Sharp anisotropic Hardy--Littlewood inequality for positive multilinear forms

TL;DR

This paper establishes sharp anisotropic Hardy-Littlewood inequalities for positive multilinear forms using elementary methods, recovering a known inequality and contributing to the understanding of multilinear analysis.

Contribution

It introduces a new proof technique for anisotropic inequalities, extending and sharpening previous results in multilinear analysis.

Findings

Proved sharp anisotropic Hardy-Littlewood inequalities for positive multilinear forms.

Recovered and generalized a 2018 inequality by F. Bayart.

Demonstrated the effectiveness of elementary techniques in this context.

Abstract

Using elementary techniques, we prove sharp anisotropic Hardy-Littlewood inequalities for positive multilinear forms. In particular, we recover an inequality proved by F. Bayart in 2018.

Equations

$$\frac{1}{p}+\frac{1}{q}<1.$$

$${\left(\sum _{j_1=1}^{\infty }{\left(\sum _{j_2=1}^{\infty }A(e_{j_1},e_{j_2}{)}^{q^{\ast }}\right)}^{\frac{1}{q^{\ast }}\times \frac{1}{1-\left(\frac{1}{p}+\frac{1}{q}\right)}}\right)}^{1-\left(\frac{1}{p}+\frac{1}{q}\right)}\le \Vert A\Vert ,$$

$$\frac{1}{p_1}+\cdots +\frac{1}{p_m}<1.$$

$${\left(\sum _{j_1,...,j_m=1}^{\infty }A(e_{j_1},...,e_{j_m}{)}^{\rho }\right)}^{1/\rho }\le \Vert A\Vert$$

$$\rho \ge \frac{1}{1-\left(\frac{1}{p_1}+\cdots +\frac{1}{p_m}\right)}.$$

$${\delta }^{s_k,...,s_m}:=\frac{1}{1-\left(\frac{1}{s_k}+\cdots +\frac{1}{s_m}\right)},$$

$${\delta }^{s_k,...,s_m}:=\infty .$$

$${\left(\sum _{j_{\sigma \left(1\right)}=1}^{\infty }{\left(\sum _{j_{\sigma \left(2\right)}=1}^{\infty }\cdots {\left(\sum _{j_{\sigma \left(m\right)}=1}^{\infty }A(e_{j_1},...,e_{j_m}{)}^{q_m}\right)}^{\frac{q_{m-1}}{q_m}}\cdots \right)}^{\frac{q_1}{q_2}}\right)}^{\frac{1}{q_1}}\le \Vert A\Vert$$

$$q_1\ge {\delta }^{p_{\sigma \left(1\right)},...,p_{\sigma \left(m\right)}},q_2\ge {\delta }^{p_{\sigma \left(2\right)},...,p_{\sigma \left(m\right)}},...,q_{m-1}\ge {\delta }^{p_{\sigma \left(m-1\right)},p_{\sigma \left(m\right)}},q_m\ge {\delta }^{p_{\sigma \left(m\right)}}.$$

$$\frac{1}{p_1}+\cdots +\frac{1}{p_m}<1.$$

$${\left(\sum _{j_1=1}^{\infty }A(e_{j_1}{)}^{q_1}\right)}^{\frac{1}{q_1}}\le \Vert A\Vert$$

$$\Vert A_n\Vert \le n^{\frac{1}{p_1^{\ast }}}.$$

$${\left(\sum _{j=1}^n{A}_n(e_j{)}^{q_1}\right)}^{\frac{1}{q_1}}=n^{\frac{1}{q_1}},$$

$$q_1\ge p_1^{\ast }={\delta }^{p_1}.$$

$$\frac{1}{p_1}+\cdots +\frac{1}{p_m}<1.$$

$$\frac{1}{p_2}+\cdots +\frac{1}{p_m}<1$$

$${\left(\sum _{j_1=1}^{\infty }{\left(\sum _{j_2=1}^{\infty }\cdots {\left(\sum _{j_m=1}^{\infty }A(e_{j_1},...,e_{j_m}{)}^{q_m}\right)}^{\frac{q_{m-1}}{q_m}}\cdots \right)}^{\frac{q_1}{q_2}}\right)}^{\frac{1}{q_1}}\le \Vert A\Vert$$

$$q_2$$

$$⋮$$

$$q_{m-1}$$

$$q_m$$

$$q_1\ge {\delta }^{p_1,...,p_m}.$$

$$B_n(x^{\left(1\right)},...,x^{\left(m\right)})=\sum _{j=1}^n{x}_j^{\left(1\right)}{x}_j^{\left(2\right)}...x_j^{\left(m\right)}.$$

$$\frac{1}{{\delta }^{p_1,\cdots ,p_m}}+\sum _{k=1}^m\frac{1}{p_k}=1\text{,}$$

$$\Vert B_n\Vert$$

$$\le \underset{\Vert x^{\left(1\right)}\Vert ,\cdots ,\Vert x^{\left(m\right)}\Vert \le 1}{\sup }\left(\prod _{k=1}^m{\left(\sum _{j=1}^n{\left|x_j^{\left(k\right)}\right|}^{p_k}\right)}^{1/p_k}{\left(\sum _{j=1}^n{\left|1\right|}^{{\delta }^{p_1\cdots p_m}}\right)}^{\frac{1}{{\delta }^{p_1,\cdots ,p_m}}}\right)$$

$$\le n^{\frac{1}{{\delta }^{p_1,\cdots ,p_m}}}.$$

$${\left(\sum _{j_1=1}^n{\left(\sum _{j_2=1}^n\cdots {\left(\sum _{j_m=1}^n{B}_n(e_{j_1},...,e_{j_m}{)}^{q_m}\right)}^{\frac{q_{m-1}}{q_m}}\cdots \right)}^{\frac{q_1}{q_2}}\right)}^{\frac{1}{q_1}}=n^{\frac{1}{q_1}},$$

$$q_1\ge {\delta }^{p_1,...,p_m}.$$

$$\Vert T\Vert$$

$$=\underset{{\Vert x^{\left(i\right)}\Vert }_{X_{p_i}}\le 1;1\le i\le m}{\sup }\left|\sum _{i_1,...,i_m=1}^{\infty }T\left(e_{i_1},...,e_{i_m}\right)x_{i_1}^{\left(1\right)}...x_{i_m}^{\left(m\right)}\right|$$

$$=\underset{{\Vert x^{\left(i\right)}\Vert }_{X_{p_i}}\le 1;1\le i\le m-1}{\sup }{\left(\sum _{i_m=1}^{\infty }{\left|\sum _{i_1,...,i_{m-1}=1}^{\infty }T\left(e_{i_1},...,e_{i_m}\right)x_{i_1}^{\left(1\right)}...x_{i_{m-1}}^{\left(m-1\right)}\right|}^{{\left(p_m\right)}^{\ast }}\right)}^{\frac{1}{{\left(p_m\right)}^{\ast }}}.$$

$${\left(\sum _{j_{{}_1}=1}^{\infty }{\left(\sum _{j_2=1}^{\infty }\cdots {\left(\sum _{j_{{}_{m-1}}=1}^{\infty }A(e_{j_1},...,e_{j_{m-1}}{)}^{{\delta }^{p_{m-1}}}\right)}^{\frac{{\delta }^{p_{m-2},p_{m-1}}}{{\delta }^{p_{m-1}}}}\cdots \right)}^{\frac{{\delta }^{p_1,...,p_{m-1}}}{{\delta }^{p_2,...,p_{m-1}}}}\right)}^{\frac{1}{{\delta }^{p_1,...,p_{m-1}}}}\le \Vert A\Vert$$

$$\frac{1}{p_1}+\cdots +\frac{1}{p_{m-1}}<1-\frac{1}{p_m}={\left({\delta }^{p_m}\right)}^{-1}$$

Full text

1. Introduction

All along this paper $\mathbb{K}:=\mathbb{R}$ or $\mathbb{C}$, $X_{p}:=\ell_{p}$ for $1\leq p<\infty$ and $X_{\infty}:=c_{0}$ and, as usual, we consider $1/\infty:=0$ and $q^{\ast}$ is the conjugate of $q$, i.e., $1/q+1/q^{\ast}=1.$ In 1934, Hardy and Littlewood [4] proved five theorems related to summability of bilinear forms. We are interested in the last one:

Theorem 1.1**.**

(See Hardy and Littlewood [4, Theorem 5]) Let $p,q\in(1,\infty]$ be such that

[TABLE]

Then

[TABLE]

for all bounded non-negative (i.e., $A(e_{j_{1}},e_{j_{2}})\geq 0$ for all $\left(j_{1},j_{2}\right)\in\mathbb{N\times N}$) bilinear forms $A:X_{p}\times X_{q}\rightarrow\mathbb{K}$.

The paper of Hardy and Littlewood was revisited in 1981 by Praciano-Pereira [5] and, recently, by several authors (see, for instance, [1] and the references therein). There are still several subtle open problems regarding the generalization of the Hardy–Littlewood inequalities to multilinear forms (see, for instance, [2]). In 2018, using a factorization result due to Schep [7], Bayart [3, Proposition 3.1] generalized Theorem 1.1 as follows:

Theorem 1.2**.**

(See Bayart [3, Proposition 3.1]) Let $m$ be a positive integer and $p_{1},\dots,p_{m}\in(1,\infty]$ with

[TABLE]

Then

[TABLE]

for all bounded non-negative $m$-linear forms $A:X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow\mathbb{K}$ if, and only if,

[TABLE]

In the present paper we prove a new generalization of Theorem 1.1, keeping its anisotropic essence. Following the notation introduced in [2], let us define $\delta^{s_{k},...,s_{m}}$ by the formula

[TABLE]

for all positive integers $m$ and $k=1,...,m.$ When $\frac{1}{s_{k}}+\cdots+\frac{1}{s_{m}}\geq 1$ it is convenient to define

[TABLE]

Also, when $q=\infty$, the notation $\left(\sum\left|x_{j}\right|^{q}\right)^{1/q}$ shall represent the supremum of $\left|x_{j}\right|.$

We prove the following:

Theorem 1.3**.**

Let $p_{1},...,p_{m}\in[1,\infty],$ $q_{1},...,q_{m}\in(0,\infty]$ and $m$ be a positive integer. Then, for any bijection $\sigma:\{1,...,m\}\rightarrow\{1,...,m\}$ we have

[TABLE]

for all bounded non-negative $m$-linear forms $A:X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow\mathbb{K}$ if, and only if

[TABLE]

Remark 1.4**.**

Note that we do not need the hypothesis

[TABLE]

The paper of Hardy and Littlewood and the recent literature just encompasses the case (2). For bilinear forms, the complementary case, called by Hardy and Littlewood  as case of spaces of type $\alpha$ was investigated in the seminal paper of M. Riesz [6].

Remark 1.5**.**

Our result recovers Theorem 1.2. In fact, if $\frac{1}{p_{1}}+\cdots+\frac{1}{p_{m}}<1,$ it is clear that $\delta^{p_{1},...,p_{m}}$ is the biggest exponent and coincides with the optimal exponent given by (1); thus the canonical inclusions of $\ell_{p}$ spaces provides the result.

2. The proof

To simplify the notation we will consider $\sigma(j)=j$ for all $j;$ the other cases are similar.

First Case. $\frac{1}{p_{1}}+\cdots+\frac{1}{p_{m}}<1.$

The proof of the direct implication is a consequence of techniques used in [2]. We present the argument for the sake of completeness. Let us suppose that $p_{1}>1$ (i.e., $\frac{1}{p_{1}}<1$) and that

[TABLE]

for all continuous non-negative linear forms $A:X_{p_{1}}\rightarrow\mathbb{K}$. For each $n$ consider the continuous non-negative linear form $A_{n}(x)=\sum\limits_{j=1}^{n}x_{j}.$ By the Hölder Inequality, we have

[TABLE]

On the other hand

[TABLE]

and, since $n$ is arbitrary,

[TABLE]

Thus the case $m=1$, is done. Now, let us proceed by induction. Suppose that the result is valid for $m-1$ and let

[TABLE]

Thus

[TABLE]

and the induction hypothesis combined with a simple argument of summability tells us that, if

[TABLE]

for all bounded non-negative $m$-linear forms $A:X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow\mathbb{K}$, then

[TABLE]

So, we must only show that

[TABLE]

For each $n$ consider the continuous non-negative $m-$linear form $B_{n}:X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow\mathbb{K}$ given by

[TABLE]

Since

[TABLE]

we use the Hölder inequality and obtain

[TABLE]

On the other hand

[TABLE]

and, since $n$ is arbitrary,

[TABLE]

Now let us prove the converse direction.

We recall that for a bounded $m$-linear form $T:X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow\mathbb{K}$, we have

[TABLE]

We denote by $X_{r}^{+}$ the set of sequences $\left(x_{j}\right)\in X_{r}$, such that $x_{j}\geq 0$ for all $j$. In the case $m=1$ the result is immediate, it holds with constant $1$ and doesn’t need the non-negative assumption. Let us show the general case $m$, supposing that the result holds for $m-1;$ so we suppose that if $p_{1},...,p_{m-1}\in(1,\infty]$ are such that $\frac{1}{p_{1}}+\cdots+\frac{1}{p_{m-1}}<1$, then

[TABLE]

for all bounded non negative $\left(m-1\right)$-linear forms $A:X_{p_{1}}\times\cdots\times X_{p_{m-1}}\rightarrow\mathbb{K}.$

Suppose that $p_{1},...,p_{m}\in(1,\infty]$ are such that $\frac{1}{p_{1}}+\cdots+\frac{1}{p_{m}}<1.$ In this case

[TABLE]

and then for all $i$ $\in\left\{1,..,m-1\right\}$, we have $p_{i}\geq\delta^{p_{m}}$.

Let $D:X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow\mathbb{K}$ be a bounded non negative $m$-linear form. We define the bounded non negative $\left(m-1\right)$-linear form $A:X_{r_{1}}\times\cdots\times X_{r_{m-1}}\rightarrow\mathbb{K}$ by

[TABLE]

with $r_{i}=p_{i}/\delta^{p_{m}}$ for each $i\in\left\{1,..,m-1\right\}$. Note that $A$ is well defined. In fact:

(i)

[TABLE]

since $D(e_{j_{1}},...,e_{j_{m-1},}\cdot):X_{p_{m}}\rightarrow\mathbb{K}$ is a linear form and $\left(e_{j}\right)$ is weakly $\left(p_{m}\right)^{\ast}$-summable.

(ii) For all positive integers $n,$ we have

[TABLE]

and by (3) we conclude that

[TABLE]

for all $n$ and all $x^{(k)}\in B_{X_{r_{k}}},$ with $k=1,...,m-1$, and we conclude that $A:X_{r_{1}}\times\cdots\times X_{r_{m-1}}\rightarrow\mathbb{K}$ is well defined.

Note that

[TABLE]

Hence, for each $i\in\left\{1,..,m-1\right\},$ a simple calculation shows that

[TABLE]

Therefore, by (4) and (6) we have

[TABLE]

By the last equality and the Induction Hypothesis we conclude that

[TABLE]

where in the last inequality we have used (5).

Second Case. $\frac{1}{p_{1}}+\cdots+\frac{1}{p_{m}}\geq 1.$

We begin by proving the direct implication.

Consider

[TABLE]

and we conclude that

[TABLE]

If

[TABLE]

for all $i$, the proof is immediate. Otherwise, at some stage $(i=2,3,...)$ we begin to have a strict inequality

[TABLE]

Denote by $i_{0}$ this index. Then

[TABLE]

and

[TABLE]

If $i_{0}=3$, we consider

[TABLE]

and we conclude that

[TABLE]

Similarly, if $i_{0}>3$, we have

[TABLE]

and

[TABLE]

Now it is simple to imitate the arguments of the previous case to complete the proof.

Now we prove the reverse implication. Recall that we are in the case

[TABLE]

If

[TABLE]

for all $i$, the proof is immediate. Otherwise, at some stage $(i=2,3,...)$ we begin to have a strict inequality

[TABLE]

Denote by $i_{0}$ this index. Then

[TABLE]

and

[TABLE]

We need to prove that

[TABLE]

for

[TABLE]

By the first case, we know that for any fixed vectors $e_{j_{1}},...,e_{j_{i_{0}-1}}$, we have

[TABLE]

for all bounded non negative $m$-linear forms $A\colon X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow\mathbb{K}$. Then,

[TABLE]

for all bounded non negative $m$-linear forms $A\colon X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow\mathbb{K}$.