Prove $f(x)$ is quadratic if $f(2x)=4f(x)$ and $f(x)$ is increasing over positive $x$ 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?

Hint :$$f(x)=ax^2+bx+c \\f(2x)=4ax^2+2bx+c \\if \space f(2x)=4f(x)\to c=0 $$so we have 
$$f(2x)=4f(x) \to 2f'(2x)=4f'(x) \\2(2a(2x)+b)=4(2ax+b) \to 8ax+2b=8ax+4b \\\implies b=0 $$so $f(x)$ is in form of $ax^2$ and finally $a>0$ because 
$$f'>0  (\forall x>0) \implies f=2ax>0 \\a>0$$ now plug your physical information into the last equation
