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