Generalized 𝑠-Convex Functions on Fractal Sets

Abstract

We introduce two kinds of generalized 𝑠-convex functions on real linear fractal sets ℝ^𝛼(0<𝛼<1). And similar to the class situation, we also study the properties of these two kinds of generalized 𝑠-convex functions and discuss the relationship between them. Furthermore, some applications are given.

1. Introduction

Let 𝑓∶𝐼⊆ℝ→ℝ. For any 𝑢,𝑣∈𝐼 and 𝑡∈[0,1], if the following inequality, 𝑓(𝑡𝑢+(1−𝑡)𝑣)≤𝑡𝑓(𝑢)+(1−𝑡)𝑓(𝑣), (1) holds, then 𝑓 is called a convex function on 𝐼.

The convexity of functions plays a significant role in many fields, such as in biological system, economy, and optimization [1, 2]. In [3], Hudzik and Maligranda generalized the definition of convex function and considered, among others, two kinds of functions which are 𝑠-convex.

Let 0<𝑠≤1 and ℝ₊=[0,∞), and then the two kinds of 𝑠-convex functions are defined, respectively, in the following way.

Definition 1. A function, 𝑓∶ℝ₊→ℝ, is said to be 𝑠-convex in the first sense if 𝑓𝛼𝑢+𝛽𝑣≤𝛼^𝑠𝑓(𝑢)+𝛽^𝑠𝑓(𝑣), (2) for all 𝑢,𝑣∈ℝ₊ and all 𝛼,𝛽≥0 with 𝛼^𝑠+𝛽^𝑠=1. One denotes this by 𝑓∈𝐾¹_𝑠.

Definition 2. A function, 𝑓∶ℝ₊→ℝ, is said to be 𝑠-convex in the second sense if 𝑓𝛼𝑢+𝛽𝑣≤𝛼^𝑠𝑓(𝑢)+𝛽^𝑠𝑓(𝑣), (3) for all 𝑢,𝑣∈ℝ₊ and all 𝛼,𝛽≥0 with 𝛼+𝛽=1. One denotes this by 𝑓∈𝐾²_𝑠.

It is obvious that the 𝑠-convexity means just the convexity when 𝑠=1, no matter whether it is in the first sense or in the second sense. In [3], some properties of 𝑠-convex functions in both senses are considered and various examples and counterexamples are given. There are many research results related to the 𝑠-convex functions; see [4–6] and so on.

In recent years, the fractal has received significantly remarkable attention from scientists and engineers. In the sense of Mandelbrot, a fractal set is the one whose Hausdorff dimension strictly exceeds the topological dimension [7–12].

The calculus on fractal set can lead to better comprehension for the various real world models from science and engineering [8]. Researchers have constructed many kinds of fractional calculus on fractal sets by using different approaches. Particularly, in [13], Yang stated the analysis of local fractional functions on fractal space systematically, which includes local fractional calculus. In [14], the authors introduced the generalized convex function on fractal sets and established the generalized Jensen inequality and generalized Hermite-Hadamard inequality related to generalized convex function. And, in [15], Wei et al. established a local fractional integral inequality on fractal space analogous to Anderson’s inequality for generalized convex functions. The generalized convex function on fractal sets ℝ^𝛼(0<𝛼<1) can be stated as follows.

Let 𝑓∶𝐼⊂ℝ→ℝ^𝛼. For any 𝑢,𝑣∈𝐼 and 𝑡∈[0,1], if the following inequality, 𝑓(𝑡𝑢+(1−𝑡)𝑣)≤𝑡^𝛼𝑓(𝑢)+(1−𝑡)^𝛼𝑓(𝑣), (4) holds, then 𝑓 is called a generalized convex on 𝐼.

Inspired by these investigations, we will introduce the generalized 𝑠-convex function in the first or second sense on fractal sets and study the properties of generalized 𝑠-convex functions.

The paper is organized as follows. In Section 2, we state the operations with real line number fractal sets and give the definitions of the local fractional calculus. In Section 3, we introduce the definitions of two kinds of generalized 𝑠-convex functions and study the properties of these functions. In Section 4, we give some applications for the two kinds of generalized 𝑠-convex functions on fractal sets.

2. Preliminaries

Let us recall the operations with real line number on fractal space and use Gao-Yang-Kang’s idea to describe the definitions of the local fractional derivative and local fractional integral [13, 16–19].

If 𝑎^𝛼,𝑏^𝛼, and 𝑐^𝛼 belong to the set ℝ^𝛼(0<𝛼≤1) of real line numbers, then one has the following:(1)𝑎^𝛼+𝑏^𝛼 and 𝑎^𝛼𝑏^𝛼 belong to the set ℝ^𝛼;(2)𝑎^𝛼+𝑏^𝛼=𝑏^𝛼+𝑎^𝛼=(𝑎+𝑏)^𝛼=(𝑏+𝑎)^𝛼;(3)𝑎^𝛼+(𝑏^𝛼+𝑐^𝛼)=(𝑎^𝛼+𝑏^𝛼)+𝑐^𝛼;(4)𝑎^𝛼𝑏^𝛼=𝑏^𝛼𝑎^𝛼=(𝑎𝑏)^𝛼=(𝑏𝑎)^𝛼;(5)𝑎^𝛼(𝑏^𝛼𝑐^𝛼)=(𝑎^𝛼𝑏^𝛼)𝑐^𝛼;(6)𝑎^𝛼(𝑏^𝛼+𝑐^𝛼)=𝑎^𝛼𝑏^𝛼+𝑎^𝛼𝑐^𝛼;(7)𝑎^𝛼+0^𝛼=0^𝛼+𝑎^𝛼=𝑎^𝛼 and 𝑎^𝛼⋅1^𝛼=1^𝛼⋅𝑎^𝛼=𝑎^𝛼.

Let us now state some definitions about the local fractional calculus on ℝ^𝛼.

Definition 3 (see [13]). A nondifferentiable function 𝑓∶ℝ→ℝ^𝛼, 𝑥→𝑓(𝑥) is called to be local fractional continuous at 𝑥₀, if, for any 𝜀>0, there exists 𝛿>0, such that 𝑓(𝑥)−𝑓𝑥₀<𝜀^𝛼 (5) holds for |𝑥−𝑥₀|<𝛿, where 𝜀,𝛿∈ℝ. If 𝑓 is local fractional continuous on the interval (𝑎,𝑏), one denotes 𝑓∈𝐶_𝛼(𝑎,𝑏).

Definition 4 (see [13]). The local fractional derivative of function 𝑓 of order 𝛼 at 𝑥=𝑥₀ is defined by 𝑓^(𝛼)𝑥₀=𝑑^𝛼𝑓(𝑥)𝑑𝑥^𝛼_𝑥=𝑥0=lim_𝑥→𝑥0Δ^𝛼𝑓(𝑥)−𝑓𝑥₀𝑥−𝑥₀^𝛼, (6) where Δ^𝛼(𝑓(𝑥)−𝑓(𝑥₀))=Γ(1+𝑎)(𝑓(𝑥)−𝑓(𝑥₀)) and the Gamma function is defined by Γ(𝑡)=∫₀^+∞𝑥^𝑡−1𝑒^−𝑥𝑑𝑥.
If there exists 𝑓^{((𝑘+1)𝛼)}(𝑥)=𝐷^𝛼_𝑥⋯𝐷^{𝑘+1times𝛼}_𝑥𝑓(𝑥) for any 𝑥∈𝐼⊆ℝ, then one denoted 𝑓∈𝐷_(𝑘+1)𝛼(𝐼), where 𝑘=0,1,2,….

Definition 5 (see [13]). Let 𝑓∈𝐶_𝛼[𝑎,𝑏]. Then the local fractional integral of the function 𝑓 of order 𝛼 is defined by _𝑎𝐼_𝑏^(𝛼)1𝑓=Γ(1+𝑎)^𝑏_𝑎𝑓(𝑡)(𝑑𝑡)^𝛼=1Γ(1+𝑎)lim^𝑁_{Δ𝑡→0𝑗=0}𝑓𝑡_𝑗Δ𝑡_𝑗^𝛼, (7) with Δ𝑡_𝑗=𝑡_𝑗+1−𝑡_𝑗, Δ𝑡=max{Δ𝑡₁,Δ𝑡₂,Δ𝑡_𝑗,…,Δ𝑡_𝑁−1}, and [𝑡_𝑗,𝑡_𝑗+1], 𝑗=0,…,𝑁−1, where 𝑡₀=𝑎<𝑡₁<⋯<𝑡_𝑖<⋯<𝑡_𝑁=𝑏 is a partition of the interval [𝑎,𝑏].

Lemma 6 (see [13]). Suppose that 𝑓,𝑔∈𝐶_𝛼[𝑎,𝑏] and 𝑓,𝑔∈𝐷_𝛼(𝑎,𝑏). If lim_𝑥→𝑥0𝑓(𝑥)=0^𝛼, lim_𝑥→𝑥0𝑔(𝑥)=0^𝛼 and 𝑔^(𝛼)(𝑥)≠0^𝛼. Suppose that lim_𝑥→𝑥0(𝑓^(𝛼)(𝑥)/𝑔^(𝛼)(𝑥))=𝐴^𝛼, and then lim_𝑥→𝑥0𝑓(𝑥)𝑔(𝑥)=𝐴^𝛼. (8)

Lemma 7 (see [13]). Suppose that 𝑓(𝑥)∈𝐶_𝛼[𝑎,𝑏]; then 𝑑^𝛼_𝑎𝐼_𝑥^(𝛼)𝑓𝑑𝑥^𝛼=𝑓(𝑥),𝑎<𝑥<𝑏. (9)

3. Generalized 𝑠-Convexity Functions

The convexity of functions plays a significant role in many fields. In this section, let us introduce two kinds of generalized 𝑠-convex functions on fractal sets. And then, we study the properties of the two kinds of generalized 𝑠-convex functions.

Definition 8. Let ℝ₊=[0,+∞). A function 𝑓∶ℝ₊→ℝ^𝛼 is said to be generalized 𝑠-convex (0<𝑠<1) in the first sense, if 𝑓𝜆₁𝑢+𝜆₂𝑣≤𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣), (10) for all 𝑢,𝑣∈ℝ₊ and all 𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂=1. One denotes this by 𝑓∈𝐺𝐾¹_𝑠.

Definition 9. A function 𝑓∶ℝ₊→ℝ^𝛼 is said to be generalized 𝑠-convex (0<𝑠<1) in the second sense, if 𝑓𝜆₁𝑢+𝜆₂𝑣≤𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣), (11) for all 𝑢,𝑣∈ℝ₊ and all 𝜆₁,𝜆₂≥0 with 𝜆₁+𝜆₂=1. One denotes this by 𝑓∈𝐺𝐾²_𝑠.

Note that, when 𝑠=1, the generalized 𝑠-convex functions in both senses are the generalized convex functions; see [14].

Theorem 10. Let 0<𝑠<1.(a)If 𝑓∈𝐺𝐾¹_𝑠, then 𝑓 is nondecreasing on (0,+∞) and 𝑓0⁺=lim_𝑢→0+𝑓(𝑢)≤𝑓(0). (12)(b)If 𝑓∈𝐺𝐾²_𝑠, then 𝑓 is nonnegative on [0,+∞).

Proof. (a) Since 𝑓∈𝐺𝐾¹_𝑠, we have, for 𝑢>0 and 𝜆∈[0,1], 𝑓𝜆^1/𝑠+(1−𝜆)^1/𝑠𝑢≤𝜆^𝛼𝑓(𝑢)+(1−𝜆)^𝛼𝑓(𝑢)=𝑓(𝑢). (13) The function ℎ(𝜆)=𝜆^1/𝑠+(1−𝜆)^1/𝑠 (14) is continuous on [0,1], decreasing on [0,1/2], and increasing on [1/2,1] and ℎ([0,1])=[ℎ(1/2),ℎ(1)]=[2^1−1/𝑠,1]. This yields that 𝑓(𝑡𝑢)≤𝑓(𝑢), (15) for 𝑢>0 and 𝑡∈[2^1−1/𝑠,1]. If 𝑡∈[2^{2(1−1/𝑠)},1], then 𝑡^1/2∈[2^1−1/𝑠,1]. Therefore, by the fact that (15) holds, we get 𝑓(𝑡𝑢)=𝑓𝑡^1/2𝑡^1/2𝑢≤𝑓𝑡^1/2𝑢≤𝑓(𝑢), (16) for all 𝑢>0. So we can obtain that 𝑓(𝑡𝑢)≤𝑓(𝑢),∀𝑢>0,𝑡∈(0,1]. (17)
So, taking 0<𝑢<𝑣, we get 𝑓(𝑢)𝑢=𝑓𝑣𝑣≤𝑓(𝑣), (18) which means that 𝑓 is nondecreasing on (0,+∞).
As for the second part, for 𝑢>0 and 𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂=1, we have 𝑓𝜆₁𝑢=𝑓𝜆₁𝑢+𝜆₂0≤𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(0). (19) And taking 𝑢→0⁺, we get lim_𝑢→0+𝑓(𝑢)=lim_𝑢→0+𝑓𝜆₁𝑢≤𝜆₁^𝑠𝛼lim_𝑢→0+𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(0). (20) So, lim_𝑢→0+𝑓(𝑢)≤𝑓(0). (21)
(b) For 𝑓∈𝐺𝐾²_𝑠, we can get that, for 𝑢∈ℝ₊, 𝑓(𝑢)𝑢=𝑓2+𝑢2≤𝑓(𝑢)2^𝑠𝛼+𝑓(𝑢)2^𝑠𝛼=2^{(1−𝑠)𝛼}𝑓(𝑢). (22)
So, (2^1−𝑠−1)^𝛼𝑓(𝑢)≥0^𝛼. This means that 𝑓(𝑢)≥0^𝛼, since 0<𝑠<1.

Remark 11. The above results do not hold, in general, in the case of generalized convex functions, that is, when 𝑠=1, because a generalized convex function, 𝑓∶ℝ₊→ℝ^𝛼, need not be either nondecreasing or nonnegative.

Remark 12. If 0<𝑠<1, then the function 𝑓∈𝐺𝐾¹_𝑠 is nondecreasing on (0,+∞) but not necessarily on [0,+∞).

Function 𝐹∶ℝ²→ℝ^𝛼 is called to be generalized convex in each variable, if 𝐹𝜆₁𝑢+𝜆₂𝑣,𝜆₁𝑟+𝜆₂𝑡≤𝜆^𝛼₁𝐹(𝑢,𝑟)+𝜆^𝛼₂𝐹(𝑣,𝑡). (23) For all (𝑢,𝑟),  (𝑣,𝑡)∈ℝ² and 𝜆₁,𝜆₂∈[0,1] with 𝜆₁+𝜆₂=1.

Theorem 13. Let 0<𝑠<1. If 𝑓,𝑔∶ℝ→ℝ and 𝑓,𝑔∈𝐾¹_𝑠 and if 𝐹∶ℝ²→ℝ^𝛼 is a generalized convex and nondecreasing function in each variable, then the function ℎ∶ℝ₊→ℝ^𝛼 defined by ℎ(𝑢)=𝐹𝑓(𝑢),𝑔(𝑢) (24) is in 𝐺𝐾¹_𝑠. In particular, if 𝑓,𝑔∈𝐾¹_𝑠, then 𝑓^𝛼+𝑔^𝛼, max{𝑓^𝛼,𝑔^𝛼}∈𝐺𝐾¹_𝑠.

Proof. If 𝑢,𝑣∈ℝ₊, then for all 𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂=1, ℎ𝜆₁𝑢+𝜆₂𝑣=𝐹𝑓𝜆₁𝑢+𝜆₂𝑣,𝑔𝜆₁𝑢+𝜆₂𝑣≤𝐹𝜆^𝑠₁𝑓(𝑢)+𝜆^𝑠₂𝑓(𝑣),𝜆^𝑠₁𝑔(𝑢)+𝜆^𝑠₂𝑔(𝑣)≤𝜆₁^𝑠𝛼𝐹𝑓(𝑢),𝑔(𝑢)+𝜆₂^𝑠𝛼𝐹𝑓(𝑣),𝑔(𝑣)=𝜆₁^𝑠𝛼ℎ(𝑢)+𝜆₂^𝑠𝛼ℎ(𝑣). (25) Thus, ℎ∈𝐺𝐾¹_𝑠.
Moreover, since 𝐹(𝑢,𝑣)=𝑢^𝛼+𝑣^𝛼,  𝐹(𝑢,𝑣)=max{𝑢^𝛼,𝑣^𝛼} are nondecreasing generalized convex functions on 𝑅², so they yield particular cases of our theorem.

Let us pay attention to the situation when the condition 𝜆^𝑠₁+𝜆^𝑠₂=1  (𝜆₁+𝜆₂=1) in the definition of 𝐺𝐾¹_𝑠(𝐺𝐾²_𝑠) can be equivalently replaced by the condition 𝜆^𝑠₁+𝜆^𝑠₂≤1  (𝜆₁+𝜆₂≤1).

Theorem 14. (a) Let 𝑓∈𝐺𝐾¹_𝑠. Then inequality (10) holds for all 𝑢,𝑣∈𝑅₊ and all 𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂<1 if and only if 𝑓(0)≤0^𝛼.
(b) Let 𝑓∈𝐺𝐾²_𝑠. Then inequality (11) holds for all 𝑢,𝑣∈𝑅₊ and all 𝜆₁,𝜆₂≥0 with 𝜆₁+𝜆₂<1 if and only if 𝑓(0)=0^𝛼.

Proof. (a) Necessity is obvious by taking 𝑢=𝑣=0 and 𝜆₁=𝜆₂=0. Let us show the sufficiency.
Assume that 𝑢,𝑣∈ℝ₊ and 𝜆₁,𝜆₂≥0 with 0<𝜆₃=𝜆^𝑠₁+𝜆^𝑠₂<1. Put 𝑎=𝜆₁𝜆₃^−1/𝑠 and 𝑏=𝜆₂𝜆₃^−1/𝑠. Then 𝑎^𝑠+𝑏^𝑠=𝜆^𝑠₁/𝜆₃+𝜆^𝑠₂/𝜆₃=1 and 𝑓𝜆₁𝑢+𝜆₂𝑣=𝑓𝑎𝜆₃^1/𝑠𝑢+𝑏𝜆₃^1/𝑠𝑣≤𝑎^𝑠𝛼𝑓𝜆₃^1/𝑠𝑢+𝑏^𝑠𝛼𝑓𝜆₃^1/𝑠𝑣=𝑎^𝑠𝛼𝑓𝜆₃^1/𝑠𝑢+1−𝜆₃^1/𝑠0+𝑏^𝑠𝛼𝑓𝜆₃^1/𝑠𝑣+1−𝜆₃^1/𝑠0≤𝑎^𝑠𝛼𝜆^𝛼₃𝑓(𝑢)+1−𝜆₃^𝛼𝑓(0)+𝑏^𝑠𝛼𝜆^𝛼₃𝑓(𝑣)+1−𝜆₃^𝛼𝑓(0)=𝑎^𝑠𝛼𝜆^𝛼₃𝑓(𝑢)+𝑏^𝑠𝛼𝜆^𝛼₃𝑓(𝑣)+1−𝜆₃^𝛼𝑓(0)≤𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣). (26)
(b) Necessity. Taking 𝑢=𝑣=𝜆₁=𝜆₂=0, we obtain 𝑓(0)≤0^𝛼. And using Theorem 10(b), we get 𝑓(0)≥0^𝛼. Therefore 𝑓(0)=0^𝛼.
Sufficiency. Let 𝑢,𝑣∈ℝ₊ and 𝜆₁,𝜆₂≥0 with 0<𝜆₃=𝜆₁+𝜆₂<1. Put 𝑎=𝜆₁/𝜆₃ and 𝑏=𝜆₂/𝜆₃, and then 𝑎+𝑏=1.
So, 𝑓𝜆₁𝑢+𝜆₂𝑣=𝑓𝑎𝜆₃𝑢+𝑏𝜆₃𝑣≤𝑎^𝑠𝛼𝑓𝜆₃𝑢+𝑏^𝑠𝛼𝑓𝜆₃𝑣=𝑎^𝑠𝛼𝑓𝜆₃𝑢+1−𝜆₃0+𝑏^𝑠𝛼𝑓𝜆₃𝑣+1−𝜆₃0≤𝑎^𝑠𝛼𝜆₃^𝑠𝛼𝑓(𝑢)+1−𝜆₃^𝑠𝛼𝑓(0)+𝑏^𝑠𝛼𝜆₃^𝑠𝛼𝑓(𝑣)+1−𝜆₃^𝑠𝛼𝑓(0)=𝑎^𝑠𝛼𝜆₃^𝑠𝛼𝑓(𝑢)+𝑏^𝑠𝛼𝜆₃^𝑠𝛼𝑓(𝑣)+1−𝜆₃^𝑠𝛼𝑓(0)=𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣). (27)

Theorem 15. (a) Let 0<𝑠≤1. If 𝑓∈𝐺𝐾²_𝑠 and 𝑓(0)=0^𝛼, then 𝑓∈𝐺𝐾¹_𝑠.
(b) Let 0<𝑠₁≤𝑠₂≤1. If 𝑓∈𝐺𝐾²_𝑠2 and 𝑓(0)=0^𝛼, then 𝑓∈𝐺𝐾²_𝑠1.
(c) Let 0<𝑠₁≤𝑠₂≤1. If 𝑓∈𝐺𝐾¹_𝑠2 and 𝑓(0)≤0^𝛼, then 𝑓∈𝐺𝐾¹_𝑠1.

Proof. (a) Assume that 𝑓∈𝐺𝐾²_𝑠 and 𝑓(0)=0^𝛼. Let 𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂=1, and we have 𝜆₁+𝜆₂≤𝜆^𝑠₁+𝜆^𝑠₂=1. From Theorem 14(b), we can get 𝑓𝜆₁𝑢+𝜆₂𝑣≤𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣), (28) for 𝑢,𝑣∈ℝ₊, and then 𝑓∈𝐺𝐾¹_𝑠.
(b) Assume that 𝑓∈𝐺𝐾²_𝑠2, 𝑢,𝑣∈ℝ₊, and 𝜆₁,𝜆₂≥0 with 𝜆₁+𝜆₂=1. Then we have 𝑓𝜆₁𝑢+𝜆₂𝑣≤𝜆^𝑠2𝛼₁𝑓(𝑢)+𝜆^𝑠2𝛼₂𝑓(𝑣)≤𝜆^𝑠1𝛼₁𝑓(𝑢)+𝜆^𝑠1𝛼₂𝑓(𝑣). (29) So 𝑓∈𝐺𝐾²_𝑠1.
(c) Assume that 𝑓∈𝐺𝐾¹_𝑠2, 𝑢,𝑣∈𝑅₊, and 𝜆₁,𝜆₂≥0 with 𝜆^𝑠1₁+𝜆^𝑠1₂=1. Then 𝜆^𝑠2₁+𝜆^𝑠2₂≤𝜆^𝑠1₁+𝜆^𝑠1₂=1. Thus, according to Theorem 14(a), we have 𝑓𝜆₁𝑢+𝜆₂𝑣≤𝜆^𝑠2𝛼₁𝑓(𝑢)+𝜆^𝑠2𝛼₂𝑓(𝑣)≤𝜆^𝑠1𝛼₁𝑓(𝑢)+𝜆^𝑠1𝛼₂𝑓(𝑣). (30) So, 𝑓∈𝐺𝐾¹_𝑠1.

Theorem 16. Let 0<𝑠<1 and 𝑝∶ℝ₊→ℝ^𝛼₊ be a nondecreasing function. Then the function 𝑓 defined for 𝑢∈ℝ₊ by 𝑓(𝑢)=𝑢^{(𝑠/(1−𝑠))𝛼}𝑝(𝑢) (31) belongs to 𝐺𝐾¹_𝑠.

Proof. Let 𝑣≥𝑢≥0 and 𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂=1. We consider two cases.
Case I. It is easy to see that 𝑓 is a nondecreasing function. Let 𝜆₁𝑢+𝜆₂𝑣≤𝑢, and then 𝑓𝜆₁𝑢+𝜆₂𝑣≤𝑓(𝑢)=𝜆₁^𝑠𝛼+𝜆₂^𝑠𝛼𝑓(𝑢)≤𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣). (32)
Case II. Let 𝜆₁𝑢+𝜆₂𝑣>𝑢, and then 𝜆₂𝑣>(1−𝜆₁)𝑢. So, 𝜆₂>0 and 𝜆₁≤𝜆^𝑠₁. Thus, 𝜆₁−𝜆₁^𝑠+1≤𝜆^𝑠₁−𝜆₁^𝑠+1; (33) that is, 𝜆₁1−𝜆₁≤𝜆^𝑠₁1−𝜆^𝑠₁=1−𝜆^𝑠₂𝜆^𝑠₂,𝜆₁𝜆₂1−𝜆₁≤𝜆₂^1−𝑠−𝜆₂. (34)
Thus, we can get that 𝜆₁𝑢+𝜆₂𝑣≤𝜆₁+𝜆₂𝑣≤𝜆^𝑠₁+𝜆^𝑠₂𝜆𝑣=𝑣,₁𝑢+𝜆₂𝜆𝑣≤¹𝜆₂𝑣1−𝜆₁+𝜆₂𝑣≤𝜆₂^1−𝑠−𝜆₂𝑣+𝜆₂𝑣=𝜆₂^1−𝑠𝑣. (35) Then, 𝜆₁𝑢+𝜆₂𝑣^{𝑠/(1−𝑠)}≤𝜆^𝑠₂𝑣^{𝑠/(1−𝑠)}. (36)
We obtain 𝑓𝜆₁𝑢+𝜆₂𝑣=𝜆₁𝑢+𝜆₂𝑣^{(𝑠/(1−𝑠))𝛼}𝑝𝜆₁𝑢+𝜆₂𝑣≤𝜆₂^𝑠𝛼𝑣^{(𝑠/(1−𝑠))𝛼}𝑝(𝑣)=𝜆₂^𝑠𝛼𝑓(𝑣)≤𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣). (37)

Theorem 17. (a) Let 𝑓∈𝐺𝐾¹_𝑠1 and 𝑔∈𝐾¹_𝑠2, where 0<𝑠₁, 𝑠₂≤1. If 𝑓 is a nondecreasing function and 𝑔 is a nonnegative function such that 𝑓(0)≤0^𝛼 and 𝑔(0)=0, then the composition 𝑓∘𝑔 of 𝑓 with 𝑔 belongs to 𝐺𝐾¹_𝑠, where 𝑠=𝑠₁𝑠₂.
(b) Let 𝑓∈𝐺𝐾¹_𝑠1 and 𝑔∈𝐺𝐾¹_𝑠2, where 0<𝑠₁, 𝑠₂≤1. Assume that 0<𝑠₁, 𝑠₂<1. If 𝑓 and 𝑔 are nonnegative functions such that either 𝑓(0)=0^𝛼 and 𝑔(0⁺)=𝑔(0), or 𝑔(0)=0^𝛼 and 𝑓(0⁺)=𝑓(0), then the product 𝑓𝑔 of 𝑓 and 𝑔 belongs to 𝐺𝐾¹_𝑠, where 𝑠=min{𝑠₁,𝑠₂}.

Proof. (a) Let 𝑢,𝑣∈ℝ₊,  𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂=1, where 𝑠=𝑠₁𝑠₂. Since 𝜆^𝑠𝑖₁+𝜆^𝑠𝑖₂≤𝜆^𝑠1𝑠2₁+𝜆^𝑠1𝑠2₂=1 for 𝑖=1,2, then according to Theorem 3(a) in [3] and Theorem 14(a) in the paper, we have 𝑓∘𝑔𝜆₁𝑢+𝜆₂𝑣=𝑓𝑔𝜆₁𝑢+𝜆₂𝑣≤𝑓𝜆^𝑠2₁𝑔(𝑢)+𝜆^𝑠2₂𝑔(𝑣)≤𝜆^{𝑠1𝑠2𝛼}₁𝑓𝑔(𝑢)+𝜆^{𝑠1𝑠2𝛼}₂𝑓𝑔(𝑣)=𝜆₁^𝑠𝛼𝑓∘𝑔(𝑢)+𝜆₂^𝑠𝛼𝑓∘𝑔(𝑣) (38) which means that 𝑓∘𝑔∈𝐺𝐾¹_𝑠.
(b) According to Theorem 10(a), 𝑓,𝑔 are nondecreasing on (0,+∞).
So, 𝑓(𝑢)−𝑓(𝑣)𝑔(𝑣)−𝑔(𝑢)≤0^𝛼, (39) or, equivalently, 𝑓(𝑢)𝑔(𝑣)+𝑓(𝑣)𝑔(𝑢)≤𝑓(𝑢)𝑔(𝑢)+𝑓(𝑣)𝑔(𝑣), (40) for all 𝑣>𝑢>0.
If 𝑣>𝑢=0, then the inequality is still true because 𝑓,𝑔 are nonnegative and either 𝑓(0)=0^𝛼 and 𝑔(0⁺)=𝑔(0) or 𝑔(0)=0^𝛼 and 𝑓(0⁺)=𝑓(0).
Now let 𝑢,𝑣∈ℝ₊ and 𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂=1, where 𝑠=min{𝑠₁,𝑠₂}. Then 𝜆^𝑠𝑖₁+𝜆^𝑠𝑖₂≤𝜆^𝑠₁+𝜆^𝑠₂=1 for 𝑖=1,2. And by Theorem 14(a), we have 𝑓𝜆₁𝑢+𝜆₂𝑣𝑔𝜆₁𝑢+𝜆₂𝑣≤𝜆^𝑠1𝛼₁𝑓(𝑢)+𝜆^𝑠1𝛼₂𝑓(𝑣)×𝜆^𝑠2𝛼₁𝑔(𝑢)+𝜆^𝑠2𝛼₂𝑔(𝑣)=𝜆^(𝑠1+𝑠2₁^)𝛼𝑓(𝑢)𝑔(𝑢)+𝜆^𝑠1𝛼₁𝜆^𝑠2𝛼₂𝑓(𝑢)𝑔(𝑣)+𝜆^𝑠2𝛼₁𝜆^𝑠1𝛼₂𝑓(𝑣)𝑔(𝑢)+𝜆^(𝑠1+𝑠2₂^)𝛼𝑓(𝑣)𝑔(𝑣)≤𝜆₁^2𝑠𝛼𝑓(𝑢)𝑔(𝑢)+𝜆₁^𝑠𝛼𝜆₂^𝑠𝛼𝑓(𝑢)𝑔(𝑣)+𝑓(𝑣)𝑔(𝑢)+𝜆₂^2𝑠𝛼𝑓(𝑣)𝑔(𝑣)≤𝜆₁^2𝑠𝛼𝑓(𝑢)𝑔(𝑢)+𝜆₁^𝑠𝛼𝜆₂^𝑠𝛼𝑓(𝑢)𝑔(𝑢)+𝑓(𝑣)𝑔(𝑣)+𝜆₂^2𝑠𝛼𝑓(𝑣)𝑔(𝑣)=𝜆₁^𝑠𝛼𝑓(𝑢)𝑔(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣)𝑔(𝑣), (41) which means that 𝑓𝑔∈𝐺𝐾¹_𝑠.

Remark 18. From the above proof, we can get that if 𝑓 is a nondecreasing function in 𝐺𝐾²_𝑠 and 𝑔 is a nonnegative convex function on [0,+∞), then the composition 𝑓∘𝑔 of 𝑓 with 𝑔 belongs to 𝐺𝐾²_𝑠.

Remark 19. Generalized convex functions on [0,+∞) need not be monotonic. However, if 𝑓 and 𝑔 are nonnegative, generalized convex and either both are nondecreasing or both are nonincreasing on [0,+∞), then the product 𝑓𝑔 is also a generalized convex function.

Let 𝑓∶ℝ₊→ℝ₊ be a continuous function. Then 𝑓 is said to be a 𝜑-function if 𝑓(0)=0 and 𝑓 is nondecreasing on ℝ₊. Similarly, we can define the 𝜑-type function on fractal sets as follows. A function 𝑓∶ℝ₊→ℝ^𝛼₊ is said to be a 𝜑-type function if 𝑓(0)=0^𝛼 and 𝑓∈𝐶_𝛼(ℝ₊) is nondecreasing.

Corollary 20. If Φ is a generalized convex 𝜑-type function and 𝑔∈𝐾¹_𝑠 is a 𝜑-function, then the composition Φ∘𝑔 belongs to 𝐺𝐾¹_𝑠. In particular, the 𝜑-type function ℎ(𝑢)=Φ(𝑢^𝑠) belongs to 𝐺𝐾¹_𝑠.

Corollary 21. If Φ is a convex 𝜑-function and 𝑓∈𝐺𝐾²_𝑠 is a 𝜑-type function, then the composition 𝑓∘Φ belongs to 𝐺𝐾²_𝑠. In particular, the 𝜑-type function ℎ(𝑢)=[Φ(𝑢)]^𝑠𝛼 belongs to 𝐺𝐾²_𝑠.

Theorem 22. If 0<𝑠<1 and 𝑓∈𝐺𝐾¹_𝑠 is a 𝜑-type function, then there exists a generalized convex 𝜑-type function Φ such that 𝑓2^−1/𝑠𝑢≤Φ𝑢^𝑠≤𝑓(𝑢), (42) for all 𝑢≥0.

Proof. By the generalized 𝑠-convexity of the function 𝑓 and by 𝑓(0)=0^𝛼, we obtain 𝑓(𝜆₁𝑢)≤𝜆₁^𝑠𝛼𝑓(𝑢) for all 𝑢≥0 and all 𝜆₁∈[0,1].
Assume now that 𝑣>𝑢>0. Then 𝑓𝑢^1/𝑠𝑢≤𝑓𝑣^1/𝑠𝑣^1/𝑠≤𝑢^𝛼𝑣^𝛼𝑓𝑣^1/𝑠; (43) that is, 𝑓𝑢^1/𝑠𝑢^𝛼≤𝑓𝑣^1/𝑠𝑣^𝛼. (44)
Inequality (44) means that the function 𝑓(𝑢^1/𝑠)/𝑢^𝛼 is a nondecreasing function on (0,+∞). And, since 𝑓 is a 𝜑-type function, thus 𝑓 is local fractional continuous [0,+∞).
Define Φ(𝑢)=0^𝛼Γ(),𝑢=0,1+𝛼₀𝐼_𝑢^(𝛼)𝑓𝑡^1/𝑠𝑡^𝛼,𝑢>0. (45)
From Lemmas 6 and 7, it is easy to see that Φ is a generalized convex 𝜑-type function and Φ𝑢^𝑠=Γ(1+𝛼)₀𝐼_𝑢^(𝛼)_𝑠𝑓𝑡^1/𝑠𝑡^𝛼≤𝑓𝑢^𝑠1/𝑠𝑢^𝑠𝛼𝑢^𝑠𝛼=𝑓(𝑢). (46) Moreover, Φ𝑢^𝑠≥Γ(1+𝛼)_(𝑢𝑠/2)𝐼_𝑢^(𝛼)_𝑠𝑓𝑡^1/𝑠𝑡^𝛼≥𝑓𝑢^𝑠/2^1/𝑠2^𝛼𝑢^−𝑠𝛼𝑢^𝑠𝛼2^𝛼=𝑓2^−1/𝑠𝑢. (47)
Therefore, 𝑓2^−1/𝑠𝑢≤Φ𝑢^𝑠≤𝑓(𝑢), (48) for all 𝑢≥0.

4. Applications

Based on the properties of the two kinds of generalized 𝑠-convex functions in the above section, some applications are given.

Example 23. Let 0<𝑠<1, and 𝑎^𝛼,𝑏^𝛼,𝑐^𝛼∈ℝ^𝛼. For 𝑢∈ℝ₊, define 𝑓(𝑢)=𝑎^𝛼𝑏,𝑢=0,_𝛼𝑢^𝑠𝛼+𝑐^𝛼,𝑢>0. (49)
We have the following conclusions.(i)If 𝑏^𝛼≥0^𝛼 and 𝑐^𝛼≤𝑎^𝛼, then 𝑓∈𝐺𝐾¹_𝑠.(ii)If 𝑏^𝛼≥0^𝛼 and 𝑐^𝛼<𝑎^𝛼, then 𝑓 is nondecreasing on (0,+∞) but not on [0,+∞).(iii)If 𝑏^𝛼≥0^𝛼 and 0^𝛼≤𝑐^𝛼≤𝑎^𝛼, then 𝑓∈𝐺𝐾²_𝑠.(iv)If 𝑏^𝛼>0^𝛼 and 𝑐^𝛼<0^𝛼, then 𝑓∉𝐺𝐾²_𝑠.

Proof. (i) Let 𝜆₁,𝜆₂≥0 with 𝜆^𝑠₁+𝜆^𝑠₂=1. Then, there are two nontrivial cases.
Case I. Let 𝑢,𝑣>0. Then 𝜆₁𝑢+𝜆₂𝑣>0.
Thus, 𝑓𝜆₁𝑢+𝜆₂𝑣=𝑏^𝛼𝜆₁𝑢+𝜆₂𝑣^𝑠𝛼+𝑐^𝛼≤𝑏^𝛼𝜆₁^𝑠𝛼𝑢^𝑠𝛼+𝜆₂^𝑠𝛼𝑣^𝑠𝛼+𝑐^𝛼=𝑏^𝛼𝜆₁^𝑠𝛼𝑢^𝑠𝛼+𝜆₂^𝑠𝛼𝑣^𝑠𝛼+𝑐^𝛼𝜆₁^𝑠𝛼+𝜆₂^𝑠𝛼=𝜆₁^𝑠𝛼𝑏^𝛼𝑢^𝑠𝛼+𝑐^𝛼+𝜆₂^𝑠𝛼𝑏^𝛼𝑣^𝑠𝛼+𝑐^𝛼=𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣). (50)
Case II. Let 𝑣>𝑢=0. We need only to consider 𝜆₂>0.
Thus, we have 𝑓𝜆₁0+𝜆₂𝑣=𝑓𝜆₂𝑣=𝑏^𝛼𝜆₂^𝑠𝛼𝑣^𝑠𝛼+𝑐^𝛼=𝑏^𝛼𝜆₂^𝑠𝛼𝑣^𝑠𝛼+𝑐^𝛼𝜆₁^𝑠𝛼+𝜆₂^𝑠𝛼=𝜆₁^𝑠𝛼𝑐^𝛼+𝜆₂^𝑠𝛼𝑏^𝛼𝑣^𝑠𝛼+𝑐^𝛼=𝜆₁^𝑠𝛼𝑐^𝛼+𝜆₂^𝑠𝛼𝑓(𝑣)≤𝜆₁^𝑠𝛼𝑎^𝛼+𝜆₂^𝑠𝛼𝑓(𝑣)=𝜆₁^𝑠𝛼𝑓(0)+𝜆₂^𝑠𝛼𝑓(𝑣). (51) So, 𝑓∈𝐺𝐾¹_𝑠.
(ii) From Theorem 10, we can see that property (ii) is true.
(iii) Let 𝜆₁,𝜆₂≥0 with 𝜆₁+𝜆₂=1. Similar to the estimate of (i), there are also two cases.
Let 𝑣,𝑣>0. Then 𝜆₁𝑢+𝜆₂𝑣>0,
Thus, 𝑓𝜆₁𝑢+𝜆₂𝑣=𝑏^𝛼𝜆₁𝑢+𝜆₂𝑣^𝑠𝛼+𝑐^𝛼<𝑏^𝛼𝜆₁^𝑠𝛼𝑢^𝑠𝛼+𝜆₂^𝑠𝛼𝑣^𝑠𝛼+𝑐^𝛼𝜆^𝛼₁+𝜆^𝛼₂≤𝑏^𝛼𝜆₁^𝑠𝛼𝑢^𝑠𝛼+𝜆₂^𝑠𝛼𝑣^𝑠𝛼+𝑐^𝛼𝜆₁^𝑠𝛼+𝜆₂^𝑠𝛼=𝜆₁^𝑠𝛼𝑏^𝛼𝑢^𝑠𝛼+𝑐^𝛼+𝜆₂^𝑠𝛼𝑏^𝛼𝑣^𝑠𝛼+𝑐^𝛼≤𝜆₁^𝑠𝛼𝑓(𝑢)+𝜆₂^𝑠𝛼𝑓(𝑣). (52)
Let 𝑣>𝑢=0. We need only to consider 𝜆₂>0.
Thus, we have 𝑓𝜆₁0+𝜆₂𝑣=𝑓𝜆₂𝑣=𝑏^𝛼𝜆₂^𝑠𝛼𝑣^𝑠𝛼+𝑐^𝛼𝜆^𝛼₁+𝜆^𝛼₂<𝑏^𝛼𝜆₂^𝑠𝛼𝑣^𝑠𝛼+𝑐^𝛼𝜆₁^𝑠𝛼+𝜆₂^𝑠𝛼=𝜆₂^𝑠𝛼𝑏^𝛼𝑣^𝑠𝛼+𝑐^𝛼+𝑐^𝛼𝜆₁^𝑠𝛼≤𝜆₂^𝑠𝛼𝑏^𝛼𝑣^𝑠𝛼+𝑐^𝛼+𝑎^𝛼𝜆₁^𝑠𝛼=𝜆₂^𝑠𝛼𝑓(𝑣)+𝜆₁^𝑠𝛼𝑓(0). (53)
So, 𝑓∈𝐺𝐾²_𝑠.
(iv) Assume that 𝑓∈𝐺𝐾²_𝑠, and then 𝑓 is nonnegative on (0,∞). On the other hand, we can take 𝑢₁>0, 𝑐₁<0 such that 𝑓(𝑢₁)=𝑏^𝛼𝑢₁^𝑠𝛼+𝑐^𝛼₁<0^𝛼, which contradict the assumption.

Example 24. Let 0<𝑠<1 and 𝑘>1. For 𝑢∈𝑅₊, define 𝑓(𝑢)=𝑢^{(𝑠/(1−𝑠))𝛼}𝑘,0≤𝑢≤1,_𝛼𝑢^{(𝑠/(1−𝑠))𝛼},𝑢>1. (54) The function 𝑓 is nonnegative, not local fractional continuous at 𝑢=1 and belongs to 𝐺𝐾¹_𝑠 but not to 𝐺𝐾²_𝑠.

Proof. From Theorem 16, we have that 𝑓∈𝐺𝐾¹_𝑠. In the following, let us show that 𝑓 is not in 𝐺𝐾²_𝑠.
Take an arbitrary 𝑎>1 and put 𝑢=1. Consider all 𝑣>1 such that 𝜆₁𝑢+𝜆₂𝑣=𝜆₁+𝜆₂𝑣=𝑎, where 𝜆₁,𝜆₂≥0 and 𝜆₁+𝜆₂=1.
If 𝑓∈𝐺𝐾²_𝑠, it must be 𝑘^𝛼𝑎^{(𝑠/(1−𝑠))𝛼}≤𝜆₁^𝑠𝛼+𝑘^𝛼1−𝜆₁^𝑠𝛼𝑎−𝜆₁1−𝜆₁^{(𝑠/(1−𝑠))𝛼}, (55) for all 𝑎>1 and all 0≤𝜆₁≤1.
Define the function 𝑓_𝜆1(𝑎)=𝜆₁^𝑠𝛼+𝑘^𝛼1−𝜆₁^𝑠𝛼𝑎−𝜆₁1−𝜆₁^{(𝑠/(1−𝑠))𝛼}−𝑘^𝛼𝑎^{(𝑠/(1−𝑠))𝛼}. (56) Then the function is local fractional continuous on the (𝜆₁,∞) and 𝑔𝜆₁=𝑓_𝜆1(1)=𝜆₁^𝑠𝛼+𝑘^𝛼1−𝜆₁^𝑠𝛼−𝑘^𝛼. (57)
It is easy to see that 𝑔 is local fractional continuous on [0,1] and 𝑔(1)=1^𝛼−𝑘^𝛼<0^𝛼. So there is a number 𝜆₁₀, 0<𝜆₁₀<1, such that 𝑔(𝜆₁₀)=𝑓_𝜆10(1)<0^𝛼. The local fractional continuity of 𝑓_𝜆10 yields that 𝑓_𝜆10(𝑎)<0^𝛼 for a certain 𝑎>1, that is, inequality (55) does not hold, which means that 𝑓∉𝐺𝐾²_𝑠.

5. Conclusion

In the paper, we introduce the definitions of two kinds of generalized 𝑠-convex function on fractal sets and study the properties of these generalized 𝑠-convex functions. When 𝛼=1, these results are the classical situation.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors would like to express their gratitude to the reviewers for their very valuable comments. And, this work is supported by the National Natural Science Foundation of China (no. 11161042).