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Let U=$R_{+}^2$, $u \in C^2(U) \cap C(\overline U)$ with $\Delta u=0$ in U. If in addition, u is bounded above on U, prove that: $sup_{U} u=sup_{\partial U} u$.

we can apply the maximum principle to the function $u(x_{1},x_{2})- \epsilon \ln \sqrt {x_{1}^2+(x_{2}+1)^2}$($\epsilon \gt 0$) on the set $U \cap$ $\brace(x_{1},x_{2}): x_{1}^2 +(x_{2}+1)^2 \lt a^2$ for large $a \gt 0$. Let $\epsilon$ approach zero.

the given set is open and bounded, so the maximum of the function will be achieved on the boundary.

Yang
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First, you show that $v_\epsilon$ is harmonic function.

Choose r such that

$$\sup u - u(0,0)-\epsilon \log r<0.$$

Drawing the set $$U_r= U\cap \{(x_1,y_1): x_1^2+(x_2+1)^2<r\}$$, you will see that its boundary contains two parts: Straight line(L) and Circle(C).

Consider $v_\epsilon$ on $U_r$, by max principle, $v_\epsilon$ has maximum on boundary of $U_r$.

By this $r$, we can show that the value of $v_\epsilon$ on Circle is smaller than $v_\epsilon(0,0)$.

That means

$$\max_{U_r}v_\epsilon =\max_{\partial U_r} v_\epsilon =\max_L v_\epsilon \geq v(0,0)> \max_Cv_\epsilon.$$

This is hold for all $R>r$. By take $r\to \infty$, we have $$\sup_U v_\epsilon = \sup_{\partial U}v_\epsilon.$$

Then take $\epsilon \to 0$, we are done.

Worawit Tepsan
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