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 $V$)



Extension of superharmonic functions
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 $u(z)=\log|z|$ 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 $\log|z-1/2|-2\log|z|$ 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.


