Periods of continuous functions

  • Thread starter vidmar
  • Start date
  • #1
11
0

Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
83
0
Proof: Suppose we have a non-constant function f:R->R which has arbitrary small period, that is: for each [tex]\delta > 0[/tex] f is periodic with some period [tex]0 < p < \delta[/tex].

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

Hence f cannot be continuous at x.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
955
In other words, there is no counter-example!
 
  • #4
1,679
3
vidmar said:
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: [tex] f(x) = \lim_{h\rightarrow{0}}sin(x/h) [/tex].

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:
  • #5
rachmaninoff
Antiphon said:
Counter-example: [tex] f(x) = \lim_{h\rightarrow{0}}sin(x/h) [/tex].

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.

o:)
 

Related Threads on Periods of continuous functions

Replies
3
Views
4K
Replies
6
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
782
Replies
2
Views
802
Replies
13
Views
16K
  • Last Post
Replies
6
Views
919
  • Last Post
Replies
4
Views
4K
Top