Double of Antiperiod is Period

Theorem

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

Then $f$ is also periodic with a period of $2A$.

Proof

Let $L_f$ be the set of all periodic elements of $f$.

By Periodic Element is Multiple of Antiperiod and Absolute Value of Real Number is not less than Divisors:

$\forall p \in L_f: A \divides p \land A \le p$

Suppose there is a $p \in L_f$ such that $p = A$.

Then:

\(\ds \map f x\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds -\map f x\)

\(\ds \leadsto \ \ \)

\(\ds \map f x\)

\(=\)

\(\ds 0\)

which contradicts Constant Function has no Period.

Therefore $\forall p \in L_f: A \divides p \land A < p$.

The smallest number $r$ such that $A < r \land A \divides r$ is $2A$.

But by Double of Antiperiodic Element is Periodic, this is a periodic element of $f$.

Hence the result.

$\blacksquare$