## 1 Hilbert’s inequalities and the operator expressions

### 1.1 Hilbert’s inequalities

In 1908, Weyl [1] published the following inequality.

If $$\{a_{m}\}_{m=1}^{\infty}$$ and $$\{b_{n}\}_{n=1}^{\infty}$$ are real sequences, satisfying $$0<\sum_{m=1}^{\infty}a_{m}^{2}<\infty$$ and $$0<\sum_{n=1}^{\infty}b_{n}^{2}<\infty$$, then

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

where the constant factor π is the best possible.

We named (1) Hilbert’s inequality; it does not contain any parameter. The best possible property of the constant factor π was proved by Schur [2] in 1911. He also gave the following integral analog of (1) at the same time.

If $$f(x)$$ and $$g(y)$$ are measurable functions, such that $$0<\int_{0}^{\infty }f^{2}(x)\, dx<\infty$$ and $$0<\int_{0}^{\infty}g^{2}(y)\, dy<\infty$$, then

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

where the constant factor π is still the best possible.

We called (2) Hilbert’s integral inequality, which still does not contain any parameter. Inequalities (1) and (2) are important in analysis and its applications. We can find a number of improvements and extensions in the vast mathematics literature, especially in [36].

### 1.2 Hilbert’s operators

We may express inequality (1) by using the form of the operator as follows.

If $$l^{2}$$ is a space of real sequences, and $$T:l^{2}\rightarrow l^{2}$$ is a linear operator, for any $$a=\{a_{m}\}_{m=1}^{\infty}\in l^{2}$$, there exists a $$c=\{c_{n}\}_{n=1}^{\infty}\in l^{2}$$, satisfying

$$c_{n}=(Ta) (n)=\sum_{m=1}^{\infty} \frac{a_{m}}{m+n},\quad n\in\mathbf{N}=\{1,2,\ldots\}.$$
(3)

Hence for any $$b=\{b_{n}\}_{n=1}^{\infty}\in l^{2}$$, we may write the inner product of Ta and b as follows:

$$(Ta,b)=(c,b)=\sum_{n=1}^{\infty} \Biggl( \sum _{m=1}^{\infty}\frac {a_{m}}{m+n} \Biggr) b_{n}=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{m+n}.$$
(4)

Expressing the norm of a as $$\|a\|_{2}=\{\sum_{n=1}^{\infty }a_{n}^{2}\}^{1/2}$$, in view of (4), inequality (1) may be rewritten as follows:

$$(Ta,b)< \pi\|a\|_{2}\|b\|_{2},$$
(5)

where $$\|a\|_{2},\|b\|_{2}>0$$. We may prove that T is a bounded operator and obtain the norm $$\|T\|=\pi$$ (cf. [7]). We call T Hilbert’s operator.

For $$\|a\|_{2}>0$$, the equivalent form of (5) is given as $$\|Ta\|_{2}<\pi\|a\|_{2}$$, e.t.

$$\sum_{n=1}^{\infty} \Biggl( \sum _{m=1}^{\infty}\frac{a_{m}}{m+n} \Biggr) ^{2}< \pi^{2}\sum_{m=1}^{\infty}a_{m}^{2},$$
(6)

where the constant factor $$\pi^{2}$$ is still the best possible. Obviously, inequalities (6) and (1) are equivalent (cf. [3]).

Similarly, if $$L^{2}(\mathbf{R}_{+})$$ is a space of real functions, we may define Hilbert’s integral operator $$\widetilde{T}:L^{2}(\mathbf {R}_{+})\rightarrow L^{2}(\mathbf{R}_{+})$$ as follows.

For any $$f\in L^{2}(\mathbf{R}_{+})$$, there exists a $$h=\widetilde {T}f\in L^{2}(\mathbf{R}_{+})$$, satisfying

$$(\widetilde{T}f) (y)=h(y)=\int_{0}^{\infty} \frac{f(x)}{x+y}\,dx,\quad y\in (0,\infty ).$$
(7)

Hence, for any $$g\in L^{2}(\mathbf{R}_{+})$$, we may still can indicate the inner product of T̃f and g as follows:

$$(\widetilde{T}f,g)=\int_{0}^{\infty} \biggl( \int _{0}^{\infty}\frac {f(x)}{x+y}\,dx \biggr) g(y)\,dy=\int _{0}^{\infty}\int_{0}^{\infty} \frac {f(x)g(y)}{x+y}\,dx\,dy.$$
(8)

Setting the norm of f as $$\|f\|_{2}= ( \int_{0}^{\infty }f^{2}(x)\,dx ) ^{\frac{1}{2}}$$, if $$\|f\|_{2},\|g\|_{2}>0$$, then (2) may be rewritten as follows:

$$(\widetilde{T}f,g)< \pi\|f\|_{2}\|g\|_{2}.$$
(9)

It follows that $$\|\widetilde{T}f\|=\pi$$ (cf. [8]), and then we have the equivalent form of (2) as $$\|\widetilde{T}f\|_{2}<\pi \|f|\|_{2}\|$$, e.t. (cf. [3]):

$$\int_{0}^{\infty} \biggl( \int_{0}^{\infty} \frac{f(x)}{x+y}\,dx \biggr) ^{2}\,dy< \pi^{2}\int _{0}^{\infty}f^{2}(x)\,dx,$$
(10)

where the constant factor $$\pi^{2}$$ is still the best possible. It is obvious that inequality (10) is the integral analog of (6).

### 1.3 A more accurate discrete Hilbert’s inequality

If we let the subscripts m, n of the double series go from 0 to infinity, then we may rewrite inequality (1) equivalently in the following form:

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{m+n+2}< \pi \Biggl( \sum _{m=0}^{\infty}a_{m}^{2}\sum _{n=0}^{\infty}b_{n}^{2} \Biggr) ^{\frac{1}{2}},$$
(11)

where the constant factor π is still the best possible. Obviously, we may raise the following question:

Is there a positive constant α (<2) that leaves the inequality still valid as we replace 2 by α in the kernel $$\frac{1}{m+n+2}?$$ The answer is positive. That is, we have the following more accurate Hilbert inequality (for short, Hilbert’s inequality) (cf. [3]):

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{m+n+1}< \pi \Biggl( \sum _{m=0}^{\infty}a_{m}^{2} \sum_{n=0}^{\infty}b_{n}^{2} \Biggr) ^{\frac{1}{2}},$$
(12)

where the constant factor π is the best possible.

Since for $$a_{m},b_{n}\geq0$$, $$\alpha\geq1$$,

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{m+n+\alpha }\leq \sum _{n=0}^{\infty}\sum_{m=0}^{\infty} \frac{a_{m}b_{n}}{m+n+1},$$

then by (12) and for $$\alpha\geq1$$, we have

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{m+n+\alpha }< \pi \Biggl( \sum _{m=0}^{\infty}a_{m}^{2} \sum_{n=0}^{\infty}b_{n}^{2} \Biggr) ^{\frac{1}{2}}.$$
(13)

For $$1\leq\alpha<2$$, inequality (13) is a refinement of (11). Obviously, we have a refinement of (6), which is equivalent to (13) as follows:

$$\sum_{n=0}^{\infty} \Biggl( \sum _{m=0}^{\infty}\frac{a_{m}}{m+n+\alpha} \Biggr) ^{2}< \pi^{2}\sum_{m=0}^{\infty}a_{m}^{2} \quad (1\leq\alpha< 2).$$
(14)

For $$0<\alpha<1$$, in 1936, Ingham [9] gave the following.

If $$\alpha\geq\frac{1}{2}$$, then

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}a_{n}}{m+n+\alpha }\leq \pi\sum _{m=0}^{\infty}a_{m}^{2};$$
(15)

if $$0<\alpha<\frac{1}{2}$$, then

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}a_{n}}{m+n+\alpha }\leq \frac{\pi}{\sin(\alpha\pi)} \sum_{m=0}^{\infty}a_{m}^{2}.$$
(16)

Note that if we set $$x=X+\frac{\alpha}{2}$$, $$y=Y+\frac{\alpha}{2}$$, $$F(X)=f(X+\frac{\alpha}{2})$$ and $$G(Y)=g(Y+\frac{\alpha}{2})$$ ($$\alpha \in \mathbf{R}=(-\infty,\infty)$$) in (2), then we obtain

$$\int_{-\frac{\alpha}{2}}^{\infty}\int_{-\frac{\alpha}{2}}^{\infty } \frac{F(X)G(Y)}{X+Y+\alpha}\, dX\, dY< \pi \biggl( \int_{-\frac{\alpha}{2}}^{\infty }F^{2}(X) \, dX\int_{-\frac{\alpha}{2}}^{\infty}G^{2}(Y)\, dY \biggr) ^{\frac {1}{2}}.$$
(17)

It is said that for $$\alpha\geq\frac{1}{2}$$, inequality (17) is an integral analog of (13) (for $$G=F$$) and for $$0<\alpha<\frac {1}{2}$$, (17) is not an integral analog of (13), since the two constant factors are different.

In recent years, by using the improved Euler-Maclaurin summation formula and introducing parameters, a few authors gave some more accurate Hilbert-type inequalities as (13) (cf. [1017]).

## 2 Hardy-Hilbert-type inequalities with a pair of conjugate exponents

### 2.1 Hardy-Hilbert’s inequalities and the operator expressions

In 1925, by introducing one pair of conjugate exponents $$(p,q)$$ with $$\frac{1}{p}+\frac{1}{q}=1$$, Hardy [18] gave an extension of (1) as follows.

If $$p>1$$, $$a_{m},b_{n}\geq0$$, such that $$0<\sum_{m=1}^{\infty }a_{m}^{p}<\infty$$ and $$0<\sum_{n=1}^{\infty}b_{n}^{q}<\infty$$, then

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\frac{\pi}{p})} \Biggl( \sum_{m=1}^{\infty}a_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}b_{n}^{q} \Biggr) ^{\frac{1}{q}},$$
(18)

where the constant factor $$\frac{\pi}{\sin(\pi/p)}$$ is the best possible.

The equivalent form of (18) is as follows:

$$\sum_{n=1}^{\infty} \Biggl( \sum _{m=1}^{\infty}\frac{a_{m}}{m+n} \Biggr) ^{p}< \biggl[ \frac{\pi}{\sin(\frac{\pi}{p})} \biggr] ^{p} \sum_{m=1}^{\infty }a_{m}^{p},$$
(19)

where the constant factor $$[\frac{\pi}{\sin(\pi/p)}]^{p}$$ is still the best possible.

In the same way, inequalities (12) and (14) (for $$\alpha=1$$) may be extended to the following equivalent forms (cf. [3]):

\begin{aligned}& \sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{m+n+1}< \frac {\pi}{\sin(\frac{\pi}{p})} \Biggl( \sum_{m=0}^{\infty}a_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=0}^{\infty}b_{n}^{q} \Biggr) ^{\frac{1}{q}}, \end{aligned}
(20)
\begin{aligned}& \sum_{n=0}^{\infty} \Biggl( \sum _{m=0}^{\infty}\frac{a_{m}}{m+n+1} \Biggr) ^{p}< \biggl[ \frac{\pi}{\sin(\frac{\pi}{p})} \biggr] ^{p}\sum _{m=0}^{\infty }a_{m}^{p}, \end{aligned}
(21)

where the constant factors $$\frac{\pi}{\sin(\pi/p)}$$ and $$[\frac{\pi }{\sin(\pi/p)}]^{p}$$ are the best possible. The equivalent integral analogs of (18) and (19) are given as follows:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \frac{\pi }{\sin(\frac{\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}
(22)
\begin{aligned}& \int_{0}^{\infty} \biggl( \int_{0}^{\infty} \frac{f(x)}{x+y}\,dx \biggr) ^{p}\,dy< \biggl[ \frac{\pi}{\sin(\frac{\pi}{p})} \biggr] ^{p}\int_{0}^{\infty }f^{p}(x) \,dx, \end{aligned}
(23)

with the best possible constant factors. We call (18) Hardy-Hilbert’s inequality and call (22) Hardy-Hilbert’s integral inequality.

Inequality (20) may be expressed in the form of the operator as follows.

If $$l^{p}$$ is a space of real sequences, $$T_{p}:l^{p}\rightarrow l^{p}$$ is a linear operator, such that for any non-negative sequence $$a=\{a_{m}\}_{m=1}\in l^{p}$$, there exists a $$T_{p}a=c=\{c_{n}\} _{n=1}^{\infty }\in l^{p}$$, satisfying

$$c_{n}=(T_{p}a) (n)=\sum_{m=0}^{\infty} \frac{a_{m}}{m+n+1},\quad n\in\mathbf {N}_{0}=\mathbf{N}\cup \{0\}.$$
(24)

For any non-negative sequence $$b=\{b_{n}\}_{n=1}\in l^{q}$$, we can indicate the formal inner product of $$T_{p}a$$ and b as follows:

$$(T_{p}a,b)=\sum_{n=0}^{\infty} \Biggl( \sum_{m=0}^{\infty}\frac {a_{m}}{m+n+1} \Biggr) b_{n}=\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac {a_{m}b_{n}}{m+n+1}.$$
(25)

Setting the norm of a as $$\|a\|_{p}= ( \sum_{n=0}^{\infty }|a_{n}|^{p} ) ^{\frac{1}{p}}$$, then inequality (20) may be rewritten as follows:

$$(T_{p}a,b)< \frac{\pi}{\sin(\frac{\pi}{p})}\|a\|_{p}\|b \|_{q},$$
(26)

where $$\|a\|_{p},\|b\|_{q}>0$$. We call $$T_{p}$$ Hardy-Hilbert’s operator.

Similarly, if $$L^{p}(\mathbf{R}_{+})$$ is a space of real functions, we may define the following Hardy-Hilbert’s integral operator $$\widetilde {T}_{p}: L^{p}(\mathbf{R}_{+})\rightarrow L^{p}(\mathbf{R}_{+})$$ as follows.

For any $$f\ (\geq0)\in L^{p}(\mathbf{R}_{+})$$, there exists a $$h=\widetilde{T}_{p}f\in L^{p}(\mathbf{R}_{+})$$, satisfying

$$(\widetilde{T}_{p}f) (y)=h(y)=\int_{0}^{\infty} \frac{f(x)}{x+y}\,dx,\quad y\in \mathbf{R}_{+}.$$
(27)

For any $$g\ (\geq0)\in L^{q}(\mathbf{R}_{+})$$, we can indicate the formal inner product of $$\widetilde{T}_{p}f$$ and g as follows:

$$(\widetilde{T}_{p}f,g)=\int_{0}^{\infty} \int_{0}^{\infty}\frac {f(x)g(y)}{x+y}\,dx\,dy.$$
(28)

Setting the norm of f as $$\|f\|_{p}= ( \int_{0}^{\infty }|f(x)|^{p}\,dx ) ^{\frac{1}{p}}$$, then inequality (22) may be rewritten as follows:

$$(\widetilde{T}_{p}f,g)< \frac{\pi}{\sin(\frac{\pi}{p})}\|f\|_{p}\|g \|_{q}.$$
(29)

### 2.2 Some kinds of Hardy-Hilbert-type inequalities

(1) For $$(p,q)$$ not being a pair of conjugate exponents, we have the following results (cf. [3], Theorem 339).

If $$p>1$$, $$q>1$$, $$\frac{1}{p}+\frac{1}{q}\geq1$$, $$0<\lambda=2-(\frac {1}{p}+\frac{1}{q})\leq1$$, then

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac {a_{m}b_{n}}{(m+n)^{\lambda}}\leq K \Biggl( \sum_{m=1}^{\infty}a_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}b_{n}^{q} \Biggr) ^{\frac{1}{q}},$$
(30)

where $$K=K(p,q)$$ relates to p, q, only for $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda=2-(\frac{1}{p}+\frac{1}{q})=1$$, the constant factor K is the best possible.

The integral analogs of (30) are given as follows:

$$\int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{(x+y)^{\lambda}}\,dx\,dy\leq K \biggl( \int_{0}^{\infty}f^{p}(x) \,dx \biggr) ^{\frac{1}{p}} \biggl( \int_{0}^{\infty}g^{q}(y) \,dy \biggr) ^{\frac{1}{q}}.$$
(31)

We also find an extension of (31) as follows (cf. [4]).

If $$p>1$$, $$q>1$$, $$\frac{1}{p}+\frac{1}{q}>1$$, $$0<\lambda=2-(\frac{1}{p}+\frac {1}{q})<1$$, then

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{f(x)g(y)}{|x+y|^{\lambda}}\,dx\,dy\leq k(p,q) \biggl( \int_{-\infty}^{\infty }f^{p}(x) \,dx \biggr) ^{\frac{1}{p}} \biggl( \int_{-\infty}^{\infty }g^{q}(y) \,dy \biggr) ^{\frac{1}{q}}.$$
(32)

For $$f(x)=g(x)=0$$, $$x\in(-\infty,0]$$, inequality (32) reduces to (31). Levin [19] also studied the expression forms of the constant factors in (30) and (31). But he did not prove their best possible property. In 1951, Bonsall [20] considered the case of (31) as regards the general kernel.

(2) If $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$h(t)>0$$, $$\phi(s)=\int_{0}^{\infty }h(t)t^{s-1}\, dt\in\mathbf{R}_{+}$$, $$f(x),g(y)\geq0$$, then we have the following integral inequalities with the non-homogeneous kernel (cf. [3], Theorem 350):

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty}h(xy)f(x)g(y) \,dx\,dy \\& \quad < \phi\biggl(\frac{1}{p}\biggr) \biggl( \int_{0}^{\infty}x^{p-2}f^{p}(x) \,dx \biggr) ^{\frac{1}{p}} \biggl( \int_{0}^{\infty}g^{q}(y) \,dy \biggr) ^{\frac{1}{q}}, \end{aligned}
(33)
\begin{aligned}& \int_{0}^{\infty} \biggl( \int_{0}^{\infty}h(xy)f(x) \,dx \biggr) ^{p}\,dy< \phi ^{p}\biggl(\frac{1}{p} \biggr)\int_{0}^{\infty}x^{p-2}f^{p}(x) \,dx, \end{aligned}
(34)
\begin{aligned}& \int_{0}^{\infty} \biggl( \int_{0}^{\infty}h(xy)f(x) \,dx \biggr) ^{p}\,dy< \phi ^{p}\biggl(\frac{1}{p} \biggr)\int_{0}^{\infty}x^{p-2}f^{p}(x) \,dx, \end{aligned}
(35)
\begin{aligned}& \int_{0}^{\infty}y^{p-2} \biggl( \int _{0}^{\infty}h(xy)f(x)\,dx \biggr) ^{p}\,dy< \phi^{p}\biggl(\frac{1}{q}\biggr)\int_{0}^{\infty}f^{p}(x) \,dx, \end{aligned}
(36)

where the integrals on the right-hand side are positive. The authors did not proved that the above constant factors are the best possible.

(3) If $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$h(t)>0$$ is decreasing with respect to $$t>0$$, $$\phi(s)=\int_{0}^{\infty}h(t)t^{s-1}\, dt\in\mathbf{R}_{+}$$, $$f(x),a_{n}\geq0$$, then we have the following half-discrete inequalities (cf. [3], Theorem 351):

\begin{aligned}& \int_{0}^{\infty} \Biggl( \sum _{n=1}^{\infty}h(nx)a_{n} \Biggr) ^{p}\,dx< \phi ^{p}\biggl(\frac{1}{p}\biggr)\sum _{n=1}^{\infty}n^{p-2}a_{n}^{p}, \end{aligned}
(37)
\begin{aligned}& \sum_{n=1}^{\infty} \biggl( \int _{0}^{\infty}h(nx)f(x)\,dx \biggr) ^{p}< \phi ^{p}\biggl(\frac{1}{p}\biggr)\int_{0}^{\infty}x^{p-2}f^{p}(x) \,dx, \end{aligned}
(38)
\begin{aligned}& \int_{0}^{\infty}x^{p-2} \Biggl( \sum _{n=1}^{\infty}h(nx)a_{n} \Biggr) ^{p}\,dx< \phi^{p}\biggl(\frac{1}{q}\biggr)\sum _{n=1}^{\infty}a_{n}^{p}, \end{aligned}
(39)
\begin{aligned}& \sum_{n=1}^{\infty}n^{p-2} \biggl( \int_{0}^{\infty}h(nx)f(x)\,dx \biggr) ^{p}< \phi^{p}\biggl(\frac{1}{q}\biggr)\int_{0}^{\infty}f^{p}(x) \,dx, \end{aligned}
(40)

where the integrals and series on the right-hand side are positive. The authors also did not prove that the above constant factors are the best possible.

### 2.3 Hardy-Hilbert-type inequalities with the general homogeneous kernel of degree −1

If $$\alpha\in\mathbf{R,}$$ the function $$k(x,y)$$ is measurable in $$\mathbf{R}_{+}^{2}$$, satisfying for any $$x,y,u>0$$, $$k(ux,uy)=u^{\alpha }k(x,y)$$, then we call $$k(x,y)$$ the homogeneous function of degree α.

In 1934, Hardy et al. published the following theorem (cf. [3], Theorem 318 and Theorem 319).

Suppose that $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$k_{1}(x,y)$$ (≥0) is a homogeneous function of degree −1 in $$\mathbf{R}_{+}^{2}$$. If $$k_{p}=\int_{0}^{\infty}k_{1}(u,1)u^{-\frac{1}{p}}\, du$$ is finite, then we have $$k_{p}=\int_{0}^{\infty}k_{1}(1,u)u^{-\frac{1}{q}}\, du$$ and the following equivalent inequalities:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty}k_{1}(x,y)f(x)g(y) \,dx\,dy \\& \quad \leq k_{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}
(41)
\begin{aligned}& \int_{0}^{\infty} \biggl( \int_{0}^{\infty}k_{1}(x,y)f(x) \,dx \biggr) ^{p}\,dy\leq k_{p}^{p}\int _{0}^{\infty}f^{p}(x)\,dx, \end{aligned}
(42)

where the constant $$k_{p}$$ is the best possible.

Moreover, if both $$k_{1}(u,1)u^{\frac{-1}{p}}$$ and $$k_{1}(1,u)u^{\frac {-1}{q}}$$ are decreasing in $$\mathbf{R}_{+}$$, then we have the following equivalent forms:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}k_{1}(m,n)a_{m}b_{n} \leq k_{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}
(43)
\begin{aligned}& \sum_{n=1}^{\infty} \Biggl( \sum _{m=1}^{\infty}k_{1}(m,n)a_{m} \Biggr) ^{p}\leq k_{p}^{p}\sum _{n=1}^{\infty}a_{n}^{p}. \end{aligned}
(44)

For $$0< p<1$$, if $$k_{p}$$ is finite, then we have the reverses of (41) and (42). (Note that we have not seen any proof of (41) and (42), and the reverse examples in the book [3].)

We name $$k_{1}(x,y)$$ the kernel of (41) and (42). If all the integrals and series in the right-hand side of inequalities (41)-(44) are positive, then we can get the following particular examples (cf. [3]):

(i) For $$k_{1}(x,y)=\frac{1}{x+y}$$, (41)-(44) deduce to (22), (23), (18), and (19).

(ii) For $$k_{1}(x,y)=\frac{1}{\max\{x,y\}}$$, (41)-(44) deduce to the following two pairs of equivalent forms:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{\max\{x,y\}}\,dx\,dy< pq \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}
(45)
\begin{aligned}& \int_{0}^{\infty} \biggl( \int_{0}^{\infty} \frac{f(x)}{\max\{x,y\}}\,dx \biggr) ^{p}\,dy< (pq)^{p} \int_{0}^{\infty}f^{p}(x)\,dx; \end{aligned}
(46)
\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\max\{m,n\}}< pq \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}
(47)
\begin{aligned}& \sum_{n=1}^{\infty} \Biggl( \sum _{m=1}^{\infty}\frac{a_{m}}{\max\{m,n\} } \Biggr) ^{p}< (pq)^{p}\sum_{n=1}^{\infty}a_{n}^{p}. \end{aligned}
(48)

(iii) For $$k_{1}(x,y)=\frac{\ln(x/y)}{x-y}$$, (41)-(44) deduce to the following two pairs of equivalent forms:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{\ln(\frac {x}{y})f(x)g(y)}{x-y}\,dx\,dy \\& \quad < \biggl[ \frac{\pi}{\sin(\frac{\pi}{p})} \biggr] ^{2} \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}
(49)
\begin{aligned}& \int_{0}^{\infty} \biggl( \int_{0}^{\infty} \frac{\ln(\frac {x}{y})f(x)}{x-y}\,dx \biggr) ^{p}\,dy< \biggl[ \frac{\pi}{\sin(\frac{\pi}{p})} \biggr] ^{2p}\int_{0}^{\infty}f^{p}(x) \,dx; \end{aligned}
(50)
\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{\ln(\frac {m}{n})a_{m}b_{n}}{m-n}< \biggl[ \frac{\pi}{\sin(\frac{\pi}{p})} \biggr] ^{2} \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}
(51)
\begin{aligned}& \sum_{n=1}^{\infty} \Biggl( \sum _{m=1}^{\infty}\frac{\ln(\frac {m}{n})a_{m}}{m-n} \Biggr) ^{p}< \biggl[ \frac{\pi}{\sin(\frac{\pi}{p})} \biggr] ^{2p}\sum _{n=1}^{\infty}a_{n}^{p}. \end{aligned}
(52)

Note that the constant factors in the above inequalities are all the best possible. We call (47) and (51) Hardy-Littlewood-Polya’s inequalities, or H-L-P inequalities. We find that the kernels in the above inequalities are all decreasing. But this is not necessary. For example, we find the following two pairs of equivalent forms with the non-decreasing kernel (cf. [21, 22]):

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{|\ln(\frac {x}{y})|f(x)g(y)}{\max\{x,y\}}\,dx\,dy \\ & \quad < \bigl(p^{2}+q^{2}\bigr) \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}
(53)
\begin{aligned}& \int_{0}^{\infty} \biggl( \int_{0}^{\infty} \frac{|\ln(\frac {x}{y})|f(x)}{\max\{x,y\}}\,dx \biggr) ^{p}\,dy< \bigl(p^{2}+q^{2} \bigr)^{p}\int_{0}^{\infty}f^{p}(x) \,dx; \end{aligned}
(54)
\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{|\ln(\frac {m}{n})|a_{m}b_{n}}{\max\{m,n\}}< \bigl(p^{2}+q^{2} \bigr) \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}
(55)
\begin{aligned}& \sum_{n=1}^{\infty} \Biggl( \sum _{m=1}^{\infty}\frac{|\ln(\frac {m}{n})|a_{m}}{\max\{m,n\}} \Biggr) ^{p}< \bigl(p^{2}+q^{2}\bigr)^{p}\sum _{n=1}^{\infty}a_{n}^{p}, \end{aligned}
(56)

where the constant factors $$p^{2}+q^{2}$$ and $$(p^{2}+q^{2})^{p}$$ are the best possible.

Other inequalities of this type with the best constants are provided as follows (cf. [23, 24]):

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{|\ln(\frac {x}{y})|f(x)g(y)}{x+y}\,dx\,dy \\& \quad < c_{0}(p) \biggl( \int_{0}^{\infty}f^{p}(x) \,dx \biggr) ^{\frac {1}{p}} \biggl( \int_{0}^{\infty}g^{p}(y) \,dy \biggr) ^{\frac{1}{q}}, \end{aligned}
(57)
\begin{aligned}& \int_{0}^{\infty} \biggl( \int_{0}^{\infty} \frac{|\ln(\frac {x}{y})|f(x)}{x+y}\,dx \biggr) ^{p}\,dy< c_{0}^{p}(p) \int_{0}^{\infty}f^{p}(x)\,dx; \end{aligned}
(58)
\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{|\ln(\frac {m}{n})|a_{m}b_{n}}{m+n}< c_{0}(2) \Biggl( \sum_{m=1}^{\infty}a_{m}^{2} \sum_{n=1}^{\infty }b_{n}^{2} \Biggr) ^{\frac{1}{2}}, \end{aligned}
(59)
\begin{aligned}& \sum_{n=1}^{\infty} \Biggl( \sum _{m=1}^{\infty}\frac{|\ln(\frac {m}{n})|a_{m}}{m+n} \Biggr) ^{2}< c_{0}^{2}(2)\sum _{n=1}^{\infty}a_{n}^{2}, \end{aligned}
(60)

where the constant factor $$c_{0}(p)$$ is indicated by

$$c_{0}(p)=2\sum_{n=1}^{\infty}(-1)^{n-1} \biggl[ \frac{1}{(n-\frac {1}{p})^{2}}-\frac{1}{(n-\frac{1}{q})^{2}} \biggr] .$$

### 2.4 Two multiple Hardy-Hilbert-type inequalities with the homogeneous kernels of degree $$-n+1$$

Suppose $$n\in\mathbf{N}\backslash\{1\}$$, n numbers $$p,q,\ldots,r$$ satisfying $$p,q,\ldots,r>1$$, $$p^{-1}+q^{-1}+\cdots+r^{-1}=1$$, $$k(x,y,\ldots,z)\geq0$$ is a homogeneous function of degree $$-n+1$$. If

$$k=\int_{0}^{\infty}\int_{0}^{\infty} \cdots\int_{0}^{\infty }k(1,y,\ldots ,z)y^{-\frac{1}{q}} \cdots z^{-\frac{1}{r}}\,dy\cdots \,dz$$

is a finite number, $$f,g,\ldots, h$$ are non-negative measurable functions in $$\mathbf{R}_{+}$$, then we have the following multiple Hilbert-type integral inequality (cf. [3], Theorem 322):

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \cdots\int_{0}^{\infty }k(x,y,\ldots ,z)f(x)g(y)\cdots h(z)\,dx\,dy\cdots \,dz \\& \quad \leq k \biggl( \int_{0}^{\infty}f^{p}(x) \,dx \biggr) ^{\frac{1}{p}} \biggl( \int_{0}^{\infty}g^{q}(y) \,dy \biggr) ^{\frac{1}{q}}\cdots \biggl( \int_{0}^{\infty}h^{r}(z) \,dz \biggr) ^{\frac{1}{r}}. \end{aligned}
(61)

Moreover, if $$a_{m},b_{n},\ldots,c_{s}\geq0$$, $$k(1,y,\ldots ,z)x^{0}y^{-\frac{1}{q}}\cdots z^{-\frac{1}{r}}$$, $$k(x,1,\ldots,z)\times x^{-\frac {1}{p}}y^{0}\cdots z^{-\frac{1}{r}}$$, … , $$k(x,y,\ldots,1)x^{-\frac {1}{p}}y^{-\frac{1}{q}}\cdots z^{0}$$ are all decreasing with respect to any single variable in $$\mathbf{R}_{+}$$, then we have

\begin{aligned}& \sum_{s=1}^{\infty}\cdots\sum _{n=1}^{\infty}\sum_{m=1}^{\infty }k(m,n, \ldots,s)a_{m}b_{n}\cdots c_{s} \\& \quad \leq k \Biggl( \sum_{m=1}^{\infty}a_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}b_{n}^{q} \Biggr) ^{\frac{1}{q}}\cdots \Biggl( \sum_{s=1}^{\infty}c_{s}^{r} \Biggr) ^{\frac{1}{r}}. \end{aligned}
(62)

For $$n=2$$, inequalities (61) and (62) reduce, respectively, to (41) and (43).

## 3 Modern research for Hilbert’s inequalities and Hardy-Hilbert’s inequalities

### 3.1 Modern research for Hilbert’s integral inequality

(1) In 1979, based on an improvement of Hölder’s inequality, Hu [25] gave a refinement of (2) (for $$f=g$$) as follows:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)f(y)}{x+y}\,dx\,dy \\& \quad < \pi \biggl[ \biggl( \int_{0}^{\infty}f^{2}(x) \,dx \biggr) ^{2}-\frac {1}{4} \biggl( \int _{0}^{\infty}f^{2}(x)\cos\sqrt{x}\,dx \biggr) ^{2} \biggr] ^{\frac{1}{2}}. \end{aligned}
(63)

Since then, he published many interesting results similar to (63) (cf. [6]).

(2) In 1998, Pachpatte [26] gave an inequality similar to (2) as follows.

For $$a,b>0$$,

\begin{aligned}& \int_{0}^{a}\int_{0}^{b} \frac{f(x)g(y)}{x+y}\,dx\,dy \\& \quad < \frac{\sqrt{ab}}{2} \biggl[ \int_{0}^{a}(a-x)f^{\prime 2}(x) \,dx\int_{0}^{b}(b-y)g^{\prime2}(y)\,dy \biggr] ^{\frac{1}{2}}. \end{aligned}
(64)

Some improvements and extensions were made by Zhao et al. [2729]. We can find other work of Pachpatte in [30].

(3) In 1998, by introducing parameters $$\lambda\in(0,1]$$ and $$a,b\in \mathbf{R}_{+}$$ ($$a< b$$), Yang [31] gave an extension of (2) as follows:

\begin{aligned}& \int_{a}^{b}\int_{a}^{b} \frac{f(x)g(y)}{(x+y)^{\lambda}}\,dx\,dy \\& \quad < B\biggl(\frac{\lambda}{2},\frac{\lambda}{2}\biggr) \biggl[ 1-\biggl( \frac{a}{b}\biggr)^{\frac {\lambda}{4}} \biggr] \biggl( \int_{a}^{b}x^{1-\lambda }f^{2}(x) \,dx\int_{a}^{b}y^{1-\lambda}g^{2}(y) \,dy \biggr) ^{\frac{1}{2}}, \end{aligned}
(65)

where $$B(u,v)$$ is the beta function. In 1999, Kuang [32] gave another extension of (2) as follows.

For $$\lambda\in(\frac{1}{2},1]$$,

$$\int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)f(y)}{x^{\lambda }+y^{\lambda}}\,dx\,dy< \frac{\pi}{\lambda\sin(\frac{\pi}{2\lambda})} \biggl( \int_{0}^{\infty}f^{2}(x) \,dx\int_{0}^{\infty}g^{2}(y)\,dy \biggr) ^{\frac {1}{2}}.$$
(66)

We can find other work of Kuang in [5] and [33].

(4) In 1999, by using the methods of algebra and analysis, Gao [34] gave an improvement of (2) as follows:

$$\int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)f(y)}{x+y}\,dx\,dy< \pi\sqrt {1-R} \biggl( \int _{0}^{\infty}f^{2}(x)\,dx\int _{0}^{\infty}g^{2}(y)\,dy \biggr) ^{\frac{1}{2}},$$
(67)

where $$R=\frac{1}{\pi}(\frac{u}{\|g\|}-\frac{v}{\|f\|})^{2}$$, $$u=\sqrt {\frac{2}{\pi}}(g,e)$$, $$v=\sqrt{2\pi}(f,e^{-x})$$, $$e(y)=\int_{0}^{\infty}\frac {e^{x}}{x+y}\,dx$$. We can find other work of Gao and Hsu in [35].

(5) In 2002, by using the operator theory, Zhang [36] gave an improvement of (2) as follows:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy \\& \quad \leq \frac{\pi}{\sqrt{2}} \biggl[ \int_{0}^{\infty }f^{2}(x) \,dx\int_{0}^{\infty}g^{2}(y)\,dy+ \biggl( \int_{0}^{\infty }f(x)g(x)\,dx \biggr) ^{2} \biggr] ^{\frac{1}{2}}. \end{aligned}
(68)

### 3.2 On the way of weight coefficients for giving a strengthened version of Hilbert’s inequality

In 1991, for giving an improvement of (1), Hsu and Wang [37] raised the way of weight coefficient as follows.

At first, by using Cauchy’s inequality in the left-hand side of (1), it follows:

\begin{aligned} I =&\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{m+n}=\sum _{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{m+n} \biggl[ \biggl(\frac {m}{n}\biggr)^{\frac{1}{4}}a_{m} \biggr] \biggl[ \biggl(\frac{n}{m}\biggr)^{\frac{1}{4}}b_{n} \biggr] \\ \leq& \Biggl\{ \sum_{m=1}^{\infty} \Biggl[ \sum_{n=1}^{\infty}\frac {1}{m+n} \biggl( \frac{m}{n} \biggr) ^{\frac{1}{2}} \Biggr] a_{m}^{2} \sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\frac{1}{m+n} \biggl( \frac{m}{n} \biggr) ^{\frac{1}{2}} \Biggr] b_{n}^{2} \Biggr\} ^{\frac{1}{2}}. \end{aligned}
(69)

Then define the weight coefficient

$$\omega(n):=\sum_{m=1}^{\infty} \frac{1}{m+n} \biggl( \frac{m}{n} \biggr) ^{\frac{1}{2}},\quad n\in \mathbf{N},$$
(70)

and rewrite (69) as follows:

$$I\leq \Biggl( \sum_{m=1}^{\infty} \omega(m)a_{m}^{2}\sum_{n=1}^{\infty } \omega(n)b_{n}^{2} \Biggr) ^{\frac{1}{2}}.$$
(71)

Afterwards, setting

$$\omega(n)=\pi-\frac{\theta(n)}{n^{1/2}},\quad n\in\mathbf{N},$$
(72)

where $$\theta(n)=(\pi-\omega(n))n^{1/2}$$, and estimating the series of $$\theta(n)$$, it follows that

$$\theta(n)= \Biggl[ \pi-\sum_{m=1}^{\infty} \frac{1}{m+n}\biggl(\frac {m}{n}\biggr)^{\frac{1}{2}} \Biggr] n^{1/2}>\theta=1.1213^{+}.$$
(73)

Then by (72), it yields

$$\omega(n)< \pi-\frac{\theta}{n^{1/2}},\quad n\in\mathbf{N},\theta=1.1213^{+}.$$
(74)

In view of (71), a strengthened version of (1) is given as follows:

$$I< \Biggl[ \sum_{m=1}^{\infty} \biggl( \pi- \frac{\theta}{m^{1/2}} \biggr) a_{m}^{2}\sum _{n=1}^{\infty} \biggl( \pi-\frac{\theta}{n^{1/2}} \biggr) b_{n}^{2} \Biggr] ^{\frac{1}{2}}.$$
(75)

Hsu also raised the open problem of obtaining the best value of (75). In 1992, Gao [38] gave the best value $$\theta_{0}=1.281669^{+}$$.

Still in 1991, by using the above method, a strengthened version of (8) was given by [39] as follows:

\begin{aligned} I < & \Biggl\{ \sum_{m=1}^{\infty} \biggl[ \frac{\pi}{\sin(\frac{\pi }{p})}-\frac{p-1}{m^{1/p}+m^{-1/q}} \biggr] a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &{}\times \Biggl\{ \sum_{n=1}^{\infty} \biggl[ \frac{\pi}{\sin(\frac{\pi }{p})}-\frac{q-1}{n^{1/q}+n^{-1/p}} \biggr] b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}
(76)

In 1997, by using the method of weight coefficients and the improved Euler-Maclaurin summation formula, Yang and Gao [40] gave

$$I < \Biggl\{ \sum_{m=1}^{\infty} \biggl[ \frac{\pi}{\sin(\frac{\pi }{p})}-\frac{1-\gamma}{m^{1/p}} \biggr] a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \Biggl\{ \sum_{n=1}^{\infty} \biggl[ \frac{\pi}{\sin(\frac{\pi }{p})}-\frac{1-\gamma}{n^{1/q}} \biggr] b_{n}^{q} \Biggr\} ^{\frac{1}{q}},$$
(77)

where $$1-\gamma=0.42278433^{+}$$ (γ is the Euler constant). We can find similar work in Gao and Yang [41].

In 1998, Yang and Debnath [42] gave another, strengthened, version of (8), which is an improvement of (76). We can find some strengthened versions of (12) and (20) in [4345].

### 3.3 Hilbert’s inequalities and Hardy-Hilbert’s inequalities with independent parameters

In 1998, by using the optimized weight coefficients and introducing an independent parameter $$\lambda\in(0,1]$$, Yang [31] gave an extension of (2) as follows.

If $$0<\int_{0}^{\infty}x^{1-\lambda}f^{2}(x)\,dx<\infty$$ and $$0<\int_{0}^{\infty}y^{1-\lambda}g^{2}(y)\,dy<\infty$$, then

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{(x+y)^{\lambda}}\,dx\,dy \\& \quad < B\biggl(\frac{\lambda}{2},\frac{\lambda}{2}\biggr) \biggl( \int _{0}^{\infty }x^{1-\lambda}f^{2}(x)\,dx\int _{0}^{\infty}y^{1-\lambda }g^{2}(y)\,dy \biggr) ^{\frac{1}{2}}, \end{aligned}
(78)

where the constant factor $$B(\frac{\lambda}{2},\frac{\lambda}{2})$$ is the best possible. The proof about the best possible property of the constant factor was given by [46], and the expressions of the beta function $$B(u,v)$$ are given in Wang and Guo [47]:

\begin{aligned} B(u,v) =&\int_{0}^{\infty}\frac{t^{u-1}\, dt}{(1+t)^{u+v}}= \int_{0}^{1}(1-t)^{u-1}t^{v-1}\, dt \\ =&\int_{1}^{\infty}\frac{(t-1)^{u-1}\, dt}{t^{u+v}}\quad (u,v>0). \end{aligned}
(79)

Some extensions of (18), (20), and (22) were given by [4850] as follows.

If $$\lambda>2-\min\{p,q\}$$, then

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{(x+y)^{\lambda}}\,dx\,dy \\& \quad < B \biggl( \frac{p+\lambda-2}{p},\frac{q+\lambda-2}{q} \biggr) \biggl( \int _{0}^{\infty}x^{1-\lambda}f^{p}(x)\,dx \biggr) ^{\frac {1}{p}} \biggl( \int_{0}^{\infty}y^{1-\lambda}g^{q}(y) \,dy \biggr) ^{\frac{1}{q}}; \end{aligned}
(80)

if $$2-\min\{p,q\}<\lambda\leq2$$, then

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac {a_{m}b_{n}}{(m+n)^{\lambda}} \\& \quad < B \biggl( \frac{p+\lambda-2}{p},\frac{q+\lambda-2}{q} \biggr) \Biggl( \sum _{m=1}^{\infty}m^{1-\lambda}a_{m}^{p} \Biggr) ^{\frac {1}{p}} \Biggl( \sum_{n=1}^{\infty}n^{1-\lambda}b_{n}^{q} \Biggr) ^{\frac{1}{q}}, \end{aligned}
(81)
\begin{aligned}& \sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac {a_{m}b_{n}}{(m+n+1)^{\lambda}} \\& \quad < B \biggl( \frac{p+\lambda-2}{p},\frac{q+\lambda-2}{q} \biggr) \Biggl[ \sum _{m=0}^{\infty} \biggl( m+\frac{1}{2} \biggr) ^{1-\lambda }a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=0}^{\infty} \biggl( n+ \frac{1}{2} \biggr) ^{1-\lambda}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}
(82)

where the constant factor $$B(\frac{p+\lambda-2}{p},\frac{q+\lambda-2}{q})$$ is the best possible.

Yang [51] also proved that (81) is valid for $$p=2$$ and $$\lambda \in(0,4]$$. Yang [52, 53] gave another extensions of (18) and (20) as follows.

If $$0<\lambda\leq\min\{p,q\}$$, then

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{m^{\lambda }+n^{\lambda}} < \frac{\pi}{\lambda\sin(\frac{\pi}{p})} \Biggl[ \sum_{m=1}^{\infty}m^{(p-1)(1-\lambda)}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{(q-1)(1-\lambda )}b_{n}^{q} \Biggr] ^{\frac{1}{q}};$$
(83)

if $$0<\lambda\leq1$$, then

\begin{aligned}& \sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{(m+\frac{1}{2} )^{\lambda}+(n+\frac{1}{2})^{\lambda}} \\& \quad < \frac{\pi}{\lambda\sin(\frac {\pi }{p})} \Biggl[ \sum_{m=0}^{\infty} \biggl( m+\frac{1}{2} \biggr) ^{p-1-\lambda }a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=0}^{\infty} \biggl( n+\frac{1}{2} \biggr) ^{q-1-\lambda}b_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}
(84)

In 2004, Yang [54] discovered the following dual form of (18):

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\frac{\pi}{p})} \Biggl( \sum_{m=1}^{\infty}m^{p-2}a_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}n^{q-2}b_{n}^{q} \Biggr) ^{\frac {1}{q}}.$$
(85)

Inequality (85) is similar to (18) but different and for $$p=2$$, both of them reduce to (1).

For $$\lambda=1$$, (84) reduces to the dual form of (20) as follows:

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{m+n+1} < \frac {\pi}{\sin(\frac{\pi}{p})} \Biggl[ \sum_{m=0}^{\infty} \biggl( m+\frac{1}{2} \biggr) ^{p-2}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=0}^{\infty} \biggl( n+\frac {1}{2} \biggr) ^{q-2}b_{n}^{q} \Biggr] ^{\frac{1}{q}}.$$
(86)

We can find some best extensions of the H-L-P inequalities such as (37)-(48) in [5561], by introducing some independent parameters.

In 2001, by introducing some parameters, Hong [62] gave a multiple integral inequality, which is an extension of (18). He et al. [63] gave a similar result for particular conjugate exponents. For making an improvement of their work, Yang [64] gave the following inequality, which is a best extension of (18).

If $$n\in\mathbf{N}\backslash\{1\}$$, $$p_{i}>1$$, $$\sum_{i=1}^{n}\frac {1}{p_{i}}=1$$, $$\lambda>n-\min_{1\leq i\leq n}\{p_{i}\}$$, $$f_{i}(t)\geq0$$, and $$0<\int_{0}^{\infty}t^{n-1-\lambda}f_{i}^{p_{i}}(t)\, dt<\infty$$ ($$i=1,2,\ldots,n$$), then we have

\begin{aligned}& \int_{0}^{\infty}\cdots\int_{0}^{\infty} \frac{\prod_{i=1}^{n}f_{i}(x_{i})}{(\sum_{i=1}^{n}x_{i})^{\lambda }}\,dx_{1}\cdots \,dx_{n} \\& \quad < \frac{1}{\Gamma(\lambda)}\prod_{i=1}^{n} \biggl( \frac{p_{i}+\lambda -n}{p_{i}} \biggr) \biggl( \int_{0}^{\infty}t^{n-1-\lambda }f_{i}^{p_{i}}(t) \,dt \biggr) ^{\frac{1}{p_{i}}}, \end{aligned}
(87)

where the constant factor $$\frac{1}{\Gamma(\lambda)}\prod_{i=1}^{n}(\frac{p_{i}+\lambda-n}{p_{i}})$$ is the best possible. In particular, for $$\lambda =n-1$$, it follows that

\begin{aligned}& \int_{0}^{\infty}\cdots\int_{0}^{\infty} \frac{\prod_{i=1}^{n}f_{i}(x_{i})}{(\sum_{i=1}^{n}x_{i})^{n-1}}\,dx_{1}\cdots \,dx_{n} \\& \quad < \frac{1}{(n-2)!}\prod_{i=1}^{n} \biggl( 1-\frac{1}{p_{i}} \biggr) \biggl( \int_{0}^{\infty}f_{i}^{p_{i}}(t) \,dt \biggr) ^{\frac{1}{p_{i}}}. \end{aligned}
(88)

In 2003, Yang and Rassias [65] introduced the method of weight coefficients and considered its applications to Hilbert-type inequalities. They summarized how to use the method of weight coefficients to obtain some new improvements and generalizations of the Hilbert-type inequalities. Since then, a number of authors discussed this problem (cf. [6686]). But how to give a best extension of inequalities (85) and (18) was solved in 2004 by introducing two pairs of conjugate exponents.

### 3.4 Hilbert-type inequalities with two conjugate exponents and multi-parameters

In 2004, by introducing an independent parameter $$\lambda>0$$ and two pairs of conjugate exponents $$(p,q)$$ and $$(r,s)$$ with $$\frac {1}{p}+\frac{1}{q}=\frac{1}{r}+\frac{1}{s}=1$$, Yang [87] gave an extension of (2) as follows.

If $$p,r>1$$, and the integrals of the right-hand side are positive, then

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x^{\lambda }+y^{\lambda}}\,dx\,dy \\& \quad < \frac{\pi}{\lambda\sin(\frac{\pi}{r})} \biggl[ \int_{0}^{\infty }x^{p(1-\frac{\lambda}{r})-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{\infty}y^{q(1-\frac{\lambda}{s})-1}g^{q}(y) \,dy \biggr] ^{\frac {1}{q}}, \end{aligned}
(89)

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

For $$\lambda=1$$, $$r=q$$, $$s=p$$, inequality (89) reduces to (22); for $$\lambda=1$$, $$r=p$$, $$s=q$$, inequality (89) reduces to the dual form of (22) as follows:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy \\& \quad < \frac{\pi}{\sin(\frac{\pi}{p})} \biggl( \int_{0}^{\infty }x^{p-2}f^{p}(x) \,dx \biggr) ^{\frac{1}{p}} \biggl( \int_{0}^{\infty }y^{q-2}g^{q}(y) \,dy \biggr) ^{\frac{1}{q}}. \end{aligned}
(90)

In 2005, by introducing an independent parameter $$\lambda>0$$, and two pairs of generalized conjugate exponents $$(p_{1},p_{2},\ldots,p_{n})$$ and $$(r_{1},r_{2},\ldots,r_{n})$$ with $$\sum_{i=1}^{n}\frac{1}{p_{i}}=\sum_{i=1}^{n}\frac{1}{r_{i}}=1$$, Yang et al. [88] gave a multiple integral inequality as follows.

For $$p_{i},r_{i}>1$$ ($$i=1,2,\ldots,n$$),

\begin{aligned}& \int_{0}^{\infty}\cdots\int_{0}^{\infty} \frac{\prod_{i=1}^{n}f_{i}(x_{i})}{(\sum_{i=1}^{n}x_{i})^{\lambda }}\,dx_{1}\cdots \,dx_{n} \\& \quad < \frac{1}{\Gamma(\lambda)}\prod_{i=1}^{n} \biggl(\frac{\lambda }{r_{i}}\biggr) \biggl[ \int_{0}^{\infty}t^{p_{i}(1-\frac{\lambda }{r_{i}})-1}f_{i}^{p_{i}}(t) \,dt \biggr] ^{\frac{1}{p_{i}}}, \end{aligned}
(91)

where the constant factor $$\frac{1}{\Gamma(\lambda)}\prod_{i=1}^{n}(\frac{\lambda}{r_{i}})$$ is the best possible. For $$r_{i}=\frac{p_{i}\lambda }{p_{i}-\lambda-n}$$ ($$i=1,2,\ldots,n$$), inequality (91) reduces to (87); for $$n=2$$, $$p_{1}=p$$, $$p_{2}=q$$, $$r_{1}=r$$, and $$r_{2}=s$$, inequality (91) reduces to the following:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{(x+y)^{\lambda}}\,dx\,dy \\& \quad < B \biggl( \frac{\lambda}{r},\frac{\lambda}{s} \biggr) \biggl[ \int _{0}^{\infty}x^{p(1-\frac{\lambda}{r})-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{\infty}y^{q(1-\frac{\lambda}{s} )-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}. \end{aligned}
(92)

It is obvious that inequality (92) is another best extension of (22).

In 2006, by using two pairs of conjugate exponents $$(p,q)$$ and $$(r,s)$$ with $$p,r>1$$, Hong [89] gave a multi-variable integral inequality as follows.

If $$\mathbf{R}_{+}^{n}=\{x=(x_{1},x_{2},\ldots ,x_{n});x_{i}>0,i=1,2,\ldots ,n\}$$, $$\alpha,\beta, \lambda>0$$, $$\|x\|_{\alpha }=(\sum_{i=1}^{n}x_{i}^{\alpha})^{\frac{1}{\alpha}}$$, $$f,g\geq0$$, $$0<\int_{\mathbf{R}_{+}^{n}}\|x\|_{\alpha}^{p(n-\frac{\beta\lambda}{r})-n}f^{p}(x)\,dx<\infty$$ and $$0<\int_{\mathbf{R}_{+}^{n}}\|y\|_{\alpha }^{q(n-\frac{\beta\lambda}{s})-n}g^{q}(y)\,dy<\infty$$, then

\begin{aligned}& \int_{\mathbf{R}_{+}^{n}}\int_{\mathbf{R}_{+}^{n}}\frac{f(x)g(y)\,dx\,dy}{(\|x\|_{\alpha}^{\beta}+\|y\|_{\alpha}^{\beta})^{\lambda}} \\& \quad < \frac {\Gamma ^{n}(\frac{1}{\alpha})}{\beta\alpha^{n-1}\Gamma(\frac{n}{\alpha})}B \biggl( \frac{\lambda}{r}, \frac{\lambda}{s} \biggr) \biggl[ \int_{\mathbf{R}_{+}^{n}}\|x \|_{\alpha}^{p(n-\frac{\beta \lambda}{r})-n}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \\& \qquad {}\times\biggl[ \int_{\mathbf{R}_{+}^{n}}\|y\|_{\alpha}^{q(n-\frac{\beta\lambda }{s})-n}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}
(93)

where the constant factor $$\frac{\Gamma^{n}(\frac{1}{\alpha})}{\beta \alpha^{n-1}\Gamma(\frac{n}{\alpha})}B(\frac{\lambda}{r},\frac {\lambda}{s})$$ is the best possible. In particular, for $$n=1$$, (93) reduces to Hong’s work in [90]; for $$n=\beta=1$$, (93) reduces to (92). In 2007, Zhong and Yang [91] generalized (93) to a general homogeneous kernel and proposed the reversion. Some other results on the multi-dimensional Hilbert-type inequalities are provided by [9295].

We can find another inequality with two parameters as follows (cf. [96]):

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{(m^{\alpha }+n^{\alpha})^{\lambda}}< \frac{1}{\alpha}B \biggl( \frac{\lambda}{r},\frac{\lambda}{s} \biggr) \Biggl[ \sum_{m=1}^{\infty}m^{p(1-\frac{\alpha\lambda}{r})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\frac{\alpha \lambda }{s})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}},$$
(94)

where $$\alpha,\lambda>0$$, $$\alpha\lambda\leq\min\{r,s\}$$. In particular, for $$\alpha=1$$, we have

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac {a_{m}b_{n}}{(m+n)^{\lambda}} < B \biggl( \frac{\lambda}{r},\frac{\lambda}{s} \biggr) \Biggl[ \sum _{m=1}^{\infty}m^{p(1-\frac{\lambda}{r})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\frac{\lambda }{s})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}.$$
(95)

For $$\lambda=1$$, $$r=q$$, (95) reduces to (18), and for $$\lambda =1$$, $$r=p$$, (95) reduces to (85). Some other results are provided by [9799].

Also we can see the reverse form as follows (cf. [100]):

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{(m+n+1)^{2}} > 2 \Biggl\{ \sum_{m=0}^{\infty} \biggl[ 1-\frac{1}{4(m+1)^{2}} \biggr] \frac{a_{m}^{p}}{2m+1} \Biggr\} ^{\frac{1}{p}} \Biggl\{ \sum_{n=0}^{\infty}\frac {b_{n}^{q}}{2n+1} \Biggr\} ^{\frac{1}{q}},$$
(96)

where $$0< p<1$$, $$\frac{1}{p}+\frac{1}{q}=1$$. The other results on the reverse of the Hilbert-type inequalities are found in [101107].

In 2006, Xin [108] gave a best extension of H-L-P integral inequality (41) as follows:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{\ln(\frac {x}{y})}{x^{\lambda }-y^{\lambda}}f(x)g(y)\,dx\,dy \\& \quad < \biggl[ \frac{\pi}{\sin(\frac{\pi}{r})} \biggr] ^{2} \biggl[ \int _{0}^{\infty}x^{p(1-\frac{\lambda}{r})-1}f^{p}(x)\,dx \biggr] ^{\frac {1}{p}} \biggl[ \int_{0}^{\infty}y^{q(1-\frac{\lambda}{s})-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}. \end{aligned}
(97)

Zhong and Yang [109] gave an extension of another H-L-P integral inequality (37) as follows:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{\max\{x^{\lambda },y^{\lambda}\}}\,dx\,dy \\& \quad < \frac{rs}{\lambda} \biggl[ \int_{0}^{\infty}x^{p(1-\frac{\lambda}{r} )-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{\infty }y^{q(1-\frac{\lambda}{s})-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}; \end{aligned}
(98)

Zhong and Yang [110] also gave the reverse form of (98).

Considering a particular kernel, Yang [111] gave

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{(\sqrt {m}+\sqrt{n})\sqrt{\max\{m,n\}}} \\& \quad < 4\ln2 \Biggl( \sum_{m=1}^{\infty}m^{\frac{p}{2}-1}a_{m}^{p} \Biggr) ^{ \frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}n^{\frac{q}{2}-1}b_{n}^{q} \Biggr) ^{\frac{1}{q}}. \end{aligned}
(99)

He also gave (cf. [112])

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{(m+an)^{2}+n^{2}} \\& \quad < \biggl( \frac{\pi}{2}-\arctan a \biggr) \Biggl( \sum _{m=1}^{\infty }\frac{a_{m}^{p}}{m} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}\frac {b_{n}^{q}}{n} \Biggr) ^{\frac{1}{q}}\quad (a\geq0). \end{aligned}
(100)

By using residue theory, Yang [113] obtained

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{(x+ay)(x+by)(x+cy)}\,dx\,dy \\& \quad < k \biggl( \int_{0}^{\infty}x^{-\frac{p}{2}-1}f^{p}(x) \,dx \biggr) ^{\frac {1}{p}} \biggl( \int_{0}^{\infty}y^{-\frac{q}{2}-1}g^{q}(y) \,dy \biggr) ^{\frac{1}{q}}, \end{aligned}
(101)

where $$k=\frac{1}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt {c})}$$ ($$a,b,c>0$$).

The constant factors in the above new inequalities are all the best possible. We can find some other new work in [114120].

In 2005, Yang [121] gave a half-discrete inequality with the kernel $$\frac{1}{(1+nx)^{\lambda}}$$ by introducing a variable and proved that the constant factor is the best possible. In 2011, Yang [122] deduced a half-discrete Hardy-Hilbert inequality with the best possible constant factor $$B(\lambda_{1},\lambda_{2})$$:

\begin{aligned}& \int_{0}^{\infty}\sum_{n=1}^{\infty} \frac{a_{n}f(x)}{(x+n)^{\lambda}}\,dx \\& \quad < B(\lambda_{1},\lambda_{2}) \biggl[ \int _{0}^{\infty}x^{p(1-\lambda _{1})-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty }n^{q(1-\lambda_{2})-1}a_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}
(102)

where $$\lambda_{1}>0$$, $$0<\lambda_{2}\leq1$$, $$\lambda_{1}+\lambda _{2}=\lambda$$.

Zhong et al. [123, 124] investigated several half-discrete Hilbert-type inequalities. A half-discrete Hilbert-type inequality with a general homogeneous kernel $$k_{\lambda}(x,n)$$ of degree $$-\lambda\in\mathbf{R}$$ and a best constant factor $$k ( \lambda _{1} )$$ was obtained, which is an extension of (102) (cf. [125]). Also a half-discrete Hilbert-type inequality with a general non-homogeneous kernel $$k_{\lambda}(1,xn)$$ and a best constant factor was given by Yang [126].

### 3.5 Modern research for Hilbert-type operators

Suppose that H is a separable Hilbert space and $$T:H\rightarrow H$$ is a bounded self-adjoint semi-positive definite operator. In 2002, Zhang [36] gave the following inequality:

$$(a,Tb)^{2}\leq\frac{\|T\|^{2}}{2}\bigl(\|a\|^{2}\|b \|^{2}+(a,b)^{2}\bigr)\quad (a,b\in H),$$
(103)

where $$(a,b)$$ is the inner product of a and b, and $$\|a\|=\sqrt{(a,a)}$$ is the norm of a. Since the Hilbert integral operator defined by (7) satisfies the condition of (103) with $$\|\widetilde{T}\|=\pi$$, inequality (2) may be improved as (68). Since the operator $$T_{p}$$ defined by (24) (for $$p=q=2$$) satisfies the condition of (103) (cf. [7]), we may improve (12) to the following form:

$$\sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{a_{m}b_{n}}{m+n+1}< \frac {\pi}{\sqrt{2}} \Biggl[ \sum_{m=0}^{\infty}a_{m}^{2} \sum_{n=0}^{\infty }b_{n}^{2}+ \Biggl(\sum_{n=0}^{\infty}a_{n}b_{n} \Biggr)^{2} \Biggr] ^{\frac{1}{2}}.$$
(104)

The key of applying (103) is to obtain the norm of the operator and to show the property of semi-definite. Now, we consider the concept and the properties of Hilbert-type integral operator as follows.

Suppose that $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$L^{r}(\mathbf{R}_{+})$$ ($$r=p,q$$) are real normal linear spaces and $$k(x,y)$$ is a non-negative symmetric measurable function in $$\mathbf{R}_{+}^{2}$$ satisfying

$$\int_{0}^{\infty}k(x,t) \biggl(\frac{x}{t} \biggr)^{\frac{1}{r}}\,dt=k_{0}(p)\in \mathbf{R}\quad (x>0).$$

We define an integral operator as

$$T:L^{r}(\mathbf{R}_{+})\rightarrow L^{r}( \mathbf{R}_{+})\quad (r=p,q),$$

for any $$f\ (\geq0)\in L^{p}(\mathbf{R}_{+})$$, there exists a $$h=Tf\in L^{p}(\mathbf{R}_{+})$$, such that

$$(Tf) (y)=h(y)=\int_{0}^{\infty}k(x,y)f(x)\,dx\quad (y>0);$$
(105)

or for any $$g\ (\geq0)\in L^{q}(\mathbf{R}_{+})$$, there exists a $$\tilde{h}=Tg\in L^{q}(\mathbf{R}_{+})$$, such that

$$(Tg) (x)=\tilde{h}(x)=\int_{0}^{\infty}k(x,y)g(y) \,dy\quad (x>0).$$
(106)

In 2006, Yang [127] proved that the operator T defined by (105) or (106) are bounded with $$\|T\|\leq k_{0}(p)$$. The following are some results in this paper.

If $$\varepsilon>0$$, is small enough and the integral $$\int_{0}^{\infty }k(x,t)(\frac{x}{t})^{\frac{1+\varepsilon}{r}}\,dt$$ ($$r=p,q$$; $$x>0$$) is convergent to a constant $$k_{\varepsilon}(p)$$ independent of x satisfying $$k_{\varepsilon}(p)=k_{0}(p)+o(1)$$ ($$\varepsilon\rightarrow 0^{+}$$), then $$\|T\|=k_{0}(p)$$. If $$\|T\|>0$$, $$f\in L^{p}(\mathbf{R}_{+})$$, $$g\in L^{q}(\mathbf{R}_{+})$$, $$\|f\|_{p},\|g\|_{q}>0$$, then we have the following equivalent inequalities:

\begin{aligned}& (Tf,g) < \|T\|\cdot\|f\|_{p}\|g\|_{q}, \end{aligned}
(107)
\begin{aligned}& \|Tf\|_{p} < \|T\|\cdot\|f\|_{p}. \end{aligned}
(108)

Some particular cases are considered in this paper.

Yang [128] also considered some properties of Hilbert-type integral operator (for $$p=q=2$$). For the homogeneous kernel of degree −1, Yang [129] considered some sufficient conditions to obtain $$\|T\|=k_{0}(p)$$. We can find some properties of the discrete Hilbert-type operator in the disperse space in Yang [130133]. A multiple integral operator is scored by Bényi and Oh [134]. In 2009, Yang [135] summarized the above part results. Some other works about Hilbert-type operators and inequalities with the general homogeneous kernel and multi-parameters were provided by [136145].

During 2009-2014, Yang published six books about the theory of Hilbert-type operators with their norms and inequalities. On January of 2009, Yang’s first book about the integral and discrete Hilbert-type operators with the general homogeneous kernels of non-negative number degree and two pairs of conjugate exponents as well as the related inequalities was published by Chinese Science Press (cf. [146]). On October of 2009, Yang’s second book about Hilbert-type integral operators with the general homogeneous kernels of real number degree and two pairs of conjugate exponents as well as their inequalities was published by Bentham Science Publishers Ltd. (cf. [147]). On February of 2011, Yang’s third book about discrete Hilbert-type operators as well as the related inequalities with the same kernels and parameters in integrals was published by Bentham Science Publishers Ltd. (cf. [148]). In 2012-2013, Yang published two books that considered multiple half-discrete Hilbert-type operators and their inequalities (cf. [149, 150]). In 2014, Yang and Debnath published a book considering general half-discrete operators and their inequalities. These six books provide an extensive account of these types of operators and inequalities successfully.

## 4 Yang-Hilbert-type inequalities with two pairs of conjugate exponents and independent parameters

### 4.1 Yang-Hilbert-type integral inequalities

In 2009, Yang [147] (Theorem 5.1.6) gave an extension of (91) as follows.

If $$n\in\mathbf{N}\backslash\{1\}$$, $$p_{i}>1$$, $$r_{i}\neq0$$ ($$i=1,2,\ldots,n$$), $$\sum_{i=1}^{n}\frac{1}{p_{i}}=\sum_{i=1}^{n}\frac{1}{r_{i}}=1$$, $$\lambda \in \mathbf{R}$$, $$k_{\lambda}(x_{1},\ldots,x_{n})$$ (≥0) is a homogeneous function of degree −λ in $$\mathbf{R}_{+}^{n}$$,

$$k_{\lambda}(r_{1},\ldots,r_{n-1}) = \int_{0}^{\infty}\cdots\int _{0}^{\infty}k_{\lambda}(u_{1},\ldots ,u_{n-1},1)\prod_{j=1}^{n-1}u_{j}^{\frac{\lambda}{r_{j}}-1} \, du_{1}\cdots \, du_{n-1}\in\mathbf{R}_{+},$$

$$f_{i}(t)\geq0$$ and $$0<\int_{0}^{\infty}t^{p_{i}(1-\frac{\lambda }{r_{i}})-1}f_{i}^{p_{i}}(t)\,dt<\infty$$ ($$i=1,2,\ldots,n$$), then we have the following inequality:

\begin{aligned}& \int_{0}^{\infty}\cdots\int_{0}^{\infty}k_{\lambda}(x_{1}, \ldots ,x_{n})\prod_{i=1}^{n}f_{i}(x_{i}) \,dx_{1}\cdots \,dx_{n} \\& \quad < k_{\lambda}(r_{1},\ldots,r_{n-1}) \biggl[ \int _{0}^{\infty }t^{p_{i}(1-\frac{\lambda}{r_{i}})-1}f_{i}^{p_{i}}(t) \,dt \biggr] ^{\frac{1}{p_{i}}}, \end{aligned}
(109)

where the constant factor $$k_{\lambda}(r_{1},\ldots,r_{n-1})$$ is the best possible.

In this reference, the equivalent form of (109), the reverses, the operator expressions, and some particular examples are provided.

In Theorem 6.14 of this book, Yang also gave the following multi-dimensional integral inequalities, an extension of (93).

If $$\lambda\in\mathbf{R}$$, $$p>1$$, $$r,s\neq0$$, $$\frac{1}{p}+\frac{1}{q}=\frac {1}{r}+\frac{1}{s}=1$$, $$k_{\lambda}(x,y)$$ (≥0) is a homogeneous function of degree −λ in $$\mathbf{R}_{+}^{2}$$,

$$k_{\lambda}(r)=\int_{0}^{\infty}k_{\lambda}(u,1)u^{\frac{\lambda}{r} -1} \, du\in\mathbf{R}_{+},$$

$$x,y\in\mathbf{R}_{+}^{n}$$, $$\alpha>0$$, $$f,g\geq0$$, $$0<\int_{\mathbf{R}_{+}^{n}}\|x\|_{\alpha}^{p(n-\frac{\lambda}{r})-n}f^{p}(x)\,dx<\infty$$, and $$0<\int_{\mathbf{R}_{+}^{n}}\|x\|_{\alpha}^{q(n-\frac{\lambda}{s})-n}g^{q}(x)\,dx<\infty$$, then we have the following inequality:

\begin{aligned}& \int_{\mathbf{R}_{+}^{n}}\int_{\mathbf{R}_{+}^{n}}k_{\lambda } \bigl(\Vert x\Vert _{\alpha },\|y\|_{\alpha}\bigr)f(x)g(y)\,dx\,dy \\& \quad < \frac{\Gamma^{n}(\frac{1}{\alpha })}{\alpha ^{n-1}\Gamma(\frac{n}{\alpha})}k_{\lambda}(r) \biggl[ \int_{\mathbf{R}_{+}^{n}} \|x\|_{\alpha}^{p(n-\frac {\lambda}{r})-n}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \\& \qquad {}\times \biggl[ \int_{\mathbf{R}_{+}^{n}}\|y\|_{\alpha}^{q(n-\frac{\lambda}{s})-n}g^{q}(y) \,dy \biggr] ^{ \frac{1}{q}}, \end{aligned}
(110)

where the constant factor $$\frac{\Gamma^{n}(\frac{1}{\alpha})}{\alpha ^{n-1}\Gamma(\frac{n}{\alpha})}k_{\lambda}(r)$$ is the best possible.

Also, the equivalent form of (110), the reverses, the Hardy-type inequalities, the operator expressions, and many particular examples are provided. Some other results of multi-dimensional Hilbert-type integral inequalities are discussed by [151, 152].

For $$n=2$$ in (109), or $$\alpha=n=1$$ in (110), we reduce the following Yang-Hilbert-type integral inequality:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda}(x,y)f(x)g(y) \,dx\,dy \\& \quad < k_{\lambda}(r) \biggl[ \int_{0}^{\infty}x^{p(1-\frac{\lambda}{r})-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{\infty }y^{q(1-\frac{\lambda}{s})-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}
(111)

where the constant factor $$k_{\lambda}(r)$$ is the best possible. The equivalent form of (111) is obtained as follows (cf. [147], Theorem 2.2.1):

$$\int_{0}^{\infty}y^{\frac{p\lambda}{s}-1} \biggl( \int _{0}^{\infty }k_{\lambda}(x,y)f(x)\,dx \biggr) ^{p}\,dy< k_{\lambda}^{p}(r)\int_{0}^{\infty }x^{p(1-\frac{\lambda}{r})-1}f^{p}(x) \,dx,$$
(112)

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

For $$\lambda=1$$, $$r=q$$, $$s=p$$, (111) and (112) reduce, respectively, to (33) and (34). Hence, Yang-Hilbert-type integral inequalities are extensions of Hardy-Hilbert-type integral inequalities.

If we replace y and $$g(\frac{1}{y})$$ to $$\frac{1}{y}$$ and $$y^{2-\lambda }g(y)$$ in (111) and (112), then we obtain the following equivalent inequalities with the non-homogeneous kernel and the best possible constant factors (cf. [153]):

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda}(xy,1)f(x)g(y) \,dx\,dy \\& \quad < k_{\lambda}(r) \biggl[ \int_{0}^{\infty}x^{p(1-\frac{\lambda}{r})-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{\infty }y^{q(1-\frac{\lambda}{r})-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}
(113)
\begin{aligned}& \int_{0}^{\infty}y^{\frac{p\lambda}{r}-1} \biggl( \int _{0}^{\infty }k_{\lambda}(xy,1)f(x)\,dx \biggr) ^{p}\,dz< k_{\lambda}^{p}(r)\int_{0}^{\infty }x^{p(1-\frac{\lambda}{r})-1}f^{p}(x) \,dx. \end{aligned}
(114)

Replacing x and $$f(\frac{1}{x})$$ to $$\frac{1}{x}$$ and $$x^{2-\lambda}f(x)$$ in (111) and (112), we also obtain the following equivalent inequalities with the non-homogeneous kernel and the best possible constant factors:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda}(1,xy)f(x)g(y) \,dx\,dy \\& \quad < k_{\lambda}(r) \biggl[ \int_{0}^{\infty}x^{p(1-\frac{\lambda}{s})-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{\infty }y^{q(1-\frac{\lambda}{s})-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}, \end{aligned}
(115)
\begin{aligned}& \int_{0}^{\infty}y^{\frac{p\lambda}{s}-1} \biggl( \int _{0}^{\infty }k_{\lambda}(1,xy)f(x)\,dx \biggr) ^{p}\,dy< k_{\lambda}^{p}(r)\int_{0}^{\infty }x^{p(1-\frac{\lambda}{s})-1}f^{p}(x) \,dx. \end{aligned}
(116)

It is evident that (111)-(116) are equivalent. In particular, if $$k_{\lambda}(x,y)$$ is symmetric, then we have

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda }(xy,1)f(x)g(y) \,dx\,dy \\& \quad < k_{\lambda}(r)\min_{a\in\{r,s\}} \biggl\{ \biggl[ \int _{0}^{\infty}x^{p(1-\frac {\lambda}{a})-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{0}^{\infty }y^{q(1-\frac{\lambda}{a})-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}} \biggr\} . \end{aligned}
(117)

The above inequalities are some refinements of (33)-(36).

### 4.2 Discrete Yang-Hilbert-type inequalities

In 2011, Yang [148] (Theorem 4.2.3) gave an extension of (35) and (36) as follows.

If $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda_{1},\lambda_{2}\in\mathbf {R}$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$k_{\lambda}(x,y)$$ (≥0) is a finite homogeneous function of degree −λ in $$\mathbf{R}_{+}^{2}$$,

$$k(\lambda_{1})=\int_{0}^{\infty}k_{\lambda}(u,1)u^{\lambda _{1}-1} \, du\in \mathbf{R}_{+},$$

$$k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{2}}}(k_{\lambda}(x,y)\frac {1}{y^{1-\lambda_{1}}})$$ is decreasing with respect to $$x(y)>0$$, $$a_{m},b_{n}\geq0$$, $$0< \sum_{m=1}^{\infty}m^{p(1-\lambda _{1})-1}a_{m}^{p}<\infty$$, $$0<\sum_{n=1}^{\infty}n^{q(1-\lambda _{2})-1}b_{n}^{q}<\infty$$, then we have the following equivalent discrete Yang-Hilbert-type inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}k_{\lambda}(m,n)a_{m}b_{n} \\& \quad < k(\lambda_{1}) \Biggl[ \sum_{m=1}^{\infty}m^{p(1-\lambda _{1})-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\lambda _{2})-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}
(118)
\begin{aligned}& \sum_{n=1}^{\infty}n^{p\lambda_{2}-1} \Biggl( \sum_{m=1}^{\infty }k_{\lambda }(m,n)a_{m} \Biggr) ^{p}< \bigl(k(\lambda_{1})\bigr)^{p}\sum _{m=1}^{\infty }m^{p(1-\lambda_{1})-1}a_{m}^{p}, \end{aligned}
(119)

where the constant factors $$k(\lambda_{1})$$ and $$(k(\lambda_{1}))^{p}$$ are the best possible.

In this reference, some extensions of (118) and (119), the reverses, the operator expressions, and some particular examples are provided.

The following multiple inequalities are considered (cf. [148], Corollary 6.1.7).

If $$n\in\mathbf{N}\backslash\{1\}$$, $$p_{i}>1$$, $$\lambda_{i}\in\mathbf{R}$$ ($$i=1,2,\ldots,n$$), $$\sum_{i=1}^{n}\frac{1}{p_{i}}=1$$, $$\sum_{i=1}^{n}\lambda _{i}=\lambda$$, $$\frac{1}{q_{n}}=1-\frac{1}{p_{n}}$$, $$k_{\lambda }(x_{1},\ldots ,x_{n})$$ (≥0) is a finite homogeneous function of degree −λ in $$\mathbf{R}_{+}^{n}$$, $$k_{\lambda}(x_{1},\ldots,x_{n})\frac{1}{x_{i}^{1-\lambda_{i}}}$$ is decreasing with respect to $$x_{i}>0$$ ($$i=1,\ldots ,n$$),

$$k(\lambda_{1},\ldots,\lambda_{n-1}) = \int_{0}^{\infty}\cdots\int _{0}^{\infty}k_{\lambda}(u_{1},\ldots ,u_{n-1},1)\prod_{j=1}^{n-1}u_{j}^{\lambda_{j}-1} \, du_{1}\cdots \, du_{n-1}\in \mathbf{R}_{+},$$

$$a_{m_{i}}^{(i)}\geq0$$ and $$0<\sum_{m_{i}=1}^{\infty }m_{i}^{p_{i}(1-\lambda_{i})-1}(a_{m_{i}}^{(i)})^{p_{i}}<\infty$$ ($$i=1,2,\ldots,n$$), then we still have the following multiple equivalent inequalities:

\begin{aligned}& \sum_{m_{n}=1}^{\infty}\cdots\sum _{m_{1}=1}^{\infty}k_{\lambda }(m_{1}, \ldots,m_{n})\prod_{i=1}^{n}a_{m_{i}}^{(i)} \\& \quad < k(\lambda_{1},\ldots,\lambda_{n-1})\prod _{i=1}^{n} \Biggl[ \sum _{m_{i}=1}^{\infty}m_{i}^{p_{i}(1-\lambda _{i})-1} \bigl(a_{m_{i}}^{(i)}\bigr)^{p_{i}} \Biggr] ^{\frac{1}{p_{i}}}, \end{aligned}
(120)
\begin{aligned}& \sum_{m_{n}=1}^{\infty}m_{n}^{q_{n}\lambda_{n}-1} \Biggl[ \sum_{m_{n-1}=1}^{\infty}\cdots\sum _{m_{1}=1}^{\infty}k_{\lambda }(m_{1}, \ldots,m_{n})\prod_{i=1}^{n-1}a_{m_{i}}^{(i)} \Biggr] ^{q_{n}} \\& \quad < \bigl(k(\lambda_{1},\ldots,\lambda_{n-1}) \bigr)^{q_{n}}\prod_{i=1}^{n-1} \Biggl[ \sum_{m_{i}=1}^{\infty}m_{i}^{p_{i}(1-\lambda _{i})-1} \bigl(a_{m_{i}}^{(i)}\bigr)^{p_{i}} \Biggr] ^{\frac{q_{n}}{p_{i}}}, \end{aligned}
(121)

where the constant factors $$k(\lambda_{1},\ldots,\lambda_{n-1})$$ and $$(k(\lambda_{1},\ldots,\lambda_{n-1}))^{q_{n}}$$ are the best possible.

In this book, the reverses of (120) and (121) are also considered. For $$n=2$$, (120) and (121) reduce, respectively, to (120) and (121); for $$\lambda=1$$, $$\lambda_{i}=1-\frac {1}{p_{i}}$$ ($$i=1,\ldots,n$$), (120) reduces to (54).

In 2014, Yang [154] (Corollary 3.2) gave the following results.

Suppose that $$i_{0},j_{0}\in\mathbf{N}$$, $$\alpha,\beta>0$$, $$\lambda _{1}< i_{0}$$, $$\lambda_{2}< j_{0}$$, $$\lambda_{1}+\lambda=\lambda$$, $$k_{\lambda }(x,y)$$ (>0) is a finite homogeneous function of degree −λ in $$\mathbf{R}_{+}^{2}$$, which is decreasing with respect to $$x(y)>0$$, there exists a constant $$0<\delta_{0}<j_{0}-\lambda_{2}$$, such that for any $$\tilde{\lambda}_{1}\in(\lambda_{1}-\delta_{0},\lambda _{1}+\delta _{0})$$, $$k(\tilde{\lambda}_{1})=\int_{0}^{\infty}k_{\lambda}(u,1)u^{ \tilde{\lambda}_{1}-1}\, du\in\mathbf{R}_{+}$$, and there exists a constant $$\delta_{1}<\lambda_{1}-\delta_{0}$$, satisfying $$k_{\lambda }(u,1)\leq\frac{L}{u^{\delta_{1}}}$$ ($$u\in(0,\infty)$$). If $$p>1$$, $$\frac {1}{p}+\frac{1}{q}=1$$, $$a(m),b(n)\geq0$$, $$0<\sum_{m\in\mathbf{N}^{i_{0}}}\|m\|_{\alpha}^{p(i_{0}-\lambda_{1})-i_{0}}a^{p}(m)<\infty$$, $$0<\sum_{n\in\mathbf{N}^{j_{0}}}\|n\|_{\beta}^{q(j_{0}-\lambda _{2})-j_{0}}b^{q}(n)<\infty$$,

$$K(\lambda_{1})= \biggl( \frac{\Gamma^{i_{0}}(\frac{1}{\alpha})}{\alpha ^{i_{0}-1}\Gamma(\frac{i_{0}}{\alpha})} \biggr) ^{\frac{1}{q}} \biggl( \frac{\Gamma^{j_{0}}(\frac{1}{\beta})}{\beta^{j_{0}-1}\Gamma(\frac{j_{0}}{ \beta})} \biggr) k(\lambda_{1}),$$

then we have the following inequality:

\begin{aligned}& \sum_{n\in\mathbf{N}^{j_{0}}}\sum_{m\in\mathbf{N}^{i_{0}}}k_{\lambda } \bigl(\Vert m\Vert _{\alpha},\|n\|_{\beta}\bigr)a(m)b(n) \\& \quad < K(\lambda_{1}) \biggl[ \sum_{m\in\mathbf{N}^{i_{0}}} \|m\|_{\alpha }^{p(i_{0}-\lambda_{1})-i_{0}}a^{p}(m) \biggr] ^{\frac{1}{p}} \biggl[ \sum_{n\in\mathbf{N}^{j_{0}}}\|n\|_{\beta}^{q(j_{0}-\lambda _{2})-j_{0}}b^{q}(n) \biggr] ^{\frac{1}{q}}, \end{aligned}
(122)
\begin{aligned}& \sum_{n\in\mathbf{N}^{j_{0}}}\|n\|_{\beta}^{p\lambda _{2}-j_{0}} \biggl( \sum_{m\in\mathbf{N}^{i_{0}}}k_{\lambda}\bigl(\Vert m \Vert _{\alpha},\|n\|_{\beta }\bigr)a(m) \biggr) ^{p} \\& \quad < K^{p}(\lambda_{1})\sum _{m\in\mathbf{N}^{i_{0}}}\|m\|_{\alpha }^{p(i_{0}-\lambda_{1})-i_{0}}a^{p}(m), \end{aligned}
(123)

where the constant factors $$K(\lambda_{1})$$ and $$K^{p}(\lambda_{1})$$ are the best possible.

For $$i_{0}=j_{0}=\alpha=\beta=1$$, (122) and (123) also reduce, respectively, to (118) and (119). In this chapter, the reverses and the operator expressions of (122) and (123) are provided. A composition formula of the operators is developed.

### 4.3 Half-discrete Yang-Hilbert-type inequalities

In 2014, Yang and Debnath [155] (Theorem 6.1) gave the following results.

Suppose that $$m\in\mathbf{N}$$, $$p_{i}>1$$, $$\lambda_{i}\in\mathbf{R}$$ ($$i=1,2,\ldots,m+1$$), $$\sum_{i=1}^{m+1}\frac{1}{p_{i}}=1$$, $$\sum_{i=1}^{m+1}\lambda_{i}=\lambda$$, $$\frac{1}{p}=1-\frac{1}{p_{m+1}}$$, $$k_{\lambda }(x_{1},\ldots,x_{n})$$ (≥0) is a finite homogeneous function of degree −λ in $$\mathbf{R}_{+}^{n}$$, there exists a constant $$\delta_{0}>0$$, such that for any $$\tilde{\lambda}_{i}\in(\lambda_{i}-\delta _{0},\lambda_{i}+\delta_{0})$$, $$\sum_{i=1}^{m+1}\tilde{\lambda}_{i}=\lambda$$, $$k_{\lambda}(x_{1},\ldots,x_{m},y)\frac {1}{y^{1-\tilde{\lambda}_{m+1}}}$$ is strictly decreasing with respect to $$y>0$$, and

$$k(\tilde{\lambda}_{m+1})=\int_{0}^{\infty} \cdots\int_{0}^{\infty }k_{\lambda}(u_{1}, \ldots,u_{m},1)\prod_{j=1}^{m}u_{j}^{\tilde {\lambda }_{j}-1} \, du_{1}\cdots\, du_{m}\in\mathbf{R}_{+}.$$

If $$f_{i}(x_{i}),a_{n}\geq0$$ and $$0<\sum_{n=1}^{\infty }n^{p_{m+1}(1-\lambda_{m+1})-1}a_{n}^{p_{m+1}}<\infty$$,

$$0< \int_{0}^{\infty}t^{p_{i}(1-\lambda_{i})-1}f_{i}^{p_{i}}(t) \,dt< \infty \quad (i=1,\ldots,m),$$

then we still have the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\int_{0}^{\infty} \cdots\int_{0}^{\infty }k_{\lambda }(x_{1}, \ldots,x_{m},n)a_{n}\prod_{i=1}^{m}f_{i}(x_{i}) \,dx_{1}\cdots \,dx_{m} \\& \quad < k(\lambda_{m+1}) \Biggl[ \sum_{n=1}^{\infty}n^{p_{m+1}(1-\lambda _{m+1})-1}a_{n}^{p_{m+1}} \Biggr] ^{\frac{1}{p_{m+1}}}\prod_{i=1}^{m} \biggl[ \int_{0}^{\infty}t^{p_{i}(1-\lambda_{i})-1}f_{i}^{p_{i}}(t) \,dt \biggr] ^{\frac{1}{p_{i}}}, \end{aligned}
(124)
\begin{aligned}& \sum_{n=1}^{\infty}n^{p\lambda_{m+1}-1} \Biggl( \int_{0}^{\infty }\cdots \int_{0}^{\infty}k_{\lambda}(x_{1}, \ldots ,x_{m},n)\prod_{i=1}^{m}f_{i}(x_{i}) \,dx_{1}\cdots \,dx_{m} \Biggr) ^{p} \\& \quad < k^{p}(\lambda_{m+1})\prod _{i=1}^{m} \biggl[ \int_{0}^{\infty }t^{p_{i}(1-\lambda_{i})-1}f_{i}^{p_{i}}(t) \,dt \biggr] ^{\frac{p}{p_{i}}}, \end{aligned}
(125)

where the constant factors

$$k(\lambda_{m+1})=\int_{0}^{\infty}\cdots\int _{0}^{\infty}k_{\lambda }(u_{1}, \ldots,u_{m},1)\prod_{j=1}^{m}u_{j}^{\lambda _{j}-1} \, du_{1}\cdots \, du_{m}$$

and $$k^{p}(\lambda_{m+1})$$ are the best possible.

In this book, the reverses, the operator expressions, and some particular examples are provided. Some other kinds of multiple half-discrete Hilbert-type inequalities are discussed in [149, 150]. The composition formula of operators are given by [156].

In Corollary 5.3 of this book, Yang also gave the following multi-dimensional half-discrete inequalities.

Suppose that $$m,s\in\mathbf{N}$$, $$\alpha,\beta>0$$, $$\lambda_{1},\lambda _{2}\in\mathbf{R}$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$k_{\lambda }(x,y)$$ (≥0) is a finite homogeneous function of degree −λ in $$\mathbf {R}_{+}^{2}$$, $$k_{\lambda}(x,y)\frac{1}{y^{s-\lambda_{2}}}$$ is decreasing with respect to $$y>0$$, and strictly decreasing in an interval $$I\subset (1,\infty )$$, $$k(\lambda_{1})=\int_{0}^{\infty}k_{\lambda}(u,1)u^{\lambda _{1}-1}\, du\in\mathbf{R}_{+}$$,

$$K(\lambda_{1})= \biggl( \frac{\Gamma^{m}(\frac{1}{\alpha})}{\alpha ^{m-1}\Gamma(\frac{m}{\alpha})} \biggr) ^{\frac{1}{q}} \biggl( \frac {\Gamma ^{s}(\frac{1}{\beta})}{\beta^{s-1}\Gamma(\frac{s}{\beta})} \biggr) k(\lambda_{1}).$$

If $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$f(x),a(n)\geq0$$, $$0<\int_{\mathbf {R}_{+}^{m}}\|x\|_{\alpha}^{p(m-\lambda_{1})-m}f^{p}(x)\,dx<\infty$$, and $$0< \sum_{n\in\mathbf{N}^{s}}\|n\|_{\beta}^{q(s-\lambda _{2})-s}a^{q}(n)<\infty$$, then we have the following equivalent inequalities:

\begin{aligned}& \int_{\mathbf{R}_{+}^{m}}\sum_{n\in\mathbf{N}^{s}}k_{\lambda } \bigl(\Vert x\Vert _{\alpha},\|n\|_{\beta}\bigr)f(x)a(n)\,dx \\& \quad < K(\lambda_{1}) \biggl[ \int_{\mathbf{R}_{+}^{m}}\|x \|_{\alpha}^{p(m-\lambda _{1})-m}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \biggl[ \sum_{n\in\mathbf{N}^{s}}\|n\|_{\beta}^{q(s-\lambda_{2})-s}a^{q}(n) \biggr] ^{\frac{1}{q}}, \end{aligned}
(126)
\begin{aligned}& \sum_{n\in\mathbf{N}^{s}}\|n\|_{\beta}^{p\lambda_{2}-s} \biggl( \int_{\mathbf{R}_{+}^{m}}k_{\lambda}\bigl(\|x\|_{\alpha},\|n \|_{\beta }\bigr)f(x)\,dx \biggr) ^{p} \\& \quad < \bigl(K(\lambda_{1})\bigr)^{p}\int _{\mathbf{R}_{+}^{m}}\|x\|_{\alpha }^{p(m-\lambda_{1})-m}f^{p}(x) \,dx, \end{aligned}
(127)
\begin{aligned}& \int_{\mathbf{R}_{+}^{m}}\|x\|_{\alpha}^{q\lambda_{1}-m} \biggl( \sum _{n\in\mathbf{N}^{s}}k_{\lambda}\bigl(\|x\|_{\alpha},\|n \|_{\beta }\bigr)a(n) \biggr) ^{q}\,dx < \bigl(K(\lambda_{1})\bigr)^{q}\sum _{n\in\mathbf{N}^{s}}\|n\|_{\beta }^{q(s-\lambda_{2})-s}a^{q}(n), \end{aligned}
(128)

where the constant factors $$K(\lambda_{1})$$, $$(K(\lambda_{1}))^{p}$$, and $$(K(\lambda_{1}))^{q}$$ are the best possible.

For $$m=s=\alpha=\beta=1$$, (126), (127), and (128) reduce to the following equivalent half-discrete Yang-Hilbert-type inequalities with the best possible constant factors:

\begin{aligned}& \int_{0}^{\infty}\sum_{n=1}^{\infty}k_{\lambda}(x,n)f(x)a_{n} \,dx \\& \quad < k(\lambda_{1}) \biggl[ \int_{0}^{\infty}x^{p(1-\lambda _{1})-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\lambda _{2})-1}a_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}
(129)
\begin{aligned}& \sum_{n=1}^{\infty}n^{p\lambda_{2}-1} \biggl( \int_{0}^{\infty }k_{\lambda }(x,n)f(x)\,dx \biggr) ^{p}< \bigl(k(\lambda_{1})\bigr)^{p}\int _{0}^{\infty }x^{p(1-\lambda_{1})-1}f^{p}(x)\,dx, \end{aligned}
(130)
\begin{aligned}& \int_{0}^{\infty}x^{q\lambda_{1}-1} \Biggl( \sum _{n=1}^{\infty }k_{\lambda }(x,n)a_{n} \Biggr) ^{q}\,dx< \bigl(k(\lambda_{1})\bigr)^{q} \sum_{n=1}^{\infty }n^{q(1-\lambda_{2})-1}a_{n}^{q}. \end{aligned}
(131)

Also, for $$m=1$$, (124) and (125) reduce, respectively, to (129) and (130).

Replacing x to $$\frac{1}{x}$$, $$x^{\lambda-2}f(\frac{1}{x})$$ to $$f(x)$$ in (129), (130), and (131), we have the following equivalent inequalities with the non-homogeneous kernel and the best possible constant factors:

\begin{aligned}& \int_{0}^{\infty}\sum_{n=1}^{\infty}k_{\lambda}(1,xn)f(x)a_{n} \,dx \\& \quad < k(\lambda_{1}) \biggl[ \int_{0}^{\infty}x^{p(1-\lambda _{2})-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\lambda _{2})-1}a_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}
(132)
\begin{aligned}& \sum_{n=1}^{\infty}n^{p\lambda_{2}-1} \biggl( \int_{0}^{\infty }k_{\lambda }(1,xn)f(x)\,dx \biggr) ^{p}< \bigl(k(\lambda_{1})\bigr)^{p}\int _{0}^{\infty }x^{p(1-\lambda_{2})-1}f^{p}(x)\,dx, \end{aligned}
(133)
\begin{aligned}& \int_{0}^{\infty}x^{q\lambda_{2}-1} \Biggl( \sum _{n=1}^{\infty }k_{\lambda }(1,xn)a_{n} \Biggr) ^{q}\,dx< \bigl(k(\lambda_{1})\bigr)^{q} \sum_{n=1}^{\infty }n^{q(1-\lambda_{2})-1}a_{n}^{q}. \end{aligned}
(134)

The above half-discrete inequalities are some refinements of (37)-(40).

### 4.4 Some simple Hilbert-type inequalities

If the Hilbert-type inequality relates to a simple symmetric homogeneous kernel of degree −1 and the best constant factor is a more brief form, which does not relate to any conjugate exponents (such as (2)), then we call it simple Hilbert-type integral inequality. Its series analog (if it exists) is called a simple Hilbert-type inequality. If the simple homogeneous kernel is of −λ degree ($$\lambda>0$$) with a parameter λ and the inequality cannot be obtained by a simple transform to a simple Hilbert-type integral inequality, then we call it a simple Hilbert-type integral inequality with a parameter.

For example, we call the following integral inequality (i.e. (2)):

$$\int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \pi \biggl( \int_{0}^{\infty}f^{2}(x) \,dx\int_{0}^{\infty}g^{2}(y)\,dy \biggr) ^{\frac {1}{2}},$$
(135)

and the following H-L-P inequalities (for $$p=2$$ in (37) and (41)):

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{\max\{x,y\}}\,dx\,dy< 4 \biggl( \int_{0}^{\infty}f^{2}(x) \,dx\int_{0}^{\infty }g^{2}(y)\,dy \biggr) ^{\frac{1}{2}}, \end{aligned}
(136)
\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{\ln(\frac {x}{y})f(x)g(y)}{x-y}\,dx\,dy< \pi^{2} \biggl( \int _{0}^{\infty}f^{2}(x)\,dx\int _{0}^{\infty }g^{2}(y)\,dy \biggr) ^{\frac{1}{2}}, \end{aligned}
(137)

simple Hilbert-type integral inequalities.

In 2006, Yang [21] gave the following simple Hilbert-type integral inequality:

$$\int_{0}^{\infty}\int_{0}^{\infty} \frac{|\ln(\frac {x}{y})|f(x)g(y)}{\max \{x,y\}}\,dx\,dy< 8 \biggl( \int_{0}^{\infty}f^{2}(x) \,dx\int_{0}^{\infty }g^{2}(y)\,dy \biggr) ^{\frac{1}{2}}.$$
(138)

In 2008, Yang [23] and [157] gave the following simple Hilbert-type integral inequalities:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{|\ln(\frac {x}{y})|f(x)g(y)}{x+y}\,dx\,dy< c_{0} \biggl( \int _{0}^{\infty}f^{2}(x)\,dx\int _{0}^{\infty }g^{2}(y)\,dy \biggr) ^{\frac{1}{2}}, \end{aligned}
(139)
\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{\arctan\sqrt{\frac {x}{y}}}{x+y}f(x)g(y)\,dx\,dy< \frac{\pi^{2}}{4} \biggl( \int _{0}^{\infty }f^{2}(x)\,dx\int _{0}^{\infty}g^{2}(y)\,dy \biggr) ^{\frac{1}{2}}, \end{aligned}
(140)

where $$c_{0}=8\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n-1)^{2}}=7.3277^{+}$$.

In 2005, Yang [158, 159] gave a simple Hilbert-type integral inequality with a parameter $$\lambda\in(0,1)$$:

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{|x-y|^{\lambda}}\,dx\,dy \\& \quad < 2B\biggl(1-\lambda,\frac{\lambda}{2}\biggr) \biggl( \int _{0}^{\infty }x^{1-\lambda }f^{2}(x)\,dx\int _{0}^{\infty}y^{1-\lambda}g^{2}(y)\,dy \biggr) ^{\frac{1}{2}}. \end{aligned}
(141)

As in (16),

\begin{aligned}& \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{(x+y)^{\lambda}}\,dx\,dy \\& \quad < B\biggl(\frac{\lambda}{2},\frac{\lambda}{2}\biggr) \biggl( \int _{0}^{\infty }x^{1-\lambda}f^{2}(x)\,dx\int _{0}^{\infty}y^{1-\lambda }g^{2}(y)\,dy \biggr) ^{\frac{1}{2}} \end{aligned}
(142)

is called a simple Hilbert-type integral inequality with the parameter $$\lambda\in(0,\infty)$$.

Also we find the following simple Hilbert-type inequality:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=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) ^{\frac{1}{2}}, \end{aligned}
(143)
\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\max\{m,n\}}< 4 \Biggl( \sum_{m=1}^{\infty}a_{m}^{2} \sum_{n=1}^{\infty }b_{n}^{2} \Biggr) ^{\frac{1}{2}}, \end{aligned}
(144)
\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{\ln(\frac {m}{n})a_{m}b_{n}}{m-n}< \pi^{2} \Biggl( \sum_{m=1}^{\infty}a_{m}^{2} \sum_{n=1}^{\infty }b_{n}^{2} \Biggr) ^{\frac{1}{2}}. \end{aligned}
(145)

In (47), for $$p=q=2$$, we have

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{|\ln(\frac {m}{n})|a_{m}b_{n}}{\max\{m,n\}}< 8 \Biggl( \sum _{m=1}^{\infty}a_{m}^{2}\sum _{n=1}^{\infty }b_{n}^{2} \Biggr) ^{\frac{1}{2}}.$$
(146)

In 2010, Xin and Yang [24] gave

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{|\ln(\frac {m}{n})|a_{m}b_{n}}{m+n}< c_{0} \Biggl( \sum_{m=1}^{\infty}a_{m}^{2} \sum_{n=1}^{\infty }b_{n}^{2} \Biggr) ^{\frac{1}{2}},$$
(147)

where $$c_{0}=8\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n-1)^{2}}=7.3277^{+}$$. Inequalities (143) and (144) are new simple Hilbert-type inequalities. We still have a simple Hilbert-type inequality with a parameter $$\lambda\in(0,4]$$ as follows (cf. [51]):

$$\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac {a_{m}b_{n}}{(m+n)^{\lambda}}< B\biggl( \frac{\lambda}{2},\frac{\lambda}{2}\biggr) \Biggl( \sum _{m=1}^{\infty }m^{1-\lambda}a_{m}^{2} \sum_{n=1}^{\infty}n^{1-\lambda }b_{n}^{2} \Biggr) ^{\frac{1}{2}}.$$
(148)

Some simple half-discrete Hilbert-type inequalities are also listed as follows:

\begin{aligned}& \int_{0}^{\infty}\sum_{n=1}^{\infty} \frac{a_{n}}{x+n}f(x)\,dx< \pi \Biggl( \int_{0}^{\infty}f^{2}(x) \,dx\sum_{n=1}^{\infty}a_{n}^{2} \Biggr) ^{\frac {1}{2}}, \end{aligned}
(149)
\begin{aligned}& \int_{0}^{\infty}\sum_{n=1}^{\infty} \frac{a_{n}}{\max\{x,n\}}f(x)\,dx< 4 \Biggl( \int_{0}^{\infty}f^{2}(x) \,dx\sum_{n=1}^{\infty }a_{n}^{2} \Biggr) ^{\frac{1}{2}}, \end{aligned}
(150)
\begin{aligned}& \int_{0}^{\infty}\sum_{n=1}^{\infty} \frac{\ln(\frac{x}{n})a_{n}}{x-n} f(x)\,dx< \pi^{2} \Biggl( \int_{0}^{\infty}f^{2}(x) \,dx\sum_{n=1}^{\infty }a_{n}^{2} \Biggr) ^{\frac{1}{2}}, \end{aligned}
(151)
\begin{aligned}& \int_{0}^{\infty}\sum_{n=1}^{\infty} \frac{|\ln(\frac {x}{n})|a_{n}}{\max \{x,n\}}f(x)\,dx< 8 \Biggl( \int_{0}^{\infty}f^{2}(x) \,dx\sum_{n=1}^{\infty }a_{n}^{2} \Biggr) ^{\frac{1}{2}}, \end{aligned}
(152)
\begin{aligned}& \int_{0}^{\infty}\sum_{n=1}^{\infty} \frac{|\ln(\frac {x}{n})|a_{n}}{x+n}f(x)\,dx< c_{0} \Biggl( \int _{0}^{\infty}f^{2}(x)\,dx\sum _{n=1}^{\infty }a_{n}^{2} \Biggr) ^{\frac{1}{2}}. \end{aligned}
(153)