# Dunford−Pettis property of $L^1(\mu)$

$\def\bs#1{\boldsymbol#1}\def\sp{\kern.4mm}$Let $\bs K$ be either the standard real or complex topological field, and let $E$ be a Hausdorff locally convex space over $\bs K\sp$. Then saying that $E$ has the Dunford−Pettis property means that if we have any Banach space $F$ over $\bs K$ and any continuous linear map $u:E\to F$ such that for all bounded sets $B$ in $E$ we have $u\sp[\,B\,]$ relatively compact in $F_\sigma\,$, then for every absolutely convex compact set $K$ in $E_\sigma$ we have $u\sp[\,K\,]$ relatively compact in $F\sp$. Here $G_\sigma$ means the underlying vector space of $G$ equipped with the weak topology $\sigma(G,G')$ against the topological dual space.

If now $\mu$ is a positive measure on some set $\Omega\sp$, it is "well-known" that $L^1(\mu)$ has the Dunford−Pettis property. The standard reference book for this kind of matters is the first volume of Dunford and Schwartz' Linear Operators which I unfortunately do not have at hand. However, according to the information given by R. E. Edwards in his Functional Analysis this matter should be in Theorem VI.8.12 on page 508 that I didn't succeed to see by Google. I have a suspicion that there is the additional restriction that $\mu$ be $\sigma$−finite. Furthermore, in the references I have seen this matter considered, there is the restrictive assumption that $\mu$ be a Radon measure on some Hausdorff locally compact topological space.

Now I am asking whether there exist any publication where this $L^1(\mu)$ having the Dunford−Pettis property is proved for more general positive measures $\mu$ than referred to above$\sp$?

In particular, I am interested to know whether $L^1(\mu)$ has the Dunford−Pettis property in the case where $\mu$ is almost decomposable in the following sense: There are a "negligible" set $N'$ and a decomposition $\mathscr A$ of $\Omega\setminus N'$ into disjoint sets $A$ of (strictly) positive finite measure such that if we have any set $N''\subseteq\Omega$ with the property that $A\cap N''$ is included in some set $N_A$ of zero measure for all $A\in\mathscr A\sp$, then $N''$ is "negligible"; this meaning that $N''\cap B$ is included in some set, depending on $B$ and $N''$, of zero measure for all $B$ having finite measure.

Added. (8.12.2017) Having slept over the night, it suddenly occurred to me how the case of an almost decomposable positive measure is "trivially reduced" to the case of a $\sigma$-finite one. It confused me when Bill Johnson said that the "general" case could be reduced the "separable" case since there are even probability measures $\pi$ with $L^1(\pi)$ not separable. The "coin tossing" measure on $\Omega={}^{\mathbb R}\,\{\sp 0\sp, 1\sp\}$ is an example. The reduction can be done by using the "convergent sequences characterization" of the Dunford−Pettis property in the case of Banach spaces by observing that $L^1(\mu)$ is linearly homeomorphic to the $\ell^{\sp 1}\,$−sum of the family $\langle\sp L^1(\mu_A):A\in\mathscr A\sp\rangle$ when $\mu$ is almost decomposable. This solves the problem in the case of an almost decomposable positive measure.

Since at the moment I still do not have access to the book of Dunford and Schwartz, I would be grateful if in a comment someone, who has access, could confirm my suspicion that there the case of a general $\sigma$-finite positive measure is fully proved, and hence that there is no "Radon restriction". Furthermore, I am still interested to know if there possibly exists a reference where the case of positive measures that are more general than "almost decomposable" ones is considered.

• It is trivial to reduce the general case to the case of separable $L_1$ spaces, and all of these are isomorphic to $L_1(0,1)$ or $\ell_1$, so what's the problem? – Bill Johnson Dec 8 '17 at 1:29
• I do not see how to do that "trivial reduction", so could you give some details, possibly in the form of an answer? – TaQ Dec 8 '17 at 2:58
• Given a separable subspace $X$ of an $L_1$ space, let $Y$ be the smallest closed sublattice that contains $X$. $Y$ is a separable abstract $L_1$ space and hence by Kakutani's representation theorem is an $L_1$ space. (You can avoid Kakutani's theorem, but it is a basic result.) See volume 2 of Lindenstrausr-Tzafriri. – Bill Johnson Dec 8 '17 at 16:15
• @BillJohnson If I understand correctly your suggestion, having a sequence $\bs x$ in the original $L^1(\mu)$ you take as $X$ the linear span of it, and then we can identify $\bs x$ as being in a separable $L^1(\mu_1)$ for some other measure $\mu_1$ for which the Dunford−Pettis result is known to hold? In the mean time, the following occurred to me: Express every member of $\bs x$ as a limit of a sequence simple functions, and let $\mu_1$ be the restriction of $\mu$ to the $\sigma$-algebra generated by the countably many "base" sets of these simple functions. Is $L^1(\mu_1)$ then separable? – TaQ Dec 8 '17 at 20:41
• @TaQ, you take the smallest closed lattice that contains your sequence; usually, this is strictly greater than the linear span but still separable. Certainly DP holds for $L_1$ (and $\ell_1$) and these are the only two Banach-space-isomorphism types of separable $L_1(\mu)$-spaces. – Tomek Kania Dec 8 '17 at 21:24