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 $mathcal{B}circmathcal{A}$ is the class of all operators of the form $BA$, where $Ainmathcal{A}$, $Binmathcal{B}$ and the codomain of $A$ coincides with the domain of $B$ (so that the composition is defined). It is well-known that $mathcal{B}circmathcal{A}$ is an operator ideal and that $mathcal{B}circmathcal{A}$ 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 $mathcal{A}^infty:= bigcap_{ninmathbb{N}}mathcal{A}^n$, 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 $nin mathbb{N}$. 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.
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