# The singular point of harmonic function

If $B$ is a unit ball in $\mathbb R^n$, then can we find a harmonic function $u$ in $B\backslash \{ 0\}$ such that $\mathop {\lim \inf }\limits_{x \to 0} u = - \infty$ while $\mathop {\lim \sup }\limits_{x \to 0} u = + \infty$ ?

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## 1 Answer

Certainly. Consider $g(z)=e^{\frac{1}{z}}$ and let $u=\text{Re}(g)$. Since $g$ has an essential singularity at $0$ we know that $g(B_r(0))$ is all of $\mathbb{C}$ except possibly for one point $p$ (this is the so-called Big Picard Theorem). Then, we know that $u(B_r(0))=\mathbb{R}$ for every $r>0$. Thus, it's easy to see that $\displaystyle \limsup_{z\to 0}u(z)=\infty$ and $\displaystyle \liminf_{z\to 0}u(z)=-\infty$.

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@Martin Indeed. I looked at the level of the other OP's questions and assumed from them that Big Picard would not be out of there reach. But yes, certainly you can use Casorati-Weierstrass. – Alex Youcis Feb 7 '13 at 23:10
I think that you can do a similar construction in arbitrary dimension, putting a Kelvin transform where now there is an inversion. This might require a bit of work, though. – Giuseppe Negro Feb 7 '13 at 23:49