# On Interpolative Prešić-Type Set-Valued Contractions

Authors: Monairah Alansari; Ghada Ali Basendwah; Muhammad Usman Ali
Journal: Journal of Mathematics
 (2022)

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## Abstract

This study aims to present the notions of interpolative Prešić-type set-valued contractions for the set-valued operators defined on product spaces. With the help of these notions, we have studied the existence of fixed points for such set-valued operators. An application of the obtained results is also discussed with the help of graph theory.


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## Body

## 1. Introduction and Preliminaries

Banach [1] initiated the study of the existence of fixed points for self-maps defined on a metric space. This study was further strengthened by Kannan and Chatterjea through their fixed point results derived in [2, 3], respectively. Following this study, Nadler [4] proposed a result to ensure the existence of fixed points for set-valued maps. Prešić [5] extended Banach contraction principle to the maps defined on product spaces, that is, Q:Rk⟶R, for any fixed k∈ℕ. Afterwards, this result was extended by Ćirić and Prešić [6]. The results of Prešić [5] and Ćirić and Prešić [6] are presented below.Theorem 1.
(see [5]). Let Q:Rk⟶R, for any fixed k∈ℕ, be a map on complete metric space R,dR and satisfies(1)dRQr1,r2,…,rk,Qr2,r3,…,rk+1≤∑j=1kβjdRrj,rj+1,for each r1,r2,…,rk,rk+1∈R, where β1,β2,…,βk≥0 with ∑j=1kβj<1. Then, there exists a unique fixed point r∈R of Q, that is, r=Qr,r,…,r︸k.Theorem 2.
(see [6]). Let Q:Rk⟶R, for any fixed k∈ℕ, be a map on complete metric space R,dR and satisfies(2)dRQr1,r2,…,rk,Qr2,r3,…,rk+1≤ζmaxdRri,ri+1:i∈1,2,…,k,for each r1,r2,…,rk,rk+1∈R, where ζ∈0,1. Then, there exists a unique fixed point r∈R of Q, that is, r=Qr,r,…,r︸k.
Karapınar [7] presented interpolative Kannan contraction by following the Kannan contraction as follows.
A mapQ:R,dR⟶R,dR is called an interpolative Kannan contraction [7] if(3)dRQr,Ql≤ζdRr,QrϑdRl,Ql1−ϑ,for each r,l∈R with r≠Qr and l≠Ql, where ζ∈0,1 and ϑ∈0,1.
The above work of Karapınar [7] is adopted by several researchers; for example, the notions of interpolative Ćirić–Reich–Rus type contractions in Branciari metric spaces, and partial metric spaces are defined by Aydi et al. [8] and Karapınar et al. [9], the notions of interpolative type F-contractions are defined by Mohammadi et al. [10] and Alansari and Ali [11], the notions of interpolative Hardy-Rogers type contractions, and set-valued interpolative Hardy–Rogers type contractions are defined by Karapınar et al. [12] and Debnath and Sen [13], the notion of interpolative Suzuki-type contraction is discussed by Fulga and Yesilkaya [14], and the notion of interpolative proximal contraction is discussed by Altun and Tasdemir [15].
Gaba and Karapınar [16] redefined the notion of interpolative Kannan contraction through modifying exponential powers in a following way.A mapQ:R,dR⟶R,dR is called an ζ,ϑ1,ϑ2-interpolative Kannan contraction if(4)dRQr,Ql≤ζdRr,Qrϑ1dRl,Qlϑ2,for each r,l∈R with r≠Qr and l≠Ql, where ϑ1,ϑ2∈0,1 with ϑ1+ϑ2<1 and ζ∈0,1.Recently, Alansari and Ali [17] defined the notion of extended interpolative Prešić-type contraction map as follows.A mapQ:Rk⟶R, for any fixed k∈ℕ, on a metric space R,dR is called extended interpolative Prešić type contraction if, for each w1,w2,…,wk, p1,p2,…,pk∈R\FixQ, we have(5)dRQw1,w2,…,wk,Qp1,p2,…,pkminγw1,p1,γw2,p2,…,γwk,pk≤ζdRwk,Qw1,w2,…,wkϑ1dRpk,Qp1,p2,…,pkϑ2,where γ:R×R⟶ℝ\0 is a map, ϑ1,ϑ2∈0,1 with ϑ1+ϑ2=1, ζ∈0,1, and FixQ=r∈R:r=Qr,r,…,r.The purpose of this study is to extend and redefine the concepts of interpolative Prešić-type contractions by introducing interpolative Prešić-type set-valued contractions for set-valued maps. We will also present a few fixed-point results to study the existence of fixed points of such maps.The literature of metric fixed-point theory contains several interesting results that are the generalizations of Banach fixed-point theorem, for example, the study of common fixed-point results for two or more maps [18] and the study of the existence of fixed points for the maps defined on generalized metric spaces, such as b-metric space [19], partial metric space [20], dislocated quasi-metric [21], hypergraphical metric space [22], and soft metric space [23, 24].Before the next section, we recall the Pompeiu–Hausdorff distance. The Pompeiu–Hausdorff distance is a mapHR:CBR×CBR⟶0,∞ defined by(6)HRJ,K=maxsupj∈JdRj,K,supk∈KdRk,J,where dRk,J=infdRk,j:j∈J and CBR represents the collection of all nonvoid closed and bounded subsets of R,dR.


## 2. Main Results

We begin this section with the following definition.Definition 1.
A mapQ:R×R⟶CBR is said to be an interpolative Prešić type-I set-valued contraction if, for all w1,w2,p1,p2∈R\FixQ, the following inequality exhibits(7)HRQw1,w2,Qp1,p2minγw1,p1,γw2,p2≤ζmaxdRw1,p1,dRw2,p2ϑ1dRw2,Qw1,w2ϑ2dRp2,Qp1,p2ϑ3,where γ:R×R⟶ℝ\0 is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1, and FixQ=r∈R:r∈Qr,r.
With the help of below stated result, we will study the existence of fixed points for the above map.Theorem 3.
LetQ:R×R⟶CBR be an interpolative Prešić type-I set-valued contraction map on a complete metric space R,dR. Also, consider that(i)
Ifminγw1,p1,γw2,p2=1, then γz1,z2=1, for all z1∈Qw1,w2andz2∈Qp1,p2(ii)
There existw1,w2∈R with minγw1,w2,γw2,z1=1, for all z1∈Qw1,w2(iii)
For each sequencerm in R with γrm,rm+1=1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r=1,∀m≥m0Then, there exists an elementr of R with r∈Qr,r.Proof.
By hypothesis (ii), we get two points inR, say r0 and r1, with(8)minγr0,r1,γr1,z=1,∀z∈Qr0,r1.
Letr2∈Qr0,r1; then, by (7), we obtain(9)dRr2,Qr1,r2≤HRQr0,r1,Qr1,r2minγr0,r1,γr1,r2≤ζmaxdRr0,r1,dRr1,r2ϑ1dRr1,Qr0,r1ϑ2dRr2,Qr1,r2ϑ3.
That is,(10)dRr2,Qr1,r21−ϑ3≤ζmaxdRr0,r1,dRr1,r2ϑ1dRr1,Qr0,r1ϑ2.
Clearly,dRr1,Qr0,r1≤maxdRr0,r1,dRr1,r2; thus, by (10), we obtain(11)dRr2,Qr1,r21−ϑ3≤ζmaxdRr0,r1,dRr1,r2ϑ1+ϑ2.
As1−ϑ3=ϑ1+ϑ2, thus, by (11), we obtain(12)dRr2,Qr1,r2≤ζmaxdRr0,r1,dRr1,r2.
The factζ∈0,1 yields the existence of some r3∈Qr1,r2 satisfying the inequality dRr2,r3≤1/ζdRr2,Qr1,r2. Thus, by the last two inequalities, we obtain(13)dRr2,r3≤ζmaxdRr0,r1,dRr1,r2.
Asr2∈Qr0,r1, r3∈Qr1,r2, and minγr0,r1,γr1,r2=1, by hypothesis (i), we get γr2,r3=1. Thus, we say that minγr1,r2,γr2,r3=1. Again, by considering (7), we obtain(14)dRr3,Qr2,r3≤HRQr1,r2,Qr2,r3minγr1,r2,γr2,r3≤ζmaxdRr1,r2,dRr2,r3ϑ1dRr2,Qr1,r2ϑ2dRr3,Qr2,r3ϑ3.
That is,(15)dRr3,Qr2,r31−ϑ3≤ζmaxdRr1,r2,dRr2,r3ϑ1+ϑ2.
As1−ϑ3=ϑ1+ϑ2, thus, by (15), we obtain(16)dRr3,Qr2,r3≤ζmaxdRr1,r2,dRr2,r3.
Asζ∈0,1, then there is some r4∈Qr2,r3 such that dRr3,r4≤1/ζdRr3,Qr2,r3. Thus, we obtain(17)dRr3,r4≤ζmaxdRr1,r2,dRr2,r3.
Continuing in that way, we reach to a sequencerm with the facts rm+1∈Qrm−1,rm for all m∈ℕ and(18)minγrm−1,rm,γrm,rm+1=1,∀m∈ℕ,and(19)dRrm+1,rm+2≤ζmaxdRrm−1,rm,dRrm,rm+1,∀m∈ℕ.
For simplicity, we usedRm=dRrm,rm+1 for each m∈ℕ∪0. We will show with induction that dRm−1≤βmM for each m∈ℕ, where β=ζ1/4 and M=maxdR0/β,dR1/β2. Trivially, dR0≤βM and dR1≤β2M. Suppose that(20)dRk−3≤βk−2M,dRk−2≤βk−1M∀k≥3.
Then,(21)dRk−1≤β2maxdRk−3,dRk−2≤β2maxβk−2M,βk−1M=βkM,for eachk≥3.
Thus,dRm−1≤βmM, for each m∈ℕ. Now, by considering this fact and the triangle inequality, for each q,n∈ℕ with q>n, we obtain(22)dRrn,rq≤∑j=nq−1dRrj,rj+1=∑j=nq−1dRj≤∑j=nq−1βj+1M.
Hence, the convergence of∑j=1∞βj, as β∈0,1, and the above inequality yields that rm is a Cauchy sequence in R. Now, the completeness of R,dR yields the existence of a point r∗∈R such that rm⟶r∗. By hypothesis (iii), we get γrm,r∗=1, as γrm,rm+1=1,∀m∈ℕ and rm⟶r∗. Now, we claim r∗∈Qr∗,r∗. If it is wrong, then, by (7), for each m∈ℕ, we obtain(23)HRQrm,rm+1,Qr∗,r∗=HRQrm,rm+1,Qr∗,r∗minγrm,r∗,γrm+1,r∗≤ζmaxdRrm,r∗,dRrm+1,r∗ϑ1dRrm+1,Qrm,rm+1ϑ2dRr∗,Qr∗,r∗ϑ3.
That is,(24)dRrm+2,Qr∗,r∗≤HRQrm,rm+1,Qr∗,r∗≤ζmaxdRrm,r∗,dRrm+1,r∗ϑ1dRrm+1,Qrm,rm+1ϑ2dRr∗,Qr∗,r∗ϑ3.
By triangle inequality and (24), we obtain(25)dRr∗,Qr∗,r∗≤dRr∗,rm+2+dRrm+2,Qr∗,r∗≤dRr∗,rm+2+ζmaxdRrm,r∗,dRrm+1,r∗ϑ1×dRrm+1,Qrm,rm+1ϑ2dRr∗,Qr∗,r∗ϑ3.
Hence, by applying the limit asm⟶∞ in (25), we get dRr∗,Qr∗,r∗=0. This shows that the claim is true and r∗∈Qr∗,r∗.Example 1.
LetR denote the set of all real numbers with a usual metric dRr,l=r−l for each r,l∈R. Define maps Q:R×R⟶CBR and γ:R×R⟶ℝ\0 by(26)Qr,l=0,r+l2,ifr,l≥0,0,otherwise,γr,l=1,ifr,l≥0,1/4,otherwise.
The hypotheses of Theorem3 can be verified on the above defined maps. Hence, there exists an element r of R with r∈Qr,r.
We now present an interpolative Prešić type-II set-valued contraction map along with fixed-point result.Definition 2.
A mapQ:R×R⟶CBR is called an interpolative Prešić type-II set-valued contraction if, for each w1,w2,p1,p2∈R\FixQ with minγw1,p1,γw2,p2≥1, we obtain(27)HRQw1,w2,Qp1,p2≤ζmaxdRw1,p1,dRw2,p2ϑ1dRw2,Qw1,w2ϑ2dRp2,Qp1,p2ϑ3,where γ:R×R⟶ℝ is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1, and FixQ=r∈R:r∈Qr,r.Theorem 4.
LetQ:R×R⟶CBR be an interpolative Prešić type-II set-valued contraction map on a complete metric space R,dR. Also, consider that(i)
Ifminγw1,p1,γw2,p2≥1, then γz1,z2≥1, for all z1∈Qw1,w2andz2∈Qp1,p2(ii)
There existw1,w2∈R with minγw1,w2,γw2,z1≥1 for all z1∈Qw1,w2(iii)
For each sequencerm in R with γrm,rm+1≥1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r≥1,∀m≥m0Then, there exists an elementr of R with r∈Qr,r.Proof.
Hypothesis (ii) makes sure the existence of two points inR, say r0 and r1, that satisfies the following:(28)minγr0,r1,γr1,z≥1,∀z∈Qr0,r1.
By defining one value ofz∈Qr0,r1 as z=r2 in the above inequality, we reach to(29)minγr0,r1,γr1,r2≥1.
Thus, by (27), we obtain(30)dRr2,Qr1,r2≤HRQr0,r1,Qr1,r2≤ζmaxdRr0,r1,dRr1,r2ϑ1dRr1,Qr0,r1ϑ2dRr2,Qr1,r2ϑ3.
That is,(31)dRr2,Qr1,r21−ϑ3≤ζmaxdRr0,r1,dRr1,r2ϑ1dRr1,Qr0,r1ϑ2≤ζmaxdRr0,r1,dRr1,r2ϑ1+ϑ2.
Since1−ϑ3=ϑ1+ϑ2, thus, by (31), we obtain(32)dRr2,Qr1,r2≤ζmaxdRr0,r1,dRr1,r2.
From the above inequality and by the fact1/ζ>1, there exists some r3∈Qr1,r2 such that(33)dRr2,r3≤1ζdRr2,Qr1,r2≤ζmaxdRr0,r1,dRr1,r2.
Sinceminγr0,r1,γr1,r2≥1 and r2∈Qr0,r1, r3∈Qr1,r2, by hypothesis (i), we get γr2,r3≥1. By proceeding the proof on the above steps, we reach to a sequence rm of the form rm+1∈Qrm−1,rm for all m∈ℕ and(34)minγrm−1,rm,γrm,rm+1≥1,∀m∈ℕ,and(35)dRrm+1,rm+2≤ζmaxdRrm−1,rm,dRrm,rm+1,∀m∈ℕ.
By viewing the above inequality and the proof of the above theorem, we conclude thatrm is a Cauchy sequence in R, and there exists a point r∗∈R with rm⟶r∗. From hypothesis (iii), we have γrm,r∗≥1 for each m∈ℕ. This implies minγrm,r∗,γrm+1,r∗≥1,∀m∈ℕ. Suppose that r∗∉Qr∗,r∗. Then, by (27), for each m∈ℕ, we obtain(36)dRrm+2,Qr∗,r∗≤HRQrm,rm+1,Qr∗,r∗≤ζmaxdRrm,r∗,dRrm+1,r∗ϑ1dRrm+1,Qrm,rm+1ϑ2dRr∗,Qr∗,r∗ϑ3.
By triangle inequality and (36), we obtain(37)dRr∗,Qr∗,r∗≤dRr∗,rm+2+dRrm+2,Qr∗,r∗≤dRr∗,rm+2+ζmaxdRrm,r∗,dRrm+1,r∗ϑ1×dRrm+1,Qrm,rm+1ϑ2dRr∗,Qr∗,r∗ϑ3.
Thus, by taking the limitm⟶∞ in (37), we get dRr∗,Qr∗,r∗=0. This shows that the supposition is wrong and r∗∈Qr∗,r∗.


### 2.1. Results for Extended Interpolative Prešić Type Set-Valued Operators

This section presents the extensions of the above listed results. Theorems5 and 6 can be considered as an extended version of Theorems 3 and 4, respectively.Theorem 5.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be an extended interpolative Prešić type-I set-valued contraction map on a complete metric space R,dR, that is, for every w1,w2,…,wk, p1,p2,…,pk∈R\FixQ, we have(38)HRQw1,w2,…,wk,Qp1,p2,…,pkminγw1,p1,γw2,p2,…,γwk,pk≤ζmaxdRwi,pi:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRpk,Qp1,p2,…,pkϑ3,where γ:R×R⟶ℝ\0 is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1, and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
Ifminγw1,p1,γw2,p2,…,γwk,pk=1, then γz1,z2=1 for all z1∈Qw1,w2,…,wk, z2∈Qp1,p2,…,pk.(ii)
There arew1,w2,…,wk∈R satisfying(39)minγw1,w2,γw2,w3,…,γwk,z=1,∀z∈Qw1,w2,…,wk.(iii)
For each sequencerm in R with γrm,rm+1=1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r=1,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.Proof.
Hypothesis (ii) says that there are pointsr1,r2,…,rk in R satisfying the condition:(40)minγr1,r2,γr2,r3,…,γrk,z=1,∀z∈Qr1,r2,…,rk.
Thus, forrk+1∈Qr1,r2,…,rk, we obtain(41)minγr1,r2,γr2,r3,…,γrk,rk+1=1.
Then, by (38), we obtain(42)dRrk+1,Qr2,r3,…,rk+1≤HRQr1,r2,…,rk,Qr2,r3,…,rk+1minγr1,r2,γr2,r3,…,γrk,rk+1≤ζmaxdRri,ri+1:i∈1,2,…,kϑ1×dRrk,Qr1,r2,…,rkϑ2dRrk+1,Qr2,r3,…,rk+1ϑ3.
That is,(43)dRrk+1,Qr2,r3,…,rk+11−ϑ3≤ζmaxdRri,ri+1:i∈1,2,…,kϑ1dRrk,Qr1,r2,…,rkϑ2.
SincedRrk,Qr1,r2,…,rk≤maxdRri,ri+1:i∈1,2,…,k, thus, by (43), we obtain(44)dRrk+1,Qr2,r3,…,rk+11−ϑ3≤ζmaxdRri,ri+1:i∈1,2,…,kϑ1+ϑ2.
Since1−ϑ3=ϑ1+ϑ2, then (44) gives(45)dRrk+1,Qr2,r3,…,rk+1≤ζmaxdRri,ri+1:i∈1,2,…,k.
As1/ζ>1, thus, there exists rk+2∈Qr2,r3,…,rk+1 of a form(46)dRrk+1,rk+2≤1ζdRrk+1,Qr2,r3,…,rk+1≤ζmaxdRri,ri+1:i∈1,2,…,k.
Hypothesis (i) implies thatγrk+1,rk+2=1, since minγr1,r2,γr2,r3,…,γrk,rk+1=1 and rk+1∈Qr1,r2,…,rk, rk+2∈Qr2,r3,…,rk+1. By the repeated application of hypothesis (i) and (38), we reach to a sequence rm with the facts rm+k∈Qrm,rm+1,…,rm+k−1 for all m∈ℕ and(47)minγrm,rm+1,γrm+1,rm+2,…,γrm+k−1,rm+k=1,∀m∈ℕ,and(48)dRrm+k,rm+k+1≤ζmaxdRrm−1+i,rm+i:i∈1,2,…,k,∀m∈ℕ.
For simplicity, takedRm=dRrm,rm+1 for each m∈ℕ; from (48), we obtain(49)dRm+k≤ζmaxdRm−1+i:i∈1,2,…,k,∀m∈ℕ.
Now, we prove by induction thatdRm≤βmM for each m∈ℕ, where β=ζ1/2k and M=maxdR1/β,dR2/β2,…,dRk/βk. Trivially, dRi≤βiM, for each i∈1,2,…,k. Suppose that dRi≤βiM for each i∈m,m+1,…,m+k−1 for some given m, as induction hypothesis. Then, by (49), we obtain(50)dRm+k≤ζmaxdRm−1+i:i∈1,2,…,k≤ζmaxβm−1+iM:i∈1,2,…,k=βkβmM=βm+kM.
Hence, it is shown by induction thatdRm≤βmM, for each m∈ℕ. This fact along with triangle inequality yield that(51)dRrn,rq≤∑j=nq−1dRrj,rj+1=∑j=nq−1dRj≤∑j=nq−1βjM,for each q,n∈ℕ with q>n. Hence, the above inequality and the convergence of ∑j=1∞βj ensure that rm is a Cauchy sequence in R. Now, the completeness of R,dR yields the existence of a point r∗∈R with rm⟶r∗. By hypothesis (iii), we get γrm,r∗=1,∀m∈ℕ, as γrm,rm+1=1,∀m∈ℕ and rm⟶r∗. Now, we claim that r∗∈Qr∗,r∗,…,r∗. Suppose it is wrong, then, by (38), for each m∈ℕ, we obtain(52)HRQrm,rm+1,…,rm+k−1,Qr∗,r∗,…,r∗=HRQrm,rm+1,…,rm+k−1,Qr∗,r∗,…,r∗minγrm,r∗,γrm+1,r∗,…,γrm+k−1,r∗≤ζmaxdRrm+i−1,r∗:i∈1,2,…,kϑ1×dRrm+k−1,Qrm,rm+1,…,rm+k−1ϑ2dRr∗,Qr∗,r∗,…,r∗ϑ3.
That is,(53)dRrm+k,Qr∗,r∗,…,r∗≤HRQrm,rm+1,…,rm+k−1,Qr∗,r∗,…,r∗≤ζmaxdRrm+i−1,r∗:i∈1,2,…,kϑ1×dRrm+k−1,Qrm,rm+1,…,rm+k−1ϑ2dRr∗,Qr∗,r∗,…,r∗ϑ3.
By triangle inequality and (53), we obtain(54)dRr∗,Qr∗,r∗,…,r∗≤dRr∗,rm+k+dRrm+k,Qr∗,r∗,…,r∗≤dRr∗,rm+k+ζmaxdRrm+i−1,r∗:i∈1,2,…,kϑ1×dRrm+k−1,Qrm,rm+1,…,rm+k−1ϑ2dRr∗,Qr∗,r∗,…,r∗ϑ3.
After applying the limit asm⟶∞ in (54), we get dRr∗,Qr∗,r∗,…,r∗=0. Hence, the claim is true and r∗∈Qr∗,r∗,…,r∗.Theorem 6.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be an extended interpolative Prešić type-II set-valued contraction map on a complete metric space R,dR; that is, for every w1,w2,…,wk, p1,p2,…,pk∈R\FixQ with minγw1,p1,γw2,p2,…,γwk,pk≥1, we have(55)HRQw1,w2,…,wk,Qp1,p2,…,pk≤ζmaxdRwi,pi:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRpk,Qp1,p2,…,pkϑ3,where γ:R×R⟶ℝ is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1, and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
Ifminγw1,p1,γw2,p2,…,γwk,pk≥1, then γz1,z2≥1, for all z1∈Qw1,w2,…,wk and z2∈Qp1,p2,…,pk.(ii)
There arew1,w2,…,wk∈R satisfying(56)minγw1,w2,γw2,w3,…,γwk,z≥1,∀z∈Qw1,w2,…,wk,(iii)
For each sequencerm in R with γrm,rm+1≥1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r≥1,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.
This result can be proved on the similar steps as the proofs of Theorems5 and 4 are obtained. By considering p1=w2, p2=w3, …, pk−1=wk and denoting pk=wk+1 in Theorems 5 and 6, we get the following results.Theorem 7.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be a set-valued map on a complete metric space R,dR such that, for every w1,w2,…,wk,wk+1∈R\FixQ, we have(57)HRQw1,w2,…,wk,Qw2,w3,…,wk+1minγw1,w2,γw2,w3,…,γwk,wk+1≤ζmaxdRwi,wi+1:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRwk+1,Qw2,w3,…,wk+1ϑ3,where γ:R×R⟶ℝ\0 is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1, and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
Ifminγw1,w2,γw2,w3,…,γwk,wk+1=1, then γz1,z2=1, for all z1∈Qw1,w2,…,wk and z2∈Qw2,w3,…,wk+1.(ii)
There arew1,w2,…,wk∈R satisfying(58)minγw1,w2,γw2,w3,…,γwk,z=1,∀z∈Qw1,w2,…,wk,(iii)
For each sequencerm in R with γrm,rm+1=1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r=1,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.Theorem 8.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be a set-valued map on a complete metric space R,dR such that, for every w1,w2,…,wk,wk+1∈R\FixQ with(59)minγw1,w2,γw2,w3,…,γwk,wk+1≥1,we have(60)HRQw1,w2,…,wk,Qw2,w3,…,wk+1≤ζmaxdRwi,wi+1:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRwk+1,Qw2,w3,…,wk+1ϑ3,where γ:R×R⟶ℝ is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1 and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
Ifminγw1,w2,γw2,w3,…,γwk,wk+1≥1, then γz1,z2≥1 for all z1∈Qw1,w2,…,wk and z2∈Qw2,w3,…,wk+1.(ii)
There arew1,w2,…,wk∈R satisfying(61)minγw1,w2,γw2,w3,…,γwk,z≥1,∀z∈Qw1,w2,…,wk,(iii)
For each sequencerm in R with γrm,rm+1≥1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r≥1,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.


## 2.1. Results for Extended Interpolative Prešić Type Set-Valued Operators

This section presents the extensions of the above listed results. Theorems5 and 6 can be considered as an extended version of Theorems 3 and 4, respectively.Theorem 5.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be an extended interpolative Prešić type-I set-valued contraction map on a complete metric space R,dR, that is, for every w1,w2,…,wk, p1,p2,…,pk∈R\FixQ, we have(38)HRQw1,w2,…,wk,Qp1,p2,…,pkminγw1,p1,γw2,p2,…,γwk,pk≤ζmaxdRwi,pi:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRpk,Qp1,p2,…,pkϑ3,where γ:R×R⟶ℝ\0 is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1, and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
Ifminγw1,p1,γw2,p2,…,γwk,pk=1, then γz1,z2=1 for all z1∈Qw1,w2,…,wk, z2∈Qp1,p2,…,pk.(ii)
There arew1,w2,…,wk∈R satisfying(39)minγw1,w2,γw2,w3,…,γwk,z=1,∀z∈Qw1,w2,…,wk.(iii)
For each sequencerm in R with γrm,rm+1=1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r=1,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.Proof.
Hypothesis (ii) says that there are pointsr1,r2,…,rk in R satisfying the condition:(40)minγr1,r2,γr2,r3,…,γrk,z=1,∀z∈Qr1,r2,…,rk.
Thus, forrk+1∈Qr1,r2,…,rk, we obtain(41)minγr1,r2,γr2,r3,…,γrk,rk+1=1.
Then, by (38), we obtain(42)dRrk+1,Qr2,r3,…,rk+1≤HRQr1,r2,…,rk,Qr2,r3,…,rk+1minγr1,r2,γr2,r3,…,γrk,rk+1≤ζmaxdRri,ri+1:i∈1,2,…,kϑ1×dRrk,Qr1,r2,…,rkϑ2dRrk+1,Qr2,r3,…,rk+1ϑ3.
That is,(43)dRrk+1,Qr2,r3,…,rk+11−ϑ3≤ζmaxdRri,ri+1:i∈1,2,…,kϑ1dRrk,Qr1,r2,…,rkϑ2.
SincedRrk,Qr1,r2,…,rk≤maxdRri,ri+1:i∈1,2,…,k, thus, by (43), we obtain(44)dRrk+1,Qr2,r3,…,rk+11−ϑ3≤ζmaxdRri,ri+1:i∈1,2,…,kϑ1+ϑ2.
Since1−ϑ3=ϑ1+ϑ2, then (44) gives(45)dRrk+1,Qr2,r3,…,rk+1≤ζmaxdRri,ri+1:i∈1,2,…,k.
As1/ζ>1, thus, there exists rk+2∈Qr2,r3,…,rk+1 of a form(46)dRrk+1,rk+2≤1ζdRrk+1,Qr2,r3,…,rk+1≤ζmaxdRri,ri+1:i∈1,2,…,k.
Hypothesis (i) implies thatγrk+1,rk+2=1, since minγr1,r2,γr2,r3,…,γrk,rk+1=1 and rk+1∈Qr1,r2,…,rk, rk+2∈Qr2,r3,…,rk+1. By the repeated application of hypothesis (i) and (38), we reach to a sequence rm with the facts rm+k∈Qrm,rm+1,…,rm+k−1 for all m∈ℕ and(47)minγrm,rm+1,γrm+1,rm+2,…,γrm+k−1,rm+k=1,∀m∈ℕ,and(48)dRrm+k,rm+k+1≤ζmaxdRrm−1+i,rm+i:i∈1,2,…,k,∀m∈ℕ.
For simplicity, takedRm=dRrm,rm+1 for each m∈ℕ; from (48), we obtain(49)dRm+k≤ζmaxdRm−1+i:i∈1,2,…,k,∀m∈ℕ.
Now, we prove by induction thatdRm≤βmM for each m∈ℕ, where β=ζ1/2k and M=maxdR1/β,dR2/β2,…,dRk/βk. Trivially, dRi≤βiM, for each i∈1,2,…,k. Suppose that dRi≤βiM for each i∈m,m+1,…,m+k−1 for some given m, as induction hypothesis. Then, by (49), we obtain(50)dRm+k≤ζmaxdRm−1+i:i∈1,2,…,k≤ζmaxβm−1+iM:i∈1,2,…,k=βkβmM=βm+kM.
Hence, it is shown by induction thatdRm≤βmM, for each m∈ℕ. This fact along with triangle inequality yield that(51)dRrn,rq≤∑j=nq−1dRrj,rj+1=∑j=nq−1dRj≤∑j=nq−1βjM,for each q,n∈ℕ with q>n. Hence, the above inequality and the convergence of ∑j=1∞βj ensure that rm is a Cauchy sequence in R. Now, the completeness of R,dR yields the existence of a point r∗∈R with rm⟶r∗. By hypothesis (iii), we get γrm,r∗=1,∀m∈ℕ, as γrm,rm+1=1,∀m∈ℕ and rm⟶r∗. Now, we claim that r∗∈Qr∗,r∗,…,r∗. Suppose it is wrong, then, by (38), for each m∈ℕ, we obtain(52)HRQrm,rm+1,…,rm+k−1,Qr∗,r∗,…,r∗=HRQrm,rm+1,…,rm+k−1,Qr∗,r∗,…,r∗minγrm,r∗,γrm+1,r∗,…,γrm+k−1,r∗≤ζmaxdRrm+i−1,r∗:i∈1,2,…,kϑ1×dRrm+k−1,Qrm,rm+1,…,rm+k−1ϑ2dRr∗,Qr∗,r∗,…,r∗ϑ3.
That is,(53)dRrm+k,Qr∗,r∗,…,r∗≤HRQrm,rm+1,…,rm+k−1,Qr∗,r∗,…,r∗≤ζmaxdRrm+i−1,r∗:i∈1,2,…,kϑ1×dRrm+k−1,Qrm,rm+1,…,rm+k−1ϑ2dRr∗,Qr∗,r∗,…,r∗ϑ3.
By triangle inequality and (53), we obtain(54)dRr∗,Qr∗,r∗,…,r∗≤dRr∗,rm+k+dRrm+k,Qr∗,r∗,…,r∗≤dRr∗,rm+k+ζmaxdRrm+i−1,r∗:i∈1,2,…,kϑ1×dRrm+k−1,Qrm,rm+1,…,rm+k−1ϑ2dRr∗,Qr∗,r∗,…,r∗ϑ3.
After applying the limit asm⟶∞ in (54), we get dRr∗,Qr∗,r∗,…,r∗=0. Hence, the claim is true and r∗∈Qr∗,r∗,…,r∗.Theorem 6.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be an extended interpolative Prešić type-II set-valued contraction map on a complete metric space R,dR; that is, for every w1,w2,…,wk, p1,p2,…,pk∈R\FixQ with minγw1,p1,γw2,p2,…,γwk,pk≥1, we have(55)HRQw1,w2,…,wk,Qp1,p2,…,pk≤ζmaxdRwi,pi:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRpk,Qp1,p2,…,pkϑ3,where γ:R×R⟶ℝ is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1, and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
Ifminγw1,p1,γw2,p2,…,γwk,pk≥1, then γz1,z2≥1, for all z1∈Qw1,w2,…,wk and z2∈Qp1,p2,…,pk.(ii)
There arew1,w2,…,wk∈R satisfying(56)minγw1,w2,γw2,w3,…,γwk,z≥1,∀z∈Qw1,w2,…,wk,(iii)
For each sequencerm in R with γrm,rm+1≥1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r≥1,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.
This result can be proved on the similar steps as the proofs of Theorems5 and 4 are obtained. By considering p1=w2, p2=w3, …, pk−1=wk and denoting pk=wk+1 in Theorems 5 and 6, we get the following results.Theorem 7.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be a set-valued map on a complete metric space R,dR such that, for every w1,w2,…,wk,wk+1∈R\FixQ, we have(57)HRQw1,w2,…,wk,Qw2,w3,…,wk+1minγw1,w2,γw2,w3,…,γwk,wk+1≤ζmaxdRwi,wi+1:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRwk+1,Qw2,w3,…,wk+1ϑ3,where γ:R×R⟶ℝ\0 is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1, and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
Ifminγw1,w2,γw2,w3,…,γwk,wk+1=1, then γz1,z2=1, for all z1∈Qw1,w2,…,wk and z2∈Qw2,w3,…,wk+1.(ii)
There arew1,w2,…,wk∈R satisfying(58)minγw1,w2,γw2,w3,…,γwk,z=1,∀z∈Qw1,w2,…,wk,(iii)
For each sequencerm in R with γrm,rm+1=1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r=1,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.Theorem 8.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be a set-valued map on a complete metric space R,dR such that, for every w1,w2,…,wk,wk+1∈R\FixQ with(59)minγw1,w2,γw2,w3,…,γwk,wk+1≥1,we have(60)HRQw1,w2,…,wk,Qw2,w3,…,wk+1≤ζmaxdRwi,wi+1:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRwk+1,Qw2,w3,…,wk+1ϑ3,where γ:R×R⟶ℝ is a map, ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1 and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
Ifminγw1,w2,γw2,w3,…,γwk,wk+1≥1, then γz1,z2≥1 for all z1∈Qw1,w2,…,wk and z2∈Qw2,w3,…,wk+1.(ii)
There arew1,w2,…,wk∈R satisfying(61)minγw1,w2,γw2,w3,…,γwk,z≥1,∀z∈Qw1,w2,…,wk,(iii)
For each sequencerm in R with γrm,rm+1≥1,∀m≥m0, for some natural number m0, and rm⟶r, we have γrm,r≥1,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.


## 3. Application

In this section, we obtain the following application of the above result through a combination of graph theory. In the following, assume thatGr=Ve,Ed be a directed graph defined on a metric space R,dR with vertex set Ve=R and edge set Ed⊂R×R contains all loops, but it has no parallel edge. From Theorem 8, by defining γw,r=1 for each w,r∈R with w,r∈Ed, for otherwise, γw,r=0, we get the following result.Theorem 9.
LetQ:Rk⟶CBR, for any fixed k∈ℕ, be a set-valued map on a complete metric space R,dR equipped with the graph Gr such that, for every w1,w2,…,wk,wk+1∈R\FixQ, with(62)w1,w2,w2,w3,…,wk,wk+1∈Ed,we have(63)HRQw1,w2,…,wk,Qw2,w3,…,wk+1≤ζmaxdRwi,wi+1:i∈1,2,…,kϑ1×dRwk,Qw1,w2,…,wkϑ2dRwk+1,Qw2,w3,…,wk+1ϑ3,where ϑ1,ϑ2,ϑ3∈0,1 with ϑ1+ϑ2+ϑ3=1, ζ∈0,1 and FixQ=r∈R:r∈Qr,r,…,r. Also, consider that(i)
For allw1,w2,…,wk,wk+1∈R with w1,w2,w2,w3,…,wk,wk+1∈Ed, we have z1,z2∈Ed, for all z1∈Qw1,w2,…,wk and z2∈Qw2,w3,…,wk+1.(ii)
There arew1,w2,…,wk∈R with(64)w1,w2,w2,w3,…,wk−1,wk,wk,z∈Ed,∀z∈Qw1,w2,…,wk.(iii)
For each sequencerm in R with rm,rm+1∈Ed,∀m≥m0, for some natural number m0, and rm⟶r, we have rm,r∈Ed,∀m≥m0.
Then, there exists an elementr of R with r∈Qr,r,…,r︸k−times.
Sarwar et al. [25] studied the existence of the solution of Caputo–Fabrizio fractional derivative of order γ, which is defined as(65)Dtγut=Nγ1−γ∫0tu′τexp−γ1−γt−τdτ,under boundary condition u0=0, where Nγ is a normalization function satisfying N0=N1=1 and a≤t≤τ≤b, by using an interpolative Dass and Gupta rational-type contraction condition. Through the work of Sarwar et al. [25], it is obvious that the existence of the solution of above defined Caputo–Fabrizio fractional derivative can also be discussed by an interpolative Kannan contraction that is a particular case of our work.


## 4. Conclusion

The notions of interpolative Prešić-type set-valued contractions for the set-valued operators defined on product spaces along with fixed-point results are presented. These notions can also be considered as an extended version of interpolative Prešić-type contractions.



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