Invertible linear substitutions preserve Schwartz space

The map $y\mapsto Ay$ is $C^{\mathrm{\infty }}$: each component ${\sum }_j{A}_{ij}{y}_j$ is a finite sum of scalar multiples of the coordinate functions, whose iterated coordinate derivatives are constant, hence continuous, so each component is $C^{\mathrm{\infty }}$ as a composite of the $C^{\mathrm{\infty }}$ identity with itself and finite sums of such [F1, F3]. It is totally differentiable at every $y$ with $D(A\cdot )(y)=A$, because $A(y+h)=Ay+Ah$ has remainder identically zero in the defining limit [F2]. Consequently $f\circ A\in C^{\mathrm{\infty }}$ [F1, F3], and for every $C^1$ map $u$ on ${\mathbb{R}}^n$ and every $j$, the chain rule gives ${\mathrm{\partial }}_j(u\circ A)(y)=(Du(Ay)\circ A)e_j$, and the matrix of $Du(Ay)$ is the Jacobian $({\mathrm{\partial }}_{i}u(Ay))$ because the partial derivatives of $u$ are continuous [F2, F4], so ${\mathrm{\partial }}_j(u\circ A)(y)={\sum }_i{A}_{ij}({\mathrm{\partial }}_{i}u)(Ay)$.

Iterating the coordinate chain rule of step 1.1 along the canonical differentiation word for $\beta$ gives a finite sum of derivatives of $f$ of order $\mathrm{\mid }\beta \mathrm{\mid }$, evaluated at $Ay$, with constant coefficients depending only on $A$. By [F7] those derivatives can be grouped by their multi-indices, giving ${\mathrm{\partial }}^{\beta }(f\circ A)(y)={\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }=\mathrm{\mid }\beta \mathrm{\mid }}{c}_{\beta \gamma }({\mathrm{\partial }}^{\gamma }f)(Ay)$. The case $\beta =0$ has its single coefficient equal to one.

Choose $K\ge 1$ with $\mathrm{\parallel }{A}^{-1}x{\mathrm{\parallel }}_2\le K\mathrm{\parallel }x{\mathrm{\parallel }}_2$ [F4], and put $N=\mathrm{\mid }\alpha \mathrm{\mid }$. For $x=Ay$ one has $\mathrm{\mid }{y}^{\alpha }\mathrm{\mid }\le K^N(1+{\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }{)}^N$. By [F6] this last power is ${\sum }_{\mathrm{\mid }\delta \mathrm{\mid }\le N}{b}_{N\delta }\mathrm{\mid }{x}^{\delta }\mathrm{\mid }$, where $b_{N\delta }:=N!\mathrm{/}((N-\mathrm{\mid }\delta \mathrm{\mid })!\delta !)$ is the multinomial coefficient with exponent tuple $(N-\mathrm{\mid }\delta \mathrm{\mid },{\delta }_1,\dots ,{\delta }_n)$. Combining this expansion with step 2.1 and taking the supremum over $y$ gives $p_{\alpha \beta }(A^{\ast }f)\le K^N{\sum }_{\mathrm{\mid }\delta \mathrm{\mid }\le N}{b}_{N\delta }{\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }=\mathrm{\mid }\beta \mathrm{\mid }}\mathrm{\mid }{c}_{\beta \gamma }\mathrm{\mid }{p}_{\delta \gamma }(f)$, hence the asserted finite-maximum bound with $C_{\alpha \beta }:=K^N{\sum }_{\mathrm{\mid }\delta \mathrm{\mid }\le N}{b}_{N\delta }{\sum }_{\mathrm{\mid }\gamma \mathrm{\mid }=\mathrm{\mid }\beta \mathrm{\mid }}\mathrm{\mid }{c}_{\beta \gamma }\mathrm{\mid }$.

Every seminorm $p_{\alpha \beta }(A^{\ast }f)$ is finite by step 3.1, so $A^{\ast }f\in \mathcal{S}({\mathbb{R}}^n)$; and the same step with $A$ replaced by $A^{-1}$ shows $(A^{-1}{)}^{\ast }f\in \mathcal{S}({\mathbb{R}}^n)$, while $(A^{-1}{)}^{\ast }(A^{\ast }f)=f=(A^{\ast })(A^{-1}{)}^{\ast }f$. For continuity, fix a basic neighbourhood $p_{{\alpha }_r{\beta }_r}(g)<{\epsilon }_r$ ($r\le m$) of $0$ in the Schwartz topology [F5]; by step 3.1 the preimage under $f\mapsto A^{\ast }f$ contains the neighbourhood of $0$ cut out by the finitely many conditions $p_{\delta \gamma }(f)<{\epsilon }_r\mathrm{/}\max (C_{{\alpha }_r{\beta }_r},1)$ over the $(\mathrm{\mid }\delta \mathrm{\mid }\le \mathrm{\mid }{\alpha }_r\mathrm{\mid },\mathrm{\mid }\gamma \mathrm{\mid }\le \mathrm{\mid }{\beta }_r\mathrm{\mid })$ appearing in the $r$-th estimate, so the map is continuous at $0$ and, being linear, everywhere. The same argument applies to $f\mapsto f\circ A^{-1}$. No choice is used: all sums, constants and maxima above range over finite index sets determined by $\alpha ,\beta$ and the fixed matrix $A$.