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A sequence $\{f_{n}\}_{n\in I}$ is a frame for a separable Hilbert space $H$ if there exists $0 such that $ A\|f\|^{2} \leq \sum_{n\in I}|\langle f,f_{n}\rangle|^{2}\leq B\|f\|^{2} $ for all $f\in H$.

Some books define a frame for just "Hilbert space" and not mentioning the "separability". Is there any difference between these two cases?

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    Sorry 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 sequences2012-10-16

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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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    Ah,$I$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