TITLE: Prove that if $\forall_{x\in\Bbb{R}} f(x)=f(x+1)$ and $f$ is continuous then there are infinitely many $c$ such that $f(c+\pi)=f(c)$
QUESTION [5 upvotes]: We have 3 assumptions about $f$:

$f: \Bbb{R} \rightarrow \Bbb{R}$
$f$ is continuous
$\forall_{x\in\Bbb{R}} f(x)=f(x+1)$

The problem asks us to prove 2 things: That $f$ reaches its supremum and infimum and also there exist infinitely many $c\in\Bbb{R}$ such that $f(\pi+c)=f(c)$.
So the first thing can be proven (I think) discovering that $f([0,1]) = f(\Bbb{R})$ (because the function is periodic) so supremum and infinimum is in $f([0,1])$ and by the extreme value theorem it can be reached.
As for the second problem I can't see why it should be true. That would mean that there exist $x\in\Bbb{R}$ such that $f(\pi +x+1) = f(x+1) = f(x)$. Wouldn't that mean that function must have 2 periods - $1$ and $\pi$? But this $\pi$ makes no sense then. Am I understanding this correctly?

REPLY [3 votes]: Since $g(x) := f(x+\pi)-f(x)$ is also periodic, we only have to show, that $g$ has a zero.
If not, we can assume that $g$ is strictly positive by the intermediate value theorem. This gives us an increasing sequence $f(n\pi)$, which converges, since $f$ is bounded.
But the sequence equals the sequence $f(n\pi-\lfloor n\pi \rfloor)$. Since $n\pi-\lfloor n\pi \rfloor$ has all points in $[0,1]$ as accumulation points, by continuity of $f$ we get that $f(n\pi-\lfloor n\pi \rfloor)$ has all points of $f([0,1])$ as accumulation points. But $f(n\pi-\lfloor n\pi \rfloor)$ converges, hence admits a unique accumulation point. We deduce that $f([0,1])$ is a singleton, hence $f$ is constant and $g=0$. Contradiction!