Weyl's lemma (Laplace equation)

Let ${\displaystyle \left(\rho _{\varepsilon }\right)_{\varepsilon >0}}$ be the standard mollifier.

Fix a compact set ${\displaystyle \Omega ^{\prime }\Subset \Omega }$ and put ${\displaystyle \varepsilon _{0}=\operatorname {dist} \left(\Omega ^{\prime },\partial \Omega \right)}$ be the distance between ${\displaystyle \Omega ^{\prime }}$ and the boundary of ${\displaystyle \Omega }$.

For each ${\displaystyle x\in \Omega ^{\prime }}$ and ${\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)}$ the function

${\displaystyle y\longmapsto \rho _{\varepsilon }(x-y)}$

belongs to test functions ${\displaystyle {\mathcal {D}}(\Omega )}$ and so we may consider

${\displaystyle \left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle }$

We assert that it is independent of ${\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)}$. To prove it we calculate ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\rho _{\varepsilon }(x-y)}$ for ${\displaystyle x,y\in \mathbb {R} ^{n}}$.

Recall that

${\displaystyle \rho _{\varepsilon }(x-y)=\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)}$

where the standard mollifier kernel ${\displaystyle \rho }$ on ${\displaystyle \mathbb {R} ^{n}}$ was defined at Mollifier#Concrete_example. If we put

${\displaystyle \theta (t)={\begin{cases}{\frac {1}{c_{n}}}\mathrm {e} ^{\frac {1}{t-1}}&{\text{ if }}t<1,\\0&{\text{ if }}t\geqslant 1,\end{cases}}}$

then ${\displaystyle \rho (x)=\theta \left(|x|^{2}\right)}$.

Clearly ${\displaystyle \theta \in \mathrm {C} ^{\infty }(\mathbb {R} )}$ satisfies ${\displaystyle \theta (t)=0}$ for ${\displaystyle t\geqslant 1}$. Now calculate

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left(\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)\right)&=-n\varepsilon ^{-n-1}\rho \left({\frac {x-y}{\varepsilon }}\right)-\varepsilon ^{-n}\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon ^{2}}}\\&=-{\frac {1}{\varepsilon ^{n+1}}}\left(n\rho \left({\frac {x-y}{\varepsilon }}\right)+\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon }}\right)\end{aligned}}}$

Put ${\displaystyle K(x)=-n\rho (x)-\nabla \rho (x)\cdot x}$ so that

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left(\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)\right)={\frac {1}{\varepsilon ^{n+1}}}K\left({\frac {x-y}{\varepsilon }}\right).}$

In terms of ${\displaystyle \rho (x)=\theta \left(|x|^{2}\right)}$ we get

${\displaystyle K(x)=-\operatorname {div} (\rho (x)x)=-\operatorname {div} \left(\theta \left(|x|^{2}\right)x\right)}$

and if we set

${\displaystyle \Theta (t)={\frac {1}{2}}\int _{t}^{\infty }\theta (s)\mathrm {d} s}$

then ${\displaystyle \Theta \in \mathrm {C} ^{\infty }(\mathbb {R} )}$ with ${\displaystyle \Theta (t)=0}$ for ${\displaystyle t\geqslant 1}$, and ${\displaystyle \Theta ^{\prime }(t)=-{\frac {1}{2}}\theta (t)}$. Consequently

${\displaystyle -\theta \left(|x|^{2}\right)x=\nabla \left(\Theta \left(|x|^{2}\right)\right)}$

and so ${\displaystyle K(x)=\operatorname {div} \nabla \left(\Theta \left(|x|^{2}\right)\right)=(\Delta \Phi )(x)}$, where ${\displaystyle \Phi (x)=\Theta \left(|x|^{2}\right)}$. Observe that ${\displaystyle \Phi \in {\mathcal {D}}\left({\overline {B_{1}(0)}}\right)}$, and

${\displaystyle {\begin{aligned}-{\frac {1}{\varepsilon ^{n+1}}}\left(n\rho \left({\frac {x-y}{\varepsilon }}\right)+\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon }}\right)&={\frac {1}{\varepsilon ^{n+1}}}\Delta _{y}\left(\Phi \left({\frac {x-y}{\varepsilon }}\right)\right)\\&=\Delta _{y}\left(\varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)\right).\end{aligned}}}$

Here ${\displaystyle y\mapsto \varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)}$ is supported in ${\displaystyle {\overline {B_{\varepsilon }(x)}}\subset \Omega }$, and so by assumption

${\displaystyle \left\langle u,\Delta _{y}\left(\varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)\right)\right\rangle =0}$.

Now by considering difference quotients we see that

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =\left\langle u,{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\rho _{\varepsilon }(x-\cdot )\right\rangle }$.

Indeed, for ${\displaystyle \varepsilon ,\varepsilon ^{\prime }>0}$ we have

${\displaystyle {\begin{aligned}{\frac {\rho _{\varepsilon +\varepsilon ^{\prime }}(x-y)-\rho _{\varepsilon }(x-y)}{\varepsilon ^{\prime }}}&{\stackrel {\mathrm {FTC} }{=}}\int _{0}^{1}{\frac {\mathrm {d} }{\mathrm {d} t}}\rho _{\varepsilon +t\varepsilon ^{\prime }}(x-y)\mathrm {d} t\\&\left.{\underset {\varepsilon ^{\prime }\searrow 0}{\longrightarrow }}{\frac {\mathrm {d} }{\mathrm {d} s}}\right|_{s=\varepsilon }\rho _{s}(x-y)\end{aligned}}}$

in ${\displaystyle {\mathcal {D}}^{\prime }(\Omega )}$ with respect to ${\displaystyle y}$, provided ${\displaystyle x\in \Omega ^{\prime }}$ and ${\displaystyle 0<\varepsilon <\varepsilon _{0}}$ (since we may differentiate both sides with respect to ${\displaystyle y)}$. But then ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =0}$, and so ${\displaystyle \left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle }$ for all ${\displaystyle \varepsilon \in \left(0,\varepsilon _{0}\right)}$, where ${\displaystyle \varepsilon _{1}\in \left(0,\varepsilon _{0}\right)}$. Now let ${\displaystyle \varphi \in {\mathcal {D}}\left(\Omega ^{\prime }\right)}$. Then, by the usual trick when convolving distributions with test functions,

${\displaystyle {\begin{aligned}\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x&=\left\langle u,\int _{\Omega ^{\prime }}\rho _{\varepsilon }(x-\cdot )\varphi (x)\mathrm {d} x\right\rangle \\&=\left\langle u,\rho _{\varepsilon }*\varphi \right\rangle \end{aligned}}}$

and so for ${\displaystyle \varepsilon \in \left(0,\varepsilon _{1}\right)}$ we have

${\displaystyle \left\langle u,\rho _{\varepsilon }*\varphi \right\rangle =\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x}$.

Hence, as ${\displaystyle \rho _{\varepsilon }*\varphi \rightarrow \varphi }$ in ${\displaystyle {\mathcal {D}}(\Omega )}$ as ${\displaystyle \varepsilon \searrow 0}$, we get

${\displaystyle \langle u,\varphi \rangle =\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x}$.

Consequently ${\displaystyle \left.u\right|_{\Omega ^{\prime }}\in \mathrm {C} ^{\infty }\left(\Omega ^{\prime }\right)}$, and since ${\displaystyle \Omega ^{\prime }}$ was arbitrary, we are done.