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Let $H$ is a Hilbert space. Let $f:H \rightarrow \mathbb{C}$ is linear and that there is a constant $C$ such that $|f(x)| \leq C ||x||$ for all $x \in H$

Suppose {$e_n$} is a complete orthonormal set for $H$.

show that

y=$\Sigma e_n \overline{f(e_n)} \in H$ and that $f(x)=<x, y>$ for all $x \in H$

I have shown that $f$ is continuous since

$|f(x)-f(y)|=|f(x-y)|\leq C||x-y||<C\delta$ by taking $\delta=\epsilon/C$

What are the next steps?

Andy
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1 Answers1

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You need to show that $\sum\overline{f(e_n)}e_n\in H$. To this end you have to prove that $\sum_n|f(e_n)|^2<\infty$. This will require using $|f(x)|\le C\|x\|$.

Angina Seng
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  • I got $||y|| = \Sigma_n |f(e_n)|^2$ using that if {$e_n$} is a complete orthonormal set for $H$ iff for all $x \in H$ $||x||^2= \Sigma_n |<x, e_k>|^2$. I'm not quite sure why $ \Sigma_n |f(e_n)|^2< \infty$ – Andy Jun 30 '17 at 05:00