Proof. First note that if $\varphi$ is a hyperstate on $\mathcal {A}$, then for all $T \in \mathcal {A}$ we have

\[ \varphi(T)= \langle T, \hat 1 \rangle_{\varphi} = \langle \mathcal{P}_\varphi(T) \hat 1, \hat 1 \rangle_\tau. \]

From this it follows that the correspondence $\varphi \mapsto \mathcal {P}_\varphi$ is one-to-one. To see that it is onto, suppose that $\mathcal {P}: \mathcal {A} \to \mathcal {B}(L^2(M, \tau ))$ is u.c.p. and $M$-bimodular. We define a state $\varphi$ on $\mathcal {A}$ by $\varphi (T) = \langle \mathcal {P}(T) \hat {1}, \hat {1} \rangle _\tau$. For all $y \in M$ we then have $\varphi (y) = \langle \mathcal {P}(y) \hat 1, \hat 1 \rangle _\tau = \tau (y)$, hence $\varphi$ is a hyperstate. Moreover, if $y, z \in M$, and $T \in \mathcal {A}$, then we have

(2)\begin{align} \langle \mathcal{P}_\varphi(T) \hat y, \hat z \rangle_\tau &= \langle \mathcal{P}_\varphi(z^* T y) \hat 1, \hat 1 \rangle_\tau \nonumber\\ &= \varphi(z^* T y)= \langle \mathcal{P}(T) \hat y, \hat z \rangle_\tau, \end{align}

hence, $\mathcal {P}_\varphi = \mathcal {P}$.

It is also easy to check that $\mathcal {P}_\varphi$ is normal if and only if $\varphi$ is.

To see that this correspondence is a homeomorphism when given the topologies above, suppose that $\varphi$ is a hyperstate, and ${\varphi _\alpha }$ is a net of hyperstates. From (2) and the fact that u.c.p. maps are contractions in norm we see that $\mathcal {P}_{{\varphi _\alpha }}$ converges in the pointwise ultraweak topology to $\mathcal {P}_\varphi$ if ${\varphi _\alpha }$ converges weak$^*$ to $\varphi$. Conversely, setting $y = z = 1$ in (2) shows that if $\mathcal {P}_{\varphi _\alpha }$ converges in the pointwise ultraweak topology to $\mathcal {P}_\varphi$, then $\varphi _\alpha$ converges weak$^*$ to $\varphi$.

Proof. By Proposition 2.1 the correspondence $\varphi \mapsto \mathcal {P}_\varphi$ is a homeomorphism from the weak$^*$ topology to the topology of pointwise ultraweak convergence. This lemma then follows easily from (3).

Proof. If we take any accumulation point $E$ of $\bigl \{ ({1}/{N}) \sum _{n = 1}^N \mathcal {P}_\varphi ^n\bigr \}_{N \in \mathbb {N}}$ in the topology of pointwise ultraweak convergence, then $E: \mathcal {B}(L^2(M, \tau ) ) \to {\rm Har}( \mathcal {B}(L^2(M, \tau ) ), \mathcal {P}_\varphi )$ gives a u.c.p. projection. As $\mathcal {B}_\varphi$ is isomorphic to ${\rm Har}( \mathcal {B}(L^2(M, \tau ) ), \mathcal {P}_\varphi )$ as an operator system, it then follows that $\mathcal {B}_\varphi$ is injective [Reference Choi and EffrosCE77, § 3].

Proof. If the support of $\mu$ generates von Neumann algebra $M_0 \subset M$ such that $M_0 \not = M$, then we have $[JxJ, e_{M_0}] = [Jx^* J, e_{M_0} ] = 0$ for each $x$ in the support of $\mu$. Hence, $\mathcal {P}_\varphi (T) = \int (J x J) T (J x^* J) \,d\mu (x) = T$, for each $T$ in the $*$-algebra generated by $M$ and $e_{M_0}$. Therefore, $\varphi$ is not generating. On the other hand, if $T \in {\rm Har}(\mathcal {B}(L^2(M, \tau )), \mathcal {P}_\varphi )$ is such that we also have $T^*T, TT^* \in {\rm Har}(\mathcal {B}(L^2(M, \tau )), \mathcal {P}_\varphi )$, then for each $a \in M$ we have

\begin{align*} & \int \| ( (J x J) T - T (J x J) ) \hat{a} \|_2^2 \,d\mu(x) \\ &\quad = \langle (T^* \mathcal{P}_\varphi(1) T - \mathcal{P}_\varphi(T^*) T - T^* \mathcal{P}_\varphi(T) + \mathcal{P}_\varphi(T^*T) ) \hat{a}, \hat{a} \rangle = 0, \end{align*}

and by symmetry we also have $\int \| ( (J x J) T^* - T^* (J x J) ) \hat {a} \|_2^2 \,d\mu (x) = 0$. Hence, $[Jx J, T] = [J x^* J, T] = 0$ for $\mu$-almost every $x \in (M)_1$. Therefore, if the support of $\mu$ generates $M$ as a von Neumann algebra, then $T \in JMJ' = M$, showing that $\varphi$ is generating, thereby proving (i).

If $y \in M$, then we have $\mathcal {P}_\varphi (JyJ) = \int J x^* y x J \,d\mu (x)$. Hence, we see that $\varphi$ is regular if and only if for all $y \in M$ we have $\tau (y) = \int \tau (x^* y x) \,d\mu (x) = \int \tau (xx^* y) \,d\mu (x)$, which is if and only if $\int xx^* \,d\mu (x) = 1$, thereby proving (ii).

If $\varphi$ is regular, then

\begin{align*} \varphi^*(T) &= \varphi( J T^* J ) = \int \langle J T^* J \widehat{x^*}, \widehat{x^*} \rangle \,d\mu(x) \\ &= \int \langle \hat{x}, T^* \hat{x} \rangle \,d\mu(x) = \int \langle T \widehat{x^*}, \widehat{x^*} \rangle \,d J_*\mu(x). \end{align*}

Therefore, if $J_*\mu = \mu$, then $\varphi$ is symmetric, thereby proving (iii).

Proof. ${\rm Har}( A, \mathcal {P})$ is clearly a self-adjoint closed subspace, thus we must show that ${\rm Har}( A, \mathcal {P})$ is an algebra. By the polarization identity it is enough to show that $x^* x \in {\rm Har}( A, \mathcal {P})$ whenever $x \in {\rm Har}( A, \mathcal {P})$. Suppose $x \in {\rm Har}(A, \mathcal {P})$. By Kadison's inequality we have $\mathcal {P}(x^*x) - x^* x = \mathcal {P}(x^*x) - \mathcal {P}(x^*) \mathcal {P}(x) \geq 0$. Also, $\varphi ( \mathcal {P}(x^*x) - x^*x) = 0$ so that by faithfulness of $\varphi$ we have $\mathcal {P}(x^*x) = x^*x$.

Proof. Let $\mathcal {P}: \mathcal {B}_\varphi \to {\rm Har}(\mathcal {B}(L^2(M, \tau )), \mathcal {P}_\varphi )$ denote the Poisson transform. If $x \in M' \cap \mathcal {B}_\varphi$, then $\mathcal {P}(x) \in M' \cap \mathcal {B}(L^2(M, \tau )) = J M J$. Since $\varphi$ is regular, $\mathcal {P}_\varphi$ preserves the trace when restricted to $J M J$. Thus, ${\rm Har}(J M J,\mathcal {P}_\varphi )$ is a von Neumann subalgebra of $J M J$ by Lemma 2.6. Since $\varphi$ is generating, $M$ is the largest von Neumann subalgebra of ${\rm Har}(\mathcal {B}(L^2(M, \tau )))$, and hence ${\rm Har}(J M J,\mathcal {P}_\varphi ) \subseteq M$, implying that ${\rm Har}(J M J,\mathcal {P}_\varphi )=\mathcal {Z}(M)$. Therefore, $\mathcal {P}(x) \in {\rm Har}(J M J, \mathcal {P}_\varphi ) = \mathcal {Z}(M)$, and hence $x \in \mathcal {Z}(M)$ since $\mathcal {P}$ is injective.

Proof. Since $A_\varphi$ is a positive trace-class operator we may write $A_\varphi = \sum _n a_n P_{y_n}$, where $a_1, a_2, \ldots$ are positive and $\{ y_n \}_n$ is an orthonormal family with $P_{y_n}$ denoting the rank-one projection onto $\mathbb {C} y_n$. For $T \in \mathcal {B}(L^2(M, \tau ))$ we then have

\[ \operatorname{Tr} (T A_\varphi)= \sum_n a_n \langle T y_n, y_n \rangle. \]

Taking $T = x^*x \in M$, we have $a_n \| x y_n \|_2^2 \leq \operatorname {Tr} (x^* x A_\varphi ) = \| x \|_2^2$, so that $y_n \in M \subset L^2(M, \tau )$ for each $n$. Hence, for $T \in \mathcal {B}(L^2(M, \tau ))$ we have

\begin{align*} \operatorname{Tr} (\mathcal{P}_\varphi(T) P_{\hat{1}} ) &= \operatorname{Tr} (T A_\varphi) = \bigg\langle \sum_{n} a_n (J y_n J) T (J y_n^* J) \hat{1}, \hat{1}\bigg\rangle \\ &= \operatorname{Tr} \bigg(\bigg(\sum_{n} a_n (J y_n J) T (J y_n^* J) \bigg) P_{\hat{1}} \bigg). \end{align*}

Since $\mathcal {P}_\varphi$ is $M$-bimodular and since $J y_n J \in M'$, it follows that for all $x, y \in M$ we have

\[ \operatorname{Tr} (\mathcal{P}_\varphi(T) x P_{\hat{1}} y) = \operatorname{Tr} \bigg(\bigg(\sum_{n} a_n (J y_n J) T (J y_n^* J)\bigg) x P_{\hat{1}} y \bigg). \]

In particular, setting $T = y = 1$, we have

\[ \tau(x) = \sum_n a_n \tau(y_n^* y_n x), \]

which shows that $\sum _n a_n y_n^* y_n = 1$.

Since the span of operators of the form $x P_{\hat {1}} y$ is dense in the space of trace-class operators, it then follows that $\mathcal {P}_\varphi (T) = \sum _{n} a_n (J y_n J) T (J y_n^* J)$ for all $T \in \mathcal {B}(L^2(M, \tau ))$. Setting $z_n = \sqrt {a_n} y_n^*$ then finishes the existence part of the proposition.

Suppose now that $\{ \tilde z_m \}_m$ is a $\tau$-orthogonal family which gives a partition of the identity $1 = \sum _n \tilde z_n^* \tilde z_n$, and set $\tilde \varphi (T) = \operatorname {Tr} ( ( \sum _n (J \tilde z_n^* J) T (J \tilde z_n J) ) P_{\hat {1}} )$. Then, the density matrix $A_{\tilde \varphi }$, corresponding to $\tilde \varphi$, is given by $A_{\tilde \varphi }=\sum _n \tilde z_n^* P_{\hat {1}} \tilde z_n$. Since $\{ \tilde z_n \}_n$ forms a $\tau$-orthogonal family it then follows easily that $\tilde z_n^*$ is an eigenvector for $A_{\tilde \varphi }$, and the corresponding eigenvalue is $\| \tilde z_n^* \|_2^2 = \| \tilde z_n \|_2^2$.

Using our notation from the first part of the proof of the proposition, we have that $A_{ \varphi }=\sum _n z_n^* P_{\hat {1}} z_n$. By the same argument as above, we get that $z_n^*$ is an eigenvector for $A_{ \varphi }$, and the corresponding eigenvalue is $\| z_n^* \|_2^2 = \| z_n \|_2^2$. Note that $\mathcal {P}_{\varphi }= \mathcal {P}_{\tilde \varphi }$ if and only if $A_{ \varphi }=A_{ \tilde \varphi }$. Since the corresponding density matrices are positive trace class operators, the moreover part of the proposition follow easily from the spectral theorem.

Proof. By considering the Poisson transform $\mathcal {P}$, it suffices to show that $\varphi$ is normal and faithful on the operator system $\operatorname {Har}(\mathcal {P}_\varphi )$. Note that here the stationary state is a vector state and hence normality follows. To see that the state is faithful fix $T \in \operatorname {Har}(\mathcal {P}_\varphi )$, with $T \geq 0$ and $\langle T \hat {1}, \hat {1} \rangle = 0$. Let $\mathcal {P}_\varphi (S)= \sum _n (Jz_n^*J) S (J z_n J)$ be the standard form of $\mathcal {P}_\varphi$. Since $T \in \operatorname {Har}(\mathcal {P}_\varphi )$, we have that $\mathcal {P}_\varphi ^k(T) =T$ for each $k \in \mathbb {N}$. Expanding the standard form gives

\[ 0 = \langle T \hat{1}, \hat{1} \rangle = \langle P_\varphi^k(T) \hat{1}, \hat{1} \rangle = \sum_{n_1, n_2, \ldots, n_k} \langle T z_{n_1} z_{n_2} \cdots z_{n_k} \hat{1}, z_{n_1} z_{n_2} \cdots z_{n_k} \hat{1} \rangle. \]

We then have $T \hat {m}=0$ for all $m$ in the unital algebra generated by $\{ z_n \}$, and as $\varphi$ is strongly generating it then follows that $T=0$.

Proof. The first condition trivially implies the second. To see that the second implies the third, suppose for each $x \in M$ that we have $xA_n-A_nx \rightarrow 0$ weakly as $n \rightarrow \infty$. Let $T \in \operatorname {Har}(\mathcal {P}_{\varphi })$. Let $x, a,b \in M$. Then, taking inner products in $L^2(M, \tau )$, we have

\begin{align*} |\langle (TJxJ- JxJ T) a\hat{1}, b\hat{1} \rangle| &= |\langle (b^*Tax^*-x^*b^*Ta) \hat{1}, \hat{1} \rangle | \\ &= | \langle \mathcal{P}_{\varphi}^n(b^*Tax^*-x^*b^*Ta) \hat{1}, \hat{1} \rangle | = | \operatorname{Tr} (A_n (b^*Tax^*-x^*b^*Ta) )| \\ &= | \operatorname{Tr} ((x^*A_n-A_nx^*)b^*Ta)| \rightarrow 0. \end{align*}

Hence $T \in JMJ' = M$.

To see that the third condition implies the first we adapt the approach of Foguel from [Reference FoguelFog75]. Suppose $\operatorname {Har}(\mathcal {P}_{\varphi })=M$. Set $\mathcal {A}_0 = \{ A \in {\rm TC}(L^2(M, \tau )) \mid \|(\mathcal {P}_{\varphi }^n)^*(A) \|_{\rm TC} \to 0 \}$. Note that since $(\mathcal {P}_{\varphi }^n)^*$ is a contraction in the trace-class norm we have that $\mathcal {A}_0$ is a closed subspace.

Since $\varphi = \frac {1}{2} \psi + \frac {1}{2} \langle \cdot \hat 1, \hat 1 \rangle$, we have $\mathcal {P}_\varphi ^* = \frac {1}{2} {\rm id} + \frac {1}{2} \mathcal {P}_\psi ^*$ and we compute

\begin{align*} (\mathcal{P}_\varphi^n)^* ({\rm id} - \mathcal{P}_\varphi^*) &= 2^{-(n + 1)} \bigg(\sum_{k = 0}^n \binom{n}{k} (\mathcal{P}_\psi^k)^* \bigg) ({\rm id} - \mathcal{P}_\psi^*) \\ &= 2^{-(n + 1)} \sum_{k = 1}^n \bigg(\binom{n}{k-1} - \binom{n}{k} \bigg) \mathcal{P}_{\psi}^*. \end{align*}

We have $\lim _{n \to \infty } 2^{-(n + 1)} \sum _{k = 1}^n | \binom {n}{k-1} - \binom {n}{k} | = 0$ (see (1.8) in [Reference Ornstein and SuchestonOS70]) hence $\| (\mathcal {P}_\varphi ^n)^*( P_{\hat 1} - \mathcal {P}_\varphi ^*(P_{\hat 1}) )\|_{\rm TC} \to 0$. Thus $P_{\hat 1} - \mathcal {P}_{\varphi }^*(P_{\hat 1}) \in \mathcal {A}_0$.

Since $\mathcal {P}_{\varphi }^*$ is $M$-bimodular we then have that $a P_{\hat 1} b - \mathcal {P}_{\varphi }^*(a P_{\hat 1} b) \in \mathcal {A}_0$ for each $a, b \in M$ and hence $B - \mathcal {P}_{\varphi }^*(B) \in \mathcal {A}_0$ for all $B \in {\rm TC}(L^2(M, \tau ))$. If $T \in \mathcal {B}(L^2(M, \tau ))$ is such that $\operatorname {Tr} (AT) = 0$ for all $A \in \mathcal {A}_0$, then for all $B \in {\rm TC}(L^2(M, \tau ))$ we have $\langle B - \mathcal {P}_{\varphi }^*(B), T \rangle = 0$ so that $T \in {\rm Har}(\mathcal {P}_{\varphi }) = M$. Hence the annihilator of $\mathcal {A}_0$ is contained in $M.$ So the pre-annihilator of $M$ must be contained in $\mathcal {A}_0$. Thus $A \in \mathcal {A}_0$ whenever $\operatorname {Tr} (A x) = 0$ for all $x \in M$. In particular, we have $x P_{\hat 1} - P_{\hat 1} x \in \mathcal {A}_0$ for all $x \in M$, which is equivalent to the fact that $\| x A_n - A_n x \|_{\rm TC} \to 0$ for each $x \in M$.

Proof. Let $A_{\varphi }$ and $A_{\psi }$ be the corresponding density operators and $\mathcal {P}_{\varphi }$ and $\mathcal {P}_{\psi }$ be the corresponding u.c.p. $M$-bimodular maps. Suppose we have the standard forms

\begin{align*} \varphi(T) &= \sum_{i \in I} \langle T \mu_i^{1/2} \widehat{a_i^*}, \mu_i^{1/2} \widehat{a_i^*} \rangle\quad \text{with }\mu_i >0,\ \|a_i^*\|_2=1, \text{ and } \tau(a_ja_i^*)=0 \text{ for all }i \neq j \in I, \\ \psi(T) &= \sum_{j \in J} \langle T \nu_j \widehat{c_j^*}, \nu_j \widehat{c_j^*} \rangle\quad \text{with }\nu_j >0,\ \|c_j^*\|_2=1, \text{ and } \tau(c_kc_l^*)=0 \text{ for all }k \neq l \in J. \end{align*}

Hence $A_{\varphi }= \sum _i \mu _i P_{\hat {a_i}}$ and $A_{\psi }= \sum _j \nu _j P_{\hat {c_j}}$.

Let $b_i=Ja_iJ$ and $d_i=Jc_iJ$ so that

\[ \mathcal{P}_{\varphi}(T)=\sum_i \mu_ib_iTb_i^{*} \quad \text{and}\quad \mathcal{P}_{\psi}(T)=\sum_j \nu_jd_j T d_j^{*}. \]

Since $\psi$ is regular we have that $\sum _i \nu _i d_i^{*}d_i=\sum _i \nu _i d_id_i^{*}=1$. Since $\varphi$ is a hyperstate we have that $\sum _i \mu _i b_ib_i^{*}=1$. Now

\[ H(\varphi \ast \psi)=-\sum_{i,j}\operatorname{Tr}[\mu_i \nu_j b_i^{*}d_j^{*}P_{\hat{1}}d_jb_i \log(A_{\varphi \ast \psi})] \]

and

\[ b_i^{*}d_j^{*}P_{\hat{1}}d_jb_i=\tau(b_ib_i^{*}d_j^{*}d_j)P_{\widehat{b_i^{*}d_j^{*}}}, \]

so that for each $k, \ell$ we have

\[ A_{\varphi \ast \psi}=\sum_{i,j} \mu_i \nu_j b_i^{*}d_j^{*}P_{\hat{1}}d_jb_i \geq \mu_k \nu_\ell \tau(b_kb_k^{*}d_\ell^{*}d_\ell)P_{\widehat{b_k^{*}d_\ell^{*}}}. \]

As $\log$ is operator monotone, for each $k, \ell$ we then have

\[ -\log(A_{\varphi \ast \psi})=-\log\bigg(\sum_{i,j}\mu_i \nu_j b_i^{*}d_j^{*}P_{\hat{1}}d_jb_i\bigg) \leq -\log((\mu_k \nu_\ell \tau(b_kb_k^{*}d_\ell^{*}d_\ell))P_{\widehat{b_k^{*}d_\ell^{*}}}). \]

Hence,

\begin{align*} H(\varphi \ast \psi) &\leq -\sum_{i,j}\operatorname{Tr}[\mu_i \nu_j\tau(b_ib_i^{*}d_j^{*}d_j) P_{\widehat{b_i^{*}d_j^{*}}}\log(\mu_i \nu_j\tau(b_ib_i^{*}d_j^{*}d_j) P_{\widehat{b_i^{*}d_j^{*}}})]\\ &= -\sum_{i,j}\operatorname{Tr}[\mu_i \nu_j\tau(b_ib_i^{*}d_j^{*}d_j)P_{\widehat{b_i^{*}d_j^{*}}} \log(\mu_i \nu_j\tau(b_ib_i^{*}d_j^{*}d_j))] \\ &\quad -\sum_{i,j}\operatorname{Tr}[\mu_i \nu_j\tau(b_ib_i^{*}d_j^{*}d_j) P_{\widehat{b_i^{*}d_j^{*}}} \log(P_{\widehat{b_i^{*}d_j^{*}}})]\\ &=-\sum_{i,j}\mu_i \nu_j\tau(b_ib_i^{*}d_j^{*}d_j) \log(\mu_i \nu_j\tau(b_ib_i^{*}d_j^{*}d_j)). \end{align*}

Now define $m$ on $I \times J$ by $m(i,j)=\mu _i \nu _j\tau (b_ib_i^{*}d_j^{*}d_j)$. Note that

\[ \sum_im(i,j)=\nu_j \tau\bigg(\sum_i \mu_i b_ib_i^{*}d_j^{*}d_j\bigg)=\nu_j\tau(d_j^{*}d_j)=\nu_j \]

and

\[ \sum_jm(i,j)=\mu_i \tau\bigg(\sum_i \nu_j b_ib_i^{*}d_j^{*}d_j\bigg)=\mu_i\tau(b_ib_i^{*})=\mu_i. \]

To finish the proof it then suffices to show

\[ H(m)=- \sum_{i,j}m(i,j)\log(m(i,j)) \leq H(\mu)+ H(\nu), \]

where $H(\mu )=-\sum _i \mu _i \log (\mu _i)$ and $H(\nu )=-\sum _i \nu _i \log (\nu _i)$. By the remark before Theorem 5.1, a direct calculation yields $H(\mu )= H(\varphi )$ and $H(\nu )= H(\psi )$.

Note that

\begin{align*} H(m) &=- \sum_{i,j}m(i,j)\log(m(i,j))\\ &= - \sum_{i,j} \mu_i \nu_j \tau( b_ib_i^{*}d_j^{*}d_j) \log(\mu_i\tau(b_ib_i^{*}d_j^{*}d_j)) -\sum_{i,j} \mu_i \nu_j \tau( b_ib_i^{*}d_j^{*}d_j) \log(\nu_j)\\ &= -\sum_{i,j} \mu_i \nu_j \tau( b_ib_i^{*}d_j^{*}d_j) \log(\mu_i) \\ &\quad -\sum_{i,j} \mu_i \nu_j \tau( b_ib_i^{*}d_j^{*}d_j) \log(\nu_j) - \sum_{i,j} \mu_i \nu_j \tau( b_ib_i^{*}d_j^{*}d_j) \log(\tau(b_ib_i^{*}d_j^{*}d_j)). \end{align*}

In the last equality above, the first summation is $H(\mu )$, since summing over $j$ yields

\[ -\sum_i \mu_i \tau(b_ib_i^{*})\log(\mu_i)=-\sum_i \mu_i \log(\mu_i), \]

while the second summation is $H(\nu )$. Hence, all that remains is to show

\[ \sum_{i,j} \mu_i \nu_j \tau( b_ib_i^{*}d_j^{*}d_j) \log(\tau(b_ib_i^{*}d_j^{*}d_j)) \geq 0. \]

Let $\eta (x)=-x\log (x)$ for $x \in [0,1]$. Note that $\eta$ is concave, and so $\eta (\sum _{i}\alpha _ix_i) \geq \sum _i \alpha _i \eta (x_i)$ whenever $\alpha _i \geq 0$ and $\sum _i \alpha _i =1$. So

\begin{align*} -\sum_{i,j} \mu_i \nu_j \tau( b_ib_i^{*}d_j^{*}d_j) \log(\tau(b_ib_i^{*}d_j^{*}d_j)) &= \sum_{i,j} \mu_i \nu_j \eta(\tau( b_ib_i^{*}d_j^{*}d_j))\\ &= \sum_i \mu_i \bigg(\sum_j \nu_j \eta(\tau( b_ib_i^{*}d_j^{*}d_j))\bigg)\\ &\leq \sum_i \mu_i \eta\bigg(\sum_j \nu_j \tau( b_ib_i^{*}d_j^{*}d_j)\bigg)\\ &= \sum_i \mu_i \eta(\tau(b_ib_i^{*}))=0. \end{align*}

Proof. The sequence $\{H(\varphi ^{*n})\}$ is subadditive by Theorem 5.1 and hence the limit exists.

Proof. As $\mathcal {A}1_{\zeta }$ forms a core for $S_{\zeta }$ we get that $z_n^*1_{\zeta } \in D(\log (\Delta _{\zeta }))$. Also, we know that

\[ \lim_{t \rightarrow 0}\frac{\Delta_{\zeta}^{it}-1}{t}\xi= i \log(\Delta_{\zeta})\xi, \]

for all $\xi \in D(\Delta _{\zeta })$. So we have that

\begin{align*} h_\varphi(M \subset \mathcal{A}, \zeta) &= -\varphi(e\log(\Delta_{\zeta})e) = -\sum_n \langle \log \Delta_\zeta z_n^* 1_\zeta, z_n^* 1_\zeta \rangle \\ &= i \sum_n \bigg\langle z_n \lim_{t \rightarrow 0} \frac{\Delta_{\zeta}^{it}-1}{t} z_n^* 1_{\zeta}, 1_{\zeta} \bigg\rangle = i \lim_{t \to 0} \frac{1}{t} \sum_n (\zeta( z_n \sigma_t^\zeta(z_n^*) )-1). \end{align*}

Proof. Let $\mathcal {P}_{\zeta }(T)=eTe$ for $T \in \mathcal {A}$. Let $\Delta _n = \int _{1/n}^{n} \lambda d\lambda$, $n \geq 1,$ denote the truncations of the modular operator $\Delta$.

Using the operator Jensen's inequality, we have

\begin{align*} h_{\varphi}(M \subset \mathcal{A}, \zeta) &= \lim_{n \rightarrow \infty}\varphi(-e \log\Delta_n e)\\ &=-\lim_{n \rightarrow \infty}\langle \mathcal{P}_{\varphi} \circ \mathcal{P}_{\zeta} (\log \Delta_n) \hat{1}, \hat{1} \rangle \geq \lim_{n \rightarrow \infty} -\langle \log (\mathcal{P}_{\varphi} \circ \mathcal{P}_{\zeta} (\Delta_n)) \hat{1}, \hat{1} \rangle \end{align*}

(recall that $\log$ is operator concave).

Notice that $e\Delta _n e \leq e\Delta e =e$. Since $\mathcal {P}_{\varphi }(e)=e$, we get $\mathcal {P}_{\varphi } \circ \mathcal {P}_{\zeta } (\Delta _n) \leq e \leq 1$. As $\log$ is operator monotone, we get that $\log (\mathcal {P}_{\varphi } \circ \mathcal {P}_{\zeta } (\Delta _n)) \leq \log (1)=0$. Hence we are done.

Proof. Suppose we have the standard forms

\begin{align*} \varphi(T) &= \sum_{i \in I} \langle T \mu_i^{1/2} \hat{a_i^*}, \mu_i^{1/2} \hat{a_i^*} \rangle\quad \text{with }\mu_i >0,\ \|a_i^*\|_2=1, \text{ and } \tau(a_ja_i^*)=0 \text{ for all }i \neq j \in I, \\ \psi(T) &= \sum_{j \in J} \langle T \nu_j \hat{b_j^*}, \nu_j \hat{b_j^*} \rangle\quad \text{with }\nu_j >0,\ \|b_j^*\|_2=1, \text{ and } \tau(b_kb_l^*)=0 \text{ for all }k \neq l \in J. \end{align*}

Let $\mathcal {P}_{\varphi }$ and $\mathcal {P}_{\psi }$ be the corresponding u.c.p. maps so that $\mathcal {P}_{\varphi }(T)= \sum _k \mu _k Ja_k^*J T Ja_kJ$ and $\mathcal {P}_{\psi }(T)= \sum _l \nu _l Jb_l^*J T Jb_lJ$. We shall denote the projection from $L^2(\mathcal {A}, \zeta )$ to $L^2(M,\tau )$ by $e$ and $\Delta _{\zeta }$ by $\Delta$. We also denote the one-parameter modular automorphism group corresponding to $\zeta$ by $\sigma _t$. We then have

\begin{align*} h_{\varphi}(M \subset \mathcal{A}, \zeta) &= i \lim_{t \rightarrow 0} \varphi\bigg(\frac{e\Delta^{it}e-1}{t}\bigg)= i \lim_{t \rightarrow 0} \frac{1}{t}\varphi(e\Delta^{it}e-1) \\ &= i \lim_{t \rightarrow 0} \frac{1}{t}\bigg(\sum_k \mu_k \langle (\Delta^{it}-1 )a_k^* 1_\zeta, a_k^* 1_{\zeta} \rangle\bigg). \end{align*}

Similarly,

\[ h_{\psi}(M \subset \mathcal{A}, \zeta)=i \lim_{t \rightarrow 0} \frac{1}{t}\bigg(\sum_l \nu_l \langle (\Delta^{it}-1 )b_l^* 1_\zeta, b_l^* 1_{\zeta} \rangle\bigg) \]

and

\begin{align*} h_{\varphi * \psi}(M \subset \mathcal{A}, \zeta) &= i \lim_{t \rightarrow 0} \frac{1}{t}\bigg(\sum_{k,l} \mu_k \nu_l \langle (\Delta^{it}-1 )a_k^*b_l^* 1_\zeta, a_k^* b_l^*1_{\zeta} \rangle\bigg) \\ &= i \lim_{t \rightarrow 0} \frac{1}{t} \bigg(\sum_{k,l} \mu_k \nu_l \langle (b_la_k\sigma_t(a_k^*b_l^*) 1_\zeta, 1_{\zeta} \rangle -1\bigg). \end{align*}

We shall now show that $\lim _{t \rightarrow 0} \frac {1}{t}\big(\sum _{k,l} \mu _k \nu _l \langle b_la_k\sigma _t(a_k^*b_l^*) 1_\zeta, 1_{\zeta } \rangle -\sum _{k,l} \mu _k \nu _l \langle b_l\sigma _t(b_l^*)\sigma _t(a_k^*) 1_\zeta,$ $1_{\zeta } \rangle \big) =0.$ Let $y_t=a_k\sigma _t(a_k^*).$ Note that $y_t \rightarrow a_ka_k^*$ as $t \rightarrow 0$, in the strong operator topology. We have

\begin{align*} y_t\sigma_t(b_l^*)-\sigma_t(b_l^*)y_t &= y_t\sigma_t(b_l^*)-y_tb_l^*+y_t b_l^*-\sigma_t(b_l^*)y_t \\ &= y_t (\sigma_t(b_l^*)- b_l^*)+ (y_t b_l^*-b_l^*y_t)+(b_l^*-\sigma_t(b_l^*))y_t. \end{align*}

Now

\begin{align*} \frac{1}{t}\bigg(\sum_{k,l} \mu_k \nu_l \langle (y_t b_l^*-b_l^*y_t) 1_{\zeta}, b_l^* 1_{\zeta} \rangle\bigg) &= \frac{1}{t}\bigg(\sum_{k,l} \mu_k \nu_l \langle b_ly_t b_l^* 1_{\zeta}, 1_{\zeta} \rangle\bigg) - \frac{1}{t}\bigg(\sum_{k,l} \mu_k \nu_l \langle y_t 1_{\zeta}, b_lb_l^* 1_{\zeta} \rangle\bigg) \\ &= \frac{1}{t} \sum_k \mu_k \bigg\langle \bigg(\sum_l \nu_l b_l y_t b_l^*\bigg) 1_{\zeta}, 1_{\zeta} \bigg\rangle - \frac{1}{t} \sum_k \mu_k \langle y_t 1_{\zeta}, 1_{\zeta} \rangle\\ &= \frac{1}{t} \langle y_t 1_{\zeta}, 1_{\zeta} \rangle- \frac{1}{t} \langle y_t 1_{\zeta}, 1_{\zeta} \rangle =0, \end{align*}

where the penultimate equality holds by $\psi$-stationarity of $\zeta$.

Also, $\lim _{t \rightarrow 0} ({1}/{t}) (y_t (\sigma _t(b_l^*)-b_l^*))$ exists, and hence

\[ \lim_{t \rightarrow 0} \frac{1}{t}\bigg(\sum_{k,l} \mu_k \nu_l \langle b_la_k\sigma_t(a_k^*b_l^*) 1_\zeta, 1_{\zeta} \rangle -\sum_{k,l} \mu_k \nu_l \langle b_l\sigma_t(b_l^*)\sigma_t(a_k^*) 1_\zeta, 1_{\zeta} \rangle\bigg) =0. \]

So we get that

\begin{align*} h_{\varphi * \psi}(M \subset \mathcal{A} , \zeta )&= i \lim_{t \rightarrow 0} \frac{1}{t}\bigg(\sum_{k,l} \mu_k \nu_l \langle (b_l\sigma_t(b_l^*)a_k\sigma_t(a_k^*)-1) 1_\zeta, 1_{\zeta} \rangle\bigg) \\ &= i \lim_{t \rightarrow 0} \frac{1}{t}\bigg(\sum_{k,l} \mu_k \nu_l[\langle(b_l\sigma_t(b_l^*)-1) 1_{\zeta}, 1_{\zeta} \rangle \\ &\quad + \langle (a_k\sigma_t(a_k^*)-1) 1_{\zeta}, 1_{\zeta} \rangle \\ &\quad\vphantom{\sum_{k,l}} + \langle (a_k\sigma_t(a_k^*)-1) 1_{\zeta}, (b_l \sigma_t(b_l^*)-1)^* 1_{\zeta} \rangle]\bigg). \end{align*}

The first term equals $h_{\varphi }(M \subset \mathcal {A}, \zeta )$, while the second term equals $h_{\psi }(M \subset \mathcal {A}, \zeta )$, and the third term equals zero, as $\lim _{t \rightarrow 0} ({1}/{t}) (a_k\sigma _t(a_k^*)-1) 1_{\zeta }$ exists, while ${\lim _{t \rightarrow 0} \sum _l \nu _l (b_l \sigma _t(b_l^*)-1)^* 1_{\zeta } =0}$.

Proof. We continue with the notation from the proof of Theorem 5.10, so that $\mathcal {P}_{\varphi }(T)= \sum _k \mu _k b_kTb_k^*$. Let $a_k=Jb_kJ \in M$. It follows from Lemma 5.7 that

\[ H(\varphi)=-\lim_{t \rightarrow 0+}\frac{\sum_{k=1}^{\infty}\mu_k\|A_{\varphi}^{t/2}a_k^* \hat{1}\|^2-\|a_k^*\hat{1}\|^2}{t}. \]

So by Corollary 5.8 it is enough to show that

\[ \lim_{t \rightarrow 0+}\frac{\sum_{k=1}^{\infty}\mu_k\|A_{\varphi}^{t/2}a_k^* \hat{1}\|^2-\|a_k^*\hat{1}\|^2}{t} \leq \lim_{t \rightarrow 0+} \frac{\sum_{k=1}^{\infty}\mu_k\|\Delta_{\varphi}^{t/2}ea_k^*\hat{1}\|^2-\|ea_k^*\hat{1}\|^2}{t}. \]

So it is enough to show that

\[ \|A_{\varphi}^{t/2}a_k\hat{1}\|^2 \leq \|\Delta_{\zeta}^{t/2}a_k1_{\zeta}\|^2. \]

Define $T: L^2(\mathcal {A},\zeta ) \rightarrow L^2(M,\tau )$ by $T(a1_{\zeta })=\mathcal {P}_{\zeta }(a)\hat {1}$. Then $\|T\| =1$, as $\|T(1_{\zeta })\|=1$ and $\|\mathcal {P}_{\zeta }\| \leq 1$. $T$ takes $\mathcal {D}(\Delta _{\zeta })$ into $\mathcal {D}(A_{\varphi })=L^2(M,\tau )$. We now denote $\Delta _{\zeta }$ by $\Delta$. By Lemma 5.6 it is enough to show that

\[ \|A_{\varphi}^{1/2}T \xi\| \leq \|\Delta^{1/2} \xi\| \quad \text{for all } \xi \in \mathcal{D}(\Delta). \]

In fact it is enough to show the above for all vectors in a core for $\mathcal {D}(\Delta )$. Recall that $\mathcal {A}1_{\zeta }$ forms a core for $\mathcal {D}(\Delta )$. So we only need to show that

\[ \|A_{\varphi}^{1/2}Ta1_{\zeta}\| \leq \|\Delta^{1/2}a1_{\zeta}\| \quad \text{for all } a \in \mathcal{A}. \]

To this end, let $a \in \mathcal {A}$. Recall that $S=J \Delta ^{1/2}$, so that $\Delta ^{1/2}=JS$. We then have

\begin{align*} \|\Delta^{1/2}a1_{\zeta}\|^2 &= \langle \Delta^{1/2} a1_{\zeta}, \Delta^{1/2} a1_{\zeta} \rangle = \langle JS a 1_{\zeta}, JS a 1_{\zeta} \rangle \\ &= \langle J a^* 1_{\zeta}, J a^* 1_{\zeta} \rangle = \langle a^* 1_{\zeta},a^* 1_{\zeta} \rangle = \zeta(aa^*) \\ &= \langle \mathcal{P}_{\zeta}(aa^*) \hat{1}, \hat{1} \rangle. \end{align*}

We also have $\mathcal {P}_{\varphi } \circ \mathcal {P}_{\zeta } = \mathcal {P}_{\zeta } \implies \varphi \circ \mathcal {P}_{\zeta } = \zeta$. Now

\begin{align*} \|A_{\varphi}^{1/2}Ta1_{\zeta}\|^2 &= \langle A_{\varphi}^{1/2} \mathcal{P}_{\zeta}(a) \hat{1}, A_{\varphi}^{1/2} \mathcal{P}_{\zeta}(a) \hat{1} \rangle = \langle A_{\varphi} \mathcal{P}_{\zeta}(a) \hat{1}, \mathcal{P}_{\zeta}(a) \hat{1} \rangle \\ &= \langle \mathcal{P}_{\zeta}(a)^* A_{\varphi} \mathcal{P}_{\zeta}(a) \hat{1}, \hat{1} \rangle \leq \operatorname{Tr}(\mathcal{P}_{\zeta}(a)^*A_{\varphi}\mathcal{P}_{\zeta}(a))\\ &= \operatorname{Tr}(A_{\varphi}\mathcal{P}_{\zeta}(a)\mathcal{P}_{\zeta}(a^*)) \leq \operatorname{Tr}(A_{\varphi}\mathcal{P}_{\zeta}(aa^*))\\ &= \langle (\varphi \circ \mathcal{P}_{\zeta})(aa^*) \hat{1},\hat{1} \rangle = \langle \mathcal{P}_{\zeta}(aa^*) \hat{1}, \hat{1} \rangle \\ &= \zeta(aa^*)=\| \Delta ^{1/2} a 1_{\zeta}\|^2. \end{align*}

Hence, we are done.

Proof. By Lemma 5.12, we have that $h_{\varphi ^{*n}}(M \subset \mathcal {A}, \zeta ) \leq H(\varphi ^{*n})$. By Corollary 5.11 we have that $h_{\varphi ^{*n}}(M \subset \mathcal {A}, \zeta ) = n h_{\varphi }(M \subset \mathcal {A}, \zeta )$. So we get

\[ h_{\varphi}(M \subset \mathcal{A}, \zeta) \leq \frac{H(\varphi^{*n})}{n} \rightarrow h(\varphi). \]

Proof. Let $\varphi$ be a standard form $\varphi (T) = \sum _k \langle T \widehat {a_k^*}, \widehat {a_k^*} \rangle$. Let $\mathcal {E}: \mathcal {A} \rightarrow M$ be a normal $\zeta$ preserving conditional expectation. Then we know that $\sigma ^{\zeta }_t(m)=m$ for all $m \in M$, where $\sigma _t^{\zeta }$ denotes the modular automorphism group corresponding to $\zeta$. Hence,

\begin{align*} h_{\varphi}(M \subset \mathcal{A}, \zeta)&= i \lim_{t \rightarrow 0} \frac{1}{t} \sum_k \langle (\Delta^{it}-1) a_k^* 1_{\zeta}, a_k^* 1_{\zeta} \rangle \\ &= i \lim_{t \rightarrow 0} \frac{1}{t} \sum_k \langle \sigma_t(a_k^*) 1_{\zeta}, a_k^* 1_{\zeta} \rangle -1 =0. \end{align*}

Conversely, suppose $h_{\varphi }(M \subset \mathcal {A}, \zeta )=0$. This part of the proof is motivated by the proof of Lemma 9.2 in [Reference Ohya and PetzOP93]. Let $\Delta _{\zeta }=\Delta$ and let $\Delta = \int _{0}^{\infty } \lambda \,d\lambda$ be its spectral resolution. Let $\Delta _n = \int _{1/n}^{n} \lambda \,d\lambda$, $n \geq 1$ be the truncations. We know that $\Delta _n$ converges to $\Delta$ in the resolvent sense. As usual, we denote by $e$ the projection from $L^2(\mathcal {A}, \zeta )$ to $L^2(M, \tau )$. We have that $e=e\Delta e \geq e \Delta _n e$ for all $n$. So $(1+t)^{-1} \leq (e\Delta _n e+ t)^{-1} \leq e(\Delta _n +t)^{-1}e$ for all $n$ and for all $t>0$. Taking limits as $n \rightarrow \infty$, we get $(1+t)^{-1} \leq e(\Delta +t)^{-1}e$. Now we shall use the integral representation of $\log$ given by

\[ \log(x)=\int_{0}^{\infty}[ (1+t)^{-1}-(x+t)^{-1}] \,dt, \]

so that

\[ h_{\varphi}(M \subset \mathcal{A}, \zeta)=-\int_{0}^{\infty}\sum_k \langle e[(1+t)^{-1}-(\Delta +t)^{-1}] e a_k^*\hat{1}, a_k^* \hat{1} \rangle \,dt. \]

From $h_{\varphi }(M \subset \mathcal {A}, \zeta )=0$ and the above discussion, we deduce that

\begin{align*} & \langle e((1+t)^{-1}-(\Delta+t)^{-1})ea_k^* \hat 1, a_k^* \hat 1 \rangle =0 \\ &\quad \Rightarrow e((1+t)^{-1}-(\Delta+t)^{-1})e a_k^* \hat 1=0 \\ &\quad \Rightarrow (1+t)^{-1} a_k^* \hat 1= e(\Delta+t)^{-1} e a_k^* \hat 1 \end{align*}

for almost all $t>0$, and hence by continuity, for all $t>0$. We now show that the last relation also holds without the compression $e$. To this end, note that by differentiating the equation $(1+t)^{-1} a_k^* \hat 1= e(\Delta +t)^{-1}a_k^* \hat 1$ with respect to $t$, we get $(1+t)^{-2}a_k^* \hat 1= e(\Delta +t)^{-2}ea_k^* \hat 1$, for all $t>0$. Therefore, by the following norm calculation in $L^2(\mathcal {A}, \zeta )$ we have

\begin{align*} \| e(\Delta+t)^{-1}ea_k^* \hat 1\|^2_2 &= \|(1+t)^{-1} a_k^* \hat 1\|_2^2= \langle (1+t)^{-2}a_k^* \hat 1, a_k^* \hat 1 \rangle \\ &= \langle e(\Delta+t)^{-2}ea_k^* \hat 1, a_k^*\hat 1 \rangle = \langle (\Delta+t)^{-2}a_k^* \hat 1, a_k^*\hat 1 \rangle =\|(\Delta+t)^{-1}a_k^* \hat 1\|^2_2. \end{align*}

So we get that $(1+t)^{-1} a_k^* \hat 1= (\Delta +t)^{-1} a_k^* \hat 1$ for all $t>0$. This implies that $\Delta ^{it}a_{k}^*1_{\zeta }=a_k^*1_{\zeta }$, which implies that $\sigma ^{\zeta }_t(a_k^*)=a_k^*$ and hence $\sigma ^{\zeta }_t(m)=m$ for all $m \in M$, as $\varphi$ is generating. Hence there exists a $\zeta$ preserving conditional expectation from $\mathcal {A}$ to $M$, which is normal, as $\zeta$ is normal.

Proof. If $h_{\varphi }(M \subset \mathcal {B_{\varphi }}, \zeta )=0$, then by Lemma 5.14 there exists a normal conditional expectation $\mathcal {E}: \mathcal {B_{\varphi }} \rightarrow M$. By Theorem 4.1, $\mathcal {E}= \rm {id}$, which implies that $\mathcal {B_{\varphi }}=M$, and hence

\[ \operatorname{Har}(\mathcal{P}_{\varphi})= \mathcal{P}(\mathcal{B}_{\varphi})=\mathcal{P}(M)=M. \]

Conversely, if $\operatorname {Har}(\mathcal {B}(L^2M, \tau ),\mathcal {P}_{\varphi })=M$, then $\Delta _{\zeta }=I$ and hence $h_{\varphi }(M \subset \mathcal {B}_{\varphi }, \zeta )=0$

Proof. Since $0 \leq h_{\varphi }(M \subset \mathcal {B}_{\varphi }, \zeta ) \leq h(\varphi )$, this result follows from Corollary 5.15.