Consider the unit circle $\mathbb{S}^1 \subset \mathbb{R}^2$ and consider the uniform measure $\nu$ on $\mathbb{S}^1$ normalised so that $\nu (\mathbb{S}^1)=1$. The collection of functions $\{1,z,\overline{z},z^2,\overline{z^2}\ldots\}$ forms an orthonormal basis for $L^2(\mathbb{S}^1,\nu)$. For each $n\geq 0$, we can think of the Bessel function $J_n$ as the Fourier transform of $z^n$, that is for $w \in \mathbb{R}^2$ $$ J_n(w) = \int_{\mathbb{S}^1} e^{-i\langle w,(\cos \theta, \sin \theta) \rangle} e^{in\theta} d\nu(\theta). $$ Note that by symmetry, $J_n(w) = J_n(\|w\|)$ and we can think of $J_n$ as just a real valued function $J_n :\mathbb{R} \to \mathbb{R}$. The Poisson representation formula for $J_n$ is known and it is as follows $$ J_{n}(r) = \frac{(r/2)^n}{\Gamma(n+\frac{1}{2}) \sqrt{\pi}} \int_{-1}^{1} e^{irt} (1-t^2)^{n-\frac{1}{2}}~ dt. $$ It is easy to see the vanishing property of $J_n$ near the origin from the above formula, that is we can conclude $$ |J_{n}(r)| \leq 10 \cdot \frac{(r/2)^n}{\sqrt{2\pi} \left(n/e\right)^n \sqrt{\pi}} \leq \left( \frac{2r}{n}\right)^n. $$
My question is: Suppose we have an arbitrary probability measure $\mu$ on $\mathbb{S}^1$ (which has no atoms) and let $\{f_1,f_2,\ldots\}$ be an orthonormal basis for $L^2(\mathbb{S}^1,\mu)$. Is it possible to conclude similar vanishing properties near the origin for the Fourier transform of $f_n$?
Thanks!