A Hardy–Hilbert-type integral inequality involving two multiple upper-limit functions

Abstract

By means of the weight functions, the idea of introducing parameters and the technique of real analysis, a new Hardy–Hilbert-type integral inequality with the homogeneous kernel $\frac{1}{(x + y)^{\lambda}}\ (\lambda > 0)$ involving two multiple upper-limit functions is obtained. The equivalent statements of the best possible constant factor related to the beta and gamma functions are considered. As applications, the equivalent forms and the case of a nonhomogeneous kernel are deduced. Some particular inequalities and the operator expressions are provided.

1 Introduction

Assuming that $p > 1,\frac{1}{p} + \frac{1}{q} = 1,a_{m},b{}_{n} \ge 0,0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty$ and $0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty$, the following Hardy–Hilbert inequality with the best possible constant factor $\pi /\sin (\frac{\pi}{p})$ was provided (cf. [1], Theorem 315):

$$\begin{aligned} \sum_{n = 1}^{\infty} \sum _{m = 1}^{\infty} \frac{a_{m}b_{n}}{m + n} < \frac{\pi}{\sin (\pi /p)} \Biggl(\sum_{m = 1}^{\infty} a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n = 1}^{\infty} b_{n}^{q} \Biggr)^{\frac{1}{q}}. \end{aligned}$$

(1)

If $f(x),g(y) \ge 0,0 < \int _{0}^{\infty} f^{p}(x)\,dx < \infty$ and $0 < \int _{0}^{\infty} g^{q}(y)\,dy < \infty$, then we still have the integral analog of (1) named in the Hardy–Hilbert integral inequality as follows (cf. [1], Theorem 316):

$$\begin{aligned} \int _{0}^{\infty} \int _{0}^{\infty} \frac{f(x)g(y)}{x + y}\,dx\,dy < \frac{\pi}{ \sin (\pi /p)} \biggl( \int _{0}^{\infty} f^{p}(x)\,dx \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{\infty} g^{q}(y)\,dy \biggr)^{\frac{1}{q}}, \end{aligned}$$

(2)

where the same constant factor $\pi /\sin (\frac{\pi}{p})$ is still the best possible. Inequalities (1) and (2) played an important role in analysis and its applications (cf. [2–13]).

In 2006, by means of the Euler–Maclaurin summation formula, Krnic et al. [14] gave an extension of (1) with the kernel $\frac{1}{(m + n)^{\lambda}}\ (0 < \lambda \le 4)$. Applying the result of [14], in 2019, Adiyasuren et al. [15] considered an extension of (1) involving partial sums, and then in 2020, Mo et al. [16] gave an extension of (2) involving two upper-limit functions. In 2016–2017, Hong et al. [17, 18] provided several equivalent statements of the extensions of (1) and (2) with multiparameters. Some similar results were given by [19–22].

In this paper, following [15] and [17], by means of the weight functions, the idea of introducing parameters and the techniques of real analysis, a new Hardy–Hilbert-type integral inequality with the kernel $\frac{1}{(x + y)^{\lambda}}\ (\lambda > 0)$ involving two multiple upper-limit functions is provided. The equivalent statements of the best possible constant factor related to the beta and gamma functions are considered. As applications, the equivalent forms, the case of a nonhomogeneous kernel, a few particular inequalities and the operator expressions are deduced. The lemmas and theorems provide an extensive account of this type of inequality.

2 Some lemmas

In what follows, we assume that $p > 1,\frac{1}{p} + \frac{1}{q} = 1,0 < \lambda _{i} < \lambda\ (i = 1,2),\hat{\lambda}_{1}: = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q},\hat{\lambda}_{2}: = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p},\ F_{0}(x)$ and $G_{0}(y)$ are nonnegative Lebesgue integrable functions in any interval $(0,b] \subset R_{ +} (b > 0)$, such that $F_{0}(u),G_{0}(u) = o(e^{tu})\ (t > 0;u \to \infty )$, the multiple upper-limit functions $\{ F_{j}(x)\}_{j = 1}^{m},\{ G_{k}(y)\}_{k = 1}^{n}$ are defined inductively by $F_{j}(x): = \int _{0}^{x} F_{j - 1}(t)\,dt (x \ge 0;j = 1,2, \ldots ,m)$, and

$$\begin{aligned} G_{k}(y): = \int _{0}^{y} G_{k - 1}(t)\,dt \quad(y \ge 0;k = 1,2, \ldots ,n), \end{aligned}$$

where $F_{1}(x) = \int _{0}^{x} F_{0} (t_{0})\,dt{}_{0}, G_{1}(y) = \int _{0}^{y} G_{0} (t_{0})\,dt{}_{0}$,

$$\begin{aligned} &F_{m}(x) = \int _{0}^{x} \biggl( \int _{0}^{t_{n - 1}} \cdots \int _{0}^{t_{1}} F_{0} (t_{0}) \,dt{}_{0} \cdots \,dt_{m - 2} \biggr)\,dt_{m - 1} \quad(t_{i},x \ge 0;m \ge 2),\quad\text{and}\\ &G_{n}(y) = \int _{0}^{y} \biggl( \int _{0}^{t_{n - 1}} \cdots \int _{0}^{t_{1}} G_{0} (t_{0}) \,dt{}_{0} \cdots \,dt_{n - 2} \biggr)\,dt_{n - 1} \quad(t_{i},y \ge 0;n \ge 2). \end{aligned}$$

For $m,n \in \mathrm{N}_{0}: = \{ 0,1, \ldots \}$, we also suppose that $F_{m}(x)$ and $G_{n}(y)$ satisfy the following inequalities:

$$\begin{aligned} 0 < \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1} - m) - 1} F_{m}^{p}(x) \,dx < \infty \quad \text{and} \quad 0 < \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy < \infty . \end{aligned}$$

We indicate the gamma function as follows:

$$\begin{aligned} \Gamma (\alpha ): = \int _{0}^{\infty} e^{ - t} t^{\alpha - 1}\,dt \quad ( \alpha > 0), \end{aligned}$$

(3)

Satisfying $\Gamma (\alpha + 1) = \alpha \Gamma (\alpha )(\alpha > 0)$, and define the following beta function (cf. [23]):

$$\begin{aligned} B(u,v): = \int _{0}^{\infty} \frac{t^{u - 1}}{(1 + t)^{u + v}} \,dt = \frac{1}{\Gamma (u + v)}\Gamma (u)\Gamma (v). \end{aligned}$$

(4)

By (3), for $\lambda ,x,y > 0$, we still have the following formula related the gamma function:

$$\begin{aligned} \frac{1}{(x + y)^{\lambda}} = \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty} t^{\lambda - 1} e^{ - (x + y)t} \,dt. \end{aligned}$$

(5)

Lemma 1

For $t > 0, m,n \in \mathrm{N}$, we have the following expressions:

$$\begin{aligned} & \int _{0}^{\infty} e^{ - tx} F_{0}(x) \,dx = t^{m} \int _{0}^{\infty} e^{ - tx} F_{m}(x) \,dx, \end{aligned}$$

(6)

$$\begin{aligned} & \int _{0}^{\infty} e^{ - ty} G_{0}(y) \,dy = t^{n} \int _{0}^{\infty} e^{ - ty} G_{n}(y) \,dy. \end{aligned}$$

(7)

Proof

For $n \in \mathrm{N}: = \{ 1,2, \ldots \}$, since $F_{1}(0) = 0$, using integration by parts, we find

$$\begin{aligned} \int _{0}^{\infty} e^{ - tx} F_{0}(x) \,dx &= \int _{0}^{\infty} e^{ - tx} \,dF_{1}(x) \\ &= e^{ - tx}F_{1}(x)|_{0}^{\infty} - \int _{0}^{\infty} F_{1} (x) \,de^{ - tx} = \lim_{x \to \infty} \frac{F_{1}(x)}{e^{tx}} + t \int _{0}^{\infty} e^{ - tx} F_{1}(x) \,dx. \end{aligned}$$

If $F{}_{1}(\infty ) =$ constant, then we have $\lim_{x \to \infty} \frac{F_{1}(x)}{e^{tx}} = 0$; if $F_{1}(\infty ) = \infty$, since $F_{0}(x) = o(e^{tx})\ (t > 0;x \to \infty )$, then we obtain that $\lim_{x \to \infty} \frac{F_{1}(x)}{e^{tx}} = \lim_{x \to \infty} \frac{F_{0}(x)}{te^{tx}} = 0$. Hence, we find

$$\begin{aligned} \int _{0}^{\infty} e^{ - tx} F_{0}(x) \,dx = t \int _{0}^{\infty} e^{ - tx} F_{1}(x) \,dx. \end{aligned}$$

In the same way, we can obtain that for $F_{i}(\infty ) = \infty\ (i = 1, \ldots ,k),\lim_{x \to \infty} \frac{F_{k}(x)}{e^{tx}} = \cdots \ =\lim_{x \to \infty} \frac{F_{0}(x)}{t^{k}e^{tx}} = 0$, and then

$$\begin{aligned} \int _{0}^{\infty} e^{ - tx} F_{k - 1}(x) \,dx = t \int _{0}^{\infty} e^{ - tx} F_{k}(x) \,dx\quad (k = 1, \ldots ,m). \end{aligned}$$

Hence, we obtain (6) inductively. In the same way, we have (7).

The lemma is proved. □

Lemma 2

Define the following weight functions:

$$\begin{aligned} &\varpi (\lambda _{2},x): = x^{\lambda + m - \lambda _{2}} \int _{0}^{\infty} \frac{t^{\lambda _{2} + n - 1}}{(x + t)^{\lambda + m + n}}\,dt \quad(x \in \mathrm{R}_{ +} ), \end{aligned}$$

(8)

$$\begin{aligned} &\omega (\lambda _{1},y): = y^{\lambda + n - \lambda _{1}} \int _{0}^{\infty} \frac{t^{\lambda _{1} + m - 1}}{(t + y)^{\lambda + m + n}} \,dt\quad (y \in \mathrm{R}_{ +} ). \end{aligned}$$

(9)

We have the following expressions:

$$\begin{aligned} &\varpi (\lambda _{2},x) = B(\lambda _{2} + n,\lambda + m - \lambda _{2})\quad (x \in \mathrm{R}_{ +} ), \end{aligned}$$

(10)

$$\begin{aligned} &\omega (\lambda _{1},y) = B(\lambda _{1} + m,\lambda + n - \lambda _{1}) \quad(y \in \mathrm{R}_{ +} ). \end{aligned}$$

(11)

Proof

Setting $u = \frac{t}{x}$, we find

$$\begin{aligned} \varpi (\lambda _{2},x) &= x^{\lambda + m - \lambda _{2}} \int _{0}^{\infty} \frac{(ux)^{\lambda _{2} + n - 1}}{(x + ux)^{\lambda + m + n}}x\,du = \int _{0}^{\infty} \frac{u^{\lambda _{2} + n - 1}}{(1 + u)^{\lambda + m + n}}\,du \\ &= B(\lambda _{2} + n,\lambda + m - \lambda _{2}), \end{aligned}$$

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

The lemma is proved. □

Lemma 3

We have the following Hardy–Hilbert integral inequality:

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{m}(x)G_{n}(y)}{(x + y)^{\lambda + m + n}} \,dx \\ &\quad< B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda {}_{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda {}_{1}) \\ &\qquad{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1} - m) - 1} F_{m}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(12)

Proof

By Hölder’s inequality (cf. [24]), we obtain

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F(x)G_{n}(y)}{(x + y)^{\lambda + m + n}} \,dx\,dy \\ &\quad = \int _{0}^{\infty} \int _{0}^{\infty} \frac{1}{(x + y)^{\lambda + m + n}} \biggl[ \frac{y^{(\lambda _{2} + n - 1)/p}}{x^{(\lambda _{1} + m - 1)/q}}F_{m}(x) \biggr] \biggl[\frac{x^{(\lambda _{1} + m - 1)/q}}{y^{(\lambda _{2} + n - 1)/p}}G_{n}(y) \biggr]\,dx\,dy \\ &\quad\le \biggl\{ \int _{0}^{\infty} \biggl[ \int _{0}^{\infty} \frac{1}{(x + y)^{\lambda + m + n}} \frac{y^{\lambda _{2} + n - 1}\,dy}{x^{(\lambda _{1} + m - 1)(p - 1)}} \biggr]F_{m}^{p}(x)\,dx \biggr\} ^{\frac{1}{p}} \\ &\qquad{}\times\biggl\{ \int _{0}^{\infty} \biggl[ \int _{0}^{\infty} \frac{1}{(x + y)^{\lambda + m + n}} \frac{x^{\lambda _{1} + m - 1}\,dx}{y^{(\lambda _{2} + n - 1)(q - 1)}} \biggr]G_{n}^{q}(y)\,dy \biggr\} ^{\frac{1}{q}} \\ &\quad= \biggl[ \int _{0}^{\infty} \varpi (\lambda {}_{2},x) x^{p(1 - \hat{\lambda}_{1} - m) - 1}F_{m}^{p}(x)\,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} \omega (\lambda _{1},y) y^{q(1 - \hat{\lambda}_{2} - n) - 1}G_{n}^{q}(y)\,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(13)

If (13) keeps the form of equality, then there exist constants A and B, such that they are not both zero, satisfying

$$\begin{aligned} A\frac{y^{\lambda _{2} + n - 1}}{x^{(\lambda _{1} + m - 1)(p - 1)}}F_{m}^{p}(x) = B\frac{x^{\lambda _{1} + m - 1}}{y^{(\lambda _{2} + n + 1)(q - 1)}}G_{n}^{q}(y)\quad \text{a.e. in }(0, \infty ) \times (0,\infty ). \end{aligned}$$

Assuming that $A \ne 0$, for fixed $a.e.y \in (0,\infty )$, we have

$$\begin{aligned} x^{p(1 - \hat{\lambda}_{1} - m) - 1}F_{m}^{p}(x) = \biggl(\frac{B}{A}y^{q(1 - \lambda _{2} - n)}G_{n}^{q}(y) \biggr)x^{ - 1 - (\lambda - \lambda _{1} - \lambda _{2})}\quad\text{a.e. in }(0,\infty ). \end{aligned}$$

Since for any $a = \lambda - \lambda _{1} - \lambda _{2} \in \mathbf{R}$, $\int _{0}^{\infty} x^{ - 1 - a}\,dx = \infty$, the above expression contradicts the fact that

$$\begin{aligned} 0 < \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1} - m) - 1} F_{m}^{p}(x) \,dx < \infty . \end{aligned}$$

Therefore, by (10), (11), and (13), we have (14).

The lemma is proved. □

3 Main results

Theorem 1

We have the following Hardy–Hilbert-type integral inequality involving two multiple upper-limit functions:

$$\begin{aligned} I: ={}& \int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{(x + y)^{\lambda}} \,dx\,dy \\ < {}& \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) \\ &{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1} - m) - 1} F_{m}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(14)

In particular, for $\lambda _{1} + \lambda _{2} = \lambda$, we reduce (14) to the following:

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{(x + y)^{\lambda}} \,dx\,dy \\ &\quad < \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B(\lambda _{1} + m,\lambda _{2} + n) \\ &\qquad{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{1} - m) - 1} F_{m}^{q}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \lambda _{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

(15)

where, the constant factor $\frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B(\lambda _{1} + m,\lambda _{2} + n)$ is the best possible.

Proof

In view of (6), (7), and Fubini’s theorem (cf. [25]), we find

$$\begin{aligned} I &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty} \int _{0}^{\infty} F_{0}(x)G_{0}(y) \biggl[ \int _{0}^{\infty} t^{\lambda - 1} e^{ - (x + y)t}\,dt \biggr]\,dx\,dy \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty} t^{\lambda - 1} \biggl( \int _{0}^{\infty} e^{ - xt}F_{0}(x) \,dx\biggr) \biggl( \int _{0}^{\infty} e^{ - yt} G_{0}(y) \,dy\biggr) \,dt \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty} t^{\lambda - 1} \biggl(t^{m} \int _{0}^{\infty} e^{ - xt}F_{m}(x)\,dx \biggr) \biggl( \int _{0}^{\infty} t^{n}e^{ - yt} G_{n}(y)\,dy\biggr) \,dt \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty} \int _{0}^{\infty} F_{m}(x)G_{n}(y) \biggl[ \int _{0}^{\infty} t^{(\lambda + m + n) - 1}e^{ - (x + y)t}\,dt \biggr] \,dx\,dy \\ &= \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )} \int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{m}(x)G_{n}(y)}{(x + y)^{\lambda + m + n}} \,dx \,dy. \end{aligned}$$

(16)

Then, by (12), we have (14).

For $\lambda _{1} + \lambda _{2} = \lambda$ in (14), by simplification, we have (15). By (3) and (4), we still have

$$\begin{aligned} \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B(\lambda _{1} + m,\lambda _{2} + n)& = \frac{1}{\Gamma (\lambda )}\Gamma (\lambda _{1} + m)\Gamma (\lambda _{2} + n) \\ &= B(\lambda _{1},\lambda _{2})\prod _{i = 1}^{m} (\lambda _{1} + i - 1) \prod _{j = 1}^{n} (\lambda _{2} + j - 1), \end{aligned}$$

(17)

where, we define $\prod_{i = 1}^{0} (\lambda _{1} + i - 1) = \prod_{j = 1}^{0} (\lambda _{2} + j - 1): = 1$.

For any $0 < \varepsilon < \min \{ p\lambda _{1},q\lambda _{2}\}$, we set the following functions:

$$\begin{aligned} \tilde{F}_{0}(x): = \textstyle\begin{cases} 0,&0 < x \le 1, \\ x^{\lambda _{1} - \frac{\varepsilon}{p} - 1},&x > 1, \end{cases}\displaystyle \qquad\tilde{G}_{0}(y): = \textstyle\begin{cases} 0,&0 < y \le 1, \\ y^{\lambda _{2} - \frac{\varepsilon}{q} - 1},& y > 1. \end{cases}\displaystyle \end{aligned}$$

We obtain that $\tilde{F}_{0}(u) = \tilde{G}_{0}(u) = o(e^{tu})\ (t > 0;u \to \infty )$,

$$\begin{aligned} &\tilde{F}_{i}(u) = \tilde{G}_{j}(u) \equiv 0\quad (0 < u \le 1;i = 1, \ldots ,m,j = 1, \ldots ,n),\\ &\tilde{F}_{1}(x) = \int _{1}^{x} t^{\lambda _{1} - \frac{\varepsilon}{p} - 1} \,dt < \frac{1}{\lambda _{1} - \frac{\varepsilon}{p}}x^{\lambda _{1} - \frac{\varepsilon}{p}}\quad (x > 1), \quad\text{and}\\ &\tilde{G}_{1}(y) = \int _{1}^{y} t^{\lambda _{2} - \frac{\varepsilon}{q} - 1} \,dt < \frac{1}{\lambda _{2} - \frac{\varepsilon}{q}}y^{\lambda _{2} - \frac{\varepsilon}{q}}\quad(y > 1). \end{aligned}$$

In general, by mathematical induction, we can show the following inequalities:

$$\begin{aligned} &\tilde{F}_{m}(x) < \frac{1}{\prod_{i = 1}^{m} (\lambda _{1} - \frac{\varepsilon}{p} + i - 1)} x^{\lambda _{1} - \frac{\varepsilon}{p} + m - 1}\quad(x > 1), \\ &\tilde{G}_{n}(y) < \frac{1}{\prod_{j = 1}^{n} (\lambda _{2} - \frac{\varepsilon}{q} + j - 1)} y^{\lambda _{2} - \frac{\varepsilon}{q} + n - 1}\quad (y > 1). \end{aligned}$$

If there exists a positive constant $M \le B(\lambda _{1},\lambda _{2})\prod_{i = 1}^{m} (\lambda _{1} + i - 1) \prod_{j = 1}^{n} (\lambda _{2} + j - 1)$, such that (15) is valid when we replace $B(\lambda _{1},\lambda _{2})\prod_{i = 1}^{m} (\lambda _{1} + i - 1) \prod_{j = 1}^{n} (\lambda _{2} + j - 1)$, by M, then in particular, since

$$\begin{aligned} \tilde{J}: = {}&\biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{1} - m) - 1} \tilde{F}_{m}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \lambda _{2} - n) - 1} \tilde{G}_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}\\ < {}& \frac{1}{\prod_{i = 1}^{m} ( \lambda _{1} - \frac{\varepsilon}{p} + i - 1)\prod_{j = 1}^{n} ( \lambda _{2} - \frac{\varepsilon}{q} + j - 1)}\biggl( \int _{1}^{\infty} x^{ - \varepsilon - 1} \,dx \biggr)^{\frac{1}{p}}\biggl( \int _{1}^{\infty} y^{ - \varepsilon - 1} \,dy \biggr)^{\frac{1}{q}} \\ ={}& \frac{1}{\varepsilon \prod_{i = 1}^{m} ( \lambda _{1} - \frac{\varepsilon}{p} + i - 1)\prod_{j = 1}^{n} ( \lambda _{2} - \frac{\varepsilon}{q} + j - 1)}, \end{aligned}$$

we have

$$\begin{aligned} \tilde{I}: = \int _{0}^{\infty} \int _{0}^{\infty} \frac{\tilde{F}_{0}(x)\tilde{G}_{0}(y)}{(x + y)^{\lambda}} \,dx\,dy < M \tilde{J} < \frac{M}{\varepsilon \prod_{i = 1}^{m} ( \lambda _{1} - \frac{\varepsilon}{p} + i - 1)\prod_{j = 1}^{n} ( \lambda _{2} - \frac{\varepsilon}{q} + j - 1)}. \end{aligned}$$

(18)

In view of Fubini’s theorem (cf. [25]), it follows that

$$\begin{aligned} \tilde{I}& = \int _{1}^{\infty} \biggl[ \int _{1}^{\infty} \frac{y^{\lambda _{2} - \frac{\varepsilon}{q} - 1}}{(x + y)^{\lambda}} \,dy \biggr]x^{\lambda _{1} - \frac{\varepsilon}{p} - 1}\,dx = \int _{1}^{\infty} x^{ - \varepsilon - 1} \biggl[ \int _{\frac{1}{x}}^{\infty} \frac{u^{\lambda _{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \biggr] \,dx\\ &= \int _{1}^{\infty} x^{ - \varepsilon - 1}\biggl[ \int _{\frac{1}{x}}^{1} \frac{u^{\lambda _{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \biggr]\,dx + \int _{1}^{\infty} x^{ - \varepsilon - 1}\biggl[ \int _{1}^{\infty} \frac{u^{\lambda _{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \biggr]\,dx \\ &= \int _{0}^{1} \biggl( \int _{\frac{1}{u}}^{\infty} x^{ - \varepsilon - 1} \,dx\biggr) \frac{u^{\lambda _{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du + \frac{1}{\varepsilon} \int _{1}^{\infty} \frac{u^{\lambda _{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \\ &= \frac{1}{\varepsilon} \biggl[ \int _{0}^{1} \frac{u^{\lambda _{2} + \frac{\varepsilon}{p} - 1}}{(1 + u)^{\lambda}} \,du + \int _{1}^{\infty} \frac{u^{\lambda _{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du\biggr]. \end{aligned}$$

Hence, by (18) and the above results, it follows that

$$\begin{aligned} \int _{0}^{1} \frac{u^{\lambda _{2} + \frac{\varepsilon}{p} - 1}}{(1 + u)^{\lambda}} \,du + \int _{1}^{\infty} \frac{u^{\lambda _{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \le \varepsilon \tilde{I} < \frac{M}{\prod_{i = 1}^{m} ( \lambda _{1} - \frac{\varepsilon}{p} + i - 1)\prod_{j = 1}^{n} ( \lambda _{2} - \frac{\varepsilon}{q} + j - 1)}. \end{aligned}$$

Putting $\varepsilon \to 0^{ +}$ in the above inequality, in view of the continuity of the beta function, we find

$$\begin{aligned} B(\lambda _{1},\lambda _{2}) \le \frac{M}{\prod_{i = 1}^{m} ( \lambda _{1} + i - 1)\prod_{j = 1}^{n} ( \lambda _{2} + j - 1)}, \end{aligned}$$

namely, $B(\lambda _{1},\lambda _{2})\prod_{i = 1}^{m} (\lambda _{1} + i - 1) \prod_{j = 1}^{n} (\lambda _{2} + j - 1) \le M$, and then

$$\begin{aligned} M = B(\lambda _{1},\lambda _{2})\prod _{i = 1}^{m} (\lambda _{1} + i - 1) \prod _{j = 1}^{n} (\lambda _{2} + j - 1) = \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B(\lambda _{1} + m,\lambda _{2} + n) \end{aligned}$$

is the best possible constant factor in (15).

The theorem is proved. □

Remark 1

For $\hat{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q},\hat{\lambda}_{2} = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}$, it follows that $\hat{\lambda}_{1} + \hat{\lambda}_{2} = \lambda$. We find $0 < \hat{\lambda}_{1} < \frac{\lambda}{p} + \frac{\lambda}{q} = \lambda$, and then $0 < \hat{\lambda}_{2} = \lambda - \hat{\lambda}_{1} < \lambda$. By Hölder’s inequality (cf. [24]), we can obtain

$$\begin{aligned} 0 &< B(\hat{\lambda}_{1} + m,\hat{\lambda}_{2} + n) = \int _{0}^{\infty} \frac{u^{\hat{\lambda}_{1} + m - 1}}{(1 + u)^{\lambda + m + n}}\,du \\ &= \int _{0}^{\infty} \frac{1}{(1 + u)^{\lambda + m + n}}u^{\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} + m - 1}\,du = \int _{0}^{\infty} \frac{1}{(1 + u)^{\lambda + m + n}} \bigl(u^{\frac{\lambda - \lambda _{2} + m - 1}{p}}\bigr) \bigl(u^{\frac{\lambda _{1} + m - 1}{q}}\bigr)\,du \\ &\le \biggl[ \int _{0}^{\infty} \frac{u^{\lambda - \lambda _{2} + m - 1}}{(1 + u)^{\lambda + m + n}}\,du \biggr]^{\frac{1}{p}}\biggl[ \int _{0}^{\infty} \frac{u^{\lambda _{1} + m - 1}}{(1 + u)^{\lambda + m + n}}\,du \biggr]^{\frac{1}{q}} \\ &= B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) < \infty . \end{aligned}$$

(19)

Theorem 2

If the constant factor

$$\begin{aligned} \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) \end{aligned}$$

in (14) is the best possible, then we have $\lambda _{1} + \lambda _{2} = \lambda$.

Proof

By (15) (for $\lambda _{i} = \hat{\lambda}_{i}\ (i = 1,2)$), since

$$\begin{aligned} \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) \end{aligned}$$

is the best possible constant factor in (15), we have

$$\begin{aligned} &\frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1})\\ &\quad \le \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B(\hat{\lambda}_{1} + m,\hat{ \lambda}_{2} + n) \quad( \in \mathrm{R} {}_{ +} ), \end{aligned}$$

namely,

$$\begin{aligned} B(\hat{\lambda}_{1} + m,\hat{\lambda}_{2} + n)\ge B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}). \end{aligned}$$

It follows that (19) keeps the form of equality.

We observe that (19) keeps the form of equality if and only if there exist constants A and B, such that they are not both zero and

$$\begin{aligned} Au^{\lambda - \lambda _{2} + m - 1} = Bu^{\lambda _{1} + m - 1}\quad \text{a.e. in }R_{ +} \end{aligned}$$

(cf. [24]). Assuming that $A \ne 0$, it follows that $u^{\lambda - \lambda _{2} - \lambda _{1}} = \frac{B}{A}\text{ a.e. in }R_{ +}$, namely, $\lambda - \lambda _{1} - \lambda _{2} = 0$, and then $\lambda _{1} + \lambda _{2} = \lambda$.

The theorem is proved. □

Theorem 3

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

• (i) Both $B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1})$ and $B(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} + m,\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p} + n)$ are independent of $p,q$;

• (ii) $B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) = B(\hat{\lambda}_{1} + m,\hat{\lambda}_{2} + n)$;

• (iii) $\lambda _{1} + \lambda _{2} = \lambda$;

• (iv) The constant factor

  $$\begin{aligned} \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) \end{aligned}$$

  in (14) is the best possible.

Proof

(i) ⇒ (ii). In view of the continuity of the beta function. we find

$$\begin{aligned} &B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) \\ &\quad= \lim_{p \to \infty} \lim_{q \to 1^{ +}} B^{\frac{1}{p}}( \lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}( \lambda _{1} + m,\lambda + n - \lambda _{1}) = B(\lambda _{1} + m,\lambda + n - \lambda _{1}), \\ &B(\hat{\lambda}_{1} + m,\hat{\lambda}_{2} + n) \\ &\quad= \lim _{p \to \infty} \lim_{q \to 1^{ +}} B \biggl( \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} + m,\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p} + n \biggr) = B( \lambda _{1} + m,\lambda + n - \lambda _{1}). \end{aligned}$$

Hence, we have

$$\begin{aligned} B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) = B(\hat{\lambda}_{1} + m,\hat{ \lambda}_{2} + n). \end{aligned}$$

(ii) ⇒ (iii). Suppose that $B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) = B(\hat{\lambda}_{1} + m,\hat{\lambda}_{2} + n)$. Then, (19) keeps the form of equality. In view of the proof of Theorem 2, we have $\lambda _{1} + \lambda _{2} = \lambda$.

(iii) ⇒ (iv). If $\lambda _{1} + \lambda _{2} = \lambda$, then by Theorem 1, the constant factor

$$\begin{aligned} &\frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})\\ &\quad\times{} B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) \biggl( = \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B(\lambda _{1} + m,\lambda _{2} + n) \biggr) \end{aligned}$$

in (14) is the best possible.

(iv) ⇒ (i). By Theorem 2, we have $\lambda _{1} + \lambda _{2} = \lambda$, and then

$$\begin{aligned} &B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) = B(\lambda _{1} + m,\lambda _{2} + n),\\ &B \biggl(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} + m,\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p} + n \biggr) = B(\lambda _{1} + m,\lambda _{2} + n). \end{aligned}$$

Both of them are independent of $p,q$.

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

The theorem is proved. □

Remark 2

(i) For $\lambda _{1} + \lambda _{2} = \lambda$ in (12), we have

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{n}(x)G_{n}(y)}{(x + y)^{\lambda + m + n}} \,dx \\ &\quad < B(\lambda _{1} + m,\lambda {}_{2} + n) \\ &\qquad{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{1} - m) - 1} F_{m}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \lambda _{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(20)

We confirm that the constant factor $B(\lambda {}_{1} + m,\lambda _{2} + n)$ in (20) is the best possible. Otherwise, we would reach a contradiction by (16) (for $\lambda _{1} + \lambda _{2} = \lambda$) that the constant factor in (15) is not the best possible.

(ii) For $m = n = 0,\lambda = 1,\lambda _{1} = \frac{1}{q},\lambda _{2} = \frac{1}{p}$, both (20) and (15) reduce to (2).

4 Equivalent forms and some particular inequalities

For $m = 0$ in (14), we have

$$\begin{aligned} I ={}& \int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{(x + y)^{\lambda}} \,dx\,dy \\ < {}& \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \\ &{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} F_{0}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(21)

In particular, for $\lambda _{1} + \lambda _{2} = \lambda$, we have

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{(x + y)^{\lambda}} \,dx\,dy \\ &\quad < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n) \\ &\qquad{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{1}) - 1} F_{0}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \lambda _{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

(22)

where the constant factor $\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n)$ is the best possible.

Theorem 4

Inequality (21) is equivalent to the following:

$$\begin{aligned} J&: = \biggl\{ \int _{0}^{\infty} x^{q\hat{\lambda} {}_{1} - 1} \biggl[ \int _{0}^{\infty} \frac{G{}_{0}(y)}{(x + y)^{\lambda}} \,dy \biggr]^{q}\,dx \biggr\} ^{\frac{1}{q}} \\ &< \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \biggl[ \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(23)

In particular, for $\lambda _{1} + \lambda _{2} = \lambda$, we reduce (23) to the equivalent form of (22) as follows:

$$\begin{aligned} &\biggl\{ \int _{0}^{\infty} x^{q\lambda {}_{1} - 1} \biggl[ \int _{0}^{\infty} \frac{G_{0}(y)}{(x + y)^{\lambda}} \,dy \biggr]^{q}\,dx \biggr\} ^{\frac{1}{q}} \\ &\quad < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n) \biggl( \int _{0}^{\infty} y^{q(1 - \lambda _{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr)^{\frac{1}{q}}, \end{aligned}$$

(24)

where the constant factor $\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n)$ is the best possible.

Proof

Suppose that (19) is valid. By Hölder’s inequality, we have

$$\begin{aligned} I = \int _{0}^{\infty} \bigl(x^{\frac{1}{q} - \hat{\lambda}_{1}}F_{0}(x) \bigr) \biggl[x^{ - \frac{1}{q} + \hat{\lambda}_{1}} \int _{0}^{\infty} \frac{G_{0}(y)}{(x + y)^{\lambda}} \,dy \biggr]\,dx\le \biggl\{ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} F_{0}^{p}(x) \,dx \biggr\} ^{\frac{1}{p}}J. \end{aligned}$$

(25)

Then, by (23), we have (21).

On the other hand, assuming that (21) is valid, we set

$$\begin{aligned} F_{0}(x): = x^{q\hat{\lambda} {}_{1} - 1} \biggl[ \int _{0}^{\infty} \frac{G_{0}(y)}{(x + y)^{\lambda}} \,dy \biggr]^{q - 1},\quad x \in \mathrm{R}_{ +}. \end{aligned}$$

If $J = 0$, then, (23) is naturally valid; if $J = \infty$, then it is impossible that (23) is valid, namely $J < \infty$. Suppose that $0 < J < \infty$. By (21), we have

$$\begin{aligned} &\int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} F_{0}^{p}(x) \,dx\\ &\quad = J^{q} = I\\ &\quad < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1})\\ &\qquad{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} F_{0}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}},\\ &J = \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} F_{0}^{p}(x) \,dx \biggr]^{\frac{1}{q}}\\ &\phantom{J }< \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \biggl[ \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

namely, (23) follows, which is equivalent to (21).

The constant factor $\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n)$ in (24) is the best possible. Otherwise, by (25) (for $\lambda _{1} + \lambda _{2} = \lambda$), we would reach a contradiction that the constant factor in (22) is not the best possible.

The theorem is proved. □

Replacing x by $\frac{1}{x}$, then replacing $x^{\lambda - 2}F_{0}(\frac{1}{x})$ by $F_{0}(x)$ in (21) and (23), by simplification, we have

Corollary 1

The following Hardy–Hilbert-type integral inequalities with a nonhomogeneous kernel are equivalent:

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{(1 + xy)^{\lambda}} \,dx\,dy \\ &\quad< \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \\ &\qquad{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{2}) - 1} F_{0}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

(26)

$$\begin{aligned} &\biggl\{ \int _{0}^{\infty} x^{q\hat{\lambda} {}_{2} - 1} \biggl[ \int _{0}^{\infty} \frac{G_{0}(y)}{(1 + xy)^{\lambda}} \,dy \biggr]^{q}\,dx \biggr\} ^{\frac{1}{q}} \\ &\quad < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \biggl[ \int _{0}^{\infty} y^{q(1 - \hat{\lambda}_{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(27)

Moreover, $\lambda _{1} + \lambda _{2} = \lambda$ if and only if the constant factor

$$\begin{aligned} \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \end{aligned}$$

in (26) and (27) is the best possible.

For $\lambda _{1} + \lambda _{2} = \lambda$, we have the following equivalent inequalities with the best possible constant factor $\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n)$:

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{(1 + xy)^{\lambda}} \,dx\,dy \\ &\quad < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n) \\ &\qquad{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{2}) - 1} F_{0}^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} y^{q(1 - \lambda _{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

(28)

$$\begin{aligned} &\biggl\{ \int _{0}^{\infty} x^{q\lambda {}_{2} - 1} \biggl[ \int _{0}^{\infty} \frac{G_{0}(y)}{(1 + xy)^{\lambda}} \,dy \biggr]^{q}\,dx \biggr\} ^{\frac{1}{q}} \\ &\quad < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n) \biggl[ \int _{0}^{\infty} y^{q(1 - \lambda _{2} - n) - 1} G_{n}^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(29)

Remark 3

(i) For $\lambda = 1,\lambda _{1} = \frac{1}{q},\lambda _{2} = \frac{1}{p}$ in (22), (24), (28), and (29), we have the following two couples of equivalent integral inequalities:

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{x + y} \,dx\,dy \\ &\quad < \frac{\pi}{\sin (\pi /p)}\prod_{i = 1}^{n} \biggl(i - \frac{1}{q} \biggr) \biggl( \int _{0}^{\infty} F_{0}^{p}(x) \,dx \biggr)^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n}} \biggr)^{q} \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

(30)

$$\begin{aligned} &\biggl[ \int _{0}^{\infty} \biggl( \int _{0}^{\infty} \frac{G_{0}(y)}{x + y} \,dy \biggr)^{q}\,dx \biggr]^{\frac{1}{q}} < \frac{\pi}{\sin (\pi /p)}\prod _{i = 1}^{n} \biggl(i - \frac{1}{q} \biggr) \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n}} \biggr)^{q} \,dy \biggr]^{\frac{1}{q}}; \end{aligned}$$

(31)

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{1 + xy} \,dx\,dy \\ &\quad < \frac{\pi}{\sin (\pi /p)}\prod_{i = 1}^{n} \biggl(i - \frac{1}{q} \biggr) \biggl( \int _{0}^{\infty} x^{p - 2} F_{0}^{p}(x) \,dx \biggr)^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n}} \biggr)^{q} \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

(32)

$$\begin{aligned} &\biggl[ \int _{0}^{\infty} x^{q - 2} \biggl( \int _{0}^{\infty} \frac{G_{0}(y)}{1 + xy} \,dy \biggr)^{q}\,dx \biggr]^{\frac{1}{q}} < \frac{\pi}{\sin (\pi /p)}\prod _{i = 1}^{n} \biggl(i - \frac{1}{q} \biggr) \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n}} \biggr)^{q} \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(33)

(ii) For $\lambda = 1,\lambda _{1} = \frac{1}{p},\lambda _{2} = \frac{1}{q}$ in (22), (24), (28), and (29), we have the dual forms of (30)–(33) as follows:

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F{}_{0}(x)G_{0}(y)}{x + y} \,dx\,dy \\ &\quad < \frac{\pi}{\sin (\pi /p)}\prod_{i = 1}^{n} \biggl(i - \frac{1}{p} \biggr) \biggl( \int _{0}^{\infty} x^{p - 2}F_{0}^{p}(x) \,dx \biggr)^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n + \frac{2}{q} - 1}} \biggr)^{q} \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

(34)

$$\begin{aligned} &\biggl[ \int _{0}^{\infty} x^{q - 2} \biggl( \int _{0}^{\infty} \frac{G_{0}(y)}{x + y} \,dy \biggr)^{q}\,dx \biggr]^{\frac{1}{q}} < \frac{\pi}{\sin (\pi /p)}\prod _{i = 1}^{n} \biggl(i - \frac{1}{p} \biggr) \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n + \frac{2}{q} - 1}} \biggr)^{q} \,dy \biggr]^{\frac{1}{q}}; \end{aligned}$$

(35)

$$\begin{aligned} &\int _{0}^{\infty} \int _{0}^{\infty} \frac{F{}_{0}(x)G_{0}(y)}{1 + xy} \,dx\,dy \\ &\quad < \frac{\pi}{\sin (\pi /p)}\prod_{i = 1}^{n} \biggl(i - \frac{1}{p} \biggr) \biggl( \int _{0}^{\infty} F_{0}^{p} (x)\,dx \biggr)^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n + \frac{2}{q} - 1}} \biggr)^{q} \,dy \biggr]^{\frac{1}{q}}, \end{aligned}$$

(36)

$$\begin{aligned} &\biggl[ \int _{0}^{\infty} \biggl( \int _{0}^{\infty} \frac{G_{0}(y)}{1 + xy} \,dy \biggr)^{q}\,dx \biggr]^{\frac{1}{q}} < \frac{\pi}{\sin (\pi /p)}\prod _{i = 1}^{n} \biggl(i - \frac{1}{p} \biggr) \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n + \frac{2}{q} - 1}} \biggr)^{q} \,dy \biggr]^{\frac{1}{q}}. \end{aligned}$$

(37)

(iii) For $p = q = 2,(2n - 1)!!: = 1 \cdot 3 \cdot \cdots \cdot (2n - 1)$, both (30) and (34) reduce to

$$\begin{aligned} \int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G_{0}(y)}{x + y} \,dx\,dy < \frac{\pi}{2^{n}}(2n - 1)!! \biggl[ \int _{0}^{\infty} F_{0}^{2}(x) \,dx \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n}} \biggr)^{2} \,dy \biggr]^{\frac{1}{2}}, \end{aligned}$$

(38)

both (31) and (35) reduce to the following equivalent inequality of (38):

$$\begin{aligned} \biggl[ \int _{0}^{\infty} \biggl( \int _{0}^{\infty} \frac{G_{0}(y)}{x + y} \,dy \biggr)^{2}\,dx \biggr]^{\frac{1}{2}} < \frac{\pi}{2^{n}}(2n - 1)!! \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n}} \biggr)^{2} \,dy \biggr]^{\frac{1}{2}}, \end{aligned}$$

(39)

both (32) and (36) reduce to

$$\begin{aligned} \int _{0}^{\infty} \int _{0}^{\infty} \frac{F_{0}(x)G{}_{0}(y)}{1 + xy} \,dx\,dy < \frac{\pi}{2^{n}}(2n - 1)!! \biggl[ \int _{0}^{\infty} F_{0}^{2}(x) \,dx \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n}} \biggr)^{2} \,dy \biggr]^{\frac{1}{2}}, \end{aligned}$$

(40)

and both (33) and (37) reduce to the following equivalent inequality of (40):

$$\begin{aligned} \biggl[ \int _{0}^{\infty} \biggl( \int _{0}^{\infty} \frac{G_{0}(y)}{1 + xy} \,dy \biggr)^{2}\,dx \biggr]^{\frac{1}{2}} < \frac{\pi}{2^{n}}(2n - 1)!! \biggl[ \int _{0}^{\infty} \biggl(\frac{G_{n}(y)}{y^{n}} \biggr)^{2} \,dy \biggr]^{\frac{1}{2}}. \end{aligned}$$

(41)

The constant factor in the above inequalities (30)–(41) are all the best possible.

5 Operator expressions

We set functions

$$\begin{aligned} \phi (x): = x^{p(1 - \hat{\lambda}_{1}) - 1},\qquad \psi (y): = y^{q(1 - \hat{\lambda}_{2} - n) - 1}, \end{aligned}$$

hence, $\phi ^{1 - q}(x) = x^{q\hat{\lambda}_{1} - 1}(x,y \in \mathrm{R}_{ +} )$. Define the following real normed spaces:

$$\begin{aligned} &L_{p,\phi} (\mathrm{R}_{ +} ): = \biggl\{ f = f(x); \Vert f \Vert _{p,\phi}: = \biggl( \int _{0}^{\infty} \phi (x) \bigl\vert f(x) \bigr\vert ^{p}\,dx \biggr)^{\frac{1}{p}} < \infty \biggr\} ,\\ &L_{q,\psi} (\mathrm{R}_{ +} ): = \biggl\{ \tilde{g} = \tilde{g}(y); \Vert \tilde{g} \Vert _{q,\psi}: = \biggl( \int _{0}^{\infty} \psi (y) \bigl\vert \tilde{g}(y) \bigr\vert ^{q}\,dy\biggr)^{\frac{1}{q}} < \infty \biggr\} , \\ &L_{q,\phi ^{1 - q}}(\mathrm{R}_{ +} ): = \biggl\{ h = h(x); \Vert h \Vert _{p,\phi ^{1 - q}}: = \biggl( \int _{0}^{\infty} \phi ^{1 - q}(x) \bigl\vert h(x) \bigr\vert ^{q}\,dx\biggr)^{\frac{1}{q}} < \infty \biggr\} . \end{aligned}$$

Assuming that $G_{0}(y)$ is a nonnegative Lebesgue integrable function in any interval $(0,b] \subset R_{ +} (b > 0)$,

$$\begin{aligned} G_{0} \in \tilde{L}(\mathrm{R}_{ +} ): = \bigl\{ g = g(y);g(y) = o \bigl(e^{ty} \bigr)\ (t > 0;y \to \infty ), G_{0} \in L_{q,\psi} (\mathrm{R}_{ +} ) \bigr\} , \end{aligned}$$

setting

$$\begin{aligned} h = h(x),\qquad h(x): = \int _{0}^{\infty} \frac{1}{(x + y)^{\lambda}} G_{0}(y) \,dy,\quad x \in \mathrm{R}_{ +}, \end{aligned}$$

(42)

we can rewrite (23) as follows:

$$\begin{aligned} \Vert h \Vert _{q,\phi ^{1 - q}} \le \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \Vert G_{n} \Vert _{q,\psi} < \infty , \end{aligned}$$

namely, $h \in L_{q,\phi ^{1 - q}}(\mathrm{R}_{ +} )$.

Definition 1

Define a Hardy–Hilbert-type operator $T:\tilde{L}(\mathrm{R}_{ +} ) \to L_{q,\phi ^{1 - q}}(\mathrm{R}_{ +} )$ as follows: For any $G_{0} \in \tilde{L}(\ \mathrm{R}_{ +} ),\ G_{0}(y)$ is a nonnegative Lebesgue integrable function in any interval $(0,b] \subset R_{ +} (b > 0)$, there exists a unique representation $h \in L_{q,\phi ^{1 - q}}(\mathrm{R}_{ +} )$, satisfying (42). Define the formal inner product of $F_{0} \in L_{p,\phi} (\mathrm{R}_{ +} )$ and $TG_{0}$, and the norm of T as follows:

$$\begin{aligned} &(F_{0},TG_{0}): = \int _{0}^{\infty} F_{0}(x) \biggl[ \int _{0}^{\infty} \frac{1}{(x + y)^{\lambda}} G_{0}(y) \,dy \biggr]\,dx = I,\\ &\Vert T \Vert : = \sup_{G_{0}( \ne 0) \in \tilde{L}(R_{ +} )}\frac{ \Vert TG_{0} \Vert _{q,\phi ^{1 - q}}}{ \Vert G_{n} \Vert _{q,\psi}}. \end{aligned}$$

By Theorem 4, we have

Theorem 5

If $F_{0} \in L_{p,\phi} (\mathrm{R}_{ +} ),G_{0} \in \tilde{L}(\mathrm{R}_{ +} ),\Vert F_{0}\Vert _{p,\phi},\Vert G_{n}\Vert _{q,\psi} > 0$, ($G_{0}(y)$ is a nonnegative Lebesgue integrable function in any interval $(0,b] \subset R_{ +} (b > 0)$), then we have the following equivalent inequalities:

$$\begin{aligned} &(F_{0},TG_{0}) < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \Vert F_{0} \Vert _{p,\phi} \Vert G_{n} \Vert _{q,\psi}, \end{aligned}$$

(43)

$$\begin{aligned} &\Vert TG_{0} \Vert _{q,\phi ^{1 - q}} < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}( \lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}( \lambda _{1},\lambda + n - \lambda _{1}) \Vert G_{n} \Vert _{q,\psi}. \end{aligned}$$

(44)

Moreover, $\lambda _{1} + \lambda _{2} = \lambda$ if and only if the constant factor $\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1})$ in (42) and (43) is the best possible, namely,

$$\begin{aligned} \Vert T \Vert = \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n). \end{aligned}$$

We also set functions $\varphi (x): = x^{p(1 - \hat{\lambda}_{2}) - 1}$, hence,

$$\begin{aligned} \varphi ^{1 - q}(x) = x^{q\hat{\lambda}_{2} - 1}\quad(x \in \mathrm{R}_{ +} ), \end{aligned}$$

and define the following real normed spaces:

$$\begin{aligned} &L_{p,\varphi} (\mathrm{R}_{ +} ): = \biggl\{ f = f(x); \Vert f \Vert _{p,\varphi}: = \biggl( \int _{0}^{\infty} \varphi (x) \bigl\vert f(x) \bigr\vert ^{p}\,dx \biggr)^{\frac{1}{p}} < \infty \biggr\} ,\\ &L_{q,\varphi ^{1 - q}}(\mathrm{R}_{ +} ): = \biggl\{ h = h(x); \Vert h \Vert _{p,\varphi ^{1 - q}}: = \biggl( \int _{0}^{\infty} \varphi ^{1 - q}(x) \bigl\vert h(x) \bigr\vert ^{q}\,dx \biggr)^{\frac{1}{q}} < \infty \biggr\} . \end{aligned}$$

Assuming that $G_{0} \in \tilde{L}(\mathrm{R}_{ +} )$, setting

$$\begin{aligned} H = H(x),\qquad H(x): = \int _{0}^{\infty} \frac{1}{(1 + xy)^{\lambda}} G_{0}(y) \,dy,\quad x \in \mathrm{R}_{ +}, \end{aligned}$$

(45)

we can rewrite (27) as follows:

$$\begin{aligned} \Vert H \Vert _{q,\varphi ^{1 - q}} \le \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \Vert G_{n} \Vert _{q,\psi} < \infty , \end{aligned}$$

namely, $H \in L_{q,\varphi ^{1 - q}}(\mathrm{R}_{ +} )$.

Definition 2

Define a Hardy–Hilbert-type operator $T_{1}:\tilde{L}(\mathrm{R}_{ +} ) \to L_{q,\varphi ^{1 - q}}(\mathrm{R}_{ +} )$ as follows: For any $G_{0} \in \tilde{L}(\ \mathrm{R}_{ +} )$, there exists a unique representation $H \in L_{q,\varphi ^{1 - q}}(\mathrm{R}_{ +} )$. Define the formal inner product of $F_{0} \in L_{p,\varphi} (\mathrm{R}_{ +} )$ and $T_{1}G_{0}$, and the norm of $T_{1}$ as follows:

$$\begin{aligned} &(F_{0},T_{1}G_{0}): = \int _{0}^{\infty} F_{0}(x) \biggl[ \int _{0}^{\infty} \frac{1}{(1 + xy)^{\lambda}} G_{0}(y) \,dy \biggr]\,dx,\\ &\Vert T_{1} \Vert : = \sup_{G_{0}( \ne \theta ) \in \tilde{L}(R_{ +} )} \frac{ \Vert TG_{0} \Vert _{q,\varphi ^{1 - q}}}{ \Vert G_{n} \Vert _{q,\psi}}. \end{aligned}$$

By Corollary 1, we have

Corollary 2

If $F_{0} \in L_{p,\varphi} (\mathrm{R}_{ +} ),G_{0} \in \tilde{L}(\mathrm{R}_{ +} ),\Vert F_{0}\Vert _{p,\varphi},\Vert G_{n}\Vert _{q,\psi} > 0$, then we have the following equivalent inequalities:

$$\begin{aligned} &(F_{0},T_{1}G_{0}) < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}( \lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}( \lambda _{1},\lambda + n - \lambda _{1}) \Vert F_{0} \Vert _{p,\varphi} \Vert G_{n} \Vert _{q,\psi}, \end{aligned}$$

(46)

$$\begin{aligned} &\Vert T_{1}G_{0} \Vert _{q,\varphi ^{1 - q}} < \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \Vert G_{n} \Vert _{q,\psi}. \end{aligned}$$

(47)

Moreover, $\lambda _{1} + \lambda _{2} = \lambda$ if and only if the constant factor

$$\begin{aligned} \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1}) \end{aligned}$$

in (44) and (45) is the best possible, namely,

$$\begin{aligned} \Vert T_{1} \Vert = \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1}, \lambda _{2} + n). \end{aligned}$$