If the operator $U^2D_2+UD_1+D_1U^*+D_2U^{*2}$ is compact then $D_1$ and $D_2$ are compact

Question

I need to show that if $U: \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$ is the right shift operator and $D_1$ and $D_2$ are multiplication operators by the sequences $(a_n)_{n \in \mathbb{Z}}$ and $(b_n)_{n \in \mathbb{Z}}$, respectively, then if the opeator $$U^2D_2+UD_1+D_1U^*+D_2U^{*2}$$ is compact, $D_1$ and $D_2$ are also compact. Could you give me any help, please?

Answer 1

For the element of the standard basis $\{e_n\}$ in $\ell^2(\mathbb{Z})$ the vectors $U^2D_2e_n,$ $UD_1e_n,$ $D_1U^*e_n$ and $D_2U^{*2}$ are pairwise orthogonal. Hence $$\|Te_n\|^2=\|U^2D_2e_n\|^2+ \|UD_1e_n\|^2+\|D_1U^*e_n\|^2+\|D_2U^{*2}\|^2\\ \ge \|D_2e_n\|^2+\|D_1e_n\|^2$$ Since $T$ is compact we get $\|Te_n\|\to 0,$ when $|n|\to \infty. $ Therefore $\|D_1e_n\|,\ \|D_2e_n\|\to 0$ when $|n|\to \infty.$ For diagonal operators the latter implies their compactness.

Answer 2

Claim 1 The multiplication operator $x\mapsto (c_n x_n)_{n\in \mathbb{Z}}$ is compact if and only if $c_n$ tends to zero.

Claim 2 A linear operator $T: \ell_2({\mathbb{Z}})\longrightarrow \ell_2({\mathbb{Z}})$ is compact if and only if is weak-norm sequentially continuous.

edit Identify $\ell_2(\mathbb{Z})$ with $\ell_2(\mathbb{N})$ for simplicity.

Now the sequence of canonical vectors $e_n$ converges weakly to zero, let $T = U^2D_2+UD_1+D_1U^\ast+D_2U^{\ast^2}$, then it must be $||Te_n||\to 0$. Computing

$T(e_n)_j=\begin{cases}b_{n-2} \quad &j=n-2\\ a_{n-1} \quad &j=n-1\\ a_n\quad &j=n+1\\ b_n\quad &j=n+2\\ 0 &\text{otherwise}\end{cases}$

therefore $||T(e_n)||^2 =b^2_{n-2}+a^2_{n-1}+a^2_n+b_n^2$, so it must be that $a_n$ and $b_n$ both tend to zero, therefore $D_1$ and $D_2$ are compact.