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Sep 23, 2013 at 19:05 answer added Yevgeny Liokumovich timeline score: 6
Sep 11, 2013 at 18:38 answer added alvarezpaiva timeline score: 3
Sep 11, 2013 at 13:40 answer added Mikhail Katz timeline score: 4
May 13, 2013 at 10:01 comment added Sergei Ivanov The question came by association from a totally different problem. I am hoping that there might be a kind of rigidity result where one concludes that the metric is round (or not far from round) by looking at some rough measures of isoperimetric profile. If there is one, I could try to apply the technique of the proof in another context.
May 13, 2013 at 7:53 comment added Mikhail Katz Just to make sure I understand what you are looking for: it seems plausible that by applying coarea to a suitable distance function one might be able to get a non-optimal bound similar to the one you asked for. Are you interested in the optimal value $2\pi$ for boundary length, or does the coarea argument get stuck on diameter issues?
May 12, 2013 at 19:01 comment added Sergei Ivanov @katz: This is just an example of statement that might be true if dividing in half fails. It might as well be an integral inequality on the isoperimetric profile. In other words, I am more interested in a affirmative answer to a slightly different question than in a counter-example to the precise one. There is nothing magic about the constant $\pi$, I just want to avoid the trivial answer "look at a small neighborhood of a point where curvature is greater than 1".
May 12, 2013 at 13:46 comment added Mikhail Katz Can elaborate why you formulated the question in terms of "area between pi and 2pi" rather than in terms of the Cheeger constant, and how this might affect the answer?
May 5, 2013 at 20:47 comment added Misha Alex Nabutovsky is the person to ask.
May 5, 2013 at 20:31 history asked Sergei Ivanov CC BY-SA 3.0