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    #1
    Hello!!!

    We say that the space $\Omega$ satisfies the exterior sphere condition at the point $x_0 \in \partial{\Omega}$ if there is a $y \notin \overline{\Omega}$ and a number $R>0$ such that $\overline{\Omega} \cap \overline{B_y(R)}=\{ x_0 \}$.

    Let the function $\phi \in C^0(\partial{\Omega})$. Let $F_{\phi}$ the set that contains the " under functions of $\phi$ in $\Omega$ ", i.e. the functions $v \in C^0(\overline{\Omega})$ that are subharmonic in $\Omega$ and in $\partial{\Omega}$ it holds that $v|_{\partial{\Omega}} \leq \phi$.


    (We say that the function $v$ is subharmonic in $\Omega$ if for each ball $B \subset \Omega$ it holds that $v \leq H_B[v]$, where $H_B[v](x)=\left\{\begin{matrix}
    \text{ harmonic for } x \in B \\
    v(x) \text{ for } x \in \Omega \setminus{B}
    \end{matrix}\right.$)



    Theorem: If $u(x)=\sup_{v \in F_{\phi}} v(x)$ and at the point $x_0$ the space $\Omega$ satisfies the exterior sphere condition, then $\lim_{x \to x_0} u(x)=\phi(x_0), x \in \Omega$.

    Proof: Let $n \geq 3$. We define the function $h(x)=\frac{1}{R^{n-2}}-\frac{1}{|x-y|^{n-2}}$.

    Obviously $h(x_0)=0$ and $h(x)>0$ in $\overline{\Omega} \setminus{\{ x_0\}}$. Since $\phi$ is continuous we have that $\forall \epsilon>0 \ \exists \delta>0$ such that

    $|\phi(x)-\phi(x_0)|< \epsilon$ for $|x-x_0|< \delta$.

    Also there is a constant $C>0$ such that $Ch(x)>2M$ for $|x-x_0| \geq \delta$ since $h(x)>0$ in $\overline{\Omega} \setminus{ \{ x_0 \}}$, here $M=\sup \phi$.

    • Could you explain to me why there is a $C$ such that $Ch(x)>2M$ ?


    In order to prove the theorem it suffices to show that $|u(x)-\phi(x_0)| \leq \epsilon+ Ch(x)$ since $Ch(x) \to 0$ while $x \to x_0$.

    So we want the folllowing inequality to hold.

    $\phi(x_0)-\epsilon-Ch(x) \leq u(x) \leq \phi(x_0)+ \epsilon+ Ch(x), \forall x \in \Omega$


    The first inequality follows from the fact that $\phi(x_0)- \epsilon-Ch(x) $ is an under function.
    Indeed, this function is harmonic and so also subharmonic and $\phi(x_0)-\epsilon-C h(x)|_{\partial{\Omega}} \leq \phi(x), \forall x \in \partial{\Omega}$.

    • I tried to show that the function is harmonic, but I got that $\Delta h=2(n-2) \frac{n-2(n-3) |x-y|}{|x-y|^{n-3}}$. Have I done something wrong? Also how do we get the inequality?
      Don't we have for $|x-x_0| \leq \delta$ that $\phi(x_0)- \epsilon> \phi(x)$ ? Or am I wrong?



    The second inequality holds since $\phi(x_0)+ \epsilon+C h(x)$ is harmonic and $u|_{\partial{\Omega}} \leq \sup \phi \leq \phi(x_0)+ Ch(x)|_{\partial{\Omega}}$.

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    #2
    Quote Originally Posted by evinda View Post
    Also there is a constant $C>0$ such that $Ch(x)>2M$ for $|x-x_0| \geq \delta$ since $h(x)>0$ in $\overline{\Omega} \setminus{ \{ x_0 \}}$, here $M=\sup \phi$.

    • Could you explain to me why there is a $C$ such that $Ch(x)>2M$ ?
    Hey evinda!!

    Can't we pick for instance:
    $$C = \frac{2M + 1}{\inf_{\xi \in\overline{\Omega} \setminus{ \{ x_0 \}}} h(\xi)}$$

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