TITLE: What else do we need to prove that $f(x)\equiv 0$?
QUESTION [5 upvotes]: The condition 
$$f^{(n)}(0)=0,\ \ n=0,1,2,\cdots$$ 
is not sufficient to conclude that $f(x)\equiv 0$. What conditions can we add to get $f(x) \equiv 0$?
Is 
$$|f^{(n)}(x)|\leqslant n!C^n,$$
where $C$ is a constant, a sufficient condition?
Can we improve it?

REPLY [2 votes]: If $f$ is analytic on a domain $D \subseteq \mathbb C$ containing $0$, and all $f^{(n)}(0)=0$, then $f(x) \equiv 0$ on $D$. 
In particular if there is $C$ such that $|f^{(n)}(x)| \le n! C^n$ for all $x$ in an interval $J$ containing $0$, then using Taylor's theorem with Lagrange remainder we find that the Taylor series of $f$ about any point $p \in J$ converges to $f$ in an interval about $p$, and thus there is such a domain
containing $J$.