Closed antilinear polar calculus and Tomita-operator closability

Put $C=c_H$ and $T=\bar{S}$. The definition of graph closure above makes $A:=\tilde{T}$ a closed densely defined complex-linear operator $\bar{H}\to H$. By [F2], write $A=UB$, where $B=(A^{\ast }A{)}^{1\mathrm{/}2}$ is positive self-adjoint with $D(B)=D(A)$ and $U$ is a partial isometry with initial space $\overline{\mathrm{ran}B}=(\ker A{)}^{\perp }$, final space $\overline{\mathrm{ran}A}$, and kernel $\ker A$. Set $J:=UC$ and $P:=C^{-1}BC$, with domain $D(P)=C^{-1}D(B)$. The composition $P$ is linear. If $E$ is the spectral measure of $B$, then $F(\mathrm{\Lambda }):=C^{-1}E(\mathrm{\Lambda })C$ is a spectral measure of orthogonal projections on $H$, and its spectral integral is $P$; hence $P$ is positive self-adjoint. More explicitly, $D(T^{\ast }T)=\{x:Cx\in D(A),ACx\in D(A^{\ast })\}=C^{-1}D(A^{\ast }A)$, and the adjoint convention [F1] gives $T^{\ast }T=C^{-1}{A}^{\ast }AC$ there. Also $D(P^2)=C^{-1}D(B^2)$ and $P^2=C^{-1}{B}^{2}C=C^{-1}{A}^{\ast }AC=T^{\ast }T$ on that domain; uniqueness of the positive self-adjoint square root gives $P=(T^{\ast }T{)}^{1\mathrm{/}2}$. For $x\in D(T)$, $JPx=UC{C}^{-1}BCx=UBCx=ACx=Tx$, so $T=JP$. The operator $J$ is antilinear, vanishes on $\ker T$, and is isometric on $(\ker T{)}^{\perp }$, since $C$ maps $\ker T$ onto $\ker A$ and preserves orthogonality. Its final space is $\overline{\mathrm{ran}T}$. If $T$ is injective and has dense range, then $A$ is injective with dense range, so $U$ is unitary by [F2] and $J=UC$ is antiunitary.

For $x,y\in M$, if $x\mathrm{\Omega }=y\mathrm{\Omega }$, then $(x-y)\mathrm{\Omega }=0$; separatingness of $\mathrm{\Omega }$ for $M$ forces $x=y$. Thus the rule $S_0(x\mathrm{\Omega })=x^{\ast }\mathrm{\Omega }$ is well defined. It is antilinear because $(ax+by{)}^{\ast }=\bar{a}{x}^{\ast }+\bar{b}{y}^{\ast }$, and its domain is dense because $\mathrm{\Omega }$ is cyclic for $M$.

Let $u_n\in D({\tilde{S}}_0)$ satisfy $u_n\to 0$ in $\bar{H}$ and ${\tilde{S}}_0{u}_n\to \eta$ in $H$. Write $u_n=C(x_n\mathrm{\Omega })$ with $x_n\in M$; then $x_n\mathrm{\Omega }\to 0$ and $x_n^{\ast }\mathrm{\Omega }=S_0(x_n\mathrm{\Omega })={\tilde{S}}_0{u}_n\to \eta$. For every $y^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$, the linear-first adjoint identity and commutation give $⟨\eta ,y^{\mathrm{\prime }}\mathrm{\Omega }⟩={\lim }_n⟨x_n^{\ast }\mathrm{\Omega },y^{\mathrm{\prime }}\mathrm{\Omega }⟩={\lim }_n⟨y^{\mathrm{\prime }\ast }\mathrm{\Omega },x_n\mathrm{\Omega }⟩=0.$ Since $\mathrm{\Omega }$ is separating for $M$, [F3] makes it cyclic for $M^{\mathrm{\prime }}$, so $M^{\mathrm{\prime }}\mathrm{\Omega }$ is dense and $\eta =0$. Thus the graph closure of the linear operator ${\tilde{S}}_0$ has zero vertical fiber: because $\bar{H}\oplus H$ is a metric space, every point of that closure is the limit of a sequence of graph points, and the argument just given excludes a nonzero limit over $0$. This closed linear subspace is therefore the graph of a closed linear extension of ${\tilde{S}}_0$. By [F4], ${\tilde{S}}_0$ is closable, and the local convention above proves that $S_0$ is closable with graph closure defined there.

Step 1.1 proves the polar decomposition for the closure of every densely-defined closable antilinear operator, and steps 1.2–2.1 prove the Tomita operator clause with its stated cyclic-separating hypotheses. The Choice assumption is inherited through the spectral theorem used in [F2].