Finding an assumption on a sequence $(a_k)_{k \in \mathbb Z}$ such that the map $t \mapsto \sum_k a_ke^{ikt}$ is entire

Question

Suppose that $(a_k)_{k \in \mathbb Z}$ is a sequence such that $\sum_k |a_k|$ is convergent. The function $f : \mathbb R \to \mathbb C$ $$ f(t) = \sum_k a_ke^{itk} $$ is then a continuous function. If we replace $t \in \mathbb R$ by $z \in \mathbb C$ then $f$ extends to function from $\mathbb C$ to $\mathbb C$. Can we find a sufficient condition on the sequence $(a_k)_k$ such that this extended function is entire?

Answer 1

A sufficient condition is that the power series

$$\sum_{k=0}^\infty a_k z^k$$

has infinite radius of convergence.

Proof:

We need compact convergence (so uniform convergence on all compact sets), because the following holds:

If $f_n:D\to\mathbb C$ is a sequence of holomorphic functions on an open set $D\subseteq \mathbb C$ which converges to $f:D\to\mathbb C$ compactly, then $f$ is holomorphic.

We can apply this to the sequence of partial sums of the series, which are all holomorphic. Since in this case $D=\mathbb C$ ($f$ is supposed to be entire), all compact subsets of $D$ are contained in a closed (and thus compact) disc centered at $z=0$. So it's sufficient to consider convergence on all closed discs. If it is uniform on all such discs, the limit will be entire.

To show uniform convergence on such discs $K$, we can show that $\sum_k a_k e^{itk}$ is Cauchy with respect to the infinity norm

$$\Vert f\Vert_\infty:=\sup_{z\in K}\vert f(z)\vert,$$

that is, $\Vert f_n-f_m\Vert_\infty$ can be made arbitrarily small by choosing $n,m$ larger than an appropriate $N\in\mathbb N$. For a series, this means

$$\left\Vert\sum_{k=m+1}^n a_k e^{izk}\right\Vert_\infty$$

can be made arbitrarily small in the same way. We can give some upper bounds to this expression, namely

$$\begin{align*}\left\Vert\sum_{k=m+1}^n a_ke^{izk}\right\Vert_\infty&\leq\sum_{k=m+1}^n\Vert a_k e^{izk}\Vert_\infty\\ &\leq\sum_{k=N+1}^\infty\Vert a_ke^{izk}\Vert_\infty\\ &=\sum_{k=N+1}^\infty\vert a_k\vert e^{Rk}, \end{align*}$$

where $R$ is the radius of the disk $K$. Remember, we want to be able to choose $N$ appropriately so that this expression is arbitrarily small. This is possible iff the series

$$\sum_{k=0}^\infty\vert a_k\vert e^{Rk}$$

converges. And we want it to converge for all $R>0$, since we want to be able to choose an arbitrarily large disc. Now notice that this series is a power series in $e^R$. So we can reduce the question for which $R$ this series converges to the question what the radius of convergence of the series $\sum_k a_k z^k$ is. If it is infinity, then the above series converges for all $R$. So we have our sufficient condition:

$\sum\limits_{k=0}^\infty a_k e^{izk}$ is entire if $\sum\limits_{k=0}^\infty a_k z^k$ has radius of convergence $\infty$, or equivalently, if this power series is entire.

You can now use the known ways to calculate radii of convergence. One of them will yield the condition Conrad mentioned in the comments.