Replicative Function of x minus Floor of x is Replicative

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

Let $f$ be a replicative function.

Let $g: \R \to \R$ be the real function defined as:
 * $\map g x = \map f {x - \floor x}$

Then $g$ is also a replicative function.

Lemma
First observe:
 * $\ds \sum_{k \mathop = 0}^{n - 1} \map g {x + \frac k n} = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n - \floor {x + \frac k n} }$

We need to show that the two final sums are equal.

Suppose:
 * $\dfrac j n \le x - \floor x < \dfrac {j + 1} n$

for some integer $j$.

By Real Number minus Floor:
 * $0 \le x - \floor x < 1$

This gives:
 * $0 \le j \le n - 1$

Let $y = x - \dfrac j n$.

Then:
 * $\floor x \le x - \dfrac j n = y$

By Number not less than Integer iff Floor not less than Integer:
 * $\floor x \le \floor y$

On the other hand:
 * $y \le x$

which gives
 * $\floor y \le \floor x$

So $\floor y = \floor x$.

We now have:
 * $\dfrac j n \le x - \floor x = y + \dfrac j n - \floor y < \dfrac {j + 1} n$

which gives:
 * $0 \le y - \floor y < \dfrac 1 n$

which satisfies the condition for the lemma.

Hence:

We also have:

Hence the result.