Let $f:\mathbb{R}^m \to \mathbb{R}$ be differentiable s.t. $f(x/2)=f(x)/2, \forall x \in \mathbb{R}^m$. Show that $f$ is linear. Basically, I am not really sure how to start. 
I thought about going through induction to show for $\mathbb{N}$ and $\mathbb{Q}$, then use the completeness of $\mathbb{R}$, but I think it is a long way there.
I am quite sure that there is a shortcut. Could you help me?
 A: Induction tells us that $f(2^k x) = 2^k f(x)$ for all $k \in \Bbb N$. Since $f$ is differentiable at the origin, for any $\epsilon > 0$ we have a $\delta > 0$ such that $$|x|<\delta \implies\frac{|f'(0)\cdot x - f(x)|}{|x|} \le \epsilon.$$ Replace $x$ with $2^k x$ and use our first fact to get 
$$|x| < 2^k \delta \implies \frac{|f'(0)\cdot x - f(x)|}{|x|} \le \epsilon.$$ 
Thus given $x$ and $\epsilon$, we can always choose $k$ large enough that the first inequality is true, and thus the second is true for all $x$ and $\epsilon$, so we have $$f(x) = f'(0) \cdot x,$$ a linear function.
A: Take partial derivative of the equation $f(x/2) = f(x)/2$ and cancel the $\frac{1}{2}$ that appears on both sides:
$$\partial_if(x/2) = \partial_if(x)$$
Now, if you also assume, that $f$ is continuously differentiable, by applying this to $x/4$, $x/8$,... you get that
$$\partial_i f(x) = \lim_{k\to\infty}\partial_if(x/2^k) = \partial_i f(0).$$
So every partial derivative is a constant. Hence $f$ is linear.
A: The assumptions imply that $f(0)=0$. Put $df(0)=:A$. The auxiliary function $g(x):=f(x)-Ax$ satisfies the same functional equation; and in addition $dg(0)=0$. Fix an arbitrary point $x\in{\mathbb R}^m$. Then
$$g(x)=2^n g\bigl(2^{-n}x\bigr)={g(2^{-n}x)-g(0)\over\bigl|2^{-n}x\bigr|}\>|x|\to0\qquad(n\to\infty)\ .$$
It follows that $g(x)=0$. As $x$ was arbitrary this proves $f=A$.
