TITLE: Show $g^{(n)}(0)=0$ for all $n$
QUESTION [5 upvotes]: Suppose $g$ is infinitely differentiable function such that for all natural $n$, there exists $c_n$ and $\delta_n$ such that if $|x|<\delta_n$, then $|g(x)|\leq c_n|x|^n$. Show  $g^{(n)}(0)=0$ for all $n$.
Could someone provide a hint on where to start? I was thinking of a proof by induction, but even the base case eludes me.

REPLY [2 votes]: Suppose you have shown that $g^{(k)} (0)=0$ for $k \leq n$. By Taylor's expansion with integral form of remainder we get $\frac 1 {n!} \int_0^{x} (x-t)^{n} g^{(n)} (t)dt =g(x)$. Suppose $g^{(n)}(0)>0$. Then we get $\frac 1 {n!} \int_0^{x} (x-t)^{n}  dt \leq D_{n+2}|x|^{n+2}$ in some neighborhood of $0$ where $D_{n+2} =2\frac {C_{n+2}} {g^{(n)}(0)}$. [ Here $x>0$ is so small that $g^{(n)}(t) >\frac {g^{(n)}(0)} 2$ for $t \in (0,x)$]. This gives $x^{n+1} \leq Cx^{n+2}$ for some constant $C$ in some interval $(0,\delta)$ which is a contradiction. Similarly, $g^{(n)}(0)<0$ leads to a contradiction.