From: ags@seaman.cc.purdue.edu (Dave Seaman) Subject: Re: Can someone provide a definition of the so-called "perfect sneak function"? Date: 27 Apr 1999 14:45:30 -0500 Newsgroups: sci.math Keywords: the Cantor function In article <3728eb3e.41504168@news.interlog.com>, Calvin Ostrum wrote: >For some reason I recently recalled a function that >a former math professor of mine told us about in >a second year calculus course. He called it >"the perfect sneak function". Well, the problem is >that I *don't* recall the function, only some >conditions that it satisfied: > >f:[0,1] -> R >f(0) = 0 >f(1) = 1 >f continuous everywhere on [0,1] >f differentiable almost everywhere on [0,1] >f'(x) = 0 wherever defined The function f is called the "Cantor function," or sometimes the "Cantor ternary function." A formal definition can be found in textbooks on real analysis. Royden, for example, discusses it in the exercises -- just look under "Cantor" in the index. Loosely speaking, the construction follows the definition of the Cantor set. You start with f(0) = 0 and f(1) = 1. Then you define f(x) = 1/2 for x in the middle third (1/3,2/3). Next you define f on the middle thirds of the two intervals that remain: f(x) = 1/4 for x in (1/9,2/9), f(x) = 3/4 for x in (7/9,8/9). You now have f defined on 3 intervals, leaving 4 gaps. Once again, define f on the middle third of each gap, using a constant value equal to the average of the values previously used on nearby intervals. That is, you define f on 4 intervals of length 1/27, with values of 1/8, 3/8, 5/8, and 7/8, respectively. Continuing this process indefinitely, you have f defined on the complement of the Cantor set C, where C is an uncountable, nowhere-dense subset of [0,1] having measure zero. It is easy to determine that there is a unique extension of f to [0,1] that is continuous on its domain, and that the resulting function has all the properties required. In particular, f' is identically zero on the complement of the Cantor set. -- Dave Seaman dseaman@purdue.edu Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal