A half-discrete Hardy-Hilbert-type inequality related to hyperbolic secant function

Abstract

By applying weight functions and technique of real analysis, a half-discrete Hardy-Hilbert-type inequality related to the kernel of hyperbolic secant function and a best possible constant factor are given. The equivalent forms, the operator expressions with the norm, the reverses, and some particular cases are also considered.

1 Introduction

If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f(x),g(y)\geq0$, $f\in L^{p}( \mathbf{R}_{+})$, $g\in L^{q}(\mathbf{R}_{+})$, $\|f\|_{p}=(\int _{0}^{\infty}f^{p}(x)\, dx)^{\frac{1}{p}}>0$, and $\|g\|_{q}>0$, 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. If $a_{m},b_{n}\geq0$, $a=\{a_{m}\}_{m=1}^{\infty}\in l^{p}$, $b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}$, $\|a\|_{p}=(\sum_{m=1}^{\infty }a_{m}^{p})^{\frac{1}{p}}>0$, and $\|b\|_{q}>0$, then we have the following Hardy-Hilbert’s inequality with the same best constant $\frac{\pi}{\sin (\pi/p)}$ [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 (see [1–5]).

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

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

(3)

Then we have the following inequality ([1], Theorem 321):

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

(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 obtain an equivalent form of (4):

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

(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 $p=q=2$. Later, Yang [5] refined [6] by giving extensions of (1) and (2) as follows.

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

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

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

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

(6)

where the constant factor $k(\lambda_{1})$ is the best possible. Moreover, if $k_{\lambda}(x,y)$ keeps 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 $a_{m},b_{n}\geq0$,

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

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

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

(7)

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

For $0<\lambda_{1},\lambda_{2}\leq1$ such that $\lambda_{1}+\lambda _{2}=\lambda$, we set

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

Then by (7) we have

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

(8)

where the constant $B(\lambda_{1},\lambda_{2})$ is the best possible, and

$$ B ( u,v ) = \int_{0}^{\infty}\frac{1}{(1+t)^{u+v}}t^{u-1} \,dt \quad (u,v>0) $$

is the beta function. Clearly, for $\lambda=1$, $\lambda_{1}=\frac{1}{q}$, $\lambda_{2}=\frac{1}{p}$, inequality (8) reduces to (2).

In 2015, by adding some conditions, Yang [7] gave an extension of (8) and (5) as follows:

$$\begin{aligned}& \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{(U_{m}+V_{n})^{\lambda}} \\& \quad < B(\lambda_{1},\lambda_{2}) \Biggl( \sum _{m=1}^{\infty}\frac{U_{m}^{p(1-\lambda_{1})-1}a_{m}^{p}}{\mu_{m}^{p-1}} \Biggr) ^{\frac {1}{p}} \Biggl( \sum_{n=1}^{\infty}\frac{V_{n}^{q(1-\lambda _{2})-1}b_{n}^{q}}{\nu _{n}^{q-1}} \Biggr) ^{\frac{1}{q}}, \end{aligned}$$

(9)

where the constant $B(\lambda_{1},\lambda_{2})$ is still the best possible.

Some other results including multidimensional Hilbert-type inequalities are provided by [8–25].

About the topic of half-discrete Hilbert-type inequalities with inhomogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1], but they did not prove that the constant factors are the best possible. However, Yang [26] gave a result with the kernel $\frac{1}{(1+nx)^{\lambda}}$ by introducing a variable and proved that the constant factor is the best possible. In 2011, Yang [27] gave the following half-discrete Hardy-Hilbert’s inequality with the best possible constant factor $B ( \lambda_{1},\lambda_{2} )$:

$$ \int_{0}^{\infty}f ( x ) \Biggl[ \sum _{n=1}^{\infty}\frac {a_{n}}{ ( x+n ) ^{\lambda}} \Biggr] \,dx< B ( \lambda_{1},\lambda _{2} ) \|f\|_{p,\phi}\|a \|_{q,\psi}, $$

(10)

where, $\lambda_{1}>0$, $0<\lambda_{2}\leq1$, $\lambda_{1}+\lambda _{2}=\lambda$. Zhong and Yang [17, 28–33] investigated several half-discrete Hilbert-type inequalities with particular kernels. Applying weight functions, a half-discrete Hilbert-type inequality with a general homogeneous kernel of degree $-\lambda\in \mathbf{R}$ with the best constant factor $k ( \lambda_{1} )$ is obtained as follows:

$$ \int_{0}^{\infty}f(x)\sum_{n=1}^{\infty}k_{\lambda }(x,n)a_{n} \,dx< k(\lambda _{1})\|f\|_{p,\phi}\|a\|_{q,\psi}, $$

(11)

which is an extension of (10) (cf. [34]). At the same time, a half-discrete Hilbert-type inequality with a general inhomogeneous kernel and the best constant factor is given by Yang [35]. In 2012-2014, Yang et al. published three books [36, 37] and [38] for building the theory of half-discrete Hilbert-type inequalities.

In this paper, by applying weight coefficients and technique of real analysis, a half-discrete Hardy-Hilbert-type inequality related to the kernel of hyperbolic secant function and the best possible constant factor is given, which is an extension of (11) for $\lambda=0$ and a particular kernel. The equivalent forms, the operator expressions with the norm, the reverses, and some particular cases are also considered.

2 Some lemmas

In the following, we make appointment that $\mu_{i},\nu_{j}>0$ ($i,j\in\mathbf{N}$), $U_{m}$ and $V_{n}$ are defined by (3), $\mu(t)$ is a positive continuous function in $\mathbf{R}_{+}=(0,\infty)$,

$$ U(x):= \int_{0}^{x}\mu(t)\,dt< \infty\quad \bigl(x\in [0, \infty )\bigr), $$

$\nu(t):=\nu_{n}$, $t\in(n-1,n]$ ($n\in\mathbf{N}$), and

$$ V(y):= \int_{0}^{y}\nu(t)\,dt\quad \bigl(y\in[0, \infty )\bigr), $$

$p\neq0,1$, $\frac{1}{p}+\frac{1}{q}=1$, $\delta\in\{-1,1\}$, $f(x),a_{n}\geq0$ ($x\in\mathbf{R}_{+}$, $n\in\mathbf{N}$), $\|f\|_{p,\Phi _{\delta}}=(\int_{0}^{\infty}\Phi_{\delta}(x)f^{p}(x)\,dx)^{\frac{1}{p}}$, $\|a\|_{q,\Psi}=(\sum_{n=1}^{\infty}\Psi(n)b_{n}^{q})^{\frac{1}{q}}$, where

$$ \Phi_{\delta}(x):=\frac{U^{p(1-\delta\sigma)-1}(x)}{\mu ^{p-1}(x)},\qquad \Psi (n):=\frac{V_{n}^{q(1-\sigma)-1}}{\nu_{n}^{q-1}} \quad (x\in\mathbf {R}_{+},n\in \mathbf{N}). $$

Example 1

For $\rho,\gamma,\sigma>0$, $\alpha>-\rho$, $\sec h(u)=\frac{2}{e^{u}+e^{-u}}$ ($u>0$) is called the hyperbolic secant function (cf. [39]), we set $h(t)=\frac{\sec h(\rho t^{\gamma})}{e^{\alpha t^{\gamma}}}$ ($t\in\mathbf{R}_{+}$).

(i) Setting $u=\rho t^{\gamma}$, we find

$$\begin{aligned} k(\sigma) : =& \int_{0}^{\infty}\frac{\sec h(\rho t^{\gamma })}{e^{\alpha t^{\gamma}}}t^{\sigma-1}\,dt \\ =& \frac{1}{\gamma\rho^{\sigma/\gamma}} \int_{0}^{\infty}\frac{\sec h(u)}{e^{\frac{\alpha}{\rho}u}}u^{\frac{\sigma}{\gamma}-1}\,du \\ =& \frac {2}{\gamma \rho^{\sigma/\gamma}} \int_{0}^{\infty}\frac{e^{-\frac{\alpha}{\rho} u}u^{\frac{\sigma}{\gamma}-1}}{e^{u}+e^{-u}}\,du \\ =& \frac{2}{\gamma\rho^{\sigma/\gamma}} \int_{0}^{\infty}\frac{e^{-( \frac{\alpha}{\rho}+1)u}u^{\frac{\sigma}{\gamma}-1}}{1+e^{-2u}}\,du \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}} \int_{0}^{\infty }\sum_{k=0}^{\infty}(-1)^{k}e^{-(2k+\frac{\alpha}{\rho}+1)u}u^{\frac{ \sigma}{\gamma}-1} \,du \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}} \int_{0}^{\infty }\sum_{k=0}^{\infty} \bigl[e^{-(4k+\frac{\alpha}{\rho}+1)u}-e^{-(4k+2+\frac {\alpha}{\rho}+1)u}\bigr]u^{\frac{\sigma}{\gamma}-1}\,du. \end{aligned}$$

By the Lebesgue term-by-term theorem (see [39]), setting $v=(2k+\frac{\alpha}{\rho}+1)u$, we have

$$\begin{aligned} k(\sigma) =& \int_{0}^{\infty}\frac{\sec h(\rho t^{\gamma })}{e^{\alpha t^{\gamma}}}t^{\sigma-1}\,dt \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}}\sum_{k=0}^{\infty } \int_{0}^{\infty}\bigl[e^{-(4k+\frac{\alpha}{\rho}+1)u}-e^{-(4k+2+\frac{\alpha}{\rho}+1)u} \bigr]u^{\frac{\sigma}{\gamma}-1}\,du \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}}\sum_{k=0}^{\infty }(-1)^{k} \int_{0}^{\infty}e^{-(2k+\frac{\alpha}{\rho}+1)u}u^{\frac {\sigma }{\gamma}-1}\,du \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}}\sum_{k=0}^{\infty} \frac {(-1)^{k}}{(2k+\frac{\alpha}{\rho}+1)^{\sigma/\gamma}} \int_{0}^{\infty }e^{-v}v^{\frac{\sigma}{\gamma}-1}\,dv \\ =&\frac{2\Gamma(\frac{\sigma}{\gamma})}{\gamma(2\rho)^{\sigma /\gamma}}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k+\frac{\alpha+\rho}{2\rho})^{\sigma/\gamma}} \\ =&\frac{2\Gamma(\frac{\sigma}{\gamma})}{\gamma(2\rho)^{\sigma /\gamma}}\xi\biggl(\frac{\sigma}{\gamma},\frac{\alpha+\rho}{2\rho}\biggr)\in \mathbf {R}_{+}, \end{aligned}$$

(12)

where $\xi(s,a):=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(k+a)^{s}}$ ($s,a>0$) and

$$ \Gamma(y):= \int_{0}^{\infty}e^{-v}v^{y-1}\,dv \quad (y>0) $$

is the gamma function (see [40]).

In particular, for $\alpha=\rho>0$ and $\gamma=\sigma$, we have $h(t)=\frac{\sec h(\rho t^{\sigma})}{e^{\rho t^{\sigma}}}$ and $k(\sigma)=\frac {\ln2}{\sigma\rho}$; for $\alpha=0$ and $\gamma=\sigma$, we find $h(t)=\sec h(\rho t^{\sigma})$ and $k(\sigma)=\frac{\pi}{2\sigma\rho}$.

(ii) We have $\frac{1}{e^{u}+e^{-u}}>0$ and $(\frac{1}{e^{u}+e^{-u}})^{\prime}=-\frac{e^{u}-e^{-u}}{(e^{u}+e^{-u})^{2}}<0$ for $u>0$. If $g(u)>0$ and $g^{\prime}(u)<0$, then for $\gamma>0$, $g(\rho t^{\gamma })>0$, $\frac{d}{\,dt}g (\rho t^{\gamma})=\rho\gamma t^{\gamma-1}g^{\prime}(\rho t^{\gamma})<0$; for $y\in(n-1,n)$, $g(V(y))>0$, $\frac{d}{dy}g(V(y))=g^{\prime }(V(y))\nu_{n}<0$ ($n\in\mathbf{N}$).

If $g_{i}(u)>0$ and $g_{i}^{\prime}(u)<0$ ($i=1,2$), then we find for $u>0$,

$$ g_{1}(u)g_{2}(u)>0,\qquad \bigl(g_{1}(u)g_{2}(u) \bigr)^{\prime}=g_{1}^{\prime }(u)g_{2}(u)+g_{1}(u)g_{2}^{\prime}(u)< 0. $$

(iii) Therefore, for $\rho,\gamma,\sigma>0$, $\alpha>-\rho$ ($\alpha \geq0$), we have $h(t)>0$ and $h^{\prime}(t)<0$ with $k(\sigma)\in\mathbf {R}_{+}$, and then for $c>0$ and $n\in\mathbf{N}$, adding $\sigma\leq1$, we have

$$ h\bigl(cV(y)\bigr)V^{\sigma-1}(y)>0, \qquad \frac{d}{dy}h\bigl(cV(y) \bigr)V^{\sigma-1}(y)< 0 \quad \bigl(y\in (n-1,n)\bigr). $$

Lemma 1

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

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

(13)

Proof

Since we have

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

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 still have

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

Hence, adding these two inequalities, we have (13). □

Lemma 2

For $\gamma,\rho>0$, $\alpha>-\rho$ ($\alpha\geq 0$), and $0<\sigma\leq1$, define the following weight coefficients:

$$\begin{aligned}& \omega_{\delta}(\sigma,x) : =\sum_{n=1}^{\infty} \frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{U^{\delta\sigma}(x)\nu_{n}}{V_{n}^{1-\sigma}},\quad x\in\mathbf{R}_{+}, \end{aligned}$$

(14)

$$\begin{aligned}& \varpi_{\delta}(\sigma,n) : = \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{\sigma}\mu(x)}{U^{1-\delta\sigma}(x)}\,dx,\quad n\in\mathbf{N}. \end{aligned}$$

(15)

Then, we have the following inequalities:

$$\begin{aligned}& \omega_{\delta}(\sigma,x) < k(\sigma)\quad (x\in\mathbf{R}_{+}), \end{aligned}$$

(16)

$$\begin{aligned}& \varpi_{\delta}(\sigma,n) \leq k(\sigma)\quad (n\in\mathbf{N}), \end{aligned}$$

(17)

where $k(\sigma)$ is defined in (12).

Proof

Since $V_{n}=V(n)$ and $V^{\prime}(t)=\nu _{n}$ for $t\in(n-1,n)$, by Example 1(iii) and the proof of Lemma 1 we have

$$\begin{aligned}& \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})\nu_{n}}{e^{\alpha (U^{\delta}(x)V_{n})^{\gamma}}V_{n}^{1-\sigma}} \\& \quad =\frac{\sec h(\rho (U^{\delta}(x)V(n))^{\gamma})\nu_{n}}{e^{\alpha(U^{\delta }(x)V(n))^{\gamma}}V^{1-\sigma}(n)} \\& \quad < \int_{n-1}^{n}\frac{\sec h(\rho(U^{\delta}(x)V(t))^{\gamma})}{e^{\alpha(U^{\delta}(x)V(t))^{\gamma}}}\frac{V^{\prime }(t)}{V^{1-\sigma }(t)} \,dt\quad (n\in\mathbf{N}), \\& \omega_{\delta}(\sigma,x) < \sum_{n=1}^{\infty} \int_{n-1}^{n}\frac {\sec h(\rho(U^{\delta}(x)V(t))^{\gamma})}{e^{\alpha(U^{\delta }(x)V(t))^{\gamma}}}\frac{U^{\delta\sigma}(x)V^{\prime }(t)}{V^{1-\sigma }(t)} \,dt \\& \hphantom{\omega_{\delta}(\sigma,x)} = \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V(t))^{\gamma})}{ e^{\alpha(U^{\delta}(x)V(t))^{\gamma}}}\frac{U^{\delta\sigma }(x)V^{\prime}(t)}{V^{1-\sigma}(t)} \,dt. \end{aligned}$$

Setting $u=U^{\delta}(x)V(t)$, by (12) we find

$$\begin{aligned} \omega_{\delta}(\sigma,x) < & \int_{0}^{U^{\delta}(x)V(\infty)}\frac {\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}\frac{U^{\delta\sigma }(x)U^{-\delta}(x)}{(uU^{-\delta}(x))^{1-\sigma}} \,du \\ \leq& \int_{0}^{\infty}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma }}}u^{\sigma-1} \,du \\ =&k(\sigma). \end{aligned}$$

Hence, (16) follows.

Setting $u=V_{n}U^{\delta}(x)$ in (15), we find $du=\delta V_{n}U^{\delta-1}(x)\mu(x)\,dx$ and

$$\begin{aligned} \varpi_{\delta}(\sigma,n) =&\frac{1}{\delta} \int_{V_{n}U^{\delta }(0)}^{V_{n}U^{\delta}(\infty)}\frac{\sec h(\rho u^{\gamma })}{e^{\alpha u^{\gamma}}}\frac{V_{n}^{\sigma}V_{n}^{-1}(V_{n}^{-1}u)^{\frac {1}{\delta}-1}}{(V_{n}^{-1}u)^{\frac{1}{\delta}-\sigma}} \,du \\ =&\frac{1}{\delta} \int_{V_{n}U^{\delta}(0)}^{V_{n}U^{\delta}(\infty )}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1} \,du. \end{aligned}$$

If $\delta=1$, then

$$ \varpi_{1}(\sigma,n)= \int_{0}^{V_{n}U(\infty)}\frac{\sec h(\rho u^{\gamma })}{e^{\alpha u^{\gamma}}}u^{\sigma-1}\,du \leq \int_{0}^{\infty}\frac {\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1}\,du; $$

if $\delta=-1$, then

$$ \varpi_{-1}(\sigma,n)=- \int_{\infty}^{V_{n}U^{-1}(\infty)}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1}\,du \leq \int_{0}^{\infty}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma }}} u^{\sigma-1} \,du. $$

Then by (12) we have (17). □

Remark 1

We do not need $\sigma\leq1$ to obtain (17). If $U(\infty)=\infty$, then we have

$$ \varpi_{\delta}(\sigma,n)=k(\sigma)\quad (n\in\mathbf{N}). $$

(18)

For example, set $\mu(t)=\frac{1}{(1+t)^{\beta}}$ ($t>0$; $0\leq\beta \leq1$). Then, for $x\geq0$, we find

$$ U(x)= \int_{0}^{x}\frac{dt}{(1+t)^{\beta}}=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{(1+x)^{1-\beta}-1}{1-\beta},& 0\leq\beta< 1, \\ \ln(1+x), &\beta=1 \end{array}\displaystyle \right . < \infty, $$

and $U(\infty)=\int_{0}^{\infty}\frac{dt}{(1+t)^{\beta}}=\infty$.

Lemma 3

If $\gamma,\rho>0$, $\alpha>-\rho$ ($\alpha\geq0$), $0<\sigma \leq1$, then there exists $n_{0}\in\mathbf{N}$ such that $\nu _{n}\geq\nu _{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots\}$), and $V(\infty )=\infty$. Moreover, then

• (i) for $x\in\mathbf{R}_{+}$, we have

  $$ k(\sigma) \bigl(1-\theta_{\delta}(\sigma,x)\bigr)< \omega_{\delta}( \sigma,x), $$

  (19)

  where $\theta_{\delta}(\sigma,x)=O((U(x))^{\delta\sigma})\in(0,1)$;

• (ii) for any $b>0$, we have

  $$ \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}= \frac{1}{b} \biggl( \frac {1}{V_{n_{0}}^{b}}+bO(1) \biggr) . $$

  (20)

Proof

By Example 1(iii) we have

$$\begin{aligned} \omega_{\delta}(\sigma,x) =&\sum_{n=1}^{\infty} \frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{U^{\delta\sigma}(x)\nu_{n}}{V_{n}^{1-\sigma}} \\ \geq&\sum_{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{\sec h(\rho (U^{\delta }(x)V(n))^{\gamma})}{e^{\alpha(U^{\delta}(x)V(n))^{\gamma}}}\frac{U^{\delta\sigma}(x)\nu_{n+1}\,dt}{(V(n))^{1-\sigma}} \\ >&\sum_{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{\sec h(\rho(U^{\delta }(x)V(t))^{\gamma})}{e^{\alpha(U^{\delta}(x)V(t))^{\gamma}}}\frac{U^{\delta\sigma}(x)V^{\prime}(t)}{(V(t))^{1-\sigma}}\,dt \\ =& \int_{n_{0}}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V(t))^{\gamma })}{e^{\alpha(U^{\delta}(x)V(t))^{\gamma}}}\frac{U^{\delta\sigma }(x)V^{\prime}(t)}{(V(t))^{1-\sigma}} \,dt. \end{aligned}$$

Setting $u=U^{\delta}(x)V(t)$, in view of $V(\infty)=\infty$, by (12) we find

$$\begin{aligned}& \omega_{\delta}(\sigma,x) > \int_{U^{\delta}(x)V_{n_{0}}}^{\infty} \frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1}\,du \\& \hphantom{\omega_{\delta}(\sigma,x)}= k(\sigma)- \int_{0}^{U^{\delta}(x)V_{n_{0}}}\frac{\sec h(\rho u^{\gamma })}{e^{\alpha u^{\gamma}}}u^{\sigma-1} \,du=k(\sigma) \bigl(1-\theta_{\delta }(\sigma,x)\bigr), \\& \theta_{\delta}(\sigma,x) : =\frac{1}{k(\sigma)} \int_{0}^{U^{\delta }(x)V_{n_{0}}}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1} \,du\in(0,1). \end{aligned}$$

Since $F(u)=\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}$ is continuous in $(0,\infty)$ and satisfies $F(u)\rightarrow1$ ($u\rightarrow 0^{+}$), $F(u)\rightarrow0$ ($u\rightarrow\infty$), there exists a constant $L>0$ such that $F(u)\leq L$, namely, $\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}\leq L$ ($u\in(0,\infty)$). Hence, we find

$$ 0< \theta_{\delta}(\sigma,x)\leq\frac{L}{k(\sigma)} \int _{0}^{U^{\delta }(x)V_{n_{0}}}u^{\sigma-1}\,du= \frac{L(U^{\delta}(x)V_{n_{0}})^{\sigma }}{k(\sigma)\sigma}, $$

and then (19) follows.

For $b>0$, we find

$$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}=\sum _{n=1}^{n_{0}}\frac {\nu _{n}}{V_{n}^{1+b}}+\sum _{n=n_{0}+1}^{\infty}\frac{\nu_{n}}{V^{1+b}(n)} \\& \hphantom{\sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}} < \sum_{n=1}^{n_{0}} \frac{\nu_{n}}{V_{n}^{1+b}}+\sum_{n=n_{0}+1}^{\infty } \int_{n-1}^{n}\frac{V^{\prime}(x)}{V^{1+b}(x)}\,dx \\& \hphantom{\sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}} = \sum_{n=1}^{n_{0}} \frac{\nu_{n}}{V_{n}^{1+b}}+ \int_{n_{0}}^{\infty} \frac{dV(x)}{V^{1+b}(x)}=\sum _{n=1}^{n_{0}}\frac{\nu_{n}}{V_{n}^{1+b}}+ \frac{1}{bV^{b}(n_{0})} \\& \hphantom{\sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}} = \frac{1}{b} \Biggl( \frac{1}{V_{n_{0}}^{b}}+b\sum _{n=1}^{n_{0}}\frac {\nu _{n}}{V_{n}^{1+b}} \Biggr) , \\& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}} \geq \sum _{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{\nu_{n+1}}{V^{1+b}(n)}\,dx>\sum _{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{V^{\prime}(x)\,dx}{V^{1+b}(x)} \\& \hphantom{\sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}} = \int_{n_{0}}^{\infty}\frac{dV(x)}{V^{1+b}(x)}=\frac {1}{bV^{b}(n_{0})}= \frac{1}{bV_{n_{0}}^{b}}. \end{aligned}$$

Hence, we have (20). □

Note

For example, $\nu_{n}=\frac{1}{n^{\beta}}$ ($n\in \mathbf{N}$; $0\leq\beta\leq1$) satisfies the conditions of Lemma 3 (for $n_{0}=1$).

3 Main results and operator expressions

Theorem 1

If $\gamma,\rho>0$, $\alpha>-\rho$ ($\alpha\geq 0$), $0<\sigma\leq1$, $k(\sigma)$ is defined in by (12), then for $p>1$, $0<\|f\|_{p,\Phi_{\delta}}, \|a\|_{q,\Psi}<\infty$, we have the following equivalent inequalities:

$$\begin{aligned}& I : =\sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n}f(x) \,dx< k(\sigma)\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

(21)

$$\begin{aligned}& J_{1} : =\sum_{n=1}^{\infty} \frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p} \\& \hphantom{J_{1}} < k(\sigma)\|f\|_{p,\Phi_{\delta}}, \end{aligned}$$

(22)

$$\begin{aligned}& J_{2} : = \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1-q\delta\sigma }(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\& \hphantom{J_{2}} < k(\sigma)\|a\|_{q,\Psi}. \end{aligned}$$

(23)

Proof

By Hölder’s inequality with weight (see [41]), we have

$$\begin{aligned}& \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p} \\& \quad = \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \biggl( \frac{U^{\frac{1-\delta\sigma}{q}}(x)f(x)}{V_{n}^{\frac{1-\sigma}{p}}\mu^{\frac {1}{q}}(x)} \biggr) \biggl( \frac{V_{n}^{\frac{1-\sigma}{p}}\mu^{\frac {1}{q}}(x)}{U^{\frac{1-\delta\sigma}{q}}(x)} \biggr) \,dx \biggr] ^{p} \\& \quad \leq \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \biggl( \frac{U^{\frac {p(1-\delta \sigma)}{q}}(x)f^{p}(x)}{V_{n}^{1-\sigma}\mu^{\frac{p}{q}}(x)} \biggr) \,dx \\& \qquad {} \times \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{ V_{n}^{(1-\sigma)(p-1)}\mu(x)}{U^{1-\delta\sigma}(x)} \,dx \biggr] ^{p-1} \\& \quad = \frac{(\varpi_{\delta}(\sigma,n))^{p-1}}{V_{n}^{p\sigma-1}\nu _{n}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx. \end{aligned}$$

(24)

In view of (17) and the Lebesgue term-by-term integration theorem (see [42]), we find

$$\begin{aligned} J_{1} \leq&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \Biggl[ \sum _{n=1}^{\infty } \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \Biggl[ \int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{U^{(1-\delta\sigma)(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \biggl[ \int_{0}^{\infty}\omega_{\delta }(\sigma,x) \frac{U^{p(1-\delta\sigma)-1}(x)}{\mu^{p-1}(x)}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$

(25)

Then by (16) we have (22).

By Hölder’s inequality (see [41]) we have

$$\begin{aligned} I =&\sum_{n=1}^{\infty} \biggl[ \frac{\nu_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\sigma}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] \biggl( \frac{V_{n}^{\frac{1}{p}-\sigma}a_{n}}{\nu_{n}^{\frac {1}{p}}} \biggr) \\ \leq&J_{1}\|a\|_{q,\Psi}. \end{aligned}$$

(26)

Then by (22) we have (21). On the other hand, assuming that (21) is valid, we set

$$ a_{n}:=\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac {\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p-1},\quad n\in\mathbf{N}. $$

Then we find $J_{1}^{p}=\|a\|_{q,\Psi}^{q}$. If $J_{1}=0$, then (22) is trivially valid; if $J_{1}=\infty$, then (22) keeps impossible. Suppose that $0< J_{1}<\infty$. By (21) we have

$$ \|a\|_{q,\Psi}^{q}=J_{1}^{p}=I< k(\sigma) \|f\|_{p,\Phi_{\delta }}\|a\|_{q,\Psi},\qquad \|a\|_{q,\Psi}^{q-1}=J_{1}< k( \sigma)\|f\|_{p,\Phi _{\delta}}, $$

and then (22) follows, which is equivalent to (21).

Again by Hölder’s inequality with weight we have

$$\begin{aligned}& \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q} \\& \quad = \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \biggl( \frac{U^{\frac{1-\delta\sigma}{q}}(x)\nu_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1-\sigma}{p}}} \biggr) \biggl( \frac{V_{n}^{\frac{1-\sigma }{p}}a_{n}}{U^{\frac{1-\delta\sigma}{q}}(x)\nu_{n}^{\frac{1}{p}}} \biggr) \Biggr] ^{q} \\& \quad \leq \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{ U^{(1-\delta\sigma)(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}} \Biggr] ^{q-1} \\& \qquad {}\times\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{V_{n}^{\frac {q(1-\sigma )}{p}}}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q} \\& \quad = \frac{(\omega_{\delta}(\sigma,x))^{q-1}}{U^{q\delta\sigma -1}(x)\mu (x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{V_{n}^{(1-\sigma )(q-1)}\mu (x)}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}. \end{aligned}$$

(27)

Then by (16) and the Lebesgue term-by-term integration theorem it follows that

$$\begin{aligned} J_{2} < &\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \int_{0}^{\infty}\sum_{n=1}^{\infty } \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha (U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta \sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \sum _{n=1}^{\infty} \int _{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}} \frac{V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta \sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \sum _{n=1}^{\infty}\varpi_{\delta }(\sigma,n) \frac{V_{n}^{q(1-\sigma)-1}}{\nu_{n}^{q-1}}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$

(28)

Then by (17) we have (23).

By Hölder’s inequality we have

$$\begin{aligned} I =& \int_{0}^{\infty} \biggl( \frac{U^{\frac{1}{q}-\delta\sigma }(x)}{\mu^{\frac{1}{q}}(x)}f(x) \biggr) \Biggl[ \frac{\mu^{\frac{1}{q}}(x)}{U^{\frac {1}{q}-\delta\sigma}(x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] \,dx \\ \leq&\|f\|_{p,\Phi_{\delta}}J_{2}. \end{aligned}$$

(29)

Then by (23) we have (21). On the other hand, assuming that (23) is valid, we set

$$ f(x):=\frac{\mu(x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum_{n=1}^{\infty } \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q-1},\quad x\in \mathbf{R}_{+}. $$

Then we find $J_{2}^{q}=\|f\|_{p,\Phi_{\delta}}^{p}$. If $J_{2}=0$, then (23) is trivially valid; if $J_{2}=\infty$, then (23) keeps impossible. Suppose that $0< J_{2}<\infty$. By (21) we have

$$ \|f\|_{p,\Phi_{\delta}}^{p}=J_{2}^{q}=I< k(\sigma) \|f\|_{p,\Phi _{\delta }}\|a\|_{q,\Psi},\qquad \|f\|_{p,\Phi_{\delta}}^{p-1}=J_{2}< k( \sigma )\|a\|_{q,\Psi}, $$

and then (23) follows, which is equivalent to (21).

Therefore, (21), (22), and (23) are equivalent. □

Theorem 2

With the assumptions of Theorem  1, if there exists $n_{0}\in\mathbf{N}$ such that $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots\}$) and $U(\infty)=V(\infty)=\infty$, then the constant factor $k(\sigma)$ in (21), (22), and (23) is the best possible.

Proof

For $\varepsilon\in(0,q\sigma)$, we set $\tilde{\sigma }=\sigma-\frac{\varepsilon}{q}$ and $\tilde{f}=\tilde{f}(x)$, $x\in \mathbf{R}_{+}$, $\tilde{a}=\{\tilde{a}_{n}\}_{n=1}^{\infty}$,

$$\begin{aligned}& \tilde{f}(x) =\left \{ \textstyle\begin{array}{l@{\quad}l} U^{\delta(\tilde{\sigma}+\varepsilon)-1}(x)\mu(x), &0< x^{\delta }\leq1, \\ 0, &x^{\delta}>0,\end{array}\displaystyle \right . \end{aligned}$$

(30)

$$\begin{aligned}& \tilde{a}_{n} =V_{n}^{\tilde{\sigma}-1}\nu _{n}=V_{n}^{\sigma-\frac{\varepsilon}{q}-1} \nu_{n}, \quad n\in\mathbf{N}. \end{aligned}$$

(31)

Then for $\delta=\pm1$, since $U(\infty)=\infty$, we find

$$ \int_{\{x>0;0< x^{\delta}\leq1\}}\frac{\mu(x)}{U^{1-\delta \varepsilon}(x)}\,dx=\frac{1}{\varepsilon}U^{\delta\varepsilon}(1). $$

(32)

By (20), (32), and (19) we obtain

$$\begin{aligned}& \|\tilde{f}\|_{p,\Phi_{\delta}}\|\tilde{a}\|_{q,\Psi} = \biggl( \int_{\{x>0;0< x^{\delta}\leq1\}}\frac{\mu(x)\,dx}{U^{1-\delta \varepsilon }(x)} \biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|\tilde{f}\|_{p,\Phi_{\delta}}\|\tilde{a}\|_{q,\Psi}} = \frac{1}{\varepsilon}U^{\frac{\delta\varepsilon}{p}}(1) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}, \\& \tilde{I} : = \int_{0}^{\infty}\sum_{n=1}^{\infty} \frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\tilde{a}_{n}\tilde{f}(x)\,dx \\& \hphantom{\tilde{I}} = \int_{\{x>0;0< x^{\delta}\leq1\}}\sum_{n=1}^{\infty} \frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{\tilde{\sigma}-1}\nu_{n}\mu(x)}{U^{1-\delta (\tilde{\sigma}+\varepsilon)}(x)}\,dx \\& \hphantom{\tilde{I}} = \int_{\{x>0;0< x^{\delta}\leq1\}}\omega_{\delta}(\tilde {\sigma},x)\frac{\mu(x)}{U^{1-\delta\varepsilon}(x)}\,dx \\& \hphantom{\tilde{I}} \geq k(\tilde{\sigma}) \int_{\{x>0;0< x^{\delta}\leq1\}}\bigl(1-\theta _{\delta}(\tilde{\sigma},x) \bigr)\frac{\mu(x)}{U^{1-\delta \varepsilon}(x)}\,dx \\& \hphantom{\tilde{I}} = k(\tilde{\sigma}) \int_{\{x>0;0< x^{\delta}\leq 1\}}\bigl(1-O\bigl(\bigl(U(x)\bigr)^{\delta(\sigma-\frac{\varepsilon}{q})}\bigr) \bigr)\frac{\mu (x)}{U^{1-\delta\varepsilon}(x)}\,dx \\& \hphantom{\tilde{I}} = k(\tilde{\sigma})\biggl[ \int_{\{x>0;0< x^{\delta}\leq1\}}\frac {\mu (x)}{U^{1-\delta\varepsilon}(x)}\,dx - \int_{\{x>0;0< x^{\delta}\leq1\}}O\biggl(\frac{\mu(x)}{U^{1-\delta (\sigma+\frac{\varepsilon}{p})}(x)}\biggr)\,dx\biggr] \\& \hphantom{\tilde{I}} = \frac{1}{\varepsilon}k\biggl(\sigma-\frac{\varepsilon}{q}\biggr) \bigl(U^{\delta \varepsilon}(1)-\varepsilon O(1)\bigr). \end{aligned}$$

(33)

If there exists a positive constant $K\leq k(\sigma)$ such that (21) is valid when replacing $k(\sigma)$ by K, then, in particular, by the Lebesgue term-by-term integration theorem we have $\varepsilon \tilde{I}<\varepsilon K\|\tilde{f}\|_{p,\Phi_{\delta}}\|\tilde {a}\|_{q,\Psi }$, namely,

$$ k\biggl(\sigma-\frac{\varepsilon}{q}\biggr) \bigl(U^{\delta\varepsilon}(1)-\varepsilon O(1)\bigr)< K\cdot U^{\frac{\delta\varepsilon}{p}}(1) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}. $$

It follows that $k(\sigma)\leq K$ ($\varepsilon\rightarrow0^{+}$). Hence, $K=k(\sigma)$ is the best possible constant factor of (21).

The constant factor $k(\sigma)$ in (22) ((23)) is still the best possible. Otherwise, we would reach a contradiction by (26) ((29)) that the constant factor in (21) is not the best possible. □

For $p>1$, we find $\Psi^{1-p}(n)=\frac{\nu_{n}}{V_{n}^{1-p\sigma}}$ ($n\in\mathbf{N}$), $\Phi _{\delta}^{1-q}(x)=\frac{\mu(x)}{U^{1-q\delta\sigma}(x)}$ ($x\in \mathbf{R}_{+}$) and define the following real normed spaces:

$$\begin{aligned}& L_{p,\Phi_{\delta}}(\mathbf{R}_{+}) = \bigl\{ f;f=f(x),x\in \mathbf{R}_{+},\|f\|_{p,\Phi_{\delta}}< \infty\bigr\} , \\& l_{q,\Psi} = \bigl\{ a;a=\{a_{n}\}_{n=1}^{\infty}, \|a\|_{q,\Psi}< \infty\bigr\} , \\& L_{q,\Phi_{\delta}^{1-q}}(\mathbf{R}_{+}) = \bigl\{ h;h=h(x),x\in\mathbf {R}_{+},\|h\|_{q,\Phi_{\delta}^{1-q}}< \infty\bigr\} , \\& l_{p,\Psi^{1-p}} = \bigl\{ c;c=\{c_{n}\}_{n=1}^{\infty}, \|c\|_{p,\Psi ^{1-p}}< \infty\bigr\} . \end{aligned}$$

Assuming that $f\in L_{p,\Phi_{\delta}}(\mathbf{R}_{+})$ and setting

$$ c=\{c_{n}\}_{n=1}^{\infty},\qquad c_{n}:= \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}f(x)\,dx, \quad n\in\mathbf{N}, $$

we can rewrite (22) as $\|c\|_{p,\Psi^{1-p}}< k(\sigma )\|f\|_{p,\Phi _{\delta}}<\infty$, namely, $c\in l_{p,\Psi^{1-p}}$.

Definition 1

Define a half-discrete Hardy-Hilbert-type operator $T_{1}:L_{p,\Phi_{\delta}}(\mathbf{R}_{+})\rightarrow l_{p,\Psi ^{1-p}}$ as follows: For any $f\in L_{p,\Phi_{\delta}}(\mathbf{R}_{+})$, the exists a unique representation $T_{1}f=c\in l_{p,\Psi^{1-p}}$. We define the formal inner product of $T_{1}f$ and $a=\{a_{n}\}_{n=1}^{\infty}\in l_{q,\Psi}$ as follows:

$$ (T_{1}f,a):=\sum_{n=1}^{\infty} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] a_{n}. $$

(34)

Then we can rewrite (21) and (22) as follows:

$$\begin{aligned}& (T_{1}f,a) < k(\sigma)\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

(35)

$$\begin{aligned}& \|T_{1}f\|_{p,\Psi^{1-p}} < k(\sigma)\|f\|_{p,\Phi_{\delta}}. \end{aligned}$$

(36)

Define the norm of operator $T_{1}$ as follows:

$$ \|T_{1}\|:=\sup_{f(\neq\theta)\in L_{p,\Phi_{\delta}}(\mathbf {R}_{+})}\frac{\|T_{1}f\|_{p,\Psi^{1-p}}}{\|f\|_{p,\Phi_{\delta}}}. $$

Then by (36) it follows that $\|T_{1}\|\leq k(\sigma)$. Since by Theorem 2 the constant factor in (36) is the best possible, we have

$$ \|T_{1}\|=k(\sigma)=\frac{2\Gamma(\frac{\sigma}{\gamma})}{\gamma (2\rho )^{\sigma/\gamma}}\xi\biggl(\frac{\sigma}{\gamma}, \frac{\alpha+\rho }{2\rho}\biggr). $$

(37)

Assuming that $a=\{a_{n}\}_{n=1}^{\infty}\in l_{q,\Psi}$ and setting

$$ h(x):=\sum_{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n}, \quad x\in\mathbf{R}_{+}, $$

we can rewrite (23) as $\|h\|_{q,\Phi_{\delta}^{1-q}}< k(\sigma )\|a\|_{q,\Psi}<\infty$, namely, $h\in L_{q,\Phi_{\delta }^{1-q}}(\mathbf{R}_{+})$.

Definition 2

Define a half-discrete Hardy-Hilbert-type operator $T_{2}:l_{q,\Psi}\rightarrow L_{q,\Phi_{\delta}^{1-q}}(\mathbf {R}_{+})$ as follows: For any $a=\{a_{n}\}_{n=1}^{\infty}\in l_{q,\Psi}$, there exists a unique representation $T_{2}a=h\in L_{q,\Phi_{\delta}^{1-q}}(\mathbf {R}_{+})$. We define the formal inner product of $T_{2}a$ and $f\in L_{p,\Phi _{\delta}}(\mathbf{R}_{+})$ as follows:

$$ (T_{2}a,f):= \int_{0}^{\infty} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}a_{n} \Biggr] f(x)\,dx. $$

(38)

Then we can rewrite (21) and (23) as follows:

$$\begin{aligned}& (T_{2}a,f) < k(\sigma)\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

(39)

$$\begin{aligned}& \|T_{2}a\|_{q,\Phi_{\delta}^{1-q}} < k(\sigma)\|a\|_{q,\Psi}. \end{aligned}$$

(40)

Define the norm of operator $T_{2}$ as follows:

$$ \|T_{2}\|:=\sup_{a(\neq\theta)\in l_{q,\Psi}}\frac{\|T_{2}a\|_{q,\Phi _{\delta}^{1-q}}}{\|a\|_{q,\Psi}}. $$

Then by (40) we find $\|T_{2}\|\leq k(\sigma)$. Since by Theorem 2 the constant factor in (40) is the best possible, we have

$$ \|T_{2}\|=k(\sigma)=\frac{2\Gamma(\frac{\sigma}{\gamma})}{\gamma (2\rho )^{\sigma/\gamma}}\xi\biggl(\frac{\sigma}{\gamma}, \frac{\alpha+\rho }{2\rho}\biggr)=\|T_{1}\|. $$

(41)

4 Some equivalent reverse inequalities

In the following, we also set

$$ \widetilde{\Phi}_{\delta}(x):=\bigl(1-\theta_{\delta}(\sigma,x) \bigr)\frac{U^{p(1-\delta\sigma)-1}(x)}{\mu^{p-1}(x)} \quad (x\in\mathbf{R}_{+}). $$

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

Theorem 3

If $\gamma,\rho>0$, $\alpha>-\rho$ ($\alpha\geq 0$), $0<\sigma\leq1$, $k(\sigma)$ is defined in (12), there exists $n_{0}\in\mathbf{N}$ such that $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots\}$), and $U(\infty)=V(\infty)=\infty$, then for $p<0$, $0<\|f\|_{p,\Phi_{\delta}}, \|a\|_{q,\Psi}<\infty$, we have the following equivalent inequalities with the best possible constant factor $k(\sigma)$:

$$\begin{aligned}& I = \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n}f(x) \,dx>k(\sigma)\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

(42)

$$\begin{aligned}& J_{1} = \sum_{n=1}^{\infty} \frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p}>k(\sigma )\|f\|_{p,\Phi_{\delta}}, \end{aligned}$$

(43)

$$\begin{aligned}& J_{2} = \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1-q\delta\sigma }(x)} \Biggl[ \sum_{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\& \hphantom{J_{2}} > k(\sigma)\|a\|_{q,\Psi}. \end{aligned}$$

(44)

Proof

By the reverse Hölder inequality with weight (see [41]), since $p<0$, similarly as obtaining (24) and (25), we have

$$\begin{aligned}& \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p} \\& \quad \leq \frac{(\varpi_{\delta}(\sigma,n))^{p-1}}{V_{n}^{p\sigma -1}\nu_{n}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx. \end{aligned}$$

Then by (18) and the Lebesgue term-by-term integration theorem it follows that

$$\begin{aligned} J_{1} \geq&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \Biggl[ \sum _{n=1}^{\infty } \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \biggl[ \int_{0}^{\infty}\omega_{\delta }(\sigma,x) \frac{U^{p(1-\delta\sigma)-1}(x)}{\mu^{p-1}(x)}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$

Then by (16) we have (43).

By the reverse Hölder inequality we have

$$\begin{aligned} I =&\sum_{n=1}^{\infty} \biggl[ \frac{\nu_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\sigma}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] \biggl( \frac{V_{n}^{\frac{1}{p}-\sigma}a_{n}}{\nu_{n}^{\frac {1}{p}}} \biggr) \\ \geq&J_{1}\|a\|_{q,\Psi}. \end{aligned}$$

(45)

Then by (43) we have (42). On the other hand, assuming that (42) is valid, we set $a_{n}$ as in Theorem 1. Then we find $J_{1}^{p}=\|a\|_{q,\Psi}^{q}$. If $J_{1}=\infty$, then (43) is trivially valid; if $J_{1}=0$, then (43) keeps impossible. Suppose that $0< J_{1}<\infty$. By (42) it follows that

$$\begin{aligned}& \|a\|_{q,\Psi}^{q} = J_{1}^{p}=I>k( \sigma)\|f\|_{p,\Phi_{\delta }}\|a\|_{q,\Psi}, \\& \|a\|_{q,\Psi}^{q-1} = J_{1}>k(\sigma)\|f \|_{p,\Phi_{\delta}}, \end{aligned}$$

and then (43) follows, which is equivalent to (42).

Still by the reverse Hölder inequality with weight, since $0< q<1$, similarly as obtaining (27) and (28), we have

$$\begin{aligned}& \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q} \\& \quad \geq \frac{(\omega_{\delta}(\sigma,x))^{q-1}}{U^{q\delta\sigma -1}(x)\mu(x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{ V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}. \end{aligned}$$

Then by (16) and the Lebesgue term-by-term integration theorem it follows that

$$\begin{aligned} J_{2} >&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \int_{0}^{\infty }\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{V_{n}^{(1-\sigma )(q-1)}\mu (x)}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \sum _{n=1}^{\infty}\varpi_{\delta }(\sigma,n) \frac{V_{n}^{q(1-\sigma)-1}}{\nu_{n}^{q-1}}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$

Then by (18) we have (44).

By the reverse Hölder inequality we have

$$\begin{aligned} I =& \int_{0}^{\infty} \biggl( \frac{U^{\frac{1}{q}-\delta\sigma }(x)}{\mu^{\frac{1}{q}}(x)}f(x) \biggr) \Biggl[ \frac{\mu^{\frac{1}{q}}(x)}{U^{\frac {1}{q}-\delta\sigma}(x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] \,dx \\ \geq&\|f\|_{p,\Phi_{\delta}}J_{2}. \end{aligned}$$

(46)

Then by (44) we have (42). On the other hand, assuming that (44) is valid, we set $f(x)$ as in Theorem 1. Then we find $J_{2}^{q}=\|f\|_{p,\Phi_{\delta}}^{p}$. If $J_{2}=\infty$, then (44) is trivially valid; if $J_{2}=0$, then (44) keeps impossible. Suppose that $0< J_{2}<\infty$. By (42) it follows that

$$ \|f\|_{p,\Phi_{\delta}}^{p}=J_{2}^{q}=I>k(\sigma) \|f\|_{p,\Phi _{\delta }}\|a\|_{q,\Psi},\qquad \|f\|_{p,\Phi_{\delta}}^{p-1}=J_{2}>k( \sigma )\|a\|_{q,\Psi}, $$

and then (44) follows, which is equivalent to (42).

Therefore, inequalities (42), (43), and (44) are equivalent.

For $\varepsilon\in(0,q\sigma)$, we set $\tilde{\sigma}=\sigma- \frac{\varepsilon}{q}$ and $\tilde{f}=\tilde{f}(x)$, $x\in \mathbf{R}_{+}$, $\tilde{ a}=\{\tilde{a}_{n}\}_{n=1}^{\infty}$,

$$\begin{aligned}& \tilde{f}(x) = \left \{ \textstyle\begin{array}{l@{\quad}l} U^{\delta(\tilde{\sigma}+\varepsilon)-1}(x)\mu(x), &0< x^{\delta }\leq1, \\ 0, &x^{\delta}>0,\end{array}\displaystyle \right . \\& \tilde{a}_{n} = \widetilde{V}_{n}^{\tilde{\sigma}-1} \nu_{n}=\widetilde{V}_{n}^{\sigma-\frac{\varepsilon}{q}-1} \nu_{n},\quad n\in \mathbf{N}. \end{aligned}$$

By (20), (32), and (16) we obtain

$$\begin{aligned}& \|\tilde{f}\|_{p,\Phi_{\delta}}\|\tilde{a}\|_{q,\Psi}=\frac {1}{\varepsilon}U^{\frac{\delta\varepsilon}{p}}(1) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}, \\& \tilde{I} = \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\tilde{a}_{n}\tilde{f}(x)\,dx \\& \hphantom{\tilde{I}} = \int_{\{x>0;0< x^{\delta}\leq1\}}\omega_{\delta}(\tilde {\sigma},x)\frac{\mu(x)}{U^{1-\delta\varepsilon}(x)}\,dx \\& \hphantom{\tilde{I}} \leq k(\tilde{\sigma}) \int_{\{x>0;0< x^{\delta}\leq1\}}\frac {\mu(x)}{U^{1-\delta\varepsilon}(x)}\,dx=\frac{1}{\varepsilon}k\biggl( \sigma-\frac{ \varepsilon}{q}\biggr)U^{\delta\varepsilon}(1). \end{aligned}$$

If there exists a positive constant $K\geq k(\sigma)$ such that (42) is valid when replacing $k(\sigma)$ by K, then, in particular, we have $\varepsilon\tilde{I}>\varepsilon K\|\tilde{f}\|_{p,\Phi _{\delta }}\|\tilde{a}\|_{q,\Psi}$, namely,

$$ k\biggl(\sigma-\frac{\varepsilon}{q}\biggr)U^{\delta\varepsilon}(1)>K\cdot U^{\frac{\delta\varepsilon}{p}}(1) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}. $$

It follows that $k(\sigma)\geq K$ ($\varepsilon\rightarrow0^{+}$). Hence, $K=k(\sigma)$ is the best possible constant factor of (42).

The constant factor $k(\sigma)$ in (43) ((44)) is still the best possible. Otherwise, we would reach a contradiction by (45) ((46)) that the constant factor in (42) is not the best possible. □

Theorem 4

With the assumptions of Theorem  3, if $0< p<1$, $0<\|f\|_{p,\Phi_{\delta}}$, and $\|a\|_{q,\Psi}<\infty$, then we have the following equivalent inequalities with the best possible constant factor $k(\sigma)$:

$$\begin{aligned}& I = \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n}f(x) \,dx>k(\sigma)\|f\|_{p,\widetilde{\Phi}_{\delta }}\|a\|_{q,\Psi}, \end{aligned}$$

(47)

$$\begin{aligned}& J_{1} = \sum_{n=1}^{\infty} \frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p}>k(\sigma )\|f\|_{p,\widetilde{\Phi}_{\delta}}, \end{aligned}$$

(48)

$$\begin{aligned}& J : = \Biggl\{ \int_{0}^{\infty}\frac{(1-\theta_{\delta}(\sigma ,x))^{1-q}\mu(x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty }\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\& \hphantom{J} > k(\sigma)\|a\|_{q,\Psi}. \end{aligned}$$

(49)

Proof

By the reverse Hölder inequality with weight, since $0< p<1$, similarly as obtaining (24) and (25), we have

$$\begin{aligned}& \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p} \\& \quad \geq \frac{(\varpi_{\delta}(\sigma,n))^{p-1}}{V_{n}^{p\sigma -1}\nu_{n}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx. \end{aligned}$$

In view of (18) and the Lebesgue term-by-term integration theorem, we find

$$\begin{aligned} J_{1} \geq&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \Biggl[ \sum _{n=1}^{\infty } \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \biggl[ \int_{0}^{\infty}\omega_{\delta }(\sigma,x) \frac{U^{p(1-\delta\sigma)-1}(x)}{\mu^{p-1}(x)}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$

Then by (19) we have (48).

By the reverse Hölder inequality we have

$$\begin{aligned} I =&\sum_{n=1}^{\infty} \biggl[ \frac{\nu_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\sigma}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] \biggl( \frac{V_{n}^{\frac{1}{p}-\sigma}a_{n}}{\nu_{n}^{\frac {1}{p}}} \biggr) \\ \geq&J_{1}\|a\|_{q,\Psi}. \end{aligned}$$

(50)

Then by (48) we have (47). On the other hand, assuming that (47) is valid, we set $a_{n}$ as in Theorem 1. Then we find $J_{1}^{p}=\|a\|_{q,\Psi}^{q}$. If $J_{1}=\infty$, then (48) is trivially valid; if $J_{1}=0$, then (48) keeps impossible. Suppose that $0< J_{1}<\infty$. By (47) it follows that

$$ \|a\|_{q,\Psi}^{q}=J_{1}^{p}=I>k(\sigma) \|f\|_{p,\widetilde{\Phi }_{\delta }}\|a\|_{q,\Psi},\qquad \|a\|_{q,\Psi}^{q-1}=J_{1}>k( \sigma )\|f\|_{p,\widetilde{\Phi}_{\delta}}, $$

and then (48) follows, which is equivalent to (47).

Again by the reverse Hölder inequality with weight, since $q<0$, we have

$$\begin{aligned}& \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q} \\& \quad \leq \frac{(\omega_{\delta}(\sigma,x))^{q-1}}{U^{q\delta\sigma -1}(x)\mu(x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{ V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}. \end{aligned}$$

Then by (19) and the Lebesgue term-by-term integration theorem it follows that

$$\begin{aligned} J >&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta \sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \sum _{n=1}^{\infty}\varpi_{\delta }(\sigma,n) \frac{V_{n}^{q(1-\sigma)-1}}{\nu_{n}^{q-1}}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$

Then by (18) we have (49).

By the reverse Hölder inequality we have

$$\begin{aligned} I =& \int_{0}^{\infty} \biggl[ \bigl(1-\theta_{\delta}( \sigma,x)\bigr)^{\frac {1}{p}}\frac{U^{\frac{1}{q}-\delta\sigma}(x)}{\mu^{\frac {1}{q}}(x)}f(x) \biggr] \\ &{}\times \Biggl[ \frac{(1-\theta_{\delta}(\sigma,x))^{\frac{-1}{p}}\mu ^{\frac{1}{q}}(x)}{U^{\frac{1}{q}-\delta\sigma}(x)}\sum_{n=1}^{\infty } \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}a_{n} \Biggr] \,dx \\ \geq&\|f\|_{p,\widetilde{\Phi }_{\delta }}J. \end{aligned}$$

(51)

Then by (49) we have (47). On the other hand, assuming that (47) is valid, we set $f(x)$ as in Theorem 1. Then we find $J^{q}=\|f\|_{p,\widetilde{\Phi}_{\delta}}^{p}$. If $J=\infty$, then (49) is trivially valid; if $J=0$, then (49) keeps impossible. Suppose that $0< J<\infty$. By (47) it follows that

$$ \|f\|_{p,\widetilde{\Phi}_{\delta}}^{p}=J^{q}=I>k(\sigma)\|f \|_{p,\widetilde{\Phi}_{\delta}}\|a\|_{q,\Psi},\qquad \|f\|_{p,\widetilde{\Phi} _{\delta}}^{p-1}=J>k( \sigma)\|a\|_{q,\Psi}, $$

and then (49) follows, which is equivalent to (47).

Therefore, inequalities (47), (48), and (49) are equivalent.

For $\varepsilon\in(0,p\sigma)$, we set $\tilde{\sigma}=\sigma+ \frac{\varepsilon}{p}$ and $\tilde{f}=\tilde{f}(x)$, $x\in \mathbf{R}_{+}$, $\tilde{a}=\{\tilde{a}_{n}\}_{n=1}^{\infty}$,

$$\begin{aligned}& \tilde{f}(x) = \left \{ \textstyle\begin{array}{l@{\quad}l} U^{\delta\tilde{\sigma}-1}(x)\mu(x),& 0< x^{\delta}\leq1, \\ 0,& x^{\delta}>0,\end{array}\displaystyle \right . \\& \tilde{a}_{n} = \widetilde{V}_{n}^{\tilde{\sigma}-\varepsilon -1} \nu_{n}=\widetilde{V}_{n}^{\sigma-\frac{\varepsilon}{q}-1}\nu _{n}, \quad n\in\mathbf{N}. \end{aligned}$$

By (19), (20), and (32) we obtain

$$\begin{aligned}& \|\tilde{f}\|_{p,\widetilde{\Phi}_{\delta}}\|\tilde {a}\|_{q,\Psi}= \biggl[ \int_{\{x>0;0< x^{\delta}\leq1\}}\bigl(1-O\bigl(\bigl(U(x)\bigr)^{\delta\sigma }\bigr) \bigr)\frac{\mu(x)\,dx}{U^{1-\delta\varepsilon}(x)} \biggr] ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+\varepsilon }} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|\tilde{f}\|_{p,\widetilde{\Phi}_{\delta}}\|\tilde {a}\|_{q,\Psi}}=\frac{1}{\varepsilon} \bigl( U^{\delta \varepsilon }(1)-\varepsilon O(1) \bigr) ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}, \\& \tilde{I} = \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\tilde{a}_{n}\tilde{f}(x)\,dx \\& \hphantom{\tilde{I}} = \sum_{n=1}^{\infty} \biggl( \int_{\{x>0;0< x^{\delta}\leq1\}}\frac {\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{\tilde{\sigma}}\mu (x)}{U^{1-\delta \tilde{\sigma}}(x)}\,dx \biggr) \frac{\nu_{n}}{V_{n}^{1+\varepsilon}} \\& \hphantom{\tilde{I}} \leq \sum_{n=1}^{\infty} \biggl( \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{\tilde{\sigma}}\mu(x)}{U^{1-\delta\tilde{\sigma }}(x)}\,dx \biggr) \frac{\nu_{n}}{V_{n}^{1+\varepsilon}} \\& \hphantom{\tilde{I}} = \sum_{n=1}^{\infty} \varpi_{\delta}(\tilde{\sigma},n)\frac {\nu_{n}}{V_{n}^{1+\varepsilon}}=k(\tilde{\sigma})\sum _{n=1}^{\infty }\frac{\nu_{n}}{V_{n}^{1+\varepsilon}} \\& \hphantom{\tilde{I}} = \frac{1}{\varepsilon}k\biggl(\sigma+\frac{\varepsilon}{p}\biggr) \biggl( \frac {1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) . \end{aligned}$$

If there exists a positive constant $K\geq k(\sigma)$ such that (42) is valid when replacing $k(\sigma)$ by K, then, in particular, we have $\varepsilon\tilde{I}>\varepsilon K\|\tilde{f}\|_{p,\widetilde {\Phi}_{\delta}}\|\tilde{a}\|_{q,\Psi}$, namely,

$$ k\biggl(\sigma+\frac{\varepsilon}{p}\biggr) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon }}+ \varepsilon O(1) \biggr) >K \bigl( U^{\delta\varepsilon}(1)-\varepsilon O(1) \bigr) ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}. $$

It follows that $k(\sigma)\geq K$ ($\varepsilon\rightarrow0^{+}$). Hence, $K=k(\sigma)$ is the best possible constant factor of (47).

The constant factor $k(\sigma)$ in (48) ((49)) is still the best possible. Otherwise, we would reach a contradiction by (50) ((51)) that the constant factor in (47) is not the best possible. □

5 Some corollaries

For $\delta=1$ in Theorems 2-4, we have the following inequalities with inhomogeneous kernel.

Corollary 1

If $\gamma,\rho>0$, $\alpha>\rho$ ($\alpha\geq 0$), $0<\sigma\leq1$, $k(\sigma)$ is indicated by (12), there exists $n_{0}\in\mathbf{N}$ such that $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots\}$), and $U(\infty)=V(\infty)=\infty$, then

• (i) for $p>1$, $0<\|f\|_{p,\Phi_{1}}$, $\|a\|_{q,\Psi}<\infty$, we have the following equivalent inequalities:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho (U(x)V_{n})^{\gamma})}{e^{\alpha(U(x)V_{n})^{\gamma}}}a_{n}f(x) \,dx< k(\sigma)\|f\|_{p,\Phi_{1}}\|a\|_{q,\Psi}, \end{aligned}$$

  (52)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U(x)V_{n})^{\gamma})}{e^{\alpha (U(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p}< k(\sigma)\|f\|_{p,\Phi_{1}}, \end{aligned}$$

  (53)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1-q\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho(U(x)V_{n})^{\gamma})}{e^{\alpha (U(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac {1}{q}}< k(\sigma )\|a \|_{q,\Psi}; \end{aligned}$$

  (54)

• (ii) for $p<0$, $0<\|f\|_{p,\Phi_{1}}$, and $\|a\|_{q,\Psi }<\infty$, we have the following equivalent inequalities:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho (U(x)V_{n})^{\gamma })}{e^{\alpha(U(x)V_{n})^{\gamma}}}a_{n}f(x) \,dx>k(\sigma)\|f\|_{p,\Phi _{1}}\|a\|_{q,\Psi}, \end{aligned}$$

  (55)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U(x)V_{n})^{\gamma})}{e^{\alpha (U(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p}>k(\sigma)\|f\|_{p,\Phi_{1}}, \end{aligned}$$

  (56)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1-q\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho(U(x)V_{n})^{\gamma})}{e^{\alpha (U(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac {1}{q}}>k(\sigma )\|a \|_{q,\Psi}; \end{aligned}$$

  (57)

• (iii) for $0< p<1$, $0<\|f\|_{p,\Phi_{1}}$, and $\|a\|_{q,\Psi }<\infty$, we have the following equivalent inequalities:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho (U(x)V_{n})^{\gamma})}{e^{\alpha(U(x)V_{n})^{\gamma}}}a_{n}f(x) \,dx>k(\sigma)\|f\|_{p,\widetilde{\Phi}_{1}}\|a\|_{q,\Psi}, \end{aligned}$$

  (58)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U(x)V_{n})^{\gamma})}{e^{\alpha (U(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p}>k(\sigma)\|f\|_{p,\widetilde {\Phi }_{1}}, \end{aligned}$$

  (59)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{(1-\theta_{1}(\sigma,x))^{1-q}\mu (x)}{U^{1-q\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho (U(x)V_{n})^{\gamma})}{e^{\alpha(U(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\& \quad > k(\sigma)\|a\|_{q,\Psi}. \end{aligned}$$

  (60)

The above inequalities are with the best possible constant factor $k(\sigma)$.

For $\delta=-1$ in Theorems 2-4, we have the following inequalities with the homogeneous kernel of degree 0.

Corollary 2

If $\gamma,\rho>0$, $\alpha>-\rho$ ($\alpha\geq 0$), $0<\sigma\leq1$, $k(\sigma)$ is defined in (12), there exists $n_{0}\in\mathbf{N}$ such that $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots\}$), and $U(\infty)=V(\infty)=\infty$, then

• (i) for $p>1$, $0<\|f\|_{p,\Phi_{-1}}$, and $\|a\|_{q,\Psi }<\infty$, we have the following equivalent inequalities:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(\frac {V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma }}}a_{n}f(x) \,dx< k(\sigma )\|f\|_{p,\Phi_{-1}}\|a\|_{q,\Psi}, \end{aligned}$$

  (61)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(\frac{V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma}}}f(x)\,dx \biggr] ^{p}< k(\sigma )\|f\|_{p,\Phi_{-1}}, \end{aligned}$$

  (62)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1+q\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho(\frac{V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{ \frac{1}{q}}< k(\sigma)\|a \|_{q,\Psi}; \end{aligned}$$

  (63)

• (ii) for $p<0$, $0<\|f\|_{p,\Phi_{-1}}$, and $\|a\|_{q,\Psi }<\infty$, we have the following equivalent inequalities:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(\frac {V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma }}}a_{n}f(x) \,dx>k(\sigma )\|f\|_{p,\Phi_{-1}}\|a\|_{q,\Psi}, \end{aligned}$$

  (64)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(\frac{V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma}}}f(x)\,dx \biggr] ^{p}>k(\sigma )\|f\|_{p,\Phi_{-1}}, \end{aligned}$$

  (65)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1+q\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho(\frac{V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{ \frac{1}{q}}>k(\sigma)\|a \|_{q,\Psi}; \end{aligned}$$

  (66)

• (iii) for $0< p<1$, $0<\|f\|_{p,\Phi_{-1}}$, and $\|a\|_{q,\Psi}<\infty$, we have the following equivalent inequalities:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(\frac {V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma }}}a_{n}f(x) \,dx>k(\sigma )\|f\|_{p,\widetilde{\Phi}_{-1}}\|a\|_{q,\Psi}, \end{aligned}$$

  (67)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(\frac{V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma}}}f(x)\,dx \biggr] ^{p}>k(\sigma )\|f\|_{p,\widetilde{\Phi}_{-1}}, \end{aligned}$$

  (68)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{(1-\theta_{-1 }(\sigma,x))^{1-q}\mu (x)}{U^{1+q\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho (\frac{V_{n}}{U(x)})^{\gamma})}{e^{\alpha(\frac{V_{n}}{U(x)})^{\gamma }}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\& \quad > k(\sigma)\|a\|_{q,\Psi}. \end{aligned}$$

  (69)

The above inequalities are with the best possible constant factor $k(\sigma)$.

For $\alpha=\rho$ and $\gamma=\sigma$ in Theorems 2-4, we have the following corollary.

Corollary 3

If $\rho>0$, $0<\sigma\leq1$, there exists $n_{0}\in\mathbf{N}$ such that $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots\}$), and $U(\infty)=V(\infty)=\infty$, then

• (i) for $p>1$, $0<\|f\|_{p,\Phi_{\delta}}$, $\|a\|_{q,\Psi }<\infty$, we have the following equivalent inequalities with the best possible constant factor $\frac{\ln2}{\sigma\rho}$:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\sigma})}{e^{\rho(U^{\delta}(x)V_{n})^{\sigma }}}a_{n}f(x) \,dx< \frac{\ln2}{\sigma\rho}\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

  (70)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\sigma})}{e^{\rho(U^{\delta}(x)V_{n})^{\sigma}}}f(x)\,dx \biggr] ^{p}< \frac{\ln 2}{\sigma\rho}\|f\|_{p,\Phi_{\delta}}, \end{aligned}$$

  (71)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\sigma})}{ e^{\rho(U^{\delta}(x)V_{n})^{\sigma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\& \quad < \frac{\ln2}{\sigma\rho}\|a \|_{q,\Psi}; \end{aligned}$$

  (72)

• (ii) for $p<0$, $0<\|f\|_{p,\Phi_{\delta}}$, $\|a\|_{q,\Psi }<\infty$, we have the following equivalent inequalities with the best possible constant factor $\frac{\ln2}{\sigma\rho}$:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\sigma})}{e^{\rho(U^{\delta}(x)V_{n})^{\sigma }}}a_{n}f(x) \,dx>\frac{\ln2}{\sigma\rho}\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

  (73)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\sigma})}{e^{\rho(U^{\delta}(x)V_{n})^{\sigma}}}f(x)\,dx \biggr] ^{p}>\frac{\ln 2}{\sigma\rho}\|f\|_{p,\Phi_{\delta}}, \end{aligned}$$

  (74)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\sigma})}{ e^{\rho(U^{\delta}(x)V_{n})^{\sigma}}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}}>\frac{\ln2}{\sigma\rho}\|a \|_{q,\Psi}; \end{aligned}$$

  (75)

• (iii) for $0< p<1$, $0<\|f\|_{p,\Phi_{\delta}}$, and $\|a\|_{q,\Psi}<\infty$, we have the following equivalent inequalities with the best possible constant factor $\frac{\ln2}{\sigma\rho}$:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\sigma})}{e^{\rho(U^{\delta}(x)V_{n})^{\sigma }}}a_{n}f(x) \,dx>\frac{\ln2}{\sigma\rho}\|f\|_{p,\widetilde{\Phi}_{\delta }}\|a\|_{q,\Psi }, \end{aligned}$$

  (76)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\sigma})}{e^{\rho(U^{\delta}(x)V_{n})^{\sigma}}}f(x)\,dx \biggr] ^{p}>\frac{\ln 2}{\sigma\rho}\|f\|_{p,\widetilde{\Phi}_{\delta}}, \end{aligned}$$

  (77)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{(1-\theta_{\delta}(\sigma ,x))^{1-q}\mu (x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\sigma})}{e^{\rho(U^{\delta}(x)V_{n})^{\sigma }}}a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\& \quad > \frac{\ln2}{\sigma\rho}\|a\|_{q,\Psi}. \end{aligned}$$

  (78)

For $\alpha=0$ and $\gamma=\sigma$ in Theorems 2-4, we have the following corollary.

Corollary 4

If $\rho>0$, $0<\sigma\leq1$, there exists $n_{0}\in\mathbf{N}$ such that $\upsilon_{n}\geq\upsilon_{n+1}$ ($n\in\{n_{0},n_{0}+1,\ldots\}$), and $U(\infty)=V(\infty)=\infty$, then

• (i) for $p>1$, $0<\|f\|_{p,\Phi_{\delta}}$, and $\|a\|_{q,\Psi}<\infty$, we have the following equivalent inequalities with the best possible constant factor $\frac{\pi}{2\sigma\rho}$:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\sec h\bigl(\rho\bigl(U^{\delta }(x)V_{n} \bigr)^{\sigma}\bigr)a_{n}f(x)\,dx< \frac{\pi}{2\sigma\rho}\|f \|_{p,\Phi _{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

  (79)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\sec h\bigl(\rho\bigl(U^{\delta}(x)V_{n} \bigr)^{\sigma }\bigr)f(x)\,dx \biggr] ^{p}< \frac{\pi}{2\sigma\rho}\|f \|_{p,\Phi_{\delta}}, \end{aligned}$$

  (80)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\sec h\bigl(\rho \bigl(U^{\delta}(x)V_{n}\bigr)^{\sigma }\bigr)a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}}< \frac{\pi}{2\sigma\rho}\|a \|_{q,\Psi}; \end{aligned}$$

  (81)

• (ii) for $p<0$, $0<\|f\|_{p,\Phi_{\delta}}$, and $\|a\|_{q,\Psi }<\infty$, we have the following equivalent inequalities with the best possible constant factor $\frac{\pi}{2\sigma\rho}$:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\sec h\bigl(\rho\bigl(U^{\delta }(x)V_{n} \bigr)^{\sigma}\bigr)a_{n}f(x)\,dx>\frac{\pi}{2\sigma\rho}\|f \|_{p,\Phi _{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

  (82)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\sec h\bigl(\rho\bigl(U^{\delta}(x)V_{n} \bigr)^{\sigma }\bigr)f(x)\,dx \biggr] ^{p}>\frac{\pi}{2\sigma\rho}\|f \|_{p,\Phi_{\delta}}, \end{aligned}$$

  (83)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{\mu(x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\sec h\bigl(\rho \bigl(U^{\delta}(x)V_{n}\bigr)^{\sigma }\bigr)a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}}>\frac{\pi}{2\sigma\rho}\|a \|_{q,\Psi}; \end{aligned}$$

  (84)

• (iii) for $0< p<1$, $0<\|f\|_{p,\Phi_{\delta}}$, and $\|a\|_{q,\Psi}<\infty$, we have the following equivalent inequalities with the best possible constant factor $\frac{\pi}{2\sigma\rho}$:

  $$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\sec h\bigl(\rho\bigl(U^{\delta }(x)V_{n} \bigr)^{\sigma}\bigr)a_{n}f(x)\,dx>\frac{\pi}{2\sigma\rho}\|f \|_{p,\widetilde{\Phi}_{\delta}}\|a\|_{q,\Psi}, \end{aligned}$$

  (85)
  $$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\sec h\bigl(\rho\bigl(U^{\delta}(x)V_{n} \bigr)^{\sigma }\bigr)f(x)\,dx \biggr] ^{p}>\frac{\pi}{2\sigma\rho}\|f \|_{p,\widetilde{\Phi}_{\delta}}, \end{aligned}$$

  (86)
  $$\begin{aligned}& \Biggl\{ \int_{0}^{\infty}\frac{(1-\theta_{\delta}(\sigma ,x))^{1-q}\mu (x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum _{n=1}^{\infty}\sec h\bigl(\rho \bigl(U^{\delta}(x)V_{n}\bigr)^{\sigma}\bigr)a_{n} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\& \quad > \frac{\pi}{2\sigma\rho}\|a\|_{q,\Psi}. \end{aligned}$$

  (87)

Remark 2

For $\mu(x)=\nu_{n}=1$ in (52), we have the following inequality with the best possible constant factor $k(\sigma)$:

$$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(x^{\delta }n)^{\gamma})}{e^{\alpha(x^{\delta}n)^{\gamma}}}a_{n}f(x) \,dx \\& \quad < k(\sigma) \biggl[ \int_{0}^{\infty}x^{p(1-\delta\sigma )-1}f^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\sigma )-1}a_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$

(88)

In particular, for $\delta=1$, we have the following inequality with inhomogeneous kernel:

$$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(xn)^{\gamma })}{e^{\alpha(xn)^{\gamma}}}a_{n}f(x) \,dx \\& \quad < k(\sigma) \biggl[ \int_{0}^{\infty}x^{p(1-\sigma)-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\sigma)-1}a_{n}^{q} \Biggr] ^{\frac{1}{q}}; \end{aligned}$$

(89)

for $\delta=-1$, we have the following inequality with inhomogeneous kernel:

$$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(\frac{n}{x})^{\gamma})}{e^{\alpha(\frac{n}{x})^{\gamma}}}a_{n}f(x) \,dx \\& \quad < k(\sigma) \biggl[ \int_{0}^{\infty}x^{p(1+\sigma)-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\sigma)-1}a_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$

(90)

We still can obtain a large number of other inequalities by using some particular parameters in theorems and corollaries.