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Volume 2007, Article ID 38752, 12 pagesdoi:10.1155/2007/38752 Research Article An Extragradient Method for Fixed Point Problems and Variational Inequality Problems Yonghong Yao, Yeong-Ch

Volume 2007, Article ID 38752, 12 pages

doi:10.1155/2007/38752

Research Article

An Extragradient Method for Fixed Point Problems and

Variational Inequality Problems

Yonghong Yao, Yeong-Cheng Liou, and Jen-Chih Yao

Received 11 September 2006; Accepted 10 December 2006

Recommended by Yeol-Je Cho

We present an extragradient method for fixed point problems and variational inequal-ity problems Using this method, we can find the common element of the set of fixed points of a nonexpansive mapping and the set of solutions of the variational inequality for monotone mapping

Copyright © 2007 Yonghong Yao 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

1 Introduction

LetC be a closed convex subset of a real Hilbert space H Recall that a mapping A of C

intoH is called monotone if

 Au − Av, u − v  ≥0, (1.1)

for allu, v ∈ C A is called α-inverse strongly monotone if there exists a positive real

num-berα such that

 Au − Av, u − v  ≥ α  Au − Av 2, (1.2)

for allu, v ∈ C It is well known that the variational inequality problem VI(A, C) is to find

u ∈ C such that

for allv ∈ C (see [1–3]) The set of solutions of the variational inequality problem is denoted byΩ The variational inequality has been extensively studied in the literature, see, for example, [4–6] and the references therein A mappingS of C into itself is called

nonexpansive if

for allu, v ∈ C We denote by F(S) the set of fixed points of S.

For finding an element ofF(S) ∩ Ω under the assumption that a set C ⊂ H is closed

and convex, a mappingS of C into itself is nonexpansive and a mapping A of C into

H is α-inverse strongly monotone, Takahashi and Toyoda [7] introduced the following iterative scheme:

x n+1 = α n x n+

1− α n



SP C



x n − λ n Ax n



for everyn =0, 1, 2, , where P Cis the metric projection ofH onto C, x0= x ∈ C, { α n }

is a sequence in (0, 1), and{ λ n }is a sequence in (0, 2α) They showed that if F(S) ∩Ω

is nonempty, then the sequence{ x n }generated by (1.5) converges weakly to somez ∈ F(S) ∩Ω Recently, Nadezhkina and Takahashi [8] introduced a so-called extragradient method motivated by the idea of Korpeleviˇc [9] for finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of a variational inequality problem They obtained the following weak convergence theorem

Theorem 1.1 (see Nadezhkina and Takahashi [8]) Let C be a nonempty closed convex sub-set of a real Hilbert space H Let A : C → H be a monotone k-Lipschitz continuous mapping, and let S : C → C be a nonexpansive mapping such that F(S) ∩Ω= ∅ Let the sequences { x n } , { y n } be generated by

x0= x ∈ H,

y n = P C



x n − λ n Ax n



,

x n+1 = α n x n+

1− α n



SP C



x n − λ n Ay n



, ∀ n ≥0,

(1.6)

where { λ n } ⊂[a, b] for some a, b ∈(0, 1/k) and { α n } ⊂[c, d] for some c, d ∈ (0, 1) Then the sequences { x n } , { y n } converge weakly to the same point P F(S) ∩Ω(x0).

Very recently, Zeng and Yao [10] introduced a new extragradient method for finding

a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of a variational inequality problem They obtained the following strong conver-gence theorem

Theorem 1.2 (see Zeng and Yao [10]) Let C be a nonempty closed convex subset of a real Hilbert space H Let A : C → H be a monotone k-Lipschitz continuous mapping, and let

S : C → C be a nonexpansive mapping such that F(S) ∩Ω= ∅ Let the sequences { x n } , { y n }

be generated by

x0= x ∈ H,

y n = P C



x n − λ n Ax n



,

x n+1 = α n x0+

1− α n



SP C



x n − λ n Ay n



, ∀ n ≥0,

(1.7)

where { λ n } and { α n } satisfy the following conditions:

(a){ λ n k } ⊂(0, 1− δ) for some δ ∈ (0, 1);

(b){ α n } ⊂ (0, 1),∞

n =0α n = ∞ , lim n →∞ α n = 0.

Then the sequences { x n } and { y n } converge strongly to the same point P F(S) ∩Ω(x0) pro-vided that

lim

Remark 1.3 The iterative scheme (1.6) inTheorem 1.1has only weak convergence The iterative scheme (1.7) inTheorem 1.2has strong convergence but imposed the assump-tion (1.8) on the sequence{ x n }

In this paper, motivated by the iterative schemes (1.6) and (1.7), we introduced a new extragradient method for finding a common element of the set of fixed points of a nonex-pansive mapping and the set of solutions of the variational inequality problem for mono-tone mapping We obtain a strong convergence theorem under some mild conditions

2 Preliminaries

LetH be a real Hilbert space with inner product ·,·and norm · and letC be a closed

convex subset ofH It is well known that for any u ∈ H, there exists unique y0∈ C such

that

 u − y :y ∈ C

We denotey0byP C u, where P Cis called the metric projection ofH onto C The metric

projectionP CofH onto C has the following basic properties:

(i) P C x − P C y  ≤  x − y , for allx, y ∈ H,

(ii) x − y, P C x − P C y  ≥  P C x − P C y 2, for everyx, y ∈ H,

(iii) x − P C x, y − P C x  ≤0, for allx ∈ H, y ∈ C,

(iv) x − y 2≥  x − P C x 2+ y − P C x 2, for allx ∈ H, y ∈ C.

Such property ofP Cwill be crucial in the proof of our main results LetA be a monotone

mapping ofC into H In the context of the variational inequality problem, it is easy to see

from (iv) that

u ∈Ω⇐⇒ u = P C(u − λAu), ∀ λ > 0. (2.2)

A set-valued mappingT : H →2H is called monotone if for allx, y ∈ H, f ∈ Tx and

g ∈ T y imply  x − y, f − g  ≥0 A monotone mappingT : H →2His maximal if its graph

G(T) is not properly contained in the graph of any other monotone mapping It is known

that a monotone mappingT is maximal if and only if for (x, f ) ∈ H × H,  x − y, f −

g  ≥ 0 for every (y, g) ∈ G(T) implies that f ∈ Tx Let A be a monotone mapping of C

intoH and let N C v be the normal cone to C at v ∈ C, that is,

N C v =w ∈ H :  v − u, w  ≥0,∀ u ∈ C

Define

Tv =

⎧

⎨

⎩

Av + N C v ifv ∈ C,

ThenT is maximal monotone and 0 ∈ Tv if and only if v ∈VI(C, A) (see [11])

Now, we introduce several lemmas for our main results in this paper

Lemma 2.1 (see [12]) Let ( E, ·,· ) be an inner product space Then, for all x, y, z ∈ E and

α, β, γ ∈ [0, 1] with α + β + γ = 1, one has

 αx + βy + γz 2= α  x 2+β  y 2+γ  z 2− αβ  x − y 2− αγ  x − z 2− βγ  y − z 2.

(2.5) Lemma 2.2 (see [13]) Let { x n } and { y n } be bounded sequences in a Banach space X and let { β n } be a sequence in [0, 1] with 0 < lim inf n →∞ β n ≤lim supn →∞ β n < 1 Suppose x n+1 =

(1− β n)y n+β n x n for all integers n ≥ 0 and lim sup n →∞( y n+1 − y n  −  x n+1 − x n )≤ 0 Then, lim n →∞  y n − x n  = 0.

Lemma 2.3 (see [14]) Assume { a n } is a sequence of nonnegative real numbers such that

a n+1 ≤1− γ n

where { γ n } is a sequence in (0, 1) and { δ n } is a sequence such that

(1)∞

n =1γ n = ∞ ;

(2) lim supn →∞ δ n /γ n ≤ 0 or∞

n =1| δ n | < ∞ Then lim n →∞ a n = 0.

3 Main results

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H Let A be a monotone L-Lipschitz continuous mapping of C into H, and let S be a nonexpansive mapping

of C into itself such that F(S) ∩Ω= ∅ For fixed u ∈ H and given x0∈ H arbitrary, let the sequences { x n } , { y n } be generated by

y n = P C

x n − λ n Ax n

,

x n+1 = α n u + β n x n+γ n SP C

x n − λ n Ay n

where { α n },{ β n } , { γ n } are three sequences in [0, 1] satisfying the following conditions:

(C1)α n+β n+γ n = 1,

(C2) limn →∞ α n = 0,∞

n =0α n = ∞ ,

(C3) 0< lim inf n →∞ β n ≤lim supn →∞ β n < 1,

(C4) limn →∞ λ n = 0.

Then { x n } converges strongly to P F(S) ∩Ωu.

Proof Let x ∗ ∈ F(S) ∩ Ω, then x ∗ = P C(x ∗ − λ n Ax ∗) Putt n = P C(x n − λ n Ay n) Substi-tutingx by x n − λ n Ay nandy by x ∗in (iv), we have

≤x n − λ n Ay n − x ∗ 2

−x n − λ n Ay n − t n 2

=x n − x ∗ 2

−2 n

Ay n,x n − x ∗

+λ2

nAy n 2

−x n − t n 2

+ 2λ n

Ay n,x n − t n

− λ2

nAy n 2

=x n − x ∗ 2

+ 2λ n

Ay n,x ∗ − t n

−x n − t n 2

=x n − x ∗ 2

−x n − t n 2

+ 2λ n

Ay n − Ax ∗,x ∗ − y n

+ 2λ n

Ax ∗,x ∗ − y n

+ 2λ n

Ay n,y n − t n

.

(3.2)

Using the fact thatA is monotonic and x ∗is a solution of the variational inequality prob-lem VI(A, C), we have

Ay n − Ax ∗,x ∗ − y n

Ax ∗,x ∗ − y n

It follows from (3.2) and (3.3) that

≤x n − x ∗ 2

−x n − t n 2

+ 2λ n

Ay n,y n − t n

=x n − x ∗ 2

−x n − y n

+

y n − t n 2

+ 2λ n

Ay n,y n − t n

=x n − x ∗ 2

−x n − y n 2

−2

x n − y n,y n − t n

−y n − t n 2

+ 2λ n

Ay n,y n − t n

=x n − x ∗ 2

−x n − y n 2

−y n − t n 2

+ 2

x n − λ n Ay n − y n,t n − y n

.

(3.4) Substitutingx by x n − λ n Ax nandy by t nin (iii), we have

x n − λ n Ax n − y n,t n − y n

It follows that

x n − λ n Ay n − y n,t n − y n

=x n − λ n Ax n − y n,t n − y n

+

λ n Ax n − λ n Ay n,t n − y n

≤λ n Ax n − λ n Ay n,t n − y n

≤ λ n Lx n − y nt n − y n.

(3.6)

By (3.4) and (3.6), we obtain

≤x n − x ∗ 2

−x n − y n 2

−y n − t n 2

+ 2λ n Lx n − y nt n − y n

≤x n − x ∗ 2

−x n − y n 2

−y n − t n 2

+λ n L2 x n − y n 2

+y n − t n 2 

≤x n − x ∗ 2

+

λ2n L2−1x n − y n 2

+

λ2n L2−1y n − t n 2

.

(3.7) Sinceλ n →0 asn → ∞, there exists a positive integerN0such thatλ2

n L2−1≤ −1/2 when

n ≥ N0 It follows from (3.7) that

By (3.1), we have

≤ α nu − x ∗+

1− α nx n − x ∗  ≤maxu − x ∗,x0− x ∗.

(3.9) Therefore,{ x n }is bounded Hence{ t n },{ St n },{ Ax n }, and{ Ay n }are also bounded For allx, y ∈ C, we get

x −I − λ n A

y 2

=(x − y) − λ n(Ax − Ay) 2

= x − y 2−2 n  x − y, Ax − Ay 

+λ2n  Ax − Ay 2≤  x − y 2+λ2n  Ax − Ay 2

≤  x − y 2+λ2n L2 x − y 2=1 +L2λ2n

 x − y 2,

(3.10) which implies that

x −I − λ n A



By (3.1) and (3.11), we have

x n+1 − λ n+1 Ay n+1

− P C

x n − λ n Ay n

≤x n+1 − λ n+1 Ay n+1

−x n − λ n Ay n

=x n+1 − λ n+1 Ax n+1

−x n − λ n+1 Ax n



+λ n+1



Ax n+1 − Ay n+1 − Ax n



+λ n Ay n

≤x n+1 − λ n+1 Ax n+1

−x n − λ n+1 Ax n

+λ n+1Ax n+1+Ay n+1+Ax n+λ nAy n

≤1 +λ n+1 Lx n+1 − x n

+λ n+1Ax n+1+Ay n+1+Ax n+λ nAy n.

(3.12)

Setx n+1 =(1− β n) n+β n x n Then, we obtain

z n+1 − z n = α n+1 u + γ n+1 St n+1

1− β n+1 − α n u + γ n St n

1− β n

1− β n+1 − α n

1− β n



u + γ n+1

1− β n+1



St n+1 − St n

1− β n+1 − γ n

1− β n



St n

(3.13)

Combining (3.12) and (3.13), we have

≤

1− β n+1 − α n

1− β n



1− β n+1



1 +λ n+1 Lx n+1 − x n

+ γ n+1

1− β n+1



λ n+1Ax n+1+Ay n+1+Ax n+λ nAy n

+

1− β n+1 − γ n

1− β n



≤

1− β n+1 − α n

1− β n



1− β n+1 λ n+1 Lx n+1 − x n

+ γ n+1

1− β n+1



λ n+1Ax n+1+Ay n+1+Ax n+λ nAy n,

(3.14)

this together with (C2) and (C4) imply that

lim sup

n →∞

Hence byLemma 2.2, we obtain z n − x n  →0 asn → ∞ Consequently,

lim

n →∞x n+1 − x n  =lim

n →∞



1− β nz n − x n  =0. (3.16) From (C4) and (3.12), we also have t n+1 − t n  →0 asn → ∞

Forx ∗ ∈ F(S) ∩Ω, fromLemma 2.1, (3.1), and (3.7), we obtain whenn ≥ N0that

=α n u + β n x n+γ n St n − x ∗ 2

≤ α nu − x ∗ 2

+β nx n − x ∗ 2

+γ nSt n − x ∗ 2

≤ α nu − x ∗ 2

+β nx n − x ∗ 2

+γ nt n − x ∗ 2

≤ α nu − x ∗ 2

+β nx n − x ∗ 2

+γ nx n − x ∗ 2

+

λ2

n L2−1x n − y n 2 

+

λ2

n L2−1y n − t n 2 

≤ α nu − x ∗ 2

+x n − x ∗ 2

−1

2x n − y n 2

,

(3.17)

which implies that

1

2x n − y n 2

≤ α nu − x ∗ 2

+x n − x ∗ 2

−x n+1 − x ∗ 2

= α nu − x ∗ 2

× x n − x ∗+x n+1 − x ∗

≤ α nu − x ∗ 2

(3.18)

Sinceα n →0 and x n − x n+1  →0, from (3.18), we have x n − y n  →0 asn → ∞ Noting that

x n − λ n Ax n

− P C

x n − λ n Ay n

≤y n − t n+α nSt n − u+β nSt n − Sx n+β nSx n − x n

≤y n − t n+α nSt n − u+β nt n − x n+β nSx n − x n.

(3.19) Consequently, from (3.19), we can infer that

≤1 +β nx n − t n+ 2t n − y n+α nSt n − u+β nSx n − x n+x n+1 − x n,

(3.20) which implies that

Also we have

Next we show that

lim sup

n →∞

u − z0,x n − z0

wherez0= P F(S) ∩Ωu.

To show it, we choose a subsequence{ t n i }of{ t n }such that

lim sup

n →∞

u − z0,St n − z0

=lim

i →∞

u − z0,St n i − z0

As{ t n i }is bounded, we have that a subsequence{ t n i j } of{ t n i }converges weakly to z.

We may assume without loss of generality thatt n i  z Since  St n − t n  →0, we obtain

St n i  z as i → ∞ Then we can obtainz ∈ F(S) ∩ Ω In fact, let us first show that z ∈Ω Let

Uv =

⎧

⎨

⎩

Av + N C v, v ∈ C,

ThenU is maximal monotone Let (v, w) ∈ G(U) Since w − Av ∈ N C v and t n ∈ C, we

have v − t n,w − Av  ≥0 On the other hand, fromt n = P C(x n − λ n Ay n), we have

v − t n,t n −x n − λ n Ay n

that is,



v − t n,t n − y n

λ n +Ay n



Therefore, we have

v − t n i,w

≥v − t n i,Av

≥v − t n i,Av

−



v − t n i,t n i − x n i

λ n i

+Ay n i



=



v − t n i,Av − Ay n i − t n i − x n i

λ n i



=v − t n i,Av − At n i

+

v − t n i,At n i − Ay n i

−



v − t n i,t n i − x n i

λ n i



≥v − t n i,At n i

−v − t n i,t n i − x n i

λ n i +Ay n i



.

(3.28)

Noting that t n i − y n i  →0 asi → ∞andA is Lipschitz continuous, hence from (3.28),

we obtain v − z, w  ≥0 asi → ∞ SinceU is maximal monotone, we have z ∈ U −10, and hencez ∈Ω

Let us show thatz ∈ F(S) Assume that z / ∈ F(S) From Opial’s condition, we have

lim inf

i →∞ t n

i − z< lim inf

i →∞ t n

i − Sz  =lim inf

i →∞ t n

i − St n i+St n i − Sz

≤lim inf

i →∞ t n

i − St n i+St n

i →∞

i − Sz

≤lim inf

i →∞ t n

This is a contradiction Thus, we obtainz ∈ F(S).

Hence, from (iii), we have

lim sup

n →∞

u − z0,x n − z0

=lim sup

n →∞

u − z0,St n − z0

=lim

i →∞

u − z0,St n i − z0

=u − z0,z − z0

=α n u + β n x n+γ n St n − z0,x n+1 − z0

= α n

u − z0,x n+1 − z0

+β n

x n − z0,x n+1 − z0

+γ n

St n − z0,x n+1 − z0

≤1

2β nx n − z0 2

+x n+1 − z0 2 

+α n

u − z0,x n+1 − z0

+1

2γ nt n − z0 2

+x n+1 − z0 2 

≤1

2



1− α nx n − z0 2

+x n+1 − z0 2 

+α n

u − z0,x n+1 − z0

, (3.31) which implies that

≤1− α nx n − z0 2

+ 2α n

u − z0,x n+1 − z0

this together with (3.30) andLemma 2.3, we can obtain the conclusion This completes

We observe that some strong convergence theorems for the iterative scheme (3.1) were established under the assumption that the mappingA is α-inverse strongly monotone in

[15]

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H Let

A be a monotone L-Lipschitz continuous mapping of C into H such thatΩ= ∅ For fixed

u ∈ H and given x0∈ H arbitrary, let the sequences { x n } , { y n } be generated by

y n = P C



x n − λ n Ax n



,

x n+1 = α n u + β n x n+γ n P C



x n − λ n Ay n



where { α n },{ β n } , { γ n } are three sequences in [0, 1] satisfying the following conditions:

(C1)α n+β n+γ n = 1,

(C2) limn →∞ α n = 0,∞

n =0α n = ∞ ,

(C3) 0< lim inf n →∞ β n ≤lim supn →∞ β n < 1,

(C4) limn →∞ λ n = 0.

Then { x n } converges strongly to PΩu.

4 Applications

A mappingT : C → C is called strictly pseudocontractive if there exists k with 0 ≤ k < 1

such that

 Tx − T y 2≤  x − y 2+k(I − T)x −(I − T)y 2

for allx, y ∈ C Put A = I − T, then we have

≤  x − y 2+k  Ax − Ay 2. (4.2)

(3.6)

By (3.4) and (3.6), we obtain

≤x... n.

(3.12)

Setx n+1 =(1−... n 2

,

(3.17)

which implies that

1

2x