AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 351372 10.1155/2014/351372 351372 Research Article Approximations for Equilibrium Problems and Nonexpansive Semigroups Wu Huan-chun Cheng Cao-zong Aydi Hassen Department of Mathematics Beijing University of Technology Beijing 100124 China bjpu.edu.cn 2014 1632014 2014 29 11 2013 04 02 2014 16 3 2014 2014 Copyright © 2014 Huan-chun Wu and Cao-zong Cheng. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We introduce a new iterative method for finding a common element of the set of solutions of an equilibrium problem and the set of all common fixed points of a nonexpansive semigroup and prove the strong convergence theorem in Hilbert spaces. Our result extends the recent result of Zegeye and Shahzad (2013). In the last part of the paper, by the way, we point out that there is a slight flaw in the proof of the main result in Shehu's paper (2012) and perfect the proof.

1. Introduction

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. A mapping T:CC is called nonexpansive if Tx-Tyx-y for all x,yC. We denote the set of fixed points of T by F(T). It is known that F(T) is closed and convex. A family 𝒮={T(t):0t<} of mappings from C into itself is called a nonexpansive semigroup on C if it satisfies the following conditions:

T(0)x=x for all xC;

T(s+t)=T(s)T(t) for all s,t0;

T(s)x-T(s)yx-y for all x,yC and s0;

for all xC,sT(s)x is continuous.

We denote by F(𝒮) the set of all common fixed points of 𝒮; that is, F(𝒮)=0t<F(T(t)). It is clear that F(𝒮) is a closed convex subset.

The equilibrium problem for f:C×C is to find x¯C such that f(x¯,y)0 for all yC. The set of such solutions is denoted by EP(f). Numerous problems in physics, optimization, and economics can be reduced to find a solution of the equilibrium problem (for instance, see ).

For solving equilibrium problem, we assume that the bifunction f satisfies the following conditions:

f(x,x)=0 for all xC;

f is monotone; that is, f(x,y)+f(y,x)0 for any x,yC;

for each x,y,zC, limsupt0f(tz+(1-t)x,y)f(x,y);

f(x,·) is convex and lower semicontinuous for each xC.

Several methods have been proposed to solve the equilibrium problem; see . For finding common fixed points of a nonexpansive semigroup, Nakajo and Takahashi  introduced a convergent sequence for nonexpansive semigroup 𝒮={T(t):0t<} as follows: (1)x0=xC,yn=αnxn+(1-αn)1tn0tnT(s)xnds,Cn={zC:yn-zxn-z},Qn={zC:xn-z,x0-xn0},xn+1=PCnQn(x0).

Some authors have paid more attention to find an element pF(𝒮)EP(f). Buong and Duong  constructed the following iterative sequence and proved the weak convergence theorem for an equilibrium problem and a nonexpansive semigroup in Hilbert spaces: (2)x0H,ukC,f(uk,y)+1rky-uk,uk-xk0yC,xk+1=μkxk+(1-μk)1tk0tkT(s)ukds.

In 2012, Shehu  studied iterative methods for fixed point problem, variational inequality, and generalized mixed equilibrium problem and introduced a new algorithm which does not involve the CQ algorithm and viscosity approximation method. However, we discover that there is a slight flaw in the proof of Theorem 3.1 in .

Motivated by Nakajo and Takahashi , Buong and Duong , and especially Shehu  and Zegeye and Shahzad , we present a new iterative method for finding a common element of the set of solutions of an equilibrium problem and the set of all common fixed points of a nonexpansive semigroup and prove the strong convergence theorem in Hilbert spaces. Our result extends the recent result of . In the last part of the paper, we perfect and simplify the proof of Theorem 3.1 in .

2. Preliminaries

Throughout this paper, let H be a real Hilbert space with inner product ·,· and norm ·, and let C be a nonempty closed convex subset of H. We write xnx to indicate that the sequence {xn} converges strongly to x. Similarly, xnx will symbolize weak convergence. It is well known that H satisfies Opial’s condition; that is, for any sequence {xn}H with xnx, we have (3)liminfnxn-x<liminfnxn-y,yx. For any xH, there exists a unique point PCxC such that (4)x-PCxx-y,yC.PC is called the metric projection of H onto C. We know that PC is a nonexpansive mapping of H onto C and PC satisfies (5)x-y,PCx-PCyPCx-PCy2,x,yH. For xH and zC, we have (6)z=PCxx-z,y-z0,for everyyC.

The following lemmas will be used in the proof of our results.

Lemma 1 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

Let C be a nonempty closed convex subset of H, and let f be a bifunction from C×C to satisfying (A1)–(A4). If r>0 and xH, then there exists zC such that (7)f(z,y)+1ry-z,z-x0,yC.

Lemma 2 (see [<xref ref-type="bibr" rid="B2">2</xref>]).

For r>0, define a mapping  Tr:H2C as follows: (8)Tr(x)={zC:f(z,y)+1ry-z,z-x0,yC}. Then the following hold:

Tr is single valued;

Tr is firmly nonexpansive; that is, for any x,yH, x-y,Trx-TryTrx-Try2;

F(Tr)=EP(f);

EP(f) is closed and convex.

Lemma 3 (see [<xref ref-type="bibr" rid="B12">12</xref>]).

Suppose that (A1)–(A4) hold. If x,yH and r1,r2>0, then (9)Tr2y-Tr1xy-x+|r2-r1|r2Tr2y-y.

Lemma 4 (see [<xref ref-type="bibr" rid="B13">13</xref>]).

Let C be a nonempty bounded closed subset of H, and let {T(s):0s<} be a nonexpansive semigroup on C. Then, for every h0, (10)limt+supxC1t0tT(s)xds-T(h)1t0tT(s)xds=0.

Lemma 5 (see [<xref ref-type="bibr" rid="B14">14</xref>]).

Let {xn} and {yn} be bounded sequences in a Banach space X, and let {βn} be a sequence in [0,1] with 0<liminfnβnlimsupnβn<1. Suppose that xn+1=(1-βn)xn+βnyn  for all integers n1 and limsupn(yn+1-yn-xn+1-xn)0. Then, limnyn-xn=0.

Lemma 6 (see [<xref ref-type="bibr" rid="B15">15</xref>]).

Let {an} be a sequence of nonnegative real numbers satisfying an+1(1-αn)an+αnβn, where

{αn}(0,1), n=1αn=;

limsupnβn0.

Then, limnan=0.

3. Strong Convergence Theorems

In this section, we introduce a new iterative method for finding a common element of the set of solutions of an equilibrium problem and the set of all common fixed points of a nonexpansive semigroup and prove the strong convergence theorem in Hilbert spaces.

Theorem 7.

Let C be a nonempty closed convex subset of a real Hilbert space H, and let f be a bifunction from C×C to satisfying (A1)–(A4). Suppose that 𝒮={T(t):0t<} is a nonexpansive semigroup on C such that F(𝒮)EP(f). For uH, let {xn}, {yn}, and {zn} be generated by (11)x1Cchosen arbitrarily,yn=αnu+(1-αn)xn,znC,such thatf(zn,y)+1rny-zn,zn-yn0yC,xn+1=(1-βn)xn+βn1tn0tnT(s)znds, where the real sequences {αn}, {βn} in (0,1) and {rn}(0,) satisfy the following conditions:

limnαn=0 and n=1αn=;

0<liminfnβnlimsupnβn<1;

0<crn<,limn|rn+1-rn|=0;

{tn}(0,), limntn= and limn(|tn+1-tn|/tn+1)=0.

Then, the sequence {xn} converges strongly to PF(𝒮)EP(f)u.

Proof.

Note that the set F(𝒮)EP(f) is closed and convex since F(𝒮) and EP(f) are closed and convex. For simplicity, we write Ω:=F(𝒮)EP(f).

From Lemmas 1 and 2, we have zn=Trnyn, and, for any pΩ, (12)zn-p=Trnyn-Trnpyn-p. Observe that (13)yn-pαnu-p+(1-αn)xn-p. It follows that (14)xn+1-p=(1-βn)xn+βn1tn0tnT(s)znds-p(1-βn)xn-p+βn1tn0tnT(s)zn-T(s)pds(1-βn)xn-p+βnzn-p(1-βn)xn-p+βn[αnu-p+(1-αn)xn-p](1-αnβn)xn-p+αnβnu-pmax{xn-p,u-p}. From a simple inductive process, one has (15)xn+1-pmax{x1-p,u-p}, which yields that {xn} is bounded. So are {yn} and {zn}.

Set σn=(1/tn)0tnT(s)znds. For any pΩ, we have (16)σn+1-σn=1tn+10tn+1T(s)zn+1ds-1tn0tnT(s)znds=1tn+10tn+1T(s)zn+1ds-1tn+10tn+1T(s)znds+1tn+10tn+1T(s)znds-1tn0tnT(s)znds=1tn+10tn+1(T(s)zn+1-T(s)zn)ds+(1tn+1-1tn)0tnT(s)znds+1tn+1tntn+1T(s)znds=1tn+10tn+1(T(s)zn+1-T(s)zn)ds+(1tn+1-1tn)0tn(T(s)zn-T(s)p)ds+(1tn+1-1tn)0tnT(s)pds+1tn+1tntn+1T(s)znds=1tn+10tn+1(T(s)zn+1-T(s)zn)ds+(1tn+1-1tn)0tn(T(s)zn-T(s)p)ds+1tn+1tntn+1(T(s)zn-T(s)p)dszn+1-zn+2|tn+1-tn|tn+1zn-p. It follows from Lemma 3 that (17)zn+1-znyn+1-yn+|rn+1-rn|rn+1zn+1-yn+1. Hence, (18)σn+1-σnyn+1-yn+|rn+1-rn|rn+1zn+1-yn+1+2|tn+1-tn|tn+1zn-pαn+1u+(1-αn+1)xn+1-αnu-(1-αn)xn+|rn+1-rn|rn+1zn+1-yn+1+2|tn+1-tn|tn+1zn-pxn+1-xn+αn+1(u+xn+1)+αn(u+xn)+|rn+1-rn|czn+1-yn+1+2|tn+1-tn|tn+1zn-p. This implies that (19)limsupn(σn+1-σn-xn+1-xn)0. It follows from Lemma 5 that limnσn-xn=0. Thus, (20)limnxn+1-xn=limnβnσn-xn=0.

For any pΩ, we have (21)zn-p2=Trnyn-Trnp2yn-p,zn-p=12[yn-p2+zn-p2-zn-yn2]. Thus (22)zn-p2yn-p2-zn-yn2. From the convexity of ·2, it follows that (23)xn+1-p2=(1-βn)(xn-p)+βn(1tn0tnT(s)znds-p)2(1-βn)xn-p2+βn1tn0tn(T(s)zn-T(s)p)ds2(1-βn)xn-p2+βnzn-p2(1-βn)xn-p2+βn[yn-p2-zn-yn2](1-βn)xn-p2+βn[αnu+(1-αn)xn-p2-zn-yn2](1-βn)xn-p2+βn[(1-αn)xn-p2+αnu-p2-zn-yn2]. Hence (24)βnzn-yn2xn-p2-xn+1-p2+αnβnu-p2xn-xn+1(xn-p+xn+1-p)+αnβnu-p2. Since limnxn+1-xn=0 and limnαn=0, one has (25)limnzn-yn=0. Observe that (26)limnyn-xn=limnαnu-xn=0. As zn-xnzn-yn+yn-xn, the following equality holds: (27)limnzn-xn=0.

Now we show that (28)limnT(s)zn-zn=0,0s<. In fact, we have (29)T(s)zn-zn=T(s)zn-T(s)1tn0tnT(s)znds=+T(s)1tn0tnT(s)znds-1tn0tnT(s)znds=+1tn0tnT(s)znds-znT(s)zn-T(s)1tn0tnT(s)znds+T(s)1tn0tnT(s)znds-1tn0tnT(s)znds+1tn0tnT(s)znds-zn21tn0tnT(s)znds-zn+T(s)1tn0tnT(s)znds-1tn0tnT(s)znds. Notice that (30)1tn0tnT(s)znds-zn1tn0tnT(s)znds-xn+xn-zn=1βnxn+1-xn+xn-zn0,asn. For any pΩ, let G={xC:x-pmax{x1-p,u-p}}. It is easy to see that G is a bounded closed convex subset and T(s)G is a subset of G. Since (31)zn-p=Trnyn-p(1-αn)xn-p+αnu-pmax{x1-p,u-p}, the sequence {zn} is contained in G. It follows from Lemma 4 that (32)limn1tn0tnT(s)znds-T(s)1tn0tnT(s)znds=0. From (29), (30), and (32), the expression (28) is obtained.

Next we prove that (33)limsupnu-x¯,xn-x¯0, where x¯=PΩu. In order to show this inequality, we can choose a subsequence {xnj} of {xn} such that (34)limsupnu-x¯,xn-x¯=limju-x¯,xnj-x¯. Due to the boundedness of {xnj}, there exists a subsequence {xnji} of {xnj} such that xnjiω. Without loss of generality, we assume that xnjω. From (27), we see that znjω. Since {znj}C and C is closed and convex, we get ωC.

We first show that ωEP(f). By zn=Trnyn, we have (35)f(zn,y)+1rny-zn,zn-yn0,yC. It follows from the monotonicity of f that (36)1rny-zn,zn-ynf(y,zn),yC. Replacing n by nj, we obtain (37)y-znj,znj-ynjrnjf(y,znj),yC. From (25), (27), and (A4), we have (38)f(y,ω)0,yC. For 0<t1, yC, set yt=ty+(1-t)ω. We have ytC and f(yt,ω)0. Hence (39)0=f(yt,yt)tf(yt,y)+(1-t)f(yt,ω)tf(yt,y). Dividing by t, we see that (40)f(yt,y)0. Letting t0 and from (A3), we get (41)f(ω,y)0,yC. That is, ωEP(f).

Second, we prove that ωF(𝒮). Note that the equality (27) implies that znjω. Suppose for contradiction that ωF(𝒮); that is, (42)there existss0>0such thatT(s0)ωω. Then from Opial’s condition and (28), we obtain (43)liminfjznj-ω<liminfjznj-T(s0)ω=liminfjznj-T(s0)znj+T(s0)znj-T(s0)ω=liminfjT(s0)znj-T(s0)ωliminfjznj-ω, which is a contradiction. Therefore, ωF(𝒮). Consequently, one gets ωΩ.

From (34) and the property of metric projection, we have (44)limsupnu-x¯,xn-x¯=limju-x¯,xnj-x¯=u-x¯,ω-x¯0. The inequality (33) arrives.

Finally we show that xnx¯. From (11), we have (45)xn+1-x¯2=(1-βn)xn+βn1tn0tnT(s)znds-x¯2(1-βn)xn-x¯2+βnzn-x¯2(1-βn)xn-x¯2+βnyn-x¯2(1-αnβn)xn-x¯2+αnβn[2(1-αn)u-x¯,xn-x¯+αnu-x¯2]. It follows from (33) and Lemma 6 that {xn} converges strongly to x¯.

Remark 8.

Let H= and C=[0,1]. Setting f(x,y)=y2-x2, we see that f(x,y) satisfies (A1)–(A4). For 0t<+, let (46)T(t)x=x1+tx,x[0,1]. Thus, it follows that 𝒮={T(t):0t<} is a nonexpansive semigroup such that F(𝒮)EP(f)={0}. If we take (47)αn=1n+1,β1=β2=12,βn=12-1n,n3,rnc>0,tn=n, then all assumptions and conditions in Theorem 7 are satisfied.

Remark 9.

Taking u=0 in Theorem 7, we obtain the iterative method for minimum-norm solution of an equilibrium problem and a nonexpansive semigroup.

As a direct consequence of Theorem 7, we obtain the following corollary.

Corollary 10.

Let C be a nonempty closed convex subset of a real Hilbert space H, and assume that 𝒮={T(t):0t<} is a nonexpansive semigroup on C such that F(𝒮). Let {αn} and {βn} be real sequences in (0,1), and let {xn} and {zn} be generated by (48)x1Cchosen arbitrarily,zn=PC[(1-αn)xn],xn+1=(1-βn)xn+βn1tn0tnT(s)znds. Suppose that the following conditions are satisfied:

limnαn=0 and n=1αn=;

0<liminfnβnlimsupnβn<1;

{tn}(0,), limntn=, and limn(|tn+1-tn|/tn+1)=0.

Then the sequence {xn} converges strongly to PF(𝒮)0.

Proof.

Letting f(x,y)=0 for all x,yC, rn=1, and u=0 in Theorem 7, we get the result.

Remark 11.

Corollary 10 extends the recent results of Zegeye and Shahzad [11, Corollaries 3.2 and 3.3] from finite family of nonexpansive mappings to a nonexpansive semigroup.

4. A Note on Shehu’s Paper “Iterative Method for Fixed Point Problem, Variational Inequality and Generalized Mixed Equilibrium Problems with Applications”

In 2012, Shehu  studied iterative methods for fixed point problem, variational inequality, and generalized mixed equilibrium problem and gave an interesting convergence theorem. However, there is a slight flaw in the proof of the main result (Theorem 3.1 in ).

Shehu obtained the following result (for more details, see ).

Theorem 12 (see [<xref ref-type="bibr" rid="B10">10</xref>]).

Let K be a closed convex subset of a real Hilbert space H, let F be a bifunction from K×K satisfying (A1)–(A4), let φ:K{+} be a proper lower semicontinuous and convex function with assumption (B1) or (B2), let A be a μ-Lipschitzian, relaxed (λ,γ)-cocoercive mapping of K into H, and let ψ be an α-inverse, strongly monotone mapping of K into H. Suppose that T:KK is a nonexpansive mapping of K into itself such that Ω:=F(T)VI(K,A)GMEP. Let {αn} and {βn} be two real sequences in (0,1) and {rn},{sn}(0,). Let {xn}, {yn}, and {un} be generated by x1K, (49)yn=PK[(1-αn)xn],un=Trn(F,φ)(yn-rnψyn),xn+1=(1-βn)xn+βnTPK(un-snAun). Suppose that the following conditions are satisfied:

limnαn=0 and n=1αn=;

0<liminfnβnlimsupnβn<1;

0<asnb<2(γ-λμ2)/μ2,limn|sn+1-sn|=0;

0<crnd<2α,limn|rn+1-rn|=0.

Then, the sequence {xn} converges strongly to an element of F(T)VI(K,A)GMEP.

This theorem is proved in  by the following steps.

Step 1.

The sequence {xn} is bounded.

Step 2.

The following equalities hold: (50)limnAun-Ax*=0,limnψyn-ψx*=0,x*Ω.

Step 3.

If ω is a weak limit of {xnj} which is a subsequence of {xn}, then ωΩ.

Step 4.

The sequence {xn} converges strongly to ω.

In Step 4, in order to show that the sequence {xn} converges strongly to ω, the author shows the inequality limsupn-ω,xn-ω0 by defining a mapping ϕ:HR as follows: ϕ(x)=μnxn-x2, where μ is a Banach limit. It is proved that the set K*={xH:ϕ(x)=minuHϕ(u)} and K*F(T). An element of K*F(T) is taken arbitrarily and is denoted by ω. Of course, the element ω is not necessarily the weak sequential cluster point of {xn}. However, in Step 3, the symbol ω stands for the weak limit of {xnj} which is a subsequence of {xn}. In the sequel, the author obtains (51)xn+1-ω2(1-βn)xn-ω2+βnyn-ω2. The meaning of the element ω in (51) is ambiguous. It is difficult to ensure consistency.

Now, we perfect and simplify the proof of Step 4. According to the equality in Step 2, limnAun-Ax*=0, for all x*Ω, we see that the set {Ax*:x*Ω} contains only one element. Since A is a relaxed (λ,γ)-cocoercive mapping of K into H, that is, there exist λ,γ>0 such that (52)Ax-Ay,x-y-λAx-Ay2+γx-y2,x,yK, it follows that the mapping A is one-to-one. Therefore, the set Ω is a singleton. By Step 3, the sequence {xn} possesses only one weak sequential cluster point. It follows from Lemma 2.38 in  that {xn} converges weakly to ω and so (53)xn+1-ω2(1-βn)xn-ω2+βnyn-ω2(1-βn)xn-ω2+βn(1-αn)xn-ω2(1-βn)xn-ω2+βn[(1-αn)xn-ω2+βn+2αn(1-αn)-ω,xn-ω+αn2ω2](1-αnβn)xn-ω2+αnβn[2(1-αn)-ω,xn-ω+αnω2]. Since {xn} converges weakly to ω, it follows from Lemma 2.2 in  or Lemma 6 in this paper that {xn} converges strongly to ωΩ.

Conflict of Interests

The authors declare that there is no conflict of interests.

Acknowledgment

The authors would like to thank referees and editors for their valuable comments and suggestions.

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