Let $V$ be a bounded open set in $\mathbb{R}^{n}$ with $n\geq2$ and $W$ be an open neighborhood of the boundary $\partial V$ of $V$. If $u$ is superharmonic on $W$, is there a way to extend $u$ to a function $\tilde{u}$ that is superharmonic on $\complement V\cup W$ and is equal to $u$ at least on $\overline{V}\cap W$? ($\complement$ means the complement in $\mathbb{R}^{n}$ and $\overline{V}$ is the closure of $V$)

1$\begingroup$ See Theorem 2.18 in amazon.com/… to this end. $\endgroup$– user64494Dec 27, 2019 at 8:35
1 Answer
In general, this is not possible. Consider the case $n=2$ take the unit disk for $V$, and some ring, for example $1/2<z<2$ for $W$. Function $u(z)=\logz$ is harmonic in $W$ but cannot be extended from any neighborhood of the unit circle to the closure of the unit disk as a superharmonic function. The obstacle is clear: $$\int_{z=1} \frac{\partial u}{\partial n} ds=\pi>0,$$ where $\partial/\partial n$ is the differentiation along the outer normal. While for superharmonic functions in the disk this integral is always $\leq 0$, which follows from the Green formula $$\int\int_V\Delta u dxdy=\int_{\partial V}\frac{\partial u}{\partial n} ds.$$ Another reason why such an extention may be impossible is that your function can blow up to $\infty$ at a boundary point of $W$ which is interior for $V$. For example, with the same $U,W$ an extension of $\logz1/22\logz$ to the unit disk is evidently impossible, though the first obstacle does not exist for this function.
A correct extension theorem of the type that you propose would be something like this: if $$\int_{\partial V}\frac{\partial u}{\partial n}ds<0$$ then $u$ extends from SOME neighborhood $W_1\subset W$ of $\partial V$ to $V$ but in general $W_1\cap V$ will be smaller than $W\cap V$.
This is just a preliminary statement requiring some smoothness assumptions on $u$ and $V$, because in general neither $\partial u/\partial n$ nor $ds$ nor $\partial/\partial n$ are defined.