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Let $H$ be a separable Hilbert space. It is known that if $\{f_{n}\}$ is a frame then it is complete, but the converse is not true. In which cases the converse will be true, i.e.

When a complete sequence $\{f_{n}\}$ becomes a frame for $H$ (or at least satisfying the lower frame bound)?


(1) A sequence $\{f_{n}\}$ is a frame for a separable Hilbert space $H$ if there exists $A,B>0$ such that for all $f$ in $H$ $$A \lVert f\rVert^2 \leq \sum |\langle f,f_n\rangle|^{2}<B\lVert f\rVert^2.$$

(2) $\{f_{n}\}$ is complete if the only element which is orthogonal to all $f_{n}$ is the zero element.

Any comments or references are welcome!

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Well, its trivially true if $f_n$ is orthonormal, by Parseval. – user12014 Apr 24 '12 at 3:20
Orthonormality is a very strong condition. – Jolie Apr 24 '12 at 12:27
Of course, I was just noting that it was true. – user12014 Apr 24 '12 at 21:38

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