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:
 * $f \left({x}\right) = \left({x - \dfrac 1 2}\right) a$

where $a \in \R$.

Proof
Let $f$ be a replicative function.

Then:
 * $\forall n > 0: f \left({n x + 1}\right) - f \left({n x}\right) = f \left({x + 1}\right) - f \left({x}\right)$

If $f$ is then also continuous:
 * $\forall x \in \R: f \left({x + 1}\right) - f \left({x}\right)$

and so:
 * $g \left({x}\right) = f \left({x}\right) - c \left \lfloor{x}\right \rfloor$

is both replicative and periodic.

We have:
 * $\displaystyle \int_0^1 e^{2 \pi i n x} g \left({x}\right) \mathrm d x = \dfrac 1 n \int_0^1 e^{2 \pi y} g \left({y}\right) \mathrm d y$

Expanding in a Fourier series shows:
 * $g \left({x}\right) = \left({x - \dfrac 1 2}\right) a$

for $0 < x < 1$.

Thus it follows that:
 * $f \left({x}\right) = \left({x - \dfrac 1 2}\right) a$