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I'm having trouble proving that the tent map

$$T_2 = 2x \ \text{for} \ 0\leq x\leq 1/2 \ \ \text{and} \ \ T_2(x) = 2(1-x) \ \text{for} \ 1/2\leq x\leq 1$$

is topologically conjugate to the quadratic map

$$F_4(x) = 4x(1-x).$$

I know that to prove topological conjugacy, one must have a homeomorphism $h$ such that

$$h\circ T_2 = F_4 \circ h.$$

Could anyone give me a hint on how to find $h$ and go on from there?

Thanks

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  • $\begingroup$ In this problem you talk about both topological conjugacy and topological equivalence. Which one do you mean? $\endgroup$ – Stella Biderman Mar 7 '17 at 5:22
  • $\begingroup$ Sorry, I meant conjugate. $\endgroup$ – sadlyfe Mar 7 '17 at 5:23
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The conjugacy equation develops as $$ h(2x) = 4h(x)(1-h(x)) $$ for $x \in [0,1/2)$. Note that the above is satisfied by functions of the form $h(x) = \sin^2(c x)$ for any $c > 0$ by the formula $\sin(2 \theta) = 2 \sin (\theta) \cos(\theta)$.

For $h$ to be a homeomorphism of the interval into itself, it should be either increasing or decreasing everywhere. If we look for homeomorphisms of the form $h(x) = \sin^2(c x)$ as above, then, it makes sense to set $c = \pi/2$ so that $h(0) = 0, h(1) = 1$, and $h$ is increasing on all of $[0,1]$.

To check that $h(x) = \sin^2(\pi x / 2)$ really is a conjugacy, it remains to verify the conjugacy equation for $x \in [1/2,1]$: $$ h(2(1- x)) = 4 h(x) (1 - h(x)) \, . $$ For this, we compute $$h(2(1-x)) = \sin^2(\pi(1 - x)) = \sin^2(\pi x) = 4 \sin^2(\pi x/2) \cos^2(\pi x/2) = 4 h(x) (1 - h(x))$$ as desired.

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  • $\begingroup$ Thanks so much. Is there any obvious reason for choosing the $h$ that you did? $\endgroup$ – sadlyfe Mar 7 '17 at 6:35
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    $\begingroup$ Sadly, nothing more than experience... $\endgroup$ – A Blumenthal Mar 7 '17 at 6:44

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