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While i am revising for exam : I am facing some problems to understand things clearly . Here are my doubts :

a) If $u$ solves $\Delta u =0 , x\in \Omega; u=g , x\in \partial \Omega$ for non constant boundary data $g$ with $g\ge0$ and $g(x_0)\> 0$ for some $x_0 \in \partial \Omega $ then why is it true that $u(x)>0$ for all $x\in \Omega$ ?

b) It's about Harnack inequality . Let $V$ be open and connected and $V\subset\subset \Omega$ ie $V$ is compactly contained in $\Omega$ then there existz a constant $C<\infty$ such that $sup_V u \le C inf_V u$ . My question is why only upto $V$ and not upto $\bar V $ ie why not upto closure ?

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Tangent: doubt and question are a bit different. –  anon Jul 11 '12 at 20:18
    
@anon : i tried looking for other possibilites but couldn't find the appropriate header. –  Theorem Jul 11 '12 at 20:21

1 Answer 1

up vote 2 down vote accepted

a) You are missing an important assumption that $g$ is continuous. Under this assumption $g(y)$ agrees with the limit of $u(x)$ as $x\to y\in\partial\Omega$. By the maximum principle $u\ge 0$ in $\Omega$, and the stronger form of the same principle says that either $u\equiv 0$ or $u>0$ everywhere. The first case is impossible, for then we would have $g(x_0)=\lim_{x\to x_0}u(x)=0 $.

b) You have my permission to replace $V$ with $\overline{V}$ in Harnack's inequality. It does not matter.

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Sir , thank you. –  Theorem Jul 11 '12 at 20:42

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