If $f:[0,1]\rightarrow \mathbb{R}$ continous and $f(0)=f(1)$. Then there exists a $c\in [0, \frac{9}{10}]$ such that $f(c)=f(c+\frac{1}{10})$ It's easy to see that we have to consider an auxiliar function $g(x)= f(x)-f(x+\frac{1}{10})$ and apply it Bolzano. But, i can't see how or why this has a change of sign. Also, I find the case where instead of using $\frac{9}{10}$ they use $\frac{1}{2}$ and that $c$ now belongs to $[0,\frac{1}{2}]$. Then, i would like to prove it in a general case: > If $f:[0,1]\rightarrow \mathbb{R}$ is continous where $f(0)=f(1)$ and $n\in\mathbb{N}$. Prove that exists $c\in [0,\frac{n-1}{n}]$ such that $f(c)=f(c+\frac{1}{n})$ However, it would be nice if i can try the first case ($n=10$). EDIT: I forgot to add the condition $f(0)=f(1)$.

Let $f:[0,1] \to \mathbb{R}$ be continuous, and assume $f(0)=f(1)$. Define $g:[0,1-\frac{1}{n}]$ by $g(x)=f(x)-f(x+\frac{1}{n})$. We want to show that there exists some $c \in [0,1-\frac{1}{n}]$ such that $g(c)=0$.

Assume by contradiction that there doesn't exist such $c$, so $g>0$ or $g<0$ (This is because $g$ is continuous; if there exist some $x_1,x_2$ such that $g(x_1)>0$ and $g(x_2)<0$, then by the intermediate value theorem there exists some $c$ such that $g(c)=0$); We assume wlog that $g>0$. Note that $0
This is a contradiction.