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Let $H$ be a separable Hilbert space.

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 $H$ $$ A\|f\|^2 \leq \sum |\langle f, f_n \rangle|^2 \leq B \|f\|^2 $$ $\{f_{n}\}$ is complete if the only element which is orthogonal to all $f_{n}$ is the zero element.

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

Of course the converse is true if $f_n$ is orthonormal. But orthogonality is a very strong condition for me.

Any comments or references are welcome!

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up vote 2 down vote accepted

Presumably you'd like some answer less tautological that "a complete sequence is a frame when it satisfies the defining condition for frames," but then it isn't clear what your rules are.

Without loss of generality, you could take $\ell^{2}$ for your $H$. Then using your $f_n$'s as rows, a sequence takes the form of a matrix (with rows in $\ell^2$). A priori, such a matrix defines an operator from $H$ to ${\Bbb C}^{\Bbb N}$. Complete means kernel $\{0\}$; frame means bounded operator, with spectrum bounded away from 0, to $\ell^2$. So I read your question as asking for a characterization of boundedness and/or spectrum bounded away from 0, directly from the appearance of the matrix coefficients. I don't believe that question admits any satisfactory general answer.

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