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On Suzuki-Wardowski type …xed point theorems

Nawab Hussain1 ; Jamshaid Ahmad2, Akbar Azam2

1Department of Mathematics, King Abdulaziz University, P.O. Box 80203,Jeddah 21589, Saudi Arabia

nhusain@kau.edu.sa

2Department of Mathematics COMSATS Institute of Information Technology, Chack Shahzad, Islamabad - 44000, Pakistan

jamshaid_jasim@yahoo.com, akbarazam@yahoo.com Abstract

Recently, Piri and Kumam [Fixed Point Theory and Applications 2014, 2014:210] improved concept ofF-contraction and proved some Wardowski and Suzuki type …xed point results in metric spaces. The aim of this article is to de…ne generalized GF-contraction and establish Wardowski and Suzuki type

…xed point results in metric and ordered metric spaces and derive main results of Piri et al. as corollaries. We also deduce certain …xed and periodic point results for orbitally continuous generalizedF-contractions and certain …xed point re- sults for integral inequalities are derived. Moreover, we discuss some illustrative examples to highlight the realized improvements.

Keywords: …xed point, -GF-contraction, - -continous function, orbitally continuous function.

1 Introduction and Preliminaries

In 1922, Banach established the most famous fundamental …xed point theorem (the so-called the Banach contraction principle [1]) which has played an impor- tant role in various …elds of applied mathematical analysis. It is known that the Banach contraction principle has been extended in many various directions by several authors (see [2]-[28]). One of the interesting results was given by Suzuki [27] which characterize the completeness of underlying metric spaces. He in- troduced a weaker notion of contraction and discussed the existence of some new …xed point theorems . Wardowski [29] introduced a new contraction called F-contraction and proved a …xed point result as a generalization of the Banach contraction principle. Abbas et al. [3] further generalized the concept of F- contraction and proved certain …xed and common …xed point results. Hussain et al. [12] introduced - -GF-contractions and obtained …xed point results in metric spaces and partially ordered metric spaces. They also established Suzuki type results for such GF-contractions. Recently Piri et al. [21] described a large class of functions by replacing condition(F30)instead of the condition(F3) in the de…ntion of F-contraction introduced by Wardowski [29] . In this paper, we improve the results of Hussain et al. [12] by replacing general conditions(F20) and(F30)instead of the conditions (F2) and(F3): We begin with some basic de…nitions and results which will be used in the sequel.

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In 2012, Samet et al. [16] introduced the concepts of - -contractive and -admissible mappings and established various …xed point theorems for such mappings de…ned on complete metric spaces. Afterwards Salimi et al. [15]

and Hussain et al. [8, 10, 11] modi…ed the notions of - -contractive and - admissible mappings and established certain …xed point theorems.

De…nition 1 [16] Let T be a self-mapping on X and :X X ![0;+1)be a function. We say thatT is an -admissible mapping if

x; y2X; (x; y) 1 =) (T x; T y) 1:

De…nition 2 [15] Let T be a self-mapping on X and ; :X X ![0;+1) be two functions. We say thatT is an -admissible mapping with respect to if

x; y2X; (x; y) (x; y) =) (T x; T y) (T x; T y):

Note that if we take (x; y) = 1 then this de…nition reduces to De…nition 1.

Also, if we take, (x; y) = 1 then we say thatT is an -subadmissible mapping.

De…nition 3 [10] Let(X; d)be a metric space. Let ; :X X![0;1)and T :X !X be functions. We say T is an - -continuous mapping on (X; d), if, for givenx2X and sequence fxng with

xn !xasn! 1; (xn; xn+1) (xn; xn+1)for alln2N=) T xn !T x:

Example 4 [10] Let X = [0;1) and d(x; y) = jx yj be a metric on X. Assume,T :X !X and ; :X X ![0;+1)be de…ned by

T x= 8<

:

x5; ifx2[0;1]

sin x+ 2; if(1;1)

; (x; y) = 8<

:

x2+y2+ 1; ifx; y2[0;1]

0; otherwise

and (x; y) = x2. Clearly, T is not continuous, but T is - -continuous on (X; d).

A mappingT :X!Xis called orbitally continuous atp2Xiflimn!1Tnx= pimplies that limn!1T Tnx=T p. The mapping T is orbitally continuous on X ifT is orbitally continuous for allp2X.

Remark 5 [10] LetT :X !X be a self-mapping on an orbitally T-complete metric spaceX. De…ne, ; :X X ![0;+1)by

(x; y) = 3; if x; y2O(w)

0; otherwise and (x; y) = 1

where O(w) is an orbit of a point w 2 X. If, T : X ! X is an orbitally continuous map on(X; d), thenT is - -continuous on(X; d).

Fixed point results for - -GF-contractions

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Wardowski [29] introduced and studied a new contraction calledF-contraction to prove a …xed point result as a generalization of the Banach contraction prin- ciple.

De…nition 6 LetF :R+!Rbe a mapping satisfying the following conditions:

(F1) F is strictly increasing;

(F2) for all sequencef ng R+,limn!1 n= 0if and only iflimn!1F( n) = 1;

(F3) there exists0< k <1 such thatlima!0+ kF( ) = 0:

Consistent with Wordowski [29], we denote by z the set of all functions F:R+!Rsatisfying conditionsF1; F2 andF3.

De…nition 7 [29] Let (X; d) be a metric space. A self-mapping T on X is called anF contraction if there exists >0 such that for x; y2X

d(T x; T y)>0 =) +F d(T x; T y) F d(x; y) whereF 2z.

Hussain et al.[12] generalized the results of Wordowski [29] by intoducing

G set of functionsG:R+4 !R+ which satisfy

(G) for all t1; tt; t3; t4 2 R+ with t1t2t3t4 = 0 there exists > 0 such that G(t1; t2; t3; t4) = :

They also given some example of such functions.

Example 8 [12] ifG(t1; t2; t3; t4) =Lminft1; t2; t3; t4g+ whereL2R+ and

>0, thenG2 G:

Example 9 [12] if G(t1; t2; t3; t4) = eLminft1; t2; t3; t4g where L 2 R+ and

>0, thenG2 G:

Example 10 [12] ifG(t1; t2; t3; t4) =Lln(minft1; t2; t3; t4g+ 1) + whereL2 R+ and >0, thenG2 G:

On the other hand Secelean [24] proved the following lemma and replaced condition(F2)by an equivalent but a more simple condition(F20):

Lemma 11 Let F :R+!Rbe an increasing map andf ng1n=1 be a sequence of positive real numbers. Then the following assertions hold:

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(a) iflimn!1F( n) = 1thenlimn!1 n = 0;

(b) ifinfF = 1and limn!1 n = 0;thenlimn!1F( n) = 1: He replaced the following condition.

(F20) infF = 1 or, also, by

(F200) there exists a sequence f ng1n=1 of positive real numbers such that limn!1F( n) = 1:

Very recently Piri et al. [21] utilized the following condition(F30)instead of (F3)in De…nition (6).

(F30)F is continuous on(0;1):

Forp 1; F( ) = 1p satis…es (F1)and (F2)but it does not satisfy(F3) while it satis…es(F30):We denote by z the family of all functionsF :R+!R which satisfy conditions(F1); (F20)and(F30):

De…nition 12 Let (X; d) be a metric space and T be a self-mapping on X:

Also suppose that ; : X X ! [0;+1) be two functions. We say T is an GF-contraction if forx; y2X with (x; T x) (x; y)andd(T x; T y)>

0, we have

G d(x; T x); d(y; T y); d(x; T y); d(y; T x) +F d(T x; T y) F d(x; y) (2.1) whereG2 G andF 2 z.

First, we prove the main result of Hussain et al. [12] by replacing conditions (F2)and(F3)with(F20)and(F30):

Theorem 13 Let (X; d) be a complete metric space. Let T : X ! X be a self-mapping satisfying the following assertions:

(i) T is -admissible mapping with respect to ; (ii) T is - -GF-contraction;

(iii) there existsx02X such that (x0; T x0) (x0; T x0);

(iv) T is - -continuous.

Then T has a …xed point. Moreover, T has a unique …xed point when (x; y) (x; x)for allx; y2F ix(T).

Proof. Letx02X such that (x0; T x0) (x0; T x0). For suchx0, we de…ne the sequencefxngbyxn=Tnx0=T xn 1:Now since,T is -admissible map- ping with respect to then, (x0; x1) = (x0; T x0) (x0; T x0) = (x0; x1).

By continuing this process we have,

(xn 1; T xn 1) = (xn 1; xn) (xn 1; xn) (2.2) for alln 2N. If there exist n0 2N such that xn0 =xn0+1; then xn0 is …xed point of T and we have nothing to prove. Hence, we assume, xn 6= xn+1 or d(T xn 1; T xn)>0for alln2N. Since,T is - -GF-contraction, so we have

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G d(xn 1; T xn 1); d(xn; T xn); d(xn 1; T xn); d(xn; T xn 1) +F d(T xn 1; T xn) F d(xn 1; xn)

which implies,

G d(xn 1; xn); d(xn; xn+1); d(xn 1; xn+1);0 +F d(xn; xn+1) F d(xn 1; xn) : (2.3) Now since,d(xn 1; xn):d(xn; xn+1):d(xn 1; xn+1):0 = 0; so from the de…nition ofG2 G;there exists >0 such that

G d(xn 1; xn); d(xn; xn+1); d(xn 1; xn+1);0 = : (2.4) From (2.4), we deduce that

F d(xn; xn+1) F d(xn 1; xn) : (2.5) Therefore

F d(xn; xn+1) F d(xn 1; xn) F d(xn 2; xn 1) 2 : : : F(d(x0; x1)) n : (2.6)

SinceF2 z;so by taking limit asn! 1in (2.6) we have,

nlim!1F d(xn; xn+1) = 1 () lim

n!1d(xn; xn+1) = 0: (2.7) Now, we claim thatfxng1n=1is a Cauchy sequence. We suppose on the contrary that fxng1n=1 is not Cauchy then we assume there exists " >0 and sequences fp(n)g1n=1andfq(n)g1n=1of natural numbers such that forp(n)> q(n)> n;we have

d(xp(n); xq(n)) ": (2.8)

Then

d(xp(n) 1; xq(n))< "

for alln2N:So, by triangle inequality and (2.8), we have

" d(xp(n); xq(n)) d(xp(n); xp(n) 1) +d(xp(n) 1; xq(n)) d(xp(n); xp(n) 1) +":

By taking the limit and using inequality (2.7), we get

nlim!1d(xp(n); xq(n)) =": (2.9) Also, from (2.7) there exists a natural numbern02Nsuch that

d(xp(n); xp(n)+1)< "

4 andd(xq(n); xq(n)+1)< "

4 (2.10)

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for alln n0:Next, we claim that

d(T xp(n); T xq(n)) =d(xp(n)+1; xq(n)+1)>0 (2.11) for alln n0:We suppose on the contrary that there existsm n0 such that

d(xp(m)+1; xq(m)+1) = 0: (2.12) Then from (2.10),(2.11) and (2.12), we have

" d(xp(m); xq(m)) d(xp(m); xp(m)+1) +d(xp(m)+1; xq(m)) d(xp(m); xp(m)+1) +d(xp(m)+1; xq(m)+1) +d(xq(m)+1; xq(m))

< "

4 + 0 +"

4 ="

2

which is a contradiction , so (2.11) holds. Thus

G d(xp(n); T xp(n)); d(xq(n); T xq(n)); d(xp(n); T xq(n)); d(xq(n); T xp(n)) +F d(T xp(n); T xq(n)) F d(xp(n); xq(n))

which implies,

G d(xp(n); xp(n)+1); d(xq(n); xq(n)+1); d(xp(n); xq(n)+1); d(xq(n); xp(n)+1

+F d(xp(n)+1; xq(n)+1) F d(xp(n); xq(n)) : (2.13)

SinceGis continuous, so from(F30);(2:9)and(2:13), we get

+F(") F("): (2.14)

Which is a contradiction. Thusfxngis a Cauchy sequence. Completeness ofX ensures that there existx 2X such that,xn!x as n! 1:Now since,T is

- -continuous and (xn 1; xn) (xn 1; xn), so

xn+1=T xn!T x as n! 1: (2.15) That is,x =T x : ThusT has a …xed point. Let x; y2F ix(T)wherex6=y:

Then from

G d(x; T x); d(y; T y); d(x; T y); d(y; T x) +F d(T x; T y) F d(x; y) we get,

+F d(x; y) F d(x; y)

which is a contradiction. Hence,x=y:Therefore,T has a unique …xed point.

Combining Theorem 13 and Example (8) we deduce the following Corollary.

Corollary 14 Let (X; d) be a complete metric space and T : X ! X be a self-mapping satisfying the following assertions:

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(i) T is -admissible mapping with respect to ;

(ii) for x; y2X with (x; T x) (x; y)andd(T x; T y)>0 we have, +F d(T x; T y) F d(x; y)

where >0 andF 2 z.

(iii) there existsx02X such that (x0; T x0) (x0; T x0);

(iv) T is - -continuous function.

Then T has a …xed point. Moreover, T has a unique …xed point when (x; y) (x; y)for allx; y2F ix(T).

Theorem 15 Let (X; d) be a complete metric space. Let T : X ! X be a self-mapping satisfying the following assertions:

(i) T is a -admissible mapping with respect to ; (ii) T is - -GF-contraction;

(iii) there existsx02X such that (x0; T x0) (x0; T x0);

(iv) iffxngis a sequence inX such that (xn; xn+1) (xn; xn+1)withxn! xasn! 1, then either

(T xn; T2xn) (T xn; x)or (T2xn; T3xn) (T2xn; x) (2.16) holds for alln2N.

Then T has a …xed point. Moreover, T has a unique …xed point whenever (x; y) (x; x) for allx; y2F ix(T).

Proof. Letx02X such that (x0; T x0) (x0; T x0). As in proof of Theorem 13 we can conclude that

(xn; xn+1) (xn; xn+1)andxn!x as n! 1 where,xn+1=T xn. So, from (iv), either

(T xn; T2xn) (T xn; x )or (T2xn; T3xn) (T2xn; x ) holds for alln2N. This implies,

(xn+1; xn+2) (xn+1; x)or (xn+2; xn+3) (xn+2; x)

holds for alln2N. Equivalently, there exists a subsequencefxnkgoffxngsuch that

(xnk; T xnk) = (xnk; xnk+1) (xnk; x ) (2.17) and so from (2.17) we deduce that,

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G d(xnk; T xnk); d(x ; T x ); d(xnk; T x ); d(x ; T xnk) +F d(T xnk; T x ) F d(xnk; x ) which implies,

F d(T xnk; T x ) F d(xnk; x ) : (2.18) From (F1) we have,

d(xnk+1; T x )< d(xnk; x ):

By taking limit ask! 1in the above inequality we get, d(x ; T x ) = 0; i:e:, x =T x :Uniqueness follows similarly as in Theorem 13.

Combining Theorem 15 and Example (8) we deduce the following Corollary.

Corollary 16 Let (X; d) be a complete metric space. Let T : X ! X be a self-mapping satisfying the following assertions:

(i) T is a -admissible mapping with respect to ;

(ii) for x; y2X with (x; T x) (x; y)andd(T x; T y)>0 we have, +F d(T x; T y) F d(x; y)

where >0 andF 2 z.

(iii) there existsx02X such that (x0; T x0) (x0; T x0);

(iv) if fxng be a sequence in X such that (xn; xn+1) (xn; xn+1) with xn!xasn! 1, then either

(T xn; T2xn) (T xn; x)or (T2xn; T3xn) (T2xn; x) holds for alln2N.

Then T has a …xed point. Moreover, T has a unique …xed point when (x; y) (x; x) for allx; y2F ix(T).

If in Corollary 16 we take (x; y) = (x; y) = 1 for allx; y 2 X, then we deduce main result of Piri et al.[21] as corollary.

Corollary 17 (Theorem 2.1 of [21]) Let(X; d)be a complete metric space and T :X !X be a self-mapping. If forx; y2X withd(T x; T y)>0 we have,

+F d(T x; T y) F d(x; y) where >0 andF 2 z. Then T has a …xed point.

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Example 18 Let X = [0;+1). We endow X with usual metric. De…ne, T :X !X, ; :X X![0;1),G:R+4 !R+ andF :R+ !Rby,

T x= 8>

<

>: 1

4e x2; ifx2[0;1]

5x ifx2(1;1)

(x; y) = 8>

><

>>

: 1

4; ifx; y2[0;1]

1

12; otherwise

and (x; y) = 16, G(t1; t2; t3; t4) = where

> 0 and F(r) = ln(r2+r): Let, (x; y) (x; y), then x; y 2 [0;1]. On the other hand, T u 2 [0;1] for all u 2 [0;1]. Then, (T x; T y) (T x; T y):

That is, T is -admissible mapping with respect to . If fxng is a sequence in X such that (xn; xn+1) (xn; xn+1) with xn ! x as n ! 1. Then, T xn; T2xn; T3xn2[0;1]for all n2N. That is,

(T xn; T2xn) (T xn; x)and (T2xn; T3xn) (T2xn; x)

hold for alln2N. Clearly, (0; T0) (0; T0). Let, (x; y) (x; T x). Now, ifx =2[0;1]or y =2[0;1], then, 1

12 1

6, which is a contradiction, sox; y2[0;1]

and hence we obtain,

d(T x; T y)(d(T x; T y) + 1) = (1

4e jx yjjx+yj)(1

4e jx yjjx+yj+ 1) e (jx yj)(e (jx yj) + 1)

e (jx yj)((jx yj) + 1)

= e d(x; y)(d(x; y) + 1) which implies,

+F(d(T x; T y)) = + ln(d(T x; T y)2+d(T x; T y)) + ln(e (d(x; y)2+d(x; y))

= ln(d(x; y)2+d(x; y)) =F(d(x; y)):

Hence,T is - -GF-contraction mapping. Thus all conditions of Corollary 16 ( and Theorem 15) hold andT has a …xed point. Letx= 0, y= 2 and >0:

Then,

+F(d(T0; T2)) F(d(T0; T2)) = ln(102+ 10)>ln(22+ 2) =F(d(0;2)):

That is Theorem 2.1 of [21] can not be applied for this example.

Recall that a self-mapping T is said to have the propertyP ifF ix(Tn) = F(T)for everyn2N.

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Theorem 19 Let (X; d) be a complete metric space and T : X ! X be an -continuous self-mapping. Assume that there exists >0such that

+F d(T x; T2x) F d(x; T x) (2.19)

holds for allx2X withd(T x; T2x)>0whereF 2 z. IfT is an -admissible mapping and there exists x0 2 X such that, (x0; T x0) 1, then T has the propertyP.

Proof. Let x0 2 X such that (x0; T x0) 1. For such x0, we de…ne the sequencefxngbyxn=Tnx0=T xn 1:Now since,T is -admissible mapping, so (x1; x2) = (T x0; T x1) 1. By continuing this process, we have

(xn 1; xn) 1

for alln2N. If there existsn02Nsuch thatxn0 =xn0+1=T xn0, thenxn0 is

…xed point ofT and we have nothing to prove. Hence, we assume, xn 6=xn+1

ord(T xn 1; T2xn 1)>0 for alln2N[ f0g. From (2.19) we have, +F d(T xn 1; T2xn 1) F d(xn 1; T xn 1) which implies,

+F d(xn; xn+1) F d(xn 1; xn) and so,

F d(xn; xn+1) F d(xn 1; xn) : Therefore,

F d(xn; xn+1) F d(xn 1; xn) F d(xn 2; xn 1) 2 : : : F(d(x0; x1)) n :

By taking limit asn! 1in above inequality, we have,limn!1F d(xn; xn+1) = 1, and since,F 2 z we obtain,

nlim!1d(xn; xn+1) = 0: (2.20) Now, we claim thatfxng1n=1is a Cauchy sequence. We suppose on the contrary that fxng1n=1 is not Cauchy then we assume there exists" >0 and sequences fp(n)g1n=1andfq(n)g1n=1of natural numbers such that forp(n)> q(n)> n;we have

d(xp(n); T xq(n) 1) =d(xp(n); xq(n)) ": (2.21) Then

d(xp(n) 1; T xq(n) 1)< " (2.22) for alln2N:So, by triangle inequality and (2.21), we have

" d(xp(n); T xq(n) 1) d(xp(n); xp(n) 1) +d(xp(n) 1; T xq(n) 1) d(xp(n); xp(n) 1) +":

(11)

By taking the limit and using inequality (2.20), we get

nlim!1d(xp(n); T xq(n) 1) =": (2.23) On the other hand, from (2.20) there exists a natural numbern02Nsuch that

d(xp(n); xp(n)+1)< "

4 andd(xq(n); xq(n)+1)< "

4 (2.24)

for alln n0:Next, we claim that

d(T xp(n); T2xq(n) 1) =d(xp(n)+1; T xq(n))>0 (2.25) for alln n0:We suppose on the contrary that there existsm n0 such that

d(T xp(m); T2xq(m) 1) =d(xp(m)+1; T xq(m)) = 0: (2.26) Then from (2.24),(2.25) and (2.26), we have

" d(xp(m); T xq(m) 1) d(xp(m); xp(m)+1) +d(xp(m)+1; T xq(m) 1) d(xp(m); xp(m)+1) +d(xp(m)+1; xq(m)+1) +d(xq(m)+1; T xq(m) 1)

= d(xp(m); xp(m)+1) +d(xp(m)+1; T xq(m)) +d(xq(m)+1; xq(m))

< "

4 + 0 +"

4 ="

2 which is a contradiction. Thus

d(T xp(n); T2xq(n) 1) =d(xp(n)+1; T xq(n))>0 (2.27) +F d(T xp(n); T2xq(n) 1) F d(xp(n); T xq(n) 1) (2.28) established. Which further implies that

+F d(xp(n)+1; xq(n)+1) F d(xp(n); xq(n)) : From(F30);(2:23)and (2:28), we get

+F(") F("): (2.29)

Which is a contradiction. Thus we proved that fxng is a Cauchy sequence.

Completeness ofX ensures that there existsx 2X such that,xn !x asn! 1:Now since,T is -continuous and (xn 1; xn) 1then,xn+1=T xn!T x asn! 1:That is,x =T x : ThusT has a …xed point andF(Tn) =F(T)for n= 1. Let n >1: Assume contrarily thatw 2F(Tn) andw =2 F(T). Then, d(w; T w)>0:Now we have,

F(d(w; T w)) = F(d(T(Tn 1w)); T2(Tn 1w))) F(d(Tn 1w); Tnw))

F(d(Tn 2w); Tn 1w)) 2 d(w; T w) n :

(12)

By taking limit asn! 1in the above inequality we have,F(d(w; T w)) = 1. Hence, by (F20) we get, d(w; T w) = 0 which is a contradictions. Therefore, F(Tn) =F(T)for alln2N.

Let(X; d; )be a partially ordered metric space. Recall thatT :X !X is nondecreasing if8x ; y2X; x y)T(x) T(y)and orderedGF-contraction if forx; y2X withx yandd(T x; T y)>0, we have

G d(x; T x); d(y; T y); d(x; T y); d(y; T x) +F d(T x; T y) F d(x; y) where G 2 G and F 2 z. Fixed point theorems for monotone operators in ordered metric spaces are widely investigated and have found various ap- plications in di¤erential and integral equations (see [2, 4, 9, 10, 13, 14] and references therein). From Theorems 13-19, we derive following new results in partially ordered metric spaces.

Theorem 20 Let (X; d; ) be a complete partially ordered metric space. As- sume that the following assertions hold true:

(i) T is nondecreasing and ordered GF-contraction;

(ii) there existsx02X such thatx0 T x0; (iii) either for a givenx2X and sequence fxng

xn !xasn! 1andxn xn+1for alln2Nwe haveT xn!T x or if fxng is a sequence such that xn xn+1 with xn ! x as n ! 1, then either

T xn x;orT2xn x holds for alln2N.

ThenT has a …xed point.

Theorem 21 Let (X; d; ) be a complete partially ordered metric space. As- sume that the following assertions hold true:

(i) T is nondecreasing and satis…es (2.19) for allx2X withd(T x; T2x)>0 whereF 2 z and >0;

(ii) there existsx02X such thatx0 T x0; (iii) for a givenx2X and sequencefxng

xn !xasn! 1andxn xn+1for alln2Nwe haveT xn!T x:

ThenT has a propertyP.

As an application of our results proved above, we deduce certain Suzuki- Wardowski type …xed point theorems.

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Theorem 22 Let(X; d)be a complete metric space andT be a continuous self- mapping on X. If for x; y2X with 12d(x; T x) d(x; y)andd(T x; T y)>0 we have,

G d(x; T x); d(y; T y); d(x; T y); d(y; T x) +F d(T x; T y) F d(x; y) (2.30) whereG2 G andF 2 z. Then T has a unique …xed point.

Proof. De…ne, ; :X X![0;1)by

(x; y) =d(x; y)and (x; y) =1 2d(x; y)

for all x; y 2 X. Now, since 12d(x; y) d(x; y) for all x; y 2 X, so (x; y) (x; y) for all x; y 2 X. That is, conditions (i) and (iii) of Theorem 13 hold true. Since T is continuous, so T is - -continuous. Let, (x; T x) (x; y) with d(T x; T y) >0: Equivalently, if 12d(x; T x) d(x; y) with d(T x; T y) > 0;

then, from (2.30) we have,

G d(x; T x); d(y; T y); d(x; T y); d(y; T x) +F d(T x; T y) F d(x; y) : That is, T is - -GF-contraction mapping. Hence, all conditions of Theorem 13 hold andT has a unique …xed point.

Combining above Corollary and Example (8) we deduce Theorem 2.2 of Piri et al. [21] as Corollary.

Corollary 23 (Theorem 2.2 [21]) Let(X; d)be a complete metric space andT be a continuous self-mapping on X. If for x; y 2 X with 12d(x; T x) d(x; y) andd(T x; T y)>0we have

+F d(T x; T y) F d(x; y) where >0 andF 2 z. Then T has a unique …xed point.

Corollary 24 Let (X; d) be a complete metric space and T be a continuous self-mapping on X. If for x; y2X with d(x; T x) d(x; y)andd(T x; T y)>0 we have,

eLminfd(x; T x); d(y; T y); d(x; T y); d(y; T x)g+F d(T x; T y) F d(x; y) where >0,L 0 andF 2 z. ThenT has a unique …xed point.

Theorem 25 Let (X; d) be a complete metric space and T be a self-mapping onX. Assume that there exists >0such that

1

2(1 + )d(x; T x) d(x; y) implies +F d(T x; T y) F d(x; y) (2.31) for x; y 2X with d(T x; T y) > 0 where F 2 z. Then T has a unique …xed point.

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Proof. De…ne, ; :X X![0;1)by

(x; y) =d(x; y)and (x; y) = 1

2(1 + )d(x; y)

for allx; y2X where >0. Now, since,2(1+ )1 d(x; y) d(x; y)for allx; y2X, so (x; y) (x; y)for allx; y2X. That is, conditions (i) and (iii) of Theorem 15 hold true. Let,fxng be a sequence with xn !x as n! 1: Assume that d(T xn; T2xn) = 0for somen. Then T xn=T2xn. That is T xn is a …xed point of T and we have nothing to prove. Hence we assume, T xn 6= T2xn for all n2 N. Since, 2(1+ )1 d(T xn; T2xn) d(T xn; T2xn)for all n 2N. Then from (2.31) we get,

F d(T2xn; T3xn) +F d(T2xn; T3xn) F d(T xn; T2xn) and so from (F1) we get,

d(T2xn; T3xn) d(T xn; T2xn): (2.32) Assume there existsn02Nsuch that,

(T xn0; T2xn0)> (T xn0; x)and (T2xn0; T3xn0)> (T2xn0; x) then,

1

2(1 + )d(T xn0; T2xn0)> d(T xn0; x)and 1

2(1 + )d(T2xn0; T3xn0)> d(T2xn0; x) so by (2.32) we have,

d(T xn0; T2xn0) d(T xn0; x) +d(T2xn0; x)

< 1

2(1 + )d(T xn0; T2xn0) + 1

2(1 + )d(T2xn0; T3xn0) 1

2(1 + )d(T xn0; T2xn0) + 1

2(1 + )d(T xn0; T2xn0)

= 2

2(1 + )d(T xn0; T2xn0) d(T xn0; T2xn0) which is a contradiction. Hence, either

(T xn; T2xn) (T xn; x)or (T2xn; T3xn) (T2xn; x)

holds for alln2N. That is condition (iv) of Theorem 15 holds. Let, (x; T x) (x; y). So, 2(1+ )1 d(x; T x) d(x; y). Then from (2.31) we get, +F d(T x; T y) F d(x; y) . Hence, all conditions of Theorem 15 hold and T has a unique …xed point.

Example 26 Consider the sequence

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S1= 1 2

S2= 1 2 + 3 4

S3= 1 2 + 3 4 + 5 6

Sn = 1 2 + 3 4 +: : :+ (2n 1)(2n) = n(n+1)(4n3 1):

Let X = fSn :n2Ng and d(x; y) = jx yj: Then (X; d) is a complete metric space. De…ne the mappingT :X!X by,

T(S1) =S1; T(Sn) =Sn 1; for alln 2:

Let us consider the mappingF(t) = t1+t, we obtain thatT isF-contraction, with = 12:

To see this, let us consider the following calculations. First observe that 1

2(1 + 12)d(Sn; T(Sn))< d(Sn; Sm) ,[(1 =n < m)_(1 =m < n)_(1< n < m)]:

For1 =n < m;we have

jT(Sm) T(S1)j=jSm 1 S1j= 3 4 + 5 6 +: : :+ (2m 3)(2m 2) (2.33) and

d(Sm; S1) =jSm S1j= 3 4 + 5 6 +: : :+ (2m 1)(2m): (2.34) Sincem >1;so we have

1

3 4 +: : :+ (2m 3)(2m 2) < 1

3 4 +: : :+ (2m 1)(2m): (2.35) From (2.35), we have

12 1

3 4 + 5 6 +: : :+ (2m 3)(2m 2) + 3 4 + +5 6 +: : :+ (2m 3)(2m 2)

< 12 1

3 4 + 5 6 +: : :+ (2m 1)(2m)+ [3 4 + +5 6 +: : :+ (2m 3)(2m 2)]

1

3 4 + 5 6 +: : :+ (2m 1)(2m)+ [3 4 + +5 6 +: : :+ (2m 3)(2m 2)] + (2m 1)(2m):

Thus from (2.33) and (2.34), we get

12 1

jT(Sm); T(S1)j+jT(Sm); T(S1)j< 1

jSm S1j +jSm S1j: (2.36) For everym; n2N withm > n >1;we have

jT(Sm) T(Sn)j= (2n 1)(2n)+(2n+1)(2n+2)+:::+(2m 3)(2m 2) (2.37) and

jSm Snj= (2n+ 1)(2n+ 2) + (2n+ 3)(2n+ 4) +:::+ (2m 1)(2m): (2.38)

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Sincem > n >1;we have

(2m 1)(2m) (2n+2)(2n+1)>(2n+2)(2n+2) = 2n(2n+2)+2(2n+2) 2n(2n+2)+12:

We know that 1

(2n 1)(2n) +:::+ (2m 3)(2m 2) < 1

(2n+ 1)(2n+ 2) +:::+ (2m 1)(2m): (2.39) From (2.39), we get

12 1

(2n 1)(2n) + (2n+ 1)(2n+ 2) +:::+ (2m 3)(2m 2) +(2n 1)(2n) + (2n+ 1)(2n+ 2) +:::+ (2m 3)(2m 2)

< 12 1

(2n+ 1)(2n+ 2) + (2n+ 3)(2n+ 4) +:::+ (2m 1)(2m) +(2n 1)(2n) + (2n+ 1)(2n+ 2) +:::+ (2m 3)(2m 2)

< 1

(2n+ 1)(2n+ 2) + (2n+ 3)(2n+ 4) +:::+ (2m 1)(2m) +(2n 1)(2n) + (2n+ 1)(2n+ 2) +:::+ (2m 3)(2m 2) +(2m 1)(2m)

= 1

(2n+ 1)(2n+ 2) + (2n+ 3)(2n+ 4) +:::+ (2m 1)(2m) +(2n 1)(2n) + (2n+ 1)(2n+ 2) +:::+ (2m 1)(2m) So from (2.37) and (2.38), we get

12 1

jT(Sm) T(S1)j+jT(Sm) T(S1)j< 1

jSm S1j+jSm S1j: Hence all the conditions of Theorem (25) are satis…ed andS1 is a unique …xed point ofT.

2 Applications to orbitally continuous mappings

Theorem 27 Let (X; d) be a complete metric space and T : X ! X be a self-mapping satisfying the following assertions:

(i) forx; y2O(w)with d(T x; T y)>0we have,

G d(x; T x); d(y; T y); d(x; T y); d(y; T x) +F d(T x; T y) F d(x; y) whereG2 G andF 2 z;

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(ii) T is an orbitally continuous function.

Then T has a …xed point. Moreover, T has a unique …xed point when F ix(T) O(w).

Proof. De…ne, ; :X X![0;+1)by (x; y) = 3; ifx; y2O(w)

0; otherwise and (x; y) = 1

where O(w) is an orbit of a point w 2 X. From Remark 5 we know that T is an - -continuous mapping. Let, (x; y) (x; y), then x; y 2 O(w).

So T x; T y 2 O(w). That is, (T x; T y) (T x; T y). Therefore, T is an - admissible mapping with respect to : Since w; T w 2O(w), then (w; T w)

(w; T w). Let, (x; y) (x; T x) andd(T x; T y)>0. Then,x; y2O(w)and d(T x; T y)>0. Therefore from (i) we have,

G d(x; T x); d(y; T y); d(x; T y); d(y; T x) +F d(T x; T y) F d(x; y) which implies, T is - -GF-contraction mapping. Hence, all conditions of Theorem 13 hold true and T has a …xed point. If F ix(T) O(w), then, (x; y) (x; y)for allx; y2F ix(T)and so from Theorem 13T has a unique

…xed point.

Corollary 28 Let (X; d) be a complete metric space and T : X ! X be a self-mapping satisfying the following assertions:

(i) forx; y2O(w)with d(T x; T y)>0we have, +F d(T x; T y) F d(x; y) where >0 andF 2 z;

(ii) T is orbitally continuous.

Then T has a …xed point. Moreover, T has a unique …xed point when F ix(T) O(w).

Corollary 29 Let (X; d) be a complete metric space and T : X ! X be a self-mapping satisfying the following assertions:

(i) forx; y2O(w)with d(T x; T y)>0we have,

eLminfd(x; T x); d(y; T y); d(x; T y); d(y; T x)g+F d(T x; T y) F d(x; y) where >0,L 0 andF 2 z;

(ii) T is orbitally continuous.

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Then T has a …xed point. Moreover, T has a unique …xed point when F ix(T) O(w).

In our next result, we prove improved version of Theorem 4 of [3].

Theorem 30 Let (X; d) be a complete metric space and T : X ! X be a self-mapping satisfying the following assertions:

(i) forx2X withd(T x; T2x)>0 we have,

+F d(T x; T2x) F d(x; T x) where >0 andF 2 z;

(ii) T is an orbitally continuous function.

ThenT has the propertyP.

Proof. De…ne, :X X ![0;+1)by

(x; y) = 1; ifx2O(w) 0; otherwise

wherew2X:Let, (x; y) 1, thenx; y2O(w). So T x; T y 2O(w). That is, (T x; T y) 1. Therefore,T is -admissible mapping. Sincew; T w2O(w), so (w; T w) 1. By Remark 5 we conclude thatT is -continuous mapping. If, x2X withd(T x; T2x)>0, then, from (i) we have,

+F d(T x; T2x) F d(x; T x) :

Thus all conditions of Theorem 19 hold true andT has the property P. We can easily deduce following results involving integral inequalities.

Theorem 31 Let (X; d) be a complete metric space and T be a continuous self-mapping onX. If forx; y2X with

Z d(x; T x)

0

(t)dt

Z d(x; y)

0

(t)dtand

Z d(T x; T y)

0

(t)dt >0 we have,

G Rd(x; T x)

0 (t)dt;Rd(y; T y)

0 (t)dt;Rd(x; T y)

0 (t)dt;Rd(y; T x)

0 (t)dt

+F Rd(T x; T y)

0 (t)dt F Rd(x; y)

0 (t)dt

where G 2 G, F 2 z and : [0;1) ! [0;1) is a Lebesgue-integrable mapping satisfyingR"

0 (t)dt >0 for" >0: ThenT has a unique …xed point.

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Theorem 32 Let (X; d) be a complete metric space and T be a self-mapping onX. Assume that there exists >0such that

1 2(1+ )

Rd(x; T x)

0 (t)dt Rd(x; y)

0 (t)dt)

+F Rd(T x; T y)

0 (t)dt F Rd(x; y)

0 (t)dt

forx; y2X withRd(T x; T y)

0 (t)dt >0where F2 z and : [0;1)![0;1) is a Lebesgue-integrable mapping satisfyingR"

0 (t)dt >0for" >0:ThenT has a unique …xed point.

Theorem 33 Let (X; d) be a complete metric space and T : X ! X be a self-mapping satisfying the following assertions:

(i) forx; y2O(w)with Rd(T x; T y)

0 (t)dt >0we have, G Rd(x; T x)

0 (t)dt;Rd(y; T y)

0 (t)dt;Rd(x; T y)

0 (t)dt;Rd(y; T x)

0 (t)dt

+F Rd(T x; T y)

0 (t)dt F Rd(x; y)

0 (t)dt

where G2 G,F 2 z and : [0;1)![0;1)is a Lebesgue-integrable mapping satisfying R"

0 (t)dt >0 for" >0:

(ii) T is an orbitally continuous function;

Then T has a …xed point. Moreover, T has a unique …xed point when F ix(T) O(w).

Theorem 34 Let (X; d) be a complete metric space and T : X ! X be a self-mapping satisfying the following assertions:

(i) forx2X withRd(T x; T2x)

0 (t)dt >0we have, +F Rd(T x; T2x)

0 (t)dt F Rd(x; T x)

0 (t)dt

where >0 andF 2 z and : [0;1)![0;1)is a Lebesgue-integrable mapping satisfying R"

0 (t)dt >0 for" >0:

(ii) T is an orbitally continuous function.

ThenT has the propertyP. Con‡ict of Interests

The authors declare that they have no competing interests.

Authors’Contribution

All authors contributed equally and signi…cantly in writing this paper. All authors read and approved the …nal paper.

Acknowledgement This article was funded by the Deanship of Scienti…c Research (DSR), King Abdulaziz University, Jeddah. Therefore, …rst author acknowledges with thanks DSR, KAU for …nancial support.

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