Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I'm posting this question again because I'm still confused about the answer!

A sequence $\{f_{n}\}_{n\in I}$ is called a Bessel sequence in a Hilbert space $H$, if there exists $B>0$ such that $$\sum_{n\in I}|\langle f,f_{n}\rangle|^{2}\leq B\|f\|^{2}$$ for all $f\in H$.

Now my questin is: if a given sequence is not Bessel sequence, does it mean that

given $B>0$, there exists (a non-zero) $f\in H$ such that $$ \sum_{n\in I}|\langle f,f_{n}\rangle|^{2}> B\|f\|^{2}$$


(old post: If a sequence is not a frame)

share|cite|improve this question
This is the negation that $\{f_{n}\}_n$ is a Bessel sequence and the negation do not make sense if $f=0$ because both sides are zero. – user29999 Jun 7 '12 at 14:04
So just to be sure: if we pick another $B$ we would find another $f\in H$ satisfying the second relation, and so forth, right! – Kim Jun 7 '12 at 14:10

The sequence $\{f_n\}$ is not a Bessel sequence if for all integer $N$, we can find $g_N\in H$ such that for all integer $N$, $$\sum_{n\in I}|\langle g_N,f_n\rangle|^2> N \lVert g_N\rVert^2,$$ exactly what you mean.

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.