Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

closed as not constructive by Andrés E. Caicedo, Michael Greinecker, William, Qiaochu Yuan Sep 9 '12 at 19:09

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

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. – Rudy the Reindeer Sep 5 '12 at 18:22
@Matt What is undeserved here? – Michael Greinecker Sep 5 '12 at 18:25
@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
@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
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.

share|cite|improve this answer

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

share|cite|improve this answer
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

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.

share|cite|improve this answer
Right. You could also use the Lefschetz fixed-point theorem... :) – t.b. Sep 5 '12 at 21:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.