Periodic Element is Multiple of Antiperiod

Theorem

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

Let $L$ be a periodic element of $f$.

Then $A \divides L$.

Proof

Consider $A + L$:

\(\ds \map f {x + \paren {A + L} }\)

\(=\)

\(\ds \map f {\paren {x + A} + L}\)

\(\ds \)

\(=\)

\(\ds \map f {x + A}\)

\(\ds \)

\(=\)

\(\ds -\map f x\)

Hence $A + L$ is an anti-periodic element of $f$.

Combining Antiperiodic Element is Multiple of Antiperiod, Divides is Reflexive, and Common Divisor Divides Difference yields:

$A \divides \paren {A + L} \land A \divides A \implies A \divides L$

Hence the result.

$\blacksquare$