The spectrum of an asymptotic operator is unbounded in both directions

Setting. Let $n \in \mathbb{N}$. 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 $f \in H$ (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 $D(A_S)$ 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 $\mu \in \mathbb{R}$, the eigenspace $E_\mu(A_S)$ is the real linear subspace $\{ f \in D(A_S) : A_S f = \mu f \}$ of $D(A_S)$; "$E_\mu(A_S) \neq 0$" means there is a nonzero $f \in D(A_S)$ with $A_S f = \mu f$.

Statement. Let $n \in \mathbb{N}$ with $n \ge 1$ (this hypothesis is assumed), let $S : \mathbb{T} \to \mathrm{Mat}_n$ be continuous with $S(t)$ symmetric for every $t \in \mathbb{T}$, and let $c \in \mathbb{R}$ be arbitrary (of any sign). Then both of the following hold:

1. there is $\mu \in \mathbb{R}$ with $\mu > c$ and a nonzero $f \in D(A_S)$ with $A_S f = \mu f$;
2. there is $\mu' \in \mathbb{R}$ with $\mu' < -c$ and a nonzero $f' \in D(A_S)$ with $A_S f' = \mu' f'$.

Since $c$ is arbitrary, this says that $A_S$ has eigenvalues strictly above every real number and eigenvalues strictly below every real number. No other hypotheses are assumed.