Let $A\subset [0,1]$ and let $\tilde A := \{ x \in \mathbb{R}^n : |x| \in A\}$ where $|\cdot|$ is the Euclidean norm. Let also $\mu$ be a radial measure on $\tilde{A}$ s.t. $|\hat \mu( \xi) | \le |\xi|^{-\alpha}$ for some $\alpha>0$, where $$ \hat \mu (\xi) = \int e^{-ix \cdot \xi} d\mu(x) $$ is the Fourier transform of $\mu$.
For every $\tau\ge 0$ we can define the translated set $A_\tau:= A+\tau$, and $\tilde{A_\tau}$ as above by rotating $A_\tau$ around the origin. Let $\mu_\tau$ be the pushforward measure of $\mu$ w.r.t. to the map $\varphi$ that, in polar coordinates, can be written as $\varphi(\rho\theta)=(\rho+\tau)\theta$.
My question is: can we conclude that, for every $\tau\ge 0$, $|\hat{\mu_\tau}(\xi)|\le |\xi|^{-\alpha}$, i.e. it has the same decay as $\mu$? If not, is there any other measure that would have this property
I would expect the answer to be yes, but I'm not sure how to prove it.
Unfortunately the transformation $\tilde A \mapsto \tilde{A_\tau}$ is not linear, so I cannot obtain the estimate on the Fourier transform of $\mu_\tau$ this way.
I tried to write $\hat{\mu_\tau}(\xi)$ in polar coordinates, writing $d\mu = d\mu^* \otimes \sigma^{n-1}$ and similarly $d\mu_\tau = d\mu^*_\tau \otimes \sigma^{n-1}$, however writing the transform in polar coordinates I only get
$$ \hat\mu(\xi) = \int_{S^{n-1}} \hat{\mu^*}(\theta\cdot \xi)d\sigma^{n-1}(\theta)$$
$$ \hat{\mu_\tau}(\xi) = \int_{S^{n-1}} \hat{\mu^*_\tau}(\theta\cdot \xi)d\sigma^{n-1}(\theta) = \int_{S^{n-1}} e^{i \tau \theta\cdot \xi}\hat{\mu^*}(\theta\cdot \xi)d\sigma^{n-1}(\theta)$$
and since I don't know the decay of $\hat{\mu^*}$ and $\hat{\mu^*_\tau}$ I don't know how to conclude.