4
$\begingroup$

I encounter the following claim in one paper:

If $(-\Delta)^{\frac14}u\in L^{2,\infty}(\mathbb{R})$, then $u\in BMO(\mathbb{R})$. Equivalently in its dual version, if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u=I_{1/2}u\in L^{2,1}(\mathbb{R})$. Here $L^{2,\infty}$ and $L^{2,1}$ are Lorentz space and $\mathcal{H}$ is the Hardy space.

I do not know how to show this fact. My knowledge of Riesz potential tells me if $u\in \mathcal{H}^1(\mathbb{R})$, then $(-\Delta)^{-\frac14}u=I_{1/2}u\in L^2(\mathbb{R})$, but why does it lie in the smaller space $L^{2,1}$? On the other hand, if $(-\Delta)^{\frac14}u\in L^2(\mathbb{R})$, then $u\in BMO$. However, this claim says we actually just need $(-\Delta)^{\frac14} u\in L^{2,\infty}$.

The paper says the first half of the claim is contained in the paper: Adams, D. R. (1975). A note on Riesz potentials. Duke Mathematical Journal. I read Adams' paper and could not figure out why.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

I suspect this is false (without additional restriction. Maybe compact support?). Indeed take $u(x)=|x|^{-1/2}$ which is in $L^{2,\infty}$, then $I_{1/2}u$ is a real constant times the convolution $u*u$ which seems to be infinite everywhere.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you for your answer. However, I believe the claim is right. It says about if $u\in \mathcal{H}^1$ then $I_{1/2}u\in L^{2,1}$. Only those $v\in L^{2,\infty}$ can be written as $(-\Delta)^{\frac14}u$ can have $I_{1/2}v\in BMO$. Do you have counter example of $u\in \mathcal{H}^1$? $\endgroup$
    – Slm2004
    Commented Sep 16, 2020 at 13:14
  • $\begingroup$ Let $v=(-\Delta)^{1/4}u$. The claim of the paper states that if $v\in L^{2,\infty}$ then $(-\Delta)^{-1/4}v\in BMO$. Is this the claim or not? $\endgroup$
    – Piero D'Ancona
    Commented Sep 17, 2020 at 4:33
  • $\begingroup$ I think you are right. Now I believe the claim of the paper should be if $(-\Delta)^{\frac14}u\in L^{2,\infty}\cap L^p$ for some $1\leq p<\infty$ then $u\in BMO$. The claim implicitly uses $I_{1/2}v$ is a locally integrable function for $v=(-\Delta)^{1/4}u$. $\endgroup$
    – Slm2004
    Commented Sep 17, 2020 at 15:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .