Proving linearity from differentiability I dont know how to prove that if $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ is differentiable and $f(x/2) = f(x)/2$ for each $x \in \mathbb{R}^n$, then $f$ is linear. Anyone could give me a hint?
 A: It is easy to see that $f(0)=0$. Put $\nabla f(0)=:a$. Then  the function $g(x):=f(x)-a\cdot x$ satisfies
$$\lim_{x\to0}{g(x)\over|x|}=\lim_{x\to0}{f(x)-a\cdot x\over|x|}=0\ ,$$
furthermore $g$ inherits the property $g(x)=2 g(x/2)$. 
Now fix an $x\in{\mathbb R}^m$. Using induction one proves that for all $n\geq0$ one has
$$g(x)=2^n g\bigl(2^{-n}x\bigr)=|x|\>{g\bigl(2^{-n}x\bigr)\over\bigl|2^{-n}x\bigr|}\ .$$
Since here the right hand side converges to $0$ when $n\to\infty$, it follows that in fact $g(x)=0$.
A: Given the fact that "f is differentiable" is important enough to state, my first thought would be to differentiate!  $\frac{1}{2}f'(x/2)= f'(x)/2$ so that $f'(x/2)= f'(x)$.  .From that, we can argue that if $f'(x_1)$ and $f'(x_2)$ are different, there exist two different sequences, $x_1, x_1/2, x_1/4, ...$, all having the same vaue, and $x_2, x_2/2, x_2/4, ...$, all having the same f value,  converging to 0.  Now, if a derivative exists, it does not have to be continuous, but it does satisfy the "intermediate value property" so that those two sequence, converging to 0, would give different values [b]at[/b] 0.  Therefore the derivative must be constant.
A: You have $f(0/2)=f(0)=f(0)/2$ implies that $f(0)=0$, 
$f(x)=Df_0(x)+O(x)$ where $lim_{\|x\|\rightarrow 0}{{O(h)}\over \|h\|}=0$.
$f(x/2^n)=Df_0(x/2^n)+O(x/2^n)=f(x)/2^n$. This implies that ${{O(x)}\over {2^n}}=O(x/2^n)$. You deduce $O(x)=2^nO(x/2^n)$.
$2^nO(x/2^n)= \|x\|{{O(x/2^n)}\over {\|x\|/2^n}}$.
$lim_{n\rightarrow +\infty}2^nO(x/2^n)= \|x\|{{O(x/2^n)}\over {\|x\|/2^n}}$ .
$lim_{n\rightarrow+\infty}{{O(x/2^n)}\over{\|x\|/2^n}}=lim_{h\rightarrow 0}O(h)/\|h\|=0$. This implies that $lim_{n\rightarrow +\infty}2^nO(x/2^n)=0$ and henceforth $O(x)=0$. We deduce that $f(x)=Df_0(x)$.
A: A function is linear if for all $a,b: f(\mathbf a+\mathbf b) = f(\mathbf a) + f(\mathbf b).$
You know that: $f(x/2) = f(x)/2$  and, $(x/2 + x/2) = f(x)/2 + f(x)/2 = f(x)$
What you will need to show:
$f(0) = 0$
Suppose that $f(\mathbf x) = c.$
Show that, 
$f(a\mathbf x) = ac$ 
You may need to show this first for inegers, then for rational, then for real scalars.
Then you will need to show that it holds for all $\mathbf a, \mathbf b$
