2

I am going through differentiable manifolds and come across a problem:
How to construct a smooth mapping $f: \mathbb R \rightarrow \mathbb R$ such that

  1. $f^{-1}(0)=0$

  2. $f^{'}(0) \neq 0$

  3. $\forall \epsilon >0,f^{-1}(-\epsilon ,\epsilon )$ is not homeomorphic to $(-\epsilon ,\epsilon )$

Any help?

talbi
  • 896
  • 4
  • 18
Barricades Mysterieuses
  • 31
  • Any thoughts do you have? – VIVID Jan 14 '21 at 16:24
  • Is there something wrong with $f(x)=\dfrac{x}{1+x^2}$? – Jean-Claude Arbaut Jan 14 '21 at 16:32
  • @VIVID I tried several times but failed. It seems that ii) implies that f is monotonous in an interval about 0, which is against iii) – Barricades Mysterieuses Jan 14 '21 at 16:45
  • @BarricadesMysterieuses There is no problem with 3. if $f$ is monotonous around $0$: you have to consider the preimage of $(-\varepsilon,\varepsilon)$. There may be small values of $f(x)$ away from $x=0$. See my example above. However, this example does not work either because $f$ is bounded (so there is an $\varepsilon>0$ that gets all of $\Bbb R$). – Jean-Claude Arbaut Jan 14 '21 at 16:52

2 Answers2

0

SUGGESTION (not a full answer): Something that oscillates a lot would be a good choice. You might want to start looking at

$$ f(x) = x^2 \sin \frac1x $$ (where $f(0)$ is defined to be 0, but I didn't want to write out cases). Now $f'(0) = 0$, so it doesnt quite work. But if $$ g(x) = f(x) + x $$ you might have something to work with...

John Hughes
  • 93,729
  • What about $f^{-1}(0) = 0$? – Paul Frost Jan 14 '21 at 16:36
  • Oh, absolutely, Paul -- I didn't say this was an answer, but rather that it was a starting point for thinking about this kind of question. Jean-Claude's answer uses oscillation, but rather differently. But I think that $g$ actually does have that property. I'm not sure it has the "not homeomorphic" property, though. – John Hughes Jan 15 '21 at 02:10
0

$$f(x)=x\left(\sin^2(x)+\frac1{1+x^2}\right)$$

  • There is only one value of $x$ such that $f(x)=0$
  • $f'(0)=1\ne0$
  • $f$ oscillates between values arbitrarily close to $0$ and arbitrarily large values.

The last property guarantees that $f^{-1}((-\varepsilon,\varepsilon))$ is made of a union of infinitely many disjoint open intervals, hence it's not homeomorphic to $(-\varepsilon,\varepsilon)$, for any $\varepsilon>0$.

enter image description here

With the example in my comment above, $f(x)=\dfrac{x}{1+x^2}$, the constraints where only partially fulfilled: for a large enough $\varepsilon>0$, $f^{-1}((-\varepsilon,\varepsilon))=\Bbb R$, because that $f$ is bounded. Here it doesn't happen.

Jean-Claude Arbaut
  • 23,277