Let $f$ be a continuous function in $[0,1]$ such that $f(1) = f(0)$. let $n$ be a natural number. I have to show that there exists $x \in [0, 1-1/n]$ such that $f(x+1/n) = f(x)$.
My question is, can it be shown for all $n$? I could show that for $n=2$ the statement is true by looking at $g(x) = f(x+1/2) - f(x)$ and then choosing $g(0)$ and $g(1/2)$ and by using the intermediate value theorem. it's easy to show that $x=1/2$ works.
I'm struggling however to prove it for all $n$. I tried doing the same thing but with $x= 1-1/n$, it didn't get me anywhere and it seems like I'm quite stuck.
i got:
$g(1-1/n) = f(1) - f(1-1/n)$
$g(0) = f(1/n) - f(0) = f(1/n) - f(1)$
if i could show that $g(0)\cdot g(1-1/n) \le 0$ then I'll be able to use the intermediate value theorem but I can't seem to succeed.
Thank you for your time!
Hint: what is $\sum_{k=0}^{n-1}{g\left(\frac{k}{n}\right)}$? Thus, can $g$ not vanish?