$G$ is dense in $X^$ in weak sense if and only if $G$ is total set

Question

I have some question on functional analysis. Recently, I'm reading an article of Coifman and Weiss, "Extensions of hardy spaces and their use in analysis".

They proved some important theorem to me by using the following functional analysis fact without proof.

Exercise (Dunford&Schwartz, p.439, #41). Let $X$ be a locally convex linear topological space and let $G$ be a linear subspace of $X^*$. Then $G$ is $X$-dense in $X^*$ if and only if $G$ is a total set of functionals on $X$.

In the case of Banach space, I can prove the one direction ($G$ is a total implies $G$ is dense in $X^*$ in the sense of weak* topology).

First, I proved that if $N$ is a subspace of $X^*$, then $(^\bot N)^\bot = \overline{N}^*$, where $-^*$ denotes the weak* closure. Here I follow Rudin's notation. So the result of one direction is proved one takes $^\bot N=\{0\}$.

But I fail to prove its reverse direction and I cannot extend to the locally convex linear topological space case.

Thank you in advance.

The dual of $(X^,weak^)$ is $X$. So the result is immediate from Hahn-Banach. — David C. Ullrich, Jul 14 '16 at 15:16
Share your proof of the direction you can show for Banach spaces, so we can see how it can (or cannot) be extended to locally convex spaces. — Aweygan, Jul 14 '16 at 15:53
@Aweygan I add proof of one direction. It seems like it does not depend on some structure of Banach space. — Will Kwon, Jul 14 '16 at 22:12
@WillKwon I fortunately have a copy of Rudin's book handy :) After reading that section, it looks like your proof still works for general locally convex spaces. — Aweygan, Jul 14 '16 at 22:20
@WillKwon No problem! always glad to help a fellow student of functional analysis — Aweygan, Jul 15 '16 at 04:46

score 3 · Accepted Answer · answered Jul 14 '16 at 15:43

3

Assume $G$ is not total. Then there is some $x\in X$ ($x\neq 0$) such that $x^*(x)=0$ for all $x^*\in G$. Thus the set $\{x^*\in X^*:x^*(x)\neq 0 \}$ is a weak$^*$ open set which is disjoint from $G$, so $G$ is not weak$^*$ dense.

answered Jul 14 '16 at 15:43

Aweygan

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$G$ is dense in $X^$ in weak sense if and only if $G$ is total set

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$G$ is dense in $X^*$ in weak* sense if and only if $G$ is total set

1 Answers1

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$G$ is dense in $X^$ in weak sense if and only if $G$ is total set