Let $\{\sigma_n\}$ be a sequence of measures on the complex unit circle $\mathbb{T}$ and let $\sigma$ also be such a measure. Suppose that $\hat{\sigma_n}(k) \rightarrow \hat{\sigma}(k)$ as $n\rightarrow\infty$ for every $k$ in $\mathbb{Z}$. Does that imply a convergence of the sequence $\{\sigma_n\}$ to $\sigma$? if so, in what sense?
[Math] Does convergence of Fourier transforms imply convergence of measures
fourier analysismeasure-theory
Best Answer
I'm assuming that the sequence $\{\sigma_n \}$ is bounded, that is $\sup_{n} \left| \sigma_n \right| (\mathbb{T}) < \infty$. I'm not sure whether this assumption is fundamental or can be deduced from the hypothesis.
Your hypothesis implies (and is actually equivalent to the fact) that the sequence $\{ \sigma_n \}$ converges weakly to $\sigma$, meaning that $\int_{\mathbb{T}} \varphi(t) \, d \sigma_n(t) \underset{n \to \infty}{\to} \int_{\mathbb{T}} \varphi(t) \, d \sigma(t)$ for all $\varphi \in \mathscr{C}(\mathbb{T})$.
This is the content of Lévy's continuity theorem, which is usually stated for probabilities on the real line, but which remains valid for complex measures on the torus.
Here is a sketch of the proof. Let denote for $k \in \mathbb{Z}$, $e_k \in \mathscr{C}(\mathbb{T})$ the function defined by $e_k(t) = e^{-ikt}$. Then, for every complex measure $\mu$ on the torus, we have $\hat{\mu}(k) = \int_{\mathbb{T}} e_k \, d\mu$.
The hypothesis implies that $\int_{\mathbb{T}} e_k \, d \sigma_n \underset{n \to \infty}{\to} \int_{\mathbb{T}} e_k \, d \sigma$ for every $k \in \mathbb{Z}$. Then, the conclusion is still valid for every trigonometric polynomial $\varphi$.
Now, if $\varphi$ is merely continuous on the torus, you can approximate it uniformly by a sequence of trigonometric polynomials, thanks to Weierstrass approximation theorem, and this implies easily the desired convergence. (this is where one uses the condition of boundedness for $\{\sigma_n\}$)
(alternatively, one can write $\varphi$ as the $L^2(\mathbb{T})$-sum of its Fourier series)