If $A$ is trace-class on $L^2(\mathbb R^2)$ then $\mathbb 1(-\Delta\leq 1)A\mathbb 1(-\Delta\leq 1)$ is finite-rank.

Question

I am trying to prove that if $A$ is trace-class on $L^2(\mathbb R^2)$ then $\mathbb 1(-\Delta\leq 1)A\mathbb 1(-\Delta\leq 1)$ is a finite-rank operator, where $\mathbb 1(-\Delta\leq 1)$ is defined on the Fourier side by $$\mathcal F[1(-\Delta\leq 1)\varphi](\xi)=1(|\xi|^2\leq 1)\widehat{\varphi}(\xi).$$

I want to prove that there exists an orthonormal basis $(\varphi_j)$ of $L^2(\mathbb R^d)$ where only finitely many $\varphi_j$ are supported in the set $\{|\xi|^2\leq 1\}$, something like $\varphi_j\approx e^{ij\cdot x}$.

Answer 1

First note that $1(-\Delta\leq 1) L^2(\mathbb{R}^2)$ is not finite dimensional. Pick $$(\varphi_j)_{j\in \mathbb{N}}\subseteq 1(-\Delta\leq 1) L^2(\mathbb{R}^2)$$ orthonormal (possible as infinite dimensional). Then define $$A=\sum_{j\in \mathbb{N}} \frac{1}{j^2} \vert \varphi_j\rangle \langle \varphi_j\vert.$$ Check that $A$ is trace-class and $A=1(-\Delta\leq 1) A 1(-\Delta\leq 1)$ is not finite rank.

Note that your statement becomes true if we would consider $L^2$ functions on the $n$-dimensional torus. In this case, the projection $1(-\Delta \leq 1)$ is a finite rank operator and hence $1(-\Delta \leq 1) A 1(-\Delta \leq 1)$ is finite rank for any bounded operator $A$.