Volume 2011, Article ID 508730,10pages doi:10.1155/2011/508730
Research Article
Fixed Point Theorems for Monotone Mappings on Partial Metric Spaces
Ishak Altun and Ali Erduran
Department of Mathematics, Faculty of Science and Arts, Kirikkale University, 71450 Yahsihan, Kirikkale, Turkey
Correspondence should be addressed to Ishak Altun,[email protected] Received 12 November 2010; Accepted 24 December 2010
Academic Editor: S. Al-Homidan
Copyrightq2011 I. Altun and A. Erduran. 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.
Matthews1994introduced a new distancepon a nonempty setX, which is called partial metric.
IfX, pis a partial metric space, thenpx, xmay not be zero forx∈X. In the present paper, we give some fixed point results on these interesting spaces.
1. Introduction
There are a lot of fixed and common fixed point results in different types of spaces. For example, metric spaces, fuzzy metric spaces, and uniform spaces. One of the most interesting is partial metric space, which is defined by Matthews 1. In partial metric spaces, the distance of a point in the self may not be zero. After the definition of partial metric space, Matthews proved the partial metric version of Banach fixed point theorem. Then, Valero 2, Oltra and Valero 3, and Altun et al. 4 gave some generalizations of the result of Matthews. Again, Romaguera 5 proved the Caristi type fixed point theorem on this space.
First, we recall some definitions of partial metric spaces and some properties of theirs.
See1–3,5–7for details.
A partial metric on a nonempty setX is a functionp :X×X → Ê such that for all x, y, z∈X:
p1xy⇔px, x px, y py, y, p2px, x≤px, y,
p3px, y py, x,
p4px, y≤px, z pz, y−pz, z.
A partial metric space is a pair X, p such that X is a nonempty set and p is a partial metric on X. It is clear that if px, y 0, then from p1 and p2 x y.
But if x y, px, y may not be 0. A basic example of a partial metric space is the pair Ê, p, where px, y max{x, y} for all x, y ∈ Ê. Other examples of partial metric spaces, which are interesting from a computational point of view, may be found in 1,8.
Each partial metricp onX generates aT0 topologyτponX, which has as a base the family openp-balls{Bpx, ε:x∈X, ε >0}, whereBpx, ε {y∈X :px, y< px, x ε}
for allx∈Xandε >0.
Ifpis a partial metric onX, then the functionps:X×X → Ê given by ps
x, y 2p
x, y
−px, x−p y, y
1.1
is a metric onX.
LetX, pbe a partial metric space, then we have the following.
iA sequence{xn}in a partial metric spaceX, pconverges to a pointx∈X if and only ifpx, x limn→ ∞px, xn.
iiA sequence{xn}in a partial metric spaceX, pis called a Cauchy sequence if there existsand is finitelimn,m→ ∞pxn, xm.
iiiA 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.
ivA mappingF :X → Xis said to be continuous atx0 ∈X, if for everyε >0, there existsδ >0 such thatFBpx0, δ⊆BpFx0, ε.
Lemma 1.1see1,3. LetX, pbe a partial metric space.
a{xn}is a Cauchy sequence inX, pif and only if it is a Cauchy sequence in the metric spaceX, ps.
bA partial metric spaceX, pis complete if and only if the metric spaceX, psis complete.
Furthermore, limn→ ∞psxn, x 0 if and only if px, x lim
n→ ∞pxn, x lim
n,m→ ∞pxn, xm. 1.2
On the other hand, existence of fixed points in partially ordered sets has been considered recently in9, and some generalizations of the result of 9 are given in10–
15in a partial ordered metric spaces. Also, in9, some applications to matrix equations are presented; in14,15, some applications to ordinary differential equations are given. Also, we can find some results on partial ordered fuzzy metric spaces and partial ordered uniform spaces in16–18, respectively.
The aim of this paper is to combine the above ideas, that is, to give some fixed point theorems on ordered partial metric spaces.
2. Main Result
Theorem 2.1. Let X,be partially ordered set, and suppose that there is a partial metric p on X such thatX, p is a complete partial metric space. SupposeF : X → X is a continuous and nondecreasing mapping such that
p
Fx, Fy
≤Ψ
max
p x, y
, px, Fx, p y, Fy
,1 2
p x, Fy
p
y, Fx 2.1
for allx, y ∈ X withy x, whereΨ :0,∞ → 0,∞is a continuous, nondecreasing function such that∞
n1Ψntis convergent for eacht >0. If there exists anx0∈Xwithx0Fx0, then there existsx∈Xsuch thatxFx. Moreover,px, x 0.
Proof. From the conditions onΨ, it is clear that limn→ ∞Ψnt 0 fort > 0 andΨt < t. If Fx0x0, then the proof is finished, so supposex0/Fx0. Now, letxnFxn−1forn1,2, . . ..
Ifxn0 xn01 for somen0 ∈ Æ, then it is clear that xn0 is a fixed point ofF. Thus, assume xn/xn1for alln∈Æ. Notice that sincex0Fx0andFis nondecreasing, we have
x0 x1 x2 · · · xnxn1 · · ·. 2.2
Now, sincexn−1xn, we can use the inequality2.1for these points, then we have pxn1, xn
pFxn, Fxn−1
≤Ψ
max
pxn, xn−1, pxn, Fxn, pxn−1, Fxn−1,1 2
pxn, Fxn−1 pxn−1, Fxn
≤Ψ
max
pxn, xn−1, pxn, xn1,1 2
pxn−1, xn pxn, xn1
Ψ max
pxn, xn−1, pxn, xn1
2.3
since
pxn, xn pxn−1, xn1≤pxn−1, xn pxn, xn1 2.4
andΨis nondecreasing. Now, if max
pxn, xn−1, pxn, xn1
pxn, xn1 2.5
for somen, then from2.3we have pxn1, xn≤Ψ
pxn, xn1
< pxn, xn1, 2.6
which is a contradiction sincepxn, xn1>0. Thus max
pxn, xn−1, pxn, xn1
pxn, xn−1 2.7
for alln. Therefore, we have
pxn1, xn≤Ψ
pxn, xn−1
, 2.8
and so
pxn1, xn≤Ψn
px1, x0
. 2.9
On the other hand, since max
pxn, xn, pxn1, xn1
≤pxn, xn1, 2.10
then from2.9we have max
pxn, xn, pxn1, xn1
≤Ψn
px1, x0
. 2.11
Therefore,
psxn, xn1 2pxn, xn1−pxn, xn−pxn1, xn1
≤2pxn, xn1 pxn, xn pxn1, xn1
≤4Ψn
px1, x0 .
2.12
This shows that limn→ ∞psxn, xn1 0. Now, we have
psxnk, xn≤psxnk, xnk−1 · · ·psxn1, xn
≤4Ψnk−1
px1, x0
· · ·4Ψn
px1, x0
. 2.13
Since∞
n1Ψntis convergent for eacht >0, then{xn}is a Cauchy sequence in the metric spaceX, ps. SinceX, pis complete, then, fromLemma 1.1, the sequence{xn}converges in the metric spaceX, ps, say limn→ ∞psxn, x 0. Again, fromLemma 1.1, we have
px, x lim
n→ ∞pxn, x lim
n,m→ ∞pxn, xm. 2.14
Moreover, since {xn} is a Cauchy sequence in the metric space X, ps, we have limn,m→ ∞psxn, xm 0, and, from2.11, we have limn→ ∞pxn, xn 0, thus, from definition ps, we have limn,m→ ∞pxn, xm 0. Therefore, from2.14, we have
px, x lim
n→ ∞pxn, x lim
n,m→ ∞pxn, xm 0. 2.15
Now, we claim thatFx x. Supposepx, Fx> 0. SinceFis continuous, then, givenε > 0, there existsδ >0 such thatFBpx, δ⊆BpFx, ε. Sincepx, x limn→ ∞pxn, x 0, then there existsk∈Æsuch thatpxn, x< px, xδfor alln≥k. Therefore, we havexn∈Bpx, δ for alln≥k. Thus,Fxn∈FBpx, δ⊆BpFx, ε, and sopFxn, Fx< pFx, Fx εfor all n≥k. This shows thatpFx, Fx limn→ ∞pxn1, Fx. Now, we use the inequality2.1for xy, then we have
pFx, Fx≤Ψ max
px, x, px, Fx Ψ
px, Fx
. 2.16
Therefore, we obtain
px, Fx≤px, xn1 pxn1, Fx−pxn1, xn1
≤px, xn1 pxn1, Fx, 2.17
and lettingn → ∞, we have
px, Fx≤ lim
n→ ∞px, xn1 lim
n→ ∞pxn1, Fx pFx, Fx
≤Ψ
px, Fx
< px, Fx,
2.18
which is a contradiction sincepx, Fx>0. Thus,px, Fx 0, and soxFx.
In the following theorem, we remove the continuity of F. Also, The contractive condition2.1does not have to be satisfied forxy, but we add a condition onX.
Theorem 2.2. LetX,be a partially ordered set, and suppose that there is a partial metricponX such thatX, pis a complete partial metric space. SupposeF :X → Xis a nondecreasing mapping such that
p
Fx, Fy
≤Ψ
max
p x, y
, px, Fx, p y, Fy
,1 2
p x, Fy
p
y, Fx 2.19
for allx, y ∈ X withy ≺ x(i.e.,y xandy /x), whereΨ : 0,∞ → 0,∞is a continuous, nondecreasing function such that∞
n1Ψntis convergent for eacht >0. Also, the condition If {xn} ⊂X is a increasing sequence withxn−→x inX, thenxn≺x, ∀n 2.20
holds. If there exists anx0 ∈Xwithx0 Fx0, then there existsx∈Xsuch thatxFx. Moreover, px, x 0.
Proof. As in the proof ofTheorem 2.1, we can construct a sequence{xn}inXbyxnFxn−1for n1,2, . . .. Also, we can assume that the consecutive terms of{xn}are different. Otherwise we are finished. Therefore, we have
x0 ≺x1 ≺x2 ≺ · · · ≺xn≺xn1≺ · · ·. 2.21
Again, as in the proof ofTheorem 2.1, we can show that{xn}is a Cauchy sequence in the metric spaceX, ps, and, therefore, there existsx∈Xsuch that
px, x lim
n→ ∞pxn, x lim
n,m→ ∞pxn, xm 0. 2.22
Now, we claim thatFxx. Supposepx, Fx>0. Since the condition2.20is satisfied, then we can use2.19foryxn. Therefore, we obtain
pFx, Fxn
≤Ψ
max
px, xn, px, Fx, pxn, Fxn,1 2
px, Fxn pxn, Fx
≤Ψ
max
px, xn, px, Fx, pxn, xn1,1 2
px, xn1 pxn, x px, Fx−px, x Ψ
max
px, xn, px, Fx, pxn, xn1, 1 2
px, xn1 pxn, x px, Fx ,
2.23
using the continuity of Ψand lettingn → ∞, we have limn→ ∞pFx, Fxn ≤ Ψpx, Fx.
Therefore, we obtain
px, Fx≤ lim
n→ ∞px, xn1 lim
n→ ∞pxn1, Fx lim
n→ ∞px, xn1 lim
n→ ∞pFxn, Fx
≤Ψ
px, Fx
< px, Fx,
2.24
which is a contradiction. Thus,px, Fx 0, and soxFx.
Example 2.3. LetX 0,∞andpx, y max{x, y}, then it is clear thatX, pis a complete partial metric space. We can define a partial order onXas follows:
xy⇐⇒xy or
x, y∈0,1withx≤y
. 2.25
LetF:X → X,
Fx
⎧⎪
⎨
⎪⎩ x2
1x, x∈0,1,
2x, x∈1,∞, 2.26
andΨ : 0,∞ → 0,∞,Ψt t2/1t. Therefore,Ψis continuous and nondecreasing.
Again we can show by induction that Ψnt ≤ tt/1 tn, and so we have ∞
n1Ψnt that is convergent. Also, F is nondecreasing with respect to , and for y ≺ x, we have
p
Fx, Fy max
x2 1x, y2
1y
x2 1x Ψ
p x, y
≤Ψ
max
p x, y
, px, Fx, p y, Fy
,1 2
p x, Fy
p
y, Fx ,
2.27
that is, the condition 2.19 of Theorem 2.2is satisfied. Also, it is clear that the condition 2.20 is satisfied, and for x0 0, we have x0 Fx0. Therefore, all conditions of Theorem 2.2are satisfied, and so F has a fixed point in X. Note that if x 1 and y 2, then
p
Fx, Fy
4/≤ 16 5 Ψ
max
p
x, y
, px, Fx, p y, Fy
,1 2
p x, Fy
p
y, Fx .
2.28
This shows that the contractive condition of Theorem 1 of4is not satisfied.
Theorem 2.4. If one uses the following condition instead of 2.1inTheorem 2.1, one has the same result.
p
Fx, Fy
≤Ψ
max
p x, y
,1 2
px, Fx p y, Fy
,1 2
p x, Fy
p y, Fx
2.29
for allx, y∈Xwithyx.
In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorem 2.4, this condition is
forx, y∈X there exists a lower bound or an upper bound. 2.30
In15, it was proved that condition2.30is equivalent to
forx, y∈X there existsz∈X which is comparable toxandy. 2.31
Theorem 2.5. Adding condition2.31to the hypotheses ofTheorem 2.4, one obtains uniqueness of the fixed point ofF.
Proof. Suppose that there existszand thaty∈Xare different fixed points ofF, thenpz, y>
0. Now, we consider the following two cases.
iIfzandyare comparable, thenFnzzandFnyyare comparable forn0,1, . . ..
Therefore, we can use the condition2.1, then we have
p z, y
p
Fnz, Fny
≤Ψ max
p
Fn−1z, Fn−1y ,1
2 p
Fn−1z, Fnz p
Fn−1y, Fny , 1
2 p
Fn−1z, Fny p
Fn−1y, Fnz Ψ
max
p
z, y ,1
2
pz, z p y, y Ψ
p z, y
< p z, y
,
2.32
which is a contradiction.
iiIfzandyare not comparable, then there existsx∈Xcomparable tozandy. Since Fis nondecreasing, thenFnxis comparable toFnzzandFnyyforn0,1, . . .. Moreover,
pz, Fnx pFnz, Fnx
≤Ψ max
p
Fn−1z, Fn−1x ,1
2 p
Fn−1z, Fnz p
Fn−1x, Fnx , 1
2 p
Fn−1z, Fnx p
Fn−1x, Fnz Ψ
max
p
z, Fn−1x ,1
2
pz, z p
Fn−1x, Fnx ,1
2
pz, Fnx p
Fn−1x, z
≤Ψ
max
p
z, Fn−1x ,1
2 p
Fn−1x, z
pz, Fnx ,1
2
pz, Fnx p
Fn−1x, z Ψ
max
p
z, Fn−1x ,1
2 p
Fn−1x, z
pz, Fnx .
2.33
Now, ifpz, Fn−1x< pz, Fnxfor somen, then we have pz, Fnx≤Ψ
pz, Fnx
< pz, Fnx, 2.34
which is a contradiction. Thus,pz, Fn−1x≥pz, Fnxfor alln, and so pz, Fnx≤Ψ
p
z, Fn−1x
< p
z, Fn−1x
. 2.35
This shows thatpz, Fnxis a nonnegative and nondecreasing sequence and so has a limit, sayα≥0. From the last inequality, we can obtain
α≤Ψα< α, 2.36
henceα0. Similarly, it can be proven that, limn→ ∞py, Fnx 0. Finally, p
z, y
≤pz, Fnx p Fnx, y
−pFnx, Fnx
≤pz, Fnx p Fnx, y
, 2.37
and taking limitn → ∞, we havepz, y 0. This contradictspz, y>0.
Consequently,Fhas no two fixed points.
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