A new Hardy-Hilbert-type inequality with multiparameters and a best possible constant factor

Abstract

By means of weight coefficients and techniques of real analysis, a new Hardy-Hilbert-type inequality with multiparameters and the best possible constant factor is given. The equivalent forms, the operator expression with the norm, and the reverse and some particular inequalities with the best possible constant factors are also considered.

1 Introduction

If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f(x),g(y)\geq0$, $f\in L^{p}(\mathbf{R}_{+})$, $g\in L^{q}(\mathbf{R}_{+})$, $\|f\|_{p}=(\int_{0}^{\infty }f^{p}(x)\,dx)^{\frac{1}{p}}>0$, $\|g\|_{q}>0$, then we have the following Hardy-Hilbert integral inequality [1]:

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

(1)

where, the constant factor $\frac{\pi}{\sin(\pi/p)}$ is the best possible. Assuming that $a_{m},b_{n}\geq0$, $a=\{a_{m}\}_{m=1}^{\infty }\in l^{p}$, $b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}$, $\|a\|_{p}=(\sum_{m=1}^{\infty }a_{m}^{p})^{\frac{1}{p}}>0$, $\|b\|_{q}>0$, we have the following Hardy-Hilbert inequality with the same best possible constant factor $\frac{\pi}{\sin(\pi/p)}$ [1]:

$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\pi/p)}\|a \|_{p}\|b\|_{q}. $$

(2)

Inequalities (1) and (2) are important in analysis and its applications [1–5].

If $\mu_{i},\upsilon_{j}>0$ ($i,j\in\mathbf{N}=\{1,2,\ldots\}$),

$$ U_{m}:=\sum_{i=1}^{m} \mu_{i},\qquad V_{n}:=\sum_{j=1}^{n} \upsilon_{j}\quad (m,n\in \mathbf{N}), $$

(3)

then we have the following inequality (see Theorem 321 of [1]):

$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{\mu_{m}^{1/q}\upsilon _{n}^{1/p}a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\pi/p)}\|a\|_{p}\|b\|_{q}. $$

(4)

Replacing $\mu_{m}^{1/q}a_{m}$ and $\upsilon_{n}^{1/p}b_{n}$ by $a_{m}$ and $b_{n}$ in (4), respectively, we have the following equivalent form of (4):

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

(5)

For $\mu_{i}=\upsilon_{j}=1$ ($i,j\in\mathbf{N}$), both (4) and (5) reduce to (2). We call (4) and (5) Hardy-Hilbert-type inequalities.

Note

The authors of [1] did not prove that (4) is valid with the best possible constant factor.

In 1998, by introducing an independent parameter $\lambda\in(0,1]$ Yang [6] gave an extension of (1) with the kernel $\frac{1}{(x+y)^{\lambda}}$ for $p=q=2$. Following the results of [6], Yang [5] gave some best extensions of (1) and (2) as follows.

If $\lambda_{1},\lambda_{2}\in\mathbf{R}$, $\lambda_{1}+\lambda _{2}=\lambda$, $k_{\lambda}(x,y)$ is a nonnegative homogeneous function of degree −λ with $k(\lambda_{1})=\int_{0}^{\infty}k_{\lambda }(t,1)t^{\lambda_{1}-1}\,dt\in\mathbf{R}_{+}$, $\phi(x)=x^{p(1-\lambda _{1})-1}$, $\psi(x)=x^{q(1-\lambda_{2})-1}$, $f(x),g(y)\geq0$,

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

$g\in L_{q,\psi}(\mathbf{R}_{+})$, $\|f\|_{p,\phi},\|g\|_{q,\psi}>0$, then

$$ \int_{0}^{\infty} \int_{0}^{\infty}k_{\lambda }(x,y)f(x)g(y)\,dx \,dy< k(\lambda _{1})\|f\|_{p,\phi}\|g\|_{q,\psi}, $$

(6)

where the constant factor $k(\lambda_{1})$ is the best possible. Moreover, if $k_{\lambda}(x,y)$ is finite and $k_{\lambda}(x,y)x^{\lambda _{1}-1}$ ($k_{\lambda}(x,y)y^{\lambda_{2}-1}$) is decreasing with respect to $x>0$ ($y>0$), then for $a_{m},b_{n}\geq0$,

$$ a\in l_{p,\phi}= \Biggl\{ a;\|a\|_{p,\phi}:= \Biggl(\sum _{n=1}^{\infty }\phi (n)|a_{n}|^{p} \Biggr)^{\frac{1}{p}}< \infty \Biggr\} , $$

$b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\psi}$, $\|a\|_{p,\phi },\|b\|_{q,\psi }>0$, it follows that

$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}k_{\lambda }(m,n)a_{m}b_{n}< k( \lambda_{1})\|a\|_{p,\phi}\|b\|_{q,\psi}, $$

(7)

where the constant factor $k(\lambda_{1})$ is still the best possible.

Clearly, for $\lambda=1$, $k_{1}(x,y)=\frac{1}{x+y}$, $\lambda _{1}=\frac{1}{q}$, $\lambda_{2}=\frac{1}{p}$, inequality (6) reduces to (1), whereas (7) reduces to (2). For $0<\lambda_{1},\lambda _{2}\leq1$, $\lambda_{1}+\lambda_{2}=\lambda$, we set

$$ k_{\lambda}(x,y)=\frac{1}{(x+y)^{\lambda}} \quad\bigl((x,y)\in\mathbf{R}_{+}^{2} \bigr). $$

Then by (7) it follows that

$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac {a_{m}b_{n}}{(m+n)^{\lambda}}< B( \lambda_{1},\lambda_{2})\|a\|_{p,\phi}\|b \|_{q,\psi}, $$

(8)

where the constant $B(\lambda_{1},\lambda_{2})$ is the best possible. Some other results including multidimensional Hilbert-type inequalities are provided in [7–24].

In this paper, by means of weight coefficients and techniques of real analysis, a new Hardy-Hilbert-type inequality with multiparameters and the best possible constant factor is given, which is with the kernel

$$ k_{\lambda}(x,y)=\frac{(\min\{x,y\})^{\alpha}}{(\max\{x,y\})^{\lambda +\alpha}} $$

similar to (4). The equivalent forms, the operator expression with the norm, the reverse and some particular inequalities with the best possible constant factors are also considered.

2 An example and some lemmas

In the following, we make appointment that $\mu_{i},\upsilon _{j}>0$ ($i,j\in\mathbf{N}$), $U_{m}$ and $V_{n}$ are defined by (3), $p\neq0,1$, $\frac{1}{p}+\frac{1}{q}=1$, $a_{m},b_{n}\geq0$ ($m,n\in \mathbf{N}$),

$$ \|a\|_{p,\Phi_{\lambda}}=\Biggl(\sum_{m=1}^{\infty} \Phi_{\lambda }(m)a_{m}^{p}\Biggr)^{\frac{1}{p}},\qquad\|b \|_{q,\Psi_{\lambda}}=\Biggl(\sum_{n=1}^{\infty} \Psi_{\lambda}(n)b_{n}^{q}\Biggr)^{\frac{1}{q}}, $$

where

$$ \Phi_{\lambda}(m):=\frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu _{m}^{p-1}},\qquad\Psi _{\lambda}(n):= \frac{V_{n}^{q(1-\lambda_{2})-1}}{\upsilon _{n}^{q-1}}\quad (m,n\in\mathbf{N}). $$

We also set

$$\begin{aligned}& \widetilde{\Phi}_{\lambda}(m) :=\bigl(1-\theta(\lambda_{2},m) \bigr)\frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu_{m}^{p-1}},\qquad \widetilde{\Psi}_{\lambda}(n) :=\bigl(1-\vartheta(\lambda_{1},n) \bigr)\frac{ V_{n}^{q(1-\lambda_{2})-1}}{\upsilon_{n}^{q-1}} \quad(m,n\in\mathbf{N}). \end{aligned}$$

Note

For $0< p<1$ or $p<0$, we still use the formal symbols $\|a\|_{p,\Phi_{\lambda}}$, $\|b\|_{q,\Psi_{\lambda}}$, $\|a\|_{p,\widetilde{\Phi}_{\lambda}}$, and $\|b\|_{q,\widetilde{\Psi }_{\lambda}}$.

Example 1

For $-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha$, $\lambda_{1}+\lambda_{2}=\lambda$, we set

$$ k_{\lambda}(x,y)=\frac{(\min\{x,y\})^{\alpha}}{(\max\{x,y\} )^{\lambda +\alpha}} \quad\bigl((x,y)\in\mathbf{R}_{+}^{2} \bigr). $$

We find

$$ \begin{aligned}[b] k(\lambda_{1}) &= \int_{0}^{\infty}k_{\lambda}(t,1)t^{\lambda _{1}-1} \,dt= \int_{0}^{\infty}\frac{(\min\{t,1\})^{\alpha}}{(\max \{t,1\})^{\lambda+\alpha}}t^{\lambda_{1}-1}\,dt\\ &= \int_{0}^{1}t^{\lambda_{1}+\alpha-1}\,dt+ \int_{1}^{\infty}\frac{1}{t^{\lambda+\alpha}}t^{\lambda_{1}-1}\,dt\\ &=\frac{1}{\lambda_{1}+\alpha}+\frac{1}{\lambda_{2}+\alpha}=\frac{\lambda+2\alpha}{(\lambda_{1}+\alpha)(\lambda_{2}+\alpha)}\in \mathbf{R}_{+}. \end{aligned} $$

(9)

Since

$$ k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{y^{\alpha+\lambda_{2}-1}}{x^{\lambda+\alpha}},& 0< y< x, \\ \frac{x^{\alpha}}{y^{1+\lambda_{1}+\alpha}},& y\geq x, \end{array}\displaystyle \right . $$

for $\lambda_{2}\leq1-\alpha^{{}}$ ($\lambda_{1}>-\alpha$), $k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}$ is decreasing for $y>0$ and strictly decreasing for y large enough. Since

$$ k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{x^{\alpha+\lambda_{1}-1}}{y^{\lambda+\alpha}},& 0< x< y, \\ \frac{y^{\alpha}}{x^{1+\lambda_{2}+\alpha}},& x\geq y, \end{array}\displaystyle \right . $$

for $\lambda_{1}\leq1-\alpha$ ($\lambda_{2}>-\alpha$), $k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}$ is decreasing for $x>0$ and strictly decreasing for x large enough.

In other words, for $-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha$, $k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}$ ($k_{\lambda }(x,y)\frac{1}{x^{1-\lambda_{1}}}$) is decreasing for $y>0$ ($x>0$) and strictly decreasing for $y(x)$ large enough, satisfying $k(\lambda_{1})\in \mathbf{R}_{+}$.

Lemma 1

If $g(t)$ (>0) is decreasing in $\mathbf{R}_{+}$, strictly decreasing in $[n_{0},\infty)$ ($n_{0}\in\mathbf{N}$), and satisfying $\int_{0}^{\infty}g(t)\,dt\in\mathbf{R}_{+}$, then we have

$$ \int_{1}^{\infty}g(t)\,dt< \sum _{n=1}^{\infty}g(n)< \int_{0}^{\infty}g(t)\,dt. $$

(10)

Proof

Since

$$\begin{aligned}& \int_{n}^{n+1}g(t)\,dt \leq g(n)\leq \int_{n-1}^{n}g(t)\,dt \quad(n=1,\ldots,n_{0}), \\& \int_{n_{0}+1}^{n_{0}+2}g(t)\,dt < g(n_{0}+1)< \int_{n_{0}}^{n_{0}+1}g(t)\,dt, \end{aligned}$$

it follows that

$$ 0< \int_{1}^{n_{0}+2}g(t)\,dt< \sum _{n=1}^{n_{0}+1}g(n)< \sum_{n=1}^{n_{0}+1} \int_{n-1}^{n}g(t)\,dt= \int_{0}^{n_{0}+1}g(t)\,dt< \infty. $$

In the same way, we have

$$ 0< \int_{n_{0}+2}^{\infty}g(t)\,dt\leq\sum _{n=n_{0}+2}^{\infty}g(n)\leq \int_{n_{0}+1}^{\infty}g(t)\,dt< \infty. $$

Adding these two inequalities, we have (10). □

Lemma 2

Let $-\alpha<\lambda_{1},\lambda_{2}\leq 1-\alpha$, $\lambda_{1}+\lambda_{2}=\lambda$, and $k(\lambda_{1})$ be as in (9). Define the following weight coefficients:

$$\begin{aligned}& \omega(\lambda_{2},m) :=\sum_{n=1}^{\infty} \frac{(\min\{U_{m},V_{n} \})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{U_{m}^{\lambda_{1}}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}},\quad m\in\mathbf{N}, \end{aligned}$$

(11)

$$\begin{aligned}& \varpi(\lambda_{1},n) :=\sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n} \})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{V_{n}^{\lambda_{2}}\mu_{m}}{U_{m}^{1-\lambda_{1}}},\quad n\in\mathbf{N}. \end{aligned}$$

(12)

Then, we have the following inequalities:

$$\begin{aligned}& \omega(\lambda_{2},m) < k(\lambda_{1})\quad (-\alpha< \lambda_{2}\leq 1-\alpha,\lambda_{1}>-\alpha;m\in \mathbf{N}), \end{aligned}$$

(13)

$$\begin{aligned}& \varpi(\lambda_{1},n) < k(\lambda_{1})\quad (-\alpha< \lambda_{1}\leq 1-\alpha,\lambda_{2}>-\alpha;n\in \mathbf{N}). \end{aligned}$$

(14)

Proof

We set $\mu(t):=\mu_{m}$, $t\in(m-1,m]$ ($m\in\mathbf{N}$); $\upsilon(t):=\upsilon_{n}$, $t\in(n-1,n]$ ($n\in\mathbf{N}$), and

$$ U(x):= \int_{0}^{x}\mu(t)\,dt\quad (x\geq0), \qquad V(y):= \int_{0}^{y}\upsilon (t)\,dt\quad (y\geq 0). $$

(15)

Then by (3) it follows that $U(m)=U_{m}$, $V(n)=V_{n}$ ($m,n\in \mathbf{N}$). For $x\in(m-1,m)$, $U^{\prime}(x)=\mu(x)=\mu_{m}$ ($m\in \mathbf{N}$); for $y\in(n-1,n)$, $V^{\prime}(y)=\upsilon(y)=\upsilon _{n}$ ($n\in\mathbf{N}$). Since $V(y)$ is strictly increasing in $(n-1,n]$, $-\alpha<\lambda_{2}\leq1-\alpha$, $\lambda_{1}>-\alpha$, in view of Example 1 and Lemma 1, we find

$$\begin{aligned} \omega(\lambda_{2},m) =&\sum_{n=1}^{\infty} \int_{n-1}^{n}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{U_{m}^{\lambda_{1}}}{V_{n}^{1-\lambda_{2}}}V^{\prime}(y) \,dy \\ < &\sum_{n=1}^{\infty} \int_{n-1}^{n}\frac{(\min\{U_{m},V(y)\} )^{\alpha}}{(\max\{U_{m},V(y)\})^{\lambda+\alpha}}\frac{U_{m}^{\lambda_{1}}}{V^{1-\lambda_{2}}(y)}V^{\prime}(y) \,dy. \end{aligned}$$

Setting $t=\frac{V(y)}{U_{m}}$, we obtain $V^{\prime}(y)\,dy=U_{m}\,dt$ and

$$\begin{aligned} \omega(\lambda_{2},m) < &\sum_{n=1}^{\infty} \int_{\frac {V(n-1)}{U_{m}}}^{\frac{V(n)}{U_{m}}}\frac{(\min\{1,t\})^{\alpha}}{(\max\{1,t\} )^{\lambda +\alpha}}t^{\lambda_{2}-1}\,dt \\ =& \int_{0}^{\frac{V(\infty)}{U_{m}}}\frac{(\min\{1,t\})^{\alpha }}{(\max \{1,t\})^{\lambda+\alpha}}t^{\lambda_{2}-1}\,dt \\ \leq& \int_{0}^{\infty}\frac{(\min\{1,t\})^{\alpha}}{(\max \{1,t\})^{\lambda+\alpha}}t^{\lambda_{2}-1} \,dt=k(\lambda_{1}). \end{aligned}$$

(16)

Hence, we have (13). In the same way, we have (14). □

Lemma 3

Let $-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha$, $\lambda_{1}+\lambda_{2}=\lambda$, $\lambda_{1}+\lambda_{2}=\lambda$, $k(\lambda_{1})$ be as in (9), $m_{0},n_{0}\in\mathbf{N}$, $\mu _{m}\geq\mu_{m+1}$ ($m\in\{m_{0},m_{0}+1,\ldots\}$), $\upsilon _{n}\geq \upsilon_{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots\}$), $U(\infty )=V(\infty)=\infty$. Then

(i) for $m,n\in\mathbf{N,}$ we have

$$\begin{aligned}& k(\lambda_{1}) \bigl(1-\theta(\lambda_{2},m)\bigr) < \omega(\lambda _{2},m) \quad(-\alpha< \lambda_{2}\leq1-\alpha, \lambda_{1}>-\alpha), \end{aligned}$$

(17)

$$\begin{aligned}& k(\lambda_{1}) \bigl(1-\vartheta(\lambda_{1},n)\bigr) < \varpi(\lambda _{1},n) \quad(-\alpha< \lambda_{1}\leq1-\alpha, \lambda_{2}>-\alpha), \end{aligned}$$

(18)

where, $\theta(\lambda_{2},m)=O(\frac{1}{U_{m}^{\lambda_{2}+\alpha }})\in (0,1)$, $\vartheta(\lambda_{1},n)=O(\frac{1}{V_{n}^{\lambda_{1}+\alpha }})\in(0,1)$;

(ii) for any $a>0$, we have

$$\begin{aligned}& \sum_{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+a}} = \frac{1}{a} \biggl( \frac{1}{U_{m_{0}}^{a}}+aO(1) \biggr) , \end{aligned}$$

(19)

$$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{1+a}} = \frac {1}{a} \biggl( \frac{1}{V_{n_{0}}^{a}}+a\widetilde{O}(1) \biggr) . \end{aligned}$$

(20)

Proof

Since $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\geq n_{0}$), $-\alpha<\lambda_{2}\leq1-\alpha$, $\lambda_{1}>-\alpha$, and $V(\infty)=\infty$, by Lemma 1 we have

$$\begin{aligned} \omega(\lambda_{2},m)&\geq\sum_{n=n_{0}}^{\infty} \frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{U_{m}^{\lambda_{1}}}{V_{n}^{1-\lambda_{2}}}\upsilon_{n+1}\\ &=\sum_{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{U_{m}^{\lambda_{1}}}{V_{n}^{1-\lambda_{2}}}V^{\prime}(y) \,dy\\ &>\sum_{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{(\min\{U_{m},V(y)\} )^{\alpha }}{(\max\{U_{m},V(y)\})^{\lambda+\alpha}}\frac{U_{m}^{\lambda_{1}}}{ V^{1-\lambda_{2}}(y)}V^{\prime}(y) \,dy\\ &=\sum_{n=n_{0}}^{\infty} \int_{\frac{V(n)}{U_{m}}}^{\frac {V(n+1)}{U_{m}}}\frac{(\min\{1,t\})^{\alpha}}{(\max\{1,t\})^{\lambda+\alpha }}t^{\lambda _{2}-1} \,dt\\ &= \int_{\frac{V(n_{0})}{U_{m}}}^{\infty}\frac{(\min\{1,t\})^{\alpha }}{(\max\{1,t\})^{\lambda+\alpha}}t^{\lambda_{2}-1} \,dt=k(\lambda _{1}) \bigl(1-\theta(\lambda_{2},m)\bigr), \end{aligned}$$

where

$$ \theta(\lambda_{2},m):=\frac{1}{k(\lambda_{1})} \int_{0}^{\frac {V(n_{0})}{U_{m}}}\frac{(\min\{1,t\})^{\alpha}}{(\max\{1,t\})^{\lambda+\alpha }}t^{\lambda_{2}-1} \,dt\in(0,1). $$

(21)

For $U_{m}>V(n_{0})$, we obtain

$$\begin{aligned} 0 < &\theta(\lambda_{2},m)=\frac{1}{k(\lambda_{1})} \int_{0}^{\frac{V(n_{0})}{U_{m}}}t^{\lambda_{2}+\alpha-1}\,dt \\ =&\frac{1}{(\lambda_{2}+\alpha)k(\lambda_{1})} \biggl( \frac {V_{n_{0}}}{U_{m}} \biggr) ^{\lambda_{2}+\alpha}, \end{aligned}$$

and then $\theta(\lambda_{2},m)=O(\frac{1}{U_{m}^{\lambda_{2}+\alpha}})$. Hence, we have (17).

In the same way, since $\mu_{m}\geq\mu_{m+1}$ ($m\geq m_{0}$), $-\alpha <\lambda_{1}\leq1-\alpha$, $\lambda_{2}>-\alpha$, and $U(\infty )=\infty$, we have

$$\begin{aligned} \varpi(\lambda_{1},n) \geq&\sum_{m=m_{0}}^{\infty} \frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{V_{n}^{\lambda_{2}}\mu_{m+1}}{U_{m}^{1-\lambda_{1}}} \\ =&\sum_{m=m_{0}}^{\infty} \int_{m}^{m+1}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{V_{n}^{\lambda_{2}}U^{\prime}(x)}{U_{m}^{1-\lambda_{1}}}\,dx \\ >&\sum_{m=m_{0}}^{\infty} \int_{m}^{m+1}\frac{(\min\{U(x),V_{n}\} )^{\alpha }}{(\max\{U(x),V_{n}\})^{\lambda+\alpha}}\frac{V_{n}^{\lambda _{2}}U^{\prime}(x)}{U^{1-\lambda_{1}}(x)}\,dx \\ \overset{t=U(x)/V_{n}}{=}&\sum_{m=m_{0}}^{\infty} \int_{\frac {U(m)}{V_{n}}}^{\frac{U(m+1)}{V_{n}}}\frac{(\min\{t,1\})^{\alpha}}{(\max \{t,1\})^{\lambda+\alpha}}t^{\lambda_{1}-1}\,dt \\ =& \int_{\frac{U(m_{0})}{V_{n}}}^{\infty}\frac{(\min\{t,1\})^{\alpha }}{(\max\{t,1\})^{\lambda+\alpha}}t^{\lambda_{1}-1} \,dt=k(\lambda _{1}) \bigl(1-\vartheta(\lambda_{1},n)\bigr), \end{aligned}$$

where

$$ \vartheta(\lambda_{1},n):=\frac{1}{k(\lambda_{1})} \int_{0}^{\frac {U(m_{0})}{V_{n}}}\frac{(\min\{t,1\})^{\alpha}}{(\max\{t,1\})^{\lambda +\alpha}}t^{\lambda_{1}-1} \,dt\in(0,1). $$

(22)

For $V_{n}>U(m_{0})$, we obtain

$$ \vartheta(\lambda_{1},n)=\frac{1}{k(\lambda_{1})} \int_{0}^{\frac {U(m_{0})}{V_{n}}}t^{\lambda_{1}+\alpha-1}\,dt= \frac{1}{(\lambda_{1}+\alpha )k(\lambda_{1})} \biggl( \frac{U(m_{0})}{V_{n}} \biggr) ^{\lambda _{1}+\alpha }. $$

Hence, we have (18).

For $a>0$, we find

$$\begin{aligned}& \begin{aligned}[b] \sum_{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+a}} &=\sum _{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+a}}+\sum _{m=m_{0}+1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+a}} \\ &=\sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+a}}+\sum _{m=m_{0}+1}^{\infty } \int_{m-1}^{m}\frac{U^{\prime}(x)}{U_{m}^{1+a}}\,dx \\ &< \sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+a}}+\sum _{m=m_{0}+1}^{\infty } \int_{m-1}^{m}\frac{U^{\prime}(x)}{U^{1+a}(x)}\,dx \\ &=\sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+a}}+ \int_{m_{0}}^{\infty} \frac{dU(x)}{U^{1+a}(x)}=\sum _{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+a}}+ \frac{1}{aU_{m_{0}}^{a}} =\frac{1}{a} \Biggl( \frac{1}{U_{m_{0}}^{a}}+a\sum _{m=1}^{m_{0}}\frac {\mu _{m}}{U_{m}^{1+a}} \Biggr) , \end{aligned}\\& \begin{aligned}[b] \sum_{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+a}} &\geq \sum_{m=m_{0}}^{\infty}\frac{\mu_{m+1}}{U_{m}^{1+a}}=\sum _{m=m_{0}}^{ \infty} \int_{m}^{m+1}\frac{U^{\prime}(x)}{U_{m}^{1+a}}\,dx \\ &>\sum_{m=m_{0}}^{\infty} \int_{m}^{m+1}\frac{U^{\prime }(x)\,dx}{U^{1+a}(x)}= \int_{m_{0}}^{\infty}\frac{dU(x)}{U^{1+a}(x)}=\frac{1}{aU_{m_{0}}^{a}}. \end{aligned} \end{aligned}$$

Hence, we have (19). In the same way, have (20). □

3 Equivalent inequalities and operator expressions

Theorem 4

If $-\alpha<\lambda_{1},\lambda_{2}\leq 1-\alpha$, $\lambda_{1}+\lambda_{2}=\lambda$, $k(\lambda_{1})$ is as in (9), then for $p>1$, $0<\|a\|_{p,\Phi_{\lambda }},\|b\|_{q,\Psi_{\lambda}}<\infty$, we have the following equivalent inequalities:

$$\begin{aligned}& I :=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}a_{m}b_{n}}{(\max\{U_{m},V_{n}\})^{\lambda +\alpha}}< k(\lambda_{1}) \|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi _{\lambda }}, \end{aligned}$$

(23)

$$\begin{aligned}& J := \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha }a_{m}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}} \Biggr] ^{p} \Biggr\} ^{\frac {1}{p}}< k( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}. \end{aligned}$$

(24)

Proof

By Hölder’s inequality with weight (see [25]) we have

$$\begin{aligned} & \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha }a_{m}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}} \Biggr] ^{p} \\ &\quad= \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha }}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}} \biggl( \frac{U_{m}^{\frac{1-\lambda _{1}}{q}}a_{m}}{V_{n}^{\frac{1-\lambda_{2}}{p}}\mu_{m}^{\frac {1}{q}}} \biggr) \biggl( \frac{V_{n}^{\frac{1-\lambda_{2}}{p}}\mu_{m}^{\frac {1}{q}}}{U_{m}^{\frac{1-\lambda_{1}}{q}}} \biggr) \Biggr] ^{p} \\ &\quad\leq\sum_{m=1}^{\infty}\frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}} \biggl( \frac{U_{m}^{(1-\lambda _{1})p/q}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p/q}}a_{m}^{p} \biggr) \\ &\qquad{}\times \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\} )^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{V_{n}^{(1-\lambda _{2})(q-1)}\mu_{m}}{U_{m}^{1-\lambda_{1}}} \Biggr] ^{p-1} \\ &\quad=\frac{V_{n}^{1-p\lambda_{2}}}{(\varpi(\lambda _{1},n))^{1-p}\upsilon _{n}}\sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{U_{m}^{(1-\lambda _{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p-1}}a_{m}^{p}. \end{aligned}$$

(25)

In view of (14), we find

$$\begin{aligned} J \leq&\bigl(k(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum _{n=1}^{\infty }\sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{U_{m}^{(1-\lambda _{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum _{m=1}^{\infty }\sum_{n=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{U_{m}^{(1-\lambda _{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum _{m=1}^{\infty}\omega (\lambda _{2},m) \frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu_{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$

(26)

Then by (13) we have (24).

By Hölder’s inequality we have

$$\begin{aligned} I =&\sum_{n=1}^{\infty} \Biggl[ \frac{\upsilon_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\lambda_{2}}}\sum_{m=1}^{\infty} \frac{(\min \{U_{m},V_{n}\})^{\alpha}a_{m}}{(\max\{U_{m},V_{n}\})^{\lambda +\alpha}} \Biggr] \biggl( \frac{V_{n}^{\frac{1}{p}-\lambda_{2}}}{\upsilon _{n}^{\frac{1}{p}}}b_{n} \biggr) \\ \leq&J\|b\|_{q,\Psi_{\lambda}}. \end{aligned}$$

(27)

Then by (24) we have (23).

On the other hand, assuming that (23) is valid, we set

$$ b_{n}:=\frac{\upsilon_{n}}{V_{n}^{1-p\lambda_{2}}} \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha}a_{m}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}} \Biggr] ^{p-1},\quad n\in\mathbf{N}. $$

Then we find $J^{p}=\|b\|_{q,\Psi_{\lambda}}^{q}$. If $J=0$, then (24) is trivially valid; if $J=\infty$, then by (26) and (13) it is impossible. Suppose that $0< J<\infty$. By (23) it follows that

$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q} =J^{p}=I< k(\lambda_{1}) \|a\|_{p,\Phi _{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$

(28)

$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q-1} =J< k_{s}( \lambda_{1})\|a\|_{p,\Phi _{\lambda}}, \end{aligned}$$

(29)

and then (24) follows, which is equivalent to (23). □

Theorem 5

With the assumptions of Theorem 4, if $m_{0},n_{0}\in \mathbf{N}$, $\mu_{m}\geq\mu_{m+1}$ ($m\in\{m_{0},m_{0}+1,\ldots\}$), $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots \}$), $U(\infty)=V(\infty)=\infty$, then the constant factor $k(\lambda _{1})$ in (23) and (24) is the best possible.

Proof

For $\varepsilon\in(0,p(\lambda_{1}+\alpha))$, we set $\widetilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon}{p}$ ($\in (-\alpha ,1-\alpha)$), $\widetilde{\lambda}_{2}=\lambda_{2}+\frac{\varepsilon }{p}$ ($>-\alpha$), and $\widetilde{a}=\{\widetilde{a}_{m}\} _{m=1}^{\infty}$, $\widetilde{b}=\{\widetilde{b}_{n}\}_{n=1}^{\infty}$,

$$ \widetilde{a}_{m}:=U_{m}^{\widetilde{\lambda}_{1}-1}\mu _{m}=U_{m}^{\lambda _{1}-\frac{\varepsilon}{p}-1}\mu_{m},\qquad\widetilde {b}_{n}=V_{n}^{\widetilde{\lambda}_{2}-\varepsilon-1}\upsilon_{n}=V_{n}^{\lambda_{2}-\frac{\varepsilon}{q}-1} \upsilon_{n}. $$

(30)

Then by (19), (20), and (18) we have

$$\begin{aligned}& \begin{aligned}[b] \|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\Psi _{\lambda }}&= \Biggl( \sum _{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}}\\ &=\frac{1}{\varepsilon} \biggl( \frac{1}{U_{m_{0}}^{\varepsilon }}+\varepsilon O(1) \biggr) ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I} &:=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}\widetilde{a}_{m}\widetilde{b}_{n} \\ &=\sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{V_{n}^{\widetilde{\lambda}_{2}}\mu_{m}}{U_{m}^{1-\widetilde{\lambda }_{1}}} \Biggr] \frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\sum_{n=1}^{\infty}\varpi(\widetilde{ \lambda}_{1},n)\frac {\upsilon_{n}}{V_{n}^{\varepsilon+1}}\geq k(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty } \bigl(1-\vartheta(\widetilde{\lambda}_{1},n)\bigr)\frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=k(\widetilde{\lambda}_{1}) \Biggl( \sum _{n=1}^{\infty}\frac{\upsilon _{n}}{V_{n}^{\varepsilon+1}}-\sum _{n=1}^{\infty}O\biggl(\frac{\upsilon _{n}}{V_{n}^{\frac{\varepsilon}{q}+\lambda_{1}+\alpha+1}}\biggr) \Biggr) \\ &=\frac{1}{\varepsilon}k(\widetilde{\lambda}_{1}) \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl(\widetilde{O}(1)-O(1)\bigr) \biggr] . \end{aligned} \end{aligned}$$

If there exists a positive constant $K\leq k(\lambda_{1})$ such that (23) is valid when replacing $k(\lambda_{1})$ to K, then, in particular, we have $\varepsilon\widetilde{I}<\varepsilon K\|\widetilde {a}\|_{p,\Phi _{\lambda}}\|\widetilde{b}\|_{q,\Psi_{\lambda}}$, namely,

$$\begin{aligned} k(\widetilde{\lambda}_{1}) \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon}}+ \varepsilon\bigl(\widetilde{O}(1)-O(1)\bigr) \biggr] < K \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{ \frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon \widetilde{O}(1) \biggr) ^{\frac{1}{q}}. \end{aligned}$$

It follows that $k(\lambda_{1})\leq K$ ($\varepsilon\rightarrow0^{+}$). Hence, $K=k(\lambda_{1})$ is the best possible constant factor of (23).

The constant factor $k(\lambda_{1})$ in (24) is still the best possible. Otherwise, we would reach a contradiction by (27) that the constant factor in (23) is not the best possible. □

For $p>1$, we find $\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon_{n}}{V_{n}^{1-p\lambda_{2}}}$ and define the following normed spaces:

$$\begin{aligned}& l_{p,\Phi_{\lambda}} :=\bigl\{ a=\{a_{m}\}_{m=1}^{\infty}; \|a\|_{p,\Phi _{\lambda}}< \infty\bigr\} , \\& l_{q,\Psi_{\lambda}} :=\bigl\{ b=\{b_{n}\}_{n=1}^{\infty}; \|b\|_{q,\Psi _{\lambda}}< \infty\bigr\} , \\& l_{p,\Psi_{\lambda}^{1-p}} :=\bigl\{ c=\{c_{n}\}_{n=1}^{\infty }; \|c\|_{p,\Psi _{\lambda}^{1-p}}< \infty\bigr\} . \end{aligned}$$

Assuming that $a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}$ and setting

$$ c=\{c_{n}\}_{n=1}^{\infty},\qquad c_{n}:=\sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}a_{m},\quad n \in\mathbf{N}, $$

we can rewrite (24) as

$$ \|c\|_{p,\Psi_{\lambda}^{1-p}}< k(\lambda_{1})\|a\|_{p,\Phi_{\lambda }}< \infty, $$

namely, $c\in l_{p,\Psi_{\lambda}^{1-p}}$.

Definition 1

Define a Hardy-Hilbert-type operator $T:l_{p,\Phi _{\lambda}}\rightarrow l_{p,\Psi_{\lambda}^{1-p}}$ as follows: For any $a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}$, there exists a unique representation $Ta=c\in l_{p,\Psi_{\lambda}^{1-p}}$. Define the formal inner product of Ta and $b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\Psi _{\lambda}}$ as follows:

$$ (Ta,b):=\sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}a_{m} \Biggr] b_{n}. $$

(31)

Then we can rewrite (23) and (24) as follows:

$$\begin{aligned}& (Ta,b) < k(\lambda_{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi _{\lambda }}, \end{aligned}$$

(32)

$$\begin{aligned}& \|Ta\|_{p,\Psi_{\lambda}^{1-p}} < k(\lambda_{1})\|a\|_{p,\Phi _{\lambda }}. \end{aligned}$$

(33)

Define the norm of the operator T as follows:

$$ \|T\|:=\sup_{a(\neq\theta)\in l_{p,\Phi_{\lambda}}}\frac {\|Ta\|_{p,\Psi _{\lambda}^{1-p}}}{\|a\|_{p,\Phi_{\lambda}}}. $$

Then by (31) we find $\|T\|\leq k(\lambda_{1})$. Since by Theorem 5 the constant factor in (31) is the best possible, we have

$$ \|T\|=k(\lambda_{1})=\frac{\lambda+2\alpha}{(\lambda_{1}+\alpha )(\lambda_{2}+\alpha)}. $$

4 Some equivalent reverse inequalities

Theorem 6

If $-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha$, $\lambda_{1}+\lambda_{2}=\lambda$, $k(\lambda_{1})$ is as in (9), $m_{0},n_{0}\in\mathbf{N}$, $\mu_{m}\geq\mu_{m+1}$ ($m\in \{m_{0},m_{0}+1, \ldots\}$), $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\in \mathbf{\{}n_{0},n_{0}+1, \ldots\}$), $U(\infty)=V(\infty)=\infty$, then for $0< p<1$, $0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi_{\lambda }}<\infty$, we have the following equivalent inequalities with the best possible constant factor $k(\lambda_{1})$:

$$\begin{aligned}& I =\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}a_{m}b_{n}>k( \lambda_{1})\|a\|_{p,\widetilde{\Phi}_{\lambda }}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$

(34)

$$\begin{aligned}& J = \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha }a_{m}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}} \Biggr] ^{p} \Biggr\} ^{\frac {1}{p}}>k( \lambda_{1})\|a\|_{p,\widetilde{\Phi}_{\lambda}}. \end{aligned}$$

(35)

Proof

By the reverse Hölder’s inequality and (14), we have the reverses of (25), (26), and (27). Then by (17) we have (35). By (35) and the reverse of (27) we have (34).

On the other hand, assuming that (34) is valid, we set $b_{n}$ as in Theorem 4. Then we find $J^{p}=\|b\|_{q,\Psi_{\lambda }}^{q}$. If $J=\infty$, then (35) is trivially valid; if $J=0$, then by reverse of (26) and (17) it is impossible. Suppose that $0< J<\infty$. By (34) it follows that

$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q} =J^{p}=I>k_{s}( \lambda_{1})\|a\|_{p,\widetilde{\Phi}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$

(36)

$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q-1} =J>k_{s}(\lambda _{1})\|a\|_{p,\widetilde{\Phi}_{\lambda}}, \end{aligned}$$

(37)

and then (35) follows, which is equivalent to (34).

For $\varepsilon\in(0,p(\lambda_{1}+\alpha))$, we set $\widetilde{\lambda}_{1}$, $\widetilde{\lambda}_{2}$, $\widetilde{a}_{m}$, and $\widetilde{b}_{n}$ as (30). Then by (19), (20), and (14) we find

$$\begin{aligned}& \begin{aligned}[b] \|a\|_{p,\widetilde{\Phi}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}&= \Biggl[ \sum _{m=1}^{\infty}\bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\mu_{m}}{U_{m}^{1+\varepsilon}} \Biggr] ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}}\\ &= \Biggl( \sum_{m=1}^{\infty} \frac{\mu_{m}}{U_{m}^{1+\varepsilon}}-\sum_{m=1}^{\infty}O \biggl(\frac{\mu_{m}}{U_{m}^{1+\lambda_{2}+\alpha +\varepsilon}}\biggr) \Biggr) ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}} \\ &=\frac{1}{\varepsilon} \biggl[ \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon \bigl(O(1)-O_{1}(1)\bigr) \biggr] ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I} &=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}\widetilde{a}_{m}\widetilde{b}_{n}\\ &=\sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{V_{n}^{\widetilde{\lambda}_{2}}\mu_{m}}{U_{m}^{1-\widetilde{\lambda }_{1}}} \Biggr] \frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\sum_{n=1}^{\infty}\varpi(\widetilde{ \lambda}_{1},n)\frac {\upsilon_{n}}{V_{n}^{\varepsilon+1}}\leq k(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\frac{1}{\varepsilon}k(\widetilde{\lambda}_{1}) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) . \end{aligned} \end{aligned}$$

If there exists a constant $K\geq k(\lambda_{1})$ such that (34) is valid when replacing $k(\lambda_{1})$ to K, then, in particular, we have $\varepsilon\widetilde{I}>\varepsilon K\|\widetilde{a}\|_{p,\widetilde {\Phi}_{\lambda}}\|\widetilde{b}\|_{q,\Psi_{\lambda}}$, namely,

$$\begin{aligned} k(\widetilde{\lambda}_{1}) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+ \varepsilon\widetilde{O}(1) \biggr) >K \biggl[ \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon \bigl(O(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon }}+\varepsilon \widetilde{O}(1) \biggr) ^{\frac{1}{q}}. \end{aligned}$$

It follows that $k(\lambda_{1})\geq K$ ($\varepsilon\rightarrow0^{+}$). Hence, $K=k(\lambda_{1})$ is the best possible constant factor of (34).

The constant factor $k(\lambda_{1})$ in (35) is still the best possible. Otherwise, we would reach a contradiction by the reverse of (27) that the constant factor in (34) is not the best possible. □

Theorem 7

With the assumptions of Theorem 6, if $p<0$, then we have the following equivalent inequalities with the best possible constant factor $k(\lambda_{1})$:

$$\begin{aligned}& I =\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}a_{m}b_{n}>k( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\widetilde {\Psi }_{\lambda}}, \end{aligned}$$

(38)

$$\begin{aligned}& \begin{aligned}[b] J_{1} &:= \Biggl\{ \sum_{n=1}^{\infty} \frac{V_{n}^{p\lambda _{2}-1}\upsilon _{n}}{(1-\vartheta(\lambda_{1},n))^{p-1}} \Biggl[ \sum_{m=1}^{\infty } \frac{(\min\{U_{m},V_{n}\})^{\alpha}a_{m}}{(\max\{U_{m},V_{n}\})^{\lambda +\alpha}} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\ &>k(\lambda_{1})\|a\|_{p,\Phi_{\lambda}}. \end{aligned} \end{aligned}$$

(39)

Proof

By the reverse Hölder inequality with weight, since $p<0$, by (18) we have

$$\begin{aligned} & \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha }}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}}a_{m} \Biggr] ^{p} \\ &\quad= \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha }}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}} \biggl( \frac{U_{m}^{(1-\lambda _{1})/q}}{V_{n}^{(1-\lambda_{2})/p}\mu_{m}^{1/q}}a_{m} \biggr) \biggl( \frac{V_{n}^{(1-\lambda_{2})/p}\mu_{m}^{1/q}}{U_{m}^{(1-\lambda _{1})/q}} \biggr) \Biggr] ^{p} \\ &\quad\leq\sum_{m=1}^{\infty}\frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}} \frac{U_{m}^{(1-\lambda_{1})p/q}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p/q}}a_{m}^{p} \\ &\qquad{}\times \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\} )^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{V_{n}^{(1-\lambda _{2})(q-1)}\mu_{m}}{U_{m}^{1-\lambda_{1}}} \Biggr] ^{p-1} \\ &\quad=\frac{V_{n}^{1-p\lambda_{2}}}{(\varpi(\lambda_{1},n))^{1-p}}\sum_{m=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{U_{m}^{(1-\lambda _{1})(p-1)}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p-1}}a_{m}^{p} \\ &\quad\leq\frac{(k(\lambda_{1}))^{p-1}V_{n}^{1-p\lambda _{2}}}{(1-\vartheta (\lambda_{1},n))^{1-p}\upsilon_{n}}\sum_{m=1}^{\infty} \frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{U_{m}^{(1-\lambda_{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu _{m}^{p-1}}a_{m}^{p}, \\ &J_{1} \geq\bigl(k(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl\{ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty}\frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}}\frac{U_{m}^{(1-\lambda _{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p-1}}a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &\hphantom{J_{1}}=\bigl(k(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl\{ \sum _{m=1}^{\infty }\sum_{n=1}^{\infty} \frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max \{U_{m},V_{n}\})^{\lambda+\alpha}}a_{m}\frac{U_{m}^{(1-\lambda _{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p-1}}a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &\hphantom{J_{1}}=\bigl(k(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl\{ \sum _{m=1}^{\infty}\omega (\lambda_{2},m) \frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu_{m}^{p-1}}a_{m}^{p} \Biggr\} ^{\frac{1}{p}}. \end{aligned}$$

(40)

Then by (13) we have (39).

By the reverse Hölder inequality we have

$$\begin{aligned} I={}&\sum_{n=1}^{\infty}\frac{V_{n}^{\lambda_{2}-\frac{1}{p}}\upsilon _{n}^{1/p}}{(1-\vartheta(\lambda_{1},n))^{1/q}} \Biggl[ \sum_{m=1}^{\infty}\frac{(\min\{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda +\alpha}}a_{m} \Biggr] \\ &{}\times \biggl[ \bigl(1-\vartheta(\lambda_{1},n)\bigr)^{\frac{1}{q}} \frac {V_{n}^{\frac{1}{p}-\lambda_{2}}}{\upsilon_{n}^{1/p}}b_{n} \biggr] \geq J_{1}\|b \|_{q,\widetilde{\Psi}_{\lambda}}. \end{aligned}$$

(41)

Then by (39) we have (38).

On the other hand, assuming that (38) is valid, we set $b_{n}$ as follows:

$$ b_{n}:=\frac{V_{n}^{p\lambda_{2}-1}\upsilon_{n}}{(1-\vartheta(\lambda _{1},n))^{p-1}} \Biggl[ \sum_{m=1}^{\infty} \frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}a_{m} \Biggr] ^{p-1},\quad n\in \mathbf{N}. $$

Then we find $J_{1}^{p}=\|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q}$. If $J_{1}=\infty$, then (39) is trivially valid; if $J_{1}=0$, then by (40) and (13) it is impossible. Suppose that $0< J_{1}<\infty$. By (38) it follows that

$$\begin{aligned}& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q} =J_{1}^{p}=I>k_{s}( \lambda _{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\widetilde{\Psi}_{\lambda}}, \\& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q-1} =J_{1}>k_{s}( \lambda _{1})\|a\|_{p,\Phi_{\lambda}}, \end{aligned}$$

and then (39) follows, which is equivalent to (38).

For $\varepsilon\in(0,q(\lambda_{2}+\alpha))$, we set $\widetilde{\lambda}_{1}=\lambda_{1}+\frac{\varepsilon}{q}$ ($>-\alpha$), $\widetilde{\lambda}_{2}=\lambda_{2}-\frac{\varepsilon}{q}$ ($\in(-\alpha ,1-\alpha)$), and

$$ \widetilde{a}_{m}:=U_{m}^{\widetilde{\lambda}_{1}-1-\varepsilon}\mu _{m}=U_{m}^{\lambda_{1}-\frac{\varepsilon}{p}-1}\mu_{m},\qquad\widetilde{b} _{n}=V_{n}^{\widetilde{\lambda}_{2}-1}\upsilon_{n}=V_{n}^{\lambda _{2}-\frac{\varepsilon}{q}-1} \upsilon_{n}. $$

Then by (19), (20), and (13) we have

$$\begin{aligned}& \begin{aligned}[b] \|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\widetilde {\Psi}_{\lambda}}&= \Biggl( \sum _{m=1}^{\infty}\frac{\mu _{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty} \bigl(1-\vartheta(\lambda _{1},n)\bigr)\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr] ^{\frac{1}{q}}\\ &= \Biggl( \sum_{m=1}^{\infty} \frac{\mu_{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}}-\sum _{n=1}^{\infty}O\biggl(\frac{\upsilon_{n}}{V_{n}^{1+\lambda_{1}+\alpha+\varepsilon}}\biggr) \Biggr) ^{\frac{1}{q}}\\ &=\frac{1}{\varepsilon} \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{p}} \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl( \widetilde{O}(1)-O_{1}(1)\bigr) \biggr] ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I} &=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha}}\widetilde{a}_{m}\widetilde{b}_{n}\\ &=\sum_{m=1}^{\infty} \Biggl[ \sum _{n=1}^{\infty}\frac{(\min \{U_{m},V_{n}\})^{\alpha}}{(\max\{U_{m},V_{n}\})^{\lambda+\alpha }}\frac{U_{m}^{\widetilde{\lambda}_{1}}\upsilon_{n}}{V_{n}^{1-\widetilde {\lambda}_{2}}} \Biggr] \frac{\mu_{m}}{U_{m}^{1+\varepsilon}}\\ &=\sum_{m=1}^{\infty}\omega(\widetilde{ \lambda}_{2},m)\frac{\mu _{m}}{U_{m}^{1+\varepsilon}}\leq k(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+\varepsilon}}\\ &=\frac{1}{\varepsilon}k(\widetilde{\lambda}_{1}) \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) . \end{aligned} \end{aligned}$$

If there exists a constant $K\geq k(\lambda_{1})$ such that (38) is valid when replacing $k(\lambda_{1})$ to K, then, in particular, we have $\varepsilon\widetilde{I}>\varepsilon K\|\widetilde{a}\|_{p,\Phi _{\lambda }}\|\widetilde{b}\|_{q,\widetilde{\Psi}_{\lambda}}$, namely,

$$\begin{aligned} k(\widetilde{\lambda}_{1}) \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+ \varepsilon O(1) \biggr) >K \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{ \frac{1}{p}} \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl(\widetilde{O}(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{q}}. \end{aligned}$$

It follows that $k(\lambda_{1})\geq K$ ($\varepsilon\rightarrow0^{+}$). Hence, $K=k(\lambda_{1})$ is the best possible constant factor of (38).

The constant factor $k(\lambda_{1})$ in (39) is still the best possible. Otherwise, we would reach a contradiction by (41) that the constant factor in (38) is not the best possible. □

Remark 1

(i) For $\alpha=0$ and $0<\lambda _{1},\lambda_{2}\leq1$ in (23) and (24), we have the following equivalent inequalities:

$$\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{(\max \{U_{m},V_{n}\})^{\lambda}}< \frac{\lambda}{\lambda_{1}\lambda_{2}}\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$

(42)

$$\begin{aligned}& \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \frac{a_{m}}{(\max\{U_{m},V_{n}\})^{\lambda }} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< \frac{\lambda}{\lambda_{1}\lambda _{2}}\|a\|_{p,\Phi_{\lambda}}; \end{aligned}$$

(43)

(ii) for $\alpha=-\lambda$ and $-1\leq\lambda_{1},\lambda_{2}<0$ in (23) and (24), we have the following equivalent inequalities:

$$\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{(\min \{U_{m},V_{n}\})^{\lambda}}< \frac{(-\lambda)}{\lambda_{1}\lambda _{2}}\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$

(44)

$$\begin{aligned}& \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \frac{a_{m}}{(\min\{U_{m},V_{n}\})^{\lambda }} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< \frac{(-\lambda)}{\lambda _{1}\lambda _{2}}\|a\|_{p,\Phi_{\lambda}}; \end{aligned}$$

(45)

(iii) for $\lambda=0$, $|\lambda_{1}|<\alpha$ ($0<\alpha\leq\frac {1}{2}$); $|\lambda_{1}|<1-\alpha$ ($\frac{1}{2}<\alpha\leq1$), $\lambda _{2}=-\lambda _{1}$ in (23) and (24), we have the following equivalent inequalities:

$$\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty} \biggl( \frac{\min\{ U_{m},V_{n}\}}{\max\{U_{m},V_{n}\}} \biggr) ^{\alpha}a_{m}b_{n}< \frac{2\alpha}{\alpha ^{2}-\lambda_{1}^{2}}\|a \|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$

(46)

$$\begin{aligned}& \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1+p\lambda _{1}}} \Biggl[ \sum_{m=1}^{\infty} \biggl( \frac{\min\{U_{m},V_{n}\}}{\max \{U_{m},V_{n}\}} \biggr) ^{\alpha}a_{m} \Biggr] ^{p} \Biggr\} ^{\frac {1}{p}}< \frac{2\alpha}{\alpha^{2}-\lambda_{1}^{2}}\|a \|_{p,\Phi_{\lambda}}. \end{aligned}$$

(47)

In view of Theorem 5, the constant factors in these inequalities with the particular kernels are all the best possible.