Dilations and their normalisations preserve Schwartz space, with scaling identities

Differentiation of a dilation. Let $f=\phi \circ A$ with $A(x)=x\mathrm{/}t$, and fix $j\le n$ and $x\in {\mathbb{R}}^n$. Writing $z=x\mathrm{/}t$ and $s=h\mathrm{/}t$, the one-variable difference quotient of the map $h\mapsto f(x+h{e}_j)$ equals $t^{-1}(\phi (z+s{e}_j)-\phi (z))\mathrm{/}s$, and $s\to 0$ exactly when $h\to 0$, so the limit exists and equals $t^{-1}{\mathrm{\partial }}_j\phi (z)$ by [L1]; there is no division by a vanishing quantity because $t>0$. Induction on $\mathrm{\mid }\beta \mathrm{\mid }$, applying the same computation to the $C^1$ function ${\mathrm{\partial }}^{\gamma }\phi$ at the point $x$ with $\gamma$ the predecessor of $\beta$, gives ${\mathrm{\partial }}^{\beta }(D_t\phi )(x)=t^{-\mathrm{\mid }\beta \mathrm{\mid }}({\mathrm{\partial }}^{\beta }\phi )(x\mathrm{/}t)\mathrm{.}$

The scaling identities. By [F1] and [L1] the functions $\phi$, $\mathrm{\mid }\phi \mathrm{\mid }$, $x^{\alpha }{\mathrm{\partial }}^{\beta }\phi$ and $\mathrm{\mid }x{\mathrm{\mid }}^m\mathrm{\mid }\phi \mathrm{\mid }$ are integrable, so the change-of-variables formula applies to them. Applying it to $\phi$ with $T(y)=ty$ and $\mathrm{\mid }\det DT\mathrm{\mid }=t^n$ gives $\int {\phi }_t(x)dx=t^{-n}\int \phi (x\mathrm{/}t)dx=\int \phi (y)dy,$ and applying it to the nonnegative integrable function $\mathrm{\mid }x{\mathrm{\mid }}^m\mathrm{\mid }\phi (x)\mathrm{\mid }$ gives $\int \mathrm{\mid }x{\mathrm{\mid }}^m\mathrm{\mid }\phi (x\mathrm{/}t)\mathrm{\mid }dx=t^{n+m}\int \mathrm{\mid }y{\mathrm{\mid }}^m\mathrm{\mid }\phi (y)\mathrm{\mid }dy$, whence the moment identity after multiplying by $t^{-n}$. The factor $t^{-n}{t}^{n+m}=t^m$ is finite for every $m\ge 0$ and $t>0$.

Membership and the seminorm identities. Substituting $y=x\mathrm{/}t$ in $\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\beta }(D_t\phi )(x)\mathrm{\mid }=t^{-\mathrm{\mid }\beta \mathrm{\mid }}\mathrm{\mid }{x}^{\alpha }({\mathrm{\partial }}^{\beta }\phi )(x\mathrm{/}t)\mathrm{\mid }$ gives $t^{\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\beta \mathrm{\mid }}\mathrm{\mid }{y}^{\alpha }{\mathrm{\partial }}^{\beta }\phi (y)\mathrm{\mid }$, valid for every $x\in {\mathbb{R}}^n$; taking suprema over $x$ is taking suprema over $y$ and proves $p_{\alpha \beta }(D_t\phi )=t^{\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\beta \mathrm{\mid }}{p}_{\alpha \beta }(\phi )<\mathrm{\infty }$. Multiplying by the scalar $t^{-n}$ proves the second identity and makes $D_t\phi ,{\phi }_t$ elements of $\mathcal{S}$, since these are finite for all $\alpha ,\beta$ by [L1] and the given. For fixed $t$ the constants $t^{\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\beta \mathrm{\mid }-n}$ are finite, so for every basic neighbourhood the finitely many relevant input seminorms of $\phi$ control the output seminorms; this is continuity of $D_t$ at zero, hence everywhere by linearity.

Conclusion. Step 2.1 gives membership, the two seminorm identities, and continuity of $D_t$ on $\mathcal{S}$; step 1.2 gives the integral and moment identities under countable choice, which is inherited from the substitution theorem. This proves the lemma.