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Let $D$ a bounded coherent space in $2$ or $3$ dimensions.

Let $u: \overline{D} \rightarrow \mathbb{R}$ a function that is continuous at $\overline{D}$ and harmonic at $D$ $\Rightarrow C^2(D)$

Then, both the maximum and minimum of $u$ at $\overline{D}$ are taken at the boundary $\partial{D}$.

$$\max_{(x,y) \in \overline{D}} u(x,y)= \max_{(x,y) \in \partial{D}} u(x,y)$$ $$$$ The proof of this theorem is the following:

Since $u$ I harmonic $\Delta u=0 \Rightarrow \Delta (-u)=0$

It's sufficient to do only the proof for the maximum.

We define the function $v=u(x,y)+\epsilon (x^2+y^2) \in C^0(\overline{D}), C^2(D)$

$\Delta v=4 \epsilon >0$, for $\epsilon >0$

We suppose that $v$ achieves its maximum at $(x_0,y_0) \in D$: $v_x(x_0,y_0)=v_y(x_0,y_0)=0, v_{xx}(x_0,y_0)<0, v_{yy}(x_0,y_0)<0$ $\Rightarrow \Delta v=v_{xx}+y_{yy}<0$

We have a contradiction, so $v$ achieves its maximum at one point at the boundary $(x_0,y_0) \in \partial{D}$.

$u(x,y) \leq v(x,y) \leq v(x_0, y_0)= u(x_0, y_0)+ \epsilon (x_0^2+y_0^2)$

$x^2+y^2<l^2, \forall (x,y) \in \overline{D}$

$\Rightarrow u(x,y) \leq u(x_0,y_0)+\epsilon l^2$

$(x_0,y_0) \in \overline{D} \Rightarrow$ $$\max_{(x,y) \in \overline{D}} u(x,y) \leq \max_{(x,y) \in \partial{D}} u(x,y) + \epsilon l^2$$ Taking the limit $\epsilon \rightarrow 0$ we conclude to the relation : $$\max_{(x,y) \in \overline{D}} u(x,y) = \max_{(x,y) \in \partial{D}} u(x,y)$$

$$$$ I haven't understood how we get from the relation: $\max\limits_{(x,y) \in \overline{D}} u(x,y) \leq \max\limits_{(x,y) \in \partial{D}} u(x,y) + \epsilon l^2$ to the relation: $\max\limits_{(x,y) \in \overline{D}} u(x,y) = \max\limits_{(x,y) \in \partial{D}} u(x,y)$. Taking the limit $\epsilon \rightarrow 0$ we get: $\max\limits_{(x,y) \in \overline{D}} u(x,y) \leq \max\limits_{(x,y) \in \partial{D}} u(x,y)$. But how do we get the equality???

gebruiker
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Mary Star
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  • I performed a rollback on this question. If you have a seperate question then it should not be added as an edit to an existing question. Just ask it as a seperate question. – gebruiker Dec 13 '15 at 12:07

1 Answers1

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$\max\limits_{(x,y)\in\bar D}u(x,y)\leq\max\limits_{(x,y)\in\partial D}u(x,y)$ says that the maximum value of $u(x,y)$ on $\bar D$ can't be greater than the maximum value of $u(x,y)$ on $\partial D$. Now since $\partial D\subseteq\bar D$ this automatically means that the maximum is attained somewhere on $\partial D$. This is why you can write $\max\limits_{(x,y)\in\bar D}u(x,y)=\max\limits_{(x,y)\in\partial D}u(x,y)$. If you're not yet fully convinced by this intuitive argument, you can prove it by contradiction.

gebruiker
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    @MaryStar Let me know if you're having trouble when you're trying the proof by contradiction part. If neccessary I can make an edit... – gebruiker May 12 '14 at 14:16
  • Good!!! I understand!!! $$$$ I have also an other question... $$u(x,y) \leq v(x,y) \leq v(x_0, y_0)= u(x_0, y_0)+ \epsilon (x_0^2+y_0^2)$$ $$\text{ Let } l>0 \text{ such that } \forall (x,y) \in \overline{D} \text{ it is: } \sqrt{x^2+y^2} \leq l. \text{ Such a } l \text{ exists, because } D \text{ is bounded.}$$ What does this mean??? Could you explain it to me?? What is this $l$?? – Mary Star Jun 08 '14 at 19:28
  • @Mary $\sqrt{x^2+y^2}\leq l$ is the equation of a circle of radius $l$. And it exists since $D$ is a bounded region. That is: because $D$ is bounded, you can always find a circle such that $\overline D$ is contained within that circle. – gebruiker Jun 09 '14 at 09:19
  • @Mary Though I'm very happy and willing to answer you questions, it is better if you ask a question like this seperately i.e. not in the comments. It also is more helpfull to you that way, because I'm basically the only one that reads this comment. A question however is visible for everyone. More people that see your question = more and reaction and quicker reactions :) – gebruiker Jun 09 '14 at 09:24