Generally in my answers I like to add sources to, in some sense, back up what I claim in said answer. I often use wikipedia, but recently have seen a few comments saying "don't trust wikipedia entirely for advanced mathematics". I personally haven't seen any fallacies from wikipedia, but then again I'm not exactly in high level math (the math I'm studying right now I can't exactly find on wikipedia ;)). The general question here is as in the title:


  Is wikipedia a valid source to use on MSE?


Recently (meaning today) I have been using wolfram pages as my backup. I do not believe the general public can edit that so I feel it may be safer. Is there some sort of regulation for this?


Question: Multiplication operator on Hilbert space
Answer: 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 $0 &lt; K &lt; +\infty$ such that
$$
 |\lambda_n - \lambda|^{-1} \leq K
$$
which holds, in turn, if and only if there exists some $0 &lt; K &lt; +\infty$ 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
$$
 \sigma(T) = \overline{\{\lambda_n\}}.
$$