Weyl's lemma for the Laplacian

The distributional Laplacian commutes with local mollification makes $h_{\epsilon }=T\ast {\rho }_{\epsilon }$ smooth and harmonic on ${\mathrm{\Omega }}_{\epsilon }$ [given].

Each $h_{\epsilon }$ is harmonic by step 1.1, so the ball and spherical mean-value properties apply under Countable Choice. Radial mollification therefore fixes it: $h_{\epsilon }\ast {\rho }_{\delta }=h_{\epsilon }$ on ${\mathrm{\Omega }}_{\epsilon +\delta }$, and likewise $h_{\delta }\ast {\rho }_{\epsilon }=h_{\delta }$. Associativity of distributional convolution with these compactly supported test kernels identifies the two double convolutions on that common shrunken domain. Hence $h_{\epsilon }=h_{\delta }$ there.

On each compact subset of $\mathrm{\Omega }$, all sufficiently small regularizations agree by step 2.1; their common value defines a smooth harmonic $h$. For a test function $\varphi$ supported in that compact subset, $⟨T_{h_{\epsilon }},\varphi ⟩=⟨T,{\check{\rho }}_{\epsilon }\ast \varphi ⟩\to ⟨T,\varphi ⟩$: the mollified tests converge to $\varphi$ with all derivatives on one fixed compact support, so distributional continuity applies. Since $h_{\epsilon }=h$ on the support for small $\epsilon$, this gives $T_h=T$.

If $T_h=T_k$, then $h=k$ almost everywhere; continuity makes $h=k$ everywhere [given]. ∎