Primitive of Periodic Function

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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.

$\blacksquare$


Also see