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单词 ClassificationOfHilbertSpaces
释义

classification of Hilbert spaces


Hilbert spacesMathworldPlanetmath can be classified, up to isometric isomorphism, according to their dimensionPlanetmathPlanetmathPlanetmath. Recall that an isometric isomorphism of Hilbert spaces is an unitary transformation, therefore it preserves the vector spaceMathworldPlanetmath structureMathworldPlanetmath along with the inner productMathworldPlanetmath structure (hence, preserving also the topological structure). Recall also that the dimension of a Hilbert space is a well defined concept, i.e. all orthonormal bases of an Hilbert space share the same cardinality.

The classification theorem we describe here states that two Hilbert spaces H1 and H2 are isometrically isomorphic if and only if they have the same dimension, i.e. if and only if an orthonormal basis of H1 has the same cardinality of an orthonormal basis of H2.

This will be achieved by proving that every Hilbert space is isometrically isomorphic to an 2(X) space (http://planetmath.org/EllpXSpace), where X has the cardinality of any orthonormal basis of the Hilbert space in consideration.

TheoremMathworldPlanetmath 1 - Suppose H is an Hilbert space and let I be a set that indexes one (and hence, any) orthonormal basis of H. Then, H is isometrically isomorphic to 2(I).

Theorem [Classification of Hilbert spaces] - Two Hilbert spaces H1 and H2 are isometrically isomorphic if and only if they have the same dimension.

Proof of Theorem 1: Let {ei}iI an orthonormal basis indexed by the set I. Let U:H2(I) be defined by

Ux(i):=x,ei

We claim that U is an isometric isomorphism. It is clear that U is linear. Using Parseval’s equality and the definition of norm in 2(I) it follows that

x2=iI|x,ei|2=iI|Ux(i)|2=Ux2(I)2

We conclude that U is isometric. It remains to see that it is surjectivePlanetmathPlanetmath (since injectivity follows from the isometric condition).

Let f2(I). By definition of the space 2(I) we must have iI|f(i)|2<, and therefore, using the Riesz-Fischer theorem, the series iIf(i)ei converges to an element x0H. We now see that

Ux0(j)=x0,ej=iIf(i)ei,ej=f(j)

or in other , Ux0=f. Hence, U is surjective.

Proof of the classification theorem :

  • () Of course, if the Hilbert spaces H1 and H2 are isometrically isomorphic, with isometric isomorphism U, then if {ei}iI is an orthonormal basis for H1 than {Uei}iI is an orthonormal basis for H2. Hence, H1 and H2 have the same dimension.

  • () If the Hilbert spaces H1 and H2 have the same dimension, then we can index any orthonormal basis of H1 and any orthonormal basis of H2 by the same set I. Using Theorem 1 we see that H1 and H2 are both isometrically isomorphic to 2(I). Hence H1 and H2 are isometrically isomorphic.

Titleclassification of Hilbert spaces
Canonical nameClassificationOfHilbertSpaces
Date of creation2013-03-22 17:56:18
Last modified on2013-03-22 17:56:18
Ownerasteroid (17536)
Last modified byasteroid (17536)
Numerical id10
Authorasteroid (17536)
Entry typeTheorem
Classificationmsc 46C15
Classificationmsc 46C05
SynonymHilbert spaces of the same dimension are isometrically isomorphic
Related topicEllpXSpace
Related topicOrthonormalBasis
Related topicClassificationOfSeparableHilbertSpaces
Related topicCategoryOfHilbertSpaces
Related topicRieszFischerTheorem
Related topicQuantumGroupsAndVonNeumannAlgebras
Definesevery Hilbert space is isometrically isomorphic to a 2(X) space
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更新时间:2025/5/4 16:04:19