Open Access

On a more accurate Hardy-Mulholland-type inequality

Journal of Inequalities and Applications20172017:163

https://doi.org/10.1186/s13660-017-1442-8

Received: 26 March 2017

Accepted: 16 June 2017

Published: 12 July 2017

Abstract

By using the way of weight coefficients, the technique of real analysis, and Hermite-Hadamard’s inequality, a more accurate Hardy-Mulholland-type inequality with multi-parameters and a best possible constant factor is given. The equivalent forms, the reverses, the operator expressions and some particular cases are considered.

Keywords

Mulholland-type inequalityweight coefficientequivalent formreverseoperator

MSC

26D1547A07

1 Introduction

Assuming that \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(a_{m},b_{n} \ge 0\), \(a = \{ a_{m}\}_{m = 1}^{\infty} \in l^{p}\), \(b = \{ b_{n}\}_{n = 1}^{\infty} \in l^{q}\), \(\Vert a \Vert _{p} = (\sum_{m = 1}^{\infty} a_{m}^{p} )^{\frac{1}{p}} > 0\), and \(\Vert b \Vert _{q} > 0\), we have the following Hardy-Hilbert’s inequality with the best possible constant \(\frac{\pi}{\sin (\pi /p)}\) (cf. [1], Theorem 315):
$$ \sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b{}_{n}}{m + n} < \frac{\pi}{\sin (\pi /p)} \Vert a \Vert _{p} \Vert b \Vert _{q}. $$
(1)
A more accurate inequality of (1) is given as follows (cf. [1], Th. 323 and [2]):
$$ \sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b{}_{n}}{m + n - \alpha} < \frac{\pi}{\sin (\pi /p)} \Vert a \Vert _{p} \Vert b \Vert _{q}\quad (0 \le \alpha \le 1), $$
(2)
where the constant factor \(\frac{\pi}{\sin (\pi /p)}\) is still the best possible.
Also we have the following Mulholland’s inequality similar to (1) with the same best possible constant factor \(\frac{\pi}{\sin (\pi /p)}\) (cf. [3] or [1], Th. 343, replacing \(\frac{a_{m}}{m}\), \(\frac{b_{n}}{n}\) by \(a_{m}\), \(b_{n}\)):
$$ \sum_{m = 2}^{\infty} \sum _{n = 2}^{\infty} \frac{a_{m}b{}_{n}}{\ln mn} < \frac{\pi}{\sin (\pi /p)} \Biggl( \sum_{m = 2}^{\infty} \frac{a_{m}^{p}}{m^{1 - p}} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 2}^{\infty} \frac{b_{n}^{q}}{n^{1 - q}} \Biggr)^{\frac{1}{q}}. $$
(3)
Inequalities (1)-(3) are important in analysis and its applications (cf. [1, 2, 418]).
Suppose that \(\mu_{i},\upsilon_{j} > 0\) (\(i,j \in \mathbb{N} = \{ 1,2, \ldots \}\)),
$$ U_{m} = \sum_{i = 1}^{m} \mu_{i},\qquad V_{n} = \sum_{j = 1}^{n} \upsilon_{j}\quad (m,n \in \mathbb{N}), $$
(4)
we have the following Hardy-Hilbert-type inequality (cf. [1], Theorem 321, replacing \(\mu_{m}^{1/q}a_{m}\) and \(\upsilon_{n}^{1/p}b_{n}\) by \(a_{m}\) and \(b_{n}\)): If \(a_{m},b_{n} \ge 0\), \(0 < \sum_{m = 1}^{\infty} \frac{a_{m}^{p}}{m^{p - 1}} < \infty\), \(0 < \sum_{n = 1}^{\infty} \frac{b_{n}^{q}}{n^{q - 1}} < \infty\), then
$$ \sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b{}_{n}}{U_{m} + V_{n}} < \frac{\pi}{\sin (\pi /p)} \Biggl( \sum_{m = 1}^{\infty} \frac{a_{m}^{p}}{\mu_{m}^{p - 1}} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 1}^{\infty} \frac{b_{n}^{q}}{\upsilon_{n}^{q - 1}} \Biggr)^{\frac{1}{q}}. $$
(5)
For \(\mu_{i} = \upsilon_{j} = 1\) (\(i,j \in \mathbb{N}\)), inequality (5) reduces to (1).
In 2015, Yang [19] gave an extension of (5) as follows: If \(0 < \lambda_{1},\lambda_{2} \le 1\), \(\lambda_{1} + \lambda_{2} = \lambda\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are positive and decreasing, with \(U_{\infty} = V_{\infty} = \infty\), then we have the following inequality with the best possible constant factor \(\pi /\sin (\frac{\pi \lambda_{1}}{\lambda} )\):
$$ \sum_{m = 1}^{\infty} \sum _{n = 1}^{\infty} \frac{a_{m}b{}_{n}}{U_{m}^{\lambda} + V_{n}^{\lambda}} < \frac{\pi}{ \lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )} \Biggl[ \sum_{m = 1}^{\infty} \frac{U_{m}^{p(1 - \lambda_{1}) - 1}a_{m}^{p}}{\mu_{m}^{p - 1}} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{n = 1}^{\infty} \frac{V_{n}^{q(1 - \lambda_{2}) - 1}b_{n}^{q}}{\upsilon_{n}^{q - 1}} \Biggr]^{\frac{1}{q}}. $$
(6)
In this paper, by using the way of weight coefficients, the technique of real analysis, and Hermite-Hadamard’s inequality, a new Hardy-Mulholland-type inequality with a best possible constant factor is given as follows: If \(\mu_{1} = \upsilon_{1} = 1\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are positive and decreasing, with \(U_{\infty} = V_{\infty} = \infty\), we have the following inequality:
$$ \sum_{m = 2}^{\infty} \sum _{n = 2}^{\infty} \frac{a_{m}b{}_{n}}{\ln U_{m}V_{n}} < \frac{\pi}{\sin (\frac{\pi}{p})} \Biggl[ \sum_{m = 2}^{\infty} \biggl( \frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{n = 2}^{\infty} \biggl(\frac{V_{n}}{\upsilon_{n + 1}}\biggr)^{q - 1}b_{n}^{q} \Biggr]^{\frac{1}{q}}, $$
(7)
which is an extension of (3). Moreover, the more accurate inequality of (7) and its extension with multi-parameters and the best possible constant factors are obtained. The equivalent forms, the reverses, the operator expressions and some particular cases are considered.

2 Some lemmas and an example

In the following, we agree that \(p \ne 0,1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(- 1 < \gamma \le 0\), \(0 < \lambda_{1},\lambda_{2} < 1\), \(\lambda_{1} + \lambda_{2} = \lambda\), \(\mu_{i},\upsilon_{j} > 0\) (\(i,j \in \mathbb{N}\)), with \(\mu_{1} = \upsilon_{1} = 1\), \(U_{m}\) and \(V_{n}\) are defined by (4),
$$\frac{1}{1 + \frac{\mu_{2}}{2}} \le \alpha \le 1,\qquad \frac{1}{1 + \frac{\upsilon_{2}}{2}} \le \beta \le 1, $$
\(a_{m},b_{n} \ge 0\), \(\Vert a \Vert _{p,\Phi_{\lambda}}: = (\sum_{m = 2}^{\infty} \Phi_{\lambda} (m)a_{m}^{p})^{\frac{1}{p}}\) and \(\Vert b \Vert _{q,\Psi_{\lambda}}: = (\sum_{n = 2}^{\infty} \Psi_{\lambda} (n)b_{n}^{q})^{\frac{1}{q}}\), where
$$ \begin{gathered} \Phi_{\lambda} (m): = \biggl(\frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}(\ln \alpha U_{m})^{p(1 - \lambda_{1}) - 1}, \\ \Psi_{\lambda} (n): = \biggl(\frac{V_{n}}{\upsilon_{n + 1}}\biggr)^{q - 1}(\ln \beta V_{n})^{q(1 - \lambda_{2}) - 1}\quad \bigl(m,n \in \mathbb{N}\backslash \{ 1\} \bigr). \end{gathered} $$
(8)

Lemma 1

If \(n \in \mathbb{N}\backslash \{ 1\}\), \(a \in (n - \frac{1}{2},n)\), \(f(x)\) is continuous in \((n - \frac{1}{2}, n + \frac{1}{2})\), and \(f'(x)\) is strictly increasing in the intervals \((n - \frac{1}{2},a)\), \((a,n)\) and \((n,n + \frac{1}{2})\), respectively, satisfying
$$ f'(a - 0) \le f'(a + 0),\qquad f'(n - 0) \le f'(n + 0), $$
(9)
then we have the following Hermite-Hadamard’s inequality (cf. [20]).

Proof

In view of \(f'(n - 0) \le f'(n + 0) = \lim_{x \to n^{ +}} f'(x)\) is finite, we set the linear function \(g(x)\) as follows:
$$g(x): = f'(n - 0) (x - n) + f(n),\quad x \in \biggl[n - \frac{1}{2}, n + \frac{1}{2}\biggr]. $$
Since \(f'(x)\) is strictly increasing in \([n - \frac{1}{2},a)\) and \((a,n)\), then for \(x \in [n - \frac{1}{2},a)\),
$$f'(x) < \lim_{x \to a^{ -}} f'(x) = f'(a - 0) \le f'(a + 0) < f'(n - 0); $$
for \(x \in (a,n)\), \(f'(x) < \lim_{x \to n^{ -}} f'(x) = f'(n - 0)\). Hence,
$$\bigl(f(x) - g(x)\bigr)' = f'(x) - f'(n - 0) < 0,\quad x \in \biggl(n - \frac{1}{2},a\biggr) \cup (a,n). $$
Since \(f(x)-g(x)\) is continuous in \((n - \frac{1}{2},n]\) with \(f(n)-g(n)=0\), it follows that
$$f(x) - g(x) > 0,\quad x \in \biggl(n - \frac{1}{2},n\biggr). $$
In the same way, since \(f'(x)\) is strictly increasing in \((n,n + \frac{1}{2})\), then for \(x \in (n,n + \frac{1}{2})\), \(f'(x) > f'(n + 0) \ge f'(n - 0)\). Hence,
$$\bigl(f(x) - g(x)\bigr)' = f'(x) - f'(n - 0) > 0,\quad x \in \biggl(n,n + \frac{1}{2}\biggr). $$
Since \(f(x)-g(x)\) is continuous in \([n,n + \frac{1}{2})\) with \(f(n)-g(n)=0\), it follows that
$$f(x) - g(x) > 0,\quad x \in \biggl(n,n + \frac{1}{2}\biggr). $$
Therefore, we have \(f(x) - g(x) > 0\), \(x \in (n - \frac{1}{2},n + \frac{1}{2})\backslash \{ n\}\). Then we find
$$\int_{n - \frac{1}{2}}^{n + \frac{1}{2}} f(x)\,dx > \int_{n - \frac{1}{2}}^{n + \frac{1}{2}} g(x)\,dx = f(n), $$
namely, (9) follows. The lemma is proved. □

Note

With the assumptions of Lemma 1, if (i) \(a \in (n,n + \frac{1}{2})\), \(f'(x)\) is strictly increasing in the intervals \((n - \frac{1}{2},n)\), \((n,a)\) and \((a,n + \frac{1}{2})\), respectively, or (ii) \(a = n\), \(f'(x)\) is strictly increasing in the intervals \((n - \frac{1}{2},n)\) and \((n,n + \frac{1}{2})\), respectively, then in the same way, we still can obtain (9).

Example 1

\(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing, we set functions \(\mu (t): = \mu_{m}\), \(t \in (m - 1,m]\) (\(m \in \mathbb{N}\)), \(\upsilon (t): = \upsilon_{n}\), \(t \in (n - 1,n]\) (\(n \in \mathbb{N}\)), and
$$ U(x): = \int_{0}^{x} \mu (t)\,dt\quad (x \ge 0),\qquad V(y): = \int_{0}^{y} \upsilon (t)\,dt\quad (y \ge 0). $$
(10)
Then it follows that \(U(m) = U_{m}\), \(V(n) = V_{n}\), \(U(\infty ) = U_{\infty}\), \(V(\infty ) = V_{\infty}\) and
$$\begin{gathered} U'(x) = \mu (x) = \mu_{m},\quad x \in (m - 1,m), \\ V'(y) = \upsilon (y) = \upsilon_{n},\quad y \in (n - 1,n)\ (x,y \in \mathbb{N}). \end{gathered} $$
For \(0 < \lambda \le 1\), \(- 1 < \gamma \le 0\), we set
$$ k_{\lambda} (x,y): = \frac{1}{x^{\lambda} + y^{\lambda} + \gamma \vert x^{\lambda} - y^{\lambda} \vert }\quad (x,y > 0). $$
(11)
We find
$$ \begin{aligned}[b] 0 &< K_{\gamma} (\lambda_{1}): = \int_{0}^{\infty} k_{\lambda} (1,t)t^{\lambda_{2} - 1} \,dt = \int_{0}^{\infty} k_{\lambda} (t,1)t^{\lambda_{1} - 1} \,dt \\ &= \int_{0}^{\infty} \frac{t^{\lambda_{1} - 1}}{t^{\lambda} + 1 + \gamma \vert t^{\lambda} - 1 \vert } \,dt = \int_{0}^{1} \frac{t^{\lambda_{1} - 1} + t^{\lambda_{2} - 1}}{1 + \gamma + (1 - \gamma )t^{\lambda}} \,dt \\ &\le \int_{0}^{1} \frac{t^{\lambda_{1} - 1} + t^{\lambda_{2} - 1}}{1 + \gamma} \,dt = \frac{1}{1 + \gamma} \biggl(\frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}\biggr) < \infty, \end{aligned} $$
(12)
namely, \(K_{\gamma} (\lambda_{1}) \in \mathbb{R}_{ +}\). In the following, we express \(K_{\gamma} (\lambda_{1})\) in other forms.
(i) For \(\gamma = 0\), we obtain
$$ K_{0}(\lambda_{1}) = \int_{0}^{\infty} \frac{t^{\lambda_{1} - 1}}{t^{\lambda} + 1} \,dt = \frac{1}{\lambda} \int_{0}^{\infty} \frac{v^{(\lambda_{1}/\lambda ) - 1}}{v + 1} \,dv = \frac{\pi}{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )}; $$
(13)
(ii) for \(- 1 < \gamma < 0\), \(0 < \frac{1 + \gamma}{1 - \gamma} < 1\), by the Lebesgue term by term integration theorem (cf. [21]), we find
$$\begin{aligned} K_{\gamma} (\lambda_{1}) &= \frac{1}{1 - \gamma} \int_{0}^{1} \frac{t^{ - \lambda_{2} - 1} + t^{ - \lambda_{1} - 1}}{\frac{1 + \gamma}{1 - \gamma} t^{ - \lambda} + 1} \,dt \\ &\mathop{ =} ^{v = \frac{1 + \gamma}{1 - \gamma} t^{ - \lambda}} \frac{1}{\lambda (1 - \gamma )} \int_{\frac{1 + \gamma}{1 - \gamma}}^{\infty} \frac{1}{v + 1} \biggl[ \biggl( \frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda} - 1}v^{\frac{\lambda_{2}}{\lambda} - 1} + \biggl( \frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda} - 1}v^{\frac{\lambda_{1}}{\lambda} - 1} \biggr] \,dv \\ &= \frac{1}{\lambda (1 - \gamma )}\biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda} - 1} \int_{0}^{\infty} \frac{v^{\frac{\lambda_{2}}{\lambda} - 1}}{v + 1} \,dv + \frac{1}{\lambda (1 - \gamma )}\biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda} - 1} \int_{0}^{\infty} \frac{v^{\frac{\lambda_{1}}{\lambda} - 1}}{v + 1} \,dv \\ &\quad {}- \frac{1}{\lambda (1 - \gamma )} \int_{0}^{\frac{1 + \gamma}{1 - \gamma}} \frac{1}{v + 1} \biggl[ \biggl( \frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda} - 1}v^{\frac{\lambda_{2}}{\lambda} - 1} + \biggl( \frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda} - 1}v^{\frac{\lambda_{1}}{\lambda} - 1} \biggr]\,dv \\ & = \frac{1}{\lambda (1 - \gamma )}\biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda} - 1} \frac{\pi}{\sin (\frac{\pi \lambda_{2}}{\lambda} )} + \frac{1}{\lambda (1 - \gamma )}\biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda} - 1}\frac{\pi}{\sin (\frac{\pi \lambda_{1}}{\lambda} )} \\ &\quad {}- \frac{1}{\lambda (1 - \gamma )} \int_{0}^{\frac{1 + \gamma}{1 - \gamma}} \sum_{k = 0}^{\infty} ( - 1)^{k}v^{k} \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda} - 1}v^{\frac{\lambda_{2}}{\lambda} - 1} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda} - 1}v^{\frac{\lambda_{1}}{\lambda} - 1} \biggr]\,dv \\ &= \frac{1}{\lambda (1 - \gamma )} \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda} - 1} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda} - 1} \biggr]\frac{\pi}{\sin (\frac{\pi \lambda_{1}}{\lambda} )} \\ &\quad {}- \frac{1}{\lambda (1 - \gamma )} \int_{0}^{\frac{1 + \gamma}{1 - \gamma}} \sum_{k = 0}^{\infty} \bigl(v^{2k} - v^{2k + 1}\bigr) \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda} - 1}v^{\frac{\lambda_{2}}{\lambda} - 1} \\ &\quad {} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda} - 1}v^{\frac{\lambda_{1}}{\lambda} - 1} \biggr]\,dv \\ & = \frac{1}{1 + \gamma} \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda}} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda}} \biggr]\frac{\pi}{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )} \\ &\quad {}- \frac{1}{\lambda (1 + \gamma )} \int_{0}^{\frac{1 + \gamma}{1 - \gamma}} \sum_{k = 0}^{\infty} \bigl(v^{2k} - v^{2k + 1}\bigr) \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda}} v^{\frac{\lambda_{2}}{\lambda} - 1} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda}} v^{\frac{\lambda_{1}}{\lambda} - 1} \biggr]\,dv \\ &= \frac{1}{1 + \gamma} \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda}} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda}} \biggr]\frac{\pi}{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )} \\ &\quad {}- \frac{1}{\lambda (1 + \gamma )}\sum_{k = 0}^{\infty} \int_{0}^{\frac{1 + \gamma}{1 - \gamma}} ( - 1)^{k} v^{k} \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda}} v^{\frac{\lambda_{2}}{\lambda} - 1} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda} - 1}v^{\frac{\lambda_{1}}{\lambda} - 1} \biggr]\,dv \\ &= \frac{1}{1 + \gamma} \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{1}}{\lambda}} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{\lambda_{2}}{\lambda}} \biggr]\frac{\pi}{\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )} \\ &\quad {}- \frac{1}{1 + \gamma} \sum_{k = 0}^{\infty} ( - 1)^{k}\biggl(\frac{1 + \gamma}{ 1 - \gamma} \biggr)^{k + 1} \biggl( \frac{1}{\lambda k + \lambda_{2}} + \frac{1}{\lambda k + \lambda_{1}} \biggr); \end{aligned}$$
(14)
(iii) for \(\lambda_{1} = \lambda_{2} = \frac{\lambda}{2}\), \(- 1 < \gamma < 0\), we find
$$ \begin{aligned}[b] K_{\gamma} \biggl(\frac{\lambda}{2} \biggr) &= 2 \int_{0}^{1} \frac{t^{(\lambda /2) - 1}}{1 + \gamma + (1 - \gamma )t^{\lambda}} \,dt\mathop{ =} ^{u = (\frac{1 - \gamma}{1 + \gamma} t^{\lambda} )^{\frac{1}{2}}} \frac{4}{\lambda (1 + \gamma )}\biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{1}{2}} \int_{0}^{(\frac{1 - \gamma}{1 + \gamma} )^{\frac{1}{2}}} \frac{du}{1 + u^{2}} \\ &= \frac{4}{\lambda (1 + \gamma )}\biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{1}{2}}\arctan \biggl(\frac{1 - \gamma}{1 + \gamma} \biggr)^{\frac{1}{2}}. \end{aligned} $$
(15)
For fixed \(m \in \mathbb{N}\backslash \{ 1\}\), we define the function \(f(y)\) as follows:
$$f(y): = k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr),\quad y \in \biggl(n - \frac{1}{2},n + \frac{1}{2}\biggr)\ \bigl(n \in \mathbb{N}\backslash \{ 1\} \bigr). $$
Then \(f(y)\) is continuous in \((n - \frac{1}{2},n + \frac{1}{2})\) (\(n \in \mathbb{N}\backslash \{ 1\} \)). There exists a unified number \(y_{0} > \frac{3}{2}\) satisfying \(V(y_{0}) = \frac{\alpha}{\beta} U_{m}\).
(i) If \(y_{0} \in (n - \frac{1}{2},n + \frac{1}{2})\), we find
$$f(y) = \textstyle\begin{cases} \frac{1}{(1 + \gamma )\ln^{\lambda} \alpha U_{m} + (1 - \gamma )\ln^{\lambda} \beta V(y)},&n - \frac{1}{2} < y < y_{0}, \\ \frac{1}{(1 - \gamma )\ln^{\lambda} \alpha U_{m} + (1 + \gamma )\ln^{\lambda} \beta V(y)},&y_{0} < y < n + \frac{1}{2}. \end{cases} $$
For \(y_{0} \ne n\), we obtain for \(y \ne n\) that
$$f'(y) = \textstyle\begin{cases} \frac{ - \lambda (1 - \gamma )V'(y)\ln^{\lambda - 1}\beta V(y)}{V(y)[(1 + \gamma )\ln^{\lambda} \alpha U_{m} + (1 - \gamma )\ln^{\lambda} \beta V(y)]^{2}},&n - \frac{1}{2} < y < y_{0}, \\ \frac{ - \lambda (1 + \gamma )V'(y)\ln^{\lambda - 1}\beta V(y)}{V(y)[(1 - \gamma )\ln^{\lambda} \alpha U_{m} + (1 + \gamma )\ln^{\lambda} \beta V(y)]^{2}},&y_{0} < y < n + \frac{1}{2}; \end{cases} $$
for \(y_{0} \ne n\), we obtain for \(y = n\) that
$$\begin{gathered} f'(n - 0) = \textstyle\begin{cases} \frac{ - \lambda (1 - \gamma )\upsilon_{n}\ln^{\lambda - 1}\beta V_{n}}{V_{n}[(1 + \gamma )\ln^{\lambda} \alpha U_{m} + (1 - \gamma )\ln^{\lambda} \beta V_{n}]^{2}},&n - \frac{1}{2} < y < y_{0}, \\ \frac{ - \lambda (1 + \gamma )\upsilon_{n}\ln^{\lambda - 1}\beta V_{n}}{V_{n}[(1 - \gamma )\ln^{\lambda} \alpha U_{m} + (1 + \gamma )\ln^{\lambda} \beta V_{n}]^{2}},&y_{0} < y < n + \frac{1}{2}, \end{cases}\displaystyle \\ f'(n + 0) = \textstyle\begin{cases} \frac{ - \lambda (1 - \gamma )\upsilon_{n + 1}\ln^{\lambda - 1}\beta V_{n}}{V_{n}[(1 + \gamma )\ln^{\lambda} \alpha U_{m} + (1 - \gamma )\ln^{\lambda} \beta V_{n}]^{2}},&n - \frac{1}{2} < y < y_{0}, \\ \frac{ - \lambda (1 + \gamma )\upsilon_{n + 1}\ln^{\lambda - 1}\beta V_{n}}{V_{n}[(1 - \gamma )\ln^{\lambda} \alpha U_{m} + (1 + \gamma )\ln^{\lambda} \beta V_{n}]^{2}},&y_{0} < y < n + \frac{1}{2}. \end{cases}\displaystyle \end{gathered} $$

Since \(0 < \lambda \le 1\), \(- 1 < \gamma \le 0\), \((1 - \gamma )\upsilon_{n} \ge (1 + \gamma )\upsilon_{n + 1}\), in view of the above results, we find \(f'(n - 0) \le f'(n + 0)\) (\(n \ne y_{0}\)), and \(f'(y)\) (<0) is strictly increasing in \((n - \frac{1}{2},y_{0})\), \((y_{0},n)\) and \((n,n + \frac{1}{2})\) for \(y_{0} < n\) or in \((n - \frac{1}{2},n)\), \((n,y_{0})\) and \((y_{0}, n + \frac{1}{2})\) for \(y_{0} > n\).

We obtain
$$\begin{aligned}& \begin{aligned} f'(y_{0} - 0) &= \frac{ - \lambda (1 - \gamma )V'(y_{0} - 0)\ln^{\lambda - 1}\beta V(y_{0})}{V(y_{0})[(1 + \gamma )\ln^{\lambda} \alpha U_{m} + (1 - \gamma )\ln^{\lambda} \beta V(y_{0})]^{2}}\\ & = \frac{ - \lambda (1 - \gamma )V'(y_{0} - 0)\ln^{\lambda - 1}\beta V(y_{0})}{V(y_{0})(2\ln^{\lambda} \alpha U_{m})^{2}}, \end{aligned} \\& \begin{aligned} f'(y_{0} + 0) &= \frac{ - \lambda (1 + \gamma )V'(y_{0} + 0)\ln^{\lambda - 1}\beta V(y_{0})}{V(y_{0})[(1 - \gamma )\ln^{\lambda} \alpha U_{m} + (1 + \gamma )\ln^{\lambda} \beta V(y_{0})]^{2}}\\ & = \frac{ - \lambda (1 + \gamma )V'(y_{0} + 0)\ln^{\lambda - 1}\beta V(y_{0})}{V(y_{0})(2\ln^{\lambda} \alpha U_{m})^{2}}. \end{aligned} \end{aligned}$$
Since for \(y_{0} = n\), \(V'(y_{0} - 0) = \upsilon_{n}\), \(V'(y_{0} + 0) = \upsilon_{n + 1}\) and for \(y_{0} \ne n\), \(V'(y_{0} - 0) = V'(y_{0})\), then we have \(\lambda (1 - \gamma )V'(y_{0} - 0) \ge \lambda (1 + \gamma )V'(y_{0} + 0)\), namely, \(f'(y_{0} - 0) \le f'(y_{0} + 0)\).
(ii) If \(y_{0} \notin (n - \frac{1}{2},n + \frac{1}{2})\), then it follows that \(f'(y) = \frac{V'(y)}{V(y)}\frac{d}{dy}k_{\lambda} (\ln \alpha U_{m},\ln \beta V(y)) < 0\), \(y \in (n - \frac{1}{2},n + \frac{1}{2})\backslash \{ n\}\). We still can find that
$$\begin{aligned} \frac{\upsilon_{n}}{V_{n}}\frac{d}{dy}k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr)\bigg\vert _{y = n} &= f'(n - 0) \\ &\le f'(n + 0) = \frac{\upsilon_{n + 1}}{V_{n}}\frac{d}{dy}k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr)\bigg\vert _{y = n}, \end{aligned} $$
and \(f'(y)\) (<0) is strictly increasing in \((n - \frac{1}{2},n)\) and \((n,n + \frac{1}{2})\).
Therefore, \(f(y)\) satisfies the conditions of Lemma 1 with Note. So does \(g(y) = \frac{f(y)}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\). Hence, by (9), we have
$$ \frac{k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})}{V_{n}\ln^{1 - \lambda_{2}}\beta V_{n}} < \int_{n - \frac{1}{2}}^{n + \frac{1}{2}} \frac{k_{\lambda} (\ln \alpha U_{m},\ln \beta V(y))}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)} \,dy\quad \bigl(n \in \mathbb{N}\backslash \{ 1\} \bigr). $$
(16)

Definition 1

Define the following weight coefficients:
$$ \begin{gathered} \omega (\lambda_{2},m): = \sum_{n = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n}) \frac{\upsilon_{n + 1}\ln^{\lambda_{1}}\alpha U_{m}}{V_{n}\ln^{1 - \lambda_{2}}\beta V_{n}},\quad m \in \mathbb{N} \backslash \{ 1\}, \\ \varpi (\lambda_{1},n): = \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n}) \frac{\mu_{m + 1}\ln^{\lambda_{2}}\beta V_{n}}{U_{m}\ln^{1 - \lambda_{1}}\alpha U_{m}},\quad n \in \mathbb{N}\backslash \{ 1\}. \end{gathered} $$
(17)

Lemma 2

If \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing and \(U_{\infty} = V_{\infty} = \infty\), then for \(m,n \in \mathbb{N}\backslash \{ 1\}\), we have the following inequalities:
$$\begin{aligned}& \omega (\lambda_{2},m) < K_{\gamma} (\lambda_{1}), \end{aligned}$$
(18)
$$\begin{aligned}& \varpi (\lambda_{1},n) < K_{\gamma} (\lambda_{1}), \end{aligned}$$
(19)
where \(K_{\gamma} (\lambda_{1})\) is determined by (12).

Proof

For \(y \in (n - \frac{1}{2},n + \frac{1}{2})\backslash \{ n\}\), \(\upsilon_{n + 1} \le V'(y)\), by (16), we find
$$\begin{aligned} \omega (\lambda_{2},m) &< \sum _{n = 2}^{\infty} \int_{n - \frac{1}{2}}^{n + \frac{1}{2}} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr) \frac{\upsilon_{n + 1}\ln^{\lambda_{1}}\alpha U_{m}}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy \\ &\le \sum_{n = 2}^{\infty} \int_{n - \frac{1}{2}}^{n + \frac{1}{2}} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr) \frac{V'(y)\ln^{\lambda_{1}}\alpha U_{m}}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy \\ &= \int_{\frac{3}{2}}^{\infty} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr)\frac{V'(y)\ln^{\lambda_{1}}\alpha U_{m}}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy. \end{aligned} $$
Setting \(t = \frac{\ln \beta V(y)}{\ln \alpha U_{m}}\) in the above, since \(\beta V(\frac{3}{2}) = \beta (1 + \frac{\upsilon_{2}}{2}) \ge 1\) and \(\frac{V'(y)}{V(y)}\,dy = (\ln \alpha U_{m})\,dt\), we find
$$\omega (\lambda_{2},m) < \int_{0}^{\infty} k_{\lambda} (1,t)t^{\lambda_{2} - 1}\,dt = K_{\gamma} (\lambda_{1}). $$
Hence, we obtain (18). In the same way, we obtain (19). □

Note

For example, \(\mu_{n} = \upsilon_{n} = \frac{1}{n^{\sigma}}\) (\(0 \le \sigma \le 1\)) satisfies the conditions of Lemma 2.

Lemma 3

With regard to the assumptions of Lemma 2, (i) for \(m,n \in \mathbb{N}\backslash \{ 1\}\), we have
$$\begin{aligned}& K_{\gamma} (\lambda_{1}) \bigl(1 - \theta ( \lambda_{2},m)\bigr) < \omega (\lambda_{2},m), \end{aligned}$$
(20)
$$\begin{aligned}& K_{\gamma} (\lambda_{1}) \bigl(1 - \vartheta ( \lambda_{1},n)\bigr) < \varpi (\lambda_{1},n), \end{aligned}$$
(21)
where
$$\begin{aligned}& \begin{aligned}[b] \theta (\lambda_{2},m) &= \frac{k_{\lambda} (1,\frac{\ln \beta (1 + \upsilon_{2}\theta (m))}{\ln \alpha U_{m}})}{\lambda_{2}K_{\gamma} (\lambda_{1})}\frac{\ln^{\lambda_{2}}\beta (1 + \upsilon_{2})}{\ln^{\lambda_{2}}\alpha U_{m}}\\ & = O \biggl(\frac{1}{\ln^{\lambda_{2}}\alpha U_{m}}\biggr) \in (0,1)\quad \biggl(\theta (m) \in \biggl( \frac{1 - \beta}{\beta \upsilon_{2}},1\biggr)\biggr), \end{aligned} \end{aligned}$$
(22)
$$\begin{aligned}& \begin{aligned}[b] \vartheta (\lambda_{1},n) &= \frac{k_{\lambda} (\frac{\ln \alpha (1 + \mu_{2}\vartheta (n))}{\ln \beta V_{n}})}{\lambda_{1}K_{\gamma} (\lambda_{1})}\frac{\ln^{\lambda_{1}}\alpha (1 + \mu_{2})}{\ln^{\lambda_{1}}\beta V_{n}}\\ & = O \biggl(\frac{1}{\ln^{\lambda_{1}}\beta V_{n}}\biggr) \in (0,1)\quad \biggl(\vartheta (n) \in \biggl( \frac{1 - \alpha}{\alpha \mu_{2}},1\biggr)\biggr); \end{aligned} \end{aligned}$$
(23)
(ii) for any \({c}>0\), we have
$$\begin{aligned}& \sum_{m = 2}^{\infty} \frac{\mu_{m + 1}}{U_{m}\ln^{1 + c}\alpha U_{m}} = \frac{1}{c} \biggl[ \frac{1}{\ln^{c}\alpha (1 + \mu_{2})} + cO(1) \biggr], \end{aligned}$$
(24)
$$\begin{aligned}& \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}\ln^{1 + c}\beta V_{n}} = \frac{1}{c} \biggl[ \frac{1}{\ln^{c}\beta (1 + \upsilon_{2})} + c\tilde{O}(1) \biggr]. \end{aligned}$$
(25)

Proof

In view of \(\beta \le 1\) and \(\beta \ge \frac{1}{1 + \upsilon_{2}/2} > \frac{1}{1 + \upsilon_{2}}\), it follows that \(1 \le \frac{1 - \beta}{\beta \upsilon_{2}} + 1 < 2\). Since, by Examples 1, \(g(y)\) is strictly decreasing in \([n,n + 1)\), then for \(m \in \mathbb{N}\backslash \{ 1\}\), we find
$$\begin{aligned} \omega (\lambda_{2},m) &> \sum _{n = 2}^{\infty} \int_{n}^{n + 1} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr) \frac{\upsilon_{n + 1}\ln^{\lambda_{1}}\alpha U_{m}}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy \\ &= \int_{2}^{\infty} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr)\frac{V'(y)\ln^{\lambda_{1}}\alpha U_{m}}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy \\ &= \int_{\frac{1 - \beta}{\beta \upsilon_{2}} + 1}^{\infty} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr)\frac{V'(y)\ln^{\lambda_{1}}\alpha U_{m}}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy \\ &\quad {}- \int_{\frac{1 - \beta}{\beta \upsilon_{2}} + 1}^{2} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr)\frac{V'(y)\ln^{\lambda_{1}}\alpha U_{m}}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy. \end{aligned} $$
Setting \(t = \frac{\ln \beta V(y)}{\ln \alpha U_{m}}\), we have \(\ln \beta V(\frac{1 - \beta}{\beta \upsilon_{2}} + 1) = \ln \beta (1 + \frac{1 - \beta}{\beta \upsilon_{2}}\upsilon_{2}) = 0\) and
$$\begin{aligned} \omega (\lambda_{2},m) &> \int_{0}^{\infty} k_{\lambda} (1,t)t^{\lambda_{2} - 1}\,dt - \int_{\frac{1 - \beta}{\beta \upsilon_{2}} + 1}^{2} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr)\frac{V'(y)\ln^{\lambda_{1}}\alpha U_{m}}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy \\ &= K_{\gamma} (\lambda_{1}) \bigl(1 - \theta ( \lambda_{2},m)\bigr), \end{aligned} $$
where
$$\theta (\lambda_{2},m): = \frac{\ln^{\lambda_{1}}\alpha U_{m}}{K_{\gamma} (\lambda_{1})} \int_{\frac{1 - \beta}{\beta \upsilon_{2}} + 1}^{2} k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V(y)\bigr)\frac{V'(y)}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy \in (0,1). $$
In view of the integral mid-value theorem, for fixed \(m \in \mathbb{N}\backslash \{ 1\}\), there exists \(\theta (m) \in (\frac{1 - \beta}{\beta \upsilon_{2}},1)\) such that
$$\begin{aligned} \theta (\lambda_{2},m) &= \frac{\ln^{\lambda_{1}}\alpha U_{m}}{K_{\gamma} (\lambda_{1})}k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V \bigl(1 + \theta (m)\bigr)\bigr) \int_{\frac{1 - \beta}{\beta \upsilon_{2}} + 1}^{2} \frac{V'(y)}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\,dy \\ &= \frac{\ln^{\lambda_{1}}\alpha U_{m}}{\lambda_{2}K_{\gamma} (\lambda_{1})}k_{\lambda} \bigl(\ln \alpha U_{m},\ln \beta V\bigl(1 + \theta (m)\bigr)\bigr)\ln^{\lambda_{2}}\beta (1 + \upsilon_{2}) \\ &= \frac{1}{\lambda_{2}K_{\gamma} (\lambda_{1})}k_{\lambda} \biggl(1,\frac{\ln \beta V(1 + \theta (m))}{\ln \alpha U_{m}}\biggr) \frac{\ln^{\lambda_{2}}\beta (1 + \upsilon_{2})}{\ln^{\lambda_{2}}\alpha U_{m}}. \end{aligned} $$
Hence, we find
$$0 < \theta (\lambda_{2},m) \le \frac{1}{\lambda_{2}K_{\gamma} (\lambda_{1})} \frac{\ln^{\lambda_{2}}\beta (1 + \upsilon_{2})}{(1 + \gamma )\ln^{\lambda_{2}}\alpha U_{m}}, $$
namely, \(\theta (\lambda_{2},m) = O(\frac{1}{\ln^{\lambda_{2}}\alpha U_{m}})\). Then we obtain (20) and (22). In the same way, we obtain (21) and (23).
For \({c}>0\), we find
$$\begin{aligned}& \begin{aligned} \sum_{m = 2}^{\infty} \frac{\mu_{m + 1}}{U_{m}\ln^{1 + c}\alpha U_{m}} &\le \sum_{m = 2}^{\infty} \frac{\mu_{m}}{U_{m}\ln^{1 + c}\alpha U_{m}} = \frac{\mu_{2}}{U_{2}\ln^{1 + c}\alpha U_{2}} + \sum_{m = 3}^{\infty} \frac{\mu_{m}}{U_{m}\ln^{1 + c}\alpha U_{m}} \\ &= \frac{\mu_{2}}{U_{2}\ln^{1 + c}\alpha U_{2}} + \sum_{m = 3}^{\infty} \int_{m - 1}^{m} \frac{U'(x)\,dx}{U_{m}\ln^{1 + c}\alpha U_{m}} \\ &< \frac{\mu_{2}}{U_{2}\ln^{1 + c}\alpha U_{2}} + \sum_{m = 3}^{\infty} \int_{m - 1}^{m} \frac{U'(x)\,dx}{U(x)\ln^{1 + c}\alpha U(x)} \\ &= \frac{\mu_{2}}{U_{2}\ln^{1 + c}\alpha U_{2}} + \int_{2}^{\infty} \frac{U'(x)\,dx}{U(x)\ln^{1 + c}\alpha U(x)}\\ & = \frac{\mu_{2}}{U_{2}\ln^{1 + c}\alpha U_{2}} + \frac{1}{c\ln^{c}\alpha (1 + \mu_{2})} \\ &= \frac{1}{c} \biggl[ \frac{1}{\ln^{c}\alpha (1 + \mu_{2})} + c\frac{\mu_{2}}{U_{2}\ln^{1 + c}\alpha U_{2}} \biggr], \end{aligned} \\& \begin{aligned} \sum_{m = 2}^{\infty} \frac{\mu_{m + 1}}{U_{m}\ln^{1 + c}\alpha U_{m}} &= \sum_{m = 2}^{\infty} \int_{m}^{m + 1} \frac{U'(x)\,dx}{U_{m}\ln^{1 + c}\alpha U_{m}} > \sum _{m = 2}^{\infty} \int_{m}^{m + 1} \frac{U'(x)\,dx}{U(x)\ln^{1 + c}\alpha U(x)} \\ &= \int_{2}^{\infty} \frac{U'(x)\,dx}{U(x)\ln^{1 + c}\alpha U(x)} = \frac{1}{c\ln^{c}\alpha (1 + \mu_{2})}. \end{aligned} \end{aligned}$$
Hence, we obtain (20). In the same way, we obtain (21). □

Lemma 4

If \(- 1 < \gamma \le 0\), \(0 < \lambda_{1},\lambda_{2} < 1\), \(\lambda_{1} + \lambda_{2} \le 1\), \(K_{\gamma} (\lambda_{1})\) is determined by (12), then for \(0 < \delta < \min \{ \lambda_{1},\lambda_{2}\}\), we have
$$ K_{\gamma} (\lambda_{1} \pm \delta ) = K_{\gamma} ( \lambda_{1}) + o(1)\quad \bigl(\delta \to 0^{ +} \bigr). $$
(26)

Proof

We find, for \(0 < \delta < \min \{ \lambda_{1},\lambda_{2}\}\),
$$\begin{aligned} \bigl\vert K_{\gamma} (\lambda_{1} + \delta ) - K_{\gamma} (\lambda_{1})\bigr\vert &\le \int_{0}^{\infty} \frac{t^{\lambda_{1} - 1}\vert t^{\delta} - 1\vert }{t^{\lambda} + 1 + \gamma \vert t^{\lambda} - 1\vert } \,dt \\ &= \int_{0}^{1} \frac{t^{\lambda_{1} - 1}(1 - t^{\delta} )}{1 + \gamma + (1 - \gamma )t^{\lambda}} \,dt + \int_{1}^{\infty} \frac{t^{\lambda_{1} - 1}(t^{\delta} - 1)}{1 - \gamma + (1 + \gamma )t^{\lambda}} \,dt \\ &\le \frac{1}{1 + \gamma} \biggl[ \int_{0}^{1} t^{\lambda_{1} - 1}\bigl(1 - t^{\delta} \bigr) \,dt + \int_{1}^{\infty} \frac{t^{\lambda_{1} - 1}(t^{\delta} - 1)}{t^{\lambda}} \,dt \biggr] \\ &= \frac{1}{1 + \gamma} \biggl( \frac{1}{\lambda_{1}} - \frac{1}{\lambda_{1} + \delta} + \frac{1}{\lambda_{2} - \delta} - \frac{1}{\lambda_{2}} \biggr) \to 0\quad \bigl(\delta \to 0^{ +} \bigr). \end{aligned} $$
In the same way, we find
$$\begin{aligned} \bigl\vert K_{\gamma} (\lambda_{1} - \delta ) - K_{\gamma} (\lambda_{1})\bigr\vert &\le \frac{1}{1 + \gamma} \biggl[ \int_{0}^{1} t^{\lambda_{1} - 1}\bigl(t^{ - \delta} - 1\bigr) \,dt + \int_{1}^{\infty} \frac{t^{\lambda_{1} - 1}(1 - t^{ - \delta} )}{t^{\lambda}} \,dt \biggr] \\ &= \frac{1}{1 + \gamma} \biggl( \frac{1}{\lambda_{1} - \delta} - \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}} - \frac{1}{\lambda_{2} + \delta} \biggr) \to 0\quad \bigl(\delta \to 0^{ +} \bigr), \end{aligned} $$
and then we have (26). □

3 Main results

In the following, we also set
$$ \begin{gathered} \tilde{\Phi}_{\lambda} (m): = \omega (\lambda_{2},m) \biggl(\frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}( \ln \alpha U_{m})^{p(1 - \lambda_{1}) - 1}\quad \bigl(m \in \mathbb{N}\backslash \{ 1\} \bigr), \\ \tilde{\Psi}_{\lambda} (n): = \varpi (\lambda_{1},n) \biggl( \frac{V_{n}}{\upsilon_{n + 1}}\biggr)^{q - 1}(\ln \beta V_{n})^{q(1 - \lambda_{2}) - 1}\quad \bigl(n \in \mathbb{N}\backslash \{ 1\} \bigr). \end{gathered} $$
(27)

Theorem 1

  1. (i)
    For \({p} >1\), we have the following equivalent inequalities:
    $$\begin{aligned}& I: = \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m}b_{n} \le \Vert a \Vert _{p,\tilde{\Phi}_{\lambda}} \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}, \end{aligned}$$
    (28)
    $$\begin{aligned}& J: = \Biggl\{ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}\ln^{p\lambda_{2} - 1}\beta V_{n}}{(\varpi (\lambda_{1},n))^{p - 1}V_{n}} \Biggl( \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} \le \Vert a \Vert _{p,\tilde{\Phi}_{\lambda}}; \end{aligned}$$
    (29)
     
  2. (ii)

    for \(0< p<1\) (or \(p<0\)), we have the equivalent reverse of (28) and (29).

     

Proof

(i) By Hölder’s inequality with weight (cf. [20]) and (17), we have
$$\begin{aligned}& \Biggl( \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr)^{p} \\& \quad = \Biggl[ \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n}) \biggl( \frac{U_{m}^{1/q}(\ln \alpha U_{m})^{(1 - \lambda_{1})/q}\upsilon_{n + 1}^{1/p}}{(\ln \beta V_{n})^{(1 - \lambda_{2})/p}\mu_{m + 1}^{1/q}}a_{m} \biggr) \\& \qquad {}\times \biggl( \frac{(\ln \beta V_{n})^{(1 - \lambda_{2})/p}\mu_{m + 1}^{1/q}}{U_{m}^{1/q}(\ln \alpha U_{m})^{(1 - \lambda_{1})/q}\upsilon_{n + 1}^{1/p}} \biggr) \Biggr]^{p} \\& \quad \le \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n}) \frac{U_{m}^{p - 1}(\ln \alpha U_{m})^{(1 - \lambda_{1})p/q}\upsilon_{n + 1}}{(\ln \beta V_{n})^{1 - \lambda_{2}}\mu_{m + 1}^{p/q}}a_{m}^{p} \\& \qquad {}\times \Biggl[ \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n}) \frac{(\ln \beta V_{n})^{(1 - \lambda_{2})(q - 1)}\mu_{m + 1}}{U_{m}(\ln \alpha U_{m})^{1 - \lambda_{1}}\upsilon_{n + 1}^{q - 1}} \Biggr]^{p - 1} \\& \quad = \frac{(\varpi (\lambda_{1},n))^{p - 1}V_{n}}{(\ln \beta V_{n})^{p\lambda_{2} - 1}\upsilon_{n + 1}}\sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n}) \frac{\upsilon_{n + 1}U_{m}^{p - 1}(\ln \alpha U_{m})^{(1 - \lambda_{1})(p - 1)}}{(\ln \beta V_{n})^{1 - \lambda_{2}}\mu_{m + 1}^{p - 1}}a_{m}^{p}. \end{aligned}$$
(30)
Then, by (16), we find
$$ \begin{aligned}[b] J &\le \Biggl[ \sum_{n = 2}^{\infty} \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n}) \frac{\upsilon_{n + 1}U_{m}^{p - 1}(\ln \alpha U_{m})^{(1 - \lambda_{1})(p - 1)}}{(\ln \beta V_{n})^{1 - \lambda_{2}}\mu_{m + 1}^{p - 1}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &= \Biggl[ \sum_{m = 2}^{\infty} \sum _{n = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n}) \frac{\upsilon_{n + 1}U_{m}^{p - 1}(\ln \alpha U_{m})^{(1 - \lambda_{1})(p - 1)}}{(\ln \beta V_{n})^{1 - \lambda_{2}}\mu_{m + 1}^{p - 1}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &= \Biggl[ \sum_{m = 2}^{\infty} \omega ( \lambda_{2},m) \biggl(\frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}(\ln \alpha U_{m})^{p(1 - \lambda_{1}) - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}}, \end{aligned} $$
(31)
and then (29) follows.
By Hölder’s inequality (cf. [20]), we have
$$ \begin{aligned}[b] I &= \sum_{n = 2}^{\infty} \Biggl[ \frac{(\ln \beta V_{n})^{\lambda_{2} - \frac{1}{p}}\upsilon_{n + 1}^{1/p}}{(\varpi (\lambda_{1},n))^{1/q}V_{n}^{1/p}}\sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr] \biggl[ \frac{(\varpi (\lambda_{1},n))^{1/q}V_{n}^{1/p}}{(\ln \beta V_{n})^{\lambda_{2} - \frac{1}{p}}\upsilon_{n + 1}^{1/p}}b_{n} \biggr] \\ &\le J \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}.\end{aligned} $$
(32)
Then, by (29), we have (28).
On the other hand, assuming that (28) is valid, we set
$$ b_{n}: = \frac{\upsilon_{n + 1}\ln^{p\lambda_{2} - 1}\beta V_{n}}{(\varpi (\lambda_{1},n))^{p - 1}V_{n}} \Biggl( \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr)^{p - 1},\quad n \in \mathbb{N}\backslash \{ 1\}. $$
(33)
Then we find \(J^{p} = \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}^{q}\). If \(J = 0\), then (29) is trivially valid; if \(J = \infty\), then by (31), (29) takes the form of equality. Suppose that \(0 < J < \infty\). By (28), it follows that
$$\begin{aligned}& \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}^{q} = J^{p} = I \le \Vert a \Vert _{p,\tilde{\Phi}_{\lambda}} \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}, \end{aligned}$$
(34)
$$\begin{aligned}& \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}^{q - 1} = J \le \Vert a \Vert _{p,\tilde{\Phi}_{\lambda}}, \end{aligned}$$
(35)
and then (29) follows, which is equivalent to (28).

(ii) For \(0<{p}<1\) (or \({p}<0\)), by the reverse Hölder’s inequality with weight (cf. [20]) and (13), we obtain the reverse of (30) (or (30)), then we have the reverse of (31), and then the reverse of (29) follows. By Hölder’s inequality (cf. [20]), we have the reverse of (32), and then by the reverse of (29), the reverse of (28) follows.

On the other hand, assuming that the reverse of (28) is valid, we set \(b_{n}\) as (33). Then we find \(J^{p} = \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}^{q}\). If \(J = \infty\), then the reverse of (29) is trivially valid; if \(J = 0\), then by the reverse of (31), (29) takes the form of equality (=0). Suppose that \(0 < J < \infty\). By the reverse of (28), it follows that the reverses of (34) and (35) are valid, and then the reverse of (29) follows, which is equivalent to the reverse of (28). □

Theorem 2

If \({p}>1\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing, \(U_{\infty} = V_{\infty} = \infty\), \(\Vert a \Vert _{p,\Phi_{\lambda}} \in \mathbb{R}_{ +}\) and \(\Vert b \Vert _{q,\Psi_{\lambda}}^{q} \in \mathbb{R}_{ +}\), then we have the following equivalent inequalities:
$$\begin{aligned}& \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m}b_{n} < K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\Phi_{\lambda}} \Vert b \Vert _{q,\Psi_{\lambda}}, \end{aligned}$$
(36)
$$\begin{aligned}& J_{1}: = \Biggl\{ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}} \ln^{p\lambda_{2} - 1}\beta V_{n} \Biggl( \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} < K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\Phi_{\lambda}}, \end{aligned}$$
(37)
where the constant factor \(K_{\gamma} (\lambda_{1})\) is the best possible.

Proof

Using (18) and (19) in (28) and (29), we obtain the equivalent inequalities (36) and (37).

For \(\varepsilon \in (0,\min \{ p\lambda_{1},p(1 - \lambda_{2})\} )\), we set \(\tilde{\lambda}_{1} = \lambda_{1} - \frac{\varepsilon}{p}\) (\(\in (0,1)\)), \(\tilde{\lambda}_{2} = \lambda_{2} + \frac{\varepsilon}{p}\) (\(\in (0,1)\)), and
$$ \begin{gathered}[b] \tilde{a}_{m}: = \frac{\mu_{m + 1}}{U_{m}}\ln^{\tilde{\lambda}_{1} - 1}\alpha U_{m} = \frac{\mu_{m + 1}}{U_{m}} \ln^{\lambda_{1} - \frac{\varepsilon}{p} - 1}\alpha U_{m}, \\ \tilde{b}_{n}: = \frac{\upsilon_{n + 1}}{V_{n}}\ln^{\tilde{\lambda}_{2} - \varepsilon - 1}\beta V_{n} = \frac{\upsilon_{n + 1}}{V_{n}}\ln^{\lambda_{2} - \frac{\varepsilon}{q} - 1}\beta V_{n}. \end{gathered} $$
(38)
Then, by (24), (25) and (21), we have
$$\begin{aligned}& \begin{aligned} \Vert \tilde{a} \Vert _{p,\Phi_{\lambda}} \Vert \tilde{b} \Vert _{q,\Psi_{\lambda}} &= \Biggl( \sum_{m = 2}^{\infty} \frac{\mu_{m + 1}}{U_{m}\ln^{1 + \varepsilon} \alpha U_{m}} \Biggr)^{\frac{1}{p}} \Biggl( \sum _{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}\ln^{1 + \varepsilon} \beta V_{n}} \Biggr)^{\frac{1}{q}} \\ &= \frac{1}{\varepsilon} \biggl[ \frac{1}{\ln^{\varepsilon} \alpha (1 + \mu_{2})} + \varepsilon O(1) \biggr]^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \tilde{O}(1) \biggr]^{\frac{1}{q}}, \end{aligned} \\& \begin{aligned} \tilde{I}&: = \sum_{n = 2}^{\infty} \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})\tilde{a}_{m} \tilde{b}_{n} \\ &= \sum_{n = 2}^{\infty} \Biggl[ \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})\frac{\mu_{m + 1}\ln^{\tilde{\lambda}_{2}}\beta V_{n}}{U_{m}\ln^{1 - \tilde{\lambda}_{1}}\alpha U_{m}} \Biggr] \frac{\upsilon_{n + 1}}{V_{n}}\ln^{ - \varepsilon - 1}\beta V_{n} \\ &= \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}\varpi (\lambda_{1},n)}{V_{n}\ln^{\varepsilon + 1}\beta V_{n}} \ge K_{\gamma} (\tilde{\lambda}_{1})\sum _{n = 2}^{\infty} \biggl( 1 - O\biggl(\frac{1}{\ln^{\tilde{\lambda}_{1}}\beta V_{n}} \biggr) \biggr)\frac{\upsilon_{n + 1}}{V_{n}\ln^{\varepsilon + 1}\beta V_{n}} \\ &= K_{\gamma} (\tilde{\lambda}_{1}) \Biggl[ \sum _{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}\ln^{\varepsilon + 1}\beta V_{n}} - \sum _{n = 2}^{\infty} O\biggl(\frac{1}{\ln^{\lambda_{1} + \frac{\varepsilon}{q} + 1}\beta V_{n}}\biggr) \frac{\upsilon_{n + 1}}{V_{n}} \Biggr] \\ &= \frac{1}{\varepsilon} K_{\gamma} (\tilde{\lambda}_{1}) \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \bigl(\tilde{O}(1) - O(1)\bigr) \biggr]. \end{aligned} \end{aligned}$$
If there exists a positive constant \(K \le K_{\gamma} (\lambda_{1})\) such that (36) is valid when replacing \(K_{\gamma} (\lambda_{1})\) by K, then, in particular, we have \(\varepsilon \tilde{I} < \varepsilon K \Vert \tilde{a} \Vert _{p,\Phi_{\lambda}} \Vert \tilde{b} \Vert _{q,\Psi_{\lambda}}\), namely,
$$\begin{aligned}& K_{\gamma} \biggl(\lambda_{1} - \frac{\varepsilon}{p}\biggr) \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \bigl( \tilde{O}(1) - O(1)\bigr) \biggr] \\ &\quad < K \biggl[ \frac{1}{\ln^{\varepsilon} \alpha (1 + \mu_{2})} + \varepsilon O(1) \biggr]^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \tilde{O}(1) \biggr]^{\frac{1}{q}}. \end{aligned} $$
In view of (26), it follows that \(K_{\gamma} (\lambda_{1}) \le K(\varepsilon \to 0^{ +} )\). Hence, \(K = K_{\gamma} (\lambda_{1})\) is the best possible constant factor of (36).
Similarly to (32), we still can find the following inequality:
$$ I \le J_{1} \Vert b \Vert _{q,\Psi_{\lambda}}. $$
(39)
Hence, we can prove that the constant factor \(K_{\gamma} (\lambda_{1})\) in (37) is the best possible. Otherwise, we would reach the contradiction by (39) that the constant factor in (36) is not the best possible. □

Remark 1

(i) For \(\alpha = \beta = 1\) in (36) and (37), setting
$$\begin{gathered} \phi_{\lambda} (m): = \biggl(\frac{U_{m}}{\mu_{m + 1}} \biggr)^{p - 1}(\ln U_{m})^{p(1 - \lambda_{1}) - 1}, \\ \psi_{\lambda} (n): = \biggl(\frac{V_{n}}{\upsilon_{n + 1}}\biggr)^{q - 1}(\ln V_{n})^{q(1 - \lambda_{2}) - 1}\quad \bigl(m,n \in \mathbb{N}\backslash \{ 1\} \bigr), \end{gathered} $$
we have the following equivalent Mulholland-type inequalities:
$$\begin{aligned}& \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \frac{a_{m}b_{n}}{\ln^{\lambda} U_{m} + \ln^{\lambda} V_{n} + \gamma \vert \ln^{\lambda} U_{m} - \ln^{\lambda} V_{n}\vert } < K_{\gamma} ( \lambda_{1}) \Vert a \Vert _{p,\phi_{\lambda}} \Vert b \Vert _{q,\psi_{\lambda}}, \end{aligned}$$
(40)
$$\begin{aligned}& \begin{aligned}[b] &\Biggl[ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}}( \ln V_{n})^{p\lambda_{2} - 1} \Biggl( \sum_{m = 2}^{\infty} \frac{a_{m}}{\ln^{\lambda} U_{m} + \ln^{\lambda} V_{n} + \gamma \vert \ln^{\lambda} U_{m} - \ln^{\lambda} V_{n} \vert } \Biggr)^{p} \Biggr]^{\frac{1}{p}} \\ & \quad < K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\phi_{\lambda}}.\end{aligned} \end{aligned}$$
(41)
(40) is an extension of (7) and the following inequality (for \(\lambda = 1\), \(\lambda_{1} = \frac{1}{q}\), \(\lambda _{2} = \frac{1}{p}\), \(\gamma = 0\)):
$$ \Biggl[ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}}( \ln V_{n})^{p\lambda_{2} - 1} \Biggl( \sum_{m = 2}^{\infty} \frac{a_{m}}{\ln U_{m}V_{n}} \Biggr)^{p} \Biggr]^{\frac{1}{p}} < \frac{\pi}{\sin (\frac{\pi}{p})} \Biggl[ \sum_{m = 2}^{\infty} \biggl(\frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}}. $$
(42)
(ii) For \(\lambda = 1\), \(\lambda_{1} = \frac{1}{q}\), \(\lambda _{2} = \frac{1}{p}\) in (36) and (37), we have the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b] &\sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \frac{a_{m}b_{n}}{\ln (\alpha \beta U_{m}V_{n}) + \gamma \vert \ln \frac{\alpha U_{m}}{\beta V_{n}} \vert } \\ &\quad < K_{1,\gamma} \biggl(\frac{1}{q}\biggr) \Biggl[ \sum_{m = 2}^{\infty} \biggl(\frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{n = 2}^{\infty} \biggl(\frac{V_{n}}{\upsilon_{n + 1}}\biggr)^{q - 1}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}$$
(43)
$$\begin{aligned}& \begin{aligned}[b] &\Biggl\{ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}} \Biggl[ \sum_{m = 2}^{\infty} \frac{a_{m}}{\ln (\alpha \beta U_{m}V_{n}) + \gamma \vert \ln \frac{\alpha U_{m}}{\beta V_{n}}\vert } \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\ &\quad < K_{1,\gamma} \biggl( \frac{1}{q}\biggr) \Biggl[ \sum_{m = 2}^{\infty} \biggl(\frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}},\end{aligned} \end{aligned}$$
(44)
where
$$ \begin{aligned}[b] K_{1,\gamma} \biggl(\frac{1}{q} \biggr)&: = \int_{0}^{1} \frac{t^{\frac{ - 1}{p} - 1} + t^{\frac{ - 1}{q} - 1}}{1 + \gamma + (1 - \gamma )t} \,dt = \frac{1}{1 + \gamma} \biggl[ \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{1}{q}} + \biggl(\frac{1 + \gamma}{1 - \gamma} \biggr)^{\frac{1}{p}} \biggr]\frac{\pi}{\sin (\frac{\pi}{p})} \\ &\quad {}- \frac{1}{1 + \gamma} \sum_{k = 0}^{\infty} ( - 1)^{k} \biggl(\frac{1 + \gamma}{ 1 - \gamma} \biggr)^{k + 1} \biggl( \frac{1}{k + \frac{1}{p}} + \frac{1}{k + \frac{1}{q}} \biggr). \end{aligned} $$
(45)
(iii) For \(\gamma = 0\), (43) reduces to the following more accurate Hardy-Mulholland-type inequality (7):
$$ \sum_{m = 2}^{\infty} \sum _{n = 2}^{\infty} \frac{a_{m}b{}_{n}}{\ln (\alpha \beta U_{m}V_{n})} < \frac{\pi}{\sin (\frac{\pi}{p})} \Biggl[ \sum_{m = 2}^{\infty} \biggl( \frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{n = 2}^{\infty} \biggl(\frac{V_{n}}{\upsilon_{n + 1}}\biggr)^{q - 1}b_{n}^{q} \Biggr]^{\frac{1}{q}}. $$
(46)
In particular, for \(\mu_{i} = \upsilon_{j} = 1\) (\(i,j \in \mathbb{N}\)), (46) reduces to the following more accurate Mulholland’s inequality (\(\frac{2}{3} \le \alpha,\beta \le 1\)):
$$ \sum_{m = 2}^{\infty} \sum _{n = 2}^{\infty} \frac{a_{m}b{}_{n}}{\ln (\alpha \beta mn)} < \frac{\pi}{\sin (\frac{\pi}{p})} \Biggl( \sum_{m = 2}^{\infty} m^{p - 1}a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl( \sum_{n = 2}^{\infty} n^{q - 1}b_{n}^{q} \Biggr)^{\frac{1}{q}}. $$
(47)
For \(p > 1\), \(\Psi_{\lambda}^{1 - p}(n) = \frac{\upsilon_{n + 1}}{V_{n}}(\ln \beta V_{n})^{p\lambda_{2} - 1}\), we define the following normed spaces:
$$\begin{gathered} l_{p,\Phi_{\lambda}}: = \bigl\{ a = \{ a_{m} \}_{m = 2}^{\infty}; \Vert a \Vert _{p,\Phi_{\lambda}} < \infty \bigr\} , \\ l_{q,\Psi_{\lambda}}: = \bigl\{ b = \{ b_{n}\}_{n = 2}^{\infty}; \Vert b \Vert _{q,\Psi_{\lambda}} < \infty \bigr\} , \\ l_{p,\Psi_{\lambda}^{1 - p}}: = \bigl\{ c = \{ c_{n}\}_{n = 2}^{\infty}; \Vert c \Vert _{p,\Psi_{\lambda}^{1 - p}} < \infty \bigr\} . \end{gathered} $$
Assuming that \(a = \{ a_{m}\}_{m = 2}^{\infty} \in l_{p,\Phi_{\lambda}}\), setting
$$c = \{ c_{n}\}_{n = 2}^{\infty},\quad c_{n}: = \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m},n \in \mathbb{N} \backslash \{ 1\}, $$
we can rewrite (37) as follows:
$$\Vert c \Vert _{p,\Psi_{\lambda}^{1 - p}} < K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\Phi_{\lambda}} < \infty, $$
namely, \(c \in l_{p,\Psi_{\lambda}^{1 - p}}\).

Definition 2

Define a Hardy-Mulholland-type operator \(T:l_{p,\Phi_{\lambda}} \to l_{p,\Psi_{\lambda}^{1 - p}}\) as follows: For any \(a = \{ a_{m}\}_{m = 2}^{\infty} \in l_{p,\Phi_{\lambda}}\), there exists a unique representation \(Ta = c \in l_{p,\Psi_{\lambda}^{1 - p}}\). Define the formal inner product of Ta and \(b = \{ b_{n}\}_{n = 2}^{\infty} \in l_{q,\Psi_{\lambda}}\) as follows:
$$ (Ta,b): = \sum_{n = 2}^{\infty} \Biggl( \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr) b_{n}. $$
(48)
Then we can rewrite (36) and (37) as follows:
$$\begin{aligned}& (Ta,b) < K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\Phi_{\lambda}} \Vert b \Vert _{q,\Psi_{\lambda}}, \end{aligned}$$
(49)
$$\begin{aligned}& \Vert Ta \Vert _{p,\Psi_{\lambda}^{1 - p}} < K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\Phi_{\lambda}}. \end{aligned}$$
(50)
Define the norm of operator T as follows:
$$\Vert T \Vert : = \sup_{a( \ne \theta ) \in l_{p,\Phi_{\lambda}}} \frac{ \Vert Ta \Vert _{p,\Psi_{\lambda}^{1 - p}}}{ \Vert a \Vert _{p,\Phi_{\lambda}}}. $$
Then, by (50), we find \(\Vert T \Vert \le K_{\gamma} (\lambda_{1})\). Since the constant factor in (50) is the best possible, we have
$$ \Vert T \Vert = K_{\gamma} (\lambda_{1}) = \int_{0}^{1} \frac{t^{\lambda_{1} - 1} + t^{\lambda_{2} - 1}}{1 + \gamma + (1 - \gamma )t^{\lambda}} \,dt. $$
(51)

4 Some reverses

In the following, we also set
$$ \begin{gathered} \tilde{\Omega}_{\lambda} (m): = \bigl(1 - \theta (\lambda_{2},m)\bigr) \biggl(\frac{U_{m}}{\mu_{m + 1}} \biggr)^{p - 1}(\ln \alpha U_{m})^{p(1 - \lambda_{1}) - 1}, \\ \tilde{\Upsilon}_{\lambda} (n): = \bigl(1 - \vartheta ( \lambda_{1},n)\bigr) \biggl(\frac{V_{n}}{\upsilon_{n + 1}}\biggr)^{q - 1}( \ln \beta V_{n})^{q(1 - \lambda_{2}) - 1}\quad \bigl(m,n \in \mathbb{N}\backslash \{ 1 \} \bigr). \end{gathered} $$
(52)
For \(0 < p < 1\) or \(p < 0\), we still use the formal symbols \(\Vert a \Vert _{p,\Phi_{\lambda}}\), \(\Vert b \Vert _{q,\Psi_{\lambda}}\), \(\Vert a \Vert _{p,\tilde{\Omega}_{\lambda}}\) and \(\Vert b \Vert _{q,\tilde{\Upsilon}_{\lambda}}\) et al.

Theorem 3

If \(0 < p < 1\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing, \(U_{\infty} = V_{\infty} = \infty\), \(\Vert a \Vert _{p,\Phi_{\lambda}} \in \mathbb{R}_{ +}\) and \(\Vert b \Vert _{q,\Psi_{\lambda}}^{q} \in \mathbb{R}_{ +}\), then we have the following equivalent inequalities with the best possible constant factor \(K_{\gamma} (\lambda_{1})\):
$$\begin{aligned}& \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m}b_{n} > K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\tilde{\Omega}_{\lambda}} \Vert b \Vert _{q,\Psi_{\lambda}}, \end{aligned}$$
(53)
$$\begin{aligned}& \Biggl\{ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}} \ln^{p\lambda_{2} - 1}\beta V_{n} \Biggl( \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} > K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\tilde{\Omega}_{\lambda}}. \end{aligned}$$
(54)

Proof

Using (20) and (19) in the reverses of (28) and (29), since
$$\begin{gathered} \bigl(\omega (\lambda_{2},m) \bigr)^{\frac{1}{p}} > \bigl(K_{\gamma} (\lambda_{1}) \bigr)^{\frac{1}{p}}\bigl(1 - \theta (\lambda_{2},m) \bigr)^{\frac{1}{p}}\quad (0 < p < 1), \\ \bigl(\varpi (\lambda_{1},n)\bigr)^{\frac{1}{q}} > \bigl(K_{\gamma} (\lambda_{1})\bigr)^{\frac{1}{q}}\quad (q < 0), \end{gathered} $$
and
$$\frac{1}{(K_{\gamma} (\lambda_{1}))^{p - 1}} > \frac{1}{(\varpi (\lambda_{1},n))^{p - 1}}\quad (0 < p < 1), $$
we obtain equivalent inequalities (53) and (54).
For \(\varepsilon \in (0,\min \{ p\lambda_{1},p(1 - \lambda_{2})\})\), we set \(\tilde{\lambda}_{1}\), \(\tilde{\lambda}_{2}\), \(\tilde{a}_{m}\) and \(\tilde{b}_{n}\) as (38). Then, by (24), (25) and (19), we find
$$\begin{aligned}& \Vert \tilde{a} \Vert _{p,\tilde{\Omega}_{\lambda}} \Vert \tilde{b} \Vert _{q,\Psi_{\lambda}} \\& \quad = \Biggl( \sum_{m = 2}^{\infty} \frac{(1 - \theta (\lambda_{2},m))\mu_{m + 1}}{U_{m}\ln^{1 + \varepsilon} \alpha U_{m}} \Biggr)^{\frac{1}{p}} \Biggl( \sum _{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}\ln^{1 + \varepsilon} \beta V_{n}} \Biggr)^{\frac{1}{q}} \\& \quad = \Biggl( \sum_{m = 2}^{\infty} \frac{\mu_{m + 1}}{U_{m}\ln^{1 + \varepsilon} \alpha U_{m}} - \sum_{m = 2}^{\infty} O \biggl(\frac{\mu_{m + 1}}{U_{m}\ln^{1 + (\lambda_{2} + \varepsilon )}\alpha U_{m}}\biggr) \Biggr)^{\frac{1}{p}} \Biggl( \sum _{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}\ln^{1 + \varepsilon} \beta V_{n}} \Biggr)^{\frac{1}{q}} \\& \quad = \frac{1}{\varepsilon} \biggl[ \frac{1}{\ln^{\varepsilon} \alpha (1 + \mu_{2})} + \varepsilon (O(1) - O_{1}(1) \biggr]^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \tilde{O}(1) \biggr]^{\frac{1}{q}}, \\& \begin{aligned} \tilde{I}&: = \sum_{n = 2}^{\infty} \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})\tilde{a}_{m} \tilde{b}_{n} \\ &= \sum_{n = 2}^{\infty} \Biggl[ \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})\frac{\mu_{m + 1}\ln^{\tilde{\lambda}_{2}}\beta V_{n}}{U_{m}\ln^{1 - \tilde{\lambda}_{1}}\alpha U_{m}} \Biggr] \frac{\upsilon_{n + 1}}{V_{n}}\ln^{ - \varepsilon - 1}\beta V_{n} \\ &= \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}\varpi (\tilde{\lambda}_{1},n)}{V_{n}\ln^{\varepsilon + 1}\beta V_{n}} \le K_{\gamma} (\tilde{\lambda}_{1})\sum _{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}\ln^{\varepsilon + 1}\beta V_{n}} \\ &= \frac{1}{\varepsilon} K_{\gamma} (\tilde{\lambda}_{1}) \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \tilde{O}(1) \biggr]. \end{aligned} \end{aligned}$$
If there exists a positive constant \(K \ge K_{\gamma} (\lambda_{1})\) such that (53) is valid when replacing \(K_{\gamma} (\lambda_{1})\) by K, then, in particular, we have \(\varepsilon \tilde{I} > \varepsilon K \Vert \tilde{a} \Vert _{p,\Phi_{\lambda}} \Vert \tilde{b} \Vert _{q,\Psi_{\lambda}}\), namely,
$$\begin{gathered} K_{\gamma} \biggl(\lambda_{1} - \frac{\varepsilon}{p}\biggr) \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \tilde{O}(1) \biggr] \\ \quad > K \biggl[ \frac{1}{\ln^{\varepsilon} \alpha (1 + \mu_{2})} + \varepsilon \bigl(O(1) - O_{1}(1) \bigr) \biggr]^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \tilde{O}(1) \biggr]^{\frac{1}{q}}. \end{gathered} $$
It follows that \(K_{\gamma} (\lambda_{1}) \ge K\) (\(\varepsilon \to 0^{ +} \)). Hence, \(K = K_{\gamma} (\lambda_{1})\) is the best possible constant factor of (53).

The constant factor \(K_{\gamma} (\lambda_{1})\) in (54) is still the best possible. Otherwise, we would reach the contradiction by the reverse of (39) that the constant factor in (53) is not the best possible. □

Remark 2

For \(\alpha = \beta = 1\), set
$$\begin{gathered} \tilde{\theta} (\lambda_{2},m) = \frac{k_{\lambda} (1,\frac{\ln (1 + \upsilon_{2}\theta (m))}{\ln U_{m}})}{\lambda_{2}K_{\gamma} (\lambda_{1})}\frac{\ln^{\lambda_{2}}(1 + \upsilon_{2})}{ln^{\lambda_{2}}U_{m}} = O\biggl(\frac{1}{ln^{\lambda_{2}}U_{m}}\biggr) \in (0,1)\quad \bigl(\theta (m) \in (0,1)\bigr), \\ \tilde{\phi}_{\lambda} (m): = \bigl(1 - \tilde{\theta} ( \lambda_{2},m)\bigr) \biggl(\frac{U_{m}}{\mu_{m + 1}}\biggr)^{p - 1}( \ln U_{m})^{p(1 - \lambda_{1}) - 1}. \end{gathered} $$
It is evident that (53) and (54) are extensions of the following equivalent inequalities:
$$\begin{aligned}& \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} k_{\lambda} (\ln U_{m},\ln V_{n})a_{m}b_{n} > K_{\gamma} ( \lambda_{1}) \Vert a \Vert _{p,\tilde{\phi}_{\lambda}} \Vert b \Vert _{q,\psi_{\lambda}}, \end{aligned}$$
(55)
$$\begin{aligned}& \Biggl\{ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}} \ln^{p\lambda_{2} - 1}V_{n} \Biggl( \sum_{m = 2}^{\infty} k_{\lambda} (\ln U_{m},\ln V_{n})a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} > K_{\gamma} ( \lambda_{1}) \Vert a \Vert _{p,\tilde{\phi}_{\lambda}}, \end{aligned}$$
(56)
where the constant factor \(K_{\gamma} (\lambda_{1})\) is the best possible.

Theorem 4

If \(p < 0\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing, \(U_{\infty} = V_{\infty} = \infty\), \(\Vert a \Vert _{p,\Phi_{\lambda}} \in \mathbb{R}_{ +}\) and \(\Vert b \Vert _{q,\Psi_{\lambda}}^{q} \in \mathbb{R}_{ +}\), then we have the following equivalent inequalities with the best possible constant factor \(K_{\gamma} (\lambda_{1})\):
$$\begin{aligned}& \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m}b_{n} > K_{\gamma} (\lambda_{1}) \Vert a \Vert _{p,\Phi_{\lambda}} \Vert b \Vert _{q,\tilde{\Upsilon}_{\lambda}}, \end{aligned}$$
(57)
$$\begin{aligned}& J_{2}: = \Biggl\{ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}\ln^{p\lambda_{2} - 1}\beta V_{n}}{(1 - \vartheta (\lambda_{1},n))^{p - 1}V_{n}} \Biggl( \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} > K_{\gamma} ( \lambda_{1}) \Vert a \Vert _{p,\Phi_{\lambda}}. \end{aligned}$$
(58)

Proof

Using (18) and (21) in the reverses of (28) and (29), since
$$\begin{gathered} \bigl(\omega (\lambda_{2},m) \bigr)^{\frac{1}{p}} > \bigl(K_{\gamma} (\lambda_{1}) \bigr)^{\frac{1}{p}}\quad (p < 0), \\ \bigl(\varpi (\lambda_{1},n)\bigr)^{\frac{1}{q}} > \bigl(K_{\gamma} (\lambda_{1})\bigr)^{\frac{1}{q}}\bigl(1 - \vartheta (\lambda_{1},n)\bigr)^{\frac{1}{q}}\quad (0 < q < 1), \end{gathered} $$
and
$$\biggl[\frac{1}{(K_{\gamma} (\lambda_{1}))^{p - 1}(1 - \vartheta (\lambda_{1},n))^{p - 1}}\biggr]^{\frac{1}{p}} > \biggl[\frac{1}{(\varpi (\lambda_{1},n))^{p - 1}} \biggr]^{\frac{1}{p}}\quad (p < 0), $$
we obtain equivalent inequalities (57) and (58).
For \(\varepsilon \in (0,\min \{ q\lambda_{2},q(1 - \lambda_{1})\})\), we set \(\tilde{\lambda}_{1} = \lambda_{1} + \frac{\varepsilon}{q}\) (\(\in (0,1)\)), \(\tilde{\lambda}_{2} = \lambda_{2} - \frac{\varepsilon}{q}\) (\(\in (0,1)\)), and
$$\begin{gathered} \tilde{a}_{m}: = \frac{\mu_{m + 1}}{U_{m}} \ln^{\tilde{\lambda}_{1} - \varepsilon - 1}\alpha U_{m} = \frac{\mu_{m + 1}}{U_{m}}\ln^{\lambda_{1} - \frac{\varepsilon}{p} - 1} \alpha U_{m}, \\ \tilde{b}_{n}: = \frac{\upsilon_{n + 1}}{V_{n}}\ln^{\tilde{\lambda}_{2} - \varepsilon - 1}\beta V_{n} = \frac{\upsilon_{n + 1}}{V_{n}}\ln^{\lambda_{2} - \frac{\varepsilon}{q} - 1}\beta V_{n}. \end{gathered} $$
Then, by (24), (25) and (18), we have
$$\begin{aligned}& \begin{gathered} \Vert \tilde{a} \Vert _{p,\Phi_{\lambda}} \Vert \tilde{b} \Vert _{q,\tilde{\Upsilon}_{\lambda}} \\ \quad = \Biggl( \sum_{m = 2}^{\infty} \frac{\mu_{m + 1}}{U_{m}\ln^{1 + \varepsilon} \alpha U_{m}} \Biggr)^{\frac{1}{p}} \Biggl( \sum _{n = 2}^{\infty} \frac{(1 - \vartheta (\lambda_{1},n))\upsilon_{n + 1}}{V_{n}\ln^{1 + \varepsilon} \beta V_{n}} \Biggr)^{\frac{1}{q}} \\ \quad = \Biggl( \sum_{m = 2}^{\infty} \frac{\mu_{m + 1}}{U_{m}\ln^{1 + \varepsilon} \alpha U_{m}} \Biggr)^{\frac{1}{p}} \Biggl( \sum _{n = 2}^{\infty} \frac{\upsilon_{n + 1}}{V_{n}\ln^{1 + \varepsilon} \beta V_{n}} - \sum _{n = 2}^{\infty} O\biggl(\frac{\upsilon_{n + 1}}{V_{n}\ln^{1 + \varepsilon} \beta V_{n}}\biggr) \Biggr)^{\frac{1}{q}} \\ \quad = \frac{1}{\varepsilon} \biggl[ \frac{1}{\ln^{\varepsilon} \alpha (1 + \mu_{2})} + \varepsilon O(1) \biggr]^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \bigl(\tilde{O}(1) - O_{1}(1)\bigr) \biggr]^{\frac{1}{q}}, \end{gathered} \\& \begin{aligned} \tilde{I} &= \sum_{n = 2}^{\infty} \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})\tilde{a}_{m} \tilde{b}_{n} \\ &= \sum_{m = 2}^{\infty} \Biggl[ \sum _{n = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})\frac{\upsilon_{n + 1}\ln^{\tilde{\lambda}_{1}}\alpha U_{m}}{V_{n}\ln^{1 - \tilde{\lambda}_{2}}\beta V_{n}} \Biggr] \frac{\mu_{m + 1}}{U_{m}\ln^{1 + \varepsilon} \alpha U_{m}} \\ &= \sum_{m = 2}^{\infty} \frac{\mu_{m + 1}\omega (\tilde{\lambda}_{2},m)}{U_{m}\ln^{\varepsilon + 1}\alpha U_{m}} \le K_{\gamma} (\tilde{\lambda}_{1})\sum _{m = 2}^{\infty} \frac{\mu_{m + 1}}{U_{m}\ln^{\varepsilon + 1}\alpha U_{m}} \\ &= \frac{1}{\varepsilon} K_{\gamma} (\tilde{\lambda}_{1}) \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon O(1) \biggr]. \end{aligned} \end{aligned}$$
If there exists a positive constant \(K \ge K_{\gamma} (\lambda_{1})\) such that (57) is valid when replacing \(K_{\gamma} (\lambda_{1})\) by K, then, in particular, we have \(\varepsilon \tilde{I} > \varepsilon K \Vert \tilde{a} \Vert _{p,\Phi_{\lambda}} \Vert \tilde{b} \Vert _{q,\tilde{\Upsilon}_{\lambda}}\), namely,
$$\begin{gathered} K_{\gamma} \biggl(\lambda_{1} + \frac{\varepsilon}{q}\biggr) \biggl[ \frac{1}{\ln^{\varepsilon} \alpha (1 + \mu_{2})} + \varepsilon O(1) \biggr] \\ \quad > K \biggl[ \frac{1}{\ln^{\varepsilon} \alpha (1 + \mu_{2})} + \varepsilon O(1) \biggr]^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon} \beta (1 + \upsilon_{2})} + \varepsilon \bigl(\tilde{O}(1) - O_{1}(1)\bigr) \biggr]^{\frac{1}{q}}. \end{gathered} $$
It follows that \(K_{\gamma} (\lambda_{1}) \ge K\) (\(\varepsilon \to 0^{ +} \)). Hence, \(K = K_{\gamma} (\lambda_{1})\) is the best possible constant factor of (57).
Similarly to the reverse of (32), we still can find that
$$ I \ge J_{2} \Vert b \Vert _{q,\tilde{\Upsilon}_{\lambda}}. $$
(59)
Hence, the constant factor \(K_{\gamma} (\lambda_{1})\) in (58) is still the best possible. Otherwise, we would reach the contradiction by (59) that the constant factor in (57) is not the best possible. □

Remark 3

For \(\alpha = \beta = 1\), set
$$\begin{gathered} \tilde{\vartheta} (\lambda_{1},n) = \frac{k_{\lambda} (\frac{\ln (1 + \mu_{2}\vartheta (n))}{\ln U_{m}},1)}{\lambda_{1}K_{\gamma} (\lambda_{1})}\frac{\ln^{\lambda_{1}}(1 + \mu_{2})}{ln^{\lambda_{1}}\beta V_{n}} = O\biggl(\frac{1}{ln^{\lambda_{2}}U_{m}}\biggr) \in (0,1)\quad \bigl(\vartheta (n) \in (0,1)\bigr), \\ \tilde{\psi}_{\lambda} (n): = \bigl(1 - \tilde{\vartheta} ( \lambda_{1},n)\bigr) \biggl(\frac{V_{n}}{\upsilon_{n + 1}}\biggr)^{q - 1}( \ln V_{n})^{q(1 - \lambda_{2}) - 1}. \end{gathered} $$
It is evident that (57) and (58) are extensions of the following equivalent inequalities:
$$\begin{aligned}& \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} k_{\lambda} (\ln U_{m},\ln V_{n})a_{m}b_{n} > K_{\gamma} ( \lambda_{1}) \Vert a \Vert _{p,\phi_{\lambda}} \Vert b \Vert _{q,\tilde{\psi}_{\lambda}}, \end{aligned}$$
(60)
$$\begin{aligned}& \Biggl\{ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}\ln^{p\lambda_{2} - 1}V_{n}}{V_{n}(1 - \tilde{\vartheta} (\lambda_{1},n))} \Biggl( \sum_{m = 2}^{\infty} k_{\lambda} (\ln U_{m},\ln V_{n})a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} > K_{\gamma} ( \lambda_{1}) \Vert a \Vert _{p,\phi_{\lambda}}, \end{aligned}$$
(61)
where the constant factor \(K_{\gamma} (\lambda_{1})\) is the best possible.

5 Conclusions

In this paper, by using the way of weight coefficients, the technique of real analysis, and Hermite-Hadamard’s inequality, a more accurate Hardy-Mulholland-type inequality with multi-parameters and a best possible constant factor is given by Theorems 1, 2, and the equivalent forms are considered. The equivalent reverses with the best possible constant factor are obtained by Theorems 3, 4. Moreover, the operator expressions and some particular cases are considered. The method of weight coefficients is very important, which helps us to prove the main inequalities with the best possible constant factor. The lemmas and theorems provide an extensive account of this type of inequalities.

Declarations

Acknowledgements

This work is supported by the National Natural Science Foundation (No. 61370186, No. 61640222), and Appropriative Researching Fund for Professors and Doctors, Guangdong University of Education (No. 2015ARF25). We are grateful for this help.

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.

Authors’ Affiliations

(1)
Department of Mathematics, Guangdong University of Education
(2)
Department of Computer Science, Guangdong University of Education

References

  1. Hardy, GH, Littlewood, JE, Pólya, G: Inequalities. Cambridge University Press, Cambridge (1934) MATHGoogle Scholar
  2. Yang, BC: Discrete Hilbert-Type Inequalities. Bentham Science Publishers Ltd., The United Arab Emirates (2011) Google Scholar
  3. Mulholland, HP: Some theorems on Dirichlet series with positive coefficients and related integrals. Proc. Lond. Math. Soc. 29(2), 281-292 (1929) View ArticleMATHGoogle Scholar
  4. Mitrinović, DS, Pečarić, JE, Fink, AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Acaremic, Boston (1991) View ArticleMATHGoogle Scholar
  5. Yang, BC: Hilbert-Type Integral Inequalities. Bentham Science Publishers Ltd., The United Arab Emirates (2009) Google Scholar
  6. Yang, BC: On Hilbert’s integral inequality. J. Math. Anal. Appl. 220, 778-785 (1998) MathSciNetView ArticleMATHGoogle Scholar
  7. Yang, BC: An extension of Mulholland’s inequality. Jordan J. Math. Stat. 3(3), 151-157 (2010) MATHGoogle Scholar
  8. Yang, BC: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijin (2009) Google Scholar
  9. Rassias, M, Yang, BC: On half-discrete Hilbert’s inequality. Appl. Comput. Math. 220, 75-93 (2013) MathSciNetMATHGoogle Scholar
  10. Rassias, M, Yang, BC: A multidimensional half-discrete Hilbert-type inequality and the Riemann zeta function. Appl. Comput. Math. 225, 263-277 (2013) MathSciNetMATHGoogle Scholar
  11. Huang, QL, Yang, BC: A more accurate half-discrete Hilbert inequality with a non-homogeneous kernel. J. Funct. Spaces Appl. 2013, 628250 (2013) MATHGoogle Scholar
  12. Huang, QL, Wang, AZ, Yang, BC: A more accurate half-discrete Hilbert-type inequality with a general non-homogeneous kernel and operator expressions. Math. Inequal. Appl. 17(1), 367-388 (2014) MathSciNetMATHGoogle Scholar
  13. Liu, T, Yang, BC: On a half-discrete reverse Mulholland-type inequality and extension. J. Inequal. Appl. 2014, 103 (2014) MathSciNetView ArticleMATHGoogle Scholar
  14. Rassias, M, Yang, BC: On a multidimensional half-discrete Hilbert-type inequality related to the hyperbolic cotangent function. Appl. Comput. Math. 242, 800-813 (2014) MathSciNetMATHGoogle Scholar
  15. Huang, QL, Wu, SH, Yang, BC: Parameterized Hilbert-type integral inequalities in the whole plane. Sci. World J. 2014, 169061 (2014) Google Scholar
  16. Chen, Q, Yang, BC: On a more accurate multidimensional Mulholland-type inequality. J. Inequal. Appl. 2014, 322 (2014) MathSciNetView ArticleMATHGoogle Scholar
  17. Rassias, M, Yang, BC: On a multidimensional Hilbert-type integral inequality associated to the gamma function. Appl. Comput. Math. 249, 408-418 (2014) MathSciNetMATHGoogle Scholar
  18. Rassias, M, Yang, BC: A Hilbert-type integral inequality in the whole plane related to the hyper geometric function and the beta function. J. Math. Anal. Appl. 428(2), 1286-1308 (2015) MathSciNetView ArticleMATHGoogle Scholar
  19. Yang, BC: An extension of a Hardy-Hilbert-type inequality. J. Guangdong Univ. Educ. 35(3), 1-7 (2015) Google Scholar
  20. Kuang, JC: Applied Inequalities. Shangdong Science Technic Press, Jinan (2004) Google Scholar
  21. Kuang, JC: Real and Functional Analysis (Continuation), vol. 2. Higher Education Press, Beijing (2015) Google Scholar

Copyright

© The Author(s) 2017