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Is there a way of endowing the unit ball $B_X$ of a Banach space $X$ (we may assume that $X$ is an AL space, if that helps) with a topology $\tau$, so that $\tau=\sigma(Y^*,Y)$ (the weak* topology) if $X=Y^*$, for some Banach space $Y$? In other words, is it possible to equip the unit ball of a Banach space $X$ with a topology that corresponds with the weak* topology if $X$ is a dual space, but is well-defined if $X$ is not a dual space?

As Nik Weaver observes in this post, "... on any dual Banach space there is no locally convex vector space topology strictly stronger than the weak* topology that makes the unit ball compact." So, given an arbitrary Banach space (or AL space) $X$, could one endow it with something like the "strongest locally convex vector space topology making the unit ball compact"?

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  • $\begingroup$ On Mathematics: An abstract characterisation of weak* topologies $\endgroup$
    – Martin Sleziak
    Commented Dec 14, 2022 at 6:46
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    $\begingroup$ Not meant seriously: Take $\tau_X=\sigma(X,Y)$ if $X$ is the dual of a Banach space $Y$ and whatever topology if $X$ does not have a predual. This nonsense indicates that should should specify further properties, e.g., functoriality of $X \mapsto (X,\tau_X)$. $\endgroup$
    – Jochen Wengenroth
    Commented Dec 14, 2022 at 8:20
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    $\begingroup$ If $X$ is a Banach space equipped with a locally convex topology $\mathcal{T}$ in which $B_X$ is compact, then $X$ is the dual space of the linear maps $X \rightarrow k$ that are continuous when restricted to $B_X$ (where $k$ is the base field $\mathbb{R}$ or \mathbb{C}$). The "locally convex" is necessary by a counterexample of J. W. Roberts. $\endgroup$
    – Robert Furber
    Commented Dec 14, 2022 at 20:13
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    $\begingroup$ Nik Weaver's answer has an extra assumption that is stated in the question and Jochen Wengenroth's comment (to Nik's answer) - that the topology on $X'$ is also coarser than $\sigma(X',X'')$ (i.e. the weak (not weak-) topology of $X'$). Without this assumption, there is the bounded weak- topology on $X'$ which is strictly finer than $\sigma(X',X)$ in the infinite-dimensional case. In fact it is the finest unique finest linear topology agreeing with $\sigma(X',X)$ on the unit ball. $\endgroup$
    – Robert Furber
    Commented Dec 14, 2022 at 20:45
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    $\begingroup$ The bounded weak-* topology is easier to characterize than the weak-* topology, therefore (if you have a locally convex topology on a normed space in which the unit ball is compact, and the topology is the finest linear topology agreeing with the topology on the ball, this space is a dual Banach space with the bounded weak-* topology). Of course, the topology on the unit ball is a piece of structure, not something intrinsic to the normed space structure, because of the existence of Banach spaces with non-isometric preduals. $\endgroup$
    – Robert Furber
    Commented Dec 14, 2022 at 20:48

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This is not possible in general. The obstruction does not come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction of the $\sigma(Y^*,Y)$-topology to the unit ball of $Y^*$ determines $Y$ uniquely : by the Krein-Smulian theorem, $Y$ coincides with the subspace of elements of $(Y^*)^*$ whose restriction to the unit ball is $\sigma(Y^*,Y)$-continuous. In particular, if $X$ admits several preduals that are not isometrically isomorphic, then there are several non-comparable maximal locally convex vector space topologies making the unit ball compact, and there is no strongest such topology.

Having a unique predual is a somewhat exceptional situation (this is the case for von Neumann algebras). The standard example of Banach space with many preduals is $\ell_1$. It has lots of very wild preduals, including the not-so-wild spaces $C(K)$ for $K$ countable and compact. Very concretely, two non-(isometrically isomorphic) preduals of $\ell_1$ are given by the space $c$ of converging sequences of complex numbers, and its subspace $c_0$ of sequences converging to $0$. See this question. See also the introduction to this article for more examples and references.

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    $\begingroup$ Actually, $c$ and $c_0$ are isomorphic, but not isometrically isomorphic. $\endgroup$
    – Dirk Werner
    Commented Dec 14, 2022 at 18:06
  • $\begingroup$ We can get a predual of $\ell^1$ not isomorphic to $c_0$ by trying $C(\omega^{\omega}+1)$, though I don't know a simple way of proving the non-isomorphism, only the Szlenk index. $\endgroup$
    – Robert Furber
    Commented Dec 14, 2022 at 20:04
  • $\begingroup$ Nonetheless, it is my favourite example where $\aleph_1$ occurs instead of $\mathfrak{c}$, there are $\aleph_1$ separable commutative C$^*$-algebras up to isomorphism (but $\mathfrak{c}$ up to isometry). $\endgroup$
    – Robert Furber
    Commented Dec 14, 2022 at 20:11
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    $\begingroup$ Thanks to both for the clarification. I know that this is clear to you, but in case somebody less familiar read these comments, let me just point out that the mere fact that $\ell_1$ has two non-isometric preduals (eg $c$ and $c_0$) is enough to say that it admits two distinct weak-* topologies.and answer the question. $\endgroup$
    – Mikael de la Salle
    Commented Dec 14, 2022 at 23:47
  • $\begingroup$ I changed the phrasing to make this clear. $\endgroup$
    – Mikael de la Salle
    Commented Dec 14, 2022 at 23:54

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