functions satisfying $f(x)=2f(2x)$ 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???
 A: I presume your function $f:\mathbb{R}^+\mapsto \mathbb{R}^+$. 
Define $g:\mathbb{R}^+\mapsto \mathbb{R}^+$ by $g(x)=x{f(x)},\quad \forall x\in\mathbb{R}^+.$ Then, you have $g(x)=g(2x).$ Now using this post to show that $g$ is constant. Consequently, you will arrive at what your wanted.
NB: You can generalize your question as follows:
Given a>1. Then a continuous function $f$ (not necessary smooth) satisfying
$$
f(x)=af(ax)
$$
implies $f(x)=k/x$ for some constant $k$.
A: Hint: Set $$r(x)=\dfrac{1}{f(x)}$$ 
then 
$$r(2x)=\dfrac{1}{f(2x)}=\dfrac{2}{f(x)}=2r(x)$$
From that you will get $r(nx)=nr(x)$ and you will find $r(x)=cx$.
A: 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. :)
