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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?

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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 $x\in [0,1-\alpha]$.

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    $\begingroup$ Exceedingly clever construction, I love it. $\endgroup$ – infinitylord Dec 17 '17 at 1:13
  • $\begingroup$ Can such examples also be given for "small" $\alpha$? $\endgroup$ – Jeppe Stig Nielsen Dec 17 '17 at 1:14
  • $\begingroup$ I suspect that the question statement is true if $\alpha\leq 1/2$. $\endgroup$ – Michael Lee Dec 17 '17 at 1:17
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    $\begingroup$ The two questions and their answers linked by dxiv in the comments give a complete answer to this. The statement is true iff $\alpha =1/n$, otherwise there are examples like this. $\endgroup$ – Torsten Schoeneberg Dec 17 '17 at 1:38
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    $\begingroup$ @MichaelLee The linked thread has an answer that claims to construct an $f$ that works to disprove your suspicion for a given $\alpha$ as long as $\alpha$ is not of the form $1/n$ for some whole number $n$. (while typing) Torsten's comment above just said it. $\endgroup$ – Jeppe Stig Nielsen Dec 17 '17 at 1:41
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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 $x\in [0, 1-\alpha]$.

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