Let $\mu$ be a finite non negative singular measure on $\mathbf{R}^d$. I would like to know if there exists some result on the infimum of the absolute value of its Fourier Transform $$\hat{\mu}(t)=\displaystyle\int \mathrm{e}^{2i\pi t\cdot x} ~\mu(dx).$$


It is known that we don't necessarily have $\hat{\mu}(t)\rightarrow0$ for singular measure. But does one have like Lebesgue measure $$\mathrm{inf}~|\hat{\mu}(t)|=0~?$$


Edit : To be more precise, $\mu$ is continuous singular, so does not have a discret part (<https://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem>)



Infimum of Fourier transform of singular measure
Yes, that's true. By Wiener's lemma we have $\sum\_{x\in \mathbb{R}^d} |\mu(\{x\})|^2 = \lim\_{R\to \infty} \frac{1}{(2R)^d} \int\_{[-R,R]^{d}} |\widehat{\mu}(t)|^2 dt$. If $\mu$ is continuous, then the left-hand side is $0$, so the infimum of $|\widehat{\mu}|$ has to be zero as well, since the right-hand side is bounded below by this infimum (squared).


