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On a multiple Hilbert-type integral inequality involving the upper limit functions

Abstract

By applying the weight functions, the idea of introducing parameters and the technique of real analysis, a new multiple Hilbert-type integral inequality involving the upper limit functions is given. The constant factor related to the gamma function is proved to be the best possible in a condition. A corollary about the case of the nonhomogeneous kernel and some particular inequalities are obtained.

Introduction

Assuming that \(0 < \sum_{m = 1}^{\infty } a_{m}^{2} < \infty \) and \(0 < \sum_{n = 1}^{\infty } b_{n}^{2} < \infty \), we have the following Hilbert’s inequality with the best possible constant factor π (cf. [1], Theorem 315):

$$ \sum_{m = 1}^{\infty } \sum _{n = 1}^{\infty } \frac{a_{m}b_{n}}{m + n} < \pi \Biggl(\sum _{m = 1}^{\infty } a_{m}^{2} \sum_{n = 1}^{\infty } b_{n}^{2} \Biggr)^{1/2}. $$
(1)

If \(0 < \int _{0}^{\infty } f^{2}(x)\,dx < \infty \) and \(0 < \int _{0}^{\infty } g^{2}(y) \,dy < \infty \), then we still have the following integral analogue of (1), named Hilbert’s integral inequality (cf. [1], Theorem 316):

$$ \int _{0}^{\infty } \int _{0}^{\infty } \frac{f(x)g(y)}{x + y}\,dx\,dy < \pi \biggl( \int _{0}^{\infty } f^{2}(x)\,dx \int _{0}^{\infty } g^{2}(y) \,dy \biggr)^{1/2}, $$
(2)

where the constant factor π is the best possible. Inequalities (1) and (2) play an important role in analysis and its applications (cf. [213]).

The following half-discrete Hilbert-type inequality was provided: If \(K(x)\ (x > 0)\) is a decreasing function, \(p > 1,\frac{1}{p} + \frac{1}{q} = 1,0 < \phi (s) = \int _{0}^{\infty } K(x)x^{s - 1} \,dx < \infty \), \(f(x) \ge 0, 0 < \int _{0}^{\infty } f^{p} (x)\,dx < \infty \), then (cf. [1], Theorem 351)

$$ \sum_{n = 1}^{\infty } n^{p - 2}\biggl( \int _{0}^{\infty } K (nx)f(x)\,dx\biggr)^{p} < \phi ^{p}\biggl(\frac{1}{q}\biggr) \int _{0}^{\infty } f^{p} (x)\,dx. $$
(3)

In recent years, some new extensions of (3) were provided by [1419].

In 2006, by using Euler–Maclaurin summation formula, Krnic et al. [20] gave an extension of (1) with the kernel \(\frac{1}{(m + n)^{\lambda }}\ (0 < \lambda \le 4)\). In 2019, following the result of [20], Adiyasuren et al. [21] considered an extension of (1) involving the partial sums. In 2016–2017, by applying the weight functions, Hong [22, 23] obtained some equivalent statements of the extensions of (1) and (2) with a few parameters. A few similar works were provided by [2438].

In this paper, following the idea of [21], by using the weight functions, the way of introducing parameters and the technique of real analysis, a new multiple Hilbert-type integral inequality with the kernel \(\frac{1}{(x{}_{1} + \cdots + x_{n})^{\lambda }}\ (\lambda > 0)\) involving the upper limit functions is given. The constant factor related to the gamma function is proved to be the best possible in a condition. A corollary about the case of the nonhomogeneous kernel and some particular inequalities are obtained.

Some lemmas

In what follows, we assume that \(n \in \mathrm{N}\backslash \{ 1\}: = \{ 2,3, \ldots \},p_{i},r_{i} > 1\ (i = 1, \ldots,n),\sum_{i = 1}^{n} \frac{1}{p_{i}} = 1,\lambda > 0\), \(c_{\lambda }: = (1 - \sum_{j = 1}^{n} \frac{1}{r_{j}} )\lambda, f_{i}(x)\) (\(i = 1, \ldots,n\)) are nonnegative measurable functions in \(R_{ +} = (0,\infty )\) such that \(f_{i}(x) = o(e^{tx})\ (t > 0;x \to \infty )\), and for any \(A = (0,a)\ (a > 0), f_{i} \in L^{1}(A)\), the upper limit functions are defined by \(F_{i}(x): = \int _{0}^{x} f_{i}(t)\,dt\ (x \ge 0)\), satisfying

$$ 0 < \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - c_{\lambda } - 1} F_{i}^{p_{i}}(x_{i})\,dx_{i} < \infty \quad (i = 1, \ldots,n). $$

By the definition of the gamma function, for \(x_{i} > 0\ (i = 1, \ldots,n)\), the following expression holds:

$$ \frac{1}{(\sum_{i = 1}^{n} x_{i} )^{\lambda }} = \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } t^{\lambda - 1} e^{ - t\sum _{i = 1}^{n} x_{i}} \,dt. $$
(4)

Lemma 1

For \(t > 0\), we have the following expressions:

$$ \int _{0}^{\infty } e^{ - tx} f_{i}(x) \,dx = t \int _{0}^{\infty } e^{ - tx} F_{i}(x) \,dx\quad (i =,1, \ldots,n), $$
(5)

Proof

In view of \(F_{i}(0) = 0\),we find

$$\begin{aligned} \int _{0}^{\infty } e^{ - tx} f_{i}(x) \,dx &= \int _{0}^{\infty } e^{ - tx} \,dF_{i}(x) \\ &= e^{ - tx}F_{i}(x)|_{0}^{\infty } - \int _{0}^{\infty } F_{i} (x) \,de^{ - tx}\\ & = \lim_{x \to \infty } \frac{F_{i}(x)}{e^{tx}} + t \int _{0}^{\infty } e^{ - tx} F_{i}(x) \,dx. \end{aligned}$$

If \(F_{i}(\infty ) = \mathrm{constant}\), then \(\lim_{x \to \infty } \frac{F_{i}(x)}{e^{tx}} = 0\) and (5) follows; if \(F_{i}(\infty ) = \infty \), since \(f_{i}(x) = o(e^{tx})\ (x \to \infty )\), we find

$$\begin{aligned} \int _{0}^{\infty } e^{ - tx} f_{i}(x) \,dx &= \lim_{x \to \infty } \frac{F'_{i}(x)}{(e^{tx})'_{x}} + t \int _{0}^{\infty } e^{ - tx} F{}_{i}(x) \,dx \\ &= \lim_{x \to \infty } \frac{f_{i}(x)}{te^{tx}} + t \int _{0}^{\infty } e^{ - tx} F_{i}(x) \,dx = 0 + t \int _{0}^{\infty } e^{ - tx} F_{i}(x) \,dx, \end{aligned}$$

and then (5) follows, too.

The lemma is proved. □

Lemma 2

For \(x_{i} > 0\ (i = 1, \ldots,n)\), the following expression holds:

$$ A: = \prod_{i = 1}^{n} \Biggl[ x_{i}^{(\frac{\lambda }{r_{i}} - 1)(1 - p_{i})}\prod_{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr]^{\frac{1}{p_{i}}} = 1. $$
(6)

Proof

We have

$$\begin{aligned} A &= \prod_{i = 1}^{n} \Biggl[ x_{i}^{(\frac{\lambda }{r_{i}} - 1)(1 - p_{i}) + 1 - \frac{\lambda }{r_{i}}}\prod_{j = 1}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr]^{\frac{1}{p_{i}}} = \prod _{i = 1}^{n} \bigl[ x_{i}^{(\frac{\lambda }{r_{i}} - 1)( - p_{i})} \bigr]^{\frac{1}{p_{i}}} \Biggl( \prod_{j = 1}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr)^{\frac{1}{p_{i}}} \\ &= \prod_{i = 1}^{n} x_{i}^{1 - \frac{\lambda }{r_{i}}} \Biggl( \prod_{j = 1}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr)^{\sum _{i = 1}^{n} \frac{1}{p_{i}}} = \prod_{i = 1}^{n} x_{i}^{1 - \frac{\lambda }{r_{i}}} \Biggl( \prod_{j = 1}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr) = 1, \end{aligned}$$

and then (6) follows.

The lemma is proved. □

Lemma 3

For \(n \in \mathrm{N}\backslash \{ 1\} \), defining the following weight functions:

$$ \omega _{\lambda }^{(i)}(x_{i}): = x_{i}^{\frac{\lambda }{r_{i}} + c_{\lambda }} \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i} )^{\lambda }} \prod _{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \,dx_{1} \cdots \,dx_{i - 1}\,dx_{i + 1} \cdots \,dx_{n}, $$
(7)

we have

$$ \omega _{\lambda }^{(i)}(x_{i}) = k_{\lambda }^{(i)}: = \frac{\Gamma (\lambda (1 - \frac{1}{r_{i}}))}{\Gamma (\sum_{j = 1(j \ne i)}^{n} \frac{\lambda }{r_{j}} )} \cdot \frac{\prod_{j = 1}^{n} \Gamma (\frac{\lambda }{r_{j}} )}{\Gamma (\lambda )} \in \mathrm{R}_{ +}\quad (i = 1, \ldots,n). $$
(8)

In particular, for \(\sum_{i = 1}^{n} \frac{1}{r_{i}} = 1\), we have

$$ k_{\lambda }^{(i)} = k_{\lambda }: = \frac{1}{\Gamma (\lambda )}\prod _{j = 1}^{n} \Gamma \biggl(\frac{\lambda }{r_{j}} \biggr)\quad (i = 1, \ldots,n). $$
(9)

Proof

For \(j \ne i\), setting \(u_{j} = \frac{x_{j}}{x_{i}}\) in (7), we have

$$\begin{aligned} \omega _{\lambda }^{(i)}(x_{i}) ={}& \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(u_{1} + \cdots u_{i - 1} + 1 + u_{i + 1} + \cdots + u{}_{n})^{\lambda }} \\ &{}\times \prod _{j = 1(j \ne i)}^{n} u_{j}^{\frac{\lambda }{r_{j}} - 1} \,du_{1} \cdots \,du_{i - 1}\,du_{i + 1} \cdots \,du_{n}. \end{aligned}$$

Then by Lemma 9.15 and (9.1.19) (cf. [2], p. 341–342), we obtain (8).

The lemma is proved. □

Lemma 4

We have the following inequality:

$$\begin{aligned} H_{\lambda }&: = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} F_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< \prod_{i = 1}^{n} \biggl[ k_{\lambda }^{(i)} \int _{0}^{\infty } x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr]^{\frac{1}{p_{i}}}. \end{aligned}$$
(10)

Proof

By (6) and Hölder’s integral inequality (cf. [39]), we obtain

$$\begin{aligned} H_{\lambda } ={}& \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} \Biggl[ x_{i}^{(\frac{\lambda }{r_{i}} - 1)(1 - p_{i})} \prod_{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr]^{\frac{1}{p_{i}}}F_{i} (x_{i})\,dx_{1} \cdots \,dx_{n} \\ \le{}& \prod_{i = 1}^{n} \Biggl\{ \int _{0}^{\infty } \Biggl[ \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} x_{i}^{\frac{\lambda }{r_{i}} + c_{\lambda }} \prod_{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \,dx_{1} \cdots \,dx_{i - 1} \,dx_{i + 1} \cdots \,dx_{n} \Biggr] \\ &{} \times x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \Biggr\} ^{\frac{1}{p_{i}}} \\ = {}&\prod_{i = 1}^{n} \biggl[ \int _{0}^{\infty } \omega _{\lambda }^{(i)}(x_{i})x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr]^{\frac{1}{p_{i}}}. \end{aligned}$$
(11)

If (11) takes the form of an equality, then there exist constants \(C_{i},C_{k}\ (i \ne k)\) such that they are not all zero and

$$\begin{aligned} &C_{i}x_{i}^{\frac{\lambda }{r_{i}} + c_{\lambda }} \prod _{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \\ &\quad = C_{k}x_{k}^{\frac{\lambda }{r_{i}} + c_{\lambda }} \prod _{j = 1(j \ne k)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} x_{k}^{p_{k}(1 - \frac{\lambda }{r_{k}}) - c_{\lambda } - 1}F_{k}^{p_{k}}(x_{k})\quad\text{a.e. in } \mathrm{R}_{ +}. \end{aligned}$$

namely, \(C_{i}x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}})}F_{i}^{p_{i}}(x_{i}) = C_{k}x_{k}^{p_{k}(1 - \frac{\lambda }{r_{k}})}F_{k}^{p_{k}}(x_{k}) = C\text{ a.e. in }\mathrm{R}_{ +} \). Assuming that \(C_{i} \ne 0\), we have

$$ x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) = \frac{C}{C_{i}}x_{i}^{ - c_{\lambda } - 1}\quad \text{a.e. in }\mathrm{R}_{ +}, $$

which contradicts the fact that \(0 < \int _{0}^{\infty } x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1} F_{i}^{p_{i}}(x_{i})\,dx_{i} < \infty \), in view of \(\int _{0}^{\infty } x_{i}^{ - c_{\lambda } - 1} \,dx_{i} = \infty \). Then by (8) and (11), we have (10).

The lemma is proved. □

Remark 1

Replacing λ (resp. \(\frac{\lambda }{r_{i}}\)) by \(\lambda + n\) (resp. \(\frac{\lambda }{r_{i}} + 1\)) in (10), we have

$$\begin{aligned} H_{\lambda + n}& = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda + n}} \prod _{i = 1}^{n} F_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< \prod_{i = 1}^{n} \biggl( \tilde{k}_{\lambda + n}^{(i)} \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr)^{\frac{1}{p_{i}}}, \end{aligned}$$
(12)

where we denote

$$ \tilde{k}_{\lambda + n}^{(i)}: = \frac{\Gamma (\lambda (1 - \frac{1}{r_{i}}) + n - 1)}{\Gamma (\sum_{j = 1(j \ne i)}^{n} (\frac{\lambda }{r_{j}} + 1))} \cdot \frac{\prod_{j = 1}^{n} \Gamma (\frac{\lambda }{r_{j}} + 1 )}{\Gamma (\lambda + n)} \in \mathrm{R}_{ +} \quad(i = 1, \ldots,n). $$

Main results and a corollary

Theorem 1

We have the following inequality:

$$\begin{aligned} I&: = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}\prod_{i = 1}^{n} \biggl( \tilde{k}_{\lambda + n}^{(i)} \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr)^{\frac{1}{p_{i}}}. \end{aligned}$$
(13)

In particular, for \(\sum_{i = 1}^{n} \frac{1}{r_{i}} = 1\), we have

$$ 0 < \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - 1} F_{i}^{p_{i}}(x_{i})\,dx_{i} < \infty\quad (i = 1, \ldots,n), $$

and the following inequality:

$$\begin{aligned} I &= \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma \biggl(\frac{\lambda }{r_{i}}\biggr) \biggl( \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr)^{\frac{1}{p_{i}}}. \end{aligned}$$
(14)

Proof

By (4) and (5), we have

$$\begin{aligned} I& = \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } \cdots \int _{0}^{\infty } \prod_{i = 1}^{n} f_{i} (x_{i}) \int _{0}^{\infty } t^{\lambda - 1} e^{ - t(x_{1} + \cdots + x_{n})}\,dt \,dx_{1} \cdots \,dx_{n} \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } t^{\lambda - 1} \prod _{i = 1}^{n} \int _{0}^{\infty } e^{ - tx_{i}} f_{i} (x_{i})\,dx_{i}\,dt \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } t^{\lambda + n - 1} \prod _{i = 1}^{n} \int _{0}^{\infty } e^{ - tx_{i}} F_{i} (x_{i})\,dx_{i}\,dt \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } \cdots \int _{0}^{\infty } \prod_{i = 1}^{n} F_{i} (x_{i}) \int _{0}^{\infty } t^{\lambda + n - 1} e^{ - t(x_{1} + \cdots + x_{n})}\,dt \,dx_{1} \cdots \,dx_{n} \\ &= \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}H_{\lambda + n}. \end{aligned}$$

Then by (12), we have (13).

The theorem is proved. □

Theorem 2

The constant factor \(\frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}})\) in (14) is the best possible.

Proof

For any \(0 < \varepsilon < \lambda \min_{1 \le i \le n}\{ \frac{p_{i}}{r_{i}}\}\), we set

$$ \tilde{f}_{i}(x_{i}): = \textstyle\begin{cases} 0,&0 < x_{i} \le 1, \\ x_{i}^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}} - 1},&x_{i} > 1, \end{cases}\displaystyle \quad(i = 1, \ldots,n). $$

We obtain that \(\tilde{f}_{i}(x_{i}) = o(e^{tx_{i}})\ (t > 0;x_{i} \to \infty )\), and \(\tilde{F}_{i}(x_{i}) \equiv 0\ (0 < x_{i} \le 1)\),

$$ \tilde{F}_{i}(x_{i}) = \int _{0}^{x_{i}} \tilde{f}_{i}(t)\,dt = \int _{1}^{x_{i}} t^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}} - 1}\,dt = \frac{x_{i}^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}}} - 1}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}}} < \frac{x_{i}^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}}}}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}}}\quad (x_{i} > 1;i = 1, \ldots,n). $$

If there exists a positive constant \(M(M \le \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}}))\) such that (14) is valid when replacing \(\frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}})\) by M, then in particular, by substitution of \(f_{i}(x_{i}) = \tilde{f}_{i}(x_{i})\text{ and }F_{i}(x_{i}) = \tilde{F}_{i}(x_{i})\), we have

$$\begin{aligned} \tilde{I}&: = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} \tilde{f}_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< M\prod_{i = 1}^{n} \biggl( \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - 1} \tilde{F}_{i}^{p_{i}}(x_{i})\,dx_{i} \biggr)^{\frac{1}{p_{i}}} \\ &= M\prod_{i = 1}^{n} \frac{1}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p{}_{i}}} \biggl( \int _{1}^{\infty } x_{i}^{ - \varepsilon - 1} \,dx_{i} \biggr)^{\frac{1}{p_{i}}} = \frac{M}{\varepsilon } \prod _{i = 1}^{n} \frac{1}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p{}_{i}}}.. \end{aligned}$$

In view of Lemma 9.1.4 (9.1.5) in [2], we find

$$ I_{\varepsilon }: = \varepsilon \tilde{I} = \varepsilon \int _{1}^{\infty } \cdots \int _{1}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} x_{i}^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}} - 1} \,dx_{1} \cdots \,dx_{n} = k{}_{\lambda } + o(1)\quad \bigl( \varepsilon \to 0^{ +} \bigr). $$

Hence, we have

$$ \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \Gamma \biggl(\frac{\lambda }{r_{i}} \biggr) + o(1) = k{}_{\lambda } + o(1) = \varepsilon \tilde{I} < M\prod_{i = 1}^{n} \frac{1}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p{}_{i}}}. $$

For \(\varepsilon \to 0^{ +} \), we find

$$ \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma \biggl(\frac{\lambda }{r_{i}} \biggr) \le M, $$

which yields that the constant factor \(M = \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}})\) in (14) is the best possible.

The theorem is proved. □

Setting \(x = \frac{1}{x_{1}},f(x) = x^{\lambda - 2}f_{1}(\frac{1}{x})\) in I of (14), we have

$$ I = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{f(x)}{(1 + \sum_{i = 2}^{n} xx_{i})^{\lambda }} \prod _{i = 2}^{n} f_{i} (x_{i})\,dx \,dx_{2} \cdots \,dx_{n}. $$

For \(f_{1}(t) = t^{\lambda - 2}f(\frac{1}{t})\), we find

$$ F_{1}(x_{1}) = \int _{0}^{x_{1}} f_{1} (t)\,dt = \int _{0}^{x_{1}} t^{\lambda - 2}f \biggl( \frac{1}{t}\biggr)\,dt. $$

Then, replacing back x (resp. \(f(x)\)) by \(x_{1}\) (resp. \(f_{1}(x_{1})\)), we have

Corollary 1

If \(\tilde{F}_{1}(x_{1}) = \int _{0}^{x_{1}} t^{\lambda - 2}f_{1} (\frac{1}{t})\,dt\),

$$ \tilde{F}_{i}(x_{i}): = \int _{0}^{x_{i}} f_{i}(t)\,dt\quad (i = 2, \ldots,n), $$

then we have the following inequality with the nonhomogeneous kernel:

$$\begin{aligned} &\int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(1 + \sum_{i = 2}^{n} x_{1}x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &\quad< \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma \biggl(\frac{\lambda }{r_{i}}\biggr) \biggl( \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - 1} \tilde{F}_{i}^{p_{i}}(x_{i})\,dx_{i} \biggr)^{\frac{1}{p_{i}}}, \end{aligned}$$
(15)

where the constant factor \(\frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}})\) in (15) is the best possible.

Remark 2

(i) For \(n = 2\), (14) reduces to (cf. [40])

$$\begin{aligned} &\int _{0}^{\infty } \int _{0}^{\infty } \frac{f_{1}(x_{1})f_{2}(x_{2})}{(x_{1} + x_{2})^{\lambda }} \,dx_{1} \,dx_{2} \\ &\quad< \frac{\lambda ^{2}}{r_{1}r_{2}}B\biggl(\frac{\lambda }{r_{1}},\frac{\lambda }{r_{2}}\biggr) \biggl( \int _{0}^{\infty } x_{1}^{ - p_{1}\frac{\lambda }{r_{1}} - 1}F_{1}^{p_{1}}(x_{1}) \,dx_{1} \biggr)^{\frac{1}{p_{1}}} \biggl( \int _{0}^{\infty } x_{2}^{ - p_{2}\frac{\lambda }{r_{2}} - 1}F_{2}^{p_{2}}(x_{2}) \,dx_{2} \biggr)^{\frac{1}{p_{2}}}, \end{aligned}$$
(16)

and (15) reduces to the following new inequality:

$$\begin{aligned} &\int _{0}^{\infty } \int _{0}^{\infty } \frac{f_{1}(x_{1})f_{2}(x_{2})}{(1 + x_{1}x_{2})^{\lambda }} \,dx_{1} \,dx_{2} \\ &\quad < \frac{\lambda ^{2}}{r_{1}r_{2}}B\biggl(\frac{\lambda }{r_{1}},\frac{\lambda }{r_{2}}\biggr) \biggl( \int _{0}^{\infty } x_{1}^{ - p_{1}\frac{\lambda }{r_{1}} - 1} \tilde{F}_{1}^{p_{1}}(x_{1})\,dx_{1} \biggr)^{\frac{1}{p_{1}}} \biggl( \int _{0}^{\infty } x_{2}^{ - p_{2}\frac{\lambda }{r_{2}} - 1} \tilde{F}_{2}^{p_{2}}(x_{2})\,dx_{2} \biggr)^{\frac{1}{p_{2}}}. \end{aligned}$$
(17)

(ii) For \(r_{i} = p_{i}\ (i = 1, \ldots,n)\), (14) reduces to

$$\begin{aligned} &\int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &\quad < \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{p_{i}}\Gamma \biggl(\frac{\lambda }{p_{i}}\biggr) \biggl( \int _{0}^{\infty } x_{i}^{ - \lambda - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr)^{\frac{1}{p_{i}}}, \end{aligned}$$
(18)

and (15) reduces to

$$\begin{aligned} &\int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(1 + \sum_{i = 2}^{n} x_{1}x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &\quad < \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{p_{i}}\Gamma \biggl(\frac{\lambda }{p_{i}}\biggr) \biggl( \int _{0}^{\infty } x_{i}^{ - \lambda - 1} \tilde{F}_{i}^{p_{i}}(x_{i})\,dx_{i} \biggr)^{\frac{1}{p_{i}}}. \end{aligned}$$
(19)

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

Conclusions

In this paper, following the idea of [21], by the use of the weight functions, the way of introducing parameters and the technique of real analysis, a new multiple Hilbert-type integral inequality with the kernel \(\frac{1}{(x_{1} + \cdots + x_{n})^{\lambda }}\ (\lambda > 0)\) involving the upper limit functions is given in Theorem 1. In a condition, the best possible constant factor related to the gamma function and a few parameters is proved in Theorem 2. A corollary about the case of nonhomogeneous kernel and some particular inequalities are obtained in Corollary 1 and Remark 2. The lemmas and theorems provide an extensive account of this type of inequalities.

Availability of data and materials

The data used to support the findings of this study are included within the article.

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Acknowledgements

The authors thank the referee for his useful propose to reform the paper.

Funding

This work is supported by the National Natural Science Foundation (No. 61772140), and Characteristic innovation project of Guangdong Provincial Colleges and universities in 2020 (No. 2020KTSCX088). We are grateful for their help.

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Authors

Contributions

BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. JZ participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.

Corresponding author

Correspondence to Jianhua Zhong.

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Zhong, J., Yang, B. On a multiple Hilbert-type integral inequality involving the upper limit functions. J Inequal Appl 2021, 17 (2021). https://doi.org/10.1186/s13660-021-02551-9

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MSC

  • 26D15

Keywords

  • Weight function
  • Hilbert-type integral inequality
  • Upper limit function
  • Parameter
  • Gamma function