Continuous Replicative Function

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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:

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

$\blacksquare$


Historical Note

Donald E. Knuth in his The Art of Computer Programming attributes this result to Nicolaas Govert de Bruijn, but gives no details as to where this result may be found.


Sources