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I'm working on the following problem:

Suppose $\{e_n\}$ is a complete orthonormal sequence in a Hilbert space $H$. Let $\{c_n\}$ be a sequence of complex numbers.

  1. Prove that there is a bounded linear operator $T$ on $H$ such that $Te_n=c_ne_n$ for each $n$, if and only if $\{c_n\}$ is bounded.
  2. Determine $\|T\|$ when $\{c_n\}$ is bounded.
  3. Determine $T^*$, the adjoint of $T$, when $\{c_n\}$ is bounded.

What I've done so far: I know that if $\{e_n\}$ is a complete orthonormal sequence then any $x\in H$ can be written as $\sum_{1}^\infty\langle x,e_n\rangle e_n$ and $\|x\|^2=\sum_1^\infty|\langle x,e_n\rangle|^2$. Then if $T$ is linear then $Tx=T(\sum\langle x,e_n\rangle e_n)=\sum\langle x,e_n\rangle c_ne_n$. But I'm not sure how to proceed.

Any help on this problem would be greatly appreciated!

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Suppose $(c_n)$ is bounded. $\|\sum c_n\langle x, e_n \rangle e_n\|^{2}=\sum |c_n\langle x, e_n \rangle |^{2} \leq M \sum |\langle x, e_n \rangle |^{2}=M^{2}\|x\|^{2}$ where $M =\sup_m |c_n|$. Hence $T$ is bounded with$\|T\| \leq \sup_m |c_n|$.

Also $\|Te_n\| =|c_n|$ and this implies (by definition of norm of an operator) that $\|T\| \geq |c_n|$. This is true for each $n$ so we conclude that $\|T\|=\sup_m |c_n|$.

Conversely, $Te_n=c_n e_n$ show that $|c_n| \leq \|T\|$ for each $n$ so boundedness of $T$ implies that $(c_n)$ is bounded.

$T^{*}$ is given by $T^{*}x=\sum \overline {c_n} \langle x, e_n \rangle e_n $ and I will let you verify it using definitiom of the adjoint.

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