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Dec 10, 2011 at 10:24 comment added André Henriques I'm perfectly happy with an argument that uses the Laplace operator. But why not use the $\bar \partial$ operator directly? Isn't this regularity thing a common feature of all elliptic differential operators?
Dec 9, 2011 at 10:45 comment added Igor Rivin I will try to find another reference (though I suspect the "regularity theory for the Laplace operator" cannot be escaped entirely, since this result is clearly a part of it...)
Dec 9, 2011 at 10:28 comment added André Henriques After having a closer look at that article, I'm not longer very satisfied by it. The argument I care about is entirely contained in the last page (one can safely disregard everything that comes before). The assumption that the boundary is $C^\infty$ is only used implicitly in the sentence "by the regularity theory of the Laplace operator". Also, I can't follow the last argument (last line of the paper): if you know that $|f(z)|^2$ is smooth, how do you conclude that $f(z)$ is smooth?
Dec 7, 2011 at 20:53 vote accept André Henriques
Dec 10, 2011 at 10:40
Dec 4, 2011 at 12:10 history answered Igor Rivin CC BY-SA 3.0