Timeline for The role of completeness in Hilbert Spaces
Current License: CC BY-SA 2.5
8 events
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| Aug 29, 2010 at 16:14 | comment | added | Jerry | @Matthew: I just ready through your post and followed the link to wikipedia. You're right about what you said, but in the context of this question, completeness is assumed. All complete inner product spaces have an orthonormal basis. Again, however, "basis" may have to be defined via convergence of nets. Your post and the ensuing discussion confused me for a minute. Thanks for that! It's good to have to go back and re-read the basics now and again. | |
| Aug 28, 2010 at 19:21 | comment | added | Matthew Daws | I asked this very question recently: see mathoverflow.net/questions/36734/… Short answer: if you're not separable, then inner-product spaces can fail to have orthonormal bases. | |
| Aug 18, 2010 at 19:59 | comment | added | Jerry |
@Nate: if one defines orthonormal basis as "maximal orthonormal set," you're right. Through you're question I realize that I'm too used to thinking of an orthonormal basis as a collection $\{x_\alpha\}_{\alpha\in A}$ of orthonormal elements such that $y = \sum_\alpha (x_\alpha, y)x_\alpha)$ for any $y$. (Equality with the sum here means convergence of nets.) In Hilbert space, the two definitions are equivalent. Having just had a quick peek at Reed-Simon, completeness is used to prove that maximal orthonormal sets have the latter property.
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| Aug 18, 2010 at 19:12 | comment | added | Nate Eldredge | Is completeness really necessary to get the existence of an orthonormal basis? It seems to me that the usual Zorn's lemma argument goes through in any inner product space. | |
| Aug 18, 2010 at 7:30 | comment | added | Olumide | I was suggesting that Cauchy convergence criterion is a means by which a pre-Hilbert space is enlarged such that it contains no "holes". What used to bother me was the statement that all Cauchy sequences converged in that space. It used to make me wonder, "How can I be really sure, without examining all possible Cauchy sequences?". I stopped worrying about this when I considered that the statement might be definition (intended to widen the space) and might not be a result to be proved. BTW, I'm not familiar with the $\varepsilon-\delta$-definition but I will look it up. | |
| Aug 17, 2010 at 13:07 | comment | added | Jerry | I'm having trouble parsing the last sentence of your comment. Let me offer the following clarification, though. If a vector space has a norm, and a sequence $(x_n)$ converges to a point $x$ with respect to that norm, i.e., using the $\varepsilon$-$\delta$-definition, <em>then</em> the sequence is Cauchy. In complete normed spaces, therefore, the characterization "convergent if and only if Cauchy" holds for sequences. More generally this is true for metric spaces; that is, the vector space structure isn't necessary. | |
| Aug 17, 2010 at 8:50 | comment | added | Olumide | I like this explanation. BTW, I've just realized that what bothered/bothers me is the fact that, regardless of the means by which the sequence is generated, every Cauchy (convergent) sequence converges to the space. My instinct was to ask why, and seek a proof. However, I'm starting to think differently. Could it be that this is condition/restriction is means by which a normed space is enlarged such that it contains the limit of any sequence, regardless of mechanism that generates the sequence. | |
| Aug 17, 2010 at 6:47 | history | answered | Jerry | CC BY-SA 2.5 |