Suppose $$f(x)=\sum\_{k=0}^\infty a\_k\frac{(i x)^k}{k!}$$ where $$a\_k=k!\int\_0^1 p\_k(y\_{k-1})\int\_0^{y\_{k-1}}p\_{k-1}(y\_{k-2})\cdots \int\_0^{y\_1}p\_1(y\_0) dy\_0\cdots dy\_{k-2}\;dy\_{k-1}$$ for functions $p\_i>0$ and $p\_i=p\_j$ if $i=j \mod 2$. Is $f$ is bounded on $\mathbb{R}$?


This is a generalization of situation when $p\_1=p\_2=p$. In this case, $$f(x)=\exp\left(ix \int\_0^1 p(y)dy\right),$$
and we know that $f(x)$ is bounded.


More generally, is there any assumptions that we can put on $a\_k$ to make sure $f(x)$ is bounded?


Any reference is appreciated.



bounded analytic function as a power series
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. 


