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. [2–4]) 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]):

$$\sum _{m=2}^{\infty }\sum _{n=2}^{\infty }\frac{a_m{b}_n}{\text{ln}mn}<\pi {\left\{\sum _{m=2}^{\infty }m{a}_m^2\sum _{n=2}^{\infty }n{b}_n^2\right\}}^{\frac{1}{2}}.$$

(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 -λ satisfying $k\left({\lambda }_{\text{1}}\right)={\displaystyle {\int }_0^{\infty }{k}_{\lambda }\left(t,1\right)t^{{\lambda }_{\text{1}}-1}dt\in R_{+,}}$, $\varphi \left(x\right)=x^{p\left(1-{\lambda }_{\text{1}}\right)-1},\psi \left(x\right)=x^{q\left(1-{\lambda }_{\text{2}}\right)-1},f\left(\ge 0\right)\in L_{p,\varphi }\left(R_{+}\right)=\left\{f\mid {\Vert f\Vert }_{p,\varphi }:={\left\{{\displaystyle {\int }_0^{\infty }\varphi \left(x\right){\left|f\left(x\right)\right|}^{p}dx}\right\}}^{\frac{1}{p}}<\infty \right\}$, $g\left(\ge 0\right)\in L_{q,\psi }\left(R_{+}\right),{\Vert f\Vert }_{p,\varphi ,}{\Vert g\Vert }_{q,\psi }>0$, then

$$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(x,y\right)f\left(x\right)g\left(y\right)dxdy<k\left({\lambda }_1\right){∥f∥}_{p,\varphi }{∥g∥}_{q,\psi },$$

(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

$$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{k}_{\lambda }\left(m,n\right)a_m{b}_n<k\left({\lambda }_1\right){∥a∥}_{p,\varphi }{∥b∥}_{q,\psi },$$

(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 [8–16].

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):

$$\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }\frac{a_n}{{\left(x+n\right)}^{\lambda }}dx<B\left({\lambda }_1,{\lambda }_2\right){∥f∥}_{p,\varphi }{\parallel a\parallel }_{q,\psi }.$$

(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:

$$\omega \left(n\right):={\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_2}\underset{\frac{1}{\sqrt{\alpha }}}{\overset{{\infty }_{}}{\int }}\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_1-1}}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}dx,n\in N\backslash \left\{1\right\},$$

(8)

$$\varpi \left(x\right):={\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_1}\sum _{n=2}^{\infty }\frac{{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_2-1}}{n{\left(\text{ln}\alpha xn\right)}^{\lambda }},x\in \left(\frac{1}{\sqrt{\alpha }},\infty \right),$$

(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

$$\begin{array}{c}\varpi \left(x\right)<{\left(\sqrt{\alpha }\text{ln}x\right)}^{{\lambda }_1}\underset{\frac{3}{2}}{\overset{\infty }{\int }}\frac{1}{y{\left(\text{ln}\alpha xy\right)}^{\lambda }}{\left(\text{ln}\sqrt{\alpha }y\right)}^{{\lambda }_2-1}dy\\ \underset{=}{t=\; (\text{ln}\sqrt{\mathit{\text{α}}}y)/(\text{ln}\sqrt{\mathit{\text{α}}}x)}\underset{\frac{\text{ln}\left(3\sqrt{\alpha }/2\right)}{\text{ln}\sqrt{\alpha }x}}{\overset{\infty }{\int }}\frac{t^{{\lambda }_2-1}dt}{{\left(1+t\right)}^{\lambda }}\le B\left({\lambda }_2,{\lambda }_1\right)B\left({\lambda }_1,{\lambda }_2\right),\end{array}$$

namely, (10) follows. □ ▪

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

$$\begin{array}{c}J:={\left\{\sum _{n=2}^{\infty }\frac{{\left(\text{ln}\sqrt{\alpha }n\right)}^{p{\lambda }_2-1}}{n}{\left[\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}dx\right]}^p\right\}}^{\frac{1}{p}}\\ \le {\left[B\left({\lambda }_1,{\lambda }_2\right)\right]}^{\frac{1}{q}}{\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\varpi \left(x\right)x^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{p\left(1-{\lambda }_1\right)}{f}^p\left(x\right)dx\right\}}^{\frac{1}{p}},\end{array}$$

(11)

$$\begin{array}{c}{L}_1:={\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{q{\lambda }_1-1}}{x{\left[\varpi \left(x\right)\right]}^{q-1}}{\left[\sum _{n=2}^{\infty }\frac{a_n}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^{q}dx\right\}}^{\frac{1}{q}}\\ \le {\left\{B\left({\lambda }_1,{\lambda }_2\right)\sum _{n=2}^{\infty }{n}^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{q\left(1-{\lambda }_2\right)-1}{a}_n^q\right\}}^{\frac{1}{q}}.\end{array}$$

(12)

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

$$\begin{array}{c}{\left[\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{f\left(x\right)dx}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^p=\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\left[\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_1\right)/q}{x}^{1/q}}{{\left(\text{ln}\sqrt{\alpha }n\right)}^{\left(1-{\lambda }_2\right)/p}{n}^{1/p}}f\left(x\right)\right]\right.\\ {\left.\times \left[\frac{\left(\text{ln}\sqrt{\alpha }{n}^{\left(1-{\lambda }_2\right)/p}{n}^{1/p}\right)}{\left(\text{ln}\sqrt{\alpha }{x}^{\left(1-{\lambda }_1\right)/q}{x}^{1/q}\right)}\right]dx\right\}}^p\le \underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{x^{p-1}}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_1\right)\left(p-1\right)}}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{1-{\lambda }_2}}{f}^p\left(x\right)dx\\ \times {\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\frac{n^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{\left(1-{\lambda }_2\right)\left(q-1\right)}}{x{\left(\text{ln}\sqrt{\alpha }x\right)}^{1-{\lambda }_1}}dx\right\}}^{p-1}\\ ={\left\{\omega \left(n\right)\frac{{\left(\text{ln}\sqrt{\alpha }n\right)}^{q\left(1-{\lambda }_2\right)-1}}{n^{1-q}}\right\}}^{p-1}\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\\ =\frac{{\left[B\left({\lambda }_1,{\lambda }_2\right)\right]}^{p-1}n}{{\left(\text{ln}\sqrt{\alpha }n\right)}^{p{\lambda }_2-1}}\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}\sqrt{\alpha }xn\right)}^{\lambda }{\left(\text{ln}\sqrt{\alpha }n\right)}^{1-{\lambda }_2}}{f}^p\left(x\right)dx.\end{array}$$

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

$$\begin{array}{c}J\le {\left[B\left({\lambda }_1,{\lambda }_2\right)\right]}^{\frac{1}{q}}{\left\{\sum _{n=2}^{\infty }\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{x^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_1\right)\left(p-1\right)}}{n{\left(\text{ln}\alpha xn\right)}^{\lambda }{\left(\text{ln}\sqrt{\alpha }n\right)}^{1-{\lambda }_2}}{f}^p\left(x\right)dx\right\}}^{\frac{1}{p}}\\ ={\left[B\left({\lambda }_1,{\lambda }_2\right)\right]}^{\frac{1}{q}}{\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\sum _{n=2}^{\infty }\frac{x^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_1\right)\left(p-1\right)}}{n{\left(\text{ln}\alpha xn\right)}^{\lambda }{\left(\text{ln}\sqrt{\alpha }n\right)}^{1-{\lambda }_2}}{f}^p\left(x\right)dx\right\}}^{\frac{1}{p}}\\ ={\left[B\left({\lambda }_1,{\lambda }_2\right)\right]}^{\frac{1}{q}}{\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\varpi \left(x\right)x^{p-1}{\left(\text{ln}\sqrt{\alpha }x\right)}^{p\left(1-{\lambda }_1\right)-1}{f}^p\left(x\right)dx\right\}}^{\frac{1}{p}},\end{array}$$

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

$$\begin{array}{c}{\left[\sum _{n=2}^{\infty }\frac{a}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^q=\left\{\sum _{n=2}^{\infty }\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\left[\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_{\text{1}}\right)/q}{x}^{1/q}}{{\left(\text{ln}\sqrt{\alpha }n\right)}^{\left(1-{\lambda }_{\text{2}}\right)/p}{n}^{1/p}}\right]\right.\\ {\left.\times \left[\frac{{\left(\text{ln}\sqrt{\alpha }n\right)}^{\left(1-{\lambda }_{\text{2}}\right)/p}{n}^{1/p}}{{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_{\text{1}}\right)/q}{x}^{1/p}}{a}_n\right]\right\}}^q\le {\left\{\sum _{n=2}^{\infty }\frac{x^{p-1}}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{\left(1-{\lambda }_1\right)\left(p-1\right)}}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{1-{\lambda }_2}}\right\}}^{q-1}\\ \times \sum _{n=2}^{\infty }\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\frac{n^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{\left(1-{\lambda }_2\right)\left(q-1\right)}}{x{\left(\text{ln}\sqrt{\alpha }x\right)}^{1-{\lambda }_1}}{a}_n^q\hfill \\ =\frac{x{\left[\varpi \left(x\right)\right]}^{q-1}}{{\left(\text{ln}\sqrt{\alpha }x\right)}^{q{\lambda }_1-1}}\sum _{n=2}^{\infty }\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_1-1}}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}{n}^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{\left(q-1\right)\left(1-{\lambda }_2\right)}{a}_n^q.\end{array}$$

Then by Beppo Levi's theorem, we have

$$\begin{array}{c}{L}_1\le {\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\sum _{n=2}^{\infty }\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_{\text{1}}-1}}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}{n}^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{\left(q-1\right)\left(1-{\lambda }_{\text{2}}\right)}{a}_n^{q}dx\right\}}^{\frac{1}{q}}\\ ={\left\{\sum _{n=2}^{\infty }\left[{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_{\text{2}}}\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_{\text{1}}-1}}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}dx\right]n^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{q\left(1-{\lambda }_{\text{2}}\right)-1}{a}_n^q\right\}}^{\frac{1}{q}}\\ ={\left\{\sum _{n=2}^{\infty }\omega \left(n\right)n^{q-1}{\left(\text{ln}\sqrt{\alpha }n\right)}^{q\left(1-{\lambda }_{\text{2}}\right)-1}{a}_n^q\right\}}^{\frac{1}{q}},\hfill \end{array}$$

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

3 Main results

We introduce two functions

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

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:

$$I:=\sum _{n=2}^{\infty }\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{a_{n}f\left(x\right)dx}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}=\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\sum _{n=2}^{\infty }\frac{f\left(x\right)a_{n}dx}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}<B\left({\lambda }_1,{\lambda }_2\right){∥f∥}_{p,\Phi }{∥a∥}_{q,\Psi },$$

(13)

$$J={\left\{\sum _{n=2}^{\infty }{\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{f\left(x\right)dx}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^p\right\}}^{\frac{1}{p}}<B\left({\lambda }_1,{\lambda }_2\right){∥f∥}_{p,\Phi },$$

(14)

$$L:={\left\{\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}{\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n=2}^{\infty }\frac{a_n}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}\right]}^{q}dx\right\}}^{\frac{1}{q}}<B\left({\lambda }_1,{\lambda }_2\right){∥a∥}_{q,\Psi },$$

(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

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

then J^p-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

$$\begin{array}{c}{∥a∥}_{q,\Psi }^q=J^p=I<B\left({\lambda }_{\text{1}},{\lambda }_2\right){∥f∥}_{p,\Psi }{∥a∥}_{q,\Psi },\text{i}\text{.e}\text{.}\hfill \\ {∥a∥}_{q,\Psi }^{q-1}=J<B\left({\lambda }_{\text{1}},{\lambda }_2\right){∥f∥}_{p,\Phi },\hfill \end{array}$$

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

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

then L^q-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

$$\begin{array}{c}{∥f∥}_{p,\Phi }^p=L^q=I<B\left({\lambda }_{\text{1}},{\lambda }_2\right){∥f∥}_{p,\Phi }{∥a∥}_{q,\Psi },\text{i}\text{.e}\text{.}\hfill \\ {∥f∥}_{p,\Phi }^{p-1}=L<B\left({\lambda }_{\text{1}},{\lambda }_2\right){∥a∥}_{q,\Psi },\hfill \end{array}$$

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

For 0 < ε <pλ₁, setting ${ã}_n=\frac{1}{n}{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_2-\frac{\in }{q}-1},n\in N\backslash \left\{1\right\}$, and

$$\tilde{f}\left(x\right):=\left\{\begin{array}{c}0,x\in \left(\frac{1}{\sqrt{\alpha }},\frac{e}{\sqrt{\alpha }}\right)\\ \frac{1}{x}{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_1-\frac{\in }{p}-1},x\in \left[\frac{e}{\sqrt{\alpha }},\infty \right)\end{array}\right.,$$

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

$$\begin{array}{ll}\hfill Ĩ:& =\sum _{n=2}^{\infty }\underset{\frac{1}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}{ã}_n\tilde{f}\left(x\right)dx<k{∥\tilde{f}∥}_{p,\Phi }{∥ã∥}_{q,\Psi }\\ =k{\left\{\underset{\frac{e}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{dx}{x{\left(\text{ln}\sqrt{\alpha }x\right)}^{\epsilon +1}}\right\}}^{\frac{1}{p}}{\left\{\frac{1}{2{\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon +1}}+\sum _{n=3}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{\epsilon +1}}\right\}}^{\frac{1}{q}}\\ <k{\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}{\left\{\frac{1}{2{\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon +1}}+\underset{2}{\overset{\infty }{\int }}\frac{1}{x{\left(\text{ln}\sqrt{\alpha }x\right)}^{\epsilon +1}}dx\right\}}^{\frac{1}{q}}\\ =\frac{k}{\epsilon }{\left\{\frac{\epsilon }{2{\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon +1}}+\frac{1}{{\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon }}\right\}}^{\frac{1}{q}},\end{array}$$

(18)

$$\begin{array}{ll}\hfill Ĩ& =\sum _{n=2}^{\infty }{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_2-\frac{\epsilon }{q}-1}\frac{1}{n}\underset{\frac{e}{\sqrt{\alpha }}}{\overset{\infty }{\int }}\frac{1}{x{\left(\text{ln}\alpha xn\right)}^{\lambda }}{\left(\text{ln}\sqrt{\alpha }x\right)}^{{\lambda }_1-\frac{\epsilon }{p}-1}dx\\ \underset{=}{t\; =\left(\text{ln}\sqrt{\mathit{\text{α}}}x\right)/\left(\text{ln}\sqrt{\mathit{\text{α}}}n\right)}\sum _{n=2}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{\epsilon +1}}\underset{1/\text{ln}\sqrt{\alpha }n}{\overset{\infty }{\int }}\frac{t^{{\lambda }_1-\frac{\epsilon }{p}-1}}{{\left(t+1\right)}^{\lambda }}dt\\ =B\left({\lambda }_1-\frac{\epsilon }{p},{\lambda }_2+\frac{\epsilon }{p}\right)\sum _{n=2}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{\epsilon +1}}-A\left(\epsilon \right)\\ >B\left({\lambda }_1-\frac{\epsilon }{p},{\lambda }_2+\frac{\epsilon }{p}\right)\underset{2}{\overset{\infty }{\int }}\frac{1}{y{\left(\text{ln}\sqrt{\alpha }y\right)}^{\epsilon +1}}dy-A\left(\epsilon \right)\\ =\frac{1}{\epsilon {\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon }}B\left({\lambda }_1-\frac{\epsilon }{p},{\lambda }_2+\frac{\epsilon }{p}\right)-A\left(\epsilon \right),\\ \hfill A\left(\epsilon \right)& :=\sum _{n=2}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{\epsilon +1}}\underset{0}{\overset{1/\text{ln}\sqrt{\alpha }n}{\int }}\frac{1}{{\left(t+1\right)}^{\lambda }}{t}^{{\lambda }_1-\frac{\epsilon }{p}-1}dt.\end{array}$$

(19)

W find

$$\begin{array}{c}0<A\left(\epsilon \right)\le \sum _{n=2}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{\epsilon +1}}\underset{0}{\overset{1/\text{ln}\sqrt{\alpha }n}{\int }}{t}^{{\lambda }_1-\frac{\epsilon }{p}-1}dt\\ =\frac{1}{{\lambda }_1-\frac{\epsilon }{p}}\sum _{n=2}^{\infty }\frac{1}{n{\left(\text{ln}\sqrt{\alpha }n\right)}^{{\lambda }_1+\frac{\epsilon }{q}+1}}<\infty ,\end{array}$$

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

$$\frac{B\left({\lambda }_1-\frac{\epsilon }{p},{\lambda }_2+\frac{\epsilon }{p}\right)}{{\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon }}-\epsilon O\left(1\right)<k{\left\{\frac{\epsilon }{2{\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon +1}}+\frac{1}{{\left(\text{ln}2\sqrt{\alpha }\right)}^{\epsilon }}\right\}}^{\frac{1}{q}},$$

(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

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

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 $\tilde{T}:l_{q,\Psi }\to L_{q,{\Phi }^{1-q}}\left(\frac{1}{\sqrt{\alpha }},\infty \right)$as follows: For a ∈ l _{q, ψ}, define $\tilde{T}a\in L_{q,{\Phi }^{1-q}}\left(\frac{1}{\sqrt{\alpha }},\infty \right)$as

$$\tilde{T}a\left(x\right)=\sum _{n=2}^{\infty }\frac{1}{{\left(\text{ln}\alpha xn\right)}^{\lambda }}{a}_n,x\in \left(\frac{1}{\sqrt{\alpha }},\infty \right).$$

Then by (15), it follows ${∥\tilde{T}a∥}_{q.{\Phi }^{1-q}}\le B\left({\lambda }_1,{\lambda }_2\right){∥a∥}_{q,\Psi }$and then $\tilde{T}$is a bounded operator with $∥\tilde{T}∥\le B\left({\lambda }_1,{\lambda }_2\right)$. Since by Theorem 1, the constant factor in (15) is the best possible, we have $∥\tilde{T}∥=B\left({\lambda }_1,{\lambda }_2\right)$.

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)