MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.