I feel like this should be wellknown, but haven't been able to find any reference so far. Consider the set of all smooth functions on $\mathbb{R}$ such that $$\sup_{x\in \mathbb{R}} e^{\alpha x} f^{(n)}(x)<\infty \qquad \mbox{for all $n\in \mathbb{N},$ $\alpha \in \mathbb{C}$. }$$ Then the twosided laplace transform of such a functions is well defined, and gives rise to a holomorphic function on $\mathbb{C}$. Is there a PaleyWiener kind of theorem that gives a characterization of the space of functions that can be generated in this way? (That is, as Laplace transforms of smooth functions with exponential decay for all its derivatives.)
We can certainly characterize these functions in terms of their Fourier transforms, but what I'm about to write down just rephrases the well known connection between holomorphic (on a strip) Fourier transforms and exponential decay, and it's not very exciting.
Let me write $F$ for the FT of $f$. Then I claim that $f$ satisfies your conditions if and only if $F$ is an entire function such that for any $n\ge 0$, $L\ge 1$, we have that $(x+iy)^nF(x+iy)$ is bounded on $L\le y\le L$ and $$ \sup_{L\le y\le L} \int_{\infty}^{\infty} x+iy^n F(x+iy)\, dx < \infty . $$
It is obvious that $F(z)=\int f(t) e^{itz}\, dt$ has these properties if $f$ is as above (note that the smoothness of $f$ gives $F$ enough decay to make it integrable). Conversely, if $F$ satisfies these conditions, then we obtain the desired estimates on $f^{(n)}(x)=\int (iz)^n F(z) e^{itz}\, dz$ by moving the path of integration into the complex plane, to pick up the required exponentially small factors.

$\begingroup$ I was not aware of this possible characterization, and it actually fits very nicely with the context I'm working on. Thank you very much! $\endgroup$ – Raul Gomez Jun 14 '17 at 17:11

$\begingroup$ @RaulGomez: Actually, I think it becomes a characterization only because you impose exponential decay at arbitrarily fast rates, I think it's not as clean when you consider the connection FT holomorphic on a strip <> function has exponential decay. $\endgroup$ – Christian Remling Jun 14 '17 at 23:23