# Closure of $L(\ell^2,\ell^2)$ in $L(\ell^2,\ell^\infty)$

Let $\ell^2$, $\ell^\infty$ denote the usual sequence spaces and let $L(\ell^2,\ell^2)$ the Banach space of bounded linear operators from $\ell^2$ to $\ell^2$ as well as $L(\ell^2,\ell^\infty)$ the Banach space of bounded linear operators from $\ell^2$ to $\ell^\infty$, equipped with the operator norm. Clearly, there holds

$$L(\ell^2,\ell^2)\subseteq L(\ell^2,\ell^\infty).$$

My question: How can I characterize the closure $\overline{L(\ell^2,\ell^2)}$ with respect to $L(\ell^2,\ell^\infty)$.

• Conjecture: the compact operators from $\ell^2$ into $c_0$. – couperin May 27 '14 at 16:14
• Certainly not, couperin. Consider the inclusion mapping from $\ell_2$ into $\c_0$. – Bill Johnson May 27 '14 at 17:03

The enclosed characterization is not completely satisfactory, but it is far from being clear that it is possible to get a better one. Observe that since $c_0$ is a closed subspace of $\ell_\infty$, the closure is in $L(\ell_2,c_0)$. It does not exhaust all of $L(\ell_2,c_0)$ as is shown in the answer to: Is $L(\ell_2,\ell_2)$ dense in $L(\ell_2,c_0)$?
Now let $B$ be an operator in the closure. It is standard that this happens if and only if we can find a sequence $B_i\in L(\ell_2,\ell_2)$ such that $||B_i||_{2\to\infty}\le 2^{-i}$ for $i\ge 2$ and $B=\sum_{i=1}^\infty B_i$ (convergence in $L(\ell_2,\ell_\infty)$). This implies that the sequence $\{Be_j\}$, where $\{e_j\}$ is the unit vector basis of $\ell_2$ can be decomposed as $Be_j=\sum_{i=1}^\infty B_ie_j$ (in $\ell_\infty$). Now let us consider sequences $\{B_ie_j\}_{j=1}^\infty$. This sequence for $i\ge 2$ has two properties: (1) As sequence in $\ell_2$ it is majorated by the orthonormal basis (in the sense that it is the image of the orthonormal basis under a bounded linear operator $\ell_2\to\ell_2$) and (2) $||\sum_{j=1}^\infty \alpha_jB_ie_j||_\infty\le 2^{-i}$ for every $\{\alpha_j\}_{j=1}^\infty\in\ell_2$.