Logistic map preserves the measure $\varphi m$

Question

This is from these lecture notes, exercise 5 on page 7.

The logistic map $$f(x) = 4x(1-x)$$ and the tent map $$g(x) = \begin{cases} 2x & x\leq \frac{1}{2}\\ 2-2x &x> \frac{1}{2} \end{cases}$$ are conjugate via $$h(x) = \frac{(1-\cos(\pi x))}{2}.$$ Prove that f preserves $\varphi m$, where $m$ is the Lebesgue measure and $$\varphi(x) = \frac{1}{\pi}\frac{1}{\sqrt{x(1-x)}}.$$

$f$ and $g$ being conjugate via $h$ means that $$f \circ h = h \circ g.$$

To show that $f$ preserves $\varphi m$, I realise I should try to show that

$$\sum_{x\in f^{-1}(y)} \frac{\varphi(x)}{|f'(x)|}=\varphi(y).$$

Substituting in, I get that what I need to show is

$$\sum_{x\in f^{-1}(y)} \frac{1}{\pi\sqrt{x(1-x)}|4-8x|}=\frac{1}{\pi\sqrt{y(1-y)}}.$$

I can't see where to go from here, or how to use conjugacy. Any hints?

Answer 1

Hint $x_{1,2}=.5*(1\pm\sqrt {1-y}) $

You need to replace x in your substitution.

Then make the sum over the two roots to get rid of the one half factor.

Answer 2

The logistic map tells us that $$y = 4x(1-x)$$ from which we get $$x_{1,2} = .5*(1\pm\sqrt{1-y}).$$

Therefore,

$$|4-8x| = 4|1-(1\pm \sqrt{1-y})|$$ $$=4|\pm \sqrt{1-y}|$$ $$=4\sqrt{1-y}.$$

Also,

$$\sqrt{x(1-x)}=\frac{\sqrt{y}}{2}.$$

So, substituting above and summing over both roots, we get the desired equality.

Thanks to Julien for the hint! This solution works, although we haven't used the given conjugacy as the lecture notes suggested.