# 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:

- $\displaystyle \int \map f x \rd x$

and:

- $\displaystyle \int \map f {x + L} \rd x$

are both primitives of the same function.

So by Primitives which Differ by Constant:

\(\displaystyle \int \map f x \rd x + C\) | \(=\) | \(\displaystyle \int \map f {x + L} \rd x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int \map f {x + L} \frac {\map \d {x + L} } {\d x} \rd x\) | as $\dfrac {\map \d {x + L} } {\d x} = 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int \map f {x + L} \map \d {x + L}\) | Integration by Substitution | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map F x + C\) | \(=\) | \(\displaystyle \map F {x + L}\) |

Aiming for a contradiction, suppose that $C \ne 0$.

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

\(\displaystyle \map F {x + \paren {n + 1} L}\) | \(=\) | \(\displaystyle \map F {x + L + n L}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map F {x + L} + n C\) | by the induction hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map F x + \paren {n + 1} C\) | by the base case |

$\Box$

Holding $x$ fixed yields:

- $\displaystyle \lim_{k \mathop \to \infty} \size {\map F {x + k L} } = \lim_{k \mathop \to \infty} \size {\map F x + k C} = \infty$

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.

$\Box$

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

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

Then:

\(\displaystyle \map F x\) | \(=\) | \(\displaystyle \map F {x + L'}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {F'} x\) | \(=\) | \(\displaystyle \map {F'} {x + L'}\) | Chain Rule for Derivatives | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map f x\) | \(=\) | \(\displaystyle \map f {x + L'}\) | Definition of Primitive of Real Function |

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

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

Hence the result.

$\blacksquare$