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Preamble: let $(X, \tau)$ be a topological vector space over $\mathbb K$, and $X'$ its (topological) dual space. Then, as I understand, the family of sets given by $$ \mathcal B = \{ f^{-1} (V) \mid f \in X' , \, V \text{ open in } \mathbb K \} $$ is a subbasis for the weak topology induced by $X'$ on $X$. This topology, of course, has the property of being the coarsest topology on $X$ for which every $f \in X'$ is $\tau$-continuous.

Question: Is it possible to make an analogous characterization of the weak-* topology induced by $X$ on $X'$? In particular, is it true that the family of sets given by $$ \mathcal E = \{ \hat x ^{-1} (V) \mid \hat x \in X'' , \, V \text{ open in } \mathbb K \} $$ is a subbasis for the weak-* topology?

Any input will be much appreciated, I am quite lost here...

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up vote 1 down vote accepted

The weak-* topology is the coarsest topology such that the maps $f_x:X'\to \mathbb{K}$ with $f_x(x') = x'(x)$ are continuous (see, e.g. Wikipedia). Hence, your familiy $\mathcal{E}$ is too large (roughly speaking: you take all $\hat x\in X''$ but the $x\in X$ are enough).

What you have written is a subbasis of the weak topology on $X'$ which may be finer than the weak-* topology on $X'$ (see Vobo's comment below).

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The weak topology $\sigma(X', X'')$ is not always finer than the weak*-topology $\sigma(X',X)$. Obviously they coincide for reflexive spaces, but there are even non-reflexive Banach spaces like somes James spaces where these topologies are identical. An easy example where they are different is $X=c_0, X'=l^1$, and $X''=l^\infty$ (consider the sequence of unit vectors in $l^1$). – Vobo Apr 6 '11 at 12:04
You may remove the vague "according to", "I think" and "seems". All this is correct, except the "finer" mentioned in my previous comment. – Vobo Apr 6 '11 at 12:30
Thanks for the comment Vobo. I edited accordingly. The example with $X' = \ell^1$ is very good. Weak convergence is equivalent to norm convergence there but weak-* convergence is only pointwise convergence, right? – Dirk Apr 6 '11 at 12:51
Let $(e_n)_n$ be the sequence of unit vectors in $l^1$. If $x=(x_i) \in c_0$, then $|<e_n,x>| = |x_n| \to 0$ as $n\to\infty$, so $(e_n)_n$ converges to $0$ in $\sigma(l^1,c_0)$. But for $y=(1,1,1,...) \in l^\infty$ you have $|<e_n,y>| = 1$, so $(e_n)_n$ does not converge to $0$ in $\sigma(l^1,l^\infty)$ – Vobo Apr 6 '11 at 12:58
I think I get it now, thank you both! – Jose L. Lykón Apr 7 '11 at 15:06

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