Spectrum of an operator

Remarks

By the bounded inverse theorem, $\lambda \in \sigma(T)$ if and only if $T - \lambda I$ is not bijective. If $X \ne \set{0}$, then $\sigma(T)$ is a nonempty compact subset of the closed disc $\set{\lambda : \abs{\lambda} \le \norm{T}}$. For $X = \C^n$, $\sigma(T)$ is the spectrum of the matrix of $T$, its set of eigenvalues.

Example: let $D(x_n) = (x_n / n)$ on $\ell^2$. For $\lambda \notin \set{1/n : n \ge 1} \cup \set{0}$, the numbers $1/(1/n - \lambda)$ are bounded, so the diagonal operator with these entries is a bounded inverse of $D - \lambda I$. Each $1/n$ is an eigenvalue, and $D$ is not surjective, so $\sigma(D) = \set{1/n : n \ge 1} \cup \set{0}$.

Non-example: the number $2$ is not in $\sigma(D)$, since $\abs{1/(1/n - 2)} \le 1$ for all $n$, so $D - 2I$ has a bounded inverse.

For a closed operator $T$ with domain $D(T)$, $\lambda$ is outside $\sigma(T)$ when $T - \lambda I \colon D(T) \to X$ is bijective with bounded inverse. The spectrum is then closed, but it can be empty or unbounded.