# Generalized Contractions of Rational Type in Ordered Partial Metric Spaces

Authors: Shujun Jiang; Zhilong Li
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Without the continuity and nondecreasing property of the comparison function, we in this paper prove some fixed point theorems of generalized contractions of rational type
in ordered partial metric spaces, which generalize and improve the corresponding results of Luong and Thuan. An example is given to support the usability of our results.


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## Body

## 1. Introduction and Preliminaries

Throughout this paper,ℝ+ and ℕ will denote the set of nonnegative real numbers and the set of all positive integer numbers.It is well known that the Banach contraction principle is one of the pivotal results of analysis. Generalizations of this principle have been obtained in several directions. In [1], Jaggi introduced a contraction of rational type in a metric space and proved the unique existence of fixed point as the contraction is continuous and the metric space is complete, which was then extended to the case of ordered metric spaces by Harjani et al. [2]. In [3], Luong and Thuan considered weak contractions of rational type in ordered metric spaces and proved the following fixed point theorem.Theorem 1 (see [3]).
Let(X,d,⪯) be a complete ordered metric space and T:X→X a nondecreasing mapping such that there exists a function η:ℝ+→ℝ+ with η-1({0})={0} such that
(1)d(Tx,Ty)≤Md·r(x,y)-η(Md·r(x,y)),
for all x,y∈X with x⪯y and x≠y, where Md·r(x,y)=max{d(x,y),((d(x,Tx)·d(y,Ty))/d(x,y))}. Assume that η is lower semicontinuous, and X has the following property:(A1)
if{xn} is a nondecreasing sequence in X such that xn→dx, then x=sup{xn}.If there exists x0∈X such that x0⪯Tx0, then T has a fixed point.In [4], Matthews introduced the partial metric space and extended the Banach contraction principle to the case of partial metric spaces, which was then improved by Oltra and Valero [5]. In [6–8], the authors studied the unique existence of fixed point of generalized contractions in partial metric spaces. Recall that a mapping T:X→X is called a Mpi-generalized contraction (Mpi-GC) if there exists a comparison function φ:ℝ+→ℝ+ such that
(2)p(Tx,Ty)≤φ(Mpi(x,y)),
where (X,p) is a partial metric space, Mp1(x,y)=max{p(x,y),p(x,Tx),p(y,Ty),(p(x,Ty)+p(y,Tx))/2}, and Mp2(x,y)=max{p(x,y),p(x,Tx),p(y,Ty)}. Under several different assumptions made on φ, they obtained the following fixed point results.Theorem 2 (see Theorem 1 of [6], Theorem 1 of [7], and Theorems 3 and 4 of [8]).
Let(X,p) be a complete partial metric space. Assume the following.(i)
T
:
X
→
X
is a
M
p
1
-GC and one of the following conditions is satisfied:
(H1)
φ
is continuous and nondecreasing, and
φ
(
t
)
<
t
for all
t
>
0
;
(H2)
φ
is nondecreasing, and the series
∑
n
=
0
∞
‍
φ
n
(
t
)
is convergent for all
t
>
0
(
φ
n
denotes the
n
th iterate of
φ
);
(H3)
φ
is upper semicontinuous from the right, and
φ
(
t
)
<
t
for all
t
>
0
.Or (ii)
T
:
X
→
X
is a
M
p
2
-GC and the following condition is satisfied:
(H4)
φ
is nondecreasing, and
lim
n
→
∞
φ
n
(
t
)
=
0
for all
t
>
0
.Then
T
has a unique fixed point.For other references concerned with various fixed point results and common fixed point results for contractions in the setting of metric-like, partial metric, and ordered partial metric spaces, we refer the readers to [9–24].In this paper, we establish some fixed point theorems for generalized contractions of rational type in ordered partial metric spaces, which generalize and improve Theorems1 and 2. An example is given to support the usability of our results. Even in the setting of metric spaces, the results presented in this paper are still new since the comparison function φ is not necessarily assumed to be upper semicontinuous from the right or nondecreasing.Following [4, 5], a partial metric on a set X is a function p:X×X→ℝ+ such that, for each x,y,z∈X,(p1)
p(x,y)=p(x,x)=p(y,y) if and only if x=y;(p2)
p(x,y)=p(y,x);(p3)
p(x,x)≤p(x,y);(p4)
p(x,y)≤p(x,z)+p(z,y)-p(z,z).Observe that, if p(x,y)=0, then x=y. A partial metric space is a pair (X,p) such that X is a set and p is a partial metric on X. Each partial metric p on X induces a T0 topology τp on X which has as a base of the family of open balls {Bp(x,ε):x∈X,ε>0}, where Bp(x,ε)={y∈X:p(x,y)<ε+p(x,x)} for all x∈X and ε>0.Let(X,p) be a partial metric space and {xn} a sequence of X. The sequence {xn} converges, with respect to τp, to a point x∈X (denoted by xn→px) if limn→∞p(xn,x)=p(x,x). Define a function ps:X×X→ℝ+ by ps(x,y)=2p(x,y)-p(x,x)-p(y,y). Then ps is a metric on X. The sequence {xn} converges, with respect to τps, to a point x∈X (denoted by xn→psx) if and only if
(3)limn→∞p(xn,x)=limm,n→∞p(xm,xn)=p(x,x).
The sequence {xn} is called a Cauchy sequence if limm,n→∞p(xm,xn) exists and is finite; (X,p) is called complete if every Cauchy sequence {xn}⊂X converges, with respect to τp, to a point x∈X such that p(x,x)=limm,n→∞p(xm,xn). In particular, {xn} is called a 0-Cauchy sequence if limm,n→∞p(xm,xn)=0;  (X,p) is called 0-complete if every 0-Cauchy sequence {xn}⊂X converges, with respect to τp, to a point x∈X such that p(x,x)=0. Every complete partial metric space (X,p) is 0-complete, but the converse may not be true; see [17].Remark 3 (see [4, 5]).
A partial metric space(X,p) is complete if and only if (X,ps) is complete.


## 2. Fixed Point Theorems

Let(X,p,⪯) be an ordered partial metric space and T:X→X. For all x,y∈X with p(x,y)>0, set
(4)Mp·r1(x,y)=max{p(x,y),p(x,Tx),p(y,Ty),p(x,Tx)·p(y,Ty)p(x,y)Mp·r1(x,y)=eemaxp(x,Tx)·p(y,Ty)p(x,y),p(x,Ty)+p(y,Tx)2},Mp·r2(x,y)=max{p(x,y),p(x,Tx),p(y,Ty),p(x,Tx)·p(y,Ty)p(x,y)Mp·r1(x,y)=eemaxp(x,Tx)·p(y,Ty)p(x,y)}.
A mapping T:X→X is said to be a Mp.ri-generalized contraction of rational type (Mp.ri-GCRT), if there exists a comparison function φ:ℝ+→ℝ+ such that
(5)p(Tx,Ty)≤φ(Mp.ri(x,y)),
for all x,y∈X with x⪯y and p(x,y)>0.Lemma 4.
Let(X,p,⪯) be an ordered partial metric space and T:X→X a nondecreasing Mp.r1-GCRT. Assume that (H5)
φ(s)<s and limsupt→s+(φ(t)/t)<1 for all s>0.For each x0∈X such that x0⪯Tx0, let xn=Tnx0 for all n∈ℕ. If xn≠xn-1 for all n∈ℕ, then
(6)limm,n→∞p(xm,xn)=0.Proof.
Note thatT is nondecreasing and x0⪯Tx0; then {xn} is nondecreasing. Since xn≠xn-1 for all n∈ℕ, then
(7)p(xn-1,xn)>0,
for all n∈ℕ. By (4) and (p4), for all n∈ℕ, we have
(8)max{p(xn-1,xn),p(xn,xn+1)}≤max{p(xn-1,xn),p(xn,xn+1),p(xn-1,xn+1)+p(xn,xn)2}=Mp·r1(xn-1,xn)≤max{p(xn-1,xn),p(xn,xn+1),p(xn-1,xn)+p(xn,xn+1)2}=max{p(xn-1,xn),p(xn,xn+1)}.
Thus by the nondecreasing property of {xn}, (7), and (8),
(9)p(xn,xn+1)≤φ(Mp·r1(xn-1,xn))=φ(max{p(xn-1,xn),p(xn,xn+1)}),
for all n∈ℕ. Now, we claim that, for all n∈ℕ,
(10)p(xn,xn+1)≤p(xn-1,xn).
Suppose, on the contrary, that there exists some n∈ℕ such that p(xn,xn+1)>p(xn-1,xn). Then by (8), (9), and φ(s)<s for all s>0, we have
(11)p(xn,xn+1)≤φ(p(xn,xn+1))<p(xn,xn+1).
This is a contradiction and hence (10) is true. Consequently, {p(xn,xn+1)} is a decreasing sequence of positive real numbers. This yields that there exists α≥0 such that
(12)p(xn,xn+1)⟶α+.
In the following, we will show that α=0. Suppose, on the contrary, that α>0. It follows from (8), (9), and (10) that, for all n∈ℕ,
(13)p(xn,xn+1)p(xn-1,xn)≤φ(p(xn-1,xn))p(xn-1,xn).
Letting n→∞ in (13), by (12), (H5), and α>0, we get a contradiction 1≤limsupt→α+(φ(t)/t)<1. Hence α=0, and consequently by (12)
(14)limn→∞p(xn,xn+1)=0.
Now, we show that conclusion (6) is true. If otherwise, there exist some β>0 and subsequences {mk}k∈ℕ and {nk}k∈ℕ in ℕ, with mk>nk≥k, such that, for all k∈ℕ,
(15)p(xnk,xmk)≥β.
From (15) we may assume that, without loss of generality, for all k∈ℕ,
(16)p(xnk,xmk-1)<β.
Then by (p4), for all k∈ℕ, we have
(17)β≤p(xnk,xmk)≤p(xnk,xmk-1)+p(xmk,xmk-1)<β+p(xmk,xmk-1).
Letting k→∞ in the above inequality, by (14), we get
(18)limk→∞p(xnk,xmk)=β.
Also by (p4), for all k∈ℕ, we have
(19)p(xnk,xmk)-p(xnk,xnk+1)-p(xmk,xmk+1)≤p(xnk+1,xmk+1)≤p(xnk,xnk+1)+p(xnk,xmk)+p(xmk,xmk+1).
Letting k→∞ in the above inequality, by (14) and (18), we get
(20)limk→∞p(xnk+1,xmk+1)=β.
Similarly, we can obtain
(21)limk→∞p(xnk,xmk+1)=limk→∞p(xnk+1,xmk)=β.
By the nondecreasing property of {xn}, (5), and (15), we get, for all k∈ℕ,
(22)p(xnk+1,xmk+1)=p(Txnk,Txmk)≤φ(Mp.r1(xnk,xmk)),
where Mp.r1(xnk,xmk)=max{p(xnk,xmk),p(xnk,xnk+1), p(xmk,xmk+1),(p((xnk,xnk+1)·p(xmk,xmk+1))/p(xnk,xmk)), (p(xnk,xmk+1)+p(xnk+1,xmk)/2)}. By (14) and (18), we have
(23)limk→∞p(xnk,xnk+1)·p(xmk,xmk+1)p(xnk,xmk)=0,
which together with (14), (18), and (21) implies that there exists k1∈ℕ such that, for all k≥k1,
(24)Mp·r1(xnk,xmk)=M1(xnk,xmk),
where M1(xnk,xmk)=max{p(xnk,xmk),(p(xnk,xmk+1)+p(xnk+1,xmk))/2}. It follows from (15), (18), and (21) that
(25)M1(xnk,xmk)⟶β+,
By (15), (22), and (24), we have, for all k≥k1,
(26)p(xnk+1,xmk+1)M1(xnk,xmk)≤φ(M1(xnk,xmk))M1(xnk,xmk).
Letting k→∞ in (26), by (20), (25), (H5), and β>0, we get a contradiction 1≤limsupt→β+(φ(t)/t)<1. Hence conclusion (6) is true. The proof is complete.Remark 5.
It is easy to see that Lemma4 is still valid for Mp·r2-GCRT and Mp1-GC.Theorem 6.
Let(X,p,⪯) be a 0-complete ordered partial metric space and T:X→X a nondecreasing Mp·r1-GCRT. Assume that(H6)
φ(s)<s and limsupt→s(φ(t)/t)<1,  for all s>0, and X has the following property:(A2)
if{xn} is a nondecreasing sequence in X such that xn→psx, then x=sup{xn}.If there exists x0∈X such that x0⪯Tx0, then T has a fixed point.Proof.
Letxn=Tnx0 for all n∈ℕ. If there exists some n∈ℕ such that Txn-1=xn=xn-1, then xn-1 is a fixed point and hence the proof is complete. Therefore, we may assume that xn≠xn-1 for all n∈ℕ. By Lemma 4, {xn} is 0-Cauchy. So by the 0-completeness of X, there exists x*∈X such that xn→px*, and consider
(27)limn→∞p(xn,x*)=p(x*,x*)=limm,n→∞p(xm,xn)=0.
That is, xn→psx*. By (A2), sup{xn}=x*, and so, for all n∈ℕ,
(28)xn⪯x*.
Now, we claim that, for all n∈ℕ,
(29)xn≠x*.
If otherwise, there exists some n* such that xn*=x*; then xn=x* for all n≥n* since {xn} is nondecreasing. This contradicts with the assumption xn≠xn-1 for all n∈ℕ, and hence (29) is true. Since T is nondecreasing, then xn+1=Txn⪯Tx*, for all n∈ℕ, and hence by (A2)
(30)x*=sup{xn}⪯Tx*.
Let yn=Tnx* for all n∈ℕ. Then {yn} is a nondecreasing sequence since T is nondecreasing. We may assume that, for all n∈ℕ,
(31)yn≠yn-1.
If otherwise, there exists some n∈ℕ such that Tyn-1=yn=yn-1; then yn-1 is a fixed point and hence the proof is complete. By Lemma 4 and the 0-completeness of X, there exists y*∈X such that yn→py*, and consider
(32)limn→∞p(yn,y*)=p(y*,y*)=limm,n→∞p(ym,yn)=0.
That is, yn→psy*. By (A2), we have, for all n∈ℕ,
(33)yn⪯y*.
It follows from xn→psx*, yn→psy*, and the continuity of the metric ps that
(34)limn→∞ps(xn,yn)=ps(x*,y*).
That is,
(35)limn→∞[2p(xn,yn)-p(xn,xn)-p(yn,yn)]=2p(x*,y*),
which together with (27) and (32) implies that
(36)limn→∞p(xn,yn)=p(x*,y*).
By (p4), we have, n∈ℕ,
(37)p(xn,yn)-p(xn,xn+1)≤p(xn+1,yn)≤p(xn,yn)+p(xn,xn+1),p(xn,yn)-p(yn,yn+1)≤p(xn,yn+1)≤p(xn,yn)+p(yn,yn+1).
Letting n→∞ in the above two inequalities, by (27), (32), and (36), we get
(38)limn→∞p(xn+1,yn)=limn→∞p(xn,yn+1)=p(x*,y*).
In what follows, we will show x*=y*. Suppose, on the contrary, that x*≠y*; then p(x*,y*)>0. Clearly, xn⪯yn and xn≠yn for all n∈ℕ by (29), (30), and (31). Consequently,
(39)p(xn,yn)>0,
for all n∈ℕ. Thus by (5), we have, for all n∈ℕ,
(40)p(xn+1,yn+1)=p(Txn,Tyn)≤φ(Mp.r1(xn,yn)),
where Mp·r1(xn,yn)=max{p(xn,yn),p(xn,xn+1),p(yn,yn+1), (p(xn,xn+1)·p(yn,yn+1))/p(xn,yn), (p(xn,yn+1)+p(yn,xn+1))/2}. By (27), (32), and (36), we have
(41)limn→∞p(xn,xn+1)·p(yn,yn+1)p(xn,yn)=0,
which together with (27), (32), (36), and (38) implies that there exists n1∈ℕ such that, for all n≥n1,
(42)Mp.r1(xn,yn)=M2(xn,yn),
where M2(xn,yn)=max{p(xn,yn),(p(xn,yn+1)+p(yn,xn+1))/2}. It follows from (36) and (38) that
(43)M2(xn,yn)⟶p(x*,y*).
By (39), (40), and (42), we have, for all n≥n1,
(44)p(xn+1,yn+1)M2(xn,yn)≤φ(M2(xn,yn))M2(xn,yn).
Letting n→∞ in (44), by (36), (43), (H6), and p(x*,y*)>0, we get a contradiction 1≤limsupt→p(x*,y*)(φ(t)/t)<1. Hence p(x*,y*)=0, and consequently x*=y*. From (30) and (33), it follows that x*⪯Tx*=y1⪯y*. This together with x*=y* yields that x*=Tx*. The proof is complete.In particular whenT is a Mp·r2-GCRT, condition (H6) could be weakened, and we have the following result.Theorem 7.
Let(X,p,⪯) be a 0-complete ordered partial metric space and T:X→X a nondecreasing Mp·r2-GCRT. Assume that (H5) is satisfied and X has property (A2). If there exists x0∈X such that x0⪯Tx0, then T has a fixed point.Proof.
Following the proof of Theorem6, we find that (27)–(36) and (39) still hold by Remark 5. Instead of (40), by (5), (6), (39), and xn⪯yn, we have, for all n∈ℕ,
(45)p(xn+1,yn+1)=p(Txn,Tyn)≤φ(Mp.r2(xn,yn))<Mp.r2(xn,yn),
where Mp·r2(xn,yn)=max{p(xn,yn),p(xn,xn+1),p(yn,yn+1), (p(xn,xn+1)·p(yn,yn+1))/p(xn,yn)}. By (28), (32), (36), and p(x*,y*)>0, there exists n2∈ℕ such that, for all n≥n2,
(46)Mp.r2(xn,yn)=p(xn,yn),
which together with (45) implies that, for all n≥n2,
(47)p(xn+1,yn+1)<p(xn,yn).
That is, {p(xn,yn)}n=n2∞ is decreasing. Moreover by (36), we have
(48)p(xn,yn)⟶p(x*,y*)+.
Instead of (44), by (39), (45), and (46), for all n≥n2,
(49)p(xn+1,yn+1)p(xn,yn)≤φ(p(xn,yn))p(xn,yn).
Letting n→∞ in (49), by (48), (H5), and p(x*,y*)>0, we get a contradiction 1≤limt→p(x*,y*)+(φ(t)/t)<1. Hence p(x*,y*)=0, and consequently x*=y*. From (30) and (33), it follows that x*⪯Tx*=y1⪯y*. This together with x*=y* yields that x*=Tx*. The proof is complete.Corollary 8.
Let(X,p,⪯) be a 0-complete ordered partial metric space and T:X→X a nondecreasing mapping such that there exists a function η:ℝ+→ℝ+ with η-1({0})={0} such that
(50)p(Tx,Ty)≤Mp·r2(x,y)-η(Mp·r2(x,y)),
for all x,y∈X with x⪯y and p(x,y)>0. Assume that, for all s>0,
(51)liminft→s+η(t)t>0,
and X has property (A2). If there exists x0∈X such that x0⪯Tx0, then T has a fixed point.Proof.
Letφ(t)=t-η(t). Clearly, φ(t)<t for all t>0 by η-1({0})={0}. For all s>0, it follows from (51) that
(52)limsupt→s+φ(t)t=limsupt→s+t-η(t)t≤1-liminft→s+η(t)t<1.
Then the conclusion follows immediately from Theorem 7. The proof is complete.Remark 9.
Ifη is lower semicontinuous from the right and η-1({0})={0}, then, for all s>0,
(53)liminft→s+η(t)t≥η(s)s>0.
Therefore Theorem 1 follows immediately from Corollary 8.Now we illustrate Theorems6 and 7 by the following example.Example 10.
LetX=ℝ+ with the partial metric p(x,y)=max{x,y} for all x,y∈X. Clearly, ps(x,y)=2p(x,y)-p(x,x)-p(y,y)=|x-y| and hence, by Remark 3, (X,p) is a complete partial metric space since (X,ps) is a complete metric space. Define a partial order ⪯ on X by
(54)x⪯y⟺{x=y,x,y∈X∖({0}∪B),x≤y,x,y∈{0}∪B,
where B={(1/n):n=2,3,…} and ≤ is the usual order of reals. Let
(55)Tx={0,x=X∖B,x1+x,x=1n,n=2,4,…,(56)φ(t)={t-t3,t∈[0,1)∖B,t1+t,t=1n,n=2,3,…,t-12,t≥1.
It is easy to see that T is nondecreasing, X has property (A2), and φ(t)<t for all t>0. Direct calculations give that
(57)limsupt→sφ(t)t={1-s2<1,s∈(0,1),1-12s<1,s≥1.
That is, (H6) is satisfied. In particular, (H5) is satisfied. For each x,y∈X∖B with x⪯y and p(x,y)>0, we must have p(Tx,Ty)=0 and hence p(Tx,Ty)≤φ(Mp.ri(x,y)),i=1,2. For each x,y∈{0}∪B with x⪯y and p(x,y)>0, we have two cases:(i)
ifx=0 andy=(1/n),n≥2, then p(Tx,Ty)=1/(n+1)≤1/n=p(x,y)=Mp.ri(x,y),i=1,2;(ii)
ify=1/n and x=1/m,m≥n≥3, then p(Tx,Ty)=1/(n+1)≤1/n=p(x,y)=Mp.ri(x,y),i=1,2.
Thus by (56), p(Tx,Ty)≤Mp.ri(x,y)=φ(Mp·ri(x,y)),i=1,2. This shows that T is a Mp·ri-GCRT(i=1,2), and consequently the existence of fixed point follows immediately from Theorem 6 or Theorem 7. In fact, 0 is a fixed point of T.In what follows, we will extend Theorem2 to the case of ordered partial metric space under weaker conditions.Theorem 11.
Let(X,p,⪯) be a 0-complete ordered partial metric space and T:X→X a nondecreasing mapping such that, for each x,y∈X with x⪯y, (2) is satisfied for Mp1. Assume that either (H4) or (H5) is satisfied and X has the following property:(A3)
if{xn} is a nondecreasing sequence in X such that xn→psx, then xn⪯x for all n∈ℕ.
If there existsx0∈X such that x0⪯Tx0, then T has a fixed point.Proof.
Case
1 ((H5) is satisfied). Following the proof of Theorem 6, we get (27) and (28) by Remark 5 and (A3). By (p4), for all n∈ℕ,
(58)p(x*,Tx*)-p(xn+1,x*)≤p(xn+1,Tx*)≤p(x*,Tx*)+p(xn+1,x*).
Letting n→∞ in (58), by (27), we get
(59)limn→∞p(xn+1,Tx*)=p(x*,Tx*).
Now, we show that x* is a fixed point of T. If otherwise, then p(x*,Tx*)>0. By (p4), (2), and (28), we have, for all n∈ℕ,
(60)p(x*,Tx*)≤p(xn+1,x*)+p(Txn,Tx*)≤p(xn+1,x*)+φ(Mp1(xn,x*)),
where Mp1(xn,x*)=max{p(xn,x*),p(xn,xn+1),p(x*,Tx*), (p(xn,Tx*)+p(x*,xn+1))/2}. It follows from (27) and (59) that
(61)limn→∞p(xn,Tx*)+p(x*,xn+1)2=p(x*,Tx*)2,
which together with (27) implies that there exist δ0∈(0,p(x*,Tx*)/2] and n0∈ℕ such that, for all n≥n0,
(62)p(xn,Tx*)+p(x*,xn+1)2≤p(x*,Tx*)2+δ0≤p(x*,Tx*),hhhhhhhhhhhhp(xn,x*)≤δ0,p(xn,xn+1)≤δ0.
Thus, for all n≥n0,
(63)Mp1(xn,x*)=p(x*,Tx*),
and hence, by (60),
(64)p(x*,Tx*)≤p(xn+1,x*)+φ(p(x*,Tx*)).
Letting n→∞ in (64), by (27), we get a contradiction
(65)p(x*,Tx*)≤φ(p(x*,Tx*))<p(x*,Tx*),
since φ(s)<s for all s>0. Hence x* is a fixed point of T.
Case
2 ((H4) is satisfied). By the same method used in Lemma 4 and [8, Theorem 4], we can show that (27) and (59) still hold. Then by analogy to Case 1, x* is a fixed point of T. The proof is complete.Remark 12.
IfT is a generalized contraction of rational type, then the method used in Theorem 11 fails to work since there exists n0∈ℕ such that Mp.ri(xn,x*)=p(x*,Tx*) for all n≥n0 unless limn→∞p(xn,xn+1)/p(xn,x*)≤1, which could not be done. Therefore the existence of fixed point of generalized contractions of rational type could not be obtained under the weaker condition (A3).By the methods used in Theorem11 and [8, Theorem 4], we have the following fixed point result.Theorem 13.
Let(X,p) be a 0-complete partial metric space and T:X→X a Mp1-GC. If either (H4) or (H5) is satisfied, then T has a unique fixed point.Remark 14.
(i) In the case that (H5) is satisfied, the continuity and nondecreasing property ofφ necessarily assumed in Theorem 2 is removed in Theorem 13. Note that (H3) implies (H5); then Theorem 13 improves Theorem1 of [6], Theorem 3 of [8], and Theorem 2.3 of [23].
(ii) In the case that (H4) is satisfied, Theorem13 is generalization of Theorem 4 of [8] to the case of Mp1-GC and hence improves Theorem 1 of [7] since (H2) implies (H4).Example 15.
LetX={0}∪{1/n:n=2,3,…} with the partial metric p(x,y)=max{x,y} for all x,y∈X. By analogy to Example 10, we find from Remark 3 that (X,p) is a complete partial metric space. Let
(66)Tx={0,x=0,x1+x,x=1n,n=2,3,…,
and φ is given by (56). It is easy to check that T is a Mp1-GC and (H5) is satisfied. Therefore the unique existence of fixed point follows immediately from Theorem 13. In fact, 0 is the unique fixed point of T.
For eachs=1/n,n≥2, we have limsupt→s+φ(t)=(n2-1)/n3>1/(n+1)=φ(s). This implies that φ is not upper semicontinuous from the right at s=1/n,n≥2. Meanwhile, it is clear that φ is decreasing on (1/3,1). Therefore, we cannot invoke Theorem 2 to show the unique existence of fixed point.Remark 16.
It is worth mentioning that, even in the setting of metric spaces, the main results in this paper are still new since the continuity and nondecreasing property of the comparison functionφ necessarily assumed in [24] and other relating references is removed.



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*Source: 928017-2013-11-17.xml*