2
$\begingroup$

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

$\endgroup$

1 Answer 1

2
$\begingroup$

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).

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – Malkoun
    Commented Apr 15, 2023 at 16:44
  • $\begingroup$ @Malkoun take $-r^{\alpha}$ for any $\alpha \in (2-n,0)$. (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. $\endgroup$
    – Malkoun
    Commented 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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