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Volume 2008, Article ID 617248, 12 pagesdoi:10.1155/2008/617248 Research Article Strong Convergence Theorem by Monotone Hybrid Algorithm for Equilibrium Problems, Hemirelatively Nonexpan

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Volume 2008, Article ID 617248, 12 pages

doi:10.1155/2008/617248

Research Article

Strong Convergence Theorem by Monotone

Hybrid Algorithm for Equilibrium Problems,

Hemirelatively Nonexpansive Mappings, and

Maximal Monotone Operators

Yun Cheng and Ming Tian

College of Science, Civil Aviation University of China, Tianjin 300300, China

Correspondence should be addressed to Yun Cheng,mathscy@126.com

Received 17 June 2008; Accepted 11 November 2008

Recommended by Wataru Takahashi

We introduce a new hybrid iterative algorithm for finding a common element of the set of fixed points of hemirelatively nonexpansive mappings and the set of solutions of an equilibrium problem and for finding a common element of the set of zero points of maximal monotone operators and the set of solutions of an equilibrium problem in a Banach space Using this theorem,

we obtain three new results for finding a solution of an equilibrium problem, a fixed point of

a hemirelatively nonexpnasive mapping, and a zero point of maximal monotone operators in a Banach space

Copyrightq 2008 Y Cheng and M Tian 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

Let E be a Banach space, let C be a closed convex subset of E, and let f be a bifunction from

C × C to R, where R is the set of real numbers The equilibrium problem is to find

x∗∈ C such that fx∗, y ≥ 0 ∀y ∈ C. 1.1

The set of such solutions x∗is denoted by EPf

In 2006, Martinez-Yanes and Xu1 obtained strong convergence theorems for finding

a fixed point of a nonexpansive mapping by a new hybrid method in a Hilbert space In particular, Takahashi and Zembayashi 2 established a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set

of fixed points of a nonexpansive mapping in a uniformly convex and uniformly smooth

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Banach space Very recently, Su et al.3 proved the following theorem by a monotone hybrid method

Theorem 1.1 see Su et al 3 Let E be a uniformly convex and uniformly smooth real Banach

space, let C be a nonempty closed convex subset of E, and let T : C → C be a closed hemirelatively nonexpansive mapping such that FT / ∅ Assume that α n is a sequence in 0, 1 such that

lim supn → ∞ α n < 1 Define a sequence x n in C by the following:

x0∈ C, chosen arbitrarily,

y n J−1α n Jx n 1 − α n JTx n ,

C n {z ∈ C n−1 ∩ Q n−1 : φz, y n ≤ φz, x n },

C0  {z ∈ C : φz, y0 ≤ φz, x0},

Q n {z ∈ C n−1 ∩ Q n−1: n − z, Jx0− Jx n ≥ 0},

Q0 C,

x n1 ΠC n∩Qn x0,

1.2

where J is the duality mapping on E Then, x n converges strongly to ΠFT x0, where ΠFT is the generalized projection from C onto FT.

In this paper, motivated by Su et al 3, we prove a strong convergence theorem for finding a common element of the set of solutions of an equilibrium problem and the set of fixed points of a hemirelatively nonexpansive mapping and for finding a common element of the set of zero points of maximal monotone operators and the set of solutions

of an equilibrium problem in a Banach space by using the monotone hybrid method Using this theorem, we obtain three new strong convergence results for finding a solution of an equilibrium problem, a fixed point of a hemirelatively nonexpnasive mapping, and a zero point of maximal monotone operators in a Banach space

2 Preliminaries

Let E be a real Banach space with dual E∗ We denote by J the normalized duality mapping from E to 2 E∗defined by

convex, then J is uniformly continuous on bounded subsets of E In this case, J is single

valued and also one to one

Let E be a smooth, strictly convex, and reflexive Banach space and let C be a nonempty closed convex subset of E Throughout this paper, we denote by φ the function defined by

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Following Alber4, the generalized projection ΠC : E → C from E onto C is defined by

ΠC x arg min

The generalized projectionΠC from E onto C is well defined and single valued, and it satisfies

x − y2≤ φy, x ≤ x y2 ∀x, y ∈ E. 2.4

If E is a Hilbert space, then φy, x y − x2andΠC is the metric projection of E onto C.

If E is a reflexive strict convex and smooth Banach space, then for x, y ∈ E, φx, y 0

if and only ifx y It is suﬃcient to show that if φx, y 0, then x y From 2.4,

we have 2  Jy2 From the definition of J, we have

Jx Jy, that is, x y.

Let C be a closed convex subset of E and let T be a mapping from C into itself.

We denote by FT the set of fixed points of T T is called hemirelatively nonexpansive if

φp, Tx ≤ φp, x for all x ∈ C and p ∈ FT.

A point p in C is said to be an asymptotic fixed point of T5 if C contains a sequence xn

which converges weakly to p such that the strong lim n → ∞ Tx n −x n 0 The set of asymptotic

fixed points of T will be denoted by FT A hemirelatively nonexpansive mapping T from C

into itself is called relatively nonexpansive1,5,6 if FT FT

We need the following lemmas for the proof of our main results

Lemma 2.1 see Alber 4 Let C be a nonempty closed convex subset of a smooth, strictly convex,

and reflexive Banach space E Then,

φ

x, Π C y

 φΠC y, y

≤ φx, y ∀x ∈ C, y ∈ E. 2.5

Lemma 2.2 see Alber 4 Let C be a nonempty closed convex subset of a smooth, strictly convex,

and reflexive Banach space, let x ∈ E, and let z ∈ C Then,

Lemma 2.3 see Kamimura and Takahashi 7 Let E be a smooth and uniformly convex Banach

space and let {x n } and {y n } be sequences in E such that either {x n } or {y n } is bounded If

limn → ∞ φx n , y n 0 Then lim n → ∞ x n − y n 0.

Lemma 2.4 see Xu 8 Let E be a uniformly convex Banach space and let r > 0 Then, there exists

a strictly increasing, continuous, and convex function g : 0, 2r → R such that g0 0 and

tx 1 − ty2≤ tx2 1 − ty2− t1 − tgx − y ∀x, y ∈ B r , t ∈ 0, 1, 2.7

where B r {z ∈ E : z ≤ r}.

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space and let r > 0 Then, there exists a strictly increasing, continuous, and convex function g :

0, 2r → R such that g0 0 and

For solving the equilibrium problem, let us assume that a bifunction f satisfies the

following conditions:

A1 fx, x 0 for all x ∈ C;

A2 f is monotone, that is, fx, y fy, x ≤ 0 for all x, y ∈ C;

A3 for all x, y, z ∈ C, lim sup t → 0 ftz 1 − tx, y ≤ fx, y;

A4 for all x ∈ C, fx, · is convex.

Lemma 2.6 see Blum and Oettli 9 Let C be a closed convex subset of a smooth, strictly convex,

and reflexive Banach space E, let f be a bifunction from C × C to R satisfying (A1)–(A4), let r > 0, and let x ∈ E Then, there exists z ∈ C such that

Lemma 2.7 see Takahashi and Zembayashi 10 Let C be a closed convex subset of a uniformly

smooth, strictly convex, and reflexive Banach space E, let f be a bifunction from C × C to R satisfying (A1)–(A4), and let x ∈ E, for r > 0 Define a mapping T r : E → 2C as follows:

T r x 

z ∈ C : fz, y 1r

∀x ∈ E. 2.10

Then, the following holds:

1 T r is single valued;

2 T r is a firmly nonexpansive-type mapping11, that is, for all x, y ∈ E,

r x − T r y, JT r x − JT r r x − T r y, Jx − Jy; 2.11

3 FT r FT r Epf;

4 Epf is closed and convex.

Lemma 2.8 see Takahashi and Zembayashi 10 Let C be a closed convex subset of a smooth,

strictly convex, and reflexive Banach space E and let f be a bifunction from C × C to R satisfying (A1)–(A4) Then, for r > 0 and x ∈ E, and q ∈ FT r ,

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Lemma 2.9 see Su et al 3 Let E be a strictly convex and smooth real Banach space, let C be

a closed convex subset of E, and let T be a hemirelatively nonexpansive mapping from C into itself Then, FT is closed and convex.

Recall that an operator T in a Banach space is called closed, if x n → x, Tx n → y, then

Tx y.

3 Strong convergence theorem

Theorem 3.1 Let E be a uniformly convex and uniformly smooth real Banach space, let C be a

nonempty closed convex subset of E, let f be a bifunction from C × C to R satisfying (A1)–(A4), and let T : C → C be a closed hemirelatively nonexpansive mapping such that FT ∩ EPf / ∅ Define

a sequence {x n } in C by the following:

y n J−1α n Jx n 1 − α n JTz n ,

z n J−1β n Jx n 1 − β n JTx n ,

u n ∈ C such that fu n , y r1

n n , Ju n − Jy n ≥ 0, ∀y ∈ C,

C n {z ∈ C n−1 ∩ Q n−1 : φz, u n ≤ φz, x n },

C0  {z ∈ C : φz, u0 ≤ φz, x0},

Q n {z ∈ C n−1 ∩ Q n−1: n − z, Jx0− Jx n ≥ 0},

Q0 C,

3.1

for every n ∈ N ∪ {0}, where J is the duality mapping on E, {α n }, {β n } are sequences in 0, 1 such

that lim inf n → ∞ 1 − α n β n 1 − β n > 0 and {r n } ⊂ a, ∞ for some a > 0 Then, {x n } converges

strongly toΠFT∩EPf x0, whereΠFT∩EPf is the generalized projection of E onto FT ∩ EPf Proof First, we can easily show that C n and Q n are closed and convex for each n≥ 0

Next, we show that FT ∩ EPf ⊂ C n for all n ≥ 0 Let u ∈ FT ∩ EPf Putting

u n T r n y n for all n ∈ N, fromLemma 2.8, we have Tr n relatively nonexpansive Since T r nare

relatively nonexpansive and T is hemirelatively nonexpansive, we have

φu, z n φu, J−1β n Jx n 1 − β n JTx n

 u2

n Jx n 1 − β n JTx n β n Jx n 1 − β n JTx n2

≤ u2− 2β n n − 21 − β n n β n x n2 1 − β n Tx n2

 β n φu, x n 1 − β n φu, Tx n

≤ φu, x n ,

φu, u n φu, T r y n ≤ φu, y n ≤ α n φu, x n 1 − α n φu, z n ≤ φu, x n .

3.2

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Hence, we have

Next, we show that FT ∩ EPf ⊂ Q n for all n ≥ 0 We prove this by induction For n 0,

we have

Suppose that FT ∩ EPf ⊂ Q n, byLemma 2.2, we have

n1 − z, Jx0− Jx n1 ≥ 0 ∀z ∈ C n ∩ Q n 3.5

As FT ∩ EPf ⊂ C n ∩ Q n, by the induction assumptions, the last inequality holds, in

particular, for all z ∈ FT ∩ EPf This, together with the definition of Q n1, implies that

FT ∩ EPf ⊂ Q n1 So,{x n} is well defined

Since x n1 ΠC n∩Qn x0and C n ∩ Q n ⊂ C n−1 ∩ Q n−1 for all n≥ 1, we have

φx n , x0 ≤ φx n1 , x0 ∀n ≥ 0. 3.6

Therefore,{φx n , x0} is nondecreasing In addition, from the definition of Q nandLemma 2.2,

x n ΠQ n x0 Therefore, for each u ∈ FT ∩ EPf, we have

φx n , x0 φΠQ n x0, x0

≤ φu, x0 − φu, x n ≤ φu, x0. 3.7

Therefore, φx n , x0 and {x n} are bounded This, together with 3.6, implies that the limit of

{φx n , x0} exists FromLemma 2.1, we have, for any positive integer m,

φx nm , x n φx nm , Π Q n x0

≤φx nm , x0−φΠQ n x0, x0

φx nm , x0−φx n , x0 ∀n ≥ 0.

3.8 Therefore,

lim

From3.9, we can prove that {xn } is a Cauchy sequence Therefore, there exists a point x ∈ C

such that{x n } converges strongly to x.

Since x n1 ∈ C n, we have

φx n1 , u n ≤ φx n1 , x n . 3.10 Therefore, we have

φx n1 , u n −→ 0. 3.11

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FromLemma 2.3, we have

lim

n → ∞ x n1 − u n lim

n → ∞ x n1 − x n 0. 3.12

So, we have

lim

Since J is uniformly norm-to-norm continuous on bounded sets, we have

lim

Let r supn∈N {x n , Tx n } Since E is a uniformly smooth Banach space, we know that E∗

is a uniformly convex Banach space Therefore, fromLemma 2.4, there exists a continuous,

strictly increasing, and convex function g with g0 0, such that

αx∗ 1 − αy∗ 2 ≤ αx∗ 2 1 − αy∗ 2− α1 − αgx∗− y∗ 3.15

for x∗, y∗∈ B r , and α ∈ 0, 1 So, we have that for u ∈ FT ∩ EPf,

φu, z n φu, J−1β n Jx n 1 − β n JTx n

 u2

n Jx n 1 − β n JTx n β n Jx n 1 − β n JTx n2

≤ φu, x n − β n 1 − β n gJx n − JTx n ,

φu, u n ≤ α n φu, x n 1 − α n φu, z n

≤ φu, x n − 1 − α n β n 1 − β n gJx n − JTx n .

3.16

Therefore, we have

1 − α n β n 1 − β n gJx n − JTx n ≤ φu, x n − φu, u n . 3.17 Since

φu, x n − φu, u n x n2−u n2

n − Ju n ≤x n −u n x n u n 2u Jx n −Ju n ,

3.18

we have

lim

n → ∞ φu, x n − φu, u n 0. 3.19 From lim infn → ∞ 1 − α n β n 1 − β n > 0, we have

lim

n → ∞ gJx n − JTx n 0. 3.20

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Therefore, from the property of g, we have

lim

Since J−1is uniformly norm-to-norm continuous on bounded sets, we have

lim

Since T is a closed operator and x n → x, then x is a fixed point of T.

On the other hand,

φu n , y n φT r n y n , y n ≤ φu, y n − φu, T r n y n ≤ φu, x n − φu, T r n y n φu, x n − φu, u n .

3.23

So, we have from3.19 that

lim

FromLemma 2.3, we have that

lim

From x n → x and x n − u n → 0, we have y n → x.

From3.25, we have

lim

From r n ≥ a, we have

lim

n → ∞

Ju n − Jy n

r n 0. 3.27

By u n T r n y n, we have

fu n , y r1

n n , Ju n − Jy n ≥ 0 ∀y ∈ C. 3.28 FromA2, we have that

1

r n n , Ju n − Jy n ≥ −fu n , y ≥ fy, u n ∀y ∈ C. 3.29

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From3.27 and A4, we have

For t with 0 < t ≤ 1 and y ∈ C, let y t ty 1 − tx We have fy t , x ≤ 0 So, from A1, we

have

0 fy t , y t ≤ tfy t , y 1 − tfy t , x ≤ tfy t , y. 3.31

Dividing by t, we have

Letting t → 0, from A3, we have

Therefore, x ∈ EPf Finally, we prove that x Π FT∩EPf x0 FromLemma 2.1, we have

φ

x, Π FT∩EPf x0

 φΠFT∩EPf x0, x0

≤ φx, x0. 3.34

Since x n1 ΠC n∩Qn x0and x ∈ FT ∩ EPf ⊂ C n ∩ Q n , for all n≥ 0, we get fromLemma 2.1 that

φ

ΠFT∩EPf x0, x n1

 φx n1 , x0 ≤ φΠFT∩EPf x0, x0

. 3.35

By the definition of φx, y, it follows that φx, x0 ≤ φΠ FT∩EPf x0, x0 and φx, x0 ≥

φΠ FT∩EPf x0, x0, whence φx, x0 φΠ FT∩EPf x0, x0 Therefore, it follows from the uniqueness ofΠFT∩EPf x0that x Π FT∩EPf x0 This completes the proof

Corollary 3.2 Let E be a uniformly convex and uniformly smooth real Banach space, let C be a

nonempty closed convex subset of E, and let f be a bifunction from C × C to R satisfying (A1)–(A4) Define a sequence {x n } in C by the following:

u n ∈ C such that fu n , y  1

r n n , Ju n − Jx n ≥ 0, ∀y ∈ C,

C0  {z ∈ C : φz, u0 ≤ φz, x0},

Q n {z ∈ C n−1 ∩ Q n−1: n − z, Jx0− Jx n ≥ 0},

Q0 C,

3.36

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for every n ∈ N ∪ {0}, where J is the duality mapping on E and {r n } ⊂ a, ∞ for some a > 0 Then, {x n } converges strongly to ΠEPfx0.

Proof Putting T I inTheorem 3.1, we obtainCorollary 3.2

nonempty closed convex subset of E, and let T : C → C be a closed hemirelatively nonexpansive mapping Define a sequence {x n } in C by the following:

u n ΠC y n ,

C n {z ∈ C n−1 ∩ Q n−1:Φz, u n ≤ φz, x n },

C0  {z ∈ C : φz, u0 ≤ φz, x0},

Q n {z ∈ C n−1 ∩ Q n−1: n − z, Jx0− Jx n ≥ 0},

Q0 C,

3.37

for every n ∈ N ∪ {0}, where J is the duality mapping on E, {α n }, {β n } are sequences in 0, 1 such

that lim inf n → ∞ 1 − α n β n 1 − β n > 0 Then, {x n } converges strongly to Π FT x0.

Proof Putting f x, y 0 for all x, y ∈ C and r n 1 for all n inTheorem 3.1, we obtain Corollary 3.3

nonempty closed convex subset of E, let f be a bifunction from C × C to R satisfying (A1)–(A4), and let T : C → C be a closed relatively nonexpansive mapping such that FT ∩ EPf / ∅ Define a sequence {x n } in C by the following:

u n ∈ C such that fu n , y  1

r n n , Ju n − Jy n ≥ 0, ∀y ∈ C,

C0  {z ∈ C : φz, u0 ≤ φz, x0},

Q n {z ∈ C n−1 ∩ Q n−1: n − z, Jx0− Jx n ≥ 0},

Q0 C,

3.38

space and. .. satisfying (A1)–(A4) Then, for r > and x ∈ E, and q ∈ FT r ,

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Lemma 2.9 see... class="text_page_counter">Trang 6

Hence, we have

Next, we show that FT ∩ EPf ⊂ Q n for all n ≥ We prove this by induction

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