Kernel operators on AL-spaces

Question

Let $E$ be an AL-space. For simplicity $E=L_{1}(X,\Sigma,\mu)$, where $(X,\Sigma,\mu)$ is a strictly localizable measure space. Let $T:E\rightarrow E$ be a bounded kernel operator on $E$ with measurable kernel $K$ defined on the product measure space $(X\times X,\Sigma\otimes\Sigma,\mu\otimes\mu)$. So $Tf(y):=\int_{X}K(x,y)f(x)d\mu(x)$. Can the closure of the range of $T$ be embedded into a sublattice $L$ of $E$, where $L$ is isometric to some $L_1$ space with $\sigma$-finite measure.

Answer 1

Your question is equivalent to asking whether $T$ has separable range. Indeed, separable subspace of a Banach lattice generates a separable sublattice.

This is indeed so, when $T$ is compact. Integral operators defined w.r.t. $\sigma$-finite measures are compact but in general they need not be (however this does not immediately ruin having potentially separable ranges).

Since you ask for strictly localisable measure spaces that behave very much as $\sigma$-finite ones, perhaps Lemma 3.3 here may help you. Then, by Lemma 4.1 such an integral operator is compact as long as certain tightness condition is satisfied.

Answer 2

The answer to that question is true for weakly compact operators $T:E\rightarrow E$, . This can be seen by the following observation: Let $(X_{i})_{i\in I}$ be a partition of $X$, where $I$ is some index set. Consider now $P_{i}:L_{1}(X)\rightarrow L_{1}(X_{i})$ is the canonical Projection onto. Suppose now that $P_{i}(R(T))\cap L_{1}(X_{i})\neq\{0\}$ for uncountable many $i$ say for simplicity again $I$. Then choose $g_{i}$ and $||f_{i}||=1$ such that $g_{i}=P_{i}Tf_{i}$. The set $T(B_{E})$ is weakly compact and hence equivalently uniformly integrable in the sense of Fremlin, see measure theory band 2, 246A, p. 185. (A set $W\subset E$ is uniformly integrable if for every $\varepsilon$ we can find a set $B\in\Sigma$ of finite measure and $M\geq0$, such that $\int_{X}(|f|-M\chi_{B})^{+}\leq\varepsilon$ for all $f\in W$. Now the set $\{g_{i}:i\in I\}$ is also uniformly integrable and hence weakly compact. This follows from the simple fact that $|g_{i}|\leq P_{i}|Tf_{i}|\leq |Tf_{i}|$. Since all $g_{i}$ have disjoint support, we see that $\overline{span}\{g_{i}/||g_{i}||\}=l_{1}(I)$. Hence $l_{1}(I)$ would be a weakly compact generated space for uncountable $I$, which is a contradiction. From this it follows that the range of $T$ is contained for countable many $i$. Hence it is contained in some $\sigma$-finite $L_1$-space.

Further remarks: There is an operator from $l_{1}(I)$ into $l_{2}(I)$ with non-separable range. Since $l_{2}(I)$ can be embedded into $L_{1}([0,1]^{I})$,( see Lacey,The ismometric theory of classical banach spaces, Thm 12, chapter 5 ) isometrically, there is an operator from $l_{1}(I)$ into $L_{1}([0,1]^{I})$ weakly compact with non-separable range. Now every weakly compact operator from an AL-space into itself is a kernel operator, see Fremlin, band 3, p. 469, 376P. So the question above is not equivalent for asking if any kernel oprator has separable range.