Weyl's lemma; doubt about the proof Weyl's lemma: A function $u \in L_{loc}^1(\Omega)$ is harmonic if
$$\int_\Omega u \Delta \varphi = 0 \quad \forall \varphi \in C_c^\infty(\Omega) $$
Here $\Omega$ is a domain i.e open and connected.
What I find confusing is the first step of the proof: Consider a radial family of mollifiers $\rho_\epsilon$, and $u_\epsilon = u \ast \rho_\epsilon$
$$\int_\Omega u_\epsilon \Delta \varphi = \int_\Omega u(\Delta \varphi 
\ast \rho_\epsilon) = \int_\Omega u \Delta (\varphi \ast \rho_\epsilon) = 0 \quad \forall \varphi \in C_c^\infty(\Omega_\epsilon)  $$
Where $\Omega_\epsilon = \{ x\in \omega: dist(x, \partial \Omega) > \epsilon \}$. In particual $\Delta u_\epsilon = 0$ in $\Omega_\epsilon$
That last part is what I don't understand, how does $$\int_\Omega u_\epsilon \Delta \varphi = 0 \quad \forall \varphi \in C_c^\infty(\Omega_\epsilon) $$ implies that $\Delta u_\epsilon = 0$ in $\Omega_\epsilon$ ?
 A: As suggested in the comments, you want to integrate by parts. From what I can tell, your worry is that $\Omega$ is only assumed to be open and bounded, and integration by parts requires some kind of regularity assumption on the boundary. You're right, but this is not an issue here because $\varphi $ has compact support in $\Omega_\varepsilon$. Indeed, for each $\varphi \in C^\infty_0(\Omega_\varepsilon)$, choose a set $U \subset \Omega$ such that $\operatorname{supp} \varphi \subset \subset U \subset \subset \Omega_\varepsilon$ and $\partial U$ is smooth. Then, by Green's identity\begin{align*}
\int_{\Omega_\varepsilon} \varphi \Delta u_\varepsilon \, dx &=\int_U \varphi \Delta u_\varepsilon \, dx \\
&= \int_U u_\varepsilon \Delta \varphi \, dx + \int_{\partial U}\bigg(u_\varepsilon \frac{\partial \varphi }{\partial \nu} - \varphi \frac{\partial u_\varepsilon}{\partial \nu} \bigg)\, d\mathcal H^{n-1}_x\\
&=0.
\end{align*} Note that all the derivative on $u_\varepsilon$ make sense because $u_\varepsilon$ is smooth. Since $\varphi$ was arbitrary, it follows that $\Delta u_\varepsilon =0$ in $\Omega_\varepsilon$ as required.
