Proof of Standard Form of a Tracial W*-Probability Space: the Commutation Theorem, Right-Bounded Vectors, Faithfulness and Self-Adjoint Vectors

Each result cited below is universally quantified over the data in its own statement.

By Tracial W*-Probability Spaces §space, $(H,M,\Omega)$ is a cyclic tracial operator algebra with $M=M''$, and $J$ is its conjugation in the sense of The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §conjugation. We apply The Tracial W*-Closure of a Cyclic Tracial Operator Algebra: the Commutant of the Right Action is a Tracial W*-Probability Space Equal to the Double Commutant with $\mathcal{A}:=M$; the conjugation used there is the conjugation of the given triple, namely $J$. Let $N=(JMJ)'$ be the algebra of that theorem. By The Tracial W*-Closure of a Cyclic Tracial Operator Algebra: the Commutant of the Right Action is a Tracial W*-Probability Space Equal to the Double Commutant §double-commutant, $N=M''=M$. By Basic Properties of Commutants: Unital Algebras Closed under Weak Limits, Order Reversal, the Triple Commutant, and Conjugation §conjugation, $JAJ\in\mathcal{L}(H)$ and $J(JAJ)J=A$ for every $A\in\mathcal{L}(H)$; and $J$ preserves norms, since $\lVert J\xi\rVert^{2}=\langle J\xi,J\xi\rangle=\langle\xi,\xi\rangle$ by Conjugation of a Complex Hilbert Space §conjugation.

Claim 1. $M=N=(JMJ)'$. By The Tracial W*-Closure of a Cyclic Tracial Operator Algebra: the Commutant of the Right Action is a Tracial W*-Probability Space Equal to the Double Commutant §commutant, $N'=JNJ$, i.e. $M'=JMJ$. Hence every element of $JM'J$ has the form $J(JTJ)J=T$ with $T\in M$, and every $T\in M$ equals $J(JTJ)J$ with $JTJ\in M'$; so $JM'J=M$.

Claim 2. By The Tracial W*-Closure of a Cyclic Tracial Operator Algebra: the Commutant of the Right Action is a Tracial W*-Probability Space Equal to the Double Commutant §right-bounded with $\mathcal{A}=M$, the hypothesis on $\zeta$ and $C$ yields exactly one $T\in N$ with $T\Omega=\zeta$, and $\lVert T\rVert_{\mathrm{op}}\le C$. Since $N=M$, this is the claim.

Claim 3. Let $\xi,\eta\in H_{\mathrm{sa}}$ and $t\in\mathbb{R}$. By Conjugation of a Complex Hilbert Space §conjugation, $J(\xi+\eta)=J\xi+J\eta=\xi+\eta$ and $J(t\xi)=\overline{t}\,J\xi=t\xi$, as $\overline{t}=t$ for real $t$. Also $\langle\xi,\eta\rangle=\langle J\xi,J\eta\rangle=\langle\eta,\xi\rangle=\overline{\langle\xi,\eta\rangle}$, the last step by conjugate symmetry of the inner product, so $\langle\xi,\eta\rangle$ is real.

Closedness: let $(\xi_{k})$ be a sequence in $H_{\mathrm{sa}}$ converging to $\xi\in H$. Since $J$ is additive, $J\xi-J\xi_{k}=J(\xi-\xi_{k})$, so

for every $k$, using $J\xi_{k}=\xi_{k}$ and that $J$ preserves norms. The right side tends to $0$, so $J\xi=\xi$ and $\xi\in H_{\mathrm{sa}}$.

Let $S\in M$ be self-adjoint. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $S$ is an adjoint of $S$, so $S^{*}=S$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique. By The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §conjugation, $J(S\Omega)=S^{*}\Omega=S\Omega$, so $S\Omega\in H_{\mathrm{sa}}$.

Density: let $\xi\in H_{\mathrm{sa}}$ and $\varepsilon>0$ real. By Cyclic Tracial Operator Algebras and Their Traces §cyclic, $M\Omega$ is dense in $H$, so there is $T\in M$ with $\lVert\xi-T\Omega\rVert<\varepsilon$. Let $S=\tfrac12(T+T^{*})$; then $S\in M$ by Cyclic Tracial Operator Algebras and Their Traces §star-algebra. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $S$ has the adjoint $\tfrac12(T^{*}+T)=S$ (as $(T^{*})^{*}=T$ and $\overline{1/2}=1/2$), so $S$ is self-adjoint by the same clause. Using $T^{*}\Omega=J(T\Omega)$ from The Trace, the Conjugation and the Right Action of a Cyclic Tracial Operator Algebra §conjugation and $\xi=\tfrac12\xi+\tfrac12J\xi$,

so $\lVert\xi-S\Omega\rVert\le\tfrac12\lVert\xi-T\Omega\rVert+\tfrac12\lVert J(\xi-T\Omega)\rVert=\lVert\xi-T\Omega\rVert<\varepsilon$.