Skewed Tent Map I have the following map: $f(x)=\begin {cases} \mu x &\text x\in[0,a] \\\mu a(1-x)/(1-a) &\text x\in [a,1]\end {cases}$
where $a\in(0,1)$. First of all I would like to check when the map $F$ is a map of $[0,1]$ into itself, i.e an endomorphism. I need to check when $f(cx+y)=cf(x)+f(y)$ holds.
Doing the calculation I see that in the first case it is always a map into itself, i.e for $x\in [0,1]$.
In the second case I end up with $1-c-y+cy+cxy=0$. What does this tell me?
Furthermore I would like to show that the fixed point is at $\bar x=\frac{\mu}{\mu+\mu_s(a)}$ where $\mu_s(\frac{1}{2})=1$. I do not see how to get this point, is this the non-trivial fixed point in all cases?
 A: The map is an endomorphism if it maps into the interval $[0,1]$. Since it's minimum value is 0 (when $x=0$ and $x=1$) and it's maximum value is $\mu a$ (when $x=a$), the map is an endomorphism if and only if $0 \le \mu a \le 1$.
For linear maps, the equation $f(c x + y) = c f(x) + f(y)$ is always true. Consequently if $0 \le x \le a$, and $0 \le y \le a$ and $0 \le c x + y \le a$, then only the left part of map (which is linear) is involved and this equation is true (both sides are equal to $\mu (c x + y)$). The equation is also true for some other specific values. For example, if $a \le x \le 1$ and $a \le y \le 1$ and $0 \le c x + y \le a$ (here we need $c<0$), then
\begin{align}
f(cx+y) &= \mu c x + \mu y \;, \\
cf(x)+f(y) &= c \mu a \frac{1-x}{1-a} + \mu a \frac{1-y}{1-a} \;.
\end{align}
By equating these, cancelling the $\mu$'s, multiplying through by $(1-a)$ and expanding, we obtain $cx + y = ac + a$,
and thus we need $c = -\frac{y-a}{x-a}$.
To find fixed points we solve $f(x) = x$. The left part of the map has a fixed point only if $\mu = 1$, in which case every value of $x$ between $0$ and $a$ is a fixed point. For the right part of the map we solve $x = \mu a \frac{1-x}{1-a}$. This gives
\begin{equation}
x = \frac{\mu a}{1 - a + \mu a} \;.
\end{equation}
