# Generalizeds-Convex Functions on Fractal Sets

Authors: Huixia Mo; Xin Sui
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We introduce two kinds of generalizeds-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.


---

## Body

## 1. Introduction

Letf
:
I
⊆
R
→
R. For any u
,
v
∈
I and t
∈
[
0,1
], if the following inequality,
(1)
f
(
t
u
+
(
1
-
t
)
v
)
≤
t
f
(
u
)
+
(
1
-
t
)
f
(
v
)
,
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.Let0
<
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
(2)
f
(
α
u
+
β
v
)
≤
α
s
f
(
u
)
+
β
s
f
(
v
)
,
for all u
,
v
∈
R
+ and all α
,
β
≥
0 with α
s
+
β
s
=
1. One denotes this by f
∈
K
s
1.Definition 2.
A function,f
:
R
+
→
R, is said to be s-convex in the second sense if
(3)
f
(
α
u
+
β
v
)
≤
α
s
f
(
u
)
+
β
s
f
(
v
)
,
for all u
,
v
∈
R
+ and all α
,
β
≥
0 with α
+
β
=
1. One denotes this by f
∈
K
s
2.It is obvious that thes-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.Letf
:
I
⊂
R
→
R
α. For any u
,
v
∈
I and t
∈
[
0,1
], if the following inequality,
(4)
f
(
t
u
+
(
1
-
t
)
v
)
≤
t
α
f
(
u
)
+
(
1
-
t
)
α
f
(
v
)
,
holds, then f is called a generalized convex on I.Inspired by these investigations, we will introduce the generalizeds-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 Section2, 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].Ifa
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
α
b
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
α
b
α
=
b
α
a
α
=
(
a
b
)
α
=
(
b
a
)
α;
(5)
a
α
(
b
α
c
α
)
=
(
a
α
b
α
)
c
α;
(6)
a
α
(
b
α
+
c
α
)
=
a
α
b
α
+
a
α
c
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 onR
α.Definition 3 (see [13]).
A nondifferentiable functionf
:
R
→
R
α, x
→
f
(
x
) is called to be local fractional continuous at x
0, if, for any ε
>
0, there exists δ
>
0, such that
(5)
|
f
(
x
)
-
f
(
x
0
)
|
<
ε
α
holds for |
x
-
x
0
|
<
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 functionf of order α at x
=
x
0 is defined by
(6)
f
(
α
)
(
x
0
)
=
d
α
f
(
x
)
d
x
α
|
x
=
x
0
=
lim
⁡
x
→
x
0
Δ
α
(
f
(
x
)
-
f
(
x
0
)
)
(
x
-
x
0
)
α
,
where Δ
α
(
f
(
x
)
-
f
(
x
0
)
)
=
Γ
(
1
+
a
)
(
f
(
x
)
-
f
(
x
0
)
) and the Gamma function is defined by Γ
(
t
)
=
∫
0
+
∞
‍
x
t
-
1
e
-
x
d
x.
If there existsf
(
(
k
+
1
)
α
)
(
x
)
=
D
x
α
⋯
D
x
α
︷
k
+
1
times
f
(
x
) for any x
∈
I
⊆
R, then one denoted f
∈
D
(
k
+
1
)
α
(
I
), where k
=
0,1
,
2
,
….Definition 5 (see [13]).
Letf
∈
C
α
[
a
,
b
]. Then the local fractional integral of the function f of order α is defined by
(7)
I
b
(
α
)
a
f
=
1
Γ
(
1
+
a
)
∫
a
b
‍
f
(
t
)
(
d
t
)
α
=
1
Γ
(
1
+
a
)
lim
⁡
Δ
t
→
0
∑
j
=
0
N
‍
f
(
t
j
)
(
Δ
t
j
)
α
,
with Δ
t
j
=
t
j
+
1
-
t
j, Δ
t
=
max
⁡
{
Δ
t
1
,
Δ
t
2
,
Δ
t
j
,
…
,
Δ
t
N
-
1
}, and [
t
j
,
t
j
+
1
], j
=
0
,
…
,
N
-
1, where t
0
=
a
<
t
1
<
⋯
<
t
i
<
⋯
<
t
N
=
b is a partition of the interval [
a
,
b
].Lemma 6 (see [13]).
Suppose thatf
,
g
∈
C
α
[
a
,
b
] and f
,
g
∈
D
α
(
a
,
b
). If lim
⁡
x
→
x
0
⁡
f
(
x
)
=
0
α, lim
⁡
x
→
x
0
⁡
g
(
x
)
=
0
α and g
(
α
)
(
x
)
≠
0
α. Suppose that lim
⁡
x
→
x
0
⁡
(
f
(
α
)
(
x
)
/
g
(
α
)
(
x
)
)
=
A
α, and then
(8)
lim
⁡
x
→
x
0
f
(
x
)
g
(
x
)
=
A
α
.Lemma 7 (see [13]).
Suppose thatf
(
x
)
∈
C
α
[
a
,
b
]; then
(9)
d
α
(
I
x
(
α
)
a
f
)
d
x
α
=
f
(
x
)
,
a
<
x
<
b
.


## 3. Generalizeds-Convexity Functions

The convexity of functions plays a significant role in many fields. In this section, let us introduce two kinds of generalizeds-convex functions on fractal sets. And then, we study the properties of the two kinds of generalized s-convex functions.Definition 8.
LetR
+
=
[
0
,
+
∞
). A function f
:
R
+
→
R
α is said to be generalized s-convex (
0
<
s
<
1
) in the first sense, if
(10)
f
(
λ
1
u
+
λ
2
v
)
≤
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
,
for all u
,
v
∈
R
+ and all λ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
=
1. One denotes this by f
∈
G
K
s
1.Definition 9.
A functionf
:
R
+
→
R
α is said to be generalized s-convex (
0
<
s
<
1
) in the second sense, if
(11)
f
(
λ
1
u
+
λ
2
v
)
≤
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
,
for all u
,
v
∈
R
+ and all λ
1
,
λ
2
≥
0 with λ
1
+
λ
2
=
1. One denotes this by f
∈
G
K
s
2.Note that, whens
=
1, the generalized s-convex functions in both senses are the generalized convex functions; see [14].Theorem 10.
Let0
<
s
<
1.(a)
Iff
∈
G
K
s
1, then f is nondecreasing on (
0
,
+
∞
) and
(12)
f
(
0
+
)
=
lim
⁡
u
→
0
+
f
(
u
)
≤
f
(
0
)
.
(b)
Iff
∈
G
K
s
2, then f is nonnegative on [
0
,
+
∞
).Proof.
(a) Sincef
∈
G
K
s
1, we have, for u
>
0 and λ
∈
[
0,1
],
(13)
f
[
(
λ
1
/
s
+
(
1
-
λ
)
1
/
s
)
u
]
≤
λ
α
f
(
u
)
+
(
1
-
λ
)
α
f
(
u
)
=
f
(
u
)
.
The function
(14)
h
(
λ
)
=
λ
1
/
s
+
(
1
-
λ
)
1
/
s
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
(15)
f
(
t
u
)
≤
f
(
u
)
,
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
(16)
f
(
t
u
)
=
f
(
t
1
/
2
(
t
1
/
2
u
)
)
≤
f
(
t
1
/
2
u
)
≤
f
(
u
)
,
for all u
>
0. So we can obtain that
(17)
f
(
t
u
)
≤
f
(
u
)
,
∀
u
>
0
,
t
∈
(
0,1
]
.
So, taking0
<
u
<
v, we get
(18)
f
(
u
)
=
f
(
(
u
v
)
v
)
≤
f
(
v
)
,
which means that f is nondecreasing on (
0
,
+
∞
).
As for the second part, foru
>
0 and λ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
=
1, we have
(19)
f
(
λ
1
u
)
=
f
(
λ
1
u
+
λ
2
0
)
≤
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
0
)
.
And taking u
→
0
+, we get
(20)
lim
⁡
u
→
0
+
f
(
u
)
=
lim
⁡
u
→
0
+
f
(
λ
1
u
)
≤
λ
1
s
α
lim
⁡
u
→
0
+
f
(
u
)
+
λ
2
s
α
f
(
0
)
.
So,
(21)
lim
⁡
u
→
0
+
f
(
u
)
≤
f
(
0
)
.
(b) Forf
∈
G
K
s
2, we can get that, for u
∈
R
+,
(22)
f
(
u
)
=
f
(
u
2
+
u
2
)
≤
f
(
u
)
2
s
α
+
f
(
u
)
2
s
α
=
2
(
1
-
s
)
α
f
(
u
)
.
So,(
2
1
-
s
-
1
)
α
f
(
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, whens
=
1, because a generalized convex function, f
:
R
+
→
R
α, need not be either nondecreasing or nonnegative.Remark 12.
If0
<
s
<
1, then the function f
∈
G
K
s
1 is nondecreasing on (
0
,
+
∞
) but not necessarily on [
0
,
+
∞
).FunctionF
:
R
2
→
R
α is called to be generalized convex in each variable, if
(23)
F
(
λ
1
u
+
λ
2
v
,
λ
1
r
+
λ
2
t
)
≤
λ
1
α
F
(
u
,
r
)
+
λ
2
α
F
(
v
,
t
)
.
For all (
u
,
r
),  (
v
,
t
)
∈
R
2 and λ
1
,
λ
2
∈
[
0,1
] with λ
1
+
λ
2
=
1.Theorem 13.
Let0
<
s
<
1. If f
,
g
:
R
→
R and f
,
g
∈
K
s
1 and if F
:
R
2
→
R
α is a generalized convex and nondecreasing function in each variable, then the function h
:
R
+
→
R
α defined by
(24)
h
(
u
)
=
F
(
f
(
u
)
,
g
(
u
)
)
is in G
K
s
1. In particular, if f
,
g
∈
K
s
1, then f
α
+
g
α, max
⁡
{
f
α
,
g
α
}
∈
G
K
s
1.Proof.
Ifu
,
v
∈
R
+, then for all λ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
=
1,
(25)
h
(
λ
1
u
+
λ
2
v
)
=
F
(
f
(
λ
1
u
+
λ
2
v
)
,
g
(
λ
1
u
+
λ
2
v
)
)
≤
F
(
λ
1
s
f
(
u
)
+
λ
2
s
f
(
v
)
,
λ
1
s
g
(
u
)
+
λ
2
s
g
(
v
)
)
≤
λ
1
s
α
F
(
f
(
u
)
,
g
(
u
)
)
+
λ
2
s
α
F
(
f
(
v
)
,
g
(
v
)
)
=
λ
1
s
α
h
(
u
)
+
λ
2
s
α
h
(
v
)
.
Thus, h
∈
G
K
s
1.
Moreover, sinceF
(
u
,
v
)
=
u
α
+
v
α,  F
(
u
,
v
)
=
max
⁡
{
u
α
,
v
α
} are nondecreasing generalized convex functions on R
2, so they yield particular cases of our theorem.Let us pay attention to the situation when the conditionλ
1
s
+
λ
2
s
=
1  (
λ
1
+
λ
2
=
1
) in the definition of G
K
s
1
(
G
K
s
2
) can be equivalently replaced by the condition λ
1
s
+
λ
2
s
≤
1  (
λ
1
+
λ
2
≤
1
).Theorem 14.
(a) Letf
∈
G
K
s
1. Then inequality (10) holds for all u
,
v
∈
R
+ and all λ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
<
1 if and only if f
(
0
)
≤
0
α.
(b) Letf
∈
G
K
s
2. Then inequality (11) holds for all u
,
v
∈
R
+ and all λ
1
,
λ
2
≥
0 with λ
1
+
λ
2
<
1 if and only if f
(
0
)
=
0
α.Proof.
(a) Necessity is obvious by takingu
=
v
=
0 and λ
1
=
λ
2
=
0. Let us show the sufficiency.
Assume thatu
,
v
∈
R
+ and λ
1
,
λ
2
≥
0 with 0
<
λ
3
=
λ
1
s
+
λ
2
s
<
1. Put a
=
λ
1
λ
3
-
1
/
s and b
=
λ
2
λ
3
-
1
/
s. Then a
s
+
b
s
=
λ
1
s
/
λ
3
+
λ
2
s
/
λ
3
=
1 and
(26)
f
(
λ
1
u
+
λ
2
v
)
=
f
(
a
λ
3
1
/
s
u
+
b
λ
3
1
/
s
v
)
≤
a
s
α
f
(
λ
3
1
/
s
u
)
+
b
s
α
f
(
λ
3
1
/
s
v
)
=
a
s
α
f
[
λ
3
1
/
s
u
+
(
1
-
λ
3
)
1
/
s
0
]
+
b
s
α
f
[
λ
3
1
/
s
v
+
(
1
-
λ
3
)
1
/
s
0
]
≤
a
s
α
[
λ
3
α
f
(
u
)
+
(
1
-
λ
3
)
α
f
(
0
)
]
+
b
s
α
[
λ
3
α
f
(
v
)
+
(
1
-
λ
3
)
α
f
(
0
)
]
=
a
s
α
λ
3
α
f
(
u
)
+
b
s
α
λ
3
α
f
(
v
)
+
(
1
-
λ
3
)
α
f
(
0
)
≤
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
.
(b) Necessity. Takingu
=
v
=
λ
1
=
λ
2
=
0, we obtain f
(
0
)
≤
0
α. And using Theorem 10(b), we get f
(
0
)
≥
0
α. Therefore f
(
0
)
=
0
α.
Sufficiency. Letu
,
v
∈
R
+ and λ
1
,
λ
2
≥
0 with 0
<
λ
3
=
λ
1
+
λ
2
<
1. Put a
=
λ
1
/
λ
3 and b
=
λ
2
/
λ
3, and then a
+
b
=
1.
So,(27)
f
(
λ
1
u
+
λ
2
v
)
=
f
(
a
λ
3
u
+
b
λ
3
v
)
≤
a
s
α
f
(
λ
3
u
)
+
b
s
α
f
(
λ
3
v
)
=
a
s
α
f
[
λ
3
u
+
(
1
-
λ
3
)
0
]
+
b
s
α
f
[
λ
3
v
+
(
1
-
λ
3
)
0
]
≤
a
s
α
[
λ
3
s
α
f
(
u
)
+
(
1
-
λ
3
)
s
α
f
(
0
)
]
+
b
s
α
[
λ
3
s
α
f
(
v
)
+
(
1
-
λ
3
)
s
α
f
(
0
)
]
=
a
s
α
λ
3
s
α
f
(
u
)
+
b
s
α
λ
3
s
α
f
(
v
)
+
(
1
-
λ
3
)
s
α
f
(
0
)
=
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
.Theorem 15.
(a) Let0
<
s
≤
1. If f
∈
G
K
s
2 and f
(
0
)
=
0
α, then f
∈
G
K
s
1.
(b) Let0
<
s
1
≤
s
2
≤
1. If f
∈
G
K
s
2
2 and f
(
0
)
=
0
α, then f
∈
G
K
s
1
2.
(c) Let0
<
s
1
≤
s
2
≤
1. If f
∈
G
K
s
2
1 and f
(
0
)
≤
0
α, then f
∈
G
K
s
1
1.Proof.
(a) Assume thatf
∈
G
K
s
2 and f
(
0
)
=
0
α. Let λ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
=
1, and we have λ
1
+
λ
2
≤
λ
1
s
+
λ
2
s
=
1. From Theorem 14(b), we can get
(28)
f
(
λ
1
u
+
λ
2
v
)
≤
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
,
for u
,
v
∈
R
+, and then f
∈
G
K
s
1.
(b) Assume thatf
∈
G
K
s
2
2, u
,
v
∈
R
+, and λ
1
,
λ
2
≥
0 with λ
1
+
λ
2
=
1. Then we have
(29)
f
(
λ
1
u
+
λ
2
v
)
≤
λ
1
s
2
α
f
(
u
)
+
λ
2
s
2
α
f
(
v
)
≤
λ
1
s
1
α
f
(
u
)
+
λ
2
s
1
α
f
(
v
)
.
So f
∈
G
K
s
1
2.
(c) Assume thatf
∈
G
K
s
2
1, u
,
v
∈
R
+, and λ
1
,
λ
2
≥
0 with λ
1
s
1
+
λ
2
s
1
=
1. Then λ
1
s
2
+
λ
2
s
2
≤
λ
1
s
1
+
λ
2
s
1
=
1. Thus, according to Theorem 14(a), we have
(30)
f
(
λ
1
u
+
λ
2
v
)
≤
λ
1
s
2
α
f
(
u
)
+
λ
2
s
2
α
f
(
v
)
≤
λ
1
s
1
α
f
(
u
)
+
λ
2
s
1
α
f
(
v
)
.
So, f
∈
G
K
s
1
1.Theorem 16.
Let0
<
s
<
1 and p
:
R
+
→
R
+
α be a nondecreasing function. Then the function f defined for u
∈
R
+ by
(31)
f
(
u
)
=
u
(
s
/
(
1
-
s
)
)
α
p
(
u
)
belongs to G
K
s
1.Proof.
Letv
≥
u
≥
0 and λ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
=
1. We consider two cases.
Case I. It is easy to see that f is a nondecreasing function. Let λ
1
u
+
λ
2
v
≤
u, and then
(32)
f
(
λ
1
u
+
λ
2
v
)
≤
f
(
u
)
=
(
λ
1
s
α
+
λ
2
s
α
)
f
(
u
)
≤
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
.
Case II. Let λ
1
u
+
λ
2
v
>
u, and then λ
2
v
>
(
1
-
λ
1
)
u. So, λ
2
>
0 and λ
1
≤
λ
1
s. Thus,
(33)
λ
1
-
λ
1
s
+
1
≤
λ
1
s
-
λ
1
s
+
1
;
that is,
(34)
λ
1
(
1
-
λ
1
)
≤
λ
1
s
(
1
-
λ
1
s
)
=
(
1
-
λ
2
s
)
λ
2
s
,
λ
1
λ
2
(
1
-
λ
1
)
≤
λ
2
1
-
s
-
λ
2
.
Thus, we can get that(35)
λ
1
u
+
λ
2
v
≤
(
λ
1
+
λ
2
)
v
≤
(
λ
1
s
+
λ
2
s
)
v
=
v
,
λ
1
u
+
λ
2
v
≤
λ
1
λ
2
v
(
1
-
λ
1
)
+
λ
2
v
≤
(
λ
2
1
-
s
-
λ
2
)
v
+
λ
2
v
=
λ
2
1
-
s
v
.
Then,
(36)
(
λ
1
u
+
λ
2
v
)
s
/
(
1
-
s
)
≤
λ
2
s
v
s
/
(
1
-
s
)
.
We obtain(37)
f
(
λ
1
u
+
λ
2
v
)
=
(
λ
1
u
+
λ
2
v
)
(
s
/
(
1
-
s
)
)
α
p
(
λ
1
u
+
λ
2
v
)
≤
λ
2
s
α
v
(
s
/
(
1
-
s
)
)
α
p
(
v
)
=
λ
2
s
α
f
(
v
)
≤
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
.Theorem 17.
(a) Letf
∈
G
K
s
1
1 and g
∈
K
s
2
1, where 0
<
s
1, s
2
≤
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 G
K
s
1, where s
=
s
1
s
2.
(b) Letf
∈
G
K
s
1
1 and g
∈
G
K
s
2
1, where 0
<
s
1, s
2
≤
1. Assume that 0
<
s
1, s
2
<
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 f
g of f and g belongs to G
K
s
1, where s
=
min
⁡
{
s
1
,
s
2
}.Proof.
(a) Letu
,
v
∈
R
+,  λ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
=
1, where s
=
s
1
s
2. Since λ
1
s
i
+
λ
2
s
i
≤
λ
1
s
1
s
2
+
λ
2
s
1
s
2
=
1 for i
=
1,2, then according to Theorem 3(a) in [3] and Theorem 14(a) in the paper, we have
(38)
f
∘
g
(
λ
1
u
+
λ
2
v
)
=
f
(
g
(
λ
1
u
+
λ
2
v
)
)
≤
f
(
λ
1
s
2
g
(
u
)
+
λ
2
s
2
g
(
v
)
)
≤
λ
1
s
1
s
2
α
f
(
g
(
u
)
)
+
λ
2
s
1
s
2
α
f
(
g
(
v
)
)
=
λ
1
s
α
f
∘
g
(
u
)
+
λ
2
s
α
f
∘
g
(
v
)
which means that f
∘
g
∈
G
K
s
1.
(b) According to Theorem10(a), f
,
g are nondecreasing on (
0
,
+
∞
).
So,(39)
(
f
(
u
)
-
f
(
v
)
)
(
g
(
v
)
-
g
(
u
)
)
≤
0
α
,
or, equivalently,
(40)
f
(
u
)
g
(
v
)
+
f
(
v
)
g
(
u
)
≤
f
(
u
)
g
(
u
)
+
f
(
v
)
g
(
v
)
,
for all v
>
u
>
0.
Ifv
>
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 letu
,
v
∈
R
+ and λ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
=
1, where s
=
min
⁡
{
s
1
,
s
2
}. Then λ
1
s
i
+
λ
2
s
i
≤
λ
1
s
+
λ
2
s
=
1 for i
=
1,2. And by Theorem 14(a), we have
(41)
f
(
λ
1
u
+
λ
2
v
)
g
(
λ
1
u
+
λ
2
v
)
≤
(
λ
1
s
1
α
f
(
u
)
+
λ
2
s
1
α
f
(
v
)
)
×
(
λ
1
s
2
α
g
(
u
)
+
λ
2
s
2
α
g
(
v
)
)
=
λ
1
(
s
1
+
s
2
)
α
f
(
u
)
g
(
u
)
+
λ
1
s
1
α
λ
2
s
2
α
f
(
u
)
g
(
v
)
+
λ
1
s
2
α
λ
2
s
1
α
f
(
v
)
g
(
u
)
+
λ
2
(
s
1
+
s
2
)
α
f
(
v
)
g
(
v
)
≤
λ
1
2
s
α
f
(
u
)
g
(
u
)
+
λ
1
s
α
λ
2
s
α
(
f
(
u
)
g
(
v
)
+
f
(
v
)
g
(
u
)
)
+
λ
2
2
s
α
f
(
v
)
g
(
v
)
≤
λ
1
2
s
α
f
(
u
)
g
(
u
)
+
λ
1
s
α
λ
2
s
α
(
f
(
u
)
g
(
u
)
+
f
(
v
)
g
(
v
)
)
+
λ
2
2
s
α
f
(
v
)
g
(
v
)
=
λ
1
s
α
f
(
u
)
g
(
u
)
+
λ
2
s
α
f
(
v
)
g
(
v
)
,
which means that f
g
∈
G
K
s
1.Remark 18.
From the above proof, we can get that iff is a nondecreasing function in G
K
s
2 and g is a nonnegative convex function on [
0
,
+
∞
), then the composition f
∘
g of f with g belongs to G
K
s
2.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 f
g is also a generalized convex function.Letf
:
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.
IfΦ is a generalized convex φ-type function and g
∈
K
s
1 is a φ-function, then the composition Φ
∘
g belongs to G
K
s
1. In particular, the φ-type function h
(
u
)
=
Φ
(
u
s
) belongs to G
K
s
1.Corollary 21.
IfΦ is a convex φ-function and f
∈
G
K
s
2 is a φ-type function, then the composition f
∘
Φ belongs to G
K
s
2. In particular, the φ-type function h
(
u
)
=
[
Φ
(
u
)
]
s
α belongs to G
K
s
2.Theorem 22.
If0
<
s
<
1 and f
∈
G
K
s
1 is a φ-type function, then there exists a generalized convex φ-type function Φ such that
(42)
f
(
2
-
1
/
s
u
)
≤
Φ
(
u
s
)
≤
f
(
u
)
,
for all u
≥
0.Proof.
By the generalizeds-convexity of the function f and by f
(
0
)
=
0
α, we obtain f
(
λ
1
u
)
≤
λ
1
s
α
f
(
u
) for all u
≥
0 and all λ
1
∈
[
0,1
].
Assume now thatv
>
u
>
0. Then
(43)
f
(
u
1
/
s
)
≤
f
(
(
u
v
)
1
/
s
v
1
/
s
)
≤
(
u
α
v
α
)
f
(
v
1
/
s
)
;
that is,
(44)
f
(
u
1
/
s
)
u
α
≤
f
(
v
1
/
s
)
v
α
.
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(45)
Φ
(
u
)
=
{
0
α
,
u
=
0
,
Γ
(
1
+
α
)
0
I
u
(
α
)
(
f
(
t
1
/
s
)
t
α
)
,
u
>
0
.
From Lemmas6 and 7, it is easy to see that Φ is a generalized convex φ-type function and
(46)
Φ
(
u
s
)
=
Γ
(
1
+
α
)
0
I
u
s
(
α
)
(
f
(
t
1
/
s
)
t
α
)
≤
(
f
(
(
u
s
)
1
/
s
)
u
s
α
)
u
s
α
=
f
(
u
)
.
Moreover,
(47)
Φ
(
u
s
)
≥
Γ
(
1
+
α
)
(
u
s
/
2
)
I
u
s
(
α
)
(
f
(
t
1
/
s
)
t
α
)
≥
(
f
(
(
u
s
/
2
)
1
/
s
)
2
α
u
-
s
α
)
u
s
α
2
α
=
f
(
2
-
1
/
s
u
)
.
Therefore,(48)
f
(
2
-
1
/
s
u
)
≤
Φ
(
u
s
)
≤
f
(
u
)
,
for all u
≥
0.


## 4. Applications

Based on the properties of the two kinds of generalizeds-convex functions in the above section, some applications are given.Example 23.
Let0
<
s
<
1, and a
α
,
b
α
,
c
α
∈
R
α. For u
∈
R
+, define
(49)
f
(
u
)
=
{
a
α
,
u
=
0
,
b
α
u
s
α
+
c
α
,
u
>
0
.
We have the following conclusions.(i)
Ifb
α
≥
0
α and c
α
≤
a
α, then f
∈
G
K
s
1.
(ii)
Ifb
α
≥
0
α and c
α
<
a
α, then f is nondecreasing on (
0
,
+
∞
) but not on [
0
,
+
∞
).
(iii)
Ifb
α
≥
0
α and 0
α
≤
c
α
≤
a
α, then f
∈
G
K
s
2.
(iv)
Ifb
α
>
0
α and c
α
<
0
α, then f
∉
G
K
s
2.Proof.
(i) Letλ
1
,
λ
2
≥
0 with λ
1
s
+
λ
2
s
=
1. Then, there are two nontrivial cases.
Case I. Let u
,
v
>
0. Then λ
1
u
+
λ
2
v
>
0.
Thus,(50)
f
(
λ
1
u
+
λ
2
v
)
=
b
α
(
λ
1
u
+
λ
2
v
)
s
α
+
c
α
≤
b
α
(
λ
1
s
α
u
s
α
+
λ
2
s
α
v
s
α
)
+
c
α
=
b
α
(
λ
1
s
α
u
s
α
+
λ
2
s
α
v
s
α
)
+
c
α
(
λ
1
s
α
+
λ
2
s
α
)
=
λ
1
s
α
(
b
α
u
s
α
+
c
α
)
+
λ
2
s
α
(
b
α
v
s
α
+
c
α
)
=
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
.
Case II. Let v
>
u
=
0. We need only to consider λ
2
>
0.
Thus, we have(51)
f
(
λ
1
0
+
λ
2
v
)
=
f
(
λ
2
v
)
=
b
α
λ
2
s
α
v
s
α
+
c
α
=
b
α
λ
2
s
α
v
s
α
+
c
α
(
λ
1
s
α
+
λ
2
s
α
)
=
λ
1
s
α
c
α
+
λ
2
s
α
(
b
α
v
s
α
+
c
α
)
=
λ
1
s
α
c
α
+
λ
2
s
α
f
(
v
)
≤
λ
1
s
α
a
α
+
λ
2
s
α
f
(
v
)
=
λ
1
s
α
f
(
0
)
+
λ
2
s
α
f
(
v
)
.
So, f
∈
G
K
s
1.
(ii) From Theorem10, we can see that property (ii) is true.
(iii) Letλ
1
,
λ
2
≥
0 with λ
1
+
λ
2
=
1. Similar to the estimate of (i), there are also two cases.
Letv
,
v
>
0. Then λ
1
u
+
λ
2
v
>
0,
Thus,(52)
f
(
λ
1
u
+
λ
2
v
)
=
b
α
(
λ
1
u
+
λ
2
v
)
s
α
+
c
α
<
b
α
(
λ
1
s
α
u
s
α
+
λ
2
s
α
v
s
α
)
+
c
α
(
λ
1
α
+
λ
2
α
)
≤
b
α
(
λ
1
s
α
u
s
α
+
λ
2
s
α
v
s
α
)
+
c
α
(
λ
1
s
α
+
λ
2
s
α
)
=
λ
1
s
α
(
b
α
u
s
α
+
c
α
)
+
λ
2
s
α
(
b
α
v
s
α
+
c
α
)
≤
λ
1
s
α
f
(
u
)
+
λ
2
s
α
f
(
v
)
.
Letv
>
u
=
0. We need only to consider λ
2
>
0.
Thus, we have(53)
f
(
λ
1
0
+
λ
2
v
)
=
f
(
λ
2
v
)
=
b
α
λ
2
s
α
v
s
α
+
c
α
(
λ
1
α
+
λ
2
α
)
<
b
α
λ
2
s
α
v
s
α
+
c
α
(
λ
1
s
α
+
λ
2
s
α
)
=
λ
2
s
α
(
b
α
v
s
α
+
c
α
)
+
c
α
λ
1
s
α
≤
λ
2
s
α
(
b
α
v
s
α
+
c
α
)
+
a
α
λ
1
s
α
=
λ
2
s
α
f
(
v
)
+
λ
1
s
α
f
(
0
)
.
So,f
∈
G
K
s
2.
(iv) Assume thatf
∈
G
K
s
2, and then f is nonnegative on (
0
,
∞
). On the other hand, we can take u
1
>
0, c
1
<
0 such that f
(
u
1
)
=
b
α
u
1
s
α
+
c
1
α
<
0
α, which contradict the assumption.Example 24.
Let0
<
s
<
1 and k
>
1. For u
∈
R
+, define
(54)
f
(
u
)
=
{
u
(
s
/
(
1
-
s
)
)
α
,
0
≤
u
≤
1
,
k
α
u
(
s
/
(
1
-
s
)
)
α
,
u
>
1
.
The function f is nonnegative, not local fractional continuous at u
=
1 and belongs to G
K
s
1 but not to G
K
s
2.Proof.
From Theorem16, we have that f
∈
G
K
s
1. In the following, let us show that f is not in G
K
s
2.
Take an arbitrarya
>
1 and put u
=
1. Consider all v
>
1 such that λ
1
u
+
λ
2
v
=
λ
1
+
λ
2
v
=
a, where λ
1
,
λ
2
≥
0 and λ
1
+
λ
2
=
1.
Iff
∈
G
K
s
2, it must be
(55)
k
α
a
(
s
/
(
1
-
s
)
)
α
≤
λ
1
s
α
+
k
α
(
1
-
λ
1
)
s
α
[
(
a
-
λ
1
)
(
1
-
λ
1
)
]
(
s
/
(
1
-
s
)
)
α
,
for all a
>
1 and all 0
≤
λ
1
≤
1.
Define the function(56)
f
λ
1
(
a
)
=
λ
1
s
α
+
k
α
(
1
-
λ
1
)
s
α
[
(
a
-
λ
1
)
(
1
-
λ
1
)
]
(
s
/
(
1
-
s
)
)
α
-
k
α
a
(
s
/
(
1
-
s
)
)
α
.
Then the function is local fractional continuous on the (
λ
1
,
∞
) and
(57)
g
(
λ
1
)
=
f
λ
1

(
1
)
=
λ
1
s
α
+
k
α
(
1
-
λ
1
)
s
α
-
k
α
.
It is easy to see thatg is local fractional continuous on [
0,1
] and g
(
1
)
=
1
α
-
k
α
<
0
α. So there is a number λ
1
0, 0
<
λ
1
0
<
1, such that g
(
λ
1
0
)
=
f
λ
1
0
(
1
)
<
0
α. The local fractional continuity of f
λ
1
0 yields that f
λ
1
0
(
a
)
<
0
α for a certain a
>
1, that is, inequality (55) does not hold, which means that f
∉
G
K
s
2.


## 5. Conclusion

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



---

*Source: 254737-2014-07-17.xml*