An operator ideal $\mathfrak J$ is a class of continuous operators. Namely, for every pair of complex Banach spaces, $\mathfrak X,\mathfrak Y$, we have that $\mathfrak J(\mathfrak X,\mathfrak Y) \subseteq \mathfrak L(\mathfrak X,\mathfrak Y)$ is a closed two-sided ideal, which means 
\begin{align\*}
1.& \ \ \ A,B\in \mathfrak J(\mathfrak X,\mathfrak Y) \Rightarrow A+B \in \mathfrak J(\mathfrak X,\mathfrak Y), \\
2.& \ \ \ \mathfrak L(\mathfrak W,\mathfrak X)\mathfrak J(\mathfrak X,\mathfrak Y)\mathfrak L(\mathfrak Y,\mathfrak Z) \subseteq \mathfrak J(\mathfrak W,\mathfrak Z),\ \textrm{and} \\
3.& \ \ \ \mathfrak J(\mathfrak X,\mathfrak Y) \supseteq \mathfrak F(\mathfrak X,\mathfrak Y), \ \textrm{the finite-rank operators}.
\end{align\*}


Now for my question: Let $\mathfrak X$ be any complex Banach space and suppose $J$ is a non-trivial closed two-sided ideal of $\mathfrak L(\mathfrak X)$. Can you always find an operator ideal $\mathfrak J$ such that $\mathfrak J(\mathfrak X,\mathfrak X) = J$?



Is every ideal part of an operator ideal?
The answer is yes; take 


$$ \mathfrak{I}(\mathfrak{Y},\mathfrak{Z} ) = {\rm span}\{ T \in \mathfrak{L}(\mathfrak{Y},\mathfrak{Z}) \mid \exists U \in \mathfrak{L}(\mathfrak{Y},\mathfrak{X}) , \exists V \in \mathfrak{L}(\mathfrak{X},\mathfrak{Z}) , \exists S \in \mathfrak{I}(\mathfrak{X}) , T= VSU \} $$


More generally, a 2003 survey article by Laustsen and Loy on closed ideals in $\mathfrak{L}(\mathfrak{X})$ defines the operator ideal generated by a set of operators; they comment that their definition has its roots in wotk of Porta.


