Proof that a periodic function is bounded and uniformly continuous. 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 $f(c)=f(c+\pi)$.

 A: 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.
A: $f$ is bounded in the compact [0,1] and as $f(x)= f(x+1), x \in \mathbb{R}, f$  is bounded.Now note that $f$ is uniformly bounde in the compacts $[0,1/2],[1/2,1]$. Then given $\epsilon>0$ there are $\delta_1, \delta_2>0$ such that
\begin{equation}
|x-y|< \delta_1 \Rightarrow |f(x)-f(y)|< \epsilon ,x, y \in [0,1]
\end{equation}
\begin{equation}
|x-y|< \delta_1 \Rightarrow |f(x)-f(y)|< \epsilon ,x, y \in [1/2,3/2]
\end{equation}
As $f(x)= f(x+1), x \in \mathbb{R}, f$ take $\delta=\min\{\delta_1,\delta_2$. Hence, $f$ is uniformily continuous. After I do the rest.
