Goldstine theorem

Goldstine theorem In functional analysis, ein Zweig der Mathematik, the Goldstine theorem, named after Herman Goldstine, is stated as follows: Goldstine theorem. Lassen {Anzeigestil 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 {Anzeigestil B^{prime prime }} of the bidual space {Anzeigestil 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 {Anzeigestil c_{0},} and its bi-dual space Lp space {displaystyle ell ^{unendlich }.} Inhalt 1 Nachweisen 1.1 Lemma 1.1.1 Proof of Lemma 1.2 Proof of Theorem 2 Siehe auch 3 References Proof Lemma For all {Anzeigestil x^{prime prime }in B^{prime prime },} {Anzeigestil Varphi _{1},Punkte ,varphi_{n}in X^{prim }} und {displaystyle delta >0,} there exists an {Anzeigestil xin (1+Delta )B} so dass {Anzeigestil Varphi _{ich}(x)=x^{prime prime }(varphi_{ich})} für alle {displaystyle 1leq ileq n.} Proof of Lemma By the surjectivity of {Anzeigestil {Start{Fälle}Phi :Xto mathbb {C} ^{n},\xmapsto left(varphi_{1}(x),cdots ,varphi_{n}(x)Rechts)Ende{Fälle}}} it is possible to find {Anzeigestil xin X} mit {Anzeigestil Varphi _{ich}(x)=x^{prime prime }(varphi_{ich})} zum {displaystyle 1leq ileq n.} Now let {Anzeigestil Y:=bigcap _{ich}ker varphi _{ich}=ker Phi .} Every element of {displaystyle zin (x+Y)cap (1+Delta )B} satisfies {displaystyle zin (1+Delta )B} und {Anzeigestil Varphi _{ich}(z)=varphi _{ich}(x)=x^{prime prime }(varphi_{ich}),} so it suffices to show that the intersection is nonempty.

Assume for contradiction that it is empty. Dann {Anzeigestil Betreibername {dist} (x,Y)geq 1+delta } and by the Hahn–Banach theorem there exists a linear form {displaystyle varphi in X^{prim }} so dass {Anzeigestil Varphi {big vert }_{Y}=0,varphi (x)geq 1+delta } und {Anzeigestil |Varphi |_{X^{prim }}=1.} Dann {displaystyle varphi in operatorname {span} links{varphi_{1},Punkte ,varphi_{n}Rechts}} [1] und deshalb {displaystyle 1+delta leq varphi (x)=x^{prime prime }(Varphi )leq |Varphi |_{X^{prim }}links|x^{prime prime }Rechts|_{X^{prime prime }}leq 1,} which is a contradiction.

Proof of Theorem Fix {Anzeigestil x^{prime prime }in B^{prime prime },} {Anzeigestil Varphi _{1},Punkte ,varphi_{n}in X^{prim }} und {displaystyle epsilon >0.} Examine the set {Anzeigestil U:=links{y^{prime prime }in X^{prime prime }:|(x^{prime prime }-y^{prime prime })(varphi_{ich})|0} there exists {displaystyle xin (1+delta )B} {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 (1+delta )J(B)cap We can scale to get {displaystyle {frac {1}{1+delta }}{text{Ev}}_{x}in J(B).} goal show sufficiently small {displaystyle delta >0,} J(B)cap Directly checking, {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 choose {displaystyle M} large {displaystyle |varphi _{i}|_{X^{prime }}leq M} {displaystyle 1leq ileq n.} [3] as well {displaystyle |x|_{X}leq ).} If {displaystyle delta } {displaystyle delta M

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