Approximate point spectrum

Remarks

Equivalently, $\lambda \in \sigma_{ap}(T)$ if and only if $T - \lambda I$ is not bounded below, that is, there is no $c > 0$ with $\norm{(T - \lambda I)x} \ge c \norm{x}$ for all $x$. It satisfies $\sigma_p(T) \cup \sigma_c(T) \subseteq \sigma_{ap}(T) \subseteq \sigma(T)$, with the point spectrum, the continuous spectrum and the spectrum. It is closed, and it contains the boundary of $\sigma(T)$, so it is nonempty when $X \ne \set{0}$.

Example: for $D(x_n) = (x_n / n)$ on $\ell^2$, the unit vectors $e_n$ satisfy $\norm{D e_n} = 1/n \to 0$, so $0 \in \sigma_{ap}(D)$.

Non-example: for the right shift $S$ on $\ell^2$, $\norm{Sx} = \norm{x}$ for all $x$, so $S$ is bounded below and $0 \notin \sigma_{ap}(S)$, although $0 \in \sigma(S)$. More generally $\norm{(S - \lambda I)x} \ge (1 - \abs{\lambda}) \norm{x}$, so $\sigma_{ap}(S)$ is the unit circle.

For a normal operator on a Hilbert space, $\sigma_{ap}(T) = \sigma(T)$.