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Well-Posedness And Data Dependence Of Strict Fixed Point For -Hardy Roger Type Contraction And Applications

Deepak Kumar

y

, Anita Tomar

z

, Ritu Sharma

x

Received 15 February 2019

Abstract

We …rst familiarise with -Hardy Roger type contraction in the frame work of metric space. Then, well-posedness, data dependence, existence and uniqueness results of strict …xed point for -Hardy Roger type contraction are presented. The obtained results generalize the existing results in the literature.

Applications to an integral inclusion equation and Fractals concludes the paper.

1 Introduction

Banach contraction principle [1] has been generalised in numerous directions and one such generalisation is due to Nadler [14], who generalised it considering set-valued contraction. There after many results are established for set-valued mappings (see for instance [3]-[8], [16]-[21]). In this paper, considering the fact that Hardy-Rogers type operator is a ´Ciri´c type operator (however the reverse need not be true), we introduce - Hardy Roger type contraction and establish strict …xed point for it using iterations of a delta distance which is not even a metric. Also, we present well-posedness and data dependence of strict …xed point problem and utilise it to solve an integral inclusion equation and in presenting a novel iterated function framework via

-Hardy-Roger type operator to obtain attractor of multifunction system.

Let(X; d)be a metric space,

P(X) =fY X :Y 6=;g; Pb(X) =fY 2P(X) :Y is boundedg; Pcl(X) =fY 2P(X) :Y is closedg and Pcp(X) =fY 2P(X) :Y is compactg:

De…ne a set-valued operator as T : X ! P(X) and T(Y) = [x2YT(x) for Y 2 P(X). Also, FT = fx 2 X : x 2 T(x)g is a set of …xed points and (SF)T = fx 2 X : Tx = fxgg is a set of strict …xed points of the set-valued operator T: Chifu and Petrusel [10] introduced the generalized functional as:

:P(X) P(X)!R+[ f1g;

(A; B) = supfd(a; b) :a2A; b2Bg:

2 Main Results

Firstly, we de…ne -Hardy Roger type contraction and establish strict …xed point making use of iterations of a delta distance which is not even a metric.

De…nition 1 If T :X !Pb(X)is a set-valued operator of a metric space(X; d) satisfying (T(x);T(y)) a1d(x; y) +a2 (x;T(x)) +a3 (y;T(y)) +a4 (x;T(y)) +a5 (y;T(x));

wherea0is2R+,P5

i=1ai<1,x; y2X, then T is called a -Hardy Roger type contraction.

Mathematics Sub ject Classi…cations: 47H10; 54H25; 46J10; 46J15.

yDepartment of Mathematics, Lovely Professional University, Phagwara, Punjab-144411, India

zDepartment of Mathematics, Government Degree College Thatyur, Tehri Garhwal, (Uttrakhand), India

xDepartment of Mathematics, V. S. K. C. Government P. G. College Dakpathar Dehradun (Uttrakhand), India

46

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Theorem 1 IfT is a -Hardy Roger type contraction of a complete metric space(X; d), then(SF)T =fu g: Proof. Letu02X. Then there existsu12 T(u0)and

(u0;T(u0)) qd(u0; u1); q >1 is arbitrary.

Now,

(u1;T(u1)) (T(u0);T(u1))

a1d(u0; u1) +a2 (u0;T(u0)) +a3 (u1;T(u1)) +a4 (u0; T(u1)) +a5 (u1;T(u0))

a1d(u0; u1) +a2qd(u0; u1) +a3 (u1;T(u1)) +a4fd(u0; u1) + (u1;T(u1))g+a5 (u1; u1) (a1+a2q+a4)d(u0; u1) + (a3+a4) (u1;T(u1)):

This implies

(1 a3 a4) (u1;T(u1)) (a1+a2q+a4)d(u0; u1);

or

(u1;T(u1)) a1+a2q+a4

1 (a3+a4) d(u0; u1):

Also,u12 T(u0); 9 u22 T(u1)and

(u1;T(u1)) qd(u1; u2):

Therefore,

(u2;T(u2)) (T(u1);T(u2))

a1d(u1; u2) +a2 (u1;T(u1)) +a3 (u2;T(u2)) +a4 (u1;T(u2)) +a5 (u2;T(u1))

a1d(u1; u2) +a2qd(u1; u2) +a3 (u2;T(u2)) +a4fd(u1; u2) + (u2;T(u2))g+a5 (u2; u2)

= (a1+a2q+a4)d(u1; u2) + (a3+a4) (u2;T(u2)):

This implies

(1 (a3+a4)) (u2;T(u2)) (a1+a4+a2q)d(u1; u2);

or

(u2;T(u2)) a1+a4+a2q

1 (a3+a4) d(u1; u2) a1+a4+a2q

1 (a3+a4) (u1;T(u1)) a1+a4+a2q

1 (a3+a4)

2

d(u0; u1):

Following the similar pattern, we construct a sequence(un)n2Nsatisfying the following properties:

(i) un2 T(un 1);

(ii) d(un; un+1) (un;T(un))

ha1+a4+a2q 1 a3 a4

in

d(u0; u1):

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Next, consider

d(un; un+p) d(un; un+1) +d(un+1; un+2) +:::+d(un+p 1; un+p) a1+a4+a2q

1 a3 a4 n

+ a1+a4+a2q 1 a3 a4

n+1

+:::+ a1+a4+a2q 1 a3 a4

n+p 1

d(u0; u1):

Assume that = a11+aa4+a2q

3 a4 . Then

d(un; un+p) n[1 + +:::+ p 1]d(u0; u1) = n

p 1

1 d(u0; u1):

Chooseq < 1 a1aa3 2a4

2 then <1:Lettingn! 1, we get d(un; un+p)!0;

i.e., (un)n2N is a Cauchy sequence. Since(X; d) is a complete metric space, there existsu 2X such that un!u as n! 1:Now, we shall demonstrate thatu 2(SF)T. Consider,

(u ;T(u )) d(u ; un) + (un;T(un)) + (T(un);T(u ))

d(u ; un) + (un;T(un)) +a1d(un; u ) +a2 (un;T(un)) +a3 (u ;T(u )) +a4 (un;T(u )) +a5 (u ;T(un)) d(u ; un) + (un;T(un)) +a1d(un; u ) +a2 (un;T(un)) +a3 (u ;T(u )) +a4(d(un; u ) + (u ;T(u )))

+a5(d(u ; un) + (un;T(un)))

= (1 +a1+a4+a5)d(u ; un) + (1 +a2+a5) (un;T(un)) +(a3+a4) (u ;T(u )):

This implies

(1 a3 a4) (u ;Tu ) (1 +a1+a4+a5)d(u ; un) + (1 +a2+a5) (un;T(un));

or

(u ;Tu ) 1 +a1+a4+a5 1 a3 a4

d(u ; un) + 1 +a2+a5 1 a3 a4

(un;T(un)):

Since (un;T(un)) nd(u0; u1);we see that (u ;Tu ) = 0. It implies thatT(u ) =fu g;i.e.,u 2(SF)T: For uniqueness, assume that there exists two distinct pointsu ; v 2(SF)T: So

d(u ; v ) = (T(u );T(v ))

a1d(u ; v ) +a2 (u ;T(u )) +a3 (v ;T(v )) +a4 (u ;T(v )) +a5 (v ;T(u ))

a1d(u ; v ) +a2 (u ;T(u )) +a3 (v ;T(v )) +a4(d(u ; v ) + (v ; T(v ))) +a5(d(v ; u ) + (u ;T(u )))

(a1+a4+a5)d(u ; v ) +a2 (u ;Tu ) +a3 (v ;Tv ) +a4 (v ;Tv ) + a5 (u ;Tu ):

This implies

(1 a1 a4 a5)d(u ; v ) a2 (u ;Tu ) +a3 (v ;Tv ) +a4 (v ;Tv ) +a5 (u ;Tu );

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or

(1 a1 a4 a5)d(u ; v ) 0;

or1 a1 a4 a5 0ora1+a4+a5 1, a contradiction to the fact thata1+a2+a3+a4+a5<1:Hence u =v .

Example 1 Let X = [0;3]be a complete metric space and the mappingT :X !Pb(X)be de…ned as

T(x) =

( [0;1]; 0 x <2;

f2g; 2 x 3:

Taking a1= 103; a2= 0; a3 = 201; a4 = 14; a5= 207; a1+a2+a3+a4+a5 = 1920 <1: Now we have following cases:

Case I: Whenx; y2[0;2);

(T(x);T(y)) = 1 a1:2 +a2:2 +a3:2 +a4:2 +a5 2:19 20: Case II: Whenx2[0;2) andy2[2;3];

(T(x);T(y)) = 2 a1:3 +a2:2 +a3:1 +a4:2 +a5:3 45 20: Case III: Whenx2[2;3]andy2[0;2);

(T(x);T(y)) = 2 a1:2 +a2:1 +a3:2 +a4:3 +a5:2 43 20: Case IV: Whenx; y2[2;3],

(T(x);T(y)) = 0 a1+a2+a3+a4+a5 19 20:

Subsequently, all the hypotheses of Theorem 1 are veri…ed and x= 2is the only strict …xed point of a discontinuous set-valued operatorT.

Next, we try to establish su¢ cient conditions for the well-posedness of a strict …xed point problem for the set-valued operator.

Theorem 2 IfT is a -Hardy Roger type contraction of a complete metric space(X; d), then the strict …xed point is well-posed forT with respect to d:

Proof. Using Theorem1,(SF)T =fu g:Supposeun2X; n2N satisfying

d(un;T(un))!0 as n! 1: We next prove thatun!u as n! 1. Now,

d(un; u ) d(un;T(un)) + d(T(un);T(u ))

d(un;T(un)) +a1d(un; u ) +a2 d(un;T(un)) +a3 d(u ;T(u )) +a4

d(un;T(u )) +a5 d(u ;T(un))

d(un;T(un)) +a1d(un; u ) +a2 d(un;T(un)) +a3 d(u ;T(u )) +a4 (d(un; u ) + d(u ;T(u )) +a5(d(u ; un) + d(un;T(un)):

This implies

(1 a1 a4 a5)d(un; u ) (1 +a2+a5) d(un;T(un));

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or

d(un; u ) 1 +a2+a5

1 a1 a4 a5 d(un;T(un))!0 as n! 1; i.e.,un!u asn! 1:

Now, we establish a data dependence result.

Theorem 3 IfT1is a -Hardy Roger type contraction of a complete metric space(X; d)andT2:X !Pb(X) is a set-valued operator such that

(i) (SF)T26=;;

(ii) 9 >0 satisfying (T1(x);T2(x)) ,x2X;

then

(u1;(SF)T2) 1 +a3+a4

1 a1 a4 a5

:

Proof. Letu22(SF)(T2). So, (u2;T2(u2)) = 0:Now, d(u1; u2) = (T1(u1);T2(u2))

(T1(u1);T1(u2)) + (T1(u2);T2(u2))

a1d(u1; u2) +a2 (u1;T1(u1)) +a3 (u2;T1(u2)) +a4 (u1;T1(u2)) + a5 (u2;T1(u1)) + (T1(u2);T2(u2))

a1d(u1; u2) +a3 (u2;T1(u2)) +a4(d(u1; u2) + (u2;T1(u2))) + a5(d(u2; u1) + (u1;T1(u1))) + (T1(u2);T2(u2)):

This implies

(1 a1 a4 a5)d(u1; u2) a3 (u2;T1(u2)) +a4 (u2;T1(u2)) + (T1(u2);T2(u2)) a3 +a4 +

(1 +a3+a4) ; or

d(u1; u2) 1 +a3+a4

1 a1 a4 a5 : By taking supremum, it follows that

(u1;(SF)T2) 1 +a3+a4 1 a1 a4 a5

:

3 Applications

3.1 Application to Volterra Integral Inclusion.

Now, we utilise Theorem1 to solve the following Volterra integral inclusion

x(t)2q(t) + Z(t)

0

k(t; s)F(s; x(s))ds; (1)

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for t 2 J, : J ! J; k : J J ! R, q : J ! E are continuous, F : J E ! C(E); E is a real Banach space with normk:kE, C(E) is the class of all non-empty closed subsets ofE and J = [0;1]in R is a closed and bounded interval. Let C(J; E) is the space of all continuous E-valued functions onJ and kxk=supt2Jjx(t)jE:We use the following de…nitions.

De…nition 2 A set-valued function :J E!2E is Caratheodory if (i) t! (t; x)is measurable, x2E and

(ii) x! (t; x)is upper semi-continuous a. e. for t2J.

De…nition 3 A Caratheodory multifunctionF(t; X)isL1-Caratheodory if for every real numberr >0 9a function hr2L1(J; R) satisfyingkF(t; x)k hrt for almost every t2J and x2E andkxkE r. Denote kF(t; x(t))k= supfkukE :u2F(t; x(t))g; TF1=fv2B(J; E) :v(t)2F(t; x(t))a.e. t2Jg;where B(J; E) is the space of all E-valued Bochner-integrable functions onJ.

Lemma 1 ([13]) If diam(E)<1andF :J E!2E is L1- Caratheodory, thenTF16= ,x2E:

Lemma 2 ([13]) Let E be a Banach space, L : L1(J; E)! C(J; E) a continuous linear mapping and F a Caratheodory set-valued mapping such that TF1 6= . Then LoTF1 : C(J; E) ! 2C(J;E) is a closed graph operator onC(J; E) C(J; E).

Theorem 4 Suppose that the following set of hypotheses hold.

(i) The functionk(t; s)is non-negative onJ J andM = supt;s2J[k(t; s)];

(ii) the set-valued function F(t; x)is Caratheodory;

(iii) the set-valued functionF(t; x)is nondecreasing inxa.e. fort2J;

(iv) jF(s; x(s)) F(s; y(s))j M1( (x; y)); s2J; x2E;where

(x; y) =a1d(x; y) +a2 (x;T(x)) +a3 (y;T(y)) +a4 (x;T(y)) +a5 (y;T(x));

a0is2R+,P5

i=1ai <1;

(v) TF16= ,x2C(J; E):

Then the integral inclusion (1) has a solution inJ.

Proof. A continuous functionx:J!Eis a solution of the integral inclusion (1), if

x(t) =q(t) + Z(t)

0

k(t; s)v(s)ds;

wherev2B(J; E)such thatv(t)2F(t; x(t)):De…ne the set-valued mappingT : [0;1]!2X as T(x) =q(t) +

Z (t) 0

k(t; s)v(s)ds;

wherev 2 TF1(x)for everyt2[0;1]. Clearly T is well-de…ned, since, from (v),TF16= . For all t2[0;1]by (ii) and (iv), we get

jT(x) T(y)j= Z (t)

0

k(t; s)v1(s) k(t; s)v2(s) ds

E

; v1; v22TF1(x):

Taking supremum on both sides

(T(x);T(y)) MjF(s; x(s)) F(s; y(s))j (x; y);

i.e., the operator T verify the hypotheses of the Theorem1 on [0;1] and consequently, the given integral inclusion has a unique solution.

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3.2 Application to Fractals.

Fixed point theory performs a signi…cant role in fractals that are the self-similar sets. Iterated function systems de…ne fractals as attractors in discrete dynamical frameworks and can be applied to wavelet analy- sis, quantum physics, computer graphics and di¤erent applied sciences. This concept was …rst introduced by Hutchinson [11] and popularized by Barnsley [2] as a natural generalization of the celebrated Banach contraction principle. Now, we present novel iterated function framework utilizing the -Hardy Rogers type operators which covers a large range of operators. The operator

T :Pcp(X)!Pcp(X);T(Y) =[mi=1Ti(Y); Y 2P(X)

is the multifractal operator generated by T = (T1; :::;Tm); such that Ti : X ! Pcp(X): A …xed point V 2Pcp(X)ofT is an attractor of the iterated multifunction systemT. Next, we establish existence of an attractor.

Theorem 5 Let Ti : X ! Pcp(X); i 2 f1; :::; mg be a …nite family of set-valued operator of a complete metric space(X; d)such that

(Ti(x);Ti(y)) A1d(x; y) +A2 (x;Ti(x)) +A3 (y;Ti(y)) +A4 (x;Ti(y)) +A5 (y;Ti(x));

A0js2R+,P5

j=1Aj<1,x; y2X. Then the operatorT :Pcp(X)!Pcp(X)de…ned byT(B) =[mi=1Ti(B) for allB2Pcp(X)satis…es:

(T(B);T(C)) A1d(B; C) +A2 (B;T(B)) +A3 (C;T(C)) +A4 (B;T(C)) +A5 (C;T(B));

where B; C 2 Pcp(X) and has attractor A in (Pcp(X); (d)) such that A = T(A) = [mi=1Ti(A); A = limn!1Ton(B)andB2Pcp(X).

Proof. Let F 2 Pcp(X), then F is a non-empty and compact in X. Clearly T(F) is non-empty. Now we establish that T(F) is compact in X. If fyng T(F), then there is a sequence fung F satisfying yn = Tun(n = 1;2; :::). Compactness of F implies that there is a subsequence fxnkg fung such that fxnkg ! x 2 F. Since T is continuous, fynkg = Txnk ! Tx 2 T(F). Hence, T(F) is compact in X. De…nition of -Hardy Roger type contraction shows thatT satis…es

(T(B);T(C)) A11d(B; C) +A21 (B;T(B)) +A31 (C;T(C)) +A41 (B;T(C)) +A51 (C;T(B)); 8 B; C 2Pcp(X):

Now, we shall use the principle of mathematical induction. The statement is clearly true form= 1. Now, form= 2;

(T(B);T(C)) = (T1(B)[ T2(B);T1(C)[ T2(C))

maxfA12g (B; C) + maxfA22g (B;T1(B)) + maxfA32g (C;T1(C)) + maxfA42g (B;T1(C)) + maxfA52g (C;T1(B)):

Hence by induction inequality is true for alli21; :::; m;i.e.,

(T(B);T(C)) A1d(B; C) +A2 (B;Tn(B)) +A3 (C;Tn(C)) +A4 (B;Tn(C)) +A5 (C;Tn(B));

where A1= maxfA1jg; A2= maxfA2jg; :::; A5= maxfA5jg, i.e., the operatorT satisfy all the hypotheses of Theorem1and consequentlyT has a attractorA=T(A) =[mi=1Ti(A)andA= limTon(B); B2Pcp(X).

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4 Remarks

Remark 1 Choosing suitably the values of constants in Theorems 1,2 and3, similar results can be estab- lished for Kannan [12], Chatterjee [9], Reich [15] and Banach [1] type set-valued contractions.

Remark 2 In Theorem 5, we established the attractors of -Hardy Roger type set-valued iterated function systems, which generalizes the celebrated Hutchinson iterated function systems.

Acknowledgment. The authors are thankful to the learned referees for the very careful reading of the manuscript and valuable suggestions.

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