TITLE: Proof that a periodic function is bounded and uniformly continuous.
QUESTION [5 upvotes]: I need to show that if $f:\mathbb{R}\to \mathbb{R}$ is continuous and $\forall x \in \mathbb R, f(x+1)=f(x)$, then:

$f$ is bounded,
$f$ is uniformly continuous,
there exists $c\in \mathbb{R}$ such that $f(c)=f(c+\pi)$.

REPLY [7 votes]: Let $$g : \begin{array}{ccc} \mathbb R & \to& \mathbb R\\ c &\mapsto &f(c+\pi) - f(c).\end{array}$$ That's a continuous function.
Now, let $c_0$ be a point where $f_{|[0,1]}$ has a minimum. Because of the periodicity, $c_0$ is a minimum for the whole of $f$. So $\forall x \in \mathbb R, f(x) \geq f(c_0)$. In particular $(x = c_0 + \pi)$, $g(c_0) \geq 0$.
The same argument with “maximum” instead of minimum gives a $c_1$ where $g(c_1) \leq0$.
The intermediate value theorem now gives a point $c$ on which $g$ vanishes, QED.