Antiperiodic Element is Multiple of Antiperiod

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Theorem

Let $f: \R \to \R$ be a real anti-periodic function with anti-period $A$.

Let $L$ be an anti-periodic element of $f$.


Then $A \divides L$.


Proof

Aiming for a contradiction, suppose that $A \nmid L$.

By the Division Theorem we have:

$\exists! q \in \Z, r \in \R: L = q A + r, 0 < r < A$

By Even and Odd Integers form Partition of Integers, it follows that $q$ must be either even or odd.


Case 1

Suppose $q$ is even.

Then:

\(\displaystyle \map f {x + L}\) \(=\) \(\displaystyle \map f {x + \paren {q A + r} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {\paren {x + r} + q A}\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {x + r}\) General Antiperiodicity Property
\(\displaystyle \) \(=\) \(\displaystyle -\map f x\)

And so $r$ is an anti-periodic element of $f$ that is less than $A$.

But then $A$ cannot be the anti-period of $f$.

Therefore by contradiction $q$ cannot be even.


Case 2

Suppose $q$ is odd.

Then:

\(\displaystyle - \map f x\) \(=\) \(\displaystyle \map f {x + L}\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {x + \paren {q A + r} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {\paren {x + r} + q A}\)
\(\displaystyle \) \(=\) \(\displaystyle - \map f {x + r}\) General Antiperiodicity Property


So:

$-\map f x = - \map f {x + r} \implies \map f x = \map f {x + r}$

It is seen that $r$ is a periodic element of $f$ such that $0 < r < A$.

But consider $0 < A - r < A$:

\(\displaystyle \map f {x + \paren {A - r} }\) \(=\) \(\displaystyle \map f {\paren {x + A} - r}\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {x + A}\)
\(\displaystyle \) \(=\) \(\displaystyle -\map f x\)


This contradicts the fact that $A$ is the anti-period of $f$.

Hence the result.

$\blacksquare$