Let $(e_n)_1^{\infty}$ be a complete orthonormal sequence in a Hilbert space $H$ and let $\lambda_n \in \mathbb{C}$ for $n \in \mathbb{N}$. Show that there is a bounded linear operator $D$ on $H$ such that $D e_n = \lambda_n e_n$ for all $n \in \mathbb{N}$, if and only if $(\lambda_n)$ is a bounded sequence. What is $||D||$ the norm of the operator $D$?
My work
I was able to prove the first part, and it is straightforward, but when it comes to the converse part, I feel I am stuck, because if I assume $(\lambda_n)$ is bounded then how can I relate it with the operator $D$ I need to prove its existence?
Also in finding the norm of the operator, I got the following
$||D|| = \sup \{ \frac{||Dx||}{||x||}, 0 \ne x \in H \} \geq \sup \{ \frac{||De_n||}{||e_n||} = \sup \{|\lambda_n |\}$ but I couldn't prove the other direction.
Best Answer
Suppose $\{\lambda_n\}_{n=1}^\infty$ is a bounded sequence. As you say, we need to show that there exists a bounded operator $D$ such that $De_n=\lambda_ne_n$. Therefore, just define $D:E\subset\mathcal{H}\rightarrow\mathcal{H}$ to be the operator such that $De_n=\lambda_ne_n$, where $E=\cup_{n=1}^\infty e_n$. Since the orthonormal basis $\{e_n\}_{n=1}^\infty$ is dense in $\mathcal{H}$, we can extend $D$ to $\tilde{D}$ onto the whole domain $\mathcal{H}$ by continuity and also preserve the operator norm.
You have already shown that $||\tilde{D}||_\mathcal{H}\geq \sup_n|\lambda_n|$. To get the other direction, note that \begin{equation} ||\tilde{D}||_\mathcal{H}=||D||_E=\sup_{x\in E\subset \mathcal{H}}||Dx||=\sup_n||\lambda_ne_n||\leq \sup_n|\lambda_n|, \end{equation} so we arrive at the desired statement.