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A non-constant real-valued continuous function (f:R->R) cannot have an arbitirarly small period!

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- Thread starter vidmar
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A non-constant real-valued continuous function (f:R->R) cannot have an arbitirarly small period!

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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.

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HallsofIvy

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In other words, there is **no** counter-example!

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vidmar said:

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.

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- #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.

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