4
$\begingroup$

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 $|x|<r<1/2$, 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 $n\geq 3$

$\endgroup$
4
$\begingroup$

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 $r$.

$\endgroup$
3
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ Ok...We have not learned about that yet. Is there a way of solving this without using this result (or without proving it?) $\endgroup$ – Adam Martens Sep 15 at 2:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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