Strong Maximum Principle for very weak supersolutions of Laplacian operator

Question

I know classical strong maximum principles for supersolutions of Laplacian operator, which says:

Suppose $ u \in C^2(\Omega)\cap C(\overline{\Omega}) $ satisfies $ -\Delta u \geq 0 $ in $ \Omega$.

If $ u \geq 0 $ on $ \partial \Omega $ then $ u \geq 0 $ in $ \Omega $. In fact, either $u > 0$ in $\Omega$ or $ u \equiv 0 $ in $\Omega$.

I am reading an article, there author use this version of strong maximum principle:

Assume $ u \in L_{\mathrm{loc}}^1(\Omega) $ is a very weak supersolutions of $ - \Delta u =0 $ in the sense of distributions, means for every non-negative $\phi \in C_0^{\infty}(\Omega) $ we have $$ \int_{\Omega} -u \, \Delta \phi \, \mathrm{d}x \geq 0 $$

Also assume $ u \geq 0 $ on $ \partial \Omega $ and $ u \not\equiv 0 $. Then
$$ u > 0 \quad \mathrm{in} \,\, \Omega $$

There is no reference for it, and I could not find any proof for it. Does anyone know a proof for it, or a reference for it's proof.

Thanks.

Answer 1

As such the statement does not make sense: if $u \in L^1_\mathrm{loc} (\Omega)$, its restriction on $\partial \Omega$ is not well-defined.

If $u \ge 0$ on $\Omega \setminus K$, where $K \subset \Omega$ is a compact subset and $-\Delta u \ge 0$ in the sense of distributions and if $\Omega$ is connected, then either $u = 0$ in $\Omega$ or $u > 0$ almost everywhere in $\Omega$. First you apply the weak maximum principle to a convolution of $u$ with a radial mollifier on a suitable subdomain of $\Omega$. This shows that $u \ge 0$. You then apply the mean value property for superharmonic functions to conclude.