Let $f:[0,1] \rightarrow\mathbb R$ be a continuous map such that $f(0)=f(1)$ Let $f:[0,1] \rightarrow \mathbb R$ be a continuous map such that $f(0)=f(1) .$ Let $n \geq 2$
Show that there is some $x \in[0,1]$ such that $f(x)=f\left(x+\frac{1}{n}\right) .$
My attempt. Assume $f(x)\neq f(x+1/n)$ for all $x$. Then either $f(x)<f(x+1/n)$ or $f(x)>f(x+1/n)$. WLOG, assume $f(x)<f(x+1/n)$. Then $f(0)

So how can I get a contradiction? May you help?

$$
0 =f(0)-f(1)\\= \left[f(0)-f\left(\frac{1}{n}\right)\right]+\left[f\left(\frac{1}{n}\right)-f\left(\frac{2}{n}\right)\right]+\left[f\left(\frac{2}{n}\right)-f\left(\frac{3}{n}\right)\right]\ldots+\left[f\left(\frac{n-1}{n}\right)-f\left(1\right)\right]
$$
You add and subtract the necessary pieces, but the terms without loss of generality are strictly negative.
