Show that there is $\xi$ s.t. $f(\xi)=f\left(\xi+\frac{1}{n}\right)$

Let $f:[0,1]\to\mathbb{R}$ be continuous and $f(0)=f(1)$. Show that for any integer $n\geqslant 2$, there is $\xi\in(0,1)$ s.t.$f(\xi)=f\left(\xi+\frac{1}{n}\right).$

I think this requires the intermediate value theorem somewhere.

Solution 1:

Fix $\delta = 1/n, n \in \mathbb{N}.$ Consider the continuous function $g_n(x) :[0,1-\delta] \to \mathbb{R} ; g_n(x) := f(x+\tfrac{1}{n}) - f(x)$ and suppose that $g_n(x) \neq 0$ on $I_0 := [0,1].$ Then by the intermediate value theorem, $g_n(x)$ does not change sign in $I_0.$ WLOG, if $g_n(x) > 0,$ then adding the $n$ inequalities, $$ g_n(1-\tfrac{1}{n}) = f(1) - f(1-\tfrac{1}{n}) > 0, $$ $$ g_n(1-\tfrac{2}{n}) = f(1-\tfrac{1}{n}) - f(1-\tfrac{2}{n}) > 0, $$ $$ \cdots $$ $$g_n(0) = f(\tfrac{1}{n}) - f(0) > 0, $$ yields $f(1) > f(0),$ which contradicts the hypothesis that $f(1) = f(0).$ Thus there exists a point $x_n$ satisfying $f(x_n + \tfrac{1}{n}) = f(x_n).$