Positive scalar multiples of self-adjoint operators: domains, resolvents, spectra and eigenbases

[Symmetry] For $x,y\in D(A)=D(cA)$, [F2] gives $⟨(cA)x,y⟩=c⟨Ax,y⟩=c⟨x,Ay⟩=⟨x,cAy⟩=⟨x,(cA)y⟩$; since $c$ is real, the scalar passes through the pairing in both slots, so $cA$ is symmetric.

[Resolvent identity] Fix $z\in \mathbb{C}$ and write $B:=cA-z$ and $B^{\mathrm{\prime }}:=A-z\mathrm{/}c$ on the common domain $D(A)$. Then $B=c{B}^{\mathrm{\prime }}$ as operators, since $c(A-z\mathrm{/}c)x=cAx-zx=(cA-z)x$ for every $x\in D(A)$. If $z\mathrm{/}c\in \rho (A)$, then $B^{\mathrm{\prime }}$ is a bijection $D(A)\to H$ with bounded inverse, so $B=c{B}^{\mathrm{\prime }}$ is a bijection with inverse $c^{-1}{B}^{\mathrm{\prime }-1}\in \mathcal{B}(H)$, which is the displayed identity; conversely if $z\in \rho (cA)$, then $c^{-1}B=B^{\mathrm{\prime }}$ is a bijection with bounded inverse $c{B}^{-1}$, so $z\mathrm{/}c\in \rho (A)$. Therefore $\rho (cA)=c\rho (A)$ and, taking complements in $\mathbb{C}$, $\sigma (cA)=c\sigma (A)$; this is (b).

[Eigenpairs] Let $\lambda \in \mathbb{R}$ and $x\in H$, $x\mathrm{\ne }0$. Since $D(cA)=D(A)$, $x\in D(A)$ if and only if $x\in D(cA)$. For such $x$, $Ax=\lambda x$ if and only if $cAx=c\lambda x$, because multiplication by $c\mathrm{\ne }0$ is injective; hence the eigenspaces coincide, which is (d).

[Adjoint domain] Let $y\in D((cA{)}^{\ast })$. Then $x\mapsto ⟨(cA)x,y⟩=c⟨Ax,y⟩$ is bounded on $D(A)$, and since $c>0$, $x\mapsto ⟨Ax,y⟩$ is bounded on $D(A)$ as well; hence $y\in D(A^{\ast })=D(A)$ and $A^{\ast }y=Ay$. For every $x\in D(A)$ the defining identity of the adjoint and [F2] then give $⟨x,(cA{)}^{\ast }y⟩=⟨(cA)x,y⟩=c⟨Ax,y⟩=c⟨x,Ay⟩=⟨x,cAy⟩$, and density of $D(A)$ in $H$ gives $(cA{)}^{\ast }y=cAy$. Hence $D((cA{)}^{\ast })\subseteq D(cA)$; the reverse inclusion holds because $cA$ is symmetric, so $D((cA{)}^{\ast })=D(cA)$ and $(cA{)}^{\ast }=cA$. This proves (a).

[All resolvents are compact] Write $R_z=(A-z{)}^{-1}$ and suppose $R_{z_0}$ is compact. The resolvent identity, obtained by subtracting the two inverses on their common operator domain, is $R_z=R_{z_0}+(z-z_0)R_{z_0}{R}_z=R_{z_0}[I+(z-z_0)R_z]$. Thus every $R_z$ is compact by [F5], irrespective of whether $z_0$ is real. Step 1.2 now gives compactness of every resolvent of $cA$.

[Finite-dimensional case] If $\dim H=N<\mathrm{\infty }$, the dense linear domain $D(A)$ equals $H$, since every finite-dimensional subspace is closed (A finite-dimensional normed subspace is closed). The operator $A$ is then bounded and compact, and its self-adjointness and [F5] give an orthonormal eigenbasis of precisely $N$ vectors, including a basis of its zero eigenspace if necessary. In that basis $A$ is the diagonal matrix with real entries ${\mu }_j$, so its spectrum is exactly those entries: off that set the diagonal inverse exists and is bounded, while at an entry there is a nonzero kernel vector. Step 1.3 gives the same family for $cA$, and step 1.2 gives the scaled spectrum.

[Infinite-dimensional eigenfamily from an arbitrary compact resolvent] Suppose $H$ is infinite dimensional. By step 2.2 both $R_i$ and $R_{-i}$ are compact; their adjoint relation $R_i^{\ast }=R_{-i}$ follows from the defining self-adjoint pairings, and the resolvent identity shows they commute. Hence $K:=R_{-i}{R}_i=R_i^{\ast }{R}_i$ is compact, self-adjoint, positive and injective. Its range lies in $D(A^2)$: $R_i$ maps into $D(A)$, $R_{-i}$ preserves $D(A)$ and commutes there with $A$, by $(A-z)R_z=I$ and $R_z(A-z)=I$ on $D(A)$. [F5] decomposes $H$ into the closed span of its finite-dimensional positive $K$-eigenspaces $E_{\nu }$, with no kernel part. For $e\in E_{\nu }$, $e={\nu }^{-1}Ke\in D(A^2)$ and $K(Ae)=A(Ke)=\nu Ae$, so $E_{\nu }$ is invariant under $A$. The restriction of $A$ there is a finite-dimensional self-adjoint operator; step 2.3 gives an orthonormal eigenbasis of that restriction. The union is a complete orthonormal eigenfamily of $A$. If $Ae=\mu e$, direct resolvent evaluation gives $Ke=({\mu }^2+1{)}^{-1}e$. There are only finitely many such vectors with $\mathrm{\mid }\mu \mathrm{\mid }\le M$ for each fixed $M$, since [F5] gives only finitely many $K$-eigenvalues above $(M^2+1{)}^{-1}$ and each has finite multiplicity. The union is countable and infinite, hence can be indexed by $\mathbb{N}$ under the retained choice assumptions, and $\mathrm{\mid }{\mu }_n\mathrm{\mid }\to \mathrm{\infty }$.

[No additional spectrum] In the eigenbasis of step 3.1, self-adjointness gives $⟨Ax,e_n⟩={\mu }_n⟨x,e_n⟩$ for $x\in D(A)$, so ${\sum }_n{\mu }_n^2\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2=\mathrm{\parallel }Ax{\mathrm{\parallel }}^2$. Conversely, if this sum is finite, the finite sums $x_N$ and $A{x}_N$ converge to $x$ and $y:={\sum }_n{\mu }_n⟨x,e_n⟩e_n$; passing to the limit in $⟨A{x}_N,v⟩=⟨x_N,Av⟩$ for every $v\in D(A)$ puts $x$ in $D(A^{\ast })=D(A)$ and gives $Ax=y$. Thus $A$ has exactly this maximal diagonal domain. Since $\mathrm{\mid }{\mu }_n\mathrm{\mid }\to \mathrm{\infty }$, a complex $z$ outside $\{{\mu }_n\}$ has positive distance from that set. The diagonal map with entries $({\mu }_n-z{)}^{-1}$ is bounded and maps $H$ into the displayed domain because ${\sup }_n\mathrm{\mid }{\mu }_n\mathrm{/}({\mu }_n-z)\mathrm{\mid }<\mathrm{\infty }$; it is the two-sided inverse of $A-z$. Every ${\mu }_n$ is an eigenvalue. This proves $\sigma (A)=\{{\mu }_n\}$, and steps 1.2 and 1.3 give the scaled spectrum and common eigenfamily, completing (c).

[Quadratic forms] If $⟨Ax,x⟩\ge \gamma \mathrm{\parallel }x{\mathrm{\parallel }}^2$ for every $x\in D(A)$, then for $x\in D(cA)=D(A)$ the same inequality scaled by $c>0$ gives $⟨(cA)x,x⟩=c⟨Ax,x⟩\ge c\gamma \mathrm{\parallel }x{\mathrm{\parallel }}^2$; this is (e), and steps 2.1, 1.2, 2.3, 4.1 and 1.3 complete the proof.