Wavefront set of a product - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:11:04Z http://mathoverflow.net/feeds/question/99404 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99404/wavefront-set-of-a-product Wavefront set of a product Rob 2012-06-13T05:26:43Z 2012-06-14T09:34:40Z <p>Let $H$ be the Heaviside function. If $f(x_1,x_2)=H(x_2)$ on $\mathbb{R}^2$, then $WF(f)=N^*\{x_2=0\}$. Similarly, if $g(x_1,x_2)=H(x_1^2-x_2)$, I think the wavefront set of $g$ is the conormal bundle to the boundary curve $x_2=x_1^2$. But what's the wavefront set of $fg$? In general, how do I determine the wavefront of a product distribution $uv$ (assuming it's well-defined) given $WF(u)$ and $WF(v)$? </p> http://mathoverflow.net/questions/99404/wavefront-set-of-a-product/99414#99414 Answer by Willie Wong for Wavefront set of a product Willie Wong 2012-06-13T08:55:51Z 2012-06-14T09:34:40Z <p>In general you can only say that $$WF(uv) \subseteq \Bigg[ WF(u) \cup WF(v) \cup \Big[WF(u)+WF(v)\Big]\Bigg]$$ where the set $$WF(u) + WF(v) = \left\lbrace (x,\xi +\eta)| (x,\xi)\in WF(u), (x,\eta)\in WF(v) \right\rbrace$$ (note that a sufficient criterion for the product to be well-defined is precisely when the above set contains no points of the form $(x,0)$). </p> <p>See, e.g. chapter 11 of Friedlander and Joshi <em>Introduction to The Theory of Distributions</em>.</p> <p>But quite obviously the $\subseteq$ is not always an equality: just take $u,v$ two compactly supported distributions with distinct supports. Note that given $WF(u)$ and $WF(v)$ you only know the singular support of $u$ and $v$ and not their actual supports. </p>