
> 
> 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 $?



Is the Poisson formula valid when the boundary condition is $ L^2 $?
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."


