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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 $f\in H$. If the sequence only satisfying the upper bound it is called a Bessel sequence.

Now my questin is: if a given sequence is Bessel sequence but not a frame, what does this mean?

My guess is that: there exists (a non-zero) $f\in H$ such that $ A\|f\|^{2} > \sum_{n\in I}|\langle f,f_{n}\rangle|^{2} $ for all $A$.

But I'm not sure if this is correct! Any help is appreciated!

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    You should make a new thread for your new questions, or edit this question in the main post. They will receive more attention if you make a new question.2012-06-03

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if a given sequence is Bessel sequence but not a frame, what does this mean?

It means that for every $A>0$ there exists $f\in H$ such that $A\|f\|^{2} > \sum_{n\in I}|\langle f,f_{n}\rangle|^{2} $

What we can say about those functions $g_m$, for example, (1) Is the sequence $g_m$ converges in $H$? Or (2) Is $|\langle f,g_m\rangle|$ bounded above? Or (3) Is $\sup_m \|g_m\|$ exists?

None of (1), (2), (3) hold generally. For example, if the subspace $V=\{f_n\}^\perp$ is nontrivial, the vectors $g_m$ can be arbitrary vectors from $V$.