Periodic Element is Multiple of Period
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Theorem
Let $f: \R \to \R$ be a real periodic function with period $P$.
Let $L$ be a periodic element of $f$.
Then $P \divides L$.
Proof
Aiming for a contradiction, suppose that $P \nmid L$.
Then by the Division Theorem we have $L = q P + r$ where $q \in \Z$ and $0 < r < P$.
And so:
| \(\ds \map f {x + L}\) | \(=\) | \(\ds \map f {x + \paren {q P + r} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f {\paren {x + r} + q P}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f {x + r}\) | General Periodicity Property | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map f x\) | Definition of Periodic Element |
But then $r$ is a periodic element of $f$ that is less than $P$.
Therefore $P$ cannot be the period of $f$.
The result follows from Proof by Contradiction.
$\blacksquare$