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?

Let $g(x)=f(x)-f(x+\frac 1 n)$. If this continuous functions is never $0$ then it is always positive or always negative. Suppose it is always positive. Write $0=f(1)-f(0)$ as $[f(\frac  1 n) -f(0)]+[f(\frac  2 n)-f(\frac 1 n)]+...+[f(\frac {n-1} n)-f(1)]$. You get  a contradiction since each term is $<0$. 
Similar argument works when $g(x) <0$ for all $x$.
