Self-adjoint operators with compact resolvent: eigenbasis and discrete spectrum

[Spectral decomposition of $R_c$] The resolvent $R_c$ is compact by hypothesis and is self-adjoint: for $x,y\in H$, put $u=R_{c}x$ and $v=R_{c}y$. Then $u,v\in D(A)$ and $y=(A-c)v$, $x=(A-c)u$ by [F1], so self-adjointness of $A-c$ gives $⟨R_{c}x,y⟩=⟨u,(A-c)v⟩=⟨(A-c)u,v⟩=⟨x,R_{c}y⟩$. It is injective: if $R_{c}x=0$, then $x=(A-c)R_{c}x=0$ by [F1]. By [F2], with $\ker R_c=\{0\}$, the nonzero eigenspaces of $R_c$ are finite dimensional with real nonzero eigenvalues, and the union of orthonormal bases of those eigenspaces is a complete orthonormal family $(e_i{)}_{i\in I}$ of eigenvectors of $R_c$ with eigenvalues ${\nu }_i\mathrm{\ne }0$. The eigenvalue set is countable, being the union over $k\ge 1$ of the finite sets $\{\nu :\mathrm{\mid }\nu \mathrm{\mid }\ge 1\mathrm{/}k\}$ (Countable unions of at most countable sets, assuming ${\mathrm{A}\mathrm{C}}_{\omega }$, Finite, countably infinite, countable, uncountable), so $I$, a union of finitely many bases over a countable set of eigenvalues, is finite or countably infinite; it is infinite because a finite complete orthonormal family would be a finite spanning set of $H$ (its span is finite dimensional, hence closed, and equals its closed span, A finite-dimensional normed subspace is closed), so that $H$ would have a finite basis and be finite dimensional (Every spanning subset of a vector space contains a basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis), contrary to hypothesis. Hence $(e_i{)}_{i\in I}$ can be indexed as $(e_n{)}_{n\in \mathbb{N}}$, and then ${\nu }_n\to 0$: for every $\epsilon >0$ only finitely many indices $n$ satisfy $\mathrm{\mid }{\nu }_n\mathrm{\mid }\ge \epsilon$ by [F2].

[From $R_c$-eigenvectors to $A$-eigenvectors] For each $n$, $e_n={\nu }_n^{-1}{R}_c{e}_n$ lies in $\mathrm{ran}{R}_c=D(A)$ by [F1], and $(A-c)e_n={\nu }_n^{-1}(A-c)R_c{e}_n={\nu }_n^{-1}{e}_n,$ so $A{e}_n=(c+{\nu }_n^{-1})e_n$. Put ${\mu }_n:=c+{\nu }_n^{-1}$. Since $\mathrm{\mid }{\nu }_n\mathrm{\mid }\to 0$ and ${\nu }_n\mathrm{\ne }0$, $\mathrm{\mid }{\mu }_n\mathrm{\mid }\ge \mathrm{\mid }{\nu }_n{\mathrm{\mid }}^{-1}-\mathrm{\mid }c\mathrm{\mid }\to \mathrm{\infty }$. The map $\lambda \mapsto c+{\lambda }^{-1}$ is injective on $\mathbb{R}\setminus \{0\}$, so distinct ${\nu }_n$ give distinct ${\mu }_n$, and every eigenvector $x$ of $A$ with eigenvalue $\mu \mathrm{\ne }c$ satisfies $R_{c}x=(\mu -c{)}^{-1}x$, so $x$ is an $R_c$-eigenvector; conversely every $R_c$-eigenvector is an $A$-eigenvector with eigenvalue $c+{\nu }^{-1}\mathrm{\ne }c$. Hence the eigenvalue set of $A$ is $\{{\mu }_n\}$ with the same eigenspaces as those of $R_c$, the associated eigenvalues are exactly ${\mu }_n$, and each eigenspace is finite dimensional; in particular the eigenspace of $A$ at ${\mu }_n$ equals the $R_c$-eigenspace at ${\nu }_n$.

[$A$ is the diagonal operator in the basis $(e_n)$] We claim $D(A)=\{x\in H:{\sum }_n{\mu }_n^2\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2<\mathrm{\infty }\}$ and $Ax={\sum }_n{\mu }_n⟨x,e_n⟩e_n$ for $x$ in this set. First let $x\in D(A)$. Then $Ax\in H$ and, by self-adjointness of $A$ and step 2.1, $⟨Ax,e_n⟩=⟨x,A{e}_n⟩={\mu }_n⟨x,e_n⟩$ for every $n$; since $(e_n)$ is a complete orthonormal family, Parseval gives ${\sum }_n{\mu }_n^2\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2=\mathrm{\parallel }Ax{\mathrm{\parallel }}^2<\mathrm{\infty }$ and $Ax={\sum }_n⟨Ax,e_n⟩e_n={\sum }_n{\mu }_n⟨x,e_n⟩e_n$. Conversely, let $x\in H$ satisfy ${\sum }_n{\mu }_n^2\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2<\mathrm{\infty }$ and put $y:={\sum }_n{\mu }_n⟨x,e_n⟩e_n\in H$; the series converges because the coefficient sequence is square summable. We claim $⟨x,A\phi ⟩=⟨y,\phi ⟩$ for every $\phi \in D(A)$: the identity holds for every $\phi =e_m$, because $⟨x,A{e}_m⟩={\mu }_m⟨x,e_m⟩=⟨y,e_m⟩$ by step 2.1 and the definition of $y$; moreover the span $Z$ of the $e_m$ is contained in $D(A)$, and it is graph-norm dense in $D(A)$: for $x^{\mathrm{\prime }}\in D(A)$ the truncations $x_N^{\mathrm{\prime }}={\sum }_{n\le N}⟨x^{\mathrm{\prime }},e_n⟩e_n$ satisfy $x_N^{\mathrm{\prime }}\to x^{\mathrm{\prime }}$ and $A{x}_N^{\mathrm{\prime }}={\sum }_{n\le N}{\mu }_n⟨x^{\mathrm{\prime }},e_n⟩e_n\to A{x}^{\mathrm{\prime }}$ by the first part of this step. Both sides of the claimed identity are continuous on $(D(A),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_A)$, since $\mathrm{\mid }⟨x,A\phi ⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }\phi {\mathrm{\parallel }}_A$ and $\mathrm{\mid }⟨y,\phi ⟩\mathrm{\mid }\le \mathrm{\parallel }y\mathrm{\parallel }\mathrm{\parallel }\phi {\mathrm{\parallel }}_A$ (Unbounded linear operators: domain, graph and extension); hence the identity extends from the dense subspace $Z$ to all of $D(A)$. Thus the functional $\phi \mapsto ⟨x,A\phi ⟩$ equals the bounded functional $\phi \mapsto ⟨y,\phi ⟩$ on $D(A)$, so $x\in D(A^{\ast })=D(A)$ with $A^{\ast }x=y$, and $Ax=A^{\ast }x=y$. This proves the claim.

[Spectrum] By step 3.1, $A$ is unitarily equivalent to the diagonal operator with entries ${\mu }_n$, so [F3] gives $\sigma (A)=\overline{\{{\mu }_n:n\in \mathbb{N}\}}=\{{\mu }_n:n\in \mathbb{N}\}$, the closure being the set itself because $\mathrm{\mid }{\mu }_n\mathrm{\mid }\to \mathrm{\infty }$ (a sequence tending to infinity has no finite accumulation point). Together with step 2.1 this is (a) and the first half of (b): $H$ has a complete orthonormal family of eigenvectors of $A$ with $\mathrm{\mid }{\mu }_n\mathrm{\mid }\to \mathrm{\infty }$, and the spectrum is exactly $\{{\mu }_n\}$.

[Essential spectrum] Every ${\mu }_n$ is an isolated point of $\sigma (A)=\{{\mu }_n\}$ (the set is discrete in $\mathbb{R}$ because $\mathrm{\mid }{\mu }_n\mathrm{\mid }\to \mathrm{\infty }$), and its eigenspace is finite dimensional by step 2.1. Hence every ${\mu }_n$ belongs to the discrete spectrum in the sense of Discrete and essential spectrum of a self-adjoint operator, and ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(A)=\sigma (A)\setminus {\sigma }_{\mathrm{d}}(A)=\mathrm{\varnothing }$. This completes (b).

[All resolvents are compact] Let $z\in \rho (A)$ and let $R_z:=(A-z{)}^{-1}$. Subtracting the inverse identities for $A-z$ and $A-c$ gives $R_z-R_c=(z-c)R_c{R}_z$, hence $R_z=R_c[I+(z-c)R_z]$. The factor $I+(z-c)R_z$ is bounded, so [F4] makes $R_z$ compact as a product with the compact $R_c$. This is (c).