I read a proof in my book on pde which I find a bit strange. Let $\Omega$ be some bounded domain. For $f\in\mathscr C^2(\Omega)\cap\mathscr C^0(\Omega)$ satisfying $\Delta f\geq0$ it holds that $\max_\bar\Omega(f)=\max_{\partial\Omega}(f)$. The proof follows by showing that it holds for $f$ with $\Delta f>0$ based on the second derivative test, and then generalizes it by considering the function $v(x)=f(x)+\epsilon|x|^2$ for $\epsilon$>0. Now since the norm function is strictly subharmonic, we have $\Delta v>0$. The author proceeds as follows:
$$\max_{\bar\Omega} (v)=\max_{\partial\Omega}(v) \Longrightarrow \max_\bar\Omega (f)+\epsilon\min_\bar\Omega(|x|^2)\leq \max_{\partial\Omega}(f)+\epsilon\max_{\partial\Omega}(|x|^2) \Longrightarrow \max_\bar\Omega(f)=\max_{\partial\Omega}(f)$$ since epsilon was arbitrary.
My question is, why would it not work to simply let epsilon be 1 in the beginning and reason as follows. $$\max_\bar\Omega(f)+\max_\bar\Omega(|x|^2)=\max_{\bar\Omega} (v)=\max_{\partial\Omega}(v)=\max_{\partial\Omega}(f)+\max_{\partial\Omega}(|x|^2)$$ which implies the result since $\max_{\bar\Omega}(|x|^2)=\max_{\partial\Omega}(|x|^2)$ since $|x|$ is strictly subharmonic. Wouldn't this be easier? What ridiculous mistake am i making? Thanks
Best Answer
$\max (f+g)\le \max f +\max g$, and equality holds only in the rare circumstance that both functions take on their max at the same point.