# Periods of continuous functions

1. Sep 4, 2005

### vidmar

I would like a proof or a counter-example for the following claim:
A non-constant real-valued continuous function (f:R->R) cannot have an arbitirarly small period!

2. Sep 4, 2005

### Timbuqtu

Proof: Suppose we have a non-constant function f:R->R which has arbitrary small period, that is: for each $$\delta > 0$$ f is periodic with some period $$0 < p < \delta$$.

Let $$x \in \mathcal{R}$$. Because f is non-constant there is a $$y \in \mathcal{R}$$ such that $$f(x) \neq f(y)$$. Now we claim that f takes the value f(y) on every neighbourhood $$\left]x-\delta,x+\delta\right[$$ of x with $$\delta > 0$$. Proof: f is periodic with some period $$0 < p < \delta$$. There exists an integer n such that $$y-np \in \left]x-\delta,x+\delta\right[$$ and $$f(y-np)=f(y)$$.

Hence f cannot be continuous at x.

3. Sep 4, 2005

### HallsofIvy

Staff Emeritus
In other words, there is no counter-example!

4. Sep 4, 2005

### Antiphon

Counter-example: $$f(x) = \lim_{h\rightarrow{0}}sin(x/h)$$.

This works for any finite h. But I don't think the limit exists, so this
is consistent with the proof given by Timbuqtu.

Last edited: Sep 4, 2005
5. Sep 4, 2005

### rachmaninoff

An incredibly important distinction must be made: between

*for arbitrarily small period p, --> there exists a function which is periodic with period P, 0<P<p

and

**there exists a function f --> for which, for arbitrarily small p, f has a period P, 0<P<p

The two statements are independent. * is true. ** is false. The OP asked about **, which is false (per Timbuqtu's proof). Of course, Antiphon successfully proves that * is true.