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

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Discussion Overview

The discussion centers around the claim that a non-constant real-valued continuous function cannot have an arbitrarily small period. Participants are exploring this concept through proofs and counter-examples, engaging in both theoretical reasoning and mathematical exploration.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests a proof or counter-example for the claim regarding non-constant continuous functions and their periods.
  • Another participant presents a proof arguing that if a non-constant function has an arbitrarily small period, it leads to a contradiction regarding continuity at a point.
  • A subsequent post asserts that there is no counter-example to the claim, reinforcing the proof provided earlier.
  • A different participant proposes a counter-example using the function f(x) = lim_{h→0} sin(x/h), suggesting that while it may work for finite h, the limit may not exist, aligning with the earlier proof.
  • Further clarification is provided distinguishing between two statements regarding periodicity and the existence of functions with arbitrarily small periods, indicating that while one statement is true, the other is false.

Areas of Agreement / Disagreement

Participants express disagreement regarding the existence of a counter-example to the original claim. Some support the proof that no such function can exist, while others propose a counter-example that raises questions about the limit's existence.

Contextual Notes

The discussion highlights the dependence on the definitions of continuity and periodicity, as well as the unresolved status of the limit in the proposed counter-example.

vidmar
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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!
 
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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.
 
In other words, there is no counter-example!
 
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:
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:)
 

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