# Does this ideal in $B(L_1)$ have a (bounded) right approximate identity?

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 $$\Vert T_\alpha S - S \Vert \to 0$$.

• Something is utterly confusing: on the one hand, the general definition is written so that the norm in $L_\infty(L_1)$ is the (essential) supremum in the first variable of $L^1$ norms in the second. Two lines later you use it as if the norm were the integral in the second variable of the $L^\infty$ norms in the first. That you swap variables in the process, doesn't make the life easier... – fedja May 14 at 17:40
• Hmm, I think my choices were consistent. I agree that the more usual way to write $L_q$ valued $L_p$ is to have a function of 2 variables which is $L_p$ in the first variable and $L_q$ in the second, but I think it is consistent to declare them to be $L_p$ in the second variable and $L_q$ in the first variable (as you say we could just do things the usual way and introduce an explicit "swapping map"). Or did I misunderstand your point? – Yemon Choi May 14 at 17:49
• Is it obvious that $f\mapsto T_f$ is 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. – Matthew Daws May 15 at 11:21
• The answer to my question follows from Egoroff's theorem. If $f$ were the uniform limit 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. – Matthew Daws May 15 at 19:13
• I missed your post of this, Yemon. I emailed you earlier today that the answer to your question is "yes". The ideal $J$ has a contractive right approximate identity consisting of idempotents. In the paper I am writing with Phillips and Schectman we show that no closed ideal in $B(L_1)$ other than the compact operators has a left approximate identity, and observe that no closed ideal of weakly compact operators in $B(L_1)$ other than the compact operators has a right approximate identity. – Bill Johnson May 16 at 23:46