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

and Applied Analysis 3 It is clear that every k-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 1 − t x t1/sy ∈ C, whenever x, y ∈ C, t ∈ 0, 1 . 2.2 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 cosA is the smallest closed s-convex set containing A. In other words, the s-convexity of the set C is equivalent to that t1x t2y ∈ C, whenever x, y ∈ C, t1, t2 ≥ 0, ts1 ts2 1. 2.3 For s 1, we obtain the usual definition of convex sets. For a subsetA ofX, the s-convex hull of A is given by


Introduction
Let X be a linear space over K K R or K 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 d p given by d p x, y x − y p for all x, y ∈ X.
Let Ω be a nonempty set, let M be a σ-algebra in Ω, and let μ : M → 0, ∞ be a positive measure. The space L p μ based on the complete measure space Ω, M, μ is an example of a p-normed space with the p-norm defined by f t p Ω f t p dμ, for f ∈ L p μ ,

1.1
Another example of a p-normed space is C p 0, 1 , the space of all continuous functions defined on the unit interval 0, 1 with the sup p-norm given by |x t | p , for x ∈ C p 0, 1 .

1.2
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 center θ 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 for s-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.
It is clear that every k-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 Let A ⊂ X. The s-convex hull of A denoted by co s A is the smallest s-convex set containing A and the closed s-convex hull of A denoted by co s A is the smallest closed s-convex set containing A.
In other words, the s-convexity of the set C is equivalent to that 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 It is easy to see that if C is a closed s-convex set, then θ ∈ C.
a The ball B θ, r is s-convex, where r > 0.
b If C ⊂ X is s-convex and α ∈ K, then αC is s-convex.
f If C is a closed s-convex set and 0 < k < s, then C is a closed k-convex set.
g If X is complete and A is a totally bounded subset of X, then co s A 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.
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 that is, 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 point x * ∈ C such that Sx * Tx * x * .
The following example shows that there are mappings which are expansive and satisfy T C ⊂ C. Then for all x, y ∈ C, we have

3.8
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; Then there exists a point x * ∈ C such that T • I − S x * x * .
Proof. Since T 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 has a unique solution x τ z ∈ T C . For any z 1 , z 2 ∈ S C , we have

3.11
Thus, 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 that is, 14 that is, This completes the proof.

3.16
Abstract and Applied Analysis 7 Proof. Let x, y ∈ C, we have

3.17
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 −1 x, F −1 y ∈ C. Now, using F −1 x, F −1 y and substituting for x, y in 3.17 , respectively, we obtain Then there exists a point x * ∈ C such that Sx * Tx * x * .
This completes the proof.

3.20
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 .

Abstract and Applied Analysis
Then there exists a point Thus by Lemma 3.5, we have y I − T −1 Sx : 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 Since T is expansive, we have It follows from 3.23 and 3.24 that Thus, by 3.25 and iv , for each x ∈ ∂D, we have which implies 3.20 . This completes the proof.
Abstract and Applied Analysis 9 Corollary 3.9. Let X, · p be a complete p-normed space and D an open s-convex subset of X, Then there exists a point x * ∈ D such that Sx * Tx * x * .
Then there exists a point x * ∈ D such that Sx * Tx * x * .
Proof. For each z ∈ S D , the mapping T z : X → X is a contraction. Thus, the equation has a unique solution x σ z ∈ X. For any z 1 , z 2 ∈ S D , from

3.29
Since T is a contraction with contractive constant α < 1, we have Thus, we have It follows from 3.31 and i that σS : D → X is compact. From 3.27 , we have 3.32

Abstract and Applied Analysis
For every x ∈ ∂D, from 3.32 and iii , we deduce that which implies 3.20 . This completes the proof.

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. Let A be a bounded subset of a metric space X, d . The Kuratowski measure of noncompactness χ A of A is defined as follows: It is easy to prove the following fundamental properties of χ, see 7 ii χ A 0 if and only if A is compact. iii vi If A 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.  and then 4.7 leads to Assume that for each ε > 0, there is a positive integer N such that 1/N < ε. For each i 0, 1, . . . , N, let Note if t ∈ 0, 1 , then t ∈ i/N, i 1 /N for some 0 ≤ i ≤ N−1. This implies that if x ∈ co s C and x / ∈ A ∪ B, then x ∈ N ε C i for some i 0, . . . , N. Thus,

4.10
By iv and 4.4 , we have

4.11
For each i, by v and vi , we deduce that

4.12
Since i N Hence, diam co s A i diam A i . Now we may assume that each A i is s-convex for each i. By 4.16 , we have Since this is true for all q > χ A then χ co s A ≤ χ A so 4.2 is proved.  As a result, we have S D S B ⊆ D and 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. 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
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 point x * ∈ C such that Sx * Tx * x * .
Proof. Let τ 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 Suppose first that S is 1-set contractive. Then 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; Then there exists a point x * ∈ C such that Sx * Tx * x * .
Proof. For each x ∈ C, by iii , there exists a y ∈ X such that y − Ty Sx.

4.29
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 −1 Sx ∈ C. Now, if A ⊂ C is bounded, then by Lemma 4.6 and 3.16 , we have Suppose first that S is 1-set contractive. Then which implies since h > 2 that I − T −1 S : C → C is a condensing map. The other case when S is condensing and h ≥ 2 also guarantees that I − T −1 S : C → C is a condensing map. The result follows from Theorem 4.4. 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 Proof. Since, for each x, y ∈ C, we have

4.33
Then F is one-to-one and the inverse of F : C → F C exists. Suppose that From the identity we have that Thus,

4.37
and so Therefore, 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 any x, y ∈ C imply Tx Sy ∈ C.
Then there exists a point x * ∈ C such that Sx * Tx * x * .
Proof. Let A ⊂ C be bounded. By Remark 4.10, Lemma 4.6 and 4.32 , we have and so I − T −1 S : 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.  iii x Tx Sy, y ∈ C ⇒ x ∈ C.
Then there exists a point x * ∈ C such that Sx * Tx * x * .
Proof. Let A ⊂ C be bounded. By Remark 4.12, Lemma 4.6 and 4.32 , we have and so I − T −1 S : 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.

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. 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 ≤ α s ρ x β s ρ y for α, β ≥ 0 and α s β s 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 modular ρ defines a corresponding modular space, that is, the space X ρ given by The ρ-diameter of A is defined by A simple example of a modular space is a p-normed space X, · p . iii f 0 is a fixed element of B. Suppose that t n , t 0 ∈ 0, 1 and t n → t 0 as n → ∞. Since T and h are continuous, then T h u is continuous at t 0 . In fact, T h u t n − T h u t 0 p ≤ Tu t n − Tu t 0 p hu t n − hu t 0 p .

5.10
Let n → ∞ so we have T h u t n − T h u t 0 p → 0, that is, T h u is continuous at t 0 .
Since t 0 ∈ 0, 1 is arbitrary, Su is continuous from 0, 1 into L p . Next, in the complete space L p , we have

5.13
Thus S T 1 h 1 . We will show that the hypotheses of Theorem 4.11 are satisfied. Now since T B h B ⊂ B we have 1 − e −t B ⊂ B, 5.14 for any t ∈ 0, 1 and u, v ∈ D. Hence T 1 D h 1 D ⊂ D. For any u, v ∈ D, we have e r−t Tu − Tv r dr.

5.15
Fix t ∈ 0, 1 . Using the same argument in the proof of Lemma 2.1 12 we now show that T 1 is a contraction. Let T {t 0 , t 1 , . . . , t n } be any subdivision of 0, t . We know that n−1 i 0 t i 1 − t i e t i −t x t i is convergent to t 0 e r−t x r dr in L p when |T | sup{|t i 1 − t i |, i 0, . . . , n − 1} → 0 as n → ∞. Therefore

5.30
Since, h B is compact, this implies that co s h B is compact. Therefore, h 1 M t is compact for all t ∈ 0, 1 . By using the Arzela-Ascoli Theorem, we obtain that h 1 M is compact and also h 1 is continuous. Thus h 1 is compact. Hence by Theorem 4.11, S has a fixed point.