TITLE: $f^{(n)}(0)=0$ and $|f^{(n)}(x)|\leq n!$ implies $f=0$
QUESTION [6 upvotes]: Let $f$ be a $C^\infty$ function such that $f^{(n)}(0)=0$ and $|f^{(n)}(x)|\leq n!$ for all integer $n$ and real $x$.
How to prove that $f(x)=0$ for all real $x$.
I proved this only for $x\in(-1,1)$ using Taylor-Lagrange Formula by showing that $$|f(x)|\leq x^{n+1}$$ for all integer $n$ and real $x$.

REPLY [5 votes]: As zhw has said in the comments, you have shown that $f \equiv 0$ on $(-1,1)$. By continuity we can strengthen this to $[-1,1]$. 
So, we have $f(1) = 0$. Note that $f^{(n)}(1)$ is the same as the $n$th left-derivative at $x=1$ since all derivatives are continuous. Hence we have $f^{(n)}(1) = 0$. By considering the Lagrange formula, we can see $|f(x)| \leq (x-1)^{n+1}$ so we have $f \equiv 0$ on $[0,2]$ as well. By induction we can see that we have $f \equiv 0$ on $[k-1,k+1]$ for all $k \in \mathbb{N}_{\geq 0}$, which implies $f \equiv 0$ for $x \geq -1$. We can use essentially the same argument for $x < -1$, but this time it's 'reverse' induction on the negative integers (e.g., now $k \in \mathbb{Z}_{<0}$).