# On Suzuki and Wardowski-Type Contraction Multivalued Mappings in Partial Symmetric Spaces

Authors: Mohammad Asim; Hassen Aydi; Mohammad Imdad
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The purpose of this paper is to provide some fixed-point results for Suzuki and Wardowski-type contraction multivalued mappings in partial symmetric spaces. We give some examples to support and substantiate the developed notions and obtained results. Also, we use one of our main results to establish the existence and uniqueness of the solution for a system of integral inclusions.


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## Body

## 1. Introduction

In 1922, Banach [1] formulated a very famous, fruitful, useful, and core fixed-point result known as the Banach contraction principle on a complete metric space. This celebrated result has been generalized and extended in several directions by various authors on some generalized metric spaces such as partial metric spaces [2], partial JS-metric spaces [3], metric-like spaces [4], b-metric spaces [5], rectangular metric spaces [6], controlled and doubled metric spaces [7, 8], generalized b-metric spaces [9], and extended rectangular b-metric spaces [10]. Sometimes, one may come across a situation wherein the full force of metric conditions is not required in the enunciation of fixed-point results. Motivated by this observation, several researchers established fixed and common fixed-point results in symmetric spaces which did not require triangular inequality, i.e., a symmetric χ on a nonempty set X is a mapping χ:X×X⟶0,∞ which satisfies dρ,ρ=0 iff ρ=ς, and dρ,ς=dς,ρ for all ρ,ς∈X.Very recently, Asim et al. [11] introduced the class of partial symmetric spaces and proved some related fixed-point results for single-valued and multivalued mappings. The theory of multivalued mappings plays an important role in many areas of mathematics due to its diverse applications, namely in differential equations, integral equations, optimization problems, game theory, control theory, economics, etc. Aydi et al. [12, 13] proved fixed and common fixed-point results for multivalued maps in partial Hausdorff metric spaces to generalize fixed-point results due to Nadler [14].In 2008, Suzuki [15] introduced a class of new types of contraction mappings and established fixed-point theorems for such mappings, which are genuine improvements of the Banach contraction principle. Thereafter, many authors attempted to extend such results for multivalued maps (employing the Pompieu-Hausdorff metric). In 2012, Wardowski [16] initiated the notion of a new kind of nonlinear contractions, namely F-contractions and proved some related fixed-point results. In recent years, the idea of F-contractions has been generalized and improved in several ways and directions. For more details, see [17–21].In our paper, based on Suzuki and Wardowski-type contractions, we consider multivalued mappings, and we establish some related fixed-point results in the setting of partial symmetric spaces. Some examples are also furnished to demonstrate the utility of our results. As an application of Theorem16, we establish the existence and uniqueness of the solution for a system of integral inclusions.


## 2. Preliminaries

In what follows, we collect some relevant definitions and auxiliary results needed in the sequel.Definition 1 [11].
LetX be a nonempty set. The function χ:X×X⟶0,∞ satisfying
χ1χρ,ς=χρ,ρ=χς,ς iff ρ=ς,
χ2χρ,ρ≤χρ,ς,
χ3χρ,ς=χς,ρis said to be a partial symmetric, and the pair X,χ is called a partial symmetric space.A partial symmetric spaceX,χ reduces to a symmetric space if χx,x=0, for all x∈X. Observe that every symmetric space is a partial symmetric space. The converse is not true as it is shown in the following examples.Example 2.
LetX=ℝand defineχ:X×X⟶0,∞by(1)χρ,ς=ρ−ςp+ρ−ςq+α,wherep,q>1andα≥0.Example 3.
LetX=0,∞and defineχ:X×X⟶0,∞by(2)χρ,ς=maxρ,ςp+ρ−ςq,wherep,q>1.Example 4.
LetX=0,πand defineχ:X×X⟶0,∞by(3)χρ,ς=sin∣ρ−ς∣+α,whereα≥0.Example 5.
LetX=0,πand defineχ:X×X⟶0,∞by(4)χρ,ς=maxρ,ςp+e∣ρ−ς∣,wherep>1.LetX,χ be a partial symmetric space. The χ-open ball with a center ρ∈X and a radius ε>0 is defined by
(5)Bχρ,ε=ς∈ρ:χρ,ς<χρ,ρ+ε.Similarly, theχ-closed ball with a center ρ∈X and a radius ε>0 is defined by
(6)Bχρ,ε=ς∈X:χρ,ς≤χρ,ρ+ε.The family ofχ-open balls denoted by
(7)Uχ=Bχρ,ε:ρ∈X,ε>0forms on X a basis of some topology τχ.Lemma 6.
LetX,τχbe a topological space andf:X⟶Xbe given mapping. If f is continuous; then, for every convergent sequenceρnto a pointρinX, the sequencefρnconverges tofρ. The converse holds ifXis metrizable.We require some more definitions, namely, a convergent sequence, a Cauchy sequence, and a complete partial symmetric space, in our forthcoming discussions.Definition 7 [11].
A sequenceρninX,χχ-converges toρ∈X, with respect toτχ, if(8)χρ,ρ=limn⟶∞χρn,ρ.Definition 8 [11].
A sequenceρninX,χisχ-Cauchy ifflimn,m⟶∞χρn,ρmexists and is finite.Definition 9 [11].
A partial symmetric spaceX,χisχ-complete if eachχ-Cauchy sequenceρninXisχ-convergent with respect toτχto a pointρ∈Xsuch that(9)χρ,ρ=limn⟶∞χρn,ρ=limn,m⟶∞χρn,ρm.LetX,χ be a partial symmetric space and CBχX be the set of all nonempty, χ-closed, and bounded subsets of X,χ. Moreover, for L,M∈CBχX and ρ∈X, we define
(10)distχρ,L=infχρ,l:l∈L,δχL,M=supdistχl,M:l∈L,δχM,L=supdistχm,L:m∈M.LetHχ.,. be the partial Pompieu-Hausdorff symmetric, that is
(11)HχL,M=maxδχL,M,δχM,L.


## 3. Main Results

Given thatf:X⟶CBχX, then for every ρ∈X, we define
(12)Sχ,f,ρ=supχρi,ρj:i,j∈ℕ,where ρi∈fρi−1 and ρj∈fρj−1.Now, we present the following lemma proved in [11], which is needed in the sequel.Lemma 10.
LetX,χbe a partial symmetric space andL,M∈CBχX. Then, for allh>1andl∈L, there ism∈Mso that(13)χl,m≤hHχL,M.Next, letψ:0,1⟶0,1 be the nonincreasing function defined by
(14)ψr=1if0≤r<121−rif12≤r<1.Theorem 11.
LetX,χ be a complete partial symmetric space and f:X⟶CBχX.
Assume the following:
Si There exists 0≤r<1 such that f satisfies the condition
(15)ψrdistχρ,fρ≤χρ,ςimpliesHχfρ,fς≤rMρ,ς,for all ρ,ς∈X, where ψ is defined by (14) and
(16)Mρ,ς=maxχρ,ς,distχρ,fρ,distχς,fς,distχρ,fς,distχς,fρ
Sii There is ρ0∈X so that Sχ,f,ρ0<∞
Siii The function ρ⟶distχρ,fρ is lower semicontinuous.Then,f has a unique fixed point.Proof.
BySi, there exists 0≤r<1 such that (15) holds. Let r1∈0,1 be such that 0≤r≤r1<1. For such ρ0∈X, define a sequence ρn∈X by ρn+1∈fρn. If we assume that ρn0∈fρn0 for some n0∈ℕ, then ρn0 is a fixed point of f, and we are done. Now, suppose that ρn∉fρn for each n∈ℕ, and so χρn,ρn+1>0. From Lemma 10 (with h=1/r1) and condition (15) (for any arbitrary n∈ℕ and for all i,j∈ℕ), we have
(17)χρn+i,ρn+j≤1r1Hχfρn−1+i,fρn−1+j.
Sinceψr≤1, we get
(18)ψrdistχρn+i,fρn+i≤distχρn+i,fρn+i≤χρn+i,ρn+1+i.
From (15) and (17), we have
(19)χρn+i,ρn+j≤1r1Hpfρn−1+i,fρn−1+j≤r1Mρn−1+i,ρn−1+j≤r1maxχρn−1+i,ρn−1+j,distχρn−1+i,fρn−1+i,distχρn−1+j,fρn−1+j,distχρn−1+i,fρn−1+j,distχρn−1+j,fρn−1+i..
The above inequality is true for alli,j∈N; therefore, by conditions Sii and (12), we have
(20)Sχ,f,fnρ0≤r1Sχ,f,fn−1ρ0.
By repeating this process, we have (for alln≥1)
(21)Sχ,f,fnρ0≤r1nSχ,f,ρ0.
Now(22)χρn,ρn+m≤Sχ,f,fnρ0≤r1nSχ,f,ρ0,∀n,m∈ℕ.
SinceSχ,f,ρ0<∞ and r1∈0,1, then
(23)limn,m⟶∞χρn,ρn+m=0,so that ρn is a χ-Cauchy sequence in X. In view of the χ-completeness of X, there exists ρ∈X such that ρnχ-converges to ρ. Thus
(24)χρ,ρ=limn⟶∞χρn,ρ=limn,m⟶∞χρn,ρn+m=0.
Assume thatρ⟶distχρ,fρ is lower semicontinuous. Then
(25)distχρ,fρ≤liminfn⟶∞distχρn,fρn≤limn⟶∞χρn,ρn+1=0.
Therefore,distχρ,fρ=0, that is, ρ∈fρ. Hence, ρ is a fixed point of f.
Next, let us show thatf has a unique fixed point. Suppose there are ρ,ς∈X such that ρ∈fρ and ς∈fς. Then, by conditions Si and χ2, we have
(26)ψrdistχρ,fρ≤distχρ,fρ≤χρ,ρ≤χρ,ς,which implies (by Lemma 10 with h=1/r1 and 0≤r≤r1<1)
(27)χρ,ς≤1r1Hχfρ,fς≤r1Mρ,ς=r1maxχρ,ς,distχρ,fρ,distχς,fς,distχρ,fς,distχρ,fς≤r1maxχρ,ς,χρ,ρ,χς,ς,χρ,ς,χς,ρ=r1χρ,ς<χρ,ς,a contradiction, so that χρ,ς=0, which implies that ρ=ς. Hence, f has a unique fixed point. This completes the proof.☐Example 12.
TakeX=0,1/10,1/5. The partial symmetricχ:X×X⟶0,∞is defined by(28)χρ,ς=12ρ−ς2+14maxρ,ς2,∀ρ,ς∈X.
Then,X,χis aχ-complete symmetric space. Note that0and1/10are bounded sets inX,χ. In fact, ifρ∈0,1/10,1/5, then(29)ρ∈0¯⟺distχρ,0=χρ,ρ⟺3ρ24=ρ24⟺ρ=0⟺ρ∈0.
Hence,0is closed. Next(30)ρ∈110¯⟺distχρ,110=χρ,ρ⟺min3ρ24,12ρ−1102+14maxρ,1102=ρ24⟺ρ∈110.
Hence,1/10is also closed. Now, definef:X⟶CBχXby(31)f0=f110=0andf15=110.
Clearly, by a routine calculation, one can easily show thatx⟶distχρ,fρis lower semicontinuous. To prove the contractive conditionSiof Theorem 11, we need the following.Case 13.
Letρ=0. Then(32)ψrdistχ0,f0=0≤χ0,ς,forallς∈X.
Forς=0 or ς=1/10, we have.
(33)Hχf0,fς=Hχ0,0=0≤rχ0,ς.
Forς=1/5, we have(34)Hχf0,f15=Hχ0,110=120−1102+14max0,1102=14120−152+14max0,152=14χ0,15.Case 14.
Letρ=1/10. Then,ψrdistχ1/10,f1/10=3/400≤χ0,ς, for all ς∈X.
Forς=0 or ς=1/10, we have(35)Hχf110,fς=0<r3ς24≤rχ110,ς.
Forς=1/5, we have(36)Hχf110,f15=Hχ0,110=120−1102+14max0,1102=1212110−152+14max110,152=12χ110,15.Case 15.
Letρ=1/5. Then(37)ψrdistχ15,f15=3200≤χ15,ς,forς∈X.
Forς=0, we have(38)Hχf15,fς=Hχ110,0≤14χ15,0.
Forς=1/10, we get(39)Hχf15,f110=Hχ110,0≤14χ15,0.
Forς=1/5, we have(40)Hχf15,f15=Hχ110,110≤14χ15,15.
Hence, the contractive conditionSiof Theorem11is satisfied. Observe thatfhas a unique fixed point (namely,ρ=0).Denote byF the class of functions F:0,∞⟶ℝ such thatF1F is increasingF2 For all αn>0,limn⟶∞αn=0ifflimn⟶∞Fαn=−∞.The related fixed-point Wardowski-type result for multivalued mappings in the setting of partial symmetric spaces is as follows.Theorem 16.
LetX,χ be a complete partial symmetric space and f:X⟶CBχX. Assume that
(i)
There areF∈F and τ>0 such that(41)τ+FHχfρ,fς≤FMρ,ς(ii)
There isρ0∈X so that Sχ,f,ρ0<∞(iii)
The functionρ⟶distχρ,fρ is lower semicontinuous.
Then,f has a unique fixed point.Proof.
For suchρ0∈X, take ρ1∈fρ0. Define the sequence ρn∈X by ρn+1∈fρn. If we assume that ρn0∈fρn0 for some n0∈ℕ, then ρn0 is a fixed point of f. Suppose ρn∉fρn for all n∈ℕ, so χρn,ρn+1>0. Since f is a generalized multivalued F-contraction, we have (for all i,j∈ℕ)
(42)Fχρn+i,ρn+j≤FHfρn−1+i,fρn−1+j≤FMρn−1+i,ρn−1+j−τ=Fmaxχρn−1+i,ρn−1+j,distχρn−1+i,fρn−1+i,distχρn−1+j,fρn−1+j,distχρn−1+i,fρn−1+j,distχρn−1+j,fρn−1+i−τ≤Fmaxχρn−1+i,ρn−1+j,χρn−1+i,ρn+i,χρn−1+j,ρn+j,χρn−1+i,ρn+j,χρn−1+j,ρn+i−τ.
The above inequality is true for alli,j∈ℕ. Therefore, by conditions ii and (12), we have
(43)FSχ,f,ρn≤FSχ,f,ρn−1−τ.
By continuing this process, we find that(44)FSχ,f,ρn≤FSχ,f,ρ0−nτ.
Now, for alln,m∈ℕ, we have
(45)χρn,ρn+m≤Sχ,f,ρnFSχ,f,ρn≤FSχ,f,ρ0−nτ.
On making the limit asn⟶∞ in (45), we have
(46)limn⟶∞FSχ,f,ρn=−∞.
Owing to conditionsii and F2, we have
(47)limn⟶∞Sχ,f,ρn=0limn,m⟶∞χρn,ρn+m=0,so that ρn is a χ-Cauchy sequence in X. In view of the χ-completeness of X, there is ρ∈X in order that ρnχ-converges to ρ. Thus, we have
(48)χρ,ρ=limn⟶∞χρn,ρ=limn,m⟶∞χρn,ρn+m=0.
Assume thatρ⟶distχρ,fρ is lower semicontinuous. Then, we have
(49)distχρ,fρ≤liminfn⟶∞distχρn,fρn≤limn⟶∞χρn,ρn+1=0.
Therefore,distχρ,fρ=0 implies that ρ∈fρ. Hence, ρ is a fixed point of f.
For the uniqueness part, suppose there existρ≠ς∈X such that ρ∈fρ and ς∈fς. Thus, by the condition i, we have
(50)FHχρ,ς=FHχfρ,fς≤FMρ,ς−τ=Fmaxχρ,ς,distχρ,fρ,distχς,fς,distχρ,fς,distχς,fρ−τ=Fmaxχρ,ς,χρ,ρ,χς,ς,χρ,ς,χς,ρ−τ=Fχρ,ς−τ,which is a contradiction. Therefore, χρ,ς=0, that is, ρ=ς. It completes the proof.☐Example 17.
LetX=0∪1/2n:n∈ℕbe equipped with the partial symmetricχ:X×X⟶0,∞defined by(51)χρ,ς=12ρ−ς2+14maxρ,ς2forallρ,ς∈X.
Then, X,χis aχ-complete symmetric space. Note that1/2n+1and0are bounded sets inX,χ. In fact(52)ρ∈12n+1¯⟺distχρ,12n+1=χρ,ρ⟺min12ρ−12n+12+14maxρ,12n+12=ρ24⟺ρ=12n+1⟺ρ∈12n+1.
Hence,1/2n+1is closed. Next(53)ρ∈0¯⟺distχρ,0=χρ,ρ⟺3ρ24=ρ24⟺ρ∈0.
Hence,0is also closed with respect to the partial symmetricχ.
Definef:X⟶CBχX and F:0,∞⟶ℝby(54)fρ=12n+1,ifρ=12n0,ifρ=0,and Fα=lnα for α>0, respectively. Then(55)distχρ,fρ=3212n+12,ifρ=12n0,ifρ=0.
Hence,ρ⟶distχρ,fρis lower semicontinuous. Now, we will show that the contractive conditioniof Theorem16is satisfied.Case 18.
Letρ,ς∈1/2n (say ρ=1/2n1 and ρ=1/2n2 for all n1,n2∈ℕ). Here(56)Hχfρ,fς=Hχ12n1+1,12n2+1=1212n1+1−12n2+12+14max12n1+1,12n2+12=141212n1−12n22+14max12n1,12n22=14χρ,ς≤14Mρ,ς.
That isforτ=ln4(57)τ+FHχfρ,fς≤FMρ,ς.Case 19.
Letρ∈1/2nandς∈0. Then(58)Hχfρ,fς=Hχ12n+1,0=maxδχ12n+1,0,δχ0,12n+1=3412n+12.
Also(59)M12n,0=maxχ12n,0,distχ12n,f12n,distχ0,0,distχ12n,0distχ0,12n+1=max3412n2,3812n2,0,3412n2,31612n2=3412n2.
Forτ=ln4, one writes(60)τ+FHχfρ,fς≤FMρ,ς.
Therefore, all the conditions of Theorem16are satisfied, andfhas a unique fixed point (namely,ς=0).


## 4. An Application to an Integral Inclusion

LetX=Ca,b,ℝ be the set of all continuous real valued functions defined on a,b. Now, we consider the following integral inclusion of Volterra type:
(61)ρt∈∫atGt,s,ρsds+ht,∀t∈a,bandh∈X,where G:a,b×a,b×ℝ⟶Kcvℝ is a multivalued operator and Kcvℝ is the set of nonempty compact and convex subsets of ℝ.Defineχ:X×X⟶ℝ+by
(62)χρt,ςt=supt∈a,bρt−ςtp+supt∈a,bρt−ςtq,q>p>1.Then,X,χ is a χ-complete partial symmetric space.Now, we are ready to present our result as follows:Theorem 20.
Suppose that for allρ,ς∈Ca,b,ℝthere exist a continuous functionj:X⟶ℝand∫atjsds≤e−τ1/pwithτ>0such that(63)HχGt,s,ρs,Gt,s,ςs≤js∣ρs−ςs∣,∀t,s∈a,b.
Then, the integral inclusion (61) has a unique solution.Proof.
Let us define the multivalued operatorf:X⟶CBχX by fρt=∈X:v∈∫01Gt,s,ρsds+ht,forallt∈0,1andh∈X.
Letρ∈X and consider Gρ=Gt,s,ρs,t,s∈a,b. For a multivalued operator Gρ:a,b×a,b⟶Kcvℝ, there exists a continuous operator gρ:a,b×a,b⟶ℝ (by Michael’s selection theorem) such that gρt,s∈Gρt,s, for all t,s∈a,b. It implies that ∫atgρt,sds+ht∈fρt. Thus, the operator fρ is nonempty. Note that fρ is closed. For more details, see [22].
Firstly, we check the conditioni of Theorem 16. Let ρ,ς∈X be such that u∈fρ. Then, there exists gρt,s∈Gρt,s, for all t,s∈a,b such thatut=∫atgρt,sds+ht,t∈a,b.
Besides, for allρ,ς∈X, we have
(64)HχGt,s,ρs,Gt,s,ςs≤js∣ρs−ςs∣,∀t,s∈a,b.
Consequently, there existsvt,s∈Gςt,s such that
(65)∣gρt,s−vt,s∣≤js∣ρs−ςs∣,∀t,s∈a,b.
Let us consider the multivalued operatorS defined by
(66)St,s=Gςt,s∩w∈ℝ:∣gρt,s−w∣≤js∣ρs−ςs∣,∀t,s∈a,b.
Sincef is a lower semicontinuous, it follows that there exists gς:a,b×a,b⟶ℝ such that gςt,s∈St,s, for all t,s∈a,b. Then, we have
(67)zt=∫atgςt,sds+ht∈∫atGt,s,ςsds+ht,t∈a,b,vt−ztp+vt−ztq=∫at∣gρt,s−gςt,s∣dsp+∫at∣gρt,s−gςt,s∣dsq≤∫atjs∣ρs−ςs∣dsp+∫atjs∣ρs−ςs∣dsq≤supt∈a,bρt−ςtpp∫atjsdsp+supt∈a,bρt−ςtqq∫atjsdsq=supt∈a,bρt−ςtp∫atjsdsp+supt∈a,bρt−ςtq∫atjsdsq≤supt∈a,bρt−ςtpe−τ+supt∈a,bρt−ςtqe−τq/p≤supt∈a,bρt−ςtp+supt∈a,bρt−ςtqe−τsince,q>p=χρ,ςe−τ.
Hence, we have(68)χv,z≤e−τχρ,ς.
On interchanging the job ofρ and ς, we have
(69)Hχfρ,fς≤e−τχρ,ς.
By usingFα=lnα, we have
(70)τ+FHχfρ,fς≤χρ,ς≤FMρ,ς.
Thus, all the hypotheses of Theorem16 are satisfied. Hence, the operator f has exactly one fixed point, that is, the Volterra integral inclusion (61) has a unique solution.☐Remark 21.
Consider the following differential inclusion (forρ∈Ca,b,ℝ):(71)ρ′t∈Gt,s,ρs,t,s∈a,b,where G:a,b×a,b×ℝ⟶Kcvℝ is a multivalued operator. Observe that, for ht=0, (71) is equivalent to (61). Also,Gsatisfies the condition (63) of Theorem20. Then, applying Theorem20, the differential inclusion (71) has a unique solution.Example 22.
For allρ,ς∈Ca,b,ℝ, let(72)Gt,s,ρs=v∈ℝ:g1t,s,ρs≤v≤g2t,s,ρs,whereg1t,s,ρs(resp.,g2t,s,ρs) is upper semicontinuous (resp., lower semicontinuous) ona,b×a,b. For ρ∈Ca,b,ℝ, assume the following:(73)ρ′t∈Gt,s,ρs,t,s∈a,b.
For allρ,ς∈Ca,b,ℝ, there exist a continuous functionj:X⟶ℝandτ>0such that∫atjsds≤e−τ1/pand(74)maxg1t,s,ρs−g2t,s,ςs≤jsρs−ςs,forallt,s∈a,b.
Hence,Gis a compact and convex valued operator. Thus,Gsatisfies the condition (63) of Theorem20(forht=0), so (74) has a unique solution.


## 5. Conclusion

We are concerned with Suzuki and Wardowski-type contraction multivalued mappings in the setting of partial symmetric spaces. We also studied a system of integral inclusions. It would be interesting to work on more generalized contraction mappings involving simulation or control functions.



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