Following is problem 8.39 from Folland.

$\mu$ is a positive Borel measure on $\mathbb{T}=[0,1)$ with $\mu(\mathbb{T})=1$, then for its Fourier transform $\hat{\mu}(k)=\int_{\mathbb{T}}e^{-2\pi ikx}d\mu(x)$, prove that $|\hat{\mu}(k)|<1$ for any $k \neq 0$ unless $\mu$ is a linear combination of point mass.

I don't even know how to start with. I was trying to use the Radon-Nikodym theorem and thus decompose the measure w.r.t Lebesgue measure but fail to push it further. Also, I can't see there's any general way to conclude to the point mass case...

Any comment and help is appreciated.


2 Answers 2


Suppose $|\hat{\mu}(k)|=1$ for some $k\ne 0$; then we can write $\hat{\mu}(k) = e^{i \alpha}$ for some real $\alpha$. This means that $$1 = e^{-i \alpha} \hat{\mu}(k) = \int_{\mathbb{T}} e^{-i(2\pi k x + \alpha)}\,d\mu(x) = \int_{\mathbb{T}} \cos(2\pi k x + \alpha)\,d\mu(x) - i \int_{\mathbb{T}} \sin(2\pi k x + \alpha)\,d\mu(x).$$ So the cosine integral must equal $1$. Argue that this can only happen if $\cos (2 \pi k x+\alpha) = 1$ for $\mu$-almost every $x$, which is to say that $\mu(A) = 1$, where $A = \{x \in \mathbb{T} : \cos(2 \pi k x + \alpha) = 1\}$. Now note that $A$ is a finite set, so write $A = \{x_1, \dots, x_n\}$. Letting $a_k = \mu(\{x_k\})$, conclude that $\mu = \sum_{k=1}^n a_k \delta_{x_k}$.

  • $\begingroup$ Very simple and clear! Thanks! $\endgroup$
    – user388493
    Commented Apr 19, 2020 at 14:34
  • $\begingroup$ it seems that you pressed the "publish" button before me! $\endgroup$ Commented Apr 19, 2020 at 14:49
  • $\begingroup$ @AdriánGonzález-Pérez: No, your answer was definitely first. I just thought of taking a slightly different approach. $\endgroup$ Commented Apr 19, 2020 at 14:59

You can use the theory of convex bodies. A closed set $D \subset \mathbb{R}^n$ is convex if, given to points $x,y \in D$, the segment $[x,y] \subset D$. Assume that the boundary $\partial D$ is closed (this is not that crucial) then for any prob measure $mu$ supported in $\partial D$, we have that $$ y = \int_{\partial D} x \, \mu(x), $$ belongs to $D$.

Case $\mathbf{k = 1}$: Assume that $\mu$ is not given by a single atom, then $\mu$ is the convex combination $\mu = p \, \mu_1 + q \, \mu_2$, with $0 < p,q$ such that $p + q = 1$. The probability measures $\mu_1$ and $\mu_2$ have support in $$ \begin{eqnarray} \mathbb{T_1} & = & \mathbb{T} \cap \{ x \in \mathbb{R}^2 : \langle \eta, x \rangle \geq 0 \}\\ \mathbb{T_2} & = & \mathbb{T} \cap \{ x \in \mathbb{R}^2 : \langle \eta, x \rangle \geq 0 \} \end{eqnarray} $$ for some $\eta \in \mathbb{T}$ such that $\mu(\{+\eta\}) = \mu(\{-\eta\}) = 0$. But now: $$ \widehat{\mu}(k) \, = \, \int_\mathbb{T} e^{2 \pi i k \theta} \mu(\theta) \, = \, p \, \int_\mathbb{T_1} z \, \mu_1(z) + q \, \int_{\mathbb{T}_2} z \, \mu_2(z). $$ So $\hat\mu(k)$ is the convex combination of two elements, The first one belongs to the convex closure of $\mathbb{T}_1$ and the second one is the convex closure of $\mathbb{T}_2$. But those are the two halves of the unit disk separated by the line spanned by $\eta$. $\hat\mu(k)$ can not be have unit norm while being a strict convex combination of two elements in two different halves of the unit disk, that gives a contradiction.

General $\mathbb{k}$ "Unwind" the integral as $$ \widehat{\mu}(k) = \int_\mathbb{T} e^{2 \pi i k \theta} \mu(\theta) = \int_{\mathbb{T}} e^{2 \pi i \theta} \, \mu_k(\theta) $$ with $\mu_k$ being the pushforward of $\mu$ under the map $z \mapsto z^k$, with $z \in \mathbb{T}$. Then you can reduce the problem to the previous case.


You must log in to answer this question.