What else do we need to prove that $f(x)\equiv 0$?

Question

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?

Where is your function defined? Do you mean $C < 1$? For which $x$ is your condition supposed to hold? — amsmath, Nov 06 '17 at 03:11
see https://math.stackexchange.com/questions/2007945/on-the-gevrey-class-of-bump-functions — Will Jagy, Nov 06 '17 at 03:19

score 2 · Answer 1 · answered Nov 06 '17 at 03:19

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 $J$.

What else do we need to prove that $f(x)\equiv 0$?

1 Answers1