REVERSED HARDY–LITTLEWOOD–PÓLYA INEQUALITIES WITH FINITE TERMS

Abstract

We prove a reversed Hardy–Littlewood–Pólya inequality with finite terms. We also give the limit of the best constant.

1 Introduction

Let $a_i, b_i \geq 0$ ( $i=1,2,\ldots $ ). The Hardy–Littlewood–Pólya inequality [3, Theorem 381, page 288] states that

(1.1) $$ \begin{align} \sum_{i,j=1,\,i \neq j}^\infty \frac{a_ib_j}{|i-j|^\lambda} \leq K_{p,q}\bigg(\sum_{i=1}^\infty a_i^p\bigg)^{1/p} \bigg(\sum_{i=1}^\infty b_i^q\bigg)^{1/q}, \end{align} $$

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 [7] 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 [4].

In 2015, Dou and Zhu in [2] 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 [5]:

$$ \begin{align*} \sum_{i,j \in \mathbb{Z}^n} \frac{|f_i||g_j|}{|i-j|^{\lambda}} +\sum_{j \in \mathbb{Z}^n}|f_j||g_j| \geq C\|f\|_{l^p}\|g\|_{l^q} \quad \mbox{for all}\ (f,g) \in l^p(\mathbb{Z}^n) \times l^q(\mathbb{Z}^n), \end{align*} $$

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

(1.2) $$ \begin{align} L_{p,q,\lambda}:=\inf\bigg\{\sum_{i,j =1}^\infty \frac{|f_i||g_j|}{|i-j|^{\lambda}} +\sum_{j =1}^\infty |f_j||g_j|: \|f\|_{l^p}=\|g\|_{l^q}=1\bigg\}. \end{align} $$

In 2011, Li and Villavert [6] proved the Hardy–Littlewood–Pólya inequality with finite terms:

(1.3) $$ \begin{align} \sum_{i,j=1,\,i \neq j}^N \frac{a_ib_j}{|i-j|} \leq K_{N}\bigg(\sum_{i=1}^N a_i^2\bigg)^{1/2} \bigg(\sum_{i=1}^N b_i^2\bigg)^{1/2}, \end{align} $$

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 [4, 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 [8]. The results in higher dimensions can be found in [1].

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

(1.4) $$ \begin{align} \sum_{i,j=1,\,i \neq j}^N \frac{a_ib_j}{|i-j|^\lambda} +\sum_{i=1}^N a_ib_i \geq L\bigg(\sum_{i=1}^N a_i^p\bigg)^{1/p} \bigg(\sum_{i=1}^N b_i^q\bigg)^{1/q}. \end{align} $$

Denote the best constant in (1.4) by

(1.5) $$ \begin{align} L_{p,q,\lambda,N}:=\min\bigg\{\sum_{i,j=1,\,i \neq j}^N\frac{a_ib_j}{|i-j|^\lambda} +\sum_{i=1}^N a_ib_i: \sum_{i=1}^N a_i^p=\sum_{i=1}^N b_i^q=1\bigg\}. \end{align} $$

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

(2.1) $$ \begin{align} \begin{cases} \displaystyle L_{p,q,\lambda,N}\ a(N)_i^{p-1} =\sum_{j=1}^N\frac{b(N)_j}{|i-j|^\lambda}+b(N)_i\\[3mm] \displaystyle L_{p,q,\lambda,N}\ b(N)_i^{q-1} =\sum_{j=1}^N\frac{a(N)_j}{|i-j|^\lambda}+a(N)_i. \end{cases} \end{align} $$

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

(2.2) $$ \begin{align} L_{p,q,\lambda,N}\sum_{i=1}^N a(N)_i^p = \sum_{i,j=1}^N \frac{a(N)_i b(N)_j}{|i-j|^\lambda}+\sum_{i=1}^N a(N)_ib(N)_i. \end{align} $$

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

(2.3) $$ \begin{align} L_{p,q,\lambda,N}=L_{p,q,\lambda,N}\sum_{i=1}^N a(N)_i^p =\sum_{i,j=1}^\infty \frac{\bar{a}(N)_i \bar{b}(N)_j}{|i-j|^\lambda} +\sum_{i=1}^\infty \bar{a}(N)_i\bar{b}(N)_i \geq L_{p,q,\lambda}. \end{align} $$

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),

(2.4) $$ \begin{align} L_{p,q,\lambda,N} &=a(N)_1^{1-p}\bigg(\sum_{j=2}^N\frac{b(N)_j}{(\,j-1)^\lambda}+b(N)_i\bigg) \notag \\ &\geq a(N)_1^{2-p}\bigg(\sum_{j=2}^N \frac{1}{(\,j-1)^\lambda} +1\bigg) =a(N)_1^{2-p}\bigg(\sum_{j=1}^{N-1} j^{-\lambda}+1\bigg) \notag \\ &\geq a(N)_1^{2-p}\bigg(\int_1^{N} (r-1)^{-\lambda}\,dr+1\bigg) =a(N)_1^{2-p}\bigg(\frac{(N-1)^{1-\lambda}}{1-\lambda}+1\bigg). \end{align} $$

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

(2.5) $$ \begin{align} L_{p,q,\lambda,N} \leq 1. \end{align} $$

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

(2.6) $$ \begin{align} \sum_{i,j=1,\,i \neq j}^\infty \frac{a_{i}^{(m),N_m}b_j^{(m),N_m}}{|i-j|^\lambda} +\sum_{i=1}^\infty a_i^{(m),N_m}b_i^{(m),N_m} \leq L_{p,q,\lambda}+\frac{2}{m} \end{align} $$

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 $(a^{(m)},b^{(m)}) \in \mathbb {S}$ ,

(2.7) $$ \begin{align} \|a^{(m),N_m}\|_{l^p}^p \geq 1-\frac{1}{m}, \quad \|b^{(m),N_m}\|_{l^q}^q \geq 1-\frac{1}{m}, \end{align} $$

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.