A Hardy-Hilbert-type inequality involving modified weight coefficients and partial sums

Xianyong Huang; Shanhe Wu; Bicheng Yang; Xianyong Huang; Shanhe Wu; Bicheng Yang

doi:10.3934/math.2022350

AIMS Mathematics

2022, Volume 7, Issue 4: 6294-6310. doi: 10.3934/math.2022350

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Research article

A Hardy-Hilbert-type inequality involving modified weight coefficients and partial sums

1.
Department of Mathematics, Guangdong University of Education, Guangzhou 510303, China
2.
Department of Mathematics, Longyan University, Longyan 364012, China
3.
Institute of Applied Mathematics, Longyan University, Longyan 364012, China

Received: 30 October 2021 Revised: 14 January 2022 Accepted: 16 January 2022 Published: 19 January 2022
MSC : 26D15, 26D10, 47A05

In this article, we construct proper weight coefficients and use them to establish a Hardy-Hilbert-type inequality involving one partial sum. Based on this inequality, the equivalent conditions of the best possible constant factor related to several parameters are discussed. We also consider the equivalent forms and the operator expressions of the obtained inequalities. At the end of the paper, we demonstrate that more new Hardy-Hilbert-type inequalities can be derived from the special cases of the present results.

Keywords:

Citation: Xianyong Huang, Shanhe Wu, Bicheng Yang. A Hardy-Hilbert-type inequality involving modified weight coefficients and partial sums[J]. AIMS Mathematics, 2022, 7(4): 6294-6310. doi: 10.3934/math.2022350

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Abstract

1. Introduction

The classical Hardy-Hilbert's inequality asserted that

$\sum\limits_{m = 1}^{\infty} \sum\limits_{n = 1}^{\infty} \frac{a_{m} b_{n}}{m+n} < \frac{\pi}{\sin (\pi / p)}\left(\sum\limits_{m = 1}^{\infty} a_{m}^{p}\right)^{\frac{1}{p}}\left(\sum\limits_{n = 1}^{\infty} b_{n}^{q}\right)^{\frac{1}{q}},$

(1)

where $p>1, \frac{1}{p}+\frac{1}{q} = 1, a_{m}, b_{n} \geq 0, 0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty ~~ { and }~~ 0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty, \frac{\pi}{\sin (\pi / p)}$ is the best possible constant factor (cf. ^[1], Theorem 315).

A sharpened inequality of (1) was included in ^[1] by Theorem 323, as follows:

$\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{m + n - 1}}} } \lt \tfrac{\pi }{{\sin (\pi /p)}}{(\sum\limits_{m = 1}^\infty {a_m^p} )^{\tfrac{1}{p}}}{(\sum\limits_{n = 1}^\infty {b_n^q} )^{\tfrac{1}{q}}} .$

(2)

In 2006, Krnić and J. Pečarić ^[2] provided an extension of (1) by introducing parameters

${\lambda _i} \in {(0, 2]^{}}(i = 1, 2), {\lambda _1} + {\lambda _2} = \lambda \in (0, 4],$ i.e.,

$\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{{{(m + n)}^\lambda }}}} } \lt B({\lambda _1}, {\lambda _2}){[\sum\limits_{m = 1}^\infty {{m^{p(1 - {\lambda _1}) - 1}}a_m^p} ]^{\tfrac{1}{p}}}{[\sum\limits_{n = 1}^\infty {{n^{q(1 - {\lambda _2}) - 1}}b_n^q} ]^{\tfrac{1}{q}}} ,$

(3)

where, the constant factor $B({\lambda _1}, {\lambda _2})$ is the best possible, and

$B(u, v) = \int_0^\infty {\tfrac{{{t^{u - 1}}}}{{{{(1 + t)}^{u + v}}}}} d{t^{}}(u, v \gt 0)$

is the beta function. For $\lambda = 1, {\lambda }_{1} = \frac{1}{q}, {\lambda }_{2} = \frac{1}{p},$ inequality (3) reduces to (1); for $p = q = 2,$ ${\lambda _1} = {\lambda _2} = \tfrac{\lambda }{2},$ (3) reduces to a generalization of Hilbert's inequality which was proved by Yang in ^[3].

Recently, by the use of inequality (3), Adiyasuren et al. ^[4] gave a Hardy-Hilbert's inequality involving partial sums, as follows:

If ${\lambda }_{i}\in \left(\mathrm{0, 1}\right]\cap (0, \lambda)\hspace{0.33em}(\lambda \in (\mathrm{0, 2}]; i = \mathrm{1, 2}), {\lambda }_{1}+{\lambda }_{2} = \lambda,$ then

$\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{{{(m + n)}^\lambda }}}} } \lt {\lambda _1}{\lambda _2}B({\lambda _1}, {\lambda _2}){(\sum\limits_{m = 1}^\infty {{m^{ - p{\lambda _1} - 1}}A_m^p} )^{\tfrac{1}{p}}}{(\sum\limits_{n = 1}^\infty {{n^{ - q{\lambda _2} - 1}}B_n^q} )^{\tfrac{1}{q}}} ,$

(4)

where the constant factor ${\lambda _1}{\lambda _2}B({\lambda _1}, {\lambda _2})$ is the best possible, and the partial sums ${A_m}: = \sum\nolimits_{i = 1}^m {{a_i}}$ and ${B_n}: = \sum\nolimits_{k = 1}^n {{b_k}}$ $(m, n \in {\text{N}} = {\text{\{ 1, 2, }} \cdots {\text{\})}}$ satisfy

$0 \lt \sum\limits_{m = 1}^\infty {{m^{ - p{\lambda _1} - 1}}A_m^p} \lt {\infty ^{}}~~an{d}~~0 \lt \sum\limits_{n = 1}^\infty {{n^{ - q{\lambda _2} - 1}}B_n^q} \lt \infty .$

(5)

Inequalities (1), (2) and the integral analogues play an important role in analysis and applications (cf. ^{[5,6,7,8,9,10,11,12,13,14,15,16,17,18]}).

In 2016, by means of the techniques of real analysis, Hong et al. ^[19] considered some equivalent statements of the extensions of (1) with the best possible constant factor related to a few parameters.

Motivated by the inequalities (2) and (4), in this paper, we establish a new Hardy-Hilbert-type inequality, which contains modified weight coefficients and partial sums. The main technical approaches are the constructing of weight coefficients and the use of Hermite-Hadamard's inequality, Euler-Maclaurin summation formula and Abel's partial summation formula. Moreover, the equivalent conditions of the best possible constant factor related to several parameters are discussed. As applications, we deal with some equivalent forms, the operator expressions and some special cases about the inequality obtained in the main result.

2. Some lemmas

In what follows, we suppose that $p > 1, \frac{1}{p}+\frac{1}{q} = 1,$ $\lambda \in (0, 2],$

${\lambda _1} \in (0, \tfrac{3}{2}] \cap (0, \lambda + 1), {\lambda _2} \in (0, \tfrac{1}{2}] \cap (0, \lambda ),$

${\hat \lambda _1}: = \tfrac{{\lambda - {\lambda _2}}}{p} + \tfrac{{{\lambda _1}}}{q}, {\hat \lambda _2}: = \tfrac{{\lambda - {\lambda _1}}}{q} + \tfrac{{{\lambda _2}}}{p},$

${\eta _i} \in {[0, \tfrac{1}{4}]^{}}(i = 1, 2), \eta : = {\eta _1} + {\eta _2}.$

We also assume that for ${a_m}, {b_n} \geqslant 0$ $(m, n \in {\text{N:}} = {\text{\{ 1, 2, }} \cdots {\text{\}), }}$ the partial sums ${B_n}: = \sum\nolimits_{k = 1}^n {{b_k}}$ satisfy ${B_n} = o({e^{t(n - {\eta _2})}})$ $(t \gt 0;n \to \infty),$ and

$0 \lt \sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{p(1 - {{\hat \lambda }_1}) - 1}}} a_m^p \lt \infty {, ^{}}0 \lt \sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - q{{\hat \lambda }_2} - 1}}} B_n^q \lt \infty .$

Lemma 1. (cf. ^[5], (2.2.3)) (i) If ${(- 1)^i}\tfrac{{{d^i}}}{{d{t^i}}}g(t) \gt 0,$ $t \in {[m, \infty)^{}}(m \in {\text{N)}}$ with ${g^{(i)}}(\infty) = 0$ $(i = 0, 1, 2, 3),$ ${P}_{i}\left(t\right), {B}_{i}(i\in N)$ are the Bernoulli functions and the Bernoulli numbers of i-order, then

$\int_m^\infty {{P_{2q - 1}}} (t)g(t)dt = - {\varepsilon _q}\tfrac{{{B_{2q}}}}{{2q}}g{(m)^{}}(0 \lt {\varepsilon _q} \lt 1;q = 1, 2, \cdots ).$

(6)

In particular, for $q = 1,$ in view of ${B_2} = \tfrac{1}{6}$ , we have

$- \tfrac{1}{{12}}g(m) \lt \int_m^\infty {{P_1}} (t)g(t)dt \lt 0;$

(7)

for $q = 2,$ in view of ${B_4} = - \tfrac{1}{{30}}$ , we have

$0 \lt \int_m^\infty {{P_3}} (t)g(t)dt \lt \tfrac{1}{{120}}g(m).$

(8)

(ii) (cf. ^[5], (2.3.2)) If $f\left(t\right)\left(>0\right)\in {C}^{3}\left[m, \infty \right), {f}^{\left(i\right)}\left(\infty \right) = 0 (i = \mathrm{0, 1}, \mathrm{2, 3})$ , then we have the following Euler-Maclaurin summation formula:

$\sum\limits_{k = m}^\infty {f(k) = \int_m^\infty {f(t)dt + \tfrac{1}{2}} } f(m) + \int_m^\infty {{P_1}} (t)f'(t)dt,$

(9)

$\int_m^\infty {{P_1}} (t)f'(t)dt = - \tfrac{1}{{12}}f'(m) + \tfrac{1}{6}\int_m^\infty {{P_3}} (t)f'''(t)dt.$

(10)

Lemma 2. Let $s \in (0, 3],$ ${s_i} \in (0, \tfrac{3}{2}] \cap (0, s),$ ${k_s}({s_i}): = B{({s_i}, s - {s_i})^{}}(i = 1, 2)$ , we define the following weight coefficient:

$\varpi ({s_2}, m): = {(m - {\eta _1})^{s - {s_2}}}{\sum\limits_{n = 1}^\infty {\tfrac{{{{(n - {\eta _2})}^{{s_2} - 1}}}}{{{{(m + n - \eta )}^s}}}} ^{}}(m \in {\text{N}}) .$

(11)

Then we have the following inequalities:

$0 \lt {k_s}({s_2})(1 - {O_1}(\tfrac{1}{{{{(m - {\eta _1})}^{{s_2}}}}})) \lt \varpi ({s_2}, m) \lt {k_s}{({s_2})^{}}(m \in {\text{N}}).$

(12)

Where

$O_1^{}(\tfrac{1}{{{{(m - {\eta _1})}^{{s_2}}}}}): = \tfrac{1}{{{k_s}({s_2})}}\int_0^{\tfrac{{1 - {\eta _2}}}{{m - \eta {}_1}}} {\tfrac{{{u^{{s_2} - 1}}}}{{{{(1 + u)}^s}}}du \gt 0} .$

Proof. For estimating the series (11), we set the following real function: For fixed $m \in {\text{N}}$ ,

$g(m, t): = {\tfrac{{{{(t - {\eta _2})}^{{s_2} - 1}}}}{{{{(m - \eta + t)}^s}}}^{}}(t \gt {\eta _2}).$

In the following we divide two cases of ${s_2} \in (0, 1) \cap (0, s)$ and ${s_2} \in [1, \tfrac{3}{2}] \cap (0, s)$ to prove (12).

(i) For ${s_2} \in (0, 1) \cap (0, s)$ , since

${( - 1)^i}{g^{(i)}}(m, t) \gt {0^{}}(t \gt {\eta _2};i = 0, 1, 2),$

by Hermite-Hadamard's inequality (cf. ^[20]), setting $u = \tfrac{{t - {\eta _2}}}{{m - {\eta _1}}}$ , we have

$\varpi ({s_2}, m) = {(m - {\eta _1})^{s - {s_2}}}\sum\limits_{n = 1}^\infty {g(m, n)} \lt {(m - {\eta _1})^{s - {s_2}}}\int_{\tfrac{1}{2}}^\infty {g(m, t)} dt$

$\begin{gathered} = {(m - {\eta _1})^{s - {s_2}}}\int_{\tfrac{1}{2}}^\infty {\tfrac{{{t^{{s_2} - 1}}}}{{{{(m - {\eta _1} + t - {\eta _2})}^s}}}} dt = \int_{\tfrac{{\tfrac{1}{2} - {\eta _2}}}{{m - {\eta _1}}}}^\infty {\tfrac{{{u^{{s_2} - 1}}}}{{{{(1 + u)}^s}}}} du \hfill \\ \leqslant \int_0^\infty {\tfrac{{{u^{{s_2} - 1}}}}{{{{(1 + u)}^s}}}} du = B({s_2}, s - {s_2}) = {k_\lambda }({s_2}). \hfill \\ \end{gathered}$

On the other hand, in view of the decreasingness property of series, setting $u = \tfrac{{t - {\eta _2}}}{{m - {\eta _1}}}$ , we obtain

$\varpi ({s_2}, m) = {(m - {\eta _1})^{s - {s_2}}}\sum\limits_{n = 1}^\infty {g(m, n)} \gt {(m - {\eta _1})^{s - {s_2}}}\int_1^\infty {g(m, t)} dt$

$\begin{gathered} = \int_{\tfrac{{1 - {\eta _2}}}{{m - {\eta _1}}}}^\infty {\tfrac{{{u^{{s_2} - 1}}}}{{{{(1 + u)}^s}}}} du = B({s_2}, s - {s_2}) - \int_0^{\tfrac{{1 - {\eta _2}}}{{m - {\eta _1}}}} {\tfrac{{{u^{{s_2} - 1}}}}{{{{(1 + u)}^s}}}} du \hfill \\ = {k_s}({s_2})(1 - {O_1}(\tfrac{1}{{{{(m - {\eta _1})}^{{s_2}}}}})) \gt 0, \hfill \\ \end{gathered}$

where

${O_1}(\tfrac{1}{{{{(m - {\eta _1})}^{{s_2}}}}}) = \tfrac{1}{{{k_s}({s_2})}}\int_0^{\tfrac{{1 - \eta {}_2}}{{m - {\eta _1}}}} {\tfrac{{{u^{{s_2} - 1}}}}{{{{(1 + u)}^s}}}} du \gt 0,$

satisfying

$0 \lt \int_0^{\tfrac{{1 - \eta {}_2}}{{m - {\eta _1}}}} {\tfrac{{{u^{{s_2} - 1}}}}{{{{(1 + u)}^s}}}} du \lt \int_0^{\tfrac{{1 - \eta {}_2}}{{m - {\eta _1}}}} {{u^{{s_2} - 1}}} du = \tfrac{1}{{{s_2}}}{(\tfrac{{1 - \eta {}_2}}{{m - {\eta _1}}})^{{s_2}}}^{}(m \in {\text{N}}).$

Hence, we obtain (12).

(ii) For ${s_2} \in [1, \tfrac{3}{2}] \cap (0, s)$ , by (9), we have

$\begin{gathered} \sum\limits_{n = 1}^\infty {g(m, n) = \int_1^\infty {g(m, t)dt + \tfrac{1}{2}} } g(m, 1) + \int_1^\infty {{P_1}(t)g'(m, t)dt} \hfill \\ \mathop {}\nolimits^{} \mathop {}\nolimits^{} \mathop {}\nolimits^{} \mathop {}\nolimits^{} = \int_{{\eta _2}}^\infty {g(m, t)dt - h(m), } \hfill \\ \end{gathered}$

where, $h(m)$ is indicated as

$\mathop {}\nolimits^{} \mathop {}\nolimits^{} h(m): = \int_{{\eta _2}}^1 {g(m, t)dt - } \tfrac{1}{2}g(m, 1) - \int_1^\infty {{P_1}(t)g'(m, t)dt.}$

We obtain $- \tfrac{1}{2}g(m, 1) = \tfrac{{ - {{(1 - {\eta _2})}^{{s_2} - 1}}}}{{2{{(m - \eta + 1)}^s}}}$ , and integrating by parts, it follows that

$\int_{\eta {}_2}^1 {g(m, t)dt} = \int_{\eta {}_2}^1 {\tfrac{{{{(t - \eta {}_2)}^{{s_2} - 1}}}}{{{{(m - \eta + t)}^s}}}dt} = \tfrac{1}{{{s_2}}}\int_{\eta {}_2}^1 {\tfrac{{d{{(t - \eta {}_2)}^{{s_2}}}}}{{{{(m - \eta + t)}^s}}}}$

$= \tfrac{1}{{{s_2}}}\tfrac{{{{(t - \eta {}_2)}^{{s_2}}}}}{{{{(m - \eta + t)}^s}}}|_{\eta {}_2}^1 + \tfrac{s}{{{s_2}}}\int_{\eta {}_2}^1 {\tfrac{{{{(t - \eta {}_2)}^{{s_2}}}dt}}{{{{(m - \eta + t)}^{s + 1}}}}} = \tfrac{1}{{{s_2}}}\tfrac{{{{(1 - \eta {}_2)}^{{s_2}}}}}{{{{(m - \eta + 1)}^s}}} + \tfrac{s}{{{s_2}({s_2} + 1)}}\int_{\eta {}_2}^1 {\tfrac{{d{{(t - \eta {}_2)}^{{s_2} + 1}}}}{{{{(m - \eta + t)}^{s + 1}}}}}$

$\begin{gathered} \gt \tfrac{1}{{{s_2}}}\tfrac{{{{(1 - {\eta _2})}^{{s_2}}}}}{{{{(m - \eta + 1)}^s}}} + \tfrac{s}{{{s_2}({s_2} + 1)}}[\tfrac{{{{(t - {\eta _2})}^{{s_2} + 1}}}}{{{{(m - \eta + t)}^{s + 1}}}}]_{{\eta _2}}^1 \hfill \\ \mathop {}\nolimits^{} \mathop {}\nolimits^{} + \tfrac{{s(s + 1)}}{{{s_2}({s_2} + 1){{(m - \eta + 1)}^{s + 2}}}}\int_{{\eta _2}}^1 {{{(t - {\eta _2})}^{{s_2} + 1}}} dt \hfill \\ \end{gathered}$

$= \tfrac{1}{{{s_2}}}\tfrac{{{{(1 - {\eta _2})}^{{s_2}}}}}{{{{(m - \eta + 1)}^s}}} + \tfrac{s}{{{s_2}({s_2} + 1)}}\tfrac{{{{(1 - {\eta _2})}^{{s_2} + 1}}}}{{{{(m - \eta + 1)}^{s + 1}}}} + \tfrac{{s(s + 1){{(1 - {\eta _2})}^{{s_2} + 2}}}}{{{s_2}({s_2} + 1)({s_2} + 2){{(m - \eta + 1)}^{s + 2}}}}.$

We find

$\begin{gathered} \mathop {}\nolimits^{} - g'(m, t) = - \tfrac{{({s_2} - 1){{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}} + \tfrac{{s{{(t - {\eta _2})}^{{s_2} - 1}}}}{{{{(m - \eta + t)}^{s + 1}}}} \hfill \\ = \tfrac{{(1 - {s_2}){{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}} + \tfrac{{s{{(t - {\eta _2})}^{{s_2} - 2}}[(m - \eta + t) - (m - {\eta _1})]}}{{{{(m - \eta + t)}^{s + 1}}}} \hfill \\ = \tfrac{{(1 - {s_2}){{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}} + \tfrac{{s{{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}} - \tfrac{{s(m - {\eta _1}){{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^{s + 1}}}} \hfill \\ = \tfrac{{(s + 1 - {s_2}){{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}} - \tfrac{{s(m - {\eta _1}){{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^{s + 1}}}}, \hfill \\ \end{gathered}$

and for ${s_2} \in [1, \tfrac{3}{2}] \cap (0, s)$ , it follows that

${( - 1)^i}\tfrac{{{d^i}}}{{d{t^i}}}[\tfrac{{{{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}}] \gt 0, {( - 1)^i}\tfrac{{{d^i}}}{{d{t^i}}}[\tfrac{{{{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^{s + 1}}}}] \gt {0^{}}(t \gt {\eta _2};i = 0, 1, 2, 3).$

By (8)–(10), for $a: = 1 - {\eta _2}(\in [\tfrac{3}{4}{, ^{}}1])$ , we obtain

$(s + 1 - {s_2})\int_1^\infty {{P_1}(t)\tfrac{{{{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}}} dt \gt - \tfrac{{s + 1 - {s_2}}}{{12{{(m - \eta + 1)}^s}}}{a^{{s_2} - 2}},$

$\begin{gathered} \mathop {}\nolimits^{} - (m - {\eta _1})s\int_1^\infty {{P_1}(t)\tfrac{{{{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^{s + 1}}}}} dt \gt \tfrac{{(m - {\eta _1})s{a^{{s_2} - 2}}}}{{12{{(m - \eta + 1)}^{s + 1}}}} - \tfrac{{(m - {\eta _1})s}}{{720}}{[\tfrac{{{{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^{s + 1}}}}]''_{t = 1}} \hfill \\ \gt \tfrac{{(m - \eta + 1)s - as}}{{12{{(m - \eta + 1)}^{s + 1}}}}{a^{{s_2} - 2}} \hfill \\ \mathop {}\nolimits^{} \mathop {}\nolimits^{} - \tfrac{{(m - \eta + 1)s}}{{720}}[\tfrac{{(s + 1)(s + 2)}}{{{{(m - \eta + 1)}^{s + 3}}}}{a^{{s_2} - 2}} + \tfrac{{2(s + 1)(2 - {s_2})}}{{{{(m - \eta + 1)}^{s + 2}}}}{a^{{s_2} - 3}} + \tfrac{{(2 - {s_2})(3 - {s_2})}}{{{{(m - \eta + 1)}^{s + 1}}}}{a^{{s_2} - 4}}] \hfill \\ = \tfrac{{s{a^{{s_2} - 2}}}}{{12{{(m - \eta + 1)}^s}}} - \tfrac{{s{a^{{s_2} - 1}}}}{{12{{(m - \eta + 1)}^{s + 1}}}} \hfill \\ \mathop {}\nolimits^{} \mathop {}\nolimits^{} - \tfrac{s}{{720}}[\tfrac{{(s + 1)(s + 2)}}{{{{(m - \eta + 1)}^{s + 2}}}}{a^{{s_2} - 2}} + \tfrac{{2(s + 1)(2 - {s_2})}}{{{{(m - \eta + 1)}^{s + 1}}}}{a^{{s_2} - 3}} + \tfrac{{(2 - {s_2})(3 - {s_2})}}{{{{(m - \eta + 1)}^s}}}{a^{{s_2} - 4}}], \hfill \\ \end{gathered}$

and then we have

$h(m) \gt \tfrac{{{a^{{s_2} - 4}}}}{{{{(m - \eta + 1)}^s}}}{h_1} + \tfrac{{s{a^{{s_2} - 3}}}}{{{{(m - \eta + 1)}^{s + 1}}}}{h_2} + \tfrac{{s(s + 1){a^{{s_2} - 2}}}}{{{{(m - \eta + 1)}^{s + 2}}}}{h_3},$

where, ${h_i}^{}(i = 1, 2, 3)$ are indicated as

$\begin{gathered} {h_1}: = \tfrac{{{a^4}}}{{{s_2}}} - \tfrac{{{a^3}}}{2} - \tfrac{{(1 - {s_2}){a^2}}}{{12}} - \tfrac{{s(2 - {s_2})(3 - {s_2})}}{{720}}, \hfill \\ {h_2}: = \tfrac{{{a^4}}}{{{s_2}({s_2} + 1)}} - \tfrac{{{a^2}}}{{12}} - \tfrac{{(s + 1)(2 - {s_2})}}{{360}}{, ^{}}and \hfill \\ \end{gathered}$

${h_3}: = \tfrac{{{a^4}}}{{{s_2}({s_2} + 1)({s_2} + 2)}} - \tfrac{{s + 2}}{{720}}.$

For $s \in (0, 3], {s_2} \in [1, \tfrac{3}{2}] \cap (0, s)$ , $a \in [\tfrac{3}{4}, 1]$ , we find

${h_1} \gt \tfrac{{{a^2}}}{{12{s_2}}}[s_2^2 - (6a + 1){s_2} + 12{a^2}] - \tfrac{1}{{120}}.$

In view of

$\tfrac{d}{{da}}[s_2^2 - (6a + 1){s_2} + 12{a^2}] = 6(4a - {s_2}) \geqslant 6(4 \cdot \tfrac{3}{4} - \tfrac{3}{2}) \gt 0,$

and

$\mathop {}\limits^{} \tfrac{d}{{d{s_2}}}[s_2^2 - (6a + 1){s_2} + 12{a^2}] = 2{s_2} - (6a + 1) \leqslant 2 \cdot \tfrac{3}{2} - (6 \cdot \tfrac{3}{4} + 1) = 3 - \tfrac{{11}}{2} \lt 0,$

we obtain

${h_1} \geqslant \tfrac{{{{(3/4)}^2}}}{{12(3/2)}}[(\tfrac{3}{2})_{}^2 - (6 \cdot \tfrac{3}{4} + 1)\tfrac{3}{2} + 12{(\tfrac{3}{4})^2}] - \tfrac{1}{{120}} = \tfrac{3}{{128}} - \tfrac{1}{{120}} \gt 0,$

${h_2} \gt {a^2}(\tfrac{{4{a^2}}}{{15}} - \tfrac{1}{{12}}) - \tfrac{1}{{90}} \geqslant {(\tfrac{3}{4})^2}[\tfrac{4}{{15}}{(\tfrac{3}{4})^2} - \tfrac{1}{{12}}] - \tfrac{1}{{90}} = \tfrac{3}{{80}} - \tfrac{1}{{90}} \gt 0,$

${h_3} \geqslant \tfrac{{8{a^4}}}{{105}} - \tfrac{5}{{720}} \geqslant \tfrac{8}{{105}}{(\tfrac{3}{4})^4} - \tfrac{1}{{144}} = \tfrac{{27}}{{1120}} - \tfrac{1}{{144}} \gt 0,$

and then $h(m) \gt 0.$

On the other hand, we also have

$\begin{gathered} \sum\limits_{n = 1}^\infty {g(m, n) = \int_1^\infty {g(m, t)dt + \tfrac{1}{2}} } g(m, 1) + \int_1^\infty {{P_1}(t)g'(m, t)dt} \hfill \\ \mathop {}\nolimits^{} \mathop {}\nolimits^{} \mathop {}\nolimits^{} \mathop {}\nolimits^{} = \int_1^\infty {g(m, t)dt + H(m), } \hfill \\ \end{gathered}$

where, $H(m)$ is indicated as

$H(m): = \tfrac{1}{2}g(m, 1) + \int_1^\infty {{P_1}(t)g'(m, t)dt.}$

We have obtained that

$\tfrac{1}{2}g(m, 1) = \tfrac{{{a^{{s_2} - 1}}}}{{2{{(m - \eta + 1)}^s}}}$

and

$g'(m, t) = - \tfrac{{(s + 1 - {s_2}){{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}} + \tfrac{{s(m - {\eta _1}){{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^{s + 1}}}}.$

For ${s}_{2}\in \left[1, \frac{3}{2}\right]\cap \left(0, s\right), 0 < s\le 3$ , by (7), we obtain

$\mathop {}\nolimits^{} - (s + 1 - {s_2})\int_1^\infty {{P_1}(t)\tfrac{{{{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^s}}}} dt \gt 0,$

$\begin{gathered} \mathop {}\nolimits^{} (m - {\eta _1})s\int_1^\infty {{P_1}(t)\tfrac{{{{(t - {\eta _2})}^{{s_2} - 2}}}}{{{{(m - \eta + t)}^{s + 1}}}}} dt \gt \tfrac{{ - (m - {\eta _1})s}}{{12{{(m - \eta + 1)}^{s + 1}}}}{a^{{s_2} - 2}} = \tfrac{{ - (m - \eta + 1)s + as}}{{12{{(m - \eta + 1)}^{s + 1}}}}{a^{{s_2} - 2}} \hfill \\ = \tfrac{{ - s}}{{12{{(m - \eta + 1)}^s}}}{a^{{s_2} - 2}} + \tfrac{s}{{12{{(m - \eta + 1)}^{s + 1}}}}{a^{{s_2} - 1}} \gt \tfrac{{ - s}}{{12{{(m - \eta + 1)}^s}}}{a^{{s_2} - 2}}. \hfill \\ \end{gathered}$

Hence, we have

$\begin{gathered} H(m) \gt \tfrac{{{a^{{s_2} - 1}}}}{{2{{(m - \eta + 1)}^s}}} - \tfrac{{s{a^{{s_2} - 2}}}}{{12{{(m - \eta + 1)}^s}}} = (\tfrac{a}{2} - \tfrac{s}{{12}})\tfrac{{{a^{{s_2} - 2}}}}{{{{(m - \eta + 1)}^s}}} \hfill \\ \mathop {}\nolimits^{} \mathop {}\nolimits^{} \geqslant (\tfrac{1}{2} \cdot \tfrac{3}{4} - \tfrac{3}{{12}})\tfrac{{{a^{{s_2} - 2}}}}{{{{(m - \eta + 1)}^s}}} = (\tfrac{3}{8} - \tfrac{3}{{12}})\tfrac{{{a^{{s_2} - 2}}}}{{{{(m - \eta + 1)}^s}}} \gt 0. \hfill \\ \end{gathered}$

Therefore, we obtain the following inequalities:

$\int_1^\infty {g(m, t)dt \lt } \sum\limits_{n = 1}^\infty {g(m, n) \lt } \int_{{\eta _2}}^\infty {g(m, t)dt} .$

In view of the the results in case (i), we have

$\varpi ({s_2}, m) > {(m - {\eta _1})^{s - {s_2}}}\int_1^\infty {g(m, t)dt} = {k_s}({s_2})(1 - {O_1}(\tfrac{1}{{{{(m - {\eta _1})}^{{s_2}}}}})) \gt 0,$

$\varpi ({s_2}, m) < {(m - {\eta _1})^{s - {s_2}}}\int_{{\eta _2}}^\infty {g(m, t)dt} \leqslant B({s_2}, s - {s_2}) = {k_\lambda }({s_2}).$

Hence, we obtain (12). The Lemma 2 is proved.

Lemma 3. Let $s \in (0, 3],$ ${s}_{i}\in \left(0, \frac{3}{2}\right]\cap \left(0, s\right)\left(i = \mathrm{1, 2}\right).$ Then we have the following Hardy-Hilbert-type inequality:

$I = \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{{{(m + n - \eta )}^s}}}} } \leqslant {(k_s^{}({s_2}))^{\tfrac{1}{p}}}{(k_s^{}({s_1}))^{\tfrac{1}{q}}}\\ \times {\{ \sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{p[1 - (\tfrac{{s - {s_2}}}{p} + \tfrac{{{s_1}}}{q})] - 1}}} a_m^p\} ^{\tfrac{1}{p}}}{\{ \sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{q[1 - (\tfrac{{s - {s_1}}}{q} + \tfrac{{{s_2}}}{p})] - 1}}} b_n^q\} ^{\tfrac{1}{q}}} .$

(13)

Proof. In the same way as the proof of inequality (12) along with the symmetry of the parameters ${s}_{1}, n, {\eta }_{2}$ and ${s}_{2}, m, {\eta }_{1}$ , we obtain the following inequalities for the next weight coefficient:

$0 \lt {k_s}({s_1})(1 - {O_2}(\tfrac{1}{{{{(n - {\eta _2})}^{{s_1}}}}}))$

$\lt \omega ({s_1}, n): = {(n - {\eta _2})^{s - {s_1}}}\sum\limits_{m = 1}^\infty {\tfrac{{{{(m - {\eta _1})}^{{s_1} - 1}}}}{{{{(m + n - \eta )}^s}}}} \lt {k_s}{({s_1})^{}}(n \in {\text{N)}},$

(14)

where

${O_2}(\tfrac{1}{{{{(n - {\eta _2})}^{{s_1}}}}}): = \tfrac{1}{{{k_s}({s_1})}}\int_0^{\tfrac{{1 - \eta {}_1}}{{n - {\eta _2}}}} {\tfrac{{{u^{{s_1} - 1}}}}{{{{(1 + u)}^s}}}} du \gt 0.$

By Hӧlder's inequality (cf. ^[20]), we obtain

$\begin{gathered} I = \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\tfrac{1}{{{{(m + n - \eta )}^s}}}} } [\tfrac{{{{(m - {\eta _1})}^{(1 - s{}_1)/q}}}}{{{{(n - {\eta _2})}^{(1 - {s_2})/p}}}}{a_m}][\tfrac{{{{(n - {\eta _2})}^{(1 - {s_2})/p}}}}{{{{(m - {\eta _1})}^{(1 - s{}_1)/q}}}}{b_n}] \\ \leqslant {[\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{1}{{{{(m + n - \eta )}^s}}}} } \tfrac{{{{(m - {\eta _1})}^{(1 - s{}_1)(p - 1)}}a_m^p}}{{{{(n - {\eta _2})}^{1 - {s_2}}}}}]^{\tfrac{1}{p}}}{[\sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\tfrac{1}{{{{(m + n - \eta )}^s}}}} } \tfrac{{{{(n - {\eta _2})}^{(1 - {s_2})(q - 1)}}b_n^q}}{{{{(m - {\eta _1})}^{1 - s{}_1}}}}]^{\tfrac{1}{q}}} \hfill \\ \end{gathered} \\ \begin{gathered} = {\{ \sum\limits_{m = 1}^\infty {\varpi ({s_2}, m)} {(m - {\eta _1})^{p[1 - (\tfrac{{s - {s_2}}}{p} + \tfrac{{s{}_1}}{q})] - 1}}a_m^p\} ^{\tfrac{1}{p}}} \hfill \\ \mathop {}\nolimits^{} \mathop {}\nolimits^{} \mathop {}\nolimits^{} \times {\{ \sum\limits_{n = 1}^\infty {\omega (s{}_1, n)} {(n - {\eta _2})^{q[1 - (\tfrac{{s - {s_1}}}{q} + \tfrac{{{s_2}}}{p})] - 1}}b_n^q\} ^{\tfrac{1}{q}}}. \hfill \\ \end{gathered}$

Then by (12) and (14), we obtain (13). The proof Lemma 3 is complete.

Remark 1. In particular, for

$s = \lambda +1\in \left(\mathrm{1, 3}\right], {s}_{1} = {\lambda }_{1}\in \left(0, \frac{3}{2}\right]\cap \left(0, \lambda +1\right), {s}_{2} = {\lambda }_{2}+1\in \left(1, \frac{3}{2}\right]\cap \left(1, \lambda +1\right),$

in (13), then

$\lambda \in \left(\mathrm{0, 2}\right], {\lambda }_{1}\in (0, \frac{3}{2}]\cap (0, \lambda +1), {\lambda }_{2}\in (0, \frac{1}{2}]\cap (0, \lambda ) ,$

replacing ${b_n}$ by ${B_n}$ , in view of the assumptions of ${a}_{m}~~and~~{B}_{n}$ , we have

$\sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\tfrac{{{a_m}{B_n}}}{{{{(m + n - \eta )}^{\lambda + 1}}}}} } \lt {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}$

$\times {[\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{p(1 - {{\hat \lambda }_1}) - 1}}} a_m^p]^{\tfrac{1}{p}}}{[\sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - q{{\hat \lambda }_2} - 1}}} B_n^q]^{\tfrac{1}{q}}} .$

(15)

Lemma 4. If $t \gt 0$ , then the following inequality holds

$\sum\limits_{n = 1}^\infty {{e^{ - t(n - {\eta _2})}}} {b_n} \leqslant t\sum\limits_{n = 1}^\infty {{e^{ - t(n - {\eta _2})}}} {B_n}.$

(16)

Proof. In view of ${B_n}{e^{ - t(n - {\eta _2})}} = o{(1)^{}}(n \to \infty)$ , by Abel's summation by parts formula, we find

$\sum\limits_{n = 1}^\infty {{e^{ - t(n - {\eta _2})}}} {b_n} = \mathop {\lim }\limits_{n \to \infty } {B_n}{e^{ - t(n - {\eta _2})}} + \sum\limits_{n = 1}^\infty {{B_n}} [{e^{ - t(n - {\eta _2})}} - {e^{ - t(n - {\eta _2} + 1)}}]$

$= \sum\limits_{n = 1}^\infty {{B_n}} [{e^{ - t(n - {\eta _2})}} - {e^{ - t(n - {\eta _2} + 1)}}] = (1 - {e^{ - t}})\sum\limits_{n = 1}^\infty {{e^{ - t(n - {\eta _2})}}} {B_n}.$

(17)

Since $1 - {e^{ - t}} \lt {t^{}}(t \gt 0)$ , by (17), we have inequality

$\sum\limits_{n = 1}^\infty {{e^{ - t(n - {\eta _2})}}} {b_n} \leqslant t\sum\limits_{n = 1}^\infty {{e^{ - t(n - {\eta _2})}}} {B_n},$

namely, (16) follows. The Lemma 4 is proved.

3. Main results

Theorem 1. Let $p > 1, \frac{1}{p}+\frac{1}{q} = 1,$ $\lambda \in (0, 2],$ ${\lambda _1} \in (0, \tfrac{3}{2}] \cap (0, \lambda + 1),$ ${\lambda _2} \in (0, \tfrac{1}{2}] \cap (0, \lambda),$ ${\hat \lambda _1}: = \tfrac{{\lambda - {\lambda _2}}}{p} + \tfrac{{{\lambda _1}}}{q},$ ${\hat \lambda _2}: = \tfrac{{\lambda - {\lambda _1}}}{q} + \tfrac{{{\lambda _2}}}{p},$ ${\eta _i} \in {[0, \tfrac{1}{4}]^{}}(i = 1, 2),$ $\eta = {\eta _1} + {\eta _2},$ ${B_n} = \sum\nolimits_{k = 1}^n {{b_k}}$ , ${B_n} = o({e^{t(n - {\eta _2})}})$ $(t > 0;n \to \infty),$ and let

$0 < \sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{p(1 - {{\hat \lambda }_1}) - 1}}} a_m^p < \infty {, ^{}}0 < \sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - q{{\hat \lambda }_2} - 1}}} B_n^q < \infty .$

Then, we have the following Hardy-Hilbert-type inequality:

$I: = \sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{{{(m + n - \eta )}^\lambda }}}} } \lt \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}} {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}$

(18)

In particular, for ${\lambda _1} + {\lambda _2} = \lambda$ ( ${\lambda _1} \in (0, \tfrac{3}{2}] \cap (0, \lambda), {\lambda _2} \in (0, \tfrac{1}{2}] \cap (0, \lambda)$ ), then we have

$\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{{{(m + n - \eta )}^\lambda }}} \lt } } {\lambda _2}B({\lambda _1}, {\lambda _2}) {[\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{p(1 - {\lambda _1}) - 1}}} a_m^p]^{\tfrac{1}{p}}}{[\sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - q{\lambda _2} - 1}}} B_n^q]^{\tfrac{1}{q}}} .$

(19)

Proof. In view of the formula that

$\tfrac{1}{{{{(m + n - \eta )}^\lambda }}} = \tfrac{1}{{\Gamma (\lambda )}}\int_0^\infty {{t^{\lambda - 1}}} {e^{ - (m + n - \eta )t}}dt ,$

by (16), it follows that

$\begin{gathered} I = \tfrac{1}{{\Gamma (\lambda )}}\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {{a_m}{b_n}\int_0^\infty {{t^{\lambda - 1}}} {e^{ - (m + n - \eta )t}}dt} } \hfill \\ \mathop {}\limits^{} = \tfrac{1}{{\Gamma (\lambda )}}\int_0^\infty {{t^{\lambda - 1}}} \sum\limits_{m = 1}^\infty {{e^{ - (m - {\eta _1})t}}{a_m}} \sum\limits_{n = 1}^\infty {{e^{ - (n - {\eta _2})t}}{b_n}} dt \hfill \\ \mathop {}\limits^{} \leqslant \tfrac{1}{{\Gamma (\lambda )}}\int_0^\infty {{t^1}} \sum\limits_{m = 1}^\infty {{e^{ - (m - {\eta _1})t}}{a_m}} \sum\limits_{n = 1}^\infty {{e^{ - (n - {\eta _2})t}}{B_n}} dt \hfill \\ \end{gathered}$

$\begin{gathered} \mathop {}\limits^{} = \tfrac{1}{{\Gamma (\lambda )}}\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {{a_m}{B_n}} } \int_0^\infty {{t^{(\lambda + 1) - 1}}} {e^{ - (m + n - \eta )t}}dt \hfill \\ \mathop {}\limits^{} = \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}}\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{B_n}}}{{{{(m + n - \eta )}^{\lambda + 1}}}}} } . \hfill \\ \end{gathered}$

Then by (15), we have (18).

For ${\lambda _1} + {\lambda _2} = {\lambda ^{}}{(\in (0, 2])^{}}({\lambda _1} \in (0, \tfrac{3}{2}] \cap (0, \lambda), {\lambda _2} \in (0, \tfrac{1}{2}] \cap (0, \lambda)),$ we have

$\begin{gathered} \mathop {}\nolimits^{} k_{\lambda + 1}^{}({\lambda _2} + 1) = k_{\lambda + 1}^{}({\lambda _1}) = B({\lambda _1}, {\lambda _2} + 1) \hfill \\ = \tfrac{{\Gamma ({\lambda _1})\Gamma ({\lambda _2} + 1)}}{{\Gamma (\lambda + 1)}} = \tfrac{{{\lambda _2}\Gamma ({\lambda _1})\Gamma ({\lambda _2})}}{{\Gamma (\lambda + 1)}} = \tfrac{{{\lambda _2}\Gamma (\lambda )}}{{\Gamma (\lambda + 1)}}B({\lambda _1}, {\lambda _2}), \hfill \\ \end{gathered}$

inequality (18) reduces to (19). The Theorem 1 is proved.

Theorem 2. Suppose that $\lambda \in (0, 2], {\lambda _1} \in (0, \tfrac{3}{2}] \cap (0, \lambda), {\lambda _2} \in (0, \tfrac{1}{2}] \cap (0, \lambda).$ If ${\lambda _1} + {\lambda _2} = \lambda,$ then the constant factor

$\tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}} {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}$

in (18) is the best possible. On the other hand, if the same constant factor in (18) is the best possible, then for $\lambda - {\lambda _1} \leqslant \tfrac{1}{2}, \lambda - {\lambda _2} \leqslant \tfrac{3}{2},$ we have ${\lambda _1} + {\lambda _2} = \lambda.$

Proof. We now prove that the constant factor ${\lambda _2}B({\lambda _1}, {\lambda _2})$ in (19) is the best possible. For any $0 \lt \varepsilon \lt \min \{ p{\lambda _1}, q{\lambda _2}\}$ , we set

${\tilde a_m}: = {m^{{\lambda _1} - \tfrac{\varepsilon }{p} - 1}}, {\tilde b_n}: = {n^{{\lambda _2} - \tfrac{\varepsilon }{q} - 1}}^{}(m, n \in {\text{N}}) .$

Since for $\lambda {}_2 \leqslant \tfrac{1}{2}$ , $g(t): = {t^{{\lambda _2} - \tfrac{\varepsilon }{q} - 1}}$ is strictly decreasing with respect to $t \gt 0$ , by the decreasingness property of series, we have

${\tilde B_n}: = \sum\limits_{k = 1}^n {{{\tilde b}_k}} = \sum\limits_{k = 1}^n {{k^{{\lambda _2} - \tfrac{\varepsilon }{q} - 1}} \lt \int_0^n {{t^{{\lambda _2} - \tfrac{\varepsilon }{q} - 1}}} } dt = \tfrac{1}{{{\lambda _2} - \tfrac{\varepsilon }{q}}}{n^{{\lambda _2} - \tfrac{\varepsilon }{q}}}.$

If there exists a positive constant $M \leqslant {\lambda _2}B({\lambda _1}, {\lambda _2})$ , such that (19) is valid when we replace ${\lambda _2}B({\lambda _1}, {\lambda _2})$ by $M$ , then in particular, for ${\eta _i} = \eta = {0^{}}(i = 1, 2)$ , substitution of ${a_m} = {\tilde a_m},$ ${b_n} = {\tilde b_n}$ and ${B_n} = {\tilde B_n}$ in (19), we have

$\tilde I: = \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^\infty {\tfrac{1}{{{{(m + n)}^\lambda }}}{{\tilde a}_m}{{\tilde b}_n}} } \lt M{[\sum\limits_{m = 1}^\infty {{m^{p(1 - {\lambda _1}) - 1}}} \tilde a_m^p]^{\tfrac{1}{p}}}{(\sum\limits_{n = 1}^\infty {{n^{ - q{\lambda _2} - 1}}} \tilde B_n^q)^{\tfrac{1}{q}}} .$

(20)

In the following, we obtain that $M \geqslant {\lambda _2}B({\lambda _1}, {\lambda _2})$ , which follows that $M = {\lambda _2}B({\lambda _1}, {\lambda _2})$ is the best possible constant factor in (19).

By (20) and the decreasingness property of series, we obtain

$\begin{gathered} \tilde I \lt M{(\sum\limits_{m = 1}^\infty {{m^{p(1 - {\lambda _1}) - 1}}} {m^{p{\lambda _1} - p - \varepsilon }})^{\tfrac{1}{p}}}\tfrac{1}{{\lambda _2^{} - \tfrac{\varepsilon }{q}}}{(\sum\limits_{n = 1}^\infty {{n^{ - q{\lambda _2} - 1}}} {n^{q{\lambda _2} - \varepsilon }})^{\tfrac{1}{q}}} \hfill \\ \mathop {}\nolimits^{} = \tfrac{M}{{\lambda _2^{} - \tfrac{\varepsilon }{q}}}(1 + \sum\limits_{m = 2}^\infty {{m^{ - \varepsilon - 1}}} ) \hfill \\ \end{gathered}$

$\mathop {}\nolimits^{} \lt \tfrac{M}{{\lambda _2^{} - \tfrac{\varepsilon }{q}}}(1 + \int_1^\infty {{x^{ - \varepsilon - 1}}dx} ) = \tfrac{M}{\varepsilon }(\tfrac{1}{{\lambda _2^{} - \tfrac{\varepsilon }{q}}})(\varepsilon + 1).$

By (14), for ${\eta _i} = \eta = {0^{}}(i = 1, 2)$ , $s = \lambda {, ^{}}{s_1} = {\lambda _1} - {\tfrac{\varepsilon }{p}^{}}(\in (0, \tfrac{3}{2}) \cap (0, \lambda)){, ^{}}{s_2} = {\lambda _2} + \tfrac{\varepsilon }{p}$ $(\in (0, \lambda)),$ we have

$\begin{gathered} \tilde I = \sum\limits_{n = 1}^\infty {[{n^{({\lambda _2} + \tfrac{\varepsilon }{p})}}\sum\limits_{m = 1}^\infty {\tfrac{1}{{{{(m + n)}^\lambda }}}{m^{({\lambda _1} - \tfrac{\varepsilon }{p}) - 1}}]{n^{ - \varepsilon - 1}}} } \hfill \\ \mathop {}\nolimits^{} = \sum\limits_{n = 1}^\infty {\omega ({\lambda _1} - \tfrac{\varepsilon }{p}, n){n^{ - \varepsilon - 1}}} \gt {k_\lambda }({\lambda _1} - \tfrac{\varepsilon }{p})\sum\limits_{n = 1}^\infty {[1 - {O_2}(\tfrac{1}{{{n^{{\lambda _1} - \tfrac{\varepsilon }{p}}}}})]{n^{ - \varepsilon - 1}}} \hfill \\ \mathop {}\nolimits^{} \gt {k_\lambda }({\lambda _1} - \tfrac{\varepsilon }{p})[\int_1^\infty {{y^{ - \varepsilon - 1}}} dy - \sum\limits_{n = 1}^\infty {{O_2}(\tfrac{1}{{{n^{{\lambda _1} + \tfrac{\varepsilon }{q} + 1}}}})]} \hfill \\ \mathop {}\nolimits^{} = \tfrac{1}{\varepsilon }B({\lambda _1} - \tfrac{\varepsilon }{p}, {\lambda _2} + \tfrac{\varepsilon }{p})(1 - \varepsilon O(1)). \hfill \\ \end{gathered}$

By (20) and the above results, we have

$B({\lambda _1} - \tfrac{\varepsilon }{p}, {\lambda _2} + \tfrac{\varepsilon }{p})(1 - \varepsilon O(1)) \lt \varepsilon \tilde I \lt M(\tfrac{1}{{\lambda _2^{} - \tfrac{\varepsilon }{q}}})(\varepsilon + 1) .$

Setting $\varepsilon \to {0^ + }$ , in view of the continuity of the beta function, we find ${\lambda _2}B({\lambda _1}, {\lambda _2}) \leqslant M$ . Hence, $M = {\lambda _2}B({\lambda _1}, {\lambda _2})$ is the best possible constant factor in (19).

On the other hand, for ${\hat \lambda _1} = \tfrac{{\lambda - {\lambda _2}}}{p} + \tfrac{{{\lambda _1}}}{q}, {\hat \lambda _2} = \tfrac{{\lambda - {\lambda _1}}}{q} + \tfrac{{{\lambda _2}}}{p},$ and $\lambda - {\lambda _2} \leqslant \tfrac{3}{2}, \lambda - {\lambda _1} \leqslant \tfrac{1}{2},$ we find

${\hat \lambda _1} + {\hat \lambda _2} = \tfrac{{\lambda - {\lambda _2}}}{p} + \tfrac{{{\lambda _1}}}{q} + \tfrac{{\lambda - {\lambda _1}}}{q} + \tfrac{{{\lambda _2}}}{p} = \lambda ,$

$0 \lt {\hat \lambda _1}, {\hat \lambda _2} \lt \lambda , {\hat \lambda _1} \leqslant \tfrac{3}{{2p}} + \tfrac{3}{{2q}} = \tfrac{3}{2}, {\hat \lambda _2} \leqslant \tfrac{1}{2} ,$

and ${\hat \lambda _2}B({\hat \lambda _1}, {\hat \lambda _2}) \in {{\text{R}}_ + } = (0, \infty)$ . By (19), we still have

$\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{{{(m + n - \eta )}^\lambda }}} \lt } } {\hat \lambda _2}B({\hat \lambda _1}, {\hat \lambda _2}) {[\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{p(1 - {{\hat \lambda }_1}) - 1}}} a_m^p]^{\tfrac{1}{p}}}{[\sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - q{{\hat \lambda }_2} - 1}}} B_n^q]^{\tfrac{1}{q}}} .$

(21)

If the constant factor $\tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda)}}{(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}$ in (18) is the best possible, then for any $M$ , when we replace $\tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda)}}{(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}$ by $M$ , we have

$\tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}}{(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}} \leqslant M ,$

and then by (21), we have the following inequality:

$\tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}}{(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}} \leqslant {\hat \lambda _2}B({\hat \lambda _1}, {\hat \lambda _2}) = \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}}{k_{\lambda + 1}}({\hat \lambda _1}).$

It follows that

${k_{\lambda + 1}}({\hat \lambda _1}) \geqslant {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}} .$

By using Hӧlder's inequality (cf. ^[20]), we obtain

$\begin{gathered} \mathop {}\nolimits^{} {k_{\lambda + 1}}({{\hat \lambda }_1}) = {k_{\lambda + 1}}(\tfrac{{\lambda - {\lambda _2}}}{p} + \tfrac{{{\lambda _1}}}{q}) \hfill \\ = \int_0^\infty {\tfrac{1}{{{{(1 + u)}^{\lambda + 1}}}}} {u^{\tfrac{{\lambda - {\lambda _2}}}{p} + \tfrac{{{\lambda _1}}}{q} - 1}}du = \int_0^\infty {\tfrac{1}{{{{(1 + u)}^{\lambda + 1}}}}(} {u^{\tfrac{{\lambda - {\lambda _2} - 1}}{p}}})({u^{\tfrac{{{\lambda _1} - 1}}{q}}})du \hfill \\ \leqslant {[\int_0^\infty {\tfrac{1}{{{{(1 + u)}^{\lambda + 1}}}}} {u^{\lambda - {\lambda _2} - 1}}du]^{\tfrac{1}{p}}}{[\int_0^\infty {\tfrac{1}{{{{(1 + u)}^{\lambda + 1}}}}} {u^{{\lambda _1} - 1}}du]^{\tfrac{1}{q}}} \hfill \\ \end{gathered}$

$= {[\int_0^\infty {\tfrac{1}{{{{(1 + v)}^{\lambda + 1}}}}} {v^{({\lambda _2} + 1) - 1}}dv]^{\tfrac{1}{p}}}{[\int_0^\infty {\tfrac{1}{{{{(1 + u)}^{\lambda + 1}}}}} {u^{{\lambda _1} - 1}}du]^{\tfrac{1}{q}}}$

$= {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}.$

(22)

Then we have

${k_{\lambda + 1}}({\hat \lambda _1}) = {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}} ,$

namely, (22) keeps the form of equality.

We observe that (22) keeps the form of equality if and only if there exist constants $A$ and $B$ , such that they are not both zero satisfying (cf. ^[20])

$A{u^{\lambda - {\lambda _2} - 1}} = B{u^{{\lambda _1} - 1}}^{}a.e.{}^{}$ in

${{\text{R}}_ + } .$

Assuming that $A \ne 0$ , we have ${u^{\lambda - {\lambda _2} - {\lambda _1}}} = {\tfrac{B}{A}^{}}a.e.{}^{}$ in ${{\text{R}}_ + }$ , and then $\lambda - {\lambda _2} - {\lambda _1} = 0$ , namely, ${\lambda _1} + {\lambda _2} = \lambda$ . This completes the proof of Theorem 2.

4. Equivalent forms and operator expressions

Theorem 3. We have the following inequality equivalent to (18):

$J: = {\left\{ {\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{q{{\hat \lambda }_1} - 1}}{{\left[ {\sum\limits_{n = 1}^\infty {\tfrac{{{b_n}}}{{{{(m + n - \eta )}^\lambda }}}} } \right]}^q}} } \right\}^{\tfrac{1}{q}}}$

$\lt \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}} {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}} {[\sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - q{{\hat \lambda }_2} - 1}}} B_n^q]^{\tfrac{1}{q}}} .$

(23)

In particular, for ${\lambda }_{1}+{\lambda }_{2} = \lambda$ ( ${\mathit{\lambda }}_{1}\in (0, \frac{3}{2}]\cap (0, \mathit{\lambda }), {\mathit{\lambda }}_{2}\in (0, \frac{1}{2}]\cap (0, \mathit{\lambda })$ ), we have the following inequality equivalent to (19):

${\left\{ {\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{q{\lambda _1} - 1}}{{\left[ {\sum\limits_{n = 1}^\infty {\tfrac{{{b_n}}}{{{{(m + n - \eta )}^\lambda }}}} } \right]}^q}} } \right\}^{\tfrac{1}{q}}} \lt {\lambda _2}B({\lambda _1}, {\lambda _2}) {[\sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - q{\lambda _2} - 1}}} B_n^q]^{\tfrac{1}{q}}} .$

(24)

Proof. Suppose that (23) is valid. By using Hӧlder's inequality (cf. ^[20]), we have

$I = \sum\limits_{m = 1}^\infty {\left[ {{{(m - {\eta _1})}^{\tfrac{1}{q} - {{\hat \lambda }_1}}}{a_m}} \right]\left[ {{{(m - {\eta _1})}^{\tfrac{{ - 1}}{q} + {{\hat \lambda }_1}}}\sum\limits_{n = 1}^\infty {\tfrac{{{b_n}}}{{{{(m + n - \eta )}^\lambda }}}} } \right]}$

$\leqslant {\left[ {\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{p(1 - {{\hat \lambda }_1}) - 1}}} a_m^p} \right]^{\tfrac{1}{p}}}J.$

(25)

Then by (23), we have (18). On the other hand, assuming that (18) is valid, we set

${a_m}: = {(m - {\eta _1})^{q{{\hat \lambda }_1} - 1}}{\left[ {\sum\limits_{n = 1}^\infty {\tfrac{{{b_n}}}{{{{(m + n - \eta )}^\lambda }}}} } \right]^{q - 1}}, m \in {\text{N}} .$

Then, it follows that $J = {\left[{\sum\nolimits_{m = 1}^\infty {{{(m - {\eta _1})}^{p(1 - {{\hat \lambda }_1}) - 1}}} a_m^p} \right]^{\tfrac{1}{q}}}.$

If $J = 0,$ then (23) is naturally valid; if $J = \infty,$ then it is impossible that makes (23) valid, namely, $J \lt \infty.$ Suppose that $0 \lt J \lt \infty.$ By (18), we have

$\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{p(1 - {{\hat \lambda }_1}) - 1}}} a_m^p = {J^q} = I$

$\lt \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}} {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}} {J^{q - 1}}{[\sum\limits_{n = 1}^\infty {{{(n - \eta {}_2)}^{ - q{{\hat \lambda }_2} - 1}}} B_n^q]^{\tfrac{1}{q}}} ,$

$J \lt \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}} {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}} {[\sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - q{{\hat \lambda }_2} - 1] - 1}}} B_n^q]^{\tfrac{1}{q}}} ,$

namely, (23) follows, which is equivalent to (18). The Theorem 3 is proved.

Theorem 4. Suppose that $\lambda \in (0, 2], {\lambda _1} \in (0, \tfrac{3}{2}] \cap (0, \lambda), {\lambda _2} \in (0, \tfrac{1}{2}] \cap (0, \lambda).$ If ${\lambda _1} + {\lambda _2} = \lambda$ , then the constant factor $\tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda)}}$ ${(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}$ in (23) is the best possible; On the other hand, if the same constant factor in (23) is the best possible, then for $\lambda - {\lambda _2} \leqslant \tfrac{3}{2}, \lambda - {\lambda _1} \leqslant \tfrac{1}{2}$ , we have ${\lambda _1} + {\lambda _2} = \lambda.$

Proof. If ${\lambda _1} + {\lambda _2} = \lambda$ , then by Theorem 2, the constant factor

$\tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}} {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}$

in (18) is the best possible. Then by (25), the constant factor in (23) is the best possible.

On the other hand, if the same constant factor in (23) is the best possible, then by the equivalency of (23) and (18), in view of ${J^q} = I$ (in the proof of Theorem 3), it follows that the same constant factor in (18) is the best possible. By Theorem 2, in view of $\lambda - {\lambda _2} \leqslant \tfrac{3}{2},$ $\lambda - {\lambda _1} \leqslant \tfrac{1}{2}$ , we have ${\lambda _1} + {\lambda _2} = \lambda.$ The theorem is proved.

Setting

$\varphi (m): = {(m - {\eta _1})^{p(1 - {{\hat \lambda }_1}) - 1}}, \psi (n): = {(n - {\eta _2})^{q(1 - {{\hat \lambda }_2}) - 1}}, \Psi (n): = {(n - {\eta _2})^{ - q{{\hat \lambda }_2} - 1}} ,$

where from,

${\varphi ^{1 - q}}(m) = {(m - {\eta _1})^{q{{\hat \lambda }_1} - 1}}^{}({\text{m}}, n \in {\text{N)}} ,$

we define the following normed linear spaces:

${l_{p, \varphi }}: = \{ a = \{ {a_m}\} _{m = 1}^\infty ;||a|{|_{p, \varphi }}: = {(\sum\limits_{m = 1}^\infty {\varphi (m)|{a_m}} {|^p})^{\tfrac{1}{p}}} \lt \infty \} ,$

${l_{q, \psi }}: = \{ b = \{ {b_n}\} _{n = 1}^\infty ;||b|{|_{q, \psi }}: = {(\sum\limits_{n = 1}^\infty {\psi (n)|{b_n}} {|^q})^{\tfrac{1}{q}}} \lt \infty \} ,$

${l_{q, \Psi }}: = \{ B = \{ {B_n}\} _{n = 1}^\infty ;||B|{|_{q, \Psi }}: = {(\sum\limits_{n = 1}^\infty {\Psi (n)|{B_n}} {|^q})^{\tfrac{1}{q}}} \lt \infty \} ,$

${l_{q, {\varphi ^{1 - q}}}}: = \{ c = \{ {c_m}\} _{m = 1}^\infty ;||c|{|_{q, {\varphi ^{1 - q}}}}: = {(\sum\limits_{m = 1}^\infty {{\varphi ^{1 - q}}(m)|{c_m}} {|^q})^{\tfrac{1}{q}}} \lt \infty \} .$

For $b = \{ {b_n}\} _{n = 1}^\infty \in {l_{q, \psi }}$ , setting $c = \{ {c_m}\} _{m = 1}^\infty :$ ${c_m}: = \sum\nolimits_{n = 1}^\infty {\tfrac{{{b_n}}}{{{{(m + n - \eta)}^\lambda }}}}$ , we can rewrite (23) as follows:

$||c|{|_{q, {\varphi ^{1 - q}}}} \lt \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}} {(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}||B|{|_{q, \Psi }} \lt \infty ,$

namely, $c \in {l_{q, {\varphi ^{1 - q}}}}$ . The proof of Theorem 4 is complete.

Definition 1. Define a more accurate Hardy-Hilbert's operator $T:{l_{q, \psi }} \to {l_{q, {\varphi ^{1 - q}}}}$ as follows: For any $b \in {l_{q, \psi }},$ there exists a unique representation $c = Tb \in {l_{q, {\varphi ^{1 - q}}}}$ , such that for any $m \in {\text{N, }}$ $Tb(m) = {c_m}$ . Define the formal inner product of $Tb$ and $a \in {l_{p, \varphi }}$ and the norm of $T$ as follows:

$(Tb, a): = \sum\limits_{m = 1}^\infty {{a_m}} \sum\limits_{n = 1}^\infty {\tfrac{{{b_n}}}{{{{(m + n - \eta )}^\lambda }}}} ,$

$||T||: = \mathop {\sup }\limits_{b( \ne 0) \in {l_{q, \psi }}} \tfrac{{||Tb|{|_{q, {\varphi ^{1 - q}}}}}}{{||B|{|_{q, \Psi }}}} .$

By Theorems 2–4, we have

Theorem 5. Suppose that

$\lambda \in (0, 2], {\lambda _1} \in (0, \tfrac{3}{2}] \cap (0, \lambda ), {\lambda _2} \in (0, \tfrac{1}{2}] \cap (0, \lambda ).$

${b^{}}( \geqslant 0) \in {l_{q, \psi }}, B \in {l_{q, \Psi }}, {a^{}}( \geqslant 0) \in {l_{p, \varphi }}, ||b|{|_{q, \psi }} \gt 0, ||B|{|_{q, \Psi }} \gt 0||a|{|_{p, \varphi }} \gt 0,$

then we have the following equivalent inequalities:

$(Tb, a) \lt \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}}{(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}||a|{|_{p, \varphi }}||B|{|_{q, \Psi }} ,$

(26)

$||Tb|{|_{q, {\varphi ^{1 - q}}}} \lt \tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda )}}{(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}||B|{|_{q, \Psi }} .$

(27)

Moreover, for ${\lambda _1} + {\lambda _2} = \lambda$ , the constant factor $\tfrac{{\Gamma (\lambda + 1)}}{{\Gamma (\lambda)}}{(k_{\lambda + 1}^{}({\lambda _2} + 1))^{\tfrac{1}{p}}}{(k_{\lambda + 1}^{}({\lambda _1}))^{\tfrac{1}{q}}}$ in (26) and (27) is the best possible, namely

$||T|| = {\lambda _2}B({\lambda _1}, {\lambda _2}) .$

On the other hand, if the same constant factor in (26) (or (27)) is the best possible, then for $\lambda - {\lambda _2} \leqslant \tfrac{3}{2},$ $\lambda - {\lambda _1} \leqslant \tfrac{1}{2}$ , we have ${\lambda _1} + {\lambda _2} = \lambda$ .

Remark 3. Taking $\lambda = 1, {\lambda _1} = {\lambda _2} = \tfrac{1}{2}$ in (19) and (24), we have the following equivalent inequalities with the best possible constant factor $\tfrac{\pi }{2}$ :

$\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{m + n - \eta }}} } \lt \tfrac{\pi }{4} {[\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{\tfrac{p}{2} - 1}}} a_m^p]^{\tfrac{1}{p}}}{[\sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - \tfrac{q}{{\text{2}}} - 1}}} B_n^q]^{\tfrac{1}{q}}} ,$

(28)

${\left\{ {\sum\limits_{m = 1}^\infty {{{(m - {\eta _1})}^{\tfrac{q}{2} - 1}}{{\left[ {\sum\limits_{n = 1}^\infty {\tfrac{{{b_n}}}{{m + n - \eta }}} } \right]}^q}} } \right\}^{\tfrac{1}{q}}} \lt \tfrac{\pi }{2} {[\sum\limits_{n = 1}^\infty {{{(n - {\eta _2})}^{ - \tfrac{q}{2} - 1}}} B_n^q]^{\tfrac{1}{q}}} .$

(29)

In particular, putting ${\eta _1} = {\eta _2} = \eta = 0$ in (28) and (29), we get

$\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{m + n}}} } \lt \tfrac{\pi }{4} {(\sum\limits_{m = 1}^\infty {{m^{\tfrac{p}{2} - 1}}} a_m^p)^{\tfrac{1}{p}}}{(\sum\limits_{n = 1}^\infty {{n^{ - \tfrac{q}{{\text{2}}} - 1}}} B_n^q)^{\tfrac{1}{q}}} ,$

(30)

${\left[ {\sum\limits_{m = 1}^\infty {{m^{\tfrac{q}{2} - 1}}{{\left( {\sum\limits_{n = 1}^\infty {\tfrac{{{b_n}}}{{m + n}}} } \right)}^q}} } \right]^{\tfrac{1}{q}}} \lt \tfrac{\pi }{2} {(\sum\limits_{n = 1}^\infty {{n^{ - \tfrac{q}{2} - 1}}} B_n^q)^{\tfrac{1}{q}}} .$

(31)

Putting ${\eta _1} = \eta {}_2 = \tfrac{1}{4}, \eta = \tfrac{1}{2}$ in (28) and (29), we obtain

$\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\tfrac{{{a_m}{b_n}}}{{m + n - \tfrac{1}{2}}}} } \lt \tfrac{\pi }{4} {[\sum\limits_{m = 1}^\infty {{{(m - \tfrac{1}{4})}^{\tfrac{p}{2} - 1}}} a_m^p]^{\tfrac{1}{p}}}{[\sum\limits_{n = 1}^\infty {{{(n - \tfrac{1}{4})}^{ - \tfrac{q}{{\text{2}}} - 1}}} B_n^q]^{\tfrac{1}{q}}} ,$

(32)

${\left\{ {\sum\limits_{m = 1}^\infty {{{(m - \tfrac{1}{4})}^{\tfrac{q}{2} - 1}}{{\left[ {\sum\limits_{n = 1}^\infty {\tfrac{{{b_n}}}{{m + n - \tfrac{1}{2}}}} } \right]}^q}} } \right\}^{\tfrac{1}{q}}} \lt \tfrac{\pi }{2} {[\sum\limits_{n = 1}^\infty {{{(n - \tfrac{1}{4})}^{ - \tfrac{q}{2} - 1}}} B_n^q]^{\tfrac{1}{q}}} .$

(33)

5. Conclusions

In this paper, by means of the weight coefficients, the idea of introduced parameters and the techniques of real analysis, using Hermite-Hadamard's inequality, the Euler-Maclaurin summation formula and Abel's summation by parts formula, a more accurate Hardy-Hilbert-type inequality involving one partial sums is given in Theorem 1. The equivalent conditions of the best possible constant factor related to several parameters are provided in Theorem 2. We also consider the equivalent forms, the operator expressions and some particular inequalities in Theorem 3, Theorem 4, Theorem 5 and Remark 3. The lemmas and theorems provide an extensive account of this type of inequalities.

Acknowledgments

This work is supported by the National Natural Science Foundation (No. 61772140, No. 12071491), the Characteristic Innovation Project of Guangdong Provincial Colleges and Universities (No. 2020KTSCX088), the Construction Project of Teaching Quality and Teaching Reform in Guangdong Undergraduate Colleges and Universities in 2018 (Specialty of Financial Mathematics), and the Natural Science Foundation of Fujian Province of China (No. 2020J01365). All authors contributed equally in the preparation of this paper.

Conflict of interest

The authors declare that they have no competing interests.

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AIMS Mathematics

A Hardy-Hilbert-type inequality involving modified weight coefficients and partial sums

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Abstract

1. Introduction

2. Some lemmas

3. Main results

4. Equivalent forms and operator expressions

5. Conclusions

Acknowledgments

Conflict of interest

References

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AIMS Mathematics

A Hardy-Hilbert-type inequality involving modified weight coefficients and partial sums

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Abstract

1. Introduction

2. Some lemmas

3. Main results

4. Equivalent forms and operator expressions

5. Conclusions

Acknowledgments

Conflict of interest

References

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