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In Ledoux and Talagrand's "Probability in Banach Spaces", for technical reasons they frequently assume that a Banach space $B$ has the property that the unit ball of $B^*$ contains a countable subset $D$ such that $$ \Vert x \Vert = \sup_{f\in D} \vert f(x) \vert$$ for every $x\in B$. Examples of such spaces $B$ include both all separable Banach spaces, and all duals of separable Banach spaces (e.g. $\ell_\infty$).

My question is, is there a standard name or alternative characterization of spaces with this property? The authors decline to discuss this at all, except to point out that separable spaces and $\ell_\infty$ have this property; they don't even point out that it extends to duals of separable spaces.

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A subset D of $B^{*}$ with the property you describe is called "norming" (wether or not it is countable).

Recall that a subset D is called "total" if if $d(x) = 0$ for all $d \in D$ implies $x = 0$. Every norming set is total. Maybe a partial answer to your question is given by proposition 5.17 in the "Handbook of the Geometry of Banach Spaces" volume 1, chapter 15 "Infinite dimensional convexity", it says when the converse is true:

Every total subset of $B^{*}$

is norming iff B is quasireflexive, that is $\frac{dim(B^{**})}{dim B} \lt \infty$.

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$l^1$ is not quasireflexive, so I am not sure that the result you quote from HGBS is particularly relevant to the question at hand... –  Yemon Choi May 14 '10 at 23:21
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Norming means that the sup defines an EQUIVALENT norm, Tim. Mark's condition is called norm determining or 1-norming. An obvious equivalent to the unit ball of $B^*$ containing a countable norming set is that B embeds isomorphically into $\ell_\infty$, while the unit ball of $B^*$ contains a countable 1-norming set iff $B$ embeds isometrically into $\ell_\infty$. The first condition is weaker than B being weak`$^*$ separable, IIRC. –  Bill Johnson May 14 '10 at 23:50
    
@Yemon: Ok, but the question "is there a standard name or alternative characterization of spaces with this property?" led me to believe that it is not about $l^1$ but more generally about which spaces have a dual with a countable 1-norming subset. @Bill: Ok, let's settle for 1-norming, the authors of the chapter in HGBS I cited defined "norming" to be r-norming for some $r \gt 0$, so we are talking about the same concept. –  Tim van Beek May 15 '10 at 5:45
    
Tim: my point is that the notion you quote holds if and only if the space if quasireflexive. $l^1$ has the property Harald is asking about, but is not quasireflexive. So the result you quote doesn't cover one of the most basic cases which Harald was asking about. –  Yemon Choi May 15 '10 at 7:16
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@Bill: Thanks, that's exactly the kind of answer I was looking for, and certainly obvious in hindsight. –  Mark Meckes May 15 '10 at 12:51
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