# Generalized Convex Functions on Fractal Sets and Two Related Inequalities

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

---

## Abstract

We introduce the generalized convex function on fractal setsRα(0<α≤1) of real line numbers and study the properties of the generalized convex function. Based on these properties, we establish the generalized Jensen’s inequality and generalized Hermite-Hadamard's inequality. Furthermore, some applications are given.


---

## Body

## 1. Introduction

Letf:I⊆R→R. For any x1,x2∈I and λ∈[0,1], if the following inequality
(1)f(λx1+(1-λ)x2)≤λf(x1)+(1-λ)f(x2)
holds, then f is called a convex function on I.The convexity of functions plays a significant role in many fields, for example, in biological system, economy, optimization, and so on [1, 2]. And many important inequalities are established for the class of convex functions. For example, Jensen’s inequality and Hermite-Hadamard’s inequality are the best known results in the literature, which can be stated as follows.Jensen’s Inequality [3]. Assume that f is a convex function on [a,b]. Then, for any xi∈[a,b] and λi∈[0,1](i=1,2,…,n) with ∑i=1nλi=1, we have
(2)f(∑i=1nλixi)≤∑i=1nλif(xi).Hermite-Hadamard’s Inequality [4]. Let f be a convex function on [a,b] with a<b. If f is integral on [a,b], then
(3)f(a+b2)≤1b-a∫abf(x)dx≤f(a)+f(b)2.In recent years, the fractal theory 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 [5–9]. Many researchers studied the properties of functions on fractal space and constructed many kinds of fractional calculus by using different approaches (see [10–14]). Particularly, in [13], Yang stated the analysis of local fractional functions on fractal space systematically, which includes local fractional calculus and the monotonicity of function.Inspired by these investigations, we will introduce the generalized convex function on fractal sets and establish the generalized Jensen’s inequality and generalized Hermite-Hadamard’s inequality related to generalized convex function. We will focus our attention on the convexity since a functionf is concave if and only if -f is convex. So, every result for the convex function can be easily restated in terms of concave functions.The paper is organized as follows. In Section2, we state the operations with real line number on fractal sets and give the definitions of the local fractional derivatives and local fractional integral. In Section 3, we introduce the definition of the generalized convex function on fractal sets and study the properties of the generalized convex functions. In Section 4, we establish the generalized Jensen’s inequality and generalized Hermite-Hadamard’s inequality on fractalsets. In Section 5, some applications are given on fractal sets by means of the generalized Jensen’s inequality.


## 2. Preliminaries

Recall the setRα of real line numbers and use Gao-Yang-Kang’s idea to describe the definitions of the local fractional derivative and local fractional integral.Recently, the theory of Yang’s fractional sets [13] was introduced as follows.For0<α≤1, we have the following α-type set of element sets.Zα: the α-type set of the integer is defined as the set {0α,±1α,±2α,…,±nα,…}.Qα: the α-type set of the rational numbers is defined as the set {mα=(p/q)α:p∈Z,q≠0}.Jα: the α-type set of the irrational numbers is defined as the set {mα≠(p/q)α:p∈Z,q≠0}.Rα: the α-type set of the real line numbers is defined as the set Rα=Qα∪Jα.Ifaα,bα, and cα belong to the set Rα of real line numbers, then one has the following:(1)
aα+bα and aαbα belong to the set Rα;(2)
aα+bα=bα+aα=(a+b)α=(b+a)α;(3)
aα+(bα+cα)=(a+b)α+cα;(4)
aαbα=bαaα=(ab)α=(ba)α;(5)
aα(bαcα)=(aαbα)cα;(6)
aα(bα+cα)=aαbα+aαcα;(7)
aα+0α=0α+aα=aα and aα1α=1αaα=aα.Let us now state some definitions about the local fractional calculus onRα.Definition 1 (see [13]).
A nondifferentiable functionf:R→Rα, x→f(x) is called local fractional continuous at x0, if, for any ε>0, there exists δ>0, such that
(4)|f(x)-f(x0)|<εα
holds for |x-x0|<δ, where ε,δ∈R. If f(x) is local fractional continuous on the interval (a,b), one denotes f(x)∈Cα(a,b).Definition 2 (see [13]).
The local fractional derivative off(x) of order α at x=x0 is defined by
(5)f(α)(x0)=dαf(x)dxα|x=x0=lim⁡x→x0Δα(f(x)-f(x0))(x-x0)α,
where Δα(f(x)-f(x0))=Γ(1+α)(f(x)-f(x0)).If there existsf((k+1)α)(x)=Dxα⋯Dxα︷k+1timesf(x) for any x∈I⊆R, then one denotes f∈D(k+1)α(I), where k=0,1,2,….Definition 3 (see [13]).
The local fractional integral of the functionf(x) of order α is defined by
(6)Iab(α)f(x)=1Γ(1+a)∫abf(t)(dt)α=1Γ(1+a)lim⁡Δt→0∑j=0N-1f(tj)(Δtj)α,
with Δtj=tj+1-tj and Δt=max⁡{Δtj∣j=1,2,…,N-1}, where [tj,tj+1], j=0,…,N-1, and t0=a<t1<⋯<ti<⋯<tN-1<tN=b is a partition of the interval [a,b].Here, it follows thatIaa(α)f(x)=0 if a=b and Iab(α)f(x)=-Iba(α)f(x) if a<b. If, for any x∈[a,b], there exists Iax(α)f(x), then it is denoted by f(x)∈Ix(α)[a,b].Lemma 4 (see [13] generalized local fractional Taylor theorem).
Suppose thatf(k+1)α(x)∈Cα(I), for interval I⊆R, k=0,1,…,n, 0<α≤1. And let x0∈[a,b]. Then, for any x∈I, there exists at least one point ξ, which lies between the points x and x0, such that
(7)f(x)=∑k=0nf(kα)(x0)Γ(1+ka)(x-x0)kα+f((n+1)α)(ξ)Γ(1+(n+1)a)(x-x0)(n+1)α.Remark 5.
WhenI⊆R is an open interval (a,b), Yang [13] has given the proof for the generalized local fractional Taylor theorem. In fact, using the generalized local fractional Lagrange’s theorem and following the proof of the class Taylor theorem, we can show that, for any interval I⊆R, the formula is also true.


## 3. Generalized Convex Functions

From an analytical point of view, we have the following definition.Definition 6.
Letf:I⊆R→Rα. For any x1, x2∈I and λ∈[0,1], if the following inequality
(8)f(λx1+(1-λ)x2)≤λαf(x1)+(1-λ)αf(x2)
holds, then f is called a generalized convex function on I.Definition 7.
Letf:I→Rα. For any x1≠x2∈I and λ∈[0,1], if the following inequality
(9)f(λx1+(1-λ)x2)<λαf(x1)+(1-λ)αf(x2)
holds, then f is called a generalized strictly convex function on I⊆R.It follows immediately, from the given definitions, that a generalized strictly convex function is also generalized convex. But, the converse is not true. And if these two inequalities are reversed, thenf is called a generalized concave function or generalized strictly concave function, respectively.Here are two basic examples of generalized strictly convex functions:(1)
f(x)=xαp, x≥0, p>1;(2)
f(x)=Eα(xα), x∈R, where Eα(xα)=∑k=0∞(xαk/Γ(1+kα)) is the Mittag-Leffler function.Note that the linear functionf(x)=aαxα+bα, x∈R is generalized convex and also generalized concave.We will focus our attention on the convexity since a functionf is concave if and only if -f is convex. So, every result for the convex function can be easily restated in terms of concave functions.In the following, we will study the properties of the generalized convex functions.Theorem 8.
Letf:I→Rα. Then f is a generalized convex function if and only if the inequality
(10)f(x1)-f(x2)(x1-x2)α≤f(x3)-f(x2)(x3-x2)α
holds, for any x1,x2,x3∈I with x1<x2<x3.Proof.
In fact, takingλ=(x3-x2)/(x3-x1), then x2=λx1+(1-λ)x3. And by the generalized convexity of f, we get
(11)f(x2)=f(λx1+(1-λ)x3)≤λαf(x1)+(1-λ)αf(x3)=(x3-x2x3-x1)αf(x1)+(x2-x1x3-x1)αf(x3).
From the above formula, it is easy to see that(12)f(x1)-f(x2)(x1-x2)α≤f(x3)-f(x2)(x3-x2)α.
Reversely, for any two pointsx1,x3(x1<x3) on I⊆R, we take x2=λx1+(1-λ)x3 for λ∈(0,1). Then x1<x2<x3 and λ=(x3-x2)/(x3-x1). Using the above inverse process, we have
(13)f(λx1+(1-λ)x3)≤λαf(x1)+(1-λ)αf(x3).
So,f is a convex function on I⊆R.
In the same way, it can be shown thatf is a generalized convex function on I⊆R if and only if
(14)f(x2)-f(x1)(x2-x1)α≤f(x3)-f(x1)(x3-x1)α≤f(x3)-f(x2)(x3-x2)α,
for any x1,x2,x3∈I with x1<x2<x3.Theorem 9.
Lettingf∈Dα(I), then the following conditions are equivalent:(1)
f is a generalized convex function on I,(2)
f(α) is an increasing function on I,(3)
for anyx1,x2∈I,
(15)f(x2)≥f(x1)+f(α)(x1)Γ(1+α)(x2-x1)α.Proof.
(
1
→
2
) Let x1,x2∈I with x1<x2. And take h>0 which is small enough such that x1-h,x2+h∈I. Since x1-h<x1<x2<x2+h, then using Theorem 8 we have
(16)Γ(1+a)f(x1)-f(x1-h)hα≤Γ(1+a)f(x2)-f(x1)(x2-x1)α≤Γ(1+a)f(x2+h)-f(x2)hα.
Sincef∈Dα(I), then letting h→0+, it follows that
(17)f(α)(x1)≤Γ(1+a)f(x2)-f(x1)(x2-x1)α≤f(α)(x2).
So,f(α) is increasing in I.
(
2
→
3
) Take x1,x2∈I. Without loss of generality, we can assume that x1<x2. Since f(α) is increasing in the interval I, then applying the generalized local fractional Taylor theorem, we have
(18)f(x2)-f(x1)=f(α)(ξ)Γ(1+a)(x2-x1)α≥f(α)(x1)Γ(1+a)(x2-x1)α,
where ξ∈(x1,x2). That is to say,
(19)f(x2)≥f(x1)+f(α)(x1)Γ(1+a)(x2-x1)α.
(
3
→
1
) For any x1, x2∈I, we let x3=λx1+(1-λ)x2, where 0<λ<1. It is easy to see that x1-x3=(1-λ)(x1-x2) and x2-x3=λ(x2-x1). Then from the third condition, we have
(20)f(x1)≥f(x3)+f(α)(x3)Γ(1+a)(x1-x3)α=f(x3)+(1-λ)αf(α)(x3)Γ(1+a)(x1-x2)α,f(x2)≥f(x3)+f(α)(x3)Γ(1+a)(x2-x3)α=f(x3)+λαf(α)(x3)Γ(1+a)(x2-x1)α.
At the above two formulas, multiply λα and (1-λ)α, respectively; then we obtain
(21)λαf(x1)+(1-λ)αf(x2)≥f(x3)=f(λx1+(1-λ)x2).
Sof is a generalized convex function on I.Corollary 10.
Letf∈D2α(a,b). Then f is a generalized convex function (or a generalized concave function) if and only if
(22)f(2α)(x)≥0(orf(2α)(x)≤0),
for any x∈(a,b).


## 4. Some Inequalities

Theorem 11 (generalized Jensen’s inequality).
Assume thatf is a generalized convex function on [a,b]. Then for any xi∈[a,b] and λi∈[0,1](i=1,2,…,n) with ∑i=1nλi=1, we have
(23)f(∑i=1nλixi)≤∑i=1nλiαf(xi).Proof.
Whenn=2, the inequality is obviously true. Assume that for n=k the inequality is also true. Then, for any x1,x2,…,xk∈[a,b] and γi>0, i=1,2,…,k, with ∑i=1kγi=1, we have
(24)f(∑i=1kγixi)≤∑i=1kγiαf(xi).
Ifx1,x2,…,xk,x(k+1)∈[a,b] and λi>0 for i=1,2,…,k+1 with ∑i=1k+1λi=1, then one sets up γi=λi/(1-λk+1), i=1,2,…,k. It is easy to see ∑i=1kγi=1.
Thus,(25)f(λ1x1+λ2x2+⋯+λkxk+λk+1xk+1)=f((1-λk+1)λ1x1+λ2x2+⋯+λkxk1-λk+1+λk+1xk+1)≤(1-λk+1)αf(γ1x1+γ2x2+⋯+γkxk)+λk+1αf(xk+1)≤(1-λk+1)α[γ1αf(x1)+γ2αf(x2)+⋯+γkαf(xk)]+λk+1αf(xk+1)=(1-λk+1)α[(λ11-λk+1)αf(x1)+(λ21-λk+1)αf(x2)ddddddddddidii+⋯+(λk1-λk+1)αf(xk)]+λk+1αf(xk+1)=∑i=1kλiαf(xi).
So, the mathematical induction gives the proof of Theorem11.Corollary 12.
Letf∈D2α[a,b] and f(2α)(x)≥0 for any x∈[a,b]. Then for any xi∈[a,b] and λi∈[0,1](i=1,2,…,n) with ∑i=1nλi=1 we have
(26)f(∑i=1nλixi)≤∑i=1nλiαf(xi).
Using the generalized Jensen’s inequality and the convexity of functions, we can also get some integral inequalities.In [13], Yang established the generalized Cauchy-Schwarz’s inequality by the estimate aα/pbα/q≤(aα/p)(bα/q), where aα,bα>0, p,q≥1, and (1/p)+(1/q)=1.Now, via the generalized Jensen’s inequality, we will give another proof for the generalized Cauchy-Schwarz’s inequality.Corollary 13 (generalized Cauchy-Schwarz’s inequality).
Let|ak|>0, |bk|>0, k=1,2,…,n. Then we have
(27)∑k=1n|ak|α|bk|α≤(∑k=1n|ak|2α)1/2(∑k=1n|bk|2α)1/2.Proof.
Takef(x)=x2α. It is easy to see that f(2α)(x)≥0 for any x∈(a,b).
Take(28)λk=|bk|2∑k=1n|bk|2,xk=|ak||bk|.
Then 0≤λk≤1(k=1,2,…,n) with ∑k=1nλk=1.
Thus, by Jensen’s inequalityf(∑k=1nλkxk)≤∑k=1nλk(α)f(xk), we have
(29)[∑k=1n|bk|2∑k=1n|bk|2|ak||bk|]2α≤∑k=1n[|bk|2∑k=1n|bk|2]α[|ak||bk|]2α.
The above formula can be reduced to
(30)[∑k=1n|bk||ak|∑k=1n|bk|2]2α≤∑k=1n[|ak|2α(∑k=1n|bk|2)α],
which implies that
(31)[∑k=1n|bk||ak|]2α≤∑k=1n|ak|2α∑k=1n|bk|2α.
Thus we have(32)∑k=1n|ak|α|bk|α≤(∑k=1n|ak|2α)1/2(∑k=1n|bk|2α)1/2.Theorem 14 (generalized Hermite-Hadamard’s inequality).
Letf(x)∈Ix(α)[a,b] be a generalized convex function on [a,b] with a<b. Then
(33)f(a+b2)≤Γ(1+α)(b-a)αIb(α)af(x)≤f(a)+f(b)2α.Proof.
Letx=a+b-y. Then
(34)∫a(a+b)/2f(x)(dx)α=∫(a+b)/2bf(a+b-y)(dy)α.
Furthermore, whenx∈[(a+b)/2,b], a+b-x∈[a,(a+b)/2]. And by the convexity of f, we have
(35)f(a+b-x)+f(x)≥2αf(a+b2).
Thus(36)∫abf(x)(dx)α=∫a(a+b)/2f(x)(dx)α+∫(a+b)/2bf(x)(dx)α=∫(a+b)/2b[f(a+b-x)+f(x)](dx)α≥∫(a+b)/2b2αf(a+b2)(dx)α=(b-a)αf(a+b2).
For another part, we first note that iff is a generalized convex function, then, for t∈[0,1], it yields
(37)f(ta+(1-t)b)≤tαf(a)+(1-t)αf(b),f((1-t)a+tb)≤(1-t)αf(a)+tαf(b).
By adding these inequalities we have(38)f(ta+(1-t)b)+f((1-t)a+tb)≤tαf(a)+(1-t)αf(b)+(1-t)αf(a)+tαf(b)=f(a)+f(b).
Then, integrating the resulting inequality with respect to t over [0,1], we obtain
(39)1Γ(1+α)∫01[f(ta+(1-t)b)+f((1-t)a+tb)](dt)α≤1Γ(1+α)∫01(f(a)+f(b))(dt)α.
It is easy to see that(40)1Γ(1+α)∫01[f(ta+(1-t)b)+f((1-t)a+tb)](dt)α=2α(b-a)αIab(α)f(x),1Γ(1+α)∫01(f(a)+f(b))(dt)α=f(a)+f(b)Γ(1+α).
So,
(41)Γ(1+α)(b-a)αIab(α)f(x)≤f(a)+f(b)2α.
Combining the inequalities (36) and (41), we have
(42)f(a+b2)≤Γ(1+α)(b-a)αIab(α)f(x)≤f(a)+f(b)2α.
Note that it will be reduced to the class Hermite-Hadamard inequality ifα=1.


## 5. Applications of Generalized Jensen’s Inequality

Using the generalized Jensen’s inequality, we can get some inequalities.Example 15.
Leta>0, b>0 and a3α+b3α≤2α. Then a+b≤2.Proof.
Letf(x)=x3α, x∈(0,+∞). It is easy to see that f is a generalized convex function.
So,(43)f(a+b2)≤f(a)+f(b)2α.
That is,
(44)(a+b)3α8α≤a3α+b3α2α≤1α.
Thus, we conclude thata+b≤2.Example 16.
Letx,y∈R. Then
(45)Eα((x+y2)α)≤12α(Eα(xα)+Eα(yα)),
where Eα(xα)=∑k=0∞(xαk/Γ(1+kα)) is the Mittag-Leffler function.Proof.
Takef(x)=Eα(xα). It is easy to see (Eα(xα))(2α)=Eα(xα)>0. So, the generalized Jensen’s inequality gives
(46)Eα((x+y2)α)≤12α(Eα(xα)+Eα(yα)).Example 17 (power mean inequality).
Leta1,a2,…,an>0 and 0<s<t or s<t<0. Denote
(47)Sr=(a1αr+a2αr+⋯+anαrnα)1/r,r∈R.ThenSs≤St. And Ss=St if and only if a1=a2=⋯=an.Proof.
Consider the following.
Case I (0<s<t). Take f(x)=x(t/s)α, x>0. Then
(48)f(2α)(x)=Γ(1+tα/s)Γ(1+(t/s-1)α)x(t/s-2)α>0.
By the generalized Jensen’s inequality, we have
(49)f(a1s+a2s+⋯+ansn)≤f(a1s)+f(a2s)+⋯+f(ans)nα.
That is,
(50)(a1s+a2s+⋯+ansn)(t/s)α≤(a1s)(t/s)α+(a2s)(t/s)α+⋯+(ans)(t/s)αnα.
From the above formula, it is easy to see
(51)(a1αs+a2αs+⋯+anαsnα)1/s≤(a1αt+a2αt+⋯+anαtnα)1/t.
So, we haveSs≤St.
Case II
(
s
<
t
<
0
). Letting bi=1/ai and applying the case for 0<-t<-s, we can get the conclusion.Example 18.
Ifa,b,c>0 and a+b+c=1, then find the minimum of
(52)(a+1a)10α+(b+1b)10α+(c+1c)10α.Solution. Note that 0<a,b,c<1. Let f(x)=(x+1/x)10α, x∈(0,1). Then, via the formula
(53)dαxkαdxα=Γ(1+kα)Γ(1+(k-1)α)x(k-1)α,
we have
(54)f(2α)(x)=Γ(1+10α)Γ(1+8α)(x+1x)8α(1-1x2)2α+Γ(1+α)Γ(1+10α)Γ(1+9α)(x+1x)9α(2x3)α>0.By the generalized Jensen’s inequality,(55)(103)10α=f(a+b+c3)≤13α[f(a)+f(b)+f(c)]=13α[(a+1a)10α+(b+1b)10α+(c+1c)10α].So, the minimum is1010α/39α, when a=b=c=1/3.Example 19.
Ifa,b,c,d>0 and c2α+d2α=(a2α+b2α)3, then
(56)a3αcα+b3αdα≥1α.Proof.
Letx1=(a3/c)1/2, x2=(b3/d)1/2, y1=(ac)1/2, y2=(bd)1/2. By the generalized Cauchy-Schwartz inequality, we have
(57)(a3αcα+b3αdα)(aαcα+bαdα)=(x12α+x22α)(y12α+y22α)≥(x1αy1α+x2αy2α)2=(a2α+b2α)2=(a2α+b2α)1/2(c2α+d2α)1/2≥aαcα+bαdα.
Cancelingaαcα+bαdα on both sides, we get the desired result.


## 6. Conclusion

In the paper, we introduce the definition of generalized convex function on fractal sets. Based on the definition, we study the properties of the generalized convex functions and establish two important inequalities: the generalized Jensen’s inequality and generalized Hermite-Hadamard’s inequality. At last, we also give some applications for these inequalities on fractal sets.



---

*Source: 636751-2014-06-15.xml*