Originally Posted by I like Serena
Not for any $\epsilon>0$...
For any given $x$ in the inside of the ball there exists an $\epsilon>0$ such that for each $\xi$ on the boundary with $|\xi - x_0|\le\delta$ we have $|\xi - x|>\epsilon$.

So for every given $x$ in the inside of the ball $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|$ is bounded.
Ah I see...

Can you also explain to me how we get the inequality for $I_2$ ?

2. Originally Posted by evinda
$P(x, \xi)=\frac{R^2-|x|^2}{w_n R |x-\xi|^n}$

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

$\forall \epsilon >0 \exists \delta>0$ 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 $x_0$.
Originally Posted by evinda
Ah I see...

Can you also explain to me how we get the inequality for $I_2$ ?
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 $\int_{|\xi-x_0|> \delta, \xi \in \partial{B_R}} P(x,\xi)ds=1$ ?

4. Originally Posted by evinda
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.

Originally Posted by I like Serena
Don't we have that:
$$\int_{\xi \in \partial B_R,\ |\xi-x_0| > \delta}dS \le \int_{\xi \in \partial B_R}dS = w_n R^{n-1}$$
Oh yes, right.

Originally Posted by I like Serena
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?
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 $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.

Originally Posted by I like Serena
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.
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 $\epsilon$ ?

6. Originally Posted by evinda
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 $\int_{\partial{B_R}} P(x,\xi) ds_{\xi}=1$.
Ok!

Quote:
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 $\epsilon$ ?
I don't think so. We cannot make $4M$ as small as we want to.

Shouldn't we pick $\xi$ sufficiently close to $x_0$ for the proof?
That is, $|\xi-x_0|<\delta$?

Originally Posted by I like Serena
Ok!

I don't think so. We cannot make $4M$ as small as we want to.

Shouldn't we pick $\xi$ sufficiently close to $x_0$ for the proof?
That is, $|\xi-x_0|<\delta$?
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 $\leq \frac{\epsilon}{2}$.

8. Originally Posted by evinda
We pick $|x-x_0|$ to be sufficiently small.
Oh yes.

Quote:
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 $\leq \frac{\epsilon}{2}$.
Good!