Fixed Point Theorems For Nonlinear Contraction In Controlled Metric Type Space
Sharafat Hussain
yzReceived 6 February 2020
Abstract
In this article, we introduced the notion of controlled comparison function and obtained some …xed point theorems involving such function in the setting of controlled metric type space. Our presented results extend, generalize and improve results on the topic in the literature. As a consequence, our …xed point results generalize the corresponding results in the Hicks and Rhoades ( Math. Japonica, 24 (1979), 327-330) and B. Samet at el. (Nonlinear Anal., 75 (2012), 2145-2165). Finally, we provide an illustratire application.
1 Introduction
Fixed point theory is an interesting subject, with a vast number of applications in various …elds of mathemat- ics. In metric spaces, this theory started with the Banach …xed-point theorem which guarantees the existence and uniqueness of …xed points of certain self-maps of complete metric spaces and provides a constructive method of …nding those …xed points. There are enormous amount of literature dealing with generalizations of this remarkable theorem (see [3, 6, 8, 10, 19] and references therein). Many researchers [3, 5, 8, 13]
generalized the structure of metric space by weakening triangle inequality and proved some …xed point for these spaces. Bourbaki [6] and Bakhtin [3] give the idea ofb-metric space. Czerwik [8] generalized Banach contraction theorem forb-metric space. Since then several articles appeared which dealt with …xed points theorems for single-valued and multi-valued mapping inb-metric space. By further weakening the triangular inequality Kamran et al. [13] introduced the concept of extended b-metric space. Recently, N. Mlaiki et al. [17] introduced the new extension of b-metric space, called controlled metric type spaces, by de…ning the triangular inequality as d(x; z) (x; y)d(x; y) + (y; z)d(y; z). They gave an example to show that controlled metric type spaces are not extended b-metric spaces in the sense of Kamran et al. They also proved the corresponding Banach …xed point theorem on controlled metric type spaces. In present work, we introduce the notion of controlled comparison function and prove some …xed point theorems involving such function in the setting of controlled metric type space. Our presented results extend, generalize and improve results on the topic in the literature. The obtained results for such setting become more useful in di¤erent avenues of applications.
2 Preliminaries
In the following, we will recall some basic de…nitions and known results needed in the sequel.
De…nition 1 ([8]) Let X be a non empty set ands 1 be a given real number. A function d:X X ! [0;1)is called b-metric if for allx; y; z2X it satis…es:
(i) d(x; y) = 0 if and only ifx=y;
Mathematics Sub ject Classi…cations: 47H10, 54H25.
yDepartment of Mathematics, Quaid i Azam University Islamabad, Pakistan
zDepartment of Mathematics, Women University of Azad Jammu and Kashmir Bagh, AJK
53
(ii) d(x; y) =d(y; x);
(iii) d(x; z) s[d(x; y) +d(y; z)]:
The pair(X; d)is called b-metric space.
The above de…nition shows that the class of metric spaces is contained in the class ofb-metric spaces. In fact, fors= 1ab-metric space is reduced into a standard metric space.
Example 1 ([13]) Let X := lp(R) with 0 < p < 1, where lp(R) := fx=fxng R:P1
n=1jxnjp<1g. Thend:X X ![0;1)is de…ned byd(x; y) = (P1
n=1jxn ynjp)1p is ab-metric onX withs= 21=p Example 2 ([13]) Let X =Lp[0;1]be the space of all functions x(t); t2[0;1]such that R1
0 jx(t)jpdt <1 with 0 < p < 1:Then (X; d) is a b-metric space with s = 21=p where d : X X ! [0;1) is de…ned by d(x; y) = R1
0 jx(t) y(t)jpdt
1 p.
De…nition 2 Let ':R+!R+. Consider the following properties:
(i)' t1 t2 implies'(t1) '(t2), for all t1; t22R+; (ii)' '(t)< t f or t >0;
(iii)' '(0) = 0;
(iv)' limn!1'n(t) = 0for all t 0;
(v)' P1
n=0'n(t)converges for allt >0.
A function 'satisfying(i)' and(iv)' is said to be a comparison function. A function'satisfying(i)'
and (v)' is known as (c)-comparison function. Any(c)-comparison function is a comparison function but converse may not be true. For example'(t) =1+tt ,t2R+is a comparison function but not a(c)-comparison function. On the other hand de…ne'(t) = 2t,0 t 1and'(t) =t 12,t >1, then'is a(c)-comparison function
Berinde [5] introduced the class of b-comparison functions and also obtained some estimations for the rate of convergence of the proposed iterative process to the …xed point.
De…nition 3 Let s 1be a …xed real number. A function':R+ !R+ is known asb-comparison function if it satis…es (i)' and the following holds;
(vi)' P1
n=0sn'n(t)converges for allt2R+.
The de…nition of b-comparison function reduces to comparison function when s = 1. Let (X; d) be a b-metric space with coe¢ cient s 1, then'(t) = t; t2R+ with0< <1s is ab-comparison function.
Recently, N. Mlaiki et al. [17] generalized b-metric space by introducing the controlled function on right side of the triangular inequality and called it a controlled metric type space. They also proved some …xed point theorems on these spaces.
De…nition 4 ([17]) Let X be a non empty set and :X X ![1;1):A functiond:X X ![0;1)is called a controlled metric type space if for allx; y; z2X it satis…es.
(i) d(x; y) = 0 if and only ifx=y;
(ii) d(x; y) =d(y; x);
(iii) d(x; z) (x; y)d(x; y) + (y; z)d(y; z):
Then the pair(X; d) is called a controlled metric type space.
Note that if for allx; y2X (x; y) =sfors 1, then triangular inequality inb-metric space is satis…ed;
however, it does not hold true in general. Thus the class of controlled metric type space is larger than that ofb-metric space. That is, everyb-metric space is a controlled metric type space, but converse may not be true. The following example shows the above remarks.
Example 3 ([17]) Let X=Nand de…ne d:X X !R+ as
d(x; y) = 8>
><
>>
:
0 if and only if x=y;
1
x if xis even andy is odd,
1
y if y is even andxis odd, 1 otherwise.
Then(X; d)is a controlled metric type space with :X X![1;1)de…ned by
(x; y) = 8<
:
x if xis even and y is odd, y if y is even and x is odd, 1 otherwise.
De…nition 5 ([17]) Let (X; d)be a controlled metric type space, x2X and let fxng be a sequence inX . Then
(i) fxng converges to x, if for every " > 0 there exists N = N(") 2 N such that d(xn; x) < "; for all n N:
(ii) fxng is a Cauchy sequence, if for every" >0 there exists N =N(")2N such that d(xm; xn)< "for all m; n N:
(iii) (X; d)is a complete extended b-metric space if every Cauchy sequence is convergent.
In the following we recollect some generalizations of Banach contraction principal in case of a controlled metric type space.
Theorem 1 ([17]) Let (X; d)be a complete controlled metric type space and supposeT :X!X satis…es:
d(T x; T y) kd(x; y)for allx; y2X wherek2(0;1)is such that for eachx02X
sup
m 1 jlim!1
(xj+1; xj+2)
(xj; xj+1) (xj+1; xm) 1 k wherexn=Tnx0. In addition, assume that, for every x2X, we have
n!lim+1 (xn; x)and lim
n!+1 (x; xn) exist and are …nite.
ThenT has a unique …xed point.
De…nition 6 Let T be a function from a subsetD of X toX. A setO(x0) =fx0; T x0; T2x0; g is called an orbit of x0 2D if O(x0) D for some x02D. A function G: D!R is said to be T-orbitally lower semi continuous ats2D iffxng O(x0)andxn!s implisesG(s) limn!1infG(xn).
Corollary 1 ([17]) Let (X; d)be a complete controlled metric type space and supposeT :X !X satis…es:
d(T x; T2x) kd (x; T x)for allx2 O(x0) wherek2(0;1)is such that for eachx02X
sup
m 1 jlim!1
(xj+1; xj+2)
(xj; xj+1) (xj+1; xm)< 1 k;
wherexn=Tnx0. Then,xn!z2X. Furthermorez is a …xed point ofT if and only ifG(x) =d(x; T x)is T-orbitally lower semi-continuous atz.
3 Main Results
In this section we introduce the notion of controlled comparison function and prove some …xed point theorems involving such function in the setting of controlled metric type space. We start this section with following de…nition.
De…nition 7 Let (X; d) be a controlled metric type space. A function ' : R+ ! R+ is called controlled comparison function if it satis…es:
(i)' t1 t2 implies'(t1) '(t2), for allt1; t22R+ (ii)' limn!+1'n(t) = 0for all t 0;
(iii)' there exists a mapping T :D X!X such that for some x02D; O(x0) D such that X1
n=1
'n(t) Yn i=1
(xi; xm)
!
(xn; xn+1)
converges for allt2R+ and for everym2N.
Here xn =Tnx0 forn= 1;2; : We say that'is a controlled comparison function forT atx0: Example 4 Let (X; d)be a controlled metric type space. De…ne ':R+!R+ by
'(t) =ktsuch that sup
m 1 jlim!1
(xj+1; xj+2)
(xj; xj+1) (xj+1; xm)< 1 k:
Then' is a controlled comparison function. Here for xo 2X , xn =Tnxo,n= 1;2;3; andT is a self map onX.
Now we proceed to establish our main result.
Theorem 2 Let (X; d)be a complete controlled metric type space. LetT :D X!X be a mapping such that O(x0) D. Suppose that
d(T x; T2x) '(d(x; T x))for each x2 O(x0); (1) where' is controlled comparison function forT atx0: ThenTnx0 !z2X (as n!+1). Furthermorez is a …xed point of T if and only ifG(x) =d(x; T x) isT-orbitally lower semi-continuous at z.
Proof. Letx02X be arbitrary, consider the sequenceTnx0=xn. By making full use of inequality (1), we yield
d(xn; xn+1) 'n(d(x0; x1)): (2)
By triangular inequality, form > nwe obtain d(xn; xm)
(xn; xn+1)d(xn; xn+1) + (xn+1; xm)(d(xn+1; xm))
(xn; xn+1)d(xn; xn+1) + (xn+1; xm) (xn+1; xn+2)d(xn+1; xn+2) + (xn+1; xm) (xn+2; xm)d(xn+2; xm)
(xn; xn+1)d(xn; xn+1) + (xn+1; xm) (xn+1; xn+2)d(xn+1; xn+2) + (xn+1; xm) (xn+2; xm) (xn+2; xn+3)d(xn+2; xn+3)
+ (xn+1; xm) (xn+2; xm) (xn+3; xm)d(xn+3; xm) ...
(xn; xn+1)d(xn; xn+1) +
mX2 i=n+1
0
@ Yi j=n+1
(xj; xm) 1
A (xi; xi+1)d(xi; xi+1)
mY1 k=n+1
(xk; xm)d(xm 1; xm):
Now by using (2) , we have
d(xn; xm) (xn; xn+1)'n(d(x0; x1)) +
mX2 i=n+1
0
@ Yi j=n+1
(xj; xm) 1
A (xi; xi+1)'i(d(x0; x1))
+
mY1 k=n+1
(xk; xm)'m 1(d(x0; x1))
(xn; xn+1)'n(d(x0; x1)) +
mX2 i=n+1
0
@ Yi j=n+1
(xj; xm) 1
A (xi; xi+1)'i(d(x0; x1))
+
mY1 k=n+1
(xk; xm)
!
(xm 1; xm)'m 1(d(x0; x1))
= (xn; xn+1)'n(d(x0; x1)) +
mX1 i=n+1
0
@ Yi j=n+1
(xj; xm) 1
A (xi; xi+1)'i(d(x0; x1))
(xn; xn+1)'n(d(x0; x1)) +
mX2 i=n+1
0
@ Yi j=0
(xj; xm) 1
A (xi; xi+1)'i(d(x0; x1)):
Since the series
X1 n=1
'n(d(x0; x1)) Yn i=1
(xi; xm)
!
(xn; xn+1)
converges for everym2N. Let S=
X1 n=1
'n(d(x0; x1)) Yn i=1
(xi; xm)
!
(xn; xn+1);
Sn= Xn i=0
'i(d(x0; x1)) 0
@ Yi j=0
(xj; xm) 1
A (xi; xi+1):
Hence form > nwe have
d(xn; xm) [ (xn; xn+1)'n(d(x0; x1)) + (Sm 1 Sn)]:
Lettingn! 1we conclude thatfxng is Cauchy sequence. SinceX is complete so there exists z2X such thatxn=Tnx0!z:Assume thatGis lower semi continuous atz2X, then
d(z; T z) lim inf
n!1d(Tnx0; Tn+1x0) lim inf
n!1'n(d(x0; x1)) = 0:
Conversely, assume thatT z=zand xn2 O(x)withxn!z. Then G(z) =d(z; T z) = 0 lim inf
n!1G(xn) =d(Tnx0; Tn+1x0):
4 Consequences
In this section,we give some consequences of our main results. We will show that many existing results in the literature can be deduced easily from our theorems. In the following we include an analogue of the theorem of Hicks and Rhoades [12], in the setting of controlled metric type space.
Theorem 3 Let (X; d)be a complete controlled metric type space and supposeT :X!X satis…es:
d(T x; T2x) kd (x; T x)for allx2 O(x0) wherek2(0;1)is such that for eachx02X
sup
m 1 jlim!1
(xj+1; xj+2)
(xj; xj+1) (xj+1; xm)< 1
k where xn=Tnx0:
Then, xn !z2X. Furthermore z is a …xed point of T if and only if G(x) =d(x:T x) isT-orbitally lower semi continuous atz.
Proof. De…ne ' : R+ ! R+ by '(t) = kt. Example 4 implies that ' is controlled comparison function.
Hence the result follows from Theorem2.
Remark 1 When (x; y) = (y; z) = 1for all x; y; z, then Theorem 3 reduces to main result of Hicks and Rhoades ([12, Theorem 1]).
De…nition 8 ([21]) Let :X X![0;1). A mapT :X!X is said to be -admissible if (x; y) 1 implies (T x; T y) 1
for allx; y2X.
Theorem 4 Let (X; d)be a complete controlled metric type space and supposeT :X!X satis…es
(x; y)d(T x; T y) '(d(x; y)for allx; y2X; (3) where'is controlled comparison function for T atx0. Assume that:
(i) T is -admissible .
(ii) (x0; T x0) 1 forx02X.
Then Tnx0 ! z 2 X. Furthermore T z = z if and only if G(x) = d(x; T x) is T-orbitally lower semi- continuous atz:
Proof. From (i) and (ii) we get
(Tnx0; Tn+1x0) 1; n= 1;2; : By using the contraction condition (4) we have
d(Tnx0; Tn+1x0) (Tnx0; Tn+1x0)d(Tnx0; Tn+1x0) '(d(Tn 1x0; Tnx0)):
The above inequality is equivalent to (1). Thus all the conditions of Theorem2 are satis…ed. Hence the result follows.
Corollary 2 ([21, Theorem 2.1]) Let(X; d)be a complete metric space and supposeT :X !X satis…es (x; y)d(T x; T y) '(d(x; y)for allx; y2X; (4) where'is controlled comparison function for T atx0. Assume that:
(i) T is -admissible .
(ii) (x0; T x0) 1 forx02X.
(iii) T is continuous ThenT has a …xed point.
Proof. The existence of …xed point follows immediately from Theorem4 by letting (x; y) = 1 = (y; z).
5 Application
In this section, we will show that how our proved results can be used to prove the exixtence of solution of nonlinear integral equation. Consider the following
x(t) =f(t) + Z t
0
K(t; u)g(u; x(u))du; t2[0;1]; (5)
wheref : [0;1]!R; g: [0;1] R!Rare two bounded continuous functions andS : [0;1] [0;1]![0;1] is a function such thatK(t; :)2L1([0;1]) for all t 2[0;1]. Before we state and prove the theorem for the existance of the solution to (5) we have the following simple lemma.
Lemma 1 Let (X; d)be a controlled metric type space,T :X !X be a self map,x02X and
sup
m 1 jlim!1
(xj+1; xj+2)
(xj; xj+1) (xj+1; xm)< 1 k
for k2(0;1). Suppose that is comparison function then '(t) =k is a controller comparison function forT atx0.
Theorem 5 Let T :X !X be the integral operator given by
T x(t) =f(t) + Z t
0
K(t; u)g(u; x(u))du:
Assume that the following conditions hold:
(i) forx2X and for everyu2[0;1], we have
0 g(u; x(u)) g(u; T x(u)) 1 4
q
ln(1 +jx(u) T x(u)j2);
(ii) for everyu2[0;1], we have
Z 1 0
K(t; u)du
1
<1:
ThenT has a …xed point.
Proof. LetX be the set of all continuous real valued functions de…ned on[0;1]. De…ned:X X ![0;1) by
d(x; y) =kx yk1= sup
t2[a;b]jx(t) y(t)j2;
then(X; d)is complete controlled metric type space with (x; y) = (y; z) = 2. Now from condition(ii), we have
T x(t) T2x(t)2 = Z t
0
K(t; u)[g(u; x(u)) g(u; T x(u))]du
2
Z t
0 jK(t; u)j2jg(u; x(u)) g(u; T x(u))j2du Z t
0 jK(t; u)j2ln(1 +jx(u) T x(u)j2)
16 du
ln(1 +kx T xk1)
16 :
Thus we obtain
T x T2x
1
ln(1 +kx T xk1)
16 :
Hence
d(T x; T2x) k'(d(x; T x)):
where k = 18 and '(r) = ln(1+r)2 . Since 'is comparison function so from Lemma 1, its follows thatk' is controlled comparison function. Moreover,
sup
m 1 jlim!1
(xj+1; xj+2)
(xj; xj+1) (xj+1; xm) = 2
for x0 2 X. Thus all the conditions of Theorem 2 are satis…ed hence T has …xed point i.e. the integral equation (5) has a solution.
Corollary 3 Let T :X!X be the integral operator given by
T x(t) =f(t) + Z t
0
1
( )(t u) 1g(u; x(u))du; t2[0;1]; 2(0;1);
where is the Euler gamma function given by ( ) =R1
0 t 1e tdt. Suppose that forx2X and for every u2[0;1], we have
0 g(u; x(u)) g(u; T x(u)) ( 1) 8
q
ln(1 +jx(u) T x(u)j2):
ThenT has a …xed point.
The above result is a special case of Theorem5for a fractional order integral equation.
Acknowledgment. The author would like to thank the referee for his/her suggestions leading to im- provements of our paper.
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