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Contents lists available atSciVerse ScienceDirect

Computers and Mathematics with Applications

journal homepage:www.elsevier.com/locate/camwa

Hybrid algorithm for generalized mixed equilibrium problems and

variational inequality problems and fixed point problems

Gang Cai

, Shangquan Bu

Department of Mathematical Sciences, Tsinghua University, 100084 Beijing, China

a r t i c l e i n f o

Article history:

Received 18 August 2011

Received in revised form 25 October 2011 Accepted 25 October 2011

Keywords:

Strong convergence

General mixed equilibrium problem Variational inequality

Fixed point

a b s t r a c t

In this paper, we introduce a new iterative algorithm by hybrid method for finding a common element of the set of solutions of finite general mixed equilibrium problems and the set of solutions of a general variational inequality problem for finite inverse strongly monotone mappings and the set of common fixed points of infinite family of strictly pseudocontractive mappings in a real Hilbert space. Then we prove strong convergence of the scheme to a common element of the three above described sets. Our result improves and extends the corresponding results announced by many others.

©2011 Elsevier Ltd. All rights reserved.

1. Introduction

In this paper, we always assume thatHis a real Hilbert space with inner product

., .

and induced norm

.

andCis a nonempty closed convex subset ofH.PCdenotes the metric projection ofHontoCandF

(

T

)

denotes the fixed points set of

a mappingT.

A mappingT

:

C

Cis said to be nonexpansive if

Tx

Ty

∥ ≤ ∥

x

y

,

x

,

y

C

.

A mappingT

:

C

His said to beL-Lipschitzian if there existsL

>

0 such that

Tx

Ty

∥ ≤

L

x

y

,

x

,

y

C

.

A mappingT

:

C

Cis said to bek-strictly pseudocontractive if there exists a constantk

∈ [

0

,

1

)

such that

Tx

Ty

2

≤ ∥

x

y

2

+

k

(

I

T

)

x

(

I

T

)

y

2

,

x

,

y

C

.

Ifk

=

0, then the mappingT is nonexpansive. Iterative approximation of fixed point ofk-strictly pseudocontractive mappings has been studied extensively by many authors (see, e.g., [1–4] and the references contained therein).

A mappingA

:

C

His said to be monotone if

Ax

Ay

,

x

y

⟩ ≥

0

,

x

,

y

C

.

A mappingA

:

C

His said to be

α

-inverse-strongly monotone if there exists a positive real number

α

such that

Ax

Ay

,

x

y

⟩ ≥

α

Ax

Ay

2

,

x

,

y

C

.

This work was supported by the NSF of China (No. 10731020, No. 11171172) and the Specialized Research Fund for the Doctoral Program of Higher

Education (No. 200800030059).

Corresponding author.

E-mail addresses:[email protected](G. Cai),[email protected](S. Bu). 0898-1221/$ – see front matter©2011 Elsevier Ltd. All rights reserved.

(2)

LetA

:

C

Hbe a nonlinear mapping. The variational inequality problem is to find ax

Csuch that

Ax

,

y

x

0

,

y

C

.

(1.1)

(see, e.g., [5,6]). The set of solution of the variational inequality problem(1.1)is denoted byVI

(

C

,

A

)

.

Let

ϕ

:

C

Ris a real-valued function andA

:

C

Ha nonlinear mapping. SupposeF

:

C

×

C

Rbe a bifunction. The generalized mixed equilibrium problem is to findx

C(see, e.g., [7,8]) such that

F

(

x

,

y

)

+

ϕ(

y

)

ϕ(

x

)

+ ⟨

Ax

,

y

x

⟩ ≥

0

,

y

C

.

(1.2) We denote the set of(1.2)byGMEP

(

F

, ϕ,

A

)

.

If

ϕ

=

0, then problem(1.2)reduces to generalized equilibrium problem studied by many authors (see, e.g., [9–11]), which is to findx

Csuch that

F

(

x

,

y

)

+ ⟨

Ax

,

y

x

⟩ ≥

0

,

y

C

.

(1.3) The set of solutions of(1.3)is denoted byEP

(

F

,

A

)

.

IfA

=

0, then problem(1.2)reduces to mixed equilibrium problem considered by many authors (see, e.g., [12–14]), which is to findx

Csuch that

F

(

x

,

y

)

+

ϕ(

y

)

ϕ(

x

)

0

,

y

C

.

(1.4) The set of solutions of(1.4)is denoted byMEP

(

F

, ϕ)

.

If

ϕ

=

0

,

A

=

0, then problem(1.2)reduces to equilibrium problem studied by many authors (see, e.g., [15–17]), which is to findx

Csuch that

F

(

x

,

y

)

0

,

y

C

.

(1.5)

The set of solutions of(1.5)is denoted byEP

(

F

)

.

The generalized mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nash equilibrium problems, and equilibrium problems as special cases. Numerous problems in Physics, optimization, and economics reduce to find a solution of problem(1.2). Several methods have been proposed to solve the fixed point problems, variational inequality problems and equilibrium problems in the literature (see, e.g., [18–30]).

LetA

,

B

:

C

Hbe two mappings. Ceng et al. [27] considered the following problem of finding

(

x

,

y

)

C

×

Csuch

that



λ

Ay

+

x

y

,

x

x

0

,

x

C

,

µ

Bx

+

y

x

,

x

y∗

0

,

x

C

,

(1.6)

which is called a general system of variational inequalities where

λ >

0 and

µ >

0 are two constants. In particular, ifA

=

B, then problem(1.6)reduces to finding

(

x

,

y

)

C

×

Csuch that



λ

Ay

+

x

y

,

x

x

0

,

x

C

,

µ

Ax

+

y

x

,

x

y

0

,

x

C

.

(1.7)

Further, if we add up the requirement thatx

=

y, the problem(1.6)reduces to the classical variational inequality problem

(1.1).

In order to find the common element of the solutions of problem(1.6)and the set of fixed points of one nonexpansive mappingT, Ceng et al. [27] studied the following algorithm:x1

=

u

Cand

yn

=

PC

(

xn

µ

Bxn

),

xn+1

=

α

nu

+

β

nxn

+

γ

nSPC

(

yn

λ

Ayn

).

(1.8)

Under appropriate conditions they obtained one strong convergence theorem.

Very recently, Cho et al. [2] introduced a hybrid projection method for finding a common element of the set of solutions of a generalized equilibrium problem, the set of solutions of a variational inequality and the set of fixed points of a strict pseudocontraction in a real Hilbert space.

In this paper, motivated and inspired by the above facts, we introduce the following system of variational inequalities in a Hilbert spaceH: LetCbe a nonempty closed convex subset of a real Hilbert spaceH. Let

{

Ai

}

Ni=1

:

C

Hbe a family of

mappings. First we consider the following problem of finding

(

x∗ 1

,

x ∗ 2

, . . . ,

xN

)

C

×

C

· · · ×

Csuch that

λ

NANxN

+

x ∗ 1

xN

,

x

x ∗ 1

0

,

x

C

,

λ

N−1AN−1xN−1

+

xN

xN−1

,

x

xN

0

,

x

C

,

...

λ

2A2x∗2

+

x ∗ 3

x ∗ 2

,

x

x ∗ 3

0

,

x

C

,

λ

1A1x∗1

+

x ∗ 2

x ∗ 1

,

x

x ∗ 2

0

,

x

C

,

(1.9)
(3)

which is called a more general system of variational inequalities in Hilbert spaces, where

λ

i

>

0 for alli

∈ {

1

,

2

, . . . ,

N

}

.

The set of solutions to(1.9)is denoted byΩ. In particular, ifN

=

2

,

A1

=

B

,

A2

=

A

, λ

1

=

µ, λ

2

=

λ,

x∗1

=

x

,

x

2

=

y, then

problem(1.9)reduces to problem(1.6). Then we study a new iterative algorithm by hybrid method for finding a common element of the set of solutions of finite general mixed equilibrium problems and the set of solutions of a general variational inequality problem for finite inverse strongly monotone mappings and the set of common fixed points of infinite family of strictly pseudocontractive mappings in a real Hilbert space. Furthermore we prove a strong convergence of the scheme to a common element of the three above described sets. Our result extends the corresponding results announced by many others.

2. Preliminaries

LetHbe a real Hilbert space. It is well known that

x

y

2

= ∥

x

2

− ∥

y

2

2

x

y

,

y

(2.1) and

λ

x

+

(

1

λ)

y

2

=

λ

x

2

+

(

1

λ)

y

2

λ(

1

λ)

x

y

2 (2.2) for allx

,

y

Hand

λ

∈ [

0

,

1

]

.

LetCbe a nonempty closed convex subset ofH. For every pointx

H, there exists a unique nearest point inC, denoted byPCx, such that

x

PCx

∥ ≤ ∥

x

y

for ally

C

.

(2.3)

Pis called a metric projection ofHontoC. It is well known thatPCis a nonexpansive mapping ofHontoCand satisfies

x

y

,

PCx

PCy

⟩ ≥ ∥

PCx

PCy

2 (2.4)

for everyx

,

y

H. Moreover,PCxis characterized by the following properties:PCx

Cand

x

PCx

,

y

PCx

⟩ ≤

0 (2.5)

x

y

2

≥ ∥

x

P

Cx

2

+ ∥

y

PCx

2 (2.6)

for allx

H

,

y

C.

In the context of the variational inequality problem, this implies that

u

VI

(

A

,

C

)

u

=

PC

(

u

λ

Au

),

for all

λ >

0

.

(2.7)

IfAis

α

-inverse-strongly monotone mapping ofCintoH, then it is obvious thatAis 1α-Lipschitz continuous. We also have that, for allu

, v

Cand

λ >

0,

(

I

λ

A

)

u

(

I

λ

A

)v

2

= ∥

(

u

v)

λ(

Au

A

v)

2

= ∥

u

v

2

2

λ

Ax

Ay

,

x

y

⟩ +

λ

2

Au

A

v

2

≤ ∥

u

v

2

+

λ(λ

2

α)

Au

A

v

2

.

(2.8) So, if

λ

2

α

, thenI

λ

Ais a nonexpansive mapping fromCtoH.

For solving the equilibrium problem, let us assume that the bifunctionFsatisfies the following conditions: (A1) F

(

x

,

x

)

=

0 for allx

C;

(A2) Fis monotone, i.e.,F

(

x

,

y

)

+

F

(

y

,

x

)

0 for anyx

,

y

C; (A3) Fis upper-semicontinuous, i.e., for eachx

,

y

,

z

C,

lim sup

t→0+

F

(

tz

+

(

1

t

)

x

,

y

)

F

(

x

,

y

)

;

(A4) F

(

x

, .)

is convex and weakly lower semicontinuous for eachx

C;

(B1) for eachx

Handr

>

0, there exists a bounded subsetDx

Candyx

Csuch that for anyz

C

\

Dx, F

(

z

,

yx

)

+

ϕ(

yx

)

ϕ(

z

)

+

1

r

yx

z

,

z

x

<

0

;

(B2) Cis a bounded set.

We recall some lemmas which will be needed in the rest of this paper.

Lemma 2.1 ([21]).Assume that F

:

C

×

C

Rsatisfies

(

A1

)

(

A4

)

and let

ϕ

:

C

Rbe a proper lower semicontinuous and convex function. Assume that either

(

B1

)

or

(

B2

)

holds. For r

>

0and x

H, define a mapping Tr(F,ϕ)

:

H

C as follows:

Tr(F,ϕ)

(

x

)

=

z

C

:

F

(

z

,

y

)

+

ϕ(

y

)

ϕ(

z

)

+

1

r

y

z

,

z

x

⟩ ≥

0

,

y

C

for all z

H. Then the following hold:

(1) for each x

H

,

Tr(F,ϕ)

(

x

)

̸= ∅

;
(4)

(3) Tr(F,ϕ)is firmly nonexpansive, that is, for any x

,

y

H,

T

r(F,ϕ)x

Tr(F,ϕ)y

2

Tr(F,ϕ)x

Tr(F,ϕ)y

,

x

y

;

(4) F

(

Tr(F,ϕ)

)

=

MEP

(

F

, ϕ)

;

(5) MEP

(

F

, ϕ)

is closed and convex.

Lemma 2.2 ([31]).Let C be a nonempty closed convex subset of a real Hilbert space H. Let T

:

C

C be a k-strict pseudo-contractive mapping such that F

(

T

)

̸= ∅

.

(1) (Demi-closed principle) T is demi-closed on C , that is, if xn

x

C and xn

Txn

0, then x

=

Tx.

(2) T satisfies the Lipschitz condition

Tx

Ty

∥ ≤

L

x

y

∥ =

1

+

k

1

k

x

y

,

x

,

y

C

.

(3) The fixed point set F

(

T

)

of T is closed and convex so that the projection PF(T)is well defined.

Lemma 2.3 ([32]).Let C be a nonempty closed convex subset of a Banach space and let

{

Tn

}

be a sequence of mappings of C into itself. Suppose

∞

n=1supxC

Tn+1x

Tnx

<

. Then for each y

C

,

{

Tny

}

converges strongly to some point of C . Moreover, let T be a mapping of C into itself defined by Ty

=

limn→∞Tny

,

y

C . Thenlimn→∞supxC

Tnx

Tx

∥ =

0.

Lemma 2.4 ([33]).Let H be a real Hilbert space, let C be a closed convex subset of H, and let T

:

C

C be a nonexpansive mapping with F

(

T

)

̸= ∅

. If

{

xn

}

is a sequence in C weakly converging to x and if

(

I

T

)

xnconverges strongly to y, then

(

I

T

)

x

=

y. Lemma 2.5. Let C be a nonempty closed convex subset of a real Hilbert space H. Let Ai

:

C

H be an

α

i-inverse-strongly accretive mapping, where i

∈ {

1

,

2

, . . . ,

N

}

. Let G

:

C

C be a mapping defined by

G

(

x

)

=

PC

(

I

λ

NAN

)

PC

(

I

λ

N−1AN−1

)

· · ·

PC

(

I

λ

2A2

)

PC

(

I

λ

1A1

)

x

,

x

C

.

If 0

< λ

i

2

α

i

,

i

=

1

,

2

, . . . ,

N, then G

:

C

C is nonexpansive.

Proof. PutΩi

=

P

C

(

I

λ

iAi

)

PC

(

I

λ

i−1Ai−1

)

· · ·

PC

(

I

λ

2A2

)

PC

(

I

λ

1A1

),

i

=

1

,

2

, . . . ,

NandΩ0

=

I, whereIis identity

mapping. ThenG

=

N. For allx

,

y

C, it follows from(2.8)that

Gx

Gy

∥ =

Nx

Ny

=

P

C

(

I

λ

NAN

)

N−1x

PC

(

I

λ

NAN

)

N−1y

(

I

λ

NAN

)

N−1x

(

I

λ

NAN

)

N−1y

N−1x

N−1y

...

Ω0x

Ω0y

,

= ∥

x

y

which impliesGis nonexpansive. This completes the proof.

Lemma 2.6. Let C be a nonempty closed convex subset of a real Hilbert space H. Let Ai

:

C

H be nonlinear mapping, where i

=

1

,

2

, . . . ,

N. For given xi

C

,

i

=

1

,

2

, . . . ,

N

, (

x ∗ 1

,

x ∗ 2

, . . . ,

x

N

)

is a solution of problem(1.9)if and only if xi

=

PC

(

I

λ

i−1Ai−1

)

xi−1

,

x

1

=

QC

(

I

λ

NAN

)

xN

,

i

=

2

, . . . ,

N

.

That is

x1

=

PC

(

I

λ

NAN

)

QC

(

I

λ

N−1AN−1

)

· · ·

PC

(

I

λ

2A2

)

PC

(

I

λ

1A1

)

x∗1

.

Proof. We can rewrite(1.9)as

x1

(

xN

λ

NANxN

),

x

x ∗ 1

0

,

x

C

,

xN

(

xN−1

λ

N−1AN−1xN−1

),

x

xN

0

,

x

C

,

...

x3

(

x2

λ

2A2x∗2

),

x

x ∗ 3

0

,

x

C

.

x2

(

x1

λ

1A1x∗1

),

x

x ∗ 2

0

,

x

C

.

(2.9)
(5)

From(2.5), we deduce that(2.9)is equivalent to xi

=

PC

(

I

λ

i−1Ai−1

)

xi−1

,

x ∗ 1

=

PC

(

I

λ

NAN

)

xN

,

i

=

2

, . . . ,

N

.

Therefore we have x1

=

PC

(

I

λ

NAN

)

PC

(

I

λ

N−1AN−1

)

· · ·

PC

(

I

λ

2A2

)

PC

(

I

λ

1A1

)

x∗1

.

This completes the proof.

3. Main results

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let

{

Fk

}

Mk=1be a family of bifunctions from C

×

C

Rsatisfying

(

A1

)

(

A4

)

,

{

ϕ

k

}

Mk=1be a family of proper lower semicontinuous functions from C intoR

∪ {+∞}

and

{

Bk

}

Mk=1be a family of

µ

k-inverse-strongly monotone mappings of C into H. Let Aibe

η

i-inverse-strongly monotone, respectively, where i

∈ {

1

,

2

, . . . ,

N

}

. Let

{

Sn

}

i∞=1be a family of

δ

i-strict pseudocontractive mappings of C into itself such that F

=

(

Mi=1 GMEP

(

Fi

, ϕ

i

,

Bi

))

F

(

G

)

(

i=1F

(

Si

))

̸= ∅

, where G is defined byLemma2.5. Assume that either

(

B1

)

or

(

B2

)

holds. Pick any x0

H and set C1

=

C . Let

{

xn

}

be a sequence generated by x1

=

PC1x0and

un

=

Tr(FMM,nM)

(

I

rM,nBM

)

T (FM−1,ϕM−1) rM−1,n

(

I

rM−1,nBM−1

)

· · ·

T (F1,ϕ1) r1,n

(

I

r1,nB1

)

xn

,

yn

=

PC

(

I

λ

NAN

)

PC

(

I

λ

N−1AN−1

)

· · ·

PC

(

I

λ

2A2

)

PC

(

I

λ

1A1

)

un

,

zn

=

α

nyn

+

(

1

α

n

)

Snyn

,

Cn+1

= {

z

Cn

: ∥

zn

z

∥ ≤ ∥

xn

z

∥}

,

xn+1

=

PCn+1x0

,

n

1

.

(3.1) Assume that

{

λ

i

} ⊂

(

0

,

2

η

i

),

i

∈ {

1

,

2

, . . . ,

N

}

,

rk,n

⊂ [

ek

,

fk

] ⊂

(

0

,

2

µ

k

)

, where k

∈ {

1

,

2

, . . . ,

M

}

. Let

δ

=

supi≥1

δ

iand assume

δ

α

n

g

<

1. Suppose

∞

n=1supxB

Sn+1x

Snx

<

for any bounded subset B of C and let S be a mapping of C into itself defined by Sx

=

limn→∞Snx for all x

C . Suppose that F

(

S

)

= ∩

n=1F

(

Sn

)

. Then

{

xn

}

converges strongly to PFx0. Proof. It is obvious thatCnis closed for eachn

N. Since

zn

z

∥ ≤ ∥

xn

z

is equivalent to

zn

xn

2

+

2

zn

xn

,

xn

z

⟩ ≤

0

,

we have thatCnis convex for eachn

N. Thus

{

xn

}

is well defined. Putting

Θk n

=

Tr(Fk,knk)

(

I

rk,nBM

)

T (Fk−1,ϕk−1) rk−1,n

(

I

rk−1,nBk−1

)

· · ·

T (F1,ϕ1) r1,n

(

I

r1,nB1

),

k

∈ {

1

,

2

, . . . ,

M

}

,

n

N

,

and Ωi

=

P C

(

I

λ

iAi

)

PC

(

I

λ

i−1Ai−1

)

· · ·

PC

(

I

λ

2A2

)

PC

(

I

λ

1A1

),

i

∈ {

1

,

2

, . . . ,

N

}

,

Θ0 n

=

0

=

I, whereIis the identity mapping onH. Then we have thatu

n

=

ΘnMxnandyn

=

Nun. FromLemma 2.1

and(2.8), it can be seen easily thatΘk

nandΩiare nonexpansive, wherek

∈ {

1

,

2

, . . . ,

M

}

,

i

∈ {

1

,

2

, . . . ,

N

}

. We divide the

proof into four steps.

Step1 We show by induction thatF

Cnfor eachn

N. It is obvious thatF

C1

=

C. Suppose thatF

Cnfor some n

N. Letp

F, byLemma 2.1, we have

un

p

∥ =

ΘnMxn

ΘnMp

≤ ∥

xn

p

.

(3.2)

It follows from(3.2)that

yn

p

∥ =

Nun

Np

≤ ∥

un

p

≤ ∥

xn

p

.

(3.3)

From(2.2)and(3.3), we have

zn

p

2

= ∥

α

n

(

yn

p

)

+

(

1

α

n

)(

Snyn

p

)

2

=

α

n

yn

p

2

+

(

1

α

n

)

Snyn

p

2

α

n

(

1

α

n

)

yn

Snyn

2

α

n

yn

p

2

+

(

1

α

n

)(

yn

p

2

+

δ

yn

Snyn

2

)

α

n

(

1

α

n

)

yn

Snyn

2

= ∥

yn

p

2

+

(

1

α

n

)(δ

α

n

)

yn

Snyn

2

≤ ∥

yn

p

2

≤ ∥

xn

p

2

.

(3.4)
(6)

Step2 We prove that limn→∞

xn

zn

∥ =

0.

Let

v

=

PFx0. Fromxn

=

PCnx0and

v

F

Cn, we obtain

xn

x0

∥ ≤ ∥

v

x0

.

(3.5)

This implies that

{

xn

}

is bounded. Sincexn+1

Cn+1

Cnandxn

=

PCnx0, we have

xn

x0

∥ ≤ ∥

xn+1

x0

,

n

N

.

Therefore limn→∞

xn

x0

exists. Fromxn

=

PCnx0

,

xn+1

Cn+1

Cnand(2.6), we obtain

xn+1

xn

2

≤ ∥

x0

xn+1

2

− ∥

x0

xn

2

,

which implies lim

n→∞

xn+1

xn

∥ =

0

.

(3.6)

It follows fromxn+1

Cn+1that

zn

xn+1

∥ ≤ ∥

xn

xn+1

, and hence

xn

zn

∥ ≤ ∥

xn

xn+1

∥ + ∥

xn+1

zn

∥ ≤

2

xn

xn+1

.

From(3.6), we have lim

n→∞

xn

zn

∥ =

0

.

(3.7)

Step3 We show that limn→∞

xn

un

∥ =

0

,

limn→∞

un

yn

∥ =

0.

First we prove that lim n→∞

Θnkxn

Θnk−1xn

=

0

,

k

=

1

,

2

, . . . ,

M

.

(3.8)

Forp

F, it follows from(2.8)that

Θnkxn

p

2

=

T (Fkk) rk,n

(

I

rk,nBk

)

Θ k−1 n xn

Tr(kF,knk)

(

I

rk,nBk

)

p

2

(

I

rk,nBk

)

Θnk−1xn

(

I

rk,nBk

)

p

2

Θnk−1xn

p

2

+

rk,n

(

rk,n

2

µ

k

)

B

kΘnk−1xn

Bkp

2

≤ ∥

xn

p

2

+

rk,n

(

rk,n

2

µ

k

)

B

kΘnk−1xn

Bkp

2

.

(3.9) By(3.3)and(3.4), we have

zn

p

2

≤ ∥

un

p

2

=

ΘnMxn

p

2

Θnkxn

p

2

.

(3.10)

Substituting(3.9)into(3.10), we have

zn

p

2

≤ ∥

xn

p

2

+

rk,n

(

rk,n

2

µ

k

)

B

kΘnk−1xn

Bkp

2

,

which implies that

rk,n

(

2

µ

k

rk,n

)

B

kΘnk−1xn

Bkp

2

≤ ∥

xn

p

2

− ∥

zn

p

2

≤ ∥

xn

zn

(

xn

p

∥ + ∥

zn

p

).

Since

rk,n

⊂ [

ek

,

fk

] ⊂

(

0

,

2

µ

k

)

, wherek

∈ {

1

,

2

, . . . ,

M

}

, and(3.7), we obtain

lim n→∞

B

kΘnk−1xn

Bkp

=

0

,

k

=

1

,

2

, . . . ,

M

.

(3.11)

ByLemma 2.1and(2.1), we have

Θnkxn

p

2

=

T (Fkk) rk,n

(

I

rk,nBk

)

Θ k−1 n xn

Tr(Fk,knk)

(

I

rk,nBk

)

p

2

(

I

rk,nBk

)

Θnk−1xn

(

I

rk,nBk

)

p

,

Θnkxn

p

(7)

=

1 2

(

I

rk,nBk

)

Θnk−1xn

(

I

rk,nBk

)

p

2

+

Θnkxn

p

2

(

I

rk,nBk

)

Θnk−1xn

(

I

rk,nBk

)

p

(

Θnkxn

p

)

2

1 2

Θk −1 n xn

p

2

+

Θnkxn

p

2

Θk −1 n xn

Θnkxn

rk,n

(

BkΘnk−1xn

Bkp

)

2

,

which implies

Θnkxn

p

2

Θnk−1xn

p

2

Θnk−1xn

Θnkxn

rk,n

(

BkΘnk−1xn

Bkp

)

2

=

Θnk−1xn

p

2

Θnk−1xn

Θnkxn

2

rk,n2

B

kΘnk−1xn

Bkp

2

+

2rk,n

Θnk−1xn

Θnkxn

,

BkΘnk−1xn

Bkp

Θk −1 n xn

p

2

Θk −1 n xn

Θnkxn

2

+

2rk,n

Θk −1 n xn

Θnkxn

B

kΘnk−1xn

Bkp

≤ ∥

xn

p

2

Θnk−1xn

Θnkxn

2

+

2rk,n

Θnk−1xn

Θnkxn

B

kΘnk−1xn

Bkp

.

(3.12)

Substituting(3.12)into(3.10), we obtain

zn

p

2

≤ ∥

xn

p

2

Θk −1 n xn

Θnkxn

2

+

2rk,n

Θk −1 n xn

Θnkxn

B

kΘnk−1xn

Bkp

,

which implies

Θk −1 n xn

Θnkxn

2

≤ ∥

xn

p

2

− ∥

zn

p

2

+

2rk,n

Θk −1 n xn

Θnkxn

B

kΘnk−1xn

Bkp

≤ ∥

xn

zn

(

xn

p

∥ + ∥

zn

p

)

+

2rk,n

Θnk−1xn

Θnkxn

B

kΘnk−1xn

Bkp

.

From(3.7)and(3.11), we have that(3.8)holds. Therefore we get

xn

un

∥ =

Θn0xn

ΘnMxn

Θn0xn

Θn1xn

+

Θn1xn

Θn2xn

+ · · · +

ΘM −1 n xn

ΘnMxn

0 asn

→ ∞

.

(3.13)

Next we show that lim n→∞

A

ii−1un

Aii−1p

=

0

,

i

=

1

,

2

, . . . ,

N

.

(3.14) By(2.8), we have

Nun

Np

2

=

P

C

(

I

λ

NAN

)

N−1un

PC

(

I

λ

NAN

)

N−1p

2

(

I

λ

NAN

)

N−1un

(

I

λ

NAN

)

N−1p

2

N −1 un

N −1 p

2

+

λ

N

N

2

η

N

)

A

NN −1 un

ANN −1 p

2

.

(3.15) By induction, we get

Nun

Np

2

≤ ∥

un

p

2

+

N

i=1

λ

i

i

2

η

i

)

A

ii−1un

Aii−1p

2

≤ ∥

xn

p

2

+

N

i=1

λ

i

i

2

η

i

)

A

ii−1un

Aii−1p

2

.

(3.16) By(3.4), we have

zn

p

2

≤ ∥

yn

p

2

=

Nun

Np

2

.

(3.17)

Substituting(3.16)into(3.17), we have

zn

p

2

≤ ∥

xn

p

2

+

N

i=1

λ

i

i

2

η

i

)

A

ii−1un

Aii−1p

2

,

(8)

which implies N

i=1

λ

i

(

2

η

i

λ

i

)

A

ii−1un

Aii−1p

2

≤ ∥

xn

p

2

− ∥

zn

p

2

≤ ∥

xn

zn

(

xn

p

∥ + ∥

zn

p

).

Since

{

λ

i

} ⊂

(

0

,

2

η

i

)

,wherei

∈ {

1

,

2

, . . . ,

N

}

and(3.7), we have(3.14)holds. From(2.4)and(2.1), we obtain

Nun

Np

2

=

P

C

(

I

λ

NAN

)

N−1un

PC

(

I

λ

NAN

)

N−1p

2

(

I

λ

NAN

)

N−1un

(

I

λ

NAN

)

N−1p

,

Nun

Np

=

1 2

(

I

λ

NAN

)

N−1un

(

I

λ

NAN

)

N−1p

2

+

Nun

Np

2

(

I

λ

NAN

)

N−1un

(

I

λ

NAN

)

N−1p

(

Nun

Np

)

2

1 2

N−1un

N−1p

2

+

Nun

Np

2

N −1 un

Nun

+

Np

N −1 p

λ

N

(

ANN −1 un

ANN −1 p

)

2

,

which implies

Nun

Np

2

N−1un

N−1p

2

N−1un

Nun

+

Np

N−1p

λ

N

(

ANN−1un

ANN−1p

)

2

=

N −1 un

N −1 p

2

N −1 un

Nun

+

Np

N −1 p

2

λ

2N

A

NN−1un

ANN−1p

2

+

2

λ

N

N−1 un

Nun

+

Np

N −1 p

,

ANN −1 un

ANN −1 p

N −1u n

N−1p

2

N −1u n

Nun

+

Np

N−1p

2

+

2

λ

N

N −1 un

Nun

+

Np

N −1 p

A

NN −1 un

ANN −1 p

.

(3.18) By induction, we have

Nun

Np

2

≤ ∥

un

p

2

N

i=1

i−1un

iun

+

ip

i−1p

2

+

N

i=1 2

λ

i

i−1un

iun

+

ip

i−1p

A

ii−1un

Aii−1p

≤ ∥

xn

p

2

N

i=1

i −1u n

iun

+

ip

i−1p

2

+

N

i=1 2

λ

i

i−1un

iun

+

ip

i−1p

A

ii−1un

Aii−1p

.

(3.19)

Substituting(3.19)into(3.17), we have

zn

p

2

≤ ∥

xn

p

2

N

i=1

i−1un

iun

+

ip

i−1p

2

+

N

i=1 2

λ

i

i−1un

iun

+

ip

i−1p

A

ii−1un

Aii−1p

,

(9)

which implies N

i=1

i−1un

iun

+

ip

i−1p

2

≤ ∥

xn

p

2

− ∥

zn

p

2

+

N

i=1 2

λ

i

i −1u n

iun

+

ip

i−1p

A

ii−1un

Aii−1p

≤ ∥

xn

zn

(

xn

p

∥ + ∥

zn

p

)

+

N

i=1 2

λ

i

i−1un

iun

+

ip

i−1p

A

ii−1un

Aii−1p

.

It follows from(3.7)and(3.14), we have lim n→∞

i−1un

iun

+

ip

i−1p

=

0

,

i

=

1

,

2

, . . . ,

N

.

(3.20) From(3.20), we obtain

un

yn

∥ =

Ω0un

Nun

N

i=1

i −1u n

iun

+

ip

i−1p

0 asn

→ ∞

.

(3.21)

From(3.13)and(3.21), we have

xn

yn

∥ ≤ ∥

xn

un

∥ + ∥

un

yn

0 asn

→ ∞

.

(3.22) We observe

zn

xn

∥ = ∥

α

n

(

yn

xn

)

+

(

1

α

n

)(

Snyn

xn

)

(

1

α

n

)

Snyn

xn

∥ −

α

n

yn

xn

,

which implies

Snyn

xn

∥ ≤

1 1

α

n

n

yn

xn

∥ + ∥

zn

xn

) .

From(3.7),(3.22)and

α

n

f

<

1, we obtain

lim

n→∞

Snyn

xn

∥ =

0

.

(3.23)

Combining(3.22)and(3.23), we have

Snyn

yn

∥ ≤ ∥

Snyn

xn

∥ + ∥

xn

yn

0 asn

→ ∞

.

(3.24)

It follows fromLemma 2.3and(3.24)that

Syn

yn

∥ ≤ ∥

Syn

Snyn

∥ + ∥

Snyn

yn

0 asn

→ ∞

.

(3.25)

Step4 We show thatxn

v

, where

v

=

PFx0.

Since

{

xn

}

is bounded, there exists a subsequence

xni

which converges weakly to

ω

. From(3.8),(3.13)and(3.21), we have thatΘk

nixni

⇀ ω,

uni

⇀ ω,

yni

⇀ ω

, wherek

∈ {

1

,

2

, . . . ,

M

}

. From(3.25)andLemma 2.2, we have that

w

F

(

S

)

=

n=1F

(

Sn

)

. From(3.21),Lemmas 2.4and2.5, we obtain

w

F

(

G

)

. Next we prove that

ω

∈ ∩

Mk=1GMEP

(

Fk

, ϕ

k

,

Bk

)

. Since

Θk nxn

=

Tr(Fk,knk)

(

I

rk,nBk

)

Θ k−1 n xn

,

n

1

,

k

∈ {

1

,

2

, . . . ,

M

}

, we have Fk

(

Θnkxn

,

y

)

+

ϕ

k

(

y

)

ϕ

k

(

Θnkxn

)

+

BkΘnk−1xn

,

y

Θnkxn

+

1 rk,n

y

Θnkxn

,

Θnkxn

Θnk−1xn

0

.

By (A2), we have

ϕ

k

(

y

)

ϕ

k

(

Θnkxn

)

+

BkΘk −1 n xn

,

y

Θnkxn

+

1 rk,n

y

Θnkxn

,

Θnkxn

Θk −1 n xn

Fk

(

y

,

Θnkxn

)

(10)

Letzt

=

ty

+

(

1

t

for allt

(

0

,

1

]

andy

C. This implies thatzt

C. Then, we have

zt

Θnkxn

,

Bkzt

ϕ

k

(

Θnkxn

)

ϕ

k

(

zt

)

+

zt

Θnkxn

,

Bkzt

zt

Θnkxn

,

BkΘnk−1xn

zt

Θnkxn

,

Θ k nxn

Θnk−1xn rk,n

+

Fk

(

zt

,

Θnkxn

)

=

ϕ

k

(

Θnkxn

)

ϕ

k

(

zt

)

+

zt

Θnkxn

,

Bkzt

BkΘnkxn

+

zt

Θnkxn

,

BkΘnkxn

BkΘk −1 n xn

zt

Θnkxn

,

Θ k nxn

Θnk−1xn rk,n

+

Fk

(

zt

,

Θnkxn

).

By(3.8), we have

B

kΘnkxn

BkΘnk−1xn

0 asn

→ ∞

. Furthermore, by the monotonicity ofBk, we obtain

zt

Θnkxn

,

Bkzt

BkΘnkxn

⟩ ≥

0. Then, by (A4) we obtain,

zt

ω,

Bkzt

⟩ ≥

ϕ

k

(ω)

ϕ

k

(

zt

)

+

Fk

(

zt

, ω).

(3.26)

Using (A1), (A4) and(3.26), we obtain 0

=

Fk

(

zt

,

zt

)

+

ϕ

k

(

zt

)

ϕ

k

(

zt

)

tFk

(

zt

,

y

)

+

(

1

t

)

Fk

(

zt

, ω)

+

t

ϕ

k

(

y

)

+

(

1

t

k

(ω)

ϕ

k

(

zt

)

t

[

Fk

(

zt

,

y

)

+

ϕ

k

(

y

)

ϕ

k

(

zt

)

] +

(

1

t

)

zt

ω,

Bkzt

=

t

[

Fk

(

zt

,

y

)

+

ϕ

k

(

y

)

ϕ

k

(

zt

)

] +

(

1

t

)

t

y

ω,

Bkzt

,

and hence 0

Fk

(

zt

,

y

)

+

ϕ

k

(

y

)

ϕ

k

(

zt

)

+

(

1

t

)

y

ω,

Bkzt

.

Lettingt

0, we have, for eachy

C,

0

Fk

(ω,

y

)

+

ϕ

k

(

y

)

ϕ

k

(ω)

+ ⟨

y

ω,

Bk

ω

.

This implies that

ω

GMEP

(

Fk

, ϕ

k

,

Bk

)

. Hence

ω

∈ ∩

Mk=1GMEP

(

Fk

, ϕ

k

,

Bk

)

. Therefore, z

F

=

(

Mi=1GMEP

(

Fi

, ϕ

i

,

Bi

))

(

F

(

G

))

(

i=1F

(

Si

)).

Since

ω

F

, v

=

PFx0,(3.5)and the lower semicontinuousness of the norm, we obtain

v

x0

∥ ≤ ∥

ω

x0

∥ ≤

lim inf i→∞

x

ni

x0

lim sup i→∞

x

ni

x0

≤ ∥

v

x0

.

Thus, we obtain

ω

=

v

and limi→∞

x

ni

x0

= ∥

ω

x0

. Fromxni

x0

⇀ ω

x0and the Kadec–Klee property ofH, we

havexni

x0

ω

x0, and hencexni

ω

. This implies thatxn

ω

=

v

. This completes the proof.

References

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