Closed adjoint sector domains

[The raising operator and its closure] For every $n$ the map $\psi \mapsto P_{+}^{(n+1)}(f\otimes {\psi }_n)$ is bounded with norm at most $\mathrm{\parallel }f\mathrm{\parallel }$ by [F2], so $D(A_{+}(f))$ is a linear subspace containing $F_0$ and $A_{+}(f)$ is a linear operator extending $S_{+}(f)$ with $(A_{+}(f)\psi {)}_n\in \mathrm{ran}{P}_{+}^{(n)}$ for all $\psi \in D(A_{+}(f))$ and ${\sum }_n\mathrm{\parallel }(A_{+}(f)\psi {)}_n{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $A_{+}(f)\psi \in F(H)$. If ${\psi }^{(k)}\in D(A_{+}(f))$ with ${\psi }^{(k)}\to \psi$ and $A_{+}(f){\psi }^{(k)}\to \eta$, then for each $m\ge 1$ continuity of $\psi \mapsto P_{+}^{(m)}(f\otimes {\psi }_{m-1})$ gives ${\eta }_m=\sqrt{m}{P}_{+}^{(m)}(f\otimes {\psi }_{m-1})$, and ${\eta }_0=0$. For every finite $N$, continuity of the norm and the finite sum gives ${\sum }_{n=0}^N(n+1)\mathrm{\parallel }{P}_{+}^{(n+1)}(f\otimes {\psi }_n){\mathrm{\parallel }}^2={\lim }_k{\sum }_{n=0}^N(n+1)\mathrm{\parallel }{P}_{+}^{(n+1)}(f\otimes {\psi }_n^{(k)}){\mathrm{\parallel }}^2\le {\lim }_k\mathrm{\parallel }{A}_{+}(f){\psi }^{(k)}{\mathrm{\parallel }}^2=\mathrm{\parallel }\eta {\mathrm{\parallel }}^2\mathrm{.}$ Taking the supremum over $N$ shows $\psi \in D(A_{+}(f))$; the coordinate formulas then give $A_{+}(f)\psi =\eta$, so $A_{+}(f)$ is closed. For $\psi \in D(A_{+}(f))$ the truncations ${\psi }^{(N)}$ lie in $F_0$, satisfy ${\psi }^{(N)}\to \psi$, and $A_{+}(f){\psi }^{(N)}\to A_{+}(f)\psi$ because the defining series of $\psi$ converges; hence $F_0$ is a core and $A_{+}(f)=\overline{S_{+}(f)}$.

[The lowering operator is the core adjoint] For $\psi \in F(H)$ set ${\eta }_n:=\sqrt{n+1}{\mathrm{\Psi }}_f^{(n)}{\psi }_{n+1}$. For $\phi \in F_0$, [F2] gives $⟨S_{+}(f)\phi ,\psi ⟩={\sum }_{n\ge 0}⟨{\phi }_n,{\eta }_n⟩$, a finite sum because $\phi$ has finite support. If ${\sum }_n\mathrm{\parallel }{\eta }_n{\mathrm{\parallel }}^2<\mathrm{\infty }$, Cauchy--Schwarz shows this functional is bounded on $F_0$ and its representing vector is $({\eta }_n{)}_n=A_{-}(f)\psi$. Conversely, if the functional is bounded with norm $C$, let ${\phi }^{(N)}$ have coordinates ${\eta }_n$ for $n\le N$ and zero thereafter. Then ${\phi }^{(N)}\in F_0$ and $S_N:={\sum }_{n\le N}\mathrm{\parallel }{\eta }_n{\mathrm{\parallel }}^2=\mathrm{\mid }⟨S_{+}(f){\phi }^{(N)},\psi ⟩\mathrm{\mid }\le C\mathrm{\parallel }{\phi }^{(N)}\mathrm{\parallel }=C\sqrt{S_N}$, so $S_N\le C^2$ for every $N$. Hence $({\eta }_n{)}_n\in F(H)$ and the functional is represented by $A_{-}(f)\psi$. Thus $\psi \in D(S_{+}(f{)}^{\ast })$ exactly when ${\sum }_n(n+1)\mathrm{\parallel }{\mathrm{\Psi }}_f^{(n)}{\psi }_{n+1}{\mathrm{\parallel }}^2<\mathrm{\infty }$, and $S_{+}(f{)}^{\ast }=A_{-}(f)$. As an adjoint $A_{-}(f)$ is closed, and it is densely defined because it contains $F_0$.

[Weighted domain and bounds] For $\psi \in D_1$, ${\sum }_n(n+1)\mathrm{\parallel }{P}_{+}^{(n+1)}(f\otimes {\psi }_n){\mathrm{\parallel }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}^2{\sum }_n(n+1)\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2<\mathrm{\infty }$ and likewise ${\sum }_n(n+1)\mathrm{\parallel }{\mathrm{\Psi }}_f^{(n)}({\psi }_{n+1}){\mathrm{\parallel }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}^2{\sum }_n(n+1)\mathrm{\parallel }{\psi }_{n+1}{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $D_1\subseteq D(A_{\pm }(f))$ and the bounds of (c) follow from the same estimates; since $F_0\subseteq D_1$, this also gives the density in (b).

[Mutual adjoints] Since $A_{+}(f)=\overline{S_{+}(f)}=S_{+}(f{)}^{\ast \ast }$ by step 1.1, [F3] gives $A_{+}(f{)}^{\ast }=S_{+}(f{)}^{\ast \ast \ast }=S_{+}(f{)}^{\ast }=A_{-}(f)$ by step 1.2. Applying the same step 1.2 with the roles of $S_{+}$ and $S_{-}$ exchanged (the adjoint relation of [F2] is symmetric in the two core operators) gives $S_{-}(f{)}^{\ast }=A_{+}(f)$, hence $A_{-}(f{)}^{\ast }=S_{+}(f{)}^{\ast \ast }=A_{+}(f)$. Consequently both operators are closed with the stated adjoint pairs, proving (a) and (b) apart from the domain inclusion.

[The weighted norm identity] Let $\psi \in D_1$ and let ${\psi }^{(N)}$ be its truncations; then ${\psi }^{(N)}\to \psi$ and, by the sector formulas and the convergence of the defining sums of $\psi$, $A_{\pm }(g){\psi }^{(N)}\to A_{\pm }(g)\psi$. The identity of Creation contraction bounds (d) applies to ${\psi }^{(N)}\in F_0$; passing to the limit in $\mathrm{\parallel }{A}_{+}(g){\psi }^{(N)}{\mathrm{\parallel }}^2=\mathrm{\parallel }{A}_{-}(g){\psi }^{(N)}{\mathrm{\parallel }}^2+\mathrm{\parallel }g{\mathrm{\parallel }}^2\mathrm{\parallel }{\psi }^{(N)}{\mathrm{\parallel }}^2$ using continuity of the norm and of addition in $F(H)$ gives the displayed identity for $\psi$. This proves (d) and completes the proof.