Suppose $f$ is continuous on $[0, 1]$ such that $f(0) = f(1)$. Let $n$ be a positive integer, prove that there exists $x$ such that

Question

Suppose $f$ is continuous on $[0, 1]$ such that $f(0) = f(1)$. Let $n$ be a positive integer, prove that there exists $x$ such that $$ 0 ≤ x < x + 1/n ≤ 1, \\f(x) = f(x+(1/n)) $$

Answer 1

For $n\in \mathbb N$ and $x\in [0,1-1/n]$ let $g(x)=f(x+1/n)-f(x).$ Then $$\sum_{j=0}^{n-1}g(j/n)=\sum_{j=0}^{n-1}f((j+1)/n)-f(j/n)=f(1)-f(0)=0.$$ So $g$ cannot take only positive or only negative values. So, since $g$ is continuous and its domain is a non-empty closed bounded interval, there exists $x\in dom(g)=[0,1-1/n]$ such that $0=g(x)=f(x+1/n)-f(x).$

Remark. If $r\in (0,1)$ and $r$ is not the reciprocal of an integer, we can construct a piece-wise-linear continuous $f:[0,1]\to \mathbb R$ with $f(0)=f(1)$, such that $f(x)\ne f(x+r)$ for all $x\in [0,1-r].$