How does inversion look like for the Fourier transform of a measure?

Question

Let $\mu$ be a probability measure, say, on the circle $\mathbb{T} = \mathbb{R} / \mathbb{Z}$. We get the coefficients $C(\mu)(n) = \int_\mathbb{T} e^{-2 \pi i n x} d\mu(x)$. How does one go back to the measure if they have the coefficients?

Answer 1

I prefer to work with the integral over $[0,2\pi]$ as opposed to over $[0,1]$ because I'm used to it. You can change variables in order to view $\mu$ over $[0,2\pi]$, and then you're considering $$ C_{\mu}(n) = \frac{1}{2\pi}\int_{0}^{2\pi}e^{-in\theta}d\mu(\theta). $$

The Poisson kernel is $$ P(r,\theta)=\frac{1}{2\pi}\frac{1-r^2}{1-2r\cos(\theta)+r^2}=\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}. $$ Therefore, $$ \int_{0}^{2\pi}P(r,\theta-\theta')d\mu(\theta') = \sum_{n=-\infty}^{\infty}r^{|n|}C_{\mu}(n)e^{in\theta}. $$ If $J_{(a,b)} = \{ e^{i\theta} : a < \theta < b \}$ and $J_{[a,b]}=\{ e^{i\theta} : a \le \theta \le b \}$, then \begin{align} &\lim_{r\uparrow 1}\int_{a}^{b}\left(\int_{0}^{2\pi}P(r,\theta-\theta')d\mu(\theta')\right)d\theta \\ &= \lim_{r\uparrow 1}\int_{0}^{2\pi}\left(\int_{a}^{b}P(r,\theta-\theta')d\theta\right)d\mu(\theta')\\ &= \int_{0}^{2\pi}\frac{1}{2}\left[\chi_{(a,b)}(\theta')+\chi_{[a,b]}(\theta')\right]d\mu(\theta')\\ &= \frac{1}{2}\mu J_{(a,b)}+\frac{1}{2}\mu J_{[a,b]}. \end{align} From that, you can obtain the $\mu$ measure of any open, closed or half-open continuous arc of the circle, which is enough to uniquely determine any Borel measure $\mu$ on $\mathbb{T}$.