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.



Modifying a superharmonic function on a neighborhood of infinity
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.


