1
$\begingroup$

Let $u(x)=\alpha+\beta U(x)$, where $U(x)=|x|^{2-m}$ ($N\geq3$) is the fondamental harmonic function, $\alpha<0$ and $\beta>0$. We know that $u$ is superharmonic on $\mathbb{R}^m$ and harmonic on $\mathbb{R}^m\setminus\{0\}$. Let $a>0$ be a given constant. Is it possible to modify $u$ on a neighborhood of infinity to obtain a function $\overline{u}$ that is $=u$ for $|x|<R$, superharmonic on $\mathbb{R}^m\setminus\{0\}$ and such that $\lim \overline{u}(x)\geq a$, as $x\to\infty$, for some $R>0$?

I tried to find by "bricolage" of adding some constant here and subtracting some other there, but in vain.

$\endgroup$

1 Answer 1

1
$\begingroup$

One way to find out that this is impossible is to note that $v = \overline u - \beta U$ is superharmonic in $\mathbb R^n$, and hence $V(r) = \int_{\partial B_r} v$ decreases with $r$. Since $V(r) = \alpha < 0$ for $r$ small enough, we have $V(r) < 0$ for all $r$, and so $v$ cannot have a positive limit at infinity. It remains to note that $U$ goes to zero, and thus $v$ and $\overline u$ have equal limit at infnity.

$\endgroup$

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.