# Bessel bound

I'm a little confuse about this problem: I have two sequences $\{a_{n}\}$ and $\{b_{n}\}$ in a Hilbert space $H$, with

$\{a_{n}\}$ is a Bessel sequence for $H$ if and only if $\{b_{n}\}$ is a Bessel sequence for $H$. The problem is that: If $\{a_{n}\}$ has Bessel bound $\geq K$, for every $K\in \mathbb N$, does this mean that $\{a_{n}\}$ is not Bessel sequence, and hence $\{b_{n}\}$ is also not Bessel sequence?

Definition: A sequence $\{a_{n}\}$ is a Bessel sequence in $H$ with Bessel bound $B>0$ if $$\sum_{n} |<f,a_{n}>|^{2}\leq B ||f||^{2}$$ for all $f\in H$

-
It maybe worthwhile defining what Bessel sequences and Bessel bounds are. (I know what the former is, but I am not sure I know what you mean by the latter.) – Willie Wong May 3 '11 at 0:33
I did, thanks.. – Sara May 3 '11 at 1:06
Then the answer to your question is "yes", $\{a_n\}$ fails the definition of a Bessel sequence. – Willie Wong May 4 '11 at 14:32
Please check the wording of your question. I can't make head nor tail of it. – Glen Wheeler May 5 '11 at 16:18
@Willie Could you share your interpretation? I really have no idea what is being asked here. – Glen Wheeler May 5 '11 at 16:19