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A relation between two simple Hardy-Mulholland-type inequalities with parameters

Abstract

By means of weight coefficients and the technique of real analysis, a new Hardy-Mulholland-type inequality with the kernel

$$ K_{\lambda}(m,n):=\frac{1}{\ln^{\lambda}U_{m}+\ln^{\lambda }V_{n}+\alpha |\ln^{\lambda}U_{m}-\ln^{\lambda}V_{n}|} $$

(\(-1<\alpha\leq1\), \(0<\lambda\leq2\); \(m,n\in\mathbf{N}\backslash\{ 1\}\)) and a best possible constant factor is provided, which is a relation between two simple Hardy-Mulholland-type inequalities with parameters. The equivalent forms, the operator expression with the norm, and some particular inequalities are studied. The lemmas and theorems of this paper provide an extensive account of this type of inequalities.

1 Introduction

Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(a_{m},b_{n}\geq 0\), \(a=\{a_{m}\}_{m=1}^{\infty}\in l^{p}\), \(b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}\), \(\|a\|_{p}=(\sum_{m=1}^{\infty}a_{m}^{p})^{\frac{1}{p}}>0\), \(\|b\|_{q}>0\). We have the following Hardy-Hilbert inequality:

$$ \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\frac{\pi}{p})}\|a \|_{p}\|b\|_{q}, $$
(1)

where the constant factor \(\frac{\pi}{\sin(\pi/p)}\) is the best possible (cf. [1]). We still have the following Hilbert-type inequality:

$$ \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\max\{m,n\}}< pq\|a \|_{p}\|b\|_{q} $$
(2)

with the best possible constant factor pq (cf. [1]). Also the following Mulholland inequality was given with the best possible constant factor \(\frac{\pi}{\sin(\pi/p)}\) (cf. [1], Theorem 343, replacing \(\frac{a_{m}}{m}\), \(\frac{b_{n}}{n}\) by \(a_{m}\), \(b_{n}\)):

$$ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln mn}< \frac {\pi }{\sin(\frac{\pi}{p})} \Biggl(\sum_{m=2}^{\infty}\frac {a_{m}^{p}}{m^{1-p}} \Biggr)^{\frac{1}{p}}\Biggl(\sum_{n=2}^{\infty} \frac{b_{n}^{q}}{n^{1-q}}\Biggr)^{\frac{1}{q}}. $$
(3)

Inequalities (1)-(3) are important in analysis and its applications (cf. [15]).

If \(\mu_{i},\upsilon_{j}>0\) (\(i,j\in\mathbf{N}=\{1,2,\ldots\}\)),

$$ U_{m}:=\sum_{i=1}^{m} \mu_{i},\qquad V_{n}:=\sum_{j=1}^{n} \upsilon_{j}\quad (m,n\in \mathbf{N}), $$
(4)

then we have the following Hardy-Hilbert-type inequality (cf. Theorem 321 of [1], replacing \(\mu_{m}^{1/q}a_{m}\) and \(\upsilon_{n}^{1/p}b_{n}\) by \(a_{m}\) and \(b_{n}\)):

$$ \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac {a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\frac{\pi}{p})} \Biggl( \sum_{m=1}^{\infty}\frac {a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty} \frac {b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}. $$
(5)

For \(\mu_{i}=\upsilon_{j}=1\) (\(i,j\in\mathbf{N}\)), inequality (5) reduces to (1).

By introducing an independent parameter \(\lambda\in(0,1]\), in 1998, Yang [6] gave an extension of the integral analogous of (1) with the kernel \(\frac{1}{(x+y)^{\lambda}}\) for \(p=q=2\). Following [6], Yang [7] gave extensions of (1) and (2) as follows.

If \(\lambda_{1},\lambda_{2}\in\mathbf{R}\), \(\lambda_{1}+\lambda _{2}=\lambda\), \(k_{\lambda}(x,y)\) is a finite non-negative homogeneous function of degree −λ, with

$$ k(\lambda_{1})= \int_{0}^{\infty}k_{\lambda}(t,1)t^{\lambda _{1}-1} \,dt\in \mathbf{R}_{+}, $$

and \(k_{\lambda}(x,y)x^{\lambda_{1}-1}\) (\(k_{\lambda}(x,y)y^{\lambda _{2}-1}\)) is decreasing with respect to \(x>0\) (\(y>0\)), \(\phi (x)=x^{p(1-\lambda _{1})-1}\), \(\psi(x)=x^{q(1-\lambda_{2})-1}\), then for \(a_{m},b_{n}\geq 0\),

$$ a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\phi}= \Biggl\{ a;\|a\|_{p,\phi }:= \Biggl( \sum_{m=1}^{\infty} \phi(m)|a_{m}|^{p} \Biggr) ^{\frac{1}{p}}< \infty \Biggr\} , $$

\(b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\psi}\), \(\|a\|_{p,\phi },\|b\|_{q,\psi }>0\), we have

$$ \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}k_{\lambda }(m,n)a_{m}b_{n}< k( \lambda_{1})\|a\|_{p,\phi}\|b\|_{q,\psi}, $$
(6)

where the constant factor \(k(\lambda_{1})\) is still the best possible. Clearly, for \(\lambda=1\), \(\lambda_{1}=\frac{1}{q}\), \(\lambda_{2}=\frac {1}{p}\), \(k_{1}(x,y)=\frac{1}{x+y}(\frac{1}{\max\{x,y\}})\), inequality (6) reduces to (1) ((2)).

Some other new results including multidimensional Hilbert-type inequalities, Hardy-Hilbert-type inequalities and Hardy-Mulholland-type inequalities are provided by [830].

In this paper, by means of weight coefficients and the technique of real analysis, a new Hardy-Mulholland-type inequality with the following kernel:

$$ K_{\lambda}(m,n):=\frac{1}{\ln^{\lambda}U_{m}+\ln^{\lambda }V_{n}+\alpha |\ln^{\lambda}U_{m}-\ln^{\lambda}V_{n}|} $$
(7)

(\(-1<\alpha\leq1\), \(0<\lambda\leq2\); \(m,n\in\mathbf{N}\backslash\{ 1\}\)) and a best possible constant factor is provided, which is a relation between two simple Hardy-Mulholland-type inequalities similarly to (2) and (3). The equivalent forms, the operator expressions with the norm and some particular inequalities are studied. The lemmas and theorems of this paper provide an extensive account of this type of inequalities.

2 An example and some lemmas

Example 1

For \(-1<\alpha\leq1\), \(0<\lambda_{1},\lambda_{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda\), we set

$$ k_{\lambda}(x,y):=\frac{1}{x^{\lambda}+y^{\lambda}+\alpha|x^{\lambda }-y^{\lambda}|}\quad \bigl((x,y)\in \mathbf{R}_{+}^{2}=\mathbf{R}_{+}\times \mathbf{R}_{+}\bigr). $$
(8)

Then by (7), it follows that \(K_{\lambda}(m,n)=k_{\lambda}(\ln U_{m},\ln V_{n})\). We find for \(-1<\alpha\leq1\), \(\lambda_{1},\lambda _{2}>0\),

$$\begin{aligned} 0 < &k_{\alpha}(\lambda_{1}):= \int_{0}^{\infty}k_{\lambda }(1,t)t^{\lambda _{2}-1} \,dt= \int_{0}^{\infty}k_{\lambda}(t,1)t^{\lambda_{1}-1} \,dt \\ =& \int_{0}^{\infty}\frac{t^{\lambda_{1}-1}\,dt}{t^{\lambda}+1+\alpha |t^{\lambda}-1|}= \int_{0}^{1}\frac{t^{\lambda_{1}-1}+t^{\lambda _{2}-1}}{1+\alpha+(1-\alpha)t^{\lambda}}\,dt \\ \leq& \int_{0}^{1}\frac{t^{\lambda_{1}-1}+t^{\lambda _{2}-1}}{1+\alpha}\,dt= \frac{1}{1+\alpha}\biggl(\frac{1}{\lambda_{1}}+\frac{1}{\lambda _{2}}\biggr)< \infty, \end{aligned}$$
(9)

namely, \(k_{\alpha}(\lambda_{1})\in\mathbf{R}_{+}\).

(1) In the following, we express \(k_{\alpha}(\lambda_{1})\) in a few cases.

(i) For \(\alpha=0\), we obtain

$$\begin{aligned} k_{0}(\lambda_{1}) =& \int_{0}^{\infty}\frac{t^{\lambda_{1}-1}}{t^{\lambda}+1}\,dt \\ =&\frac{1}{\lambda} \int_{0}^{\infty}\frac{u^{(\lambda_{1}/\lambda )-1}}{u+1}\,du= \frac{\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}. \end{aligned}$$
(10)

(ii) For \(\alpha=1\), we obtain

$$\begin{aligned} k_{1}(\lambda_{1}) =& \int_{0}^{\infty}\frac{t^{\lambda_{1}-1}}{t^{\lambda}+1+|t^{\lambda}-1|}\,dt= \int_{0}^{\infty}\frac{t^{\lambda _{1}-1}}{2\max\{t^{\lambda},1\}}\,dt \\ =&\frac{1}{2} \int_{0}^{1}\bigl(t^{\lambda_{1}-1}+t^{\lambda _{2}-1} \bigr)\,dt=\frac{\lambda}{2\lambda_{1}\lambda_{2}}. \end{aligned}$$
(11)

(iii) For \(0<\alpha<1\), \(0<\frac{1-\alpha}{1+\alpha}<1\), in view of (9) and the Lebesgue term by term integration theorem (cf. [31]), we find

$$\begin{aligned} k_{\alpha}(\lambda_{1}) =&\frac{1}{1+\alpha} \int_{0}^{1}\frac {t^{\lambda _{1}-1}+t^{\lambda_{2}-1}}{1+\frac{1-\alpha}{1+\alpha}t^{\lambda }}\,dt \\ =&\frac{1}{1+\alpha} \int_{0}^{1}\bigl(t^{\lambda_{1}-1}+t^{\lambda _{2}-1} \bigr)\sum_{k=0}^{\infty}(-1)^{k} \biggl( \frac{1-\alpha}{1+\alpha} \biggr) ^{k}t^{\lambda k}\,dt \\ =&\frac{1}{1+\alpha} \int_{0}^{1}\bigl(t^{\lambda_{1}-1}+t^{\lambda _{2}-1} \bigr)\sum_{j=0}^{\infty} \biggl( \frac{1-\alpha}{1+\alpha} \biggr) ^{2j} \biggl( 1-\frac{1-\alpha}{1+\alpha}t^{\lambda} \biggr) t^{2\lambda j}\,dt \\ =&\frac{1}{1+\alpha}\sum_{j=0}^{\infty} \int_{0}^{1}\bigl(t^{\lambda _{1}-1}+t^{\lambda_{2}-1} \bigr) \biggl( \frac{1-\alpha}{1+\alpha} \biggr) ^{2j} \biggl( 1- \frac{1-\alpha}{1+\alpha}t^{\lambda} \biggr) t^{2\lambda j}\,dt \\ =&\frac{1}{1+\alpha}\sum_{k=0}^{\infty}(-1)^{k} \biggl( \frac{1-\alpha }{1+\alpha} \biggr) ^{k} \int_{0}^{1}\bigl(t^{\lambda_{1}-1}+t^{\lambda _{2}-1} \bigr)t^{\lambda k}\,dt \\ =&\frac{1}{1+\alpha}\sum_{k=0}^{\infty}(-1)^{k} \biggl( \frac{1-\alpha }{1+\alpha} \biggr) ^{k} \biggl( \frac{1}{\lambda k+\lambda_{1}}+ \frac{1}{\lambda k+\lambda_{2}} \biggr). \end{aligned}$$
(12)

(iv) For \(-1<\alpha<0\), \(0<\frac{1+\alpha}{1-\alpha}<1\), by (9) and the Lebesgue term by term integration theorem (cf. [31]), we find

$$\begin{aligned} k_{\alpha}(\lambda_{1}) =&\frac{1}{1+\alpha} \int_{0}^{1}\frac {t^{\lambda _{1}-1}+t^{\lambda_{2}-1}}{1+\frac{1-a}{1+a}t^{\lambda}}\,dt \\ \stackrel{ v=\frac{1+\alpha}{1-\alpha}t^{-\lambda} }{=}&\frac {1}{\lambda (1+\alpha)} \int_{\frac{1+\alpha}{1-\alpha}}^{\infty}\frac{1}{v+1} \\ &{}\times \biggl[ \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac {\lambda _{1}}{\lambda}}v^{\frac{-\lambda_{1}}{\lambda}}+ \biggl( \frac {1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}}v^{\frac{-\lambda _{2}}{\lambda}} \biggr] \,dv \\ =&\frac{1}{\lambda(1+\alpha)} \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{1}}{\lambda}} \int_{0}^{\infty}\frac{1}{v+1}v^{(1-\frac {\lambda_{1}}{\lambda})-1}\,dv \\ &{}+\frac{1}{\lambda(1+\alpha)} \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}} \int_{0}^{\infty}\frac{1}{v+1}v^{(1-\frac {\lambda_{2}}{\lambda})-1}\,dv \\ &{}-\frac{1}{\lambda(1+\alpha)} \int_{0}^{\frac{1+\alpha}{1-\alpha }}\frac{1}{v+1} \\ &{}\times \biggl[ \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac {\lambda _{1}}{\lambda}}v^{\frac{-\lambda_{1}}{\lambda}}+ \biggl( \frac {1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}}v^{\frac{-\lambda _{2}}{\lambda}} \biggr] \,dv \\ =&\frac{1}{\lambda(1+\alpha)}\biggl(\frac{1+\alpha}{1-\alpha}\biggr)^{\frac {\lambda _{1}}{\lambda}} \frac{\pi}{\sin(\frac{\pi\lambda_{2}}{\lambda})} \\ &{}+\frac{1}{\lambda(1+\alpha)}\biggl(\frac{1+\alpha}{1-\alpha}\biggr)^{\frac {\lambda _{2}}{\lambda}} \frac{\pi}{\sin(\frac{\pi\lambda_{1}}{\lambda})} \\ &{}-\frac{1}{\lambda(1+\alpha)} \int_{0}^{\frac{1+\alpha}{1-\alpha}}\sum_{k=0}^{\infty}(-1)^{k}v^{k} \\ &{}\times \biggl[ \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac {\lambda _{1}}{\lambda}}v^{\frac{-\lambda_{1}}{\lambda}}+ \biggl( \frac {1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}}v^{\frac{-\lambda _{2}}{\lambda}} \biggr] \,dv \\ =&\frac{1}{\lambda(1+\alpha)} \biggl[ \biggl( \frac{1+\alpha}{1-\alpha } \biggr) ^{\frac{\lambda_{1}}{\lambda}}+ \biggl( \frac{1+\alpha }{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}} \biggr] \frac{\pi}{\sin(\frac {\pi \lambda_{1}}{\lambda})} \\ &{}-\frac{1}{\lambda(1+\alpha)} \int_{0}^{\frac{1+\alpha}{1-\alpha}}\sum_{k=0}^{\infty}(-1)^{k}v^{k} \\ &{}\times \biggl[ \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac {\lambda _{1}}{\lambda}}v^{\frac{-\lambda_{1}}{\lambda}}+ \biggl( \frac {1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}}v^{\frac{-\lambda _{2}}{\lambda}} \biggr] \,dv \\ =&\frac{1}{\lambda(1+\alpha)} \biggl[ \biggl( \frac{1+\alpha}{1-\alpha } \biggr) ^{\frac{\lambda_{1}}{\lambda}}+ \biggl( \frac{1+\alpha }{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}} \biggr] \frac{\pi}{\sin(\frac {\pi \lambda_{1}}{\lambda})} \\ &{}-\frac{1}{\lambda(1+\alpha)} \int_{0}^{\frac{1+\alpha}{1-\alpha}}\sum_{k=0}^{\infty} \bigl(v^{2k}-v^{2k+1}\bigr) \\ &{}\times \biggl[ \biggl(\frac{1+\alpha}{1-\alpha}\biggr)^{\frac{\lambda _{1}}{\lambda}}v^{\frac{-\lambda_{1}}{\lambda}}+ \biggl(\frac{1+\alpha}{1-\alpha}\biggr)^{\frac {\lambda_{2}}{\lambda}}v^{\frac{-\lambda_{2}}{\lambda}} \biggr] \,dv \\ =&\frac{1}{\lambda(1+a)} \biggl[ \biggl(\frac{1+a}{1-a}\biggr)^{\frac{\lambda_{1}}{ \lambda}}+ \biggl(\frac{1+a}{1-a}\biggr)^{\frac{\lambda_{2}}{\lambda}} \biggr] \frac{\pi}{\sin(\frac{\pi\lambda_{1}}{\lambda})} \\ &{}-\frac{1}{\lambda(1+a)}\sum_{k=0}^{\infty} \int_{0}^{\frac{1+a}{1-a}}\bigl(v^{2k}-v^{2k+1} \bigr) \\ &{}\times \biggl[ \biggl(\frac{1+a}{1-a}\biggr)^{\frac{\lambda_{1}}{\lambda }}v^{\frac{-\lambda_{1}}{\lambda}}+ \biggl(\frac{1+a}{1-a}\biggr)^{\frac{\lambda _{2}}{\lambda}}v^{\frac{-\lambda_{2}}{\lambda}} \biggr] \,dv \\ =&\frac{1}{\lambda(1+\alpha)} \biggl[ \biggl( \frac{1+\alpha}{1-\alpha } \biggr) ^{\frac{\lambda_{1}}{\lambda}}+ \biggl( \frac{1+\alpha }{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}} \biggr] \frac{\pi}{\sin(\frac {\pi \lambda_{1}}{\lambda})} \\ &{}-\frac{1}{\lambda(1+\alpha)}\sum_{k=0}^{\infty} \int_{0}^{\frac {1+\alpha }{1-\alpha}}(-1)^{k}v^{k} \\ &{}\times \biggl[ \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac {\lambda _{1}}{\lambda}}v^{\frac{\lambda_{2}}{\lambda}-1}+ \biggl( \frac {1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}}v^{\frac{\lambda _{1}}{\lambda}-1} \biggr] \,dv \\ =&\frac{1}{1+\alpha} \biggl[ \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{1}}{\lambda}}+ \biggl( \frac{1+\alpha}{1-\alpha} \biggr) ^{\frac{\lambda_{2}}{\lambda}} \biggr] \frac{\pi}{\lambda\sin(\frac {\pi \lambda_{1}}{\lambda})} \\ &{}-\frac{1}{1+\alpha}\sum_{k=0}^{\infty}(-1)^{k} \biggl( \frac{1+\alpha }{1-\alpha} \biggr) ^{k+1} \biggl( \frac{1}{\lambda k+\lambda_{2}}+ \frac {1}{\lambda k+\lambda_{1}} \biggr) . \end{aligned}$$
(13)

(v) For \(\lambda_{1}=\lambda_{2}=\frac{\lambda}{2}\in(0,1]\), \(-1<\alpha <1\), in view of (9), we find

$$\begin{aligned} k_{\alpha}\biggl(\frac{\lambda}{2}\biggr) =&\frac{2}{1+\alpha} \int_{0}^{1}\frac {t^{\frac{\lambda}{2}-1}}{(1+\frac{1-\alpha}{1+\alpha})t^{\lambda}}\,dt \\ \stackrel{ u=(\frac{1-\alpha}{1+\alpha}t^{\lambda})^{\frac{1}{2}} }{=}& \frac{4(\frac{1+\alpha}{1-\alpha})^{\frac{1}{2}}}{\lambda(1+\alpha)} \int_{0}^{(\frac{1-\alpha}{1+\alpha})^{\frac{1}{2}}}\frac {1}{1+u^{2}}\,du \\ =&\frac{4}{\lambda(1-\alpha^{2})^{\frac{1}{2}}}\arctan \biggl( \frac{1-\alpha}{1+\alpha} \biggr) ^{\frac{1}{2}}. \end{aligned}$$

We still have \(k_{1}(\frac{\lambda}{2})=\lim_{\alpha\rightarrow 1^{-}}k_{\alpha}(\frac{\lambda}{2})=\frac{2}{\lambda}\).

(2) For fixed \(x>0\), in view of \(-1<\alpha\leq1\), \(\lambda>0\), we find that

$$\begin{aligned} k_{\lambda}(x,y) =&\frac{1}{x^{\lambda}+y^{\lambda}+\alpha |x^{\lambda }-y^{\lambda}|} \\ =&\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{(1+\alpha)x^{\lambda}+(1-\alpha)y^{\lambda}},&0< y< x, \\ \frac{1}{(1-\alpha)x^{\lambda}+(1+\alpha)y^{\lambda}},&y\geq x,\end{array}\displaystyle \right . \end{aligned}$$

is decreasing for \(y>0\) and strictly decreasing for \(y\in{}[ x,\infty)\). In the same way, for fixed \(y>0\), \(k_{\lambda}(x,y)\) is decreasing for \(x>0\) and strictly decreasing for \(x\in{}[ y,\infty)\).

Lemma 1

(cf. [29])

Suppose that \(g(t)\) (>0) is decreasing in \(\mathbf{R}_{+}\) and strictly decreasing in \([n_{0},\infty)\) (\(n_{0}\in \mathbf{N}\)), satisfying \(\int_{0}^{\infty}g(t)\,dt\in\mathbf{R}_{+}\). We have

$$ \int_{1}^{\infty}g(t)\,dt< \sum _{n=1}^{\infty}g(n)< \int_{0}^{\infty}g(t)\,dt. $$
(14)

Lemma 2

Suppose that \(U_{m}\) and \(V_{n}\) are defined by (4) with \(\mu_{1},\upsilon_{1}\geq1\), \(-1<\alpha\leq1\), \(0<\lambda_{1},\lambda_{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda \), \(K_{\lambda}(m,n)\), and \(k_{\alpha}(\lambda_{1})\) are indicated by (7) and (9). Define the following weight coefficients:

$$\begin{aligned}& \omega(\lambda_{2},m) : =\sum_{n=2}^{\infty}K_{\lambda}(m,n) \frac{\upsilon_{n}\ln^{\lambda_{1}}U_{m}}{V_{n}\ln^{1-\lambda _{2}}V_{n}},\quad m\in \mathbf{N}\backslash\{1\}, \end{aligned}$$
(15)
$$\begin{aligned}& \varpi(\lambda_{1},n) : =\sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac {\mu _{m}\ln^{\lambda_{2}}V_{n}}{U_{m}\ln^{1-\lambda_{1}}U_{m}},\quad n\in \mathbf{N}\backslash\{1\}. \end{aligned}$$
(16)

Then we have the following inequalities:

$$\begin{aligned}& \omega(\lambda_{2},m) < k_{\alpha}(\lambda_{1}) \quad \bigl(0< \lambda_{2}\leq 1,\lambda_{1}>0;m\in\mathbf{N} \backslash\{1\}\bigr), \end{aligned}$$
(17)
$$\begin{aligned}& \varpi(\lambda_{1},n) < k_{\alpha}(\lambda_{1}) \quad \bigl(0< \lambda_{1}\leq 1,\lambda_{2}>0;n\in\mathbf{N} \backslash\{1\}\bigr). \end{aligned}$$
(18)

Proof

We set \(\mu(t):=\mu_{m}\), \(t\in(m-1,m]\) (\(m\in\mathbf{N}\)); \(\upsilon(t):=\upsilon_{n}\), \(t\in(n-1,n]\) (\(n\in\mathbf{N}\)), and

$$ U(x):= \int_{0}^{x}\mu(t)\,dt\quad (x\geq0),\qquad V(y):= \int_{0}^{y}\upsilon (t)\,dt\quad (y\geq 0). $$
(19)

Then it follows that \(U(m)=U_{m}\), \(V(n)=V_{n}\) (\(m,n\in\mathbf {N}\)). \(U^{\prime }(x)=\mu(x)=\mu_{m}\), for \(x\in(m-1,m)\) (\(m\in\mathbf{N}\)); \(V^{\prime }(y)=\upsilon(y)=\upsilon_{n} \), for \(y\in(n-1,n)\) (\(n\in\mathbf{N}\)). Since \(V(y)\) is strictly increasing in \((n-1,n]\), \(0<\lambda_{2}\leq 1\), \(\lambda_{1}>0\), in view of Example 1(2) and Lemma 1, we find

$$\begin{aligned} \omega(\lambda_{2},m) =&\sum_{n=2}^{\infty} \int_{n-1}^{n}k_{\lambda }(\ln U_{m}, \ln V_{n})\frac{\ln^{\lambda_{1}}U_{m}}{V_{n}\ln ^{1-\lambda _{2}}V_{n}}V^{\prime}(y)\,dy \\ < &\sum_{n=2}^{\infty} \int_{n-1}^{n}k_{\lambda}\bigl(\ln U_{m},\ln V(y)\bigr)\frac{\ln^{\lambda_{1}}U_{m}}{V(y)\ln^{1-\lambda_{2}}V(y)}V^{\prime }(y)\,dy \\ =& \int_{1}^{\infty}k_{\lambda}\bigl(\ln U_{m},\ln V(y)\bigr)\frac{\ln^{\lambda _{1}}U_{m}}{V(y)\ln^{1-\lambda_{2}}V(y)}V^{\prime}(y)\,dy. \end{aligned}$$

Setting \(t=\frac{\ln V(y)}{\ln U_{m}}\), we obtain \(\frac {1}{V(y)}V^{\prime }(y)\,dy=\ln U_{m}\,dt\) and

$$ \omega(\lambda_{2},m)< \int_{\frac{\ln V(1)}{\ln U_{m}}}^{\frac{\ln V(\infty)}{\ln U_{m}}}k_{\lambda}(1,t)t^{\lambda_{2}-1} \,dt\leq \int_{0}^{\infty}k_{\lambda}(1,t)t^{\lambda_{2}-1} \,dt=k_{\alpha }(\lambda _{1}). $$
(20)

Hence, we have (17). In the same way, we have (18). □

Note

We do not need the condition \(\lambda_{1}\leq1\) (\(\lambda _{2}\leq1\)) to obtain (17) ((18)).

Lemma 3

As regards the assumptions of Lemma  2, if \(U(\infty )=V(\infty)=\infty\), there exist \(m_{0},n_{0}\in\mathbf{N}\backslash \{1\}\), such that \(\mu_{m}\geq\mu_{m+1}\) (\(m\in \{m_{0},m_{0}+1,\ldots\}\)), \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\in \{n_{0},n_{0}+1,\ldots\}\)), then:

  1. (i)

    For \(m,n\in\mathbf{N}\backslash \{1\}\), we have

    $$\begin{aligned}& k_{\alpha}(\lambda_{1}) \bigl(1-\theta(\lambda_{2},m) \bigr) < \omega(\lambda _{2},m)\quad (0< \lambda_{2}\leq1, \lambda_{1}>0), \end{aligned}$$
    (21)
    $$\begin{aligned}& k_{\alpha}(\lambda_{1}) \bigl(1-\vartheta( \lambda_{1},n)\bigr) < \varpi (\lambda _{1},n)\quad (0< \lambda_{1}\leq1,\lambda_{2}>0), \end{aligned}$$
    (22)

    where

    $$\begin{aligned}& \theta(\lambda_{2},m) : =\frac{1}{k_{\alpha}(\lambda_{1})} \int _{0}^{\frac{\ln V_{n_{0}}}{\ln U_{m}}}\frac{t^{\lambda_{2}-1}\,dt}{1+t^{\lambda }+\alpha|1-t^{\lambda}|} \\& \hphantom{\theta(\lambda_{2},m)} = O\biggl(\frac{1}{\ln^{\lambda_{2}}U_{m}}\biggr)\in(0,1), \end{aligned}$$
    (23)
    $$\begin{aligned}& \vartheta(\lambda_{1},n) : =\frac{1}{k_{\alpha}(\lambda_{1})} \int _{0}^{\frac{\ln U_{m_{0}}}{\ln V_{n}}}\frac{t^{\lambda_{2}-1}\,dt}{1+t^{\lambda }+\alpha|1-t^{\lambda}|} \\& \hphantom{\vartheta(\lambda_{1},n)} = O\biggl(\frac{1}{\ln^{\lambda_{1}}V_{n}}\biggr)\in(0,1). \end{aligned}$$
    (24)
  2. (ii)

    For any \(b>0\), we have

    $$\begin{aligned}& \sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}} = \frac{1}{b} \biggl( \frac{1}{\ln^{b}U_{m_{0}}}+bO_{1}(1) \biggr) , \end{aligned}$$
    (25)
    $$\begin{aligned}& \sum_{n=2}^{\infty}\frac{\upsilon_{n}}{V_{n}\ln^{1+b}V_{n}} = \frac {1}{b} \biggl( \frac{1}{\ln^{b}V_{n_{0}}}+bO_{2}(1) \biggr) . \end{aligned}$$
    (26)

Proof

Since \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\geq n_{0}\)), \(0<\lambda_{2}\leq1\), \(\lambda_{1}>0\), and \(V(\infty)=\infty\), by Example 1(2), Lemma 1, and (23), we find

$$\begin{aligned} \omega(\lambda_{2},m) \geq&\sum_{n=n_{0}}^{\infty}K_{\lambda }(m,n) \frac{\upsilon_{n+1}\ln^{\lambda_{1}}U_{m}}{V_{n}\ln^{1-\lambda_{2}}V_{n}} \\ =&\sum_{n=n_{0}}^{\infty} \int_{n}^{n+1}k_{\lambda}(\ln U_{m}, \ln V_{n})\frac{\ln^{\lambda_{1}}U_{m}}{V_{n}\ln^{1-\lambda _{2}}V_{n}}V^{\prime }(y)\,dy \\ >&\sum_{n=n_{0}}^{\infty} \int_{n}^{n+1}k_{\lambda}\bigl(\ln U_{m},\ln V(y)\bigr)\frac{\ln^{\lambda_{1}}U_{m}}{V(y)\ln^{1-\lambda_{2}}V(y)}V^{\prime }(y) \,dy \\ =& \int_{n_{0}}^{\infty}k_{\lambda}\bigl(\ln U_{m},\ln V(y)\bigr)\frac{\ln ^{\lambda _{1}}U_{m}}{V(y)\ln^{1-\lambda_{2}}V(y)}V^{\prime}(y)\,dy \\ \stackrel{ t=\frac{\ln V(y)}{\ln U_{m}} }{=}& \int_{\frac{\ln V_{n_{0}}}{\ln U_{m}}}^{\infty}\frac{t^{\lambda_{2}-1}\,dt}{1+t^{\lambda}+\alpha |1-t^{\lambda}|}=k_{\alpha}( \lambda_{1}) \bigl(1-\theta(\lambda_{2},m)\bigr). \end{aligned}$$

We obtain

$$ 0< \theta(\lambda_{2},m)\leq\frac{1}{k_{\alpha}(\lambda_{1})} \int _{0}^{\frac{\ln V_{n_{0}}}{\ln U_{m}}}t^{\lambda_{2}-1}\,dt= \frac{1}{\lambda _{2}k_{\alpha}(\lambda_{1})} \biggl( \frac{\ln V_{n_{0}}}{\ln U_{m}} \biggr) ^{\lambda_{2}}, $$

namely, \(\theta(\lambda_{2},m)=O(\frac{1}{\ln^{\lambda_{2}}U_{m}})\). Hence we have (21). In the same way, we obtain (22).

For \(b>0\), we find

$$\begin{aligned}& \sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}=\sum _{m=2}^{m_{0}}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}+ \sum_{m=m_{0}+1}^{\infty}\frac {\mu _{m}}{U_{m}\ln^{1+b}U_{m}} \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}} = \sum_{m=2}^{m_{0}} \frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}+\sum_{m=m_{0}+1}^{\infty} \int_{m-1}^{m}\frac{U^{\prime}(x)}{U_{m}\ln ^{1+b}U_{m}}\,dx \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}} < \sum_{m=2}^{m_{0}} \frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}+\sum_{m=m_{0}+1}^{\infty} \int_{m-1}^{m}\frac{U^{\prime}(x)}{U(x)\ln ^{1+b}U(x)}\,dx \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}} = \sum_{m=2}^{m_{0}} \frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}+ \int_{m_{0}}^{\infty}\frac{dU(x)}{U(x)\ln^{1+b}U(x)} \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}}= \frac{1}{b} \Biggl( \frac{1}{\ln^{b}U_{m_{0}}}+b\sum _{m=2}^{m_{0}}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}} \Biggr) , \\& \sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}\geq \sum _{m=m_{0}}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}\geq \sum _{m=m_{0}}^{\infty}\frac{\mu_{m+1}}{U_{m}\ln^{1+b}U_{m}} \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}} = \sum_{m=m_{0}}^{\infty} \int_{m}^{m+1}\frac{U^{\prime }(x)\,dx}{U_{m}\ln ^{1+b}U_{m}}>\sum _{m=m_{0}}^{\infty} \int_{m}^{m+1}\frac{U^{\prime }(x)\,dx}{U(x)\ln^{1+b}U(x)} \\& \hphantom{\sum_{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+b}U_{m}}} = \int_{m_{0}}^{\infty}\frac{dU(x)}{U(x)\ln^{1+b}U(x)}=\frac{1}{b\ln ^{b}U_{m_{0}}}. \end{aligned}$$

Hence we have (25). In the same way, we have (26). □

Lemma 4

If \(-1<\alpha\leq1\), \(0<\lambda_{1},\lambda_{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(k_{\alpha}(\lambda_{1})\) is indicated in (9), then for \(0<\delta<\min\{\lambda _{1},\lambda _{2}\}\), we have

$$ k_{\alpha}(\lambda_{1}\pm\delta)=k_{\alpha}(\lambda _{1})+o(1)\quad \bigl(\delta \rightarrow0^{+}\bigr). $$
(27)

Proof

We find for \(0<\delta<\min\{\lambda_{1},\lambda_{2}\}\),

$$\begin{aligned}& \bigl\vert k_{\alpha}(\lambda_{1}+\delta)-k_{\alpha}( \lambda_{1})\bigr\vert \\& \quad \leq \int_{0}^{\infty}\frac{t^{\lambda_{1}-1}|t^{\delta}-1|}{t^{\lambda }+1+\alpha|t^{\lambda}-1|}\,dt \\& \quad = \int_{0}^{1}\frac{t^{\lambda_{1}-1}(1-t^{\delta})\,dt}{1+\alpha +(1-\alpha)t^{\lambda}}+ \int_{1}^{\infty}\frac{t^{\lambda _{1}-1}(t^{\delta}-1)\,dt}{1-\alpha+(1+\alpha)t^{\lambda}} \\& \quad \leq \frac{1}{1+\alpha} \biggl[ \int_{0}^{1}t^{\lambda _{1}-1}\bigl(1-t^{\delta } \bigr)\,dt+ \int_{1}^{\infty}\frac{t^{\lambda_{1}-1}(t^{\delta }-1)}{t^{\lambda}}\,dt \biggr] \\& \quad = \frac{1}{1+\alpha} \biggl( \frac{1}{\lambda_{1}}-\frac{1}{\lambda _{1}+\delta}+ \frac{1}{\lambda_{2}-\delta}-\frac{1}{\lambda _{2}} \biggr) \rightarrow0\quad \bigl(\delta \rightarrow0^{+}\bigr). \end{aligned}$$

In the same way, we find

$$\begin{aligned}& \bigl\vert k_{\alpha}(\lambda_{1}-\delta)-k_{\alpha}( \lambda_{1})\bigr\vert \\& \quad \leq \frac{1}{1+\alpha} \biggl[ \int_{0}^{1}t^{\lambda _{1}-1}\bigl(t^{-\delta }-1 \bigr)\,dt+ \int_{1}^{\infty}\frac{t^{\lambda_{1}-1}(1-t^{-\delta})}{t^{\lambda}}\,dt \biggr] \\& \quad \leq \frac{1}{1+\alpha} \biggl( \frac{1}{\lambda_{1}-\delta}-\frac {1}{\lambda_{1}}+ \frac{1}{\lambda_{2}}-\frac{1}{\lambda_{2}+\delta } \biggr) \rightarrow0\quad \bigl(\delta \rightarrow0^{+}\bigr), \end{aligned}$$

and then we have (27). □

3 Main results

In the following, we agree that \(\mu_{1},\upsilon_{1}\geq1\), \(\mu _{i},\upsilon_{j}>0\) (\(i,j\in\mathbf{N}\backslash\{1\}\)), \(U_{m}\) and \(V_{n}\) are defined by (4), \(-1<\alpha\leq1\), \(0<\lambda _{1},\lambda_{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(K_{\lambda }(m,n)\), and \(k_{\alpha}(\lambda_{1})\) are indicated by (7) and (9), \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(a_{m},b_{n}\geq0 \) (\(m,n\in\mathbf{N}\backslash\{1\}\)),

$$ \|a\|_{p,\Phi_{\lambda}}:=\Biggl(\sum_{m=2}^{\infty} \Phi_{\lambda }(m)a_{m}^{p}\Biggr)^{\frac{1}{p}},\qquad \|b\|_{q,\Psi_{\lambda}}:=\Biggl(\sum_{n=2}^{\infty } \Psi_{\lambda}(n)b_{n}^{q}\Biggr)^{\frac{1}{q}}, $$

where

$$\begin{aligned}& \Phi_{\lambda}(m) : =\biggl(\frac{U_{m}}{\mu_{m}}\biggr)^{p-1}(\ln U_{m})^{p(1-\lambda_{1})-1}, \\& \Psi_{\lambda}(n) : =\biggl(\frac{V_{n}}{\upsilon_{n}}\biggr)^{q-1}(\ln V_{n})^{q(1-\lambda_{2})-1}\quad \bigl(m,n\in\mathbf{N}\backslash\{1\}\bigr). \end{aligned}$$

Theorem 1

If \(0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi _{\lambda}}<\infty\), then we have the following equivalent inequalities:

$$\begin{aligned}& I : =\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}K_{\lambda }(m,n)a_{m}b_{n}< k_{\alpha}( \lambda_{1})\|a\|_{p,\Phi_{\lambda }}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(28)
$$\begin{aligned}& J : = \Biggl[ \sum_{n=2}^{\infty} \frac{\upsilon_{n}}{V_{n}}(\ln V_{n})^{p\lambda_{2}-1}\Biggl(\sum _{m=2}^{\infty}K_{\lambda}(m,n)a_{m} \Biggr)^{p} \Biggr] ^{\frac{1}{p}}< k_{\alpha}( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}. \end{aligned}$$
(29)

In particular, for \(\lambda_{1}=\lambda_{2}=\frac{\lambda}{2}\in(0,1]\), the constant factor \(k_{\alpha}(\lambda_{1})\) in the above inequalities is expressed in the following form:

$$ k_{\alpha}\biggl(\frac{\lambda}{2}\biggr)=\frac{4}{\lambda(1-\alpha^{2})^{\frac {1}{2}}}\arctan \biggl( \frac{1-\alpha}{1+\alpha} \biggr) ^{\frac{1}{2}}. $$

Proof

By Hölder’s inequality with weight (cf. [32]) and (16), we have

$$\begin{aligned} \Biggl( \sum_{m=2}^{\infty}K_{\lambda}(m,n)a_{m} \Biggr) ^{p} =&\Biggl[ \sum_{m=2}^{\infty}K_{\lambda}(m,n) \biggl( \frac{U_{m}^{1/q}(\ln U_{m})^{(1-\lambda _{1})/q}}{(\ln V_{n})^{(1-\lambda_{2})/p}\mu_{m}^{1/q}}a_{m} \biggr) \biggl( \frac {(\ln V_{n})^{(1-\lambda_{2})/p}\mu_{m}^{1/q}}{U_{m}^{1/q}(\ln U_{m})^{(1-\lambda_{1})/q}} \biggr) \Biggr] ^{p} \\ \leq&\sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac{U_{m}^{p-1}(\ln U_{m})^{(1-\lambda_{1})p/q}}{(\ln V_{n})^{1-\lambda_{2}}\mu _{m}^{p/q}}a_{m}^{p} \\ &{}\times \Biggl[ \sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac{(\ln V_{n})^{(1-\lambda_{2})(q-1)}\mu_{m}}{U_{m}(\ln U_{m})^{1-\lambda _{1}}} \Biggr] ^{p-1} \\ =&\frac{(\varpi(\lambda_{1},n))^{p-1}V_{n}}{(\ln V_{n})^{p\lambda _{2}-1}\upsilon_{n}}\sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac{\upsilon _{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda_{1})(p-1)}a_{m}^{p}}{V_{n}(\ln V_{n})^{1-\lambda_{2}}\mu_{m}^{p-1}}. \end{aligned}$$
(30)

Then by (18), we obtain

$$\begin{aligned} J \leq&\bigl(k_{\alpha}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{n=2}^{\infty }\sum _{m=2}^{\infty}K_{\lambda}(m,n)\frac{\upsilon_{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda_{1})(p-1)}a_{m}^{p}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}\mu_{m}^{p-1}} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{\alpha}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=2}^{\infty }\sum _{n=2}^{\infty}K_{\lambda}(m,n)\frac{\upsilon_{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda_{1})(p-1)}a_{m}^{p}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}\mu_{m}^{p-1}} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{\alpha}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=2}^{\infty }\omega(\lambda_{2},m) \frac{(\ln U_{m})^{p(1-\lambda_{1})-1}}{U_{m}^{1-p}\mu_{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$
(31)

Hence, by (17), we have (29).

By Hölder’s inequality (cf. [32]), we have

$$\begin{aligned} I =&\sum_{n=2}^{\infty} \Biggl[ \biggl( \frac{\upsilon_{n}}{V_{n}}\biggr)^{1/p}(\ln V_{n})^{\lambda_{2}-\frac{1}{p}} \sum_{m=2}^{\infty}K_{\lambda }(m,n)a_{m} \Biggr] \\ &{}\times \biggl[ \biggl(\frac{\upsilon_{n}}{V_{n}}\biggr)^{-1/p}(\ln V_{n})^{\frac {1}{p}-\lambda_{2}}b_{n} \biggr] \leq J\|b \|_{q,\Psi_{\lambda}}, \end{aligned}$$
(32)

and then by (29), we have (28).

On the other hand, assuming that (29) is valid, setting

$$ b_{n}:=\frac{\upsilon_{n}}{V_{n}}(\ln V_{n})^{p\lambda_{2}-1} \Biggl( \sum_{m=2}^{\infty}K_{\lambda}(m,n)a_{m} \Biggr) ^{p-1},\quad n\in\mathbf{ N}\backslash\{1\}, $$
(33)

we find \(J^{p}=\|b\|_{q,\Psi_{\lambda}}^{q}\). If \(J=0\), then (29) is trivially valid; if \(J=\infty\), then by (31) and (17), it is impossible; if \(0< J<\infty\), then by (28), it follows that

$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q} = J^{p}=I< k_{\alpha}( \lambda _{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(34)
$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q-1} = J< k_{\alpha}(\lambda _{1})\|a\|_{p,\Phi _{\lambda}}, \end{aligned}$$
(35)

namely, (29) follows, which is equivalent to (28). □

Theorem 2

With regards the assumptions of Theorem  1, if \(U(\infty )=V(\infty)=\infty\), there exist \(m_{0},n_{0}\in\mathbf{N}\backslash \{1\}\), such that \(\mu_{m}\geq\mu_{m+1}\) (\(m\in \{m_{0},m_{0}+1,\ldots\}\)), \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\in \{n_{0},n_{0}+1,\ldots\}\)), then the constant factor \(k_{\alpha }(\lambda_{1})\) in (28) and (29) is the best possible.

Proof

For \(\varepsilon\in(0,p\lambda_{1})\), we set \(\tilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon}{p}\) (\(\in(0,1)\)), \(\tilde{ \lambda}_{2}=\lambda_{2}+\frac{\varepsilon}{p}\) (>0), \(\tilde {a}=\{\tilde{a}_{m}\}_{m=2}^{\infty}\), \(\tilde{b}=\{\tilde{b}_{n}\}_{n=2}^{\infty}\), where

$$ \begin{aligned} &\tilde{a}_{m} : =\frac{\mu_{m}}{U_{m}} \ln^{\tilde{\lambda}_{1}-1}U_{m}=\frac{\mu_{m}}{U_{m}}\ln^{\lambda_{1}-\frac{\varepsilon }{p}-1}U_{m}, \\ &\tilde{b}_{n} : =\frac{\upsilon_{n}}{V_{n}}\ln^{\tilde {\lambda}_{2}-\varepsilon-1}V_{n}= \frac{\upsilon_{n}}{V_{n}}\ln^{\lambda_{2}- \frac{\varepsilon}{q}-1}V_{n}. \end{aligned} $$
(36)

Then by (25), (26), and (22), we find

$$\begin{aligned}& \|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi_{\lambda}}= \Biggl( \sum _{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+\varepsilon }U_{m}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=2}^{\infty} \frac{\upsilon_{n}}{V_{n}\ln^{1+\varepsilon}V_{n}} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi_{\lambda}}} = \frac{1}{\varepsilon} \biggl( \frac{1}{\ln^{\varepsilon}U_{m_{0}}}+ \varepsilon O_{1}(1) \biggr) ^{\frac{1}{p}} \biggl( \frac{1}{\ln ^{\varepsilon }V_{n_{0}}}+\varepsilon O_{2}(1) \biggr) ^{\frac{1}{q}}, \\& \tilde{I} : =\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}K_{\lambda }(m,n)\tilde{a}_{m}\tilde{b}_{n} \\& \hphantom{\tilde{I} } = \sum_{n=2}^{\infty} \Biggl( \sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac {\mu _{m}\ln^{\tilde{\lambda}_{2}}V_{n}}{U_{m}\ln^{1-\tilde {\lambda}_{1}}U_{m}} \Biggr) \frac{\upsilon_{n}}{V_{n}\ln^{\varepsilon +1}V_{n}} \\& \hphantom{\tilde{I} } = \sum_{n=2}^{\infty} \frac{\varpi(\tilde{\lambda }_{1},n)\upsilon_{n}}{V_{n}\ln^{\varepsilon+1}V_{n}}\geq k_{\alpha}(\tilde{\lambda}_{1}) \sum_{n=2}^{\infty} \biggl( 1-O\biggl( \frac{1}{\ln^{\tilde{\lambda}_{1}}V_{n}}\biggr) \biggr) \frac{\upsilon_{n}}{V_{n}\ln^{\varepsilon +1}V_{n}} \\& \hphantom{\tilde{I} } = k_{\alpha}(\tilde{\lambda}_{1}) \Biggl[ \sum _{n=2}^{\infty}\frac {\upsilon_{n}}{V_{n}\ln^{\varepsilon+1}V_{n}}-\sum _{n=2}^{\infty }O \biggl( \frac{\upsilon_{n}}{V_{n}(\ln V_{n})^{(\frac{\varepsilon}{q}+\lambda _{1})+1}} \biggr) \Biggr] \\& \hphantom{\tilde{I} } = \frac{1}{\varepsilon}k_{\alpha}(\tilde{ \lambda}_{1}) \biggl[ \frac{1}{\ln^{\varepsilon}V_{n_{0}}}+\varepsilon \bigl(O_{2}(1)-O(1)\bigr) \biggr] . \end{aligned}$$

If there exists a positive constant \(K\leq k_{\alpha}(\lambda_{1})\), such that (28) is valid when replacing \(k_{\alpha}(\lambda_{1})\) by K, then, in particular, we have \(\varepsilon\tilde{I}<\varepsilon K\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi_{\lambda}}\), namely

$$\begin{aligned}& k_{\alpha}\biggl(\lambda_{1}-\frac{\varepsilon}{p}\biggr) \biggl[ \frac{1}{\ln ^{\varepsilon}V_{n_{0}}}+\varepsilon\bigl(O_{2}(1)-O(1)\bigr) \biggr] \\& \quad < K \biggl( \frac{1}{\ln^{\varepsilon}U_{m_{0}}}+\varepsilon O_{1}(1) \biggr) ^{\frac{1}{p}} \biggl( \frac{1}{\ln^{\varepsilon }V_{n_{0}}}+\varepsilon O_{2}(1) \biggr) ^{\frac{1}{q}}. \end{aligned}$$

By (27), it follows that \(k_{\alpha}(\lambda_{1})\leq K\) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k_{\alpha}(\lambda_{1})\) is the best possible constant factor of (28).

The constant factor \(k_{\alpha}(\lambda_{1})\) in (29) is still the best possible. Otherwise, we would reach a contradiction by (32) that the constant factor in (28) is not the best possible. □

For \(p>1\), we find \(\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon _{n}}{V_{n}}(\ln V_{n})^{p\lambda_{2}-1}\) (\(n\in\mathbf{N}\backslash\{1\}\)) and define the following normed spaces:

$$\begin{aligned}& l_{p,\Phi_{\lambda}} : =\bigl\{ a=\{a_{m}\}_{m=2}^{\infty}; \|a\|_{p,\Phi _{\lambda}}< \infty\bigr\} , \\& l_{q,\Psi_{\lambda}} : =\bigl\{ b=\{b_{n}\}_{n=2}^{\infty}; \|b\|_{q,\Psi _{\lambda}}< \infty\bigr\} , \\& l_{p,\Psi_{\lambda}^{1-p}} : =\bigl\{ c=\{c_{n}\}_{n=2}^{\infty }; \|c\|_{p,\Psi _{\lambda}^{1-p}}< \infty\bigr\} . \end{aligned}$$

Assuming that \(a=\{a_{m}\}_{m=2}^{\infty}\in l_{p,\Phi_{\lambda}}\), setting

$$ c=\{c_{n}\}_{n=2}^{\infty},\qquad c_{n}:= \sum_{m=2}^{\infty}K_{\lambda }(m,n)a_{m}, \quad n\in\mathbf{N}, $$

we can rewrite (29) as \(\|c\|_{p,\Psi_{\lambda }^{1-p}}< k_{\alpha }(\lambda_{1})\|a\|_{p,\Phi_{\lambda}}<\infty\), namely, \(c\in l_{p,\Psi _{\lambda}^{1-p}}\).

Definition 1

Define a Hardy-Mulholland-type operator \(T:l_{p,\Phi _{\lambda}}\rightarrow l_{p,\Psi_{\lambda}^{1-p}}\) as follows: for any \(a=\{a_{m}\}_{m=2}^{\infty}\in l_{p,\Phi_{\lambda}}\), there exists a unique representation \(Ta=c\in l_{p,\Psi_{\lambda}^{1-p}}\). Define the formal inner product of Ta and \(b=\{b_{n}\}_{n=2}^{\infty}\in l_{q,\Psi_{\lambda}}\) as follows:

$$ (Ta,b):=\sum_{n=2}^{\infty} \Biggl( \sum _{m=2}^{\infty}K_{\lambda }(m,n)a_{m} \Biggr) b_{n}. $$
(37)

Then we can rewrite (28) and (29) as follows:

$$\begin{aligned}& (Ta,b) < k_{\alpha}(\lambda_{1})\|a\|_{p,\Phi_{\lambda }}\|b \|_{q,\Psi _{\lambda}}, \end{aligned}$$
(38)
$$\begin{aligned}& \|Ta\|_{p,\Psi_{\lambda}^{1-p}} < k_{\alpha}(\lambda _{1})\|a \|_{p,\Phi _{\lambda}}. \end{aligned}$$
(39)

Define the norm of operator T as follows:

$$ \|T\|:=\sup_{a\, (\neq\theta)\in l_{p,\Phi_{\lambda}}}\frac {\|Ta\|_{p,\Psi _{\lambda}^{1-p}}}{\|a\|_{p,\Phi_{\lambda}}}. $$

Then by (39), it follows that \(\|T\|\leq k_{\alpha}(\lambda_{1})\). In view of Theorem 2, the constant factor in (39) is the best possible, we have

$$ \|T\|=k_{\alpha}(\lambda_{1})= \int_{0}^{1}\frac{t^{\lambda _{1}-1}+t^{\lambda_{2}-1}}{1+\alpha+(1-\alpha)t^{\lambda}}\,dt. $$
(40)

Remark 1

(i) For \(\lambda=1\), \(\lambda_{1}=\frac{1}{q}\), \(\lambda_{2}=\frac{1}{p}\), (28) reduces to

$$ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln U_{m}V_{n}+\alpha|\ln\frac{U_{m}}{V_{n}}|}< k_{\alpha} \biggl(\frac {1}{q}\biggr) \Biggl( \sum_{m=2}^{\infty} \frac{a_{m}^{p}}{\mu_{m}^{p-1}} \Biggr) ^{\frac {1}{p}} \Biggl( \sum _{n=2}^{\infty}\frac{b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}. $$
(41)

In particular, for \(\alpha=0\), we have the following simple Hardy-Mulholland-type inequality:

$$ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln U_{m}V_{n}}< \frac{\pi}{\sin(\pi/p)} \Biggl( \sum_{m=2}^{\infty} \frac {a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum _{n=2}^{\infty}\frac {b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}, $$
(42)

which is an extension of (3); for \(\alpha=1\), we have another simple Hardy-Mulholland-type inequality:

$$ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\max\{\ln U_{m},\ln V_{n}\}}< pq \Biggl( \sum _{m=2}^{\infty}\frac{a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=2}^{\infty} \frac {b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}. $$
(43)

Hence, inequality (28) is a relation between (42) and (43).

(ii) For \(\alpha=1\) in (28) and (29), in view of (9), we have the following equivalent Hardy-Mulholland-type inequalities with parameters:

$$\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\max\{\ln ^{\lambda}U_{m},\ln^{\lambda}V_{n}\}}< \frac{\lambda}{\lambda _{1}\lambda _{2}}\|a \|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(44)
$$\begin{aligned}& \Biggl[ \sum_{n=2}^{\infty}\frac{\upsilon_{n}}{V_{n}}( \ln V_{n})^{p\lambda_{2}-1} \Biggl( \sum_{m=2}^{\infty} \frac{a_{m}}{\max\{ \ln ^{\lambda}U_{m},\ln^{\lambda}V_{n}\}} \Biggr) ^{p} \Biggr] ^{\frac{1}{p}} \\& \quad < \frac{\lambda}{\lambda_{1}\lambda_{2}}\|a\|_{p,\Phi_{\lambda}}. \end{aligned}$$
(45)

(iii) For \(\alpha=0\) in (28) and (29), in view of (11), we have another equivalent Hardy-Mulholland-type inequalities with parameters:

$$\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }U_{m}+\ln^{\lambda}V_{n}}< \frac{\pi}{\lambda\sin(\frac{\pi \lambda_{1}}{\lambda})}\|a \|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(46)
$$\begin{aligned}& \Biggl[ \sum_{n=2}^{\infty}\frac{\upsilon_{n}}{V_{n}}( \ln V_{n})^{p\lambda_{2}-1} \Biggl( \sum_{m=2}^{\infty} \frac{a_{m}}{\ln ^{\lambda}U_{m}+\ln^{\lambda}V_{n}} \Biggr) ^{p} \Biggr] ^{\frac{1}{p}} \\& \quad < \frac{\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}\|a\|_{p,\Phi_{\lambda}}. \end{aligned}$$
(47)

In view of Theorem 2, the constant factors in the above inequalities are all the best possible. Inequality (28) is also a relation between the two Hardy-Mulholland-type inequalities (46) and (44) with parameters.

References

  1. Hardy, GH, Littlewood, JE, Pólya, G: Inequalities. Cambridge University Press, Cambridge (1934)

    Google Scholar 

  2. Mitrinović, DS, Pečarić, JE, Fink, AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic, Boston (1991)

    Book  MATH  Google Scholar 

  3. Yang, BC: Hilbert-Type Integral Inequalities. Bentham Science Publishers, Sharjah (2009)

    Google Scholar 

  4. Yang, BC: Discrete Hilbert-Type Inequalities. Bentham Science Publishers, Sharjah (2011)

    Google Scholar 

  5. Chen, Q, Yang, BC: A survey on the study of Hilbert-type inequalities. J. Inequal. Appl. 2015, 302 (2015)

    Article  Google Scholar 

  6. Yang, BC: On Hilbert’s integral inequality. J. Math. Anal. Appl. 220, 778-785 (1998)

    Article  MathSciNet  MATH  Google Scholar 

  7. Yang, BC: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing (2009)

    Google Scholar 

  8. Yang, BC, Brnetić, I, Krnić, M, Pečarić, JE: Generalization of Hilbert and Hardy-Hilbert integral inequalities. Math. Inequal. Appl. 8(2), 259-272 (2005)

    MathSciNet  MATH  Google Scholar 

  9. Krnić, M, Pečarić, JE: Hilbert’s inequalities and their reverses. Publ. Math. (Debr.) 67(3-4), 315-331 (2005)

    MATH  Google Scholar 

  10. Yang, BC, Rassias, TM: On the way of weight coefficient and research for Hilbert-type inequalities. Math. Inequal. Appl. 6(4), 625-658 (2003)

    MathSciNet  MATH  Google Scholar 

  11. Yang, BC, Rassias, TM: On a Hilbert-type integral inequality in the subinterval and its operator expression. Banach J. Math. Anal. 4(2), 100-110 (2010)

    Article  MathSciNet  MATH  Google Scholar 

  12. Azar, L: On some extensions of Hardy-Hilbert’s inequality and applications. J. Inequal. Appl. 2009, Article ID 546829 (2009)

    MathSciNet  Google Scholar 

  13. Bényi, Á, Oh, C: Best constant for certain multilinear integral operator. J. Inequal. Appl. 2006, Article ID 28582 (2006)

    Article  Google Scholar 

  14. Kuang, JC, Debnath, L: On Hilbert’s type inequalities on the weighted Orlicz spaces. Pac. J. Appl. Math. 1(1), 95-103 (2007)

    MathSciNet  MATH  Google Scholar 

  15. Zhong, WY: The Hilbert-type integral inequalities with a homogeneous kernel of −λ-degree. J. Inequal. Appl. 2008, Article ID 917392 (2008)

    Article  Google Scholar 

  16. Hong, Y: On Hardy-Hilbert integral inequalities with some parameters. J. Inequal. Pure Appl. Math. 6(4), Article 92 (2005)

    MathSciNet  Google Scholar 

  17. Zhong, WY, Yang, BC: On a multiple Hilbert-type integral inequality with the symmetric kernel. J. Inequal. Appl. 2007, Article ID 27962 (2007). doi:10.1155/2007/27962

    Article  MathSciNet  Google Scholar 

  18. Yang, BC, Krnić, M: On the norm of a multi-dimensional Hilbert-type operator. Sarajevo J. Math. 7(20), 223-243 (2011)

    MathSciNet  Google Scholar 

  19. Krnić, M, Pečarić, JE, Vuković, P: On some higher-dimensional Hilbert’s and Hardy-Hilbert’s type integral inequalities with parameters. Math. Inequal. Appl. 11, 701-716 (2008)

    MathSciNet  MATH  Google Scholar 

  20. Krnic, M, Vuković, P: On a multidimensional version of the Hilbert-type inequality. Anal. Math. 38, 291-303 (2012)

    Article  MathSciNet  MATH  Google Scholar 

  21. Rassias, MT, Yang, BC: On half-discrete Hilbert’s inequality. Appl. Math. Comput. 220, 75-93 (2013)

    Article  MathSciNet  Google Scholar 

  22. Rassias, MT, Yang, BC: A multidimensional half-discrete Hilbert-type inequality and the Riemann zeta function. Appl. Math. Comput. 225, 263-277 (2013)

    Article  MathSciNet  Google Scholar 

  23. Rassias, MT, Yang, BC: On a multidimensional half-discrete Hilbert-type inequality related to the hyperbolic cotangent function. Appl. Math. Comput. 242, 800-813 (2014)

    Article  MathSciNet  Google Scholar 

  24. Rassias, MT, Yang, BC: On a multidimensional Hilbert-type integral inequality associated to the gamma function. Appl. Math. Comput. 249, 408-418 (2014)

    Article  MathSciNet  Google Scholar 

  25. Shi, YP, Yang, BC: On a multidimensional Hilbert-type inequality with parameters. J. Inequal. Appl. 2015, 371 (2015)

    Article  Google Scholar 

  26. Yang, BC: An extension of a Hardy-Hilbert-type inequality. J. Guangdong Univ. Educ. 35(3), 1-7 (2015)

    Google Scholar 

  27. Huang, QL, Yang, BC: A more accurate Hardy-Hilbert-type inequality. J. Guangdong Univ. Educ. 35(5), 27-35 (2015)

    Google Scholar 

  28. Wang, AZ, Huang, QL, Yang, BC: A strengthened Mulholland-type inequality with parameters. J. Inequal. Appl. 2015, 329 (2015)

    Article  MathSciNet  Google Scholar 

  29. Yang, BC, Chen, Q: On a Hardy-Hilbert-type inequality with parameters. J. Inequal. Appl. 2015, 339 (2015)

    Article  Google Scholar 

  30. Shi, YP, Yang, BC: A new Hardy-Hilbert-type inequality with multiparameters and a best possible constant factor. J. Inequal. Appl. 2015, 380 (2015)

    Article  Google Scholar 

  31. Kuang, JC: Real and Functional Analysis (Continuation), vol. 2. Higher Education Press, Beijing (2015)

    Google Scholar 

  32. Kuang, JC: Applied Inequalities. Shandong Science Technic Press, Jinan (2004)

    Google Scholar 

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Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 61370186), Appropriative Researching Fund for Professors and Doctors, Guangdong University of Education (No. 2015ARF25) and Hunan Province Natural Science Foundation (No. 2015JJ4041).

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Correspondence to Qiang Chen.

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BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. QC and YS participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.

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Chen, Q., Shi, Y. & Yang, B. A relation between two simple Hardy-Mulholland-type inequalities with parameters. J Inequal Appl 2016, 75 (2016). https://doi.org/10.1186/s13660-016-1020-5

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