Discrete and essential spectrum of a self-adjoint operator

Assume the Axiom of Choice. Let $A$ be a self-adjoint operator on a complex Hilbert space $H$ with spectral projection valued measure $E$ on $\mathbb{R}$ (Spectral theorem for unbounded self-adjoint operators (PVM form)). The discrete spectrum ${\sigma }_{\mathrm{d}}(A)$ is the set of eigenvalues of $A$ that are isolated points of $\sigma (A)$ and whose eigenspace is finite dimensional; the essential spectrum is ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A):=\sigma (A)\setminus {\sigma }_{\mathrm{d}}(A),$ with $\sigma (A)$ as in Resolvent and spectrum of an unbounded operator.

For $H=\{0\}$ use its unique PVM directly: the spectrum and both parts are empty and every projection has rank zero. Below suppose $H\mathrm{\ne }\{0\}$, as required by the cited spectral theorem.

Spectral-projection description, with proofs. Fix $\lambda \in \mathbb{R}$ and write $P_{\epsilon }:=E((\lambda -\epsilon ,\lambda +\epsilon ))$ for $\epsilon >0$.

Here rank means the algebraic dimension of the range when finite; rank $=\mathrm{\infty }$ means the range is not finite dimensional. The calculus Unbounded Borel functional calculus: domains, products, spectral mapping gives the support facts: $E(\mathbb{R}\setminus \sigma (A))=0$, and every open interval about a spectral point has nonzero projection. Projections on disjoint sets have orthogonal ranges, and $E(B)E(C)=E(B\cap C)$ Projection valued measure.

1. $E(\{\lambda \})$ is the projection onto $\ker (A-\lambda )$. If $x=E(\{\lambda \})x$, its scalar measure $E_x$ is supported on $\{\lambda \}$ by the projection identity. Thus $\int {\mu }^{2}d{E}_x={\lambda }^2\mathrm{\parallel }x{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $x\in D(A)$, and $\mathrm{\parallel }(A-\lambda )x{\mathrm{\parallel }}^2=\int \mathrm{\mid }\mu -\lambda {\mathrm{\mid }}^{2}d{E}_x=0.$ The domain and norm identities are supplied by the unbounded calculus and The unbounded PVM integral is densely defined, closed and normal. Hence $Ax=\lambda x$. Conversely, for an eigenvector (or the zero vector) the same norm identity gives zero integral. On $\{\mathrm{\mid }\mu -\lambda \mathrm{\mid }\ge 1\mathrm{/}n\}$ this bounds the measure by $n^2$ times that zero integral; taking the countable union shows $E_x(\mathbb{R}\setminus \{\lambda \})=0$. Since $\mathrm{\parallel }(I-E(\{\lambda \}))x{\mathrm{\parallel }}^2$ equals this scalar measure, $E(\{\lambda \})x=x$.
2. A finite-rank interval contains only finitely many spectral points. If $P_{\epsilon }$ has rank $r<\mathrm{\infty }$ and its interval contained $r+1$ distinct spectral points, choose disjoint small open intervals about those finitely many points, all contained in the given interval. Each has a nonzero projection by support. Choose one unit vector in each range. They are orthonormal vectors in $\mathrm{ran}{P}_{\epsilon }$, hence linearly independent (take inner products with each vector), contradicting its dimension $r$. Thus there are at most $r$ spectral points there. In particular any $\lambda \in \sigma (A)$ with such a finite-rank $P_{\epsilon }$ is isolated: take a smaller interval around $\lambda$ excluding the other finitely many points. For that interval the projection is $E(\{\lambda \})$ by support, is nonzero by support, and has finite rank since its range lies in $\mathrm{ran}{P}_{\epsilon }$. By item 1, $\lambda$ is an eigenvalue of finite multiplicity.
3. The two rank characterizations. If $\lambda \in {\sigma }_{\mathrm{d}}(A)$, an isolating interval has projection $E(\{\lambda \})$, of finite rank by item 1. Conversely, if $\lambda$ is an eigenvalue and some $P_{\epsilon }$ has finite rank, item 2 proves it discrete. Hence $\lambda \in {\sigma }_{\mathrm{d}}(A)⟺\lambda \text{ is an eigenvalue and some }{P}_{\epsilon }\text{ has finite rank}\mathrm{.}$ If $\lambda \in {\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$, item 2 excludes every finite-rank interval. Conversely, if every $P_{\epsilon }$ has infinite rank, each is nonzero, so support puts $\lambda$ in $\sigma (A)$, and the just-proved discrete characterization excludes it from ${\sigma }_{\mathrm{d}}(A)$. Therefore $\lambda \in {\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)⟺\mathrm{rank}{P}_{\epsilon }=\mathrm{\infty }\text{ for every }\epsilon >0.$
4. ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$ is closed. If ${\lambda }_n\in {\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$ and ${\lambda }_n\to \lambda$, then for every $\epsilon >0$ some $n$ has $({\lambda }_n-\epsilon \mathrm{/}2,{\lambda }_n+\epsilon \mathrm{/}2)\subseteq (\lambda -\epsilon ,\lambda +\epsilon )$, so $\mathrm{rank}{P}_{\epsilon }\ge \mathrm{rank}{P}_{\epsilon \mathrm{/}2}^{(n)}=\mathrm{\infty }$ and $\lambda \in {\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$ by item 3.

The closure conclusion is in $\mathbb{R}$, and also in $\mathbb{C}$ since $\mathbb{R}$ is closed there. A spectral accumulation point has infinitely many spectral points in every surrounding interval, so item 2 forces infinite rank. An isolated eigenvalue of infinite multiplicity has its infinite-dimensional eigenspace inside every interval range by item 1. Consequently an accumulation point of $\sigma (A)$ and an isolated eigenvalue of infinite multiplicity both lie in ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$, and $\sigma (A)$ is the disjoint union of ${\sigma }_{\mathrm{d}}(A)$ and ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)$.