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

Abstract

By applying the weight functions and the idea of introduced parameters we give a new Hilbert-type integral inequality involving the upper limit functions and the beta and gamma functions. We consider equivalent statements of the best possible constant factor related to a few parameters. As applications, we obtain a corollary in the case of a nonhomogeneous kernel and some particular inequalities.

1 Introduction

If $0 < \sum_{m = 1}^{\infty} a_{m}^{2} < \infty$ and $0 < \sum_{n = 1}^{\infty} b_{n}^{2} < \infty$, then we have the following discrete Hilbert inequality with the best possible constant factor π ([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)

Assuming that $0 < \int_{0}^{\infty} f^{2}(x)\,dx < \infty$ and $0 < \int_{0}^{\infty} g^{2}(y) \,dy < \infty$, we still have the following integral analogue of (1) ([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) are playing an important role in analysis and its applications [2–13].

The following half-discrete Hilbert-type inequality was provided in 1934 ([1], Theorem 351): If $K(x)$ ($x > 0$) is a decreasing function, $p > 1$, $\frac{1}{p} + \frac{1}{q} = 1$, $0 < \phi (s) = \int_{0}^{\infty} K(x)x^{s - 1} \,dx < \infty$, $f(x) \ge 0$, and $0 < \int_{0}^{\infty} f^{p} (x)\,dx < \infty$, then

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

(3)

In recent years, some new extensions of (3) were given by [14–19].

In 2006, using the Euler–Maclaurin summation formula, Krnic et al. [20] gave an extension of (1) with the kernel $\frac{1}{(m + n)^{\lambda}}$ ($0 < \lambda \le 14$). In 2019, following [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] considered some equivalent statements of the extensions of (1) and (2) with a few parameters. Some similar works were provided in [24–26].

In this paper, following [21, 22], by the use of the weight functions and the idea of introduced parameters, we give a new Hilbert-type integral inequality with the kernel $\frac{1}{(x + y)^{\lambda}}$ ($\lambda > 0$) involving the upper limit functions and the beta and gamma functions. We consider the equivalent statements of the best possible constant factor related to a few parameters. As applications, we obtain a corollary in the case of nonhomogeneous kernel and some particular inequalities.

2 Some lemmas

In what follows, we assume that $p > 1$, $\frac{1}{p} + \frac{1}{q} = 1$, $\lambda > 0$, $\lambda_{1},\lambda_{2} \in (0,\lambda + 1)$, $f(x)$ and $g(y)$ are nonnegative measurable functions in $R_{ +} = (0,\infty )$, $f(x) = o(e^{x})$, $g(y) = o(e^{y})$ ($x,y \to \infty$), such that for any $A = (0,a)$ ($a > 0$), $f,g \in L^{1}(A)$, and the upper limit functions are defined by

$$F(x): = \int_{0}^{x} f(t)\,dt\quad (x \ge 0)\quad \mbox{and} \quad G(y): = \int_{0}^{y} g(t)\,dt\quad (y \ge 0), $$

satisfying

$$0 < \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx < \infty \quad \mbox{and}\quad 0 < \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy < \infty. $$

By the definition of the gamma function, for $\lambda,x,y > 0$, the following expression holds:

$$ \frac{1}{(x + y)^{\lambda}} = \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} t^{\lambda - 1} e^{ - (x + y)t}\,dt. $$

(4)

Lemma 1

For $t > 0$, we have the following expressions:

$$\begin{aligned}& \int_{0}^{\infty} e^{ - tx} f(x)\,dx = t \int_{0}^{\infty} e^{ - tx} F(x) \,dx, \end{aligned}$$

(5)

$$\begin{aligned}& \int_{0}^{\infty} e^{ - ty} g(y)\,dy = t \int_{0}^{\infty} e^{ - ty} G(y) \,dy. \end{aligned}$$

(6)

Proof

We find

$$\begin{aligned} \int_{0}^{\infty} e^{ - tx} f(x)\,dx &= \int_{0}^{\infty} e^{ - tx}\, dF(x) \\ &= e^{ - tx}F(x)|_{0}^{\infty} - \int_{0}^{\infty} F (x)\,de^{ - tx} \\ &= \lim _{x \to \infty} \frac{F(x)}{e^{tx}} + t \int_{0}^{\infty} e^{ - tx} F(x)\,dx. \end{aligned}$$

If $F(\infty ) =\mbox{constant}$, then $\lim_{x \to \infty} \frac{F(x)}{e^{tx}} = 0$, and (5) follows; if $F(\infty ) = \infty$, in view of $f(x) = o(e^{x})$ ($x \to \infty$), we find

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

and then (5) follows. In the same way, we have (6).

The lemma is proved. □

Lemma 2

For $s > 0,\mu,\sigma \in (0,s)$, define the following weight functions:

$$\begin{aligned}& \varpi (\sigma,x): = x^{s - \sigma} \int_{0}^{\infty} \frac{t^{\sigma - 1}}{(x + t)^{s}}\,dt\quad (x \in \mathrm{R}_{ +} ), \end{aligned}$$

(7)

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

(8)

We have the following expressions:

$$\begin{aligned}& \varpi (\sigma,x) = B(\sigma,s - \sigma )\quad (x \in \mathrm{R}_{ +} ), \end{aligned}$$

(9)

$$\begin{aligned}& \omega (\mu,y) = B(\mu,s - \mu )\quad (y \in \mathrm{R}_{ +} ), \end{aligned}$$

(10)

where $B(u,v): = \int_{0}^{\infty} \frac{t^{u - 1}}{(1 + t)^{u + v}} \,dt$ ($u,v > 0$) is the beta function satisfying

$$B(u,v) = \frac{1}{\varGamma (u + v)}\varGamma (u)\varGamma (v). $$

Proof

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

$$\varpi (\sigma,x) = x^{s - \sigma} \int_{0}^{\infty} \frac{(ux)^{\sigma - 1}}{(x + ux)^{s}}x\,du = \int_{0}^{\infty} \frac{u^{\sigma - 1}}{(1 + u)^{s}}\,du = B\quad ( \sigma,s - \sigma ), $$

namely, (9) follows. In the same way, we have (10).

The lemma is proved. □

Lemma 3

Suppose that $s > 0,\mu,\sigma \in (0,s)$. We have the following inequality:

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{s}} \,dx \le B^{\frac{1}{p}}(\sigma,s - \sigma )B^{\frac{1}{q}}(\mu,s - \mu ) \\& \quad {} \times \biggl[ \int_{0}^{\infty} x^{p(1 - \mu ) - (s - \mu - \sigma ) - 1} f^{p}(x)\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} y^{q(1 - \sigma ) - (s - \mu - \sigma ) - 1} g^{q}(y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$

(11)

For $\lambda > 0$, $s = \lambda + 2( > 2)$, $\lambda_{1} = \mu - 1 \in (0,\lambda + 1)$, $\lambda_{2} = \sigma - 1 \in (0,\lambda + 1)$, by the substitution $f(x) = F(x)$and $g(y) = G(y)$in (11) we can reduce it to the following:

$$\begin{aligned}& \begin{gathered}[b] \int_{0}^{\infty} \int_{0}^{\infty} \frac{F(x)G(y)}{(x + y)^{\lambda + 2}} \,dx\,dy < B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\ \quad {} \times \biggl[ \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy\biggr]^{\frac{1}{q}}. \end{gathered} \end{aligned}$$

(12)

Proof

By Hölder’s inequality (see [27]) we obtain

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{s}} \,dx\,dy = \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{(x + y)^{s}} \biggl[ \frac{y^{(\sigma - 1)/p}}{x^{(\mu - 1)/q}}f(x)\biggr] \biggl[\frac{x^{(\mu - 1)/q}}{y^{(\sigma - 1)/p}}g(y)\biggr]\,dx\,dy \\& \quad \le \biggl\{ \int_{0}^{\infty} \biggl[ \int_{0}^{\infty} \frac{1}{(x + y)^{s}} \frac{y^{\sigma - 1}\,dy}{x^{(\mu - 1)(p - 1)}}\biggr]f^{p}(x)\,dx\biggr\} ^{\frac{1}{p}} \\& \qquad {}\times\biggl\{ \int_{0}^{\infty} \biggl[ \int_{0}^{\infty} \frac{1}{(x + y)^{s}} \frac{x^{\mu - 1}\,dx}{y^{(\sigma - 1)(q - 1)}}\biggr]g^{q}(y)\,dy\biggr\} ^{\frac{1}{q}} \\& \quad = \biggl[ \int_{0}^{\infty} \varpi (\sigma,x) x^{p(1 - \mu ) - (\lambda - \mu - \sigma ) - 1}f^{p}(x)\,dx\biggr]^{\frac{1}{p}} \\& \qquad {}\times \biggl[ \int_{0}^{\infty} \omega (\mu,y) y^{q(1 - \sigma ) - (\lambda - \mu - \sigma ) - 1}g^{q}(y) \,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$

(13)

Then by (9) and (10) we have (11).

By simplifications of (11) we have

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{F(x)G(y)}{(x + y)^{\lambda + 2}} \,dx\,dy \le B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\& \quad {}\times \biggl[ \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$

(14)

If (14) keeps the form of equality, then, in view of the proof of (13), there exist constants A and B such that they are not all zero, satisfying for $s = \lambda + 2$, $\lambda_{1} = \mu - 1$, $\lambda_{2} = \sigma - 1$,

$$\begin{aligned}& Ax^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1}x^{(\lambda - \lambda_{1} - \lambda_{2}) + 1}F^{p}(x) \\& \quad = By^{ - q\lambda_{2}}G^{q}(y)\quad \mbox{a.e. in }(0,\infty ) \times (0,\infty ). \end{aligned}$$

Without loss of generality, we assume that $A \ne 0$. Then for fixed $y \in (0,\infty )$, we have

$$x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1}F^{p}(x) = \biggl(\frac{B}{A}y^{ - q\lambda_{2}}G^{q}(y) \biggr)x^{ - 1 - (\lambda - \lambda_{1} - \lambda_{2})}\quad \mbox{a.e. in }(0,\infty ), $$

which contradicts the fact that

$$0 < \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx < \infty, $$

since for any $\lambda - \lambda_{1} - \lambda_{2} \in\mathbf{R}$, $\int_{0}^{\infty} x^{ - 1 - (\lambda - \lambda_{1} - \lambda_{2})}\,dx = \infty$. Therefore inequality (12) follows.

The lemma is proved. □

3 Main results

Theorem 1

We have the following inequality:

$$\begin{aligned} I : =& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\ &{} \times \biggl[ \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$

(15)

In particular, for $\lambda_{1} + \lambda_{2} = \lambda$ ($\lambda_{1},\lambda_{2} \in (0,\lambda )$), we reduce it to the following inequality:

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < & \lambda_{1}\lambda_{2}B(\lambda_{1}, \lambda_{2}) \biggl( \int_{0}^{\infty} x^{ - p\lambda_{1} - 1} F^{p}(x)\,dx\biggr)^{\frac{1}{p}} \\ &{}\times\biggl( \int_{0}^{\infty} y^{ - q\lambda_{2} - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}, \end{aligned}$$

(16)

where the constant factor $\lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})$is the best possible.

Proof

Using (4), (5), and (6), we find

$$\begin{aligned} I =& \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} \int_{0}^{\infty} f(x)g(y) \biggl( \int_{0}^{\infty} t^{\lambda - 1} e^{ - (x + y)t}\,dt\biggr)\,dx\,dy \\ =& \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} t^{\lambda - 1} \biggl( \int_{0}^{\infty} e^{ - xt}f(x)\,dx\biggr) \biggl( \int_{0}^{\infty} e^{ - yt} g(y)\,dy\biggr) \,dt \\ =& \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} t^{\lambda + 1} \biggl( \int_{0}^{\infty} e^{ - xt}F(x)\,dx\biggr) \biggl( \int_{0}^{\infty} e^{ - yt} G(y)\,dy\biggr) \,dt \\ =& \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} \int_{0}^{\infty} F(x)G(y) \biggl[ \int_{0}^{\infty} t^{\lambda + 1}e^{ - (x + y)t} \,dt\biggr] \,dx\,dy \\ =& \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )} \int_{0}^{\infty} \int_{0}^{\infty} \frac{F(x)G(y)}{(x + y)^{\lambda + 2}} \,dx\,dy. \end{aligned}$$

(17)

In view of (12), we have (15).

In the case of $\lambda_{1} + \lambda_{2} = \lambda$ ($\lambda_{1},\lambda_{2} \in (0,\lambda )$), we find

$$\begin{aligned}& \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\& \quad = \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1, \lambda_{1} + 1)B^{\frac{1}{q}}(\lambda_{1} + 1, \lambda_{2} + 1) \\& \quad = \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B(\lambda_{1} + 1, \lambda_{2} + 1) = \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}\frac{\varGamma (\lambda_{1} + 1)\varGamma (\lambda_{2} + 1)}{\varGamma (\lambda + 2)} \\& \quad = \lambda_{1}\lambda_{2}\frac{\varGamma (\lambda_{1})\varGamma (\lambda_{2})}{\varGamma (\lambda )} = \lambda_{1}\lambda_{2}B(\lambda_{1}, \lambda_{2}), \end{aligned}$$

and then (16) follows.

For any $0 < \varepsilon < \min \{ p\lambda_{1},q\lambda_{2}\}$, we set

$$\tilde{f}(t): = \textstyle\begin{cases} 0,&0 < t \le 1, \\ t^{\lambda_{1} - \frac{\varepsilon}{p} - 1},&t > 1, \end{cases}\displaystyle \qquad \tilde{g}(t): = \textstyle\begin{cases} 0,&0 < t \le 1, \\ t^{\lambda_{2} - \frac{\varepsilon}{q} - 1},&t > 1. \end{cases} $$

We obtain that $\tilde{f}(x) = o(e^{x})$, $\tilde{g}(y) = o(e^{y})$ ($x,y \to \infty$), and $\tilde{F}(x) = \tilde{G}(y) \equiv 0$ ($0 < x,y \le 1$), where

$$\begin{aligned}& \tilde{F}(x) = \int_{0}^{x} \tilde{f}(t)\,dt = \int_{1}^{x} t^{\lambda_{1} - \frac{\varepsilon}{p} - 1}\,dt = \frac{x^{\lambda_{1} - \frac{\varepsilon}{p}} - 1}{\lambda_{1} - \frac{\varepsilon}{p}} < \frac{x^{\lambda_{1} - \frac{\varepsilon}{p}}}{\lambda_{1} - \frac{\varepsilon}{p}}\quad (x > 1), \\& \tilde{G}(y) = \int_{0}^{y} \tilde{g}(t)\,dt = \int_{1}^{y} t^{\lambda_{2} - \frac{\varepsilon}{q} - 1}\,dt = \frac{y^{\lambda_{2} - \frac{\varepsilon}{q}} - 1}{\lambda_{2} - \frac{\varepsilon}{q}} < \frac{y^{\lambda_{2} - \frac{\varepsilon}{q}}}{\lambda_{2} - \frac{\varepsilon}{q}}\quad (y > 1). \end{aligned}$$

If there exists a positive constant M ($M \le \lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})$) such that (16) is valid when replacing $\lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})$ by M, then, in particular, by substitution of $f(x) = \tilde{f}(x)$ and $g(y) = \tilde{g}(y)$ we have

$$\tilde{I}: = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\tilde{f}(x)\tilde{g}(y)}{(x + y)^{\lambda}} \,dx\,dy < M \biggl( \int_{0}^{\infty} x^{ - p\lambda_{1} - 1} \tilde{F}^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - q\lambda_{2} - 1} \tilde{G}^{q}(y)\,dy\biggr)^{\frac{1}{q}}. $$

We find

$$\begin{aligned} \tilde{J} : =& \biggl( \int_{0}^{\infty} x^{ - p\lambda_{1} - 1} \tilde{F}^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - q\lambda_{2} - 1} \tilde{G}^{q}(y)\,dy\biggr)^{\frac{1}{q}} \\ < & \frac{1}{(\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})}\biggl[ \int_{1}^{\infty} x^{ - p\lambda_{1} - 1} \bigl(x^{\lambda_{1} - \frac{\varepsilon}{p}}\bigr)^{p}\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{1}^{\infty} y^{ - q\lambda_{2} - 1} \bigl(y^{\lambda_{2} - \frac{\varepsilon}{q}}\bigr)^{q}\,dy\biggr]^{\frac{1}{q}} \\ =& \frac{1}{(\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})}\biggl( \int_{1}^{\infty} x^{ - \varepsilon - 1} \,dx \biggr)^{\frac{1}{p}}\biggl( \int_{1}^{\infty} y^{ - \varepsilon - 1} \,dy \biggr)^{\frac{1}{q}} \\ =& \frac{1}{(\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})} \int_{1}^{\infty} x^{ - \varepsilon - 1} \,dx = \frac{1}{\varepsilon (\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})}. \end{aligned}$$

In view of the Fubini theorem (see [28]), it follows that

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

So we obtain

$$\int_{0}^{1} \frac{u^{\lambda_{2} + \frac{\varepsilon}{p} - 1}}{(1 + u)^{\lambda}} \,du + \int_{1}^{\infty} \frac{u^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \le \varepsilon \tilde{I}< \varepsilon M\tilde{J} < \frac{M}{(\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})}. $$

As $\varepsilon \to 0^{ +}$ in this inequality, in view of the continuity of the beta function, we find $B(\lambda_{1},\lambda_{2}) \le \frac{M}{\lambda_{1}\lambda_{2}}$, namely $\lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2}) \le M$. Hence $M = \lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})$ is the best possible constant factor of (14).

The theorem is proved. □

Remark 1

We set $\hat{\lambda}_{1}: = \lambda_{1} + \frac{\lambda - \lambda_{1} - \lambda_{2}}{p}$, $\hat{\lambda}_{2}: = \lambda_{2} + \frac{\lambda - \lambda_{1} - \lambda_{2}}{q}$. It follows that $\hat{\lambda}_{1} + \hat{\lambda}_{2} = \lambda$. For $\lambda - \lambda_{1} - \lambda_{2}\ \in ( - p\lambda_{1},p(\lambda - \lambda_{1}))$, we find

$$\hat{\lambda}_{1} > \lambda_{1} + \frac{ - p\lambda_{1}}{p} = 0,\qquad \hat{\lambda}_{1} < \lambda_{1} + \frac{p(\lambda - \lambda_{1})}{p} = \lambda, $$

namely, $0 < \hat{\lambda}_{1} < \lambda$, and then $0 < \hat{\lambda}_{2} < \lambda$. So we reduce (15) as follows:

$$\begin{aligned} I : =& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\ &{}\times \biggl( \int_{0}^{\infty} x^{ - p\hat{\lambda}_{1} - 1} F^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - q\hat{\lambda}_{2} - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}. \end{aligned}$$

(18)

Theorem 2

If $\lambda - \lambda_{1} - \lambda_{2} \in ( - p\lambda_{1},p(\lambda - \lambda_{1}))$and the constant factor

$$\frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) $$

in (18) is the best possible, then $\lambda_{1} + \lambda_{2} = \lambda$with $\lambda_{1},\lambda_{2} \in (0,\lambda )$.

Proof

As regards to the assumptions, we find $0 < \hat{\lambda}_{1},\hat{\lambda}_{2} < \lambda$. By (16) the unified best possible constant factor in (18) must be of the form

$$\hat{\lambda}_{1}\hat{\lambda}_{2}B(\hat{ \lambda}_{1},\hat{\lambda}_{2}) \biggl(= \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B(\hat{\lambda}_{1} + 1,\hat{ \lambda}_{2} + 1)\biggr), $$

namely, it follows that

$$B(\hat{\lambda}_{1} + 1,\hat{\lambda}_{2} + 1)= B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}). $$

By Hölder’s inequality (see [27]) we obtain

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

(19)

We observe that (19) becomes equality if and only if there exist constants A and B such that they are not all zero and

$$Au^{\lambda - \lambda_{2}} = Bu^{\lambda_{1}}\quad \mbox{a.e. in }R_{ +} $$

(see [26]). Without loss of generality, we suppose $A \ne 0$. It follows that $u^{\lambda - \lambda_{2} - \lambda_{1}} = \frac{B}{A}$ a.e. in $R_{ +}$, namely, $\lambda - \lambda_{1} - \lambda_{2} = 0$, and then $\lambda_{1} + \lambda_{2} = \lambda$ with $\lambda_{1},\lambda_{2} \in (0,\lambda )$.

The theorem is proved. □

Theorem 3

The following statements are equivalent:

• (i) $B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1})$is independent ofp, q;

• (ii) $B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1})$is expressible as a single integral;

• (iii) If $\lambda - \lambda_{1} - \lambda_{2} \in ( - p\lambda_{1},p(\lambda - \lambda_{1}))$, then $\lambda_{1} + \lambda_{2} = \lambda$ ($\lambda_{1},\lambda_{2} \in (0,\lambda )$);

• (iv) The constant factor

  $$\frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) $$

  in (15) is the best possible.

Proof

(i) ⇒ (ii). Since $B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1})$ is independent of p, q, we find

$$\begin{aligned}& B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\& \quad = \lim_{p \to \infty} \lim_{q \to 1^{ +}} B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\& \quad = B(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) = \int_{0}^{\infty} \frac{u^{\lambda_{1}}}{(1 + u)^{\lambda + 2}} \,du, \end{aligned}$$

which is a single integral.

(ii) ⇒ (iii). Suppose that $B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1})$ is expressible as a single integral $\int_{0}^{\infty} \frac{1}{(1 + u)^{\lambda + 2}}u^{\frac{\lambda + 1 - \lambda_{2}}{p} + \frac{\lambda_{1} + 1}{q} - 1}\,du$. Then (19) keeps the form of equality. By the proof of Theorem 2 we have $\lambda_{1} + \lambda_{2} = \lambda$ ($\lambda_{1},\lambda_{2} \in (0,\lambda )$).

(iii) ⇒ (iv). If $\lambda_{1} + \lambda_{2} = \lambda$ ($\lambda_{1},\lambda_{2} \in (0,\lambda )$), then by Theorem 1 the constant factor

$$\frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \bigl( = \lambda_{1} \lambda_{2}B(\lambda_{1},\lambda_{2})\bigr) $$

in (13) is the best possible.

(iv) ⇒ (i). In this case, by Theorem 2 we have $\lambda_{1} + \lambda_{2} = \lambda$, and

$$B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) = B(\lambda_{1} + 1, \lambda_{2} + 1) $$

is independent of p, q.

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

The theorem is proved. □

Remark 2

If $\mu + \sigma = s$ ($\mu,\sigma \in (0,s)$), then inequality (11) reduces to

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{s}} \,dx\,dy \le& B(\mu,\sigma )\biggl[ \int_{0}^{\infty} x^{p(1 - \mu ) - 1} f^{p}(x)\,dx\biggr]^{\frac{1}{p}} \\ &{}\times\biggl[ \int_{0}^{\infty} y^{q(1 - \sigma ) - 1}g^{q} (y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$

(20)

We confirm that the constant factor $B(\mu,\sigma )$ in (20) is the best possible. Otherwise, we would reach a contradiction by (17) that the constant factor in (16) is not the best possible.

Replacing x by $\frac{1}{x}$ and then $x^{s - 2}f(\frac{1}{x})$ by $f(x)$ in (20), we have the following Hardy–Hilbert’s integral inequality with a nonhomogeneous kernel and the best possible constant factor $B(s - \sigma,\sigma )$:

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{s}} \,dx\,dy \le& B(s - \sigma,\sigma )\biggl[ \int_{0}^{\infty} x^{p(1 - \sigma ) - 1} f^{p}(x)\,dx\biggr]^{\frac{1}{p}} \\ &{}\times\biggl[ \int_{0}^{\infty} y^{q(1 - \sigma ) - 1}g^{q} (y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$

(21)

4 A corollary and some particular cases

Replacing x by $\frac{1}{x}$ in (15) and setting $\hat{f}(x) = x^{\lambda - 2}f(\frac{1}{x})$, we define

$$F_{\lambda} (x): = \int_{\frac{1}{x}}^{\infty} t^{ - \lambda} \hat{f}(t)\,dt \biggl( = \int_{\frac{1}{x}}^{\infty} f\biggl(\frac{1}{u} \biggr)\frac{1}{u^{2}}\,du = \int_{0}^{x} f(t)\,dt \biggr). $$

Then replacing $\hat{f}(x)$ by $f(x)$, we have $F_{\lambda} (x) = \int_{\frac{1}{x}}^{\infty} t^{ - \lambda} f(t)\,dt$ and the following Hilbert-type integral inequality with nonhomogeneous kernel:

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < & \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\ &{}\times \biggl[ \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F_{\lambda}^{p}(x)\,dx\biggr]^{\frac{1}{p}} \\ &{}\times \biggl[ \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy\biggr]^{\frac{1}{q}}, \end{aligned}$$

(22)

which is equivalent to (15).

In view of Theorem 3, we have the following:

Corollary 1

Assuming that $\lambda - \lambda_{1} - \lambda_{2} \in ( - p\lambda_{1},p(\lambda - \lambda_{1}))$, the constant factor

$$\frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) $$

in (22) is the best possible if and only if $\lambda_{1} + \lambda_{2} = \lambda$ ($\lambda_{1},\lambda_{2} \in (0,\lambda )$).

In the case of $\lambda_{1} + \lambda_{2} = \lambda$, (22) reduces to the following Hilbert-type integral inequality with nonhomogeneous kernel and the best possible constant factor $\lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})$:

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < & \lambda_{1}\lambda_{2}B(\lambda_{1}, \lambda_{2}) \\ &{}\times \biggl\{ \int_{0}^{\infty} x^{ - p\lambda_{1} - 1} F_{\lambda}^{p}(x)\,dx\biggr\} ^{\frac{1}{p}}\biggl\{ \int_{0}^{\infty} y^{ - q\lambda_{2} - 1} G^{q}(y)\,dy\biggr\} ^{\frac{1}{q}}, \end{aligned}$$

(23)

which is equivalent to (16).

Remark 3

In (16) and (23), for $\lambda_{1} = \frac{\lambda}{ q}$, $\lambda_{2} = \frac{\lambda}{p}$, we have the following equivalent inequalities:

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < \frac{\lambda^{2}}{pq}B\biggl(\frac{\lambda}{p},\frac{\lambda}{q}\biggr) \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy< } {}\times \biggl( \int_{0}^{\infty} x^{(1 - p)\lambda - 1} F^{p}(x)\,dx\biggr)^{\frac{1}{p}} \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < } {}\times \biggl( \int_{0}^{\infty} y^{(1 - q)\lambda - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}, \end{aligned}$$

(24)

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < \frac{\lambda^{2}}{pq}B\biggl(\frac{\lambda}{p},\frac{\lambda}{q}\biggr) \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < } {}\times \biggl( \int_{0}^{\infty} x^{(1 - p)\lambda - 1} F_{\lambda}^{p}(x)\,dx\biggr)^{\frac{1}{p}} \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < } {}\times \biggl( \int_{0}^{\infty} y^{(1 - q)\lambda - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}, \end{aligned}$$

(25)

and for $\lambda_{1} = \frac{\lambda}{p}$, $\lambda_{2} = \frac{\lambda}{q}$, we have the following equivalent inequalities:

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < \frac{\lambda^{2}}{pq}B\biggl(\frac{\lambda}{p},\frac{\lambda}{q}\biggr) \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < }{}\times \biggl( \int_{0}^{\infty} x^{ - \lambda - 1} F^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - \lambda - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}, \end{aligned}$$

(26)

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < \frac{\lambda^{2}}{pq}B\biggl(\frac{\lambda}{p},\frac{\lambda}{q}\biggr) \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < } {}\times \biggl( \int_{0}^{\infty} x^{ - \lambda - 1} F_{\lambda}^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - \lambda - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}. \end{aligned}$$

(27)

In particular, for $p = q = 2$, both inequalities (24) and (26) reduce to

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < & \frac{\lambda^{2}}{4}B\biggl(\frac{\lambda}{2},\frac{\lambda}{2}\biggr) \\ &{}\times \biggl( \int_{0}^{\infty} x^{ - \lambda - 1} F^{2}(x)\,dx \int_{0}^{\infty} y^{ - \lambda - 1} G^{2}(y)\,dy\biggr)^{\frac{1}{2}}, \end{aligned}$$

(28)

and both (25) and (27) reduce to the following equivalent form of (25):

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < & \frac{\lambda^{2}}{4}B\biggl(\frac{\lambda}{2},\frac{\lambda}{2}\biggr) \\ &{}\times \biggl( \int_{0}^{\infty} x^{ - \lambda - 1} F_{\lambda}^{2}(x)\,dx \int_{0}^{\infty} y^{ - \lambda - 1} G^{2}(y)\,dy\biggr)^{\frac{1}{2}}. \end{aligned}$$

(29)

The constant factors in the inequalities of Remark 3 are the best possible.

5 Conclusions

In this paper, following [21, 22], using the weight functions and the idea of introduced parameters, we give a new Hilbert-type integral inequality with the kernel $\frac{1}{(x + y)^{\lambda}}$ ($\lambda > 0$) involving the upper limit functions and the beta and gamma functions (Theorem 1). The preliminaries and the equivalent statements of the best possible constant factor related to a few parameters are considered in Theorems 2 and 3. As applications, we obtain a corollary in the case of nonhomogeneous kernel and some particular inequalities (Corollary 1 and Remark 3). The lemmas and theorems provide an extensive account of inequalities of this type.