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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.

Rebellos
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    I believe you need to add some assumption on the function $r$. If $r$ could be any function, take it to be the Dirichlet function, that is, the indicator function of the rational numbers. It is periodic and the period can be arbitrarily small. – Amitai Yuval Feb 24 '19 at 17:22
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    Why is there a differential geometry tag on this? – Ted Shifrin Feb 24 '19 at 23:49

1 Answers1

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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

Thomas Davis
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