# On Ćirić-Prešić Operators in Metric Spaces

Authors: Narongsuk Boonsri; Satit Saejung; Kittipong Sitthikul
Journal: Journal of Function Spaces
 (2021)

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## Abstract

We show that the Prešić type operators of several variables can be regarded as an operator of a single variable and the fixed point problem of Prešić type can be regarded as a classical fixed point problem. We extend the recent result of Ćirić and Prešić by using the classical approach of Prešić. The key of the proof is based on the mappings introduced by Kada, Suzuki, and Takahashi. We also discuss the convergence problems of recursive real sequences and the Volterra integral equations as an application of our result.


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## Body

## 1. Introduction

LetX,d be a metric space and k≥1 be a fixed integer. Suppose that T:Xk⟶X is given. The following two problems were studied by Prešić [1]:(i)
Findx^∈X such that(1)x^=Tx^,x^,⋯,x^.(ii)
For any givenx1,x2,⋯,xk∈X, if we define(2)xn+k=Txn,xn+1,⋯,xn+k−1foralln≥1,then what can we say about the convergence of the sequence xn?This problem is very interesting and was further studied by many authors, for example, see [2–4].In 1965, Prešić [1] proved the following interesting theorem which can be regarded as a generalization of the classical fixed point theorem proposed by Banach [5]. Let k be a positive integer and let q1,q2,⋯,qk be nonnegative real numbers such that q1+q2+⋯+qk<1. Let Pq1,q2,⋯,qk be a family of mappings T:Xk⟶X such that(3)dTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤∑i=1kqidui,ui+1forallu1,u2,⋯,uk+1∈X.Theorem P1. Let X,d be a complete metric space. Let αnn=1∞ be a sequence of positive real numbers such that ∑n=1∞αn<∞ and liminfn⟶∞αn+1/αn=1. Let Tnn=1∞ be a sequence of operators in Pq1,q2,⋯,qk such that(4)dTnu1,u2,⋯,uk,Tn+1u1,u2,⋯,uk≤αnforallu1,u2,⋯,uk∈Xandforalln≥1.Let xnn=1∞ be a sequence in X such that x1,x2,⋯,xk∈X are arbitrary and(5)xn+k=Tnxn,xn+1,⋯,xn+k−1foralln≥1.Then, the following statements are true.(a)
There exists an elementx^∈X such that x^=limn⟶∞xn(b)
There exists an operatorT∈Pq1,q2,⋯,qk such that Tnn=1∞ converges to T uniformly; that is, limn⟶∞supx∈XdTnx,Tx=0(c)
The elementx^ is the only one satisfying Equation (1)Theorem P1, wherek=1, is nothing but Banach’s fixed point theorem [5].Remark 1.
It was shown in [6] that the condition liminfn⟶∞αn+1/αn=1 is superfluous.As a consequence of Theorem P1, we have the following result.Theorem P2. Let X,d be a complete metric space and let k be a positive integer. Let q1,q2,⋯,qk be nonnegative real numbers such that q1+q2+⋯+qk<1. Let T∈Pq1,q2,⋯,qk. Let xnn=1∞ be a sequence in X such that x1,x2,⋯,xk∈X are arbitrary and Equation (2) holds. Then, the following statements are true.(a)
There exists an elementx^∈X such that x^=limn⟶∞xn(b)
The elementx^ is the only one satisfying Equation (1).Ćirić and Prešić [2] proved the following improvement of Theorem P2.Theorem CP. Let X,d be a complete metric space. Let k be a positive integer and let q be a positive real number such that q<1. Let T:Xk⟶X be such that(6)dTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤qmaxdui,ui+1:i=1,2,⋯,kforallu1,u2,⋯,uk+1∈X.Let xnn=1∞ be a sequence in X such that x1,x2,⋯,xk∈X are arbitrary and Equation (2) holds. Then, the following statements are true.(a)
There exists an elementx^∈X such that x^=limn⟶∞xn and Equation (1) holds.(b)
If, in addition(7)dTu,u,⋯,u,Tv,v,⋯,v<du,vforallu,v∈X,then the element x^ is the only one satisfying Equation (1).Remark 2.
It is clear that ifT:Xk⟶X satisfies Expression (3), then it satisfies Expressions (6) and (7).Proof.
Suppose thatT:Xk⟶X satisfies Expression (3) with nonnegative constants q1,q2,⋯,qk such that q1+q2+⋯+qk<1. To show that T satisfies Expression (6) with q≔∑i=1kqi, let u1,u2,⋯,uk+1∈X. It follows that(8)dTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤∑i=1kqidui,ui+1≤qmaxdui,ui+1:i=1,2,⋯,k.
To see thatT satisfies Equation (7), let u,v∈X. It follows that(9)dTu,u,⋯,u,Tv,v,⋯,v≤dTu,u,⋯,u,Tv,u,⋯,u+dTv,u,⋯,u,Tv,v,u,⋯,u+⋯+dTv,v,⋯,v,u,Tv,v,⋯,v≤q1+q2+⋯+qkdu,v<du,v.
This completes the proof.☐It is natural to ask the following:Question:Is it possible to generalize Theorem CP by using the approach of Theorem P1?In this paper, we answer the question above by considering a wider class of mappings than those satisfying Expression (6). The class of mappings studied in this paper is motivated by the one in work of Kada et al. [7]. Some progress on these mappings can be found in [8].Theorem KST (see [7]). Let X,d be a complete metric space. Let r∈0,1 be given and let T:X⟶X be a mapping such that the following conditions hold:(1)
(KST1) dTx,T2x≤rdx,Tx for all x∈X(2)
(KST2) If y≠Ty, then infdx,Tx+dx,y:x∈X>0Then, every sequencexnn=1∞ with arbitrary x1 and xn+1≔Txn for all n≥1 converges to a fixed point of T.


## 2. Main Results

### 2.1. On the Fixed Point Problem of Prešić Type

We first show that the fixed point problem of Prešić type is equivalent to the classical fixed point problem. Suppose thatX is a nonempty set and k≥1 is an integer. The fixed point problem of Prešić type for a given mapping T:Xk⟶X is to find u∈X such that(10)u=Tu,u,⋯,u.We denote byPFixT the set of all solutions of the problem above. We show that this problem is connected to the classical fixed point problem. To simplify the notation of the following result, we write X≔Xk and ẋ≔x,x,⋯,x∈X where x∈X.Theorem 3.
Suppose thatX is a nonempty set and k≥1 is an integer. Suppose that T:Xk⟶X is given. Then, there exists a mapping T:X⟶X such that the following statements hold.(a)
Ifu∈PFixT, then u̇∈FixT(b)
Ifu1,u2,⋯,uk∈FixT, then u1=u2=⋯=uk∈PFixTProof.
DefineT:X⟶X by(11)Tu≔u2,⋯,uk,Tu1,⋯,uk,where u≔u1,⋯,uk∈X.
The statement (a) is trivial. We prove the statement (b). Suppose thatu1,u2,⋯,uk∈FixT. It follows that(12)u1,u2,⋯,uk=u2,⋯,uk,Tu1,⋯,uk.
This implies that(13)u1=u2=⋯=uk=Tu1,⋯,uk.
This completes the proof.☐


### 2.2. On the Class of Mappings

In this subsection, we discuss the following classes of mappings. The first two classes are from Theorem P and Theorem CP and the last one from Theorem KST.Suppose thatX≔X,d is a metric space and k≥1 is an integer. Suppose that r,q,q1,⋯,qk are nonnegative real numbers. Define(14)FX,d,q1,⋯,qk≔T:Xk⟶Xsatisfies3,GX,d,q≔T:Xk⟶Xsatisfies6,HX,d,r≔T:X⟶XsatisfiesConditionsKST1andKST2.Remark 4.
Suppose thatX≔X,d is a metric space and k≥1 is an integer. The following statements are true.(a)
Ifq1,⋯,qk are nonnegative real numbers, then FX,d,q1,⋯,qk⊂GX,d,q, where q≔∑i=1kqi(b)
Ifq is a nonnegative real number and T∈GX,d,q, then T∈HX,d,r, where X≔Xk, r=q1/k, T is defined by Equation (11) and(15)dx,y≔maxq−i/kdxi,yi:i=1,⋯,kforallx≔x1,⋯,xk∈Xandy≔y1,⋯,yk∈X.Proof.
The statement (a) is trivial. We prove the statement (b). LetT∈GX,d,q. It is clear that d is a metric on X. We now prove that T∈HX,d,r. To see that T satisfies Condition (KST1), let u≔u1,⋯,uk∈X. We write uk+1≔Tu1,⋯,uk and uk+2≔Tu2,⋯,uk+1. It follows from T∈GX,d,q that(16)duk+1,uk+2≤qmaxdui,ui+1:i=1,⋯,k.
Note that(17)Tu=u2,⋯,uk+1,T2u=u3,⋯,uk+2.
Putr≔1/s∈0,1 where s=q−1/k. It follows from Equation (6) that(18)dTu,T2u=maxsidui+1,ui+2:i=1,⋯,k=maxmaxsidui+1,ui+2:i=1,⋯,k−1,skduk+1,uk+2=maxmaxsidui+1,ui+2:i=1,⋯,k−1,skdTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤maxmaxsidui+1,ui+2:i=1,⋯,k−1,skqmaxdui,ui+1:i=1,⋯,k=maxmaxsidui+1,ui+2:i=1,⋯,k−1,maxdui,ui+1:i=1,⋯,k=maxmaxsidui+1,ui+2:i=1,⋯,k−1,du1,u2,maxdui,ui+1:i=2,⋯,k=maxdu1,u2,maxsidui+1,ui+2:i=1,⋯,k−1=1smaxsdu1,u2,smaxsidui+1,ui+2:i=1,⋯,k−1=rmaxsdu1,u2,maxsidui,ui+1:i=2,⋯,k=rmaxsidui,ui+1:i=1,⋯,k=rmaxq−i/kdui,ui+1:i=1,⋯,k=rdu,Tu.
Finally, we show thatT satisfies Condition (KST2). To see this, let un be a sequence in X and let u∈X be such that limn⟶∞dun,Tun=limn⟶∞dun,u=0. For each n∈ℕ, we write un≔u1n,u2n,⋯,ukn and u≔u1,u2,⋯,uk. It follows from the definition of d that(19)limn⟶∞duin,ui+1n=limn⟶∞duin,ui=0,foralli=1,2,⋯,k,whereuk+1n≔Tu1n,u2n,⋯,uknforalln≥1.In particular, we have u1=u2=⋯=uk. This implies that u=u1,u1,⋯,u1 and limn⟶∞du1,Tu1n,u2n,⋯,ukn=0. We consider(20)dTu1n,u2n,⋯,ukn,Tu1,u1,⋯,u1≤dTu1n,u2n,⋯,ukn,Tu2n,u3n,⋯,ukn,u1+dTu2n,u3n,⋯,ukn,u1,Tu3n,⋯,ukn,u1,u1+⋯+dTukn,u1,⋯,u1,Tu1,u1,⋯,u1≤qmaxdukn,u1,duin,ui+1n:i=1,⋯,k−1+qmaxdukn,u1,duin,ui+1n:i=2,⋯,k−1+⋯+qdukn,u1.
Hence,limn⟶∞dTu1n,u2n,⋯,ukn,Tu1,u1,⋯,u1=0. Since(21)limn⟶∞dTu1n,u2n,⋯,ukn,u1=limn⟶∞duk+1n,u1=0,we have u1=Tu1,u1,⋯,u1. Hence, u=Tu. This completes the proof.☐Remark 5.
The classesFX,d,q1,⋯,qk and GX,d,q are closed under the pointwise convergence, that is, if Tn is a sequence in FX,d,q1,⋯,qk (GX,d,q, respectively) and there exists a mapping T:Xk⟶X such that(22)limn⟶∞dTnu1,u2,⋯,uk,Tu1,u2,⋯,uk=0forallu1,u2,⋯,uk∈Xk,then T∈FX,d,q1,⋯,qk (T∈GX,d,q, respectively).Proof.
Suppose thatTn is a sequence in GX,d,q and there exists a mapping T:X⟶X such that(23)limn⟶∞dTnu1,u2,⋯,uk,Tu1,u2,⋯,uk=0forallu1,u2,⋯,uk∈Xk.
We prove thatT∈GX,d,q. To see this, let u1,u2,⋯,uk,uk+1∈X. Note that(24)limn⟶∞dTu1,u2,⋯,uk,Tnu1,u2,⋯,uk=limn⟶∞dTnu2,u3,⋯,uk+1,Tu2,u3,⋯,uk+1=0.
It follows that(25)dTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤dTu1,u2,⋯,uk,Tnu1,u2,⋯,uk+dTnu1,u2,⋯,uk,Tnu2,u3,⋯,uk+1+dTnu2,u3,⋯,uk+1,Tu2,u3,⋯,uk+1≤dTu1,u2,⋯,uk,Tnu1,u2,⋯,uk+qmaxdui,ui+1:i=1,2,⋯,k+dTnu2,u3,⋯,uk+1,Tu2,u3,⋯,uk+1.
In particular,(26)dTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤qmaxdui,ui+1:i=1,2,⋯,k.
This implies thatT∈GX,d,q. The case that Tn is a sequence in FX,d,q1,q2,⋯,qk can be done similarly.☐


### 2.3. A Generalization of Theorem KST and Its Consequences

The following result is analogous to Theorem P1 with a wider class of mappings.Theorem 6.
LetX,d be a complete metric space and r∈0,1 be given. Let αnn=1∞ be a sequence of positive real numbers such that ∑n=1∞αn<∞. Let Tnn=1∞ be a sequence of operators in HX,d,r such that(27)dTnx,Tn+1x≤αnforallx∈X.Let xnn=1∞ be a sequence in X such that x1∈X is arbitrary and xn+1≔Tnxn for all n≥1. Then, the following statements are true.(a)
There exists an elementx^∈X such that x^=limn⟶∞xn(b)
There exists an operatorT:X⟶X such that Tn converges to T uniformly(c)
IfT satisfies Condition (KST2), then x^∈FixTLemma 7.
Letsnn=1∞ and αnn=1∞ be two sequences of nonnegative real numbers such that ∑n=1∞αn<∞. Let r∈0,1 be given. If(28)sn+1≤rsn+αnforalln≥1,then ∑n=1∞sn<∞.Proof.
Note that, for eachm≥1, we have ∑n=2m+1sn≤r∑n=1msn+∑n=1mαn. In particular, 1−r∑n=2msn≤1−r∑n=2msn+sm+1≤rs1+∑n=1mαn≤rs1+∑n=1∞αn for all m≥2. Hence, the conclusion follows.☐Proof of Theorem 6.
First, we note that(29)dxn+1,xn+2=dTnxn,Tn+1xn+1≤dTnxn,Tnxn+1+dTnxn+1,Tn+1xn+1=dTnxn,Tn2xn+dTnxn+1,Tn+1xn+1≤rdxn,Tnxn+αn=rdxn,xn+1+αn.
It follows from Lemma7 that ∑n=1∞dxn,xn+1<∞. In particular, xn is a Cauchy sequence and hence limn⟶∞xn=x^ for some x^∈X by the completeness of X. Hence, (a) is proved.
We now prove (b). For eachx∈X and for each n,k≥1, we note that(30)dTnx,Tn+kx≤∑j=nn+k−1dTjx,Tj+1x≤∑j=nn+k−1αj≤∑j=n∞αj.
It follows thatTnxn=1∞ is a Cauchy sequence and hence limn⟶∞Tnx exists. We then define T:X⟶X by(31)Tx≔limn⟶∞Tnx.
It follows then that(32)dTnx,Tx≤limk⟶∞dTnx,Tn+kx≤limk⟶∞∑j=nn+k−1dTjx,Tj+1x≤limk⟶∞∑j=nn+k−1αj=∑j=n∞αjforallx∈Xandforalln≥1.To see that limn⟶∞supx∈XdTnx,Tx=0, let ε>0. Since ∑n=1∞αn<∞, there exists an integer M such that ∑j=n∞αj<ε for all n≥M. For each x∈X and n≥M, we have(33)dTnx,Tx≤∑j=n∞αj<ε.
That is,supx∈XdTnx,Tx≤ε for all n≥M. Hence,(34)limn⟶∞supx∈XdTnx,Tx=0.
Finally, we assume thatT satisfies (KST2). Note that limn⟶∞dxn,x^=0. Moreover,(35)limn⟶∞dxn,Txn≤limn⟶∞dxn,Tnxn+dTnxn,Txn≤limn⟶∞dxn,xn+1+limn⟶∞∑j=n∞αj=0.
It follows from Condition (KST2) thatx^=Tx^.We are now ready to give an affirmative answer of the problem in the introduction. In fact, we can generalize Theorem CP by using the approach of Theorem P1.Theorem 8.
LetX,d be a complete metric space and k≥1 be a fixed integer. Let q be a positive real number such that q<1. Let αnn=1∞ be a sequence of positive real numbers such that ∑n=1∞αn<∞. Let Tnn=1∞ be a sequence of operators in GX,d,q such that(36)dTnu1,u2,⋯,uk,Tn+1u1,u2,⋯,uk≤αnforallu1,u2,⋯,uk∈Xandforalln≥1.Let xnn=1∞ be a sequence in X such that x1,x2,⋯,xk∈X are arbitrary and(37)xn+k≔Tnxn,xn+1,⋯,xn+k−1foralln≥1.
Then, the following statements are true.(a)
There exists an elementx^∈X such that x^=limn⟶∞xn(b)
There exists an operatorT∈GX,d,q such that Tnn=1∞ converges uniformly to T and x^∈PFixT(c)
PFixT=x^, that is, x^ is the only solution of the fixed point problem of Prešić type for T, provided that(38)dTu,u,⋯,u,Tv,v,⋯,v<du,vforallu,v∈X.Proof.
Suppose thatαn, Tn, and xn are given as in the statement of the theorem. Let X and d be defined as in Theorem 3 and Remark 4. For each n≥1, we define Tn:X⟶X by(39)Tnu≔u2,⋯,uk,Tnu1,⋯,uk,where u≔u1,⋯,uk∈X. By Remark 4, we have Tn∈HX,d,r for all n≥1. Define xnn=1∞ be a sequence in X by(40)xn≔xn,xn+1,⋯,xn+k−1foralln≥1.
We prove thatxn+1≔Tnxn for each n≥1. To see this, we have(41)xn+1=xn+1,xn+2,⋯,xn+k=xn+1,xn+2,⋯,Tnxn,xn+1,⋯,xn+k−1=Tnxn.
Next, we prove that(42)dTnu,Tn+1u≤αnqforalln≥1andu∈X.
In fact, for eachn≥1 and u≔u1,⋯,uk∈X, we have(43)dTnu,Tn+1u=du2,u3,⋯,uk,Tnu1,⋯,uk,u2,u3,⋯,uk,Tn+1u1,⋯,uk=q−1dTnu1,⋯,uk,Tn+1u1,⋯,uk≤αnq.
It follows from Theorem6 that there exists an element x^≔x^1,x^2,⋯,x^k∈X such that x^=limn⟶∞xn. In particular, we have limn⟶∞xn=x^1=x^2=⋯=x^k. It also follows from Theorem 6 that there exists an operator S:X⟶X such that Tn converges uniformly to S. Note that, for each u1,u2,⋯,uk∈X, we have(44)Su1,u2,⋯,uk=limn⟶∞Tnu1,u2,⋯,uk=limn⟶∞u2,u3,⋯,uk,Tnu1,u2,⋯,uk=u2,u3,⋯,uk,limn⟶∞Tnu1,u2,⋯,uk.
We now defineT:X⟶X (from S) by(45)Tu1,u2,⋯,uk≔vk⟺defSu1,u2,⋯,uk≔v1,v2,⋯,vk.
In particular, we have(i)
Tu1,u2,⋯,uk=limn⟶∞Tnu1,u2,⋯,uk,(ii)
Su1,u2,⋯,uk=u2,u3,⋯,uk,Tu1,u2,⋯,uk for all u1,u2,⋯,uk∈X.
It follows from eachTn∈GX,d,q and Remark 5 that T∈GX,d,q. Hence S∈HX,d,r. In particular, S satisfies Condition KST2 and hence x^=Sx^ by Theorem 6. This implies that(46)x^1,x^1,⋯,x^1=Sx^1,x^1⋯,x^1=x^1,x^1,⋯,x^1,Tx^1,x^1⋯,x^1.
Hence,(47)Tx^1,x^1⋯,x^1=x^1.
The uniqueness of the fixed point of Prešić type is obvious if the additional hypothesis is assumed.☐


## 2.1. On the Fixed Point Problem of Prešić Type

We first show that the fixed point problem of Prešić type is equivalent to the classical fixed point problem. Suppose thatX is a nonempty set and k≥1 is an integer. The fixed point problem of Prešić type for a given mapping T:Xk⟶X is to find u∈X such that(10)u=Tu,u,⋯,u.We denote byPFixT the set of all solutions of the problem above. We show that this problem is connected to the classical fixed point problem. To simplify the notation of the following result, we write X≔Xk and ẋ≔x,x,⋯,x∈X where x∈X.Theorem 3.
Suppose thatX is a nonempty set and k≥1 is an integer. Suppose that T:Xk⟶X is given. Then, there exists a mapping T:X⟶X such that the following statements hold.(a)
Ifu∈PFixT, then u̇∈FixT(b)
Ifu1,u2,⋯,uk∈FixT, then u1=u2=⋯=uk∈PFixTProof.
DefineT:X⟶X by(11)Tu≔u2,⋯,uk,Tu1,⋯,uk,where u≔u1,⋯,uk∈X.
The statement (a) is trivial. We prove the statement (b). Suppose thatu1,u2,⋯,uk∈FixT. It follows that(12)u1,u2,⋯,uk=u2,⋯,uk,Tu1,⋯,uk.
This implies that(13)u1=u2=⋯=uk=Tu1,⋯,uk.
This completes the proof.☐


## 2.2. On the Class of Mappings

In this subsection, we discuss the following classes of mappings. The first two classes are from Theorem P and Theorem CP and the last one from Theorem KST.Suppose thatX≔X,d is a metric space and k≥1 is an integer. Suppose that r,q,q1,⋯,qk are nonnegative real numbers. Define(14)FX,d,q1,⋯,qk≔T:Xk⟶Xsatisfies3,GX,d,q≔T:Xk⟶Xsatisfies6,HX,d,r≔T:X⟶XsatisfiesConditionsKST1andKST2.Remark 4.
Suppose thatX≔X,d is a metric space and k≥1 is an integer. The following statements are true.(a)
Ifq1,⋯,qk are nonnegative real numbers, then FX,d,q1,⋯,qk⊂GX,d,q, where q≔∑i=1kqi(b)
Ifq is a nonnegative real number and T∈GX,d,q, then T∈HX,d,r, where X≔Xk, r=q1/k, T is defined by Equation (11) and(15)dx,y≔maxq−i/kdxi,yi:i=1,⋯,kforallx≔x1,⋯,xk∈Xandy≔y1,⋯,yk∈X.Proof.
The statement (a) is trivial. We prove the statement (b). LetT∈GX,d,q. It is clear that d is a metric on X. We now prove that T∈HX,d,r. To see that T satisfies Condition (KST1), let u≔u1,⋯,uk∈X. We write uk+1≔Tu1,⋯,uk and uk+2≔Tu2,⋯,uk+1. It follows from T∈GX,d,q that(16)duk+1,uk+2≤qmaxdui,ui+1:i=1,⋯,k.
Note that(17)Tu=u2,⋯,uk+1,T2u=u3,⋯,uk+2.
Putr≔1/s∈0,1 where s=q−1/k. It follows from Equation (6) that(18)dTu,T2u=maxsidui+1,ui+2:i=1,⋯,k=maxmaxsidui+1,ui+2:i=1,⋯,k−1,skduk+1,uk+2=maxmaxsidui+1,ui+2:i=1,⋯,k−1,skdTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤maxmaxsidui+1,ui+2:i=1,⋯,k−1,skqmaxdui,ui+1:i=1,⋯,k=maxmaxsidui+1,ui+2:i=1,⋯,k−1,maxdui,ui+1:i=1,⋯,k=maxmaxsidui+1,ui+2:i=1,⋯,k−1,du1,u2,maxdui,ui+1:i=2,⋯,k=maxdu1,u2,maxsidui+1,ui+2:i=1,⋯,k−1=1smaxsdu1,u2,smaxsidui+1,ui+2:i=1,⋯,k−1=rmaxsdu1,u2,maxsidui,ui+1:i=2,⋯,k=rmaxsidui,ui+1:i=1,⋯,k=rmaxq−i/kdui,ui+1:i=1,⋯,k=rdu,Tu.
Finally, we show thatT satisfies Condition (KST2). To see this, let un be a sequence in X and let u∈X be such that limn⟶∞dun,Tun=limn⟶∞dun,u=0. For each n∈ℕ, we write un≔u1n,u2n,⋯,ukn and u≔u1,u2,⋯,uk. It follows from the definition of d that(19)limn⟶∞duin,ui+1n=limn⟶∞duin,ui=0,foralli=1,2,⋯,k,whereuk+1n≔Tu1n,u2n,⋯,uknforalln≥1.In particular, we have u1=u2=⋯=uk. This implies that u=u1,u1,⋯,u1 and limn⟶∞du1,Tu1n,u2n,⋯,ukn=0. We consider(20)dTu1n,u2n,⋯,ukn,Tu1,u1,⋯,u1≤dTu1n,u2n,⋯,ukn,Tu2n,u3n,⋯,ukn,u1+dTu2n,u3n,⋯,ukn,u1,Tu3n,⋯,ukn,u1,u1+⋯+dTukn,u1,⋯,u1,Tu1,u1,⋯,u1≤qmaxdukn,u1,duin,ui+1n:i=1,⋯,k−1+qmaxdukn,u1,duin,ui+1n:i=2,⋯,k−1+⋯+qdukn,u1.
Hence,limn⟶∞dTu1n,u2n,⋯,ukn,Tu1,u1,⋯,u1=0. Since(21)limn⟶∞dTu1n,u2n,⋯,ukn,u1=limn⟶∞duk+1n,u1=0,we have u1=Tu1,u1,⋯,u1. Hence, u=Tu. This completes the proof.☐Remark 5.
The classesFX,d,q1,⋯,qk and GX,d,q are closed under the pointwise convergence, that is, if Tn is a sequence in FX,d,q1,⋯,qk (GX,d,q, respectively) and there exists a mapping T:Xk⟶X such that(22)limn⟶∞dTnu1,u2,⋯,uk,Tu1,u2,⋯,uk=0forallu1,u2,⋯,uk∈Xk,then T∈FX,d,q1,⋯,qk (T∈GX,d,q, respectively).Proof.
Suppose thatTn is a sequence in GX,d,q and there exists a mapping T:X⟶X such that(23)limn⟶∞dTnu1,u2,⋯,uk,Tu1,u2,⋯,uk=0forallu1,u2,⋯,uk∈Xk.
We prove thatT∈GX,d,q. To see this, let u1,u2,⋯,uk,uk+1∈X. Note that(24)limn⟶∞dTu1,u2,⋯,uk,Tnu1,u2,⋯,uk=limn⟶∞dTnu2,u3,⋯,uk+1,Tu2,u3,⋯,uk+1=0.
It follows that(25)dTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤dTu1,u2,⋯,uk,Tnu1,u2,⋯,uk+dTnu1,u2,⋯,uk,Tnu2,u3,⋯,uk+1+dTnu2,u3,⋯,uk+1,Tu2,u3,⋯,uk+1≤dTu1,u2,⋯,uk,Tnu1,u2,⋯,uk+qmaxdui,ui+1:i=1,2,⋯,k+dTnu2,u3,⋯,uk+1,Tu2,u3,⋯,uk+1.
In particular,(26)dTu1,u2,⋯,uk,Tu2,u3,⋯,uk+1≤qmaxdui,ui+1:i=1,2,⋯,k.
This implies thatT∈GX,d,q. The case that Tn is a sequence in FX,d,q1,q2,⋯,qk can be done similarly.☐


## 2.3. A Generalization of Theorem KST and Its Consequences

The following result is analogous to Theorem P1 with a wider class of mappings.Theorem 6.
LetX,d be a complete metric space and r∈0,1 be given. Let αnn=1∞ be a sequence of positive real numbers such that ∑n=1∞αn<∞. Let Tnn=1∞ be a sequence of operators in HX,d,r such that(27)dTnx,Tn+1x≤αnforallx∈X.Let xnn=1∞ be a sequence in X such that x1∈X is arbitrary and xn+1≔Tnxn for all n≥1. Then, the following statements are true.(a)
There exists an elementx^∈X such that x^=limn⟶∞xn(b)
There exists an operatorT:X⟶X such that Tn converges to T uniformly(c)
IfT satisfies Condition (KST2), then x^∈FixTLemma 7.
Letsnn=1∞ and αnn=1∞ be two sequences of nonnegative real numbers such that ∑n=1∞αn<∞. Let r∈0,1 be given. If(28)sn+1≤rsn+αnforalln≥1,then ∑n=1∞sn<∞.Proof.
Note that, for eachm≥1, we have ∑n=2m+1sn≤r∑n=1msn+∑n=1mαn. In particular, 1−r∑n=2msn≤1−r∑n=2msn+sm+1≤rs1+∑n=1mαn≤rs1+∑n=1∞αn for all m≥2. Hence, the conclusion follows.☐Proof of Theorem 6.
First, we note that(29)dxn+1,xn+2=dTnxn,Tn+1xn+1≤dTnxn,Tnxn+1+dTnxn+1,Tn+1xn+1=dTnxn,Tn2xn+dTnxn+1,Tn+1xn+1≤rdxn,Tnxn+αn=rdxn,xn+1+αn.
It follows from Lemma7 that ∑n=1∞dxn,xn+1<∞. In particular, xn is a Cauchy sequence and hence limn⟶∞xn=x^ for some x^∈X by the completeness of X. Hence, (a) is proved.
We now prove (b). For eachx∈X and for each n,k≥1, we note that(30)dTnx,Tn+kx≤∑j=nn+k−1dTjx,Tj+1x≤∑j=nn+k−1αj≤∑j=n∞αj.
It follows thatTnxn=1∞ is a Cauchy sequence and hence limn⟶∞Tnx exists. We then define T:X⟶X by(31)Tx≔limn⟶∞Tnx.
It follows then that(32)dTnx,Tx≤limk⟶∞dTnx,Tn+kx≤limk⟶∞∑j=nn+k−1dTjx,Tj+1x≤limk⟶∞∑j=nn+k−1αj=∑j=n∞αjforallx∈Xandforalln≥1.To see that limn⟶∞supx∈XdTnx,Tx=0, let ε>0. Since ∑n=1∞αn<∞, there exists an integer M such that ∑j=n∞αj<ε for all n≥M. For each x∈X and n≥M, we have(33)dTnx,Tx≤∑j=n∞αj<ε.
That is,supx∈XdTnx,Tx≤ε for all n≥M. Hence,(34)limn⟶∞supx∈XdTnx,Tx=0.
Finally, we assume thatT satisfies (KST2). Note that limn⟶∞dxn,x^=0. Moreover,(35)limn⟶∞dxn,Txn≤limn⟶∞dxn,Tnxn+dTnxn,Txn≤limn⟶∞dxn,xn+1+limn⟶∞∑j=n∞αj=0.
It follows from Condition (KST2) thatx^=Tx^.We are now ready to give an affirmative answer of the problem in the introduction. In fact, we can generalize Theorem CP by using the approach of Theorem P1.Theorem 8.
LetX,d be a complete metric space and k≥1 be a fixed integer. Let q be a positive real number such that q<1. Let αnn=1∞ be a sequence of positive real numbers such that ∑n=1∞αn<∞. Let Tnn=1∞ be a sequence of operators in GX,d,q such that(36)dTnu1,u2,⋯,uk,Tn+1u1,u2,⋯,uk≤αnforallu1,u2,⋯,uk∈Xandforalln≥1.Let xnn=1∞ be a sequence in X such that x1,x2,⋯,xk∈X are arbitrary and(37)xn+k≔Tnxn,xn+1,⋯,xn+k−1foralln≥1.
Then, the following statements are true.(a)
There exists an elementx^∈X such that x^=limn⟶∞xn(b)
There exists an operatorT∈GX,d,q such that Tnn=1∞ converges uniformly to T and x^∈PFixT(c)
PFixT=x^, that is, x^ is the only solution of the fixed point problem of Prešić type for T, provided that(38)dTu,u,⋯,u,Tv,v,⋯,v<du,vforallu,v∈X.Proof.
Suppose thatαn, Tn, and xn are given as in the statement of the theorem. Let X and d be defined as in Theorem 3 and Remark 4. For each n≥1, we define Tn:X⟶X by(39)Tnu≔u2,⋯,uk,Tnu1,⋯,uk,where u≔u1,⋯,uk∈X. By Remark 4, we have Tn∈HX,d,r for all n≥1. Define xnn=1∞ be a sequence in X by(40)xn≔xn,xn+1,⋯,xn+k−1foralln≥1.
We prove thatxn+1≔Tnxn for each n≥1. To see this, we have(41)xn+1=xn+1,xn+2,⋯,xn+k=xn+1,xn+2,⋯,Tnxn,xn+1,⋯,xn+k−1=Tnxn.
Next, we prove that(42)dTnu,Tn+1u≤αnqforalln≥1andu∈X.
In fact, for eachn≥1 and u≔u1,⋯,uk∈X, we have(43)dTnu,Tn+1u=du2,u3,⋯,uk,Tnu1,⋯,uk,u2,u3,⋯,uk,Tn+1u1,⋯,uk=q−1dTnu1,⋯,uk,Tn+1u1,⋯,uk≤αnq.
It follows from Theorem6 that there exists an element x^≔x^1,x^2,⋯,x^k∈X such that x^=limn⟶∞xn. In particular, we have limn⟶∞xn=x^1=x^2=⋯=x^k. It also follows from Theorem 6 that there exists an operator S:X⟶X such that Tn converges uniformly to S. Note that, for each u1,u2,⋯,uk∈X, we have(44)Su1,u2,⋯,uk=limn⟶∞Tnu1,u2,⋯,uk=limn⟶∞u2,u3,⋯,uk,Tnu1,u2,⋯,uk=u2,u3,⋯,uk,limn⟶∞Tnu1,u2,⋯,uk.
We now defineT:X⟶X (from S) by(45)Tu1,u2,⋯,uk≔vk⟺defSu1,u2,⋯,uk≔v1,v2,⋯,vk.
In particular, we have(i)
Tu1,u2,⋯,uk=limn⟶∞Tnu1,u2,⋯,uk,(ii)
Su1,u2,⋯,uk=u2,u3,⋯,uk,Tu1,u2,⋯,uk for all u1,u2,⋯,uk∈X.
It follows from eachTn∈GX,d,q and Remark 5 that T∈GX,d,q. Hence S∈HX,d,r. In particular, S satisfies Condition KST2 and hence x^=Sx^ by Theorem 6. This implies that(46)x^1,x^1,⋯,x^1=Sx^1,x^1⋯,x^1=x^1,x^1,⋯,x^1,Tx^1,x^1⋯,x^1.
Hence,(47)Tx^1,x^1⋯,x^1=x^1.
The uniqueness of the fixed point of Prešić type is obvious if the additional hypothesis is assumed.☐


## 3. Applications

We finally discuss some applications of our result.


### 3.1. Some Convergence Problem of Recursive Real Sequences

We reconsider the following example studied by Ćirić and Prešić [2, Example 1] and give some remark.Example 9.
LetX≔0,1∪2,3 be a metric space endowed with the usual metric d and T:X2⟶X be defined by(48)Tx,y≔x+y4ifx,y∈0,12,1+x+y4ifx,y∈2,32,x+y4−12ifx,y∈0,1×2,3∪2,3×1,2.
It was proved in [2] that(49)dTx,y,Ty,z≤12maxdx,y,dy,zforallx,y∈Xandz≔Tx,y.The author of [2] claimed that T∈GX,d,1/2. We note that T∉GX,d,q for all 0<q<1. In fact, let x≔3, y≔2, and z≔1; it follows that(50)dTx,y,Ty,z=d94,14=2whilemaxdx,y,dy,z=1.
Using our approach, letX≔X2 and(51)dx,y≔max2dx1,x2,2dy1,y2,where x≔x1,x2∈X and y≔y1,y2∈X. Moreover, let T:X⟶X be defined by(52)Tx≔x2,Tx1,x2,where x≔x1,x2∈X. We can follow the proof in Remark 4 to show that(53)dTx,T2x≤12dx,Txforallx∈X.
Next, we show thatT satisfies Condition (KST2). Note that PFixT=0,2 so FixT=0,0,2,2. Let un be a sequence in X and let u∈X be such that limn⟶∞dun,Tun=limn⟶∞dun,u=0. For each n∈ℕ, we write un≔u1n,u2n and u≔u1,u2. It follows from the definition of d that(54)limn⟶∞du1n,u2n=limn⟶∞du2n,Tu1n,u2n=limn⟶∞du1n,u1=limn⟶∞du2n,u2=0.
In particular, we haveu1=u2. This implies that u=u1,u1 and limn⟶∞du1,Tu1n,u2n=0. We consider the following two cases.Case 1.
u1∈0,1. In this case, we may assume that u1n and u2n are sequences in 0,1 and hence Tu1n,u2n=u1n+u2n/4. This implies that u1=u1+u2/4=u1/2 and hence u1=0. Then, u≔u1,u2=u1,u1=0,0∈FixT.Case 2.
u1∈2,3. In this case, we may assume that u1n and u2n are sequences in 2,3 and hence Tu1n,u2n=1+u1n+u2n/4. This implies that u1=1+u1+u2/4=1+u1/2 and hence u1=2. Then u≔u1,u2=u1,u1=2,2∈FixT.Hence,T∈HX,d,1/2. In particular, we can apply this example to our Theorem 8. Note that, since PFixT=0,2, the condition (7) cannot be omitted for the uniqueness of the solution as claimed in [2].We now discuss the following convergence problem of real sequences inspired by [6].Example 10.
Suppose thatxn is a real sequence satisfying the following recursive relation: x1,x2∈0,1∪2,3 and for each n≥1(55)xn+2≔xn+xn+14ifxn,xn+1∈0,12,1+xn+xn+14ifxn,xn+1∈2,32,xn+xn+14−12ifxn,xn+1∈0,1×2,3∪2,3×0,1.
It is clear that this example is related to the preceding one. Table1 shows the numerical experiment regarding to the problem with respect to the initial inputs x1 and x2. For example, if x1≔0.5 and x2≔1, then limn⟶∞xn=0; if x1≔2 and x2≔3, then limn⟶∞xn=2.Table 1
A numerical experiment for Example10.
x10.51123x211233x30.3750.50.252.252.5x40.343750.3750.06252.31252.375x50.17968750.218750.0781252.1406252.21875x60.1308593750.14843750.035156252.113281252.1484375x70.0776367190.0917968750.0283203132.0634765632.091796875x80.0521240230.0600585940.0158691412.0441894532.060058594x90.0324401860.0379638670.0110473632.0269165042.037963867x100.0211410520.0245056150.0067291262.0177764892.024505615x110.0133953090.0156173710.0044441222.0111732482.015617371x120.008634090.0100307460.0027933122.0072374342.010030746x130.005507350.0064120290.0018093592.0046026712.006412029x140.003535360.0041106940.0011506682.0029600262.004110694x150.0022606780.0026306810.0007400072.0018906742.002630681x160.0014490090.0016853440.0004726692.0012126752.001685344x170.0009274220.0010790060.0003031692.0007758372.001079006x180.0005941080.0006910870.0001939592.0004971282.000691087x190.0003803820.0004425230.0001242822.0003182412.000442523x200.0002436230.0002834037.95603E−052.0002038422.000283403


### 3.2. Volterra Type Integral Equations

We discuss a further application of our Theorem8 in the context of Volterra type integral equations.Theorem 12.
Suppose thatX≔C0,T,ℝ is the space of continuous real-valued functions defined on an interval 0,T, where T>0, equipped with the supremum metric d defined by(56)du,v≔suput−vt:t∈0,Tforallu,v∈X.
Suppose thatλ,μ:0,T×ℝ⟶0,∞ and f,g:0,T×ℝ⟶ℝ are continuous functions such that(i)
f and g satisfy the Lipschitz condition for the second argument with constants α and β, respectively, that is,(57)ft,s−ft,s′≤αs−s′and∣gt,s−gt,s′∣≤β∣s−s′∣forallt∈0,Tandforalls,s′∈ℝ,(ii)
q≔αmaxt∈0,T∫0Tλt,sds+βmaxt∈0,T∫0Tμt,sds<1.
Suppose thathn is a sequence in X such that ∑n=1∞αn<∞ where αn≔dhn,hn+1 for all n≥1. If u1,u2∈X and for each n≥1(58)un+2t≔hnt+∫0Tλt,sfs,unsds+∫0Tμt,sgs,un+1sdst∈0,T,then un∈X for all n≥3 and there exists an element u^∈X such that limn⟶∞dun,u^=0 and u^ is a unique solution of the following Volterra type integral equation(59)ut=ht+∫0Tλt,sfs,usds+∫0Tμt,sgs,usdst∈0,T,where ht≔limn⟶∞hnt.Proof.
Now, for eachn≥1, we define Fn:X×X⟶X, by for each u,v∈X,(60)Fnu,vt≔hnt+∫0Tλt,sfs,usds+∫0Tμt,sgs,vsdst∈0,T.
To apply our Theorem8, it is sufficient to prove that(a)
Fn∈GX,d,q for all n≥1(b)
dFnu1,u2,Fn+1u1,u2=αn for all n≥1 and for all u1,u2∈X
We now prove the statements (a) and (b).(a)
Letn≥1 and let u1,u2,u3∈X. Then,(61)dFnu1,u2,Fnu2,u3=supt∈0,T∣Fnu1,u2t−Fnu2,u3t∣≤supt∈0,T∫0Tλt,s∣fs,u1s−fs,u2s∣ds+supt∈0,T∫0Tμt,s∣gs,u2s−gs,u3s∣ds≤αmaxt∈0,T∫0Tλt,sdsdu1,u2+βmaxt∈0,T∫0Tμt,sdsdu2,u3≤qmaxdu1,u2,du2,u3.(b)
Letn≥1 and let u1,u2∈X. Then,(62)dFnu1,u2,Fn+1u1,u2=supt∈0,T∣Fnu1,u2t−Fn+1u1,u2t∣=supt∈0,T∣hnt−hn+1t∣=dhn,hn+1=αn.
Note that(63)un+2t=Fnunt,un+1tt∈0,T.
It follows from our Theorem8, where k=2, there exists an element u^∈X such that(64)limn⟶∞dun,u^=0.
Moreover,(65)u^=Fu^,u^⟺u^isasolutionof11,where F:X×X⟶X is defined by(66)Fu,vt≔ht+∫0Tλt,sfs,usds+∫0Tμt,sgs,vsdst∈0,T,and ht=limn⟶∞hnt. Finally, to show that u^ is a unique solution of the Volterra type integral Equation (59), we show that dFu,u,Fv,v<du,v for all u,v∈X. Let u,v∈X. Then,(67)dFu,u,Fv,v=supt∈0,T∣Fu,ut−Fv,vt∣≤supt∈0,T∫0Tλt,s∣fs,us−fs,vs∣ds+supt∈0,T∫0Tμt,s∣gs,us−gs,vs∣ds≤αmaxt∈0,T∫0Tλt,sdsdu,v+βmaxt∈0,T∫0Tμt,sdsdu,v=qdu,v<du,v.
This completes the proof.☐


## 3.1. Some Convergence Problem of Recursive Real Sequences

We reconsider the following example studied by Ćirić and Prešić [2, Example 1] and give some remark.Example 9.
LetX≔0,1∪2,3 be a metric space endowed with the usual metric d and T:X2⟶X be defined by(48)Tx,y≔x+y4ifx,y∈0,12,1+x+y4ifx,y∈2,32,x+y4−12ifx,y∈0,1×2,3∪2,3×1,2.
It was proved in [2] that(49)dTx,y,Ty,z≤12maxdx,y,dy,zforallx,y∈Xandz≔Tx,y.The author of [2] claimed that T∈GX,d,1/2. We note that T∉GX,d,q for all 0<q<1. In fact, let x≔3, y≔2, and z≔1; it follows that(50)dTx,y,Ty,z=d94,14=2whilemaxdx,y,dy,z=1.
Using our approach, letX≔X2 and(51)dx,y≔max2dx1,x2,2dy1,y2,where x≔x1,x2∈X and y≔y1,y2∈X. Moreover, let T:X⟶X be defined by(52)Tx≔x2,Tx1,x2,where x≔x1,x2∈X. We can follow the proof in Remark 4 to show that(53)dTx,T2x≤12dx,Txforallx∈X.
Next, we show thatT satisfies Condition (KST2). Note that PFixT=0,2 so FixT=0,0,2,2. Let un be a sequence in X and let u∈X be such that limn⟶∞dun,Tun=limn⟶∞dun,u=0. For each n∈ℕ, we write un≔u1n,u2n and u≔u1,u2. It follows from the definition of d that(54)limn⟶∞du1n,u2n=limn⟶∞du2n,Tu1n,u2n=limn⟶∞du1n,u1=limn⟶∞du2n,u2=0.
In particular, we haveu1=u2. This implies that u=u1,u1 and limn⟶∞du1,Tu1n,u2n=0. We consider the following two cases.Case 1.
u1∈0,1. In this case, we may assume that u1n and u2n are sequences in 0,1 and hence Tu1n,u2n=u1n+u2n/4. This implies that u1=u1+u2/4=u1/2 and hence u1=0. Then, u≔u1,u2=u1,u1=0,0∈FixT.Case 2.
u1∈2,3. In this case, we may assume that u1n and u2n are sequences in 2,3 and hence Tu1n,u2n=1+u1n+u2n/4. This implies that u1=1+u1+u2/4=1+u1/2 and hence u1=2. Then u≔u1,u2=u1,u1=2,2∈FixT.Hence,T∈HX,d,1/2. In particular, we can apply this example to our Theorem 8. Note that, since PFixT=0,2, the condition (7) cannot be omitted for the uniqueness of the solution as claimed in [2].We now discuss the following convergence problem of real sequences inspired by [6].Example 10.
Suppose thatxn is a real sequence satisfying the following recursive relation: x1,x2∈0,1∪2,3 and for each n≥1(55)xn+2≔xn+xn+14ifxn,xn+1∈0,12,1+xn+xn+14ifxn,xn+1∈2,32,xn+xn+14−12ifxn,xn+1∈0,1×2,3∪2,3×0,1.
It is clear that this example is related to the preceding one. Table1 shows the numerical experiment regarding to the problem with respect to the initial inputs x1 and x2. For example, if x1≔0.5 and x2≔1, then limn⟶∞xn=0; if x1≔2 and x2≔3, then limn⟶∞xn=2.Table 1
A numerical experiment for Example10.
x10.51123x211233x30.3750.50.252.252.5x40.343750.3750.06252.31252.375x50.17968750.218750.0781252.1406252.21875x60.1308593750.14843750.035156252.113281252.1484375x70.0776367190.0917968750.0283203132.0634765632.091796875x80.0521240230.0600585940.0158691412.0441894532.060058594x90.0324401860.0379638670.0110473632.0269165042.037963867x100.0211410520.0245056150.0067291262.0177764892.024505615x110.0133953090.0156173710.0044441222.0111732482.015617371x120.008634090.0100307460.0027933122.0072374342.010030746x130.005507350.0064120290.0018093592.0046026712.006412029x140.003535360.0041106940.0011506682.0029600262.004110694x150.0022606780.0026306810.0007400072.0018906742.002630681x160.0014490090.0016853440.0004726692.0012126752.001685344x170.0009274220.0010790060.0003031692.0007758372.001079006x180.0005941080.0006910870.0001939592.0004971282.000691087x190.0003803820.0004425230.0001242822.0003182412.000442523x200.0002436230.0002834037.95603E−052.0002038422.000283403


## 3.2. Volterra Type Integral Equations

We discuss a further application of our Theorem8 in the context of Volterra type integral equations.Theorem 12.
Suppose thatX≔C0,T,ℝ is the space of continuous real-valued functions defined on an interval 0,T, where T>0, equipped with the supremum metric d defined by(56)du,v≔suput−vt:t∈0,Tforallu,v∈X.
Suppose thatλ,μ:0,T×ℝ⟶0,∞ and f,g:0,T×ℝ⟶ℝ are continuous functions such that(i)
f and g satisfy the Lipschitz condition for the second argument with constants α and β, respectively, that is,(57)ft,s−ft,s′≤αs−s′and∣gt,s−gt,s′∣≤β∣s−s′∣forallt∈0,Tandforalls,s′∈ℝ,(ii)
q≔αmaxt∈0,T∫0Tλt,sds+βmaxt∈0,T∫0Tμt,sds<1.
Suppose thathn is a sequence in X such that ∑n=1∞αn<∞ where αn≔dhn,hn+1 for all n≥1. If u1,u2∈X and for each n≥1(58)un+2t≔hnt+∫0Tλt,sfs,unsds+∫0Tμt,sgs,un+1sdst∈0,T,then un∈X for all n≥3 and there exists an element u^∈X such that limn⟶∞dun,u^=0 and u^ is a unique solution of the following Volterra type integral equation(59)ut=ht+∫0Tλt,sfs,usds+∫0Tμt,sgs,usdst∈0,T,where ht≔limn⟶∞hnt.Proof.
Now, for eachn≥1, we define Fn:X×X⟶X, by for each u,v∈X,(60)Fnu,vt≔hnt+∫0Tλt,sfs,usds+∫0Tμt,sgs,vsdst∈0,T.
To apply our Theorem8, it is sufficient to prove that(a)
Fn∈GX,d,q for all n≥1(b)
dFnu1,u2,Fn+1u1,u2=αn for all n≥1 and for all u1,u2∈X
We now prove the statements (a) and (b).(a)
Letn≥1 and let u1,u2,u3∈X. Then,(61)dFnu1,u2,Fnu2,u3=supt∈0,T∣Fnu1,u2t−Fnu2,u3t∣≤supt∈0,T∫0Tλt,s∣fs,u1s−fs,u2s∣ds+supt∈0,T∫0Tμt,s∣gs,u2s−gs,u3s∣ds≤αmaxt∈0,T∫0Tλt,sdsdu1,u2+βmaxt∈0,T∫0Tμt,sdsdu2,u3≤qmaxdu1,u2,du2,u3.(b)
Letn≥1 and let u1,u2∈X. Then,(62)dFnu1,u2,Fn+1u1,u2=supt∈0,T∣Fnu1,u2t−Fn+1u1,u2t∣=supt∈0,T∣hnt−hn+1t∣=dhn,hn+1=αn.
Note that(63)un+2t=Fnunt,un+1tt∈0,T.
It follows from our Theorem8, where k=2, there exists an element u^∈X such that(64)limn⟶∞dun,u^=0.
Moreover,(65)u^=Fu^,u^⟺u^isasolutionof11,where F:X×X⟶X is defined by(66)Fu,vt≔ht+∫0Tλt,sfs,usds+∫0Tμt,sgs,vsdst∈0,T,and ht=limn⟶∞hnt. Finally, to show that u^ is a unique solution of the Volterra type integral Equation (59), we show that dFu,u,Fv,v<du,v for all u,v∈X. Let u,v∈X. Then,(67)dFu,u,Fv,v=supt∈0,T∣Fu,ut−Fv,vt∣≤supt∈0,T∫0Tλt,s∣fs,us−fs,vs∣ds+supt∈0,T∫0Tμt,s∣gs,us−gs,vs∣ds≤αmaxt∈0,T∫0Tλt,sdsdu,v+βmaxt∈0,T∫0Tμt,sdsdu,v=qdu,v<du,v.
This completes the proof.☐


## 4. Conclusion

We show that the fixed point problem of Prešić type (with respect to several variables) can be regarded a fixed point problem (of a single variable) by using a product space approach. With an appropriate metric on the product space, the Ćirić-Prešić operator can be regarded a mapping studied by Kada et al. in the product space. In particular, we deduce the fixed point result under a weaker assumption. We apply our result for the convergence problems of recursive real sequences and the Volterra type integral equations.



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*Source: 5758032-2021-07-31.xml*