arXiv:1704.08190v1 [math.CA] 15 Apr 2017
Generalized Convex Functions and Their Applications
Adem Kili¸ cman a,1 and Wedad Saleh b a
Institute for Mathematical Research, University Putra Malaysia, Malaysia e-mail :
[email protected] b Department of Mathematics, Putra University of Malaysia, Malaysia e-mail : wed 10
[email protected] Abstract : This study focuses on convex functions and their generalized. Thus, we start this study by giving the definition of convex functions and some of their properties and discussing a simple geometric property. Then we generalize E-convex functions and establish some their properties. Moreover, we give generalized s-convex functions in the second sense and present some new inequalities of generalized Hermite-Hadamard type for the class of functions whose second local fractional derivatives of order α in absolute value at certain powers are generalized s-convex functions in the second sense. At the end, some examples that these inequalities are able to be applied to some special means are showed.
1
Introduction
Let M ⊆ R be an interval. A function ϕ : M ⊆ R −→ R is called a convex if for any y1 , y2 ∈ M and η ∈ [0, 1], ϕ(ηy1 + (1 − η)y2 ) ≤ ηϕ(y1 ) + (1 − η)ϕ(y2 ).
(1.1)
If the inequality (1.1) is the strict inequality, then ϕ is called a strict convex function. From a geometrical point of view, a function ϕ is convex provided that the line segment connecting any two points of its graph lies on or above the graph. The function ϕ is strictly convex provided that the line segment 1
Corresponding author email:
[email protected] ( Adem Kili¸ cman)
connecting any two points of its graph lies above the graph. If −ϕ is convex (resp. strictly convex), then ϕ is called concave (resp. strictly concave). The convexity of functions have been widely used in many branches of mathematics, for example in mathematical analysis, function theory, functional analysis, optimization theory and so on. For aproduction function x = ϕ(L), concacity of ϕ is expressed economically by saying that ϕ exhibits diminishing returns. While if ϕ is convex, then it exhibits increasing returns. Due to its applications and significant importance, the concept of convexity has been extended and generalized in several directions, see([2, 12, 23]. Recently, 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[10, 19]. Many researchers studied the properties of functions on fractal space and constructed many kinds of fractional calculus by using different approaches [2, 6, 31]. Particularly, in [27], Yang stated the analysis of local fractional functions on fractal space systematically, which includes local fractional calculus and the monotonicity of function. Throughout this chapter Rα will be denoted a real linear fractal set. Definition 1.1. [15] A function ϕ : M ⊂ R −→ Rα is called generalized convex if ϕ(ηy1 + (1 − η)y2 ) ≤ η α ϕ(y1 ) + (1 − η)α ϕ(y2 ) (1.2) for all y1 , y2 ∈ M , η ∈ [0, 1] and α ∈ ( 0, 1] . It is called strictly generalized convex if the inequality (1.2) holds strictly whenever y1 and y2 are distinct points and η ∈ (0, 1). If −ϕ is generalized convex (respectively, strictly generalized convex), then ϕ is generalized concave (respectively, strictly generalized concave). In α = 1, we have a convex function ,i.e, (1.1) is obtained. Let f ∈ Then,
f
(α) a1 Ia2
a1 + a2 2
be a generalized convex function on [a1 , a2 ] with a1 < a2 .
≤
Γ(1 + α) (a2 − a1 )α
(α) a1 Ia2 f (x)
2
≤
f (a1 ) + f (a2 ) . 2α
(1.3)
is known as generalized Hermite-Hadmard’s inequality [14]. Many authers paid attention to the study of generalized Hermite-Hadmard’s inequality and generalized convex function, see [4, 16]. If α = 1 in (1.3), then [8] Z a2 a1 + a2 f (a1 ) + f (a2 ) 1 f , (1.4) f (x)dx ≤ ≤ 2 a2 − a1 a1 2 which is known as classical Hermite-Hadamard inequality, for more properties about this inequality we refer the interested readers to [9, 13].
2
Generalized E-convex Functions
In 1999, Youness [30] introduced E-convexity of sets and functions, which have some important applications in various branches of mathematical sciences [1, 20]. However, Yang [26] showed that some results given by Youness [30] seem to be incorrect. Chen [7] extended E-convexity to a semi E-convexity and discussed some of its properties. For more results on E-convex function or semi E-convex function see [3, 11, 17, 18, 25]. Definition 2.1. [30] (i) A set B ⊆ Rn is called a E-convex iff there exists E : Rn −→ Rn such that ηE(r1 ) + (1 − η)E(r2 ) ∈ B, ∀r1 , r2 ∈ B, η ∈ [0, 1]. (ii) A function g : Rn −→ R is called E-convex (ECF) on a set B ⊆ Rn iff there exists E : Rn −→ Rn and g(ηE(r1 )+(1−η)E(r2 ) ≤ ηg(E(r1 ))+(1−η)g(E(r2 )), ∀r1 , r2 ∈ B, η ∈ [0, 1]. The following propositions were proved in [30]: Proposition 2.2. E(B) ⊆ B.
(i) Suppose that a set B ⊆ Rn is E-convex, then
(ii) Assume that E(B) is convex and E(B) ⊆ B, then B is E-convex. Definition 2.3. A function g : Rn −→ Rα is called a generalized E-convex function (gECF) on a set B ⊆ Rn iff there exists a map E : Rn −→ Rn such that B is an E-convex set and g(ηE(r1 ) + (1 − η)E(r2 )) ≤ η α g(E(r1 )) + (1 − η)α g(E(r2 )),
3
(2.1)
∀r1 , r2 ∈ B, η ∈ (0, 1) and α ∈ ( 0, 1] On the other hand, if g(ηE(x1 ) + (1 − η)E(x2 )) ≥ η α g(E(x1 )) + (1 − η)α g(E(x1 )), ∀x1 , x2 ∈ B, η ∈ (0, 1) and α ∈ ( 0, 1] , then g is called generalized E-concave on B. If the inequality sings in the previous two inequality are strict, then g is called generalized strictly E-convex and generalized strictly E-concave, respectively. Proposition 2.4. E = I.
(i) Every ECF on a convex set B is gECF , where
(ii) If α = 1 in equation (2.1), then g is called ECF on a set B. (iii) If α = 1 and E = I in equation (2.1), then g is called a convex function The following two examples show that generalized E-convex function which are not necessarily generalized convex. Example 2.5. Assume that B ⊆ R2 is given as B = (x1 , x2 ) ∈ R2 : µ1 (0, 0) + µ2 (0, 3) + µ3 (2, 1) , with µi > 0,
3 X
µi = 1 and define a map E : R2 −→ R2 such as E(x1 , x2 ) =
i=1
(0, x2 ). The function g : R2 −→ Rα defined by ( x3α 1 ; x2 < 1, g(x1 , x2 ) = α x1 x3α 2 ; x2 ≥ 1 The function g is gECF on B, but is not generalized convex. Remark 2.6. If α −→ 0 in the above example, then g goes to generalized convex function. Example 2.7. Assume that g : R −→ Rα is defined as ( 1α ; r > 0, g(r) = (−r)α ; r ≤ 0 and assume that E : R −→ R is defined as E(r) = −r 2 . Hence, R is an E-convex set and g is gECF, but is not generalized convex.
4
Theorem 2.1. Assume that B ⊆ Rn is an E-convex set and g1 : B −→ R is an ECF. If g2 : U −→ Rα is non-decreasing generalized convex function such that the rang g1 ⊂ U , then g2 og1 is a gECF on B. Proof. Since g1 is ECF, then g1 (ηE(r1 ) + (1 − η)E(r2 )) ≤ ηg1 (E(r1 )) + (1 − η)g1 (E(r2 )), ∀r1 , r2 ∈ B and η ∈ [0, 1]. Also, since g2 is non-decreasing generalized convex function, then g2 og1 (ηE(r1 ) + (1 − η)E(r2 )) ≤ g2 [ηg1 (E(r1 )) + (1 − η)g1 (E(r2 ))] ≤ η α g2 (g1 (E(r1 ))) + (1 − η)α g2 (g1 (E(r2 ))) = η α g2 og1 (E(r1 )) + (1 − η)α g2 og1 (E(r2 )) which implies that g2 og1 is a gECF on B. Similarly, g2 og1 is a strictly gECF if g2 is a strictly non-decreasing generalized convex function. Theorem 2.2. Assume that B ⊆ Rn is an E-convex set, and gi : B −→ Rα , i = 1, 2, ..., l are generalized E-convex function. Then, g=
l X
kiα gi
i=1
is a generalized E-convex on B for all kiα ∈ Rα Proof. Since gi , i = 1, 2, ..., l are gECF, then gi (ηE(r1 ) + (1 − η)E(r2 )) ≤ η α gi (E(r1 )) + (1 − η)α gi (E(r2 )), ∀r1 , r2 ∈ B , η ∈ [0, 1] and α ∈ ( 0, 1] . Then, l X
kiα gi (ηE(r1 ) + (1 − η)E(r2 ))
i=1
≤ ηα
l X
kiα gi (E(r1 )) + (1 − η)α
l X i=1
i=1
= η α g(E(r1 )) + (1 − η)α g(E(r2 )). Thus, g is a gECF.
5
kiα gi (E(r2 ))
Definition 2.8. Assume that B ⊆ Rn is a convex set. A function g : B −→ Rα is called generalized quasi convex if g(ηr1 + (1 − η)r2 ) ≤ max {g(r1 ), g(r2 )} , ∀r1 , r2 ∈ B and η ∈ [0, 1]. Definition 2.9. Assume that B ⊆ Rn is an E-convex set. A function g : B −→ Rα is called (i) Generalized E-quasiconvex function iff g(ηE(r1 ) + (1 − η)E(r2 ) ≤ max {g(E(r1 )), E(g(r2 ))} , ∀r1 , r2 ∈ B and η ∈ [0, 1]. (ii) Strictly generalized E-quasiconcave function iff g(ηE(r1 ) + (1 − η)E(r2 ) > min {g(E(r1 )), E(g(r2 ))} , ∀r1 , r2 ∈ B and η ∈ [0, 1]. Theorem 2.3. Assume that B ⊆ Rn is an E-convex set, and gi : B −→ Rα , i = 1, 2, ..., l are gECF. Then, (i) The function g : B −→ Rα which is defined by g(r) = supi∈I gi (r), r ∈ B is a gECF on B. (ii) If gi , i = 1, 2, ..., l are generalized E-quasiconvex functions on B, then the function g is a generalized E-quasiconvex function on B. Proof.
(i) Due to gi , i ∈ I be gECF on B, then g(ηE(r1 ) + (1 − η)E(r2 )) = sup gi (ηE(r1 ) + (1 − η)E(r2 )) i∈I
≤ η α sup gi (E(r1 )) + (1 − η)α sup gi (E(r2 )) i∈I
i∈I
= η α g(E(r1 )) + (1 − η)α g(E(r2 )). Hence, g is a gECF on B.
6
(ii) Since gi , i ∈ I are generalized E-quasiconvex functions on B, then g(ηE(x1 ) + (1 − η)E(x2 )) = sup gi (ηE(x1 ) + (1 − η)E(x2 )) i∈I
≤ sup max {gi (E(x1 )), gi (E(x2 ))} i∈I = max sup gi (E(x1 )), sup gi (E(x2 )) i∈I
i∈I
= max {g(E(x1 )), g(E(x2 ))} . Hence, g is a generalized E-quasiconvex function on B.
Considering B ⊆ Rn is a nonempty E-convex set. From Propostion2.2(i), we get E(B) ⊆ B.Hence, for any g : B −→ Rα , the restriction g˜ : E(B) −→ Rα of g to E(B) defined by g˜(˜ x) = g(˜ x), ∀˜ x ∈ E(B) is well defined. Theorem 2.4. Assume that B ⊆ Rn , and g : B −→ Rα is a generalized E-quasiconvex function on B. Then, the restriction g˜ : U −→ Rα of g to any nonempty convex subset U of E(B) is a generalized quasiconvex on U . Proof. Assume that x1 , x2 ∈ U ⊆ E(B), then there exist x∗1 , x∗2 ∈ B such that x1 = E(x∗1 ) and x2 = E(x∗2 ). Since U is a convex set, we have ηx1 + (1 − η)x2 = ηE(x∗1 ) + (1 − η)E(x∗2 ) ∈ U, ∀η ∈ [0, 1]. Therefore, we have g˜(ηx1 + (1 − η)x2 ) = g˜(ηE(x∗1 ) + (1 − η)E(x∗2 )) ≤ max {g(E(x∗1 )), g(E(x∗2 ))} = max {g(x1 ), g(x2 )} = max {˜ g(x1 ), g˜(x2 )} .
Theorem 2.5. Assume that B ⊆ Rn is an E-convex set, and E(B) is a convex set. Then, g : B −→ Rα is a generalized E-quasiconvex on B iff its restriction g˜ = g|E(B) is a generalized quasiconvex function on E(B).
7
Proof. Due to Theorem2.4, the if condition is true. Conversely, suppose that x1 , x2 ∈ B, then E(x1 ), E(x2 ) ∈ E(B) and ηE(x1 ) + (1 − η)E(x2 ) ∈ E(B) ⊆ B, ∀η ∈ [0, 1]. Since E(B) ⊆ B, then g(ηE(x1 ) + (1 − η)E(x2 )) = g˜(ηE(x1 ) + (1 − η)E(x2 )) ≤ max {˜ g(E(x1 )), g˜(E(x2 ))} = max {g(E(x1 )), g(E(x2 ))} .
An analogous result to Theorem2.4 for the generalized E-convex case is as follows: Theorem 2.6. Assume that B ⊆ Rn is an E-convex set, and g : B −→ Rα is a gECF on B. Then, the restriction g˜ : U −→ Rα of g to any nonempty convex subset U of E(B) is a gCF. An analogous result to Theorem2.5 for the generalized E-convex case is as follows: Theorem 2.7. Assume that B ⊆ Rn is an E-convex set, and E(B) is a convex set. Then, g : B −→ Rα is a gECF on B iff its restriction g˜ = g|E(B) is a gCF on E(B). The lower level set of goE : B −→ Rα is defined as Lrα (goE) = {x ∈ B : (goE)(x) = g(E(x)) ≤ r α , r α ∈ Rα } . The lower level set of g˜ : E(B) −→ Rα is defined as Lrα (˜ g ) = {˜ x ∈ E(B) : g˜(˜ x) = g(˜ x ) ≤ r α , r α ∈ Rα } . Theorem 2.8. Supose that E(B) be a convex set. A function g : B −→ Rα is a generalized E-quasiconvex iff Lrα (˜ g ) of its restriction g˜ : E(B) −→ Rα is a convex set for each r α ∈ Rα . Proof. Due to E(B) be a convex set, then for each E(x1 ), E(x2 ) ∈ E(B), we have ηE(x1 ) + (1 − η)E(x2 ) ∈ E(B) ⊆ B. Let x ˜1 = E(x1 ) and x ˜2 = E(x2 ).
8
If x ˜1 , x ˜2 ∈ Lrα (˜ g ), then g(˜ x1 ) ≤ r α and g(˜ x2 ) ≤ r α . Thus, g˜(η˜ x1 + (1 − η)˜ x2 ) = g(η˜ x1 + (1 − η)˜ x2 ) = g(ηE(x1 ) + (1 − η)E(x2 )) ≤ max {g(E(x1 )), g(E(x2 ))} = max {g(˜ x1 ), g(˜ x2 )} = max {˜ g(˜ x1 ), g˜(˜ x2 )} ≤ rα. which show that η˜ x1 + (1 − η)˜ x2 ∈ Lrα (˜ g ). Hence, Lrα (˜ g ) is a convex set. Conversely, let Lrα (˜ g ) be a convex set for each r α ∈ Rα , i.e., η˜ x1 + (1 − η)˜ x2 ∈ Lrα (˜ g ), ∀˜ x1 , x ˜2 ∈ Lrα (˜ g ) and r α = max {g(˜ x1 ), g(˜ x2 )}. Thus, g(ηE(x1 ) + (1 − η)E(x2 )) = g˜(ηE(x1 ) + (1 − η)E(x2 )) = g˜(η˜ x1 + (1 − η)˜ x2 ) ≤ rα = max {g(˜ x1 ), g(˜ x2 )} = max {g(E(x1 )), g(E(x2 ))} . Hence, g is a generalized E-quasiconvex. Theorem 2.9. Let B ⊆ Rn be a nonempty E-convex set and let g1 : B −→ Rα be a generalized E-quasiconvex on B. Suppose that g2 : Rα −→ Rα is a non-decreasing function. Then, g2 og1 is a generalized E-quasiconvex. Proof. Since g1 : B −→ Rα is generalized E-quasiconvex on B and g2 : Rα −→ Rα is a non-decreasing function, then (g2 og1 )(ηE(x1 ) + (1 − η)E(x2 )) = g2 (g1 (ηE(x1 ) + (1 − η)E(x2 ))) ≤ g2 (max {g1 (E(x1 )), g1 (E(x2 ))}) = max {(g2 og1 )(E(x1 )), (g2 og1 )(E(x2 ))} which shows that g2 og1 is a generalized E-quasiconvex on B.
9
Theorem 2.10. If the function g is a gECF on B ⊆ Rn , then g is a generalized E-quasiconvex on B. Proof. Assume that g is a gECF on B. Then, g(ηE(r1 ) + (1 − η)E(r2 )) ≤ η α g(E(r1 )) + (1 − η)α g(E(r2 )) ≤ η α max {g(E(r1 )), g(E(r2 ))} +(1 − η)α max {g(E(r1 )), g(E(r2 ))} = max {g(E(r1 )), g(E(r2 ))} .
E α -epigraph
3
Definition 3.1. Assume that B ⊆ Rn × Rα and E : Rn −→ Rn , then the set B is called E α -convex set iff (ηE(x1 ) + (1 − η)E(x2 ), η α r1α + (1 − η)α r2α ) ∈ B ∀(x1 , r1α ), (x2 , r2α ) ∈ B, η ∈ [0, 1] and α ∈ ( 0, 1] . Now, the E α - epigraph of g is given by epiE α (g) = {(E(x), r α ) : x ∈ B, r α ∈ Rα , g(E(x)) ≤ r α } . A sufficient condition for g to be a gECF is given by the following theorem: Theorem 3.1. Let E : Rn −→ Rn be an idempoted map. Assume that B ⊆ Rn is an E-convex set and epiE α (g) is an E α -convex set where g : B −→ Rα , then g is a gECF on B. Proof. Assume that r1 , r2 ∈ B and (E(r1 ), g(E(r1 ))), (E(r2 ), g(E(r2 ))) ∈ epiE α (g). Since epiE α (g) is E α - convex set, we have (ηE(E(r1 )) + (1 − η)E(E(r2 )), η α g(E(r1 )) + (1 − η)α g(E(r2 ))) ∈ epiE α (g), then g(E(ηE(r1 )) + (1 − η)E(E(r2 ))) ≤ η α g(E(r1 )) + (1 − η)α g(E(r2 )). Due to E : Rn −→ Rn be an idempotent map, then g(ηE(r1 ) + (1 − η)E(r2 )) ≤ η α g(E(r1 )) + (1 − η)α g(E(r2 )). Hence, g is a gECF.
10
Theorem 3.2. Assume that {Bi }i∈I is a family of E α -convex sets. Then, their intersection ∩i∈I Bi is an E α -convex set. Proof. Considering (x1 , r1α ), (x2 , r2α ) ∈ ∩i∈I Bi , then (x1 , r1α ), (x2 , r2α ) ∈ Bi , ∀i ∈ I. By E α -convexity of Bi , ∀i ∈ I, then we have (ηE(x1 ) + (1 − η)E(x2 ), η α r1α + (1 − η)α r2α ) ∈ Bi , ∀η ∈ [0, 1] and α ∈ ( 0, 1] . Hence, (ηE(x1 ) + (1 − η)E(x2 ), η α r1α + (1 − η)α r2α ) ∈ ∩i∈I Bi .
Rn
The following theorem is a special case of Theorem 2.3(i) where E : Rn −→ is an idempotent map.
Theorem 3.3. Assume that E : Rn −→ Rn is an idempotent map, and B ⊆ Rn is an E-convex set. Let {gi }i∈I be a family function which have bounded from above. If epiE α (gi ) are E α -convex sets, then the function g which defined by g(x) = supi∈I gi (x), x ∈ B is a gECF on B. Proof. Since epiE α (gi ) = {(E(x), r α ) : x ∈ B, r α ∈ Rα , gi (E(x)) ≤ r α , i ∈ I} are E α - convex set in B × Rα , then ∩i∈I epiE α (gi ) = {(E(x), r α ) : x ∈ B, r α ∈ Rα , gi (E(x)) ≤ r α , i ∈ I} = {(E(x), r α ) : x ∈ B, r α ∈ Rα , g(E(x)) ≤ r α } ,
(3.1)
where g(E(x)) = supi∈I gi (E(x)), also is E α -convex set. Hence, ∩i∈I epiE α (gi ) is an E α -epigraph, then by Theorem3.2, g is a generalized E-convex function on B.
4
Generalized s-convex functions
There are many researchers studied the properties of functions on fractal space and constructed many kinds of fractional calculus by using different approaches see [5, 24, 28] In [14], two kinds of generlized s-convex functions on fractal sets are introduced as follows:
11
Definition 4.1. (i) A function ϕ : R+ −→ Rα , is called a generalized s-convex (0 < s < 1)in the first sense if ϕ(η1 y1 + η2 y2 ) ≤ η1sα ϕ(y1 ) + η2sα ϕ(y2 )
(4.1)
for all y1 , y2 ∈ R+ and all η1 , η2 ≥ 0 with η1s + η2s = 1, this class of functions is denoted by GKs1 . (ii) A function ϕ : R+ −→ Rα , is called a generalized s-convex (0 < s < 1) in the second sense if (4.1) holds for all y1 , y2 ∈ R+ and all η1 , η2 ≥ 0 with η1 + η2 = 1, this class of functions is denoted by GKs2 . In the same paper,[14], Mo and Sui proved that all functions which are generalized s-convex in the second sense, for s ∈ (0, 1), are non-negative. If α = 1 in Definition 4.1 , then we have the classical s-convex functions in the first sense (second sense) see [8]. Also, in [8], Dragomir and Fitzatrick demonstrated a variation of Hadamard’s inequality which holds for s-convex functions in the second sense. Theorem 4.2. Assume that ϕ : R+ −→ R+ is a s-convex function in the second sense, 0 < s < 1 and y1 , y2 ∈ R+ , y1 < y2 . If ϕ ∈ L1 ([y1 , y2 ]), then Z y2 y1 + y2 ϕ(y1 ) + ϕ(y2 ) 1 s−1 2 ϕ . (4.2) ϕ(z)dz ≤ ≤ 2 y 2 − y 1 y1 s+1 If we set k = in (4.2).
1 , then it is the best possible in the second inequality s+1
A varation of generalized Hadamard’s inequality which holds for generalized s-convex functions in the second sense [2]. Theorem 4.3. Assume that ϕ : R+ −→ Rα+ is a generalized s-convex function in the second sense where 0 < s < 1 and y1 , y2 ∈ R+ with y1 < y2 . If ϕ ∈ L1 ([y1 , y2 ]), then Γ(1 + α) y1 + y2 (α) α(s−1) ϕ(x) ≤ 2 ϕ y I 2 (y2 − y1 )α 1 y2 Γ(1 + sα)Γ(1 + α) (ϕ(y1 ) + ϕ(y2 )) . (4.3) ≤ Γ(1 + (s + 1)α)
12
Proof. We know that ϕ is generalized s-convex in the second sense,which lead to ϕ(ηy1 + (1 − η)y2 ) ≤ η αs ϕ(y1 ) + (1 − η)αs ϕ(y2 ), ∀η ∈ [0, 1]. Then the following inequality can be written: (α)
(α)
Γ(1 + α) 0 I1 ϕ(ηy1 + (1 − η)y2 ) ≤ ϕ(y1 )Γ(1 + α) 0 I1 η αs (α)
+ϕ(y2 )Γ(1 + α) 0 I1 (1 − η)αs Γ(1 + sα)Γ(1 + α) (ϕ(y2 ) + y2 ) Γ(1 + (s + 1)α)
=
By considering z = ηy1 + (1 − η)y2 . Then (α)
Γ(1 + α) 0 I1 ϕ(ηy1 + (1 − η)y2 ) = =
Γ(1 + α) (y1 − y2 )α Γ(1 + α) (y2 − y1 )α
(α) y2 Iy1 ϕ(x) (α) y1 Iy2 ϕ(z).
Here Γ(1 + α) (y2 − y1 )α
(α) y1 Iy2 ϕ(x)
Γ(1 + sα)Γ(1 + α) (ϕ(y1 ) + ϕ(y2 )). Γ(1 + (s + 1)α)
≤
Then, the second inequality in (4.3) is given. Now ϕ
≤
y1 + y2 2
z1 + z2 2
ϕ
ϕ(z1 ) + ϕ(z2 ) , ∀z1 , z2 ∈ I. 2αs
(4.4)
Let z1 = ηy1 + (1 − η)y2 and z2 = (1 − η)y1 + ηy2 with η ∈ [0, 1]. Hence, by applying (4.4), the next inequalty holds y1 + y2 ϕ(ηy1 + (1 − η)y2 ) + ϕ((1 − η)y1 + ηy2 ) ϕ , ∀η ∈ [0, 1]. ≤ 2 2αs So 1 Γ(1 + α)
Z
0
1
(dη)α ≤
1 2α(s−1) (y
2
− y1 )α
(α) y1 Iy2 ϕ(z)
Then, α(s−1)
2
ϕ
y1 + y2 2
≤
Γ(1 + α) (y2 − y1 )α
13
(α) y1 Iy2 ϕ(z).
Lemma 4.4. Assume that ϕ : [y1 , y2 ] ⊂ R −→ Rα is a local fractional derivative of order α (ϕ ∈ Dα ) on (y1 , y2 ) with y1 < y2 . If ϕ(2α) ∈ Cα [y1 , y2 ], then the following equality holds: y1 + y2 Γ(1 + 2α) Γ(1 + 2α)[Γ(1 + α)]2 (α) ϕ y1 Iy2 ϕ(x) − 2α (y2 − y1 ) 2α 2 2α (y2 − y1 ) y1 + y2 (α) 2α (2α) = + (1 − γ)y1 γ 0 I1 γ ϕ 16α 2 y1 + y2 (α) 2α (2α) + 0 I1 (γ − 1) ϕ γy2 + (1 − γ) 2 Proof. From the local fractional integration by parts, we get B1
Z 1 1 y1 + y2 2α (2α) = + (1 − γ)y1 (dγ)α γ ϕ γ Γ(1 + α) 0 2 α 2 (α) y1 + y2 ϕ = y2 − y1 2 2α 1 y1 + y2 2 α γ ϕ γ + (1 − γ)y1 −Γ(1 + 2α) y2 − y1 2 0 2α Z 1 2 y1 + y2 +Γ(1 + 2α)Γ(1 + α) + (1 − γ)y1 (dγ)α ϕ γ b−a 2 0 α 2α y1 + y2 2 y + y 2 1 2 (α) ϕ ϕ − Γ(1 + 2α) = y2 − y1 2 y2 − y1 2 2α Z 1 2 y1 + y2 +Γ(1 + 2α)Γ(1 + α) ϕ γ + (1 − γ)y1 (dγ)α y2 − y1 2 0
y1 + y2 + (1 − γ)y1 , for γ ∈ [0, 1] and multiply the both sides 2 (y2 − y1 )2α in the last equation by , we get 16α
Setting x = γ
B1 = =
(y2 − y1 )2α (α) 2α (2α) y1 + y2 + (1 − γ)y1 I1 γ ϕ γ 16α 2 0 y1 + y2 Γ(1 + 2α) (y2 − y1 )α (α) y1 + y2 ϕ ϕ − 8α 2 4α 2 Z y1 +y2 2 Γ(1 + 2α)Γ(1 + α) + ϕ(x)(dx)α . α α 2 (y2 − y1 ) y1
14
By the similar way, also we have
B2
(y2 − y1 )2α (α) y1 + y2 2α (2α) = I1 (γ − 1) ϕ γy2 + (1 − γ) 16α 2 0 α (y2 − y1 ) (α) y1 + y2 y1 + y2 Γ(1 + 2α) = − ϕ ϕ − 8α 2 4α 2 Z Γ(1 + 2α)Γ(1 + α) y2 + ϕ(x)(dx)α . y1 +y2 2α (y2 − y1 )α 2
Thus, adding B1 and B2 , we get the desired result.
Theorem 4.5. Assume that ϕ : U ⊂ [ 0, ∞) −→ Rα such that ϕ ∈ Dα on Int(U ) (Int(U ) is the interior of U ) and ϕ(2α) ∈ Cα [y1 , y2 ], where y1 , y2 ∈ U with y1 < y1 . If |ϕ| is generalized s-convex on [y1 , y2 ], for some fixed 0 < s ≤ 1, then the following inequality holds:
Γ(1 + 2α) y1 + y2 Γ(1 + 2α)[Γ(1 + α)]2 (α) ϕ I ϕ(x) − y y 1 2 2α 2 2α (y2 − y1 )α Γ(1 + sα) (y2 − y1 )2α 2α Γ(1 + (s + 2)α) (2α) y1 + y2 ϕ ≤ + Γ(1 + (s + 1)α) 16α Γ(1 + (s + 3)α) 2 i Γ(1 + (s + 2)α) h (2α) (2α) α Γ(1 + (s + 1)α) −2 + (y2 ) (y1 ) + ϕ (4.5) ϕ Γ(1 + (s + 2)α) Γ(1 + (s + 3)α) ( Γ(1 + sα) (y2 − y1 )2α 2α(2−s) Γ(1 + (s + 2)α) Γ(1 + sα)Γ(1 + α) + ≤ α 16 Γ(1 + (s + 3)α) Γ(1 + (s + 1)α) Γ(1 + (s + 1)α) i 2α Γ(1 + (s + 1)α) Γ(1 + (s + 2)α) h (2α) − + (4.6) (y1 ) + ϕ(2α) (y2 ) . ϕ Γ(1 + (s + 2)α) Γ(1 + (s + 3)α)
15
Proof. From Lemma 4.4, we have
Γ(1 + 2α) y1 + y2 Γ(1 + 2α)[Γ(1 + α)]2 (α) ϕ I ϕ(x) − y y 1 2 2α 2 2α (y2 − y1 )α (y2 − y1 )2α y1 + y2 (α) 2α (2α) ≤ + (1 − γ)y1 ) ϕ (γ 0 I1 γ α 16 2 y1 + y2 (α) 2α (2α) + 0 I1 (γ − 1) ϕ γy2 + (1 − γ) 2 (y2 − y1 )2α (α) 2α αs (2α) αs (2α) y1 + y2 (y1 ) γ ϕ ≤ 0 I1 γ + (1 − γ) ϕ 16α 2 (y2 − y1 )2α (α) 2α αs (2α) αs (2α) y1 + y2 + γ ϕ (y2 ) + (1 − γ) ϕ 0 I1 (γ − 1) α 16 2 (y2 − y1 )2α Γ(1 + (s + 2)α) (2α) y1 + y2 = ϕ 16α Γ(1 + (s + 3)α) 2 Γ(1 + (s + 2)α) (2α) Γ(1 + αs) α Γ(1 + (s + 1)α) −2 + (y1 ) + ϕ Γ(1 + (s + 1)α) Γ(1 + (s + 2)α) Γ(1 + (s + 3)α) (y2 − y1 )2α Γ(1 + (s + 2)α) (2α) y1 + y2 + ϕ 16α Γ(1 + (s + 3)α) 2 Γ(1 + (s + 2)α) (2α) Γ(1 + αs) α Γ(1 + (s + 1)α) −2 + (a2 ) + ϕ Γ(1 + (s + 1)α) Γ(1 + (s + 2)α) Γ(1 + (s + 3)α) Γ(1 + (s + 2)α) (2α) y1 + y2 + ϕ Γ(1 + (s + 3)α) 2 (y2 − y1 )2α 2α Γ(1 + (s + 2)α) (2α) y1 + y2 Γ(1 + sα) = ϕ + α 16 Γ(1 + (s + 3)α) 2 Γ(1 + (s + 1)α) h i Γ(1 + (s + 1)α) (2α) (2α) α Γ(1 + (s + 1)α) −2 (y1 ) + ϕ (y2 ) . + ϕ Γ(1 + (s + 2)α) Γ(1 + (s + 3)α) This proves inequality (4.5). Since
α(s−1) (2α)
2
ϕ
y1 + y2 2
≤
Γ(1 + sα)Γ(1 + α) (2α) ϕ (y1 ) + ϕ(2α) (y2 ) , Γ(1 + (s + 1)α)
16
then 2 Γ(1 + 2α) y + y Γ(1 + 2α)[Γ(1 + α)] 1 2 (α) ϕ − y1 Iy2 ϕ(x) α α α 2 2 2 (y2 − y1 ) ( i (y2 − y1 )2α 2α Γ(1 + (s + 2)α) 2−α(s−1) Γ(1 + sα)Γ(1 + α) h (2α) (2α) ≤ (y2 ) (y1 ) + ϕ ϕ 16α Γ(1 + (s + 3)α) Γ(1 + (s + 1)α) i Γ(1 + sα) 2α Γ(1 + (s + 1)α) Γ(1 + (s + 2)α) h (2α) (2α) (y1 ) + ϕ − + (y2 ) + ϕ Γ(1 + (s + 1)α) Γ(1 + (s + 2)α) Γ(1 + (s + 3)α) ( α(2−s) 2α Γ(1 + sα) 2 Γ(1 + (s + 2)α) Γ(1 + sα)Γ(1 + α) (y2 − y1 ) + = α 16 Γ(1 + (s + 3)α) Γ(1 + (s + 1)α) Γ(1 + (s + 1)α) i 2α Γ(1 + (s + 1)α) Γ(1 + (s + 2)α) h (2α) (2α) − (y2 ) (y1 ) + ϕ + ϕ Γ(1 + (s + 2)α) Γ(1 + (s + 3)α) Thus, we get the inequality (4.6)and the proof is complete.
1. When α = 1, Theorem 4.5 reduce to Theorem 2 in[21].
Remark 4.6.
2. If s = 1 in Theorem 4.5 , then Γ(1 + 2α) y1 + y2 Γ(1 + 2α)[Γ(1 + α)]2 (α) ϕ I ϕ(x) − y 1 y 2 2α 2 2α (y2 − y1 )α α 2α Γ(1 + α) 2 Γ(1 + 3α) (2α) y1 + y2 (y2 − y1 ) ≤ ϕ + α 16 Γ(1 + 4α) 2 Γ(1 + 2α) h i Γ(1 + 3α) (2α) (2α) α Γ(1 + 2α) (y2 ) + (y1 ) + ϕ −2 ϕ Γ(1 + 3α) Γ(1 + 4α) ( (y2 − y1 )2α 2α Γ(1 + 3α) [Γ(1 + α)]2 Γ(1 + α) ≤ + α 16 Γ(1 + 4α) Γ(1 + 2α) Γ(1 + 2α) i 2α Γ(1 + 2α) Γ(1 + 3α) h (2α) (2α) . (y1 ) + ϕ + (y2 ) (4.7) − ϕ Γ(1 + 3α) Γ(1 + 4α) 3. If s = 1 and α = 1 in Theorem 4.5, then Z y2 1 ϕ y1 + y2 − ϕ(x)dx 2 y 2 − y 1 y1 ′′ ′′ y1 + y2 ′′ (y2 − y1 )2 ≤ 6 ϕ + ϕ (y1 ) + ϕ (y2 ) 192 2 ≤
(y2 − y1 )2 ′′ ϕ (y1 ) + ϕ′′ (y2 ) 48
17
We give a new upper bound of the left generalized Hadamard’s inequality for generalized s-convex functions in the following theorem: Theorem 4.7. Assume that ϕ : U ⊂ [ 0, ∞) −→ Rα such that ϕ ∈ Dα on Int(U ) and ϕ(2α) ∈ Cα [y1 , y2 ], where y1 , y2 ∈ U with y1 < y2 . If |ϕ(2α) |p2 is generalized s-convex on [y1 , y2 ], for some fixed 0 < s ≤ 1 and p2 > 1 with 1 1 p1 + p2 = 1, then the following inequality holds: 2 Γ(1 + 2α) y + y Γ(1 + 2α)[Γ(1 + α)] 1 2 (α) ϕ I ϕ(x) − y y 1 2 2α 2 2α (y2 − y1 )α 1 1 p2 p1 Γ(1 + 2p1 α) Γ(1 + sα) (y2 − y1 )2α ≤ α 16 Γ(1 + (s + 1)α) Γ(1 + (2p1 + 1)α) " p2 p2 p12 (2α) y1 + y2 (2α) × ϕ (y1 ) + ϕ 2 # p2 p12 (2α) y1 + y2 p2 (2α) + ϕ + ϕ (y2 ) . (4.8) 2
Proof. Let p1 > 1, then from Lemma 4.4 and using generalized H¨older’s inequality [27], we obtain 2 Γ(1 + 2α) y + y Γ(1 + 2α)[Γ(1 + α)] 1 2 (α) ϕ − y1 Iy2 ϕ(x) α α α 2 2 2 (y2 − y1 ) (y2 − y1 )2α y1 + y2 (α) 2α (2α) ≤ ϕ γ + (1 − γ)y1 0 I1 γ α 16 2 y1 + y2 (α) +0 I1 (γ − 1)2α ϕ(2α) γy2 + (1 − γ) 2 (y2 − y1 )2α (α) 2p1 α p11 ≤ 0 I1 γ 16α p2 1 p2 y1 + y2 (α) (2α) γ × 0 I1 ϕ + (1 − γ)y1 2 1 2α (y2 − y1 ) (α) 2p1 α p1 + I (1 − γ) 0 1 16α 1 y1 + y2 p2 p2 (α) (2α) . × 0 I1 ϕ γy2 + (1 − γ) 2
18
p2 Since ϕ(2α) is generalized s-convex, then p2 y1 + y2 (α) (2α) ϕ I γ + (1 − γ)y 0 1 1 2 p2 (2α) y1 + y2 p2 Γ(1 + sα) Γ(1 + sα) (2α) + ϕ (y ) ≤ ϕ 1 , Γ(1 + (s + 1)α) 2 Γ(1 + (s + 1)α) which means
(α) 0 I1
(2α) y1 + y2 p2 ϕ γy + (1 − γ) 2 2 p2 Γ(1 + sα) (2α) (y2 ) ≤ ϕ Γ(1 + (s + 1)α) (2α) y1 + y2 p2 Γ(1 + sα) ϕ . + Γ(1 + (s + 1)α) 2
Hence Γ(1 + 2α) y1 + y2 Γ(1 + 2α)[Γ(1 + α)]2 (α) ϕ I ϕ(x) − y y 1 2 2α 2 2α (y2 − y1 )α 1 1 p2 p1 Γ(1 + sα) (y2 − y1 )2α Γ(1 + 2p1 α) ≤ α 16 Γ(1 + (s + 1)α) Γ(1 + (2p1 + 1)α) ( p2 p2 p12 (2α) y1 + y2 (2α) + ϕ (y1 ) × ϕ 2 ) p2 p12 (2α) y1 + y2 p2 (2α) + ϕ (y2 ) . + ϕ 2
The proof is complete.
Remark 4.8. If s = 1 in Theorem 4.7, then 2 Γ(1 + 2α) y + y Γ(1 + 2α)[Γ(1 + α)] 1 2 (α) ϕ − y1 Iy2 ϕ(x) α α α 2 2 2 (y2 − y1 ) 1 1 p1 (y2 − y1 )2α Γ(1 + α) p2 Γ(1 + 2p1 α) ≤ 16α Γ(1 + 2α) Γ(1 + (2p1 + 1)α) ( p2 p2 p12 (2α) y1 + y2 (2α) + ϕ (y1 ) × ϕ 2 ) p2 p12 (2α) y1 + y2 p2 (2α) + ϕ . (4.9) + ϕ (y2 ) 2
19
Corollary 4.9. Assume that ϕ : U ⊂ [ 0, ∞) −→ Rα such that ϕ ∈ Dα on Int(U ) and ϕ(2α) ∈ Cα [y1 , y2 ], where y1 , y2 ∈ U with y1 < y1 . If |ϕ(2α) |p2 is generalized s-convex on [y1 , y2 ], for some fixed 0 < s ≤ 1 and p2 > 1 with 1 1 p1 + p2 = 1, then the following inequality holds: Γ(1 + 2α) y1 + y2 Γ(1 + 2α)[Γ(1 + α)]2 (α) ϕ − y1 Iy2 ϕ(x) α α α 2 2 2 (y2 − y1 ) 1 1 p1 Γ(1 + 2p1 α) (y2 − y1 )2α [Γ(1 + sα)] p2 ≤ 2 16α [Γ(1 + (s + 1)α)] p2 Γ(1 + (2p1 + 1)α) 1 p 2α(1−s) Γ(1 + sα)Γ(1 + α) + Γ(1 + (s + 1)α) 2 × h i α(1−s) 1 1 (2α) (2α) p p p (y2 ) (y1 ) + ϕ +2 2 [Γ(1 + α)] 2 [Γ(1 + α)] 2 ϕ
p2 Proof. Since ϕ(2α) is generalized s-convex, then α(s−1) (2α)
2
ϕ
y1 + y2 2
≤
Γ(1 + sα)Γ(1 + α) (2α) ϕ (y1 ) + ϕ(2α) (y2 ) . Γ(1 + (s + 1)α)
Hence using (4.8), we get Γ(1 + 2α) y1 + y2 Γ(1 + 2α)[Γ(1 + α)]2 (α) ϕ I ϕ(x) − y y 1 2 2α 2 2α (y2 − y1 )α 1 1 p1 [Γ(1 + sα)] p2 (y2 − y1 )2α Γ(1 + 2p1 α) ≤ 2 α 16 [Γ(1 + (s + 1)α)] p2 Γ(1 + (2p1 + 1)α) nh × 2α(1−s) Γ(1 + sα)Γ(1 + α) + Γ(1 + (s + 1)α) |ϕ(2α) (y1 )|p2 i1 p +2α(1−s) Γ(1 + sα)Γ(1 + α)|ϕ(2α) (y2 )|p2 2 h + 2α(1−s) Γ(1 + sα)Γ(1 + α)|ϕ(2α) (y1 )|p2 i1 α(1−s) (2α) p2 q + 2 Γ(1 + sα)Γ(1 + α) + Γ(1 + (s + 1)α) |ϕ (y2 )| k k k X X X αn αn ziαn xi + (xi + zi ) ≤ and since i=1
i=1
i=1
20
for 0 < n < 1, xi , zi ≥ 0; ∀1 ≤ i ≤ k, then we have Γ(1 + 2α) y1 + y2 Γ(1 + 2α)[Γ(1 + α)]2 (α) ϕ I ϕ(x) − y 1 y 2 2α 2 2α (y2 − y1 )α 1 1 p1 Γ(1 + 2p1 α) (y2 − y1 )2α [Γ(1 + sα)] p2 ≤ 2 16α [Γ(1 + (s + 1)α)] p2 Γ(1 + (2p1 + 1)α) 1 p 2α(1−s) Γ(1 + sα)Γ(1 + α) + Γ(1 + (s + 1)α) 2 ϕ(2α) (y1 ) × α(1−s) 1 1 (2α) p2 p2 p2 (y2 ) [Γ(1 + sα)] [Γ(1 + α)] ϕ +2 α(1−s) 1 1 + 2 p2 [Γ(1 + sα)] p2 [Γ(1 + α)] p2 ϕ(2α) (y1 ) 1 p2 (2α) α(1−s) (y2 ) + 2 Γ(1 + sα)Γ(1 + α) + Γ(1 + (s + 1)α) ϕ
where 0 < p12 < 1 for p2 > 1. By a simple calculation, we obtain the required result.
Now, the generalized Hadamard’s type inequality for generalized sconcave functions. Theorem 4.10. Assume that ϕ : U ⊂ [ 0, ∞) −→ Rα such that ϕ ∈ Dα on Int(U ) and ϕ(2α) ∈ Cα [y1 , y2 ], where y1 , y2 ∈ U with y1 < y1 . If |ϕ(2α) |p2 is generalized s-convex on [y1 , y2 ], for some fixed 0 < s ≤ 1 and p2 > 1 with 1 1 p1 + p2 = 1, then the following inequality holds: Γ(1 + 2α) y + y Γ(1 + 2α) [Γ(1 + α)]2 1 2 ϕ − 2α 2 2α (y2 − y1 )α α(s−1) p2
1 p1 Γ(1 + 2p1 α) ≤ 1 16α (Γ(1 + α)) p2 Γ(1 + (2p1 + 1)α) (2α) 3y1 + y2 (2α) y1 + 3y2 × ϕ + ϕ 4 4 2
(y2 − y1 )2α
21
(α) y1 Iy2 ϕ(x)
Proof. From Lemma 4.4 and using the generalized H¨older inequality for p2 > 1 and p11 + p12 = 1, we get Γ(1 + 2α) y + y Γ(1 + 2α) [Γ(1 + α)]2 1 2 (α) ϕ − y1 Iy2 ϕ(x) 2α 2 2α (a2 − a1 )α y1 + y2 (y2 − y1 )2α (α) 2α (2α) ϕ γ + (1 − γ)y1 ≤ 0 I1 γ α 16 2 y1 + y2 (α) 2α (2α) + 0 I1 (γ − 1) ϕ γy2 + (1 − γ) 2 p2 1 p2 y1 + y2 (y2 − y1 )2α (α) 2p1 α p11 (α) (2α) γ ≤ + (1 − γ)y1 0 I1 ϕ 0 I1 γ α 16 2 1 1 (y2 − y1 )2α (α) y1 + y2 p2 p2 (α) (2α) 2p1 α p1 + γy2 + (1 − γ) . 0 I1 (γ − 1) 0 I1 ϕ 16α 2 p2 Since ϕ(2α) is generalized s-concave, then (α) 0 I1
also (α) 0 I1
p2 p2 α(s−1) (2α) 3y + y 2 y + y 1 2 1 2 (2α) (4.10) ϕ ϕ + (1 − γ)y1 ≤ γ 2 Γ(1 + α) 4
(2α) y1 + y2 p2 2α(s−1) (2α) y1 + 3y2 p2 ϕ γy2 + (1 − γ) . ≤ Γ(1 + α) ϕ (4.11) 2 4
From (4.10) and (4.11), we observe that Γ(1 + 2α) y + y Γ(1 + 2α) [Γ(1 + α)]2 1 2 ϕ − 2α 2 2α (y2 − y1 )α
(α) y1 Iy2 ϕ(x)
α(s−1) 1 p1 (2α) 3y1 + y2 2 p2 (y2 − y1 )2α Γ(1 + 2p1 α) ≤ 1 ϕ 16α Γ(1 + (2p1 + 1)α) 4 p2 (Γ(1 + α)) α(s−1) 1 p1 (2α) y1 + 3y2 2 p2 (y2 − y1 )2α Γ(1 + 2p1 α) + 1 ϕ 16α Γ(1 + (2p1 + 1)α) 4 p2 (Γ(1 + α)) α(s−1) 1 p1 (2α) 3y1 + y2 (2α) y1 + 3y2 2 p2 (y2 − y1 )2α Γ(1 + 2p1 α) + ϕ ϕ = 1 4 4 16α (Γ(1 + α)) p2 Γ(1 + (2p1 + 1)α)
the proof is complete.
22
Remark 4.11. 1. If α = 1 in Theorem 4.10, then Z y2 y + y 1 1 2 ϕ ϕ(x)dx − 2 y 2 − y 1 y1 s−1 1 p1 ′′ 3y1 + y2 ′′ y1 + 3y2 1 2 q (y2 − y1 )2 ≤ + ϕ ϕ 16 Γ(2p1 + 1) 4 4
5
h i1 p1 Γ(1+2p1 α) 2. If s = 1 and 13 < Γ(1+(2p < 1, p1 > 1 in Theorem 4.10, 1 +1)α) then Γ(1 + 2α) y + y Γ(1 + 2α) [Γ(1 + α)]2 1 2 (α) ϕ I ϕ(x) − y 1 y2 α α α 2 2 2 (y2 − y1 ) (2α) 3y1 + y2 (2α) y1 + 3y2 (y2 − y1 )2α ≤ ϕ + ϕ 1 4 4 16α (Γ(1 + α)) p2
Applications to special means
As in [22], some generalized means are considered such as : y α +y α A(y1 , y2 ) = 1 2α 2 , y1 , y2 ≥ 0; i 1 h n (n+1)α (n+1)α Γ(1+nα) y2 − y1 , n ∈ Z{−1, 0}, y1 , y2 ∈ Ln (y1 , y2 ) = Γ(1+(n+1)α) R, y1 6= y2 . In [14], the following example was given: let 0 < s < 1 and y1α , y2α , y3α ∈ Rα . Defining for x ∈ R+ , ( b = 0, y1α ϕ(b) = sα α α y2 b + y3 b > 0. If y2α ≥ 0α and 0α ≤ y3α ≤ y1α , then ϕ ∈ GKs2 . Proposition 5.1. Let 0 < y1 < y2 and s ∈ (0, 1). Then Γ(1 + 2α) 2 Γ(1 + 2α) [Γ(1 + α)] s s A (y , y ) − L (y , y ) 1 2 s 1 2 2α 2α (y2 − y1 )α ( (y2 − y1 )2α Γ(1 + sα) 2α Γ(1 + 3α) [Γ(1 + α)]2 ≤ Γ(1 + (s − 2)α) 16α Γ(1 + 4α)Γ(1 + 2α) i Γ(1 + α) 2α Γ(1 + 2α) Γ(1 + 3α) h (s−2)α + − + |y1 | + |y2 |(s−2)α Γ(1 + 2α) Γ(1 + 3α) Γ(1 + 4α)
23
Proof. The result follows from Remark 4.7 (2) with ϕ : [0, 1] −→ [0α , 1α ], ϕ(x) = xsα . and when α = 1, we have the following inequalitly: (y2 − y1 )2 |s(s − 1)| s−2 s 1 s ≤ A (y1 , y2 ) − L (y , y ) |y1 | + |y2 |s−2 .(5.1) 1 2 s y2 − y1 48 Proposition 5.2. Let 0 < y1 < y2 and s ∈ (0, 1). Then Γ(1 + 2α) 2 Γ(1 + 2α) [Γ(1 + α)] s s A (y , y ) − L (y , y ) 1 2 1 2 s 2α 2α (y2 − y1 )α 1 1 p1 (y2 − y1 )2α Γ(1 + α) p2 Γ(1 + sα) Γ(1 + 2p1 α) ≤ α 16 Γ(1 + 2α) Γ(1 + (s − 2)α) Γ(1 + (2p1 + 1)α) !1 !1 (s−2)p2 α p2 p2 y1 + y2 (s−2)p2 α (s−2)p2 α (s−2)p2 α y1 + y2 , + × + |y | + |y | 1 2 2 2 where p2 > 1 and
1 p1
+
1 p2
= 1.
Proof. The result follows (4.9) with ϕ : [0, 1] −→ [0α , 1α ], ϕ(x) = xsα , and when α = 1, we have the following inequalitly: s 1 s A (y1 , y2 ) − L (y , y ) 1 2 s y2 − y1 !1 p2 (y2 − y1 )2 |s(s − 1)| y1 + y2 (s−2)p2 (s−2)p2 + |y | ≤ 1 1 1 2 2 p2 16(2p1 + 1) p1 !1 (s−2)p2 p2 y1 + y2 (s−2)p2 + |y | + . (5.2) 2 2 Where A(y1 , y2 ) and Ln (y1 , y2 ) in (5.1) and (5.2) are known as 1. Arithmetic mean: A(y1 , y2 ) =
y1 + y2 , y1 , y2 ∈ R+ ; 2
24
2. Logarithmic mean : L(y1 , y2 ) =
y1 − y2 , |y1 | = 6 y2 , y1 , a2 6= 0, y1 , y2 ∈ R+ ; ln |y1 | − ln |y2 |
Generalized Log-mean: y2n+1 − y1n+1 Ln (y1 , y2 ) = (n + 1)(y2 − y1 )
n1
, n ∈ Z \ {−1, 0} , y1 , y2 ∈ R+ .
Now, we give application to wave equation on Cantor sets: the wave equation on Cantor sets (local fractional wave equation) was given by [27] 2α ∂ 2α f (x, t) 2α ∂ f (x, t) = A (5.3) ∂t2α ∂x2α Following (5.3), a wave equation on Cantor sets was proposed as follows [29]: ∂ 2α f (x, t) ∂ 2α f (x, t) x2α = , 0≤α≤1 (5.4) ∂t2α Γ(1 + 2α) ∂x2α where f (x, t) is a fractal wave function and the initial value is given by xα . The solution of (5.4) is given as f (x, 0) = Γ(1 + α) f (x, t) =
xα t2α + . Γ(1 + α) Γ(1 + 2α)
By using (4.4), we have Z y1 + y2 Γ(1 + 2α) Γ(1 + 2α)Γ(1 + α) y2 α f x, f (x, t)(dt) − 2α (y2 − y1 )α 2α 2 y " 1 2α 2α (y2 − y1 )α 2 (α) 2α 2α ∂ f (x, t) = α y1 I y2 +y1 (t − y1 ) x 8 Γ(1 + 2α) y2 − y1 ∂xα 2 # 2α 2α 2(t − y1 ) (α) 2α ∂ f (x, t) x −1 + y1 I y2 +y1 y2 − y1 ∂xα 2
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