Let $f:[0,1] \to \mathbf {R}$, continuous and $f(0)=f(1)$. Prove $f(x)=f(x+\frac{1}{2})$ has a solution $x_0 \in [0,\frac{1}{2}]$
$\mathbf {EDIT}$, I know this must have been asked before but I didn't specific reference to the edge case where $x_0 = 0,\frac{1}{2}$
So I wrote proof and I would like to run it by another set of eyes, especially since I'm missing reference to case where $x_0=0,\frac{1}{2}$.
First let $g:[0,\frac{1}{2}] \to \mathbf {R}$ such that $g(x)=f(x)-f(x+\frac{1}{2})$
Notice, $g(0)=f(0)-f(\frac{1}{2})$ and $g(\frac{1}{2})=f(\frac{1}{2})-f(1)$. Denote, since $f(0)=f(1)$ we have $g(\frac{1}{2})=-g(0)$.
$g(x)$ is continuous by arithmetic of continuous functions and therefore we can apply intermediate value theorem, thus implementing we have $x_0 \in (0,\frac{1}{2})$ such that $g(x_0)=0$ and specifically, $f(x_0)=f(x_0+\frac{1}{2})$.