elliptic operators on compact manifolds in [41, Theorem 5.2].

######  Lemma 3.4.5.

Let $g$ be a metric on $\mathbb{T}^{p}$, and let $D_{g}$ denote the
corresponding Dirac operator. Then there is a constant $C_{g}>0$ such that for
all $v\in L_{2}\Omega(\mathbb{T}^{p},g)$ such that $D_{g}v\in
L_{2}\Omega(\mathbb{T}_{p},g)$, we have

$\|v\|_{H^{1}\Omega(\mathbb{T}^{p},g)}\leq
C_{g}(\|v\|_{L_{2}\Omega(\mathbb{T}^{p})}+\|D_{g}v\|_{L_{2}\Omega(\mathbb{T}^{p},g)}).$

The following lemma is the essential technical result allowing us to transfer
the Cwikel-type estimates for $\mathbb{T}^{p}$ with the flat metric to
$\mathbb{T}^{p}$ with an arbitrary metric.

######  Lemma 3.4.6.

Let $g$ be an arbitrary metric on $\mathbb{T}^{p}$, and let $g_{0}$ be the
flat metric on $\mathbb{T}^{p}$. Denote by $D_{g}$ and $D_{0}$ the Hodge-Dirac
operators corresponding to $g$ and $g_{0}$ respectively. Then the operator

$D_{0}(D_{g}+i)^{-1}$

defined initially on $\Omega(\mathbb{T}^{p})$ has bounded extension to
$L_{2}\Omega(\mathbb{T}^{p},g)$ (or equivalently
$L_{2}\Omega(\mathbb{T}^{p},g_{0})$).

###### Proof.

By the definition of the Sobolev norm
$\|\cdot\|_{H^{1}\Omega(\mathbb{T}^{p},g)}$, for all $v\in
H^{1}\Omega(\mathbb{T}^{p},g)$ we have:

$\|D_{0}v\|_{L_{2}\Omega(\mathbb{T}^{p},g)}^{2}\leq\|v\|_{H^{1}\Omega(\mathbb{T}^{p},g)}^{2}.$

Thus by Lemma 3.4.5, there is a constant $C$ such that,

$\|D_{0}v\|_{L_{2}\Omega(\mathbb{T}^{p},g)}\leq
C(\|D_{g}v\|_{L_{2}\Omega(\mathbb{T}^{p},g)}+\|v\|_{L_{2}\Omega(\mathbb{T}^{p},g)}).$

Hence, if $v\in\mathrm{dom}(D_{g})$ then $v\in\mathrm{dom}(D_{0})$, and so if
$u\in L_{2}\Omega(\mathbb{T}^{p},g)$ and $v=(D_{g}+i)^{-1}u$ then
$v\in\mathrm{dom}(D_{0})$. So substituting $v=(D_{g}+i)^{-1}u$ we obtain:

$\|D_{0}(D_{g}+i)^{-1}u\|_{L_{2}\Omega(\mathbb{T}^{p},g)}\leq
C(1+\|(D_{g}+i)^{-1}\|_{\mathcal{L}_{\infty}(L_{2}\Omega(\mathbb{T}^{p},g)})\|u\|_{L_{2}\Omega(\mathbb{T}^{p},g)}.$

However since $(D_{g}+i)^{-1}$ is bounded, we get:

$\|D_{0}(D_{g}+i)^{-1}u\|_{L_{2}\Omega(\mathbb{T}^{p},g)}\leq
C\|u\|_{L_{2}\Omega(\mathbb{T}^{p},g)}.$

∎

As a consequence of Lemmas 3.4.2 and 3.4.6, we get:

######  Corollary 3.4.7.

Let $g$ be an arbitrary metric on $\mathbb{T}^{p}$ and let $D_{g}$ be the
corresponding Hodge-Dirac operator. Then,

  1. (i)

We have,

$(D_{g}+i)^{-1}\in\mathcal{L}_{p,\infty}(L_{2}\Omega(\mathbb{T}^{p},g)).$

  2. (ii)

For $\lambda\to\infty$, we have

$\|(D_{g}+i\lambda)^{-1}\|_{\mathcal{L}_{p+1}(L_{2}\Omega(\mathbb{T}^{p},g))}=O(\lambda^{-\frac{1}{p+1}}).$

###### Proof.

Part (i) follows immediately from Lemma 3.4.2 and Lemma 3.4.6.i.

Now we prove (ii). First we compute $(D_{0}+i\lambda)(D_{g}+i\lambda)^{-1}$
(working on the dense domain $\Omega(\mathbb{T}^{p})$):

$\displaystyle(D_{0}+i\lambda)(D_{g}+i\lambda)^{-1}$
$\displaystyle=(D_{0}-D_{g}+D_{g}+i\lambda)(D_{g}+i\lambda)^{-1}$
$\displaystyle=1+(D_{0}-D_{g})(D_{g}+i\lambda)^{-1}$
$\displaystyle=1+D_{0}(D_{g}+i)^{-1}\frac{D_{g}+i}{D_{g}+i\lambda}-\frac{D_{g}}{D_{g}+i\lambda}.$

By Lemma 3.4.6, the operator $D_{0}(D_{g}+i)^{-1}$ has bounded extension.
Moreover, by functional calculus the operators
$\frac{D_{g}+i}{D_{g}+i\lambda}$ and $\frac{D_{g}}{D_{g}+i\lambda}$ have
bounded extension with norm bounded by a constant independent of $\lambda$.
Thus,

$\|(D_{0}+i\lambda)(D_{g}+i\lambda)^{-1}\|_{\mathcal{L}_{\infty}(L_{2}\Omega(\mathbb{T}^{p},g)}=O(1).$

Now Lemma 3.4.2.(ii) yields the result. ∎

With the above results in hand we are able to establish our first Cwikel-type
result for $(X,g).$ A similar result to Lemma 3.4.8 can also be found in [55,
Proposition 13]. The result in [55] is very similar in nature (despite
applying to the somewhat different situation of Riemannian spin manifolds).
The method of proof here is different: we use reduce the problem to
$\mathbb{T}^{p}$ by using local coordinates, rather than a doubling
construction as employed in [55].

######  Lemma 3.4.8.

Let $f\in C^{\infty}_{c}(X).$ We have

$M_{f}(D_{g}+i)^{-1}\in\mathcal{L}_{p,\infty}(L_{2}\Omega(X,g)).$

###### Proof.

By using a partition of unity if necessary, we may assume without loss of
generality that $f$ is supported in a single chart $(U,h)$ where
$h:U\to\mathbb{R}^{p}$ is a homeomorphism onto its image, and since $f$ has
compact support we may further assume without loss of generality that $h(U)$
is bounded. Since $h(U)$ is bounded, there is a sufficiently large box
$[-N,N]^{p}$ with $h(U)$ in the interior of $[-N,N]^{p}$. By applying a
translation and dilation if necessary, we may assume without loss of
generality that $h(U)$ is contained within the interior of the box
$[0,1]^{p}$. By identifying the edges of $[0,1]^{p}$, we may view $h$ as a
continuous function $h:U\to\mathbb{T}^{p}$.

We define three smooth “cut-off” functions $\phi_{1},\phi_{2},\phi_{3}$
compactly supported in $h(U)$, defined so that for each $j=1,2,3$ we have
$0\leq\phi_{j}\leq 1$, and

  1. (a)

for all $x\in U$, $\phi_{1}(h(x))f(x)=f(x)$,

  2. (b)

we have $\phi_{2}\phi_{1}=\phi_{1}$,

  3. (c)

we have $\phi_{3}\phi_{2}=\phi_{2}$.

In other words, on $\mathrm{supp}(f\circ h^{-1})$ we have $\phi_{1}=1$, and on
$\mathrm{supp}(\phi_{1})$ we have $\phi_{2}=1$ and on
$\mathrm{supp}(\phi_{2})$ we have $\phi_{3}=1$.

For $j=1,2,3$, we also define the function $\psi_{j}$ by pulling back
$\phi_{j}$ to $X$:

$\psi_{j}(x)=\begin{cases}(\phi_{j}\circ h)(x),\quad x\in U.\\\ 0,\quad
x\notin U.\end{cases}$

Since $\phi_{j}$ is compactly supported in $h(U)$, the function $\psi_{j}$ is
smooth and compactly supported in $U$.

Let $g_{0}$ denote the flat metric on $\mathbb{T}^{p}$. The metric $g$ can be
pushed forward by $h$ to a metric $h^{*}g$ on $h(U)$. We then define a new
metric $g_{1}$ on $\mathbb{T}^{p}$ by:

$g_{1}:=(h^{*}g)\phi_{3}+g_{0}(1-\phi_{3}).$

Since $\phi_{3}$ is compactly supported in $h(U)$, the metric $g_{1}$ is well
defined. Moreover, on $\mathrm{supp}(\phi_{2})$ we have $\phi_{3}=1$ so on
$\mathrm{supp}(\phi_{2})$ the metric $g_{1}$ is identical to $h^{*}g$.

We define a partial isometry $V:L_{2}\Omega(X,g)\to
L_{2}\Omega(\mathbb{T}^{p},g_{1})$ on $\xi\in L_{2}\Omega(X,g)$,
$z\in\mathbb{T}^{p}$ by:

$V\xi(z)=\begin{cases}\xi\circ h^{-1}(z),\quad z\in\mathrm{supp}(\phi_{2}),\\\
0,\quad z\notin\mathrm{supp}(\phi_{2}).\end{cases}$

By construction, $V$ induces an isometry from
$L_{2}\Omega(\mathrm{supp}(\psi_{2}),g)\to
L_{2}\Omega(\mathrm{supp}(\phi_{2}),g_{1})$, and

$\displaystyle VV^{*}$ $\displaystyle=M_{\chi_{\mathrm{supp}{\phi_{2}}}},$
$\displaystyle V^{*}V$ $\displaystyle=M_{\chi_{\mathrm{supp}{\psi_{2}}}}.$

We also have that if $j=1,2$ then

$M_{\phi_{j}}V=VM_{\psi_{j}}.$

We use the important fact that for $j=1,2$, we have an equality on
$\Omega_{c}(X)$:

$V^{*}D_{g_{1}}M_{\phi_{j}}=D_{g}M_{\psi_{j}}V^{*}.$

Next we consider the following two operators on
$L_{2}\Omega(\mathbb{T}^{p},g_{1})$.

$\displaystyle P:=M_{\phi_{1}}(D_{g_{1}}+i)^{-1}M_{\phi_{1}}$ $\displaystyle
Q:=M_{\phi_{2}}(D_{g_{1}}+i)^{-1}M_{\phi_{2}}.$

By Lemma 3.4.7, the operators $P$ and $Q$ are in
$\mathcal{L}_{p,\infty}(L_{2}\Omega(\mathbb{T}^{p},g_{1}))$.

We now consider the operator $(D_{g}+i)M_{\psi_{2}}V^{*}PV$. We note that this
operator is well defined on $\Omega_{c}(X)$, since if $u\in\Omega_{c}(X)$,
then $M_{\psi_{2}}V^{*}PV$ is smooth and supported in
$\mathrm{supp}{\psi_{2}}$. Hence (working on $\Omega_{c}(X)$):

$\displaystyle(D_{g}+i)M_{\psi_{2}}V^{*}PV$
$\displaystyle=V^{*}(D_{g_{1}}+i)M_{\phi_{2}}PV$
$\displaystyle=V^{*}(D_{g_{1}}+i)M_{\phi_{1}}(D_{g_{1}}+i)^{-1}M_{\phi_{1}}V$
$\displaystyle=V^{*}([D_{g_{1}},M_{\phi_{1}}](D_{g_{1}}+i)^{-1}M_{\phi_{1}}+M_{\phi_{1}}^{2})V.$

Now recalling that $D_{g_{1}}$ is a local operator, we have
$[D_{g_{1}},M_{\phi_{1}}]=[D_{g_{1}},M_{\phi_{1}}]M_{\phi_{2}}$.

Moreover, $V^{*}M_{\phi_{1}}^{2}V=M_{\psi_{1}}^{2}$ and so

$(D_{g}+i)M_{\psi_{2}}V^{*}PV=V^{*}[D_{g_{1}},M_{\phi_{1}}]QM_{\phi_{1}}V+M_{\psi_{1}}^{2}.$

Now multiplying on the left by $(D_{g}+i)^{-1}$, we arrive at:

$M_{\psi_{2}}V^{*}PV=(D_{g}+i)^{-1}V^{*}[D_{g_{1}},M_{\phi_{1}}]QM_{\phi_{1}}V+(D_{g}+i)^{-1}M_{\psi_{1}}^{2}.$

The final step is to use the fact that $\psi_{2}=1$ on the support of $f$, so
we may use $M_{\psi_{2}}^{2}M_{f}=M_{f}$, and multiply on the right by $M_{f}$
to obtain:

$M_{\psi_{2}}V^{*}PVM_{f}=(D_{g}+i)^{-1}V^{*}[D_{1},M_{\phi_{1}}]QM_{\phi_{1}}VM_{f}+(D_{g}+i)^{-1}M_{f}.$

Since both $P$ and $Q$ are in
$\mathcal{L}_{p,\infty}(L_{2}\Omega(\mathbb{T}^{p},g_{1}))$, we finally obtain
that $(D_{g}+i)^{-1}M_{f}\in\mathcal{L}_{p,\infty}(L_{2}\Omega(X,g))$. ∎

######  Lemma 3.4.9.

Let $f\in C^{\infty}_{c}(X).$ Then:

$\|M_{f}(D_{g}+i\lambda)^{-1}\|_{\mathcal{L}_{p+1}(L_{2}\Omega(X,g))}=O(\lambda^{-\frac{1}{p+1}}),\quad\lambda\to\infty.$

###### Proof.

This proof proceeds along similar lines to Lemma 3.4.8. We again assume
without loss of generality that $f$ is supported in a single chart $(U,h)$,
and construct the metric $g_{1}$ on $\mathbb{T}^{p}$ and the partial isometry
$V$ identically to the proof of Lemma 3.4.8. We also use the same cut-off
functions $\phi_{1}$, $\phi_{2}$ and $\phi_{3}$, and
$\psi_{1},\psi_{2},\psi_{3}$.

In place of the operators $P$ and $Q$, we introduce $P_{\lambda}$ and
$Q_{\lambda}$ given by:

$\displaystyle P_{\lambda}$
$\displaystyle=M_{\phi_{1}}(D_{g_{1}}+i\lambda)^{-1}M_{\phi_{1}}$
$\displaystyle Q_{\lambda}$
$\displaystyle=M_{\phi_{2}}(D_{g_{1}}+i\lambda)^{-1}M_{\phi_{2}}.$

Following the argument of Lemma 3.4.8 with $P_{\lambda}$ and $Q_{\lambda}$ in
place of $P$ and $Q$, we arrive at:

$M_{\psi_{2}}V^{*}P_{\lambda}VM_{f}=(D_{g}+i\lambda)^{-1}V^{*}[D_{1},M_{\phi_{1}}]Q_{\lambda}M_{\phi_{1}}VM_{f}+(D_{g}+i\lambda)^{-1}M_{f}.$

Due to Lemma 3.4.7.(i), we have
$\|P_{\lambda}\|_{p+1}=O(\lambda^{-\frac{1}{p+1}})$ and similarly for
$Q_{\lambda}$. Thus,

$\|(D_{g}+i\lambda)^{-1}\|_{\mathcal{L}_{p+1}(L_{2}\Omega(X,g))}=O(\lambda^{-\frac{1}{p+1}}).$

∎

We now finally have the results necessary to prove Theorem 3.4.1.

###### Proof of Theorem 3.4.1.

We will instead work with Hypothesis 3.2.7, as justified by Theorem 3.2.9.

First, we show that 3.2.7.(ii) holds for the triple
$(C^{\infty}_{c}(X),L_{2}\Omega(X,g),D_{g})$.

To this end let $f\in C^{\infty}_{c}(X)$. We will prove by induction that for
all $j\geq 1$, we have

(3.11)
$M_{f}(D_{g}+i)^{-j}\in\mathcal{L}_{\frac{p}{j},\infty}(L_{2}\Omega(X,g)).$

The case $j=1$ is already established by Lemma 3.4.8.

Suppose now that (3.11) holds for $j\geq 1$. Choose $\phi\in
C^{\infty}_{c}(X)$ such that $f\phi=f$. Then,

$\displaystyle M_{f}(D_{g}+i)^{-j-1}$
$\displaystyle=M_{\phi}M_{f}(D_{g}+i)^{-1}\cdot(D_{g}+i)^{-j}$
$\displaystyle=-M_{\phi}[(D_{g}+i)^{-1},M_{f}](D_{g}+i)^{-j}+M_{\phi}(D_{g}+i)^{-1}M_{f}(D_{g}+i)^{-j}$
$\displaystyle=M_{\phi}(D_{g}+i)^{-1}[D,M_{f}](D_{g}+i)^{-j-1}+M_{\phi}(D_{g}+i)^{-1}M_{f}(D_{g}+i)^{-j}$
$\displaystyle=M_{\phi}(D_{g}+i)^{-1}[D,M_{f}]M_{\phi}(D_{g}+i)^{-j-1}+M_{\phi}(D_{g}+i)^{-1}M_{f}(D_{g}+i)^{-j}.$

Due to Lemma 3.4.8, we have $M_{\phi}(D_{g}+i)^{-1}\in\mathcal{L}_{p,\infty}$,
and by the inductive assumption we also have
$M_{\phi}(D_{g}+i)^{-j}\in\mathcal{L}_{\frac{p}{j},\infty}$. Then
$M_{f}(D_{g}+i)^{-j-1}\in\mathcal{L}_{p,\infty}\cdot\mathcal{L}_{\frac{p}{j},\infty}$,
so applying the Hölder inequality we arrive at
$M_{f}(D_{g}+i)^{-j-1}\in\mathcal{L}_{\frac{p}{j+1},\infty}$. Taking $j=p$, we
get that $M_{f}(D_{g}+i)^{-p}\in\mathcal{L}_{1,\infty}(L_{2}\Omega(X,g))$.

Similarly, since $D_{g}$ is a local operator, we have that
$[D_{g},M_{f}]=[D_{g},M_{f}]M_{\phi}$. Hence,
$[D_{g},M_{f}](D_{g}+i)^{-p}\in\mathcal{L}_{1,\infty}(L_{2}\Omega(X,g))$. This
completes the proof of Hypothesis 3.2.7.(ii) in the case $k=0$.

What remains is to show that Hypothesis 3.2.7.(iii) holds. We will first deal
with the $k=0$ case. To that end, we will show by induction that for all
$j\geq 1$:

(3.12)
$\|M_{f}(D_{g}+i\lambda)^{-j}\|_{\frac{p+1}{j}}=O(\lambda^{-\frac{j}{p+1}}),\lambda\to\infty.$

The base case $j=1$ is the result of Lemma 3.4.9, and the case $j=p+1$ is what
is required for Hypothesis 3.2.7.

Suppose now that (3.12) holds for $j\geq 1$, and again choose $\phi\in
C^{\infty}_{c}(X)$ such that $f\phi=f$. Then,

$\displaystyle M_{f}(D_{g}+i\lambda)^{-j-1}$ $\displaystyle=M_{\phi}\cdot
M_{f}(D_{g}+i\lambda)^{-1}(D_{g}+i\lambda)^{-j}$
$\displaystyle=-M_{\phi}[(D_{g}+i\lambda)^{-1},M_{f}](D_{g}+i\lambda)^{-j}$
$\displaystyle\quad+M_{\phi}(D_{g}+i\lambda)^{-1}M_{f}(D_{g}+i\lambda)^{-j}$
$\displaystyle=M_{\phi}(D_{g}+i\lambda)^{-1}[D,M_{f}]M_{\phi}(D_{g}+i\lambda)^{-j-1}$
$\displaystyle\quad+M_{\phi}(D_{g}+i\lambda)^{-1}M_{f}(D_{g}+i\lambda)^{-j}.$

By Lemma 3.4.9, we have
$\|M_{\phi}(D_{g}+i\lambda)^{-1}\|_{p+1}=O(\lambda^{-\frac{1}{p+1}})$, and by
the inductive assumption we also have
$\|M_{\phi}(D_{g}+i\lambda)^{-k}\|_{\frac{p+1}{j}}=O(\lambda^{-\frac{j}{p+1}})$.
Then by the Holder inequality,

$\|M_{f}(D_{g}+i\lambda)^{-j-1}\|_{\frac{p+1}{j+1}}\leq
O(\lambda^{-\frac{j}{p+1}})\cdot O(\lambda^{-\frac{1}{p+1}}).$

So
$\|M_{f}(D_{g}+i\lambda)^{-j-1}\|_{\frac{p+1}{j+1}}=O(\lambda^{-\frac{j+1}{p+1}})$.
To conclude the same for $\partial(f)$ in place of $f$, we once more use the
fact that $[D_{g},M_{f}]M_{\phi}=[D_{g},M_{f}]$. This completes the proof of
the $k=0$ case of Hypothesis 3.2.7.(iii).

For $k>0$, if $\phi f=f$, then we have

$\Lambda^{k}(M_{f})=\Lambda^{k}(M_{f})M_{\phi}$

so we may apply the $k=0$ case to $\phi$ to deduce the result.

That $(C_{c}^{\infty}(X),L_{2}\Omega(X,g),D_{g})$ satisfies 3.2.7.(i) follows
from similar reasoning. ∎

##  Chapter 4 Asymptotic of the heat trace

In this chapter we complete the proof of Theorem 1.2.2. This will require some
delicate computations exploiting Hochschild homology.

For the remainder of this chapter, we assume that $(\mathcal{A},H,D)$ is a
spectral triple satisfying Hypothesis 1.2.1. Furthermore we will need the
following auxiliary assumption:

######  Hypothesis 4.0.1.

The spectral triple $(\mathcal{A},H,D)$ satisfies the following:

  1. (i)

$(\mathcal{A},H,D)$ has the same parity as $p$: this means that
$(\mathcal{A},H,D)$ is even when $p$ is even (with grading $\Gamma$) and odd
when $p$ is odd.

  2. (ii)

$D$ has a spectral gap at $0$.

We will show at the end of this chapter how Hypothesis 4.0.1 can be removed.

Recall from Definition 2.2.15 the fundamental mappings $\mathrm{ch}$ and
$\Omega$, given by:

$\displaystyle\mathrm{ch}(a_{0}\otimes a_{1}\otimes\cdots\otimes a_{p})$
$\displaystyle:=\Gamma F[F,a_{0}][F,a_{1}]\cdots[F,a_{p}],$
$\displaystyle\Omega(a_{0}\otimes a_{1}\otimes\cdots\otimes a_{p})$
$\displaystyle:=\Gamma
a_{0}\partial(a_{1})\partial(a_{2})\cdots\partial(a_{p})$

and that Theorem 1.2.2 states that for all Hochschild cycles
$c\in\mathcal{A}^{\otimes(p+1)}$,

$\mathrm{Tr}(\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}D^{2}})=\frac{p}{2}\mathrm{Tr}(\mathrm{ch}(c))s^{-2}+O(s^{-1}),\quad
s\to 0.$

The computations in Sections 4.1 and 4.2 are inspired by those in [12, 15].
Since we do not assume that the algebra $\mathcal{A}$ is unital, the
computations are more delicate than those of [12, 15].

###  4.1. Combinatorial expression for
$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})$

We begin this section with the introduction of a new set of multilinear maps:

######  Definition 4.1.1.

Let $\mathscr{A}\subseteq\\{1,2,\ldots,p\\}$. We define the multilinear map
$\mathcal{W}_{\mathscr{A}}:\mathcal{A}^{\otimes(p+1)}\to\mathcal{L}_{\infty}$
by:

$\mathcal{W}_{\mathscr{A}}(a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p})=\Gamma a_{0}\prod_{k=1}^{p}b_{k}(a_{k})$

where for $a\in\mathcal{A}$ and each $k$ we define:

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

In the case where $\mathscr{A}=\\{m\\}$, a single number, $1\leq m\leq p$,
then write $\mathcal{W}_{m}$ for $\mathcal{W}_{\\{m\\}}$.

Since by assumption $(\mathcal{A},H,D)$ is smooth, the operators
$\delta(a_{k})$ are defined and bounded, so that $\mathcal{W}_{\mathscr{A}}$
is well defined as a bounded operator.

The two extreme cases, $\mathcal{W}_{\emptyset}$ and
$\mathcal{W}_{\\{1,2,\ldots,p\\}}$ are easily described as:

$\mathcal{W}_{\emptyset}(a_{0}\otimes a_{1}\otimes\cdots\otimes a_{p})=\Gamma
a_{0}\prod_{k=1}^{p}[F,a_{k}]$

It will be important to observe that if $(\mathcal{A},H,D)$ has the same
parity as $p$, then
$\mathrm{ch}(c)=\mathcal{W}_{\emptyset}(c)+F\mathcal{W}_{\emptyset}(c)F$ (this
is Lemma 4.3.6).

On the other extreme.

$\mathcal{W}_{\\{1,2,\ldots,p\\}}(a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p})=\Gamma a_{0}\prod_{k=1}^{p}\delta(a_{k}).$

Associated to a subset $\mathscr{A}\subseteq\\{1,2,\ldots,p\\}$ we have the
number,

$n_{\mathscr{A}}=|\\{(j,k)\in\\{1,2,\ldots,p\\}^{2}\;:\;j<k\text{ and
}j\in\mathscr{A},k\neq\mathscr{A}\\}|,$

where $|\cdot|$ denotes the cardinality of a set.

Theorem 4.1.7 (to be stated below) is the main result of this section. Roughly
speaking, it shows that one can replace $\Omega(c)$ in
$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})$ with a sum of
$\mathcal{W}_{\mathscr{A}}(c)$ over all subsets
$\mathscr{A}\subseteq\\{1,\ldots,p\\}$. However first we need a Lemma which
constitutes the core of the proof of Theorem 4.1.7. Most of this section is
devoted to the proof of the following lemma, which is split into various
parts.

######  Lemma 4.1.2.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple where $D$ has a spectral
gap at $0.$ For all $c\in(\mathcal{A}+\mathbb{C})^{\otimes(p+1)},$ the
operator

$\left(\Omega(c)|D|^{2-p}-\sum_{\mathscr{A}\subseteq\\{1,\cdots,p\\}}(-1)^{n_{\mathscr{A}}}\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}\right)\cdot|D|^{p-1}.$

has bounded extension.

Before proving Lemma 4.1.2 above we initially need:

######  Lemma 4.1.3.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple, where $D$ has a spectral
gap at $0$. For all $c\in(\mathcal{A}+\mathbb{C})^{\otimes(p+1)},$ the
operator

$\mathcal{W}_{\mathscr{A}}(c)\cdot|D|^{p-|\mathscr{A}|}$

has bounded extension.

###### Proof.

By linearity it suffices to prove the assertion for elementary tensors. Let
$c=a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p}\in(\mathcal{A}+\mathbb{C})^{\otimes(p+1)}$ and let
$c^{\prime}=a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p-1}\in(\mathcal{A}+\mathbb{C})^{\otimes p}.$

We will prove this assertion by induction on $p.$ If $p=1$ and
$\mathscr{A}=\emptyset,$ then

$\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}=\Gamma a_{0}[F,a_{p}]|D|$

and recall that $[F,a_{1}]|D|$ has a bounded extension $L(a_{p})$. On the
other hand, if $p=1$ and $\mathscr{A}=\\{1\\},$ then

$\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}=\Gamma
a_{0}\delta(a_{1})\in\mathcal{L}_{\infty}.$

This proves the base of induction.

Next, let $p\geq 2$ and assume that the statement is true for $p-1$. To this
end, let $\mathscr{B}\subseteq\\{1,\cdots,p-1\\}$ be defined by the formula
$\mathscr{B}=\mathscr{A}\setminus\\{p\\}.$ There are two distinct cases: when
$p\in\mathscr{A}$ and $p\notin\mathscr{A}$.

First suppose $p\in\mathscr{A}.$ Here we have $|\mathscr{B}|=|\mathscr{A}|-1$,
and

$\displaystyle\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}$
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})\delta(a_{p})|D|^{p-1-|\mathscr{B}|}$
$\displaystyle=\left(\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-1-|\mathscr{B}|}\right)\left(|D|^{-p+1+|\mathscr{B}|}\delta(a_{p})|D|^{p-1-|\mathscr{B}|}\right)$

The first factor in the right hand side has bounded extension by the inductive
assumption. Moreover, the second factor on the right hand side has bounded
extension by Lemma A.1.2. This proves step of induction for the case
$p\in\mathscr{A}.$

Now we deal with the case where $p\notin\mathscr{A}$. Then
$\mathscr{B}=\mathscr{A}$, and

$\displaystyle\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}$
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})L(a_{p})|D|^{p-1-|\mathscr{B}|}$
$\displaystyle=\left(\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-1-|\mathscr{B}|}\right)\left(|D|^{-p+1+|\mathscr{B}|}L(a_{p})|D|^{p-1-|\mathscr{B}|}\right)$

The first factor in the above right hand side is bounded by the inductive
assumption. For the second factor, we use the expression
$L(a_{p})=\partial(a_{p})-F\delta(a_{p})$. Since
$\partial(a_{p}),\delta(a_{p})\in\mathcal{B}$ then it follows from Lemma A.1.2
that the second factor on the right hand side has bounded extension. Hence,
$\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}$ extends to a bounded
linear operator.

∎

In order to prove Lemma 4.1.2, we introduce two more classes of multilinear
functionals.

######  Definition 4.1.4.

Let $\mathscr{A}\subseteq\\{1,2,\ldots,p\\}$. We define the multilinear map
$\mathcal{R}_{\mathscr{A}}:\mathcal{A}^{\otimes(p+1)}\to\mathcal{L}_{\infty}$
by

$\mathcal{R}_{\mathscr{A}}(a_{0}\otimes\cdots\otimes a_{p}):=\Gamma
a_{0}\prod_{k=1}^{p}x_{k}(a_{k}),$

where for each $1\leq k\leq p$ and $a\in\mathcal{A}$,

$x_{k}(a):=\begin{cases}F\delta(a),\,k\in\mathscr{A},\\\
L(a),k\notin\mathscr{A}.\end{cases}$

We also define the multilinear map
$\mathcal{P}_{\mathscr{A}}:\mathcal{A}^{\otimes(p+1)}\to\mathcal{L}_{\infty}$
by

$\mathcal{P}_{\mathscr{A}}(a_{0}\otimes\cdots\otimes a_{p}):=\Gamma
a_{0}\prod_{k=1}^{p}y_{k}(a_{k})$

where for each $1\leq k\leq p$ and $a\in\mathcal{A}$,

$y_{k}(a):=\begin{cases}\delta(a),\,k\in\mathscr{A},\\\
L(a),\,k\notin\mathscr{A}.\end{cases}$

######  Lemma 4.1.5.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple, where $D$ has a spectral
gap at $0$. For all $c\in(\mathcal{A}+\mathbb{C})^{\otimes(p+1)}$ and all
$\mathscr{A}\subseteq\\{1,2,\ldots,p\\}$, the operator

$\left(\mathcal{R}_{\mathscr{A}}(c)-(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{A}}(c)\cdot
F^{|\mathscr{A}|}\right)\cdot|D|$

has bounded extension.

###### Proof.

This proof is similar to that of 4.1.3. Once again it suffices to prove the
result for an elementary tensor $c=a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p}$, and we prove the statement by induction on $p$. Denote, for brevity,
$c^{\prime}=a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p-1}\in(\mathcal{A}+\mathbb{C})^{\otimes p}.$

For the base of induction, when $p=1$, we deal with the two possibilities
$\mathscr{A}=\\{1\\}$ and $\mathscr{A}=\emptyset$. If $\mathscr{A}=\\{1\\}$,
then

$\displaystyle\mathcal{R}_{\mathscr{A}}(c)$ $\displaystyle=\Gamma
a_{0}F\delta(a_{1})$
$\displaystyle\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|}$
$\displaystyle=\Gamma a_{0}\delta(a_{1})F.$

So,

$\displaystyle\left(\mathcal{R}_{\mathscr{A}}(c)-(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{A}}(c)\cdot
F^{|\mathscr{A}|}\right)\cdot|D|$ $\displaystyle=\Gamma
a_{0}(F\delta(a_{1})-\delta(a_{1})F)|D|$ $\displaystyle=\Gamma
a_{0}[F,\delta(a_{1})]|D|$ $\displaystyle=\Gamma a_{0}L(\delta(a_{1})).$

since $L(\delta(a_{1}))$ has bounded extension, so does the above left hand
side.

Now in the case that $p=1$ and $\mathscr{A}=\emptyset$,

$\displaystyle\mathcal{R}_{\mathscr{A}}(c)$ $\displaystyle=\Gamma
a_{0}L(a_{1})$ $\displaystyle\mathcal{P}_{\mathscr{A}}(c)$
$\displaystyle=\Gamma a_{0}L(a_{1})$

and $|\mathscr{A}|=0$, so
$\mathcal{R}_{\mathscr{A}}(c)-\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|}=0$.
This proves the $p=1$ case.

Now suppose that $p>1$ and the assertion is proved for $p-1$. For this
purpose, let $\mathscr{B}\subset\\{1,\cdots,p-1\\}$ be defined by
$\mathscr{B}=\mathscr{A}\backslash\\{p\\}.$

If $p\in\mathscr{A}$, then $n_{\mathscr{A}}=n_{\mathscr{B}}$, and if
$p\notin\mathscr{A}$ then $n_{\mathscr{A}}=n_{\mathscr{B}}+|\mathscr{B}|$.

Now we consider separately the cases $p\in\mathscr{A}$ and
$p\notin\mathscr{A}$.

First, if $p\in\mathscr{A}$, then,

$\displaystyle\mathcal{R}_{\mathscr{A}}(c)$
$\displaystyle=\mathcal{R}_{\mathscr{B}}(c^{\prime})F\delta(a_{p}),$
$\displaystyle\mathcal{P}_{\mathscr{A}}(c)$
$\displaystyle=\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta(a_{p}).$

Hence, since $|\mathscr{A}|=|\mathscr{B}|+1$ and
$n_{\mathscr{A}}=n_{\mathscr{B}}$ in this case:

$(\mathcal{R}_{\mathscr{A}}(c)-(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|})|D|=(\mathcal{R}_{\mathscr{B}}(c^{\prime})F\delta(a_{p})-(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta(a_{p})F^{|\mathscr{B}|+1})|D|=$
$=\mathcal{R}_{\mathscr{B}}(c^{\prime})[F,\delta(a_{p})]|D|+(\mathcal{R}_{\mathscr{B}}(c^{\prime})\delta(a_{p})F-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta(a_{p})F\cdot
F^{|\mathscr{B}|})|D|.$
$=\mathcal{R}_{\mathscr{B}}(c^{\prime})L(a_{p})+(\mathcal{R}_{\mathscr{B}}(c^{\prime})\delta(a_{p})-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta(a_{p})\cdot
F^{|\mathscr{B}|})D.$

In the case that $|\mathscr{B}|$ is even, we have $F^{|\mathscr{B}|}=1$ and
so:

$(\mathcal{R}_{\mathscr{A}}(c)-(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|})|D|=$
$=\mathcal{R}_{\mathscr{B}}(c^{\prime})L(a_{p})+(\mathcal{R}_{\mathscr{B}}(c^{\prime})-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})\cdot
F^{|\mathscr{B}|})\cdot\delta(a_{p})D.$
$=\mathcal{R}_{\mathscr{B}}(c^{\prime})L(a_{p})+(\mathcal{R}_{\mathscr{B}}(c^{\prime})-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})\cdot
F^{|\mathscr{B}|})D\cdot\delta(a_{p})-$
$-(\mathcal{R}_{\mathscr{B}}(c^{\prime})-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})\cdot
F^{|\mathscr{B}|})\cdot\partial(\delta(a_{p})).$

Since $\delta(a_{p}),$ $\partial(\delta(a_{p}))$ and $L(a_{p})$ are bounded,
by the inductive hypothesis the above has bounded extension, completing the
proof of the case where $p\in\mathscr{A}$ and $|\mathscr{B}|$ is even.

On the other hand, if $p\in\mathscr{A}$ and $|\mathscr{B}|$ is odd, then:

$(\mathcal{R}_{\mathscr{A}}(c)-(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|})|D|=$
$=\mathcal{R}_{\mathscr{B}}(c^{\prime})L(a_{p})+\mathcal{R}_{\mathscr{B}}(c^{\prime})\delta(a_{p})D-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta(a_{p})|D|=$
$=\mathcal{R}_{\mathscr{B}}(c^{\prime})L(a_{p})+(\mathcal{R}_{\mathscr{B}}(c^{\prime})-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})F^{|\mathscr{B}|})D\cdot\delta(a_{p})-$
$-\mathcal{R}_{\mathscr{B}}(c^{\prime})\delta(a_{p})D+(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta^{2}(a_{p}).$

Since $\delta(a_{p}),$ $\delta^{2}(a_{p})$ and $L(a_{p})$ are bounded, by the
inductive hypothesis the above has bounded extension, completing the proof of
the case where $p\in\mathscr{A}$ and $|\mathscr{B}|$ is odd.

Now assume that $p\notin\mathscr{A}$. Then,

$\displaystyle\mathcal{R}_{\mathscr{A}}(c)$
$\displaystyle=\mathcal{R}_{\mathscr{B}}(c^{\prime})L(a_{p})$
$\displaystyle\mathcal{P}_{\mathscr{A}}(c)$
$\displaystyle=\mathcal{P}_{\mathscr{B}}(c^{\prime})L(a_{p}).$

Focusing on $\mathcal{P}_{\mathscr{A}}(c)$:

$\displaystyle\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|}$
$\displaystyle=\mathcal{P}_{\mathscr{B}}(c^{\prime})L(a_{p})F^{|\mathscr{B}|}.$

So:

$(\mathcal{R}_{\mathscr{A}}(c)-(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|})|D|=(\mathcal{R}_{\mathscr{B}}(c^{\prime})L(a_{p})-(-1)^{n_{\mathscr{B}}+|\mathscr{B}|}\mathcal{P}_{\mathscr{B}}(c^{\prime})L(a_{p})F^{|\mathscr{B}|})|D|.$

Note that since $F$ anticommutes with $[F,a_{p}]$, $F$ also anticommutes with
$L(a_{p})$. Hence,

$L(a_{p})F^{|\mathscr{B}|}=(-1)^{|\mathscr{B}|}F^{|\mathscr{B}|}L(a_{p})$

and so

$\displaystyle(\mathcal{R}_{\mathscr{A}}(c)-(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|})|D|$
$\displaystyle=(\mathcal{R}_{\mathscr{B}}(c^{\prime})-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})F^{|\mathscr{B}|})L(a_{p})|D|$
$\displaystyle=(\mathcal{R}_{\mathscr{B}}(c^{\prime})-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})F^{|\mathscr{B}|})(|D|L(a_{p})+L(\delta(a_{p}))).$

By the inductive assumption, the operator
$(\mathcal{R}_{\mathscr{B}}(c^{\prime})-(-1)^{n_{\mathscr{B}}}\mathcal{P}_{\mathscr{B}}(c^{\prime})F^{|\mathscr{B}|})|D|$
has bounded extension. This completes the proof of the $p\notin\mathscr{A}$
case.

Hence, the statement is true for $p$ and this completes the induction. ∎

######  Lemma 4.1.6.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple. Suppose $D$ has a
spectral gap at $0.$ For all $c\in(\mathcal{A}+\mathbb{C})^{\otimes(p+1)}$ and
for all $\mathscr{A}\subseteq\\{1,\ldots,p\\}$ the operator

$\left(\mathcal{P}_{\mathscr{A}}(c)-\mathcal{W}_{\mathscr{A}}(c)\cdot|D|^{p-|\mathscr{A}|}\right)\cdot|D|.$

has bounded extension.

###### Proof.

This proof is again similar to the proofs of Lemmas 4.1.3 and 4.1.5.

Once more, it suffices to prove the assertion for an elementary tensor
$c=a_{0}\otimes a_{1}\otimes\cdots\otimes
a_{p}\in(\mathcal{A}+\mathbb{C})^{\otimes(p+1)}$. Let
$c^{\prime}:=a_{0}\otimes\cdots\otimes
a_{p-1}\in(\mathcal{A}+\mathbb{C})^{\otimes p}$. The proof proceeds by
induction on $p$.

First, if $p=1$, then either $\mathscr{A}=\\{1\\}$ or $\mathscr{A}=\emptyset$.
If $\mathscr{A}=\\{1\\}$, then

$\displaystyle\mathcal{P}_{\mathscr{A}}(c)$ $\displaystyle=\Gamma
a_{0}\delta(a_{1}),$
$\displaystyle\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}$
$\displaystyle=\Gamma a_{0}\delta(a_{1}).$

and if $\mathscr{A}=\emptyset$, then

$\displaystyle\mathcal{P}_{\mathscr{A}}(c)$ $\displaystyle=\Gamma
a_{0}L(a_{1}),$
$\displaystyle\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}$
$\displaystyle=\Gamma a_{0}[F,a_{1}]|D|.$

Since $L(a_{1})=[F,a_{1}]|D|$ on the dense subspace $H_{\infty}$, it follows
that in all cases with $p=1$ the difference
$\mathcal{P}_{\mathscr{A}}(c)-\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}$
is identically zero on $H_{\infty}$ and therefore has trivial bounded
extension to $H$. This establishes the $p=1$ case.

Now suppose that $p>1$ and the assertion has been proved for $p-1$. Let
$\mathscr{B}=\mathscr{A}\setminus\\{p\\}$, and we consider the two cases of
$p\in\mathscr{A}$ and $p\notin\mathscr{A}$.

Suppose that $p\in\mathscr{A}$. Then,

$\displaystyle\mathcal{P}_{\mathscr{A}}(c)$
$\displaystyle=\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta(a_{p}),$
$\displaystyle\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|}$
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})\delta(a_{p})|D|^{p-1-|\mathscr{B}|}.$

So,

$\displaystyle\mathcal{P}_{\mathscr{A}}(c)|D|$
$\displaystyle=\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta(a_{p})|D|$ (4.1)
$\displaystyle=\mathcal{P}_{\mathscr{B}}(c^{\prime})|D|-\mathcal{P}_{\mathscr{B}}(c^{\prime})\delta^{2}(a_{p}).$

and

$\displaystyle\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|+1}$
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})\delta(a_{p})|D|^{p-|\mathscr{B}|}$
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-|\mathscr{B}|}\delta(a_{p})+\mathcal{W}_{\mathscr{B}}(c^{\prime})[\delta(a_{p}),|D|^{p-|\mathscr{B}|}]$
(4.2)
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-|\mathscr{B}|}\delta(a_{p})-\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-|\mathscr{B}|-1}|D|^{|\mathscr{B}|-p+1}[|D|^{p-|\mathscr{B}|},\delta(a_{p})].$

Applying Lemma A.1.2, the operator
$|D|^{|\mathscr{B}|-p+1}[|D|^{p-|\mathscr{B}|},\delta(a_{p})]$ has bounded
extension. So combining (4.1) and (4.1):

$\displaystyle(\mathcal{P}_{\mathscr{A}}(c)-\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|})|D|-(\mathcal{P}_{\mathscr{B}}(c^{\prime})-\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-1-|\mathscr{B}|})|D|$

has bounded extension. So by the inductive hypothesis,
$(\mathcal{P}_{\mathscr{A}}(c)-\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|})|D|$
has bounded extension in the case that $p\in\mathscr{A}$.

Now suppose that $p\notin\mathscr{A}$. In this case, we have:

$\displaystyle\mathcal{P}_{\mathscr{A}}(c)$
$\displaystyle=\mathcal{P}_{\mathscr{B}}(c^{\prime})L(a_{p})$
$\displaystyle\mathcal{W}_{\mathscr{A}}(c)$
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})[F,a_{p}].$

Multiplying by $|D|$, we have

(4.3)
$\mathcal{P}_{\mathscr{A}}(c)|D|=\mathcal{P}_{\mathscr{B}}(c^{\prime})|D|L(a_{p})-\mathcal{P}_{\mathscr{B}}(c^{\prime})L(\delta(a_{p})).$

Note that $\mathcal{P}_{\mathscr{B}}(c^{\prime})L(\delta(a_{p}))$ is bounded.

Also,

$\displaystyle\mathcal{W}_{\mathscr{A}}(c)|D|^{p+1-|\mathscr{A}|}$
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})L(a_{p})|D|^{p-|\mathscr{A}|}$
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-|\mathscr{A}|}L(a_{p})-\mathcal{W}_{\mathscr{B}}(c^{\prime})[|D|^{p-|\mathscr{A}|},L(a_{p})]$
(4.4)
$\displaystyle=\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-|\mathscr{A}|}L(a_{p})-\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-1-|\mathscr{A}|}|D|^{-p+1+|\mathscr{A}|}[|D|^{p-|\mathscr{A}|},L(a_{p})].$

By Lemma A.1.2, the operator
$|D|^{-p+1+|\mathscr{A}|}[|D|^{p-|\mathscr{A}|},L(a_{p})]$ has bounded
extension. So combining (4.3) and (4.1), it follows that

$(\mathcal{P}_{\mathscr{A}}(c)-\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|})|D|-(\mathcal{P}_{\mathscr{B}}(c^{\prime})-\mathcal{W}_{\mathscr{B}}(c^{\prime})|D|^{p-1-|\mathscr{B}|})|D|L(a_{p})$

has bounded extension. So by the inductive hypothesis, it follows that
$(\mathcal{P}_{\mathscr{A}}(c)-\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|})|D|$
has bounded extension in the case $p\notin\mathscr{A}$. ∎

The main idea used in the proof of Lemma 4.1.2 is the algebraic identity,

(4.5)
$\prod_{k=1}^{p}(x_{k}+y_{k})=\sum_{\mathscr{A}\subseteq\\{1,\cdots,p\\}}z_{\mathscr{\mathscr{A}}}$

where $z_{\mathscr{\mathscr{A}}}$ is given by the product $z_{1}z_{2}\cdots
z_{p}$, where $z_{k}=x_{k}$ for $k\in\mathscr{A}$ and $z_{k}=y_{k}$ for
$k\notin\mathscr{A}$.

Now we are ready to complete the proof of Lemma 4.1.2:

###### Proof of Lemma 4.1.2.

Since

$[D,a]=F[|D|,a]+[F,a]|D|,$

as an equality of operators on $H_{\infty}$, it follows that we have
$\partial(a)=F\delta(a)+L(a)$. Now using (4.5):

$\displaystyle\Omega(c)$ $\displaystyle=\Gamma
a_{0}\prod_{k=1}^{p}(F\delta(a_{k})+L(a_{k}))$
$\displaystyle=\sum_{\mathscr{A}\subseteq\\{1,\cdots,p\\}}\mathcal{R}_{\mathscr{A}}(c).$

So on $H_{\infty}$:

$\Omega(c)|D|=\sum_{\mathscr{A}\subseteq\\{1,\cdots,p\\}}\mathcal{R}_{\mathscr{A}}(c)|D|.$

We may now apply Lemma 4.1.5 to each summand to conclude that

$\Omega(c)|D|-\sum_{\mathscr{A}\subseteq\\{1,\cdots,p\\}}(-1)^{n_{\mathscr{A}}}\mathcal{P}_{\mathscr{A}}(c)F^{|\mathscr{A}|}|D|$

has bounded extension. Now applying Lemma 4.1.6 to each summand, we have that
the operator

$\Omega(c)|D|-\sum_{\mathscr{A}\subseteq\\{1,\cdots,p\\}}(-1)^{n_{\mathscr{A}}}\mathcal{W}_{\mathscr{A}}(c)|D|^{p-|\mathscr{A}|+1}F^{|\mathscr{A}|}$

has bounded extension.

Equivalently,

$\left(\Omega(c)|D|^{2-p}-\sum_{\mathscr{A}\subseteq\\{1,\ldots,p\\}}(-1)^{n_{\mathscr{A}}}\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}\right)|D|^{p-1}$

has bounded extension. ∎

We now prove the main result of this section.

######  Theorem 4.1.7.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. For all $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\displaystyle\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=\sum_{\mathscr{A}\subseteq\\{1,2,\ldots,p\\}}(-1)^{n_{\mathscr{A}}}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})+O(s^{-1})$

as $s\to 0$.

###### Proof.

As in the preceding lemmas it suffices to prove the result for an elementary
tensor $c=a_{0}\otimes\cdots\otimes a_{p}\in\mathcal{A}^{\otimes(p+1)}$. Let
$c^{\prime}=1\otimes a_{1}\otimes\cdots\otimes a_{p}$. By the cyclicity of the
trace and the fact that $\mathcal{A}$ commutes with $\Gamma$, we have:

$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=\mathrm{Tr}(\Omega(c^{\prime})|D|^{2-p}e^{-s^{2}D^{2}}a_{0})$

and for all $\mathscr{A}\subseteq\\{1,2,\ldots,p\\}$:

$\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})=\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c^{\prime})D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}}a_{0}).$

Thus,

$\displaystyle\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})-\sum_{\mathscr{A}\subseteq\\{1,\ldots,p\\}}(-1)^{n_{\mathscr{A}}}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})$
$\displaystyle=\mathrm{Tr}\left(\left(\Omega(c^{\prime})|D|^{2-p}-\sum_{\mathscr{A}\subseteq\\{1,\ldots,p\\}}(-1)^{n_{\mathscr{A}}}\mathcal{W}_{\mathscr{A}}(c^{\prime})D^{2-|\mathscr{A}|}\right)|D|^{p-1}|D|^{1-p}e^{-s^{2}D^{2}}a_{0}\right)$

Hence,

$\displaystyle\left|\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}}-\sum_{\mathscr{A}\subseteq\\{1,\ldots,p\\}}(-1)^{n_{\mathscr{A}}}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})\right|$
$\displaystyle\leq\left\|\left(\Omega(c^{\prime})D^{2-p}-\sum_{\mathscr{A}\subseteq\\{1,\ldots,p\\}}(-1)^{n_{\mathscr{A}}}\mathcal{W}_{\mathscr{A}}(c^{\prime})D^{2-|\mathscr{A}|}\right)|D|^{p-1}\right\|_{\infty}\||D|^{1-p}e^{-s^{2}D^{2}}a_{0}\|_{1}$

The first factor is finite, by Lemma 4.1.2, and the second factor is
$O(s^{-1})$, by Lemma A.1.4. This completes the proof. ∎

###  4.2. Auxiliary commutator estimates

This section is a slight detour from the main task of this chapter. Here we
establish bounds on the $\mathcal{L}_{1}$-norm of commutators of the form
$[f(s|D|),x]$, where $x\in\mathcal{B}$, $s>0$ and $f$ is the square of a
Schwartz class function on $\mathbb{R}.$ These bounds are used everywhere in
the subsequent sections of this chapter.

Recall that the algebra $\mathcal{B}$ is defined in Definition 2.2.7.

The following lemma serves the same purpose as [12, Lemma 18], but the right
hand sides are different here due to the fact that we deal with non-unital
spectral triples.

######  Lemma 4.2.1.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple. Let $h$ be a Schwartz
class function on $\mathbb{R}$ and let $f=h^{2}$. Then for all
$x\in\mathcal{B}$, we have

$\left\|[f(s|D|),x]-\frac{s}{2}\\{f^{\prime}(s|D|),\delta(x)\\}\right\|_{1}\leq\frac{1}{2}s^{2}\|\widehat{h^{\prime\prime}}\|_{1}\cdot\left(\|\delta^{2}(x)h(s|D|)\|_{1}+\|h(s|D|)\delta^{2}(x)\|_{1}\right)$

Here, $\\{\cdot,\cdot\\}$ denotes the anti-commutator.

###### Proof.

Since $f^{\prime}(s|D|)=2h^{\prime}(s|D|)h(s|D|)$, by the Leibniz rule we
have:

$[f(s|D|),x]-\frac{s}{2}\\{f^{\prime}(s|D|),\delta(x)\\}=[h(s|D|)^{2},x]-s\\{h^{\prime}(s|D|)h(s|D|),\delta(x)\\}=$
$=h(s|D|)\big{(}[h(s|D|),x]-sh^{\prime}(s|D|)\delta(x)\big{)}+\big{(}[h(s|D|),x]-s\delta(x)h^{\prime}(s|D|)\big{)}h(s|D|).$

Applying Lemma A.2.3, we have

$[h(s|D|),x]-sh^{\prime}(s|D|)\delta(x)=-s^{2}\int_{-\infty}^{\infty}\int_{0}^{1}\widehat{h^{\prime\prime}}(u)(1-v)e^{ius(1-v)|D|}\delta^{2}(x)e^{iusv|D|}\,dvdu$
$[h(s|D|),x]-s\delta(x)h^{\prime}(s|D|)=-s^{2}\int_{-\infty}^{\infty}\int_{0}^{1}\widehat{h^{\prime\prime}}(u)(1-v)e^{iusv|D|}\delta^{2}(x)e^{ius(1-v)|D|}\,dvdu.$

Therefore,

$\displaystyle[f(s|D|),x]-\frac{s}{2}$
$\displaystyle\\{f^{\prime}(s|D|),\delta(x)\\}=$
$\displaystyle-s^{2}\int_{-\infty}^{\infty}\int_{0}^{1}\widehat{h^{\prime\prime}}(u)(1-v)e^{iu(1-v)s|D|}h(s|D|)\delta^{2}(x)e^{iuvs|D|}\,dvdu$
$\displaystyle-s^{2}\int_{-\infty}^{\infty}\int_{0}^{1}\widehat{h^{\prime\prime}}(u)(1-v)e^{iuvs|D|}\delta^{2}(x)h(s|D|)e^{iu(1-v)s|D|}\,dvdu.$

Now applying Lemma 2.3.2 to each integral, we have

$\displaystyle\|[f(s|D|),x]$
$\displaystyle-\frac{s}{2}\\{f^{\prime}(s|D|),\delta(x)\\}\|_{1}$
$\displaystyle\leq
s^{2}\int_{-\infty}^{\infty}\int_{0}^{1}\left\|\widehat{h^{\prime\prime}}(u)(1-v)e^{iu(1-v)s|D|}h(s|D|)\delta^{2}(x)e^{iuvs|D|}\right\|_{1}dvdu$
$\displaystyle+s^{2}\int_{-\infty}^{\infty}\int_{0}^{1}\left\|\widehat{h^{\prime\prime}}(u)(1-v)e^{iuvs|D|}\delta^{2}(x)h(s|D|)e^{iu(1-v)s|D|}\right\|_{1}dvdu$
$\displaystyle=s^{2}\|\widehat{h^{\prime\prime}}\|_{1}(\|\delta^{2}(x)h(s|D|)\|_{1}+\|h(s|D|)\delta^{2}(x)\|_{1})\int_{0}^{1}(1-v)dv$
$\displaystyle=\frac{1}{2}s^{2}\|\widehat{h^{\prime\prime}}\|_{1}(\|\delta^{2}(x)h(s|D|)\|_{1}+\|h(s|D|)\delta^{2}(x)\|_{1}).$

∎

######  Lemma 4.2.2.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple and assume that $D$ has a
spectral gap at $0$. Let $h$ be a Schwartz function on $\mathbb{R}$ and let
$f=h^{2}.$ Then for every $x\in\mathcal{B},$ we have

$\left\||D|^{m}[f(s|D|),x]\right\|_{1}\leq
s\|\widehat{h^{\prime}}\|_{1}\left(\left\||D|^{m}h(s|D|)\delta(x)\right\|_{1}+\left\||D|^{m}\delta(x)h(s|D|)\right\|_{1}\right).$

###### Proof.

Since $f=h^{2}$, by the Leibniz rule, we have

$[f(s|D|),x]=h(s|D|)[h(s|D|),x]+[h(s|D|),x]h(s|D|).$

Using Lemma A.2.2, we have

$[h(s|D|),x]=s\int_{\mathbb{R}}\int_{0}^{1}\widehat{h^{\prime}}(u)e^{ius(1-v)|D|}\delta(x)e^{iusv|D|}dvdu.$

Thus,

$\displaystyle|D|^{m}[f(s|D|),x]$
$\displaystyle=s\int_{\mathbb{R}}\int_{0}^{1}\widehat{h^{\prime}}(u)e^{ius(1-v)|D|}|D|^{m}h(s|D|)\delta(x)e^{iusv|D|}\,dvdu$
$\displaystyle\quad\quad+s\int_{\mathbb{R}}\int_{0}^{1}\widehat{h^{\prime}}(u)e^{ius(1-v)|D|}|D|^{m}\delta(x)h(s|D|)e^{iusv|D|}\,dvdu.$

Bounding the $\mathcal{L}_{1}$ norm using Lemma 2.3.2, we have

$\displaystyle\left\||D|^{m}[f(s|D|),x]\right\|_{1}$ $\displaystyle\leq
s\int_{\mathbb{R}}\int_{0}^{1}|\widehat{h^{\prime}}(u)|\||D|^{m}h(s|D|)\delta(x)\|_{1}\,dvdu$
$\displaystyle\quad\quad+s\int_{\mathbb{R}}\int_{0}^{1}|\widehat{h^{\prime}}(u)|\||D|^{m}\delta(x)h(s|D|)\|_{1}\,dvdu$
$\displaystyle=s\|\widehat{h^{\prime}}\|_{L_{1}(\mathbb{R})}(\||D|^{m}h(s|D|)\delta(x)\|_{1}+\||D|^{m}\delta(x)h(s|D|)\|_{1}).$

∎

######  Lemma 4.2.3.

Let $(\mathcal{A},H,D)$ be a smooth spectral triple satisfying Hypothesis
1.2.1 and Hypothesis 4.0.1. For all $x\in\mathcal{B}$ and for all integers
$m>-p,$ we have

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

###### Proof.

Let $h(t)=e^{-\frac{1}{2}t^{2}},$ $t\in\mathbb{R}.$ By Lemma 4.2.2, we have

$\||D|^{m}[e^{-s^{2}D^{2}},x]\|_{1}\leq
s\|\widehat{h^{\prime}}\|_{1}(\||D|^{m}e^{-\frac{1}{2}s^{2}D^{2}}\delta(x)\|_{1}+\||D|^{m}\delta(x)e^{-\frac{1}{2}s^{2}D^{2}}\|_{1}).$

If $m\leq 0$, then the assertion now follows from applying Lemma A.1.4 to the
terms $\||D|^{m}e^{-\frac{1}{2}s^{2}D^{2}}\delta(x)\|_{1}$ and
$\||D|^{m}\delta(x)e^{-\frac{1}{2}s^{2}D^{2}}\|_{1}$. Assume now $m>0$. Using
Lemma A.1.3 with the Schwartz function $t\mapsto t^{m}e^{-\frac{1}{2}t^{2}}$,
we obtain

$s\left\||D|^{m}e^{-\frac{1}{2}s^{2}D^{2}}\delta(x)\right\|_{1}=O(s^{1-p-m}),\quad
s\to 0.$

By Lemma A.1.1, we have

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

Now we apply Lemma A.1.3 to each summand, using the function $t\mapsto
t^{k}e^{-\frac{1}{2}s^{2}D^{2}}$ for the $k$th summand. So,

$\displaystyle s\||D|^{m}\delta(x)e^{-\frac{1}{2}s^{2}D^{2}}\|_{1}$
$\displaystyle\leq s\sum_{k=0}^{m}\binom{m}{k}O(s^{1-p-k})$
$\displaystyle=O(s^{1-p-m}),\quad s\to 0.$

∎

The following lemma is used in the proof of Theorem 4.4.2.

######  Lemma 4.2.4.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. Let $f(t)=e^{-t^{2}},$ $t\in\mathbb{R}.$

  1. (i)

for every $a\in\mathcal{A},$ we have

$\Big{\|}[f(s|D|),a]-s\delta(a)f^{\prime}(s|D|)\Big{\|}_{\infty}=O(s^{2}),\quad
s\downarrow 0.$

  2. (ii)

for every $a\in\mathcal{A},$ we have

$\Big{\|}[f(s|D|),a]-s\delta(a)f^{\prime}(s|D|)\Big{\|}_{1}=O(s^{2-p}),\quad
s\downarrow 0.$

  3. (iii)

for every $a\in\mathcal{A},$ we have

$\Big{\|}[f(s|D|),a]-s\delta(a)f^{\prime}(s|D|)\Big{\|}_{p,1}=O(s),\quad
s\downarrow 0.$

###### Proof.

First we prove (i): this is a simple combination of Lemma A.2.3 and the
triangle inequality:

$\|[f(s|D|),a]-sf^{\prime}(s|D|)\delta(a)\|_{\infty}\leq
s^{2}\|\widehat{f^{\prime\prime}}\|_{1}\|\delta^{2}(a)\|_{\infty}.$

Now we prove (ii). Let $h(t)=e^{-t^{2}/2}$, $t\in\mathbb{R}$, so that
$f=h^{2}$. By Lemma 4.2.1, for all $a\in\mathcal{A}$ we have

(4.6)
$\left\|[f(s|D|),a]-\frac{s}{2}\\{f^{\prime}(s|D|),\delta(a)\\}\right\|_{1}\leq\frac{1}{2}s^{2}\|\hat{h^{\prime\prime}}\|_{1}(\|\delta^{2}(a)h(s|D|)\|_{1}+\|h(s|D|)\delta^{2}(a)\|_{1}).$

Using Lemma A.1.3, we have

$\displaystyle\|\delta^{2}(a)h(s|D|)\|_{1}$ $\displaystyle=O(s^{-p})\text{
and, }$ (4.7) $\displaystyle\|h(s|D|)\delta^{2}(a)\|_{1}$
$\displaystyle=O(s^{-p}).$

Combining (4.6) and (4.2), we arrive at:

(4.8)
$\|[f(s|D|),a]-\frac{s}{2}\\{f^{\prime}(s|D|),\delta(a)\\}\|_{1}=O(s^{2-p}).$

On the other hand,

$\displaystyle\|[f^{\prime}(s|D|),\delta(a)]\|_{1}$
$\displaystyle=2s\|[|D|e^{-s^{2}D^{2}},\delta(a)]\|_{1}$ $\displaystyle\leq
2s\|\delta^{2}(a)e^{-s^{2}D^{2}}\|_{1}+2s\||D|[e^{-s^{2}D^{2}},\delta(a)]\|_{1}.$

Due to Lemma A.1.3, we have
$2s\|\delta^{2}(a)e^{-s^{2}D^{2}}\|_{1}=O(s^{1-p})$, and by Lemma 4.2.3 we
also have $2s\||D|[e^{-s^{2}D^{2}},\delta(a)]\|_{1}=O(s^{1-p})$. Therefore,

(4.9) $\|[f^{\prime}(s|D|),\delta(a)]\|_{1}=O(s^{1-p}),\quad s\downarrow 0.$

By combining (4.8) and (4.9), we obtain (ii).

Finally, to prove (iii), we use the inequality

$\|T\|_{p,1}\leq\|T\|_{1}^{\frac{1}{p}}\|T\|_{\infty}^{1-\frac{1}{p}}$

and write

$\|[f(s|D|),a]-s\delta(a)f^{\prime}(s|D|)\|_{p,1}\leq$
$\leq\|[f(s|D|),a]-s\delta(a)f^{\prime}(s|D|)\|_{1}^{\frac{1}{p}}\|[f(s|D|,a)]-s\delta(a)f^{\prime}(s|D|)\|_{\infty}^{1-\frac{1}{p}}=$
$=O(s^{2-p})^{\frac{1}{p}}\cdot O(s^{2})^{1-\frac{1}{p}}=O(s).$

∎

The following Lemma is used in Lemma 4.3.2, Lemma 4.3.3 and Lemma 4.5.2.

######  Lemma 4.2.5.

Let $(\mathcal{A},H,D)$ be a smooth $p-$dimensional spectral triple satisfying
Hypothesis 1.2.1 and Hypothesis 4.0.1. For every $0\leq m\leq p,$ and
$x\in\mathcal{B}$ we have

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

###### Proof.

By the Leibniz rule,

$[D^{2-m}e^{-s^{2}D^{2}},x]=[D^{2-m},a]e^{-s^{2}D^{2}}+D^{2-m}[e^{-s^{2}D^{2}},x].$

Thus,

$\|D^{m-p}[D^{2-m}e^{-s^{2}D^{2}},x]\|_{1}\leq\|D^{2-p}[e^{-s^{2}D^{2}},x]\|_{1}+\|D^{m-p}[D^{2-m},x]e^{-s^{2}D^{2}}\|_{1}.$

By Lemma 4.2.3, we have $\|D^{2-p}[e^{-s^{2}D^{2}},x]\|_{1}=O(s^{-1})$, so we
now focus on the second summand. First, for the case when $m>2$ we apply the
Leibniz rule

$\displaystyle[D^{2-m},x]$ $\displaystyle=-D^{2-m}[D^{m-2},x]D^{2-m}$
$\displaystyle=-\sum_{k+l=m-3}D^{k+2-m}\partial(x)D^{l+2-m}.$

Now using the triangle inequality:

$\|D^{m-p}[D^{2-m},x]e^{-s^{2}D^{2}}\|_{1}\leq\sum_{k+l=m-3}\||D|^{k+2-p}\partial(x)|D|^{l+2-m}e^{-s^{2}|D|^{2}}\|_{1}.$

Applying Lemma A.1.4 to each summand, we then conclude that

${\|D^{m-p}[D^{2-m},x]e^{-s^{2}D^{2}}\|_{1}=O(s^{-1})},$

thus proving the claim for $m>2$.

We now deal with the remaining cases $m=0,1,2$ individually. In the case
$m=2$, we have $D^{m-p}[D^{2-m},x]e^{-s^{2}D^{2}}=0$, and so the claim follows
trivially in this case.

For $m=1$, we have

$\|D^{m-p}[D^{2-m},x]e^{-s^{2}D^{2}}\|_{1}=\||D|^{1-p}\partial(x)e^{-s^{2}D^{2}}\|_{1}$

So by Lemma A.1.4, we also have in this case that the above is $O(s^{-1})$.

Finally, for $m=0$,

$\displaystyle[D^{2},x]$ $\displaystyle=[|D|^{2},x]$
$\displaystyle=|D|\delta(x)+\delta(x)|D|$
$\displaystyle=2|D|\delta(x)-\delta^{2}(x).$

So by the triangle inequality:

$\|D^{m-p}[D^{2-m},x]e^{-s^{2}D^{2}}\|_{1}\leq
2\||D|^{1-p}\delta(x)e^{-s^{2}D^{2}}\|_{1}+\||D|^{-p}\delta^{2}(x)e^{-s^{2}D^{2}}\|_{1}$

so an application of Lemma A.1.4 to each of the above summands yields the
result. ∎

###  4.3. Exploiting Hochschild homology

Recall the multilinear mapping $\mathcal{W}_{p}$ from Definition 4.1.1. In
this section, we prove the following:

######  Theorem 4.3.1.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. For every Hochschild cycle $c\in\mathcal{A}^{\otimes(p+1)}$
we have:

$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})-p\mathrm{Tr}(\mathcal{W}_{p}(c)De^{-s^{2}D^{2}})=O(s^{-1}),\quad
s\downarrow 0.$

We achieve this using the commutator estimates of the preceding section.

Our strategy to prove Theorem 4.3.1 is to start from Theorem 4.1.7 and then
show that:

  1. (1)

all of the terms with $|\mathscr{A}|\geq 2$ are $O(s^{-1})$ (see Lemma 4.3.4)

  2. (2)

the $\mathscr{A}=\emptyset$ term is $O(s^{-1})$ (see Lemma 4.3.7

  3. (3)

finally we complete the proof by showing that the terms with $|\mathscr{A}|=1$
are all equal to the $\mathscr{A}=\\{p\\}$ term up to terms of size
$O(s^{-1})$.

The proofs in this section rely crucially on the assumption that $c$ is a
Hochschild cycle.

First, we show that terms in Theorem 4.1.7 such that there is some $m$ with
$m-1,m\in\mathscr{A}$ are $O(s^{-1})$.

######  Lemma 4.3.2.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. Let $m\in\\{1,\ldots,p\\}$ and suppose that
$m-1,m\in\mathscr{A}$ (so necessarily we have $|\mathscr{A}|\geq 2$). For
every Hochschild cycle $c\in\mathscr{A}^{\otimes(p+1)},$ we have

$\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})=O(s^{-1}),\quad
s\downarrow 0.$

###### Proof.

For $s>0$, consider the multilinear mapping $\theta_{s}:\mathcal{A}^{\otimes
p}\to\mathbb{C}$ defined by:

$\theta_{s}(a_{0}\otimes\cdots\otimes a_{p-1})=\mathrm{Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-2}b_{k}(a_{k})\right)\delta^{2}(a_{m-1})\left(\prod_{k=m}^{p-1}b_{k+1}(a_{k})\right)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}}\right)$

where $b_{k}$ is as in Definition 4.1.1. Then, from the computation in
Appendix A.3, we have that the Hochschild coboundary is:

$\displaystyle(b\theta_{s})(a_{0}\otimes$ $\displaystyle\cdots\otimes a_{p})$
$\displaystyle=(-1)^{p}\mathrm{Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-2}[b_{k},a_{k}]\right)\delta^{2}(a_{m-1})\left(\prod_{k=m}^{p-1}[b_{k+1},a_{k}]\right)[D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}},a_{p}]\right)$
$\displaystyle+2(-1)^{m-1}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(a_{0}\otimes\cdots\otimes
a_{p})D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}}).$

We now claim that the first summand is $O(s^{-1})$ as $s\downarrow 0$. Indeed,
dividing and multiplying by $D^{|\mathscr{A}|-p}$:

$\displaystyle\Big{\|}\Gamma a_{0}\left(\prod_{k=1}^{m-2}[b_{k},a_{k}]\right)$
$\displaystyle\delta^{2}(a_{m-1})\left(\prod_{k=m}^{p-1}[b_{k+1},a_{k}]\right)[D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}},a_{p}]\Big{\|}_{1}$
$\displaystyle\leq\left\|D^{|\mathscr{A}|-p}[D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}},a_{p}]\right\|_{1}$
$\displaystyle\times\left\|\Gamma
a_{0}\left(\prod_{k=1}^{m-2}[b_{k},a_{k}]\right)\delta^{2}(a_{m-1})\left(\prod_{k=m}^{p-1}[b_{k+1},a_{k}]\right)|D|^{p-|\mathscr{A}|}\right\|_{\infty}$

The first factor is $O(s^{-1})$ by Lemma 4.2.5, and the second factor is
finite by Lemma A.4.2 and has no dependence on $s$.

To summarise, so far we that if $c\in\mathcal{A}^{\otimes(p+1)}$:

$(b\theta_{s})(c)=2(-1)^{m-1}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0.$

If $c$ is a Hochschild cycle, then $(b\theta_{s})(c)=\theta_{s}(bc)=0$, and so

$2(-1)^{m-1}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})=O(s^{-1})$

as required. ∎

######  Lemma 4.3.3.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. Let
$\mathscr{A}_{1},\mathscr{A}_{2}\subseteq\\{1,\ldots,p\\}$, with
$|\mathscr{A}_{1}|=|\mathscr{A}_{2}|$ and that the symmetric difference
$\mathscr{A}_{1}\Delta\mathscr{A}_{2}=\\{m-1,m\\}$ for some $m.$ Then for
every Hochschild cycle $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{1}}(c)D^{2-|\mathscr{A}_{1}|}e^{-s^{2}D^{2}})+\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{2}}(c)D^{2-|\mathscr{A}_{2}|}e^{-s^{2}D^{2}})=O(s^{-1}),\quad
s\downarrow 0.$

###### Proof.

This proof is similar to that of Lemma 4.3.2. For $s>0$ we consider the
multilinear mapping $\theta_{s}:\mathcal{A}^{\otimes p}\to\mathbb{C}$ given by

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

Here, as in Lemma 4.3.2, the operators $b_{k}$ are defined as in Definition
4.1.1, relative to the set $\mathscr{A}=\mathscr{A}_{1}$. From the computation
in Appendix A.3,

$\displaystyle(b\theta_{s})(a_{0}$ $\displaystyle\otimes\cdots\otimes a_{p})$
$\displaystyle=(-1)^{p}\mathrm{Tr}\left(\Gamma
a_{0}\left(\prod_{k=1}^{m-2}[b_{k},a_{k}]\right)[F,\delta(a_{m-1})]\left(\prod_{k=m}^{p-1}[b_{k+1},a_{k}]\right)[D^{2-|\mathscr{A}_{1}|}e^{-s^{2}D^{2}},a_{p}]\right)$
$\displaystyle+(-1)^{m-1}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{1}}(a_{0}\otimes\cdots\otimes
a_{p})D^{2-|\mathscr{A}_{1}|}e^{-s^{2}D^{2}})$
$\displaystyle+(-1)^{m-1}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{2}}(a_{0}\otimes\cdots\otimes
a_{p})D^{2-|\mathscr{A}_{2}|}e^{-s^{2}D^{2}}).$

We first show that the first summand above is $O(s^{-1})$. Indeed,

$\displaystyle\Big{\|}\Gamma a_{0}\left(\prod_{k=1}^{m-2}[b_{k},a_{k}]\right)$
$\displaystyle[F,\delta(a_{m-1})]\left(\prod_{k=m}^{p-1}[b_{k+1},a_{k}]\right)[D^{2-|\mathscr{A}_{1}|}e^{-s^{2}D^{2}},a_{p}]\Big{\|}_{1}$
$\displaystyle\leq\left\|D^{|\mathscr{A}_{1}|-p}[D^{2-|\mathscr{A}_{1}|}e^{-s^{2}D^{2}},a_{p}]\right\|_{1}$
$\displaystyle\times\left\|\Gamma
a_{0}\prod_{k=1}^{m-2}[b_{k},a_{k}][F,\delta(a_{m-1})]\prod_{k=m}^{p-1}[b_{k+1},a_{k}]\cdot|D|^{p-|\mathscr{A}_{1}|}\right\|_{\infty}.$

The first factor above is $O(s^{-1})$ due to Lemma 4.2.5, and the second
factor is finite by Lemma A.4.3 and has no dependence on $s$.

Summarising the above, if $c\in\mathcal{A}^{\otimes(p+1)}$ we have

$\displaystyle(b\theta_{s})(c)$
$\displaystyle=(-1)^{m-1}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{1}}(c)D^{2-|\mathscr{A}_{1}|}e^{-s^{2}D^{2}})$
$\displaystyle+(-1)^{m-1}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{2}}(c)D^{2-|\mathscr{A}_{2}|}e^{-s^{2}D^{2}})+O(s^{-1})$

as $s\downarrow 0$. Hence, if $c$ is a Hochschild cycle then
$(b\theta_{s})(c)=\theta_{s}(bc)=0$, and this completes the proof. ∎

######  Lemma 4.3.4.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. For every Hochschild cycle $c\in\mathcal{A}^{\otimes(p+1)}$
and for every $\mathscr{A}\subset\\{1,\cdots,p\\}$ with $|\mathscr{A}|\geq 2,$
we have

$\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})=O(s^{-1}),\quad
s\downarrow 0.$

###### Proof.

Let $m$ be the maximum element in $\mathscr{A}$ and let $n$ be the maximal
element in $\mathscr{A}\backslash\\{m\\}.$ If $n=m-1,$ then the assertion is
already proved in Lemma 4.3.2. If $n<m-1,$ then $m-n>1$, and hence
$n+j\notin\mathscr{A}$ for all $1\leq j<m-n$. Now for each $0\leq j<m-n$ we
define $\mathscr{A}_{j}$ to be $\mathscr{A}$ with $n$ replaced with $n+j$.
That is:

$\mathscr{A}_{j}:=(\mathscr{A}\backslash\\{n\\})\cup\\{j+n\\},\quad 0\leq
j<m-n.$

Then by construction:

  1. (1)

$|\mathscr{A}_{j}|=|\mathscr{A}|$ and
$\mathscr{A}_{j}\Delta\mathscr{A}_{j-1}=\\{n+j,n+j-1\\}$ for all $1\leq
j<m-n.$

  2. (2)

$\mathscr{A}_{0}=\mathscr{A}$ and $m-1,m\in\mathscr{A}_{m-n-1}.$

Hence if $1\leq j<m-n$ the subsets $\mathscr{A}_{j}$ and $\mathscr{A}_{j-1}$
satisfy the conditions of Lemma 4.3.3. So for all Hochschild cycles
$c\in\mathcal{A}^{\otimes(p+1)}$:

$\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{j-1}}(c)D^{2-|\mathscr{A}_{i-1}|}e^{-s^{2}D^{2}})=-\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{j}}(c)D^{2-|\mathscr{A}_{j}|}e^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0.$

So by induction, we have:

$\displaystyle\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{0}}(c)D^{2-|\mathscr{A}_{0}|}e^{-s^{2}D^{2}})$
$\displaystyle=(-1)^{m-n-1}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{m-n-1}}(c)D^{2-|\mathscr{A}_{m-n-1}|}e^{-s^{2}D^{2}})$
(4.10) $\displaystyle\quad+O(s^{-1}),\quad s\downarrow 0.$

On the other hand, since $m-1,m\in\mathscr{A}_{m-n-1}$, we may apply Lemma
4.3.2 to $\mathscr{A}_{m-n-1}$ to obtain:

$\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{m-n-1}}(c)D^{2-|\mathscr{A}_{m-n-1}|}e^{-s^{2}D^{2}})=O(s^{-1}),\quad
s\downarrow 0.$

Combining (4.3) and (4.3), we get

$\mathrm{Tr}(\mathcal{W}_{\mathscr{A}_{0}}(c)D^{2-|\mathscr{A}_{0}|}e^{-s^{2}D^{2}})=O(s^{-1}).$

since $\mathscr{A}_{0}=\mathscr{A}$, the proof is complete. ∎

Recall the mapping $\mathrm{ch}$ from Definition 2.2.15.

######  Lemma 4.3.5.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. For every $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\displaystyle\|\mathrm{ch}(c)D^{2}e^{-s^{2}D^{2}}\|_{1}=O(s^{-1}),\quad
s\downarrow 0.$

###### Proof.

Recall that on $H_{\infty}$ we have $[F,a_{p}]|D|=L(a_{p})$. So on
$H_{\infty}$:

$\displaystyle[F,a_{p-1}][F,a_{p}]|D|^{2}$ $\displaystyle=[F,a_{p-1}]\cdot
L(a_{p})\cdot|D|$
$\displaystyle=[F,a_{p-1}]\cdot|D|L(a_{p})-[F,a_{p-1}]\cdot[|D|,L(a_{p})]$
$\displaystyle=[F,a_{p-1}]|D|\cdot L(a_{p})-[F,a_{p-1}]\cdot\delta(L(a_{p}))$
$\displaystyle=L(a_{p-1})\cdot L(a_{p})-[F,a_{p-1}]\cdot L(\delta(a_{p})).$

So for $c=a_{0}\otimes\cdots\otimes a_{p}\in\mathcal{A}^{\otimes(p+1)}$ we
have

$\displaystyle\mathrm{ch}(c)\cdot D^{2}e^{-s^{2}D^{2}}$
$\displaystyle=\Gamma\left(\prod_{k=0}^{p-2}[F,a_{k}]\right)|D|^{p-1}\cdot|D|^{1-p}L(a_{p-1})L(a_{p})e^{-s^{2}D^{2}}$
$\displaystyle\quad-\Gamma\left(\prod_{k=0}^{p-1}[F,a_{k}]\right)|D|^{p-1}\cdot|D|^{1-p}L(\delta(a_{p}))e^{-s^{2}D^{2}}.$

Using Lemma A.4.4, the operators
$\left(\prod_{k=0}^{p-2}[F,a_{k}]\right)|D|^{p-1}$ and
$\left(\prod_{k=0}^{p-1}[F,a_{k}]\right)|D|^{p-1}$ both have bounded extension
and no dependence on $s$. From Lemma A.1.4, we have that:

$\displaystyle\||D|^{1-p}L(a_{p-1})L(a_{p})e^{-s^{2}D^{2}}\|_{1}$
$\displaystyle=O(s^{-1})$
$\displaystyle\||D|^{1-p}L(\delta(a_{p}))e^{-s^{2}D^{2}}\|_{1}$
$\displaystyle=O(s^{-1}).$

So by the triangle inequality:
$\|\mathrm{ch}(c)D^{2}e^{-s^{2}D^{2}}\|_{1}=O(s^{-1})$ as $s\downarrow 0$. ∎

######  Lemma 4.3.6.

Let $(\mathcal{A},H,D)$ be a spectral triple of dimension $p$, where $p$ has
the same parity as $(\mathcal{A},H,D)$. If $c\in\mathcal{A}^{\otimes(p+1)}$
then

$\mathrm{ch}(c)=\mathcal{W}_{\emptyset}(c)+F\mathcal{W}_{\emptyset}(c)F.$

###### Proof.

Let $c=a_{0}\otimes\cdots\otimes a_{p}\in\mathcal{A}^{\otimes(p+1)}$. Recall
that

$\mathcal{W}_{\emptyset}(c)=\Gamma a_{0}\prod_{k=1}^{p}[F,a_{k}].$

Using the fact that $F$ anticommutes with $[F,a_{k}]$ for all $k$, we have:

$\displaystyle\mathrm{ch}(c)$ $\displaystyle=\Gamma
F[F,a_{0}]\prod_{k=1}^{p}[F,a_{k}]$ $\displaystyle=\Gamma
a_{0}\prod_{k=1}^{p}[F,a_{k}]-\Gamma Fa_{0}F\prod_{k=1}^{p}[F,a_{k}].$
$\displaystyle=\mathcal{W}_{\emptyset}(c)+(-1)^{p+1}\Gamma
Fa_{0}\left(\prod_{k=1}^{p}[F,a_{k}]\right)F.$

Since $\Gamma^{2}=1$,

$\mathrm{ch}(c)=\mathcal{W}_{\emptyset}(c)+(-1)^{p+1}\Gamma
F\Gamma\mathcal{W}_{\emptyset}(c)F.$

Since the parity of $p$ matches the parity of $\Gamma$, we have $\Gamma
F=(-1)^{p+1}F\Gamma$. This completes the proof. ∎

######  Lemma 4.3.7.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. For every $c\in\mathcal{A}^{\otimes(p+1)}$, we have

$\mathrm{Tr}(\mathcal{W}_{\emptyset}(c)D^{2}e^{-s^{2}D^{2}})=O(s^{-1}),\quad
s\downarrow 0.$

###### Proof.

By Lemma 4.3.6, we have:

$\mathrm{Tr}(\mathrm{ch}(c)D^{2}e^{-s^{2}D^{2}})=\mathrm{Tr}(\mathcal{W}_{\emptyset}(c)D^{2}e^{-s^{2}D^{2}})+\mathrm{Tr}(F\mathcal{W}_{\emptyset}(c)FD^{2}e^{-s^{2}D^{2}}).$

However since $F$ commutes with $D^{2}e^{-s^{2}D^{2}}$ and $F^{2}=1$ we have:

$2\mathrm{Tr}(\mathcal{W}_{\emptyset}(c)D^{2}e^{-s^{2}D^{2}})=\mathrm{Tr}(\mathrm{ch}(c)D^{2}e^{-s^{2}D^{2}}).$

However by Lemma 4.3.5,

$|\mathrm{Tr}(\mathrm{ch}(c)D^{2}e^{-s^{2}D^{2}})|=O(s^{-1}).$

Hence $\mathrm{Tr}(\mathcal{W}_{\emptyset}(c)D^{2}e^{-s^{2}D^{2}})=O(s^{-1})$.
∎

We are now ready to prove the main result of this section.

###### Proof of Theorem 4.3.1.

Let $c\in\mathcal{A}^{\otimes(p+1)}$. Then using Theorem 4.1.7:

$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=\sum_{\mathscr{A}\subseteq\\{1,\ldots,p\\}}(-1)^{n_{\mathscr{A}}}\mathrm{Tr}(\mathcal{W}_{\mathscr{A}}(c)D^{2-|\mathscr{A}|}e^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0.$

Applying Lemma 4.3.4 to every summand with $|\mathscr{A}|\geq 2$, and Lemma
4.3.7 to the summand $\mathscr{A}=\emptyset$, it follows that:

$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=\sum_{k=1}^{p}(-1)^{n_{\\{k\\}}}\mathrm{Tr}(\mathcal{W}_{k}(c)De^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0.$

Recall that
$n_{\mathscr{A}}=|\\{(j,k)\in\\{1,\ldots,p\\}^{2}\;:\;j\in\mathscr{A},k\notin\mathscr{A}\\}|$.
So in particular, $n_{\\{k\\}}=p-k$. Hence:

(4.11)
$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=\sum_{k=1}^{p}(-1)^{p-k}\mathrm{Tr}(\mathcal{W}_{k}(c)De^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0.$

For any $1\leq k\leq p-1$, the sets $\mathscr{A}_{1}=\\{k\\}$ and
$\mathscr{A}_{2}=\\{k+1\\}$ satisfy the conditions of Lemma 4.3.3 with
$m=k+1$. So we have

$\mathrm{Tr}(\mathcal{W}_{k}(c)De^{-s^{2}D^{2}})=-\mathrm{Tr}(\mathcal{W}_{k+1}(c)De^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0,$

Hence by induction, for any $1\leq k\leq p$:

(4.12)
$\mathrm{Tr}(\mathcal{W}_{k}(c)De^{-s^{2}D^{2}})=(-1)^{p-k}\mathrm{Tr}(\mathcal{W}_{p}De^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0.$

Substituting (4.12) into each summand of (4.11) we finally get:

$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=p\mathrm{Tr}(\mathcal{W}_{p}(c)De^{-s^{2}D^{2}})+O(s^{-1}).$

∎

###  4.4. Preliminary heat semigroup asymptotic

In this section, we move closer to proving Theorem 1.2.2. We will show that if
$(\mathcal{A},H,D)$ satisfies Hypothesis 1.2.1 and Hypothesis 4.0.1 (in
particular, $D$ has a spectral gap at $0$), then for a Hochschild cycle
$c\in\mathcal{A}^{\otimes(p+1)}$ we have

(4.13)
$\mathrm{Tr}(\mathcal{W}_{p}(c)|D|^{2-p}e^{-s^{2}D^{2}})=\frac{1}{4}\mathrm{Tr}(\mathrm{ch}(c))s^{-2}+O(s^{-1}),s\downarrow
0.$

By the Theorem 4.3.1, this is a formula very close to Theorem 1.2.2: the only
difference is the assumption of Hypothesis 4.0.1 and that the occurance of
$|D|$ should be replaced with $(1+D^{2})^{1/2}.$

We start with the following asymptotic result.

######  Lemma 4.4.1.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. For all $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\mathrm{Tr}(\mathrm{ch}(c)e^{-s^{2}D^{2}})=\mathrm{Tr}(\mathrm{ch}(c))+O(s),\quad
s\downarrow 0.$

###### Proof.

We wish to show that

$\mathrm{Tr}(\mathrm{ch}(c)(1-e^{-s^{2}D^{2}}))=O(s)$

so it suffices to prove

$\|\mathrm{ch}(c)(1-e^{-s^{2}D^{2}})\|_{1}=O(s).$

Let $c=a_{0}\otimes\cdots\otimes a_{p}\in\mathcal{A}^{\otimes(p+1)}$. Then
using $[F,a_{p}]=|D|^{-1}\partial(a_{p})-|D|^{-1}\delta(a_{p})F$, we have (on
$H_{\infty}$)

$\displaystyle\mathrm{ch}(c)(1-e^{-s^{2}D^{2}})$ $\displaystyle=\Gamma
F\left(\prod_{k=0}^{p-1}[F,a_{k}]\right)(|D|^{p}|D|^{-p})[F,a_{p}](1-e^{-s^{2}D^{2}})$
$\displaystyle=\Gamma
F\left(\prod_{k=0}^{p-1}[F,a_{k}]\right)|D|^{p}\cdot|D|^{-p-1}\partial(a_{p})(1-e^{-s^{2}D^{2}})$
$\displaystyle\quad-\Gamma
F\left(\prod_{k=0}^{p-1}[F,a_{k}]\right)|D|^{p}\cdot|D|^{-p-1}\delta(a_{p})(1-e^{-s^{2}D^{2}})F.$

Thus,

$\displaystyle\|\mathrm{ch}(c)(1-e^{-s^{2}D^{2}})\|_{1}$
$\displaystyle\leq\left\|\left(\prod_{k=0}^{p-1}[F,a_{k}]\right)|D|^{p}\right\|_{\infty}\cdot\left\||D|^{-p-1}\partial(a_{p})(1-e^{-s^{2}D^{2}})\right\|_{1}$
$\displaystyle\quad+\left\|\left(\prod_{k=0}^{p-1}[F,a_{k}]\right)|D|^{p}\right\|_{\infty}\cdot\left\||D|^{-p-1}\delta(a_{p})(1-e^{-s^{2}D^{2}})\right\|_{1}.$

In both summands, the first factor is finite by Lemma A.4.4, the second factor
is $O(s)$ by Lemma A.1.6. ∎

######  Theorem 4.4.2.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1 and
Hypothesis 4.0.1. For every Hochschild cycle $c\in\mathcal{A}^{\otimes(p+1)},$
we have

(4.14)
$\mathrm{Tr}(\mathcal{W}_{p}(c)De^{-s^{2}D^{2}})=\frac{\mathrm{Ch}(c)}{2}s^{-2}+O(s^{-1}),\quad
s\downarrow 0.$

###### Proof.

Let $c\in\mathcal{A}^{\otimes(p+1)}$ be a Hochschild cycle. By Lemma 4.4.1, we
have

$\mathrm{Tr}(\mathrm{ch}(c)e^{-s^{2}D^{2}})=\mathrm{Tr}(\mathrm{ch}(c))+O(s),\quad
s\downarrow 0.$

Since the spectral triple and $p$ both have the same parity, we may apply
Lemma 4.3.6 to get:

(4.15)
$2\mathrm{Tr}(\mathcal{W}_{\emptyset}(c)e^{-s^{2}D^{2}})=\mathrm{Tr}(\mathrm{ch}(c))+O(s),\quad
s\downarrow 0,$

for all $c\in\mathcal{A}^{\otimes(p+1)}$

Define the multilinear mappings
$\mathcal{K}_{s},\,\mathcal{H}_{s}:\mathcal{A}^{\otimes(p+1)}\to\mathbb{C}$ by
setting

$\displaystyle\mathcal{K}_{s}(a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle=\mathrm{Tr}(\Gamma
a_{0}\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)[Fe^{-s^{2}D^{2}},a_{p}]),$
$\displaystyle\mathcal{H}_{s}(a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle=\mathrm{Tr}(\Gamma
a_{0}\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)F[e^{-s^{2}D^{2}},a_{p}]).$

By the Leibniz rule, we have

$[F,a_{p}]e^{-s^{2}D^{2}}=[Fe^{-s^{2}D^{2}},a_{p}]-F[e^{-s^{2}D^{2}},a_{p}].$

Therefore:

(4.16)
$\mathrm{Tr}(\mathcal{W}_{\emptyset}(c)e^{-s^{2}D^{2}})=\mathcal{K}_{s}(c)-\mathcal{H}_{s}(c).$

Now combining (4.15) and (4.16), we arrive at

(4.17)
$2\mathcal{K}_{s}(c)-2\mathcal{H}_{s}(c)=\mathrm{Tr}(\mathrm{ch}(c))+O(s),\quad
s\downarrow 0,$

However it is shown in Appendix A.3 that $\mathcal{K}_{s}$ is a Hochschild
coboundary, and thus since $c$ is a Hochschild cycle we have
$\mathcal{K}_{s}(c)=0$.

Using (4.17), we obtain

(4.18) $-2\mathcal{H}_{s}(c)=\mathrm{Tr}(\mathrm{ch}(c))+O(s),\quad
s\downarrow 0,$

Define the multilinear mapping
$\mathcal{V}_{s}:\mathcal{A}^{\otimes(p+1)}\to\mathbb{C}$ by setting

$\mathcal{V}_{s}(a_{0}\otimes\cdots\otimes a_{p})=\mathrm{Tr}(\Gamma
a_{0}\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)F\delta(a_{p})|D|e^{-s^{2}D^{2}}).$

Let $\frac{1}{q}=1-\frac{1}{p}$. By the Hölder inequality in the form (2.7):

$\displaystyle|(\mathcal{H}_{s}+2s^{2}\mathcal{V}_{s})$
$\displaystyle(a_{0}\otimes\cdots\otimes a_{p})|$
$\displaystyle=\left|\mathrm{Tr}\Big{(}\Gamma
a_{0}\Big{(}\prod_{k=1}^{p-1}[F,a_{k}]\Big{)}F\cdot\Big{(}[e^{-s^{2}D^{2}},a_{p}]+2s^{2}\delta(a_{p})|D|e^{-s^{2}D^{2}}\Big{)}\Big{)}\right|$
$\displaystyle\leq\Big{\|}\Gamma
a_{0}\prod_{k=1}^{p-1}[F,a_{k}]F\Big{\|}_{q,\infty}\Big{\|}[e^{-s^{2}D^{2}},a_{p}]+2s^{2}\delta(a_{p})|D|e^{-s^{2}D^{2}}\Big{\|}_{p,1}.$

The first factor on the above right hand side is finite by Proposition 3.1.5,
and the second factor above is $O(s)$ by Lemma 4.2.4.(iii). Therefore, we have

(4.19) $(\mathcal{H}_{s}+2s^{2}\mathcal{V}_{s})(c)=O(s),\quad s\downarrow 0,$

Combining (4.19) and (4.18), we arrive at

(4.20) $4s^{2}\mathcal{V}_{s}(c)=\mathrm{Tr}(\mathrm{ch}(c))+O(s),\quad
s\downarrow 0,$

for all Hochschild cycles $c\in\mathcal{A}^{\otimes(p+1)}.$

From the definition of $\mathcal{W}_{p}$, if $a_{0}\otimes\cdots\otimes
a_{p}\in\mathcal{A}^{\otimes(p+1)}$:

$\displaystyle\mathrm{Tr}(\mathcal{W}_{p}(a_{0}\otimes\cdots\otimes
a_{p})De^{-s^{2}D^{2}})$ $\displaystyle=\mathrm{Tr}(\Gamma
a_{0}\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)\delta(a_{p})De^{-s^{2}D^{2}})$
$\displaystyle=\mathrm{Tr}(\Gamma
a_{0}\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)\delta(a_{p})F|D|e^{-s^{2}D^{2}}).$

Thus,

$\displaystyle\Big{|}\mathcal{V}_{s}(a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle-\mathrm{Tr}(\mathcal{W}_{p}(a_{0}\otimes\cdots\otimes
a_{p})De^{-s^{2}D^{2}})\Big{|}$ $\displaystyle=\Big{|}\mathrm{Tr}(\Gamma
a_{0}\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)[F,\delta(a_{p})]|D|e^{-s^{2}D^{2}})\Big{|}$
$\displaystyle=\Big{|}\mathrm{Tr}(\Gamma\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)[F,\delta(a_{p})]|D|e^{-s^{2}D^{2}}a_{0})\Big{|}$
$\displaystyle\leq\Big{\|}\Gamma\left(\prod_{k=1}^{p-1}[F,a_{k}]\right)[F,\delta(a_{p})]|D|^{p}\Big{\|}_{\infty}\cdot\Big{\|}|D|^{1-p}e^{-s^{2}D^{2}}a_{0}\Big{\|}_{1}.$

By Lemma A.1.4, we have that this is $O(s^{-1})$ and therefore, we have

(4.21)
$\mathcal{V}_{s}(c)=\mathrm{Tr}(\mathcal{W}_{p}(c)De^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0,$

for all $c\in\mathcal{A}^{\otimes(p+1)}.$

Combining (4.20) and (4.21), we arrive at

$4s^{2}\mathrm{Tr}(\mathcal{W}_{p}(c)De^{-s^{2}D^{2}})=\mathrm{Tr}(\mathrm{ch}(c))+O(s),\quad
s\downarrow 0$

for all Hochschild cycles $c\in\mathcal{A}^{\otimes(p+1)}.$ Dividing by
$4s^{2}$,

$\mathrm{Tr}(\mathcal{W}_{p}(c)De^{-s^{2}D^{2}})=\frac{1}{4}s^{-2}\mathrm{Tr}(\mathrm{ch}(c))+O(s^{-1}).$

Since $\mathrm{Ch}(c)=\frac{1}{2}\mathrm{Tr}(\mathrm{ch}(c))$, this formula
coincides with (4.14). ∎

We remark that (4.13) follows as a simple combination of Theorems 4.3.1 and
4.4.2.

###  4.5. Heat semigroup asymptotic: the proof of the first main result

In this section, we finally complete the proof of Theorem 1.2.2. We start by
removing the assumption of Hypothesis 4.0.1.

The following two lemmas show that if the parity of $p$ does not match
$(\mathcal{A},H,D)$, then the statement of (4.13) becomes trivial.

######  Lemma 4.5.1.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1.
Suppose that $D$ has a spectral gap at $0$. Suppose that the dimension $p$ is
odd but $(\mathcal{A},H,D)$ is even. Then:

  1. (i)

for every $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=0,\quad s>0.$

  2. (ii)

for every $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\mathrm{Tr}(\mathrm{ch}(c))=0.$

###### Proof.

Let us prove (i). Since $\Gamma D=-D\Gamma$ and $\Gamma$ commutes with
$a\in\mathcal{A}$, we have $\Gamma[D,a]=-[D,a]\Gamma$ on $H_{\infty}$. Hence
$\Gamma\partial(a)=-\partial(a)\Gamma$ for all $a\in\mathcal{A}$. Thus for
$c=a_{0}\otimes\cdots\otimes a_{p}\in\mathcal{A}^{\otimes(p+1)}$ since $p$ is
odd we have:

$\Omega(c)=\Gamma
a_{0}\prod_{k=1}^{p}\partial(a_{k})=-a_{0}\left(\prod_{k=1}^{p}\partial(a_{k})\right)\Gamma.$

However $\Gamma D^{2}=D^{2}\Gamma$, so by the spectral theorem we have $\Gamma
e^{-s^{2}D^{2}}=e^{-s^{2}D^{2}}$. So multiplying by $e^{-s^{2}D^{2}}$ on the
right and taking the trace,

$\displaystyle\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})$
$\displaystyle=-\mathrm{Tr}(a_{0}\left(\prod_{k=1}^{p}\partial(a_{k})\right)|D|^{2-p}e^{-s^{2}D^{2}}\Gamma)$
$\displaystyle=-\mathrm{Tr}(\Gamma
a_{0}\left(\prod_{k=1}^{p}\partial(a_{k})\right)|D|^{2-p}e^{-s^{2}D^{2}}).$

This proves (i)

The argument for (ii) is similar. Note that we have $\Gamma[F,a]=-[F,a]\Gamma$
for every $a\in\mathcal{A}.$ Thus since $p+1$ is even:

$\displaystyle\mathrm{ch}(c)$ $\displaystyle=\Gamma F\prod_{k=0}^{p}[F,a_{k}]$
$\displaystyle=-F\Gamma\prod_{k=0}^{p}[F,a_{k}]$
$\displaystyle=-F\cdot\prod_{k=0}^{p}[F,a_{k}]\Gamma.$

Thus,

$\displaystyle\mathrm{Tr}(\mathrm{ch}(c))$ $\displaystyle=-\mathrm{Tr}(\Gamma
F\prod_{k=0}^{p}[F,a_{k}])$ $\displaystyle=-\mathrm{Tr}(\mathrm{ch}(c)).$

This proves the second assertion. ∎

Now, we deal with the other case where the parity of $(\mathcal{A},H,D)$ does
not match $p$.

######  Lemma 4.5.2.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1.
Suppose $D$ has a spectral gap at $0$ and $(\mathcal{A},H,D)$ is odd but $p$
is even.

  1. (i)

for every Hochschild cycle $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=O(s^{-1}),\quad s\downarrow
0.$

  2. (ii)

for every $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\mathrm{Tr}(\mathrm{ch}(c))=0.$

###### Proof.

First, we prove (i). Consider the multilinear mapping
$\theta_{s}:\mathcal{A}^{\otimes p}\to\mathbb{C}$ defined by the formula

$\theta_{s}(a_{0}\otimes\cdots\otimes
a_{p-1})=\mathrm{Tr}(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)|D|^{2-p}e^{-s^{2}D^{2}}).$

The Hochschild coboundary $b\theta_{s}$ is computed in Section A.3 by the
formula:

$\displaystyle(b\theta_{s})(a_{0}\otimes\cdots\otimes a_{p})$
$\displaystyle=2\mathrm{Tr}(a_{0}\left(\prod_{k=1}^{p}\partial(a_{k})\right)|D|^{2-p}e^{-s^{2}D^{2}})$
$\displaystyle\quad+\mathrm{Tr}(a_{0}\left(\prod_{k=1}^{p-1}\partial(a_{k})\right)[|D|^{2-p}e^{-s^{2}D^{2}},\partial(a_{p})])$
$\displaystyle\quad+\mathrm{Tr}(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)[|D|^{2-p}e^{-s^{2}D^{2}},a_{p}]).$

Using the Hölder inequality, we have

$\displaystyle\Big{|}\mathrm{Tr}(a_{0}\left(\prod_{k=1}^{p-1}\partial(a_{k})\right)$
$\displaystyle[|D|^{2-p}e^{-s^{2}D^{2}},\partial(a_{p})])\Big{|}$
$\displaystyle\leq\|a_{0}\|_{\infty}\prod_{k=1}^{p-1}\|\partial(a_{k})\|_{\infty}\cdot\Big{\|}[|D|^{2-p}e^{-s^{2}D^{2}},\partial
a_{p}])\Big{\|}_{1}.$

By Lemma 4.2.5, we have

(4.22)
$\mathrm{Tr}(a_{0}\left(\prod_{k=1}^{p-1}\partial(a_{k})\right)[|D|^{2-p}e^{-s^{2}D^{2}},\partial(a_{p})])=O(s^{-1}),\quad
s\downarrow 0.$

Similarly, we have

$\Big{|}\mathrm{Tr}(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)[|D|^{2-p}e^{-s^{2}D^{2}},a_{p}])\Big{|}\leq\prod_{k=0}^{p-1}\|\partial(a_{k})\|_{\infty}\cdot\Big{\|}[|D|^{2-p}e^{-s^{2}D^{2}},a_{p}])\Big{\|}_{1}.$

By Lemma 4.2.5, we have

(4.23)
$\mathrm{Tr}(\left(\prod_{k=0}^{p-1}\partial(a_{k})\right)[|D|^{2-p}e^{-s^{2}D^{2}},a_{p}])=O(s^{-1}),\quad
s\downarrow 0.$

For every $c\in\mathcal{A}^{\otimes(p+1)},$ it follows from (4.22) and (4.23)
that

$(b\theta_{s})(c)=2\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})+O(s^{-1}),\quad
s\downarrow 0.$

If $c$ is a Hochschild cycle, then $(b\theta_{s})(c)=0.$ Thus,

$\mathrm{Tr}(\Omega(c)|D|^{2-p}e^{-s^{2}D^{2}})=O(s^{-1}),\quad s\downarrow
0.$

for every Hochschild cycle $c.$ This completes the proof of (i).

The proof of (ii) is similar to Lemma 4.5.1.(ii). For all $a\in\mathcal{A}$,
we have $F[F,a]=-[F,a]F.$ Since $p+1$ is odd,

$F\cdot\prod_{k=0}^{p}[F,a_{k}]=-\left(\prod_{k=0}^{p}[F,a_{k}]\right)F.$

Hence,

$\mathrm{Tr}(F\prod_{k=0}^{p}[F,a_{k}])=-\mathrm{Tr}(F\prod_{k=0}^{p}[F,a_{k}]).$

This proves (ii). ∎

The preceding two lemmas show how to remove the assumption that the parities
of $p$ and $(\mathcal{A},H,D)$ match. It remains to remove the assumption that
$D$ has a spectral gap at $0.$ For this purpose, we use the doubling trick.
The ”doubling trick” in this form follows [12, Definition 6].

Let $\mu>0.$ We define another spectral triple
$(\pi(\mathcal{A}),H_{0},D_{\mu}),$ where

$\displaystyle H_{0}$ $\displaystyle=\mathbb{C}^{2}\otimes H,$ $\displaystyle
D_{\mu}$ $\displaystyle=\begin{pmatrix}D&\mu\\\ \mu&-D\end{pmatrix}.$

and $\pi$ is the same representation of $\mathcal{A}$ as in Definition 2.2.17.
That is:

$\pi(a):=\begin{pmatrix}a&0\\\ 0&0\end{pmatrix}.$

For a tensor $c\in\mathcal{A}^{\otimes(p+1)}$, we denote $\pi(c)$ for the
corresponding element of $(\pi(\mathcal{A}))^{\otimes(p+1)}$ obtained by
applying the map $\pi^{\otimes(p+1)}$ to $c$. The spectral triple
$(\pi(\mathcal{A}),H_{0},D_{\mu})$ is equipped with grading operator

$\Gamma_{0}=\begin{pmatrix}\Gamma&0\\\ 0&(-1)^{\rm deg}\Gamma\end{pmatrix}$

where $\Gamma$ is the (possibly trivial) grading of $(\mathcal{A},H,D)$ (see
Definition 2.2.17).

Let $\Omega_{\mu}$ and ${\rm ch}_{\mu}$ be the multilinear mappings $\Omega$
and ${\rm ch}$ (just as in Definition 2.2.15) as applied to the spectral
triple $(\pi(\mathcal{A}),H_{0},D_{\mu}).$

######  Lemma 4.5.3.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1. Let
$F_{0}$ be as in Definition 2.2.17. If $a\in\mathcal{A},$ then as
$\mu\downarrow 0$ we have:

$[{\rm sgn}(D_{\mu}),\pi(a)]\to[F_{0},\pi(a)]$

in $\mathcal{L}_{p+1}$.

###### Proof.

We have

$\mathrm{sgn}(D_{\mu})=\begin{pmatrix}\frac{D}{(D^{2}+\mu^{2})^{1/2}}&\frac{\mu}{(D^{2}+\mu^{2})^{1/2}}\\\
\frac{\mu}{(D^{2}+\mu^{2})^{1/2}}&-\frac{D}{(D^{2}+\mu^{2})^{1/2}}\end{pmatrix}.$

Hence,

$[\mathrm{sgn}(D_{\mu}),\pi(a)]=\begin{pmatrix}\left[\frac{D}{(D^{2}+\mu^{2})^{1/2}},a\right]&-a\frac{\mu}{(D^{2}+\mu^{2})^{1/2}}\\\
\frac{\mu}{(D^{2}+\mu^{2})^{1/2}}a&0.\end{pmatrix}$

On the other hand, we have

$[F_{0},\pi(a)]=\begin{pmatrix}[\mathrm{sgn}(D),a]&-aP\\\ Pa&0\end{pmatrix}.$

Therefore:

$\displaystyle\Big{\|}[{\rm sgn}(D_{\mu}),\pi(a)$
$\displaystyle]-[F_{0},\pi(a)]\Big{\|}_{p+1}$
$\displaystyle\leq\Big{\|}\Big{(}{\rm
sgn}(D)-\frac{D}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}a\Big{\|}_{p+1}+\Big{\|}a\Big{(}{\rm
sgn}(D)-\frac{D}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}\Big{\|}_{p+1}$
$\displaystyle\quad+\Big{\|}\Big{(}P-\frac{\mu}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}a\Big{\|}_{p+1}+\Big{\|}a\Big{(}P-\frac{\mu}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}\Big{\|}_{p+1}$
$\displaystyle=\Big{\|}a^{*}\Big{(}{\rm
sgn}(D)-\frac{D}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}a\Big{\|}_{\frac{p+1}{2}}^{\frac{1}{2}}+\Big{\|}a\Big{(}{\rm
sgn}(D)-\frac{D}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}a^{*}\Big{\|}_{\frac{p+1}{2}}^{\frac{1}{2}}$
$\displaystyle\quad+\Big{\|}a^{*}\Big{(}P-\frac{\mu}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}a\Big{\|}_{\frac{p+1}{2}}^{\frac{1}{2}}+\Big{\|}a\Big{(}P-\frac{\mu}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}a^{*}\Big{\|}_{\frac{p+1}{2}}^{\frac{1}{2}}.$

By functional calculus, as $\mu\downarrow 0$ we have:

$\displaystyle\Big{(}{\rm
sgn}(D)-\frac{D}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}$
$\displaystyle\downarrow 0,$
$\displaystyle\Big{(}P-\frac{\mu}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}$
$\displaystyle\downarrow 0$

in the weak operator topology. Hence,

$a^{*}\Big{(}{\rm
sgn}(D)-\frac{D}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}a\downarrow 0,\quad
a\Big{(}{\rm
sgn}(D)-\frac{D}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}a^{*}\downarrow
0,\quad\mu\downarrow 0,$
$a^{*}\Big{(}P-\frac{\mu}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}a\downarrow
0,\quad
a\Big{(}P-\frac{\mu}{(D^{2}+\mu^{2})^{\frac{1}{2}}}\Big{)}^{2}a^{*}\downarrow
0,\quad\mu\downarrow 0.$

also in the weak operator topology. The assertion follows now from the order
continuity of the $\mathcal{L}_{\frac{p+1}{2}}$ norm. ∎

######  Lemma 4.5.4.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1. If
$c\in\mathcal{A}^{\otimes(p+1)},$ then

$\lim_{\mu\to 0}{\rm ch}_{\mu}(\pi(c))={\rm ch}_{0}(c)$

in the $\mathcal{L}_{1}$-norm.

###### Proof.

It suffices to prove the assertion for $c=a_{0}\otimes\cdots\otimes a_{p}.$
Then we have

$\displaystyle{\rm ch}_{0}(c)-{\rm ch}_{\mu}(\pi(c))$
$\displaystyle=\Gamma_{0}(F_{0}-{\rm
sgn}(D_{\mu}))\prod_{k=0}^{p}[F_{0},\pi(a_{k})]$
$\displaystyle-\Gamma_{0}\mathrm{sgn}(D_{\mu})\Big{(}\prod_{k=0}^{p}[{\rm
sgn}(D_{\mu}),\pi(a_{k})]-\prod_{k=0}^{p}[F_{0},\pi(a_{k})]\Big{)}.$

Next we use the fact that if $q\geq 1$ and if $A_{\mu}\to A$ in the strong
operator topology, $B_{\mu}\to B\in\mathcal{L}_{q}$ in the $\mathcal{L}_{q}$
norm, and $\sup_{\mu\downarrow 0}\|A_{\mu}\|_{\infty}<\infty$ then
$A_{\mu}B_{\mu}\to AB$ in $\mathcal{L}_{1}$ (see [60, Chapter 2, Example 3]).

We have that $\mathrm{sgn}(D_{\mu})-F_{0}\to 0$ in the strong operator
topology, and since $\prod_{k=0}^{p}[F_{0},\pi(a_{k})]\in\mathcal{L}_{1}$, it
follows that the first summand converges to $0$ in the $\mathcal{L}_{1}$ norm.

For the second summand, we apply Lemma 4.5.3. Since for each $k$ we have that
$[\mathrm{sgn}(D_{\mu}),\pi(a_{k})]\to[F_{0},\pi(a_{k})]$ in
$\mathcal{L}_{p+1}$, it follows that the second summand converges to $0$ in
$\mathcal{L}_{1}$. ∎

######  Lemma 4.5.5.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1. For
every $c\in\mathcal{A}^{\otimes(p+1)},$ we have

$\Big{\|}\Omega_{\mu}(\pi(c))(1\otimes(1+D^{2})^{-\frac{p}{2}}e^{-s^{2}D^{2}})\Big{\|}_{1}=O(s^{-1}),\quad
s\downarrow 0.$

###### Proof.

Once more it suffices to prove the assertion for an elementary tensor
$c=a_{0}\otimes\cdots\otimes a_{p}.$ We have

$\displaystyle\Big{\|}\Omega_{\mu}$
$\displaystyle(\pi(c))(1\otimes(1+D^{2})^{-\frac{p}{2}}e^{-s^{2}D^{2}})\Big{\|}_{1}$
$\displaystyle\leq\Big{\|}\Gamma_{0}\pi(a_{0})\prod_{k=1}^{p-1}[D_{\mu},\pi(a_{k})]\Big{\|}_{\infty}\Big{\|}[D_{\mu},\pi(a_{p})](1\otimes(1+D^{2})^{-\frac{p}{2}}e^{-s^{2}D^{2}})\Big{\|}_{1}$
$\displaystyle\leq\|a_{0}\|_{\infty}\prod_{k=1}^{p-1}\|[D_{\mu},\pi(a_{k})]\|_{\infty}\Big{\|}[D,a_{p}](1+D^{2})^{-\frac{p}{2}}e^{-s^{2}D^{2}}\Big{\|}_{1}$
$\displaystyle\quad+2\mu\|a_{0}\|_{\infty}\prod_{k=1}^{p-1}\|[D_{\mu},\pi(a_{k})]\|_{\infty}\Big{\|}a_{p}(1+D^{2})^{-\frac{p}{2}}e^{-s^{2}D^{2}}\Big{\|}_{1}.$

The assertion follows by applying Lemma A.1.4 (with $m_{1}=0$ and $m_{2}=p-1$)
to the odd spectral triple $(\mathcal{A},H,F(1+D^{2})^{\frac{1}{2}}).$ ∎

We note that since $\pi$ is an algebra homomorphism, if
$c\in\mathcal{A}^{\otimes(p+1)}$ is a Hochschild cycle then so is $\pi(c)$.

######  Lemma 4.5.6.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1. Let
$c\in\mathcal{A}^{\otimes(p+1)}$ be a Hochschild cycle. We have

$(\mathrm{Tr}_{2}\otimes\mathrm{Tr})(\Omega_{\mu}(\pi(c))(1\otimes(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}D^{2}}))=\frac{p}{2}\mathrm{Ch}_{\mu}(\pi(c))s^{-2}+O(s^{-1}),\quad
s\downarrow 0.$

###### Proof.

For $\mu>0,$ the spectral triple $(\pi(\mathcal{A}),H_{0},D_{\mu})$ satisfies
Hypothesis 1.2.1 and the spectral gap assumption. This allows us to apply
Theorems 4.3.1 and 4.4.2 (or Lemmas 4.5.1 and 4.5.2) to the Hochschild cycle
$\pi(c)\in(\pi(\mathcal{A}))^{\otimes(p+1)}.$

A combination of Theorems 4.3.1 and 4.4.2 (if the parities of $p$ and
$(\pi(\mathcal{A}),H_{0},D_{\mu})$ match) or one of Lemmas 4.5.1 and 4.5.2 (if
parities of $p$ and $(\pi(A),H_{0},D_{\mu})$ do not match) yields

$(\mathrm{Tr}_{2}\otimes\mathrm{Tr})(\Omega_{\mu}(\pi(c))|D_{\mu}|^{2-p}e^{-s^{2}D_{\mu}^{2}})=\frac{p}{2}\mathrm{Ch}_{\mu}(\pi(c))s^{-2}+O(s^{-1}),\quad
s\downarrow 0.$

Noting that $D_{\mu}^{2}=D_{0}^{2}+\mu^{2},$ and $e^{-s^{2}\mu^{2}}=O(1)$ we
obtain

$(\mathrm{Tr}_{2}\otimes\mathrm{Tr})(\Omega_{\mu}(\pi(c))|D_{\mu}|^{2-p}e^{-s^{2}D_{0}^{2}})=\frac{p}{2}\mathrm{Ch}_{\mu}(\pi(c))s^{-2}+O(s^{-1}),\quad
s\downarrow 0.$

By Lemma 4.5.5, we have

$\displaystyle\Big{\|}\Omega_{\mu}(\pi(c))$
$\displaystyle|D_{\mu}|^{2-p}e^{-s^{2}D_{0}^{2}}-\Omega_{\mu}(\pi(c))|D_{1}|^{2-p}e^{-s^{2}D_{0}^{2}}\Big{\|}_{1}$
$\displaystyle\leq\Big{\|}\Omega_{\mu}(\pi(c))|D_{1}|^{-p}e^{-s^{2}D_{0}^{2}}\Big{\|}_{1}\Big{\|}(|D_{\mu}|^{2-p}-|D_{1}|^{2-p})\cdot|D_{1}|^{p-2}\Big{\|}_{\infty}$
$\displaystyle=O(s^{-1}).$

∎

We are now ready to prove the main result of this chapter.

###### Proof of Theorem 1.2.2.

For $\mathscr{A}\subseteq\\{1,\cdots,p\\},$ we define the multilinear
functional on $a_{0}\otimes\cdots a_{p}\in\mathcal{A}^{\otimes(p+1)}$ by:

$\mathcal{T}_{\mathscr{A}}(a_{0}\otimes\cdots\otimes
a_{p})=\Gamma_{0}\pi(a_{0})\prod_{k=1}^{p}y_{k}(a_{k}).$

Here,

$y_{k}(a):=\begin{cases}\begin{pmatrix}\partial(a)&0\\\
0&0\end{pmatrix},k\notin\mathscr{A}\\\ \begin{pmatrix}0&-a\\\
a&0\end{pmatrix},k\in\mathscr{A}.\end{cases}$

In particular,

$\mathcal{T}_{\emptyset}(a_{0}\otimes\cdots
a_{p})=\begin{pmatrix}\Omega(a_{0}\otimes\cdots\otimes a_{p})&0\\\
0&0\end{pmatrix}.$

For $c\in\mathcal{A}^{\otimes(p+1)},$ we apply (4.5) to get

$\Omega_{\mu}(\pi(c))=\sum_{\mathscr{A}\subseteq\\{1,\cdots,p\\}}\mu^{|\mathscr{A}|}\mathcal{T}_{\mathscr{A}}(c).$

For each $0\leq k\leq p,$ we set

$f_{k}(s)=\sum_{|\mathscr{A}|=k}(\mathrm{Tr}_{2}\otimes\mathrm{Tr})(\mathcal{T}_{\mathscr{A}}(c)\cdot(1\otimes(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}D^{2}})).$

That is, $f_{k}(s)$ is the coefficient of $\mu^{k}$ in
$(\mathrm{Tr}_{2}\otimes\mathrm{\mathrm{Tr}})(\Omega_{\mu}(\pi(c))(1\otimes(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}D^{2}}))$.

Now if $c\in\mathcal{A}^{\otimes(p+1)}$ is a Hochschild cycle, then Lemma
4.5.6 yields

(4.24)
$\sum_{k=0}^{p}\mu^{k}f_{k}(s)=\frac{p}{2}\mathrm{Ch}_{\mu}(\pi(c))s^{-2}+O(s^{-1}),\quad
s\downarrow 0.$

Select a set $\\{\mu_{0},\ldots,\mu_{p}\\}$ of distinct positive numbers, and
for each $0\leq l\leq p$ we may take $\mu=\mu_{l}$ in (4.24) to arrive at:

$\sum_{k=0}^{p}\mu_{l}^{k}f_{k}(s)=\frac{p}{2}\mathrm{Ch}_{\mu_{l}}(\pi(c))s^{-2}+O(s^{-1}),\quad
s\downarrow 0,\quad 0\leq l\leq p.$

Since the Vandermonde matrix $\\{\mu_{l}^{k}\\}_{0\leq l,k\leq p}$ is
invertible, it follows that there exist $\\{\alpha_{0},\ldots,\alpha_{k}\\}$
such that:

$f_{k}(s)=\frac{\alpha_{k}}{s^{2}}+O(s^{-1}),\quad s\downarrow 0,\quad 0\leq
k\leq p.$

Substituting this back to (4.24), we obtain

$\sum_{k=0}^{p}\mu^{k}\alpha_{k}=\frac{p}{2}\mathrm{Ch}_{\mu}(\pi(c)).$

In particular

$\alpha_{0}=\frac{p}{2}\lim_{\mu\to 0}\mathrm{Ch}_{\mu}(\pi(c))$

So by Lemma 4.5.4,

$\alpha_{0}=\frac{p}{2}{\rm Ch}(c).$

Hence,

$f_{0}(s)=\frac{p}{2}\mathrm{Ch}(c)s^{-2}+O(s^{-1}),\quad s\downarrow 0.$

The assertion follows now from the definition of $f_{0}.$ ∎

##  Chapter 5 Residue of the $\zeta-$function and the Connes character formula

In this chapter we complete the proofs of Theorem 1.2.3 and Theorem 1.2.5.

For a spectral triple $(\mathcal{A},H,D)$ satisfying Hypothesis 1.2.1, we
define the zeta function of a Hochschild cycle
$c\in\mathcal{A}^{\otimes(p+1)}$ by the formula

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

Indeed, by Hypothesis 1.2.1.(iii) if $\Re(z)\geq p+1$ then the operator
$\Omega(c)(1+D^{2})^{-z/2}$ is trace class, and so $\zeta_{c,D}$ is well
defined when $\Re(z)>p+1$. In Section 5.1 we prove that $\zeta_{c,D}$ is
holomorphic and has analytic continuation to the set

$\\{z\in\mathbb{C}\;:\;\Re(z)>p-1\\}\setminus\\{p\\}.$

We also show that the point $p$ is a simple pole for $\zeta_{c,D}$, and that
the corresponding residue of $\zeta_{c,D}$ at $p$ is equal to
$p\mathrm{Ch}(c)$, thus completing the proof of Theorem 1.2.3.

Then we undertake the more difficult task of proving Theorem 1.2.5. We achieve
this by a new characterisation of universal measurability in Section 5.5,
which allows us to deduce Theorem 1.2.5 as a corollary of Theorem 1.2.3.

The most novel feature of this chapter, and of this manuscript as a whole, is
a certain integral representation of the difference
$B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}$ for two positive bounded
operators $A$ and $B$ and $z\in\mathbb{C}$ with $\Re(z)>0$ (see Lemma 5.2.1).
This result first appeared in [25, Lemma 5.2] for the special case where $z$
is real and positive. With this integral representation we are able to prove
the analyticity of the function

$z\mapsto\mathrm{Tr}(XB^{z}A^{z})-\mathrm{Tr}(X(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z})$

for a bounded operator $X$ and for $z$ in a certain domain in the complex
plane, under certain assumptions on $A$ and $B$. This result is stated in full
in Section 5.4.

In sections 5.6 and 5.7 we complete the proof of Theorem 1.2.5.

###  5.1. Analyticity of the $\zeta$-function for $\Re(z)>p-1$, $z\neq p$

This section contains the proof of Theorem 1.2.3. The proof is relatively
short, since we are able to use Theorem 1.2.2.

######  Lemma 5.1.1.

Let $h\in L_{\infty}(0,1)$ and $u\in L_{\infty}(1,\infty)$. Then,

  1. (i)

the function

$F(z):=\int_{0}^{1}s^{z-1}h(s)ds,\quad\Re(z)>0,$

is analytic.

  2. (ii)

the function

$G(z):=\int_{1}^{\infty}s^{z-1}u(s)e^{-s}ds,\quad z\in\mathbb{C},$

is analytic.

###### Proof.

Let us prove (i). Define

$F_{n}(z)=\int_{\frac{1}{n}}^{1}s^{z-1}h(s)ds,\quad\Re(z)>0.$

Then for $\Re(z)>0$ we have:

$\displaystyle|F(z)-F_{n}(z)|$
$\displaystyle=\left|\int_{0}^{\frac{1}{n}}s^{z-1}h(s)ds\right|$
$\displaystyle\leq\int_{0}^{\frac{1}{n}}s^{\Re(z)-1}|h(s)|ds$
$\displaystyle\leq\|h\|_{\infty}\int_{0}^{\frac{1}{n}}s^{\Re(z)-1}ds$
$\displaystyle=\frac{\|h\|_{\infty}}{\Re(z)}n^{-\Re(z)}.$

So for every $\varepsilon>0$, we have that $F_{n}$ converges uniformly to $F$
on the set $\\{z\;:\;\Re(z)>\varepsilon\\}$.

We now show that for each $n$ the function $F_{n}$ is entire. Indeed, we have
the power series expansion

$\displaystyle s^{z-1}$ $\displaystyle=e^{(z-1)\log(s)}$
$\displaystyle=\sum_{k\geq 0}\frac{(\log(s))^{k}}{k!}(z-1)^{k}.$

which converges uniformly on compact subsets of $\mathbb{C}$. Therefore,
interchanging the integral and summation, we have that for all
$z\in\mathbb{C}$

(5.1) $F_{n}(z)=\sum_{k\geq
0}\frac{1}{k!}\left(\int_{\frac{1}{n}}^{1}(\log(s))^{k}h(s)ds\right)(z-1)^{k},\quad
z\in\mathbb{C}.$

This power series has infinite radius of convergence, since

$\left|\int_{\frac{1}{n}}^{1}(\log(s))^{k}h(s)ds\right|\leq\|h\|_{\infty}(\log(n))^{k}.$

So each $F_{n}$ is entire.

In summary, the sequence $\\{F_{n}\\}_{n\geq 1}$ of entire functions converges
to $F$ uniformly on the half-plane $\\{z\;:\;\Re(z)>\varepsilon\\}$. Since
$\varepsilon$ is arbitrary, the sequence $\\{F_{n}\\}_{n\geq 0}$ converges
uniformly to $F$ on compact subsets of the half plane $\\{z\;:\;\Re(z)>0\\}$.
Thus, $F$ is holomorphic on this half-plane.

To prove (ii), we consider the functions

$G_{n}(z):=\int_{1}^{n}s^{z-1}u(s)e^{-s}ds,\quad z\in\mathbb{C}$

Exactly the same argument as above shows that each $G_{n}$ is entire. For all
$n\geq 1$, we have:

$\displaystyle|G(z)-G_{n}(z)|$
$\displaystyle\leq\int_{n}^{\infty}s^{\Re(z)-1}|u(s)|e^{-s}\,ds$
$\displaystyle\leq\|u\|_{\infty}\int_{n}^{\infty}s^{\Re(z)-1}e^{-s}\,ds$

So for any $N>0$, we have that $G_{n}$ converges uniformly to $G$ in the set
$\\{z\in\mathbb{C}\;:\;\Re(z)<N\\}$, and therefore on compact subsets of the
plane. Hence, $G$ is entire. ∎

We are now able to prove Theorem 1.2.3.

######  Theorem.

Let $(\mathcal{A},H,D)$ be a spectral triple satisfying Hypothesis 1.2.1, and
let $c\in\mathcal{A}^{\otimes(p+1)}$ be a Hochschild cycle. Then the function

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

is analytic, and has analytic continuation to the set
$\\{z\in\mathbb{C}\;:\;\Re(z)>p-1\\}\setminus\\{p\\}$ and a simple pole at $p$
with residue $p\mathrm{Ch}(c)$.

###### Proof of Theorem 1.2.3.

Let $z\in\mathbb{C}$ with $\Re(z)>1$. Then for all $x>0,$ we have

$\displaystyle\int_{0}^{\infty}s^{z-1}e^{-s^{2}x^{2}}\,ds$
$\displaystyle=x^{-z}\int_{0}^{\infty}t^{z-1}e^{-t^{2}}\,dt$
$\displaystyle=x^{-z}\int_{0}^{\infty}u^{\frac{z-1}{2}}e^{-u}\frac{u^{-\frac{1}{2}}}{2}\,du$
$\displaystyle=\frac{x^{-z}}{2}\Gamma\left(\frac{z}{2}\right).$

Thus,

$x^{2-z}=\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{0}^{\infty}s^{z-1}x^{2}e^{-x^{2}s^{2}}\,ds$

So by the functional calculus, for $\Re(z)>2$ we have an integral in the weak
operator topology:

$(1+D^{2})^{1-\frac{z}{2}}=\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{0}^{\infty}s^{z-1}(1+D^{2})e^{-s^{2}(1+D^{2})}\,ds.$

We now multiply on the left by the bounded operator
$\Omega(c)(1+D^{2})^{-p/2}$ to arrive at:

$\Omega(c)(1+D^{2})^{1-\frac{z+p}{2}}=\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{0}^{\infty}s^{z-1}\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})}\,ds.$

We claim that this integral converges in $\mathcal{L}_{1}$. First, if $p=1$
then consider the function $h_{1}(t)=te^{-t^{2}}$. Applying Lemma A.1.3 with
the function $h_{1}$, we have:

$\displaystyle\|s^{\Re(z)-1}\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})}\|_{1}$
$\displaystyle=s^{\Re(z)-2}\|\Omega(c)h_{1}(s(1+D^{2})^{1/2})\|_{1}$
$\displaystyle=O(s^{\Re(z)-3}),\quad s\downarrow 0.$

On the other hand, if $p>1$ then define
$h_{p}(t)=(1+t^{2})^{1-p/2}e^{-t^{2}}.$ Now applying Lemma A.1.3 with the
function $h_{p}$:

$\displaystyle\|s^{z-1}\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})}\|_{1}$
$\displaystyle\leq\left\|\left(\frac{1+D^{2}}{1+s^{2}D^{2}}\right)^{1-\frac{p}{2}}\right\|_{\infty}\|s^{z-1}\Omega(c)h_{p}(s|D|)\|_{1}$
$\displaystyle=s^{\Re(z)+p-3}\|\Omega(c)h_{p}(s|D|)\|_{1}$
$\displaystyle=O(s^{\Re(z)-3}),\quad s\downarrow 0.$

So in both cases, since $\Re(z)>2$, the function
$s\mapsto\|s^{z-1}\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})}\|_{1}$
is integrable on $[0,1].$

For $s>1,$ we have

$e^{-s^{2}D^{2}}\leq e^{-D^{2}}\leq(1+D^{2})^{-3/2}.$

so we have that

$\|s^{z-1}\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})}\|_{1}\leq
s^{\Re(z)-1}\|\Omega(c)(1+D^{2})^{-\frac{p+1}{2}}\|_{1}.$

Hence,
$s\mapsto\|s^{z-1}\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})}\|_{1}$
is integrable on the interval $(1,\infty)$.

By Lemma 2.3.2, for $\Re(z)>2$ we therefore have:

$\displaystyle\|\Omega(c)(1+D^{2})^{1-\frac{z+p}{2}}\|_{1}$
$\displaystyle\leq\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{0}^{\infty}\|s^{z-1}\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})}\|_{1}\,ds$
$\displaystyle<\infty$

and

(5.2)
$\mathrm{Tr}(\Omega(c)(1+D^{2})^{1-\frac{z+p}{2}})=\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{0}^{\infty}s^{z-1}\mathrm{Tr}(\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})})\,ds.$

We will now apply the result of Theorem 1.2.2 to the integrand. First we
define a function $h$ on $(0,\infty)$ by:

$h(s):=\begin{cases}e^{s}\mathrm{Tr}(\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})}),\quad
s\geq 1\\\
s\mathrm{Tr}(\Omega(c)(1+D^{2})^{1-\frac{p}{2}}e^{-s^{2}(1+D^{2})})-\frac{p}{2}\mathrm{Ch}(c)s^{-1},\quad
0<s<1.\end{cases}$

By Theorem 1.2.2, the function $h$ is bounded on the interval $(0,1).$ For
$s>1$, we have a constant $c$ such that.

$|h(s)|\leq Ce^{s-s^{2}}$

Hence, $h$ is bounded on $[0,\infty)$. Substituting $h$ in (5.2):

$\displaystyle\mathrm{Tr}(\Omega(c)(1+D^{2})^{1-\frac{z+p}{2}})$
$\displaystyle=\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{0}^{1}s^{z-1}(s^{-1}h(s)+\frac{p}{2}\mathrm{Ch}(c)s^{-2})\,ds$
$\displaystyle\quad+\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{1}^{\infty}s^{z-1}e^{-s}h(s)\,ds.$

By Lemma 5.1.1.(ii), the second term in the above sum has extension to an
entire function, and so we focus on the first term. We have,

$\displaystyle\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{0}^{1}s^{z-1}(s^{-1}h(s)+\frac{p}{2}\mathrm{Ch}(c)s^{-2})\,ds$
$\displaystyle=\frac{2}{\Gamma\left(\frac{z}{2}\right)}\int_{0}^{1}s^{z-2}h(s)\,ds$
$\displaystyle\quad+\frac{p}{\Gamma\left(\frac{z}{2}\right)}\mathrm{Ch}(c)\int_{0}^{1}s^{z-3}\,ds.$

Due to Lemma 5.1.1.(i), the first term in the above sum has extension to an
analytic function for $\Re(z-1)>0$. That is, for $\Re(z)>1$.

As for the second term, since we are still working with $\Re(z)>2$ we may
compute:

$\int_{0}^{1}s^{z-3}\,ds=(z-2)^{-1}.$

So in summary, the function initially defined for $\Re(z)>2$ given by:

$z\mapsto\mathrm{Tr}(\Omega(c)(1+D^{2})^{1-\frac{p+z}{2}})-\frac{p}{\Gamma\left(\frac{z}{2}\right)}(z-2)^{-1}$

is analytic on the set $\Re(z)>2$, and has analytic continuation to the set
$\Re(z)>1$. Since the function $\frac{1}{\Gamma\left(\frac{z}{2}\right)}$ is
entire, and $\Gamma(1)=1$, we may equivalently say that the function

$z\mapsto\zeta_{c,D}(z+p-2)-p\mathrm{Ch}(c)(z-2)^{-1},\quad\Re(z)>2$

has analytic continuation to the set $\Re(z)>1$. In other words, for
$\Re(z)>p$

$\zeta_{c,D}(z)-p\mathrm{Ch}(c)(z-p)^{-1}$

has analytic continuation to the set $\Re(z)>p-1$. This is exactly the
statement of the theorem. ∎

###  5.2. An Integral Representation for
$B^{z}A^{z}-(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}$

In this section, we follow the convention that for all $s\in\mathbb{R}$ we
have $0^{is}=0$, so in particular we have the unusual convention that
$0^{i0}=0$. This section is devoted to the proof of Theorem 5.2.1 (stated
below). Theorem 5.2.1 is a strengthening of [25, Lemma 5.2] (which corresponds
to the special case where $z$ is real and $B$ is compact). Theorem 5.2.1 also
substantially strengthens [10, Proposition 4.4].

Here we work with abstract operators on a separable Hilbert space $H$. Given a
positive bounded operator $A$ on $H$, and a complex number $z$ with
$\Re(z)>0$, the operator $A^{z}$ may be defined by continuous functional
calculus.

######  Theorem 5.2.1.

Let $A$ and $B$ be bounded, positive operators on $H$, and let
$z\in\mathbb{C}$ with $\Re(z)>1$. Let $Y:=A^{1/2}BA^{1/2}$. We define the
mapping $T_{z}:\mathbb{R}\to\mathcal{L}_{\infty}$ by,

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

We also define the function $g_{z}:\mathbb{R}\to\mathbb{C}$ by:

$\displaystyle g_{z}(0)$ $\displaystyle:=1-\frac{z}{2},$ $\displaystyle
g_{z}(t)$
$\displaystyle:=1-\frac{e^{\frac{z}{2}t}-e^{-\frac{z}{2}t}}{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})(e^{\left(\frac{z-1}{2}\right)t}+e^{-\left(\frac{z-1}{2}\right)t})}.$

Then:

  1. (i)

The mapping $T_{z}:\mathbb{R}\to\mathcal{L}_{\infty}$ is continuous in the
weak operator topology.

  2. (ii)

We have:

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

######  Remark 5.2.2.

For $\Re(z)>1$, the function $g_{z}$ is Schwartz, and hence so is the
(rescaled) Fourier transform $\widehat{g}_{z}$.

###### Proof.

For $t\neq 0$, we may rewrite $g_{z}(t)$ as:

$g_{z}(t)=\frac{1}{2}\left(\frac{\tanh((z-1)t/2)}{\tanh(t/2)}-1\right).$

Letting $s=t/2$ and $w=z-1$, it then suffices to show that for $\Re(w)>0$ the
function

$f_{w}(s)=\frac{\tanh(ws)}{\tanh(s)}-1,\quad s\neq 0$

with $f_{w}(0)=w-1$ is Schwartz. Since $\lim_{s\to 0}\frac{\tanh(s)}{s}=1$, it
is evident that $f_{w}$ is continuous at $0$ and that $f_{w}$ is smooth in
$[-1,1]$.

It suffices now to show that the function $\tanh(ws)-\tanh(s)$ is Schwartz,
since for $|s|>1$ the function $\frac{1}{\tanh(s)}$ is smooth and bounded with
all derivatives bounded. For $s>1$, we can write,

$\tanh(ws)-\tanh(s)=\tanh(ws)-1+(1-\tanh(s))$

and then note that since $\Re(w)>0$, the functions $\tanh(ws)-1$ and
$1-\tanh(s)$ have rapid decay as $s\to\infty$, with all derivatives to all
orders also of rapid decay as $s\to\infty$. Similarly, for $s<-1$, we can
write $\tanh(ws)-\tanh(s)=\tanh(ws)+1-(\tanh(s)+1)$ and then use the fact that
$\tanh(s)+1$ and $\tanh(ws)+1$ have rapid decay, with all derivatives of rapid
decay, as $s\to-\infty$. ∎

######  Lemma 5.2.3.

Let $A_{k},X_{k}\in\mathcal{L}_{\infty},$ $1\leq k\leq n,$ and let $X_{k}\geq
0,$ $1\leq k\leq n.$ The function from $\mathbb{R}$ to $\mathcal{L}_{\infty}$
given by:

$s\mapsto\prod_{k=1}^{n}A_{k}X_{k}^{is},s\in\mathbb{R}$

is continuous in the strong operator topology (and in particular in the weak
operator topology).

###### Proof.

If uniformly bounded nets $\\{A_{i}\\}_{i\in\mathbb{I}}$ and
$\\{B_{i}\\}_{i\in\mathbb{I}}$ converge in the strong operator topology to $A$
and $B$ respectively, then the net $\\{A_{i}B_{i}\\}_{i\in\mathbb{I}}$
converges to $AB$ in strong operator topology. This fact is standard and can
be found e.g. in [8, Proposition 2.4.1]. Therefore it suffices to show that
for each $k=1,\ldots,n$ that the function $s\mapsto X_{k}^{is}$ is continuous
in the strong operator topology.

We note if $X_{k}$ has a spectral gap at $0$, then $\log(X_{k})$ is well
defined by continuous functional calculus and $X_{k}^{is}=\exp(is\log(X_{k}))$
is strongly continuous by the Stone-von Neumann theorem.

If $X_{k}$ does not necessarily have a spectral gap, then instead we use the
Borel function:

$\log_{0}(t):=\begin{cases}\log(t),\quad t>0\\\ 0,\quad t=0.\end{cases}$

Hence, for $s\in\mathbb{R}$ and $t\geq 0$,

$\exp(is\log_{0}(t))=\begin{cases}t^{is},\quad t>0,\\\ 1,\quad
t=0.\end{cases}$

Recalling our convention stated at the start of this section, that $0^{is}=0$
for all $t\geq 0$, we have:

$t^{is}=\exp(is\log_{0}(t))(1-\chi_{\\{0\\}}(t)).$

Let $P_{k}$ be the support projection of $X_{k}$ (i.e., the projection onto
the orthogonal complement of the kernel of $X_{k}$). Then since
$P_{k}=1-\chi_{\\{0\\}}(X_{k})$ by Borel functional calculus we have:

$X_{k}^{is}=P_{k}\exp(is\log_{0}(X_{k})).$

Since the operator $\log_{0}(X_{k})$ is self-adjoint, by the Stone-von Neumann
theorem it follows that $s\mapsto\exp(is\log_{0}(X_{k}))$ is strongly
continuous. Hence, $s\mapsto X_{k}^{is}$ is strongly continuous and the proof
is complete. ∎

######  Lemma 5.2.4.

Let $X$ and $Y$ be positive bounded operators and $z\in\mathbb{C}$ with
$\Re(z)>1$. Set $V_{z}:=X^{z-1}(X-Y)+(X-Y)Y^{z-1}$. Then,

$X^{z}-Y^{z}=V_{z}-\int_{\mathbb{R}}X^{is}V_{z}Y^{-is}\widehat{g}_{z}(s)ds.$

The integral is understood in the weak operator topology sense. The function
$g_{z}$ is the same as in the statement of Theorem 5.2.1.

###### Proof.

We define the function $\phi_{1,z}$ on $[0,\infty)\times[0,\infty)$ by:

$\displaystyle\phi_{1,z}(\lambda,\mu)$
$\displaystyle:=g_{z}(\log(\frac{\lambda}{\mu}))\quad\lambda,\mu>0,$
$\displaystyle\phi_{1,z}(0,\mu)$ $\displaystyle:=0,\quad\mu\geq 0,$
$\displaystyle\phi_{1,z}(\lambda,0)$ $\displaystyle:=0,\quad\lambda\geq 0.$

We caution the reader that $\phi_{1,z}$ is not continuous at $(0,0)$ unless
$z=2$ (indeed, $\phi_{1,z}(\lambda,\lambda)=g_{z}(0)=1-\frac{z}{2}$ for all
$\lambda>0$). If we rewrite the definition of $g_{z}$ in terms of
exponentials, then for $t\neq 0$ we get

$g_{z}(t)=1-\frac{e^{\frac{z}{2}t}-e^{-\frac{z}{2}t}}{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})(e^{\left(\frac{z-1}{2}\right)t}+e^{-\left(\frac{z-1}{2}\right)t})},$

and therefore:

(5.3)
$\phi_{1,z}(\lambda,\mu)=1-\frac{\lambda^{z}-\mu^{z}}{(\lambda-\mu)(\lambda^{z-1}+\mu^{z-1})},\quad\lambda,\mu>0,\lambda\neq\mu.$

We claim that

(5.4)
$\phi_{1z}(\lambda,\mu)=\int_{\mathbb{R}}\widehat{g}_{z}(s)\lambda^{is}\mu^{-is}ds,\quad\lambda,\mu\geq
0.$

Indeed, since $g_{z}$ is Schwartz we can use the Fourier inversion theorem:

$g_{z}(t)=\int_{\mathbb{R}}\widehat{g}_{z}(s)e^{ist}ds,\quad t\in\mathbb{R}.$

If $\lambda,\mu>0$, we simply substitute $t=\log(\lambda/\mu)$. For
$\lambda=0$ or $\mu=0$, then the right hand side of (5.4) vanishes, as does
$\phi_{1,z}$ by definition. Hence (5.4) is valid for all $\lambda,\mu\geq 0$.

Thus, by the definition of the double operator integral (2.14), we have:

(5.5)
$T_{\phi_{1,z}}^{X,Y}(A)=\int_{\mathbb{R}}\widehat{g}_{z}(s)X^{is}AY^{-is}ds.$

Indeed, since $g_{z}$ is a Schwartz function, it follows that
$\widehat{g}_{z}\in L_{1}(\mathbb{R})$ and so the condition (2.13) holds.
Therefore, (5.5) follows as a consequence of (2.14). Here, the integral on the
right hand side of (5.5) is understood in the weak operator topology sense.

Measurability of the function $s\mapsto X^{is}AY^{-is}$ in the weak operator
topology is guaranteed by Lemma 5.2.3 and condition (2.10) follows from the
inequality

$\|\widehat{g}_{z}(s)X^{is}AY^{-is}\|_{\infty}\leq|\widehat{g}_{z}(s)|\cdot\|A\|_{\infty},\quad
s\in\mathbb{R},$

and from the fact that $\widehat{g}_{z}$ is a Schwartz (and in particular
integrable) function. So it follows that $T^{X,Y}_{\phi_{1,z}}$ is bounded in
the operator norm from $\mathcal{L}_{\infty}$ to $\mathcal{L}_{\infty}.$

We introduce two more functions on $[0,\infty)\times[0,\infty)$. First,

$\phi_{2,z}(\lambda,\mu)=(\lambda^{z-1}+\mu^{z-1})(\lambda-\mu),\quad\lambda,\mu\geq
0$

and secondly,

$\phi_{3,z}(\lambda,\mu)=(\lambda^{z-1}+\mu^{z-1})(\lambda-\mu)-(\lambda^{z}-\mu^{z}),\quad\lambda,\mu\geq
0.$

Both functions are bounded on compact subsets of $[0,\infty)^{2}$, and so in
particular on $\mathrm{Spec}(X)\times\mathrm{Spec}(Y)$, since by assumption
both $X$ and $Y$ are bounded.

The equality $\phi_{3,z}=\phi_{1,z}\phi_{2,z}$ holds on
$[0,\infty)\times[0,\infty)$. Indeed this follows from (5.3) for
$\lambda,\mu>0$, $\lambda\neq\mu.$ For $\lambda=\mu>0$ one has
$\phi_{1,z}(\lambda,\lambda)=1-\frac{z}{2},$ $\phi_{2,z}(\lambda,\lambda)=0$
and $\phi_{3,z}(\lambda,\lambda)=0.$ If $\lambda=0$ or $\mu=0$ one has
$\phi_{1,z}(\lambda,\mu)=0$ and $\phi_{3,z}(\lambda,\mu)=0.$

Using formulae (2.16) and (2.14), we obtain that
$T^{X,Y}_{\phi_{2,z}}:\mathcal{L}_{\infty}\to\mathcal{L}_{\infty}$ and

$T^{X,Y}_{\phi_{2,z}}(A)=X^{z}A-X^{z-1}AY+XAY^{z-1}-AY^{z}.$

Since $\phi_{3,z}$ bounded on $\mathrm{Spec}(X)\times\mathrm{Spec}(Y)$, we
also get that $T^{X,Y}_{\phi_{3,z}}$ is bounded in the operator norm from
$\mathcal{L}_{\infty}$ to $\mathcal{L}_{\infty}$, and

$T^{X,Y}_{\phi_{3,z}}(A)=(X^{z}A-X^{z-1}AY+XAY^{z-1}-AY^{z})-(X^{z}A-AY^{z}).$

We note at this point that $T^{X,Y}_{\phi_{3,z}}(1)=V_{z}$.

We have $\phi_{3,z}=\phi_{1,z}\phi_{2,z}$ on
$\mathrm{Spec}(X)\times\mathrm{Spec}(Y)$, and thus by (2.15):

$\displaystyle T^{X,Y}_{\phi_{1,z}}(V_{z})$
$\displaystyle=T^{X,Y}_{\phi_{1,z}}(T^{X,Y}_{\phi_{2,z}}(1))$
$\displaystyle=T^{X,Y}_{\phi_{3,z}}(1)$ $\displaystyle=V_{z}-(X^{z}-Y^{z}).$

The assertion follows now from (5.5). ∎

We are now able to prove Theorem 5.2.1 in the special case where the spectrum
of $B$ is a finite set.

######  Lemma 5.2.5.

Theorem 5.2.1 holds under the additional assumption that the spectrum of $B$
consists of a finite set of points.

###### Proof.

Suppose that $\mathrm{Spec}(B)=\\{\lambda_{1},\ldots,\lambda_{n}\\}$, where
each $\lambda_{k}\geq 0$ is distinct. By the spectral theorem there exists $n$
mutually orthogonal projections $\\{P_{k}\\}_{k=1}^{n}$ such that

$B=\sum_{k=1}^{n}\lambda_{k}P_{k}$

and $\sum_{k=1}^{n}P_{k}=1$. We have,

$B^{z}=\sum_{k=1}^{n}\lambda_{k}^{z}P_{k}.$

Therefore,

$\displaystyle B^{z}A^{z}-Y^{z}$
$\displaystyle=\sum_{k=1}^{n}(P_{k}\lambda_{k}^{z}A^{z}-P_{k}Y^{z})$ (5.6)
$\displaystyle=\sum_{k=1}^{n}P_{k}((\lambda_{k}A)^{z}-Y^{z}).$

Applying Lemma 5.2.4 to each term in the above sum, with $X=\lambda_{k}A$, if

$V_{k,z}=(\lambda_{k}A)^{z-1}(\lambda_{k}A-Y)+(\lambda_{k}A-Y)Y^{z-1}$

then

$(\lambda_{k}A)^{z}-Y^{z}=V_{k,z}-\int_{\mathbb{R}}(\lambda_{k}A)^{is}V_{k,z}Y^{-is}\widehat{g}_{z}(s)\,ds.$

Now substituting into (5.6), we have:

$\displaystyle B^{z}A^{z}-Y^{z}$
$\displaystyle=\sum_{k=1}^{n}P_{k}\left(V_{k,z}-\int_{\mathbb{R}}(\lambda_{k}A)^{is}V_{k,z}Y^{-is}\widehat{g}_{z}(s)\,ds\right)$
$\displaystyle=\sum_{k=1}^{n}P_{k}V_{k,z}-\int_{\mathbb{R}}\left(\sum_{k=1}^{n}P_{k}(\lambda_{k}A)^{is}V_{k,z}\right)Y^{-is}\,\widehat{g}_{z}(s)\,ds.$

By the definition of $V_{k,z}$:

$V_{k,z}=(\lambda_{k}A)^{z}-(\lambda_{k}A)^{z-1}Y+\lambda_{k}AY^{z-1}-Y^{z}$

and so

$\displaystyle\sum_{k=1}^{n}P_{k}V_{k,z}$
$\displaystyle=B^{z}A^{z}-B^{z-1}A^{z-1}Y+BAY^{z-1}-Y^{z}$
$\displaystyle=B^{z-1}(BA^{z}-A^{z-1}Y)+(BA-Y)Y^{z-1}$
$\displaystyle=B^{z-1}(BA^{z}-A^{z-1}A^{1/2}BA^{1/2})+(BA-A^{1/2}BA^{1/2})Y^{z-1}$
$\displaystyle=[BA^{1/2},A^{z-1/2}]+[BA^{1/2},A^{1/2}]Y^{z-1}$
$\displaystyle=T_{z}(0).$

We may also compute the sum in the integrand:

$\displaystyle\sum_{k=1}^{n}P_{k}(\lambda_{k}A)^{is}V_{k,z}$
$\displaystyle=\sum_{k=1}^{n}P_{k}\left((\lambda_{k}A)^{z+is}-(\lambda_{k}A)^{z-1+is}Y+(\lambda_{k}A)^{1+is}Y^{z-1}-(\lambda_{k}A)^{is}Y^{z}\right)$
$\displaystyle=\sum_{k=1}^{n}jP_{k}\lambda_{k}^{z+is}\cdot
A^{p+is}-\sum_{k=1}^{n}P_{k}\lambda_{k}^{z-1+is}\cdot A^{z-1+is}Y$
$\displaystyle\quad+\sum_{k=1}^{n}P_{k}\lambda_{k}^{1+is}A^{1+is}Y^{z-1}-\sum_{k=1}^{n}P_{k}\lambda_{k}^{is}\cdot
A^{is}Y^{z}.$

By functional calculus, we have

$\displaystyle\sum_{k=1}^{n}P_{k}(\lambda_{k}A)^{is}V_{k,z}$
$\displaystyle=B^{z+is}A^{z+is}-B^{z-1+is}A^{z-1+is}Y+B^{1+is}A^{1+is}Y^{z-1}-B^{is}A^{is}Y^{z}$
$\displaystyle=B^{z-1+is}(BA^{z+is}-A^{z-1+is}Y)+B^{is}(BA^{1+is}-A^{is}Y)Y^{z-1}$
$\displaystyle=B^{z-1+is}[BA^{\frac{1}{2}},A^{z-\frac{1}{2}+is}]+B^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{z-1}.$

So multiplying on the right by $Y^{-is}$,

$\left(\sum_{k=1}^{n}P_{k}(\lambda_{k}A)^{is}V_{k,z}\right)Y^{-is}=B^{z-1+is}[BA^{\frac{1}{2}},A^{z-\frac{1}{2}+is}]Y^{-is}+B^{is}[BA^{\frac{1}{2}},A^{\frac{1}{2}+is}]Y^{z-1-is}.$

We recognise this right hand side as exactly $T_{z}(s)$, and so

$B^{z}A^{z}-Y^{z}=T_{z}(0)-\int_{\mathbb{R}}T_{z}(s)\widehat{g}_{z}(s)\,ds$

as required. ∎

We now explain how to deduce the general version of Lemma 5.2.1 from the
special case of Lemma 5.2.5 (i.e, when $\mathrm{Spec}(B)$ is a finite set). To
do this we will select a sequence $\\{B_{n}\\}_{n=1}^{\infty}$ with $B_{n}\to
B$ in the uniform norm and such that the spectrum of each $B_{n}$ is finite.

The following lemma shows that under certain conditions, if $B_{n}\to B$ in
the uniform norm, then $B_{n}^{is}\to B^{is}$ in the weak operator topology
for each fixed $s\in\mathbb{R}$.

For a bounded operator $T$ we denote $\mathrm{supp}(T)$ for the projection
onto the orthogonal complement of $\ker(T)$ (this is the support projection of
$T$).

######  Lemma 5.2.6.

Let $C$ be a positive bounded operator, and let $\\{C_{n}\\}_{n=1}^{\infty}$
be a sequence of positive bounded operators such that $C_{n}\to C$ in the
operator norm, and for each $n$ we have
$\mathrm{supp}(C_{n})=\mathrm{supp}(C)$. Then for all $s\in\mathbb{R}$, we
have that $C_{n}^{is}\to C^{is}$ in the weak operator topology.

###### Proof.

By definition, we need to show that for all $s\in\mathbb{R}$ and $\xi,\eta\in
H$ we have

$\lim_{n\to\infty}\langle C_{n}^{is}\xi,\eta\rangle=\langle
C^{is}\xi,\eta\rangle.$

By assumption, $\mathrm{supp}(C_{n})=\mathrm{supp}(C)$ for every $n\geq 0.$ By
taking a quotient by the closed subspace $\ker(C)$ if necessary, we may assume
without loss of generality that $\mathrm{supp}(C_{n})=\mathrm{supp}(C)=1$ for
every $n\geq 0.$ Also without loss of generality, we assume that
$\|C\|_{\infty}\leq 1$ and $\sup_{n\geq 0}\|C_{n}\|_{\infty}\leq 1$. Let
$\xi,\eta\in H$ be such that $\|\xi\|=\|\eta\|=1$.

Fix $\varepsilon>0$. Since $\ker(C)=\\{0\\}$, we may select $m>1$ such that

$\|\chi_{[0,\frac{1}{m}]}(C))\xi\|<\varepsilon.$

Let $0\leq\phi\leq 1$ be a smooth function supported on the interval
$[\frac{1}{m+1},2]$ such that $\phi=1$ on the interval $[\frac{1}{m},1].$ We
note that therefore

$\displaystyle\|(1-\phi(C))\xi\|$
$\displaystyle=\|(1-\phi(C))\chi_{[0,\frac{1}{m})}\xi\|$
$\displaystyle\leq\|\chi_{[0,\frac{1}{m})}\xi\|$ $\displaystyle<\varepsilon.$

Let $\psi(t):=t^{is}\phi(t)$. Since $\phi$ and $\psi$ are smooth and compactly
supported, it follows that their first and second derivatives are in
$L_{2}(\mathbb{R})$. These conditions are sufficient for $\phi$ and $\psi$ to
be operator Lipschitz (see [51, Lemma 6, Lemma 7]): i.e., there are constants
$C_{\phi}$ and $C_{\psi}$ such that

$\displaystyle\|\phi(C_{n})-\phi(C)\|_{\infty}$ $\displaystyle\leq
C_{\phi}\|C_{n}-C\|_{\infty},$ $\displaystyle\|\psi(C_{n})-\psi(C)\|_{\infty}$
$\displaystyle\leq C_{\psi}\|C_{n}-C\|_{\infty}.$

Select $N>0$ such that for all $n>N$ we thus have,

$\displaystyle\|\phi(C_{n})-\phi(C)\|_{\infty}$
$\displaystyle\leq\epsilon\text{ and, }$
$\displaystyle\|\phi(C_{n})C_{n}^{is}-\phi(C)C^{is}\|_{\infty}$
$\displaystyle\leq\epsilon.$

For $n>N$ we have:

$\displaystyle\langle C_{n}^{is}\xi,\eta\rangle-\langle C^{is}\xi,\eta\rangle$
$\displaystyle=\langle C_{n}^{is}(1-\phi(C_{n}))\xi,\eta\rangle+\langle
C^{is}(\phi(C)-1)\xi,\eta\rangle$ (5.7)
$\displaystyle\quad+\langle(C_{n}^{is}\phi(C_{n})-C^{is}\phi(C))\xi,\eta\rangle.$

For the first term in (5.7) above, we have:

$\displaystyle|\langle C_{n}^{is}(1-\phi(C_{n}))\xi,\eta\rangle|$
$\displaystyle\leq\|(1-\phi(C_{n}))\xi\|$
$\displaystyle\leq\|(1-\phi(C))\xi\|+\|\phi(C_{n})-\phi(C)\|_{\infty}$
$\displaystyle<2\varepsilon.$

Next, for the second term in (5.7), we have

$\displaystyle|\langle C^{is}(\phi(C)-1)\xi,\eta\rangle|$
$\displaystyle\leq\|(1-\phi(C))\xi\|_{\infty}$ $\displaystyle<\varepsilon.$

Finally, for the third term in (5.7),

$\displaystyle|\langle(C_{n}^{is}\phi(C_{n})-C^{is}\phi(C))\xi,\eta\rangle|$
$\displaystyle\leq\|\psi(C_{n})-\psi(C)\|_{\infty}\leq\varepsilon.$

So in summary, for $n\geq N$ we have:

$|\langle C_{n}^{is}\xi,\eta\rangle-\langle C^{is}\xi,\eta\rangle|\leq
4\varepsilon.$

Since $\varepsilon>0$ is arbitrarily small, the assertion follows. ∎

We are now ready to prove Theorem 5.2.1.

###### Proof of Theorem 5.2.1.

Without loss of generality, $\|B\|_{\infty}=1$ (if not, then we replace the
couple $(A,B)$ with a couple $(cA,c^{-1}B)$ with a suitable constant $c>0$).
In this case we have $\mathrm{Spec}(B)\subseteq[0,1]$ and
$1\in\mathrm{Spec}(B)$. For every $n\geq 1,$ set

$B_{n}=\sum_{m=1}^{n}\frac{m}{n}\chi_{(\frac{m-1}{n},\frac{m}{n}]}(B).$

Recall that $Y:=A^{1/2}BA^{1/2}$, and let $Y_{n}:=A^{1/2}B_{n}A^{1/2}$, and
let $T_{n,z}(s)$ be defined as $T_{z}(s)$ with the occurances of $B$ replaced
with $B_{n}$ and $Y$ replaced with $Y_{n}$.

By construction, the spectrum of $B$ consists of at most $n$ points, indeed by
the spectral mapping theorem:

$\mathrm{Spec}(B_{n})\subseteq\left\\{\frac{m}{n}\right\\}_{m=1}^{n}.$

Since $1\in\mathrm{Spec}(B)$, we always have that
$\chi_{(\frac{n-1}{n},1]}(B)\neq 0$, so $1\in\mathrm{Spec}(B_{n})$ and
$\|B_{n}\|_{\infty}=1$. We also have that
$\mathrm{supp}(B_{n})=\mathrm{supp}(B)$, and
$\mathrm{supp}(Y_{n})=\mathrm{supp}(Y)$.

Moreover, $B_{n}$ converges in norm to $B$, since
$\|B-B_{n}\|_{\infty}\leq\frac{1}{n}$. Thus by Lemma 5.2.6, for any
$s\in\mathbb{R}$ we also have that $B_{n}^{is}\to B^{is}$ in the weak operator
topology. Similarly, $Y_{n}\to Y$ in the norm topology and $Y_{n}^{is}\to
Y^{is}$ in the weak operator topology.

It follows now that for each $s\in\mathbb{R}$ and $z\in\mathbb{C}$ with
$\Re(z)>1$, we have that $T_{n,z}(s)\to T_{z}(s)$ in the weak operator
topology. One can also see that $\sup_{s\in\mathbb{R}}\sup_{n\geq
1}\|T_{n,z}(s)\|_{\infty}<\infty$.

In other words, for every $\xi,\eta\in H,$ we have

$\langle T_{n,z}(s)\xi,\eta\rangle\to\langle T_{z}(s)\xi,\eta\rangle.$

Since $|\langle T_{n,z}(s)\xi,\eta\rangle|\leq\sup_{n\geq
1}\|T_{n,z}(s)\|_{\infty}$, and this is bounded in $s$, we may use the
Dominated Convergence theorem to obtain

$\int_{\mathbb{R}}\langle
T_{n}(s)\xi,\eta\rangle\widehat{g}_{z}(s)ds\to\int_{\mathbb{R}}\langle
T(s)\xi,\eta\rangle\widehat{g}_{z}(s)ds$

By Lemma 5.2.5, for all $n\geq 1$ and $\xi,\eta\in H$ we have:

(5.8) $\langle(B_{n}^{z}A^{z}-Y_{n}^{z})\xi,\eta\rangle=\langle
T_{n,z}(0)\xi,\eta\rangle-\int_{\mathbb{R}}\langle
T_{n,z}(s)\xi,\eta\rangle\widehat{g}_{z}(s)\,ds.$

As already discussed, $B_{n}^{z}\to B^{z}$ in the weak operator topology, and
similarly $Y_{n}^{z}\to Y^{z}$ in the weak operator topology. Hence both sides
of (5.8) converge, and:

$\langle(B^{z}A^{z}-Y^{z})\xi,\eta\rangle=\langle
T_{z}(0)\xi,\eta\rangle-\int_{\mathbb{R}}\langle
T_{z}(s)\xi,\eta\rangle\widehat{g}_{z}(s)\,ds.$

Since $\xi$ and $\eta$ are arbitrary, this completes the proof. ∎

###  5.3. Analyticity of the mapping $z\mapsto g_{z}$

So far we have considered the function,

$g_{z}(t):=1-\frac{e^{\frac{z}{2}t}-e^{-\frac{z}{2}t}}{(e^{\frac{t}{2}}-e^{-\frac{t}{2}})(e^{\left(\frac{z-1}{2}\right)t}+e^{-\left(\frac{z-1}{2}\right)t})},t\neq
0$

with $g_{z}(0):=1-\frac{z}{2}$ as a Schwartz function of $t$ with a fixed
parameter $z\in\mathbb{C}$, with $\Re(z)>1$.

We may equally well consider $g$ as a function of $z$. That is, the mapping
$z\mapsto g_{z}$ defines a function:

$\\{z\in\mathbb{C}\;:\;\Re(z)>1\\}\to\mathcal{S}(\mathbb{R}).$

As a matter of fact, the function $z\mapsto g_{z}$ is holomorphic with values
in the Hilbert-Sobolev space:

$H^{2}(\mathbb{R}):=\\{f\in
L_{2}(\mathbb{R})\;:\;f^{\prime},f^{\prime\prime}\in L_{2}(\mathbb{R})\\},$

equipped with the Sobolev norm:

$\|f\|_{H^{2}(\mathbb{R})}^{2}:=\|f\|_{L_{2}(\mathbb{R})}^{2}+\|f^{\prime}\|_{L_{2}(\mathbb{R})}^{2}+\|f^{\prime\prime}\|_{L_{2}(\mathbb{R})}^{2}.$

We remind the reader of the meaning of Banach space valued holomorphy. If
$D\subseteq\mathbb{C}$ is a domain, $X$ is a Banach space and $f:D\to X$ then
the following two conditions are equivalent:

  1. (a)

For any continuous linear functional $\varpi\in X^{*}$, the function
$\varpi\circ f:D\to\mathbb{C}$ is holomorphic.

  2. (b)

For any $z\in\mathbb{C}$, the limit in the norm topology of $X$

$f^{\prime}(z)=\lim_{\zeta\to z}\frac{f(z)-f(\zeta)}{z-\zeta}$

exists.

The equivalence of these conditions is well known, see e.g. [57, Theorem
3.31].

We work with the first condition. Since $H^{2}(\mathbb{R})$ is a Hilbert
space, for any continuous linear functional $\varpi$ on $H^{2}(\mathbb{R})$
there exists $h\in H^{2}(\mathbb{R})$ such that:

$\varpi(g_{z})=\int_{\mathbb{R}}g_{z}(t)h(t)\,dt+\int_{\mathbb{R}}g_{z}^{\prime}(t)h^{\prime}(t)\,dt+\int_{\mathbb{R}}g_{z}^{\prime\prime}(t)h^{\prime\prime}(t)\,dt.$

So we focus on proving that for each $h\in H^{2}(\mathbb{R})$, the mapping
$z\mapsto\varpi(g_{z})$ is holomorphic.

######  Lemma 5.3.1.

Let $G:\\{z\in\mathbb{C}\;:\;\Re(z)>1\\}\to H^{2}(\mathbb{R})\\}$ be the
function given by $G(z)=g_{z}$. Then $G$ is continuous on its domain.

###### Proof.

It suffices to prove that the mappings $G:\\{z\in\mathbb{C}\;:\;\Re(z)>1\\}\to
H^{2}(-1,1)\\},$ $G:\\{z\in\mathbb{C}\;:\;\Re(z)>1\\}\to H^{2}(1,\infty)\\}$
and $G:\\{z\in\mathbb{C}\;:\;\Re(z)>1\\}\to H^{2}(-\infty,-1)\\}$ are
continuous on their domains.

We write the first function as follows:

$g_{z}=1-a_{z}bc_{z},\quad\Re(z)>1,$

where

$a_{z}(t)=\frac{\sinh(\frac{zt}{2})}{t},\quad
b(t)=\frac{t}{2\sinh(\frac{t}{2})},\quad
c_{z}(t)=\frac{1}{\cosh(\left(\frac{z-1}{2}\right)t)}.$

Since $z\to a_{z}$ and $z\to c_{z}$ are continuous $C^{2}[-1,1]-$valued
mappings, the first assertion follows.

We rewrite our function as follows.

$g_{z}=-\frac{\sinh(\frac{(z-2)t}{2})}{2\sinh(\frac{t}{2})\cosh(\frac{(z-1)t}{2})},\quad
t\in\mathbb{R}.$

Equivalently,

$g_{z}=-a_{z}bc_{z},$

where

$a_{z}=e^{-\frac{(z+1)t}{2}}-e^{-\frac{(3z-3)t}{2}},\quad
b=\frac{e^{-t}}{1-e^{-t}},\quad c_{z}=\frac{1}{1+e^{-(z-1)t}}.$

Since $z\to a_{z}$ and $z\to c_{z}$ are continuous $C^{2}(1,\infty)-$valued
mappings and $b\in H^{2}(1,\infty),$ the second assertion follows. ∎

######  Theorem 5.3.2.

Let $G:\\{z\in\mathbb{C}\;:\;\Re(z)>1\\}\to H^{2}(\mathbb{R})\\}$ be the
function given by $G(z)=g_{z}$. Then $G$ is holomorphic on its domain.

###### Proof.

To show that $G$ is holomorphic with values in $H^{2}(\mathbb{R})$, it
suffices to show for all continuous linear functionals $\varpi$ on
$H^{2}(\mathbb{R})$ that $z\mapsto\varpi(G(z))$ is holomorphic.

Since $H^{2}(\mathbb{R})$ is a Hilbert space, it suffices to show for all
$h\in H^{2}(\mathbb{R})$ that:

(5.9)
$z\mapsto\int_{\mathbb{R}}g_{z}(t)h(t)+g_{z}^{\prime}(t)h^{\prime}(t)+g_{z}^{\prime\prime}(t)h^{\prime\prime}(t)\,dt,\quad\Re(z)>1$

is holomorphic.

Let $\gamma$ be a simple closed curve contained in $\\{z\;:\;\Re(z)>1\\}$. By
Lemma 5.3.1, the function

$(z,t)\mapsto
g_{z}(t)h(t)+g_{z}^{\prime}(t)h^{\prime}(t)+g_{z}^{\prime\prime}(t)h^{\prime\prime}(t),\quad
t\in\mathbb{R},z\in\gamma$

is integrable on $\gamma\times\mathbb{R}.$ Indeed,

$\int_{\gamma}\Big{(}\int_{\mathbb{R}}|g_{z}(t)h(t)+g_{z}^{\prime}(t)h^{\prime}(t)+g_{z}^{\prime\prime}(t)h^{\prime\prime}(t)|dt\Big{)}|dz|\leq$
$\leq\int_{\gamma}\|g_{z}\|_{H^{2}(\mathbb{R})}|dz|\leq{\rm
length}(\gamma)\cdot\sup_{z\in\gamma}\|g_{z}\|_{H^{2}(\mathbb{R})}<\infty.$

We may apply Fubini’s theorem to conclude that:

$\int_{\gamma}\varpi(G_{z})\,dz=\int_{\mathbb{R}}\Big{(}\int_{\gamma}(g_{z}(t)h(t)+g_{z}^{\prime}(t)h^{\prime}(t)+g_{z}^{\prime\prime}(t)h^{\prime\prime}(t))dz\Big{)}dt.$

For each fixed $t$ it follows from the definition that the functions
$g_{z}(t)$, $g_{z}^{\prime}(t)$ and $g_{z}^{\prime\prime}(t)$ are holomorphic
in $z$. Hence $\int_{\gamma}\varpi(G(z))\,dz=0$ for all simple closed curves
$\gamma$ contained in $\\{z\;:\;\Re(z)>1\\}$. Since $z\mapsto\varpi(G(z))$ is
continuous, by Morera’s theorem $z\mapsto\varpi(G(z))$ is holomorphic in the
domain $\\{z\;:\;\Re(z)>1\\}$. Since $\varpi$ is arbitrary, $G$ is
$H^{2}(\mathbb{R})$-valued holomorphic. ∎

###  5.4. The function $z\to{\rm Tr}(XB^{z}A^{z})-{\rm
Tr}(X(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z})$ admits analytic continuation to
$\\{\Re(z)>p-1\\}$

As indicated in the title, in this section we prove (under certain assumptions
on $A$ and $B$) that for all $X\in\mathcal{L}_{\infty}$ the mapping

$z\mapsto\mathrm{Tr}(XB^{z}A^{z})-\mathrm{Tr}(X(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z})$

defined initially on $\\{z\in\mathbb{C}\;:\;\Re(z)>p\\}$ is holomorphic, and
admits analytic continuation to the set $\\{z\;:\;\Re(z)>p-1\\}$. The precise
assumptions on $A$ and $B$ are as follows:

######  Condition 5.4.1.

Let $p>2$ and let $0\leq A,B\in\mathcal{L}_{\infty}$ satisfy the conditions

  1. (i)

$B^{p}A\in\mathcal{L}_{1,\infty}$

  2. (ii)

$B^{q-2}[B,A]\in\mathcal{L}_{1}$ for every $q>p$

  3. (iii)

$A^{\frac{1}{2}}BA^{\frac{1}{2}}\in\mathcal{L}_{p,\infty}$

  4. (iv)

$[B,A^{\frac{1}{2}}]\in\mathcal{L}_{\frac{p}{2},\infty}$.

The main result of this section is the following:

######  Theorem 5.4.2.

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

$z\to{\rm Tr}(XB^{z}A^{z})-{\rm
Tr}(X(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{z}),\quad\Re(z)>p,$

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

######  Lemma 5.4.3.

Assume that $p\geq 1$ and that $A,B$ satisfy Condition 5.4.1. Then for all
$r\geq 1$ we have $B^{\frac{p}{r}}A\in\mathcal{L}_{r,\infty}$. More precisely,
we have the following norm bound:

$\|B^{\frac{p}{r}}A\|_{r,\infty}\leq
e\|A\|_{\infty}^{1-\frac{1}{r}}\|B^{p}A\|_{1,\infty}^{\frac{1}{r}}.$

###### Proof.

We show this is a consequence of the Araki-Lieb-Thirring inequality (2.1) and
(2.4).

Fix $r\geq 1$. Then by the Araki-Lieb-Thirring inequality:

$|B^{p/r}A|^{r}\prec\prec_{\log}B^{p}A^{r}.$

Now using (2.4):

$\displaystyle\|B^{p/r}A\|_{r,\infty}$
$\displaystyle=\||B^{p/r}A|^{r}\|_{1,\infty}$ $\displaystyle\leq
e\|B^{p}A^{r}\|_{1,\infty}$ $\displaystyle\leq
e\|A\|_{\infty}^{r-1}\|B^{p}A\|_{1,\infty}.$

∎

The next lemma provides a sufficient condition for a function to be
holomorphic with values in a Banach ideal of $\mathcal{L}_{\infty}$.

######  Lemma 5.4.4.

Assume that $D\subseteq\mathbb{C}$ is a domain (i.e., a connected open set)
and that $F:D\to\mathcal{L}_{\infty}$ is a holomorphic function. If
$\mathcal{I}$ is a Banach-normed ideal of $\mathcal{L}_{\infty}$ such that $F$
takes values in $\mathcal{I}$ and $F:D\to\mathcal{I}$ is continuous, then $F$
is also an $\mathcal{I}$-valued holomorphic function.

###### Proof.

Fix a contour $\gamma\subset D.$ such that the interior of $\gamma$ is
contained in $D$. Then for all $z$ in the interior of $\gamma$, we have

$F(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{F(w)}{w-z}\,dw$

A priori, the integral is a weak integral. However, $F:D\to\mathcal{I}$ is
continuous and, therefore, the integral is Bochner in $\mathcal{I}.$

In order to show that $F$ is holomorphic, it suffices to show that it is
differentiable. Let

$G(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{F(w)}{(w-z)^{2}}\,dw$

Once more, since $F:D\to\mathcal{I}$ is continuous, then this integral is
defined as an $\mathcal{I}$-valued Bochner integral. If $F$ were holomorphic,
then $G$ would be the derivative of $F$. The proof will be completed upon
showing that $G$ is indeed the derivative of $F$ considered as an
$\mathcal{I}$-valued mapping.

For every $z$ and $z_{0}$ in the interior of $\gamma$, we have:

$\frac{F(z)-F(z_{0})}{z-z_{0}}-G(z_{0})=\frac{z-z_{0}}{2\pi
i}\int_{\gamma}\frac{F(w)}{(w-z)(w-z_{0})^{2}}\,dw$

Again, this integral is an $\mathcal{I}$-valued Bochner integral.

Thus,

$\displaystyle\Big{\|}\frac{F(z)-F(z_{0})}{z-z_{0}}-G(z_{0})\Big{\|}_{\mathcal{I}}$
$\displaystyle\leq\frac{|z-z_{0}|}{2\pi}\int_{\gamma}\|F(w)\|_{\mathcal{I}}\frac{1}{|w-z|\cdot|w-z_{0}|^{2}}\,|dw|$
$\displaystyle\leq\sup_{w\in\gamma}\|F(w)\|_{\mathcal{I}}\cdot\frac{|z-z_{0}|}{2\pi}\int_{\gamma}\frac{1}{|w-z|\cdot|w-z_{0}|^{2}}\,|dw|.$

Since $f$ is continuous, the right hand side tends to $0$ as $z\to z_{0}$.
Hence, $G=F^{\prime}$ and so $F$ is holomorphic. ∎

######  Lemma 5.4.5.

Let $0\leq A\in\mathcal{L}_{\infty}.$ The function $z\to A^{z}$ is
$\mathcal{L}_{\infty}-$valued holomorphic on the half-plane
$\\{z\;:\;\Re(z)>0\\}.$

###### Proof.

For $z_{0},z\in\mathbb{C}$ with $\Re(z_{0})>0$ and $\Re(z)>0,$ we define the
operator $A^{z_{0}}\log(A)$ by means of functional calculus (the convention
$0^{z_{0}}\log(0)=0$ is used).

Hence,

$\displaystyle\Big{\|}\frac{A^{z}-A^{z_{0}}}{z-z_{0}}-A^{z_{0}}\log(A)\Big{\|}_{\infty}$
$\displaystyle\leq\sup_{0\leq\lambda\leq\|A\|_{\infty}}\Big{|}\frac{\lambda^{z}-\lambda^{z_{0}}}{z-z_{0}}-\lambda^{z_{0}}\log(\lambda)\Big{|}$
$\displaystyle=O(z-z_{0}),\quad z_{0}\to z.$

Hence, for all $z$ with $\Re(z)>0$

$A^{z}-A^{z_{0}}=A^{z_{0}}\log(A)(z-z_{0})+o(1),\quad z_{0}\to z$

and so $A^{z}$ is $\mathcal{L}_{\infty}$-valued holomorphic with derivative
$A^{z}\log(A)$. ∎

######  Lemma 5.4.6.

Let $p>2$ and assume $A$ and $B$ satisfy condition 5.4.1. The mapping

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

is a holomorphic $\mathcal{L}_{\frac{p}{2},\infty}-$valued function.

###### Proof.

We take care to note that since $p>2$, the ideal $\mathcal{L}_{p/2,\infty}$
can be equipped with a norm generating the same topology as that of the
canonical quasi-norm (see [3, Chapter 4, Lemma 4.5]). Denote such a norm as
$\|\cdot\|_{\mathcal{L}_{p/2,\infty}}^{\prime}$.

Denote

$F(z)=[B,A^{z}].$

Since $F(z)=BA^{z}-A^{z}B$ it now follows from Lemma 5.4.5 that $F$ is
$\mathcal{L}_{\infty}$-valued holomorphic. Due to Lemma 5.4.4, it now suffices
to show that $F$ is $\mathcal{L}_{p/2,\infty}$-valued and
$\mathcal{L}_{p/2,\infty}$-continuous.

Let $\phi$ be a compactly supported smooth function on $\mathbb{R}$, such that
$0\leq\phi\leq 1$ and $\phi=1$ on the interval $[0,\|A\|_{\infty}]$. Define

$\phi_{z}(t)=|t|^{z}\phi(t),\quad t\in\mathbb{R},\Re(z)>1.$

Since $\phi=1$ on the spectrum of $A$, we have that $\phi_{z}(A)=A^{z}$ and so
for all $z,z_{1},z_{2}$ with $\Re(z),\Re(z_{1}),\Re(z_{2})>1$,

$\displaystyle F(z)$ $\displaystyle=[B,\phi_{z}(A)]$ $\displaystyle
F(z_{1})-F(z_{2})$ $\displaystyle=[B,(\phi_{z_{1}}-\phi_{z_{2}})(A)].$

Now we refer to [50], where it is proved that if $A$ and $B$ are self-adjoint
operators and $r>1$ is such that $[A,B]\in\mathcal{L}_{r,\infty}$, then for
all Lipschitz functions $f$, there is a constant $c_{r}$ such that:

$\|[f(A),B]\|_{\mathcal{L}_{r,\infty}}\leq
c_{r}\|f^{\prime}\|_{L_{\infty}(\mathbb{R})}\|[A,B]\|_{\mathcal{L}_{r,\infty}}.$

Since $p>2$, we may apply this result with $r=p/2$ and since $\phi$ is smooth
and compactly supported, we may take $f=\phi_{z}$. Thus,

$\|F(z)\|_{\mathcal{L}_{p/2,\infty}}^{\prime}\leq
c_{p/2}\|\phi_{z}^{\prime}\|_{L_{\infty}(\mathbb{R})}\|[B,A]\|_{\mathcal{L}_{p/2,\infty}}^{\prime}$

and similarly taking $f=\phi_{z_{1}}-\phi_{z_{2}}$,

$\|F(z_{1})-F(z_{2})\|_{\mathcal{L}_{p/2,\infty}}^{\prime}\leq
c_{p/2}\|\phi_{z_{1}}^{\prime}-\phi_{z_{2}}^{\prime}\|_{L_{\infty}(\mathbb{R})}\|[B,A]\|_{\mathcal{L}_{p/2,\infty}}^{\prime}.$

Note that for $z_{1}\to z_{2}$ we have:

$\|\phi_{z_{1}}^{\prime}-\phi_{z_{2}}^{\prime}\|_{L_{\infty}(\mathbb{R})}\to
0$

and hence $F$ is $\mathcal{L}_{p/2,\infty}$-valued continuous. The result now
follows. ∎

######  Lemma 5.4.7.

Let $p>2$ and let $A$ and $B$ satisfy Condition 5.4.1. Then:

  1. (i)

The mapping
$F_{0}(z):=B^{z-1}[B,A^{z-\frac{1}{2}}]A^{\frac{1}{2}}+[B,A^{\frac{1}{2}}]A^{\frac{1}{2}}Y^{z-1}$
is an $\mathcal{L}_{1}-$valued holomorphic function for the domain
$\Re(z)>p-1.$

  2. (ii)

The mapping $F_{1}(z):=B^{z-1}A^{z-1}$ is an
$\mathcal{L}_{\frac{p}{p-2},1}-$valued holomorphic function for the domain
$\Re(z)>p-1.$

  3. (iii)

The mapping $F_{2}(z):=B^{z-1}[BA,A^{z-1}]$ is an $\mathcal{L}_{1}-$valued
holomorphic function for the domain $\Re(z)>p-1.$

  4. (iv)

The mapping $F_{3}(z):=Y^{z-1}$ is an $\mathcal{L}_{\frac{p}{p-2},1}-$valued
holomorphic function for the domain $\Re(z)>p-1.$ (Recall that
$Y=A^{1/2}BA^{1/2}$).

###### Proof.

We first prove (ii). Fix $q\in(p,p+2).$ If $\Re(z)>q-1,$ then

$B^{z-1}A^{z-1}=B^{z-q+1}B^{q-2}A\cdot A^{z-2}.$

By Lemma 5.4.3, we have that

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

and correspondingly by Lemma 5.4.5 the mapping $z\mapsto B^{z-1}A^{z-1}$ is
continuous in the $\mathcal{L}_{p/(p-2),1}$ norm.

Moreover Lemma 5.4.5 implies that the mappings $z\to B^{z-q+1}$ and $z\to
A^{z-2}$ are $\mathcal{L}_{\infty}-$valued holomorphic for $\Re(z)>q-1.$ Thus
by Lemma 5.4.4 the mapping $z\to B^{z-1}A^{z-1}$ is
$\mathcal{L}_{\frac{p}{p-2},1}-$valued holomorphic for $\Re(z)>q-1.$ Since
$q>p$ is arbitrary, (ii) follows.

Now we prove (iii). Again fix $q\in(p,p+2)$ and let $z$ satisfy $\Re(z)>q-1.$
We rewrite $F_{2}$ as:

$\displaystyle F_{2}(z)$ $\displaystyle=B^{z}A^{z}-B^{z-1}A^{z-1}BA$
$\displaystyle=B^{z-1}A^{z-1}\cdot[A,B]+[B,B^{z-1}A^{z}]$ (5.10)
$\displaystyle=B^{z-1}A^{z-1}\cdot[A,B]+B^{z-1}\cdot[B,A]\cdot
A^{z-1}+B^{z-1}A\cdot[B,A^{z-1}].$

The first summand in (5.10) can be written as

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

Due to Lemma 5.4.5, the mappings $z\to B^{z-q+1}$ and $z\to A^{z-2}$ are
$\mathcal{L}_{\infty}-$valued holomorphic for $\Re(z)>q-1.$ By Lemma 5.4.3, we
have that

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

The element $[A,B]=A^{1/2}[A^{1/2},B]+[A^{1/2},B]A^{1/2}$ belongs to