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Let $\mu$ be a Borel probability measure with compact support on $\mathbb{R}$. Let $\hat{\mu}(\xi)=\int e^{-2\pi i\xi x}d\mu(x)$ denote the Fourier transform of $\mu$. Is the following claim true?

$\mu$ is not a Dirac measure if and ony if $|\hat{\mu}(\xi)|\not\equiv1, \xi\in \mathbb{R}.$

We know that if $\mu$ is a Dirac measure on $\mathbb{R}$, then $|\hat{\mu}(\xi)|\equiv1, \xi\in \mathbb{R},$ So, the sufficiency is true. Is the other hand true?

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2 Answers 2

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Assume that $|\int e^{-2\pi i \xi x} \mu(dx)| = 1$.

Fix $\xi$. Thanks to the equality case in the integral triangular inequality Equality in the triangle inequality for integrals, it means that there is $\theta$ such that $e^{-2\pi i \xi x} = e^{i\theta}$ for $\mu$-almost every $x$.

That means that $2\pi \xi x \in \theta + 2\pi \mathbb Z$, in other words, $x \in \dfrac {\theta}{\xi} + \dfrac 1 \xi \mathbb Z$ for $\mu$-almost every $x$.

Now consider $\xi_1, \xi_2$such that $\dfrac {\xi_1}{\xi_2} \notin \mathbb Q$. For $\mu$-a.e. $x$ we have $$x \in (\dfrac {\theta_1}{\xi_1} + \dfrac 1 {\xi_1} \mathbb Z)\cap(\dfrac {\theta_2}{\xi_2} + \dfrac 1 {\xi_2} \mathbb Z)$$

But this latter set cannot have more than one point (easy to show, if two distinct points, $\dfrac {\xi_1}{\xi_2} \in \mathbb Q$). This just says that the support of $\mu$ is a single point.

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  • $\begingroup$ Many thanks! @justt $\endgroup$
    – ljjpfx
    Commented Sep 27, 2022 at 12:23
  • $\begingroup$ Thank you, happy learning. The other answer is much more elegant in my opinion though ! $\endgroup$
    – justt
    Commented Sep 27, 2022 at 16:22
  • $\begingroup$ How to understand the last statement? i.e, You can do that for 2 different ξ which are not commensurable. This implies that x is constant for μ-almost every x. Would you like to give some further comments?Thanks! $\endgroup$
    – ljjpfx
    Commented Sep 28, 2022 at 4:37
  • $\begingroup$ Edited to answer $\endgroup$
    – justt
    Commented Sep 28, 2022 at 10:05
  • $\begingroup$ Many thanks for your kind reply! I think the constant $\theta$ depends on $\xi$, and $x\in \frac{\theta}{2\pi \xi}+\frac{1}{\xi}\mathbb{Z}$. How to prove that the set $(\frac{\theta_{\xi_1}}{2\pi \xi_1}+\frac{1}{\xi_1}\mathbb{Z})\cap (\frac{\theta_{\xi_2}}{2\pi \xi_2}+\frac{1}{\xi_2}\mathbb{Z})$ cannot have more than one point? $\endgroup$
    – ljjpfx
    Commented Sep 28, 2022 at 23:18
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Yes. Here is a probabilistic argument: Let $X,Y$ be i.i.d with law $\mu$. Then the characteristic function of $X-Y$ is $|\hat {\mu}|^{2}=1$. This implies that $X=Y$ a.s. But $X$ and $Y$ are independent , so $X$ must be a constant.

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  • $\begingroup$ Many thanks! @geentha290krm $\endgroup$
    – ljjpfx
    Commented Sep 27, 2022 at 12:22

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