Knowing when the Logistic map sequence converges for all initial guesses Let $4 \geq \alpha \geq 0$ and $T_{\alpha}:[0,1] \rightarrow [0,1]$ be the function defined by $$T_{\alpha}(x) =\alpha x(1-x)$$ for all $x\in [0,1]$. For which values of $\alpha$ the sequences given by $y^{(x,\alpha)}_n = {T_{\alpha}^{n}(x)}$ converges for all $x \in (0,1)$?

Notation:
Inductively define $T^{n+1} = T^{n} \circ T $ and $T^{0} = I$, the identity function.

Observations:
It looks that this is the Logistic map. I think I need only find a proof that these sequences converges for $1-1/\alpha$ for $1 \leq  \alpha \leq 3$ and diverges for some $x \in (0,1)$ with $ \alpha >3.$
 A: $\textbf{Case 2:}$ $1 < \alpha \leq 2$
Let, first, $x\in (0,1/2]$, we have that
\begin{align*}
T_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) =  & \alpha x (1-x)-\left(1-\dfrac{1}{\alpha}\right) \\
=  & \alpha x \left( 1 - \dfrac{1}{\alpha} -x + \dfrac{1}{\alpha}\right)-\left(1-\dfrac{1}{\alpha} - x + x\right) \\
=  & \left(\alpha x- 1 \right) \left(\left(1-\dfrac{1}{\alpha}\right) - x\right).
\end{align*}
Now, since $| \alpha x- 1 | < 1, $ for $0<x\leq 1/2$, we get
\begin{equation*}
 \left| T_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) \right|<\left| x-\left(1-\dfrac{1}{\alpha}\right) \right|, \tag{1}
\end{equation*} whenever $x \not = 1-\dfrac{1}{\alpha} $. Noting that $T_{\alpha} \left( 1- \dfrac{1}{\alpha}\right) = \left( 1- \dfrac{1}{\alpha}\right)$ and (1) holds, $\inf_{k\in\mathbb{N}} \left| x^{k} -\left( 1- \dfrac{1}{\alpha}\right) \right|$ cannot be greater than zero. Once the sequence $\left\lbrace \left| x^{k} -\left( 1- \dfrac{1}{\alpha}\right) \right|\right\rbrace_{k\in\mathbb{N}} $ strictly decreasing, $x^{k}$ converges to $\left(1- \dfrac{1}{\alpha}\right).$
On the other hand, if $ 1/2 < x \leq 1$, $0<T_{\alpha}(x) \leq 1/2$ and the argument below applies.
$\textbf{Case 3:}$ $2<\alpha \leq 3$
This case is slightly harder and its proof works for the other, but since is more complicated I have decided to split both cases. We need to prove:

*

*The sequence defined by $T^{2 k}_{\alpha} (y)$ converges for all  initial guess $y \in \left(0,\dfrac{\alpha}{4}\right]$;


*Take into account that for $1 > x > \dfrac{\alpha}{4}$ its true $ 0 < T(x)\leq \dfrac{\alpha}{4}$;


*Conclude that $x_{2k}$ converges to $1-1/\alpha$ and $x_{2k+1}$ converges to $1-1/\alpha$ as well and get the desired result;
The same argument applied to the $\textbf{Case 1}$, implies that for all $x \in (0,1),$
\begin{align*}
T_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) = \left(\alpha x- 1 \right) \left(\left(1-\dfrac{1}{\alpha}\right) - x\right),
\end{align*} then
\begin{align*}
T^{2}_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) = & \left(\alpha T_{\alpha}(x) - 1 \right) \left(\left(1-\dfrac{1}{\alpha}\right) - T_{\alpha}(x) \right) \\
& \left(\alpha T_{\alpha}(x) - 1 \right) \left( 1 - \alpha x \right) \left(\left(1-\dfrac{1}{\alpha}\right) - x\right) \\
= & \left(\alpha^2 x (1-x) - 1 \right) \left( 1 - \alpha x \right) \left(\left(1-\dfrac{1}{\alpha}\right) - x\right). \\
\end{align*}
Since problem
\begin{equation*}
\begin{array}{c c}
\text{maximize}_{(x,\alpha) \in \mathbb{R}^{2}} &  \left| \left(\alpha^2 x (1-x) - 1 \right) \left( 1 - \alpha x \right)\right| \\
\text{subject to} &  0 \leq x \leq \dfrac{\alpha}{4}\\
& 1 \leq \alpha \leq 3
\end{array}
\end{equation*} have maximum less or igual than $1$ and additionally
$$ \left| \left(\alpha^2 x (1-x) - 1 \right) \left( 1 - \alpha x \right)\right| \not = 1$$ for all $ 0 < x \leq \dfrac{\alpha}{4}$ and $x \not = 1 -  \dfrac{1}{\alpha},$ we have
\begin{equation*}
\left| T^{2}_{\alpha}(x)-\left(1-\dfrac{1}{\alpha}\right) \right|  < \left| \left(\left(1-\dfrac{1}{\alpha}\right) - x\right)\right|, \tag{2}
\end{equation*} for all $ 0 < x \leq \dfrac{\alpha}{4}$ and $x \not = 1 -  \dfrac{1}{\alpha}. $
Since $0 <T_{\alpha} (y) \leq \dfrac{\alpha}{4}$ for all $ 0 < y < 1$, given  $0 < y < 1$ we have
$$
\left| T^{2(k+1)}_{\alpha}(y) - \left(1-\dfrac{1}{\alpha} \right) \right| = \left| T^{2}_{\alpha}( T_{\alpha}^{2k} (y)) - \left(1-\dfrac{1}{\alpha} \right) \right| < \left|  T_{\alpha}^{2k} (y) - \left(1-\dfrac{1}{\alpha} \right) \right|.
$$ Hence, the sequence $\left\lbrace \left| T^{2k}_{\alpha}(y) - \left( 1-\dfrac{1}{\alpha}\right) \right| \right\rbrace_{k\in\mathbb{N}}$ is decreasing and
\begin{equation}
\inf_{k\in\mathbb{N}} \left| T^{2k}_{\alpha}(y) - \left(1-\dfrac{1}{\alpha} \right) \right|
\end{equation} cannot be greater than zero for all $ 0 < y <1 $, which means that
\begin{equation*}
\lim_{k\in\mathbb{N}} T^{2k}_{\alpha}(y) = 1-\dfrac{1}{\alpha} \tag{3}
\end{equation*} for all $ 0 < y < 1$. With the aim to prove it, is needed to put all the sequence into a compact that does not contain $0$, $1-\dfrac{1}{\alpha}$ and $1$ and note that the inequality (2) would remain with a sufficiently small positive constant added in the LHS for all $x$ in that compact, which would would lead to a contradiction.
Now, for any $x \in \left(0,1\right)$, we have that $0 < T_{\alpha}(x) \leq \dfrac{\alpha}{4}$, hence
\begin{align*}
\left| T^{2k + 3}_{\alpha}(x) - \left(1-\dfrac{1}{\alpha} \right) \right| = & \left| T^{2(k+1)}_{\alpha}( T_{\alpha} (x) ) - \left(1-\dfrac{1}{\alpha} \right) \right| \\
 = & \left| T^{2}_{\alpha}( T_{\alpha}^{2k} (  T_{\alpha} (x) )) - \left(1-\dfrac{1}{\alpha} \right) \right| \\ 
 <  &\left|  T_{\alpha}^{2k} (  T_{\alpha} (x) ) - \left(1-\dfrac{1}{\alpha} \right) \right| \rightarrow 0.
\end{align*}
On the other hand, the same argument implies that
\begin{align*}
\left| T^{2k+4}_{\alpha}(x) - \left(1-\dfrac{1}{\alpha} \right) \right| = &\left| T^{2(k +1)}_{\alpha}( T^{2}_{\alpha} (x) ) - \left(1-\dfrac{1}{\alpha} \right) \right| \\
= & \left| T^{2}_{\alpha}( T_{\alpha}^{2(k+1)} ( x )) - \left(1-\dfrac{1}{\alpha} \right) \right| \\ 
<  &\left|  T_{\alpha}^{2(k+1)} ( x ) - \left(1-\dfrac{1}{\alpha} \right) \right| \\
=  &\left|  T_{\alpha}^{2k} ( T_{\alpha}^2 ( x ) ) - \left(1-\dfrac{1}{\alpha} \right) \right| \rightarrow 0.
\end{align*} Hence the sequence $\left\lbrace T^{k}(x)\right\rbrace_{k\in\mathbb{N}}$ converges to the same number for odd and even numbers. Finally, it must be that $\left\lbrace T^{k}(x)\right\rbrace_{k\in\mathbb{N}}$ converges.
A: Partial answer
The sequence $y_n$ is constructed as follows
\begin{cases}
y_{n+1} = \alpha y_n(1-y_n) \\
y_0=x \in(0,1)
\end{cases}
Case 1: $0 \le \alpha \le 1$
It's easy to notice that $y_n >0$ for all $n$ and the sequence is decreasing as
$$y_{n+1}-y_n = (\alpha-1) y_n - \alpha y_n^2 <0$$
So, the sequence converges for all values of $y_0 = x \in(0,1)$
$$$$
Case 2: $1 < \alpha < 2$
Let's denote $r$ the solution of $r = T_{\alpha}(r)$ then $r = 1-\frac{1}{\alpha}$. We notive that $$T'_{\alpha}(r) = \alpha(1-2r) = 2-\alpha<1$$
According to this result, as $x_{n+1} = T_{\alpha}(x_n)$ and the point $r = 1-\frac{1}{\alpha}$ is stable, the sequence converges to $r$.
Case 3: $2 \le \alpha \le 3$
Not yet
$$$$
Case 4: $3 < \alpha \le 4$
$\qquad$ First, we will find $(x,y)$ such that $x \ne y$, $x \in (0,1)$ and
\begin{cases}
y = \alpha x(1-x) \\\tag{1}
x = \alpha y(1-y)
\end{cases}
From the 2 equations $(1)$, we deduce that $x+y=1+\frac{1}{\alpha}$. And then $x$ is the solution of
$$1+\frac{1}{\alpha}-r=\alpha r(1-r) \tag{2}$$
The equation $(2)$ has 2 distinct real roots in $(0,1)$
$$r_{1,2} = \frac{(\alpha+1) \color{red}{\pm} \sqrt{(\alpha-1)^2-4}}{2\alpha}$$
$\qquad$ Second, take $y_0 = r1_1$, we will have $y_1 = r_2$ and in general
\begin{cases}
y_{2n} = r_1 \\
y_{2n+1} = r_2
\end{cases}
So, with $3 < \alpha \le 4$, the sequence $y_n$ doesn't converge for all $y_0 \in (0,1)$
$$$$
A: I have conjectured that the problem
\begin{equation*}
\begin{array}{c c}
\text{maximize}_{(x,\alpha) \in \mathbb{R}^{2}} &  \left| \left(\alpha^2 x (1-x) - 1 \right) \left( 1 - \alpha x \right)\right| \\
\text{subject to} &  \epsilon \leq x \leq \dfrac{\alpha}{4}\\
& 1 \leq \alpha \leq 3 \\
&  \epsilon^2 \leq \left(x-\left(1-\dfrac{1}{\alpha}\right) \right)^{2} \\
\end{array}
\end{equation*} has minimum strictly less than 1 for all $\epsilon>0$. If someone prove it, the prove above will be complete. =)
