For a probability measure $\mu$, does $|1-\hat{\mu} (\xi)| \geq c \xi^2$ hold near zero?

Question

Let $\mu$ be a probability measure on $\mathbb{R}$, with $\mu \neq \delta_0$.

Does there exists $c>0$ and a neighborhood of $0$ on which $|1-\hat{\mu} (\xi)| \geq c \xi^2$?

This is true if $\mu$ has a finite second moment (then $\hat{\mu}$ is $\mathcal{C}^2$ with non-zero second derivative at $0$), but I neither have a general argument nor a counter-example.

Answer 1

Yes, this is true. $|1-\hat {\mu} (\xi)| \geq \int_{-r}^{r} [1-\cos \xi x] d\mu (x)$. We can choose $s$ so that $1-\cos u \geq \frac {u^{2}} 3$ for $|u| \leq s$. So we get $|1-\hat {\mu} (\xi)| \geq c\xi^{2}$ for $|\xi| \leq \frac s r$ where $c=\frac 1 3 \int_{-r}^{r} x^{2} d\mu(x)$ which cannot be $0$ for every $r$ unless $\mu = \delta_0$.