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 $f\in L^2(\mathbb{R}^{d})$
\begin{equation}
A||Hf||\_2^2 \leq \sum\_{n=1}^q |\langle Hf, \pi(x\_n,\omega\_n)g \rangle|^2 \leq B||Hf||\_2^2.
\end{equation}
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:
\begin{equation}
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.
\end{equation}
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?



Insights about a frame-like inequality
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.


Finally, it is not too hard to see that a self adjoint operator with finite rank has finite spectrum.


