# On Complex-Valued Triple Controlled Metric Spaces and Applications

Authors: Nabil Mlaiki; Thabet Abdeljawad; Wasfi Shatanawi; Hassen Aydi; Yaé Ulrich Gaba
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this manuscript, we introduce the concept of complex-valued triple controlled metric spaces as an extension of rectangular metric type spaces. To validate our hypotheses and to show the usability of the Banach and Kannan fixed point results discussed herein, we present an application on Fredholm-type integral equations and an application on higher degree polynomial equations.


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## Body

## 1. Introduction

Since the breakthrough of Banach [1] in 1922, where he was able to show that a contractive mapping on a complete metric space has a unique fixed point, the field of fixed point theory has become an important research focus in the field of mathematics; see [2–6]. Due to the fact that fixed point theory has many applications in many fields of science, many researchers have been working on generalizing his result by either generalizing the type of contractions [7–10] or by extending the metric space itself (b-metric spaces [11, 12], controlled metric spaces [13], double controlled metric spaces [14], etc.). On the other hand, Azam et al. [15] defined complex-valued metric spaces and gave common fixed point results. Rao et al. [16] introduced the complex-valued b-metric spaces in the year 2013. Going in the same direction, recently, Ullah et al. [17] presented complex-valued extended b-metric spaces to extend the idea of extended b-metric spaces.In this manuscript, following the path of the work done in [18], we extend complex-valued rectangular extended b-metric spaces [19] to complex-valued triple controlled metric spaces. The layout of our manuscript is as follows. In the second section, we present some backgrounds along with the definition of complex-valued triple controlled metric spaces. In the third section, we prove some fixed point results in such spaces. In the fourth section, we present an application for our findings. In closing, we present two open questions.


## 2. Preliminaries

In what follows, owing to Azam et al. [15], we recall several notations and definitions which will be used in the sequel.Letℂ be the set of all complex numbers and s1,s2∈ℂ. The partial order on ℂ is defined as s1≺¯s2 if and only if Res1≤Res2 and Ims1≤Ims2. This implies that s1≺¯s2 if one of the below conditions is fulfilled:
(i)
Res1=Res2,Ims1<Ims2(ii)
Res1<Res2, Ims1=Ims2(iii)
(Res1<Res2, Ims1<Ims2(iv)
Res1=Res2, Ims1=Ims2Following [15], the authors in [17] developed the notion of complex-valued extended b-metric spaces.Definition 1 (see [17]).
LetX be a nonempty set and ξ:X×X⟶1,∞ be a function. Then, Le:X2⟶ℂ is known as a complex-valued extended b-metric space if the following are satisfied for all s,κ,u∈X:
(1)
0≺¯Les,κ and Les,κ=0 if and only if s=κ(2)
Les,κ=Leκ,s(3)
Les,κ≺¯ξs,κLes,u+Leu,κThen, the pairX,Le is known as a complex-valued extended b-metric space.As an extension of complex-valued extendedb-metric spaces, Ullah et al. in [19] introduced the concept of complex-valued rectangular extended b-metric spaces.Definition 2 (see [19]).
LetX be a nonempty set and ξ:X2⟶1,∞ and Lr:X2⟶ℂ. We say that X,Lr is a complex-valued rectangular extended b-metric space if for all a,b∈X each of which is different from κ,v∈X, we have
(1)
Lrs,κ=0 if and only if s=κ(2)
Lrs,κ=Lrκ,s(3)
Lra,b≺¯ξa,bLra,κ+Lrκ,v+Lrv,bThe authors in [20] have recently introduced the idea of triple controlled metric type spaces as follows.Definition 3 (see [20]).
LetX be a nonempty set. Given three functions ξ,ρ,ς:X2⟶1,∞ and LT:X2⟶0,∞. We say that X,LT is a triple controlled metric type space if for all a,b,κ,v∈X, we have
(1)
LTs,κ=0 if and only if s=κ(2)
LTs,κ=LTκ,s(3)
LTa,b≤ξa,κLTa,κ+ρκ,vLTκ,v+ςv,bLTv,bHighly motivated by the abovementioned concepts, we now present the definition of complex-valued triple controlled metric spaces.Definition 4.
LetX be a nonempty set. Given three functions ξ,ρ,ς:X2⟶1,∞ and Lt:X2⟶ℂ. We say that X,Lt is a complex-valued triple controlled metric space if for all a,b∈X, each of which is different from κ,v∈X, we have
(1)
Lts,κ=0 if and only if s=κ(2)
Lts,κ=Ltκ,s(3)
Lta,b≺¯ξa,κLta,κ+ρκ,vLtκ,v+ςv,bLtv,bThroughout the rest of this paper, we will denote a complex-valued triple controlled metric space by (CV-TCMS). Next, we present the topology of (CV-TCMSs).Definition 5.
LetX,Lt be a (CV-TCMS).
(1)
We say that a sequencean is Lt-convergent to some a∈X if Ltan,a⟶0asn⟶∞(2)
We say that a sequencean is Lt-Cauchy if and only if limn,m⟶∞∣Ltan,am∣=0(3)
We say thatX,Lt is Lt-complete if for every Lt-Cauchy sequence is Lt-convergent(4)
Letx∈X. An open ball of center x and radius η>0 in the (CV-TCMS) X,Lt is Bξx,η=b∈X∣Ltx,b≺¯ηNote that a CV rectangular metric space is a CV-TCMS. The converse is not true. Next, we present an example that confirms this statement.Example 1.
LetX=Y∪Z where Y=1/k∣k∈ℕ and Z is the set of positive integers. We define Lt:X2⟶ℂ by
(1)Lta,b=0,⇔a=b,2iβ,ifa,b∈Y,iβ2,otherwise,where β>0. Now, define ξ:X2⟶1,∞ by ξa,b=4β. Given ρ:X2⟶1,∞ as ρa,b=3β and ς:X2⟶1,∞ as ςa,b=maxa,b+2β.Note thatX,Lt is a CV-TCMS. On the other hand, X,Lt is not a CV rectangular metric space. Indeed,
(2)Lt12,13=2iβ≻Lt12,2+Lt2,3+Lt3,13=3iβ2.In this paper, we prove the Banach and Kannan fixed point results in the setting of CV-TCMSs. Two related applications are also investigated.


## 3. Main Results

Theorem 1.
LetX,Lt be a Lt-complete CV-TCMS. Let T:X⟶X satisfy LtTx,Ty≺¯δLtx,y where 0<δ<1. Assume that there exists x0∈X such that the sequence xn defined by xn=Tnx0 satisfies the following:
(3)limn⟶∞ξxn,xn+1≤1δ,limn⟶∞ρxn,xn+1≤1δ,limn⟶∞ςxn,y<∞,limn⟶∞ξy,xn<∞foranyy∈X,supm≥1limn⟶∞ξxn,xn+1ςxn,xm≤1δ,supm≥1limn⟶∞ρxn,xn+1ςxn,xm≤1δ.Then,T has a unique fixed point in X.Proof.
First, we haveLtxn,xn+1≺¯δLtxn−1,xn≺¯δ2Ltxn−2,xn−1≺¯⋯≺¯δnLtx0,x1. Then,
(4)Ltxn,xn+1⟶0asn⟶∞.Now, letLi=Ltxn+i,xn+i+1. We need to consider the following two cases.Case 1.
Letxn=xm for some natural numbers n and m with n≠m. Without loss of generality, take m>n. If Tm−nxn=xn; then, by choosing y=xn and p=m−n, we get Tpy=y, which implies that y is a periodic point of T. Hence, Lty,Ty=LtTpy,Tp+1y≺¯δpLty,Ty. Since δ∈0,1, we get ∣Lty,Ty∣=0, so y=Ty, that is, T has a fixed point.From now on, we consider the following case.Case 2.
Assume that for all natural numbersn≠m, we have xn=Tnx0≠Tmx0=xm. Let n<m. To prove that xn is a Lt-Cauchy sequence, we need to consider the following two subcases.Subcase 1.
Ifm=n+2p+1 (where p≥1 is a fixed natural number), then by the rectangle inequality of the CV-TCMS, we have
(5)∣Ltxn,xn+2p+1∣≤ξxn,xn+1∣Ltxn,xn+1∣+ρxn+1,xn+2∣Ltxn+1,xn+2∣+ςxn+2,xn+2p+1∣Ltxn+2,xn+2p+1∣≤ξxn,xn+1∣Ltxn,xn+1∣+ρxn+1,xn+2∣Ltxn+1,xn+2∣+ςxn+2,xn+2p+1ξxn+2,xn+3∣Ltxn+2,xn+3∣+ρxn+3,xn+4∣Ltxn+3,xn+4∣+ςxn+4,xn+2p+1∣Ltxn+4,xn+2p+1∣≤ξxn,xn+1δn∣L0∣+ρxn+1,xn+2δn+1∣L0∣+ςxn+2,xn+2p+1ξxn+2,xn+3δn+2∣L0∣+ρxn+3,xn+4δn+3∣L0∣+ςxn+4,xn+2p+1∣Ltxn+4,xn+2p+1∣≤ξxn,xn+1δn∣L0∣+ρxn+1,xn+2δn+1∣L0∣+ςxn+2,xn+2p+1ξxn+2,xn+3δn+2∣L0∣+ςxn+2,xn+2p+1ρxn+3,xn+4δn+3∣L0∣+ςxn+2,xn+2p+1ςxn+4,xn+2p+1ξxn+4,xn+5∣Ltxn+4,xn+5∣+ρxn+5,xn+6∣Ltxn+5,xn+6∣+ςxn+6,xn+2p+1∣Ltxn+6,xn+2p+1∣≤ξxn,xn+1δn∣L0∣+ρxn+1,xn+2δn+1∣L0∣+ςxn+2,xn+2p+1ξxn+2,xn+3δn+2∣L0∣+ςxn+2,xn+2p+1ρxn+3,xn+4δn+3∣L0∣+ςxn+2,xn+2p+1ςxn+4,xn+2p+1ξxn+4,xn+5δn+4∣L0∣+ςxn+2,xn+2p+1ςxn+4,xn+2p+1ρxn+5,xn+6δn+5∣L0∣+ςxn+2,xn+2p+1ςxn+4,xn+2p+1ςxn+6,xn+2p+1∣Ltxn+6,xn+2p+1∣≤⋯=ξxn,xn+1δn∣L0∣+ρxn+1,xn+2δn+1∣L0∣+∑l=1p∏i=1lξxn+2l,xn+2l+1ςxn+2i,xn+2p+1δn+2l∣L0+∑l=1p∏i=1lρxn+2l+1,xn+2l+2ςxn+2i,xn+2p+1δn+2l+1L0.Now, given that(6)supm≥1limn⟶∞ξxn,xn+1ςxn,xm≤1δ,supm≥1limn⟶∞ρxn,xn+1ςxn,xm≤1δ,we can easily deduce that
(7)∣Ltxn,xn+2p+1∣≤δn−1+δn−1+∑l=1pδn+l+∑l=1pδn+l+1L0.Sincelimn⟶∞δn=0, the last right-hand side goes to zero at the limit n⟶∞ (for any integer p≥1). Therefore, ∣Ltxn,xn+2p+1∣n is convergent.Subcase 2.
Letm=n+2p (where p≥1 is a fixed integer). First, notice the following:
(8)Ltxn,xn+2≺¯δLtxn−1,xn+1≺¯δ2Ltxn−2,xn≺¯⋯≺¯δnLtx0,x2,which leads us to conclude that
(9)Ltxn,xn+2⟶0asn⟶∞.Thus, by Subcase1 and using the rectangular inequality of the complex-valued triple controlled metric, we have
(10)Ltxn,xn+2p≤ξxn,xn+2p−3Ltxn,xn+2p−3+ρxn+2p−3,xn+2p−2Ltxn+2p−3,xn+2p−2+ςxn+2p−2,xn+2pLtxn+2p−2,xn+2p≤ξxn,xn+1δnL0+ρxn+1,xn+2δn+1∣L0∣+∑l=1p−1∏i=1lξxn+2l,xn+2l+1ςxn+2i,xn+2p−3δn+2lL0+∑l=1p−1∏i=1lρxn+2l+1,xn+2l+2ςxn+2i,xn+2p−3δn+2l+1L0+ςxn+2p−2,xn+2pδnLtx0,x2.Now, similar to Subcase1, one can easily deduce that Ltxn,xn+2pn is a convergent sequence as n⟶∞ (for any integer p≥1). Hence, by Subcases 1 and 2, we conclude that xn is a Lt-Cauchy sequence. Since X,Lt is a Lt-complete CV-TCMS, there is ν∈X such that xn⟶ν as n⟶∞.Now, if there existsN∈ℕ such that xN=ν, then since we deal with Case 2, one writes xn=Tnx0≠ν for all n>N. Also, xn=Tnx0≠Tν for all n>N. Next, assume that there exists N∈ℕ with xN=TNx0=Tν. Once again, we confirm that xn=Tnx0∈ν,Tν for all n>N. Thus, without loss of generality, we may assume xn∈ν,Tν for all natural numbers n. We have
(11)Ltν,Tν≺¯ξν,xnLtν,xn+ρxn,xn+1Ltxn,xn+1+ςxn+1,TνLtxn+1,Tν≺¯ξν,xnLtν,xn+ρxn,xn+1Ltxn,xn+1+ςxn+1,TνδLtxn,ν,which implies
(12)∣Ltν,Tν∣≤ξν,xn∣Ltν,xn∣+ρxn,xn+1∣Ltxn,xn+1∣+ςxn+1,TνδLtxn,ν.Therefore, in view of the assumptions in the theorem, asn⟶∞, we deduce that ∣Ltν,Tν∣=0 and that is Tν=ν as required.In closing, assume there exist two fixed points ofT, say ν and μ where ν≠μ. Thus,
(13)Ltν,μ=LtTν,Tμ≺¯δLtν,μ≺Ltν,μ,which is a contradiction. Therefore, the fixed point of T is unique.Theorem 2.
LetX,Lt be a Lt-complete CV-TCMS and T be a self mapping on X satisfying the following condition: for all a,b∈X, there exists 0<δ<1/2 such that
(14)LtTa,Tb≺¯δLta,Ta+Ltb,Tb,and there exists x0∈X in order that the sequence xn defined by xn=Tnx0 satisfies the following:
(15)limn⟶∞ξy,xn≤1δ,limn⟶∞ςxn,y<1δforanyy∈X,limn⟶∞ρxn,xn+1≤1δ.Then,T has a unique fixed point in X.Proof.
First of all, note that for alln≥1, we have
(16)Ltxn,xn+1≺¯δLtxn−1,xn+Ltxn,xn+1.Consequently,(17)Ltxn,xn+1≺¯δ1−δLtxn−1,xn.Since0<δ<1/2, one has 0<δ/1−δ<1. Set μ=δ/1−δ. One writes
(18)∣Ltxn,xn+1∣≤μ∣Ltxn−1,xn∣≤μ2∣Ltxn−2,xn−1∣≤⋯≤μnLtx0,x1.Therefore,(19)Ltxn,xn+1⟶0asn⟶∞.Also, for alln,m≥1, we have
(20)Ltxn,xm≺¯δLtxn−1,xn+Ltxm−1,xm.By (19), we deduce that ∣Ltxn,xm∣⟶0 as n,m⟶∞. Hence, xn is a Lt-Cauchy sequence. Since X,Lt is a Lt-complete CV-TCMS, the sequence xn converges to some ν∈X.By the argument of the proof of Theorem1, assume that for all n≥1, we have xn∈ν,Tν. Thus,
(21)Ltν,Tν≺¯ξν,xnLtν,xn+ρxn,xn+1Ltxn,xn+1+ςxn+1,TνLtxn+1,Tν≺¯ξν,xnLtν,xn+ρxn,xn+1Ltxn,xn+1+ςxn+1,TνδLtxn,Txn+Ltν,Tν.Asn⟶∞, we obtain
(22)∣Ltν,Tν∣≤0+0+0+limsupn⟶∞ςxn+1,Tνδ∣Ltν,Tν∣<Ltν,Tν.At the limitn⟶∞, we find that ∣Ltν,Tν∣=0 and that is Tν=ν as required. Now, assume that we have two fixed points of T, say ν and s. Therefore,
(23)∣Ltν,s∣=∣LtTν,Ts∣≤δ∣Ltν,ν∣+∣Lts,s∣=0.Hence,ν=s, as desired.


## 4. Applications

### 4.1. A Fredholm-Type Integral Equation

Consider the setX=C0,1,ℝ. Given the following Fredholm-type integral equation
(24)a′t=∫01Mt,s,a′tds,fort,s∈0,1,where Mt,s,a′t is a continuous function from 0,12 into ℝ. Now, define
(25)Lt:X×X⟶ℂa,b↦isupt∈0,1∣at∣+∣bt∣2.Note thatX,Lt is a complete CV-TCMS, where
(26)ξa,b=2,ρa,b=1andςa,b=3.Theorem 3.
Assume that for alla,b∈X(1)
∣Mt,s,a′t∣+∣Mt,s,bt∣≤δ∣a′t∣+∣bt∣, for some δ∈0,1/4(2)
Mt,s,∫01Mt,s,a′tds<Mt,s,a′t for all t,sThen, the above integral equation has a unique solution.Proof.
LetT:X⟶X be defined by Ta′t=∫01Mt,s,a′tds. Then,
(27)LtTa′,Tb=isupt∈0,1∣Ta′t∣+∣Tbt∣2.Now, we have(28)LtTa′t,Tbt=i∣Ta′t∣+∣Tbt∣2=i∣∫01Mt,s,a′tds∣+∣∫01Mt,s,btds∣2≺¯i∫01∣Mt,s,a′t∣ds+∫01∣Mt,s,bt∣ds2=i∫01∣Mt,s,a′t∣+∣Mt,s,bt∣ds2≺¯i∫01δ∣a′t∣+∣bt∣ds2≺¯δLta′t,bt.Thus,LtTa′,Tb≺¯δLta′,b. Since δ∈0,1/4, one gets
(29)ξa,b<1δ,ρa,b<1δ,ςa,b<1δ.Therefore, all the hypotheses of Theorem1 are satisfied, and hence, equation (24) has a unique solution.


### 4.2. A Polynomial Equation of a Degree Greater or Equal to 3

The following is an application on higher degree polynomial equations.Theorem 4.
For any natural numberβ≥3 and real ∣α∣≤1, the following equation
(30)αβ+1=β4−1αβ+1+β4αhas a unique real solution.Proof.
It is not difficult to see that if∣α∣>1, equation (30) does not have a solution. So, let X=−1,1 and for all α,r∈X, let Ltα,r=∣α−r∣+i∣α−r∣ and ξu,v=3,ρu,v=4 and ςu,v=maxu,v+2. Note that X,Lt is a Lt-complete CV-TCMS. Now, let
(31)Tα=αβ+1β4−1αβ+β4.Notice that, sinceβ≥2, we can deduce that β4≥6. Thus,
(32)LtTα,Tr=∣αβ+1β4−1αβ+β4−rβ+1β4−1rβ+β4∣+i∣αβ+1β4−1αβ+β4−rβ+1β4−1rβ+β4∣=∣αβ−rββ4−1αβ+β4β4−1rβ+β4∣+i∣αβ−rββ4−1αβ+β4β4−1rβ+β4∣≺¯∣α−r∣β4+i∣α−r∣β4≺¯∣α−r∣6+i∣α−r∣6=16Ltα,r.Hence,(33)LtTα,Tr≺¯δLtα,r,whereδ=16.Moreover, it is easy to see that for allα0∈X, we have
(34)αn=Tnα0≤2β4.Note that all the conditions of Theorem1 are satisfied. Thus, T possesses a unique fixed point in X, and equation (30) has a unique real solution.


## 4.1. A Fredholm-Type Integral Equation

Consider the setX=C0,1,ℝ. Given the following Fredholm-type integral equation
(24)a′t=∫01Mt,s,a′tds,fort,s∈0,1,where Mt,s,a′t is a continuous function from 0,12 into ℝ. Now, define
(25)Lt:X×X⟶ℂa,b↦isupt∈0,1∣at∣+∣bt∣2.Note thatX,Lt is a complete CV-TCMS, where
(26)ξa,b=2,ρa,b=1andςa,b=3.Theorem 3.
Assume that for alla,b∈X(1)
∣Mt,s,a′t∣+∣Mt,s,bt∣≤δ∣a′t∣+∣bt∣, for some δ∈0,1/4(2)
Mt,s,∫01Mt,s,a′tds<Mt,s,a′t for all t,sThen, the above integral equation has a unique solution.Proof.
LetT:X⟶X be defined by Ta′t=∫01Mt,s,a′tds. Then,
(27)LtTa′,Tb=isupt∈0,1∣Ta′t∣+∣Tbt∣2.Now, we have(28)LtTa′t,Tbt=i∣Ta′t∣+∣Tbt∣2=i∣∫01Mt,s,a′tds∣+∣∫01Mt,s,btds∣2≺¯i∫01∣Mt,s,a′t∣ds+∫01∣Mt,s,bt∣ds2=i∫01∣Mt,s,a′t∣+∣Mt,s,bt∣ds2≺¯i∫01δ∣a′t∣+∣bt∣ds2≺¯δLta′t,bt.Thus,LtTa′,Tb≺¯δLta′,b. Since δ∈0,1/4, one gets
(29)ξa,b<1δ,ρa,b<1δ,ςa,b<1δ.Therefore, all the hypotheses of Theorem1 are satisfied, and hence, equation (24) has a unique solution.


## 4.2. A Polynomial Equation of a Degree Greater or Equal to 3

The following is an application on higher degree polynomial equations.Theorem 4.
For any natural numberβ≥3 and real ∣α∣≤1, the following equation
(30)αβ+1=β4−1αβ+1+β4αhas a unique real solution.Proof.
It is not difficult to see that if∣α∣>1, equation (30) does not have a solution. So, let X=−1,1 and for all α,r∈X, let Ltα,r=∣α−r∣+i∣α−r∣ and ξu,v=3,ρu,v=4 and ςu,v=maxu,v+2. Note that X,Lt is a Lt-complete CV-TCMS. Now, let
(31)Tα=αβ+1β4−1αβ+β4.Notice that, sinceβ≥2, we can deduce that β4≥6. Thus,
(32)LtTα,Tr=∣αβ+1β4−1αβ+β4−rβ+1β4−1rβ+β4∣+i∣αβ+1β4−1αβ+β4−rβ+1β4−1rβ+β4∣=∣αβ−rββ4−1αβ+β4β4−1rβ+β4∣+i∣αβ−rββ4−1αβ+β4β4−1rβ+β4∣≺¯∣α−r∣β4+i∣α−r∣β4≺¯∣α−r∣6+i∣α−r∣6=16Ltα,r.Hence,(33)LtTα,Tr≺¯δLtα,r,whereδ=16.Moreover, it is easy to see that for allα0∈X, we have
(34)αn=Tnα0≤2β4.Note that all the conditions of Theorem1 are satisfied. Thus, T possesses a unique fixed point in X, and equation (30) has a unique real solution.


## 5. Conclusion

Finally, we would like to leave the following questions.Question 1.
LetX,Lt be a CV-TCMS and T:X⟶X. Given a function ς:X2⟶1,∞. Suppose there exists δ∈0,1 such that, for all s,r∈X,
(35)LtTs,Tr≺¯δςs,rLts,r.Under what conditions doesT have a unique fixed point in X?Question 2.
LetX,Lt be a CV-TCMS, and T:X⟶X. Given a function ς:X2⟶1,∞. Suppose there exists δ∈0,1/2 such that, for all s,r∈X,
(36)LtTs,Tr≺¯δςs,rLts,Ts+Ltr,Tr.Under what conditions doesT have a unique fixed point in X?



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