Let $f$ be an *infinitely differentiable function* on $[0,1]$ and suppose that for each $x \in [0,1]$ there is an integer $n \in \mathbb{N}$ such that $f^{(n)}(x)=0$. Then does $f$ coincide on $[0,1]$ with some polynomial? If yes then how.


I thought of using Weierstrass approximation theorem, but couldn't succeed.



If $f$ is infinitely differentiable then $f$ coincides with a polynomial
The proof is by contradiction. Assume $f$ is not a polynomial.


Consider the following closed sets:
$$
S\_n = \{x: f^{(n)}(x) = 0\} 
$$
and
$$
X = \{x: \forall (a,b)\ni x: f\restriction\_{(a,b)}\text{ is not a polynomial} \}. 
$$


It is clear that $X$ is a non-empty closed set without isolated points. Applying Baire category theorem to the covering $\{X\cap S\_n\}$ of $X$ we get that there exists an interval $(a,b)$ such that $(a,b)\cap X$ is non-empty and
$$
(a,b)\cap X\subset S\_n
$$
for some $n$. Since every $x\in (a,b)\cap X$ is an accumulation point we also have that $x\in S\_m$ for all $m\ge n$ and $x\in (a,b)\cap X$. 


Now consider any maximal interval $(c,e)\subset ((a,b)-X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$. Therefore $f^{(d)}=\mathrm{const}\neq 0$ on $[c,e]$. Hence $d< n$. (Since either $c$ or $e$ is in $X$.)


So we get that $f^{(n)}=0$ on $(a,b)$ which is in contradiction with $(a,b)\cap X$ being non-empty.


