How to show that $f\equiv 0$? [closed]

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 $f\equiv0$?

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

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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 $x \in [0,1]$

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

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 $f(x)=0,\ x\in[0,1]$

• Who says we're differentiable? – Mathmo123 Feb 8 '15 at 13:38
• 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 $[0, 1]$. – anomaly Feb 9 '15 at 15:04