# 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.

 $\displaystyle \int \map f x \rd x + C$ $=$ $\displaystyle \int \map f {x + L} \rd x$ $\displaystyle$ $=$ $\displaystyle \int \map f {x + L} \frac {\map \d {x + L} } {\d x} \rd x$ 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.

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.

$\blacksquare$