Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2012, Article ID 597074, 11 pages
doi:10.1155/2012/597074
Research Article
A Fixed Point Result for Boyd-Wong Cyclic
Contractions in Partial Metric Spaces
Hassen Aydi1 and Erdal Karapinar2
1
Institut Supérieur d’Informatique et des Technologies de Communication de Hammam Sousse,
Université de Sousse, Route GP1-4011, Hammam Sousse, 4002 Sousse, Tunisia
2
Department of Mathematics, Atilim University, 06836 İncek, Turkey
Correspondence should be addressed to Hassen Aydi, hassen.aydi@isima.rnu.tn
Received 31 March 2012; Accepted 31 May 2012
Academic Editor: Billy Rhoades
Copyright q 2012 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 Matthews 1 to the study of denotational semantics
of data networks. In particular, he proved a partial metric version of the Banach contraction
principle 2. Subsequently, many fixed points results in partial metric spaces appeared see,
e.g., 1, 3–19 for more details.
Throughout this paper, the letters R and N∗ 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 set X is a function p : X × X → 0, ∞ such
that for all x, y, z ∈ X
p1 x y ⇔ px, x px, y py, y,
p2 px, x ≤ px, y,
p3 px, y py, x,
p4 px, y ≤ px, z pz, y − pz, z.
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A partial metric space is a pair X, p such that X is a nonempty set and p is a partial metric
on X.
If p is a partial metric on X, then the function dp : X × X → 0, ∞ given by
dp x, y 2p x, y − px, x − p y, y
1.1
is a metric on X.
Example 1.2 see, e.g., 1, 3, 11, 12. Consider X 0, ∞ with px, y max{x, y}. Then,
X, p is a partial metric space. It is clear that p is not a usual metric. Note that in this case
dp x, y |x − y|.
Example 1.3 see, e.g., 1. Let X {a, b : a, b, ∈ R, a ≤ b}, and define pa, b, c, d max{b, d} − min{a, c}. Then, X, p is a partial metric space.
Example 1.4 see, e.g., 1, 20. Let X : 0, 1 ∪ 2, 3, and define p : X × X → 0, ∞ by
p x, y max x, y
x − y
if x, y ∩ 2, 3 / ∅,
if x, y ⊂ 0, 1.
1.2
Then, X, p is a complete partial metric space.
Each partial metric p on X generates a T0 topology τp on X, which has as a base the
family of open p-balls {Bp x, ε, x ∈ X, ε > 0}, where Bp x, ε {y ∈ X : px, y < px, x ε}
for all x ∈ X and ε > 0.
Definition 1.5. Let X, p be a partial metric space and {xn } a sequence in X. Then,
i {xn } converges to a point x ∈ X if and only if px, x limn → ∞ px, xn ,
ii {xn } is called a Cauchy sequence if limn,m → ∞ pxn , xm exists 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 1.7 see, e.g., 3, 11, 12. Let X, p be a partial metric space. Then,
a {xn } is a Cauchy sequence in X, p if and only if it is a Cauchy sequence in the metric
space X, dp ,
b X, p is complete if and only if the metric space X, dp is complete. Furthermore,
limn → ∞ dp xn , x 0 if and only if
px, x lim pxn , x n → ∞
lim pxn , xm .
n,m → ∞
1.3
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3
Lemma 1.8 see, e.g., 3, 11, 12. Let X, p be a partial metric space. Then,
a if px, y 0, then x y,
b if x / y, then px, y > 0.
Remark 1.9. If x y, px, y may not be 0.
Lemma 1.10 see, e.g., 3, 11, 12. Let xn → z as n → ∞ in a partial metric space X, p where
pz, z 0. Then, limn → ∞ pxn , y pz, y for every y ∈ X.
Let Φ be the set of functions φ : 0, ∞ → 0, ∞ such that
i φ is upper semicontinuous i.e., for any sequence {tn } in 0, ∞ such that tn → t as
n → ∞, we have lim supn → ∞ φtn ≤ φt,
ii φt < t for each t > 0.
Recently, Romaguera 21 obtained the following fixed point theorem of Boyd-Wong
type 22.
Theorem 1.11. Let X, p be a complete partial metric space, and let T : X → X be a map such that
for all x, y ∈ X
p T x, T y ≤ φ M x, y ,
1.4
1 .
M x, y max p x, y , px, T x, p y, T y , p x, T y p y, T x
2
1.5
where φ ∈ Φ and
Then, T has a unique fixed point.
In 2003, Kirk et al. 23 introduced the following definition.
Definition 1.12 see 23. Let X be a nonempty set, m a positive integer, and T : X → X a
mapping. X m
i1 Ai is said to be a cyclic representation of X with respect to T if
i Ai , i 1, 2, . . . , m are nonempty closed sets,
ii T A1 ⊂ A2 , . . . , T Am−1 ⊂ Am , T Am ⊂ A1 .
Recently, fixed point theorems involving a cyclic representation of X with respect to a
self-mapping T have appeared in many papers see, e.g., 24–28.
Very recently, Abbas et al. 24 extended Theorem 1.11 to a class of cyclic mappings
and proved the following result, but with φ ∈ Φ being a continuous map.
Theorem 1.13. Let X, p be a complete partial metric space. Let m be a positive integer,
A1 , A2 , . . . , Am nonempty closed subsets of X, dp , and Y m
i1 Ai . Let T : Y → Y be a mapping
such that
i Y m
i1 Ai is a cyclic representation of Y with respect to T ,
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ii there exists φ : 0, ∞ → 0, ∞ such that φ is continuous and φt < t for each t > 0,
satisfying
p T x, T y ≤ φ M x, y ,
1.6
for any x ∈ Ai , y ∈ Ai1 , i 1, 2, . . . , m, where Am1 A1 and Mx, y is defined by
1.5.
Then, T has a unique fixed point z ∈ m
i1 Ai .
In the following example, φ ∈ Φ, but it is not continuous.
Example 1.14. Define φ : 0, ∞ → 0, ∞ by φt t/2 for all t ∈ 0, 1 and φt nn 1/n 2 for t ∈ n, n 1, n ∈ N∗ . Then, φ is upper semicontinuous on 0, ∞ with φt < t
for all t > 0. However, it is not continuous at t n for all n ∈ N.
Following Example 1.14, the main aim of this paper is to present the analog of
Theorem 1.13 for a weaker hypothesis on φ, that is, with φ ∈ Φ. Our proof is simpler than
that in 24. 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 , . . . , Am nonempty closed subsets of X, dp , and Y m
i1 Ai . Let T : Y → Y be a mapping
such that
1 Y m
i1 Ai is a cyclic representation of Y with respect to T ,
2 there exists φ ∈ Φ such that
p T x, T y ≤ φ M x, y ,
2.1
for any x ∈ Ai , y ∈ Ai1 , i 1, 2, . . . , m, where Am1 A1 and Mx, y is defined by
1.5.
Then, T has a unique fixed point z ∈ m
i1 Ai .
Proof. Let x0 ∈ Y m
i1 Ai . Consider the Picard iteration {xn } given by T xn xn1 for n 0, 1, 2, . . . . If there exists n0 such that xn0 1 xn0 , then xn0 1 T xn0 xn0 and the existence of
the fixed point is proved.
xn1 , for each n ≥ 0. Having in mind that Y m
Assume that xn /
i1 Ai , so for each
n ≥ 0, there exists in ∈ {1, 2, . . . , m} such that xn ∈ Ain and xn1 T xn ∈ T Ain ⊆ Ain 1 . Then,
by 2.1
pxn1 , xn2 pT xn , T xn1 ≤ φMxn , xn1 ,
2.2
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5
where
Mxn , xn1 max pxn , xn1 , pxn , T xn , pxn1 , T xn1 ,
pxn , T xn1 pxn1 , T xn 2
pxn , xn2 pxn1 , xn1 max pxn , xn1 , pxn1 , xn2 ,
2
pxn , xn1 pxn1 , xn2 max pxn , xn1 , pxn1 , xn2 ,
2
max pxn , xn1 , pxn1 , xn2 .
2.3
Therefore,
Mxn , xn1 max pxn , xn1 , pxn1 , xn2 ∀n ≥ 0.
2.4
If for some k ∈ N, we have Mxk , xk1 pxk1 , xk2 , so by 2.2
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, from 2.2, 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
exists γ ≥ 0 such that limn → ∞ pxn , xn1 γ. Assume that γ > 0. Letting n → ∞ in the above
inequality, we get using the upper semicontinuity of φ
0 < γ ≤ lim sup φpxn1 , xn2 ≤ φ γ < γ,
n→∞
2.8
which is a contradiction, so that γ 0, that is,
lim pxn , xn1 0.
n→∞
2.9
By 1.1, we have dp x, y ≤ 2px, y for all x, y ∈ X, and then from 2.9
lim dp xn , xn1 0.
n→∞
2.10
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Also, by p2,
lim pxn , xn 0.
n→∞
2.11
In the sequel, we will prove that {xn } is a Cauchy sequence in the partial metric space Y m
i1 Ai , p. By Lemma 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 exists ε > 0 for which we can find subsequences {xmk } and {xnk } of {xn } with
nk > mk ≥ k such that
dp xnk , xmk ≥ ε.
2.12
Further, corresponding to mk, we can choose nk in such a way that it is the smallest
integer with nk > mk and satisfying 2.12. Then,
dp xnk−1 , xmk < ε.
2.13
We use 2.13 and the triangular inequality
ε ≤ dp xnk , xmk ≤ dp xnk , xnk−1 dp xnk−1 , xmk
< ε dp xnk , xnk−1 .
2.14
Letting k → ∞ in 2.14 and using 2.10, we find
lim dp xnk , xmk ε.
k→∞
2.15
On the other hand
dp xnk , xmk ≤ dp xnk , xnk1 dp xnk1 , xmk1 dp xmk1 , xmk ,
dp xnk1 , xmk1 ≤ dp xnk1 , xnk dp xnk , xmk dp xmk , xmk1 .
2.16
Letting k → ∞ in the two above inequalities and using 2.10 and 2.15,
lim dp xnk1 , xmk1 ε.
2.17
lim dp xnk , xmk1 lim dp xmk , xnk1 ε.
2.18
k→∞
Similarly, we have
k→∞
k → ∞
International Journal of Mathematics and Mathematical Sciences
7
Also, by 1.1, 2.11, and 2.15–2.18, we may find
lim p xnk , xmk lim p xnk , xmk1 ,
k→∞
2
lim p xnk1 , xmk1 lim p xmk , xnk1 .
k→∞
k→∞
2
k→∞
2.19
On the other hand, for all k, there exists jk, 0 ≤ jk ≤ p, such that nk − mk jk ≡ 1p.
Then, xmk−jk for k large enough, mk > jk and xnk lie in different adjacently labeled
sets Ai and Ai1 for certain i 1, . . . , p. Using the contractive condition 2.1, we get
p xnk1 , xmk−jk1 p T xnk , T xmk−jk
≤ φ M xnk , xmk−jk ,
2.20
where
M xnk , xmk−jk max p xnk , xmk−jk , p xnk , T xnk , p xmk−jk , T xmk−jk ,
p xnk , T xmk−jk p xmk−jk , T xnk
2
max p xnk , xmk−jk , p xnk , xnk1 , p xmk−jk , xmk−jk1 ,
p xnk , xmk−jk1 p xmk−jk , xnk1
.
2
2.21
As 2.19, using 2.9, we may get
lim p xnk , xmk−jk lim p xnk1 , xmk−jk1 ,
k→∞
k→∞
2
lim p xnk , xmk−jk1 lim p xnk1 , xmk−jk .
k→∞
k→∞
2
2.22
2.23
By 2.22 and 2.23, we get that
lim M xnk , xmk−jk .
k→∞
2
2.24
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International Journal of Mathematics and Mathematical Sciences
Letting n → ∞ in 2.20, we get using 2.22, 2.24, and the upper semicontinuity of φ
0<
≤ lim sup φ M xnk , xmk−jk ≤ φ
< ,
2
2
2
k→∞
2.25
which is a contradiction.
This shows that {xn } is a Cauchy sequence in the complete subspace Y m
i1 Ai
equipped with the metric dp . Thus, there exists u limn → ∞ xn ∈ Y, dp . Notice that the
sequence {xn }n∈N has an infinite number of terms in each Ai , i 1, . . . , m, so since Y, dp is
complete, from each Ai , i 1, . . . , m one can extract a subsequence of {xn } that converges to
u. Because Ai , i 1, . . . , m are closed in Y, dp , it follows that
u∈
m
Ai .
2.26
i1
m
Ai / ∅.
For simplicity, set A m
i1 Ai . Clearly, A is also closed in Y, dp , so it is a complete
subspace of Y, dp and then A, p is a complete partial metric space. Consider the restriction
of T on A, that is, T/A : A → A. Then, T/A satisfies the assumptions of Theorem 1.11, and
thus T/A has a unique fixed point in Z.
Thus,
i1
3. Examples
We give some examples illustrating our results.
Example 3.1. Let X R and px, y max{|x|, |y|}. It is obvious that X, p is a complete
partial metric space.
Set A1 −8, 0, A2 0, 8, and Y A1 ∪ A2 −8, 8. Define T : T → Y by
⎧
⎨ −x
4
Tx ⎩0
if x ∈ −1, 1,
otherwise.
3.1
Notice that T −8, −1 0 and T −1, 0 0, 1/4, and hence T A1 ⊆ A2 .
Analogously, T 1, 8 0 and T 0, 1 −1/4, 0, and hence T A2 ⊆ A1 .
Take
⎧
t
⎪
⎨
φt 3 n2
⎪
⎩
n2 1
if t ∈ 0, 1,
if t ∈ n, n 1, n ∈ N∗ .
Clearly, T satisfies condition 2.1. Indeed, we have the following cases.
3.2
International Journal of Mathematics and Mathematical Sciences
9
Case 1. x ∈ −8, −1 and y ∈ 1, 8. Inequality 2.1 turns into
p T x, T y max{0, 0} 0 ≤ φ M x, y ,
3.3
which is necessarily true.
Case 2. x ∈ −8, −1 and y ∈ 0, 1. Inequality 2.1 becomes
y
y
≤ φ M x, y
p T x, T y max 0,
4
4
1 φ max p x, y , px, T x, p T y, y , p x, T y p T x, y
2
1
φ max |x|, |x|, y, |x| y
2
3.4
φ|x|.
It is clear that 1/2 ≤ φt < 1 for all t > 1. Hence, 3.4 holds.
Case 3. x ∈ −1, 0 and y ∈ 1, 8. Inequality 2.1 turns into
|x|
|x|
p T x, T y max
,0 ≤ φ M x, y
4
4
1 φ max p x, y , px, T x, p T y, y , p x, T y p T x, y
2
1
φ max y, |x|, y, |x| y
2
φ y φ y ,
3.5
which is true again by the fact that 1/2 ≤ φt < 1 for all t > 1.
Case 4. x ∈ −1, 0 and y ∈ 0, 1. Inequality 2.1 becomes
|x| y
p T x, T y max
,
≤ φ M x, y
4 4
1 φ max p x, y , px, T x, p T y, y , p x, T y p T x, y
3.6
2
y
1
|x| , y max |x|,
.
φ max max |x|, y , |x|, y , max
2
4
4
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International Journal of Mathematics and Mathematical Sciences
Let use examine all possibilities:
p T x, T y ⎧
|x|
⎪
⎪
⎪
⎪ 4 ⎪
⎨ y
⎪
4
⎪
⎪
⎪
⎪
⎩ y
4
⎧
⎪
|x|
⎪
⎪
⎪
⎨
M x, y ≤ y
⎪
⎪
⎪ ⎪
⎩y
if |x| ≥ y,
y
if
≤ |x| ≤ y,
4 y
if |x| ≤
,
4
if |x|
≥ y ,
y
if
≤ |x| ≤ y,
4 y
if |x| ≤
.
4
3.7
Thus, 2.1 holds for φt t/3.
The rest of the assumptions of Theorem 2.1 are also satisfied. The function T has 0 as a
unique fixed point.
However, since φ is not a continuous function, we could not apply Theorem 1.13.
Example 3.2. Let X 0, 1 and px, y max{x, y} for all x, y ∈ X. Then, X, p is a complete
partial metric space. Take A1 · · · Ap X. Define T : X → X by T x x/2. Consider
φ : 0, ∞ → 0, ∞ given by Example 1.14.
For all x, y ∈ X, we have
x y
φ p x, y ≤ φ M x, y .
p T x, T y max ,
2 2
3.8
Then all the assumptions of Theorem 2.1 are satisfied. The function T has 0 as a unique fixed
point.
Similarly, Theorem 1.13 is not applicable.
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