The asymptotic operator has an orthonormal eigenbasis with discrete spectrum

Setting. Let $n \in \mathbb{N}$ be arbitrary ($n = 0$ is allowed). Let $I = \{1,\dots,n\} \sqcup \{1,\dots,n\}$ (a first and a second copy), $V_n = \mathbb{R}^I \cong \mathbb{R}^{2n}$ with the standard Euclidean inner product, and $\mathrm{Mat}_n$ the real $I \times I$ (i.e. $2n \times 2n$) matrices. In block form with respect to the two copies, $J_0 = \begin{pmatrix} 0 & -\mathrm{Id}_n \\ \mathrm{Id}_n & 0 \end{pmatrix}$. Let $\mathbb{T} = \mathbb{R}/\mathbb{Z}$ with its Haar probability measure $dt$ (total mass $1$), and let $H = L^2(\mathbb{T}, V_n)$ be the real Hilbert space of a.e.-classes of square-integrable functions, with $\langle f, g \rangle = \int_{\mathbb{T}} \langle f(t), g(t) \rangle \, dt$. For $n = 0$, $V_0 = \{0\}$ and $H = \{0\}$.

For $f \in H$ (any representative) and $k \in \mathbb{Z}$, the Fourier coefficient is the vector $\hat f_k = \int_{\mathbb{T}} e^{-2\pi i k t} f(t)\, dt \in \mathbb{C}^I$, where $f(t)$ is regarded as a complex vector componentwise; for a continuous map $S : \mathbb{T} \to \mathrm{Mat}_n$, $\widehat{(Sf)}_k = \int_{\mathbb{T}} e^{-2\pi i k t} S(t) f(t)\, dt \in \mathbb{C}^I$ (with $S(t)$ regarded as a complex matrix). These are Bochner integrals (which by convention are $0$ when the integrand is not integrable). Say $f \in W^{1,2}$ if the family $\big((1 + k^2) \sum_{i \in I} |(\hat f_k)_i|^2\big)_{k \in \mathbb{Z}}$ is summable. Let $G_S \subseteq H \times H$ be the set of pairs $(f, g)$ with $f \in W^{1,2}$ and

The operator $A_S$ is defined as follows: if there exists a partially defined linear map on $H$ (a linear subspace $D \subseteq H$ and a linear map $T : D \to H$) whose graph $\{(x, Tx) : x \in D\}$ equals $G_S$, then $A_S$ is such a map, chosen by the axiom of choice (its domain is then $\{f : \exists g,\ (f,g) \in G_S\}$ and $A_S f$ is the corresponding $g$); if no such map exists, $A_S$ is the zero map with domain all of $H$. The statement does not itself assert which case occurs.

For a partially defined linear map $A$ on $H$ with domain $D(A)$, "$A$ has a discrete eigenbasis" means: there exist an index set $\iota$, a Hilbert basis $(b_i)_{i \in \iota}$ of $H$ (an orthonormal family with dense linear span), and real numbers $(\lambda_i)_{i \in \iota}$ such that every $b_i$ lies in $D(A)$ with $A b_i = \lambda_i b_i$, and $|\lambda_i| \to \infty$ along the cofinite filter on $\iota$, i.e. for every $C \in \mathbb{R}$ the set $\{ i \in \iota : |\lambda_i| \le C \}$ is finite. (When $\iota$ is finite this last condition holds automatically.)

Statement. For every $n \in \mathbb{N}$ and every continuous $S : \mathbb{T} \to \mathrm{Mat}_n$ such that $S(t)$ is symmetric ($S(t)^{\mathsf T} = S(t)$) for every $t \in \mathbb{T}$, the operator $A_S$ (as defined above, including the fallback) has a discrete eigenbasis in the above sense: there are an index set $\iota$, a Hilbert basis $(b_i)$ of $L^2(\mathbb{T}, \mathbb{R}^{2n})$, and reals $\lambda_i$ with $b_i \in D(A_S)$, $A_S b_i = \lambda_i b_i$ for all $i$, and $\{i : |\lambda_i| \le C\}$ finite for every $C$. No other hypotheses are assumed. For $n = 0$ the space is $\{0\}$ and the statement is satisfied with an empty index set.