When is a Fourier series analytic?

Question

By Fourier theory, every continuously differentiable function $f : S^1 \to \mathbf C$ admits a unique, uniformly convergent Fourier expansion

$$f(\theta) = \sum_{n\in \mathbf Z} a_n e^{in\theta}.$$

Does there exist a condition on the sequence $(a_n)$ equivalent to $f$ being analytic in $\theta$ in a neighborhood of $\theta=0$ (meaning that it has a convergent Taylor expansion $\sum_{m\geq 0}\frac{\theta^m}{m!} b_m$ in a neighborhood of $0$)?

Writing $e^{in\theta} = \sum_{m\geq 0} \frac{(in)^m}{m!} \theta^m$, formally switching the two sums yields

$$f(\theta) = \sum_{n\in \mathbf Z} \sum_{m\geq 0} a_n \frac{(in)^m}{m!} \theta^m = \sum_{m\geq 0} \sum_{n\in \mathbf Z} a_n \frac{(in)^m}{m!} \theta^m = \sum_{m\geq 0}\frac{\theta^m}{m!} b_m$$

where $$b_m = i^m\sum_{n \in \mathbf Z} a_n n^m.$$

(Remark that there is the formal "double-sided" Dirichlet series

$$L(f, s) = \sum_{n \in \mathbf Z} a_n n^{-s}$$

and formally, we can write $b_m = i^mF(-m)$.)

First question: is it true that $f(\theta)$ is analytic at $0$ if and only if:

1. the sums defining the $b_m$'s converge absolutely, and
2. the series $\sum_{m\geq 0}\frac{\theta^m}{m!} b_m$ converges in a neighborhood of $0$, in which case it coincides with $f$ around $0$?

Second question: Can we determine whether $f$ is analytic around $0$ directly by looking at the sequence $(a_n)$?

Answer 1

I begin with the second question. By looking at the coefficients $a_n$, you can easily see analyticity on all of $S^1$: it holds iff $|a_n|$ decay exponentially in both directions. This is Exercise I.4.4 from An Introduction to Harmonic Analysis by Y. Katznelson, 3rd edition:

Show that $f$ is analytic on $\mathbb T$ if and only if there exist constants $K>0$ and $a>0$ such that $|\hat f(j)|\le K e^{-a|j|}$.

Indeed, think in terms of complex plane, replacing $\exp(i\theta )$ by $z$. Two-sided exponential decay gives you convergent Laurent series in some annulus containing the circle. Conversely, analytic function on the circle can be extended to analytic function on some annulus; such a function is represented by a convergent Laurent series, which gives exponential decay of coefficients.

At the same time, Fourier coefficients are poorly suited to detect analyticity at a point, since they take into account the whole function. If you have a break in second derivative somewhere on the circle, that is going to dominate the behavior of coefficients; the contribution of analytic piece somewhere will be tiny in comparison.

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First question: the conditions 1-2 are sufficient for analyticity in a neighborhood of $0$, but are not necessary. To begin with, the series $$b_m = i^m\sum_{n \in \mathbf Z} a_n n^m \tag{1}$$ converges for all $m$ iff $|a_n|$ decay faster than any power of $n$. This kind of decay is equivalent to $f$ being in $C^\infty(\mathbb T)$ (see section I.4 in Katznelson's book). So, if you have an analytic function in a neighborhood of $0$ that is not $C^\infty$ on the entire circle, the condition for $b_m$ does not make sense*.

(*) You could try to regularize the divergent series (1), but I think the regularization would amount to summing the Fourier series and then differentiating the sum.

Sufficiency can be established without rearranging the series. As noted above, the convergence of every series in (1) implies $f\in C^\infty(\mathbb T)$. We evaluate the derivatives of $f$ at $0$ just by plugging in $\theta=0$ into the differentiated Fourier series. This gives Taylor coefficients, in terms of which the analyticity at $0$ is characterized.