0
$\begingroup$

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 $V$)

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.