Re: f is constant?

From: Robert Israel (israel_at_math.ubc.ca)
Date: 03/02/05 

Date: 2 Mar 2005 00:48:20 GMT

In article <d01q9n$kvs$1@mailhub227.itcs.purdue.edu>,
Dave Seaman <dseaman@no.such.host> wrote:

>I suspect that's where the difference lies. The function I mean by
>"Cantor function" is the one described in Royden (exercise 46 on p. 48 of
>the 2nd ed.). That function has the properties f(0) = 0, f(1) = 1, f is
>monotone, and f is constant on each interval contained in the complement
>of the Cantor set. Therefore f' is zero wherever it is defined.

I don't think the "therefore" is at all obvious, though it may be true
for the case of the Cantor set.
It's certainly false for some functions with the same properties except
that the Cantor set is replaced by another set that is homeomorphic to it.
Thus there are increasing homeomorphisms g and h of [0,1] to itself such
that g(f(h(x))) has a nonzero derivative at, say, x = 1/2, because
we can make, say, x <= g(f(h(x))) <= x + (x-1/2)^2.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
 

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