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Let $u_k$ be continuous on $\overline \Omega$ and harmonic in $\Omega$. Suppose $u_k$ converges uniformly on $\partial \Omega$. Can we conclude that $u_k$ converges uniformly on $\Omega$?

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  • $\begingroup$ You need some assumptions on $\Omega$. On a ball, the Poisson's kernel gives you a way to go. $\endgroup$ May 25, 2015 at 22:33
  • $\begingroup$ @GiuseppeNegro In class, the teacher says by maximum principle. I don't know how it follows from the maximum principle. $\endgroup$
    – Sherry
    May 25, 2015 at 22:36

1 Answer 1

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Careful: This is not true for any $\Omega.$ For example let $u_k(x,y) = (-1)^ky$ in the upper half plane $\Omega $ of $\mathbb {R}^2.$ Then all $u_k$ vanish on $\partial \Omega,$ hence converge uniformly there, but the $u_k$ converge nowhere in $\Omega.$

This will be true for all bounded $\Omega$ however. There it's just the maximum principle:

$$\sup_{\overline {\Omega}} |u_k-u_j|\le \sup_{ \partial \Omega}|u_k-u_j|.$$

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