Proving that the tent map is topologically conjugate to the quadratic map

Question

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

Answer 1

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.