Does zero Fourier dimension imply there is no Rajchman measure?

Question

Let $K\subset R^d$ be a compact set. It is well known that its Fourier dimension is defined by $$\dim_F K=\sup\{s\ge 0: \exists \mu \in M_1(K) s.t. \hat{\mu}(x)=O(|x|^{-s/2})\}(|x|\to\infty),$$ where $M_1(K)$ denotes the set of probability measures with support in $K.$

My question is: Does the condition $\dim_FK=0$ imply that $K$ can not support a measure $\mu$ with decaying Fourier transformation? In other word, there is no Rajchman measure spported on $K?$

Maybe there exists measure with decaying Fourier transformation according other ``speed" rather than polynomial speed. Can somebody give an example? Many thanks.

Answer 1

No, this does not follow. Since the Hausdorff dimension dominates the Fourier dimension, it suffices to establish the existence of a compact set $K$ of Hausdorff dimension zero that supports a Rajchman measure.

The last theorem of section 3 (attributed to Ivashev-Musatov 1962) of Lyons's survey gives a considerably stronger version of this statement (note that $\mathcal U_0$ is defined in the introduction of the paper as the collection of sets that are annihilated by all Rajchman measures).