Fourier transform of a tempered distribution

Assume Countable Choice, exactly as required by the published Schwartz Fourier theorem. For $\phi \in \mathcal{S}({\mathbb{R}}^n)$ put

$\widehat{\phi }(\xi )=\mathcal{F}\phi (\xi )={\int }_{{\mathbb{R}}^n}\phi (x)e^{-2\pi ix\cdot \xi }dx\mathrm{.}$

For $u\in {\mathcal{S}}^{\mathrm{\prime }}({\mathbb{R}}^n)$ define its Fourier transform by

$⟨\mathcal{F}u,\phi ⟩=⟨u,\mathcal{F}\phi ⟩(\phi \in \mathcal{S}({\mathbb{R}}^n))\mathrm{.}$

The published automorphism Fourier transform is a topological automorphism of Schwartz space sends Schwartz tests continuously to Schwartz tests, so composition with the continuous functional of Tempered distribution is again a continuous complex-linear functional. Hence $\mathcal{F}u\in {\mathcal{S}}^{\mathrm{\prime }}$. The pairing is bilinear: there is no conjugation and no inverse transform on the right-hand side. Countable Choice is used only through the cited published Fourier construction; transposition itself uses no additional choice.