Let $f : \mathbb{R} \to \mathbb{R}$ be a continuous periodic function of period 1. Show that there exists $x_0 \in \mathbb{R}$ such that $f(x_0 + π) = f(x_0).$
This is visually pretty clear but I cannot think of a complete proof. What I thought was to take a function $g(x)=f(x+k)-f(x)$ where $k=π-3$ and need to show $g(x)$ takes positive values somewhere and negative values somewhere.
But the problem is at $x=0,\\ f(k)-f(0)$ is positive does not imply at $x=1-k, \\f(1)-f(1-k)=f(0)-f(-k)$ is negative. i.e. I cannot use the trick which we do for 1/2*period case.
Best Answer
Hint. Good start, but there is nothing special about taking $x=0$. Instead, suppose that $f$ has a minimum when $x=a$ and consider $$g(a)\quad\hbox{and}\quad g(a-\pi)\ ,$$ where $g(x)=f(x+\pi)-f(x)$.