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Which of the following statements is true?

1.If $f\in C^\infty$ and $f^{(k)}(0) = 0$ for all integers $k\geq 0$, then $f=0$.

2.$f : [0,\infty] \rightarrow [0,\infty ]$ is continuous and bounded, then $f$ has a fixed point.

for 1 completely out of idea. i tried to find some counter example but could not find it. for 2.this result is true for closed bounded interval.but what is the case here?

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    @MichaelHardy: not even continuous? Surely you can just e$x$tend the function to be constant in the imaginar$y$ part...2012-09-05

3 Answers 3

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For the second question: there is a fixed point. Indeed, define

$g:[0,+\infty[\to \mathbb{R};\quad g(x)=f(x)-x$.

$g$ is continuous. $g(0)\geq 0$, and since $f$ is bounded, $g(N)<0$ for $N$ large enough. Now use the intermediate value theorem.

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For 1, a counter-example is the canonical non-analytic smooth function: $ f(x)=\begin{cases}\exp(-1/x)&\text{if }x>0\\ 0&\text{if }x\le0\end{cases} $

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    @MichaelHardy, the wikipedia page mentioned in my answer has a proof.2012-09-05
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The second statement is also a consequence of Brouwer's fixed point theorem.

Let $M \in \mathbb{R}_{+}$ be such that $f \leq M$ by boundedness so that $\operatorname{image}(f) \in [0, M]$ and restrict $f$ to this interval. Then

$f_{M}: [0, M] \rightarrow [0, M]$

has a fixed point by Brouwer's theorem.

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    Right. You could also use the [Lefschetz fixed-point theorem](http://en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem)... :)2012-09-05