A half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor

Abstract

By introducing two pairs of conjugate exponents and using the improved Euler-Maclaurin summation formula, we estimate the weight functions and obtain a half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor. We also consider its equivalent forms.

MSC:26D15.

1 Introduction

If $a_n,b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_n^2<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{b}_n^2<\mathrm{\infty }$, then we have the famous Hilbert’s inequality as follows (cf. [1]):

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{m+n}<\pi {\left(\sum _{n=1}^{\mathrm{\infty }}{a}_n^2\sum _{n=1}^{\mathrm{\infty }}{b}_n^2\right)}^{\frac{1}{2}},$$

(1)

where the constant factor π is the best possible.

Under the same condition of (1), Xin et al. [2] gave the following inequality:

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{|ln(m/n)|}{m+n}{a}_m{b}_n<c_0{\left(\sum _{n=1}^{\mathrm{\infty }}{a}_n^2\sum _{n=1}^{\mathrm{\infty }}{b}_n^2\right)}^{\frac{1}{2}},$$

(2)

where the constant factor $c_0=8{\sum }_{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{{(2n-1)}^2}={7.3277}^{+}$ is the best possible. And Yang [3] gave the integral analogues of (2).

In 1934, Hardy et al. [1] established a few results on the half-discrete Hilbert-type inequalities with the non-homogeneous kernel (see Theorem 351). But they did not prove that the constant factors are the best possible. However, Yang [4] gave a result by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang et al. [5–9] gave some half-discrete Hilbert-type inequalities and their reverses with the monotone kernels and best constant factors.

Recently, Yang [10] gave the following half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor 8:

$$\sum _{n=1}^{\mathrm{\infty }}{\int }_1^{\mathrm{\infty }}\frac{|ln(x/n)|a_n}{max\{x,n\}}f(x)dx<8{(\sum _{n=1}^{\mathrm{\infty }}{a}_n^2{\int }_1^{\mathrm{\infty }}{f}^2(x)dx)}^{\frac{1}{2}}.$$

(3)

Obviously, for a half-discrete Hilbert-type inequality with the monotone kernel, it is easy to build the relating inequality by estimating the series form and the integral form of weight functions. However, for a half-discrete Hilbert-type inequality with the non-monotone kernel, it is much more difficult to prove.

In this paper, by using the way of weight functions, we give a new half-discrete Hilbert-type inequality with the non-monotone kernel as follows:

$$\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_1^{\mathrm{\infty }}\frac{|ln(\frac{x}{n})|f(x)}{x+n}dx<8\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^k}{{(2k+1)}^2}{(\sum _{n=1}^{\mathrm{\infty }}{a}_n^2{\int }_1^{\mathrm{\infty }}{f}^2(x))}^{\frac{1}{2}},$$

(4)

where the constant factor $8{\sum }_{k=0}^{\mathrm{\infty }}\frac{{(-1)}^k}{{(2k+1)}^2}$ is the best possible. The main objective of this paper is to build the best extension of (4) with parameters and equivalent forms.

2 Some lemmas

Lemma 2.1 If$x_1\in \mathbf{R}$, $n_1\in \mathbf{Z}$ (Z is the set of non-negative integers), $[x_1]=n_1$, $\rho (y)=y-[y]-\frac{1}{2}$ ($y\in \mathbf{R}$) is the Bernoulli function of first order[11], then we have (cf. [10])

$${\int }_{n_1}^{x_1}\rho (y)dy=-\frac{{\epsilon }_1}{8}({\epsilon }_1\in [0,1]).$$

(5)

Lemma 2.2 If$r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $f(x,y):=\frac{|ln(\frac{x}{y})|}{x+y}{(\frac{x}{y})}^{\frac{1}{r}}$ ($x,y\in (0,\mathrm{\infty })$), N is the set of positive integers, define the weight functions as follows:

(6)

(7)

Then we have

$$\omega (n)<c_r:=\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k[\frac{1}{{(k+\frac{1}{r})}^2}+\frac{1}{{(k+\frac{1}{s})}^2}],\varpi (x)<c_r.$$

(8)

Proof Setting $u=\frac{x}{n}$ in (6), we have

$$\omega (n)={\int }_{\frac{1}{n}}^{\mathrm{\infty }}\frac{|lnu|}{u+1}{u}^{-\frac{1}{s}}du<{\int }_0^{\mathrm{\infty }}\frac{|lnu|}{u+1}{u}^{-\frac{1}{s}}du=c_r.$$

Setting $u=\frac{y}{x}$, then it follows

$${\int }_0^{1}f(x,y)dy={\int }_0^{\frac{1}{x}}\frac{(-lnu)}{u+1}{u}^{-\frac{1}{r}}du>s{\int }_0^{\frac{1}{x}}\frac{(-lnu)}{\frac{1}{x}+1}d{u}^{-\frac{1}{r}+1}=\frac{x^{\frac{1}{r}}}{x+1}(slnx+s^2).$$

(9)

For $1\le y<x$, $f(x,y)=-\frac{ln(\frac{y}{x})}{x+y}{(\frac{x}{y})}^{\frac{1}{r}}$, we have

$$\begin{array}{rcl}{f}_y^{\prime }(x,y)& =& x^{\frac{1}{r}}[-\frac{1}{(y+x)y^{1+\frac{1}{r}}}+\frac{ln(\frac{y}{x})}{{(y+x)}^2{y}^{\frac{1}{r}}}+\frac{ln(\frac{y}{x})}{r(y+x)y^{1+\frac{1}{r}}}]\\ =& -x^{\frac{1}{r}}[\frac{1}{(y+x)y^{1+\frac{1}{r}}}+\frac{2xln(\frac{y}{x})}{(y-x){(y+x)}^2{y}^{\frac{1}{r}}}+\frac{2xln(\frac{y}{x})}{r(y-x)(y+x)y^{1+\frac{1}{r}}}]\\ +x^{\frac{1}{r}}[\frac{ln(\frac{y}{x})}{(y-x)(y+x)y^{\frac{1}{r}}}+\frac{ln(\frac{y}{x})}{r(y-x)y^{1+\frac{1}{r}}}].\end{array}$$

For $y\ge x$, $f(x,y)=\frac{ln(\frac{y}{x})}{x+y}{(\frac{x}{y})}^{\frac{1}{r}}$, we find

$$\begin{array}{rcl}{f}_y^{\prime }(x,y)& =& x^{\frac{1}{r}}[\frac{1}{(y+x)y^{1+\frac{1}{r}}}-\frac{ln(\frac{y}{x})}{{(y+x)}^2{y}^{\frac{1}{r}}}-\frac{ln(\frac{y}{x})}{r(y+x)y^{1+\frac{1}{r}}}]\\ =& x^{\frac{1}{r}}[\frac{1}{(y+x)y^{1+\frac{1}{r}}}+\frac{2xln(\frac{y}{x})}{(y-x){(y+x)}^2{y}^{\frac{1}{r}}}+\frac{2xln(\frac{y}{x})}{r(y-x)(y+x)y^{1+\frac{1}{r}}}]\\ -x^{\frac{1}{r}}[\frac{ln(\frac{y}{x})}{(y-x)(y+x)y^{\frac{1}{r}}}+\frac{ln(\frac{y}{x})}{r(y-x)y^{1+\frac{1}{r}}}].\end{array}$$

Define two functions as follows:

Then we have $-f_y^{\prime }(x,y)=g(y)-h(y)$. Setting $a=\frac{1}{2x^2}$, $b=-\frac{1}{2x^2}$, then $a-b=\frac{1}{x^2}$. Define two functions as follows:

$$\tilde{g}(y)=\{\begin{array}{cc}{g}_1(y)-a,\hfill & y<x;\hfill \\ g_2(y),\hfill & y\ge x,\hfill \end{array}\tilde{h}(y)=\{\begin{array}{cc}{h}_1(y)-b,\hfill & y<x;\hfill \\ h_2(y),\hfill & y\ge x.\hfill \end{array}$$

Since $g_1(x-0)-a=g_2(x)$, $h_1(x-0)-b=h_2(x)$, then both $\tilde{g}(y)$ and $\tilde{h}(y)$ ($y\in [1,\mathrm{\infty })$) are decreasing and continuous. Besides $y=x$, we have ${(-1)}^i{\tilde{g}}^{(i)}(y)\ge 0$, ${(-1)}^i{\tilde{h}}^{(i)}(y)\ge 0$ ($i=0,1$), and $\tilde{g}(\mathrm{\infty })=\tilde{h}(\mathrm{\infty })=0$. By the improved Euler-Maclaurin summation formula (cf. [11], Theorem 2.2.2) and (5), for ${\epsilon }_1\in [0,1]$, ${\epsilon }_i\in (0,1)$ ($i=2,3$), it follows

Since $g_1(1)-a\ge g_1(x-0)-a=g_2(x)>0$, $h_1(1)-b\ge h_1(x-0)-b=h_2(x)>0$, then we have

$$\begin{array}{rcl}-{\int }_1^{\mathrm{\infty }}\rho (y)f_y^{\prime }(x,y)dy& >& -\frac{1}{8}(g_1(1)-a)-\frac{1}{8}(a-b)\\ =& -\frac{x^{\frac{1}{r}}}{8(x+1)}-\frac{x^{1+\frac{1}{r}}lnx}{4{(x+1)}^2(x-1)}-\frac{x^{1+\frac{1}{r}}lnx}{4r(x+1)(x-1)}-\frac{1}{16x^2}.\end{array}$$

(10)

By the improved Euler-Maclaurin summation formula [11], we have

$$\begin{array}{rcl}\varpi (x)& =& \sum _{k=1}^{\mathrm{\infty }}f(x,k)={\int }_1^{\mathrm{\infty }}f(x,y)dy+\frac{1}{2}f(x,1)+{\int }_1^{\mathrm{\infty }}\rho (y)f_y^{\prime }(x,y)dy\\ =& {\int }_0^{\mathrm{\infty }}f(x,y)dy-({\int }_0^{1}f(x,y)dy-\frac{1}{2}f(x,1)-{\int }_1^{\mathrm{\infty }}\rho (y)f_y^{\prime }(x,y)dy)=c_r-\theta (x),\end{array}$$

where

$$\theta (x):={\int }_0^{1}f(x,y)dy-\frac{1}{2}f(x,1)-{\int }_1^{\mathrm{\infty }}\rho (y)f_y^{\prime }(x,y)dy.$$

Since $-\frac{1}{2}f(x,1)=-\frac{x^{\frac{1}{r}}lnx}{2(x+1)}$, in view of (9), (10), (i) for $1\le x<2$, $-\frac{lnx}{x-1}\ge -1$, we have

$$\begin{array}{rcl}\theta (x)& >& \frac{x^{\frac{1}{r}}}{x+1}(slnx+s^2)-\frac{x^{\frac{1}{r}}lnx}{2(x+1)}-\frac{x^{\frac{1}{r}}}{8(x+1)}\\ -\frac{x^{1+\frac{1}{r}}lnx}{4{(x+1)}^2(x-1)}-\frac{x^{1+\frac{1}{r}}lnx}{4r(x+1)(x-1)}-\frac{1}{16x^2}\\ \ge & \frac{x^{\frac{1}{r}}lnx}{(x+1)}(s-\frac{1}{2})+\frac{x^{\frac{1}{r}}}{x+1}[s^2-\frac{1}{8}-\frac{x}{4(x+1)}-\frac{x}{4r}-\frac{x+1}{16x^{2+\frac{1}{r}}}]\\ >& \frac{x^{\frac{1}{r}}}{(x+1)}(1-\frac{1}{8}-\frac{1}{4}-\frac{1}{2}-\frac{1}{8})=0;\end{array}$$

(ii) for $x\ge 2$, $-\frac{1}{x-1}\ge -\frac{2}{x}$, we have

$$\begin{array}{rcl}\theta (x)& \ge & \frac{x^{\frac{1}{r}}lnx}{(x+1)}[s-\frac{1}{2}-\frac{1}{2(x+1)}-\frac{1}{2r}]+\frac{x^{\frac{1}{r}}}{x+1}(s^2-\frac{1}{8}-\frac{x+1}{16x^{2+\frac{1}{r}}})\\ >& \frac{x^{\frac{1}{r}}lnx}{(x+1)}(s-\frac{1}{2}-\frac{1}{6}-\frac{1}{2r})=\frac{(6s+2r-3)x^{\frac{1}{r}}lnx}{6r(x+1)}>0.\end{array}$$

Hence, for $x\ge 1$, we have $\theta (x)>0$, it follows $\varpi (x)<c_r$. The lemma is proved. □

Lemma 2.3 As the assumption of Lemma  2.2, if$0<\epsilon <\frac{p}{r}$, $R={(\frac{1}{r}-\frac{\epsilon }{p})}^{-1}$, $S={(\frac{1}{s}+\frac{\epsilon }{p})}^{-1}$, then we have

$$\overline{I}:=\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{1}{S}-\epsilon -1}{\int }_1^{\mathrm{\infty }}\frac{|ln(\frac{x}{n})|}{x+n}{x}^{\frac{1}{R}-1}dx\ge \frac{c_R}{\epsilon }-O(1).$$

(11)

Proof It is obvious that $R>1$, $\frac{1}{R}+\frac{1}{S}=1$. Setting $u=\frac{x}{n}$, we have

$$\begin{array}{rcl}\overline{I}& =& \sum _{n=1}^{\mathrm{\infty }}{n}^{-\epsilon -1}{\int }_{\frac{1}{n}}^{\mathrm{\infty }}\frac{|lnu|}{u+1}{u}^{\frac{1}{R}-1}du\\ =& \sum _{n=1}^{\mathrm{\infty }}{n}^{-\epsilon -1}[{\int }_0^{\mathrm{\infty }}\frac{|lnu|}{u+1}{u}^{\frac{1}{R}-1}du+{\int }_0^{\frac{1}{n}}\frac{lnu}{u+1}{u}^{\frac{1}{R}-1}du]\\ >& c_R{\int }_1^{\mathrm{\infty }}{x}^{-\epsilon -1}dx+R\sum _{n=1}^{\mathrm{\infty }}{n}^{-\epsilon -1}{\int }_0^{\frac{1}{n}}lnud{u}^{\frac{1}{R}}\\ =& \frac{c_R}{\epsilon }-[R\sum _{n=1}^{\mathrm{\infty }}\frac{lnn}{n^{1+\epsilon +\frac{1}{R}}}+R^2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{1+\epsilon +\frac{1}{R}}}]\\ =& \frac{c_R}{\epsilon }-O(1)(\epsilon \to 0^{+}).\end{array}$$

The lemma is proved. □

Lemma 2.4 If$p,r>1$, $\frac{1}{r}+\frac{1}{s}=\frac{1}{p}+\frac{1}{q}=1$, $a_n\ge 0$, $f(x)$is a non-negative measurable function, then we have

(12)

(13)

Proof By Hölder’s inequality [12], in view of (6) and (7), we have

Hence we have (12). Still by Hölder’s inequality [12], (6) and (7), we have

Then we have (13). The lemma is proved. □

3 Main results and applications

Theorem 3.1 If$p,r>1$, $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+\frac{1}{s}=1$, $a_n,f(x)\ge 0$such that$0<{\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^p(x)dx<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_n^q<\mathrm{\infty }$, then we have the following equivalent inequalities:

(14)

(15)

(16)

where the constant factors$c_r={\sum }_{k=0}^{\mathrm{\infty }}{(-1)}^k[\frac{1}{{(k+\frac{1}{r})}^2}+\frac{1}{{(k+\frac{1}{s})}^2}]$, $c_r^p$, $c_r^q$are the best possible.

Proof By the Lebesgue term-by-term integration theorem [13], there are two kinds of representation in (14). By the conditions of Theorem 3.1, (12) takes the form of a strict inequality, and we have (15). By Hölder’s inequality [12], we have

$$I=\sum _{n=1}^{\mathrm{\infty }}\left[n^{\frac{1}{s}-\frac{1}{p}}{\int }_1^{\mathrm{\infty }}\frac{|ln(\frac{x}{n})|f(x)}{x+n}dx\right]\left[n^{-\frac{1}{s}+\frac{1}{p}}{a}_n\right]\le J_1^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_n^q\right\}}^{\frac{1}{q}}.$$

(17)

By (15), we have (14). On the other hand, suppose that (14) is valid. Setting $a_n:=n^{\frac{p}{s}-1}{[{\int }_1^{\mathrm{\infty }}\frac{|ln(\frac{x}{n})|f(x)}{x+n}dx]}^{p-1}$, $n\in \mathbf{N}$, then it follows $J_1={\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_n^q$. By (12), we have $J<\mathrm{\infty }$. If $J=0$, then (15) is obvious value; if $0<J<\mathrm{\infty }$, then by (14), we obtain

$$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_n^q=J_1=I<c_r{\{{\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_n^q\right\}}^{\frac{1}{q}},\\ J_1^{\frac{1}{p}}={\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_n^q\right\}}^{\frac{1}{p}}<c_r{\{{\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^p(x)dx\}}^{\frac{1}{p}}.\end{array}$$

(18)

Hence we have (15), which is equivalent to (14).

By Hölder’s inequality [12], we have

$$I={\int }_1^{\mathrm{\infty }}[x^{\frac{1}{r}-\frac{1}{q}}f(x)]\left[x^{-\frac{1}{r}+\frac{1}{q}}\sum _{n=1}^{\mathrm{\infty }}\frac{|ln(\frac{x}{n})|a_n}{x+n}\right]dx\le {\{{\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^p(x)dx\}}^{\frac{1}{p}}{J}_2^{\frac{1}{q}}.$$

(19)

By (16), we have (14). On the other hand, suppose that (14) is valid. Setting $f(x):=x^{\frac{q}{r}-1}{[{\sum }_{n=1}^{\mathrm{\infty }}\frac{|ln(\frac{x}{n})|a_n}{x+n}]}^{q-1}$, $x\in [1,\mathrm{\infty })$, then it follows $J_2={\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^p(x)dx$. By (13), we have $J_2<\mathrm{\infty }$. If $J_2=0$, then (16) is obvious value; if $0<J_2<\mathrm{\infty }$, then by (14), we obtain

$$\begin{array}{r}{\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^p(x)dx=J_2=I<c_r{\{{\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_n^q\right\}}^{\frac{1}{q}},\\ J_2^{\frac{1}{q}}={\{{\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{f}^p(x)dx\}}^{\frac{1}{q}}<c_r{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{a}_n^q\right\}}^{\frac{1}{q}}.\end{array}$$

(20)

Hence we have (16), which is equivalent to (14). Therefore (14), (15) and (16) are equivalent.

If the constant factor $c_r$ in (14) is not best possible, then there exists a positive number K, with $0<K<c_r$, such that (14) is still valid if we replace $c_r$ by K. For $0<\epsilon <{\epsilon }_0$, setting $\overline{f}(x)=n^{\frac{1}{r}-\frac{\epsilon }{p}-1}$, ${\overline{a}}_n=n^{\frac{1}{s}-\frac{\epsilon }{q}-1}$ ($n\in N$), we have

$$\begin{array}{rcl}\overline{I}& =& \sum _{n=1}^{\mathrm{\infty }}{\overline{a}}_n{\int }_1^{\mathrm{\infty }}\frac{|ln(\frac{x}{n})|\overline{f}(x)}{x+n}dx<K{\{{\int }_1^{\mathrm{\infty }}{x}^{\frac{p}{s}-1}{\overline{f}}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{\overline{a}}_n^q\right\}}^{\frac{1}{q}}\\ =& K{\left({\int }_1^{\mathrm{\infty }}{x}^{-1-\epsilon }dx\right)}^{\frac{1}{p}}{(1+\sum _{n=2}^{\mathrm{\infty }}{n}^{-1-\epsilon })}^{\frac{1}{q}}\\ <& K{\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}{(1+{\int }_1^{\mathrm{\infty }}{x}^{-1-\epsilon }dx)}^{\frac{1}{q}}=\frac{K}{\epsilon }{(\epsilon +1)}^{\frac{1}{q}}.\end{array}$$

(21)

By (11) and (21), we have $c_R-\epsilon O(1)<K{(\epsilon +1)}^{\frac{1}{q}}$ and for $\epsilon \to 0^{+}$, by Fatou lemma [13], we have $c_r\le {lim}_{\epsilon \to 0^{+}}(c_R-\epsilon O(1))\le K$. This is a contradiction. Hence we can conclude that the constant $c_r$ in (14) is the best possible. If the constant factors in (15) and (16) are not the best possible, then we can imply a contradiction that the constant factor in (14) is not the best possible by (17) and (19). The theorem is proved. □

Remark For $p=q=r=s=2$, (14) reduces to (4). Inequality (4) is a new basic half-discrete Hilbert-type inequality with the non-monotone kernel.