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Consider a radial weight $w=|x|^2 \geq 0$ for all $x\in \mathbb{R}^n$ and consider the operator $$Lu= \frac{1}{w}\operatorname{div}(w\nabla u).$$

Then if $-Lu\leq 0$ on a smooth bounded domain $\Omega$ and $u=0$ on the boundary of $\Omega$, does this imply that $u\leq 0$ on all $\Omega$?

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    $\begingroup$ If you are working e.g. in the class of functions $u$ such that $w|\nabla u|^2$ is integrable, then yes (use $w\max\{u,\,0\}$ as a test function). $\endgroup$
    – Connor Mooney
    Commented Mar 30, 2023 at 23:45
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    $\begingroup$ If you care about this operator on a ball centered at the origin (or the full space) you can write this out with spherical harmonics and get Euler ode's and then you can probably analyze the operator exactly on weighted spaces... $\endgroup$
    – Math604
    Commented Mar 31, 2023 at 0:30
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    $\begingroup$ web.math.princeton.edu/~const/maxhar.pdf $\endgroup$
    – Alexandre Eremenko
    Commented Mar 31, 2023 at 13:36

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