Can a Non-Constant Continuous Function Have an Arbitrarily Small Period?

[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!

 

[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.

 

[HallsofIvy]

In other words, there is no counter-example!

 

[Antiphon]

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!

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.

 

[rachmaninoff]

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.

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.