Prove topological conjugacy of an affine and a quadratic map using a linear map

Question

Let $Q_c(x)=x^2+c$. Prove that if $c<\frac14$, there is a unique $\mu>1$ such that $Q_c$ is topologically conjugate to $F_\mu(x)=\mu(1-x)$ via a map of the form $h(x)=\alpha x+\beta$.

Interpretation: $h(x)$ is a linear map. $Q_c(x)$ and $F_\mu(x)$ are a quadratic and affine map respectively. The claims is that if $c<\frac14$, then $\exists$ $\mu>1$ and the quadratic and affine maps are conjugate to one another.

Definition. (Topological conjugacy). Let $Q_c:X\to X$ and $F_\mu: Y\to Y$, and let $x_1\ne x_2$. Then $Q_c$ and $F_\mu$ are topologically conjugate if $\exists$ a homeomorphism $h:X\to Y$, $\ni$ $$h\circ Q_c = F_\mu\circ h$$ or $$h(Q_c(x_1)) = F_{\mu}(h(x_2)).$$

Now, we form, with $x_1\ne x_2$ $$h(Q_c(x_1)) = \alpha (x_1^2+c)+\beta$$and $$F_{\mu}(h(x_2))= \mu(1-\alpha x_2+\beta) $$ hence, $$\alpha (x_1^2+c)+\beta=\mu(1-\alpha x_2-\beta)$$ which gives \begin{equation} c = -\frac{1}{\alpha}(\beta \mu + \beta - \mu + \alpha x_1^2 + \alpha \mu x_2) \ \ \ \text{where}\ \alpha\ne 0\ \text{and}\ \beta + \alpha x_2\ne1 \end{equation} or \begin{equation} \mu=\frac{\alpha c+\beta +\alpha x_1^2}{-x_2 \alpha + \beta + 1} \end{equation} Insert for $c<\frac14$, i.e. $c=\frac15$: \begin{equation} \mu=\frac{\alpha \frac15+\beta +\alpha x_1^2}{-x_2 \alpha + \beta + 1} \end{equation} then we insert for $\alpha=1$ and $\beta=0$ for a simple case map $h(x)=x$: \begin{equation} \mu=\frac{\frac15+x_1^2}{1-x_2} \end{equation} Here we see that $\mu$ will be non-negative only in the unit interval, hence when $0\leq x_2<1$. Furthermore, we see $\mu>1$, when $x_2<x_1$ within the unit interval, which is the case for the quadratic family on the unit interval. So the claim holds for any mapping $h$, only in the unit interval.

But is it valid as proof when it only holds for the unit interval and that $x_1>x_2$ strictly within the unit interval?

Thanks

UPDATE:

By Lutz Lehmanns point, we have the new solution:

Insert for $c<\frac14$, i.e. $c=\frac15$: \begin{equation} \mu=\frac{\alpha \frac15+\beta +\alpha p^2}{- \alpha p + \beta + 1} \end{equation} By the claim, we set: \begin{equation} \frac{\beta + \alpha (p^2 + \frac15)}{\beta -\alpha p + 1}>1 \end{equation}

which is satisfied only when by \begin{equation} \begin{split} &p\in\mathbb{R}\\ &\frac{1}{10}\bigg( \sqrt{5} \sqrt{\frac{\alpha + 20}{\alpha}} - 5\bigg)<p<\frac{\beta+1}{\alpha}\\ &\beta<-1\\ &\alpha\leq-20, \end{split} \end{equation} which gives with the given conditions for $h(x)=\alpha x+\beta$ that the periodic point is always located in the unit interval $[0,1]$, and the quadratic and affine maps are topologicaly conjugated.

If $c<\frac14$, i.e. $c=\frac13$, then we obtain that the periodic point is outside the unit interval, i.e. $p = -\frac16, \beta<\frac16 (-\alpha - 6)$, hence there exists no topological conjugacy between the quadratic and affine maps.

Answer 1

There are lots of quadratic recursion sequences $$ z_{n+1}=az_n^2+bz_n+c. $$ Inserting a linear transformation $z=\alpha x+\beta$ results in another quadratic recursion $$ x_{n+1}=a'x_n^2+b'x_n+c' $$ with the same qualitative properties. Now one can ask what are the most simple, most easy to compare examples in such a transformation class?

The linear transformation has 2 free parameters. That allows generically to pose 2 conditions on the transformed coefficients $a',b',c'$.

The condition combinations that have "won" as being popular are $a'=1$, $b'=0$ resulting in the Mandelbrot iteration, and $c'=0$, $a'+b'=0$ giving the Feigenbaum/logistic map.

These conditions are themselves quadratic equations in the transformation parameters. Thus it is unsurprising that they can have complex solutions. So it can be of interest to ask when, given real coefficients in the original recursion, the transformation parameters are also real.

The task asks when the the Mandelbrot map with real $c$ can be transformed into the logistic map using only real parameters.

So insert the transformation $$ h(x_{n+1})=h(x_n)^2+c\\ αx_{n+1}+β=α^2x_n^2+2βαx_n+β^2+c\\ a'=α\\ b'=2β\\ c'=(β^2-β+c)/α $$ So $α=-2β$ and $β$ is a root of $(2β-1)^2+4c-1=0$.