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!
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.