Proposition: Dilation identity for W (μ) ρ / zero-moment extensions

Let ${\varpi }_{\mu }(z)=\varpi ((\mu /{\mu }_0)z)$ and let ${W^{(\mu )}}_{\rho }$, ${w^{(\rho ,\mu )}}_{\alpha }$, ${\mathfrak{R}}_{\mathrm{L},\mu }$ be as in XVI-L, Definition def:RLmu. Write $E=z^{\rho }{\partial }_{\rho }$.

1. For every $\rho$, $\mu {\partial }_{\mu }{\varpi }_{\mu }=E{\varpi }_{\mu }$.

2. For $\rho =0$ (so $N=0$, ${w^{(0,\mu )}}_0={\varpi }_{\mu }$):

   $$⟨A_{\mu }(u),\phi ⟩=-\phi (0)⟨u,E{\varpi }_{\mu }⟩,A_{\mu }(u)=-⟨u,E{\varpi }_{\mu }⟩\delta .$$

3. For $u\in {\mathcal{H}}_4$ homogeneous of degree $-4$ on ${\mathbb{R}}^4\setminus \{0\}$, the dilation defect satisfies

   $$\mathcal{A}(u):=(E+4){\mathfrak{R}}_{\mathrm{L},\mu }(u)=A_{\mu }(u).$$

Proof. (1) Chain rule: ${\partial }_{\mu }\varpi ((\mu /{\mu }_0)z)=(z/{\mu }_0)\cdot (\nabla \varpi )((\mu /{\mu }_0)z)$, hence $\mu {\partial }_{\mu }{\varpi }_{\mu }=E{\varpi }_{\mu }$. (2) For $\rho =0$, $⟨{\mathfrak{R}}_{\mathrm{L},\mu }(u),\phi ⟩=⟨u,\phi -\phi (0){\varpi }_{\mu }⟩$; differentiate in $\mu$ and use (1). (3) $⟨(E+4)\mathfrak{R},\phi ⟩=-⟨\mathfrak{R},E\phi ⟩$. Since $(E\phi )(0)=0$, $W_0(E\phi )=E\phi$, so $⟨\mathfrak{R},E\phi ⟩=⟨u,E\phi ⟩$. Write $\phi =W_0\phi +\phi (0){\varpi }_{\mu }$ and $E\phi =E(W_0\phi )+\phi (0)E{\varpi }_{\mu }$. The continuous extension of $u$ to $\{\psi :\psi (0)=0\}$ obeys $⟨u,E(W_0\phi )⟩=0$ (homogeneity of degree $-4$; polar identity on the Euclidean sphere), whence $⟨\mathcal{A}(u),\phi ⟩=-\phi (0)⟨u,E{\varpi }_{\mu }⟩=⟨A_{\mu }(u),\phi ⟩$.