TITLE: Continuous $f:[0,1]\to\mathbb{R}$ such that $f(0)=f(1)$ and $\forall\alpha\in(0,1)\exists c\in[0,1-\alpha]|f(c)=f(c+\alpha)$?
QUESTION [13 upvotes]: Let $f:[0,1]\to\mathbb{R}$ continuous such that $f(0)=f(1)$. Is it true that $\forall\alpha\in(0,1)\exists c\in[0,1-\alpha]|f(c)=f(c+\alpha)$?
At first I tried to find a counterexample but my intuition says it's true.
Then I've got the idea of applying Bolzano's Theorem to $g(x)=f(x)-f(x+\alpha)$ defined on $[0,1-\alpha]$  but I didn't get anything. What can I do?

REPLY [13 votes]: If you choose $$f(x):= \begin{cases} x &: x < \frac{1}{4} \\
                       \frac{1}{2}-x &:  \frac{1}{4}\leqslant x\leqslant \frac{3}{4} \\
                     x-1 &: \frac{3}{4}\leqslant x \leqslant 1\end{cases}$$ and $\alpha=\frac{3}{4}$ then $f(x)\geq 0$ and $f(x+\alpha)\leqslant 0$ for all $x\in [0,1-\alpha]$.

REPLY [6 votes]: Consider $f(x) = \sin(2\pi x)$ and $1/2 < \alpha < 1$. Then, $f(x)\geq 0$ and $f(x+\alpha)\leq 0$ for all $x\in [0, 1-\alpha]$.