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

  2. The operator $$T:l^2 \rightarrow H, \;\; (c_k)_{k\in \mathbb{N}} \mapsto \sum_{k\in \mathbb{N}}c_kf_k$$ is a well-defined bounded operator from $l_2$ onto $H$ and $\|T\|\leq\sqrt{B}$, called synthesis operator.

I want to show that these two statements are equivalent.

From 2. to 1., I started with the Cauchy-Schwarz inequality:

$$\sum_{k\in \mathbb{N}}|\langle f,f_{k}\rangle|^{2}\leq \|f\|^2\sum_{k\in \mathbb{N}}\|f_k\|^2$$

Then $B = \sum_{k\in \mathbb{N}}\|f_k\|^2$. Now I'd like to show that this series indeed converges. However I fail to continue from here on because I don't know how to bring the boundedness of the operator $T$ or it's existance into effect.

For the converse implication I'm also rather clueless. Any help appreciated!

user161913
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1 Answers1

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Hints: For 2. $\to$ 1. note that for the standard unit vectors $e_k \in \ell^2$ we have $Te_k = f_k$ $\def\<#1>{\left<#1\right>}\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$, hence $$ \sum_k\abs{\<f, f_k>}^2 = \sum_k\abs{\<T^*f,e_k>}^2 $$ Now use that $e_k$ is an orthogonal system and Bessel's inequality will do the rest.

For 1. $\to$ 2. we aim to let $T$ given by $Te_k = f_k$. To show that this gives a well-defined operator, note that for $n < m$ \begin{align*} \norm{\sum_{k=n+1}^m c_kf_k} &= \sup_{\norm f = 1} \abs{\<\sum_{k=n+1}^m c_kf_k, f>}\\ &\le \sup_{\norm f = 1} \sum_{k=n+1}^m \abs{c_k}\abs{\<f_k,f>}\\ &\le \sup_{\norm f = 1} \left(\sum_{k=n+1}^m \abs{c_k}^2\right)^{1/2} \cdot \left(\sum_{k=n+1}^m \abs{\<f_k,f>}^2\right)^{1/2}\\ &\le \sqrt B \cdot \left(\sum_{k=n+1}^m \abs{c_k}^2\right)^{1/2} \end{align*} Hence $T(c_k) = \sum_k c_k f_k$ is well-defined. The statement about its norm follows along the lines of the above.

martini
  • 84,101