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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?

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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)

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Thanks. Can you please give me a reference for this theorem? –  user25640 Sep 17 '12 at 12:03
    
Also, Is $\mathscr{C}(\mathbb{T})$ the space of continuous functions from $\mathbb{T}$ to $\mathbb{C}$ ? –  user25640 Sep 17 '12 at 12:13
    
Yes, it is. I'm not able to locate a precise reference, so I'll give some elements of the proof, which is a straightforward adaptation of the standard proof in the case of probability measures on the real line. I'm not sure now whether one should also assume the sequence $\{\sigma_n\}$ to be bounded ... –  Ahriman Sep 17 '12 at 12:17
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The total mass of a non-negative measure is given by the zeroth Fourier coefficient, so provided $\{\sigma_n\}$ are non-negative, they are already bounded. –  Byron Schmuland Sep 17 '12 at 12:36
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This is used in proofs of the central limit theorem in probability theory. If $X_1,X_2,X_3,\ldots$ is a sequence of independent identically distributed real-valued random variables and $\mathbb{E}(X_1^2)<\infty$, then the probability distribution of the random variable $$ \frac{(X_1+\cdots+X_n)-n\mathbb{E}(X_1)}{\sqrt{n\operatorname{var}(X_1)}} $$ approaches the standard normal distribution, whose probability density function is $$ x\mapsto \frac{1}{\sqrt{2\pi}} e^{-x^2/2},\tag{1} $$ as $n\to\infty$.

Proof: Show that the sequence of Fourier transforms of this sequence of probability distributions approaches the Fourier transform of the stated limit.

This of course relies on an affirmative answer to the question posed here, except that it deals with probability measures on the line rather than on the compact circle. I don't think it's hard to deduce the case for the circle from the case for the line, since as a measure space, the circle is merely a subset of finite measure, of the line.

In what sense? The sense of convergence used in the statement of this theorem in probability theory is usually stated something like this: Let $F$ be the cumulative probability distribution function $F(x) = \Pr(X_1\le x)$. Then the sequence of cumulative probability distribution functions whose convergence to $F$ is assserted, converges pointwise to $F$ on the set of points of continuity of $F$. Now notice that the cumulative distribution function corresponding to $(1)$ is continuous everywhere, so that last phrase doesn't matter.

(The reason for not requiring convergence at points where $F$ is not continuous can be seen by thinking about the probability measure assigning probability $1$ to every set containing $1/n$ as a member and $0$ to every set disjoint from $\{1/n\}$. Under this definition, this approaches the measure concentrating all probability at $0$, as $n\to\infty$. One wants the definition to imply that.)

See the Wikipedia article: Lévy's continuity theorem.

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