Throughout, $X$ and $Y$ will denote Banach spaces with $T\in\mathcal{L}(X,Y)$ (the space of continuous linear operators between $X$ and $Y$). We define the operator $\overline{T}\in\mathcal{L}(X^{**}/X,Y^{**}/Y)$ by the rule
\begin{equation}\overline{T}(x^{**}+X)=T^{**}x^{**}+Y.\end{equation}
Suppose $\mathcal{J}$ is an operator ideal (see here for the definition). We define the **co-ideal** of $\mathcal{J}$ via components
\begin{equation}\mathcal{J}^{co}(X,Y):=\left\{T\in\mathcal{L}(X,Y):\overline{T}\in\mathcal{J}(X^{**}/X,Y^{**}/Y)\right\}\end{equation}
It is clear that $\mathcal{J}^{co}$ is an operator ideal. Furthermore, it's easy to check that if $\mathcal{J}(X,Y)$ is closed (as a subspace of $\mathcal{L}(X,Y)$ endowed with the operator norm) then so is $\mathcal{J}^{co}(X,Y)$. (I do not know whether the converse is true in general.)

Denote by $\mathcal{F}$, $\mathcal{K}$, and $\mathcal{X}$, the respective ideals of finite-rank operators, compact operators, and operators with separable range. We define the ideal of **quasi-weakly compact** operators as the co-ideal $\mathcal{F}^{co}$. This may seem like an odd name, but it makes sense given that an operator is quasi-weakly compact if and only if it factors through a quasireflexive space. (Similarly, an operator is weakly compact if and only if it factors through a reflexive space.)

The following is probably already known, but if not then it is easy to prove by using weighted $\ell_2$-direct sums of identity operators acting on quasireflexive spaces of order $n=1,2,3,\cdots$.

**Proposition 1.** The ideal of quasi-weakly compact operators is not norm-closed. In particular, there exists a component space $\mathcal{F}^{co}(X)$ which is not norm-closed.

As the ideal of quasi-weakly compact operators is not closed, then what exactly *is* its closure? I suspect the following.

**Conjecture 2.** The closure of the quasi-weakly compact operators is just the co-ideal of the approximable operators. In symbols: $\overline{\mathcal{F}^{co}}=(\overline{\mathcal{F}})^{co}$.

*Proof idea.* It is clear that $\overline{\mathcal{F}^{co}}\subseteq(\overline{\mathcal{F}})^{co}$, so it remains only to show that $(\overline{\mathcal{F}})^{co}\subseteq\overline{\mathcal{F}^{co}}$. Suppose $T\in(\overline{\mathcal{F}})^{co}(X,Y)$. Then we can find finite-rank operators $\widehat{F}_n\in\mathcal{L}(X^{**}/X,Y^{**}/Y)$ with $\widehat{F}_n\to\overline{T}$. To complete the proof, it is sufficient to find $F_n\in\mathcal{L}(X,Y)$ such that $\overline{F}_n=\widehat{F}_n$ and $\|F_n-T\|\leq\|\widehat{F}_n-\overline{T}\|$. However, I do not immediately see how to construct $F_n$.

Regardless of whether or not conjecture 2 is true, we might be more interested in co-compact operators, i.e. the ideal $\mathcal{K}^{co}$, than we would in co-approximable operators. (I suppose there is a chance that these ideals coincide, i.e. that $\mathcal{K}^{co}=(\overline{\mathcal{F}})^{co}$, although I doubt it very much.)

**Question 3.** Is there any literature investigating the properties of $\mathcal{K}^{co}$?

As operators factoring through quasi-reflexive spaces fail to form a closed operator ideal, we could instead consider operators factoring through a Banach space $E$ such that $E^{**}/E$ is separable. By Example 2.5 and Theorem 2.6 in [Go93], this ideal coincides with the co-ideal $\mathcal{X}^{co}$, and hence is norm-closed.

Note that $E^{**}/E$ being separable is a very natural generalization of quasireflexivity, and as such has probably already been investigated. This property probably has a special name, too. However, I can find no information on this.

**Question 4.** Is there any literature discussing the property of $E^{**}/E$ being separable, and/or discussing the ideal $\mathcal{X}^{co}$ of operators factoring through such spaces?

Thanks!