$\alpha_{t}^{\psi_{n}}\hskip 2.84544pt=\hskip
2.84544ptu(t)^{*}\alpha_{t}^{\hat{\phi}}u(t)=\text{ad}_{u(t)}\alpha_{t}^{\hat{\phi}}$

for the unitary $u(t)=e^{-ita_{n}}$. Note that $\hat{\Phi}$ commute with
$\alpha_{t}^{\hat{\phi}}$ by $\hat{\phi}$-symmetry, and also commutes with
$\text{ad}_{u(t)}$ because $u(t)\in\lambda(G)^{\prime\prime}$ is in
$\hat{\Phi}$’s multiplicative domain. Thus $\hat{\Phi}$ commutes with
$\alpha_{t}^{\psi_{n}}$ and hence the conditional expectation
$E_{{\mathcal{M}}_{n}}=2^{-n}\int_{0}^{2^{-n}}\alpha_{t}^{\psi_{n}}$. This
proves i).

For ii), we note that for $x,y\in{\mathcal{M}}_{n}$

$\psi_{n}(x\Phi_{n}(y))=\hat{\phi}(e^{-a_{n}}x\hat{\Phi}(y))=\hat{\phi}(\hat{\Phi}(e^{-a_{n}}x)y)=\hat{\phi}(e^{-a_{n}}\hat{\Phi}(x)y)=\psi_{n}(\Phi_{n}(x)y)\hskip
2.84544pt,$

where we used the fact that $\hat{\Phi}$ is $\hat{\phi}$-symmetric and
$e^{-a_{n}}\in\lambda(G)^{\prime\prime}$ is in the fixed point subspace of
$\hat{\Phi}$. Finally, iii) follows from applying i) and ii) to $\hat{E}$. ∎

######  Remark 4.6.

For the restriction state $\phi_{n}=\hat{\phi}_{{\mathcal{M}}_{n}}$ with
density $d_{\phi_{n}}$ w.r.t to $\tau_{n}$, the above implies the KMS symmetry
for each $n$

$\displaystyle\langle\hat{\Phi}(x),y\rangle_{\phi_{n}}=\tau_{n}(d_{\phi_{n}}^{1/2}\hat{\Phi}(x^{*})d_{\phi_{n}}^{1/2}y)=\tau_{n}(d_{\phi_{n}}^{1/2}x^{*}d_{\phi_{n}}^{1/2}\hat{\Phi}(y))=\langle\hat{\Phi}(x),y\rangle_{\phi_{n}}\hskip
2.84544pt.$

Hence the density $d_{\phi_{n}}^{1/2}$ is in the multiplicative domain of
$\hat{\Phi}$.

To summarize the lemma above, we have the following commuting diagrams
${{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\hat{l}}{\mathcal{M}}}$${\hat{{\mathcal{M}}}}$${{\mathcal{M}}_{n}{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\hat{l}}}$${{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\hat{l}}{\mathcal{M}}}$${\hat{{\mathcal{M}}}}$${{\mathcal{M}}_{n}{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\hat{l}}}$$\Phi$$E_{\mathcal{M}}$$\hat{\Phi}$$E_{{\mathcal{M}}_{n}}$$\Phi_{n}$$E_{\mathcal{M}}$$E_{{\mathcal{M}}_{n}}$
${{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\hat{l}}{\mathcal{M}}}$${\hat{{\mathcal{M}}}}$${{\mathcal{M}}_{n}{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\hat{l}}}$${{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\hat{l}}{\mathcal{N}}}$${\hat{{\mathcal{N}}}}$${{\mathcal{N}}_{n}{\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}\hat{l}}}$$E$$E_{\mathcal{M}}$$\hat{E}$$E_{{\mathcal{M}}_{n}}$$E_{n}$$E_{\mathcal{N}}$$E_{{\mathcal{N}}_{n}}$
Basically, $\Phi_{n}$ is a family of trace symmetric channel which
approximates $\hat{\Phi}$ as a natural extension of $\Phi$. The same picture
holds for the conditional expectation $E,\hat{E}$ and $E_{n}$

###  4.3. Entropy Decay via CB return time

We are shall now discuss the entropy decay result for GNS-symmetric channels.
The first step is to extend the entropy difference Lemma 2.1. Define the state
space that are bounded w.r.t $\phi$,

$S_{b}({\mathcal{M}},\phi)=\\{\rho\in S({\mathcal{M}})\hskip 2.84544pt|\hskip
2.84544ptc^{-1}\phi\leq\rho\leq c\phi\hskip 2.84544pt,\hskip
2.84544pt\text{for some }c>0\\}\hskip 2.84544pt.$

For all $\rho\in S_{b}({\mathcal{M}},\phi)$, $D(\rho||\phi)<\infty$ is finite.
Such $S_{b}({\mathcal{M}},\phi)$ is a dense subset of $S({\mathcal{M}})$
because for any $\rho$ and $0<\varepsilon<1$,
$\rho_{\varepsilon}=(1-\varepsilon)\rho+\varepsilon\phi\in
S_{b}({\mathcal{M}})$. For $\rho\in S_{b}({\mathcal{M}})$, we define the
entropy difference for a $\phi$-quantum channel $\Phi_{*}$ as

$D_{\Phi_{*}}(\rho):=D(\rho||E_{*}(\rho))-D(\Phi_{*}(\rho)||\Phi_{*}\circ
E_{*}(\rho))\hskip 2.84544pt.$

Recall the chain rule [77, Theorem 2]

$D(\rho||\psi)=D(\rho||E_{*}(\rho))+D(\rho||E_{*}(\rho))$

Thus for $\rho\in S_{b}({\mathcal{M}})$,

$D(\Phi_{*}(\rho)||\Phi_{*}\circ E_{*}(\rho))\leq D(\rho||E_{*}(\rho))\leq
D(\rho||\psi)<\infty\hskip 2.84544pt.$

Then $D_{\Phi_{*}}(\rho)$ is well-defined and finite. Also by the chain rule,
we have the alternative expression
$D_{\Phi_{*}}(\rho)=D(\rho||\psi)-D(\Phi_{*}(\rho)||\psi)$. In the trace
symmetric case,
$D_{\Phi_{*}}(\rho)=D(\rho||1)-D(\Phi_{*}(\rho)||1)=H(\rho)-H(\Phi_{*}(\rho))$
as in Section 2.2.

######  Lemma 4.7.

Let $\Phi_{*}$ be a $\phi$-symmetric quantum channel. For any state
$\rho,\omega\in S_{b}({\mathcal{M}},\phi)$,

$D(\rho||\Phi_{*}^{2}(\omega))\hskip 2.84544pt\leq\hskip
2.84544ptD_{\Phi_{*}}(\rho)+D(\rho||\omega)\hskip 2.84544pt.$

###### Proof.

Recall that we use
$\rho_{n}=\hat{\rho}|_{{\mathcal{M}}_{n}}=E_{{\mathcal{M}}_{n},*}(\hat{\rho})$
and
$\omega_{n}=\hat{\omega}|_{{\mathcal{M}}_{n}}=E_{{\mathcal{M}}_{n},*}(\hat{\omega})$,
as the states in the finite von Neumann algebra
${\mathcal{M}}_{n}\subset\hat{{\mathcal{M}}}$ obtained from the Haagerup
reduction. By Lemma 4.5, we know that
$\Phi_{n}=\hat{\Phi}|_{{\mathcal{M}}_{n}}$ is a quantum Markov map symmetric
with respect to the tracial state $\tau_{n}$. Thus by Lemma 2.1 in the tracial
case,

$D(\rho_{n}||\Phi_{n}^{2}(\omega_{n}))\hskip 2.84544pt\leq\hskip
2.84544ptD_{\Phi_{n}}(\rho_{n})+D(\rho_{n}||\omega_{n})\hskip 2.84544pt,$

where we identify $\Phi_{n}=\Phi_{n,*}$ by trace symmetry. Here,

$D_{\Phi_{n}}(\rho_{n})=D(\rho_{n}||\hat{E}_{n}(\rho_{n}))-D(\Phi_{n}(\rho_{n})||\Phi_{n}\hat{E}_{n}(\rho_{n})),$

where $\hat{E}_{n}:{\mathcal{M}}_{n}\to{\mathcal{N}}_{n}$ is the
$\tau_{n}$-preserving conditional expectation. By the definition of $\Phi_{n}$
and $\rho_{n}$, recalling the hat ”$\hat{}$” notation for states on
$\hat{{\mathcal{M}}}$,

$\displaystyle\Phi_{n}(\rho_{n})=\hat{\rho}|_{{\mathcal{M}}_{n}}\circ\Phi_{n}=\hat{\rho}\circ\hat{\Phi}|_{{\mathcal{M}}_{n}}=\widehat{\Phi_{*}(\rho)}|_{{\mathcal{M}}_{n}}=\Phi_{*}(\rho)_{n}\hskip
2.84544pt,$
$\displaystyle\Phi_{n}^{2}(\rho_{n})=\Phi_{n}(\Phi_{*}(\rho)_{n})=\Phi_{*}^{2}(\rho)_{n}\hskip
2.84544pt.$

Apply this to $E_{n}=\hat{E}|_{{\mathcal{M}}_{n}}$, we also have

$\displaystyle E_{n}(\rho_{n})=E_{*}(\rho)_{n}\hskip 2.84544pt,\hskip
2.84544pt\Phi_{n}\hat{E}_{n}(\rho_{n})=\Phi_{n}(E_{*}(\rho)_{n})=\Phi_{*}E_{*}(\rho)_{n}\hskip
2.84544pt.$

Then by Lemma 4.4, we can approximate every entropic term

$\displaystyle\lim_{n}D(\rho_{n}||(\Phi_{n})^{2}(\omega_{n}))=$
$\displaystyle\lim_{n}D(\rho_{n}||\Phi_{*}^{2}(\omega)_{n})=D(\rho||\Phi_{*}^{2}(\omega))$
$\displaystyle\lim_{n}D_{\Phi_{n}}(\rho_{n})=$
$\displaystyle\lim_{n}D(\rho_{n}||E_{n}(\rho_{n}))-D(\Phi_{n}(\rho_{n})||\Phi_{n}E_{n}(\rho_{n}))$
$\displaystyle=$
$\displaystyle\lim_{n}D(\rho_{n}||E_{*}(\rho)_{n}))-D(\Phi_{*}(\rho)_{n}||\Phi_{*}E_{*}(\rho)_{n})=\lim_{n}D_{\Phi}(\rho)$
$\displaystyle\lim_{n}D(\rho_{n}||\omega_{n})=$ $\displaystyle
D(\rho||\omega)\qed$

The next lemma shows the CB-return time is also compatible with Haargerup
reduction.

######  Lemma 4.8.

Let $\Psi:{\mathcal{M}}\to{\mathcal{M}}$ be a $\phi$-symmetric quantum Markov
map and $E$ be the conditional expectation on its multiplicative domain.

$(1-\varepsilon)E\leq_{cp}\Psi\leq_{cp}(1+\varepsilon)E\hskip 2.84544pt.$ (54)

Then for all $n\in{\mathbb{N}}$,

$(1-\varepsilon)E_{n}\leq_{cp}\Psi_{n}\leq_{cp}(1+\varepsilon)E_{n}\hskip
2.84544pt.$

In particular, if $0.9E\leq_{cp}\Psi\leq_{cp}1.1E$ and $\Psi\circ E=E$, then
for any $\rho\in S_{b}({\mathcal{M}},\phi)$

$\frac{1}{2}D(\rho||E_{*}(\rho))\leq D(\rho||\Psi_{*}(\rho))\hskip 2.84544pt.$

###### Proof.

The CP order inequality follows from the fact that both maps $E_{n}$ and
$\Psi_{n}$ are the restriction of $E\otimes\operatorname{\operatorname{id}}$
and $\Psi\otimes\operatorname{\operatorname{id}}$ on the subalgebra
${\mathcal{M}}_{n}\subset\hat{{\mathcal{M}}}\subset{\mathcal{M}}\otimes
B(\ell_{2}(G))$. Then the entropy inequality can be obtained by the tracial
case Lemma 2.4 and the approximation as in Lemma 4.7. ∎

Recall that we define

$\alpha(\Phi)=\sup_{\rho}\frac{D(\Phi_{*}(\rho)||\Phi_{*}\circ
E_{*}(\rho))}{D(\rho||E_{*}(\rho))}\hskip 2.84544pt,$

as the entropy contraction coefficient and
$\alpha_{c}(\Phi):=\sup_{{\mathcal{Q}}}\alpha(\operatorname{\operatorname{id}}_{\mathcal{Q}}\otimes\Phi)$
as the complete bounded coefficient. We obtain the identical estimate as in
the tracical case.

######  Theorem 4.9.

Let $\Phi:{\mathcal{M}}\to{\mathcal{M}}$ be a $\phi$-symmetric quantum Markov
map and $E$ be the $\phi$-preserving conditional expectation onto its
multiplicative domain ${\mathcal{N}}$. Denote

$k_{cb}(\Phi)\hskip 2.84544pt=\hskip
2.84544pt\inf\\{k\in{\mathbb{N}}^{+}\hskip 2.84544pt|\hskip
2.84544pt0.9E\leq_{cp}\Phi^{2k}\leq_{cp}1.1E\hskip 2.84544pt\\}$

Then

$\alpha(\Phi)\leq\alpha_{c}(\Phi)\leq\Big{(}1-\frac{1}{2k_{cb}(\Phi)}\Big{)}.$

As a consequence, for any finite von Neumann algebra ${\mathcal{Q}}$, state
$\rho\in S({\mathcal{M}}\overline{\otimes}{\mathcal{Q}})$ and $n\geq 1$,

$D(\Phi_{*}^{n}\otimes\operatorname{\operatorname{id}}(\rho)||\Phi_{*}^{n}E_{*}\otimes\operatorname{\operatorname{id}}(\rho))\hskip
2.84544pt\leq\hskip
2.84544pt\Big{(}1-\frac{1}{2k_{cb}(\Phi)}\Big{)}^{n}D(\rho||E_{*}\otimes\operatorname{\operatorname{id}}(\rho)\hskip
2.84544pt.$

###### Proof.

For $\rho\in S_{b}({\mathcal{M}},\phi)$, the proof is same as the tracial case
Theorem 2.6 by using Lemma 4.7 and Lemma 4.8 above. The general case $\rho\in
S({\mathcal{M}})$ can be approximated by
$\rho_{\varepsilon}=(1-\varepsilon)\rho+\varepsilon\phi$. ∎

######  Remark 4.10.

The above theorem gives substantial improvement to [37, Theorem 4.1] which is
almost an existence proof of $\alpha_{c}(\Phi)<1$ given the spectral gap

$\lambda(\Phi):=\parallel\\!\Phi(\operatorname{\operatorname{id}}-E):L_{2}({\mathcal{M}},\phi)\to
L_{2}({\mathcal{M}},\phi)\\!\parallel<1$

and finite index $C_{cb}(E)<\infty$. Our results yield explicit bound
depending on $\lambda$ and $C_{cb}(E)$ and also applies to infinite
dimensional setting even with $C_{cb}(E)=\infty$.

For a $\phi$-symmetric quantum Markov semigroup
$T_{t}=e^{-tL}:{\mathcal{M}}\to{\mathcal{M}}$, we define the modified log-
Sobolev (MLSI) constant $\alpha_{1}(L)$ as the largest $\alpha$ such that

$\displaystyle D(T_{t,*}(\rho)||E_{*}(\rho))\hskip 2.84544pt\leq\hskip
2.84544pte^{-2\alpha
t}D(\rho||E_{*}\otimes\operatorname{\operatorname{id}}(\rho))\hskip
2.84544pt,\hskip 2.84544pt\forall\rho\in S({\mathcal{M}})\hskip 2.84544pt,$
(55)

where $E$ is the $\phi$-preserving conditional expectation onto the fixed
point subalgebra ${\mathcal{N}}$ The complete MLSI constant is then defined as
$\alpha_{c}(L):=\sup_{\mathcal{{\mathcal{Q}}}}\alpha(\operatorname{\operatorname{id}}_{\mathcal{Q}}\otimes
L)$.

######  Theorem 4.11.

Denote

$t_{cb}(L)\hskip 2.84544pt=\hskip 2.84544pt\inf\\{t>0\hskip 2.84544pt|\hskip
2.84544pt0.9E\leq_{cp}T_{t}\leq_{cp}1.1E\\}$

Then

$\alpha_{1}(L)\leq\alpha_{c}(L)\leq\frac{1}{2t_{cb}(L)}.$

Namely, for any finite von Neumann algebra ${\mathcal{Q}}$ and state $\rho\in
S({\mathcal{M}}\overline{\otimes}{\mathcal{Q}})$, we have the exponential
decay of relative entropy

$D(T_{t,*}\otimes\operatorname{\operatorname{id}}(\rho)||E_{*}\otimes\operatorname{\operatorname{id}}(\rho))\hskip
2.84544pt\leq\hskip
2.84544pte^{-\frac{t}{t_{cb}(L)}}D(\rho||E_{*}\otimes\operatorname{\operatorname{id}}(\rho))\hskip
2.84544pt,\hskip 2.84544ptt\geq 0\hskip 2.84544pt.$

###### Proof.

This can be approximated using the tracial case Theorem 2.6 as Lemma 4.7
above. ∎

######  Remark 4.12.

In finite dimensions, the MLSI constant of a GNS-symmetric semigroup is
defined as

$2\alpha D(\rho||E_{*}(\rho))\leq
I_{L}(\rho):=\tau(L_{*}(\rho)(\ln\rho-\ln\phi))\hskip 2.84544pt,$

The right hand side $I_{L}(\rho)$ is the entropy production, and the
equivalence to entropy decay rely on the de Bruijn identity

$\displaystyle
I_{L}(\rho)=-\frac{d}{dt}D(T_{*}(\rho)||E_{*}(\rho))|_{t=0}\hskip 2.84544pt,$
(56)

In infinite dimensions, the de Bruijn identity (56) is less justified even in
$B(H)$ with $\dim(H)=+\infty$ (see discussions in [48, 57]). Here, our
definition of MLSI (55) via exponential entropy decay avoids this issue, as
the relative entropy is always well-defined even on Type III algebra via
Araki’s definition (11). We note that this definition of MLSI also does not
depend on any choice of reference state $\phi$.

The next proposition shows that the GNS-symmetry is also independent of the
choice of invariant state $\phi$.

######  Lemma 4.13.

Let $T_{t}:{\mathcal{M}}\to{\mathcal{M}}$ be a $\phi$-symmetric quantum Markov
semigroup for a normal faithful state $\phi$. Denote
$E:{\mathcal{M}}\to{\mathcal{N}}$ as the $\phi$-preserving conditional
expectation onto the fixed point space. Suppose $\psi$ is an another normal
faithful state invariant under $E$, i.e. $\psi\circ E=\psi$. Then
$T_{t}:{\mathcal{M}}\to{\mathcal{M}}$ is also $\psi$-symmetric.

###### Proof.

Without loss of generosity we assume $\psi\leq C\phi$ for some $C>0$. We first
view them as the state on the subalgebra ${\mathcal{N}}$ by restriction. By
[93, Theorem 3.17], there exists $h\in{\mathcal{N}}$ such that

$\psi(x)=\phi(h^{*}xh)\hskip 2.84544pt,\hskip 2.84544pt\forall
x\in{\mathcal{N}}\hskip 2.84544pt.$

This identity actually also holds for $y\in{\mathcal{M}}$. Indeed, because of
$\phi\circ E=\phi$ and $\psi\circ E=\psi$,

$\psi(y)=\psi(E(y))=\phi(h^{*}E(y)h)=\phi(E(h^{*}yh))=\phi(h^{*}yh)\hskip
2.84544pt,\hskip 2.84544pt\forall y\in{\mathcal{M}}\hskip 2.84544pt.$

Thus the GNS-symmetry with respect to $\psi$ follows that for
$x,y\in{\mathcal{M}}$,

$\psi(xT_{t}(y))=\phi(h^{*}xT_{t}(y)h)=\phi(h^{*}xT_{t}(yh))=\phi(T_{t}(h^{*}x)yh)=\phi(h^{*}T_{t}(x)yh)=\psi(T_{t}(x)y)\hskip
2.84544pt,$

where we used the bimodule property of $T_{t}(axb)=aT_{t}(x)b$ for
$a,b\in{\mathcal{N}}$. The general case can be obtained via
$\psi_{\varepsilon}=(1-\varepsilon)\psi+\varepsilon\phi$. ∎

We remark that under spectral gap or MLSI, one have $\lim_{t}T_{t}=E$ in
$L_{2}$ or $L_{1}$ norm. Then the above $E$-invariant condition $\phi\circ
E=\phi$ can be replaced by $\phi=T_{t}\circ\phi$ for any/some $t>0$.

######  Remark 4.14.

In the above Haagerup’s reduction, both $\hat{\Phi}$ and $\Phi_{n}$ are always
non-primitive even given $\Phi$ is primitive. From this point of view, our
consideration for non-primitive cases is essential even for primitive $\Phi$.
It also indicates that Haagerup’s reduction does not work for
LSI/hypercontractivity.

As we have seen in the tracial case, a combination of heat kernel estimates
and spectral gap allow us to bound CB return time. This remains true in the
GNS-symmetric case. Here we briefly illustrate the idea, as a careful
treatment in finite dimenisonal cases is already given in [6]. Let
$E:{\mathcal{M}}\to{\mathcal{N}}$ be a $\phi$-preserving conditional
expectation. Our technical tool, following [35], is the use of the module
basis $x_{\alpha},\alpha\in I$ such that

$E(x_{\alpha}^{*}x_{\beta})\hskip 2.84544pt=\hskip
2.84544pt\delta_{\alpha,\beta}e_{\alpha}$

holds for a family of projections $e_{\alpha}\in{\mathcal{N}}$. For each
$a\in{\mathcal{M}}$, there is a unique decomposition
$a=\sum_{\alpha}x_{\alpha}a_{\alpha}$ with $a_{\alpha}=E(x_{\alpha}^{*}a)\in
L_{2}({\mathcal{N}},\phi)$. In full generality the elements $x_{\alpha}$ may
no longer be in ${\mathcal{M}}$, but belong to the strong closure of
${\mathcal{M}}\subset B(L_{2}({\mathcal{M}},\phi))$ (see [49, 52]). For an
${\mathcal{N}}$-bimodule map $T:{\mathcal{M}}\to{\mathcal{M}}$, the module
Choi operator is defined as

$\chi_{T}=\sum_{\alpha,\beta\in I}|{\alpha}\rangle\langle{\beta}|\otimes
T(x_{\alpha}^{*}x_{\beta})\in B(\ell_{2}(I))\overline{\otimes}{\mathcal{M}}$

$T$ and $\chi_{T}$ determine each other because for a family
$a_{i}=\sum_{\alpha}x_{\alpha}a_{\alpha,i}$

$\displaystyle\operatorname{\operatorname{id}}\otimes
T(\sum_{i,j}|{i}\rangle\langle{j}|\otimes a_{i}^{*}a_{j})=$
$\displaystyle\sum_{i,j}|{i}\rangle\langle{j}|\otimes T(a_{i}^{*}a_{j})$
$\displaystyle=$ $\displaystyle\sum_{i,j}|{i}\rangle\langle{j}|\otimes
T(a_{i}^{*}a_{j})$ $\displaystyle=$
$\displaystyle\sum_{i,j}|{i}\rangle\langle{j}|\otimes\sum_{\alpha,\beta}a_{\alpha,i}^{*}T(x_{\alpha}^{*}x_{\beta})a_{\beta,j}$
$\displaystyle=$
$\displaystyle(\sum_{\alpha,i}|{i}\rangle\langle{\alpha}|\otimes
a_{\alpha,i}^{*})\chi_{T}(\sum_{\beta,i}|{\beta}\rangle\langle{j}|\otimes
a_{\beta,i})\hskip 2.84544pt.$

This also shows that $T$ is completely positive if and only if $\chi_{T}$ is
positive.

For $1\leq p\leq\infty$, we define the $\phi$-weighted conditional
$L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)$ space as the
completion of ${\mathcal{M}}$ under the norm

$\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}=\sup\\{\parallel\\!axb\\!\parallel_{p,\phi}\hskip
2.84544pt|\hskip 2.84544pta,b\in{\mathcal{N}},\hskip
2.84544pt\parallel\\!aa^{*}\\!\parallel_{p,\phi}=\parallel\\!b^{*}b\\!\parallel_{p,\phi}=1\hskip
2.84544pt\\}\hskip 2.84544pt.$

At $p=1$, we have for $x\geq 0$,

$\parallel\\!x\\!\parallel_{L_{\infty}^{1}({\mathcal{N}}\subset{\mathcal{M}},\phi)}=\parallel\\!E(x)\\!\parallel_{\infty}\hskip
2.84544pt.$

For a bimodule map $T$,

$\displaystyle\parallel\\!\chi_{T}\\!\parallel_{B(\ell_{2}(I))\overline{\otimes}{\mathcal{M}}}=\parallel\\!T:L_{\infty}^{1}({\mathcal{N}}\subset{\mathcal{M}},\phi)\to{\mathcal{M}}\\!\parallel_{cb}\hskip
2.84544pt,$ (57)

Here, we only need the easy direction ”$\leq$”, which stems from

$\displaystyle\parallel\\!\sum_{\alpha,\beta}|{\alpha}\rangle\langle{\beta}|\otimes
x_{\alpha}^{*}x_{\beta}\\!\parallel_{B(\ell_{2}(I))\underset{min}{\otimes}L_{\infty}^{1}({\mathcal{N}}\subset{\mathcal{M}})}=$
$\displaystyle\parallel\\!\sum_{\alpha,\beta}|{\alpha}\rangle\langle{\beta}|\otimes
x_{\alpha}^{*}x_{\beta}\\!\parallel_{L_{\infty}^{1}(B(\ell_{2}(I))\overline{\otimes}{\mathcal{N}}\subset
B(\ell_{2}(I))\overline{\otimes}{\mathcal{M}})}$ $\displaystyle=$
$\displaystyle\parallel\\!\operatorname{\operatorname{id}}\otimes
E(\sum_{\alpha,\beta}|{\alpha}\rangle\langle{\beta}|\otimes
x_{\alpha}^{*}x_{\beta})\\!\parallel_{\infty}$ $\displaystyle=$
$\displaystyle\parallel\\!\sum_{\alpha}|{\alpha}\rangle\langle{\alpha}|\otimes
E_{\alpha}\\!\parallel_{\infty}\leq 1\hskip 2.84544pt.$

Denote $\chi_{E}=\sum_{\alpha,\beta}|{\alpha}\rangle\langle{\beta}|\otimes
x_{\alpha}^{*}x_{\beta}$ as the Choi operator of $E$,

$\displaystyle\parallel\\!\chi_{T}\\!\parallel_{B(\ell_{2}(I))\overline{\otimes}{\mathcal{M}}}=$
$\displaystyle\parallel\\!\operatorname{\operatorname{id}}\otimes
T(\chi_{E})\\!\parallel_{B(\ell_{2}(I))\overline{\otimes}{\mathcal{M}}}$
$\displaystyle\leq$
$\displaystyle\parallel\\!T:L_{\infty}^{1}(B(\ell_{2}(I))\overline{\otimes}{\mathcal{N}}\subset
B(\ell_{2}(I))\overline{\otimes}{\mathcal{M}})\to
B(\ell_{2}(I))\overline{\otimes}{\mathcal{M}})\\!\parallel$
$\displaystyle\cdot\parallel\\!\chi_{0}\\!\parallel_{L_{\infty}^{1}(B(\ell_{2}(I))\overline{\otimes}{\mathcal{N}}\subset
B(\ell_{2}(I))\overline{\otimes}{\mathcal{M}})}$ $\displaystyle\leq$
$\displaystyle\parallel\\!T:L_{\infty}^{1}({\mathcal{N}}\subset{\mathcal{M}},\phi)\to{\mathcal{M}}\\!\parallel_{cb}$

As a consequence, we have

$\displaystyle\parallel\\!T-E:L_{\infty}^{1}({\mathcal{N}}\subset{\mathcal{M}},\phi)\to{\mathcal{M}}\\!\parallel_{cb}\leq\varepsilon\Longrightarrow(1-\varepsilon)E\leq_{cp}T\leq_{cp}(1+\varepsilon)E\hskip
2.84544pt.$ (58)

It is clear that the right hand side only rely on positivity structure.
Indeed, the $L_{\infty}^{1}({\mathcal{N}}\subset{\mathcal{M}},\phi)$ norm is
independent of the choice of the invariant state $\phi=\phi\circ E$.

######  Lemma 4.15.

Let $\phi$ be faithful normal state and $E:{\mathcal{M}}\to{\mathcal{N}}$ be a
$\phi$-preserving conditional expectation. Suppose $\psi=\psi\circ E$ is
another normal faithful state preserved by $E$. Then for any
$x\in{\mathcal{M}}$,

$\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}=\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\psi)}$

The identity extends to all $x\in
L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)$.

###### Proof.

Note that if both $\phi$ and $\psi$ are trace-preserving, then
$d_{\psi}^{-\frac{1}{2p}}d_{\phi}^{\frac{1}{2p}}$ is affiliated to
${\mathcal{N}}$. Indeed, as argued in Lemma 4.13, if $\psi\leq C\phi$, then
$d_{\psi}=hd_{\phi}h^{*}$ for some $h\in{\mathcal{N}}$, and the general case
follows from approximation
$\psi\leq\frac{1}{\varepsilon}((1-\varepsilon)\phi+\varepsilon\psi)$. Then we
have

$\parallel\\!aa^{*}\\!\parallel_{\phi,p}=\parallel\\!d_{\phi}^{\frac{1}{2p}}aa^{*}d_{\phi}^{\frac{1}{2p}}\\!\parallel_{p}=\parallel\\!d_{\psi}^{-\frac{1}{2p}}d_{\phi}^{\frac{1}{2p}}aa^{*}d_{\phi}^{\frac{1}{2p}}d_{\psi}^{-\frac{1}{2p}}\\!\parallel_{\psi,2p}$

Denote $a_{1}=d_{\psi}^{-\frac{1}{2p}}d_{\phi}^{\frac{1}{2p}}a$ and
$b_{1}=bd_{\phi}^{\frac{1}{2p}}d_{\psi}^{-\frac{1}{2p}}$. For
$x\in{\mathcal{M}}$,

$\displaystyle\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}=$
$\displaystyle\sup_{\parallel\\!\hskip 2.84544ptaa^{*}\hskip
2.84544pt\\!\parallel_{\phi,2p}=\parallel\\!\hskip 2.84544ptb^{*}b\hskip
2.84544pt\\!\parallel_{\phi,2p}=1}\parallel\\!axb\\!\parallel_{\phi,p}=\sup_{\parallel\\!\hskip
2.84544ptaa^{*}\hskip 2.84544pt\\!\parallel_{\phi,p}=\parallel\\!\hskip
2.84544ptb_{1}b_{1}\hskip
2.84544pt\\!\parallel_{\phi,p}=1}\parallel\\!d_{\psi}^{-\frac{1}{2p}}d_{\phi}^{\frac{1}{2p}}axbd_{\phi}^{\frac{1}{2p}}d_{\psi}^{-\frac{1}{2p}}\\!\parallel_{\psi,p}$
$\displaystyle=$ $\displaystyle\sup_{\parallel\\!\hskip
2.84544pta_{1}a_{1}^{*}\hskip 2.84544pt\\!\parallel_{2p}=\parallel\\!\hskip
2.84544ptb_{1}b_{1}^{*}\hskip
2.84544pt\\!\parallel_{2p}=1}\parallel\\!a_{1}xb_{1}\\!\parallel_{\psi,p}=\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\psi)}$

where the supremum are for $a,b\in{\mathcal{N}}$. ∎

Based on above we have obtain

######  Proposition 4.16.

The Proposition 2.12 and 3.4 remains valid in the GNS-symmetric cases by
replacing $L_{2}({\mathcal{M}},\tau)$ by $L_{2}({\mathcal{M}},\phi)$ and
$L_{\infty}^{1}({\mathcal{N}}\subset{\mathcal{M}})$ and
$L_{\infty}^{1}({\mathcal{N}}\subset{\mathcal{M}},\phi)$.

###### Proof.

The argument is identical to the tracial cases by using the property of
$L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)$ for general von
Neumann algebra established [51]. See also [6, Section 5] for argument in
finite dimensional GNS-symmetric cases. ∎

######  Remark 4.17.

For finite ${\mathcal{M}}$, one particular invariant state of $E$ used in [9,
6] is

$\phi_{{\rm\text{tr}}}=E_{*}(1)\hskip 2.84544pt.$

This state is convenient because $\phi_{{\rm\text{tr}}}|_{\mathcal{N}}$ is a
tracial state. Then by Lemma 4.15, we have

$\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}=\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi_{\rm\text{tr}})}=\sup\\{\parallel\\!axb\\!\parallel_{p,\phi_{\rm\text{tr}}}\hskip
2.84544pt|a,b\in{\mathcal{N}},\hskip
2.84544pt\parallel\\!a\\!\parallel_{p,\phi_{\rm\text{tr}}}=\parallel\\!b\\!\parallel_{p,\phi_{\rm\text{tr}}}=1\hskip
2.84544pt\\}$

where we used the fact $L_{p}({\mathcal{N}},\phi_{\rm\text{tr}})$ is a tracial
$L_{p}$-space. We will use this point to simplify the discussion in Section
4.6.

###  4.4. Applications to finite Quantum Markov chains

Let $T_{t}=e^{-Lt}:{\mathbb{M}}_{d}\to{\mathbb{M}}_{d}$ be a quantum Markov
semigroup on matrix algebra ${\mathbb{M}}_{d}$. Its generator $L$ admits the
following Lindbladian form ([38, 63])

$L(x)=i[h,x]+\sum_{j}\gamma_{j}(V_{j}^{*}[x,V_{j}]+[V_{j}^{*},x]V_{j})\hskip
2.84544pt,$

where $h,V_{j}\in{\mathbb{M}}_{d}$ and $h=h^{*}$ is Hermitian. When $T_{t}$ is
GNS-symmetric, Carlen and Maas proved the following simplified form

$L(x)=\sum_{j}e^{-w_{j}/2}\Big{(}V_{j}^{*}[x,V_{j}]+[V_{j}^{*},x]V_{j}\Big{)}\hskip
2.84544pt,$

where $\\{V_{j}\\}=\\{V_{j}\\}^{*}$ is an orthonormal set with respect to
trace inner product and the eigenvector of modular group

$\alpha_{t}^{\phi}(V_{j})=e^{-iw_{j}t}V_{j}\hskip 2.84544pt.$

Gao and Rouzé in [37, Theorem 3.3] proved that in finite dimensions,

$\displaystyle
0<\frac{\lambda}{2C_{cb}(E)}\leq\alpha_{c}\leq\alpha_{1}\leq\lambda\hskip
2.84544pt,$ (59)

where $E$ is the conditional expectation onto the fixed point subalgebra.
Here, we obtain an exponential improvement on the dependence of the dimension
constant $C_{cb}(E)$.

######  Theorem 4.18.

Let $\phi$ be a faithful state. For a $\phi$-symmetric quantum Makrov
semigroup $T_{t}=e^{-tL}:{\mathbb{M}}_{d}\to{\mathbb{M}}_{d}$,

$\displaystyle\frac{\lambda}{2\ln(10C_{cb}(E))}\leq\alpha_{c}(L)\leq\alpha_{1}(L)\leq\lambda(L)\hskip
2.84544pt.$ (60)

###### Proof.

It follows from combining Theorem 4.11 and Proposition 4.16. ∎

######  Remark 4.19.

In the primitive case ${\mathcal{N}}=\mathbb{C}1$, the conditional expectation
$E_{\phi}(x)=\phi(x)1$ has index

$C(E_{\phi})=\parallel\\!\phi^{-1}\\!\parallel_{\infty}\hskip
2.84544pt,C_{cb}(E_{\phi})\leq\parallel\\!\phi^{-1}\\!\parallel_{\infty}^{2}\hskip
2.84544pt.$

The above theorem gives

$\alpha_{1}\geq\alpha_{c}\geq\frac{\lambda}{2\ln
10+4\ln\parallel\\!\phi^{-1}\\!\parallel_{\infty}}\hskip 2.84544pt.$

It can be compared to the bound

$\displaystyle\frac{2(1-\frac{2}{\parallel\\!\phi^{-1}\\!\parallel_{\infty}})\lambda}{\ln(\parallel\\!\phi^{-1}\\!\parallel_{\infty}-1)}\leq\alpha_{2}\leq\alpha_{1}\leq\lambda\hskip
2.84544pt.$ (61)

This bound was proved Diaconis and Saloff-Coste [30] for symmetric classical
Markov semigroups. In the quantum case, it is only partially obtained for
unital semigroups [55] and $d=2$ [11]. In both classical and quantum
depolarizing semigroup $L(x)=x-\phi(x)1$, this bound is known to be tight for
$\alpha_{2}$, which upper bounds $\alpha_{1}$. In this sense, our results
gives a general
$\mathcal{O}(\frac{\lambda}{\parallel\\!\phi^{-1}\\!\parallel_{\infty}})$
lower bound for $\alpha_{1}$, not only for the primitive case, but also for
non-primitive cases and complete constant $\alpha_{c}$. We also note that for
both classical and quantum depolarizing semigroup $L(x)=x-\phi(x)1$, this
bound is known to be tight for $\alpha_{2}$, which upper bounds $\alpha_{1}$.

######  Remark 4.20.

The Corollary 3.9 shows that the CMLSI constant $\alpha_{c}$ for classical
Markov semigroup is lower bounded by LSI constant $\alpha_{2}$ up to a
$O(\ln\ln\parallel\\!\mu^{-1}\\!\parallel_{\infty})$ term. This argument does
not work for Quantum Markov semigroup
$T_{t}:\mathbb{M}_{d}\to{\mathbb{M}}_{d}$ on matrix algebras, despite relation
(41) remain valid for primitive quantum Markov semigroup. The difference is
that for matrix algebra, the bounded return time

$t_{b}:=\frac{1}{2}\inf\\{t>0\hskip 2.84544pt|\hskip
2.84544pt\parallel\\!T_{t}-E:L_{1}({\mathbb{M}}_{d},\phi)\to
L_{\infty}({\mathbb{M}}_{d})\\!\parallel<1/e^{2}\\}$

and the CB return time of completely bounded norm

$\displaystyle t_{cb}=\inf\\{t>0\hskip 2.84544pt|\hskip
2.84544pt\parallel\\!T_{t}-E_{\mu}:L_{1}({\mathbb{M}}_{d},\phi)\to
L_{\infty}({\mathbb{M}}_{d})\\!\parallel_{cb}<1/10\hskip 2.84544pt.\\}$

are quite different. In the classical setting, we used the fact

$\parallel\\!T:L_{1}(\Omega)\to
L_{\infty}(\Omega)\\!\parallel=\parallel\\!T:L_{1}(\Omega)\to
L_{\infty}(\Omega)\\!\parallel_{cb}\hskip 2.84544pt.$

So the $t_{b}$ and $t_{cb}$ are comparable by absolute constants. In the
noncommutative setting, we can only have

$\parallel\\!T_{t}-E_{\mu}:L_{1}({\mathbb{M}}_{d},\phi)\to
L_{\infty}({\mathbb{M}}_{d})\\!\parallel_{cb}\leq
d\parallel\\!T_{t}-E_{\mu}:L_{1}({\mathbb{M}}_{d})\to
L_{\infty}({\mathbb{M}}_{d})\\!\parallel.$

In the trace symmetric case, $\parallel\\!\mu^{-1}\\!\parallel_{\infty}=d$ and
$t_{cb}\leq 3t_{b}+\ln d$,

$\alpha_{c}\geq\frac{1}{4t_{cb}}\geq\frac{1}{12t_{b}+4\ln d}\sim
O(\frac{\alpha_{2}}{\ln d})\hskip 2.84544pt,$

which is worse than the lower bound in Theorem 4.18 as
$\alpha_{2}\leq\lambda$.

###  4.5. An example of noncommutative Birth-Death Process

We illustrate our estimate with a noncommutative Birth-Death process. This
example is an extension of usual graph Laplacians to matrix algebras (see [62,
50] for similar constructions). To fix the notation, let $\mathcal{G}=(V,E)$
be an undirected graph with $n=|V|$ vertices and edge set $E$. For each edge
$(r,s)$, we introduce the edge Lindbladian on ${\mathbb{M}}_{n}$,

$\displaystyle L_{rs}(x)=$ $\displaystyle
e^{\beta_{rs}/2}L_{e_{rs}}(x)+e^{-\beta_{rs}}L_{e_{sr}}(x)$
$\displaystyle=e^{\beta_{rs}/2}(e_{ss}x+xe_{ss}-2e_{sr}xe_{rs})+e^{-\beta_{rs}/2}(e_{rr}x+xe_{rr}-2e_{sr}xe_{sr})\hskip
2.84544pt,$

where $e_{rs}\in{\mathbb{M}}_{n}$ is the matrix unit with $1$ at the $(r,s)$
position. The total Lindbladian is a weighted sum over the edge set $E$,

$L\hskip 2.84544pt=\hskip 2.84544pt\sum_{(r,s)\in E}w(r,s)L_{rs}$

where we assume $\beta_{rs}=-\beta_{sr}$ and $w(r,s)=w(s,r)>0$ for the GNS
symmetry condition. Note that for $j\neq k$,

$\displaystyle L(e_{jk})=2(\sum_{(r,k)\in
E}w(r,k)e^{\beta_{rk}/2}+\sum_{(r,j)\in E}w(r,j)e^{\beta_{r,j}/2})e_{jk}$
$\displaystyle L(e_{jj})=4\sum_{(r,j)\in
E}w(r,j)(e^{\beta_{r,j}/2}e_{jj}-e^{-\beta_{r,j}/2}e_{rr})\hskip 2.84544pt.$

Let us collect some relevant facts of such Lindbladian $L$ as noncommutative
extension of graph Laplacian.

  1. i)

Denote $\ell_{\infty}(V)\subset{\mathbb{M}}_{n}$ as the diagonal subalgebra.
$L(\ell_{\infty}(V))\subset\ell_{\infty}(V)$, and $L|_{\ell_{\infty}(V)}$ is a
weighted graph Laplacian;

  2. ii)

For $r\neq s$, the matrix unit $e_{rs}$ is an eigenvector of $L$

$L(e_{rs})\hskip 2.84544pt=\hskip 2.84544pt\gamma_{rs}e_{rs}$

where $\gamma_{rs}=2(\sum_{(r,j)\in E}w(r,j)e^{\beta_{rj}/2}+\sum_{(k,s)\in
E}w(k,s)e^{-\beta_{ks}/2})$.

  3. iii)

$\ker(L)\subset\ell_{\infty}(V)$, and $\ker(L)=\mathbb{C}1$ if
$\mathcal{G}=(V,E)$ is connected.

  4. iv)

Let $\mu=(\mu_{k})\in\ell_{\infty}(V)$ be density operator in the diagonal
subalgebra. Then $L$ is $\mu$-symmetric if $e^{\beta_{rs}}=\mu_{s}/\mu_{r}$
for any $s\neq r$.

Assume $L=\sum_{(s,r)\in E}L_{sr}$ is a primitive graph Lindbladian satisfying
$\mu$-symmetric condition for a diagonal density operator $\mu$. Denote
$E_{d}$ as the projection onto diagonal subalgebra. We can decompose the
semigroup $T_{t}=e^{-tL}$ on the diagonal part and off diagonal part.

$\displaystyle
T_{t}=T_{t}E_{d}+T_{t}(\operatorname{\operatorname{id}}-E_{d}):=T_{t}^{diag}+T_{t}^{off}\hskip
2.84544pt.$

It is clear from i) and ii) that $T_{t}E_{d}$ is a classical graph random walk
and $T_{t}(\operatorname{\operatorname{id}}-E_{d})$ is a Schur multiplier on
${\mathbb{M}}_{n}$. Using this decomposition, we have the CB-return time of
the semigroup

$t_{cb}(\varepsilon):=\inf\\{t>0\hskip 2.84544pt|\hskip
2.84544pt\parallel\\!T_{t}:L_{1}({\mathbb{M}}_{n},\mu)\to{\mathbb{M}}_{n}\\!\parallel\\}$

satisfies

$t_{cb}(2\varepsilon)\leq
t^{diag}_{cb}(\varepsilon)+t^{off}_{cb}(\varepsilon)$

where $t^{diag}_{cb}$ and $t^{off}_{cb}$ are the CB-return time for the
diagonal part $T_{t}E_{d}$ and off diagonal part
$T_{t}(\operatorname{\operatorname{id}}-E_{d})$ respectively. For the diagonal
part, this is a classical bounded mixing time, i.e. the smallest $t$ such that

$\parallel\\!T_{t}E_{d}:L_{1}(V,\mu)\to
L_{\infty}(V)\\!\parallel\leq\varepsilon$

For the off-diagonal term, we have by Effros-Ruan isomorphism that a Schur
multiplier map

$\displaystyle\parallel\\!T_{t}(\operatorname{\operatorname{id}}-E_{d}):L_{1}({\mathbb{M}}_{n},\mu)\to{\mathbb{M}}_{n}\\!\parallel_{cb}=$
$\displaystyle\parallel\\!\sum_{r\neq
s}\mu_{r}^{-1/2}e^{-\gamma_{rs}t}\mu_{s}^{-1/2}e_{rs}\otimes
e_{rs}\\!\parallel_{\infty}$ $\displaystyle=$
$\displaystyle\parallel\\!\sum_{r\neq
s}\mu_{r}^{-1/2}e^{-\gamma_{rs}t}\mu_{s}^{-1/2}e_{rs}\\!\parallel_{\infty}\hskip
2.84544pt.$

Note that for each $t$,

$A_{t}=\sum_{r\neq s}\mu_{r}^{-1/2}e^{-\gamma_{rs}t}\mu_{s}^{-1/2}e_{rs}$

is a symmetric matrix with positive entry. A standard application of Schur’s
lemma for matrices with positive entries implies

$\displaystyle\parallel\\!A_{t}\\!\parallel_{\infty}\leq\sup_{r}\Big{(}\sum_{s}\mu_{r}^{-1/2}e^{-\gamma_{rs}t}\mu_{s}^{-1/2}\Big{)}\hskip
2.84544pt,$

which gives us an estimate for the off diagonal term.

Now we consider the birth-death process on a finite state space
$V=\\{1,\cdots,n\\}$, which we denote as $L^{BD}_{n}$. The corresponding edge
$E$ set consists of only successive vertices $E=\\{(j,j+1)|1\leq j\leq
n-1\\}$. The simplest case chooses the uniform weight $w(r,s)=1$ for $(r,s)\in
E$ and allows only one Bohr frequency $e^{-\beta}=\frac{\mu_{j}}{\mu_{j+1}}$,
and the resulting stationary measure is the well-studied thermal state

$\mu=Z_{\beta}^{-1}(e^{-\beta j})_{j=1}^{n}$

where $Z_{\beta}=\sum_{j=1}^{n}e^{-\beta j}$ is the normalization constant. In
this case, $\gamma_{rs}=8(\cosh\beta)t$, and the off diagonal CB norm can be
estimated by

$\displaystyle\parallel\\!A_{t}\\!\parallel_{\infty}\leq$
$\displaystyle\sup_{r}\Big{(}\sum_{s=1}^{n}e^{\beta r/2}e^{\beta
s/2}\Big{)}Z_{\beta}e^{-8(\cosh\beta)t}$ $\displaystyle\leq$ $\displaystyle
e^{\beta\frac{n-1}{2}}e^{\beta/2}e^{-\beta}\frac{1-e^{n\beta/2}}{1-e^{\beta/2}}\frac{1-e^{-n\beta}}{1-e^{-\beta}}e^{-8(\cosh\beta)t}$

Thus $t_{cb}^{off}(\varepsilon)\leq C_{1}(\beta)n$ for some constant
$C_{1}(\beta)$ depending on $\beta$. For the classic part, We refer to [68]
and [24] for the fact that the spectral gap is of order $O(1)$, i.e.

$c(\beta)\leq\lambda(L^{diag}_{n})\leq C(\beta)$

for all $n\in{\mathbb{N}}$. For the commutative system on the diagonal part
this implies (see also [30])

$t^{diag}_{cb}(\varepsilon)\hskip 2.84544pt\leq\hskip
2.84544pt2c(\beta)^{-1}(2+|\ln\mu_{n}|)\hskip 2.84544pt\leq\hskip
2.84544ptC_{2}(\beta)n\hskip 2.84544pt,$

(for $\varepsilon=e^{-2}$, but here the actual value of $\varepsilon$ does not
matter the asymptotic). On the other hand, Miclo [68] proves that

$t^{diag}_{cb}\geq\alpha_{1}(L^{diag}_{n})^{-1}\hskip 2.84544pt\geq\hskip
2.84544ptc(\beta)n\hskip 2.84544pt.$

Combining the diagonal and off diagonal part, we know
$t^{diag}_{cb}(L^{BD}_{n})\sim\frac{1}{n}$. It turns out the same asymptotic
holds for CMLSI constant $\alpha_{c}(L^{BD}_{n})$, which indicates our
$\alpha_{c}\geq\frac{1}{2t_{cb}}$ estimate is asymptotically tight.

######  Theorem 4.21.

For $\beta>0$, there exist constants $c(\beta),C(\beta)>0$ such that the CMLSI
constant of noncommutative birth-death process $L_{n}^{BD}$ satisfies

$\frac{c(\beta)}{n}\hskip 2.84544pt\leq\hskip
2.84544pt\alpha_{c}(L_{n}^{BD})\leq\alpha_{1}(L^{BD}_{n})\hskip
2.84544pt\leq\hskip 2.84544pt\frac{C(\beta)}{n}\hskip 2.84544pt.$

The same $\Theta(\frac{1}{n})$ asymptotic holds for the CB return time
$t_{cb}(L^{BD}_{n})$.

###### Proof.

It suffices to show that

$\alpha_{c}(L^{BD_{n}})\leq\alpha_{1}(L^{BD}_{n})\leq c(\beta)/n\hskip
2.84544pt.$

For this, we consider the function in the commutative system on the diagonal

$f(k)=\frac{Z(\beta)}{n}e^{\beta k}\text{ and
}\sum_{k=1}^{n}f(k)\mu(k)=\sum_{k=1}^{n}\frac{Z(\beta)}{n}e^{\beta
k}\frac{1}{Z(\beta)}e^{-\beta k}=1$

so that $f\mu:=\rho$ represents a probability density. The relative entropy
term satisfies

$D(\rho||\mu)=D(f\mu||\mu)\hskip 2.84544pt=\hskip
2.84544pt\sum_{k}\frac{e^{-\beta k}}{Z(\beta)}f(k)(\beta k+\ln Z(\beta)-\ln
n)\ \hskip 2.84544pt=\hskip 2.84544pt\ln Z(\beta)-\ln
n+\beta\frac{n+1}{2}\hskip 2.84544pt.$

Our density is $\rho\equiv(\frac{1}{Z(\beta)n})$ and the reference density if
$\mu(k)=\frac{e^{-\beta k}}{Z(\beta)}$. Denote $a_{k}=|k\rangle\langle k+1|$.
On the diagonal, we have

$\displaystyle\frac{1}{2}L^{*}(f)$
$\displaystyle=\sum_{k}e^{\beta/2}(a_{k}a_{k}^{*}f-a_{k}^{*}fa_{k})+e^{-\beta/2}(a_{k}^{*}a_{k}f-a_{k}fa_{k}^{*})$
$\displaystyle=\sum_{k}e^{\beta/2}(e_{k}f(k)-f(k)e_{k+1})+e^{-\beta/2}(f(k+1)e_{k+1}-f(k+1)e_{k})$
$\displaystyle=\frac{1}{Z(\beta)n}(e^{\beta/2}(e_{0}-e_{n})+e^{-\beta/2}(e_{n}-e_{0}))\hskip
2.84544pt.$

Note that

$\ln\rho-\ln\mu\hskip 2.84544pt=\hskip 2.84544pt\ln f\hskip 2.84544pt=\hskip
2.84544pt\big{(}\beta k-\ln(Z(\beta)n)\big{)}_{k=1}^{n}\hskip 2.84544pt.$

Since there are only two terms, we have

$EP(\rho)\sim c(\beta)\hskip 2.84544pt$

for some constant $c(\beta)$ only depending on $\beta$. This holds for
$n\hskip 2.84544pt\geq\hskip 2.84544ptn_{0}$ large enough. ∎

######  Remark 4.22.

When $\beta>0$, $\sum_{k=1}^{n}e^{-\beta k}=O(1)$ as a geometric series. In
the case that $\beta=0$, Theorem 4.21 no longer yields with
$\alpha_{c}(L_{n})\sim\frac{1}{n}$. For $\beta=0$, the birth-death process
reduces to a “broken” version of the cyclic graph as in Example 3.11 with
$\alpha_{c}(L_{n})\sim 1/n^{2}$.

###  4.6. Noncommutative concentration inequality

One important application of modified Log-Sobolev inequality is to derive
concentration inequalities. Such implication is discovered by Otto and Villani
[75] for classical Markov semigroups (see also the work of Erbar and Maas [33]
in the discrete case), and recently extended to the noncommutative setting in
[82, 35, 23]. In this section we will show that CMLSI of a GNS symmetric
semigroup also implies concentration inequalities. A key quantity in the
discussion of concentration inequalities is the Lipschitz norm

$\|x\|_{\text{Lip}}^{2}\hskip 2.84544pt=\hskip
2.84544pt\max\\{\parallel\\!\Gamma_{L}(x,x)\\!\parallel\hskip 2.84544pt,\hskip
2.84544pt\parallel\\!\Gamma_{L}(x^{*},x^{*})\\!\parallel\\}$

where the _gradient form_ (or Carré du Champ operator) is

$\Gamma_{L}(x,y)\hskip 2.84544pt=\hskip
2.84544pt\frac{1}{2}\Big{(}L(x^{*})y+x^{*}L(y)-L(x^{*}y)\Big{)}\hskip
2.84544pt,\hskip 2.84544pt\forall x,y\in\text{dom}(L)\hskip 2.84544pt.$

Note that $\|\cdot\|_{\text{Lip}}$ is a semi-norm (satisfying triangle
inequality) because $\Gamma_{L}$ is completely positive bilinear form. Our
first lemma is to show that $\|x\|_{\text{Lip}}$ can be approximated by
Haagerup reduction.

######  Lemma 4.23.

Let $x\in{\mathcal{M}}$. Then for all $n\in{\mathbb{N}}$

$\|E_{{\mathcal{M}}_{n}}(x)\|_{Lip}\hskip 2.84544pt\leq\hskip
2.84544pt\|x\|_{Lip}\hskip 2.84544pt.$

###### Proof.

Recall the conditional expectation
$E_{{\mathcal{M}}_{n}}:\hat{{\mathcal{M}}}\to{\mathcal{M}}_{n}$ is given by

$E_{{\mathcal{M}}_{n}}(x)=2^{n}\int_{0}^{2^{-n}}\alpha_{t}^{\psi_{n}}(x)dt$

Note that $\alpha_{t}^{\psi_{n}}$ is an inner automorphism on
${\mathcal{M}}\rtimes_{\alpha}2^{-n}\mathbb{Z}$. We note that for a modular
group $\alpha_{t}$ such that $L\alpha_{t}=\alpha_{t}L$,

$\Gamma_{L}(\alpha_{t}(x),\alpha_{t}(y))\hskip 2.84544pt=\hskip
2.84544pt\alpha_{t}(\Gamma_{L}(x,y))\hskip 2.84544pt,$

which implies
$\parallel\\!x\\!\parallel_{\text{Lip}}=\parallel\\!\alpha_{t}(x)\\!\parallel_{\text{Lip}}$.
Here both $\alpha_{t}^{\hat{\phi}}$ and $\alpha_{t}^{\psi_{n}}$ commutes with
$\hat{L}=\operatorname{\operatorname{id}}_{\mathbb{T}}\otimes L$. Then by
triangle inequality,

$\displaystyle\parallel\\!E_{{\mathcal{M}}_{n}}(x)\\!\parallel_{\text{Lip}}$
$\displaystyle=\Big{\|}2^{n}\int_{0}^{2^{-n}}\alpha_{t}^{\psi_{n}}(x)dt\Big{\|}_{\text{Lip}}\leq
2^{n}\int_{0}^{2^{-n}}\parallel\\!x\\!\parallel_{\text{Lip}}dt=\parallel\\!x\\!\parallel_{\text{Lip}}\hskip
2.84544pt.\quad\qed$

######  Lemma 4.24.

Let ${\mathcal{M}}_{0},{\mathcal{N}}\subset{\mathcal{M}}$ be two subalgebras
and $\phi$ be a normal faithful state $\phi$. Suppose
$E_{0}:{\mathcal{M}}\to{\mathcal{M}}_{0}$ and
$E:{\mathcal{M}}\to{\mathcal{N}}$ are $\phi$-preserving conditional
expectations onto ${\mathcal{M}}_{0}$ and ${\mathcal{N}}$ respectively.
Suppose $E\circ E_{0}=E_{0}\circ E$ satisfy the commuting square condition

${{\mathcal{M}}}$${{\mathcal{M}}_{0}}$${{\mathcal{N}}}$${{\mathcal{N}}_{0}}$$E$$E_{0}$$E_{0}$$E$

Then for any $p\in[1,\infty]$ and any $x\in{\mathcal{M}}$,

$\parallel\\!E_{0}(x)\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{0}\subset{\mathcal{M}}_{0},\phi)}=\parallel\\!E(x)\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}\leq\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}$

In other words,
$L_{\infty}^{p}({\mathcal{N}}_{0}\subset{\mathcal{M}}_{0},\phi)\subset
L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)$ is a $1$-complemented
subspace with projection $E_{0}$.

###### Proof.

We can assume $\phi=\phi_{\rm\text{tr}}$ in the Remark 4.17. Using commuting
square assumption, we know $E_{0}(a)\in{\mathcal{N}}_{0}$ for
$a\in{\mathcal{N}}$. By definition,

$\displaystyle\parallel\\!E_{0}(x)\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{0}\subset{\mathcal{M}}_{0},\phi)}=$
$\displaystyle\sup_{a,b\in\hskip
2.84544pt{\mathcal{N}}_{0}}\parallel\\!aE_{0}(x)b\\!\parallel_{\phi,p}\leq\sup_{a,b\in\hskip
2.84544pt{\mathcal{N}}_{0}}\parallel\\!E_{0}(axb)\\!\parallel_{\phi,p}$
$\displaystyle\leq$ $\displaystyle\sup_{a,b\in\hskip
2.84544pt{\mathcal{N}}_{0}}\parallel\\!axb\\!\parallel_{\phi,p}\leq\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}$

where the first supremum is for all $a,b$ in the corresponding subalgebra with
$\parallel\\!a\\!\parallel_{\phi,p}=\parallel\\!b\\!\parallel_{\phi,p}=1$. Now
it suffices to show the other direction

$\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{0}\subset{\mathcal{M}}_{0},\phi)}\geq\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{0}\subset{\mathcal{M}}_{0},\phi)}\hskip
2.84544pt,$

for $x\in{\mathcal{M}}_{0}$. For that, we revoke that for
$\frac{1}{p}+\frac{1}{q}$,
$L_{1}^{p^{\prime}}({\mathcal{N}}\subset{\mathcal{M}})\subset
L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)^{*}$ is as a
weak∗-dense subspace. Here for $x\in{\mathcal{M}}$,

$\parallel\\!y\\!\parallel_{L_{1}^{q}({\mathcal{N}}\subset{\mathcal{M}})}=\inf_{y=azb}\parallel\\!a\\!\parallel_{2p,\phi}\parallel\\!y\\!\parallel_{q,\phi}\parallel\\!b\\!\parallel_{2p,\phi}$

where the infimum is over all factorization
$a,b\in{\mathcal{N}},z\in{\mathcal{M}}$. The duality pairing is given by the
KMS inner product,

$\langle x,y\rangle=\tau(x^{*}d_{\phi}^{1/2}yd_{\phi}^{1/2})=\langle
x,y\rangle_{\phi}$

Indeed, it was proved in [51, Corollary 3.13] that

$E_{0}:L_{1}^{q}({\mathcal{N}}\subset{\mathcal{M}})\to
L_{1}^{q}({\mathcal{N}}_{0}\subset{\mathcal{M}}_{0})$

is a contraction by the commuting square condition. Therefore, for
$x\in{\mathcal{M}}_{0}$, by the $\phi$-symmetry of $E_{0}$,

$\displaystyle\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}=$
$\displaystyle\sup_{\parallel\\!\hskip 2.84544pty\hskip
2.84544pt\\!\parallel_{L_{1}^{q}({\mathcal{N}}\subset{\mathcal{M}})}=1}\langle
x,y\rangle_{\phi}$ $\displaystyle=$ $\displaystyle\sup_{\parallel\\!\hskip
2.84544pty\hskip
2.84544pt\\!\parallel_{L_{1}^{q}({\mathcal{N}}\subset{\mathcal{M}})}=1}\langle
x,E_{0}(y)\rangle_{\phi}$ $\displaystyle\leq$
$\displaystyle\sup_{\parallel\\!\hskip 2.84544ptz\hskip
2.84544pt\\!\parallel_{L_{1}^{q}({\mathcal{N}}_{0}\subset{\mathcal{M}}_{0})}=1}\langle
x,z\rangle_{\phi}=\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{0}\subset{\mathcal{M}}_{0},\phi)}\hskip
2.84544pt.\qed$

######  Lemma 4.25.

For $x\in{\mathcal{M}}$,
$\lim_{n}\|E_{{\mathcal{M}}_{n}}(x)\|_{L_{\infty}^{p}({\mathcal{N}}_{n}\subset{\mathcal{M}}_{n},\psi_{n})}=\|x\|_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}$

###### Proof.

Recall the commuting square condition
$E_{{\mathcal{M}}_{n}}\circ\hat{E}=\hat{E}\circ E_{{\mathcal{M}}_{n}}$. Then
by Lemma 4.15 & 4.24,

$\displaystyle\parallel\\!E_{{\mathcal{M}}_{n}}(x)\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{n}\subset{\mathcal{M}}_{n},\psi_{n})}=$
$\displaystyle\parallel\\!E_{{\mathcal{M}}_{n}}(x)\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{n}\subset{\mathcal{M}}_{n},\hat{\phi})}=\parallel\\!E_{{\mathcal{M}}_{n}}(x)\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\hat{\phi})}\leq\parallel\\!x\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\hat{\phi})}\hskip
2.84544pt.$

The other direction follows from the weak∗-convergence
$E_{{\mathcal{M}}_{n}}(x)\to x$. Fix $\frac{1}{q}+\frac{1}{p}=1$. For any
$\varepsilon>0$, there exists $a_{0},b_{0}\in\hat{{\mathcal{N}}}$ and
$y_{0}\in\hat{{\mathcal{M}}}$ such that

$\displaystyle\parallel\\!aa^{*}\\!\parallel_{p,\hat{\phi}}=\parallel\\!b^{*}b\\!\parallel_{p,\hat{\phi}}=\parallel\\!y\\!\parallel_{\hat{\phi},q}=1$
$\displaystyle\hat{\tau}(d_{\hat{\phi}}^{1/2}axbd_{\hat{\phi}}^{1/2}y)\geq\parallel\\!x\\!\parallel_{L_{\infty}^{p}(\hat{{\mathcal{N}}}\subset\hat{{\mathcal{M}}},\hat{\phi})}-\varepsilon$

By the weak∗-density, we can choose $n_{1},n_{2},n_{3}$ and
$n_{4}\geq\max\\{n_{1},n_{2},n_{3}\\}$ inductively such that

$\displaystyle\tau(d_{\phi}^{1/2}E_{{\mathcal{M}}_{n_{1}}}(a)E_{{\mathcal{M}}_{n_{4}}}(x)E_{{\mathcal{M}}_{n_{2}}}(b)d_{\phi}^{1/2}E_{{\mathcal{M}}_{n_{3}}}(y))>\tau(d_{\phi}^{1/2}axbd_{\phi}^{1/2}y)-\varepsilon>\parallel\\!x\\!\parallel_{L_{\infty}^{p}(\hat{{\mathcal{N}}}\subset\hat{{\mathcal{M}}},\hat{\phi})}-2\varepsilon\hskip
2.84544pt.$

Since $E_{{\mathcal{M}}_{n}}(\hat{{\mathcal{N}}})={\mathcal{N}}_{n}$ (see the
commuting diagram after Lemma 4.5), we have

$\displaystyle\parallel\\!E_{{\mathcal{M}}_{n_{1}}}(a)E_{{\mathcal{M}}_{n_{1}}}(a)^{*}\\!\parallel_{\hat{\phi},p}\leq\parallel\\!E_{{\mathcal{M}}_{n_{1}}}(aa^{*})\\!\parallel_{\hat{\phi},p}\leq\parallel\\!aa^{*}\\!\parallel_{\hat{\phi},p}=1$
$\displaystyle\parallel\\!E_{{\mathcal{M}}_{n_{3}}}(b^{*})E_{{\mathcal{M}}_{n_{3}}}(b)\\!\parallel_{\hat{\phi},p}\leq\parallel\\!b^{*}b\\!\parallel_{\hat{\phi},p}=1\hskip
2.84544pt,$
$\displaystyle\parallel\\!E_{{\mathcal{M}}_{n_{4}}}(y)\\!\parallel_{\hat{\phi},q}\leq\parallel\\!y\\!\parallel_{\hat{\phi},q}=1$

by the $\hat{\phi}$-symmetry of $E_{{\mathcal{M}}_{n}}$. Then, for
$n=n_{4}=\max\\{n_{1},n_{2},n_{3},n_{4}\\}$,

$\displaystyle\parallel\\!x\\!\parallel_{L_{\infty}^{p}(\hat{{\mathcal{N}}}_{n}\subset\hat{{\mathcal{M}}}_{n},\hat{\phi})}$
$\displaystyle\geq\parallel\\!E_{{\mathcal{M}}_{n}}(x)\\!\parallel_{L_{\infty}^{p}(\hat{{\mathcal{N}}}_{n}\subset\hat{{\mathcal{M}}}_{n},\hat{\phi})}$
$\displaystyle\geq\tau(d_{\hat{\phi}}^{1/2}E_{{\mathcal{M}}_{n_{1}}}(a)xE_{{\mathcal{M}}_{n_{2}}}(b)d_{\hat{\phi}}^{1/2}E_{{\mathcal{M}}_{n_{3}}}(y))$
$\displaystyle\geq\parallel\\!x\\!\parallel_{L_{\infty}^{p}(\hat{{\mathcal{N}}}\subset\hat{{\mathcal{M}}},\hat{\phi})}-2\varepsilon$

which proves

$\lim_{n}\|E_{{\mathcal{M}}_{n}}(x)\|_{L_{\infty}^{p}({\mathcal{N}}_{n}\subset{\mathcal{M}}_{n},\psi_{n})}=\|x\|_{L_{\infty}^{p}(\hat{{\mathcal{N}}}\subset\hat{{\mathcal{M}}},\hat{\phi})}\hskip
2.84544pt.$

Finally, the assertion follows from

$\|x\|_{L_{\infty}^{p}(\hat{{\mathcal{N}}}\subset\hat{{\mathcal{M}}},\hat{\phi})}=\|x\|_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}\hskip
2.84544pt,$

as a consequence of $E_{0}\circ\hat{E}=\hat{E}\circ E_{0}$ by Lemma 4.24. ∎

######  Theorem 4.26.

Let ${\mathcal{M}}$ be a finite von Neumann algebra and $T_{t}=e^{-tL}$ be a
$\phi$-symmetric quantum Markov semigroup. Suppose $T_{t}$ satisfies MLSIwith
parameter $\alpha>0$. There exists an universal constant $c$ such that for
$2\leq p<\infty$

$\alpha\|x-E(x)\|_{L_{p}({\mathcal{M}},\phi)}\leq\alpha\|x-E(x)\|_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}\hskip
2.84544pt\leq\hskip
2.84544ptc\sqrt{p}\parallel\\!x\\!\parallel_{\text{Lip}}\hskip 2.84544pt.$

###### Proof.

We first show that if $T_{t}$ satisfying $\alpha$-MLSI, so does the
approximation semigroup.

$T_{n,t}=\hat{T}_{t}:{\mathcal{M}}_{n}\to{\mathcal{M}}_{n}$

Indeed, as we see in the discussion above,
${\mathcal{M}}_{n}\subset{\mathcal{M}}\rtimes_{\alpha^{\phi}_{t}}2^{-n}\mathbb{Z}\cong
L_{\infty}(\mathbb{T},{\mathcal{M}})$ and the extension
$\hat{T}_{t}=T_{t}\otimes\operatorname{\operatorname{id}}_{\mathbb{T}}$ has
$\alpha$-MLSI (because $L_{\infty}(\mathbb{T})$ is a commutative space). Note
that since
${\mathcal{M}}_{n}\subset{\mathcal{M}}\rtimes_{\alpha^{\phi}_{t}}2^{-n}\mathbb{Z}\subset{\mathcal{M}}\rtimes_{\alpha^{\phi}_{t}}G$,
the restriction
$E_{{\mathcal{M}}_{n}}:{\mathcal{M}}\rtimes_{\alpha^{\phi}_{t}}2^{-n}\mathbb{Z}\to{\mathcal{M}}_{n}$
is also a conditional expectation. Then for any $\rho,\sigma\in
S({\mathcal{M}}_{n})$, we have

$D(E_{{\mathcal{M}}_{n},*}\rho||E_{{\mathcal{M}}_{n},*}\sigma)\leq
D(\rho||\sigma)=D(\rho|_{{\mathcal{M}}_{n}}||\sigma|_{{\mathcal{M}}_{n}})\leq
D(E_{{\mathcal{M}}_{n},*}\rho||E_{{\mathcal{M}}_{n},*}\sigma)\hskip
2.84544pt.$

Now using the commutation relation $\hat{T}_{t}\circ
E_{{\mathcal{M}}_{n}}=E_{{\mathcal{M}}_{n}}\circ\hat{T}_{t}$ and
$E_{{\mathcal{M}}_{n}}\circ\hat{E}=E_{n}\circ E_{{\mathcal{M}}_{n}}$, we have

$\displaystyle D(T_{t,n,*}\rho||E_{n,*}\rho)=$ $\displaystyle
D(E_{{\mathcal{M}}_{n},*}T_{t,n,*}\rho||E_{{\mathcal{M}}_{n},*}E_{n,*}\rho)=D(T_{t,n,*}E_{{\mathcal{M}}_{n},*}\rho||\hat{E}_{*}E_{{\mathcal{M}}_{n},*}\rho)$
$\displaystyle\leq$ $\displaystyle e^{-2\alpha
t}D(E_{{\mathcal{M}}_{n},*}\rho||\hat{E}_{*}E_{{\mathcal{M}}_{n},*}\rho)$
$\displaystyle=$ $\displaystyle e^{-2\alpha
t}D(E_{{\mathcal{M}}_{n},*}\rho||E_{{\mathcal{M}}_{n},*}E_{n,*}\rho)=e^{-2\alpha
t}D(\rho||E_{n,*}\rho)\hskip 2.84544pt.$

Thus $T_{n,t}$ has $\alpha$-MLSI on ${\mathcal{M}}_{n}$. Note that $T_{n,t}$
is both $\hat{\phi}$-symmetric for the extension state $\hat{\phi}$ and also
$\psi_{n}$-symmetric for the trace $\psi_{n}$. Now, we may use the tracial
version of the concentration inequality [35, Theorem 6.10] that for
$x\in{\mathcal{M}}_{n}$

$\alpha\parallel\\!x-E_{n}(x)\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{n}\subset{\mathcal{M}}_{n})}\leq
C\sqrt{p}\parallel\\!x\\!\parallel_{\text{Lip}}$

Now by the approximation of Lemma 4.25 and independence of $L_{\infty}^{p}$ on
the reference state, for $x\in{\mathcal{M}}$,

$\displaystyle\parallel\\!{x-E(x)}\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}=$
$\displaystyle\lim_{n}\parallel\\!E_{{\mathcal{M}}_{n}}(x-E(x))\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{n}\subset{\mathcal{M}}_{n},\psi_{n})}$
$\displaystyle=$
$\displaystyle\lim_{n}\parallel\\!E_{{\mathcal{M}}_{n}}(x)-E_{n}E_{{\mathcal{M}}_{n}}(x))\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}_{n}\subset{\mathcal{M}}_{n},\psi_{n})}$
$\displaystyle\leq$ $\displaystyle
C\sqrt{p}\parallel\\!E_{{\mathcal{M}}_{n}}(x)\\!\parallel_{\text{Lip}}\leq
C\sqrt{p}\parallel\\!x\\!\parallel_{\text{Lip}}\hskip 2.84544pt.$

The other inequality

$\parallel\\!y\\!\parallel_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)}\geq\parallel\\!y\\!\parallel_{L_{p}({\mathcal{M}},\phi)}$

is clear from definition of
$L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\phi)$. ∎

For Gaussian type concentration property, we need the following definition.

######  Definition 4.27.

For an operator $O$ we say that

$\text{Prob}_{\phi}(|O|>t)\hskip 2.84544pt\leq\hskip 2.84544pt\varepsilon$

if there exist a projection $e$ such that

$\|eOe\|_{\infty}\leq t\quad\mbox{and}\quad\phi(1-e)\leq\varepsilon\hskip
2.84544pt.$

The next lemma is a Cheybev inequality for $\phi$-weighted $L_{p}$ norm.

######  Lemma 4.28.

Let $x\in L_{p}({\mathcal{M}},\phi)$ and $1<p<\infty$. Then

$\text{Prob}_{\phi}(|x|>t)\hskip 2.84544pt\leq\hskip
2.84544pt2\Big{(}\frac{t}{4}\Big{)}^{-p}\|x\|_{p,\phi}\hskip 2.84544pt.$

###### Proof.

We start with a positive element $x=y^{2}$ and assume $\|x\|_{p,\phi}=M$. Then
we have

$M=\|x\|_{p,\phi}\hskip 2.84544pt=\hskip
2.84544pt\|d_{\phi}^{1/2p}xd_{\phi}^{1/2p}\|_{p}\hskip 2.84544pt=\hskip
2.84544pt\|yd_{\phi}^{1/2p}\|_{2p}^{2}\hskip 2.84544pt.$

Recall the asymmetric Kosaki $L_{p}$-space

$\parallel\\!y\\!\parallel_{L_{2p}^{c}({\mathcal{M}},\phi)}:=\|yd_{\phi}^{1/2p}\|_{2p}\hskip
2.84544pt,$

and the complex interpolation relation.

$L_{2p}^{c}({\mathcal{M}},\phi)=[{\mathcal{M}},L_{2}^{c}({\mathcal{M}},\phi)]_{1/p}\hskip
2.84544pt.$

By the relation between real and complex interpolation

$L_{2p}^{c}({\mathcal{M}},\phi)=[{\mathcal{M}},L_{2}^{c}({\mathcal{M}},\phi)]_{1/p}\subset[N,L_{2}^{c}({\mathcal{M}},\phi)]_{1/p,\infty}\hskip
2.84544pt,$

for every $s>0$ we have a decomposition $y\hskip 2.84544pt=\hskip
2.84544pty_{1}+y_{2}$ such that

$\|y_{1}\|_{\infty}+s\parallel\\!y_{2}\\!\parallel_{L_{2}^{c}({\mathcal{M}},\phi)}\hskip
2.84544pt\leq\hskip 2.84544pts^{1/p}M^{1/2}\hskip 2.84544pt.$

The by Chebychev’s inequality for the spectral projection
$e=e_{[0,a]}(y_{2}^{*}y_{2})$ of $y_{2}^{*}y_{2}$, we have

$a\phi(1-e)\hskip 2.84544pt\leq\hskip 2.84544pt\phi(y_{2}^{*}y_{2})\hskip
2.84544pt\leq\hskip
2.84544pts^{2/p-2}M\quad\mbox{and}\quad\|y_{2}e\|_{\infty}^{2}\hskip
2.84544pt=\hskip 2.84544pt\|ey_{2}^{*}y_{2}e\|_{\infty}\hskip
2.84544pt\leq\hskip 2.84544pta\hskip 2.84544pt.$

Choose $a=s^{2/p}$ and deduce that

$\|exe\|_{\infty}=\|ye\|_{\infty}^{2}\hskip 2.84544pt\leq\hskip
2.84544pt(\|y_{1}e\|_{\infty}+\|y_{2}e\|_{\infty})^{2}\hskip
2.84544pt\leq\hskip 2.84544pt(s^{1/p}+s^{1/p})^{2}\hskip 2.84544pt\leq\hskip
2.84544pt4s^{2/p}M\hskip 2.84544pt.$

Then for $t=4s^{2/p}M$ and

$\phi(1-e)\hskip 2.84544pt\leq\hskip 2.84544pta^{-1}s^{2/p-2}\hskip
2.84544pt=\hskip 2.84544pts^{-2}\hskip 2.84544pt=\hskip
2.84544pt(\frac{t}{4M})^{-p}=(\frac{t}{4})^{-p}M^{p}\hskip 2.84544pt.$

For an arbitrary $x$, we may write $x=x_{1}x_{2}$ such that

$\|d_{\phi}^{1/2p}x_{1}\|_{2p}\hskip 2.84544pt=\hskip
2.84544pt\|x_{2}d_{\phi}^{1/2p}\|_{2p}\hskip 2.84544pt=\hskip
2.84544pt\parallel\\!x\\!\parallel_{p,\phi}=1\hskip 2.84544pt.$

Then for each $s>0$, we have decompotision

$x_{1}\hskip 2.84544pt=\hskip 2.84544ptx_{11}+x_{21}\hskip 2.84544pt,\hskip
2.84544ptx_{2}=x_{21}+x_{22}\hskip 2.84544pt.$

with

$\displaystyle\|x_{11}\|_{\infty}+s\parallel\\!x_{21}\\!\parallel_{L_{2}^{c}({\mathcal{M}},\phi)}\hskip
2.84544pt\leq\hskip 2.84544pts^{1/p}M^{1/2}\hskip
2.84544pt,\|x_{12}\|_{\infty}+s\parallel\\!x_{21}\\!\parallel_{L_{2}^{c}({\mathcal{M}},\phi)}\hskip
2.84544pt\leq\hskip 2.84544pts^{1/p}M^{1/2}\hskip 2.84544pt.$

We then use the Chebychev inequality for

$\phi(x_{21}^{*}x_{21}+x_{22}^{*}x_{22})\hskip 2.84544pt\leq\hskip
2.84544pt2s^{2/p-2}M\hskip 2.84544pt.$

Then, using the choice of $a=s^{2/p}$ and the spectral projection,

$\displaystyle\|exe\|_{\infty}$
$\displaystyle=\|e(x_{1}x_{2})e\|_{\infty}\hskip 2.84544pt=\hskip
2.84544pt\|e(x_{11}+x_{12})(x_{21}+x_{22})e\|_{\infty}$ $\displaystyle\hskip
2.84544pt\leq\hskip
2.84544pt\|x_{11}x_{22}\|_{\infty}+\|ex_{12}x_{21}\|_{\infty}\|x_{11}x_{22}e\|_{\infty}+\|ex_{12}x_{22}e\|_{\infty}$
$\displaystyle\hskip 2.84544pt\leq\hskip 2.84544pt4s^{2/p}M\hskip 2.84544pt.$

Thus for $t=4s^{2/p}M$,

$\displaystyle\phi(1-e)$ $\displaystyle\leq a^{-1}2s^{2/p-2}\hskip
2.84544pt=\hskip 2.84544pt2s^{-2}\hskip 2.84544pt=\hskip
2.84544pt2(\frac{t}{4M})^{-p}=2(\frac{t}{4})^{-p}M^{p}\hskip 2.84544pt.\qed$

######  Corollary 4.29.

Let $T_{t}=e^{-tL}$ be a $\phi$-symmetric quantum Markov semigroup. Suppose
$T_{t}$ satisfies $\alpha$-MLSI. Then for any $x\in{\mathcal{M}}$ and $t>0$,

$\text{Prob}_{\phi}(|x-E_{fix}(x)|>t)\hskip 2.84544pt\leq\hskip
2.84544pt2\exp\Big{(}-\frac{1}{e}\Big{(}\frac{\alpha
t}{4c\parallel\\!x\\!\parallel_{\text{Lip}}}\Big{)}^{2}\Big{)}$

where $c$ is an universal constant.

###### Proof.

By Lemma 4.28, we have

$\text{Prob}_{\phi}(|x-E(x)|>t)\leq
2(t/4)^{-p}\parallel\\!x-E(x)\\!\parallel_{L_{p}({\mathcal{M}},\phi)}^{p}\leq
2\big{(}\frac{4C\parallel\\!x\\!\parallel_{\text{Lip}}\sqrt{p}}{\alpha
t}\big{)}^{p}$

The standard optimization gives $p=\frac{1}{e}(\frac{\alpha
t}{4C\parallel\\!x\\!\parallel_{\text{Lip}}})^{2}$, which implies

$\displaystyle\text{Prob}_{\phi}(|x-E(x)|>t)\leq
2\exp(-\frac{\alpha^{2}t^{2}}{16ec^{2}\parallel\\!x\\!\parallel_{\text{Lip}}^{2}})\qed$

We remark that our argument for the above concentration inequality applies to
general von Neumann algebras. The details are left to readers.

######  Remark 4.30.

In the primitive case, the above results can be compared to [82, Theorem 8],
which states that for self-adjoint $x=x*$

$\displaystyle\phi(e_{\\{|x-E(x)|>t\\}})\leq\exp\Big{(}-\frac{\alpha
t^{2}}{8\parallel\\!d_{\phi}^{-1/2}xd_{\phi}^{1/2}\\!\parallel_{\tilde{\text{Lip}}}^{2}}\Big{)}$

with a different Lipschitz norm. Our results here use a more natural
definition of Lipschitz norm and apply to non-primitive cases. Nevertheless,
the projection we have for

$\text{Prob}_{\phi}(|x-E(x)|>t)$

is not necessarily a spectral projection $e_{\\{|x-E(x)|>t\\}}$ and depends on
the state $\phi.$

######  Remark 4.31.

In the operator valued setting, let ${\mathcal{Q}}$ be any finite von Neumann
algebra and $T_{t}\otimes\operatorname{\operatorname{id}}_{{\mathcal{Q}}}$ be
the amplification semigroup on ${\mathcal{Q}}\overline{\otimes}{\mathcal{M}}$.
The conditional expectation for
$T_{t}\otimes\operatorname{\operatorname{id}}_{{\mathcal{Q}}}$ is
$E\otimes\operatorname{\operatorname{id}}_{{\mathcal{Q}}}$. Note that by Lemma
4.15, $T_{t}\otimes\operatorname{\operatorname{id}}_{{\mathcal{Q}}}$ is GNS-
symmetric to the product state $\phi\otimes\sigma$, for any state $\sigma\in
S({\mathcal{Q}})$ and any invariant state $\phi\in E_{*}(S({\mathcal{M}}))$.
This means we obtain

$\text{Prob}_{\phi\otimes\sigma}(|x-E_{fix}(x)|>t)\leq
2e^{-\frac{\alpha^{2}t^{2}}{C\parallel\\!x\\!\parallel_{\text{Lip}}^{2}}}$

for any product state $\phi\otimes\sigma$ of this specific form. The
projection depends, of course, on $\phi$ and $\sigma$.

We illustrate our result with a special case as matrix concentration
inequalities.

######  Example 4.32 (Matrix concentration inequality).

Let $S_{1},\cdots,S_{n}$ be an independent sequence of random $d\times
d$-matrices $S_{1},\cdots,S_{n}$ such that

$\parallel\\!S_{i}-\mathbb{E}S_{i}\\!\parallel_{\infty}\leq M\hskip
2.84544pt,\hskip 2.84544pta.e.$

It was obtained in [96, Corollary 6.1.2] by Tropp the matrix Bernstein
inequality that for the sum $Z=\sum_{k=1}^{n}S_{k}$

${\mathbb{E}}\parallel\\!Z-\mathbb{E}Z\\!\parallel_{\infty}\leq\sqrt{2v(Z)\ln(2d)}+\frac{1}{3}M\ln(2d)$

and the matrix Chernoff bound

$P(|Z-\mathbb{E}Z|>t)\leq
2d\exp\big{(}-\frac{t^{2}}{v(Z)+\frac{t}{3}M}\big{)}$

where

$v(Z)=\max\\{\parallel\\!{\mathbb{E}}((Z-\mathbb{E}Z)^{*}(Z-\mathbb{E}Z))\\!\parallel_{\hskip
2.84544pt},\parallel\\!{\mathbb{E}}((Z-\mathbb{E}Z)^{*}(Z-\mathbb{E}Z))\\!\parallel\\}\hskip
2.84544pt.$

Now to apply our results, we recall the depolarizing semigroup with generator
$L(f):=(I-E_{\mu})(f)=f-\mu(f)1_{\Omega}$ on any probability space
$(\Omega,\mu)$ has $\alpha_{c}\geq\frac{1}{2}$. For a random matrix
$f:\Omega\to M_{d}$, the Lipschitz norm is

$\displaystyle\parallel\\!f\\!\parallel_{\text{Lip}}^{2}=$
$\displaystyle\frac{1}{2}\max\\{\parallel\\!\hat{f}^{*}\hat{f}+E(\hat{f}^{*}\hat{f})\\!\parallel_{\infty}\hskip
2.84544pt,\hskip
2.84544pt\parallel\\!\hat{f}\hat{f}^{*}+E(\hat{f}\hat{f}^{*})\\!\parallel_{\infty}\\}$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}(\parallel\\!f\\!\parallel_{\infty}^{2}+v(f)),$

where $\hat{f}=f-E_{\mu}(f)$ is the mean zero part. Then we consider
$S_{k}:\Omega_{k}\to M_{d}$ as a random matrix on $(\Omega_{k},\mu_{k})$. Then
on the product space
$(\Omega,\mu)=(\Omega_{1},\mu_{1})\times\cdots\times(\Omega_{n},\mu_{n})$, we
have for $Z=\sum_{k}S_{k}$

$\displaystyle{\mathbb{E}}\parallel\\!Z-{\mathbb{E}}Z\\!\parallel_{\infty}\leq\Big{(}\frac{1}{d}{\mathbb{E}}\parallel\\!Z-{\mathbb{E}}Z\\!\parallel_{p}^{p}\Big{)}^{1/p}\leq
d^{1/p}\parallel\\!Z-{\mathbb{E}}Z\\!\parallel_{L_{\infty}(M_{d},L_{p}(\Omega))}\leq
2cd^{1/p}\sqrt{p}\parallel\\!Z\\!\parallel_{\text{Lip}}\hskip 2.84544pt,$

where $\parallel\\!\cdot\\!\parallel_{p}$ is the $p$-norm for the normalized
trace, and

$\parallel\\!Z\\!\parallel_{\text{Lip}}^{2}\leq\frac{1}{2}(v(Z)+M^{2})\hskip
2.84544pt.$

Optimizing $p$ gives

${\mathbb{E}}\parallel\\!Z-{\mathbb{E}}Z\\!\parallel_{\infty}\leq
2ce^{-1/2}\sqrt{(v(Z)+M^{2})\ln d}$

And for the matrix Chernoff bound

$P(|Z-\mathbb{E}Z|>t)\leq
d\text{Prob}_{\mu\otimes\frac{{\rm\text{tr}}}{d}}(|Z-\mathbb{E}Z|>t)\leq
2\exp\Big{(}-\frac{t^{2}}{64ec^{2}(v(Z)+M^{2})}\Big{)}\hskip 2.84544pt.$

We will now prove a second concentration inequality under more restrictive
assumptions. Our starting point is a finite von Neumann algebra
$({\mathcal{M}},\tau)$ and a normal faithful weight $\phi$ whose density
operator has finite spectrum

$d_{\phi}\hskip 2.84544pt=\hskip 2.84544pt\sum_{k=1}^{m}\lambda_{k}e_{k}\hskip
2.84544pt,\hskip 2.84544pt\alpha_{t}^{\phi}(x)\hskip 2.84544pt=\hskip
2.84544ptd_{\phi}^{-it}xd_{\phi}^{it}\hskip 2.84544pt,$

where $\lambda_{k}>0$ and $\sum{e_{k}}=1$. By re-scaling $\phi$, we can always
assume $\lambda_{k}\geq 1$. Let $T_{t}=e^{-Lt}:{\mathcal{M}}\to{\mathcal{M}}$
be a $\phi$-symmetric quantum Markov semigroup. We now analyze Haagerup’s
construction for a fixed $n\in{\mathbb{N}}$. We work with the subgroup
$2^{-n}\mathbb{Z}\subset{\mathbb{R}}$ of the modular automorphism group
$\alpha_{t},t\in\mathbb{R}$. By the duality $L(\mathbb{Z})\cong
L_{\infty}(\mathbb{T})$, the crossed product can be represented as

${\mathcal{M}}\rtimes_{\alpha}2^{-n}\mathbb{Z}\cong
L_{\infty}(\mathbb{T},{\mathcal{M}})\hskip 2.84544pt$

with the embedding

$\displaystyle\pi:{\mathcal{M}}\to L_{\infty}(\mathbb{T},{\mathcal{M}})\hskip
2.84544pt,\hskip 2.84544pt\pi(x)=1\otimes x$
$\displaystyle\lambda:2^{-n}\mathbb{Z}\to
L_{\infty}(\mathbb{T},{\mathcal{M}})\hskip 2.84544pt,\hskip
2.84544pt\lambda(g)=\exp(i2^{n}gs)\otimes d_{\phi}^{ig}$

Under this identification we write

$w(1):=\lambda(2^{-n})=\exp(is)\otimes d_{\phi}^{\frac{i}{2^{n}}}\hskip
2.84544pt,\hskip 2.84544pts\in[0,2\pi]$

as the generator of the group $\lambda(2^{-n}\mathbb{Z})$. Then the canonical
extension state is

$\hat{\phi}(x)=\int_{0}^{2\pi}\phi(x(s))\frac{ds}{2\pi}\hskip 2.84544pt,\hskip
2.84544ptx\in L_{\infty}(\mathbb{T},{\mathcal{M}})$

Note that the representation $\pi$ and $\lambda$ above is different with the
one we used in (4.2). Let us also recall the rest of the construction under
this identification

$\displaystyle b_{n}(s)$ $\displaystyle\hskip 2.84544pt=\hskip
2.84544pt-i\text{Log}(w(1))\hskip 2.84544pt=\hskip
2.84544pt\sum_{k=1}^{m}[s+2\frac{\ln\lambda_{k}}{2^{n}}]e_{k}\hskip
2.84544pt,$ $\displaystyle a_{n}(s)$ $\displaystyle\hskip 2.84544pt=\hskip
2.84544pt2^{n}\sum_{k=1}^{m}[s+\frac{\ln\lambda_{k}}{2^{n}}]e_{k}\hskip
2.84544pt,$ $\displaystyle\psi_{n}(x)$ $\displaystyle\hskip 2.84544pt=\hskip
2.84544pt\hat{\phi}(e^{-a_{n}(s)}x)\hskip 2.84544pt.$

where $[s]$ is the number in $[0,2\pi)$ such that $s-[s]=2\pi l$ for some
$l\in\mathbb{Z}$. The key point here is that for $n$ large enough we may
assume that $0\leq\frac{\ln\lambda_{k}}{2^{n}}<\pi$ and hence

$\displaystyle a_{n}(s)\hskip 2.84544pt=\hskip
2.84544pt\sum_{k}(2^{n}s+\ln\lambda_{k})e_{k}\hskip 2.84544pt\hskip
2.84544pt\text{for}\hskip 2.84544pt0\leq s\leq\pi\hskip 2.84544pt.$ (62)

######  Lemma 4.33.

Suppose $2^{n}\pi>\max\\{\ln\lambda_{j}\\}$. Then for $x\in{\mathcal{M}}$,

  1. i)

$E_{{\mathcal{M}}_{n}}(x)(s)=x$ for $s\in[0,\pi]$;

  2. ii)

$\|E_{{\mathcal{M}}_{n}}(x)\|_{\text{Lip}}=\|x\|_{\text{Lip}}$;

###### Proof.

It suffices to consider the element $x=e_{k}xe_{j}$. Then the cocyle chain
rule tells us that for $s\in[0,\pi]$, we have by (62)

$\displaystyle\alpha_{t}^{\psi_{n}}(x)(s)$ $\displaystyle\hskip
2.84544pt=\hskip
2.84544pte^{ita_{n}}(s)\alpha_{t}^{\hat{\phi}}(e_{k}xe_{j})e^{-ita_{n}}(s)$
$\displaystyle\hskip 2.84544pt=\hskip
2.84544pte^{ita_{n}}(s)\lambda_{k}^{it}\lambda_{j}^{-it}(e_{k}xe_{j})e^{-ita_{n}}(s)$

Because $e_{k}$ and $e_{k}$ are projectors, they leave only terms in the
Taylor expansion for $e^{ita_{n}}(s)$ that contain only powers of themselves.
Hence we may rewrite the sums in the exponential to obtain

$\displaystyle\alpha_{t}^{\psi_{n}}(x)(s)$ $\displaystyle\hskip
2.84544pt=\hskip
2.84544pt\lambda_{k}^{it}\lambda_{j}^{-it}e^{-it2^{n}[s+2^{-n}\ln\lambda_{k}]}e_{k}xe_{j}e^{it2^{n}[s+2^{-n}\ln\lambda_{j}]}$
$\displaystyle\hskip 2.84544pt=\hskip
2.84544pt\lambda_{k}^{it}\lambda_{j}^{-it}\lambda_{k}^{-it}\lambda_{j}^{it}e^{-its2^{n}}e^{its2^{n}}e_{k}xe_{j}\hskip
2.84544pt=\hskip 2.84544pte_{k}xe_{j}\hskip 2.84544pt.$

For the second assertion, we recall that the upper estimate holds thanks to
Lemma 4.23. For the other direction follows, we have by i) that for
$s\in[0,\pi]$

$\Gamma_{\hat{L}}(E_{{\mathcal{M}}_{n}}(x),E_{{\mathcal{M}}_{n}}(x))(s)=\Gamma_{L}(E_{{\mathcal{M}}_{n}}(x)(s),E_{{\mathcal{M}}_{n}}(x)(s))=\Gamma_{L}(x,x)\hskip
2.84544pt,$

Therefore

$\displaystyle\parallel\\!\Gamma_{\hat{L}}(E_{{\mathcal{M}}_{n}}(x),E_{{\mathcal{M}}_{n}}(x))\\!\parallel=\sup_{s\in[0,2\pi]}\parallel\\!\Gamma_{L}(E_{{\mathcal{M}}_{n}}(x)(s),E_{{\mathcal{M}}_{n}}(x)(s))\\!\parallel\geq\parallel\\!\Gamma_{L}(x,x)\\!\parallel,$

which implies
$\parallel\\!E_{{\mathcal{M}}_{n}}(x)\\!\parallel_{\text{Lip}}\geq\parallel\\!x\\!\parallel_{\text{Lip}}$.∎

######  Theorem 4.34.

Let ${\mathcal{M}}$ be a finite von Neumann algebra with finite trace $\tau$
and let $\phi$ be a faithful density operator with finite spectrum. Suppose
$L$ is a $\phi$-symmetric generator satisfying CMLSI with parameter $\alpha$.
For $2\leq p<\infty$, we have

$\alpha\|x-E(x)\|_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\tau)}\hskip
2.84544pt\leq\hskip 2.84544ptc\parallel\\!x\\!\parallel_{\text{Lip}}\hskip
2.84544pt.$

Note that here
$\|x-E(x)\|_{L_{\infty}^{p}({\mathcal{N}}\subset{\mathcal{M}},\tau)}$ is the
$L_{\infty}^{p}$ norm depending on the trace.

###### Proof.

Let $n$ be as in Lemma 4.33. Now, we consider the subalgebra

${\mathcal{M}}_{n}\hskip 2.84544pt=\hskip
2.84544pt\\{x|\alpha_{t}^{\psi_{n}}(x)=x\\}\subset
L_{\infty}(\mathbb{T},{\mathcal{M}})\hskip 2.84544pt.$

By definition of CMLSI, $\operatorname{\operatorname{id}}_{\mathbb{T}}\otimes
L$ also satisfies $\alpha$-CMLSI on $L_{\infty}(\mathbb{T},{\mathcal{M}})$. By
restriction we have $\alpha$-CMLSI for each ${\mathcal{M}}_{n}$. On
${\mathcal{M}}_{n}$ we have two natural traces. Indeed, the Haagerup
construction leads to $\tau_{n}(x)\hskip 2.84544pt=\hskip
2.84544pt\psi_{n}(x)$ and this is compatible with the restriction of
$\hat{\phi}$ to ${\mathcal{M}}_{n}$. The other trace is $\mu\otimes\tau$,
where $\mu$ can be probability measure on $\mathbb{T}$. By the independence of
tracial state for CMLSI, we use the tracial version [35, Theorem 6.10] for
$\mu\otimes\tau$ restricted on ${\mathcal{M}}_{n}$ that

$\alpha\|x\|_{L_{\infty}^{p}({\mathcal{N}}_{n}\subset{\mathcal{M}}_{n},\mu\otimes\tau)}\hskip
2.84544pt\leq\hskip 2.84544ptc_{0}\sqrt{p}\|x\|_{Lip}\hskip 2.84544pt.$

Applying this to $E_{{\mathcal{M}}_{n}}(x)|_{[0,\pi]}=1_{[0,\pi]}\otimes x$
yields the assertion, as
$E_{{\mathcal{M}}_{n}}(E(x))=\hat{E}E_{{\mathcal{M}}_{n}}(x)$ is also of the
form $1_{[0,\pi]}\otimes E(x)$. ∎

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