Open AccessArticle

A New Hardy–Hilbert-Type Integral Inequality Involving One Multiple Upper Limit Function and One Derivative Function of Higher Order

Bicheng Yang

¹ and

Michael Th. Rassias

^2,3,*

Department of Mathematics, Guangdong University of Education, Guangzhou 510303, China

Department of Mathematics and Engineering Sciences, Hellenic Military Academy, 16673 Vari Attikis, Greece

Institute for Advanced Study, Program in Interdisciplinary Studies, 1 Einstein Dr, Princeton, NJ 08540, USA

Author to whom correspondence should be addressed.

Axioms 2023, 12(5), 499; https://doi.org/10.3390/axioms12050499

Submission received: 8 April 2023 / Revised: 11 May 2023 / Accepted: 16 May 2023 / Published: 19 May 2023

(This article belongs to the Special Issue Fractional Calculus - Theory and Applications II)

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Abstract

Using weight functions and parameters, as well as applying real analytic techniques, we derive a new Hardy–Hilbert-type integral inequality with the homogeneous kernel

\frac{1}{{(x + y)}^{λ + n}}

involving one multiple upper limit function and one derivative function of higher order. Certain equivalent statements of the optimal constant factor related to some parameters are considered. A few particular inequalities and the case of reverses are also provided.

Keywords:

weight function; Hardy–Hilbert-type integral inequality; multiple upper limit function; derivative function of higher order; parameter; gamma function; reverse

MSC:

26D15; 47A05

1. Introduction

Assuming that

p > 1, \frac{1}{p} + \frac{1}{q} = 1, a_{m}, b_{n} \geq 0,

0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty

and

0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty,

the following Hardy–Hilbert inequality with the optimal constant factor

π / sin (\frac{π}{p})

has been proven (cf. [1], Theorem 315):

\sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{a_{m} b_{n}}{m + n} < \frac{π}{sin (π / p)} {(\sum_{m = 1}^{\infty} a_{m}^{p})}^{\frac{1}{p}} {(\sum_{n = 1}^{\infty} b_{n}^{q})}^{\frac{1}{q}} .

(1)

f (x), g (y) \geq 0,

0 < \int_{0}^{\infty} f (x) d x < \infty

and

0 < \int_{0}^{\infty} g (y) d y < \infty,

then we still have the following integral analogue of (1) known as Hardy–Hilbert integral inequality (cf. [1], Theorem 316):

\int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{x + y} d x d y < \frac{π}{sin (π / p)} {(\int_{0}^{\infty} f^{p} (x) d x)}^{\frac{1}{p}} {(\int_{0}^{\infty} g^{q} (y) d y)}^{\frac{1}{q}},

(2)

where the identical constant factor

π / sin (\frac{π}{p})

remains optimal. Inequalities (1) and (2) have proven to be essential in various applications of mathematical analysis (cf. [2,3,4,5,6,7,8,9,10,11,12,13]).

In 2006, applying the Euler–Maclaurin summation formula, Krnic et al. [14] established an extension of (1) with the kernel

\frac{1}{{(m + n)}^{λ}} (0 < λ \leq 4)

. Making use of the result of [14], in 2019, Adiyasuren et al. [15] considered an extension of (1), which involved two partial sums, and subsequently, in 2020, Mo et al. [16] proved an extension of (2), which involved two upper-limit functions. In 2016–2017, Hong et al. [17,18] presented several equivalent statements of the extensions of (1) and (2) with the best possible constant factors and multi-parameters. Some similar results were established in [19,20,21,22,23,24,25,26,27].

In the present paper, following the methods of [15,17], using weight functions and parameters, as well as applying real analytic techniques, we prove a new Hardy–Hilbert-type integral inequality with the kernel

\frac{1}{{(x + y)}^{λ + n}}

involving one multiple upper limit function and one derivative function of higher order. Equivalent statements of the best possible constant factor related to the parameters are considered. Some particular inequalities and the case of reverses are obtained. The lemmas and theorems provide an extensive account of this type of inequalities.

2. Some Lemmas

In what follows, we suppose that

p > 0

(p \neq 1), \frac{1}{p} + \frac{1}{q} = 1

, m, n \in N = {1, 2, \dots},

λ > - min {m, n},

- m < λ_{1} < λ,

0 < λ_{2} < λ + m,

{\hat{λ}}_{1} : = \frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q},

{\hat{λ}}_{2} : = \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p}

and the following

Assumption (I):

For

F_{0} (x) : = f (x)

, being a non-negative continuous function, except for finitely many points in

R_{+} = (0, \infty),

satisfying

f (x) = o (e^{t x}) (t > 0; x \to \infty),

F_{k} (x) : = \int_{0}^{x} F_{k - 1} (x) d x (k = 1, \dots, m),

g^{(n)} (y)

is a non-negative continuous function, except for finite points in

R_{+},

satisfying

g^{(n)} (y) = o (e^{t y}) (t > 0; y \to \infty),

g^{(0)} (y) = g (y),

g^{(j - 1)} (y)

is a non-negative differentiable function in

R_{+}

with

g^{(j - 1)} (0^{+}) = 0 (j = 1, \dots, n) .

We also assume that

0 < \int_{0}^{\infty} x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) d x < \infty and 0 < \int_{0}^{\infty} y^{q (1 - {\hat{λ}}_{2}) - 1} {(g^{(n)} (y))}^{q} d y < \infty .

Note: According to Assumption (I), since

f (x) \geq 0,

F_{k} (x)

is increasing and

F_{k} (\infty) = \infty

or constant. If there exists a last constant

k_{0} \leq k

such that

F_{k_{0}} (\infty) =

constant, then

lim_{x \to \infty} \frac{F_{k} (x)}{e^{t x}} = \dots = \frac{1}{t^{k - k_{0}}} lim_{x \to \infty} \frac{F_{k_{0}} (x)}{e^{t x}} = 0;

otherwise, we still have

lim_{x \to \infty} \frac{F_{k} (x)}{e^{t x}} = \dots = \frac{1}{t^{k}} lim_{x \to \infty} \frac{f (x)}{e^{t x}} = 0,

namely,

F_{k} (x) = o (e^{t x}) (t > 0; x \to \infty) (k = 1, \dots, m) .

In the same way, we still can show that

g^{(j - 1)} (y) = o (e^{t y}) (t > 0; y \to \infty) (j = 1, \dots, n) .

We define the gamma function as follows:

Γ (α) : = \int_{0}^{\infty} e^{- t} t^{α - 1} d t^{} (α > 0),

(3)

satisfying

Γ (α + 1) = α Γ {(α)}^{} (α > 0)

, and define the following beta function (cf. [28]):

B (u, v) : = \int_{0}^{\infty} \frac{t^{u - 1}}{{(1 + t)}^{u + v}} d t = \frac{Γ (u) Γ (v)}{Γ (u + v)} (u, v > 0) .

(4)

According to (3), for

λ, x, y > 0,

we still have the following formula related to the gamma function:

\frac{1}{{(x + y)}^{λ}} = \frac{1}{Γ (λ)} \int_{0}^{\infty} e^{- (x + y) t} t^{λ - 1} d t .

(5)

Lemma 1.

For

t > 0, m, n \in N,

we have the following expressions:

\begin{matrix} \int_{0}^{\infty} e^{- t x} f (x) d x & = & t^{m} \int_{0}^{\infty} e^{- t x} F_{m} (x) d x, \end{matrix}

(6)

\begin{matrix} \int_{0}^{\infty} e^{- t y} g (y) d y & = & t^{- n} \int_{0}^{\infty} e^{- t y} g^{(n)} (y) d y . \end{matrix}

(7)

Proof.

According to Assumption (I), for

k = 1, \dots, m,

we get that

e^{- t x} F_{k} {(x) |}_{0}^{\infty} = 0

, then integration by parts,

\begin{matrix} \int_{0}^{\infty} e^{- t x} F_{k - 1} (x) d x \\ = & \int_{0}^{\infty} e^{- t x} d F_{k} (x) = e^{- t x} F_{k} {(x) |}_{0}^{\infty} - \int_{0}^{\infty} F_{k} (x) d e^{- t x} \\ = & t \int_{0}^{\infty} e^{- t x} F_{k} (x) (x) d x . \end{matrix}

Substituting

k = 1, \dots, m

in the above expression, by simplification, we obtain (6).

According to Assumption (I), for

j = 1, \dots, n,

we get that

e^{- t y} g^{(j - 1)} {(y) |}_{0}^{\infty} = 0,

and

\begin{matrix} \int_{0}^{\infty} e^{- t y} g^{(j)} (y) d y \\ = & \int_{0}^{\infty} e^{- t y} d g^{(j - 1)} (y) = e^{- t y} g^{(j - 1)} {(y) |}_{0}^{\infty} - \int_{0}^{\infty} g^{(j - 1)} (y) d e^{- t y} \\ = & t \int_{0}^{\infty} e^{- t y} g^{(j - 1)} (y) d y . \end{matrix}

Substituting

j = 1, \dots, n

in the above expression, we obtain (7).

This completes the proof of the lemma. □

Note: (6) (resp. (7)) is naturally the value for

m = 0

(resp.

n = 0) .

Lemma 2.

For

0 < s_{1}, s_{2} < s,

define the following weight functions:

\begin{matrix} ω_{s} (s_{2}, x) & : & = x^{s - s_{2}} \int_{0}^{\infty} \frac{t^{s_{2} - 1}}{{(x + t)}^{s}} d t^{} (x \in R_{+}), \end{matrix}

(8)

\begin{matrix} ω_{s} (s_{1}, y) & : & = y^{s - s_{1}} \int_{0}^{\infty} \frac{t^{s_{1} - 1}}{{(x + t)}^{s}} d t^{} (y \in R_{+}) . \end{matrix}

(9)

We have the following expressions:

\begin{matrix} ω_{s} (s_{2}, x) & : = & B {(s_{2}, s - s_{2})}^{} (x \in R_{+}), \end{matrix}

(10)

\begin{matrix} ω_{s} (s_{1}, y) & : = & B {(s_{1}, s - s_{1})}^{} (y \in R_{+}) . \end{matrix}

(11)

Proof.

Setting

u = \frac{t}{x}

, we derive

\begin{matrix} ω_{s} (s_{2}, x) & = & x^{s - s_{2}} \int_{0}^{\infty} \frac{{(u x)}^{s_{2} - 1}}{{(x + u x)}^{s}} x d u \\ = & \int_{0}^{\infty} \frac{u^{s_{2} - 1} d u}{{(1 + u)}^{s}} = B (s_{2}, s - s_{2}), \end{matrix}

namely, (10) follows. In the same way, we obtain (11).

This completes the proof of the lemma. □

Lemma 3.

Suppose that

λ > - m, - m < λ_{1} < λ, 0 < λ_{2} < λ + m .

(i): For $p > 1,$ we have the following extended Hardy–Hilbert integral inequality:

$\begin{matrix} I_{λ + m} & : = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{F_{m} (x) g^{(n)} (y)}{{(x + y)}^{λ + m}} d x d y \\ < & B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ \times {[\int_{0}^{\infty} x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - {\hat{λ}}_{2}) - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}}; \end{matrix}$

(12)
(ii): for $0 < p < 1,$ we have the reverse of (13).

Proof.

(i) By Hölder’s inequality (cf. [29]), we obtain

\begin{matrix} I_{λ + m} & = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{{(x + y)}^{λ + m}} [\frac{y^{(λ_{2} - 1) / p}}{x^{(λ_{1} + m - 1) / q}} F_{m} (x)] \\ \times [\frac{x^{(λ_{1} + m - 1) / q}}{y^{(λ_{2} - 1) / p}} g^{(n)} (y)] d x d y \\ \leq & {\{\int_{0}^{\infty} [\int_{0}^{\infty} \frac{1}{{(x + y)}^{λ + m}} \frac{y^{λ_{2} - 1}}{x^{(λ_{1} + m - 1) (p - 1)}} d y] F_{m}^{p} (x) d x\}}^{\frac{1}{p}} \\ \times {\{\int_{0}^{\infty} [\int_{0}^{\infty} \frac{1}{{(x + y)}^{λ + m}} \frac{x^{λ_{1} + m - 1}}{y^{(λ_{2} - 1) (q - 1)}} d x] {(g^{(n)} (y))}^{q} d y\}}^{\frac{1}{q}} \\ = & {[\int_{0}^{\infty} ω_{λ + m} (λ_{2}, x) x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} \\ \times {[\int_{0}^{\infty} ω_{λ + m} (λ_{1} + m, y) y^{q (1 - {\hat{λ}}_{2}) - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}} . \end{matrix}

(13)

If (14) retains the form of equality, then, there exist constants A and B such that they are not both zero, satisfying

A \frac{y^{λ_{2} - 1}}{x^{(λ_{1} + m - 1) (p - 1)}} F_{m}^{p} (x) = B \frac{x^{λ_{1} + m - 1}}{y^{(λ_{2} - 1) (q - 1)}} {(g^{(n)} (y))}^{q} a . e . in R_{+}^{2} .

Assuming that

A \neq 0

, for fixed

a, e,

y \in R_{+},

we have

x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) = [\frac{B}{A} y^{q (1 - {\hat{λ}}_{2})} {(g^{(n)} (y))}^{q}] x^{- 1 - (λ - λ_{1} - λ_{2})} a . e . in R_{+} .

Since for any

a : = λ - λ_{1} - λ_{2} \in R

\int_{0}^{\infty} x^{- 1 - a} d x = \infty

, the above expression contradicts the fact that

0 < \int_{0}^{\infty} x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) d x < \infty .

Therefore, by (10) and (11), setting

s = λ + m

(> 0), s_{1} = λ_{1} + m

(\in (0, λ + m)),

s_{2} = λ_{2}

(\in (0, λ + m))

, in view of (14), we have (13).

(ii) Similarly, according to the reverse Hölder inequality, we obtain the reverse of (13).

This completes the proof of the lemma. □

3. Main Results

Theorem 1.

Suppose that

λ > - min {m, n},

- m < λ_{1} < λ, 0 < λ_{2} < λ + m .

(i): For $p > 1,$ we have the following Hardy–Hilbert-type integral inequality involving one multiple upper limit function and one derivative function of higher order:

$\begin{matrix} I & : = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + n}} d x d y \\ < & \frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ \times {[\int_{0}^{\infty} x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - {\hat{λ}}_{2}) - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}} . \end{matrix}$

(14)
(ii): For $0 < p < 1,$ we obtain the reverse of (14).

In particular, for

λ_{1} + λ_{2} = λ

(0 < λ_{1}, λ_{2} < λ),

we reduce (14) to the following:

\begin{matrix} I & = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + n}} d x d y < \frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2}) \\ \times {[\int_{0}^{\infty} x^{p (1 - λ_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - λ_{2}) - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}}, \end{matrix}

(15)

where the constant factor

\frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2})

is the best possible.

Proof.

(i) In view of (6), (7) and Fubini’s theorem (cf. [30]), we obtain

\begin{matrix} I & = & \frac{1}{Γ (λ + n)} \int_{0}^{\infty} \int_{0}^{\infty} f (x) g (y) [\int_{0}^{\infty} t^{λ + n - 1} e^{- (x + y) t} d t] d x d y \\ = & \frac{1}{Γ (λ + n)} \int_{0}^{\infty} t^{λ + n - 1} (\int_{0}^{\infty} e^{- x t} f (x) d x) (\int_{0}^{\infty} g (y) e^{- y t} d y) d t \\ = & \frac{1}{Γ (λ + n)} \int_{0}^{\infty} t^{λ + n - 1} (t^{m} \int_{0}^{\infty} e^{- x t} F_{m} (x) d x) (t^{- n} \int_{0}^{\infty} g^{(n)} (y) e^{- y t} d y) d t \\ = & \frac{1}{Γ (λ + n)} \int_{0}^{\infty} \int_{0}^{\infty} F_{m} (x) g^{(n)} (y) [\int_{0}^{\infty} t^{λ + m - 1} e^{- (x + y) t} d t] d x d y \\ = & \frac{Γ (λ + m)}{Γ (λ + n)} \int_{0}^{\infty} \int_{0}^{\infty} \frac{F_{m} (x) g^{(n)} (y)}{{(x + y)}^{λ + m}} d x d y = \frac{Γ (λ + m) I_{λ + m}}{Γ (λ + n)} . \end{matrix}

(16)

Then, according to (13), we obtain (14).

(ii) According to (17) and the reverse of (13), we derive the reverse of (14).

For

λ_{1} + λ_{2} = λ

(0 < λ_{1}, λ_{2} < λ)

in (14), we deduce (15).

For any

0 < ε < min {p λ_{1}, q λ_{2}},

we set the following functions:

\begin{matrix} {\tilde{F}}_{0} (x) & = & \tilde{f} (x) : = \{\begin{matrix} 0, 0 < x < 1, \\ x^{λ_{1} - \frac{ε}{p} - 1}, x \geq 1 \end{matrix}, \\ {\tilde{F}}_{k} (x) & = & \int_{0}^{x} (\int_{0}^{t_{k - 1}} \dots \int_{0}^{t_{2}} \tilde{f} (t_{1}) d t_{1} \dots d t_{k - 1}) d t_{k} (k = 1, \dots, m), \end{matrix}

and then

{\tilde{F}}_{m} (x) = 0, 0 < x < 1,

\begin{matrix} {\tilde{F}}_{m} (x) & = & \int_{1}^{x} (\int_{1}^{t_{m - 1}} \dots \int_{1}^{t_{2}} t_{1}^{λ_{1} - \frac{ε}{p} - 1} d t_{1} \dots d t_{m - 1}) d t_{m} \\ = & {[\prod_{i = 0}^{m - 1} (λ_{1} - \frac{ε}{p} + i)]}^{- 1} (x^{λ_{1} - \frac{ε}{p} + m - 1^{}} - O_{1} (x^{m - 1})) \\ \leq & {[\prod_{i = 0}^{m - 1} (λ_{1} - \frac{ε}{p} + i)]}^{- 1} x^{λ_{1} - \frac{ε}{p} + m - 1^{}} (x \geq 1), \end{matrix}

where

O_{1} (x^{m - 1})

(x \geq 1)

is a positive polynomial of

(m - 1)

-degree. We also set

\begin{matrix} {\tilde{g}}^{(n)} (y) & = & \{\begin{matrix} 0, 0 < y \leq 1, \\ \prod_{i = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}) y^{λ_{2} - \frac{ε}{q} - 1}, y > 1 \end{matrix}, \\ {\tilde{g}}^{(l)} (y) & = & \prod_{j = 0}^{l - 1} (λ_{2} + j - \frac{ε}{q}) \int_{0}^{y} (\int_{0}^{t_{n - l - 1}} \dots \int_{0}^{t_{2}} {\tilde{g}}^{(n)} (t_{1}) d t_{1} \dots d t_{n - l - 1}) d t_{n - l} \\ (l & = & 1, \dots, n), \end{matrix}

and

\tilde{g} (y) = 0

(0 < y \leq 1),

\begin{matrix} \tilde{g} (y) & = & \prod_{j = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}) \int_{1}^{y} (\int_{1}^{t_{n - 1}} \dots \int_{1}^{t_{2}} t_{1}^{λ_{2} - \frac{ε}{q} - 1} d t_{1} \dots d t_{n - 1}) d t_{n} \\ = & y^{λ_{2} - \frac{ε}{q} + n - 1} - O_{2} (y^{n - 1}) \leq y^{λ_{2} - \frac{ε}{q} + n - 1} (y > 1), \end{matrix}

where

O_{2} (y^{n - 1})

(y > 1)

is a positive polynomial of

n - 1

-degree.

We observe that

{\tilde{F}}_{k} (x)

(k = 0, \dots, m)

and

{\tilde{g}}^{(l)} (y)

(l = 0, \dots, n)

all satisfy Assumption (I) on

F_{k}, g^{(l)}

If there exists a positive constant

M \leq \frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2})

such that (15) is valid when we replace

\frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2})

with M, then, in particular, since

\begin{matrix} \tilde{J} & : & = {[\int_{0}^{\infty} x^{p (1 - λ_{1} - m) - 1} {\tilde{F}}_{m}^{p} (x)) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - λ_{2}) - 1} {({\tilde{g}}^{(n)} (y))}^{q} d y]}^{\frac{1}{q}} \\ \leq & {[\prod_{i = 0}^{m - 1} (λ_{1} + i - \frac{ε}{p})]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}) \int_{1}^{\infty} x^{- ε - 1} d x \\ = & \frac{1}{ε} {[\prod_{i = 0}^{m - 1} (λ_{1} + i - \frac{ε}{p})]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}), \end{matrix}

we have

\begin{matrix} \tilde{I} & : = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{\tilde{f} (x) \tilde{g} (y)}{{(x + y)}^{λ + n}} d x d y \\ < & M \tilde{J} = \frac{M}{ε} {[\prod_{i = 0}^{m - 1} (λ_{1} + i - \frac{ε}{p})]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}) . \end{matrix}

In view of Fubini’s theorem (cf. [30]), it follows that

\tilde{I} = \int_{1}^{\infty} [\int_{1}^{\infty} \frac{y^{λ_{2} + n - \frac{ε}{q} - 1} - O_{2} (y^{n - 1})}{{(x + y)}^{λ + n}} d y] x^{λ_{1} - \frac{ε}{p} - 1} d x = I_{0} - I_{1},

where

\begin{matrix} I_{0} & = & \int_{1}^{\infty} [\int_{1}^{\infty} \frac{y^{λ_{2} + n - \frac{ε}{q} - 1}}{{(x + y)}^{λ + n}} d y] x^{λ_{1} + m - \frac{ε}{p} - 1} d x \\ = & \int_{1}^{\infty} x^{- ε - 1} [\int_{\frac{1}{x}}^{\infty} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u] d x \\ = & \int_{1}^{\infty} x^{- ε - 1} [\int_{\frac{1}{x}}^{1} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u] d x + \frac{1}{ε} \int_{1}^{\infty} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u \\ = & \int_{0}^{1} (\int_{\frac{1}{u}}^{1} x^{- ε - 1} d x) \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u + \frac{1}{ε} \int_{1}^{\infty} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u \\ = & \frac{1}{ε} [\int_{0}^{1} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u + \int_{1}^{\infty} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u], \end{matrix}

\begin{matrix} 0 & < & I_{1} : = \int_{1}^{\infty} [\int_{1}^{\infty} \frac{O_{2} (y^{n - 1})}{{(x + y)}^{λ + n}} d y] x^{λ_{1} - \frac{ε}{p} - 1} d x \\ = & \int_{1}^{\infty} [\int_{1}^{\infty} \frac{O_{2} (y^{n - 1})}{{(x + y)}^{(λ_{2} / 2) + n}} d y] \frac{x^{λ_{1} - \frac{ε}{p} - 1}}{{(x + y)}^{λ_{1} + (λ_{2} / 2)}} d x \\ \leq & \int_{1}^{\infty} [\int_{1}^{\infty} \frac{O_{2} (y^{n - 1})}{y^{(λ_{2} / 2) + n}} d y] \frac{x^{λ_{1} - \frac{ε}{p} - 1}}{x^{λ_{1} + (λ_{2} / 2)}} d x \\ = & \int_{1}^{\infty} x^{- \frac{λ_{2}}{2} - \frac{ε}{p} - 1} d x [\int_{1}^{\infty} O (y^{- \frac{λ_{2}}{2} - 1}) d y] \leq M_{1} < \infty . \end{matrix}

According to the above results, it follows that

\begin{matrix} \int_{0}^{1} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u + \int_{1}^{\infty} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u - ε I_{1} \\ = & ε \tilde{I} < M {[\prod_{i = 0}^{m - 1} (λ_{1} + i - \frac{ε}{p})]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}) . \end{matrix}

Letting

ε \to 0^{+}

in the above inequality, in view of the continuity of the beta function, we obtain

B (λ_{1}, λ_{2} + n) \leq M {[\prod_{i = 0}^{m - 1} (λ_{1} + i)]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j),

namely,

\frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2}) = B (λ_{1}, λ_{2} + n) \prod_{i = 0}^{m - 1} (λ_{1} + i) {[\prod_{j = 0}^{n - 1} (λ_{2} + j)]}^{- 1} \leq M

and then

M = \frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2})

is the best possible constant factor of (15).

This completes the proof of the theorem. □

Remark 1.

For

{\hat{λ}}_{1} = \frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q},

{\hat{λ}}_{2} = \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p},

it follows that

{\hat{λ}}_{1} +

{\hat{λ}}_{2} = λ .

We find

0 < {\hat{λ}}_{1} < \frac{λ}{p} + \frac{λ}{q} = λ

, then

0 < {\hat{λ}}_{2} = λ - {\hat{λ}}_{1} < λ

. According to Hölder’s inequality (cf. [29]), it follows that

\begin{matrix} 0 & < & B ({\hat{λ}}_{1} + m, {\hat{λ}}_{2}) = \int_{0}^{\infty} \frac{u^{\frac{λ + m - λ_{2}}{p} + \frac{λ_{1} + m}{q} - 1}}{{(1 + u)}^{λ + m}} d u \\ = & \int_{0}^{\infty} \frac{1}{{(1 + u)}^{λ + m}} (u^{\frac{λ + m - λ_{2} - 1}{p}}) (u^{\frac{λ_{1} + m - 1}{q}}) d u \\ \leq & {[\int_{0}^{\infty} \frac{u^{λ + m - λ_{2} - 1}}{{(1 + u)}^{λ + m}} d u]}^{\frac{1}{p}} {[\int_{0}^{\infty} \frac{u^{λ_{1} + m - 1}}{{(1 + u)}^{λ + m}} d u]}^{\frac{1}{q}} \\ = & B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) < \infty . \end{matrix}

(17)

Theorem 2.

For

p > 1,

if the constant factor

\frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})

in (14) is the best possible, then for

0 < λ_{1}, λ_{2} < λ,

we have

λ_{1} + λ_{2} = λ

Proof.

According to (15) (for

λ_{i} = {\hat{λ}}_{i}^{} (i = 1, 2)

), since

\frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})

is the best possible constant factor in (14), we have

\begin{matrix} \frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ \leq & \frac{Γ (λ + m)}{Γ (λ + n)} B ({\hat{λ}}_{1} + m, {\hat{λ}}_{2}) (\in R_{+}), \end{matrix}

namely

B ({\hat{λ}}_{1} + m, {\hat{λ}}_{2}) \geq B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) .

It follows that (18) retains the form of equality.

We observe that (18) retains the form of equality if and only if there exist constants A and B such that they are not both zero and

A u^{λ + m - λ_{2} - 1} = B u^{λ_{1} + m - 1}

a . e .

R_{+}

(cf. [29]). Assuming that

A \neq 0

, it follows that

u^{λ - λ_{2} - λ_{1}} = B / A

a . e .

R_{+}

, namely,

λ - λ_{2} - λ_{1} = 0

. Hence, we have

λ_{1} + λ_{2} = λ

This completes the proof of the theorem. □

Theorem 3.

The following statements ((i), (ii), (iii) and (iv)) are equivalent:

(i): Both

$B^{\frac{1}{p}} (λ_{2}, λ + m + - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})$

and

$B (\frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q} + m, \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p})$

are independent of $p, q$ ;
(ii): $B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) = B (\frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q} + m, \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p})$ ;
(iii): For $0 < λ_{1}, λ_{2} < λ,$ we have $λ_{1} + λ_{2} = λ$ ;
(iv): The constant factor

$\frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})$

is the best possible in (14).

Proof.

(i) \Rightarrow (i i)

. In view of the continuity of the beta function, we derive

\begin{matrix} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ = & lim_{p \to \infty} lim_{q \to 1^{+}} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ = & B (λ_{1} + m, λ - λ_{1}), \\ B (\frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q} + m, \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p}) \\ = & lim_{p \to \infty} lim_{q \to 1^{+}} B (\frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q} + m, \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p}) \\ = & B (λ_{1} + m, λ - λ_{1}) . \end{matrix}

Hence, we have

\begin{matrix} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ = & B (\frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q} + m, \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p}) . \end{matrix}

(i i) \Rightarrow (i i i)

. In view of

B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) = B ({\hat{λ}}_{1}, {\hat{λ}}_{2}),

(17) retains the form of equality. In view of the proof of Theorem 2, we have

λ_{1} + λ_{2} = λ .

(i i i) \Rightarrow (i v)

. If

λ_{1} + λ_{2} = λ

(0 < λ_{1}, λ_{2} < λ)

, then according to Theorem 1, the constant factor

\frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})

in (14) is the best possible.

(i v) \Rightarrow (i)

. According to Theorem 2, we have

λ_{1} + λ_{2} = λ

; then,

\begin{matrix} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) & = & B (λ_{1} + m, λ_{2}), \\ B (\frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q} + m, \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p}) & = & B (λ_{1} + m, λ_{2}), \end{matrix}

both of which are independent of

p, q

Hence, statements (i), (ii), (iii) and (iv) are equivalent.

This completes the proof of the theorem. □

For

n = m,

we have:

Corollary 1.

For

p > 1,

we have the following Hardy–Hilbert-type integral inequality involving one multiple upper limit function and one derivative function of m-order:

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + m}} d x d y \\ < & B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ \times {[\int_{0}^{\infty} x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - {\hat{λ}}_{2}) - 1} {(g^{(m)} (y))}^{q} d y]}^{\frac{1}{q}} . \end{matrix}

(18)

Moreover, for

0 < λ_{1}, λ_{2} < λ,

the constant factor

B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})

in (19) is the best possible if and only if

λ_{1} + λ_{2} = λ .

For

λ_{1} + λ_{2} = λ (0 < λ_{1}, λ_{2} < λ),

we reduce (19) to the following:

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + m}} d x d y < B (λ_{1} + m, λ_{2}) \\ \times & {[\int_{0}^{\infty} x^{p (1 - λ_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - λ_{2}) - 1} {(g^{(m)} (y))}^{q} d y]}^{\frac{1}{q}}, \end{matrix}

(19)

where the constant factor

B (λ_{1} + m, λ_{2})

is the best possible factor.

Remark 2.

(i) For

λ_{1} + λ_{2} = λ

in (13), we have

\begin{matrix} I_{λ + m} & = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{F_{m} (x) g^{(n)} (y)}{{(x + y)}^{λ + m}} d x d y < B (λ_{1} + m, λ_{2}) \\ \times {[\int_{0}^{\infty} x^{p (1 - λ_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - λ_{2}) - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}} . \end{matrix}

(20)

We confirm that the constant factor

B (λ_{1} + m, λ_{2})

in (20) is the best possible. Otherwise, we would reach a contradiction according to (17) (for

λ_{1} + λ_{2} = λ

) that the constant factor in (15) is not the best possible.

(ii) In view of the note of Lemma 1, Theorem 1, Theorem 2 and Theorem 3 are valid for

m = 0

n = 0 .

For

m = n = 0, λ = 1, λ_{1} = \frac{1}{q}, λ_{2} = \frac{1}{p},

both (15) and (20) reduce to (2).

Remark 3.

(i) For

λ = 1, λ_{1} = \frac{1}{q}, λ_{2} = \frac{1}{p}

in (15), we have

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{1 + n}} d x d y < \frac{π}{n! sin (\frac{π}{p})} \prod_{i = 0}^{m - 1} (\frac{1}{q} + i) \\ \times & {(\int_{0}^{\infty} x^{- p m} F_{m}^{p} (x) d x)}^{\frac{1}{p}} {[\int_{0}^{\infty} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}}; \end{matrix}

(21)

(ii) For

λ = 1, λ_{1} = \frac{1}{p}, λ_{2} = \frac{1}{q}

in (15), we have the dual form of (21) as follows:

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{1 + n}} d x d y < \frac{π}{n! sin (\frac{π}{p})} \prod_{i = 0}^{m - 1} (\frac{1}{p} + i) \\ \times & {(\int_{0}^{\infty} x^{p (1 - m) - 2} F_{m}^{p} (x) d x)}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{p - 2} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}}; \end{matrix}

(22)

(iii) For

p = q = 2,

both (15) and (21) reduce to

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{1 + n}} d x d y \\ < & \frac{π (2 m - 1)!!}{n! 2^{m}} {[\int_{0}^{\infty} x^{- 2 m} F_{m}^{2} (x) d x \int_{0}^{\infty} {(g^{(n)} (y))}^{2} d y]}^{\frac{1}{2}} . \end{matrix}

(23)

The constant factors in the above inequalities are all the best possible.

4. The Reverses

According to Lemma 3 and Theorem 1 (ii), we have:

Theorem 4.

For

0 < p < 1

(q < 0),

we have the following reverse Hardy–Hilbert-type integral inequality involving one multiple upper limit function and one derivative function of higher order:

\begin{matrix} I & = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + n}} d x d y \\ > & \frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ \times {[\int_{0}^{\infty} x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - {\hat{λ}}_{2}) - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}} . \end{matrix}

(24)

In particular, for

λ_{1} + λ_{2} = λ

(0 < λ_{1}, λ_{2} < λ),

we reduce (24) to the following:

\begin{matrix} I & = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + n}} d x d y > \frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2}) \\ \times {[\int_{0}^{\infty} x^{p (1 - λ_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - λ_{2}) - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}}, \end{matrix}

(25)

where the constant factor

\frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2})

is the best possible.

Proof.

We only prove that the constant factor

\frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2})

in (25) is the best possible.

For any

0 < ε < λ_{1} min {p, | q |},

we consider the functions

{\tilde{F}}_{k} (x)

(k = 0, \dots, m)

and

{\tilde{g}}^{(n - l)} (y)

(l = 0, \dots, n)

as in Theorem 1. If there exists a positive constant

M \geq \frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2}),

such that (25) is valid when we replace

\frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2})

with M, then in particular, since

\begin{matrix} \tilde{J} & : = & {[\int_{0}^{\infty} x^{p (1 - λ_{1} - m) - 1} {\tilde{F}}_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - λ_{2}) - 1} {({\tilde{g}}^{(n)} (y))}^{p} d y]}^{\frac{1}{q}} \\ = & {[\int_{1}^{\infty} x^{- ε - 1} d x - \int_{1}^{\infty} O_{1} (x^{- p λ_{1} - 1}) d x]}^{\frac{1}{p}} {(\int_{1}^{\infty} y^{- ε - 1} d y)}^{\frac{1}{q}} \\ = & \frac{1}{ε} {[\prod_{i = 0}^{m - 1} (λ_{1} + i - \frac{ε}{p})]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}) {(1 - ε O (1))}^{\frac{1}{p}}, \end{matrix}

we have

\begin{matrix} \tilde{I} & : = & \int_{0}^{\infty} \int_{0}^{\infty} \frac{\tilde{f} (x) \tilde{g} (y)}{{(x + y)}^{λ}} d x d y \\ > & M \tilde{J} = \frac{M}{ε} {[\prod_{i = 0}^{m - 1} (λ_{1} + i - \frac{ε}{p})]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}) {(1 - ε O (1))}^{\frac{1}{p}} . \end{matrix}

In view of the proof of the results of Theorem 1, it follows that

\tilde{I} = \int_{1}^{\infty} [\int_{1}^{\infty} \frac{y^{λ_{2} + n - \frac{ε}{q} - 1} - O_{2} (y^{n - 1})}{{(x + y)}^{λ + n}} d y] x^{λ_{1} - \frac{ε}{p} - 1} d x = I_{0} - I_{1},

where

\begin{matrix} I_{0} & = & \int_{1}^{\infty} [\int_{1}^{\infty} \frac{y^{λ_{2} + n - \frac{ε}{q} - 1}}{{(x + y)}^{λ + n}} d y] x^{λ_{1} - \frac{ε}{p} - 1} d x \\ = & \frac{1}{ε} [\int_{0}^{1} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u + \int_{1}^{\infty} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u] . \end{matrix}

0 < I_{1} = \int_{1}^{\infty} [\int_{1}^{\infty} \frac{O_{2} (y^{n - 1})}{{(x + y)}^{λ + n}} d y] x^{λ_{1} - \frac{ε}{p} - 1} d x \leq M_{2} < \infty,

According to the above results, it follows that

\begin{matrix} \int_{0}^{1} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u + \int_{1}^{\infty} \frac{u^{λ_{2} + n - \frac{ε}{q} - 1}}{{(1 + u)}^{λ + n}} d u - ε I_{1} \\ = & ε \tilde{I} > M {[\prod_{i = 0}^{m - 1} (λ_{1} + i - \frac{ε}{p})]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j - \frac{ε}{q}) {(1 - ε O (1))}^{\frac{1}{p}} . \end{matrix}

Letting

ε \to 0^{+}

in the above inequality, in view of the continuity of the beta function, we obtain

B (λ_{1}, λ_{2} + n) \geq M {[\prod_{i = 0}^{m - 1} (λ_{1} + i)]}^{- 1} \prod_{j = 0}^{n - 1} (λ_{2} + j),

namely

\frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2}) = B (λ_{1}, λ_{2} + n) \prod_{i = 0}^{m - 1} (λ_{1} + i) {[\prod_{j = 0}^{n - 1} (λ_{2} + j)]}^{- 1} \geq M

and then,

M = \frac{Γ (λ + m)}{Γ (λ + n)} B (λ_{1} + m, λ_{2})

is the best possible constant factor of (25).

This completes the proof of the theorem. □

Remark 4.

For

{\hat{λ}}_{1} = \frac{λ - λ_{2}}{p} + \frac{λ_{1}}{q} = \frac{λ - λ_{1} - λ_{2}}{p} + λ_{1},

{\hat{λ}}_{2} = \frac{λ - λ_{1}}{q} + \frac{λ_{2}}{p} = \frac{λ - λ_{1} - λ_{2}}{q} + λ_{2},

it follows that

{\hat{λ}}_{1} +

{\hat{λ}}_{2} = λ .

We find that for

- p λ_{1} < λ - λ_{1} - λ_{2} < p (λ - λ_{1}),

0 < {\hat{λ}}_{1} < λ;

then,

0 < {\hat{λ}}_{2} = λ - {\hat{λ}}_{1} < λ .

According to the reverse Hölder inequality (cf. [29]), we obtain

\begin{matrix} \infty & > & B ({\hat{λ}}_{1} + m, {\hat{λ}}_{2}) = \int_{0}^{\infty} \frac{u^{\frac{λ + m - λ_{2}}{p} + \frac{λ_{1} + m}{q} - 1}}{{(1 + u)}^{λ + m}} d u \\ = & \int_{0}^{\infty} \frac{1}{{(1 + u)}^{λ + m}} (u^{\frac{λ + m - λ_{2} - 1}{p}}) (u^{\frac{λ_{1} + m - 1}{q}}) d u \\ \geq & {[\int_{0}^{\infty} \frac{u^{λ + m - λ_{2} - 1}}{{(1 + u)}^{λ + m}} d u]}^{\frac{1}{p}} {[\int_{0}^{\infty} \frac{u^{λ_{1} + m - 1}}{{(1 + u)}^{λ + m}} d u]}^{\frac{1}{q}} \\ = & B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) > 0 . \end{matrix}

(26)

Theorem 5.

For

0 < p < 1,

0 < λ_{1}, λ_{2} < λ,

if the constant factor

\frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})

in (24) is the best possible, then for

- p λ_{1} < λ - λ_{1} - λ_{2} < p (λ - λ_{1}),

we have

λ_{1} + λ_{2} = λ

Proof.

For

- p λ_{1} < λ - λ_{1} - λ_{2} < p (λ - λ_{1}),

by (25) (for

λ_{i} = {\hat{λ}}_{i}^{} (i = 1, 2)

), since

\frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})

is the best possible constant factor in (24), we have

\begin{matrix} \frac{Γ (λ + m)}{Γ (λ + n)} B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ \geq & \frac{Γ (λ + m)}{Γ (λ + n)} B ({\hat{λ}}_{1} + m, {\hat{λ}}_{2}) (\in R_{+}), \end{matrix}

namely,

B ({\hat{λ}}_{1} + m, {\hat{λ}}_{2}) \leq B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) .

It follows that (27) retains the form of equality.

We observe that (27) retains the form of equality if and only if there exist constants A and B such that they are not both zero and

A u^{λ + m - λ_{2} - 1} = B u^{λ_{1} + m - 1}

a . e .

R_{+}

(cf. [29]). Assuming that

A \neq 0

, it follows that

u^{λ - λ_{2} - λ_{1}} = B / A

a . e .

R_{+}

, namely

λ - λ_{2} - λ_{1} = 0

; then,

λ_{1} + λ_{2} = λ

This completes the proof of the theorem. □

For

n = m,

we have:

Corollary 2.

For

0 < p < 1,

we have the following reverse Hardy–Hilbert-type integral inequality involving one multiple upper limit function and one derivative function of m-order:

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + m}} d x d y \\ > & B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1}) \\ \times {[\int_{0}^{\infty} x^{p (1 - {\hat{λ}}_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - {\hat{λ}}_{2}) - 1} {(g^{(m)} (y))}^{q} d y]}^{\frac{1}{q}} . \end{matrix}

(27)

Moreover, if

0 < λ_{1}, λ_{2} < λ,

the constant factor

B^{\frac{1}{p}} (λ_{2}, λ + m - λ_{2}) B^{\frac{1}{q}} (λ_{1} + m, λ - λ_{1})

in (28) is the best possible, then for

- p λ_{1} < λ - λ_{1} - λ_{2} < p (λ - λ_{1}),

we have

λ_{1} + λ_{2} = λ .

For

λ_{1} + λ_{2} = λ

(0 < λ_{1}, λ_{2} < λ),

we reduce (19) to the following:

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + m}} d x d y > B (λ_{1} + m, λ_{2}) \\ \times & {[\int_{0}^{\infty} x^{p (1 - λ_{1} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - λ_{2}) - 1} {(g^{(m)} (y))}^{q} d y]}^{\frac{1}{q}}, \end{matrix}

(28)

where the constant factor

B (λ_{1} + m, λ_{2})

is the best possible.

Remark 5.

For

r > 1, \frac{1}{r} + \frac{1}{s} = 1, λ_{1} = \frac{λ}{r}, λ_{2} = \frac{λ}{s}

in (25), we have the following reverse inequality

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ + n}} d x d y \\ > & \prod_{i = 0}^{m - 1} (\frac{λ}{r} + i) {[\prod_{j = 0}^{n - 1} (λ + j)]}^{- 1} B (\frac{λ}{r}, \frac{λ}{s}) \\ \times {[\int_{0}^{\infty} x^{p (1 - \frac{λ}{r} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{q (1 - \frac{λ}{s}) - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}}, \end{matrix}

(29)

where the constant factor

\prod_{i = 0}^{m - 1} (\frac{λ}{r} + i) {[\prod_{j = 0}^{n - 1} (λ + j)]}^{- 1} B (\frac{λ}{r}, \frac{λ}{s})

is the best possible. In particular, for

λ = 1,

we have

\begin{matrix} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{1 + n}} d x d y \\ > & \frac{π}{n! sin (\frac{π}{r})} {[\int_{0}^{\infty} x^{p (\frac{1}{s} - m) - 1} F_{m}^{p} (x) d x]}^{\frac{1}{p}} {[\int_{0}^{\infty} y^{\frac{q}{r} - 1} {(g^{(n)} (y))}^{q} d y]}^{\frac{1}{q}}, \end{matrix}

(30)

where the constant factor

\frac{π}{n! sin (\frac{π}{r})}

is still the best possible.

5. Conclusions

In the present paper, we followed the methods of [15,17], used weight functions and introduced parameters in order to prove a new Hardy–Hilbert-type integral inequality with the kernel

\frac{1}{{(x + y)}^{λ + n}}

involving one multiple upper limit function and one derivative function of higher order. In this study, we also considered equivalent statements of the best possible constant factor related to the parameters and obtained some particular inequalities, in addition to considering the case of reverses. The lemmas and theorems presented in this work provide an extensive account of this type of inequalities.

Author Contributions

Writing—original draft preparation, B.Y. and M.T.R.; writing—review and editing, B.Y. and M.T.R. Both authors contributed equally in the preparation of this work. All authors have read and agreed to the published version of the manuscript.

Funding

This work was supported by the National Natural Science Foundation (No. 61772140), Science and Technology Projects in Guangzhou (No. 202103010004) and the Characteristic Innovation Project of Guangdong Provincial Colleges and Universities (No. 2020KTSCX088).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

Conflicts of Interest

The authors have no conflict of interests.

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Yang, B.; Rassias, M.T. A New Hardy–Hilbert-Type Integral Inequality Involving One Multiple Upper Limit Function and One Derivative Function of Higher Order. Axioms 2023, 12, 499. https://doi.org/10.3390/axioms12050499

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Yang B, Rassias MT. A New Hardy–Hilbert-Type Integral Inequality Involving One Multiple Upper Limit Function and One Derivative Function of Higher Order. Axioms. 2023; 12(5):499. https://doi.org/10.3390/axioms12050499

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Yang, Bicheng, and Michael Th. Rassias. 2023. "A New Hardy–Hilbert-Type Integral Inequality Involving One Multiple Upper Limit Function and One Derivative Function of Higher Order" Axioms 12, no. 5: 499. https://doi.org/10.3390/axioms12050499

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A New Hardy–Hilbert-Type Integral Inequality Involving One Multiple Upper Limit Function and One Derivative Function of Higher Order

Abstract

1. Introduction

2. Some Lemmas

3. Main Results

4. The Reverses

5. Conclusions

Author Contributions

Funding

Institutional Review Board Statement

Informed Consent Statement

Data Availability Statement

Acknowledgments

Conflicts of Interest

References

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