Derivative of Periodic Function

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Theorem

Let $f: \R \to \R$ be a real function.

Let $f$ be differentiable on all of $\R$.


Then if $f$ is periodic with period $L$, then its derivative is also periodic with period $L$.


Proof

Let $f$ be differentiable on all of $\R$.

Let $f$ be periodic with period $L$.


Then taking the derivative of both sides using the Chain Rule yields:

$\map f x = \map f {x + L} \implies \map {f'} x = \map {f'} {x + L}$


Let $L'$ be the period of $f'$.

Suppose that $\size {L'} < \size L$.


$f$ is differentiable and therefore continuous, by Differentiable Function is Continuous.

From Image of Closed Real Interval is Bounded, it follows that $f$ is bounded on $\closedint 0 {\size L}$.

But from the General Periodicity Property, it follows that $f$ is bounded on all of $\R$.

Then from Primitive of Periodic Function, it follows that $L'$ is the period of $f$.


But we had previously established that $L$ was the period of $f$.

This is a contradiction, therefore $L' = L$.

Hence the result.

$\blacksquare$


Also see