monge-ampere operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:17:14Z http://mathoverflow.net/feeds/question/110633 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110633/monge-ampere-operator monge-ampere operator digital 2012-10-25T08:40:34Z 2013-05-11T01:22:00Z <p>hello everybody, i'm studying the article of Bedford-Taylor "Fine topology silov boundary..." but i don't understand the proof of the following proposition.</p> <p>Let u,v plurisubharmonic function defined on $\Omega$ open subset of $\mathbb{C}^n$ such that $u,v\in L^{\infty}_{loc}(\Omega)$. </p> <p>Let $O$ be a fine-open subset of $\Omega$ where $u>v$. Then $(dd^{c}\max(u,v))^{n}|O=(dd^{n}u)^{n}|O$.</p> <p>They say that if $O$ is open then it is obvious. while if $O$ is just fine open then using a decreasing approximation sequence $u_k$ of smooth functions for $u$.</p> <p>Then since $O=\bigcap_{k}(u_k>v)$ and since $(dd^{c}\max(u_k,v))^{n}|O_k=(dd^{n}u_k)^{n}|O_k$ (this dont'understand),</p> <p>where $O_k=(u_k>v)$, the statement holds.</p> <p>Maybe it is trivial but i'm not able to see it. thanks </p> http://mathoverflow.net/questions/110633/monge-ampere-operator/110736#110736 Answer by digital for monge-ampere operator digital 2012-10-26T08:13:19Z 2012-10-26T08:50:09Z <p>yes i see it, but even on $O$ $u=\max(u,v)$ so i don't understand why he passes to an approximation sequence. </p> http://mathoverflow.net/questions/110633/monge-ampere-operator/110765#110765 Answer by bonho for monge-ampere operator bonho 2012-10-26T16:15:39Z 2012-10-26T16:15:39Z <p>Hi digital: I can not give comments, hence give answer here for your answer. If O is not open, then u=max(u,v) on O does not neccessarily imply that dd^cu=dd^cmax (u,v) on a neighborhood of O, so that when restricting to O you get the equality. (If you want to do derivatives, you need to do it on an open set).</p>