑛𝜛_𝑛+1×ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−1−𝛼_𝑛𝑣_𝑛≤1−𝛼_𝑛+1𝑣_𝑛+1−𝑣_𝑛+𝛼_𝑛+1−𝛼_𝑛𝑣_𝑛+𝜛_𝑛+1−𝜛_𝑛𝜂₁×ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−1−𝛼_𝑛𝑣_𝑛. (21) From (10), we have 𝑣_𝑛+1−𝑣_𝑛=proj_ℂΨ𝑢_𝑛+1−𝜇_𝑛+1𝔸𝑢_𝑛+1−proj_ℂΨ𝑢_𝑛−𝜇_𝑛𝔸𝑢_𝑛≤Ψ𝑢_𝑛+1−𝜇_𝑛+1𝔸𝑢_𝑛+1−Ψ𝑢_𝑛−𝜇_𝑛+1𝔸𝑢_𝑛+𝜇_𝑛+1−𝜇_𝑛𝔸𝑢_𝑛≤Ψ𝑢_𝑛+1−Ψ𝑢_𝑛+𝜇_𝑛+1−𝜇_𝑛𝔸𝑢_𝑛. (22) Therefore, 𝑧_𝑛+1−𝑧_𝑛≤1−𝛼_𝑛+1Ψ𝑢_𝑛+1−Ψ𝑢_𝑛+𝛼_𝑛+1−𝛼_𝑛𝑣_𝑛+𝜇_𝑛+1−𝜇_𝑛𝔸𝑢_𝑛+1𝜂₁𝜛_𝑛+1−𝜛_𝑛ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−1−𝛼_𝑛𝑣_𝑛. (23) It follows that 𝑧_𝑛+1−𝑧_𝑛−Ψ𝑢_𝑛+1−Ψ𝑢_𝑛≤𝛼_𝑛+1−𝛼_𝑛𝑣_𝑛+𝜇_𝑛+1−𝜇_𝑛𝔸𝑢_𝑛+1𝜂₁𝜛_𝑛+1−𝜛_𝑛ϝ_𝜛𝑛+11−𝛼_𝑛𝑣_𝑛−1−𝛼_𝑛𝑣_𝑛. (24) Since lim_𝑛→∞𝛼_𝑛=0, lim_𝑛→∞(𝜇_𝑛+1−𝜇_𝑛)=0, and lim_𝑛→∞(𝜛_𝑛+1−𝜛_𝑛)=0 and the sequences {Ψ(𝑢_𝑛)}, {𝑧_𝑛}, {𝑣_𝑛}, and {𝔸𝑢_𝑛} are bounded, we deduce that limsup_𝑛→∞𝑧_𝑛+1−𝑧_𝑛−Ψ𝑢_𝑛+1−Ψ𝑢_𝑛≤0. (25) Applying Lemma 4, we obtain lim_𝑛→∞𝑧_𝑛−Ψ𝑢_𝑛=0. (26) Thus, lim_𝑛→∞Ψ𝑢_𝑛+1−Ψ𝑢_𝑛=lim_𝑛→∞1−𝜅_𝑛𝑧_𝑛−Ψ𝑢_𝑛=0. (27) This together with the 𝜁-strong monotonicity of Ψ implies that lim_𝑛→∞𝑢_𝑛+1−𝑢_𝑛=0. (28) From (13) and (16), we derive Ψ𝑢_𝑛+1−Ψ(𝑥̆)²≤𝜅_𝑛Ψ𝑢_𝑛−Ψ(𝑥̆)²+1−𝜅_𝑛𝑧_𝑛−Ψ(𝑥̆)²≤𝜅_𝑛Ψ𝑢_𝑛−Ψ(𝑥̆)²+1−𝜅_𝑛×1−𝛼_𝑛𝑣_𝑛−Ψ(𝑥̆)²+𝛼_𝑛‖Ψ(𝑥̆)‖²≤1−𝜅_𝑛×1−𝛼_𝑛Ψ𝑢_𝑛−𝜇_𝑛𝔸𝑢_𝑛−Ψ(𝑥̆)−𝜇_𝑛𝔸𝑥̆²+𝛼_𝑛‖Ψ(𝑥̆)‖²+𝜅_𝑛Ψ𝑢_𝑛−Ψ(𝑥̆)²≤𝜅_𝑛Ψ𝑢_𝑛−Ψ(𝑥̆)²+1−𝜅_𝑛1−𝛼_𝑛×Ψ𝑢_𝑛−Ψ(𝑥̆)²+𝜇_𝑛𝜇_𝑛−2𝜍𝔸𝑢_𝑛−𝔸𝑥̆²+𝛼_𝑛‖Ψ(𝑥̆)‖²≤Ψ𝑢_𝑛−Ψ(𝑥̆)²+1−𝜅_𝑛1−𝛼_𝑛𝜇_𝑛𝜇_𝑛−2𝜍𝔸𝑢_𝑛−𝔸𝑥̆²+𝛼_𝑛‖Ψ(𝑥̆)‖². (29) Hence, 1−𝜅_𝑛1−𝛼_𝑛𝜇_𝑛2𝜍−𝜇_𝑛𝔸𝑢_𝑛−𝔸𝑥̆²≤Ψ𝑢_𝑛−Ψ(𝑥̆)²−Ψ𝑢_𝑛+1−Ψ(𝑥̆)²+𝛼_𝑛‖Ψ(𝑥̆)‖²≤Ψ𝑢_𝑛−Ψ(𝑥̆)+Ψ𝑢_𝑛+1(𝑥̆×Ψ𝑢−Ψ)_𝑛+1−Ψ𝑢_𝑛+𝛼_𝑛‖Ψ(𝑥̆)‖². (30) Since 𝛼_𝑛→0, ‖Ψ(𝑢_𝑛+1)−Ψ(𝑢_𝑛)‖→0, and liminf_𝑛→∞(1−𝜅_𝑛)(1−𝛼_𝑛)𝜇_𝑛(2𝜍−𝜇_𝑛)>0, we obtain lim_𝑛→∞𝔸𝑢_𝑛−𝔸𝑥̆=0. (31) Set 𝑦_𝑛=Ψ(𝑢_𝑛)−𝜇_𝑛𝔸𝑢_𝑛−(Ψ(𝑥̆)−𝜇_𝑛𝔸𝑥̆) for all 𝑛. By using the firm nonexpansivity of projection, we get 𝑣_𝑛−Ψ(𝑥̆)²=proj_ℂΨ𝑢_𝑛−𝜇_𝑛𝔸𝑢_𝑛²−proj_ℂΨ(𝑥̆)−𝜇_𝑛𝔸𝑥̆²≤𝑦_𝑛,𝑣_𝑛(𝑥̆=1−Ψ)2𝑦_𝑛²+𝑣_𝑛−Ψ(𝑥̆)²−𝑦_𝑛−𝑣_𝑛+Ψ(𝑥̆)²≤12Ψ𝑢_𝑛−Ψ(𝑥̆)²+𝑣_𝑛−Ψ(𝑥̆)²−Ψ𝑢_𝑛−𝑣_𝑛−𝜇_𝑛𝔸𝑢_𝑛−𝔸𝑥̆²=12Ψ𝑢_𝑛−Ψ(𝑥̆)²+𝑣_𝑛−Ψ(𝑥̆)²−Ψ𝑢_𝑛−𝑣_𝑛²−𝜇²_𝑛𝔸𝑢_𝑛−𝔸𝑥̆+2𝜇_𝑛Ψ𝑢_𝑛−𝑣_𝑛,𝔸𝑢_𝑛−𝔸𝑥̆. (32) It follows that 𝑣_𝑛−Ψ(𝑥̆)²≤Ψ𝑢_𝑛−Ψ(𝑥̆)²−Ψ𝑢_𝑛−𝑣_𝑛²+2𝜇_𝑛Ψ𝑢_𝑛−𝑣_𝑛𝔸𝑢_𝑛−𝔸𝑥̆. (33) From (29) and (32), we have Ψ𝑢_𝑛+1−Ψ(𝑥̆)²≤𝜅_𝑛Ψ𝑢_𝑛−Ψ(𝑥̆)²+1−𝜅_𝑛×1−𝛼_𝑛𝑣_𝑛−Ψ(𝑥̆)²+𝛼_𝑛‖Ψ(𝑥̆)‖²≤𝜅_𝑛Ψ𝑢_𝑛−Ψ(𝑥̆)²+1−𝛼_𝑛1−𝜅_𝑛×Ψ𝑢_𝑛−Ψ(𝑥̆)²−1−𝜅_𝑛Ψ𝑢_𝑛−𝑣_𝑛²+1−𝜅_𝑛𝛼_𝑛‖Ψ(𝑥̆)‖²+2𝜇_𝑛1−𝜅_𝑛×Ψ𝑢_𝑛−𝑣_𝑛𝔸𝑢_𝑛𝑥̆≤Ψ𝑢−𝔸_𝑛−Ψ(𝑥̆)²−1−𝜅_𝑛Ψ𝑢_𝑛−𝑣_𝑛²+2𝜇_𝑛Ψ𝑢_𝑛−𝑣_𝑛𝔸𝑢_𝑛−𝔸𝑥̆+𝛼_𝑛‖Ψ(𝑥̆)‖². (34) Then, we obtain 1−𝜅_𝑛Ψ𝑢_𝑛−𝑣_𝑛²≤Ψ𝑢_𝑛−Ψ(𝑥̆)+Ψ𝑢_𝑛+1(𝑥̆×Ψ𝑢−Ψ)_𝑛+1−Ψ𝑥_𝑛+2𝜇_𝑛Ψ𝑢_𝑛−𝑣_𝑛𝔸𝑢_𝑛−𝔸𝑥̆+𝛼_𝑛‖Ψ(𝑥̆)‖². (35) Since lim_𝑛→∞𝛼_𝑛=0, lim_𝑛→∞‖Ψ(𝑢_𝑛+1)−Ψ(𝑢_𝑛)‖=0, and lim_𝑛→∞‖𝔸𝑢_𝑛−𝔸𝑥̆‖=0, we deduce that lim_𝑛→∞Ψ𝑢_𝑛−𝑣_𝑛=0. (36) Next, we prove limsup_𝑛→∞⟨Ψ(𝑥^∗),𝑣_𝑛−Ψ(𝑥^∗)⟩≥0, where 𝑥^∗ satisfies (GVI): ⟨Ψ(𝑥^∗),Ψ(𝑥)−Ψ(𝑥^∗)⟩≥0, for all 𝑥∈Υ (note that Ψ is 𝜁-strongly monotone; we can easily deduce that the solution of (GVI) is unique). We take a subsequence {𝑣_𝑛𝑖} of {𝑣_𝑛} such that limsup_𝑛→∞Ψ𝑥^∗,𝑣_𝑛−Ψ𝑥^∗=lim_𝑖→∞Ψ𝑥^∗,𝑣_𝑛𝑖−Ψ𝑥^∗=lim_𝑖→∞Ψ𝑥^∗,Ψ𝑢_𝑛𝑖−Ψ𝑥^∗. (37)
By the boundedness of {𝑢_𝑛𝑖}, we can choose a subsequence {𝑢_𝑛𝑖𝑗} of {𝑢_𝑛𝑖} such that 𝑢_𝑛𝑖𝑗→𝑧 weakly. For the convenience, we may assume that 𝑢_𝑛𝑖⇀𝑧. This implies that Ψ(𝑢_𝑛𝑖)⇀Ψ(𝑧) due to the weak continuity of Ψ. Now, we show 𝑧∈Υ. We firstly show Ψ(𝑧)∈EP(𝜚,ℂ).
Note that 𝜛_𝑛∈(𝜂₁,𝜂₂). Then we choose a subsequence {𝜛_𝑛𝑖} of {𝜛_𝑛} such that lim_𝑖→∞𝜛_𝑛𝑖=𝜛∈(𝜂₁,𝜂₂). From (26) and (36), we deduce that ‖𝑧_𝑛−𝑣_𝑛‖=‖ϝ_𝜛𝑛(1−𝛼_𝑛)𝑣_𝑛−𝑣_𝑛‖→0. Thus, ‖𝑧_𝑛𝑖−𝑣_𝑛𝑖‖=‖ϝ_𝜛𝑛𝑖(1−𝛼_𝑛𝑖)𝑣_𝑛𝑖−𝑣_𝑛𝑖‖→0. From Lemma 2, we know that ϝ_𝜛 is nonexpansive. By demiclosed principle (Lemma 5), we get immediately that Ψ(𝑧)∈Fix(ϝ_𝜛)=EP(𝜚,ℂ).
Next we prove 𝑧∈VI(𝔸,Ψ). Set 𝑅𝑣=𝔸𝑣+𝑁_𝐶(𝑣),𝑣∈𝐶,∅,𝑣∉𝐶. (38)
By [27], we know that 𝑅 is maximal Ψ-monotone. Let (𝑣,𝑤)∈𝐺(𝑅). Since 𝑤−𝔸𝑣∈𝑁_𝐶(𝑣) and 𝑢_𝑛∈𝐶, we have ⟨Ψ(𝑣)−Ψ(𝑢_𝑛),𝑤−𝔸𝑣⟩≥0. Noting that 𝑣_𝑛=proj_ℂ(Ψ(𝑢_𝑛)−𝜇_𝑛𝔸𝑢_𝑛), we get Ψ(𝑣)−𝑣_𝑛,𝑣_𝑛−Ψ𝑢_𝑛−𝜇_𝑛𝔸𝑢_𝑛≥0. (39) It follows that Ψ(𝑣)−𝑣_𝑛,𝑣_𝑛−Ψ𝑢_𝑛𝜇_𝑛+𝔸𝑢_𝑛≥0. (40) Then, Ψ(𝑣)−Ψ𝑢_𝑛𝑖≥Ψ(𝑣)𝑢,𝑤−Ψ_𝑛𝑖≥Ψ(𝑣)𝑢,𝔸𝑣−Ψ_𝑛𝑖−Ψ(𝑣),𝔸𝑣−𝑣_𝑛𝑖,𝑣_𝑛𝑖−Ψ𝑢_𝑛𝑖𝜇_𝑛𝑖−Ψ(𝑣)−𝑣_𝑛𝑖,𝔸𝑢_𝑛𝑖=Ψ(𝑣)−Ψ𝑢_𝑛𝑖,𝔸𝑣−𝔸𝑢_𝑛𝑖+Ψ(𝑣)−Ψ𝑢_𝑛𝑖,𝔸𝑢_𝑛𝑖−Ψ(𝑣)−𝑣_𝑛𝑖,𝑣_𝑛𝑖−Ψ𝑢_𝑛𝑖𝜇_𝑛𝑖−Ψ(𝑣)−𝑣_𝑛𝑖,𝔸𝑢_𝑛𝑖≥−Ψ(𝑣)−𝑣_𝑛𝑖,𝑣_𝑛𝑖−Ψ𝑢_𝑛𝑖𝜇_𝑛𝑖−Ψ𝑢_𝑛𝑖−𝑣_𝑛𝑖,𝔸𝑢_𝑛𝑖. (41) Since ‖Ψ(𝑢_𝑛𝑖)−𝑣_𝑛𝑖‖→0 and Ψ(𝑢_𝑛𝑖)⇀Ψ(𝑧), we deduce that ⟨Ψ(𝑣)−Ψ(𝑧),𝑤⟩≥0 by taking 𝑖→∞ in (41). Thus, 𝑧∈𝑅⁻¹0 by the maximal Ψ-monotonicity of 𝑅. Hence, 𝑧∈VI(𝔸,Ψ). Therefore, 𝑧∈Υ. From (37), we obtain limsup_𝑛→∞Ψ𝑥^∗,𝑣_𝑛−Ψ𝑥^∗=lim_𝑖→∞Ψ𝑥^∗,Ψ𝑢_𝑛𝑖−Ψ𝑥^∗=Ψ𝑥^∗,Ψ(𝑧)−Ψ𝑥^∗≥0. (42) From (12), we have Ψ𝑢_𝑛+1−Ψ𝑥^∗2≤𝜅_𝑛Ψ𝑢_𝑛−Ψ𝑥^∗2+1−𝜅_𝑛1−𝛼_𝑛𝑣_𝑛−Ψ𝑥^∗2≤𝜅_𝑛Ψ𝑢_𝑛−Ψ𝑥^∗2+1−𝜅_𝑛×1−𝛼_𝑛𝑣_𝑛−Ψ𝑥^∗2−2𝛼_𝑛1−𝛼_𝑛Ψ𝑥^∗,𝑣_𝑛−Ψ𝑥^∗+𝛼²_𝑛Ψ𝑥^∗2≤𝜅_𝑛Ψ𝑢_𝑛−Ψ𝑥^∗2+1−𝜅_𝑛×1−𝛼_𝑛Ψ𝑢_𝑛−Ψ𝑥^∗2−2𝛼_𝑛1−𝛼_𝑛×Ψ𝑥^∗,𝑣_𝑛−Ψ𝑥^∗+𝛼²_𝑛Ψ𝑥^∗2=1−1−𝜅_𝑛𝛼_𝑛Ψ𝑢_𝑛−Ψ𝑥^∗2+1−𝜅_𝑛𝛼_𝑛×21−𝛼_𝑛−Ψ𝑥^∗,𝑣_𝑛−Ψ𝑥^∗+𝛼_𝑛Ψ𝑥^∗2. (43) Using Lemma 6, we conclude that Ψ(𝑢_𝑛)→Ψ(𝑥^∗), and hence 𝑢_𝑛→𝑥^∗. This completes the proof.