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Let $X$ be a Banach space over the complex field. Let $X^*$ denote its topological dual. If $S$ is a subspace of $X$, write $$\mathrm{ann}_L(S)= \{ \varphi \in X^* : \varphi S = 0\}$$ and note the result is a weak-star closed subspace of $X^*$. Similarly, if $S$ is a subspace of $X^*$, write $$\mathrm{ann}_R (S) = \{ x \in X : S x = 0 \}$$ and note the result is a weakly closed subspace of $X$. It is obvious that $\mathrm{ann}_L$ and $\mathrm{ann}_R$ are order-reversing. Also, if $S, T$ are subspaces of $X, X^*$ respectively, we have $$ S \subset \mathrm{ann}_R(T) \Leftrightarrow T \subset \mathrm{ann}_L(S) \Leftrightarrow TS = 0$$ so that $\mathrm{ann}_L$ and $\mathrm{ann}_R$ set up an (antitone) Galois connection between the posets of subspaces of $X$ and $X^*$. Various things then follow by abstract nonsense. For instance, $\mathrm{ann}_R \circ \mathrm{ann}_L$ and $\mathrm{ann}_L \circ \mathrm{ann}_R$ are abstract closure operators and the associated "closed subspaces" of $X$ and $X^*$ are put into order-reversing bijection by $\mathrm{ann}_L$ and $\mathrm{ann}_R$. In light of the fact that the range of $\mathrm{ann}_L$ consists of weak-star closed subspaces and the range of $\mathrm{ann}_R$ consists of weakly closed subspaces, it is natural to wonder whether these closure operators are, in fact, equal to the weak and weak-star closure operators. In essence, I am asking the following.

Question 1 (answered): Let $S$ be a subspace of $X$ and let $x \in X$. If $\varphi S = 0$ implies $\varphi(x) =0$ for all $\varphi \in X^*$, does it follow that $x$ is in the weak closure of $S$?

Question 2: Let $S$ be a subspace of $X^*$ and let $\varphi \in X^*$. If $S x= 0$ implies $\varphi(x) =0$ for all $x \in X$, does it follow that $\varphi$ is in the weak-star closure of $S$?

Thank you in advance for any answers or clarification on surrounding issues.

Added: I've managed to answer Question 1 affirmatively. First recall that the weak closure of $S$ is the same as the norm closure of $S$. More generally, the weak closure and norm closure coincide for convex subsets of $X$ (Conway, A Course in Functional Analysis, Theorem V.1.4). The Hahn-Banach Theorem implies the following statement: If $S \subset X$ is a subspace and $x \in X$ is a positive distance $d$ away from $S$, then there exists a functional $\varphi \in X^*$ with $\|\varphi\| = 1$ and $\varphi(x) = d$ and $\varphi S = 0$. It is easy to answer Question 1 using these two facts.

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So, in fact, if $S \subset X$ is a subspace then $\mathrm{ann}_R(\mathrm{ann}_L(S))$ equals the norm closure of $S$ equals the weak closure of $S$. –  Mike F Dec 13 '12 at 4:46
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1 Answer 1

up vote 3 down vote accepted

The key for question 2 is to observe that the weak-star continuous functionals on $X^\ast$ are precisely the evaluation functionals at points $x \in X$. From there the answer proceeds similarly to what you did in your answer for question 1:

If $\varphi$ is not in the weak-star closure $\overline{S}^{w\ast}$ of $S$ then Hahn-Banach applied to $X^\ast$ with the weak-star topology ensures that we can separate $\varphi$ from $\overline{S}^{w\ast}$, so there must be $x \in X$ such that $\varphi(x) \gt s(x)$ for all $s \in \overline{S}^{w\ast}$. Since $\overline{S}^{w\ast}$ is a linear subspace, we must have that $x \in \mathop{{\rm ann}_R} S$ and hence $\varphi \notin \mathop{{\rm ann}_L} \mathop{{\rm ann}_R} S$. This shows $\mathop{{\rm ann}_L} \mathop{{\rm ann}_R} S \subseteq \overline{S}^{w\ast}$ while $S \subseteq \mathop{{\rm ann}_L} \mathop{{\rm ann}_R}S$ and weak-star closedness of the latter shows the other inclusion.

A good reference for this and related results is Rudin's Functional Analysis, chapter IV on duality of Banach spaces.

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OK this sounds good. I'm aware that the weak-star continuous functionals on $X^*$ are precisely the point evaluations - in fact I checked this earlier today :). What's new to me is applying Hahn-Banach in anything more general than a Banach space (here, $X^*$ in the weak-star topology). Do you just use the formulation in terms of sublinear functionals or is there something more subtle at play? –  Mike F Dec 13 '12 at 5:03
    
Nothing much more subtle: In an arbitrary topological vector space you can separate an open convex set $U$ from an arbitrary convex set disjoint from it by using the Minkowski-functional $p_U$ (which is continuous by openness of $U$). This entails that you can separate a compact convex set from a disjoint closed convex set in a locally convex space. This is also explained in Rudin, chapter III if I remember correctly. Edit: The Wikipedia article contains the precise statements: en.wikipedia.org/wiki/Hahn-Banach_theorem –  Martin Dec 13 '12 at 5:09
    
Alright thanks, you've put your finger on it. The need for an application of Hahn-Banach certainly explains why I wasn't getting anywhere with this. I'll hold off on accepting for the moment to give your answer a chance to attract more votes - but yes I think this settles things for me. –  Mike F Dec 13 '12 at 5:21
    
Great, glad to hear that. By the way, your analogy with orthogonal complements is also emphasized by the common notation $S^\bot = \mathop{{\rm ann}_R} S \subseteq X^\ast$ for $S \subseteq X$ and ${}^\bot T = \mathop{{\rm ann}_L} T \subseteq X$ for $T \subseteq X^\ast$ as well as the "pairing notation" $\langle x, \varphi \rangle_{X,X^\ast} = \varphi(x)$ for the duality pairing. Your identities then become ${}^\bot(S^\bot) = \overline{S} = \overline{S}^w$ and $({}^\bot T)^\bot = \overline{T}^{w\ast}$ –  Martin Dec 13 '12 at 5:29
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