The ideal generated by the discrete multiplication operators is closed

Question

I have been trying to show that the ideal of $B(\ell^2,\ell^2)$ given by $$I=\{\alpha_1M_{x_1}\beta_1+\cdots+\alpha_nM_{x_n}\beta_n:\alpha_j,\beta_j\in B(\ell^2,\ell^2),x_j\in c_0,n<\infty\}$$ is closed, where $M_x:(v_n)_n\mapsto(x_nv_n)_n$ for any $v\in\ell^2$. Here $c_0$ denotes the sequences in $\textbf{C}$ that vanish (indexed by the natural numbers), and $\ell^2$ denotes the square-summable sequences in $\textbf{C}$ (indexed by the natural numbers). I considered an arbitrary Cauchy sequence $$\left(\sum_{j\in\mathcal{J}_n}\alpha_jM_{x_j}\beta_j\right)_n$$ in $I$ where $(\mathcal{J}_n)_n$ is some sequence in $\mathcal{P}(\textbf{N})$. The problem I have is that $\#\mathcal{J}_n$ is not necessarily constant, nor does it necessarily converge to a finite number. Is there a better way to approach this problem?

Answer 1

Note that, since each $M_x$ is a compact operator, $I$ is contained in the ideal $K(\ell^2)$ of $B(\ell^2)$ consisting of compact operators on $\ell^2$, which is a closed ideal. Now given any compact operator $T\in K(\ell^2)$, write $T_r=\frac12(T+T^*)$, $T_i=\frac{1}{2i}(T-T^*)$. Then $T_r$ and $T_i$ are compact, self-adjoint operators on $\ell^2$, so the spectral theorem tells us that we can find unitaries $u_r,u_i\in B(\ell^2)$ and $c_0$-sequences $x_r,x_i$ such that $T_r=u_rM_{x_r}u_r^*$ and $T_i=u_iM_{x_i}u_i^*$. Now $$T=T_r+iT_i=u_rM_{x_r}u_r^*+(iu_i)M_{x_i}u_i^*\in I.$$ Therefore $I=K(\ell^2)$ is a closed ideal of $B(\ell^2)$.