Let $C_k^\infty$ be the space of functions $k$ times continuously differentiabl on an open set $\Omega,$ with compact support, i.e., which vanish outside some compact set $K\subset \Omega.$ The space $\ L_{loc}^p(\Omega)$ consists of Borel measurable functions defined on all of $\Omega$ with the property that $||f||_{L^p(K)}<\infty.$ We want to prove the Weyl's lemma:
Let $u \in L_{loc}^p(\Omega)$ and suppose that for all $\varphi \in C_0^\infty,$ $$\int_{\Omega}u\Delta\varphi=0.$$ Then, $u$ is harmonic and in particular smooth in $\Omega.$
For the proof, we use a mollifier $\rho,$ which is a radial function in $C_0^\infty(\mathbb{R}^n)$ such that $\int_{\mathbb{R}^n}\rho=1.$ Theh, we consider the mollifications $$u_r(x)=\dfrac{1}{r^n}\int_{\Omega}\rho\left(\dfrac{|x-y|}{r}\right)u(y)dy), \ \ r>0.$$
For $\varphi \in C_0^\infty(\Omega)$ and $0<r<\text{dist}(\text{supp } (\varphi), \partial\Omega),$ we obtain \begin{equation} \begin{split} \int_{\Omega} u_r(x)\Delta\varphi(x)&=\int_{\Omega} \left(\dfrac{1}{r^n}\int_{\Omega}\rho\left(\dfrac{|x-y|}{r}\right)u(y)dy\right)\Delta\varphi(x)dx\\ &=\int_{\Omega}u(y)\varphi_r(y)dy=0, \end{split} \end{equation} since $\varphi_r \in C_0^\infty(\Omega).$ But here, we've used that $\Delta(\varphi_r)=(\Delta\varphi)_r,$ which is apparently true but I don't know how to prove it. Please, any ideas or hints about how to show this equallity. Thanks.
