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 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 , {\displaystyle c_{0},} and its bi-dual space Lp space ℓ ∞ . {\displaystyle \ell ^{\infty }.}
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.}
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 .}
Every element of z ∈ ( x + Y ) ∩ ( 1 + δ ) B {\displaystyle z\in (x+Y)\cap (1+\delta )B} satisfies z ∈ ( 1 + δ ) B {\displaystyle z\in (1+\delta )B} and φ i ( z ) = φ i ( x ) = x ′ ′ ( φ i ) , {\displaystyle \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 dist ( x , Y ) ≥ 1 + δ {\displaystyle \operatorname {dist} (x,Y)\geq 1+\delta } and by the Hahn–Banach theorem there exists a linear form φ ∈ X ′ {\displaystyle \varphi \in X^{\prime }} such that φ | Y = 0 , φ ( x ) ≥ 1 + δ {\displaystyle \varphi {\big \vert }_{Y}=0,\varphi (x)\geq 1+\delta } and ‖ φ ‖ X ′ = 1. {\displaystyle \|\varphi \|_{X^{\prime }}=1.} Then φ ∈ span { φ 1 , … , φ n } {\displaystyle \varphi \in \operatorname {span} \left\{\varphi _{1},\ldots ,\varphi _{n}\right\}} [ 1] and therefore 1 + δ ≤ φ ( x ) = x ′ ′ ( φ ) ≤ ‖ φ ‖ X ′ ‖ x ′ ′ ‖ X ′ ′ ≤ 1 , {\displaystyle 1+\delta \leq \varphi (x)=x^{\prime \prime }(\varphi )\leq \|\varphi \|_{X^{\prime }}\left\|x^{\prime \prime }\right\|_{X^{\prime \prime }}\leq 1,} which is a contradiction.
Fix 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 ϵ > 0. {\displaystyle \epsilon >0.} Examine the set U := { y ′ ′ ∈ X ′ ′ : | ( x ′ ′ − y ′ ′ ) ( φ i ) | < ϵ , 1 ≤ i ≤ n } . {\displaystyle U:=\left\{y^{\prime \prime }\in X^{\prime \prime }:|(x^{\prime \prime }-y^{\prime \prime })(\varphi _{i})|<\epsilon ,1\leq i\leq n\right\}.}
Let J : X → X ′ ′ {\displaystyle J:X\rightarrow X^{\prime \prime }} be the embedding defined by J ( x ) = Ev x , {\displaystyle J(x)={\text{Ev}}_{x},} where Ev x ( φ ) = φ ( x ) {\displaystyle {\text{Ev}}_{x}(\varphi )=\varphi (x)} is the evaluation at x {\displaystyle x} map. Sets of the form U {\displaystyle U} form a base for the weak* topology,[ 2] so density follows once it is shown J ( B ) ∩ U ≠ ∅ {\displaystyle J(B)\cap U\neq \varnothing } for all such U . {\displaystyle U.} The lemma above says that for any δ > 0 {\displaystyle \delta >0} there exists a x ∈ ( 1 + δ ) B {\displaystyle x\in (1+\delta )B} such that x ′ ′ ( φ i ) = φ i ( x ) , {\displaystyle x^{\prime \prime }(\varphi _{i})=\varphi _{i}(x),} 1 ≤ i ≤ n , {\displaystyle 1\leq i\leq n,} and in particular Ev x ∈ U . {\displaystyle {\text{Ev}}_{x}\in U.} Since J ( B ) ⊂ B ′ ′ , {\displaystyle J(B)\subset B^{\prime \prime },} we have Ev x ∈ ( 1 + δ ) J ( B ) ∩ U . {\displaystyle {\text{Ev}}_{x}\in (1+\delta )J(B)\cap U.} We can scale to get 1 1 + δ Ev x ∈ J ( B ) . {\displaystyle {\frac {1}{1+\delta }}{\text{Ev}}_{x}\in J(B).} The goal is to show that for a sufficiently small δ > 0 , {\displaystyle \delta >0,} we have 1 1 + δ Ev x ∈ J ( B ) ∩ U . {\displaystyle {\frac {1}{1+\delta }}{\text{Ev}}_{x}\in J(B)\cap U.}
Directly checking, one has | [ x ′ ′ − 1 1 + δ Ev x ] ( φ i ) | = | φ i ( x ) − 1 1 + δ φ i ( x ) | = δ 1 + δ | φ i ( x ) | . {\displaystyle \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)|.}
Note that one can choose M {\displaystyle M} sufficiently large so that ‖ φ i ‖ X ′ ≤ M {\displaystyle \|\varphi _{i}\|_{X^{\prime }}\leq M} for 1 ≤ i ≤ n . {\displaystyle 1\leq i\leq n.} [ 3] Note as well that ‖ x ‖ X ≤ ( 1 + δ ) . {\displaystyle \|x\|_{X}\leq (1+\delta ).} If one chooses δ {\displaystyle \delta } so that δ M < ϵ , {\displaystyle \delta M<\epsilon ,} then δ 1 + δ | φ i ( x ) | ≤ δ 1 + δ ‖ φ i ‖ X ′ ‖ x ‖ X ≤ δ ‖ φ i ‖ X ′ ≤ δ M < ϵ . {\displaystyle {\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 .}
Hence one gets 1 1 + δ Ev x ∈ J ( B ) ∩ U {\displaystyle {\frac {1}{1+\delta }}{\text{Ev}}_{x}\in J(B)\cap U} as desired.
^ Rudin, Walter. Functional Analysis (Second ed.). Lemma 3.9. pp. 63– 64. {{cite book }}
: CS1 maint: location (link ) ^ Rudin, Walter. Functional Analysis (Second ed.). Equation (3) and the remark after. p. 69. {{cite book }}
: CS1 maint: location (link ) ^ Folland, Gerald. Real Analysis: Modern Techniques and Their Applications (Second ed.). Proposition 5.2. pp. 153– 154. {{cite book }}
: CS1 maint: location (link )
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