Continuous Replicative Function

Theorem
Let $f: \R \to \R$ be a real function.

Let $f$ be continuous on $\R$.

Let $f$ also be a replicative function.

Then $f$ is of the form:
 * $\map f x = \paren {x - \dfrac 1 2} a$

where $a \in \R$.

Proof
Let $f$ be a replicative function.

Then:
 * $\forall n > 0: \map f {n x + 1} - \map f {n x} = \map f {x + 1} - \map f x$

If $f$ is then also continuous:
 * $\forall x \in \R: \map f {x + 1} - \map f x$

and so:
 * $\map g x = \map f x - c \floor x$

is both replicative and periodic.

We have:
 * $\ds \int_0^1 e^{2 \pi i n x} \map g x \rd x = \dfrac 1 n \int_0^1 e^{2 \pi y} \map g y \rd y$

Expanding in a Fourier series shows:
 * $\map g x = \paren {x - \dfrac 1 2} a$

for $0 < x < 1$.

Thus it follows that:
 * $\map f x = \paren {x - \dfrac 1 2} a$