# A Note on the Górnicki-Proinov Type Contraction

Authors: Erdal Karapınar; Manuel De La Sen; Andreea Fulga
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we propose a notion of the Górnicki-Proinov type contraction. Then, we prove the uniqueness and existence of the fixed point for such mappings in the framework of the complete metric spaces. Some illustrative examples are also expressed to strengthen the observed results.


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## Body

## 1. Introduction and Preliminaries

The history of the fixed point theory goes back about a century. Banach’s result initiated the metric fixed point theory in 1922 [1]. The first outstanding extension of this initial theorem was given by Kannan [2] in 1968. In this first generalization, Kannan [2] removed the necessity of the continuity of the contraction mapping. Recently, Górnicki [3] expressed an extension of Kannan type of contraction but the continuity condition was assumed. After then, Bisht [4] refined the result of Górnicki [3] by replacing the continuity condition for the considered mapping with orbitally continuity or p-continuity. Very recently, Górnicki [5] improved these two mentioned results by introducing new contractions, “Geraghty-Kannan type” and “ϕ-Kannan type.” He proved the existence of a fixed point for such mappings. On the other hand, Proinov [6] discussed some existing results and noted that these results are particular cases of Skof [7]. He also proposed a very general fixed point theorem that also contains the result of Skof [7].We first recall the pioneer theorem of Banach [1] and Kannan [2]. On a complete metric space X,d, a mapping T:X→X admits a unique fixed point if there exists 0≤K<1 such that
(1)dTu,Tv≤k·du,v,and
(2)dTu,Tv≤K·du,Tu+dv,Tv,for all u,v∈X. The inequality (1) belongs to Banach [1] and (2) belongs to Kannan [2]. By using the “asymptotic regularity” concept, Górnicki [3] proved an extension of Kannan Theorem 1.2. Before giving this interesting result, we recollect the interesting concepts:LetT be a self-mapping on a metric space X,d and Tnu be the Picard iterative sequence, for an initial point u∈X.(o) The setOT,u=Tnu:n=0,1,2,⋯ is called the orbit of the mappingT at u.The mappingT is said to be [3, 5]:(o-c)orbitally continuous at a pointw∈X if for any sequence un in OT,u for some u∈X, limn→∞dun,w=0 implies limn→∞dTun,Tw=0.(p-c)p-continuous at a pointw∈X (p=1,2,3,⋯) if for any sequence un in Xlimn→∞dTp−1un,w=0 implies limn→∞dTpun,Tw=0.(a-r)asymptotically regular at a pointu∈X if limn→∞dTnu,Tn+1u=0. If T is asymptotically regular at each point of X, we say that it is asymptotically regular.Remark 1.
In [8], it is shown that p-continuity of T and the continuity of Tp are independent conditions for the case p>1.Theorem 2 (see [3, 5]).
On a complete metric spaceX,d, a continuous asymptotically regular mapping T:X→X admits a unique fixed point if there exist 0≤k<1 and 0≤K<+∞ such that
(3)dTu,Tv≤k·du,v+K·du,Tu+dv,Tv,for all u,v∈X.Later, the assumption of continuity of the mappingT was replaced with weaker notions of continuity.Theorem 3 (see [4]).
On a complete metric spaceX,d and a mapping T:X→X. Suppose that there exists 0≤K<1 such that
(4)dTu,Tv≤k·du,v+K·du,Tu+dv,Tv,for all u,v∈X. Then, T admits a unique fixed point if either T is (o-c) or (p-c) for p≥1.In [5], some generalizations of Theorems 2 and 3 are considered, by replacing the constant k with some real-valued functions.Theorem 4 (see [5]).
LetX,d be a complete metric space and T:X→X be an (a-r) mapping such that there exist ψ:0,∞→0,∞ and 0≤K<∞ such that
(5)dTu,Tv≤ϕdu,v+K·du,Tu+dv,Tv,for all u,v∈X. Suppose that:
(i)
ϕθ<θ for all θ>0 and ϕ is upper semicontinuous(ii)
eitherT is (o-c) or T is (p-c) for some p≥1Then,T has a unique fixed point u∗∈X and for each u∈X, Tnu→u∗ as n→∞.Theorem 5 (see [5]).
LetX,d be a complete metric space and T:X→X be an (a-r) mapping such that there exist ς:0,∞→0,1 and 0≤K<∞ such that
(6)dTu,Tv≤ςdu,v·du,v+K·du,Tu+dv,Tv,for all u,v∈X. Suppose that:
(1)
ςθn→1⇒θn→0;(2)
eitherT is (o-c) or T is (p-c) for some p≥1Then,T has a unique fixed point u∗∈X and for each u∈X, Tnu→u∗ as n→∞.On the other hand, very recently, Proinov announced some results which unify many known results [6].Theorem 6 (see [6]).
LetX,d be a complete metric space and T:X→X be a mapping such that
(7)ψdTu,Tv≤ϕdu,v,for all u,v∈X with dTu,Tv>0, where the functions ψ,ϕ:0,∞→ℝ are such that the following conditions are satisfied:
(p1) ϕθ<ψθ for any θ>0;
(p2) ψ is nondecreasing;
(p3) limsupθ→e+ϕθ<ψe+ for any e>0.Then,T admits a unique fixed point.Theorem 7 (see [6]).
LetX,d be a complete metric space and T:X→X be a mapping such that
(8)ψdTu,Tv≤ϕdu,v,for all u,v∈X with dTu,Tv>0, where ψ,ϕ:0,∞→ℝ are two functions such that the following conditions are satisfied:
(p1) ϕθ<ψθ for any θ>0;
(p4) infθ>eψθ>−∞ for any e;
(p5) limsupθ→e+ϕθ<liminfθ→eψθ or limsupθ→eϕθ<liminfθ→e+ψθ for any e>0;
(p6) limsupθ→0+ϕθ<liminfθ→eψθ for any e>0;
(p7) if the sequences ψθn and ϕθn are convergent with the same limit and ψθn is strictly decreasing, then θn→0 as n→∞.Then,T admits a unique fixed point.Lemma 8 (see [6]).
Letun be a sequence in a metric space X,d such that dun,un+1→0 as n→∞. If the sequence un is not Cauchy, then there exist e>0 and two subsequences sk, rk of positive integers such that
(9)limk→∞dusk+1,urk+1=e+,limk→∞dusk,urk=limk→∞dusk+1,urk=limk→∞dusk,urk+1=eLemma 9 (see [6]).
Letun be a sequence in a metric space X,d such that dun,un+1→0 as n→∞. If the sequence un is not Cauchy, then there exist e>0 and two subsequences sk, rk of positive integers such that
(10)limk→∞dusk,urk=e+,limk→∞dusk+1,urk+1=limk→∞dusk+1,urk=limk→∞dusk,urk+1=eIn the end of this section, we recall the notions ofα-orbital admissible and triangular α-orbital admissible mappings [9] with mention that these notions were extended in many directions, see, e.g., [10] and it could be potentially extended also to several approaches of recent developments in fixed point theory. See, for instance, [11–21].On a metric spaceX,d, a self-mapping T is called
(i)
α-orbital admissible if(11)αu,Tu≥1⇒αTu,T2u≥1,for any u,v∈X, where α:X×X→0,∞(ii)
triangularα-orbital admissible if it is α-orbital admissible and the following condition is satisfied(12)αu,v≥1andαv,Tv≥1⇒αu,Tv≥1,for any u,v,w∈XLemma 10.
If for an triangularα-orbital admissible mapping T:X→X there exists u0∈X such that αu0,Tu0≥1, then
(13)αun,up≥1,foralln,p∈ℕ,where the sequence un is defined as un+1=Tun.LetX,d be a metric space and the function α:X×X→0,∞. The following conditions will be used further:R If for a sequence un in X such that un→u and αun,un+1≥1 for all n∈ℕ, then there exists a subsequence upk of un such that αupk,u≥1.U For all u,v∈FixXT=z∈X:Tz=z, we have αu,v≥1.


## 2. Main Results

LetΛ be the set of all functions ϕ:0,∞→ℝ. For ϕ,ψ∈Λ, we are considering the following conditions:(a1) ϕθ<ψθ for θ>0(a2) limsupθ→e+ϕθ<liminfθ→eψθ, for any e>0(a3) limsupθ→eϕθ<liminfθ→e+ψθ, for any e>0(a4) limsupθ→e+ϕθ<ψe+, for any e>0Definition 11.
LetX,d be a metric space, the functions ψ,ϕ∈Λ and α:X×X→0,∞. An (a-r) mapping T:X→X is said to be α,ψ,ϕ-contraction if there exists 0≤K<∞ such that
(14)αu,vψdu,v≤ϕdu,v+K·du,Tu+dv,Tv,for each u,v∈X with dTu,Tv>0.Theorem 12.
On a complete metric spaceX,d an α,ψ,ϕ-contraction T:X→X has a fixed point provided that
(1)
the functionsψ,ϕ∈Λ satisfy (a1) and either (a2) or (a3)(2)
T is triangular α-orbital admissible and there exists u0∈X such that αu0,Tu0≥1(3)
eitherT is (o-c) or T is (p-c), for some p≥1Moreover, if propertyU is satisfied, then the fixed point of T is unique.Proof.
Letu be any point (but fixed) in X and we build the sequence un, where u0=u and un=Tnu for any n∈ℕ. If there exists m0∈ℕ such that Tm0u=Tm0+1u=TTm0u, then Tm0u is a fixed point of T. For this reason, we can suppose that Tnu≠Tn+1u, for every n∈ℕ∪0 and we claim that un is Cauchy sequence. Assuming the contrary, that the sequence un is not Cauchy, from Lemma 1, it follows that we can find e and two subsequences sk and rk of positive integers such that (9) holds. Letting u=usk and v=urk in (14), we have αusk,urk≥1 (taking into account (1.8)), and then,
(15)ψdusk+1,urk+1≤αusk,urkψdusk+1,urk+1=αusk,urkψdTusk,Turk≤ϕdusk,urk+K·dusk,usk+1+durk,urk+1,or denoting ξk=dusk+1,urk+1 and ζk=dusk,urk(16)ψξk≤ϕζk+K·dusk,usk+1+durk,urk+1.
Taking into account the asymptotically regularity ofT, from (9), it follows that
(17)ξk→e+andζk→e.
Thus, letting the limit in (16), we have
(18)liminfθ→e+ψθ≤liminfk→∞ψξk≤limsupk→∞ϕζk≤limsupθ→eϕθ.
This contradicts the assumption (a2).
Similarly, if we consider that the functionsψ,ϕ satisfy (a3), the conclusion follows in the same way, but taking into account Lemma 2.
Therefore,un is a Cauchy sequence, and because the space X,d is complete, there exists u∗ such that
(19)limn→∞un=u∗.
We claim thatu∗ is a fixed point of T.
IfT is orbitally continuous, then since un∈OT,u and un→u∗, we have un+1=Tun→Tu∗ as n→∞. The uniqueness of the limit gives Tu∗=u∗.
IfT is p-continuous, for some p≥1, by (19), we have limn→∞Tp−1un=u∗ which implies limn→∞Tpun=Tu∗ (because T is p-continuous). Therefore, by uniqueness of the limit, we have Tu∗=u∗.
Now, supposing that there existsv∗∈X such that Tv∗=v∗≠u∗=Tu∗, from (14) and taking into account the property U, we have
(20)ψdu∗,v∗≤αu∗,v∗ψdTu∗,Tv∗≤ϕdu∗,v∗+K·du∗,Tu∗+dv∗,Tv∗=ϕdu∗,v∗<ψdu∗,v∗,which is a contradiction. Therefore, u∗=v∗.Lettingαu,v=1 in Theorem 12, we get the following:Corollary 13.
LetX,d be a complete metric space and an (a-r) mapping T:X→X. Suppose that there exists 0≤K<∞ such that
(21)ψdu,v≤ϕdu,v+K·du,Tu+dv,Tv,for each u,v∈X with dTu,Tv>0, where ψ,ϕ∈Λ. Suppose also that:
(1)
the functionsψ,ϕ∈Λ satisfy (a1) and either (a2) or (a3)(2)
eitherT is (o-r) or T is (p-o), for some p≥1Then,T has a unique fixed point.Corollary 14.
LetT,d be a complete metric space and T:X→X be an (a-r) mapping such that
(22)dTu,Tv≤ςdu,vdu,v+K·du,Tu+dv,Tv,for each u,v∈X, where 0≤K<∞ and the function ς:0,∞→0,1 is such that limsupθ→e+ςθ<1 for any e>0. If T is either (o-c) or (p-c) for some p≥1, then T has a unique fixed point.Proof.
Letψθ=θ in Corollary 1.Takingψθ=θ and ϕθ=k·θ, with k∈0,1 Corollary 1 becomes:Corollary 15.
LetT,d be a complete metric space and T:X→X be an (a-r) mapping. If there exist k∈0,1 and 0≤K<∞ such that
(23)dTu,Tv≤kdu,v+K·du,Tu+dv,Tvfor each u,v∈X, then T admits a unique fixed point provided that T is (o-c) or (p-c) for some p≥1.Theorem 16.
LetX,d be a complete metric space, α:X×X→0,∞, ψ,φ∈Λ such that (a1) and (a2) are satisfied. Let T:X→X be an (a-r) mapping. Suppose that there exists 0≤K<∞ such that
(24)ψαu,vdTu,Tv≤φdu,v+K·du,Tu+dv,Tv,for each u,v∈X with dTu,Tv>0. Suppose also that
(i)
ψ is nondecreasing and limsupθ→e+<ψe+ for any e>0(ii)
T is triangular α-orbital admissible and there exists u0∈X such that αu0,Tu0≥1(iii)
the mappingT is either (o-c) or (p-c)Then, the mappingT possesses a fixed point. Moreover, the fixed point is unique, provided that property U is satisfied.Proof.
Letun be the sequence defined as in the previous theorem, as un=Tnu, where u∈X is arbitrary but fixed. Letting u=usk and v=urk in (2.7), we have
(25)ψαusk,urkdusk+1,urk+1≤ϕdusk,urk+K·dusk,usk+1+dusk,usk+1,and taking into account the assumptions (i), (ii), and Lemma 3, we get
(26)ψdusk+1,urk+1≤ϕdusk,urk+K·dusk,usk+1+durk,urk+1,
Settingξk=dusk+1,urk+1 and ζk=dusk,urk and since ϕθ<ψθ, we get
(27)ψξk≤ϕζk+K·dusk,usk+1+durk,urk+1<ψζk+K·dusk,usk+1+durk,urk+1.
On the other hand, from 1.5 thatξk→e+, ζk→e+ and then, letting the limit as k→∞ in the above inequality, since T is an (a-r) mapping and taking into account the second part of the assumption (i), we have
(28)ψe+=limk→∞ψξk≤limsupk→∞ϕζk≤limsupθ→e+ϕθ<ψe+,which is a contradiction. Thus, the sequence un is Cauchy on a metric space, so there exists u∗ such that un→u∗ as n→∞ and following the lines of the previous proof, we get that u∗ is the unique fixed point of T.Again, lettingαu,v=1 for any u,v∈X we get the following:Theorem 17.
LetX,d be a complete metric space, and two functions ψ,φ∈Λ such that (a1) is satisfied. Let T:X→X be an (a-r) mapping. Suppose that there exists 0≤K<∞ such that
(29)ψdTu,Tv≤ϕdu,v+K·du,Tu+dv,Tv,for each u,v∈X with dTu,Tv>0, where ψ,ϕ∈Λ. Suppose also that
(i)
ψ is nondecreasing and limsupθ→e+<ψe+ for any e>0(ii)
the mappingT is either (o-c) or (p-c)Then, the mappingT possesses a unique fixed point.Theorem 18.
LetX,d be a complete metric space, and two functions ψ,φ∈Λ such that (a1) is satisfied. Let T:X→X be an (a-r) mapping. Suppose that there exists 0≤K<∞ such that
(30)ψdTu,Tv≤ςdu,vψdu,v+K·du,Tu+dv,Tv,for each u,v∈X with dTu,Tv>0, where ψ∈Λ and ς:0,∞→0,1. Suppose also that
(i)
ψ is nondecreasing and limsupθ→e+ςθ<1 for any e>0(ii)
the mappingT is either (o-c) or (p-c)Then, the mappingT possesses a unique fixed point.Proof.
Takeϕθ=αθψθ, for θ>0 in Theorem 17.Next, we consider mappings that satisfy a similar condition as (14), but for which the asymptotic regularity condition is not necessary.Definition 19.
LetX,d be a complete metric space, α:X×X→0,∞ and ψ,φ∈Λ. A mapping T:X→X is called α,ψ,ϕ-contraction of type 2 if there exists 0≤K<∞ such that
(31)αu,vψdTu,Tv≤ϕdu,v+K·du,Tu+dv,Tvdu,Tvdv,Tu,for each u,v∈X with dTu,Tv>0.Theorem 20.
On a complete metric spaceX,d, an α,ψ,ϕ-contraction of type 2, T:X→X has a fixed point provided that property (R) and the following conditions hold:
(A)
T is triangular α-orbital admissible and there exists u0∈X such that αu0,Tu0≥1(B)
ψ,ϕ satisfy the assumptions (a1) and (a4)(C)
ψ is nondecreasing(D)
limsupθ→0+ϕθ<liminfθ→eψθ, for any e>0Moreover, if the propertyU holds, the fixed point of T is unique.Proof.
Letu be a sequence in X defined as
(32)un=Tnu0,foreveryn∈ℕ,where u0 is an arbitrary but fixed point in X. Replacing in (31) and taking into account (11), we have
(33)ψdun,un+1≤αun−1,unψdTun−1,Tun≤ϕdun−1,un++K·dun−1,Tun−1+dun,Tundun−1,Tundun,Tun−1=ϕdun−1,un+K·dun−1,un+dun,un+1dun−1,un+1dun,un=ϕdun−1,un,or setting xn=dun−1,un (we can suppose that xn>0) and taking into account the condition (a1) for any θ>0, we get
(34)ψxn≤ϕxn−1<ψxn−1.
If the condition (C) holds, from the above inequality, we getxn<xn−1, for every n∈ℕ. Consequently, being positive and strictly decreasing, the sequence xn is convergent and there is x≥0∈X such that xn→x. If we assume that x>0, then from the above inequality, we have
(35)ϕx+=limn→∞ψxn≤limsupn→∞ϕxn≤limsupθ→x+ϕθ<ϕx+,which is a contradiction. Thus,
(36)limn→∞xn=x=0.
The aim for the next step is to prove that the sequenceun is Cauchy. Supposing by contradiction, the sequence un is not Cauchy, by (36), and taking into account Lemma 1, we can find e>0 and two subsequences usk and urk of un such that (9) holds. Taking u=usk and v=urk in (14) and keeping in mind (1.7), we have
(37)ψdusk+1,urk+1≤αusk,urkψdTusk,Turk≤ϕdusk,urk++K·dusk,usk+1+durk,urk+1dusk,urk+1durk,usk+1.
Letting the limit ask→∞ in the previous inequality (since dusk+1,urk+1→e+ and dusk,urk→e and using (2.11), we get
(38)liminfθ→e+ψθ≤liminfk→∞ψdusk+1,urk+1≤limsupk→∞ϕdusk,urk≤limsupθ→eϕθ.
This is a contradiction to (a4). Thus, un is a Cauchy sequence on a complete metric space, so it is convergent. Let u∗=limn→∞un and we claim that u∗ is a fixed point of T. From (31) and R, for u=un and v=u∗, we have
(39)ψdun+1,Tu∗≤αun,u∗ψdTun,Tu∗≤ϕdun,u∗++K·dun,un+1+du∗,Tu∗dun,Tu∗du∗,un+1.
Sincelimn→∞dun+1,Tu∗=du∗,Tu∗ and limn→∞dun,u∗=0 if we suppose that du∗,Tu∗>0, the above inequality yields
(40)liminfθ→du∗,Tu∗ψθ≤liminfn→∞ψdun+1,Tu∗≤limsupn→∞ϕdun,u∗≤limsupθ→0ϕθ,which is a contradiction to (D). Therefore, du∗,Tu∗=0, that is u∗ is a fixed point of T. As in the Theorem 12, adding the condition U to the statement of Theorem 20, we are able to prove that the fixed point is unique. Indeed, if we suppose that v∗∈X is such that Tv∗=v∗≠u∗=Tu∗, from (2.10), we have
(41)ψdu∗,v∗≤αu∗,v∗ψdTu∗,Tv∗≤ϕdu∗,v∗++K·du∗,Tu∗+dv∗,Tv∗du∗,Tv∗+dv∗,Tu∗.
Lettingn→∞ in the above inequality and keeping in mind (a1), we have
(42)ψdu∗,v∗≤ϕdu∗,v∗<ψdu∗,v∗,which is a contradiction.Example 21.
Let the setX=A1,A2,A3,A4,A5 endowed with the distance d:X×X→0,∞, where du,u=0, du,v=dv,u for any u,v∈X and
(43)dA1,A2=dA2,A3=dA3,A4=1,dA1,A3=dA2,A4=dA4,A5=2,dA1,A4=dA2,A5=dA3,A5=3,dA1,A5=4.Let the mappingT:X→X defined by
(44)TA1=A1,TA2=A3,TA3=A5,TA4=A2,TA5=A2.Let also the functionα:X×X→0,∞, with
(45)αu,v=2,ifu,v=Ai,A1,fori=1,2,3,4,51,ifu,v∈A3,A4,A4,A30,otherwiseThen,T does not satisfy Banach, neither Kannan type condition. Indeed, letting for example u=A1,v=A3,
(46)dTA1,TA3=dA1,A5=4>2k=kdA1,A3forany0≤k<1,dTA1,TA3=dA1,A5=4>3K=K·dA1,A1+dA3,A5=K·dA1,TA1+dA3,TA3,forany0≤K<12.On the other hand,T is not (a-r), so Theorem 3 cannot be applied. Let the functions ψ,ϕ∈Λ, ϕθ=θ, ψθ=θ/2, for θ>0 and K=8. For an easier reading, we will set
(47)Au,v=ϕdu,v+K·du,Tu+dv,Tvdu,Tvdv,Tu=du,v2+8·du,Tu+dv,Tvdu,Tvdv,Tu.Let us check that the mappingT is an α,ψ,ϕ-contraction of type 2. For this purpose, we must consider the following cases:
(i)
u=A1,v=A2,(48)αA1,A2ψdTA1,TA2=2dA1,A3=4<332=AA1,A2(ii)
u=A1,v=A3,(49)αA1,A3ψdTA1,TA3=2dA1,A5=8<193=AA1,A3(iii)
u=A1,v=A4,(50)αA1,A4ψdTA1,TA4=2dA1,A2=2<992=AA1,A4(iv)
u=A1,v=A5,(51)αA1,A5ψdTA1,TA5=2dA1,A2=2<98=AA1,A5(v)
u=A3,v=A4,(52)αA3,A4ψdTA3,TA4=dA5,A2=3<972=AA3,A4Moreover, it is easy to see that all the assumptions of Theorem20 are satisfied, so that T has a unique fixed point.Example 22.
Let the setX=0,∞ be endowed with the usual distance d on ℝ. Consider the mapping T:X→X defined by
(53)Tu=1−u,if0≤u≤1ln1+eu,ifu>1.Then,Tis neither continuous, a contraction, nor (a-r).Define the functionα:X×X→0,∞ by
(54)αu,av=2,ifu∈14,12,1,v=121,ifu=2,v=10,otherwise.Consider also, the functionsψ,ϕ∈Λ, where ψθ=eθ and ϕθ=θ+1, for θ>0. Let, for example, K=64. Using the same notation as in Example 1, taking into account the definition of the function α, we have the following:
(i)
u=1/4, v=1/2(55)α14,12ψdT14,T12=2e14≤174=ϕd14,12+A14,12(ii)
u=1, v=1/2(56)α1,12ψdT1,T12=2e12≤352=ϕd1,12+A1,12(iii)
u=2, v=1(57)α2,1ψdT2,T1=eln1+e2=1+e2≤ϕd2,1+A2,1.Since it easy to check that all the assumptions of Theorem20 are verified, we can conclude that T has a unique fixed point.Corollary 23.
LetX,d be a complete metric space and a mapping T:X→X such that for all u,v∈X with dTu,Tv>0,
(58)ψdTu,Tv≤ϕdu,v+K·du,Tu+dv,Tvdu,Tvdv,Tu,where 0≤K<1 and the functions ψ,ϕ∈Λ are such that
(a)
ψ,ϕ satisfy (a1) and (a4)(b)
ψ is not decreasingThen,T admits a unique fixed point.Corollary 24 (Theorem6).
LetX,d be a complete metric space and a mapping T:X→X such that for all u,v∈X with dTu,Tv>0,
(59)ψdTu,Tv≤ϕdu,v,where the functions ψ,ϕ∈Λ are such that
(a)
ψ,ϕ satisfy the assumptions (a1) and (a4)(b)
ψ is not decreasingThen,T admits a unique fixed point.Proof.
Letαu,v=0 and K=0 in Theorem 20.Corollary 25.
LetT,d be a complete metric space and T:X→X be a mapping such that
(60)ψdTu,Tv≤ςdu,vψdu,v+K·du,Tu+dv,Tvdu,Tvdv,Tu,for each u,v∈X with dTu,Tv>0, where 0≤K<∞ and the functions ς:0,∞→0,1, ψ:0,∞→0,1 are such that
(i)
limsupθ→e+ςθ<1 for any e>0(ii)
ψ is nondecreasingThen,T has a unique fixed point.Proof.
Letϕθ=ςdu,vψdu,v in Corollary 4.Corollary 26.
LetX,d be a complete metric space and a mapping T:X→X. Suppose that there exist 0≤k<1 and 0≤K<∞ such that for all u,v∈X,
(61)dTu,Tv≤kdu,v+K·du,Tu+dv,Tvdu,Tvdv,Tv.Then,T admits a unique fixed point.Proof.
Letαu,v=0, ψθ=θ and ϕθ=k·θ, with 0≤k<1 in Theorem 20.



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*Source: 6686644-2021-01-28.xml*