TITLE: Prove $f(x)$ is quadratic if $f(2x)=4f(x)$ and $f(x)$ is increasing over positive $x$
QUESTION [5 upvotes]: The problem arose in the context of kinetic energy, where it can be proven from symmetry principles that $E(2v)=4E(v)$ without assuming $E=mv^2/2$ (see e.g. physics stackexchange).
While one may do further physics from this point to prove the desired result (that $E$ is quadratic in $v$) -- consider a system with other prime numbers of balls, then do algebra to prove the result for rational scaling in $v$, then use the fact that there are rational numbers between any two real numbers and assume the function is increasing to prove it for all real scaling -- it seems intuitively obvious from this point immediately, that if $E$ is increasing in $v$, $E=kmv^2$.
How would one prove this functional equation?

REPLY [4 votes]: This is false.
Take an arbitrary increasing function $f$ defined on $[1,2]$ taking values in $[1,4]$.
Then define $f(x) = 4^{-n}f(2^nx)$ whenever $x\in[2^{-n},2^{1-n}]$. Then $f$ is increasing and $f(2x) = 4f(x)$, but $f$ is not in general quadratic.
Moreover, if we define $f(0)=0$, then $f$ is continuous at $0$, since $f(x) \leq 4^{-n}f(2) $ whenever $x \in [2^{-n},2^{1-n}]$.