Let $D$ be a differential operator with smooth coefficients in $\mathbb{R}^n$. Suppose $E$ is a bounded open set in $\mathbb{R}^n$. Suppose $u$ is a function that is smooth up to the boundary of $E$. If $D(u\cdot\chi\_{E})$, as a distribution in $\mathbb{R}^n$, has a finite support, is it true that $D(u\cdot\chi\_{E})$ is actually zero?


The answer is no in the case $n=1$. An example is $D=d/dx$, $E=(0,1)$ and $u(x)=1$ for all $x$. In this case, the support of $D(u\cdot\chi\_{(0,1)})$ is $\{0,1\}$. I hope that the answer is yes if $n\geq 2$. If the answer is yes, what weaker conditions in u are actually needed to guarantee that the answer is yes? Any suggestion for appropriate references would be greatly appreciated.



When is a distribution having a finite support actually zero?
Let $E$ be the square $(0,1)^2$ in $R^2$, $D=\partial\_x\partial\_y$ and $u=1$. The support of $D(\chi\_E u)$ is the set of corners of $E$.


