(ProQuest: ... denotes non-US-ASCII text omitted.)
Academic Editor:Hari M. Srivastava
School of Science, Beijing University of Posts and Telecommunications, Beijing 100876, China
Received 5 May 2014; Revised 27 June 2014; Accepted 27 June 2014; 17 July 2014
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let f : I ⊆ R [arrow right] R . For any u , v ∈ I and t ∈ [ 0,1 ] , if the following inequality, [figure omitted; refer to PDF] 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 ...4; 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 + [arrow right] R , is said to be s -convex in the first sense if [figure omitted; refer to PDF] for all u , v ∈ R + and all α , β ...5; 0 with α s + β s = 1 . One denotes this by f ∈ K s 1 .
Definition 2.
A function, f : R + [arrow right] R , is said to be s -convex in the second sense if [figure omitted; refer to PDF] for all u , v ∈ R + and all α , β ...5; 0 with α + β = 1 . One denotes this by f ∈ K s 2 .
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 [arrow right] R α . For any u , v ∈ I and t ∈ [ 0,1 ] , if the following inequality, [figure omitted; refer to PDF] 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 ...4; 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 on R α .
Definition 3 (see [13]).
A nondifferentiable function f : R [arrow right] R α , x [arrow right] f ( x ) is called to be local fractional continuous at x 0 , if, for any [straight epsilon] > 0 , there exists δ > 0 , such that [figure omitted; refer to PDF] holds for | x - x 0 | < δ , where [straight epsilon] , 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 = x 0 is defined by [figure omitted; refer to PDF] 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 exists f ( ( 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]).
Let f ∈ C α [ a , b ] . Then the local fractional integral of the function f of order α is defined by [figure omitted; refer to PDF] 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 that f , g ∈ C α [ a , b ] and f , g ∈ D α ( a , b ) . If lim ... x [arrow right] x 0 ... f ( x ) = 0 α , lim ... x [arrow right] x 0 ... g ( x ) = 0 α and g ( α ) ( x ) ...0; 0 α . Suppose that lim ... x [arrow right] x 0 ... ( f ( α ) ( x ) / g ( α ) ( x ) ) = A α , and then [figure omitted; refer to PDF]
Lemma 7 (see [13]).
Suppose that f ( x ) ∈ C α [ a , b ] ; then [figure omitted; refer to PDF]
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 + [arrow right] R α is said to be generalized s -convex ( 0 < s < 1 ) in the first sense, if [figure omitted; refer to PDF] for all u , v ∈ R + and all λ 1 , λ 2 ...5; 0 with λ 1 s + λ 2 s = 1 . One denotes this by f ∈ G K s 1 .
Definition 9.
A function f : R + [arrow right] R α is said to be generalized s -convex ( 0 < s < 1 ) in the second sense, if [figure omitted; refer to PDF] for all u , v ∈ R + and all λ 1 , λ 2 ...5; 0 with λ 1 + λ 2 = 1 . One denotes this by f ∈ G K s 2 .
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 ∈ G K s 1 , then f is nondecreasing on ( 0 , + ∞ ) and [figure omitted; refer to PDF]
(b) If f ∈ G K s 2 , then f is nonnegative on [ 0 , + ∞ ) .
Proof.
(a) Since f ∈ G K s 1 , we have, for u > 0 and λ ∈ [ 0,1 ] , [figure omitted; refer to PDF] The function [figure omitted; refer to PDF] 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 [figure omitted; refer to PDF] 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 [figure omitted; refer to PDF] for all u > 0 . So we can obtain that [figure omitted; refer to PDF]
So, taking 0 < u < v , we get [figure omitted; refer to PDF] which means that f is nondecreasing on ( 0 , + ∞ ) .
As for the second part, for u > 0 and λ 1 , λ 2 ...5; 0 with λ 1 s + λ 2 s = 1 , we have [figure omitted; refer to PDF] And taking u [arrow right] 0 + , we get [figure omitted; refer to PDF] So, [figure omitted; refer to PDF]
(b) For f ∈ G K s 2 , we can get that, for u ∈ R + , [figure omitted; refer to PDF]
So, ( 2 1 - s - 1 ) α f ( u ) ...5; 0 α . This means that f ( u ) ...5; 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 + [arrow right] R α , need not be either nondecreasing or nonnegative.
Remark 12.
If 0 < s < 1 , then the function f ∈ G K s 1 is nondecreasing on ( 0 , + ∞ ) but not necessarily on [ 0 , + ∞ ) .
Function F : R 2 [arrow right] R α is called to be generalized convex in each variable, if [figure omitted; refer to PDF] For all ( u , r ) , ( v , t ) ∈ R 2 and λ 1 , λ 2 ∈ [ 0,1 ] with λ 1 + λ 2 = 1 .
Theorem 13.
Let 0 < s < 1 . If f , g : R [arrow right] R and f , g ∈ K s 1 and if F : R 2 [arrow right] R α is a generalized convex and nondecreasing function in each variable, then the function h : R + [arrow right] R α defined by [figure omitted; refer to PDF] 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.
If u , v ∈ R + , then for all λ 1 , λ 2 ...5; 0 with λ 1 s + λ 2 s = 1 , [figure omitted; refer to PDF] Thus, h ∈ G K s 1 .
Moreover, since F ( 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 condition λ 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 ...4; 1 ( λ 1 + λ 2 ...4; 1 ) .
Theorem 14.
(a) Let f ∈ G K s 1 . Then inequality (10) holds for all u , v ∈ R + and all λ 1 , λ 2 ...5; 0 with λ 1 s + λ 2 s < 1 if and only if f ( 0 ) ...4; 0 α .
(b) Let f ∈ G K s 2 . Then inequality (11) holds for all u , v ∈ R + and all λ 1 , λ 2 ...5; 0 with λ 1 + λ 2 < 1 if and only if f ( 0 ) = 0 α .
Proof.
(a) Necessity is obvious by taking u = v = 0 and λ 1 = λ 2 = 0 . Let us show the sufficiency.
Assume that u , v ∈ R + and λ 1 , λ 2 ...5; 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 [figure omitted; refer to PDF]
(b) Necessity. Taking u = v = λ 1 = λ 2 = 0 , we obtain f ( 0 ) ...4; 0 α . And using Theorem 10(b), we get f ( 0 ) ...5; 0 α . Therefore f ( 0 ) = 0 α .
Sufficiency. Let u , v ∈ R + and λ 1 , λ 2 ...5; 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, [figure omitted; refer to PDF]
Theorem 15.
(a) Let 0 < s ...4; 1 . If f ∈ G K s 2 and f ( 0 ) = 0 α , then f ∈ G K s 1 .
(b) Let 0 < s 1 ...4; s 2 ...4; 1 . If f ∈ G K s 2 2 and f ( 0 ) = 0 α , then f ∈ G K s 1 2 .
(c) Let 0 < s 1 ...4; s 2 ...4; 1 . If f ∈ G K s 2 1 and f ( 0 ) ...4; 0 α , then f ∈ G K s 1 1 .
Proof.
(a) Assume that f ∈ G K s 2 and f ( 0 ) = 0 α . Let λ 1 , λ 2 ...5; 0 with λ 1 s + λ 2 s = 1 , and we have λ 1 + λ 2 ...4; λ 1 s + λ 2 s = 1 . From Theorem 14(b), we can get [figure omitted; refer to PDF] for u , v ∈ R + , and then f ∈ G K s 1 .
(b) Assume that f ∈ G K s 2 2 , u , v ∈ R + , and λ 1 , λ 2 ...5; 0 with λ 1 + λ 2 = 1 . Then we have [figure omitted; refer to PDF] So f ∈ G K s 1 2 .
(c) Assume that f ∈ G K s 2 1 , u , v ∈ R + , and λ 1 , λ 2 ...5; 0 with λ 1 s 1 + λ 2 s 1 = 1 . Then λ 1 s 2 + λ 2 s 2 ...4; λ 1 s 1 + λ 2 s 1 = 1 . Thus, according to Theorem 14(a), we have [figure omitted; refer to PDF] So, f ∈ G K s 1 1 .
Theorem 16.
Let 0 < s < 1 and p : R + [arrow right] R + α be a nondecreasing function. Then the function f defined for u ∈ R + by [figure omitted; refer to PDF] belongs to G K s 1 .
Proof.
Let v ...5; u ...5; 0 and λ 1 , λ 2 ...5; 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 ...4; u , and then [figure omitted; refer to PDF]
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 ...4; λ 1 s . Thus, [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF]
Thus, we can get that [figure omitted; refer to PDF] Then, [figure omitted; refer to PDF]
We obtain [figure omitted; refer to PDF]
Theorem 17.
(a) Let f ∈ G K s 1 1 and g ∈ K s 2 1 , where 0 < s 1 , s 2 ...4; 1 . If f is a nondecreasing function and g is a nonnegative function such that f ( 0 ) ...4; 0 α and g ( 0 ) = 0 , then the composition f [composite function] g of f with g belongs to G K s 1 , where s = s 1 s 2 .
(b) Let f ∈ G K s 1 1 and g ∈ G K s 2 1 , where 0 < s 1 , s 2 ...4; 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) Let u , v ∈ R + , λ 1 , λ 2 ...5; 0 with λ 1 s + λ 2 s = 1 , where s = s 1 s 2 . Since λ 1 s i + λ 2 s i ...4; 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 [figure omitted; refer to PDF] which means that f [composite function] g ∈ G K s 1 .
(b) According to Theorem 10(a), f , g are nondecreasing on ( 0 , + ∞ ) .
So, [figure omitted; refer to PDF] or, equivalently, [figure omitted; refer to PDF] 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 λ 1 , λ 2 ...5; 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 ...4; λ 1 s + λ 2 s = 1 for i = 1,2 . And by Theorem 14(a), we have [figure omitted; refer to PDF] which means that f g ∈ G K s 1 .
Remark 18.
From the above proof, we can get that if f is a nondecreasing function in G K s 2 and g is a nonnegative convex function on [ 0 , + ∞ ) , then the composition f [composite function] 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.
Let f : R + [arrow right] R + be a continuous function. Then f is said to be a [straight phi] -function if f ( 0 ) = 0 and f is nondecreasing on R + . Similarly, we can define the [straight phi] -type function on fractal sets as follows. A function f : R + [arrow right] R + α is said to be a [straight phi] -type function if f ( 0 ) = 0 α and f ∈ C α ( R + ) is nondecreasing.
Corollary 20.
If Φ is a generalized convex [straight phi] -type function and g ∈ K s 1 is a [straight phi] -function, then the composition Φ [composite function] g belongs to G K s 1 . In particular, the [straight phi] -type function h ( u ) = Φ ( u s ) belongs to G K s 1 .
Corollary 21.
If Φ is a convex [straight phi] -function and f ∈ G K s 2 is a [straight phi] -type function, then the composition f [composite function] Φ belongs to G K s 2 . In particular, the [straight phi] -type function h ( u ) = [ Φ ( u ) ] s α belongs to G K s 2 .
Theorem 22.
If 0 < s < 1 and f ∈ G K s 1 is a [straight phi] -type function, then there exists a generalized convex [straight phi] -type function Φ such that [figure omitted; refer to PDF] for all u ...5; 0 .
Proof.
By the generalized s -convexity of the function f and by f ( 0 ) = 0 α , we obtain f ( λ 1 u ) ...4; λ 1 s α f ( u ) for all u ...5; 0 and all λ 1 ∈ [ 0,1 ] .
Assume now that v > u > 0 . Then [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF]
Inequality (44) means that the function f ( u 1 / s ) / u α is a nondecreasing function on ( 0 , + ∞ ) . And, since f is a [straight phi] -type function, thus f is local fractional continuous [ 0 , + ∞ ) .
Define [figure omitted; refer to PDF]
From Lemmas 6 and 7, it is easy to see that Φ is a generalized convex [straight phi] -type function and [figure omitted; refer to PDF] Moreover, [figure omitted; refer to PDF]
Therefore, [figure omitted; refer to PDF] for all u ...5; 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 [figure omitted; refer to PDF]
We have the following conclusions.
(i) If b α ...5; 0 α and c α ...4; a α , then f ∈ G K s 1 .
(ii) If b α ...5; 0 α and c α < a α , then f is nondecreasing on ( 0 , + ∞ ) but not on [ 0 , + ∞ ) .
(iii): If b α ...5; 0 α and 0 α ...4; c α ...4; a α , then f ∈ G K s 2 .
(iv) If b α > 0 α and c α < 0 α , then f ∉ G K s 2 .
Proof.
(i) Let λ 1 , λ 2 ...5; 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, [figure omitted; refer to PDF]
Case II. Let v > u = 0 . We need only to consider λ 2 > 0 .
Thus, we have [figure omitted; refer to PDF] So, f ∈ G K s 1 .
(ii) From Theorem 10, we can see that property (ii) is true.
(iii) Let λ 1 , λ 2 ...5; 0 with λ 1 + λ 2 = 1 . Similar to the estimate of (i), there are also two cases.
Let v , v > 0 . Then λ 1 u + λ 2 v > 0 ,
Thus, [figure omitted; refer to PDF]
Let v > u = 0 . We need only to consider λ 2 > 0 .
Thus, we have [figure omitted; refer to PDF]
So, f ∈ G K s 2 .
(iv) Assume that f ∈ 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.
Let 0 < s < 1 and k > 1 . For u ∈ R + , define [figure omitted; refer to PDF] 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 Theorem 16, 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 arbitrary a > 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 ...5; 0 and λ 1 + λ 2 = 1 .
If f ∈ G K s 2 , it must be [figure omitted; refer to PDF] for all a > 1 and all 0 ...4; λ 1 ...4; 1 .
Define the function [figure omitted; refer to PDF] Then the function is local fractional continuous on the ( λ 1 , ∞ ) and [figure omitted; refer to PDF]
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 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 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.
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).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Abstract
We introduce two kinds of generalized s -convex functions on real linear fractal sets [superscript] R α [/superscript] ( 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.
You have requested "on-the-fly" machine translation of selected content from our databases. This functionality is provided solely for your convenience and is in no way intended to replace human translation. Show full disclaimer
Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer