13
$\begingroup$

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?

$\endgroup$
2

2 Answers 2

13
$\begingroup$

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

$\endgroup$
5
  • 4
    $\begingroup$ Exceedingly clever construction, I love it. $\endgroup$ Dec 17, 2017 at 1:13
  • $\begingroup$ Can such examples also be given for "small" $\alpha$? $\endgroup$ Dec 17, 2017 at 1:14
  • $\begingroup$ I suspect that the question statement is true if $\alpha\leq 1/2$. $\endgroup$
    – Michael L.
    Dec 17, 2017 at 1:17
  • 4
    $\begingroup$ 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. $\endgroup$ Dec 17, 2017 at 1:38
  • 3
    $\begingroup$ @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. $\endgroup$ Dec 17, 2017 at 1:41
6
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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