Is fourier series of a function with $e^{j\theta}$ replaced with a complex variable $z$ holomorphic on the unit disc?

Question

Consider any continuous $2\pi$ periodic function (of bounded variation) $f : \mathbb{R} \to \mathbb{R}$ and its fourier series given as

$f(\theta) = \frac{a_o}{2} + \sum\limits_{n = 1}^{\infty} (a_n\cos{n\theta} + b_n\sin{n\theta})$

The above series is the real part of the series

$F(z) = \frac{a_o}{2} + \sum\limits_{n=1}^{\infty} (a_n-jb_n)z^n$

It is stated in this link that $F(z)$ is holomorphic on the unit disc without any restrictions on $f(\theta)$. I suspect that $F(z)$ is not holomorphic for any $f(\theta)$ and request you to clarify the validity of this statement.

Answer 1

In the given link it is only stated that $F(z)$ is a "power series along the unit circle" which says no more than that the coordinate $\theta$ on $\partial D$ has been replaced by the new variable $z$.

Now to your question: If the function $f(\theta)$ is continuous and of bounded variation on ${\mathbb R}/(2\pi{\mathbb Z})$ then its complex Fourier coefficients $c_k={1\over 2}(a_k-i b_k)$ $\ (k\geq0)$ are of order $O({1\over k})$. This implies that the radius of convergence of the power series $\sum_{k=0}^\infty c_k z^k$ is at least $1$. Therefore the function $F(z)$ is holomorphic in $D$.

Answer 2

It can be easily seen that $F(z)$ is always holomorphic because it is defined by a Taylor expansion and Taylor-expanded functions are always holomorphic within the disk where the Taylor expansion converges: the only thing we have to avoid are singularities.

And the radius of convergence is surely bigger than $1$ because $f(\theta)$ which is the value of $F(z)$ on the unit circle is, by assumption, convergent and well-defined. (If the real part of Taylor expansion is convergent, the imaginary part has to be as well.) Because the Taylor expansion converges on the unit circle, it must also converge for all smaller values of $|z|$ i.e. inside the unit disk.

The statement would actually work even for distributions, not just "genuine" functions $f(\theta)$. For example, if $f(\theta)=\delta(\theta)$, then the Fourier transform is proportional to $1+2\sum_{n\in Z^+} \cos n\theta$ and the corresponding Taylor expansion for $F(z)$ will be $\sum_{n\in Z+} z^n$ which is geometric series converging to $1/(1-z)$.

One picks the delta-function using the usual rules for the distributions, $1/(\theta+i\epsilon)$ is equal to the principal value minus $i\pi$ times $\delta(\theta)$ as long as the imaginary value is picked from the expansion. So distributions will produce singularities of $F(z)$ at the unit circle - but nothing strictly inside the unit disk.