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The Gaussian isoperimetric inequality (Tsirelson,Sudakov, Borell) states that among all sets of given Gaussian measure in the n-dimensional Euclidean space, half-spaces have the minimal Gaussian boundary measure. Suppose we put an additional restriction on the set, that it should be symmetric about the origin. Then can we conclude that quarter-spaces (intuitively the first and third quadrant in 2-dimensions, say) have the minimal Gaussian boundary measure?

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Also, if possible, could someone please suggest a more readable version of the Gaussian isoperimetry proofs, and possibly other related references on Gaussian measures (like lecture notes or surveys available free on the internet)? –  BharatRam May 23 '11 at 17:10
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A bit off topic, but you could find this helpful: a nice review connecting concentration (and isoperimetric inequalities) to Markov chains. This allows to study discrete analogues of the picture. Yann Ollivier, A survey of Ricci curvature for metric spaces and Markov chains (pdf) yann-ollivier.org/rech/publs/surveycurvmarkov.pdf –  Leonid Petrov May 24 '11 at 5:28
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A little bit of computation shows that 'quarter-spaces' (or a symmetrization of halfspaces around the origin) is clearly not the best we can do. For example in two dimensions, just a circle of measure 1/2 has smaller boundary measure than the above set. But the question remains open. Also, thanks everyone, for the references. –  BharatRam May 24 '11 at 12:39
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Following up on Ryan's suggestions, here's a paper by Barthe (subscription probably required) whose introduction suggests that the problem as you stated, and the analogous problem on the sphere, are open and difficult: journals.cambridge.org/action/… –  Mark Meckes May 24 '11 at 15:01
    
Nice find, Mark. –  Ryan O'Donnell May 25 '11 at 2:59

2 Answers 2

My guess is that the optimizer is actually a "strip"; i.e., a set of the form {$x : -t \leq x_1 \leq t$}. But I'm somewhat sure that the solution to this problem is not known. You might take a look at the discussion surrounding after Corollary 3.6 in this paper by Klartag and Regev:

http://eccc.hpi-web.de/report/2010/140/

Barthe may also have some relevant papers.

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Suppose we are looking at sets of measure 1/2 say. Then t as above is root(2) erfinverse(.5) which is roughly .67449 The boundary measure in this case is .635553 Compared to a boundary measure of .588 in the case of a circle of measure 1/2. Assuming my calculations are correct. –  BharatRam May 24 '11 at 13:11
    
Yeah, that's very possible. So maybe a sphere is best in general? Another good question (which might suggest the answer) is whether the analogous isoperimetric problem on the surface of the sphere is solved. –  Ryan O'Donnell May 24 '11 at 13:46

Answering more @Bratt's comment than the original question: Talagrand's book

people.math.jussieu.fr/~talagran/book.ps.gz

Seems quite nice.

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