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I am reading Bollobás' Linear Analysis. Chapter 8. Theorem 6., as the title says:

$B(X^{**})$ is the $w^*$-closure of $B(X)$ in $X^{**}$

The proof starts by saying that i) $B(X^{**}$) is $w^*$-closed, and ii) $B \subset B(X^{**})$ where $B$ is the $w^*$-closure of $B(X)$ in $X^{**}$. Then, suppose that there is $x_0^{**} \in B(X^{**}) \setminus B$. By the separation theorem, there is a bounded linear functional $x_0^{***}$ on $X^{**}$ such that $$\langle x_0^{***}, b \rangle \leq 1 < \langle x_0^{***}, x_0^{**} \rangle$$ for every $b \in B$. By considering $x_0^{*}$, the restriction of $x_0^{***}$ onto the subspace $X$ of $X^{**}$, we see that $\langle x_0^{*}, x \rangle \leq 1$ for every $x \in B(X)$, thus $\| x_0^{*} \| \leq 1$. This far everything is clear. To finish up the proof, the book argues as follows: $$1 \geq \langle x_0^{*}, x_0^{**} \rangle = \langle x_0^{***}, x_0^{**} \rangle > 1$$ claiming contradiction.

I don't really understand why $\langle x_0^{*}, x_0^{**} \rangle = \langle x_0^{***}, x_0^{**} \rangle$. I see that we can view $x_0^*$ as an element of $X^{***}$, but what guarantees, that this is the same element as $x_0^{***}$?

stomfaig
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  • Is it possible that the book argued that (i) $B(X^{})$ is $w^{*}$-closed in $X^{}$ and (ii) that the $w^{}$-closure of the embedded image of $B(X)$ into $X^{}$ is contained in $B(X^{})$? If so, the result follows from a separation argument on $(X^{}, w^{})$. If not, then it is not obvious as to how the result follows. – Dean Miller Mar 17 '24 at 03:25
  • You’re right about i) and ii), but I still don’t see why that helps – stomfaig Mar 17 '24 at 09:48

1 Answers1

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Note that as $B$ is a closed convex subset of $(X^{**}, w^{*})$ and as $x_{0}^{**}\not\in B$, it follows from a separation argument that there exists some $x_{0}^{***}$ in the dual of $(X^{**}, w^{*})$ which strictly separates $B$ and $x_{0}^{**}$. But as continuous linear functionals on $(X^{**}, w^{*})$ are elements of the form $x^{***}\in X^{***}$ where there is some $x^{*}\in X^{*}$ such that $\langle x^{***}, x^{**} \rangle = \langle x^{**}, x^{*} \rangle$ for all $x^{**} \in X^{**}$, it follows that there is some $x_{0}^{*}\in X^{*}$ such that $\langle x_{0}^{***}, x^{**} \rangle = \langle x^{**}, x_{0}^{*} \rangle$ for all $x^{**} \in X^{**}$. Consequently, $x_{0}^{*}$ is the desired element.

It is also worth mentioning that this theorem is known as Goldstine's theorem.

Dean Miller
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    Ah okay. So the part I was missing is that $x_0^{*}$ being a linear functional on $(X^{}, w^)$, it is the canonical embedding of an element $x^ \in X^{}$ into $X^{**}$? – stomfaig Mar 17 '24 at 10:39
  • Yes, that's correct if you have that $x_{0}^{***}$ is continuous as in the question. – Dean Miller Mar 17 '24 at 10:40
  • And I guess this statement follows from the fact that $w^$ is defined to be the weakest topology making the induced linear functionals continuous, so if $x_0^{**}$ wasn't like that, it couldn't be continuous. Thanks for your answer. – stomfaig Mar 17 '24 at 10:50
  • No problem. For the characterisation of the dual, see this post and this post. – Dean Miller Mar 17 '24 at 10:56