Goldstine theorem explained

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 ^{\prime\prime} of the bidual space X ^{\prime\prime} is a weak*-dense subset.c _{0,} and its bi-dual space Lp space \ell ^{infty} .

Proof

Lemma

For all

x ^{\prime\prime} \in B ^{\prime\prime} , \varphi _{1,} \ldots, \varphi _{n} \in X ^{\prime} and \delta > 0 , there exists an x \in ( 1+ \delta)B such that \varphi _{i(x)} = x ^{\prime\prime} (\varphi _{i)} for all 1 \leq i \leq n.

Proof of lemma

By the surjectivity of$$\backslash begin\backslash Phi\; :\; X\; \backslash to\; \backslash Complex^,\; \backslash \backslash \; x\; \backslash mapsto\; \backslash left(\backslash varphi\_1(x),\; \backslash cdots,\; \backslash varphi\_n(x)\; \backslash right)\backslash end$$it is possible to find

x \in X with \varphi _{i(x)} = x ^{\prime\prime} (\varphi _{i)} for 1 \leq i \leq n.

Now let$$Y\; :=\; \backslash bigcap\_i\; \backslash ker\; \backslash varphi\_i\; =\; \backslash ker\; \backslash Phi.$$

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 ^{\prime\prime} (\varphi _{i),} so it suffices to show that the intersection is nonempty.

Assume for contradiction that it is empty. Then

\operatorname{dist}(x, Y) \geq 1 + \delta and by the Hahn–Banach theorem there exists a linear form \varphi \in X ^{\prime} such that \varphi \ vert _{Y} = 0 , \varphi(x) \geq 1 + \delta and \|\varphi\| _{X ^{\prime}} = 1 . Then \varphi \in \operatorname{span} \left\{ \varphi _{1,} \ldots, \varphi _{n} \right\} ^[1] and therefore$$1+\backslash delta\; \backslash leq\; \backslash varphi(x)\; =\; x^(\backslash varphi)\; \backslash leq\; \backslash |\backslash varphi\backslash |\_\; \backslash left\backslash |x^\backslash right\backslash |\_\; \backslash leq\; 1,$$which is a contradiction.

Proof of theorem

Fix

x ^{\prime\prime} \in B ^{\prime\prime} , \varphi _{1,} \ldots, \varphi _{n} \in X ^{\prime} and \epsilon > 0 . Examine the set$$U\; :=\; \backslash left\backslash .$$

Let

J : X → X ^{\prime\prime} be the embedding defined by J(x) = Ev _{x,} where Ev _{x(\varphi)} = \varphi(x) is the evaluation at x map. Sets of the form U form a base for the weak* topology,^[2] so density follows once it is shown J(B) \cap U ≠ \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 ^{\prime\prime} (\varphi _{i)=\varphi} _{i(x),} 1 \leq i\leq n, and in particular Ev _{x} \in U. Since J(B) \subset B ^{\prime\prime} , we have Ev _{x} \in ( 1+ \delta)J(B) \cap U. We can scale to get (1)/(1+ \delta) Ev _{x} \in J(B). The goal is to show that for a sufficiently small \delta > 0 , we have (1)/(1+ \delta) Ev _{x} \in J(B) \cap U.

Directly checking, one has$$\backslash left|\backslash left[x^\{\backslash prime\backslash prime\}\; -\; \backslash frac\{1\}\{1+\backslash delta\}\; \backslash text\{Ev\}\_x\backslash right](\backslash varphi\_i)\backslash right|\; =\; \backslash left|\backslash varphi\_i(x)\; -\; \backslash frac\backslash varphi\_i(x)\backslash right|\; =\; \backslash frac\; |\backslash varphi\_i(x)|.$$

Note that one can choose

M sufficiently large so that \|\varphi _{i\|} ^{X ^{\prime}} \leq M for 1 \leq i \leq n. ^[3] Note as well that \|x\| _{X} \leq ( 1+ \delta). If one chooses \delta so that \delta M < \epsilon, then

Hence one gets

(1)/(1+ \delta) Ev _{x} \in J(B) \cap U as desired.

Notes and References

1. Book: Rudin. Walter . Functional Analysis. Lemma 3.9. 63–64. Second.
2. Book: Rudin. Walter. Functional Analysis. Equation (3) and the remark after. 69. Second.
3. Book: Folland. Gerald. Real Analysis: Modern Techniques and Their Applications. Proposition 5.2. 153–154. Second.