Continuous function on $[0,1]$, $f(0)=f(1)$

Question

I came across this very interesting question, which seems to be partially answered in a couple posts around here:
Let $f:[0,1]\rightarrow\mathbb{R}$ continuous such that $f(0)=f(1)$. Then for all $n>1\in \mathbb{N}$, there exists $x_n\in [0,1-1/n]$ such that $f(x_n)=f(x_n+1/n)$. The case $n=2$ is simple and has been answered in this forum already. But for $n>2$ I can't seem to use the same logic (intermediate value theorem). Any ideas?

Answer 1

Consider the function $$g(x) = f(x+\frac{1}{n}) - f(x)$$ Then $$g(0)+g(1/n)+\cdots+g((n-1)/n) = f(1)-f(0)=0$$ Hence for some $i\neq j$, we have $g(i/n) \geq 0$ and $g(j/n) \leq 0$. The intermediate value says there exist $\xi$ such that $g(\xi) = 0$.

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On the other hand, for any $\alpha$ not of the form $1/n$, you can construct a continuous function $f_\alpha: [0,1]\to \mathbb{R}$, with $f_\alpha(0)=f_\alpha(1)$ such that $f_\alpha(x+\alpha) = f_\alpha(x)$ has no solution.