When can one extend a function on the unit circle to an analytic function? Suppose I have a function $f$ defined on the unit circle. Under what conditions can I define a new function $g$ defined on a subset of the complex plane containing the unit disk such that $f = g$ on the unit circle?
 A: A necessary condition would be that $f$ is analytic, in the sense that $f(e^{it})=e^{iF(t)}$, where $F(t)=\sum_{n \geq 0} a_n t^n$ and $F: \mathbb R \to \mathbb C$ satisfies $F(t+2\pi)=F(t)+2k\pi$ for some integer constant $k$.
Additionnally, since you want your extension to be defined near the unit disk (and not just near the unit circle), you'll want $F$ to extend to a holomorphic function on the upper half-plane. That condition is also clearly sufficient.
Off the top of my head, I don't see any simple necessary and sufficient condition on the $a_n$ to ensure this condition; for example, it is clearly sufficient that $F$ is entire. 
A: If $f:\>S^1\to{\mathbb C}$ is a reasonable function then
$$f(e^{it})=\sum_{k=-\infty}^\infty c_k e^{ikt}\ ,$$
whereby this series is convergent in some sense. On the other hand, if $$g:\quad\bar D\to{\mathbb C},\quad z\mapsto \sum_{k=0}^\infty a_kz^k$$
is analytic in a neighborhood of $\bar D$ then
$$g(e^{it})=\sum_{k=0}^\infty a_k e^{ik t}\ .$$
It follows that for $f=g\restriction\partial D$ it is necessary all Fourier coefficients $$c_k={1\over2\pi}\int_0^{2\pi} f(t)e^{-ikt}\>dt\qquad(k<0)$$  are zero. When $f$ is sufficiently smooth this is also sufficient.
See also this answer:  Connection between Fourier transform and Taylor series
