Multilinear Fractional Integral Operators: A counter-example

TL;DR

This paper presents a counter-example demonstrating that multilinear fractional operators are not bounded from a product of Hardy spaces into a Hardy space, challenging previous assumptions about their boundedness properties.

Contribution

The paper provides the first explicit counter-example showing the unboundedness of multilinear fractional operators on Hardy spaces.

Findings

Multilinear fractional operators are not bounded from a product of Hardy spaces into a Hardy space.

Counter-example disproves previous conjectures about boundedness.

Highlights limitations of multilinear fractional operators in harmonic analysis.

Abstract

By means of a counter-example we show that the multilinear fractional operator is not bounded from a product of Hardy spaces into a Hardy space.

Equations

$${\mathcal{I}}_{\gamma }(f_1,...,f_m)(x)={\int }_{({\mathbb{R}}^n{)}^m}\frac{f_1(y_1)\cdot \cdot \cdot f_m(y_m)}{(|x-y_1|+\cdot \cdot \cdot +|x-y_m|{)}^{nm-\gamma }}d{y}_1\cdot \cdot \cdot d{y}_m,x\in {\mathbb{R}}^n.$$

$$\Vert {\mathcal{I}}_{\gamma }(f_1,...,f_m){\Vert }_{L^q}\le C\Vert f_1{\Vert }_{H^{p_1}}\cdot \cdot \cdot \Vert f_m{\Vert }_{H^{p_m}}.$$

$${\mathcal{I}}_{\alpha +1}(f_1,f_2)(x)={\iint }_{{\mathbb{R}}^2}\frac{f_1(s)f_2(t)}{(|x-s|+|x-t|{)}^{1-\alpha }}dsdt,x\in \mathbb{R}.$$

$$({\mathcal{M}}_{\phi }f)(x)=\underset{t>0}{\sup }|({\phi }_t\ast f)(x)|$$

$${\int }_{\mathbb{R}}{\mathcal{I}}_{\alpha +1}(a_1,a_2)(x)dx=\frac{\alpha -1}{\alpha }{\iint }_{{\mathbb{R}}^2}{a}_1(s)a_2(t)|t-s{|}^{\alpha }dsdt,$$

$$K_{s,t}^{\alpha }(x)={\left(|x-s|+|x-t|\right)}^{\alpha -1},x\in \mathbb{R}.$$

$$\widehat{K_{s,t}^{\alpha }}(\xi )=-2^{\alpha }\Gamma (\alpha )\sin \left(\frac{(\alpha -1)\pi }{2}\right)e^{-i\xi (\frac{s+t}{2})}|\xi {|}^{-\alpha }+|t-s{|}^{\alpha -1}\mathrm{sgn}(t-s){\int }_s^t{e}^{-ix\xi }dx$$

$$-\frac{|t-s{|}^{\alpha }}{\alpha }{e}^{-i\frac{(s+t)}{2}\xi }\cos \left(\frac{|t-s|\xi }{2}\right)+\frac{i{2}^{\alpha }\xi e^{-i\xi (\frac{s+t}{2})}}{\alpha }{\int }_0^{\frac{|t-s|}{2}}{x}^{\alpha }\sin (x\xi )dx,$$

$$\left(\widehat{K_{s,t}^{\alpha }},\varphi \right)=\left(K_{s,t}^{\alpha },\hat{\varphi }\right)={\int }_{\mathbb{R}}{K}_{s,t}^{\alpha }(x)\hat{\varphi }(x)dx$$

$$={\int }_t^{+\infty }{K}_{s,t}^{\alpha }(x)\hat{\varphi }(x)dx+{\int }_s^t{K}_{s,t}^{\alpha }(x)\hat{\varphi }(x)dx+{\int }_{-\infty }^s{K}_{s,t}^{\alpha }(x)\hat{\varphi }(x)dx=I+II+III.$$

$$I={\int }_{\mathbb{R}}{\chi }_{(t,+\infty )}(x){\left(2x-(s+t)\right)}^{\alpha -1}\hat{\varphi }(x)dx$$

$$=2^{\alpha -1}{\int }_{\mathbb{R}}{x}^{\alpha -1}{\chi }_{(\frac{t-s}{2},+\infty )}(x)(e^{-i(\cdot )\frac{(s+t)}{2}}\varphi )\hat{}(x)dx$$

$$=2^{\alpha -1}{\int }_{\mathbb{R}}{x}_{+}^{\alpha -1}(e^{-i(\cdot )\frac{(s+t)}{2}}\varphi )\hat{}(x)dx-2^{\alpha -1}{\int }_{\mathbb{R}}{x}^{\alpha -1}{\chi }_{(0,\frac{t-s}{2})}(x)(e^{-i(\cdot )\frac{(s+t)}{2}}\varphi )\hat{}(x)dx$$

$$=2^{\alpha -1}{\int }_{\mathbb{R}}{x}_{+}^{\alpha -1}(e^{-i(\cdot )\frac{(s+t)}{2}}\varphi )\hat{}(x)dx$$

$$-2^{\alpha -1}{\int }_{\mathbb{R}}\left(\frac{(t-s{)}^{\alpha }{e}^{-i\xi \frac{(t-s)}{2}}}{2^{\alpha }\alpha }+\frac{i\xi }{\alpha }{\int }_0^{\frac{t-s}{2}}{x}^{\alpha }{e}^{-ix\xi }dx\right)e^{-i\xi \frac{(s+t)}{2}}\varphi (\xi )d\xi ,$$

$$III=2^{\alpha -1}{\int }_{\mathbb{R}}{x}_{-}^{\alpha -1}(e^{-i(\cdot )\frac{(s+t)}{2}}\varphi )\hat{}(x)dx$$

$$-2^{\alpha -1}{\int }_{\mathbb{R}}\left(\frac{(t-s{)}^{\alpha }}{2^{\alpha }\alpha }{e}^{i\xi (\frac{t-s}{2})}-\frac{i\xi }{\alpha }{\int }_0^{\frac{t-s}{2}}{x}^{\alpha }{e}^{ix\xi }dx\right)e^{-i\xi \frac{(s+t)}{2}}\varphi (\xi )d\xi ,$$

$$I+III=2^{\alpha -1}{\int }_{\mathbb{R}}|x{|}^{\alpha -1}(e^{-i(\cdot )\frac{(s+t)}{2}}\varphi )\hat{}(x)dx$$

$$-{\int }_{\mathbb{R}}\frac{(t-s{)}^{\alpha }}{\alpha }{e}^{-i\frac{(s+t)}{2}\xi }\cos \left(\frac{(t-s)\xi }{2}\right)\varphi (\xi )d\xi$$

$$+{\int }_{\mathbb{R}}\left(\frac{i{2}^{\alpha }\xi e^{-i\xi (\frac{s+t}{2})}}{\alpha }{\int }_0^{\frac{(t-s)}{2}}{x}^{\alpha }\sin (x\xi )dx\right)\varphi (\xi )d\xi .$$

$$II={\int }_{\mathbb{R}}{\chi }_{(s,t)}(x)(t-s{)}^{\alpha -1}\hat{\varphi }(x)dx={\int }_{\mathbb{R}}\left((t-s{)}^{\alpha -1}{\int }_s^t{e}^{-ix\xi }dx\right)\varphi (\xi )d\xi .$$

$$\widehat{|x{|}^{\alpha -1}}(\xi )=-2\Gamma (\alpha )\sin \left(\frac{(\alpha -1)\pi }{2}\right)|\xi {|}^{-\alpha }$$

$${\int }_{\mathbb{R}}{\mathcal{I}}_{\alpha +1}(a_1,a_2)(x)dx=\frac{\alpha -1}{\alpha }{\iint }_{{\mathbb{R}}^2}{a}_1(s)a_2(t)|t-s{|}^{\alpha }dsdt.$$

$${\int }_{\mathbb{R}}{\mathcal{I}}_{\alpha +1}(a_1,a_2)(x)dx=\underset{ϵ\to 0^{+}}{\lim }{\int }_{\mathbb{R}}{\mathcal{I}}_{\alpha +1}(a_1,a_2)(x){\phi }_{ϵ}(x)dx,$$

$$\underset{ϵ\to 0^{+}}{\lim }{\int }_{\mathbb{R}}{\mathcal{I}}_{\alpha +1}(a_1,a_2)(x){\phi }_{ϵ}(x)dx=\underset{ϵ\to 0^{+}}{\lim }{\iint }_{{\mathbb{R}}^2}{a}_1(s)a_2(t)\left({\int }_{\mathbb{R}}{K}_{s,t}^{\alpha }(x){\phi }_{ϵ}(x)dx\right)dsdt$$

$$=\underset{ϵ\to 0^{+}}{\lim }{\iint }_{{\mathbb{R}}^2}{a}_1(s)a_2(t)\left({\int }_{\mathbb{R}}\widehat{K_{s,t}^{\alpha }}(\xi )\widehat{{\phi }_{ϵ}}(\xi )d\xi \right)dsdt$$

$$={\iint }_{{\mathbb{R}}^2}{a}_1(s)a_2(t)\underset{ϵ\to 0^{+}}{\lim }\left({\int }_{\mathbb{R}}\left(\widehat{K_{s,t}^{\alpha }}(ϵ\xi )+2^{\alpha }\Gamma (\alpha )\sin \left(\frac{(\alpha -1)\pi }{2}\right)|ϵ\xi {|}^{-\alpha }\right)\hat{\phi }(\xi )d\xi \right)dsdt$$

$$=\frac{\alpha -1}{\alpha }{\iint }_{{\mathbb{R}}^2}{a}_1(s)a_2(t)|t-s{|}^{\alpha }dsdt,$$

$$\iint a_1(s)a_2(t)|t-s{|}^{\alpha }dsdt=\frac{4\cdot 3^{\alpha +2}-4^{\alpha +2}-6\cdot 2^{\alpha +2}+4}{(\alpha +1)(\alpha +2)}\ne 0.$$

$$\left|{\int }_{\mathbb{R}}{\mathcal{I}}_{\alpha +1}(a_1,b)(x)dx\right|\ge \left|{\int }_{\mathbb{R}}{\mathcal{I}}_{\alpha +1}(a_1,a_2)(x)dx\right|-{\int }_{\mathbb{R}}|{\mathcal{I}}_{\alpha +1}(a_1,a_2-b)(x)|dx$$

Full text

Key words and phrases: Multilinear Fractional Operators, Hardy Spaces.

2.010 Math. Subject Classification: 42B20, 42B30.

1. Introduction

Given positive integers $m,n$ and a real number $0<\gamma<mn$, it is define the multilinear fractional operator $\mathcal{I}_{\gamma}$ by

[TABLE]

Y. Lin and S. Lu in [3] proved Hardy space estimates for the multilinear fractional operator $\mathcal{I}_{\gamma}$. More precisely, they proved that if $0<\gamma<n$, $0<p_{1},...,p_{m},q\leq 1$, and $q$ such that $\frac{1}{q}=\frac{1}{p_{1}}+\cdot\cdot\cdot+\frac{1}{p_{m}}-\frac{\gamma}{n}>0$, then

[TABLE]

Recently, D. Cruz-Uribe, K. Moen and H. van Nguyen in [1] generalized the result of Lin and Lu to weighted Hardy spaces on the full range $0<\gamma<nm$.

The purpose of this note is to give a counter-example to show that the multilinear fractional operator $\mathcal{I}_{\gamma}$ is not bounded from a product of Hardy spaces into a Hardy space. For them, we consider $n=1$, $m=2$, $\gamma=\alpha+1$ with $0<\alpha<1$, so $1<\gamma<2$, $2-\gamma=1-\alpha$ and the multilinear fractional operator $\mathcal{I}_{\alpha+1}$ in this case is given by

[TABLE]

We will prove that the operator $\mathcal{I}_{\alpha+1}$ is not bounded from $H^{1}(\mathbb{R})\times H^{p}(\mathbb{R})$ into $H^{q}(\mathbb{R})$, for $0<p\leq(\alpha+1)^{-1}$ and $\frac{1}{q}=\frac{1}{p}-\alpha$.

We briefly recall the definition of Hardy space on $\mathbb{R}^{n}$. The Hardy space $H^{p}(\mathbb{R}^{n})$ (for $0<p<\infty$) consists of tempered distributions $f\in\mathcal{S}^{\prime}(\mathbb{R}^{n})$ such that for some Schwartz function $\varphi$ with $\int\varphi=1$, the maximal operator

[TABLE]

is in $L^{p}(\mathbb{R}^{n})$, where $\varphi_{t}(x):=\frac{1}{t^{n}}\varphi(\frac{x}{t})$. In this case we define $\left\|f\right\|_{H^{p}}:=\left\|\mathcal{M}_{\varphi}f\right\|_{p}$ as the $H^{p}$ “norm”. It can be shown that this definition does not depend on the choice of the function $\varphi$. For $1<p<\infty$, it is well known that $H^{p}(\mathbb{R}^{n})\cong L^{p}(\mathbb{R}^{n})$, $H^{1}(\mathbb{R}^{n})\subset L^{1}(\mathbb{R}^{n})$ strictly, and for $0<p<1$ the spaces $H^{p}(\mathbb{R}^{n})$ and $L^{p}(\mathbb{R}^{n})$ are not comparable.

The following sentences hold in Hardy spaces $H^{p}(\mathbb{R}^{n})$ for $0<p\leq 1$ (see pp. 128-129 in [4]):

(S1) A bounded compactly supported function $f$ belongs to $H^{p}(\mathbb{R}^{n})$ if and only if it satisfies the moment conditions $\int x^{\beta}f(x)dx=0$ for all $|\beta|\leq n(p^{-1}-1)$.

(S2) If $f\in L^{1}(\mathbb{R}^{n})\cap H^{p}(\mathbb{R}^{n})$ then $\int x^{\beta}f(x)dx=0$, whenever $|\beta|\leq n(p^{-1}-1)$ and the function $x^{\beta}f(x)$ is in $L^{1}(\mathbb{R}^{n})$.

To obtain our result we will compute explicitly in Section 2 the Fourier transform of the kernel $\left(|x-s|+|x-t|\right)^{\alpha-1}$ in the $x$ variable, this allows us to get the following identity

[TABLE]

valid for bounded functions $a_{1}$ and $a_{2}$ having compact support with $\int a_{1}=0$ or $\int a_{2}=0$. Then, from (S2), the counter-example will follow to consider $a_{1}\in H^{1}(\mathbb{R})$ and $a_{2}\in H^{p}(\mathbb{R})$ such that $\iint_{\mathbb{R}^{2}}a_{1}(s)\,a_{2}(t)\,|t-s|^{\alpha}\,dsdt\neq 0$.

Notation: We use the following convention for the Fourier transform in $\mathbb{R}$ $\widehat{f}(\xi)=\int f(x)e^{-ix\xi}\,dx$. As usual we denote with $\mathcal{S}(\mathbb{R})$ the Schwartz space on $\mathbb{R}$.

2. Preliminaries

We start with the following lemma.

Lemma 1**.**

For $0<\alpha<1$ and $s\neq t\in\mathbb{R}$ fixed, let $K_{s,t}^{\alpha}$ be the function defined in $\mathbb{R}$ by

[TABLE]

Then

[TABLE]

[TABLE]

in the distributional sense.

Proof.

First we assume that $s<t$. Then for each $\phi\in\mathcal{S}(\mathbb{R})$ fixed, we have

[TABLE]

[TABLE]

Let us now proceed to compute each one of these integrals,

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[TABLE]

[TABLE]

[TABLE]

[TABLE]

to compute $III$ we proceed as in $I$, thus

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[TABLE]

so

[TABLE]

[TABLE]

[TABLE]

Now $II$ is easy, indeed

[TABLE]

Since

[TABLE]

(see equation (12), pp. 173, in [2]), the lemma follows for the case $s<t$. Finally, exchanging the roles of $s$ and $t$ we obtain the statement of the lemma. ∎

Corollary 2**.**

If $a_{1}$ and $a_{2}$ are two bounded functions on $\mathbb{R}$ with compact support and such that $\int a_{1}=0$ or $\int a_{2}=0$, then

[TABLE]

Proof.

It is easy to check that $\mathcal{I}_{\alpha+1}(a_{1},a_{2})(\cdot)\in L^{1}(\mathbb{R})$. Let $\varphi\in\mathcal{S}(\mathbb{R})$ be an even function such that $\varphi(0)=1$ and for $\epsilon>0$ let $\varphi_{\epsilon}(x)=\varphi(\epsilon x)$. Since

[TABLE]

we will proceed to compute this limit.

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[TABLE]

[TABLE]

[TABLE]

where the third equality follows from the moment condition of $a_{1}$ (or $a_{2}$) and the last one from Lemma 1 and that $\varphi(0)=1$. ∎

3. A Counter-example

We take $a_{1}(s)=\chi_{(-1,0)}(s)-\chi_{(0,1)}(s)$ and $a_{2}(t)=a_{1}(t-2)$. From (S1) it follows that $a_{1}\in H^{1}(\mathbb{R})$ and $a_{2}\in H^{(\alpha+1)^{-1}}(\mathbb{R})$ for each $0<\alpha<1$. A computation gives

[TABLE]

From (S2) and corollary 2 it obtains that $\mathcal{I}_{\alpha+1}(a_{1},a_{2})(\cdot)\notin H^{1}(\mathbb{R})$, for each $0<\alpha<1$. For $0<p<(\alpha+1)^{-1}$ and $\frac{1}{q}=\frac{1}{p}-\alpha$, we take $N$ as any fixed integer with $N>p^{-1}-1$, then the set of all bounded, compactly supported functions for which $\int x^{\beta}f(x)dx=0$, for all $|\beta|\leq N$ is dense in $H^{r}(\mathbb{R})$ for each $p\leq r\leq 1$ (see 5.2 b), pp. 128, in [4]). In particular, there exists $b\in H^{p}(\mathbb{R})$ such that $\|a_{1}\|_{H^{1}}\|a_{2}-b\|_{H^{(\alpha+1)^{-1}}}<\left|\int_{\mathbb{R}}\mathcal{I}_{\alpha+1}(a_{1},a_{2})(x)dx\right|/2C$. Then

[TABLE]

[TABLE]

[TABLE]

where the second inequality follows from Theorem 1.1 in [1] with $p_{1}=1$, $p_{2}=(\alpha+1)^{-1}$ and $q=1$. But then the operator $\mathcal{I}_{\alpha+1}$ is not bounded from $H^{1}(\mathbb{R})\times H^{p}(\mathbb{R})$ into $H^{q}(\mathbb{R})$ for $0<p\leq(\alpha+1)^{-1}$ and $\frac{1}{q}=\frac{1}{p}-\alpha$, since $\int_{\mathbb{R}}\mathcal{I}_{\alpha+1}(a_{1},b)(x)dx\neq 0$.

We conclude this note by summarizing our main result in the following theorem.

Theorem 3**.**

For $1<\gamma<2$, let $\mathcal{I}_{\gamma}$ be the multilinear fractional integral operator given by

[TABLE]

Then the operator $\mathcal{I}_{\gamma}$ is not bounded from $H^{1}(\mathbb{R})\times H^{p}(\mathbb{R})$ into $H^{q}(\mathbb{R})$ for            $0<p\leq\gamma^{-1}$ and $\frac{1}{q}=1+\frac{1}{p}-\gamma$.