TITLE: Function with derivatives with "fixed" values at $0$
QUESTION [5 upvotes]: If $ \{a_n \}_{n \in \mathbb{N}} \subset \mathbb{R}$ is a non-constant sequence of real numbers, is possible to find a function $f \in \mathcal{C}^{\infty}_{b}(\mathbb{R})$ ( = space of bounded smooth functions) such that and $f^{(n)}(0)=a_n$ for every $n \in \mathbb{N}$?

I am struggling with this problem, arisen from my sleepless nights, since a couple of days, but I have no idea of how to prove/disprove it. I only noticed that if I remove the hypothesis of boundedness and I assume $\limsup_{n} \sqrt[n]{a_n / n!}=0$ then $f(x) = \sum_{n \ge 1} \frac{a_n}{n!} x^n$ should work, but this doesn't help so much.
Thanks in advance!

REPLY [3 votes]: Outline: It can be done. We use "bump" functions, please see Wikipedia. Let $f_0(x)=a_0$ on the interval $(-1/2,1/2)$ and make it decay to identically $0$  beyond $(-1,1)$ by multiplying $a_0$ by a suitable bump function which is $1$ on $(-1/2,1/2)$ and decays to $0$ beyond $(-1,1)$.
Let $f_1(x)=a_1x$ on $(-1/4,1/4)$, and let it decay to $0$ by multiplying $a_1x$ by a bump function which is $1$ on $(-1/4,1/4)$  and decays to identically $0$ beyond $(-1/2,1/2)$. Let $f_2(x)=\frac{a_2x^2}{2!}$ on $(-1/8,1/8)$ and decay to identically $0$ beyond $(-1/4,1/4)$. And so on. Let $f=\sum_0^\infty f_n(x)$.