The space of bounded operators $L(X \rightarrow X)$ where $X=\ell_1$ is NOT seperable

Question

How can one prove that the space of bounded operators $L(X \rightarrow X)$ where $X=\ell_1$ is NOT seperable? I am trying to think what can contradict seperability, but I didn't have any progress.

Answer 1

Consider the canonical Schrauder Basis $\{e_n\}_{n\in \mathbb{N}}$ of $\ell^1(\mathbb{N})$ given by $e_n(k)=\delta_{kn}$. Each permutation $\sigma: \mathbb{N}\rightarrow \mathbb{N}$ induces an isometric linear map $T_\sigma: X \rightarrow X$ given by $T_{\sigma}(e_n)=e_{\sigma(n)}$. Now, for two distinct permutations $\sigma_1$ and $\sigma_2$ there exists $k\in \mathbb{N}$ such that $\sigma_1(k)\neq \sigma_2(k)$. Thus, $$\|T_{\sigma_1}-T_{\sigma_2}\|\geq \|T_{\sigma_1}(e_k)-T_{\sigma_2}(e_k)\|=\|e_{\sigma_1(k)}-e_{\sigma_2(k)}\|=2.$$ Then $\{T_\sigma: \sigma \text{ is a permutation on }\mathbb{N}\}$ is an uncountable subset of $\ell^1(\mathbb{N})$ whose elements are all a distance of at least $2$ apart.

Answer 2

I prefer the following. Let $E\subset\mathbb{N}.$ Define $T_E:X\to X$ by $$T_E(e_n)=\begin{cases} e_n & n\in E\\ 0 & n\notin E\end{cases}$$ where $\{e_n\}_{n=1}^\infty$ denotes the standard topological basis in $\ell^1.$

Then $\|T_E\|=1$ and $\|T_E-T_F\|=2$ for $E\neq F.$ The cardinality of $\{T_E\}_{E\subset \mathbb{N}}$ is continuum.

According to suggestion of @RobertFurber, one can as well consider $$S_E(x)=\left (\sum_{n\in E} x_n\right )e_1$$ Again $\|S_E\|=1,$ $\|S_E-S_F\|=2$ for $E\neq F.$

Remark The first method can be applied to any $\ell^p$ space.