*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.



Strong Maximum Principle for very weak supersolutions of Laplacian operator
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.


