Primitive of Periodic Function

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

Let $F$ be a primitive of $f$ that is bounded on all of $\R$.

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

Then $F$ is also periodic with period $L$.

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

Let $f$ have a primitive $F$ that is bounded on all of $\R$.

By definition of a periodic function, it is seen that:
 * $\map f x = \map f {x + L}$.

Then:
 * $\ds \int \map f x \rd x$

and:
 * $\ds \int \map f {x + L} \rd x$

are both primitives of the same function.

So by Primitives which Differ by Constant:

$C \ne 0$.

Then it is to be shown that for all $k \in \N_{> 0}$:
 * $\map F x + k C = \map F {x + k L}$

The case for $k = 1$ has already been proven, so this will be proven by induction.

Suppose that for some $n \in \N_{> 0}$ we have:
 * $\map F x + n C = \map F {x + n L}$

Then

Holding $x$ fixed yields:
 * $\ds \lim_{k \mathop \to \infty} \size {\map F {x + k L} } = \lim_{k \mathop \to \infty} \size {\map F x + k C} = \infty$

and so $\map F x$ is unbounded.

But we had previously established that $\map F x$ was bounded.

This is a contradiction.

Therefore our assumption that $C \ne 0$ was false.

Hence $C = 0$ and so:
 * $\map F x = \map F {x + L}$

It has been shown that $F$ is periodic.

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

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

Then:

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

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

Hence the result.

Also see

 * Derivative of Periodic Function