Bedford-Taylor theory - MathOverflow most recent 30 from http://mathoverflow.net2013-05-20T20:57:02Zhttp://mathoverflow.net/feeds/question/91125http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/91125/bedford-taylor-theoryBedford-Taylor theory Vamsi2012-03-13T21:39:53Z2012-03-14T18:29:17Z
<p>The Dirichlet problem for the Complex Monge-Ampere equation on a bounded pseudoconvex domain in $\mathbb{C}^n$ was studied in Bedford-Taylor's seminal paper wherein they defined $(dd^{c} u)^n$ for locally bounded plurisubharmonic $u$. But, they don't seem to use it in their Perron-type method, instead using a convex-measure-theoretic construction claiming that the upper envelope is not well-behaved. Now that we know more about psh functions, have people studied the Dirichlet problem without using the measure-theoretic construction of Goffman and Serrin? (and reproved Bedford-Taylor's results)</p>
http://mathoverflow.net/questions/91125/bedford-taylor-theory/91201#91201Answer by Margaret Friedland for Bedford-Taylor theory Margaret Friedland2012-03-14T18:29:17Z2012-03-14T18:29:17Z<p>There is a result in the paper</p>
<p>Cegrell, Urban
On the Dirichlet problem for the complex Monge-Ampère operator.
Math. Z. 185 (1984), no. 2, 247–251.</p>
<p>Theorem. Assume that $\Omega$ is strictly pseudoconvex and that $H(t, z)$ is a measurable,
bounded and non-negative function on $(-\infty,, \max h] \times \Omega$, where $h$ is a
continuous function on $\partial \Omega$.. If $H(t,z)$ is continuous on $(-\infty, \max h]$ for every
fixed $z \in \partial \Omega$, then the Dirichlet problem $\phi \in P(\Omega),
(dd^c \varphi)^n =H(\varphi, z)dV$ on $\partial \Omega$,
$\lim_{z \to \zeta)\varphi(z) = h(\zeta)$ on $\partial \Omega$
has a solution. (Here, $dV$ denotes the Lebesgue measure and $P(\omega)$ is the class of bounded plurisubharmonic functions in $\Omega$.)</p>
<p>This theorem is a generalization of Theorem A in Bedford and Taylor
where $H^{1/n}$ is also required to be convex and increasing in $t$.(These properties are due to the fact that $H$ comes from the Goffman-Serrin operator.)</p>
<p>The proof uses fixed point methods.</p>
<p>Further results of this type were obtained (by Cegrell as well as his students and collaborators), mainly in the setting of hyperconvex domains.</p>