Predual of $l^1(\Gamma)$

Question

Let $\Gamma$ be an uncountable index set. For example $\Gamma=\mathbb R$. Let $l^1(\Gamma)$ be the set of functions with countable support and finite sum: $$ \sum_{a\in\Gamma}|f(a)|<\infty. $$ The space $l^1(\Gamma)$ is a Banach space with the norm $\|f\|:=\sum_{a\in\Gamma}|f(a)|$. It is not separable.

My question is: does $l^1(\Gamma)$ have a separable pre-dual space?

I have seen statements that $l^1(\Gamma)$ is isometric to the dual space of $c_0(\Gamma)$, however I did not found a definition of $c_0(\Gamma)$ nor a statement about its separability. The usual Krein-Milman based argument does not fail as in the $L^1$ case (the unit ball of $l^1(\Gamma)$ is the closed convex hull of its extreme points).

Answer 1

We have $$ c_0(\Gamma) := \{x \in \mathbf K^\Gamma \mid \forall \epsilon > 0 \,\exists \Gamma' \subseteq \Gamma, |\Gamma'| < \infty \land \sup_{\gamma \in \Gamma - \Gamma'} |x(\gamma)| < \epsilon \} $$ which is a Banach space with respect to the norm $$ \def\norm#1{\left\|#1\right\|}\norm{x}_\infty := \sup_{\gamma \in \Gamma} \def\abs#1{\left|#1\right|}\abs{x(\gamma)} $$ $c_0(\Gamma)^*$ is isometric to $\ell^1(\Gamma)$ via $$ T \colon \ell^1(\Gamma) \to c_0(\Gamma)^*, \quad (Ty)(x) := \sum_{\gamma\in \Gamma} y(\gamma)x(\gamma), \qquad x \in c_0(\Gamma), y \in \ell^1(\Gamma) $$ Regarding seperability: $c_0(\Gamma)$ is not separable: For $\delta \in \Gamma$, consider $e_\delta \colon \Gamma \to \mathbf K$ given by $e_\delta(\gamma) = 0$ for $\gamma \ne \delta$ and $e_\delta(\delta) = 1$. Then $e_\delta \in c_0(\Gamma)$ and $$ \norm{e_\delta - e_{\delta'}} = 1, \qquad \delta \ne \delta' $$ So, if $\Gamma$ is uncountable, $c_0(\Gamma)$ is not separable.

Answer 2

When $\Gamma$ is uncountable, $c_0(\Gamma)$ is not separable. It consists of all functions $g : \Gamma \to \mathbb R$ (with countable support) such that for every $\epsilon>0$, the set $\{\gamma \in \Gamma : |g(\gamma)|>\epsilon\}$ is finite.

But (you may ask) even if this particular pre-dual is nonseparable, might there not be some other pre-dual that is separable? The answer to that is "no". (But the proof I have in mind is a theorem of Stegall about "Asplund space" ... Israel J. Math. 29 (1978), 408–412. Maybe there is a more conventional proof?)

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Stegall:

Stegall, Charles, The duality between Asplund spaces and spaces with the Radon-Nikodym property, Isr. J. Math. 29, 408-412 (1978). ZBL0374.46015.

If $X$ is a Banach space, and $X^*$ has the Radon-Nikodym property, then every separable subspace of $X$ has separable dual.

Consequence: since $l^1(\Gamma)$ has the Radon-Nikodym property, if $X$ is any predual of $l^1(\Gamma)$, then every separable subspace of $X$ has separable dual. In particular, if $X$ itself were separable, then $X^* \cong l^1(\Gamma)$ would be separable, so $\Gamma$ would be countable.