# Insights about a frame-like inequality

I'm a graduate student doing research on time-frequency analysis. I am considering the existence of a certain frame-like inequality. Let $H: L^2(\mathbb{R}^d) \rightarrow L^2(\mathbb{R}^d)$ be a compact self-adjoint operator, and for each $(x,\omega)\in \mathbb{R}^{d} \times \mathbb{R}^d,$ let $\pi(x,\omega): L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R}^{d})$ be the time-frequency shift operator $\pi(x,\omega)f =f(t-x) e^{2\pi i \langle \omega,t-x \rangle}$. Fix a window function $g\in L^2(\mathbb{R}^{d}),$ and suppose there exist finite points $\{(x_n,\omega_{n})\}_{n=1}^q$ such that there are $A,B>0$ where for all $f\in L^2(\mathbb{R}^{d})$ $$A||Hf||_2^2 \leq \sum_{n=1}^q |\langle Hf, \pi(x_n,\omega_n)g \rangle|^2 \leq B||Hf||_2^2.$$ You might recognize that the inequality above looks like the Gabor Frame inequality except we only have finite 'atoms' and we have a compact self-adjoint operator inside. Let us review the spectral theorem for compact operators:

Let $H$ be a Hilbert space and $T \in K(H)$ where $K(H)$ denotes the set of compact operators from $H$ to $H$. Let $T$ be normal or self adjoint depending on whether the underlying field is $\mathbb R$ or $\mathbb C$, then there exists an orthonormal set $\{e_i \mathrel| i\in I\}$ where $I$ is either $\mathbb N$ or $\{1,2,\dotsc,k\}$ and a sequence $(\lambda_i)_{i\in I} \in \mathbb K$ which converges to $0$ such that

$$\operatorname{span}{(x_i : i ∈ I})^{⊥} = \ker(T)$$

and also $\forall x \in H$ : $Tx= \sum_{i\in I} \lambda_i \langle x, e_i \rangle e_i$ with unconditional convergence.

Now denote the eigenvalue-eigenfunction pairs of $H$ by $(\lambda_i, e_i)_{i \in \mathcal{I}}$. For each $N\in \mathbb{N}$ let $V_N = \operatorname{span}\{e_1,e_2,...,e_N \} \subset L^2(\mathbb{R}^d)$, and $P_N$ be the orthogonal projection operator from $L^2(\mathbb{R}^d)$ onto $V_N.$ Since orthogonal projections are self-adjoint and idempotent, then for all $f\in L^2(\mathbb{R}^d)$ the inequality above implies: $$A||P_NHf||_2^2 \leq \sum_{n=1}^q |\langle P_NHf, P_N\pi(x_n,\omega_n)g \rangle|^2 \leq B||P_NHf||_2^2.$$ The above is also due to the fact that $H$ commutes with $P_N,$ which should not be hard to prove. From frame theory, the inequality above implies that for all $N\in\mathbb{N},$ the finite set $\{P_N\pi(x_n,\omega_n)g \}_{n=1}^q$ is a spanning set of $\operatorname{Im}P_NH,$ and therefore $\sup_{N\in \mathbb{N}}(\operatorname{dim}\operatorname{Im}P_NH) = \sup_{N\in \mathbb{N}}(\operatorname{rank}P_NH) \leq q < \infty$ It should not also be that hard to show that $V_N = \operatorname{Im}HP_N.$ So finally we have $\sup_{N\in \mathbb{N}}\operatorname{dim}V_N \leq q.$ Am I right in thinking that this should imply that $H$ should only have a finite spectrum (and hence finite rank)?

Does this mean that our starting inequality (provided it is possible to construct one) is only valid for compact self-adjoint operators whose spectrum is finite?

Yes, your starting inequality can only hold if $H$ has finite dimensional range. In fact, the dimension can be at most $q$.
To see this, let $Y$ denote the range of $H$. Your inequality implies (why?) that if $y \in Y$ with $\langle y, \pi(x_n, \omega_n) g\rangle =0$, then $y =0$. In other words, the linear map $$Y \to \Bbb{C}^q, y \mapsto (\langle y, \pi(x_n, \omega_n) g\rangle )_{i=1,\dots,q}$$ is injective. By elementary linear algebra, this means that $Y$ has dimension at most $q$.
Note that we did not use any properties of the operator $H$, except that its range is a linear space and that it satisfies your inequality.
• Your mapping is injective since norm 'norm' equivalences are also equivalence relations (i.e we can swap the inequality above so that $A^{\prime} \sum |\langle Hf, \pi(x_n,\omega_n)g \rangle|^2 \leq ||Hf||_2^2 \leq B^{\prime} \sum |\langle Hf, \pi(x_n,\omega_n)g \rangle|^2$). So your hypothesis would imply $Hf = 0$ from the axioms of a normed linear space. Anyway, did you use the rank-nullity theorem for your conclusion? It's kinda confusing for me since AFAIK it can only be used if we know a priori that our mapping's domain is finite dimensional, which is kind of circular. – Kurome Jun 25 '18 at 2:00
• @Kurmo: Let's call the map from my proof $\Phi$. Since it is injective, I we denote by $Z$ the range of $\Phi$, then $\Phi : Y \to Z$ is an isomorphism. Since $Z$ is finite dimensional as a subspace of the finite dimensional space $\Bbb{C}^q$, also $Y$ is finite dimensional [if $z_1,...,z_N$ is a basis for $Z$, then $\Phi^{-1}(z_1),...,\Phi^{-1}(z_N)$ is a basis for $Y$]. – PhoemueX Jun 25 '18 at 5:30
• I see, I think we can also simply argue that: $Y$ cannot be isomorphic to a space of finite dimension if it is of infinite dimension. Thank you, that's illuminating. – Kurome Jun 25 '18 at 6:03