HARDY-LITTLEWOOD TYPE INEQUALITY FOR ψ-RIEMANN-LIOUVILLE FRACTIONAL INTEGRALS

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HARDY-LITTLEWOOD TYPE INEQUALITY FOR ψ-RIEMANN-LIOUVILLE FRACTIONAL

INTEGRALS

César Ledesma, J. Vanterler da C. Sousa, Amado Cruz

To cite this version:

César Ledesma, J. Vanterler da C. Sousa, Amado Cruz. HARDY-LITTLEWOOD TYPE INEQUAL-

ITY FOR ψ-RIEMANN-LIOUVILLE FRACTIONAL INTEGRALS. 2021. �hal-03189239�

ψ-RIEMANN-LIOUVILLE FRACTIONAL INTEGRALS

C ´ ESAR T. LEDESMA, J. VANTERLER DA C. SOUSA, AND AMADO M. CRUZ

Abstract. In this present paper, we present a new extension for Hardy-Littlewood type inequality using the Riemann-Liouville fractional integral with respect to another function in the spaces

L^p

[a, b] and

L^q

[a, b]. The result is investigated under certain conditions for

p, q

and via Holder inequality.

1. Introduction

In 1928, Hardy and Littlewood [1] first established boundedness properties for Riemann- Liouville fractional integrals. The classical Hardy inequality for fractional integrals states that

x

^β−α

Z

x

0

f (y) y

^β

(x − y)

^1−α

dy

L^p(0,b)

≤ C kfk

_Lp(0,b)

with 0 < α < 1, where α −

¹_p

< β <

¹_q

,

¹_q

+

¹_p

= 1 and 0 < b ≤ ∞.

On the other hand, 10 years later, classical inequality was discussed, duo to Hardy, Littlewood and Sobolev, states that [1]

Z

Rⁿ

Z

Rⁿ

f (x)|x − y|

^−λ

g(y)dxdy

≤ C

_n,p,q⁰

||f ||

_Lq0

||g||

_L^p

for all f ∈ L

^q0

( R

ⁿ

), g ∈ L

^p

( R

ⁿ

), 1 < p, q

⁰

< ∞,

¹_p

+

_q¹0

+

^λ_n

= 2, 0 < λ < n and q

⁰

=

_q−1^q

. In this sense, Lie and Loss [12], discussed an estimate for the constant C

_n,p,q⁰

.

The Hardy–Littlewood–Sobolev inequality was extended by Stein and Weiss to the following Stein–Weiss inequalities [19]

Z

Rⁿ

Z

Rⁿ

|x|

^−α

|x − y|

^−λ

f (x)g(y)|y|

^−β

dxdy

≤ C

_n,α,β,p,q⁰

||f ||

_Lq0

||g||

_L^p

where 1 < p, q

⁰

< ∞, β, α and λ satisfying the following conditions.

_q¹0

+

¹_p

+

^α+β+λ_n

= 2,

1

q⁰

+

¹_p

≥ 1, α + β ≥ 0, α <

ⁿ_q

, β <

_pⁿ0

and 0 < λ < n.

Over these 92 years since the first ideas of Hardy-Littlewood inequality, numerous results have been addressed, in particular, applications. We also highlight new extensions to the Hardy-Littlewood inequality, in particular, with the fractional Poisson kernel [10, 11, 13–15].

We also highlight the extent to the Hardy-Littlewood-Polya inequality [16]. It is noted that the importance and impact of the first result on the inequality of Hardy and Littlewood, caused in theory. Other inequalities involving Hardy and Littlewood can be found in the articles [17,18,20].

2010

Mathematics Subject Classification.

Primary 26A33; Secondary 34B15, 35J20, 58E05.

Key words and phrases. ψ-fractional derivative space,

variational structure, fractional differential equations,boundary value problem, mountain pass theorem.

C´ esar T. Ledesma was partially supported by CONCYTEC, Peru, 379-2019-FONDECYT “ASPECTOS CUALITATIVOS DE ECUACIONES NO-LOCALES Y APLICACIONES”..

1

As seen above, the first results of the Hardy-Littlewood inequality in the sense of fractional integrals, was discussed only for the case of Riemann-Liouville. As is known, by means of the Riemann-Liouville fractional integral, the fractional derivatives of Caputo and Riemann- Liouville are defined. However, over the decades, new versions of fractional integrals have emerged. Then, there was a need to propose a new general operator that contained as particular cases, a wide class of existing fractional integrals [2, 6]. So, we have the Riemann-Liouville fractional integral call with respect to another function, that is, ψ(·), as will be presented below:

Let α ∈ (0, 1) and ξ ∈ L

¹

[a, b] , [a, b] be a finite or infinite interval of the real line R (−∞ ≤ a < b ≤ ∞). Also let ψ : [a, b] → R be an increasing and positive continuous function, having a continuous and non decreasing derivatives ψ

⁰

(x) 6= 0 on (a, b). The left-sided fractional integrals of a function f with respect to another function ψ on [a, b], is defined as [2–4]

I

^α;ψ_a+,x

ξ (x) = 1 Γ (α)

Z

x a

ψ

⁰

(t) (ψ (x) − ψ (t))

^α−1

ξ (t) dt

The right-sided fractional integrals of a function f with respect to another function ψ on [a, b], is defined in an analogous way I

^α;ψ_x,b−

(·).

Let α > 0 and δ > 0.Then, we have the following semigroup property given by [2,3]

I

^α;ψ_a+,x

I

^δ;ψ_a+,x

ξ (x) = I

^α+δ;ψ_a+,x

ξ (x) .

Since we have a fractional integration operator that contains and a unique (operator) a wide class of possible particular cases, then the following question arises: what conditions about the ψ-Riemann-Liouville fractional integral and about Hardy’s inequality-Littlewood is necessary and enough, so that we can extend it?

So, motivated by these issues and the work above, in this paper we consider the boundedness of ψ−Riemann-Liouville fractional integrals. More precisely we are going to prove the following result:

H-L Theorem 1.1. Let 0 < α < 1 and p ∈ [1, ∞], then the following statements are satisfied:

(1) If 1 ≤ p <

_α¹

, then the operators I

^α;ψ_a+

, I

^α;ψ_b−

: L

^p

[a, b] → L

^q

[a, b] are continuous for every q ∈ h

1,

_1−αp^p

,

(2) If α =

¹_p

, then the operators I

^α;ψ_a+

, I

^α;ψ_b−

: L

^p

[a, b] → L

^q

[a, b] are continuous for every q ∈ [1, ∞),

(3) If α ∈

1 p

, 1

, then the operators I

^α;ψ_a+

, I

^α;ψ_b−

: L

^p

[a, b] → L

^q

[a, b] are continuous for every p ≤ q ≤ ∞.

(4) If p = ∞, then the operators I

^α;ψ_a+

, I

^α;ψ_b−

: L

^p

[a, b] → C[a, b] ∩ L

^∞

[a, b] are continuous.

2. Hardy-Littlewood type result

In this section, we are going to prove Theorem 1.1. We start our analysis with the following remark. Let α ∈ (0, 1) and ψ be an increasing and positive continuous function on (a, b] with ψ

⁰

∈ C(a, b), ψ

⁰

6= 0 and ψ

⁰

increasing. Then, for x ∈ (a, b] we have

HL01

HL01 (2.1)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

dt ≤ (ψ(b) − ψ(a))

^α

α .

In fact, by doing the change of variable u = ψ(x) − ψ(t) and since ψ is increasing then Z

x

a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

dt =

Z

ψ(x)−ψ(a) 0

u

^α−1

du

= (ψ(x) − ψ(a))

^α

α ≤ (ψ(b) − ψ(a))

^α

α .

Next, we consider the boundedness of I

^α;ψ_a+

in L

^p

spaces.

H-L1 Lemma 2.1. Let p ∈ [1, ∞] and u ∈ L

^p

(a, b). Then Z

x

a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|dt < ∞ for a.e. x ∈ (a, b].

Moreover, I

^α;ψ_a+

u ∈ L

^p

(a, b) with

HL02

HL02 (2.2) kI

^α;ψ_a+

uk

_Lp(a,b)

≤ (ψ(b) − ψ(a))

^α

Γ(α + 1) kuk

_Lp(a,b)

. Proof. Let p = ∞, then by (2.1) we obtain

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|dt ≤ kuk

_L∞(a,b)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

dt

≤ (ψ(b) − ψ(a))

^α

α kuk

_L^∞_(a,b)

, x ∈ (a, b].

Let p = 1, next Z

b

a

|I

^α;ψ_a+

u(x)|dx = Z

b

a

1 Γ(α)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

u(t)dt

dx

≤ 1 Γ(α)

Z

b a

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|dtdx

≤ 1 Γ(α)

Z

b a

Z

x a

ψ

⁰

(x)(ψ(x) − ψ(t))

^α−1

|u(t)|dtdx

= 1

Γ(α) Z

b

a

|u(t)|

Z

b t

ψ

⁰

(x)(ψ(x) − ψ(t))

^α−1

dxdt

= 1

Γ(α + 1) Z

b

a

(ψ(b) − ψ(t))

^α

|u(t)|dt

≤ (ψ(b) − ψ(a))

^α

Γ(α + 1)

Z

b a

|u(t)|dt.

Now, we consider the case 1 < p < ∞. Let q > 0 such that 1

p + 1

q = 1.

Next, by H¨ older inequality we get

|I

^α;ψ_a+

u(x)| =

1 Γ(α)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

u(t)dt

≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^1q

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^1p

|u(t)|dt

≤ 1 Γ(α)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

dt

1/q

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|

^p

dt

1/p

≤ 1 Γ(α)

(ψ(b) − ψ(a))

^α

α

1/q

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|

^p

dt

1/p

. So, since ψ

⁰

is increasing we obtain

Z

b a

|I

^α;ψ_a+

u(x)|

^p

dx ≤ 1 [Γ(α)]

^p

(ψ(b) − ψ(a))

^α

α

p/q

Z

b a

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|

^p

dtdx

≤ 1 [Γ(α)]

^p

(ψ(b) − ψ(a))

^α

α

p/q

Z

b a

Z

x a

ψ

⁰

(x)(ψ(x) − ψ(t))

^α−1

|u(t)|

^p

dtdx

= 1

[Γ(α)]

^p

(ψ(b) − ψ(a))

^α

α

p/q

Z

b a

|u(t)|

^p

Z

b

t

ψ

⁰

(x)(ψ(x) − ψ(t))

^α−1

dxdt

≤ 1 [Γ(α)]

^p

(ψ(b) − ψ(a))

^α

α

p/q

(ψ(b) − ψ(a))

^α

α

Z

b a

|u(t)|

^p

dt

= 1

[Γ(α)]

^p

(ψ(b) − ψ(a))

^α

α

^p

q+1

Z

b a

|u(t)|

^p

dt.

Therefore

kI

^α;ψ_a+

uk

_Lp(a,b)

≤ (ψ(b) − ψ(a))

^α

Γ(α + 1) kuk

_Lp(a,b)

.

Now, we are ready to give the proof of Theorem 1.1.

Proof of Theorem 1.1:

(1) The case p = q ∈ [1, ∞] was proved in Lemma 2.1, were we obtain HL03

HL03 (2.3) kI

^α;ψ_a+

uk

_Lp(a,b)

≤ (ψ(b) − ψ(a))

^α

Γ(α + 1) kuk

_Lp(a,b)

.

Now we consider the case p = 1 and 1 ≤ q <

_1−α¹

. The case q = 1 is given by (2.3), therefore we consider the case 1 < q <

_1−α¹

.

By H¨ older inequality we have

|I

^α;ψ_a+

u(x)| ≤ 1 Γ(α)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|dt

= 1

Γ(α) Z

x

a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|

^1q

|u(t)|

^1−1q

dt

≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^q

|u(t)|dt

1/q

Z

x a

|u(t)|dt

1−¹_q

.

Hence HL04

HL04 (2.4) |I

^α;ψ_a+

u(x)|

^q

≤ 1 [Γ(α)]

^q

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α

− 1]

^q

|u(t)|dtkuk

^q−1_L1(a,b)

.

Next,

k|I

^α;ψ_a+

uk

^q_Lq(a,b)

≤ 1 [Γ(α)]

^q

Z

b a

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^q

|u(t)|dtdxkuk

^q−1_L1(a,b)

≤ 1 [Γ(α)]

^q

Z

b a

|u(t)|

Z

b t

[ψ

⁰

(x)(ψ(x) − ψ(t))

^α−1

]

^q

dxdtkuk

^q−1_L1(a,b)

= 1

[Γ(α)]

^q

Z

b

a

|u(t)|[ψ

⁰

(b)]

^q−1

Z

b

t

ψ

⁰

(x)(ψ(x) − ψ(t))

^(α−1)q

dxdtkuk

^q−1_L1(a,b)

= [ψ

⁰

(b)]

^q−1

[Γ(α)]

^q

Z

b a

|u(t)| (ψ(b) − ψ(t))

^{1−(1−α)q}

1 − (1 − α)q dtkuk

^q−1_L1(a,b)

≤ [ψ

⁰

(b)]

^q−1

(ψ(b) − ψ(a))

^{1−(1−α)q}

(1 − (1 − α)q)[Γ(α)]

^q

kuk

^q_L1(a,b)

. Therefore

HL05

HL05 (2.5) k|I

^α;ψ_a+

uk

_Lq(a,b)

≤ [ψ

⁰

(b)]

^q−1

(ψ(b) − ψ(a))

^{1−(1−α)q}

(1 − (1 − α)q)[Γ(α)]

^q

!

1/q

kuk

_L1(a,b)

, ∀1 ≤ q < 1 1 − α Continuing with our analysis, we consider the case 1 < p <

_α¹

and p ≤ q <

_1−αp^p

. The case q = p was considered in (2.3), so we only consider the case

p < q < p 1 − αp . In fact, let

HL06

HL06 (2.6) δ = 1

p − 1 q , then 0 < δ < α < 1. Define r as

HL07

HL07 (2.7) 1

r = 1 + 1 q − 1

p = 1 − δ = ⇒ r = 1 1 − δ . Note that r < q, next by H¨ older inequality we have

|I

^α;ψ_a+

u(x)| ≤ 1 Γ(α)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|dt

= 1

Γ(α) Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^rq

|u(t)|

^pq

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^1−rq

|u(t)|

^p(1p−1q)

dt

≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt

1/q

Z

x a

|u(t)|

^p

dt

¹_p−¹_q

× Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

dt

1−¹

p

= 1

Γ(α)

[ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

1−¹_p

× Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt

1/q

Z

x a

|u(t)|

^p

dt

¹

p−¹

q

. Note that the last expression is well defined since

1

r = 1 − δ > 1 − α = ⇒ (1 − α)r < 1.

Therefore, we have

kI

^α;ψ_a+

uk

^q_Lq(a,b)

= Z

b

a

|I

^α;ψ_a+

u(x)|

^q

dx

≤ 1 [Γ(α)]

^q

Z

b a

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dtkuk

^q−p_Lp(a,b)

[ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q(1−¹_p)



 dx

≤

kuk

^q−p_Lp(a,b)

[Γ(α)]

^q

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q(1−¹

p)

Z

b a

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dtdx

≤ kuk

^q−p_Lp(a,b)

[Γ(α)]

^q

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q(1−¹

p)

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r kuk

^p_Lp(a,b)

= 1

[Γ(α)]

^q

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q(1−_p¹)+1

kuk

^q_Lp(a,b)

.

Hence HL08

HL08 (2.8)

kI

^α;ψ_a+

uk

_Lq(a,b)

≤ 1 Γ(α)

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

1−¹

p+¹_q

kuk

_Lp(a,b)

∀p ≤ q < p 1 − αp .

(2) Now we consider the case p =

_α¹

and

_α¹

≤ q < ∞. Just we consider the case

_α¹

< q < ∞, since by (2.2) we have the case q = p.

Define δ and r as

δ = 1 p − 1

q = α − 1 q and

1

r = 1 − δ = 1 + 1 q − 1

p .

Note that 0 < δ < α < 1 and r <

_1−α¹

. Next by H¨ older inequality HL09

HL09 (2.9)

|I

^α;ψ_a+

u(x)| ≤ 1 Γ(α)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|dt

= 1

Γ(α) Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^rq

|u(t)|

^pq

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r(1−1p)

|u(t)|

^1−pq

dt

≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt

1/q

×

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r(1−1p)θ

|u(t)|

^(1−pq)θ

dt

1/θ

, 1 q + 1

θ = 1.

By other side, by H¨ older inequality we derive

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r(1−1p)θ

|u(t)|

^(1−pq)θ

dt

≤ Z

x

a

[φ

⁰

(t)(φ(x) − φ(t))

^α−1

]

^r

1−¹_p

θs

dt

1/s

Z

x a

|u(t)|

1−^p_q

θm

dt

1/m

, 1 s + 1

m = 1,

note that

m = p(q − 1)

q − p , θ = q

q − 1 and s = p(q − 1)

q(p − 1) = ⇒ r

1 − 1 p

θs = r and

1 − p q

θm = p.

Therefore

HL10

HL10 (2.10)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r(1−1p)θ

|u(t)|

^(1−pq)θ

dt

≤ Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

dt

^q(p−1)_p(q−1)

Z

x a

|u(t)|

^p

dt

_p(q−1)^q−p

= [ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

^q(p−1)_p(q−1)

Z

x

a

|u(t)|

^p

dt

_p(q−1)^q−p

.

Now, replacing (2.10) in (2.9) we get

|I

^α;ψ_a+

u(x)| ≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt

1/q

[ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

^p−1_p

× Z

x

a

|u(t)|

^p

dt

^q−p_pq

.

So HL11

HL11 (2.11)

|I

^α;ψ_a+

u(x)|

^q

≤ 1

[Γ(α)]

^q

[ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q

1−¹

p

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt

× kuk

^q−p_Lp(a,b)

.

Therefore HL12

HL12 (2.12)

kI

^α;ψ_a+

uk

^q_Lq(a,b)

= Z

b

a

|I

^α;ψ_a+

u(x)|

^q

dx

≤ 1

[Γ(α)]

^q

kuk

^q−p_Lp(a,p)

Z

b

a

[ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q 1−¹_p

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dtdx

≤ kuk

^q−p_Lp(a,b)

[Γ(α)]

^q

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q 1−¹

p

Z

b a

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dtdx

≤ kuk

^q−p_Lp(a,b)

[Γ(α)]

^q

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q 1−¹_p

Z

b a

Z

b t

[ψ

⁰

(x)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dxdt

≤

kuk

^q−p_Lp(a,b)

[Γ(α)]

^q

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

q

1−¹

p

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r kuk

^p_Lp(a,b)

=

kuk

^q_Lp(a,b)

[Γ(α)]

^q

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

1+q 1−¹_q

.

That is

kI

^α;ψ_a+

uk

_Lq(a,b)

≤ 1

Γ(α) [ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

1−¹

p+¹_q

kuk

_Lp(a,b)

∀p = 1

α ≤ q < ∞.

(3) Now we consider the case 1

α < p < ∞ and p ≤ q ≤ ∞.

The case q = p was considered in (2.3), so we only consider the case p < q ≤ ∞.. WE start considering the case p < q < ∞. Let

δ = 1 p − 1

q and 1

r = 1 − δ = 1 + 1 q − 1

p . So

0 < δ < α and 1 < r < 1 1 − α . Next

|I

^α;ψ_a+

u(x)| ≤ 1 Γ(α)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|dt

= 1

Γ(α) Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^rq

|u(t)|

^pq

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^1−rq

|u(t)|

^p

1 p−¹

q

dt.

Since

1 − r q = r

1 − 1

p

,

by H¨ older inequality we derive HL13

HL13 (2.13)

|I

^α;ψ_a+

u(x)| ≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^rq

|u(t)|

^pq

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

1−¹_p

|u(t)|

^p

1 p−¹_q

dt

≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt

1/q

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^rθ

1−¹

p

|u(t)|

^pθ

1 p−¹

q

dt

¹_θ

,

where

¹_q

+

¹_θ

= 1. On the other hand HL14

HL14 (2.14) Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^rθ

1−¹_p

|u(t)|

^pθ

q−p pq dt

dt ≤ Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^rθm

1−¹_p

dt

_m¹

× Z

x

a

|u(t)|

^pθn

1 p−¹_q

dt

1/n

, 1 m + 1

n = 1.

Note that

θ = q

q − 1 , m = p(q − 1)

q(p − 1) and n = p(q − 1) q − p . So, replacing (2.14) into (2.13) we get

|I

^α;ψ_a+

u(x)| ≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt

1/q

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

^p−1_p

× Z

x

a

|u(t)|

^p

dt

^q−p_pq

≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt

1/q

[ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

^p−1_p

× kuk

q−p q

L^p(a,b)

. Therefore

HL15

HL15 (2.15)

|I

^α;ψ

a⁺

u(x)|

^q

≤ 1 [Γ(α)]

^q

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

|u(t)|

^p

dt [ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

^q(p−1)_p

× kuk

^q−p_Lp(a,b)

. Finally

kI

^α;ψ_a+

uk

^q_Lq(a,b)

= Z

b

a

|I

^α;ψ_a+

u(x)|

^q

dx

≤

kuk

^q−p_Lp(a,b)

[Γ(α)]

^q

Z

b a

[ψ

⁰

(x)]

^r−1

(ψ(x) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

^q(p−1)

p

Z

x

a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]|u(t)|

^p

dtdx

≤ 1

[Γ(α)]

^q

[ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

1+^q(p−1)_p

kuk

^q_Lp(a,b)

.

That is

HL16

HL16 (2.16) kI

^α;ψ_a+

uk

_Lq(a,b)

≤ 1

Γ(α) [ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

1−¹

p+¹_q

kuk

_Lp(a,b)

.

Now we consider the case 1

α < p < ∞ and q = ∞.

By H¨ older inequality we derive

|I

^α;ψ_a+

u(x)| ≤ 1 Γ(α)

Z

x a

ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

|u(t)|dt

≤ 1 Γ(α)

Z

x a

[ψ

⁰

(t)(ψ(x) − ψ(t))

^α−1

]

^r

dt

1/r

Z

x a

|u(t)|

^p

dt

1/p

≤ 1

Γ(α) [ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

1/r

kuk

_Lp(a,b)

.

Therefore

HL17

HL17 (2.17) kI

^α;ψ_a+

uk

_L^∞_(a,b)

≤ 1

Γ(α) [ψ

⁰

(b)]

^r−1

(ψ(b) − ψ(a))

^{1−(1−α)r}

1 − (1 − α)r

!

1−¹

p

kuk

_Lp(a,b)

.

(4) Finally, the case p = q = ∞ was proved in (2.3).

Remark 1. Here, we present proof of the results only for the left operator I

^α;ψ_a+

(·), in an analogous way, follows to the operator the right I

^α;ψ_b−

(·).

References

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Chen [12] Chen, W., and C. Li. The best constant in a weighted Hardy-Littlewood-Sobolev inequality. Proc. Amer.

Math. Soc. (2008): 955-962.

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Love [13] Love, E. R. Generalizations of a classical inequality. Applicable Anal. 8.1 (1978): 47-59.

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Everitt [14] Everitt, W. N. Some Examples of Hardy-Littlewood Type Integral Inequalities. General Inequalities 4.

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4885-4896.

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Khajeh-Khalili [16] Khalili, P. Generalization of a Hardy-Littlewood-Polya inequality. J. Approx. Theory. 66.2 (1991): 115-124.

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Smith [17] Smith, K. T. A generalization of an inequality of Hardy and Littlewood. Canadian J. Math. 8 (1956):

157-170.

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De [18] De N´ apoli, Pablo L., I. Drelichman, and R. G. Dur´ an. On weighted inequalities for fractional integrals of radial functions. Illinois J. Math. 55.2 (2011): 575-587.

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Chen1 [19] Chen, L., G. Lu, and C. Tao. Hardy—Littlewood—Sobolev Inequalities with the Fractional Poisson Kernel and Their Applications in PDEs. Acta Mathematica Sinica, English Series. 35.6 (2019): 853-875.

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1

(C´ esar T. Ledesma)

Departamento de Matem´ aticas Universidad Nacional de Trujillo Av. Juan Pablo II s/n. Trujillo-Per´ u

Email address

:

ctl 576@yahoo.es

(J. Vanterler da C. Sousa)

Centro de Matem´ atica, Computac ¸˜ ao e Cognic ¸˜ ao Universidade Federal do ABC-UFABC

Santo Andr´ e, SP, Brazil

Email address

:

jose.vanterler@ufabc.edu.br

(Amado M. Cruz)

Departamento de Matem´ aticas

Universidad Nacional de Trujillo

Av. Juan Pablo II s/n. Trujillo-Per´ u

Email address

:

gamc55@hotmail.com