# Infinitely times differentiable function

Let $f$ belong to $C^{\infty}[0,1]$ and for each $x \in [0,1]$ there exists $n \in \mathbb{N}$ so that $f^{(n)}(x)=0$. Prove that $f$ is a polynomial in $[0,1]$.

I am trying to use Baire Category Theorem , but cannot perfectly complete the proof .

Thanks for any help.

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Dear Timothy - Given that this is a relatively "short" question (in terms of mathematical notation), perhaps this would be a good question to start learning how to use LaTeX to format your questions? –  amWhy Nov 8 '12 at 20:35
See also this MathJax LaTeX tutorial. –  amWhy Nov 8 '12 at 20:51
(I voted to close, but perhaps I shouldn't have. This question is a slight generalization of the "duplicate", but there are no correct answers (except in the comments) to the duplicate question. The answers in the comments lead to mathoverflow, which answers this more general question.) –  Jason DeVito Nov 8 '12 at 23:03
The solution is given in this MO thread: mathoverflow.net/questions/34059/… –  Lukas Geyer Nov 11 '12 at 16:24
@Timothy What's wrong with the solution provided at math.overflow? –  Davide Giraudo Nov 15 '12 at 21:29

In the latest, we argue by contradiction. The set $X$ is the points $x$ such that for all $a<b$, the restriction of $f$ to $(a,b)$ is not an polynomial if $x\in (a,b)$. It doesn't contain an isolated point (if $x_0$ were such a point, there would be $r>0$ such that if $|x-x_0|<r$, and $x\neq x_0$, then $x\notin X$). Let $x$ such a point, and $f$ restricted to $(a,b)$ is not a polynomial, $x\in (a,b)$. But there is an open interval $I$ containing $x_0$ and $(a,b)$, and since $x_0\in X$, $f$ restricted to this interval is a polynomial, a contradiction.
$X$ is closed, as if $x_n\to x$, $x_n\in X$ for all $n$, take $(a,b)$ containing $x$. Then it contains a $x_n$ for some $n$. So $x_n\in (a,b)$ and $f$ restricted to this interval is a polynomial.