**Motivation and background** This question is motivated by the problem of classifying the (two-sided) closed ideals of the Banach algebra $\mathcal{B}(L\_\infty)$ of all (bounded, linear) operators on $L\_\infty$ $(=L\_\infty[0,1])$. As far as I am aware, the only known nontrivial ideals in $\mathcal{B}(L\_\infty)$ are the ideals $\mathcal{K}(L\_\infty)$ of compact operators and $\mathcal{W}(L\_\infty)$ of weakly compact operators. Most of the other well-known closed operator ideals not containing the identity operator of $L\_\infty$ seem to coincide with one of these two operator ideals on $L\_\infty$. Let me mention explicitly the following further relevant pieces of background information:


* Any nontrivial closed ideal $\mathcal{J}$ of $\mathcal{B}(L\_\infty)$ must satisfy $\mathcal{K}(L\_\infty)\subseteq \mathcal{J} \subseteq \mathcal{W}(L\_\infty)$.
* We have $\mathcal{S}(L\_\infty) = \mathcal{W}(L\_\infty)$, where $\mathcal{S}$ denotes the (closed) operator ideal of strictly singular operators. Moreover, the ideal of operators $L\_\infty \longrightarrow L\_\infty$ that factor through $L\_2$ $(=L\_2[0,1])$ is a norm dense subset of $\mathcal{W}(L\_\infty)$. With regards to this latter fact, we have that if $\mathcal{J}$ is any closed ideal of $\mathcal{B}(L\_\infty)$ satisfying $\mathcal{K}(L\_\infty)\subsetneq \mathcal{J} \subsetneq \mathcal{W}(L\_\infty)$, then there exists an element of $\mathcal{W}(L\_\infty)\setminus \mathcal{J}$ that factors through $L\_2$.


One possible avenue towards discovering more closed ideals in $\mathcal{B}(L\_\infty)$ is to consider products of closed operator ideals, and my question below concerns but one approach along these lines. We recall now that for operator ideals $\mathcal{A}$ and $\mathcal{B}$, their product $\mathcal{B}\circ\mathcal{A}$ is the class of all operators of the form $BA$, where $A\in\mathcal{A}$, $B\in\mathcal{B}$ and the codomain of $A$ coincides with the domain of $B$ (so that the composition is defined). It is well-known that $\mathcal{B}\circ\mathcal{A}$ is an operator ideal and that $\mathcal{B}\circ\mathcal{A}$ is a closed operator ideal whenever $\mathcal{A}$ and $\mathcal{B}$ are, the latter fact being due to Stefan Heinrich. For $n$ a natural number, one may define the power of an operator ideal $\mathcal{A}^n$ as the product $\mathcal{A}\circ \ldots \circ \mathcal{A}$ with $n$ factors, and $\mathcal{A}^n$ is closed for all $n$ whenever $\mathcal{A}$ is closed. Moreover, we may define $\mathcal{A}^\infty:= \bigcap\_{n\in\mathbb{N}}\mathcal{A}^n$, with $\mathcal{A}^\infty$ being a closed operator ideal whenever $\mathcal{A}$ is.


It is well-known that every operator from $L\_\infty$ to $L\_2$ $(=L\_2[0,1])$ is strictly singular. However, the answer to the following question is not so clear to me:


**Question:** *Is $\mathcal{B} (L\_\infty, L\_2) = \mathcal{S}^\infty (L\_\infty, L\_2)$? If no, what is the least $n$ for which $\mathcal{B} (L\_\infty, L\_2) \neq \mathcal{S}^n (L\_\infty, L\_2)$?*


**Further comments:** A negative answer to my question above will yield through easy arguments that $\mathcal{K}(L\_\infty) \subsetneq \mathcal{S}^\infty (L\_\infty) \subsetneq \mathcal{W}(L\_\infty)$ (note that $\mathcal{K}(L\_\infty) \subsetneq \mathcal{S}^\infty (L\_\infty)$ in any case since there are noncompact operators $L\_\infty \longrightarrow L\_\infty$ that have the formal inclusion operator $\ell\_2 \hookrightarrow \ell\_\infty$ as a factor, and this inclusion operator belongs to $\mathcal{S}^\infty$).


It seems to me that the best chance of obtaining an element of $\mathcal{B} (L\_\infty, L\_2) \setminus \mathcal{S}^\infty (L\_\infty, L\_2)$ would be to consider a surjective element of $\mathcal{B} (L\_\infty, L\_2)$, for instance the adjoint of an isomorphic embedding of $L\_2$ into $L\_1$. A further possibility along these lines would be to factor such an embedding as the product of $n$ operators whose adjoints are strictly singular, for each $n\in \mathbb{N}$. However, it seems to me that such an approach will require a far deeper knowledge of the subspace structure of $L\_1$ than I have at the present time. Interpolation methods may work too, but I am also too ignorant of that theory to know whether it's a genuinely feasible approach.



Factoring operators $L_\infty \longrightarrow L_2$ as the composition of $n$ strictly singular operators, $n\in \mathbb{N}$
I think this is an outline of a proof, Phil. It is enough to factor $I\_{\infty,2}$ as the product of strictly singular operators, where $I\_{\infty,2}$ is the identity from $L\_\infty(\mu)$ to $L\_2(\mu)$ with $\mu$ a probability (since every operator from $L\_\infty$ to $L\_2$ is $2$-summing). I guess it can be assumed that $\mu$ is a separable measure (if not, I think using Maharam's theorem works) and certainly it is enough to look at $\mu$ purely non atomic. But then $I\_{\infty,2}$ can be regarded as the identity from $L\_\infty([0,1])$ to $L\_2([0,1])$ since every separable non atomic probability space is measure isometric to $[0,1]$ with Lebesgue measure. Now $I\_{\infty,2}$ factors through $L\_\Phi([0,1])$ for every fast growing Orlicz function $\Phi$. Consider $\Phi\_p(t)=e^{t^p} -1$ for $p>$. Rodin and Semenov showed that the identity from $L\_{\Phi\_p}$ to $L\_2$ 
is strictly singular and the injection from $L\_\infty$ to $L\_\Phi$ is strictly singular
always. This takes care of the case $n=2$. 


Schechtman suggests that to do general $n$ to show that the injection from $L\_{\Phi\_p}$ to $L\_{\Phi\_r}$ is strictly singular for 
$2<r<p$. At least this map is not an isomorphism on the span of the Rademachers. 


