A sequence $\{f_{n}\}_{n\in I}$ is a frame for a separable Hilbert space $H$ if there exists $0
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!
If a sequence is not a frame
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hilbert-spaces
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0Welcome to math.stackexchange kim! You almost had it, but your logical ordering is incorrect. Currently it says : There exists a single non-zero $f$ such that blah blah holds for all A,B, while the actual negation would be: Choose any A,B. Then there exists a non-zero $f$ such that blah blah... (EDIT: You updated your question a bit, but I think you will still see my point.) – 2012-06-02
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0Ok, so $f$ will depends on $A$ and $B$ ? – 2012-06-02
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0Yes. $ $ $ $ $ $ – 2012-06-02
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0So, if this is the case, then we have: given $A_{m}>0$, we can find $g_{m}\in H$ such that $$A_{m}\|g_{m}\|^{2}> \sum_{n\in I}|\langle f,g_{m}\rangle|^{2}.$$ 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? – 2012-06-02
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0You 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
1 Answers
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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 $V$.