Continuous probability measures on the unit circle Is there a continuous probability measure on the unit circle in the complex plane - $\sigma$  with full support, such that $\hat{\sigma}(n_k)\rightarrow1$ as $k\rightarrow\infty$ for some increasing sequence of integers $\ n_k$ 
 A: Define a sequence $(X_m)_{m\geq 1}$ of independent $\{0,1\}$-valued random variables by 
$$\mathbf{P}(X_m = 0) = p_m > 1/2,$$
where the sequence $p_m\to1$ slowly. Define
$$X = X_1/2 + X_2/4 + \cdots.$$
Then the distribution $\sigma$ of $X$ is continuous in $[0,1]$ provided only $\prod_{m\geq1} p_m = 0$. Moreover $\sigma$ has full supported provided $0<p_m<1$ for all $m$.
Define
$$Y_m = 2^m X \text{(mod $1$)} = X_{m+1}/2 + X_{m+2}/4 + \cdots.$$
Then $\mathbf{P}(Y_m\leq 1/2^k) = p_{m+1}\cdots p_{m+k}\to 1$ as $m\to\infty$, for each fixed $k$. Thus $Y_m$ tends to $0$ in distribution. It follows that
$$\hat{\sigma}(2^m) = \int e^{-i2\pi 2^m x} d\sigma(x) = \mathbf{E}(e^{-i2\pi Y_m}) \to 1.$$
A: Define the sequence $n_k$, a sequence of positive reals $r_k$ and a sequence of nested subsets $A_k$ of the circle $\mathbb T$ as follows.  Each $A_k$ will be the union of $2^k$ open intervals of length $r_k$ on which $|e^{in_k t} - 1| < 2^{-k}$, and each of these intervals will contain two intervals of $A_{k+1}$.  This can be done inductively: all we need is to take $n_{k+1}$ large enough so that each interval of $A_k$ contains at least two points where $e^{i n_{k+1} t} = 1$, and take intervals of small enough length $r_k$ around two of those to form $A_{k+1}$.  I will also choose $n_{k+1}$ to be a multiple of $n_k$.
Now let $\mu_k$ be normalized Lebesgue measure on $A_k$, and $\mu$ a weak limit point of $\mu_k$.  Then $|\hat{\mu}(n_k) - 1| \le 2^{-k}$ for all $k$.  $\mu$ is a singular continuous probability measure.  But you wanted one with full support.
OK, take $\sigma = \sum_{j = 1}^\infty 2^{-j} T_{t_j}\mu$ where $T_t$ is translation by $t \in [0,2 \pi)$, choosing $t_j$ a sequence dense in $[0,2 \pi]$ such that $n_j t_j/(2 \pi)$ is an integer. This is again a singular continuous probability measure, but with full support.  For $k \ge j$, $n_k$ is a multiple of $n_j$ and so $\widehat{T_{t_j}\mu}(n_k) = \hat{\mu}(n_k)$.  Thus $$|\hat{\sigma}(n_k) - 1| \le \sum_{j=1}^k 2^{-j} |\hat{\mu}(n_k) - 1| + \sum_{j=k+1}^\infty 2^{-j} |\widehat{T_{t_j}\mu}(n_k) - 1|
\le 2^{2-k}$$
A: If I'm not mistaken, the Riemann–Lebesgue lemma rules this out if "continuous" means absolutely continuous with respect to Lebesgue measure.  I'm not sure what happens if you mix in a "continuous singular" measure.
(If you concentrate measure $1$ at a point, you get every coefficient of the Fourier series equal to $1$, so your hypothesis of full support certainly can't be dropped.)
