Let $f:\mathbb{I} \to \mathbb{R}$ continuous function such that $f(0)=f(1)$.

Question

$\mathbb{I} = [0,1]$

Let $f:\mathbb{I} \to \mathbb{R}$ continuous function such that $f(0)=f(1)$. Prove that for all $n \in \mathbb{N} $ there $ x \in \mathbb{I}$ such that $ x + \frac{1}{n} \in \mathbb{I}$ and $f( x + \frac{1}{n})=f(x)$

Could you help me by giving me an idea of ​​how to do it?

Answer 1

Extend $f$ to $\mathbb{R}$ periodically (and so continuously by $f(0) = f(1)$). Let $g(x): = f(x+1/n) - f(x)$. Then we have $$ g(x) + g\left(x+\frac{1}{n}\right) + \cdots + g\left(x + \frac{n-1}{n}\right) = 0 $$ This implies that $g(x)$ can't have always same sign, and so $g$ has a zero by intermediate value theorem.

I'm sure that there's better way to show that $f(x+\alpha) = f(x)$ has root for any $0<\alpha < 1$, since this method only works for $\alpha\in \mathbb{Q}$. But I don't have any idea for this now.

Answer 2

Suppose there is no such $x$. Then either $f(x+\frac 1 n ) >f(x)$ for all $x$ or $f(x+\frac 1 n ) <f(x)$ for all $x$ (by IVP applied to the continuous function $f(x+\frac 1 n ) -f(x))$. Assume that $f(x+\frac 1 n ) >f(x)$ for all $x$. (the proof is similar in the other case). Then $f(0)<f(\frac 1 n) <f(\frac 2 n)<\cdots <f(1)$ which is a contradiction.