# Weyl's lemma (Laplace equation)

In mathematics, Weyl's lemma, named after Hermann Weyl, states that every weak solution of Laplace's equation is a smooth solution. This contrasts with the wave equation, for example, which has weak solutions that are not smooth solutions. Weyl's lemma is a special case of elliptic or hypoelliptic regularity.

## Statement of the lemma

Let $\Omega$ be an open subset of $n$-dimensional Euclidean space $\mathbb {R} ^{n}$, and let $\Delta$ denote the usual Laplace operator. Weyl's lemma states that if a locally integrable function $u\in L_{\mathrm {loc} }^{1}(\Omega )$ is a weak solution of Laplace's equation, in the sense that
    $\int _{\Omega }u(x)\,\Delta \varphi (x)\,dx=0$
for every test function (smooth function with compact support) $\varphi \in C_{c}^{\infty }(\Omega )$, then (up to redefinition on a set of measure zero) $u\in C^{\infty }(\Omega )$ is smooth and satisfies $\Delta u=0$ pointwise in $\Omega$.
This result implies the interior regularity of harmonic functions in $\Omega$, but it does not say anything about their regularity on the boundary $\partial \Omega$.

## Idea of the proof

To prove Weyl's lemma, one convolves the function $u$ with an appropriate mollifier $\varphi _{\varepsilon }$ and shows that the mollification $u_{\varepsilon }=\varphi _{\varepsilon }\ast u$ satisfies Laplace's equation, which implies that $u_{\varepsilon }$ has the mean value property. Taking the limit as $\varepsilon \to 0$ and using the properties of mollifiers, one finds that $u$ also has the mean value property, which implies that it is a smooth solution of Laplace's equation. Alternative proofs use the smoothness of the fundamental solution of the Laplacian or suitable a priori elliptic estimates.
Let $\left(\rho _{\varepsilon }\right)_{\varepsilon >0}$ be the standard mollifier.
Fix a compact set $\Omega ^{\prime }\Subset \Omega$ and put $\varepsilon _{0}=\operatorname {dist} \left(\Omega ^{\prime },\partial \Omega \right)$ be the distance between $\Omega ^{\prime }$ and the boundary of $\Omega$.
For each $x\in \Omega ^{\prime }$ and $\varepsilon \in \left(0,\varepsilon _{0}\right)$ the function
    $y\longmapsto \rho _{\varepsilon }(x-y)$
belongs to test functions ${\mathcal {D}}(\Omega )$ and so we may consider
    $\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle$
We assert that it is independent of $\varepsilon \in \left(0,\varepsilon _{0}\right)$. To prove it we calculate ${\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\rho _{\varepsilon }(x-y)$ for $x,y\in \mathbb {R} ^{n}$.
Recall that
    $\rho _{\varepsilon }(x-y)=\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)$
where the standard mollifier kernel $\rho$ on $\mathbb {R} ^{n}$ was defined at Mollifier#Concrete_example. If we put
    $\theta (t)={\begin{cases}{\frac {1}{c_{n}}}\mathrm {e} ^{\frac {1}{t-1}}&{\text{ if }}t<1,\\0&{\text{ if }}t\geqslant 1,\end{cases}}$
then $\rho (x)=\theta \left(|x|^{2}\right)$.
Clearly $\theta \in \mathrm {C} ^{\infty }(\mathbb {R} )$ satisfies $\theta (t)=0$ for $t\geqslant 1$. Now calculate
    ${\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left(\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)\right)&=-n\varepsilon ^{-n-1}\rho \left({\frac {x-y}{\varepsilon }}\right)-\varepsilon ^{-n}\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon ^{2}}}\\&=-{\frac {1}{\varepsilon ^{n+1}}}\left(n\rho \left({\frac {x-y}{\varepsilon }}\right)+\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon }}\right)\end{aligned}}$
Put $K(x)=-n\rho (x)-\nabla \rho (x)\cdot x$ so that
    ${\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left(\varepsilon ^{-n}\rho \left({\frac {x-y}{\varepsilon }}\right)\right)={\frac {1}{\varepsilon ^{n+1}}}K\left({\frac {x-y}{\varepsilon }}\right).$
In terms of $\rho (x)=\theta \left(|x|^{2}\right)$ we get
    $K(x)=-\operatorname {div} (\rho (x)x)=-\operatorname {div} \left(\theta \left(|x|^{2}\right)x\right)$
and if we set
    $\Theta (t)={\frac {1}{2}}\int _{t}^{\infty }\theta (s)\mathrm {d} s$
then $\Theta \in \mathrm {C} ^{\infty }(\mathbb {R} )$ with $\Theta (t)=0$ for $t\geqslant 1$, and $\Theta ^{\prime }(t)=-{\frac {1}{2}}\theta (t)$. Consequently
    $-\theta \left(|x|^{2}\right)x=\nabla \left(\Theta \left(|x|^{2}\right)\right)$
and so $K(x)=\operatorname {div} \nabla \left(\Theta \left(|x|^{2}\right)\right)=(\Delta \Phi )(x)$, where $\Phi (x)=\Theta \left(|x|^{2}\right)$. Observe that $\Phi \in {\mathcal {D}}\left({\overline {B_{1}(0)}}\right)$, and
    ${\begin{aligned}-{\frac {1}{\varepsilon ^{n+1}}}\left(n\rho \left({\frac {x-y}{\varepsilon }}\right)+\nabla \rho \left({\frac {x-y}{\varepsilon }}\right)\cdot {\frac {x-y}{\varepsilon }}\right)&={\frac {1}{\varepsilon ^{n+1}}}\Delta _{y}\left(\Phi \left({\frac {x-y}{\varepsilon }}\right)\right)\\&=\Delta _{y}\left(\varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)\right).\end{aligned}}$
Here $y\mapsto \varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)$ is supported in ${\overline {B_{\varepsilon }(x)}}\subset \Omega$, and so by assumption
    $\left\langle u,\Delta _{y}\left(\varepsilon ^{1-n}\Phi \left({\frac {x-y}{\varepsilon }}\right)\right)\right\rangle =0$.
Now by considering difference quotients we see that
    ${\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =\left\langle u,{\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\rho _{\varepsilon }(x-\cdot )\right\rangle$.
Indeed, for $\varepsilon ,\varepsilon ^{\prime }>0$ we have
    ${\begin{aligned}{\frac {\rho _{\varepsilon +\varepsilon ^{\prime }}(x-y)-\rho _{\varepsilon }(x-y)}{\varepsilon ^{\prime }}}&{\stackrel {\mathrm {FTC} }{=}}\int _{0}^{1}{\frac {\mathrm {d} }{\mathrm {d} t}}\rho _{\varepsilon +t\varepsilon ^{\prime }}(x-y)\mathrm {d} t\\&\left.{\underset {\varepsilon ^{\prime }\searrow 0}{\longrightarrow }}{\frac {\mathrm {d} }{\mathrm {d} s}}\right|_{s=\varepsilon }\rho _{s}(x-y)\end{aligned}}$
in ${\mathcal {D}}^{\prime }(\Omega )$ with respect to $y$, provided $x\in \Omega ^{\prime }$ and $0<\varepsilon <\varepsilon _{0}$ (since we may differentiate both sides with respect to $y)$. But then ${\frac {\mathrm {d} }{\mathrm {d} \varepsilon }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =0$, and so $\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle =\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle$ for all $\varepsilon \in \left(0,\varepsilon _{0}\right)$, where $\varepsilon _{1}\in \left(0,\varepsilon _{0}\right)$. Now let $\varphi \in {\mathcal {D}}\left(\Omega ^{\prime }\right)$. Then, by the usual trick when convolving distributions with test functions,
    ${\begin{aligned}\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon }(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x&=\left\langle u,\int _{\Omega ^{\prime }}\rho _{\varepsilon }(x-\cdot )\varphi (x)\mathrm {d} x\right\rangle \\&=\left\langle u,\rho _{\varepsilon }*\varphi \right\rangle \end{aligned}}$
and so for $\varepsilon \in \left(0,\varepsilon _{1}\right)$ we have
    $\left\langle u,\rho _{\varepsilon }*\varphi \right\rangle =\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x$.
Hence, as $\rho _{\varepsilon }*\varphi \rightarrow \varphi$ in ${\mathcal {D}}(\Omega )$ as $\varepsilon \searrow 0$, we get
    $\langle u,\varphi \rangle =\int _{\Omega ^{\prime }}\left\langle u,\rho _{\varepsilon _{1}}(x-\cdot )\right\rangle \varphi (x)\mathrm {d} x$.
Consequently $\left.u\right|_{\Omega ^{\prime }}\in \mathrm {C} ^{\infty }\left(\Omega ^{\prime }\right)$, and since $\Omega ^{\prime }$ was arbitrary, we are done.

## Generalization to distributions

More generally, the same result holds for every distributional solution of Laplace's equation: If $T\in D'(\Omega )$ satisfies $\langle T,\Delta \varphi \rangle =0$ for every $\varphi \in C_{c}^{\infty }(\Omega )$, then $T$ is a regular distribution associated with a smooth solution $u\in C^{\infty }(\Omega )$ of Laplace's equation.

## Connection with hypoellipticity

Weyl's lemma follows from more general results concerning the regularity properties of elliptic or hypoelliptic operators. A linear partial differential operator $P$ with smooth coefficients is hypoelliptic if the singular support of $Pu$ is equal to the singular support of $u$ for every distribution $u$. The Laplace operator is hypoelliptic, so if $\Delta u=0$, then the singular support of $u$ is empty since the singular support of $0$ is empty, meaning that $u\in C^{\infty }(\Omega )$. In fact, since the Laplacian is elliptic, a stronger result is true, and solutions of $\Delta u=0$ are real-analytic.