# 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?

• No quadratic function is increasing. Commented Sep 16, 2017 at 7:21
• Sorry -- increasing over positive $v$. Commented Sep 16, 2017 at 7:22
• Increasing $\implies$ discontinuous at most at countably many points Commented Sep 16, 2017 at 7:23

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:

$$f(x) = ax^2, \quad a > 0$$

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 $x \in [2^{-n},2^{1-n}]$.

• @EwanDelanoy This is not true. According this answer, we can make $f$ increasing, continuous on all of $\mathbb{R}$, but not quadratic. Commented Sep 16, 2017 at 8:01
• @pisco125 You're right, what we really need is $\frac{f(x)}{x^2}$ to have a limit at $0$. I'll correct my comment on Khosrotash's answer Commented Sep 16, 2017 at 8:03
• @EwanDelanoy See the answer by mechandroid. I think that might be the most natural "physical" assumption. Commented Sep 16, 2017 at 8:04

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 $0$.

• "Increasing for positive $x$" is not sufficient: Try $$f(x)=x^2\,\exp\left(\frac{\ln2}{\pi}\sin\frac{2\pi\ln x}{\ln2}\right).$$
– user436658
Commented Sep 16, 2017 at 7:46
• @ProfessorVector What is your $f(0)$? And is it continuous at $0$? Commented Sep 16, 2017 at 7:52
• Of course, I'd assume $f(0)=0.$ Then, it is continuous at $0$, because the second factor is periodic, continuous and thus bounded for $x>0$. It's easy to check that $f'(x)\ge0$ for $x>0$.
– user436658
Commented Sep 16, 2017 at 7:55
• @velutluna Ignore "Professor Vector", for some time now he's been spamming randomly this website with irrevelant and/or stupid comments. What you say is correct, but I think you should add some more details : say that $h(v)=\frac{E(v)}{v^2}$, and that since $0$ is the limit of $\frac{v}{2^n}$ when $n\to\infty$, if $h$ is continuous at $0$ we have $h(0)=h(v)$. Commented Sep 16, 2017 at 7:56
• @EwanDelanoy Thanks! Added. Is that what you want? Commented Sep 16, 2017 at 8:05

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