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

Journal of Inequalities and Applications20162016:75

https://doi.org/10.1186/s13660-016-1020-5

• Accepted: 12 February 2016
• Published:

## 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.

## Keywords

• Hardy-Mulholland-type inequality
• weight coefficient
• equivalent form
• operator
• norm

• 26D15
• 47A07

## 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. ). 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. ). Also the following Mulholland inequality was given with the best possible constant factor $$\frac{\pi}{\sin(\pi/p)}$$ (cf. , 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. ).
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 , 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  gave an extension of the integral analogous of (1) with the kernel $$\frac{1}{(x+y)^{\lambda}}$$ for $$p=q=2$$. Following , Yang  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 .

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. ), 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. ), 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. )

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. ) 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. ), 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.

## Declarations

### 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).

## Authors’ Affiliations

(1)
Department of Computer Science, Guangdong University of Education, Guangzhou, Guangdong, 51003, P.R. China
(2)
Normal College of Jishou University, Jishou, Hunan, 416000, P.R. China
(3)
Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong, 51003, P.R. China

## References 