- #1

vidmar

- 11

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter vidmar
- Start date

- #1

vidmar

- 11

- 0

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

- #2

Timbuqtu

- 83

- 0

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

- 43,021

- 970

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

- #4

Antiphon

- 1,683

- 3

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.

Last edited:

- #5

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.

Share:

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 10

- Views
- 5K

- Replies
- 6

- Views
- 1K

- Replies
- 3

- Views
- 5K

- Last Post

- Replies
- 7

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 11K

- Last Post

- Replies
- 3

- Views
- 958