Goldstine theorem

Goldstine theorem In functional analysis, a branch of mathematics, the Goldstine theorem, named after Herman Goldstine, is stated as follows: Goldstine theorem. Let {displaystyle X} be a Banach space, then the image of the closed unit ball {displaystyle Bsubseteq X} under the canonical embedding into the closed unit ball {displaystyle B^{prime prime }} of the bidual space {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 {displaystyle c_{0},} and its bi-dual space Lp space {displaystyle ell ^{infty }.} Contents 1 Proof 1.1 Lemma 1.1.1 Proof of Lemma 1.2 Proof of Theorem 2 See also 3 References Proof Lemma For all {displaystyle x^{prime prime }in B^{prime prime },} {displaystyle varphi _{1},ldots ,varphi _{n}in X^{prime }} and {displaystyle delta >0,} there exists an {displaystyle xin (1+delta )B} such that {displaystyle varphi _{i}(x)=x^{prime prime }(varphi _{i})} for all {displaystyle 1leq ileq n.} Proof of Lemma By the surjectivity of {displaystyle {begin{cases}Phi :Xto mathbb {C} ^{n},\xmapsto left(varphi _{1}(x),cdots ,varphi _{n}(x)right)end{cases}}} it is possible to find {displaystyle xin X} with {displaystyle varphi _{i}(x)=x^{prime prime }(varphi _{i})} for {displaystyle 1leq ileq n.} Now let {displaystyle Y:=bigcap _{i}ker varphi _{i}=ker Phi .} Every element of {displaystyle zin (x+Y)cap (1+delta )B} satisfies {displaystyle zin (1+delta )B} and {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 {displaystyle operatorname {dist} (x,Y)geq 1+delta } and by the Hahn–Banach theorem there exists a linear form {displaystyle varphi in X^{prime }} such that {displaystyle varphi {big vert }_{Y}=0,varphi (x)geq 1+delta } and {displaystyle |varphi |_{X^{prime }}=1.} Then {displaystyle varphi in operatorname {span} left{varphi _{1},ldots ,varphi _{n}right}} [1] and therefore {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.

Proof of Theorem Fix {displaystyle x^{prime prime }in B^{prime prime },} {displaystyle varphi _{1},ldots ,varphi _{n}in X^{prime }} and {displaystyle epsilon >0.} Examine the set {displaystyle U:=left{y^{prime prime }in X^{prime prime }:|(x^{prime prime }-y^{prime prime })(varphi _{i})|0} there exists a {displaystyle xin (1+delta )B} such that {displaystyle x^{prime prime }(varphi _{i})=varphi _{i}(x),} {displaystyle 1leq ileq n,} and in particular {displaystyle {text{Ev}}_{x}in U.} Since {displaystyle J(B)subset B^{prime prime },} we have {displaystyle {text{Ev}}_{x}in (1+delta )J(B)cap U.} We can scale to get {displaystyle {frac {1}{1+delta }}{text{Ev}}_{x}in J(B).} The goal is to show that for a sufficiently small {displaystyle delta >0,} we have {displaystyle {frac {1}{1+delta }}{text{Ev}}_{x}in J(B)cap U.} Directly checking, we have {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 we can choose {displaystyle M} sufficiently large so that {displaystyle |varphi _{i}|_{X^{prime }}leq M} for {displaystyle 1leq ileq n.} [3] Note as well that {displaystyle |x|_{X}leq (1+delta ).} If we choose {displaystyle delta } so that {displaystyle delta M

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