Let $V$ be a bounded open set in $\mathbb{R}^{n}$ with $n\geq2$, and $u$ a subharmonic function on an open set containing the closure $ \overline{V} $ of $V$. Suppose we have $u<M$ on $V$ ($M$ is a constant). Can we conclude that $u\leq M$ on $ \overline{V} $?
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$\begingroup$ Do you allow your subharmonic functions to attain the value $-\infty$? $\endgroup$ – Daniel Fischer♦ Jul 21 '17 at 15:07
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$\begingroup$ Yes, but cannot be identically $-\infty$. $\endgroup$ – M. Rahmat Jul 22 '17 at 0:52
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I have a partial answer to this question: if $u$ is continuous at each point of the boundary $\partial V$ of $V$, the answer is yes! In fact, we have by continuity of $u$ and the maximum principle, $$u(\zeta)=\limsup_{\substack{x\rightarrow\zeta\\(x\in V)}}u(x)\leq M$$ for all point $\zeta\in\partial V$, which proves the claim!
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$\begingroup$ This actually works for all functions that are continuous on the boundary, as $u(x)<M$ for all $x\in V$ by assumption. No need for the maximum principle. $\endgroup$ – Giuseppe Negro Jul 27 '17 at 11:29
