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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 $$ \sum_{n} ||^{2}\leq B ||f||^{2} $$ for all $f\in H$

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    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.)2011-05-03
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    I did, thanks..2011-05-03
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    Then the answer to your question is "yes", $\{a_n\}$ fails the definition of a Bessel sequence.2011-05-04
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    Please check the wording of your question. I can't make head nor tail of it.2011-05-05
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    @Willie Could you share your interpretation? I really have no idea what is being asked here.2011-05-05
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    @Glen Let $P(A)$ be the statement that $A = \{a_n\}$ is a Bessel sequence. Assume you know that for some fixed $A,B$ sequences in $H$, $P(A)\iff P(B)$. I interpret the statement $A$ has Bessel bound $\geq K \forall K\in \mathbb{N}$ to mean that $\forall K \exists f\in H$ such that $\sum\langle f,a_n\rangle^2 \geq K\|f\|^2_H$. The OP asks: does this imply $\neg P(A)$ and therefore also $\neg P(B)$. I answer: yes, by definition $\neg P(A)$. By assumed proposition $P(A)\iff P(B)$ we have $\neg P(B)$.2011-05-05

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