I am learning the theory of standard von Neumann algebras by following Chapter $10$ from the book 'Lectures on von Neumann algebras' by Strătilă and Zsidó. But I got stuck with an argument provided in Section $6^{\circ}$ of the Chapter $10$. Let me elaborate the notations and prerequisites before posing my problems.
Let $\mathscr{M}\subseteq\mathcal{B}(\mathcal{H})$ be a von Neumann algebra which has cyclic and separating vector $\xi_0\in\mathcal{H}$. Consider the conjugate linear operator $S_0:\mathscr{M}\xi_0\ni x\xi_0\mapsto x^*\xi_0\in\mathscr{M}\xi_0\subseteq\mathcal{H};\,\mathscr{D}_{S_0}=\mathscr{M}\xi_0$. Then $S_0$ is closable and denote its closure by $S$. Consider the modular operator $\Delta=S^*S$ and the polar decomposition $S=J\Delta^{1/2}$ where $J=J^*=J^{-1}$. See Section $1^{\circ}$ for the details.
Now let me recall a few results from the book which will be used later.
Proposition 9.23: Let $A$ and $B$ be positive self-adjoint operators in $\mathcal{H}$, such that $\text{Kernel}(A)=\text{Kernel}(B)=\{0\}$ and $\omega\in\mathbb{C},\,|\omega|=1,\,\omega\neq -1$. For any $x\in\mathcal{B}(\mathcal{H})$, there exists a unique $y\in\mathcal{B}(\mathcal{H})$, such that $$\langle x\eta |\xi \rangle = \omega \langle yB^{-1/2}\eta | A^{1/2}\xi\rangle +\langle yB^{1/2}\eta | A^{-1/2}\xi\rangle ,$$ $$\xi\in\mathscr{D}_{(A^{1/2})}\cap \mathscr{D}_{(A^{-1/2})},\;\eta\in\mathscr{D}_{(B^{1/2})}\cap \mathscr{D}_{(B^{-1/2})},$$ and it is given by $y=\int_{-\infty}^{\infty}\frac{{\omega}^{it-\frac{1}{2}}}{e^{\pi t}+e^{-\pi t}}A^{it}xB^{-it}dt$.
Corollary 9.23: Let $A$ be a positive self-adjoint operator in $\mathcal{H}$, such that $\text{Kernel}(A)=\{0\}$ and $\omega\in\mathbb{C},\,|\omega|=1,\,\omega\neq -1$. Then $$A^{-1/2}(\omega + A^{-1})^{-1}=\int_{-\infty}^{\infty}\frac{{\omega}^{it-\frac{1}{2}}}{e^{\pi t} + e^{-\pi t}}A^{it}dt.$$
Lemma $3^{\circ}$: For $\omega\in\mathbb{C},\,|\omega|=1,\,\omega\neq -1$, we have $(\Delta +\omega)^{-1}\mathscr{M}'\xi_0\subseteq \mathscr{M}\xi_0$.
Lemma $5^{\circ}$: Let $x'\in\mathscr{M}'$ and $\omega\in\mathbb{C},\,|\omega|=1,\,\omega\neq -1$. Then $$\langle x'\eta | \xi \rangle = \langle Jx^*J{\Delta}^{-1/2}\eta | {\Delta}^{1/2}\zeta\rangle + \omega\langle Jx^*J{\Delta}^{1/2}\eta | {\Delta}^{-1/2}\zeta\rangle $$ for any $\eta,\zeta\in\mathscr{D}_{{\Delta}^{1/2}}\cap \mathscr{D}_{{\Delta}^{-1/2}}$ where $x\in\mathscr{M}$ is given by $x'\xi_0=(\Delta +\omega)x\xi_0$ (by Lemma $3^{\circ}$).
Now Section $6^{\circ}$ goes as follows. Given $x'\in\mathscr{M}'$ and $\omega\in\mathbb{C},\,|\omega|=1,\,\omega\neq -1$, we obtain the equation of Lemma $5^{\circ}$ with $x\in\mathscr{M}$ such that $x'\xi_0=(\Delta + \omega)x\xi_0$. Consider now another arbitrary $y'\in\mathscr{M}'$. By Proposition $9.23$ from Lemma $5^{\circ}$, we get $$Jx^*J=\int_{-\infty}^{\infty}\frac{{\omega}^{it-\frac{1}{2}}}{e^{\pi t} + e^{-\pi t}}{\Delta}^{-it}x'{\Delta}^{it}dt,$$ $$\text{hence } x^*=\int_{-\infty}^{\infty}\frac{{\omega}^{it-\frac{1}{2}}}{e^{\pi t} + e^{-\pi t}}J{\Delta}^{-it}x'{\Delta}^{it}J dt,$$ $$\text{i.e. } x^*y'\xi_0=\int_{-\infty}^{\infty}\frac{{\omega}^{it-\frac{1}{2}}}{e^{\pi t} + e^{-\pi t}}J{\Delta}^{-it}x'{\Delta}^{it}Jy'\xi_0 dt.$$ In contrast, by using the Corollary $9.23$, $$x^*y'\xi_0=y'x^*\xi_0=y'Sx\xi_0=y'J{\Delta}^{1/2}(\Delta + \omega)^{-1}x'\xi_0=\int_{-\infty}^{\infty}\frac{{\omega}^{it-\frac{1}{2}}}{e^{\pi t} + e^{-\pi t}}y'J{\Delta}^{-it}x'\xi_0 dt.$$ It follows that $$F(\omega):=\int_{-\infty}^{\infty}\frac{{\omega}^{it-\frac{1}{2}}}{e^{\pi t} + e^{-\pi t}}(J{\Delta}^{-it}x'{\Delta}^{it}Jy'\xi_0 - y'J{\Delta}^{-it}x'\xi_0) dt=0$$ for every $\omega\in\mathbb{C},\,|\omega|=1,\,\omega\neq -1$. Now I have the following problems.
Problem 1: The authors say that the above equality extends by analyticity to all $\omega\in\mathbb{C}\setminus (-\infty,0)$. I did not get this reasoning. How can I conclude that $F$ is analytic on $\mathbb{C}\setminus (-\infty,0)$?
Problem 2: Why is the function $t\mapsto \frac{1}{e^{\pi t} + e^{-\pi t}}(J{\Delta}^{-it}x'{\Delta}^{it}Jy'\xi_0 - y'J{\Delta}^{-it}x'\xi_0)$ an $L^1$ function?
Thanks in advance for any help.
You have $$ \|J{\Delta}^{-it}x'{\Delta}^{it}Jy'\xi_0 - y'J{\Delta}^{-it}x'\xi_0\|\leq 2\|x'\|\,\|y'\|\,\|\xi_0\|. $$ Then $$ \int_{-\infty}^\infty\Big\|\frac{1}{e^{\pi t} + e^{-\pi t}}(J{\Delta}^{-it}x'{\Delta}^{it}Jy'\xi_0 - y'J{\Delta}^{-it}x'\xi_0)\Big\|\,dt\leq\int_{-\infty}^\infty\frac{2\|x'\|\,\|y'\|\,\|\xi_0\|}{e^{\pi t}+e^{-\pi t}}\,dt<\infty, $$ since $t\mapsto \frac{1}{e^{\pi t} + e^{-\pi t}}$ is in $L^1(\mathbb R)$.
Then the function $F$ is defined for any complex number with argument other than $\pi$. Moreover, because the integrand is integrable, Dominated Convergence gives you that $$ \lim_{h\to0}\frac{F(\omega+h)-F(\omega)}h=\int_{-\infty}^{\infty}\frac{\big(-\tfrac12+it\big){\omega}^{it-\frac{3}{2}}}{e^{\pi t} + e^{-\pi t}}(J{\Delta}^{-it}x'{\Delta}^{it}Jy'\xi_0 - y'J{\Delta}^{-it}x'\xi_0) dt. $$ So $F$ is analytic on $\mathbb C\setminus(-\infty,0]$.