# An Alternative Regularization Method for

Authors: Shuo Sun
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We introduce a new regularization iterative algorithm for equilibrium and fixed point problems of nonexpansive mapping. Then, we prove a strong convergence theorem for nonexpansive mappings to solve a unique solution of the variational inequality and the unique sunny nonexpansive retraction. Our results extend beyond the results of S. Takahashi and W. Takahashi (2007), and many others.


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## Body

## 1. Introduction

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, respectively. let C be a nonempty closed convex subset of H. Let ϕ be a bifunction of C×C into ℝ, where ℝ is the set of real numbers. The equilibrium problem for ϕ:C×C→ℝ is to find x∈C such that
(1.1)ϕ(x,y)≥0,∀y∈C.
The set of solutions of (1.1) is denoted by EP(ϕ). Given a mapping T:C→H, let ϕ(x,y)=〈Tx,y-x〉 for all x,y∈C. Then, z∈EP(ϕ) if and only if 〈Tz,y-z〉≥0 for all y∈C, that is, z is a solution of the variational inequality. Numerous problems in physics, optimization, and economics reduce to find a solution of (1.1). Some methods have been proposed to solve the equilibrium problem; see, for instance, [1–6].A mappingS of C into H is said to be nonexpansive if
(1.2)‖Sx-Sy‖≤‖x-y‖∀x,y∈C.We denote byF(S) the set of fixed points of S. The fixed point equation Tx=x is ill-posed (it may fail to have a solution, nor uniqueness of solution) in general. Regularization therefore is needed. Contractions can be used to regularize nonexpansive mappings. In fact, the following regularization has widely been implemented ([7–9]). Fixing a point u∈C and for each t∈(0,1), one defines a contraction Tt:C→C by
(1.3)Ttx=tu+(1-t)Tx,x∈C.
In this paper we provide an alternative regularization method. Our idea is to shrink x first and then apply T to the convex combination of the shrunk x and the anchor u (this idea appeared implicitly in [10] where iterative methods for finding zeros of maximal monotone operators were investigated). In other words, we fix an anchor u∈C and t∈(0,1) and define a contraction Tt:C→C by
(1.4)Ttx=T(tu+(1-t)x),x∈C.
Compared with (1.1), (1.4) looks slightly more compact in the sense that the mapping T is more directly involved in the regularization and thus may be more convenient in manipulations since the nonexpansivity of T is utilized first.In 2000, Moudafi [11] proved the following strong convergence theorem.Theorem 1.1 (Moudafi [11]).
LetC be a nonempty closed convex subset of a Hilbert space  H and let  S be a nonexpansive mapping of  C into itself such that  F(S) is nonempty. Let  f be a contraction of  C into itself and let  {xn} be a sequence defined as follows:  x1=x∈C and
(1.5)xn+1=11+εnSxn+εn1+εnf(xn)
for all n∈N, where {εn}⊂(0,1) satisfies
(1.6)lim⁡n→∞εn=0,∑n=1∞εn=∞,lim⁡n→∞|1εn+1-1εn|=0.
Then, {xn} converges strongly to z∈F(S), where z=PF(S)f(z) and PF(S) is the metric projection of H onto F(S).Such a method for approximation of fixed points is called the viscosity approximation method.In 2007, S. Takahashi and W. Takahashi [5] introduced and considered the following iterative algorithm by the viscosity approximation method in the Hilbert space:(1.7)x1∈H,ϕ(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C,xn+1=αnf(xn)+(1-αn)Sun,
for all n∈N, where {αn}⊂[0,1] and {rn}⊂(0,∞) satisfy some appropriate conditions. Furthermore, they proved that {xn} and {un} converge strongly to z∈F(S)∩EP(ϕ), where z=PF(S)∩EP(ϕ)f(z).In this paper, motivated and inspired by the above results, we introduce an iterative scheme by the general iterative method for finding a common element of the set of solutions of (1.1) and the set of fixed points of a nonexpansive mapping in Hilbert space.LetS:C→C be a nonexpansive mapping. Starting with an arbitrary x1,u∈H, define sequences {xn} and {un} by
(1.8)ϕ(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C,xn+1=βnf(xn)+(1-βn)T(αnu+(1-αn)un),n≥1.
We will prove in Section 3 that if the sequences {αn}, {βn}, and {γn} of parameters satisfy appropriate conditions, then the sequence {xn} and {un} generated by (1.8) converges strongly to the unique solution of the variational inequality
(1.9)〈(I-f)z,x-z〉≥0,∀x∈F(S)∩EP⁡(ϕ),
which is the optimality condition for the minimization problem
(1.10)min⁡x∈F(S)∩EP⁡(ϕ)12〈x,x〉-h(x),
where h is a potential function for f and at the same time, the sequence {xn} and {un} generated by (1.8) converges in norm to Q(u), where Q:C→Fix(T) is the sunny nonexpansive retraction.


## 2. Preliminaries

Throughout this paper, we considerH as the Hilbert space with inner product 〈·,·〉 and norm ∥·∥, respectively, C is a nonempty closed convex subset of H. Consider a subset D of C and a mapping Q:C→D. Then we say that(i)
Q is a retraction provided Qx=xforx∈D;(ii)
Q is a nonexpansive retraction provided Q is a retraction that is also nonexpansive;(iii)
Q is a sunny nonexpansive retraction provided Q is a nonexpansive retraction with the additional property: Q(x+t(x-Qx))=Qx whenever x+t(x-Qx)∈C, where x∈C and t≥0.Let nowT:C→C be a nonexpansive mapping. For a fixed anchor u∈C and each t∈(0,1) recall that zt∈C is the unique fixed point of the contraction C∋x↦T(tu+(1-t)x). Namely, zt∈C is the unique solution in C to the fixed point equation
(2.1)zt=T(tu+(1-t)zt).
In the Hilbert space (either uniformly smooth or reflexive with a weakly continuous duality map), then zt is strongly convergent should it is bounded as t→0+.We also know that for any sequence{xn}⊂H with xn⇀x, the inequality
(2.2)liminf⁡n→∞‖xn-x‖<liminf⁡n→∞‖xn-y‖
holds for every y∈H with x≠y, (we usually call it Opial’s condition); see [12, 13] for more details.For solving the equilibrium problem for a bifunctionϕ:C×C→ℝ, let us assume that ϕ satisfies the following conditions:(A1)
ϕ(x,x)=0,forallx∈C;(A2)
ϕ is monotone, that is, ϕ(x,y)+ϕ(y,x)≤0,forallx,y∈C;(A3)
For eachx,y,z∈C, limt↓0ϕ(tz+(1-t)x,y)≤ϕ(x,y);(A4)
For eachx∈C, the function y↦ϕ(x,y) is convex and lower semicontinuous.The following lemma appeared implicitly in [14].Lemma 2.1 (see [14]).
LetC be a nonempty closed convex subset of  H and let  ϕ:C×C→ℝ be a bifunction satisfying  (A1)–(A4). Let  r>0 and  x∈H, then, there exists  z∈C such that
(2.3)ϕ(z,y)+1r〈y-z,z-x〉≥0,∀y∈C.Lemma 2.2 (see [6]).
Assume thatϕ:C×C→ℝ satisfies  (A1)–(A4). For  r>0 and  x∈H, define a mapping  Tr:H→C as follows:
(2.4)Tr(x)={z∈C:ϕ(z,y)+1r〈y-z,z-x〉≥0,∀y∈C},
for all z∈H. Then, the following hold:(1)
Tr is single-valued;(2)
Tr is firmly nonexpansive, that is, for any x,y∈H,(2.5)‖Trx-Try‖2≤〈Trx-Try,x-y〉;(3)
F(Tr)=EP(ϕ);(4)
EP(ϕ) is closed and convex.Lemma 2.3 (see [15]).
Let{an}⊂[0,∞),  {bn}⊂[0,∞) and  {cn}⊂[0,1) be sequences of real numbers such that
(2.6)an+1≤(1-cn)an+bn,∀n∈N,∑n=1∞cn=∞,∑n=1∞bn<∞(orlimsup⁡n→∞bncn≤0),
then, limn→∞an=0.Lemma 2.4 (see [9]).
Suppose thatX is a smooth Banach space. Then a retraction  Q:C→D is sunny nonexpansive if and only if
(2.7)〈x-Qx,J(y-Qx)〉≤0,x∈C,y∈D.Lemma 2.5.
LetX be a uniformly smooth Banach space,  C a nonempty closed convex subset of  X, and  T:C→C a nonexpansive mapping. Let  zt be defined by (2.1). Then  (zt) remains bounded as  t→0 if and only if  Fix(T)≠. Moreover, if  Fix(T)≠, then  (zt) converges  in norm, as  t→0+, to a fixed point of T; and if one sets
(2.8)Q(u):=lim⁡t→0zt,
then Q defines the unique sunny nonexpansive retraction from C onto Fix(T).Lemma 2.6.
In the Hilbert space, the following inequalities always hold(i)
∥x+y∥2≤∥x∥2+2〈y,x+y〉;(ii)
∥tx+(1-t)y∥2≤t∥x∥2+(1-t)∥y∥2.


## 3. Main Results

Theorem 3.1.
LetC be a nonempty closed convex subset of  H,  ϕ:C×C→∞ be a bifunction satisfying  (A1)–(A4) and  T:C→C be a nonexpansive mapping of  C into  H such that  F(T)∩EP(ϕ)≠∅. Let  f be a contraction of  H into itself with  α∈(0,1), initially give an arbitrary element  x1∈H and let  {xn} and  {un} be sequences generated by (1.8), where  {αn}⊂[0,1] and  {rn}⊂(0,∞) satisfy the following conditions: (I)
limn→∞αn=0,∑n=1∞αn=∞and∑n=1∞|αn+1-αn|<∞;(II)
liminfnrn>0and∑n=1∞|rn+1-rn|<∞;(III)
limn→∞βn=0,∑n=1∞βn=∞and∑n=1∞|βn+1-βn|<∞;(IV)
limn→∞(αn/βn)=0. Then, the sequences {xn} and {un} converge strongly to z∈F(T)∩EP(ϕ), where z=PF(T)∩EP(ϕ)f(z) and converge in norm to Q(u), where Q:C→Fix(T) is the sunny nonexpansive retraction.Proof.
LetQ=PF(S)∩EP(ϕ). Then Qf is a contraction of H into itself. In fact, there exists a∈[0,1) such that ∥f(x)-f(y)∥≤a∥x-y∥ for all x,y∈H. So, we have that
(3.1)‖Qf(x)-Qf(y)‖≤‖f(x)-f(y)‖≤a‖x-y‖
for all x,y∈H. So, Qf is a contraction of H into itself. Since H is complete, there exists a unique element z∈H such that z=Qf(z), such a z∈H is an element of C. We divide the proof into serval steps.
Step 1.
{xn} and {un} are all bounded. Let p∈F(T)∩EP(ϕ), Then from un=Trnxn, we have
(3.2)‖un-p‖=‖Trnxn-Trnp‖≤‖xn-p‖
for all n∈N. Put yn=αnu+(1-αn)un, so {xn+1} can be rewritten as
(3.3)xn+1=βnf(xn)+(1-βn)Tyn,‖yn-p‖=‖αnu+(1-αn)un-p‖=‖αn(u-p)+(1-αn)(un-p)‖≤αn‖u-p‖+(1-αn)‖un-p‖.
Therefore, from (3.2) we get
(3.4)‖yn-p‖≤αn‖u-p‖+(1-αn)‖xn-p‖≤max⁡⁡{‖u-p‖,‖xn-p‖}.
If ∥xn-p∥≤∥u-p∥, then {xn} is bounded. So, we assume that ∥xn-p∥≥∥u-p∥.
Therefore∥yn-p∥≤∥xn-p∥,
(3.5)‖xn+1-p‖=‖βnf(xn)+(1-βn)Tyn-p‖=‖βn(f(xn)-p)+(1-βn)(Tyn-p)‖≤βn‖f(xn)-p‖+(1-βn)‖yn-p‖≤βnα‖xn-p‖+(1-βn)‖xn-p‖+βn‖f(p)-p‖=(1-(1-α)βn)‖xn-p‖+βn(1-α)‖f(p)-p‖1-α≤max⁡⁡{‖xn-p‖,‖‖f(p)-p‖1-α‖}.
So, by induction, we have
(3.6)‖xn-p‖≤max⁡⁡{‖x1-p‖,‖‖f(p)-p‖1-α‖},
hence {xn} is bounded. we also obtain that {un}, {Tun}, {Txn}, {f(xn)} and {yn} are bounded.Step 2.
∥xn+1-xn∥→0 as n→∞,
(3.7)‖xn+1-xn‖=‖βnf(xn)+(1-βn)Tyn-βn-1f(xn-1)-(1-βn-1)Tyn-1‖=‖βn(f(xn)-f(xn-1))+(βn-βn-1)f(xn-1)+Tyn-Tyn-1-βn(Tyn-Tyn-1)-(βn-βn-1)Tyn-1‖≤βnα‖xn-xn-1‖+|βn-βn-1|‖f(xn-1)-Tyn-1‖+(1-βn)‖yn-yn-1‖,(3.8)‖yn-yn-1‖=‖αnu+(1-αn)un-αn-1u-(1-αn-1)un-1‖=‖(αn-αn-1)u+un-un-1-αn(un-un-1)-(αn-αn-1)un-1‖≤|αn-αn-1|‖u-un-1‖+(1-αn)‖un-un-1‖.
On the other hand, fromun=Trnxn and un+1=Trn+1xn+1, we have
(3.9)ϕ(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C,(3.10)ϕ(un+1,y)+1rn+1〈y-un+1,un+1-xn+1〉≥0,∀y∈C.
Putting y=un+1 in (3.9) and y=un in (3.10), we have
(3.11)ϕ(un,un+1)+1rn〈un+1-un,un-xn〉≥0,ϕ(un+1,un)+1rn+1〈un-un+1,un+1-xn+1〉≥0.
So, from (A2) we have
(3.12)〈un+1-un,un-xnrn-un+1-xn+1rn+1〉≥0,
and hence
(3.13)〈un+1-un,un-un+1+un+1-xn-rnrn+1(un+1-xn+1)〉≥0.
Without loss of generality, let us assume that there exists a real number b such that rn>b>0 for all n∈N. Then, we have
(3.14)‖un+1-un‖2≤〈un+1-un,xn+1-xn+(1-rnrn+1)(un+1-xn+1)〉≤‖un+1+un‖{‖xn+1-xn‖+|1-rnrn+1|‖un+1-xn+1‖},
and hence
(3.15)‖un+1-un‖≤‖xn+1-xn‖+1rn+1|rn+1-rn|‖un+1-xn+1‖≤‖xn+1-xn‖+1b|rn+1-rn|L,
where L=sup{∥un-xn∥:n∈N}. Then we obtain
(3.16)‖un-un-1‖≤‖xn-xn-1‖+1b|rn-rn-1|L.
So, put (3.8) and (3.16) into (3.7) we have
(3.17)‖xn+1-xn‖≤βnα‖xn-xn-1‖+|βn-βn-1|‖f(xn-1)-Tyn-1‖+(1-βn)(|αn-αn-1|‖u-un-1‖+(1-αn)(‖xn+1-xn‖+1b|rn+1-rn|L))≤(1-(1-α)βn)‖xn-xn-1‖+1b|rn+1-rn|L+|αn-αn-1|K1+|βn-βn-1|K2,
where K1:=sup{∥u-un∥,∀n≥1} is a constant; K2:=sup{∥f(xn-1)∥+∥Tyn-1∥,∀n≥1} is a constant.
Using Lemma2.3 and conditions (I), (II), (III) we have
(3.18)lim⁡n→∞‖xn+1-xn‖=0.
From (3.15) and |rn+1-rn|→0, we have
(3.19)lim⁡n→∞‖un+1-un‖=0.
Since yn=αnu+(1-αn)un,xn=βnf(xn-1)+(1-βn)Tyn-1, we have
(3.20)‖yn-un‖=αn‖u-un‖⟶0,asn⟶∞,(3.21)‖xn-Tun‖≤‖xn-Tyn-1‖+‖Tyn-1-Tun-1‖+‖Tun-1-Tun‖≤βn-1‖f(xn-1)+Tyn-1‖+‖yn-1-un-1‖+‖un-un-1‖⟶0,asn⟶∞.
For p∈F(T)∩EP(ϕ), we have
(3.22)‖un-p‖2=‖Trnxn-Trnp‖2≤〈Trnxn-Trnp,xn-p〉=〈un-p,xn-p〉=12(‖un-p‖2+‖xn-p‖2-‖un-xn‖2),‖un-p‖2≤‖xn-p‖2-‖un-xn‖2.
Therefore, we have
(3.23)‖xn+1-p‖2=‖(1-βn)(Tyn-p)+βn(f(xn)-p)‖2≤(1-βn)2‖Tyn-p‖2+2βn〈f(xn)-p,xn+1-p〉≤(1-βn)2(‖yn-un‖+‖un-p‖)2+2βnα‖xn-p‖‖xn+1-p‖+2βn‖f(p)-p‖‖xn+1-p‖=(1-βn)2‖un-p‖2+‖yn-un‖(1-βn)2(‖yn-un‖+2‖un-p‖)+2βnα‖xn-p‖‖xn+1-p‖+2βn‖f(p)-p‖‖xn+1-p‖=(1-βn)2(‖xn-p‖2-‖un-xn‖2)+‖yn-un‖(1-βn)2(‖yn-un‖+2‖un-p‖)+2βnα‖xn-p‖‖xn+1-p‖+2βn‖f(p)-p‖‖xn+1-p‖=(1-2βn+βn2)‖xn-p‖2-(1-βn)2‖un-xn‖2+‖yn-un‖(1-βn)2(‖yn-un‖+2‖un-p‖)+2βnα‖xn-p‖‖xn+1-p‖+2βn‖f(p)-p‖‖xn+1-p‖,(1-βn)2‖un-xn‖2≤‖xn-p‖2-‖xn+1-p‖2+βn(βn‖xn-p‖2-2‖xn-p‖2)-(1-βn)2‖un-xn‖2+‖yn-un‖(1-βn)2(‖yn-un‖+2‖un-p‖)+2βnα‖xn-p‖‖xn+1-p‖+2βn‖f(p)-p‖‖xn+1-p‖≤‖xn-xn+1‖{‖xn-p‖+‖xn+1-p‖}+βn(βn‖xn-p‖2-2‖xn-p‖2)-(1-βn)2‖un-xn‖2+‖yn-un‖(1-βn)2(‖yn-un‖+2‖un-p‖)+2βnα‖xn-p‖‖xn+1-p‖+2βn‖f(p)-p‖‖xn+1-p‖.
By the above of what we have and the condition of limn→∞βn=0, we get limn→∞∥xn-un∥=0. Since ∥Tun-un∥≤∥Tun-xn∥+∥xn-un∥, it follows that ∥Tun-un∥→0.Step 3.
we show that(3.24)limsup⁡n→∞〈f(z)-z,xn-z〉≤0,
where z=PF(S)∩EP(ϕ)f(z). To show this inequality, we choose a subsequence {uni} of {un} such that
(3.25)limsup⁡i→∞〈f(z)-z,xni-z〉=limsup⁡n→∞〈f(z)-z,xn-z〉.
Since {uni} is bounded, there exists a subsequence {unij} of {uni} which converges weakly to w. Without loss of generality, we can assume that uni⇀w. From ∥Tun-un∥→0, we obtain Tuni⇀w. Let us show w∈EP(ϕ). By un=Trnxn, we have
(3.26)ϕ(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C.
From (A2), we also have
(3.27)1rn〈y-un,un-xn〉≥ϕ(y,un),
and hence
(3.28)〈y-uni,uni-xnirni〉≥ϕ(y,uni).
Since uni-xni/rni→0 and uni⇀w, from (A4) we have 0≥ϕ(y,w) for all y∈C. For t with 0<t<1 and y∈C, let yt=ty+(1-t)w. Since y∈C and w∈C, we have yt∈C and hence ϕ(yt,w)≤0. So, from (A1) and A4 we have
(3.29)0=ϕ(yt,yt)≤tϕ(yt,y)+(1-t)ϕ(yt,w)≤tϕ(yt,y),
and hence 0≤ϕ(yt,y). From (A3), we have 0≤ϕ(w,y) for all y∈C, and hence w∈EP(ϕ). We will show that w∈F(T). Assume that w∉F(T). Since uni⇀w and w≠Tw, from Opial’s theorem we have
(3.30)liminf⁡i→∞‖uni-w‖<liminf⁡i→∞‖uni-Tw‖≤liminf⁡i→∞{‖uni-Tuni‖+‖Tuni-Tw‖}≤liminf⁡i→∞‖uni-w‖.
This is a contradiction. So, we get w∈F(T). Therefore, w∈F(T)∩EP(ϕ). Since z=PF(T)∩EP(ϕ)f(z), we have
(3.31)limsup⁡n→∞〈f(z)-z,xn-z〉=lim⁡i→∞〈f(z)-z,xni-z〉=〈f(z)-z,w-z〉≤0.
Fromxn+1-z=(1-βn)(Tyn-z)+βn(f(xn)-z), we have
(3.32)‖xn+1-z‖2≤(1-βn)2‖yn-z‖2+2βn〈f(xn)-z,xn+1-z〉≤(1-βn)2‖αn(u-z)+(1-αn)(xn-z)‖2+2βnα‖xn-z‖‖xn+1-z‖+2βn〈f(z)-z,xn+1-z〉≤(1-βn)2(αn‖u-z‖2+(1-αn)‖xn-z‖2)+βnα(‖xn-z‖2+‖xn+1-z‖2)+2βn〈f(z)-z,xn+1-z〉≤[(1-βn)2(1-αn)+βnα]‖xn-z‖2+βnα‖xn+1-z‖2+(1-βn)2αn‖u-z‖2+2βn〈f(z)-z,xn+1-z〉,‖xn+1-z‖2≤((1-βn)2+βnα)1-αβn‖xn-z‖2+(1-βn)2αn1-αβn‖u-z‖2+2βn1-αβn〈f(z)-z,xn+1-z〉≤[1-2(1-α)βn1-αβn]‖xn-z‖2+βn21-αβn‖xn-z‖2+(1-βn)2αn1-αβn‖u-z‖2+2βn1-αβn〈f(z)-z,xn+1-z〉≤[1-2(1-α)βn1-αβn]‖xn-z‖2+2(1-α)βn1-αβn×{βn2(1-α)M+(1-βn)2αn2(1-α)βn‖u-z‖2+11-α〈f(z)-z,xn+1-z〉}=(1-δn)‖xn-z‖2+δnζn,
where M=sup{∥xn-z∥2:n∈ℕ},   δn=2(1-α)βn/1-αβn and ζn:=βn/2(1-α)M+(1-βn)2αn/2(1-α)βn∥u-z∥2+1/1-α〈f(z)-z,xn+1-z〉. It is easy to see that δn→0,Σn=1∞δn=∞ and limsupn→∞ζn/δn≤0 by (3.31) and the conditions. Hence, by Lemma 2.3, the sequence {xn} converges strongly to z.
Ifzt is definition as (2.1), then, from Lemma 2.5, we have ∥zt-q∥→0 as t→0, and if we set Q(u):=limt→0zt, then Q defines the unique sunny nonexpansive retraction from C onto Fix(T). So, if we replace t with αn, the corollary still holds. and it is that zn=T(αnu+(1-αn)zn) is a fixed point sequence and ∥zn-q∥→0 as n→∞, and if we set Q(u):=limn→∞zn, then Q defines the unique sunny nonexpansive retraction from C onto Fix(T). In the iterative algorithm of Theorem 3.1, we can take zn to replace Tyn in particular. Then, we have xn+1=βnf(xn)+(1-βn)zn, so ∥xn+1-zn∥=βn∥f(xn)-zn∥→0 as n→∞. By the uniqueness of limit, we have z=q, that is, z=Q(u), where Q defines the unique sunny nonexpansive retraction from C onto Fix(T).Remark 3.
We notice thatun=Trnxn has not influence on xn,un→z=PF(T)∩EP(ϕ)f(z).As direct consequences of Theorem3.1, we obtain corollary.Corollary 3.2.
LetC be a nonempty closed convex subset of H, S:C→C be a nonexpansive mapping of C into H such that F(S)≠∅. Let f be a contraction of H into itself and let {xn} and {un} be sequences generated initially by an arbitrary elements x1∈H and then by
(3.33)xn+1=βnf(xn)+(1-βn)S(αnu+(1-αn)PCxn)
for all n∈N, where {αn}⊂(0,∞) satisfies the following conditions:(I)
limn→∞αn=0,∑n=1∞αn=∞and∑n=1∞|αn+1-αn|<∞;(II)
limn→∞βn=0,∑n=1∞βn=∞and∑n=1∞|βn+1-βn|<∞;(III)
limn→∞αn/βn=0. Then, the sequences {xn} converge strongly to z∈F(S), where z=PF(S)f(z).Proof.
Putϕ(x,y)=0, for all x,y∈C and rn=1, for all n∈ℕ in Theorem 3.1.
Then we haveun=PCxn. So, from Theorem 3.1, the sequence xn generated by x1∈H and
(3.34)xn+1=βnf(xn)+(1-βn)S(αnu+(1-αn)PCxn)
for all n∈ℕ converges strongly to z∈F(S), where z=PF(S)f(z).


## 4. Application for Zeros of Maximal Monotone Operators

We adapt in this section the iterative algorithm (3.1) to find zeros of maximal monotone operators and find EP(ϕ). Let us recall that an operator A with domain D(A) and range R(A) in a real Hilbert space H with inner product 〈·,·〉 and norm ∥·∥ is said to be monotone if the graph of A,
(4.1)G(T):={(x,y)∈H×H:x∈D(T),y∈Tx}
is a monotone set. Namely,
(4.2)〈x-x′,y-y′〉≥0,(x,y),(x′,y′)∈G(A).A monotone operatorA is said to be maximal monotone of the graph G(T) is not properly contained in the graph of any other monotone defined in H. See Brezis [16] for more details on maximal monotone operators.In this section we always assume thatA is maximal monotone and the set of zeros of A, N(A)={x∈D(A):0∈Ax}, is nonempty so that the metric projection PN(A) from H onto N(A) is well-defined.One of the major problems in the theory of maximal operators is to find a point in the zero setN(A) because various problems arising from economics, convex programming, and other applied areas can be formulated as finding a zero of maximal monotone operators. The proximal point algorithm (PPA) of Rockafellar [17] is commonly recognized as the most powerful algorithm in finding a zero of maximal monotone operators. This (PPA) generates, starting with any initial guess x0∈H, a sequence {xn} according to the inclusion:
(4.3)xn+en∈xn+1+cnA(xn+1),
where {en} is a sequence of errors and {cn} is a sequence of positive regularization parameters. Equivalently, we can write
(4.4)xn+1=JcnA(xn+en),
where for c>0,JcA denotes the resolvent of A,
(4.5)JcA=(I+cA)-1,
with I being the identity operator on the space H.Rockafellar [17] proved the weak convergence of his algorithm (4.4) provided the regularization sequence {cn} remains bounded away from zero and the error sequence {en} satisfies the condition
(4.6)∑n=0∞‖en‖<∞.The aim of this section is to combine algorithm (3.1) with algorithm (4.4). Our algorithm generates a sequence {xn} and {un} be sequences generated initially by an arbitrary elements x1∈H and then by
(4.7)ϕ(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C,xn+1=βnf(xn)+(1-βn)JcnA(αnu+(1-αn)un+en),
where αn and cn are sequences of positive real numbers. Furthermore, we prove that {xn} and {un} converge strongly to z∈N(A)∩EP(ϕ), where z=PN(A)∩EP(ϕ)f(z).Before stating the convergence theorem of the algorithm (4.7), we list some properties of maximal monotone operators.Proposition 4.1.
LetA be a maximal monotone operator in  H and let  JcA=(I+cA)-1 denote the resolvent, where  c>0, (a)
JcA is nonexpansive for all c>0;(b)
N(A)=F(Jc) for all c>0;(c)
Forc>c′>0,∥Jc′A-x∥≤2∥JcA-x∥forx∈H;(d)
(The Resolvent Identity) Forλ,μ>0, there holds the identity:(4.8)JλTx=JμT(μλx+(1-μλ)JλTx),x∈H.Theorem 4.2.
LetC be a nonempty closed convex subset of  H,  ϕ:C×C→∞ be a bifunction satisfying  (A1)–(A4) and  A be a maximal monotone operator such that  N(A)∩EP(ϕ)≠∅. Let  f be a contraction of  H into itself and let  {xn} and  {un} be sequences generated initially by an arbitrary elements  x0∈H and then by
(4.9)ϕ(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈C,xn+1=βnf(xn)+(1-βn)JcnA(αnu+(1-αn)un+en),
for all n∈N, where {αn}⊂[0,1] and {rn}⊂(0,∞) satisfy the following conditions:(I)
limn→∞αn=0, Σn=1∞αn=∞ and Σn=1∞|αn+1-αn|<∞;(II)
liminfnrn>0 and Σn=1∞|rn+1-rn|<∞;(III)
limn→∞(cn+1/cn)=1;(IV)
limn→∞βn=0, Σn=1∞βn=∞, Σn=1∞|βn+1-βn|<∞, and limn→∞(αn/βn)=0. Then, the sequences {xn} and {un} converge strongly to z∈N(A)∩EP(ϕ), where z=PN(A)∩EP(ϕ)f(z).Proof.
Below we writeJc=JcA for simplicity. Setting
(4.10)wn=αnu+(1-αn)un+en,yn=Jcnwn,
we rewrite xn+1 of (4.7) as
(4.11)xn+1=βnf(xn)+(1-βn)Jcnwn=βnf(xn)+(1-βn)yn.
Because the proof is similar to Theorem 3.1, here we just give the main steps as follows:(1)
{xn} is bounded;(2)
∥xn+1-xn∥→0, as n→0;(3)
∥un-Jcnun∥→0, as n→0;(4)
∥xn-un∥→0, as n→0;(5)
limsupn→∞〈f(z)-z,xn-z〉≤0;(6)
xn,un→z, as n→z.


## 5. Application for Optimization Problem

In this section, we study a kind of optimization problem by using the result of this paper. That is, we will give an iterative algorithm of solution for the following optimization problem with nonempty set of solutions(5.1)min⁡h(x),x∈C,
where h(x) is a convex and lower semicontinuous functional defined on a closed convex subset C of a Hilbert space H. We denoted by B the set of solutions of (5.1). Let ϕ be a bifunction from C×C to R defined by ϕ(x,y)=h(y)-h(x). We consider the following equilibrium problem, that is, to find x∈C such that
(5.2)ϕ(x,y)≥0,∀y∈C.It is obvious thatEP(ϕ)=B, where EP(ϕ) denotes the set of solutions of equilibrium problem (5.2). In addition, it is easy to see that ϕ(x,y) satisfies the conditions (A1)–(A4) in the Section 2. Therefore, from Theorem 3.1, we know that the following iterative algorithm:
(5.3)h(y)-h(un)+1rn〈y-un,un-xn〉≥0,∀y∈C,xn+1=βnf(xn)+(1-βn)T(αnu+(1-αn)un),
for any initial guess x1, converges strongly to a solution z=PBf(z) of optimization problem (5.1), where {αn}⊂[0,1], {βn}⊂[0,1], and {rn}⊂[0,∞) satisfy(5.4)lim⁡n→∞αn=0,∑n=1∞αn=∞,∑n=1∞|αn+1-αn|<∞,lim⁡n→∞βn=0,∑n=1∞βn=∞,∑n=1∞|βn+1-βn|<∞,lim⁡n→∞αnβn=0lim⁡inf⁡⁡rnn→∞>0,∑n=1∞|rn+1-rn|<∞.,For a special case, we pickf(x)=η, for all η∈H, and rn=1,  βn=1/2 and αn=0 for all n≥1, then xn+1=(1/2)Tun+(1/2)η, from (5.3), we get the special iterative algorithm as follows:
(5.5)h(y)-h(un)+〈y-un,un-(12η+12Tun)〉≥0,∀y∈C,n≥2,h(y)-h(u1)+〈y-u1,u1-(12η+12Tu1)〉≥0,∀y∈C.
Then {un} converges strongly to a solution z=PBη of optimization problem (5.1).In fact,z is the minimum norm point from η onto the B, furthermore, if η=0, then z is the minimum norm point on the B.



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*Source: 202860-2012-05-13.xml*