Modular automorphisms and physical-time KMS conversion

Write $z=t+is$ with $0\le s\le 1$. The spectral multiplier $f_z(\lambda )=\exp (-iz\log (\lambda )\mathrm{/}2)$ has $\mathrm{\mid }{f}_z(\lambda ){\mathrm{\mid }}^2={\lambda }^s$. Since ${\lambda }^s\le 1+\lambda$ for every $\lambda >0$ and $s\in [0,1]$, [F2] gives $\mathrm{\parallel }{A}^{-iz\mathrm{/}2}\xi {\mathrm{\parallel }}^2=\int {\lambda }^{s}d{\mu }_{\xi }\le \int (1+\lambda )d{\mu }_{\xi }=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2+\mathrm{\parallel }{A}^{1\mathrm{/}2}\xi {\mathrm{\parallel }}^2\mathrm{.}$ This also proves that every vector named in the Statement is in the domain of the indicated multiplier.

If $z_n=t_n+i{s}_n\to z=t+is$ in $S$, then $f_{z_n}(\lambda )\to f_z(\lambda )$ pointwise and $\mathrm{\mid }{f}_{z_n}(\lambda )-f_z(\lambda ){\mathrm{\mid }}^2\le 2{\lambda }^{s_n}+2{\lambda }^s\le 4(1+\lambda )\mathrm{.}$ The majorant is integrable by $\xi \in D(A^{1\mathrm{/}2})$. Dominated convergence applied to the scalar spectral measure therefore gives $\mathrm{\parallel }{A}^{-i{z}_n\mathrm{/}2}\xi -A^{-iz\mathrm{/}2}\xi {\mathrm{\parallel }}^2\to 0$, proving strong continuity on the closed strip.

Fix $z_0$ with $s_0=\mathrm{Im}{z}_0\in (0,1)$. Choose $\delta >0$ so that $2\delta <s_0<1-2\delta$. For $\mathrm{\mid }h\mathrm{\mid }\le \delta$, the scalar difference quotient of $f_z(\lambda )$ along the segment from $z_0$ to $z_0+h$ is bounded by $\frac{1}{2}\mathrm{\mid }\log \lambda \mathrm{\mid }{\sup }_{0\le r\le 1}{\lambda }^{\mathrm{Im}(z_0+rh)\mathrm{/}2}\mathrm{.}$ When $0<\lambda \le 1$ this is bounded by $\frac{1}{2}\mathrm{\mid }\log \lambda \mathrm{\mid }{\lambda }^{\delta \mathrm{/}2}$, and when $\lambda \ge 1$ it is bounded by $\frac{1}{2}\mathrm{\mid }\log \lambda \mathrm{\mid }{\lambda }^{(1-\delta )\mathrm{/}2}$. The first expression has a finite supremum and the second is bounded by a constant depending on $\delta$ times ${\lambda }^{1\mathrm{/}2}$. Thus the difference quotients are dominated in $L^2({\mu }_{\xi })$ by a constant times $1+{\lambda }^{1\mathrm{/}2}$. Their pointwise limit is $(-i\mathrm{/}2)(\log \lambda )f_{z_0}(\lambda )$. Dominated convergence in the spectral norm proves that the vector difference quotient converges in $H$ to the spectral integral of this limit. Hence the vector map is complex-differentiable at $z_0$.

Step 2.2 applies at every point of $S^{\circ }$, so the vector map is norm-holomorphic throughout the open strip. The estimate and closed-strip continuity were proved in steps 1.1 and 2.1.

Let $t_n\to t$. For every $\lambda >0$, ${\lambda }^{i{t}_n}\to {\lambda }^{it}$, while $\mathrm{\mid }{\lambda }^{i{t}_n}-{\lambda }^{it}{\mathrm{\mid }}^2\le 4$. Since ${\mu }_{\xi }((0,\mathrm{\infty }))=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2<\mathrm{\infty }$, [F1] and [F2] give $\mathrm{\parallel }{A}^{i{t}_n}\xi -A^{it}\xi {\mathrm{\parallel }}^2={\int }_{(0,\mathrm{\infty })}\mathrm{\mid }{\lambda }^{i{t}_n}-{\lambda }^{it}{\mathrm{\mid }}^{2}d{\mu }_{\xi }(\lambda )⟶0.$ Thus the orbit is sequentially continuous.

The real parameter space and Hilbert-space norm topology are metric, so sequential continuity is equivalent to continuity. Therefore $t\mapsto A^{it}\xi$ is norm-continuous for every $\xi \in H$.

If $x\mathrm{\Omega }=y\mathrm{\Omega }$ for $x,y\in M$, then $(x-y)\mathrm{\Omega }=0$ and separatingness gives $x=y$. Thus $S_0$ is well defined; it is antilinear by $(ax+by{)}^{\ast }=\bar{a}{x}^{\ast }+\bar{b}{y}^{\ast }$, and its domain is dense by cyclicity. The same reasoning makes $F_0$ well defined; [F4] makes $M^{\mathrm{\prime }}\mathrm{\Omega }$ dense. Also $S_0^2(x\mathrm{\Omega })=x\mathrm{\Omega }$ and $F_0^2(x^{\mathrm{\prime }}\mathrm{\Omega })=x^{\mathrm{\prime }}\mathrm{\Omega }$.

By [F4], $\mathrm{\Omega }$ is cyclic for $M^{\mathrm{\prime }}$ because it is separating for $M$, and separating for $M^{\mathrm{\prime }}$ because it is cyclic for $M$. Applying [F2] to the Tomita operator $F_0$ of $M^{\mathrm{\prime }}$ therefore proves directly that $F_0$ is closable; applying [F2] to $M$ proves that $S_0$ is closable. For $x\in M$ and $x^{\mathrm{\prime }}\in M^{\mathrm{\prime }}$, the adjoint identity and commutation give $(x^{\mathrm{\prime }}\mathrm{\Omega }\mathrm{\mid }{S}_{0}x\mathrm{\Omega })=(x^{\mathrm{\prime }}\mathrm{\Omega }\mathrm{\mid }{x}^{\ast }\mathrm{\Omega })=(x\mathrm{\Omega }\mathrm{\mid }{x}^{\mathrm{\prime }\ast }\mathrm{\Omega })=(x\mathrm{\Omega }\mathrm{\mid }{F}_0{x}^{\mathrm{\prime }}\mathrm{\Omega })\mathrm{.}$ By [F3] this says $F_0\subseteq S_0^{\ast }$ and $S_0\subseteq F_0^{\ast }$.

Let $\xi \in D(S)$ and choose ${\xi }_n\in M\mathrm{\Omega }$ with ${\xi }_n\to \xi$ and $S_0{\xi }_n\to S\xi$, using the definition of graph closure. Since $S_0^2{\xi }_n={\xi }_n$, the graph points $(S{\xi }_n,S(S{\xi }_n))=(S_0{\xi }_n,{\xi }_n)$ converge to $(S\xi ,\xi )$. Closedness of $S$ implies $S\xi \in D(S)$ and $S^2\xi =\xi$. Hence $S$ is an involution, so it is injective and its range is the dense domain $D(S)$.

Apply the closed-antilinear polar result [F2] to $S$. Injectivity and dense range make its partial anti-isometry $J$ antiunitary. The inverse of $S=J{\mathrm{\Delta }}^{1\mathrm{/}2}$ is ${\mathrm{\Delta }}^{-1\mathrm{/}2}{J}^{\ast }=J^{\ast }(J{\mathrm{\Delta }}^{-1\mathrm{/}2}{J}^{\ast })$, whose second factor is positive self-adjoint. Since $S^{-1}=S$, uniqueness of the polar decomposition gives $J=J^{\ast }$ and ${\mathrm{\Delta }}^{1\mathrm{/}2}=J{\mathrm{\Delta }}^{-1\mathrm{/}2}J$. Thus $J^2=I$ and $J\mathrm{\Delta }J={\mathrm{\Delta }}^{-1}$. For antiunitary conjugation the functional-calculus rule is $Jf(\mathrm{\Delta })J=\bar{f}(J\mathrm{\Delta }J)$. Taking $f(s)=s^{it}$ gives $J{\mathrm{\Delta }}^{it}J=({\mathrm{\Delta }}^{-1}{)}^{-it}={\mathrm{\Delta }}^{it}$, so $J$ commutes with every ${\mathrm{\Delta }}^{it}$.

Because $S$ is the graph closure of $S_0$, their adjoints agree: the adjoint identity for $S_0$ extends along convergent graph pairs to $S$, and restriction gives the converse. Hence $F=S^{\ast }=S_0^{\ast }$ and $F_0\subseteq F$. From $S=J{\mathrm{\Delta }}^{1\mathrm{/}2}$, the antilinear adjoint identity [F3] and self-adjointness of ${\mathrm{\Delta }}^{1\mathrm{/}2}$ give $F={\mathrm{\Delta }}^{1\mathrm{/}2}J$. Functional calculus applied to $J\mathrm{\Delta }J={\mathrm{\Delta }}^{-1}$ gives ${\mathrm{\Delta }}^{1\mathrm{/}2}J=J{\mathrm{\Delta }}^{-1\mathrm{/}2}$ on their common domain.

Since $S_0\mathrm{\Omega }=F_0\mathrm{\Omega }=\mathrm{\Omega }$ and $F_0\subseteq F$, steps 3.1 and 4.1 give ${\mathrm{\Delta }}^{1\mathrm{/}2}\mathrm{\Omega }=J\mathrm{\Omega }={\mathrm{\Delta }}^{-1\mathrm{/}2}\mathrm{\Omega }$. For the spectral measure of $\mathrm{\Delta }$, this equality has squared norm difference ${\int }_{(0,\mathrm{\infty })}\mathrm{\mid }{\lambda }^{1\mathrm{/}2}-{\lambda }^{-1\mathrm{/}2}{\mathrm{\mid }}^{2}d{\mu }_{\mathrm{\Omega }}(\lambda )=0$. The integrand vanishes only at $\lambda =1$, so $\mathrm{\Delta }\mathrm{\Omega }=\mathrm{\Omega }$ and then $J\mathrm{\Omega }=\mathrm{\Omega }$.

Let $M=\{L_A:A\in M_n(\mathbb{C})\}$ on ${\mathcal{H}}_{\mathrm{H}\mathrm{S}}$, where $L_{A}X=AX$. The Hilbert-Schmidt adjoint satisfies $L_A^{\ast }=L_{A^{\ast }}$, so $M$ is a unital star subalgebra. The ambient operator space $\mathcal{B}({\mathcal{H}}_{\mathrm{H}\mathrm{S}})$ is finite dimensional; hence $M$ is weak-operator closed. By [F2], $M$ is a concrete von Neumann algebra. Since ${\rho }^{1\mathrm{/}2}$ is invertible, every $X\in {\mathcal{H}}_{\mathrm{H}\mathrm{S}}$ equals $L_{X{\rho }^{-1\mathrm{/}2}}\mathrm{\Omega }$, so $\mathrm{\Omega }$ is cyclic. If $L_A\mathrm{\Omega }=0$, then $A{\rho }^{1\mathrm{/}2}=0$ and invertibility gives $A=0$, so $\mathrm{\Omega }$ is separating. Finally, $⟨L_A\mathrm{\Omega },\mathrm{\Omega }{⟩}_{\mathrm{H}\mathrm{S}}=\mathrm{Tr}({\rho }^{1\mathrm{/}2}A{\rho }^{1\mathrm{/}2})=\mathrm{Tr}(\rho A),$ using finite trace cyclicity.

For $A,B\in M_n(\mathbb{C})$ define $F_{A,B}(z)=\mathrm{Tr}\text{\hspace{0.17em}⁣}\left(\rho A{e}^{izH\mathrm{/}\mathrm{\hslash }}B{e}^{-izH\mathrm{/}\mathrm{\hslash }}\right)\mathrm{.}$ Finite matrix exponentials are entire, so $F_{A,B}$ is entire. If $0\le \mathrm{Im}z\le \beta \mathrm{\hslash }$, the real-time factors are unitary and the two imaginary-time factors have norms at most $e^{\beta \mathrm{\parallel }H\mathrm{\parallel }}$ each. Since $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1=1$, $\mathrm{\mid }{F}_{A,B}(z)\mathrm{\mid }\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }B\mathrm{\parallel }{e}^{2\beta \mathrm{\parallel }H\mathrm{\parallel }}$ uniformly on that strip. The group ${\alpha }_s$ is norm-continuous because its matrix entries are continuous.

Choose an orthonormal eigenbasis of $\rho$, with eigenvalues $p_j>0$, and write $E_{ij}=\mathrm{\mid }i⟩⟨j\mathrm{\mid }$. Then $E_{ij}\mathrm{\Omega }=\sqrt{p_j}{E}_{ij}$ and $E_{ij}^{\ast }\mathrm{\Omega }=\sqrt{p_i}{E}_{ji}$, whence $S(E_{ij})=\sqrt{p_i\mathrm{/}{p}_j}{E}_{ji}$. The maps $JX=X^{\ast }$ and $\mathrm{\Delta }X=\rho X{\rho }^{-1}$ satisfy $J{\mathrm{\Delta }}^{1\mathrm{/}2}{E}_{ij}=\sqrt{p_i\mathrm{/}{p}_j}{E}_{ji}$ and $\mathrm{\Delta }{E}_{ij}=(p_i\mathrm{/}{p}_j)E_{ij}$. Thus $S=J{\mathrm{\Delta }}^{1\mathrm{/}2}$ on a basis, so uniqueness of the polar decomposition gives the stated modular data.

For $X\in {\mathcal{H}}_{\mathrm{H}\mathrm{S}}$, ${\mathrm{\Delta }}^{it}{L}_A{\mathrm{\Delta }}^{-it}X={\rho }^{it}A{\rho }^{-it}X,$ because ${\mathrm{\Delta }}^{-it}X={\rho }^{-it}X{\rho }^{it}$ and the final right factor ${\rho }^{it}$ cancels ${\rho }^{-it}$. Hence ${\sigma }_t(A)={\rho }^{it}A{\rho }^{-it}$.

For $\rho =e^{-\beta H}\mathrm{/}Z$, functional calculus gives ${\rho }^{it}=Z^{-it}{e}^{-i\beta tH}$; the scalar factor cancels in conjugation. Substituting $t=-s\mathrm{/}(\beta \mathrm{\hslash })$ into step 3.1 yields ${\sigma }_{-s\mathrm{/}(\beta \mathrm{\hslash })}(A)=e^{isH\mathrm{/}\mathrm{\hslash }}A{e}^{-isH\mathrm{/}\mathrm{\hslash }}={\alpha }_s(A)$. Here $\beta \mathrm{\hslash }$ has units seconds, so the modular parameter is dimensionless.

At $z=t+i\beta \mathrm{\hslash }$, write $U=e^{itH\mathrm{/}\mathrm{\hslash }}$. Since $U$ commutes with $e^{\pm \beta H}$, $F_{A,B}(t+i\beta \mathrm{\hslash })=Z^{-1}\mathrm{Tr}\text{\hspace{0.17em}⁣}(e^{-\beta H}A{e}^{-\beta H}{\alpha }_t(B)e^{\beta H})=Z^{-1}\mathrm{Tr}\text{\hspace{0.17em}⁣}(A{e}^{-\beta H}{\alpha }_t(B))=\mathrm{Tr}\text{\hspace{0.17em}⁣}(\rho {\alpha }_t(B)A)\mathrm{.}$ The middle equality cycles $e^{\beta H}$ past the trace, and the last cycles $A$ past the trace. The lower boundary is $F_{A,B}(t)=\mathrm{Tr}(\rho A{\alpha }_t(B))$. Thus [F3] gives the physical-time $\beta$-KMS condition with width $\beta \mathrm{\hslash }$.

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 $1\mathrm{/}2$.