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Article

# Hilfer-Polya, ψ-Hilfer Ostrowski and ψ-Hilfer-Hilbert-Pachpatte Fractional Inequalities

by
George A. Anastassiou
Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA
Symmetry 2021, 13(3), 463; https://doi.org/10.3390/sym13030463
Submission received: 26 February 2021 / Revised: 11 March 2021 / Accepted: 11 March 2021 / Published: 12 March 2021
(This article belongs to the Special Issue Functional Equations and Analytic Inequalities)

## Abstract

:
Here we present Hilfer-Polya, $ψ$-Hilfer Ostrowski and $ψ$-Hilfer-Hilbert-Pachpatte types fractional inequalities. They are univariate inequalities involving left and right Hilfer and $ψ$-Hilfer fractional derivatives. All estimates are with respect to norms $· p$, $1 ≤ p ≤ ∞$. At the end we provide applications.

## 1. Introduction

We are motivated by the following famous Polya’s integral inequality, see [1], (p. 62, [2]), [3] and (p. 83, [4]).
Theorem 1.
Let $f x$ be a differentiable and not identically a constant on $a , b$ with $f a = f b = 0 .$ Then there exists at least one point $ξ ∈ a , b$ such that
$f ′ ξ > 4 b − a 2 ∫ a b f x d x .$
We are inspired also by the related first fractional Polya Inequality, see Chapter 2, p. 9, [5].
In this article, we establish fractional integral inequalities using the Hilfer and $ψ$-Hilfer fractional derivatives. These are of Polya, Ostrowski and Hilbert-Pachpatte types.

## 2. Background

Let $− ∞ < a < b < ∞$, the left and right Riemann-Liouville fractional integrals of order $α ∈ C$ ($R α > 0$) are defined by
$I a + α f x = 1 Γ α ∫ a x x − t α − 1 f t d t ,$
$x > a$; where $Γ$ stands for the gamma function,
And
$I b − α f x = 1 Γ α ∫ x b t − x α − 1 f t d t ,$
$x < b$.
The Riemann-Liouville left and right fractional derivatives of order $α ∈ C$ ($R α ≥ 0$) are defined by
$Δ a + α y x = d d x n I a + n − α y x = 1 Γ n − α d d x n ∫ a x x − t n − α − 1 y t d t$
($n = R α$, $·$ means ceiling of the number; $x > a$)
$Δ b − α y x = − 1 n d d x n I b − n − α y x =$
$− 1 n Γ n − α d d x n ∫ x b t − x n − α − 1 y t d t$
($n = R α$; $x < b$), respectively, where $R α$ is the real part of $α$.
In particular, when $α = n ∈ Z +$, then
$Δ a + 0 y x = Δ b − 0 y x = y x ;$
$Δ a + n y x = y n x , and Δ b − n y x = − 1 n y n x , n ∈ N ,$
see [6].
Let $α > 0$, $I = a , b ⊂ R$, f an integrable function defined on I and $ψ ∈ C 1 I$ an increasing function such that $ψ ′ x ≠ 0$, for all $x ∈ I$. Left fractional integrals and left Riemann-Liouville fractional derivatives of a function f with respect to another function $ψ$ are defined as ([6,7])
$I a + α , ψ f x = 1 Γ α ∫ a x ψ ′ t ψ x − ψ t α − 1 f t d t ,$
and
$Δ a + α , ψ f x = 1 ψ ′ x d d x n I a + n − α , ψ f x =$
$1 Γ n − α 1 ψ ′ x d d x n ∫ a x ψ ′ t ψ x − ψ t n − α − 1 f t d t ,$
respectively, where $n = α$.
Similarly, we define the right ones:
$I b − α , ψ f x = 1 Γ α ∫ x b ψ ′ t ψ t − ψ x α − 1 f t d t ,$
and
$Δ b − α , ψ f x = − 1 ψ ′ x d d x n I b − n − α , ψ f x =$
$1 Γ n − α − 1 ψ ′ x d d x n ∫ x b ψ ′ t ψ t − ψ x n − α − 1 f t d t .$
The following semigroup property holds; if $α , β > 0$, $f ∈ C I$, then
$I a + α , ψ I a + β , ψ f = I a + α + β , ψ f and I b − α , ψ I b − β , ψ f = I b − α + β , ψ f .$
Next let again $α > 0$, $n = α$, $I = a , b$, $f , ψ ∈ C n I : ψ$ is increasing and $ψ ′ x ≠ 0$, for all $x ∈ I$. The left $ψ$-Caputo fractional derivative of f of order $α$ is given by ([8])
$C D a + α , ψ f x = I a + n − α , ψ 1 ψ ′ x d d x n f x ,$
and the right $ψ$-Caputo fractional derivative ([8])
$C D b − α , ψ f x = I b − n − α , ψ − 1 ψ ′ x d d x n f x .$
We set
$f ψ n x : = f ψ n f x : = 1 ψ ′ x d d x n f x .$
Clearly, when $α = m ∈ N$ we have
$C D a + α , ψ f x = f ψ m x and C D b − α , ψ f x = − 1 m f ψ m x ,$
and if $α ∉ N$, then
$C D a + α , ψ f x = 1 Γ n − α ∫ a x ψ ′ t ψ x − ψ t n − α − 1 f ψ n t d t ,$
and
$C D b − α , ψ f x = − 1 n Γ n − α ∫ x b ψ ′ t ψ t − ψ x n − α − 1 f ψ n t d t .$
If $ψ x = x$, then we get the usual left and right Caputo fractional derivatives
$C D a + m f x = f m x , C D b − m f x = − 1 m f m x ,$
for $m ∈ N$, and ($α ∉ N$)
$D ∗ a α f x = C D a + α f x = 1 Γ n − α ∫ a x x − t n − α − 1 f n t d t ,$
$D b − α x = C D b − α f x = − 1 n Γ n − α ∫ x b t − x n − α − 1 f n t d t .$
Also we set
$C D a + 0 , ψ f x = C D b − 0 , ψ f x = f x .$
Next we will deal with the $ψ$-Hilfer fractional derivative.
Definition 1.
([9]) Let $n − 1 < α < n$, $n ∈ N$, $I = a , b ⊂ R$ and $f , ψ ∈ C n a , b$, ψ is increasing and $ψ ′ x ≠ 0$, for all $x ∈ I$. The ψ-Hilfer fractional derivative (left-sided and right-sided) $H D a + b − α , β ; ψ f$ of order α and type $0 ≤ β ≤ 1$, respectively, are defined by
$H D a + α , β ; ψ f x = I a + β n − α ; ψ 1 ψ ′ x d d x n I a + 1 − β n − α ; ψ f x ,$
and
$H D b − α , β ; ψ f x = I b − β n − α ; ψ − 1 ψ ′ x d d x n I b − 1 − β n − α ; ψ f x , x ∈ a , b .$
The original Hilfer fractional derivatives ([10]) come from $ψ x = x$, and are denoted by $H D a + α , β f x$ and $H D b − α , β f x$.
When $β = 0$, we get Riemann-Liouville fractional derivatives, while when $β = 1$ we have Caputo type fractional derivatives.
We define $γ = α + β n − α$. We notice that $n − 1 < α ≤ α + β n − α ≤ α + n − α = n$, hence $γ = n$. We can easily write that ([9])
$H D a + α , β ; ψ f x = I a + γ − α ; ψ Δ a + γ ; ψ f x ,$
and
$H D b − α , β ; ψ f x = I b − γ − α ; ψ Δ b − γ ; ψ f x , x ∈ a , b .$
We have that ([9])
$Δ a + γ , ψ f x = 1 ψ ′ x d d x n I a + 1 − β n − α ; ψ f x ,$
and
$Δ b − γ , ψ f x = − 1 ψ ′ x d d x n I b − 1 − β n − α ; ψ f x .$
In particular, when $0 < α < 1$ and $0 ≤ β ≤ 1$; $γ = α + β 1 − α$, we have that
$H D a + α , β ; ψ f x = 1 Γ γ − α ∫ a x ψ ′ t ψ x − ψ t γ − α − 1 Δ a + γ ; ψ f t d t ,$
and
$H D b − α , β ; ψ f x = 1 Γ γ − α ∫ x b ψ ′ t ψ t − ψ x γ − α − 1 Δ b − γ ; ψ f t d t ,$
$x ∈ a , b .$
Remark 1.
([9]) Let $μ = n 1 − β + β α$, then $μ = n .$
Assume that $g x = I a + 1 − β n − α ; ψ f x ∈ C n a , b$, we have that
$H D a + α , β ; ψ f x = I a + n − μ ; ψ 1 ψ ′ x d d x n g x .$
Thus,
$H D a + α , β ; ψ f = C D a + μ ; ψ g x = C D a + μ ; ψ I a + 1 − β n − α ; ψ f x .$
Assume that $w x = I b − 1 − β n − α ; ψ f x ∈ C n a , b$. Hence
$H D b − α , β ; ψ f x = I b − β n − α ; ψ − 1 ψ ′ x d d x n w x = I b − n − μ ; ψ − 1 ψ ′ x d d x n w x .$
Thus,
$H D b − α , β ; ψ f = C D b − μ ; ψ w x = C D b − μ ; ψ I b − 1 − β n − α ; ψ f x .$
We mention the simplified $ψ$-Hilfer fractional Taylor formulae:
Theorem 2.
(see also [9]) Let $ψ , f ∈ C n a , b$, with ψ being increasing such that $ψ ′ x ≠ 0$ over $a , b$, where $n − 1 < α < n$, $0 ≤ β ≤ 1$, and $γ = α + β n − α$, $x ∈ a , b$. Then
$f x − ∑ k = 1 n − 1 ψ x − ψ a γ − k Γ γ − k + 1 f ψ n − k I a + 1 − β n − α ; ψ f a =$
$1 Γ α ∫ a x ψ ′ t ψ x − ψ t α − 1 H D a + α , β ; ψ f t d t ,$
and
$f x − ∑ k = 1 n − 1 − 1 k ψ b − ψ x γ − k Γ γ − k + 1 f ψ n − k I b − 1 − β n − α ; ψ f b =$
$1 Γ α ∫ x b ψ ′ t ψ t − ψ x α − 1 H D b − α , β ; ψ f t d t .$
Here notice that $I a + 1 − β n − α ; ψ f a = I b − 1 − β n − α ; ψ f b = 0$.
We also mention the following alternative $ψ$-Hilfer fractional Taylor formulae:
Theorem 3.
([11]) Let $f , ψ ∈ C n a , b$, with ψ being increasing, $ψ ′ x ≠ 0$ over $a , b ⊂ R ,$$α > 0 : α = n$, $0 ≤ β ≤ 1$, $μ = n 1 − β + β α$. Assume that $g x = I a + 1 − β n − α ; ψ f x ,$$w x = I b − 1 − β n − α ; ψ f x ∈ C n a , b$.
Then
(1)
$I a + μ ; ψ H D a + α , β ; ψ f x = g x − ∑ k = 0 n − 1 g ψ k a k ! ψ x − ψ a k ,$
where
$g ψ k x = 1 ψ ′ x d d x k g x , k = 0 , 1 , … , n − 1 ,$
and
(2)
$I b − μ ; ψ H D b − α , β ; ψ f x = w x − ∑ k = 0 n − 1 − 1 k w ψ k b k ! ψ b − ψ x k ,$
where
$w ψ k x = 1 ψ ′ x d d x k w x , k = 0 , 1 , … , n − 1 ; x ∈ a , b .$
Next we list two Hilfer fractional derivatives representation formulae:
Theorem 4.
([11]) Let $α > 0$, $α ∉ N$, $α = n$, $0 < β < 1$; $f ∈ C n a , b$, $a , b ⊂ R$; and set $γ = α + β n − α$. Assume further that $Δ a + γ f ∈ C a , b : Δ a + γ − j f a = 0$, for $j = 1 , … , n$. Let also $α ¯ > 0 : α ¯ = n ¯$, with $γ ¯ = α ¯ + β n ¯ − α ¯$, and assume that $α > α ¯$ and $γ > γ ¯$. Then
$H D a + α ¯ , β f x = 1 Γ α − α ¯ ∫ a x x − t α − α ¯ − 1 H D a + α , β f t d t ,$
$x ∈ a , b ,$
Furthermore, $H D a + α ¯ , β f ∈ A C a , b$ (absolutely continuous functions) if $α − α ¯ ≥ 1$ and $H D a + α ¯ , β f ∈ C a , b$ if $α − α ¯ ∈ 0 , 1$.
Theorem 5.
([11]) Let $α > 0$, $α ∉ N$, $α = n$, $0 < β < 1$; $f ∈ C n a , b$, $a , b ⊂ R$; and set $γ = α + β n − α$. Assume further that $Δ b − γ f ∈ C a , b : Δ b − γ − j f b = 0$, $j = 1 , … , n$. Let also $α ¯ > 0 : α ¯ = n ¯$, with $γ ¯ = α ¯ + β n ¯ − α ¯$, and assume that $α > α ¯$ and $γ > γ ¯$. Then
$H D b − α ¯ , β f x = 1 Γ α − α ¯ ∫ x b t − x α − α ¯ − 1 H D b − α , β f t d t ,$
$x ∈ a , b ,$
Furthermore, $H D b − α ¯ , β f ∈ A C a , b$ if $α − α ¯ ≥ 1$ and $H D b − α ¯ , β f ∈ C a , b$ if $α − α ¯ ∈ 0 , 1$.

## 3. Main Results

We present the following Hilfer-Polya type fractional inequalities:
Theorem 6.
Let $α > 0$, $α ∉ N$, $α = n$, $0 < β < 1$; $f ∈ C n a , b$, $a , b ⊂ R$; and set $γ = α + β n − α$. Assume further that $Δ a + γ f ∈ C a , b : Δ a + γ − j f a = 0 ,$ for $j = 1 , … , n ;$ and $Δ b − γ f ∈ C a , b : Δ b − γ − j f b = 0$, $j = 1 , … , n$. Let also $α ¯ > 0 : α ¯ = n ¯$, with $γ ¯ = α ¯ + β n ¯ − α ¯$, and assume that $α > α ¯$ and $γ > γ ¯$.
Set
$H D α ¯ , β f x : = H D a + α ¯ , β f x , x ∈ a , a + b 2 , H D b − α ¯ , β f x , x ∈ a + b 2 , b ,$
and
$M 1 : = max H D a + α , β f ∞ , a , a + b 2 , H D b − α , β f ∞ , a + b 2 , b .$
Then
$∫ a b H D α ¯ , β f x d x ≤ ∫ a b H D α ¯ , β f x d x ≤ M 1 b − a α − α ¯ + 1 2 α − α ¯ Γ α − α ¯ + 2 .$
Proof.
From (33) we have
$H D a + α ¯ , β f x = 1 Γ α − α ¯ ∫ a x x − t α − α ¯ − 1 H D a + α , β f t d t ,$
$x ∈ a , a + b 2 .$
By (34), we get
$H D b − α ¯ , β f x = 1 Γ α − α ¯ ∫ x b t − x α − α ¯ − 1 H D b − α , β f t d t ,$
$x ∈ a + b 2 , b .$
We derive that
$H D a + α ¯ , β f x ≤ H D a + α , β f ∞ , a , a + b 2 Γ α − α ¯ + 1 x − a α − α ¯ ,$
$x ∈ a , a + b 2 ,$ and similarly,
$H D b − α ¯ , β f x ≤ H D b − α , β f ∞ , a + b 2 , b Γ α − α ¯ + 1 b − x α − α ¯ ,$
$x ∈ a + b 2 , b .$
We notice that:
$∫ a b H D α ¯ , β f x d x = ∫ a a + b 2 H D α ¯ , β f x d x + ∫ a + b 2 b H D α ¯ , β f x d x =$
$∫ a a + b 2 H D a + α ¯ , β f x d x + ∫ a + b 2 b H D b − α ¯ , β f x d x .$
We further derive that
$∫ a a + b 2 H D a + α ¯ , β f x d x ≤ H D a + α , β f ∞ , a , a + b 2 Γ α − α ¯ + 1 ∫ a a + b 2 x − a α − α ¯ d x =$
$H D a + α , β f ∞ , a , a + b 2 Γ α − α ¯ + 2 a + b 2 − a α − α ¯ + 1 = H D a + α , β f ∞ , a , a + b 2 Γ α − α ¯ + 2 b − a 2 α − α ¯ + 1 .$
That is, it holds
$∫ a a + b 2 H D a + α ¯ , β f x d x ≤ H D a + α , β f ∞ , a , a + b 2 Γ α − α ¯ + 2 b − a 2 α − α ¯ + 1 .$
Similarly, it holds
$∫ a + b 2 b H D b − α ¯ , β f x d x ≤ H D b − α , β f ∞ , a + b 2 , b Γ α − α ¯ + 2 b − a 2 α − α ¯ + 1 .$
Therefore, we obtain
$∫ a b H D α ¯ , β f x d x ≤ ∫ a b H D α ¯ , β f x d x =$
$∫ a a + b 2 H D a + α ¯ , β f x d x + ∫ a + b 2 b H D b − α ¯ , β f x d x ≤$
$H D a + α , β f ∞ , a , a + b 2 Γ α − α ¯ + 2 b − a 2 α − α ¯ + 1 + H D b − α , β f ∞ , a + b 2 , b Γ α − α ¯ + 2 b − a 2 α − α ¯ + 1 =$
$b − a 2 α − α ¯ + 1 Γ α − α ¯ + 2 H D a + α , β f ∞ , a , a + b 2 + H D b − α , β f ∞ , a + b 2 , b ≤$
$2 M 1 b − a α − α ¯ + 1 2 α − α ¯ + 1 Γ α − α ¯ + 2 = M 1 b − a α − α ¯ + 1 2 α − α ¯ Γ α − α ¯ + 2 .$
□
We continue with the $L 1$-variant:
Theorem 7.
All as in Theorem 6 with $α − α ¯ > 1$ (i.e., $α > α ¯ + 1$). Call
$M 2 : = max H D a + α , β f 1 , a , a + b 2 , H D b − α , β f 1 , a + b 2 , b .$
Then
$∫ a b H D α ¯ , β f x d x ≤ ∫ a b H D α ¯ , β f x d x ≤ M 2 b − a α − α ¯ 2 α − α ¯ − 1 Γ α − α ¯ + 1 .$
Proof.
By (38) we have
$H D a + α ¯ , β f x ≤ 1 Γ α − α ¯ ∫ a x x − t α − α ¯ − 1 H D a + α , β f t d t ≤$
$x − a α − α ¯ − 1 Γ α − α ¯ H D a + α , β f 1 , a , a + b 2 ,$
$x ∈ a , a + b 2 .$
Similarly, from (39) we find that
$H D b − α ¯ , β f x ≤ b − x α − α ¯ − 1 Γ α − α ¯ H D b − α , β f 1 , a + b 2 , b ,$
$x ∈ a + b 2 , b .$
Furthermore, we obtain
$∫ a a + b 2 H D a + α ¯ , β f x d x ≤ H D a + α , β f 1 , a , a + b 2 Γ α − α ¯ ∫ a a + b 2 x − a α − α ¯ − 1 d x =$
$H D a + α , β f 1 , a , a + b 2 Γ α − α ¯ + 1 b − a 2 α − α ¯ .$
Similarly, we derive
$∫ a + b 2 b H D b − α ¯ , β f x d x ≤ H D b − α , β f 1 , a + b 2 , b Γ α − α ¯ + 1 b − a 2 α − α ¯ .$
Therefore we obtain
$∫ a b H D α ¯ , β f x d x ≤ ∫ a b H D α ¯ , β f x d x =$
$∫ a a + b 2 H D a + α ¯ , β f x d x + ∫ a + b 2 b H D b − α ¯ , β f x d x ≤$
$H D a + α , β f 1 , a , a + b 2 + H D b − α , β f 1 , a + b 2 , b Γ α − α ¯ + 1 b − a 2 α − α ¯ ≤$
$2 M 2 Γ α − α ¯ + 1 b − a α − α ¯ 2 α − α ¯ = M 2 Γ α − α ¯ + 1 b − a α − α ¯ 2 α − α ¯ − 1 .$
□
Next comes the $L q$-variant of Hilfer-Polya fractional inequality:
Theorem 8.
All as in Theorem 6 with $α − α ¯ > 1 q$, where $p , q > 1 : 1 p + 1 q = 1$. Call
$M 3 : = max H D a + α , β f q , a , a + b 2 , H D b − α , β f q , a + b 2 , b .$
Then
$∫ a b H D α ¯ , β f x d x ≤ ∫ a b H D α ¯ , β f x d x ≤$
$M 3 Γ α − α ¯ p α − α ¯ − 1 + 1 1 p α − α ¯ + 1 p b − a α − α ¯ + 1 p 2 α − α ¯ − 1 q .$
Proof.
By (38) we have
$H D a + α ¯ , β f x ≤ 1 Γ α − α ¯ ∫ a x x − t α − α ¯ − 1 H D a + α , β f t d t ≤$
$1 Γ α − α ¯ ∫ a x x − t p α − α ¯ − 1 d t 1 p H D a + α , β f q , a , a + b 2 =$
$x − a α − α ¯ − 1 q Γ α − α ¯ p α − α ¯ − 1 + 1 1 p H D a + α , β f q , a , a + b 2 ,$
$x ∈ a , a + b 2$, with $α − α ¯ > 1 q .$
And, by (39), similarly we derive
$H D b − α ¯ , β f x ≤ b − x α − α ¯ − 1 q Γ α − α ¯ p α − α ¯ − 1 + 1 1 p H D b − α , β f q , a + b 2 , b ,$
$x ∈ a + b 2 , b$, with $α − α ¯ > 1 q .$
Consequently, we obtain that
$∫ a a + b 2 H D a + α ¯ , β f x d x ≤ H D a + α , β f q , a , a + b 2 Γ α − α ¯ p α − α ¯ − 1 + 1 1 p ∫ a a + b 2 x − a α − α ¯ − 1 q d x =$
$H D a + α , β f q , a , a + b 2 Γ α − α ¯ p α − α ¯ − 1 + 1 1 p α − α ¯ + 1 p b − a 2 α − α ¯ + 1 p .$
Similarly, we derive
$∫ a + b 2 b H D b − α ¯ , β f x d x ≤ H D b − α , β f q , a + b 2 , b Γ α − α ¯ p α − α ¯ − 1 + 1 1 p α − α ¯ + 1 p b − a 2 α − α ¯ + 1 p .$
Therefore, we obtain
$∫ a b H D α ¯ , β f x d x ≤ ∫ a b H D α ¯ , β f x d x =$
$H D a + α , β f q , a , a + b 2 + H D b − α , β f q , a + b 2 , b Γ α − α ¯ p α − α ¯ − 1 + 1 1 p α − α ¯ + 1 p b − a 2 α − α ¯ + 1 p ≤$
$M 3 Γ α − α ¯ p α − α ¯ − 1 + 1 1 p α − α ¯ + 1 p b − a α − α ¯ + 1 p 2 α − α ¯ − 1 q ,$
proving the claim. □
Next come $ψ$-Hilfer-Ostrowski type inequalities for several functions involved.
For basic $ψ$-Hilfer-Ostrowski type inequalities involving one function see [11].
We make
Remark 2.
Our setting here follows: Let $f i ∈ C n a , b$, $α ∉ N$, $n = α$, $α > 0$; $i = 1 , … , r ∈ N − { 1 }$, $x 0 ∈ a , b$. Assume that $g 1 i x = I x 0 + 1 − β n − α ; ψ f i x ∈ C n x 0 , b$ and $w 1 i x = I x 0 − 1 − β n − α ; ψ f i x ∈ C n a , x 0$, for all $i = 1 , … , r .$
Define
$φ i x 0 x : = g 1 i x , x ∈ x 0 , b w 1 i x , x ∈ [ a , x 0 ) .$
Notice that if $β = 1$, we get $g 1 i x 0 = w 1 i x 0 = φ i x 0 x 0 = f i x 0$, all $i = 1 , … , r .$
In general, for $f ∈ C a , b$ we have
$I a + α , ψ f x ≤ 1 Γ α ∫ a x ψ ′ t ψ x − ψ t α − 1 f t d t ≤$
$f ∞ , a , b Γ α + 1 ψ x − ψ a α , ∀ x ∈ a , b .$
Hence $I a + α , ψ f a = 0 .$
Similalry, we have
$I b − α , ψ f x ≤ 1 Γ α ∫ x b ψ ′ t ψ t − ψ x α − 1 f t d t ≤$
$f ∞ , a , b Γ α + 1 ψ b − ψ x α , ∀ x ∈ a , b .$
That is $I b − α , ψ f b = 0 .$
So when $0 ≤ β < 1$, by the above we obtain $g 1 i x 0 = w 1 i x 0 = φ i x 0 x 0 = 0$, for all $i = 1 , … , r .$
Thus, it is always true that $g 1 i x 0 = w 1 i x 0$, $i = 1 , … , r .$
We present
Theorem 9.
Let $ψ , f i ∈ C n a , b$, $α ∉ N$, $n = α$, $α > 0$; $i = 1 , … , r ∈ N − { 1 }$, $x 0 ∈ a , b$. Here ψ is increasing, $ψ ′ x ≠ 0$ over $a , b ⊂ R$, $0 ≤ β ≤ 1$, $μ = n 1 − β + β α$. Assume that $g 1 i x = I x 0 + 1 − β n − α ; ψ f i x ∈ C n x 0 , b$ and $w 1 i x = I x 0 − 1 − β n − α ; ψ f i x ∈ C n a , x 0$, for all $i = 1 , … , r ,$ and $φ i x 0 x$ is as in (61). Assume also that $g 1 i ψ k x 0 = w 1 i ψ k x 0 = 0$, for all $k = 1 , … , n − 1 .$
Then
(1)
$θ ψ f 1 , … , f r x 0 : =$
$r ∫ a b ∏ λ = 1 r φ λ x 0 x d x − ∑ i = 1 r φ i x 0 x 0 ∫ a b ∏ j = 1 j ≠ i r φ j x 0 x d x =$
$∑ i = 1 r ∫ a x 0 ∏ j = 1 j ≠ i r φ j x 0 x I x 0 − μ ; ψ H D x 0 − α , β ; ψ f i x d x +$
$∫ x 0 b ∏ j = 1 j ≠ i r φ j x 0 x I x 0 + μ ; ψ H D x 0 + α , β ; ψ f i x d x ,$
and in case of $0 ≤ β < 1$, we have that
$θ ψ f 1 , … , f r x 0 = r ∫ a b ∏ λ = 1 r φ λ x 0 x d x ,$
(2) furthermore, it holds
$θ ψ f 1 , … , f r x 0 ≤ 1 Γ μ + 1$
$∑ i = 1 r H D x 0 − α , β ; ψ f i ∞ , a , x 0 ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 ψ x 0 − ψ a μ +$
$∑ i = 1 r H D x 0 + α , β ; ψ f i ∞ , x 0 , b ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b ψ b − ψ x 0 μ .$
It follows the $L 1$-variant.
Theorem 10.
All as in Theorem 9, with $α > 1$. Then
$θ ψ f 1 , … , f r x 0 ≤ 1 Γ μ$
$∑ i = 1 r H D x 0 − α , β ; ψ f i L 1 a , x 0 , ψ ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 ψ x 0 − ψ a μ − 1 +$
$∑ i = 1 r H D x 0 + α , β ; ψ f i L 1 x 0 , b , ψ ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b ψ b − ψ x 0 μ − 1 .$
Next we have the $L q$-variant.
Theorem 11.
All as in Theorem 9. Let also $p , q > 1 : 1 p + 1 q = 1$ with $α > 1 q$. Then
$θ ψ f 1 , … , f r x 0 ≤ 1 Γ μ p μ − 1 + 1 1 p$
$∑ i = 1 r H D x 0 − α , β ; ψ f i L q a , x 0 , ψ ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 ψ x 0 − ψ a μ − 1 q +$
$∑ i = 1 r H D x 0 + α , β ; ψ f i L q x 0 , b , ψ ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b ψ b − ψ x 0 μ − 1 q .$
Proof of Theorems 9–11.
By Theorem 3 we have
$g 1 i x − g 1 i x 0 = I x 0 + μ ; ψ H D x 0 + α , β ; ψ f i x , ∀ x ∈ x 0 , b , and w 1 i x − w 1 i x 0 = I x 0 − μ ; ψ H D x 0 − α , β ; ψ f i x , ∀ x ∈ a , x 0 ,$
for all $i = 1 , … , r .$
That is
$φ i x 0 x − φ i x 0 x 0 = I x 0 + μ ; ψ H D x 0 + α , β ; ψ f i x , ∀ x ∈ x 0 , b , and φ i x 0 x − φ i x 0 x 0 = I x 0 − μ ; ψ H D x 0 − α , β ; ψ f i x , ∀ x ∈ [ a , x 0 ) ,$
for all $i = 1 , … , r .$
Multiplying (70) by $∏ j = 1 j ≠ i r φ j x 0 x$ we get, respectively,
$∏ λ = 1 r φ λ x 0 x − ∏ j = 1 j ≠ i r φ j x 0 x φ i x 0 x 0 = ∏ j = 1 j ≠ i r φ j x 0 x I x 0 + μ ; ψ H D x 0 + α , β ; ψ f i x ,$
$x ∈ x 0 , b ,$
And
$∏ λ = 1 r φ λ x 0 x − ∏ j = 1 j ≠ i r φ j x 0 x φ i x 0 x 0 = ∏ j = 1 j ≠ i r φ j x 0 x I x 0 − μ ; ψ H D x 0 − α , β ; ψ f i x ,$
$x ∈ [ a , x 0 ) ,$ for all $i = 1 , … , r .$
Adding (71) and (72), separately, we obtain
$r ∏ λ = 1 r φ λ x 0 x − ∑ i = 1 r ∏ j = 1 j ≠ i r φ j x 0 x φ i x 0 x 0 =$
$∑ i = 1 r ∏ j = 1 j ≠ i r φ j x 0 x I x 0 + μ ; ψ H D x 0 + α , β ; ψ f i x ,$
$x ∈ x 0 , b ,$
$r ∏ λ = 1 r φ λ x 0 x − ∑ i = 1 r ∏ j = 1 j ≠ i r φ j x 0 x φ i x 0 x 0 =$
$∑ i = 1 r ∏ j = 1 j ≠ i r φ j x 0 x I x 0 − μ ; ψ H D x 0 − α , β ; ψ f i x ,$
$x ∈ [ a , x 0 ) .$
Next integrate (73) and (74) with respect to $x ∈ a , b .$ We have
$r ∫ x 0 b ∏ λ = 1 r φ λ x 0 x d x − ∑ i = 1 r φ i x 0 x 0 ∫ x 0 b ∏ j = 1 j ≠ i r φ j x 0 x d x =$
$∑ i = 1 r ∫ x 0 b ∏ j = 1 j ≠ i r φ j x 0 x I x 0 + μ ; ψ H D x 0 + α , β ; ψ f i x d x ,$
and
$r ∫ a x 0 ∏ λ = 1 r φ λ x 0 x d x − ∑ i = 1 r φ i x 0 x 0 ∫ a x 0 ∏ j = 1 j ≠ i r φ j x 0 x d x =$
$∑ i = 1 r ∫ a x 0 ∏ j = 1 j ≠ i r φ j x 0 x I x 0 − μ ; ψ H D x 0 − α , β ; ψ f i x d x .$
Finally adding (75) and (76) we obtain the useful and nice identity (64).
Identity (64) implies
$θ ψ f 1 , … , f r x 0 ≤ ∑ i = 1 r ∫ a x 0 ∏ j = 1 j ≠ i r φ j x 0 x I x 0 − μ ; ψ H D x 0 − α , β ; ψ f i x d x +$
$∫ x 0 b ∏ j = 1 j ≠ i r φ j x 0 x I x 0 + μ ; ψ H D x 0 + α , β ; ψ f i x d x =$
$∑ i = 1 r ∫ a x 0 ∏ j = 1 j ≠ i r φ j x 0 x 1 Γ μ$
$∫ x x 0 ψ ′ t ψ t − ψ x μ − 1 H D x 0 − α , β ; ψ f i t d t d x +$
$∫ x 0 b ∏ j = 1 j ≠ i r φ j x 0 x 1 Γ μ$
$∫ x 0 x ψ ′ t ψ x − ψ t μ − 1 H D x 0 + α , β ; ψ f i t d t d x ≤$
$1 Γ μ + 1 ∑ i = 1 r H D x 0 − α , β ; ψ f i ∞ , a , x 0 ∫ a x 0 ψ x 0 − ψ x μ ∏ j = 1 j ≠ i r φ j x 0 x d x$
$+ H D x 0 + α , β ; ψ f i ∞ , x 0 , b ∫ x 0 b ψ x − ψ x 0 μ ∏ j = 1 j ≠ i r φ j x 0 x d x ≤$
$1 Γ μ + 1 ∑ i = 1 r H D x 0 − α , β ; ψ f i ∞ , a , x 0 ψ x 0 − ψ a μ ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 +$
$H D x 0 + α , β ; ψ f i ∞ , x 0 , b ψ b − ψ x 0 μ ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b =$
$1 Γ μ + 1 ∑ i = 1 r H D x 0 − α , β ; ψ f i ∞ , a , x 0 ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 ψ x 0 − ψ a μ +$
$∑ i = 1 r H D x 0 + α , β ; ψ f i ∞ , x 0 , b ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b ψ b − ψ x 0 μ ,$
proving (66).
If $α ∉ N$ and $α > 1$, then $n = α > 1$, and $n − 1 ≥ 1 > β n − α$. Hence $n − β n − α > 1$ and $μ > 1$. So we have
$θ ψ f 1 , … , f r x 0 ≤ ∑ i = 1 r ∫ a x 0 ∏ j = 1 j ≠ i r φ j x 0 x 1 Γ μ$
$∫ x x 0 ψ ′ t ψ t − ψ x μ − 1 H D x 0 − α , β ; ψ f i t d t d x +$
$∫ x 0 b ∏ j = 1 j ≠ i r φ j x 0 x 1 Γ μ$
$∫ x 0 x ψ ′ t ψ x − ψ t μ − 1 H D x 0 + α , β ; ψ f i t d t d x ≤$
$1 Γ μ ∑ i = 1 r H D x 0 − α , β ; ψ f i L 1 a , x 0 , ψ ∫ a x 0 ψ x 0 − ψ x μ − 1 ∏ j = 1 j ≠ i r φ j x 0 x d x$
$+ H D x 0 + α , β ; ψ f i L 1 x 0 , b , ψ ∫ x 0 b ψ x − ψ x 0 μ − 1 ∏ j = 1 j ≠ i r φ j x 0 x d x ≤$
$1 Γ μ ∑ i = 1 r H D x 0 − α , β ; ψ f i L 1 a , x 0 , ψ ψ x 0 − ψ a μ − 1 ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 +$
$H D x 0 + α , β ; ψ f i L 1 x 0 , b , ψ ψ b − ψ x 0 μ − 1 ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b =$
$1 Γ μ ∑ i = 1 r H D x 0 − α , β ; ψ f i L 1 a , x 0 , ψ ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 ψ x 0 − ψ a μ − 1 +$
$∑ i = 1 r H D x 0 + α , β ; ψ f i L 1 x 0 , b , ψ ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b ψ b − ψ x 0 μ − 1 ,$
proving (67).
Let $α > 0$ with $α = n ∈ N$, and let $p , q > 1 : 1 p + 1 q = 1$, with $α > 1 q$. Clearly $n > 1 q$. Let $0 < β ≤ 1$, then $α β > β q$, furthermore $μ = n 1 − β + β α > β q + n 1 − β ≥ β q + 1 − β = β q + 1 p + 1 q − β p − β q = 1 q + 1 p 1 − β ≥ 1 q$. That is $μ > 1 q .$
From (81), by using Hölder’s inequality twice, we have
$θ ψ f 1 , … , f r x 0 ≤ 1 Γ μ$
$∑ i = 1 r ∫ a x 0 ∏ j = 1 j ≠ i r φ j x 0 x ψ x 0 − ψ x p μ − 1 + 1 p p μ − 1 + 1 1 p H D x 0 − α , β ; ψ f i L q a , x 0 , ψ d x +$
$∫ x 0 b ∏ j = 1 j ≠ i r φ j x 0 x ψ x − ψ x 0 p μ − 1 + 1 p p μ − 1 + 1 1 p H D x 0 + α , β ; ψ f i L q x 0 , b , ψ d x =$
$1 Γ μ p μ − 1 + 1 1 p$
$∑ i = 1 r ∫ a x 0 ∏ j = 1 j ≠ i r φ j x 0 x ψ x 0 − ψ x μ − 1 q H D x 0 − α , β ; ψ f i L q a , x 0 , ψ d x +$
$∫ x 0 b ∏ j = 1 j ≠ i r φ j x 0 x ψ x − ψ x 0 μ − 1 q H D x 0 + α , β ; ψ f i L q x 0 , b , ψ d x ≤$
$1 Γ μ p μ − 1 + 1 1 p$
$∑ i = 1 r H D x 0 − α , β ; ψ f i L q a , x 0 , ψ ψ x 0 − ψ a μ − 1 q ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 +$
$H D x 0 + α , β ; ψ f i L q x 0 , b , ψ ψ b − ψ x 0 μ − 1 q ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b =$
$1 Γ μ p μ − 1 + 1 1 p$
$∑ i = 1 r H D x 0 − α , β ; ψ f i L q a , x 0 , ψ ∏ j = 1 j ≠ i r φ j x 0 1 , a , x 0 ψ x 0 − ψ a μ − 1 q +$
$∑ i = 1 r H D x 0 + α , β ; ψ f i L q x 0 , b , ψ ∏ j = 1 j ≠ i r φ j x 0 1 , x 0 , b ψ b − ψ x 0 μ − 1 q ,$
proving (68). □
Next we present a $ψ$-Hilfer-Hilbert-Pachpatte left fractional inequality:
Theorem 12.
Let $i = 1 , 2 ;$$ψ i ,$$f i ∈ C n i a i , b i$, with $ψ i$ being strictly increasing over $a i , b i$, where $n i − 1 < α i < n i$, $0 ≤ β i ≤ 1$, and $γ i = α i + β i n i − α i$, $x i ∈ a i , b i$. Assume that $f i ψ i n i − k i I a i + 1 − β i n i − α i ; ψ i f i a i = 0$, for $k i = 1 , … , n i − 1$. Let also $p , q > 1 : 1 p + 1 q = 1$, such that $α 1 > 1 q$ and $α 2 > 1 p$. Then
$∫ a 1 b 1 ∫ a 2 b 2 f 1 x 1 f 2 x 2 d x 1 d x 2 ψ 1 x 1 − ψ 1 a 1 p α 1 − 1 + 1 p p α 1 − 1 | + 1 + ψ 2 x 2 − ψ 2 a 2 q α 2 − 1 + 1 q q α 2 − 1 + 1 ≤$
$b 1 − a 1 b 2 − a 2 Γ α 1 Γ α 2 H D a 1 + α 1 , β 1 ; ψ 1 f 1 L q a 1 , b 1 , ψ 1 H D a 2 + α 2 , β 2 ; ψ 2 f 2 L p a 2 , b 2 , ψ 2 .$
Proof.
By Theorem 2 we have
$f i x i = 1 Γ α i ∫ a i x i ψ i ′ t i ψ i x i − ψ i t i α i − 1 H D a i + α i , β i ; ψ i f i t i d t i ,$
$x i ∈ a i , b i$, $i = 1 , 2 .$
Then
$f i x i ≤ 1 Γ α i ∫ a i x i ψ i ′ t i ψ i x i − ψ i t i α i − 1 H D a i + α i , β i ; ψ i f i t i d t i ,$
$i = 1 , 2 ,$$x i ∈ a i , b i .$
By Hölder’s inequality we obtain
$f 1 x 1 ≤ 1 Γ α 1 ψ 1 x 1 − ψ 1 a 1 p α 1 − 1 + 1 p p α 1 − 1 + 1 1 p H D a 1 + α 1 , β 1 ; ψ 1 f 1 L q a 1 , b 1 , ψ 1 ,$
$x 1 ∈ a 1 , b 1 ,$ and
$f 2 x 2 ≤ 1 Γ α 2 ψ 2 x 2 − ψ 2 a 2 q α 2 − 1 + 1 q q α 2 − 1 + 1 1 q H D a 2 + α 2 , β 2 ; ψ 2 f 2 L p a 2 , b 2 , ψ 2 ,$
$x 2 ∈ a 2 , b 2 .$
Hence we have
$f 1 x 1 f 2 x 2 ≤ 1 Γ α 1 Γ α 2 p α 1 − 1 + 1 1 p q α 2 − 1 + 1 1 q$
$ψ 1 x 1 − ψ 1 a 1 p α 1 − 1 + 1 p ψ 2 x 2 − ψ 2 a 2 q α 2 − 1 + 1 q$
$H D a 1 + α 1 , β 1 ; ψ 1 f 1 L q a 1 , b 1 , ψ 1 H D a 2 + α 2 , β 2 ; ψ 2 f 2 L p a 2 , b 2 , ψ 2 ≤$
(using Young’s inequality for $a , b ≥ 0$, $a 1 p b 1 q ≤ a p + b q$)
$1 Γ α 1 Γ α 2 ψ 1 x 1 − ψ 1 a 1 p α 1 − 1 + 1 p p α 1 − 1 + 1 + ψ 2 x 2 − ψ 2 a 2 q α 2 − 1 + 1 q q α 2 − 1 + 1$
$H D a 1 + α 1 , β 1 ; ψ 1 f 1 L q a 1 , b 1 , ψ 1 H D a 2 + α 2 , β 2 ; ψ 2 f 2 L p a 2 , b 2 , ψ 2 ,$
$x i ∈ a i , b i$; $i = 1 , 2 .$
So far we have
$f 1 x 1 f 2 x 2 ψ 1 x 1 − ψ 1 a 1 p α 1 − 1 + 1 p p α 1 − 1 + 1 + ψ 2 x 2 − ψ 2 a 2 q α 2 − 1 + 1 q q α 2 − 1 + 1 ≤$
$H D a 1 + α 1 , β 1 ; ψ 1 f 1 L q a 1 , b 1 , ψ 1 H D a 2 + α 2 , β 2 ; ψ 2 f 2 L p a 2 , b 2 , ψ 2 Γ α 1 Γ α 2 ,$
$x i ∈ a i , b i$; $i = 1 , 2 .$
The denominator in (94) can be zero only when $x 1 = a 1$ and $x 2 = a 2$.
Therefore we obtain (87), by integrating (94) over $a 1 , b 1 × a 2 , b 2 .$ □
It follows the right side analog of last theorem.
Theorem 13.
Let $i = 1 , 2 ;$$ψ i ,$$f i ∈ C n i a i , b i$, with $ψ i$ being strictly increasing over $a i , b i$, where $n i − 1 < α i < n i$, $0 ≤ β i ≤ 1$, and $γ i = α i + β i n i − α i$, $x i ∈ a i , b i$. Let also $p , q > 1 : 1 p + 1 q = 1$, such that $α 1 > 1 q$ and $α 2 > 1 p$. Assume that $f i ψ i n i − k i I b i − 1 − β i n i − α i ; ψ i f i b i = 0$, for $k i = 1 , … , n i − 1$. Then
$∫ a 1 b 1 ∫ a 2 b 2 f 1 x 1 f 2 x 2 d x 1 d x 2 ψ 1 b 1 − ψ 1 x 1 p α 1 − 1 + 1 p p α 1 − 1 | + 1 + ψ 2 b 2 − ψ 2 x 2 q α 2 − 1 + 1 q q α 2 − 1 + 1 ≤$
$b 1 − a 1 b 2 − a 2 Γ α 1 Γ α 2 H D b 1 − α 1 , β 1 ; ψ 1 f 1 L q a 1 , b 1 , ψ 1 H D b 2 − α 2 , β 2 ; ψ 2 f 2 L p a 2 , b 2 , ψ 2 .$
Proof.
Similar to Theorem 12, by the use of (30). □
We continue with other Hilfer-Hilbert-Pachpatte fractional inequalities.
Theorem 14.
Let $i = 1 , 2 ;$$α i > 0$, $α i ∉ N$, $α i = n i ,$$0 < β i < 1 ,$$f i ∈ C n i a i , b i$, $a i , b i ⊂ R$ and set $γ i = α i + β i n i − α i$. Assume further that $Δ a i + γ i f i ∈ C a i , b i : Δ a i + γ i − j i f i a i = 0$, for $j i = 1 , … , n i$. Let also $α ¯ i > 0 : α ¯ i = n ¯ i$, with $γ ¯ i = α ¯ i + β i n ¯ i − α ¯ i$, and assume that $α i > α ¯ i$ and $γ i > γ ¯ i$. Furthermore, let $p , q > 1 : 1 p + 1 q = 1$, such that $α 1 > 1 q$ and $α 2 > 1 p$. Then
$∫ a 1 b 1 ∫ a 2 b 2 H D a 1 + α ¯ 1 , β 1 f 1 x 1 H D a 2 + α ¯ 2 , β 2 f 2 x 2 d x 1 d x 2 x 1 − a 1 p α 1 − α ¯ 1 − 1 + 1 p p α 1 − α ¯ 1 − 1 | + 1 + x 2 − a 2 q α 2 − α ¯ 2 − 1 + 1 q q α 2 − α ¯ 2 − 1 + 1 ≤$
$b 1 − a 1 b 2 − a 2 Γ α 1 − α ¯ 1 Γ α 2 − α ¯ 2 H D a 1 + α 1 , β 1 f 1 L q a 1 , b 1 H D a 2 + α 2 , β 2 f 2 L p a 2 , b 2 .$
Proof.
Similar to Theorem 12, by the use of Theorem 4. □
It follows
Theorem 15.
Let $i = 1 , 2 ;$$α i > 0$, $α i ∉ N$, $α i = n i ,$$0 < β i < 1 ,$$f i ∈ C n i a i , b i$, $a i , b i ⊂ R$ and set $γ i = α i + β i n i − α i$. Assume further that $Δ b i − γ i f i ∈ C a i , b i : Δ b i − γ i − j i f i b i = 0$, for $j i = 1 , … , n i$. Let also $α ¯ i > 0 : α ¯ i = n ¯ i$, with $γ ¯ i = α ¯ i + β i n ¯ i − α ¯ i$, and assume that $α i > α ¯ i$ and $γ i > γ ¯ i$. Furthermore, let $p , q > 1 : 1 p + 1 q = 1$, such that $α 1 > 1 q$ and $α 2 > 1 p$. Then
$∫ a 1 b 1 ∫ a 2 b 2 H D b 1 − α ¯ 1 , β 1 f 1 x 1 H D b 2 − α ¯ 2 , β 2 f 2 x 2 d x 1 d x 2 b 1 − x 1 p α 1 − α ¯ 1 − 1 + 1 p p α 1 − α ¯ 1 − 1 | + 1 + b 2 − x 2 q α 2 − α ¯ 2 − 1 + 1 q q α 2 − α ¯ 2 − 1 + 1 ≤$
$b 1 − a 1 b 2 − a 2 Γ α 1 − α ¯ 1 Γ α 2 − α ¯ 2 H D b 1 − α 1 , β 1 f 1 L q a 1 , b 1 H D b 2 − α 2 , β 2 f 2 L p a 2 , b 2 .$
Proof.
Similar to Theorem 12, by the use of Theorem 5. □
We finish with two applications:
Corollary 1.
All as in Theorem 12, with $ψ 1 x 1 = e x 1$, $ψ 2 x 2 = e x 2$. Then
$∫ a 1 b 1 ∫ a 2 b 2 f 1 x 1 f 2 x 2 d x 1 d x 2 e x 1 − e a 1 p α 1 − 1 + 1 p p α 1 − 1 | + 1 + e x 2 − e a 2 q α 2 − 1 + 1 q q α 2 − 1 + 1 ≤$
$b 1 − a 1 b 2 − a 2 Γ α 1 Γ α 2 H D a 1 + α 1 , β 1 ; e x 1 f 1 L q a 1 , b 1 , e x 1 H D a 2 + α 2 , β 2 ; e x 2 f 2 L p a 2 , b 2 , e x 2 .$
Proof.
By Theorem 12. □
Corollary 2.
All as in Theorem 13, with $a i , b i ⊂ 0 , + ∞$, $i = 1 , 2 ;$ and $ψ 1 x 1 = ln x 1$, $ψ 2 x 2 = ln x 2$. Then
$∫ a 1 b 1 ∫ a 2 b 2 f 1 x 1 f 2 x 2 d x 1 d x 2 ln b 1 x 1 p α 1 − 1 + 1 p p α 1 − 1 | + 1 + ln b 2 x 2 q α 2 − 1 + 1 q q α 2 − 1 + 1 ≤$
$b 1 − a 1 b 2 − a 2 Γ α 1 Γ α 2 H D b 1 − α 1 , β 1 ; ln x 1 f 1 L q a 1 , b 1 , ln x 1 H D b 2 − α 2 , β 2 ; ln x 2 f 2 L p a 2 , b 2 , ln x 2 .$
Proof.
By Theorem 13. □

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## Conflicts of Interest

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Anastassiou, G.A. Hilfer-Polya, ψ-Hilfer Ostrowski and ψ-Hilfer-Hilbert-Pachpatte Fractional Inequalities. Symmetry 2021, 13, 463. https://doi.org/10.3390/sym13030463

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Anastassiou GA. Hilfer-Polya, ψ-Hilfer Ostrowski and ψ-Hilfer-Hilbert-Pachpatte Fractional Inequalities. Symmetry. 2021; 13(3):463. https://doi.org/10.3390/sym13030463

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Anastassiou, George A. 2021. "Hilfer-Polya, ψ-Hilfer Ostrowski and ψ-Hilfer-Hilbert-Pachpatte Fractional Inequalities" Symmetry 13, no. 3: 463. https://doi.org/10.3390/sym13030463

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