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    #1
    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 $I_2$ ?

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    #2
    Hey evinda!!

    Quote Originally Posted by evinda View Post
    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 $\left| u(x)- \phi(x_0)\right|=\left| \int_{\partial{B_R}} ( P(x, \xi) \phi(\xi)-\phi(x_0) ) dS\right|$ ?
    Isn't:
    $$\int_{\partial{B_R}} \phi(x_0)dS = w_nR^{n-1}\phi(x_0) \ne \phi(x_0)$$


    While:
    $$\phi(x_0) = u(x_0) = \int_{\partial{B_R}} P(x_0,\xi) \phi(x_0)dS
    $$

    Quote Quote:
    Also why does it hold that $\int_{\partial{B_R}} P(x, \xi) |\phi(\xi)-\phi(x_0)| dS= \int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS+\int_{|\xi-x_0| > \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS$ ?
    We can split up the domain of an integral into 2 subsets that combine together to the domain.
    Compare for instance:
    $$\int_0^1 f(x)dx = \int_0^\delta f(x)dx + \int_\delta^1 f(x)dx$$
    Note that we're still constrained to $\partial{B_R}$.

    Quote Quote:
    Could you also explain to me how we get the inequalities for $I_1$ and $I_2$ ?
    Let's start with $I_1$.

    $$|I_1|=\left|\int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS\right| \\
    \le \sup_{|\xi-x_0| \leq \delta} |P(x,\xi)| \cdot \sup_{|\xi-x_0| \leq \delta} |\phi(\xi)-\phi(x_0)| \cdot \int_{\partial{B_R}}dS \\
    \le M \cdot \epsilon' \cdot w_nR^{n-1} = \frac\epsilon 2
    $$
    where $\epsilon'$ is introduced since $\phi$ is given to be continuous.
    So given an $\epsilon$ we can find an $\epsilon'$ and from the continuity of $\phi$, we can find the appropriate $\delta$.

    Note that we can only guarantee the bound on $|P(x, \xi)|$ if $x\notin \partial{B_R}$, which is given since $x\in B_R$.

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    #3 Thread Author
    Quote Originally Posted by I like Serena View Post
    Isn't:
    $$\int_{\partial{B_R}} \phi(x_0)dS = w_nR^{n-1}\phi(x_0) \ne \phi(x_0)$$


    While:
    $$\phi(x_0) = u(x_0) = \int_{\partial{B_R}} P(x_0,\xi) \phi(x_0)dS
    $$
    Oh yes, right...

    Quote Originally Posted by I like Serena View Post
    We can split up the domain of an integral into 2 subsets that combine together to the domain.
    Compare for instance:
    $$\int_0^1 f(x)dx = \int_0^\delta f(x)dx + \int_\delta^1 f(x)dx$$
    Note that we're still constrained to $\partial{B_R}$.
    Ok...

    Quote Originally Posted by I like Serena View Post

    Let's start with $I_1$.

    $$|I_1|=\left|\int_{|\xi-x_0| \leq \delta} P(x, \xi)|\phi(\xi)-\phi(x_0)| dS\right|
    \le \sup_{|\xi-x_0| \leq \delta} |P(x,\xi)| \cdot \sup_{|\xi-x_0| \leq \delta} |\phi(\xi)-\phi(x_0)| \cdot \int_{\partial{B_R}}dS
    \le M \cdot \epsilon' \cdot w_nR^{n-1} = \frac\epsilon 2
    $$
    where $\epsilon'$ is introduced since $\phi$ is given to be continuous.
    I see.

    Quote Originally Posted by I like Serena View Post
    So given an $\epsilon$ we can find an $\epsilon'$ and from the continuity of $\phi$, we can find the appropriate $\delta$.
    Don't we have from the continuity of $\phi$ that $\forall \epsilon'>0 \exists \delta>0$ such that if $|\xi-x_0| \leq \delta$ then $|\phi(\xi)-\phi(x_0)| \leq \epsilon'$ and then we set $\epsilon=2M \epsilon' w_n R^{n-1}$ ? Or am I wrong?

    How can find the appropriate $\delta$?

    Quote Originally Posted by I like Serena View Post

    Note that we can only guarantee the bound on $|P(x, \xi)|$ if $x\notin \partial{B_R}$, which is given since $x\in B_R$.
    Could you explain to me further why $|P(x, \xi)|$ is only bounded if $x\notin \partial{B_R}$?

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    #4
    Quote Originally Posted by evinda View Post
    Don't we have from the continuity of $\phi$ that $\forall \epsilon'>0 \exists \delta>0$ such that if $|\xi-x_0| \leq \delta$ then $|\phi(\xi)-\phi(x_0)| \leq \epsilon'$ and then we set $\epsilon=2M \epsilon' w_n R^{n-1}$ ? Or am I wrong?

    How can find the appropriate $\delta$?
    We just did.

    For any $\epsilon>0$ we define $\epsilon'=\frac{\epsilon}{2M w_n R^{n-1}}$.
    From the continuity of $\phi$ we know that for this $\epsilon'>0$ we can find a $\delta>0$, such that if $|\xi-x_0| <\delta$ then $|\phi(\xi)-\phi(x_0)| < \epsilon'$.
    It follows that for this same $\delta$ we now also have $|I_1|<\frac\epsilon 2$.


    Quote Quote:
    Could you explain to me further why $|P(x, \xi)|$ is only bounded if $x\notin \partial{B_R}$?
    Because $P(x,\xi)=\frac{R^2-|x|^2}{w_nR |x-\xi|^n}$, which is not defined if $x=\xi$.
    But that can't be because $x$ is an element of the interior of the ball, and $\xi$ is on the boundary of the ball.

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    #5 Thread Author
    Quote Originally Posted by I like Serena View Post
    We just did.

    For any $\epsilon>0$ we define $\epsilon'=\frac{\epsilon}{2M w_n R^{n-1}}$.
    From the continuity of $\phi$ we know that for this $\epsilon'>0$ we can find a $\delta>0$, such that if $|\xi-x_0| <\delta$ then $|\phi(\xi)-\phi(x_0)| \leq \epsilon'$.
    It follows that for this same $\delta$ we now also have $|I_1|<\frac\epsilon 2$.
    I understand.

    Quote Originally Posted by I like Serena View Post
    Because $P(x,\xi)=\frac{R^2-|x|^2}{w_nR |x-\xi|^n}$, which is not defined if $x=\xi$.
    But that can't be because $x$ is an element of the interior of the ball, and $\xi$ is on the boundary of the ball.

    Do we assume that $x$ is an element of the interior of the ball, and $\xi$ is on the boundary of the ball? Or how do we know it?

    I thought that we would have at the boundary $|\xi-x|=R$. So do we have $|\xi-x|=0$ ?

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    #6
    Quote Originally Posted by evinda View Post
    Do we assume that $x$ is an element of the interior of the ball, and $\xi$ is on the boundary of the ball? Or how do we know it?
    The given formula for $u(x)$ has an integral in it that is not defined if $x$ is on the boundary.
    That's because it has $|x-\xi|$ in the denominator, which would become zero when $\xi=x$ ($\xi$ is always a point on the boundary).
    Therefore we have to conclude that $x$ is intended to be a point in the interior of the ball.


    Quote Quote:
    I thought that we would have at the boundary $|\xi-x|=R$. So do we have $|\xi-x|=0$ ?
    $x$ is an arbitrary point inside $B_R$, which is the ball with center $0$ and radius $R$, while $\xi$ is a point on the boundary.
    So we have $|\xi - 0|=R$ and $0<|\xi-x|<2R$.

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    #7 Thread Author
    Quote Originally Posted by I like Serena View Post
    The given formula for $u(x)$ has an integral in it that is not defined if $x$ is on the boundary.
    That's because it has $|x-\xi|$ in the denominator, which would become zero when $\xi=x$ ($\xi$ is always a point on the boundary).
    Therefore we have to conclude that $x$ is intended to be a point in the interior of the ball.

    I see.

    Quote Originally Posted by I like Serena View Post
    $x$ is an arbitrary point inside $B_R$, which is the ball with center $0$ and radius $R$, while $\xi$ is a point on the boundary.So we have $|\xi - 0|=R$ and $0<|\xi-x|<2R$.

    Ok. So we have

    $$\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|=\sup_{|\xi-x_0| \leq \delta} \left| \frac{R^2-|x|^2}{w_n R |x-\xi|^n} \right|=\frac{R}{w_n} \sup_{|\xi-x_0| \leq \delta} \frac{1}{|x-\xi|^n}$$

    Since $|x-\xi|>0$ we have that $\frac{1}{|x-\xi|^n} \to 0$ while $n \to +\infty$ and so $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)| \to 0$ and so it is bounded. Right?

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    #8
    Quote Originally Posted by evinda View Post
    Ok. So we have

    $$\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|=\sup_{|\xi-x_0| \leq \delta} \left| \frac{R^2-|x|^2}{w_n R |x-\xi|^n} \right|=\frac{R}{w_n} \sup_{|\xi-x_0| \leq \delta} \frac{1}{|x-\xi|^n}$$

    Since $|x-\xi|>0$ we have that $\frac{1}{|x-\xi|^n} \to 0$ while $n \to +\infty$ and so $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)| \to 0$ and so it is bounded. Right?
    Erm... shouldn't we pick $n$ to be some unknown constant? I thought it wasn't supposed to go to infinity. Or is it?

    We do have that there is some $\epsilon > 0$ such that $|x-\xi|>\epsilon >0$.
    That's because $x$ is supposed to be inside the ball, while $\xi$ is supposed to be on the boundary of the ball.
    Isn't therefore:
    $$\sup_{|\xi-x_0| \leq \delta} \frac{1}{|x-\xi|^n} < \frac{1}{\epsilon^n}$$

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    #9 Thread Author
    Quote Originally Posted by I like Serena View Post
    Erm... shouldn't we pick $n$ to be some unknown constant? I thought it wasn't supposed to go to infinity. Or is it?
    Oh yes, you are right.

    Quote Originally Posted by I like Serena View Post
    We do have that there is some $\epsilon > 0$ such that $|x-\xi|>\epsilon >0$.
    That's because $x$ is supposed to be inside the ball, while $\xi$ is supposed to be on the boundary of the ball.
    Isn't therefore:
    $$\sup_{|\xi-x_0| \leq \delta} \frac{1}{|x-\xi|^n} < \frac{1}{\epsilon^n}$$
    I see... But $\frac{1}{\epsilon^n}$ isn't bounded for any $\epsilon>0$. So how can we show that $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|$ is bounded?

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    #10
    Quote Originally Posted by evinda View Post
    I see... But $\frac{1}{\epsilon^n}$ isn't bounded for any $\epsilon>0$. So how can we show that $\sup_{|\xi-x_0| \leq \delta} |P(x,\xi)|$ is bounded?
    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.

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