A Fixed Point Result for Boyd Wong Cyclic Contractions in Partial Metric Spaces

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Volume 2012, Article ID 597074,11pages doi:10.1155/2012/597074

Research Article

A Fixed Point Result for Boyd-Wong Cyclic

Contractions in Partial Metric Spaces

Hassen Aydi

1

_{and Erdal Karapinar}

2

1_{Institut Sup´erieur d’Informatique et des Technologies de Communication de Hammam Sousse,}

Universit´e de Sousse, Route GP1-4011, Hammam Sousse, 4002 Sousse, Tunisia

2_{Department of Mathematics, Atilim University, 06836 ˙Incek, Turkey}

Correspondence should be addressed to Hassen Aydi,[email protected]

Received 31 March 2012; Accepted 31 May 2012

Academic Editor: Billy Rhoades

Copyrightq2012 H. Aydi and E. Karapinar. 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.

A fixed point theorem involving Boyd-Wong-type cyclic contractions in partial metric spaces is proved. We also provide examples to support the concepts and results presented herein.

1. Introduction and Preliminaries

Partial metric spaces were introduced by Matthews1to the study of denotational semantics of data networks. In particular, he proved a partial metric version of the Banach contraction principle2. Subsequently, many fixed points results in partial metric spaces appearedsee,

e.g.,1,3–19for more details.

Throughout this paper, the lettersRandN∗ will denote the sets of all real numbers and positive integers, respectively. We recall some basic definitions and fixed point results of partial metric spaces.

Definition 1.1. A partial metric on a nonempty setXis a functionp :X×X → 0,∞such that for allx, y, z∈X

p1xy⇔px, x px, y py, y,

p2px, x≤px, y,

p3px, y py, x,

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A partial metric space is a pairX, psuch thatX is a nonempty set andpis a partial metric onX.

Ifpis a partial metric onX, then the functiondp:X×X → 0,∞given by

dpx, y2px, y−px, x−py, y 1.1

is a metric onX.

Example 1.2see, e.g., 1,3,11,12. ConsiderX 0,∞with px, y max{x, y}. Then, X, pis a partial metric space. It is clear thatpis not ausualmetric. Note that in this case

dpx, y |x−y|.

Example 1.3see, e.g., 1. LetX {a, b : a, b,∈ R, a ≤ b}, and definepa, b,c, d

max{b, d} −min{a, c}. Then,X, pis a partial metric space.

Example 1.4see, e.g.,1,20. LetX: 0,1∪2,3, and definep:X×X → 0,∞by

px, y

maxx, y ifx, y∩2,3/∅,

_x₋_y _if

x, y⊂0,1. 1.2

Then,X, pis a complete partial metric space.

Each partial metricponX generates aT0 topologyτponX, which has as a base the family of openp-balls{Bpx, ε, x∈X, ε >0}, whereBpx, ε {y∈X:px, y< px, x ε}

for allx∈Xandε >0.

Definition 1.5. LetX, pbe a partial metric space and{xn}a sequence inX. Then,

i{xn}converges to a pointx∈Xif and only ifpx, x limn→∞px, xn,

ii{xn}is called a Cauchy sequence if limn,m→∞pxn, xmexists and is finite.

Definition 1.6. A partial metric space X, p is said to be complete if every Cauchy sequence {xn} in X converges, with respect to τp, to a point x ∈ X, such that px, x

limn,m→∞pxn, xm.

Lemma 1.7see, e.g.,3,11,12. LetX, pbe a partial metric space. Then,

a{xn}is a Cauchy sequence inX, pif and only if it is a Cauchy sequence in the metric spaceX, dp,

b X, p is complete if and only if the metric space X, dp is complete. Furthermore,

limn→∞dpxn, x 0 if and only if

px, x lim

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Lemma 1.8see, e.g.,3,11,12. LetX, pbe a partial metric space. Then,

aifpx, y 0, thenxy,

bif_{x /}y, thenpx, y>0.

Remark 1.9. Ifxy,px, ymay not be 0.

Lemma 1.10see, e.g.,3,11,12. Letxn → zasn → ∞in a partial metric spaceX, pwhere

pz, z 0. Then, limn→ ∞pxn, y pz, yfor everyy∈X.

LetΦbe the set of functionsφ:0,∞ → 0,∞such that

iφis upper semicontinuousi.e., for any sequence{tn}in0,∞such thattn → tas

n → ∞, we have lim sup_n_{→ ∞}φtn≤φt,

iiφt< tfor eacht >0.

Recently, Romaguera21obtained the following fixed point theorem of Boyd-Wong type22.

Theorem 1.11. LetX, pbe a complete partial metric space, and letT :X → Xbe a map such that

for allx, y∈X

pTx, Ty≤φMx, y, 1.4

whereφ∈Φand

Mx, ymax

px, y, px, Tx, py, Ty,1

2 p

x, Typy, Tx. 1.5

Then,T has a unique fixed point.

In 2003, Kirk et al.23introduced the following definition.

Definition 1.12see23. LetX be a nonempty set,ma positive integer, andT : X → X a mapping.X m_i1Aiis said to be a cyclic representation ofXwith respect toTif

iAi, i1,2, . . . , mare nonempty closed sets,

iiTA1⊂A2, . . . , TAm−1⊂Am, TAm⊂A1.

Recently, fixed point theorems involving a cyclic representation ofXwith respect to a self-mappingThave appeared in many paperssee, e.g.,24–28.

Very recently, Abbas et al.24extendedTheorem 1.11to a class of cyclic mappings and proved the following result, but withφ∈Φbeing a continuous map.

Theorem 1.13. Let X, p be a complete partial metric space. Let m be a positive integer,

A1, A2, . . . , Amnonempty closed subsets ofX, dp, andY m_i1Ai. LetT :Y → Y be a mapping

such that

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iithere existsφ : 0,∞ → 0,∞such thatφis continuous andφt < tfor eacht >0,

satisfying

pTx, Ty≤φMx, y, 1.6

for anyx∈ Ai,y ∈Ai1,i 1,2, . . . , m, whereAm1 A1 andMx, yis defined by

1.5.

Then,T has a unique fixed pointz∈m_i1Ai.

In the following example,φ∈Φ, but it is not continuous.

Example 1.14. Defineφ : 0,∞ → 0,∞byφt t/2 for allt ∈ 0,1 andφt nn

1/n2fort∈n, n1,n∈N∗. Then,φis upper semicontinuous on0,∞withφt< t

for allt >0. However, it is not continuous attnfor alln∈N.

Following Example 1.14, the main aim of this paper is to present the analog of

Theorem 1.13for a weaker hypothesis onφ, that is, withφ ∈ Φ. Our proof is simpler than

that in24. Also, some examples are given.

2. Main Results

Our main result is the following.

Theorem 2.1. Let X, p be a complete partial metric space. Let m be a positive integer,

A1, A2, . . . , Amnonempty closed subsets ofX, dp, andY m_i1Ai. LetT :Y → Y be a mapping

such that

1Y m_i1Aiis a cyclic representation ofY with respect toT,

2there existsφ∈Φsuch that

pTx, Ty≤φMx, y, 2.1

for anyx∈ Ai,y ∈Ai1,i 1,2, . . . , m, whereAm1 A1 andMx, yis defined by

1.5.

Then,T has a unique fixed pointz∈m_i1Ai.

Proof. Letx0 ∈ Y m_i1Ai. Consider the Picard iteration{xn}given byTxn xn1forn

0,1,2, . . . .If there existsn0such thatxn01 xn0, thenxn01 Txn0 xn0and the existence of

the fixed point is proved.

Assume thatxn/xn1, for eachn ≥ 0. Having in mind thatY m_i1Ai, so for each

n≥0, there existsin ∈ {1,2, . . . , m}such thatxn∈Ainandxn1Txn∈TAin⊆Ain1. Then,

by2.1

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where

Mxn, xn1 max

pxn, xn1, pxn, Txn, pxn1, Txn1,

pxn, Txn1 pxn1, Txn

2

max

pxn, xn1, pxn1, xn2,pxn, xn2 pxn1, xn1

2

max

pxn, xn1, pxn1, xn2,pxn, xn1 pxn1, xn2

2

maxpxn, xn1, pxn1, xn2.

2.3

Therefore,

Mxn, xn1 maxpxn, xn1, pxn1, xn2 ∀n≥0. 2.4

If for somek∈N, we haveMxk, xk1 pxk1, xk2, so by2.2

0< pxk1, xk2≤φpxk1, xk2< pxk1, xk2, 2.5

which is a contradiction. It follows that

Mxn, xn1 pxn, xn1 ∀n≥0. 2.6

Thus, from2.2, we get that

0< pxn1, xn2≤φpxn1, xn2< pxn1, xn2. 2.7

Hence,{pxn, xn1}is a decreasing sequence of positive real numbers. Consequently, there existsγ≥0 such that limn→ ∞pxn, xn1 γ. Assume thatγ >0. Lettingn → ∞in the above

inequality, we get using the upper semicontinuity ofφ

0< γ≤lim sup

n→ ∞ φpxn1, xn2≤φ

γ< γ, _2.8

which is a contradiction, so thatγ 0, that is,

lim

n→ ∞pxn, xn1 0. 2.9

By1.1, we havedpx, y≤2px, yfor allx, y∈X, and then from2.9

lim

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Also, byp2,

lim

n→ ∞pxn, xn 0. 2.11

In the sequel, we will prove that{xn}is a Cauchy sequence in the partial metric spaceY

_m

i1Ai, p. ByLemma 1.7, it suﬃces to prove that{xn}is Cauchy sequence in the metric space

Y, dp. We argue by contradiction. Assume that {xn}is not a Cauchy sequence in Y, dp.

Then, there existsε >0 for which we can find subsequences{xmk}and{xnk}of{xn}with

nk> mk≥ksuch that

dpxnk, xmk≥ε. 2.12

Further, corresponding to mk, we can choosenk in such a way that it is the smallest integer withnk> mkand satisfying2.12. Then,

dpxnk−1, xmk< ε. 2.13

We use2.13and the triangular inequality

ε≤dpxnk, xmk≤dpxnk, xnk−1

dpxnk−1, xmk

< εdpxnk, xnk−1

. 2.14

Lettingk → ∞in2.14and using2.10, we find

lim

k→ ∞dp

xnk, xmkε. _2.15

On the other hand

dpxnk, xmk≤dpxnk, xnk1dpxnk1, xmk1dpxmk1, xmk,

dpxnk1, xmk1≤dpxnk1, xnkdpxnk, xmkdpxmk, xmk1. 2.16

Lettingk → ∞in the two above inequalities and using2.10and2.15,

lim

k→ ∞dp

xnk1, xmk1ε. _2.17

Similarly, we have

lim

k→ ∞dp

xnk, xmk1 lim

k→∞dp

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Also, by1.1,2.11, and2.15–2.18, we may find

lim

k→ ∞p

xnk, xmk lim

k→ ∞p

xnk, xmk1

2,

lim

k→ ∞p

xnk1, xmk1 lim

k→ ∞p

xmk, xnk1

2.

2.19

On the other hand, for allk, there existsjk, 0≤jk≤p, such thatnk−mk jk≡1p.

Then,xmk−jkforklarge enough,mk> jkandxnklie in diﬀerent adjacently labeled

setsAiandAi1for certaini1, . . . , p. Using the contractive condition2.1, we get

pxnk1, xmk−jk1pTxnk, Txmk−jk

≤φMxnk, xmk−jk,

2.20

where

Mxnk, xmk−jk max

pxnk, xmk−jk, pxnk, Txnk, pxmk−jk, Txmk−jk,

pxnk, Txmk−jkpxmk−jk, Txnk

2

max

pxnk, xmk−jk, pxnk, xnk1, pxmk−jk, xmk−jk1,

pxnk, xmk−jk1pxmk−jk, xnk1

2

.

2.21

As2.19, using2.9, we may get

lim

k→ ∞p

xnk, xmk−jk lim

k→ ∞p

xnk1, xmk−jk1 ₂, 2.22

lim

k→ ∞p

xnk, xmk−jk1 lim

k→ ∞p

xnk1, xmk−jk ₂. 2.23

By2.22and2.23, we get that

lim

k→ ∞M

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Lettingn → ∞in2.20, we get using2.22,2.24, and the upper semicontinuity ofφ

0<

2 ≤lim sup_k_{→ ∞} φ

Mxnk, xmk−jk≤φ

2

<

2, 2.25

which is a contradiction.

This shows that {xn} is a Cauchy sequence in the complete subspace Y m_i1Ai

equipped with the metric dp. Thus, there existsu limn→ ∞xn ∈ Y, dp. Notice that the

sequence{xn}_n_∈_Nhas an infinite number of terms in eachAi,i 1, . . . , m, so sinceY, dpis complete, from eachAi,i1, . . . , mone can extract a subsequence of{xn}that converges to

u. BecauseAi,i1, . . . , mare closed inY, dp, it follows that

u∈

m

i1

Ai. 2.26

Thus,m_i1Ai/∅.

For simplicity, setA m_i1Ai. Clearly,Ais also closed inY, dp, so it is a complete

subspace ofY, dpand thenA, pis a complete partial metric space. Consider the restriction ofT onA, that is,T/A : A → A. Then,T/A satisfies the assumptions ofTheorem 1.11, and thusT/Ahas a unique fixed point inZ.

3. Examples

We give some examples illustrating our results.

Example 3.1. LetX Rand px, y max{|x|,|y|}. It is obvious thatX, pis a complete partial metric space.

SetA1 −8,0,A2 0,8, andY A1∪A2 −8,8. DefineT :T → Yby

Tx

⎧ ⎨ ⎩

−x

4 if x∈−1,1,

0 otherwise. 3.1

Notice that T−8,−1 0 and T−1,0 0,1/4, and hence TA1 ⊆ A2. Analogously,T1,8 0 andT0,1 −1/4,0, and henceTA2⊆A1.

Take

φt

⎧ ⎪ ⎨ ⎪ ⎩ t

3 if t∈0,1,

n2

n2₁ if t∈n, n1, n∈N∗.

3.2

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Case 1.x∈−8,−1andy∈1,8. Inequality2.1turns into

pTx, Tymax{0,0}0≤φMx, y, 3.3

which is necessarily true.

Case 2.x∈−8,−1andy∈0,1. Inequality2.1becomes

pTx, Ty max

0,y

4

y 4 ≤φ

Mx, y

φ

max

px, y, px, Tx, pTy, y,1

2 p

x, TypTx, y

φ

max

|x|,|x|,y,1

2 |x|y

φ|x|.

3.4

It is clear that 1/2≤φt<1 for allt >1. Hence,3.4holds.

Case 3.x∈−1,0andy∈1,8. Inequality2.1turns into

pTx, Ty max

_|x|

4 ,0

|x|

4 ≤φ

Mx, y

φ

max

2 p

x, TypTx, y

φ

maxy,|x|,y,1

2 |x|y

φyφy,

3.5

which is true again by the fact that 1/2≤φt<1 for allt >1.

Case 4.x∈−1,0andy∈0,1. Inequality2.1becomes

pTx, Ty max

|x| 4 , _y 4

≤φMx, y

φ

max

2 p

x, TypTx, y

φ

max

max|x|,y,|x|,y,1

2

max

_|x|

4 ,y

max

|x|,y

4

.

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Let use examine all possibilities:

pTx, Ty

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ |x|

4 if|x| ≥y,

_y

4 if

_y

4 ≤ |x| ≤y,

_y

4 if|x| ≤

_y

4 ,

Mx, y≤

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

|x| if |x| ≥y,

_y if y

4 ≤ |x| ≤y,

_y if |x| ≤ y

4 .

3.7

Thus,2.1holds forφt t/3.

The rest of the assumptions ofTheorem 2.1are also satisfied. The functionT has 0 as a unique fixed point.

However, sinceφis not a continuous function, we could not applyTheorem 1.13.

Example 3.2. LetX 0,1andpx, y max{x, y}for allx, y∈X. Then,X, pis a complete partial metric space. TakeA1 · · · Ap X. DefineT : X → X byTx x/2. Consider

φ:0,∞ → 0,∞given byExample 1.14.

For allx, y∈X, we have

pTx, Tymaxx

2,

y

2

φpx, y ≤φMx, y. 3.8

Then all the assumptions ofTheorem 2.1are satisfied. The functionT has 0 as a unique fixed point.

Similarly,Theorem 1.13is not applicable.

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