3
$\begingroup$

A probability measure $\mu$ on $\mathbb{R}^d$ is said to be a Frostman measure if $$\mu(B)\lesssim r(B)^\alpha \ \ \ \ (1)$$ for all open ball $B$, where $r(B)$ denotes the radius and $\alpha>0$. If $\mu$ is a Frostman measure, then so is $\mu*\mu$ since $$\mu*\mu(B):=\int\mu(B-x)d\mu(x)\lesssim r(B)^\alpha \ \ \ \ (2)$$

My question is whether the converse is true, i.e. if $\mu*\mu$ satisfies $(2)$, is it necessarily true that $\mu$ satisfies $(1)$. Thank you!

$\endgroup$
1
  • $\begingroup$ You could try to describe Frostman measures in terms of the fourier transform of the measure, and then translate the problem in terms of the square of the transform. Limitation has relations with regularity, so maybe that can help. $\endgroup$
    – guaraqe
    Commented May 29, 2012 at 2:01

1 Answer 1

Reset to default
2
$\begingroup$

Let $d=1$ and $2/3<\alpha\le 1$. The measure $\mu =(2/3) x^{-1/3}\chi_{[0,1]}\,dx$ does not satisfy the Frostman condition. But the convolution $\mu*\mu$ has bounded density, because $\int_0^1 t^{-1/3}(x-t)^{-1/3}\,dt\le \int_0^1 t^{-2/3}\,dt$ by the Cauchy-Schwarz inequality. Therefore, $\mu*\mu$ is a Frostman measure.

$\endgroup$
2
  • $\begingroup$ Is it possible to find a counterexample in which $\mu$ is supported on a set of fractional dimension? $\endgroup$
    – Syang Chen
    Commented Jun 13, 2012 at 15:56
  • $\begingroup$ @SyangChen Probably yes, but I will leave the details to you. Note that the measure $\mu$ above is the pushforward of the Lebesgue measure on $[0,1]$ under the map $f(x)=x^{3/2}$. If you take the standard probability measure on the Cantor set and push it forward by $f(x)=x^p$, $p>1$, the behavior of the pushforward should be similar. $\endgroup$
    – user31373
    Commented Jun 13, 2012 at 16:21

You must log in to answer this question.

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