A sequence $\{f_{n}\}_{n\in I}$ is a frame for a separable Hilbert space $H$ if there exists $0
Some books define a frame for just "Hilbert space" and not mentioning the "separability". Is there any difference between these two cases?
Different definitions of frames for Hilbert spaces
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real-analysis
functional-analysis
hilbert-spaces
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0That's strange! No one have any idea about this problem! – 2012-10-15
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1All that I can add is: some books take it as the definition that a Hilbert Space is separable, this being the most interesting case. Perhaps this helps?! – 2012-10-15
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0@Sebastian: Thank you. I was wondering why a "separable" Hilbert space is most interesting than just Hilbert space? – 2012-10-16
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0Sorry this is a bit late. You can prove that any two separable Hilbert Spaces are isomorphic, by the appropriate sense of isomorphism. This gives nice results such as $L^2([0,1])$ being equivalent to $l^2$ the space of square bounded sequences – 2012-10-16
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If the Hilbert space is not separable and $\{f_n\}_{n\in I}$ is a frame in this sense, then take $f\neq 0$ orthogonal to all the $f_n$'s: it's possible, otherwise the Hilbert space would be separable. By the first inequality, we would have that $f=0$, which is not possible.
But it can make sense if we deal with an arbitrary set. For example, take $\ell²(0,1)$, which is not separable and $f_i(k):=\delta_{ik}$ for $i,k\in (0,1)$.
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0Is $I$ assumed to be countable? – 2012-10-17
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0I assumed that as in the OP we talked about a sequence. – 2012-10-18
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1Ah, I missed the word "sequence". The concept would make sense for an arbitrary index set $I$ and this would remove the restriction to separable Hilbert spaces, maybe that generalization led to the OP's confusion? – 2012-10-18