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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... –  Michael Greinecker Sep 5 '12 at 18:17
    
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. –  Matt N. Sep 5 '12 at 18:22
    
@Matt What is undeserved here? –  Michael Greinecker Sep 5 '12 at 18:25
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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. –  rschwieb Sep 5 '12 at 18:48
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@MichaelHardy: not even continuous? Surely you can just extend the function to be constant in the imaginary part... –  Ben Millwood Sep 5 '12 at 21:07
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closed as not constructive by Andres Caicedo, Michael Greinecker, William, Qiaochu Yuan Sep 9 '12 at 19:09

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3 Answers

up vote 6 down vote accepted

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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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... :) –  t.b. Sep 5 '12 at 21:07
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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 $C^\infty$. –  Michael Hardy Sep 5 '12 at 18:42
    
@MichaelHardy, the wikipedia page mentioned in my answer has a proof. –  lhf Sep 5 '12 at 19:32
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