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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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    Using the shift-key, you can write capital letters too...2012-09-05
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    How about you texify your post? I'm feeling too lazy to do that for you. If you do it I'll plus one your post to undo the undeserved downvote you got.2012-09-05
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    @Matt What is undeserved here?2012-09-05
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    What's "not constructive" about the question?2012-09-05
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    @MichaelGreinecker Maybe half an hour grace period should be allowed for the author to improve the question, rather than near-instant demolition.2012-09-05
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    @rschwieb I think the 30 day grace period should be enough to learn how to post a question and what the community norms are. This goes on for quite some time now with no improvements in sight.2012-09-05
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    @MichaelGreinecker Seems like a nonnative speaker. 30 days is a pretty short time to master a language :S Let me assure you I acknowledge your frustration, though.2012-09-05
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    The existence of non-zero functions satisfying $(1)$ is a sharp contrast between real and complex functions. "lhf"'s answer gives an example that is zero on $(-\infty,0]$ and positive on $(0,\infty)$. There is no way to extend it to a complex-valued function that is differentiable, or even continuous, in an open neighborhood of $0$. Functions like this are building blocks of "test functions" that have bounded support and yet are in $C^\infty$. Those are used in the definition of "generalized functions".2012-09-05
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    @rschwieb The poster has never stated anything about his or her background, so that is just speculation. But the questions do look like copy and paste, which either means that the poster can translate questions as long as they are in the imperative, studies using an English text, or takes a course taught in English.2012-09-05
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    @MichaelGreinecker "look like" I notice you answered speculation with speculation of your own :) I'm not going to continue this: I can see you won't consider any excuse for this particular poster. I'll just trust that you won't let this zeal spill onto sincere posters with an English language handicap.2012-09-05
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    It is of course [fine to ask questions in languages other than English](http://meta.math.stackexchange.com/questions/1617/what-is-the-site-etiquette-about-i-asking-and-ii-answering-questions-in-a-la).2012-09-05
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    @MichaelHardy: not even continuous? Surely you can just extend the function to be constant in the imaginary 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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    If I were feeling energetic and didn't have other things to do, I'd think about posting a proof that this function is in $C^\infty$.2012-09-05
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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