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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.

  • 2
    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](http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto)?2012-11-08
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    See also this [MathJax LaTeX tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference).2012-11-08
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    @amWhy I shall definitely try it next time .Sorry for the inconvenience2012-11-08
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    No worries. Note that when you ask, and/or answer, a question, you can view what you are formatting, as it would appear on the site...before actually posting.2012-11-08
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    possible duplicate of [A question on Taylor Series and polynomial](http://math.stackexchange.com/questions/93452/a-question-on-taylor-series-and-polynomial)2012-11-08
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    (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.)2012-11-08
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    The solution is given in this MO thread: http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial2012-11-11
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    @Timothy What's wrong with the solution provided at math.overflow?2012-11-15
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    Below, you asked about some step in the proofs of the link. Maybe you can write them in the OP.2012-11-19

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Look at Andrea Ferretti's solution or Andrey Gogolev's one at MathOverflow.

In the latest, we argue by contradiction. The set $X$ is the points $x$ such that for all $a, 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|, 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.

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    I posted it as CW as it's not the idea of mine and I just provide a link. And the question won't remain unanswered. But I don't deserve the bounty, of course, so if nobody gives an answer, I don't know what will happen.2012-11-17
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    @DavideI put it on bounty as it does not explain certain steps like X is closed without isolated points2012-11-18
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    Thank you very much Davide2012-11-19