Theorem 1
Inequality (9) is equivalent to the following inequalities:
$$\begin{aligned} J_{1}&: = \Biggl\{ \sum_{n = 1}^{\infty } n^{p(\frac{\sigma }{p} + \frac{\sigma _{1}}{q}) - 1} \biggl[ \int _{0}^{\infty } \frac{f(x)}{(1 + xn)^{\lambda }}\,dx \biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\ &< k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}( \sigma _{1}) \biggl\{ \int _{0}^{\infty } x^{p[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}}, \end{aligned}$$
(13)
$$\begin{aligned} J_{2}&: = \Biggl\{ \int _{0}^{\infty } x^{q(\frac{\sigma }{p} + \frac{\sigma _{1}}{q}) - 1} \Biggl[\sum _{n = 1}^{\infty } \frac{a_{n}}{(1 + xn)^{\lambda }} \Biggr]^{q} \,dx \Biggr\} ^{\frac{1}{q}} \\ &< k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}( \sigma _{1}) \Biggl\{ \sum_{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1} a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(14)
If the constant factor in (9) is the best possible, then so is the constant factor in (13) and (14).
Proof
Suppose that (13) (or (14)) is valid. By Hölder’s inequality, we have
$$\begin{aligned} I &= \sum_{n = 1}^{\infty } \biggl[n^{\frac{ - 1}{p} + (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})} \int _{0}^{\infty } \frac{f(x)}{(1 + xn)^{\lambda }}\,dx \biggr] \bigl[n^{\frac{1}{p} - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})}a_{n} \bigr] \\ &\le J_{1} \Biggl\{ \sum_{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1} a_{n}^{q} \Biggr\} ^{\frac{1}{q}}, \end{aligned}$$
(15)
$$\begin{aligned} I &= \int _{0}^{\infty } \bigl[x^{\frac{1}{q} - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})}f(x) \bigr] \Biggl[x^{\frac{ - 1}{q} + (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})}\sum_{n = 1}^{\infty } \frac{1}{(1 + xn)^{\lambda }} a_{n} \Biggr]\,dx \\ &\le \biggl\{ \int _{0}^{\infty } x^{p[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}}J_{2}. \end{aligned}$$
(16)
Then, by (13) (or (14)), we have (9). On the other hand, assuming (9) is valid, we set
$$ a_{n}: = n^{p(\frac{\sigma }{p} + \frac{\sigma _{1}}{q}) - 1} \biggl[ \int _{0}^{\infty } \frac{f(x)}{(1 + xn)^{\lambda }}\,dx \biggr]^{p - 1}\quad (n \in \mathbb{N}). $$
If \(J_{1} = 0\), then (13) is naturally valid; if \(J_{1} = \infty \), then it is impossible that it makes (13) valid. Suppose that \(0 < J_{1} < \infty \). By (9) we have
$$\begin{aligned}& \sum_{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1}a_{n}^{q} \\& \quad = J_{1}^{p} = I < k_{\lambda }^{\frac{1}{p}}( \sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1}) \biggl\{ \int _{0}^{\infty } x^{p[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}} \Biggl\{ \sum_{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1} a_{n}^{q} \Biggr\} ^{\frac{1}{q}}, \\& \Biggl\{ \sum_{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1} a_{n}^{q} \Biggr\} ^{\frac{1}{p}} = J_{1} < k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1}) \biggl\{ \int _{0}^{\infty } x^{p[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}}, \end{aligned}$$
namely, (13) follows.
In the same way, assuming (9) is valid, we set
$$ f(x): = x^{q(\frac{\sigma }{p} + \frac{\sigma _{1}}{q}) - 1} \Biggl[\sum_{n = 1}^{\infty } \frac{1}{(1 + xn)^{\lambda }} a_{n} \Biggr]^{q - 1}\quad (x \in \mathbb{R}). $$
If \(J_{2} = 0\), then (14) is naturally valid; if \(J_{2} = \infty \), then it is impossible that makes (14) valid. Suppose that \(0 < J_{2} < \infty \). By (9) we have
$$\begin{aligned}& \int _{0}^{\infty } x^{p[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1}f^{p}(x)\,dx \\& \quad =J_{2}^{q} = I< k_{\lambda }^{\frac{1}{p}}( \sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1}) \biggl\{ \int _{0}^{\infty } x^{p[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}} \Biggl\{ \sum_{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1} a_{n}^{q} \Biggr\} ^{\frac{1}{q}}, \\& \biggl\{ \int _{0}^{\infty } x^{p[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{q}} = J_{2} < k_{\lambda }^{\frac{1}{p}}( \sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1}) \Biggl\{ \sum _{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})] - 1} a_{n}^{q} \Biggr\} ^{\frac{1}{q}}, \end{aligned}$$
namely, (14) follows. Hence, inequalities (9), (13) and (14) are equivalent.
If the constant factor in (9) is the best possible, then so is constant factor in (13) (or (14)). Otherwise, by (15) (or (16)), we would reach the contradiction that the constant factor in (9) is not the best possible. □
Theorem 2
The statements (i), (ii), (iii) and (iv) are equivalent:
-
(i)
\(k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1})\)
is independent of
\(p,q\);
-
(ii)
\(k_{\lambda }^{\frac{1}{p}}(\sigma)k_{\lambda }^{\frac{1}{q}}(\sigma _{1})\)
is expressed as a single integral;
-
(iii)
\(k_{\lambda }^{\frac{1}{p}}(\sigma)k_{\lambda }^{\frac{1}{q}}(\sigma _{1})\)
in (9) is the best possible constant;
-
(iv)
\(\sigma _{1} = \sigma\).
If the statement (iv) follows, then we have the following equivalent inequalities with the best possible constant factor
\(B(\sigma ,\lambda - \sigma )\):
$$\begin{aligned}& \int _{0}^{\infty } \sum_{n = 1}^{\infty } \frac{a_{n}}{(1 + xn)^{\lambda }} f(x)\,dx \\& \quad < B(\sigma ,\lambda - \sigma ) \biggl[ \int _{0}^{\infty } x^{p(1 - \sigma ) - 1}f^{p}(x)\,dx \biggr]^{\frac{1}{p}} \Biggl[\sum_{n = 1}^{\infty } n^{q(1 - \sigma ) - 1} a_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(17)
$$\begin{aligned}& \Biggl[\sum_{n = 1}^{\infty } n^{p\sigma - 1} \biggl( \int _{0}^{\infty } \frac{f(x)}{(1 + xn)^{\lambda }}\,dx \biggr)^{p} \Biggr]^{\frac{1}{p}} \\& \quad < B(\sigma ,\lambda - \sigma ) \biggl[ \int _{0}^{\infty } x^{p(1 - \sigma ) - 1}f^{p}(x)\,dx \biggr]^{\frac{1}{p}}, \end{aligned}$$
(18)
$$\begin{aligned}& \Biggl[ \int _{0}^{\infty } x^{q\sigma - 1} \Biggl(\sum _{n = 1}^{\infty } \frac{a_{n}}{(1 + xn)^{\lambda }} \Biggr)^{q} \,dx \Biggr]^{\frac{1}{q}} \\& \quad < B(\sigma ,\lambda - \sigma ) \Biggl[\sum_{n = 1}^{\infty } n^{q(1 - \sigma ) - 1} a_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(19)
Proof
(i)⇒(ii). By (i) we have
$$ k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1}) = \lim_{p \to 1^{ +}} k_{\lambda }^{\frac{1}{p}}( \sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1}) = k_{\lambda } (\sigma ), $$
namely, \(k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1})\) is expressed as a single integral.
(ii)⇒(iv). In (12), if \(k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1})\) is expressed as a single integral \(k_{\lambda } (\frac{\sigma }{p} + \frac{\sigma _{1}}{q})\), then (12) keeps the form of equality. In view of the proof of Lemma 4, if and only if \(\sigma _{1} = \sigma \), (12) keeps the form of equality.
(iv)⇒(i). If \(\sigma _{1} = \sigma \), then \(k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1}) = k_{\lambda } (\sigma )\), which is independent of \(p,q\). Hence, we have (i) ↔ (ii) ⇔ (iv).
(iii)⇒(iv). By Lemma 4, we have \(\sigma _{1} = \sigma \). (iv)⇒(iii). By Lemma 3, \(k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\sigma _{1}) = k_{\lambda } (\sigma )\) in (9) (for \(\sigma _{1} = \sigma \)) is the best possible constant. Therefore, we have (iii) ⇔ (iv).
Hence, the statements (i), (ii), (iii) and (iv) are equivalent. □
Replacing x by \(\frac{1}{x}\), and then \(x^{\lambda - 2}f(\frac{1}{x})\) by \(f(x)\) in Theorem 1, setting \(\sigma _{1} = \lambda - \mu \), we have the following.
Corollary 1
The following inequalities with the homogeneous kernel are equivalent:
$$\begin{aligned} &\int _{0}^{\infty } \sum_{n = 1}^{\infty } \frac{a_{n}}{(x + n)^{\lambda }} f(x)\,dx \\ &\quad < k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}( \lambda - \mu ) \biggl\{ \int _{0}^{\infty } x^{p[1 - (\frac{\lambda - \sigma }{p} + \frac{\mu }{q})] - 1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}} \Biggl\{ \sum_{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\lambda - \mu }{q})] - 1} a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(20)
$$\begin{aligned} &\Biggl\{ \sum_{n = 1}^{\infty } n^{p(\frac{\sigma }{p} + \frac{\lambda - \mu }{q}) - 1} \biggl[ \int _{0}^{\infty } \frac{f(x)}{(x + n)^{\lambda }}\,dx \biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\ &\quad < k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}( \lambda - \mu ) \biggl\{ \int _{0}^{\infty } x^{p[1 - (\frac{\lambda - \sigma }{p} + \frac{\mu }{q})] - 1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}}, \end{aligned}$$
(21)
$$\begin{aligned} &\Biggl\{ \int _{0}^{\infty } x^{q(\frac{\lambda - \sigma }{p} + \frac{\mu }{q}) - 1} \Biggl[\sum _{n = 1}^{\infty } \frac{a_{n}}{(x + n)^{\lambda }} \Biggr]^{q} \,dx \Biggr\} ^{\frac{1}{q}} \\ &\quad < k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}( \lambda - \mu ) \Biggl\{ \sum_{n = 1}^{\infty } n^{q[1 - (\frac{\sigma }{p} + \frac{\lambda - \mu }{q})] - 1} a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(22)
If the constant factor in (20) is the best possible, then so is the constant factor in (21) and (22).
Corollary 2
The statements (I), (II), (III) and (IV) are equivalent:
-
(I)
\(k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\lambda - \mu )\)
is independent of
\(p,q\);
-
(II)
\(k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\lambda - \mu )\)
is expressed as a single integral;
-
(III)
\(k_{\lambda }^{\frac{1}{p}}(\sigma )k_{\lambda }^{\frac{1}{q}}(\lambda - \mu )\)
in (20) is the best possible constant;
-
(IV)
\(\mu + \sigma = \lambda\).
If the statement (IV) follows, then we have the following equivalent inequalities with the best possible constant factor
\(B(\mu ,\sigma )\):
$$ \int _{0}^{\infty } \sum_{n = 1}^{\infty } \frac{a_{n}}{(x + n)^{\lambda }} f(x)\,dx< B(\mu ,\sigma ) \biggl[ \int _{0}^{\infty } x^{p(1 - \mu ) - 1}f^{p}(x)\,dx \biggr]^{\frac{1}{p}} \Biggl[\sum_{n = 1}^{\infty } n^{q(1 - \sigma ) - 1} a_{n}^{q} \Biggr]^{\frac{1}{q}}. $$
(23)
$$ \Biggl\{ \sum_{n = 1}^{\infty } n^{p\sigma - 1} \biggl[ \int _{0}^{\infty } \frac{f(x)}{(x + n)^{\lambda }}\,dx \biggr]^{p} \Biggr\} ^{\frac{1}{p}}< B(\mu ,\sigma ) \biggl[ \int _{0}^{\infty } x^{p(1 - \sigma ) - 1}f^{p}(x)\,dx \biggr]^{\frac{1}{p}}, $$
(24)
$$ \Biggl\{ \int _{0}^{\infty } x^{q\mu - 1} \Biggl[\sum _{n = 1}^{\infty } \frac{a_{n}}{(x + n)^{\lambda }} \Biggr]^{q} \,dx \Biggr\} ^{\frac{1}{q}}< B(\mu ,\sigma ) \Biggl[\sum _{n = 1}^{\infty } n^{q(1 - \sigma ) - 1} a_{n}^{q} \Biggr]^{\frac{1}{q}}. $$
(25)
Remark 1
(i) For \(\sigma = \frac{1}{p}( < \lambda )\) in (17), (18) and (19), we have the following equivalent inequalities with the nonhomogeneous kernel and the best possible constant factor \(B(\frac{1}{p},\lambda - \frac{1}{p})\):
$$ \int _{0}^{\infty } \sum_{n = 1}^{\infty } \frac{a_{n}}{(1 + xn)^{\lambda }} f(x)\,dx< B \biggl(\frac{1}{p},\lambda - \frac{1}{p} \biggr) \biggl( \int _{0}^{\infty } x^{p - 2}f^{p}(x)\,dx \biggr)^{\frac{1}{p}} \Biggl(\sum_{n = 1}^{\infty } a_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(26)
$$ \Biggl\{ \sum_{n = 1}^{\infty } \biggl[ \int _{0}^{\infty } \frac{f(x)}{(1 + xn)^{\lambda }}\,dx \biggr]^{p} \Biggr\} ^{\frac{1}{p}}< B \biggl(\frac{1}{p},\lambda - \frac{1}{p} \biggr) \biggl( \int _{0}^{\infty } x^{p - 2}f^{p}(x)\,dx \biggr)^{\frac{1}{p}}, $$
(27)
$$ \Biggl\{ \int _{0}^{\infty } x^{q - 2} \Biggl[\sum _{n = 1}^{\infty } \frac{a_{n}}{(1 + xn)^{\lambda }} \Biggr]^{q} \,dx \Biggr\} ^{\frac{1}{q}}< B \biggl(\frac{1}{p},\lambda - \frac{1}{p} \biggr) \Biggl(\sum_{n = 1}^{\infty } a_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(28)
(ii) For \(\sigma = \frac{1}{q}( < \lambda )\) in (17), (18) and (19), we have the following equivalent inequalities with the best possible constant factor \(B(\frac{1}{q},\lambda - \frac{1}{q})\):
$$ \int _{0}^{\infty } \sum_{n = 1}^{\infty } \frac{a_{n}}{(1 + xn)^{\lambda }} f(x)\,dx< B \biggl(\frac{1}{q},\lambda - \frac{1}{q} \biggr) \biggl( \int _{0}^{\infty } f^{p}(x)\,dx \biggr)^{\frac{1}{p}} \Biggl(\sum_{n = 1}^{\infty } n^{q - 2}a_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(29)
$$ \Biggl\{ \sum_{n = 1}^{\infty } n^{p - 2} \biggl[ \int _{0}^{\infty } \frac{f(x)}{(1 + xn)^{\lambda }}\,dx \biggr]^{p} \Biggr\} ^{\frac{1}{p}}< B \biggl(\frac{1}{q},\lambda - \frac{1}{q} \biggr) \biggl( \int _{0}^{\infty } f^{p}(x)\,dx \biggr)^{\frac{1}{p}}, $$
(30)
$$ \Biggl\{ \int _{0}^{\infty } \Biggl[ \sum _{n = 1}^{\infty } \frac{a_{n}}{(1 + xn)^{\lambda }} \Biggr]^{q} \,dx \Biggr\} ^{\frac{1}{q}}< B \biggl(\frac{1}{q},\lambda - \frac{1}{q} \biggr) \Biggl(\sum_{n = 1}^{\infty } n^{q - 2}a_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(31)
(iii) For \(\lambda = 1\), \(\sigma = \frac{1}{p}\), \(\mu = \frac{1}{q}\) in (23), (24) and (25), we have the following equivalent inequalities with the homogeneous kernel and the best possible constant factor \(\frac{\pi }{\sin (\frac{\pi }{p})}\):
$$ \int _{0}^{\infty } \sum_{n = 1}^{\infty } \frac{a_{n}}{x + n} f(x)\,dx< \frac{\pi }{\sin (\frac{\pi }{p})} \biggl( \int _{0}^{\infty } f^{p}(x)\,dx \biggr)^{\frac{1}{p}} \Biggl(\sum_{n = 1}^{\infty } a_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(32)
$$ \Biggl[\sum_{n = 1}^{\infty } \biggl( \int _{0}^{\infty } \frac{f(x)}{x + n}\,dx \biggr)^{p} \Biggr]^{\frac{1}{p}}< \frac{\pi }{\sin (\frac{\pi }{p})} \biggl( \int _{0}^{\infty } f^{p}(x)\,dx \biggr)^{\frac{1}{p}}, $$
(33)
$$ \Biggl[ \int _{0}^{\infty } \Biggl( \sum _{n = 1}^{\infty } \frac{a_{n}}{x + n} \Biggr)^{q} \,dx \Biggr]^{\frac{1}{q}}< \frac{\pi }{\sin (\frac{\pi }{p})} \Biggl(\sum _{n = 1}^{\infty } a_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(34)
(iv) For \(\lambda = 1\), \(\sigma = \frac{1}{q}\), \(\mu = \frac{1}{p}\) in (23), (24) and (25), we have the following equivalent inequalities with the best possible constant factor \(\frac{\pi }{\sin (\frac{\pi }{p})}\):
$$ \int _{0}^{\infty } \sum_{n = 1}^{\infty } \frac{a_{n}}{x + n} f(x)\,dx< \frac{\pi }{\sin (\frac{\pi }{p})} \biggl( \int _{0}^{\infty } x^{p - 2}f^{p}(x)\,dx \biggr)^{\frac{1}{p}} \Biggl(\sum_{n = 1}^{\infty } n^{q - 2}a_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(35)
$$ \Biggl[\sum_{n = 1}^{\infty } n^{p - 2} \biggl( \int _{0}^{\infty } \frac{f(x)}{x + n}\,dx \biggr)^{p} \Biggr]^{\frac{1}{p}}< \frac{\pi }{\sin (\frac{\pi }{p})} \biggl( \int _{0}^{\infty } x^{p - 2}f^{p}(x)\,dx \biggr)^{\frac{1}{p}}, $$
(36)
$$ \Biggl[ \int _{0}^{\infty } x^{q - 2} \Biggl( \sum _{n = 1}^{\infty } \frac{a_{n}}{x + n} \Biggr)^{q} \,dx \Biggr]^{\frac{1}{q}}< \frac{\pi }{\sin (\frac{\pi }{p})} \Biggl(\sum _{n = 1}^{\infty } n^{q - 2}a_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(37)