Periodic fluid Fourier Sobolev spaces

Under $x=2\pi y$, normalized measure on ${\mathbb{T}}^d$ pushes forward to $(2\pi {)}^{-d}dx$ on $Q_d$, while $e^{2\pi ik\cdot y}$ becomes $e^{ik\cdot x}$. Thus Fourier coefficients and the componentwise $L^2$ inner product are unchanged by this rescaling.

Apply [F4] to each of the $N$ components after this rescaling. The coefficient map identifies scalar $L^2(Q_d)$ isometrically with ${\mathrm{\ell }}^2({\mathbb{Z}}^d)$; summing the component identities gives the vector-valued coefficient isometry. The Countable Choice hypothesis is exactly [F1].

Since $w_k^{2s}>0$, symmetric truncations to $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}\le M$ converge in the weighted ${\mathrm{\ell }}^2$ norm by convergence of the defining nonnegative series. The conjugate-symmetry condition is preserved by these cutoffs. To verify completeness of the coefficient space without Countable Choice, let $(a_n)$ be a Cauchy sequence in its weighted norm. For each $k$, $\mathrm{\mid }{a}_n(k)-a_m(k)\mathrm{\mid }\le w_k^{-s}\mathrm{\parallel }{a}_n-a_m{\mathrm{\parallel }}_{H^s},$ so scalar completeness [F5] applied to the finitely many real and imaginary coordinates gives a unique limit $a(k)$. Conjugate symmetry passes to these coordinate limits. Given $\epsilon >0$, choose $n_0$ so that $\mathrm{\parallel }{a}_n-a_m{\mathrm{\parallel }}_{H^s}\le \epsilon$ for $n,m\ge n_0$. For fixed $n\ge n_0$ and every finite box $F_M=\{k:\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}\le M\}$, passing $m\to \mathrm{\infty }$ in the finite sum gives ${\sum }_{k\in F_M}{w}_k^{2s}\mathrm{\mid }{a}_n(k)-a(k){\mathrm{\mid }}^2\le {\epsilon }^2\mathrm{.}$ The inequality $\mathrm{\mid }a(k){\mathrm{\mid }}^2\le 2\mathrm{\mid }{a}_n(k){\mathrm{\mid }}^2+2\mathrm{\mid }a(k)-a_n(k){\mathrm{\mid }}^2$ bounds the finite-box weighted sums for $a$ by $2\mathrm{\parallel }{a}_n{\mathrm{\parallel }}_{H^s}^2+2{\epsilon }^2$, so $a$ lies in the weighted coefficient space. Taking the supremum over $M$ in the preceding difference bound gives $\mathrm{\parallel }{a}_n-a{\mathrm{\parallel }}_{H^s}\le \epsilon$. Thus the coefficient space is complete, and the explicit cutoff density already proved identifies it with the completion of finite coefficient arrays. The coordinate limits are unique and the cutoffs are specified, so this argument makes no choice of function representatives and uses no Countable Choice. For $s\ge 0$ under Countable Choice, $w_k^{2s}\ge 1$, hence the coefficient sequence is in ${\mathrm{\ell }}^2$ and [F4] gives its unique initially complex $L^2$ class. For each component $f$ of this class, complex conjugation preserves $L^2$ and the normalized coefficient convention gives $\widehat{\bar{f}}(k)=\overline{\widehat{f}(-k)}=\overline{a_{-k}}=a_k=\widehat{f}(k),$ where $a_k$ is its coefficient coordinate. Uniqueness of Fourier synthesis in [F4] therefore gives $f=\bar{f}$ almost everywhere, so $\mathrm{Re}f$ is a real representative with the same coefficients. Applying this to the finitely many components gives a real ${\mathbb{R}}^N$-valued representative, unique up to equality almost everywhere by the coefficient isometry in step 2.1. For $s<0$ the construction gives only the weighted coefficient completion and makes no $L^2$ claim.