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Let $B=\{x \in \mathbb{R}^n : |x|<1\}$ and $u,h \in C^2(B) \cap C^0(\overline{B})$ such that $\Delta u \ge 0=\Delta h $ in $B$, $ u \le h $ on $\partial B$ and $u \ge 0=h(0)$ in $B$. Which value does $u(1,0,\dots,0)$ attend?

$h$ is harmonic and $u$ is subharmonic in $B$. The function $u-h$ is subharmonic and $u-h \le 0$ in $\partial B$.

Case 1: Assume it exists $x_0 \in B: (u-h)(x_0):= max_{x \in \overline{B}} (u-h) (x) \Rightarrow u-h $ constant after the strong max principle. Since $u-h \le 0 \Rightarrow c \le 0$ but on the other hand $c=(u-h)(0)=u(0)\ge 0$ and so $c=0$.

Case 2: Otherwise (it doesn't exists such a $x_0$) we have $u-h<0$ in $B$.

In both cases $0 \le u \le h$ in B and $h(0)=0$ a mimimum point in the iterior. Since h is continous up to the boundary we also have $0 \le h $ in $\overline{B}$. So $h\equiv 0 $.

In the first case: $u(1,0,...,0)=h(1,0,..,0)=0$

In the second case: $u=u-0=u-h <0 $ in B. So we have a contradiction to $u \ge 0$ in B.

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    Use the strong maximum principle instead of weak - either the inequality is strict or it is an equality. – 2017-01-08
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    Ok, but how you can calculate with this information the value $u(1,0,..0)$? – 2017-01-08
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    If you already know that $u=h$ is a nonnegative harmonic function and equals zero at some interior point, then it is zero everywhere. – 2017-01-08
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    But for this we also need that $h=u\ge 0$ on $\partial B$, we only have this information in B. – 2017-01-09
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    The functions are continuous up to the boundary. – 2017-01-10
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    I editet all your suggestions. Is the proof now correct? – 2017-01-10

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