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I believe that the following result is true:

Let $u,v \in H^{1}(\Omega)$, where $\Omega$ is open bounded set of $R^{n} $. Supoose that $u,v$ is harmonic in $\Omega$ in the weak sense, that is

$$ \int_{\Omega} \nabla u . \nabla \varphi \ dx = \int_{\Omega} \nabla v . \nabla \varphi \ dx = 0, \forall \varphi \in C^{\infty}_{0}(\Omega).$$

Supoose that $lim \ sup_{y \rightarrow x } u(y) \leq lim inf_{y \rightarrow x} v(y) $ for almost everywhere $x \in \partial \Omega$. Then $u\leq v $ in $\Omega$.

If the result is true someone can say to me a reference for a proof ?

I know that if $lim \ sup_{y \rightarrow x } u(y) \leq lim inf_{y \rightarrow x} v(y) $ hold for all $x \in \partial \Omega$, then $u\leq v $ in $\Omega$. But i am not finding in books a proof for the case when the inequality holds almost everywhere.

math student
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  • If we can show that $(u-v)_+\in H_0^1(\Omega)$ we are done. – Tomás Feb 25 '14 at 23:02
  • Where do you found this problem? – Tomás Feb 26 '14 at 00:03
  • you're right. I am studying a article and if my affirmation is true then it will help me a lot. I believe the affirmation is true. But i have no idea to how to prove this. (sorry for my english, my english is terrible) – math student Feb 26 '14 at 03:13
  • I don't think this is true. Assume for instance that $u,v\in C^2(\overline{\Omega})$ are harmonic functions and hence weakly harmonic functions. If there is some point $x$ in the boundary, where $u(x)>v(x)$ then $u(y)\ge v(y)$ in a neighbourhood of $x$. Therefore your statement cannot be true, because the point $x$ is a zero measure set. – Tomás Feb 26 '14 at 12:40

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