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

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 in p-normed spaces.

1. Introduction

Let 𝑋 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 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=_Ω𝑓(𝑡)^𝑝𝑑𝜇,for𝑓∈𝐿^𝑝𝜇,𝐿^𝑝𝜇=𝑓∶𝑓∶Ω→𝐾ismeasurable,_Ω𝑓(𝑡)^𝑝𝑑𝜇<+∞. (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‖_𝑝=sup_0≤𝑡≤1|𝑥(𝑡)|^𝑝,for𝑥∈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.

2. Preliminaries

Throughout this paper, we denote the closure and the boundary of a subset 𝐴 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,𝑑_𝑋) and (𝑌,𝑑_𝑌) be two metric spaces and 𝑇∶𝑋→𝑌. The mapping 𝑇 is said to be 𝑘-Lipschitz, where 𝑘 is a positive constant, if 𝑑_𝑌𝑇𝑥,𝑇𝑦≤𝑘𝑑_𝑋𝑥,𝑦,∀𝑥,𝑦∈𝑋. (2.1)𝑇 is said to be nonexpansive if 𝑘=1, and to be a contraction if 𝑘<1.

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)^1/𝑠𝑥+𝑡^1/𝑠𝑦∈𝐶,whenever𝑥,𝑦∈𝐶,𝑡∈[0,1]. (2.2) Let 𝐴⊂𝑋. The 𝑠-convex hull of 𝐴 denoted by 𝑐𝑜_𝑠𝐴 is the smallest 𝑠-convex set containing 𝐴 and the closed 𝑠-convex hull of 𝐴 denoted by 𝑐𝑜_𝑠𝐴 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,whenever𝑥,𝑦∈𝐶,𝑡₁,𝑡₂≥0,𝑡^𝑠₁+𝑡^𝑠₂=1. (2.3) 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 𝑐𝑜_𝑠𝐴=^𝑛_𝑖=1𝑡_𝑖𝑥_𝑖∶𝑡_𝑖≥0,^𝑛_𝑖=1𝑡^𝑠_𝑖=1,𝑥_𝑖∈𝐴,𝑛≥2. (2.4) 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 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. (c)If 𝐶₁,𝐶₂⊂𝑋 are 𝑠-convex, then 𝐶₁+𝐶₂ is 𝑠-convex.(d)If 𝐶_𝑖⊂𝑋,𝑖=1,2,… are all 𝑠-convex, then ⋂^∞_𝑖=1𝐶_𝑖 is 𝑠-convex.(e) If 𝐴⊂𝑋 and 𝜃∈𝐴, then 𝑐𝑜_𝑠𝐴⊂𝑐𝑜𝐴, where 𝑐𝑜𝐴 is the convex hull of 𝐴.(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_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. Let 𝑄=𝑐𝑜_𝑠(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 𝑑𝑇𝑥,𝑇𝑦≥ℎ𝑑𝑥,𝑦,∀𝑥,𝑦∈𝐶. (2.5)

Theorem 2.7 (see [4]). Let 𝐶 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 point 𝑥^∗∈𝐶 such that 𝑆𝑥^∗+𝑇𝑥^∗=𝑥^∗.

Proof. Let 𝑧∈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 𝑇𝑥+𝑧=𝑥 (3.1) has a unique solution 𝑥=𝜏(𝑧)∈𝐶. For any 𝑧₁,𝑧₂∈𝑆(𝐶), we have 𝑇𝜏𝑧₁+𝑧₁=𝜏𝑧₁,𝑇𝜏𝑧₂+𝑧₂=𝜏𝑧₂, (3.2) and so 𝑇𝜏𝑧₁−𝑇𝜏𝑧_2𝑝=𝜏𝑧₁−𝑧₁−𝜏𝑧₂+𝑧_2𝑝≤𝑧₁−𝑧_2𝑝+𝜏𝑧₁−𝜏𝑧_2𝑝. (3.3) Since 𝑇 is expansive, there exists a constant ℎ>1 such that 𝑇𝜏𝑧₁−𝑇𝜏𝑧_2𝑝≥ℎ𝜏𝑧₁−𝜏𝑧_2𝑝. (3.4) As a result 𝜏𝑧₁−𝜏𝑧_2𝑝≤1𝑧ℎ−1¹−𝑧_2𝑝. (3.5) 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 𝑇𝜏𝑆𝑥^∗+𝑆𝑥^∗=𝜏𝑆𝑥^∗, (3.6) that is, 𝑇𝑥^∗+𝑆𝑥^∗=𝑥^∗. (3.7) 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.

Example 3.3. Let 𝑋=𝐶=ℝ with the usual metric and consider 𝑇𝑥=𝑥³+2𝑥+1 for 𝑥∈𝐶. Then for all 𝑥,𝑦∈𝐶, we have 𝑇𝑥−𝑇𝑦=𝑥³−𝑦³=𝑥+2𝑥−𝑦𝑥−𝑦²+𝑥𝑦+𝑦²=𝑥+2𝑥−𝑦₂+𝑥𝑦+𝑦².+2𝑥−𝑦≥2𝑥−𝑦 (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;(iii)𝑧∈𝑆(𝐶) implies 𝐶+𝑧⊂𝑇(𝐶)⊂𝐶 where 𝐶+𝑧={𝑦+𝑧∶𝑦∈𝐶}.
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 𝑇⁻¹𝑥+𝑧=𝑥 (3.9) has a unique solution 𝑥=𝜏(𝑧)∈𝑇(𝐶). For any 𝑧₁,𝑧₂∈𝑆(𝐶), we have 𝑇⁻¹𝜏𝑧₁+𝑧₁=𝜏𝑧₁,𝑇⁻¹𝜏𝑧₂+𝑧₂=𝜏𝑧₂, (3.10) so that 𝜏𝑧₁−𝜏𝑧_2𝑝≤𝑧₁−𝑧_2𝑝+𝑇⁻¹𝜏𝑧₁−𝑇⁻¹𝜏𝑧_2𝑝≤𝑧₁−𝑧_2𝑝+1ℎ𝜏𝑧₁−𝜏𝑧_2𝑝. (3.11) Thus, 𝜏𝑧₁−𝜏𝑧_2𝑝≤ℎ𝑧ℎ−1¹−𝑧_2𝑝. (3.12) 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 𝑇⁻¹𝜏𝑆𝑥^∗+𝑆𝑥^∗=𝜏𝑆𝑥^∗, (3.13) that is, 𝑇⁻¹𝑥^∗+𝑆𝑥^∗=𝑥^∗, (3.14) that is, 𝑇∘(𝐼−𝑆)𝑥^∗=𝑥^∗. (3.15) 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 𝐹⁻¹𝑥−𝐹⁻¹𝑦_𝑝≤1ℎ−1𝑥−𝑦_𝑝,𝑥,𝑦∈𝐹(𝐶). (3.16)

Proof. Let 𝑥,𝑦∈𝐶, we have 𝐹𝑥−𝐹𝑦_𝑝=𝑇𝑥−𝑇𝑦−𝑥−𝑦_𝑝≥𝑇𝑥−𝑇𝑦_𝑝−𝑥−𝑦_𝑝≥(ℎ−1)𝑥−𝑦_𝑝. (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⁻¹𝑥,𝐹⁻¹𝑦∈𝐶. Now, using 𝐹⁻¹𝑥,𝐹⁻¹𝑦 and substituting for 𝑥,𝑦 in (3.17), respectively, we obtain 𝐹⁻¹𝑥−𝐹⁻¹𝑦_𝑝≤1ℎ−1𝑥−𝑦_𝑝. (3.18)

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 point 𝑥^∗∈𝐶 such that 𝑆𝑥^∗+𝑇𝑥^∗=𝑥^∗.

Proof. From (iii), for each 𝑦∈𝐶 since 𝑆(𝐶)⊂(𝐼−𝑇)(𝑋) or 𝑆(𝐶)⊂(𝐼−𝑇)(𝐶) there is an 𝑥∈𝑋 such that 𝑥−𝑇𝑥=𝑆𝑦. (3.19) 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 ‖𝑥−𝑆𝑥‖_𝑝≥‖𝑆𝑥‖_𝑝,∀𝑥∈𝜕𝐷. (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].

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 point 𝑥^∗∈𝐷 such that 𝑆𝑥^∗+𝑇𝑥^∗=𝑥^∗.

Proof. From (iii), for each 𝑥∈𝐷, there is a 𝑦∈𝑋 such that 𝑦−𝑇𝑦=𝑆𝑥. (3.21) 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 𝑇(𝐺𝑆𝑥)+𝑆𝑥=𝐺𝑆𝑥, (3.22) so ‖𝑇(𝐺𝑆𝑥)−𝑇𝜃‖_𝑝≤‖𝐺𝑆𝑥‖_𝑝+‖𝑆𝑥+𝑇𝜃‖_𝑝. (3.23) Since 𝑇 is expansive, we have ‖𝑇(𝐺𝑆𝑥)−𝑇𝜃‖_𝑝≥ℎ‖𝐺𝑆𝑥‖_𝑝. (3.24) It follows from (3.23) and (3.24) that ‖𝐺𝑆𝑥‖_𝑝≤1‖‖ℎ−1𝑆𝑥+𝑇𝜃_𝑝. (3.25) Thus, by (3.25) and (iv), for each 𝑥∈𝜕𝐷, we have ‖𝐺𝑆𝑥‖²_𝑝−‖𝐺𝑆𝑥‖_𝑝−‖𝑥‖_𝑝²=‖𝑥‖_𝑝2‖𝐺𝑆𝑥‖_𝑝−‖𝑥‖_𝑝≤‖𝑥‖_𝑝2‖‖ℎ−1𝑆𝑥+𝑇𝜃_𝑝−‖𝑥‖_𝑝≤0, (3.26) 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 point 𝑥^∗∈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 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 𝑇𝑥+𝑧=𝑥 (3.27) has a unique solution 𝑥=𝜎(𝑧)∈𝑋. For any 𝑧₁,𝑧₂∈𝑆(𝐷), from 𝑇𝜎𝑧₁+𝑧₁=𝜎𝑧₁,𝑇𝜎𝑧₂+𝑧₂=𝜎𝑧₂, (3.28) it follows that 𝜎𝑧₁−𝜎𝑧_2𝑝=𝑇𝜎𝑧₁+𝑧₁−𝑇𝜎𝑧₂−𝑧_2𝑝≤𝑇𝜎𝑧₁−𝑇𝜎𝑧_2𝑝+𝑧₁−𝑧_2𝑝. (3.29) Since 𝑇 is a contraction with contractive constant 𝛼<1, we have 𝑇𝜎𝑧₁−𝑇𝜎𝑧_2𝑝≤𝛼𝜎𝑧₁−𝜎𝑧_2𝑝. (3.30) Thus, we have 𝜎𝑧₁−𝜎𝑧_2𝑝≤1𝑧1−𝛼¹−𝑧_2𝑝,𝑧₁,𝑧₂∈𝑆𝐷. (3.31) It follows from (3.31) and (i) that 𝜎𝑆∶𝐷→𝑋 is compact. From (3.27), we have ‖𝜎𝑆𝑥‖_𝑝≤1‖‖1−𝛼𝑆𝑥+𝑇𝜃_𝑝,𝑥∈𝐷. (3.32) For every 𝑥∈𝜕𝐷, from (3.32) and (iii), we deduce that ‖𝜎𝑆𝑥‖²_𝑝−‖𝜎𝑆𝑥‖_𝑝−‖𝑥‖_𝑝²=‖𝑥‖_𝑝2‖𝜎𝑆𝑥‖_𝑝−‖𝑥‖_𝑝≤‖𝑥‖_𝑝2‖‖1−𝛼𝑆𝑥+𝑇𝜃_𝑝−‖𝑥‖_𝑝≤0, (3.33) 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. 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: 𝜒(𝐴)=inf𝛿>0∶thereisafinitenumberofsubsets𝐴_𝑖⋃⊂𝐴suchthat𝐴⊆_𝑖𝐴_𝑖anddiam𝐴_𝑖≤𝛿, (4.1) where diam(𝐴_𝑖) denotes the diameter of set 𝐴_𝑖.

It is easy to prove the following fundamental properties of 𝜒, 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) If 𝐴,𝐵 are bounded, then 𝜒(𝐴+𝐵)≤𝜒(𝐴)+𝜒(𝐵). (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.

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 𝜒𝑐𝑜_𝑠(𝐴)=𝜒(𝐴), (4.2) where 0<𝑠≤𝑝.

Proof. Let 𝜀>0 and 𝑈(𝑥;𝜀) the open ball with center 𝑥 and radius 𝜀. Note diam𝑁_𝜀(𝑆)≤diam(𝑆)+2𝜀, (4.3) holds for any bounded set 𝑆 in a metric space; here 𝑁_𝜀(𝐴)=⋃_𝑥∈𝐴𝑈(𝑥;𝜀). Then, if 𝐴 is a bounded set, we have that 𝜒𝑁_𝜀(𝐴)≤𝜒(𝐴)+2𝜀. (4.4)
Assume first that 𝐴 and 𝐵 are bounded 𝑠-convex sets. Let 𝐶=𝐴∪𝐵. We now prove that 𝜒𝑐𝑜_𝑠(𝐶)≤max𝜒(𝐴),𝜒(𝐵). (4.5) To do it, suppose that 𝑥∈𝑐𝑜_𝑠(𝐶) and 𝑥∉𝐴∪𝐵. Then there exist 𝑥_𝑖∈𝐶 and 𝑡_𝑖≥0, 𝑖=1,…,𝑛,∑^𝑛_𝑖=1𝑡^𝑠_𝑖=1, such that 𝑥=∑^𝑛_𝑖=1𝑡_𝑖𝑥_𝑖. Let 𝑥_𝑖∈𝐴 for 1≤𝑖≤𝑚 and 𝑥_𝑖∈𝐵 for 𝑚+1≤𝑖≤𝑛. Then, we have 𝑥=^𝑚_𝑖=1𝑡^𝑠_𝑖^1/𝑠∑^𝑚_𝑖=1𝑡_𝑖𝑥_𝑖∑^𝑚_𝑖=1𝑡^𝑠_𝑖^1/𝑠+^𝑛_𝑖=𝑚+1𝑡^𝑠_𝑖^1/𝑠∑^𝑛_𝑖=𝑚+1𝑡_𝑖𝑥_𝑖∑^𝑛_𝑖=𝑚+1𝑡^𝑠_𝑖^1/𝑠. (4.6) Consequently 𝑥=^𝑚_𝑖=1𝑡^𝑠_𝑖^1/𝑠𝑢+^𝑛_𝑖=𝑚+1𝑡^𝑠_𝑖^1/𝑠𝑣,where𝑢∈𝐴,𝑣∈𝐵, (4.7) and then (4.7) leads to 𝑥=𝑡^1/𝑠𝑢+(1−𝑡)^1/𝑠𝑣,where𝑢∈𝐴,𝑣∈𝐵,𝑡∈[0,1]. (4.8) Assume that for each 𝜀>0, there is a positive integer 𝑁 such that 1/𝑁<𝜀. For each 𝑖=0,1,…,𝑁, let 𝐶_𝑖=𝑥∈𝑐𝑜_𝑠(𝐶)𝑖∶𝑥=𝑁^1/𝑠𝑖𝑢+1−𝑁^1/𝑠𝑣forsome𝑢∈𝐴,𝑣∈𝐵. (4.9) 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, 𝑐𝑜_𝑠(𝐶)⊆^𝑛_𝑖=1𝑁_𝜀𝐶_𝑖∪𝐴∪𝐵. (4.10) By (iv) and (4.4), we have 𝜒𝑐𝑜_𝑠(𝐶)≤max𝜒(𝐴),𝜒(𝐵),max_{1≤𝑖≤𝑁}𝜒𝑁_𝜀𝐶_𝑖≤max𝜒(𝐴),𝜒(𝐵),max_{1≤𝑖≤𝑁}𝜒𝐶_𝑖+2𝜀. (4.11) For each 𝑖, by (v) and (vi), we deduce that 𝜒𝐶_𝑖𝑖=𝜒𝑁^1/𝑠𝑖𝐴+1−𝑁^1/𝑠𝐵≤𝑖𝑁^1/𝑠𝜒(𝐴)+𝑖1−𝑁^1/𝑠𝜒(𝐵)≤𝑖𝑁^1/𝑠+𝑖1−𝑁^1/𝑠max𝜒(𝐴),𝜒(𝐵). (4.12) Since 𝑖𝑁^1/𝑠+𝑖1−𝑁^1/𝑠≤1, (4.13) we have 𝜒𝐶_𝑖≤max𝜒(𝐴),𝜒(𝐵). (4.14) Hence, 𝜒𝑐𝑜_𝑠(𝐶)≤max𝜒(𝐴),𝜒(𝐵)+2𝜀. (4.15) Since 𝜀>0 is arbitrary we obtain (4.5).
Consequently, if 𝐶=⋃^𝑛_𝑖=1𝐴_𝑖, where each 𝐴_𝑖 is bounded 𝑠-convex, we have that 𝜒𝑐𝑜_𝑠(𝐶)≤max_{1≤𝑖≤𝑁}𝜒𝐴_𝑖. (4.16)
Now, to prove (4.2), let 𝑞>𝜒(𝐴). Then 𝐴⊆⋃^𝑛_𝑖=1𝐴_𝑖, where diam(𝐴_𝑖)≤𝑞. We now claim that diam(𝑐𝑜_𝑠(𝐴_𝑖))=diam(𝐴_𝑖). In fact, since 𝐴_𝑖⊆𝑐𝑜_𝑠(𝐴_𝑖), we have diam𝐴_𝑖≤diam𝑐𝑜_𝑠𝐴_𝑖, (4.17) for each 𝑖. Also, let 𝑥,𝑦∈𝑐𝑜_𝑠(𝐴_𝑖), such that 𝑥=∑^𝑛_𝑖=1𝑡_𝑖𝑥_𝑖,𝑦=∑^𝑛_𝑖=1𝑡_𝑖𝑦_𝑖,𝑥_𝑖,𝑦_𝑖∈𝐴_𝑖 and ∑^𝑛_𝑖=1𝑡^𝑠_𝑖=1. Then sup_{𝑥,𝑦∈𝑐𝑜𝑠𝐴𝑖}𝑥−𝑦_𝑝≤^𝑛_𝑖=1𝑡^𝑝_𝑖𝑥_𝑖−𝑦_𝑖𝑝≤^𝑛_𝑖=1𝑡^𝑠_𝑖𝑥_𝑖−𝑦_𝑖𝑝≤^𝑛_𝑖=1𝑡^𝑠_𝑖diam𝐴_𝑖, (4.18) that is, diam𝑐𝑜_𝑠𝐴_𝑖≤diam𝐴_𝑖. (4.19) 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 𝜒𝑐𝑜_𝑠(𝐴)≤max_{1≤𝑖≤𝑁}𝜒𝐴_𝑖≤max_{1≤𝑖≤𝑁}diam𝐴_𝑖≤𝑞. (4.20) Since this is true for all 𝑞>𝜒(𝐴) then 𝜒(𝑐𝑜_𝑠(𝐴))≤𝜒(𝐴) so (4.2) is proved.

Definition 4.3. Let 𝑋, 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 𝜒(𝑇(𝐴))≤𝑘𝜒(𝐴). (4.21)𝑇 is strictly 𝑘-set contractive if 𝑇 is 𝑘-set contractive and 𝜒(𝑇(𝐴))<𝑘𝜒(𝐴) (4.22) 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 𝜒(𝑇(𝐴))<𝜒(𝐴) (4.23) for all bounded sets 𝐴⊂Ω with 𝜒(𝐴)≠0.

Notice that 𝑇 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 fixed 𝑥∈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 𝐵=_𝐴∈𝔅𝐴,𝐷=𝑐𝑜_𝑠{𝑆(𝐵)∪{𝑥}}. (4.24) 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 have 𝑆(𝐷)=𝑆(𝐵)⊆𝐷 and 𝜒(𝐷)=𝜒𝑐𝑜_𝑠{𝑆(𝐵)∪{𝑥}}=𝜒({𝑆(𝐵)∪{𝑥}})=𝜒(𝑆(𝐵))=𝜒(𝑆(𝐷)). (4.25) 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 point 𝑥^∗∈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, 𝑇𝑥−𝑇𝑦_𝑝≤𝑘𝑥−𝑦_𝑝,𝑥,𝑦∈𝐶. (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 𝜒(𝜏(𝑆(Ω≤1)))𝜒(𝑆(Ω.ℎ−1)) (4.27) Suppose first that 𝑆 is 1-set contractive. Then 𝜒Ω≤1((𝜏𝑆)())𝜒(Ω),ℎ−1 (4.28) 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 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)⁻¹𝑆𝑥∈𝐶. Now, if 𝐴⊂𝐶 is bounded, then by Lemma 4.6 and (3.16), we have 𝜒(𝐼−𝑇)⁻¹𝑆(𝐴≤1)𝜒(𝑆(𝐴.ℎ−1)) (4.30) Suppose first that 𝑆 is 1-set contractive. Then 𝜒(𝐼−𝑇)⁻¹𝑆(𝐴≤1)𝜒(𝐴),ℎ−1 (4.31) 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 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 𝐹⁻¹𝑥−𝐹⁻¹𝑦_𝑝≤11−𝛼𝑥−𝑦_𝑝,𝑥,𝑦∈𝐹(𝐶). (4.32)

Proof. Since, for each 𝑥,𝑦∈𝐶, we have 𝐹𝑥−𝐹𝑦_𝑝=𝑥−𝑦−𝑇𝑥−𝑇𝑦_𝑝≥𝑥−𝑦_𝑝−𝑇𝑥−𝑇𝑦_𝑝≥(1−𝛼)𝑥−𝑦_𝑝. (4.33) Then 𝐹 is one-to-one and the inverse of 𝐹∶𝐶→𝐹(𝐶) exists. Suppose that 𝐺∶=𝐹⁻¹−𝐼∶𝐹(𝐶)→𝑋. (4.34) From the identity 𝐼=𝐹∘𝐹⁻¹=(𝐼−𝑇)∘(𝐼+𝐺)=𝐼+𝐺−𝑇∘(𝐼+𝐺), (4.35) we have that 𝐺=𝑇∘(𝐼+𝐺). (4.36) Thus, 𝐺𝑥−𝐺𝑦_𝑝≤𝛼(𝐼+𝐺)𝑥−(𝐼+𝐺)𝑦_𝑝≤𝛼𝑥−𝑦_𝑝+𝐺𝑥−𝐺𝑦_𝑝, (4.37) and so 𝐺𝑥−𝐺𝑦_𝑝≤𝛼1−𝛼𝑥−𝑦_𝑝,𝑥,𝑦∈𝐹(𝐶). (4.38) Therefore, 𝐹⁻¹𝑥−𝐹⁻¹𝑦_𝑝≤𝐺𝑥−𝐺𝑦_𝑝+𝑥−𝑦_𝑝≤11−𝛼𝑥−𝑦_𝑝. (4.39)

Remark 4.10. If 𝑇∶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 𝑇𝑥+𝑆𝑦∈𝐶,forany𝑥,𝑦∈𝐶 (4.40) 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, 𝑇𝑥+𝑧=𝑥 (4.41) 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 Theorem 4.4, 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)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 𝜒(𝐼−𝑇)⁻¹𝑆(𝐴≤1)𝜒(𝑆(𝐴,1−𝛼)) (4.42) 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 𝑥=𝑇𝑥+𝑆𝑦,𝑦∈𝐶⟹𝑥∈𝐶 (4.43) 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 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 𝜒(𝐼−𝑇)⁻¹𝑆(𝐴≤1)𝜒(𝑆(𝐴,1−𝛼)) (4.44) 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]). Let 𝑋 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)≤𝛼^𝑠𝜌(𝑥)+𝛽^𝑠𝜌(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 𝑋_𝜌=𝑥∈𝑋∶𝜌(𝜆𝑥)⟶0as𝜆⟶0. (5.1)

The subset 𝐴 of 𝑋_𝜌 is 𝜌-bounded if sup_{𝑥,𝑦∈𝐴}𝜌𝑥−𝑦<+∞. (5.2) The 𝜌-diameter of 𝐴 is defined by 𝛿_𝜌(𝐴)=sup_{𝑥,𝑦∈𝐴}𝜌𝑥−𝑦. (5.3)

A simple example of a modular space is a 𝑝-normed space (𝑋,‖⋅‖_𝑝).

Now, we discuss the existence of solutions for the following perturbed integral equation 𝑢(𝑡)=𝑒^−𝑡𝑓₀+^𝑡₀𝑒^𝑟−𝑡(𝑇+ℎ)𝑢(𝑟)𝑑𝑟, (5.4) in the modular space (𝑝-normed space) 𝐶_𝑝=𝐶([0,1],𝐿^𝑝), where 𝐿^𝑝=𝐿^𝑝[]=[]𝑓(𝑡0,1𝑓∶𝑓∶0,1→𝐾ismeasurable,and)_𝑝=𝑓(𝑡)^𝑝𝑑𝑡<+∞, (5.5) and 𝐶_𝑝=𝐶([0,1],𝐿^𝑝) is the space of all continuous functions from [0,1] to 𝐿^𝑝[0,1] under the modular (norm) 𝑔_𝑝𝑐=sup_0≤𝑡≤1𝑔(𝑡)_𝑝. (5.6) Consider the following assumptions:

(i) 𝐵 is a bounded closed 𝑠-convex subset of 𝐿^𝑝 with 𝜃∈𝐵, where 0<𝑝≤1,0<𝑠≤𝑝;

(ii) 𝑇∶𝐵→𝐵 is a mapping satisfying 𝑇𝑥−𝑇𝑦_𝑝≤𝑘𝑥−𝑦_𝑝, (5.7)

for all 𝑥,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∈𝐶_𝑝.

Proof. Consider 𝐶_𝑝 with ‖𝑢‖_𝑝𝑐=sup_0≤𝑡≤1‖𝑢(𝑡)‖_𝑝, (5.8) for 𝑢∈𝐷 with 𝐷=𝐶([0,1],𝐵). First we show 𝑆∶𝐷→𝐷 where 𝑆 is given by 𝑆𝑢(𝑡)=𝑒^−𝑡𝑓₀+^𝑡₀𝑒^𝑟−𝑡(𝑇+ℎ)𝑢(𝑟)𝑑𝑟. (5.9) Suppose that 𝑡_𝑛,𝑡₀∈[0,1] and 𝑡_𝑛→𝑡₀ as 𝑛→∞. Since 𝑇 and ℎ are continuous, then (𝑇+ℎ)𝑢 is continuous at 𝑡₀. In fact, (𝑇+ℎ)𝑢𝑡_𝑛−(𝑇+ℎ)𝑢𝑡_0𝑝≤𝑇𝑢𝑡_𝑛−𝑇𝑢𝑡_0𝑝+ℎ𝑢𝑡_𝑛−ℎ𝑢𝑡_0𝑝. (5.10) Let 𝑛→∞ so we have ‖(𝑇+ℎ)𝑢(𝑡_𝑛)−(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 𝐿^𝑝. Next, in the complete space 𝐿^𝑝, we have ^𝑡₀𝑒^𝑟−𝑡(𝑇+ℎ)𝑢(𝑟)𝑑𝑟∈^𝑡₀𝑒^𝑟−𝑡𝑑𝑟𝑐𝑜_𝑠(𝑇+ℎ)𝑢(𝑟), (5.11) where 0≤𝑟≤𝑡. Since (𝑇+ℎ)(𝐵)⊆𝐵, it follows that ^𝑡₀𝑒^𝑟−𝑡(𝑇+ℎ)𝑢(𝑟)𝑑𝑟∈1−𝑒^−𝑡𝑐𝑜_𝑠(𝐵), (5.12) and since 𝐵 is closed 𝑠-convex, we have 𝑐𝑜_𝑠(𝐵)=𝐵=𝐵. Therefore, 𝑆𝑢(𝑡)∈𝑒^−𝑡𝐵+(1−𝑒^−𝑡)𝐵⊆𝐵 for all 𝑡∈[0,1]. Thus 𝑆∶𝐷→𝐷.
Let 𝑇₁𝑢(𝑡)=𝑒^−𝑡𝑓₀+^𝑡₀𝑒^𝑟−𝑡(𝑟)ℎ𝑇𝑢𝑑𝑟,₁𝑢(𝑡)=^𝑡₀𝑒^𝑟−𝑡ℎ𝑢(𝑟)𝑑𝑟. (5.13) Thus 𝑆=𝑇₁+ℎ₁. We will show that the hypotheses of Theorem 4.11 are satisfied.
Now since 𝑇(𝐵)+ℎ(𝐵)⊂𝐵 we have 𝑇₁𝑢(𝑡)+ℎ₁𝑣(𝑡)∈𝑒^−𝑡𝐵+1−𝑒^−𝑡𝐵⊂𝐵, (5.14) for any 𝑡∈[0,1] and 𝑢,𝑣∈𝐷. Hence 𝑇₁(𝐷)+ℎ₁(𝐷)⊂𝐷. For any 𝑢,𝑣∈𝐷, we have 𝑇₁𝑢(𝑡)−𝑇₁𝑣(𝑡)=^𝑡₀𝑒^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 𝑇₁ is a contraction. Let 𝑇={𝑡₀,𝑡₁,…,𝑡_𝑛} be any subdivision of [0,𝑡]. We know that ∑^𝑛−1_𝑖=0(𝑡_𝑖+1−𝑡_𝑖)𝑒^{𝑡𝑖−𝑡}𝑥(𝑡_𝑖) is convergent to ∫^𝑡₀𝑒^𝑟−𝑡𝑥(𝑟)𝑑𝑟 in 𝐿^𝑝 when |𝑇|=sup{|𝑡_𝑖+1−𝑡_𝑖|,𝑖=0,…,𝑛−1}→0 as 𝑛→∞. Therefore ^𝑡₀𝑒^𝑟−𝑡𝑥(𝑟)𝑑𝑟_𝑝≤lim^𝑛−1_𝑖=0𝑡_𝑖+1−𝑡_𝑖𝑒^{𝑡𝑖−𝑡}𝑥𝑡_𝑖𝑝. (5.16) On the other hand, since ^𝑛−1_𝑖=0𝑡_𝑖+1−𝑡_𝑖𝑒^{𝑡𝑖−𝑡}≤^𝑡₀𝑒^𝑟−𝑡𝑑𝑟=1−𝑒^−𝑡≤1, (5.17) then we have ^𝑛−1_𝑖=0𝑡_𝑖+1−𝑡_𝑖𝑒^{𝑡𝑖−𝑡}𝑥𝑡_𝑖𝑝≤^𝑛−1_𝑖=0𝑡_𝑖+1−𝑡_𝑖𝑒^{𝑡𝑖−𝑡𝑝}𝑥𝑡_𝑖𝑝≤𝑥𝑡_𝑖𝑝≤sup𝑥𝑡_𝑖𝑝≤‖𝑥‖_𝑝𝑐, (5.18) that is, ^𝑡₀𝑒^𝑟−𝑡𝑥(𝑟)𝑑𝑟_𝑝≤‖𝑥‖_𝑝𝑐. (5.19) This implies that 𝑇₁𝑢(𝑡)−𝑇₁𝑣(𝑡)_𝑝=^𝑡₀𝑒^𝑟−𝑡(𝑇𝑢−𝑇𝑣)(𝑟)𝑑𝑟_𝑝≤‖𝑇𝑢−𝑇𝑣‖_𝑝𝑐. (5.20) Note also ‖𝑇𝑢−𝑇𝑣‖_𝑝𝑐=sup_[]𝑡∈0,1‖𝑇𝑢(𝑡)−𝑇𝑣(𝑡)‖_𝑝≤𝑘‖𝑢−𝑣‖_𝑝𝑐. (5.21) Hence, 𝑇₁𝑢(𝑡)−𝑇₁𝑣(𝑡)_𝑝≤𝑘‖𝑢−𝑣‖_𝑝𝑐, (5.22) and so 𝑇₁𝑢−𝑇₁𝑣_𝑝𝑐≤𝑘‖𝑢−𝑣‖_𝑝𝑐, (5.23) for any 𝑢,𝑣∈𝐷. Therefore, 𝑇₁ is a contraction.
Now we show that ℎ₁ is compact. Let 𝑀⊂𝐷. Then ℎ₁(𝑀) is equicontinuous. To see this let 𝑢∈𝑀. Then ℎ₁𝑢(𝑡)−ℎ₁𝑢𝑡^∗=^𝑡₀𝑒^𝑟−𝑡ℎ𝑢(𝑟)𝑑𝑟−^𝑡∗₀𝑒^{𝑟−𝑡∗}ℎ𝑢(𝑟)𝑑𝑟=𝑒^−𝑡𝑡₀𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟−𝑒^{−𝑡∗𝑡∗}₀𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟=𝑒^−𝑡𝑡₀𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟−𝑒^{−𝑡∗𝑡}₀𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟+𝑒^{−𝑡∗𝑡}₀𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟−𝑒^{−𝑡∗𝑡∗}₀𝑒^𝑟(𝑟)=𝑒ℎ𝑢𝑑𝑟_−𝑡−𝑒^{−𝑡∗𝑡}₀𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟+𝑒^{−𝑡∗𝑡}_𝑡∗𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟. (5.24) Hence, ℎ₁𝑢(𝑡)−ℎ₁𝑢𝑡^∗_𝑝≤𝑒^−𝑡−𝑒^{−𝑡∗𝑡}₀𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟_𝑝+𝑒^{−𝑡∗𝑡}_𝑡∗𝑒^𝑟ℎ𝑢(𝑟)𝑑𝑟_𝑝=𝑒^−𝑡−𝑒^{−𝑡∗𝑝1}₀𝑒^𝑟𝑑𝑟^𝑝𝛿_{‖⋅‖𝑝}(𝐵)+𝑒^{−𝑡∗𝑝𝑡}_𝑡∗𝑒^𝑟𝑑𝑟^𝑝𝛿_{‖⋅‖𝑝}(𝐵)≤2𝑒^−𝑡−𝑒^{−𝑡∗𝑝}𝛿_{‖⋅‖𝑝}(𝐵)+^𝑡_𝑡∗𝑒^𝑟𝑑𝑟^𝑝𝛿_{‖⋅‖𝑝}(𝐵)=2𝑒^−𝑡−𝑒^{−𝑡∗𝑝}𝛿_{‖⋅‖𝑝}(𝐵)+𝑒^𝑡−𝑒^𝑡∗𝑝𝛿_{‖⋅‖𝑝}(𝐵). (5.25) Recall the functions 𝑡↦𝑒^−𝑡 and 𝑡↦𝑒^𝑡 are uniformly continuous on [0,1]. Therefore, for 𝜀>0, there exists 𝜂₁>0 such that if |𝑡−𝑡^∗|<𝜂₁ then 𝑒^−𝑡−𝑒^−𝑡∗≤𝜀4𝛿_{‖⋅‖𝑝}(𝐵)^1/𝑝, (5.26) and there exists 𝜂₂>0 such that if |𝑡−𝑡^∗|<𝜂₂ then 𝑒^𝑡−𝑒^𝑡∗≤𝜀2𝛿_{‖⋅‖𝑝}(𝐵)^1/𝑝. (5.27) As a result with 𝜂=min(𝜂₁,𝜂₂) note if |𝑡−𝑡^∗|<𝜂 then ℎ₁𝑢(𝑡)−ℎ₁𝑢𝑡^∗_𝑝≤𝜀, (5.28) for any 𝑢∈𝑀. Thus, ℎ₁(𝑀) is equicontinuous. Moreover, ℎ₁𝑢(𝑡)=^𝑡₀𝑒^𝑟−𝑡ℎ𝑢(𝑟)𝑑𝑟∈1−𝑒^−𝑡𝑐𝑜_𝑠{(𝑟)}⊂ℎ𝑢,0≤𝑟≤𝑡1−𝑒^−𝑡𝑐𝑜_𝑠(ℎ(𝐵)). (5.29) Hence, for all 𝑡∈[0,1], we have ℎ₁(𝑀(𝑡))⊂1−𝑒^−𝑡𝑐𝑜_𝑠(ℎ(𝐵)). (5.30) Since, ℎ(𝐵) is compact, this implies that 𝑐𝑜_𝑠(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.

Acknowledgment

The research of the first and third authors was partially supported by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia.