Modular resolvent invariance and commutant equality

Fix $\mathrm{\mid }\lambda \mathrm{\mid }=1$ with $\lambda \mathrm{\ne }-1$. Write $\lambda =e^{i\theta }$ with $-\pi <\theta <\pi$, set $\alpha =(1+\lambda {)}^{-1}$, and note $\alpha +\bar{\alpha }=1$. Because $\mathrm{\Delta }$ is positive, its spectrum lies in $[0,\mathrm{\infty })$; $-\lambda \mathrm{\notin }[0,\mathrm{\infty })$, so $(\mathrm{\Delta }+\lambda {)}^{-1}$ is a bounded spectral multiplier.

First let $x^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$ be a positive contraction. In the linear-second notation $(u\mathrm{\mid }v)=⟨v,u⟩$, define $\psi (y)=(x^{\mathrm{\prime }}\mathrm{\Omega }\mathrm{\mid }y\mathrm{\Omega })=\omega (x^{\mathrm{\prime }}y)$ and ${\psi }_x(y)=\omega (\alpha xy+\bar{\alpha }yx)$ for $y\in M$ and $x=x^{\ast }\in M$. Since $x^{\mathrm{\prime }}$ commutes with $M$, $0\le \psi \le \omega$, and $\psi$ is normal by its rank-one trace-class representation. Each ${\psi }_x$ is self-adjoint. For fixed $y$, both $x\mapsto \omega (xy)$ and $x\mapsto \omega (yx)$ are normal matrix-coefficient functionals, so $x\mapsto {\psi }_x$ is continuous from the weak-star topology of $M_{\mathrm{s}\mathrm{a}}$ to the pointwise topology of $M_{\ast }^{\mathrm{s}\mathrm{a}}$. The self-adjoint unit ball of $M$ is weak-star closed; [F3]–[F4] make it compact. Hence $V=\{{\psi }_x:x=x^{\ast },\mathrm{\parallel }x\mathrm{\parallel }\le 1\}$ is compact and convex.

Suppose $\psi \mathrm{\notin }V$. The pointwise topology on $M_{\ast }^{\mathrm{s}\mathrm{a}}$ has a base of finite evaluation conditions, so there are self-adjoint $a_1,\dots ,a_m\in M$ for which $q=(\psi (a_j){)}_{j=1}^m$ is outside the compact convex set $K=\{(\varphi (a_j){)}_{j=1}^m:\varphi \in V\}\subset {\mathbb{R}}^m$. Choose $0<r<\mathrm{dist}(q,K)$, put $U=K+B(0,r)$, choose $u_0\in U$, and set $C=U-u_0$. The open convex neighborhood $C$ of $0$ is absorbing; its gauge $p(v)=\inf \{s>0:v\in sC\}$ is finite and sublinear, and convexity and openness give $C=\{v:p(v)<1\}$. With $v_0=q-u_0\mathrm{\notin }C$, $p(v_0)\ge 1$. On the line $\mathbb{R}{v}_0$ define $f(t{v}_0)=tp(v_0)$. Positive homogeneity gives $f\le p$ for $t\ge 0$; for $t<0$ the inequality $p(v_0)+p(-v_0)\ge 0$ gives the same domination. By [F5], extend $f$ to a real linear $L$ on ${\mathbb{R}}^m$ with $L\le p$. Then $L(k-u_0)<1\le L(q-u_0)$ for every $k\in K$. Writing $L(z)={\sum }_j{c}_j{z}_j$ and $y_0={\sum }_j{c}_j{a}_j\in M_{\mathrm{s}\mathrm{a}}$ gives $\psi (y_0)>{\sup }_{\mathrm{\parallel }x\mathrm{\parallel }\le 1,x=x^{\ast }}{\psi }_x(y_0)$.

Fix $\mathrm{\mid }\lambda \mathrm{\mid }=1$, $\lambda =e^{i\theta }$, $-\pi <\theta <\pi$. For $n\ge 1$ let $E_n=1_{[1\mathrm{/}n,n]}(\mathrm{\Delta })$. On $E_{n}H$, $\mathrm{\Delta }$ and its inverse are bounded. Apply [F8] to each scalar matrix coefficient of $f_n(z)=\frac{e^{i\theta z}}{\sin (\pi z)}{\mathrm{\Delta }}^{-z}{E}_{n}x{E}_n{\mathrm{\Delta }}^z$ around the rectangle $-1\mathrm{/}2\le \mathrm{Re}z\le 1\mathrm{/}2$, $\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }\le R$. On $E_{n}H$, $\log \mathrm{\Delta }$ is bounded, so $f_n$ is norm-meromorphic with simple poles at integers; the rectangle encloses only $0$, with residue $E_{n}x{E}_n\mathrm{/}\pi$. On the horizontal sides, the norm is at most $2e^{\mathrm{\mid }\theta \mathrm{\mid }R}{n}^2\mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}(e^{\pi R}-e^{-\pi R})$, which tends to zero. On the right vertical side, $z=it+1\mathrm{/}2$ gives $\sin (\pi z)=\cosh (\pi t)$, $e^{i\theta z}=e^{-\theta t}{e}^{i\theta \mathrm{/}2}$, and ${\mathrm{\Delta }}^{-z}{E}_{n}x{E}_n{\mathrm{\Delta }}^z={\mathrm{\Delta }}^{-1\mathrm{/}2}{\mathrm{\Delta }}^{-it}{E}_{n}x{E}_n{\mathrm{\Delta }}^{it}{\mathrm{\Delta }}^{1\mathrm{/}2}$. On the left side, $z=it-1\mathrm{/}2$ gives $\sin (\pi z)=-\cosh (\pi t)$ and the corresponding factors ${\mathrm{\Delta }}^{1\mathrm{/}2}{\mathrm{\Delta }}^{-it}{E}_{n}x{E}_n{\mathrm{\Delta }}^{it}{\mathrm{\Delta }}^{-1\mathrm{/}2}$ and $e^{-\theta t}{e}^{-i\theta \mathrm{/}2}$. Substitution in the oriented contour formula therefore gives $E_{n}x{E}_n={\mathrm{\Delta }}^{1\mathrm{/}2}{E}_n\tilde{x}{E}_n{\mathrm{\Delta }}^{-1\mathrm{/}2}+\lambda {\mathrm{\Delta }}^{-1\mathrm{/}2}{E}_n\tilde{x}{E}_n{\mathrm{\Delta }}^{1\mathrm{/}2},\tilde{x}=e^{-i\theta \mathrm{/}2}{\int }_{\mathbb{R}}\frac{e^{-\theta t}}{e^{\pi t}+e^{-\pi t}}{\mathrm{\Delta }}^{-it}x{\mathrm{\Delta }}^{it}dt\mathrm{.}$ The scalar weight is integrable because $\mathrm{\mid }\theta \mathrm{\mid }<\pi$. For fixed $\xi \in H$, the orbit $t\mapsto {\mathrm{\Delta }}^{-it}x{\mathrm{\Delta }}^{it}\xi$ is norm-continuous: writing $U_t={\mathrm{\Delta }}^{-it}$, $\mathrm{\parallel }{U}_{t}x{U}_t^{\ast }\xi -U_{s}x{U}_s^{\ast }\xi \mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }(U_t^{\ast }-U_s^{\ast })\xi \mathrm{\parallel }+\mathrm{\parallel }(U_t-U_s)x{U}_s^{\ast }\xi \mathrm{\parallel },$ and each term tends to zero as $t\to s$ by [F10]. Its norm is at most $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }\xi \mathrm{\parallel }$. Define the integral as the norm limit of ordinary vector integrals over finite intervals; the integrable scalar weight makes the tails Cauchy. The resulting linear map has norm at most $\mathrm{\parallel }x\mathrm{\parallel }$ times the weight's $L^1$ norm, so it is a bounded operator. This contour identity has the normalization fixed by the residue $E_{n}x{E}_n\mathrm{/}\pi$.

The last strict inequality is impossible. Positivity and $0\le \psi \le \omega$ imply $\psi (y_0)\le \psi (\mathrm{\mid }{y}_0\mathrm{\mid })\le \omega (\mathrm{\mid }{y}_0\mathrm{\mid })$. For $n\ge 1$, let $x_n=f_n(y_0)$ by [F6], where $f_n(s)=s\mathrm{/}(\mathrm{\mid }s\mathrm{\mid }+1\mathrm{/}n)$. Each $x_n$ is self-adjoint with $\mathrm{\parallel }{x}_n\mathrm{\parallel }\le 1$ and commutes with $y_0$. The positive functions $g_n(s)=s^2\mathrm{/}(\mathrm{\mid }s\mathrm{\mid }+1\mathrm{/}n)$ increase to $\mathrm{\mid }s\mathrm{\mid }$, so $x_n{y}_0=y_0{x}_n=g_n(y_0)\uparrow \mathrm{\mid }{y}_0\mathrm{\mid }$. Normality gives ${\psi }_{x_n}(y_0)=\omega (g_n(y_0))\uparrow \omega (\mathrm{\mid }{y}_0\mathrm{\mid })$, contradicting the strict separator. Therefore $\psi ={\psi }_x$ for some self-adjoint contraction $x$.

Apply the same residue calculation to $g_n(z)=e^{i\theta z}{\mathrm{\Delta }}^{-z}{E}_n\mathrm{/}\sin (\pi z)$, whose residue at $0$ is $E_n\mathrm{/}\pi$. On the right vertical side its multiplier is $2e^{-\theta t}{e}^{i\theta \mathrm{/}2}{\mathrm{\Delta }}^{-1\mathrm{/}2}{\mathrm{\Delta }}^{-it}{E}_n\mathrm{/}(e^{\pi t}+e^{-\pi t})$; on the left it is $-2e^{-\theta t}{e}^{-i\theta \mathrm{/}2}{\mathrm{\Delta }}^{1\mathrm{/}2}{\mathrm{\Delta }}^{-it}{E}_n\mathrm{/}(e^{\pi t}+e^{-\pi t})$. Including the opposite orientations of the two vertical sides and using $\lambda =e^{i\theta }$ yields ${\mathrm{\Delta }}^{1\mathrm{/}2}(\mathrm{\Delta }+\lambda {)}^{-1}{E}_n=e^{-i\theta \mathrm{/}2}\left({\int }_{\mathbb{R}}\frac{e^{-\theta t}}{e^{\pi t}+e^{-\pi t}}{\mathrm{\Delta }}^{-it}dt\right)E_n\mathrm{.}$ This follows by factoring ${\mathrm{\Delta }}^{-1\mathrm{/}2}(\mathrm{\Delta }+\lambda )={\mathrm{\Delta }}^{1\mathrm{/}2}+\lambda {\mathrm{\Delta }}^{-1\mathrm{/}2}$ from the vertical-side integral. Define the strong integral vectorwise as in step 1.4. The bounded spectral multiplier on the left and the integrable weight on the right allow $n\to \mathrm{\infty }$ strongly, so the identity holds with $E_n$ omitted. In particular the scalar factor has no additional $i$: at $\lambda =1$ and $\mathrm{\Delta }=I$ both sides equal $1\mathrm{/}2$.

For every $y\in M$, $\psi ={\psi }_x$ and $\alpha +\bar{\alpha }=1$ give $(x^{\mathrm{\prime }}\mathrm{\Omega }-\bar{\alpha }x\mathrm{\Omega }\mathrm{\mid }y\mathrm{\Omega })=\omega (x^{\mathrm{\prime }}y)-\alpha \omega (xy)=\bar{\alpha }\omega (yx)=(S_{0}y\mathrm{\Omega }\mathrm{\mid }\bar{\alpha }x\mathrm{\Omega })\mathrm{.}$ The antilinear-adjoint definition therefore puts $\bar{\alpha }x\mathrm{\Omega }$ in $D(F)$ and gives $F(\bar{\alpha }x\mathrm{\Omega })=x^{\mathrm{\prime }}\mathrm{\Omega }-\bar{\alpha }x\mathrm{\Omega }$. Since $\bar{\alpha }\mathrm{\ne }0$ and $D(F)$ is linear, $x\mathrm{\Omega }\in D(F)$ and antilinearity gives $\alpha F(x\mathrm{\Omega })=x^{\mathrm{\prime }}\mathrm{\Omega }-\bar{\alpha }x\mathrm{\Omega }$. Since $x=x^{\ast }$, $Sx\mathrm{\Omega }=x\mathrm{\Omega }$; the polar identity $\mathrm{\Delta }=S^{\ast }S=FS$ then gives $x\mathrm{\Omega }\in D(\mathrm{\Delta })$ and $F(x\mathrm{\Omega })=\mathrm{\Delta }x\mathrm{\Omega }$. Thus $x^{\mathrm{\prime }}\mathrm{\Omega }=(\alpha \mathrm{\Delta }+\bar{\alpha })x\mathrm{\Omega }=(\mathrm{\Delta }+\lambda )x\mathrm{\Omega }\mathrm{/}(1+\lambda )$. Replacing $x$ by $x\mathrm{/}(1+\lambda )$ proves $(\mathrm{\Delta }+\lambda {)}^{-1}{x}^{\mathrm{\prime }}\mathrm{\Omega }\in M\mathrm{\Omega }$. For a general $x^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$, write $x^{\mathrm{\prime }}=h_1+i{h}_2$ with $h_j=h_j^{\ast }$. If $c\ge \max (1,\mathrm{\parallel }{h}_1\mathrm{\parallel },\mathrm{\parallel }{h}_2\mathrm{\parallel })$, each $(I\pm h_j\mathrm{/}c)\mathrm{/}2$ is a positive contraction, so $x^{\mathrm{\prime }}$ is a complex linear combination of positive contractions. Linearity extends the inclusion to all of $M^{\mathrm{\prime }}\mathrm{\Omega }$. Uniqueness follows because $\mathrm{\Delta }+\lambda$ is injective and $\mathrm{\Omega }$ is separating.

Let $x^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$ and $x\in M$ satisfy $x^{\mathrm{\prime }}\mathrm{\Omega }=(\mathrm{\Delta }+\lambda )x\mathrm{\Omega }$. For $y_1,y_2\in M$, commutation, $\mathrm{\Delta }=FS$, $S(y\mathrm{\Omega })=y^{\ast }\mathrm{\Omega }$, and $S=J{\mathrm{\Delta }}^{1\mathrm{/}2}={\mathrm{\Delta }}^{-1\mathrm{/}2}J$ give $(y_1\mathrm{\Omega }\mathrm{\mid }{x}^{\mathrm{\prime }}{y}_2\mathrm{\Omega })=(y_2^{\ast }{y}_1\mathrm{\Omega }\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\Omega })=(Sx\mathrm{\Omega }\mathrm{\mid }S(y_2^{\ast }{y}_1\mathrm{\Omega }))+\lambda (y_1\mathrm{\Omega }\mathrm{\mid }{y}_{2}x\mathrm{\Omega })=(y_1{x}^{\ast }\mathrm{\Omega }\mathrm{\mid }{y}_2\mathrm{\Omega })+\lambda (y_1\mathrm{\Omega }\mathrm{\mid }S(x^{\ast }S(y_2\mathrm{\Omega })))=({\mathrm{\Delta }}^{-1\mathrm{/}2}JxJ{\mathrm{\Delta }}^{1\mathrm{/}2}{y}_1\mathrm{\Omega }\mathrm{\mid }{y}_2\mathrm{\Omega })+\lambda (y_1\mathrm{\Omega }\mathrm{\mid }{\mathrm{\Delta }}^{-1\mathrm{/}2}J{x}^{\ast }J{\mathrm{\Delta }}^{1\mathrm{/}2}{y}_2\mathrm{\Omega })\mathrm{.}$ If $y_1\mathrm{\Omega },y_2\mathrm{\Omega }\in (\mathrm{\Delta }+1{)}^{-1}{M}^{\mathrm{\prime }}\mathrm{\Omega }$, then both lie in $D({\mathrm{\Delta }}^{1\mathrm{/}2})\cap D({\mathrm{\Delta }}^{-1\mathrm{/}2})$. Since $JxJ$ is bounded with adjoint $J{x}^{\ast }J$, $({\mathrm{\Delta }}^{-1\mathrm{/}2}JxJ{\mathrm{\Delta }}^{1\mathrm{/}2}{y}_1\mathrm{\Omega }\mathrm{\mid }{y}_2\mathrm{\Omega })=(JxJ{\mathrm{\Delta }}^{1\mathrm{/}2}{y}_1\mathrm{\Omega }\mathrm{\mid }{\mathrm{\Delta }}^{-1\mathrm{/}2}{y}_2\mathrm{\Omega })=({\mathrm{\Delta }}^{1\mathrm{/}2}{y}_1\mathrm{\Omega }\mathrm{\mid }J{x}^{\ast }J{\mathrm{\Delta }}^{-1\mathrm{/}2}{y}_2\mathrm{\Omega }),$ and self-adjointness of ${\mathrm{\Delta }}^{-1\mathrm{/}2}$ gives $(y_1\mathrm{\Omega }\mathrm{\mid }{\mathrm{\Delta }}^{-1\mathrm{/}2}J{x}^{\ast }J{\mathrm{\Delta }}^{1\mathrm{/}2}{y}_2\mathrm{\Omega })=({\mathrm{\Delta }}^{-1\mathrm{/}2}{y}_1\mathrm{\Omega }\mathrm{\mid }J{x}^{\ast }J{\mathrm{\Delta }}^{1\mathrm{/}2}{y}_2\mathrm{\Omega })\mathrm{.}$ Therefore $(y_1\mathrm{\Omega }\mathrm{\mid }{x}^{\mathrm{\prime }}{y}_2\mathrm{\Omega })=({\mathrm{\Delta }}^{1\mathrm{/}2}{y}_1\mathrm{\Omega }\mathrm{\mid }J{x}^{\ast }J{\mathrm{\Delta }}^{-1\mathrm{/}2}{y}_2\mathrm{\Omega })+\lambda ({\mathrm{\Delta }}^{-1\mathrm{/}2}{y}_1\mathrm{\Omega }\mathrm{\mid }J{x}^{\ast }J{\mathrm{\Delta }}^{1\mathrm{/}2}{y}_2\mathrm{\Omega })\mathrm{.}$

The space ${\mathrm{\Delta }}^{-1\mathrm{/}2}{M}^{\mathrm{\prime }}\mathrm{\Omega }$ is dense: $F(y^{\mathrm{\prime }}\mathrm{\Omega })=y^{\mathrm{\prime }\ast }\mathrm{\Omega }$ and $F=J{\mathrm{\Delta }}^{-1\mathrm{/}2}$ give ${\mathrm{\Delta }}^{-1\mathrm{/}2}{M}^{\mathrm{\prime }}\mathrm{\Omega }=J{M}^{\mathrm{\prime }}\mathrm{\Omega }$, which is dense by cyclicity of $\mathrm{\Omega }$ for $M^{\mathrm{\prime }}$. In particular each $y^{\mathrm{\prime }}\mathrm{\Omega }$ belongs to $D({\mathrm{\Delta }}^{-1\mathrm{/}2})$. Fix $\xi \in D({\mathrm{\Delta }}^{1\mathrm{/}2})\cap D({\mathrm{\Delta }}^{-1\mathrm{/}2})$ and put $\eta =({\mathrm{\Delta }}^{1\mathrm{/}2}+{\mathrm{\Delta }}^{-1\mathrm{/}2})\xi \in H$. Choose $y_n^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$ so that ${\eta }_n:={\mathrm{\Delta }}^{-1\mathrm{/}2}{y}_n^{\mathrm{\prime }}\mathrm{\Omega }\to \eta$ in $H$. Step 3.1 at $\lambda =1$ supplies $y_n\in M$ with $y_n\mathrm{\Omega }=(\mathrm{\Delta }+1{)}^{-1}{y}_n^{\mathrm{\prime }}\mathrm{\Omega }$. The spectral calculus gives $y_n\mathrm{\Omega }=q(\mathrm{\Delta }){\eta }_n,{\mathrm{\Delta }}^{1\mathrm{/}2}{y}_n\mathrm{\Omega }=a(\mathrm{\Delta }){\eta }_n,{\mathrm{\Delta }}^{-1\mathrm{/}2}{y}_n\mathrm{\Omega }=b(\mathrm{\Delta }){\eta }_n,$ where $q(\lambda )=\sqrt{\lambda }\mathrm{/}(1+\lambda )$, $a(\lambda )=\lambda \mathrm{/}(1+\lambda )$ and $b(\lambda )=1\mathrm{/}(1+\lambda )$ for $\lambda >0$. These three multipliers are bounded. The identities follow by multiplying their scalar functions on the domain of ${\mathrm{\Delta }}^{-1\mathrm{/}2}$; moreover ${\lambda }^{1\mathrm{/}2}q(\lambda )=a(\lambda )$ and ${\lambda }^{-1\mathrm{/}2}q(\lambda )=b(\lambda )$ are bounded, so $q(\mathrm{\Delta }){\eta }_n$ belongs to both required domains. Applied to $\eta$, the same identities give $q(\mathrm{\Delta })\eta =\xi$, $a(\mathrm{\Delta })\eta ={\mathrm{\Delta }}^{1\mathrm{/}2}\xi$ and $b(\mathrm{\Delta })\eta ={\mathrm{\Delta }}^{-1\mathrm{/}2}\xi$. Thus convergence of ${\eta }_n$ in $H$ implies $y_n\mathrm{\Omega }\to \xi$ and ${\mathrm{\Delta }}^{\pm 1\mathrm{/}2}{y}_n\mathrm{\Omega }\to {\mathrm{\Delta }}^{\pm 1\mathrm{/}2}\xi$. This proves density of the core from step 4.1 in the joint graph norm, without treating ${\mathrm{\Delta }}^{-1\mathrm{/}2}$ as a bounded operator.

Both sides of the identity in step 4.1 are continuous in those graph norms: the left side is bounded by $\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }{\xi }_1\mathrm{\parallel }\mathrm{\parallel }{\xi }_2\mathrm{\parallel }$, and the right side by $\mathrm{\parallel }x\mathrm{\parallel }$ times the products of the indicated ${\mathrm{\Delta }}^{\pm 1\mathrm{/}2}$ norms. Passing to the approximating sequences of step 5.1 proves the displayed form identity for every ${\xi }_1,{\xi }_2\in D({\mathrm{\Delta }}^{1\mathrm{/}2})\cap D({\mathrm{\Delta }}^{-1\mathrm{/}2})$.

Fix $x^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$ and $x\in M$ as in step 3.1, so $x^{\mathrm{\prime }}\mathrm{\Omega }=(\mathrm{\Delta }+\lambda )x\mathrm{\Omega }$. Put $B:=J{x}^{\ast }J$ and let $C:=\widetilde{x^{\mathrm{\prime }}}$ be the bounded operator defined by the integral in step 1.4 with input $x^{\mathrm{\prime }}$. The form identity in step 6.1 says that $x^{\mathrm{\prime }}$ and $B$ satisfy $x^{\mathrm{\prime }}={\mathrm{\Delta }}^{1\mathrm{/}2}B{\mathrm{\Delta }}^{-1\mathrm{/}2}+\lambda {\mathrm{\Delta }}^{-1\mathrm{/}2}B{\mathrm{\Delta }}^{1\mathrm{/}2}$ in matrix coefficients on $D({\mathrm{\Delta }}^{1\mathrm{/}2})\cap D({\mathrm{\Delta }}^{-1\mathrm{/}2})$; step 1.4 gives the same identity with $C$ in place of $B$. Thus, on $E_{n}H$, $v_n:=E_n(B-C)E_n$ satisfies ${\mathrm{\Delta }}_n{v}_n+\lambda v_n{\mathrm{\Delta }}_n=0$, where ${\mathrm{\Delta }}_n=\mathrm{\Delta }{\mathrm{\mid }}_{E_{n}H}$, by multiplying the difference on the left and right by ${\mathrm{\Delta }}_n^{1\mathrm{/}2}$. If $E_{n}H=\{0\}$ then $v_n=0$; otherwise the spectra of ${\mathrm{\Delta }}_n$ and $-\lambda {\mathrm{\Delta }}_n$ are disjoint compact sets because $\sigma ({\mathrm{\Delta }}_n)\subset [1\mathrm{/}n,n]$ and $-\lambda$ is not positive real. Choose a positively oriented contour $\mathrm{\Gamma }$ enclosing $\sigma ({\mathrm{\Delta }}_n)$ and excluding $\sigma (-\lambda {\mathrm{\Delta }}_n)$. For $z\in \mathrm{\Gamma }$, multiplying the Sylvester equation by $(z-{\mathrm{\Delta }}_n{)}^{-1}$ on the left and $(z+\lambda {\mathrm{\Delta }}_n{)}^{-1}$ on the right gives $(z-{\mathrm{\Delta }}_n{)}^{-1}{v}_n=v_n(z+\lambda {\mathrm{\Delta }}_n{)}^{-1}\mathrm{.}$ By the continuous functional calculus, the two contour integrals are the operators associated to the scalar functions $s\mapsto (2\pi i{)}^{-1}{\int }_{\mathrm{\Gamma }}(z-s{)}^{-1}dz$ and $s\mapsto (2\pi i{)}^{-1}{\int }_{\mathrm{\Gamma }}(z+\lambda s{)}^{-1}dz$ on $\sigma ({\mathrm{\Delta }}_n)$. Uniform convergence of the contour Riemann sums permits this functional-calculus interchange. The scalar residue theorem gives the constant functions $1$ and $0$, respectively, so $(2\pi i{)}^{-1}{\int }_{\mathrm{\Gamma }}(z-{\mathrm{\Delta }}_n{)}^{-1}dz=I$ and $(2\pi i{)}^{-1}{\int }_{\mathrm{\Gamma }}(z+\lambda {\mathrm{\Delta }}_n{)}^{-1}dz=0$. Integrating the resolvent identity yields $v_n=0$.

The projections $E_n$ increase strongly to $I$ because $\ker \mathrm{\Delta }=0$. Step 7.1 gives $E_n(B-C)E_n=0$ for all $n$. For any $\xi ,\eta \in H$, $⟨(B-C)\xi ,\eta ⟩={\lim }_{n\to \mathrm{\infty }}⟨E_n(B-C)E_n\xi ,\eta ⟩=0,$ since $E_n\xi \to \xi$ and $E_n\eta \to \eta$ and $B-C$ is bounded. Thus $B=C$, which is the resolvent-core formula $J{x}^{\ast }J=e^{-i\theta \mathrm{/}2}{\int }_{\mathbb{R}}\frac{e^{-\theta t}}{e^{\pi t}+e^{-\pi t}}{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}dt\mathrm{.}$ The scalar kernel has integral $1\mathrm{/}2$ at $\theta =0$, matching the normalization checked in step 2.2.

Fix $x^{\mathrm{\prime }},y^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$, let $\lambda =e^{i\theta }$ with $-\pi <\theta <\pi$, and use step 3.1 to choose $x\in M$ with $x^{\mathrm{\prime }}\mathrm{\Omega }=(\mathrm{\Delta }+\lambda )x\mathrm{\Omega }$. Since $(\mathrm{\Delta }+\lambda {)}^{-1}{x}^{\mathrm{\prime }}\mathrm{\Omega }=x\mathrm{\Omega }$, step 2.2 gives $y^{\mathrm{\prime }}J{\mathrm{\Delta }}^{1\mathrm{/}2}(\mathrm{\Delta }+\lambda {)}^{-1}{x}^{\mathrm{\prime }}\mathrm{\Omega }=e^{i\theta \mathrm{/}2}{\int }_{\mathbb{R}}\frac{e^{-\theta t}}{e^{\pi t}+e^{-\pi t}}{y}^{\mathrm{\prime }}J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}\mathrm{\Omega }dt\mathrm{.}$ The left side is $y^{\mathrm{\prime }}J{\mathrm{\Delta }}^{1\mathrm{/}2}x\mathrm{\Omega }=y^{\mathrm{\prime }}{x}^{\ast }\mathrm{\Omega }=x^{\ast }{y}^{\mathrm{\prime }}\mathrm{\Omega }$, because $Sx\mathrm{\Omega }=x^{\ast }\mathrm{\Omega }$ and $S=J{\mathrm{\Delta }}^{1\mathrm{/}2}$. Conjugating the operator identity of step 8.1 by the antiunitary $J$ gives $x^{\ast }{y}^{\mathrm{\prime }}\mathrm{\Omega }=e^{i\theta \mathrm{/}2}{\int }_{\mathbb{R}}\frac{e^{-\theta t}}{e^{\pi t}+e^{-\pi t}}J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}J{y}^{\mathrm{\prime }}\mathrm{\Omega }dt\mathrm{.}$ Equating these expressions and cancelling their common nonzero scalar gives, for each $-\pi <\theta <\pi$, ${\int }_{\mathbb{R}}\frac{e^{-\theta t}}{e^{\pi t}+e^{-\pi t}}(J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}J{y}^{\mathrm{\prime }}\mathrm{\Omega }-y^{\mathrm{\prime }}J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}\mathrm{\Omega })dt=0.$ The first summand is bounded by $\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }{y}^{\mathrm{\prime }}\mathrm{\parallel }$ and norm-continuous by the conjugation-orbit estimate in step 1.4 applied to $x^{\mathrm{\prime }}$ and $J{y}^{\mathrm{\prime }}\mathrm{\Omega }$; the second is bounded by the same constant and norm-continuous by [F10] applied to $x^{\mathrm{\prime }}\mathrm{\Omega }$, followed by the fixed bounded maps $J$ and $y^{\mathrm{\prime }}$. For any $\eta \in H$, let $c(t)$ be its scalar coefficient. If $\mathrm{\mid }\mathrm{Re}z\mathrm{\mid }\le r<\pi$, then $\mathrm{\mid }c(t)e^{-zt}\mathrm{/}(e^{\pi t}+e^{-\pi t})\mathrm{\mid }\le C{e}^{-(\pi -r)\mathrm{\mid }t\mathrm{\mid }}$ for a fixed $C$. For difference quotients with $\mathrm{\mid }h\mathrm{\mid }\le \delta <\pi -r$, the integral formula for $e^{-ht}-1$ gives the integrable bound $C\mathrm{\mid }t\mathrm{\mid }{e}^{-(\pi -r-\delta )\mathrm{\mid }t\mathrm{\mid }}$. By [F11], dominated convergence differentiates the integral in $z$ on every such compact substrip, so its Laplace transform is holomorphic on $\mathrm{\mid }\mathrm{Re}z\mathrm{\mid }<\pi$. It vanishes on the real interval $(-\pi ,\pi )$. A scalar holomorphic function with zeros accumulating at an interior point is identically zero: its Taylor series at that point either has a first nonzero coefficient, which would make the zero isolated, or all coefficients vanish and it is zero in a neighborhood; the set of points with a zero neighborhood is then open and closed in the connected strip. Evaluating at $z=is$ shows the Fourier transform of that $L^1$ coefficient is zero for every $s\in \mathbb{R}$.

By [F9] the scalar coefficient in step 9.1 vanishes almost everywhere; continuity makes it vanish everywhere. Since this holds for every $\eta \in H$, $J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}J{y}^{\mathrm{\prime }}\mathrm{\Omega }=y^{\mathrm{\prime }}J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}\mathrm{\Omega }(t\in \mathbb{R})\mathrm{.}$ For fixed $t,x^{\mathrm{\prime }}$, apply this identity with $y_2^{\mathrm{\prime }}$ and then with $y_1^{\mathrm{\prime }}{y}_2^{\mathrm{\prime }}$. It gives $(y_1^{\mathrm{\prime }}J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}J)y_2^{\mathrm{\prime }}\mathrm{\Omega }=y_1^{\mathrm{\prime }}{y}_2^{\mathrm{\prime }}J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}\mathrm{\Omega }=(J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}J)y_1^{\mathrm{\prime }}{y}_2^{\mathrm{\prime }}\mathrm{\Omega }\mathrm{.}$ The vectors $M^{\mathrm{\prime }}\mathrm{\Omega }$ are dense, so $J{\mathrm{\Delta }}^{-it}{x}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}J$ commutes with every $y_1^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$. Thus it lies in $M^{\mathrm{\prime }\mathrm{\prime }}=M$. At $t=0$, $J{M}^{\mathrm{\prime }}J\subseteq M$, hence $M^{\mathrm{\prime }}\subseteq JMJ$.

To prove the reverse inclusion, fix $x,y\in M$ and $x^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$. Step 10.1 gives $J{x}^{\mathrm{\prime }}J\in M$. Using commutation of $x^{\mathrm{\prime }}$ and $y$, the identities $F=S^{\ast }=JSJ$, $S(x\mathrm{\Omega })=x^{\ast }\mathrm{\Omega }$, and $(JxJ{)}^{\ast }=J{x}^{\ast }J$, we obtain $(x^{\mathrm{\prime }\ast }\mathrm{\Omega }\mathrm{\mid }yJx\mathrm{\Omega })=(y^{\ast }\mathrm{\Omega }\mathrm{\mid }{x}^{\mathrm{\prime }}Jx\mathrm{\Omega })=(y^{\ast }\mathrm{\Omega }\mathrm{\mid }J(J{x}^{\mathrm{\prime }}J)x\mathrm{\Omega })=(y^{\ast }\mathrm{\Omega }\mathrm{\mid }JS(x^{\ast }J{x}^{\mathrm{\prime }\ast }J\mathrm{\Omega }))=(y^{\ast }\mathrm{\Omega }\mathrm{\mid }F(J{x}^{\ast }J{x}^{\mathrm{\prime }\ast }\mathrm{\Omega }))=(J{x}^{\ast }J{x}^{\mathrm{\prime }\ast }\mathrm{\Omega }\mathrm{\mid }S{y}^{\ast }\mathrm{\Omega })=(x^{\mathrm{\prime }\ast }\mathrm{\Omega }\mathrm{\mid }JxJy\mathrm{\Omega })\mathrm{.}$ Since $x^{\mathrm{\prime }\ast }\mathrm{\Omega }$ ranges over the dense subspace $M^{\mathrm{\prime }}\mathrm{\Omega }$, this implies $yJx\mathrm{\Omega }=JxJy\mathrm{\Omega }$. For $y_1,y_2\in M$, it follows that $y_{1}JxJ{y}_2\mathrm{\Omega }=y_1{y}_{2}Jx\mathrm{\Omega }=JxJ{y}_1{y}_2\mathrm{\Omega }$. Density of $M\mathrm{\Omega }$ then gives $JxJ\in M^{\mathrm{\prime }}$. Therefore $JMJ\subseteq M^{\mathrm{\prime }}$, and step 10.1 yields $JMJ=M^{\mathrm{\prime }}$.

The inclusions $J{\mathrm{\Delta }}^{-it}{M}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}J\subseteq M$ from step 10.1 and $JMJ=M^{\mathrm{\prime }}$ from step 11.1 imply ${\mathrm{\Delta }}^{-it}{M}^{\mathrm{\prime }}{\mathrm{\Delta }}^{it}\subseteq M^{\mathrm{\prime }}$ for every $t$. Replacing $t$ by $-t$ and conjugating gives equality. Taking commutants and using $M^{\mathrm{\prime }\mathrm{\prime }}=M$ gives ${\mathrm{\Delta }}^{it}M{\mathrm{\Delta }}^{-it}=M$.

For $x^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$, write $x^{\mathrm{\prime }}=JxJ$ with $x\in M$ by step 11.1. Since $J\mathrm{\Omega }=\mathrm{\Omega }$ and $S_0(x\mathrm{\Omega })=x^{\ast }\mathrm{\Omega }$, direct substitution gives $J{S}_{0}J(x^{\mathrm{\prime }}\mathrm{\Omega })=x^{\mathrm{\prime }\ast }\mathrm{\Omega }=F_0(x^{\mathrm{\prime }}\mathrm{\Omega })$. Thus $F_0=J{S}_{0}J$, so graph closure gives $\overline{F_0}=JSJ=F=S^{\ast }$. Steps 3.1, 11.1–12.1, and this closure identity establish all stated conclusions.