Find $g:[0,1]\rightarrow [0,1]$ such that $g(f(x))+g(x)=1$ Let $f:[0,1]\rightarrow [0,1]$ be a continuous function such that $f_\alpha= \frac{1-x}{1+\alpha x}$ , $\alpha \gt-1$.
Can you help me to find a function $g:[0,1]\rightarrow [0,1]$ such that $g(f(x))+g(x)=1$

Consider a transform $g$ of the form $g(x) = \frac{a+bx}{c+dx}$. Then we must find $a, b, c, d$ satisfying
\begin{align*}
\frac{a + b\frac{1-x}{1+\alpha x}}{c + d\frac{1-x}{1+\alpha x}} + \frac{a + bx}{c + dx} = 1
\end{align*}
Or in other words
\begin{align*}
\frac{(a+b) + (a\alpha-b)x}{(c+d) + (c\alpha-d)x} = \frac{(c-a) + (d-b)x}{c + dx}
\end{align*}
which gives us the system of equations
\begin{align*}
a + b &= \lambda(c-a) \\
a\alpha - b &= \lambda(d-b) \\
c+d &= \lambda c \\
c\alpha - d &= \lambda d \\
\end{align*}
We have four equations for five variables ($\alpha$ is assumed to be known). In fact, one of the equations turns out to be redundant, giving us two free parameters in the end. We may solve for the positive roots of $c, d$ with a shameless application of your favorite computer algebra program as
\begin{align*}
c = \frac{\sqrt{\alpha+1}a + (a+b)}{\sqrt{\alpha+1}}, \quad d = \frac{(\sqrt{\alpha+1}-1)(\sqrt{\alpha+1}a + (a+b))}{\sqrt{\alpha+1}}
\end{align*}
Now, let's find sufficient conditions on $a, b$ such that $g([0, 1]) \subseteq [0,1]$. If we allow $a > 0$ and $b = 0$, then we have $c \ge a$ and $d > 0$. Clearly $g(x)\ge 0$. Also, for $0 \le x \le 1$,
\begin{align*}
g(x) = \frac{a}{c+dx} + \frac{b}{c+dx}x \le \frac{a}{c} \le 1
\end{align*}
And we are done. 
