MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Wavefront set of a product

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 $WF(v)$?

-

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 $(x,0)$).
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.
 If $WF(g)=N^*\{x_1^2=x_2\}$ (is it?), then is $fg$ well-defined? If $WF(g)=N^*\{x_1^2=x_2\}$, then e.g. $(x_1,x_2,\xi_1,\xi_2)=(0,0,0,1)\in WF(g)$, but $(0,0,0,-1)\in N^*\{x_2=0\}=WF(f)$. – Rob Jun 14 2012 at 0:51 In your specific case if you use additional information which is not available in your wave front set, then you can define $fg$. In particular, you have that $f,g\in L^\infty\subset L^1_{\text{loc}}$ and using that $L^\infty$ is an algebra you have point-wise product which is locally integrable and hence can be viewed as distributions. But if you are only given the two wave-front sets, in general two distributions with those wave front sets need not admit a well-defined product. – Willie Wong Jun 14 2012 at 10:20