Exercise :
Let $r(t)$ be a periodic and non constant function. Prove that there exists a minimum $T \in \mathbb R$ with $T>0$ such that $r(t)$ is $T$-periodic, meaning that $T$ is the fundamental period of $r(t)$.
Thoughts :
So, since $r(t)$ is periodic, that means that there exists a positive $P>0$ such that : $$r(t+P) = r(t)$$ Also, $r(t)$ is non constant, which means that : $$\frac{\mathrm{d}r(t)}{\mathrm{d}t} \neq 0, \; \; \forall t \in D_r$$ Other than these, though, I don't see how one would come up with something that shows that there exists a least positive $T$ such that $r(t) = r(t+T)$.
Any help will be greatly appreciated.
Let $r:\mathbb{R}\rightarrow \mathbb{R}$ be periodic, non-constant, and continuous*. Then there exists at least one $P\in \mathbb{R}$ such that for any $t\in \mathbb{R}, r(t + P) = r(t)$.
Let $T = \{P\in \mathbb{R} : P\geq 0, r(t+P) = r(t), \forall t \in \mathbb{R}\}$.
$T$ is non-empty (since $r$ is periodic), and is a subset of $\mathbb{R}$. Show that $\inf T$ exists, and you’re done.
Note: Since $r$ is non-constant, we can show that $0$ is a lower bound for $T$, but not a greatest lower bound.
*Added continuous because of @Amitai Yuval’s comment on the question