Let $V$ be an infinite-dimensional vector space and $V^*$ its dual.
For a linear subspace $W\subset V$ define $W^ \circ\subset V^*$ as the subspace of linear forms on $V$ vanishing on $W$.
Dually, for $\Gamma\subset V^*$ define $\Gamma^\diamond \subset V$ as the set of vectors $v\in V$ such that $\gamma(v)=0$ for all linear forms $\gamma\in \Gamma$.
It is slightly surprising but not too difficult to show that that we have for all subspaces $W\subset V$ the equality $(W^\circ) ^\diamond=W$.
But is it true that for all $\Gamma\subset V^*$ we have $(\Gamma^\diamond)^\circ=\Gamma$ ?
And is there a reference (article, book, lecture notes,...) where this problem is mentioned?
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$\begingroup$ A maybe-more accessible question that keeps it all on the dual side: since $((\Gamma^\diamond)^\circ)^\diamond$ equals $\Gamma^\diamond$, we can ask: if $\Gamma$ is contained in $\Lambda$ and $\Gamma^\diamond$ equals $\Lambda^\diamond$, then under what circumstances does $\Gamma$ equal $\Lambda$? $\endgroup$– LSpiceCommented Aug 16, 2020 at 14:00
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2$\begingroup$ In the infinite dimensional case, you can’t get by without topology, in this case the weak topologies $\sigma(V,V^\ast)$ and $\sigma(V^\ast,V)$. It is well known that the bipolars of subspaces of either of your spaces are precisely their closures for the corresponding topologies. So if they are not closed, then your claims will fail. $\endgroup$– user131781Commented Aug 16, 2020 at 14:12
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2$\begingroup$ I think @user131781's point is less that the problem can't be stated, or even answered, without topology, and more that it is probably most useful to think of the problem in topological terms even if one is not required to do so (for example, this reveals 'heuristically' that the answer must be no, even if one is left to, as you and I did, construct a specific non-closed subspace). $\endgroup$– LSpiceCommented Aug 16, 2020 at 14:54
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1$\begingroup$ For example, your surprising true fact is the statement that all subspace are closed in the weak topology (with respect to the full algebraic dual). $\endgroup$– LSpiceCommented Aug 16, 2020 at 15:00
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1$\begingroup$ Dear LSpice, thank you for your explanations. I know nothing about topological vector spaces but I appreciate your point that if I did I would have immediately realized that the question is quite easy, whereas in reality I spent much time coming up with a solution. $\endgroup$– lefunesteCommented Aug 16, 2020 at 15:02
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2 Answers
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No, $(\Gamma^\diamond)^\circ$ need not always equal $\Gamma$. Let $\mathcal B$ be a basis for $V$, and let $\Gamma$ be the span of the 'dual' set $\{e_b \mathrel: b \in \mathcal B\}$, so $e_b(c)$ is the Iverson bracket $[b = c]$ for all $b, c \in \mathcal B$. Then $\Gamma^\diamond$ is $0$, so $(\Gamma^\diamond)^\circ$ is all of $V^*$; but $\Gamma$ itself does not contain, for example, the element $\sum_{b \in \mathcal B} e_b$ of $V^*$.
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$\begingroup$ Your symbol $\sum_{b \in \mathcal B} e_b$ does not make sense because you cannot sum infinitely many non-zero vectors in a vector space. $\endgroup$ Commented Aug 16, 2020 at 14:36
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2$\begingroup$ Sure it does: $\bigl(\sum_{b \in \mathcal B} e_b\bigr)(v) = \sum_{b \in \mathcal B} e_b(v)$ is a finitely supported sum for each $v \in V$. I think that, as @user131781 points out, this is a convergent sum in a weak topology; but anyway it can be defined without direct reference to topology. $\endgroup$– LSpiceCommented Aug 16, 2020 at 14:47
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1$\begingroup$ OK, I agree, what you write makes sense. However I don't know about weak topology and I thank you for your definition which doesn't make reference to topology. $\endgroup$ Commented Aug 16, 2020 at 14:54
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1$\begingroup$ The topology @user131781 mentions is the weak* topology, weakest making all eval. fncls. continuous. Since $\lim_{F \subseteq \mathcal B} \bigl(\sum_{b \in F} e_b\bigr)(v)$ equals $\bigl(\sum_{b \in \mathcal B} e_b\bigr)(v)$ for all $v \in V$, this topology must declare that $\lim_{F \subseteq \mathcal B} \sum_{b \in F} e_b$ equals $\sum_{b \in \mathcal B} e_b$ (the limit taken over the finite subsets $F$ of $\mathcal B$—the so called "unordered sum"). $\endgroup$– LSpiceCommented Aug 16, 2020 at 15:01
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1$\begingroup$ Thanks again for your clear explanations, LSpice. I have upvoted your answer and your comments. $\endgroup$ Commented Aug 16, 2020 at 15:09
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The equality is false in general.
Here is a counterexample: fix a basis $v_i, i\in I$ of $V$ and consider the set of coordinate linear forms $v^*_i, i\in I$.
These forms are linearly independant but never form a basis since $V$ is infinite-dimensional.
So complete these forms to a basis $(v^*_j), j\in J$ with $J\setminus I\neq\emptyset$.
Choose $l\in J\setminus I$ and put $J'=J\setminus \{l\}$
If you define $\Gamma \subset V^*$ as the vector space generated by the $v_j^*, j\in J'$, then $\Gamma^\diamond =0$ (since already the subspace of $V^*$ generated by the $v_i^*, i\in I$ kill all vectors in $V$) so that $\Gamma\subsetneq (\Gamma^\diamond)^\circ=\{0\}^\circ=V^*$ yielding the required counterexample.
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1$\begingroup$ Yes, it is the same idea, but I posted my answer without seeing yours. The technical details are different and I find your formulation indeed simpler. Also I wrote a comment to your post which you should take into account. $\endgroup$ Commented Aug 16, 2020 at 14:50