Solutions to $f'(x)-f(\alpha x)=0$ Looking at this question:
Solution of recursion $T(n) = T(n-2) + 2T(\frac{n}{2})$
I was interested to know what are/is the solution/s to the (functional) differential equation:
$f'(x)-f(\alpha x)=0, \quad f(0)=0,f:\mathrm{R}^+\rightarrow\mathrm{R},0<\alpha<1$
I am not sure that the differential equation is related to the linked problem, but I got interested in the equation by itself.
TRIAL: I just noticed that defining $g(x) \equiv f(e^x),g:[-\infty,+\infty]\rightarrow \mathrm{R}$, this satistfies $g'(x)=e^xg(x+ln(\alpha))$ so that now we have a time delay $\tau=ln(\alpha)<0$, $g(-\infty)=0$ and maybe standard techinques can be used? But actually I do not know which are the standard techniques, if they exists, for this type of equations.
 A: If we assume a series solution
$$ f(x) = \sum_{k=0}^\infty c_k x^k$$
we get $$ k c_k = \alpha^{k-1} c_{k-1} \ \text{for}\ k \ge 1$$
so that
$$ c_k = \frac{c_0 \alpha^{k(k-1)/2}}{k!} $$
The series converges for all $z$ if $|\alpha| \le 1$.
A: The only differentiable function $f: \Bbb R^+ \to \Bbb R$ satisfying
$$
 f'(x) = f(\alpha x) \, , \, f(0) = 0
$$
for a given $\alpha \in (0, 1)$ is the “zero function” $f(x) = 0$. 
One can proceed as in $|f'(x)|\le|f(x)|$ and $f(0)=0$, prove that $\forall x\in[0,\frac 1 2]:f(x)=0$ and prove that
$$ \tag{*}
 f(x_0) = 0 \implies f(x) = 0 \text{ for } x_0 \le x \le x_0 + \frac 12 \, .
$$
It is clear that this – together with $f(0) = 0$ – implies that $f(x) = 0$ for all $x \ge 0$.
So it remains to prove $(*)$: Assume that $f(x_0) = 0$ and define
$$
 M = \max \{ |f(x) | : x_0 \le x \le x_0 + \frac 12 \} \, .
$$
Then 
$$
f(x) = \int_{x_0}^x f'(t) \, dt = \int_{x_0}^x f(\alpha t) \, dt \\
\implies |f(x)| \le \int_{x_0}^x |f(\alpha t)| \, dt \le x M \le \frac 12 M 
$$
for $x_0 \le x \le x_0 + \frac 12$. It follows that
$$
 0 \le M \le \frac 12 M \implies M = 0\, .
$$
