We can write:
$$|I_2|=\left|\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS\right| \\
\le \sup_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} |P(x, \xi)| \cdot \sup_{\xi \in \partial B_R,\ |\xi-x_0| > \delta} |\phi(\xi)-\phi(x_0)| \cdot \left|\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS\right|
$$
How might we continue?
How can we get a relation for the integral $\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS$ ?
Also we have that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$. Does this imply that $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds=1$ ?
Don't we have that:
$$\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS \le \int_{\xi \in \partial B_R}dS = w_n R^{n-1}$$
Btw, I don't understand yet why it's supposed to hold that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.
Do you?
Is it somewhere in your notes?
Anyway, if it holds, then, since $P(x,\xi) > 0$, we have:
$$\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds \le \int_{\partial{B_R}} P(x,\xi) ds_{\xi} = 1$$
It seems we don't need it though.
Oh yes, right.
We have that if $u \in C^2(B_R(0)) \cap C^1(\overline{B_R}(0))$ is harmonic in $B_R(0)$, then the following formula holds:
$u(x)=\frac{R^2-|x|^2}{w_n R} \int_{\partial{B_R}} \frac{u}{|x-\xi|^n} ds_{\xi}$.
For $u=1$ we get that $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.
So then we have $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| dS \leq \sup_{\xi \in \partial{B_R}, |\xi-x_0|> \delta} |\phi(\xi)-\phi(x_0)| \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} P(x, \xi) dS \leq 2M \int_{\partial{B_R}} P(x,\xi) dS=2M $
So we have to pick $4M$ as $\epsilon$ ?
We pick $|x-x_0|$ to be sufficiently small.
I think that I understood the inequality of my notes.
Since $\phi$ is bounded, there is a $M>0$ such that $|\phi| \leq M$.
So $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi) |\phi(\xi)-\phi(x_0)| ds \leq 2M \int_{|\xi-x_0|>\delta, \xi \in \partial{B_R}} \frac{R^2-|x|^2}{w_n R |x-\xi|^n} ds < 2M \frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1}, \text{ since } |\xi-x_0|> \delta$.
Since $|x-x_0|$ is sufficiently small , $|x| \to |x_0|=|R|$ and so $\frac{R^2-|x|^2}{R} \frac{1}{\delta^n} R^{n-1} \to 0$ and so it is $\leq \frac{\epsilon}{2}$.