TITLE: functions satisfying $f(x)=2f(2x)$
QUESTION [6 upvotes]: How can I prove every smooth real function f satisfying $f(x)=2f(2x)$ is of the from $f(x)=k/x$ where $k$ is a constant? I have tried this for integers that can be divided by $2^n$, but then I cannot proceed anymore. Can anybody give me some help???

REPLY [2 votes]: We note that one solution is $g(x) = 1/x$. Suppose that another solution is $h(x)$.
Then
$$g(x)/h(x) = 2g(2x)/2h(2x) = g(2x)/h(2x)$$
But looking at it the other way around, we have
$$g(x)/h(x) = g(x/2)/h(x/2)$$
continuing this inductively we have
$$g(x)/h(x) = g(x/2)/h(x/2) = g(x/4)/h(x/4) = g(x/8)/h(x/8) = g(x/16)/h(x/16) = \dots$$
Thus we have for all $a$ and $b \in R$
$$g(a)/h(a) = \lim_{n\to\infty} g(a/2^n)/h(a/2^n) = \lim_{t\to 0} g(t)/h(t) = \lim_{n\to\infty} g(b/2^n)/h(b/2^n) = g(b)/h(b)$$
Thus $g(x)/h(x)$ evaluates to a constant for all $x$, which implies that $h(x)=kg(x)=k/x$ for some $k\in R$
I hope this helps. :)