# On the Fourier transform of Dirac measure

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?

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.

• Many thanks! @justt Commented Sep 27, 2022 at 12:23
• Thank you, happy learning. The other answer is much more elegant in my opinion though ! Commented Sep 27, 2022 at 16:22
• 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! Commented Sep 28, 2022 at 4:37
• Edited to answer Commented Sep 28, 2022 at 10:05
• 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? Commented Sep 28, 2022 at 23:18

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.

• Many thanks! @geentha290krm Commented Sep 27, 2022 at 12:22