# Strong Convergence Theorems for Maximal Monotone Operators, Fixed-Point Problems, and Equilibrium Problems

Authors: Huan-chun Wu; Cao-zong Cheng; De-ning Qu
Journal: ISRN Applied Mathematics
 (2013)

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## Abstract

We present a new iterative method for finding a common element of the set of fixed points of a nonexpansive mapping, the set of solutions to an equilibrium problem, and the set of zeros of the sum of maximal monotone operators and prove the strong convergence theorems in the Hilbert spaces. We also apply our results to variational inequality and optimization problems.


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## Body

## 1. Introduction

LetC be a nonempty closed convex subset of a real Hilbert space H. A mapping S:C→C is nonexpansive if ∥Sx-Sy∥≤∥x-y∥forallx,y∈C. The set of fixed points of S is denoted by Fix(S). It is well known that Fix(S) is closed and convex. There are two iterative methods for approximating fixed points of a nonexpansive mapping. One is introduced by Mann in [1] and the other by Halpern in [2]. The iteration procedure of Mann's type for approximating fixed points of a nonexpansive mapping S is the following: x1∈C and
(1)xn+1=αnxn+(1-αn)Sxn,
where {αn} is a sequence in [0,1]. The iteration procedure of Halpern's type is the following: u∈C,x1∈C and
(2)xn+1=αnu+(1-αn)Sxn,
where {αn} is a sequence in [0,1].Letf be a bifunction from C×C to ℝ, where ℝ is the set of real numbers. The equilibrium problem is to find x¯∈C such that f(x¯,y)≥0 for all y∈C. The set of such solutions is denoted by EP(f). Numerous problems in physics, optimization, and economics reduce to finding a solution to the equilibrium problem (e.g., see [3]). For solving the equilibrium problem, we assume that the bifunction f satisfies the following conditions:(A1)
f(x,x)=0 for all x∈C,(A2)
f is monotone, that is, f(x,y)+f(y,x)≤0 for all x,y∈C,(A3)
for everyx,y,z∈C, limsupt↓0f(tz+(1-t)x,y)≤f(x,y),(A4)
f(x,·) is convex and lower semicontinuous for each x∈C. Equilibrium problems have been studied extensively; see [3–9].LetB be a mapping of H into 2H. The effective domain of B is denoted by dom(B), that is, dom(B)={x∈H:Bx≠∅}. A multivalued mapping B is said to be monotone if
(3)〈x-y,u-v〉≥0∀x,y∈dom(B),u∈Bx,v∈By.
A monotone operator B is said to be maximal if its graph is not properly contained in the graph of any other monotone operator. For a maximal monotone operator B on H and r>0, the operator Jr=(I+rB)-1:H→dom(B) is called the resolvent of B for r. It is known that Jr is firmly nonexpansive. Given a positive constant α, a mapping A:C→H is said to be α-inverse strongly monotone if
(4)〈x-y,Ax-Ay〉≥α∥Ax-Ay∥2∀x,y∈C.
Some authors have paid more attention to finding an element in the set of zeros of A+B. For a mapping A from C into H, we know that (A+B)-10=Fix(Jλ(I-λA)); see [10].  Takahashi et al. [11] constructed the following iterative sequence. Let u∈C,x1=x∈C, and let {xn} be a sequence generated by
(5)xn+1=αnu+(1-αn)Jλn(xn-λnAxn).
Under appropriate conditions they proved that the sequence {xn} converges strongly to a point z0∈(A+B)-10. Lin and Takahashi [12] introduced an iterative sequence that converges strongly to an element of (A+B)-10∩F-10, where F is another maximal monotone operator. Takahashi et al. [13] presented a new iterative sequence converging strongly to an element of (A+B)-10∩Fix(S).Motivated by the above results, in this paper, we introduce a new iterative algorithm for finding a common element of the set of fixed points of a nonexpansive mapping, the set of solutions to an equilibrium problem, and the set of zeros of the sum of maximal monotone operators and prove the strong convergence theorems in the Hilbert spaces. Finally, we give the applications to the variational inequality and optimization problems.


## 2. Preliminaries

Throughout this paper, letH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, and let C be a nonempty closed convex subset of H. We write xn→x to indicate that the sequence {xn} converges strongly to x. Similarly,xn⇀x  will mean weak convergence. It is well known that H satisfies Opial's condition; that is, for any sequence {xn}⊂H with xn⇀x, we have
(6)liminfn→∞∥xn-x∥<liminfn→∞∥xn-y∥∀y≠x.
For any x∈H, there exists a unique point PCx∈C such that
(7)∥x-PCx∥≤∥x-y∥,∀y∈C.PC  is called the metric projection of H onto C. Note that PC  is a nonexpansive mapping of H onto C. For x∈H and z∈C, we have
(8)z=PCx⟺〈z-y,x-z〉≥0foreveryy∈C.Forγ->0, a mapping V on H is called γ--strongly monotone if
(9)〈x-y,Vx-Vy〉≥γ-∥x-y∥2∀x,y∈H.
Taking L>0, a mapping T on H is said to be L-Lipschitzian continuous if
(10)∥Tx-Ty∥≤L∥x-y∥∀x,y∈H.
It is easy to see that A is γ-/L2-inverse strongly monotone whenever A is γ--strongly monotone and L-Lipschitzian continuous. Now we consider inverse strongly monotone. Let α>0, and let A:C→H be an α-inverse strongly monotone operator. If 0<λ≤2α, then I-λA is a nonexpansive mapping. Indeed, for x,y∈C  and  0<λ≤2α, we get
(11)∥(I-λA)x-(I-λA)y∥2=∥(x-y)-λ(Ax-Ay)∥2=∥x-y∥2-2λ〈x-y,Ax-Ay〉+λ2∥Ax-Ay∥2≤∥x-y∥2-2λα∥Ax-Ay∥2+λ2∥Ax-Ay∥2≤∥x-y∥2+λ(λ-2α)∥Ax-Ay∥2≤∥x-y∥2.
Therefore, the operator I-λA is a nonexpansive mapping of C into H.We need the following lemmas.Lemma 1 (see [3]).
Let C be a nonempty closed convex subset ofH, and let f be a bifunction from C×C to ℝ satisfying (A1)–(A4). If r>0 and x∈H, then there exists z∈C such that
(12)f(z,y)+1r〈y-z,z-x〉≥0∀y∈C.Lemma 2 (see [7]).
Let C be a nonempty closed convex subset ofH, and let f be a bifunction from C×C to ℝ satisfying (A1)–(A4). For r>0, define a mapping Tr:H→2C as follows:
(13)Tr(x)={z∈C:f(z,y)+1r〈y-z,z-x〉≥0∀y∈C}.
Then the following hold:  (i)
Tr is single valued,(ii)
Tr is firmly nonexpansive; that is, for any x,y∈H,
(14)〈x-y,Trx-Try〉≥∥Trx-Try∥2,(iii)
Fix(Tr)=EP(f),(iv)
EP(f) is closed and convex.Lemma 3.
LetV be a γ--strongly monotone and L-Lipschitzian continuous operator on a real Hilbert space H with γ-,L>0 and 2γ--1<L2<2γ-. Suppose that {βn} is a sequence in (0,1). For all x,y∈H, one has
(15)∥(I-βnV)x-(I-βnV)y∥≤(1-βnτ)∥x-y∥,
where τ=γ--L2/2.Proof.
Observe that(16)∥(I-βnV)x-(I-βnV)y∥2=∥x-y∥2+βn2∥Vx-Vy∥2-2βn〈x-y,Vx-Vy〉≤∥x-y∥2+βn2L2∥x-y∥2-2βnγ-∥x-y∥2=(1+βn2L2-2βnγ-)∥x-y∥2=[1+βn2L2-2βn(τ+L22)]∥x-y∥2=(1-2βnτ-βnL2+βn2L2)∥x-y∥2≤(1-2βnτ+βn2τ2-βnL2+βn2L2)∥x-y∥2=[(1-βnτ)2-βn(L2-βnL2)]∥x-y∥2.
Since the sequence {βn}⊂(0,1) and 2γ--1<L2<2γ-, we obtain
(17)∥(I-βnV)x-(I-βnV)y∥≤(1-βnτ)∥x-y∥.Lemma 4 (see [8]).
Suppose that (A1)–(A4) hold. Ifx,y∈H and r1,r2>0, then
(18)∥Tr2y-Tr1x∥≤∥y-x∥+|r2-r1|r2∥Tr2y-y∥.Lemma 5 (see [13]).
LetH be a real Hilbert space, and let B be a maximal monotone operator on H. Then the following holds:
(19)s-ts〈Jsx-Jtx,Jsx-x〉≥∥Jsx-Jtx∥2
for all s,t>0 and x∈H.Lemma 6 (see [14, 15]).
Let{sn} be a sequence of nonnegative real numbers satisfying
(20)sn+1≤(1-αn)sn+αnβn+γn,n≥0,
where {αn}, {βn}, and {γn} satisfy the following conditions:  (i)
{αn}⊂[0,1],  ∑n=1∞αn=∞,(ii)
limsupn→∞βn≤0,(iii)
γn≥0,∑n=1∞γn<∞. Then limn→∞sn=0.The following lemma is an immediate consequence of the inner product onH.Lemma 7.
For allx,y∈H, the inequality ∥x+y∥2≤∥x∥2+2〈y,x+y〉 holds.Lemma 8 (see [16] (demiclosedness principle)).
LetC be a nonempty closed convex subset of H, S:C→H a nonexpansive mapping, and x a point in H, the sequence {xn} in C. Suppose that xn⇀x and that xn-Sxn→0. Then x∈Fix(S).


## 3. Strong Convergence Theorems

In this section, we present a new iterative method for finding a common element of the set of fixed points of a nonexpansive mapping, the set of solutions to an equilibrium problem, and the set of zeros of the sum of maximal monotone operators.Theorem 9.
LetC be a nonempty closed convex subset of a real Hilbert space H and A an α-inverse strongly monotone operator of C into H. Let B be a maximal monotone operator on H such that the domain of B is included in C. Let Jλ=(I+λB)-1 be the resolvent of B for λ>0, and let S be a nonexpansive mapping of C into itself. Suppose that V is a γ--strongly monotone and L-Lipschitzian continuous operator on H with γ-,L>0 and 2γ--1<L2<2γ-. Assume that f:C×C→ℝ satisfies (A1)–(A4). Suppose that (A+B)-10∩Fix(S)∩EP(f)≠∅. Let ω∈C and x1∈C, and let {xn} be a sequence generated by
(21)un∈C,suchthatf(un,y)+1rn〈y-un,un-xn〉≥0∀y∈C,xn+1=βnω+(I-βnV)×[αnxn+(1-αn)Jλn(I-λnA)Sun],
where the sequences {αn},  {βn},  {λn},  and  {rn} satisfy the following conditions:  (1)
{αn}⊂(0,1)  and  ∑n=1∞αn<∞,(2)
{βn}⊂(0,1),βn→0,∑n=1∞βn=∞,  and  ∑n=1∞|βn+1-βn|<∞,(3)
0<a≤λn≤b<2α  and  ∑n=1∞|λn+1-λn|<∞,(4)
0<c≤rn  and  ∑n=1∞|rn+1-rn|<∞. Then the sequence {xn} converges strongly to an element of (A+B)-10∩Fix(S)∩EP(f).Proof.
The proof will be completed by eight steps.
Step  1. Show that the sequences {xn} and {un} are bounded.
Note that(A+B)-10∩Fix(S)∩EP(f) is a closed convex subset of H since (A+B)-10, Fix(S), and  EP(f) are closed and convex. For simplicity, we write
(22)Ω:=(A+B)-10∩Fix(S)∩EP(f).
From Lemmas1 and 2, we have un=Trnxn, and for any z∈Ω,
(23)∥un-z∥=∥Trnxn-Trnz∥≤∥xn-z∥.
Set yn=αnxn+(1-αn)Jλn(I-λnA)Sun. It follows that
(24)∥yn-z∥=∥αnxn+(1-αn)Jλn(I-λnA)Sun-z∥=∥αn(xn-z)+(1-αn)[Jλn(I-λnA)Sun-z]∥≤αn∥xn-z∥+(1-αn)∥xn-z∥=∥xn-z∥.
Lemma 3 implies that
(25)∥xn+1-z∥=∥Jλnβnω+(I-βnV)×[αnxn+(1-αn)Jλn(I-λnA)Sun]-z∥=∥βn(ω-Vz)+(I-βnV)yn-(I-βnV)z∥≤βn∥ω-Vz∥+∥(I-βnV)yn-(I-βnV)z∥≤βn∥ω-Vz∥+(1-βnτ)∥yn-z∥≤βn∥ω-Vz∥+(1-βnτ)∥xn-z∥≤βnτ∥ω-Vz∥τ+(1-βnτ)∥xn-z∥≤max{∥xn-z∥,∥ω-Vz∥τ},whereτ=γ--L22.
From a simple inductive process, it follows that
(26)∥xn+1-z∥≤max{∥x1-z∥,∥ω-Vz∥τ},
which yields that {xn} is bounded, so is the sequence {un}.
Step  2. Show that limn→∞∥xn+1-xn∥=0.
Since(27)∥xn+2-xn+1∥=∥[αnxn+(1-αn)Jλn(I-λnA)Sun]βn+1ω+(I-βn+1V)×[αn+1xn+1+(1-αn+1)Jλn+1(I-λn+1A)Sun+1]-βnω-(I-βnV)×[αnxn+(1-αn)Jλn(I-λnA)Sun]∥=∥(βn+1-βn)ω+(I-βn+1V)yn+1-(I-βn+1V)yn+(I-βn+1V)yn-(I-βnV)yn∥≤|βn+1-βn|∥ω∥+(1-βn+1τ)∥yn+1-yn∥+|βn+1-βn|∥Vyn∥≤|βn+1-βn|(∥ω∥+∥Vyn∥)+(1-βn+1τ)×∥αn+1xn+1+(1-αn+1)Jλn+1(I-λn+1A)Sun+1-αnxn-(1-αn)Jλn(I-λnA)Sun∥≤|βn+1-βn|(∥ω∥+∥Vyn∥)+(1-βn+1τ)×[αn+1∥xn+1-Jλn+1(I-λn+1A)Sun+1∥+αn∥xn-Jλn(I-λnA)Sun∥+∥Jλn+1(I-λn+1A)Sun+1-Jλn(I-λnA)Sun∥]≤|βn+1-βn|(∥ω∥+∥Vyn∥)+(1-βn+1τ)×[αn+1∥xn+1-Jλn+1(I-λn+1A)Sun+1∥+αn∥xn-Jλn(I-λnA)Sun∥+∥Jλn+1(I-λn+1A)Sun+1-Jλn+1(I-λn+1A)Sun+Jλn+1(I-λn+1A)Sun-Jλn+1(I-λnA)Sun+Jλn+1(I-λnA)Sun-Jλn(I-λnA)Sun∥]≤|βn+1-βn|(∥ω∥+∥Vyn∥)+(1-βn+1τ)×[αn+1∥xn+1-Jλn+1(I-λn+1A)Sun+1∥+αn∥xn-Jλn(I-λnA)Sun∥+∥un+1-un∥+|λn+1-λn|∥ASun∥+∥Jλn+1(I-λnA)Sun-Jλn(I-λnA)Sun∥],
it follows from Lemmas 4 and 5 that
(28)∥xn+2-xn+1∥≤|βn+1-βn|(∥ω∥+∥Vyn∥)+(1-βn+1τ)×[|λn+1-λn|λn+1αn+1∥xn+1-Jλn+1(I-λn+1A)Sun+1∥+αn∥xn-Jλn(I-λnA)Sun∥+|λn+1-λn|∥ASun∥+∥xn+1-xn∥+|rn+1-rn|rn+1∥un+1-xn+1∥+|λn+1-λn|λn+1×∥Jλn+1(I-λnA)Sun-(I-λnA)Sun∥+|λn+1-λn|λn+1]≤|βn+1-βn|(∥ω∥+∥Vyn∥)+(1-βn+1τ)×[|λn+1-λn|aαn+1∥xn+1-Jλn+1(I-λn+1A)Sun+1∥+αn∥xn-Jλn(I-λnA)Sun∥+|λn+1-λn|∥ASun∥+∥xn+1-xn∥+|rn+1-rn|c∥un+1-xn+1∥+|λn+1-λn|a×∥Jλn+1(I-λnA)Sun-(I-λnA)Sun∥+|λn+1-λn|a]≤(1-βn+1τ)∥xn+1-xn∥+αn+1∥xn+1-Jλn+1(I-λn+1A)Sun+1∥+αn∥xn-Jλn(I-λnA)Sun∥+|λn+1-λn|∥ASun∥+|rn+1-rn|c∥un+1-xn+1∥+|λn+1-λn|a∥Jλn+1(I-λnA)Sun-(I-λnA)Sun∥+|βn+1-βn|(∥ω∥+∥Vyn∥).
SetM=supn∈ℕ{∥xn-Jλn(I-λnA)Sun∥,∥ASun∥,∥un+1-xn+1∥,∥Jλn+1(I-λnA)Sun-(I-λnA)Sun∥,(∥ω∥+∥Vyn∥)}. We have
(29)∥xn+2-xn+1∥≤(1-βn+1τ)∥xn+1-xn∥+M(|rn+1-rn|cαn+1+αn+|λn+1-λn|+|rn+1-rn|c+|λn+1-λn|a+|βn+1-βn|).
By the assumptions ∑n=1∞βn=∞,  ∑n=1∞αn<∞,  ∑n=1∞|λn+1-λn|<∞,  ∑n=1∞|rn+1-rn|<∞,   and ∑n=1∞|βn+1-βn|<∞,   it follows from Lemma 6 that
(30)limn→∞∥xn+1-xn∥=0.
Step  3. Show that limn→∞∥xn-un∥=0.
For anyz∈Ω, we have
(31)∥un-z∥2=∥Trnxn-Trnz∥2≤〈xn-z,un-z〉=12[∥xn-z∥2+∥un-z∥2-∥xn-un∥2],
which implies that
(32)∥un-z∥2≤∥xn-z∥2-∥xn-un∥2.
With the help of Lemma7, we get
(33)∥xn+1-z∥2=∥Jλnβn(ω-Vz)+(I-βnV)×[αnxn+(1-αn)Jλn(I-λnA)Sun]-(I-βnV)zJλn∥2≤∥(I-βnV)[αnxn+(1-αn)Jλn(I-λnA)Sun]-(I-βnV)zJλn∥2+2〈βn(ω-Vz),xn+1-z〉≤(1-βnτ)∥Jλnαn(xn-z)+(1-αn)×(Jλn(I-λnA)Sun-z)∥2+2βn〈ω-Vz,xn+1-z〉.
Consequently,
(34)∥xn+1-z∥2≤(1-βnτ)[αn∥xn-z∥2+(1-αn)∥un-z∥2]+2βn〈ω-Vz,xn+1-z〉≤(1-βnτ)[αn∥xn-z∥2+(1-αn)(∥xn-z∥2-∥xn-un∥2)]+2βn〈ω-Vz,xn+1-z〉≤(1-βnτ)[∥xn-z∥2-(1-αn)∥xn-un∥2]+2βn〈ω-Vz,xn+1-z〉≤∥xn-z∥2-(1-βnτ)(1-αn)∥xn-un∥2+2βn〈ω-Vz,xn+1-z〉.
Hence,
(35)(1-βnτ)(1-αn)∥xn-un∥2≤∥xn-z∥2-∥xn+1-z∥2+2βn〈ω-Vz,xn+1-z〉≤∥xn-xn+1∥(∥xn-z∥+∥xn+1-z∥)+2βn∥ω-Vz∥∥xn+1-z∥.
Since limn→∞∥xn+1-xn∥=0 and limn→∞βn=0, we get
(36)limn→∞∥xn-un∥=0.
Step  4. Show that limn→∞∥ASun-Az∥=0, for all z∈Ω.
Forz∈Ω, we get
(37)∥Jλn(I-λnA)Sun-z∥2=∥Jλn(I-λnA)Sun-Jλn(I-λnA)Sz∥2≤∥(Sun-Sz)-λn(ASun-ASz)∥2≤∥un-z∥2-2λn〈Sun-Sz,ASun-ASz〉+λn2∥ASun-ASz∥2≤∥xn-z∥2-2λnα∥ASun-ASz∥2+λn2∥ASun-ASz∥2≤∥xn-z∥2+λn(λn-2α)∥ASun-ASz∥2.
This together with (33) deduces that
(38)∥xn+1-z∥2≤(1-βnτ)×[αn∥xn-z∥2+(1-αn)×(∥xn-z∥2+λn(λn-2α)∥ASun-ASz∥2)]+2βn〈ω-Vz,xn+1-z〉≤(1-βnτ)×[∥xn-z∥2+(1-αn)λn(λn-2α)∥ASun-ASz∥2]+2βn〈ω-Vz,xn+1-z〉≤∥xn-z∥2+(1-βnτ)(1-αn)λn(λn-2α)×∥ASun-Az∥2+2βn〈ω-Vz,xn+1-z〉.
Thus,
(39)(1-βnτ)(1-αn)λn(2α-λn)∥ASun-ASz∥2≤∥xn-z∥2-∥xn+1-z∥2+2βn∥ω-Vz∥∥xn+1-z∥≤∥xn-xn+1∥(∥xn-z∥+∥xn+1-z∥)+2βn∥ω-Vz∥∥xn+1-z∥.
Since 0<a≤λn≤b<2α  and  ∑n=1∞|λn+1-λn|<∞, the sequence {λn} is a Cauchy sequence. Assume that λn→λ0∈[a,b]. It follows that
(40)limn→∞∥ASun-Az∥=0.
Step  5. Show that limn→∞∥Jλn(I-λnA)Sun-Sun∥=0.
Sethn=Jλn(I-λnA)Sun. For z∈Ω, we have
(41)∥hn-z∥2=∥Jλn(I-λnA)Sun-Jλn(I-λnA)Sz∥2≤〈(I-λnA)Sun-(I-λnA)Sz,hn-z〉=12[∥(I-λnA)Sun-(I-λnA)Sz∥2+∥hn-z∥2-∥(I-λnA)Sun-(I-λnA)Sz-(hn-z)∥2]≤12[∥un-z∥2+∥hn-z∥2-∥(Sun-hn)-λn(ASun-ASz)∥2].
Therefore,(42)∥hn-z∥2≤∥un-z∥2-∥(Sun-hn)-λn(ASun-ASz)∥2≤∥xn-z∥2-∥Sun-hn∥2-λn2∥ASun-ASz∥2+2λn〈Sun-hn,ASun-ASz〉.
Using (33) again, we obtain that
(43)∥xn+1-z∥2≤(1-βnτ)×[αn∥xn-z∥2+(1-αn)∥Jλn(I-λnA)Sun-z∥2]+2βn〈ω-Vz,xn+1-z〉≤(1-βnτ)[αn∥xn-z∥2+(1-αn)×(∥xn-z∥2-∥Sun-hn∥2-λn2∥ASun-ASz∥2+2λn〈Sun-hn,ASun-ASz〉∥xn-z∥2)∥xn-z∥2]+2βn〈ω-Vz,xn+1-z〉≤(1-βnτ)[∥xn-z∥2-(1-αn)∥Sun-hn∥2-(1-αn)λn2∥ASun-ASz∥2+2(1-αn)λn〈Sun-hn,ASun-ASz〉[xn-z]2]+2βn〈ω-Vz,xn+1-z〉≤∥xn-z∥2-(1-βnτ)(1-αn)∥Sun-hn∥2+2λn(1-αn)∥Sun-hn∥∥ASun-ASz∥+2βn∥ω-Vz∥∥xn+1-z∥.
Thus,
(44)(1-βnτ)(1-αn)∥Sun-hn∥2≤∥xn-z∥2-∥xn+1-z∥2+2λn∥Sun-hn∥∥ASun-ASz∥+2βn∥ω-Vz∥∥xn+1-z∥≤∥xn-xn+1∥(∥xn-z∥+∥xn+1-z∥)+2λn∥Sun-hn∥∥ASun-Az∥+2βn∥ω-Vz∥∥xn+1-z∥.
It follows from (30), (40), and limn→∞βn=0 that
(45)limn→∞∥Jλn(I-λnA)Sun-Sun∥=0.
Step  6. Show that limn→∞∥un-Sun∥=0.
Since(46)∥xn+1-Sun∥=∥βn(ω-VSun)+(I-βnV)×[αnxn+(1-αn)Jλn(I-λnA)Sun]-(I-βnV)Sun∥≤βn∥ω-VSun∥+(1-βnτ)×∥αnxn+(1-αn)Jλn(I-λnA)Sun-Sun∥≤βn∥ω-VSun∥+(1-βnτ)×[Jλnαn∥xn-Sun∥+(1-αn)×∥Jλn(I-λnA)Sun-Sun∥],
equality (45) implies that
(47)limn→∞∥xn+1-Sun∥=0.
As
(48)∥un-Sun∥≤∥un-xn∥+∥xn-xn+1∥+∥xn+1-Sun∥,
it follows from (30), (36), and (47) that
(49)limn→∞∥un-Sun∥=0.
Step  7. Show that limsupn→∞〈ω-Vz0,xn+1-z0〉≤0, where z0=PΩ[ω+(I-V)z0].
Observe that the mappingx↦PΩ[ω+(I-V)x] is a contraction. Indeed, for any x,y∈H,
(50)∥PΩ[ω+(I-V)x]-PΩ[ω+(I-V)y]∥2≤∥(I-V)x-(I-V)y∥2=∥(x-y)-(Vx-Vy)∥2=∥x-y∥2-2〈x-y,Vx-Vy〉+∥Vx-Vy∥2≤∥x-y∥2-2γ-∥x-y∥2+L2∥x-y∥2=(1-2γ-+L2)∥x-y∥2.
As 2γ--1<L2<2γ-, we have 0<1-2γ-+L2<1. The Banach contraction mapping principle guarantees that the mapping x→PΩ[ω+(I-V)x] has a unique fixed point z0; that is, z0=PΩ[ω+(I-V)z0].
In order to show this inequality, we can choose a subsequence{xni} of {xn} such that
(51)limsupn→∞〈ω-Vz0,xn+1-z0〉=limi→∞〈ω-Vz0,xni-z0〉.
In view of the boundedness of{xni}, there exists a subsequence {xnij} of {xni} such that xnij⇀p. Without loss of generality, we assume that xni⇀p. It follows from (36) that uni⇀p. Since {uni}⊂C and C is closed and convex, we get p∈C. Now we show that p∈Ω.
First we prove thatp∈EP(f). By (21),
(52)f(un,y)+1rn〈y-un,un-xn〉≥0∀y∈C.
The monotonicity of f implies that
(53)1rn〈y-un,un-xn〉≥f(y,un)∀y∈C.
Replacing n by ni, we obtain
(54)〈y-uni,uni-xnirni〉≥f(y,uni)∀y∈C.
Applying (36) and (A4), we have
(55)f(y,p)≤0∀y∈C.
For 0<t≤1, y∈C,   set  yt=ty+(1-t)p. Then yt∈C  and  f(yt,p)≤0. Thus,
(56)0=f(yt,yt)≤tf(yt,y)+(1-t)f(yt,p)≤tf(yt,y).
Dividing by t, we see that
(57)f(yt,y)≥0.
Letting t↓0, we get
(58)f(p,y)≥0∀y∈C.
That is, p∈EP(f).
Now we prove thatp∈Fix(S). Otherwise, assume that p∉Fix(S), that is, p≠Sp. Opial's condition and (49) imply that
(59)liminfi→∞∥uni-p∥<liminfi→∞∥uni-Sp∥=liminfi→∞∥uni-Suni+Suni-Sp∥=liminfi→∞∥Suni-Sp∥≤liminfi→∞∥uni-p∥.
This is a contradiction. Thus, p∈Fix(S).
Next we will show thatp∈(A+B)-10.
In fact, letλn→λ0∈[a,b], and let vn=Sun. It follows from Lemma 5 that
(60)∥Jλ0(I-λ0A)vn-Jλn(I-λnA)vn∥=∥Jλ0(I-λ0A)vn-Jλ0(I-λnA)vn+Jλ0(I-λnA)vn-Jλn(I-λnA)vn∥≤|λn-λ0|∥Avn∥+|λ0-λn|λ0×∥Jλ0(I-λnA)vn-(I-λnA)vn∥.
Thus,
(61)limn→∞∥Jλ0(I-λ0A)vn-Jλn(I-λnA)vn∥=0.
Since
(62)∥vn-Jλ0(I-λ0A)vn∥≤∥vn-Jλn(I-λnA)vn∥+∥Jλn(I-λnA)vn-Jλ0(I-λ0A)vn∥,
equalities (45) and (61) imply that
(63)limn→∞∥vn-Jλ0(I-λ0A)vn∥=0.
Therefore,
(64)limi→∞∥vni-Jλ0(I-λ0A)vni∥=0.
It follows from uni⇀p and (49) that vni⇀p. As Jλ0(I-λ0A) is nonexpansive, Lemma 8 implies that p=Jλ0(I-λ0A)p. That is, p∈(A+B)-10. Hence, p∈Ω. By (51) and the property of metric projection, we have
(65)limsupn→∞〈ω-Vz0,xn+1-z0〉=limi→∞〈ω-Vz0,xni-z0〉=〈ω-Vz0,p-z0〉=〈[ω+(I-V)z0]-z0,p-z0〉≤0.
Step  8. Show that xn→z0, where z0=PΩ[ω+(I-V)z0].
According to (21), we get
(66)∥xn+1-z0∥2=∥Jλnβn(ω-Vz0)+(I-βnV)×[αnxn+(1-αn)Jλn(I-λnA)Sun]-(I-βnV)z0Jλn∥2≤(1-βnτ)∥Jλnαn(xn-z0)+(1-αn)(Jλn(I-λnA)Sun-z0)∥2+2βn〈ω-Vz0,xn+1-z0〉≤(1-βnτ)[αn∥xn-z0∥2+(1-αn)∥xn-z0∥2]+2βn〈ω-Vz0,xn+1-z0〉≤(1-βnτ)∥xn-z0∥2+2βn〈ω-Vz0,xn+1-z0〉.
It follows from (65) and Lemma 6 that {xn} converges strongly to z0∈Ω.Remark 10.
By an examination of the proof of Theorem9, the conclusion still holds in the case that αn≡0.Remark 11.
Consider the following quadratic optimization problem:(67)minx∈H12〈Vx,x〉-〈x,ω〉,
where H is a real Hilbert space, V is a self-adjoint bounded linear operator on H such that
(68)〈Vx,x〉≥γ¯∥x∥2,∀x∈Handsomeγ¯>0.
Letting  A=0,  B=∂δC (i.e., the subdifferential of the indicator function of C),  C=H,f(x,y)≡0,   and  αn≡0, algorithm (21) reduces to
(69)xn+1=βnω+(I-βnV)xn.
Xu [17] showed that the sequence in algorithm (69) converges strongly to the solution of problem (67).Remark 12.
Consider the setting of Theorem9 with f(x,y)≡0,αn≡0,  and  V=S=I. Then algorithm (21) corresponds to the algorithm in [11, Theorem 9].The corollaries below are the direct consequences of Theorem9.Corollary 13.
LetC be a nonempty closed convex subset of a real Hilbert space H and A an α-inverse strongly monotone operator of C into H. Let B be a maximal monotone operator on H such that the domain of B is included in C. Let Jλ=(I+λB)-1 be the resolvent of B for λ>0, and let S be a nonexpansive mapping of C into itself. Suppose that V is a γ--strongly monotone and L-Lipschitzian continuous operator on H with γ-,L>0 and 2γ--1<L2<2γ-. Suppose that (A+B)-10∩Fix(S)≠∅. Let ω∈C and x1∈C, and let {xn} be a sequence generated by
(70)un=PCxn,xn+1=βnω+(I-βnV)[αnxn+(1-αn)Jλn(I-λnA)Sun],
where the sequences {αn}, {βn}, and {λn} satisfy the following conditions:  (1)
{αn}⊂(0,1)  and  ∑n=1∞αn<∞,(2)
{βn}⊂(0,1),βn→0,∑n=1∞βn=∞,  and  ∑n=1∞|βn+1-βn|<∞,(3)
0<a≤λn≤b<2α  and  ∑n=1∞|λn+1-λn|<∞. Then the sequence {xn} converges strongly to an element of  (A+B)-10∩Fix(S).Proof.
Lettingf(x,y)≡0 for all x,y∈C and rn=1 in Theorem 9, we get the result.Corollary 14.
LetC be a nonempty closed convex subset of a real Hilbert space H and A an α-inverse strongly monotone operator of C into H. Let B be a maximal monotone operator on H such that the domain of B is included in C. Let Jλ=(I+λB)-1 be the resolvent of B for λ>0. Assume that f:C×C→ℝ satisfies (A1)–(A4). Suppose that (A+B)-10∩EP(f)≠∅. Let ω∈C and x1∈C, and let {xn} be a sequence generated by
(71)un∈C,suchthatf(un,y)+1rn〈y-un,un-xn〉≥0∀y∈C,xn+1=βnω+(1-βn)[αnxn+(1-αn)Jλn(I-λnA)un],
where the sequences {αn}, {βn}, {λn}, and {rn} satisfy the following conditions:  (1)
{αn}⊂(0,1)  and  ∑n=1∞αn<∞,(2)
{βn}⊂(0,1),βn→0,∑n=1∞βn=∞,  and  ∑n=1∞|βn+1-βn|<∞,(3)
0<a≤λn≤b<2α  and  ∑n=1∞|λn+1-λn|<∞,(4)
0<c≤rn  and  ∑n=1∞|rn+1-rn|<∞. Then the sequence {xn} converges strongly to an element of (A+B)-10∩EP(f).Proof.
PuttingV=S=I in Theorem 9, we can obtain the desired result.


## 4. Applications

In this section, we apply the results in the preceding section to variational inequality and optimization problems. Now we consider the variational inequality problem. LetH be a real Hilbert space, and let f be a proper lower semicontinuous convex function of H into (-∞,+∞]. Then the subdifferential ∂f of f is defined as
(72)∂f(x)={z∈H:f(y)-f(x)≥〈z,y-x〉,∀y∈H}
for all x∈H. Rockafellar [18] claimed that ∂f is a maximal monotone operator. Let C be a nonempty closed convex subset of H, and let δC be the indicator function of C. That is,
(73)δC(x)={0x∈C,+∞x∉C.
Since δC is a proper lower semicontinuous convex function on H, the subdifferential ∂δC of δC is a maximal monotone operator. The resolvent Jλ of ∂δC for λ>0 is defined by
(74)Jλx=(I+λ∂δC)-1x∀x∈H.
We have
(75)u=Jλx=(I+λ∂δC)-1x⟺x∈u+λ∂δCu⟺x∈u+λNCu⟺x-u∈λNCu⟺1λ〈x-u,y-u〉≤0∀y∈C⟺〈x-u,y-u〉≤0∀y∈C⟺u=PCx,
where NCu={z∈H:〈z,y-u〉≤0,∀y∈C}. The variational inequality problem for nonlinear operator A is to find z∈C such that
(76)〈Az,y-z〉≥0,∀y∈C.
The set of its solutions is denoted by VI(C,A). Then we have
(77)z∈VI(C,A)⟺〈Az,y-z〉≥0,∀y∈C⟺〈-Az,y-z〉≤0∀y∈C⟺-Az∈NCz⟺0∈Az+NCz⟺0∈Az+∂δCz⟺z∈(A+∂δC)-10.
Using Theorem 9, we obtain the strong convergence theorem for the variational inequality problem.Theorem 15.
LetC be a nonempty closed convex subset of a real Hilbert space H and A an α-inverse strongly monotone operator of C into H, and let V be a γ--strongly monotone and L-Lipschitzian continuous operator on H with γ-,L>0 and 2γ--1<L2<2γ-. Suppose that VI(C,A)≠∅. Let ω∈C and x1∈C, and let {xn} be a sequence generated by
(78)un=PCxn,xn+1=βnω+(I-βnV)[αnxn+(1-αn)PC(I-λnA)un],
where the sequences {αn}, {βn}, and {λn} satisfy the following conditions:  (1)
{αn}⊂(0,1)  and  ∑n=1∞αn<∞,(2)
{βn}⊂(0,1),βn→0,∑n=1∞βn=∞,  and  ∑n=1∞|βn+1-βn|<∞,(3)
0<a≤λn≤b<2α  and  ∑n=1∞|λn+1-λn|<∞. Then the sequence {xn} converges strongly to an element of VI(C,A).Proof.
Notice thatVI(C,A)=(A+∂δC)-10. Letting f(x,y)≡0 for all x,y∈C,  rn=1, and S=I, Theorem 9 yields that the sequence {xn} converges strongly to an element of VI(C,A).Next we study the optimization problem(79)ming(x)x∈C,
where g(x) is a proper lower semicontinuous convex function of H into (-∞,+∞] such that C is included in domg={x∈H:g(x)<+∞}. We denote by Sol(g,C) the set of solutions to problem (79). Let f:C×C→ℝ be a bifunction defined by
(80)f(x,y)=g(y)-g(x).
It is clear that f(x,y) satisfies (A1)–(A4) and EP(f)=Sol(g,C). Therefore, by Theorem 9, the following result is obtained.Theorem 16.
Letg(x) be a proper lower semicontinuous convex function of H into (-∞,+∞] and C a nonempty closed convex subset of H such that C is included in domg, and let V be a γ--strongly monotone and L-Lipschitzian continuous operator on H with γ-,L>0 and 2γ--1<L2<2γ-. Suppose that Sol(g,C)≠∅. Let ω∈C and x1∈C, and let {xn} be a sequence generated by
(81)un=argminy∈C{g(y)+12rn∥y-xn∥2},xn+1=βnω+(I-βnV)[αnxn+(1-αn)un],
where the sequences {αn}, {βn}, and {rn} satisfy the following conditions:  (1)
{αn}⊂(0,1)  and  ∑n=1∞αn<∞,(2)
{βn}⊂(0,1),βn→0,∑n=1∞βn=∞,  and  ∑n=1∞|βn+1-βn|<∞,(3)
0<c≤rn  and  ∑n=1∞|rn+1-rn|<∞. Then the sequence {xn} converges strongly to an element of Sol(g,C).Proof.
LettingS=I, A=0, B=∂δC, and f(x,y)=g(y)-g(x) in Theorem 9, we get the conclusion.



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