# Extension of superharmonic functions

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 $$V$$)

• See Theorem 2.18 in amazon.com/… to this end. Dec 27, 2019 at 8:35

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.