Let $f$ differentiable at 0. If $f(\frac{x}{2}) = \frac{f(x)}{2}$, show that there exists $k \in \mathbb{R}$, such that $f(x) = kx$. Problem:
Let $f: \mathbb{R} \to \mathbb{R}$ be a function differentiable at 0. If $f(\frac{x}{2}) = \frac{f(x)}{2}$, show that there exists $k \in \mathbb{R}$, such that $f(x) = kx$.
This is part of my analysis I course, so I'm only allowed to use differentiation. I've been able to derive this facts but nothing more:

*

*$f(0) = 0$

*$f'(\frac{x}{2}) = f'(x)$
Other than this I have not been able to come up with anything useful. I would appreciate any help.

 A: Proof sketch:
Assume for contradiction that there are two non-zero $x_1, x_2$ such that $\frac{f(x_1)}{x_1}\neq \frac{f(x_2)}{x_2}$. Show that $f$ turns out to not be differentiable at $0$.
A: Step by step.
All possible functions that obey $f(\frac{x}{2})=\frac{f(x)}{2}$ are completely defined by two possibly independent functions: one, $u_p$, in the interval $[1,2)$ another, $u_n$, in the interval $(-2,-1]$.
The rest we get by copying the half of the function, that is $\frac{u_p}{2},\frac{u_p}{4}...$ to the interval $[\frac{1}{2},1),[\frac{1}{4},\frac{1}{2}),...$ and doubles that is $2u_p,4u_p$ into $[2,4),[4,8),...$. The similar in negative direction for $u_n$.
The functions $u_p$ and $u_n$ must be bounded otherwise we would not even have a continuous function at $0$, let alone differentiable one because we would get an infinite value at any distance close to $0$.
Since they are bounded it means:
$$f(0)=\lim_{n \to \infty}\frac{f(1)}{2^n}=0$$
Suppose that there is a discontinuity at any point for $u_p$ or $u_n$. This would mean that by definition
$$f'(0)=\frac{f(0+h)-f(0)}{h}=\frac{f(h)}{h}$$ would potentially have two different values as we would have a cloud of values around $0$. At one discontinuity with value $d_1$ we would approach $0$ at one rate while we would have another rate at another discontinuity.
However, since $f(x)=c, c \neq 0$ is not the solution, as it has a discontinuity at $1$, we do have at least two different distant values. What happens with these two different values then?
So let us have two values between $1$ and $2$ $f(1+a)=x$, $f(1+b)=y$
The chain will be
$$f(1-\frac{a}{2})=\frac{x}{2},f(1-\frac{b}{2})=\frac{y}{2}$$
and so on to get in general:
$$f(\frac{1}{2^n}-\frac{a}{2^{n+1}})=\frac{x}{2^{n+1}},f(\frac{1}{2^n}-\frac{b}{2^{n+1}})=\frac{y}{2^{n+1}}$$
Let us find the first derivative now as we are closing towards $0$:
$$f(0)'=\lim_{n \to \infty}\frac{\frac{x}{2^{n+1}}-\frac{y}{2^{n+1}}}{\frac{1}{2^n}-\frac{a}{2^{n+1}}-(\frac{1}{2^n}-\frac{b}{2^{n+1}})}=\frac{x-y}{b-a}$$
Apart from the interval $[1,2)$ this, obviously, must be valid for any two different points, so in order all to have the same differential at $0$ it must be:
$$f(a)-f(b) \sim a-b$$
or in your vernacular
$$f(a)-f(b) = k (a-b)$$
Since $f(0)=0$ it follows that it must be $f(a) = ka$
A: The idea is that by iterating the condition $f(x/2) = {1 \over 2} f(x)$, you will get that for any $x_0 \neq 0$, there is a line $y = kx$ that contains the graph of $f(x)$ above $x_0, {x_0\over 2}, {x_0 \over 4}, {x_0 \over 8},...$
Since $f(x)$ is differentiable at $x = 0$, $k = f'(0)$ regardless of which $x_0$ you use. Hence $f(x)$ is just the function $kx$.
A: Expand in a Taylor series and note that differentiating more times you see that
$$ \frac{1}{2^{n-1}} f^{(n)}\left(\frac{x}{2} \right) = f^{(n)}(x). $$
Thus setting $x = 0$ we see that $f^{(n)}(0) = 0$ for all $n \ge 2$. Hence $f(x) = f(0) + f'(0)x = k x$.
