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Periods of continuous functions

  1. Sep 4, 2005 #1
    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. jcsd
  3. Sep 4, 2005 #2
    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.
  4. Sep 4, 2005 #3


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    In other words, there is no counter-example!
  5. Sep 4, 2005 #4
    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: Sep 4, 2005
  6. Sep 4, 2005 #5
    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


    **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.

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