Dirichlet problem for Laplace equation as follows
\begin{eqnarray}
\Delta{u}&=&0\text{ in }B_r(0)\\
u&=&g\text{ on }\partial B_{r}(0),
\end{eqnarray}
where $ g $ is continuous.
It is already known that $ u(x)=C_n\int_{\partial B_{r}(0)}\frac{r^2-\lvert x\rvert^2}{\lvert x-y\rvert^n}g(y)dS(y) $ by the construction of Green function in balls, where $ C_n $ is a constant depending only on $ n $. From this formula, we can see that $ u(x)\in C^{\infty}(B_r(0)) $ and $ u(x)\rightarrow g(\xi) $ if $ x\rightarrow \xi $ with $ \xi\in\partial B_r(0) $. I am thinking about the problem when $ g $ is not continuous but $ g\in L^2(\partial B_{r}(0)) $. In that case the integration $ u(x)=C_n\int_{\partial B_{r}(0)}\frac{r^2-\lvert x\rvert^2}{\lvert x-y\rvert^n}g(y)dS(y) $ still makes sense and $ u(x) $ is still smooth. Can I say that $ u(x) $ actually solves the Dirichlet problem? In other words, can I show that for a.e. $ \xi\in\partial B_{r}(0) $, $ u(x)\rightarrow g(\xi) $ if $ x\rightarrow \xi $?
Best Answer
This is clearly false as stated, since a necessary condition is that $g(x)\to g(\xi)$ as $x\to\xi$ a. e. in the sphere, but if $g$ is merely $L^2$, this may well fail for every point.
The convergence holds in a weaker sense. For example, it is true that if $g_\rho(x)=g(\rho x)$ for $x\in\partial B_r(0)$ and $\rho<1$, then $g_\rho\to g$ a. e. and in $L^2$ as $\rho\to 1.$ Moreover, non-tangential limits (i. e., limits when $x\to\xi$ staying in the cone $\mathrm{dist}(x;[0,\xi])<\alpha |x-\xi|$ for some fixed $\alpha<1$) exist almost everywhere. See, for example, Axler, Boudron, Ramey, "Harmonic function theory."