# 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 $$\Omega_\epsilon$$ ?

• Did you try to integrate by parts?
– fwd
Feb 11, 2023 at 21:28
• I don't know the lemma, but it seems like you want the integral $=0?$ Or do you want the integral to just exist? Feb 11, 2023 at 21:46
• ... indeed : look at the way you have written the first "equation". Feb 11, 2023 at 21:49
• I don't think one can integrate by parts, at least in the book, the set $\Omega$ is just assumed to be open and connected, there's no regularity assumptions on $\partial \Omega$ Feb 12, 2023 at 0:42
• I think it's possible to integrate by parts over $\Omega_\epsilon$ Feb 12, 2023 at 0:46

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.