limit involving harmonic function

Question

Let $u$ harmonic function in $\mathbb{R}^3 -\{0\}$. I know that $$\lim_{x\to0} \sqrt{|x|} \cdot u(x)=k< \infty$$

I'm trying to show that $k=0$. I tried by contradiction, but I failed and I'm beginning to suspect that $k$ may be different from $0$.

Answer 1

Using $\lim_{x\to0} \sqrt{|x|} \cdot u(x)=k< \infty$:

$$\lim_{x \to 0}|x|u(x)=\lim_{x\to0} \sqrt{|x|} \cdot \sqrt{|x|} \cdot u(x)=0 \cdot k=0$$

then $0$ is removable point. Let $v:\mathbb{R}^3 \to \mathbb{R}$ harmonic extension of $u $ so:

$$\lim_{x\to0} \sqrt{|x|} \cdot u(x)= \lim_{x\to0} \sqrt{|x|} \cdot v(x)=0 \cdot v(0)=0=k$$

Answer 2

Hint: Let $\epsilon>0$ and consider

$$u(x)-P[u|_S](x) + \frac{\epsilon}{|x|},\,\,\text{for }0<|x|<1.$$

Here $P$ is the Poisson integral for the unit ball and $S$ is the unit sphere. This function is harmonic on $\{0<|x|<1\}$ with boundary values $\ge 0.$