Solving a functional equation $ g(x) = \frac{1}{1-x} g\left(\frac{ax}{1-x}\right) $ Consider a following functional equation
$$ g(x) = \frac{1}{1-x} g\left(\frac{ax}{1-x}\right) $$
for a given fixed real parameter $a$.
I don't know whether the solution for $g$ is unique. However, assuming it has a Taylor series around $x=0$, we get, expanding the both sides and comparing each power of $x$,
$$ \frac{1}{n!} g^{(n)}(0) = \sum_{k=0}^n \frac{a^k}{k!} \binom{n}{k} g^{(k)}(0)$$
Upto a constant factor $g(0)$, we see clearly that all of the derivatives of $g$ at zero are uniquely determined.
On the other hand, note that a simple guess $$g(x) = \frac{1-a}{1-a-x}g(0)$$ satisfies the functional equation for $g$. Together with the uniqueness of the Taylor coefficients, we then deduce it is the solution of the initial problem.
However, is there a generic approach how to arrive to this solution? Based on manipulation of the functional equation for example?
 A: I will assume that $g$ is continuous at $0$ and $a\in (0,1)$.
Let's prove by induction the following: for $n\in \Bbb N$,
\begin{align}
g(x) &= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}g\left(\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}\right)
\end{align}
For $n=0$ the statement is equivalement to the functional equation. Assume it is true for $n\in \Bbb N$,
\begin{align}
g(x) &= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}g\left(\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}\right)\\
&= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x} \frac{1}{1-\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}}g\left(\frac{a\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}}{1-\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}}\right)\\
&= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x} \frac{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}{1 - \left(\sum\limits_{k=0}^{n+1}a^k\right)x}g\left(\frac{a^{n+2}x}{1-\left(\sum\limits_{k=0}^{n+1}a^k\right)x}\right)\\
&= \frac{1}{1 - \left(\sum\limits_{k=0}^{n+1}a^k\right)x}g\left(\frac{a^{n+2}x}{1 - \left(\sum\limits_{k=0}^{n+1}a^k\right)x}\right)
\end{align}
So for fixed $x$, $$g(x) = \lim\limits_{n\to \infty} \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}g\left(\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}\right) = \frac{1}{1 - \left(\sum\limits_{k=0}^{\infty}a^k\right)x}g\left(0\right) = \frac{1}{1-\frac{x}{1-a}} g(0)$$
