Proposition 18

The restriction to $\mathscr{S}(\mathbf{R}^n)$ of the Fourier transform is an automorphism of topological vector spaces whose inverse is the restriction of the cotransform of Fourier.

Let $\varphi \in \mathscr{S}(\mathbf{R}^n)$. By the preceding lemma, the Fourier transform of $\varphi$ belong to $\mathscr{C}^{\infty}(\mathbf{R}^n)$. Moreover, for $\alpha \in \mathbf{N}^n$ and $\beta \in \mathbf{N}^n$, one has

In particular, the function $X^{\beta}\partial^{\alpha}(\mathscr{F}(\varphi ))$ is bounded. Since $\alpha$ and $\beta$ are arbitrary in $\mathbf{N}^n$, this means that $\mathscr{F}(\varphi )$ belongs to $\mathscr{S}(\mathbf{R}^n)$. Moreover, this calculation implies

for $(\alpha , \beta )\in \mathbf{N}^n\times \mathbf{N}^n$ and $\varphi \in \mathscr{S}(\mathbf{R}^n)$.

Since the inclusion of the space $\mathscr{S}(\mathbf{R}^n)$ in $L^1(\mathbf{R}^n)$ is continuous (prop. 13 of IV, p. 213), the mapping $q_{\alpha ,\beta}\circ \mathscr{F}$ from $\mathscr{S}(\mathbf{R}^n)$ into $\mathbf{R}$ is continuous. It follows that the Fourier transform is continuous from the space $\mathscr{S}(\mathbf{R}^n)$ into itself (cf. EVT, II, p. 7, prop. 5, c)). One verifies analogously that the Fourier cotransform is continuous from the space $\mathscr{S}(\mathbf{R}^n)$ into itself. By the fourier inversion formula (theorem 3 of II, p. 222), the Fourier transform and the Fourier cotransform are reciprocal isomorphisms of one another.