# Some Variants of Krasnoselskii-Type Fixed Point Results for Equiexpansive Mappings with Applications

Authors: Niaz Ahmad; Nayyar Mehmood; Ahmed Al-Rawashdeh
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we investigate the Krasnoselskii-type fixed point results for the operatorF of two variables by assuming that the family Fx,.:x is equiexpansive. The results may be considered as variants of the Krasnoselskii fixed point theorem in a general setting. We use our main results to obtain the existence of solutions of a fractional evolution differential equation. An example of a controlled system is given to illustrate the application.


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## Body

## 1. Introduction

Fractional evolution equations give a unique way to evaluate the well-posedness of many complicated systems. Differential models with fractional derivatives provide a great tool for the description of memory and hereditary properties. The fractional-order models of real physical systems are always more appropriate than the classical integer order systems. Many fractional order controlled problems and fractional evolution differential equations are recently studied. Their existence results can be seen in [1–5]. Most of the results involve contractive operators with more restrictive conditions. This is the reason that many existence results cover a restrictive class of physical problems.Krasnoselskii fixed point theorem is a generalized form of Schauder and Banach fixed point theorems. While studying the solutions of delay and neutral differential/integral equations, it has been noticed that the solution can be expressed as a sum of contractive and compact operators. This theorem plays an important role in the existence of solutions of delay integral equations and neutral functional equations. Many generalizations and modifications of the Krasnoselskii fixed point theorem have appeared; for example, see [6–16] and the references therein. The Krasnoselskii fixed point theorem [17] may be stated as follows:Theorem 1.
LetV be a Banach space and Ω be a closed convex nonempty subset of V. Suppose ξ and ζ map Ω into V such that
(1)
ξν+ζυ∈Ω for all ν,υ in Ω(2)
ξ is compact and continuous(3)
ζ is contraction mapping
Then, there is aυ in Ω such that ξυ+ζυ=υ.In Krasnoselskii fixed point theorem, there are two operators in whichξ is compact and continuous and ζ is a contraction mapping. It is important to note that there is a very restrict and small class of operators that are contractions; due to this reason, many theorems involving contractive operators are less applicable. Therefore, it is required to cover more applications with different types of operators; in this regard, one of the suitable choices is the class of expansive operators. Also, there are a number of equations in which we cannot decompose the operator as a sum of two or more operators. To overcome this situation, the authors in [9–11] established Krasnoselskii-type results in a more general setting consisting of a single map Γ depending upon two variables. The following result may be seen in [11]:Theorem 2.
LetΩ be a bounded closed convex subset of a given Banach space X, and Γ be a mapping of Ω×Ω into Ω such that
(1)
Γν,υ1−Γν,υ2≤γυ1−υ2 for some γ∈0,1, and for all υ∈X(2)
Γν1,υ−Γν2,υ≤ξν1−ξν2where ξ is completely continuous and maps Ω into Ω. Then, there is an element υ∈Ω such that Γυ,υ=υ.If we defineΓν,υ=ξν+ζυ where ξ is compact and continuous and B is contraction then, Theorem 1 is the special case of Theorem 2.In most fixed point results, there is a condition such thatTΩ⊆Ω, where Ω⊆X, X is a metric space and T is an operator on Ω into Ω. No much attention has been given to the case Ω⊆TΩ. Xiang and Yuan [15] investigated the case Ω⊆TΩ and obtained the following important results in this direction.Theorem 3 (see [15]).
LetΩ be a nonempty closed subset of a complete metric space X and ζ is an expansive mapping of Ω into X such that Ω⊆ζΩ. Then, there is a unique s∈Ω such that ζs=s.Theorem 4 (see [15]).
LetX be a Banach space and Ω be a nonempty closed convex subset of X. Suppose ξ and ζ are mappings of Ω into X such that
(1)
ξ is continuous such that ξΩ lies in a compact subset of X(2)
ζ is an expansive mapping(3)
Ω⊆υ+ζΩ for υ∈ξΩ
Then, there iss∈Ω with ξs+ζs=s.Definition 5 (see [9]).
LetX be a Banach space and Γ:X×X⟶X. The family Γ.,υ:υ∈X is called equicontractive if there is a k∈0,1 such that
(1)Γν1,υ−Γν2,υ≤kν1−ν2,for all ν1,υ,ν2,υ in the domain of Γ.
In a similar way, we define the following.Definition 6.
LetX be a Banach space and Γ:X×X⟶X. The family Γν,.:ν∈X is called equiexpansive if there is h>1 such that
(2)Γν,υ1−Γν,υ2≥hυ1−υ2,for all ν,υ1,ν,υ2 in the domain of Γ.


## 2. Main Results

Inspired by the above results (Theorems1, 2, 3, and 4), we study the Krasnoselskii-type fixed point results for the operator Γ with conditions:
(i)
The familyΓν,.:ν∈X is equiexpansive(ii)
Ω⊆Γν,Ω for ν∈Ω or Ω⊆Γυ,Ω for υ∈ξΩwhereΩ is a subset of a Banach space X and ξ is an operator of Ω into Ω or X.Remark 7.
In Theorem2, the mapping Γ is equicontractive and Γν,υ∈Ω for ν,υ in Ω while in (i) and (ii), the family Γν,.:ν is equiexpansive and Ω⊆Γυ,Ω for υ∈ξΩ.Theorem 8.
SupposeΩ be a nonempty closed convex subset of a Banach space X and ξ be a continuous mapping of Ω into X such that ξΩ resides in a compact subset of X. Let Γ be a mapping of ξΩ×Ω into X such that the family Γυ,.:υ∈ξΩ is equiexpansive and Ω⊆Γυ,Ω for υ∈ξΩ with
(3)Γυ1,υ−Γυ2,υ≤υ1−υ2.
Then, there iss∈Ω such that Γξs,s=s.Proof.
Foru∈ξΩ, define a mapping H of Ω into X by Hυ=Γu,υ. The mapping H is expansive because
(4)Hυ1−Hυ2=Γu,υ1−Γu,υ2≥hυ1−υ2,and Ω⊆Γυ,Ω=HΩ. By Theorem 3, there exists a unique point in s∗∈Ω such that Hs∗=s∗, since for each u∈ξΩ, there is a unique s∗∈Ω, so we can define ϝ:ξΩ⟶Ω to obtain ϝυ=s∗ for some υ∈ξΩ); therefore,
we haveϝυ=Γυ,ϝυ. Now, for υ1,υ2∈ξΩ,
(5)ϝυ1=ϝυ2=Γυ1,ϝυ1−Γυ2,ϝυ2=Γυ2,ϝυ1−Γυ2,ϝυ2−Γυ2,ϝυ1−Γυ1,ϝυ1≥Γυ2,ϝυ1−Γυ2,ϝυ2−Γυ2,ϝυ1−Γυ1,ϝυ1≥hϝυ1−ϝυ2−υ1−υ2.
This means that(6)ϝυ1−ϝυ2≤1h−1υ1−υ2,h>1,which shows that ϝ is a continuous function of ξΩ into Ω. Also, ϝ∘ξ is a continuous function of Ω into Ω. Since ξΩ⊆C, where C is any compact set; therefore, ξΩ¯⊆C¯=C and ξΩ¯ is compact by [18] (page 37). By [19] (page 412), ϝξΩ¯ is compact because ξΩ is relatively compact, and ϝξΩ resides in a compact set ϝξΩ¯. Since ϝξΩ⊆Ω, therefore, ϝξΩ¯⊆Ω¯=Ω. Hence, ϝ∘ξΩ=ϝξΩ⊆ϝξΩ¯=ϝ∘ξΩ¯⊆Ω and ϝ∘ξ is a continuous function of Ω into a compact set ϝ∘ξΩ¯⊆Ω. By Schauder second theorem [17], there is s∈Ω such that ϝ∘ξs=s or ϝξs=s. Since for ξs∈ξΩ there is unique ϝξs∈Ω such that Γξs,ϝξs=ϝξs and also ϝξs=s, therefore, Γξs,s=s.Corollary 9 (see [15]).
LetX be a Banach space and Ω be a nonempty closed convex subset of X. Suppose ξ and ζ are mappings of Ω into X such that
(1)
ξ is continuous such that ξΩ lies in a compact subset of X(2)
ζ is an expansive mapping(3)
Ω⊆υ+ζΩ for υ∈ξΩ
Then, there iss∈Ω with ξs+ζs=s.Proof.
DefineΓυ,υ=υ+ζυ with υ=ξν∈ξΩ (Theorem 8). Since Γυ,Ω=υ+ζΩ and from (3) Ω⊆υ+ζΩ, therefore, Ω⊆Γυ,Ω. Now, since ζ is expansive mapping, therefore,
(7)Γυ,υ1−Γυ,υ2=ζυ1−ζυ2≥hυ1−υ2.
Also,(8)Γυ1,υ−Γυ2,υ=υ1−υ2≤υ1−υ2.
Hence, there iss∈Ω such that s=Γξs,s=ξs+ζs.Remark 10.
The above corollary is Theorem 2.2 in [15] or above Theorem 4. If we take ξ=0, zero operator, then, we obtain Theorem 3 or Theorem 2.1 in [15].Corollary 11.
LetX be a Banach space and Ω be a nonempty closed convex subset of X. Suppose ξ and ζ are mappings of Ω into X such that
(1)
ξ nonexpansive and ξΩ⊆Ω lies in a compact subset of X(2)
ζ is expansive mapping(3)
Ω⊆ξυ+ζΩ for υ∈ξΩ
Then, there iss∈Ω with ξ∘ξs+ζs=s.Proof.
DefineΓυ,υ=ξυ+ζυ with υ=ξν∈ξΩ (Theorem 8). Since Γυ,Ω=ξυ+ζΩ and from (3) Ω⊆ξυ+ζΩ for υ∈ξΩ, therefore, Ω⊆Γυ,Ω. Now, since ζ is expansive mapping, therefore,
(9)Γυ,υ1−Γυ,υ2=ζυ1−ζυ2≥hυ1−υ2.
Also, from (1),ξ is continuous and
(10)Γυ1,υ−Γυ2,υ=ξυ1−ξυ2≤υ1−υ2.
Hence, there iss∈Ω such that s=Γξs,s=ξ∘ξs+ζs.Theorem 12.
SupposeΩ be a nonempty closed convex set such that Ω lies in a compact subset of a Banach space X. Γ be a mapping of Ω×Ω into X such that the family Γν,.:ν∈Ω is equiexpansive and Ω⊆Γν,Ω for ν∈Ω. Let ξ be a continuous mapping of Ω into a metric space Y with
(11)Γν1,υ−Γν2,υ≤dυξν1,ξν2.
Then, there iss∈Ω such that Γs,s=s.Proof.
Forν∈Ω, define a mapping H of Ω into X by Hυ=Γν,υ. The mapping H is expansive and Ω⊆Γν,Ω=HΩ. By (Theorem 3), there exists a unique point in Ω say ϝν such that ϝν=Γν,ϝν. Now,
(12)ϝν1−ϝν2=Γν1,ϝν1−Γν2,ϝν2=Γν2,ϝν1−Γν2,ϝν2−Γν2,ϝν1−Γν1,ϝν1≥Γν2,ϝν1−Γν2ϝ−Γν2,ϝν1−Γν1,ϝν1≥hϝν1−ϝν2−dυξν1,ξν2.
This means that(13)ϝν1−ϝν2≤1h−1dυξν1,ξν2,h>1.
In the above expression, the continuity ofξ implies the continuity of ϝ. Since Ω lies in a compact subset of X and ϝ is continuous, therefore, ϝΩ lies in a compact subset of Ω. So by Schauder second theorem [17], there is s∈Ω such that ϝs=s. Since for s∈Ω there is unique ϝs∈Ω such that Γs,ϝs=ϝs and also ϝs=s, therefore, Γs,s=s.Corollary 13.
LetΩ be a nonempty closed convex set such that Ω lies in a compact subset of a Banach space X. Γ be a mapping of Ω×Ω into X such that
(1)
Ω⊆Γν,Ω for ν∈Ω(2)
Γν,υ1−Γν,υ2≥hυ1−υ2, h>1(3)
Γν1,υ−Γν2,υ≤ξν1−ξν2where ξ be a continuous mapping of Ω into X; then, there is s∈Ω such that Γs,s=s.Proof.
Takeυ=X (where X is a Banach space) in the above theorem.Remark 14.
The above corollary is a variant of Theorem 2 or Theorem 8 in [11].Corollary 15.
LetX be a Banach space and Ω be a nonempty closed convex set such that Ω lies in a compact subset of X. Suppose ξ and ζ are mappings of Ω into X such that
(1)
ξ is a continuous mapping(2)
ζ is an expansive mapping(3)
Ω⊆ξν+ζΩ for ν∈Ω
Then, there iss∈Ω such that ξs+ζs=s.Proof.
DefineΓν,υ=ξν+ζυ (in the above Corollary 13). Since Γν,Ω=ξν+ζΩ and from (3) Ω⊆ξν+ζΩ, therefore, Ω⊆Γν,Ω. Now, since ζ is expansive mapping, therefore,
(14)Γν,υ1−Γν,υ2=ζυ1−ζυ2≥hυ1−υ2.
Also,(15)Γν1,υ−Γν2,υ=ξν1−ξν2≤ξν1−ξν2.
Hence, there iss∈Ω such that s=Γs,s=ξs+ζs.Theorem 16.
SupposeΩ be a nonempty closed convex subset of a Banach space X and ξ be a continuous mapping of Ω into Ω such that ξΩ is contained in a compact subset of X. Let Γ be a mapping of ξΩ×Ω into X such that the family Γυ,.:υ∈ξΩ equiexpansive and Ω⊆Γυ,Ω for υ∈ξΩ with
(16)Γυ1,υ−Γυ2,υ≤ξυ1−ξυ2.
Then, there existss∈Ω such that Γξs,s=s.Proof.
Foru∈ξΩ, define a mapping H of Ω into X by Hυ=Γu,υ. The mapping H is expansive because
(17)Hυ1−Hυ2=Γυ,υ1−Γυ,υ2≥hυ1−υ2.
Also,Ω⊆Γυ,Ω=HΩ. By Theorem 3, there exists a unique point say ϝυ such that ϝυ=Γυ,ϝυ. Now,
(18)ϝυ1−ϝυ2=Γυ1,ϝυ1−Γυ2,ϝυ2=Γυ2,ϝυ1−Γυ2,ϝυ2−Γυ2,ϝυ1−Γυ1,ϝυ1≥Γυ2,ϝυ1−Γυ2,ϝυ2−Γυ2,ϝυ1−Γυ1,ϝυ1≥hϝυ1−ϝυ2−ξυ1−ξυ2.
This means that(19)ϝυ1−ϝυ2≤1h−1ξυ1−ξυ2,h>1.
Sinceξ is continuous, so ϝ is a continuous function of ξΩ into Ω. Also, ϝ∘ξ is a continuous function of Ω into Ω and ϝ∘ξΩ resides in a compact subset of Ω. So by Schauder second theorem [17], there is s∈Ω such that ϝ∘ξs=s or ϝξs=s. Since for ξs∈ξΩ there is unique ϝξs∈Ω such that Γξs,ϝξs=ϝξs and also ϝξs=s, therefore, Γξs,s=s.Remark 17.
In the above theorem,ξ is a mapping of Ω into Ω while in Theorem 8, ξ is a mapping of Ω into X.Corollary 18.
LetX be a Banach space and Ω be a nonempty closed convex subset of X. Suppose ξ is a mapping of Ω into Ω and ζ is mapping of Ω into X such that
(1)
ξ is continuous such that ξΩ lies in a compact subset of X(2)
ζ is an expansive mapping(3)
υ∈ξΩ implies Ω⊆ξυ+ζΩ
Then, there existss∈Ω with ξ∘ξs+ζs=s.Proof.
TakingΓu,υ=ξu+ζυ with υ=ξu∈ξΩ (in the above Theorem 16), we can obtain the required result.Theorem 19.
SupposeΩ be a nonempty closed convex subset of a Banach space X and ξ be a continuous mapping of Ω into Y such that ξΩ is a precompact subset of a metric space Y. Let Γ be a continuous mapping of ξΩ¯×Ω into X such that the family Γυ,.:υ∈ξΩ¯ equiexpansive and Ω⊆Γυ,Ω for υ∈ξΩ¯. Then, there is s∈Ω such that Γξs,s=s.Proof.
Foru∈ξΩ¯, define a mapping H of Ω into X by Hυ=Γu,υ. The mapping H is expansive because
(20)Hυ1−Hυ2=Γu,υ1−Γu,υ2≥hυ1−υ2.
Also,Ω⊆Γu,Ω=HΩ. By Theorem 3, there is a unique point in Ω say ϝυ such that ϝυ=Γυ,ϝυ. We show that the mapping ϝ of ξΩ¯ into Ω is continuous. Let υn be a sequence in ξΩ¯ such that υn converges to υ0 in ξΩ¯. Then,
(21)ϝυn−ϝυ0=Γυn,ϝυn−Γυ0,ϝυ0=Γυn,ϝυn−Γυn,ϝυ0−Γυ0,ϝυ0−Γυn,ϝυ0≥Γυn,ϝυn−Γυn,ϝυ0−Γυ0,ϝυ0−Γυn,ϝυ0≥hϝυn−ϝυ0−Γυ0,ϝυ0−Γυn,ϝυ0,and therefore,
(22)ϝυn−ϝυ0≤1h−1Γυ0,ϝυ0−Γυn,ϝυ0.
SinceΓ is continuous, therefore, ϝ is continuous. The operator ϝ maps ξΩ¯ into Ω and is continuous. Since ξΩ is relatively compact, therefore, by [19] (page 412), ϝξΩ is relatively compact. Let coϝξΩ be the convex hull of ϝξΩ. By [20] (page 195), coϝξΩ is relatively compact. Since Ω is closed and convex, so coϝξΩ¯⊆Ω, and by [21] (page 415), co¯ϝξΩ⊆Ω. Let P=co¯ϝξΩ; then, P is a subset of Ω and ϝ∘ξP⊆P. Since P=coϝξΩ¯ also, therefore, P is compact and convex. So by first Schauder theorem [17], there is a point s∈P such that ϝξs=s. Thus, s=Γξs,s by similar arguments given in the proof of Theorem 5.Remark 20.
The above theorem is a variant of Theorem 2.2 in [9].


## 3. Application

LetX be a Banach space with norm ·, for ξ>0 denote J=0,ξ a closed interval in ℝ. Let CJ,X be the Banach space of all continuous functions from J into X with norm ν=supρ∈Jνρ, where ν∈CJ,X. Let U be a closed and convex subset of X. Consider the following controlled system:
(23)Dανϱ=−ξνϱ+Γϱ,νϱ+Cϱυϱ,ϱ∈J≔0,ξ,υ∈U,andα∈0,1,ν0=ν0.The mild solution corresponding toυ of ξ is given by [22]:
(24)νυϱ=ℵϱν0+∫0ϱρ−sα−1Iϱ−sΓs,νsds+∫0ϱρ−sα−1Iρ−sCsυsds,where
(25)ℵϱ=∫0∞ξαθTϱαθdθ,Iϱ=α∫0∞θξαθTϱαθdθ,ξα is the probability density function defined on 0,∞; for more details about the notations, we refer the readers to [22]. Also,
(26)Tϱ≤ΩforsomeΩ>1,ℵϱν≤Ων,Iϱν≤αΩΓα+1ν,and ℵϱ and Iϱ are strongly continuous for ϱ≥0 and are compact for each ϱ>0.We can writeζ in the form
(27)Γν,υϱ=ℵϱν0+∫0ϱϱ−sα−1Iϱ−sΓs,νsds+Qυ,where
(28)Qυϱ=∫0ϱϱ−sα−1Iϱ−sCsυsds.Assume that the following conditions hold:H1Q:ζδ⟶X is continuous and expansive, i.e., Qυ−Qv≥ℓυ−v for all υ,v∈ζ and some ℓ>1H2Γ:Qζδ¯×ζδ⟶X is continuousH3ζδ⊆Γυ,ζδ for each υ∈Qζδ¯whereζδ is a closed ball with radius δ and center at 0 in X.We show thatΓ is equiexpansive. For a fixed ν, and any υ,u∈Qζδ¯, consider
(29)Γν,υ−Γν,u=ℵϱν0+∫0ϱϱ−sα−1Iϱ−sΓs,νsds+Qυ−ℵϱν0−∫0ϱϱ−sα−1Iϱ−sΓs,νsds−Qu=Qυ−Qu≥ℓυ−u,which shows that Γv,·:v∈Qζδ¯ is equiexpansive.We prove compactness ofQ. Continuity of I and C implies the continuity of Q; we show that Q is uniformly bounded on ζδ. For υ∈ζδ, consider
(30)Qυϱ=∫0ϱϱ−sα−1Iϱ−sCsυsds≤∫0ϱϱ−sα−1ξqIϱ−sCsυsds≤CΩqαΓ2−qΓ1+α1−qδ∫0ϱϱ−sα−1ds≤CΩqαΓ2−qΓ1+α1−qδTα1−qα1−q≔ϱ,which implies that Q is uniformly bounded. To prove the equicontinuity of Q, we consider for 0<ϱ1<ϱ2<T,(31)Qυϱ2−Qυϱ1=∫0ϱ2ϱ2−sα−1Iϱ2−sCsυsds−∫0ϱ1ϱ1−sα−1Iϱ1−sCsυsds≤CΩqαΓ2−qΓ1+α1−qδ∫0ϱ2ϱ2−sα−1ds−∫0ϱ1ϱ1−sα−1ds≤CΩqΓ2−qΓ1+α1−qδϱ2α−ϱ1α⟶0asϱ2⟶ϱ1,which is independent of υ, and Qυϱ2−Qυϱ1⟶0 as ϱ2⟶ϱ1. Therefore, Q is equicontinuous. By Arzela-Ascoli theorem, Q is a compact operator. From H3,ζδ⊆Γυ,ζδ for each υ∈Qζδ¯. All conditions of Theorem 19 are satisfied to obtain the z∈X such that ΓQz,z=z, which is the solution of controlled problem ξ.Finally, we summarize the above discussion in the following theorem.Theorem 21.
LetX be a Banach space and ζδ be a closed ball with center at 0 and radius δ. Assume that Q:ζδ⟶X and Γ:Qζδ¯×ζδ⟶X satisfy H1–H3. Then, there exists a mild solution of controlled problem ξ.Example 22.
Consider the following controlled system:(32)∂αυϱ,ν∂ϱα=∂2υϱ,ν∂ν2+Γϱ,υϱ,ν+∫0ϱϝs,υs,νds,υϱ,0=υϱ,π=0,υ′ϱ,0=υ′ϱ,π=0,υ0,ν+Σξiυϱi,ν=υ0ν,
whereϱ∈J=0,1,ν∈0,π,0<α<1, let X=L20,π and consider the operator ξ:Dξ⊆X⟶X defined by
(33)ξυ=∂2υ∂ν2,with domain
(34)Dξ=υ,∂υ∂ν,∂2υ∂ν2∈X.Clearly,ξ is densely defined in X and is the infinitesimal generator of a resolvent family Tαϱρ≥0 on X [4] and let υ,v∈CJ,X. Define the operators Γ,ϝ:J×U⟶X by
(35)Γϱ,υ=1+eϱ1+υϱ,νe−ϱυϱ,ν,ϝϱ,υ=2υϱ,ν.Clearly,(36)Γϱ,υ−Γϱ,v≥1+eϱv−υ,which shows that Γ is equiexpansive. Note that ξ and ϝ are continuous and compact. All conditions of Theorem 21 are satisfied to obtain the solution of the given controlled problem.


## 4. Conclusion

Some variants of the Krasnoselskii-type fixed point theorem are presented using the notion of newly defined equiexpansive mappings. As an application, we present an existence result for controlled problem with less conditions. This problem is also a generalized kind of Cauchy problem for evolution differential equations.



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*Source: 2648057-2021-09-11.xml*