Multiplication operator on Hilbert space i looked to the question Spectrum and point spectrum of this operator. I will go further with asking. We know that $T$ is well-defined iff $(\lambda_n)\in\ell^{\infty}$. But if $(\lambda_n)\notin\ell^{\infty}$ then we can defined this operator also on a new domain $D(T)$ given by $D(T)=\{\sum_{n=0}^{\infty}{x_ne_n}\in\ell^2: \sum_{n=0}^{\infty}{|\lambda_nx_n|^2}<\infty\}$. My first question is: Is this $D(T)$ dense in $\ell^2$? 
I observe that $T$ is well-defined on $D(T)$. Thus we can also ask about the spectrum of $T$. We have seen in Spectrum and point spectrum of this operator that the spectrum is $\sigma(T)=\overline{\{\lambda_n:n\in\mathbb{N}\}}$. Can we say this also for $T$ defined on $D(T)$? Is it the same argumentation?
Thank you for help.
 A: Yes, $D(T)$ is dense in $\ell^2$, since it contains all finite linear combinations of the $e_n$, which are dense in $\ell^2$ since $\{e_n\}$ is an orthonormal basis for $\ell^2$.
Let us turn to $\sigma(T)$. It is still the case that $T e_n = \lambda_n e_n$ for each $n$, so each $\lambda_n$ is an honest-to-goodness eigenvalue of $T$, and hence, $\overline{\{\lambda_n\}} \subseteq \sigma(T)$. On the other hand, observe that for any $\lambda \in \mathbb{C} \setminus \{\lambda_n\}$,
$$
 (T - \lambda \operatorname{Id})\left(\sum_n x_n e_n \right) = \sum_n (\lambda_n - \lambda)x_n e_n,
$$
so that $T - \lambda \operatorname{Id}$ is invertible at least on the dense subspace of finite linear combinations of the $e_n$, with inverse
$$
 (T - \lambda \operatorname{Id})^{-1}\left(\sum_n x_n e_n \right) = \sum_n (\lambda_n - \lambda)^{-1}x_n e_n.
$$
Hence, by the result you cite, $(T - \lambda \operatorname{Id})^{-1}$ extends to a bounded operator on $\ell^2$ if and only if
$$
 \left( (\lambda_n - \lambda)^{-1} \right) \in \ell^\infty,
$$
if and only if there exists some $0 < K < +\infty$ such that
$$
 |\lambda_n - \lambda|^{-1} \leq K
$$
which holds, in turn, if and only if there exists some $0 < K < +\infty$ such that
$$
 |\lambda_n - \lambda| \geq K^{-1},
$$
which is equivalent, if I'm not mistaken, to $\lambda \notin \overline{\{\lambda_n\}}$. Hence, one should still have that
$$
 \sigma(T) = \overline{\{\lambda_n\}}.
$$
