# Fourier transform of measure on $\mathbb{T}$

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.

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 $$\mu = \sum_{k=1}^n a_k \delta_{x_k}$$.

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