I will take a roundabout way to defining this ideal, because (a) this route is how my collaborators and I came to it (b) this alternative definition, rather than the standard one, may suggest a direct attack on the question stated in the title.

**Notational conventions:** $L_p$ is short-hand for $L_p([0,1])$ with the Lebesgue measure on $[0,1]$. If $E$ is a Banach space then $L_\infty(E)$ denotes the space of essentially bounded, strongly measurable functions $[0,1]\to E$ (modulo equivalence a.e.).

In the case where $E=L_1$, we shall regard elements of the space $A=L_\infty(L_1)$ as functions of two variables which are $L_\infty$ in the *second* variable and $L_1$ in the *first* variable. So here the norm would be given by
$$
\Vert f \Vert = \operatorname{ess.sup}_{t\in [0,1]} \Vert f( \cdot, t)\Vert_1
$$
This convention has the advantage that given $f\in A$ and $\xi \in L_1$ we can define $T_f(\xi) \in L_1$ by
$$
T_f(\xi)(s) = \int_0^1 f(s,y)\xi(y)\,dy
$$
The map $f\mapsto T_f$ is an isometric embedding of $A$ as a closed subalgebra of $B(L_1)$. For given $f,g\in A$, we may define
$$
(f\bullet g)(s,t) = \int_0^1 f(s,x) g(x,t)\,dt
$$
Then $f\bullet g\in A$ and $T_fT_g = T_{f\bullet g}$. The image of $A$ in $B(L_1)$ under this embedding, which we will denote by $J$, turns out to be a much studied object in the theory of operators on $L_1$: it is the set of *representable* operators from $L_1$ to itself.

QUESTION: does $J$ (or equivalently $(A,\bullet)$) have a bounded right approximate identity? What if we drop the requirement of boundedness?

Note that the naive attempt of taking simple functions in $L_\infty(L_1)$ that approximate the "Dirac" measure concentrated on the diagonal in $[0,1]^2$ won't work, because such functions get sent by our embedding to elements of $K(L_1)$, which is properly contained in $J$ (see below).

*Some remarks for background context, which might be relevant to a solution.*

It can be shown that $J$ has no left approximate identity (bounded or otherwise). I would like to thank W. B. Johnson for indicating why this is the case; an expanded and paraphrased version of his explanation is given below.

From some vector measure theory, we know that $J$ contains the ideal $W(L_1)$ of weakly compact operators. Moreover, $J$ is contained in the ideal $CC(L_1)$ of completely continuous operators; "completely continuous" means that weakly convergent sequences are mapped to norm convergent sequences).

(The containment $J\subseteq CC(L_1)$ follows from a theorem of Lewis and Stegall, which characterizes operators in $J$ as those which factor through $\ell_1$. This also shows that $J$ is a *$2$-sided ideal* in $B(L_1)$, not just a subalgebra.)

It follows that if $S\in W(L_1)$ and $T\in J$, then $TS\in K(L_1)$. Since there exist weakly compact operators $S$ on $L_1$ which are not compact (e.g. take any non-compact map $L_1\to \ell_2$ and then compose with an isometric embedding $\ell_2\to L_1$), it follows that there is no net $(T_\alpha)$ in $J$ such that $\Vert T_\alpha S - S \Vert \to 0$.

isometric? If $f$ is a simple function, this is clear (looking actually at the adjoint $T_f^*$ acting on $L_\infty$). But in general $f$ is only the pointwise (a.e.) limit of simples. $\endgroup$ – Matthew Daws May 15 at 11:21uniformlimit of simple functions, then clearly $\|f\|=\|T_f\|$. As $L_1$ is separable (more generally, one could appeal to the Pettis Measurability Theorem) it follows that for any $\epsilon>0$ there is a set $B$ of measure smaller than $\epsilon$, and off $B$ we do have uniform convergence. Letting $\epsilon\rightarrow 0$ gives the result. $\endgroup$ – Matthew Daws May 15 at 19:13