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I know that weak* topology is the weakest topology so that $Jx$ is continuous for $\forall x\in X$, where $J$ is the isometry from $X$ to $X''$. But what exactly is this topology? What is the open set in general look like?

And moreover, I want to prove another topology induced by metric is exactly the weak* topology. How can I prove this topology is weaker than weak* topology, so that it is exactly the weak* topology?

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Concerning your second question: can you make it a bit more precise, please? From context, I assume that you're working on the closed unit ball of the dual space $X'$ for $X$ separable, is that it? [Note also that the weak$^\ast$-topology on all $X'$ is never metrizable if $X$ is infinite-dimensional.] –  t.b. May 2 '12 at 2:46
    
@t.b. yes, you're right. But I'd really like to know more about weak* topology. And also, what is the relationship between this topology and weak* convergence? –  henryforever14 May 2 '12 at 2:49
    
What do you mean? A sequence (or net) is weak$^\ast$-convergent in the sense that $\varphi_i(x) \to \varphi(x)$ for all $x$ if and only if $\varphi_i \to \varphi$ in the weak$^\ast$-topology (it is simply the topology of pointwise convergence of linear functionals on $X$). For more detail you'll have to wait for a full-fledged answer because I don't have time right now. –  t.b. May 2 '12 at 2:56
    
@t.b. Thanks anyway. I am aware of the definition for weak* convergence. It's just that they share the a common part "weak*" in their names. Is it just a coincidence? –  henryforever14 May 2 '12 at 2:59
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@henryforever14 The weak* topology is defined by specifying neighborhood bases of the form $\{\Lambda \in X^* : |\Lambda(x_j)-\Lambda_0(x_j)|<\epsilon$ for some finite collection $j=1,...,n$ and $x_j \in X \}$. –  Chris Janjigian May 2 '12 at 3:05

1 Answer 1

Weak*-open (non-empty) sets contain infinite-dimensional subspaces, so in this sense they are huge.

Roughly, we have less open sets, hence more compact setes. This is reflected in the Banach-Alaoglu theorem which says that a polar set of any nbhd of the origin in $X$ is weak* compact in $X^*$.

You work with Banach spaces with separable duals. In this case, the dual ball (which is weak*-compact) is metrisable. Thus, you cannot find any other compact Hausdorff topology (in particular, any other metric giving topology which is compact) which is weaker that the (relative) weak* topology since compact Hausdorff topologies are minimal.

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You don't need the dual to be separable, only the space itself. This guarantees the metrizability of the unit ball of the dual space. –  Vobo May 5 '12 at 7:00

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