$\mathcal{L}_{\frac{p}{2},\infty}.$ Hence, the first summand is
$\mathcal{L}_{1}$-valued holomorphic for $\Re(z)>q-1.$

The second summand in (5.10) can be written as

$z\to B^{z-q+1}\cdot B^{q-2}[B,A]\cdot A^{z-1}.$

By our assumption of Condition 5.4.1, the operator $B^{q-2}[B,A]$ belongs to
$\mathcal{L}_{1}$ and accordingly the map $z\to B^{z-1}\cdot[B,A]\cdot
A^{z-1}$ is $\mathcal{L}_{1}$-continuous. Once again due to Lemma 5.4.5, the
mappings $z\to B^{z-q+1}$ and $z\to A^{z-1}$ are $\mathcal{L}_{\infty}-$valued
holomorphic for $\Re(z)>q-1.$ Hence, the second summand of (5.10) is
$\mathcal{L}_{1}$-valued holomorphic for $\Re(z)>q-1.$

We now treat the third summand of (5.10). By Lemma 5.4.6, the mapping

$z\to[B,A^{z-1}],\quad\Re(z)>q-1,$

is a holomorphic $\mathcal{L}_{\frac{p}{2},\infty}-$valued function. Due to
Condition 5.4.1, we have that

$B^{q-2}A\in\mathcal{L}_{\frac{p}{q-2},\infty}\subset\mathcal{L}_{\frac{p}{p-2},1}.$

Hence

$z\to B^{z-q+1}\cdot B^{q-2}A\cdot[B,A^{z-1}],\quad\Re(z)>q-1,$

is an $\mathcal{L}_{1}$-valued holomorphic function.

So, all three summands in (5.10) are holomorphic for $\Re(z)>q-1.$ Thus, $z\to
F_{2}(z)$ is holomorphic for $\Re(z)>q-1.$ Since $q>p$ is arbitrary, it
follows that $z\to F_{2}(z)$ is holomorphic for $\Re(z)>p-1.$ This proves
(iii).

For (iv), we note that this is an immediate consequence of the assumption of
Condition 5.4.1.(iii) and Lemma 5.4.5.

Finally we prove (i). We write $F_{0}(z)$ as:

$\displaystyle F_{0}(z)$
$\displaystyle=B^{z-1}[B,A^{z-1}A^{1/2}]A^{1/2}+[B,A^{1/2}]A^{1/2}Y^{z-1}$
$\displaystyle=F_{2}(z)+B^{z-1}A^{z-1}[B,A^{1/2}]A^{1/2}+[B,A^{1/2}]A^{1/2}F_{3}(z)$
$\displaystyle=F_{2}(z)+F_{1}(z)[B,A^{1/2}]A^{1/2}+[B,A^{1/2}]F_{3}(z).$

Hence, by (ii), (iii) and (iv) and Condition 5.4.1.(iv),

$F_{0}(z)\in\mathcal{L}_{1}+\mathcal{L}_{p/(p-2),1}\cdot\mathcal{L}_{p/2,\infty}+\mathcal{L}_{p/2,\infty}\mathcal{L}_{p/(p-2),\infty}.$

so by the Hölder-type inequality (2.7), $F_{0}(z)\in\mathcal{L}_{1}$, and it
continuous in the $\mathcal{L}_{1}$-norm. So by Lemma 5.4.4, $F_{0}$ is
$\mathcal{L}_{1}$-valued holomorphic. ∎

######  Lemma 5.4.8.

Let $s\to H(s)$ be a bounded $\mathcal{L}_{\infty}-$valued function measurable
in the weak operator topology (see Definition 2.3.1). Let $g_{z}$ be as in
Theorem 5.2.1. Define

$G(z):=\int_{\mathbb{R}}H(s)\widehat{g}_{z}(s)ds,\quad\Re(z)>1$

as a weak operator topology integral.

  1. (i)

$G$ is an $\mathcal{L}_{\infty}-$valued holomorphic function for the domain
$\Re(z)>1.$

  2. (ii)

if there is $r>1$ such that for all $s\in\mathbb{R}$ we have
$\|H(s)\|_{r,\infty}\leq 1+|s|$, then $G$ is an
$\mathcal{L}_{r,\infty}-$valued holomorphic function for the domain
$\Re(z)>1.$

###### Proof.

Define:

$g_{1,z}=\frac{\partial}{\partial z}g_{z},\quad z\in\mathbb{C},\quad\Re(z)>1.$

Define the $\mathcal{L}_{\infty}$-valued function $G_{1}$ by:

$G_{1}(z)=\int_{\mathbb{R}}H(s)\hat{g}_{1,z}(s)ds.$

We will show that $G$ is $\mathcal{L}_{\infty}$-valued holomorphic by showing
that $G_{1}$ is the derivative of $G$.

Let $z,z_{0}\in\mathbb{C}$ have real part greater than $1$. Then we have

$\frac{G(z)-G(z_{0})}{z-z_{0}}-G_{1}(z_{0})=\int_{\mathbb{R}}H(s)\Big{(}\frac{\hat{g}_{z}(s)-\hat{g}_{z_{0}}(s)}{z-z_{0}}-\hat{g}_{1,z_{0}}(s)\Big{)}ds.$

So by the triangle inequality,

$\Big{\|}\frac{G(z)-G(z_{0})}{z-z_{0}}-G^{\prime}(z_{0})\Big{\|}_{\infty}\leq\operatorname*{ess\,sup}_{s\in\mathbb{R}}\|H(s)\|_{\mathcal{L}_{\infty}}\int_{\mathbb{R}}\Big{|}\frac{\hat{g}_{z}(s)-\hat{g}_{z_{0}}(s)}{z-z_{0}}-\hat{g}_{1,z_{0}}(s)\Big{|}ds.$

By [51, Lemma 7], we have an absolute constant $c_{\mathrm{abs}}$ such that:

$\int_{\mathbb{R}}\Big{|}\frac{\hat{g}_{z}(s)-\hat{g}_{z_{0}}(s)}{z-z_{0}}-\hat{g}_{1,z_{0}}(s)\Big{|}ds\leq
c_{abs}\Big{(}\Big{\|}\frac{g_{z}-g_{z_{0}}}{z-z_{0}}-g_{1,z_{0}}\Big{\|}_{2}+\Big{\|}\frac{g_{z}^{\prime}-g_{z_{0}}^{\prime}}{z-z_{0}}-g_{1,z_{0}}^{\prime}\Big{\|}_{2}\Big{)}.$

The assertion (i) follows now from Theorem 5.3.2, and hence moreover we have
that $G_{1}=G^{\prime}$.

Let us now establish (ii). Indeed, by Lemma 2.3.3, we have

$\displaystyle\|G(z)\|_{r,\infty}$
$\displaystyle\leq\frac{r}{r-1}\int_{\mathbb{R}}(1+|s|)|\hat{g}_{z}(s)|ds,$
$\displaystyle\|G^{\prime}(z)\|_{r,\infty}$
$\displaystyle\leq\frac{r}{r-1}\int_{\mathbb{R}}(1+|s|)|\hat{g}_{1,z}(s)|ds,$
$\displaystyle\Big{\|}\frac{G(z)-G(z_{0})}{z-z_{0}}-G^{\prime}(z_{0})\Big{\|}_{r,\infty}$
$\displaystyle\leq\frac{r}{r-1}\int_{\mathbb{R}}(1+|s|)\Big{|}\frac{\hat{g}_{z}(s)-\hat{g}_{z_{0}}(s)}{z-z_{0}}-\hat{g}_{1,z_{0}}(s)\Big{|}ds.$

Once more using [51, Lemma 7], we have

$\int_{\mathbb{R}}(1+|s|)|\hat{g}_{z}(s)|ds=\|\hat{g}_{z}\|_{1}+\|\hat{g^{\prime}}_{z}\|_{1}\leq
c_{\mathrm{abs}}\|g_{z}\|_{W^{2,2}}.$

Similarly,

$\int_{\mathbb{R}}(1+|s|)|\hat{g}_{1,z}(s)|ds=\|\hat{g}_{1,z}\|_{1}+\|\hat{g^{\prime}}_{1,z}\|_{1}\leq
c_{abs}\|g_{1,z}\|_{W^{2,2}}$

and

$\int_{\mathbb{R}}(1+|s|)\Big{|}\frac{\hat{g}_{z}(s)-\hat{g}_{z_{0}}(s)}{z-z_{0}}-\hat{g}_{1,z_{0}}(s)\Big{|}ds=$
$=\Big{\|}\frac{\hat{g}_{z}-\hat{g}_{z_{0}}}{z-z_{0}}-\hat{g}_{1,z_{0}}\Big{\|}_{1}+\Big{\|}\frac{\hat{g^{\prime}}_{z}-\hat{g^{\prime}}_{z_{0}}}{z-z_{0}}-\hat{g^{\prime}}_{1,z_{0}}\Big{\|}_{1}\leq$
$\leq
c_{abs}\Big{\|}\frac{g_{z}-g_{z_{0}}}{z-z_{0}}-g_{1,z_{0}}\Big{\|}_{W^{2,2}}.$

We now deduce (ii) from Theorem 5.3.2. ∎

######  Lemma 5.4.9.

Let $p>2$ and let $A$ and $B$ satisfy Condition 5.4.1. If
$X\in\mathcal{L}_{\infty},$ then

  1. (i)

the mapping

$G_{1}(z):=\int_{\mathbb{R}}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{-is}XB^{is}\hat{g}_{z}(s)ds,$

is an $\mathcal{L}_{\frac{p}{2},\infty}-$valued holomorphic function for
$\Re(z)>1.$

  2. (ii)

the mapping

$G_{2}(z):=\int_{\mathbb{R}}A^{is}Y^{-is}XB^{is}\hat{g}_{z}(s)ds,$

is $\mathcal{L}_{\infty}-$valued holomorphic for $\Re(z)>1.$

  3. (iii)

the mapping

$G_{3}(z):=\int_{\mathbb{R}}Y^{-is}XB^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]\hat{g}_{z}(s)ds,$

is $\mathcal{L}_{\frac{p}{2},\infty}-$valued holomorphic for $\Re(z)>1.$

###### Proof.

By Lemma 5.4.8 (i), the functions $G_{1},$ $G_{2}$ and $G_{3}$ are
$\mathcal{L}_{\infty}-$valued holomorphic. In particular, this proves (ii). We
now prove the first and third assertions.

Set

$H_{1}(s)=[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{-is}XB^{is},\quad
s\in\mathbb{R},$

and

$H_{2}(s)=Y^{-is}XB^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}],\quad
s\in\mathbb{R}.$

For every $s\in\mathbb{R},$ we have for $j=1,2$,

$\Big{\|}H_{j}(s)\Big{\|}_{\frac{p}{2},\infty}\leq\|X\|_{\infty}\cdot\Big{\|}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]\Big{\|}_{\frac{p}{2},\infty},$

Let $\phi_{s}(t):=|t|^{1+2is},$ $t\in\mathbb{R}.$ We now write

$[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]=[BA^{\frac{1}{2}},\phi_{s}(A^{\frac{1}{2}})].$

We again refer to [50]. Since $p>2$ and $\phi_{s}$ is a Lipschitz function, we
have

$\displaystyle\|[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]\|_{\frac{p}{2},\infty}$
$\displaystyle\leq
c_{p}\|\phi_{s}^{\prime}\|_{\infty}\|[BA,A^{\frac{1}{2}}]\|_{\frac{p}{2},\infty}$
$\displaystyle\leq c_{p}(1+|s|)\|[BA,A^{\frac{1}{2}}]\|_{\frac{p}{2},\infty}.$

Hence for $j=1,2$ we have:

$\|H_{j}(s)\|_{\frac{p}{2},\infty}\leq
c_{p}(1+|s|)\|[BA,A^{\frac{1}{2}}]\|_{\frac{p}{2},\infty}\|X\|_{\infty}.$

The assertions (i) and (iii) now follow from Lemma 5.4.8.(ii) upon taking
$j=1$ for (i) and $j=2$ for (iii). ∎

######  Lemma 5.4.10.

Let $p>2$ and let $A$ and $B$ satisfy Condition 5.4.1. Let $T_{z}(s)$,
$s\in\mathbb{R}$ be defined as in Theorem 5.2.1. Then if $\Re(z)>p-1$ we have:

$\int_{\mathbb{R}}\|T_{z}(s)\|_{1}\cdot|\hat{g}_{z}(s)|ds<\infty.$

###### Proof.

We recall the definition of $T_{z}(s)$:

$T_{z}(s)=B^{z-1+is}[BA^{\frac{1}{2}},A^{z-\frac{1}{2}+is}]Y^{-is}+B^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{z-1-is},\quad
s\in\mathbb{R}.$

Consider the first summand in the definition of $T_{z}(s)$. Using the Leibniz
rule:

$\displaystyle B^{z-1+is}[BA^{\frac{1}{2}},A^{z-\frac{1}{2}+is}]Y^{-is}$
$\displaystyle=B^{z-1+is}[BA^{\frac{1}{2}},A^{z-1}A^{\frac{1}{2}+is}]Y^{-is}$
$\displaystyle=B^{is}B^{z-1}A^{z-1}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{-is}$
$\displaystyle\quad+B^{is}B^{z-1}[BA,A^{z-1}]A^{is}Y^{-is}.$

By the $\mathcal{L}_{1}$-triangle inequality, we have

$\displaystyle\|B^{z-1+is}[BA^{\frac{1}{2}}$
$\displaystyle,A^{z-\frac{1}{2}+is}]Y^{-is}\|_{1}$
$\displaystyle\leq\|B^{z-1}A^{z-1}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]\|_{1}+\|B^{z-1}[BA,A^{z-1}]\|_{1}.$

We now apply the Hölder-type inequality (2.7),

$\displaystyle\|B^{z-1+is}[BA^{\frac{1}{2}}$
$\displaystyle,A^{z-\frac{1}{2}+is}]Y^{-is}\|_{1}$ (5.11)
$\displaystyle\leq\|B^{z-1}A^{z-1}\|_{\frac{p}{p-2},1}\|[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]\|_{\frac{p}{2},\infty}+\|B^{z-1}[BA,A^{z-1}]\|_{1}.$

Consider the function $\phi_{s}(t)=|t|^{1+2is}$, $t\in\mathbb{R}$.
Immediately, $\phi_{s}$ is Lipschitz and
$\|\phi_{s}^{\prime}\|_{L_{\infty}}\leq 2(1+|s|).$ Since $p>2$, we may apply
the result of [50] to obtain:

$\|[BA^{1/2},\phi(A^{1/2})]\|_{p/2,\infty}\leq
C_{p}(1+|s|)\|[BA^{1/2},A^{1/2}]\|_{p/2,\infty}.$

Therefore,

(5.12) $\|[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]\|_{\frac{p}{2},\infty}\leq
C_{p}(1+|s|)\|[BA^{\frac{1}{2}},A^{\frac{1}{2}}]\|_{\frac{p}{2},\infty}.$

Combining (5.11) and (5.12), we have

$\displaystyle\|B^{z-1+is}[BA^{\frac{1}{2}}$
$\displaystyle,A^{z-\frac{1}{2}+is}]Y^{-is}\|_{1}$ (5.13) $\displaystyle\leq
C_{p}(1+|s|)\|[BA^{\frac{1}{2}},A^{\frac{1}{2}}]\|_{\frac{p}{2},\infty}\|B^{z-1}A^{z-1}\|_{\frac{p}{p-2},1}+\|B^{z-1}[BA,A^{z-1}]\|_{1}.$

Let us now consider the second summand in $T_{z}(s).$ Using the Hölder
inequality in the form of (2.7), we obtain

$\displaystyle\|B^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{z-1-is}\|_{1}$
$\displaystyle\leq\|[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{z-1}\|_{1}$ (5.14)
$\displaystyle\leq\|[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]\|_{\frac{p}{2},\infty}\|Y^{z-1}\|_{\frac{p}{p-2},1}.$

Now combining (5.14) with (5.12), we arrive at:

$\displaystyle\|B^{is}[BA^{\frac{1}{2}}$
$\displaystyle,A^{\frac{1}{2}+is}]Y^{z-1-is}\|_{1}$ (5.15) $\displaystyle\leq
C_{p}(1+|s|)\|[BA^{\frac{1}{2}},A^{\frac{1}{2}}]\|_{\frac{p}{2},\infty}\|Y^{z-1}\|_{\frac{p}{p-2},1}.$

Now we may combine (5.13) and (5.15):

$\displaystyle\|T_{z}(s)\|_{1}$ $\displaystyle\leq\|B^{z-1}[BA,A^{z-1}]\|_{1}$
$\displaystyle\quad+c_{abs}(1+|s|)\|[BA^{\frac{1}{2}},A^{\frac{1}{2}}]\|_{\frac{p}{2},\infty}\cdot\Big{(}\|B^{z-1}A^{z-1}\|_{\frac{p}{p-2},1}+\|Y^{z-1}\|_{\frac{p}{p-2},1}\Big{)}.$

By Lemma 5.4.7,

$\|B^{z-1}[BA,A^{z-1}]\|_{1},\|B^{z-1}A^{z-1}\|_{\frac{p}{p-2},1},\|Y^{z-1}\|_{\frac{p}{p-2},1}<\infty,\quad\Re(z)>p-1.$

Hence,

$\|T_{z}(s)\|_{1}\leq C_{A,B,z}\cdot(1+|s|).$

We now have

$\int_{\mathbb{R}}\|T_{z}(s)\|_{1}\cdot|\hat{g}_{z}(s)|ds\leq
C_{A,B,z}\int_{\mathbb{R}}(1+|s|)|\hat{g}_{z}(s)|ds.$

By [51, Lemma 7], we have

$\int_{\mathbb{R}}(1+|s|)|\hat{g}_{z}(s)|ds=\|g_{z}\|_{2}+\|g_{z}^{\prime}\|_{2}\leq
c_{abs}\|g_{z}\|_{W^{2,2}}.$

Thus,

$\int_{\mathbb{R}}\|T_{z}(s)\|_{1}|\hat{g}_{z}(s)|ds\leq
C_{A,B,z}\|g_{z}\|_{W^{2,2}}.$

The assertion follows now from Theorem 5.3.2. ∎

######  Corollary 5.4.11.

If $p>2$ and $A$ and $B$ satisfy condition 5.4.1, then for all $\Re(z)>p-1$ we
have:

$B^{z}A^{z}-(A^{1/2}BA^{1/2})^{z}\in\mathcal{L}_{1}.$

###### Proof.

The integral formula from Lemma 5.2.1 is valid for $\Re(z)>1$. Since $p>2$, we
therefore have a weak operator topology integral:

$B^{z}A^{z}-(A^{1/2}BA^{1/2})^{z}=T_{z}(0)-\int_{\mathbb{R}}T_{z}(s)\widehat{g}_{z}(s)\,ds.$

From Lemma 5.4.10, we have that

$\int_{\mathbb{R}}\|T_{z}(s)\|_{1}\cdot|\hat{g}_{z}(s)|ds<\infty.$

Thus by Lemma 2.3.2, we have that
$\int_{\mathbb{R}}T_{z}(s)\widehat{g}_{z}(s)\in\mathcal{L}_{1}$.

Recalling the definition of $T_{z}(0)$:

$T_{z}(0):=B^{z-1}[BA^{\frac{1}{2}},A^{z-\frac{1}{2}}]+[BA^{\frac{1}{2}},A^{\frac{1}{2}}]Y^{z-1}$

So by (5.11) and (5.14), $T_{z}(0)\in\mathcal{L}_{1}$. Therefore,
$B^{z}A^{z}-(A^{1/2}BA^{1/2})^{z}\in\mathcal{L}_{1}$. ∎

######  Lemma 5.4.12.

Assume that $p>2$ and let $A$ and $B$ satisfy Condition 5.4.1. If
$X\in\mathcal{L}_{\infty},$ then for $\Re(z)>p$:

${\rm
Tr}\Big{(}X\Big{(}B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}\Big{)}\Big{)}={\rm
Tr}(XF_{0}(z))-\sum_{k=1}^{3}{\rm Tr}(G_{k}(z)F_{k}(z))$

Here, the functions $F_{k}$ are as in Lemma 5.4.7 and the functions $G_{k}$
are as in Lemma 5.4.9.

###### Proof.

By Lemma 5.4.10,

$\int_{\mathbb{R}}\|T_{z}(s)\|_{1}|\widehat{g}_{z}(s)|\,ds<\infty.$

So by Lemma 2.3.2, we have:

$\int_{\mathbb{R}}T_{z}(s)\widehat{g}_{z}(s)ds\in\mathcal{L}_{1}$

and

${\rm
Tr}\Big{(}X\int_{\mathbb{R}}T_{z}(s)\widehat{g}_{z}(s)ds\Big{)}=\int_{\mathbb{R}}{\rm
Tr}(XT_{z}(s))\widehat{g}_{z}(s)ds.$

By Theorem 5.2.1, we have:

(5.16) ${\rm
Tr}\Big{(}X\Big{(}B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}\Big{)}\Big{)}={\rm
Tr}(XT_{z}(0))-\int_{\mathbb{R}}{\rm Tr}(XT_{z}(s))\hat{g}_{z}(s)ds$

for $\Re(z)>p.$

Observing that $F_{0}(z)=T_{z}(0)$, we have:

(5.17) ${\rm
Tr}\Big{(}X\Big{(}B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}\Big{)}\Big{)}={\rm
Tr}(X\cdot F_{0}(z))-\int_{\mathbb{R}}{\rm Tr}(XT_{z}(s))\hat{g}_{z}(s)ds.$

By (5.11) and (5.14), we have

$\displaystyle B^{z-1}[BA^{\frac{1}{2}},A^{z-\frac{1}{2}+is}]$
$\displaystyle\in\mathcal{L}_{1},$
$\displaystyle[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{z-1}$
$\displaystyle\in\mathcal{L}_{1}.$

Now by the definition of $T_{z}(s),$ we have

$\displaystyle\mathrm{Tr}(XT_{z}(s))$
$\displaystyle=\mathrm{Tr}(Y^{-is}XB^{is}B^{z-1}[BA^{\frac{1}{2}},A^{z-\frac{1}{2}+is}])$
$\displaystyle\quad+\mathrm{Tr}(Y^{-is}XB^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{z-1}).$

By an application of the Leibniz rule, we have

$B^{z-1}[BA^{\frac{1}{2}},A^{z-\frac{1}{2}+is}]=B^{z-1}A^{z-1}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]+B^{z-1}[BA,A^{z-1}]A^{is}.$

Each of the above terms is $\mathcal{L}_{1}$, by (5.11) and Lemma 5.4.7.(iii)
respectively. Therefore,

$\displaystyle{\rm Tr}(X\cdot T_{z}(s))$ $\displaystyle={\rm
Tr}([BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{-is}XB^{is}B^{z-1}A^{z-1})$
$\displaystyle+{\rm Tr}(A^{is}Y^{-is}XB^{is}B^{z-1}[BA,A^{z-1}])+{\rm
Tr}(Y^{-is}XB^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{z-1}).$

Thus,

$\displaystyle\int_{\mathbb{R}}{\rm Tr}$
$\displaystyle(XT_{z}(s))\hat{g}_{z}(s)ds$ $\displaystyle={\rm
Tr}\Big{(}\int_{\mathbb{R}}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{-is}XB^{is}\hat{g}_{z}(s)ds\cdot
B^{z-1}A^{z-1}\Big{)}$ $\displaystyle\quad+{\rm
Tr}\Big{(}\int_{\mathbb{R}}A^{is}Y^{-is}XB^{is}\hat{g}_{z}(s)ds\cdot
B^{z-1}[BA,A^{z-1}]\Big{)}$ $\displaystyle\quad+{\rm
Tr}\Big{(}\int_{\mathbb{R}}Y^{-is}XB^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]\hat{g}_{z}(s)ds\cdot
Y^{z-1}\Big{)}.$

Using the notations from Lemma 5.4.7 and Lemma 5.4.9, we may summarise the
above equality as:

(5.18) $\int_{\mathbb{R}}{\rm
Tr}(XT_{z}(s))\hat{g}_{z}(s)ds=\sum_{k=1}^{3}{\rm Tr}(G_{k}(z)F_{k}(z)).$

Combining (5.17) and (5.18) completes the proof. ∎

We are now ready to complete the proof of Theorem 5.4.2.

###### Proof of Theorem 5.4.2.

We will show that the function

$A(z):={\rm Tr}(X\cdot F_{0}(z))-\sum_{k=1}^{3}{\rm Tr}(G_{k}(z)F_{k}(z))$

is analytic for $\Re(z)>p-1$.

For the $F_{0}$ term, we use Lemma 5.4.7.(i): the mapping $z\to XF_{0}(z)$ is
$\mathcal{L}_{1}-$valued analytic for $\Re(z)>p-1.$ For the $G_{1}F_{1}$ term,
we use Lemma 5.4.7.(ii) and Lemma 5.4.9.(i) to see that the mapping $z\to
G_{1}(z)F_{1}(z)$ is $\mathcal{L}_{1}-$valued analytic for $\Re(z)>p-1.$ For
the $G_{2}F_{2}$ term, we use Lemma 5.4.7.(iii) and Lemma 5.4.9.(ii) to see
that the mapping $z\to G_{2}(z)F_{2}(z)$ is $\mathcal{L}_{1}-$valued analytic
for $\Re(z)>p-1.$ Finally, for the $G_{3}F_{3}$ term, we use Lemma 5.4.7. (iv)
and Lemma 5.4.9.(iii) to see that mapping $z\to G_{3}(z)F_{3}(z)$ is
$\mathcal{L}_{1}-$valued analytic for $\Re(z)>p-1.$

Hence, $A$ is holomorphic in the set $\Re(z)>p-1$. By Lemma 5.4.12,

$A(z)={\rm
Tr}\Big{(}X\Big{(}B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}\Big{)}\Big{)}$

and so the proof is complete. ∎

###  5.5. Criterion for universal measurability in terms of a $\zeta-$function

In this section we provide a sufficient condition for universal measurability
of operators in $\mathcal{L}_{1,\infty}$.

We recall that a linear functional $\varphi$ on the weak Schatten ideal
$\mathcal{L}_{1,\infty}$ is called a trace if for all unitary operators $U$
and $T\in\mathcal{L}_{1,\infty}$, we have $\varphi(U^{*}TU)=\varphi(T)$.
Equivalently, for all bounded operators $A$ we have $\varphi(AT)=\varphi(TA)$.
We say that $\varphi$ is normalised if

$\varphi\left(\mathrm{diag}\left\\{\frac{1}{n+1}\right\\}_{n\geq 0}\right)=1.$

An operator $T\in\mathcal{L}_{1,\infty}$ is called _universally measurable_ if
all normalised traces take the same value on $T$.

In this section we prove Theorem 1.2.7, which provides a sufficient condition
for operators of the form $AV$, $A\in\mathcal{L}_{\infty}$,
$V\in\mathcal{L}_{1,\infty}$ to be universally measurable. This result is new,
and is sufficiently powerful to allow us to prove Theorem 1.2.5. A similar
characterisation is provided in [63, Theorem 4.13], but the result provided
here is stronger. A previously known characterisation of universal
measurability in terms of a heat trace can be found in [15, Proposition 6].

Let $b$ be a signed Borel measure on $[0,\infty)$. Recall that $b$ can be
written as a difference of two positive measures, $b=b_{+}-b_{-}$ such that
$b_{+}$ and $b_{-}$ are mutually singular to each other. Given a Borel set
$S$, the total variation of $b$ on $S$ is defined to be
$\mathrm{Var}_{S}(b):=b_{+}(S)+b_{-}(S)$.

We will consider measures $b$ which satisfy,

(5.19) $\sup_{x\geq 0}\mathrm{Var}_{[x,x+1]}(b)=c_{b}<\infty.$

Clearly any measure of finite total variation will satisfy this condition, as
will some measures with infinite total variation such as Lebesgue measure and
the measure $d\nu(t)=\sin(t)dt$.

######  Lemma 5.5.1.

Let $b$ be a signed Borel measure satisfying the condition (5.19) and let $f$
be the Laplace transform of $b$, that is,

$f(z):=\int_{0}^{\infty}e^{-tz}\,db(t),\quad\Re(z)>0.$

The function $f$ is analytic on the half-plane $\\{\Re(z)>0\\}.$

###### Proof.

For every $n\geq 0,$ let

$f_{n}(z)=\int_{0}^{n}e^{-tz}db(t),\quad z\in\mathbb{C}.$

Each function $f_{n},$ $n\geq 0,$ is entire. Let $\varepsilon>0.$ For
$\Re(z)>\varepsilon,$ we have

$|f(z)-f_{n}(z)|=|\int_{n}^{\infty}e^{-tz}db(t)|\leq\sum_{k\geq
n}e^{-k\varepsilon}\mathrm{Var}_{[k,k+1]}(b)\leq
c_{b}\frac{e^{-n\varepsilon}}{1-e^{-\varepsilon}}.$

Therefore, $f_{n}\to f$ uniformly on the half-plane
$\\{\Re(z)>\varepsilon\\}.$ Since $\varepsilon>0$ is arbitrary, it follows
that $f_{n}\to f$ uniformly on the compact subsets of the half-plane
$\\{\Re(z)>0\\}.$ Thus, $f$ is analytic on the half-plane $\\{\Re(z)>0\\}.$ ∎

For $x\geq 0$, we denote

$b(x):=b([0,x]).$

We caution the reader that $b(x)$ does not denote $b(\\{x\\})$ nor the Radon-
Nikodym derivative of $b$ at $x$.

The following lemma is very similar to [61, Lemma 2.1.3]. However we require a
slightly different formulation of that result and we were unable to find an
English-language version of its proof, and so we include a self-contained
proof here.

######  Lemma 5.5.2.

Let $b$ be a signed Borel measure on $[0,\infty)$ satisfying (5.19), and and
let $f$ be the Laplace transform of $b.$ For $x\geq 1$, $x\to\infty$, we have

$b(x)=\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}f(\frac{1}{x}+it)e^{(\frac{1}{x}+it)x}dt+O(1),\quad
x\geq 1.$

###### Proof.

Let $x\geq 1$. By definition we have:

$\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}f(\frac{1}{x}+it)e^{(\frac{1}{x}+it)x}dt=\int_{-1}^{1}\int_{0}^{\infty}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}e^{(\frac{1}{x}+it)(x-s)}db(s)dt.$

Examining the integrand, we see that the function

$(s,t)\mapsto\frac{(1-t^{2})^{2}}{x^{-1}+t}e^{(\frac{1}{x}+it)(x-s)}$

is bounded above in absolute value by

$(s,t)\mapsto exe^{-s/x}.$

Since $x\geq 0$ the function $s\to e^{-s/x}$ is in $L_{1}([0,\infty),b)$ we
may interchange the integrals to get:

$\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}f(\frac{1}{x}+it)e^{(\frac{1}{x}+it)x}\,dt=\int_{0}^{\infty}\left(\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}e^{(\frac{1}{x}+it)(x-s)}\,dt\right)\,db(s)$

Now we refer to Proposition A.5.1, where it is proved that:

$\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}e^{(\frac{1}{x}+it)(x-s)}dt=(1+x^{-2})^{2}\chi_{[0,x]}(s)+\min\\{1,(x-s)^{-2}\\}\cdot
O(1).$

Thus,

$\displaystyle\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}f(\frac{1}{x}+it)e^{(\frac{1}{x}+it)x}dt$
$\displaystyle=\int_{0}^{x}(1+x^{-2})^{2}db(s)$
$\displaystyle+\int_{0}^{\infty}\min\\{1,(x-s)^{-2}\\}\cdot O(1)\,db(s).$

We let $h(s)$ be the total variation of $b$ on $[0,s]$. That is,
$h(s):=\mathrm{Var}_{[0,s]}(b)$. Then by the triangle inequality, there is a
positive constant $c_{\mathrm{abs}}$ such that

$\displaystyle\Big{|}\int_{0}^{\infty}\min\\{1,(x-s)^{-2}\\}\cdot O(1)\cdot
db(s)\Big{|}$ $\displaystyle\leq
c_{abs}\cdot\int_{0}^{\infty}\min\\{1,(x-s)^{-2}\\}dh(s)$
$\displaystyle=c_{abs}\cdot\sum_{k\geq
0}\int_{k}^{k+1}\min\\{1,(x-s)^{-2}\\}dh(s)$ $\displaystyle\leq
c_{abs}\cdot\sum_{k\geq
0}\sup_{s\in[k,k+1]}\min\\{1,(x-s)^{-2}\\}\cdot\int_{k}^{k+1}dh(s)$
$\displaystyle\leq c_{abs}\cdot c_{b}\cdot\sum_{k\geq
0}\sup_{s\in[k,k+1]}\min\\{1,(x-s)^{-2}\\}$ $\displaystyle=O(1)$

Thus,

(5.20)
$\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}f(\frac{1}{x}+it)e^{(\frac{1}{x}+it)x}dt=\int_{0}^{x}(1+x^{-2})^{2}db(s)+O(1).$

By the definition of $b(x)$,

$|b(x)-b(0)|\leq\mathrm{Var}_{[0,x]}(b)\leq c_{b}(1+x),\quad x>0.$

Therefore,

$\int_{0}^{x}(1+x^{-2})^{2}db(s)=(1+x^{-2})^{2}\cdot(b(x)-b(0))=b(x)+O(1),\quad
x\geq 1.$

A combination of the latter inequality with (5.20) completes the proof. ∎

The following Lemma is similar in spirit to (but much stronger than) the well-
known Wiener-Ikehara Tauberian theorem [59, Theorem 14.1].

######  Lemma 5.5.3.

Let $b$ be a signed Borel measure on $[0,\infty)$ satisfying (5.19), and let
$f$ be the Laplace transform of $b.$ If there exists $\epsilon>0$ such that
$f$ has analytic continuation to a half-plane $\\{z\;:\;\Re(z)>-\epsilon\\},$
then for $x\geq 1$

$b(x)=O(1),\quad x\to\infty.$

###### Proof.

We write $f(z)=f(0)+zf_{0}(z),$ where $f_{0}$ is an analytic function on the
half-plane $\\{z\;:\;\Re(z)>-\epsilon\\}.$ In particular, since the (closure
of the) set

$\Big{\\{}\frac{1}{x}+it:\ x\geq 1,\ t\in[-1,1]\Big{\\}},$

is a compact subset in $\\{\Re(z)>-\epsilon\\},$ it follows that

(5.21) $\sup_{x\geq 1}\sup_{t\in[-1,1]}|f_{0}(\frac{1}{x}+it)|<\infty.$

The assertion of Lemma 5.5.2 is now written as follows.

$\displaystyle\frac{1}{2\pi}\int_{-1}^{1}(1-t^{2})^{2}f_{0}(\frac{1}{x}+it)$
$\displaystyle
e^{(\frac{1}{x}+it)x}dt+\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}f(0)e^{(\frac{1}{x}+it)x}dt$
(5.22) $\displaystyle=b(x)+O(1),\quad x\to\infty.$

The first summand in the left hand side is bounded for $x\geq 1.$ due to
(5.21). By Proposition A.5.1, we have

(5.23)
$\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{\frac{1}{x}+it}e^{(\frac{1}{x}+it)x}dt=1+O(x^{-2}),\quad
x\to\infty.$

Combining (5.22) and (5.23), we get:

$b(x)+O(1)=f(0)+O(1),\quad x\to\infty.$

So $b(x)=O(1)$ as $x\to\infty$. ∎

In order to continue our discussion of measurability we refer to the concept
of a modulated operator. This theory was introduced in [37] and is developed
extensively in [45, Section 11.2]. If $V\in\mathcal{L}_{1,\infty}$ is
positive, and $T\in\mathcal{L}_{\infty}$, we say that $T$ is $V$-modulated if

$\sup_{t>0}t^{1/2}\|T(1+tV)^{-1}\|_{\mathcal{L}_{2}}<\infty.$

It can be easily seen that if $T$ is $V$-modulated, and
$A\in\mathcal{L}_{\infty}$, then $AT$ is $V$-modulated (this is also [45,
Proposition 11.2.2]). It is proved in [45, Lemma 11.2.8] that $V$ is
$V$-modulated, and therefore that $AV$ is $V$-modulated.

The relevance of the notion of a $V$-modulated operator to measurability comes
from [45, Theorem 11.2.3], which states that if $V\geq 0$ is in
$\mathcal{L}_{1,\infty}$, $\ker(V)=0$, $T$ is $V$-modulated and
$\\{e_{n}\\}_{n\geq 0}$ is an eigenbasis for $V$ ordered so that
$Ve_{n}=\mu(n,V)e_{n}$ for $n\geq 0$, then

  1. (a)

$T\in\mathcal{L}_{1,\infty}$ and $\mathrm{diag}\\{\langle
Te_{n},e_{n}\rangle\\}_{n=0}^{\infty}\in\mathcal{L}_{1,\infty}$

  2. (b)

We have:

$\sum_{k=0}^{n}\lambda(k,T)-\sum_{k=0}^{n}\langle
Te_{k},e_{k}\rangle=O(1),\quad n\to\infty.$

Recall that $\\{\lambda(k,T)\\}_{k=0}^{\infty}$ denotes an eigenvalue sequence
for $T$, ordered with non-increasing absolute value.

######  Lemma 5.5.4.

Let $0\leq V\in\mathcal{L}_{1,\infty}$ satisfy $\ker(V)=0$ and let
$A\in\mathcal{L}_{\infty}.$ Let $\\{e_{k}\\}_{k=0}^{\infty}$ be an eigenbasis
for $V$ ordered such that $Ve_{k}=\mu(k,V)e_{k}$. We have

$\sum_{k=0}^{n}\lambda(k,AV)=\sum_{\mu(k,V)>\frac{\|V\|_{1,\infty}}{n}}\langle
Ae_{k},e_{k}\rangle\mu(k,V)+O(1).$

Here, $e_{k}$ is the eigenvector of $V$ corresponding to the eigenvalue
$\mu(k,V).$

###### Proof.

We have that $AV$ is $V-$modulated. [45, Theorem 11.2.3] now states that as
$n\to\infty$

$\displaystyle\sum_{k=0}^{n}\lambda(k,AV)$
$\displaystyle=\sum_{k=0}^{n}\langle AVe_{k},e_{k}\rangle+O(1)$
$\displaystyle=\sum_{k=0}^{n}\langle Ae_{k},e_{k}\rangle\mu(k,V)+O(1).$

For $n\geq 1$, let

$m(n)=\max\\{k\in\mathbb{N}:\ \mu(k,V)>\frac{\|V\|_{1,\infty}}{n}\\}.$

Using this notation, we write

$\sum_{\mu(k,V)>\frac{\|V\|_{1,\infty}}{n}}\langle
Ae_{k},e_{k}\rangle\mu(k,V)=\sum_{k=0}^{m(n)}\langle
Ae_{k},e_{k}\rangle\mu(k,V).$

We have $\mu(k,V)\leq\frac{\|V\|_{1,\infty}}{k+1}$ for every $k\geq 0$ and,
therefore,

$m(n)\leq\max\Big{\\{}k\in\mathbb{Z}_{+}:\
\frac{\|V\|_{1,\infty}}{k+1}>\frac{\|V\|_{1,\infty}}{n}\Big{\\}}=n-2<n.$

On the other hand, we have $\mu(k,V)\leq\frac{\|V\|_{1,\infty}}{n}$ for all
$k>m(n)$. Thus,

$\displaystyle\Big{|}\sum_{k=m(n)+1}^{n}\langle
Ae_{k},e_{k}\rangle\mu(k,V)\Big{|}$
$\displaystyle\leq\|A\|_{\infty}\sum_{k=m(n)+1}^{n}\mu(k,V)$
$\displaystyle\leq\frac{\|A\|_{\infty}\|V\|_{1,\infty}}{n}\sum_{k=m(n)+1}^{n}1$
$\displaystyle=O(1).$

Finally, we have

$\displaystyle\sum_{k=0}^{n}\lambda(k,AV)$
$\displaystyle=\sum_{k=0}^{n}\langle Ae_{k},e_{k}\rangle\mu(k,V)+O(1)$
$\displaystyle=\sum_{k=0}^{m(n)}\langle Ae_{k},e_{k}\rangle\mu(k,V)+O(1)$
$\displaystyle=\sum_{\mu(k,V)>\frac{\|V\|_{1,\infty}}{n}}\langle
Ae_{k},e_{k}\rangle\mu(k,V)+O(1).$

∎

We now conclude with the proof of Theorem 1.2.7, a sufficient condition for
universal measurability in terms of a $\zeta$-function. Recall that if $f$ is
a meromorphic function of a complex variable $z$ with a simple pole at $z=0$,
then $\mathrm{Res}_{z=0}f(z)$ denotes the coefficient of $z^{-1}$ in the
Laurent expansion of $f$, or equivalently the value of $zf(z)$ at $z=0$.

######  Theorem.

Let $0\leq V\in\mathcal{L}_{1,\infty}$ and let $A\in\mathcal{L}_{\infty}.$
Define the $\zeta$-function:

$\zeta_{A,V}(z):=\mathrm{Tr}(AV^{1+z}),\quad\Re(z)>0.$

If there exists $\varepsilon>0$ such that $\zeta_{A,V}$ admits an analytic
continuation to the set $\\{z\;:\;\Re(z)>-\varepsilon\\}\setminus\\{0\\}$ with
a simple pole at $0$, then for every normalised trace $\varphi$ on
$\mathcal{L}_{1,\infty}$ we have:

$\varphi(AV)=\mathrm{Res}_{z=0}\zeta_{A,V}(z).$

In particular, $AV$ is universally measurable.

###### Proof.

Assume without loss of generality that $\ker(V)=0$. Select an orthonormal
basis $\\{e_{k}\\}_{k=0}^{\infty}$ such that $Ve_{k}=\mu(k,V)e_{k}$.

We define:

$b(t):=-t\mathrm{Res}_{w=0}\zeta_{A,V}(w)+\sum_{\mu(k,V)>e^{-t}}\langle
Ae_{k},e_{k}\rangle\mu(k,V),\quad t\geq 0.$

Since $b$ is a linear combination of monotone functions, it is of locally
bounded variation. Hence there is a signed Borel measure $b$ such that
$b([0,x])=b(x)$, $x\geq 0$.

We first prove that $b$ satisfies (5.19). If $x\geq 0$, then:

$\displaystyle\mathrm{Var}_{[x,x+1]}b$
$\displaystyle\leq|\mathrm{Res}_{w=0}\zeta_{A,V}(w)|+\sum_{\mu(k,V)\in(e^{-1-x},e^{-x}]}|\langle
Ae_{k},e_{k}\rangle|\mu(k,V)$
$\displaystyle\leq|\mathrm{Res}_{w=0}\zeta_{A,V}(w)|+2\|A\|_{\infty}e^{-x}\sum_{\mu(k,V)\in(e^{-1-x},e^{-x}]}1$
$\displaystyle\leq|\mathrm{Res}_{w=0}\zeta_{A,V}(w)|+2\|A\|_{\infty}e^{-x}\sum_{\mu(k,V)\in(e^{-1-x},\infty]}1.$

Since $V\in\mathcal{L}_{1,\infty},$ we have that

$\sum_{\mu(k,V)\in(e^{-1-x},\infty]}1\leq e^{1+x}\|V\|_{1,\infty},\quad
x\in\mathbb{R}.$

Therefore,

$\mathrm{Var}_{[x,x+1]}b\leq|\mathrm{Res}_{w=0}\zeta_{A,V}(w)|+2e\|A\|_{\infty}\|V\|_{1,\infty}.$

So $b$ indeed satisfies (5.19).

Let $\alpha_{k}:=\log(\frac{1}{\mu(k,V)})$ and $b_{k}:=\langle
Ae_{k},e_{k}\rangle\mu(k,V),$ then the function $b$ has a jump discontinuity
at the point $\alpha_{k}$ of magnitude $b_{k}.$ Let $\Re(z)>0$. Using the
identity $e^{-\alpha_{k}z}=\mu(k,V)^{z}$, we have:

$\displaystyle\int_{0}^{\infty}e^{-zt}db(t)$ $\displaystyle=\sum_{k\geq
0}e^{-\alpha_{k}z}\cdot
b_{k}-\mathrm{Res}_{w=0}\zeta_{A,V}(w)\int_{0}^{\infty}e^{-zt}dt$
$\displaystyle=\sum_{k\geq 0}\mu(k,V)^{z}\cdot\langle
Ae_{k},e_{k}\rangle\mu(k,V)-\mathrm{Res}_{w=0}\zeta_{A,V}(w)\int_{0}^{\infty}e^{-zt}dt$
$\displaystyle=\sum_{k\geq 0}\langle
AV^{1+z}e_{k},e_{k}\rangle-\mathrm{Res}_{w=0}\zeta_{A,V}(w)\int_{0}^{\infty}e^{-zt}dt$
$\displaystyle=\mathrm{Tr}(AV^{1+z})-\frac{1}{z}\mathrm{Res}_{w=0}\zeta_{A,V}(w).$

By assumption, the above right hand side has analytic continuation to the set
$\\{z\;:\;\Re(z)>-\varepsilon\\}$.

Thus, since $b$ satisfies (5.19) the assumptions of Lemma 5.5.3 are satisfied,
and we may then conclude that $b(t)=O(1),$ for $t\to\infty$. Thus by the
definition of $b$:,

$\sum_{\mu(k,V)>e^{-t}}\langle
Ae_{k},e_{k}\rangle\mu(k,V)=t\cdot\mathrm{Res}_{w=0}\zeta_{A,V}(w)+O(1),\quad
t\to\infty.$

Setting $e^{-t}=\frac{\|V\|_{1,\infty}}{n},$ we obtain

$\sum_{\mu(k,V)>\frac{\|V\|_{1,\infty}}{n}}\langle
Ae_{k},e_{k}\rangle\mu(k,V)=\log(n)\cdot\mathrm{Res}_{w=0}\zeta_{A,V}(w)+O(1),\quad
t\to\infty.$

By Lemma 5.5.4, we have

$\sum_{k=0}^{n}\lambda(k,AV)=\log(n)\cdot\mathrm{Res}_{w=0}\zeta_{A,V}(w)+O(1),\quad
t\to\infty.$

The assertion follows now from Theorem 2.1.5. ∎

###  5.6. Proof of Theorem 1.2.5, $p>2$

In this section we complete the proof of Theorem 1.2.5 under the restriction
that $p>2$. We require this restriction in order to directly apply the results
of Section 5.4. We will handle the $p=1$ and $p=2$ cases separately in the
next section.

######  Lemma 5.6.1.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1. Let
$0\leq a\in\mathcal{A}.$ If $p>2,$ then the operators $A=a^{2}$ and
$B=(1+D^{2})^{-\frac{1}{2}}$ satisfy Condition 5.4.1.

###### Proof.

Let $D_{0}=F(1+D^{2})^{\frac{1}{2}}.$ By Lemma 3.1.4, the spectral triple
$(\mathcal{A},H,D_{0})$ satisfies Hypothesis 1.2.1. Since $|D_{0}|\geq 1$, we
have that $\||D_{0}|^{-1}\|_{\infty}\leq 1$.

Let us establish Condition 5.4.1.(i). We have

$B^{p}A=|D_{0}|^{-p}a^{2}.$

Since $(\mathcal{A},H,D)$ is $p$-dimensional, we have that
$|D_{0}|^{-p}a\in\mathcal{L}_{1,\infty}$, and so
$B^{p}A\in\mathcal{L}_{1,\infty}$.

Next let us prove Condition 5.4.1.(ii). Let $q>p.$ We have

$\displaystyle B^{q-2}[B,A]$ $\displaystyle=|D_{0}|^{2-q}[|D_{0}|^{-1},a^{2}]$
$\displaystyle=-|D_{0}|^{1-q}\delta_{0}(a^{2})|D_{0}|^{-1}$
$\displaystyle=-|D_{0}|^{1-q}\delta_{0}(a)a|D_{0}|^{-1}-|D_{0}|^{1-q}a\delta_{0}(a)|D_{0}|^{-1}.$

Referring to Lemma 3.1.3, we have

$|D_{0}|^{1-q}\delta_{0}(a)\in\mathcal{L}_{p/(q-1),\infty},\quad|D_{0}|^{1-q}a\in\mathcal{L}_{p/(q-1),\infty},$
$\delta_{0}(a)|D_{0}|^{-1}\in\mathcal{L}_{p,\infty},\quad
a|D_{0}|^{-1}\in\mathcal{L}_{p,\infty}.$

Therefore,

$\displaystyle B^{q-2}[B,A]$
$\displaystyle\in\mathcal{L}_{p/(q-1),\infty}\cdot\mathcal{L}_{p,\infty}+\mathcal{L}_{p/(q-1),\infty}\cdot\mathcal{L}_{p,\infty}$

So by the Hölder inequality, $B^{q-2}[B,A]\in\mathcal{L}_{p/q,\infty}$. Since
$q>p$, it then follows that $B^{q-2}[B,A]\in\mathcal{L}_{1}$.

Now we establish Condition 5.4.1.(iii). We have

$A^{\frac{1}{2}}BA^{\frac{1}{2}}=a|D_{0}|^{-1}a.$

Thus,

$\|A^{\frac{1}{2}}BA^{\frac{1}{2}}\|_{p,\infty}\leq\|a\|_{\infty}\|a|D_{0}|^{-1}\|_{p,\infty}.$

The above right hand side is finite by Lemma 3.1.3.

Finally we prove Condition 5.4.1.(iv). We recall the notation that
$\delta_{0}(a)$ denotes the bounded extension of $[|D_{0}|,a]$. Using (2.2),
we have:

$[B,A^{\frac{1}{2}}]=[|D_{0}|^{-1},a]=-|D_{0}|^{-1}\delta_{0}(a)|D_{0}|^{-1}.$

By Theorem 9 in [62], we have

$|D_{0}|^{-1}\delta_{0}(a)|D_{0}|^{-1}\prec\prec\delta_{0}(a)|D_{0}|^{-2}.$

By Lemma 3.1.3, we have

$\delta_{0}(a)|D_{0}|^{-2}\in\mathcal{L}_{\frac{p}{2},\infty}.$

Since the norm in the space $\mathcal{L}_{\frac{p}{2},\infty}$ is monotone
with respect to the Hardy-Littlewood submajorisation (recall that $p>2$), it
follows that also

$[B,A^{\frac{1}{2}}]=-|D_{0}|^{-1}\delta_{0}(a)|D_{0}|^{-1}\in\mathcal{L}_{\frac{p}{2},\infty}.$

∎

Now we may prove Theorem 1.2.5 for the case $p>2$.

######  Theorem (1.2.5, $p>2$ case).

Assume $p>2$ and let $(\mathcal{A},H,D)$ be a spectral triple satisfying
Hypothesis 1.2.1. If $c\in\mathcal{A}^{\otimes(p+1)}$ is a local Hochschild
cycle, then for every normalised trace $\varphi$ on $\mathcal{L}_{1,\infty}$
we have:

$\varphi(\Omega(c)(1+D^{2})^{-\frac{p}{2}})=\mathrm{Ch}(c).$

###### Proof.

By Theorem 1.2.3, the function

$\zeta_{c,D}(z):=\mathrm{Tr}(\Omega(c)(1+D^{2})^{-z/2}),\quad\Re(z)>p$

admits an analytic continuation to the set
$\\{z\;:\;\Re(z)>p-1\\}\setminus\\{p\\}$, and $p$ is a simple pole for
$\zeta_{c,D}$ with residue $p\mathrm{Ch}(c)$.

Let $c=\sum_{j=1}^{m}a^{j}_{0}\otimes a^{j}_{1}\otimes\cdots\otimes
a^{j}_{p}$. We assume that $c$ is local, i.e. that there exists $0\leq
a\in\mathcal{A}$ such that for all $j$ we have $aa^{j}_{0}=a^{j}_{0}$.
Equivalently, $(1-a)a^{j}_{0}=0$. So
$\mathrm{im}(a^{j}_{0})\subseteq\ker(1-a)$. Since the support projection
$\mathrm{supp}(1-a)$ is the projection onto the orthogonal complement of the
kernel, we have:

$\mathrm{supp}(1-a)a^{j}_{0}=0.$

By functional calculus, $\mathrm{supp}(1-a)=1-\chi_{\\{1\\}}(a)$, so moreover
we have that $\chi_{\\{1\\}}(a)a^{j}_{0}=a^{j}_{0}$. Therefore, for all
$z\in\mathbb{C}$ with $\Re(z)>0$:

$a^{2z}a_{0}^{j}=a^{2z}\chi_{\\{1\\}}(a)a_{0}^{j}=1^{2z}a_{0}^{j}=a_{0}^{j}.$

Recall that $\Omega(c)=\sum_{j=0}^{m}\Gamma
a_{0}^{j}\partial(a_{1}^{j})\cdots\partial(a_{p}^{j})$. Since $a$ commutes
with $\Gamma$, we have for all $\Re(z)>0$,

(5.24) $a^{2z}\Omega(c)=\Omega(c).$

Let $A=a^{2}$ and $B=(1+D^{2})^{-\frac{1}{2}}$, as in Lemma 5.6.1. Then
$B^{z}A^{z}=(1+D^{2})^{-z/2}a^{2z}$, and hence

$\displaystyle\mathrm{Tr}(\Omega(c)B^{z}A^{z})$
$\displaystyle=\mathrm{Tr}(\Omega(c)(1+D^{2})^{-\frac{z}{2}}a^{2z})$
$\displaystyle=\mathrm{Tr}(a^{2z}\Omega(c)(1+D^{2})^{-\frac{z}{2}})$
$\displaystyle=\mathrm{Tr}(\Omega(c)(1+D^{2})^{-\frac{z}{2}}).$

By Theorem 1.2.3, it then follows that
$z\mapsto\mathrm{Tr}(\Omega(c)B^{z}A^{z})$ admits an analytic continuation to
the set $\\{\Re(z)>p-1\\}\setminus\\{p\\}$ with a simple pole at $z=p$ and
residue $p\mathrm{Ch}(c)$.

Condition 5.4.1 holds for $A$ and $B$ by Lemma 5.6.1. Hence we may apply
Theorem 5.4.2 to conclude that

$z\to{\rm
Tr}\Big{(}\Omega(c)\Big{(}B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}\Big{)}\Big{)}$

admits an analytic continuation to the set $\\{\Re(z)>p-1\\}.$

By Lemma 5.6.1, $A^{\frac{1}{2}}BA^{\frac{1}{2}}\in\mathcal{L}_{p,\infty}.$
Hence, the function (defined a priori for $\Re(z)>p$)

$z\to{\rm Tr}(\Omega(c)(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z})$

admits an analytic continuation to the set $\\{\Re(z)>p-1\\}\setminus\\{p\\}$,
with $z=p$ being a simple pole with residue $p\mathrm{Ch}(c).$ Consider
$V=(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{p}\in\mathcal{L}_{1,\infty}.$ It has
just been demonstrated that

$z\to{\rm Tr}(\Omega(c)V^{z})$

admits an analytic continuation to the set
$\\{\Re(z)>1-\frac{1}{p}\\}\setminus\\{1\\},$ with a simple pole at $z=1$ and
the corresponding residue being $\mathrm{Ch}(c).$

By Theorem 1.2.7, we therefore have

$\varphi(\Omega(c)V)=\mathrm{Ch}(c)$

for every normalised trace $\varphi$ on $\mathcal{L}_{1,\infty}.$

By Corollary 5.4.11 we have:

$V-B^{p}A^{p}=(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{p}-B^{p}A^{p}\in\mathcal{L}_{1}.$

Since $\varphi$ vanishes on $\mathcal{L}_{1},$ it follows that

$\varphi(\Omega(c)B^{p}A^{p})=\varphi(\Omega(c)V)-\varphi(\Omega(c)(V-B^{p}A^{p}))=\varphi(\Omega(c)V)$

for every normalised trace $\varphi$ on $\mathcal{L}_{1,\infty}.$ Now using
(5.24) with $z=p$:

$\varphi(\Omega(c)(1+D^{2})^{-\frac{p}{2}})=\varphi(a^{2p}\Omega(c)(1+D^{2})^{-\frac{p}{2}})=\varphi(\Omega(c)B^{p}A^{p})=\mathrm{Ch}(c)$

for every normalised trace $\varphi$ on $\mathcal{L}_{1,\infty}.$ ∎

###  5.7. Proof of Theorem 1.2.5, $p=1,2$

In this final section we complete the proof of Theorem 1.2.5 by dealing with
the remaining cases of $p=1$ and $p=2$. We require adjustment for these cases
since Theorem 5.4.2 is inapplicable for $p\leq 2$.

######  Lemma 5.7.1.

Let $(\mathcal{A},H,D)$ satisfy Hypothesis 1.2.1. Suppose that spectrum of the
operator $D$ does not intersect the interval $(-1,1).$ Then for all
$x\in\mathrm{dom}(\delta)$ we have an absolute constant $c_{\mathrm{abs}}$
such that

$\|[|D|^{\frac{1}{3}},x]\|_{1}\leq c_{\mathrm{abs}}\|[|D|,x]\|_{1},$

and for all $r\in(1,\infty)$ a constant $c_{r}>0$ such that

$\|[|D|^{\frac{1}{3}},x]\|_{r,\infty}\leq c_{r}\|[|D|,x]\|_{r,\infty}.$

These inequalities are understood to be trivially true if the right hand side
is infinite.

###### Proof.

We only prove the first assertion. One can prove the second inequality by an
identical argument, with the $\mathcal{L}_{r,\infty}$ quasi-norm in place of
the $\mathcal{L}_{1}$ norm.

Let $f$ be a smooth function on $\mathbb{R}$ such that
$f(t)=|t|^{\frac{1}{3}}$ for $|t|>1.$ For $\varepsilon>0$, set
$f_{\varepsilon}(t)=f(t)e^{-\varepsilon^{2}t^{2}},$ $t\in\mathbb{R}.$ Then,

$\displaystyle f_{\varepsilon}^{\prime}(t)$
$\displaystyle=(f^{\prime}(t)-2\varepsilon^{2}tf(t))e^{-\varepsilon^{2}t^{2}},$
$\displaystyle f_{\varepsilon}^{\prime\prime}(t)$
$\displaystyle=(f^{\prime\prime}(t)-4\varepsilon^{2}tf^{\prime}(t)+(4t\varepsilon^{4}-2\varepsilon^{2})f(t))e^{-\varepsilon^{2}t^{2}}$

Since for $|t|>1$ we have $f^{\prime}(t)=\frac{1}{3}|t|^{-2/3}$, we have that
as $\varepsilon\to 0$ the $L_{2}$-norm
$\|f_{\varepsilon}^{\prime}\|_{L_{2}(\mathbb{R})}$ is uniformly bounded.
Similarly, since for $|t|>1$, $f^{\prime\prime}(t)=-\frac{2}{9}|t|^{-5/3}$, we
also have that $\|f_{\varepsilon}^{\prime\prime}(t)\|_{L_{2}(\mathbb{R})}$ is
uniformly bounded.

We have that (see e.g. [51, Lemma 7]):

$\|\widehat{f_{\epsilon}^{\prime}}\|_{1}\leq
c_{abs}\Big{(}\|f_{\epsilon}^{\prime}\|_{2}+\|f_{\epsilon}^{\prime\prime}\|_{2}\Big{)}.$

and so if $\varepsilon\in(0,1)$, $\|\widehat{f_{\varepsilon}^{\prime}}\|_{1}$
is uniformly bounded.

By Lemma A.2.2 (taken with $s=1$), we have the identity:

$[f_{\epsilon}(|D|),x]=\int_{-\infty}^{\infty}\Big{(}\int_{0}^{1}\hat{f_{\epsilon}^{\prime}}(u)e^{iu(1-v)|D|}\delta(x)e^{iuv|D|}dv\Big{)}du.$

So taking the $\mathcal{L}_{1}$-norm, we conclude from Lemma 2.3.3 that

$\displaystyle\Big{\|}[f_{\epsilon}(|D|),x]\Big{\|}_{1}$
$\displaystyle\leq\|\hat{f_{\epsilon}^{\prime}}\|_{1}\|\delta(x)\|_{1}$
$\displaystyle\leq c_{\mathrm{abs}}\|\delta(x)\|_{1}.$

Fix $N>0.$ We have

$\Big{\|}\chi_{[0,N]}(|D|)[f_{\epsilon}(|D|),x]\chi_{[0,N]}(|D|)\Big{\|}_{r,\infty}\leq\frac{c_{abs}r}{r-1}\|\delta(x)\|_{r,\infty}.$

Since as $\varepsilon\to 0$, we have that $f_{\varepsilon}$ converges
uniformly to $f$ on the set $[0,N]$, we have that: As $\epsilon\to 0,$ we have

$\chi_{[0,N]}(|D|)[f_{\epsilon}(|D|),x]\chi_{[0,N]}(|D|)\to\chi_{[0,N]}(|D|)[|D|^{\frac{1}{3}},x]\chi_{[0,N]}(|D|)$

in the operator norm. By the Fatou property of the $\mathcal{L}_{1}$-norm:

$\Big{\|}\chi_{[0,N]}(|D|)\cdot[|D|^{\frac{1}{3}},x]\cdot\chi_{[0,N]}(|D|)\Big{\|}_{1}\leq
c_{\mathrm{abs}}\|\delta(x)\|_{1}.$

Since the above inequality is true for arbitrary $N>0,$ we may take the limit
$N\to\infty$ and again using the Fatou property of the $\mathcal{L}_{1}$ norm,
we arrive at

$\Big{\|}[|D|^{\frac{1}{3}},x]\Big{\|}_{1}\leq
c_{\mathrm{abs}}\|\delta(x)\|_{1}.$

∎

As a replacement for Lemma 5.6.1 in the $p=1$ case we use the following:

######  Lemma 5.7.2.

Let $(\mathcal{A},H,D)$ be a $1-$dimensional spectral triple satisfying
Hypothesis 1.2.1. If $0\leq a\in\mathcal{A},$ then the operators $A=a^{4}$ and
$B=(1+D^{2})^{-\frac{1}{6}}$ satisfy Condition 5.4.1 with $p=3.$

###### Proof.

This proof is similar to that of Lemma 5.6.1.

Let $D_{0}=F(1+D^{2})^{\frac{1}{2}}.$ By Lemma 3.1.4, the $1-$dimensional
spectral triple $(\mathcal{A},H,D_{0})$ satisfies Hypothesis 1.2.1. Since
$|D_{0}|\geq 1$, we have that $\||D_{0}|^{-1}\|_{\infty}\leq 1$.

Let us establish Condition 5.4.1.(i). We have

$B^{p}A=|D_{0}|^{-1}a^{4}\in\mathcal{L}_{1,\infty}$

since by assumption $(\mathcal{A},H,D)$ is $1$-dimensional.

Next we establish Condition 5.4.1.(ii). Let $q\in(3,4).$ Using (2.2), we have
on $H_{\infty}$:

$\displaystyle B^{q-2}[B,A]$
$\displaystyle=|D_{0}|^{\frac{2-q}{3}}[|D_{0}|^{-\frac{1}{3}},a^{4}]$
$\displaystyle=-|D_{0}|^{\frac{1-q}{3}}[|D_{0}|^{\frac{1}{3}},a^{4}]|D_{0}|^{-\frac{1}{3}}$
$\displaystyle=-[|D_{0}|^{\frac{1}{3}},|D_{0}|^{\frac{1-q}{3}}a^{4}|D_{0}|^{-\frac{1}{3}}].$

By Lemma 5.7.1, we have

$\|B^{q-2}[B,A]\|_{1}\leq
c_{abs}\|[|D_{0}|,|D_{0}|^{\frac{1-q}{3}}a^{4}|D_{0}|^{-\frac{1}{3}}]\|_{1}.$

Still working on $H_{\infty}$, we also have:

$\displaystyle[|D_{0}|,|D_{0}|^{\frac{1-q}{3}}a^{4}|D_{0}|^{-\frac{1}{3}}]$
$\displaystyle=|D_{0}|^{\frac{1-q}{3}}\delta_{0}(a^{4})|D_{0}|^{-\frac{1}{3}}$
$\displaystyle=|D_{0}|^{\frac{1-q}{3}}\delta_{0}(a^{2})a^{2}|D_{0}|^{-\frac{1}{3}}+|D_{0}|^{\frac{1-q}{3}}a^{2}\delta_{0}(a^{2})|D_{0}|^{-\frac{1}{3}}.$

Applying Lemma 3.1.3, we have

$[|D_{0}|,|D_{0}|^{\frac{1-q}{3}}a^{4}|D_{0}|^{-\frac{1}{3}}]\in\mathcal{L}_{\frac{3}{q-1},\infty}\cdot\mathcal{L}_{3,\infty}\subset\mathcal{L}_{\frac{3}{q},\infty}$

by the Holder inequality, since $q>3$,
$\mathcal{L}_{3/q,\infty}\subset\mathcal{L}_{1}$, and so
$B^{q-2}[B,A]\in\mathcal{L}_{1}$.

Now we establish Condition 5.4.1.(iii). We may compute:

$A^{\frac{1}{2}}BA^{\frac{1}{2}}=a^{2}|D_{0}|^{-\frac{1}{3}}a^{2}.$

Thus,

$\|A^{\frac{1}{2}}BA^{\frac{1}{2}}\|_{3,\infty}\leq\|a\|_{\infty}^{3}\|a|D_{0}|^{-\frac{1}{3}}\|_{3,\infty}.$

The right hand side is finite by Lemma 3.1.3.

Finally we verify Condition 5.4.1.(iv). We have:

$\displaystyle[B,A^{\frac{1}{2}}]$
$\displaystyle=[|D_{0}|^{-\frac{1}{3}},a^{2}]$
$\displaystyle=-|D_{0}|^{-\frac{1}{3}}[|D_{0}|^{\frac{1}{3}},a^{2}]|D_{0}|^{-\frac{1}{3}}$
$\displaystyle=-[|D_{0}|^{\frac{1}{3}},|D_{0}|^{-\frac{1}{3}}a^{2}|D_{0}|^{-\frac{1}{3}}].$

By Lemma 5.7.1, we have

$\|[B,A^{\frac{1}{2}}]\|_{\frac{3}{2},\infty}\leq
c_{\mathrm{abs}}\|[|D_{0}|,|D_{0}|^{-\frac{1}{3}}a^{2}|D_{0}|^{-\frac{1}{3}}]\|_{\frac{3}{2},\infty}.$

Moreover, by the Leibniz rule

$\displaystyle[|D_{0}|,|D_{0}|^{-\frac{1}{3}}a^{2}|D|^{-\frac{1}{3}}]$
$\displaystyle=|D_{0}|^{-\frac{1}{3}}\delta_{0}(a^{2})|D_{0}|^{-\frac{1}{3}}$
$\displaystyle=|D_{0}|^{-\frac{1}{3}}\delta_{0}(a)a|D_{0}|^{-\frac{1}{3}}+|D_{0}|^{-\frac{1}{3}}a\cdot\delta_{0}(a)|D_{0}|^{-\frac{1}{3}}.$

By Lemma 3.1.3 and the Hölder inequality, we have

$[|D_{0}|,|D_{0}|^{-\frac{1}{3}}a^{2}|D_{0}|^{-\frac{1}{3}}]\in\mathcal{L}_{3,\infty}\cdot\mathcal{L}_{3,\infty}\subset\mathcal{L}_{\frac{3}{2},\infty}.$

∎

For the case $p=2$, we instead use:

######  Lemma 5.7.3.

Let $(\mathcal{A},H,D)$ be a $2-$dimensional spectral triple satisfying
Hypothesis 1.2.1. If $0\leq a\in\mathcal{A},$ then the operators $A=a^{4}$ and
$B=(1+D^{2})^{-\frac{1}{6}}$ satisfy Condition 5.4.1 with $p=6.$

###### Proof.

This proof is similar to those of Lemmas 5.6.1 and 5.7.2.

Let $D_{0}=F(1+D^{2})^{\frac{1}{2}}.$ By Lemma 3.1.4, the $2-$dimensional
spectral triple $(\mathcal{A},H,D_{0})$ satisfies Hypothesis 1.2.1. By
rescaling $D$ if necessary, we may assume without loss of generality that the
spectrum of $D_{0}$ does not intersect the interval $(-1,1)$.

Let us establish Condition 5.4.1.(i). We have

$B^{p}A=|D_{0}|^{-2}a^{4}\in\mathcal{L}_{1,\infty}$

by Hypothesis 1.2.1.

Next we establish Condition 5.4.1.(ii). Let $q\in(6,7).$ We have

$\displaystyle B^{q-2}[B,A]$
$\displaystyle=|D_{0}|^{\frac{2-q}{3}}[|D_{0}|^{-\frac{1}{3}},a^{4}]$
$\displaystyle=-|D_{0}|^{\frac{1-q}{3}}[|D_{0}|^{\frac{1}{3}},a^{4}]|D_{0}|^{-\frac{1}{3}}$
$\displaystyle=-[|D_{0}|^{\frac{1}{3}},|D_{0}|^{\frac{1-q}{3}}a^{4}|D_{0}|^{-\frac{1}{3}}].$

By Lemma 5.7.1, we have

$\|B^{q-2}[B,A]\|_{1}\leq
c_{abs}\|[|D_{0}|,|D_{0}|^{\frac{1-q}{3}}a^{4}|D_{0}|^{-\frac{1}{3}}]\|_{1}.$

However,

$\displaystyle[|D_{0}|,|D_{0}|^{\frac{1-q}{3}}a^{4}|D_{0}|^{-\frac{1}{3}}]$
$\displaystyle=|D_{0}|^{\frac{1-q}{3}}\delta_{0}(a^{4})|D_{0}|^{-\frac{1}{3}}$
$\displaystyle=|D_{0}|^{\frac{1-q}{3}}\delta_{0}(a^{2})a^{2}|D_{0}|^{-\frac{1}{3}}+|D_{0}|^{\frac{1-q}{3}}a^{2}\delta_{0}(a^{2})|D_{0}|^{-\frac{1}{3}}.$

By Lemma 3.1.3, we have by the Hölder inequality:

$[|D_{0}|,|D_{0}|^{\frac{1-q}{3}}a^{4}|D|^{-\frac{1}{3}}]\in\mathcal{L}_{\frac{6}{q-1},\infty}\cdot\mathcal{L}_{6,\infty}\subset\mathcal{L}_{\frac{6}{q},\infty}$

Since $q>6$, we have that $\mathcal{L}_{6/q,\infty}\subset\mathcal{L}_{1}$,
and so $B^{q-2}[B,A]\in\mathcal{L}_{1}$.

Now we prove Condition 5.4.1.(iii). We have

$A^{\frac{1}{2}}BA^{\frac{1}{2}}=a^{2}|D_{0}|^{-\frac{1}{3}}a^{2}.$

Thus,

$\|A^{\frac{1}{2}}BA^{\frac{1}{2}}\|_{6,\infty}\leq\|a\|_{\infty}^{3}\|a|D_{0}|^{-\frac{1}{3}}\|_{6,\infty}.$

The above right hand side is finite by Lemma 3.1.3.

Finally, let us establish Condition 5.4.1.(iv). We may compute on
$H_{\infty}$:

$\displaystyle[B,A^{\frac{1}{2}}]$
$\displaystyle=[|D_{0}|^{-\frac{1}{3}},a^{2}]$
$\displaystyle=-|D_{0}|^{-\frac{1}{3}}[|D_{0}|^{\frac{1}{3}},a^{2}]|D_{0}|^{-\frac{1}{3}}$
$\displaystyle=-[|D_{0}|^{\frac{1}{3}},|D_{0}|^{-\frac{1}{3}}a^{2}|D_{0}|^{-\frac{1}{3}}].$

Therefore using Lemma 5.7.1, we have

$\|[B,A^{\frac{1}{2}}]\|_{3,\infty}\leq
c_{abs}\|[|D_{0}|,|D_{0}|^{-\frac{1}{3}}a^{2}|D_{0}|^{-\frac{1}{3}}]\|_{3,\infty}.$

Applying the Leibniz rule,

$\displaystyle[|D_{0}|,|D_{0}|^{-\frac{1}{3}}a^{2}|D_{0}|^{-\frac{1}{3}}]$
$\displaystyle=|D_{0}|^{-\frac{1}{3}}\delta_{0}(a^{2})|D_{0}|^{-\frac{1}{3}}$
$\displaystyle=|D_{0}|^{-\frac{1}{3}}\delta_{0}(a)a|D_{0}|^{-\frac{1}{3}}+|D_{0}|^{-\frac{1}{3}}a\delta_{0}(a)|D_{0}|^{-\frac{1}{3}}.$

By Lemma 3.1.3, we then have from the Hölder inequality:

$[D_{0},|D_{0}|^{-\frac{1}{3}}a^{2}|D_{0}|^{-\frac{1}{3}}]\in\mathcal{L}_{6,\infty}\cdot\mathcal{L}_{6,\infty}\subset\mathcal{L}_{3,\infty}.$

∎

We may now at last complete the proof of Theorem 1.2.5.

######  Theorem.

Assume $p=1$ or $p=2$ and let $(\mathcal{A},H,D)$ be a spectral triple
satisfying Hypothesis 1.2.1. If $c\in\mathcal{A}^{\otimes(p+1)}$ is a local
Hochschild cycle, then for every normalised trace $\varphi$ on
$\mathcal{L}_{1,\infty}$ we have:

$\varphi(\Omega(c)(1+D^{2})^{-\frac{p}{2}})=\mathrm{Ch}(c).$

###### Proof.

By Theorem 1.2.3, the function

$\zeta_{c,D}(z)=\mathrm{Tr}(\Omega(c)(1+D^{2})^{-z/2}),\quad\Re(z)>p$

admits an analytic continuation to the set
$\\{z\;:\;\Re(z)>p-1\\}\setminus\\{p\\}$, and the point $p$ is a simple pole
with corresponding residue $p\mathrm{Ch}(c)$.

Let $c=\sum_{j=1}^{m}a_{0}^{j}\otimes\cdots\otimes a_{p}^{j}$. Since $c$ is
local, we may choose $0\leq a\in\mathcal{A}$ such that $aa_{0}^{j}=a_{0}^{j}$
for all $j$.

By exactly the same argument as in the $p>2$ case, we can show that for all
$\Re(z)>0$:

(5.25) $a^{4z}\Omega(c)=\Omega(c).$

We let $A=a^{4}$ and $B=(1+D^{2})^{-\frac{1}{6}}$ as in Lemmas 5.7.2 and
5.7.3.

We have that:

$\displaystyle\mathrm{Tr}(\Omega(c)B^{z}A^{z})$
$\displaystyle=\mathrm{Tr}(a^{4z}\Omega(c)(1+D^{2})^{-\frac{z}{6}})$
$\displaystyle=\mathrm{Tr}(\Omega(c)(1+D^{2})^{-\frac{z}{6}}),\quad\Re(z)>3p.$

We recognise the above function as being precisely $z\mapsto\zeta_{c,D}(z/3)$.
Hence by Theorem 1.2.3, the function
$z\mapsto\mathrm{Tr}(\Omega(c)B^{z}A^{z})$ admits an analytic continuation to
the set $\\{z\;:\;\Re(z)>3(p-1)\\}\setminus\\{3p\\}$, with a simple pole at
$3p$ with corresponding residue $3p\mathrm{Ch}(c)$.

Assume now that $p=1.$ By Lemma 5.7.2, Condition 5.4.1 holds for $A$ and $B$
(with $p=3$). By Theorem 5.4.2 the function

$z\to{\rm
Tr}\Big{(}\Omega(c)\Big{(}B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}\Big{)}\Big{)}$

admits an analytic continuation to the set $\\{\Re(z)>2\\}.$

By Lemma 5.7.2, $A^{\frac{1}{2}}BA^{\frac{1}{2}}\in\mathcal{L}_{3,\infty}.$
Hence, the function (defined a priori for $\Re(z)>3$)

$z\to{\rm Tr}(\Omega(c)(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z})$

admits an analytic continuation to the set $\\{\Re(z)>2\\}\setminus\\{3\\}.$
with a pole at $z=3$ and corresponding residue $3\mathrm{Ch}(c).$ Define
$V_{1}:=(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{3}\in\mathcal{L}_{1,\infty}.$ Then,

$\mathrm{Tr}(\Omega(c)(A^{1/2}BA^{1/2})^{z})=\mathrm{Tr}(\Omega(c)V_{1}^{z/3}).$

We now know that function

$z\mapsto{\rm Tr}(\Omega(c)V_{1}^{z/3})$

admits an analytic continuation to the set $\\{\Re(z)>2\\}\setminus\\{3\\}$
with a simple pole at $3$ and corresponding residue $3\mathrm{Ch}(c)$. So by
rescaling the argument, we can equivalently say that the function

$z\mapsto\mathrm{Tr}(\Omega(c)V_{1}^{z})$

has analytic continuation to the set $\\{z\;:\;\Re(z)>2/3\\}\setminus\\{1\\}$
with a simple pole at $1$ with corresponding residue $\mathrm{Ch}(c)$.

Thus by Theorem 1.2.7, for any continuous normalised trace $\varphi$ on
$\mathcal{L}_{1,\infty}$ we have

$\varphi(\Omega(c)V_{1})=\mathrm{Ch}(c).$

Due to Lemma 5.7.2, we have that $V_{1}-B^{3}A^{3}\in\mathcal{L}_{1}$, and
since $\varphi$ vanishes on $\mathcal{L}_{1}$ it follows that

$\varphi(\Omega(c)B^{3}A^{3})=\mathrm{Ch}(c).$

So

$\varphi(a^{12}\Omega(c)(1+D^{2})^{-1/2})=\mathrm{Ch}(c).$

By taking $z=3$ in (5.25), we have that $a^{12}\Omega(c)=\Omega(c)$, this
completes the proof in the case $p=1$.

Now assume that $p=2$. By Lemma 5.7.3, Condition 5.4.1 holds for $A$ and $B$
(with $p=6$). By Theorem 5.4.2 the function

$z\to{\rm
Tr}\Big{(}\Omega(c)\Big{(}B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}\Big{)}\Big{)}$

admits an analytic extension to the set $\\{\Re(z)>5\\}.$

By Lemma 5.7.3, $A^{\frac{1}{2}}BA^{\frac{1}{2}}\in\mathcal{L}_{6,\infty}.$
Hence, the function (defined a priori for $\Re(z)>6$)

$z\to{\rm Tr}(\Omega(c)(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z})$

admits an analytic extension to the set $\\{\Re(z)>5\\}\setminus\\{6\\}.$ The
point $z=6$ is a simple pole with corresponding residue $6\mathrm{Ch}(c).$
Consider
$V_{2}=(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{6}\in\mathcal{L}_{1,\infty}.$ We
have so far shown that the function

$z\to{\rm Tr}(\Omega(c)V_{2}^{z})$

admits an analytic extension to the set
$\\{z\;:\;\Re(z)>\frac{5}{6}\\}\setminus\\{1\\}.$ The point $z=1$ is a simple
pole with corresponding residue $\mathrm{Ch}(c)$.

Hence, by Theorem 1.2.7, for any continuous normalised trace $\varphi$ on
$\mathcal{L}_{1,\infty}$, we have

$\varphi(\Omega(c)V_{2})=\mathrm{Ch}(c).$

By Lemma 5.7.3, the operator $V_{2}-B^{6}A^{6}$ is trace class. Thus,

$\varphi(\Omega(c)B^{6}A^{6})=\mathrm{Ch}(c).$

So $\varphi(a^{12}\Omega(c)(1+D^{2})^{-1})=\mathrm{Ch}(c)$. Since
$a^{12}\Omega(c)=\Omega(c)$, this completes the proof for the case $p=2$. ∎

##  Appendix A Appendix

###  A.1. Properties of the algebra $\mathcal{B}$

For this section, $(\mathcal{A},H,D)$ is a smooth spectral triple. Recall from
Definition 2.2.7 that $\mathcal{B}$ is the $*-$algebra generated by all
elements of the form $\delta^{k}(a)$ or $\partial(\delta^{k}(a)),$ $k\geq 0,$
$a\in\mathcal{A}.$ Recall that we define $H_{\infty}:=\bigcap_{k\geq
1}\mathrm{dom}(D^{k})$, and that for all $T\in\mathcal{B}$, we have
$T:H_{\infty}\to H_{\infty}$, and for all $k\geq 0$ we have
$D^{k},|D|^{k}:H_{\infty}\to H_{\infty}$.

The following should be compared with [13, Lemma 6.2]. See also the discussion
following [33, Lemma 10.22].

######  Lemma A.1.1.

For every $x\in\mathcal{B}$ and for every $m\geq 0,$ we have the following
equalities of linear (potentially unbounded) operators on $H_{\infty}$:

$\displaystyle|D|^{m}x$
$\displaystyle=\sum_{k=0}^{m}\binom{m}{k}\delta^{m-k}(x)|D|^{k}\text{ and, }$
$\displaystyle x|D|^{m}$
$\displaystyle=\sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}|D|^{k}\delta^{m-k}(x).$

###### Proof.

We prove only the first equality as the proof of the second one follows by an
identical argument.

This formula can be seen by induction on $m.$ Indeed, for $m=1$, this is
simply the claim that since
$\mathcal{B}\subseteq\mathrm{dom}_{\infty}(\delta)$ we have an equality of
operators on $H_{\infty}$:

$|D|x=\delta(x)+x|D|.$

Now suppose that the claim is true for $m-1$. Then on $H_{\infty}$ we have

$\displaystyle|D|^{m}x$ $\displaystyle=|D|\cdot|D|^{m-1}x$
$\displaystyle=|D|\cdot\sum_{k=0}^{m-1}\binom{m-1}{k}\delta^{m-1-k}(x)|D|^{k}$
$\displaystyle=\sum_{k=0}^{m-1}\binom{m-1}{k}|D|\delta^{m-1-k}(x)\cdot|D|^{k}$
$\displaystyle=\sum_{k=0}^{m-1}\binom{m-1}{k}\delta^{m-1-k}(x)|D|^{k+1}+\sum_{k=0}^{m-1}\binom{m-1}{k}\delta^{m-k}(x)|D|^{k}$
$\displaystyle=\sum_{k=1}^{m}\binom{m-1}{k-1}\delta^{m-k}(x)|D|^{k}+\sum_{k=0}^{m-1}\binom{m-1}{k}\delta^{m-k}(x)|D|^{k}$
$\displaystyle=\sum_{k=0}^{m}\binom{m}{k}\delta^{m-k}(x)|D|^{k}.$

and so the statement follows for $m$. ∎

######  Lemma A.1.2.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple, and assume that $D$ has a
spectral gap at $0.$ Then for all $x\in\mathcal{B}$ and $m\geq 0$ we have

  1. (i)

The operators $|D|^{-m}x|D|^{m},|D|^{m}x|D|^{-m}:H_{\infty}\to H_{\infty}$
have bounded extension

  2. (ii)

$|D|^{1-m}[|D|^{m},x]:H_{\infty}\to H_{\infty}$ has bounded extension.

###### Proof.

By Lemma A.1.1, on $H_{\infty}$ we have:

$\displaystyle|D|^{m}x|D|^{-m}$
$\displaystyle=\sum_{k=0}^{m}\binom{m}{k}\delta^{m-k}(x)|D|^{k-m},$
$\displaystyle|D|^{-m}x|D|^{m}$
$\displaystyle=\sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}|D|^{k-m}\delta^{m-k}(x).$

Clearly, the expressions on the right hand side have bounded extension. This
proves the first assertion.

By Lemma A.1.1, we have

$\displaystyle[|D|^{m},x]$
$\displaystyle=|D|^{m}x-\sum_{k=0}^{m}(-1)^{m-k}\binom{m}{k}|D|^{k}\delta^{m-k}(x)$
$\displaystyle=\sum_{k=0}^{m-1}(-1)^{m-k-1}\binom{m}{k}|D|^{k}\delta^{m-k}(x).$

Therefore,

$|D|^{1-m}[|D|^{m},x]=\sum_{k=0}^{m-1}(-1)^{m-k-1}\binom{m}{k}|D|^{k+1-m}\delta^{m-k}(x).$

Since $x\in\mathrm{dom}_{\infty}(\delta)$, for each $0\leq k\leq m-1$ the
operator $|D|^{k+1-m}\delta^{m-k}(x)$ has bounded extension. Hence,
$|D|^{1-m}[|D|^{m},x]$ has bounded extension. ∎

######  Lemma A.1.3.

Assume that $(\mathcal{A},H,D)$ satisfies Hypothesis 1.2.1. Let $h$ be a Borel
function on $\mathbb{R}$ such that

$t\mapsto(1+t^{2})^{\frac{p+1}{2}}h(t),\quad t\in\mathbb{R}.$

is bounded. Then for all $x\in\mathcal{B}$ and $s>0$ the operator $xh(sD)$ is
in $\mathcal{L}_{1}$, and:

$\|xh(sD)\|_{1}=O(s^{-p}),\quad s\downarrow 0.$

###### Proof.

Let $s>0$. Clearly,

$(1+s^{2}D^{2})^{-\frac{p+1}{2}}=|(1-isD)^{-p-1}|.$

Setting $\lambda=\frac{1}{s},$ we obtain from Hypothesis 1.2.1 that:

$\displaystyle\|x(1+s^{2}D^{2})^{-\frac{p+1}{2}}\|_{1}$
$\displaystyle=s^{-p-1}\|x(D+\frac{i}{s})^{-p-1}\|_{1}$
$\displaystyle=s^{-p-1}\cdot O(s)$ $\displaystyle=O(s^{-p}),\quad s\downarrow
0.$

Since the operator $(1+s^{2}D^{2})^{\frac{p+1}{2}}h(sD)$ is bounded, with

$\|(1+s^{2}D^{2})^{\frac{p+1}{2}}h(sD)\|_{\infty}\leq\sup_{t\in\mathbb{R}}(1+t^{2})^{\frac{p+1}{2}}|h(t)|$

We can conclude that:

$\displaystyle\|xh(sD)\|_{1}$
$\displaystyle\leq\|x(1+s^{2}D^{2})^{-\frac{p+1}{2}}\|_{1}\|(1+s^{2}D^{2})^{\frac{p+1}{2}}h(sD)\|_{\infty}$
$\displaystyle=O(s^{-p}),\quad s\downarrow 0.$

∎

We note in particular that the assumption on $h$ in Lemma A.1.3 is satisfied
if $h$ is a Schwartz function.

######  Lemma A.1.4.

Let $(\mathcal{A},H,D)$ satisfy Hypothesis 1.2.1, assume that $D$ has a
spectral gap at $0$ and let $x\in\mathcal{B}$. For all non-negative integers
$m_{1},m_{2}\geq 0$ with $m_{1}+m_{2}<p,$ we have

$\||D|^{-m_{1}}x|D|^{-m_{2}}e^{-s^{2}D^{2}}\|_{1}=O(s^{m_{1}+m_{2}-p}),\quad
s\downarrow 0.$

###### Proof.

Suppose first that $m_{1}>0.$ Using the triangle inequality:

$\displaystyle\||D|^{-m_{1}}x|D|^{-m_{2}}e^{-s^{2}D^{2}}\|_{1}$
$\displaystyle\leq\sum_{k,l=0}^{\infty}\|\chi_{[2^{k},2^{k+1}]}(|D|)|D|^{-m_{1}}x|D|^{-m_{2}}e^{-s^{2}D^{2}}\chi_{[2^{l},2^{l+1}]}(|D|)\|_{1}$
$\displaystyle\leq\sum_{k,l=0}^{\infty}2^{-km_{1}-lm_{2}}\cdot
e^{-2^{2l}s^{2}}\cdot\|\chi_{[2^{k},2^{k+1}]}(|D|)x\chi_{[2^{l},2^{l+1}]}(|D|)\|_{1}$
$\displaystyle\leq\sum_{k,l=0}^{\infty}2^{-km_{1}-lm_{2}}\cdot
e^{-2^{2l}s^{2}}\cdot\|\chi_{[0,2^{k+1}]}(|D|)x\chi_{[0,2^{l+1}]}(|D|)\|_{1}.$

If $m=\min\\{k,l\\},$ then

$\displaystyle\|\chi_{[0,2^{k+1}]}(|D|)x\chi_{[0,2^{l+1}]}(|D|)\Big{\|}_{1}$
$\displaystyle\leq\min\Big{\\{}\Big{\|}x\chi_{[0,2^{m+1}]}(|D|)\Big{\|}_{1},\Big{\|}\chi_{[0,2^{m+1}]}(|D|)x\Big{\|}_{1}\Big{\\}}$
$\displaystyle\leq
e\min\Big{\\{}\Big{\|}xe^{-2^{-2(m+1)}|D|^{2}}\Big{\|}_{1},\Big{\|}x^{*}e^{-2^{-2(m+1)}|D|^{2}}\Big{\|}_{1}\Big{\\}}$
$\displaystyle=O(2^{mp}).$

Thus,

$\||D|^{-m_{1}}x|D|^{-m_{2}}e^{-s^{2}D^{2}}\|_{1}\leq\Big{(}\sum_{k,l=0}^{\infty}2^{-km_{1}-lm_{2}}\cdot
e^{-2^{2l}s^{2}}\cdot 2^{p\cdot\min\\{k,l\\}}\Big{)}\cdot O(1).$

Since $p-m_{1}>0,$ it follows that

$\displaystyle\sum_{\begin{subarray}{c}k,l\geq 0\\\ k\leq
l\end{subarray}}2^{-km_{1}-lm_{2}}\cdot e^{-2^{2l}s^{2}}\cdot
2^{p\cdot\min\\{k,l\\}}$ $\displaystyle=\sum_{\begin{subarray}{c}k,l\geq 0\\\
k\leq l\end{subarray}}2^{(p-m_{1})k}\cdot 2^{-lm_{2}}\cdot e^{-2^{2l}s^{2}}$
$\displaystyle\leq 2\cdot\sum_{l=0}^{\infty}2^{(p-m_{1})l}\cdot
2^{-lm_{2}}\cdot e^{-2^{2l}s^{2}}.$

Now due to our assumption that $m_{1}>0,$ it follows that

$\displaystyle\sum_{\begin{subarray}{c}k,l\geq 0\\\ k\geq
l\end{subarray}}2^{-km_{1}-lm_{2}}\cdot e^{-2^{2l}s^{2}}\cdot
2^{p\cdot\min\\{k,l\\}}$ $\displaystyle=\sum_{\begin{subarray}{c}k,l\geq 0\\\
k\geq l\end{subarray}}2^{-km_{1}}\cdot 2^{(p-m_{2})l}\cdot e^{-2^{2l}s^{2}}$
$\displaystyle\leq 2\cdot\sum_{l=0}^{\infty}2^{-lm_{1}}\cdot
2^{(p-m_{2})l}\cdot e^{-2^{2l}s^{2}}.$

Note that for $m>0$, we have

$\sum_{l=0}^{\infty}2^{lm}e^{-2^{2l}s^{2}}=O(s^{-m}),\quad s\downarrow 0.$

Therefore,

$\displaystyle\||D|^{-m_{1}}x|D|^{-m_{2}}e^{-s^{2}D^{2}}\|_{1}$
$\displaystyle\leq\Big{(}\sum_{l=0}^{\infty}2^{l(p-m_{1}-m_{2})}\cdot
e^{-2^{2l}s^{2}}\Big{)}\cdot O(1)$ $\displaystyle=O(s^{m_{1}+m_{2}-p}).$

This completes the proof for the case $m_{1}>0.$

To complete the proof we now deal with the case where $m_{1}=0.$ We have

$\displaystyle\||D|^{-m_{1}}x|D|^{-m_{2}}e^{-s^{2}D^{2}}\|_{1}$
$\displaystyle\leq\sum_{l=0}^{\infty}\Big{\|}x|D|^{-m_{2}}e^{-s^{2}D^{2}}\chi_{[2^{l},2^{l+1}]}(|D|)\Big{\|}_{1}$
$\displaystyle\leq\sum_{l=0}^{\infty}2^{-lm_{2}}\cdot
e^{-2^{2l}s^{2}}\cdot\Big{\|}x\chi_{[2^{l},2^{l+1}]}(|D|)\Big{\|}_{1}$
$\displaystyle\leq\sum_{l=0}^{\infty}2^{-lm_{2}}\cdot
e^{-2^{2l}s^{2}}\cdot\Big{\|}x\chi_{[0,2^{l+1}]}(|D|)\Big{\|}_{1}.$

Thus,

$\displaystyle\||D|^{-m_{1}}x|D|^{m_{2}}e^{-s^{2}D^{2}}\|_{1}$
$\displaystyle\leq\Big{(}\sum_{l=0}^{\infty}2^{l(p-m_{2})}\cdot
e^{-2^{2l}s^{2}}\Big{)}\cdot O(1)$ $\displaystyle=O(s^{m_{2}-p}).$

This completes the proof for the case $m_{1}=0.$ ∎

######  Lemma A.1.5.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
assume that $D$ has spectral gap at $0.$ For all $x\in\mathcal{B}$ we have

$\|x|D|^{-p-1}(1-e^{-s^{2}D^{2}})\|_{1}=O(s),\quad s\downarrow 0.$

###### Proof.

By the triangle inequality, we have

$\displaystyle\|x|D|^{-p-1}(1-e^{-s^{2}D^{2}})\|_{1}$
$\displaystyle\leq\|x|D|^{-p-1}(1-e^{-s^{2}D^{2}})\chi_{(\frac{1}{s},\infty)}(|D|)\|_{1}$
(A.1)
$\displaystyle\quad+\|x|D|^{-p-1}(1-e^{-s^{2}D^{2}})\chi_{[0,\frac{1}{s}]}(|D|)\|_{1}.$

Let us estimate the first summand of (A.1). Since for $t>1$ we have:

$t^{-p-1}(1-e^{-t^{2}})\leq t^{-p-1}\leq
2^{\frac{p+1}{2}}\cdot(t^{2}+1)^{-\frac{p+1}{2}},$

it follows that

$|D|^{-p-1}(1-e^{-s^{2}D^{2}})\chi_{(\frac{1}{s},\infty)}(|D|)\leq
s^{p+1}\cdot 2^{\frac{p+1}{2}}\cdot(1+s^{2}D^{2})^{-\frac{p+1}{2}}.$

So we may estimate the first summand by:

$\displaystyle\|x|D|^{p-1}(1-e^{-s^{2}D^{2}})\chi_{(\frac{1}{s},\infty)}(|D|)\|_{1}$
$\displaystyle\leq
2^{\frac{p+1}{2}}\Big{\|}x(D^{2}+s^{-2})^{-\frac{p+1}{2}}\Big{\|}_{1}$
$\displaystyle=2^{\frac{p+1}{2}}\Big{\|}x(D+\frac{i}{s})^{-p-1}\Big{\|}_{1}$
$\displaystyle=O(s)$

Where the final step follows from Hypothesis 1.2.1.(iii).

Let us estimate the second summand of (A.1). We have

$t^{-p-1}(1-e^{-t^{2}})\leq t^{1-p}\leq et^{1-p}\cdot e^{-t^{2}},\quad
t\in[0,1],$

so it follows that

$|D|^{-p-1}(1-e^{-s^{2}D^{2}})\chi_{[0,\frac{1}{s}]}(|D|)\leq
es^{p+1}\cdot(s|D|)^{1-p}\cdot e^{-s^{2}D^{2}}.$

So, the second summand in (A.1) can be estimated by:

$\|x|D|^{-p-1}(1-e^{-s^{2}D^{2}})\chi_{[0,\frac{1}{s})]}(|D|)\|_{1}\leq
es^{2}\Big{\|}x|D|^{1-p}e^{-s^{2}D^{2}}\Big{\|}_{1}.$

From Lemma A.1.4, this is $O(s)$ as $s\downarrow 0$. ∎

######  Lemma A.1.6.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
assume that $D$ has a spectral gap at $0.$ For every $x\in\mathcal{B},$ we
have

$\||D|^{-p-1}x(1-e^{-s^{2}D^{2}})\|_{1}=O(s),\quad s\downarrow 0.$

###### Proof.

On the subspace $H_{\infty}$, we have:

$\displaystyle|D|^{-p-1}x$ $\displaystyle=x|D|^{-p-1}+[|D|^{-p-1},x]$
$\displaystyle=x|D|^{-p-1}-|D|^{-p-1}[|D|^{p+1},x]|D|^{-p-1}.$

By Lemma A.1.1, we have (again on $H_{\infty}$):

$\displaystyle[|D|^{p+1},x]$ $\displaystyle=|D|^{p+1}x-x|D|^{p+1}$
$\displaystyle=\left(\sum_{k=0}^{p+1}\binom{p+1}{k}\delta^{p+1-k}(x)|D|^{k}\right)-x|D|^{p+1}$
$\displaystyle=\sum_{k=0}^{p}\binom{p+1}{k}\delta^{p+1-k}(x)|D|^{k}.$

Thus on $H_{\infty}$:

$|D|^{-p-1}x=x|D|^{-p-1}-\sum_{k=0}^{p}\binom{p+1}{k}|D|^{-p-1}\delta^{p+1-k}(x)|D|^{k-p-1}.$

Multiplying on the right by $(1-e^{-s^{2}D^{2}})$, on $H_{\infty}$ we have:

(A.2)
$|D|^{-p-1}x(1-e^{-s^{2}D^{2}})=x|D|^{-p-1}(1-e^{-s^{2}D^{2}})-\sum_{k=0}^{p}\binom{p+1}{k}|D|^{-p-1}\delta^{p+1-k}(x)|D|^{k-p-1}(1-e^{-s^{2}D^{2}})$

From Lemma A.1.5, we have that $x|D|^{-p-1}(1-e^{-s^{2}D^{2}})$ has bounded
extension to an operator in $\mathcal{L}_{1}$ with norm bounded by $O(s)$.

For $0\leq k\leq p,$ we have

$\displaystyle\Big{\|}|D|^{-p-1}\delta^{p+1-k}$
$\displaystyle(x)|D|^{k-p-1}(1-e^{-s^{2}D^{2}})\Big{\|}_{1}$
$\displaystyle\leq\Big{\|}|D|^{-p-1}\delta^{p+1-k}(x)\Big{\|}_{1}\cdot\Big{\|}|D|^{k-p}\Big{\|}_{\infty}$
$\displaystyle\quad\cdot
s\Big{\|}(s|D|)^{-1}(1-e^{-s^{2}D^{2}})\Big{\|}_{\infty}.$

The first factor here is finite by Remark 3.1.1. The second factor is finite
because $k\leq p.$ The third factor is $O(s)$ by the functional calculus.
Thus,

$\||D|^{-p-1}\delta^{p+1-k}(x)|D|^{k-p-1}(1-e^{-s^{2}D^{2}})\|_{1}=O(s),\quad
s\downarrow 0.$

Hence, each summand in (A.2) extends to a trace class operator with norm
bounded by $O(s)$, $s\downarrow 0$. By the triangle inequality, this completes
the proof. ∎

###  A.2. Integral formulae for commutators

In this section of the appendix, we collect results concerning formulae for
commutators with functions of $D$. Many of the results of this section will be
known to the expert reader, but since they are scattered around various
sources we provide them here with short and self-contained proofs.

In this section, $(\mathcal{A},H,D)$ is a smooth $p$-dimensional spectral
triple.

The following is essentially a consequence of the classical Duhamel formula.

######  Lemma A.2.1.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple. If $x\in\mathcal{B}$ then
for all $t\in\mathbb{R}$:

$[e^{it|D|},x]=it\int_{0}^{1}e^{it(1-v)|D|}\delta(x)e^{itv|D|}\,dv.$

Here, the integral is understood in the weak operator topology sense.

###### Proof.

For $n\geq 0$, we define the projection

$p_{n}:=\chi_{[0,n]}(|D|).$

Since $|D|$ is non-negative, as $n\to\infty$ the sequence of projections
$\\{p_{n}\\}_{n\geq 0}$ converges in the strong operator topology to the
identity. We define the functions $\xi$ and $\eta$ by:

$\displaystyle\xi(v)$ $\displaystyle:=p_{n}\exp(it(1-v)|D|),$
$\displaystyle\eta(v)$ $\displaystyle:=x\exp(itv|D|)p_{n},\quad
v\in\mathbb{R}.$

Since $p_{n}|D|\leq n$, the operator valued functions $\xi$ and $\eta$ are
continuous and differentiable in the uniform norm.

Since $\xi$ and $\eta$ are continuous and differentiable, we have:

$\xi(1)\eta(1)-\xi(0)\eta(0)=\int_{0}^{1}\xi^{\prime}(v)\eta(v)+\xi(v)\eta^{\prime}(v)\,dv.$

where since $\xi,$ $\xi^{\prime},$ $\eta$ and $\eta^{\prime}$ are continuous,
this may be considered as a Bochner integral. Therefore in particular, this is
an integral in the weak operator topology.

We can compute the terms in the integrand as:

$\displaystyle\xi^{\prime}(v)\eta(v)$
$\displaystyle=-itp_{n}\exp(it(1-v)|D|)|D|x\exp(itv|D|)p_{n},$
$\displaystyle\xi(v)\eta^{\prime}(v)$
$\displaystyle=itp_{n}\exp(it(1-v)|D|)x|D|\exp(itv|D|)p_{n}.$

Thus,

$\xi(1)\eta(1)-\xi(0)\eta(0)=-it\int_{0}^{1}p_{n}\exp(it(1-v)|D|)\delta(x)\exp(itv|D|)p_{n}\,dv.$

Since the operator $\exp(it(1-v)|D|)\delta(x)\exp(itv|D|)$ is a continuous
function of $v$ (in weak operator topology), the weak operator topology
integral

$\int_{0}^{1}\exp(it(1-v)|D|)\delta(x)\exp(itv|D|)\,dv$

exists, and we have:

$p_{n}\int_{0}^{1}\exp(it(1-v)|D|)\delta(x)\exp(itv|D|)\,dv\cdot
p_{n}=\int_{0}^{1}p_{n}\exp(it(1-v)|D|)\delta(x)\exp(itv|D|)p_{n}\,dv.$

Therefore:

$\xi(1)\eta(1)-\xi(0)\eta(0)=p_{n}\int_{0}^{1}\exp(it(1-v)|D|)\delta(x)\exp(itv|D|)\,dv\cdot
p_{n}.$

On the other hand, we can compute $\xi(1),\eta(1),\xi(0)$ and $\eta(0)$
directly:

$\displaystyle\xi(1)\eta(1)-\xi(0)\eta(0)$
$\displaystyle=p_{n}x\exp(it|D|)p_{n}-p_{n}\exp(it|D|)xp_{n}$
$\displaystyle=p_{n}[x,\exp(it|D|)]p_{n}.$

Thus,

$p_{n}[\exp(it|D|),x]p_{n}=itp_{n}\int_{0}^{1}\exp(it(1-v)|D|)\delta(x)\exp(itv|D|)\,dv\cdot
p_{n}.$

Since $n\geq 0$ is arbitrary, we may take $n\to\infty$ and since $p_{n}$
converges in the strong operator topology to the identity we get the desired
equality. ∎

Combining Lemma A.2.1 with the Fourier inversion theorem yields a formula for
$[f(s|D|),x]$ for quite general functions $f$. The following formula is well
known and appears in many places, for example [8, Theorem 3.2.32].

######  Lemma A.2.2.

If $\widehat{f},\widehat{f^{\prime}}\in L_{1}(\mathbb{R}),$ then for all
$x\in\mathcal{B}$, and $s>0$,

$[f(s|D|),x]=s\int_{-\infty}^{\infty}\left(\int_{0}^{1}\widehat{f^{\prime}}(u)e^{ius(1-v)|D|}\delta(x)e^{iusv|D|}dv\right)\,du.$

Here, the integral is understood in a weak operator topology sense.

###### Proof.

Indeed, by the Fourier inversion formula, by functional calculus we have a
weak operator topology integral:

$f(s|D|)=\int_{\mathbb{R}}\widehat{f}(u)e^{ius|D|}du.$

Therefore,

$[f(s|D|),x]=\int_{-\infty}^{\infty}\widehat{f}(u)[e^{ius|D|},x]du.$

By Lemma A.2.1, we have a weak operator topology integral:

$[e^{ius|D|},x]=ius\int_{0}^{1}e^{ius(1-v)|D|}\delta(x)e^{iusv|D|}dv.$

Thus,

$[f(s|D|),x]=s\int_{-\infty}^{\infty}iu\widehat{f}(u)\Big{(}\int_{0}^{1}e^{ius(1-v)|D|}\delta(x)e^{iusv|D|}dv\Big{)}du.$

Since $iu\widehat{f}(u)=\widehat{f^{\prime}}(u),$ the assertion follows. ∎

######  Lemma A.2.3.

If $\widehat{f},\widehat{f^{\prime}},\widehat{f^{\prime\prime}}\in
L_{1}(\mathbb{R}),$ then for all $x\in\mathcal{B}$ and $s>0$ we have:

$[f(s|D|),x]-sf^{\prime}(s|D|)\delta(x)=-s^{2}\int_{-\infty}^{\infty}\left(\int_{0}^{1}\widehat{f^{\prime\prime}}(u)(1-v)e^{ius(1-v)|D|}\delta^{2}(x)e^{iusv|D|}dv\right)du.$
$[f(s|D|),x]-s\delta(x)f^{\prime}(s|D|)=-s^{2}\int_{-\infty}^{\infty}\left(\int_{0}^{1}\widehat{f^{\prime\prime}}(u)(1-v)e^{iusv|D|}\delta^{2}(x)e^{ius(1-v)|D|}dv\right)du.$

Here once again the integrals are understood in the weak operator topology.

###### Proof.

We only prove the first equality as the proof of the second one is similar.

By the Fourier inversion theorem and functional calculus we have a weak
operator topology integral representation:

$f^{\prime}(s|D|)=\int_{\mathbb{R}}\widehat{f^{\prime}}(u)e^{ius|D|}du.$

So multiplying on the left by the bounded operator $\delta(x)$,

$\displaystyle sf^{\prime}(s|D|)\delta(x)$
$\displaystyle=s\int_{\mathbb{R}}\widehat{f^{\prime}}(u)e^{ius|D|}\delta(x)du$
$\displaystyle=s\int_{-\infty}^{\infty}\left(\int_{0}^{1}\widehat{f^{\prime}}(u)e^{ius|D|}\delta(x)dv\right)du.$

Now representing $[f(s|D|),x]$ by the integral representation given by Lemma
A.2.2, we infer that

$\displaystyle[f(s|D|),x]-sf^{\prime}(s|D|)\delta(x)$
$\displaystyle=s\int_{-\infty}^{\infty}\Big{(}\int_{0}^{1}\widehat{f^{\prime}}(u)\Big{(}e^{ius(1-v)|D|}\delta(x)e^{iusv|D|}-e^{ius|D|}\delta(x)\Big{)}dv\Big{)}du$
(A.3)
$\displaystyle=s\int_{-\infty}^{\infty}\Big{(}\int_{0}^{1}\widehat{f^{\prime}}(u)\Big{(}e^{ius(1-v)|D|}[\delta(x),e^{iusv|D|}]\Big{)}dv\Big{)}du.$

Applying Lemma A.2.1 to $\delta(x)\in\mathcal{B}$, we have:

(A.4)
$[\delta(x),e^{iusv|D|}]=-iusv\int_{0}^{1}e^{iusv(1-w)|D|}\delta^{2}(x)e^{iusvw|D|}dw.$

Combining (A.3) and (A.4), we obtain

(A.5)
$[f(s|D|),x]-sf^{\prime}(s|D|)\delta(x)=-s^{2}\int_{-\infty}^{\infty}\Big{(}\int_{0}^{1}\Big{(}\int_{0}^{1}\widehat{f^{\prime\prime}}(u)ve^{ius(1-vw)|D|}\delta^{2}(x)e^{iusvw|D|}dw\Big{)}dv\Big{)}du.$

We focus on the inner integral. Performing a linear change of variables,
$w_{0}=vw$, we get:

$\int_{0}^{1}ve^{ius(1-vw)|D|}\delta^{2}(x)e^{iusvw|D|}\,dw=\int_{0}^{v}e^{ius(1-w_{0})|D|}\delta^{2}(x)e^{iusw_{0}|D|}\,dw_{0},$

and therefore:

$\int_{0}^{1}\int_{0}^{1}ve^{ius(1-vw)|D|}\delta^{2}(x)e^{iusvw|D|}\,dwdv=\int_{0}^{1}\int_{0}^{v}e^{ius(1-w)|D|}\delta^{2}(x)e^{iusw|D|}\,dwdv.$

Since the integrand is continuous, we may apply Fubini’s theorem to
interchange the integrals:

$\displaystyle\int_{0}^{1}\int_{0}^{1}ve^{ius(1-vw)|D|}\delta^{2}(x)e^{iusvw|D|}\,dwdv$
$\displaystyle=\int_{0}^{1}\int_{w}^{1}e^{ius(1-w)|D|}\delta^{2}(x)e^{iusw|D|}\,dvdw$
$\displaystyle=\int_{0}^{1}(1-w)e^{ius(1-w)|D|}\delta^{2}(x)e^{iusw|D|}\,dw.$

So from (A.5):

$[f(s|D|),x]-sf^{\prime}(s|D|)\delta(x)=-s^{2}\int_{-\infty}^{\infty}\int_{0}^{1}(1-w)e^{ius(1-w)|D|}\delta^{2}(x)e^{iusw|D|}\,dwdu.$

∎

###  A.3. Hochschild coboundary computations

In this part of the appendix we include some of the lengthy algebraic
computations required for Sections 4.3 and 4.4. Recall that for a multilinear
functional $\theta:\mathcal{A}^{\otimes p}\to\mathbb{C}$ the Hochschild
coboundary $b\theta:\mathcal{A}^{\otimes(p+1)}\to\mathbb{C}$ is defined in
terms of the Hochschild boundary $b$ by $b\theta(c)=\theta(bc)$.

Explicitly, for an elementary tensor $a_{0}\otimes\cdots\otimes
a_{p}\in\mathcal{A}^{\otimes(p+1)}$ we have:

$\displaystyle(b\theta)(a_{0}\otimes\cdots a_{p})$
$\displaystyle=\theta(a_{0}a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{p})+\sum_{k=1}^{p-1}(-1)^{k}\theta(a_{0}\otimes\cdots\otimes
a_{k}a_{k+1}\otimes\cdots\otimes a_{p})$
$\displaystyle\quad+(-1)^{p}\theta(a_{p}a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p-1}).$

####  A.3.1. Coboundaries in Section 4.3

Let $\mathscr{A}\subseteq\\{1,\ldots,p\\}$. Let
$T:=D^{2-|\mathscr{A}|}|D|^{p+1}e^{-s^{2}D^{2}}.$ Following the notation of
Definition 4.1.1, we define for $a\in\mathcal{A}$,

$b_{k}(a):=\begin{cases}\delta(a),\quad k\in\mathscr{A},\\\ [F,a],\quad
k\notin\mathscr{A}.\end{cases}$

Fix $1\leq m\leq p-1$. We introduce a pair of multilinear mappings,
$\theta_{s}^{1}$ and $\theta_{s}^{2}$, defined on $a_{0}\otimes\cdots\otimes
a_{p-1}\in\mathcal{A}^{\otimes p}$ by:

$\theta_{s}^{1}(a_{0}\otimes\cdots\otimes a_{p-1}):=\mathrm{Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)\delta^{2}(a_{m-1})\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)T\right).$

and

$\theta_{s}^{2}(a_{0}\otimes\cdots\otimes a_{p-1}):=\mathrm{Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)[F,\delta(a_{m-1})]\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)T\right).$

The mapping that here is denoted $\theta_{s}^{1}$ is exactly the multilinear
mapping $\theta_{s}$ introduced in Lemma 4.3.2, and similarly the multilinear
mapping $\theta_{s}^{2}$ is the multilinear mapping $\theta_{s}$ introduced in
Lemma 4.3.3. For $1\leq k<m$ we also introduce $X_{k}^{1}$ and $X_{k}^{2}$
defined by:

$\displaystyle X_{k}^{1}$ $\displaystyle:={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{k-1}b_{l}(a_{l})\right)a_{k}\left(\prod_{l=k}^{m-2}b_{l}(a_{l+1})\right)\delta^{2}(a_{m})\left(\prod_{l=m+1}^{p}b_{l}(a_{l})\right)\cdot
T\right),$ $\displaystyle X_{k}^{2}$ $\displaystyle:={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{k-1}b_{l}(a_{l})\right)a_{k}\left(\prod_{l=k}^{m-2}b_{l}(a_{l+1})\right)[F,\delta(a_{m})]\left(\prod_{l=m+1}^{p}b_{l}(a_{l})\right)\cdot
T\right).$

Now if $m\leq k\leq p$, we define $Y^{1}_{k}$ and $Y^{2}_{k}$ by:

$\displaystyle Y_{k}^{1}$ $\displaystyle:={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{k-1}b_{l+1}(a_{l})\right)a_{k}\left(\prod_{l=k+1}^{p}b_{l}(a_{l})\right)\cdot
T\right),$ $\displaystyle Y_{k}^{2}$ $\displaystyle:={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)[F,\delta(a_{m-1})]\left(\prod_{l=m}^{k-1}b_{l+1}(a_{l})\right)a_{k}\left(\prod_{l=k+1}^{p}b_{l}(a_{l})\right)\cdot
T\right).$

######  Lemma A.3.1.

For $j=1,2$ and $a_{0}\otimes\cdots\otimes
a_{p}\in\mathcal{A}^{\otimes(p+1)}$. we have:

$\theta^{j}_{s}(a_{0}a_{1}\otimes a_{2}\otimes\cdots\otimes a_{p})=X_{1}^{j}$

###### Proof.

This follows immediately from the definition. ∎

######  Lemma A.3.2.

For $j=1,2$, $1\leq k<m-1$ and $a_{0}\otimes\cdots\otimes
a_{p}\in\mathcal{A}^{\otimes(p+1)}$ we have

$\theta_{s}^{j}(a_{0}\otimes\cdots\otimes a_{k}a_{k+1}\otimes\cdots\otimes
a_{p})=X_{k}^{j}+X_{k+1}^{j}.$

###### Proof.

We will only describe the $j=1$ case, since the $j=2$ case is identical. By
the definition of $\theta_{s}^{j},$ we have

$\displaystyle\theta_{s}^{1}(a_{0}\otimes$ $\displaystyle
a_{1}\otimes\cdots\otimes a_{k-1}\otimes a_{k}a_{k+1}\otimes
a_{k+2}\otimes\cdots\otimes a_{p})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{k-1}b_{l}(a_{l})\right)b_{k}(a_{k}a_{k+1})\left(\prod_{l=k+1}^{m-2}b_{l}(a_{l+1})\right)\delta^{2}(a_{m})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l+1})\right)\cdot
T\right).$

Now applying the Leibniz rule to $b_{k}(a_{k}a_{k+1})$,

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{k-1}\otimes a_{k}a_{k+1}\otimes
a_{k+2}\otimes\cdots\otimes a_{p})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{k-1}b_{l}(a_{l})\right)a_{k}b_{k}(a_{k+1})\left(\prod_{l=k+1}^{m-2}b_{l}(a_{l+1})\right)\delta^{2}(a_{m})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l+1})\right)\cdot
T\right)$ $\displaystyle\quad+{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{k-1}b_{l}(a_{l})\right)b_{k}(a_{k})a_{k+1}\left(\prod_{l=k+1}^{m-2}b_{l}(a_{l+1})\right)\delta^{2}(a_{m})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l+1})\right)\cdot
T\right)$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{k-1}b_{l}(a_{l})\right)a_{k}\left(\prod_{l=k}^{m-2}b_{l}(a_{l+1})\right)\delta^{2}(a_{m})\left(\prod_{l=m+1}^{p}b_{l}(a_{l})\right)T\right)$
$\displaystyle\quad+{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{k}b_{l}(a_{l})\right)a_{k+1}\left(\prod_{l=k+1}^{m-2}b_{l}(a_{l+1})\right)\delta^{2}(a_{m})\left(\prod_{l=m+1}^{p}b_{l}(a_{l})\right)\cdot
T\right)$ $\displaystyle=X_{k}^{1}+X_{k+1}^{1}$

as required. ∎

######  Lemma A.3.3.

For $m\leq k<p,$ $j=1,2$ and $a_{0}\otimes\cdots\otimes
a_{p}\in\mathcal{A}^{\otimes(p+1)}$ we have

$\theta_{s}^{j}(a_{0}\otimes a_{1}\otimes\cdots\otimes a_{k-1}\otimes
a_{k}a_{k+1}\otimes a_{k+2}\otimes\cdots\otimes a_{p})=Y_{k}^{j}+Y_{k+1}^{j}.$

###### Proof.

Again we demonstrate only the $j=1$ case since the $j=2$ case is identical. By
definition we have:

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{k-1}\otimes a_{k}a_{k+1}\otimes
a_{k+2}\otimes\cdots\otimes a_{p})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{k-1}b_{l+1}(a_{l})\right)b_{k+1}(a_{k}a_{k+1})\left(\prod_{l=m+1}^{p-1}b_{l+1}(a_{l+1})\right)\cdot
T\right).$

Applying the Leibniz rule to $b_{k+1}(a_{k}a_{k+1})$ we have:

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{k-1}\otimes a_{k}a_{k+1}\otimes
a_{k+2}\otimes\cdots\otimes a_{p})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{k-1}b_{l+1}(a_{l})\right)a_{k}b_{k+1}(a_{k+1})\left(\prod_{l=k+1}^{p-1}b_{l+1}(a_{l+1})\right)\cdot
T\right)$ $\displaystyle\quad+{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{k-1}b_{l+1}(a_{l})\right)b_{k+1}(a_{k})a_{k+1}\left(\prod_{l=k+1}^{p-1}b_{l+1}(a_{l+1})\right)\cdot
T\right)$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{k-1}b_{l+1}(a_{l}\right)a_{k}\left(\prod_{l=k+1}^{p}b_{l}(a_{l})\right)\cdot
T\right)$ $\displaystyle\quad+{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{k}b_{l+1}(a_{l})\right)a_{k+1}\left(\prod_{l=k+2}^{p}b_{l}(a_{l})\right)\cdot
T\right)$ $\displaystyle=Y_{k}^{1}+Y_{k+1}^{1}.$

∎

We recall also the multilinear maps $\mathcal{W}_{\mathscr{A}}$ from
Definition 4.1.1.

######  Lemma A.3.4.

Let $c=a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p}\in\mathcal{A}^{\otimes(p+1)}$. If we assume that $m-1,m\in\mathscr{A}$,
then we have:

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{m-2}\otimes a_{m-1}a_{m}\otimes
a_{m+1}\otimes\cdots\otimes a_{p})$ $\displaystyle=X_{m-1}^{1}+Y_{m}^{1}+2{\rm
Tr}(\mathcal{W}_{\mathscr{A}}(c)\cdot T).$

Now if $\mathscr{B}\subseteq\\{1,\ldots,p\\}$ is such that
$|\mathscr{B}|=|\mathscr{A}|$ and $\mathscr{A}\Delta\mathscr{B}=\\{m-1,m\\}$
(where $\Delta$ denotes the symmetric difference) then

$\displaystyle\theta_{s}^{2}($ $\displaystyle a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{m-2}\otimes a_{m-1}a_{m}\otimes
a_{m+1}\otimes\cdots\otimes a_{p})$ $\displaystyle=X_{m-1}^{2}+Y_{m}^{2}+{\rm
Tr}(\mathcal{W}_{\mathscr{A}}(c)\cdot T)+{\rm
Tr}(\mathcal{W}_{\mathscr{B}}(c)\cdot T).$

###### Proof.

Using the repeated Leibniz rule, we obtain

$\displaystyle\delta^{2}(a_{m-1}a_{m})$
$\displaystyle=\delta^{2}(a_{m-1})a_{m}+2\delta(a_{m-1})\delta(a_{m})+a_{m-1}\delta^{2}(a_{m}),$
$\displaystyle[F,\delta(a_{m-1}a_{m})]$
$\displaystyle=[F,\delta(a_{m-1})]a_{m}+[F,a_{m-1}]\delta(a_{m})+\delta(a_{m-1})[F,a_{m}]+a_{m-1}[F,\delta(a_{m})].$

Let us focus on proving the assertion relating to $\theta^{1}_{s}$, since the
other assertion is identical.

By the definition of $\theta_{s}^{1},$ we have

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{m-2}\otimes a_{m-1}a_{m}\otimes
a_{m+1}\otimes\cdots\otimes a_{p})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1}a_{m})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l+1})\right)\cdot
T\right)$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1}a_{m})\left(\prod_{l=m+1}^{p}b_{l}(a_{l})\right)\cdot
T\right).$

Applying the Leibniz rule to the $\delta^{2}(a_{m-1}a_{m})$ term:

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{m-2}\otimes a_{m-1}a_{m}\otimes
a_{m+1}\otimes\cdots\otimes a_{p})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)a_{m-1}\delta^{2}(a_{m})\left(\prod_{l=m+1}^{p}b_{l}(a_{l})\right)\cdot
T\right)$ $\displaystyle\quad+{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)2\delta(a_{m-1})\delta(a_{m})\left(\prod_{l=m+1}^{p}b_{l}(a_{l})\right)\cdot
T\right)$ $\displaystyle\quad+{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})a_{m}\left(\prod_{l=m+1}^{p}b_{l}(a_{l})\right)\cdot
T\right)$ $\displaystyle=X_{m-1}+2{\rm Tr}(\mathcal{W}_{\mathscr{A}}(c)\cdot
T)+Y_{m}.$

∎

######  Lemma A.3.5.

For $a_{0}\otimes\cdots\otimes a_{p}\in\mathcal{A}^{\otimes(p+1)}$, we have

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{p}a_{0}\otimes a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{p-1})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l})\right)[T,a_{p}]\right)+Y^{1}_{p}.$

We also have:

$\displaystyle\theta_{s}^{2}($ $\displaystyle a_{p}a_{0}\otimes a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{p-1})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)[F,\delta(a_{m-1})]\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l})\right)[T,a_{p}]\right)+Y_{p}^{2}.$

###### Proof.

We prove only the assertion involving $\theta_{s}^{1}$, since one can prove
the other result by an identical argument. Since $\Gamma$ commutes with
$a_{p}$, we have:

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{p}a_{0}\otimes a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{p-1})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{p}a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l})\right)\cdot
T\right)$ $\displaystyle={\rm Tr}\left(a_{p}\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l})\right)\cdot
T\right).$

Using the cyclicity of the trace, we have

$\displaystyle\theta_{s}^{1}($ $\displaystyle a_{p}a_{0}\otimes a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{p-1})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l})\right)\cdot
Ta_{p}\right)$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l})\right)\cdot[T,a_{p}]\right)+Y_{p}^{1}.$

∎

Note that for $j=1,2$, the telescopic sum

$\displaystyle X_{1}^{j}+$
$\displaystyle\sum_{k=1}^{m-1}(-1)^{k}(X_{k}^{j}+X_{k+1}^{j})+(-1)^{m-1}(X_{m-1}^{j}+Y_{m}^{j})$
$\displaystyle+\sum_{k=m}^{p-1}(-1)^{k}(Y_{k}^{j}+Y_{k+1}^{j})+(-1)^{p}Y_{p}^{j}.$

vanishes.

Therefore, combining Lemmas A.3.1, A.3.2, A.3.3, A.3.4 and A.3.5 we have:

$\displaystyle(b\theta_{s}^{1})(a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle=2\cdot(-1)^{m-1}\cdot{\rm Tr}(\mathcal{W}_{\mathscr{A}}(c)\cdot
T)$ $\displaystyle\quad+(-1)^{p}\cdot{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)\delta^{2}(a_{m-1})\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l})\right)\cdot[T,a_{p}]\right).$

Similarly, if $\mathscr{B}$ is such that $|\mathscr{A}|=|\mathscr{B}|$ and
$\mathscr{A}\Delta\mathscr{B}=\\{m-1,m\\}$ then:

$\displaystyle(b\theta_{s}^{2})(a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle=(-1)^{m-1}\cdot{\rm Tr}(\mathcal{W}_{\mathscr{A}}(c)\cdot
T)+(-1)^{m-1}\cdot{\rm Tr}(\mathcal{W}_{\mathscr{B}}(c)\cdot T)$
$\displaystyle\quad+(-1)^{p}\cdot{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{l=1}^{m-2}b_{l}(a_{l})\right)[F,\delta(a_{m-1})]\left(\prod_{l=m}^{p-1}b_{l+1}(a_{l})\right)\cdot[T,a_{p}]\right).$

This completes the computation of the coboundaries for $\theta^{1}_{s}$ and
$\theta^{2}_{s}$.

####  A.3.2. Coboundaries in Section 4.4

Again in this subsection, $(\mathcal{A},H,D)$ is a smooth spectral triple
where $D$ has a spectral gap at $0$. Let $T=Fe^{-s^{2}D^{2}}$. Note that this
is different to $T$ in the preceding section.

We define the multilinear mapping $\mathcal{L}_{s}:\mathcal{A}^{\otimes
p}\to\mathbb{C}$ on $a_{0}\otimes\cdots\otimes a_{p-1}\in\mathcal{A}^{\otimes
p}$ by:

$\mathcal{L}_{s}(a_{0}\otimes\cdots\otimes a_{p-1}):={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)\cdot T\right).$

We also define the multilinear mapping
$\mathcal{K}_{s}:\mathcal{A}^{\otimes(p+1)}\to\mathbb{C}$ by:

$\mathcal{K}_{s}(a_{0}\otimes\cdots\otimes a_{p}):={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)\cdot[T,a_{p}]\right).$

By definition, $\mathcal{K}_{s}$ is exactly the mapping from Theorem 4.4.2.

For $1\leq m\leq p$, we define $X_{m}$ by:

$X_{m}:={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-1}[F,a_{k}]\right)a_{m}\left(\prod_{k=m+1}^{p}[F,a_{k}]\right)T\right).$

We have

$\displaystyle\mathcal{L}_{s}(a_{0}a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{p})$ $\displaystyle={\rm Tr}(\Gamma a_{0}a_{1}\prod_{k=2}^{p}[F,a_{k}]\cdot
T)$ $\displaystyle=X_{1}.$

Applying the Leibniz rule to $[F,a_{m-1}a_{m}]$:

$\displaystyle\mathcal{L}_{s}($ $\displaystyle a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{m-2}\otimes a_{m-1}a_{m}\otimes
a_{m+1}\otimes\cdots\otimes a_{p})$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-2}[F,a_{k}]\right)[F,a_{m-1}a_{m}]\left(\prod_{k=m+1}^{p}[F,a_{k}]\right)\cdot
T\right)$ $\displaystyle={\rm Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-2}[F,a_{k}]\right)a_{m-1}\left(\prod_{k=m}^{p}[F,a_{k}]\right)\cdot
T\right)$ $\displaystyle\quad+{\rm Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-1}[F,a_{k}]\right)a_{m}\left(\prod_{k=m+1}^{p}[F,a_{k}]\right)\cdot
T\right)$ $\displaystyle=X_{m}+X_{m+1}.$

Finally,

$\displaystyle\mathcal{L}_{s}(a_{p}a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p-1})$ $\displaystyle={\rm Tr}(\Gamma
a_{p}a_{0}\prod_{k=1}^{p-1}[F,a_{k}]\cdot T)$ $\displaystyle={\rm Tr}(\Gamma
a_{0}\prod_{k=1}^{p-1}[F,a_{k}]\cdot Ta_{p})$ $\displaystyle={\rm Tr}(\Gamma
a_{0}\prod_{k=1}^{p-1}[F,a_{k}]\cdot[T,a_{p}])+{\rm Tr}(\Gamma
a_{0}\prod_{k=1}^{p-1}[F,a_{k}]\cdot a_{p}T)$
$\displaystyle=\mathcal{K}_{s}(a_{0}\otimes\cdots\otimes a_{p})+X_{p}.$

Thus,

$\displaystyle(b\mathcal{L}_{s})(a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle=\mathcal{K}_{s}(a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle\quad+X_{1}+\Big{(}\sum_{m=1}^{p-1}(-1)^{m}(X_{m}+X_{m+1})\Big{)}+(-1)^{p}X_{p}.$

The latter sum telescopes and indeed vanishes, so it follows that
$b\mathcal{L}_{s}=\mathcal{K}_{s}.$

####  A.3.3. Coboundaries in Section 4.5

In this section, $(\mathcal{A},H,D)$ is a smooth spectral triple where $D$ has
a spectral gap at $0$, and $T:=|D|^{2-p}e^{-s^{2}D^{2}}$. We define the
multilinear mapping $\theta_{s}:\mathcal{A}^{\otimes p}\to\mathbb{C}$ on
$a_{0}\otimes\cdots a_{p-1}\in\mathcal{A}^{\otimes p}$ by:

$\theta_{s}(a_{0}\otimes\cdots\otimes a_{p-1})={\rm
Tr}\left(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)T\right).$

For $0\leq k\leq p$ we also define:

$X_{k}={\rm
Tr}\left(\left(\prod_{l=0}^{k-1}\partial(a_{l})\right)a_{k}\left(\prod_{l=k+1}^{p}\partial(a_{l})\right)T\right).$

So in particular,

$X_{0}=\mathrm{Tr}\left(a_{0}\left(\prod_{l=1}^{p}\partial(a_{l})\right)T\right).$

Applying the Leibniz rule to $\partial(a_{k}a_{a+1})$ we get:

$\displaystyle\theta_{s}($ $\displaystyle a_{0}\otimes\cdots\otimes
a_{k-1}\otimes a_{k}a_{k+1}\otimes a_{k+2}\otimes\cdots\otimes a_{p})$
$\displaystyle={\rm
Tr}\left(\left(\prod_{l=0}^{k-1}\partial(a_{l})\right)\partial(a_{k}a_{k+1})\left(\prod_{l=k+2}^{p}\partial(a_{l})\right)\cdot
T\right)$ $\displaystyle={\rm
Tr}\left(\left(\prod_{l=0}^{k-1}\partial(a_{l})\right)a_{k}\left(\prod_{l=k+1}^{p}\partial(a_{l})\right)\cdot
T\right)$ $\displaystyle\quad+{\rm
Tr}\left(\left(\prod_{l=0}^{k}\partial(a_{l})\right)a_{k+1}\left(\prod_{l=k+2}^{p}\partial(a_{l})\right)\cdot
T\right)$ $\displaystyle=X_{k}+X_{k+1}.$

We also have

$\displaystyle\theta_{s}($ $\displaystyle a_{p}a_{0}\otimes
a_{1}\otimes\cdots\otimes a_{p-1})$ $\displaystyle={\rm
Tr}\left(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)\cdot Ta_{p}\right)$
$\displaystyle\quad+{\rm
Tr}\left(a_{0}\left(\prod_{l=1}^{p-1}\partial(a_{k})\right)\cdot
T\partial(a_{p})\right)$ $\displaystyle=X_{p}+{\rm
Tr}\left(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)\cdot[T,a_{p}]\right)$
$\displaystyle\quad+X_{0}+{\rm
Tr}\left(a_{0}\left(\prod_{k=1}^{p-1}\partial(a_{k})\right)\cdot[T,\partial(a_{p})]\right).$

If we assume that $p$ is even, then:

$\displaystyle(b\theta_{s})($ $\displaystyle a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle=\left(\sum_{k=0}^{p-1}(-1)^{k}(X_{k}+X_{k+1})\right)+(X_{p}+X_{0})$
$\displaystyle\quad+{\rm
Tr}\left(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)\cdot[T,a_{p}]\right)+{\rm
Tr}\left(a_{0}\left(\prod_{k=1}^{p-1}\partial(a_{k})\right)\cdot[T,\partial(a_{p})]\right)$
$\displaystyle=2X_{0}+{\rm
Tr}\left(a_{0}\left(\prod_{k=1}^{p-1}\partial(a_{k})\right)[T,\partial
a_{p}]\right)+{\rm
Tr}\left(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)[T,a_{p}]\right)$
$\displaystyle=2{\rm
Tr}\left(a_{0}\left(\prod_{k=1}^{p}\partial(a_{k})\right)\cdot T\right)+{\rm
Tr}\left(a_{0}\left(\prod_{k=1}^{p-1}\partial(a_{k})\right)[T,\partial
a_{p}]\right)$ $\displaystyle\quad+{\rm
Tr}\left(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)[T,a_{p}]\right).$

This completes the computation of the coboundaries.

###  A.4. Technical estimates for Section 4.4.1

For this section, $(\mathcal{A},H,D)$ is assumed to be a spectral triple
satisfying Hypothesis 1.2.1 and we assume that $D$ has a spectral gap at $0$.

The results of this section are very similar to that of Lemma 4.1.3. However
additional technicalities make the proofs more involved and therefore are
included here in the appendix. We make use of the mappings $b_{k}$ from
Definition 4.1.1, defined in terms of $m\geq 1$ and a subset
$\mathscr{B}\subseteq\\{1,\ldots,m\\}$ on $a\in\mathcal{A}$

$b_{k}(a):=\begin{cases}\delta(a),\quad k\in\mathscr{B},\\\ [F,a],\quad
k\notin\mathscr{B}.\end{cases}$

######  Lemma A.4.1.

Let $m\geq 1$ and Let $n\in\mathbb{Z}$. Then for all
$\mathscr{B}\subseteq\\{1,\ldots,m\\}$ the operator on $H_{\infty}$ given by:

$|D|^{-n}\left(\prod_{k=1}^{m}b_{k}(a_{k})\right)|D|^{n+m-|\mathscr{B}|}$

has bounded extension, where $b_{k}$ is defined in terms of $\mathscr{B}$.

###### Proof.

We prove the assertion by induction on $m.$ If $m=1$ then there are two
possible cases, $\mathscr{B}=\emptyset$ and $\mathscr{B}=\\{1\\}$.

If $\mathscr{B}=\emptyset$, then on $H_{\infty}$ we have:

$\displaystyle|D|^{-n}\left(\prod_{k=1}^{m}b_{k}(a_{k})\right)|D|^{n+m-|\mathscr{B}|}$
$\displaystyle=|D|^{-n}[F,a_{1}]|D|^{n+1}$
$\displaystyle=|D|^{-n}L(a_{1})|D|^{n}$
$\displaystyle=|D|^{-n}\partial(a_{1})|D|^{n}-F|D|^{-n}\delta(a_{1})|D|^{n}.$

By Lemma A.1.2, the operators $|D|^{-n}\partial(a_{1})|D|^{n}$ and
$|D|^{-n}\delta(a_{1})|D|^{n}$ have bounded extension, so this proves the
$\mathscr{B}=\emptyset$ case. On the other hand, if $\mathscr{B}=\\{1\\},$
then on $H_{\infty}$:

$|D|^{-n}\left(\prod_{k=1}^{m}b_{k}(a_{k})\right)|D|^{n+m-|\mathscr{B}|}=|D|^{-n}\delta(a_{1})|D|^{n}.$

Again by Lemma A.1.2, the operator $|D|^{-n}\delta(a_{1})|D|^{n}$ has bounded
extension. This completes the proof of the $\mathcal{B}=\\{1\\}$ case.

Now assume that $m>1$ and the assertion is true for $m-1$. Define
$\mathscr{C}:=\mathscr{B}\setminus\\{m\\}.$ and let
$n_{1}=n+(m-1)-|\mathscr{C}|$. Then on $H_{\infty}$:

$\displaystyle|D|^{-n}$
$\displaystyle\left(\prod_{k=1}^{m}b_{k}(a_{k})\right)|D|^{n+m-|\mathscr{B}|}$
(A.6)
$\displaystyle=\left(|D|^{-n}\left(\prod_{k=1}^{m-1}b_{k}(a_{k})\right)|D|^{n+(m-1)-|\mathscr{C}|}\right)\left(|D|^{-n_{1}}b_{m}(a_{m})|D|^{n_{1}+1-|\mathscr{B}\cap\\{m\\}|}\right).$

By the inductive assumption, the first factor has bounded extension.

If $m\in\mathscr{B},$ then the second factor in (A.6) is:

$|D|^{-n_{1}}\delta(a_{m})|D|^{n_{1}}$

which has bounded extension by Lemma A.1.2.(i). On the other hand, if
$m\notin\mathscr{B},$ then the second factor in (A.6) is

$|D|^{-n_{1}}F(a_{m})|D|^{n_{1}+1}=|D|^{-n_{1}}(\partial(a_{m})-F\delta(a_{m}))|D|^{n_{1}}$

which also has bounded extension by Lemma A.1.2.(i). In either case, the
second factor of (A.6) has bounded extension. So the assertion is proved for
$m$, completing the proof by induction. ∎

######  Lemma A.4.2.

Let $\mathscr{A}\subseteq\\{1,\cdots,p\\}$ Assume that there is $m>1$ be such
that $m-1,m\in\mathscr{A}.$ The operator on $H_{\infty}$ given by:

$\Gamma
a_{0}\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)\delta^{2}(a_{m-1})\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)|D|^{p-|\mathscr{A}|}$

has bounded extension.

###### Proof.

Let $n_{1}=|\\{1,\cdots,m-2\\}\setminus\mathscr{A}|$ and
$n_{2}=|\\{m+1,\cdots,p\\}\setminus\mathscr{A}|$ so that immediately
$n_{1}\leq m-2-|\mathscr{A}|$ and $n_{2}\leq p-m-|\mathscr{A}|$. On
$H_{\infty}$ we have

$\Gamma
a_{0}\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)\delta^{2}(a_{m-1})\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)|D|^{p-|\mathscr{A}|}=\mathrm{I}\cdot\mathrm{II}\cdot\mathrm{III}.$

Here,

$\displaystyle\mathrm{I}$ $\displaystyle=\Gamma
a_{0}\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)|D|^{n_{1}},$
$\displaystyle\mathrm{II}$
$\displaystyle=|D|^{-n_{1}}\delta^{2}(a_{m-1})|D|^{n_{1}},$
$\displaystyle\mathrm{III}$
$\displaystyle=|D|^{-n_{1}}\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)|D|^{n_{1}+n_{2}}.$

The operators $\mathrm{I}$ and $\mathrm{III}$ have bounded extension by Lemma
A.4.1. On the other hand, $\mathrm{II}$ has bounded extension due to Lemma
A.1.2. ∎

######  Lemma A.4.3.

Let $\mathscr{A}\subset\\{1,\cdots,p\\}$ and assume that there is $m>1$ be
such that $m-1\in\mathscr{A}$ and $m\notin\mathscr{A}.$ The operator on
$H_{\infty}$ given by

$\displaystyle\Gamma
a_{0}\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)[F,\delta(a_{m-1})]\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)|D|^{p-|\mathscr{A}|}$

has bounded extension.

###### Proof.

Let $n_{1}=|\\{1,\cdots,m-2\\}\setminus\mathscr{A}|$ and
$n_{2}=|\\{m+1,\cdots,p\\}\setminus\mathscr{A}|,$ so that we immediately have
$n_{1}\leq m-2-|\mathscr{A}|$ and $n_{2}\leq p-m-|\mathscr{A}|$ as in Lemma
A.4.2. We have

$\Gamma
a_{0}\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)[F,\delta(a_{m-1})]\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)|D|^{p-|\mathscr{A}|}=\Gamma
a_{0}\cdot\mathrm{I}\cdot\mathrm{II}\cdot\mathrm{III}.$

Here,

$\displaystyle\mathrm{I}$
$\displaystyle=\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)|D|^{n_{1}},$
$\displaystyle\mathrm{II}$
$\displaystyle=|D|^{-n_{1}}[F,\delta(a_{m-1})]|D|^{1+n_{1}},$
$\displaystyle\mathrm{III}$
$\displaystyle=|D|^{-n_{1}-1}\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)|D|^{1+n_{1}+n_{2}}.$

The operators $\mathrm{I}$ and $\mathrm{III}$ have bounded extension by Lemma
A.4.1. On the other hand,

$\mathrm{II}=|D|^{-n_{1}}(\partial(\delta(a_{m-1}))-F\delta^{2}(a_{m-1}))|D|^{n_{1}}$

which has bounded extension by Lemma A.1.2.(i). ∎

######  Lemma A.4.4.

For every $m\geq 1,$ the operator on $H_{\infty}$ given by

$\Big{(}\prod_{k=0}^{m}[F,a_{k}]\Big{)}|D|^{m+1}$

has bounded extension.

###### Proof.

We prove the assertion by induction on $m.$ For $m=0,$ on $H_{\infty}$ we have

$[F,a_{0}]|D|=\partial(a_{0})-F\delta(a_{0})$

which has has bounded extension.

Now let $m>1$ and assume that the assertion is true for $m-1$. On $H_{\infty}$
we write

$\Big{(}\prod_{k=0}^{m}[F,a_{k}]\Big{)}|D|^{m+1}=\Big{(}\Big{(}\prod_{k=0}^{m-1}[F,a_{k}]\Big{)}|D|^{m}\Big{)}\cdot\Big{(}|D|^{-m}[F,a_{m}]|D|^{m+1}\Big{)}.$

The first factor has bounded extension by the induction assumption. The second
factor is

$|D|^{-m}L(a_{m})|D|^{m}=|D|^{-m}\partial(a_{m})|D|^{m}-F\cdot|D|^{-m}\delta(a_{m})|D|^{m}$

which has bounded extension by Lemma A.1.2.(i). Hence, the assertion holds for
$m$, completing the inductive proof. ∎

###  A.5. Subkhankulov’s computation

The following assertion is identical to [61, Lemma 2.1.1]. However to the best
of our knowledge there is no published proof in English, and [61] is not
easily accessible. For the convenience of the reader we include a proof here.

######  Proposition A.5.1.

For all $u\in(0,1)$ and $v\in\mathbb{R},$ we have

$\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{u+it}e^{(u+it)v}dt=(1+u^{2})^{2}\chi_{[0,\infty]}(v)+e^{uv}\cdot\min\\{1,v^{-2}\\}\cdot
O(1).$

###### Proof.

We will deal separately with the $v\geq 0$ and $v<0$ cases. First, assume that
$v\geq 0$.

Let $\gamma_{0}$ be a smooth curve in $\mathbb{C}$ without self-intersections
such that

  1. (1)

$\gamma_{0}$ starts at $i$ and ends at $-i.$

  2. (2)

$\gamma_{0}$ lies in the half-plane $\\{\Re(z)\leq 0\\},$

  3. (3)

The distance between $\gamma_{0}$ and the interval $[-1,0]$ is greater than or
equal to $1$,

  4. (4)

the length of $\gamma_{0}$ is at most $10.$

  5. (5)

$\gamma_{0}$ is contained in the disc $\\{z\;:\;|z|\leq 10\\}$.

Let $\gamma_{1}$ be the interval starting at $-i$ and ending at $i$ and let
the contour $\gamma$ be the concatenation of $\gamma_{0}$ and $\gamma_{1}.$

Define

$f(z):=\frac{(1+z^{2})^{2}}{u+z},\quad z\in\mathbb{C}\setminus\\{-u\\}.$

So that by definition:

(A.7)
$\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{u+it}e^{(u+it)v}\,dt=\frac{1}{2\pi
i}e^{uv}\int_{\gamma_{1}}f(z)e^{zv}\,dz.$

Since $z\mapsto e^{zv}$ is entire, the function $z\to f(z)e^{zv}$ is
holomorphic in the set $\mathbb{C}\setminus\\{-u\\}$ and has a simple pole at
at $z=-u$ with corresponding residue $(1+u^{2})^{2}e^{-uv}.$ By construction,
the point $-u$ is in the interior of the curve $\gamma$ and so by the Cauchy
integral formula we have:

$\frac{1}{2\pi i}\int_{\gamma}f(z)e^{zv}dz=(1+u^{2})^{2}e^{-uv}.$

Since $\gamma=\gamma_{0}\cup\gamma_{1}$:

(A.8) $\frac{1}{2\pi i}\int_{\gamma_{0}}f(z)e^{zv}dz+\frac{1}{2\pi
i}\int_{\gamma_{1}}f(z)e^{zv}dz=(1+u^{2})^{2}e^{-uv}.$

By definition $\gamma_{0}$ has length at most $10$, so by the triangle
inequality we have the bound:

$\left|\int_{\gamma_{0}}f(z)e^{zv}dz\right|\leq
10\sup_{z\in\gamma_{0}}|f(z)||e^{zv}|.$

Since $\gamma_{0}$ is contained in the half-plane $\\{z\;:\;\Re(z)\leq 0\\}$
we also have that $\sup_{z\in\gamma_{0}}|e^{zv}|\leq 1$ and therefore:

$\left|\int_{\gamma_{0}}f(z)e^{zv}dz\right|\leq
10\sup_{z\in\gamma_{0}}|f(z)|.$

For $z\in\gamma$ we have that $|z|\leq 10$ and $|u+z|\geq 1$ so it follows
that

$\sup_{z\in\gamma_{0}}|f(z)|\leq(1+10^{2})^{2}\leq 10^{5}.$

Therefore, we have

(A.9) $\int_{\gamma_{0}}f(z)e^{zv}dz=O(1).$

Using integration by parts twice and taking into account that $f(\pm
i)=f^{\prime}(\pm i)=0,$ we obtain

$\int_{\gamma_{0}}f(z)e^{zv}dz=v^{-2}\int_{\gamma_{0}}f^{\prime\prime}(z)e^{vz}dz.$

Thus,

$v^{2}\left|\int_{\gamma_{0}}f(z)e^{zv}dz\right|\leq
10\sup_{z\in\gamma_{0}}|f^{\prime\prime}(z)|.$

We may compute $f^{\prime\prime}(z)$ directly as:

$f^{\prime\prime}(z)=(4+12z)\frac{1}{u+z}-8z(1+z^{2})\frac{1}{(u+z)^{2}}+(1+z^{2})^{2}\frac{2}{(u+z)^{3}}.$

Since $|z|\leq 10$ and $|u+z|\geq 1$ for every $z\in\gamma_{0},$ it follows
that

$\displaystyle\sup_{z\in\gamma_{0}}|f^{\prime\prime}(z)|$
$\displaystyle\leq(4+12\cdot 10)+8\cdot
10\cdot(1+10^{2})+2\cdot(1+10^{2})^{2}$ $\displaystyle\leq 10^{5}.$

Therefore, we have:

(A.10) $\int_{\gamma_{0}}f(z)e^{zv}dz=O(v^{-2}),$

Hence,

(A.11) $\int_{\gamma_{0}}f(z)e^{zv}dz=\min\\{1,v^{-2}\\}\cdot O(1).$

Combining (A.8) and (A.11), we obtain

$\frac{1}{2\pi
i}\int_{\gamma_{1}}f(z)e^{zv}dz=(1+u^{2})^{2}e^{-uv}+\min\\{1,v^{-2}\\}\cdot
O(1).$

So using (A.7) and (A.8):

$\frac{1}{2\pi}\int_{-1}^{1}\frac{(1-t^{2})^{2}}{u+it}e^{(u+it)v}dt=(1+u^{2})^{2}+e^{uv}\cdot\min\\{1,v^{-2}\\}\cdot
O(1).$

This completes the proof of the $v\geq 0$.

Now assume that $v<0$. This proof is similar, but instead we consider a
contour in the half plane $\\{z\;:\;\Re(z)\geq 0\\}$. Let $\gamma_{2}$ be a
smooth curve without self-intersections such that

  1. (1)

$\gamma_{2}$ starts at $-i$ and ends at $i.$

  2. (2)

$\gamma_{2}$ lies in the half-plane $\\{\Re(z)\geq 0\\}.$

  3. (3)

the distance between $\gamma_{2}$ and $[-1,0]$ is greater than or equal to
$1$.

  4. (4)

the length of $\gamma_{2}$ is at most $10.$

  5. (5)

$\gamma_{2}$ is contained in the disc $\\{z\;:\;|z|\leq 10\\}$.

As in the $v\geq 0$ case, $\gamma_{1}$ denotes the interval joining $-i$ and
$i$, and now write $\gamma^{\prime}$ for the concatenation of $\gamma_{1}$ and
$\gamma_{2}$.

Since $f$ is holomorphic in the half-plane $\Re(z)\geq 0$, we have:

$\int_{\gamma^{\prime}}f(z)e^{zv}dz=0.$

Since $\gamma^{\prime}=\gamma_{1}\cup\gamma_{2}$ we have:

(A.12) $\frac{1}{2\pi i}\int_{\gamma_{1}}f(z)e^{zv}dz+\frac{1}{2\pi
i}\int_{\gamma_{2}}f(z)e^{zv}dz=0.$

Since by definition $\gamma_{2}$ has length at most $10$, and we are assuming
$v<0$ we have:,

$\displaystyle\left|\int_{\gamma_{2}}f(z)e^{zv}dz\right|$ $\displaystyle\leq
10\sup_{z\in\gamma_{2}}|f(z)e^{zv}|$ $\displaystyle\leq
10\sup_{z\in\gamma_{2}}|f(z)|$ $\displaystyle\leq 10(1+10^{2})^{2}.$
$\displaystyle=O(1).$

Using integration by parts and taking into account once again that $f(\pm
i)=f^{\prime}(\pm i)=0,$ we obtain in also in the $v<0$ case that:

$\int_{\gamma_{2}}f(z)e^{zv}dz=v^{-2}\int_{\gamma_{2}}f^{\prime\prime}(z)e^{vz}dz.$

Thus,

$\displaystyle v^{2}\left|\int_{\gamma_{2}}f(z)e^{zv}dz\right|$
$\displaystyle\leq 10\sup_{z\in\gamma_{2}}|f^{\prime\prime}(z)||e^{zv}|$
$\displaystyle\leq 10^{6}$ $\displaystyle=O(1).$

Therefore,

(A.13) $\int_{\gamma_{2}}f(z)e^{zv}dz=\min\\{1,v^{-2}\\}\cdot O(1).$

Combining (A.12) and (A.13), we obtain

$\frac{1}{2\pi i}\int_{\gamma_{1}}f(z)e^{zv}dz=\min\\{1,v^{-2}\\}\cdot O(1).$

Hence, by (A.7) we conclude the proof for the $v<0$ case. ∎

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