The algebras of compact operators on $\ell_p$ as direct limits of matrix algebras

Question

Consider $M_n(\mathbb{C})$ as $B(\ell_p^n)$ for $n\in\mathbb{N}$ where $p\in[1,\infty)$, and include $M_n(\mathbb{C})$ in $M_{n+1}(\mathbb{C})$ as the upper left corner. Is it true that $\overline{\bigcup_n M_n(\mathbb{C})}=K(\ell_p)$ where $K(\ell_p)$ denotes the compact operators on $\ell_p$?

I know it is true when $p=2$. Also, I know that for $p\in[1,\infty)$, every $L_p$-space has the approximation property so every compact operator is approximable but does this imply the above?

Answer 1

Yes for $p\in (1,\infty)$. The reason is that the canonical basis of $\ell_p$ in this case is shrinking (a basis in a Banach space is shrinking if the coordinate functionals form a basis of the dual space; every basis in a reflexive space is shrinking).

Let $(e_n)_{n=1}^\infty$ be a shrinking basis in a Banach space $E$ and let $(e_n^*)_{n=1}^\infty$ be the associated coordinate functionals. Set $P_n = e_1\otimes e_1^* + \ldots e_n\otimes e_n^*$ ($n\in \mathbb{N}$). Then for a bounded linear operator $T\colon E\to E$ the following are equiverdicial:

1. $T$ is compact,
2. $\|T-TP_n\|\to 0$ as $n\to \infty$.

Note that the ranges of $P_n$ can be identified with $\mathscr{B}(\ell_p^n)$, so the conclusion follows as in this case we have $\|T-P_nTP_n\|\to 0$. As for the proof of the above result you may consult Lemma 2.1 here.