Continuous and residual spectrum

Remarks

By construction, $\sigma(T)$ is the disjoint union $\sigma_p(T) \cup \sigma_c(T) \cup \sigma_r(T)$. For $\lambda \in \sigma_c(T)$, the image of $T - \lambda I$ is dense but not all of $X$, since a bounded bijection between Banach spaces has a bounded inverse.

Example: for $D(x_n) = (x_n / n)$ on $\ell^2$, $0 \in \sigma_c(D)$. Indeed $D$ is injective, it is not surjective, and its image contains every finitely supported sequence, a dense set.

Non-example: for the right shift $S(x_1, x_2, \ldots) = (0, x_1, x_2, \ldots)$ on $\ell^2$, $0$ is not in $\sigma_c(S)$. The image of $S$ is the closed subspace $\set{x : x_1 = 0}$, which is not dense, so $0 \in \sigma_r(S)$.

For the right shift, $\sigma_p(S) = \emptyset$, $\sigma_r(S)$ is the open unit disc and $\sigma_c(S)$ is the unit circle. A self-adjoint operator on a Hilbert space has empty residual spectrum.