If $f(x)$ is bounded, the Laplace transform

$${\mathscr L}f(s) = \int_0^\infty f(x) e^{-sx}\; dx$$
is analytic in the open right half plane, and the same goes for the Laplace transform of $\widetilde{f}(x) = f(-x)$.
On the other hand, $|a_k| \le C^k$ implies that $\sum_{k=0}^\infty a_k i^k s^{-k-1}$ converges absolutely to an analytic function $g(s)$ for $|s|> C$, with
$|g(s)| \le 1/(|s|-|C|)$ and
this agrees with $\mathscr L f(s)$ and $- \mathscr L \widetilde{f}(-s)$ in the intersections of this region with the open right and left half planes.
Thus $\mathscr L f(s)$ can be analytically continued to a function analytic in
$\mathbb C \backslash I$ where $I$ is the closed line segment from $-Ci$ to $Ci$.

Conversely, if $g(s)$ is an analytic function in $\mathbb C \backslash I$ with $\lim_{|s| \to \infty} g(s) = 0$, the Bromwich integral defines $f(x)$ on $\mathbb R$ that has this Laplace transform. However, this is not necessarily bounded. A sufficient condition is that
$g(s) = \int_{-C}^C (s-it)^{-1} d\mu(t)$ where $\mu$ is a signed measure on $[-C,C]$, which translates to $f(x) = \int_{-C}^C \exp(itx)\; d\mu(t)$.
This is also necessary in the case that $g(s)$ is a rational function: in that case it means that the only singularities of $g$ are simple poles.

absolutelyfor all $x$, then the necessary and sufficient condition is that $\forall k>0: a_k = 0$. $\endgroup$onceto make this clear to all (and then please stop editing your question for a while...). $\endgroup$11more comments