# 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 $f \left({x}\right) = f \left({x + L}\right)$.

Then:

- $\displaystyle \int f \left({x}\right) \, \mathrm d x$

and:

- $\displaystyle \int f \left({x + L}\right) \, \mathrm d x$

are both primitives of the same function.

So by Primitives which Differ by Constant:

\(\displaystyle \int f \left({x}\right) \, \mathrm d x + C\) | \(=\) | \(\displaystyle \int f \left({x + L}\right) \, \mathrm d x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int f \left({x + L}\right) \, \frac {\mathrm d \left({x + L}\right)}{\mathrm d x} \, \mathrm d x\) | $\dfrac {\mathrm d\left({x + L}\right)}{\mathrm d x} = 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int f \left({x + L}\right) \, \mathrm d \left({x + L}\right)\) | Integration by Substitution | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle F \left({x}\right) + C\) | \(=\) | \(\displaystyle F \left({x + L}\right)\) |

Suppose that $C \ne 0$.

Then it is to be shown that for all $k \in \N_{\gt 0}$:

- $F \left({x}\right) + k C = F \left({x + k L}\right)$

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

Suppose that for some $n \in \N_{\gt 0}$ we have:

- $F \left({x}\right) + n C = F \left({x + n L}\right)$

Then

\(\displaystyle F \left({x + \left({n + 1}\right) L}\right)\) | \(=\) | \(\displaystyle F \left({x + L + n L}\right)\) | |||||||||||

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

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

$\Box$

Holding $x$ fixed yields:

- $\displaystyle \lim_{k \mathop \to \infty} \left\lvert{F \left({x + k L}\right)}\right\rvert = \lim_{k \mathop \to \infty} \left\lvert{F \left({x}\right) + k C}\right \rvert = \infty$

and so $F \left({x}\right)$ is unbounded.

But we had previously established that $F \left({x}\right)$ was bounded.

This is a contradiction, therefore $C = 0$ and so:

- $F \left({x}\right) = F \left({x + L}\right)$

It has been shown that $F$ is periodic.

$\Box$

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

Suppose that $\left\lvert{L'}\right\rvert \lt \left\lvert{L}\right\rvert$.

Then:

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

\(\displaystyle \implies \ \ \) | \(\displaystyle F' \left({x}\right)\) | \(=\) | \(\displaystyle F' \left({x + L'}\right)\) | Chain Rule | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle f \left({x}\right)\) | \(=\) | \(\displaystyle f \left({x + L'}\right)\) | Definition of Definition: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.

$\blacksquare$