6
$\begingroup$

Let $f:[0;1]\to \mathbb{R}$ be a continuous function satisfying

$$f\left(\frac{x}{2}\right) + f\left(\frac{x+1}{2}\right)=3f(x).$$

How to show that $f\equiv0$?

$\endgroup$

closed as off-topic by dustin, colormegone, Lord_Farin, Najib Idrissi, N. F. Taussig Feb 17 '15 at 20:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – dustin, colormegone, Lord_Farin, Najib Idrissi, N. F. Taussig
If this question can be reworded to fit the rules in the help center, please edit the question.

12
$\begingroup$

Hint:
Show that $f$ does not get maximum and minimum in the interval, which leads to that it's constant and conclude that it is the $0$ function

Update:
Basically the idea is to let $x_0$ be the point where $f$ gets its maximum.
Then we have: $$ 3f(x_0) = f(\frac{x_0}{2}) + f(\frac{x_0+1}{2}) \le 2f(x_0) \implies f(x_0) \le 0 \implies f(x) \le 0, x \in [0,1] $$ Similarly, if $f$ gets its minimum at $x_1$ we get $f(x) \ge 0$ for $x \in [0,1]$

$\endgroup$
  • $\begingroup$ Very nice answer. +1 $\endgroup$ – Timbuc Feb 8 '15 at 13:43
  • $\begingroup$ Thank you. I tried but I didn't manage to show that can't get a maximum in the interval... $\endgroup$ – wase5 Feb 8 '15 at 14:14
  • $\begingroup$ Thank you very much @benji ! $\endgroup$ – wase5 Feb 8 '15 at 14:32
  • $\begingroup$ @wase5 please see the update $\endgroup$ – benji Feb 8 '15 at 14:53
1
$\begingroup$

Since $f$ is continuous on $[0,1]$ we can see after a little effort that $$f(0)=f(1)=1/2f(1/2)$$ Thus, $|f(x)|<\infty$ for $x\in [0,1]$. Let $\max_{x\in [0,1]}|f(x)|=M$. Then, we have $$|f(x)|=\left|\frac{1}{3^n}\sum_{k=1}^{n}\left[f\left(\frac{x+2^k-2}{2^n}\right)+f\left(\frac{x+2^k-1}{2^n}\right)\right]\right|\le \frac{2Mn}{3^n},\ \forall n\in \mathbb{N}$$ Thus $f(x)=0,\ x\in[0,1]$

$\endgroup$
  • 3
    $\begingroup$ Who says we're differentiable? $\endgroup$ – Mathmo123 Feb 8 '15 at 13:38
  • $\begingroup$ What do the values of $f(0), f(1)$ have to do with the fact that $f$ is bounded? That's automatic from the compactness of $[0, 1]$. $\endgroup$ – anomaly Feb 9 '15 at 15:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.