Hardy-type inequalities for generalized fractional ...

DOI 10.1515/tmj-2017-0005

Hardy-type inequalities for generalized fractional integral operators ˇ Sajid Iqbal1 , Josip Peˇcarić2 , Muhammad Samraiz3 and Zivorad Tomovski4 1 SIqbal-Department of Mathematics, University of Sargodha (Sub-Campus Bhakkar), Bhakkar, Pakistan 2 SJ. Peˇ cari´ c-Faculty of Textile Technology, University of Zagreb, Prilaz baruna Filipovi´ ca 28a, 10000 Zagreb, Croatia 3 SM. Samraiz-Department of Mathematics, University of Sargodha, Sargodha, Pakistan 4 SFaculty of Mathematics and Natural Sciences, Gazi Baba bb, 1000 Skopje, Macedonia; Zivorad Tomovski Depart-

ment of Mathematics, University of Rijeka Radmile Matejcic 2, 51000 Rijeka, Croatia E-mail: sajid [email protected] , [email protected] , [email protected] , [email protected] , [email protected]

Abstract The aim of this research paper is to establish the Hardy-type inequalities for Hilfer fractional derivative and generalized fractional integral involving Mittag-Leffler function in its kernel using convex and increasing functions. 2010 Mathematics Subject Classification. 26D15. 26D10, 26A33 Keywords. Hilfer fractional derivative; Mittag-Leffler function; Fractional integral.

1

Introduction

The Hardy inequality states that:  x p p Z∞ Z∞ Z 1 p  f (y)dy  ≤ f p (x)dx, p > 1, x p−1 0

0

0

where equality holds only if f ≡ 0 introduced by G. H. Hardy in [1]. It is one of the most important inequality of analysis. Such an inequality is broadly applied to various fascinating problems in partial differential equations such as eigenvalue and boundary value problems. It also been studied for vector field as well. A variety of mathematicians [2–7] awarded the generalizations and improvements of Hardy’s inequality. In this paper, we establish some more general inequalities of G. H. Hardy given in [6, 7] and applications of such inequalities for Hilfer fractional derivative and generalized fractional integral containing Mittag-Leffler function in its kernel via convex and increasing functions. We first need the following basic definition of convex function and elementary information about a particular class of function. The following definition is given in [8]. Definition 1.1. Let I be an interval in R. A function ϕ : I → R is called convex if ϕ(λx + (1 − λ)y) ≤ λϕ(x) + (1 − λ)ϕ(y),

(1.1)

for all points x, y ∈ I and all λ ∈ [0, 1]. The function ϕ is strictly convex if inequality (1.1) holds strictly for all distinct points in I and λ ∈ (0, 1).

Tbilisi Mathematical Journal 10(1) (2017), pp. 75–90. Tbilisi Centre for Mathematical Sciences. Received by the editors: 11 May 2015. Accepted for publication: 14 June 2016.

Authenticated | [email protected] author's copy Download Date | 2/12/17 7:18 AM

ˇ Tomovski S. Iqbal, J. Peˇ cari´ c, M. Samraiz, Z.

76

Let (Σ1 , Ω1 , µ1 ) and (Σ2 , Ω2 , µ2 ) be measure spaces with positive σ-finite measures. Let U (f ) denote the class of functions g : Ω1 → R with the representation Z g(x) := k(x, y)f (y)dµ2 (y), Ω2

and Ak be an integral operator defined by g(x) 1 Ak f (x) := = K(x) K(x)

Z k(x, y)f (y)dµ2 (y), Ω2

where k : Ω1 × Ω2 → R is measurable and non-negative kernel, f : Ω2 → R is measurable function and Z 0 < K(x) := k(x, y)dµ2 (y), x ∈ Ω1 . (1.2) Ω2

The upcoming result is given in [6]. Theorem 1.2. Let u be a weight function on Ω1 , k a non-negative measurable function on Ω1 × Ω2 and K is defined on Ω1 by (1.2). Assume that the function x 7→ u(x) k(x,y) K(x) is integrable on Ω1 for each fixed y ∈ Ω2 . Define v on Ω2 by Z k(x, y) v(y) := u(x) dµ1 (x) < ∞. K(x) Ω1 If ϕ : (0, ∞) → R is convex and increasing function, then the inequality Z Z g(x) dµ1 (x) ≤ u(x)ϕ v(y)ϕ(|f (y)|)dµ2 (y) K(x) Ω1 Ω2

(1.3)

holds for all measurable functions f : Ω2 → R. f1 , where fi : Ω2 → R, (i = 1, 2) are measurable f2 functions in Theorem 1.2, then the following result is obtained (see [9, p. 220]). By substituting k(x, y) by k(x, y)f2 (y) and f by

Theorem 1.3. Let fi : Ω2 → R be measurable functions, gi ∈ U (fi ), (i = 1, 2), where g2 (x) > 0 for every x ∈ Ω1 . Let u be a weight function on Ω1 and k a non-negative measurable function on is integrable on Ω1 for each fixed y ∈ Ω2 . Ω1 × Ω2 . Assume that the function x 7→ u(x) f2 (y)k(x,y) g2 (x) Define v on Ω2 by Z u(x)k(x, y) v(y) := f2 (y) dµ1 (x) < ∞. g2 (x) Ω1

If ϕ : (0, ∞) → R is convex and increasing function, then the inequality Z Z g1 (x) f1 (y) dµ2 (y) u(x)ϕ dµ1 (x) ≤ v(y)ϕ g2 (x) f2 (y) Ω1

Ω2

holds.


Hardy-type inequalities for generalized fractional integral operators

77

The upcoming result is given in [7]. Theorem 1.4. Let (Σ1 , Ω1 , µ1 ) and (Σ2 , Ω2 , µ2 ) be measure spaces with σ-finite measures, u be a weight function on Ω1 , k a non-negative measurable function on Ω1 × Ω2 . Let 0 < p ≤ q < ∞, K be defined on Ω1 and the function x 7→ u(x) k(x,y) K(x) is integrable on Ω1 for each fixed y ∈ Ω2 , then v is defined as:  pq  pq Z k(x, y) dµ1 (x) < ∞. v(y) :=  u(x) K(x) Ω1

If ϕ is non-negative convex function on the interval I ⊆ R, then the inequality   q1   p1 Z Z q  u(x) (ϕ (Ak f (x))) p dµ1 (x) ≤  v(y)ϕ (f (y)) dµ2 (y) Ω1

Ω2

holds for all measurable functions f : Ω2 → R such that Imf ⊆ I. Next result is represented in [7]. Theorem 1.5. Let gi ∈ U (fi ), (i = 1, 2, 3), where g2 (x) > 0 for every x ∈ Ω1 . Let u be a weight function on Ω1 , k a non-negative measurable function on Ω1 × Ω2 , then v is defined by Z u(x)k(x, y) dx < ∞. v(y) := f2 (y) g2 (x) Ω1

If ϕ : (0, ∞) × (0, ∞) → R is a convex and increasing function, then the inequality Z Z g1 (x) g3 (x) f1 (y) f3 (y) dµ2 (y) u(x)ϕ , , dµ1 (x) ≤ v(y)ϕ g2 (x) g2 (x) f2 (y) f2 (y) Ω1

Ω2

holds. Next we give the well known definition of Riemann-Liouville fractional integrals (see [10, p. 69-71]). Definition 1.6. Let [a, b] be a finite interval on R. The left and right sided Riemann-Liouville fractional integrals Iaα+ f and Ibα− f of order α > 0 are defined as: Iaα+ f (x)

1 = Γ(α)

Zx

(x − y)α−1 f (y)dy, x > a,

a

and Ibα− f (x)

1 = Γ(α)

Zb

(y − x)α−1 f (y)dy, x < b,

x

respectively. Here Γ represents Gamma function. The paper is organized as follows: After introduction in Section 2, we present the Hardy-type inequalities for Hilfer fractional derivative. Section 3 consists of results for generalized fractional integral which involve the Mittag-Leffler function in its kernel.



78

2

Hardy-type Inequalities for Hilfer fractional derivative

In this section, we give the basic definitions of Hilfer fractional derivative, then we present Hardytype inequalities for the said derivative. But before this, we first recall the well known definition of absolutely continuous function (see [7, p. 9]). Definition 2.1. Let 0 < a < b ≤ ∞. By C n [a, b] we denote the space of all functions on [a, b] which have continuous derivatives up to order n and AC[a, b] is the space of all absolutely continuous functions on [a, b]. By AC n [a, b], we denote the space of all functions f ∈ C n−1 [a, b] with f (n−1) ∈ AC[a, b]. Let us now recall the definition of Hilfer fractional derivative given in [11]. Definition 2.2. Let f ∈ L1 [a, b], f ∗ K(1−ν)(1−µ) ∈ AC 1 [a, b]. The fractional derivative operator µ,ν Da+ of order 0 < µ < 1 and type 0 < ν ≤ 1 with respect to x ∈ [a, b] is defined by ν(1−µ) d (1−ν)(1−µ) µ,ν Da+ f (x) := Ia+ f (x) , Ia+ dx

(2.1)

whenever the right hand side exists. The derivative (2.1) is usually called Hilfer fractional derivative. The more general integral representation of equation (2.1) given in [12] define as: Let f ∈ L1 [a, b] , f ∗ K(1−ν)(n−µ) ∈ AC n [a, b] , n − 1 < µ < n, 0 < ν ≤ 1, n ∈ N, then the following equation holds true: n ν(n−µ) d (1−ν)(n−µ) µ,ν Da+ f (x) = Ia+ f (x) . (2.2) I dxn a+ µ,0 µ Specially for ν = 0, Da+ f = Da+ f is a Riemann- Liouvile fractional derivative of order µ and for µ,1 µ ν = 1 it is a Caputo fractional derivative Da+ f =C Da+ f of order µ. Applying the properties of Riemann-Liouvile integral the relation (2.2) can be rewritten in the form: ν(n−µ) n−(1−ν)(n−µ) µ,ν f (x) f (x) = Ia+ Da+ Da+

=

1 Γ (ν (n − µ))

Zx

ν(n−µ)−1

(x − y)

µ+ν(n−µ)

Da+

f (y) dy.

a

Our first result is given in upcoming theorem. Theorem 2.3. Let n − 1 < µ < n, 0 < ν ≤ 1, n ∈ N, 0 < p < Lq (a, b) , then the following inequality holds true: Zb

µ,ν Da+ f (x) q dx ≤ C

a

where

1 p

+

1 q

= 1 and C =

1 1−ν(n−µ) ,

Zb q µ+ν(n−µ) f (y) dy, Da+

µ+ν(n−µ)

q > 1. If Da+

f∈

(2.3)

a (b−a)qν(n−µ) 1 (Γ(ν(n−µ)))q ((ν(n−µ)−1)p+1)q/p (qν(n−µ)) .



79

Proof. Since 1 Γ (ν (n − µ))

µ,ν Da+ f (x) ≤

Zx (x − y)

ν(n−µ)−1

µ+ν(n−µ) f (y) dy. Da+

a

Using Hölder’s inequality for {p, q} the above inequality becomes µ,ν Da+ f (x)  ≤

1  Γ (ν (n − µ))

1/p 

Zx (x − y)

(ν(n−µ)−1)p

1/q Zx q µ+ν(n−µ)  Da+ f (y) dy 

dy 

a

a ν(n−µ)−1+ 1

≤

p (x − a) 1 Γ (ν (n − µ)) ((ν (n − µ) − 1) p + 1)1/p

 b 1/q Z q µ+ν(n−µ)  Da+ f (y) dy  . a

Thus we have µ,ν Da+ f (x) q ≤

q[ν(n−µ)−1]+ q

p 1 (x − a) q (Γ (ν (n − µ))) ((ν (n − µ) − 1) p + 1)q/p

Zb q µ+ν(n−µ) f (y) dy. Da+ a

Integrating both sides from a to b gives inequality (2.3). If in particular we take ν = 1, we obtain the upcoming result given in [6]. Remark 2.4. Let n − 1 < µ < n, n ∈ N , 0 < p < following inequality holds true: Zb

C

q µ Da+ f (x)

1 p

+

1 q

= 1 and C =

q > 1. If f (n) ∈ Lq (a, b) , then the

Zb q dx ≤ C f (n) (y) dy,

a

where

1 1−(n−µ) ,

a

(b−a)q(n−µ) 1 (Γ(n−µ))q (((n−µ)−1)p+1)q/p (q(n−µ)) .

The upcoming corollary is a special case of Theorem 1.2 for Hilfer fractional derivative. Corollary 2.5. Let u be a weight function on (a, b) and let n − 1 < µ < n, 0 < ν ≤ 1, n ∈ N. Let f ∈ L (a, b) and define v on (a, b) by Zb v (y) = ν (n − µ)

u (x) y

(x − y)

ν(n−µ)−1 ν(n−µ)

(x − a)

dx < ∞.



80

If ϕ : (0, ∞) → R is a convex and increasing function, then the inequality Zb u (x) ϕ

Γ (ν (n − µ) + 1) ν(n−µ)

(x − a)

a

Zb ≤

! µ,ν Da+ f (x)

dx

µ+ν(n−µ) f (y) dy v (y) ϕ Da+

(2.4)

a

holds true. Proof. Applying Theorem 1.2 with Ω1 = Ω2 = (a, b), dµ1 (x) = dx, dµ2 (y) = dy ( (x−y)ν(n−µ)−1 Γ(ν(n−µ)) , a ≤ y ≤ x; k (x, y) = 0, x < y ≤ b, µ+ν(n−µ) (x−a)ν(n−µ) µ,ν f and g (x) = Da+ f (x) , we obtain inequality and K (x) = Γ(ν(n−µ)+1) . Replace f by Da+ (2.4). Specially for ν = 1, we obtain Corollary 2.9 of [6].

Remark 2.6. Choose u(x) = (x − a)ν(n−µ) a particular weight on (a, b) in Corollary 2.5, then we obtain the following inequality: Zb

(x − a)ν(n−µ) ϕ

Γ(ν(n − µ) + 1) µ,ν | D f (x)| dx a+ (x − a)ν(n−µ)

a

Zb ≤

µ+ν(n−µ) f (y) dy. (b − y)ν(n−µ) ϕ Da+

(2.5)

a

Although (1.3) holds for all convex and increasing functions but some choices of ϕ are of particular interest. Namely, we shall consider power function. Let q > 1 and the function ϕ : R+ → R be defined by ϕ(x) = xq , then (2.5) reduces to Zb

ν(n−µ)

(x − a)

q Γ(ν(n − µ) + 1) µ,ν | Da+ f (x)| dx (x − a)ν(n−µ)

a

Zb ≤

q µ+ν(n−µ) (b − y)ν(n−µ) Da+ f (y) dy.

(2.6)

a



81

Since x ∈ (a, b) and ν(n − µ)(1 − q) < 0, then we obtain that the left-hand side of (2.6) satisfies Zb

ν(n−µ)

(x − a)

Γ (ν (n − µ) + 1) ν(n−µ)

(x − a)

a

ν(n−µ)(1−q)

≥ (b − a)

!q µ,ν Da+ f (x)

(Γ(ν(n − µ)) + 1)

dx

q

Zb

µ,ν Da+ f (x) q dx,

(2.7)

a

and the right-hand side of (2.6) satisfies Zb

q µ+ν(n−µ) (b − y)ν(n−µ) Da+ f (y) dy

a ν(n−µ)

≤ (b − a)

Zb q µ+ν(n−µ) f (y) dy. Da+

(2.8)

a

Combining (2.6), (2.7) and (2.8) we get Zb

µ,ν Da+ f (x) q dx ≤

ν(n−µ)

(b − a) Γ (ν(n − µ) + 1)

!q Zb

a

Taking power

q µ+ν(n−µ) f (y) dy. Da+

a 1 q

on both sides, we obtain ν(n−µ)

µ,ν kDa+ f kq

≤

(b − a) Γ (ν(n − µ) + 1)

! µ+ν(n−µ)

kDa+

f kq .

Remark 2.7. In particular if ν = 0 inequality (2.6) represents inequality of G. H. Hardy for Riemann-Liouvile fractional derivative of order µ and for ν = 1 it becomes inequality of G. H. Hardy for Caputo fractional derivative of order µ. Corollary 2.8. Let u be a weight function on (a, b) and let n − 1 < µ < n, 0 < ν ≤ 1, n ∈ N. Define v on (a, b) as: µ+ν(n−µ)

(Da+ f2 )(y) v(y) = Γ(ν(n − µ))

Zb

ν(n−µ)−1

u(x) y

(x − y) dx < ∞. (Daµ,ν + f2 )(x)

If ϕ : (0, ∞) → R is convex and increasing function, then the inequality     µ+ν(n−µ) Dµ,ν f (x) Zb Zb f1 )(y) (Da+ a+ 1  dx ≤ v(y)ϕ   dy u(x)ϕ  µ,ν µ+ν(n−µ) (Da+ f2 ) (x) f2 )(y) (Da+ a

a

holds true for all fi ∈ L1 [a, b].


(2.9)


82

Proof. Applying Theorem 1.3 with Ω1 = Ω2 = (a, b), dµ1 (x) = dx, dµ2 (y) = dy ( (x−y)ν(n−µ)−1 Γ(ν(n−µ)) , a ≤ y ≤ x k(x, y) = 0, x ≤ y ≤ b, µ+ν(n−µ)

µ,ν and replace fi by Da+ fi , (i = 1, 2) and gi = Da+ fi , (i = 1, 2), we get inequality (2.9). In particular if we take ν = 1, we obtain Corollary 3.14 of [13]. The upcoming corollary is the generalization of Corollary 2.5.

Corollary 2.9. Let 0 < p ≤ q < ∞, u be a weight function on (a, b), n − 1 < µ < n, 0 < ν ≤ 1, µ,ν n ∈ N. Let Da+ be the left sided Hilfer fractional derivative and v is defined on (a, b) by  pq  b q Z ν(n−µ)−1 p (x − y) dx < ∞. v(y) := ν(n − µ)  u(x) (x − a)ν(n−µ) y

If ϕ is a non-negative increasing convex function on an interval I ⊆ R, then the following inequality  b  q1 pq Z µ,ν  u(x) ϕ Γ(ν(n − µ) + 1) Da+ dx f (x) (x − a)ν(n−µ) a

 b  p1 Z µ+ν(n−µ) ≤  v(y)ϕ Da+ f (y) dy 

(2.10)

a µ+ν(n−µ)

holds true for all measurable functions f : (a, b) → R such that ImDa+

f ⊆ I.

Proof. Applying Theorem 1.4 with Ω1 = Ω2 = (a, b), dµ1 (x) = dx, dµ2 (y) = dy ( (x−y)ν(n−µ)−1 Γ(ν(n−µ)) , a ≤ y ≤ x; k (x, y) = 0, x < y ≤ b, and K (x) = (2.10).

(x−a)ν(n−µ) Γ(ν(n−µ)+1) .

µ+ν(n−µ)

Replace f by Da+

µ,ν f and g (x) = Da+ f (x) , we obtain inequality

ν(n−µ)q

µ,ν Example 2.10. If Da+ is the Hilfer fractional derivative, u(x) = (x − y) p is a particular s weight and ϕ(x) = x , s ≥ 1, x > 0 is convex, then after some calculation we obtain the following



83

inequality:  b  q1 Z sq µ,ν  Da+ f (x) p dx a q(ν(n−µ)s−1)+p

1

≤

pq (ν(n − µ)) p (b − a) q1 s (ν(n − µ) pq + 1) (Γ(ν(n − µ) + 1)) p

 b  p1 Z s µ+ν(n−µ) ×  Da+ f (y) dy  . a µ,ν Theorem 2.11. Let u be a weight function, n − 1 < µ < n, 0 < ν ≤ 1,n ∈ N and let Da+ f be the µ+ν(n−µ)

Hilfer fractional derivative. If fi ∈ L[a, b], 0 < a < b < ∞ and x 7→

u(x) Da+

f2 (y)(x−y)ν(n−µ)−1

µ,ν Γ(ν(n−µ))(Da+ f2 )(x)

is integrable over (a, b), then v(y) is defined by b µ+ν(n−µ) Da+ f2 (y) Z (x − y)ν(n−µ)−1 v(y) = dx. u(x) µ,ν Γ(ν(n − µ)) Da+ f2 (x) y

If ϕ : (0, ∞) × (0, ∞) → R is a convex and increasing function, then the inequality Zb a

! Dµ,ν f (x) Dµ,ν f (x) a+ 1 a+ 3 u(x)ϕ µ,ν , µ,ν dx Da+ f2 (x) Da+ f2 (x) µ+ν(n−µ)   µ+ν(n−µ) D D Zb f (y) f (y) 1 3 a+ a+ ,  dy ≤ v(y)ϕ  µ+ν(n−µ) µ+ν(n−µ) Da+ f2 (y) Da+ f2 (y)

(2.11)

a

holds true. Proof. Applying Theorem 1.5 with with Ω1 = Ω2 = (a, b), dµ1 (x) = dx, dµ2 (y) = dy ( (x−y)ν(n−µ)−1 Γ(ν(n−µ)) , a ≤ y ≤ x; k (x, y) = 0, x < y ≤ b, and K (x) = (2.11).

(x−a)ν(n−µ) Γ(ν(n−µ)+1) .

µ+ν(n−µ)

Replace fi by Da+

µ,ν fi (x) to obtain inequality fi and gi (x) = Da+

Remark 2.12. In particular if we choose ν = 1 in Theorem 2.11, then v(y) can be written as: (n)

f2 (y) v(y) = Γ((n − µ))

Zb u(x)

(x − y)(n−µ)−1 dx, µ (D∗a f2 ) (x)

y



84

and inequality (2.11) represents results for Caputo fractional derivative which is given as: Zb

µ µ (D∗a f1 ) (x) (D∗a , µ f3 ) (x) dx u(x)ϕ µ (D∗a f2 ) (x) (D∗a f2 ) (x)

a

Zb ≤ a

3

! f (n) (y) f (n) (y) 1 3 v(y)ϕ (n) , (n) dy. f (y) f (y) 2 2

Hardy-type Inequalities for generalized fractional derivative involving Mittag-Leffler function in its kernel

First, we survey some facts about fractional integral operator which contain 6 parameter MittagLeffler function in the kernal (see [14, p. 1-13]), then we present Hardy-type inequalities for that integral operator. Definition 3.1. Let α, β, γ, δ ∈ C; min{R(α), R(β), R(γ), R(δ)} > 0; p, q > 0 and q < Rα + p, then the integral operator defined by Zx γ,δ,q γ,δ,q εα,β,p,ω,a+ g (x) = (x − y)β−1 Eα,β,p (ω(x − y)α )g(y)dy, a

which contains the generalized Mittag-Leffler function γ,δ,q Eα,β,p (z) =

∞ X

(γ)qn zn , Γ(αn + β) (δ)pn n=0

(3.1)

in its kernel is investigated and its boundedness is proved under certain conditions. The function (3.1) represents all the previous generalizations of Mittag-Leffler function by setting ∞ P (γ)n γ,δ zn • p = q = 1, it reduces to Eα,β (z) = Γ(αn+β) (δ)n defined by Salim in [15]. n=0

γ,q

• δ = p = 1, it represents Eα,β (z) =

∞ P n=0

(γ)qn z n Γ(αn+β) n!

which was introduced by A. K. Shukla and

J. C. Prajapati in [16]. In [17] H. M. Srivastava and Z. Tomovski investigated the properties of this function and its existence for a wider set of parameters. γ

• δ = p = q = 1, the operator (3.1) is defined by Prabhakar in [18] and is denoted as: Eα,β (z) = ∞ P

n=0

(γ)n zn Γ(αn+β) n! .

• γ = δ = p = q = 1, it reduces to Wiman’s function presented in [19], moreover if β = 1,

Mittag-Leffler function Eα (z) will be the result. β−1 γ,δ,q Lemma 3.2. Let α, β, γ, δ, ω > 0; p, q > 0, q < Rα + p and take eγ,δ,q Eα,β,p (ωxα ) . α,β,p,ω (x) = x Then the following integral holds true: Zx γ,δ,q eγ,δ,q α,β,p,ω (x − y) dy = eα,β+1,p,ω (x − a) . a



85

Proof. Consider Zx

eγ,δ,q α,β,p,ω

Zx (x − y) dy

=

γ,δ,q (x − y)β−1 Eα,β,p (ω(x − y)α )dy

a

a

Zx =

(x − y)β−1

a ∞ X

∞ n X (γ)qn (w(x − y)α ) dy Γ(αn + β)(δ)pn n=0 n

(γ)qn (w) = Γ(αn + β)(δ)pn n=0 =

β

(x − a)

Zx

a γ,δ,q Eα,β+1,p (ω(x

(x − y)αn+β−1 dy − a)α )

= eγ,δ,q α,β+1,p,ω (x − a) . γ This completes the lemma. It also holds for 3 parameter functions eγα,β,ω (x) = xβ−1 Eα,β (ωxα ) . The upcoming corollary represents Theorem 1.2 for fractional integral involving Mittag-Leffler function in its kernel.

Corollary 3.3. Let u be a weight function on (a, b) and α, β, γ > 0. Let f ∈ L (a, b) and v is defined on (a, b) by Zb v (y) =

u (x) y

(x − y)

β−1 β

α

γ Eα,β (ω (x − y) ) α

γ (x − a) Eα,β+1 (ω (x − a) )

dx < ∞.

If ϕ : (0, ∞) → R is a convex and increasing function, then the following inequality   γ Zb Zb εα,β,ω,a+ f (x)  dx ≤ v (y) ϕ (|f (y)|) dy u (x) ϕ  β γ α (x − a) Eα,β+1 (ω (x − a) )

(3.2)

a

a

holds. Proof. Applying Theorem 1.2 with Ω1 = Ω2 = (a, b), dµ1 (x) = dx, dµ2 (y) = dy β−1 γ α (x − y) Eα,β (ω (x − y) ) , a ≤ y ≤ x; k (x, y) = 0, x < y ≤ b, β γ α K (x) = (x − a) Eα,β+1 (ω (x − a) ) . Taking g(x) = εγα,β,ω,a+ f (x), we get inequality (3.2). Next result is an extension of Corollary 3.3. Corollary 3.4. Let u be a weight function on (a, b) and let α, β, γ, δ, ω be positive real numbers. Also p, q > 0 and q < Rα + p. Let f ∈ L (a, b) and define v on (a, b) by Zb v (y) =

β−1

u (x) y

(x − y)

β

α

γ,δ,q Eα,β,p (ω (x − y) ) α

γ,δ,q (x − a) Eα,β+1,p (ω (x − a) )

dx < ∞.



86

If ϕ : (0, ∞) → R is a convex and increasing function, then the inequality   γ,δ,q Zb Zb εα,β,p,ω,a+ f (x)   u (x) ϕ dx ≤ v (y) ϕ (|f (y)|) dy β γ,δ,q α (x − a) Eα,β+1,p (ω (x − a) ) a

(3.3)

a

holds true. Proof. Applying Theorem 1.2 with Ω1 = Ω2 = (a, b) , dµ1 (x) = dx, dµ2 (y) = dy β−1 γ,δ,q α (x − y) Eα,β,p (ω (x − y) ) , a ≤ y ≤ x; k(x, y) = 0, x ≤ y ≤ b, β γ,δ,q α and K (x) = (x − a) Eα,β+1,p (ω (x − a) ) . Taking g (x) = εγ,δ,q f (x) , we get inequality α,β,p,ω,a+ (3.3). Corollary 3.5. Let the assumptions of Corollary 3.4 be satisfied and v on (a, b) is defined by Zb v(y) = f2 (y)

u(x)

γ,δ,q (x − y)β−1 Eα,β,p (w(x − y)α )

y

εγ,δ,q α,β,p,ω,a+ f2 (x)

dx.

If ϕ : (0, ∞) → R is convex and increasing function, then the following inequality   γ,δ,q ε Zb Zb α,β,p,ω,a+ f1 (x) f1 (y)   u(x)ϕ dx ≤ v(y)ϕ f2 (y) dy εγ,δ,q α,β,p,ω,a+ f2 (x) a a

(3.4)

holds for all measurable functions fi : Ω2 → R, (i = 1, 2). Proof. Applying Theorem 1.3 with Ω1 = Ω2 = (a, b), dµ1 (x) = dx, dµ2 (y) = dy γ,δ,q (x − y)β−1 Eα,β,p (w(x − y)α ) , a ≤ y ≤ x k(x, y) = 0, x ≤ y ≤ b, β γ,δ,q α K (x) = (x − a) Eα,β+1,p (ω (x − a) ) . Taking gi (x) = εγ,δ,q f (x), we get inequality (3.4). + i α,β,p,ω,a Remark 3.6. If we choose ω = 0 in Corollary 3.5, then we obtained Corollary 3.11 of [7, p. 44] for left sided Riemann-Liouville fractional integral operator. We next give Hardy-type inequality for generalized fractional integral operator involving generalized Mittag-Leffler function in its kernel. Theorem 3.7. Let p, q > 1 such that p1 + 1q = 1 and α, β, γ, δ, ω > 0. If f ∈ Lq (a, b) , 0 < a < b < ∞, then the inequality Zb Zb q γ,δ,q q εα,β,p,ω,a+ f (x) dx ≤ C |f (x)| dx a

holds true, where C =

h

eγ,δ,q α,β+2,p,ω

a

iq (b − a) .



87

Proof. Applying H¨ older’s inequality, we have Zx γ,δ,q γ,δ,q eα,β,p,ω (x − y) |f (y)| dy εα,β,p,ω,a+ f (x) ≤ a

1/p  x 1/q  x Z Z p   |f (y)|q dy  ≤  eγ,δ,q α,β,p,ω (x − y) dy a

a

 x  b 1/q Z Z   |f (x)|q dx ≤  eγ,δ,q α,β,p,ω (x − y) dy a

=

h

a

eγ,δ,q α,β+1,p,ω

 1/q i Zb q (x − a)  |f (x)| dx . a

Thus we have   q h iq Zb γ,δ,q q γ,δ,q εα,β,p,ω,a+ f (x) ≤ eα,β+1,p,ω (x − a)  |f (x)| dx , a

for every x ∈ [a, b] . Integrating on both sides from a to b, we get  b  b  Zb Z h Z q iq γ,δ,q q dx  |f (x)| dx eγ,δ,q εα,β,p,ω,a+ f (x) dx ≤  α,β+1,p,ω (x − a) a

a

a

 b q  b  Z Z q   |f (x)| dx . ≤  eγ,δ,q α,β+1,p,ω (x − a) dx a

a

Applying Lemma 3.2, we obtain   Zb q q Zb γ,δ,q q γ,δ,q εα,β,p,ω,a+ f (x) dx ≤ eα,β+2,p,ω (b − a)  |f (x)| dx . a

a

Corollary 3.8. Let u(x) be a weight function on (a, b) and let α, β, γ, δ, ω > 0; p, q > 0, q < Rα+p. Define v on (a, b) by p

 b Z  v (y) = u (x) y

(x − (x −

β−1

γ,δ,q y) Eα,β,p β γ,δ,q a) Eα,β+1,p

! pq  q (ω (x − y) ) dx  < ∞. α (ω (x − a) ) α



88

If ϕ : I → R is a convex and increasing function, then the inequality    Zb   u(x) ϕ  a

 pq

εγ,δ,q f (x) α,β,p,ω,a+

β

(x − a)

γ,δ,q Eα,β+1,p

α

(ω (x − a) )

 q1

 dx 

 b  p1 Z ≤  v (y) ϕ (|f (y)|) dy 

(3.5)

a

holds true for all measurable functions f : (a.b) → R such that Imf ⊆ I. Proof. Applying Theorem 1.4 with Ω1 = Ω2 = (a, b), dµ1 (x) = dx, dµ2 (y) = dy γ,δ,q (x − y)β−1 Eα,β,p (w(x − y)α ) , a ≤ y ≤ x k(x, y) = 0, x ≤ y ≤ b, g = εγ,δ,q α,β,p,ω,a+ f, we get inequality (3.5). The upcoming result is an application of Theorem 1.5. Corollary 3.9. Let u be a weight function and α, β, γ, δ, ω be the positive real numbers. If fi ∈ L (a, b) , 0 < a < b < ∞, p, q > 0, then v(y) is defined by Zb v(y) = f2 (y)

γ,δ,q u(x)(x − y)β−1 Eα,β,p (ω(x − y)α )

εγ,δ,q α,β,p,ω,a+ f2 (x)

y

dx.

If ϕ : (0, ∞) × (0, ∞) → R is a convex and increasing function, then the inequality Zb a

γ,δ,q γ,δ,q ! Zb ε f1 (y) f3 (y) α,β,p,ω,a+ f1 (x) εα,β,p,ω,a+ f3 (x) , dy u(x)ϕ γ,δ,q , γ,δ,q dx ≤ ϕ ε ε f2 (y) f2 (y) α,β,p,ω,a+ f2 (x) α,β,p,ω,a+ f2 (x)

(3.6)

a

holds true. Remark 3.10. If in particular we choose ω = 0 in Corollary 3.9, then v(y) becomes f2 (y) v(y) = Γ(β)

Zb

u(x)(x − y)β−1

y

Iaβ+ f2 (x)

dx,

and inequality (3.6) can be written as: Zb a

! Zb I β f (x) I β f (x) f1 (y) f3 (y) a+ 1 a+ 3 dy. , u(x)ϕ β , dx ≤ ϕ I + f2 (x) I β+ f2 (x) f2 (y) f2 (y) a a a



89

Acknowledgements. The research of J. Peˇcarić has been fully supported by Croatian Science Foundation under the project 5435. The author Zivorad Tomovski was supported under the European Commission and the Croatian Ministry of Science, Education and Sports Co-Financing agreement No. 291823. In particular Z. Tomovski acknowledge project financing from the Maria Curie FP-7-PEOPLE-2011-COFUND program NEWFELPRO Grant Agreement No. 37 (Anomalous diffusion).

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