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]: 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]: Inequalities (1) and (2) are important in analysis and its applications (see [1‐5]).
$$ \int_{0}^{\infty} \int_{0}^{\infty}\frac{f(x)g(y)}{x+y}\,dx\,dy< \frac{\pi }{\sin(\pi/p)}\|f\|_{p}\|g\|_{q}, $$
(1)
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\pi/p)}\|a \|_{p}\|b\|_{q}. $$
(2)
Suppose that \(\mu_{i},\upsilon_{j}>0\) (\(i,j\in\mathbf{N}=\{1,2,\ldots \}\)), Then we have the following inequality ([1], Theorem 321): 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): 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.
$$ U_{m}:=\sum_{i=1}^{m} \mu_{i},\qquad V_{n}:=\sum_{j=1}^{n} \nu_{j}\quad (m,n\in \mathbf{N}). $$
(3)
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{\mu_{m}^{1/q}\nu _{n}^{1/p}a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\pi/p)}\|a\|_{p}\|b\|_{q}. $$
(4)
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac {a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\frac{\pi}{p})} \Biggl( \sum_{m=1}^{\infty}\frac {a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty} \frac {b_{n}^{q}}{\nu_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}. $$
(5)
Anzeige
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 \(k(\lambda_{1})=\int_{0}^{\infty}k_{\lambda }(t,1)t^{\lambda_{1}-1}\, dt\in\mathbf{R}_{+}\), \(\phi(x)=x^{p(1-\lambda _{1})-1}\), \(\psi(x)=x^{q(1-\lambda_{2})-1}\), \(f(x),g(y)\geq0\),
\(g\in L_{q,\psi}(\mathbf{R}_{+})\), \(\|f\|_{p,\phi},\|g\|_{q,\psi}>0\), we have 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 \(a_{m},b_{n}\geq0\),
\(b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\psi}\), \(\|a\|_{p,\phi },\|b\|_{q,\psi }>0\), we have where the constant factor \(k(\lambda_{1})\) is still the best possible.
$$ f\in L_{p,\phi}(\mathbf{R}_{+})= \biggl\{ f;\|f \|_{p,\phi }:=\biggl\{ \int_{0}^{\infty}\phi(x)\bigl\vert f(x)\bigr\vert ^{p}\,dx\biggr\} ^{\frac{1}{p}}< \infty \biggr\} , $$
$$ \int_{0}^{\infty} \int_{0}^{\infty}k_{\lambda }(x,y)f(x)g(y)\,dx \,dy< k(\lambda _{1})\|f\|_{p,\phi}\|g\|_{q,\psi}, $$
(6)
$$ a\in l_{p,\phi}= \Biggl\{ a;\|a\|_{p,\phi}:=\Biggl(\sum _{n=1}^{\infty}\phi (n)|a_{n}|^{p} \Biggr)^{\frac{1}{p}}< \infty \Biggr\} , $$
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}k_{\lambda }(m,n)a_{m}b_{n}< k( \lambda_{1})\|a\|_{p,\phi}\|b\|_{q,\psi}, $$
(7)
For \(0<\lambda_{1},\lambda_{2}\leq1\) such that \(\lambda_{1}+\lambda _{2}=\lambda\), we set Then by (7) we have where the constant \(B(\lambda_{1},\lambda_{2})\) is the best possible, and is the beta function. Clearly, for \(\lambda=1\), \(\lambda_{1}=\frac{1}{q}\), \(\lambda_{2}=\frac{1}{p}\), inequality (8) reduces to (2).
$$ k_{\lambda}(x,y)=\frac{1}{(x+y)^{\lambda}}\quad \bigl((x,y)\in \mathbf{R}_{+}^{2}\bigr). $$
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac {a_{m}b_{n}}{(m+n)^{\lambda}}< B( \lambda_{1},\lambda_{2})\|a\|_{p,\phi}\|b \|_{q,\psi}, $$
(8)
$$ B ( u,v ) = \int_{0}^{\infty}\frac{1}{(1+t)^{u+v}}t^{u-1} \,dt \quad (u,v>0) $$
Anzeige
In 2015, by adding some conditions, Yang [7] gave an extension of (8) and (5) as follows: where the constant \(B(\lambda_{1},\lambda_{2})\) is still the best possible.
$$\begin{aligned}& \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{(U_{m}+V_{n})^{\lambda}} \\& \quad < B(\lambda_{1},\lambda_{2}) \Biggl( \sum _{m=1}^{\infty}\frac{U_{m}^{p(1-\lambda_{1})-1}a_{m}^{p}}{\mu_{m}^{p-1}} \Biggr) ^{\frac {1}{p}} \Biggl( \sum_{n=1}^{\infty}\frac{V_{n}^{q(1-\lambda _{2})-1}b_{n}^{q}}{\nu _{n}^{q-1}} \Biggr) ^{\frac{1}{q}}, \end{aligned}$$
(9)
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 \(B ( \lambda_{1},\lambda_{2} ) \): 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: 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.
$$ \int_{0}^{\infty}f ( x ) \Biggl[ \sum _{n=1}^{\infty}\frac {a_{n}}{ ( x+n ) ^{\lambda}} \Biggr] \,dx< B ( \lambda_{1},\lambda _{2} ) \|f\|_{p,\phi}\|a \|_{q,\psi}, $$
(10)
$$ \int_{0}^{\infty}f(x)\sum_{n=1}^{\infty}k_{\lambda }(x,n)a_{n} \,dx< k(\lambda _{1})\|f\|_{p,\phi}\|a\|_{q,\psi}, $$
(11)
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 \(\mathbf{R}_{+}=(0,\infty)\),
\(\nu(t):=\nu_{n}\), \(t\in(n-1,n]\) (\(n\in\mathbf{N}\)), and
\(p\neq0,1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\delta\in\{-1,1\}\), \(f(x),a_{n}\geq0\) (\(x\in\mathbf{R}_{+}\), \(n\in\mathbf{N}\)), \(\|f\|_{p,\Phi _{\delta}}=(\int_{0}^{\infty}\Phi_{\delta}(x)f^{p}(x)\,dx)^{\frac{1}{p}}\), \(\|a\|_{q,\Psi}=(\sum_{n=1}^{\infty}\Psi(n)b_{n}^{q})^{\frac{1}{q}}\), where
$$ U(x):= \int_{0}^{x}\mu(t)\,dt< \infty\quad \bigl(x\in [0, \infty )\bigr), $$
$$ V(y):= \int_{0}^{y}\nu(t)\,dt\quad \bigl(y\in[0, \infty )\bigr), $$
$$ \Phi_{\delta}(x):=\frac{U^{p(1-\delta\sigma)-1}(x)}{\mu ^{p-1}(x)},\qquad \Psi (n):=\frac{V_{n}^{q(1-\sigma)-1}}{\nu_{n}^{q-1}} \quad (x\in\mathbf {R}_{+},n\in \mathbf{N}). $$
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 By the Lebesgue term-by-term theorem (see [39]), setting \(v=(2k+\frac{\alpha}{\rho}+1)u\), we have where \(\xi(s,a):=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(k+a)^{s}}\) (\(s,a>0\)) and is the gamma function (see [40]).
$$\begin{aligned} k(\sigma) : =& \int_{0}^{\infty}\frac{\sec h(\rho t^{\gamma })}{e^{\alpha t^{\gamma}}}t^{\sigma-1}\,dt \\ =& \frac{1}{\gamma\rho^{\sigma/\gamma}} \int_{0}^{\infty}\frac{\sec h(u)}{e^{\frac{\alpha}{\rho}u}}u^{\frac{\sigma}{\gamma}-1}\,du \\ =& \frac {2}{\gamma \rho^{\sigma/\gamma}} \int_{0}^{\infty}\frac{e^{-\frac{\alpha}{\rho} u}u^{\frac{\sigma}{\gamma}-1}}{e^{u}+e^{-u}}\,du \\ =& \frac{2}{\gamma\rho^{\sigma/\gamma}} \int_{0}^{\infty}\frac{e^{-( \frac{\alpha}{\rho}+1)u}u^{\frac{\sigma}{\gamma}-1}}{1+e^{-2u}}\,du \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}} \int_{0}^{\infty }\sum_{k=0}^{\infty}(-1)^{k}e^{-(2k+\frac{\alpha}{\rho}+1)u}u^{\frac{ \sigma}{\gamma}-1} \,du \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}} \int_{0}^{\infty }\sum_{k=0}^{\infty} \bigl[e^{-(4k+\frac{\alpha}{\rho}+1)u}-e^{-(4k+2+\frac {\alpha}{\rho}+1)u}\bigr]u^{\frac{\sigma}{\gamma}-1}\,du. \end{aligned}$$
$$\begin{aligned} k(\sigma) =& \int_{0}^{\infty}\frac{\sec h(\rho t^{\gamma })}{e^{\alpha t^{\gamma}}}t^{\sigma-1}\,dt \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}}\sum_{k=0}^{\infty } \int_{0}^{\infty}\bigl[e^{-(4k+\frac{\alpha}{\rho}+1)u}-e^{-(4k+2+\frac{\alpha}{\rho}+1)u} \bigr]u^{\frac{\sigma}{\gamma}-1}\,du \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}}\sum_{k=0}^{\infty }(-1)^{k} \int_{0}^{\infty}e^{-(2k+\frac{\alpha}{\rho}+1)u}u^{\frac {\sigma }{\gamma}-1}\,du \\ =&\frac{2}{\gamma\rho^{\sigma/\gamma}}\sum_{k=0}^{\infty} \frac {(-1)^{k}}{(2k+\frac{\alpha}{\rho}+1)^{\sigma/\gamma}} \int_{0}^{\infty }e^{-v}v^{\frac{\sigma}{\gamma}-1}\,dv \\ =&\frac{2\Gamma(\frac{\sigma}{\gamma})}{\gamma(2\rho)^{\sigma /\gamma}}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k+\frac{\alpha+\rho}{2\rho})^{\sigma/\gamma}} \\ =&\frac{2\Gamma(\frac{\sigma}{\gamma})}{\gamma(2\rho)^{\sigma /\gamma}}\xi\biggl(\frac{\sigma}{\gamma},\frac{\alpha+\rho}{2\rho}\biggr)\in \mathbf {R}_{+}, \end{aligned}$$
(12)
$$ \Gamma(y):= \int_{0}^{\infty}e^{-v}v^{y-1}\,dv \quad (y>0) $$
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 it follows that In the same way, we still have Hence, adding these two inequalities, we have (13). □
$$\begin{aligned}& \int_{n}^{n+1}g(t)\,dt \leq g(n)\leq \int_{n-1}^{n}g(t)\,dt\quad (n=1, \ldots,n_{0}), \\& \int_{n_{0}+1}^{n_{0}+2}g(t)\,dt < g(n_{0}+1)< \int_{n_{0}}^{n_{0}+1}g(t)\,dt, \end{aligned}$$
$$ 0< \int_{1}^{n_{0}+2}g(t)\,dt< \sum _{n=1}^{n_{0}+1}g(n)< \sum_{n=1}^{n_{0}+1} \int_{n-1}^{n}g(t)\,dt= \int_{0}^{n_{0}+1}g(t)\,dt< \infty. $$
$$ 0< \int_{n_{0}+2}^{\infty}g(t)\,dt\leq\sum _{n=n_{0}+2}^{\infty}g(n)\leq \int_{n_{0}+1}^{\infty}g(t)\,dt< \infty. $$
Lemma 2
For
\(\gamma,\rho>0\), \(\alpha>-\rho\) (\(\alpha\geq 0\)), and
\(0<\sigma\leq1\), define the following weight coefficients:
Then, we have the following inequalities:
where
\(k(\sigma)\)
is defined in (12).
$$\begin{aligned}& \omega_{\delta}(\sigma,x) : =\sum_{n=1}^{\infty} \frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{U^{\delta\sigma}(x)\nu_{n}}{V_{n}^{1-\sigma}},\quad x\in\mathbf{R}_{+}, \end{aligned}$$
(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)
$$\begin{aligned}& \omega_{\delta}(\sigma,x) < k(\sigma)\quad (x\in\mathbf{R}_{+}), \end{aligned}$$
(16)
$$\begin{aligned}& \varpi_{\delta}(\sigma,n) \leq k(\sigma)\quad (n\in\mathbf{N}), \end{aligned}$$
(17)
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 Setting \(u=U^{\delta}(x)V(t)\), by (12) we find Hence, (16) follows.
$$\begin{aligned}& \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})\nu_{n}}{e^{\alpha (U^{\delta}(x)V_{n})^{\gamma}}V_{n}^{1-\sigma}} \\& \quad =\frac{\sec h(\rho (U^{\delta}(x)V(n))^{\gamma})\nu_{n}}{e^{\alpha(U^{\delta }(x)V(n))^{\gamma}}V^{1-\sigma}(n)} \\& \quad < \int_{n-1}^{n}\frac{\sec h(\rho(U^{\delta}(x)V(t))^{\gamma})}{e^{\alpha(U^{\delta}(x)V(t))^{\gamma}}}\frac{V^{\prime }(t)}{V^{1-\sigma }(t)} \,dt\quad (n\in\mathbf{N}), \\& \omega_{\delta}(\sigma,x) < \sum_{n=1}^{\infty} \int_{n-1}^{n}\frac {\sec h(\rho(U^{\delta}(x)V(t))^{\gamma})}{e^{\alpha(U^{\delta }(x)V(t))^{\gamma}}}\frac{U^{\delta\sigma}(x)V^{\prime }(t)}{V^{1-\sigma }(t)} \,dt \\& \hphantom{\omega_{\delta}(\sigma,x)} = \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V(t))^{\gamma})}{ e^{\alpha(U^{\delta}(x)V(t))^{\gamma}}}\frac{U^{\delta\sigma }(x)V^{\prime}(t)}{V^{1-\sigma}(t)} \,dt. \end{aligned}$$
$$\begin{aligned} \omega_{\delta}(\sigma,x) < & \int_{0}^{U^{\delta}(x)V(\infty)}\frac {\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}\frac{U^{\delta\sigma }(x)U^{-\delta}(x)}{(uU^{-\delta}(x))^{1-\sigma}} \,du \\ \leq& \int_{0}^{\infty}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma }}}u^{\sigma-1} \,du \\ =&k(\sigma). \end{aligned}$$
Setting \(u=V_{n}U^{\delta}(x)\) in (15), we find \(du=\delta V_{n}U^{\delta-1}(x)\mu(x)\,dx\) and If \(\delta=1\), then if \(\delta=-1\), then Then by (12) we have (17). □
$$\begin{aligned} \varpi_{\delta}(\sigma,n) =&\frac{1}{\delta} \int_{V_{n}U^{\delta }(0)}^{V_{n}U^{\delta}(\infty)}\frac{\sec h(\rho u^{\gamma })}{e^{\alpha u^{\gamma}}}\frac{V_{n}^{\sigma}V_{n}^{-1}(V_{n}^{-1}u)^{\frac {1}{\delta}-1}}{(V_{n}^{-1}u)^{\frac{1}{\delta}-\sigma}} \,du \\ =&\frac{1}{\delta} \int_{V_{n}U^{\delta}(0)}^{V_{n}U^{\delta}(\infty )}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1} \,du. \end{aligned}$$
$$ \varpi_{1}(\sigma,n)= \int_{0}^{V_{n}U(\infty)}\frac{\sec h(\rho u^{\gamma })}{e^{\alpha u^{\gamma}}}u^{\sigma-1}\,du \leq \int_{0}^{\infty}\frac {\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1}\,du; $$
$$ \varpi_{-1}(\sigma,n)=- \int_{\infty}^{V_{n}U^{-1}(\infty)}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1}\,du \leq \int_{0}^{\infty}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma }}} u^{\sigma-1} \,du. $$
Remark 1
We do not need \(\sigma\leq1\) to obtain (17). If \(U(\infty)=\infty\), then we have For example, set \(\mu(t)=\frac{1}{(1+t)^{\beta}}\) (\(t>0\); \(0\leq\beta \leq1\)). Then, for \(x\geq0\), we find and \(U(\infty)=\int_{0}^{\infty}\frac{dt}{(1+t)^{\beta}}=\infty\).
$$ \varpi_{\delta}(\sigma,n)=k(\sigma)\quad (n\in\mathbf{N}). $$
(18)
$$ U(x)= \int_{0}^{x}\frac{dt}{(1+t)^{\beta}}=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{(1+x)^{1-\beta}-1}{1-\beta},& 0\leq\beta< 1, \\ \ln(1+x), &\beta=1 \end{array}\displaystyle \right . < \infty, $$
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
where
\(\theta_{\delta}(\sigma,x)=O((U(x))^{\delta\sigma})\in(0,1)\);
$$ k(\sigma) \bigl(1-\theta_{\delta}(\sigma,x)\bigr)< \omega_{\delta}( \sigma,x), $$
(19)
(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 Setting \(u=U^{\delta}(x)V(t)\), in view of \(V(\infty)=\infty\), by (12) we find
$$\begin{aligned} \omega_{\delta}(\sigma,x) =&\sum_{n=1}^{\infty} \frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{U^{\delta\sigma}(x)\nu_{n}}{V_{n}^{1-\sigma}} \\ \geq&\sum_{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{\sec h(\rho (U^{\delta }(x)V(n))^{\gamma})}{e^{\alpha(U^{\delta}(x)V(n))^{\gamma}}}\frac{U^{\delta\sigma}(x)\nu_{n+1}\,dt}{(V(n))^{1-\sigma}} \\ >&\sum_{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{\sec h(\rho(U^{\delta }(x)V(t))^{\gamma})}{e^{\alpha(U^{\delta}(x)V(t))^{\gamma}}}\frac{U^{\delta\sigma}(x)V^{\prime}(t)}{(V(t))^{1-\sigma}}\,dt \\ =& \int_{n_{0}}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V(t))^{\gamma })}{e^{\alpha(U^{\delta}(x)V(t))^{\gamma}}}\frac{U^{\delta\sigma }(x)V^{\prime}(t)}{(V(t))^{1-\sigma}} \,dt. \end{aligned}$$
$$\begin{aligned}& \omega_{\delta}(\sigma,x) > \int_{U^{\delta}(x)V_{n_{0}}}^{\infty} \frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1}\,du \\& \hphantom{\omega_{\delta}(\sigma,x)}= k(\sigma)- \int_{0}^{U^{\delta}(x)V_{n_{0}}}\frac{\sec h(\rho u^{\gamma })}{e^{\alpha u^{\gamma}}}u^{\sigma-1} \,du=k(\sigma) \bigl(1-\theta_{\delta }(\sigma,x)\bigr), \\& \theta_{\delta}(\sigma,x) : =\frac{1}{k(\sigma)} \int_{0}^{U^{\delta }(x)V_{n_{0}}}\frac{\sec h(\rho u^{\gamma})}{e^{\alpha u^{\gamma}}}u^{\sigma-1} \,du\in(0,1). \end{aligned}$$
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 and then (19) follows.
$$ 0< \theta_{\delta}(\sigma,x)\leq\frac{L}{k(\sigma)} \int _{0}^{U^{\delta }(x)V_{n_{0}}}u^{\sigma-1}\,du= \frac{L(U^{\delta}(x)V_{n_{0}})^{\sigma }}{k(\sigma)\sigma}, $$
For \(b>0\), we find Hence, we have (20). □
$$\begin{aligned}& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}=\sum _{n=1}^{n_{0}}\frac {\nu _{n}}{V_{n}^{1+b}}+\sum _{n=n_{0}+1}^{\infty}\frac{\nu_{n}}{V^{1+b}(n)} \\& \hphantom{\sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}} < \sum_{n=1}^{n_{0}} \frac{\nu_{n}}{V_{n}^{1+b}}+\sum_{n=n_{0}+1}^{\infty } \int_{n-1}^{n}\frac{V^{\prime}(x)}{V^{1+b}(x)}\,dx \\& \hphantom{\sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}} = \sum_{n=1}^{n_{0}} \frac{\nu_{n}}{V_{n}^{1+b}}+ \int_{n_{0}}^{\infty} \frac{dV(x)}{V^{1+b}(x)}=\sum _{n=1}^{n_{0}}\frac{\nu_{n}}{V_{n}^{1+b}}+ \frac{1}{bV^{b}(n_{0})} \\& \hphantom{\sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}} = \frac{1}{b} \Biggl( \frac{1}{V_{n_{0}}^{b}}+b\sum _{n=1}^{n_{0}}\frac {\nu _{n}}{V_{n}^{1+b}} \Biggr) , \\& \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}} \geq \sum _{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{\nu_{n+1}}{V^{1+b}(n)}\,dx>\sum _{n=n_{0}}^{\infty} \int_{n}^{n+1}\frac{V^{\prime}(x)\,dx}{V^{1+b}(x)} \\& \hphantom{\sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+b}}} = \int_{n_{0}}^{\infty}\frac{dV(x)}{V^{1+b}(x)}=\frac {1}{bV^{b}(n_{0})}= \frac{1}{bV_{n_{0}}^{b}}. \end{aligned}$$
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 Then by (16) we have (22).
$$\begin{aligned} J_{1} \leq&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \Biggl[ \sum _{n=1}^{\infty } \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \Biggl[ \int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{U^{(1-\delta\sigma)(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \biggl[ \int_{0}^{\infty}\omega_{\delta }(\sigma,x) \frac{U^{p(1-\delta\sigma)-1}(x)}{\mu^{p-1}(x)}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
(25)
By Hölder’s inequality (see [41]) we have Then by (22) we have (21). On the other hand, assuming that (21) is valid, we set 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 and then (22) follows, which is equivalent to (21).
$$\begin{aligned} I =&\sum_{n=1}^{\infty} \biggl[ \frac{\nu_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\sigma}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] \biggl( \frac{V_{n}^{\frac{1}{p}-\sigma}a_{n}}{\nu_{n}^{\frac {1}{p}}} \biggr) \\ \leq&J_{1}\|a\|_{q,\Psi}. \end{aligned}$$
(26)
$$ a_{n}:=\frac{\nu_{n}}{V_{n}^{1-p\sigma}} \biggl[ \int_{0}^{\infty}\frac {\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p-1},\quad n\in\mathbf{N}. $$
$$ \|a\|_{q,\Psi}^{q}=J_{1}^{p}=I< k(\sigma) \|f\|_{p,\Phi_{\delta }}\|a\|_{q,\Psi},\qquad \|a\|_{q,\Psi}^{q-1}=J_{1}< k( \sigma)\|f\|_{p,\Phi _{\delta}}, $$
Again by Hölder’s inequality with weight we have Then by (16) and the Lebesgue term-by-term integration theorem it follows that Then by (17) we have (23).
$$\begin{aligned}& \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q} \\& \quad = \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \biggl( \frac{U^{\frac{1-\delta\sigma}{q}}(x)\nu_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1-\sigma}{p}}} \biggr) \biggl( \frac{V_{n}^{\frac{1-\sigma }{p}}a_{n}}{U^{\frac{1-\delta\sigma}{q}}(x)\nu_{n}^{\frac{1}{p}}} \biggr) \Biggr] ^{q} \\& \quad \leq \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{ U^{(1-\delta\sigma)(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}} \Biggr] ^{q-1} \\& \qquad {}\times\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{V_{n}^{\frac {q(1-\sigma )}{p}}}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q} \\& \quad = \frac{(\omega_{\delta}(\sigma,x))^{q-1}}{U^{q\delta\sigma -1}(x)\mu (x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{V_{n}^{(1-\sigma )(q-1)}\mu (x)}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}. \end{aligned}$$
(27)
$$\begin{aligned} J_{2} < &\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \int_{0}^{\infty}\sum_{n=1}^{\infty } \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha (U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta \sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \sum _{n=1}^{\infty} \int _{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}} \frac{V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta \sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \sum _{n=1}^{\infty}\varpi_{\delta }(\sigma,n) \frac{V_{n}^{q(1-\sigma)-1}}{\nu_{n}^{q-1}}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(28)
By Hölder’s inequality we have Then by (23) we have (21). On the other hand, assuming that (23) is valid, we set 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 and then (23) follows, which is equivalent to (21).
$$\begin{aligned} I =& \int_{0}^{\infty} \biggl( \frac{U^{\frac{1}{q}-\delta\sigma }(x)}{\mu^{\frac{1}{q}}(x)}f(x) \biggr) \Biggl[ \frac{\mu^{\frac{1}{q}}(x)}{U^{\frac {1}{q}-\delta\sigma}(x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] \,dx \\ \leq&\|f\|_{p,\Phi_{\delta}}J_{2}. \end{aligned}$$
(29)
$$ f(x):=\frac{\mu(x)}{U^{1-q\delta\sigma}(x)} \Biggl[ \sum_{n=1}^{\infty } \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q-1},\quad x\in \mathbf{R}_{+}. $$
$$ \|f\|_{p,\Phi_{\delta}}^{p}=J_{2}^{q}=I< k(\sigma) \|f\|_{p,\Phi _{\delta }}\|a\|_{q,\Psi},\qquad \|f\|_{p,\Phi_{\delta}}^{p-1}=J_{2}< k( \sigma )\|a\|_{q,\Psi}, $$
Theorem 2
Proof
For \(\varepsilon\in(0,q\sigma)\), we set \(\tilde{\sigma }=\sigma-\frac{\varepsilon}{q}\) and \(\tilde{f}=\tilde{f}(x)\), \(x\in \mathbf{R}_{+}\), \(\tilde{a}=\{\tilde{a}_{n}\}_{n=1}^{\infty}\),
Then for \(\delta=\pm1\), since \(U(\infty)=\infty\), we find By (20), (32), and (19) we obtain
$$\begin{aligned}& \tilde{f}(x) =\left \{ \textstyle\begin{array}{l@{\quad}l} U^{\delta(\tilde{\sigma}+\varepsilon)-1}(x)\mu(x), &0< x^{\delta }\leq1, \\ 0, &x^{\delta}>0,\end{array}\displaystyle \right . \end{aligned}$$
(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)
$$ \int_{\{x>0;0< x^{\delta}\leq1\}}\frac{\mu(x)}{U^{1-\delta \varepsilon}(x)}\,dx=\frac{1}{\varepsilon}U^{\delta\varepsilon}(1). $$
(32)
$$\begin{aligned}& \|\tilde{f}\|_{p,\Phi_{\delta}}\|\tilde{a}\|_{q,\Psi} = \biggl( \int_{\{x>0;0< x^{\delta}\leq1\}}\frac{\mu(x)\,dx}{U^{1-\delta \varepsilon }(x)} \biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|\tilde{f}\|_{p,\Phi_{\delta}}\|\tilde{a}\|_{q,\Psi}} = \frac{1}{\varepsilon}U^{\frac{\delta\varepsilon}{p}}(1) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}, \\& \tilde{I} : = \int_{0}^{\infty}\sum_{n=1}^{\infty} \frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\tilde{a}_{n}\tilde{f}(x)\,dx \\& \hphantom{\tilde{I}} = \int_{\{x>0;0< x^{\delta}\leq1\}}\sum_{n=1}^{\infty} \frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{\tilde{\sigma}-1}\nu_{n}\mu(x)}{U^{1-\delta (\tilde{\sigma}+\varepsilon)}(x)}\,dx \\& \hphantom{\tilde{I}} = \int_{\{x>0;0< x^{\delta}\leq1\}}\omega_{\delta}(\tilde {\sigma},x)\frac{\mu(x)}{U^{1-\delta\varepsilon}(x)}\,dx \\& \hphantom{\tilde{I}} \geq k(\tilde{\sigma}) \int_{\{x>0;0< x^{\delta}\leq1\}}\bigl(1-\theta _{\delta}(\tilde{\sigma},x) \bigr)\frac{\mu(x)}{U^{1-\delta \varepsilon}(x)}\,dx \\& \hphantom{\tilde{I}} = k(\tilde{\sigma}) \int_{\{x>0;0< x^{\delta}\leq 1\}}\bigl(1-O\bigl(\bigl(U(x)\bigr)^{\delta(\sigma-\frac{\varepsilon}{q})}\bigr) \bigr)\frac{\mu (x)}{U^{1-\delta\varepsilon}(x)}\,dx \\& \hphantom{\tilde{I}} = k(\tilde{\sigma})\biggl[ \int_{\{x>0;0< x^{\delta}\leq1\}}\frac {\mu (x)}{U^{1-\delta\varepsilon}(x)}\,dx - \int_{\{x>0;0< x^{\delta}\leq1\}}O\biggl(\frac{\mu(x)}{U^{1-\delta (\sigma+\frac{\varepsilon}{p})}(x)}\biggr)\,dx\biggr] \\& \hphantom{\tilde{I}} = \frac{1}{\varepsilon}k\biggl(\sigma-\frac{\varepsilon}{q}\biggr) \bigl(U^{\delta \varepsilon}(1)-\varepsilon O(1)\bigr). \end{aligned}$$
(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, It follows that \(k(\sigma)\leq K\) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k(\sigma)\) is the best possible constant factor of (21).
$$ k\biggl(\sigma-\frac{\varepsilon}{q}\biggr) \bigl(U^{\delta\varepsilon}(1)-\varepsilon O(1)\bigr)< K\cdot U^{\frac{\delta\varepsilon}{p}}(1) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}. $$
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 we can rewrite (22) as \(\|c\|_{p,\Psi^{1-p}}< k(\sigma )\|f\|_{p,\Phi _{\delta}}<\infty\), namely, \(c\in l_{p,\Psi^{1-p}}\).
$$ c=\{c_{n}\}_{n=1}^{\infty},\qquad c_{n}:= \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}f(x)\,dx, \quad n\in\mathbf{N}, $$
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: 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
$$ \|T_{1}\|:=\sup_{f(\neq\theta)\in L_{p,\Phi_{\delta}}(\mathbf {R}_{+})}\frac{\|T_{1}f\|_{p,\Psi^{1-p}}}{\|f\|_{p,\Phi_{\delta}}}. $$
$$ \|T_{1}\|=k(\sigma)=\frac{2\Gamma(\frac{\sigma}{\gamma})}{\gamma (2\rho )^{\sigma/\gamma}}\xi\biggl(\frac{\sigma}{\gamma}, \frac{\alpha+\rho }{2\rho}\biggr). $$
(37)
Assuming that \(a=\{a_{n}\}_{n=1}^{\infty}\in l_{q,\Psi}\) and setting we can rewrite (23) as \(\|h\|_{q,\Phi_{\delta}^{1-q}}< k(\sigma )\|a\|_{q,\Psi}<\infty\), namely, \(h\in L_{q,\Phi_{\delta }^{1-q}}(\mathbf{R}_{+})\).
$$ h(x):=\sum_{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n}, \quad x\in\mathbf{R}_{+}, $$
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: 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
$$ \|T_{2}\|:=\sup_{a(\neq\theta)\in l_{q,\Psi}}\frac{\|T_{2}a\|_{q,\Phi _{\delta}^{1-q}}}{\|a\|_{q,\Psi}}. $$
$$ \|T_{2}\|=k(\sigma)=\frac{2\Gamma(\frac{\sigma}{\gamma})}{\gamma (2\rho )^{\sigma/\gamma}}\xi\biggl(\frac{\sigma}{\gamma}, \frac{\alpha+\rho }{2\rho}\biggr)=\|T_{1}\|. $$
(41)
4 Some equivalent reverse inequalities
In the following, we also set For \(0< p<1\) or \(p<0\), we still use the formal symbols \(\|f\|_{p,\Phi_{\delta}}\), \(\|f\|_{p,\widetilde{\Phi}_{\delta}}\), and \(\|a\|_{q,\Psi}\).
$$ \widetilde{\Phi}_{\delta}(x):=\bigl(1-\theta_{\delta}(\sigma,x) \bigr)\frac{U^{p(1-\delta\sigma)-1}(x)}{\mu^{p-1}(x)} \quad (x\in\mathbf{R}_{+}). $$
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 Then by (18) and the Lebesgue term-by-term integration theorem it follows that Then by (16) we have (43).
$$\begin{aligned}& \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p} \\& \quad \leq \frac{(\varpi_{\delta}(\sigma,n))^{p-1}}{V_{n}^{p\sigma -1}\nu_{n}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx. \end{aligned}$$
$$\begin{aligned} J_{1} \geq&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \Biggl[ \sum _{n=1}^{\infty } \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \biggl[ \int_{0}^{\infty}\omega_{\delta }(\sigma,x) \frac{U^{p(1-\delta\sigma)-1}(x)}{\mu^{p-1}(x)}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
By the reverse Hölder inequality we have 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 and then (43) follows, which is equivalent to (42).
$$\begin{aligned} I =&\sum_{n=1}^{\infty} \biggl[ \frac{\nu_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\sigma}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] \biggl( \frac{V_{n}^{\frac{1}{p}-\sigma}a_{n}}{\nu_{n}^{\frac {1}{p}}} \biggr) \\ \geq&J_{1}\|a\|_{q,\Psi}. \end{aligned}$$
(45)
$$\begin{aligned}& \|a\|_{q,\Psi}^{q} = J_{1}^{p}=I>k( \sigma)\|f\|_{p,\Phi_{\delta }}\|a\|_{q,\Psi}, \\& \|a\|_{q,\Psi}^{q-1} = J_{1}>k(\sigma)\|f \|_{p,\Phi_{\delta}}, \end{aligned}$$
Still by the reverse Hölder inequality with weight, since \(0< q<1\), similarly as obtaining (27) and (28), we have Then by (16) and the Lebesgue term-by-term integration theorem it follows that Then by (18) we have (44).
$$\begin{aligned}& \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q} \\& \quad \geq \frac{(\omega_{\delta}(\sigma,x))^{q-1}}{U^{q\delta\sigma -1}(x)\mu(x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{ V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}. \end{aligned}$$
$$\begin{aligned} J_{2} >&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \int_{0}^{\infty }\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{V_{n}^{(1-\sigma )(q-1)}\mu (x)}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \sum _{n=1}^{\infty}\varpi_{\delta }(\sigma,n) \frac{V_{n}^{q(1-\sigma)-1}}{\nu_{n}^{q-1}}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
By the reverse Hölder inequality we have 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 and then (44) follows, which is equivalent to (42).
$$\begin{aligned} I =& \int_{0}^{\infty} \biggl( \frac{U^{\frac{1}{q}-\delta\sigma }(x)}{\mu^{\frac{1}{q}}(x)}f(x) \biggr) \Biggl[ \frac{\mu^{\frac{1}{q}}(x)}{U^{\frac {1}{q}-\delta\sigma}(x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] \,dx \\ \geq&\|f\|_{p,\Phi_{\delta}}J_{2}. \end{aligned}$$
(46)
$$ \|f\|_{p,\Phi_{\delta}}^{p}=J_{2}^{q}=I>k(\sigma) \|f\|_{p,\Phi _{\delta }}\|a\|_{q,\Psi},\qquad \|f\|_{p,\Phi_{\delta}}^{p-1}=J_{2}>k( \sigma )\|a\|_{q,\Psi}, $$
For \(\varepsilon\in(0,q\sigma)\), we set \(\tilde{\sigma}=\sigma- \frac{\varepsilon}{q}\) and \(\tilde{f}=\tilde{f}(x)\), \(x\in \mathbf{R}_{+}\), \(\tilde{ a}=\{\tilde{a}_{n}\}_{n=1}^{\infty}\), By (20), (32), and (16) we obtain
$$\begin{aligned}& \tilde{f}(x) = \left \{ \textstyle\begin{array}{l@{\quad}l} U^{\delta(\tilde{\sigma}+\varepsilon)-1}(x)\mu(x), &0< x^{\delta }\leq1, \\ 0, &x^{\delta}>0,\end{array}\displaystyle \right . \\& \tilde{a}_{n} = \widetilde{V}_{n}^{\tilde{\sigma}-1} \nu_{n}=\widetilde{V}_{n}^{\sigma-\frac{\varepsilon}{q}-1} \nu_{n},\quad n\in \mathbf{N}. \end{aligned}$$
$$\begin{aligned}& \|\tilde{f}\|_{p,\Phi_{\delta}}\|\tilde{a}\|_{q,\Psi}=\frac {1}{\varepsilon}U^{\frac{\delta\varepsilon}{p}}(1) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}, \\& \tilde{I} = \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\tilde{a}_{n}\tilde{f}(x)\,dx \\& \hphantom{\tilde{I}} = \int_{\{x>0;0< x^{\delta}\leq1\}}\omega_{\delta}(\tilde {\sigma},x)\frac{\mu(x)}{U^{1-\delta\varepsilon}(x)}\,dx \\& \hphantom{\tilde{I}} \leq k(\tilde{\sigma}) \int_{\{x>0;0< x^{\delta}\leq1\}}\frac {\mu(x)}{U^{1-\delta\varepsilon}(x)}\,dx=\frac{1}{\varepsilon}k\biggl( \sigma-\frac{ \varepsilon}{q}\biggr)U^{\delta\varepsilon}(1). \end{aligned}$$
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, It follows that \(k(\sigma)\geq K\) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k(\sigma)\) is the best possible constant factor of (42).
$$ k\biggl(\sigma-\frac{\varepsilon}{q}\biggr)U^{\delta\varepsilon}(1)>K\cdot U^{\frac{\delta\varepsilon}{p}}(1) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}. $$
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 In view of (18) and the Lebesgue term-by-term integration theorem, we find Then by (19) we have (48).
$$\begin{aligned}& \biggl[ \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] ^{p} \\& \quad \geq \frac{(\varpi_{\delta}(\sigma,n))^{p-1}}{V_{n}^{p\sigma -1}\nu_{n}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx. \end{aligned}$$
$$\begin{aligned} J_{1} \geq&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \Biggl[ \sum _{n=1}^{\infty } \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}} \frac{U^{(1-\delta\sigma )(p-1)}(x)\nu_{n}}{V_{n}^{1-\sigma}\mu^{p-1}(x)}f^{p}(x)\,dx \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{q}} \biggl[ \int_{0}^{\infty}\omega_{\delta }(\sigma,x) \frac{U^{p(1-\delta\sigma)-1}(x)}{\mu^{p-1}(x)}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
By the reverse Hölder inequality we have 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 and then (48) follows, which is equivalent to (47).
$$\begin{aligned} I =&\sum_{n=1}^{\infty} \biggl[ \frac{\nu_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\sigma}} \int_{0}^{\infty}\frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}f(x)\,dx \biggr] \biggl( \frac{V_{n}^{\frac{1}{p}-\sigma}a_{n}}{\nu_{n}^{\frac {1}{p}}} \biggr) \\ \geq&J_{1}\|a\|_{q,\Psi}. \end{aligned}$$
(50)
$$ \|a\|_{q,\Psi}^{q}=J_{1}^{p}=I>k(\sigma) \|f\|_{p,\widetilde{\Phi }_{\delta }}\|a\|_{q,\Psi},\qquad \|a\|_{q,\Psi}^{q-1}=J_{1}>k( \sigma )\|f\|_{p,\widetilde{\Phi}_{\delta}}, $$
Again by the reverse Hölder inequality with weight, since \(q<0\), we have Then by (19) and the Lebesgue term-by-term integration theorem it follows that Then by (18) we have (49).
$$\begin{aligned}& \Biggl[ \sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma })}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}a_{n} \Biggr] ^{q} \\& \quad \leq \frac{(\omega_{\delta}(\sigma,x))^{q-1}}{U^{q\delta\sigma -1}(x)\mu(x)}\sum_{n=1}^{\infty} \frac{\sec h(\rho(U^{\delta }(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta}(x)V_{n})^{\gamma}}}\frac{ V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta\sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}. \end{aligned}$$
$$\begin{aligned} J >&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{(1-\sigma)(q-1)}\mu(x)}{U^{1-\delta \sigma}(x)\nu_{n}^{q-1}}a_{n}^{q}\,dx \Biggr\} ^{\frac{1}{q}} \\ =&\bigl(k(\sigma)\bigr)^{\frac{1}{p}} \Biggl\{ \sum _{n=1}^{\infty}\varpi_{\delta }(\sigma,n) \frac{V_{n}^{q(1-\sigma)-1}}{\nu_{n}^{q-1}}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
By the reverse Hölder inequality we have 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 and then (49) follows, which is equivalent to (47).
$$\begin{aligned} I =& \int_{0}^{\infty} \biggl[ \bigl(1-\theta_{\delta}( \sigma,x)\bigr)^{\frac {1}{p}}\frac{U^{\frac{1}{q}-\delta\sigma}(x)}{\mu^{\frac {1}{q}}(x)}f(x) \biggr] \\ &{}\times \Biggl[ \frac{(1-\theta_{\delta}(\sigma,x))^{\frac{-1}{p}}\mu ^{\frac{1}{q}}(x)}{U^{\frac{1}{q}-\delta\sigma}(x)}\sum_{n=1}^{\infty } \frac{\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}a_{n} \Biggr] \,dx \\ \geq&\|f\|_{p,\widetilde{\Phi }_{\delta }}J. \end{aligned}$$
(51)
$$ \|f\|_{p,\widetilde{\Phi}_{\delta}}^{p}=J^{q}=I>k(\sigma)\|f \|_{p,\widetilde{\Phi}_{\delta}}\|a\|_{q,\Psi},\qquad \|f\|_{p,\widetilde{\Phi} _{\delta}}^{p-1}=J>k( \sigma)\|a\|_{q,\Psi}, $$
For \(\varepsilon\in(0,p\sigma)\), we set \(\tilde{\sigma}=\sigma+ \frac{\varepsilon}{p}\) and \(\tilde{f}=\tilde{f}(x)\), \(x\in \mathbf{R}_{+}\), \(\tilde{a}=\{\tilde{a}_{n}\}_{n=1}^{\infty}\), By (19), (20), and (32) we obtain
$$\begin{aligned}& \tilde{f}(x) = \left \{ \textstyle\begin{array}{l@{\quad}l} U^{\delta\tilde{\sigma}-1}(x)\mu(x),& 0< x^{\delta}\leq1, \\ 0,& x^{\delta}>0,\end{array}\displaystyle \right . \\& \tilde{a}_{n} = \widetilde{V}_{n}^{\tilde{\sigma}-\varepsilon -1} \nu_{n}=\widetilde{V}_{n}^{\sigma-\frac{\varepsilon}{q}-1}\nu _{n}, \quad n\in\mathbf{N}. \end{aligned}$$
$$\begin{aligned}& \|\tilde{f}\|_{p,\widetilde{\Phi}_{\delta}}\|\tilde {a}\|_{q,\Psi}= \biggl[ \int_{\{x>0;0< x^{\delta}\leq1\}}\bigl(1-O\bigl(\bigl(U(x)\bigr)^{\delta\sigma }\bigr) \bigr)\frac{\mu(x)\,dx}{U^{1-\delta\varepsilon}(x)} \biggr] ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty}\frac{\nu_{n}}{V_{n}^{1+\varepsilon }} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|\tilde{f}\|_{p,\widetilde{\Phi}_{\delta}}\|\tilde {a}\|_{q,\Psi}}=\frac{1}{\varepsilon} \bigl( U^{\delta \varepsilon }(1)-\varepsilon O(1) \bigr) ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}, \\& \tilde{I} = \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\tilde{a}_{n}\tilde{f}(x)\,dx \\& \hphantom{\tilde{I}} = \sum_{n=1}^{\infty} \biggl( \int_{\{x>0;0< x^{\delta}\leq1\}}\frac {\sec h(\rho(U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{\tilde{\sigma}}\mu (x)}{U^{1-\delta \tilde{\sigma}}(x)}\,dx \biggr) \frac{\nu_{n}}{V_{n}^{1+\varepsilon}} \\& \hphantom{\tilde{I}} \leq \sum_{n=1}^{\infty} \biggl( \int_{0}^{\infty}\frac{\sec h(\rho (U^{\delta}(x)V_{n})^{\gamma})}{e^{\alpha(U^{\delta }(x)V_{n})^{\gamma}}}\frac{V_{n}^{\tilde{\sigma}}\mu(x)}{U^{1-\delta\tilde{\sigma }}(x)}\,dx \biggr) \frac{\nu_{n}}{V_{n}^{1+\varepsilon}} \\& \hphantom{\tilde{I}} = \sum_{n=1}^{\infty} \varpi_{\delta}(\tilde{\sigma},n)\frac {\nu_{n}}{V_{n}^{1+\varepsilon}}=k(\tilde{\sigma})\sum _{n=1}^{\infty }\frac{\nu_{n}}{V_{n}^{1+\varepsilon}} \\& \hphantom{\tilde{I}} = \frac{1}{\varepsilon}k\biggl(\sigma+\frac{\varepsilon}{p}\biggr) \biggl( \frac {1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) . \end{aligned}$$
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, It follows that \(k(\sigma)\geq K\) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k(\sigma)\) is the best possible constant factor of (47).
$$ k\biggl(\sigma+\frac{\varepsilon}{p}\biggr) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon }}+ \varepsilon O(1) \biggr) >K \bigl( U^{\delta\varepsilon}(1)-\varepsilon O(1) \bigr) ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{q}}. $$
5 Some corollaries
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
\(k(\sigma)\).
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)
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: 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)
$$\begin{aligned}& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{\sec h(\rho(\frac{n}{x})^{\gamma})}{e^{\alpha(\frac{n}{x})^{\gamma}}}a_{n}f(x) \,dx \\& \quad < k(\sigma) \biggl[ \int_{0}^{\infty}x^{p(1+\sigma)-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty}n^{q(1-\sigma)-1}a_{n}^{q} \Biggr] ^{\frac{1}{q}}. \end{aligned}$$
(90)
We still can obtain a large number of other inequalities by using some particular parameters in theorems and corollaries.
Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 61370186) and 2013 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2013KJCX0140).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. QC participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.