1
$\begingroup$

Let $f$ and $g: \mathbb{R} \to \mathbb{R}$ be given by $f(x)= \frac{1}{2}x$ and $g(x)= 3x$. I want to determine if they are topologically conjugate, i.e., that there exists a homeomorphism such that $\phi(f(x))=g(\phi(x))$. A necessary condition for conjugacy is that they have the same number of fixed points, which turns out to be just $x=0$ and implies that $\phi(0)=0$. But how can I find a sufficient condition? I've tries $\phi(x)=x^a$ and similar but I'm not sure if I actually need to find $\phi(x)$ to prove conjugacy.

$\endgroup$

1 Answer 1

1
$\begingroup$

The maps are not topologically conjugate. If they were, both fixed points would have the same type of stability, but one is stable and the other is unstable.

$\endgroup$
3
  • $\begingroup$ How would I prove that? If $\phi(f(x))=g(\phi(x))$, $\phi '(f(x))f'(x)=g'(\phi(x)) \phi '(x)$, and since the fixed point is at $x=\phi(0)=0$ hence $\phi '(0)f'(0)=g'(0) \phi '(0)$. Is there some way I can say for certain that $\phi '(0) \neq 0$? $\endgroup$
    – Freelunch
    Oct 4, 2016 at 17:12
  • 1
    $\begingroup$ You complicate things a lot. Do read carefully what I wrote and think, without complicating. Also, you seem to be taking the derivative of a homeomorphism... $\endgroup$
    – John B
    Oct 6, 2016 at 1:48
  • 1
    $\begingroup$ $|f'(0)|<1$ so it is stable and $|g'(0)|>1$ hence unstable, I was trying to figure out why there is a relation. $\endgroup$
    – Freelunch
    Oct 6, 2016 at 6:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .