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,
1M. Derafshpour,
1and N. Shahzad
21Department 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 :A∪B → A∪Bis a map satisfyingTA⊆B,TB⊆Aand there existsk ∈0,1such thatdTx, Ty≤kdx, yfor allx∈Aandy∈B. Then,Thas a unique fixed point inA∩B.
LetAandB be nonempty closed subsets of a metric spaceX, dand ϕ : 0,∞ →
0,∞a strictly increasing map. We say thatT :A∪B → A∪Bis a cyclicϕ-contraction map
2wheneverTA⊆B,TB⊆Aand
dTx, Ty≤dx, y−ϕdx, yϕdA, B 1.1
for allx∈Aandy∈B, wheredA, B:inf{dx, y:x∈A, y∈B}. Also,x∈A∪Bis 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
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 :A∪B → A∪Ba cyclicϕ-contraction map,x0 ∈A∪B
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
M−ϕdTx0, T2n0x0
ϕdx0, T2n0−1x0
2ϕdA, B.
SincedTx, Ty≤dx, yfor allx∈Aandy∈B, 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 :A∪B → A∪Ba cyclicϕ-contraction map,x0 ∈A∪B, 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
i1x0≤dT2x0, 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
nmn1kmm−k11. 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}
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, Ty≤dx, yfor allx∈Aandy∈B, 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,kmm ≤μ 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
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:A∪B → A∪Ba cyclicϕ-contraction map. Then there existsx, y∈A×Bsuch that
x−ydA, 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 thatx2nk →w x∈A. As{x2nk1}is bounded andBis weakly closed,
we can say, without loss of generality, thatx2nk1 →w y∈Bask → ∞. Sincex2nk−x2nk1 →w
x−y /0 ask → ∞, there exists a bounded linear functionalf:X → 0,∞such thatf1 andfx−y x−y. For eachk≥1, we have
fx2nk−x2nk1 ≤fx2nk−x2nk1x2nk−x2nk1. 2.17
Since limk→ ∞fx2nk−x2nk1 fx−y x−y, by using2, Theorem 3we obtain
x−y lim
k→ ∞
fx2nk −x2nk1 ≤ lim
k→ ∞x2nk−x2nk1x2nk−x2nk1dA, B. 2.18
Hence,x−ydA, B.
Definition 2.4. see2LetAandBbe nonempty subsets of a normed spaceX,T :A∪B → A∪B,TA⊆B, andTB⊆A. We say thatT satisfies the proximal property if
xn−→w x∈A∪B, xn−Txn −→dA, B ⇒ x−TxdA, 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 :A∪B → A∪Ba cyclic
ϕ-contraction map. Then, there existsx∈Asuch thatx−TxdA, 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
thatx2nk →w x∈Aask → ∞.
aSinceT is weakly continuous onAandTA ⊆ B, we havex2nk1 →w Tx ∈ Bas
k → ∞. Sox2nk −x2nk1 →w x−Tx /0 ask → ∞. The rest of the proof is similar to
bBy2, Theorem 3, we have
x2nk−Tx2nkx2nk−x2nk1 −→dA, B 2.20
ask → ∞. SinceT satisfies the proximal property, we havex−TxdA, 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 :A∪B → A∪B
a cyclicϕ-contraction map. IfA−A∩B−B {0}, then there exists a uniquex∈Asuch that
T2xxandx−TxdA, 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, y∈A×B
withx−y dA, B. To show the uniqueness ofx, y, suppose that there exists another
x, y ∈ A×Bwith x−y dA, B. SinceA−A∩B−B {0}, we conclude that
x−y /x−y. As bothAandBare convex, by the strict convexity ofX, we have
x2x− yy
2
x−y
2
x−y
2
< dA, B, 2.21
which is a contradiction. SinceTy−TxTx−Tyx−ydA, 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 :
A∪B → A∪Ba cyclicϕ-contraction map. Then, there exists a uniquex∈Asuch thatT2xxand x−TxdA, 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 existsx∈Awith
x−Tx dA, B. Thus,T2x x. Indeed, if we assume thatT2x−Tx /x−Tx. Then from
the convexity ofAand the strict convexity ofX, we have
T
2xx
2 −Tx
T
2x−Tx
2
x−Tx
2
< dA, B, 2.22
which is a contradiction. The uniqueness ofxfollows as in the proof of2, Theorem 8.
Acknowledgment
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.