Approximation argument for Poisson integral formula

Question

In the PDE book,

for a harmonic function $u \in C^2(B_R(0))\cap C^1(\overline B_R(0)),$ we have the following Poisson integral formula $$ u(y)=\frac{R^2-|y|^2}{n w_n R}\int_{\partial B_R(y)}\frac{u(x)}{|x-y|^n}ds_x $$ An approximation argument shows that the Poisson integral formula continues to hold for $u \in C^2(B_R(0))\cap C(\overline B_R(0)).$

What is the approximation argument? For a harmonic function $u \in C(\overline B_R(0))$, is there a sequence of harmonic functions $u_k \in C^1(\overline B_R(0))$ such that $u_k \to u$ in $C(\overline B_R(0))$? I have no idea for harmonic functions. May I use convolution?

Please let me know if you have any hint or comment for it. Thanks in advance!

Answer 1

A typical way to approximate a function defined on a ball $B_R(0)$ is $$u_k(x) = u(\lambda_k x),\quad \lambda_k\nearrow 1$$ For any $u \in C(\overline B_R(0))$, this results in $u_k \to u$ in $C(\overline B_R(0))$ by virtue of the uniform continuity of $u$. If $u$ is smooth (harmonic, holomorphic...) in the open ball $B_R(0)$, then $u_k$ has these properties on a larger ball.

Answer 2

There is a sequence of polynomials converging to $u$ in $C(\overline {B}_R(y))$ by Stone Wierstrass Theorem.

Answer 3

For $t\in(0,1)$, consider $\lambda: \overline{B_r(0)}\rightarrow B_r(0)$, $\lambda(y)=ty$, then $\lambda\in C^{\infty}(\overline{B_r(0)})$. As $u\in C^2(B_r(0))$ follows that $\hat{u}(y)=u(ty)\in C^2(\overline{B_r(0)})$. Moreover $$ \frac{u(ty)}{|x-y|^n}\rightarrow \frac{u(y)}{|x-y|^n}, \text{ as } t\rightarrow 1, \text{ and } \frac{|u(ty)|}{|x-y|^n} \leq \frac{\displaystyle\max_{\overline{B_r(0)}}|u|}{|x-y|^n}\in L^1(\partial B_r(0)). $$ Therefore, from the dominated convergence theorem, for all $y\in B_r(0)$, \begin{eqnarray*} u(y)&=&\lim\limits_{t\rightarrow 1}u(ty) =\lim\limits_{t\rightarrow 1}\hat{u}(y)\\ &=&\frac{r^2-|y|^2}{nw_nr} \lim\limits_{t\rightarrow 1}\int_{\partial B_r(0)}\frac{u(tx)}{|x-y|^n}ds_x =\frac{r^2-|y|^2}{nw_nr} \int_{\partial B_r(0)}\lim\limits_{t\rightarrow 1}\frac{u(tx)}{|x-y|^n}ds_x\\ &=&\frac{r^2-|y|^2}{nw_nr} \int_{\partial B_r(0)}\frac{u(x)}{|x-y|^n}ds_x. \end{eqnarray*}