# Mollifier of a subharmonic function

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?

• I think yes... Shouldn't the same proof work? what's the issue – mathworker21 Jul 5 '19 at 1:11
• 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? – M. Rahmat Jul 5 '19 at 4:11
• 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 $\phi_n$, $x-t \not \in F$. – mathworker21 Jul 5 '19 at 4:18
• 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? – M. Rahmat Jul 5 '19 at 4:27
• 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... – mathworker21 Jul 5 '19 at 4:34