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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.

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    $\begingroup$ I think the answer is no even in higher dimension, since the gradient of any characteristic function $\chi_\Omega$, $\Omega\Subset\Bbb R^n$ as exactly the same properties of $D=d/dx$, i.e. $\nabla \chi_\Omega\subseteq\partial\Omega$ as it is shown in this Q&A. And if you need a single PDE satisfying this property, you can take the divergence of the gradient and form the laplacian of the characteristic function. $\endgroup$
    – Daniele Tampieri
    Commented Jun 23, 2020 at 4:49

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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 $E$.

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    $\begingroup$ Up to a factor $1/\sqrt\pi $. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 23, 2020 at 5:46
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    $\begingroup$ $E$ seems unbounded in you example. $\endgroup$
    – Zamanyan
    Commented Jun 23, 2020 at 6:08
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    $\begingroup$ Perhaps you should edit your answer and only keep the $\partial_x\partial_y$ example? the first part is wrong not only because the domain is unbounded, but more importantly because the fundamental solution is NOT smooth up to the boundary $\endgroup$
    – leo monsaingeon
    Commented Jun 23, 2020 at 8:25
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    $\begingroup$ If $E$ has a smooth boundary, we can assume it is a half space $x_n>0$ near a point (say the origin) of the support of $D(\chi_Eu)$. Then $\chi_E=H(x_n)$ and we can write $D(\chi_Eu)=\Sigma_{j=0}^k a_j(x)H^{(j)}(x_n)$. Then $a_k$ must vanish in a punctured nbh of the origin in $x_n=0$ and so also at the origin by continuity (assuming n>1). Using that $x_nH^{(k)}(x_n)$ is a multiple of $H^{(k-1)}(x_n)$ we can eliminate the highest order term in the sum and inductively continue to eliminate all (I guess). $\endgroup$
    – NJK
    Commented Jun 23, 2020 at 13:06
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    $\begingroup$ In the smooth case I believe you are right, and it might be sufficient to consider the case $E$ is a half space (EDIT: see the previous comment :) $\endgroup$
    – Piero D'Ancona
    Commented Jun 23, 2020 at 13:36

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