Theorem 3.1.11 (Weyl's lemma)C9127FBA9AE8

Proof. Recall from section 2.5 that $(T*\phi )(y)=\langle T,\phi (y-\cdot )\rangle $ for a test function $\phi $ on $\mathbb R^ d$ supported in $\overline{B(0,s)}$ and $y\in \Omega _ s$, and that $T_\varepsilon =T*\rho _\varepsilon $. By part (1) of proposition 2.5.3, $T*\phi $ is smooth on $\Omega _ s$ and $\partial ^\alpha (T*\phi )=T*\partial ^\alpha \phi $; moreover $\Delta T_\varepsilon =(\Delta T)_\varepsilon =0$, so $T_\varepsilon $ is a smooth harmonic function on $\Omega _\varepsilon $. We show that $T_\varepsilon $ converges to the required $h$, locally uniformly with all derivatives, by writing $T_\varepsilon =T*(\rho _\varepsilon *\rho _ r)$ near each point for a fixed $r>0$.

Fix $x\in \Omega $ and $r>0$ with $\overline{B(x,3r)}\subset \Omega $, and let $0<\varepsilon <r$. Since $\operatorname {dist}(x,\mathbb R^ d\setminus \Omega )>3r$, every point of $B(x,2r)$ has distance greater than $r$ from $\mathbb R^ d\setminus \Omega $, and every point of $B(x,r)$ has distance greater than $2r$; thus $B(x,2r)\subset \Omega _\varepsilon $ and $B(x,r)\subset \Omega _{2r}$. Let $y\in B(x,r)$. The closed ball $\overline{B(y,r)}$ lies in $B(x,2r)$, where $T_\varepsilon $ is harmonic, so the last assertion of theorem 3.1.7, applied to $T_\varepsilon $ on $B(x,2r)$, gives $T_\varepsilon (y)=\int T_\varepsilon (z)\rho _ r(y-z)\, \mathrm dz$. The test function $\chi =\rho _ r(y-\cdot )$ is supported in $\overline{B(y,r)}\subset \Omega _\varepsilon $, so (2.5.4) gives $T_\varepsilon (y)=\langle T,\chi *\rho _\varepsilon \rangle $. Substituting $z=w+v$ and using that $\rho _\varepsilon $ is even, we find

The function $\rho _\varepsilon *\rho _ r$ is a test function supported in $\overline{B(0,\varepsilon +r)}\subset \overline{B(0,2r)}$, so we have shown that

Since $\rho _\varepsilon *\rho _ r=\rho _ r*\rho _\varepsilon $, part (3) of lemma 2.5.1, applied on $\mathbb R^ d$ to $\rho _ r$, shows that $\partial ^\alpha (\rho _\varepsilon *\rho _ r)\to \partial ^\alpha \rho _ r$ uniformly for every multi-index $\alpha $ as $\varepsilon \to 0$. Let $C$ and $m$ be the constants of definition 2.4.1 for $T$ and the compact set $\overline{B(x,3r)}$, which contains the supports of $(\rho _\varepsilon *\rho _ r)(y-\cdot )$ and $\rho _ r(y-\cdot )$ for $y\in B(x,r)$. For $y\in B(x,r)$ and a multi-index $\alpha $, the difference $\partial ^\alpha T_\varepsilon (y)-\partial ^\alpha (T*\rho _ r)(y)$ is the pairing of $T$ with $w\mapsto \psi (y-w)$, where $\psi =\partial ^\alpha (\rho _\varepsilon *\rho _ r-\rho _ r)$, and the derivatives of order $\beta $ of this test function are $\pm (\partial ^\beta \psi )(y-w)$. Hence

which tends to zero. Thus $T_\varepsilon \to T*\rho _ r$ on $B(x,r)$, uniformly with all derivatives.

Every point of $\Omega $ is the center of such a ball. Therefore $h(y)=\lim _{\varepsilon \to 0}T_\varepsilon (y)$ exists for every $y\in \Omega $, and on each ball $B(x,r)$ as above it equals $T*\rho _ r$ and is the limit of $T_\varepsilon $ uniformly with all derivatives. Thus $h$ is smooth and $\Delta h=\lim _{\varepsilon \to 0}\Delta T_\varepsilon =0$. Every compact subset of $\Omega $ is covered by finitely many of these balls, so $T_\varepsilon \to h$ uniformly on compact sets, and part (2) of proposition 2.5.3 gives $\int h\chi \, \mathrm dx=\lim _{\varepsilon \to 0}\int T_\varepsilon \chi \, \mathrm dx=\langle T,\chi \rangle $ for every test function $\chi $. If two continuous functions define the same distribution, they agree almost everywhere by part (4) of proposition 2.5.3, and hence everywhere, since nonempty open sets have positive measure. This proves uniqueness. Finally, if $T$ is given by a locally integrable function $u$, then $u$ and $h$ define the same distribution, so $u=h$ almost everywhere by the same proposition. $\square$