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A sequence $\{f_{k}\}_{k=1}^{\infty}$ is called a Bessel sequence in a Hilbert space $H$, if there exists $B>0$ such that $$\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}\leq B\|f\|^{2}$$ for all $f\in H$.

But my question is: is this an equivalent definition for if $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}<\infty$ for all $f\in H$, then $\{f_{k}\}_{k=1}^{\infty}$ is a Bessel sequence.

If yes, how to show that if $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}<\infty$ for all $f\in H$, then $\{f_{k}\}_{k=1}^{\infty}$ is a Bessel sequence.

How to relate the infinity with $||f||^2$?

I was thinking can we use method of contradiction?

Suppose $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}> B\|f\|^{2}$, so we have $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}=\infty$ (is this conclusion right?), so a contradiction. So we must have $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}\leq B\|f\|^{2}$.

Does this make sense?

Thanks in advance.

glS
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Ian
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1 Answers1

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Let us define the operators $T_n:H\rightarrow \mathcal{l}_2$, $T_nf = (\langle f_1,f\rangle,\langle f_2,f\rangle,\ldots,\langle f_n,f\rangle,0,0,\ldots )$.

We will prove that if $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}<\infty$ for all $f\in H$ then $\{f_{k}\}_{k=1}^{\infty}$ is a Bessel sequence. The condition $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}<\infty$ for all $f\in H$ implies that $\forall_{f\in H} \sup_n\| T_nf\| \leq \infty$. Now using Banach-Steinhaus theorem (see e.g. http://en.wikipedia.org/wiki/Banach-Steinhaus_theorem) we conclude that $\sup_n\|T_n\|\leq \infty$ and thus $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}\leq (\sup_n\|T_n\|)^2\|f\|^2$, which shows that $\{f_{k}\}_{k=1}^{\infty}$ is a Bessel sequence.

user72829
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