# Harmonic in punctured ball and bounded implies harmonic in ball.

If $$u$$ is harmonic and bounded in $$B_1(0)\setminus\{0\}$$, then can we say that $$u$$ is harmonic in $$B_1(0)$$? I believe the answer is yes and I think the way to show it is by the Mean Value Property... but there is a problem. For $$|x|<1/2$$ and $$|x|, how do we know that $$\frac{1}{|B_r|}\int_{B_r(x)} u(y) dy =u(x)$$? Since $$u$$ is harmonic in the punctured ball, the MVP holds only when $$B_r(x)\subset B_1(0)\setminus\{0\}$$. How does one get around this problem?

Note we are working in $$\mathbb{R}^n$$ for some $$n\geq 3$$

Define $$\newcommand{\dashint}{\mathchoice{\rlap{\,\,-}\int}{\rlap{\,-}\int}{\rlap{\ -}\int}{\rlap{\,-}\int}} \dashint_Af(x)\,\mathrm{d}x=\frac1{|A|}\int_Af(x)\,\mathrm{d}x$$
Using $$(1)$$ and $$(2)$$ from this answer and $$|S(r,p)|=\omega_{n-1}r^{n-1}$$, we get that \begin{align} &r_2^{n-1}\frac\partial{\partial r}\dashint_{S(r_2,p_2)}u(x)\,\mathrm{d}\sigma-r_1^{n-1}\frac\partial{\partial r}\dashint_{S(r_1,p_1)}u(x)\,\mathrm{d}\sigma\\ &=\frac1{\omega_{n-1}}\int_{B(r_2,p_2)\setminus B(r_1,p_1)}\Delta u(x)\,\mathrm{d}x\\[6pt] &=0\tag1 \end{align} where $$r_k\gt|p_k|$$ (the origin is inside both spheres).
Thus, for some constant $$C$$, independent of $$p$$ (as long as $$r\gt|p|$$), $$r^{n-1}\frac\partial{\partial r}\dashint_{S(r,p)}u(x)\,\mathrm{d}\sigma=C\tag2$$ Therefore, $$\dashint_{S(r,p)}u(x)\,\mathrm{d}\sigma =\left\{\begin{array}{} A(p)-\frac{C}{(n-2)\,r^{n-2}}&\text{if }n\ge3\\ A(p)+C\log(r)&\text{if }n=2 \end{array}\right. \tag3$$ If $$u$$ is bounded, then by considering $$p=0$$, we get $$C=0$$, and therefore, for $$r\gt|p|$$, $$\dashint_{S(r,p)}u(x)\,\mathrm{d}\sigma=A(p)\tag4$$ Since $$u$$ is harmonic away from $$0$$, the Mean Value Property says that for $$r\lt|p|$$, $$\dashint_{S(r,p)}u(x)\,\mathrm{d}\sigma=u(p)\tag5$$ However, because $$u$$ is smooth away from $$0$$, and bounded, $$\dashint_{S(r,p)}u(x)\,\mathrm{d}\sigma$$ is a continuous function of $$r$$. That is, $$A(p)=u(p)$$ and we have $$(5)$$ for all $$r$$.
The function $$u$$ is harmonic on a neighborhood of $$0$$ with $$u(x) |x|^{n-2} \to 0$$ as $$x\to 0$$, so it extends to a harmonic function across $$0$$. (This is a standard result about harmonic functions, but the proof is basically solving the Dirichlet problem on a small punctured ball around $$0$$ with boundary value $$u$$, then using the maximum principle and the condition above to ensure that this new solution must just be $$u$$ itself.)