Iterative Methods for the Sum of Two Monotone Operators

Abstract

We introduce an iterative for finding the zeros point of the sum of two monotone operators. We prove that the suggested method converges strongly to the zeros point of the sum of two monotone operators.

1. Introduction

Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let 𝐴∶𝐶→𝐻 be a single-valued nonlinear mapping and let 𝐵∶𝐻→2^𝐻 be a multivalued mapping. The “so-called” quasi-variational inclusion problem is to find a 𝑢∈2^𝐻 such that0∈𝐴𝑥+𝐵𝑥.(1.1) The set of solutions of (1.1) is denoted by (𝐴+𝐵)⁻¹(0). A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions; see, for instance, [1–4]. The problem (1.1) includes many problems as special cases.(1)If 𝐵=𝜕𝜙∶𝐻→2^𝐻, where 𝜙∶𝐻→𝑅∪+∞ is a proper convex lower semicontinuous function and 𝜕𝜙 is the subdif and if onlyerential of 𝜙, then the variational inclusion problem (1.1) is equivalent to find 𝑢∈𝐻 such that ⟨𝐴𝑢,𝑦−𝑢⟩+𝜙(𝑦)−𝜙(𝑢)≥0,∀𝑦∈𝐻,(1.2) which is called the mixed quasi-variational inequality (see, Noor [5]).(2)If 𝐵=𝜕𝛿_𝐶, where 𝐶 is a nonempty closed convex subset of 𝐻 and 𝛿_𝐶∶𝐻→[0,∞] is the indicator function of 𝐶, that is, 𝛿_𝐶=0+,∞,𝑥𝑥∈∉𝐶𝐶,,(1.3) then the variational inclusion problem (1.1) is equivalent to find 𝑢∈𝐶 such that ⟨𝐴𝑢,𝑣−𝑢⟩≥0,∀𝑣∈𝐶.(1.4)

This problem is called Hartman-Stampacchia variational inequality (see, e.g., [6]).

Recently, Zhang et al. [7] introduced a new iterative scheme for finding a common element of the set of solutions to the inclusion problem, and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [8] introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping. For some related works, please see [9–27] and the references therein.

Inspired and motivated by the works in the literature, in this paper, we introduce an iterative for solving the problem (1.1). We prove that the suggested method converges strongly to the zeros point of the sum of two monotone operators 𝐴+𝐵.

2. Preliminaries

Let 𝐻 be a real Hilbert space with inner product ⟨⋅,⋅⟩ and norm ‖⋅‖, respectively. Let 𝐶 be a nonempty closed convex subset of 𝐻. Recall that a mapping 𝐴∶𝐶→𝐻 is said to be 𝛼-inverse strongly-monotone if and if only ⟨𝐴𝑥−𝐴𝑦,𝑥−𝑦⟩≥𝛼‖𝐴𝑥−𝐴𝑦‖²(2.1) for some 𝛼>0 and for all 𝑥,𝑦∈𝐶. It is known that if 𝐴 is 𝛼-inverse strongly monotone, then ‖𝐴𝑥−𝐴𝑦‖≤1𝛼‖𝑥−𝑦‖(2.2) for all 𝑥,𝑦∈𝐶.

Let 𝐵 be a mapping of 𝐻 into 2^𝐻. The effective domain of 𝐵 is denoted by dom(𝐵), that is,dom(𝐵)={𝑥∈𝐻∶𝐵𝑥≠∅}.(2.3) A multivalued mapping 𝐵 is said to be a monotone operator on 𝐻 if and if only ⟨𝑥−𝑦,𝑢−𝑣⟩≥0(2.4) for all 𝑥,𝑦∈dom(𝐵), 𝑢∈𝐵𝑥, and 𝑣∈𝐵𝑦. A monotone operator 𝐵 on 𝐻 is said to be maximal if and if only its graph is not strictly contained in the graph of any other monotone operator on 𝐻. Let 𝐵 be a maximal monotone operator on 𝐻 and let 𝐵⁻¹0={𝑥∈𝐻∶0∈𝐵𝑥}.

For a maximal monotone operator 𝐵 on 𝐻 and 𝜆>0, we may define a single-valued operator: 𝐽^𝐵𝜆=(𝐼+𝜆𝐵)⁻¹∶𝐻→dom(𝐵),(2.5) which is called the resolvent of 𝐵 for 𝜆. It is known that the resolvent 𝐽^𝐵𝜆 is firmly nonexpansive, that is, ‖‖𝐽^𝐵𝜆𝑥−𝐽^𝐵𝜆𝑦‖‖²≤𝐽^𝐵𝜆𝑥−𝐽^𝐵𝜆𝑦,𝑥−𝑦(2.6) for all 𝑥,𝑦∈𝐶 and 𝐵⁻¹0=𝐹(𝐽^𝐵𝜆) for all 𝜆>0.

The following resolvent identity is well known: for 𝜆>0 and 𝜇>0, there holds the following identity:𝐽^𝐵𝜆𝑥=𝐽^𝐵𝜇𝜇𝜆𝑥+1−𝜇𝜆𝐽^𝐵𝜆𝑥,𝑥∈𝐻.(2.7)

We use the following notation: (i)𝑥_𝑛⇀𝑥 stands for the weak convergence of (𝑥_𝑛) to 𝑥; (ii)𝑥_𝑛→𝑥 stands for the strong convergence of (𝑥_𝑛) to 𝑥.

We need the following lemmas for the next section.

Lemma 2.1 (see [28]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻. Let the mapping 𝐴∶𝐶→𝐻 be 𝛼-inverse strongly monotone and let 𝜆>0 be a constant. Then, one has ‖(𝐼−𝜆𝐴)𝑥−(𝐼−𝜆𝐴)𝑦‖²≤‖𝑥−𝑦‖²+𝜆(𝜆−2𝛼)‖𝐴𝑥−𝐴𝑦‖²,∀𝑥,𝑦∈𝐶.(2.8) In particular, if 0≤𝜆≤2𝛼, then 𝐼−𝜆𝐴 is nonexpansive.

Lemma 2.2 (see [29]). Let {𝑥_𝑛} and {𝑦_𝑛} be bounded sequences in a Banach space 𝑋 and let {𝛽_𝑛} be a sequence in [0,1] with 0<liminf_𝑛→∞𝛽_𝑛≤limsup_𝑛→∞𝛽_𝑛<1.(2.9) Suppose that 𝑥_𝑛+1=1−𝛽_𝑛𝑦_𝑛+𝛽_𝑛𝑥_𝑛(2.10) for all 𝑛≥0 and limsup_𝑛→∞‖‖𝑦_𝑛+1−𝑦_𝑛‖‖−‖‖𝑥_𝑛+1−𝑥_𝑛‖‖≤0.(2.11) Then, lim_𝑛→∞‖𝑦_𝑛−𝑥_𝑛‖=0.

Lemma 2.3 (see [30]). Assume that {𝑎_𝑛} is a sequence of nonnegative real numbers such that 𝑎_𝑛+1≤1−𝛾_𝑛𝑎_𝑛+𝛿_𝑛𝛾_𝑛,(2.12) where {𝛾_𝑛} is a sequence in (0,1) and {𝛿_𝑛} is a sequence such that (1)∑^∞𝑛=1𝛾_𝑛=∞; (2)limsup_𝑛→∞𝛿_𝑛≤0 or ∑^∞𝑛=1|𝛿_𝑛𝛾_𝑛|<∞. Then lim_𝑛→∞𝑎_𝑛=0.

3. Main Results

In this section, we will prove our main result.

Theorem 3.1. Let 𝐶 be a nonempty closed and convex subset of a real Hilbert space 𝐻. Let 𝐴 be an 𝛼-inverse strongly monotone mapping of 𝐶 into H and let 𝐵 be a maximal monotone operator on 𝐻, such that the domain of 𝐵 is included in 𝐶. Let 𝐽^𝐵𝜆=(𝐼+𝜆𝐵)⁻¹ be the resolvent of 𝐵 for 𝜆>0. Suppose that (𝐴+𝐵)⁻¹0≠∅. For 𝑢∈𝐶 and given 𝑥₀∈𝐶, let {𝑥_𝑛}⊂𝐶 be a sequence generated by 𝑥_𝑛+1=𝛽_𝑛𝑥_𝑛+1−𝛽_𝑛𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛(3.1) for all 𝑛≥0, where {𝜆_𝑛}⊂(0,2𝛼), {𝛼_𝑛}⊂(0,1), and {𝛽_𝑛}⊂(0,1) satisfy (i)lim_𝑛→∞𝛼_𝑛=0 and ∑_𝑛𝛼_𝑛=∞; (ii)0<liminf_𝑛→∞𝛽_𝑛≤limsup_𝑛→∞𝛽_𝑛<1; (iii)𝑎≤𝜆_𝑛≤𝑏 where [𝑎,𝑏]⊂(0,2𝛼) and lim_𝑛→∞(𝜆_𝑛+1−𝜆_𝑛)=0. Then {𝑥_𝑛} generated by (3.1) converges strongly to 𝑥̃=𝑃_(𝐴+𝐵)⁻¹₀(𝑢).

Proof. First, we choose any 𝑧∈(𝐴+𝐵)⁻¹0. Note that 𝑧=𝐽^𝐵𝜆_𝑛𝑧−𝜆_𝑛1−𝛼_𝑛𝐴𝑧=𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑧+1−𝛼_𝑛𝑧−𝜆_𝑛𝐴𝑧(3.2) for all 𝑛≥0. Since 𝐽^𝐵𝜆 is nonexpansive for all 𝜆>0, we have ‖‖𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑧‖‖²=‖‖𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑧+1−𝛼_𝑛𝑧−𝜆_𝑛𝐴𝑧‖‖²≤‖‖𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝛼_𝑛𝑧+1−𝛼_𝑛𝑧−𝜆_𝑛𝐴𝑧‖‖²=‖‖1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑧−𝜆_𝑛𝐴𝑧+𝛼_𝑛(𝑢−𝑧)‖‖².(3.3) Since 𝐴 is 𝛼-inverse strongly monotone, we get ‖‖1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑧−𝜆_𝑛𝐴𝑧+𝛼_𝑛(𝑢−𝑧)‖‖²≤1−𝛼_𝑛‖‖𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑧−𝜆_𝑛𝐴𝑧‖‖²+𝛼_𝑛‖𝑢−𝑧‖²=1−𝛼_𝑛‖‖𝑥_𝑛−𝑧−𝜆_𝑛𝐴𝑥_𝑛−𝐴𝑧‖‖²+𝛼_𝑛‖𝑢−𝑧‖²=1−𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²−2𝜆_𝑛⟨𝐴𝑥_𝑛−𝐴𝑧,𝑥_𝑛−𝑧⟩+𝜆^2𝑛‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²+𝛼^𝑛‖𝑢−𝑧‖²≤1−𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²−2𝛼𝜆_𝑛‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²+𝜆^2𝑛‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²+𝛼^𝑛‖𝑢−𝑧‖²=1−𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²+𝜆_𝑛𝜆_𝑛−2𝛼‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²+𝛼^𝑛‖𝑢−𝑧‖².(3.4) By (3.3) and (3.4), we obtain ‖‖𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑧‖‖²≤1−𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²+𝜆_𝑛𝜆_𝑛−2𝛼‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²+𝛼^𝑛‖𝑢−𝑧‖²≤1−𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²+𝛼_𝑛‖𝑢−𝑧‖².(3.5) It follows from (3.1) and (3.5) that ‖‖𝑥_𝑛+1−𝑧‖‖²=‖‖𝛽_𝑛𝑥_𝑛−𝑧+1−𝛽_𝑛𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑧‖‖²≤𝛽_𝑛‖‖𝑥_𝑛−𝑧‖‖²+1−𝛽_𝑛‖‖𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑧‖‖²≤=𝛽1^𝑛‖‖−𝑥_𝑛1−−𝑧‖‖𝛽²_𝑛+1−𝛽_𝑛1−𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²+𝛼_𝑛‖𝑢−𝑧‖²𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²+1−𝛽_𝑛𝛼_𝑛‖𝑢−𝑧‖²≤max‖‖𝑥_𝑛−𝑧‖‖²,‖𝑢−𝑧‖².(3.6) By induction, we have ‖‖𝑥_𝑛+1−𝑧‖‖≤max‖‖𝑥₀−𝑧‖‖,‖𝑢−𝑧‖.(3.7) Therefore, {𝑥_𝑛} is bounded. We deduce immediately that {𝐴𝑥_𝑛} is also bounded. Set 𝑢_𝑛=𝛼_𝑛𝑢+(1−𝛼_𝑛)(𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛) for all 𝑛. Then {𝑢_𝑛} and {𝐽^𝐵𝜆_𝑛𝑢_𝑛} are bounded.
Next, we estimate ‖𝐽^𝐵𝜆_𝑛+1𝑢_𝑛+1−𝐽^𝐵𝜆_𝑛𝑢_𝑛‖. In fact, we have ‖‖𝐽=^𝐵𝜆_𝑛+1𝑢‖‖𝐽_𝑛+1−_{𝐵𝜆𝑛+1}𝐽^𝐵𝜆_𝑛𝑢_𝑛‖‖≤‖‖𝐽𝛼_𝑛+1𝑢+1−𝛼_𝑛+1𝑥_𝑛+1−𝜆_𝑛+1𝐴𝑥_𝑛+1−𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛_{𝐵𝜆𝑛+1}‖‖+‖‖𝐽𝛼_𝑛+1𝑢+1−𝛼_𝑛+1𝑥_𝑛+1−𝜆_𝑛+1𝐴𝑥_𝑛+1−𝐽^𝐵𝜆_𝑛+1𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛_{𝐵𝜆𝑛+1}‖‖≤‖‖𝛼𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛_𝑛+1‖‖+‖‖𝐽𝑢+1−𝛼_𝑛+1𝑥_𝑛+1−𝜆_𝑛+1𝐴𝑥_𝑛+1−𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛_{𝐵𝜆𝑛+1}‖‖≤‖‖𝐼−𝜆𝑢_𝑛−_𝑛+1𝐽^𝐵𝜆_𝑛𝑢_𝑛‖‖+𝛼_𝑛+1𝐴𝑥_𝑛+1−𝐼−𝜆_𝑛+1𝐴𝑥_𝑛‖‖+||𝜆_𝑛+1−𝜆_𝑛||‖‖𝐴𝑥_𝑛‖‖‖𝑢‖+‖‖𝑥_𝑛+1‖‖+𝜆_𝑛+1‖‖𝐴𝑥_𝑛+1‖‖+𝛼_𝑛‖𝑢‖+‖‖𝑥_𝑛‖‖+𝜆_𝑛‖‖𝐴𝑥_𝑛‖‖+‖‖𝐽^𝐵𝜆_𝑛+1𝑢_𝑛−𝐽^𝐵𝜆_𝑛𝑢_𝑛‖‖.(3.8)
Since 𝐼−𝜆_𝑛+1𝐴 is nonexpansive for 𝜆_𝑛+1∈(0,2𝛼), we have ‖(𝐼−𝜆_𝑛+1𝐴)𝑥_𝑛+1−(𝐼−𝜆_𝑛+1𝐴)𝑥_𝑛‖≤‖𝑥_𝑛+1−𝑥_𝑛‖. By the resolvent identity (2.7), we have 𝐽^𝐵𝜆_𝑛+1𝑢_𝑛=𝐽^𝐵𝜆_𝑛𝜆𝜆^𝑛_𝑛+1𝑢_𝑛+1−𝜆^𝑛𝜆_𝑛+1𝐽^{𝐵𝜆𝑛+1}𝑢_𝑛.(3.9) It follows that ‖‖𝐽^𝐵𝜆_𝑛+1𝑢_𝑛−𝐽^𝐵𝜆_𝑛𝑢_𝑛‖‖=‖‖‖𝐽^𝐵𝜆_𝑛𝜆𝜆^𝑛_𝑛+1𝑢_𝑛+1−𝜆^𝑛𝜆_𝑛+1𝐽^{𝐵𝜆𝑛+1}𝑢_𝑛−𝐽^𝐵𝜆𝑛𝑢_𝑛≤‖‖‖𝜆_𝑛‖‖‖𝜆_𝑛+1𝑢_𝑛+1−𝜆^𝑛𝜆_𝑛+1𝐽^{𝐵𝜆𝑛+1}𝑢_𝑛−𝑢^𝑛≤||𝜆_𝑛+1‖‖‖−𝜆_𝑛||𝜆_𝑛+1‖‖𝑢_𝑛−𝐽^𝐵𝜆_𝑛+1𝑢_𝑛‖‖.(3.10) So, ‖‖𝐽^𝐵𝜆_𝑛+1𝑢_𝑛+1−𝐽^𝐵𝜆_𝑛𝑢_𝑛‖‖≤+‖‖𝛼𝑥_𝑛+1𝑛−𝑥_𝑛‖‖+||𝜆_𝑛+1−𝜆_𝑛||‖‖𝐴𝑥_𝑛‖‖+𝛼_𝑛+1‖𝑢‖+‖‖𝑥_𝑛+1‖‖+𝜆_𝑛+1‖‖𝐴𝑥_𝑛+1‖‖‖𝑢‖+‖‖𝑥_𝑛‖‖+𝜆_𝑛‖‖𝐴𝑥_𝑛‖‖+||𝜆^𝑛+1−𝜆_𝑛||𝜆_𝑛+1‖‖𝑢_𝑛−𝐽^𝐵𝜆_𝑛+1𝑢_𝑛‖‖.(3.11) Thus, limsup_𝑛→∞‖‖𝐽^𝐵𝜆_𝑛+1𝑢_𝑛+1−𝐽^𝐵𝜆_𝑛𝑢_𝑛‖‖−‖‖𝑥_𝑛+1−𝑥_𝑛‖‖≤0.(3.12) From Lemma 2.2, we get lim_𝑛→∞‖‖𝐽^𝐵𝜆_𝑛𝑢_𝑛−𝑥_𝑛‖‖=0.(3.13) Consequently, we obtain lim_𝑛→∞‖‖𝑥_𝑛+1−𝑥_𝑛‖‖=lim_𝑛→∞1−𝛽_𝑛‖‖𝐽^𝐵𝜆_𝑛𝑢_𝑛−𝑥_𝑛‖‖=0.(3.14) From (3.5) and (3.6), we have ‖‖𝑥_𝑛+1−𝑧‖‖²≤𝛽_𝑛‖‖𝑥_𝑛−𝑧‖‖²+1−𝛽_𝑛‖‖𝐽^𝐵𝜆_𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑧‖‖²≤+1𝛽_𝑛−𝛽_𝑛1−𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²+𝜆_𝑛𝜆_𝑛−2𝛼‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²+𝛼^𝑛‖𝑢−𝑧‖²‖‖𝑥_𝑛−𝑧‖‖²=1−1−𝛽_𝑛𝛼_𝑛‖‖𝑥_𝑛−𝑧‖‖²+1−𝛽_𝑛𝜆_𝑛𝜆_𝑛−2𝛼‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²+1−𝛽_𝑛𝛼_𝑛‖𝑢−𝑧‖²≤‖‖𝑥_𝑛−𝑧‖‖²+1−𝛽_𝑛𝜆_𝑛𝜆_𝑛−2𝛼‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²+1−𝛽_𝑛𝛼_𝑛‖𝑢−𝑧‖².(3.15) It follows that 1−𝛽_𝑛𝜆_𝑛2𝛼−𝜆_𝑛‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖²≤‖‖𝑥_𝑛−𝑧‖‖²−‖‖𝑥_𝑛+1−𝑧‖‖²+1−𝛽_𝑛𝛼_𝑛‖𝑢−𝑧‖²≤‖‖𝑥_𝑛−𝑧‖‖−‖‖𝑥_𝑛+1−𝑧‖‖‖‖𝑥_𝑛+1−𝑥_𝑛‖‖+1−𝛽_𝑛𝛼_𝑛‖𝑢−𝑧‖².(3.16) Since lim_𝑛→∞𝛼_𝑛=0, lim_𝑛→∞‖𝑥_𝑛+1−𝑥_𝑛‖=0, and liminf_𝑛→∞(1−𝛽_𝑛)𝜆_𝑛(2𝛼−𝜆_𝑛)>0, we have lim_𝑛→∞‖‖𝐴𝑥_𝑛−𝐴𝑧‖‖=0.(3.17) Put 𝑥̃=𝑃_(𝐴+𝐵)⁻¹₀(𝑢). Set 𝑣_𝑛=𝑥_𝑛−(𝜆_𝑛/(1−𝛼_𝑛))(𝐴𝑥_𝑛−𝐴𝑥̃) for all 𝑛. Take 𝑧=𝑥̃ in (3.17) to get ‖𝐴𝑥_𝑛−𝐴𝑥̃‖→0. First, we prove limsup_𝑛→∞⟨𝑢−𝑥̃,𝑣_𝑛−𝑥̃⟩≤0. We take a subsequence {𝑣_𝑛𝑖} of {𝑣_𝑛} such that limsup_𝑛→∞⟨𝑢−𝑥̃,𝑣_𝑛−𝑥̃⟩=lim_𝑖→∞𝑢−𝑥̃,𝑣_𝑛𝑖−𝑥̃.(3.18) It is clear that {𝑣_𝑛𝑖} is bounded due to the boundedness of {𝑥_𝑛} and ‖𝐴𝑥_𝑛−𝐴𝑥̃‖→0. Then, there exists a subsequence {𝑣_𝑛𝑖𝑗} of {𝑣_𝑛𝑖} which converges weakly to some point 𝑤∈𝐶. Hence, {𝑥_𝑛𝑖𝑗} also converges weakly to 𝑤 because of ‖𝑣_𝑛𝑖𝑗−𝑥_𝑛𝑖𝑗‖→0. By the similar argument as that in [31], we can show that 𝑤∈(𝐴+𝐵)⁻¹0. This implies that limsup_𝑛→∞⟨𝑢−𝑥̃,𝑣_𝑛−𝑥̃⟩=lim_𝑗→∞𝑢−𝑥̃,𝑣^𝑛_𝑖𝑗−𝑥̃=⟨𝑢−𝑥̃,𝑤−𝑥̃⟩.(3.19) Note that 𝑥̃=𝑃_(𝐴+𝐵)⁻¹₀(𝑢). Then, ⟨𝑢−𝑥̃,𝑤−𝑥̃⟩≤0,𝑤∈(𝐴+𝐵)⁻¹0. Therefore, limsup_𝑛→∞⟨𝑢−𝑥̃,𝑣_𝑛−𝑥̃⟩≤0.(3.20) Finally, we prove that 𝑥_𝑛→𝑥̃. From (3.1), we have ‖‖𝑥_𝑛+1−𝑥̃‖‖²≤𝛽_𝑛‖‖𝑥_𝑛−𝑥̃‖‖²+1−𝛽_𝑛‖‖𝐽^𝐵𝜆_𝑛𝑢_𝑛−𝑥̃‖‖²=𝛽_𝑛‖‖𝑥_𝑛−𝑥̃‖‖²+1−𝛽_𝑛‖‖𝐽^𝐵𝜆_𝑛𝑢_𝑛−𝐽^𝐵𝜆_𝑛𝑥̃−1−𝛼_𝑛𝜆_𝑛𝐴𝑥̃‖‖²≤𝛽_𝑛‖‖𝑥_𝑛−𝑥̃‖‖²+1−𝛽_𝑛‖‖𝑢_𝑛−𝑥̃−1−𝛼_𝑛𝜆_𝑛𝐴𝑥̃‖‖²=𝛽_𝑛‖‖𝑥_𝑛−𝑥̃‖‖²+1−𝛽_𝑛‖‖𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑥̃−1−𝛼_𝑛𝜆_𝑛𝐴𝑥̃‖‖²=𝛽_𝑛‖‖𝑥_𝑛−𝑥̃‖‖²+1−𝛽_𝑛‖‖1−𝛼_𝑛𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑥̃−𝜆_𝑛𝐴𝑥̃+𝛼_𝑛(𝑢−𝑥̃)‖‖²=×𝛽_𝑛‖‖𝑥1_𝑛−−𝑥𝛼̃_𝑛‖‖²+1−𝛽_𝑛²‖‖𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑥̃−𝜆_𝑛𝐴𝑥̃‖‖²≤𝛽_𝑛+2𝛼_𝑛1−𝛼_𝑛𝑢−𝑥̃,𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝑥̃−𝜆_𝑛𝐴𝑥̃+𝛼^2𝑛‖𝑢−𝑥̃‖²×‖‖𝑥1_𝑛−−𝑥𝛼̃_𝑛‖‖²+1−𝛽_𝑛+𝛼_2𝑛‖‖𝑥_𝑛−𝑥̃‖‖²+2𝛼_𝑛1−𝛼_𝑛𝑢−𝑥̃,𝑥_𝑛−𝜆_𝑛𝐴𝑥_𝑛−𝐴𝑥̃−𝑥̃≤1−1‖𝑢−−𝛽_𝑛𝑥̃‖²𝛼_𝑛‖‖𝑥_𝑛−𝑥̃‖‖²+1−𝛽_𝑛𝛼_𝑛21−𝛼_𝑛⟨𝑢−𝑥̃,𝑣_𝑛−𝑥̃⟩+𝛼_𝑛‖𝑢−𝑥̃‖².(3.21) It is clear that ∑_𝑛(1−𝛽_𝑛)𝛼_𝑛=∞ and limsup_𝑛→∞(2(1−𝛼_𝑛)⟨𝑢−𝑥̃,𝑣_𝑛−𝑥̃⟩+𝛼_𝑛‖𝑢−𝑥̃‖²)≤0. We can therefore apply Lemma 2.3 to conclude that 𝑥_𝑛→𝑥̃. This completes the proof.

4. Applications

Next, we consider the problem for finding the minimum norm solution of a mathematical model related to equilibrium problems. Let 𝐶 be a nonempty, closed, and convex subset of a Hilbert space and let 𝐺∶𝐶×𝐶→𝑅 be a bifunction satisfying the following conditions:(E1)𝐺(𝑥,𝑥)=0 for all 𝑥∈𝐶;(E2)𝐺 is monotone, that is, 𝐺(𝑥,𝑦)+𝐺(𝑦,𝑥)≤0 for all 𝑥,𝑦∈𝐶;(E3)for all 𝑥,𝑦,𝑧∈𝐶, limsup_𝑡↓0𝐺(𝑡𝑧+(1−𝑡)𝑥,𝑦)≤𝐺(𝑥,𝑦);(E4)for all 𝑥∈𝐶, 𝐺(𝑥,⋅) is convex and lower semicontinuous.

Then, the mathematical model related to equilibrium problems (with respect to 𝐶) is to find 𝑥̃∈𝐶 such that𝐺(𝑥̃,𝑦)≥0(4.1) for all 𝑦∈𝐶. The set of such solutions 𝑥̃ is denoted by 𝐸𝑃(𝐺). The following lemma appears implicitly in Blum and Oettli [32].

Lemma 4.1. Let 𝐶 be a nonempty, closed, and convex subset of 𝐻 and let 𝐺 be a bifunction of 𝐶×𝐶 into 𝑅 satisfying (E1)–(E4). Let 𝑟>0 and 𝑥∈𝐻. Then, there exists 𝑧∈𝐶 such that 𝐺(𝑧,𝑦)+1𝑟⟨𝑦−𝑧,𝑧−𝑥⟩≥0,∀𝑦∈C.(4.2)

The following lemma was given by Combettes and Hirstoaga [33].

Lemma 4.2. Assume that 𝐺∶𝐶×𝐶→𝑅 satisfies (E1)–(E4). For 𝑟>0 and 𝑥∈𝐻, define a mapping 𝑇_𝑟∶𝐻→𝐶 as follows: 𝑇_𝑟(𝑥)=𝑧∈𝐶∶𝐺(𝑧,𝑦)+1𝑟⟨𝑦−𝑧,𝑧−𝑥⟩≥0,∀𝑦∈𝐶(4.3) for all 𝑥∈𝐻. Then, the following holds: (1)𝑇_𝑟 is single valued; (2)𝑇_𝑟 is a firmly nonexpansive mapping, that is, for all 𝑥,𝑦∈𝐻, ‖‖𝑇_𝑟𝑥−𝑇_𝑟𝑦‖‖²≤⟨𝑇_𝑟𝑥−𝑇_𝑟𝑦,𝑥−𝑦⟩;(4.4)(3)𝐹(𝑇_𝑟)=𝐸𝑃(𝐺); (4)𝐸𝑃(𝐺) is closed and convex.

We call such 𝑇_𝑟 the resolvent of 𝐺 for 𝑟>0. Using Lemmas 4.1 and 4.2, we have the following lemma. See [34] for a more general result.

Lemma 4.3. Let 𝐻 be a Hilbert space and let 𝐶 be a nonempty, closed, and convex subset of 𝐻. Let 𝐺∶𝐶×𝐶→𝑅 satisfy (E1)–(E4). Let 𝐴_𝐺 be a multivalued mapping of 𝐻 into itself defined by 𝐴_𝐺𝑥={∅𝑧,∈𝐻∶𝐺(𝑥,𝑦)≥⟨𝑦−𝑥,𝑧⟩,∀𝑦∈𝐶},𝑥𝑥∈∉𝐶𝐶,.(4.5) Then, 𝐸𝑃(𝐺)=𝐴^𝐺−1(0) and 𝐴_𝐺 is a maximal monotone operator with dom(𝐴_𝐺)⊂𝐶. Further, for any 𝑥∈𝐻 and 𝑟>0, the resolvent 𝑇_𝑟 of 𝐺 coincides with the resolvent of 𝐴_𝐺; that is, 𝑇_𝑟𝑥=(𝐼+𝑟𝐴_𝐺)⁻¹𝑥.(4.6)

Form Lemma 4.3 and Theorems 3.1, we have the following result.

Theorem 4.4. Let 𝐶 be a nonempty, closed, and convex subset of a real Hilbert space 𝐻. Let 𝐺 be a bifunction from 𝐶×𝐶→𝑅 satisfying (E1)–(E4) and let 𝑇_𝜆 be the resolvent of 𝐺 for 𝜆>0. Suppose 𝐸𝑃(𝐺)≠∅. For 𝑢∈𝐶 and given 𝑥₀∈𝐶, let {𝑥_𝑛}⊂𝐶 be a sequence generated by 𝑥_𝑛+1=𝛽_𝑛𝑥_𝑛+1−𝛽_𝑛𝑇_𝜆𝑛𝛼_𝑛𝑢+1−𝛼_𝑛𝑥_𝑛(4.7) for all 𝑛≥0, where {𝜆_𝑛}⊂(0,∞), {𝛼_𝑛}⊂(0,1), and {𝛽_𝑛}⊂(0,1) satisfy (i)lim_𝑛→∞𝛼_𝑛=0 and ∑_𝑛𝛼_𝑛=∞;(ii)0<liminf_𝑛→∞𝛽_𝑛≤limsup_𝑛→∞𝛽_𝑛<1;(iii)𝑎≤𝜆_𝑛≤𝑏 where [𝑎,𝑏]⊂(0,∞) and lim_𝑛→∞(𝜆_𝑛+1−𝜆_𝑛)=0.Then {𝑥_𝑛} converges strongly to a point 𝑥̃=𝑃_{𝐸𝑃(𝐺)}(𝑢).

Acknowledgment

The author was supported in part by NSC 100-2221-E-230-012.