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For a Hilbert space $H$, the linear span of the matrix-coefficient functionals $T\mapsto ⟨Tx,y⟩$ is a predual of $\mathcal{B}(H)$. The unit ball is compact in the resulting weak operator topology by the Tychonoff theorem, and the compact norming dual-pair criterion identifies $\mathcal{B}(H)$ with the dual of that span.

The dense commutant orbit makes $\Omega$ separating for $M$, so $S_0(a\Omega )=a^{\ast }\Omega$ is well defined on the dense subspace $M\Omega$. Put $\tau (a)=⟨a\Omega ,\Omega ⟩$. The assumed identity says that this positive vector functional is a tracial positive functional. HenceIt extends to an antiunitary involution $J$ on $H$, with $J\Omega =\Omega$. This is the modular conjugation, and the modular operator in this tracial case is $\Delta =I$.

On the dense subspace $M\Omega$,Thus it acts by right multiplication and commutes with all left multiplications; $JMJ\subseteq M^{\prime }$. We prove the reverse inclusion rather than assuming the general Tomita commutation theorem.

Let $T\in M^{\prime }$, $C=\Vert T\Vert$, and $\eta =T\Omega$. The setis weakly compact and convex. Indeed, trace-class duality and the Banach-Alaoglu theorem make the bounded-operator ball compact in a topology stronger than the weak operator topology. The algebra $M$ is weak-operator closed, and $a\mapsto a\Omega$ is weakly continuous, so its image is weakly compact.

For $b\in M$ use the polar decomposition of a bounded operator $b=uh$, $h=|b|$. Both $h$ and $u$ belong to $M$: continuous functional calculus gives $h$, and $b(h+\epsilon I{)}^{-1}$ tends strongly to $u$. The polar decomposition is proved independently in Question 8. Since $T$ commutes with $M$,The trace identity givesTherefore $|⟨\eta ,b\Omega ⟩|\le C\tau (h)$. Meanwhilewith the supremum attained by $a=Cu$. For the upper bound, traciality gives $\tau (b^{\ast }a)=\tau (h^{1/2}{u}^{\ast }a{h}^{1/2})$; evaluating this on $\Omega$ bounds its absolute value by $\Vert a\Vert \tau (h)$.

These support inequalities imply $\eta \in \mathcal{B}$. Otherwise the Hahn-Banach separation theorem would separate $\eta$ from this closed convex set by a real inner-product functional. Since $M\Omega$ is dense, its separating vector could be approximated by some $b\Omega$, preserving strict separation, contrary to the displayed inequality. Thus $T\Omega =a\Omega$ for an $a\in M$ of norm at most $C$.

The bounded operator $R_a=J{a}^{\ast }J$ commutes with $M$ and sends $\Omega$ to $a\Omega$. It agrees with $T$ on the cyclic vector and therefore on all of $M\Omega$, hence on $H$. This proves $T\in JMJ$, and so

For the regular representation, take $\Omega ={\delta }_e$. It is cyclic for the left action and for the commuting right action. On finite sums of left translations, its vector functional is the identity coefficient, so $\tau (ab)=\tau (ba)$. This extends to the whole group von Neumann algebra by the bounded strong-star approximations of Question 6, taking limits of the vector functional and products. The conjugation is explicitlyConsequentlyThus the commutant in (b) is exactly the Von Neumann algebra of right regular translations, with the inverse in $\rho (g)$ fixing the representation convention.

The Von Neumann double commutant theorem states that for a unital star-subalgebra $A\subseteq B(H)$,The commutant of an operator algebra consists of operators commuting with every element. Each commutation equation is closed in the weak operator topology, so $A^{\prime \prime }$ is weakly closed and contains both indicated closures.

Let $T\in A^{\prime \prime }$ and fix ${\xi }_1,\dots ,{\xi }_m\in H$. On $H^m$ let $A$ act diagonally and put $\xi =({\xi }_1,\dots ,{\xi }_m)$. The subspace $K=\overline{\{(a{\xi }_1,\dots ,a{\xi }_m):a\in A\}}$ reduces the diagonal star-algebra. Its orthogonal projection $P$ therefore commutes with every diagonal $a$. Writing $P$ as an $m\times m$ operator matrix shows that each entry belongs to $A^{\prime }$. Since $T\in A^{\prime \prime }$, the diagonal $T$ commutes with $P$. Unitality gives $\xi \in K$, so $P\xi =\xi$ and $P(T\xi )=T\xi$. Thus $T\xi \in K$. By definition of $K$, for any $\epsilon >0$ a single $a\in A$ satisfies ${\sum }_j\Vert (T-a){\xi }_j{\Vert }^2<{\epsilon }^2$. These are precisely the finite-vector neighbourhoods of the strong operator topology. Hence $T$ belongs to the strong closure, completing the proof. Unitality, or the corresponding nondegeneracy hypothesis, is essential to the theorem's stated form.

The estimateshows $\Vert {\mu }_{x,y}\Vert \le \Vert x\Vert \Vert y\Vert$. For nonzero $x,y$, the rank-one operatorhas norm one and attains equality. The zero cases are immediate.

Embed the operator unit ball intoby $T\mapsto (⟨Tx,y⟩{)}_{x,y}$. The product is compact by Tychonoff theorem. A pointwise limit of these coordinates is a bilinear form $\theta$ satisfyingBy the stated representation theorem for bounded bilinear forms, $\theta (x,y)=⟨Tx,y⟩$ for a unique operator with $\Vert T\Vert \le 1$. The image is therefore closed and compact. Its product topology is precisely the weak operator topology $\sigma (\mathcal{B}(H),M)$.

The linear span $Z=\mathrm{span}M$ separates operators, and the preceding compactness lets part (a) identify $\mathcal{B}(H)$ isometrically with $Z^{\ast }$. Thus $\mathcal{B}(H)$ is a dual Banach space. This is the operator predual from matrix coefficients.

Here functional calculus convergence means that for every $f\in C(\mathbb{T})$,

Embed ${\mathcal{H}}_n$ in $\mathcal{H}$ and write ${\tilde{A}}_n=P_n^{\ast }{A}_n{P}_n$. The hypothesis gives ${\tilde{A}}_n\rightharpoonup A$ in the weak operator topology. Since $A_n$ and $A$ are unitary operators,Thus ${\tilde{A}}_n\to A$ in the strong operator topology. Applying the same argument to the adjoints gives ${\tilde{A}}_n^{\ast }\to A^{\ast }$ strongly.

Products of uniformly bounded strongly convergent operators converge strongly, so for every integer $k$,strongly, with negative $k$ interpreted through adjoints. Therefore convergence holds for every Laurent polynomial. The Stone-Weierstrass theorem says that Laurent polynomials are uniformly dense in $C(\mathbb{T})$. Since the continuous functional calculus is contractive, uniform approximation finishes the proof for every $f\in C(\mathbb{T})$.

A unital star-subalgebra of the bounded operators on a Hilbert space that is closed in the weak operator topology, equivalently in the strong operator topology. The Von Neumann double commutant theorem makes these conditions equivalent to equality with the double commutant of an operator algebra. It is norm closed and hence a C-star algebra.

For a unital star-subalgebra $A$ of the bounded operators on a complex Hilbert space, its strong operator topology and weak operator topology closures equal its double commutant of an operator algebra. To approximate $T\in A^{\prime \prime }$ on finitely many vectors, take their tuple in $H^m$. The projection onto its diagonal-$A$ cyclic subspace has matrix entries in $A^{\prime }$, so it commutes with the diagonal action of $T$. Therefore the image tuple under $T$ lies in that cyclic subspace and can be approximated by a single $a\in A$.