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.

  • $\begingroup$ If you also require that the sum defining $f(x)$ converges absolutely for all $x$, then the necessary and sufficient condition is that $\forall k>0: a_k = 0$. $\endgroup$ Feb 25, 2016 at 20:01
  • $\begingroup$ @MarkFischler Sorry? I thought the power series always converges absolutely... $\endgroup$
    – Qijun Tan
    Feb 25, 2016 at 20:07
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    $\begingroup$ Okay, so could you please edit your answer once to make this clear to all (and then please stop editing your question for a while...). $\endgroup$ Feb 25, 2016 at 21:05
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    $\begingroup$ Anyway, you realize that the newer version is essentially equivalent to the original formulation (by taking real and imaginary parts), don't you? Since you haven't explained in which terms (algebraic, analytic,...) you want a necessary and sufficient condition, it seems difficult to answer the question as it stands. $\endgroup$ Feb 25, 2016 at 21:17
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    $\begingroup$ @QijunTan May be this link would be some how related to the question: en.wikipedia.org/wiki/… $\endgroup$ Feb 26, 2016 at 19:46

1 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.


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