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Suppose $\mu$ is a finite measure on $\mathbb{R}^2$ and consider the Fourier transform of $\mu$, denoted by $\hat{\mu}$ \begin{align} \hat{\mu} (x,y):= \int_{\mathbb{R}^2} e^{-i\langle (x,y),(s,t)\rangle} d\mu (s,t). \end{align} I have the following question regarding $\hat{\mu}$:

Suppose we know $\hat{\mu}$ on the unit circle $\mathbb{S}^1$, does it completely determine $\mu$? In other words is it possible to find two different measures $\mu_1$ and $\mu_2$ such that $\hat{\mu_1} = \hat{\mu_2}$ on $\mathbb{S}^1$?

Thanks!

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Clearly not. Say $\mu_1$ and $\mu_2$ are both rotation-invariant measures. Then $\hat\mu_j$ is constant on $S^1$. Wlog (adding $\delta_0$, say) both constants are non-zero, so there exist $a_j>0$ such that $a_1\hat\mu_1=a_2\hat\mu_2$ on $S^1$.

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  • $\begingroup$ That answers my question, thanks. $\endgroup$
    – April
    Commented Jan 29, 2020 at 16:39

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