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Suppose one is given the weak maximum principle for subharmonic functions, if $u \in C^2(\Omega)\cap C^0(\overline{\Omega})$ with $\Omega$ open, bounded and $\Delta u \geqslant 0$ in $\Omega$, then $\text{max}_{\overline{\Omega}} \ u = \text{max}_{\delta \Omega} \ u$ . Apparently, a weak maximum principle for the Helmholtz equation should be an easy consequence: if $\Delta u +\lambda u=0$ in $\Omega$, with $\lambda < 0$ and it is known that $\text{max}_{\overline{\Omega}} \ u > 0$, then $\text{max}_{\overline{\Omega}} \ u = \text{max}_{\delta \Omega} \ u$. One should apply the weak maximum priniciple to $u$ in $\Omega_{0}=\left \{ z \in \Omega \mid u(z)>0 \right \}$, where $u$ is subharmonic. However, it would appear to me that $u\equiv 0$ on $\delta \Omega_0$ so I cannot see how that will be of any use (leads to an immediate contradiction). Any thoughts?

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  • $\begingroup$ Not so fast - if $z \in \partial \Omega_0$ then it's not necessarily true that $u(z)=0$. What's the other possibility? $\endgroup$ – Anthony Carapetis Apr 14 '17 at 12:03
  • $\begingroup$ That $z \in \delta \Omega$ if $u(z)>0$. So, it really is thát simple? $\endgroup$ – Romanda de Gore Apr 14 '17 at 13:51
  • $\begingroup$ Well, there's a little more work to do. If $z \in \partial\Omega_0$ then either $u(z)=0$ or $z \in \partial \Omega$, but the former is (as you deduced) a contradiction. Thus $\partial \Omega_0 \subset \partial \Omega$, so $\Omega_0$ is either empty or all of $\Omega$. Hopefully you can finish it from there. $\endgroup$ – Anthony Carapetis Apr 15 '17 at 1:07

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