On a half-discrete Hilbert-type inequality related to hyperbolic functions

Abstract

By the introduction of a new half-discrete kernel which is composed of several exponent functions, and using the method of weight coefficient, a Hilbert-type inequality and its equivalent forms involving multiple parameters are established. In addition, it is proved that the constant factors of the newly obtained inequalities are the best possible. Furthermore, by the use of the rational fraction expansion of the tangent function and introducing the Bernoulli numbers, some interesting and special half-discrete Hilbert-type inequalities are presented at the end of the paper.

1 Introduction

Let $a_{n},\mu _{n}>0$, $a=\{a_{n}\}_{n=1}^{\infty }$, and $p>1$. Define

$$\begin{aligned} l_{p,\mu }:= \Biggl\{ a: \Vert a \Vert _{p,\mu }:= \Biggl(\sum_{k=1}^{\infty }\mu _{k}a_{k}^{p} \Biggr)^{\frac{1}{p}}< \infty \Biggr\} . \end{aligned}$$

In particular, if $\mu _{n}=1$, then we have the abbreviated notations $\|a\|_{p}:=\|a\|_{p,\mu }$ and $l_{p}:=l_{p,\mu }$.

Let $p>1$ and consider measurable functions $f(x),\nu (x)>0$. Define

$$\begin{aligned} L_{p,\nu }\bigl(\mathbb{R}^{+}\bigr):= \biggl\{ f: \Vert f \Vert _{p,\nu }:= \biggl({ \int _{ \mathbb{R}^{+}}\nu (x)f^{p}(x)\,\mathrm{d}x} \biggr)^{\frac{1}{p}}< \infty \biggr\} . \end{aligned}$$

For $\nu (x)=1$, we have the abbreviations as follows: $\|f\|_{p}:=\|f\|_{p,\nu }$ and $L_{p}(\mathbb{R}^{+}):=L_{p,\nu }(\mathbb{R}^{+})$.

Consider two sequences of real numbers, $a=\{a_{n}\}_{n=1}^{\infty }\in l_{2}$ and $b=\{b_{n}\}_{n=1}^{\infty }\in l_{2}$. Then

$$\begin{aligned} \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \pi \Vert a \Vert _{2} \Vert b \Vert _{2}, \end{aligned}$$

(1.1)

where the constant factor π is the best possible. Inequality (1.1) was first proved by the German mathematician D. Hilbert in 1908 in his lectures on integral equations, and is usually known as Hilbert’s double series inequality [1]. Three years later, Schur established the integral analogue of (1.1), that is[1],

$$\begin{aligned} \int _{0}^{\infty } \int _{0}^{\infty }\frac{f(x)g(y)}{x+y} \,\mathrm{d}x \, \mathrm{d}y< \pi \Vert f \Vert _{2} \Vert g \Vert _{2}, \end{aligned}$$

(1.2)

where $f, g\geq 0$ are two real-valued functions, and $f, g\in L_{2}(\mathbb{R}^{+})$.

In 1925, by the introduction of a pair of conjugate parameters p and q, $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, Hardy and Riesz generalized (1.1) as follows:

$$\begin{aligned} \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{\pi }{\sin \frac{\pi }{p}} \Vert a \Vert _{p} \Vert b \Vert _{q}, \end{aligned}$$

(1.3)

where $a=\{a_{n}\}_{n=1}^{\infty }\in l_{p}$ and $b=\{b_{n}\}_{n=1}^{\infty }\in l_{q}$.

Since the 1990s, by the introduction of parameters and special functions, researchers established quite a few generalizations of (1.3) (see [2–11]).

It should be noted that there is a sharper form of (1.3) (see [1, Theorem 323]):

$$\begin{aligned} \sum_{n=0}^{\infty }\sum _{m=0}^{\infty }\frac{a_{m}b_{n}}{m+n+1}< \frac{\pi }{\sin \frac{\pi }{p}} \Vert a \Vert _{p} \Vert b \Vert _{q}. \end{aligned}$$

(1.4)

Regarding extensions of (1.4), we can refer to [12–16]. In addition, some extensions of (1.2) were also established in the past 20 years (see [10, 11, 17–20]). Furthermore, by constructing new kernel functions, introducing parameters, and considering coefficient refinement, reverse form, and multi-dimensional extension, a large number of new inequalities similar to (1.1) and (1.2) were established in the past several decades (see [21–31]). These newly constructed inequalities are generally called Hilbert-type inequalities.

In addition to integral and discrete forms, Hilbert-type inequalities also appear in half-discrete form. The first half-discrete Hilbert-type inequality was proved by Hardy et al. [1, Theorem 351], but the constant factor was not proved to be the best possible. In 2011, Yang [32] proved that

$$\begin{aligned} \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty } \frac{a_{n}}{n+x}\,\mathrm{d}x< \pi \Vert f \Vert _{2} \Vert a \Vert _{2}, \end{aligned}$$

(1.5)

where the constant factor π is the best possible. With regard to the related results of half-discrete Hilbert inequality, we can refer to [33–40].

The main objective of this paper is to establish a half-discrete Hilbert-type inequality with the kernel function, in particular exponential or hyperbolic function. We first present some relevant results in the literature. For example, Yang [41] proved that

$$\begin{aligned} \int _{0}^{\infty } \int _{0}^{\infty }e^{-\frac{x}{y}}f(x)g(y)\,\mathrm{d}y< \Vert f \Vert _{2,\nu _{1}} \Vert g \Vert _{2,\nu _{2}}, \end{aligned}$$

(1.6)

where $\nu _{1}(x)=x^{-1}$ and $\nu _{2}(y)=y^{3}$.

In 2013, Liu [42] established the following inequality involving hyperbolic secant function:

$$\begin{aligned} \int _{0}^{\infty } \int _{0}^{\infty }\operatorname{sech}(xy) f(x)g(y) \, \mathrm{d}x\,\mathrm{d}y< 2c_{0} \Vert f \Vert _{2,\nu } \Vert g \Vert _{2, \nu }, \end{aligned}$$

(1.7)

where $\operatorname{sech} (u)=\frac{2}{e^{u}+e^{-u}}$, $\nu (x)=x^{-3}$, and $c_{0}=\sum_{k=0}^{\infty }\frac{(-1)^{k}}{(2k+1)^{2}}=0.915965^{+}$ is the Catalan constant.

Another Hilbert-type inequality with a half-discrete kernel involving hyperbolic secant function was established by Zhong [43] in 2012. It reads

$$\begin{aligned} \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty }a_{n} \operatorname{sech} \biggl(\frac{n}{x} \biggr) \,\mathrm{d}x< \frac{\pi }{2} \Vert f \Vert _{2,\nu } \Vert a \Vert _{2, \mu }, \end{aligned}$$

(1.8)

where $\nu (x)=x^{3}$, $\mu _{n}=\frac{1}{n}$.

In this work, we will establish the following half-discrete Hilbert-type inequalities with the kernels involving hyperbolic tangent and cotangent functions:

$$\begin{aligned} &\int _{0}^{\infty }f(x)\sum_{n=1}^{\infty }a_{n} \bigl[\coth { \sqrt[2m]{nx}}-1 \bigr]\,\mathrm{d}x< B_{m}\pi ^{2m} \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }, \end{aligned}$$

(1.9)

$$\begin{aligned} &\int _{0}^{\infty }f(x)\sum_{n=1}^{\infty }a_{n} \bigl[1-\tanh { \sqrt[2m]{nx} } \bigr]\,\mathrm{d}x< \biggl(1-\frac{1}{2^{2m-1}} \biggr)B_{m}\pi ^{2m} \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }, \end{aligned}$$

(1.10)

where $\nu (x)=\frac{1}{x}$, $\mu _{n}=\frac{1}{n}$, $m\in \mathbb{N}^{+}$.

It is of interest that we also present some other half-discrete inequalities involving hyperbolic functions. More generally, we construct a kernel function with multiple parameters, which unifies the homogeneous and nonhomogeneous kernels, and then a half-discrete Hilbert-type inequality and its equivalent forms are established.

2 Some lemmas

Lemma 2.1

Let $\eta _{1},\eta _{2}\in \{1,-1\}$, and $\eta _{2}\neq -1$ for $\eta _{1}=1$. Let $\gamma _{1}\in \mathbb{R}^{+},\gamma _{2},\gamma _{3}\in \mathbb{R}$. Suppose that $\gamma _{3}\leq \gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$ for $\eta _{2}=1$, and $\gamma _{3}<\gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$ for $\eta _{2}=-1$. Define

$$\begin{aligned} \kappa (u):= \frac{ e^{\gamma _{2}u}+\eta _{2}e^{\gamma _{3}u}}{ e^{\gamma _{1}u}+\eta _{1}e^{-\gamma _{1}u}},\quad u>0. \end{aligned}$$

(2.1)

Then $\kappa (u)$ is decreasing on $\mathbb{R}^{+}$.

Proof

It is easy to show that

$$\begin{aligned} \frac{\mathrm{d} \kappa }{\mathrm{d} u}= {}&\bigl[ (\gamma _{2}- \gamma _{1} )e^{(\gamma _{2}+\gamma _{1})u}+\eta _{1} ( \gamma _{2}+ \gamma _{1} )e^{(\gamma _{2}-\gamma _{1})u} \\ &{} + \eta _{2} (\gamma _{3}-\gamma _{1} )e^{(\gamma _{3}+ \gamma _{1})u}+\eta _{1}\eta _{2} (\gamma _{3}+\gamma _{1} )e^{(\gamma _{3}-\gamma _{1})u} \bigr] \\ &{} \times \bigl( e^{\gamma _{1}u}+\eta _{1}e^{-\gamma _{1}u} \bigr)^{-2}:=\kappa _{1}(u) \bigl( e^{\gamma _{1}u}+\eta _{1}e^{- \gamma _{1}u} \bigr)^{-2} \end{aligned}$$

(2.2)

and

$$\begin{aligned} \frac{\mathrm{d} \kappa _{1}}{\mathrm{d} u}={}& \bigl\{ \bigl(\gamma _{2}^{2}- \gamma _{1}^{2} \bigr) \bigl[e^{(\gamma _{2}+\gamma _{1})u}+\eta _{1}e^{( \gamma _{2}-\gamma _{1})u} \bigr] \\ &{} + \eta _{2} \bigl(\gamma _{3}^{2}- \gamma _{1}^{2} \bigr) \bigl[e^{(\gamma _{3}+\gamma _{1})u}+\eta _{1}e^{(\gamma _{3}- \gamma _{1})u} \bigr] \bigr\} . \end{aligned}$$

(2.3)

For $\eta _{2}=1,\eta _{1}=\pm 1$, since $\gamma _{3}\leq \gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$, we can obtain $\gamma _{2}^{2}-\gamma _{1}^{2}\leq 0$ and $\gamma _{3}^{2}-\gamma _{1}^{2}\leq 0$. In addition, it is obvious that $e^{(\gamma _{2}+\gamma _{1})u}+\eta _{1}e^{(\gamma _{2}-\gamma _{1})u}>0$ and $e^{(\gamma _{3}+\gamma _{1})u}+\eta _{1}e^{(\gamma _{3}-\gamma _{1})u}>0$. Therefore $\frac{\mathrm{d} \kappa _{1}}{\mathrm{d} u}\leq 0$ for $\eta _{2}=1,\eta _{1}=\pm 1$.

For $\eta _{2}=-1,\eta _{1}=-1$, write $\kappa _{2}(t):=e^{(t+\gamma _{1})u}-e^{(t-\gamma _{1})u}$, $t\in \mathbb{R}$, $u>0$. Then

$$\begin{aligned} \frac{\mathrm{d}\kappa _{2}}{\mathrm{d}t}= ue^{(t+\gamma _{1})u}-ue^{(t- \gamma _{1})u}>0. \end{aligned}$$

Therefore $\kappa _{2}(t)$ is decreasing on $\mathbb{R}$, and it follows that

$$\begin{aligned} \frac{\mathrm{d} \kappa _{1}}{\mathrm{d} u}\leq{}& \bigl\{ \bigl( \gamma _{2}^{2}-\gamma _{1}^{2} \bigr) \bigl[e^{(\gamma _{2}+\gamma _{1})u}-e^{( \gamma _{2}-\gamma _{1})u} \bigr] \\ & {}- \bigl(\gamma _{3}^{2}-\gamma _{1}^{2} \bigr) \bigl[e^{( \gamma _{2}+\gamma _{1})u}-e^{(\gamma _{2}-\gamma _{1})u} \bigr] \bigr\} \\ = {}&\bigl(\gamma _{2}^{2}-\gamma _{3}^{2} \bigr) \bigl[e^{(\gamma _{2}+ \gamma _{1})u}-e^{(\gamma _{2}-\gamma _{1})u} \bigr]\leq 0. \end{aligned}$$

(2.4)

Based on the above discussions, it follows that $\kappa _{1}(u)$ is a decreasing function on $\mathbb{R}^{+}$. Therefore, for $\eta _{1}=1, \eta _{2}=1$, $\kappa _{1}(u)<\kappa _{1}(0)=2(\gamma _{2}+\gamma _{3})\leq 0$. Similarly, we can obtain that $\kappa _{1}(u)<\kappa _{1}(0)=-4\gamma _{1}<0$ for $\eta _{1}=-1, \eta _{2}=1$, and $\kappa _{1}(u)<\kappa _{1}(0)=0$ for $\eta _{1}=-1, \eta _{2}=-1$. Applying $\kappa _{1}(u)<0$ to (2.2), we get $\frac{\mathrm{d} \kappa }{\mathrm{d} u}<0$, and it follows that $\kappa (u)$ is decreasing on $\mathbb{R}^{+}$. □

Lemma 2.2

Let $\eta _{1},\eta _{2}\in \{1,-1\}$ and $\eta _{2}\neq -1$ for $\eta _{1}=1$. Let $\beta _{1}\in \mathbb{R}^{+}$, $\beta _{2}\in \mathbb{R}\setminus \{ 0 \}$. Assume that $\beta \beta _{1}\leq 1$, $\beta \geq 1$ and $\beta \neq 1$ for $\eta _{1}=-1$, $\eta _{2}=1$. Let $\gamma _{1}\in \mathbb{R}^{+},\gamma _{2},\gamma _{3}\in \mathbb{R}$, and $\gamma _{2},\gamma _{3}\neq \gamma _{1}$. Suppose that $\gamma _{3}\leq \gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$ for $\eta _{2}=1$, and $\gamma _{3}<\gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$ for $\eta _{2}=-1$. Define

$$\begin{aligned} K(n,x):= \frac{ e^{\gamma _{2}n^{\beta _{1}}x^{\beta _{2}}}+\eta _{2}e^{\gamma _{3}n^{\beta _{1}}x^{\beta _{2}}}}{ e^{\gamma _{1}n^{\beta _{1}}x^{\beta _{2}}}+\eta _{1}e^{-\gamma _{1}n^{\beta _{1}}x^{\beta _{2}}}},\quad n\in \mathbb{N}^{+}, x\in \mathbb{R}^{+}, \end{aligned}$$

(2.5)

and

$$\begin{aligned} C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta )=\sum _{j=0}^{\infty } \biggl[ \frac{(-\eta _{1})^{j}}{(2\gamma _{1}j-\gamma _{2}+\gamma _{1})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{j}}{(2\gamma _{1}j-\gamma _{3}+\gamma _{1})^{\beta }} \biggr]. \end{aligned}$$

(2.6)

Then

$$\begin{aligned} & \omega (n):= \int _{0}^{\infty }K(n,x)x^{\beta \beta _{2}-1}\,\mathrm{d}x= \frac{n^{-\beta \beta _{1}}}{ \vert \beta _{2} \vert }\Gamma (\beta )C (\gamma _{1},\gamma _{2}, \gamma _{3},\eta _{1},\eta _{2},\beta ), \end{aligned}$$

(2.7)

$$\begin{aligned} &\varpi (x):=\sum_{n=1}^{\infty }K(n,x)n^{\beta \beta _{1}-1}< \frac{x^{-\beta \beta _{2}}}{\beta _{1}}\Gamma (\beta )C ( \gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta ). \end{aligned}$$

(2.8)

Proof

Setting $n^{\beta _{1}}x^{\beta _{2}}=u$, we obtain

$$\begin{aligned} \omega (n)=\frac{n^{-\beta \beta _{1}}}{ \vert \beta _{2} \vert } \int _{0}^{ \infty }\kappa (u)u^{\beta -1}\, \mathrm{d}u. \end{aligned}$$

(2.9)

Expanding $\kappa (u)$ into a power series of $e^{u}$, we have

$$\begin{aligned} \kappa (u)&= \frac{ e^{(\gamma _{2}-\gamma _{1})u}+\eta _{2}e^{(\gamma _{3}-\gamma _{1})u}}{ 1+\eta _{1}e^{-2\gamma _{1}u}} \\ &=\sum_{j=0}^{\infty }(-\eta _{1})^{j} \bigl[ e^{(-2\gamma _{1}j+ \gamma _{2}-\gamma _{1})u}+\eta _{2} e^{(-2\gamma _{1}j+\gamma _{3}- \gamma _{1})u} \bigr]. \end{aligned}$$

(2.10)

Therefore, by the use of Lebesgue term-by-term integration theorem, we obtain

$$\begin{aligned} \int _{0}^{\infty }\kappa (u)u^{\beta -1}\, \mathrm{d}u ={}&\sum_{j=0}^{\infty } \biggl\{ (-\eta _{1})^{j} \biggl[ \int _{0}^{\infty }e^{(-2\gamma _{1}j+ \gamma _{2}-\gamma _{1})u}u^{\beta -1}\, \mathrm{d}u \biggr] \\ &{} +\eta _{2}(-\eta _{1})^{j} \biggl[ \int _{0}^{\infty }e^{(-2 \gamma _{1}j+\gamma _{3}-\gamma _{1})u}u^{\beta -1}\, \mathrm{d}u \biggr] \biggr\} . \end{aligned}$$

(2.11)

Setting $(-2\gamma _{1}j+\gamma _{2}-\gamma _{1})u=-t$, it follows that

$$\begin{aligned} \int _{0}^{\infty }e^{(-2\gamma _{1}j+\gamma _{2}-\gamma _{1})u}u^{ \beta -1}\, \mathrm{d}u= \frac{\int _{0}^{\infty }e^{-t}t^{\beta -1}\,\mathrm{d}t}{(2\gamma _{1}j-\gamma _{2}+\gamma _{1})^{\beta }} = \frac{\Gamma (\beta )}{(2\gamma _{1}j-\gamma _{2}+\gamma _{1})^{\beta }}. \end{aligned}$$

(2.12)

Similarly, we can obtain

$$\begin{aligned} \int _{0}^{\infty }e^{(-2\gamma _{1}j+\gamma _{3}-\gamma _{1})u}u^{ \beta -1}\, \mathrm{d}u= \frac{\int _{0}^{\infty }e^{-t}t^{\beta -1}\,\mathrm{d}t}{(2\gamma _{1}j-\gamma _{3}+\gamma _{1})^{\beta }} = \frac{\Gamma (\beta )}{(2\gamma _{1}j-\gamma _{3}+\gamma _{1})^{\beta }}. \end{aligned}$$

(2.13)

Applying (2.12) and (2.13) to (2.11), and using (2.6), we have

$$\begin{aligned} \int _{0}^{\infty }\kappa (u)u^{\beta -1}\, \mathrm{d}u =\Gamma (\beta )C (\gamma _{1},\gamma _{2}, \gamma _{3},\eta _{1},\eta _{2},\beta ). \end{aligned}$$

(2.14)

Plugging (2.14) into (2.9), we have (2.7).

Furthermore, observing that $\beta _{1}>0$ and $\beta \beta _{1}-1<0$, by Lemma 2.1, it’s easy to see that $K(n,x)n^{\beta \beta _{1}-1}$ is monotonically decreasing with respect to n. Therefore,

$$\begin{aligned} \sum_{n=1}^{\infty }K(n,x)n^{\beta \beta _{1}-1}< \int _{0}^{\infty }K(u,x)u^{ \beta \beta _{1}-1}\,\mathrm{d}u. \end{aligned}$$

(2.15)

Setting $u^{\beta _{1}}x^{\beta _{2}}=t$, and using (2.14), we obtain

$$\begin{aligned} \int _{0}^{\infty }K(u,x)u^{\beta \beta _{1}-1}\,\mathrm{d}u&= \frac{x^{-\beta \beta _{2}}}{\beta _{1}} \int _{0}^{\infty }\kappa (t)t^{ \beta -1}\, \mathrm{d}t \\ &=\frac{x^{-\beta \beta _{2}}}{\beta _{1}}\Gamma (\beta )C ( \gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta ). \end{aligned}$$

(2.16)

Combining (2.15) and (2.16), we have (2.8). The proof of Lemma 2.2 is completed. □

Lemma 2.3

Let $\eta _{1},\eta _{2}\in \{1,-1\}$ and $\eta _{2}\neq -1$ for $\eta _{1}=1$. Let $\beta _{1}\in \mathbb{R}^{+}$, $\beta _{2}\in \mathbb{R}\setminus \{ 0 \}$. Assume that $\beta \beta _{1}\leq 1$, $\beta \geq 1$ and $\beta \neq 1$ for $\eta _{1}=-1$, $\eta _{2}=1$. Let $\gamma _{1}\in \mathbb{R}^{+},\gamma _{2},\gamma _{3}\in \mathbb{R}$, and $\gamma _{2},\gamma _{3}\neq \gamma _{1}$. Suppose that $\gamma _{3}\leq \gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$ for $\eta _{2}=1$, and $\gamma _{3}<\gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$ for $\eta _{2}=-1$. Let $K(n,x)$ and $C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta )$ be defined via (2.5) and (2.6), respectively. For a sufficiently small positive number ε, setting

$$\begin{aligned} a(\varepsilon ):= \bigl\{ a_{n}(\varepsilon ) \bigr\} _{n=1}^{\infty }:= \bigl\{ n^{\frac{q(\beta \beta _{1}-1)-\beta _{1}\varepsilon }{q}} \bigr\} _{n=1}^{\infty }, \end{aligned}$$

(2.17)

and

$$\begin{aligned} f_{\varepsilon }(x)&:= \textstyle\begin{cases} x^{\frac{p(\beta \beta _{2}-1)+\beta _{2}\varepsilon }{p}}, &x\in \Omega, \\ 0, & x\in \mathbb{R^{+}}\setminus \Omega, \end{cases}\displaystyle \end{aligned}$$

(2.18)

where $\Omega = \{ x: x>0, x^{\frac{\beta _{2}}{ \vert \beta _{2} \vert }}<1 \}$. Then

$$\begin{aligned} \tilde{J} &:=\sum_{n=1}^{\infty } a_{n}(\varepsilon) \int _{\Omega }K(n,x) f_{\varepsilon}(x)\,\mathrm{d}x = \int _{\Omega }f_{\varepsilon}(x)\sum_{n=1}^{\infty } a_{n}(\varepsilon)K(n,x)\,\mathrm{d}x \\ &>\frac{1}{ \vert \beta _{1}\beta _{2} \vert } \bigl[\Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1},\eta _{2},\beta )+o(1) \bigr]. \end{aligned}$$

(2.19)

Proof

By Lemma 2.1, we have

$$\begin{aligned} \tilde{J} &> \int _{\Omega }x^{ \frac{p(\beta \beta _{2}-1)+\beta _{2}\varepsilon }{p}} \biggl[ \int _{1}^{ \infty } y^{\frac{q(\beta \beta _{1}-1)-\beta _{1}\varepsilon }{q}} \frac{ e^{\gamma _{2}y^{\beta _{1}}x^{\beta _{2}}}+\eta _{2}e^{\gamma _{3}y^{\beta _{1}}x^{\beta _{2}}}}{ e^{\gamma _{1}y^{\beta _{1}}x^{\beta _{2}}}+\eta _{1}e^{-\gamma _{1}y^{\beta _{1}}x^{\beta _{2}}}} \,\mathrm{d}y \biggr]\,\mathrm{d}x. \end{aligned}$$

(2.20)

Setting $y^{\beta _{1}}x^{\beta _{2}}=u$, we get

$$\begin{aligned} \tilde{J}>{}&\frac{1}{\beta _{1}} \int _{\Omega }x^{\beta _{2} \varepsilon -1} \biggl[ \int _{x^{\beta _{2}}}^{\infty }\kappa (u)u^{ \beta -\frac{\varepsilon }{q}-1}\, \mathrm{d}u \biggr]\,\mathrm{d}x \\ ={}&\frac{1}{\beta _{1}} \int _{\Omega }x^{\beta _{2}\varepsilon -1} \biggl[ \int _{1}^{\infty }\kappa (u)u^{\beta -\frac{\varepsilon }{q}-1} \, \mathrm{d}u \biggr]\,\mathrm{d}x \\ &{}+ \frac{1}{\beta _{1}} \int _{\Omega }x^{\beta _{2}\varepsilon -1} \biggl[ \int _{x^{\beta _{2}}}^{1}\kappa (u)u^{\beta - \frac{\varepsilon }{q}-1}\, \mathrm{d}u \biggr]\,\mathrm{d}x \\ ={}&\frac{1}{ \vert \beta _{1}\beta _{2} \vert \varepsilon } \int _{1}^{ \infty }\kappa (u)u^{\beta -\frac{\varepsilon }{q}-1}\, \mathrm{d}u + \frac{1}{\beta _{1}} \int _{\Omega }x^{\beta _{2}\varepsilon -1} \biggl[ \int _{x^{\beta _{2}}}^{1}\kappa (u)u^{\beta -\frac{\varepsilon }{q}-1} \, \mathrm{d}u \biggr]\,\mathrm{d}x. \end{aligned}$$

(2.21)

For $\beta _{2}>0$ or $\beta _{2}<0$, by Fubini’s theorem, it follows that

$$\begin{aligned} \int _{\Omega }x^{\beta _{2}\varepsilon -1} \biggl[ \int _{x^{\beta _{2}}}^{1} \kappa (u)u^{\beta -\frac{\varepsilon }{q}-1}\, \mathrm{d}u \biggr] \,\mathrm{d}x= \frac{1}{ \vert \beta _{2} \vert \varepsilon } \int _{0}^{1} \kappa (u)u^{\beta +\frac{\varepsilon }{p}-1}\, \mathrm{d}u. \end{aligned}$$

(2.22)

Applying (2.22) to (2.21), and using (2.14), we obtain (2.19). The proof of Lemma 2.3 is completed. □

Lemma 2.4

Let $-1< z<1$, $\psi _{1}(u)=\tan u$, $\psi _{2}(u)=\sec u$, and $m\in \mathbb{N}$. Then

$$\begin{aligned} &\psi _{1}^{(2m)} \biggl(\frac{z\pi }{2} \biggr)= \frac{2^{2m+1}(2m)!}{\pi ^{2m+1}} \sum_{j=0}^{\infty } \biggl[ \frac{1}{(2j+1-z)^{2m+1}}-\frac{1}{(2j+1+z)^{2m+1}} \biggr], \end{aligned}$$

(2.23)

$$\begin{aligned} & \psi _{1}^{(2m+1)} \biggl(\frac{z\pi }{2} \biggr)= \frac{2^{2m+2}(2m+1)!}{\pi ^{2m+2}}\sum_{j=0}^{\infty } \biggl[ \frac{1}{(2j+1-z)^{2m+2}}+\frac{1}{(2j+1+z)^{2m+2}} \biggr], \end{aligned}$$

(2.24)

$$\begin{aligned} &\psi _{2}^{(2m)} \biggl(\frac{z\pi }{2} \biggr)= \frac{2^{2m+1}(2m)!}{\pi ^{2m+1}}\sum_{j=0}^{\infty } \biggl[ \frac{(-1)^{j}}{(2j+1-z)^{2m+1}}+\frac{(-1)^{j}}{(2j+1+z)^{2m+1}} \biggr]. \end{aligned}$$

(2.25)

Proof

The rational fraction expansion of $\psi _{1}(u)=\tan u$ can be written as follows [44, 45]:

$$\begin{aligned} \psi _{1}(u)=\tan u=2\sum _{j=0}^{\infty } \biggl[ \frac{1}{(2j+1)\pi -2u}- \frac{1}{(2j+1)\pi +2u} \biggr]. \end{aligned}$$

(2.26)

Finding the $(2m)$th derivative of $\psi _{1}(u)=\tan u$, we have

$$\begin{aligned} \psi _{1}^{(2m)}(u)=2^{2m+1}(2m)! \sum _{j=0}^{\infty } \biggl\{ \frac{1}{[(2j+1)\pi -2u]^{2m+1}} - \frac{1}{[(2j+1)\pi +2u]^{2m+1}} \biggr\} . \end{aligned}$$

(2.27)

Letting $u=\frac{z\pi }{2}$ in (2.27), we obtain (2.23). Finding the first derivative of (2.27) and letting $u=\frac{z\pi }{2}$, we arrive at (2.24). In view of

$$\begin{aligned} 2\psi _{2}(2u)=\psi _{1}\biggl( \frac{\pi }{4}+u\biggr)+\psi _{1}\biggl(\frac{\pi }{4}-u \biggr), \end{aligned}$$

(2.28)

and finding the $(2m)$th derivative of (2.28), we obtain

$$\begin{aligned} 2^{2m+1}\psi _{2}^{(2m)}(2u)=\psi _{1}^{(2m)}\biggl(\frac{\pi }{4}+u\biggr)+\psi _{1}^{(2m)}\biggl( \frac{\pi }{4}-u\biggr). \end{aligned}$$

(2.29)

Letting $u=\frac{z\pi }{4}$ in (2.29), and using (2.23), we have

$$\begin{aligned} \psi _{2}^{(2m)} \biggl(\frac{z\pi }{2} \biggr)={}& \frac{2^{2m+1}(2m)!}{\pi ^{2m+1}}\sum_{j=0}^{\infty } \biggl[ \frac{1}{(4j+1-z)^{2m+1}}-\frac{1}{(4j+3+z)^{2m+1}} \biggr] \\ &{} +\frac{2^{2m+1}(2m)!}{\pi ^{2m+1}} \sum_{j=0}^{\infty } \biggl[ \frac{1}{(4j+1+z)^{2m+1}}-\frac{1}{(4j+3-z)^{2m+1}} \biggr] \\ ={}& \frac{2^{2m+1}(2m)!}{\pi ^{2m+1}}\sum_{j=0}^{\infty } \biggl[ \frac{(-1)^{j}}{(2j+1-z)^{2m+1}}+\frac{(-1)^{j}}{(2j+1+z)^{2m+1}} \biggr]. \end{aligned}$$

(2.30)

Lemma 2.4 is proved. □

3 Main results

Theorem 3.1

Let $\eta _{1},\eta _{2}\in \{1,-1\}$ and $\eta _{2}\neq -1$ for $\eta _{1}=1$. Let $\beta _{1}\in \mathbb{R}^{+}$, $\beta _{2}\in \mathbb{R}\setminus \{ 0 \}$. Assume that $\beta \beta _{1}\leq 1$, $\beta \geq 1$ and $\beta \neq 1$ for $\eta _{1}=-1$, $\eta _{2}=1$. Let $\gamma _{1}\in \mathbb{R}^{+},\gamma _{2},\gamma _{3}\in \mathbb{R}$, and $\gamma _{2},\gamma _{3}\neq \gamma _{1}$. Suppose that $\gamma _{3}\leq \gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$ for $\eta _{2}=1$, and $\gamma _{3}<\gamma _{2}\leq -\gamma _{3}\leq \gamma _{1}$ for $\eta _{2}=-1$. Let $\mu _{n}= n^{q(1-\beta \beta _{1})-1}$ and $\nu (x)= x^{p(1-\beta \beta _{2})-1}$. Let $a_{n}$, $f(x)\geq 0$ with $a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\mu }$ and $f(x)\in L_{p,\nu }(\mathbb{R}^{+})$. Consider $K(n,x)$ and $C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta )$ defined via (2.5) and (2.6), respectively. Then the following equivalent inequalities hold true:

$$\begin{aligned} &I_{1}:=\sum_{n=1}^{\infty }n^{p\beta \beta _{1}-1} \biggl[ \int _{0}^{ \infty }K(n,x)f(x)\,\mathrm{d}x \biggr]^{p} \end{aligned}$$

(3.1)

$$\begin{aligned} &\phantom{I_{1}:}< \bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{p} \Vert f \Vert _{p,\nu }^{p}, \\ &I_{2}: = \int _{0}^{\infty }x^{q\beta \beta _{2}-1} \Biggl[\sum _{n=1}^{ \infty }K(n,x)a_{n} \Biggr]^{q}\,\mathrm{d}x \end{aligned}$$

(3.2)

$$\begin{aligned} &\phantom{I_{2}:}< \bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{q} \Vert a \Vert _{q, \mu }^{q}, \\ & J:=\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }K(n,x)f(x)\,\mathrm{d}x = \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty }K(n,x)a_{n} \,\mathrm{d}x \\ &\phantom{J:}< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}}\Gamma ( \beta )C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta ) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }, \end{aligned}$$

(3.3)

where the constant factors

$$\begin{aligned} &\bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{p}, \\ &\bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{q}, \end{aligned}$$

and $\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}}\Gamma ( \beta )C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta )$ are the best possible.

Proof

By Hölder’s inequality and (2.7), we have

$$\begin{aligned} & \biggl[ \int _{0}^{\infty }K(n,x)f(x)\,\mathrm{d}x \biggr]^{p} \\ &\quad= \biggl[ \int _{0}^{\infty }K(n,x) \biggl( \frac{n^{\frac{\beta \beta _{1}-1}{p}}}{x^{\frac{\beta \beta _{2}-1}{q}}}f(x) \biggr) \biggl( \frac{x^{\frac{\beta \beta _{2}-1}{q}}}{n^{\frac{\beta \beta _{1}-1}{p}}} \biggr)\,\mathrm{d}x \biggr]^{p} \\ &\quad\leq \int _{0}^{\infty }K(n,x) \biggl[ \frac{n^{\beta \beta _{1}-1}}{x^{\frac{p(\beta \beta _{2}-1)}{q}}}f^{p}(x) \biggr]\,\mathrm{d}x \biggl\{ \int _{0}^{\infty }K(n,x) \biggl[ \frac{x^{\beta \beta _{2}-1}}{n^{\frac{q(\beta \beta _{1}-1)}{p}}} \biggr]\,\mathrm{d}x \biggr\} ^{p-1} \\ &\quad= \int _{0}^{\infty }K(n,x) \biggl[ \frac{n^{\beta \beta _{1}-1}}{x^{\frac{p(\beta \beta _{2}-1)}{q}}}f^{p}(x) \biggr]\,\mathrm{d}x \bigl[ \omega (n)n^{ \frac{q(1-\beta \beta _{1})}{p}} \bigr]^{p-1} \\ &\quad= \frac{n^{-p\beta \beta _{1}+\beta \beta _{1}}}{ \vert \beta _{2} \vert ^{p-1}} \bigl[\Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3}, \eta _{1},\eta _{2},\beta ) \bigr]^{p-1} \int _{0}^{\infty }K(n,x)x^{ \frac{p(1-\beta \beta _{2})}{q}}f^{p}(x) \,\mathrm{d}x. \end{aligned}$$

(3.4)

Therefore, by Lebesgue term-by-term integration theorem and (2.8), we obtain

$$\begin{aligned} I_{1}&\leq \sum_{n=1}^{\infty } \frac{n^{\beta \beta _{1}-1}}{ \vert \beta _{2} \vert ^{p-1}} \bigl[ \Gamma (\beta )C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1}, \eta _{2}, \beta ) \bigr]^{p-1} \int _{0}^{\infty }K(n,x)x^{ \frac{p(1-\beta \beta _{2})}{q}}f^{p}(x) \,\mathrm{d}x \\ &=\frac{1}{ \vert \beta _{2} \vert ^{p-1}} \bigl[\Gamma (\beta )C ( \gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1},\eta _{2},\beta ) \bigr]^{p-1} \int _{0}^{\infty }x^{\frac{p(1-\beta \beta _{2})}{q}}f^{p}(x) \varpi (x) \,\mathrm{d}x \\ &< \frac{1}{\beta _{1} \vert \beta _{2} \vert ^{p-1}} \bigl[\Gamma ( \beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta ) \bigr]^{p} \int _{0}^{\infty }x^{p(1-\beta \beta _{2})-1}f^{p}(x) \, \mathrm{d}x \\ &= \bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{p} \Vert f \Vert _{p,\nu }^{p}. \end{aligned}$$

(3.5)

Hence (3.1) is proved. Similarly, by Hölder’s inequality in the form of series and (2.8), we have

$$\begin{aligned} & \Biggl[\sum_{n=1}^{\infty }K(n,x)a_{n} \Biggr]^{q} \\ &\quad= \Biggl[\sum_{n=1}^{\infty }K(n,x) \biggl( \frac{n^{\frac{\beta \beta _{1}-1}{p}}}{x^{\frac{\beta \beta _{2}-1}{q}}} \biggr) \biggl( \frac{x^{\frac{\beta \beta _{2}-1}{q}}}{n^{\frac{\beta \beta _{1}-1}{p}}}a_{n} \biggr) \Biggr]^{q} \\ &\quad\leq \bigl[\varpi (x)x^{\frac{p(1-\beta \beta _{2})}{q}} \bigr]^{q-1} \sum _{n=1}^{\infty }K(n,x)n^{\frac{q(1-\beta \beta _{1})}{p}}a_{n}^{q}x^{ \beta \beta _{2}-1} \\ & \quad< \frac{x^{-q\beta \beta _{2}+\beta \beta _{2}}}{\beta _{1}^{q-1}} \bigl[\Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3}, \eta _{1},\eta _{2},\beta ) \bigr]^{q-1}\sum_{n=1}^{\infty }K(n,x)n^{ \frac{q(1-\beta \beta _{1})}{p}}a_{n}^{q}. \end{aligned}$$

(3.6)

Plugging (3.6) into the left-hand side of (3.2), and using Lebesgue term-by-term integration theorem again, as well as (2.7), we obtain

$$\begin{aligned} I_{2}&< \frac{1}{\beta _{1}^{q-1}} \bigl[\Gamma (\beta )C ( \gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2},\beta ) \bigr]^{q-1} \int _{0}^{\infty }x^{\beta \beta _{2}-1}\sum _{n=1}^{ \infty }K(n,x)n^{\frac{q(1-\beta \beta _{1})}{p}}a_{n}^{q} \,\mathrm{d}x \\ &= \frac{1}{\beta _{1}^{q-1}} \bigl[\Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1},\eta _{2},\beta ) \bigr]^{q-1} \sum_{n=1}^{\infty }n^{\frac{q(1-\beta \beta _{1})}{p}}a_{n}^{q} \omega (n) \\ &= \frac{1}{\beta _{1}^{q-1} \vert \beta _{2} \vert } \bigl[\Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1},\eta _{2},\beta ) \bigr]^{q}\sum_{n=1}^{\infty }n^{q(1-\beta \beta _{1})-1}a_{n}^{q} \\ &= \bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{q} \Vert a \Vert _{q, \mu }^{q}. \end{aligned}$$

(3.7)

The proof of (3.2) is completed.

By the use of Lebesgue term-by-term integration theorem, it is obvious that there exist two forms of J. Using Hölder’s inequality and (3.1), we have

$$\begin{aligned} J&=\sum_{n=1}^{\infty } \biggl\{ \biggl[n^{\beta \beta _{1}-\frac{1}{p}} \int _{0}^{\infty }K(n,x)f(x)\,\mathrm{d}x \biggr] \bigl(a_{n}n^{-\beta \beta _{1}+\frac{1}{p}} \bigr) \biggr\} \\ &\leq I_{1}^{\frac{1}{p}} \Vert a \Vert _{q, \mu } < \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}}\Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1},\eta _{2},\beta ) \Vert f \Vert _{p, \nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

Therefore, we obtained (3.3) via (3.1). It will be proved that (3.3) can also be obtained via (3.2). In fact, it follows from Hölder’s inequality and (3.2) that

$$\begin{aligned} J&= \int _{0}^{\infty } \Biggl\{ \Biggl[x^{\beta \beta _{2}-\frac{1}{q}} \sum _{n=1}^{\infty }a_{n}K(n,x) \Biggr] \bigl[x^{-\beta \beta _{2}+ \frac{1}{q}}f(x) \bigr] \Biggr\} \,\mathrm{d}x \\ & \leq I_{2}^{\frac{1}{q}} \Vert f \Vert _{p, \nu } < \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}}\Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1},\eta _{2},\beta ) \Vert f \Vert _{p, \nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

In order to prove the equivalence of (3.1), (3.2), and (3.3), we will show that both (3.1) and (3.2) hold when (3.3) is true. Let

$$\begin{aligned} a_{n}:=n^{p\beta \beta _{1}-1} \biggl[ \int _{0}^{\infty }K(n,x)f(x) \,\mathrm{d}x \biggr]^{p-1}. \end{aligned}$$

It follows from (3.3) that

$$\begin{aligned} 0&< \bigl( \Vert a \Vert _{q, \mu } \bigr)^{pq}= \Biggl\{ \sum_{n=1}^{\infty }n^{p \beta \beta _{1}-1} \biggl[ \int _{0}^{\infty }K(n,x)f(x)\,\mathrm{d}x \biggr]^{p} \Biggr\} ^{p} \\ &= \Biggl[\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }K(n,x)f(x) \,\mathrm{d}x \Biggr]^{p} \\ &< \bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{p} \Vert f \Vert _{p,\nu }^{p} \Vert a \Vert _{q, \mu }^{p}. \end{aligned}$$

Therefore,

$$\begin{aligned} 0&< \sum_{n=1}^{\infty }n^{p\beta \beta _{1}-1} \biggl[ \int _{0}^{\infty }K(n,x)f(x) \,\mathrm{d}x \biggr]^{p}\\ &= \bigl( \Vert a \Vert _{q, \mu } \bigr)^{q} \\ &< \bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{p} \Vert f \Vert _{p,\nu }^{p}. \end{aligned}$$

(3.8)

Hence, (3.1) is proved via (3.3). Similarly, let

$$\begin{aligned} f(x):=x^{q\beta \beta _{2}-1} \Biggl[\sum_{n=1}^{\infty }K(n,x)a_{n} \Biggr]^{q-1}. \end{aligned}$$

By the use of (3.3), we have

$$\begin{aligned} 0&< \bigl( \Vert f \Vert _{p, \nu } \bigr)^{pq}= \Biggl\{ \int _{0}^{\infty }x^{q \beta \beta _{2}-1} \Biggl[\sum _{n=1}^{\infty }K(n,x)a_{n} \Biggr]^{q} \,\mathrm{d}x \Biggr\} ^{q} \\ &= \Biggl[ \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty }K(n,x)a_{n} \,\mathrm{d}x \Biggr]^{q} \\ &< \bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{q} \Vert f \Vert _{p,\nu }^{q} \Vert a \Vert _{q, \mu }^{q}. \end{aligned}$$

(3.9)

It follows from (3.9) that

$$\begin{aligned} 0&< \int _{0}^{\infty }x^{q\beta \beta _{2}-1} \Biggl[\sum _{n=1}^{\infty }K(n,x)a_{n} \Biggr]^{q}\,\mathrm{d}x= \bigl( \Vert f \Vert _{p, \nu } \bigr)^{p} \\ &< \bigl[\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \Gamma (\beta )C (\gamma _{1}, \gamma _{2},\gamma _{3},\eta _{1}, \eta _{2},\beta ) \bigr]^{q} \Vert a \Vert _{q,\nu }^{q}. \end{aligned}$$

(3.10)

Inequality (3.2) is also proved via (3.3). According to the above discussions, (3.1), (3.2), and (3.3) are equivalent.

At last, it will be proved that the constant factors on the right-hand side of (3.1), (3.2), and (3.3) are the best possible. Assuming that the constant factor $\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}}\Gamma ( \beta )C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1}, \eta _{2}, \beta )$ in (3.3) is not the best possible, there must be a positive number c such that (3.3) still holds if $\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}}\Gamma ( \beta )C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta )$ is replaced with c. That is,

$$\begin{aligned} \sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }K(n,x)f(x)\,\mathrm{d}x = \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty }a_{n}K(n,x) \,\mathrm{d}x< c \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(3.11)

Replacing $a_{n}$ and $f(x)$ with $a_{n}(\varepsilon )$ and $f_{\varepsilon }(x)$ defined in Lemma 2.3, respectively, and using (2.19), we have

$$\begin{aligned} &\frac{1}{ \vert \beta _{1}\beta _{2} \vert }\Gamma (\beta )C ( \gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta )+o(1) \\ &\quad< \varepsilon c \Vert f_{\varepsilon } \Vert _{p,\nu } \bigl\Vert a(\varepsilon ) \bigr\Vert _{q, \mu }= \varepsilon c \biggl( \int _{\Omega }x^{\beta _{2}\varepsilon -1} \,\mathrm{d}x \biggr)^{\frac{1}{p}} \Biggl(1+\sum_{n=2}^{\infty }n^{- \beta _{1}\varepsilon -1} \Biggr)^{\frac{1}{q}} \\ &\quad< \varepsilon c \biggl( \int _{\Omega }x^{\beta _{2}\varepsilon -1} \,\mathrm{d}x \biggr)^{\frac{1}{p}} \biggl(1+ \int _{1}^{\infty }x^{-\beta _{1} \varepsilon -1}\,\mathrm{d}x \biggr)^{\frac{1}{q}} \\ &\quad= c \vert \beta _{2} \vert ^{-\frac{1}{p}} \bigl( \varepsilon +\beta _{1}^{-1} \bigr)^{\frac{1}{q}}. \end{aligned}$$

(3.12)

Let $\varepsilon \to 0^{+}$, then we obtain $\beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}}\Gamma ( \beta )C (\gamma _{1},\gamma _{2},\gamma _{3},\eta _{1},\eta _{2}, \beta )\leq c$, which contradicts the assumption, obviously. Therefore, the constant factor on the right-hand side of inequality (3.3) is the best possible. From the equivalence of (3.1), (3.2), and (3.3), the constant factors in (3.1) and (3.2) are obviously the best possible. Theorem 3.1 is proved. □

4 Corollaries

Let $\eta _{1}=\eta _{2}=-1$, $\gamma _{3}=-\gamma _{2}$ and $\beta =2m+1$ $(m\in \mathbb{N})$ in Theorem 3.1. By the use of (2.23), we obtain the following corollary.

Corollary 4.1

Let $\beta _{1}\in \mathbb{R}^{+}$, $\beta _{2}\in \mathbb{R}\setminus \{ 0 \}$, and $(2m+1)\beta _{1}\leq 1$ $(m\in \mathbb{N})$. Let $0<\gamma _{2}<\gamma _{1}$. Suppose that $\mu _{n}= n^{q[1-(2m+1)\beta _{1}]-1}$, $\nu (x)= x^{p[1-(2m+1)\beta _{2}]-1}$, and $\psi _{1}(u)=\tan u$. Let $a_{n}$, $f(x)\geq 0$ with $a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\mu }$ and $f(x)\in L_{p,\nu }(\mathbb{R}^{+})$. Then

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\sinh { \bigl(\gamma _{2}n^{ \beta _{1}}x^{\beta _{2}} \bigr)}\mathrm{csch} { \bigl( \gamma _{1}n^{ \beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad = \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty }a_{n} \sinh { \bigl( \gamma _{2}n^{\beta _{1}}x^{\beta _{2}} \bigr)} \mathrm{csch} { \bigl( \gamma _{1}n^{\beta _{1}}x^{\beta _{2}} \bigr)}\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \biggl( \frac{\pi }{2\gamma _{1}} \biggr)^{2m+1}\psi _{1}^{(2m)} \biggl( \frac{\gamma _{2}\pi }{2\gamma _{1}} \biggr) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.1)

In particular, let $\gamma _{1}=2\gamma$, $\gamma _{2}=\gamma$ $(\gamma >0)$ in (4.1), then it follows that

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{sech} { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \frac{1}{2^{4m+1}} \biggl(\frac{\pi }{\gamma } \biggr)^{2m+1}\psi _{1}^{(2m)} \biggl( \frac{\pi }{4} \biggr) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.2)

Setting $\gamma =1$, $\beta _{1}=1$, $\beta _{2}=-1$, $m=0$ in (4.2), we obtain (1.8).

In addition, let $\gamma _{1}=4\gamma$, $\gamma _{2}=\gamma$ $(\gamma >0)$ in (4.1), then we have

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{sech} { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}\mathrm{sech} { \bigl(2 \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \frac{1}{2^{6m+1}} \biggl(\frac{\pi }{\gamma } \biggr)^{2m+1}\psi _{1}^{(2m)} \biggl( \frac{\pi }{8} \biggr) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.3)

Setting $\beta _{1}=\beta _{2}=\gamma =1$, $m=0$ in (4.2) and (4.3), we obtain

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{sech} {(nx)}f(x) \, \mathrm{d}x< \frac{\pi }{2} \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }, \end{aligned}$$

(4.4)

$$\begin{aligned} & \sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{sech}(nx) \mathrm{sech}(2nx)f(x)\,\mathrm{d}x< \frac{(\sqrt{2}-1)\pi }{2} \Vert f \Vert _{p, \nu } \Vert a \Vert _{q, \mu }, \end{aligned}$$

(4.5)

where $\mu _{n}=\frac{1}{n}$, $\nu (x)=\frac{1}{x}$.

Let $\eta _{1}=-1$, $\eta _{2}=1$, $\gamma _{3}=-\gamma _{2}$ and $\beta =2m+2$ $(m\in \mathbb{N})$ in Theorem 3.1. By the use of (2.24), we obtain the following corollary.

Corollary 4.2

Let $\beta _{1}\in \mathbb{R}^{+}$, $\beta _{2}\in \mathbb{R}\setminus \{ 0 \}$, and $(2m+2)\beta _{1}\leq 1$ $(m\in \mathbb{N})$. Let $0\leq \gamma _{2}<\gamma _{1}$. Suppose that $\mu _{n}= n^{q[1-(2m+2)\beta _{1}]-1}$, $\nu (x)= x^{p[1-(2m+2)\beta _{2}]-1}$, and $\psi _{1}(u)=\tan u$. Let $a_{n}$, $f(x)\geq 0$ with $a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\mu }$ and $f(x)\in L_{p,\nu }(\mathbb{R}^{+})$. Then

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\cosh { \bigl(\gamma _{2}n^{ \beta _{1}}x^{\beta _{2}} \bigr)}\mathrm{csch} { \bigl( \gamma _{1}n^{ \beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad = \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty } \cosh { \bigl(\gamma _{2}n^{ \beta _{1}}x^{\beta _{2}} \bigr)} \mathrm{csch} { \bigl(\gamma _{1}n^{ \beta _{1}}x^{\beta _{2}} \bigr)}a_{n}\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \biggl( \frac{\pi }{2\gamma _{1}} \biggr)^{2m+2}\psi _{1}^{(2m+1)} \biggl( \frac{\gamma _{2}\pi }{2\gamma _{1}} \biggr) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.6)

Particularly, let $\gamma _{1}=\gamma >0$, $\gamma _{2}=0$ in (4.6), then it follows that

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{csch} { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \biggl( \frac{\pi }{2\gamma } \biggr)^{2m+2}\psi _{1}^{(2m+1)} (0 ) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.7)

Additionally, letting $\gamma _{1}=2\gamma$, $\gamma _{2}=\gamma$ $(\gamma >0)$ in (4.6), we have

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{csch} { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \frac{1}{2^{4m+3}} \biggl(\frac{\pi }{\gamma } \biggr)^{2m+2}\psi _{1}^{(2m+1)} \biggl( \frac{\pi }{4} \biggr) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.8)

Compare (4.7) with (4.8). By (2.24), it is easy to show that $\psi _{1}^{(2m+1)} (0 )=\frac{1}{2^{2m+1}}\psi _{1}^{(2m+1)} (\frac{\pi }{4} )$. Therefore, (4.7) and (4.8) are equivalent. Setting $\beta _{1}=\beta _{2}=\frac{1}{2}$, $\gamma =1$, $m=0$ in (4.7), we obtain

$$\begin{aligned} \sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{csch} { (\sqrt{nx} )}f(x)\, \mathrm{d}x< \frac{\pi ^{2}}{2} \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }, \end{aligned}$$

(4.9)

where $\mu _{n}=\frac{1}{n}$, $\nu (x)=\frac{1}{x}$.

Furthermore, letting $\gamma _{1}=4\gamma$, $\gamma _{2}=\gamma$ $(\gamma >0)$ in (4.6), we have

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{csch} { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}\mathrm{sech} { \bigl(2 \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \frac{1}{2^{6m+3}} \biggl(\frac{\pi }{\gamma } \biggr)^{2m+2}\psi _{1}^{(2m+1)} \biggl( \frac{\pi }{8} \biggr) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.10)

Let $\eta _{1}=-1$, $\eta _{2}=1$, $\gamma _{1}=\gamma$, $\gamma _{2}=\gamma _{3}=-\gamma$, and $\beta =2m+2$ $(m\in \mathbb{N})$ in Theorem 3.1. By the following equality

$$\begin{aligned} \sum_{j=0}^{\infty } \frac{1}{(j+1)^{2m+2}}=\frac{2^{2m+1}}{(2m+2)!} \pi ^{2m+2}B_{m+1}, \end{aligned}$$

where $B_{m+1}$ is Bernoulli number, $B_{1}=\frac{1}{6}, B_{2}=\frac{1}{30}, B_{3}=\frac{1}{42},\dots$ Then, the following corollary holds.

Corollary 4.3

Let $\gamma,\beta _{1}\in \mathbb{R}^{+}$, $\beta _{2}\in \mathbb{R}\setminus \{ 0 \}$, and $(2m+2)\beta _{1}\leq 1$ $(m\in \mathbb{N})$. Suppose that $\mu _{n}= n^{q[1-(2m+2)\beta _{1}]-1}$, $\nu (x)= x^{p[1-(2m+2)\beta _{2}]-1}$, and $\psi _{2}(u)=\sec u$. Let $a_{n}$, $f(x)\geq 0$ with $a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\mu }$ and $f(x)\in L_{p,\nu }(\mathbb{R}^{+})$. Then

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty } \bigl[\coth { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}-1 \bigr]f(x)\,\mathrm{d}x \\ &\quad = \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty } \bigl[\coth { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}-1 \bigr]a_{n}\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \frac{B_{m+1}}{2m+2} \biggl(\frac{\pi }{\gamma } \biggr)^{2m+2} \Vert f \Vert _{p, \nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.11)

Setting $\beta _{1}=\beta _{2}=\frac{1}{2m+2}$, $\gamma =1$, and replacing $m+1$ with m, we obtain (1.9).

Let $\eta _{1}=\eta _{2}=1$, $\gamma _{3}=-\gamma _{2}$ and $\beta =2m+1$ $(m\in \mathbb{N})$ in Theorem 3.1. By the use of (2.25), we obtain Corollary 4.4.

Corollary 4.4

Let $\beta _{1}\in \mathbb{R}^{+}$, $\beta _{2}\in \mathbb{R}\setminus \{ 0 \}$, and $(2m+1)\beta _{1}\leq 1$ $(m\in \mathbb{N})$. Let $0\leq \gamma _{2}<\gamma _{1}$. Suppose that $\mu _{n}= n^{q[1-(2m+1)\beta _{1}]-1}$, $\nu (x)= x^{p[1-(2m+1)\beta _{2}]-1}$, and $\psi _{2}(u)=\sec u$. Let $a_{n}$, $f(x)\geq 0$ with $a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\mu }$ and $f(x)\in L_{p,\nu }(\mathbb{R}^{+})$. Then

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\cosh { \bigl(\gamma _{2}n^{ \beta _{1}}x^{\beta _{2}} \bigr)}\mathrm{sech} { \bigl( \gamma _{1}n^{ \beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad = \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty } \cosh { \bigl(\gamma _{2}n^{ \beta _{1}}x^{\beta _{2}} \bigr)} \mathrm{sech} { \bigl(\gamma _{1}n^{ \beta _{1}}x^{\beta _{2}} \bigr)}a_{n}\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \biggl( \frac{\pi }{2\gamma _{1}} \biggr)^{2m+1}\psi _{2}^{(2m)} \biggl( \frac{\gamma _{2}\pi }{2\gamma _{1}} \biggr) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.12)

Let $\gamma _{1}=\gamma >0$, $\gamma _{2}=0$ in (4.12), we have

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{sech} { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \biggl( \frac{\pi }{2\gamma } \biggr)^{2m+1}\psi _{2}^{(2m)} (0 ) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.13)

Comparing (4.2) with (4.13), it can be shown that (4.2) and (4.13) are equivalent. In fact, in view of

$$\begin{aligned} \sum_{j=0}^{\infty } \biggl[\frac{1}{(4j+1)^{2m+1}}- \frac{1}{(4j+3)^{2m+1}} \biggr]=\sum_{j=0}^{\infty } \frac{(-1)^{j}}{(2j+1)^{2m+1}}, \end{aligned}$$

(4.14)

and using (2.23) and (2.25), we have $\psi _{1}^{(2m)} (\frac{\pi }{4} )=2^{2m}\psi _{2}^{(2m)}(0)$. Therefore, (4.2) is equivalent to (4.13).

Additionally, letting $\gamma _{1}=2\gamma$, $\gamma _{2}=\gamma$ $(\gamma >0)$ in (4.12), we have

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty }\mathrm{csch} { \bigl( \gamma _{1}n^{\beta _{1}}x^{\beta _{2}} \bigr)}\mathrm{tanh} { \bigl(2 \gamma _{1}n^{\beta _{1}}x^{\beta _{2}} \bigr)}f(x)\,\mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \frac{1}{2^{4m+1}} \biggl(\frac{\pi }{\gamma } \biggr)^{2m+1}\psi _{1}^{(2m)} \biggl( \frac{\pi }{4} \biggr) \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.15)

Let $\eta _{1}=\eta _{2}=1$, $\gamma _{1}=\gamma$, $\gamma _{2}=\gamma _{3}=-\gamma$, and $\beta =2m+2$ $(m\in \mathbb{N})$. Due to the following equality [44, 45]:

$$\begin{aligned} \sum_{j=0}^{\infty } \frac{(-1)^{j}}{(j+1)^{2m+2}}= \frac{2^{2m+1}-1}{(2m+2)!}\pi ^{2m+2}B_{m+1}, \end{aligned}$$

the following corollary holds.

Corollary 4.5

Let $\gamma,\beta _{1}\in \mathbb{R}^{+}$, $\beta _{2}\in \mathbb{R}\setminus \{ 0 \}$, and $(2m+2)\beta _{1}\leq 1$ $(m\in \mathbb{N})$. Suppose that $\mu _{n}= n^{q[1-(2m+2)\beta _{1}]-1}$, $\nu (x)= x^{p[1-(2m+2)\beta _{2}]-1}$, and $\psi _{2}(u)=\sec u$. Let $a_{n}$, $f(x)\geq 0$ with $a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\mu }$ and $f(x)\in L_{p,\nu }(\mathbb{R}^{+})$. Then

$$\begin{aligned} &\sum_{n=1}^{\infty }a_{n} \int _{0}^{\infty } \bigl[1-\tanh { \bigl( \gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)} \bigr]f(x)\,\mathrm{d}x \\ &\quad = \int _{0}^{\infty }f(x)\sum_{n=1}^{\infty }a_{n} \bigl[1-\tanh { \bigl(\gamma n^{\beta _{1}}x^{\beta _{2}} \bigr)} \bigr]\, \mathrm{d}x \\ &\quad< \beta _{1}^{-\frac{1}{p}} \vert \beta _{2} \vert ^{-\frac{1}{q}} \frac{2^{2m+1}-1}{m+1} \biggl(\frac{\pi }{2\gamma } \biggr)^{2m+2}B_{m+1} \Vert f \Vert _{p,\nu } \Vert a \Vert _{q, \mu }. \end{aligned}$$

(4.16)

Setting $\beta _{1}=\beta _{2}=\frac{1}{2m+2}$, $\gamma =1$, and replacing $m+1$ with m, we arrive at (1.10).