# On a more accurate half-discrete multidimensional Hilbert-type inequality involving one derivative function of m-order

## Abstract

By means of the weight functions, the idea of introduced parameters, using the transfer formula and Hermite–Hadamard’s inequality, a more accurate half-discrete multidimensional Hilbert-type inequality with the homogeneous kernel as $$\frac{1}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}}\ (x,\lambda > 0)$$ involving one derivative function of m-order is given. The equivalent conditions of the best possible constant factor related to several parameters are considered. The equivalent forms. the operator expressions and some particular inequalities are obtained.

## 1 Introduction

Suppose that $$p > 1,\frac{1}{p} + \frac{1}{q} = 1,a_{m},b_{n} \ge 0,0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty$$ and $$0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty$$. We have the following discrete Hardy–Hilbert’s inequality with the best possible constant factor $$\pi /\sin (\frac{\pi}{ p})$$ (cf. , Theorem 315):

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

The integral analogues of (1) named in Hardy–Hilbert’s integral inequality was provided as follows (cf. , Theorem 316):

\begin{aligned} \int _{0}^{\infty} \int _{0}^{\infty} \frac{f(x)g(y)}{x + y} \,dx,\,dy < \frac{\pi}{\sin (\pi /p)} \biggl( \int _{0}^{\infty} f^{p}(x)\,dx \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{\infty} g^{q}(y)\,dy \biggr)^{\frac{1}{q}}, \end{aligned}
(2)

with the same best possible factor. The more accurate form of (1) was given as follows (cf. , Theorem 323):

\begin{aligned} \sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b_{n}}{m + n - 1} < \frac{\pi}{\sin (\pi /p)} \Biggl( \sum_{m = 1}^{\infty} a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 1}^{\infty} b_{n}^{q} \Biggr)^{\frac{1}{q}}. \end{aligned}
(3)

Inequalities (1)–(3) with their extensions played an important role in analysis and its applications (cf. ).

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

\begin{aligned} \sum_{n = 1}^{\infty} n^{p - 2}\biggl( \int _{0}^{\infty} K (nx)f(x)\,dx\biggr)^{p} < \phi ^{p}\biggl(\frac{1}{q}\biggr) \int _{0}^{\infty} f^{p} (x)\,dx. \end{aligned}
(4)

Some new extensions of (3) were given by .

In 2006, by using Euler–Maclaurin summation formula, Krnic et al.  gave an extension of (1) with the kernel as $$\frac{1}{(m + n)^{\lambda}}\ (0 < \lambda \le 4)$$. In 2019–2020, following the results of , Adiyasuren et al.  provided an extension of (1) involving partial sums, and Mo et al.  gave an extension of (2) involving the upper limit functions, which a new application of the way in . In 2016–2017, Hong et al. [23, 24] considered some equivalent statements of the extensions of (1) and (2) with a few parameters. Some further results were provided by . In 2023, Hong et. al.  published a more accurate half-discrete multidimensional Hilbert-type inequality involving one multiple upper limit function.

In this paper, following the way of , by means of the weight functions, the idea of introduced parameters, using the transfer formula and Hermite–Hadamard’s inequality, a more accurate half-discrete multidimensional Hilbert-type inequality with the homogeneous kernel as $$\frac{1}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}}\ (x,\lambda > 0)$$ involving one derivative function of m-order and the beta function is given. The equivalent conditions of the best possible constant factor related to several parameters are considered. The equivalent forms, the operator expressions and some particular inequalities are obtained. Our new work is different to , which is involving one higher-order derivative function but not involving one multiple upper limit function.

## 2 Some formulas and lemmas

Hereinafter in this paper, we suppose that $$p > 1,\frac{1}{p} + \frac{1}{q} = 1,\lambda > 0,\lambda _{1},\lambda _{2} \in (0,\lambda ), m,n \in \mathrm{N} = \{ 1,2, \ldots \}, \alpha \in (0,1],\xi \in [0,\frac{1}{2}],\hat{\lambda}_{1}: = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q},\hat{\lambda}_{2}: = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}, \Vert y\Vert _{\alpha}: = (\sum_{k = 1}^{n} |y_{i}|^{\alpha} )^{\frac{1}{\alpha}} (y = (y_{1}, \ldots,y_{n}) \in \mathrm{R}^{\mathrm{n}})$$. We also assume that $$f(x)\ ( \ge 0)$$ is a differentiable function of m-order unless at finite points in $$\mathrm{R}_{ +} = (0,\infty )$$,

\begin{aligned} &f^{(k - 1)}(x) = o\bigl(e^{tx}\bigr)\quad (t > 0;x \to \infty ),\qquad f^{(k - 1)}\bigl(0^{ +} \bigr) = 0\quad (k = 1, \ldots,m),\\ &f^{(m)}(x),\ a_{k} = (a_{k_{1}}, \ldots,a_{k_{n}}) \ge 0\ \bigl(x \in \mathrm{R}_{ +} = (0,\infty ),\ k = (k_{1}, \ldots,k_{n}) \in \mathrm{N}^{\mathrm{n}} \bigr), \text{ such that}\\ &\quad 0 < \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx < \infty\quad \text{and} \quad 0 < \sum _{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n} a_{k}^{q} < \infty. \end{aligned}

For $$M > 0,\psi (u)\ (u > 0)$$ is a nonnegative measurable function, we have the following transfer formula (cf. , (9.3.3)):

\begin{aligned} &\int \cdots \int _{\{ y \in R_{ +}^{n};0 < \sum _{i = 1}^{n} (\frac{y_{i}}{M})^{\alpha} \le 1\}} \psi \Biggl(\sum_{i = 1}^{n} \biggl(\frac{y_{i}}{M}\biggr)^{\alpha} \Biggr) \,dy_{1} \cdots \,dy_{n} \\ &\quad= \frac{M^{n}\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n}\Gamma (\frac{n}{\alpha} )} \int _{0}^{1} \psi (u)u^{\frac{n}{\alpha} - 1} \,du. \end{aligned}
(5)

In particular, (i) in view of $$\Vert y\Vert _{\alpha} = M[\sum_{i = 1}^{n} (\frac{y_{i}}{M})^{\alpha} ]^{\frac{1}{\alpha}}$$, by (5), we have

\begin{aligned} &\int _{R_{ +}^{n}} \phi \bigl( \Vert y \Vert _{\alpha} \bigr)\,dy \\ &\quad= \lim_{M \to \infty} \int \cdots \int _{\{ y \in R_{ +}^{n};0 < \sum _{i = 1}^{n} (\frac{y_{i}}{M})^{\alpha} \le 1\}} \phi \Biggl(M\Biggl[\sum _{i = 1}^{n} \biggl(\frac{y_{i}}{M} \biggr)^{\alpha} \Biggr]^{\frac{1}{\alpha}} \Biggr) \,dy_{1} \cdots \,dy_{n} \\ &\quad= \lim_{M \to \infty} \frac{M^{n}\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n}\Gamma (\frac{n}{\alpha} )} \int _{0}^{1} \phi \bigl(Mu^{\frac{1}{\alpha}} \bigr)u^{\frac{n}{\alpha} - 1} \,du\mathop{ =} ^{v = Mu^{\frac{1}{\alpha}}} \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \int _{0}^{\infty} \phi (v)v^{n - 1} \,dv; \end{aligned}
(6)

(ii) for $$\psi (u) = \phi (Mu^{\frac{1}{\alpha}} ) = 0.u < \frac{b^{\alpha}}{ M^{\alpha}} (b > 0)$$, by (5), we have

\begin{aligned} \int _{\{ y \in R_{ +}^{n}, \Vert y \Vert _{\alpha} \ge b\}} \phi \bigl( \Vert y \Vert _{\alpha} \bigr)\,dy &=\lim_{M \to \infty} \frac{M^{n}\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n}\Gamma (\frac{n}{\alpha} )} \int _{\frac{b^{\alpha}}{M^{\alpha}}}^{1} \phi \bigl(Mu^{\frac{1}{\alpha}} \bigr)u^{\frac{n}{\alpha} - 1} \,du \\ & =\frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \int _{b}^{\infty} \phi (v)v^{n - 1} \,dv. \end{aligned}
(7)

### Lemma 1

For $$s > 0,\alpha \in (0,1],\xi \in [0,\frac{1}{2}],A_{\xi}: = \{ y = \{ y_{1}, \ldots,y_{n}\};y_{i} > \xi\ (i = 1, \ldots,n)\}$$, define the following function:

\begin{aligned} g_{x}(y): = \frac{1}{(x + \Vert y - \xi \Vert _{\alpha} )^{s}} = \frac{1}{\{ x + [\sum_{i = 1}^{n} ( y_{i} - \xi )^{\alpha} ]^{1/\alpha} \}^{s}}\quad \bigl(x > 0,y = (y_{1}, \ldots,y_{n}) \in A_{\xi} \bigr). \end{aligned}

Then we have $$\frac{\partial}{\partial y_{j}}g_{x}(y) < 0,\frac{\partial ^{2}}{\partial y_{j}^{2}}g_{x}(y) > 0\ (y \in A_{\xi};j = 1, \ldots,n)$$.

### Proof

We obtain that for $$s > 0,\alpha \in (0,1],\xi \in [0,\frac{1}{2}],y \in A_{\xi}$$,

\begin{aligned} & \frac{\partial}{\partial y_{j}}g_{x}(y) = \frac{ - s[\sum_{i = 1}^{n} ( y_{i} - \xi )^{\alpha} ]^{\frac{1}{\alpha} - 1}(y_{j} - \xi )^{\alpha - 1}}{\{ x + [\sum_{i = 1}^{n} ( y_{i} - \xi )^{\alpha} ]^{1/\alpha} \}^{s + 1}} < 0, \\ &\frac{\partial ^{2}}{\partial y_{j}^{2}}g_{x}(y) = \frac{s(s + 1)[\sum_{i = 1}^{n} ( y_{i} - \xi )^{\alpha} ]^{\frac{2}{\alpha} - 2}(y_{j} - \xi )^{2\alpha - 2}}{\{ x + [\sum_{i = 1}^{n} ( y_{i} - \xi )^{\alpha} ]^{1/\alpha} \}^{s + 2}} \\ &\phantom{\frac{\partial ^{2}}{\partial y_{j}^{2}}g_{x}(y) =}{}+ \frac{s(1 - \alpha )[\sum_{i = 1}^{n} ( y_{i} - \xi )^{\alpha} ]^{\frac{1}{\alpha} - 2}(y_{j} - \xi )^{\alpha - 2}}{\{ x + [\sum_{i = 1}^{n} ( y_{i} - \xi )^{\alpha} ]^{1/\alpha} \}^{s + 1}} \Biggl[ \sum_{i = 1}^{n} (y_{i} - \xi )^{\alpha} - (y_{j} - \xi )^{\alpha} \Biggr] > 0. \end{aligned}

The lemma is proved. □

Note. In the same way, for $$s_{2} \le n,\alpha \in (0,1],\xi \in [0,\frac{1}{2}],y \in A_{\xi}$$, we can prove that

\begin{aligned} \frac{\partial}{\partial y_{j}} \Vert y - \xi \Vert _{\alpha}^{s_{2} - n} \le 0, \frac{\partial ^{2}}{\partial y_{j}^{2}} \Vert y - \xi \Vert _{\alpha}^{s_{2} - n} \ge 0\quad (j = 1, \ldots,n), \end{aligned}
(8)

and then for $$s_{2} \le n,\alpha \in (0,1],\xi \in [0,\frac{1}{2}], h_{x}(y): = g_{x}(y)\Vert y - \xi \Vert _{\alpha}^{s_{2} - n}\ (x > 0,y \in A_{\xi} )$$, by Lemma 1, we obtain

\begin{aligned} &\frac{\partial}{\partial y_{j}}h_{x}(y) = \Vert y - \xi \Vert _{\alpha}^{s_{2} - n}\frac{\partial}{\partial y_{j}}g_{x}(y) + g_{x}(y)\frac{\partial}{ \partial y_{j}} \Vert y - \xi \Vert _{\alpha}^{s_{2} - n} < 0, \\ &\frac{\partial ^{2}}{\partial y_{j}^{2}}h_{x}(y) = \frac{\partial}{\partial y_{j}} \Vert y - \xi \Vert _{\alpha}^{s_{2} - n}\frac{\partial}{\partial y_{j}}g_{x}(y) + \Vert y - \xi \Vert _{\alpha}^{s_{2} - n}\frac{\partial ^{2}}{\partial y_{j}^{2}}g_{x}(y) \\ &\phantom{\frac{\partial ^{2}}{\partial y_{j}^{2}}h_{x}(y) = }{}+ \frac{\partial}{\partial y_{j}}g_{x}(y)\frac{\partial}{\partial y_{j}} \Vert y - \xi \Vert _{\alpha}^{s_{2} - n} + g_{x}(y)\frac{\partial ^{2}}{\partial y^{2}} \Vert y - \xi \Vert _{\alpha}^{s_{2} - n} > 0,\quad (j = 1, \ldots,n). \end{aligned}
(9)

### Lemma 2

For $$c > 0$$, we have the following inequalities:

\begin{aligned} \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{c\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} < \sum_{k} \Vert k \Vert _{\alpha}^{ - c - n} < \frac{2^{c}\Gamma ^{n}(\frac{1}{\alpha} )}{c\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}, \end{aligned}
(10)

where, $$\sum_{k} G(k) = \sum_{k_{n} = 1}^{\infty} \cdots \sum_{k_{1} = 1}^{\infty} G(k_{1}, \ldots,k_{n})$$.

### Proof

By (8) (for $$\xi = 0$$), in view of $$- c - n < 0$$, we find that

\begin{aligned} \frac{\partial}{\partial y_{j}} \Vert y \Vert _{\alpha}^{ - c - n} < 0, \frac{\partial ^{2}}{\partial y_{j}^{2}} \Vert y \Vert _{\alpha}^{ - c - n} > 0\quad (j = 1, \ldots,n), \end{aligned}

and then by Hermite–Hadamard’s inequality (cf. ) and (7), we have

\begin{aligned} \sum_{k} \Vert k \Vert _{\alpha}^{ - c - n} < \int _{\{ y \in R_{ +}^{n}, \Vert y \Vert _{\alpha} \ge \frac{1}{2}\}} \Vert y \Vert _{\alpha}^{ - c - n} \,dy =\frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \int _{\frac{1}{2}}^{\infty} v^{ - c - n}v^{n - 1} \,dv = \frac{2^{c}\Gamma ^{n}(\frac{1}{\alpha} )}{c\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}. \end{aligned}

By the decreasingness property of series and (7), it follows that

\begin{aligned} \sum_{k} \Vert k \Vert _{\alpha}^{ - c - n} > \int _{\{ y \in R_{ +}^{n}, \Vert y \Vert _{\alpha} \ge 1\}} \Vert y \Vert _{\alpha}^{ - c - n} \,dy =\frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \int _{1}^{\infty} v^{ - c - 1} \,dv = \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{c\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}, \end{aligned}

namely, inequalities (10) follow.

The lemma is proved. □

### Lemma 3

For $$s > 0$$, we fine the following weight functions:

\begin{aligned} &\varpi _{s}(s_{2},x): = x^{s - s_{2}}\sum _{k} \frac{ \Vert k - \xi \Vert _{\alpha}^{s_{2} - n}}{(x + \Vert k - \xi \Vert _{\alpha} )^{s}} \quad (x \in \mathrm{R}_{ +} ), \end{aligned}
(11)
\begin{aligned} &\omega _{s}(s_{1},k): = \Vert k - \xi \Vert _{\alpha}^{s - s_{1}} \int _{0}^{\infty} \frac{x^{s_{1} - 1}}{(x + \Vert k - \xi \Vert _{\alpha} )^{s}} \,dx\quad \bigl(k \in \mathrm{N}^{\mathrm{n}}\bigr), \end{aligned}
(12)
1. (i)

For $$0 < s_{2} < s, s_{2} \le n$$, we have the following inequality:

\begin{aligned} \varpi _{s}(s_{2},x) < \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(s_{2},s - s_{2}) \quad(x \in \mathrm{R}_{ +} ); \end{aligned}
(13)
2. (ii)

for $$0 < s_{1} < s$$, the following expression follows:

\begin{aligned} \omega _{s}(s_{1},k) = B(s_{1},s - s_{1})\quad \bigl(y \in \mathrm{R}_{ +}^{\mathrm{n}}\bigr), \end{aligned}
(14)

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

### Proof

(i) For $$0 < s_{2} < s, s_{2} \le n$$, by (9) and Hermite–Hadamard’s inequality, we have

\begin{aligned} \varpi _{s}(s_{2},x) &< x^{s - s_{2}} \int _{A_{1/2}} \frac{ \Vert y - \xi \Vert _{\alpha}^{s_{2} - n}}{(x + \Vert y - \xi \Vert _{\alpha} )^{s}} \,dy \le x^{s - s_{2}} \int _{A_{\xi}} \frac{ \Vert y - \xi \Vert _{\alpha}^{s_{2} - n}}{(x + \Vert y - \xi \Vert _{\alpha} )^{s}} \,dy \\ &= x^{s - s_{2}} \int _{R_{ +}^{n}} \frac{ \Vert y \Vert _{\alpha}^{s_{2} - n}}{(x + \Vert y \Vert _{\alpha} )^{s}} \,dy. \end{aligned}

Setting $$\phi (v): = \frac{v^{s_{2} - n}}{(x + v)^{s}}$$, by (6), it follows that

\begin{aligned} \varpi _{s}(s_{2},x) &< x^{s - s_{2}} \int _{R_{ +}^{n}} \phi \bigl( \Vert y \Vert _{\alpha} \bigr) \,dy = x^{s - s_{2}}\frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \int _{0}^{\infty} \phi (v)v^{n - 1} \,dv\\ & = x^{s - s_{2}}\frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \int _{0}^{\infty} \frac{v^{s_{2} - 1}}{(x + v)^{s}} \,dv\mathop{ =} ^{u = v/x}\frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \int _{0}^{\infty} \frac{u^{s_{2} - 1}}{(1 + u)^{s}} \,du \\ &= \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(s_{2},s - s_{2}), \end{aligned}

namely, (13) follows.

(ii) Setting $$u = \frac{x}{\Vert k - \xi \Vert _{\alpha}}$$ in (12), we find

\begin{aligned} \omega _{s}(s_{1},k) &= \Vert k - \xi \Vert _{\alpha}^{s - s_{1}} \int _{0}^{\infty} \frac{(u \Vert k - \xi \Vert _{\alpha} )^{s_{1} - 1} \Vert k - \xi \Vert _{\alpha}}{(u \Vert k - \xi \Vert _{\alpha} + \Vert k - \xi \Vert _{\alpha} )^{s}} \,du\\ & = \int _{0}^{\infty} \frac{u^{s_{1} - 1}}{(u + 1)^{s}} \,du = B(s_{1},s - s_{1}), \end{aligned}

and then we have (17).

The lemma is proved. □

We indicate the following gamma function (cf. ): $$\Gamma (\alpha ): = \int _{0}^{\infty} e^{ - t} t^{\alpha - 1}\,dt\ (\alpha > 0)$$, satisfying $$\Gamma (\alpha + 1) = \alpha \Gamma (\alpha )$$ ($$\alpha > 0$$) and $$B(u,v) = \frac{1}{\Gamma (u + v)}\Gamma (u)\Gamma (v)\ (u,v > 0)$$. By the definition of the gamma function, for $$\lambda,x > 0$$, the following expression holds:

\begin{aligned} \frac{1}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} = \frac{1}{\Gamma (\lambda + m)} \int _{0}^{\infty} t^{\lambda + m - 1} e^{ - (x + \Vert k - \xi \Vert _{\alpha} )t} \,dt. \end{aligned}
(15)

### Lemma 4

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

\begin{aligned} \int _{0}^{\infty} e^{ - tx} f(x)\,dx = t^{ - m} \int _{0}^{\infty} e^{ - tx} f^{(m)}(x) \,dx. \end{aligned}
(16)

### Proof

Since $$f(0) = 0, f(x) = o(e^{tx})\ (t > 0;x \to \infty )$$, we find

\begin{aligned} \int _{0}^{\infty} e^{ - tx} f'(x) \,dx &= \int _{0}^{\infty} e^{ - tx} \,df(x)= e^{ - tx}f(x)|_{0}^{\infty} - \int _{0}^{\infty} f (x)\,de^{ - tx} \\ &= \lim_{x \to \infty} \frac{f(x)}{e^{tx}} + t \int _{0}^{\infty} e^{ - tx} f(x)\,dx = t \int _{0}^{\infty} e^{ - tx} f(x)\,dx. \end{aligned}

Inductively, for $$f^{(i)}(0^{ +} ) = 0,f^{(i)}(x) = o(e^{tx})$$ ($$t > 0,i = 1, \ldots,m;x \to \infty$$), we still have

\begin{aligned} \int _{0}^{\infty} e^{ - tx} f(x)\,dx = t^{ - 1} \int _{0}^{\infty} e^{ - tx} f'(x) \,dx = \cdots = t^{ - i} \int _{0}^{\infty} e^{ - tx} f^{(i)}(x) \,dx, \end{aligned}

namely, expression (17) follows.

The lemma is proved. □

### Lemma 5

We have the following inequality:

\begin{aligned} I_{\lambda}: ={}& \sum_{k} \int _{0}^{\infty} \frac{f^{(m)}(x)a_{k}}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda}} \,dx \\ < {}& \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1}) \\ &{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \biggl[ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n} a_{k}^{q} \biggr]^{\frac{1}{q}}. \end{aligned}
(17)

### Proof

By Hölder’s inequality (cf. ), and Lebesgue term by term integration theorem (cf. ), we obtain

\begin{aligned} I_{\lambda} = {}&\sum_{k} \int _{0}^{\infty} \frac{1}{(x + \Vert k - \xi \Vert _{\alpha} |)^{\lambda}} \biggl[ \frac{ \Vert k - \xi \Vert _{\alpha}^{(\lambda _{2} - n)/p}}{x^{(\lambda _{1} - 1)/q}}f^{(m)}(x) \biggr] \biggl[ \frac{x^{(\lambda _{1} - 1)/q}}{ \Vert k - \xi \Vert _{\alpha}^{(\lambda _{2} - n)/p}}a_{k} \biggr]\,dx \\ \le{}& \biggl\{ \int _{0}^{\infty} \biggl[ \sum _{k} \frac{1}{(x + \Vert k - \xi \Vert _{\alpha} |)^{\lambda}} \frac{ \Vert k - \xi \Vert _{\alpha}^{\lambda _{2} - n}}{x^{(\lambda _{1} - 1)(p - 1)}} \biggr] \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr\} ^{\frac{1}{p}} \\ &{}\times \biggl\{ \sum_{k} \biggl[ \int _{0}^{\infty} \frac{1}{(x + \Vert k - \xi \Vert _{\alpha} |)^{\lambda}} \frac{x^{\lambda _{1} - 1}}{ \Vert k - \xi \Vert _{\alpha}^{(\lambda _{2} - n)(q - 1)}} \,dx \biggr] a_{k}^{q} \biggr\} ^{\frac{1}{q}} \\ ={}& \biggl[ \int _{0}^{\infty} \varpi _{\lambda} (\lambda _{2},x) x^{p(1 - \hat{\lambda}_{1}) - 1}\bigl(f^{(m)}(x)\bigr)^{p} \,dx \biggr]^{\frac{1}{p}}\\ &{}\times \biggl[ \sum_{k} \omega _{\lambda} ( \lambda _{1},k) \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n}a_{k}^{q} \biggr]^{\frac{1}{q}}. \end{aligned}

Therefore, by (13) and (15) (for $$s = \lambda,s_{1} = \lambda _{1},s_{2} = \lambda _{2}$$), we have (18).

The lemma is proved. □

## 3 Main results

### Theorem 1

We have the following more accurate half-discrete multidimensional Hilbert-type inequality involving one derivative function of m-order:

\begin{aligned} I: ={}& \sum_{k} \int _{0}^{\infty} \frac{f(x)a_{k}}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} \,dx \\ &{}< \Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2}, \lambda - \lambda _{2})\biggr) )^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1}) \\ &{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \biggl[ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n} a_{k}^{q} \biggr]^{\frac{1}{q}}. \end{aligned}
(18)

In particular, for $$\lambda _{1} + \lambda _{2} = \lambda$$, we reduce (19) to the following:

\begin{aligned} I = {}&\sum_{k} \int _{0}^{\infty} \frac{f(x)a_{k}}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} \,dx \\ < {}&\biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}} \Biggl[ \prod _{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1}B( \lambda _{1},\lambda _{2}) \\ &{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \biggl[ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \lambda _{2}) - n} a_{k}^{q} \biggr]^{\frac{1}{q}}. \end{aligned}
(19)

where, the constant factor $$( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} )^{\frac{1}{p}}[\prod_{i = 0}^{m - 1} (\lambda + i) ]^{ - 1}B(\lambda _{1},\lambda _{2})$$ is the best possible.

### Proof

Using (16) and (17), in view of Lebesgue term by term integration theorem (cf. ), we find

\begin{aligned} I &= \frac{1}{\Gamma (\lambda + m)}\sum_{k} \int _{0}^{\infty} f(x)a_{k} \biggl[ \int _{0}^{\infty} t^{\lambda + m - 1} e^{ - (x + \Vert k - \xi \Vert _{\alpha} )t}\,dt \biggr]\,dx\\ &= \frac{1}{\Gamma (\lambda + m)} \int _{0}^{\infty} t^{\lambda + m - 1} \biggl( \int _{0}^{\infty} e^{ - xt}f(x)\,dx \biggr) \biggl( \sum_{k} e^{ - \Vert k - \xi \Vert _{\alpha} t}a_{k} \biggr)\,dt \\ &= \frac{1}{\Gamma (\lambda + m)} \int _{0}^{\infty} t^{\lambda + m - 1} \biggl( t^{ - m} \int _{0}^{\infty} e^{ - xt}f^{(m)}(x) \,dx \biggr) \biggl( \sum_{k} e^{ - \Vert k - \xi \Vert _{\alpha} t} a_{k} \biggr)\,dt \\ &= \frac{1}{\Gamma (\lambda + m)}\sum_{k} \int _{0}^{\infty} f^{(m)}(x)a_{k} \biggl[ \int _{0}^{\infty} t^{\lambda - 1} e^{ - (x + \Vert k - \xi \Vert _{\alpha} )t}\,dt \biggr]\,dx\\ &= \frac{\Gamma (\lambda )}{\Gamma (\lambda + m)}\sum_{k} \int _{0}^{\infty} \frac{f^{(m)}(x)a_{k}}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda}} \,dx = \Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i)\Biggr] ]^{ - 1}I_{\lambda}. \end{aligned}

Then by (18), we have (19).

For $$\lambda _{1} + \lambda _{2} = \lambda$$ in (19), we have (20). For any $$0 < \varepsilon < p\lambda _{1}$$, we set $$\tilde{a}_{k}: = \Vert k\Vert _{\alpha}^{\lambda _{2} - \frac{\varepsilon}{q} - n}(k \in N^{n}), \tilde{f}(x): = 0,0 < x < 1$$,

\begin{aligned} \tilde{f}(x)&: = \prod_{i = 0}^{m - 1} \biggl(\lambda _{1} + i - \frac{\varepsilon}{p}\biggr) \int _{1}^{x} \biggl( \int _{1}^{t_{m}} \cdots \int _{1}^{t_{2}} t_{1}^{\lambda _{1} - \frac{\varepsilon}{p} - 1} \,dt_{1} \cdots \,dt_{m - 1} \biggr) \,dt_{m} \\ &= x^{\lambda _{1} - \frac{\varepsilon}{p} + m - 1} - O\bigl(x^{m - 1}\bigr),\quad x \ge 1, \end{aligned}

where, for $$m \in \mathrm{N}, O(x^{m - 1})$$ is indicated a nonnegative polynomial of ($$m - 1$$)-order.

If there exists a positive constant $$M( \le (\frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )})^{\frac{1}{p}}[\prod_{i = 0}^{m - 1} (\lambda + i)]^{ - 1} B(\lambda _{1},\lambda _{2}))$$, such that (20) is valid when we replace $$( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} )^{\frac{1}{p}}[\prod_{i = 0}^{m - 1} (\lambda + i) ]^{ - 1}B(\lambda _{1},\lambda _{2})$$ by M, then in particular, for $$\xi = 0$$, we still have

\begin{aligned} \tilde{I}: ={}& \sum_{k} \int _{0}^{\infty} \frac{\tilde{f}(x)\tilde{a}_{k}}{(x + \Vert k \Vert _{\alpha} )^{\lambda + m}} \,dx < M \biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{1}) - 1} \bigl( \tilde{f}^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \\ &{}\times\biggl[ \sum_{k} \Vert k \Vert _{\alpha}^{q(n - \lambda _{2}) - n} \tilde{a}_{k}^{q} \biggr]^{\frac{1}{q}}. \end{aligned}
(20)

By (10), we obtain

\begin{aligned} \tilde{J}&: = \biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{1}) - 1} \bigl( \tilde{f}^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \biggl[ \sum_{k} \Vert k \Vert _{\alpha}^{q(n - \lambda _{2}) - n} \tilde{a}_{k}^{q} \biggr]^{\frac{1}{q}} \\ &< \prod_{i = 0}^{m - 1} \biggl(\lambda _{1} + i - \frac{\varepsilon}{p}\biggr) \biggl( \int _{1}^{\infty} x^{ - \varepsilon - 1} \,dx \biggr)^{\frac{1}{p}} \biggl(\sum_{k} \Vert k \Vert _{\alpha}^{ - \varepsilon - n} \biggr)^{\frac{1}{q}} \\ &= \frac{1}{\varepsilon} \prod_{i = 0}^{m - 1} \biggl(\lambda _{1} + i - \frac{\varepsilon}{p}\biggr) \biggl( \frac{2^{\varepsilon} \Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{q}}. \end{aligned}
(21)

We obtain

\begin{aligned} \tilde{I} = \sum_{k} \int _{1}^{\infty} \frac{(x^{\lambda _{1} - \frac{\varepsilon}{p} + m - 1} - O(x^{m - 1}))}{(x + \Vert k \Vert _{\alpha} )^{\lambda + m}} \Vert k \Vert _{\mathrm{A}}^{\lambda _{2} - \frac{\varepsilon}{q} - n}\,dx = I_{0} - I_{1}, \end{aligned}

where, $$I_{0}: = \sum_{k} \int _{1}^{\infty} \frac{x^{\lambda _{1} - \frac{\varepsilon}{p} + m - 1}}{(x + \Vert k\Vert _{\alpha} )^{\lambda + m}} \Vert k\Vert _{\mathrm{A}}^{\lambda _{2} - \frac{\varepsilon}{q} - n}\,dx, I_{1}: = \sum_{k} \int _{1}^{\infty} \frac{O(x^{m - 1})}{(x + \Vert k\Vert _{\alpha} )^{\lambda + m}} \Vert k\Vert _{\mathrm{A}}^{\lambda _{2} - \frac{\varepsilon}{q} - n}\,dx$$.

By (10), we also find that $$\frac{1}{c - \varepsilon} \sum_{k} \Vert k\Vert _{\alpha}^{ - c - n} = O(1)\ (c = \lambda _{1} + m + \frac{\varepsilon}{ q})$$. For $$s = \lambda + m > 0,s_{1} = \lambda _{1} + m - \frac{\varepsilon}{p} \in (0,s)$$ in (12) and (15), by (10), we find

\begin{aligned} I_{0} &= \sum_{k} \Vert k \Vert _{\alpha}^{ - \varepsilon - n} \biggl[ \Vert k \Vert _{\alpha}^{(\lambda _{2} + \frac{\varepsilon}{p})} \int _{1}^{\infty} \frac{x^{(\lambda _{1} + m - \frac{\varepsilon}{p}) - 1}}{(x + \Vert k \Vert _{\alpha} )^{\lambda + m}} \,dx \biggr] \\ &= \sum_{k} \Vert k \Vert _{\alpha}^{ - \varepsilon - n} \biggl[ \Vert k \Vert _{\alpha}^{(\lambda _{2} + \frac{\varepsilon}{p})} \int _{0}^{\infty} \frac{x^{(\lambda _{1} + m - \frac{\varepsilon}{p}) - 1}}{(x + \Vert k \Vert _{\alpha} )^{\lambda + m}} \,dx - \Vert k \Vert _{\alpha}^{(\lambda _{2} + \frac{\varepsilon}{p})} \int _{0}^{1} \frac{x^{(\lambda _{1} + m - \frac{\varepsilon}{p}) - 1}}{(x + \Vert k \Vert _{\alpha} )^{\lambda + m}} \,dx \biggr] \\ &\ge \sum_{k} \Vert k \Vert _{\alpha}^{ - \varepsilon - n} \biggl[ \omega _{\lambda + m}\biggl(\lambda _{1} + m - \frac{\varepsilon}{p},k\biggr) - \Vert k \Vert _{\alpha}^{(\lambda _{2} + \frac{\varepsilon}{p})} \int _{0}^{1} \frac{x^{(\lambda _{1} + m - \frac{\varepsilon}{p}) - 1}}{ \Vert k \Vert _{\alpha}^{\lambda + m}} \,dx \biggr] \\ &= \sum_{k} \Vert k \Vert _{\alpha}^{ - \varepsilon - n} \omega _{\lambda + m}\biggl(\lambda _{1} + m - \frac{\varepsilon}{p},k \biggr) - \frac{1}{\lambda _{1} + m - \frac{\varepsilon}{p}}\sum_{k} \Vert k \Vert _{\alpha}^{ - (\lambda _{1} + m + \frac{\varepsilon}{q}) - n} \\ & = B\biggl(\lambda _{1} + m - \frac{\varepsilon}{p},\lambda _{2} + \frac{\varepsilon}{p}\biggr)\sum_{k} \Vert k \Vert _{\alpha}^{ - \varepsilon - n} - O(1)\\ &> \frac{1}{\varepsilon} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B\biggl(\lambda _{1} + m - \frac{\varepsilon}{p},\lambda _{2} + \frac{\varepsilon}{p}\biggr) - \varepsilon O(1) \biggr). \end{aligned}

We still find that

\begin{aligned} 0 &< I_{1} = \sum _{k} \frac{ \Vert k \Vert _{\mathrm{A}}^{\lambda _{2} - \frac{\varepsilon}{q} - n}}{(x + \Vert k \Vert _{\alpha} )^{\lambda _{2}}} \int _{1}^{\infty} \frac{O(x^{m - 1})}{(x + \Vert k \Vert _{\alpha} )^{\lambda _{1} + m}} \,dx \\ &\le \sum_{k} \frac{ \Vert k \Vert _{\mathrm{A}}^{\lambda _{2} - \frac{\varepsilon}{q} - n}}{ \Vert k \Vert _{\alpha}^{\lambda _{2}}} \int _{1}^{\infty} \frac{O(x^{m - 1})}{x^{\lambda _{1} + m}} \,dx \le M_{1} < \infty. \end{aligned}

Hence, by (21) and the above results, we have the following inequality

\begin{aligned} &\frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B\biggl(\lambda _{1} + m - \frac{\varepsilon}{p}, \lambda _{2} + \frac{\varepsilon}{p}\biggr) - \varepsilon O(1) - \varepsilon I_{1} \\ &\quad < \varepsilon \tilde{I} < \varepsilon M\tilde{J}\le M\prod _{i = 0}^{m - 1} \biggl(\lambda _{1} + i - \frac{\varepsilon}{p}\biggr) \biggl( \frac{2^{\varepsilon} \Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{q}}. \end{aligned}
(22)

For $$\varepsilon \to 0^{ +}$$ in (23), in view of the continuity of the beta function, we find

\begin{aligned} \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{1} + m,\lambda _{2})\le M \prod_{i = 0}^{m - 1} (\lambda _{1} + i) \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{q}}, \end{aligned}

namely, $$( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} )^{\frac{1}{p}}[\prod_{i = 0}^{m - 1} (\lambda + i)]^{ - 1} B(\lambda _{1},\lambda _{2}) \le M$$. It follows that

\begin{aligned} M = \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}}\Biggl[\prod _{i = 0}^{m - 1} (\lambda + i)\Biggr]^{ - 1} B( \lambda _{1},\lambda _{2}) \end{aligned}

is the best possible constant factor of (20).

The theorem is proved. □

### Remark 1

For $$\hat{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q},\hat{\lambda}_{2} = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p} = \lambda _{2} + \frac{\lambda - \lambda _{1} - \lambda _{2}}{q}$$, we find $$\hat{\lambda}_{1} + \hat{\lambda}_{2} = \lambda$$,

\begin{aligned} 0 < \hat{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} < \frac{\lambda}{p} + \frac{\lambda}{q} = \lambda,\qquad 0 < \hat{\lambda}_{2} = \lambda - \hat{\lambda}_{1} < \lambda. \end{aligned}

For $$\lambda - \lambda _{1} - \lambda _{2} \le q(n - \lambda _{2})$$, we still can find $$\hat{\lambda}_{2} \le n$$. In this case, we can rewrite (20) as follows:

\begin{aligned} &\sum_{k} \int _{0}^{\infty} \frac{f(x)a_{k}}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} \,dx \\ &\quad < \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}} \Biggl[ \prod _{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1}B( \hat{\lambda}_{1},\hat{\lambda}_{2}) \\ &\qquad{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \biggl[ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n} a_{k}^{q} \biggr]^{\frac{1}{q}}. \end{aligned}
(23)

### Theorem 2

If $$\lambda - \lambda _{1} - \lambda _{2} \le q(n - \lambda _{2})$$, the constant factor

\begin{aligned} \Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1}) \end{aligned}

in (19) is the best possible, then we have $$\lambda - \lambda _{1} - \lambda _{2} = 0,\lambda _{1} + \lambda _{2} = \lambda$$.

### Proof

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

\begin{aligned} B(\hat{\lambda}_{1},\hat{ \lambda}_{2})& = \int _{0}^{\infty} \frac{u^{\hat{\lambda}_{1} - 1}}{(1 + u)^{\lambda}} \,du = \int _{0}^{\infty} \frac{1}{(1 + u)^{\lambda}} u^{\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} - 1} \,du \\ &= \int _{0}^{\infty} \frac{1}{(1 + u)^{\lambda}} u^{\frac{\lambda - \lambda _{2} - 1}{p}}u^{\frac{\lambda _{1} - 1}{q}}\,du \le \biggl[ \int _{0}^{\infty} \frac{u^{\lambda - \lambda _{2} - 1}}{(1 + u)^{\lambda}} \,du \biggr]^{\frac{1}{p}} \biggl[ \int _{0}^{\infty} \frac{u^{\lambda _{1} - 1}}{(1 + u)^{\lambda}} \,du \biggr]^{\frac{1}{q}} \\ &= B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}( \lambda _{1},\lambda - \lambda _{1}). \end{aligned}
(24)

In view of the assumption, compare with the constant factors in (19) and (24), we have the following inequality:

\begin{aligned} &\Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}( \lambda _{1},\lambda - \lambda _{1})\\ &\quad\le \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}} \Biggl[ \prod _{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1}B( \hat{\lambda}_{1},\hat{\lambda}_{2}), \end{aligned}

namely, $$B(\hat{\lambda}_{1},\hat{\lambda}_{2})\ge B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})$$, which follows that (25) retains the form of equality. We observe that (25) retains the form of equality if and only if there exist constants A and B, such that they are not both zero and $$Au^{\lambda - \lambda _{2} - 1} = Bu^{\lambda _{1} - 1}\text{ a.e. in }\mathrm{R}_{ +}$$ (cf. ). Assuming that $$A \ne 0$$, we have $$u^{\lambda - \lambda _{2} - \lambda _{1}} = \frac{B}{A}\text{ a.e.in }\mathrm{R}_{ +}$$, namely, $$\lambda - \lambda _{1} - \lambda _{2} = 0$$ and then $$\lambda _{1} + \lambda _{2} = \lambda$$.

The theorem is proved. □

## 4 Equivalent forms and operator expressions

### Theorem 3

Inequality (19) is equivalent to the following inequality:

\begin{aligned} J: ={}& \biggl\{ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{p\hat{\lambda}_{2} - n} \biggl[ \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} \,dx \biggr]^{p} \biggr\} ^{\frac{1}{p}} \\ < {}& \Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1}) \\ &{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}}. \end{aligned}
(25)

In particular, for $$\lambda _{1} + \lambda _{2} = \lambda$$, we reduce (26) to the equivalent form of (20) as follows:

\begin{aligned} &\biggl\{ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{p\lambda _{2} - n} \biggl[ \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} \,dx \biggr]^{p} \biggr\} ^{\frac{1}{p}} \\ &\quad< \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}} \Biggl[ \prod _{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1}B( \lambda _{1},\lambda _{2}) \biggl[ \int _{0}^{\infty} x^{p(1 - \lambda _{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}}, \end{aligned}
(26)

where, the constant factor $$( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} )^{\frac{1}{p}}[\prod_{i = 0}^{m - 1} (\lambda + i) ]^{ - 1}B(\lambda _{1},\lambda _{2})$$ is the best possible.

### Proof

Suppose that (26) is valid. By Hölder’s inequality, we have

\begin{aligned} I &= \sum_{k} \biggl[ \Vert k - \xi \Vert _{\alpha}^{\frac{ - n}{p} + \hat{\lambda}_{2}} \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} \,dx \biggr] \bigl[ \Vert k - \xi \Vert _{\alpha}^{\frac{n}{p} - \hat{\lambda}_{2}}a_{k} \bigr] \\ &\le J \biggl[ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n} a_{k}^{q} \biggr]^{\frac{1}{q}}. \end{aligned}
(27)

Then by (26), we have (19).

On the other hand, assuming that (19) is valid, we set

\begin{aligned} a_{k}: = \Vert k - \xi \Vert _{\alpha}^{p\hat{\lambda}_{2} - n} \biggl[ \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} \,dx \biggr]^{p - 1},\quad k \in N^{n}. \end{aligned}

If $$J = 0$$, then (26) is naturally valid; if $$J = \infty$$, then it is impossible to make (26) valid, namely $$J < \infty$$. Suppose that $$0 < J < \infty$$. By (19), we have

\begin{aligned} &\sum_{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n} a_{k}^{q} \\ &\quad= J^{p} = I < \Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\\ &\phantom{\quad= J^{p} = I < }{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \biggl[ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n} a_{k}^{q} \biggr]^{\frac{1}{q}},\\ &\biggl[ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n} a_{k}^{q} \biggr]^{\frac{1}{p}}\\ &\quad = J< \Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\\ &\phantom{\quad = J< }{}\times \biggl[ \int _{0}^{\infty} x^{p(1 - \hat{\lambda}_{1}) - 1} \bigl(f^{(m)}(x)\bigr)^{p}\,dx \biggr]^{\frac{1}{p}}, \end{aligned}

namely, (26) follows, which is equivalent to (19).

The constant factor $$( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} )^{\frac{1}{p}}[\prod_{i = 0}^{m - 1} (\lambda + i) ]^{ - 1}B(\lambda _{1},\lambda _{2})$$ in (27) is the best possible. Otherwise, by (28) (for $$\lambda _{1} + \lambda _{2} = \lambda$$), we would reach a contradiction that the constant factor in (20) is not the best possible.

The theorem is proved. □

We set functions $$\phi (x): = x^{p(1 - \hat{\lambda}_{1}) - 1},\psi (k): = \Vert k - \xi \Vert _{\alpha}^{q(n - \hat{\lambda}_{2}) - n}$$, then,

\begin{aligned} \psi ^{1 - p}(k) = ( \Vert k - \xi \Vert _{\alpha}^{p\hat{\lambda}_{2} - n}\quad \bigl(x \in \mathrm{R}_{ +},k \in \mathrm{N}^{\mathrm{n}}\bigr). \end{aligned}

Define the following real normed spaces:

\begin{aligned} &L_{p,\phi} (\mathrm{R}_{ +} ): = \biggl\{ f = f(x); \Vert f \Vert _{p,\phi}: = \biggl( \int _{0}^{\infty} \phi (x) \bigl\vert f(x) \bigr\vert ^{p}\,dx\biggr)^{\frac{1}{p}} < \infty \biggr\} ,\\ &l_{q,\psi}: = \biggl\{ a = \{ a_{k_{1}, \ldots,k_{n}} \}; \Vert a \Vert _{q,\psi}: = \biggl(\sum _{k} \psi (k) \vert a_{k} \vert ^{q} \biggr)^{\frac{1}{q}} < \infty \biggr\} , \\ &l_{p,\psi ^{1 - p}}: = \biggl\{ b = \{ b_{k_{1}, \ldots,k_{n}}\}; \Vert b \Vert _{q,\psi}: = \biggl(\sum_{k} \psi ^{1 - p}(k) \vert b_{k} \vert ^{p} \biggr)^{\frac{1}{p}} < \infty \biggr\} , \end{aligned}

and $$\tilde{L}(R_{ +} ): = \{ f \in L_{p,\phi} (R_{ +} ); f(x)$$ is a nonnegative differentiable function of m-order, unless at finite points in $$\mathrm{R}_{ +}, f^{(k - 1)}(x) = o(e^{tx})\ (t > 0;x \to \infty ),f^{(k - 1)}(0^{ +} ) = 0\ (k = 1, \ldots,m)\}$$.

For any $$f \in \tilde{L}(R_{ +} )$$, setting $$b_{k}: = \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}} \,dx,k \in \mathrm{N}^{\mathrm{n}}$$, we can rewrite (26) as follows:

\begin{aligned} \Vert b \Vert _{p,\psi ^{1 - p}} \le \Biggl[ \prod _{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1}, \lambda - \lambda _{1}) \bigl\Vert f^{(m)} \bigr\Vert _{p,\phi} < \infty, \end{aligned}

namely, $$b \in l_{p,\psi ^{1 - p}}$$.

### Definition 1

Define a Hilbert-type operator $$T:\tilde{L}(\mathrm{R}_{ +} ) \to l_{p,\psi ^{1 - p}}$$ as follows: For any $$f \in \tilde{L}( \mathrm{R}_{ +} )$$, there exists a unique representation $$Tf = b \in l_{p,\psi ^{1 - p}}$$, satisfying $$Tf(k) = b_{k}\ (k \in \mathrm{N}^{\mathrm{n}})$$. Define the formal inner product of Tf and $$a \in l_{q,\psi}$$, and the norm of T as follows:

\begin{aligned} &(Tf,a): =\sum_{k} a_{k} \biggl[ \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}}\,dx \biggr] = I,\\ &\Vert T \Vert : = \sup_{f( \ne 0) \in L_{p,\phi} (R_{ +} )}\frac{ \Vert Tf \Vert _{p,\psi ^{1 - p}}}{ \Vert f^{(m)} \Vert _{p,\phi}}. \end{aligned}

By Theorem 1-3, we have

### Theorem 4

If $$f \in \tilde{L}(\mathrm{R}_{ +} ),a \in l_{q,\psi},\Vert f^{(m)}\Vert _{p,\phi},\Vert a\Vert _{q,\psi} > 0$$, then we have the following equivalent inequalities:

\begin{aligned} &(Tf,a) < \Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2}, \lambda - \lambda _{2}) \biggr)^{\frac{1}{p}} \\ & \phantom{(Tf,a) < }{}\times B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1}) \bigl\Vert f^{(m)} \bigr\Vert _{p,\phi} \Vert a \Vert _{q,\psi}, \end{aligned}
(28)
\begin{aligned} &\Vert Tf \Vert _{p,\psi ^{1 - p}} < \Biggl[ \prod _{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1}, \lambda - \lambda _{1}) \bigl\Vert f^{(m)} \bigr\Vert _{p,\phi}. \end{aligned}
(29)

Moreover, if $$\lambda _{1} + \lambda _{2} = \lambda$$ then the constant factor $$[\prod_{i = 0}^{m - 1} ( \lambda + i)]^{ - 1} ( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) )^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})$$ in (29) and (30) is the best possible, namely, $$\Vert T\Vert = ( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} )^{\frac{1}{p}}[\prod_{i = 0}^{m - 1} ( \lambda + i)]^{ - 1}B(\lambda _{1},\lambda _{2})$$. On the other hand, if $$\lambda - \lambda _{1} - \lambda _{2} \le q(n - \lambda _{2})$$, the constant factor

\begin{aligned} \Biggl[ \prod_{i = 0}^{m - 1} (\lambda + i) \Biggr]^{ - 1} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )}B(\lambda _{2},\lambda - \lambda _{2}) \biggr)^{\frac{1}{p}}B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1}) \end{aligned}

in (29) or (30) is the best possible, then we have $$\lambda - \lambda _{1} - \lambda _{2} = 0$$, namely, $$\lambda _{1} + \lambda _{2} = \lambda$$.

### Remark 2

(i) For $$\lambda = 1,\lambda _{1} = \frac{1}{q},\lambda _{2} = \frac{1}{p}$$ in (20) and (27), we have the following equivalent Hilbert-type inequalities:

\begin{aligned} &\sum_{k} \int _{0}^{\infty} \frac{f(x)a_{k}}{(x + \Vert k - \xi \Vert _{\alpha} )^{1 + m}} \,dx \\ &\quad< \frac{1}{m!} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}} \frac{\pi}{\sin (\pi /p)}\biggl[ \int _{0}^{\infty} \bigl(f^{(m)} (x) \bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \biggl[ \sum _{k} \Vert k - \xi \Vert _{\alpha}^{(q - 1)(n - 1)} a_{k}^{q} \biggr]^{\frac{1}{q}}, \end{aligned}
(30)
\begin{aligned} &\biggl\{ \sum_{k} |k - \xi \Vert _{\alpha}^{1 - n} \biggl[ \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{1 + m}} \,dx \biggr]^{p} \biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{1}{m!} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}} \frac{\pi}{\sin (\pi /p)} \biggl[ \int _{0}^{\infty} \bigl(f^{(m)} (x) \bigr)^{p}\,dx \biggr]^{\frac{1}{p}}; \end{aligned}
(31)

(ii) for $$\lambda = 1,\lambda _{1} = \frac{1}{p},\lambda _{2} = \frac{1}{q}$$ in (20) and (27), we have the following equivalent dual forms of (32) and (33):

\begin{aligned} &\sum_{k} \int _{0}^{\infty} \frac{f(x)a_{k}}{(x + \Vert k - \xi \Vert _{\alpha} )^{1 + m}} \,dx \\ &\quad < \frac{1}{m!} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}} \frac{\pi}{\sin (\pi /p)} \biggl[ \int _{0}^{\infty} x^{p - 2} \bigl(f^{(m)}(x) \bigr)^{p}\,dx \biggr]^{\frac{1}{p}} \\ &\qquad{}\times \biggl[ \sum _{k} \Vert k - \xi \Vert _{\alpha}^{(q - 1)n - 1} a_{k}^{q} \biggr]^{\frac{1}{q}}, \end{aligned}
(32)
\begin{aligned} &\biggl\{ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{p - 1 - n} \biggl[ \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{1 + m}} \,dx \biggr]^{p} \biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{1}{m!} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{p}} \frac{\pi}{\sin (\pi /p)} \biggl[ \int _{0}^{\infty} x^{p - 2} \bigl(f^{(m)}(x) \bigr)^{p}\,dx \biggr]^{\frac{1}{p}}; \end{aligned}
(33)

(iii) for $$p = q = 2$$, both (31) and (33) reduce to

\begin{aligned} &\sum_{k} \int _{0}^{\infty} \frac{f(x)a_{k}}{(x + \Vert k - \xi \Vert _{\alpha} )^{1 + m}} \,dx \\ &\quad < \frac{\pi}{m!} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{2}} \biggl[ \int _{0}^{\infty} \bigl(f^{(m)} (x) \bigr)^{2}\,dx \sum_{k} \Vert k - \xi \Vert _{\alpha}^{n - 1} a_{k}^{2} \biggr]^{\frac{1}{2}}, \end{aligned}
(34)

and both (32) and (34) reduce to the equivalent form of (35) as follows:

\begin{aligned} &\biggl\{ \sum_{k} \Vert k - \xi \Vert _{\alpha}^{1 - n} \biggl[ \int _{0}^{\infty} \frac{f(x)}{(x + \Vert k - \xi \Vert _{\alpha} )^{1 + m}} \,dx \biggr]^{2} \biggr\} ^{\frac{1}{2}} \\ &\quad< \frac{\pi}{m!} \biggl( \frac{\Gamma ^{n}(\frac{1}{\alpha} )}{\alpha ^{n - 1}\Gamma (\frac{n}{\alpha} )} \biggr)^{\frac{1}{2}} \biggl[ \int _{0}^{\infty} \bigl(f^{(m)} (x) \bigr)^{2}\,dx \biggr]^{\frac{1}{2}}, \end{aligned}
(35)

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

### Remark 3

For $$\alpha > 0$$, we only obtain $$\frac{\partial}{ y_{j}}h_{x}(y) < 0\ (j = 1, \ldots,n)$$ in (9). In this case, we can’t use Hermite–Hadamard’s inequality to obtain (11). But for $$\xi = 0$$, we still can obtain (11), and then the equivalent inequalities (19) and (26) for $$\xi = 0$$ with the best possible constant factor were proved.

## 5 Conclusions

In this paper, following the way of , by means of the weight functions, the idea of introduced parameters and the transfer formula, a more accurate half-discrete multidimensional Hilbert-type inequality with the homogeneous kernel as $$\frac{1}{(x + \Vert k - \xi \Vert _{\alpha} )^{\lambda + m}}\ (x,\lambda > 0)$$ involving one derivative function of m-order and the beta function is given in Theorem 1. The equivalent conditions of the best possible constant factor related to several parameters are considered in Theorem 2. The equivalent forms, the operator expressions and some particular Hilbert-type inequalities are obtained Theorem 3, Theorem 4 and Remark 2. The lemmas and theorems provide an extensive account of this type of inequalities.

## Availability of data and materials

We declare that the data and material in this paper can be used publicly.

## References

1. Hardy, G.H., Littlewood, J.E., Polya, G.: Inequalities. Cambridge University Press, Cambridge (1934)

2. Yang, B.C.: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing (2009)

3. Yang, B.C.: Hilbert-Type Integral Inequalities. Bentham Science Publishers Ltd., The United Arab Emirates (2009)

4. Yang, B.C.: On the norm of an integral operator and applications. J. Math. Anal. Appl. 321, 182–192 (2006)

5. Xu, J.S.: Hardy–Hilbert’s inequalities with two parameters. Adv. Math. 36(2), 63–76 (2007)

6. Yang, B.C.: On the norm of a Hilbert’s type linear operator and applications. J. Math. Anal. Appl. 325, 529–541 (2007)

7. Xie, Z.T., Zeng, Z., Sun, Y.F.: A new Hilbert-type inequality with the homogeneous kernel of degree-2. Adv. Appl. Math. Sci. 12(7), 391–401 (2013)

8. Zhen, Z., Raja Rama Gandhi, K., Xie, Z.T.: A new Hilbert-type inequality with the homogeneous kernel of degree-2 and with the integral. Bull. Math. Sci. Appl. 3(1), 11–20 (2014)

9. Xin, D.M.: A Hilbert-type integral inequality with the homogeneous kernel of zero degree. Mathematical Theory and Applications 30(2), 70–74 (2010)

10. Azar, L.E.: The connection between Hilbert and Hardy inequalities. J. Inequal. Appl. 2013, 452 (2013)

11. Batbold, T., Sawano, Y.: Sharp bounds for m-linear Hilbert-type operators on the weighted Morrey spaces. Math. Inequal. Appl. 20, 263–283 (2017)

12. Adiyasuren, V., Batbold, T., Krnic, M.: Multiple Hilbert-type inequalities involving some differential operators. Banach J. Math. Anal. 10, 320–337 (2016)

13. Adiyasuren, V., Batbold, T., Krni’c, M.: Hilbert-type inequalities involving differential operators, the best constants and applications. Math. Inequal. Appl. 18, 111–124 (2015)

14. Rassias, M., Yang, B.C.: On half-discrete Hilbert’s inequality. Appl. Math. Comput. 220, 75–93 (2013)

15. Yang, B.C., Krnic, M.: A half-discrete Hilbert-type inequality with a general homogeneous kernel of degree 0. J. Math. Inequal. 6(3), 401–417 (2012)

16. Rassias, M., Yang, B.C.: A multidimensional half-discrete Hilbert-type inequality and the Riemann zeta function. Appl. Math. Comput. 225, 263–277 (2013)

17. Rassias, M., Yang, B.C.: On a multidimensional half-discrete Hilbert-type inequality related to the hyperbolic cotangent function. Appl. Math. Comput. 242, 800–813 (2013)

18. Huang, Z.X., Yang, B.C.: On a half-discrete Hilbert-type inequality similar to Mulholland’s inequality. J. Inequal. Appl. 2013, 290 (2013)

19. Yang, B.C., Lebnath, L.: Half-Discrete Hilbert-Type Inequalities. World Scientific, Singapore (2014)

20. Krnic, M., Pecaric, J.: Extension of Hilbert’s inequality. J. Math. Anal. Appl. 324(1), 150–160 (2006)

21. Adiyasuren, V., Batbold, T., Azar, L.E.: A new discrete Hilbert-type inequality involving partial sums. J. Inequal. Appl. 2019, 127 (2019)

22. Mo, H.M., Yang, B.C.: On a new Hilbert-type integral inequality involving the upper limit functions. J. Inequal. Appl. 2020, 5 (2020)

23. Hong, Y., Wen, Y.: A necessary and sufficient condition of that Hilbert type series inequality with homogeneous kernel has the best constant factor. Ann. Math. 37A(3), 329–336 (2016)

24. Hong, Y.: On the structure character of Hilbert’s type integral inequality with homogeneous kernel and applications. J. Jilin Univ. Sci. Ed. 55(2), 189–194 (2017)

25. Hong, Y., Huang, Q.L., Yang, B.C., Liao, J.L.: The necessary and sufficient conditions for the existence of a kind of Hilbert-type multiple integral inequality with the non-homogeneous kernel and its applications. J. Inequal. Appl. 2017, 316 (2017)

26. Xin, D.M., Yang, B.C., Wang, A.Z.: Equivalent property of a Hilbert-type integral inequality related to the beta function in the whole plane. J. Funct. Spaces 2018, Article ID 2691816 (2018)

27. Hong, Y., He, B., Yang, B.C.: Necessary and sufficient conditions for the validity of Hilbert type integral inequalities with a class of quasi-homogeneous kernels and its application in operator theory. J. Math. Inequal. 12(3), 777–788 (2018)

28. Liao, J.Q., Wu, S.H., Yang, B.C.: On a new half-discrete Hilbert-type inequality involving the variable upper limit integral and the partial sum. Mathematics 8, Article ID 229 (2020). https://doi.org/10.3390/math8020229

29. Hong, Y., Zhong, Y.R., Yang, B.C.: A more accurate half-discrete multidimensional Hilbert-type inequality involving one multiple upper limit function. Axioms 12, Article ID 211 (2023). https://doi.org/10.3390/axioms12020211

30. Kuang, J.C.: Applied Inequalities. Shangdong Science and Technology Press, Jinan (2004)

31. Wang, Z.X., Guo, D.R.: Introduction to Special Functions. Science Press, Beijing (1979)

32. Kuang, J.C.: Real and Functional Analysis (Continuation), vol. 2. Higher Education Press, Beijing (2015)

## Acknowledgements

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

## Funding

This work is supported by the National Natural Science Foundation of China (No. 62166011), and the Innovation Key Project of Guangxi Province (No. 222068071). We are grateful for this help.

## Author information

Authors

### Contributions

BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. YH and YZ participated in the design of the study and performed the numerical analysis. All authors reviewed the manuscript.

### Corresponding author

Correspondence to Yanru Zhong.

## Ethics declarations

### Competing interests

The authors declare no competing interests. 