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?
 A: If we assume $f \in C^2$ (or at least that $f$ is twice differentiable and $f''$ is continuous at $0$), we can prove uniqueness. Taking the derivative twice, we obtain:
$$f(2x) = 4f(x) \implies f''(2x) = f''(x)$$
Now we have, using continuity of $f''$ at $0$:
$$f''(x) = f''\left(\frac{x}{2}\right) = f''\left(\frac{x}{4}\right) = \ldots = f''\left(\frac{x}{2^n}\right) \xrightarrow{n\to\infty} f''(0)$$
Hence, $f''$ is constant so $f(x) = ax^2 + bx + c$.
Now using the answer of @Khosrotash, you can deduce that:
$$f(x) = ax^2, \quad a > 0$$
A: 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 $x \in [2^{-n},2^{1-n}]$.
A: I guess you will need to assume continuity of $E/v^2$ at $v=0$ in order to prove uniqueness.
Consider the function
$$h(v)=\frac{E(v)}{v^2}.$$
We have
$$h(2v)=h(v)$$
and $h$ continuous at $0$.
The only function satisfying $h(2v)=h(v)$ and $h(0)=c$ and is continuous at $v=0$ is $h(v)=c$.
Proof:
Suppose $h(b)\ne c$ for some non-zero $b$. Then $h(b/2^N)=h(b)$. For all $\epsilon$, $\delta$, there exists $N$ such that $|b/2^N|<\delta$ but $|h(b/2^N)-c|>\epsilon$, contradicting the continuity of $h(v)$ at $0$.
A: 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
