On a composition of two Hilbert-Hardy-type integral operators and related inequalities

Abstract

By applying the way of real and functional analysis and estimating the weight functions, we build some lemmas and deduce some Hilbert-type and Hilbert-Hardy-type integral inequalities with the best possible constant factors. The equivalent forms, the reverses and the operator expressions are all considered. The composition formula of two Hilbert-Hardy-type integral operators and some examples are given.

1 Introduction

Assuming that $f(x),g(y)\geq0$, $f,g\in L^{2}(\mathbf{R}_{+})=\{f;\|f\| _{2}=(\int_{0}^{\infty}|f(x)|^{2}\, dx)^{\frac{1}{2}}<\infty\}$, $\|f\|_{2},\|g\|_{2}>0$, we have the following well-known Hilbert integral inequality and the equivalent form (cf. [1]):

$$\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \pi \|f\|_{2}\|g\|_{2}, \end{aligned}$$

(1)

$$\begin{aligned}& \biggl[ \int_{0}^{\infty} \biggl( \int _{0}^{\infty}\frac {f(x)}{x+y}\,dx \biggr) ^{2}\,dy \biggr] ^{\frac{1}{2}}< \pi\|f\|_{2}, \end{aligned}$$

(2)

where the constant factor π is the best possible.

In 1925, by introducing a pair of conjugate exponents $( p,q )$ ($\frac{1}{p}+\frac{1}{q}=1$), Hardy [2] gave some extensions of (1) and (2) as follows: For $p>1$, $f(x),g(y)\geq0$, $f\in L^{p}(\mathbf{R}_{+})$, $g\in L^{q}(\mathbf{R}_{+})$, $\|f\|_{p},\|g\|_{q}>0$, we have the following Hardy-Hilbert integral inequality and the equivalent form:

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

(3)

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

(4)

where the constant factor $\frac{\pi}{\sin(\pi/p)}$ is the best possible. For $p=q=2$, inequalities (3) and (4) reduce respectively to (1) and (2).

Definition 1

If $\lambda\in\mathbf{R}=(-\infty,\infty)$, $\mathbf{R} _{+}=(0,\infty)$, $k_{\lambda}(x,y)$ is a measurable function in $\mathbf {R}_{+}^{2}=\mathbf{R}_{+}\times\mathbf{R}_{+}$, satisfying for any $t,x,y\in \mathbf{R}_{+}$, $k_{\lambda}(tx,ty)=t^{-\lambda}k_{\lambda}(x,y)$, then we call $k_{\lambda}(x,y)$ homogeneous function of degree −λ.

In 1934, by using a general non-negative homogeneous function of degree −1 as $k_{1}(x,y)$, Hardy et al. [3] gave some extensions of (3) and (4) as follows: For $p>1$, $k_{p}=\int_{0}^{\infty }k_{1}(u,1)u^{\frac{-1}{p}}\, du\in\mathbf{R}_{+}$, $f(x),g(y)\geq0$, $f\in L^{p}(\mathbf{R}_{+})$, $g\in L^{q}(\mathbf{R}_{+})$, $\|f\|_{p},\|g\|_{q}>0$, we have the following Hardy-Hilbert-type integral inequality and the equivalent form:

$$\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty }k_{1}(x,y)f(x)g(y) \,dx\,dy< k_{p}\|f\|_{p}\|g\|_{q}, \end{aligned}$$

(5)

$$\begin{aligned}& \biggl[ \int_{0}^{\infty} \biggl( \int _{0}^{\infty}k_{1}(x,y)f(x)\,dx \biggr) ^{p}\,dy \biggr] ^{\frac{1}{p}}< k_{p}\|f\|_{p}, \end{aligned}$$

(6)

where the constant factor $k_{p}$ is the best possible. Some applications of (5) and (6) are provided in [4].

In 1998, by introducing an independent parameter $\lambda\in(0,1]$, Yang [5] gave an extension of (3) with the homogeneous kernel of degree −λ as $\frac{1}{(x+y)^{\lambda}}$. In 2009, by using a general non-negative homogeneous function of degree −λ as $k_{\lambda}(x,y)$ and adding another pair of conjugate exponents $( r,s )$ ($\frac{1}{r}+\frac{1}{s}=1$), Yang [6] gave some extensions of (5) and (6) as follows: For $p,r>1$, $\Phi (x)=x^{p(1-\frac{\lambda}{r})-1}$, $\Psi(y)=y^{q(1-\frac{\lambda}{s})-1}$ ($x,y\in\mathbf{R}_{+}$), $k_{\lambda}(r)=\int_{0}^{\infty}k_{\lambda }(u,1)u^{\frac{\lambda}{r}-1}\,du\in\mathbf{R}_{+}$, $f(x),g(y)\geq0$,

$$ f\in L_{p,\Phi}(\mathbf{R}_{+})=\biggl\{ f;\|f \|_{p,\Phi}=\biggl(\int_{0}^{\infty }\Phi (x) \bigl\vert f(x)\bigr\vert ^{p}\,dx\biggr)^{\frac{1}{p}}< \infty\biggr\} , $$

$g\in L_{q,\Psi}(\mathbf{R}_{+})$, $\|f\|_{p,\Phi},\|g\|_{q,\Psi }>0$, we have the following Yang-Hilbert-type integral inequality and the equivalent form:

$$\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda }(x,y)f(x)g(y) \,dx\,dy< k_{\lambda }(r)\|f\|_{p,\Phi}\|g\|_{q,\Psi}, \end{aligned}$$

(7)

$$\begin{aligned}& \biggl[ \int_{0}^{\infty}y^{\frac{p\lambda}{s}-1} \biggl( \int_{0}^{\infty }k_{\lambda}(x,y)f(x)\,dx \biggr) ^{p}\,dy \biggr] ^{\frac{1}{p}}< k_{\lambda }(r)\|f \|_{p,\Phi}, \end{aligned}$$

(8)

where the constant factor $k_{\lambda}(r)$ is the best possible.

Remark 1

(i) When $\lambda=1$, $r=q$, $s=p$, (7) and (8) reduce respectively to (5) and (6). (ii) By (8), setting $y=\frac{1}{z}$, we have the following Yang-Hilbert-type inequality with the best possible constant factor and a non-homogeneous kernel:

$$ \biggl[ \int_{0}^{\infty}z^{\frac{p\lambda}{r}-1} \biggl( \int_{0}^{\infty }k_{\lambda}(xz,1)f(x)\,dx \biggr) ^{p}\, dz \biggr] ^{\frac{1}{p}}< k_{\lambda }(r)\|f\|_{p,\Phi}. $$

(9)

Using (2), we may define Hilbert’s integral operator $T:L^{2}(\mathbf{R}_{+})\rightarrow L^{2}(\mathbf{R}_{+})$ as follows (cf. [7]): For any $f\in L^{2}(\mathbf{R}_{+})$, there exists $Tf\in L^{2}(\mathbf{R}_{+})$ satisfying

$$ Tf(y)=\int_{0}^{\infty}\frac{f(x)}{x+y}\,dx\quad (y \in\mathbf{R}_{+}). $$

Then by (2) we have $\|Tf\|_{2}\leq\pi\|f\|_{2}$, and T is a bounded linear operator satisfying $\|T\|\leq\pi$. Since the constant factor in (2) is the best possible, we have $\|T\|=\pi$.

About the discrete forms of (1) and (2), in 1950, Wilhelm [8] gave an operator expression. In 2002, by using the operator theory, Zhang [9] gave some improvements of (2) and the discrete form. In 2006 to 2009, [10] considered a new Hilbert-type operator and its applications, and [11] and [12] gave some multiple Hilbert-type operator expressions.

By using (8), we can define the Yang-Hilbert-type integral operator $T:L_{p,\Phi}(\mathbf{R}_{+})\rightarrow L_{p,\Phi}(\mathbf{R}_{+})$ as follows (cf. [6]): For any $f\in L_{p,\Phi}(\mathbf{R}_{+})$, there exists $Tf\in L_{p,\Phi}(\mathbf{R}_{+})$ satisfying

$$ Tf(y)=y^{\lambda-1}\int_{0}^{\infty} \frac{f(x)}{x+y}\,dx\quad (y\in\mathbf{R} _{+}). $$

Then by (8) we have $\|Tf\|_{p,\Phi}\leq k_{\lambda }(r)\|f\|_{p,\Phi}$, and T is a bounded linear operator satisfying $\|T\|\leq k_{\lambda}(r)$. Since the constant factor in (8) is the best possible, we have $\|T\|=k_{\lambda}(r)$.

About the composition of two Hilbert-type operators, the main objective is to build the expression $\|T_{1}T_{2}\|=\|T_{1}\|\cdot\|T_{2}\|$. Recently, [13] published a composition of two discrete Hilbert-Hardy-type operators with particular kernels. Adiyasuren et al. [14] published a composition of two half-discrete Hilbert-Hardy-type operators with some particular kernels, and [15] and [16] published some composition of two Hilbert-Hardy-type integral operators with particular kernels. These works are hard and interesting.

In this paper, applying the way of real and functional analysis and estimating the weight functions, we build some lemmas and deduce some Hilbert-type and Hilbert-Hardy-type integral inequalities with the best possible constant factors. The equivalent forms, the reverses and the operator expressions are all considered. The composition formulas of two Hilbert-Hardy-type integral operators and some examples are given, which are some extensions of the results of [15] and [16].

2 Some lemmas

In the following, we agree on that $p>0$ ($p\neq1$), $\frac{1}{p}+\frac{1}{q}=1$.

Lemma 1

(cf. [17], Lemma 2.2.5)

Suppose that $\lambda\in A=(0,c)$ ($0< c\leq\infty$), $k_{\lambda}^{(s)}(x,y)$ are non-negative homogeneous functions of degree −λ in $\mathbf{R}_{+}^{2}$,

$$ k^{(s)}\biggl(\frac{\lambda}{2}\biggr):=\int_{0}^{\infty}k_{\lambda }^{(s)}(u,1)u^{\frac{\lambda}{2}-1} \,du\quad (s=1,2,3), $$

(10)

there exists a constant $\delta_{0}\in(0,\frac{\lambda}{2})$ such that $k^{(s)}(\frac{\lambda}{2}\pm\delta_{0})\in\mathbf{R}_{+}$. Then, for any $\delta\in[0,\delta_{0})$, we have $k^{(s)}(\frac{\lambda }{2}\pm \delta)\in\mathbf{R}_{+}$ and

$$ \lim_{\delta\rightarrow0^{+}}k^{(s)}\biggl(\frac{\lambda}{2}\pm\delta \biggr)=k^{(s)}\biggl(\frac{\lambda}{2}\biggr)\quad (s=1,2,3). $$

With the assumptions of Lemma 1, we set the following conditions.

Condition (i)

For $\lambda\in A$, there exist constants $\delta _{1}\in(0,\delta_{0})$ and $L_{1}>0$ such that

$$ k_{\lambda}^{(s)}(u,1)u^{\frac{\lambda}{2}+\delta_{1}}\leq L_{1}\quad \bigl(u\in (1,\infty);s=2,3\bigr). $$

(11)

Condition (ii)

For $\lambda\in(0,1)\cap A$, there exists a constant $L_{2}>0$ such that

$$ k_{\lambda}^{(s)}(u,1) (u-1)^{\lambda}\leq L_{2} \quad \bigl(u\in(1,\infty);s=2,3\bigr). $$

(12)

Condition (iii)

For $\lambda\in(0,1)\cap A$, there exist constants $a\in(0,\lambda)$ and $L_{3}>0$ such that

$$ k_{\lambda}^{(1)}(u,1)u^{a}\leq L_{3}\quad \bigl(u\in(0,\infty)\bigr). $$

(13)

Example 1

For $\lambda\in A=(0,\infty)$, $s=1,2,3$, the functions

$$ k_{\lambda}^{(s)}(u,1)=\frac{1}{(u+1)^{\lambda}},\frac{1}{u^{\lambda }+1},\frac{\ln u}{u^{\lambda}-1},\frac{|\ln u|^{\beta-1}}{(\max \{u,1\})^{\lambda}}\quad (\beta\geq1) $$

satisfy Conditions (i) and (iii). In fact, for $b=\frac{\lambda }{2}+\delta _{1}$ or $b=a\in(0,\lambda)$, we find

$$ \lim_{u\rightarrow0^{+}}k_{\lambda}^{(s)}(u,1)u^{b}= \lim_{u\rightarrow \infty}k_{\lambda}^{(s)}(u,1)u^{b}=0. $$

In view of the continuity, $k_{\lambda}^{(s)}(u,1)u^{b}$ ($s=1,2,3$) are bounded in $(0,\infty)$ and then satisfy (11) and (13).

It is evident that for $\lambda\in A=(0,1)$, the functions

$$ k_{\lambda}^{(s)}(u,1)=\frac{1}{(u-1)^{\lambda}}\quad \bigl(u\in(1, \infty);s=2,3\bigr) $$

satisfy Condition (ii).

Definition 2

With the assumptions of Lemma 1 and Condition (i), we define the following two sequences of real functions:

$$ \begin{aligned} &\widetilde{F}_{k}(y):=\left \{ \begin{array}{l@{\quad}l} y^{\lambda-1}\int_{0}^{1}k_{\lambda}^{(2)}(x,y)x^{\frac{\lambda }{2}+\frac{1}{pk}-1}\,dx,&y\in(0,1), \\ 0,&y\in{}[1,\infty),\end{array} \right . \\ &\widetilde{G}_{k}(x):=\left \{ \begin{array}{l@{\quad}l} x^{\lambda-1}\int_{1}^{\infty}k_{\lambda}^{(3)}(x,y)y^{\frac{\lambda }{2}-\frac{1}{qk}-1}\,dy,&x\in(1,\infty), \\ 0,&x\in(0,1],\end{array} \right . \end{aligned} $$

(14)

where $k>\max\{\frac{1}{|q|\delta_{1}},\frac{1}{p\delta_{1}}\}$ ($k\in \mathbf{N}=\{1,2,\ldots\}$).

Setting $u=x/y$ ($0< y<1$), we find

$$\begin{aligned}& \widetilde{F}_{k}(y) = y^{\frac{\lambda}{2}+\frac{1}{pk}-1}\int_{0}^{\frac{1}{y}}k_{\lambda}^{(2)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,du \\& \hphantom{\widetilde{F}_{k}(y)} = y^{\frac{\lambda}{2}+\frac{1}{pk}-1} \biggl( \int_{0}^{\infty }k_{\lambda }^{(2)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,du-\int_{\frac{1}{y}}^{\infty}k_{\lambda}^{(2)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,du \biggr) \\& \hphantom{\widetilde{F}_{k}(y)} = y^{\frac{\lambda}{2}+\frac{1}{pk}-1}k^{(2)}\biggl(\frac{\lambda}{2}+ \frac {1}{pk}\biggr)-F(y), \\& F(y) : =y^{\frac{\lambda}{2}+\frac{1}{pk}-1}\int_{\frac{1}{y}}^{\infty }k_{\lambda}^{(2)}(u,1)u^{\frac{\lambda}{2}+\frac {1}{pk}-1} \,du\quad \bigl(y\in (0,1)\bigr). \end{aligned}$$

(15)

(a) If $k_{\lambda}^{(2)}(u,1)$ satisfies Condition (i) (for $\lambda \in A$), then by (11) we have

$$ 0\leq F(y)\leq L_{1}y^{\frac{\lambda}{2}+\frac{1}{pk}-1}\int_{\frac {1}{y}}^{\infty}u^{-\frac{\lambda}{2}-\delta_{1}}u^{\frac{\lambda }{2}+\frac{1}{pk}-1} \,du=\frac{L_{1}y^{\frac{\lambda}{2}+\delta_{1}-1}}{\delta _{1}-\frac{1}{pk}}\quad \bigl(y\in(0,1)\bigr). $$

(b) If $k_{\lambda}^{(2)}(u,1)$ satisfies Condition (ii) (for $\lambda \in(0,1)\cap A$), then by (12) we have

$$\begin{aligned} 0 \leq&F(y)\leq L_{2}y^{\frac{\lambda}{2}+\frac{1}{pk}-1}\int_{\frac {1}{y}}^{\infty} \frac{u^{\frac{\lambda}{2}+\frac{1}{pk}-1}}{(u-1)^{\lambda }}\,du\overset{v=1/(yu)}{=}L_{2}y^{\lambda-1} \int_{0}^{1}\frac{v^{\frac {\lambda}{2}-\frac{1}{pk}-1}}{(1-yv)^{\lambda}}\,dv \\ \leq&\frac{L_{2}y^{\lambda-1}}{(1-y)^{\lambda}}\int_{0}^{1}v^{\frac{ \lambda}{2}-\frac{1}{pk}-1} \,dv=\frac{L_{2}}{\frac{\lambda}{2}-\frac {1}{pk}}\frac{y^{\lambda-1}}{(1-y)^{\lambda}}\quad \bigl(y\in(0,1)\bigr). \end{aligned}$$

Still setting $u=x/y$ ($x>1$), we obtain

$$\begin{aligned}& \widetilde{G}_{k}(x) = x^{\frac{\lambda}{2}-\frac{1}{qk}-1}\int_{0}^{x}k_{\lambda}^{(3)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{qk}-1} \,du \\& \hphantom{\widetilde{G}_{k}(x)} = x^{\frac{\lambda}{2}-\frac{1}{qk}-1} \biggl( \int_{0}^{\infty }k_{\lambda }^{(3)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{qk}-1} \,du-\int_{x}^{\infty }k_{\lambda}^{(3)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{qk}-1} \,du \biggr) \\& \hphantom{\widetilde{G}_{k}(x)} = x^{\frac{\lambda}{2}-\frac{1}{qk}-1}k^{(3)}\biggl(\frac{\lambda}{2}+ \frac {1}{qk}\biggr)-G(x), \\& G(x) : =x^{\frac{\lambda}{2}-\frac{1}{qk}-1}\int_{x}^{\infty }k_{\lambda }^{(3)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{qk}-1} \,du\quad \bigl(x\in(1,\infty)\bigr). \end{aligned}$$

(16)

(c) If $k_{\lambda}^{(3)}(u,1)$ satisfies Condition (i) (for $\lambda \in A$), then by (11) we have

$$ 0\leq G(x)\leq L_{1}x^{\frac{\lambda}{2}-\frac{1}{qk}-1}\int_{x}^{\infty }u^{-\frac{\lambda}{2}-\delta_{1}}u^{\frac{\lambda}{2}+\frac {1}{qk}-1} \,du=\frac{L_{1}x^{\frac{\lambda}{2}-\delta_{1}-1}}{\delta_{1}-\frac {1}{qk}}\quad \bigl(x\in(1,\infty)\bigr). $$

(d) If $k_{\lambda}^{(3)}(u,1)$ satisfies Condition (ii) (for $\lambda \in(0,1)\cap A$), then by (12) we have

$$\begin{aligned} 0 \leq&G(x)\leq L_{2}x^{\frac{\lambda}{2}-\frac{1}{qk}-1}\int_{x}^{\infty } \frac{u^{\frac{\lambda}{2}+\frac{1}{qk}-1}}{(u-1)^{\lambda }}\,du\overset{v=x/u}{=}L_{2}x^{\lambda-1} \int_{0}^{1}\frac{v^{\frac{\lambda }{2}-\frac{1}{qk}-1}}{(x-v)^{\lambda}}\,dv \\ \leq&\frac{L_{2}x^{\lambda-1}}{(x-1)^{\lambda}}\int_{0}^{1}v^{\frac{ \lambda}{2}-\frac{1}{qk}-1} \,dv=\frac{L_{2}x^{\lambda-1}}{(\frac {\lambda}{2}-\frac{1}{qk})(x-1)^{\lambda}}\quad \bigl(x\in(1,\infty)\bigr). \end{aligned}$$

Remark 2

In view of the results of (a)-(d), there exists a large constant $L>0$ such that

• (a) $F(y)\leq Ly^{\frac{\lambda}{2}+\delta_{1}-1}$ ($y\in(0,1)$; $\lambda \in A$);

• (b) $F(y)\leq L\frac{y^{\lambda-1}}{(1-y)^{\lambda}}$ ($y\in (0,1)$; $\lambda \in(0,1)\cap A$);

• (c) $G(x)\leq Lx^{\frac{\lambda}{2}-\delta_{1}-1}$ ($x\in(1,\infty )$; $\lambda \in A$);

• (d) $G(x)\leq L\frac{x^{\lambda-1}}{(x-1)^{\lambda}}$ ($x\in(1,\infty )$; $\lambda\in(0,1)\cap A$).

Lemma 2

With the assumptions of Lemma  1, (1) $k_{\lambda }^{(2)}(u,1)$ ($k_{\lambda}^{(3)}(u,1)$) satisfies Condition (i) or Condition (ii); (2) if $k_{\lambda}^{(s)}(u,1)$ ($s=2,3$) only satisfy Condition (i), then $\lambda\in A$; otherwise, $\lambda\in(0,1)\cap A$. Then we have

$$ \widetilde{L}_{k} :=\frac{1}{k}\int_{0}^{\infty} \int_{0}^{\infty }k_{\lambda}^{(1)}(xy,1) \widetilde{F}_{k}(y)\widetilde{G}_{k}(x)\, dy\, dx \geq\prod_{i=1}^{3}k_{\lambda}^{(i)} \biggl(\frac{\lambda}{2}\biggr)+o(1)\quad (k\rightarrow\infty). $$

(17)

Proof

In view of (15) and (16), we have

$$\begin{aligned} \widetilde{L}_{k} =&\frac{1}{k}\int_{1}^{\infty} \int_{0}^{1}k_{\lambda }^{(1)}(xy,1) \biggl[ y^{\frac{\lambda}{2}+\frac{1}{pk}-1}k_{\lambda }^{(2)}\biggl(\frac{\lambda}{2}+\frac{1}{pk}\biggr)-F(y) \biggr] \\ &{}\times \biggl[ x^{\frac{\lambda}{2}-\frac{1}{qk}-1}k_{\lambda }^{(3)}\biggl( \frac{\lambda}{2}+\frac{1}{qk}\biggr)-G(x) \biggr] \,dy \,dx=I_{1}-I_{2}-I_{3}+I_{4}, \end{aligned}$$

(18)

where we define

$$\begin{aligned}& I_{1} : =\frac{1}{k}k_{\lambda}^{(2)}\biggl( \frac{\lambda}{2}+\frac{1}{pk}\biggr)k_{\lambda}^{(3)} \biggl(\frac{\lambda}{2}+\frac{1}{qk}\biggr) \\& \hphantom{I_{1} : =}\times\int_{1}^{\infty} \biggl( \int _{0}^{1}k_{\lambda }^{(1)}(xy,1)y^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,dy \biggr) x^{\frac{\lambda}{2}-\frac {1}{qk}-1}\,dx, \\& I_{2} : =\frac{1}{k}k_{\lambda}^{(3)}\biggl( \frac{\lambda}{2}+\frac{1}{qk}\biggr)\int_{1}^{\infty} \biggl( \int_{0}^{1}k_{\lambda }^{(1)}(xy,1)F(y) \,dy \biggr) x^{\frac{\lambda}{2}-\frac{1}{qk}-1}\,dx, \\& I_{3} : =\frac{1}{k}k_{\lambda}^{(2)}\biggl( \frac{\lambda}{2}+\frac{1}{pk}\biggr)\int_{1}^{\infty} \biggl( \int_{0}^{1}k_{\lambda}^{(1)}(xy,1)y^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,dy \biggr) G(x)\,dx, \\& I_{4} : =\frac{1}{k}\int_{1}^{\infty} \biggl( \int_{0}^{1}k_{\lambda }^{(1)}(xy,1)F(y) \,dy \biggr) G(x)\,dx. \end{aligned}$$

It is evident that

$$ I_{1}-I_{2}-I_{3}\leq\widetilde{L}_{k} \leq I_{1}+I_{4}. $$

(19)

By Fubini’s theorem, we obtain that (cf. [18])

$$\begin{aligned}& \int_{1}^{\infty} \biggl( \int_{0}^{1}k_{\lambda}^{(1)}(xy,1)y^{\frac{ \lambda}{2}+\frac{1}{pk}-1} \,dy \biggr) x^{\frac{\lambda}{2}-\frac{1}{qk}-1}\,dx \\& \quad \overset{u=xy}{=} \int_{1}^{\infty} \biggl( \int_{0}^{x}k_{\lambda }^{(1)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,du \biggr) x^{-\frac {1}{k}-1}\,dx \\& \quad = \int_{1}^{\infty} \biggl( \int _{0}^{1}k_{\lambda}^{(1)}(u,1)u^{\frac{ \lambda}{2}+\frac{1}{pk}-1} \,du \biggr) x^{-\frac{1}{k}-1}\,dx \\& \qquad {} +\int_{1}^{\infty} \biggl( \int _{1}^{x}k_{\lambda}^{(1)}(u,1)u^{\frac{ \lambda}{2}+\frac{1}{pk}-1} \,du \biggr) x^{-\frac{1}{k}-1}\,dx \\& \quad = k\int_{0}^{1}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}+\frac {1}{pk} -1} \,du \\& \qquad {} + \int_{1}^{\infty} \biggl( \int _{u}^{\infty}x^{-\frac{1}{k}-1}\,dx \biggr) k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1}\,du \\& \quad = k \biggl( \int_{0}^{1}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda }{2}+\frac{1}{pk}-1} \,du+\int_{1}^{\infty}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}- \frac{1}{qk}-1} \,du \biggr) . \end{aligned}$$

Since $\{k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1}\}_{k=1}^{\infty}$ ($u\in(0,1)$) is increasing, by Levi’s theorem (cf. [18]), it follows that

$$ \int_{0}^{1}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,du\rightarrow\int_{0}^{1}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda }{2}-1} \,du\quad (k\rightarrow\infty). $$

Since $k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}-\frac{1}{qk}-1}\leq k_{\lambda}^{(1)}(u,1)u^{\mu+\delta_{1}-1}$ ($u\in(1,\infty)$) and

$$ 0\leq\int_{1}^{\infty}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda }{2}+\delta _{1}-1} \,du\leq k^{(1)}\biggl(\frac{\lambda}{2}+\delta_{1}\biggr)< \infty, $$

then by the Lebesgue convergence control theorem (cf. [18]), we have

$$ \int_{1}^{\infty}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}-\frac {1}{qk}-1} \,du\rightarrow\int_{1}^{\infty}k_{\lambda}^{(1)}(u,1)u^{\frac {\lambda}{2}-1} \,du\quad (k\rightarrow\infty). $$

Hence, by Lemma 1, we find, for $k\rightarrow\infty$,

$$\begin{aligned} I_{1} =&k^{(2)}\biggl(\frac{\lambda}{2}+\frac{1}{pk} \biggr)k^{(3)}\biggl(\frac{\lambda }{2}+\frac{1}{qk} \biggr) \\ &{}\times \biggl( \int_{0}^{1}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda }{2}+\frac{1}{pk}-1} \,du+\int_{1}^{\infty}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda }{2}-\frac{1}{qk}-1} \,du \biggr) \rightarrow\prod_{s=1}^{3}k^{(s)} \biggl(\frac {\lambda}{2}\biggr). \end{aligned}$$

(20)

(1) We estimate $I_{2}$.

(a) If $k_{\lambda}^{(2)}(u,1)$ satisfies Condition (i) for $\lambda \in A$, then by Remark 2(a) we have

$$\begin{aligned} 0 \leq&J_{2}:=\int_{1}^{\infty} \biggl( \int_{0}^{1}k_{\lambda }^{(1)}(xy,1)F(y) \,dy \biggr) x^{\frac{\lambda}{2}-\frac{1}{qk}-1}\,dx \\ \leq&L\int_{1}^{\infty} \biggl( \int _{0}^{\infty}k_{\lambda }^{(1)}(xy,1)y^{\frac{\lambda}{2}+\delta_{1}-1} \,dy \biggr) x^{\frac {\lambda }{2}-\frac{1}{qk}-1}\,dx \\ \overset{u=xy}{=}&L\int_{1}^{\infty} \biggl( \int _{0}^{\infty }k_{\lambda }^{(1)}(u,1)u^{\frac{\lambda}{2}+\delta_{1}-1} \,du \biggr) x^{-\delta _{1}-\frac{1}{qk}-1}\,dx \\ =&\frac{L\cdot k_{\lambda}^{(1)}(\frac{\lambda}{2}+\delta _{1})}{\delta _{1}+\frac{1}{qk}}< \infty. \end{aligned}$$

(b) If $k_{\lambda}^{(2)}(u,1)$ satisfies Condition (ii) for $\lambda \in (0,1)\cap A$, then by Remark 2(b) we have

$$\begin{aligned} 0 \leq&J_{2}\leq L\int_{0}^{1} \biggl( \int_{0}^{\infty}k_{\lambda }^{(1)}(xy,1)x^{\frac{\lambda}{2}-\frac{1}{qk}-1} \,dx \biggr) \frac {y^{\lambda -1}}{(1-y)^{\lambda}}\,dy \\ \overset{u=xy}{=}&L\int_{0}^{1} \biggl( \int _{0}^{\infty}k_{\lambda }^{(1)}(u,1)u^{\frac{\lambda}{2}-\frac{1}{qk}-1} \,du \biggr) \frac {y^{\frac{\lambda}{2}+\frac{1}{qk}-1}}{(1-y)^{\lambda}}\,dy \\ =&L\cdot k^{(1)}\biggl(\frac{\lambda}{2}-\frac{1}{qk}\biggr)B \biggl(1-\lambda,\frac {\lambda }{2}+\frac{1}{qk}\biggr)< \infty. \end{aligned}$$

Therefore, in view of (a) and (b), we have $I_{2}\rightarrow 0$ ($k\rightarrow \infty$).

(2) We estimate $I_{3}$.

(c) If $k_{\lambda}^{(3)}(u,1)$ satisfies Condition (i) for $\lambda \in A$, then by Remark 2(c) we have

$$\begin{aligned} 0 \leq&J_{3}:=\int_{1}^{\infty} \biggl( \int_{0}^{1}k_{\lambda }^{(1)}(xy,1)y^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,dy \biggr) G(x)\,dx \\ \leq&L\int_{1}^{\infty} \biggl( \int _{0}^{\infty}k_{\lambda }^{(1)}(xy,1)y^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,dy \biggr) x^{\frac {\lambda }{2}-\delta_{1}-1}\,dx \\ \overset{u=xy}{=}&L\int_{1}^{\infty} \biggl( \int _{0}^{\infty }k_{\lambda }^{(1)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,du \biggr) x^{-\delta _{1}-\frac{1}{pk}-1}\,dx \\ =&\frac{L\cdot k^{(1)}(\frac{\lambda}{2}+\frac{1}{pk})}{\delta _{1}+\frac{1}{pk}}< \infty. \end{aligned}$$

(d) If $k_{\lambda}^{(3)}(u,1)$ satisfies Condition (ii) for $\lambda \in (0,1)\cap A$, then by Remark 2(d), we have

$$\begin{aligned} 0 \leq&J_{3}\leq L\int_{1}^{\infty} \biggl( \int_{0}^{\infty }k_{\lambda }^{(1)}(xy,1)y^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,dy \biggr) \frac {x^{\lambda -1}}{(x-1)^{\lambda}}\,dx \\ \overset{u=xy}{=}&L\int_{1}^{\infty} \biggl( \int _{0}^{\infty }k_{\lambda }^{(1)}(u,1)u^{\frac{\lambda}{2}+\frac{1}{pk}-1} \,du \biggr) \frac {x^{\frac{\lambda}{2}-\frac{1}{pk}-1}}{(x-1)^{\lambda}}\,dx \\ =&L\cdot k_{\lambda}^{(1)}\biggl(\frac{\lambda}{2}+ \frac {1}{pk}\biggr)B\biggl(1-\lambda,\frac{\lambda}{2}+ \frac{1}{pk}\biggr)< \infty. \end{aligned}$$

Therefore, in view of (c) and (d), we have $I_{3}\rightarrow 0$ ($k\rightarrow \infty$).

By (19) and the above results, we have (17). □

Lemma 3

Suppose that (1) $\lambda\in A=(0,c)$ ($0< c\leq\infty$), $k_{\lambda}^{(s)}(x,y)$ are non-negative homogeneous functions of degree −λ in $\mathbf{R}_{+}^{2}$,

$$ k^{(s)}\biggl(\frac{\lambda}{2}\biggr)=\int_{0}^{\infty}k_{\lambda}^{(s)}(u,1)u^{ \frac{\lambda}{2}-1} \,du\quad (s=1,2,3), $$

there exists a constant $\delta_{0}\in(0,\frac{\lambda}{2})$ such that $k^{(s)}(\frac{\lambda}{2}\pm\delta_{0})\in\mathbf{R}_{+}$; (2) $k_{\lambda}^{(2)}(u,1)$ ($k_{\lambda}^{(3)}(u,1)$) satisfies Condition (i) or Condition (ii); (3) if both $k_{\lambda}^{(2)}(u,1)$ and $k_{\lambda }^{(3)}(u,1)$ satisfy Condition (ii), then $k_{\lambda}^{(1)}(u,1)$ satisfies Condition (iii); (4) if $k_{\lambda}^{(s)}(u,1)$ ($s=2,3$) only satisfy Condition (i), then $\lambda\in A$; otherwise, $\lambda\in (0,1)\cap A$. Then we have the reverse of (17), namely

$$\begin{aligned} \widetilde{L}_{k} =&\frac{1}{k}\int_{0}^{\infty} \int_{0}^{\infty }k_{\lambda}^{(1)}(xy,1) \widetilde{F}_{k}(y)\widetilde{G}_{k}(x)\,dy\,dx \\ =&\prod_{s=1}^{3}k^{(s)}\biggl( \frac{\lambda}{2}\biggr)+o(1)\quad (k\rightarrow\infty). \end{aligned}$$

(21)

Proof

We have four cases to show that in any case, $I_{4}\rightarrow0$ ($k\rightarrow\infty$).

Case (i). $\lambda\in A$, $F(y)\leq Ly^{\frac{\lambda}{2}+\delta _{1}-1}$ ($y\in (0,1)$), $G(x)\leq Lx^{\frac{\lambda}{2}-\delta_{1}-1}$ ($x\in(1,\infty)$). We have

$$\begin{aligned} J_{4} :=&\int_{1}^{\infty} \biggl( \int _{0}^{1}k_{\lambda }^{(1)}(xy,1)F(y) \,dy \biggr) G(x)\,dx \\ \leq&L^{2}\int_{1}^{\infty} \biggl( \int _{0}^{\infty}k_{\lambda }^{(1)}(xy,1)y^{\frac{\lambda}{2}+\delta_{1}-1} \,dy \biggr) x^{\frac {\lambda }{2}-\delta_{1}-1}\,dx \\ \overset{u=xy}{=}&L^{2}\int_{1}^{\infty} \biggl( \int_{0}^{\infty }k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}+\delta_{1}-1} \,du \biggr) x^{-2\delta_{1}-1}\,dx \\ =&\frac{L^{2}}{2\delta_{1}}k^{(1)}\biggl(\frac{\lambda}{2}+\delta _{1}\biggr)< \infty. \end{aligned}$$

Case (ii). $\lambda\in(0,1)\cap A$, $F(y)\leq Ly^{\frac{\lambda }{2}+\delta _{1}-1}$ ($y\in(0,1)$), $G(x)\leq L\frac{x^{\lambda-1}}{(x-1)^{\lambda}}$ ($x\in(1,\infty)$). We have

$$\begin{aligned} J_{4} \leq&L^{2}\int_{1}^{\infty} \biggl( \int_{0}^{\infty}k_{\lambda }^{(1)}(xy,1)y^{\frac{\lambda}{2}+\delta_{1}-1} \,dy \biggr) \frac {x^{\lambda -1}}{(x-1)^{\lambda}}\,dx \\ \overset{u=xy}{=}&L^{2}\int_{1}^{\infty} \biggl( \int_{0}^{\infty }k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}+\delta_{1}-1} \,du \biggr) \frac{x^{\frac{\lambda}{2}-\delta_{1}-1}}{(x-1)^{\lambda}}\,dx \\ =&L^{2}k^{(1)}\biggl(\frac{\lambda}{2}+ \delta_{1}\biggr)B\biggl(1-\lambda,\frac {\lambda}{2}+ \delta_{1}\biggr)< \infty. \end{aligned}$$

Case (iii). $\lambda\in(0,1)\cap A$, $F(y)\leq L\frac{y^{\lambda-1}}{(1-y)^{\lambda}}$ ($y\in(0,1)$), $G(x)\leq Lx^{\frac{\lambda}{2}-\delta _{1}-1}$ ($x\in(1,\infty)$). We have

$$\begin{aligned} J_{4} \leq& L^{2}\int_{0}^{1} \biggl( \int_{0}^{\infty}k_{\lambda }^{(1)}(xy,1)x^{\frac{\lambda}{2}-\delta_{1}-1} \,dx \biggr) \frac {y^{\lambda -1}}{(1-y)^{\lambda}}\,dy \\ \overset{u=xy}{=}&L^{2}\int_{0}^{1} \biggl( \int_{0}^{\infty}k_{\lambda }^{(1)}(u,1)u^{\frac{\lambda}{2}-\delta_{1}-1} \,du \biggr) \frac{y^{\frac {\lambda}{2}+\delta_{1}-1}}{(1-y)^{\lambda}}\,dy \\ =&L^{2}k^{(1)}\biggl(\frac{\lambda}{2}- \delta_{1}\biggr)B\biggl(1-\lambda,\frac {\lambda}{2}+ \delta_{1}\biggr)< \infty. \end{aligned}$$

Case (iv). $\lambda\in(0,1)\cap A$, $F_{k}(y)\leq L\frac{y^{\lambda-1}}{ (1-y)^{\lambda}}$ ($y\in(0,1)$), $G_{k}(x)\leq L\frac{x^{\lambda-1}}{(x-1)^{\lambda}}$ ($x\in(1,\infty)$), $k_{\lambda}^{(1)}(u,1)$ satisfies Condition (iii). We have

$$\begin{aligned} J_{4} \leq&L^{2}L_{3}\int_{1}^{\infty} \biggl( \int_{0}^{1}(xy)^{-a} \frac{y^{\lambda-1}}{(1-y)^{\lambda}}\,dy \biggr) \frac{x^{\lambda-1}}{(x-1)^{\lambda}}\,dx \\ =&L^{2}L_{3}\int_{1}^{\infty} \biggl( \int_{0}^{1}\frac{y^{\lambda -a-1}}{(1-y)^{\lambda}}\,dy \biggr) \frac{x^{\lambda-a-1}}{(x-1)^{\lambda}}\,dx \\ =&L^{2}L_{3}B(1-\lambda,\lambda-a)B(1-\lambda,a)< \infty. \end{aligned}$$

Hence, in any case, $I_{4}=\frac{1}{k}J_{4}\rightarrow0$ ($k\rightarrow \infty$).

Therefore, by (19) and (20), we have the reverse of (17), and then (21) follows. □

3 Some equivalent Hilbert-type inequalities

We set functions $\varphi(x):=x^{p(1-\frac{\lambda}{2})-1}$, $\psi (y):=y^{q(1-\frac{\lambda}{2})-1}$ ($x,y\in\mathbf{R}_{+}$) in the following theorem.

Theorem 1

Suppose that (1) $\lambda\in A=(0,c)$ ($0< c\leq \infty$), $k_{\lambda}^{(s)}(x,y)$ are non-negative homogeneous functions of degree −λ in $\mathbf{R}_{+}^{2}$,

$$ k^{(s)}\biggl(\frac{\lambda}{2}\biggr)=\int_{0}^{\infty}k_{\lambda}^{(s)}(u,1)u^{ \frac{\lambda}{2}-1} \,du\quad (s=1,2,3), $$

there exists a constant $\delta_{0}\in(0,\frac{\lambda}{2})$ such that $k^{(s)}(\frac{\lambda}{2}\pm\delta_{0})\in\mathbf{R}_{+}$; (2) $k_{\lambda}^{(2)}(u,1)$ ($k_{\lambda}^{(3)}(u,1)$) satisfies Condition (i) or Condition (ii); (3) if both $k_{\lambda}^{(2)}(u,1)$ and $k_{\lambda }^{(3)}(u,1)$ satisfy Condition  (ii), then $k_{\lambda}^{(1)}(u,1)$ satisfies Condition (iii); (4) if $k_{\lambda}^{(s)}(u,1)$ ($s=2,3$) only satisfy Condition (i), then $\lambda\in A$; otherwise, $\lambda\in (0,1)\cap A$. For $p>1$, $f(x),G(y)\geq0$, $f\in L_{p,\varphi}(\mathbf{R}_{+})$, $G\in L_{q,\psi}(\mathbf{R}_{+})$, $\|f\|_{p,\varphi},\|G\| _{q,\psi }>0$, and

$$ F_{\lambda}(y):=\left \{ \begin{array}{l@{\quad}l} y^{\lambda-1}\int_{0}^{\infty}k_{\lambda}^{(2)}(x,y)f(x)\,dx,&y\in \{y>0;f(y)>0\}, \\ 0,&y\in\{y>0;f(y)=0\},\end{array} \right . $$

(22)

we have the following equivalent inequalities:

$$\begin{aligned}& I:=\int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda }^{(1)}(xy,1)F_{\lambda }(y)G(x) \,dy\,dx< \prod_{s=1}^{2}k^{(s)} \biggl(\frac{\lambda}{2}\biggr)\|f\|_{p,\varphi }\|G\|_{q,\psi}, \end{aligned}$$

(23)

$$\begin{aligned}& J:= \biggl[ \int_{0}^{\infty}x^{\frac{p\lambda}{2}-1} \biggl( \int_{0}^{\infty }k_{\lambda}^{(1)}(xy,1)F_{\lambda}(y) \,dy \biggr) ^{p}\,dx \biggr] ^{\frac {1}{p}}< \prod _{s=1}^{2}k^{(s)}\biggl(\frac{\lambda}{2} \biggr)\|f\|_{p,\varphi}, \end{aligned}$$

(24)

where the constant factor $\prod_{s=1}^{2}k^{(s)}(\frac{\lambda}{2})$ is the best possible.

In particular, for $g(y)\geq0$, $g\in L_{q,\psi}(\mathbf{R}_{+})$, $\|g\|_{q,\psi}>0$, and

$$ G(x)=G_{\lambda}(x):=\left \{ \begin{array}{l@{\quad}l} x^{\lambda-1}\int_{0}^{\infty}k_{\lambda}^{(3)}(x,y)g(y)\,dy,&x\in \{x>0;g(x)>0\}, \\ 0,&x\in\{x>0;g(x)=0\},\end{array} \right . $$

(25)

we have the following inequality:

$$ \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda}^{(1)}(xy,1)F_{\lambda }(y)G_{\lambda}(x) \,dy\,dx< \prod_{s=1}^{3}k^{(s)} \biggl(\frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}\|g \|_{q,\psi}, $$

(26)

where the constant factor $\prod_{s=1}^{3}k^{(s)}(\frac{\lambda}{2})$ is still the best possible.

Proof

By (9) and (8) (for $r=s=2$), we have

$$\begin{aligned}& J= \biggl[ \int_{0}^{\infty}y^{\frac{p\lambda}{2}-1} \biggl( \int_{0}^{\infty }k_{\lambda}^{(1)}(xy,1)F_{\lambda}(x) \,dx \biggr) ^{p}\,dy \biggr] ^{\frac {1}{p}}\leq k^{(1)} \biggl(\frac{\lambda}{2}\biggr)\|F_{\lambda}\|_{p,\varphi}, \end{aligned}$$

(27)

$$\begin{aligned}& \|F_{\lambda}\|_{p,\varphi}= \biggl[ \int_{0}^{\infty}y^{p(1-\frac {\lambda }{2})-1} \biggl( y^{\lambda-1}\int_{0}^{\infty}k_{\lambda }^{(2)}(x,y)f(x) \,dx \biggr) ^{p}\,dy \biggr] ^{\frac{1}{p}} \\& \hphantom{\|F_{\lambda}\|_{p,\varphi}}= \biggl[ \int_{0}^{\infty}y^{\frac{p\lambda}{2}-1} \biggl( \int_{0}^{\infty }k_{\lambda}^{(2)}(x,y)f(x) \,dx \biggr) ^{p}\,dy \biggr] ^{\frac {1}{p}} \\& \hphantom{\|F_{\lambda}\|_{p,\varphi}}< k^{(2)} \biggl(\frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}. \end{aligned}$$

(28)

Then we have (24).

By Hölder’s inequality (cf. [19]), we have

$$ I=\int_{0}^{\infty} \biggl( x^{\frac{\lambda}{2}-\frac{1}{p}}\int _{0}^{\infty}k_{\lambda}^{(1)}(xy,1)F_{\lambda}(y) \,dy \biggr) \bigl(x^{-\frac{\lambda}{2}+\frac{1}{p}}G(x)\bigr)\,dx\leq J\|G\|_{q,\psi}. $$

(29)

Then by (24) we have (23).

On the other hand, suppose that (23) is valid. Setting

$$ G(x):=y^{\frac{p\lambda}{2}-1} \biggl( \int_{0}^{\infty}k_{\lambda }^{(1)}(xy,1)F_{\lambda}(x) \,dx \biggr) ^{p-1}\quad (x\in\mathbf{R}_{+}), $$

we find $\|G\|_{q,\psi}^{q}=J^{p}$. If $J=0$, then (24) is trivially valid; if $J=\infty$, then by (27) we have $\|F_{\lambda }\|_{p,\varphi}=\infty$, which contradicts the fact of (28). Assuming that $0< J<\infty$, then by (23) we have

$$\begin{aligned}& \|G\|_{q,\psi}^{q} = J^{p}=I< \prod _{s=1}^{2}k^{(s)}\biggl(\frac{\lambda}{2} \biggr)\|f\|_{p,\varphi}\|G\|_{q,\psi}, \\& \|G\|_{q,\psi}^{q-1} = J<\prod_{s=1}^{2}k^{(s)} \biggl(\frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}, \end{aligned}$$

then we have (24), which is equivalent to (23).

Since we find similar to (28) that

$$\begin{aligned} \|G_{\lambda}\|_{q,\psi} =& \biggl[ \int_{0}^{\infty}x^{q(1-\frac {\lambda}{2})-1} \biggl( x^{\lambda-1}\int_{0}^{\infty}k_{\lambda }^{(3)}(x,y)g(y) \,dy \biggr) ^{q}\,dx \biggr] ^{\frac{1}{q}} \\ =& \biggl[ \int_{0}^{\infty}x^{\frac{q\lambda}{2}-1} \biggl( \int_{0}^{\infty }k_{\lambda}^{(3)}(x,y)g(y) \,dy \biggr) ^{q}\,dy \biggr] ^{\frac {1}{q}}< k^{(3)} \biggl(\frac{\lambda}{2}\biggr)\|g\|_{q,\Psi}, \end{aligned}$$

setting $G(x)=G_{\lambda}(x)$ in (23), we have (26).

For any $k>\max\{\frac{1}{|q|\delta_{1}},\frac{1}{p\delta_{1}}\}$ ($k\in \mathbf{N}$), we set

$$ \widetilde{f}(x)=\left \{ \begin{array}{l@{\quad}l} x^{\frac{\lambda}{2}+\frac{1}{pk}-1},&x\in(0,1), \\ 0,&x\in{}[1,\infty),\end{array} \right .\qquad \widetilde{g}(y)=\left \{ \begin{array}{l@{\quad}l} 0,&y\in(0,1], \\ y^{\frac{\lambda}{2}-\frac{1}{qk}-1},&y\in(1,\infty). \end{array} \right . $$

Then we have

$$\begin{aligned}& \widetilde{F}_{k}(y) =\left \{ \begin{array}{l@{\quad}l} y^{\lambda-1}\int_{0}^{1}k_{\lambda}^{(2)}(x,y)x^{\frac{\lambda }{2}+\frac{1}{pk}-1}\,dx,&y\in(0,1), \\ 0,&y\in{}[1,\infty)\end{array} \right . \\& \hphantom{\widetilde{F}_{k}(y)}=\left \{ \begin{array}{l@{\quad}l} y^{\lambda-1}\int_{0}^{\infty}k_{\lambda}^{(2)}(x,y)\widetilde{f}(x)\,dx,&y\in\{y>0;\widetilde{f}(y)>0\}, \\ 0,&y\in\{y>0;\widetilde{f}(y)=0\},\end{array} \right . \\& \widetilde{G}_{k}(x) :=\left \{ \begin{array}{l@{\quad}l} x^{\lambda-1}\int_{1}^{\infty}k_{\lambda}^{(3)}(x,y)y^{\frac{\lambda }{2}-\frac{1}{qk}-1}\,dy,&x\in(1,\infty), \\ 0,&x\in(0,1] \end{array} \right . \\& \hphantom{\widetilde{G}_{k}(x)}=\left \{ \begin{array}{l@{\quad}l} x^{\lambda-1}\int_{0}^{\infty}k_{\lambda}^{(3)}(x,y)\widetilde{g}(y)\,dy,&x\in\{x>0;\widetilde{g}(x)>0\}, \\ 0,&x\in\{x>0;\widetilde{g}(x)=0\}.\end{array} \right . \end{aligned}$$

If there exists a positive constant $K\leq\prod_{s=1}^{3}k^{(s)}(\frac{ \lambda}{2})$ such that (26) is valid when replacing $\prod_{s=1}^{3}k^{(s)}(\frac{\lambda}{2})$ by K, then, in particular, we have

$$ \widetilde{L}_{k}=\frac{1}{k}\int_{0}^{\infty} \int_{0}^{\infty }k_{\lambda }^{(1)}(xy,1) \widetilde{F}_{k}(y)\widetilde{G}_{k}(x)\,dy\,dx< \frac{1}{k}K\| \widetilde{f}\|_{p,\varphi}\|\widetilde{g} \|_{q,\psi}=K. $$

By (17), we find $\prod_{s=1}^{3}k^{(s)}(\frac{\lambda }{2})+o(1)\leq \widetilde{L}_{k}< K$, and then $\prod_{s=1}^{3}k^{(s)}(\frac{\lambda }{2})\leq K$ ($k\rightarrow\infty$). Hence $K=\prod_{s=1}^{3}k^{(s)}(\frac{\lambda}{2})$ is the best possible constant factor of (26).

The constant factor in (23) is the best possible. Otherwise, setting $G(x)=\widetilde{G}_{\lambda}(x)$, we would reach a contradiction that the constant factor in (26) is not the best possible. By the equivalency, if the constant factor in (24) is not the best possible, then by (29) we would reach a contradiction that the constant factor in (23) is not the best possible. □

Theorem 2

Suppose that (1) $\lambda\in A=(0,c)$ ($0< c\leq\infty$), $k_{\lambda}^{(s)}(x,y)$ are non-negative homogeneous functions of degree −λ in $\mathbf{R}_{+}^{2}$,

$$ k^{(s)}\biggl(\frac{\lambda}{2}\biggr)=\int_{0}^{\infty}k_{\lambda}^{(s)}(u,1)u^{ \frac{\lambda}{2}-1} \,du\quad (s=1,2,3), $$

there exists a constant $\delta_{0}\in(0,\frac{\lambda}{2})$ such that $k^{(s)}(\frac{\lambda}{2}\pm\delta_{0})\in\mathbf{R}_{+}$; (2) $k_{\lambda}^{(2)}(u,1)$ ($k_{\lambda}^{(3)}(u,1)$) satisfies Condition (i) or Condition (ii); (3) if both $k_{\lambda}^{(2)}(u,1)$ and $k_{\lambda }^{(3)}(u,1)$ satisfy Condition (ii), then $k_{\lambda}^{(1)}(u,1)$ satisfies Condition (iii); (3) if $k_{\lambda}^{(s)}(u,1)$ ($s=2,3$) only satisfy Condition (i), then $\lambda\in A$; otherwise, $\lambda\in (0,1)\cap A$. For $0< p<1$, $f(x),G(y)\geq0$, $f\in L_{p,\varphi}(\mathbf {R}_{+})$, $G\in L_{q,\psi}(\mathbf{R}_{+})$, $\|f\|_{p,\varphi}$, $\|G\|_{q,\psi}>0$, and $F_{\lambda}(y)$ being as (22), we have the equivalent reverses of (24) and (25) with the best possible constant factor $\prod_{s=1}^{2}k^{(s)}(\frac{\lambda}{2})$.

In particular, for $g(y)\geq0$, $g\in L_{q,\psi}(\mathbf{R}_{+})$, $\|g\|_{q,\psi}>0$, and $G(x)=G_{\lambda}(x)$ as (25), we have the reverse of (26) with the best possible constant factor $\prod_{s=1}^{3}k^{(s)}(\frac{\lambda}{2})$.

Proof

By the reverse Hölder inequality (cf. [19]), we obtain the reverses of (27) and (28). Then we deduce the reverse of (24).

By the reverse Hölder inequality, we have

$$ I=\int_{0}^{\infty} \biggl( x^{\frac{\lambda}{2}-\frac{1}{p}}\int _{0}^{\infty}k_{\lambda}^{(1)}(xy,1)F_{\lambda}(y) \,dy \biggr) \bigl(x^{-\frac{\lambda}{2}+\frac{1}{p}}G(x)\bigr)\,dx\geq J\|G\|_{q,\psi}. $$

(30)

Then by the reverse of (24), we obtain the reverse of (23).

On the other hand, suppose that the reverse of (23) is valid. Setting $G(x)$ as (25), we find $\|G\|_{q,\psi}^{q}=J^{p}$. If $J=\infty$, then the reverse of (24) is trivially valid; if $J=0$, then by the reverse of (27), we have $\|F_{\lambda}\|_{p,\varphi}=0$, which contradicts the reverse of (28). Assuming that $0< J<\infty$, by the reverse of (23), we have

$$\begin{aligned}& \|G\|_{q,\psi}^{q} = J^{p}=I>\prod _{s=1}^{2}k^{(s)}\biggl(\frac{\lambda}{2} \biggr)\|f\|_{p,\varphi}\|G\|_{q,\psi}, \\& \|G\|_{q,\psi}^{q-1} = J>\prod_{s=1}^{2}k^{(s)} \biggl(\frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}, \end{aligned}$$

and then the reverse of (24) follows, which is equivalent to the reverse of (23).

For $q<0$, since we find similar to the reverse of (28) that

$$\begin{aligned} \|G_{\lambda}\|_{q,\psi} =& \biggl[ \int_{0}^{\infty}x^{q(1-\frac {\lambda}{2})-1} \biggl( x^{\lambda-1}\int_{0}^{\infty}k_{\lambda }^{(3)}(x,y)g(y) \,dy \biggr) ^{q}\,dx \biggr] ^{\frac{1}{q}} \\ =& \biggl[ \int_{0}^{\infty}x^{\frac{q\lambda}{2}-1} \biggl( \int_{0}^{\infty }k_{\lambda}^{(3)}(x,y)g(y) \,dy \biggr) ^{q}\,dy \biggr] ^{\frac {1}{q}}>k^{(3)} \biggl(\frac{\lambda}{2}\biggr)\|g\|_{q,\Psi}, \end{aligned}$$

setting $G(x)=G_{\lambda}(x)$ in the reverse of (23), we have the reverse of (26).

For any $k>\max\{\frac{1}{|q|\delta_{1}},\frac{1}{p\delta_{1}}\}$ ($k\in \mathbf{N}$), we set $\widetilde{f}(x)$, $\widetilde{g}(y)$ as Theorem 1. If there exists a positive constant $K\geq\prod_{s=1}^{3}k^{(s)}(\frac {\lambda}{2})$ such that the reverse of (26) is valid when replacing $\prod_{s=1}^{3}k^{(s)}(\frac{\lambda}{2})$ by K, then, in particular, we have

$$ \widetilde{L}_{k}=\frac{1}{k}\int_{0}^{\infty} \int_{0}^{\infty }k_{\lambda }^{(1)}(xy,1) \widetilde{F}_{k}(y)\widetilde{G}_{k}(x)\,dy\,dx> \frac{1}{k}K\| \widetilde{f}\|_{p,\varphi}\|\widetilde{g} \|_{q,\psi}=K. $$

By (21), we find $\prod_{s=1}^{3}k^{(s)}(\frac{\lambda}{2})+o(1)= \widetilde{L}_{k}>K$, and then $\prod_{s=1}^{3}k^{(s)}(\frac{\lambda }{2})\geq K$ ($k\rightarrow\infty$). Hence $K=\prod_{s=1}^{3}k^{(s)}(\frac{\lambda}{2})$ is the best possible constant factor of the reverse of (26).

The constant factor in the reverse of (23) is the best possible. Otherwise, setting $G(x)=\widetilde{G}_{\lambda}(x)$, we would reach a contradiction that the constant factor in the reverse of (26) is not the best possible. By the equivalency, if the constant factor in the reverse of (24) is not the best possible, then by (30) we would reach a contradiction that the constant factor in the reverse of (23) is not the best possible. □

4 Some corollaries on Hilbert-Hardy-type inequalities

In the following sections, if the best possible constant factor in a Hilbert-type inequality is related to $k_{j}^{(s)}(\frac{\lambda}{2})$ ($s=1,2,3$, $j=0,1,2$) defined as follows, then we call this inequality Hilbert-Hardy-type inequality. The related operator is called Hilbert-Hardy-type operator.

Assuming that $k_{\lambda}^{(1)}(xy,1)=0$ ($0<\frac{1}{x}\leq y$), we find $k_{\lambda}^{(1)}(u,1)=0$ ($u\geq1$), and

$$ k^{(1)}\biggl(\frac{\lambda}{2}\biggr)=k_{1}^{(1)} \biggl(\frac{\lambda}{2}\biggr):=\int_{0}^{1}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}-1} \,du. $$

(31)

By Theorem 1 and Theorem 2, we have the following.

Corollary 1

With the assumptions of Theorem  1, for $p>1$, $k_{1}^{(1)}(\frac{\lambda}{2})\in\mathbf{R}_{+}$, we have the following equivalent inequalities:

$$\begin{aligned}& \int_{0}^{\infty}G(x)\int_{0}^{\frac{1}{x}}k_{\lambda }^{(1)}(xy,1)F_{\lambda}(y) \,dy\,dx< k_{1}^{(1)}\biggl(\frac{\lambda}{2} \biggr)k^{(2)}\biggl( \frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}\|G \|_{q,\psi}, \end{aligned}$$

(32)

$$\begin{aligned}& \biggl[ \int_{0}^{\infty}x^{\frac{p\lambda}{2}-1} \biggl( \int_{0}^{\frac {1}{x}}k_{\lambda}^{(1)}(xy,1)F_{\lambda}(y) \,dy \biggr) ^{p}\,dx \biggr] ^{\frac {1}{p}}< k_{1}^{(1)} \biggl(\frac{\lambda}{2}\biggr)k^{(2)}\biggl(\frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}, \end{aligned}$$

(33)

where the constant factor $k_{1}^{(1)}(\frac{\lambda}{2})k^{(2)}(\frac{ \lambda}{2})$ is the best possible.

In particular, for $g(y)\geq0$, $g\in L_{q,\psi}(\mathbf{R}_{+})$, $\|g\|_{q,\psi}>0$, $G(x)=G_{\lambda}(x)$ as (25), we have the following inequality:

$$ \int_{0}^{\infty}\int_{0}^{\frac{1}{x}}k_{\lambda }^{(1)}(xy,1)F_{\lambda }(y)G_{\lambda}(x) \,dy\,dx< k_{1}^{(1)}\biggl(\frac{\lambda}{2}\biggr)\prod_{s=2}^{3}k^{(s)} \biggl(\frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}\|g\| _{q,\psi}, $$

(34)

where the constant factor $k_{1}^{(1)}(\frac{\lambda}{2})\prod_{s=2}^{3}k^{(s)}(\frac{\lambda}{2})$ is still the best possible.

Corollary 2

With the assumptions of Theorem  2, for $0< p<1$, $k_{1}^{(1)}(\frac{\lambda}{2})\in\mathbf{R}_{+}$, we have the equivalent reverses of (32) and (33), where the constant factor $k_{1}^{(1)}(\frac{\lambda}{2})k^{(2)}(\frac{\lambda}{2})$ is the best possible.

In particular, for $g(y)\geq0$, $g\in L_{q,\psi}(\mathbf{R}_{+})$, $\|g\|_{q,\psi}>0$, $G(x)=G_{\lambda}(x)$ as (25), we have the reverse of (34) with the best value $k_{1}^{(1)}(\frac{\lambda }{2})\prod_{s=2}^{3}k^{(s)}(\frac{\lambda}{2})$.

Assuming that $k_{\lambda}^{(1)}(xy,1)=0$ ($0< y\leq\frac{1}{x}$), then we find $k_{\lambda}^{(1)}(u,1)=0$ ($0< u\leq1$), and

$$ k^{(1)}\biggl(\frac{\lambda}{2}\biggr)=k_{2}^{(1)} \biggl(\frac{\lambda}{2}\biggr):=\int_{1}^{\infty}k_{\lambda}^{(1)}(u,1)u^{\frac{\lambda}{2}-1} \,du. $$

(35)

By Theorem 1 and Theorem 2, we have the following.

Corollary 3

With the assumptions of Theorem  1, for $p>1$, $k_{2}^{(1)}(\frac{\lambda}{2})\in\mathbf{R}_{+}$, we have the following equivalent inequalities:

$$\begin{aligned}& \int_{0}^{\infty}G(x)\int_{\frac{1}{x}}^{\infty}k_{\lambda }^{(1)}(xy,1)F_{\lambda}(y) \,dy\,dx< k_{2}^{(1)}\biggl(\frac{\lambda}{2} \biggr)k^{(2)}\biggl( \frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}\|G \|_{q,\psi}, \end{aligned}$$

(36)

$$\begin{aligned}& \biggl[ \int_{0}^{\infty}x^{\frac{p\lambda}{2}-1} \biggl( \int_{\frac {1}{x}}^{\infty}k_{\lambda}^{(1)}(xy,1)F_{\lambda}(y) \,dy \biggr) ^{p}\,dx \biggr] ^{\frac{1}{p}}< k_{2}^{(1)} \biggl(\frac{\lambda}{2}\biggr)k^{(2)}\biggl(\frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}, \end{aligned}$$

(37)

where the constant factor $k_{2}^{(1)}(\frac{\lambda}{2})k^{(2)}(\frac{ \lambda}{2})$ is the best possible.

In particular, for $g(y)\geq0$, $g\in L_{q,\psi}(\mathbf{R}_{+})$, $\|g\|_{q,\psi}>0$, and $G(x)=G_{\lambda}(x)$ as (25), we have the following inequality:

$$ \int_{0}^{\infty}G_{\lambda}(x)\int _{\frac{1}{x}}^{\infty}k_{\lambda }^{(1)}(xy,1)F_{\lambda}(y) \,dy\,dx< k_{2}^{(1)}\biggl(\frac{\lambda}{2}\biggr) \prod_{s=2}^{3}k^{(s)}\biggl( \frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}\|g\| _{q,\psi}, $$

(38)

where the constant factor $k_{2}^{(1)}(\frac{\lambda}{2})\prod_{s=2}^{3}k^{(s)}(\frac{\lambda}{2})$ is still the best possible.

Corollary 4

With the assumptions of Theorem  2, if $0< p<1$, $k_{2}^{(1)}(\frac{\lambda}{2})\in\mathbf{R}_{+}$, we have the equivalent reverses of (36) and (37), where the constant factor $k_{2}^{(1)}(\frac{\lambda}{2})k^{(2)}(\frac{\lambda}{2})$ is the best possible.

In particular, for $g(y)\geq0$, $g\in L_{q,\psi}(\mathbf{R}_{+})$, $\|g\|_{q,\psi}>0$, and $G(x)=G_{\lambda}(x)$ as (25), we have the reverse of (38) with the best value $k_{2}^{(1)}(\frac{\lambda }{2})\prod_{s=2}^{3}k^{(s)}(\frac{\lambda}{2})$.

Remark 3

For $x>0$, we set $A_{x,0}:=(0,\infty )$, $A_{x,1}:=(0,\frac{1}{x})$, $A_{x,2}:=(\frac{1}{x},\infty)$. By (24), (33) and (37), putting $k_{0}^{(1)}(\frac{\lambda}{2}):=k^{(1)}(\frac{\lambda }{2})$, for $i=0,1,2$, we have the following Hilbert-Hardy-type inequalities:

$$ \biggl[ \int_{0}^{\infty}x^{\frac{p\lambda}{2}-1} \biggl( \int_{A_{x,i}}k_{\lambda}^{(1)}(xy,1)F_{\lambda}(y) \,dy \biggr) ^{p}\,dx \biggr] ^{\frac{1}{p}}< k_{i}^{(1)} \biggl(\frac{\lambda}{2}\biggr)k^{(2)}\biggl(\frac{\lambda}{2} \biggr) \|f\|_{p,\varphi}, $$

(39)

where the constant factor $k_{i}^{(1)}(\frac{\lambda}{2})k^{(2)}(\frac{ \lambda}{2})$ ($i=0,1,2$) is still the best possible.

For $x>0$, we set some sets $B_{x,0}:=(0,\infty)$, $B_{x,1}:=(x,\infty )$, $B_{x,2}:=(0,x)$. If $k_{\lambda}^{(2)}(x,y)=0$ ($y\in\mathbf{R}_{+}\backslash B_{x,1}$), then we find $k_{\lambda}^{(2)}(u,1)=0$ ($u\geq1$), and

$$ k^{(2)}\biggl(\frac{\lambda}{2}\biggr)=k_{1}^{(2)} \biggl(\frac{\lambda}{2}\biggr):=\int_{0}^{1}k_{\lambda}^{(2)}(u,1)u^{\frac{\lambda}{2}-1} \,du; $$

if $k_{\lambda}^{(2)}(x,y)=0$ ($y\in\mathbf{R}_{+}\backslash B_{x,2}$), then we find $k_{\lambda}^{(2)}(u,1)=0$ ($0< u\leq1$), and

$$ k^{(2)}\biggl(\frac{\lambda}{2}\biggr)=k_{2}^{(2)} \biggl(\frac{\lambda}{2}\biggr):=\int_{1}^{\infty}k_{\lambda}^{(2)}(u,1)u^{\frac{\lambda}{2}-1} \,du. $$

Assuming that $k_{0}^{(2)}(\frac{\lambda}{2}):=k^{(2)}(\frac{\lambda}{2})$, $k_{i}^{(1)}(\frac{\lambda}{2}),k_{j}^{(2)}(\frac{\lambda}{2})\in \mathbf{R}_{+}$, for $i,j=0,1,2$, setting

$$ F_{\lambda,j}(y):=\left \{ \begin{array}{l@{\quad}l} y^{\lambda-1}\int_{B_{x,j}}k_{\lambda}^{(2)}(x,y)f(x)\,dx,&y\in\{y\in \mathbf{R}_{+};f(y)>0\}, \\ 0,&y\in\{y\in\mathbf{R}_{+};f(y)=0\},\end{array} \right . $$

then it follows that $F_{\lambda,0}(y)=F_{\lambda}(y)$, and by (39) we have the following united expression of Hilbert-Hardy-type inequalities:

$$ \biggl[ \int_{0}^{\infty}x^{\frac{p\lambda}{2}-1} \biggl( \int_{A_{x,i}}k_{\lambda}^{(1)}(xy,1)F_{\lambda,j}(y) \,dy \biggr) ^{p}\,dx \biggr] ^{\frac{1}{p}}< k_{i}^{(1)} \biggl(\frac{\lambda}{2}\biggr)k_{j}^{(2)}\biggl( \frac{ \lambda}{2}\biggr)\|f\|_{p,\varphi}, $$

(40)

where the constant factor $k_{i}^{(1)}(\frac{\lambda }{2})k_{j}^{(2)}(\frac{\lambda}{2})$ ($i,j=0,1,2$) is the best possible.

In the same way, we still can find by (27) and (28) that

$$\begin{aligned}& \biggl[ \int_{0}^{\infty}x^{\frac{p\lambda}{2}-1} \biggl( \int_{A_{x,i}}k_{\lambda}^{(1)}(xy,1)F_{\lambda,j}(y) \,dy \biggr) ^{p}\,dx \biggr] ^{\frac{1}{p}}\leq k_{i}^{(1)}\biggl(\frac{\lambda}{2}\biggr) \|F_{\lambda ,j}\|_{p,\varphi}, \end{aligned}$$

(41)

$$\begin{aligned}& \|F_{\lambda,j}\|_{p,\varphi}< k_{j}^{(2)}\biggl( \frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}\quad (i,j=0,1,2), \end{aligned}$$

(42)

where the constant factors $k_{i}^{(1)}(\frac{\lambda}{2})$ and $k_{j}^{(2)}(\frac{\lambda}{2})$ are the best possible.

Example 2

(i) For $k_{\lambda}^{(s)}(x,y)=\frac{|\ln x/y|^{\beta -1}}{(\max\{x,y\})^{\lambda}}$ ($\lambda>0,\beta\geq1$; $s=1,2,3$), we find

$$\begin{aligned}& k_{0}^{(s)}\biggl(\frac{\lambda}{2}\biggr)=\int _{0}^{\infty}k_{\lambda }^{(s)}(u,1)u^{\frac{\lambda}{2}-1} \,du=\int_{0}^{\infty}\frac{|\ln u|^{\beta -1}}{(\max \{u,1\})^{\lambda}}u^{\frac{\lambda}{2}-1} \,du \\& \hphantom{k_{0}^{(s)}\biggl(\frac{\lambda}{2}\biggr)}=2\int_{0}^{1}(-\ln u)^{\beta-1}u^{\frac{\lambda}{2}-1}\,du \\& \hphantom{k_{0}^{(s)}\biggl(\frac{\lambda}{2}\!}\overset {v=-\ln u}{=}2\int _{0}^{\infty}e^{-\frac{\lambda}{2}v}v^{\beta-1}\,dv=2 \biggl(\frac{2}{ \lambda}\biggr)^{\beta}\Gamma(\beta), \\& k_{i}^{(s)}\biggl(\frac{\lambda}{2}\biggr)=\biggl( \frac{2}{\lambda}\biggr)^{\beta}\Gamma (\beta )\quad (i=1,2). \end{aligned}$$

(ii) For $k_{\lambda}^{(s)}(x,y)=\frac{1}{|x-y|^{\lambda}}$ ($0<\lambda <1$; $s=1,2,3$), we find

$$\begin{aligned}& k_{0}^{(s)}\biggl(\frac{\lambda}{2}\biggr) = \int _{0}^{\infty}k_{\lambda }^{(s)}(u,1)u^{\frac{\lambda}{2}-1} \,du=\int_{0}^{\infty}\frac{1}{|u-1|^{\lambda}}u^{\frac{\lambda}{2}-1} \,du \\& \hphantom{k_{0}^{(s)}\biggl(\frac{\lambda}{2}\biggr)} = 2\int_{0}^{1} \frac{1}{(1-u)^{\lambda}}u^{\frac{\lambda}{2}-1}\,du=2B\biggl(1-\lambda,\frac{\lambda}{2} \biggr), \\& k_{i}^{(s)}\biggl(\frac{\lambda}{2}\biggr) = B\biggl(1- \lambda,\frac{\lambda}{2}\biggr)\quad (i=1,2). \end{aligned}$$

5 A composition of two Hilbert-Hardy-type operators

For $F\in L_{p,\varphi}(\mathbf{R}_{+})$, we set $h_{i}(x):=x^{\lambda -1}\int_{A_{x,i}}k_{\lambda}^{(1)}(xy,1)F(y)\,dy$ ($x\in\mathbf{R}_{+}$; $i=0,1,2$). Then by (41) we have

$$ \|h_{i}\|_{p,\varphi}\leq k_{i}^{(1)} \biggl(\frac{\lambda}{2}\biggr)\|F\| _{p,\varphi}. $$

(43)

Definition 3

With the assumptions of Theorem 1, for any $i=0,1,2$, $k_{i}^{(1)}(\frac{\lambda}{2})\in\mathbf{R}_{+}$, we define a Hilbert-Hardy-type operator $T_{1}^{(i)}:L_{p,\varphi}(\mathbf{R}_{+})\rightarrow L_{p,\varphi}(\mathbf{R}_{+})$ as follows: For any $F\in L_{p,\varphi}(\mathbf{R}_{+})$, there exists a unified expression $T_{1}^{(i)}F=h_{i}\in L_{p,\varphi}(\mathbf{R}_{+})$ such that for any $x\in\mathbf{R}_{+}$, $T_{1}^{(i)}F(x)=h_{i}(x)$.

By (43), we have $\|T_{1}^{(i)}F\|_{p,\varphi}\leq k_{i}^{(1)}(\frac{\lambda}{2})\|F\|_{p,\varphi}$. Hence, $T_{1}^{(i)}$ is a bounded linear operator with

$$ \bigl\Vert T_{1}^{(i)}\bigr\Vert := \sup _{F(\neq\theta)\in L_{p,\varphi}(\mathbf {R}_{+})}\frac{\|T_{1}^{(i)}F\|_{p,\varphi}}{\|F\|_{p,\varphi}}\leq k_{i}^{(1)} \biggl(\frac{\lambda}{2}\biggr). $$

Since the constant factor in (43) is the best possible, we have $\|T_{1}^{(i)}\|=k_{i}^{(1)}(\frac{\lambda}{2})$.

Definition 4

With the assumptions of Theorem 1, for any $j=0,1,2$, $k_{j}^{(2)}(\frac{\lambda}{2})\in\mathbf{R}_{+}$, we define a Hilbert-Hardy-type operator $T_{2}^{(j)}:L_{p,\varphi}(\mathbf{R}_{+})\rightarrow L_{p,\varphi}(\mathbf{R}_{+})$ as follows: For any $f\in L_{p,\varphi}(\mathbf{R}_{+})$, there exists a unified expression $T_{2}^{(j)}f=F_{\lambda,j}\in L_{p,\varphi}(\mathbf{R}_{+})$ such that for any $y\in\mathbf{R}_{+}$, $T_{2}^{(j)}f(y)=F_{\lambda,j}(y)$.

By (42), we have $\|T_{2}^{(j)}f\|_{p,\varphi}=\|F_{\lambda ,j}\|_{p,\varphi}\leq k_{j}^{(2)}(\frac{\lambda}{2})\|f\|_{p,\varphi}$. Hence, $T_{2}^{(j)}$ is a bounded linear operator with

$$ \bigl\Vert T_{2}^{(j)}\bigr\Vert = \sup _{f(\neq\theta)\in L_{p,\varphi}(\mathbf {R}_{+})}\frac{\|T_{2}^{(j)}f\|_{p,\varphi}}{\|f\|_{p,\varphi}}\leq k_{j}^{(2)} \biggl(\frac{\lambda}{2}\biggr). $$

Since the constant in (42) is the best possible, we have $\|T_{2}^{(j)}\|=k_{j}^{(2)}(\frac{\lambda}{2})$.

Definition 5

With the assumptions of Theorem 1, for any $i,j\in \{0,1,2\}$, $k_{i}^{(1)}(\frac{\lambda}{2}),k_{j}^{(2)}(\frac{\lambda }{2})\in\mathbf{R}_{+}$, we define a Hilbert-Hardy-type operator $T_{i,j}:L_{p,\varphi}(\mathbf{R}_{+})\rightarrow L_{p,\varphi }(\mathbf{R}_{+})$ as follows: For any $f\in L_{p,\varphi}(\mathbf{R}_{+})$, there exists a unified expression $T_{i,j}f=T_{1}^{(i)}F_{\lambda,j}\in L_{p,\varphi}(\mathbf{R}_{+})$ such that for any $x\in\mathbf{R}_{+}$,

$$ T_{i,j}f(x)=T_{1}^{(i)}F_{\lambda,j}(x)=x^{\lambda -1} \int_{A_{x,i}}k_{\lambda}^{(1)}(xy,1)F_{\lambda,j}(y) \,dy. $$

It is evident that $T_{i,j}f=T_{1}^{(i)}F_{\lambda ,j}=T_{1}^{(i)}(T_{2}^{(j)}f)=(T_{1}^{(i)}T_{2}^{(j)})f$, and then $T_{i,j}=T_{1}^{(i)}T_{2}^{(j)}$. Hence, $T_{i,j}$ is the composition of $T_{1}^{(i)}$ and $T_{2}^{(j)}$, and (cf. [20])

$$ \|T_{i,j}\|=\bigl\Vert T_{1}^{(i)}T_{2}^{(j)} \bigr\Vert \leq\bigl\Vert T_{1}^{(i)}\bigr\Vert \cdot \bigl\Vert T_{2}^{(j)}\bigr\Vert =k_{i}^{(1)} \biggl(\frac{\lambda}{2}\biggr)k_{j}^{(2)}\biggl( \frac {\lambda}{2}\biggr). $$

By (40), we have

$$ \|T_{i,j}f\|_{p,\varphi}=\bigl\Vert T_{1}^{(i)}F_{\lambda,j} \bigr\Vert _{p,\varphi}\leq k_{i}^{(1)}\biggl( \frac{\lambda}{2}\biggr)k_{j}^{(2)}\biggl( \frac{\lambda}{2}\biggr)\|f\|_{p,\varphi}. $$

Since the constant factor in (40) is the best possible, then the theorem follows.

Theorem 3

With the assumptions of Theorem  1, if for any $i,j\in \{0,1,2\}$, $k_{i}^{(1)}(\frac{\lambda}{2}),k_{j}^{(2)}(\frac{\lambda }{2})\in\mathbf{R}_{+}$, then we have the composition formula of two Hilbert-Hardy-type operators as follows:

$$ \|T_{i,j}\|=\bigl\Vert T_{1}^{(i)}T_{2}^{(j)} \bigr\Vert =\bigl\Vert T_{1}^{(i)}\bigr\Vert \cdot \bigl\Vert T_{2}^{(j)}\bigr\Vert =k_{i}^{(1)} \biggl(\frac{\lambda}{2}\biggr)k_{j}^{(2)}\biggl( \frac {\lambda}{2}\biggr). $$

(44)

Example 3

For $k_{\lambda}^{(1)}(xy,1)=\frac {1}{|xy-1|^{\lambda}}$, $k_{\lambda}^{(2)}(x,y)=\frac{|\ln(x/y)|^{\beta-1}}{(\max \{x,y\})^{\lambda}}$ ($\beta\geq1$), $\lambda\in(0,1)$, by Example 2 and (44), we have

$$\begin{aligned}& \Vert T_{0,0}\Vert = \bigl\Vert T_{1}^{(0)}T_{2}^{(0)} \bigr\Vert =\bigl\Vert T_{1}^{(0)}\bigr\Vert \cdot \bigl\Vert T_{2}^{(0)}\bigr\Vert =4B\biggl(1-\lambda, \frac{\lambda}{2}\biggr) \biggl(\frac{2}{\lambda }\biggr)^{\beta }\Gamma( \beta), \\& \Vert T_{0,j}\Vert = \bigl\Vert T_{1}^{(0)}T_{2}^{(j)} \bigr\Vert =\bigl\Vert T_{1}^{(0)}\bigr\Vert \cdot \bigl\Vert T_{2}^{(j)}\bigr\Vert =2B\biggl(1-\lambda, \frac{\lambda}{2}\biggr) \biggl(\frac{2}{\lambda }\biggr)^{\beta }\Gamma( \beta)\quad (j=1,2), \\& \Vert T_{i,0}\Vert = \bigl\Vert T_{1}^{(i)}T_{2}^{(0)} \bigr\Vert =\bigl\Vert T_{1}^{(i)}\bigr\Vert \cdot \bigl\Vert T_{2}^{(0)}\bigr\Vert =2B\biggl(1-\lambda, \frac{\lambda}{2}\biggr) \biggl(\frac{2}{\lambda }\biggr)^{\beta }\Gamma( \beta)\quad (i=1,2), \\& \Vert T_{i,j}\Vert = \bigl\Vert T_{1}^{(i)}T_{2}^{(j)} \bigr\Vert =\bigl\Vert T_{1}^{(i)}\bigr\Vert \cdot \bigl\Vert T_{2}^{(j)}\bigr\Vert =B\biggl(1-\lambda, \frac{\lambda}{2}\biggr) \biggl(\frac{2}{\lambda }\biggr)^{\beta }\Gamma( \beta)\quad (i,j=1,2). \end{aligned}$$