From: grubb@math.niu.edu (Daniel Grubb)
Subject: Re: function strictly increasing with a derivative equal to zero on a   dense set
Date: 27 Oct 1999 12:38:05 GMT
Newsgroups: sci.math

OK, I'll give the example taken out of 'An introduction to Classical
Real Analysis' by Karl Stromberg,  Wadsworth International Mathematics
Series:

Let {r_n} be dense in the real line.

Define

h(x)=x + \sum (x-r_n )^(1/3) /(n^2 (1+ \abs{r_n})^(1/3)

This function is strictly increasing, has (possibly infinite) derivative
everywhere with h'(x)>1 for all x. Furthermore, h'(r_n)=\infty for all n.

Now let H be the inverse function of h. Then H is strictly increasing,
is differentiable and has H'=0 on a dense set of the reals.

Proving that h has a derivative everywhere is a bit tricky, but if you use
the fact that for \phi(x)=x^(1/3), we have

0<= [ \phi(x)-\phi(t)]/(x-t) <= \phi '(t)

for all x and t, it is essentially a matter of looking at difference
quotients.


--Dan Grubb