Injective resolvents need not be bounded below

On $H={\mathrm{\ell }}^2({\mathbb{N}}_{>0})=\{x:{\mathbb{N}}_{>0}\to \mathbb{C}:{\sum }_{n\ge 1}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<\mathrm{\infty }\}$, with inner product $⟨x,y⟩={\sum }_{n\ge 1}{x}_n\overline{y_n}$ and standard basis $e_n$ given by $(e_n{)}_m=1$ if $m=n$ and $0$ otherwise, all coordinate indices below range over ${\mathbb{N}}_{>0}$. Define $D(A)=\{x=(x_n):{\sum }_{n\ge 1}{n}^2\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<\mathrm{\infty }\}$ and $(Ax{)}_n=n{x}_n$. Finite sequences lie in $D(A)$ and are dense. For $x\in D(A)$, Cauchy–Schwarz [F2] gives $\sum n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\le (\sum n^2\mathrm{\mid }{x}_n{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}(\sum \mathrm{\mid }{x}_n{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$, so $⟨Ax,x⟩=\sum n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\ge 0$ and $A$ is nonnegative.

The operator is self-adjoint. If $y\in D(A^{\ast })$, there is $z\in {\mathrm{\ell }}^2$ with $⟨Ax,y⟩=⟨x,z⟩$ for every finite sequence $x$; choosing $x=e_n$ gives $z_n=n{y}_n$, so $\sum n^2\mathrm{\mid }{y}_n{\mathrm{\mid }}^2=\mathrm{\parallel }z{\mathrm{\parallel }}^2<\mathrm{\infty }$ and $y\in D(A)$. Conversely, if $y\in D(A)$ then $z=(n{y}_n)\in {\mathrm{\ell }}^2$ satisfies the adjoint identity for every $x\in D(A)$: both inner-product series converge by [F2] because $Ax,y,x,z\in {\mathrm{\ell }}^2$, and their terms agree. Thus $D(A^{\ast })=D(A)$ and $A^{\ast }=A$.

The inverse of $I+A$ is the bounded diagonal operator $B{e}_n=(n+1{)}^{-1}{e}_n$. Indeed for any $x\in {\mathrm{\ell }}^2$, $y_n=x_n\mathrm{/}(n+1)$ satisfies $\sum n^2\mathrm{\mid }{y}_n{\mathrm{\mid }}^2\le \sum \mathrm{\mid }{x}_n{\mathrm{\mid }}^2$, and $(I+A)y=x$. The operator $B$ is positive and injective, but it is not bounded below because $\mathrm{\parallel }B{e}_n\mathrm{\parallel }=(n+1{)}^{-1}\to 0$ for unit vectors $e_n$. The vector $y=(1\mathrm{/}n{)}_{n\ge 1}$ belongs to ${\mathrm{\ell }}^2$: for $n\ge 2$, $n^{-2}\le (n-1{)}^{-1}-n^{-1}$, so ${\sum }_{n\ge 1}{n}^{-2}\le 2$. It is not in $\mathrm{Ran}B$, since $Bx=y$ would require $x_n=(n+1)\mathrm{/}n>1$ for every $n$, which is not square-summable. Hence $B$ is not surjective and $0\in \sigma (B)$.

Define $E_B(S)x={\sum }_{n:(n+1{)}^{-1}\in S}{x}_n{e}_n$ for Borel $S\subseteq \sigma (B)$. Its values are orthogonal coordinate projections, and the coordinate definition gives $E_B(\mathrm{\varnothing })=0$, $E_B(\sigma (B))=I$, and $E_B(S)E_B(T)=E_B(S\cap T)$. For disjoint Borel $S_k$, each coordinate $n$ occurs in at most one projection, so the partial sums ${\sum }_{k\le N}{E}_B(S_k)x$ converge in ${\mathrm{\ell }}^2$ to $E_B({\bigcup }_k{S}_k)x$: the squared norm of the omitted coordinates is a tail of the convergent nonnegative series ${\sum }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2$. Thus $E_B$ is strongly countably additive. It integrates the coordinate function to $B$, because its integral acts diagonally by $(n+1{)}^{-1}$ on each $e_n$. For every $x$, its finite scalar measure is ${\mu }_x(S):=⟨E_B(S)x,x⟩={\sum }_{n:(n+1{)}^{-1}\in S}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2$. For any Borel $S\subseteq \sigma (B)$, the finite set $K_N:=\{(n+1{)}^{-1}:n\le N,(n+1{)}^{-1}\in S\}$ is compact and contained in $S$, and ${\mu }_x(K_N)\uparrow {\mu }_x(S)$ because ${\sum }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<\mathrm{\infty }$. Hence ${\mu }_x$ is inner regular on every Borel set. Since $\sigma (B)$ is compact, for any Borel $S$ and $\epsilon >0$, inner regularity of $S^c$ gives a compact $K\subseteq S^c$ with ${\mu }_x(S^c\setminus K)<\epsilon$; then the open set $\sigma (B)\setminus K$ contains $S$ and its excess measure is less than $\epsilon$. Thus ${\mu }_x$ is also outer regular, so $E_B$ is a regular PVM. Hence [F1] identifies it with the bounded-resolvent spectral PVM. Since no atom is at zero, $E_B(\{0\})=0$, although step 3.1 proves $0\in \sigma (B)$. This refutes the asserted implication and shows why a zero spectral projection must be distinguished from a spectral value.