In functional analysis, a branch of mathematics, the Goldstine theorem, named after Herman Goldstine, is stated as follows:
Goldstine theorem. Let X {\displaystyle X} be a Banach space, then the image of the closed unit ball B ⊆ X {\displaystyle B\subseteq X} under the canonical embedding into the closed unit ball B ′ ′ {\displaystyle B^{\prime \prime }} of the bidual space X ′ ′ {\displaystyle X^{\prime \prime }} is a weak*-dense subset. The conclusion of the theorem is never true for the norm topology if X {\displaystyle X} is not reflexive; indeed, the image of X {\displaystyle X} in its bidual X ′ ′ {\displaystyle X^{\prime \prime }} is always norm closed, being the continuous image of a complete metric space under an isometry.
Proof
Lemma For all x ′ ′ ∈ B ′ ′ , {\displaystyle x^{\prime \prime }\in B^{\prime \prime },} φ 1 , … , φ n ∈ X ′ {\displaystyle \varphi _{1},\ldots ,\varphi _{n}\in X^{\prime }} and δ > 0 , {\displaystyle \delta >0,} there exists an x ∈ ( 1 + δ ) B {\displaystyle x\in (1+\delta )B} such that φ i ( x ) = x ′ ′ ( φ i ) {\displaystyle \varphi _{i}(x)=x^{\prime \prime }(\varphi _{i})} for all 1 ≤ i ≤ n . {\displaystyle 1\leq i\leq n.}
Proof of lemma By the surjectivity of
{ Φ : X → C n , x ↦ ( φ 1 ( x ) , ⋯ , φ n ( x ) ) {\displaystyle {\begin{cases}\Phi :X\to \mathbb {C} ^{n},\\x\mapsto \left(\varphi _{1}(x),\cdots ,\varphi _{n}(x)\right)\end{cases}}}
it is possible to find x ∈ X {\displaystyle x\in X} with φ i ( x ) = x ′ ′ ( φ i ) {\displaystyle \varphi _{i}(x)=x^{\prime \prime }(\varphi _{i})} for 1 ≤ i ≤ n . {\displaystyle 1\leq i\leq n.}
Now let
Y := ⋂ i ker φ i = ker Φ . {\displaystyle Y:=\bigcap _{i}\ker \varphi _{i}=\ker \Phi .}
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