# Super harmonic function

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 $$\liminf_{|y|\rightarrow 0}u(y)>0$$?

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 $$\liminf_{x\to 0}u(x)>0.$$
• @Malkoun take $-r^{\alpha}$ for any $\alpha \in (2-n,0)$. (The subharmonic extension is allowed to take $-\infty$ as a value.) Commented Apr 15, 2023 at 17:11
• Thank you @WillieWong. What about the case $n = 2$ please? Yes, I had forgotten that $-\infty$ was allowed... Thank you for that. Commented Apr 15, 2023 at 18:26
• @Malkoun: when $n = 2$ Liouville's theorem states that any subharmonic function that is bounded above is constant. Commented Apr 15, 2023 at 20:05