On a more accurate half-discrete mulholland's inequality and an extension

Abstract

By using the way of weight functions and Jensen-Hadamard's inequality, a more accurate half-discrete Mulholland's inequality with a best constant factor is given. The extension with multi-parameters, the equivalent forms as well as the operator expressions are considered.

Mathematics Subject Classication 2000: 26D15; 47A07.

1 Introduction

Assuming that $f,g\in {L}^{2}\left({R}_{+}\right),∥f∥={\left\{{\int }_{0}^{\infty }{f}^{2}\left(x\right)dx\right\}}^{\frac{1}{2}}<0,∥g∥<0$, we have the following Hilbert's integral inequality (cf. [1]):

$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{x+y}dxdy<\pi ∥f∥∥g∥,$
(1)

where the constant factor π is the best possible. Moreover, for $a={\left\{{a}_{m}\right\}}_{m=1}^{\infty }\in {l}^{2},b={\left\{{b}_{n}\right\}}_{n=1}^{\infty }\in {l}^{2},∥a∥={\left\{\sum _{m=1}^{\infty }{a}_{m}^{2}\right\}}^{\frac{1}{2}}<0,∥b∥>0$, we still have the following discrete Hilbert's inequality

$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{{a}_{m}{b}_{n}}{m+n}<\pi ∥a∥∥b∥,$
(2)

with the same best constant factor π. Inequalities (1) and (2) are important in analysis and its applications (cf. [24]) and they still represent the field of interest to numerous mathematicians. Also we have the following Mulholland's inequality with the same best constant factor (cf. [1, 5]):

$\sum _{m=2}^{\infty }\sum _{n=2}^{\infty }\frac{{a}_{m}{b}_{n}}{\text{ln}\phantom{\rule{2.77695pt}{0ex}}mn}<\pi {\left\{\sum _{m=2}^{\infty }m{a}_{m}^{2}\sum _{n=2}^{\infty }n{b}_{n}^{2}\right\}}^{\frac{1}{2}}.$
(3)

In 1998, by introducing an independent parameter λ (0, 1], Yang [6] gave an extension of (1). By generalizing the results from [6], Yang [7] gave some best extensions of (1) and (2) as follows: If $p>1,\frac{1}{p}+\frac{1}{q}=1,{\lambda }_{\text{1}}+{\lambda }_{\text{2}}=\lambda \text{,}{k}_{\lambda }\left(x,y\right)$ is a non-negative homogeneous function of degree satisfying $k\left({\lambda }_{\text{1}}\right)={\int }_{0}^{\infty }{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{\text{1}}-1}dt\in {R}_{+,}$, $\varphi \left(x\right)={x}^{p\left(1-{\lambda }_{\text{1}}\right)-1},\psi \left(x\right)={x}^{q\left(1-{\lambda }_{\text{2}}\right)-1},f\left(\ge 0\right)\in {L}_{p,\varphi }\left({R}_{+}\right)=\left\{f\mid {‖f‖}_{p,\varphi }:={\left\{{\int }_{0}^{\infty }\varphi \left(x\right){|f\left(x\right)|}^{p}dx\right\}}^{\frac{1}{p}}<\infty \right\}$, $g\left(\ge 0\right)\in {L}_{q,\psi }\left({R}_{+}\right),{‖f‖}_{p,\varphi ,}{‖g‖}_{q,\psi }>0$, then

$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(x,y\right)f\left(x\right)g\left(y\right)dxdy
(4)

where the constant factor k(λ1) is the best possible. Moreover if k λ (x, y) is finite and ${k}_{\lambda }\left(x,y\right){x}^{{\lambda }_{1}-1}\left({k}_{\lambda }\left(x,y\right){y}^{{\lambda }_{2}-1}\right)$ is decreasing for x > 0(y > 0), then for $a={\left\{{a}_{m}\right\}}_{m=1}^{\infty }\in {l}_{p,\varphi }:=\left\{a|{∥a∥}_{p,\varphi }:={\left\{\sum _{n=1}^{\infty }\varphi \left(n\right){\left|{a}_{n}\right|}^{p}\right\}}^{\frac{1}{p}}<\infty \right\},b={\left\{{b}_{n}\right\}}_{n=1}^{\infty }\in {l}_{q,\psi },{∥a∥}_{p,\varphi },{∥b∥}_{q,\psi }>0$ we have

$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{k}_{\lambda }\left(m,n\right){a}_{m}{b}_{n}
(5)

where, k(λ1) is still the best value. Clearly, for $p=q=2,\lambda =1,{k}_{1}\left(x,y\right)=\frac{1}{x+y},{\lambda }_{1}={\lambda }_{2}=\frac{1}{2}$, inequality (4) reduces to (1), while (5) reduces to (2). Some other results about Hilbert-type inequalities are provided by [816].

On half-discrete Hilbert-type inequalities with the general non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the the constant factors in the inequalities are the best possible. However Yang [17] gave a result with the kernel $\frac{1}{{\left(1+nx\right)}^{\lambda }}$ by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang [18] gave the following half-discrete Hilbert's inequality with the best constant factor B(λ1, λ2)(λ1 > 0, 0 < λ2 1, λ1 + λ2 = λ):

$\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left(x+n\right)}^{\lambda }}dx
(6)

In this article, by using the way of weight functions and Jensen-Hadamard's inequality, a more accurate half-discrete Mulholland's inequality with a best constant factor similar to (6) is given as follows:

$\underset{\frac{3}{2}}{\overset{\infty }{\int }}f\left(x\right)\sum _{n=2}^{\infty }\frac{{a}_{n}}{\text{ln}\frac{4}{9}xn}dx<\pi {\left\{\underset{\frac{3}{2}}{\overset{\infty }{\int }}x{f}^{2}\left(x\right)dx\sum _{n=2}^{\infty }n{a}_{n}^{2}\right\}}^{\frac{1}{2}}.$
(7)

Moreover, a best extension of (7) with multi-parameters, some equivalent forms as well as the operator expressions are also considered.

2 Some lemmas

Lemma 1 If ${\lambda }_{1}>0,0<{\lambda }_{2}\le 1,{\lambda }_{1}+{\lambda }_{2}=\lambda ,\alpha \ge \frac{4}{9}$, setting weight functions ω(n) and ϖ(x) as follows:

$\omega \left(n\right):={\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_{2}}\underset{\frac{1}{\sqrt{\alpha }}}{\overset{{\infty }_{}}{\int }}\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_{1}-1}}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}dx,\phantom{\rule{1em}{0ex}}n\in N\\left\{1\right\},$
(8)
$\varpi \left(x\right):={\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_{1}}\sum _{n=2}^{\infty }\frac{{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_{2}-1}}{n{\left(\text{ln}\alpha xn\right)}^{\lambda }},\phantom{\rule{1em}{0ex}}x\in \left(\frac{1}{\sqrt{\alpha }},\infty \right),$
(9)

then we have

$\varpi \left(x\right)<\omega \left(n\right)=B\left({\lambda }_{1},{\lambda }_{2}\right).$
(10)

Proof. Applying the substitution $t=\frac{\text{ln}\sqrt{\alpha }x}{\text{ln}\sqrt{\alpha }n}$ to (8), we obtain

$\omega \left(n\right)=\underset{0}{\overset{\infty }{\int }}\frac{1}{{\left(1+t\right)}^{\lambda }}{t}^{{\lambda }_{1}-1}dt=B\left({\lambda }_{1},{\lambda }_{2}\right).$

Since by the conditions and for fixed $x\ge \frac{1}{\sqrt{\alpha }}$,

$h\left(x,y\right):=\frac{{\left(\text{ln}\sqrt{\alpha }y\right)}^{{\lambda }_{2}-1}}{y{\left(\text{ln}\alpha xy\right)}^{\lambda }}=\frac{1}{y{\left(\text{ln}\sqrt{\alpha }x+\text{ln}\sqrt{\alpha }y\right)}^{\lambda }{\left(\text{ln}\sqrt{\alpha }y\right)}^{1-{\lambda }_{2}}}$

is decreasing and strictly convex in $y\in \left(\frac{3}{2},\infty \right)$, then by Jensen-Hadamard's inequality (cf. [1]), we find

$\begin{array}{c}\varpi \left(x\right)<{\left(\sqrt{\alpha }\text{ln}x\right)}^{{\lambda }_{1}}\underset{\frac{3}{2}}{\overset{\infty }{\int }}\frac{1}{y{\left(\text{ln}\alpha xy\right)}^{\lambda }}{\left(\text{ln}\sqrt{\alpha }y\right)}^{{\lambda }_{2}-1}dy\\ \underset{=}{t= \left(\text{ln}\sqrt{\mathit{\text{α}}}y\right)/\left(\text{ln}\sqrt{\mathit{\text{α}}}x\right)}\underset{\frac{\text{ln}\left(3\sqrt{\alpha }/2\right)}{\text{ln}\sqrt{\alpha }x}}{\overset{\infty }{\int }}\frac{{t}^{{\lambda }_{2}-1}dt}{{\left(1+t\right)}^{\lambda }}\le B\left({\lambda }_{2},{\lambda }_{1}\right)B\left({\lambda }_{1},{\lambda }_{2}\right),\end{array}$

namely, (10) follows. □

Lemma 2 Let the assumptions of Lemma 1 be fulfilled and additionally, $p>1,\frac{1}{p}+\frac{1}{q}=1,{a}_{n}\ge 0,\in N\\left\{\text{1}\right\},f\left(x\right)$is a non-negative measurable function in $\left(\frac{1}{\sqrt{\alpha }},\infty \right)$. Then we have the following inequalities:

$\begin{array}{c}J\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}:\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}={\left\{\sum _{n=2}^{\infty }\frac{{\left(\text{ln}\sqrt{\alpha }n\right)}^{p{\lambda }_{2}-1}}{n}{\left[\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}dx\right]}^{p}\right\}}^{\frac{1}{p}}\\ \phantom{\rule{1em}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}{\left[B\left({\lambda }_{1},{\lambda }_{2}\right)\right]}^{\frac{1}{q}}{\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\varpi \left(x\right){x}^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{p\left(1-{\lambda }_{1}\right)}{f}^{p}\left(x\right)dx\right\}}^{\frac{1}{p}},\end{array}$
(11)
$\begin{array}{c}{L}_{1}:\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}={\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{q{\lambda }_{1}-1}}{x{\left[\varpi \left(x\right)\right]}^{q-1}}{\left[\sum _{n=2}^{\infty }\frac{{a}_{n}}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^{q}dx\right\}}^{\frac{1}{q}}\\ \phantom{\rule{1em}{0ex}}\le \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{\left\{B\left({\lambda }_{1},{\lambda }_{2}\right)\sum _{n=2}^{\infty }{n}^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}}.\end{array}$
(12)

Proof. By Hälder's inequality cf. [1] and (10), it follows

Then by Beppo Levi's theorem (cf. [19]), we have

$\begin{array}{c}J\le {\left[B\left({\lambda }_{1},{\lambda }_{2}\right)\right]}^{\frac{1}{q}}{\left\{\sum _{n=2}^{\infty }\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{{x}^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{n{\left(\text{ln}\alpha xn\right)}^{\lambda }{\left(\text{ln}\sqrt{\alpha }n\right)}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)dx\right\}}^{\frac{1}{p}}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}={\left[B\left({\lambda }_{1},{\lambda }_{2}\right)\right]}^{\frac{1}{q}}{\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\sum _{n=2}^{\infty }\frac{{x}^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{n{\left(\text{ln}\alpha xn\right)}^{\lambda }{\left(\text{ln}\sqrt{\alpha }n\right)}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)dx\right\}}^{\frac{1}{p}}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.6em}{0ex}}\phantom{\rule{0.6em}{0ex}}={\left[B\left({\lambda }_{1},{\lambda }_{2}\right)\right]}^{\frac{1}{q}}{\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\varpi \left(x\right){x}^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)dx\right\}}^{\frac{1}{p}},\end{array}$

that is, (11) follows. Still by Hölder's inequality, we have

Then by Beppo Levi's theorem, we have

$\begin{array}{c}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}{L}_{1}\le {\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\sum _{n=2}^{\infty }\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_{\text{1}}-1}}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}{n}^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{\left(q-1\right)\left(1-{\lambda }_{\text{2}}\right)}{a}_{n}^{q}dx\right\}}^{\frac{1}{q}}\\ ={\left\{\sum _{n=2}^{\infty }\left[{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_{\text{2}}}\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_{\text{1}}-1}}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}dx\right]{n}^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{q\left(1-{\lambda }_{\text{2}}\right)-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}}\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}={\left\{\sum _{n=2}^{\infty }\omega \left(n\right){n}^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{q\left(1-{\lambda }_{\text{2}}\right)-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}},\hfill \end{array}$

and then in view of (10), inequality (12) follows. □ ▪

3 Main results

We introduce two functions

$\begin{array}{c}\Phi \left(x\right):\phantom{\rule{1em}{0ex}}={x}^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{p\left(1-{\lambda }_{\text{1}}\right)-1}\left(x\in \left(\frac{1}{\sqrt{\alpha }},\infty \right)\right),\phantom{\rule{2.77695pt}{0ex}}\text{and}\hfill \\ \Psi \left(n\right):\phantom{\rule{1em}{0ex}}={n}^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{q\left(1-{\lambda }_{\text{2}}\right)-1}\left(n\in N\\left\{\text{1}\right\}\right),\hfill \end{array}$

wherefrom, ${\left[\Phi \left(x\right)\right]}^{1-q}=\frac{1}{x}{\left(\text{ln}\sqrt{\alpha }x\right)}^{q{\lambda }_{\text{1}}-1}$, and ${\left[\Psi \left(n\right)\right]}^{1-p}=\frac{1}{n}{\left(\text{ln}\sqrt{\alpha }n\right)}^{p{\lambda }_{\text{2}}-1}$.

Theorem 3 If $p>1,\frac{1}{p}+\frac{1}{q}=1,{\lambda }_{\text{1}}>0,0<{\lambda }_{2}\le 1,{\lambda }_{1}+{\lambda }_{2}=\lambda ,\alpha \ge \frac{4}{9},f\left(x\right),{a}_{n}\ge 0,f\in {L}_{p,\Phi }\left(\frac{1}{\sqrt{\alpha }},\infty \right),a={\left\{{a}_{n}\right\}}_{n=2}^{\infty }\in {l}_{q,\Psi },{∥f∥}_{p,\Phi }>0$, then we have the following equivalent inequalities:

$I:=\sum _{n=2}^{\infty }\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{{a}_{n}f\left(x\right)dx}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}=\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\sum _{n=2}^{\infty }\frac{f\left(x\right){a}_{n}dx}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}
(13)
$J={\left\{\sum _{n=2}^{\infty }{\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{f\left(x\right)dx}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^{p}\right\}}^{\frac{1}{p}}
(14)
$L:={\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}{\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n=2}^{\infty }\frac{{a}_{n}}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^{q}dx\right\}}^{\frac{1}{q}}
(15)

where the constant B(λ1, λ2) is the best possible in the above inequalities.

Proof. By Beppo Levi's theorem (cf. [19]), there are two expressions for I in (13). In view of (11), for ϖ(x) < B(λ1, λ2), we have (14). By Hälder's inequality, we have

$I=\sum _{n=2}^{\infty }\left[{\Psi }^{\frac{-1}{q}}\left(n\right)\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}f\left(x\right)dx\right]\left[{\Psi }^{\frac{1}{q}}\left(n\right){a}^{n}\right]\le J{∥a∥}_{q,\Psi }.$
(16)

Then by (14), we have (13). On the other-hand, assuming that (13) is valid, setting

${a}_{n}:={\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}f\left(x\right)dx\right]}^{p-1},n\in N\\left\{\text{1}\right\},$

then Jp-1 = ||a|| q , Ψ. By (11), we find J < . If J = 0, then (14) is valid trivially; if J > 0, then by (13), we have

$\begin{array}{c}{∥a∥}_{q,\Psi }^{q}={J}^{p}=I

that is, (14) is equivalent to (13). By (12), since [ϖ(x)]1-q>[B(λ1, λ2)]1-q, we have (15). By Hälder's inequality, we find

$I=\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\left[{\Phi }^{\frac{1}{p}}\left(x\right)f\left(x\right)\right]\left[{\Phi }^{\frac{-1}{p}}\left(x\right)\sum _{n=2}^{\infty }\frac{{a}_{n}}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]dx\le {∥f∥}_{p,\Phi }L.$
(17)

Then by (15), we have (13). On the other-hand, assuming that (13) is valid, setting

$f\left(x\right):={\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n=2}^{\infty }\frac{{a}_{n}}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^{q-1},\phantom{\rule{1em}{0ex}}x\in \left(\frac{1}{\sqrt{\alpha }},\infty \right),$

then Lq-1 = ║f║ p , Φ.. By (12), we find L < . If L = 0, then (15) is valid trivially; if L > 0, then by (13), we have

$\begin{array}{c}{∥f∥}_{p,\Phi }^{p}\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}{L}^{q}=I

That is, (15) is equivalent to (13). Hence inequalities (13), (14) and (15) are equivalent.

For 0 < ε <pλ1, setting ${ã}_{n}=\frac{1}{n}{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_{2}-\frac{\in }{q}-1},n\in N\\left\{1\right\}$, and

$\stackrel{̃}{f}\left(x\right):=\left\{\begin{array}{c}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}0,x\in \left(\frac{1}{\sqrt{\alpha }},\frac{e}{\sqrt{\alpha }}\right)\\ \frac{1}{x}{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_{1}-\frac{\in }{p}-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}x\in \left[\frac{e}{\sqrt{\alpha }},\infty \right)\end{array}\right\,$

if there exists a positive number k(≤ B(λ1, λ2)), such that (13) is valid as we replace B(λ1, λ2) with k, then in particular, it follows

$\begin{array}{ll}\hfill Ĩ:& =\sum _{n=2}^{\infty }\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}{ã}_{n}\stackrel{̃}{f}\left(x\right)dx
(18)
$\begin{array}{ll}\hfill \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}Ĩ& =\sum _{n=2}^{\infty }{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{1}{n}\underset{\frac{e}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}dx\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\underset{=}{t =\left(\text{ln}\sqrt{\mathit{\text{α}}}x\right)/\left(\text{ln}\sqrt{\mathit{\text{α}}}n\right)}\sum _{n=2}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{\epsilon +1}}\underset{1/\text{ln}\sqrt{\alpha }n}{\overset{\infty }{\int }}\frac{{t}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}}{{\left(t+1\right)}^{\lambda }}dt\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=B\left({\lambda }_{1}-\frac{\epsilon }{p},{\lambda }_{2}+\frac{\epsilon }{p}\right)\sum _{n=2}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{\epsilon +1}}-A\left(\epsilon \right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}>B\left({\lambda }_{1}-\frac{\epsilon }{p},{\lambda }_{2}+\frac{\epsilon }{p}\right)\underset{2}{\overset{\infty }{\int }}\frac{1}{y{\left(\text{ln}\sqrt{\alpha }y\right)}^{\epsilon +1}}dy-A\left(\epsilon \right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\frac{1}{\epsilon {\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon }}B\left({\lambda }_{1}-\frac{\epsilon }{p},{\lambda }_{2}+\frac{\epsilon }{p}\right)-A\left(\epsilon \right),\phantom{\rule{2em}{0ex}}\\ \hfill A\left(\epsilon \right)& :=\sum _{n=2}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{\epsilon +1}}\underset{0}{\overset{1/\text{ln}\sqrt{\alpha }n}{\int }}\frac{1}{{\left(t+1\right)}^{\lambda }}{t}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}dt.\phantom{\rule{2em}{0ex}}\end{array}$
(19)

W find

$\begin{array}{c}0

that is, A(ε) = O(1) (ε → 0+). Hence by (18) and (19), it follows

$\frac{B\left({\lambda }_{1}-\frac{\epsilon }{p},{\lambda }_{2}+\frac{\epsilon }{p}\right)}{{\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon }}-\epsilon O\left(1\right)
(20)

and B(λ1, λ2) ≤ k(ε → 0+). Hence, k = B(λ1, λ2) is the best value of (13).

Due to the equivalence, the constant factor B(λ1, λ2) in (14) and (15) is the best possible. Otherwise, we can imply a contradiction by (16) and (17) that the constant factor in (13) is not the best possible. □

Remark 1 (i) Define the first type half-discrete Mulholland's operator $T:{L}_{p,\Phi }\left(\frac{1}{\sqrt{\alpha }},\infty \right)\to {l}_{p,{\Psi }^{1-p}}$as follows: for $f\in {L}_{p,\Phi }\left(\frac{1}{\sqrt{\alpha }},\infty \right)$, we define $Tf\in {l}_{p,{\Psi }^{1-p}}$as

$Tf\left(n\right)=\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}f\left(x\right)dx,\phantom{\rule{1em}{0ex}}n\in N\\left\{\text{1}\right\}\text{.}$

Then by (14), it follows ${∥Tf∥}_{p,{\Psi }^{1-p}}\le B\left({\lambda }_{1},{\lambda }_{2}\right){∥f∥}_{p,\Phi }$and then T is a bounded operator with ║T║ ≤ B(λ1, λ2). Since by Theorem 1, the constant factor in (14) is the best possible, we have ║T║ = B(λ1, λ2).

(ii) Define the second type half-discrete Mulholland's operator $\stackrel{̃}{T}:{l}_{q,\Psi }\to {L}_{q,{\Phi }^{1-q}}\left(\frac{1}{\sqrt{\alpha }},\infty \right)$as follows: For a l q, ψ, define $\stackrel{̃}{T}a\in {L}_{q,{\Phi }^{1-q}}\left(\frac{1}{\sqrt{\alpha }},\infty \right)$as

$\stackrel{̃}{T}a\left(x\right)=\sum _{n=2}^{\infty }\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}{a}_{n},\phantom{\rule{1em}{0ex}}x\in \left(\frac{1}{\sqrt{\alpha }},\infty \right).$

Then by (15), it follows ${∥\stackrel{̃}{T}a∥}_{q.{\Phi }^{1-q}}\le B\left({\lambda }_{1},{\lambda }_{2}\right){∥a∥}_{q,\Psi }$and then $\stackrel{̃}{T}$is a bounded operator with $∥\stackrel{̃}{T}∥\le B\left({\lambda }_{1},{\lambda }_{2}\right)$. Since by Theorem 1, the constant factor in (15) is the best possible, we have $∥\stackrel{̃}{T}∥=B\left({\lambda }_{1},{\lambda }_{2}\right)$.

Remark 2 We set $p=q=2,\lambda =1,{\lambda }_{1}={\lambda }_{2}=\frac{1}{2}$in (13), (14) and (15). (i) if $\alpha =\frac{4}{9}$, then we deduce (7) and the following equivalent inequalities:

$\sum _{n=2}^{\infty }\frac{1}{n}{\left[\underset{\frac{3}{2}}{\overset{\infty }{\int }}\frac{f\left(x\right)}{\text{ln}\frac{4}{9}xn}dx\right]}^{2}<{\pi }^{\text{2}}\underset{\frac{3}{2}}{\overset{\infty }{\int }}x{f}^{2}\left(x\right)dx,$
(21)
$\underset{\frac{3}{2}}{\overset{\infty }{\int }}\frac{1}{x}{\left[\sum _{n=2}^{\infty }\frac{{a}_{n}}{\text{ln}\frac{4}{9}xn}\right]}^{2}dx<{\pi }^{\text{2}}\sum _{n=2}^{\infty }n{a}_{n}^{2};$
(22)

(ii) if α = 1, then we have the following half-discrete Mulholland's inequality and its equivalent forms:

$\underset{1}{\overset{\infty }{\int }}{f\left(x\right)\sum _{n=2}^{\infty }\frac{{a}_{n}}{\text{ln}xn}dx<\pi \left\{\underset{\text{1}}{\overset{\infty }{\int }}x{f}^{2}\left(x\right)dx\sum _{n=2}^{\infty }n{a}_{n}^{2}\right\}}^{\frac{1}{2}},$
(23)
$\sum _{n=2}^{\infty }{\frac{1}{n}\left[\underset{1}{\overset{\infty }{\int }}\frac{f\left(x\right)}{\text{ln}xn}dx\right]}^{2}<{\pi }^{\text{2}}\underset{1}{\overset{\infty }{\int }}x{f}^{2}\left(x\right)dx,$
(24)
$\underset{1}{\overset{\infty }{\int }}\frac{1}{x}{\left[\sum _{n=2}^{\infty }\frac{{a}_{n}}{\text{ln}xn}\right]}^{2}dx<{\pi }^{\text{2}}\sum _{n=2}^{\infty }n{a}_{n}^{2}.$
(25)

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Acknowledgements

This work is supported by the Guangdong Science and Technology Plan Item (No. 2010B010600018), and the Guangdong Modern Information Service industry Develop Particularly item 2011 (No. 13090).

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Correspondence to Bicheng Yang.

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QC conceived of the study, and participated in its design and coordination. BY wrote and reformed the article. All authors read and approved the final manuscript.

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Chen, Q., Yang, B. On a more accurate half-discrete mulholland's inequality and an extension. J Inequal Appl 2012, 70 (2012). https://doi.org/10.1186/1029-242X-2012-70