The Gaussian isoperimetric inequality (Tsirelson,Sudakov, Borell) states that among all sets of given Gaussian measure in the ndimensional Euclidean space, halfspaces 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 quarterspaces (intuitively the first and third quadrant in 2dimensions, say) have the minimal Gaussian boundary measure?

1$\begingroup$ 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)? $\endgroup$ – BharatRam May 23 '11 at 17:10

1$\begingroup$ 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) yannollivier.org/rech/publs/surveycurvmarkov.pdf $\endgroup$ – Leonid Petrov May 24 '11 at 5:28

1$\begingroup$ A little bit of computation shows that 'quarterspaces' (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. $\endgroup$ – BharatRam May 24 '11 at 12:39

2$\begingroup$ 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/… $\endgroup$ – Mark Meckes May 24 '11 at 15:01

$\begingroup$ Nice find, Mark. $\endgroup$ – Ryan O'Donnell May 25 '11 at 2:59
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.hpiweb.de/report/2010/140/
Barthe may also have some relevant papers.

$\begingroup$ 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. $\endgroup$ – BharatRam May 24 '11 at 13:11

$\begingroup$ 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. $\endgroup$ – 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.