Lemma: Taylor reduction and first-order operators

1. For $c\in C^{\infty }(V_{+})$ and $\gamma \in {\mathbb{N}}^5$,

   $$c(x^{\prime },n){{\partial }_n}^{\gamma }\delta (n)=\sum _{{\gamma }^{\prime }\le \gamma }(-1{)}^{|\gamma -{\gamma }^{\prime }|}(\genfrac{}{}{0ex}{}{\gamma }{{\gamma }^{\prime }})({{\partial }_n}^{\gamma -{\gamma }^{\prime }}c)(x^{\prime },0){{\partial }_n}^{{\gamma }^{\prime }}\delta (n).$$

   Hence $c\cdot {{\mathcal{J}}^k}_{\infty }\subseteq {{\mathcal{J}}^k}_{\infty }$.

2. Base derivatives ${\partial }_{{x^{\prime }}_i}$ map ${{\mathcal{J}}^k}_{\infty }\to {{\mathcal{J}}^k}_{\infty }$; normal derivatives ${\partial }_{n_j}$ map ${{\mathcal{J}}^k}_{\infty }\to {{\mathcal{J}}^{k+1}}_{\infty }$.

3. Every first-order differential operator with coefficients in $C^{\infty }(V_{+})$ maps ${{\mathcal{J}}^k}_{\infty }\to {{\mathcal{J}}^{k+1}}_{\infty }$, and every such operator with no normal-derivative component maps ${{\mathcal{J}}^k}_{\infty }\to {{\mathcal{J}}^k}_{\infty }$.

Proof. (i) Pair with $\varphi$: $⟨c{\partial }^{\gamma }\delta ,\varphi ⟩=(-1{)}^{|\gamma |}\int {{\partial }_n}^{\gamma }(c\varphi )(x^{\prime },0)\mathrm{d}{x}^{\prime }$; expand by Leibniz and use $⟨{\partial }^{{\gamma }^{\prime }}\delta ,\psi ⟩=(-1{)}^{|{\gamma }^{\prime }|}{\partial }^{{\gamma }^{\prime }}\psi (0)$. The coefficients $({\partial }^{\gamma -{\gamma }^{\prime }}c)(x^{\prime },0)$ are smooth on $S^{+}$. (ii) ${\partial }_{{x^{\prime }}_i}(b\otimes {\partial }^{\gamma }\delta )=({\partial }_{{x^{\prime }}_i}b)\otimes {\partial }^{\gamma }\delta$ and ${\partial }_{n_j}(b\otimes {\partial }^{\gamma }\delta )=b\otimes {\partial }^{\gamma +e_j}\delta$. (iii) follows from (i), (ii).