Solve $f(x)=c \times f(\frac{x}{2})$ for $c$ Given:


*

*Function $f(x)$ is infinitely differentiable 

*equation (1)
$f(x)=c \times f(\frac{x}{2})$
We have to find all $c$, for which the (1) has non-zero solutions 
Any hints on theorems to apply here, I reckon it's somehow related to ODEs

If
$ f(x)=c*f(x/2)
$
then
$\begin{array}\\
 f(x)
&=cf(x/2)\\
&=c^2f(x/4)\\
&=c^3f(x/8)\\
&...\\
&=c^nf(x/2^n)\\
\end{array}
$
If $|c| < 1$ then
$f(x) \to 0$
so
$f(x) = 0$
for all $x$.
If $f(0) \ne 0$,
$\dfrac{f(x)}{c^n}
\to f(0)
$.
If $|c| > 1$,
$\dfrac{f(x)}{c^n}
\to 0
$
which contradicts
$f(0) \ne 0$.
If $f(0) = 0$,
then,
for small $x$,
$f(x) = xf'(0)+O(x*2)
$
so
$f(x/2^n)
=xf'(0)/2^n+O(x^2/4^n)
$
so
$f(x)
=c^n(xf'(0)/2^n+O(x^2/4^n))
=xf'(0)(c/2)^n+O(x^2(c/4)^n))
$.
This only works if
$c=2$;
it goes to zero if
$|c| < 2$
and to $\infty$ is
$|c| > 2$.
Therefore we must have
$c = 2$.
