I dont know how to prove that if $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ is diferentiable and $f(x/2) = f(x)/2$ for all $x \in \mathbb{R}^n$ then $f$ is linear. Anyone could give me a hint?

  • $\begingroup$ list out all the definitions for diffrentiable, linear. and it should help $\endgroup$ – shai horowitz May 23 '16 at 1:48
  • $\begingroup$ Maybe take the gradient and you can show each component is a constant function. Just a thought, I'm not certain that solves it. $\endgroup$ – jdods May 23 '16 at 1:49
  • $\begingroup$ 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 $\endgroup$ – user247327 May 23 '16 at 1:54
  • $\begingroup$ i got that $f^{'}(x/2) = f^{'}(x)$ but what can i argue about $f$????? $\endgroup$ – Eduardo Silva May 23 '16 at 2:02
  • $\begingroup$ I initially posted this as an answer, but I'm less sure of it now so I'm posting it as a comment. If we take the gradient of both sides, we get that $\frac{1}{2} \nabla f(x/2) = \nabla f(x)/2$, so $\nabla f(x/2) = \nabla f(x)$. So now I think the problem basically reduces to the following problem: if g is a differentiable function from R to R, and g'(t/2) = g'(t) for all t, then show that g' is constant. Well, for any integer n, it's easy to show that g'(t*2^-n) = g'(t). So now I think we should try to prove g'(t) = g'(0). $\endgroup$ – Keshav Srinivasan May 23 '16 at 2:08

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 $g(x)=0$.

  • $\begingroup$ that is good but why if $g(x) = 2 g(2^{-1}x)$ i can assume the this is true for all $n \in \mathbb{N}$? $\endgroup$ – Eduardo Silva May 24 '16 at 2:17

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.

  • $\begingroup$ Could you spell out your reasoning at the end concerning the intermediate value property? $\endgroup$ – Keshav Srinivasan May 23 '16 at 2:14

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 $f(x)=Df_0(x)$.

  • $\begingroup$ What does $Df_0(x)$ mean? $\endgroup$ – Keshav Srinivasan May 23 '16 at 2:19
  • $\begingroup$ The differential of $f$ at $0$. $\endgroup$ – Tsemo Aristide May 23 '16 at 2:25
  • $\begingroup$ What do you mean by the differential? You clearly don't mean the gradient, so what do you mean? $\endgroup$ – Keshav Srinivasan May 23 '16 at 2:27
  • $\begingroup$ $Df_0:R^n\rightarrow 0$ is a linear function such that $f(x)=Df_0(x)+O(x)$ where $lim_{\|x\|\rightarrow 0}{{O(x)}\over{\|x\|}}=0$. $\endgroup$ – Tsemo Aristide May 23 '16 at 2:31
  • $\begingroup$ OK, so you mean the tangent hyperplane/local linear approximation. Thanks for clarifying. $\endgroup$ – Keshav Srinivasan May 23 '16 at 2:35

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 $\mathbf a, \mathbf b$


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