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

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?

This is definitely helpful to you https://math.stackexchange.com/questions/1536042/if-a-continuous-function-takes-equal-values-at-endpoints-in-cannot-be-injective — Edumaths555, Dec 17 '17 at 01:04
See Construct a continuous function $f$ over $[0,1]$ satisfying $f(0) = f(1)$ but $f(x) \neq f(x+a)$ and the linked Show that for any positive integer $n$ there is $x \in [0,1-\frac{1}{n}]$ for which $f(x) = f(x+\frac{1}{n})$. — dxiv, Dec 17 '17 at 01:08

score 13 · Accepted Answer · answered Dec 17 '17 at 01:10

13

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

answered Dec 17 '17 at 01:10

Nightgap

1,261

4

Exceedingly clever construction, I love it. – infinitylord Dec 17 '17 at 01:13
Can such examples also be given for "small" $\alpha$? – Jeppe Stig Nielsen Dec 17 '17 at 01:14
I suspect that the question statement is true if $\alpha\leq 1/2$. – Michael L. Dec 17 '17 at 01:17
4

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. – Torsten Schoeneberg Dec 17 '17 at 01:38
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@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. – Jeppe Stig Nielsen Dec 17 '17 at 01:41

score 6 · Answer 2 · answered Dec 17 '17 at 01:15

6

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

answered Dec 17 '17 at 01:15

Michael L.

5,521

Continuous $f:[0,1]\to\mathbb{R}$ such that $f(0)=f(1)$ and $\forall\alpha\in(0,1)\exists c\in[0,1-\alpha]|f(c)=f(c+\alpha)$?

2 Answers2