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Suppose $D$ is a bounded open set in $\mathbb{R}^{m}$ with $m\geq2$, and $u(x)$ is locally integrable on $D$. We know that if $\omega$ is an open set that is included with its closure to $D$, the mollifier $u_{n}$ of $u$ (https://en.wikipedia.org/wiki/Mollifier) exists on $\omega$ for $n$ sufficiently big. We know also that if $u$ is subharmonic on $D$, then the mollifier is subharmonic on $\omega$ for $n$ sufficiently big.

My question is: suppose $u$ is locally integrable on $D$ but subharmonic on $D\setminus F$, where $F$ is a closed set with empty interior. Can we say that $u_{n}$ is subharmonic on $\omega\setminus F$ for $n$ big enough?

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    $\begingroup$ I think yes... Shouldn't the same proof work? what's the issue $\endgroup$ – mathworker21 Jul 5 '19 at 1:11
  • $\begingroup$ The sequence is defined by $$u_{n}(x)=\int u(x-t)\phi_{n}(t)dt.$$ In the regular case, where $u$ is subharmonic on $D$, $x-t$ belongs also to $\omega$ and so $x\mapsto u(x-t)$ is subharmonic on $\omega$. But now, what if $x\in \omega$ but $x-t\in F$? How can we have subharmonicity of $x\mapsto u(x-t)$, or circumvent it? $\endgroup$ – M. Rahmat Jul 5 '19 at 4:11
  • $\begingroup$ For any fixed $x$, since $F$ is closed, if we take $\phi$ to have compact support (which we may do), then for $n$ large enough and $t$ in the support of $\phi_n$, $x-t \not \in F$. $\endgroup$ – mathworker21 Jul 5 '19 at 4:18
  • $\begingroup$ Consider $x\in F$ and a sequence $x_{m}$ outside $F$ that converges to $x$. For a fixed $n$ I cannot say that $u_{n}$ is subharmonic at a neighborhood of each $x_{m}$, can I? $\endgroup$ – M. Rahmat Jul 5 '19 at 4:27
  • $\begingroup$ Does the closure of $\omega$ avoid $F$? If so, then my previous comment applies uniformly to $x \in \omega$, and the answer to your last comment is "yes, if each $x_m$ is in $\omega$". If the closure of $\omega$ intersects $F$, then I'm not sure... $\endgroup$ – mathworker21 Jul 5 '19 at 4:34

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