Weighted Fourier candidate norm on Schwartz space

Assume Countable Choice. Fix $n\ge 1$ and $s\in \mathbb{R}$, and let $\widehat{u}=\mathcal{F}u$ use the repository's negative-sign $2\pi$ Fourier transform. For $u,v\in \mathcal{S}({\mathbb{R}}^n)$ define $Q_s(u,v)={\int }_{{\mathbb{R}}^n}⟨\xi {⟩}^{2s}\widehat{u}(\xi )\overline{\widehat{v}(\xi )}d\xi ,q_s(u)=\parallel ⟨\xi {⟩}^s\widehat{u}{\parallel }_{L^2}\mathrm{.}$

The integral is finite and $Q_s$ is linear in its first variable. Indeed, Fourier transformation preserves Schwartz space (Fourier transform is a topological automorphism of Schwartz space), the bracket multiplier preserves it (Real powers of the Japanese bracket act on Schwartz space), and Schwartz functions define $L^2$ classes (Schwartz space is dense in L2); the complex $L^2$ pairing and Cauchy–Schwarz are those of Complex completeness, density, and inner product: the consumer interface. Since the weight is real and positive, $Q_s(u,v)$ is the corresponding $L^2$ pairing of $⟨\xi {⟩}^s\widehat{u}$ and $⟨\xi {⟩}^s\widehat{v}$. At this stage $q_s$ is only the candidate seminorm; the next item proves that its kernel is zero.

The Countable Choice assumption (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) is inherited from the cited Fourier and complex $L^2$ interfaces. The integral definition itself makes no selection, and no full Axiom of Choice is used. With the repository convention $\mathcal{F}({\mathrm{\partial }}_{j}u)=2\pi i{\xi }_j\mathcal{F}u$, this definition asserts no equality at integer order with a derivative-sum norm or the norm defined by the symbol $(1+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{k\mathrm{/}2}$.