Let $f:\Bbb R\to \Bbb R$, $f$ in $C^{\infty}$. Suppose that for all $x \in \Bbb R$, there exists an integer $n$ (which depends on $x$) such that
$f^{n}(x) = 0$ ($f^{n}$ is composing $f$ $n$ times)
Prove that there exists a non-empty open set in $\Bbb R$ where $f$ is a polynomial.
My professor told me to use Baire theorem, but I can't see how!
Could you help me, please?
On
If you mean by $f^n$ the $n$th derivative.
For $n\in \Bbb N$ set $F_n=\{x\in \Bbb R \ \ ,\ f^{(n)}(x)=0\}$.
For each $n$, $F_n$ is a closed subset of $\Bbb R$.
We have $\cup_{n\in \Bbb N}F_n=\Bbb R$, since $\Bbb R$ is complete and the interior of $\cup_{n\in \Bbb N}F_n$ is not empty , (By Bair Theorem ) there exits $N\in \Bbb N$ such the interior of $F_N$ is not empty , That is $F_N$ contain an open interval $]a,b[$ ($a<b$).
For all $x\in ]a,b[$, $f^{(N)}(x)=0$ hence $f$ is polynomial in $]a,b[$.
This is a very difficult problem, assuming that $f^{(n)}(x)=0$, instead of $f^n(x)=0$.
For a proof see here.
However, if you further assume that $f$ is real analytical, then it is not hard to show that it is a polynomial, using a standard cardinals argument.