Thursday, November 10, 2005

plotting my own doom ..

Or as it's more commonly known here, writing a syllabus of topics for my Prelim Exam. A good bit of them I've seen before, but it still seems a scary task to remember it all .. or at least enough to keep my committee happy.

If you're curious, here is a tentative breakdown of the subjects:
  1. Preliminaries
  2. Lips¢hitz Functions and C0nvex Functions
  3. Sob01ev Spaces and Fine Properties
  4. Highlights from Ge0metric Mea$ure The0ry
Believe me, you can say a lot about each heading. I only hope that I can, too. Wish me luck!



As I was browsing through books for syllabus items, I found this theorem in a book of P. Mattila. The result is credited to P. Jones, which generalises a previous result of G. David.

Maybe I am rather dull and ignorant, but it seems a little unbelievable.

Let Q be the unit cube in Rn and let m ≤ n. Given ε > 0, there is a number Nε where for all 1-Lipschitz functions f mapping Q into Rm, there are balls {Bi : i = 1 to N} with N < Nε, where Hn(f(Q \ Ui Bi)) < ε and f|Bi is Nε-biLipschitz!

Essentially, this means that off a small set set of small image, 1-Lipschitz functions from the unit cube to a higher-dimensional Euclidean space are locally bi-Lipschitz (and probably with very large biLipschitz constant). It's not even clear to me that a 1-Lipschitz function should be locally invertible, much less locally biLipschitz!

Edit: perhaps thinking of $ard's Lemm@ gives plenty of motivation. Now that I think about it, I'm not as impressed. Since n ≥ m, then the Hausdorff content given by Hn doesn't detect the image that well, since in some ways, the image is an "m-dimensional set." I'd be much more impressed if the theorem used Hm instead.

Anyways, back to work.

4 comments:

Anonymous said...

Not off a small set, but off a set with small image. Think of Sard's lemma.

janus said...

Good point. I think I made the standard mistake of misinterpreting a Sard-type lemma.

Thanks for the correction; I've edited the post to reflect this.

Anonymous said...

You have a wrong inequality between m and n. This is clear even without the book: if m was less than n, then f could not be bi-Lipschitz on any ball.

You can be impressed again. :)

janus said...

Argh! You're right. Let me re-post the theorem with greater caution ..