Hello!!!

Let $(\star)\left\{\begin{matrix}

\Delta u=0 & \text{ in } B_R \\

u|_{\partial{B_R}}=\phi &

\end{matrix}\right.$.

Theorem: If $\phi \in C^0(\partial{B_R})$ then there is a unique solution of the problem $(\star)$ and $u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{\phi(\xi) dS}{|x- \xi|^n}$.

Proof: if $x_0 \in \partial{B_R}$ then it has to hold that $\lim_{x \to x_0 } u(x)=\phi(x_0)$.

$P(x, \xi)=\frac{R^2-|x|^2}{w_n R |x-\xi|^n}$

It holds that $\int_{\partial{B_R}} P(x, \xi) dS=1$.

$$\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} P(x, \xi) (\phi(\xi)-\phi(x_0)) dS\right| \leq \int_{\partial{B_R}} P(x, \xi) |\phi(\xi)-\phi(x_0)| dS= \int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS+\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS $$

Let $I_1=\int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$ and $I_2=\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$.

$\forall \epsilon >0 \exists \delta>0$ so that if $|\xi-x_0| \leq \delta$ then $|I_1|< \frac{\epsilon}{2}$.

$I_2 \leq 2M \frac{R^2-|x|^2}{w_n R \delta^n} \int_{|\xi-x_0|=R} dS=2M \frac{R^2-|x|^2}{\delta^n} R^{n-2} \leq \frac{\epsilon}{2}$ if $x$ is near to $ x_0 $.

So we have the following: $\forall \epsilon>0 \exists \delta>0$ such that $|u(x)-\phi(x_0)| \leq \epsilon \Leftrightarrow \lim_{x \to x_0} u(x)=\phi(x_0)$.

First of all, why does it hold that $\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} P(x, \xi) (\phi(\xi)-\phi(x_0)) dS\right| $ and not $\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} ( P(x, \xi) \phi(\xi)-\phi(x_0) ) dS\right|$ ?

Also why does it hold that $\int_{\partial{B_R}} P(x, \xi) |\phi(\xi)-\phi(x_0)| dS= \int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS+\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$ ?

Could you also explain to me how we get the inequalities for $I_1$ and $I_2$ ?