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!


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.