Let $u_k$ be continuous on $\overline\Omega$, $u_k$ harmonic in $\Omega$. Suppose $u_k|\partial\Omega$ converge uniformly. Then $u_k$ converge uniformly in $\Omega$.
The hint is using Maximum Principle.
My attempt:
By Poisson Integral foumula:
$u_k(x)=\int_{\partial\Omega}\frac{\partial G}{\partial n}(x,y)u_k(y)ds(y)$. Then let $k$ goes to infinity. The RHS is harmonic so $u(x)$ is harmonic in $\Omega$.
My question: I didn't use maximum principle.
I have been thinking about it for several hours and I didn't figure a thing about maximum principle？
Can anyone help me with some answers or clues? That will be really helpful!
Thanks a lot!:)
Since $u_k|_{\partial \Omega}$ is uniformly convergent it is uniformly Cauchy: if $$M_{j,k} = \max_{x \in \partial \Omega} |u_j(x) - u_k(x)|$$ then $M_{j,k} \to 0$ as $j,k \to \infty$.
Given two indices $j,k$ the difference $u_j - u_k$ is harmonic. For any $x_0 \in \Omega$ you have by the maximum principle $$u_j(x_0) - u_k(x_0) \le \max_{x \in \overline \Omega} (u_j(x) - u_k(x)) = \max_{x \in \partial \Omega} (u_j(x) - u_k(x)) \le M_{j,k}.$$ Interchange the roles of $j$ and $k$ to obtain also $$u_k(x_0) - u_j(x_0) \le M_{j,k}.$$ Thus $|u_j(x_0) - u_k(x_0)| \le M_{j,k}$ for all $x_0 \in \Omega$. This means that $\{u_k\}$ is uniformly Cauchy in $\Omega$, hence uniformly convergent.