Hindawi Publishing CorporationJournal of Function Spaces and ApplicationsVolume 2013 Article ID 143686 9 pageshttpdxdoiorg1011552013143686

Research ArticleSuzuki-Type Fixed Point Results in Metric-Like Spaces

Nabiollah Shobkolaei1 Shaban Sedghi2 Jamal Rezaei Roshan2 and Nawab Hussain3

1 Department of Mathematics Babol Branch Islamic Azad University Babol Iran2Department of Mathematics Qaemshahr Branch Islamic Azad University Qaemshahr Iran3Department of Mathematics King Abdulaziz University PO Box 80203 Jeddah 21589 Saudi Arabia

Correspondence should be addressed to Nawab Hussain nhusainkauedusa

Received 16 May 2013 Accepted 17 July 2013

Academic Editor Pei De Liu

Copyright copy 2013 Nabiollah Shobkolaei et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We demonstrate a fundamental lemma for the convergence of sequences in metric-like spaces and by using it we prove someSuzuki-type fixed point results in the setup of metric-like spaces As an immediate consequence of our results we obtain certainrecent results in partial metric spaces as corollaries Finally three examples are presented to verify the effectiveness and applicabilityof our main results

1 Introduction

There are a lot of generalizations of Banach fixed-point prin-ciple in the literature So far several authors have studiedthe problem of existence and uniqueness of a fixed point formappings satisfying different contractive conditions (eg [1ndash20]) In 2008 Suzuki introduced an interesting generalizationof Banach fixed-point principle This interesting fixed-pointresult is as follows

Theorem 1 (see [19]) Let (119883 119889) be a complete metric spaceand let 119879 be a mapping on119883 Define a nonincreasing function120579 from [0 1] into [12 1] by

120579 (119903) =

1 0 le 119903 leradic5 minus 1

2

1 minus 119903

1199032radic5 minus 1

2le 119903 le

1

radic2

1

1 + 1199031

radic2le 119903 lt 1

(1)

Assume that there exists 119903 isin [0 1] such that

120579 (119903) 119889 (119909 119879119909) le 119889 (119909 119910) 997904rArr 119889 (119879119909 119879119910) le 119903119889 (119909 119910) (2)

for all 119909 119910 isin 119883 then there exists a unique fixed-point 119911 of 119879Moreover lim

119899rarrinfin119879119899119909 = 119911 for all 119909 isin 119883

Suzuki proved also the following version of Edelsteinrsquosfixed point theorem

Theorem2 Let (119883 119889) be a compactmetric space Let119879 119883 rarr119883 be a self-map satisfying for all 119909 119910 isin 119883 119909 = 119910 the condition

1

2119889 (119909 119879119909) le 119889 (119909 119910) 997904rArr 119889 (119879119909 119879119910) lt 119889 (119909 119910) (3)

Then 119879 has a unique fixed point in119883

This theorem was generalized in [3]In addition to the above results Kikkawa and Suzuki [8]

provided a Kannan type version of the theorems mentionedbefore In [14] Chatterjea type version is provided Popescu[15] presented a Ciric type version Recently Kikkawa andSuzuki also provided multivalued versions which can befound in [9 10]

Very recently Hussain et al [4] have extended SuzukirsquosTheorems 1 and 2 as well as Popescursquos results from [15] to thecase of metric type spaces and cone metric type spaces (seealso [5ndash7 11])

The aim of this paper is to generalize the above-men-tioned results Indeed we prove a fixed point theorem in theset up of metric-like spaces and derive certain new results ascorollaries Finally three examples are presented to verify theeffectiveness and applicability of our main results

2 Journal of Function Spaces and Applications

In the rest of this section we recall some definitionsand facts which will be used throughout the paper First wepresent some known definitions and propositions in partialmetric and metric-like spaces

A partial metric on a nonempty set 119883 is a mapping 119901 119883 times 119883 rarr R+ such that for all 119909 119910 119911 isin 119883

(p1) 119909 = 119910 if and only if 119901(119909 119909) = 119901(119909 119910) = 119901(119910 119910)

(p2) 119901(119909 119909) le 119901(119909 119910)

(p3) 119901(119909 119910) = 119901(119910 119909)

(p4) 119901(119909 119910) le 119901(119909 119911) + 119901(119911 119910) minus 119901(119911 119911)

A partial metric space is a pair (119883 119901) such that 119883 is anonempty set and 119901 is a partial metric on 119883 It is clear thatif 119901(119909 119910) = 0 then from (p

1) and (p

2) 119909 = 119910 But if 119909 = 119910

119901(119909 119910) may not be 0 A basic example of a partial metricspace is the pair (R+ 119901) where 119901(119909 119910) = max119909 119910 for all119909 119910 isin R+

Lemma 3 (see [17]) Let (119883 119889) and (119883 119901) be a metric spaceand partial metric space respectively Then

(i) the function 120588 119883 times 119883 rarr R+ defined by 120588(119909 119910) =119889(119909 119910) + 119901(119909 119910) is a partial metric

(ii) let 120588 119883 times 119883 rarr R+ be defined by 120588(119909 119910) = 119889(119909 119910) +max120596(119909) 120596(119910) then120588 is a partialmetric on119883 where120596 119883 rarr R+ is an arbitrary function

(iii) Let 120588 R times R rarr R be defined by 120588(119909 119910) = max21199092119910 then 120588 is a partial metric on R

(iv) Let 120588 119883times119883 rarr R+ be defined by 120588(119909 119910) = 119889(119909 119910)+119886then 120588 is a partial metric on 119883 where 119886 ge 0

Other examples of the partial metric spaces which areinteresting from a computational point of viewmay be foundin [7 11 12 18]

Each partial metric 119901 on119883 generates a 1198790topology 120591

119901on

119883 which has as a base the family of open 119901-balls 119861119901(119909 120576)

119909 isin 119883 120576 gt 0 where 119861119901(119909 120576) = 119910 isin 119883 119901(119909 119910) lt 119901(119909 119909) +

120576 for all 119909 isin 119883 and 120576 gt 0Let (119883 119901) be a partial metricA sequence 119909

119899 in a partial metric space (119883 119901) converges

to a point 119909 isin 119883 if and only if 119901(119909 119909) = lim119899rarrinfin

119901(119909 119909119899)

A sequence 119909119899 in a partial metric space (119883 119901) is

called a Cauchy sequence if there exists (and is finite)lim119899119898rarrinfin

119901(119909119899 119909119898)

A partialmetric space (119883 119901) is said to be complete if everyCauchy sequence 119909

119899 in119883 converges with respect to 120591

119901 to a

point 119909 isin 119883 such that 119901(119909 119909) = lim119899119898rarrinfin

119901(119909119899 119909119898)

Suppose that 119909119899 is a sequence in partial metric space

(119883 119901) then we define 119871(119909119899) = 119909 | 119909

119899rarr 119909

The following example shows that every convergentsequence 119909

119899 in a partial metric space (119883 119901) may not be a

Cauchy sequence In particular it shows that the limit is notunique

Example 4 (see [17]) Let 119883 = [0infin) and 119901(119909 119910) =max119909 119910 Let

119909119899= 0 119899 = 2119896

1 119899 = 2119896 + 1(4)

Then clearly it is a convergent sequence and for every 119909 ge 1we have lim

119899rarrinfin119901(119909119899 119909) = 119901(119909 119909) hence 119871(119909

119899) = [1infin)

But lim119899119898rarrinfin

119901(119909119899 119909119898) does not exist that is it is not a

Cauchy sequence

Definition 5 (see [2]) A metric-like on a nonempty set119883 is amapping 120590 119883 times 119883 rarr R+ such that for all 119909 119910 119911 isin 119883

(1205901) 120590(119909 119910) = 0 rArr 119909 = 119910(1205902) 120590(119909 119910) = 120590(119910 119909)(1205903) 120590(119909 119910) le 120590(119909 119911) + 120590(119911 119910)

The pair (119883 120590) is called a metric-like space Then ametric-like on 119883 satisfies all of the conditions of a metricexcept that 120590(119909 119909) may be positive for 119909 isin 119883 Each metric-like 120590 on 119883 generates a topology 120591

120590on 119883 whose base is the

family of open 120590-ball 119861120590(119909 120576) 119909 isin 119883 120576 gt 0 where

119861120590(119909 120576) = 119910 isin 119883 |120590(119909 119910) minus 120590(119909 119909)| lt 120576 for all 119909 isin 119883

and 120576 gt 0A sequence 119909

119899 in ametric-like space (119883 120590) converges to

a point 119909 isin 119883 if and only if lim119899rarrinfin

120590(119909 119909119899) = 120590(119909 119909)

A sequence 119909119899 in a metric-like space (119883 120590) is

called a 120590-Cauchy sequence if there exists (and is finite)lim119899119898rarrinfin

120590(119909119899 119909119898)

A metric-like space (119883 120590) is said to be complete if every120590-Cauchy sequence 119909

119899 in 119883 converges with respect to 120591

120590

to a point 119909 isin 119883 such that

lim119899rarrinfin120590 (119909119899 119909) = 120590 (119909 119909) = lim

119899119898rarrinfin120590 (119909119899 119909119898) (5)

Every partial metric space is a metric-like space Belowwe give some examples of a metric-like space

Example 6 Let 119883 = [0 1] then mapping 1205901 119883 times 119883 rarr R+

defined by 1205901(119909 119910) = 119909 + 119910 minus 119909119910 is a metric-like on119883

Example 7 Let119883 = R then mappings 120590119894 119883times119883 rarr R+ (119894 isin

2 3 4) defined by

1205902(119909 119910) = |119909| +

10038161003816100381610038161199101003816100381610038161003816 + 119886

1205903(119909 119910) = |119909 minus 119887| +

1003816100381610038161003816119910 minus 1198871003816100381610038161003816

1205904(119909 119910) = 119909

2+ 1199102

(6)

are metric-like space on119883 where 119886 ge 0 and 119887 isin R

2 Main Results

We start our work by proving the following crucial lemma

Lemma 8 Let (119883 120590) be a metric-like space and suppose that119909119899 is 120590-convergent to 119909 Then for every 119910 isin 119883 one has

120590 (119909 119910) minus 120590 (119909 119909) le lim inf119899rarrinfin

120590 (119909119899 119910)

Journal of Function Spaces and Applications 3

le lim sup119899rarrinfin

120590 (119909119899 119910)

le 120590 (119909 119910) + 120590 (119909 119909)

(7)

In particular if 120590(119909 119909) = 0 then one has lim119899rarrinfin

120590(119909119899 119910) =

120590(119909 119910)

Proof Using the triangle inequality in a metric-like space itis easy to see that

120590 (119909119899 119910) le 120590 (119909

119899 119909) + 120590 (119909 119910)

120590 (119909 119910) le 120590 (119909 119909119899) + 120590 (119909

119899 119910)

(8)

Taking the upper limit as 119899 rarr infin in the first inequalityand the lower limit as 119899 rarr infin in the second inequality weobtain the desired result

Theorem 9 Let (119883 120590) be a complete metric-like space Let 119879 119883 rarr 119883 be a self-map and let 120579 = [0 1) rarr (12 1] bedefined by

120579 (119903) =

1 0 le 119903 leradic5 minus 1

2

1 minus 119903

1199032radic5 minus 1

2le 119903 le

1

radic2

1

1 + 1199031

radic2le 119903 lt 1

(9)

If there exists 119903 isin [0 1) such that for each 119909 119910 isin 119883

120579 (119903) 120590 (119909 119879119909) le 120590 (119909 119910) 997904rArr 120590 (119879119909 119879119910) le 119903120590 (119909 119910) (10)

Then 119879 has a unique fixed point 119911 isin 119883 and for each 119909 isin 119883the sequence 119879119899119909 converges to 119911

Proof Putting 119910 = 119879119909 in (10) hence from

120579 (119903) 120590 (119909 119879119909) le 120590 (119909 119879119909) (11)

it follows

120590 (119879119909 1198792119909) le 119903120590 (119909 119879119909) (12)

for every 119909 isin 119883 Let 1199090isin 119883 be arbitrary and form the

sequence 119909119899 by 119909

1= 1198791199090and 119909

119899+1= 119879119909119899for 119899 isin N cup 0

By (12) we have

120590 (119909119899 119909119899+1) = 120590 (119879119909

119899minus1 1198792119909119899minus1)

le 119903120590 (119909119899minus1 119879119909119899minus1)

= 119903120590 (119909119899minus1 119909119899)

le 119903119899120590 (1199090 1199091)

(13)

Also by the condition 1205903 of the definition of metric-likespace for all119898 ge 119899 we have

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2)

+ sdot sdot sdot + 120590 (119909119898minus1 119909119898)

le 119903119899120590 (1199090 1199091) + 119903119899+1120590 (1199090 1199091)

+ sdot sdot sdot + 119903119898minus1120590 (1199090 1199091)

=119903119899 minus 119903119898

1 minus 119903120590 (1199090 1199091)

lt119903119899

1 minus 119903120590 (1199090 1199091) 997888rarr 0 as 119899 997888rarr infin

(14)

Hence 119909119899 is a 120590-Cauchy sequence

Since119883 is 120590-complete there exists 119911 isin 119883 such that

lim119899rarrinfin120590 (119909119899 119911) = 120590 (119911 119911) = lim

119898119899rarrinfin120590 (119909119898 119909119899) = 0 (15)

That is lim119899rarrinfin

119909119899+1= lim119899rarrinfin

119879119909119899= 119911 We prove that 119879119911 =

119911 Putting 119909 = 119879119899minus1119911 in (12) we get that

120590 (119879119899119911 119879119899+1119911) le 119903120590 (119879

119899minus1119911 119879119899119911) (16)

holds for each 119899 isin N (where 1198790119911 = 119911) It follows by inductionthat

120590 (119879119899119911 119879119899+1119911) le 119903

119899120590 (119911 119879119911) (17)

Let us prove now that

120590 (119911 119879119909) le 119903120590 (119911 119909) (18)

holds for each 119909 = 119911 Since 120590(119909119899 119879119909119899) rarr 0 and by

Lemma 8 120590(119909119899 119909) rarr 120590(119911 119909) = 0 it follows that there exists

1198990isin N such that

120579 (119903) 120590 (119909119899 119879119909119899) le 120590 (119909

119899 119909) (19)

holds for every 119899 ge 1198990 Assumption (10) implies that for such

119899120590(119879119909119899 119879119909) le 119903120590(119909

119899 119909) thus as 119899 rarr infin we get that

120590 (119911 119879119909) le 119903120590 (119911 119909) (20)

We will prove that

120590 (119879119899119911 119911) le 120590 (119879119911 119911) (21)

for each 119899 isin N For 119899 = 1 this relation is obvious Supposethat it holds for some 119898 isin N If 119879119898119911 = 119911 then 119879119898+1119911 =119879119911 and 120590(119879119898+1119911 119911) = 120590(119879119911 119911) le 120590(119879119911 119911) If 119879119898119911 = 119911 thenapplying (18) and the induction hypothesis we get that

120590 (119879119898+1119911 119911) le 119903120590 (119879

119898119911 119911)

le 119903120590 (119879119911 119911) le 120590 (119879119911 119911)

(22)

and (21) is proved by induction

4 Journal of Function Spaces and Applications

In order to prove that 119879119911 = 119911 we consider two possiblecases

Case I 0 le 119903 lt 1radic2 (and hence 120579(119903) le (1 minus 119903)1199032) We willprove first that

120590 (119879119899119911 119879119911) le 119903120590 (119879119911 119911) (23)

for 119899 ge 2 For 119899 = 2 it follows from (16) Suppose that (23)holds for some 119899 gt 2 Then

120590 (119879119911 119911) le 120590 (119911 119879119899119911) + 120590 (119879

119899119911 119879119911)

le 120590 (119911 119879119899119911) + 119903120590 (119911 119879119911)

(24)

which implies (1 minus 119903)120590(119911 119879119911) le 120590(119911 119879119899119911) Using (17) weobtain

120579 (119903) 120590 (119879119899119911 119879119899+1119911) le

1 minus 119903

119903119899120590 (119879119899119911 119879119899+1119911)

le1 minus 119903

119903119899sdot 119903119899120590 (119911 119879119911)

= (1 minus 119903) 120590 (119911 119879119911) le 120590 (119911 119879119899119911)

(25)

Assumption (10) and relation (21) imply that

120590 (119879119911 119879119899+1119911) le 119903120590 (119911 119879

119899119911)

le 119903120590 (119911 119879119911)

(26)

So relation (23) is proved by inductionNow 119879119911 = 119911 and (23) implies that 119879119899119911 = 119911 for each 119899 isin N

Hence (18) imply that

120590 (119911 119879119899+1119911) le 119903120590 (119911 119879

119899119911)

le 1199032120590 (119911 119879

119899minus1119911)

le 119903119899120590 (119911 119879119911)

(27)

Hence lim119899rarrinfin

120590(119911 119879119899+1119911) = 0 = 120590(119911 119911) thus 119879119899119911 rarr 119911

and using Lemma 8 in (23) we have 120590(119911 119879119911) le 119903120590(119879119911 119911) as119899 rarr infin which implies that 120590(119911 119879119911) = 0 a contradiction

Case II 1radic2 le 119903 lt 1 (and so 120579(119903) = 1(1+119903)) We will provethat there exists a subsequence 119909

119899119896 of 119909

119899 such that

120579 (119903) 120590 (119909119899119896 119879119909119899119896) = 120579 (119903) 120590 (119909

119899119896 119909119899119896+1) le 120590 (119909

119899119896 119911) (28)

holds for each 119896 isin N From (12) we know that 120590(119909119899 119909119899+1) le

119903120590(119909119899minus1 119909119899) holds for each 119899 isin N Suppose that

1

1 + 119903120590 (119909119899minus1 119909119899) gt 120590 (119909

119899minus1 119911)

1

1 + 119903120590 (119909119899 119909119899+1) gt 120590 (119909

119899 119911)

(29)

hold for some 119899 isin N Then120590 (119909119899minus1 119909119899) le 120590 (119909

119899minus1 119911) + 120590 (119911 119909

119899)

lt1

1 + 119903120590 (119909119899minus1 119909119899) + 120590 (119909

119899 119911)

lt1

1 + 119903120590 (119909119899minus1 119909119899) +

1

1 + 119903120590 (119909119899 119909119899+1)

le1

1 + 119903120590 (119909119899minus1 119909119899) +

119903

1 + 119903120590 (119909119899minus1 119909119899)

= 120590 (119909119899minus1 119909119899)

(30)

which is impossibleHence one of the following holds for each119899

120579 (119903) 120590 (119909119899minus1 119909119899) le 120590 (119909

119899minus1 119911) (31)

or

120579 (119903) 120590 (119909119899 119909119899+1) le 120590 (119909

119899 119911) (32)

In particular

120579 (119903) 120590 (1199092119899minus1 1199092119899) le 120590 (119909

2119899minus1 119911) (33)

or

120579 (119903) 120590 (1199092119899 1199092119899+1) le 120590 (119909

2119899 119911) (34)

In other words there is a subsequence 119909119899119896 of 119909

119899 such that

(28) holds for each 119896 isin N But then assumption (10) impliesthat

120590 (119879119909119899119896 119879119911) le 119903120590 (119909

119899119896 119911) (35)

or

120590 (119879119909119899119896minus1 119879119911) le 119903120590 (119909

119899119896minus1 119911) (36)

Passing to the limit when 119896 rarr infin we get that 120590(119911 119879119911) le 0which is possible only if 119879119911 = 119911

Thus we have proved that 119911 is a fixed point of 119879 Theuniqueness of the fixed point follows easily from (10) Indeedif 119910 and 119911 are two fixed points of 119879 such that 119910 = 119911 then from(18) we have

120590 (119910 119911) = 120590 (119910 119879119911)

le 119903120590 (119910 119911) (37)

which is a contradiction

According toTheorem 9 we get the following result

Corollary 10 (see [19]) Let (119883 119889) be a complete metric spaceand let 119879 be a mapping on119883 Define a nonincreasing function120579 from [0 1] into [12 1] by

120579 (119903) =

1 0 le 119903 leradic5 minus 1

2

1 minus 119903

1199032radic5 minus 1

2le 119903 le

1

radic2

1

1 + 1199031

radic2le 119903 lt 1

(38)

Journal of Function Spaces and Applications 5

Assume that there exists 119903 isin [0 1] such that

120579 (119903) 119889 (119909 119879119909) le 119889 (119909 119910) 997904rArr 119889 (119879119909 119879119910) le 119903119889 (119909 119910) (39)

for all 119909 119910 isin 119883 then there exists a unique fixed-point 119911 of 119879Moreover lim

119899rarrinfin119879119899119909 = 119911 for all 119909 isin 119883

Proof Using a similar argument given in Theorem 9 for120590(119909 119910) = 119889(119909 119910) the desired result is obtained

Now in order to support the useability of our results letus introduce the following example

Example 11 Let119883 = [0infin) Define 120590 119883 times 119883 rarr R+ by

120590 (119909 119910) = 119909 + 119910 (40)

for all 119909 119910 isin 119883 Then (119883 120590) is a complete metric-like spaceDefine a map 119879 119883 rarr 119883 by

119879 (119909) = ln(1 + 1radic2119909) (41)

for 119909 isin 119883 Then for each 119909 119910 isin 119883 we have

1

1 + 1radic2120590 (119909 119879119909) =

radic2

radic2 + 1(119909 + ln(1 + 1

radic2119909))

leradic2

radic2 + 1(119909 +

1

radic2119909) = 119909

le 119909 + 119910 = 120590 (119909 119910)

(42)

On the other hand we have

120590 (119879119909 119879119910) = ln(1 + 1radic2119909) + ln(1 + 1

radic2119910)

le1

radic2119909 +

1

radic2119910

=1

radic2120590 (119909 119910)

(43)

Thus 119879 satisfies all the hypotheses ofTheorem 9 and hence 119879has a unique fixed point Indeed 119903 = 1radic2 120579(119903) = 1(1 + 119903)and 0 is the unique fixed point of 119879

Theorem 12 Let (119883 120590) be a complete metric-like space Let 119878119879 119883 rarr 119883 be two self-mappings Suppose that there exists119903 isin [0 1) such that

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))(44)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(45)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary and define a sequence 119909

119899 by

119909119899= 119878 (119909

119899minus1) if 119899 is odd

= 119879 (119909119899minus1) if 119899 is even

(46)

Then if 119899 isin N is odd we have

120590 (119909119899 119909119899+1)

= 120590 (119878 (119909119899minus1) 119879 (119909

119899))

= 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le max 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119878 (119909119899minus1)) 120590 (119909

119899minus1 119879 (119909119899minus1))

le 119903120590 (119909119899minus1 119878 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(47)

If 119899 is even then by (44) we have

120590 (119909119899 119909119899+1)

= 120590 (119879 (119909119899minus1) 119878 (119909

119899))

= 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le max 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119879 (119909119899minus1)) 120590 (119909

119899minus1 119878 (119909119899minus1))

le 119903120590 (119909119899minus1 119879 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(48)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899 119909119899+1) le 119903120590 (119909

119899minus1 119909119899) (49)

Repeating (49) we obtain

120590 (119909119899 119909119899+1) le 119903119899120590 (1199090 1199091) (50)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [119903119899+ 119903119899+1+ sdot sdot sdot + 119903

119898minus1] 120590 (119909

0 1199091)

le119903119899

1 minus 119903120590 (1199090 1199091)

(51)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0

That is 119909119899 is a 120590-Cauchy sequence in the metric-like

space (119883 120590) Since (119883 120590) is 120590-complete there exist 119911 isin 119883such that120590 (119911 119911) = lim

119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (52)

6 Journal of Function Spaces and Applications

Assume that 119911 is not a common fixed point of 119878 and 119879Then by hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 120590 (119909

119899 119909119899+1) 119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 119903119899120590 (1199090 1199091) 119899 isin N = 0

(53)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = max 120590 (119878 (V) 119879119878 (V)) 120590 (119879 (V) 119878119879 (V))

le 119903min 120590 (V 119878 (V)) 120590 (V 119879 (V))

= 119903min 120590 (V V) 120590 (V V)

= 119903120590 (V V)

(54)

which gives that 120590(V V) = 0

Example 13 Let (119883 120590) be a metric-like space where 119883 =1119899infin

119899=1cup 0 and 120590(119909 119910) = 119909 + 119910 Define 119878 119883 rarr 119883 by

119878(0) = 0 119878(12119899) = 1(4119899 + 3) and 119878(1(2119899 minus 1)) = 0 and119879(0) = 0 119879(1(2119899 minus 1)) = 1(4119899 + 4) and 119879(12119899) = 0 Thenfor 119909 = 12119899 we have

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

= max 120590 (119878 ( 12119899) 119879 (119878 (

1

2119899)))

120590 (119879(1

2119899) 119878 (119879(

1

2119899)))

= max 1

4119899 + 3+

1

8119899 + 12 0 =

1

4119899 + 3+

1

8119899 + 12

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

= 119903min 12119899+1

4119899 + 31

2119899+ 0 = 119903

1

2119899

(55)

It is easy to see that the above inequality is true for119909 = 1(2119899minus1) and for 34 le 119903 lt 1 Also

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(56)

for every 119910 isin 119883 with y is not a common fixed point of 119878 and119879 This shows that all conditions of Theorem 12 are satisfiedand 0 is a common fixed point for 119878 and 119879

Corollary 14 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be a mapping Suppose that there exists 119903 isin[0 1) such that

120590 (119879 (119909) 1198792(119909)) le 119903120590 (119909 119879 (119909)) (57)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) + 120590 (119909 119879 (119909)) 119909 isin 119883 gt 0 (58)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 12 the conclusion of the cor-ollary follows

Theorem 15 Let (119883 120590) be a complete metric-like space Let119878 119879 be mappings from 119883 onto itself Suppose that there exists119903 gt 1 such that

min 120590 (119879119878 (119909) 119878 (119909)) 120590 (119878119879 (119909) 119879 (119909))

ge 119903max 120590 (119878119909 119909) 120590 (119879119909 119909)(59)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(60)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary Since 119878 is onto there is an

element 1199091satisfying 119909

1isin 119878minus1(119909

0) Since 119879 is also onto there

is an element 1199092satisfying 119909

2isin 119879minus1(119909

1) Proceeding in the

same way we can find that 1199092119899+1

isin 119878minus1(1199092119899) and 119909

2119899+2isin

119879minus1(1199092119899+1) for 119899 = 1 2 3 Therefore 119909

2119899= 1198781199092119899+1

and1199092119899+1= 1198791199092119899+2

for 119899 = 0 1 2 If 119899 = 2119898 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898minus1 1199092119898)

= 120590 (1198791199092119898 1198781199092119898+1)

= 120590 (1198791198781199092119898+1 1198781199092119898+1)

ge min 120590 (119879119878 (1199092119898+1) 119878 (119909

2119898+1))

120590 (119878119879 (1199092119898+1) 119879 (119909

2119898+1))

ge 119903max 120590 (1198781199092119898+1 1199092119898+1) 120590 (119879119909

2119898+1 1199092119898+1)

ge 119903120590 (1198781199092119898+1 1199092119898+1)

= 119903120590 (1199092119898 1199092119898+1)

= 119903120590 (119909119899 119909119899+1)

(61)

Journal of Function Spaces and Applications 7

If 119899 = 2119898 + 1 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898 1199092119898+1)

= 120590 (1198781199092119898+1 1198791199092119898+2)

= 120590 (1198781198791199092119898+2 1198791199092119898+2)

ge min 120590 (119879119878 (1199092119898+2) 119878 (119909

2119898+2))

120590 (119878119879 (1199092119898+2) 119879 (119909

2119898+2))

ge 119903max 120590 (1198781199092119898+2 1199092119898+2) 120590 (119879119909

2119898+2 1199092119898+2)

ge 119903120590 (1198791199092119898+2 1199092119898+2)

= 119903120590 (1199092119898+1 1199092119898+2)

= 119903120590 (119909119899 119909119899+1)

(62)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899minus1 119909119899) ge 119903120590 (119909

119899 119909119899+1) (63)

which implies that

120590 (119909119899 119909119899+1) le1

119903120590 (119909119899minus1 119909119899) le sdot sdot sdot le (

1

119903)119899

120590 (1199090 1199091) (64)

Let 120572 = 1119903 then 0 lt 120572 lt 1 since 119903 gt 1Now (64) becomes

120590 (119909119899 119909119899+1) le 120572119899120590 (1199090 1199091) (65)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [120572119899+ 120572119899+1+ sdot sdot sdot + 120572

119898minus1] 120590 (119909

0 1199091)

le120572119899

1 minus 120572120590 (1199090 1199091)

(66)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0 That is 119909

119899 is a 120590-Cauchy

sequence in the metric-like space (119883 120590) Since (119883 120590) is 120590-complete there exists 119911 isin 119883 such that

120590 (119911 119911) = lim119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (67)

Assume that 119911 is not a common fixed point of 119878 and 119879 Thenby hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120590 (119909

119899minus1 119909119899) 119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120572119899minus1120590 (1199090 1199091) 119899 isin N = 0

(68)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)

If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = min 120590 (119879119878 (V) 119878 (V)) 120590 (119878119879 (V) 119879 (V))ge 119903max 120590 (119878 (V) V) 120590 (119879 (V) V)= 119903max 120590 (V V) 120590 (V V)= 119903120590 (V V)

(69)

which gives that 120590(V V) = 0

Corollary 16 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be an onto mapping Suppose that there exists119903 isin [0 1) such that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (70)

for every 119909 isin 119883 and that120572 (119910) = inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (71)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 15we have the desired result

Definition 17 Let (119883 120590) and (119884 120591) bemetric-like spacesThen119891 119883 rarr 119884 is said to be a continuous mapping iflim119899rarrinfin

119909119899= 119909 implies that lim

119899rarrinfin119891(119909119899) = 119891(119909)

Corollary 18 Let (119883 120590) be a complete metric-like space andlet 119879 be a mapping of 119883 into itself If there is a real number 119903with 119903 gt 1 satisfying

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (72)

for every 119909 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Assume that there exists 119910 isin 119883 with 119910 = 119879(119910) and

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 = 0 (73)

Then there exists a sequence 119909119899 such that

lim119899rarrinfin

120590 (119909119899 119910) + 120590 (119879 (119909

119899) 119909119899) = 0 (74)

So we have 120590(119909119899 119910) rarr 0 and 120590(119879(119909

119899) 119909119899) rarr 0 as 119899 rarr infin

Since 120590(119910 119910) le 120590(119910 119909119899) + 120590(119909

119899 119910) hence 120590(119910 119910) rarr 0 as

119899 rarr infin Now120590 (119879 (119909

119899) 119910) le 120590 (119879 (119909

119899) 119909119899) + 120590 (119909

119899 119910) 997888rarr 0

as 119899 997888rarr infin(75)

Since 119879 is continuous we have

119879 (119910) = 119879 ( lim119899rarrinfin119909119899) = lim119899rarrinfin119879 (119909119899) = 119910 (76)

This is a contradiction Hence if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (77)

which is condition (71) of Corollary 16 By Corollary 16 thereexists 119911 isin 119883 such that 119911 = 119879(119911)

8 Journal of Function Spaces and Applications

Now we give an example to support our result

Example 19 Let 119883 = [0infin) and 120590(119909 119910) = 119909 + 119910 Define 119879 119883 rarr 119883 by 119879(119909) = 2119909

Obviously 119879 is onto and continuous Also for each 119909 119910 isin119883 we have

120590 (1198792119909 119879119909) = 4119909 + 2119909 = 6119909 ge 1199033119909 = 119903120590 (119879119909 119909) (78)

where 119903 = 2 Thus 119879 satisfies the conditions given inCorollary 18 and 0 is the fixed point of 119879

Corollary 20 Let (119883 120590) be a complete metric-like space and119879 be a mapping of119883 into itself If there is a real number 119903 with119903 gt 1 satisfying

120590 (119879 (119909) 119879 (119910)) ge 119903min 120590 (119909 119879 (119909)) 120590 (119879 (119910) 119910) 120590 (119909 119910)(79)

for every 119909 119910 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Replacing 119910 by 119879(119909) in (79) we obtain

120590 (119879 (119909) 1198792(119909))

ge 119903min 120590 (119909 119879 (119909)) 120590 (1198792 (119909) 119879 (119909)) 120590 (119909 119879 (119909))(80)

for all 119909 isin 119883Without loss of generality we may assume that 119879(119909) =

1198792(119909) Otherwise 119879 has a fixed point Since 119903 gt 1 it followsfrom (80) that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (81)

for every 119909 isin 119883 By the argument similar to that used inCorollary 18 we can prove that if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (82)

which is condition (71) of Corollary 16 So Corollary 16applies to obtain a fixed point of 119879

According toTheorem 12 we get the following result

Corollary 21 (see [17 Theorem 1]) Let (119883 119901) be a completepartial metric space Let 119878 119879 119883 rarr 119883 be two self-mappingsSuppose that there exists 119903 isin [0 1) such that

max 119901 (119878 (119909) 119879119878 (119909)) 119901 (119879 (119909) 119878119879 (119909))

le 119903min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))(83)

for every 119909 isin 119883 and that

120572 (119910) = inf 119901 (119909 119910) +min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))

119909 isin 119883 gt 0

(84)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 119901(V V) = 0

Proof Using a similar argument given in the Theorem 12 for120590(119909 119910) = 119901(119909 119910) the desired result is obtained where 119901 is apartial metric on119883

Also according to Theorem 15 we get Theorem 2 from[17]

References

[1] A Aghajani S Radenovic and J R Roshan ldquoCommon fixedpoint results for four mappings satisfying almost generalized(119878 119879)-contractive condition in partially ordered metric spacesrdquoAppliedMathematics andComputation vol 218 no 9 pp 5665ndash5670 2012

[2] A Amini-Harandi ldquoMetric-like spaces partial metric spacesand fixed pointsrdquo Fixed PointTheory and Applications vol 2012article 204 10 pages 2012

[3] D ETHoric Z Kadelburg and S Radenovic ldquoEdelstein-Suzuki-type fixed point results in metric and abstract metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 75 no4 pp 1927ndash1932 2012

[4] N Hussain D ETHoric Z Kadelburg and S Radenovic ldquoSuzuki-type fixed point results in metric type spacesrdquo Fixed PointTheory and Applications vol 2012 article 126 12 pages 2012

[5] N Hussain and M H Shah ldquoKKMmappings in cone 119887-metricspacesrdquoComputersampMathematics withApplications vol 62 no4 pp 1677ndash1684 2011

[6] N Hussain M H Shah A Amini-Harandi and Z AkhtarldquoCommon fixed point theorems for generalized contractivemappings with applicationsrdquo Fixed Point Theory and Applica-tions vol 2013 article 169 17 pages 2013

[7] N Hussain Z Kadelburg S Radenovic and F Al-SolamyldquoComparison functions and fixed point results in partial metricspacesrdquo Abstract and Applied Analysis vol 2012 Article ID605781 15 pages 2012

[8] M Kikkawa and T Suzuki ldquoSome similarity between con-tractions and Kannan mappingsrdquo Fixed Point Theory andApplications vol 2008 Article ID 649749 8 pages 2008

[9] MKikkawa andT Suzuki ldquoSomenotes on fixed point theoremswith constantsrdquo Bulletin of the Kyushu Institute of Technologyno 56 pp 11ndash18 2009

[10] M Kikkawa and T Suzuki ldquoThree fixed point theorems forgeneralized contractions with constants in complete metricspacesrdquoNonlinearAnalysisTheoryMethodsampApplications vol69 no 9 pp 2942ndash2949 2008

[11] M A Kutbi J Ahmad N Hussain and M Arshad ldquoCommonfixed point results for mappings with rational expressionsrdquoAbstract and Applied Analysis vol 2013 Article ID 549518 11pages 2013

[12] S G Matthews ldquoPartial metric topologyrdquo in Papers on GeneralTopology and Applications vol 728 of Annals of the NewYork Academy of Sciences pp 183ndash197 New York Academy ofSciences New York NY USA 1994

[13] V Parvaneh J R Roshan and S Radenovic ldquoExistence oftripled coincidence points in ordered b-metric spaces and anapplication to a systemof integral equationsrdquo Fixed PointTheoryand Applications vol 2013 article 130 19 pages 2013

[14] O Popescu ldquoFixed point theorem in metric spacesrdquo Bulletinof the Transilvania University of Brasov vol 150 pp 479ndash4822008

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Journal of Function Spaces and Applications

In the rest of this section we recall some definitionsand facts which will be used throughout the paper First wepresent some known definitions and propositions in partialmetric and metric-like spaces

A partial metric on a nonempty set 119883 is a mapping 119901 119883 times 119883 rarr R+ such that for all 119909 119910 119911 isin 119883

(p1) 119909 = 119910 if and only if 119901(119909 119909) = 119901(119909 119910) = 119901(119910 119910)

(p2) 119901(119909 119909) le 119901(119909 119910)

(p3) 119901(119909 119910) = 119901(119910 119909)

(p4) 119901(119909 119910) le 119901(119909 119911) + 119901(119911 119910) minus 119901(119911 119911)

A partial metric space is a pair (119883 119901) such that 119883 is anonempty set and 119901 is a partial metric on 119883 It is clear thatif 119901(119909 119910) = 0 then from (p

1) and (p

2) 119909 = 119910 But if 119909 = 119910

119901(119909 119910) may not be 0 A basic example of a partial metricspace is the pair (R+ 119901) where 119901(119909 119910) = max119909 119910 for all119909 119910 isin R+

Lemma 3 (see [17]) Let (119883 119889) and (119883 119901) be a metric spaceand partial metric space respectively Then

(i) the function 120588 119883 times 119883 rarr R+ defined by 120588(119909 119910) =119889(119909 119910) + 119901(119909 119910) is a partial metric

(ii) let 120588 119883 times 119883 rarr R+ be defined by 120588(119909 119910) = 119889(119909 119910) +max120596(119909) 120596(119910) then120588 is a partialmetric on119883 where120596 119883 rarr R+ is an arbitrary function

(iii) Let 120588 R times R rarr R be defined by 120588(119909 119910) = max21199092119910 then 120588 is a partial metric on R

(iv) Let 120588 119883times119883 rarr R+ be defined by 120588(119909 119910) = 119889(119909 119910)+119886then 120588 is a partial metric on 119883 where 119886 ge 0

Other examples of the partial metric spaces which areinteresting from a computational point of viewmay be foundin [7 11 12 18]

Each partial metric 119901 on119883 generates a 1198790topology 120591

119901on

119883 which has as a base the family of open 119901-balls 119861119901(119909 120576)

119909 isin 119883 120576 gt 0 where 119861119901(119909 120576) = 119910 isin 119883 119901(119909 119910) lt 119901(119909 119909) +

120576 for all 119909 isin 119883 and 120576 gt 0Let (119883 119901) be a partial metricA sequence 119909

119899 in a partial metric space (119883 119901) converges

to a point 119909 isin 119883 if and only if 119901(119909 119909) = lim119899rarrinfin

119901(119909 119909119899)

A sequence 119909119899 in a partial metric space (119883 119901) is

called a Cauchy sequence if there exists (and is finite)lim119899119898rarrinfin

119901(119909119899 119909119898)

A partialmetric space (119883 119901) is said to be complete if everyCauchy sequence 119909

119899 in119883 converges with respect to 120591

119901 to a

point 119909 isin 119883 such that 119901(119909 119909) = lim119899119898rarrinfin

119901(119909119899 119909119898)

Suppose that 119909119899 is a sequence in partial metric space

(119883 119901) then we define 119871(119909119899) = 119909 | 119909

119899rarr 119909

The following example shows that every convergentsequence 119909

119899 in a partial metric space (119883 119901) may not be a

Cauchy sequence In particular it shows that the limit is notunique

Example 4 (see [17]) Let 119883 = [0infin) and 119901(119909 119910) =max119909 119910 Let

119909119899= 0 119899 = 2119896

1 119899 = 2119896 + 1(4)

Then clearly it is a convergent sequence and for every 119909 ge 1we have lim

119899rarrinfin119901(119909119899 119909) = 119901(119909 119909) hence 119871(119909

119899) = [1infin)

But lim119899119898rarrinfin

119901(119909119899 119909119898) does not exist that is it is not a

Cauchy sequence

Definition 5 (see [2]) A metric-like on a nonempty set119883 is amapping 120590 119883 times 119883 rarr R+ such that for all 119909 119910 119911 isin 119883

(1205901) 120590(119909 119910) = 0 rArr 119909 = 119910(1205902) 120590(119909 119910) = 120590(119910 119909)(1205903) 120590(119909 119910) le 120590(119909 119911) + 120590(119911 119910)

The pair (119883 120590) is called a metric-like space Then ametric-like on 119883 satisfies all of the conditions of a metricexcept that 120590(119909 119909) may be positive for 119909 isin 119883 Each metric-like 120590 on 119883 generates a topology 120591

120590on 119883 whose base is the

family of open 120590-ball 119861120590(119909 120576) 119909 isin 119883 120576 gt 0 where

119861120590(119909 120576) = 119910 isin 119883 |120590(119909 119910) minus 120590(119909 119909)| lt 120576 for all 119909 isin 119883

and 120576 gt 0A sequence 119909

119899 in ametric-like space (119883 120590) converges to

a point 119909 isin 119883 if and only if lim119899rarrinfin

120590(119909 119909119899) = 120590(119909 119909)

A sequence 119909119899 in a metric-like space (119883 120590) is

called a 120590-Cauchy sequence if there exists (and is finite)lim119899119898rarrinfin

120590(119909119899 119909119898)

A metric-like space (119883 120590) is said to be complete if every120590-Cauchy sequence 119909

119899 in 119883 converges with respect to 120591

120590

to a point 119909 isin 119883 such that

lim119899rarrinfin120590 (119909119899 119909) = 120590 (119909 119909) = lim

119899119898rarrinfin120590 (119909119899 119909119898) (5)

Every partial metric space is a metric-like space Belowwe give some examples of a metric-like space

Example 6 Let 119883 = [0 1] then mapping 1205901 119883 times 119883 rarr R+

defined by 1205901(119909 119910) = 119909 + 119910 minus 119909119910 is a metric-like on119883

Example 7 Let119883 = R then mappings 120590119894 119883times119883 rarr R+ (119894 isin

2 3 4) defined by

1205902(119909 119910) = |119909| +

10038161003816100381610038161199101003816100381610038161003816 + 119886

1205903(119909 119910) = |119909 minus 119887| +

1003816100381610038161003816119910 minus 1198871003816100381610038161003816

1205904(119909 119910) = 119909

2+ 1199102

(6)

are metric-like space on119883 where 119886 ge 0 and 119887 isin R

2 Main Results

We start our work by proving the following crucial lemma

Lemma 8 Let (119883 120590) be a metric-like space and suppose that119909119899 is 120590-convergent to 119909 Then for every 119910 isin 119883 one has

120590 (119909 119910) minus 120590 (119909 119909) le lim inf119899rarrinfin

120590 (119909119899 119910)

Journal of Function Spaces and Applications 3

le lim sup119899rarrinfin

120590 (119909119899 119910)

le 120590 (119909 119910) + 120590 (119909 119909)

(7)

In particular if 120590(119909 119909) = 0 then one has lim119899rarrinfin

120590(119909119899 119910) =

120590(119909 119910)

Proof Using the triangle inequality in a metric-like space itis easy to see that

120590 (119909119899 119910) le 120590 (119909

119899 119909) + 120590 (119909 119910)

120590 (119909 119910) le 120590 (119909 119909119899) + 120590 (119909

119899 119910)

(8)

Taking the upper limit as 119899 rarr infin in the first inequalityand the lower limit as 119899 rarr infin in the second inequality weobtain the desired result

Theorem 9 Let (119883 120590) be a complete metric-like space Let 119879 119883 rarr 119883 be a self-map and let 120579 = [0 1) rarr (12 1] bedefined by

120579 (119903) =

1 0 le 119903 leradic5 minus 1

2

1 minus 119903

1199032radic5 minus 1

2le 119903 le

1

radic2

1

1 + 1199031

radic2le 119903 lt 1

(9)

If there exists 119903 isin [0 1) such that for each 119909 119910 isin 119883

120579 (119903) 120590 (119909 119879119909) le 120590 (119909 119910) 997904rArr 120590 (119879119909 119879119910) le 119903120590 (119909 119910) (10)

Then 119879 has a unique fixed point 119911 isin 119883 and for each 119909 isin 119883the sequence 119879119899119909 converges to 119911

Proof Putting 119910 = 119879119909 in (10) hence from

120579 (119903) 120590 (119909 119879119909) le 120590 (119909 119879119909) (11)

it follows

120590 (119879119909 1198792119909) le 119903120590 (119909 119879119909) (12)

for every 119909 isin 119883 Let 1199090isin 119883 be arbitrary and form the

sequence 119909119899 by 119909

1= 1198791199090and 119909

119899+1= 119879119909119899for 119899 isin N cup 0

By (12) we have

120590 (119909119899 119909119899+1) = 120590 (119879119909

119899minus1 1198792119909119899minus1)

le 119903120590 (119909119899minus1 119879119909119899minus1)

= 119903120590 (119909119899minus1 119909119899)

le 119903119899120590 (1199090 1199091)

(13)

Also by the condition 1205903 of the definition of metric-likespace for all119898 ge 119899 we have

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2)

+ sdot sdot sdot + 120590 (119909119898minus1 119909119898)

le 119903119899120590 (1199090 1199091) + 119903119899+1120590 (1199090 1199091)

+ sdot sdot sdot + 119903119898minus1120590 (1199090 1199091)

=119903119899 minus 119903119898

1 minus 119903120590 (1199090 1199091)

lt119903119899

1 minus 119903120590 (1199090 1199091) 997888rarr 0 as 119899 997888rarr infin

(14)

Hence 119909119899 is a 120590-Cauchy sequence

Since119883 is 120590-complete there exists 119911 isin 119883 such that

lim119899rarrinfin120590 (119909119899 119911) = 120590 (119911 119911) = lim

119898119899rarrinfin120590 (119909119898 119909119899) = 0 (15)

That is lim119899rarrinfin

119909119899+1= lim119899rarrinfin

119879119909119899= 119911 We prove that 119879119911 =

119911 Putting 119909 = 119879119899minus1119911 in (12) we get that

120590 (119879119899119911 119879119899+1119911) le 119903120590 (119879

119899minus1119911 119879119899119911) (16)

holds for each 119899 isin N (where 1198790119911 = 119911) It follows by inductionthat

120590 (119879119899119911 119879119899+1119911) le 119903

119899120590 (119911 119879119911) (17)

Let us prove now that

120590 (119911 119879119909) le 119903120590 (119911 119909) (18)

holds for each 119909 = 119911 Since 120590(119909119899 119879119909119899) rarr 0 and by

Lemma 8 120590(119909119899 119909) rarr 120590(119911 119909) = 0 it follows that there exists

1198990isin N such that

120579 (119903) 120590 (119909119899 119879119909119899) le 120590 (119909

119899 119909) (19)

holds for every 119899 ge 1198990 Assumption (10) implies that for such

119899120590(119879119909119899 119879119909) le 119903120590(119909

119899 119909) thus as 119899 rarr infin we get that

120590 (119911 119879119909) le 119903120590 (119911 119909) (20)

We will prove that

120590 (119879119899119911 119911) le 120590 (119879119911 119911) (21)

for each 119899 isin N For 119899 = 1 this relation is obvious Supposethat it holds for some 119898 isin N If 119879119898119911 = 119911 then 119879119898+1119911 =119879119911 and 120590(119879119898+1119911 119911) = 120590(119879119911 119911) le 120590(119879119911 119911) If 119879119898119911 = 119911 thenapplying (18) and the induction hypothesis we get that

120590 (119879119898+1119911 119911) le 119903120590 (119879

119898119911 119911)

le 119903120590 (119879119911 119911) le 120590 (119879119911 119911)

(22)

and (21) is proved by induction

4 Journal of Function Spaces and Applications

In order to prove that 119879119911 = 119911 we consider two possiblecases

Case I 0 le 119903 lt 1radic2 (and hence 120579(119903) le (1 minus 119903)1199032) We willprove first that

120590 (119879119899119911 119879119911) le 119903120590 (119879119911 119911) (23)

for 119899 ge 2 For 119899 = 2 it follows from (16) Suppose that (23)holds for some 119899 gt 2 Then

120590 (119879119911 119911) le 120590 (119911 119879119899119911) + 120590 (119879

119899119911 119879119911)

le 120590 (119911 119879119899119911) + 119903120590 (119911 119879119911)

(24)

which implies (1 minus 119903)120590(119911 119879119911) le 120590(119911 119879119899119911) Using (17) weobtain

120579 (119903) 120590 (119879119899119911 119879119899+1119911) le

1 minus 119903

119903119899120590 (119879119899119911 119879119899+1119911)

le1 minus 119903

119903119899sdot 119903119899120590 (119911 119879119911)

= (1 minus 119903) 120590 (119911 119879119911) le 120590 (119911 119879119899119911)

(25)

Assumption (10) and relation (21) imply that

120590 (119879119911 119879119899+1119911) le 119903120590 (119911 119879

119899119911)

le 119903120590 (119911 119879119911)

(26)

So relation (23) is proved by inductionNow 119879119911 = 119911 and (23) implies that 119879119899119911 = 119911 for each 119899 isin N

Hence (18) imply that

120590 (119911 119879119899+1119911) le 119903120590 (119911 119879

119899119911)

le 1199032120590 (119911 119879

119899minus1119911)

le 119903119899120590 (119911 119879119911)

(27)

Hence lim119899rarrinfin

120590(119911 119879119899+1119911) = 0 = 120590(119911 119911) thus 119879119899119911 rarr 119911

and using Lemma 8 in (23) we have 120590(119911 119879119911) le 119903120590(119879119911 119911) as119899 rarr infin which implies that 120590(119911 119879119911) = 0 a contradiction

Case II 1radic2 le 119903 lt 1 (and so 120579(119903) = 1(1+119903)) We will provethat there exists a subsequence 119909

119899119896 of 119909

119899 such that

120579 (119903) 120590 (119909119899119896 119879119909119899119896) = 120579 (119903) 120590 (119909

119899119896 119909119899119896+1) le 120590 (119909

119899119896 119911) (28)

holds for each 119896 isin N From (12) we know that 120590(119909119899 119909119899+1) le

119903120590(119909119899minus1 119909119899) holds for each 119899 isin N Suppose that

1

1 + 119903120590 (119909119899minus1 119909119899) gt 120590 (119909

119899minus1 119911)

1

1 + 119903120590 (119909119899 119909119899+1) gt 120590 (119909

119899 119911)

(29)

hold for some 119899 isin N Then120590 (119909119899minus1 119909119899) le 120590 (119909

119899minus1 119911) + 120590 (119911 119909

119899)

lt1

1 + 119903120590 (119909119899minus1 119909119899) + 120590 (119909

119899 119911)

lt1

1 + 119903120590 (119909119899minus1 119909119899) +

1

1 + 119903120590 (119909119899 119909119899+1)

le1

1 + 119903120590 (119909119899minus1 119909119899) +

119903

1 + 119903120590 (119909119899minus1 119909119899)

= 120590 (119909119899minus1 119909119899)

(30)

which is impossibleHence one of the following holds for each119899

120579 (119903) 120590 (119909119899minus1 119909119899) le 120590 (119909

119899minus1 119911) (31)

or

120579 (119903) 120590 (119909119899 119909119899+1) le 120590 (119909

119899 119911) (32)

In particular

120579 (119903) 120590 (1199092119899minus1 1199092119899) le 120590 (119909

2119899minus1 119911) (33)

or

120579 (119903) 120590 (1199092119899 1199092119899+1) le 120590 (119909

2119899 119911) (34)

In other words there is a subsequence 119909119899119896 of 119909

119899 such that

(28) holds for each 119896 isin N But then assumption (10) impliesthat

120590 (119879119909119899119896 119879119911) le 119903120590 (119909

119899119896 119911) (35)

or

120590 (119879119909119899119896minus1 119879119911) le 119903120590 (119909

119899119896minus1 119911) (36)

Passing to the limit when 119896 rarr infin we get that 120590(119911 119879119911) le 0which is possible only if 119879119911 = 119911

Thus we have proved that 119911 is a fixed point of 119879 Theuniqueness of the fixed point follows easily from (10) Indeedif 119910 and 119911 are two fixed points of 119879 such that 119910 = 119911 then from(18) we have

120590 (119910 119911) = 120590 (119910 119879119911)

le 119903120590 (119910 119911) (37)

which is a contradiction

According toTheorem 9 we get the following result

Corollary 10 (see [19]) Let (119883 119889) be a complete metric spaceand let 119879 be a mapping on119883 Define a nonincreasing function120579 from [0 1] into [12 1] by

120579 (119903) =

1 0 le 119903 leradic5 minus 1

2

1 minus 119903

1199032radic5 minus 1

2le 119903 le

1

radic2

1

1 + 1199031

radic2le 119903 lt 1

(38)

Journal of Function Spaces and Applications 5

Assume that there exists 119903 isin [0 1] such that

120579 (119903) 119889 (119909 119879119909) le 119889 (119909 119910) 997904rArr 119889 (119879119909 119879119910) le 119903119889 (119909 119910) (39)

for all 119909 119910 isin 119883 then there exists a unique fixed-point 119911 of 119879Moreover lim

119899rarrinfin119879119899119909 = 119911 for all 119909 isin 119883

Proof Using a similar argument given in Theorem 9 for120590(119909 119910) = 119889(119909 119910) the desired result is obtained

Now in order to support the useability of our results letus introduce the following example

Example 11 Let119883 = [0infin) Define 120590 119883 times 119883 rarr R+ by

120590 (119909 119910) = 119909 + 119910 (40)

for all 119909 119910 isin 119883 Then (119883 120590) is a complete metric-like spaceDefine a map 119879 119883 rarr 119883 by

119879 (119909) = ln(1 + 1radic2119909) (41)

for 119909 isin 119883 Then for each 119909 119910 isin 119883 we have

1

1 + 1radic2120590 (119909 119879119909) =

radic2

radic2 + 1(119909 + ln(1 + 1

radic2119909))

leradic2

radic2 + 1(119909 +

1

radic2119909) = 119909

le 119909 + 119910 = 120590 (119909 119910)

(42)

On the other hand we have

120590 (119879119909 119879119910) = ln(1 + 1radic2119909) + ln(1 + 1

radic2119910)

le1

radic2119909 +

1

radic2119910

=1

radic2120590 (119909 119910)

(43)

Thus 119879 satisfies all the hypotheses ofTheorem 9 and hence 119879has a unique fixed point Indeed 119903 = 1radic2 120579(119903) = 1(1 + 119903)and 0 is the unique fixed point of 119879

Theorem 12 Let (119883 120590) be a complete metric-like space Let 119878119879 119883 rarr 119883 be two self-mappings Suppose that there exists119903 isin [0 1) such that

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))(44)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(45)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary and define a sequence 119909

119899 by

119909119899= 119878 (119909

119899minus1) if 119899 is odd

= 119879 (119909119899minus1) if 119899 is even

(46)

Then if 119899 isin N is odd we have

120590 (119909119899 119909119899+1)

= 120590 (119878 (119909119899minus1) 119879 (119909

119899))

= 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le max 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119878 (119909119899minus1)) 120590 (119909

119899minus1 119879 (119909119899minus1))

le 119903120590 (119909119899minus1 119878 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(47)

If 119899 is even then by (44) we have

120590 (119909119899 119909119899+1)

= 120590 (119879 (119909119899minus1) 119878 (119909

119899))

= 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le max 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119879 (119909119899minus1)) 120590 (119909

119899minus1 119878 (119909119899minus1))

le 119903120590 (119909119899minus1 119879 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(48)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899 119909119899+1) le 119903120590 (119909

119899minus1 119909119899) (49)

Repeating (49) we obtain

120590 (119909119899 119909119899+1) le 119903119899120590 (1199090 1199091) (50)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [119903119899+ 119903119899+1+ sdot sdot sdot + 119903

119898minus1] 120590 (119909

0 1199091)

le119903119899

1 minus 119903120590 (1199090 1199091)

(51)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0

That is 119909119899 is a 120590-Cauchy sequence in the metric-like

space (119883 120590) Since (119883 120590) is 120590-complete there exist 119911 isin 119883such that120590 (119911 119911) = lim

119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (52)

6 Journal of Function Spaces and Applications

Assume that 119911 is not a common fixed point of 119878 and 119879Then by hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 120590 (119909

119899 119909119899+1) 119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 119903119899120590 (1199090 1199091) 119899 isin N = 0

(53)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = max 120590 (119878 (V) 119879119878 (V)) 120590 (119879 (V) 119878119879 (V))

le 119903min 120590 (V 119878 (V)) 120590 (V 119879 (V))

= 119903min 120590 (V V) 120590 (V V)

= 119903120590 (V V)

(54)

which gives that 120590(V V) = 0

Example 13 Let (119883 120590) be a metric-like space where 119883 =1119899infin

119899=1cup 0 and 120590(119909 119910) = 119909 + 119910 Define 119878 119883 rarr 119883 by

119878(0) = 0 119878(12119899) = 1(4119899 + 3) and 119878(1(2119899 minus 1)) = 0 and119879(0) = 0 119879(1(2119899 minus 1)) = 1(4119899 + 4) and 119879(12119899) = 0 Thenfor 119909 = 12119899 we have

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

= max 120590 (119878 ( 12119899) 119879 (119878 (

1

2119899)))

120590 (119879(1

2119899) 119878 (119879(

1

2119899)))

= max 1

4119899 + 3+

1

8119899 + 12 0 =

1

4119899 + 3+

1

8119899 + 12

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

= 119903min 12119899+1

4119899 + 31

2119899+ 0 = 119903

1

2119899

(55)

It is easy to see that the above inequality is true for119909 = 1(2119899minus1) and for 34 le 119903 lt 1 Also

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(56)

for every 119910 isin 119883 with y is not a common fixed point of 119878 and119879 This shows that all conditions of Theorem 12 are satisfiedand 0 is a common fixed point for 119878 and 119879

Corollary 14 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be a mapping Suppose that there exists 119903 isin[0 1) such that

120590 (119879 (119909) 1198792(119909)) le 119903120590 (119909 119879 (119909)) (57)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) + 120590 (119909 119879 (119909)) 119909 isin 119883 gt 0 (58)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 12 the conclusion of the cor-ollary follows

Theorem 15 Let (119883 120590) be a complete metric-like space Let119878 119879 be mappings from 119883 onto itself Suppose that there exists119903 gt 1 such that

min 120590 (119879119878 (119909) 119878 (119909)) 120590 (119878119879 (119909) 119879 (119909))

ge 119903max 120590 (119878119909 119909) 120590 (119879119909 119909)(59)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(60)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary Since 119878 is onto there is an

element 1199091satisfying 119909

1isin 119878minus1(119909

0) Since 119879 is also onto there

is an element 1199092satisfying 119909

2isin 119879minus1(119909

1) Proceeding in the

same way we can find that 1199092119899+1

isin 119878minus1(1199092119899) and 119909

2119899+2isin

119879minus1(1199092119899+1) for 119899 = 1 2 3 Therefore 119909

2119899= 1198781199092119899+1

and1199092119899+1= 1198791199092119899+2

for 119899 = 0 1 2 If 119899 = 2119898 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898minus1 1199092119898)

= 120590 (1198791199092119898 1198781199092119898+1)

= 120590 (1198791198781199092119898+1 1198781199092119898+1)

ge min 120590 (119879119878 (1199092119898+1) 119878 (119909

2119898+1))

120590 (119878119879 (1199092119898+1) 119879 (119909

2119898+1))

ge 119903max 120590 (1198781199092119898+1 1199092119898+1) 120590 (119879119909

2119898+1 1199092119898+1)

ge 119903120590 (1198781199092119898+1 1199092119898+1)

= 119903120590 (1199092119898 1199092119898+1)

= 119903120590 (119909119899 119909119899+1)

(61)

Journal of Function Spaces and Applications 7

If 119899 = 2119898 + 1 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898 1199092119898+1)

= 120590 (1198781199092119898+1 1198791199092119898+2)

= 120590 (1198781198791199092119898+2 1198791199092119898+2)

ge min 120590 (119879119878 (1199092119898+2) 119878 (119909

2119898+2))

120590 (119878119879 (1199092119898+2) 119879 (119909

2119898+2))

ge 119903max 120590 (1198781199092119898+2 1199092119898+2) 120590 (119879119909

2119898+2 1199092119898+2)

ge 119903120590 (1198791199092119898+2 1199092119898+2)

= 119903120590 (1199092119898+1 1199092119898+2)

= 119903120590 (119909119899 119909119899+1)

(62)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899minus1 119909119899) ge 119903120590 (119909

119899 119909119899+1) (63)

which implies that

120590 (119909119899 119909119899+1) le1

119903120590 (119909119899minus1 119909119899) le sdot sdot sdot le (

1

119903)119899

120590 (1199090 1199091) (64)

Let 120572 = 1119903 then 0 lt 120572 lt 1 since 119903 gt 1Now (64) becomes

120590 (119909119899 119909119899+1) le 120572119899120590 (1199090 1199091) (65)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [120572119899+ 120572119899+1+ sdot sdot sdot + 120572

119898minus1] 120590 (119909

0 1199091)

le120572119899

1 minus 120572120590 (1199090 1199091)

(66)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0 That is 119909

119899 is a 120590-Cauchy

sequence in the metric-like space (119883 120590) Since (119883 120590) is 120590-complete there exists 119911 isin 119883 such that

120590 (119911 119911) = lim119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (67)

Assume that 119911 is not a common fixed point of 119878 and 119879 Thenby hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120590 (119909

119899minus1 119909119899) 119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120572119899minus1120590 (1199090 1199091) 119899 isin N = 0

(68)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)

If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = min 120590 (119879119878 (V) 119878 (V)) 120590 (119878119879 (V) 119879 (V))ge 119903max 120590 (119878 (V) V) 120590 (119879 (V) V)= 119903max 120590 (V V) 120590 (V V)= 119903120590 (V V)

(69)

which gives that 120590(V V) = 0

Corollary 16 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be an onto mapping Suppose that there exists119903 isin [0 1) such that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (70)

for every 119909 isin 119883 and that120572 (119910) = inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (71)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 15we have the desired result

Definition 17 Let (119883 120590) and (119884 120591) bemetric-like spacesThen119891 119883 rarr 119884 is said to be a continuous mapping iflim119899rarrinfin

119909119899= 119909 implies that lim

119899rarrinfin119891(119909119899) = 119891(119909)

Corollary 18 Let (119883 120590) be a complete metric-like space andlet 119879 be a mapping of 119883 into itself If there is a real number 119903with 119903 gt 1 satisfying

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (72)

for every 119909 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Assume that there exists 119910 isin 119883 with 119910 = 119879(119910) and

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 = 0 (73)

Then there exists a sequence 119909119899 such that

lim119899rarrinfin

120590 (119909119899 119910) + 120590 (119879 (119909

119899) 119909119899) = 0 (74)

So we have 120590(119909119899 119910) rarr 0 and 120590(119879(119909

119899) 119909119899) rarr 0 as 119899 rarr infin

Since 120590(119910 119910) le 120590(119910 119909119899) + 120590(119909

119899 119910) hence 120590(119910 119910) rarr 0 as

119899 rarr infin Now120590 (119879 (119909

119899) 119910) le 120590 (119879 (119909

119899) 119909119899) + 120590 (119909

119899 119910) 997888rarr 0

as 119899 997888rarr infin(75)

Since 119879 is continuous we have

119879 (119910) = 119879 ( lim119899rarrinfin119909119899) = lim119899rarrinfin119879 (119909119899) = 119910 (76)

This is a contradiction Hence if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (77)

which is condition (71) of Corollary 16 By Corollary 16 thereexists 119911 isin 119883 such that 119911 = 119879(119911)

8 Journal of Function Spaces and Applications

Now we give an example to support our result

Example 19 Let 119883 = [0infin) and 120590(119909 119910) = 119909 + 119910 Define 119879 119883 rarr 119883 by 119879(119909) = 2119909

Obviously 119879 is onto and continuous Also for each 119909 119910 isin119883 we have

120590 (1198792119909 119879119909) = 4119909 + 2119909 = 6119909 ge 1199033119909 = 119903120590 (119879119909 119909) (78)

where 119903 = 2 Thus 119879 satisfies the conditions given inCorollary 18 and 0 is the fixed point of 119879

Corollary 20 Let (119883 120590) be a complete metric-like space and119879 be a mapping of119883 into itself If there is a real number 119903 with119903 gt 1 satisfying

120590 (119879 (119909) 119879 (119910)) ge 119903min 120590 (119909 119879 (119909)) 120590 (119879 (119910) 119910) 120590 (119909 119910)(79)

for every 119909 119910 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Replacing 119910 by 119879(119909) in (79) we obtain

120590 (119879 (119909) 1198792(119909))

ge 119903min 120590 (119909 119879 (119909)) 120590 (1198792 (119909) 119879 (119909)) 120590 (119909 119879 (119909))(80)

for all 119909 isin 119883Without loss of generality we may assume that 119879(119909) =

1198792(119909) Otherwise 119879 has a fixed point Since 119903 gt 1 it followsfrom (80) that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (81)

for every 119909 isin 119883 By the argument similar to that used inCorollary 18 we can prove that if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (82)

which is condition (71) of Corollary 16 So Corollary 16applies to obtain a fixed point of 119879

According toTheorem 12 we get the following result

Corollary 21 (see [17 Theorem 1]) Let (119883 119901) be a completepartial metric space Let 119878 119879 119883 rarr 119883 be two self-mappingsSuppose that there exists 119903 isin [0 1) such that

max 119901 (119878 (119909) 119879119878 (119909)) 119901 (119879 (119909) 119878119879 (119909))

le 119903min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))(83)

for every 119909 isin 119883 and that

120572 (119910) = inf 119901 (119909 119910) +min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))

119909 isin 119883 gt 0

(84)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 119901(V V) = 0

Proof Using a similar argument given in the Theorem 12 for120590(119909 119910) = 119901(119909 119910) the desired result is obtained where 119901 is apartial metric on119883

Also according to Theorem 15 we get Theorem 2 from[17]

References

[1] A Aghajani S Radenovic and J R Roshan ldquoCommon fixedpoint results for four mappings satisfying almost generalized(119878 119879)-contractive condition in partially ordered metric spacesrdquoAppliedMathematics andComputation vol 218 no 9 pp 5665ndash5670 2012

[2] A Amini-Harandi ldquoMetric-like spaces partial metric spacesand fixed pointsrdquo Fixed PointTheory and Applications vol 2012article 204 10 pages 2012

[3] D ETHoric Z Kadelburg and S Radenovic ldquoEdelstein-Suzuki-type fixed point results in metric and abstract metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 75 no4 pp 1927ndash1932 2012

[4] N Hussain D ETHoric Z Kadelburg and S Radenovic ldquoSuzuki-type fixed point results in metric type spacesrdquo Fixed PointTheory and Applications vol 2012 article 126 12 pages 2012

[5] N Hussain and M H Shah ldquoKKMmappings in cone 119887-metricspacesrdquoComputersampMathematics withApplications vol 62 no4 pp 1677ndash1684 2011

[6] N Hussain M H Shah A Amini-Harandi and Z AkhtarldquoCommon fixed point theorems for generalized contractivemappings with applicationsrdquo Fixed Point Theory and Applica-tions vol 2013 article 169 17 pages 2013

[7] N Hussain Z Kadelburg S Radenovic and F Al-SolamyldquoComparison functions and fixed point results in partial metricspacesrdquo Abstract and Applied Analysis vol 2012 Article ID605781 15 pages 2012

[8] M Kikkawa and T Suzuki ldquoSome similarity between con-tractions and Kannan mappingsrdquo Fixed Point Theory andApplications vol 2008 Article ID 649749 8 pages 2008

[9] MKikkawa andT Suzuki ldquoSomenotes on fixed point theoremswith constantsrdquo Bulletin of the Kyushu Institute of Technologyno 56 pp 11ndash18 2009

[10] M Kikkawa and T Suzuki ldquoThree fixed point theorems forgeneralized contractions with constants in complete metricspacesrdquoNonlinearAnalysisTheoryMethodsampApplications vol69 no 9 pp 2942ndash2949 2008

[11] M A Kutbi J Ahmad N Hussain and M Arshad ldquoCommonfixed point results for mappings with rational expressionsrdquoAbstract and Applied Analysis vol 2013 Article ID 549518 11pages 2013

[12] S G Matthews ldquoPartial metric topologyrdquo in Papers on GeneralTopology and Applications vol 728 of Annals of the NewYork Academy of Sciences pp 183ndash197 New York Academy ofSciences New York NY USA 1994

[13] V Parvaneh J R Roshan and S Radenovic ldquoExistence oftripled coincidence points in ordered b-metric spaces and anapplication to a systemof integral equationsrdquo Fixed PointTheoryand Applications vol 2013 article 130 19 pages 2013

[14] O Popescu ldquoFixed point theorem in metric spacesrdquo Bulletinof the Transilvania University of Brasov vol 150 pp 479ndash4822008

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces and Applications 3

le lim sup119899rarrinfin

120590 (119909119899 119910)

le 120590 (119909 119910) + 120590 (119909 119909)

(7)

In particular if 120590(119909 119909) = 0 then one has lim119899rarrinfin

120590(119909119899 119910) =

120590(119909 119910)

Proof Using the triangle inequality in a metric-like space itis easy to see that

120590 (119909119899 119910) le 120590 (119909

119899 119909) + 120590 (119909 119910)

120590 (119909 119910) le 120590 (119909 119909119899) + 120590 (119909

119899 119910)

(8)

Taking the upper limit as 119899 rarr infin in the first inequalityand the lower limit as 119899 rarr infin in the second inequality weobtain the desired result

Theorem 9 Let (119883 120590) be a complete metric-like space Let 119879 119883 rarr 119883 be a self-map and let 120579 = [0 1) rarr (12 1] bedefined by

120579 (119903) =

1 0 le 119903 leradic5 minus 1

2

1 minus 119903

1199032radic5 minus 1

2le 119903 le

1

radic2

1

1 + 1199031

radic2le 119903 lt 1

(9)

If there exists 119903 isin [0 1) such that for each 119909 119910 isin 119883

120579 (119903) 120590 (119909 119879119909) le 120590 (119909 119910) 997904rArr 120590 (119879119909 119879119910) le 119903120590 (119909 119910) (10)

Then 119879 has a unique fixed point 119911 isin 119883 and for each 119909 isin 119883the sequence 119879119899119909 converges to 119911

Proof Putting 119910 = 119879119909 in (10) hence from

120579 (119903) 120590 (119909 119879119909) le 120590 (119909 119879119909) (11)

it follows

120590 (119879119909 1198792119909) le 119903120590 (119909 119879119909) (12)

for every 119909 isin 119883 Let 1199090isin 119883 be arbitrary and form the

sequence 119909119899 by 119909

1= 1198791199090and 119909

119899+1= 119879119909119899for 119899 isin N cup 0

By (12) we have

120590 (119909119899 119909119899+1) = 120590 (119879119909

119899minus1 1198792119909119899minus1)

le 119903120590 (119909119899minus1 119879119909119899minus1)

= 119903120590 (119909119899minus1 119909119899)

le 119903119899120590 (1199090 1199091)

(13)

Also by the condition 1205903 of the definition of metric-likespace for all119898 ge 119899 we have

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2)

+ sdot sdot sdot + 120590 (119909119898minus1 119909119898)

le 119903119899120590 (1199090 1199091) + 119903119899+1120590 (1199090 1199091)

+ sdot sdot sdot + 119903119898minus1120590 (1199090 1199091)

=119903119899 minus 119903119898

1 minus 119903120590 (1199090 1199091)

lt119903119899

1 minus 119903120590 (1199090 1199091) 997888rarr 0 as 119899 997888rarr infin

(14)

Hence 119909119899 is a 120590-Cauchy sequence

Since119883 is 120590-complete there exists 119911 isin 119883 such that

lim119899rarrinfin120590 (119909119899 119911) = 120590 (119911 119911) = lim

119898119899rarrinfin120590 (119909119898 119909119899) = 0 (15)

That is lim119899rarrinfin

119909119899+1= lim119899rarrinfin

119879119909119899= 119911 We prove that 119879119911 =

119911 Putting 119909 = 119879119899minus1119911 in (12) we get that

120590 (119879119899119911 119879119899+1119911) le 119903120590 (119879

119899minus1119911 119879119899119911) (16)

holds for each 119899 isin N (where 1198790119911 = 119911) It follows by inductionthat

120590 (119879119899119911 119879119899+1119911) le 119903

119899120590 (119911 119879119911) (17)

Let us prove now that

120590 (119911 119879119909) le 119903120590 (119911 119909) (18)

holds for each 119909 = 119911 Since 120590(119909119899 119879119909119899) rarr 0 and by

Lemma 8 120590(119909119899 119909) rarr 120590(119911 119909) = 0 it follows that there exists

1198990isin N such that

120579 (119903) 120590 (119909119899 119879119909119899) le 120590 (119909

119899 119909) (19)

holds for every 119899 ge 1198990 Assumption (10) implies that for such

119899120590(119879119909119899 119879119909) le 119903120590(119909

119899 119909) thus as 119899 rarr infin we get that

120590 (119911 119879119909) le 119903120590 (119911 119909) (20)

We will prove that

120590 (119879119899119911 119911) le 120590 (119879119911 119911) (21)

for each 119899 isin N For 119899 = 1 this relation is obvious Supposethat it holds for some 119898 isin N If 119879119898119911 = 119911 then 119879119898+1119911 =119879119911 and 120590(119879119898+1119911 119911) = 120590(119879119911 119911) le 120590(119879119911 119911) If 119879119898119911 = 119911 thenapplying (18) and the induction hypothesis we get that

120590 (119879119898+1119911 119911) le 119903120590 (119879

119898119911 119911)

le 119903120590 (119879119911 119911) le 120590 (119879119911 119911)

(22)

and (21) is proved by induction

4 Journal of Function Spaces and Applications

In order to prove that 119879119911 = 119911 we consider two possiblecases

Case I 0 le 119903 lt 1radic2 (and hence 120579(119903) le (1 minus 119903)1199032) We willprove first that

120590 (119879119899119911 119879119911) le 119903120590 (119879119911 119911) (23)

for 119899 ge 2 For 119899 = 2 it follows from (16) Suppose that (23)holds for some 119899 gt 2 Then

120590 (119879119911 119911) le 120590 (119911 119879119899119911) + 120590 (119879

119899119911 119879119911)

le 120590 (119911 119879119899119911) + 119903120590 (119911 119879119911)

(24)

which implies (1 minus 119903)120590(119911 119879119911) le 120590(119911 119879119899119911) Using (17) weobtain

120579 (119903) 120590 (119879119899119911 119879119899+1119911) le

1 minus 119903

119903119899120590 (119879119899119911 119879119899+1119911)

le1 minus 119903

119903119899sdot 119903119899120590 (119911 119879119911)

= (1 minus 119903) 120590 (119911 119879119911) le 120590 (119911 119879119899119911)

(25)

Assumption (10) and relation (21) imply that

120590 (119879119911 119879119899+1119911) le 119903120590 (119911 119879

119899119911)

le 119903120590 (119911 119879119911)

(26)

So relation (23) is proved by inductionNow 119879119911 = 119911 and (23) implies that 119879119899119911 = 119911 for each 119899 isin N

Hence (18) imply that

120590 (119911 119879119899+1119911) le 119903120590 (119911 119879

119899119911)

le 1199032120590 (119911 119879

119899minus1119911)

le 119903119899120590 (119911 119879119911)

(27)

Hence lim119899rarrinfin

120590(119911 119879119899+1119911) = 0 = 120590(119911 119911) thus 119879119899119911 rarr 119911

and using Lemma 8 in (23) we have 120590(119911 119879119911) le 119903120590(119879119911 119911) as119899 rarr infin which implies that 120590(119911 119879119911) = 0 a contradiction

Case II 1radic2 le 119903 lt 1 (and so 120579(119903) = 1(1+119903)) We will provethat there exists a subsequence 119909

119899119896 of 119909

119899 such that

120579 (119903) 120590 (119909119899119896 119879119909119899119896) = 120579 (119903) 120590 (119909

119899119896 119909119899119896+1) le 120590 (119909

119899119896 119911) (28)

holds for each 119896 isin N From (12) we know that 120590(119909119899 119909119899+1) le

119903120590(119909119899minus1 119909119899) holds for each 119899 isin N Suppose that

1

1 + 119903120590 (119909119899minus1 119909119899) gt 120590 (119909

119899minus1 119911)

1

1 + 119903120590 (119909119899 119909119899+1) gt 120590 (119909

119899 119911)

(29)

hold for some 119899 isin N Then120590 (119909119899minus1 119909119899) le 120590 (119909

119899minus1 119911) + 120590 (119911 119909

119899)

lt1

1 + 119903120590 (119909119899minus1 119909119899) + 120590 (119909

119899 119911)

lt1

1 + 119903120590 (119909119899minus1 119909119899) +

1

1 + 119903120590 (119909119899 119909119899+1)

le1

1 + 119903120590 (119909119899minus1 119909119899) +

119903

1 + 119903120590 (119909119899minus1 119909119899)

= 120590 (119909119899minus1 119909119899)

(30)

which is impossibleHence one of the following holds for each119899

120579 (119903) 120590 (119909119899minus1 119909119899) le 120590 (119909

119899minus1 119911) (31)

or

120579 (119903) 120590 (119909119899 119909119899+1) le 120590 (119909

119899 119911) (32)

In particular

120579 (119903) 120590 (1199092119899minus1 1199092119899) le 120590 (119909

2119899minus1 119911) (33)

or

120579 (119903) 120590 (1199092119899 1199092119899+1) le 120590 (119909

2119899 119911) (34)

In other words there is a subsequence 119909119899119896 of 119909

119899 such that

(28) holds for each 119896 isin N But then assumption (10) impliesthat

120590 (119879119909119899119896 119879119911) le 119903120590 (119909

119899119896 119911) (35)

or

120590 (119879119909119899119896minus1 119879119911) le 119903120590 (119909

119899119896minus1 119911) (36)

Passing to the limit when 119896 rarr infin we get that 120590(119911 119879119911) le 0which is possible only if 119879119911 = 119911

Thus we have proved that 119911 is a fixed point of 119879 Theuniqueness of the fixed point follows easily from (10) Indeedif 119910 and 119911 are two fixed points of 119879 such that 119910 = 119911 then from(18) we have

120590 (119910 119911) = 120590 (119910 119879119911)

le 119903120590 (119910 119911) (37)

which is a contradiction

According toTheorem 9 we get the following result

Corollary 10 (see [19]) Let (119883 119889) be a complete metric spaceand let 119879 be a mapping on119883 Define a nonincreasing function120579 from [0 1] into [12 1] by

120579 (119903) =

1 0 le 119903 leradic5 minus 1

2

1 minus 119903

1199032radic5 minus 1

2le 119903 le

1

radic2

1

1 + 1199031

radic2le 119903 lt 1

(38)

Journal of Function Spaces and Applications 5

Assume that there exists 119903 isin [0 1] such that

120579 (119903) 119889 (119909 119879119909) le 119889 (119909 119910) 997904rArr 119889 (119879119909 119879119910) le 119903119889 (119909 119910) (39)

for all 119909 119910 isin 119883 then there exists a unique fixed-point 119911 of 119879Moreover lim

119899rarrinfin119879119899119909 = 119911 for all 119909 isin 119883

Proof Using a similar argument given in Theorem 9 for120590(119909 119910) = 119889(119909 119910) the desired result is obtained

Now in order to support the useability of our results letus introduce the following example

Example 11 Let119883 = [0infin) Define 120590 119883 times 119883 rarr R+ by

120590 (119909 119910) = 119909 + 119910 (40)

for all 119909 119910 isin 119883 Then (119883 120590) is a complete metric-like spaceDefine a map 119879 119883 rarr 119883 by

119879 (119909) = ln(1 + 1radic2119909) (41)

for 119909 isin 119883 Then for each 119909 119910 isin 119883 we have

1

1 + 1radic2120590 (119909 119879119909) =

radic2

radic2 + 1(119909 + ln(1 + 1

radic2119909))

leradic2

radic2 + 1(119909 +

1

radic2119909) = 119909

le 119909 + 119910 = 120590 (119909 119910)

(42)

On the other hand we have

120590 (119879119909 119879119910) = ln(1 + 1radic2119909) + ln(1 + 1

radic2119910)

le1

radic2119909 +

1

radic2119910

=1

radic2120590 (119909 119910)

(43)

Thus 119879 satisfies all the hypotheses ofTheorem 9 and hence 119879has a unique fixed point Indeed 119903 = 1radic2 120579(119903) = 1(1 + 119903)and 0 is the unique fixed point of 119879

Theorem 12 Let (119883 120590) be a complete metric-like space Let 119878119879 119883 rarr 119883 be two self-mappings Suppose that there exists119903 isin [0 1) such that

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))(44)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(45)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary and define a sequence 119909

119899 by

119909119899= 119878 (119909

119899minus1) if 119899 is odd

= 119879 (119909119899minus1) if 119899 is even

(46)

Then if 119899 isin N is odd we have

120590 (119909119899 119909119899+1)

= 120590 (119878 (119909119899minus1) 119879 (119909

119899))

= 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le max 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119878 (119909119899minus1)) 120590 (119909

119899minus1 119879 (119909119899minus1))

le 119903120590 (119909119899minus1 119878 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(47)

If 119899 is even then by (44) we have

120590 (119909119899 119909119899+1)

= 120590 (119879 (119909119899minus1) 119878 (119909

119899))

= 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le max 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119879 (119909119899minus1)) 120590 (119909

119899minus1 119878 (119909119899minus1))

le 119903120590 (119909119899minus1 119879 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(48)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899 119909119899+1) le 119903120590 (119909

119899minus1 119909119899) (49)

Repeating (49) we obtain

120590 (119909119899 119909119899+1) le 119903119899120590 (1199090 1199091) (50)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [119903119899+ 119903119899+1+ sdot sdot sdot + 119903

119898minus1] 120590 (119909

0 1199091)

le119903119899

1 minus 119903120590 (1199090 1199091)

(51)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0

That is 119909119899 is a 120590-Cauchy sequence in the metric-like

space (119883 120590) Since (119883 120590) is 120590-complete there exist 119911 isin 119883such that120590 (119911 119911) = lim

119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (52)

6 Journal of Function Spaces and Applications

Assume that 119911 is not a common fixed point of 119878 and 119879Then by hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 120590 (119909

119899 119909119899+1) 119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 119903119899120590 (1199090 1199091) 119899 isin N = 0

(53)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = max 120590 (119878 (V) 119879119878 (V)) 120590 (119879 (V) 119878119879 (V))

le 119903min 120590 (V 119878 (V)) 120590 (V 119879 (V))

= 119903min 120590 (V V) 120590 (V V)

= 119903120590 (V V)

(54)

which gives that 120590(V V) = 0

Example 13 Let (119883 120590) be a metric-like space where 119883 =1119899infin

119899=1cup 0 and 120590(119909 119910) = 119909 + 119910 Define 119878 119883 rarr 119883 by

119878(0) = 0 119878(12119899) = 1(4119899 + 3) and 119878(1(2119899 minus 1)) = 0 and119879(0) = 0 119879(1(2119899 minus 1)) = 1(4119899 + 4) and 119879(12119899) = 0 Thenfor 119909 = 12119899 we have

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

= max 120590 (119878 ( 12119899) 119879 (119878 (

1

2119899)))

120590 (119879(1

2119899) 119878 (119879(

1

2119899)))

= max 1

4119899 + 3+

1

8119899 + 12 0 =

1

4119899 + 3+

1

8119899 + 12

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

= 119903min 12119899+1

4119899 + 31

2119899+ 0 = 119903

1

2119899

(55)

It is easy to see that the above inequality is true for119909 = 1(2119899minus1) and for 34 le 119903 lt 1 Also

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(56)

for every 119910 isin 119883 with y is not a common fixed point of 119878 and119879 This shows that all conditions of Theorem 12 are satisfiedand 0 is a common fixed point for 119878 and 119879

Corollary 14 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be a mapping Suppose that there exists 119903 isin[0 1) such that

120590 (119879 (119909) 1198792(119909)) le 119903120590 (119909 119879 (119909)) (57)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) + 120590 (119909 119879 (119909)) 119909 isin 119883 gt 0 (58)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 12 the conclusion of the cor-ollary follows

Theorem 15 Let (119883 120590) be a complete metric-like space Let119878 119879 be mappings from 119883 onto itself Suppose that there exists119903 gt 1 such that

min 120590 (119879119878 (119909) 119878 (119909)) 120590 (119878119879 (119909) 119879 (119909))

ge 119903max 120590 (119878119909 119909) 120590 (119879119909 119909)(59)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(60)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary Since 119878 is onto there is an

element 1199091satisfying 119909

1isin 119878minus1(119909

0) Since 119879 is also onto there

is an element 1199092satisfying 119909

2isin 119879minus1(119909

1) Proceeding in the

same way we can find that 1199092119899+1

isin 119878minus1(1199092119899) and 119909

2119899+2isin

119879minus1(1199092119899+1) for 119899 = 1 2 3 Therefore 119909

2119899= 1198781199092119899+1

and1199092119899+1= 1198791199092119899+2

for 119899 = 0 1 2 If 119899 = 2119898 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898minus1 1199092119898)

= 120590 (1198791199092119898 1198781199092119898+1)

= 120590 (1198791198781199092119898+1 1198781199092119898+1)

ge min 120590 (119879119878 (1199092119898+1) 119878 (119909

2119898+1))

120590 (119878119879 (1199092119898+1) 119879 (119909

2119898+1))

ge 119903max 120590 (1198781199092119898+1 1199092119898+1) 120590 (119879119909

2119898+1 1199092119898+1)

ge 119903120590 (1198781199092119898+1 1199092119898+1)

= 119903120590 (1199092119898 1199092119898+1)

= 119903120590 (119909119899 119909119899+1)

(61)

Journal of Function Spaces and Applications 7

If 119899 = 2119898 + 1 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898 1199092119898+1)

= 120590 (1198781199092119898+1 1198791199092119898+2)

= 120590 (1198781198791199092119898+2 1198791199092119898+2)

ge min 120590 (119879119878 (1199092119898+2) 119878 (119909

2119898+2))

120590 (119878119879 (1199092119898+2) 119879 (119909

2119898+2))

ge 119903max 120590 (1198781199092119898+2 1199092119898+2) 120590 (119879119909

2119898+2 1199092119898+2)

ge 119903120590 (1198791199092119898+2 1199092119898+2)

= 119903120590 (1199092119898+1 1199092119898+2)

= 119903120590 (119909119899 119909119899+1)

(62)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899minus1 119909119899) ge 119903120590 (119909

119899 119909119899+1) (63)

which implies that

120590 (119909119899 119909119899+1) le1

119903120590 (119909119899minus1 119909119899) le sdot sdot sdot le (

1

119903)119899

120590 (1199090 1199091) (64)

Let 120572 = 1119903 then 0 lt 120572 lt 1 since 119903 gt 1Now (64) becomes

120590 (119909119899 119909119899+1) le 120572119899120590 (1199090 1199091) (65)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [120572119899+ 120572119899+1+ sdot sdot sdot + 120572

119898minus1] 120590 (119909

0 1199091)

le120572119899

1 minus 120572120590 (1199090 1199091)

(66)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0 That is 119909

119899 is a 120590-Cauchy

sequence in the metric-like space (119883 120590) Since (119883 120590) is 120590-complete there exists 119911 isin 119883 such that

120590 (119911 119911) = lim119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (67)

Assume that 119911 is not a common fixed point of 119878 and 119879 Thenby hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120590 (119909

119899minus1 119909119899) 119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120572119899minus1120590 (1199090 1199091) 119899 isin N = 0

(68)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)

If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = min 120590 (119879119878 (V) 119878 (V)) 120590 (119878119879 (V) 119879 (V))ge 119903max 120590 (119878 (V) V) 120590 (119879 (V) V)= 119903max 120590 (V V) 120590 (V V)= 119903120590 (V V)

(69)

which gives that 120590(V V) = 0

Corollary 16 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be an onto mapping Suppose that there exists119903 isin [0 1) such that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (70)

for every 119909 isin 119883 and that120572 (119910) = inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (71)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 15we have the desired result

Definition 17 Let (119883 120590) and (119884 120591) bemetric-like spacesThen119891 119883 rarr 119884 is said to be a continuous mapping iflim119899rarrinfin

119909119899= 119909 implies that lim

119899rarrinfin119891(119909119899) = 119891(119909)

Corollary 18 Let (119883 120590) be a complete metric-like space andlet 119879 be a mapping of 119883 into itself If there is a real number 119903with 119903 gt 1 satisfying

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (72)

for every 119909 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Assume that there exists 119910 isin 119883 with 119910 = 119879(119910) and

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 = 0 (73)

Then there exists a sequence 119909119899 such that

lim119899rarrinfin

120590 (119909119899 119910) + 120590 (119879 (119909

119899) 119909119899) = 0 (74)

So we have 120590(119909119899 119910) rarr 0 and 120590(119879(119909

119899) 119909119899) rarr 0 as 119899 rarr infin

Since 120590(119910 119910) le 120590(119910 119909119899) + 120590(119909

119899 119910) hence 120590(119910 119910) rarr 0 as

119899 rarr infin Now120590 (119879 (119909

119899) 119910) le 120590 (119879 (119909

119899) 119909119899) + 120590 (119909

119899 119910) 997888rarr 0

as 119899 997888rarr infin(75)

Since 119879 is continuous we have

119879 (119910) = 119879 ( lim119899rarrinfin119909119899) = lim119899rarrinfin119879 (119909119899) = 119910 (76)

This is a contradiction Hence if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (77)

which is condition (71) of Corollary 16 By Corollary 16 thereexists 119911 isin 119883 such that 119911 = 119879(119911)

8 Journal of Function Spaces and Applications

Now we give an example to support our result

Example 19 Let 119883 = [0infin) and 120590(119909 119910) = 119909 + 119910 Define 119879 119883 rarr 119883 by 119879(119909) = 2119909

Obviously 119879 is onto and continuous Also for each 119909 119910 isin119883 we have

120590 (1198792119909 119879119909) = 4119909 + 2119909 = 6119909 ge 1199033119909 = 119903120590 (119879119909 119909) (78)

where 119903 = 2 Thus 119879 satisfies the conditions given inCorollary 18 and 0 is the fixed point of 119879

Corollary 20 Let (119883 120590) be a complete metric-like space and119879 be a mapping of119883 into itself If there is a real number 119903 with119903 gt 1 satisfying

120590 (119879 (119909) 119879 (119910)) ge 119903min 120590 (119909 119879 (119909)) 120590 (119879 (119910) 119910) 120590 (119909 119910)(79)

for every 119909 119910 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Replacing 119910 by 119879(119909) in (79) we obtain

120590 (119879 (119909) 1198792(119909))

ge 119903min 120590 (119909 119879 (119909)) 120590 (1198792 (119909) 119879 (119909)) 120590 (119909 119879 (119909))(80)

for all 119909 isin 119883Without loss of generality we may assume that 119879(119909) =

1198792(119909) Otherwise 119879 has a fixed point Since 119903 gt 1 it followsfrom (80) that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (81)

for every 119909 isin 119883 By the argument similar to that used inCorollary 18 we can prove that if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (82)

which is condition (71) of Corollary 16 So Corollary 16applies to obtain a fixed point of 119879

According toTheorem 12 we get the following result

Corollary 21 (see [17 Theorem 1]) Let (119883 119901) be a completepartial metric space Let 119878 119879 119883 rarr 119883 be two self-mappingsSuppose that there exists 119903 isin [0 1) such that

max 119901 (119878 (119909) 119879119878 (119909)) 119901 (119879 (119909) 119878119879 (119909))

le 119903min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))(83)

for every 119909 isin 119883 and that

120572 (119910) = inf 119901 (119909 119910) +min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))

119909 isin 119883 gt 0

(84)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 119901(V V) = 0

Proof Using a similar argument given in the Theorem 12 for120590(119909 119910) = 119901(119909 119910) the desired result is obtained where 119901 is apartial metric on119883

Also according to Theorem 15 we get Theorem 2 from[17]

References

[1] A Aghajani S Radenovic and J R Roshan ldquoCommon fixedpoint results for four mappings satisfying almost generalized(119878 119879)-contractive condition in partially ordered metric spacesrdquoAppliedMathematics andComputation vol 218 no 9 pp 5665ndash5670 2012

[2] A Amini-Harandi ldquoMetric-like spaces partial metric spacesand fixed pointsrdquo Fixed PointTheory and Applications vol 2012article 204 10 pages 2012

[3] D ETHoric Z Kadelburg and S Radenovic ldquoEdelstein-Suzuki-type fixed point results in metric and abstract metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 75 no4 pp 1927ndash1932 2012

[4] N Hussain D ETHoric Z Kadelburg and S Radenovic ldquoSuzuki-type fixed point results in metric type spacesrdquo Fixed PointTheory and Applications vol 2012 article 126 12 pages 2012

[5] N Hussain and M H Shah ldquoKKMmappings in cone 119887-metricspacesrdquoComputersampMathematics withApplications vol 62 no4 pp 1677ndash1684 2011

[6] N Hussain M H Shah A Amini-Harandi and Z AkhtarldquoCommon fixed point theorems for generalized contractivemappings with applicationsrdquo Fixed Point Theory and Applica-tions vol 2013 article 169 17 pages 2013

[7] N Hussain Z Kadelburg S Radenovic and F Al-SolamyldquoComparison functions and fixed point results in partial metricspacesrdquo Abstract and Applied Analysis vol 2012 Article ID605781 15 pages 2012

[8] M Kikkawa and T Suzuki ldquoSome similarity between con-tractions and Kannan mappingsrdquo Fixed Point Theory andApplications vol 2008 Article ID 649749 8 pages 2008

[9] MKikkawa andT Suzuki ldquoSomenotes on fixed point theoremswith constantsrdquo Bulletin of the Kyushu Institute of Technologyno 56 pp 11ndash18 2009

[10] M Kikkawa and T Suzuki ldquoThree fixed point theorems forgeneralized contractions with constants in complete metricspacesrdquoNonlinearAnalysisTheoryMethodsampApplications vol69 no 9 pp 2942ndash2949 2008

[11] M A Kutbi J Ahmad N Hussain and M Arshad ldquoCommonfixed point results for mappings with rational expressionsrdquoAbstract and Applied Analysis vol 2013 Article ID 549518 11pages 2013

[12] S G Matthews ldquoPartial metric topologyrdquo in Papers on GeneralTopology and Applications vol 728 of Annals of the NewYork Academy of Sciences pp 183ndash197 New York Academy ofSciences New York NY USA 1994

[13] V Parvaneh J R Roshan and S Radenovic ldquoExistence oftripled coincidence points in ordered b-metric spaces and anapplication to a systemof integral equationsrdquo Fixed PointTheoryand Applications vol 2013 article 130 19 pages 2013

[14] O Popescu ldquoFixed point theorem in metric spacesrdquo Bulletinof the Transilvania University of Brasov vol 150 pp 479ndash4822008

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

Submit your manuscripts athttpwwwhindawicom

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Journal of Function Spaces and Applications

In order to prove that 119879119911 = 119911 we consider two possiblecases

Case I 0 le 119903 lt 1radic2 (and hence 120579(119903) le (1 minus 119903)1199032) We willprove first that

120590 (119879119899119911 119879119911) le 119903120590 (119879119911 119911) (23)

for 119899 ge 2 For 119899 = 2 it follows from (16) Suppose that (23)holds for some 119899 gt 2 Then

120590 (119879119911 119911) le 120590 (119911 119879119899119911) + 120590 (119879

119899119911 119879119911)

le 120590 (119911 119879119899119911) + 119903120590 (119911 119879119911)

(24)

which implies (1 minus 119903)120590(119911 119879119911) le 120590(119911 119879119899119911) Using (17) weobtain

120579 (119903) 120590 (119879119899119911 119879119899+1119911) le

1 minus 119903

119903119899120590 (119879119899119911 119879119899+1119911)

le1 minus 119903

119903119899sdot 119903119899120590 (119911 119879119911)

= (1 minus 119903) 120590 (119911 119879119911) le 120590 (119911 119879119899119911)

(25)

Assumption (10) and relation (21) imply that

120590 (119879119911 119879119899+1119911) le 119903120590 (119911 119879

119899119911)

le 119903120590 (119911 119879119911)

(26)

So relation (23) is proved by inductionNow 119879119911 = 119911 and (23) implies that 119879119899119911 = 119911 for each 119899 isin N

Hence (18) imply that

120590 (119911 119879119899+1119911) le 119903120590 (119911 119879

119899119911)

le 1199032120590 (119911 119879

119899minus1119911)

le 119903119899120590 (119911 119879119911)

(27)

Hence lim119899rarrinfin

120590(119911 119879119899+1119911) = 0 = 120590(119911 119911) thus 119879119899119911 rarr 119911

and using Lemma 8 in (23) we have 120590(119911 119879119911) le 119903120590(119879119911 119911) as119899 rarr infin which implies that 120590(119911 119879119911) = 0 a contradiction

Case II 1radic2 le 119903 lt 1 (and so 120579(119903) = 1(1+119903)) We will provethat there exists a subsequence 119909

119899119896 of 119909

119899 such that

120579 (119903) 120590 (119909119899119896 119879119909119899119896) = 120579 (119903) 120590 (119909

119899119896 119909119899119896+1) le 120590 (119909

119899119896 119911) (28)

holds for each 119896 isin N From (12) we know that 120590(119909119899 119909119899+1) le

119903120590(119909119899minus1 119909119899) holds for each 119899 isin N Suppose that

1

1 + 119903120590 (119909119899minus1 119909119899) gt 120590 (119909

119899minus1 119911)

1

1 + 119903120590 (119909119899 119909119899+1) gt 120590 (119909

119899 119911)

(29)

hold for some 119899 isin N Then120590 (119909119899minus1 119909119899) le 120590 (119909

119899minus1 119911) + 120590 (119911 119909

119899)

lt1

1 + 119903120590 (119909119899minus1 119909119899) + 120590 (119909

119899 119911)

lt1

1 + 119903120590 (119909119899minus1 119909119899) +

1

1 + 119903120590 (119909119899 119909119899+1)

le1

1 + 119903120590 (119909119899minus1 119909119899) +

119903

1 + 119903120590 (119909119899minus1 119909119899)

= 120590 (119909119899minus1 119909119899)

(30)

which is impossibleHence one of the following holds for each119899

120579 (119903) 120590 (119909119899minus1 119909119899) le 120590 (119909

119899minus1 119911) (31)

or

120579 (119903) 120590 (119909119899 119909119899+1) le 120590 (119909

119899 119911) (32)

In particular

120579 (119903) 120590 (1199092119899minus1 1199092119899) le 120590 (119909

2119899minus1 119911) (33)

or

120579 (119903) 120590 (1199092119899 1199092119899+1) le 120590 (119909

2119899 119911) (34)

In other words there is a subsequence 119909119899119896 of 119909

119899 such that

(28) holds for each 119896 isin N But then assumption (10) impliesthat

120590 (119879119909119899119896 119879119911) le 119903120590 (119909

119899119896 119911) (35)

or

120590 (119879119909119899119896minus1 119879119911) le 119903120590 (119909

119899119896minus1 119911) (36)

Passing to the limit when 119896 rarr infin we get that 120590(119911 119879119911) le 0which is possible only if 119879119911 = 119911

Thus we have proved that 119911 is a fixed point of 119879 Theuniqueness of the fixed point follows easily from (10) Indeedif 119910 and 119911 are two fixed points of 119879 such that 119910 = 119911 then from(18) we have

120590 (119910 119911) = 120590 (119910 119879119911)

le 119903120590 (119910 119911) (37)

which is a contradiction

According toTheorem 9 we get the following result

Corollary 10 (see [19]) Let (119883 119889) be a complete metric spaceand let 119879 be a mapping on119883 Define a nonincreasing function120579 from [0 1] into [12 1] by

120579 (119903) =

1 0 le 119903 leradic5 minus 1

2

1 minus 119903

1199032radic5 minus 1

2le 119903 le

1

radic2

1

1 + 1199031

radic2le 119903 lt 1

(38)

Journal of Function Spaces and Applications 5

Assume that there exists 119903 isin [0 1] such that

120579 (119903) 119889 (119909 119879119909) le 119889 (119909 119910) 997904rArr 119889 (119879119909 119879119910) le 119903119889 (119909 119910) (39)

for all 119909 119910 isin 119883 then there exists a unique fixed-point 119911 of 119879Moreover lim

119899rarrinfin119879119899119909 = 119911 for all 119909 isin 119883

Proof Using a similar argument given in Theorem 9 for120590(119909 119910) = 119889(119909 119910) the desired result is obtained

Now in order to support the useability of our results letus introduce the following example

Example 11 Let119883 = [0infin) Define 120590 119883 times 119883 rarr R+ by

120590 (119909 119910) = 119909 + 119910 (40)

for all 119909 119910 isin 119883 Then (119883 120590) is a complete metric-like spaceDefine a map 119879 119883 rarr 119883 by

119879 (119909) = ln(1 + 1radic2119909) (41)

for 119909 isin 119883 Then for each 119909 119910 isin 119883 we have

1

1 + 1radic2120590 (119909 119879119909) =

radic2

radic2 + 1(119909 + ln(1 + 1

radic2119909))

leradic2

radic2 + 1(119909 +

1

radic2119909) = 119909

le 119909 + 119910 = 120590 (119909 119910)

(42)

On the other hand we have

120590 (119879119909 119879119910) = ln(1 + 1radic2119909) + ln(1 + 1

radic2119910)

le1

radic2119909 +

1

radic2119910

=1

radic2120590 (119909 119910)

(43)

Thus 119879 satisfies all the hypotheses ofTheorem 9 and hence 119879has a unique fixed point Indeed 119903 = 1radic2 120579(119903) = 1(1 + 119903)and 0 is the unique fixed point of 119879

Theorem 12 Let (119883 120590) be a complete metric-like space Let 119878119879 119883 rarr 119883 be two self-mappings Suppose that there exists119903 isin [0 1) such that

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))(44)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(45)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary and define a sequence 119909

119899 by

119909119899= 119878 (119909

119899minus1) if 119899 is odd

= 119879 (119909119899minus1) if 119899 is even

(46)

Then if 119899 isin N is odd we have

120590 (119909119899 119909119899+1)

= 120590 (119878 (119909119899minus1) 119879 (119909

119899))

= 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le max 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119878 (119909119899minus1)) 120590 (119909

119899minus1 119879 (119909119899minus1))

le 119903120590 (119909119899minus1 119878 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(47)

If 119899 is even then by (44) we have

120590 (119909119899 119909119899+1)

= 120590 (119879 (119909119899minus1) 119878 (119909

119899))

= 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le max 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119879 (119909119899minus1)) 120590 (119909

119899minus1 119878 (119909119899minus1))

le 119903120590 (119909119899minus1 119879 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(48)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899 119909119899+1) le 119903120590 (119909

119899minus1 119909119899) (49)

Repeating (49) we obtain

120590 (119909119899 119909119899+1) le 119903119899120590 (1199090 1199091) (50)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [119903119899+ 119903119899+1+ sdot sdot sdot + 119903

119898minus1] 120590 (119909

0 1199091)

le119903119899

1 minus 119903120590 (1199090 1199091)

(51)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0

That is 119909119899 is a 120590-Cauchy sequence in the metric-like

space (119883 120590) Since (119883 120590) is 120590-complete there exist 119911 isin 119883such that120590 (119911 119911) = lim

119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (52)

6 Journal of Function Spaces and Applications

Assume that 119911 is not a common fixed point of 119878 and 119879Then by hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 120590 (119909

119899 119909119899+1) 119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 119903119899120590 (1199090 1199091) 119899 isin N = 0

(53)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = max 120590 (119878 (V) 119879119878 (V)) 120590 (119879 (V) 119878119879 (V))

le 119903min 120590 (V 119878 (V)) 120590 (V 119879 (V))

= 119903min 120590 (V V) 120590 (V V)

= 119903120590 (V V)

(54)

which gives that 120590(V V) = 0

Example 13 Let (119883 120590) be a metric-like space where 119883 =1119899infin

119899=1cup 0 and 120590(119909 119910) = 119909 + 119910 Define 119878 119883 rarr 119883 by

119878(0) = 0 119878(12119899) = 1(4119899 + 3) and 119878(1(2119899 minus 1)) = 0 and119879(0) = 0 119879(1(2119899 minus 1)) = 1(4119899 + 4) and 119879(12119899) = 0 Thenfor 119909 = 12119899 we have

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

= max 120590 (119878 ( 12119899) 119879 (119878 (

1

2119899)))

120590 (119879(1

2119899) 119878 (119879(

1

2119899)))

= max 1

4119899 + 3+

1

8119899 + 12 0 =

1

4119899 + 3+

1

8119899 + 12

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

= 119903min 12119899+1

4119899 + 31

2119899+ 0 = 119903

1

2119899

(55)

It is easy to see that the above inequality is true for119909 = 1(2119899minus1) and for 34 le 119903 lt 1 Also

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(56)

for every 119910 isin 119883 with y is not a common fixed point of 119878 and119879 This shows that all conditions of Theorem 12 are satisfiedand 0 is a common fixed point for 119878 and 119879

Corollary 14 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be a mapping Suppose that there exists 119903 isin[0 1) such that

120590 (119879 (119909) 1198792(119909)) le 119903120590 (119909 119879 (119909)) (57)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) + 120590 (119909 119879 (119909)) 119909 isin 119883 gt 0 (58)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 12 the conclusion of the cor-ollary follows

Theorem 15 Let (119883 120590) be a complete metric-like space Let119878 119879 be mappings from 119883 onto itself Suppose that there exists119903 gt 1 such that

min 120590 (119879119878 (119909) 119878 (119909)) 120590 (119878119879 (119909) 119879 (119909))

ge 119903max 120590 (119878119909 119909) 120590 (119879119909 119909)(59)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(60)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary Since 119878 is onto there is an

element 1199091satisfying 119909

1isin 119878minus1(119909

0) Since 119879 is also onto there

is an element 1199092satisfying 119909

2isin 119879minus1(119909

1) Proceeding in the

same way we can find that 1199092119899+1

isin 119878minus1(1199092119899) and 119909

2119899+2isin

119879minus1(1199092119899+1) for 119899 = 1 2 3 Therefore 119909

2119899= 1198781199092119899+1

and1199092119899+1= 1198791199092119899+2

for 119899 = 0 1 2 If 119899 = 2119898 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898minus1 1199092119898)

= 120590 (1198791199092119898 1198781199092119898+1)

= 120590 (1198791198781199092119898+1 1198781199092119898+1)

ge min 120590 (119879119878 (1199092119898+1) 119878 (119909

2119898+1))

120590 (119878119879 (1199092119898+1) 119879 (119909

2119898+1))

ge 119903max 120590 (1198781199092119898+1 1199092119898+1) 120590 (119879119909

2119898+1 1199092119898+1)

ge 119903120590 (1198781199092119898+1 1199092119898+1)

= 119903120590 (1199092119898 1199092119898+1)

= 119903120590 (119909119899 119909119899+1)

(61)

Journal of Function Spaces and Applications 7

If 119899 = 2119898 + 1 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898 1199092119898+1)

= 120590 (1198781199092119898+1 1198791199092119898+2)

= 120590 (1198781198791199092119898+2 1198791199092119898+2)

ge min 120590 (119879119878 (1199092119898+2) 119878 (119909

2119898+2))

120590 (119878119879 (1199092119898+2) 119879 (119909

2119898+2))

ge 119903max 120590 (1198781199092119898+2 1199092119898+2) 120590 (119879119909

2119898+2 1199092119898+2)

ge 119903120590 (1198791199092119898+2 1199092119898+2)

= 119903120590 (1199092119898+1 1199092119898+2)

= 119903120590 (119909119899 119909119899+1)

(62)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899minus1 119909119899) ge 119903120590 (119909

119899 119909119899+1) (63)

which implies that

120590 (119909119899 119909119899+1) le1

119903120590 (119909119899minus1 119909119899) le sdot sdot sdot le (

1

119903)119899

120590 (1199090 1199091) (64)

Let 120572 = 1119903 then 0 lt 120572 lt 1 since 119903 gt 1Now (64) becomes

120590 (119909119899 119909119899+1) le 120572119899120590 (1199090 1199091) (65)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [120572119899+ 120572119899+1+ sdot sdot sdot + 120572

119898minus1] 120590 (119909

0 1199091)

le120572119899

1 minus 120572120590 (1199090 1199091)

(66)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0 That is 119909

119899 is a 120590-Cauchy

sequence in the metric-like space (119883 120590) Since (119883 120590) is 120590-complete there exists 119911 isin 119883 such that

120590 (119911 119911) = lim119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (67)

Assume that 119911 is not a common fixed point of 119878 and 119879 Thenby hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120590 (119909

119899minus1 119909119899) 119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120572119899minus1120590 (1199090 1199091) 119899 isin N = 0

(68)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)

If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = min 120590 (119879119878 (V) 119878 (V)) 120590 (119878119879 (V) 119879 (V))ge 119903max 120590 (119878 (V) V) 120590 (119879 (V) V)= 119903max 120590 (V V) 120590 (V V)= 119903120590 (V V)

(69)

which gives that 120590(V V) = 0

Corollary 16 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be an onto mapping Suppose that there exists119903 isin [0 1) such that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (70)

for every 119909 isin 119883 and that120572 (119910) = inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (71)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 15we have the desired result

Definition 17 Let (119883 120590) and (119884 120591) bemetric-like spacesThen119891 119883 rarr 119884 is said to be a continuous mapping iflim119899rarrinfin

119909119899= 119909 implies that lim

119899rarrinfin119891(119909119899) = 119891(119909)

Corollary 18 Let (119883 120590) be a complete metric-like space andlet 119879 be a mapping of 119883 into itself If there is a real number 119903with 119903 gt 1 satisfying

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (72)

for every 119909 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Assume that there exists 119910 isin 119883 with 119910 = 119879(119910) and

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 = 0 (73)

Then there exists a sequence 119909119899 such that

lim119899rarrinfin

120590 (119909119899 119910) + 120590 (119879 (119909

119899) 119909119899) = 0 (74)

So we have 120590(119909119899 119910) rarr 0 and 120590(119879(119909

119899) 119909119899) rarr 0 as 119899 rarr infin

Since 120590(119910 119910) le 120590(119910 119909119899) + 120590(119909

119899 119910) hence 120590(119910 119910) rarr 0 as

119899 rarr infin Now120590 (119879 (119909

119899) 119910) le 120590 (119879 (119909

119899) 119909119899) + 120590 (119909

119899 119910) 997888rarr 0

as 119899 997888rarr infin(75)

Since 119879 is continuous we have

119879 (119910) = 119879 ( lim119899rarrinfin119909119899) = lim119899rarrinfin119879 (119909119899) = 119910 (76)

This is a contradiction Hence if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (77)

which is condition (71) of Corollary 16 By Corollary 16 thereexists 119911 isin 119883 such that 119911 = 119879(119911)

8 Journal of Function Spaces and Applications

Now we give an example to support our result

Example 19 Let 119883 = [0infin) and 120590(119909 119910) = 119909 + 119910 Define 119879 119883 rarr 119883 by 119879(119909) = 2119909

Obviously 119879 is onto and continuous Also for each 119909 119910 isin119883 we have

120590 (1198792119909 119879119909) = 4119909 + 2119909 = 6119909 ge 1199033119909 = 119903120590 (119879119909 119909) (78)

where 119903 = 2 Thus 119879 satisfies the conditions given inCorollary 18 and 0 is the fixed point of 119879

Corollary 20 Let (119883 120590) be a complete metric-like space and119879 be a mapping of119883 into itself If there is a real number 119903 with119903 gt 1 satisfying

120590 (119879 (119909) 119879 (119910)) ge 119903min 120590 (119909 119879 (119909)) 120590 (119879 (119910) 119910) 120590 (119909 119910)(79)

for every 119909 119910 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Replacing 119910 by 119879(119909) in (79) we obtain

120590 (119879 (119909) 1198792(119909))

ge 119903min 120590 (119909 119879 (119909)) 120590 (1198792 (119909) 119879 (119909)) 120590 (119909 119879 (119909))(80)

for all 119909 isin 119883Without loss of generality we may assume that 119879(119909) =

1198792(119909) Otherwise 119879 has a fixed point Since 119903 gt 1 it followsfrom (80) that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (81)

for every 119909 isin 119883 By the argument similar to that used inCorollary 18 we can prove that if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (82)

which is condition (71) of Corollary 16 So Corollary 16applies to obtain a fixed point of 119879

According toTheorem 12 we get the following result

Corollary 21 (see [17 Theorem 1]) Let (119883 119901) be a completepartial metric space Let 119878 119879 119883 rarr 119883 be two self-mappingsSuppose that there exists 119903 isin [0 1) such that

max 119901 (119878 (119909) 119879119878 (119909)) 119901 (119879 (119909) 119878119879 (119909))

le 119903min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))(83)

for every 119909 isin 119883 and that

120572 (119910) = inf 119901 (119909 119910) +min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))

119909 isin 119883 gt 0

(84)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 119901(V V) = 0

Proof Using a similar argument given in the Theorem 12 for120590(119909 119910) = 119901(119909 119910) the desired result is obtained where 119901 is apartial metric on119883

Also according to Theorem 15 we get Theorem 2 from[17]

References

[1] A Aghajani S Radenovic and J R Roshan ldquoCommon fixedpoint results for four mappings satisfying almost generalized(119878 119879)-contractive condition in partially ordered metric spacesrdquoAppliedMathematics andComputation vol 218 no 9 pp 5665ndash5670 2012

[2] A Amini-Harandi ldquoMetric-like spaces partial metric spacesand fixed pointsrdquo Fixed PointTheory and Applications vol 2012article 204 10 pages 2012

[3] D ETHoric Z Kadelburg and S Radenovic ldquoEdelstein-Suzuki-type fixed point results in metric and abstract metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 75 no4 pp 1927ndash1932 2012

[4] N Hussain D ETHoric Z Kadelburg and S Radenovic ldquoSuzuki-type fixed point results in metric type spacesrdquo Fixed PointTheory and Applications vol 2012 article 126 12 pages 2012

[5] N Hussain and M H Shah ldquoKKMmappings in cone 119887-metricspacesrdquoComputersampMathematics withApplications vol 62 no4 pp 1677ndash1684 2011

[6] N Hussain M H Shah A Amini-Harandi and Z AkhtarldquoCommon fixed point theorems for generalized contractivemappings with applicationsrdquo Fixed Point Theory and Applica-tions vol 2013 article 169 17 pages 2013

[7] N Hussain Z Kadelburg S Radenovic and F Al-SolamyldquoComparison functions and fixed point results in partial metricspacesrdquo Abstract and Applied Analysis vol 2012 Article ID605781 15 pages 2012

[8] M Kikkawa and T Suzuki ldquoSome similarity between con-tractions and Kannan mappingsrdquo Fixed Point Theory andApplications vol 2008 Article ID 649749 8 pages 2008

[9] MKikkawa andT Suzuki ldquoSomenotes on fixed point theoremswith constantsrdquo Bulletin of the Kyushu Institute of Technologyno 56 pp 11ndash18 2009

[10] M Kikkawa and T Suzuki ldquoThree fixed point theorems forgeneralized contractions with constants in complete metricspacesrdquoNonlinearAnalysisTheoryMethodsampApplications vol69 no 9 pp 2942ndash2949 2008

[11] M A Kutbi J Ahmad N Hussain and M Arshad ldquoCommonfixed point results for mappings with rational expressionsrdquoAbstract and Applied Analysis vol 2013 Article ID 549518 11pages 2013

[12] S G Matthews ldquoPartial metric topologyrdquo in Papers on GeneralTopology and Applications vol 728 of Annals of the NewYork Academy of Sciences pp 183ndash197 New York Academy ofSciences New York NY USA 1994

[13] V Parvaneh J R Roshan and S Radenovic ldquoExistence oftripled coincidence points in ordered b-metric spaces and anapplication to a systemof integral equationsrdquo Fixed PointTheoryand Applications vol 2013 article 130 19 pages 2013

[14] O Popescu ldquoFixed point theorem in metric spacesrdquo Bulletinof the Transilvania University of Brasov vol 150 pp 479ndash4822008

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces and Applications 5

Assume that there exists 119903 isin [0 1] such that

120579 (119903) 119889 (119909 119879119909) le 119889 (119909 119910) 997904rArr 119889 (119879119909 119879119910) le 119903119889 (119909 119910) (39)

for all 119909 119910 isin 119883 then there exists a unique fixed-point 119911 of 119879Moreover lim

119899rarrinfin119879119899119909 = 119911 for all 119909 isin 119883

Proof Using a similar argument given in Theorem 9 for120590(119909 119910) = 119889(119909 119910) the desired result is obtained

Now in order to support the useability of our results letus introduce the following example

Example 11 Let119883 = [0infin) Define 120590 119883 times 119883 rarr R+ by

120590 (119909 119910) = 119909 + 119910 (40)

for all 119909 119910 isin 119883 Then (119883 120590) is a complete metric-like spaceDefine a map 119879 119883 rarr 119883 by

119879 (119909) = ln(1 + 1radic2119909) (41)

for 119909 isin 119883 Then for each 119909 119910 isin 119883 we have

1

1 + 1radic2120590 (119909 119879119909) =

radic2

radic2 + 1(119909 + ln(1 + 1

radic2119909))

leradic2

radic2 + 1(119909 +

1

radic2119909) = 119909

le 119909 + 119910 = 120590 (119909 119910)

(42)

On the other hand we have

120590 (119879119909 119879119910) = ln(1 + 1radic2119909) + ln(1 + 1

radic2119910)

le1

radic2119909 +

1

radic2119910

=1

radic2120590 (119909 119910)

(43)

Thus 119879 satisfies all the hypotheses ofTheorem 9 and hence 119879has a unique fixed point Indeed 119903 = 1radic2 120579(119903) = 1(1 + 119903)and 0 is the unique fixed point of 119879

Theorem 12 Let (119883 120590) be a complete metric-like space Let 119878119879 119883 rarr 119883 be two self-mappings Suppose that there exists119903 isin [0 1) such that

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))(44)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(45)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary and define a sequence 119909

119899 by

119909119899= 119878 (119909

119899minus1) if 119899 is odd

= 119879 (119909119899minus1) if 119899 is even

(46)

Then if 119899 isin N is odd we have

120590 (119909119899 119909119899+1)

= 120590 (119878 (119909119899minus1) 119879 (119909

119899))

= 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le max 120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119878 (119909119899minus1)) 120590 (119909

119899minus1 119879 (119909119899minus1))

le 119903120590 (119909119899minus1 119878 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(47)

If 119899 is even then by (44) we have

120590 (119909119899 119909119899+1)

= 120590 (119879 (119909119899minus1) 119878 (119909

119899))

= 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

le max 120590 (119879 (119909119899minus1) 119878119879 (119909

119899minus1))

120590 (119878 (119909119899minus1) 119879119878 (119909

119899minus1))

le 119903min 120590 (119909119899minus1 119879 (119909119899minus1)) 120590 (119909

119899minus1 119878 (119909119899minus1))

le 119903120590 (119909119899minus1 119879 (119909119899minus1))

= 119903120590 (119909119899minus1 119909119899)

(48)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899 119909119899+1) le 119903120590 (119909

119899minus1 119909119899) (49)

Repeating (49) we obtain

120590 (119909119899 119909119899+1) le 119903119899120590 (1199090 1199091) (50)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [119903119899+ 119903119899+1+ sdot sdot sdot + 119903

119898minus1] 120590 (119909

0 1199091)

le119903119899

1 minus 119903120590 (1199090 1199091)

(51)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0

That is 119909119899 is a 120590-Cauchy sequence in the metric-like

space (119883 120590) Since (119883 120590) is 120590-complete there exist 119911 isin 119883such that120590 (119911 119911) = lim

119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (52)

6 Journal of Function Spaces and Applications

Assume that 119911 is not a common fixed point of 119878 and 119879Then by hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 120590 (119909

119899 119909119899+1) 119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 119903119899120590 (1199090 1199091) 119899 isin N = 0

(53)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = max 120590 (119878 (V) 119879119878 (V)) 120590 (119879 (V) 119878119879 (V))

le 119903min 120590 (V 119878 (V)) 120590 (V 119879 (V))

= 119903min 120590 (V V) 120590 (V V)

= 119903120590 (V V)

(54)

which gives that 120590(V V) = 0

Example 13 Let (119883 120590) be a metric-like space where 119883 =1119899infin

119899=1cup 0 and 120590(119909 119910) = 119909 + 119910 Define 119878 119883 rarr 119883 by

119878(0) = 0 119878(12119899) = 1(4119899 + 3) and 119878(1(2119899 minus 1)) = 0 and119879(0) = 0 119879(1(2119899 minus 1)) = 1(4119899 + 4) and 119879(12119899) = 0 Thenfor 119909 = 12119899 we have

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

= max 120590 (119878 ( 12119899) 119879 (119878 (

1

2119899)))

120590 (119879(1

2119899) 119878 (119879(

1

2119899)))

= max 1

4119899 + 3+

1

8119899 + 12 0 =

1

4119899 + 3+

1

8119899 + 12

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

= 119903min 12119899+1

4119899 + 31

2119899+ 0 = 119903

1

2119899

(55)

It is easy to see that the above inequality is true for119909 = 1(2119899minus1) and for 34 le 119903 lt 1 Also

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(56)

for every 119910 isin 119883 with y is not a common fixed point of 119878 and119879 This shows that all conditions of Theorem 12 are satisfiedand 0 is a common fixed point for 119878 and 119879

Corollary 14 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be a mapping Suppose that there exists 119903 isin[0 1) such that

120590 (119879 (119909) 1198792(119909)) le 119903120590 (119909 119879 (119909)) (57)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) + 120590 (119909 119879 (119909)) 119909 isin 119883 gt 0 (58)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 12 the conclusion of the cor-ollary follows

Theorem 15 Let (119883 120590) be a complete metric-like space Let119878 119879 be mappings from 119883 onto itself Suppose that there exists119903 gt 1 such that

min 120590 (119879119878 (119909) 119878 (119909)) 120590 (119878119879 (119909) 119879 (119909))

ge 119903max 120590 (119878119909 119909) 120590 (119879119909 119909)(59)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(60)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary Since 119878 is onto there is an

element 1199091satisfying 119909

1isin 119878minus1(119909

0) Since 119879 is also onto there

is an element 1199092satisfying 119909

2isin 119879minus1(119909

1) Proceeding in the

same way we can find that 1199092119899+1

isin 119878minus1(1199092119899) and 119909

2119899+2isin

119879minus1(1199092119899+1) for 119899 = 1 2 3 Therefore 119909

2119899= 1198781199092119899+1

and1199092119899+1= 1198791199092119899+2

for 119899 = 0 1 2 If 119899 = 2119898 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898minus1 1199092119898)

= 120590 (1198791199092119898 1198781199092119898+1)

= 120590 (1198791198781199092119898+1 1198781199092119898+1)

ge min 120590 (119879119878 (1199092119898+1) 119878 (119909

2119898+1))

120590 (119878119879 (1199092119898+1) 119879 (119909

2119898+1))

ge 119903max 120590 (1198781199092119898+1 1199092119898+1) 120590 (119879119909

2119898+1 1199092119898+1)

ge 119903120590 (1198781199092119898+1 1199092119898+1)

= 119903120590 (1199092119898 1199092119898+1)

= 119903120590 (119909119899 119909119899+1)

(61)

Journal of Function Spaces and Applications 7

If 119899 = 2119898 + 1 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898 1199092119898+1)

= 120590 (1198781199092119898+1 1198791199092119898+2)

= 120590 (1198781198791199092119898+2 1198791199092119898+2)

ge min 120590 (119879119878 (1199092119898+2) 119878 (119909

2119898+2))

120590 (119878119879 (1199092119898+2) 119879 (119909

2119898+2))

ge 119903max 120590 (1198781199092119898+2 1199092119898+2) 120590 (119879119909

2119898+2 1199092119898+2)

ge 119903120590 (1198791199092119898+2 1199092119898+2)

= 119903120590 (1199092119898+1 1199092119898+2)

= 119903120590 (119909119899 119909119899+1)

(62)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899minus1 119909119899) ge 119903120590 (119909

119899 119909119899+1) (63)

which implies that

120590 (119909119899 119909119899+1) le1

119903120590 (119909119899minus1 119909119899) le sdot sdot sdot le (

1

119903)119899

120590 (1199090 1199091) (64)

Let 120572 = 1119903 then 0 lt 120572 lt 1 since 119903 gt 1Now (64) becomes

120590 (119909119899 119909119899+1) le 120572119899120590 (1199090 1199091) (65)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [120572119899+ 120572119899+1+ sdot sdot sdot + 120572

119898minus1] 120590 (119909

0 1199091)

le120572119899

1 minus 120572120590 (1199090 1199091)

(66)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0 That is 119909

119899 is a 120590-Cauchy

sequence in the metric-like space (119883 120590) Since (119883 120590) is 120590-complete there exists 119911 isin 119883 such that

120590 (119911 119911) = lim119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (67)

Assume that 119911 is not a common fixed point of 119878 and 119879 Thenby hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120590 (119909

119899minus1 119909119899) 119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120572119899minus1120590 (1199090 1199091) 119899 isin N = 0

(68)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)

If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = min 120590 (119879119878 (V) 119878 (V)) 120590 (119878119879 (V) 119879 (V))ge 119903max 120590 (119878 (V) V) 120590 (119879 (V) V)= 119903max 120590 (V V) 120590 (V V)= 119903120590 (V V)

(69)

which gives that 120590(V V) = 0

Corollary 16 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be an onto mapping Suppose that there exists119903 isin [0 1) such that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (70)

for every 119909 isin 119883 and that120572 (119910) = inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (71)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 15we have the desired result

Definition 17 Let (119883 120590) and (119884 120591) bemetric-like spacesThen119891 119883 rarr 119884 is said to be a continuous mapping iflim119899rarrinfin

119909119899= 119909 implies that lim

119899rarrinfin119891(119909119899) = 119891(119909)

Corollary 18 Let (119883 120590) be a complete metric-like space andlet 119879 be a mapping of 119883 into itself If there is a real number 119903with 119903 gt 1 satisfying

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (72)

for every 119909 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Assume that there exists 119910 isin 119883 with 119910 = 119879(119910) and

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 = 0 (73)

Then there exists a sequence 119909119899 such that

lim119899rarrinfin

120590 (119909119899 119910) + 120590 (119879 (119909

119899) 119909119899) = 0 (74)

So we have 120590(119909119899 119910) rarr 0 and 120590(119879(119909

119899) 119909119899) rarr 0 as 119899 rarr infin

Since 120590(119910 119910) le 120590(119910 119909119899) + 120590(119909

119899 119910) hence 120590(119910 119910) rarr 0 as

119899 rarr infin Now120590 (119879 (119909

119899) 119910) le 120590 (119879 (119909

119899) 119909119899) + 120590 (119909

119899 119910) 997888rarr 0

as 119899 997888rarr infin(75)

Since 119879 is continuous we have

119879 (119910) = 119879 ( lim119899rarrinfin119909119899) = lim119899rarrinfin119879 (119909119899) = 119910 (76)

This is a contradiction Hence if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (77)

which is condition (71) of Corollary 16 By Corollary 16 thereexists 119911 isin 119883 such that 119911 = 119879(119911)

8 Journal of Function Spaces and Applications

Now we give an example to support our result

Example 19 Let 119883 = [0infin) and 120590(119909 119910) = 119909 + 119910 Define 119879 119883 rarr 119883 by 119879(119909) = 2119909

Obviously 119879 is onto and continuous Also for each 119909 119910 isin119883 we have

120590 (1198792119909 119879119909) = 4119909 + 2119909 = 6119909 ge 1199033119909 = 119903120590 (119879119909 119909) (78)

where 119903 = 2 Thus 119879 satisfies the conditions given inCorollary 18 and 0 is the fixed point of 119879

Corollary 20 Let (119883 120590) be a complete metric-like space and119879 be a mapping of119883 into itself If there is a real number 119903 with119903 gt 1 satisfying

120590 (119879 (119909) 119879 (119910)) ge 119903min 120590 (119909 119879 (119909)) 120590 (119879 (119910) 119910) 120590 (119909 119910)(79)

for every 119909 119910 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Replacing 119910 by 119879(119909) in (79) we obtain

120590 (119879 (119909) 1198792(119909))

ge 119903min 120590 (119909 119879 (119909)) 120590 (1198792 (119909) 119879 (119909)) 120590 (119909 119879 (119909))(80)

for all 119909 isin 119883Without loss of generality we may assume that 119879(119909) =

1198792(119909) Otherwise 119879 has a fixed point Since 119903 gt 1 it followsfrom (80) that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (81)

for every 119909 isin 119883 By the argument similar to that used inCorollary 18 we can prove that if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (82)

which is condition (71) of Corollary 16 So Corollary 16applies to obtain a fixed point of 119879

According toTheorem 12 we get the following result

Corollary 21 (see [17 Theorem 1]) Let (119883 119901) be a completepartial metric space Let 119878 119879 119883 rarr 119883 be two self-mappingsSuppose that there exists 119903 isin [0 1) such that

max 119901 (119878 (119909) 119879119878 (119909)) 119901 (119879 (119909) 119878119879 (119909))

le 119903min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))(83)

for every 119909 isin 119883 and that

120572 (119910) = inf 119901 (119909 119910) +min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))

119909 isin 119883 gt 0

(84)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 119901(V V) = 0

Proof Using a similar argument given in the Theorem 12 for120590(119909 119910) = 119901(119909 119910) the desired result is obtained where 119901 is apartial metric on119883

Also according to Theorem 15 we get Theorem 2 from[17]

References

[1] A Aghajani S Radenovic and J R Roshan ldquoCommon fixedpoint results for four mappings satisfying almost generalized(119878 119879)-contractive condition in partially ordered metric spacesrdquoAppliedMathematics andComputation vol 218 no 9 pp 5665ndash5670 2012

[2] A Amini-Harandi ldquoMetric-like spaces partial metric spacesand fixed pointsrdquo Fixed PointTheory and Applications vol 2012article 204 10 pages 2012

[3] D ETHoric Z Kadelburg and S Radenovic ldquoEdelstein-Suzuki-type fixed point results in metric and abstract metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 75 no4 pp 1927ndash1932 2012

[4] N Hussain D ETHoric Z Kadelburg and S Radenovic ldquoSuzuki-type fixed point results in metric type spacesrdquo Fixed PointTheory and Applications vol 2012 article 126 12 pages 2012

[5] N Hussain and M H Shah ldquoKKMmappings in cone 119887-metricspacesrdquoComputersampMathematics withApplications vol 62 no4 pp 1677ndash1684 2011

[6] N Hussain M H Shah A Amini-Harandi and Z AkhtarldquoCommon fixed point theorems for generalized contractivemappings with applicationsrdquo Fixed Point Theory and Applica-tions vol 2013 article 169 17 pages 2013

[7] N Hussain Z Kadelburg S Radenovic and F Al-SolamyldquoComparison functions and fixed point results in partial metricspacesrdquo Abstract and Applied Analysis vol 2012 Article ID605781 15 pages 2012

[8] M Kikkawa and T Suzuki ldquoSome similarity between con-tractions and Kannan mappingsrdquo Fixed Point Theory andApplications vol 2008 Article ID 649749 8 pages 2008

[9] MKikkawa andT Suzuki ldquoSomenotes on fixed point theoremswith constantsrdquo Bulletin of the Kyushu Institute of Technologyno 56 pp 11ndash18 2009

[10] M Kikkawa and T Suzuki ldquoThree fixed point theorems forgeneralized contractions with constants in complete metricspacesrdquoNonlinearAnalysisTheoryMethodsampApplications vol69 no 9 pp 2942ndash2949 2008

[11] M A Kutbi J Ahmad N Hussain and M Arshad ldquoCommonfixed point results for mappings with rational expressionsrdquoAbstract and Applied Analysis vol 2013 Article ID 549518 11pages 2013

[12] S G Matthews ldquoPartial metric topologyrdquo in Papers on GeneralTopology and Applications vol 728 of Annals of the NewYork Academy of Sciences pp 183ndash197 New York Academy ofSciences New York NY USA 1994

[13] V Parvaneh J R Roshan and S Radenovic ldquoExistence oftripled coincidence points in ordered b-metric spaces and anapplication to a systemof integral equationsrdquo Fixed PointTheoryand Applications vol 2013 article 130 19 pages 2013

[14] O Popescu ldquoFixed point theorem in metric spacesrdquo Bulletinof the Transilvania University of Brasov vol 150 pp 479ndash4822008

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Function Spaces and Applications

Assume that 119911 is not a common fixed point of 119878 and 119879Then by hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 120590 (119909

119899 119909119899+1) 119899 isin N

le inf 119903119899

1 minus 119903120590 (1199090 1199091) + 119903119899120590 (1199090 1199091) 119899 isin N = 0

(53)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = max 120590 (119878 (V) 119879119878 (V)) 120590 (119879 (V) 119878119879 (V))

le 119903min 120590 (V 119878 (V)) 120590 (V 119879 (V))

= 119903min 120590 (V V) 120590 (V V)

= 119903120590 (V V)

(54)

which gives that 120590(V V) = 0

Example 13 Let (119883 120590) be a metric-like space where 119883 =1119899infin

119899=1cup 0 and 120590(119909 119910) = 119909 + 119910 Define 119878 119883 rarr 119883 by

119878(0) = 0 119878(12119899) = 1(4119899 + 3) and 119878(1(2119899 minus 1)) = 0 and119879(0) = 0 119879(1(2119899 minus 1)) = 1(4119899 + 4) and 119879(12119899) = 0 Thenfor 119909 = 12119899 we have

max 120590 (119878 (119909) 119879119878 (119909)) 120590 (119879 (119909) 119878119879 (119909))

= max 120590 (119878 ( 12119899) 119879 (119878 (

1

2119899)))

120590 (119879(1

2119899) 119878 (119879(

1

2119899)))

= max 1

4119899 + 3+

1

8119899 + 12 0 =

1

4119899 + 3+

1

8119899 + 12

le 119903min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

= 119903min 12119899+1

4119899 + 31

2119899+ 0 = 119903

1

2119899

(55)

It is easy to see that the above inequality is true for119909 = 1(2119899minus1) and for 34 le 119903 lt 1 Also

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(56)

for every 119910 isin 119883 with y is not a common fixed point of 119878 and119879 This shows that all conditions of Theorem 12 are satisfiedand 0 is a common fixed point for 119878 and 119879

Corollary 14 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be a mapping Suppose that there exists 119903 isin[0 1) such that

120590 (119879 (119909) 1198792(119909)) le 119903120590 (119909 119879 (119909)) (57)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) + 120590 (119909 119879 (119909)) 119909 isin 119883 gt 0 (58)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 12 the conclusion of the cor-ollary follows

Theorem 15 Let (119883 120590) be a complete metric-like space Let119878 119879 be mappings from 119883 onto itself Suppose that there exists119903 gt 1 such that

min 120590 (119879119878 (119909) 119878 (119909)) 120590 (119878119879 (119909) 119879 (119909))

ge 119903max 120590 (119878119909 119909) 120590 (119879119909 119909)(59)

for every 119909 isin 119883 and that

120572 (119910) = inf 120590 (119909 119910) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909))

119909 isin 119883 gt 0

(60)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 120590(V V) = 0

Proof Let 1199090isin 119883 be arbitrary Since 119878 is onto there is an

element 1199091satisfying 119909

1isin 119878minus1(119909

0) Since 119879 is also onto there

is an element 1199092satisfying 119909

2isin 119879minus1(119909

1) Proceeding in the

same way we can find that 1199092119899+1

isin 119878minus1(1199092119899) and 119909

2119899+2isin

119879minus1(1199092119899+1) for 119899 = 1 2 3 Therefore 119909

2119899= 1198781199092119899+1

and1199092119899+1= 1198791199092119899+2

for 119899 = 0 1 2 If 119899 = 2119898 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898minus1 1199092119898)

= 120590 (1198791199092119898 1198781199092119898+1)

= 120590 (1198791198781199092119898+1 1198781199092119898+1)

ge min 120590 (119879119878 (1199092119898+1) 119878 (119909

2119898+1))

120590 (119878119879 (1199092119898+1) 119879 (119909

2119898+1))

ge 119903max 120590 (1198781199092119898+1 1199092119898+1) 120590 (119879119909

2119898+1 1199092119898+1)

ge 119903120590 (1198781199092119898+1 1199092119898+1)

= 119903120590 (1199092119898 1199092119898+1)

= 119903120590 (119909119899 119909119899+1)

(61)

Journal of Function Spaces and Applications 7

If 119899 = 2119898 + 1 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898 1199092119898+1)

= 120590 (1198781199092119898+1 1198791199092119898+2)

= 120590 (1198781198791199092119898+2 1198791199092119898+2)

ge min 120590 (119879119878 (1199092119898+2) 119878 (119909

2119898+2))

120590 (119878119879 (1199092119898+2) 119879 (119909

2119898+2))

ge 119903max 120590 (1198781199092119898+2 1199092119898+2) 120590 (119879119909

2119898+2 1199092119898+2)

ge 119903120590 (1198791199092119898+2 1199092119898+2)

= 119903120590 (1199092119898+1 1199092119898+2)

= 119903120590 (119909119899 119909119899+1)

(62)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899minus1 119909119899) ge 119903120590 (119909

119899 119909119899+1) (63)

which implies that

120590 (119909119899 119909119899+1) le1

119903120590 (119909119899minus1 119909119899) le sdot sdot sdot le (

1

119903)119899

120590 (1199090 1199091) (64)

Let 120572 = 1119903 then 0 lt 120572 lt 1 since 119903 gt 1Now (64) becomes

120590 (119909119899 119909119899+1) le 120572119899120590 (1199090 1199091) (65)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [120572119899+ 120572119899+1+ sdot sdot sdot + 120572

119898minus1] 120590 (119909

0 1199091)

le120572119899

1 minus 120572120590 (1199090 1199091)

(66)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0 That is 119909

119899 is a 120590-Cauchy

sequence in the metric-like space (119883 120590) Since (119883 120590) is 120590-complete there exists 119911 isin 119883 such that

120590 (119911 119911) = lim119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (67)

Assume that 119911 is not a common fixed point of 119878 and 119879 Thenby hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120590 (119909

119899minus1 119909119899) 119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120572119899minus1120590 (1199090 1199091) 119899 isin N = 0

(68)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)

If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = min 120590 (119879119878 (V) 119878 (V)) 120590 (119878119879 (V) 119879 (V))ge 119903max 120590 (119878 (V) V) 120590 (119879 (V) V)= 119903max 120590 (V V) 120590 (V V)= 119903120590 (V V)

(69)

which gives that 120590(V V) = 0

Corollary 16 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be an onto mapping Suppose that there exists119903 isin [0 1) such that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (70)

for every 119909 isin 119883 and that120572 (119910) = inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (71)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 15we have the desired result

Definition 17 Let (119883 120590) and (119884 120591) bemetric-like spacesThen119891 119883 rarr 119884 is said to be a continuous mapping iflim119899rarrinfin

119909119899= 119909 implies that lim

119899rarrinfin119891(119909119899) = 119891(119909)

Corollary 18 Let (119883 120590) be a complete metric-like space andlet 119879 be a mapping of 119883 into itself If there is a real number 119903with 119903 gt 1 satisfying

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (72)

for every 119909 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Assume that there exists 119910 isin 119883 with 119910 = 119879(119910) and

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 = 0 (73)

Then there exists a sequence 119909119899 such that

lim119899rarrinfin

120590 (119909119899 119910) + 120590 (119879 (119909

119899) 119909119899) = 0 (74)

So we have 120590(119909119899 119910) rarr 0 and 120590(119879(119909

119899) 119909119899) rarr 0 as 119899 rarr infin

Since 120590(119910 119910) le 120590(119910 119909119899) + 120590(119909

119899 119910) hence 120590(119910 119910) rarr 0 as

119899 rarr infin Now120590 (119879 (119909

119899) 119910) le 120590 (119879 (119909

119899) 119909119899) + 120590 (119909

119899 119910) 997888rarr 0

as 119899 997888rarr infin(75)

Since 119879 is continuous we have

119879 (119910) = 119879 ( lim119899rarrinfin119909119899) = lim119899rarrinfin119879 (119909119899) = 119910 (76)

This is a contradiction Hence if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (77)

which is condition (71) of Corollary 16 By Corollary 16 thereexists 119911 isin 119883 such that 119911 = 119879(119911)

8 Journal of Function Spaces and Applications

Now we give an example to support our result

Example 19 Let 119883 = [0infin) and 120590(119909 119910) = 119909 + 119910 Define 119879 119883 rarr 119883 by 119879(119909) = 2119909

Obviously 119879 is onto and continuous Also for each 119909 119910 isin119883 we have

120590 (1198792119909 119879119909) = 4119909 + 2119909 = 6119909 ge 1199033119909 = 119903120590 (119879119909 119909) (78)

where 119903 = 2 Thus 119879 satisfies the conditions given inCorollary 18 and 0 is the fixed point of 119879

Corollary 20 Let (119883 120590) be a complete metric-like space and119879 be a mapping of119883 into itself If there is a real number 119903 with119903 gt 1 satisfying

120590 (119879 (119909) 119879 (119910)) ge 119903min 120590 (119909 119879 (119909)) 120590 (119879 (119910) 119910) 120590 (119909 119910)(79)

for every 119909 119910 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Replacing 119910 by 119879(119909) in (79) we obtain

120590 (119879 (119909) 1198792(119909))

ge 119903min 120590 (119909 119879 (119909)) 120590 (1198792 (119909) 119879 (119909)) 120590 (119909 119879 (119909))(80)

for all 119909 isin 119883Without loss of generality we may assume that 119879(119909) =

1198792(119909) Otherwise 119879 has a fixed point Since 119903 gt 1 it followsfrom (80) that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (81)

for every 119909 isin 119883 By the argument similar to that used inCorollary 18 we can prove that if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (82)

which is condition (71) of Corollary 16 So Corollary 16applies to obtain a fixed point of 119879

According toTheorem 12 we get the following result

Corollary 21 (see [17 Theorem 1]) Let (119883 119901) be a completepartial metric space Let 119878 119879 119883 rarr 119883 be two self-mappingsSuppose that there exists 119903 isin [0 1) such that

max 119901 (119878 (119909) 119879119878 (119909)) 119901 (119879 (119909) 119878119879 (119909))

le 119903min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))(83)

for every 119909 isin 119883 and that

120572 (119910) = inf 119901 (119909 119910) +min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))

119909 isin 119883 gt 0

(84)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 119901(V V) = 0

Proof Using a similar argument given in the Theorem 12 for120590(119909 119910) = 119901(119909 119910) the desired result is obtained where 119901 is apartial metric on119883

Also according to Theorem 15 we get Theorem 2 from[17]

References

[1] A Aghajani S Radenovic and J R Roshan ldquoCommon fixedpoint results for four mappings satisfying almost generalized(119878 119879)-contractive condition in partially ordered metric spacesrdquoAppliedMathematics andComputation vol 218 no 9 pp 5665ndash5670 2012

[2] A Amini-Harandi ldquoMetric-like spaces partial metric spacesand fixed pointsrdquo Fixed PointTheory and Applications vol 2012article 204 10 pages 2012

[3] D ETHoric Z Kadelburg and S Radenovic ldquoEdelstein-Suzuki-type fixed point results in metric and abstract metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 75 no4 pp 1927ndash1932 2012

[4] N Hussain D ETHoric Z Kadelburg and S Radenovic ldquoSuzuki-type fixed point results in metric type spacesrdquo Fixed PointTheory and Applications vol 2012 article 126 12 pages 2012

[5] N Hussain and M H Shah ldquoKKMmappings in cone 119887-metricspacesrdquoComputersampMathematics withApplications vol 62 no4 pp 1677ndash1684 2011

[6] N Hussain M H Shah A Amini-Harandi and Z AkhtarldquoCommon fixed point theorems for generalized contractivemappings with applicationsrdquo Fixed Point Theory and Applica-tions vol 2013 article 169 17 pages 2013

[7] N Hussain Z Kadelburg S Radenovic and F Al-SolamyldquoComparison functions and fixed point results in partial metricspacesrdquo Abstract and Applied Analysis vol 2012 Article ID605781 15 pages 2012

[8] M Kikkawa and T Suzuki ldquoSome similarity between con-tractions and Kannan mappingsrdquo Fixed Point Theory andApplications vol 2008 Article ID 649749 8 pages 2008

[9] MKikkawa andT Suzuki ldquoSomenotes on fixed point theoremswith constantsrdquo Bulletin of the Kyushu Institute of Technologyno 56 pp 11ndash18 2009

[10] M Kikkawa and T Suzuki ldquoThree fixed point theorems forgeneralized contractions with constants in complete metricspacesrdquoNonlinearAnalysisTheoryMethodsampApplications vol69 no 9 pp 2942ndash2949 2008

[11] M A Kutbi J Ahmad N Hussain and M Arshad ldquoCommonfixed point results for mappings with rational expressionsrdquoAbstract and Applied Analysis vol 2013 Article ID 549518 11pages 2013

[12] S G Matthews ldquoPartial metric topologyrdquo in Papers on GeneralTopology and Applications vol 728 of Annals of the NewYork Academy of Sciences pp 183ndash197 New York Academy ofSciences New York NY USA 1994

[13] V Parvaneh J R Roshan and S Radenovic ldquoExistence oftripled coincidence points in ordered b-metric spaces and anapplication to a systemof integral equationsrdquo Fixed PointTheoryand Applications vol 2013 article 130 19 pages 2013

[14] O Popescu ldquoFixed point theorem in metric spacesrdquo Bulletinof the Transilvania University of Brasov vol 150 pp 479ndash4822008

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces and Applications 7

If 119899 = 2119898 + 1 then using (59)

120590 (119909119899minus1 119909119899)

= 120590 (1199092119898 1199092119898+1)

= 120590 (1198781199092119898+1 1198791199092119898+2)

= 120590 (1198781198791199092119898+2 1198791199092119898+2)

ge min 120590 (119879119878 (1199092119898+2) 119878 (119909

2119898+2))

120590 (119878119879 (1199092119898+2) 119879 (119909

2119898+2))

ge 119903max 120590 (1198781199092119898+2 1199092119898+2) 120590 (119879119909

2119898+2 1199092119898+2)

ge 119903120590 (1198791199092119898+2 1199092119898+2)

= 119903120590 (1199092119898+1 1199092119898+2)

= 119903120590 (119909119899 119909119899+1)

(62)

Thus for any positive integer 119899 it must be the case that

120590 (119909119899minus1 119909119899) ge 119903120590 (119909

119899 119909119899+1) (63)

which implies that

120590 (119909119899 119909119899+1) le1

119903120590 (119909119899minus1 119909119899) le sdot sdot sdot le (

1

119903)119899

120590 (1199090 1199091) (64)

Let 120572 = 1119903 then 0 lt 120572 lt 1 since 119903 gt 1Now (64) becomes

120590 (119909119899 119909119899+1) le 120572119899120590 (1199090 1199091) (65)

So if119898 gt 119899 then

120590 (119909119899 119909119898) le 120590 (119909

119899 119909119899+1)

+ 120590 (119909119899+1 119909119899+2) + sdot sdot sdot + 120590 (119909

119898minus1 119909119898)

le [120572119899+ 120572119899+1+ sdot sdot sdot + 120572

119898minus1] 120590 (119909

0 1199091)

le120572119899

1 minus 120572120590 (1199090 1199091)

(66)

Thus lim119899119898rarrinfin

120590(119909119899 119909119898) = 0 That is 119909

119899 is a 120590-Cauchy

sequence in the metric-like space (119883 120590) Since (119883 120590) is 120590-complete there exists 119911 isin 119883 such that

120590 (119911 119911) = lim119899rarrinfin120590 (119909119899 119911) = lim

119899119898rarrinfin120590 (119909119899 119909119898) = 0 (67)

Assume that 119911 is not a common fixed point of 119878 and 119879 Thenby hypothesis

0 lt inf 120590 (119909 119911) +min 120590 (119909 119878 (119909)) 120590 (119909 119879 (119909)) 119909 isin 119883

le inf 120590 (119909119899 119911) +min 120590 (119909

119899 119878 (119909119899)) 120590 (119909

119899 119879 (119909119899))

119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120590 (119909

119899minus1 119909119899) 119899 isin N

le inf 120572119899

1 minus 120572120590 (1199090 1199091) + 120572119899minus1120590 (1199090 1199091) 119899 isin N = 0

(68)

which is a contradiction Therefore 119911 = 119878(119911) = 119879(119911)

If V = 119878(V) = 119879(V) for some V isin 119883 then

120590 (V V) = min 120590 (119879119878 (V) 119878 (V)) 120590 (119878119879 (V) 119879 (V))ge 119903max 120590 (119878 (V) V) 120590 (119879 (V) V)= 119903max 120590 (V V) 120590 (V V)= 119903120590 (V V)

(69)

which gives that 120590(V V) = 0

Corollary 16 Let (119883 120590) be a complete metric-like space andlet 119879 119883 rarr 119883 be an onto mapping Suppose that there exists119903 isin [0 1) such that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (70)

for every 119909 isin 119883 and that120572 (119910) = inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (71)

for every 119910 isin 119883 with 119910 = 119879(119910) Then there exists 119911 isin 119883 suchthat 119911 = 119879(119911) Moreover if V = 119879(V) then 120590(V V) = 0

Proof Taking 119878 = 119879 inTheorem 15we have the desired result

Definition 17 Let (119883 120590) and (119884 120591) bemetric-like spacesThen119891 119883 rarr 119884 is said to be a continuous mapping iflim119899rarrinfin

119909119899= 119909 implies that lim

119899rarrinfin119891(119909119899) = 119891(119909)

Corollary 18 Let (119883 120590) be a complete metric-like space andlet 119879 be a mapping of 119883 into itself If there is a real number 119903with 119903 gt 1 satisfying

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (72)

for every 119909 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Assume that there exists 119910 isin 119883 with 119910 = 119879(119910) and

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 = 0 (73)

Then there exists a sequence 119909119899 such that

lim119899rarrinfin

120590 (119909119899 119910) + 120590 (119879 (119909

119899) 119909119899) = 0 (74)

So we have 120590(119909119899 119910) rarr 0 and 120590(119879(119909

119899) 119909119899) rarr 0 as 119899 rarr infin

Since 120590(119910 119910) le 120590(119910 119909119899) + 120590(119909

119899 119910) hence 120590(119910 119910) rarr 0 as

119899 rarr infin Now120590 (119879 (119909

119899) 119910) le 120590 (119879 (119909

119899) 119909119899) + 120590 (119909

119899 119910) 997888rarr 0

as 119899 997888rarr infin(75)

Since 119879 is continuous we have

119879 (119910) = 119879 ( lim119899rarrinfin119909119899) = lim119899rarrinfin119879 (119909119899) = 119910 (76)

This is a contradiction Hence if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (77)

which is condition (71) of Corollary 16 By Corollary 16 thereexists 119911 isin 119883 such that 119911 = 119879(119911)

8 Journal of Function Spaces and Applications

Now we give an example to support our result

Example 19 Let 119883 = [0infin) and 120590(119909 119910) = 119909 + 119910 Define 119879 119883 rarr 119883 by 119879(119909) = 2119909

Obviously 119879 is onto and continuous Also for each 119909 119910 isin119883 we have

120590 (1198792119909 119879119909) = 4119909 + 2119909 = 6119909 ge 1199033119909 = 119903120590 (119879119909 119909) (78)

where 119903 = 2 Thus 119879 satisfies the conditions given inCorollary 18 and 0 is the fixed point of 119879

Corollary 20 Let (119883 120590) be a complete metric-like space and119879 be a mapping of119883 into itself If there is a real number 119903 with119903 gt 1 satisfying

120590 (119879 (119909) 119879 (119910)) ge 119903min 120590 (119909 119879 (119909)) 120590 (119879 (119910) 119910) 120590 (119909 119910)(79)

for every 119909 119910 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Replacing 119910 by 119879(119909) in (79) we obtain

120590 (119879 (119909) 1198792(119909))

ge 119903min 120590 (119909 119879 (119909)) 120590 (1198792 (119909) 119879 (119909)) 120590 (119909 119879 (119909))(80)

for all 119909 isin 119883Without loss of generality we may assume that 119879(119909) =

1198792(119909) Otherwise 119879 has a fixed point Since 119903 gt 1 it followsfrom (80) that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (81)

for every 119909 isin 119883 By the argument similar to that used inCorollary 18 we can prove that if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (82)

which is condition (71) of Corollary 16 So Corollary 16applies to obtain a fixed point of 119879

According toTheorem 12 we get the following result

Corollary 21 (see [17 Theorem 1]) Let (119883 119901) be a completepartial metric space Let 119878 119879 119883 rarr 119883 be two self-mappingsSuppose that there exists 119903 isin [0 1) such that

max 119901 (119878 (119909) 119879119878 (119909)) 119901 (119879 (119909) 119878119879 (119909))

le 119903min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))(83)

for every 119909 isin 119883 and that

120572 (119910) = inf 119901 (119909 119910) +min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))

119909 isin 119883 gt 0

(84)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 119901(V V) = 0

Proof Using a similar argument given in the Theorem 12 for120590(119909 119910) = 119901(119909 119910) the desired result is obtained where 119901 is apartial metric on119883

Also according to Theorem 15 we get Theorem 2 from[17]

References

[1] A Aghajani S Radenovic and J R Roshan ldquoCommon fixedpoint results for four mappings satisfying almost generalized(119878 119879)-contractive condition in partially ordered metric spacesrdquoAppliedMathematics andComputation vol 218 no 9 pp 5665ndash5670 2012

[2] A Amini-Harandi ldquoMetric-like spaces partial metric spacesand fixed pointsrdquo Fixed PointTheory and Applications vol 2012article 204 10 pages 2012

[3] D ETHoric Z Kadelburg and S Radenovic ldquoEdelstein-Suzuki-type fixed point results in metric and abstract metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 75 no4 pp 1927ndash1932 2012

[4] N Hussain D ETHoric Z Kadelburg and S Radenovic ldquoSuzuki-type fixed point results in metric type spacesrdquo Fixed PointTheory and Applications vol 2012 article 126 12 pages 2012

[5] N Hussain and M H Shah ldquoKKMmappings in cone 119887-metricspacesrdquoComputersampMathematics withApplications vol 62 no4 pp 1677ndash1684 2011

[6] N Hussain M H Shah A Amini-Harandi and Z AkhtarldquoCommon fixed point theorems for generalized contractivemappings with applicationsrdquo Fixed Point Theory and Applica-tions vol 2013 article 169 17 pages 2013

[7] N Hussain Z Kadelburg S Radenovic and F Al-SolamyldquoComparison functions and fixed point results in partial metricspacesrdquo Abstract and Applied Analysis vol 2012 Article ID605781 15 pages 2012

[8] M Kikkawa and T Suzuki ldquoSome similarity between con-tractions and Kannan mappingsrdquo Fixed Point Theory andApplications vol 2008 Article ID 649749 8 pages 2008

[9] MKikkawa andT Suzuki ldquoSomenotes on fixed point theoremswith constantsrdquo Bulletin of the Kyushu Institute of Technologyno 56 pp 11ndash18 2009

[10] M Kikkawa and T Suzuki ldquoThree fixed point theorems forgeneralized contractions with constants in complete metricspacesrdquoNonlinearAnalysisTheoryMethodsampApplications vol69 no 9 pp 2942ndash2949 2008

[11] M A Kutbi J Ahmad N Hussain and M Arshad ldquoCommonfixed point results for mappings with rational expressionsrdquoAbstract and Applied Analysis vol 2013 Article ID 549518 11pages 2013

[12] S G Matthews ldquoPartial metric topologyrdquo in Papers on GeneralTopology and Applications vol 728 of Annals of the NewYork Academy of Sciences pp 183ndash197 New York Academy ofSciences New York NY USA 1994

[13] V Parvaneh J R Roshan and S Radenovic ldquoExistence oftripled coincidence points in ordered b-metric spaces and anapplication to a systemof integral equationsrdquo Fixed PointTheoryand Applications vol 2013 article 130 19 pages 2013

[14] O Popescu ldquoFixed point theorem in metric spacesrdquo Bulletinof the Transilvania University of Brasov vol 150 pp 479ndash4822008

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Function Spaces and Applications

Now we give an example to support our result

Example 19 Let 119883 = [0infin) and 120590(119909 119910) = 119909 + 119910 Define 119879 119883 rarr 119883 by 119879(119909) = 2119909

Obviously 119879 is onto and continuous Also for each 119909 119910 isin119883 we have

120590 (1198792119909 119879119909) = 4119909 + 2119909 = 6119909 ge 1199033119909 = 119903120590 (119879119909 119909) (78)

where 119903 = 2 Thus 119879 satisfies the conditions given inCorollary 18 and 0 is the fixed point of 119879

Corollary 20 Let (119883 120590) be a complete metric-like space and119879 be a mapping of119883 into itself If there is a real number 119903 with119903 gt 1 satisfying

120590 (119879 (119909) 119879 (119910)) ge 119903min 120590 (119909 119879 (119909)) 120590 (119879 (119910) 119910) 120590 (119909 119910)(79)

for every 119909 119910 isin 119883 and 119879 is onto and continuous then 119879 has afixed point

Proof Replacing 119910 by 119879(119909) in (79) we obtain

120590 (119879 (119909) 1198792(119909))

ge 119903min 120590 (119909 119879 (119909)) 120590 (1198792 (119909) 119879 (119909)) 120590 (119909 119879 (119909))(80)

for all 119909 isin 119883Without loss of generality we may assume that 119879(119909) =

1198792(119909) Otherwise 119879 has a fixed point Since 119903 gt 1 it followsfrom (80) that

120590 (1198792(119909) 119879 (119909)) ge 119903120590 (119879 (119909) 119909) (81)

for every 119909 isin 119883 By the argument similar to that used inCorollary 18 we can prove that if 119910 = 119879(119910) then

inf 120590 (119909 119910) + 120590 (119879 (119909) 119909) 119909 isin 119883 gt 0 (82)

which is condition (71) of Corollary 16 So Corollary 16applies to obtain a fixed point of 119879

According toTheorem 12 we get the following result

Corollary 21 (see [17 Theorem 1]) Let (119883 119901) be a completepartial metric space Let 119878 119879 119883 rarr 119883 be two self-mappingsSuppose that there exists 119903 isin [0 1) such that

max 119901 (119878 (119909) 119879119878 (119909)) 119901 (119879 (119909) 119878119879 (119909))

le 119903min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))(83)

for every 119909 isin 119883 and that

120572 (119910) = inf 119901 (119909 119910) +min 119901 (119909 119878 (119909)) 119901 (119909 119879 (119909))

119909 isin 119883 gt 0

(84)

for every 119910 isin 119883 with 119910 that is not a common fixed point of 119878and 119879 Then there exists 119911 isin 119883 such that 119911 = 119878(119911) = 119879(119911)Moreover if V = 119878(V) = 119879(V) then 119901(V V) = 0

Proof Using a similar argument given in the Theorem 12 for120590(119909 119910) = 119901(119909 119910) the desired result is obtained where 119901 is apartial metric on119883

Also according to Theorem 15 we get Theorem 2 from[17]

References

[1] A Aghajani S Radenovic and J R Roshan ldquoCommon fixedpoint results for four mappings satisfying almost generalized(119878 119879)-contractive condition in partially ordered metric spacesrdquoAppliedMathematics andComputation vol 218 no 9 pp 5665ndash5670 2012

[2] A Amini-Harandi ldquoMetric-like spaces partial metric spacesand fixed pointsrdquo Fixed PointTheory and Applications vol 2012article 204 10 pages 2012

[3] D ETHoric Z Kadelburg and S Radenovic ldquoEdelstein-Suzuki-type fixed point results in metric and abstract metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 75 no4 pp 1927ndash1932 2012

[4] N Hussain D ETHoric Z Kadelburg and S Radenovic ldquoSuzuki-type fixed point results in metric type spacesrdquo Fixed PointTheory and Applications vol 2012 article 126 12 pages 2012

[5] N Hussain and M H Shah ldquoKKMmappings in cone 119887-metricspacesrdquoComputersampMathematics withApplications vol 62 no4 pp 1677ndash1684 2011

[6] N Hussain M H Shah A Amini-Harandi and Z AkhtarldquoCommon fixed point theorems for generalized contractivemappings with applicationsrdquo Fixed Point Theory and Applica-tions vol 2013 article 169 17 pages 2013

[7] N Hussain Z Kadelburg S Radenovic and F Al-SolamyldquoComparison functions and fixed point results in partial metricspacesrdquo Abstract and Applied Analysis vol 2012 Article ID605781 15 pages 2012

[8] M Kikkawa and T Suzuki ldquoSome similarity between con-tractions and Kannan mappingsrdquo Fixed Point Theory andApplications vol 2008 Article ID 649749 8 pages 2008

[9] MKikkawa andT Suzuki ldquoSomenotes on fixed point theoremswith constantsrdquo Bulletin of the Kyushu Institute of Technologyno 56 pp 11ndash18 2009

[10] M Kikkawa and T Suzuki ldquoThree fixed point theorems forgeneralized contractions with constants in complete metricspacesrdquoNonlinearAnalysisTheoryMethodsampApplications vol69 no 9 pp 2942ndash2949 2008

[11] M A Kutbi J Ahmad N Hussain and M Arshad ldquoCommonfixed point results for mappings with rational expressionsrdquoAbstract and Applied Analysis vol 2013 Article ID 549518 11pages 2013

[12] S G Matthews ldquoPartial metric topologyrdquo in Papers on GeneralTopology and Applications vol 728 of Annals of the NewYork Academy of Sciences pp 183ndash197 New York Academy ofSciences New York NY USA 1994

[13] V Parvaneh J R Roshan and S Radenovic ldquoExistence oftripled coincidence points in ordered b-metric spaces and anapplication to a systemof integral equationsrdquo Fixed PointTheoryand Applications vol 2013 article 130 19 pages 2013

[14] O Popescu ldquoFixed point theorem in metric spacesrdquo Bulletinof the Transilvania University of Brasov vol 150 pp 479ndash4822008

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces and Applications 9

[15] O Popescu ldquoTwo fixed point theorems for generalized contrac-tions with constants in complete metric spacerdquo Central Euro-pean Journal of Mathematics vol 7 no 3 pp 529ndash538 2009

[16] J R Roshan V Parvaneh S Sedghi N Shobkolaei and WShatanawi ldquoCommon fixed points of almost generalized(120595 120593)

119904-contractivemappings in ordered b-metric spacesrdquo Fixed

Point Theory and Applications vol 2013 article 159 23 pages2013

[17] S Sedghi and N Shobkolaei ldquoCommon fixed point of mapsin complete partial metric spacesrdquo East Asian MathematicalJournal vol 29 no 1 pp 1ndash12 2013

[18] N Shobkolaei S Sedghi J R Roshan and I Altun ldquoCommonfixed point of mappings satisfying almost generalized (119878 119879)-contractive condition in partially ordered partialmetric spacesrdquoApplied Mathematics and Computation vol 219 no 2 pp 443ndash452 2012

[19] T Suzuki ldquoA generalized Banach contraction principle thatcharacterizesmetric completenessrdquo Proceedings of the AmericanMathematical Society vol 136 no 5 pp 1861ndash1869 2008

[20] T Suzuki ldquoA new type of fixed point theorem in metric spacesrdquoNonlinear Analysis Theory Methods amp Applications vol 71 no11 pp 5313ndash5317 2009

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of