In functional analysis, a branch of mathematics, the '''Goldstine theorem''', named after Herman Goldstine, is stated as follows:
:'''Goldstine theorem.''' Let <math>X</math> be a Banach space, then the image of the closed unit ball <math>B \subseteq X</math> under the canonical embedding into the closed unit ball <math>B^{\prime\prime}</math> of the bidual space <math>X^{\prime\prime}</math> 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 <math>c_0,</math> and its bi-dual space Lp space <math>\ell^{\infty}.</math>
== Proof ==
=== Lemma ===
For all <math>x^{\prime\prime} \in B^{\prime\prime},</math> <math>\varphi_1, \ldots, \varphi_n \in X^{\prime}</math> and <math>\delta > 0,</math> there exists an <math>x \in (1+\delta)B</math> such that <math>\varphi_i(x) = x^{\prime\prime}(\varphi_i)</math> for all <math>1 \leq i \leq n.</math>
==== Proof of lemma ====
By the surjectivity of <math display="block">\begin{cases} \Phi : X \to \Complex^{n}, \\ x \mapsto \left(\varphi_1(x), \cdots, \varphi_n(x) \right) \end{cases}</math> it is possible to find <math>x \in X</math> with <math>\varphi_i(x) = x^{\prime\prime}(\varphi_i)</math> for <math>1 \leq i \leq n.</math>
Now let <math display="block">Y := \bigcap_i \ker \varphi_i = \ker \Phi.</math>
Every element of <math>z \in (x + Y) \cap (1 + \delta)B</math> satisfies <math>z \in (1+\delta)B</math> and <math>\varphi_i(z) = \varphi_i(x)= x^{\prime\prime}(\varphi_i),</math> so it suffices to show that the intersection is nonempty.
Assume for contradiction that it is empty. Then <math>\operatorname{dist}(x, Y) \geq 1 + \delta</math> and by the Hahn–Banach theorem there exists a linear form <math>\varphi \in X^{\prime}</math> such that <math>\varphi\big\vert_Y = 0, \varphi(x) \geq 1 + \delta</math> and <math>\|\varphi\|_{X^{\prime}} = 1.</math> Then <math>\varphi \in \operatorname{span} \left\{ \varphi_1, \ldots, \varphi_n \right\}</math><ref>{{cite book|last1=Rudin|first1=Walter |title=Functional Analysis|location=Lemma 3.9|pages=63–64|edition=Second}}</ref> and therefore <math display="block">1+\delta \leq \varphi(x) = x^{\prime\prime}(\varphi) \leq \|\varphi\|_{X^{\prime}} \left\|x^{\prime\prime}\right\|_{X^{\prime\prime}} \leq 1,</math> which is a contradiction.
=== Proof of theorem ===
Fix <math>x^{\prime\prime} \in B^{\prime\prime},</math> <math>\varphi_1, \ldots, \varphi_n \in X^{\prime}</math> and <math>\epsilon > 0.</math> Examine the set <math display="block">U := \left\{ y^{\prime\prime} \in X^{\prime\prime} : |(x^{\prime\prime} - y^{\prime\prime})(\varphi_i)| < \epsilon, 1 \leq i \leq n \right\}.</math>
Let <math>J : X \rightarrow X^{\prime\prime}</math> be the embedding defined by <math>J(x) = \text{Ev}_x,</math> where <math>\text{Ev}_x(\varphi) = \varphi(x)</math> is the evaluation at <math>x</math> map. Sets of the form <math>U</math> form a base for the weak* topology,<ref>{{cite book|last1=Rudin|first1=Walter|title=Functional Analysis|location=Equation (3) and the remark after|page=69|edition=Second}}</ref> so density follows once it is shown <math>J(B) \cap U \neq \varnothing</math> for all such <math>U.</math> The lemma above says that for any <math>\delta > 0</math> there exists a <math>x \in (1+\delta)B</math> such that <math>x^{\prime\prime}(\varphi_i)=\varphi_i(x),</math> <math>1\leq i\leq n,</math> and in particular <math>\text{Ev}_x \in U.</math> Since <math>J(B) \subset B^{\prime\prime},</math> we have <math>\text{Ev}_x \in (1+\delta)J(B) \cap U.</math> We can scale to get <math>\frac{1}{1+\delta} \text{Ev}_x \in J(B).</math> The goal is to show that for a sufficiently small <math>\delta > 0,</math> we have <math>\frac{1}{1+\delta} \text{Ev}_x \in J(B) \cap U.</math>
Directly checking, one has <math display="block">\left|\left[x^{\prime\prime} - \frac{1}{1+\delta} \text{Ev}_x\right](\varphi_i)\right| = \left|\varphi_i(x) - \frac{1}{1+\delta}\varphi_i(x)\right| = \frac{\delta}{1+\delta} |\varphi_i(x)|.</math>
Note that one can choose <math>M</math> sufficiently large so that <math>\|\varphi_i\|_{X^{\prime}} \leq M</math> for <math>1 \leq i \leq n.</math><ref>{{cite book|last1=Folland|first1=Gerald|title=Real Analysis: Modern Techniques and Their Applications|location=Proposition 5.2|pages=153–154|edition=Second}}</ref> Note as well that <math>\|x\|_{X} \leq (1+\delta).</math> If one chooses <math>\delta</math> so that <math>\delta M < \epsilon,</math> then <math display="block">\frac{\delta}{1+\delta} \left|\varphi_i(x)\right| \leq \frac{\delta}{1+\delta} \|\varphi_i\|_{X^{\prime}} \|x\|_{X} \leq \delta \|\varphi_i\|_{X^{\prime}} \leq \delta M < \epsilon.</math>
Hence one gets <math>\frac{1}{1+\delta} \text{Ev}_x \in J(B) \cap U</math> as desired.
== See also ==
* {{annotated link|Banach–Alaoglu theorem}} * {{annotated link|Bishop–Phelps theorem}} * {{annotated link|Eberlein–Šmulian theorem}} * {{annotated link|James' theorem}} * {{annotated link|Mazur's lemma}}
== References ==
{{reflist}}
* {{Rudin Walter Functional Analysis|edition=2}} <!-- {{sfn | Rudin | 1991 | p=}} -->
{{Banach spaces}} {{Functional Analysis}}
Category:Banach spaces Category:Theorems in functional analysis