If $u>0$ in $\mathbb{R}^n\backslash\{0\}$ ($n\geq 2$) and $\Delta u>0$ in $\mathbb{R}^n\backslash\{0\}$, is it true that $\liminf_{y\rightarrow 0}u(y)>0$?
1 Answer
Yes, this is true. Let me consider $v=u$ for convenience, so $v$ is subharmonic and negative. For a subharmonic function bounded from above, an isolated point is removable, so v is actually subharmonic in $R^n$, if we define $v(0)=\limsup_{x\to 0} v(x).$ Now since it is strictly negative, we have by the average property $$v(0)\leq c_n\int_{x=1} v(x)dx<0,$$ that is $\liminf_{x\to 0}u(x)>0.$
Reference for removable singularity:
Carleson, Lennart Selected problems on exceptional sets. Zbl 0189.10903 Princeton, N.J.Toronto, Ont.London: D. Van Nostrand Co., Inc. V, 151 p. (1967).

$\begingroup$ Many thanks! I‘m not sure about the details of proofs in the removable of the isolated point, would you give more details. Thanks again! $\endgroup$ Commented Apr 15, 2023 at 14:48

$\begingroup$ By the way, is there such a function (strictly subharmonic and negative)? I cannot easily picture it. $\endgroup$– MalkounCommented Apr 15, 2023 at 16:44

$\begingroup$ @Malkoun take $r^{\alpha}$ for any $\alpha \in (2n,0)$. (The subharmonic extension is allowed to take $\infty$ as a value.) $\endgroup$ Commented Apr 15, 2023 at 17:11

$\begingroup$ Thank you @WillieWong. What about the case $n = 2$ please? Yes, I had forgotten that $\infty$ was allowed... Thank you for that. $\endgroup$– MalkounCommented Apr 15, 2023 at 18:26

$\begingroup$ @Malkoun: when $n = 2$ Liouville's theorem states that any subharmonic function that is bounded above is constant. $\endgroup$ Commented Apr 15, 2023 at 20:05