1 Introduction
Let
$a_i, b_i \geq 0$
(
$i=1,2,\ldots $
). The Hardy–Littlewood–Pólya inequality [Reference Hardy, Littlewood and Pólya3, Theorem 381, page 288] states that
$$ \begin{align} \sum_{i,j=1,\,i \neq j}^\infty \frac{a_ib_j}{|i-j|^\lambda} \leq K_{p,q}\bigg(\sum_{i=1}^\infty a_i^p\bigg)^{1/p} \bigg(\sum_{i=1}^\infty b_i^q\bigg)^{1/q}, \end{align} $$
where
$p,q>1$
,
$1/p+1/q>1$
,
$\lambda =2-(1/p+1/q)$
. In 2015, Huang, Li and Yin used the Hardy–Littlewood–Sobolev inequality [Reference Lieb7] to generalise (1.1) to the case of higher dimensions. In addition, they also proved that the best constant can be approximated by the corresponding functional with finite terms [Reference Huang, Li and Yin4].
In 2015, Dou and Zhu in [Reference Dou and Zhu2] established a reversed Hardy–Littlewood–Sobolev inequality:
$$ \begin{align*} \bigg|\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\frac{f(x)g(y)\,dx\,dy}{|x-y|^\lambda} \bigg| \geq C\|f\|_{L^p(\mathbb{R}^n)}\|g\|_{L^q(\mathbb{R}^n)}\quad \mbox{for all } f \in L^p(\mathbb{R}^n), g \in L^q(\mathbb{R}^n), \end{align*} $$
where
$n \geq 1$
and
$p,q \in (n/(n-\lambda ),1)$
satisfy
$1/p+1/q+\lambda /n=2$
. In addition, they proved the best constant is attained. From this inequality, a reversed discrete inequality in higher dimensions was also deduced in [Reference Lei, Li and Tang5]:
$$ \begin{align*} \sum_{i,j \in \mathbb{Z}^n} \frac{|f_i||g_j|}{|i-j|^{\lambda}} +\sum_{j \in \mathbb{Z}^n}|f_j||g_j| \geq C\|f\|_{l^p}\|g\|_{l^q} \quad \mbox{for all}\ (f,g) \in l^p(\mathbb{Z}^n) \times l^q(\mathbb{Z}^n), \end{align*} $$
where
$f=(f_i)_{i \in \mathbb {Z}^n}$
,
$g=(g_j)_{j \in \mathbb {Z}^n}$
,
$\lambda <0$
,
$n/(n-\lambda )<p,q<1$
and
$1/p+1/q+\lambda /n \leq 2$
. When
$n=1$
and replacing
$\mathbb {Z}^n$
by
$\mathbb {N}$
, we denote the best constant by
$$ \begin{align} L_{p,q,\lambda}:=\inf\bigg\{\sum_{i,j =1}^\infty \frac{|f_i||g_j|}{|i-j|^{\lambda}} +\sum_{j =1}^\infty |f_j||g_j|: \|f\|_{l^p}=\|g\|_{l^q}=1\bigg\}. \end{align} $$
In 2011, Li and Villavert [Reference Li and Villavert6] proved the Hardy–Littlewood–Pólya inequality with finite terms:
$$ \begin{align} \sum_{i,j=1,\,i \neq j}^N \frac{a_ib_j}{|i-j|} \leq K_{N}\bigg(\sum_{i=1}^N a_i^2\bigg)^{1/2} \bigg(\sum_{i=1}^N b_i^2\bigg)^{1/2}, \end{align} $$
where the constant
$K_N$
satisfies
$$ \begin{align*} 2\ln N-2 \leq K_N \leq 2(\ln N-\ln 2)+2. \end{align*} $$
Comparing (1.3) with (1.1) for
$p=q=2$
shows that (1.3) is an inequality in the critical case. In contrast to the estimate of
$K_N$
above, the best constant for the upper-critical inequality is bounded with respect to N [Reference Huang, Li and Yin4, Lemma 2.2]. The bounds for the best constant are helpful in giving a better understanding of the Coulomb energy in the Thomas–Fermi model describing electron gas and N-body systems [Reference Lieb, Feng, Klauder and Strayer8]. The results in higher dimensions can be found in [Reference Cheng and Li1].
In this paper, we always assume
$a_i,b_i \geq 0$
(
$i=1,2,\ldots ,N$
). We will prove the following reversed Hardy–Littlewood–Pólya inequality with finite terms.
Theorem 1.1. Let
$\lambda <0$
and
$p,q \in (0,1)$
satisfy
$1/p+1/q \leq 2-\lambda $
. Then we can find a constant
$L>0$
which only depends on
$p,q,\lambda ,N$
such that
$$ \begin{align} \sum_{i,j=1,\,i \neq j}^N \frac{a_ib_j}{|i-j|^\lambda} +\sum_{i=1}^N a_ib_i \geq L\bigg(\sum_{i=1}^N a_i^p\bigg)^{1/p} \bigg(\sum_{i=1}^N b_i^q\bigg)^{1/q}. \end{align} $$
Denote the best constant in (1.4) by
$$ \begin{align} L_{p,q,\lambda,N}:=\min\bigg\{\sum_{i,j=1,\,i \neq j}^N\frac{a_ib_j}{|i-j|^\lambda} +\sum_{i=1}^N a_ib_i: \sum_{i=1}^N a_i^p=\sum_{i=1}^N b_i^q=1\bigg\}. \end{align} $$
Theorem 1.2. Let
$\lambda <0$
and
$p,q \in ((1-\lambda )^{-1},1)$
satisfy
$1/p+1/q \leq 2-\lambda $
. Then
$L_{p,q,\lambda ,N} \to L_{p,q,\lambda }$
when
$N \to \infty $
.
2 Proof of Theorems 1.1 and 1.2
Proof of Theorem 1.1
Write
$a=(a_1,a_2,\ldots ,a_N)$
and
$b=(b_1,b_2,\ldots ,b_N)$
. Set
$$ \begin{align*} J(a,b)=\sum_{i,j=1,\,i \neq j}^N\frac{a_ib_j}{|i-j|^\lambda} +\sum_{i=1}^N a_ib_i -L_{p,q,\lambda,N}\bigg(\sum_{i=1}^N a_i^p\bigg)^{1/p} \bigg(\sum_{i=1}^N b_i^q\bigg)^{1/q}. \end{align*} $$
Clearly,
$J(a,b) \geq 0$
for all
$a,b \in \mathbb {R}_+^N:= \{x=(x_1,x_2,\ldots ,x_N) : x_i \geq 0, i=1,2,\ldots ,N\}$
. However,
$$ \begin{align*} \mathbb{S}(N):=\bigg\{(a,b): \sum_{i=1}^N a_i^p =\sum_{i=1}^N b_i^q=1\bigg\} \end{align*} $$
is compact in
$\mathbb {R}_+^N \times \mathbb {R}_+^N$
and hence the minimisation problem (1.5) has solutions in
$\mathbb {S}(N)$
. Thus, we can find
$(a(N),b(N)) \in \mathbb {S}(N)$
such that
$J(a(N),b(N)) =0$
. We call
$(a(N),b(N))$
the minimiser of J. Therefore, both the partial derivatives of J are equal to zero at
$(a(N),b(N))$
. Namely,
$$ \begin{align*} \bigg[\frac{d}{dt} J(a(N)+ta,b(N))\bigg]_{t=0}= \bigg[\frac{d}{dt} J(a(N),b(N)+tb)\bigg]_{t=0}=0 \end{align*} $$
for any
$(a,b) \in \mathbb {R}_+^N \times \mathbb {R}_+^N$
. From this result, by simple calculation, we see that
$$ \begin{align} \begin{cases} \displaystyle L_{p,q,\lambda,N}\ a(N)_i^{p-1} =\sum_{j=1}^N\frac{b(N)_j}{|i-j|^\lambda}+b(N)_i\\[3mm] \displaystyle L_{p,q,\lambda,N}\ b(N)_i^{q-1} =\sum_{j=1}^N\frac{a(N)_j}{|i-j|^\lambda}+a(N)_i. \end{cases} \end{align} $$
Noting
$(a(N),b(N))\neq (0,0)$
(because
$(a(N),b(N))\in \mathbb {S}(N)$
), from (2.1) we see that
$$ \begin{align*} \min\{a(N)_i,b(N)_i\}>0 \quad \mbox{for}\ 1 \leq i \leq N. \end{align*} $$
Therefore,
$L_{p,q,\lambda ,N}>0$
.
Next, we prove that
$L_{p,q,\lambda ,N}$
has a positive lower bound which is independent of N. Multiplying (2.1)
$_1$
by
$a(N)_i$
and summing from
$1$
to N gives
$$ \begin{align} L_{p,q,\lambda,N}\sum_{i=1}^N a(N)_i^p = \sum_{i,j=1}^N \frac{a(N)_i b(N)_j}{|i-j|^\lambda}+\sum_{i=1}^N a(N)_ib(N)_i. \end{align} $$
Write
$$ \begin{align*}\begin{cases} \bar{a}(N)=(a(N)_1,a(N)_2,\ldots,a(N)_N,0,\ldots),\\ \bar{b}(N)=(b(N)_1,b(N)_2,\ldots,b(N)_N,0,\ldots). \end{cases} \end{align*} $$
Since
$(a(N),b(N)) \in \mathbb {S}(N)$
,
$$ \begin{align*} (\bar{a}(N),\bar{a}(N)) \in \mathbb{S}:=\{(a,b): \|a\|_{l^p}=\|b\|_{l^q}=1\}. \end{align*} $$
From (2.2) and (1.2), it follows that
$$ \begin{align} L_{p,q,\lambda,N}=L_{p,q,\lambda,N}\sum_{i=1}^N a(N)_i^p =\sum_{i,j=1}^\infty \frac{\bar{a}(N)_i \bar{b}(N)_j}{|i-j|^\lambda} +\sum_{i=1}^\infty \bar{a}(N)_i\bar{b}(N)_i \geq L_{p,q,\lambda}. \end{align} $$
Therefore,
$L_{p,q,\lambda ,N}>0$
with the lower bound (2.3). This proves (1.4).
Remark 2.1. We claim that
$$ \begin{align*} \lim_{N \to \infty}\min_{1 \leq i \leq N} \{a(N)_i,b(N)_i\}=0. \end{align*} $$
Without loss of generality, we can assume
$a(N)_1=\min _{1 \leq i \leq N} \{a(N)_i,b(N)_i\}$
. From (2.1),
$$ \begin{align} L_{p,q,\lambda,N} &=a(N)_1^{1-p}\bigg(\sum_{j=2}^N\frac{b(N)_j}{(\,j-1)^\lambda}+b(N)_i\bigg) \notag \\ &\geq a(N)_1^{2-p}\bigg(\sum_{j=2}^N \frac{1}{(\,j-1)^\lambda} +1\bigg) =a(N)_1^{2-p}\bigg(\sum_{j=1}^{N-1} j^{-\lambda}+1\bigg) \notag \\ &\geq a(N)_1^{2-p}\bigg(\int_1^{N} (r-1)^{-\lambda}\,dr+1\bigg) =a(N)_1^{2-p}\bigg(\frac{(N-1)^{1-\lambda}}{1-\lambda}+1\bigg). \end{align} $$
However, since
$\mathbb {S}(N) \subset \mathbb {S}(N+1)$
, it follows that
$L_{p,q,\lambda ,N}$
is nonincreasing with respect to N. Therefore,
$L_{p,q,\lambda ,N} \leq L_{p,q,\lambda ,1}$
. Taking
$a_1=b_1=1$
, we see that
$(a_1,b_1) \in \mathbb {S}(1)$
and hence
$L_{p,q,\lambda ,1} \leq a_1b_1=1$
. This gives the upper bound
$$ \begin{align} L_{p,q,\lambda,N} \leq 1. \end{align} $$
Combining (2.5) with (2.4) yields
$$ \begin{align*} a(N)_1^{2-p} = O(N^{\lambda-1}) \quad (N \to \infty). \end{align*} $$
This implies our claim.
Proof of Theorem 1.2
By (1.2), we can find a minimising sequence
$(a^{(m)},b^{(m)}) \in \mathbb {S}$
such that
$$ \begin{align*} \sum_{i,j=1,\,i \neq j}^\infty \frac{a_i^{(m)}b_j^{(m)}}{|i-j|^\lambda} +\sum_{i=1}^\infty a_i^{(m)}b_i^{(m)} \leq L_{p,q,\lambda}+\frac{1}{m}. \end{align*} $$
The convergence of this series implies
$$ \begin{align} \sum_{i,j=1,\,i \neq j}^\infty \frac{a_{i}^{(m),N_m}b_j^{(m),N_m}}{|i-j|^\lambda} +\sum_{i=1}^\infty a_i^{(m),N_m}b_i^{(m),N_m} \leq L_{p,q,\lambda}+\frac{2}{m} \end{align} $$
when
$N_m>m$
is sufficiently large. Here,
$$ \begin{align*} \begin{cases} a_{i}^{(m),N_m}=a_{i}^{(m)} & \mbox{when} \ i \leq N_m,\\ a_{i}^{(m),N_m}=0 & \mbox{when} \ i> N_m, \end{cases} \end{align*} $$
and
$b_{i}^{(m),N_m}$
is defined by the same truncation. Since
$(a^{(m)},b^{(m)}) \in \mathbb {S}$
,
$$ \begin{align} \|a^{(m),N_m}\|_{l^p}^p \geq 1-\frac{1}{m}, \quad \|b^{(m),N_m}\|_{l^q}^q \geq 1-\frac{1}{m}, \end{align} $$
when
$N_m>m$
is sufficiently large. Therefore, noting that
$$ \begin{align*} \bigg(\frac{a^{(m),N_m}}{\|a^{(m),N_m}\|_{l^p}}, \frac{b^{(m),N_m}}{\|b^{(m),N_m}\|_{l^q}}\bigg) \in \mathbb{S}(N_m), \end{align*} $$
from (1.5), (2.6) and (2.7), we deduce
$$ \begin{align*} L_{p,q,\lambda,N_m} \leq \bigg(L_{p,q,\lambda}+\frac{2}{m}\bigg) \bigg(1-\frac{1}{m}\bigg)^{-({1}/{p}+{1}/{q})} \end{align*} $$
for large
$N_m$
. Letting
$m \to \infty $
and combining with (2.3) completes the proof.
Acknowledgement
The authors thank the anonymous referee very much for the useful suggestions.
