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I had just recently picked up Functional Analysis so my problem may sound trivial. But I appreciate any help.

I am having trouble to reconcile the two statements (said to be true in my notes):
let $B \subset$ X*, dual space of X and define $B^o$ and $B^z$ to be the set of its annihilators and pre-annihilators respectively.

1) $(B^z)^o$ is the weak* closure of $R:=$ convex hull of $B$.
2) The norm closure of spanB is a strict subset of $(B^z)^o$.

I feel one of the two is wrong because 1) seems to contradict 2). Here's my reasoning.
$R$ is a subset of spanB, hence norm closure of $R$ is contained in the norm closure of spanB. But the weak* topology is contained in the weak topology of X* i.e. the smallest topology to have X** to be continuous. So weak* closure of R is weak closed implying it's also normed closed, as R is convex. So the norm closure of $R$ is in norm closure of spanB, which conflicts with 2) if 1) is true.

Thank you for any clarifications.

user2319
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1 Answers1

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The statements are certainly contradictory. Note that nothing precludes you from starting with $B$ a weak* closed subspace of $X^*$. In that case, 1) implies that $(B^z)^o=R=B$, making 2) false.

It is not hard to see that 2) is false in general. Let $X=C[0,1]$, $B=\{\text{Lebesgue measure}\}$. Then $$ B^z=\{f\in C[0,1]:\ \int_0^1f=0\}. $$ It is not hard to see that $(B^z)^o=\mathbb C\,B=\text{span}\,B$; indeed, if $\mu f=0$ for all $f$ in $B^z$ it is not hard to see that $\mu$ is necessarily a scalar multiple of Lebesgue measure. So we see that for this $B$ 1) holds while 2) is false.

A maybe simpler example can be obtained by making $X=H$, a Hilbert space. We can take $B=\{v\}$ for a certain vector $v$. Then $$ B^z=\{y\in H:\ \langle y,v\rangle=0\}=\{v\}^\perp, $$ $$ (B^z)^o=\{z\in H:\ \langle z,y\rangle=0\ \text{for all }y\in \{v\}^\perp\}=\{v\}^{\perp\perp}=\mathbb Cv. $$

Martin Argerami
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