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Fixed Point Theory and Applications Volume 2010, Article ID 946178,7pages doi:10.1155/2010/946178

Research Article

Best Proximity Points of Cyclic

ϕ

-Contractions on

Reflexive Banach Spaces

Sh. Rezapour,

1

M. Derafshpour,

1

and N. Shahzad

2

1Department of Mathematics, Azarbaidjan University of Tarbiat Moallem, Azarshahr, Tabriz, Iran

2Department of Mathematics, King AbdulAziz University, P.O. Box 80203, Jeddah 21859, Saudi Arabia

Correspondence should be addressed to N. Shahzad,nshahzad@kau.edu.sa

Received 8 July 2009; Revised 4 January 2010; Accepted 12 January 2010

Academic Editor: Juan Jose Nieto

Copyrightq2010 Sh. Rezapour 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 provide a positive answer to a question raised by Al-Thagafi and ShahzadNonlinear Analysis, 702009, 3665-3671 about best proximity points of cyclic ϕ-contractions on reflexive Banach spaces.

1. Introduction

As a generalization of Banach contraction principle, Kirk et al. proved, in 2003, the following fixed point result; see1.

Theorem 1.1. LetAandBbe nonempty closed subsets of a complete metric spaceX, d. Suppose thatT :ABABis a map satisfyingTAB,TBAand there existsk ∈0,1such thatdTx, Tykdx, yfor allxAandyB. Then,Thas a unique fixed point inAB.

LetAandB be nonempty closed subsets of a metric spaceX, dand ϕ : 0,∞ →

0,∞a strictly increasing map. We say thatT :ABABis a cyclicϕ-contraction map

2wheneverTAB,TBAand

dTx, Tydx, yϕdx, yϕdA, B 1.1

for allxAandyB, wheredA, B:inf{dx, y:xA, yB}. Also,xABis called a best proximity point ifdx, Tx dA, B. As a special case, whenϕt 1−αt, in which

α∈0,1is a constant,T is called cyclic contraction.

In 2005, Petrusel proved some periodic point results for cyclic contraction maps

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contraction maps in 20064. They raised a question about the existence of a best proximity point for a cyclic contraction map in a reflexive Banach space. In 2009, Al-Thagafi and Shahzad gave a positive answer to the question2. More precisely, they proved some results on the existence and convergence of best proximity points of cyclic contraction maps defined on reflexiveand strictly convexBanach spaces2, Theorems 9, 10, 11, and 12. They also introduced cyclicϕ-contraction maps and raised the following question in2.

Question 1. It is interesting to ask whether Theorems 9 and 10 resp., Theorems 11 and 12 hold for cyclic ϕ-contraction maps where the space is only reflexive resp., reflexive and strictly convexBanach space.

In this paper, we provide a positive answer to the above question. For obtaining the answer, we use some results of2.

2. Main Results

First, we give the following extension of4, Proposition 3.3for cyclicϕ-contraction maps, whereϕis unbounded.

Theorem 2.1. Letϕ:0,∞ → 0,be a strictly increasing unbounded map. Also, letAandBbe nonempty subsets of a metric spaceX, d,T :ABABa cyclicϕ-contraction map,x0 ∈AB

andxn1Txnfor alln≥0. Then, the sequences{x2n}and{x2n1}are bounded.

Proof. Suppose that x0 ∈ A the proof when x0 ∈ B is similar. By 2, Theorem 3,

dx2n, x2n1 → dA, B. Hence, it is sufficient to prove that{x2n1}is bounded. Sinceϕis

unbounded, there existsM >0 such that

ϕM> dx0, Tx0 ϕdA, B. 2.1

If{x2n1}is not bounded, then there exists a natural numbern0such that

dT2x0, T2n01x0

> M, dT2x0, T2n0−1x0

M. 2.2

Then, we have

M < dT2x0, T2n01x0

dTx0, T2n0x0

ϕdTx0, T2n0x0

ϕdA, B

dx0, T2n0−1x0

ϕdTx0, T2n0x0

ϕdx0, T2n0−1x0

2ϕdA, B

dx0, T2x0

dT2x0, T2n0−1x0

ϕdTx0, T2n0x0

ϕdx0, T2n0−1x0

2ϕdA, B

dx0, Tx0 d

Tx0, T2x0

MϕdTx0, T2n0x0

ϕdx0, T2n0−1x0

2ϕdA, B.

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SincedTx, Tydx, yfor allxAandyB, we obtain

M <2dx0, Tx0 M

ϕdTx0, T2n0x0

ϕdx0, T2n0−1x0

2ϕdA, B. 2.4

SincedTx0, T2n0x0≤dx0, T2n0−1x0, we have

ϕdTx0, T2n0x0

ϕdx0, T2n0−1x0

. 2.5

Thus, we obtainϕdTx0, T2n0x0 < dx0, Tx0 ϕdA, B. Since

M < dT2x0, T2n01x0

dTx0, T2n0x0

, 2.6

ϕM < ϕdTx0, T2n0x0. Hence, ϕM < dx0, Tx0 ϕdA, B. This contradiction

completes the proof.

Since the proof of last result was classic, we presented it separately. Here, we provide our key result via a special proof which is a general case ofTheorem 2.1.

Theorem 2.2. Letϕ:0,∞ → 0,be a strictly increasing map. Also, letAandBbe nonempty subsets of a metric spaceX, d,T :ABABa cyclicϕ-contraction map,x0 ∈AB, and

xn1Txnfor alln≥0. Then, the sequences{x2n}and{x2n1}are bounded.

Proof. Suppose that x0 ∈ A the proof when x0 ∈ B is similar. By 2, Theorem 3,

dx2n, x2n1 → dA, B. Hence, either{x2n1}and{x2n}are bounded or both are unbounded.

Suppose that both sequences are unbounded. Fixn1∈Nand define

en,kd

T2nx0, T2n1k1x0

2.7

for all n, k ≥ 1. Since {x2n1} is unbounded, lim supk→ ∞en,k ∞ for all n ≥ 1. Thus,

we can choose a strictly increasing subsequence {e1,k1

i}i≥1 of the sequence{e1,k}k≥1. Since

dT2x

0, T2n1k

1

i1x0dT2x0, T4x0 dT4x0, T2n1k1i1x0, we have

lim sup

i→ ∞ e2,k

1

i. 2.8

Again, we can choose a strictly increasing subsequence {e2,k2

i}i≥1 of the sequence {e2,ki1}i≥1

such that lim supi→ ∞e2,k2

i ∞. By continuing this process, for each natural number n,

we can choose a strictly increasing subsequence {en,kn

i}i≥1 of the sequence{en,kin−1}i≥1 such

that lim supi→ ∞en,kn

i ∞. By the construction, if we consider the sequence {k i

i}i≥1, then

limi→ ∞kii∞,{en,ki

i}i≥1is a strictly increasing subsequence of{en,kin}i≥1and lim supi→ ∞en,ki i ∞for alln≥1. Now, definen2n1k22−k11. Also, by induction define the sequence{nm}m≥1by

nmn1kmmk11. Note that, the sequence{nm}m≥1is strictly increasing and lim supm→ ∞nm ∞. Since T is a cyclic ϕ-contraction map, {dT2nmx

0, T2nmk

1 11x0}

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sequence. Hence by the construction of the sequence{nm}m≥1,{dT2nmx0, T2n1k m m1x

0}m≥1

is a decreasing sequence. Letm≥1 be given. Sinceenm,k1 1≤enm,k

m

m, we have

dT2nmx0, T2n1k

1 11x0

dT2nmx0, T2n1k m m1x

0

. 2.9

Thus,

dT2nmx

0, T2n1k

1 11x0

dT2n1x

0, T2n1k

1 11x0

2.10

for allm≥1. Hence, we have

dT2n1kmm1x0, T2n1k111x

0

dT2nmx0, T2n1k

1 11x0

dT2nmx0, T2n1k m m1x0

dT2nmx0, T2n1k

1 11x0

dT2nmx0, T2nmk

1 11x0

dT2n1x

0, T2n1k

1 11x0

dT2nmx0, T2nmk

1 11x0

2.11

for allm≥1. SincedTx, Tydx, yfor allxAandyB, we obtain

dT2n1kmm1x

0, T2n1k

1 11x0

dT2n1x

0, T2n1k

1 11x0

dT2nm−1x0, T2nmk

1 1x0

dT2n1x

0, T2n1k

1 11x0

dx0, T2k

1 11x0

2.12

for allm≥1. Consequently

dT2n1kmm1x

0, T2n1k

1 1x0

dT2n1kmm1x

0, T2n1k

1 11x0

dT2n1k111x0, T2n1k11x0

dT2n1x

0, T2n1k

1 11x0

dx0, T2k

1 11x0

dT2n1k111x0, T2n1k11x0

2.13

for allm≥1. This implies that

en1k1

1,kmmμ 2.14

for allm≥1, where

μdT2n1x

0, T2n1k

1 11x0

dx0, T2k

1 11x0

dT2n1k111x0, T2n1k11x0

2.15

is a constant. But, lim supi→ ∞en,ki

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Now by using this key result, we provide our main results which give positive answer to the question. Their proofs are basically due to Al-Thagafi and Shahzad2. However, the crucial role is played by our key result. Weak convergence of{xn}toxis denoted byxn−→w x.

Theorem 2.3. Letϕ:0,∞ → 0,be a strictly increasing map. Also, letAandBbe nonempty weakly closed subsets of a reflexive Banach space andT:ABABa cyclicϕ-contraction map. Then there existsx, yA×Bsuch that

xydA, B. 2.16

Proof. IfdA, B 0, the result follows from2, Theorem 1. So, we assume thatdA, B >

0. For x0 ∈ A, definexn1 Txn for all n ≥ 1. By Theorem 2.2, the sequences {x2n}and {x2n1}are bounded. SinceX is reflexive andAis weakly closed, the sequence{x2n}has a

subsequence{x2nk}such thatx2nkw xA. As{x2nk1}is bounded andBis weakly closed,

we can say, without loss of generality, thatx2nk1 →w yBask → ∞. Sincex2nkx2nk1 →w

xy /0 ask → ∞, there exists a bounded linear functionalf:X → 0,∞such thatf1 andfxy xy. For eachk≥1, we have

fx2nkx2nk1 fx2nkx2nk1x2nkx2nk1. 2.17

Since limk→ ∞fx2nkx2nk1 fxy xy, by using2, Theorem 3we obtain

xy lim

k→ ∞

fx2nk x2nk1 lim

k→ ∞x2nkx2nk1x2nkx2nk1dA, B. 2.18

Hence,xydA, B.

Definition 2.4. see2LetAandBbe nonempty subsets of a normed spaceX,T :ABAB,TAB, andTBA. We say thatT satisfies the proximal property if

xn−→w xAB, xnTxn −→dA, BxTxdA, B. 2.19

Theorem 2.5. Letϕ:0,∞ → 0,be a strictly increasing map. Also, letAandBbe nonempty subsets of a reflexive Banach spaceX such thatAis weakly closed andT :ABABa cyclic

ϕ-contraction map. Then, there existsxAsuch thatxTxdA, Bprovided that one of the following conditions is satisfied

aT is weakly continuous onA.

bT satisfies the proximal property.

Proof. IfdA, B 0, the result follows from2, Theorem 1. So, we assume thatdA, B>0. Forx0 ∈A, definexn1 Txn for alln≥1. ByTheorem 2.2, the sequence{x2n}is bounded.

SinceXis reflexive andAis weakly closed, the sequence{x2n}has a subsequence{x2nk}such

thatx2nkw xAask → ∞.

aSinceT is weakly continuous onAandTAB, we havex2nk1 →w TxBas

k → ∞. Sox2nkx2nk1 →w xTx /0 ask → ∞. The rest of the proof is similar to

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bBy2, Theorem 3, we have

x2nkTx2nkx2nkx2nk1 −→dA, B 2.20

ask → ∞. SinceT satisfies the proximal property, we havexTxdA, B.

Theorem 2.6. Letϕ:0,∞ → 0,be a strictly increasing map. Also, letAandBbe nonempty closed and convex subsets of a reflexive and strictly convex Banach space andT :ABAB

a cyclicϕ-contraction map. IfAABB {0}, then there exists a uniquexAsuch that

T2xxandxTxdA, B.

Proof. IfdA, B 0, the result follows from2, Theorem 1. So, we assume thatdA, B>0. SinceAis closed and convex, it is weakly closed. ByTheorem 2.3, there existsx, yA×B

withxy dA, B. To show the uniqueness ofx, y, suppose that there exists another

x, yA×Bwith xy dA, B. SinceAABB {0}, we conclude that

xy /xy. As bothAandBare convex, by the strict convexity ofX, we have

x2xyy

2

xy

2

xy

2

< dA, B, 2.21

which is a contradiction. SinceTyTxTxTyxydA, B, we obtain, from the uniqueness ofx, y, thatTy, Tx x, y. HenceTxy,TyxandT2xx.

Theorem 2.7. Letϕ:0,∞ → 0,be a strictly increasing map. Also, letAandBbe nonempty subsets of a reflexive and strictly convex Banach spaceX such thatAis closed and convex andT :

ABABa cyclicϕ-contraction map. Then, there exists a uniquexAsuch thatT2xxand xTxdA, Bprovided that one of the following conditions is satisfied

aT is weakly continuous onA.

bT satisfies the proximal property.

Proof. IfdA, B 0, the result follows from2, Theorem 1. So, we assume thatdA, B>0. SinceAis closed and convex, it is weakly closed. ByTheorem 2.5that there existsxAwith

xTx dA, B. Thus,T2x x. Indeed, if we assume thatT2xTx /xTx. Then from

the convexity ofAand the strict convexity ofX, we have

T

2xx

2 −Tx

T

2xTx

2

xTx

2

< dA, B, 2.22

which is a contradiction. The uniqueness ofxfollows as in the proof of2, Theorem 8.

Acknowledgment

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References

1 W. A. Kirk, P. S. Srinivasan, and P. Veeramani, “Fixed points for mappings satisfying cyclical contractive conditions,”Fixed Point Theory, vol. 4, no. 1, pp. 79–89, 2003.

2 M. A. Al-Thagafi and N. Shahzad, “Convergence and existence results for best proximity points,”

Nonlinear Analysis. Theory, Methods & Applications, vol. 70, no. 10, pp. 3665–3671, 2009.

3 G. Petrus¸el, “Cyclic representations and periodic points,”Universitatis Babes¸-Bolyai. Studia. Mathemat-ica, vol. 50, no. 3, pp. 107–112, 2005.

References

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