$W_{loc}^{1,2}$ regularity of nonnegative subharmonic functions I'm trying to solve the following excercise:
Let $v \in C(D)$ be a nonnegative subharmonic function in an open set $D$. Prove that $v \in W_{loc}^{1,2}(D)$
The problem has the following hint:
Mollifications $v_\epsilon$ of $v$ satisfy the inequality
$$ \int_D \nabla v_\epsilon \cdot \nabla \phi \;dx \leq 0 $$
for any nonnegative $\phi \in C_0^\infty(D)$. Take $\phi = v_\epsilon \zeta^2$ with $\zeta \in C_0^\infty(D)$ and let $\epsilon \to 0+$.
I got confused trying to integrate by parts and use the Poincaré inequality to prove a uniform bound, but I guess and don't really understand how to use the hint. I know that under the hypotheses the convergence $v_\epsilon \to v$ is in $L_{loc}^2(D)$ and uniform on compact subsets of $D$, but not more than that.
I will really appreciate any comment.
 A: I don't know if you're still interested in the answer for this since the questions was posted some time ago. In any case, I was looking at this exercise from Petrosyan's book a couple of days ago so here goes a sketch of my solution.
We basically want a uniform bound (on $\varepsilon$) for the family $\{\nabla v_\varepsilon\}_{\varepsilon>0}$. If we get it, then there exists a vector valued function $g$ in $L^2$ and a subsequence $\varepsilon_j$ such that $\nabla v_{\varepsilon_j}$ converges weakly to $g$ as $\varepsilon_j\rightarrow0$. It's easy to check that $g=\nabla v$.  
Following the hint and plugging in the test function suggested we have
\begin{align*}
\int \zeta^2|\nabla v_\varepsilon|^2 \:dx & \leq -\int 2\zeta v_\varepsilon\nabla v_\varepsilon\cdot\nabla\zeta\\
                                          & \leq 2\int \zeta v_\varepsilon|\nabla v_\varepsilon\nabla \zeta|\:dx \\
                                          & \leq 2\|v\|_{L^1}\|\zeta\nabla v_\varepsilon\|_{L^2}\|\nabla \zeta\|_{L^2}
\end{align*}
where the $L^1$ norm of $v$ comes from the subharmonicity (is a weak Harnack inequality, if you don't now it it's a nice exercise to prove it). Therefore
$$
\|\zeta\nabla v_\varepsilon\|_{L^2}\leq C
$$
for a constant $C$ which is independent of $\varepsilon$. Now take your favorite set $V\subset\subset B_1$ and take $\zeta\in C^\infty_c(B_1), \zeta\equiv 1$ in $V$ and you are done. 
PS: I know this is sort of sketchy, but I hope that the overall idea is clear.  
A: We follow the hint. Set $\phi=v_{\varepsilon}\zeta^{2}$, with the extra condition that $\zeta=1$ on $B_{\rho} \subset \mathrm{supp} \zeta=:K$. Since $v$ is subharmonic, $v_{\varepsilon}$ satisfy
$$ \int_{D}|\nabla v_{\varepsilon}|^{2}\zeta^{2}+\int_{D}v_{\varepsilon} \nabla v_{\varepsilon} \cdot \nabla \zeta^{2} \leq 0.$$
Since $\nabla(v_{\varepsilon}^{2}/2)=v_{\varepsilon} \nabla v_{\varepsilon}$, the previous calculation is equal to
$$\int_{D}|\nabla v_{\varepsilon}|^{2}\zeta^{2}+\int_{D}\nabla\left(\frac{v_{\varepsilon}^{2}}{2}\right)\nabla \zeta^{2}\leq 0, $$
and using the Green identity we obtain
$$\int_{D}|\nabla v_{\varepsilon}|^{2}\zeta^{2}-\int_{D}\frac{v_{\varepsilon}^{2}}{2}\Delta \zeta^{2}\leq 0.$$
Notice that the boundary term vanishes because $\zeta$ vanishes. Then,
$$\int_{B_{\rho}}|\nabla v_{\varepsilon}|^{2}\leq \int_{K}|\nabla v_{\varepsilon}|^{2}\zeta^{2}\leq C(\zeta)\int_{K}v_{\varepsilon}^{2},$$
where $C$ only depends on $\zeta$ and not from $\varepsilon$.
Then, letting $\varepsilon \to 0^{+}$,
$$\|v\|_{W^{1,2}(B_{\rho})}\leq C\|v\|_{L^{2}(B_{\rho})} \leq  C_{1},$$
were in the last inequality I used that $v \in C(D)$. So, $v \in W_{\mathrm{loc}}^{1,2}(D)$.
