Strong Convergence Theorems for a Generalized Equilibrium Problem with a Relaxed Monotone Mapping and a Countable Family of Nonexpansive Mappings in a Hilbert Space

Volume 2010, Article ID 230304,22pages doi:10.1155/2010/230304

Research Article

Strong Convergence Theorems for a Generalized Equilibrium Problem with a Relaxed Monotone Mapping and a Countable Family of Nonexpansive Mappings in a Hilbert Space

Shenghua Wang,

Giuseppe Marino,

and Fuhai Wang

1School of Applied Mathematics and Physics, North China Electric Power University, Baoding 071003, China

2Dipartimento di Matematica, Universit´a della Calabria, 87036 Arcavacata di Rende, Italy

Correspondence should be addressed to Shenghua Wang,sheng-huawang@hotmail.com Received 15 March 2010; Accepted 20 June 2010

Academic Editor: Naujing Jing Huang

Copyrightq2010 Shenghua Wang et al. 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 a generalized equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings in a Hilbert space and then prove that the sequence converges strongly to a common element of the two sets. Using this result, we prove several new strong convergence theorems in fixed point problems, variational inequalities, and equilibrium problems.

1. Introduction

Throughout this paper, letRdenote the set of all real numbers, let Ndenote the set of all positive integer numbers, letHbe a real Hilbert space, and letCbe a nonempty closed convex subset ofH. LetS:C → Cbe a mapping. We callSnonexpansive if

Sx−Sy≤x−y, ∀x, y∈C. 1.1 The set of fixed points ofSis denoted by FixS. We know that the set FixSis closed and convex. LetΦ:C×C → Rbe a bifunction. The equilibrium problem forΦis to findz∈C such that

Φ z, y

≥0, ∀y∈C. 1.2

The set of all solutions of the equilibrium problem is denoted by EPΦ, that is,

EPΦ

z∈C:Φ z, y

≥0, ∀y∈C

. 1.3

Some iterative methods have been proposed to find an element of EPΦ∩FixS; see1,2.

A mappingA:C → His called inverse-strongly monotone if there existsλ >0 such that

x−y, Ax−Ay

≥λAx−Ay², ∀x, y∈C. 1.4 Such a mappingAis also calledλ-inverse-strongly monotone. It is known that each nonex- pansive mapping is 1/2-inverse-strongly monotone and eachκ-strictly pseudocontraction is 1−κ/2-inverse-strongly monotone; see3,4. If there existsu∈Csuch that

v−u, Au ≥0, ∀v∈C, 1.5

then u is called a solution of the variational inequality. The set of all solutions of the variational inequality is denoted by VIC, A. It is known that VIC, Ais closed and convex.

Recently Takahashi and Toyoda5introduced an iterative method for finding an element of FixS∩VIC, A; see also6. On the other hand, Plubtieng and Punpaeng7introduced an iterative method for finding an element of FixS∩EPΦ∩VIC, A; see also8.

Consider a general equilibrium problem:

Findz∈Csuch thatΦ z, y

Az, y−z

≥0, ∀y∈C. 1.6

The set of all solutions of the equilibrium problem is denoted by EP, that is, EP

z∈C:Φ z, y

Az, y−z

≥0, ∀y∈C

. 1.7

In the case ofA≡0, EP coincides with EPΦ. In the caseΦ≡0, EP coincides with VIC, A.

Recently, S. Takahashi and W. Takahashi9introduced an iterative method to find an element of EP∩FixS. More precisely, they introduced the following iterative scheme:u∈C,x1∈C, and

Φ zn, y

Axn, y−zn

λn

y−zn, zn−xn

≥0, ∀y∈C, xn1βnxn

1−βn

Sαnu 1−αnzn, n∈N, 1.8

where{αn} ⊂0,1,{βn} ⊂0,1, and{λn} ⊂0,2λare three control sequences. They proved that{xn}converges strongly tozPFixS∩EPu.

A mappingT :C → H is said to be relaxedη-αmonotone if there exist a mapping η : C×C → H and a functionα : H → Rpositively homogeneous of degree p, that is, αtz t^pαzfor allt >0 andz∈Hsuch that

Tx−Ty, η x, y

≥α x−y

, ∀x, y∈C, 1.9

wherep > 1 is a constant; see10. In the case of ηx, y x−yfor allx, y ∈ C,T is said to be relaxedα-monotone. In the case ofηx, y x−yfor allx, y ∈ Candαz kz^p, wherep >1 andk >0,T is said to bep-monotone; see11–13. In fact, in this case, ifp2, then T is ak-strongly monotone mapping. Moreover, every monotone mapping is relaxed η-αmonotone withηx, y x−yfor allx, y∈Candα≡0.

In this paper, we consider a new general equilibrium problem with a relaxed monotone mapping:

Find z∈Csuch thatΦ z, y

Tz, η

y, z

Az, y−z

≥0, ∀y∈C. 1.10 The set of all solutions of the equilibrium problem is denoted by GEPΦ, T, that is,

GEPΦ, T

z∈C:Φ z, y

Tz, η

y, z

Az, y−z

≥0, ∀y∈C

. 1.11

In the case ofA≡0,1.10is deduced to Findz∈Csuch thatΦ

z, y

Tz, η y, z

≥0, ∀y∈C. 1.12 The set of all solutions of1.12is denoted by EPΦ, T, that is,

EPΦ, T

z∈C:Φ z, y

Tz, η

y, z

≥0, ∀y∈C

. 1.13

In the case ofT ≡0, GEPΦ, Tcoincides with EP. In the case ofT ≡0 andA≡0, GEPΦ, T coincides with EPΦ.

In this paper, we introduce a new iterative scheme for finding a common element of the set of solutions of a general equilibrium problem with a relaxed monotone mapping and the set of common fixed points of a countable family of nonexpansive mappings and then obtain a strong convergence theorem. More precisely, we introduce the following iterative scheme:

x1∈Cchosen arbitrarily, Φ

un, y

Tun, η y, un

Axn, y−un

1 λn

y−un, un−xn

≥0, ∀y∈C,

ynαnxn ⁿ

i1

αi−1−αiβnSixn 1−αn 1−βn

un,

Cn

z∈C:yn−z≤ xn−z , Dnⁿ

j1

Cj,

xn1PDnx1, n≥1,

1.14

whereT :C → His a relaxedη-αmonotone mapping,A:C → His aλ-inverse-strongly monotone mapping, and{Sn}^∞_n1 :C → Cis a countable family of nonexpansive mappings

such thatF_∞

n1FixSn∩GEPΦ, T/∅,α01, and{αn}^∞_n1,{βn}^∞_n1, and{λn}^∞_n1are three control sequences. We prove that{xn} defined by1.14converges strongly tox^∗ PFx1. Using the main result in this paper, we also prove several new strong convergence theorems for finding the elements of FixS∩EP, FixS∩EPΦ, FixS∩EPΦ, T, and FixS∩VIC, A, respectively, whereS:C → Cis a nonexpansive mapping.

2. Preliminaries

LetA : C → H be aλ-inverse-strongly monotone mapping and letI denote the identity mapping ofH. For allx, y∈Candk >0, one has6

I−kAx−I−kAy²≤x−y²kk−2λAx−Ay². 2.1 Hence, ifk∈0,2λ, thenI−kAis a nonexpansive mapping ofCintoH.

For each pointx∈H, there exists a unique nearest point ofC, denoted byPCx, such thatx−PCx ≤ x−yfor ally∈C. Such aPCis called the metric projection fromHontoC.

The well-known Browder’s characterization ofPC ensures thatPC is a firmly nonexpansive mapping fromHontoC, that is,

PCx−PCy²≤

PCx−PCy, x−y

, x, y∈H. 2.2

Further, we know that for anyx∈Handz∈C,zPCxif and only if x−z, z−y

≥0, ∀y∈C. 2.3

LetSbe a nonexpansive mapping ofCinto itself such that FixS/∅. Then we have

x∈FixS⇐⇒ Sx−x²≤2x−Sx, x−x , ∀x∈C, 2.4

which is obtained directly from the following:

x−x ²≥ Sx−Sx ²Sx−x ²Sx−x x−x ²

Sx−x²x−x ²2Sx−x, x−x . 2.5 This inequality is a very useful characterization of FixS. Observe what is more that it immediately yields that FixSis a convex closed set.

LetΦbe a bifunction ofC×CintoRsatisfying the following conditions:

A1 Φx, x 0 for allx∈C;

A2 Φis monotone, that is,Φx, y Φy, x≤0 for allx, y∈C;

A3for eachx, y, z∈C, limt↓0Φtz 1−tx, y≤Φx, y;

A4for eachx∈C,y→Φx, yis convex and lower semicontinuous.

Definition 2.1 see10. Let E be a Banach space with the dual space E^∗ and let K be a nonempty subset of E. Let T : K → E^∗ and η : K ×K → E be two mappings. The mappingT :K → E^∗is said to beη-hemicontinuous if, for any fixedx, y ∈K, the function f:0,1 → −∞,∞defined byft T1−txty, ηx, y is continuous at 0.

Lemma 2.2. LetH be a Hilbert space and letCbe a nonempty closed convex subset ofH. LetT : C → Hbe an η-hemicontinuous and relaxedη-αmonotone mapping. LetΦbe a bifunction from C×CtoRsatisfying (A1) and (A4). Letr >0 andz∈C. Assume that

iηx, x 0 for allx∈C;

iifor any fixedu, v∈C, the mappingx→ Tv, ηx, u is convex.

Then the following problems2.6and2.7are equivalent:

Findx∈Csuch thatΦ x, y

Tx, η

y, x 1

r

y−x, x−z

≥0, ∀y∈C; 2.6 Findx∈Csuch thatΦ

x, y

Ty, η y, x

1 r

y−x, x−z

≥α y−x

, ∀y∈C. 2.7

Proof. Letx∈Cbe a solution of the problem2.6. SinceTis relaxedη-αmonotone, we have Φ

x, y

Ty, η y, x

1 r

y−x, x−z

≥Φ x, y

α y−x

1 r

y−x, x−z

Tx, η y, x

≥α y−x

, ∀y∈C.

2.8

Thusx∈Cis a solution of the problem2.7.

Conversely, letx∈Cbe a solution of the problem2.7. Letting

yt 1−txty, ∀t∈0,1, 2.9

thenyt∈C. Sincex∈Cis a solution of the problem2.7, it follows that Φ

x, yt

Tyt, η

yt, x 1

r

yt−x, x−z

≥α yt−x

t^pα y−x

. 2.10

The conditionsi,ii,A1, andA4imply that Tyt, η

yt, x

≤1−t

Tyt, ηx, x t

Tyt, η y, x t

T xt

y−x , η

y, x , Φ

x, yt

≤1−tΦx, x tΦ x, y

tΦ x, y

.

2.11

It follows from2.10-2.11that

Φ x, y

T

xt y−x

, η y, x

1 r

y−x, x−z

≥t^p−1α y−x

, ∀y∈C. 2.12

SinceT isη-hemicontinuous andp >1, lettingt → 0 in2.12, we get

Φ x, y

Tx, η

y, x 1

r

y−x, x−z

≥0 2.13

for all y ∈ C. Therefore, x ∈ Cis also a solution of the problem2.6. This completes the proof.

Definition 2.3 see14. Let E be a Banach space with the dual space E^∗ and let K be a nonempty subset of E. A mapping F : K → 2^E is called a KKM mapping if, for any {x1, . . . , xn} ⊂K, co{x1, . . . , xn} ⊂_n

i1Fxi, where 2^Edenotes the family of all the nonempty subsets ofE.

Lemma 2.4see14. LetMbe a nonempty subset of a Hausdorfftopological vector spaceXand letF:M → 2^Xbe a KKM mapping. IfFxis closed inXfor allx∈XinKand compact for some x∈K, then

x∈MFx/∅.

Next we use the concept of KKM mapping to prove two basic lemmas for our main result. The idea of the proof of the next lemma is contained in the paper of Fang and Huang 10.

Lemma 2.5. LetH be a real Hilbert space and Cbe a nonempty bounded closed convex subset of H. LetT : C → H be anη-hemicontinuous and relaxedη-αmonotone mapping, and letΦbe a bifunction fromC×CtoRsatisfying (A1) and (A4). Letr >0. Assume that

iηx, y ηy, x 0 for allx, y∈C;

iifor any fixed u, v ∈ C, the mapping x → Tv, ηx, u is convex and lower semicontinuous;

iiiα:H → Ris weakly lower semicontinuous; that is, for any net{xβ},xβconverges toxin σH, Hwhich implies thatαx≤lim infαxβ.

Then problem2.6is solvable.

Proof. Letz∈C. Define two set-valued mappingsFz, Gz:C → 2^Has follows:

Fz

y

x∈C:Φ x, y

Tx, η

y, x 1

r

y−x, x−z

≥0

, ∀y∈C,

Gz

y

x∈C:Φ x, y

Ty, η

y, x 1

r

y−x, x−z

≥α

y−x

, ∀y∈C.

2.14

We claim that Fz is a KKM mapping. If Fz is not a KKM mapping, then there exist {y1, . . . , yn} ⊂Candti>0,i1, . . . , n, such that

n i1

ti1, y ⁿ

i1

tiyi/∈ⁿ

i1

Fz

yi

. 2.15

By the definition ofF, we have Φ

y, yi

Ty, η

yi, y 1

r

yi−y, y−z

<0, ∀i1, . . . , n. 2.16

It follows fromA1,A4, andiithat 0 Φ

y, y Φ

y,

n i1

tiyi

Ty, η

_n

i1

tiyi, y

≤ ⁿ

i1

tiΦ y, yi

ⁿ

i1

ti

Ty, η yi, y

<

n i1

ti1 r

y−yi, y−z

0,

2.17

which is a contradiction. This implies thatFzis a KKM mapping.

Now, we prove that

Fz

y

⊂Gz

y

, ∀y∈C. 2.18

For any giveny∈C, takingx∈Fzy, then Φ

x, y

Tx, η y, x

1 r

y−x, x−z

≥0. 2.19

SinceT is relaxedη-αmonotone, we have Φ

x, y

Ty, η y, x

1 r

y−x, x−z

≥Φ x, y

Tx, η

y, x α

y−x 1

r

y−x, x−z

≥α y−x

.

2.20

It follows thatx∈Gzyand so Fz

y

⊂Gz

y

, ∀y∈C. 2.21

This implies thatGzis also a KKM mapping. Now, sincex→ Ty, ηx, y is a convex lower- semicontinuous function, we know that it is weakly lower semicontinuous. Thus from the definition ofGzand the weak lower semicontinuity ofα, it follows thatGzyis weakly closed for ally∈C. SinceCis bounded closed and convex, we know thatCis weakly compact, and soGzyis weakly compact inCfor eachy∈C. It follows from Lemmas2.2and2.4that

y∈C

Fz

y

y∈C

Gz

y

/∅. 2.22

Hence there existsx∈Csuch that Φ

x, y

Tx, η y, x

1 r

y−x, x−z

≥0, ∀y∈C. 2.23

This completes the proof.

Lemma 2.6. LetHbe a real Hilbert space and letCbe a nonempty bounded closed convex subset ofH.

LetT :C → Hbe anη-hemicontinuous and relaxedη-αmonotone mapping and letΦbe a bifunction fromC×CtoRsatisfying (A1), (A2), and (A4). Letr >0 and define a mappingTr : H → Cas follows:

Trx

z∈C:Φ z, y

Tz, η

y, z 1

r

y−z, z−x

≥0, ∀y∈C

2.24

for allx∈H. Assume that

iηx, y ηy, x 0, for allx, y∈C;

iifor any fixedu, v∈C, the mappingx→ Tv, ηx, u is convex and lower semicontinuous and the mappingx→ Tu, ηv, x is lower semicontinuous;

iiiα:H → Ris weakly lower semicontinuous;

ivfor anyx, y∈C,αx−y αy−x≥0.

Then, the following holds:

1Tr is single-valued;

2Tr is a firmly nonexpansive mapping, that is, for allx, y∈H, Trx−Try²≤

Trx−Try, x−y

; 2.25

3FTr EPΦ, T;

4EPΦ, Tis closed and convex.

Proof. The fact thatTr is nonempty is exactly the thesis of the previous lemma. We claim that Tr is single-valued. Indeed, forx∈Handr >0, letz1, z2∈Trx. Then,

Φz1, z2

Tz1, ηz2, z1 1

rz2−z1, z1−x ≥0, Φz2, z1

Tz2, ηz1, z2 1

rz1−z2, z2−x ≥0.

2.26

Adding the two inequalities, fromiwe have

Φz1, z2 Φz2, z1

Tz1−Tz2, ηz2, z1 1

rz1−z2, z2−z1 ≥0. 2.27 FromA2, we have

Tz1−Tz2, ηz2, z1 1

rz1−z2, z2−z1 ≥0, 2.28 that is,

1

rz1−z2, z2−z1 ≥

Tz2−Tz1, ηz2, z1

. 2.29

SinceT is relaxedη-αmonotone andr >0, one has

z1−z2, z2−z1 ≥rαz2−z1. 2.30

In2.29exchanging the position ofz1andz2, we get 1

rz2−z1, z1−z2 ≥

Tz1−Tz2, ηz1, z2

≥αz1−z2, 2.31

that is,

z1−z2, z2−z1 ≥rαz1−z2. 2.32

Now, adding the inequalities2.30and2.32, by usingivwe have

−2z1−z2²2z1−z2, z2−z1 ≥0. 2.33

Hence,z1z2.

Next we show thatTr is firmly nonexpansive. Indeed, forx, y∈H, we have

Φ

Trx, Try

TTrx, η

Try, Trx 1

r

Try−Trx, Trx−x

≥0, Φ

Try, Trx

TTry, η

Trx, Try 1

r

Trx−Try, Try−y

≥0.

2.34

Adding the two inequalities and byiandA2, we get TTrx−TTry, η

Try, Trx 1

r

Try−Trx, Trx−Try−xy

≥0, 2.35

that is,

1 r

Try−Trx, Trx−Try−xy

≥

TTry−TTrx, η

Try, Trx

≥α

Try−Trx .

2.36

In2.36exchanging the position ofTrxandTry, we get 1

r

Trx−Try, Try−Trx−yx

≥α

Trx−Try

. 2.37

Adding the inequalities2.36and2.37, we have

2

Trx−Try, Try−Trx−yx

≥r α

Trx−Try α

Try−Trx

. 2.38

It follows fromivthat

Trx−Try, Try−Trx−yx

≥0, 2.39

that is,

Trx−Try²≤

Trx−Try, x−y

. 2.40

This shows thatTris firmly nonexpansive.

Next, we claim thatFTr EPΦ, T. Indeed, we have the following:

u∈FTr⇐⇒uTru

⇐⇒Φ u, y

Tu, η

y, u 1

r

y−u, u−u

≥0, ∀y∈C

⇐⇒Φ u, y

Tu, η

y, u

≥0, ∀y∈C

⇐⇒u∈EPΦ, T.

2.41

Finally, we prove that EPΦ, T is closed and convex. Indeed, Since every firm nonexpansive mapping is nonexpansive, we see thatTr is nonexpansive from2. On the other hand, since the set of fixed points of every nonexpansive mapping is closed and convex, we have that EPΦ, T is closed and convex from 2 and 3. This completes the proof.

3. Main Results

In this section, we prove a strong convergence theorem which is our main result.

Theorem 3.1. Let C be a nonempty bounded closed convex subset of a real Hilbert spaceH and letΦ : C×C → Rbe a bifunction satisfying (A1), (A2), (A3), and (A4). LetT : C → H be anη-hemicontinuous and relaxedη-αmonotone mapping, letA :C → Hbe aλ-inverse-strongly monotone mapping, and let{Sn}^∞_n1 : C → C be a countable family of nonexpansive mappings such thatF _∞

n1FixSn∩GEPΦ, T/∅. Assume that the conditions (i)–(iv) ofLemma 2.6are satisfied. Letα0 1 and assume that{αn}^∞_n1 ⊂0,1is a strictly decreasing sequence. Assume that {βn}^∞_n1⊂c, dwith somec, d∈0,1and{λn}^∞_n1⊂a, bwith somea, b∈0,2λ. Then, for any x1 ∈ C, the sequence{xn}generated by1.14converges strongly tox^∗ PFx1. In particular, ifC contains the origin 0, takingx10, then the sequence{xn}generated by1.14converges strongly to the minimum norm element inF.

Proof. We split the proof into following steps.

Step 1. F is closed and convex, the sequence{xn}generated by1.14is well defined, and F⊂Dnfor alln≥1.

First, we prove thatFis closed and convex. It suffices to prove that GEPΦ, Tis closed and convex. Indeed, it is easy to prove the conclusion by the following fact:

∀p∈GEPΦ, T⇐⇒Φ p, y

Tp, η

y, p 1

λn

y−p, λnAp

≥0, ∀y∈C

⇐⇒Φ p, y

Tp, η

y, p 1

λn

y−p, p−

p−λnAp

≥0, ∀y∈C

⇐⇒pTλnI−λnAp.

3.1

This implies that GEPΦ, T FixTλnI−λnA. Noting thatTλnI−λnAis a nonexpansive mapping forλn < 2λand the set of fixed points of a nonexpansive mapping is closed and convex, we have that GEPΦ, Tis closed and convex.

Next we prove that the sequence{xn}generated by1.14is well defined andF ⊂Dn

for alln≥1. It is easy to see thatCnis closed and convex for alln∈Nfrom the construction ofCn. Hence,Dnis closed and convex for alln∈N. For anyp∈F, sinceunTλnxn−λnAxn andI−λnAis nonexpansive, we havenote that{αn}is strictly decreasing

yn−p αn

xn−p ⁿ

i1

α_i−1−αiβn

Sixn−p

1−αn 1−βn

un−p

≤αnxn−p ⁿ

i1

αi−1−αiβnSixn−p 1−αn

1−βnun−p

≤αnxn−p ⁿ

i1

αi−1−αiβnxn−p 1−αn

1−βnTλnxn−λnAxn−Tλn

p−λnAp

≤αnxn−p 1−αnβnxn−p 1−αn

1−βnxn−λnAxn−

p−λnAp

≤αnxn−p 1−αnβnxn−p 1−αn

1−βnxn−p xn−p.

3.2

So,F ⊂Cnfor alln∈N. HenceF ⊂ _n

j1Cj, that is,F ⊂Dnfor alln∈N. SinceDnis closed, convex, and nonempty, the sequence{xn}is well defined.

Step 2. lim_{n→ ∞}x_n1−xn0 and there existsx^∗∈Csuch thatxn → x^∗asn → ∞.

From the definition ofDn, we see thatDn1⊂Dnfor alln∈Nand hence

x_n2PDn1x1∈D_n1⊂Dn. 3.3

Noting thatx_n1 PDnx1, we get

xn1−x1 ≤ xn2−x1 3.4

for alln ≥ 1. This shows that{xn−x1} is increasing. SinceCis bounded,{xn −x1}is bounded. So, we have that lim_{n→ ∞}xn−x1exists.

Noting thatxn1PDnx1andxm1PDmx1∈Dm⊂Dnfor allm≥n, we have

x_n1−x1, x_m1−x_n1 ≥0. 3.5

It follows from3.5that xm1−xn1²

xm1−x1−xn1−x1²

xm1−x1²xn1−x1²−2xm1−x1, xn1−x1

xm1−x1²xn1−x1²−2xn1−x1, xm1−xn1xn1−x1

xm1−x1²− xn1−x1²−2xn1−x1, xm1−xn1

≤ xm1−x1²− xn1−x1².

3.6

By takingmn1 in3.6, we get

xn2−xn1 ≤ xn2−x1²− xn1−x1². 3.7

Since the limits ofxn−x1exists we get

nlim→ ∞xn2−xn10, 3.8

that is,x_n1−xn → 0 asn → ∞. Moreover, from3.6we also have

m,nlim→ ∞xm1−xn10. 3.9

This shows that{xn}is a Cauchy sequence. Hence, there existsx^∗∈Csuch that

xn−→x^∗∈C, asn−→ ∞. 3.10 Step 3. lim_{n→ ∞}xn−un0.

Sincexn1 ∈Cnandxn−xn1 → 0 asn → ∞, we have

yn−x_n1≤ xn−x_n1 −→0 as n−→ ∞, 3.11 and hence

yn−xn≤yn−xn1xn−xn1 −→0 asn−→ ∞. 3.12

Note that un can be rewritten as un Tλnxn −λnAxn for all n ≥ 1. Take p ∈ F. Since pTλnp−λnAp,Aisλ-inverse-strongly monotone, and 0< λn <2λ, we know that, for all n∈N,

un−p²Tλnxn−λnAxn−Tλnp−λnAp²

≤xn−λnAxn−p−λnAp² xn−p−λnAxn−Ap² xn−p²−2λn

xn−p, Axn−Ap

λ²_nAxn−Ap²

≤xn−p²−2λnλAxn−Ap²λ²_nAxn−Ap² xn−p²λnλn−2λAxn−Ap²

≤xn−p².

3.13

Using1.14and3.13, we havenote that{αn}is strictly decreasing

yn−p²

αnxn−p ⁿ

i1

αi−1−αiβnSixn−p 1−αn1−βnun−p

2

≤αnxn−p^{2 n}

i1

α_i−1−αiβnSixn−p² 1−αn

1−βnun−p²

≤αnxn−p^{2 n}

i1

α_i−1−αiβnxn−p² 1−αn

1−βnxn−p²λnλn−2λAxn−Ap² xn−p² 1−αn

1−βn

λnλn−2λAxn−Ap²,

3.14

and hence

1−αn1−da2λ−bAxn−Ap²≤1−αn 1−βn

λn2λ−λnAxn−Ap²

≤xn−p²−yn−p²

≤xn−ynxn−pyn−p.

3.15

Since{xn}and{yn}are both bounded,αn → 0, andxn−yn → 0, we have

Axn−Ap−→0 asn−→ ∞. 3.16

UsingLemma 2.6, we get

un−p²Tλnxn−λnAxn−Tλnp−λnAp²

≤

xn−λnAxn−

p−λnAp

, un−p 1

2

xn−λnAxn−p−λnAp²un−p²

−xn−λnAxn−p−λnAp−un−p²

≤ 1 2

xn−p²un−p²−xn−un−λnAxn−Ap² 1

2

xn−p²un−p²− xn−un² 2λn

xn−un, Axn−Ap

−λ²_nAxn−Ap² .

3.17

So, we have

un−p²≤xn−p²− xn−un²2λn

xn−un, Axn−Ap

−λ²_nAxn−Ap². 3.18

From3.18, we have yn−p²

αnxn−p ⁿ

i1

α_i−1−αiβnSixn−p 1−αn1−βnun−p

2

≤αnxn−p^{2 n}

i1

αi−1−αiβnxn−p² 1−αn

1−βnun−p²

≤αnxn−p² 1−αnβnxn−p² 1−αn 1−βn

×xn−p²− xn−un²2λn

xn−un, Axn−Ap

−λ²_nAxn−Ap²

≤xn−p²−1−αn 1−βn

xn−un²21−αn 1−βn

λn

xn−un, Axn−Ap , 3.19

and hence

1−d1−αnxn−un²≤ 1−βn

1−αnxn−un²

≤xn−ynxn−pyn−p 21−αn

1−βn

λnxn−unAxn−Ap.

3.20

By usingxn−yn → 0 and3.16, we have

xn−un −→0 asn−→ ∞. 3.21 Step 4. lim_{n→ ∞}xn−Sixn0, for alli0,1, . . . .

It follows from the definition of scheme1.14that yn ⁿ

i1

αi−1−αiβnxn−Sixn−1−αnβnxnαnxn 1−αn 1−βn

un, 3.22

that is,

n i1

αi−1−αiβnxn−Sixn xn−yn−xnαnxn 1−αnβnxn 1−αn 1−βn

un

xn−yn 1−αn βn−1

xn 1−αn 1−βn

un

xn−yn 1−αn 1−βn

un−xn.

3.23

Hence, for anyp∈F, one has

n i1

α_i−1−αiβn

xn−Sixn, xn−p

1−αn 1−βn

un−xn, xn−p

xn−yn, xn−p . 3.24

Since eachSiis nonexpansive, by2.4we have Sixn−xn²≤2

xn−Sixn, xn−p

. 3.25

Hence, combining this inequality with3.24, we get 1

2

n i1

αi−1−αiβnSixn−xn²≤1−αn 1−βn

un−xn, xn−p

xn−yn, xn−p

, 3.26

that isnoting that{αn}is strictly decreasing, Sixn−xn²≤ 21−αn

1−βn

αi−1−αiβn

un−xn, xn−p

2 αi−1−αiβn

xn−yn, xn−p

≤ 21−αn 1−βn

α_i−1−αiβn un−xnxn−p 2 α_i−1−αiβn

xn−ynxn−p. 3.27

Sinceun−xn → 0 andxn−yn → 0, we have

nlim→ ∞Sixn−xn0, ∀i1,2, . . . . 3.28

Step 5. xn → x^∗PFx1. First we provex^∗ ∈_∞

i1FixSi. Indeed, sincexn → x^∗andSixn−xn → 0, we have x^∗∈FixSifor eachi1,2, . . .. Hence,x^∗∈_∞

i1FixSi.

Next, we show thatx^∗∈GEPΦ, T. Noting thatunTλnxn−λnAxn, one obtains Φ

un, y

Tun, η y, un

Axn, y−un

1 λn

y−un, un−xn

≥0, ∀y∈C. 3.29

Pututty 1−tx^∗for allt∈0,1andy∈C. Then, we haveut∈C. So, fromA2,i, and 3.29we have

ut−un, Aut ≥ ut−un, Aut − ut−un, Axn −

ut−un,un−xn

λn

Φut, un

Tun, ηun, ut

ut−un, Aut−Aun ut−un, Aun−Axn −

ut−un,un−xn

λn

Φut, un

Tun, ηun, ut .

3.30

Sincexn−un → 0, we haveAun−Axn → 0. Further, from monotonicity ofA, we have ut−un, Aut−Aun ≥0. So, fromA4,ii, andη-hemicontinuity ofT we have

ut−x^∗, Aut ≥Φut, x^∗

Tx^∗, ηx^∗, ut

. 3.31

FromA1,A4,ii, and3.31we also have 0 Φut, ut

Tx^∗, ηut, ut

≤t Φ

ut, y

Tx^∗, η y, ut

1−t

Φut, x^∗

Tx^∗, ηx^∗, ut

≤t Φ

ut, y

Tx^∗, η y, ut

1−tut−x^∗, Aut

t Φ

ut, y

Tx^∗, η y, ut

1−tt

y−x^∗, Aut

,

3.32

and hence

0≤Φ ut, y

Tx^∗, η y, ut

1−t

y−x^∗, Aut

. 3.33

Lettingt → 0, fromA3andiiwe have, for eachy∈C, 0≤Φ

x^∗, y

Tx^∗, η y, x^∗

y−x^∗, Ax^∗

. 3.34

This implies thatx^∗∈GEPΦ, T. Hence, we getx^∗∈F_∞

n1FixSn∩GEPΦ, T.

Finally, we show thatx^∗PFx. Indeed, fromxn1PDnxandF⊂Dn, we have x−xn1, xn1−v ≥0, ∀v∈F. 3.35

Taking the limit in3.35and noting thatxn → x^∗asn → ∞, we get

x−x^∗, x^∗−v ≥0, ∀v∈F. 3.36

In view of2.3, one sees thatx^∗PFx. This completes the proof.

Corollary 3.2. Let C be a nonempty bounded closed convex subset of a Hilbert space H and let Φ : C×C → Rbe a bifunction satisfying (A1), (A2), (A3), and (A4). LetT : C → H be anη- hemicontinuous and relaxedη-αmonotone mapping and letS:C → Cbe a nonexpansive mapping such that FixS∩ EPΦ, T/∅. Assume that the conditions (i)–(iv) of Lemma 2.6 are satisfied.

Assume that{αn}^∞_n1 ⊂ 0,1with lim sup_{n→ ∞}αn < 1,{βn}^∞_n1 ⊂ c, dwith somec, d ∈ 0,1 and{λn}^∞_n1⊂a,∞witha∈0,∞. Letx1∈Cand let{xn}be generated by

Φ un, y

Tun, η

y, un

1 λn

y−un, un−xn

≥0, ∀y∈C,

ynαnxn 1−αnβnSxn 1−αn 1−βn

un, Cn

z∈C:yn−z≤ xn−z , Dnⁿ

j1

Cj,

xn1PDnx1, n≥1.

3.37

Then the sequence {xn} converges strongly to x^∗ PFixS∩EPΦ,Tx1. In particular, if C contains the origin 0, takingx1 0, the sequence{xn}converges strongly to the minimum norm element in FixS∩EPΦ, T.

Proof. InTheorem 3.1, putA≡0,S1· · ·Sn· · ·S. Then, we have

ynαnxn ⁿ

i1

α_i−1−αiβnSixn 1−αn 1−βn

un

αnxn 1−αnβnSxn 1−αn 1−βn

un,

Sxn−xn² ≤ 21−αn 1−βn

1−αnβn un−xnxn−p 2 1−αnβn

xn−ynxn−p. 3.38

On the other hand, for allλ∈0,∞, we have that x−y, Ax−Ay

≥λAx−Ay², ∀x, y∈C. 3.39