Ideal structure and factorization properties of the regular kernel operators

Abstract

We consider the structure of the lattice of (order and algebra) ideals of the band of regular kernel operators on $L^p$-spaces. We show, in particular, that for any $L^p(\mu )$ space, with $\mu$ $\sigma$-finite and $1<p<\infty$, the norm-closure of the ideal of finite-rank operators on $L^p(\mu )$, is the only non-trivial proper closed (order and algebra) ideal of this band. Key to our results in the $L^p$ setting is the fact that every regular kernel operator on an $L^p(\mu )$ space ($\mu$ and p as before) factors with regular factors through $\ell _p$. We show that a similar but weaker factorization property, where $\ell _p$ is replaced by some reflexive purely atomic Banach lattice, characterizes the regular kernel operators from a reflexive Banach lattice with weak order unit to a KB-space with weak order unit.

1 Introduction

Let X be a Banach lattice and let ${{\mathcal {L}}}^r(X)$ be the space of all regular maps (i.e., linear combinations of positive maps) from X to itself. It is well known that ${{\mathcal {L}}}^r(X)$, endowed with the regular norm and with composition as product, is a Banach algebra and an ordered vector space whose positive cone is closed under multiplication.

There has been interest in understanding the structure of the lattice of closed ideals of ${{\mathcal {L}}}^r(X)$, where by ideal, here and henceforth, we shall mean an order and two-sided algebra ideal. For instance, if $X = \ell _p$ $(1<p<\infty )$ then ${{\mathcal {A}}}^r(\ell _p)$ (:= the closure of the ideal of finite-rank operators in ${{\mathcal {L}}}^r(\ell _p)$) is the only non-trivial proper closed ideal of ${{\mathcal {L}}}^r(\ell _p)$ [3, 4]. If $X = c_0$ or $\ell _1$, then ${{\mathcal {L}}}^r(X) = {{\mathcal {B}}}(X)$ (:= the Banach algebra of all bounded operators on X) and the same conclusion follows from a classical result of Gohberg, Markus and Feldman. It is also known that if $X = L^p[0,1]$ ($1<p<\infty$) then the structure of the lattice of closed ideals of ${{\mathcal {L}}}^r(X)$ significantly departs from that of the lattice of closed (two-sided) algebra ideals of ${{\mathcal {B}}}(X)$, e.g., for every $1<p<\infty$, there are in ${{\mathcal {L}}}^r(L^p[0,1])$ at least four distinct well-documented closed ideals apart from the trivial ones, namely, ${{\mathcal {A}}}^r(L^p[0,1])$, ${{\mathcal {I}}}^r(L^p[0,1])$ (see below for definitions), the span of the positive compact operators and the AM-compact operators (this can be easily deduced, for instance, from [22, Theorem 125.5], [17, Theorem 3.3] and [20, Theorem 3.4]).

In this note, our main concern will be with the band ${{\mathcal {I}}}^r(X)$ of all kernel (or integral) operators in ${{\mathcal {L}}}^r(X)$. When X is order continuous, the latter is a closed ideal (in the sense of the note) of ${{\mathcal {L}}}^r(X)$. If X is in addition atomic then ${{\mathcal {I}}}^r(X) = {{\mathcal {L}}}^r(X)$ and one can think of the results of [3, 4], mentioned above, as results about the ideal structure of ${{\mathcal {I}}}^r(X)$. Arendt and Sourour [3] asked whether ${{\mathcal {A}}}^r(L^2[0,1])$ is likewise the only non-trivial proper closed ideal of ${{\mathcal {I}}}^r(L^2[0,1])$. Here, we answer this question in the affirmative. In fact, we will show that for any $L^p(\mu )$ space, with $\mu$ a $\sigma$-finite measure and $1<p<\infty$, ${{\mathcal {A}}}^r(L^p(\mu ))$ is the only non-trivial proper closed ideal of ${{\mathcal {I}}}^r(L^p(\mu ))$.

Key to establishing the result from the previous paragraph will be the fact that the kernel operators in ${{\mathcal {L}}}^r(L^p(\mu ))$ (with $\mu$ and p as above) are precisely the operators that factor through $\ell _p$ with regular factors. It is thus natural to ask whether a similar characterization is possible for arbitrary regular kernel operators. In this direction, our main result will be that a regular operator from a reflexive Banach lattice with a weak order unit into a KB-space also with a weak order unit is a kernel operator if and only if it factors through a reflexive purely atomic Banach lattice with both factors regular. Unfortunately, our path towards this more general but weaker result will be different from the one followed to establish the factorization result in the case of $L^p$-spaces. For this reason, it will be discussed separately.

The note has been organized as follows. In the next section we gather some notation and terminology, as well as some basic facts needed throughout the note. In Sect. 3, we first give an extension of the factorization result for regular kernel operators on $L^p(\mu )$ spaces, mentioned above, and then prove our main results on the ideal structure of the band of regular kernel operators on such spaces. Whenever possible, we shall work and present our results in a slightly more general framework, which we explain below. Finally, in Sect. 4, we investigate the connection between regular kernel operators and factorization of regular maps (with regular factors) through purely atomic Banach lattices. We prove, amongst other things, the result stated at the end of the previous paragraph.

2 Some background and notation

Given a Banach lattice X, we shall write $X_+$ for its positive cone and $X_{[\rho ]}$ for the closed ball in X centered at the origin and of radius $\rho$. If S is a subset of X, we shall denote by $\mathrm {sp}(S)$ its linear span and by $S^{dd}$ the band in X generated by S. In the case of an indexed set $\{x_i : i\in I\}$, we shall write $[x_i : i\in I]$ (or just $[x_i]$ if I is clear from context) for the norm-closure of its linear span in X.

The topological dual of X will be denoted by $X'$, and we shall always identify X with its image in $X''$ under the canonical embedding. Recall a Banach lattice X is said to be a KB-space if every increasing norm bounded sequence $(x_n)\subset X$ is norm convergent. Recall also every KB-space is a band in its second dual.

A linear subspace ${{\mathcal {I}}}$ of an ordered vector space V shall be called an order ideal if (i) $\pm v\le w\in {{\mathcal {I}}}$ $\Rightarrow$ $v\in {{\mathcal {I}}}$; and (ii) for every $v\in {{\mathcal {I}}}$ there is $w\in {{\mathcal {I}}}$ such that $\pm v\le w$. One easily checks that, when V is a vector lattice, the latter is equivalent to the standard definition of ideal in the vector lattice setting.

Given a linear map $T:X\rightarrow Y$ and a subspace $E\subset X$ (resp. $F\subset Y$), we shall write $T|_E$ (resp. $T|^F$) for its restriction (resp. corestriction) to E (resp. F). Moreover, we shall denote by $T'$ its (topological) adjoint and by T(X) its image. The identity map on X shall be denoted by $\mathrm {id}_X$ (or just $\mathrm {id}$ if X is clear from context).

As customary, a linear map T between Banach lattices X and Y shall be said to be positive if $T(X_+)\subseteq Y_+$ and regular if it can be written as a linear combination of positive maps. We shall write ${{\mathcal {L}}}^r(X,Y)$ for the linear space of all regular maps from X to Y. The latter is an ordered vector space and becomes a Banach space when endowed with the regular norm $\Vert T\Vert _r := \big \{\Vert S\Vert : S\in {{\mathcal {B}}}(X,Y),\, |Tx|\le S|x| \;(x\in X)\big \}$ $\,(T\in {{\mathcal {L}}}^r(X,Y))$, where $\Vert \cdot \Vert$ stands for the operator norm on ${{\mathcal {B}}}(X,Y) :=$ the Banach space of all bounded linear operators from X to Y. When Y is Dedekind complete, ${{\mathcal {L}}}^r(X,Y)$ is also a Banach lattice. Given subspaces ${{\mathcal {I}}}\subseteq {{\mathcal {L}}}^r(Y,Z)$ and ${{\mathcal {J}}}\subseteq {{\mathcal {L}}}^r(X,Y)$, we shall write ${{\mathcal {I}}}\circ {{\mathcal {J}}}$ for the set $\{ST : S\in {{\mathcal {I}}}\text { and } T\in {{\mathcal {J}}}\}\subseteq {{\mathcal {L}}}^r(X,Z)$.

We shall write ${{\mathcal {F}}}(X,Y)$ for the linear space of all continuous finite-rank operators from X to Y and ${{\mathcal {A}}}^r(X,Y)$ for its norm closure in ${{\mathcal {L}}}^r(X,Y)$. We shall call the elements of ${{\mathcal {A}}}^r(X,Y)$ approximable regular operators. Two well-known facts about ${{\mathcal {A}}}^r(X,Y)$ that we shall use repeatedly later on are that it is always a Banach lattice and that $X'_+\otimes Y_+ := \big \{\sum _{i=1}^n x_i'\otimes y_i : x_i'\in X'_+,\, y_i\in Y_+ \text { and } n\in {\mathbb {N}}\big \}$ is norm-dense in ${{\mathcal {A}}}^r(X,Y)_+$ (here, of course, we are identifying $X'\otimes Y$ with ${{\mathcal {F}}}(X,Y)$ in the usual way).

Given Banach lattices X and Y, we shall say $T\in {{\mathcal {L}}}^r(X,Y)$ is a regular kernel operator if it belongs to the band $(X'_{oc}\otimes Y)^{dd}$, where we have written $X'_{oc}$ for the order continuous dual of X, and $X'_{oc}\otimes Y$ for the linear space of all order continuous finite-rank operators in ${{\mathcal {L}}}^r(X,Y)$. It is well known that if X and Y are Banach function spaces on $\sigma$-finite measure spaces $(\varOmega _1,\varSigma _1,\mu _1)$ and $(\varOmega _2,\varSigma _2,\mu _2)$, respectively, then $T\in (X'_{oc}\otimes Y)^{dd}$ if and only if there is a $\mu _1\times \mu _2$-measurable function on $\varOmega _1\times \varOmega _2$ such that $(Tf)(x) := \int _{\varOmega _1} k(t,x)f(t)\,d\mu _1(t)$ for $\mu _2$-almost every $x\in \varOmega _2$ and the function $\int _{\varOmega _1} |k(t,\cdot )||f(t)|\, d\mu _1(t)$ belongs to Y $\,(f\in X)$. We shall let ${{\mathcal {I}}}^r(X,Y) := (X'_{oc}\otimes Y)^{dd}$. When X is order continuous, $X'_{oc} = X'$ and ${{\mathcal {I}}}^r(X,Y) = {{\mathcal {F}}}(X,Y)^{dd}$. It is known that if X (resp. Y) is order continuous then for every Banach lattice E, ${{\mathcal {I}}}^r(X,Y)\circ {{\mathcal {L}}}^r(E,X)\subseteq {{\mathcal {I}}}^r(E,Y)$ (resp. ${{\mathcal {L}}}^r(Y,E)\circ {{\mathcal {I}}}^r(X,Y)\subseteq {{\mathcal {I}}}^r(X,E)$), and in general, the inclusion may fail. Furthermore, if either X is order continuous and atomic, or if Y is atomic, then ${{\mathcal {F}}}(X,Y)^{dd} = {{\mathcal {L}}}^r(X,Y)$.

A known fact connecting the above classes of regular operators, which we shall use throughout the note, is that if $X'$ and Y are order continuous Banach lattices then the operators in ${{\mathcal {I}}}^r(X,Y)$ whose modules are compact are precisely the elements of ${{\mathcal {A}}}^r(X,Y)$ (see for instance [17, Theorem 3.3]).

To simplify, all Banach lattices of the note will be assumed to be real. Other facts and terminology will be introduced and explained as they are needed. For unexplained material about Banach lattices, we refer the reader to [2, 16].

3 The ideal structure of ${{\mathcal {I}}}^r(L^p)$ $(1\le p<\infty )$

Mimicking the Banach space situation, given Banach lattices E, F, X and Y, and $L\in {{\mathcal {L}}}^r(E,F)$, we shall say an operator $T\in {{\mathcal {L}}}^r(X,Y)$ r-factors through L if there are operators $R\in {{\mathcal {L}}}^r(F,Y)$ and $S\in {{\mathcal {L}}}^r(X,E)$ such that $T = RLS$. We then define

$$\begin{aligned} \varGamma _{L,r}(X,Y) := \mathrm {sp}\big \{T\in {{\mathcal {L}}}^r(X,Y) : T \text { { r}-factors through } L\big \}, \end{aligned}$$

and

$$\begin{aligned} \varGamma _{L,r}^\mathrm {o}(X,Y) := \big \{T\in {{\mathcal {L}}}^r(X,Y) : \pm T\le S \text { for some } S\in \varGamma _{L,r}(X,Y)\big \}. \end{aligned}$$

If $X = Y$ we simply write $\varGamma _{L,r}(X)$ (resp. $\varGamma _{L,r}^\mathrm {o}(X)$) for $\varGamma _{L,r}(X,X)$ (resp. $\varGamma _{L,r}^\mathrm {o}(X,X)$). Clearly, $\varGamma _{L,r}^\mathrm {o}(X,Y)$ is an order ideal of ${{\mathcal {L}}}^r(X,Y)$ and one can easily verify that for every pair of Banach lattices E, F, ${{\mathcal {L}}}^r(Y,F)\circ \varGamma _{L,r}(X,Y)\circ {{\mathcal {L}}}^r(E,X)\subseteq \varGamma _{L,r}(E,F)$ and ${{\mathcal {L}}}^r(Y,F)\circ \varGamma _{L,r}^\mathrm {o}(X,Y)\circ {{\mathcal {L}}}^r(E,X)\subseteq \varGamma _{L,r}^\mathrm {o}(E,F)$. If $L = \mathrm {id}_Z$ for some Banach lattice Z then we shall simply say that T r-factors through Z. (Note the latter simply amounts to T factoring through Z, in the usual sense, but with both factors being regular.) In this case, we shall write $\varGamma _{Z,r}$ and $\varGamma _{Z,r}^\mathrm {o}$ for $\varGamma _{\mathrm {id}_Z,r}$ and $\varGamma _{\mathrm {id}_Z,r}^\mathrm {o}$, respectively.

Remark 1

In general, $\varGamma _{Z,r}(X,Y)\ne \{T\in {{\mathcal {L}}}^r(X,Y) : T\text { r-factors through } Z\}$. However, if Z admits a decomposition $Z_1\oplus Z_2\oplus Z_3$ with $Z_1$ and $Z_2$ isomorphic as Banach lattices to Z and each ‘coordinate’ projection $P_i:Z\rightarrow Z_i$ $(i = 1,2,3)$ regular, then equality holds. (The proof of this is the same as in the Banach space situation. Clearly, one only needs to check that the set on the right, call it ${{\mathcal {V}}}$, is closed under addition. So let $T_1,T_2\in {{\mathcal {V}}}$ be arbitrary and let $R_1S_1$ and $R_2S_2$ be r-factorizations through Z of $T_1$ and $T_2$, respectively. For each $i\in \{1,2\}$, let $\phi _i:Z\rightarrow Z_i$ and $\imath _i:Z_i\rightarrow Z$ be a Banach lattice isomorphism and the corresponding coordinate embedding, respectively, and let $R := R_1 \phi _1^{-1} P_1 + R_2 \phi _2^{-1} P_2$ and $S := \imath _1 \phi _1 S_1 + \imath _2 \phi _2 S_2$. Then $RS = T_1+T_2$.) The spaces $c_0$, $\ell _p$ and $L^p[0,1]$, $1\le p\le \infty$, all have this property.

Remark 2

It may be worth pointing out that, unless L is positive, $\varGamma _{L,r}^o(X)$ need not coincide with the order and algebra ideal of ${{\mathcal {L}}}^r(X)$ generated by L. For instance, if T is a compact operator on $\ell _p$ $(1<p<\infty )$ whose module is not compact, then there is no $S\in \varGamma _{T,r}(\ell _p)$ such that $|T|\le S$. One can easily see that the smallest order and algebra ideal of ${{\mathcal {L}}}^r(\ell _p)$ that contains T is precisely $\varGamma _{|T|,r}^o(\ell _p)$. So, in the study of (order and algebra) ideals of ${{\mathcal {L}}}^r(X)$, r-factorization through positive operators has a special significance.

It is known [see [18, Theorems 3.1 and 4.1] (resp. [18, Corollary 3.3])] that if $(\varOmega _1,\varSigma _1,\mu _1)$ and $(\varOmega _2,\varSigma _2,\mu _2)$ are finite measure spaces and $T:L^p(\mu _1)\rightarrow L^p(\mu _2)$, $1<p\le \infty$ (resp. $1\le p<\infty$), is a positive kernel operator, then for every $\varepsilon >0$ there is $A\subseteq \varOmega _2$ (resp. $A\subseteq \varOmega _1$) of measure $\ge \mu _2(\varOmega _2)-\varepsilon$ (resp. $\ge \mu _1(\varOmega _1)-\varepsilon$) such that the map $f\mapsto \chi _A T(f)$, $L^p(\mu _1)\rightarrow L^p(\mu _2)$ (resp. $f\mapsto T(\chi _A f)$, $L^p(\mu _1)\rightarrow L^p(\mu _2)$), is approximable regular. (We should mention here that a similar result to [18, Corollary 3.3] for $p=1$, was first established by Lewis and Stegall in the case where $\mu _1$ is a Radon measure and $\varOmega _1$ is a compact set (see [11, Lemma]).) An easy consequence of these results is that every regular kernel operator $T:L^p(\mu _1)\rightarrow L^p(\mu _2)$ r-factors through $\ell _p$ $(1\le p\le \infty )$. The first result of the section will be a slightly more general version of this factorization result in the case where $1<p<\infty$. Recall first a linear operator T from a Banach lattice X to a Banach lattice Y is said to be p-convex (resp. p -concave) for some $1\le p<\infty$ if for some constant C and every finite sequence $x_1,\ldots ,x_n\in X$,

$$\begin{aligned} \Bigg \Vert \bigg (\sum _{i=1}^n |Tx_i|^p\bigg )^{\frac{1}{p}}\Bigg \Vert\le & {} C\Bigg (\sum _{i=1}^n \Vert x_i\Vert ^p\Bigg )^{\frac{1}{p}} \quad \Bigg (\text {resp.} \Bigg (\sum _{i=1}^n \Vert Tx_i\Vert ^p\Bigg )^{\frac{1}{p}}\\\le & {} C\Bigg \Vert \bigg (\sum _{i=1}^n |x_i|^p\bigg )^{\frac{1}{p}}\Bigg \Vert \Bigg ). \end{aligned}$$

In turn, X is said to be p-convex (resp. p-concave) for some $1\le p<\infty$ if $\mathrm {id}_X$ is p-convex (resp. p-concave).

The result on r-factorization can now be stated as follows.

Theorem 1

Let X and Y be Banach lattices and let $T\in {{\mathcal {L}}}^r(X,Y)$. Suppose there is a $\sigma$-finite measure space $(\varOmega ,\varSigma ,\mu )$ such that, for some $1<p<\infty$, T r-factors through $L^p(\mu )$ with factors $R:L^p(\mu )\rightarrow Y$ and $S:X\rightarrow L^p(\mu )$. If S is a regular kernel operator and X is p-convex, or if R is a regular kernel operator and Y is p-concave, then T r-factors through $\ell _p$.

Our proof of Theorem 1 shall require the following version of [18, Theorem 3.1].

Lemma 1

Let X be a p-convex Banach lattice, let $(\varOmega ,\varSigma ,\mu )$ be a finite-measure space and let $T:X\rightarrow L^p(\mu )$ $(1<p<\infty )$ be a regular kernel operator. Then, for every $\varepsilon >0$ there exists $A\in \varSigma$ such that $\mu (\varOmega {\setminus } A)< \varepsilon$ and $M_A\circ T$ is approximable regular, where $M_A:L^p(\mu )\rightarrow L^p(\mu )$, $f\mapsto \chi _A f$.

Proof

Consider first the case where T is positive. Let $\varepsilon >0$ arbitrary, let $\imath :L^p(\mu )\rightarrow L^1(\mu )$ be the natural embedding, and let ${{\mathcal {T}}} := \imath \circ T$, which is also a kernel operator (because $p<\infty$). Since X is p-convex, $s(X)>1$, so ${{\mathcal {T}}}$ is approximable regular (see [17, Corollary 3.4(iii)]). Let $(U_n)$ be a sequence in ${{\mathcal {F}}}(X,L^p(\mu ))$ such that $\sum _n \imath \circ U_n = {{\mathcal {T}}}$ and $\sum _n n\Vert \imath \circ U_n\Vert _r< \infty$, and set $\sum _n n|\imath \circ U_n| =: {{\mathcal {S}}}:X\rightarrow L^1(\mu )$. Since X is p-convex and ${{\mathcal {S}}}\ge 0$, ${{\mathcal {S}}}$ is p-convex too (see [10, Proposition 1.d.9]), and so, it follows from [12, Théorème 2] that there is a strictly positive function $g\in L^r(\mu )$ $(r = p/(p-1))$ such that $g^{-1}{{\mathcal {S}}}(x)\in L^p(\mu )$ $(x\in X)$.

Let $A_t := \{\omega \in \varOmega : g(\omega )\ge t\}$ $(t\in {\mathbb {R}}_+)$ and choose $t_0$ small enough so that $\mu (\varOmega {\setminus } A_{t_0})<\varepsilon$. Set $A := A_{t_0}$. It is clear that $\chi _A{{\mathcal {S}}}(x)\in L^p(\mu )$ $(x\in X)$, so ${\widetilde{S}}:X\rightarrow L^p(\mu )$, $x\mapsto \chi _A{{\mathcal {S}}}(x)$, is a well-defined positive (and therefore continuous) operator. To see $M_A\circ T:X\rightarrow L^p(\mu )$ is regular approximable, first note that $(M_A\circ T)(x) = \sum _n \chi _A U_n(x)$ $(x\in X)$. Then note that, for every $N\in {\mathbb {N}}$ and $x\in X_+$,

$$\begin{aligned} \begin{aligned} \Big \Vert \sum \limits _{n\ge N} \chi _A U_n(x)\Big \Vert&\le \Big \Vert \sum \limits _{n\ge N} \chi _A \big |\imath \circ U_n\big |(x)\Big \Vert \\&\le \frac{1}{N} \Big \Vert \sum \limits _{n\ge N} n\chi _A \big |\imath \circ U_n\big |(x)\Big \Vert \le \frac{1}{N}\Vert {\widetilde{S}}\Vert \Vert x\Vert , \end{aligned} \end{aligned}$$

so $M_A\circ T$ is positive compact. Since $M_A\circ T$ is also a kernel operator, it must be approximable regular.

In general, if $T = T_+ - T_-$, then, by the previous part of the argument, for every $\varepsilon >0$ there are sets $A_1,A_2\in \varSigma$ such that $\mu (\varOmega {\setminus } A_i)< \varepsilon /2$ $\,(i = 1,2)$, and both $M_{A_1}\circ T_+$ and $M_{A_2}\circ T_-$ are approximable regular. One then easily verifies that $A := A_1\cap A_2$ meets all the requirements. $\square$

Proof of Theorem 1

Suppose first X is p-convex and $S:X\rightarrow L^p(\mu )$ is a positive kernel operator. Clearly, it will suffice to produce an r-factorization of S through $\ell _p$.

By the previous lemma, there is a partition $\{\varOmega _i : i\in {\mathbb {N}}\}$ of $\varOmega$ such that, if $M_i:L^p(\mu )\rightarrow L^p(\mu )$ is defined by $M_i(f) := \chi _{\varOmega _i} f$ $(f\in L^p(\mu ))$, then $M_i\circ S:X\rightarrow L^p(\mu )$ is approximable regular $(i\in {\mathbb {N}})$. Let $\varepsilon >0$ arbitrary, and for each $i\in {\mathbb {N}}$, choose $V_i = \sum _{j=1}^{n_i} x_{ij}'\otimes f_{ij}\in {{\mathcal {F}}}(X,L^p(\mu ))_+$ so that $\Vert V_i - M_iS\Vert _r\le \varepsilon /2^i$. Let ${\widetilde{S}}:X\rightarrow L^p(\mu )$ be the strong operator limit of the sequence $(\sum _{i=1}^n V_i)$.

Since the linear space $S(\mu )$ of all simple integrable functions on $(\varOmega ,\varSigma ,\mu )$ is dense in $L^p(\mu )$, we can further assume $\{f_{ij} : 1\le j\le n_i\}\subset M_i(S(\mu ))$ $(i\in {\mathbb {N}})$. For each i, let $E_i$ be the smallest vector sublattice of $L^p(\mu )$ such that $[f_{ij} : 1\le j\le n_i]\subseteq E_i$, and let $\imath _i:E_i\rightarrow L_p(\mu )$ be the natural inclusion. Let $E := (\bigoplus _i E_i)_p$ (which is clearly isomorphic, as a Banach lattice, to $\ell _p$). The maps $S_2:X\rightarrow E$, $x\mapsto (V_ix)$, and $S_1:E\rightarrow L^p(\mu )$, $(z_i)\mapsto \sum _i \imath _i(z_i)$, are both well-defined, for

$$\begin{aligned} \begin{aligned} \Vert S_2x\Vert _E&= \Big (\sum \limits _i \Vert V_ix\Vert ^p_{E_i}\Big )^{1/p}\\&\le \Big (\sum \limits _i \Vert M_iSx\Vert _{E_i}^p\Big )^{1/p} + \Big (\sum \limits _i \Vert (V_i-M_iS)x\Vert _{E_i}^p\Big )^{1/p} \\&\le \Vert {\widetilde{S}}x\Vert +\varepsilon \Big (\sum \limits _i 2^{-ip}\Big )^{1/p}\Vert x\Vert \le (\Vert {\widetilde{S}}\Vert +\varepsilon )\Vert x\Vert \quad (x\in X), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \Vert S_1((z_i))\Vert _{L^p} = \Big \Vert \sum \limits _i \imath _i(z_i)\Big \Vert _{L^p} = \Big (\sum \limits _i \Vert \imath _i(z_i)\Vert _{L^p}^p\Big )^{1/p} = \Vert (z_i)\Vert _E \quad ((z_i)\in E). \end{aligned}$$

Clearly, they provide an r-factorization of ${\widetilde{S}}$ through $\ell _p$ with positive factors.

To obtain an r-factorization of S it just remains to find an r-factorization for $S_0 := S-{\widetilde{S}}$. To this end, first choose a sequence $(U_i)\subset {{\mathcal {F}}}(X,S(\mu )){\setminus } \{0\}$ such that $\sum _i U_i = S_0$ and $\sum _i \Vert U_i\Vert _r =: M<\infty$ (one such sequence exists because $S_0 = \sum _i (M_iS-V_i)\in {{\mathcal {A}}}^r(X,L^p(\mu ))$). For each $i\in {\mathbb {N}}$, let $F_i$ be the smallest Riesz subspace of $L^p(\mu )$ that contains $U_i(X)$, and let $\jmath _i:F_i\rightarrow L^p(\mu )$ be the natural embedding. Let $F := (\bigoplus _i F_i)_p$, let $r_i := \Vert U_i\Vert _r^{1/q}$ $(i\in {\mathbb {N}})$, where $q = p/(p-1)$, let $S_{02}:X\rightarrow F$, $x\mapsto (r_i^{-1} U_i(x))$, and let $S_{01}:F\rightarrow L^p(\mu )$, $(z_i)\mapsto \sum _i r_i\jmath _i(z_i)$. Then

$$\begin{aligned} \Vert S_{02}x\Vert _F= & {} \Big (\sum \limits _i r_i^{-p}\Vert U_i x\Vert _{F_i}^p\Big )^{1/p} \le M^{1/p}\Vert x\Vert \quad (x\in X), \\ \Vert S_{01}((z_i))\Vert _{L^p}= & {} \Big \Vert \sum \limits _i r_i\jmath _i(z_i)\Big \Vert _{L^p}\le \sum \limits _i r_i\Vert z_i\Vert _{F_i} \le M^{1/q}\Vert (z_i)\Vert _F \quad ((z_i)\in F), \end{aligned}$$

and $S_{01}S_{02} = S_0$. The rest is clear.

Now suppose S is a p-convex regular kernel operator. If $S_+^{(1)}S_+^{(2)}$ and $S_-^{(1)}S_-^{(2)}$ are r-factorizations of $S_+$ and $S_-$, respectively, through $\ell _p$, then noting that $\ell _p\oplus _p \ell _p\cong \ell _p$ as Banach lattices, one readily sees that $T_1:X\rightarrow \ell _p\oplus _p\ell _p$, $x\mapsto (S_+^{(1)}x,S_-^{(1)}x)$, and $T_2:\ell _p\oplus _p \ell _p\rightarrow L^p(\mu )$, $(\xi _1,\xi _2)\mapsto RS_+^{(2)}\xi _1 - RS_-^{(2)}\xi _2$, provide the required r-factorization of T through $\ell _p$.

In the case where Y is p-concave and $R:L^p(\mu )\rightarrow Y$ is a regular kernel operator, one simply applies the previous part of the argument to the adjoint $R':Y'\rightarrow L^p(\mu )' = L^q(\mu )$ to obtain regular maps $R_1:\ell _q\rightarrow L^q(\mu )$ and $R_2:Y'\rightarrow \ell _q$ such that $R' = R_1R_2$ (by [10, Proposition 1.d.4(iii)], Y is p-concave $\iff$ $Y'$ is q-convex). Since Y is p-concave and $p<\infty$, Y is a KB-space and therefore a band in $Y''$. Let $P:Y''\rightarrow Y$ be the corresponding band projection. Then $R = (P\circ R_2')\circ R_1'$, which is an r-factorization through $\ell _p$. $\square$

Remark 3

If $p=\infty$ and $\mu$ and $\nu$ are $\sigma$-finite measures, then an argument similar to (but simpler than) the one used in the first part of the proof of Theorem 1, gives that every kernel operator $T:L^\infty (\mu )\rightarrow L^\infty (\nu )$ r-factors, or equivalently, factors (see [1, Theorem 9]) through $\ell _\infty$.

As a first consequence of Theorem 1, we now have the following.

Corollary 1

Let X (resp. Y) be a p-convex (resp. p-concave) Banach lattice, with $1<p<\infty$, and let $\mu$ be a $\sigma$-finite measure. Then

$$\begin{aligned} {{\mathcal {I}}}^r(X,L^p(\mu ))= & {} \varGamma _{\ell _p,r}^\text {o}(X,L^p(\mu )) = \varGamma _{\ell _p,r}(X,L^p(\mu )) \\ (\text {resp. } {{\mathcal {I}}}^r(L^p(\mu ),Y)= & {} \varGamma _{\ell _p,r}^\text {o}(L^p(\mu ),Y) = \varGamma _{\ell _p,r}(L^p(\mu ),Y)). \end{aligned}$$

In particular, ${{\mathcal {I}}}^r(L^p(\mu )) = \varGamma _{\ell _p,r}^\text {o}(L^p(\mu )) = \varGamma _{\ell _p,r}(L^p(\mu ))$ $\,(1<p<\infty )$.

Proof

Suppose first X is p-convex. By Theorem 1, ${{\mathcal {I}}}^r(X,\ell _p)\subseteq \varGamma _{\ell _p,r}(X,L^p(\mu ))$, and since ${{\mathcal {I}}}^r(X,\ell _p)$ is a band, it will suffice to show that $\varGamma _{\ell _p,r}(X,L^p(\mu ))\subseteq {{\mathcal {I}}}^r(X,\ell _p)$, for then one has that $\varGamma _{\ell _p,r}^\text {o}(X,L^p(\mu ))\subseteq {{\mathcal {F}}}(X,L^p(\mu ))^{dd}$, and clearly, $\varGamma _{\ell _p,r}(X,L^p(\mu ))\subseteq \varGamma _{\ell _p,r}^\text {o}(X,L^p(\mu ))$. In turn, by Remark 1, we only need to check that if an operator $T:X\rightarrow L^p(\mu )$ r-factors through $\ell _p$, then $T\in {{\mathcal {I}}}^r(X,\ell _p)$. But $\ell _p$ is order continuous and atomic, so ${{\mathcal {I}}}^r(X,\ell _p) = {{\mathcal {L}}}^r(X,\ell _p)$, and therefore ${{\mathcal {L}}}^r(\ell _p,L^p(\mu ))\circ {{\mathcal {L}}}^r(X,\ell _p) = {{\mathcal {L}}}^r(\ell _p,L^p(\mu ))\circ {{\mathcal {I}}}^r(X,\ell _p) \subseteq {{\mathcal {I}}}^r(X,L^p(\mu ))$.

The proof that ${{\mathcal {I}}}^r(L^p(\mu ),Y) = \varGamma _{\ell _p,r}^\text {o}(L^p(\mu ),Y) = \varGamma _{\ell _p,r}(L^p(\mu ),Y)$ if Y is p-concave, is completely analogous. The rest is clear. $\square$

Remark 4

Since $\ell _1$ is order continuous and, as noted at the beginning of this section, if $\mu$ and $\nu$ are finite measures, every kernel operator $T:L^1(\mu )\rightarrow L^1(\nu )$ r-factors (or equivalently, factors) through $\ell _1$ (see [1, Theorem 9]), the last part of Corollary 1 holds true also for $p=1$ when $\mu$ is finite.

It should be clear now, from Corollary 1, that in order to show ${{\mathcal {A}}}^r(L^p(\mu ))$ is the only non-zero proper closed ideal of ${{\mathcal {I}}}^r(L^p(\mu ))$ (with $\mu$ as in Corollary 1) it would suffice to show that $\mathrm {id}_{\ell _p}$ r-factors through any linear map in ${{\mathcal {I}}}^r(L^p(\mu )){\setminus }{{\mathcal {A}}}^r(L^p(\mu ))$. We shall prove this below for any $1<p<\infty$. In fact, for $p\ne 2$, slightly more can be said.

Before stating our next result, we recall a few things. First, a sequence $(x_i)\subset X$ is said to be complemented by a projection, or just complemented, if $[x_i]$ is the image of a bounded projection on X. Second, an order continuous Banach lattice X is said to have the subsequence splitting property (SSP in short) if for every bounded sequence $(x_n)\subset X$ there is an increasing sequence $(n_i)\subset {\mathbb {N}}$, an equi-integrable sequence $(f_i)\subset X$ (i.e., a sequence such that for some $u\in X_+$, $\lim _{t\rightarrow \infty } \sup _k \Vert |f_k|-|f_k|\wedge tu\Vert = 0$) and a disjoint sequence $(g_i)\subset X$ such that $|f_i|\wedge |g_i| = 0$ and $x_{n_i} = f_i+g_i$ $(i\in {\mathbb {N}})$. (This is the same as [19, Definition 2.1], except for the fact that it has not been restricted to order continuous Banach lattices with weak order units as in [19]). In view of [10, Proposition 1.a.9] and the results of [19], it is clear that any q-concave Banach lattice, with $q<\infty$, has the SSP. Third, the lower index, s(X), of a Banach lattice X is defined as $s(X) := \sup \big \{p\in [1,\infty ] : X \text { satisfies an upper p-estimate}\big \}$, where X is said to satisfy an upper p-estimate if for some constant C and every finite disjoint sequence $x_1,\ldots ,x_n\in X$, $\Vert \sum _{i=1}^n x_i\Vert \le C(\sum _{i=1}^n \Vert x_i\Vert ^p)^{1/p}$. Lastly, recall $T\in {{\mathcal {L}}}^r(X,Y)$ is AM-compact if it maps order bounded sets to precompact ones. It is easy to see (and well known) that an AM-compact operator must also map semicompact sets to precompact ones, where by a semicompact set we mean a subset $A\subset X$ with the property that for every $\varepsilon >0$ there exists a $u\in X_+$ such that $\Vert (|\xi |-u)_+\Vert < \varepsilon$ $(\xi \in A)$. If Y is order continuous, then the space $\mathcal {AM}(X,Y)$ of all AM-compact operators from X to Y is a band in ${{\mathcal {L}}}^r(X,Y)$ [7, Theorem 4.7], and hence, ${{\mathcal {I}}}^r(X,Y)\subseteq {\mathcal {AM}}(X,Y)$.

Proposition 1

Let X and Y be Banach lattices. Suppose there are positive disjoint normalized sequences $(\xi _i)\subset X$ and $(\eta _i)\subset Y$ satisfying:

• For every positive disjoint normalized sequence ${\mathbf {x}} = (x_i)\subset X$ there is a constant $C = C({\mathbf {x}})$ such that $\Vert \sum _i \alpha _i x_i\Vert \le C\Vert \sum _i \alpha _i\xi _i\Vert$ $((\alpha _i)\in c_{00})$;

• For every positive disjoint normalized sequence ${\mathbf {y}} = (y_i)\subset Y$ there is a constant $C = C({\mathbf {y}})$ such that $\Vert \sum _i \alpha _i\eta _i\Vert \le C\Vert \sum _i \alpha _i y_i\Vert$ $((\alpha _i)\in c_{00})$;

• There is a constant M such that $\Vert \sum _i \alpha _i\eta _i\Vert \le M\Vert \sum _i \alpha _i\xi _i\Vert$ $((\alpha _i)\in c_{00})$.

Furthermore, suppose X is order continuous, has the SSP, $s(X)>2$ and every positive disjoint normalized sequence in Y contains a subsequence complemented by a regular projection. Then $K:[\xi _i]\rightarrow [\eta _i]$, $\sum _i \alpha _i\xi _i\mapsto \sum _i \alpha _i\eta _i$, r-factors through every positive noncompact AM-compact operator $T:X\rightarrow Y$.

Proof

Let X and Y be Banach lattices as in the hypotheses and let $T\in {{\mathcal {L}}}^r(X,Y)_+$ be a noncompact AM-compact operator. Since T is noncompact, there must be a sequence $(x_k)$ in $X_+$ such that $(Tx_k)$ has no Cauchy subsequences, and since X has the SSP, $(x_k)$ has a subsequence $(x_{k_n})$ such that $(x_{k_n}) = (f_n)+(g_n)$, with $(f_n)\subset X$ equi-integrable, $(g_n)\subset X$ disjoint and $f_n\wedge g_n = 0$ $(n\in {\mathbb {N}})$. Next, as T is AM-compact, $\{Tf_n : n\in {\mathbb {N}}\}$ must be precompact (recall T must map semicompact sets to precompact ones), and so, $(g_n)$ must contain a subsequence $(g_{n_i})$ such that $\inf _i \Vert Tg_{n_i}\Vert =: \delta >0$.

Set $y_i := Tg_{n_i}$ $(i\in {\mathbb {N}})$. By [10, Proposition 1.c.10], either $(y_i)$ contains an asymptotically disjoint subsequence (i.e., a subsequence $(y_{i_k})$ such that $\lim _k \Vert y_{i_k}-\phi _k\Vert = 0$ for some disjoint sequence $(\phi _k)\subset Y$) or there is a constant C such that

$$\begin{aligned} \begin{aligned} \Bigg (\sum _{i=1}^N |\alpha _i|^2\Bigg )^\frac{1}{2}&\le C 2^{-N} \sum _{\varepsilon _i = \pm 1} \Bigg \Vert \sum _{i=1}^N \varepsilon _i\alpha _i y_i\Bigg \Vert \le C M\Vert T\Vert \bigg \Vert \sum _{i=1}^N \alpha _i g_{n_i}\bigg \Vert , \end{aligned} \end{aligned}$$

for every set of scalars $\alpha _1,\ldots ,\alpha _N$ and $N\in {\mathbb {N}}$, where M stands for the unconditionality constant of $(g_{n_i})$. But $s(X)>2$, so $(y_i)$ must contain an asymptotically disjoint subsequence $(y_{i_k})$. In view of our hypotheses on Y, by further passing to a subsequence if necessary, we can assume there is a positive disjoint sequence $(\psi _k)\subset Y$, complemented by a regular projection Q and such that $\sum _k \Vert y_{i_k}-\psi _k\Vert <\delta /2$.

Since $(\psi _k)$ is a 1-unconditional basis for $[\psi _k]$, each biorthogonal functional $\psi _k^*$ has norm $\le \Vert \psi _k\Vert ^{-1}< 2/\delta$ (for $\inf _i \Vert y_i\Vert = \delta$ and $\Vert y_{i_k}-\psi _k\Vert <\delta /2$ $(k\in {\mathbb {N}})$), and by [6, Chapter V, Theorem 9], $(y_{i_k})$ is a basic sequence equivalent to $(\psi _k)$, and hence a basis for $[y_{i_k}]$. For each $k\in {\mathbb {N}}$, let $z_k^*$ be any norm-preserving extension to Y, of the k-th biorthogonal functional on $[y_{i_k}]$ corresponding to the basis $(y_{i_k})$. Note $(z_k^*)$ is norm-bounded, for if $L:[y_{i_k}]\rightarrow [\psi _k]$, $\sum _k \alpha _k y_{i_k}\mapsto \sum _k \alpha _k\psi _k$, then $y_{i_k}^* = \psi _k^*\circ L$ $(k\in {\mathbb {N}})$.

To finish, let $R_1:[\xi _k]\rightarrow X$, $\sum _k \alpha _k\xi _k\mapsto \sum _k \alpha _k g_{n_{i_k}}$, let $S_1:Y\rightarrow Y$, $y\mapsto y +\sum _k z_k^*(y)(\psi _k-y_{i_k})$ and let $S_2:[\psi _k]\rightarrow [\eta _k]$, $\sum _k \alpha _k\psi _k \mapsto \sum _k \alpha _k\eta _k$. All the maps just defined are readily seen to be well-defined and regular (in the case of $S_1$ simply note that it is a nuclear perturbation of $\mathrm {id}_Y$) and one easily checks that $S_2 Q S_1 T R_1 = K$. $\square$

Corollary 2

Let p, q be such that either $1<p\le q<2$ or $2<p\le q<\infty$, let $\imath _{pq}:\ell _p\rightarrow \ell _q$ be the natural embedding (of course, $\imath _{pp}$ is just $\mathrm {id}_{\ell _p})$, and let $\mu$ and $\nu$ be $\sigma$-finite measures. If $L^p(\mu )$ and $L^q(\nu )$ are infinite dimensional, then ${{\mathcal {I}}} := \varGamma _{\imath _{pq},r}^{o}(L^p(\mu ),L^q(\nu ))$ is the smallest order ideal of ${{\mathcal {I}}}^r(L^p(\mu ),L^q(\nu ))$, not contained in ${{\mathcal {A}}}^r(L^p(\mu ),L^q(\nu ))$ and with the property that ${{\mathcal {L}}}^r(L^q(\nu ))\circ {{\mathcal {I}}} \circ {{\mathcal {L}}}^r(L^p(\mu ))\subseteq {{\mathcal {I}}}$.

Proof

Suppose first $2<p\le q<\infty$. Let $(\xi _i)$ and $(\eta _i)$ be positive disjoint normalized sequences in $L^p(\mu )$ and $L^q(\nu )$, respectively. Let $\phi :[\xi _i]\rightarrow \ell _p$, $\sum _i \alpha _i\xi _i\mapsto (\alpha _i)$, and $\psi :\ell _q\rightarrow [\eta _i]$, $(\beta _i)\mapsto \sum _i \beta _i\eta _i$. By Proposition 1, if $T\in {{\mathcal {L}}}^r(L^p(\mu ),L^q(\nu ))_+$ is noncompact AM-compact then there are $V\in {{\mathcal {L}}}^r(L^q(\nu ),[\eta _i])$ and $W\in {{\mathcal {L}}}^r([\xi _i],L^p(\mu ))$ so that $VTW = \psi \imath _{pq}\phi$. In turn, $\imath _{pq} = \psi ^{-1}VTW\phi ^{-1}$, and therefore, $\varGamma _{\imath _{pq},r}(L^p(\mu ),L^q(\nu ))\subseteq \varGamma _{T,r}(L^p(\mu ),L^q(\nu ))$.

Now let $T\in {{\mathcal {I}}}^r(L^p(\mu ),L^q(\nu )){\setminus }{{\mathcal {A}}}^r(L^p(\mu ),L^q(\nu ))$ arbitrary. Then |T| is a non-compact AM-compact operator. The smallest order ideal ${{\mathcal {J}}}$ of ${{\mathcal {L}}}^r(L^p(\mu ),L^q(\nu ))$ that contains T and satisfies ${{\mathcal {L}}}^r(L^q(\nu ))\circ {{\mathcal {J}}}\circ {{\mathcal {L}}}^r(L^p(\mu ))\subseteq {{\mathcal {J}}}$ is clearly $\varGamma _{|T|,r}^o(L^p(\mu ),L^q(\nu ))$, and by the result from the previous paragraph, $\varGamma _{\imath _{pq},r}(L^p(\mu ),L^q(\nu ))\subseteq \varGamma _{|T|,r}(L^p(\mu ),L^q(\nu ))$. From this, one readily obtains that ${{\mathcal {I}}}\subseteq \varGamma _{|T|,r}^o(L^p(\mu ),L^q(\nu ))$. Since the last holds for any $T\in {{\mathcal {I}}}^r(L^p(\mu ),L^q(\nu )){\setminus }{{\mathcal {A}}}^r(L^p(\mu ),L^q(\nu ))$ and ${{\mathcal {I}}}$ is an order ideal not contained in ${{\mathcal {A}}}^r(L^p(\mu ),L^q(\nu ))$ and such that ${{\mathcal {L}}}^r(L^q(\nu ))\circ {{\mathcal {I}}}\circ {{\mathcal {L}}}^r(L^p(\mu ))\subseteq {{\mathcal {I}}}$, the desired conclusion follows for $2<p\le q<\infty$.

Second, suppose $1<p\le q<2$ and let $T\in {{\mathcal {L}}}^r(L^p(\mu ),L^q(\nu ))_+$ be a noncompact AM-compact operator. Then $2<q'\le p'<\infty$ and $T'\in {{\mathcal {L}}}^r(L^{q'}(\nu ),L^{p'}(\mu ))_+$ is also noncompact AM-compact (see [22, Theorem 125.6]), where $p'$ and $q'$ stand for the conjugates of p and q, respectively. As in the previous case, $\imath _{q'p'} = RT'S$ for some $R\in {{\mathcal {L}}}^r(L^{p'}(\mu ),\ell _{p'})$ and $S\in {{\mathcal {L}}}^r(\ell _{q'},L^{q'}(\nu ))$. Taking adjoints on the last identity, one obtains that $\imath _{pq} = S'TR'$, and in turn, that $\varGamma _{\imath _{pq},r}(L^p(\mu ),L^q(\nu ))\subseteq \varGamma _{T,r}(L^p(\mu ),L^q(\nu ))$. From this, arguing exactly as in the previous paragraph, one obtains the desired conclusion for $1<p\le q<2$. $\square$

Unfortunately, except for the case where $p=2=q$, which we discuss next, we are unable to shed any more light into what happens when $1<p\le 2\le q<\infty$. Our next proposition strongly suggests the need for a different approach in these cases.

Proposition 2

Let $L^2(\mu )$ be infinite-dimensional, with $\mu$ $\sigma$-finite, let $(\chi _i)$ be a disjoint sequence of non-zero characteristic functions and let P be the average projection defined by $P(f) := \sum _i \Vert \chi _i\Vert ^{-2} (\int f\chi _i\,d\mu ) \chi _i$ $\,(f\in L^2(\mu ))$. If $T\in {{\mathcal {L}}}^r(L^2(\mu ))$ is a positive noncompact kernel operator then $\varGamma _{T,r}^o(L^2(\mu ))\ni P$.

Proof

Let $X := L^2(\mu )$ and let $T\in {{\mathcal {L}}}^r(X)_+$ be a noncompact kernel operator. Let $\delta := \inf \big \{\Vert T-S\Vert : S\in {{\mathcal {A}}}^r(X)\cap [0,T]\big \}$. Note $\delta >0$ since otherwise T would be compact. Let ${\mathfrak {M}}\subset X$ be a complete orthonormal system for X, and for every $E\subset {\mathfrak {M}}$, let $P_E$ be the natural projection onto $\mathrm {sp}(E)$ with respect to ${\mathfrak {M}}$.

Given $\varepsilon > 0$, we construct positive sequences $(\xi _i)\subset X$ and $(\phi _i)\subset X'$, and an increasing sequence of finite subsets of ${\mathfrak {M}}$, as follows. First, set $T_1 := T$ and choose $\xi _1\in X_{[1]}\cap X_+$ and $\phi _1\in X'_{[1]}\cap X'_+$ so that $\phi _1(T_1\xi _1)\ge \Vert T_1\Vert -\varepsilon /2$. Next, choose $F_1\subset {\mathfrak {M}}$ finite so that

$$\begin{aligned} \max \big \{\Vert \xi _1 - P_{F_1}\xi _1\Vert , \Vert \phi _1 - \phi _1P_{F_1}\Vert \big \}\le \frac{\varepsilon }{2\Vert T_1\Vert }, \end{aligned}$$

and then choose $S_1\in {{\mathcal {A}}}^r(X)\cap [0,T_1]$ so that

$$\begin{aligned} \max \big \{\Vert P_{F_1}(T_1-S_1)\Vert , \Vert (T_1-S_1)P_{F_1}\Vert \big \}\le \frac{\varepsilon }{2}. \end{aligned}$$

(To see the latter is possible, recall ${{\mathcal {A}}}^r(X)$ is an order ideal whenever X and $X'$ are both order continuous, and so, $T = \sup ([0,T]\cap {{\mathcal {A}}}^r(X))\,$ [13, Proposition 1.2.6].) In turn, $Tx = \sup \{Sx : S\in [0,T]\cap {{\mathcal {A}}}^r(X)\}$ $(x\in X_+)$, and since X is order-continuous, $\lim _{S\in {\mathfrak {S}}} \Vert Tx-Sx\Vert = 0$ $(x\in X_+)$, where we have written ${\mathfrak {S}}$ for the upwards directed set $[0,T]\cap {{\mathcal {A}}}^r(X)$. If $\varDelta$ is an $\bar{\varepsilon }$-net for $P_{F_1}(X_{[1]})$, then, by the previous discussion, there is an $S\in {{\mathcal {A}}}^r(X)$ so that $\Vert (T-S)(x)\Vert < \bar{\varepsilon }$ $(x\in \varDelta )$, and hence, so that $\Vert (T-S)P_{F_1}\Vert < (1+\Vert T\Vert )\bar{\varepsilon }$. As for the other inequality, note $T'$ is a kernel operator ([22, Theorem 97.3]), so, by the previous part, there is for every $\bar{\varepsilon } >0$ an operator $R\in {{\mathcal {A}}}^r(X')$ so that $\Vert (T'-R)P'_{F_1}\Vert < \bar{\varepsilon }$. Then note that $\Vert P_{F_1}(T-R')\Vert = \Vert (T'-R)P'_{F_1}\Vert$ and $R'\in {{\mathcal {A}}}^r(X)$.) Set $T_2 := T_1-S_1$. In general, if $T_i\in {{\mathcal {L}}}^r(X)_+$ has been defined and $F_{i-1}\subset {\mathfrak {M}}$ has been chosen, choose $\xi _i\in X_{[1]}\cap X_+$ and $\phi _i\in X'_{[1]}\cap X'_+$ so that $\phi _i(T_i\xi _i)\ge \Vert T_i\Vert -\varepsilon /2^i$. Then choose $F_i\supset F_{i-1}$ finite so that

$$\begin{aligned} \max \big \{\Vert \xi _j - P_{F_{i}}\xi _j\Vert , \Vert \phi _j - \phi _jP_{F_{i}}\Vert \big \}\le \frac{\varepsilon }{2^i \Vert T_i\Vert } \quad (1\le j\le i), \end{aligned}$$

choose $S_i\in {{\mathcal {A}}}^r(X)\cap [0,T_i]$ so that

$$\begin{aligned} \max \big \{\Vert P_{F_i}(T_i-S_i)\Vert , \Vert (T_i-S_i)P_{F_{i}}\Vert \big \}\le \frac{\varepsilon }{2^i}, \end{aligned}$$

and set $T_{i+1} := T_i-S_i$. Further, set $P_{F_0} = T_0 = S_0 = 0$, and to simplify notations set $Q_i := P_{F_i}-P_{F_{i-1}}$ $(i\in {\mathbb {N}})$.

Let $F := \bigcup _i F_i$. Then, for every $i\in {\mathbb {N}}$,

$$\begin{aligned} \begin{aligned} \Vert T_i\Vert ^2\big (1 - \Vert (P_F - Q_i )\xi _i\Vert ^2\big )&\ge \Vert T_i\Vert ^2\Vert Q_i\xi _i\Vert ^2 \ge \Vert T_i Q_i\xi _i\Vert ^2 \\&\ge \big (\Vert T_iP_{F_i}\xi _i\Vert - \Vert (T_{i-1}-S_{i-1})P_{F_{i-1}}\Vert \big )^2 \\&\ge \Big (\Vert T_i\Vert - \frac{\varepsilon }{2^{i-2}}\Big )^2, \end{aligned} \end{aligned}$$

and so,

$$\begin{aligned} \Vert (P_F-Q_i)\xi _i\Vert ^2 \le 1-\bigg (1-\frac{\varepsilon }{2^{i-2}\Vert T_i\Vert }\bigg )^2 \le \frac{\varepsilon }{2^{i-3}\delta }. \end{aligned}$$

In turn, $\sum _i \Vert \xi _i-Q_i\xi _i\Vert < 8\sqrt{\varepsilon /\delta }$ (note that $P_F\xi _i = \xi _i$ $(i\in {\mathbb {N}})$). On the other hand, it follows from our construction of $(\xi _i)$ that $\Vert T\Vert \Vert \xi _i\Vert \ge \Vert T_i\Vert -\varepsilon /2^i\ge \delta /2$ $\,(i\in {\mathbb {N}})$, and hence, that $\Vert Q_i\xi _i\Vert \ge \delta /(2\Vert T\Vert )-2\sqrt{\varepsilon /\delta }$ $(i\in {\mathbb {N}})$.

Write F as a sequence $(x_i)$ in such a way that if $|F_{k-1}|< i\le |F_k|$ then $x_i\in F_k{\setminus } F_{k-1}$ $(k\in {\mathbb {N}})$, with $F_0 := \emptyset$. Let us suppose $\delta /(2\Vert T\Vert )> 2\sqrt{\varepsilon /\delta }$, so $Q_i\xi _i\ne 0$ $(i\in {\mathbb {N}})$. In this situation, it is clear that $(Q_i\xi _i)$ is a block basic sequence with respect to $(x_i)$, whose unconditionality constant equals 1. If $f_i$ denotes the i-th biorthogonal functional on $[Q_i\xi _i]$, corresponding to the basis $(Q_i\xi _i)$, then $\Vert f_i\Vert \le 1/\Vert Q_i\xi _i\Vert$ $(i\in {\mathbb {N}})$ and from our previous estimates we obtain that

$$\begin{aligned} \sum _i \Vert f_i\Vert \Vert \xi _i-Q_i\xi _i\Vert \le \frac{16\Vert T\Vert \sqrt{\varepsilon }}{\delta ^{\frac{3}{2}} - 4\Vert T\Vert \sqrt{\varepsilon }}. \end{aligned}$$

So provided $\varepsilon$ is small enough, $\sum _i \Vert f_i\Vert \Vert \xi _i-Q_i\xi _i\Vert < 1$ and $(\xi _i)$ is a basic sequence in X, equivalent to $(Q_i\xi _i)$ (see [6, Chapter V, Theorem 9]). Analogously, one shows $(\phi _i)$ is equivalent to $(Q_i'\phi _i)$ for every $\varepsilon >0$ sufficiently small.

Suppose $\varepsilon$ has been chosen so that $(\xi _i)$ and $(\phi _i)$ are basic sequences equivalent to $(Q_i\xi _i)$ and $(Q_i'\phi _i)$, respectively. Define $U:X\rightarrow X$ by $Uf := \sum _i \Vert \chi _i\Vert ^{-1} (\int f\chi _i\,d\mu ) \xi _i$ $(f\in X)$ and $V:X\rightarrow X$ by $Vf := \sum _i \Vert \chi _i\Vert ^{-1}\phi _i(f)\chi _i$ $(f\in X)$. It is easy to see that both $(Q_i\xi _i)$ and $(Q_i'\phi _i)$ are equivalent to the unit vector basis of $\ell _2$, and hence, to $(\Vert \chi _i\Vert ^{-1}\chi _i)$. In turn, $(\xi _i)$ and $(\phi _i)$ are equivalent to $(\Vert \chi _i\Vert ^{-1}\chi _i)$, and one readily sees U and V are well-defined. Clearly, U and V are positive (hence, regular). To finish the proof, simply note that $\phi _i(T\xi _i)\ge \phi _i(T_i\xi _i)\ge \Vert T_i\Vert -\varepsilon /2^i> \delta /2$ $(i\in {\mathbb {N}})$, so

$$\begin{aligned} VTUf\ge \sum _i \phi _i(T\xi _i) \Vert \chi _i\Vert ^{-2}\bigg (\int f\chi _i\,d\mu \bigg ) \chi _i\ge \frac{\delta }{2} Pf \quad (f\in X_+). \end{aligned}$$

The rest is clear. $\square$

The bimonotonicity of the basic sequence $(x_i)$, in the proof of Proposition 2, although (admittedly) not entirely apparent, played a key role. For $p\ne 2$, no basis for a separable non-atomic $L^p$-space can be bimonotone (recall the Haar system in $L^p[0,1]$ is precisely reproducible [10, Theorem 2.c.8]; and is bimonotone only if $p=2$ [see [14] for a much stronger result)]. This appears to be the main obstacle preventing the above argument from working for non-atomic $L^p$-spaces when $p\ne 2$.

We can now prove the following result, announced in the abstract:

Corollary 3

If $\mu$ is $\sigma$-finite then ${{\mathcal {A}}}^r(L^p(\mu ))$ is the only non-zero proper closed ideal (i.e., order and algebra ideal) of ${{\mathcal {I}}}^r(L^p(\mu ))$ for every $1<p<\infty$.

Proof

First, by Corollary 1, for every $1<p<\infty$, ${{\mathcal {I}}}^r(L^p(\mu )) = \varGamma _{\ell _p,r}^o(L^p(\mu ))$. If either $1<p<2$ or $2<p<\infty$, then, by Corollary 2, $\varGamma _{\ell _p,r}^o(L^p(\mu ))$ is the smallest ideal of ${{\mathcal {I}}}^r(L^p(\mu ))$ not contained in ${{\mathcal {A}}}^r(L^p(\mu ))$, and the desired conclusion follows. If on the other hand, $p = 2$, then, by Proposition 2, the order and algebra ideal of ${{\mathcal {L}}}^r(L^2(\mu ))$ generated by an operator $T\in {{\mathcal {I}}}^r(L^2(\mu )){\setminus } {{\mathcal {A}}}^r(L^2(\mu ))$ must contain a positive projection P whose range is the span of a normalized disjoint sequence in $L^2(\mu )_+$, and therefore, isometrically Riesz isomorphic to $\ell _2$. It follows readily from this that any operator in ${{\mathcal {L}}}^r(L^2(\mu ))$, r-factoring through $\ell _2$, r-factors through P, and hence, that $\varGamma _{\ell _2,r}^o(L^2(\mu ))\subseteq \varGamma _{T,r}^o(L^2(\mu ))$. The rest is clear. $\square$

The result of Corollary 3 fails for non-atomic $L^1$-spaces since, in this case, the approximable regular operators no longer form an order ideal but just a Banach lattice. However, it is still true that if X is an $L^1(\mu )$-space, with $\mu$-finite, then there is only one closed non-trivial proper (order and algebra) ideal in the band of kernel operators. We suspect this is probably known, although we do not have a reference for it. Clearly, it does not follow from our previous results, so, for completeness, we provide next a sketched proof of it.

To fix ideas, let X be an $L^1(\mu )$-space with $\mu$ finite. Recall first that the algebra ideal of weakly compact operators on X, ${{\mathcal {W}}}(X)$, is an order ideal (see [2, Corollary 5.36]) and satisfies ${{\mathcal {W}}}(X)\subsetneq {{\mathcal {I}}}^r(X)$ (see for instance [13, Theorem 3.5.16]). It turns out that

${{\mathcal {W}}}(X)$ is the only closed non-trivial proper ideal of ${{\mathcal {I}}}^r(X)$.

Proof

We show first that ${{\mathcal {W}}}(X)$ is the smallest norm-closed order ideal of ${{\mathcal {I}}}^r(X)$ that contains ${{\mathcal {F}}}(X)$. Clearly, it will suffice to show that every $T\in {{\mathcal {W}}}(X)_+$ belongs to the closed order ideal generated by ${{\mathcal {F}}}(X)$. So let $T\in {{\mathcal {W}}}(X)_+$ and $\varepsilon >0$ arbitrary. Now, as T is weakly compact, by [9, Theorem III.2.12], there is $k_T\in L^\infty (\mu ,X)$ such that $Tf = \int _\varOmega k_Tf\,d\mu$ $(f\in X)$ and $\{k_T(t) : t\in \varOmega \}$ is relatively weakly compact in X (where $\varOmega$ stands for the underlying set of the measure space). Furthermore, since relatively weakly compact subsets of X are semicompact (see [13, Theorem 2.5.4]), there exists $u\in X_+$ such that $\{k_T(t) : t\in \varOmega \}\subseteq [-u,u]+\varepsilon X_{[1]}$. Define $Sf := (\int _\varOmega f\,d\mu )u$ $(f\in X)$. Then $S\in {{\mathcal {F}}}(X)$ and for every $f\in X_+$,

$$\begin{aligned} \begin{aligned} \Vert (T-S)_+(f)\Vert _1&\le \bigg \Vert \int _\varOmega (k_T(t)-u)_+ f(t)\,d\mu (t)\bigg \Vert _1 \\&\le \int _\varOmega \Vert (k_T(t)-u)_+\Vert _1 f(t)\,d\mu (t)\le \varepsilon \Vert f\Vert _1 \end{aligned} \end{aligned}$$

(the first inequality follows easily from the fact that $|(T-S)|f = \sup _{|\xi |\le f} |(T-S)\xi |\le \int |k_T(t)-u|f(t)\,d\mu$). One concludes readily from this that $\Vert (T-S)_+\Vert \le \varepsilon$, and since $T = T\wedge S+(T-S)_+$ and $\varepsilon >0$ was arbitrary, the desired conclusion follows.

To see ${{\mathcal {W}}}(X)$ is also the only closed ideal of ${{\mathcal {I}}}^r(X)$, let $T\in {{\mathcal {I}}}^r(X){\setminus }{{\mathcal {W}}}(X)$ be arbitrary. Since $T(X_{[1]})$ is not relatively weakly compact, it must contain a sequence $(y_i)$ equivalent to the unit vector basis of $\ell _1$ (see [21, III.C.12]). Let Y be a separable closed Riesz subspace of X containing $[y_i]$. The space Y must be isomorphic, as a Banach space, to either $\ell _1$ or $L^1[0,1]$, and so, $[y_i]$ must contain a sequence $(z_i)$ equivalent to the unit vector basis of $\ell _1$ and such that $[z_i]$ is complemented in Y (see for instance [5, Propositions IV.1.2 & VI.2.8]). In turn, as Y is complemented in X, so is $[z_i]$. Choose a bounded sequence $(x_i)\subset X$ such that $Tx_i = z_i$ $(i\in {\mathbb {N}})$ (that such a sequence exists, follows on noting that if $(\xi _i)\subset X_{[1]}$ is such $y_i = T\xi _i$ $(i\in {\mathbb {N}})$, then $(\xi _i)$ is equivalent to $(y_i)$ and the restriction of T to $[\xi _i]$ is an isomorphism onto $[y_i]$). Let $P:X\rightarrow [z_i]$ be a bounded projection, let $(e_i)$ stand for the unit vector basis of $\ell _1$, let $R:\ell _1\rightarrow X$, $\sum _i \alpha _ie_i\mapsto \sum _i \alpha _ix_i$, and let $S:[z_i]\rightarrow \ell _1$, $\sum _i \alpha _iz_i\mapsto \sum _i \alpha _ie_i$. Then SPTR is a factorization of $\mathrm{id}_{\ell _1}$ through T. Since every integral operator on $L^1(\mu )$ must factor through $\ell _1$ (see Remark 4), the algebra ideal generated by T must be the whole of ${{\mathcal {I}}}^r(X)$. $\square$

Recall an operator $T\in {{\mathcal {L}}}^r(X)$ is said to be semicompact if $T(X_{[1]})$ is semicompact. The linear span $\mathcal {SK}^r(X)$ of the positive semicompact operators on a Dedekind complete Banach lattice X is always a norm-closed ideal of ${{\mathcal {L}}}^r(X)$ (this is well-known and easy to prove). Now let X be an $L^p(\mu )$-space for some $1\le p<\infty$. If $p=1$ then ${\mathcal {SK}}^r(X) = {{\mathcal {W}}}(X)$ (see [13, Theorem 2.5.4] and [2, Theorem 5.35]), while if $1<p<\infty$ then ${{\mathcal {I}}}^r(X)\cap {\mathcal {SK}}^r(X) = {{\mathcal {A}}}^r(X)$ [17, Theorem 3.3]. One can thus restate the result of Corollary 3 for finite $\mu$ and the above result, in a unified way, as follows:

Corollary 4

For every $1\le p<\infty$ and $\mu$ finite, ${{\mathcal {I}}}^r(L^p(\mu ))\cap {\mathcal {SK}}^r(L^p(\mu ))$ is the only closed non-trivial proper ideal of ${{\mathcal {I}}}^r(L^p(\mu ))$.

4 r-Factorization properties of regular kernel operators

As explained in the introduction, the main purpose of this section is to shed further light into the relationship between factorization properties of the kind encountered in the previous section and the property of being a (regular) kernel operator. In particular, we wish to know to which extent such properties determine the kernel operators within the regular ones. Our main findings in this direction are collected in the next theorem.

Theorem 2

Let X and Y be Banach lattices and let $T\in {{\mathcal {L}}}^r(X,Y)$.

• (i) Suppose both $X'$ and Y are order continuous, both have weak order units and there is a regular projection $P:Y''\rightarrow Y$. Under these assumptions, if $T\in {{\mathcal {F}}}(X,Y)^{dd}$ then T r-factors through a purely atomic Banach lattice. Moreover, if T is positive then at least one of the factors (any one of them) can be chosen to be positive.

• (ii) If T r-factors through an order continuous purely atomic Banach lattice, then T belongs to ${{\mathcal {F}}}(X,Y)^{dd}$.

• (iii) Let X be reflexive, let Y be a KB space and suppose both X and Y have weak order units. Then all the following are equivalent:

  • (a) T is a kernel operator.

  • (b) T r-factors through a purely atomic Banach lattice Z and the factors $S:X\rightarrow Z$ and $R:Z\rightarrow Y$ can be chosen so that for every atom $z\in S(X)^{dd}$, $Rz\ne 0$.

  • (c) T r-factors through a reflexive purely atomic Banach lattice.

  Furthermore, in parts (b) and (c), if T is positive then at least one of the factors (any one of them) can be chosen to be positive.

In proving the theorem, we shall make use of the following simple fact.

Lemma 2

Let X and Y be Banach lattices and let $T\in {{\mathcal {A}}}^r(X,Y)$. Then, for every $\varepsilon >0$, there are sequences $(f_i)\subset X'_+$, $(y_i)\subset Y_+$ and $(\eta _i)\subset \{\pm 1\}^{{\mathbb {N}}}$ so that $T = \sum _i \eta _i f_i\otimes y_i$ and $\sum _i f_i(x)g(y_i)\le (2\Vert T\Vert _r + \varepsilon ) \Vert x\Vert \Vert g\Vert$ $(x\in X_+,\,g\in Y'_+)$.

Proof

Let $\varepsilon >0$ arbitrary, let H be the family of all words in the digits 0 and 1, and let $H_n := \{h\in H : |h| = n\}$ $(n\in {\mathbb {N}})$, where |h| stands for the length of the word $h\in H$. Let $T\in {{\mathcal {A}}}^r(X,Y)$ arbitrary. We shall construct a family $\{T_h : h\in H\}\subset X_+'\otimes Y_+$ as follows. First, let e stand for the empty word and set $S_e := T$. Then choose $T_0,\,T_1\in X_+'\otimes Y_+$ so that $\max \big \{\Vert (S_e)_+-T_0\Vert _r, \Vert (S_e)_- - T_1\Vert _r\big \}\le \varepsilon /4$, and set $S_0 := (S_e)_+-T_0$ and $S_1 := (S_e)_--T_1$. In general, if $T_h$ and $S_h$ are known, choose $T_{h0},\,T_{h1}\in X_+'\otimes Y_+$ so that

$$\begin{aligned} \max \big \{\Vert (S_h)_+-T_{h0}\Vert _r, \Vert (S_h)_--T_{h1}\Vert _r\big \}\le \frac{\varepsilon }{4^{|h|+1}}, \end{aligned}$$

and set $S_{h0} := (S_h)_+-T_{h0}$ and $S_{h1} := (S_h)_- - T_{h1}$.

Let s(h) $(h\in H)$ be the number of ones in the word h. It is then easy to see that

$$\begin{aligned} \Bigg \Vert T - \sum _{n=1}^m \sum _{h\in H_n} (-1)^{s(h)} T_h\Bigg \Vert _r = \Bigg \Vert \sum _{h\in H_m} (-1)^{s(h)} S_h\Bigg \Vert _r \le \frac{\varepsilon }{2^m} \quad (m\in {\mathbb {N}}). \end{aligned}$$

Furthermore, since

$$\begin{aligned} \sum _n \sum _{h\in H_n} \Vert T_h\Vert _r\le 2\Vert T\Vert _r + \frac{\varepsilon }{2} + \sum _{n>1} 2^n\bigg (\frac{\varepsilon }{4^{n-1}} + \frac{\varepsilon }{4^n}\bigg ) \le 2\Vert T\Vert _r + 5\varepsilon , \end{aligned}$$

(1)

convergence of $\sum _{h\in H} (-1)^{s(h)} T_h$ is absolute. For each $h\in H$, let $\{f_{i,h} : 1\le i\le k(h)\}\subset X_+'$ and $\{y_{i,h} : 1\le i\le k(h)\}\subset Y_+$ be so that $\sum _{i=1}^{k(h)} f_{i,h}\otimes y_{i,h} = T_h$. Then

$$\begin{aligned} T = \sum _h \sum _{i=1}^{k(h)} (-1)^{s(h)} \,f_{i,h}\otimes y_{i,h}, \end{aligned}$$

and since $0\le \sum _{i\in S} f_{i,h}\otimes y_{i,h}\le T_h$ for every $S\subseteq \{1,\ldots ,k(h)\}$ $\,(h\in H)$, also the series $\sum _{h,i} (-1)^{s(h)} f_{i,h}\otimes y_{i,h}$ converges absolutely to T. Both claims of the lemma now follow easily from this last observation and (1) above. $\square$

Proof of Theorem 2

(i) It will suffice to consider the case where $T\ge 0$, for if $T\in {{\mathcal {L}}}^r(X,Y)$ is such that $T_+$ r-factors through a Banach lattice $Z_1$ with factors $R_1\in {{\mathcal {L}}}^r(Z_1,Y)$ and $S_1\in {{\mathcal {L}}}^r(X,Z_1)$, and $T_-$ r-factors through a Banach lattice $Z_2$ with factors $R_2\in {{\mathcal {L}}}^r(Z_2,Y)$ and $S_2\in {{\mathcal {L}}}^r(X,Z_2)$, then T r-factors through $Z_1\oplus _2 Z_2$ with factors $R:Z_1\oplus _2 Z_2\rightarrow Y$, $(z_1,z_2)\mapsto R_1z_1-R_2z_2$, and $S:X\rightarrow Z_1\oplus _2 Z_2$, $x\mapsto (S_1x,S_2x)$ (here, $Z_1\oplus _2 Z_2$ stands for the $\ell _2$-sum of $Z_1$ and $Z_2$, with the natural order).

So let $T\ge 0$ and let $\xi$ and e be weak order units for $X'$ and Y, respectively. Set $S_n := T\wedge (n\xi \otimes e)$ $(n\in {\mathbb {N}})$, and let $T_n := S_{n+1}-S_n$ $(n\in {\mathbb {N}})$. Note that since $X'$ and Y are order continuous, $S_n\in {{\mathcal {A}}}^r(X,Y)$ $(n\in {\mathbb {N}})$. Clearly, $\{\xi \otimes e\}^{dd} = {{\mathcal {F}}}(X,Y)^{dd} = {{\mathcal {A}}}^r(X,Y)^{dd}$, and so, for every $R\in {{\mathcal {A}}}^r(X,Y)\cap [0,T]$, $R\wedge (n\xi \otimes e)\uparrow R$. It follows that $\sup _n S_nx\ge Rx$ for every $R\in {{\mathcal {A}}}^r(X,Y)\cap [0,T]$ and $x\in X_+$, and hence, that $\sup _n S_n x\ge \sup \{Rx : R\in {{\mathcal {A}}}^r(X,Y)\cap [0,T]\} = Tx$. Since Y is order continuous and $S_n x\uparrow Tx$ $(x\in X_+)$, $(S_n)$ converges to T in the strong operator topology.

Let $\varepsilon >0$ arbitrary. For each $n\in {\mathbb {N}}$, choose ${\widetilde{T}}_n\in X_+'\otimes Y_+$ so that $\Vert T_n - {\widetilde{T}}_n\Vert _r <2^{-n}\varepsilon$, and choose sequences $(f_{i,n})_{i=1}^{k(n)}\subset X'_+$ and $(y_{i,n})_{i=1}^{k(n)}\subset Y_+$ such that ${\widetilde{T}}_n = \sum _{i=1}^{k(n)} f_{i,n}\otimes y_{i,n}$. Since $\sum _n T_nx = Tx$ $(x\in X)$, the series $\sum _n {\widetilde{T}}_n$ converges in the strong operator topology to some positive linear operator ${\widetilde{T}}$. Clearly, $\sum _n (T_n-{\widetilde{T}}_n) =: T_0\in {{\mathcal {A}}}^r(X,Y)$, so, by the previous lemma, there are sequences $(f_{i,0})\subset X'_+$, $(y_{i,0})\subset Y_+$ and $(\eta _i)\subset \{\pm 1\}^{{\mathbb {N}}}$ such that $T_0(x) = \sum _i \eta _i f_{i,0}(x)y_{i,0}$ and

$$\begin{aligned} \sum \limits _i f_{i,0}(x)g(y_{i,0})\le (2\Vert T_0\Vert _r + \varepsilon )\Vert x\Vert \Vert g\Vert \le 3\varepsilon \Vert x\Vert \Vert g\Vert \quad (x\in X_+,\,g\in Y'_+). \end{aligned}$$

One readily sees that $Tx = \sum _{n\in {\mathbb {N}}} \sum _{i=1}^{k(n)} f_{i,n}(x)y_{i,n} + \sum _i \eta _i f_{i,0}(x)y_{i,0}$ $(x\in X)$, and so, that there are sequences $(f_i)\subset X_+'$, $(y_i)\subset Y_+$ and $(\tau _i)\subset \{\pm 1\}^{{\mathbb {N}}}$ such that $Tx = \sum _i \tau _i f_i(x)y_i$ $(x\in X)$ and

$$\begin{aligned} \begin{aligned} \sum \limits _i f_i(x)g(y_i)&\le (3\varepsilon + \Vert {\widetilde{T}}\Vert _r)\Vert x\Vert \Vert g\Vert \\&\le (\Vert T\Vert _r + 4\varepsilon )\Vert x\Vert \Vert g\Vert \quad (x\in X_+,\,g\in Y'_+). \end{aligned} \end{aligned}$$

(2)

Now, let $\varGamma$ be the solid hull of the set $\big \{(f_i(x)) : x\in X_{[1]}\big \}\subset {\mathbb {R}}^{{\mathbb {N}}}$, where ${\mathbb {R}}^{{\mathbb {N}}}$ is endowed with the canonical order, and let $Z_0 := \mathrm{sp}(\varGamma )$. Then $\varGamma$ is an absolutely convex, absorbent and radially bounded subset of $Z_0$. (That $\varGamma$ is absolutely convex is obvious; that it is absorbent follows on noting that if $\xi = \sum _{i=1}^n \alpha _i\xi _i$ $(\xi \ne 0)$, with $\xi _1,\ldots ,\xi _n\in \varGamma$, then $(\sum _{j=1}^n |\alpha _j|)^{-1} |\xi |\le (\sum _{j=1}^n |\alpha _j|)^{-1}\sum _{i=1}^n |\alpha _i| |\xi _i|\in \varGamma$; to see $\varGamma$ is radially bounded simply note that for every $\xi \in Z_0{\setminus } \{0\}$, if k is such that $\xi (k)\ne 0$, then $t\xi \in \varGamma$ $\Rightarrow$ $|t\xi (k)|\le f_k(x)$ for some $x\in X_{[1]}\cap X_+$ $\Rightarrow$ $|t|\le \Vert f_k\Vert /|\xi (k)|$.) The Minkowski functional of $\varGamma$ is thus a lattice norm on $Z_0$, and we shall define Z as the completion of $Z_0$ with respect to such norm. Also note that, by (2),

$$\begin{aligned} \sum \limits _i \xi _i g(y_i)\le 2\Vert T\Vert _r + 8\varepsilon \quad ((\xi _i)\in \varGamma ,\, g\in Y'_{[1]}), \end{aligned}$$

for $(\xi _i)\in \varGamma$ $\Rightarrow$ $(|\xi _i|)\le (f_i(x))$ for some $x\in X_{[1]}\cap X_+$, and $g\in Y'_{[1]}$ $\Rightarrow$ $g_-,g_+\in Y'_{[1]}$.

Define $S:X\rightarrow Z$ by $Sx := (\tau _i f_i(x))$ $(x\in X)$, and $U:Y'\rightarrow Z'$ by $Ug := (g(y_i))$ $(g\in Y')$. By the previous paragraph, it is clear that both U and S are well-defined. Furthermore, both are regular and therefore bounded (in fact, it is clear from the previous paragraph that their norms are bounded by $2\Vert T\Vert _r + 8\varepsilon$ and 1, respectively). Also, for every $x\in X$ and $g\in Y'$,

$$\begin{aligned} g(Tx) = \sum \limits _i \tau _i f_i(x)g(y_i) = Ug(Sx) = (U'(Sx))(g), \end{aligned}$$

so $Tx = U'(Sx)$ $(x\in X)$. Let $R := P\circ U'\big |_Z$. Then $T = RS$ is a factorization of the kind claimed by the theorem. Note that if $z_i$ is the sequence (or atom) in Z with 1 in the i-th entry and 0 in all the others, then $(U'z_i)(g) = z_i(Ug) = g(y_i)$ $(g\in Y')$, i.e., $U'z_i = y_i$, and therefore $Rz_i = y_i\ne 0$ $(i\in {\mathbb {N}})$.

As for the final assertion of part (i), simply note that above, we could have defined $S{:}X\rightarrow Z$ by $Sx:= (f_i(x))$ $(x\in X)$, and $U{:}Y'\rightarrow Z'$ by $Ug := (\tau _i g(y_i))$ $(g\in Y')$.

(ii) Here, we could argue as in Corollary 1. Instead, though, we shall provide an alternative, more elementary and self-contained argument. To this end, suppose T r-factors through an atomic Banach lattice Z, i.e., there are $R\in {{\mathcal {L}}}^r(Z,Y)$ and $S\in {{\mathcal {L}}}^r(X,Z)$ so that $RS = T$. Consider first the case where R and S (and therefore T) are positive. Let $\{z_i : i\in I\}$ be the collection of all normalized atoms of Z whose linear span, since Z is order continuous, must be dense in Z. Let $\{z_i^* : i\in I\}$ be the corresponding set of biorthogonal functionals in $Z'$, and define $f_i := S'z_i^*$ and $y_i := Rz_i$ $(i\in I)$. Then

$$\begin{aligned} \begin{aligned} Tx = RSx&= R\Big (\sum \limits _i z_i^*(Sx)z_i\Big ) = \sum \limits _i z_i^*(Sx)Rz_i = \sum \limits _i f_i(x)y_i \quad (x\in X). \end{aligned} \end{aligned}$$

We thus have that $0\le \sum _{i\in F} f_i\otimes y_i\uparrow T$, where F runs over all finite subsets of I, and hence, $T\in {{\mathcal {F}}}(X,Y)^{dd}$.

In the general case, $T = S_+R_+ + S_-R_- - S_+R_- - S_-R_+$, and by the previous discussion, $S_+R_+$, $S_-R_-$, $S_+R_-$ and $S_-R_+$ belong to ${{\mathcal {F}}}(X,Y)^{dd}$.

(iii) (a) $\Rightarrow$ (b): Let X and Y be as in the hypotheses and let $T:X\rightarrow Y$ be a regular kernel operator. Since X is order continuous and has a weak order unit, $X'$ has a weak order unit too (see for instance [10, Theorem 1.b.14]); and since Y is a KB space, Y is a band in $Y''$. Thus, arguing as in the proof of part (i), one can construct a purely atomic Banach lattice Z and operators $R\in {{\mathcal {L}}}^r(Z,Y)$ and $S\in {{\mathcal {L}}}^r(X,Z)$, such that $RS = T$, $S(X)^{dd} = Z$ and $Rz\ne 0$ for every atom $z\in Z$.

It is also clear, from the proof of part (i), that if $T\ge 0$, at least one of the two factors (any one of them) can be chosen to be positive.

(b) $\Rightarrow$ (c): Let Z be a purely atomic Banach lattice such that there is an r-factorization of T through Z with factors R and S as in (iii)(b). Let $S_1\in {{\mathcal {L}}}^r(X,Z)_+$ and $R_1\in {{\mathcal {L}}}^r(Z,Y)_+$ be such that $\pm S\le S_1$ and $\pm R\le R_1$. What follows now goes more or less along the lines of the argument of the proof of [2, Theorem 5.45], which under the given assumptions, yields an r-factorization of T through a reflexive purely atomic Banach lattice. Before starting with the argument, and for the reader’s convenience, we recall a few facts.

Given a Banach lattice E and a convex solid norm-bounded subset of it, W say, we write $E_W$ for the Banach lattice whose underlying vector space is $\{x\in E : \sum _{i=1}^\infty \Vert x\Vert _i^2< \infty \}$ (where for each $i\in {\mathbb {N}}$, $\Vert \cdot \Vert _i$ is the (equivalent) norm on E given by the Minkowski functional of the set $2^i W + 2^{-i}E_{[1]}$), endowed with the norm $|||x|||:= (\sum _{i=1}^\infty \Vert x\Vert _i^2)^{\frac{1}{2}}$ $(x\in E_W)$. From [8, Lemma 1], we know that $W\subseteq (E_W)_{[1]}$ and that if W is relatively weakly compact then $E_W$ is reflexive. Furthermore, $E_W$ is a Banach lattice and an order ideal of E such that the natural inclusion $\imath :E_W\rightarrow E$ and its adjoint $\imath ':E'\rightarrow E_W'$ are both lattice homomorphisms ([2, Theorem 5.41]). Lastly, if W is the convex solid hull of a relatively weakly compact set, then $E_W'$ is order continuous, and therefore, a KB space ([2, Theorem 5.43]).

We now go through the argument. Consider first the weakly compact operator S. Let U be the solid hull of $S_1(X_{[1]})$ in Z and let $\varPsi := Z_U$ (so $S(X_{[1]})\cup S_1(X_{[1]})\subseteq U\subseteq \varPsi _{[1]}$). Then $\varPsi$ is a Banach lattice and an order ideal of Z, hence purely atomic, whose dual $\varPsi '$ is a KB space. Let ${\widetilde{S}}:X\rightarrow \varPsi$, $x\mapsto Sx$, let ${\widetilde{S}}_1:X\rightarrow \varPsi$, $x\mapsto S_1x$, and let $\imath :\varPsi \rightarrow Z$ be the natural inclusion (which is a lattice homomorphism). Then $\pm {\widetilde{S}}\le {\widetilde{S}}_1$, so ${\widetilde{S}}$ is regular, and clearly, $S = \imath \circ {\widetilde{S}}$.

Consider next the map $L := R\circ \imath :\varPsi \rightarrow Y$. Let $L_1 := R_1\circ \imath :\varPsi \rightarrow Y$. Since $\varPsi '$ is order continuous and Y is a KB space, both L and $L_1$ are weakly compact ([2, Theorem 5.27]). Thus, $L_1'$ is weakly compact, and since $\varPsi '$ is a KB space, the solid hull of $L_1'(Y'_{[1]})$, V say, is relatively weakly compact ([2, Theorem 4.39(2)]). Let $\varUpsilon := (\varPsi ')_V$. Then $\varUpsilon$ is a reflexive Banach lattice and the natural inclusion $\jmath :\varUpsilon \rightarrow \varPsi '$ and its adjoint $\jmath ':\varPsi ''\rightarrow \varUpsilon '$ are both lattice homomorphisms. Like above, note that $L'(Y'_{[1]})\cup L_1'(Y'_{[1]})\subseteq V\subseteq \varUpsilon _{[1]}$. So $H:Y'\rightarrow \varUpsilon$, $y'\mapsto L'y'$, and $H_1:Y'\rightarrow \varUpsilon$, $y'\mapsto L_1'y'$, are well-defined. Furthermore, $L' = \jmath \circ H$ and $\pm H\le H_1$.

Let $\varPhi$ be the closure of $\jmath '(\varPsi )$ in $\varUpsilon '$. Clearly $\varPhi$ is reflexive and we will show next it is also atomic. For this, let $\{e_i : i\in I\}$ be the set of all normalized atoms of $\varPsi$, let $\gamma _i := \jmath '(e_i)$ $(i\in I)$ and let $\varGamma := \{\gamma _i : i\in I\}$. Note that $0\notin \varGamma$, for $H'(\gamma _i) = L''(e_i) = L(e_i) = R(e_i)\ne 0$ $(i\in I)$, where we have identified $\varPsi$ with $\imath (\varPsi )$. Furthermore, each $\gamma _i$ is an atom in $\varPhi$. To see it, fix i and let $\phi \in \varPhi$ be such that $0<\phi \le \gamma _i$. Let $(\psi _n)\subset \varPsi _+$ be such that $\phi = \lim _n \jmath '(\psi _n)$. Without loss of generality, we can assume $0< \jmath '(\psi _n\wedge e_i)\le \gamma _i$ $(n\in {\mathbb {N}})$. Since X is atomic $\psi _n\wedge e_i$ is a supremum of elements of the form $\sum _{j\in F} t_j e_j$, with $F\subset I$ finite and $t_j\ge 0$, $j\in F$. But $0\le \sum _{j\in F} t_j e_j\le \psi _n\wedge e_i$ $\Rightarrow$ $0\le \sum _{j\in F} t_j\gamma _j\le \gamma _i$ $\Rightarrow$ $t_j = 0$ $(j\in F{\setminus }\{i\})$. Thus, $\psi _n\wedge e_i\in {\mathbb {R}} e_i$ $(n\in {\mathbb {N}})$, and in turn, $\phi \in {\mathbb {R}}\gamma _i$. It remains to show $\varGamma ^{dd} = \varPhi$. Clearly, it will suffice to show $\varGamma ^d = \{0\}$. For this, suppose towards a contradiction there is a $b\in \varGamma ^d{\setminus } \{0\}$. Since $\varGamma ^d$ is a band, we can further assume, by replacing b if necessary, that $0< b =\jmath '(\psi )$ for some $\psi \in \varPsi _+$. Since $\varPsi$ is atomic, we would have that $0< te_i\le \psi$ for some $t>0$ and $i\in I$, and hence, that $\gamma _i = \gamma _i\wedge t^{-1}b = 0$, which is clearly impossible. Thus, $\varPhi$ is atomic.

Finally, letting $P:Y''\rightarrow Y$ be the corresponding band projection, and noting that $L'' = H'\circ \jmath '$ and that $L''(\varPsi '')\subseteq Y$, it is easy to see that the maps $\jmath '\big |_{\varPsi }^{\varPhi }$ and $P\circ H'\big |_{\varPhi }$ provide a factorization of L through $\varPhi$ with regular factors. In turn, $\jmath '\big |_{\varPsi }^{\varPhi }\circ {\widetilde{S}}$ and $P\circ H'\big |_{\varPhi }$ provide the desired factorization of T through $\varPhi$.

If $T\ge 0$, then we can assume R (resp. S) to be positive. If R (resp. S) is positive it should be clear from the above argument that the same will be true about $P\circ H'\big |_{\varPhi }$ (resp. $\jmath '\big |_{\varPsi }^{\varPhi }\circ {\widetilde{S}}$).

(c) $\Rightarrow$ (a): This is immediate from (ii). $\square$

Remark 5

It should be pointed out that it is not possible, in general, for a positive T in parts (i) and (iii) of the theorem, to produce an r-factorization with both factors positive (see the example at the end of [15]). This appears to be the main obstacle in extending parts (i) and (iii) to arbitrary non-separable Banach lattices. It is not hard to see (from the proof of the theorem), though, that if $T\ge 0$ then for every $\varepsilon >0$ there exist an atomic Banach lattice Z and operators $R\in {{\mathcal {L}}}^r(Z,Y)_+$ (resp. $R\in {{\mathcal {L}}}^r(Z,Y)_+ + {{\mathcal {A}}}^r(Z,Y)_{[\varepsilon ]}$) and $S\in {{\mathcal {L}}}^r(X,Z)_+ + {{\mathcal {A}}}^r(X,Z)_{[\varepsilon ]}$ (resp. $S\in {{\mathcal {L}}}^r(X,Z)_+$) such that $RS = T$.

It follows easily from Theorem 2 that for every reflexive Banach lattice X with a weak order unit there is a reflexive purely atomic Banach lattice Z such that ${{\mathcal {I}}}^r(X) = \{T\in {{\mathcal {L}}}^r(X) : T \text { r-factors through } Z\}$. Information on the lattice of ideals of ${{\mathcal {L}}}^r(Z)$ may then lead to information on that of ${{\mathcal {I}}}^r(X)$ via the map that associates to every ideal ${{\mathcal {J}}}$ of ${{\mathcal {L}}}^r(Z)$ the ideal $\{T\in {{\mathcal {L}}}^r(X) : \pm T\le S\in \mathrm{sp}({{\mathcal {L}}}^r(Z,X)\circ {{\mathcal {J}}} \circ {{\mathcal {L}}}^r(X,Z))\}$ of ${{\mathcal {I}}}^r(X)$. Of course, the usefulness of this approach, for a given X, will depend on the properties of the latter map, and in turn, on our choice of Z.