# Some Generalized Fixed Point Results with Applications to Dynamic Programming

Authors: Hamed H. Al-Sulami; Nawab Hussain; Jamshaid Ahmad
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The aim of this paper is to introduce some generalized contractions and prove certain new fixed point results for self-mappings satisfying these contractions in the setting ofF-metric space. As an application of our results, we investigate the problem of dynamic programming related to the multistage process which formulates the problems of computer programming and mathematical optimization. We also provide an example to support the validity of our main results.


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## Body

## 1. Introduction

Because of utility and applications of metric fixed point theory in mathematics and related fields like social sciences, physical sciences, computer sciences, and engineering have grown in many directions. More general theorems have been introduced along with the provision of many useful tools to solve problems arising in several diverse areas of research.Recently, Jlei and Samet [1] initiated a generalized metric space named as F-metric space and showed a generalization of the Banach contraction principle. Meanwhile, researchers have picked keen interests in extending results in this generalized metric space; see for instance, [2–5]. In this paper, we define some generalized contractions and establish some results in the context of F-metric spaces.


## 2. Preliminaries

Here, we record some requisites definitions and results for the purpose of the next sections.Definition 1 (see [1]).
LetF be the class of functions f:0,+∞⟶ℝ satisfying these assertions:
F10<s<ι=⟹fs≤fι
F2∀ιn⊆R+, limn→∞ιn=0⇔limn→∞fιn=−∞Example 2.
Iff:0,+∞⟶ℝ are defined as.
(1)
fϱ=lnϱ(2)
fϱ=ϱ+lnϱ(3)
fϱ=−1/ϱ(4)
fϱ=lnϱ2+ϱfor ϱ>0, then f∈F.Definition 3 (see [1]).
LetM≠∅, and dF:M×M⟶0,+∞. Assume that ∃f,τ∈F×0,+∞ such that
(D1) for all ζ,ξ∈M, dFζ,ξ=0⇔ζ=ξ
(D2) dFζ,ξ=dFξ,ζ, for all ζ,ξ∈M×M
(D3) for every ζ,ξ∈M×M, for every N∈ℕ, N≥2, and wii=1N⊂M, with w1,wN=ζ,ξ, we have
(1)dFζ,ξ>0⟹fdFζ,ξ≤f∑i=1N−1dFζi,ζi+1+τ.Then,M,dF is called an F-metric space.Example 4 (see [1]).
LetdF:ℝ×ℝ⟶ℝ+ be defined by
(2)dFζ,ξ=ζ−ξ2ifζ,ξ∈0,3×0,3,ζ−ξifζ,ξ∈0,3×0,3,with fϱ=lnϱ and τ=ln3. Then, (ℝ,dF) is an F-metric space.Theorem 5.
[1].
LetM,dF be an F-metric space and J:M⟶M. Assume that these conditions are satisfied:
(i)
M,dF is F-complete(ii)
∃0<λ<1 such that(3)dFJζ,Jξ≤λdFζ,ξ.Then,∃ζ∗∈M such that Jζ∗=ζ∗. Furthermore, for any ζ0∈M, ζn⊂M defined by
(4)ζn+1=Jζn,n∈ℕ,is F-convergent to ζ∗.Afterwards, many researchers [2–9] worked in this space.In this article, we give the notions of twistedα,β-admissible and twisted α,β−φ-rational contractions in the setting of F-metric spaces and prove some new theorems.


## 3. Main Results

In 2012, Samet et al. [10] introduced the concepts of α-admissibility mappings and α-φ-contraction in complete metric spaces.Definition 6 (see [10]).
LetJ:M⟶M and α:M×M⟶0,+∞. Then, J is said to be α-admissible if ζ,ξ∈M,
(5)αζ,ξ≥1⟹αJζ,Jξ≥1.According to Samet et al. [10], Ψ represents the class of all nondecreasing functions φ:0,+∞⟶0,+∞ such that ∑n=1∞φnι<+∞ for all ι>0, where φn is the nth iterate of φ.Lemma 7 (see [10]).
Ifφ∈Ψ, then
(i)
φnιn∈ℕ⟶0 as n⟶∞,∀ι∈0,+∞(ii)
φι<ι for all ι>0(iii)
φι=0 iff ι=0Now, we give the concept of twistedα,β-admissible in the F-metric space as follows.Definition 8.
LetM,dF be an F-metric space, J:M⟶M and α,β:M×M⟶0,+∞. Then, J is said to be twisted α,β-admissible if
(6)αζ,ξ≥1βζ,ξ≥1⟹αJζ,Jξ≥1,βJζ,Jξ≥1,for ζ,ξ∈M.Now, we state our main result.Theorem 9.
LetM,dF be an F-metric space and J:M⟶M be twisted α,β-admissible. Suppose that the following assertions are satisfied:
(a)
M,dF is F-complete(b)
there existsζ0∈M such that αζ0,Jζ0≥1 and βζ0,Jζ0≥1(c)
J is continuous. If any one of these assertions hold:(7)αζ,ξβζ,ξdFJζ,Jξ≤φRζ,ξ(i)
∃0<l≤1 such that(8)αζ,ξβζ,ξ+ldFJζ,Jξ≤1+lφRζ,ξ(ii)
∃l≥1 such that(9)dFJζ,Jξ+lαζ,ξβζ,ξ≤φRζ,ξ+lwhere
(10)Rζ,ξ=maxdFζ,ξ,dFζ,JζdFξ,Jξ1+dFζ,ξ,for all ζ,ξ∈M, then ∃ζ∗∈M such that Jζ∗=ζ∗.Proof.
Letζ0∈M such that αζ0,Jζ0≥1 and βζ0,Jζ0≥1. Generate ζn in M by ζn+1=Jζn,∀n∈ℕ. If ζn+1=ζn for some n∈ℕ, then ζ∗=ζn is a fixed point of J. So we suppose that ζn+1≠,∀n∈ℕ. Then as J is twisted α,β-admissible, we get αζ0,ζ1=αζ0,Jζ0≥1 implies αζ1,ζ2=αJζ0,Jζ1≥1 and βζ1,ζ2=βJζ0,Jζ1≥1. By induction, we get αζn,ζn+1≥1 and βζn,ζn+1≥1 for all n∈ℕ. Suppose the inequality (7) holds. So with ζ=ζn−1 and ξ=ζn, we have
(11)dFζn,ζn+1=dFJζn−1,Jζn≤αζn−1,ζnβζn−1,ζndFJζn−1,Jζn≤φRζn−1,ζn,where
(12)Rζn−1,ζn=maxdFζn−1,ζn,dFζn−1,Jζn−1dFζn,Jζn1+dFζn−1,ζn=maxdFζn−1,ζn,dFζn−1,ζndFζn,ζn+11+dFζn−1,ζn,
IfmaxdFζn−1,ζn,dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn=dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn, then from (11), we obtain
(13)dFζn,ζn+1≤φdFζn−1,ζndFζn,ζn+11+dFζn−1,ζn<dFζn−1,ζndFζn,ζn+11+dFζn−1,ζn≤dFζn,ζn+1,a contradiction. Hence, maxdFζn−1,ζn,dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn=dFζn−1,ζn; therefore, (11) becomes
(14)dFζn,ζn+1≤φdFζn−1,ζn.
Consequently, we get(15)dFζn,ζn+1≤φndFζ0,ζ1,
∀n∈ℕ.
Assume inequality (8) holds and ∃0<l≤1 such that
(16)1+ldFζn,ζn+1≤αζn−1,ζnβζn−1,ζn+ldFJζn−1,Jζn≤1+lφRζn−1,ζn,which implies that
(17)dFζn,ζn+1≤φRζn−1,ζn,where
(18)Rζn−1,ζn=maxdFζn−1,ζn,dFζn−1,Jζn−1dFζn,Jζn1+dFζn−1,ζn=maxdFζn−1,ζn,dFζn−1,ζndFζn,ζn+11+dFζn−1,ζn.
IfmaxdFζn−1,ζn,dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn=dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn, then from (17), we obtain
(19)dFζn,ζn+1≤φdFζn−1,ζndFζn,ζn+11+dFζn−1,ζn<dFζn−1,ζndFζn,ζn+11+dFζn−1,ζn≤dFζn,ζn+1,a contradiction. Thus, maxdFζn−1,ζn,dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn=dFζn−1,ζn; therefore, (17) becomes
(20)dFζn,ζn+1≤φdFζn−1,ζn.
Consequently, we get(21)dFζn,ζn+1≤φndFζ0,ζ1.
Assume inequality (9) holds and ∃l≤1 such that
(22)dFζn,ζn+1+l≤dFJζn−1,Jζn+lαζn−1,ζnβζn−1,ζn≤φRζn−1,ζn+l,which implies that
(23)dFζn,ζn+1≤φRζn−1,ζn,where
(24)Rζn−1,ζn=maxdFζn−1,ζn,dFζn−1,Jζn−1dFζn,Jζn1+dFζn−1,ζn=maxdFζn−1,ζn,dFζn−1,ζndFζn,ζn+11+dFζn−1,ζn,
IfmaxdFζn−1,ζn,dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn=dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn, then from (23), we obtain
(25)dFζn,ζn+1≤φdFζn−1,ζndFζn,ζn+11+dFζn−1,ζn<dFζn−1,ζndFζn,ζn+11+dFζn−1,ζn≤dFζn,ζn+1,
a contradiction. Thus,maxdFζn−1,ζn,dFζn−1,ζndF(ζn,ζn+1)/1+dFζn−1,ζn=dFζn−1,ζn; therefore, (23) becomes
(26)dFζn,ζn+1≤φdFζn−1,ζn.
Consequently, we get(27)dFζn,ζn+1≤φndFζ0,ζ1,
Letf,τ∈F×0,+∞ be such that (D3) is satisfied. Let ε>0 be fixed. By (F2), ∃δ>0 such that
(28)0<ι<δ⟹fι<fδ−τ.
Letnε∈ℕ such that 0<∑n≥nεφndFζ0,ζ1<δ. Hence, by (27), F1, and (F2), we have
(29)f∑i=nm−1dFζi,ζi+1≤f∑i=nm−1φidFζ0,ζ1≤f∑n≥nεφndFζ0,ζ1<fε−τ,for m>n≥nε. Using (D3) and (29), we get dFζn,ζm>0,m>n≥nε implies
(30)fdFζn,ζm≤f∑i=nm−1dFζi,ζi+1+τ<fε,which implies by (F1) that dFζn,ζm<ε,m>n≥nε. This shows that ζn is F-Cauchy. As M,dF is F-complete, ∃ζ∗∈M such that ζn is F-convergent to ζ∗. As J is continuous, so we have Jζ∗=limn→∞Jζn≪limn→∞ζn+1=ζ∗. Thus, ∃ζ∗∈M such that Jζ∗=ζ∗.In the next result, we omit the continuity ofJ and use an adjunctive condition on M.Theorem 10.
LetM,dF be an F-metric space and J:M⟶M be twisted α,β-admissible. Suppose that the following assertions are satisfied:
(a)
M,dF is F-complete(b)
∃ζ0∈M such that αζ0,Jζ0≥1 and βζ0,Jζ0≥1(c)
Ifζn is a sequence in M such that αζn,ζn+1≥1 and βζn,ζn+1≥1,∀n and ζn⟶ζ∗∈M as n⟶∞, then αζn,ζ∗≥1 and βζn,ζ∗≥1,∀n∈ℕ
If any one of these assertions hold:(31)αζ,ξβζ,ξdFJζ,Jξ≤φRζ,ξ,(i)
∃0<l≤1 such that(32)αζ,ξβζ,ξ+ldFJζ,Jξ≤1+lφRζ,ξ(ii)
∃l≥1 such that(33)dFJζ,Jξ+lαζ,ξβζ,ξ≤φRζ,ξ+lwhere
(34)Rζ,ξ=maxdFζ,ξ,dFζ,JζdFξ,Jξ1+dFζ,ξ,for all ζ,ξ∈M, then ∃ζ∗∈M such that Jζ∗=ζ∗.Proof.
Letζ0∈M such that αζ0,Jζ0≥1 and βζ0,Jζ0≥1. Proceeding as in the proof of Theorem 9, we have ζ∗∈M such that ζn is F-convergent to ζ∗, i.e.,
(35)limn→∞dFζn,ζ∗=0.
Suppose thatdFJζ∗,ζ∗>0 and inequality (31) holds. By F1 and (D3), we have
(36)dFJζ∗,ζn+1=dFJζ∗,Jζn≤αζ∗,ζnβζ∗,ζndFJζ∗,Jζn≤φmaxdFζ∗,ζn,dFζ∗,Jζ∗dFζn,Jζn1+dFζ∗,ζn.
Similarly, if inequality (32) holds. So ∃0<l≤1 such that
(37)1+ldFJζ∗,ζn+1=1+ldFJζ∗,Jζn≤αζ∗,ζnβζ∗,ζn+ldFJζ∗,Jζn≤1+lφRζ∗,ζn,which implies that
(38)dFJζ∗,ζn+1≤φRζ∗,ζn,that is
(39)dFJζ∗,ζn+1≤φmaxdFζ∗,ζn,dFζ∗,Jζ∗dFζn,Jζn1+dFζ∗,ζn.
Also if inequality (33) holds, then ∃l≥1 such that
(40)dFJζ∗,ζn+1+l≤dFJζ∗,Jζn+lαζ∗,ζnβζ∗,ζn≤φRζ∗,ζn+ldFJζ∗,ζn+1≤φRζ∗,ζn,that is
(41)dFJζ∗,ζn+1≤φmaxdFζ∗,ζn,dFζ∗,Jζ∗dFζn,Jζn1+dFζ∗,ζn.
Thus, for all cases, byF1 and (D3), we have
(42)fdFJζ∗,ζ∗≤fdFJζ∗,Jζn+dFJζn,ζ∗+τ≤fφmaxdFζ∗,ζn,dFζ∗,Jζ∗dFζn,Jζn1+dFζ∗,ζn+dFζn+1,ζ∗+τ,<fmaxdFζ∗,ζn,dFζ∗,Jζ∗dFζn,ζn+11+dFζ∗,ζn+dFζn+1,ζ∗+τ,for n∈ℕ. If maxdFζ∗,ζn,dFζ∗,Jζ∗dFζn,ζn+1/1+dFζ∗,ζn=dFζ∗,ζn, then
(43)fdFJζ∗,ζ∗≤fdFζ∗,ζn+dFζn+1,ζ∗+τ,
Lettingn⟶∞ and utilizing (F2) and (35), we get
(44)limn→∞fdFJζ∗,ζ∗≤limn→∞fdFζ∗,ζn+dFζn+1,ζ∗+τ=−∞,which implies that dFJζ∗,ζ∗=0, a contradiction.
IfmaxdFζ∗,ζn,dF(ζ∗,Jζ∗)dF(ζn,ζn+1)/1+dFζ∗,ζn=dF(ζ∗,Jζ∗)dF(ζn,ζn+1)/1+dFζ∗,ζn, then
(45)fdFJζ∗,ζ∗≤fdFζ∗,Jζ∗dFζn,ζn+11+dFζ∗,ζn+dFζn+1,ζ∗+τ.
Lettingn⟶∞ and utilizing (F2) and (35), we get
(46)limn→∞fdFJζ∗,ζ∗≤limn→∞fdFζ∗,Jζ∗dFζn,ζn+11+dFζ∗,ζn+dFζn+1,ζ∗+τ=−∞,which implies that dFJζ∗,ζ∗=0, a contradiction. Thus, we have dFJζ∗,ζ∗=0, i.e., Jζ∗=ζ∗.For the uniqueness of the fixed point, we take the following property:(P)αζ,ξ≥1 and βζ,ξ≥1 for all fixed points ζ,ξ∈M of JTheorem 11.
If we add the property (P) in supposition of Theorem10,then we get that the fixed point of the mapping J is unique.Proof.
Letζ∗,ζ^∈M be such that Jζ∗=ζ∗ and Jζ^=ζ^ such that ζ∗=ζ^. Then, by hypothesis (P), we have αζ∗,ζ^≥1 and βζ∗,ζ^≥1. Suppose (i) holds. Then,
(47)dFζ∗,ζ^=dFJζ∗,Jζ^≤αζ∗,ζ^βζ∗,ζ^dFJζ∗,Jζ^≤φmaxdFζ∗,ζ^,dFζ∗,Jζ∗dFζ^,Jζ^1+dFζ∗,ζ^=φdFζ∗,ζ^<dFζ∗,ζ^,a contradiction. Hence, ∃ζ∗∈M such that Jζ∗=ζ∗. Suppose (ii) holds. Then, there exists 0<l≤1 such that
(48)αζ∗,ζ^βζ∗,ζ^+ldFζ∗,ζ^=αζ∗,ζ^βζ∗,ζ^+ldFJζ∗,Jζ^≤1+lφRζ∗,ζ^,which implies that
(49)dFζ∗,ζ^≤φmaxdFζ∗,ζ^,dFζ∗,Jζ∗dFζ^,Jζ^1+dFζ∗,ζ^=φdFζ∗,ζ^<dFζ∗,ζ^,a contradiction. Hence, ∃ζ∗∈M such that Jζ∗=ζ∗. Suppose (iii) holds. Then, there exists l≥1 such that
(50)dFζ∗,ζ^+l≤dFJζ∗,Jζ^+lαζ∗,ζ^βζ∗,ζ^≤φRζ∗,ζ^+l,which implies that
(51)dFζ∗,ζ^≤φmaxdFζ∗,ζ^,dFζ∗,Jζ∗dFζ^,Jζ^1+dFζ∗,ζ^=φdFζ∗,ζ^<dFζ∗,ζ^,a contradiction. Hence, ∃ζ∗∈M such that Jζ∗=ζ∗.Corollary 12.
LetM,dF be an F-metric space and J:M⟶M be α-admissible. Assume that these assertions hold:
(a)
M,dF is F-complete(b)
∃ζ0∈M such that αζ0,Jζ0≥1(c)
J is continuous or if ζn is a sequence in M such that αζn,ζn+1≥1 for all n and ζn⟶ζ∗∈M as n⟶∞, then αζn,ζ∗≥1,∀n∈ℕ. If any one of these assertions hold:(52)αζ,ξdFJζ,Jξ≤φRζ,ξ(i)
∃0<l≤1 such that(1)
(53)αζ,ξ+ldFJζ,Jξ≤1+lφRζ,ξ(ii)
∃l≥1 such that(2)
(54)dFJζ,Jξ+lαζ,ξ≤φRζ,ξ+lwhere
(55)Rζ,ξ=maxdFζ,ξ,dFζ,JζdFξ,Jξ1+dFζ,ξ
∀ζ,ξ∈M, then ∃ζ∗∈M such that Jζ∗=ζ∗.Proof.
Takingβζ,ξ=1 for all ζ,ξ∈M in Theorem 10.Corollary 13.
LetM,dF be an F-metric space and J:M⟶M be α-admissible mapping such that
(56)αζ,ξdFJζ,Jξ≤φRζ,ξ,where
(57)Rζ,ξ=maxdFζ,ξ,dFζ,JζdFξ,Jξ1+dFζ,ξ,
∀ζ,ξ∈M. Suppose that the following assertions are satisfied:
(a)
M,dF is F-complete(b)
∃ζ0∈M such that αζ0,Jζ0≥1(c)
J is continuous or if ζn is a sequence in M such that αζn,ζn+1≥1 for all n and ζn⟶ζ∗∈M as n⟶∞, then αζn,ζ∗≥1,∀n∈ℕ. Then ∃ζ∗∈M such that Jζ∗=ζ∗Proof.
If only (i) holds in Corollary12.Corollary 14.
LetM,dF be an F-metric space and J:M⟶M be α,β-admissible mapping such that
(58)αζ,ξβζ,ξdFJζ,Jξ≤φdFζ,ξ,∀ζ,ξ∈M. Assume that these assertions hold:
(a)
M,dF is F-complete(b)
∃ζ0∈M such that αζ0,Jζ0≥1 and βζ0,Jζ0≥1(c)
J is continuous or if ζn is a sequence in M such that αζn,ζn+1≥1 and βζn,ζn+1≥1,∀n and ζn⟶ζ∗∈M as n⟶∞, then αζn,ζ∗≥1 and βζn,ζ∗≥1,∀n∈ℕ. Then, ∃ζ∗∈M such that Jζ∗=ζ∗Proof.
If only hypothesis (i) holds in Theorem10 and maxdFζ,ξ,dF(ζ,Jζ)dF(ξ,Jξ)/1+dFζ,ξ=dFζ,ξ.Following is Boyd and Wong type in the setting ofF-metric space which is a consequence of Corollary 14.Corollary 15.
LetM,dF be an F-metric space and J:M⟶M be a self-mapping such that
(59)dFJζ,Jξ≤φdFζ,ξ
∀ζ,ξ∈M. Assume that these assertions hold:
(a)
M,dF is F-complete(b)
J is continuous. Then, ∃ζ∗∈M such that Jζ∗=ζ∗Proof.
Takingαζ,ξ=βζ,ξ=1 for all ζ,ξ∈M in Corollary 14.Example 16.
LetM=ℝand F-metric dF given by
(60)dFζ,ξ=2ζ−ξ,ifζ≠ξ,0,ifζ=ξ.
Takefϱ=−1/ϱ and τ=1. Define J:M⟶M by
(61)Jζ=2ζ,ifζ>1,ζ2,if0≤ζ≤1,0,otherwise.
Now, we defineα,β:M×M⟶0,+∞ by
(62)αζ,ξ=βζ,ξ=1ifζ,ξ∈0,1,0,ifζ,ξ∈0,1.
Evidently,J is twisted α,β-φ rational contraction of type (i) with φι=kι,∀ι≥0 and k∈0,1. Actually, ∀ζ,ξ∈M, we have
(63)dFJζ,Jξ≤kmaxdFζ,ξ,dFζ,JζdFξ,Jξ1+dFζ,ξ.All the conditions of Theorem9 are satisfied. Hence, J0=0 which is unique.


## 4. Applications in Dynamic Programming

In this section, we now establish the solution of functional equations arising from dynamic programming related to multistage process [11, 12] as an application of Theorem 10. Recall that a dynamic programming problem is a decision-making problem in n variables in which the problem being subdivided into n subproblems (stages), each being a decision-making problem in one variable only. The decision is the “goodness” of a selected alternative depending on satisfying the optimal policy of the problem. The state of the system at any stage is regarded as the information that links the stages together, such that the optimal decisions for the remaining stages can be made. The state allows us to consider each stage separately and guarantees that the solution is feasible for all the stages. This setting formulates the problems of mathematical optimization and computer programming which are converted into the problems of functional equations
(64)μζ=supξ∈Dfζ,ξ+Kζ,ξ,μgζ,ξ,ζ∈W,where f:W×D⟶ℝ and K:W×D×ℝ⟶ℝ and g:W×D→W,W and D are Banach spaces, precisely, W is a state space and D is a decision space.LetBW represent the set of all bounded real-valued functions defined on W. For h∈BW, consider
(65)h=supζ∈Whζ.Clearly, (BW,h) is a Banach space. We endow BW with the F-metric (with fϱ=lnϱ and a=0) defined by
(66)dFh,k=supι∈Whι−kι.Evidently, (BW,dF) is a F-complete F-metric space. To show the existence of a solution of (64), we define J:BW⟶BW by
(67)Jhζ=supξ∈Dfζ,ξ+Kζ,ξ,ζgζ,ξ,∀h∈BW and ζ∈W. Clearly, J is well-defined as f and K are bounded.Theorem 17.
LetJ:BW⟶BW is given by (67) and assume that these assertions are satisfied:
Assume that there existsΘ:BW×BW⟶ℝ such that
(i)
Θh,k≥0⇒ΘJk,Jh≥0, where h,k∈BW(ii)
Kζ,ξ,hζ−Kζ,ξ,kζ≤φhζ−kζ, where φ∈Ψ, h,k∈BW, Θh,k≥0, ζ∈W and ξ∈D(iii)
ifhn is a sequence in BW such that Θhn,hn+1≥0 for all n∈ℕ∪0 and hn⟶h∗ as n⟶+∞, then Θhn,h∗≥0 for all n∈ℕ∪0(iv)
there existsh0∈BW such that Θh0,Jh0≥0Then, functional equation (64) has a unique and bounded solution.Proof.
Note thatBW,dF is a F-complete F- metric space. Let ε be an arbitrary positive number and h1,h2∈BW such that Θh1,h2≥0, then there exist ξ1,ξ2∈D such that
(68)Jh1ζ<fζ,ξ1+Kζ,ξ1,h1gζ,ξ1+ε,(69)Jh2ζ<fζ,ξ2+Kζ,ξ2,h2gζ,ξ2+ε,(70)Jh2ζ≥fζ,ξ2+Kζ,ξ2,h2gζ,ξ2,(71)Jh1ζ≥fζ,ξ1+Kζ,ξ1,h1gζ,ξ1.
Now, from (68) and (71), it follows easily that
(72)Jh1ζ−Jh2ζ<Kζ,ξ1,h1gζ,ξ1−Kζ,ξ1,h2gζ,ξ1+≤Kζ,ξ1,h1gζ,ξ1−Kζ,ξ1,h2gζ,ξ1+≤φh1ζ−h2ζ+ε.
Hence, we get(73)Jh1ζ−Jh2ζ<φh1ζ−h2ζ+ε.
Similarly, from (69) and (70), we obtain
(74)Jh2ζ−Jh1ζ<φh1ζ−h2ζ+ε.
Therefore, from (73) and (74), we have
(75)Jh1ζ−Jh2ζ<φh1ζ−h2ζ+ε.that implies,
(76)dJh1,Jh2≤φdh1,h2+ε.
Sinceε>0 is arbitrary, then
(77)dJh1,Jh2≤φdh1,h2.
Define(78)αh,k=βh,k=1ifΘh,k≥0,whereh,k∈BW,0,otherwise.
Consequently, we have(79)αh1,h2βh1,h2dJh1,Jh2≤φdh1,h2,that is, J is a twisted α,β−φ-contractive mapping of type (i) with αh,k=βh,k=1 for all h,k∈BW. Thus, by Theorem 10, J has a fixed point. Thus, all the suppositions of Theorem 10 are satisfied. Therefore, there exists h, such that Jh=h, which is the bounded solution of the functional Equation (64).


## 5. Conclusion

In the present paper, we have defined the notions of twistedα,β-admissible and twisted α,β−φ-rational contractions in the context of F-metric spaces and established some generalized fixed point results. As an application of our main results, we investigate the problem of dynamic programming related to multistage process which reduces to the problem of solving the functional equation.



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*Source: 8130764-2020-09-26.xml*