# proving that $\mathcal{A}_\infty(X)$ is or is not norm-closed in $\mathcal{L}(X)$ for each Banach space $X$

Fix any $1\leq p\leq\infty$. If $X$ is a Banach space and $C\in(0,\infty)$, we say that $T\in\mathcal{A}_C(X)$ whenever, for each $(x_n)_{n=1}^\infty\subset B_X$ (where $B_X$ is the closed unit ball in $X$), and each $\epsilon>0$, there exists $(a_n)_{n=1}^\infty\in c_{00}$ such that $$\|\sum_{n=1}^\infty a_nTx_n\|_X<\epsilon\|(a_n)_{n=1}^\infty\|_p\;\;\;\text{ and }\;\;\;\|\sum_{n=1}^\infty a_nx_n\|_X\leq C\|(a_n)_{n=1}^\infty\|_p$$ where $\|\cdot\|_p$ denotes the $\ell_p$-norm. We then define $$\mathcal{A}_\infty(X):=\bigcup_{0<C<\infty}\mathcal{A}_C(X).$$

Proposition. The set $\mathcal{A}_\infty(X)$ is a linear space and a two-sided ideal in $\mathcal{L}(X)$, which contains the compact operators.

Question #1. Is $\mathcal{A}_\infty(X)$ norm-closed in $\mathcal{L}(X)$?

The answer to question #1 above is yes in whenever $p=1$ or $p=\infty$. In these cases, $\mathcal{A}_\infty(X)=\mathcal{R}(X)$ or $\mathcal{A}_\infty(X)=\mathcal{K}(X)$, respectively, where $\mathcal{R}$ is the class of Rosenthal operators and $\mathcal{K}$ the compact operators, each norm-closed operator ideals. However, it is not clear whether $\mathcal{A}_\infty(X)$ is norm-closed in $\mathcal{L}(X)$ when $1<p<\infty$.

The next proposition outlines an idea for giving an affirmative answer to question #1. The proof is based on a simple application of the Baire Category Theorem.

Proposition. Suppose there exists a norm $\|\cdot\|_{\mathcal{A}_\infty(X)}$ on the space $\mathcal{A}_\infty(X)$ which satisfies the following two properties:

(i) There exists $K\in(0,\infty)$ such that for all $C\in[K,\infty)$, the set $\mathcal{A}_C(X)$ is $\|\cdot\|_{\mathcal{A}_\infty(X)}$-closed in $\mathcal{A}_\infty(X)$.

(ii) The space $\mathcal{A}_\infty(X)$ is complete under the norm $\|\cdot\|_{\mathcal{A}_\infty(X)}$.

Then $\mathcal{A}_\infty(X)$ is a closed subspace of $\mathcal{L}(X)$ under the operator norm.

I'm not sure whether that helps or not. Or maybe it is a lost cause, and there is some choice of $X$ for which $\mathcal{A}_\infty(X)$ is not norm-closed in $\mathcal{L}(X)$.

What do y'all think? Ideas?

Thanks!