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Let $ \displaystyle{ U \subset \mathbb R ^n }$ open, bounded and connented.

If $ u \in C(U) \cap C(\bar U)$ such that:

$$ \Delta u =0 \quad \text { in U}$$

$$u=g \geq 0 \quad \text {in} \quad \partial U $$

and $ g >0 $ on a point in $ \partial U$ then $ u>0 $ in whole $U$.

I know that I have to use the maximum principle someway but for some reason I cant't solve it.

Any help ?

Thank's in advance!

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    That doesn't seem right. $u(x) = x_1$ on $B_1(0)$ is positive on some point of $\partial U$, but not on all.2012-10-12
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    @martini: I made a typo. It is $ g \geq 0 $and $g \neq 0$2012-10-12

2 Answers 2

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(After the edit): Suppose you had $u(x) \le 0$ for some $x \in U$. Then, as $-u$ is harmonic in $U$ by the maximum principle: \[ 0 \le \sup_{U} -u = \max_{\partial U} -u = \max_{\partial U} -g \le 0 \] So $\sup_U -u = -u(x) = 0$, and again by the maximum principle $u = 0$, which contradicts $g \ne 0$ and $u \in C(\bar U)$.

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    why it is $ 0 \leq \sup_{U} (-u) $?2012-10-12
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    Since there is $x \in U$ with $u(x) \le 0$, we have \\[ \sup_U (-u) \ge -u(x) \ge 0 \\]2012-10-12
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    O.K it was obvious...sorry... Thank you very much for your time!2012-10-12
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I dont think this is true. Take the function $x^2-y^2$ defined on the square $[-1,1]\times [-1,1]$.

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    At the point $(0,1)$ of the boundary of your domain, the function takes the negative value $-1$. The question asks that $u|_{\partial U} \geq 0$.2012-10-12
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    Because he edited the question... Take a look in first post2012-10-12