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

Abstract

Here we present Hilfer-Polya, $\psi$-Hilfer Ostrowski and $\psi$-Hilfer-Hilbert-Pachpatte types fractional inequalities. They are univariate inequalities involving left and right Hilfer and $\psi$-Hilfer fractional derivatives. All estimates are with respect to norms ${∥·∥}_p$, $1\le p\le \infty$. 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 $f\left(x\right)$ be a differentiable and not identically a constant on $\left[a,b\right]$ with $f\left(a\right)=f\left(b\right)=0.$ Then there exists at least one point $\xi \in \left[a,b\right]$ such that

$$\left|f^{\prime }\left(\xi \right)\right|>\frac{4}{{\left(b-a\right)}^2}{\int }_a^{b}f\left(x\right)dx.$$

(1)

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 $\psi$-Hilfer fractional derivatives. These are of Polya, Ostrowski and Hilbert-Pachpatte types.

2. Background

Let $-\infty <a<b<\infty$, the left and right Riemann-Liouville fractional integrals of order $\alpha \in \mathbb{C}$ ($\mathcal{R}\left(\alpha \right)>0$) are defined by

$$\left(I_{a+}^{\alpha }f\right)\left(x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\left(x-t\right)}^{\alpha -1}f\left(t\right)dt,$$

(2)

$x>a$; where $\Gamma$ stands for the gamma function,

And

$$\left(I_{b-}^{\alpha }f\right)\left(x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\left(t-x\right)}^{\alpha -1}f\left(t\right)dt,$$

(3)

$x<b$.

The Riemann-Liouville left and right fractional derivatives of order $\alpha \in \mathbb{C}$ ($\mathcal{R}\left(\alpha \right)\ge 0$) are defined by

$$\left({\Delta }_{a+}^{\alpha }y\right)\left(x\right)={\left(\frac{d}{dx}\right)}^n\left(I_{a+}^{n-\alpha }y\right)\left(x\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{d}{dx}\right)}^n{\int }_a^x{\left(x-t\right)}^{n-\alpha -1}y\left(t\right)dt$$

(4)

($n=⌈\mathcal{R}\left(\alpha \right)⌉$, $⌈·⌉$ means ceiling of the number; $x>a$)

$$\left({\Delta }_{b-}^{\alpha }y\right)\left(x\right)={\left(-1\right)}^n{\left(\frac{d}{dx}\right)}^n\left(I_{b-}^{n-\alpha }y\right)\left(x\right)=$$

$$\frac{{\left(-1\right)}^n}{\Gamma \left(n-\alpha \right)}{\left(\frac{d}{dx}\right)}^n{\int }_x^b{\left(t-x\right)}^{n-\alpha -1}y\left(t\right)dt$$

(5)

($n=⌈\mathcal{R}\left(\alpha \right)⌉$; $x<b$), respectively, where $\mathcal{R}\left(\alpha \right)$ is the real part of $\alpha$.

In particular, when $\alpha =n\in {\mathbb{Z}}_{+}$, then

$$\left({\Delta }_{a+}^{0}y\right)\left(x\right)=\left({\Delta }_{b-}^{0}y\right)\left(x\right)=y\left(x\right);$$

$$\left({\Delta }_{a+}^{n}y\right)\left(x\right)=y^{\left(n\right)}\left(x\right),\mathrm{and}\left({\Delta }_{b-}^{n}y\right)\left(x\right)={\left(-1\right)}^n{y}^{\left(n\right)}\left(x\right),n\in \mathbb{N},$$

see [6].

Let $\alpha >0$, $I=\left[a,b\right]\subset \mathbb{R}$, f an integrable function defined on I and $\psi \in C^1\left(I\right)$ an increasing function such that ${\psi }^{\prime }\left(x\right)\ne 0$, for all $x\in I$. Left fractional integrals and left Riemann-Liouville fractional derivatives of a function f with respect to another function $\psi$ are defined as ([6,7])

$$I_{a+}^{\alpha ,\psi }f\left(x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\psi }^{\prime }\left(t\right){\left(\psi \left(x\right)-\psi \left(t\right)\right)}^{\alpha -1}f\left(t\right)dt,$$

(6)

and

$${\Delta }_{a+}^{\alpha ,\psi }f\left(x\right)={\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^n{I}_{a+}^{n-\alpha ,\psi }f\left(x\right)=$$

(7)

$$\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^n{\int }_a^x{\psi }^{\prime }\left(t\right){\left(\psi \left(x\right)-\psi \left(t\right)\right)}^{n-\alpha -1}f\left(t\right)dt,$$

respectively, where $n=⌈\alpha ⌉$.

Similarly, we define the right ones:

$$I_{b-}^{\alpha ,\psi }f\left(x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\psi }^{\prime }\left(t\right){\left(\psi \left(t\right)-\psi \left(x\right)\right)}^{\alpha -1}f\left(t\right)dt,$$

(8)

and

$${\Delta }_{b-}^{\alpha ,\psi }f\left(x\right)={\left(-\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^n{I}_{b-}^{n-\alpha ,\psi }f\left(x\right)=$$

$$\frac{1}{\Gamma \left(n-\alpha \right)}{\left(-\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^n{\int }_x^b{\psi }^{\prime }\left(t\right){\left(\psi \left(t\right)-\psi \left(x\right)\right)}^{n-\alpha -1}f\left(t\right)dt.$$

(9)

The following semigroup property holds; if $\alpha ,\beta >0$, $f\in C\left(I\right)$, then

$$I_{a+}^{\alpha ,\psi }{I}_{a+}^{\beta ,\psi }f=I_{a+}^{\alpha +\beta ,\psi }f\mathrm{and}{I}_{b-}^{\alpha ,\psi }{I}_{b-}^{\beta ,\psi }f=I_{b-}^{\alpha +\beta ,\psi }f.$$

Next let again $\alpha >0$, $n=⌈\alpha ⌉$, $I=\left[a,b\right]$, $f,\psi \in C^n\left(I\right):\psi$ is increasing and ${\psi }^{\prime }\left(x\right)\ne 0$, for all $x\in I$. The left $\psi$-Caputo fractional derivative of f of order $\alpha$ is given by ([8])

$${}^C{D}_{a+}^{\alpha ,\psi }f\left(x\right)=I_{a+}^{n-\alpha ,\psi }{\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^{n}f\left(x\right),$$

(10)

and the right $\psi$-Caputo fractional derivative ([8])

$${}^C{D}_{b-}^{\alpha ,\psi }f\left(x\right)=I_{b-}^{n-\alpha ,\psi }{\left(-\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^{n}f\left(x\right).$$

(11)

We set

$$f_{\psi }^{\left[n\right]}\left(x\right):=f_{\psi }^{\left(n\right)}f\left(x\right):={\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^{n}f\left(x\right).$$

(12)

Clearly, when $\alpha =m\in \mathbb{N}$ we have

$${}^C{D}_{a+}^{\alpha ,\psi }f\left(x\right)=f_{\psi }^{\left[m\right]}\left(x\right)\mathrm{and}{}^C{D}_{b-}^{\alpha ,\psi }f\left(x\right)={\left(-1\right)}^m{f}_{\psi }^{\left[m\right]}\left(x\right),$$

and if $\alpha \notin \mathbb{N}$, then

$${}^C{D}_{a+}^{\alpha ,\psi }f\left(x\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\int }_a^x{\psi }^{\prime }\left(t\right){\left(\psi \left(x\right)-\psi \left(t\right)\right)}^{n-\alpha -1}{f}_{\psi }^{\left[n\right]}\left(t\right)dt,$$

(13)

and

$${}^C{D}_{b-}^{\alpha ,\psi }f\left(x\right)=\frac{{\left(-1\right)}^n}{\Gamma \left(n-\alpha \right)}{\int }_x^b{\psi }^{\prime }\left(t\right){\left(\psi \left(t\right)-\psi \left(x\right)\right)}^{n-\alpha -1}{f}_{\psi }^{\left[n\right]}\left(t\right)dt.$$

(14)

If $\psi \left(x\right)=x$, then we get the usual left and right Caputo fractional derivatives

$${}^C{D}_{a+}^{m}f\left(x\right)=f^{\left(m\right)}\left(x\right),{}^C{D}_{b-}^{m}f\left(x\right)={\left(-1\right)}^m{f}^{\left(m\right)}\left(x\right),$$

for $m\in \mathbb{N}$, and ($\alpha \notin \mathbb{N}$)

$$D_{\ast a}^{\alpha }f\left(x\right)={}^C{D}_{a+}^{\alpha }f\left(x\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\int }_a^x{\left(x-t\right)}^{n-\alpha -1}{f}^{\left(n\right)}\left(t\right)dt,$$

(15)

$$D_{b-}^{\alpha }\left(x\right)={}^C{D}_{b-}^{\alpha }f\left(x\right)=\frac{{\left(-1\right)}^n}{\Gamma \left(n-\alpha \right)}{\int }_x^b{\left(t-x\right)}^{n-\alpha -1}{f}^{\left(n\right)}\left(t\right)dt.$$

(16)

Also we set

$${}^C{D}_{a+}^{0,\psi }f\left(x\right)={}^C{D}_{b-}^{0,\psi }f\left(x\right)=f\left(x\right).$$

Next we will deal with the $\psi$-Hilfer fractional derivative.

Definition 1.

([9]) Let $n-1<\alpha <n$, $n\in \mathbb{N}$, $I=\left[a,b\right]\subset \mathbb{R}$ and $f,\psi \in C^n\left(\left[a,b\right]\right)$, ψ is increasing and ${\psi }^{\prime }\left(x\right)\ne 0$, for all $x\in I$. The ψ-Hilfer fractional derivative (left-sided and right-sided) ${}^H{\mathbb{D}}_{a+\left(b-\right)}^{\alpha ,\beta ;\psi }f$ of order α and type $0\le \beta \le 1$, respectively, are defined by

$${}^H{\mathbb{D}}_{a+}^{\alpha ,\beta ;\psi }f\left(x\right)=I_{a+}^{\beta \left(n-\alpha \right);\psi }{\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^n{I}_{a+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right),$$

(17)

and

$${}^H{\mathbb{D}}_{b-}^{\alpha ,\beta ;\psi }f\left(x\right)=I_{b-}^{\beta \left(n-\alpha \right);\psi }{\left(-\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^n{I}_{b-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right),x\in \left[a,b\right].$$

(18)

The original Hilfer fractional derivatives ([10]) come from $\psi \left(x\right)=x$, and are denoted by ${}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f\left(x\right)$ and ${}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f\left(x\right)$.

When $\beta =0$, we get Riemann-Liouville fractional derivatives, while when $\beta =1$ we have Caputo type fractional derivatives.

We define $\gamma =\alpha +\beta \left(n-\alpha \right)$. We notice that $n-1<\alpha \le \alpha +\beta \left(n-\alpha \right)\le \alpha +n-\alpha =n$, hence $⌈\gamma ⌉=n$. We can easily write that ([9])

$${}^H{\mathbb{D}}_{a+}^{\alpha ,\beta ;\psi }f\left(x\right)=I_{a+}^{\gamma -\alpha ;\psi }{\Delta }_{a+}^{\gamma ;\psi }f\left(x\right),$$

(19)

and

$${}^H{\mathbb{D}}_{b-}^{\alpha ,\beta ;\psi }f\left(x\right)=I_{b-}^{\gamma -\alpha ;\psi }{\Delta }_{b-}^{\gamma ;\psi }f\left(x\right),x\in \left[a,b\right].$$

(20)

We have that ([9])

$${\Delta }_{a+}^{\gamma ,\psi }f\left(x\right)={\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^n{I}_{a+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right),$$

(21)

and

$${\Delta }_{b-}^{\gamma ,\psi }f\left(x\right)={\left(-\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^n{I}_{b-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right).$$

(22)

In particular, when $0<\alpha <1$ and $0\le \beta \le 1$; $\gamma =\alpha +\beta \left(1-\alpha \right)$, we have that

$${}^H{\mathbb{D}}_{a+}^{\alpha ,\beta ;\psi }f\left(x\right)=\frac{1}{\Gamma \left(\gamma -\alpha \right)}{\int }_a^x{\psi }^{\prime }\left(t\right){\left(\psi \left(x\right)-\psi \left(t\right)\right)}^{\gamma -\alpha -1}{\Delta }_{a+}^{\gamma ;\psi }f\left(t\right)dt,$$

(23)

and

$${}^H{\mathbb{D}}_{b-}^{\alpha ,\beta ;\psi }f\left(x\right)=\frac{1}{\Gamma \left(\gamma -\alpha \right)}{\int }_x^b{\psi }^{\prime }\left(t\right){\left(\psi \left(t\right)-\psi \left(x\right)\right)}^{\gamma -\alpha -1}{\Delta }_{b-}^{\gamma ;\psi }f\left(t\right)dt,$$

(24)

$x\in \left[a,b\right].$

Remark 1.

([9]) Let $\mu =n\left(1-\beta \right)+\beta \alpha$, then $⌈\mu ⌉=n.$

Assume that $g\left(x\right)=I_{a+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right)\in C^n\left(\left[a,b\right]\right)$, we have that

$${}^H{\mathbb{D}}_{a+}^{\alpha ,\beta ;\psi }f\left(x\right)=I_{a+}^{n-\mu ;\psi }{\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^{n}g\left(x\right).$$

(25)

Thus,

$${}^H{\mathbb{D}}_{a+}^{\alpha ,\beta ;\psi }f={}^C{D}_{a+}^{\mu ;\psi }g\left(x\right)={}^C{D}_{a+}^{\mu ;\psi }\left[I_{a+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right)\right].$$

(26)

Assume that $w\left(x\right)=I_{b-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right)\in C^n\left(\left[a,b\right]\right)$. Hence

$${}^H{\mathbb{D}}_{b-}^{\alpha ,\beta ;\psi }f\left(x\right)=I_{b-}^{\beta \left(n-\alpha \right);\psi }{\left(-\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^{n}w\left(x\right)=I_{b-}^{n-\mu ;\psi }{\left(-\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^{n}w\left(x\right).$$

(27)

Thus,

$${}^H{\mathbb{D}}_{b-}^{\alpha ,\beta ;\psi }f={}^C{D}_{b-}^{\mu ;\psi }w\left(x\right)={}^C{D}_{b-}^{\mu ;\psi }\left(I_{b-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right)\right).$$

(28)

We mention the simplified $\psi$-Hilfer fractional Taylor formulae:

Theorem 2.

(see also [9]) Let $\psi ,f\in C^n\left(\left[a,b\right]\right)$, with ψ being increasing such that ${\psi }^{\prime }\left(x\right)\ne 0$ over $\left[a,b\right]$, where $n-1<\alpha <n$, $0\le \beta \le 1$, and $\gamma =\alpha +\beta \left(n-\alpha \right)$, $x\in \left[a,b\right]$. Then

$$f\left(x\right)-\sum _{k=1}^{n-1}\frac{{\left(\psi \left(x\right)-\psi \left(a\right)\right)}^{\gamma -k}}{\Gamma \left(\gamma -k+1\right)}{f}_{\psi }^{\left[n-k\right]}\left(I_{a+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\right)\left(a\right)=$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\psi }^{\prime }\left(t\right){\left(\psi \left(x\right)-\psi \left(t\right)\right)}^{\alpha -1}{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta ;\psi }f\left(t\right)dt,$$

(29)

and

$$f\left(x\right)-\sum _{k=1}^{n-1}\frac{{\left(-1\right)}^k{\left(\psi \left(b\right)-\psi \left(x\right)\right)}^{\gamma -k}}{\Gamma \left(\gamma -k+1\right)}{f}_{\psi }^{\left[n-k\right]}\left(I_{b-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\right)\left(b\right)=$$

$$\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\psi }^{\prime }\left(t\right){\left(\psi \left(t\right)-\psi \left(x\right)\right)}^{\alpha -1}{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta ;\psi }f\left(t\right)dt.$$

(30)

Here notice that $\left(I_{a+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\right)\left(a\right)=\left(I_{b-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\right)\left(b\right)=0$.

We also mention the following alternative $\psi$-Hilfer fractional Taylor formulae:

Theorem 3.

([11]) Let $f,\psi \in C^n\left(\left[a,b\right]\right)$, with ψ being increasing, ${\psi }^{\prime }\left(x\right)\ne 0$ over $\left[a,b\right]\subset \mathbb{R},$$\alpha >0:⌈\alpha ⌉=n$, $0\le \beta \le 1$, $\mu =n\left(1-\beta \right)+\beta \alpha$. Assume that $g\left(x\right)=I_{a+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right),$$w\left(x\right)=I_{b-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }f\left(x\right)\in C^n\left(\left[a,b\right]\right)$.

Then

(1)

$$I_{a+}^{\mu ;\psi }{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta ;\psi }f\left(x\right)=g\left(x\right)-\sum _{k=0}^{n-1}\frac{g_{\psi }^{\left[k\right]}\left(a\right)}{k!}{\left(\psi \left(x\right)-\psi \left(a\right)\right)}^k,$$

(31)

where

$$g_{\psi }^{\left[k\right]}\left(x\right)={\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^{k}g\left(x\right),k=0,1,\dots ,n-1,$$

and

(2)

$$I_{b-}^{\mu ;\psi }{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta ;\psi }f\left(x\right)=w\left(x\right)-\sum _{k=0}^{n-1}\frac{{\left(-1\right)}^k{w}_{\psi }^{\left[k\right]}\left(b\right)}{k!}{\left(\psi \left(b\right)-\psi \left(x\right)\right)}^k,$$

(32)

where

$$w_{\psi }^{\left[k\right]}\left(x\right)={\left(\frac{1}{{\psi }^{\prime }\left(x\right)}\frac{d}{dx}\right)}^{k}w\left(x\right),k=0,1,\dots ,n-1;x\in \left[a,b\right].$$

Next we list two Hilfer fractional derivatives representation formulae:

Theorem 4.

([11]) Let $\alpha >0$, $\alpha \notin \mathbb{N}$, $⌈\alpha ⌉=n$, $0<\beta <1$; $f\in C^n\left(\left[a,b\right]\right)$, $\left[a,b\right]\subset \mathbb{R}$; and set $\gamma =\alpha +\beta \left(n-\alpha \right)$. Assume further that ${\Delta }_{a+}^{\gamma }f\in C\left(\left[a,b\right]\right):{\Delta }_{a+}^{\gamma -j}f\left(a\right)=0$, for $j=1,\dots ,n$. Let also $\overline{\alpha }>0:⌈\overline{\alpha }⌉=\overline{n}$, with $\overline{\gamma }=\overline{\alpha }+\beta \left(\overline{n}-\overline{\alpha }\right)$, and assume that $\alpha >\overline{\alpha }$ and $\gamma >\overline{\gamma }$. Then

$${}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)=\frac{1}{\Gamma \left(\alpha -\overline{\alpha }\right)}{\int }_a^x{\left(x-t\right)}^{\alpha -\overline{\alpha }-1}{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f\left(t\right)dt,$$

(33)

∀$x\in \left[a,b\right],$

Furthermore, ${}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\in AC\left(\left[a,b\right]\right)$ (absolutely continuous functions) if $\alpha -\overline{\alpha }\ge 1$ and ${}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\in C\left(\left[a,b\right]\right)$ if $\alpha -\overline{\alpha }\in \left(0,1\right)$.

Theorem 5.

([11]) Let $\alpha >0$, $\alpha \notin \mathbb{N}$, $⌈\alpha ⌉=n$, $0<\beta <1$; $f\in C^n\left(\left[a,b\right]\right)$, $\left[a,b\right]\subset \mathbb{R}$; and set $\gamma =\alpha +\beta \left(n-\alpha \right)$. Assume further that ${\Delta }_{b-}^{\gamma }f\in C\left(\left[a,b\right]\right):{\Delta }_{b-}^{\gamma -j}f\left(b\right)=0$, $j=1,\dots ,n$. Let also $\overline{\alpha }>0:⌈\overline{\alpha }⌉=\overline{n}$, with $\overline{\gamma }=\overline{\alpha }+\beta \left(\overline{n}-\overline{\alpha }\right)$, and assume that $\alpha >\overline{\alpha }$ and $\gamma >\overline{\gamma }$. Then

$${}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)=\frac{1}{\Gamma \left(\alpha -\overline{\alpha }\right)}{\int }_x^b{\left(t-x\right)}^{\alpha -\overline{\alpha }-1}{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f\left(t\right)dt,$$

(34)

∀$x\in \left[a,b\right],$

Furthermore, ${}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\in AC\left(\left[a,b\right]\right)$ if $\alpha -\overline{\alpha }\ge 1$ and ${}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\in C\left(\left[a,b\right]\right)$ if $\alpha -\overline{\alpha }\in \left(0,1\right)$.

3. Main Results

We present the following Hilfer-Polya type fractional inequalities:

Theorem 6.

Let $\alpha >0$, $\alpha \notin \mathbb{N}$, $⌈\alpha ⌉=n$, $0<\beta <1$; $f\in C^n\left(\left[a,b\right]\right)$, $\left[a,b\right]\subset \mathbb{R}$; and set $\gamma =\alpha +\beta \left(n-\alpha \right)$. Assume further that ${\Delta }_{a+}^{\gamma }f\in C\left(\left[a,b\right]\right):{\Delta }_{a+}^{\gamma -j}f\left(a\right)=0,$ for $j=1,\dots ,n;$ and ${\Delta }_{b-}^{\gamma }f\in C\left(\left[a,b\right]\right):{\Delta }_{b-}^{\gamma -j}f\left(b\right)=0$, $j=1,\dots ,n$. Let also $\overline{\alpha }>0:⌈\overline{\alpha }⌉=\overline{n}$, with $\overline{\gamma }=\overline{\alpha }+\beta \left(\overline{n}-\overline{\alpha }\right)$, and assume that $\alpha >\overline{\alpha }$ and $\gamma >\overline{\gamma }$.

Set

$${}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right):=\left\{\begin{array}{c}{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right),x\in \left[a,\frac{a+b}{2}\right],\hfill \\ \\ {}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right),x\in \left(\frac{a+b}{2},b\right],\hfill \end{array}\right.$$

(35)

and

$$M_1:=max\left\{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{\infty ,\left[a,\frac{a+b}{2}\right]},{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{\infty ,\left[\frac{a+b}{2},b\right]}\right\}.$$

(36)

Then

$$\left|{\int }_a^b{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx\right|\le {\int }_a^b\left|{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{M_1{\left(b-a\right)}^{\alpha -\overline{\alpha }+1}}{2^{\alpha -\overline{\alpha }}\Gamma \left(\alpha -\overline{\alpha }+2\right)}.$$

(37)

Proof. 

From (33) we have

$${}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)=\frac{1}{\Gamma \left(\alpha -\overline{\alpha }\right)}{\int }_a^x{\left(x-t\right)}^{\alpha -\overline{\alpha }-1}{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f\left(t\right)dt,$$

(38)

∀$x\in \left[a,\frac{a+b}{2}\right].$

By (34), we get

$${}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)=\frac{1}{\Gamma \left(\alpha -\overline{\alpha }\right)}{\int }_x^b{\left(t-x\right)}^{\alpha -\overline{\alpha }-1}{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f\left(t\right)dt,$$

(39)

∀$x\in \left[\frac{a+b}{2},b\right].$

We derive that

$$\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|\le \frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{\infty ,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }+1\right)}{\left(x-a\right)}^{\alpha -\overline{\alpha }},$$

(40)

∀$x\in \left[a,\frac{a+b}{2}\right],$ and similarly,

$$\left|{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)\right|\le \frac{{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{\infty ,\left[\frac{a+b}{2},b\right]}}{\Gamma \left(\alpha -\overline{\alpha }+1\right)}{\left(b-x\right)}^{\alpha -\overline{\alpha }},$$

(41)

∀$x\in \left[\frac{a+b}{2},b\right].$

We notice that:

$${\int }_a^b{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx={\int }_a^{\frac{a+b}{2}}{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx+{\int }_{\frac{a+b}{2}}^b{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx=$$

$${\int }_a^{\frac{a+b}{2}}{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)dx+{\int }_{\frac{a+b}{2}}^b{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)dx.$$

(42)

We further derive that

$${\int }_a^{\frac{a+b}{2}}\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{\infty ,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }+1\right)}{\int }_a^{\frac{a+b}{2}}{\left(x-a\right)}^{\alpha -\overline{\alpha }}dx=$$

(43)

$$\frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{\infty ,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }+2\right)}{\left(\frac{a+b}{2}-a\right)}^{\alpha -\overline{\alpha }+1}=\frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{\infty ,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }+2\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+1}.$$

That is, it holds

$${\int }_a^{\frac{a+b}{2}}\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{\infty ,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }+2\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+1}.$$

(44)

Similarly, it holds

$${\int }_{\frac{a+b}{2}}^b\left|{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{\infty ,\left[\frac{a+b}{2},b\right]}}{\Gamma \left(\alpha -\overline{\alpha }+2\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+1}.$$

(45)

Therefore, we obtain

$$\left|{\int }_a^b{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx\right|\le {\int }_a^b\left|{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)\right|dx=$$

$${\int }_a^{\frac{a+b}{2}}\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|dx+{\int }_{\frac{a+b}{2}}^b\left|{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le$$

$$\frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{\infty ,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }+2\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+1}+\frac{{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{\infty ,\left[\frac{a+b}{2},b\right]}}{\Gamma \left(\alpha -\overline{\alpha }+2\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+1}=$$

(46)

$$\frac{{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+1}}{\Gamma \left(\alpha -\overline{\alpha }+2\right)}\left[{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{\infty ,\left[a,\frac{a+b}{2}\right]}+{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{\infty ,\left[\frac{a+b}{2},b\right]}\right]\le$$

$$\frac{2M_1{\left(b-a\right)}^{\alpha -\overline{\alpha }+1}}{2^{\alpha -\overline{\alpha }+1}\Gamma \left(\alpha -\overline{\alpha }+2\right)}=\frac{M_1{\left(b-a\right)}^{\alpha -\overline{\alpha }+1}}{2^{\alpha -\overline{\alpha }}\Gamma \left(\alpha -\overline{\alpha }+2\right)}.$$

 □

We continue with the $L_1$-variant:

Theorem 7.

All as in Theorem 6 with $\alpha -\overline{\alpha }>1$ (i.e., $\alpha >\overline{\alpha }+1$). Call

$$M_2:=max\left\{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{1,\left[a,\frac{a+b}{2}\right]},{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{1,\left[\frac{a+b}{2},b\right]}\right\}.$$

(47)

Then

$$\left|{\int }_a^b{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx\right|\le {\int }_a^b\left|{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{M_2{\left(b-a\right)}^{\alpha -\overline{\alpha }}}{2^{\alpha -\overline{\alpha }-1}\Gamma \left(\alpha -\overline{\alpha }+1\right)}.$$

(48)

Proof. 

By (38) we have

$$\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|\le \frac{1}{\Gamma \left(\alpha -\overline{\alpha }\right)}{\int }_a^x{\left(x-t\right)}^{\alpha -\overline{\alpha }-1}\left|{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f\left(t\right)\right|dt\le$$

$$\frac{{\left(x-a\right)}^{\alpha -\overline{\alpha }-1}}{\Gamma \left(\alpha -\overline{\alpha }\right)}{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{1,\left[a,\frac{a+b}{2}\right]},$$

(49)

∀$x\in \left[a,\frac{a+b}{2}\right].$

Similarly, from (39) we find that

$$\left|{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)\right|\le \frac{{\left(b-x\right)}^{\alpha -\overline{\alpha }-1}}{\Gamma \left(\alpha -\overline{\alpha }\right)}{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{1,\left[\frac{a+b}{2},b\right]},$$

(50)

∀$x\in \left[\frac{a+b}{2},b\right].$

Furthermore, we obtain

$${\int }_a^{\frac{a+b}{2}}\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{1,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }\right)}{\int }_a^{\frac{a+b}{2}}{\left(x-a\right)}^{\alpha -\overline{\alpha }-1}dx=$$

$$\frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{1,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }+1\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }}.$$

(51)

Similarly, we derive

$${\int }_{\frac{a+b}{2}}^b\left|{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{1,\left[\frac{a+b}{2},b\right]}}{\Gamma \left(\alpha -\overline{\alpha }+1\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }}.$$

(52)

Therefore we obtain

$$\left|{\int }_a^b{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx\right|\le {\int }_a^b\left|{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)\right|dx=$$

$${\int }_a^{\frac{a+b}{2}}\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|dx+{\int }_{\frac{a+b}{2}}^b\left|{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le$$

$$\frac{\left[{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{1,\left[a,\frac{a+b}{2}\right]}+{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{1,\left[\frac{a+b}{2},b\right]}\right]}{\Gamma \left(\alpha -\overline{\alpha }+1\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }}\le$$

(53)

$$\frac{2M_2}{\Gamma \left(\alpha -\overline{\alpha }+1\right)}\frac{{\left(b-a\right)}^{\alpha -\overline{\alpha }}}{2^{\alpha -\overline{\alpha }}}=\frac{M_2}{\Gamma \left(\alpha -\overline{\alpha }+1\right)}\frac{{\left(b-a\right)}^{\alpha -\overline{\alpha }}}{2^{\alpha -\overline{\alpha }-1}}.$$

 □

Next comes the $L_q$-variant of Hilfer-Polya fractional inequality:

Theorem 8.

All as in Theorem 6 with $\alpha -\overline{\alpha }>\frac{1}{q}$, where $p,q>1:\frac{1}{p}+\frac{1}{q}=1$. Call

$$M_3:=max\left\{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{q,\left[a,\frac{a+b}{2}\right]},{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{q,\left[\frac{a+b}{2},b\right]}\right\}.$$

(54)

Then

$$\left|{\int }_a^b{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx\right|\le {\int }_a^b\left|{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le$$

$$\frac{M_3}{\Gamma \left(\alpha -\overline{\alpha }\right){\left(p\left(\alpha -\overline{\alpha }-1\right)+1\right)}^{\frac{1}{p}}\left(\alpha -\overline{\alpha }+\frac{1}{p}\right)}\frac{{\left(b-a\right)}^{\alpha -\overline{\alpha }+\frac{1}{p}}}{2^{\alpha -\overline{\alpha }-\frac{1}{q}}}.$$

(55)

Proof. 

By (38) we have

$$\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|\le \frac{1}{\Gamma \left(\alpha -\overline{\alpha }\right)}{\int }_a^x{\left(x-t\right)}^{\alpha -\overline{\alpha }-1}\left|{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f\left(t\right)\right|dt\le$$

$$\frac{1}{\Gamma \left(\alpha -\overline{\alpha }\right)}{\left({\int }_a^x{\left(x-t\right)}^{p\left(\alpha -\overline{\alpha }-1\right)}dt\right)}^{\frac{1}{p}}{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{q,\left[a,\frac{a+b}{2}\right]}=$$

$$\frac{{\left(x-a\right)}^{\left(\alpha -\overline{\alpha }-\frac{1}{q}\right)}}{\Gamma \left(\alpha -\overline{\alpha }\right){\left(p\left(\alpha -\overline{\alpha }-1\right)+1\right)}^{\frac{1}{p}}}{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{q,\left[a,\frac{a+b}{2}\right]},$$

(56)

∀$x\in \left[a,\frac{a+b}{2}\right]$, with $\alpha -\overline{\alpha }>\frac{1}{q}.$

And, by (39), similarly we derive

$$\left|{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)\right|\le \frac{{\left(b-x\right)}^{\left(\alpha -\overline{\alpha }-\frac{1}{q}\right)}}{\Gamma \left(\alpha -\overline{\alpha }\right){\left(p\left(\alpha -\overline{\alpha }-1\right)+1\right)}^{\frac{1}{p}}}{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{q,\left[\frac{a+b}{2},b\right]},$$

(57)

∀$x\in \left[\frac{a+b}{2},b\right]$, with $\alpha -\overline{\alpha }>\frac{1}{q}.$

Consequently, we obtain that

$${\int }_a^{\frac{a+b}{2}}\left|{}^H{\mathbb{D}}_{a+}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{q,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }\right){\left(p\left(\alpha -\overline{\alpha }-1\right)+1\right)}^{\frac{1}{p}}}{\int }_a^{\frac{a+b}{2}}{\left(x-a\right)}^{\left(\alpha -\overline{\alpha }-\frac{1}{q}\right)}dx=$$

$$\frac{{∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{q,\left[a,\frac{a+b}{2}\right]}}{\Gamma \left(\alpha -\overline{\alpha }\right){\left(p\left(\alpha -\overline{\alpha }-1\right)+1\right)}^{\frac{1}{p}}\left(\alpha -\overline{\alpha }+\frac{1}{p}\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+\frac{1}{p}}.$$

(58)

Similarly, we derive

$${\int }_{\frac{a+b}{2}}^b\left|{}^H{\mathbb{D}}_{b-}^{\overline{\alpha },\beta }f\left(x\right)\right|dx\le \frac{{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{q,\left[\frac{a+b}{2},b\right]}}{\Gamma \left(\alpha -\overline{\alpha }\right){\left(p\left(\alpha -\overline{\alpha }-1\right)+1\right)}^{\frac{1}{p}}\left(\alpha -\overline{\alpha }+\frac{1}{p}\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+\frac{1}{p}}.$$

(59)

Therefore, we obtain

$$\left|{\int }_a^b{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)dx\right|\le {\int }_a^b\left|{}^H{\mathbb{D}}^{\overline{\alpha },\beta }f\left(x\right)\right|dx=$$

$$\frac{\left({∥{}^H{\mathbb{D}}_{a+}^{\alpha ,\beta }f∥}_{q,\left[a,\frac{a+b}{2}\right]}+{∥{}^H{\mathbb{D}}_{b-}^{\alpha ,\beta }f∥}_{q,\left[\frac{a+b}{2},b\right]}\right)}{\Gamma \left(\alpha -\overline{\alpha }\right){\left(p\left(\alpha -\overline{\alpha }-1\right)+1\right)}^{\frac{1}{p}}\left(\alpha -\overline{\alpha }+\frac{1}{p}\right)}{\left(\frac{b-a}{2}\right)}^{\alpha -\overline{\alpha }+\frac{1}{p}}\le$$

(60)

$$\frac{M_3}{\Gamma \left(\alpha -\overline{\alpha }\right){\left(p\left(\alpha -\overline{\alpha }-1\right)+1\right)}^{\frac{1}{p}}\left(\alpha -\overline{\alpha }+\frac{1}{p}\right)}\frac{{\left(b-a\right)}^{\alpha -\overline{\alpha }+\frac{1}{p}}}{2^{\alpha -\overline{\alpha }-\frac{1}{q}}},$$

proving the claim. □

Next come $\psi$-Hilfer-Ostrowski type inequalities for several functions involved.

For basic $\psi$-Hilfer-Ostrowski type inequalities involving one function see [11].

We make

Remark 2.

Our setting here follows: Let $f_i\in C^n\left(\left[a,b\right]\right)$, $\alpha \notin \mathbb{N}$, $n=⌈\alpha ⌉$, $\alpha >0$; $i=1,\dots ,r\in \mathbb{N}-\left\{1\right\}$, $x_0\in \left[a,b\right]$. Assume that $g_{1i}\left(x\right)=I_{x_0+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }{f}_i\left(x\right)\in C^n\left(\left[x_0,b\right]\right)$ and $w_{1i}\left(x\right)=I_{x_0-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }{f}_i\left(x\right)\in C^n\left(\left[a,x_0\right]\right)$, for all $i=1,\dots ,r.$

Define

$${\phi }_{i{x}_0}\left(x\right):=\left\{\begin{array}{c}{g}_{1i}\left(x\right),x\in \left[x_0,b\right]\hfill \\ \\ w_{1i}\left(x\right),x\in [a,x_0)\hfill \end{array}\right\}.$$

(61)

Notice that if $\beta =1$, we get $g_{1i}\left(x_0\right)=w_{1i}\left(x_0\right)={\phi }_{i{x}_0}\left(x_0\right)=f_i\left(x_0\right)$, all $i=1,\dots ,r.$

In general, for $f\in C\left(\left[a,b\right]\right)$ we have

$$\left|I_{a+}^{\alpha ,\psi }f\left(x\right)\right|\le \frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\psi }^{\prime }\left(t\right){\left(\psi \left(x\right)-\psi \left(t\right)\right)}^{\alpha -1}\left|f\left(t\right)\right|dt\le$$

$$\frac{{∥f∥}_{\infty ,\left[a,b\right]}}{\Gamma \left(\alpha +1\right)}{\left(\psi \left(x\right)-\psi \left(a\right)\right)}^{\alpha },\forall x\in \left[a,b\right].$$

(62)

Hence $I_{a+}^{\alpha ,\psi }f\left(a\right)=0.$

Similalry, we have

$$\left|I_{b-}^{\alpha ,\psi }f\left(x\right)\right|\le \frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\psi }^{\prime }\left(t\right){\left(\psi \left(t\right)-\psi \left(x\right)\right)}^{\alpha -1}\left|f\left(t\right)\right|dt\le$$

$$\frac{{∥f∥}_{\infty ,\left[a,b\right]}}{\Gamma \left(\alpha +1\right)}{\left(\psi \left(b\right)-\psi \left(x\right)\right)}^{\alpha },\forall x\in \left[a,b\right].$$

(63)

That is $I_{b-}^{\alpha ,\psi }f\left(b\right)=0.$

So when $0\le \beta <1$, by the above we obtain $g_{1i}\left(x_0\right)=w_{1i}\left(x_0\right)={\phi }_{i{x}_0}\left(x_0\right)=0$, for all $i=1,\dots ,r.$

Thus, it is always true that $g_{1i}\left(x_0\right)=w_{1i}\left(x_0\right)$, $i=1,\dots ,r.$

We present

Theorem 9.

Let $\psi ,f_i\in C^n\left(\left[a,b\right]\right)$, $\alpha \notin \mathbb{N}$, $n=⌈\alpha ⌉$, $\alpha >0$; $i=1,\dots ,r\in \mathbb{N}-\left\{1\right\}$, $x_0\in \left[a,b\right]$. Here ψ is increasing, ${\psi }^{\prime }\left(x\right)\ne 0$ over $\left[a,b\right]\subset \mathbb{R}$, $0\le \beta \le 1$, $\mu =n\left(1-\beta \right)+\beta \alpha$. Assume that $g_{1i}\left(x\right)=I_{x_0+}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }{f}_i\left(x\right)\in C^n\left(\left[x_0,b\right]\right)$ and $w_{1i}\left(x\right)=I_{x_0-}^{\left(1-\beta \right)\left(n-\alpha \right);\psi }{f}_i\left(x\right)\in C^n\left(\left[a,x_0\right]\right)$, for all $i=1,\dots ,r,$ and ${\phi }_{i{x}_0}\left(x\right)$ is as in (61). Assume also that $g_{1i\psi }^{\left[k\right]}\left(x_0\right)=w_{1i\psi }^{\left[k\right]}\left(x_0\right)=0$, for all $k=1,\dots ,n-1.$

Then

(1)

$${\theta }^{\psi }\left(f_1,\dots ,f_r\right)\left(x_0\right):=$$

(64)

$$r{\int }_a^b\left(\prod _{\lambda =1}^r{\phi }_{\lambda x_0}\left(x\right)\right)dx-\sum _{i=1}^r\left[{\phi }_{i{x}_0}\left(x_0\right){\int }_a^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)dx\right]=$$

$$\sum _{i=1}^r\left[\left[{\int }_a^{x_0}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)\left(I_{x_0-}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\left(x\right)\right)dx\right]+\right.$$

$$\left.\left[{\int }_{x_0}^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)\left(I_{x_0+}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\left(x\right)\right)dx\right]\right],$$

and in case of $0\le \beta <1$, we have that

$${\theta }^{\psi }\left(f_1,\dots ,f_r\right)\left(x_0\right)=r{\int }_a^b\left(\prod _{\lambda =1}^r{\phi }_{\lambda x_0}\left(x\right)\right)dx,$$

(65)

(2) furthermore, it holds

$$\left|{\theta }^{\psi }\left(f_1,\dots ,f_r\right)\left(x_0\right)\right|\le \frac{1}{\Gamma \left(\mu +1\right)}$$

$$\left\{\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{\infty ,\left[a,x_0\right]}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right){\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu }+\right.$$

$$\left.\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{\infty ,\left[x_0,b\right]}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right){\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu }\right\}.$$

(66)

It follows the $L_1$-variant.

Theorem 10.

All as in Theorem 9, with $\alpha >1$. Then

$$\left|{\theta }^{\psi }\left(f_1,\dots ,f_r\right)\left(x_0\right)\right|\le \frac{1}{\Gamma \left(\mu \right)}$$

$$\left\{\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_1\left(\left[a,x_0\right],\psi \right)}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right){\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu -1}+\right.$$

$$\left.\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_1\left(\left[x_0,b\right],\psi \right)}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right){\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu -1}\right\}.$$

(67)

Next we have the $L_q$-variant.

Theorem 11.

All as in Theorem 9. Let also $p,q>1:\frac{1}{p}+\frac{1}{q}=1$ with $\alpha >\frac{1}{q}$. Then

$$\left|{\theta }^{\psi }\left(f_1,\dots ,f_r\right)\left(x_0\right)\right|\le \frac{1}{\Gamma \left(\mu \right){\left(p\left(\mu -1\right)+1\right)}^{\frac{1}{p}}}$$

$$\left\{\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[a,x_0\right],\psi \right)}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right){\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu -\frac{1}{q}}+\right.$$

$$\left.\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[x_0,b\right],\psi \right)}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right){\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu -\frac{1}{q}}\right\}.$$

(68)

Proof of Theorems 9–11. 

By Theorem 3 we have

$$\begin{array}{c}{g}_{1i}\left(x\right)-g_{1i}\left(x_0\right)=I_{x_0+}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\left(x\right),\forall x\in \left[x_0,b\right],\hfill \\ \mathrm{and}\hfill \\ w_{1i}\left(x\right)-w_{1i}\left(x_0\right)=I_{x_0-}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\left(x\right),\forall x\in \left[a,x_0\right],\hfill \end{array}$$

(69)

for all $i=1,\dots ,r.$

That is

$$\begin{array}{c}{\phi }_{i{x}_0}\left(x\right)-{\phi }_{i{x}_0}\left(x_0\right)=I_{x_0+}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\left(x\right),\forall x\in \left[x_0,b\right],\hfill \\ \mathrm{and}\hfill \\ {\phi }_{i{x}_0}\left(x\right)-{\phi }_{i{x}_0}\left(x_0\right)=I_{x_0-}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\left(x\right),\forall x\in [a,x_0),\hfill \end{array}$$

(70)

for all $i=1,\dots ,r.$

Multiplying (70) by $\left({\displaystyle \prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r}{\phi }_{j{x}_0}\left(x\right)\right)$ we get, respectively,

$$\prod _{\lambda =1}^r{\phi }_{\lambda x_0}\left(x\right)-\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right){\phi }_{i{x}_0}\left(x_0\right)=\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)I_{x_0+}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\left(x\right),$$

(71)

∀$x\in \left[x_0,b\right],$

And

$$\prod _{\lambda =1}^r{\phi }_{\lambda x_0}\left(x\right)-\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right){\phi }_{i{x}_0}\left(x_0\right)=\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)I_{x_0-}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\left(x\right),$$

(72)

∀$x\in [a,x_0),$ for all $i=1,\dots ,r.$

Adding (71) and (72), separately, we obtain

$$r\left(\prod _{\lambda =1}^r{\phi }_{\lambda x_0}\left(x\right)\right)-\sum _{i=1}^r\left[\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right){\phi }_{i{x}_0}\left(x_0\right)\right]=$$

(73)

$$\sum _{i=1}^r\left[\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)I_{x_0+}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\left(x\right)\right],$$

∀$x\in \left[x_0,b\right],$

In addition,

$$r\left(\prod _{\lambda =1}^r{\phi }_{\lambda x_0}\left(x\right)\right)-\sum _{i=1}^r\left[\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right){\phi }_{i{x}_0}\left(x_0\right)\right]=$$

(74)

$$\sum _{i=1}^r\left[\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)I_{x_0-}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\left(x\right)\right],$$

∀$x\in [a,x_0).$

Next integrate (73) and (74) with respect to $x\in \left[a,b\right].$ We have

$$r{\int }_{x_0}^b\left(\prod _{\lambda =1}^r{\phi }_{\lambda x_0}\left(x\right)\right)dx-\sum _{i=1}^r\left[{\phi }_{i{x}_0}\left(x_0\right){\int }_{x_0}^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)dx\right]=$$

$$\sum _{i=1}^r\left[{\int }_{x_0}^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)\left(I_{x_0+}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\left(x\right)\right)dx\right],$$

(75)

and

$$r{\int }_a^{x_0}\left(\prod _{\lambda =1}^r{\phi }_{\lambda x_0}\left(x\right)\right)dx-\sum _{i=1}^r\left[{\phi }_{i{x}_0}\left(x_0\right){\int }_a^{x_0}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)dx\right]=$$

$$\sum _{i=1}^r\left[{\int }_a^{x_0}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}\left(x\right)\right)\left(I_{x_0-}^{\mu ;\psi }{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\left(x\right)\right)dx\right].$$

(76)

Finally adding (75) and (76) we obtain the useful and nice identity (64).

Identity (64) implies

$$\left|{\theta }^{\psi }\left(f_1,\dots ,f_r\right)\left(x_0\right)\right|\le \sum _{i=1}^r\left[\left[{\int }_a^{x_0}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)\left(I_{x_0-}^{\mu ;\psi }\left|{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\right|\left(x\right)\right)dx\right]+\right.$$

$$\left.\left[{\int }_{x_0}^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)\left(I_{x_0+}^{\mu ;\psi }\left|{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\right|\left(x\right)\right)dx\right]\right]=$$

(77)

$$\sum _{i=1}^r\left[\left[{\int }_a^{x_0}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)\frac{1}{\Gamma \left(\mu \right)}\right.\right.$$

$$\left.\left({\int }_x^{x_0}{\psi }^{\prime }\left(t\right){\left(\psi \left(t\right)-\psi \left(x\right)\right)}^{\mu -1}\left|\left({}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\right)\left(t\right)\right|dt\right)dx\right]+$$

$$\left[{\int }_{x_0}^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)\frac{1}{\Gamma \left(\mu \right)}\right.$$

$$\left.\left.\left({\int }_{x_0}^x{\psi }^{\prime }\left(t\right){\left(\psi \left(x\right)-\psi \left(t\right)\right)}^{\mu -1}\left|\left({}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\right)\left(t\right)\right|dt\right)dx\right]\right]\le$$

$$\frac{1}{\Gamma \left(\mu +1\right)}\sum _{i=1}^r\left[\left[{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{\infty ,\left[a,x_0\right]}{\int }_a^{x_0}{\left(\psi \left(x_0\right)-\psi \left(x\right)\right)}^{\mu }\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)dx\right]\right.$$

(78)

$$\left.+\left[{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{\infty ,\left[x_0,b\right]}{\int }_{x_0}^b{\left(\psi \left(x\right)-\psi \left(x_0\right)\right)}^{\mu }\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)dx\right]\right]\le$$

$$\frac{1}{\Gamma \left(\mu +1\right)}\sum _{i=1}^r\left[\left[{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{\infty ,\left[a,x_0\right]}{\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu }{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right]+\right.$$

$$\left.\left[{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{\infty ,\left[x_0,b\right]}{\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu }{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right]\right]=$$

(79)

$$\frac{1}{\Gamma \left(\mu +1\right)}\left\{\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{\infty ,\left[a,x_0\right]}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right){\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu }+\right.$$

$$\left.\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{\infty ,\left[x_0,b\right]}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right){\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu }\right\},$$

(80)

proving (66).

If $\alpha \notin \mathbb{N}$ and $\alpha >1$, then $n=⌈\alpha ⌉>1$, and $n-1\ge 1>\beta \left(n-\alpha \right)$. Hence $n-\beta \left(n-\alpha \right)>1$ and $\mu >1$. So we have

$$\left|{\theta }^{\psi }\left(f_1,\dots ,f_r\right)\left(x_0\right)\right|\le \sum _{i=1}^r\left[\left[{\int }_a^{x_0}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)\frac{1}{\Gamma \left(\mu \right)}\right.\right.$$

$$\left.\left({\int }_x^{x_0}{\psi }^{\prime }\left(t\right){\left(\psi \left(t\right)-\psi \left(x\right)\right)}^{\mu -1}\left|\left({}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i\right)\left(t\right)\right|dt\right)dx\right]+$$

$$\left[{\int }_{x_0}^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)\frac{1}{\Gamma \left(\mu \right)}\right.$$

$$\left.\left.\left({\int }_{x_0}^x{\psi }^{\prime }\left(t\right){\left(\psi \left(x\right)-\psi \left(t\right)\right)}^{\mu -1}\left|\left({}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i\right)\left(t\right)\right|dt\right)dx\right]\right]\le$$

(81)

$$\frac{1}{\Gamma \left(\mu \right)}\sum _{i=1}^r\left[\left[{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_1\left(\left[a,x_0\right],\psi \right)}{\int }_a^{x_0}{\left(\psi \left(x_0\right)-\psi \left(x\right)\right)}^{\mu -1}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)dx\right]\right.$$

$$\left.+\left[{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_1\left(\left[x_0,b\right],\psi \right)}{\int }_{x_0}^b{\left(\psi \left(x\right)-\psi \left(x_0\right)\right)}^{\mu -1}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)dx\right]\right]\le$$

$$\frac{1}{\Gamma \left(\mu \right)}\sum _{i=1}^r\left[\left[{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_1\left(\left[a,x_0\right],\psi \right)}{\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu -1}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right]+\right.$$

(82)

$$\left.\left[{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_1\left(\left[x_0,b\right],\psi \right)}{\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu -1}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right]\right]=$$

$$\frac{1}{\Gamma \left(\mu \right)}\left\{\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_1\left(\left[a,x_0\right],\psi \right)}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right){\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu -1}+\right.$$

(83)

$$\left.\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_1\left(\left[x_0,b\right],\psi \right)}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right){\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu -1}\right\},$$

proving (67).

Let $\alpha >0$ with $⌈\alpha ⌉=n\in \mathbb{N}$, and let $p,q>1:\frac{1}{p}+\frac{1}{q}=1$, with $\alpha >\frac{1}{q}$. Clearly $n>\frac{1}{q}$. Let $0<\beta \le 1$, then $\alpha \beta >\frac{\beta }{q}$, furthermore $\mu =n\left(1-\beta \right)+\beta \alpha >\frac{\beta }{q}+n\left(1-\beta \right)\ge \frac{\beta }{q}+1-\beta =\frac{\beta }{q}+\frac{1}{p}+\frac{1}{q}-\frac{\beta }{p}-\frac{\beta }{q}=\frac{1}{q}+\frac{1}{p}\left(1-\beta \right)\ge \frac{1}{q}$. That is $\mu >\frac{1}{q}.$

From (81), by using Hölder’s inequality twice, we have

$$\left|{\theta }^{\psi }\left(f_1,\dots ,f_r\right)\left(x_0\right)\right|\le \frac{1}{\Gamma \left(\mu \right)}$$

$$\sum _{i=1}^r\left[\left[{\int }_a^{x_0}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)\frac{{\left(\psi \left(x_0\right)-\psi \left(x\right)\right)}^{\frac{p\left(\mu -1\right)+1}{p}}}{{\left(p\left(\mu -1\right)+1\right)}^{\frac{1}{p}}}{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[a,x_0\right],\psi \right)}dx\right]+\right.$$

$$\left.\left[{\int }_{x_0}^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right)\frac{{\left(\psi \left(x\right)-\psi \left(x_0\right)\right)}^{\frac{p\left(\mu -1\right)+1}{p}}}{{\left(p\left(\mu -1\right)+1\right)}^{\frac{1}{p}}}{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[x_0,b\right],\psi \right)}dx\right]\right]=$$

(84)

$$\frac{1}{\Gamma \left(\mu \right){\left(p\left(\mu -1\right)+1\right)}^{\frac{1}{p}}}$$

$$\sum _{i=1}^r\left[\left[{\int }_a^{x_0}\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right){\left(\psi \left(x_0\right)-\psi \left(x\right)\right)}^{\mu -\frac{1}{q}}{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[a,x_0\right],\psi \right)}dx\right]+\right.$$

$$\left.\left[{\int }_{x_0}^b\left(\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r\left|{\phi }_{j{x}_0}\left(x\right)\right|\right){\left(\psi \left(x\right)-\psi \left(x_0\right)\right)}^{\mu -\frac{1}{q}}{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[x_0,b\right],\psi \right)}dx\right]\right]\le$$

$$\frac{1}{\Gamma \left(\mu \right){\left(p\left(\mu -1\right)+1\right)}^{\frac{1}{p}}}$$

$$\sum _{i=1}^r\left[\left[{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[a,x_0\right],\psi \right)}{\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu -\frac{1}{q}}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right]+\right.$$

(85)

$$\left.\left[{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[x_0,b\right],\psi \right)}{\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu -\frac{1}{q}}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right]\right]=$$

$$\frac{1}{\Gamma \left(\mu \right){\left(p\left(\mu -1\right)+1\right)}^{\frac{1}{p}}}$$

$$\left\{\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0-}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[a,x_0\right],\psi \right)}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[a,x_0\right]}\right){\left(\psi \left(x_0\right)-\psi \left(a\right)\right)}^{\mu -\frac{1}{q}}+\right.$$

(86)

$$\left.\left(\sum _{i=1}^r{∥{}^H{\mathbb{D}}_{x_0+}^{\alpha ,\beta ;\psi }{f}_i∥}_{L_q\left(\left[x_0,b\right],\psi \right)}{∥\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^r{\phi }_{j{x}_0}∥}_{1,\left[x_0,b\right]}\right){\left(\psi \left(b\right)-\psi \left(x_0\right)\right)}^{\mu -\frac{1}{q}}\right\},$$

proving (68). □

Next we present a $\psi$-Hilfer-Hilbert-Pachpatte left fractional inequality:

Theorem 12.

Let $i=1,2;$${\psi }_i,$$f_i\in C^{n_i}\left(\left[a_i,b_i\right]\right)$, with ${\psi }_i$ being strictly increasing over $\left[a_i,b_i\right]$, where $n_i-1<{\alpha }_i<n_i$, $0\le {\beta }_i\le 1$, and ${\gamma }_i={\alpha }_i+{\beta }_i\left(n_i-{\alpha }_i\right)$, $x_i\in \left[a_i,b_i\right]$. Assume that $f_{i{\psi }_i}^{\left[n_i-k_i\right]}\left(I_{a_i+}^{\left(1-{\beta }_i\right)\left(n_i-{\alpha }_i\right);{\psi }_i}{f}_i\right)\left(a_i\right)=0$, for $k_i=1,\dots ,n_i-1$. Let also $p,q>1:\frac{1}{p}+\frac{1}{q}=1$, such that ${\alpha }_1>\frac{1}{q}$ and ${\alpha }_2>\frac{1}{p}$. Then

$${\int }_{a_1}^{b_1}{\int }_{a_2}^{b_2}\frac{\left|f_1\left(x_1\right)\right|\left|f_2\left(x_2\right)\right|d{x}_{1}d{x}_2}{\left(\frac{{\left({\psi }_1\left(x_1\right)-{\psi }_1\left(a_1\right)\right)}^{p\left({\alpha }_1-1\right)+1}}{p\left(p\left({\alpha }_1-1\right)|+1\right)}+\frac{{\left({\psi }_2\left(x_2\right)-{\psi }_2\left(a_2\right)\right)}^{q\left({\alpha }_2-1\right)+1}}{q\left(q\left({\alpha }_2-1\right)+1\right)}\right)}\le$$

$$\frac{\left(b_1-a_1\right)\left(b_2-a_2\right)}{\Gamma \left({\alpha }_1\right)\Gamma \left({\alpha }_2\right)}{∥{}^H{\mathbb{D}}_{a_1+}^{{\alpha }_1,{\beta }_1;{\psi }_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right],{\psi }_1\right)}{∥{}^H{\mathbb{D}}_{a_2+}^{{\alpha }_2,{\beta }_2;{\psi }_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right],{\psi }_2\right)}.$$

(87)

Proof. 

By Theorem 2 we have

$$f_i\left(x_i\right)=\frac{1}{\Gamma \left({\alpha }_i\right)}{\int }_{a_i}^{x_i}{\psi }_i^{\prime }\left(t_i\right){\left({\psi }_i\left(x_i\right)-{\psi }_i\left(t_i\right)\right)}^{{\alpha }_i-1}{}^H{\mathbb{D}}_{a_i+}^{{\alpha }_i,{\beta }_i;{\psi }_i}{f}_i\left(t_i\right)d{t}_i,$$

(88)

∀$x_i\in \left[a_i,b_i\right]$, $i=1,2.$

Then

$$\left|f_i\left(x_i\right)\right|\le \frac{1}{\Gamma \left({\alpha }_i\right)}{\int }_{a_i}^{x_i}{\psi }_i^{\prime }\left(t_i\right){\left({\psi }_i\left(x_i\right)-{\psi }_i\left(t_i\right)\right)}^{{\alpha }_i-1}\left|{}^H{\mathbb{D}}_{a_i+}^{{\alpha }_i,{\beta }_i;{\psi }_i}{f}_i\left(t_i\right)\right|d{t}_i,$$

(89)

$i=1,2,$∀$x_i\in \left[a_i,b_i\right].$

By Hölder’s inequality we obtain

$$\left|f_1\left(x_1\right)\right|\le \frac{1}{\Gamma \left({\alpha }_1\right)}\frac{{\left({\psi }_1\left(x_1\right)-{\psi }_1\left(a_1\right)\right)}^{\frac{p\left({\alpha }_1-1\right)+1}{p}}}{{\left(p\left({\alpha }_1-1\right)+1\right)}^{\frac{1}{p}}}{∥{}^H{\mathbb{D}}_{a_1+}^{{\alpha }_1,{\beta }_1;{\psi }_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right],{\psi }_1\right)},$$

(90)

∀$x_1\in \left[a_1,b_1\right],$ and

$$\left|f_2\left(x_2\right)\right|\le \frac{1}{\Gamma \left({\alpha }_2\right)}\frac{{\left({\psi }_2\left(x_2\right)-{\psi }_2\left(a_2\right)\right)}^{\frac{q\left({\alpha }_2-1\right)+1}{q}}}{{\left(q\left({\alpha }_2-1\right)+1\right)}^{\frac{1}{q}}}{∥{}^H{\mathbb{D}}_{a_2+}^{{\alpha }_2,{\beta }_2;{\psi }_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right],{\psi }_2\right)},$$

(91)

∀$x_2\in \left[a_2,b_2\right].$

Hence we have

$$\left|f_1\left(x_1\right)\right|\left|f_2\left(x_2\right)\right|\le \frac{1}{\Gamma \left({\alpha }_1\right)\Gamma \left({\alpha }_2\right){\left(p\left({\alpha }_1-1\right)+1\right)}^{\frac{1}{p}}{\left(q\left({\alpha }_2-1\right)+1\right)}^{\frac{1}{q}}}$$

$${\left({\psi }_1\left(x_1\right)-{\psi }_1\left(a_1\right)\right)}^{\frac{p\left({\alpha }_1-1\right)+1}{p}}{\left({\psi }_2\left(x_2\right)-{\psi }_2\left(a_2\right)\right)}^{\frac{q\left({\alpha }_2-1\right)+1}{q}}$$

(92)

$${∥{}^H{\mathbb{D}}_{a_1+}^{{\alpha }_1,{\beta }_1;{\psi }_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right],{\psi }_1\right)}{∥{}^H{\mathbb{D}}_{a_2+}^{{\alpha }_2,{\beta }_2;{\psi }_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right],{\psi }_2\right)}\le$$

(using Young’s inequality for $a,b\ge 0$, $a^{\frac{1}{p}}{b}^{\frac{1}{q}}\le \frac{a}{p}+\frac{b}{q}$)

$$\frac{1}{\Gamma \left({\alpha }_1\right)\Gamma \left({\alpha }_2\right)}\left(\frac{{\left({\psi }_1\left(x_1\right)-{\psi }_1\left(a_1\right)\right)}^{p\left({\alpha }_1-1\right)+1}}{p\left(p\left({\alpha }_1-1\right)+1\right)}+\frac{{\left({\psi }_2\left(x_2\right)-{\psi }_2\left(a_2\right)\right)}^{q\left({\alpha }_2-1\right)+1}}{q\left(q\left({\alpha }_2-1\right)+1\right)}\right)$$

(93)

$${∥{}^H{\mathbb{D}}_{a_1+}^{{\alpha }_1,{\beta }_1;{\psi }_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right],{\psi }_1\right)}{∥{}^H{\mathbb{D}}_{a_2+}^{{\alpha }_2,{\beta }_2;{\psi }_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right],{\psi }_2\right)},$$

∀$x_i\in \left[a_i,b_i\right]$; $i=1,2.$

So far we have

$$\frac{\left|f_1\left(x_1\right)\right|\left|f_2\left(x_2\right)\right|}{\left(\frac{{\left({\psi }_1\left(x_1\right)-{\psi }_1\left(a_1\right)\right)}^{p\left({\alpha }_1-1\right)+1}}{p\left(p\left({\alpha }_1-1\right)+1\right)}+\frac{{\left({\psi }_2\left(x_2\right)-{\psi }_2\left(a_2\right)\right)}^{q\left({\alpha }_2-1\right)+1}}{q\left(q\left({\alpha }_2-1\right)+1\right)}\right)}\le$$

(94)

$$\frac{{∥{}^H{\mathbb{D}}_{a_1+}^{{\alpha }_1,{\beta }_1;{\psi }_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right],{\psi }_1\right)}{∥{}^H{\mathbb{D}}_{a_2+}^{{\alpha }_2,{\beta }_2;{\psi }_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right],{\psi }_2\right)}}{\Gamma \left({\alpha }_1\right)\Gamma \left({\alpha }_2\right)},$$

∀$x_i\in \left[a_i,b_i\right]$; $i=1,2.$

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 $\left[a_1,b_1\right]\times \left[a_2,b_2\right].$ □

It follows the right side analog of last theorem.

Theorem 13.

Let $i=1,2;$${\psi }_i,$$f_i\in C^{n_i}\left(\left[a_i,b_i\right]\right)$, with ${\psi }_i$ being strictly increasing over $\left[a_i,b_i\right]$, where $n_i-1<{\alpha }_i<n_i$, $0\le {\beta }_i\le 1$, and ${\gamma }_i={\alpha }_i+{\beta }_i\left(n_i-{\alpha }_i\right)$, $x_i\in \left[a_i,b_i\right]$. Let also $p,q>1:\frac{1}{p}+\frac{1}{q}=1$, such that ${\alpha }_1>\frac{1}{q}$ and ${\alpha }_2>\frac{1}{p}$. Assume that $f_{i{\psi }_i}^{\left[n_i-k_i\right]}\left(I_{b_i-}^{\left(1-{\beta }_i\right)\left(n_i-{\alpha }_i\right);{\psi }_i}{f}_i\right)\left(b_i\right)=0$, for $k_i=1,\dots ,n_i-1$. Then

$${\int }_{a_1}^{b_1}{\int }_{a_2}^{b_2}\frac{\left|f_1\left(x_1\right)\right|\left|f_2\left(x_2\right)\right|d{x}_{1}d{x}_2}{\left(\frac{{\left({\psi }_1\left(b_1\right)-{\psi }_1\left(x_1\right)\right)}^{p\left({\alpha }_1-1\right)+1}}{p\left(p\left({\alpha }_1-1\right)|+1\right)}+\frac{{\left({\psi }_2\left(b_2\right)-{\psi }_2\left(x_2\right)\right)}^{q\left({\alpha }_2-1\right)+1}}{q\left(q\left({\alpha }_2-1\right)+1\right)}\right)}\le$$

$$\frac{\left(b_1-a_1\right)\left(b_2-a_2\right)}{\Gamma \left({\alpha }_1\right)\Gamma \left({\alpha }_2\right)}{∥{}^H{\mathbb{D}}_{b_1-}^{{\alpha }_1,{\beta }_1;{\psi }_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right],{\psi }_1\right)}{∥{}^H{\mathbb{D}}_{b_2-}^{{\alpha }_2,{\beta }_2;{\psi }_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right],{\psi }_2\right)}.$$

(95)

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;$${\alpha }_i>0$, ${\alpha }_i\notin \mathbb{N}$, $⌈{\alpha }_i⌉=n_i,$$0<{\beta }_i<1,$$f_i\in C^{n_i}\left(\left[a_i,b_i\right]\right)$, $\left[a_i,b_i\right]\subset \mathbb{R}$ and set ${\gamma }_i={\alpha }_i+{\beta }_i\left(n_i-{\alpha }_i\right)$. Assume further that ${\Delta }_{a_i+}^{{\gamma }_i}{f}_i\in C\left(\left[a_i,b_i\right]\right):{\Delta }_{a_i+}^{{\gamma }_i-j_i}{f}_i\left(a_i\right)=0$, for $j_i=1,\dots ,n_i$. Let also ${\overline{\alpha }}_i>0:⌈{\overline{\alpha }}_i⌉={\overline{n}}_i$, with ${\overline{\gamma }}_i={\overline{\alpha }}_i+{\beta }_i\left({\overline{n}}_i-{\overline{\alpha }}_i\right)$, and assume that ${\alpha }_i>{\overline{\alpha }}_i$ and ${\gamma }_i>{\overline{\gamma }}_i$. Furthermore, let $p,q>1:\frac{1}{p}+\frac{1}{q}=1$, such that ${\alpha }_1>\frac{1}{q}$ and ${\alpha }_2>\frac{1}{p}$. Then

$${\int }_{a_1}^{b_1}{\int }_{a_2}^{b_2}\frac{\left|{}^H{\mathbb{D}}_{a_1+}^{{\overline{\alpha }}_1,{\beta }_1}{f}_1\left(x_1\right)\right|\left|{}^H{\mathbb{D}}_{a_2+}^{{\overline{\alpha }}_2,{\beta }_2}{f}_2\left(x_2\right)\right|d{x}_{1}d{x}_2}{\left(\frac{{\left(x_1-a_1\right)}^{p\left({\alpha }_1-{\overline{\alpha }}_1-1\right)+1}}{p\left(p\left({\alpha }_1-{\overline{\alpha }}_1-1\right)|+1\right)}+\frac{{\left(x_2-a_2\right)}^{q\left({\alpha }_2-{\overline{\alpha }}_2-1\right)+1}}{q\left(q\left({\alpha }_2-{\overline{\alpha }}_2-1\right)+1\right)}\right)}\le$$

$$\frac{\left(b_1-a_1\right)\left(b_2-a_2\right)}{\Gamma \left({\alpha }_1-{\overline{\alpha }}_1\right)\Gamma \left({\alpha }_2-{\overline{\alpha }}_2\right)}{∥{}^H{\mathbb{D}}_{a_1+}^{{\alpha }_1,{\beta }_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right]\right)}{∥{}^H{\mathbb{D}}_{a_2+}^{{\alpha }_2,{\beta }_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right]\right)}.$$

(96)

Proof. 

Similar to Theorem 12, by the use of Theorem 4. □

It follows

Theorem 15.

Let $i=1,2;$${\alpha }_i>0$, ${\alpha }_i\notin \mathbb{N}$, $⌈{\alpha }_i⌉=n_i,$$0<{\beta }_i<1,$$f_i\in C^{n_i}\left(\left[a_i,b_i\right]\right)$, $\left[a_i,b_i\right]\subset \mathbb{R}$ and set ${\gamma }_i={\alpha }_i+{\beta }_i\left(n_i-{\alpha }_i\right)$. Assume further that ${\Delta }_{b_i-}^{{\gamma }_i}{f}_i\in C\left(\left[a_i,b_i\right]\right):{\Delta }_{b_i-}^{{\gamma }_i-j_i}{f}_i\left(b_i\right)=0$, for $j_i=1,\dots ,n_i$. Let also ${\overline{\alpha }}_i>0:⌈{\overline{\alpha }}_i⌉={\overline{n}}_i$, with ${\overline{\gamma }}_i={\overline{\alpha }}_i+{\beta }_i\left({\overline{n}}_i-{\overline{\alpha }}_i\right)$, and assume that ${\alpha }_i>{\overline{\alpha }}_i$ and ${\gamma }_i>{\overline{\gamma }}_i$. Furthermore, let $p,q>1:\frac{1}{p}+\frac{1}{q}=1$, such that ${\alpha }_1>\frac{1}{q}$ and ${\alpha }_2>\frac{1}{p}$. Then

$${\int }_{a_1}^{b_1}{\int }_{a_2}^{b_2}\frac{\left|{}^H{\mathbb{D}}_{b_1-}^{{\overline{\alpha }}_1,{\beta }_1}{f}_1\left(x_1\right)\right|\left|{}^H{\mathbb{D}}_{b_2-}^{{\overline{\alpha }}_2,{\beta }_2}{f}_2\left(x_2\right)\right|d{x}_{1}d{x}_2}{\left(\frac{{\left(b_1-x_1\right)}^{p\left({\alpha }_1-{\overline{\alpha }}_1-1\right)+1}}{p\left(p\left({\alpha }_1-{\overline{\alpha }}_1-1\right)|+1\right)}+\frac{{\left(b_2-x_2\right)}^{q\left({\alpha }_2-{\overline{\alpha }}_2-1\right)+1}}{q\left(q\left({\alpha }_2-{\overline{\alpha }}_2-1\right)+1\right)}\right)}\le$$

$$\frac{\left(b_1-a_1\right)\left(b_2-a_2\right)}{\Gamma \left({\alpha }_1-{\overline{\alpha }}_1\right)\Gamma \left({\alpha }_2-{\overline{\alpha }}_2\right)}{∥{}^H{\mathbb{D}}_{b_1-}^{{\alpha }_1,{\beta }_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right]\right)}{∥{}^H{\mathbb{D}}_{b_2-}^{{\alpha }_2,{\beta }_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right]\right)}.$$

(97)

Proof. 

Similar to Theorem 12, by the use of Theorem 5. □

We finish with two applications:

Corollary 1.

All as in Theorem 12, with ${\psi }_1\left(x_1\right)=e^{x_1}$, ${\psi }_2\left(x_2\right)=e^{x_2}$. Then

$${\int }_{a_1}^{b_1}{\int }_{a_2}^{b_2}\frac{\left|f_1\left(x_1\right)\right|\left|f_2\left(x_2\right)\right|d{x}_{1}d{x}_2}{\left(\frac{{\left(e^{x_1}-e^{a_1}\right)}^{p\left({\alpha }_1-1\right)+1}}{p\left(p\left({\alpha }_1-1\right)|+1\right)}+\frac{{\left(e^{x_2}-e^{a_2}\right)}^{q\left({\alpha }_2-1\right)+1}}{q\left(q\left({\alpha }_2-1\right)+1\right)}\right)}\le$$

$$\frac{\left(b_1-a_1\right)\left(b_2-a_2\right)}{\Gamma \left({\alpha }_1\right)\Gamma \left({\alpha }_2\right)}{∥{}^H{\mathbb{D}}_{a_1+}^{{\alpha }_1,{\beta }_1;e^{x_1}}{f}_1∥}_{L_q\left(\left[a_1,b_1\right],e^{x_1}\right)}{∥{}^H{\mathbb{D}}_{a_2+}^{{\alpha }_2,{\beta }_2;e^{x_2}}{f}_2∥}_{L_p\left(\left[a_2,b_2\right],e^{x_2}\right)}.$$

(98)

Proof. 

By Theorem 12. □

Corollary 2.

All as in Theorem 13, with $\left[a_i,b_i\right]\subset \left(0,+\infty \right)$, $i=1,2;$ and ${\psi }_1\left(x_1\right)=ln{x}_1$, ${\psi }_2\left(x_2\right)=ln{x}_2$. Then

$${\int }_{a_1}^{b_1}{\int }_{a_2}^{b_2}\frac{\left|f_1\left(x_1\right)\right|\left|f_2\left(x_2\right)\right|d{x}_{1}d{x}_2}{\left(\frac{{\left(ln\frac{b_1}{x_1}\right)}^{p\left({\alpha }_1-1\right)+1}}{p\left(p\left({\alpha }_1-1\right)|+1\right)}+\frac{{\left(ln\frac{b_2}{x_2}\right)}^{q\left({\alpha }_2-1\right)+1}}{q\left(q\left({\alpha }_2-1\right)+1\right)}\right)}\le$$

$$\frac{\left(b_1-a_1\right)\left(b_2-a_2\right)}{\Gamma \left({\alpha }_1\right)\Gamma \left({\alpha }_2\right)}{∥{}^H{\mathbb{D}}_{b_1-}^{{\alpha }_1,{\beta }_1;ln{x}_1}{f}_1∥}_{L_q\left(\left[a_1,b_1\right],ln{x}_1\right)}{∥{}^H{\mathbb{D}}_{b_2-}^{{\alpha }_2,{\beta }_2;ln{x}_2}{f}_2∥}_{L_p\left(\left[a_2,b_2\right],ln{x}_2\right)}.$$

(99)

Proof. 

By Theorem 13. □