Primitive of Periodic Real 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
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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:
| \(\ds \int \map f x \rd x + C\) | \(=\) | \(\ds \int \map f {x + L} \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map f {x + L} \frac {\map \d {x + L} } {\d x} \rd x\) | as $\dfrac {\map \d {x + L} } {\d x} = 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map f {x + L} \map \d {x + L}\) | Integration by Substitution | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map F x + C\) | \(=\) | \(\ds \map F {x + L}\) |
Aiming for a contradiction, suppose $C \ne 0$.
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
| \(\ds \map F {x + \paren {n + 1} L}\) | \(=\) | \(\ds \map F {x + L + n L}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map F {x + L} + n C\) | by the induction hypothesis | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map F x + \paren {n + 1} C\) | by the base case |
$\Box$
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.
$\Box$
Let $L'$ be the period of $F$.
Then:
| \(\ds \map F x\) | \(=\) | \(\ds \map F {x + L'}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {F'} x\) | \(=\) | \(\ds \map {F'} {x + L'}\) | Chain Rule for Derivatives | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map f x\) | \(=\) | \(\ds \map f {x + L'}\) | Definition of Primitive of Real Function |
But it was previously established that $L$ was the period of $f$.
Hence it follows that $L' = L$.
Hence the result.
$\blacksquare$