Goldstine theorem

In functional analysis, a branch of mathematics, the Goldstine theorem, named after Herman Goldstine, is stated as follows:

:Goldstine theorem. Let $X$ be a Banach space, then the image of the closed unit ball $B \subseteq X$ under the canonical embedding into the closed unit ball $B^$ of the bidual space $X^$ is a weak*-dense subset.

The conclusion of the theorem is not true for the norm topology, which can be seen by considering the Banach space of real sequences that converge to zero, c0 space $c_0,$ and its bi-dual space Lp space $\ell^.$

Proof

Lemma

For all $x^ \in B^,$ $\varphi_1, \ldots, \varphi_n \in X^$ and $\delta > 0,$ there exists an $x \in (1+\delta)B$ such that $\varphi_i(x) = x^(\varphi_i)$ for all $1 \leq i \leq n.$

Proof of lemma

By the surjectivity of
$\begin \Phi : X \to \Complex^, \\ x \mapsto \left(\varphi_1(x), \cdots, \varphi_n(x) \right) \end$
it is possible to find $x \in X$ with $\varphi_i(x) = x^(\varphi_i)$ for $1 \leq i \leq n.$

Now let
$Y := \bigcap_i \ker \varphi_i = \ker \Phi.$

Every element of $z \in (x + Y) \cap (1 + \delta)B$ satisfies $z \in (1+\delta)B$ and $\varphi_i(z) = \varphi_i(x)= x^(\varphi_i),$ so it suffices to show that the intersection is nonempty.

Assume for contradiction that it is empty. Then $\operatorname(x, Y) \geq 1 + \delta$ and by the Hahn–Banach theorem there exists a linear form $\varphi \in X^$ such that $\varphi\big\vert_Y = 0, \varphi(x) \geq 1 + \delta$ and $\, \varphi\, _ = 1.$ Then $\varphi \in \operatorname \left\$ and therefore
$1+\delta \leq \varphi(x) = x^(\varphi) \leq \, \varphi\, _ \left\, x^\right\, _ \leq 1,$
which is a contradiction.

Proof of theorem

Fix $x^ \in B^,$ $\varphi_1, \ldots, \varphi_n \in X^$ and $\epsilon > 0.$ Examine the set
$U := \left\.$

Let $J : X \rightarrow X^$ be the embedding defined by $J(x) = \text_x,$ where $\text_x(\varphi) = \varphi(x)$ is the evaluation at $x$ map. Sets of the form $U$ form a base for the weak* topology, so density follows once it is shown $J(B) \cap U \neq \varnothing$ for all such $U.$ The lemma above says that for any $\delta > 0$ there exists a $x \in (1+\delta)B$ such that $x^(\varphi_i)=\varphi_i(x),$ $1\leq i\leq n,$ and in particular $\text_x \in U.$ Since $J(B) \subset B^,$ we have $\text_x \in (1+\delta)J(B) \cap U.$ We can scale to get $\frac \text_x \in J(B).$ The goal is to show that for a sufficiently small $\delta > 0,$ we have $\frac \text_x \in J(B) \cap U.$

Directly checking, one has
\left, \left ^ - \frac \text_x\right\varphi_i)\ = \left, \varphi_i(x) - \frac\varphi_i(x)\ = \frac , \varphi_i(x), .

Note that one can choose $M$ sufficiently large so that $\, \varphi_i\, _ \leq M$ for $1 \leq i \leq n.$ Note as well that $\, x\, _ \leq (1+\delta).$ If one chooses $\delta$ so that $\delta M < \epsilon,$ then
$\frac \left, \varphi_i(x)\ \leq \frac \, \varphi_i\, _ \, x\, _ \leq \delta \, \varphi_i\, _ \leq \delta M < \epsilon.$

Hence one gets $\frac \text_x \in J(B) \cap U$ as desired.

See also

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References

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{{Functional Analysis

Banach spaces
Theorems in functional analysis

de:Schwach-*-Topologie#Eigenschaften