Show that $f(a)=f(a+\frac{1}{n})$ for some $a \in [0, 1-\frac{1}{n}]$, given that: Show that $f(a)=f(a+\frac{1}{n})$ for some $a \in [0, 1-\frac{1}{n}]$, given that:


*

*$f$ is continuous on $[0,1]$ and $f(0)=f(1)$. 

*$f(a)=f(a+\frac{1}{2})$ for some $a \in [0, 1/2]$.

*$n\in \Bbb N$ and $n>2$
I tried using the intermediate value theorem on $g(x)=f(x+\frac{1}{n})-f(x)=0$ on $[0, 1-\frac{1}{n}]$
and got that $g(0)=f(\frac{1}{n})-f(0)$ and $g(1-\frac{1}{n})=f(1)-f(1-\frac{1}{n})$
but I'm stuck and not sure if I'm on the right track...

Yes, you're on the right track. Consider that you can show that:
$$g\left(\frac{0}n\right)+g\left(\frac{1}n\right)+g\left(\frac{2}n\right)+\ldots+g\left(\frac{n-2}n\right)+g\left(\frac{n-1}n\right)=f(1)-f(0)=0$$
which implies that not all the $g$ can be positive.
