# Krasnosel’skii Type Fixed Point Theorems for Mappings on Nonconvex Sets

Authors: Maryam A. Alghamdi; Donal O'Regan; Naseer Shahzad
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We prove Krasnosel'skii type fixed point theorems in situations where the domain is not necessarily convex. As an application, the existence of solutions for perturbed integral equation is considered inp-normed spaces.


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## Body

## 1. Introduction

LetX be a linear space over  K(K=ℝorK=ℂ)  with the origin  θ. A functional ∥·∥p:X→[0,∞)  with  0<p≤1  is called a p-norm on  X  if the following conditions hold(a)
∥x∥p=0  if and only if  x=θ;(b)
∥λx∥p=|λ|p∥x∥p, for all x∈X,λ∈K;(c)
∥x+y∥p≤∥x∥p+∥y∥p, for all  x,y∈X.The pair(X,∥·∥p)  is called a p-normed space. If  p=1, then X is a usual normed space. A p-normed space is a metric linear space with a translation invariant metric   dp  given by  dp(x,y)=∥x-y∥p  for all  x,y∈X.LetΩ  be a nonempty set, let  M  be a  σ-algebra in  Ω,  and let  μ:M→[0,+∞)  be a positive measure. The space  Lp(μ) based on the complete measure space  (Ω,M,μ)  is an example of  a p-normed space with the  p-norm defined by
(1.1)∥f(t)∥p=∫Ω|f(t)|pdμ,forf∈Lp(μ),Lp(μ)={f:f:Ω→Kismeasurable,∫Ω|f(t)|pdμ<+∞}.
Another example of a p-normed space is Cp[0,1], the space of all continuous functions defined on the unit interval [0,1] with the sup  p-norm given by
(1.2)∥x∥p=sup0≤t≤1|x(t)|p,forx∈Cp[0,1].
The class of p-normed spaces (0<p≤1) is a significant generalization of the class of usual normed spaces. For more details about  p-normed spaces, we refer the reader to [1, 2].It is noted that most fixed point theorems are concerned with convex sets. As we know, there exists nonconvex sets also, for example, the unit ball with centerθ in a  p-normed space (0<p<1) is not a convex set. It is a natural question whether the well-known fixed point theorems could be extended to nonconvex sets.  Xiao and Zhu [3] established the existence of fixed points of mappings on s-convex sets in p-normed spaces, where 0<p≤1,0<s≤p.Theorem 1.1 (see [3] (Krasnosel’skii-type)).
Let(X,∥·∥p)  be a complete  p-normed space and  C  a bounded closed s-convex subset of X,  where 0<p≤1,0<s≤p.  Let  T:C→X be a contraction mapping and S:C→X a completely continuous mapping. If Tx+Sy∈C for all x,y∈C,  then there exists  x*∈C such that  Sx*+Tx*=x*.In this paper, we investigate the fixed point problem of the sum of an expansive mapping and a compact mapping. Our results extend and complement the classical Krasnosel’skii fixed point theorem. We also prove the Sadovskii theorem fors-convex sets in p-normed spaces, where 0<p≤1,0<s≤p, and from it we obtain some fixed point theorems for the sum of two mappings. In the last section, as an application of a Krasnosel’skii-type theorem, the existence of solutions for perturbed integral equation is considered in p-normed spaces.


## 2. Preliminaries

Throughout this paper, we denote the closure and the boundary of a subsetA  of  X  by  A- and  ∂A, respectively. B(x,r) will be the open ball of  X  with center  x∈X and radius  r>0.Definition 2.1.
Let(X,dX) and (Y,dY)  be two metric spaces and  T:X→Y.  The mapping  T  is said to be k-Lipschitz, where  k  is a positive constant, if
(2.1)dY(Tx,Ty)≤kdX(x,y),∀x,y∈X.T is said to be nonexpansive if  k=1, and to be a contraction if  k<1.It is clear that everyk-Lipschitz mapping is continuous. Moreover, the Banach contraction principle holds for a closed subset in a complete  p-normed space.Definition 2.2 (see [3]).
Let(X,∥·∥p)(0<p≤1) be a  p-normed space and 0<s≤p. A set C⊂X  is said to be s-convex if the following condition is satisfied
(2.2)(1-t)1/sx+t1/sy∈C,wheneverx,y∈C,t∈[0,1].
Let  A⊂X. The  s-convex hull of A denoted by  cosA  is the smallest  s-convex set containing A and the closed s-convex hull of A denoted by cos¯A is the smallest closed s-convex set containing  A.In other words, thes-convexity of the set C is equivalent to that
(2.3)t1x+t2y∈C,wheneverx,y∈C,t1,t2≥0,t1s+t2s=1.
For  s=1, we obtain the usual definition of convex sets. For a subset  A  of  X, the s-convex hull of  A  is given by
(2.4)cosA={∑i=1ntixi:ti≥0,∑i=1ntis=1,xi∈A,n≥2}.
It is easy to see that if  C  is a closed  s-convex set, then θ∈C.Lemma 2.3 (see [3]).
Let(X,∥·∥p)(0<p≤1)  be a  p-normed space and  0<s≤p.(a)
The ballB(θ,r)  is  s-convex, where  r>0.(b)
IfC⊂X  is  s-convex and α∈K, then αC  is  s-convex.(c)
IfC1,C2⊂X  are  s-convex, then C1+C2  is  s-convex.(d)
IfCi⊂X,i=1,2,…  are all  s-convex, then ⋂i=1∞Ci  is   s-convex.(e)
IfA⊂X  and  θ∈A,  then  cosA⊂coA, where coA  is the convex hull of A.(f)
IfC is a closed  s-convex set and 0<k<s, then C is a closed  k-convex set.(g)
IfX is complete and A is a totally bounded subset of X, then co¯sA(0<s≤p) is compact.Theorem 2.4 (see [3] (Schauder-type)).
Let(X,∥·∥p)  be a complete  p-normed space and C  a compact  s-convex subset of  X, where  0<p≤1,0<s≤p. If  S:C→C is continuous, then  S  has a fixed point  (i.e., there exists x*∈C such that  Sx*=x*).Theorem 2.5.
Let(X,∥·∥p)  be a complete  p-normed space and C a closed  s-convex subset of X, where 0<p≤1,0<s≤p. If  S:C→C is a continuous compact map (i.e., the image of  C under  S  is compact), then  S  has a fixed point.Proof.
LetQ=cos¯(S(C)¯). Note Q is a closed compact s-convex subset of X and S(Q)⊆S(C)⊆Q. The result follows from Theorem 2.4.We will need the following definition.Definition 2.6 (see [4]).
Let(X,d)  be a metric space and C a subset of  X. The mapping  T:C→X  is said to be expansive, if there exists a constant  h>1  such that
(2.5)d(Tx,Ty)≥hd(x,y),∀x,y∈C.Theorem 2.7 (see [4]).
LetC  be a closed subset of a complete metric space (X,d). Assume that the mapping T:C→X is expansive and T(C)⊃C,  then there exists a unique point x*∈C such that Tx*=x*.Recently, Xiang and Yuan [4] established a Krasnosel’skii type fixed point theorem when the mapping  T  is expansive. For other related results, see also [5, 6].


## 3. Main Results

Theorem 3.1.
Let(X,∥·∥p)  be a complete  p-normed space and C a closed s-convex subset of X, where 0<p≤1,0<s≤p.  Suppose that(i)
S:C→X is a continuous compact mapping (i.e., the image of C under S is compact);(ii)
T:C→X is an expansive mapping;(iii)
z∈S(C)  implies  T(C)+z⊃C  where  T(C)+z={y+z:y∈T(C)}.
Then there exists a pointx*∈C such that Sx*+Tx*=x*.Proof.
Letz∈S(C). Then the mapping T+z:C→X satisfies the assumptions of Theorem 2.7 by virtue of (ii) and (iii), which guarantees that the equation
(3.1)Tx+z=x
has a unique solution x=τ(z)∈C. For any z1,z2∈S(C), we have
(3.2)T(τ(z1))+z1=τ(z1),T(τ(z2))+z2=τ(z2),
and so
(3.3)∥T(τ(z1))-T(τ(z2))∥p=∥τ(z1)-z1-τ(z2)+z2∥p≤∥z1-z2∥p+∥τ(z1)-τ(z2)∥p.
Since  T  is expansive, there exists a constant  h>1  such that
(3.4)∥T(τ(z1))-T(τ(z2))∥p≥h∥τ(z1)-τ(z2)∥p.
As a result
(3.5)∥τ(z1)-τ(z2)∥p≤1h-1∥z1-z2∥p.
This implies that  τ:S(C)→C is continuous. Since  S  is continuous on  C, it follows that τS:C→C  is also continuous. Since S is compact, so is τS. By Theorem 2.5, there exists x*∈C, such that  τ(S(x*))=x*. From (3.1), we have
(3.6)T(τ(S(x*)))+S(x*)=τ(S(x*)),
that is,
(3.7)Tx*+Sx*=x*.
This completes the proof.Corollary 3.2.
Let(X,∥·∥p)  be a complete  p-normed space and C a closed  s-convex subset of  X, where 0<p≤1,0<s≤p.  Suppose that(i)
S:C→X  is a continuous compact mapping;(ii)
T:C→X is an expansive and onto mapping.
Then there exists a pointx*∈C such that Sx*+Tx*=x*.The following example shows that there are mappings which are expansive and satisfyT(C)⊂C.Example 3.3.
LetX=C=ℝ with the usual metric and consider Tx=x3+2x+1 for x∈C.  Then for all x,y∈C,  we have
(3.8)|Tx-Ty|=|(x3-y3)+2(x-y)|=|(x-y)(x2+xy+y2)+2(x-y)|=|(x2+xy+y2+2)||x-y|≥2|x-y|.
Thus  T  is expansive with  T(C)⊂C.Theorem 3.4.
Let(X,∥·∥p)  be a complete  p-normed space and C a closed  s-convex subset of  X, where 0<p≤1,0<s≤p.  Suppose that(i)
S:C→X is a continuous compact mapping;(ii)
T:C→X is an expansive mapping;(iii)
z∈S(C) implies C+z⊂T(C)⊂C where C+z={y+z:y∈C}.
Then there exists a pointx*∈C such that T∘(I-S)x*=x*.Proof.
SinceT  is expansive, it follows that the inverse of  T:C→T(C) exists,  T-1:T(C)→C  is a contraction and hence continuous. Thus  T(C)  is a closed set. Then, for each fixed z∈S(C), the equation
(3.9)T-1x+z=x
has a unique solution x=τ(z)∈T(C). For any z1,z2∈S(C), we have
(3.10)T-1(τ(z1))+z1=τ(z1),T-1(τ(z2))+z2=τ(z2),
so that
(3.11)∥τ(z1)-τ(z2)∥p≤∥z1-z2∥p+∥T-1(τ(z1))-T-1(τ(z2))∥p≤∥z1-z2∥p+1h∥τ(z1)-τ(z2)∥p.
Thus,
(3.12)∥τ(z1)-τ(z2)∥p≤hh-1∥z1-z2∥p.
This shows that  τ:S(C)→T(C)  is continuous. Since S is continuous on C, it follows that τS:C→T(C)⊂C is also continuous and since S is compact, so is τS. Then, by Theorem 2.5, there exists x*∈C, such that τ(S(x*))=x*. From (3.9) we have
(3.13)T-1(τ(S(x*)))+S(x*)=τ(S(x*)),
that is,
(3.14)T-1x*+Sx*=x*,
that is,
(3.15)T∘(I-S)x*=x*.
This completes the proof.Lemma 3.5.
Let(X,∥·∥p)(0<p≤1) be a  p-normed space and C⊂X.  Suppose that the mapping T:C→X  is expansive with constant h>1. Then the inverse of F=I-T:C→(I-T)(C) exists and
(3.16)∥F-1x-F-1y∥p≤1h-1∥x-y∥p,x,y∈F(C).Proof.
Letx,y∈C, we have
(3.17)∥Fx-Fy∥p=∥(Tx-Ty)-(x-y)∥p≥∥Tx-Ty∥p-∥x-y∥p≥(h-1)∥x-y∥p.
From (3.17) we see that F is one-to-one. Therefore, the inverse of F:C→F(C) exists. Thus, for x,y∈F(C), we have F-1x,F-1y∈C. Now, using F-1x,F-1y and substituting for x,y in (3.17), respectively, we obtain
(3.18)∥F-1x-F-1y∥p≤1h-1∥x-y∥p.Theorem 3.6.
Let(X,∥·∥p) be a complete  p-normed space and C a closed  s-convex subset of  X, where 0<p≤1,0<s≤p. Suppose that(i)
S:C→X is a continuous compact mapping;(ii)
T:X→X (or  T:C→X) is an expansive mapping with constant  h>1;(iii)
S(C)⊂(I-T)(X) and [x=Tx+Sy,y∈C implies x∈C] or S(C)⊂(I-T)(C).
Then there exists a pointx*∈C  such that  Sx*+Tx*=x*.Proof.
From (iii), for eachy∈C  since  S(C)⊂(I-T)(X) or S(C)⊂(I-T)(C) there is an x∈X such that
(3.19)x-Tx=Sy.
If  S(C)⊂(I-T)(C) then x∈C  whereas if S(C)⊂(I-T)(X) then Lemma 3.5 and (iii) imply x=(I-T)-1Sy∈C. Now (I-T)-1 is continuous, and so (I-T)-1S is a continuous mapping of C into C. Since S is compact, so is (I-T)-1S. By Theorem 2.5, (I-T)-1S has a fixed point x*∈C with x*=(I-T)-1Sx*, that is, Sx*+Tx*=x*. This completes the proof.Theorem 3.7 (Petryshyn-type).
Let(X,∥·∥p) be a complete  p-normed space and D an open  s-convex subset of  X with θ∈D, where 0<p≤1,0<s≤p. Let S:D¯→X be a continuous compact mapping and
(3.20)∥x-Sx∥p≥∥Sx∥p,∀x∈∂D.
Then there exists z∈D¯ such that Sz=z.Proof.
The proof is exactly the same as the proof of Theorem  2.20 (b) [3]. Here we use Theorem 2.5 instead of Theorem  2.14 [3].Theorem 3.8.
Let(X,∥·∥p) be a complete p-normed space and D an open  s-convex subset of  X, where 0<p≤1,0<s≤p. Suppose that(i)
S:D¯→X is a continuous compact mapping;(ii)
T:X→Xis an expansive map with constant h>1;(iii)
S(D¯)⊂(I-T)(X);(iv)
∥Sx+Tθ∥p≤((h-1)/2)∥x∥p    for each x∈∂D.
Then there exists a pointx*∈D¯ such that Sx*+Tx*=x*.Proof.
From (iii), for eachx∈D¯, there is a y∈X such that
(3.21)y-Ty=Sx.
Thus by Lemma 3.5,  we have y=(I-T)-1Sx:=GSx∈X. Again by Lemma 3.5 and (i), we see that GS:D¯→X is compact. We now prove that (3.20) holds with S replaced by GS. In fact, for each x∈D¯,  from (3.21), we have
(3.22)T(GSx)+Sx=GSx,
so
(3.23)∥T(GSx)-Tθ∥p≤∥GSx∥p+∥Sx+Tθ∥p.
Since  T  is expansive, we have
(3.24)∥T(GSx)-Tθ∥p≥h∥GSx∥p.
It follows from (3.23) and (3.24) that
(3.25)∥GSx∥p≤1h-1∥Sx+Tθ∥p.
Thus, by (3.25) and (iv), for each x∈∂D, we have
(3.26)∥GSx∥p2-(∥GSx∥p-∥x∥p)2=∥x∥p(2∥GSx∥p-∥x∥p)≤∥x∥p(2h-1∥Sx+Tθ∥p-∥x∥p)≤0,
which implies (3.20). This completes the proof.Corollary 3.9.
Let(X,∥·∥p)  be a complete  p-normed space and D an open  s-convex subset of  X, where 0<p≤1,0<s≤p.  Suppose that(i)
S:D¯→X is a continuous compact mapping;(ii)
T:X→X is an expansive map with constant h>1;(iii)
z∈S(D¯) implies T(X)+z=X where T(X)+z={y+z:y∈T(X)};(iv)
∥Sx+Tθ∥p≤((h-1)/2)∥x∥p  for each x∈∂D.
Then there exists a pointx*∈D¯ such that Sx*+Tx*=x*.Theorem 3.10.
Let(X,∥·∥p)  be a complete  p-normed space and D an open  s-convex subset of  X, where 0<p≤1,0<s≤p.  Suppose that(i)
S:D¯→X is a continuous compact mapping;(ii)
T:X→X is a contraction with contractive constant α<1;(iii)
∥Sx+Tθ∥p≤((1-α)/2)∥x∥p for each x∈∂D.
Then there exists a pointx*∈D¯ such that Sx*+Tx*=x*.Proof.
For eachz∈S(D¯),  the mapping T+z:X→X  is  a  contraction. Thus, the equation
(3.27)Tx+z=x
has a unique solution x=σ(z)∈X. For any z1,z2∈S(D¯), from
(3.28)T(σ(z1))+z1=σ(z1),T(σ(z2))+z2=σ(z2),
it follows that
(3.29)∥σ(z1)-σ(z2)∥p=∥T(σ(z1))+z1-T(σ(z2))-z2∥p≤∥T(σ(z1))-T(σ(z2))∥p+∥z1-z2∥p.
Since  T is a contraction with contractive constant α<1, we have
(3.30)∥T(σ(z1))-T(σ(z2))∥p≤α∥σ(z1)-σ(z2)∥p.
Thus, we have
(3.31)∥σ(z1)-σ(z2)∥p≤11-α∥z1-z2∥p,z1,z2∈S(D¯).
It follows from (3.31) and (i) that σS:D¯→X is compact. From (3.27), we have
(3.32)∥σSx∥p≤11-α∥Sx+Tθ∥p,x∈D¯.
For every  x∈∂D,  from (3.32) and (iii), we deduce that
(3.33)∥σSx∥p2-(∥σSx∥p-∥x∥p)2=∥x∥p(2∥σSx∥p-∥x∥p)≤∥x∥p(21-α∥Sx+Tθ∥p-∥x∥p)≤0,
which implies (3.20). This completes the proof.


## 4. Condensing Mappings

Now, we extend the above results to a class of condensing mappings. For convenience, we recall some definitions, see [4, 7].Definition 4.1.
LetA  be a bounded subset of a metric space (X,d). The Kuratowski measure of noncompactness χ(A) of A is defined as follows:
(4.1)χ(A)=inf{δ>0:thereisafinitenumberofsubsetsAi⊂AsuchthatA⊆⋃iAianddiam(Ai)≤δ},
where  diam(Ai) denotes the diameter of set Ai.It is easy to prove the following fundamental properties ofχ, see [7](i)
χ(A)=χ(A¯).(ii)
χ(A)=0  if and only if  A¯  is compact.(iii)
A⊆B⇒χ(A)≤χ(B).(iv)
χ(A∪B)=max{χ(A),χ(B)}.(v)
IfA,B are bounded,  then χ(A+B)≤χ(A)+χ(B).(vi)
IfA is bounded  and  λ∈ℝ, then χ(λA)=|λ|χ(A).For (ii) one should remember that a set is compact if and only if it is closed and totally bounded.Proposition 4.2.
Let(X.∥·∥p)(0<p≤1) be a complete  p-normed space and let  A, B be two bounded subsets of X. Then
(4.2)χ(cos(A))=χ(A),
where 0<s≤p.Proof.
Letε>0  and  U(x;ε) the open ball with center x and radius ε. Note
(4.3)diam(Nε(S))≤diam(S)+2ε,
holds for any bounded set S in a metric space; here Nε(A)=⋃x∈AU(x;ε).  Then, if A is a bounded set, we have that
(4.4)χ(Nε(A))≤χ(A)+2ε.
Assume first thatA and B are bounded s-convex sets. Let C=A∪B. We now prove that
(4.5)χ(cos(C))≤max{χ(A),χ(B)}.
To do it, suppose that x∈cos(C) and  x∉A∪B. Then there exist xi∈C and ti≥0, i=1,…,n,∑i=1ntis=1,  such that x=∑i=1ntixi. Let  xi∈A for 1≤i≤m and xi∈B for m+1≤i≤n. Then, we have
(4.6)x=(∑i=1mtis)1/s∑i=1mtixi(∑i=1mtis)1/s+(∑i=m+1ntis)1/s∑i=m+1ntixi(∑i=m+1ntis)1/s.
Consequently
(4.7)x=(∑i=1mtis)1/su+(∑i=m+1ntis)1/sv,whereu∈A,v∈B,
and then (4.7) leads to
(4.8)x=t1/su+(1-t)1/sv,whereu∈A,v∈B,t∈[0,1].
Assume that for each ε>0, there is a positive integer  N  such that  1/N<ε. For each i=0,1,…,N, let
(4.9)Ci={x∈cos(C):x=(iN)1/su+(1-iN)1/svforsomeu∈A,v∈B}.
Note if t∈(0,1),  then  t∈[i/N,(i+1)/N]  for some 0≤i≤N-1. This implies that if x∈cos(C) and  x∉A∪B, then x∈Nε(Ci)  for some  i=0,…,N. Thus,
(4.10)cos(C)⊆(⋃i=1nNε(Ci))∪A∪B.
By (iv) and (4.4), we have
(4.11)χ(cos(C))≤max{χ(A),χ(B),max1≤i≤Nχ(Nε(Ci))}≤max{χ(A),χ(B),max1≤i≤N[χ(Ci)+2ε]}.
For each i,  by (v) and (vi), we deduce that
(4.12)χ(Ci)=χ((iN)1/sA+(1-iN)1/sB)≤(iN)1/sχ(A)+(1-iN)1/sχ(B)≤[(iN)1/s+(1-iN)1/s]max{χ(A),χ(B)}.
Since
(4.13)(iN)1/s+(1-iN)1/s≤1,
we have
(4.14)χ(Ci)≤max{χ(A),χ(B)}.
Hence,
(4.15)χ(cos(C))≤max{χ(A),χ(B)}+2ε.
Since  ε>0 is arbitrary we obtain (4.5).
Consequently, ifC=⋃i=1nAi,  where each Ai  is bounded  s-convex, we have that
(4.16)χ(cos(C))≤max1≤i≤Nχ(Ai).
Now, to prove  (4.2), let q>χ(A).  Then A⊆⋃i=1nAi,  where diam(Ai)≤q.  We now claim that  diam(cos(Ai))=diam(Ai).  In fact, since Ai⊆cos(Ai),  we have
(4.17)diam(Ai)≤diam(cos(Ai)),
for each i.  Also, let  x,y∈cos(Ai),  such that x=∑i=1ntixi,y=∑i=1ntiyi,xi,yi∈Ai  and ∑i=1ntis=1.  Then
(4.18)supx,y∈cos(Ai)∥x-y∥p≤∑i=1ntip∥xi-yi∥p≤∑i=1ntis∥xi-yi∥p≤∑i=1ntisdiam(Ai),
that is,
(4.19)diam(cos(Ai))≤diam(Ai).
Hence, diam(cos(Ai))=diam(Ai).
Now we may assume that eachAi is  s-convex for each i.  By (4.16), we have
(4.20)χ(cos(A))≤max1≤i≤Nχ(Ai)≤max1≤i≤Ndiam(Ai)≤q.
Since this is true for all q>χ(A) then χ(cos(A))≤χ(A) so (4.2) is proved.Definition 4.3.
LetX, Y be two metric spaces and Ω a subset of X. A bounded continuous map T:Ω→Y is k-set contractive if for any bounded set A⊂Ω, we have
(4.21)χ(T(A))≤kχ(A).T  is strictly  k-set contractive if  T  is  k-set  contractive and
(4.22)χ(T(A))<kχ(A)
for all bounded sets A⊂Ω with χ(A)≠0. We say T is a condensing map if T is a bounded continuous 1-set contractive map and
(4.23)χ(T(A))<χ(A)
for all bounded sets A⊂Ω with χ(A)≠0.Notice thatT is a compact map if and only if T is a 0-set contractive map.Now, we extend Sadovskii theorem in [7] to a map, that is, defined on a  p-normed space.Theorem 4.4 (Sadovskii type).
Let(X.∥·∥p) be a complete  p-normed space and  C  a closed  s-convex subset of X, where 0<p≤1,0<s≤p. If S:C→C is condensing and S(C) is bounded, then S has a fixed point.Proof.
For each fixedx∈C, let 𝔅 be the family of all closed  s-convex subsets A of C, such that x∈A and S:A→A.  Suppose that
(4.24)B=⋂A∈𝔅A,D=cos¯{S(B)∪{x}}.
Since x∈B and S:B→B, it follows that D⊆B. This implies that S(D)⊆S(B)⊆D. On the other hand, since x∈D, we have D∈𝔅 and B⊆D. Therefore, B=D.
As a result, we haveS(D)=S(B)⊆D  and
(4.25)χ(D)=χ(cos¯{S(B)∪{x}})=χ({S(B)∪{x}})=χ(S(B))=χ(S(D)).
Since S  is condensing, it follows that χ(D)=0, that is, D is compact. Therefore S is a compact mapping of  s-convex set D into itself. By Theorem 2.5,  S has a fixed point.We next make use of the main ideas established in [5, 6, 8] to obtain a Krasnosel’skii fixed point theorem in a  p-normed space.Theorem 4.5.
Let(X,∥·∥p)  be a complete p-normed space and C a closed  s-convex subset of X, where 0<p≤1,0<s≤p.  Suppose that S:C→X and(i)
T:C→X is such that the inverse of (I-T):C→(I-T)(C) exists;(ii)
S(C)⊆(I-T)(C);(iii)
(I-T)-1S:C→X is condensing and (I-T)-1S(C) is bounded.
Then there exists a pointx*∈C  such that  Sx*+Tx*=x*.Proof.
From (ii), we have(I-T)-1S:C→C.  Thus, (I-T)-1S  is a condensing map of C into itself.  By Theorem 4.4, (I-T)-1S has a fixed point. This completes the proof.The following lemma is easy to prove.Lemma 4.6.
Let(X,∥·∥p)(0<p≤1) be a complete  p-normed space and  C⊂X. Assume that T:C→X is a  k-Lipschitizian map, that is,
(4.26)∥Tx-Ty∥p≤k∥x-y∥p,x,y∈C.
Then for each bounded subset Ω of C, we have χ(T(Ω))≤kχ(Ω).Theorem 4.7.
Let(X,∥·∥p) be a complete  p-normed space and C a closed  s-convex subset of X, where 0<p≤1,0<s≤p. Suppose that(i)
S:C→X is a 1-set contractive map (condensing) and S(C) is bounded;(ii)
T:C→X is an expansive map with constant h>2(h≥2);(iii)
z∈S(C) implies T(C)+z⊃C where T(C)+z={y+z:y∈T(C)}.
Then there exists a pointx*∈C such that Sx*+Tx*=x*.Proof.
Letτ  be the function defined as in Theorem 3.1. We will show that τS:C→C is a condensing map. Let Ω be bounded in  C.  From (3.5) and Lemma 4.6, it follows that
(4.27)χ(τ(S(Ω)))≤1h-1χ(S(Ω)).
Suppose first that  S  is  1-set contractive. Then
(4.28)χ((τS)(Ω))≤1h-1χ(Ω),
which implies that τS:C→C is a condensing map. The other case when S is condensing and h≥2 also guarantees that τS:C→C is a condensing map. The result follows from Theorem 4.4. This completes the proof.Theorem 4.8.
Let(X,∥·∥p) be a complete p-normed space and C a closed s-convex subset of X, where 0<p≤1,0<s≤p. Suppose that(i)
S:C→X is a 1-set contractive map (condensing) and S(C) is bounded;(ii)
T:X→X (or T:C→X) is an expansive map with constant h>2(h≥2);(iii)
S(C)⊂(I-T)(X) and [x=Tx+Sy,y∈C implies x∈C] or S(C)⊂(I-T)(C).
Then there exists a pointx*∈C such that Sx*+Tx*=x*.Proof.
For eachx∈C, by (iii), there exists a y∈X such that
(4.29)y-Ty=Sx.
If  S(C)⊂(I-T)(C) then y∈C whereas if S(C)⊂(I-T)(X) then it follows from Lemma 3.5 and (iii), that y=(I-T)-1Sx∈C. Now, if A⊂C is bounded, then by Lemma 4.6 and (3.16), we have
(4.30)χ(((I-T)-1S)(A))≤1h-1χ(S(A)).
Suppose first that S   is  1-set contractive. Then
(4.31)χ(((I-T)-1S)(A))≤1h-1χ(A),
which implies since h>2 that (I-T)-1S:C→C is a condensing map. The other case when S is condensing and h≥2 also guarantees that (I-T)-1S:C→C is a condensing map.
The result follows from Theorem4.4. This completes the proof.Lemma 4.9.
Let(X,∥·∥p)(0<p≤1) be a  p-normed space and C⊂X. Suppose that the mapping T:C→X is a contraction with contractive constant α<1. Then the inverse of F=I-T:C→(I-T)(C)  exists and
(4.32)∥F-1x-F-1y∥p≤11-α∥x-y∥p,x,y∈F(C).Proof.
Since, for eachx,y∈C, we have
(4.33)∥Fx-Fy∥p=∥(x-y)-(Tx-Ty)∥p≥∥x-y∥p-∥Tx-Ty∥p≥(1-α)∥x-y∥p.
Then F is one-to-one and the inverse of F:C→F(C) exists. Suppose that
(4.34)G:=F-1-I:F(C)→X.
From the identity
(4.35)I=F∘F-1=(I-T)∘(I+G)=I+G-T∘(I+G),
we have that
(4.36)G=T∘(I+G).
Thus,
(4.37)∥Gx-Gy∥p≤α∥(I+G)x-(I+G)y∥p≤α(∥x-y∥p+∥Gx-Gy∥p),
and so
(4.38)∥Gx-Gy∥p≤α1-α∥x-y∥p,x,y∈F(C).
Therefore,
(4.39)∥F-1x-F-1y∥p≤∥Gx-Gy∥p+∥x-y∥p≤11-α∥x-y∥p.Remark 4.10.
IfT:C→X is a contraction with contractive constant α<1, by Lemma 4.9, it follows that (I-T)-1 exists and is continuous. If in addition
(4.40)Tx+Sy∈C,foranyx,y∈C
holds, then (I-T)-1S:C→C. To see this first note, since S(C)+T(C)⊆C, for each z∈S(C), the mapping T+z:C→C is a contraction. Thus,
(4.41)Tx+z=x
has a unique solution x∈C. Hence, z∈(I-T)(C) and we have S(C)⊆(I-T)(C). This shows that (I-T)-1S:C→C.
From Theorem4.4, we generalize a Krasnosel’skii type theorem (Theorem 1.1) as follows.Theorem 4.11.
Let(X,∥·∥p)  be a complete  p-normed space and C a closed  s-convex subset of X, where 0<p≤1,0<s≤p. Suppose that(i)
S:C→X is a strictly (1-α)-set contractive map (or a β-set contractive map with β<1-α) and S(C) is bounded;(ii)
T:C→X is a contraction with contractive constant α<1;(iii)
anyx,y∈C imply Tx+Sy∈C.
Then there exists a pointx*∈C such that Sx*+Tx*=x*.Proof.
LetA⊂C be bounded. By Remark 4.10,  Lemma 4.6 and (4.32), we have
(4.42)χ(((I-T)-1S)(A))≤11-αχ(S(A)),
and so (I-T)-1S:C→C is a condensing map. Hence, from Theorem 4.4, there exists a point x*∈C such that Sx*+Tx*=x*. This completes the proof.Remark 4.12.
If(I-T)-1 exists and is continuous on  X and if in addition
(4.43)x=Tx+Sy,y∈C⟹x∈C
holds, then (I-T)-1S:C→C. To see this first note if z∈S(C), then there exists y∈C such that z=S(y). Let x=(I-T)-1z, so x-Tx=z. This implies x=Tx+Sy, and then x∈C. Hence, z∈(I-T)(C) and we have S(C)⊆(I-T)(C). This shows that (I-T)-1S:C→C.Theorem 4.13.
Let(X,∥·∥p)  be a complete  p-normed space and C a closed  s-convex subset of  X, where 0<p≤1,0<s≤p. Suppose that(i)
S:C→X is a strictly (1-α)-set contractive map (or a β-set contractive map with β<1-α) and S(C) is bounded;(ii)
T:X→X is a contraction with contractive constant α<1;(iii)
[x=Tx+Sy,y∈C]⇒x∈C.
Then there exists a pointx*∈C such that Sx*+Tx*=x*.Proof.
LetA⊂C be bounded. By Remark 4.12, Lemma 4.6 and (4.32), we have
(4.44)χ(((I-T)-1S)(A))≤11-αχ(S(A)),
and so (I-T)-1S:C→C is a condensing map. Hence, from Theorem 4.4, there exists a point x*∈C such that Sx*+Tx*=x*. This completes the proof.


## 5. Application

In [9], Hajji and Hanebaly presented a modular version of Krasnosel’skii fixed point theorem and applied their result to the existence of solutions to perturbed integral equations in modular spaces. In this section, we use the same argument as in [9] to give an application of Krasnosel’skii fixed point theorem to a p-normed space. For more details about modular spaces, we refer the reader to [10, 11]. Now, we recall some definitions.Definition 5.1 (see [12]).
LetX be an arbitrary linear space over K, where K=ℝ or K=ℂ. A functional ρ:X→[0,+∞) is called a modular if (i)
ρ(x)=0 if and only if x=θ;(ii)
ρ(αx)=ρ(x) if α∈K with |α|=1, for all x∈X;(iii)
ρ(αx+βy)≤ρ(x)+ρ(y) if α, β≥0 with α+β=1, for all x,y∈X.If in place of (iii), we have(iv)
ρ(αx+βy)≤αsρ(x)+βsρ(y) for α, β≥0 and αs+βs=1 with an s∈(0,1],then the modular ρ is called an s-convex modular, and if s=1, ρ is called convex modular.A modularρ defines a corresponding modular space, that is, the space Xρ given by
(5.1)Xρ={x∈X:ρ(λx)⟶0asλ⟶0}.The subsetA of Xρ is ρ-bounded if
(5.2)supx,y∈Aρ(x-y)<+∞.
The ρ-diameter of A is defined by
(5.3)δρ(A)=supx,y∈Aρ(x-y).A simple example of a modular space is ap-normed space (X,∥·∥p).Now, we discuss the existence of solutions for the following perturbed integral equation(5.4)u(t)=e-tf0+∫0ter-t(T+h)u(r)dr,
in the modular space (p-normed space) Cp=C([0,1],Lp), where
(5.5)Lp=Lp[0,1]={f:f:[0,1]→Kismeasurable,and∥f(t)∥p=∫|f(t)|pdt<+∞},
and Cp=C([0,1],Lp) is the space of all continuous functions from [0,1] to Lp[0,1] under the modular (norm)
(5.6)∥g∥pc=sup0≤t≤1∥g(t)∥p.
Consider the following assumptions:(i)B is a bounded closed  s-convex subset of  Lp with θ∈B,  where 0<p≤1,0<s≤p;(ii)T:B→B is a mapping satisfying
(5.7)∥Tx-Ty∥p≤k∥x-y∥p,for allx,y∈B where 0<k<1, and h:B→B is a continuous compact mapping such that T(B)+h(B)⊆B;(iii)f0 is a fixed element of B.Theorem 5.2.
Suppose that (i)–(iii) are satisfied then the integral equation (5.4) has a solution u∈Cp.Proof.
ConsiderCp with
(5.8)∥u∥pc=sup0≤t≤1∥u(t)∥p,
for u∈D with D=C([0,1],B). First we show S:D→D where S is given by
(5.9)Su(t)=e-tf0+∫0ter-t(T+h)u(r)dr.
Suppose that tn,t0∈[0,1] and tn→t0 as n→∞. Since T and h are continuous, then (T+h)u is continuous at t0. In fact,
(5.10)∥(T+h)u(tn)-(T+h)u(t0)∥p≤∥Tu(tn)-Tu(t0)∥p+∥hu(tn)-hu(t0)∥p.
Let n→∞ so we have ∥(T+h)u(tn)-(T+h)u(t0)∥p→0, that is, (T+h)u is continuous at t0. Since t0∈[0,1] is arbitrary, Su is continuous from [0,1] into Lp. Next, in the complete space Lp, we have
(5.11)∫0ter-t(T+h)u(r)dr∈(∫0ter-tdr)cos¯(T+h)u(r),
where 0≤r≤t. Since (T+h)(B)⊆B, it follows that
(5.12)∫0ter-t(T+h)u(r)dr∈(1-e-t)cos¯(B),
and since B is closed s-convex, we have cos¯(B)=B¯=B. Therefore, Su(t)∈e-tB+(1-e-t)B⊆B for all t∈[0,1]. Thus S:D→D.
Let(5.13)T1u(t)=e-tf0+∫0ter-tTu(r)dr,h1u(t)=∫0ter-thu(r)dr.
Thus S=T1+h1. We will show that the hypotheses of Theorem 4.11 are satisfied.
Now sinceT(B)+h(B)⊂B we have
(5.14)T1u(t)+h1v(t)∈e-tB+(1-e-t)B⊂B,
for any t∈[0,1] and u,v∈D. Hence T1(D)+h1(D)⊂D. For any u,v∈D, we have
(5.15)T1u(t)-T1v(t)=∫0ter-t(Tu-Tv)(r)dr.
Fix t∈[0,1]. Using the same argument in the proof of Lemma  2.1 [12] we now show that T1 is a contraction. Let T={t0,t1,…,tn} be any subdivision of [0,t]. We know that ∑i=0n-1(ti+1-ti)eti-tx(ti) is convergent to ∫0ter-tx(r)dr in Lp when |T|=sup{|ti+1-ti|,i=0,…,n-1}→0 as n→∞. Therefore
(5.16)∥∫0ter-tx(r)dr∥p≤lim∥∑i=0n-1(ti+1-ti)eti-tx(ti)∥p.
On the other hand, since
(5.17)∑i=0n-1(ti+1-ti)eti-t≤∫0ter-tdr=1-e-t≤1,
then we have
(5.18)∥∑i=0n-1(ti+1-ti)eti-tx(ti)∥p≤|∑i=0n-1(ti+1-ti)eti-t|p∥x(ti)∥p≤∥x(ti)∥p≤sup∥x(ti)∥p≤∥x∥pc,
that is,
(5.19)∥∫0ter-tx(r)dr∥p≤∥x∥pc.
This implies that
(5.20)∥T1u(t)-T1v(t)∥p=∥∫0ter-t(Tu-Tv)(r)dr∥p≤∥Tu-Tv∥pc.
Note also
(5.21)∥Tu-Tv∥pc=supt∈[0,1]∥Tu(t)-Tv(t)∥p≤k∥u-v∥pc.
Hence,
(5.22)∥T1u(t)-T1v(t)∥p≤k∥u-v∥pc,
and so
(5.23)∥T1u-T1v∥pc≤k∥u-v∥pc,
for any u,v∈D. Therefore, T1 is a contraction.
Now we show thath1 is compact. Let M⊂D. Then h1(M) is equicontinuous. To see this let u∈M. Then
(5.24)h1u(t)-h1u(t*)=∫0ter-thu(r)dr-∫0t*er-t*hu(r)dr=e-t∫0terhu(r)dr-e-t*∫0t*erhu(r)dr=e-t∫0terhu(r)dr-e-t*∫0terhu(r)dr+e-t*∫0terhu(r)dr-e-t*∫0t*erhu(r)dr=(e-t-e-t*)∫0terhu(r)dr+e-t*∫t*terhu(r)dr.
Hence,
(5.25)∥h1u(t)-h1u(t*)∥p≤∥(e-t-e-t*)∫0terhu(r)dr∥p+∥e-t*∫t*terhu(r)dr∥p=|e-t-e-t*|p|∫01erdr|pδ∥·∥p(B)+|e-t*|p|∫t*terdr|pδ∥·∥p(B)≤2|e-t-e-t*|pδ∥·∥p(B)+|∫t*terdr|pδ∥·∥p(B)=2|e-t-e-t*|pδ∥·∥p(B)+|et-et*|pδ∥·∥p(B).
Recall the functions t↦e-t and t↦et are uniformly continuous on [0,1]. Therefore, for ε>0, there exists η1>0 such that if |t-t*|<η1 then
(5.26)|e-t-e-t*|≤[ε4δ∥·∥p(B)]1/p,
and there exists η2>0 such that if |t-t*|<η2 then
(5.27)|et-et*|≤[ε2δ∥·∥p(B)]1/p.
As a result with η=min(η1,η2) note if |t-t*|<η then
(5.28)∥h1u(t)-h1u(t*)∥p≤ε,
for any u∈M. Thus, h1(M) is equicontinuous. Moreover,
(5.29)h1u(t)=∫0ter-thu(r)dr∈(1-e-t)cos¯{hu(r),0≤r≤t}⊂(1-e-t)cos¯(h(B)).
Hence, for all t∈[0,1], we have
(5.30)h1(M(t))⊂(1-e-t)cos¯(h(B)).
Since, h(B) is compact, this implies that cos¯(h(B)) is compact. Therefore, h1(M(t))¯ is compact for all t∈[0,1]. By using the Arzela-Ascoli Theorem, we obtain that h1(M)¯ is compact and also h1 is continuous. Thus h1 is compact. Hence by Theorem 4.11, S has a fixed point.



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*Source: 267531-2012-10-03.xml*