Is the Essential Spectrum the same as the Continuous Spectrum and Residual Spectrum

Question

Given a linear operator $A:D(A)\rightarrow X$ (where $D(A)$ is a dense subset of $X$ and $X$ is a Banach space or Hilbert Space), we define the spectrum to be $$ \sigma(A)=\{\lambda\in\mathbb{C}: A-\lambda \text{is not invertible}\} $$ We note that $\sigma(A)$ will be a closed subset of $\mathbb{C}$. Now there are a few ways that one may break up the spectrum.

Firstly, we can decompose it into the discrete spectrum and the essential spectrum that is $\sigma(A)=\sigma_d(A)\cup \sigma_{ess}(A)$. Where we have that the discrete spectrum, $\sigma_d(A)$, consists of all eigenvalues of $A$ such that $A-\lambda$ has finite algebraic multiplicity. Then we define the essential spectrum $\sigma_{ess}(A)$ to be the complement of $\sigma_d(A)$ inside of $\sigma(A)$. It is clear that this is a partition of the spectrum.

Now there is another way in which we may break up the spectrum. We can decompose it as $\sigma(A)=\sigma_{eigenvalues}(A)\cup \sigma_{cont}(A)\cup\sigma_{res}(A)$. In this decomposition, we have that $\sigma_{eigenvalues}(A)$ are the eigenvalues of $A$ consisting of those $\lambda\in\mathbb{C}$ such that $\ker(A-\lambda)\neq 0$. Thus, the remaining cases what happens if $\ker(A-\lambda)=0$, then this splits up into two cases depending upon the range of $A-\lambda$. If $\text{Ran}(A-\lambda)$ is a dense subset of $X$, then we have that $A-\lambda$ will have a densely defined inverse (which necessarily must be unbounded since if it were bounded then $A-\lambda$ would be invertible and $\lambda$ would not be in the spectrum) such $\lambda$ are a part of the continuous spectrum $\sigma_{cont}(A)$. Then the last case is if $\ker(A-\lambda)=0$, but $\text{Ran}(A-\lambda)$ is not dense, then put such $\lambda$ in the residual spectrum $\sigma_{res}(A)$.

Thus, we have these two different decompositions of the spectrum. My question is how related are these two decompositions? Since intuitively I would like to say that $\sigma_{eigenvalues}(A)=\sigma_d(A)$, and then $\sigma_{ess}(A)=\sigma_{cont}(A)\cup\sigma_{res}(A)$, but this seems wrong after thinking about it since there may be an eigenvalue with infinite algebraic multiplicity, so is all that we can say is that $\sigma_d(A)\subset\sigma_{eigenvalues}(A)$ and hence $\sigma_{cont}(A)\cup\sigma_{res}(A)\subset\sigma_{ess}(A)$, or is there some sort of further decomposition relating these? Or maybe there is some property of $A$ (like maybe being normal or self-adjoint or something like that) which might make this true?

Answer 1

In general, these spectra do not coincide. In fact, assume for simplicity that $A: H \to H$ is a bounded and self-adjoint operator, where $H$ is a Hilbert space. Then on one hand, $\sigma(A) = \sigma_p(A) \cup \sigma_{\textit{cont}} (A)$, as the residual spectrum is empty. On the other hand, we indeed have$^{(*)}$ the decomposition $\sigma(A) = \sigma_{\textit{disc}}(A) \cup \sigma_{\textit{ess}} (A)$. For both decompositions, the union is disjoint.

The inclusion $\sigma_{\textit{disc}}(A) \subset \sigma_p (A)$ (or equivalently $\sigma_{\textit{cont}} (A) \subset \sigma_{\textit{ess}} (A)$) always holds, since $\sigma_p(A)$ consists of all eigenvalues, whereas $\sigma_{\textit{disc}}(A)$ consists of isolated eigenvalues of finite multiplicity.

However, the above inclusions are strict, a useful example being that of finite-rank operators. For instance, consider an orthonormal basis $\{w_n, n \geq 0\}$ of $H$ (w.r.t. the scalar product $\langle \cdot, \cdot \rangle_H$), and let $A$ be the orthogonal projection onto $\operatorname{Span}(w_0)$, that is: $$ \forall u \in H, \quad Au := \langle u, w_0 \rangle_H\ w_0. $$ It can be easily checked that $\sigma_p (A) = \{0, \|w_0\|^2_H = 1\}$, where $1$ is a simple eigenvalue (with eigenvector $w_0$), and where $0$ is an eigenvalue of infinite multiplicity ($\operatorname{Ker}(A) = \operatorname{Span}(\{w_n, n \neq 0\})$. Therefore, $\sigma_{\textit{disc}}(A) = \{1\}$ is strictly included in $\sigma_p(A)$. Similarly, from the spectral theorem, $\sigma(A) = \sigma_p(A)$, so that $\sigma_{\textit{cont}}(A) = \varnothing$, while $\sigma_{\textit{ess}}(A) = \{0\}$.

$~^{(*)}$ Beware that the essential spectrum is defined without ambiguity only if $A$ is self-adjoint. For non self-adjoint operators, there are many definitions (5 on Wikipedia) of the essential spectrum, which are not equivalent.