bounded analytic function as a power series

Question

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.

Answer 1

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.