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
 A: 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
$c = 2$.
A: $f(x) = c^{\ln(x)/\ln(2) - 1}$
By graphing $f(x)$ for different values of $c$, you can observe that $f(x)$ is non-zero for all $c > 0$.
A: We have $f(x)=c^nf(\frac{x}{2^n})$.
If $|c|<1$, then by taking $n\to\infty$, we get that $f(x)=\lim_{n\to\infty}c^nf(\frac{x}{2^n})=0f(0)=0.$ So, if $|c|<1$ the only solution is $f=0$.
Ideas:(I don't know if this helps in solving the problem)
Note that since $f(x)=cf(\frac{x}{2})$, we have that $f$ is determined by the values of the function on $[1,2]$ and $[-2,-1]$. So, we may define $f$ to be a bump function on these intervals and extend it to the entire real line by using $f(x)=cf(\frac{x}{2})$ (does this work for $|c|>1$?)
A: $$f(x)=c \times f\left(\frac{x}{2}\right)$$
Since $f(x)$ is infinitely differentiable, let us consider the Maclaurin series expansion:
$$f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f^{(2)}(0) + \dots$$
Also, 
$$f^{(n)}(x) = \frac{c}{2^n}f^{(n)}\left(\frac{x}{2}\right)$$
$$\implies f(x) = f(0) + x\frac{c}{2}f'(0) + x^2\frac{c^2}{2^22!}f^{(2)}(0)+\dots$$
If the two series converge, we can equate the coefficients of $x^n$:
$$\implies \frac{c^n}{2^nn!} = \frac{1}{n!}$$
$$\implies c = 2$$
