The domain of the asymptotic operator is $W^{1,2}(S^1,\mathbb{R}^{2n})$ and is dense

Statement. For every $n \in \mathbb{N}$ (including $n = 0$) and every continuous map $S : S^1 \to \mathrm{Mat}_n$ — no symmetry or other hypothesis on $S$ — both of the following hold for the operator $\mathcal A_S$ defined below:

1. for every $f \in L^2_n$: $f$ lies in the domain of $\mathcal A_S$ if and only if $f$ is in $W^{1,2}$;
2. the domain of $\mathcal A_S$ is dense in $L^2_n$ (for the $L^2$ norm).

Meaning of the symbols.

• $S^1 = \mathbb{R}/\mathbb{Z}$ with its Haar probability measure $\lambda$; $\mathrm{Mat}_n$ is the space of real $2n\times 2n$ matrices indexed by $\{1,\dots,n\}\sqcup\{1,\dots,n\}$; $J_0 = \begin{pmatrix} 0 & -I_n \\ I_n & 0\end{pmatrix}$.
• $V_n = \mathbb{R}^{2n}$ with the Euclidean inner product, and $L^2_n = L^2(S^1,\lambda;V_n)$, the real Hilbert space of a.e.-classes of square-integrable maps with $\langle f, g\rangle = \int \langle f(t), g(t)\rangle\, d\lambda(t)$ and norm $\|f\| = (\int |f|^2 d\lambda)^{1/2}$. Pointwise values of an element of $L^2_n$ are those of an arbitrary measurable representative.
• Fourier coefficients: $\hat f_k = \int_{S^1} e^{-2\pi i k t} f(t)\, d\lambda(t) \in \mathbb{C}^{2n}$ and $\widehat{(Sf)}_k = \int_{S^1} e^{-2\pi i k t} S(t) f(t)\, d\lambda(t)$, $k \in \mathbb{Z}$ (real entries viewed in $\mathbb{C}$; Bochner integral, value $0$ if not integrable).
• $f$ is "in $W^{1,2}$" iff $\sum_{k\in\mathbb{Z}} (1+k^2)\sum_i |(\hat f_k)_i|^2$ is summable, $|\cdot|$ the complex modulus, $i$ over all $2n$ coordinates.
• $\Gamma_S$ is the set of pairs $(f, g) \in L^2_n\times L^2_n$ with $f$ in $W^{1,2}$ and $\hat g_k = -(2\pi k\, i) J_0 \hat f_k - \widehat{(Sf)}_k$ for all $k \in \mathbb{Z}$.
• $\mathcal A_S$ is defined by cases: if some partially defined real-linear operator on $L^2_n$ has graph exactly $\Gamma_S$, then $\mathcal A_S$ is such an operator (picked by choice); otherwise $\mathcal A_S$ is the zero operator with domain all of $L^2_n$. The theorem is about this $\mathcal A_S$, whichever case occurs (it does not itself assume or assert which case occurs). In the fallback case the domain is all of $L^2_n$, so claim 1 would then say that every $f \in L^2_n$ is in $W^{1,2}$, and claim 2 holds trivially.
• For $f \in L^2_n$ (on a probability space) and continuous $S$, both Fourier integrands are integrable, so the value-$0$ convention for non-integrable integrands does not arise for elements of $L^2_n$.

For $n = 0$, $L^2_0 = \{0\}$ and both claims concern the single element $0$.