# Krein’s trace theorem revisited

D. Potapov  School of Mathematics and Statistics, University of New South
Wales, Kensington, NSW, Australia<EMAIL_ADDRESS>,  F. Sukochev
<EMAIL_ADDRESS>and  D. Zanin<EMAIL_ADDRESS>
###### Abstract.

We supply the first proof of Krein’s Trace Theorem which does not use complex
analysis. Our proof holds for $\sigma$-finite von Neumann algebras
$\mathcal{M}$ of type II and unbounded perturbations from the predual of
$\mathcal{M}$.

###### 2000 Mathematics Subject Classification:

47C15

##  1\. Introduction

The first attempt to deliver a proof of the existence of the Krein Spectral
Shift Function (KrSSF) without using complex analytical facts was made by
M.Sh. Birman and M.Z. Solomyak in 1972 (see [3] and also [2]). Their method
was based on the theory of double operator integrals developed by those
authors in [4, 5, 6]. That attempt led to introducing of an important notion
of the spectral averaging measure (see also [17]), but was not successful
since the authors of [3] failed to prove directly the absolute continuity of
that measure with respect to Lebesgue measure. The second attempt to deliver
such a proof is due to D. Voiculescu [20], whose method is based on the usage
of classical Weyl-von Neumann theorem. However, his attempt also fails to
recover the full generality of Krein’s original result. In our present paper,
we combine methods drawn from the double operator integration theory with
Voiculescu’s ideas and deliver a rather short and straightforward new proof of
Krein’s result in full generality. The advantage of our present approach is
seen from the following extension.

We deliver the complete proof of the existence of the spectral shift function
$\xi_{A,B}$ in the setting when $A$ and $B$ are self-adjoint operators
affiliated with a $\sigma$-finite semifinite von Neumann algebra
$\mathcal{M}$, whose difference $A-B$ belongs to the predual of $\mathcal{M}$.
If $\mathcal{M}$ is a type I factor, then this is precisely the Krein’s
result. For a general semifinite von Neumann algebra, the earlier attempt
based on emulating the Krein’s complex-analytical proof only yielded the
result for the special case when $V$ was necessarily a bounded trace class
perturbation [1].

##  2\. Preliminaries

Let $H$ be an infinite dimensional Hilbert space and let $\mathcal{L}(H)$ be
the algebra of all bounded operators in $H$. In what follows, $\mathcal{M}$ is
a von Neumann algebra on $H$, that is a $*$-subalgebra of $\mathcal{L}(H)$
closed in the weak operator topology. The identity in $\mathcal{M}$ is denoted
by $1$. We are only interested in semifinite von Neumann algebras, that is,
those which admit a faithful normal semifinite trace $\tau$. We fix a couple
$(\mathcal{M},\tau)$. A von Neumann algebra is said to be $\sigma$-finite if
it admits at most countably many orthogonal projections.

An (unbounded) operator is said to be affiliated with $\mathcal{M}$ if it
commutes with every operator in the commutant $\mathcal{M}^{\prime}$ of
$\mathcal{M}$. Closed densely defined operator $A$ affiliated with
$\mathcal{M}$ is said to be $\tau$-measurable if, for every $\varepsilon>0$,
there exists a projection $p\in\mathcal{M}$ such that $\tau(p)<\varepsilon$
and such that $(1-p)H\subset{\rm dom}(A)$. The collection of all
$\tau$-measurable operators is denoted by $S(\tau)$. The real vector space
$S_{h}\left(\tau\right)=\left\\{A\in S\left(\tau\right):A=A^{\ast}\right\\}$
is a partially ordered vector space with the ordering defined by setting
$A\geq 0$ if and only if $\left\langle A\xi,\xi\right\rangle\geq 0$ for all
$\xi\in\mathcal{D}\left(A\right)$. The positive cone in
$S_{h}\left(\tau\right)$ will be denoted by $S\left(\tau\right)^{+}$. The
positive part $A_{+}$ and negative part $A_{-}$ of an operator $A\in
S_{h}\left(\tau\right)$ are defined by

$A_{+}:=\int_{\mathbb{R}}\lambda^{+}dE_{A}(-\infty,\lambda]\text{ \ \ \ and \
\ \ }A_{-}:=\int_{\mathbb{R}}\lambda^{-}dE_{A}(-\infty,\lambda]$

respectively, where $\lambda^{+}=\max\left(\lambda,0\right)$ and
$\lambda^{-}=\max\left(-\lambda,0\right)$ and $E_{A}(-\infty,\lambda]$ is the
spectral projection of the self-adjoint operator $A$ corresponding to the
interval $(-\infty,\lambda]$. It follows immediately from the spectral theorem
that $A=A_{+}-A_{-}$.

The notions of the distribution function $n_{A}$, $A\in
S_{h}\left(\tau\right)$ and that of the singular value function $\mu(A)$,
$A\in S\left(\tau\right)$ are defined as follows

$n_{A}(t):=\tau(E_{A}(t,\infty)),\ t\in\mathbb{R}\ \ \text{and}\ \
\mu(t;A):=\inf\\{s:n_{|A|}(s)\leq t\\},\quad t\geq 0.$

It follows directly that the singular value function $\mu(A)$ is a decreasing,
right-continuous function on the positive half-line $[0,\infty)$. The trace
$\tau$ extends to $S\left(\tau\right)^{+}$ as a non-negative extended real-
valued functional which is positively homogeneous, additive, unitarily
invariant and normal. This extension is given by

$\tau\left(A\right)=\int_{0}^{\infty}\mu\left(t;A\right)dt,\ \ \ A\in
S\left(\tau\right)^{+},$

and satisfies $\tau\left(A^{\ast}A\right)=\tau\left(AA^{\ast}\right)$ for all
$A\in S\left(\tau\right)$. If ${\mathcal{M}}={\mathcal{L}}({\mathcal{H}})$ and
$\tau$ is the standard trace, then $S(\tau)={\mathcal{M}}$. In this case, an
operator $A\in S(\tau)$ is compact if and only if
$\lim_{t\to\infty}\mu(t;A)=0$ and

$\mu_{n}(A)=\mu\left(t;A\right),\quad t\in[n,n+1),\quad n=0,1,2,\dots,$

and the sequence $\\{\mu_{n}(A)\\}_{{}_{n=0}}^{\infty}$ is just the sequence
of eigenvalues of $|A|$ in non-increasing order and counted according to
multiplicity.

The noncommutative space $\mathcal{L}_{p}=\mathcal{L}_{p}(\mathcal{M},\tau)$,
$1\leq p\leq\infty$ is defined as follows

$\mathcal{L}_{p}=\\{A\in S(\tau):\ \mu(|A|)\in L_{p}:=L_{p}(0,\infty)\\},$

where $L_{p}(0,\infty)$ is the usual Lebesgue space. The space
$\mathcal{L}_{p}$ is a linear subspace of $S\left(\tau\right)$ and the
functional
$A\longmapsto\left\|A\right\|_{p}:=\left(\tau(|A|^{p})\right)^{1/p}$,
$A\in\mathcal{L}_{p}$, $1\leq p<\infty$ is a norm. For convenience, we set
$(\mathcal{L}_{\infty},\|\cdot\|_{\infty})=\mathcal{M}$ equipped with the
uniform operator norm. We have, in particular,
$\left\|A\right\|_{p}=\left\|\mu\left(A\right)\right\|_{p}$ for all
$A\in\mathcal{L}_{p}$, $1\leq p\leq\infty$ (we denote the norm on
$\mathcal{L}_{p}$ and the standard norms on Lebesgue spaces $L_{p}(0,\infty)$
and $L_{p}(\mathbb{R})$ by the same symbol $\|\cdot\|_{p}$ and this should not
cause any confusion). We recall the following useful formula

$\|n_{A}\|_{1}=\int_{0}^{\infty}n_{A}(s)ds=sn_{A}(s)|_{0}^{\infty}-\int_{0}^{\infty}sdn_{A}(s)=\tau(A)=\|A\|_{1},$

which holds for every $A\in S\left(\tau\right)^{+}$. Equipped with the norm
$\|\cdot\|_{1}$, the space $\mathcal{L}_{1}$ is a Banach space. It is well
known (see e.g. [14]) that $\mathcal{L}_{1}$ is isometric to a predual
$\mathcal{M}_{*}$ of the von Neumann algebra $\mathcal{M}$. In what follows,
we will need the following result whose proof follows verbatim from that of
[19, Lemmas 15 and 16], where this result stated for $s,t>0$.

######  Lemma 1.

Suppose that $\mathcal{M}$ is a finite von Neumann algebra and that
$\tau(1)<\infty$. If operators $A,B\in S_{h}\left(\tau\right)$ satisfy $A\geq
B$, then $n_{A}\geq n_{B}$ and the inequality

$n_{A+B}(s+t)\leq n_{A}(s)+n_{B}(t)$ (1)

holds for all $s,t\in\mathbb{R}$.

The following Weyl-von Neumann type theorem is at the core of our approach in
this paper. It is (implicitly) proved in [16]. For the classical Weyl-von
Neumann theorem we refer the reader to [8].

######  Theorem 2.

Let $\mathcal{M}$ be a $\sigma$-finite von Neumann algebra equipped with a
faithful normal semifinite trace $\tau$. For every $A\in\mathcal{M}$, there
exists a sequence of $\tau$-finite projections $p_{n}\uparrow 1$, $n\geq 0$,
such that $\|[A,p_{n}]\|_{2}\to 0$.

###### Proof.

It follows from Lemma 6.4 in [16] that there exists a net $p_{i}$,
$i\in\mathbb{I}$, of orthogonal projections such that the algebra
$p_{i}\mathcal{M}p_{i}$ admits a $\tau$-finite generating projection. Since
$\mathcal{M}$ is $\sigma$-finite, it follows that the set $\mathbb{I}$ is, at
most, countable. The assertion follows now from Lemma 6.3 in [16]. ∎

The following two lemmas constitute a small complement to Theorem 2. These
lemmas are at the core of Voiculescu’s approach [20]. If $A\in S_{h}(\tau)$,
then the projection onto the closure of the range of $|A|$ is called the
support of $A$ and is denoted by ${\rm supp}(A)$.

######  Lemma 3.

Let $\mathcal{M}$ be a von Neumann algebra equipped with a faithful normal
semifinite trace $\tau$. If $p_{n}\uparrow 1$ and if $C\in\mathcal{M}$,
$C^{*}=C$ is such that $\tau({\rm supp}(C))<\infty$, then
$\|[C,p_{n}]\|_{2}\to 0$.

###### Proof.

Define $A_{n}={\rm supp}(C)p_{n}{\rm supp}(C)$. We have $A_{n}\uparrow{\rm
supp}(C)$ and

$\|[C,p_{n}]\|_{2}^{2}=-\tau([C,p_{n}]^{2})=2\tau(A_{n}C^{2})-2\tau((A_{n}C)^{2}).$

The sequence $A_{n}$ strongly converges to ${\rm supp}(C)$. Therefore,
$A_{n}C^{2}\to C^{2}$ and $A_{n}C\to C$ strongly. By [7, Proposition 2.4.1],
we have $(A_{n}C)^{2}\to C^{2}$ strongly. Since the trace $\tau$ is strongly
continuous (see e.g. [18, Lemma 1.2, Theorem 1.10]) and the algebra ${\rm
supp}(C)\mathcal{M}{\rm supp}(C)$ is finite, it follows that
$\tau(A_{n}C^{2})\to\tau(C^{2})$ and $\tau((A_{n}C)^{2})\to\tau(C^{2})$. This
suffices to conclude the proof. ∎

######  Lemma 4.

Let $\mathcal{M}$ be a von Neumann algebra equipped with a faithful normal
semifinite trace $\tau$. Let $A\in\mathcal{M}$ and let $\\{p_{n}\\}_{n\geq
0}\subset\mathcal{M}$ be a sequence of $\tau$-finite projections. If
$\|[A,p_{n}]\|_{2}\to 0$, then, for every $m\geq 1$, we have

$\tau((p_{n}Ap_{n})^{m}-A^{m}p_{n})\to 0.$

###### Proof.

For $m=1$ the assertion is obvious. For every $m\geq 2$, we have

$|\tau((p_{n}Ap_{n})^{m}-A^{m}p_{n})|=|\sum_{k=1}^{m-1}p_{n}A^{k}(1-p_{n})(Ap_{n})^{m-k}|\leq\\\
\sum_{k=1}^{m-1}\|p_{n}A^{k}(1-p_{n})\|_{2}\|(1-p_{n})Ap_{n}\|_{2}\|A\|_{\infty}^{m-k-1}.$

Since $(1-p_{n})Ap_{n}=[A,p_{n}]p_{n}$ and
$p_{n}A^{k}(1-p_{n})=p_{n}[p_{n},A^{k}]$, and since

$[p_{n},A^{k}]=\sum_{l=0}^{k-1}A^{l}[p_{n},A]A^{k-1-l},$

it follows that

$|\tau((p_{n}Ap_{n})^{m}-A^{m}p_{n})|\leq\\\
\sum_{k=1}^{m-1}k\|[p_{n},A]\|_{2}\|A\|_{\infty}^{k-1}\cdot\|[A,p_{n}]\|_{2}\cdot\|A\|_{\infty}^{m-k-1}\\\
=\frac{m(m-1)}{2}\|[A,p_{n}]\|_{2}^{2}\|A\|_{\infty}^{m-2}.$

∎

### The class $W_{1}$

Originally Theorem 7 is proved for the function of class $W_{1}$. That is,

$W_{1}=\left\\{f\in S^{\prime}(\mathbb{R}):\quad\mathcal{F}(f^{\prime})\in
L_{1}(\mathbb{R})\right\\},$

where $S^{\prime}(\mathbb{R})$ is the class of all tempered distributions on
$\mathbb{R}$, $\mathcal{F}$ is the Fourier transform and $L_{1}(\mathbb{R})$
is the Lebesgue space of all integrable functions on $\mathbb{R}$. The class
$W_{1}$ is equipped with the semi-norm

$\left\|f\right\|_{W_{1}}=\left\|\mathcal{F}(f^{\prime})\right\|_{1}.$

We need the following simple observation which directly follows from the fact
that the class $S(\mathbb{R})$ of all Schwartz functions is dense in
$L_{1}(\mathbb{R})$.

######  Lemma 5.

The class of primitives of functions in $S(\mathbb{R})$ is dense in $W_{1}$,
that is for every $f\in W_{1}$, there exists a sequence $f_{n}\in
C^{1}_{b}(\mathbb{R})$ such that $f_{n}^{\prime}\in S(\mathbb{R})$ and

$\lim_{n\rightarrow\infty}\left\|f-f_{n}\right\|_{W_{1}}=0.$

######  Theorem 6.

Let $\mathcal{M}$ be a von Neumann algebra equipped with a faithful normal
semifinite trace $\tau$. If $A,B$ are unbounded self-adjoint operators
affiliated with $\mathcal{M}$ such that $A-B\in\mathcal{L}_{1}$, then
$f(A)-f(B)\in\mathcal{L}_{1}$ for every $f\in W_{1}$. Moreover, we have

$\|f(A)-f(B)\|_{1}\leq\|f\|_{W_{1}}\|A-B\|_{1},$
$\tau(f(A)-f(B))=\int_{0}^{1}\tau(f^{\prime}((1-z)A+zB)(A-B))dz$

for every $f\in W_{1}.$

###### Proof.

The first inequality is proved by Widom [21]. We present it here for
convenience of the reader. It follows from the equality

$e^{iC}-e^{iD}=i\int_{0}^{1}e^{itC}(C-D)e^{i(1-t)D}dt$

that

$\|e^{iC}-e^{iD}\|_{1}\leq\|C-D\|_{1}.$

Hence,

$\|e^{isA}-e^{isB}\|_{1}\leq|s|\|A-B\|_{1}.$

Recall the obvious equality

$f(A)-f(B)=\int_{\mathbb{R}}(\mathcal{F}f)(s)(e^{isA}-e^{isB})ds.$

It follows that

$\|f(A)-f(B)\|_{1}\leq\int_{\mathbb{R}}|(\mathcal{F}f)(s)|\cdot\|e^{isA}-e^{isB}\|_{1}ds\leq$
$\leq\int_{\mathbb{R}}|s(\mathcal{F}f)(s)|\cdot\|A-B\|_{1}ds=\|f\|_{W_{1}}\cdot\|A-B\|_{1}.$

The second equality is proved in [3] (see equation (2.1) there). ∎

##  3\. Krein’s theorem in semifinite setting

The present section proves the following theorem in complete generality. In
the setting ${\mathcal{M}}={\mathcal{L}}({\mathcal{H}})$ the result is
originally due to M.G. Krein.

######  Theorem 7.

Let $\mathcal{M}$ be a $\sigma$-finite von Neumann algebra with a faithful
normal semifinite trace $\tau$. If self-adjoint operators $A,B$ affiliated
with $\mathcal{M}$ are such that $A-B\in\mathcal{L}_{1}$, then there is a
function $\xi=\xi_{A,B}\in L_{1}(\mathbb{R})$ such that

$\tau(f(A)-f(B))=\int_{\mathbb{R}}f^{\prime}(s)\xi(s)ds$ (2)

for every $f\in W_{1}.$ In addition, we have

$\int_{\mathbb{R}}|\xi(t)|\,dt\leq\left\|V\right\|_{1},\quad\int_{\mathbb{R}}\xi(t)dt=\tau(V).$

### Proof of Theorem 7

We shall approach the proof of Theorem 7 via step by step relaxing of
conditions on the trace $\tau$, function $f$ and the operators $A$ and $B$.
This is presented as a series of lemmas from Lemma 8 to Lemma 12. The final
extension to the class $W_{1}$ is given in Lemma 13. For reader’s convenience,
we denote the function $\xi_{A,B}$ at different stages with different indices.

We start with rather restrictive case as in the following lemma. The lemma was
noted yet by Krein [9, 10] (see also p.360 in [1]).

######  Lemma 8.

Suppose that the assumptions of Theorem 7 hold.

  1. (a)

If $\tau$ is finite and if $A,B\in\mathcal{M}$, then (2) holds with
$\xi=\xi^{(1)}_{A,B}=n_{A}-n_{B}$. The function $\xi$ is integrable and is
supported on $[0,\max\\{\|A\|_{\infty},\|B\|_{\infty}\\}]$.

  2. (b)

Furthermore,

$\|\xi^{(1)}_{A,B}\|_{\infty}\leq\tau({\rm supp}(A-B))\ \text{and}\ \
\|\xi^{(1)}_{A,B}\|_{1}\leq\|A-B\|_{1}.$

  3. (c)

If $A\geq B$, then $\|\xi^{(1)}_{A,B}\|_{1}=\|A-B\|_{1}$.

###### Proof.

(a) Indeed, it follows from the functional calculus that

$f(A)=-\int_{-\infty}^{\infty}f(s)dE_{A}(s,\infty).$

Taking the trace and integrating by parts, we obtain

$\tau(f(A))=-\int_{-\infty}^{\infty}f(s)dn_{A}(s)=\int_{-\infty}^{\infty}f^{\prime}(s)n_{A}(s)ds.$

The equation (2) follows immediately. Since $n_{A}$ and $n_{B}$ are
integrable, then so is $\xi^{(1)}_{A,B}$.

(b) It is clear that $A\leq B+|A-B|$ and, therefore,

$n_{A}(t)\leq
n_{A+|B-A|}(t)\stackrel{{\scriptstyle(\ref{nsum})}}{{\leq}}n_{B}(t)+n_{|A-B|}(0).$

Similarly, we have $B\leq A+|A-B|$ and, therefore,

$n_{B}(t)\leq
n_{B+|B-A|}(t)\stackrel{{\scriptstyle(\ref{nsum})}}{{\leq}}n_{A}(t)+n_{|A-B|}(0).$

Combining these inequalities, we obtain $|n_{A}-n_{B}|\leq n_{|A-B|}(0)$,
which proves the first assertion. In order to prove the second inequality,
observe that

$|B-A|=(B-A)_{+}+(B-A)_{-}\ \ \text{and}\ \ B-A=(B-A)_{+}-(B-A)_{-}.$

Set $C:=A+(B-A)_{+}=B-(B-A)_{-}$. We then have $C\geq A$ and $C\geq B$. In
particular, $n_{C}\geq n_{A}$ and $n_{C}\geq n_{B}$. Therefore,

$\|n_{C}-n_{A}\|_{1}=\|n_{C}\|_{1}-\|n_{A}\|_{1}=\tau(C)-\tau(A)=\tau((B-A)_{+}),$
$\|n_{C}-n_{B}\|_{1}=\|n_{C}\|_{1}-\|n_{B}\|_{1}=\tau(C)-\tau(B)=\tau((B-A)_{-}).$

Hence,

$\|n_{A}-n_{B}\|_{1}\leq\|n_{C}-n_{A}\|_{1}+\|n_{C}-n_{B}\|_{1}\leq\tau((B-A)_{+})+\tau((B-A)_{-})=\|A-B\|_{1}.$

∎

The key step in our approach is the extension of Lemma 8 to the following
lemma via approximation process set out in Theorem 2 and Lemmas 3 and 4.

######  Lemma 9.

Suppose that the assumptions of Theorem 7 hold. If $A\geq B\geq 0$, and if
$\tau({\rm supp}(A-B))<\infty$, then (2) holds for every $f(s)=s^{m}$,
$m\in\mathbb{N}$, and for the compactly supported positive function
$\xi=\xi^{(2)}_{A,B}$. If the conditions of Lemma 8 are also met, then
$\xi^{(1)}_{A,B}=\xi^{(2)}_{A,B}$.

###### Proof.

By Theorem 2 and Lemma 3, there exists a family of $\tau$-finite projections
$p_{n}$, $n\geq 0$, such that $p_{n}\uparrow 1$ and such that
$\|[A,p_{n}]\|_{2}\to 0$ and $\|[B,p_{n}]\|_{2}\to 0$ as $n\to\infty$. It
follows from Lemma 4 that

$\tau((p_{n}Ap_{n})^{m}-A^{m}p_{n})\to
0,\quad\tau((p_{n}Bp_{n})^{m}-B^{m}p_{n})\to 0.$

Hence,

$\tau((p_{n}Ap_{n})^{m}-(p_{n}Bp_{n})^{m})\to\tau(A^{m}-B^{m}).$

Set $a:=\|A\|_{\infty}$. Since $A\geq B\geq 0$ in $\mathcal{M}$ it follows
that $p_{n}Ap_{n}\geq p_{n}Bp_{n}\geq 0$ in the algebra
$p_{n}\mathcal{M}p_{n}$. In particular, we have $n_{p_{n}Ap_{n}}\geq
n_{p_{n}Bp_{n}}$ for all $n\geq 0$ in the algebra $p_{n}\mathcal{M}p_{n}$. By
Lemma 8 (a), for every $n\geq 0$, there exists a positive function
$\xi_{n}=\xi^{(1)}_{p_{n}Ap_{n},p_{n}Bp_{n}}$ supported on $[0,a]$ such that

$\tau((p_{n}Ap_{n})^{m}-(p_{n}Bp_{n})^{m})=\int_{0}^{a}ms^{m-1}\xi_{n}(s)ds.$

By Lemma 8 (b), we have

$\|\xi_{n}\|_{\infty}\leq\tau({\rm
supp}(A-B)),\quad\|\xi_{n}\|_{1}\leq\|A-B\|_{1}.$

Since $L_{\infty}[0,a]$ is a Banach dual for $L_{1}[0,a]$, it follows from the
Banach-Alaoglu theorem that there exists a directed set $\mathbb{I}$ and the
mapping $\psi:\mathbb{I}\to\mathbb{Z}_{+}$ such that

  1. (1)

for every $n\in\mathbb{Z}_{+}$, there exists $i(n)\in\mathbb{I}$ such that
$\psi(i)>n$ for $i>i(n)$.

  2. (2)

the net $\xi_{\psi(i)}$, $i\in\mathbb{I}$ converges in weak∗ topology.

We set

$\xi^{(2)}_{A,B}=\lim_{i\in\mathbb{I}}\xi_{\psi(i)}.$

Therefore,

$\int_{0}^{b}ms^{m-1}\xi^{(2)}_{A,B}(s)ds=\lim_{i\in
I}\int_{0}^{b}ms^{m-1}\xi_{i}(s)ds=\\\ =\lim_{i\in
I}\tau((p_{\psi(i)}Ap_{\psi(i)})^{m}-(p_{\psi(i)}Bp_{\psi(i)})^{m})=\tau(A^{m}-B^{m}).$
(3)

This shows (2) with $f(s)=s^{m}$. The function $\xi^{(2)}_{A,B}$ is positive
as a weak∗-limit of positive functions.

In order to see that $\xi^{(1)}_{A,B}=\xi^{(2)}_{A,B}$, we simply write the
trace formula (2)

$\int_{\mathbb{R}}ms^{m-1}\xi^{(1)}_{A,B}(s)ds\stackrel{{\scriptstyle Lemma\
\ref{fin
case}}}{{=}}\tau(A^{m}-B^{m})\stackrel{{\scriptstyle(\ref{tbmd})}}{{=}}\int_{\mathbb{R}}ms^{m-1}\xi^{(2)}_{A,B}(s)ds.$

Identification now follows from the fact that the polynomials are a separating
family of functionals on $L_{1}([0,a])$. ∎

The following lemma removes the positivity assumption on the operators $A$ and
$B$ and the assumption that $\tau({\rm supp}(A-B))$ is finite.

######  Lemma 10.

Suppose that the assumptions of Theorem 7 hold. If $A,B\in\mathcal{M}$, then
(2) holds for every $f\in C^{2}(\mathbb{R})$ and for the unique compactly
supported function $\xi=\xi^{(3)}_{A,B}$. In addition,

$\|\xi\|_{1}\leq\|A-B\|_{1},\ \ \text{and}\ \
\int_{\mathbb{R}}\xi(s)\,ds=\tau(A-B).$

If, in addition, the conditions of Lemma 9 are met, then
$\xi^{(2)}_{A,B}=\xi^{(3)}_{A,B}$.

###### Proof.

Set $C:=A+(B-A)_{+}=B-(B-A)_{-}$. Observe that $C\geq A$, $C\geq B$ and that
$C-A,C-B\in\mathcal{L}_{1}$. Since

$\tau(f(A)-f(B))=\tau(f(C)-f(B))-\tau(f(C)-f(A)),$

it is sufficient to consider only the case $A\geq B\geq 0$.

Clearly, the left hand side of (2) does not depend on the values of $f$
outside of the interval $[-\|A\|_{\infty},\|A\|_{\infty}]$. For every $f\in
C^{2}(\mathbb{R})$, there exists $g\in W_{1}$ such that $f=g$ on the interval
$[-\|A\|_{\infty},\|A\|_{\infty}]$. Thus, we may assume without loss of
generality that $f\in W_{1}$.

Let $0\leq D_{n}\leq A-B$ be such that $D_{n}\uparrow A-B$ and such that
$\tau({\rm supp}(D_{n}))<\infty$. It follows from Theorem 6 that

$|\tau(f(B+D_{n}))-f(A))|\leq{\rm
const}\cdot\|f\|_{W_{1}}\cdot\|B+D_{n}-A\|_{1}\to 0.$ (4)

Since the functions which are polynomials on
$[-\|A\|_{\infty},\|A\|_{\infty}]$ are dense in $W_{1}$, it follows from Lemma
9 that

$\tau(f(B+D_{n})-f(B))=\int_{0}^{\infty}f^{\prime}(s)\xi^{(2)}_{B+D_{n},B}(s)ds$
(5)

for every $f\in W_{1}$. Hence, we have

$\int_{0}^{\infty}f^{\prime}(s)\xi_{B+D_{n},B+D_{m}}(s)ds\stackrel{{\scriptstyle
Lemma\ \ref{main lemma}}}{{=}}\tau(f(B+D_{n})-f(B+D_{m}))=$
$=\tau(f(B+D_{n})-f(B))-\tau(f(B+D_{m})-f(B))=$
$\stackrel{{\scriptstyle(\ref{tbd2})}}{{=}}\int_{0}^{\infty}f^{\prime}(s)(\xi_{B+D_{n},B}(s)-\xi_{B+D_{m},B}(s))ds.$

Since $f^{\prime}$ is an arbitrary $C^{1}$ function on
$[-\|A\|_{\infty},\|A\|_{\infty}]$, it follows that

$\xi^{(2)}_{B+D_{n},B}-\xi^{(2)}_{B+D_{m},B}=\xi^{(2)}_{B+D_{n},B+D_{m}}\geq
0,\quad n\geq m.$

Setting $f(s)=s$ in Lemma 9, we obtain that
$\|\xi^{(2)}_{B+D_{n},B}\|_{1}=\|D_{n}\|_{1}$. Since
$\|D_{n}\|_{1}\leq\|A-B\|_{1}$, it follows from the Monotone Convergence
Principle, that the sequence $\xi^{(2)}_{B+D_{n},B}$ converges in
$L_{1}(\mathbb{R})$. Denote its limit by $\xi^{(3)}_{A,B}$. Therefore,

$\tau(f(A)-f(B))\stackrel{{\scriptstyle(\ref{tbd1})}}{{=}}\lim_{n\to\infty}\tau(f(B+D_{n})-f(B))=$
$\stackrel{{\scriptstyle(\ref{tbd2})}}{{=}}\lim_{n\to\infty}\int_{0}^{\infty}f^{\prime}(s)\xi^{(2)}_{B+D_{n},B}(s)ds=\int_{0}^{\infty}f^{\prime}(s)\xi^{(3)}_{A,B}(s)ds.$

The identification $\xi^{(2)}_{A,B}=\xi^{(3)}_{A,B}$ can be established
similarly to that in Lemma 9. ∎

The next step in our approach is removing the assumption that the operators
$A$ and $B$ are bounded. We remove this assumption in two steps: (i) first we
show our construction of a locally integrable $\xi$ for unbounded pair of $A$
and $B$ (see Lemma 11); and (ii) we show that this new $\xi$ is integrable
(and positive when $A\geq B$) (see Lemma 12).

######  Lemma 11.

Suppose that the assumptions of Theorem 7 hold. The equality (2) holds for
$f\in C_{b}^{1}(\mathbb{R})$ such that $f^{\prime}\in S(\mathbb{R})$ with the
unique locally integrable function

$\xi^{(4)}_{A,B}=\xi^{(3)}_{h(A),h(B)}\circ h.$

Here, $h:\mathbb{R}\to(0,1)$ is a $C^{2}$-bijection such that

$(h^{-1})^{\prime}(t),(h^{-1})^{\prime\prime}(t)=O((h^{-1}(t))^{n}),\quad t\to
0,1,$ (6)

for some $n<\infty$. If $A,B$ are bounded, then
$\xi^{(4)}_{A,B}=\xi^{(3)}_{A,B}$.

###### Proof.

As in Lemma 10, we may assume that $A\geq B$. Applying Theorem 6 to the
operators $A$, $B$ and to the function $h$, we obtain

$\|h(A)-h(B)\|_{1}\leq{\rm const}\cdot\|A-B\|_{1}.$

Define the function $g$ by setting $g:=f\circ h^{-1}$. It is a composition of
two $C^{2}$ functions and, therefore, $g\in C^{2}(-1,1)$. The assumption
$f^{\prime}\in S(\mathbb{R})$ guarantees that $f^{\prime},f^{\prime\prime}$
decrease faster than any power function. This together with (6) implies that

$g^{\prime}(+0)=g^{\prime}(1-0)=0,\quad
g^{\prime\prime}(+0)=g^{\prime\prime}(1-0)=0.$

Hence, we have $g\in C^{2}[0,1]$. Clearly, $g$ extends to a function $g\in
C^{2}_{b}(\mathbb{R})$. Applying Lemma 10 to the operators $h(A)$ and $h(B)$,
we now obtain

$\tau(f(A)-f(B))=\tau(g(h(A))-g(h(B)))=\\\
=\int_{\mathbb{R}}g^{\prime}(s)\xi^{(3)}_{h(A),h(B)}(s)ds=\int_{\mathbb{R}}f^{\prime}(u)\xi^{(3)}_{h(A),h(B)}(h(u))du,$
(7)

where in the last step we substituted $s=h(u)$. This proves (2). Since the
left hand side does not depend on $h$ and since $f^{\prime}$ is an arbitrary
Schwartz function, the uniqueness follows. The local integrability of
$\xi^{(4)}_{A,B}$ follows from the integrability of $\xi^{(3)}_{h(A),h(B)}$
combined with the assumption that $h$ is a $C^{2}$-bijection.

If $A$ and $B$ are bounded, then by using the trace formula for the left hand
side of (7), we obtain

$\int_{\mathbb{R}}f^{\prime}(u)\xi^{(3)}_{A,B}(u)du=\tau\left(f(A)-f(B)\right)=\int_{\mathbb{R}}f^{\prime}(u)\xi^{(4)}_{A,B}(u)du.$

The latter implies that $\xi^{(3)}_{A,B}=\xi^{(4)}_{A,B}$ in this case. ∎

In what follows, $C_{b}(\mathbb{R})$ is the space of all bounded functions in
$C(\mathbb{R})$. The spaces $C_{b}^{1}(\mathbb{R})$ and
$C_{b}^{2}(\mathbb{R})$ are defined in an obvious manner.

######  Lemma 12.

Suppose that the assumptions of Theorem 7 hold. The equality (2) holds with
$\xi^{(4)}_{A,B}$ of Lemma 11, for every $f\in C^{1}_{b}(\mathbb{R})$ such
that $f^{\prime}\in S(\mathbb{R})$. Moreover, $\xi^{(4)}_{A,B}\in
L_{1}(\mathbb{R})$. If $A\geq B$, then we also have $\xi^{(4)}_{A,B}\geq 0$.

###### Proof.

Without loss of generality, $A\geq B$. We show that the function of Lemma 11
is positive and integrable.

We show the positivity first. Let $\alpha>0$ and let $a<b\in\mathbb{R}$.
Define the function $h_{a,b,\alpha}$ by setting $h_{a,b,\alpha}(-\infty)=0$
and

$h_{a,b,\alpha}^{\prime}(t)=\begin{cases}1,\quad t\in[a,b]\\\
\frac{\alpha^{3}}{(\alpha^{2}+(t-b)^{2})^{3/2}},\quad t>b\\\
\frac{\alpha^{3}}{(\alpha^{2}+(t-a)^{2})^{3/2}},\quad t<a\end{cases}$

The function $h_{a,b,\alpha}$ is a $C^{2}$-bijection $\mathbb{R}\mapsto(0,c)$
where $c=h_{a,b,\alpha}(+\infty)$. A simple computation shows that
$h_{a,b,\alpha}$ satisfies the condition (6) in Lemma 11. If $\xi^{(4)}_{A,B}$
is from Lemma 11, then

$\xi^{(4)}_{A,B}=\xi^{(3)}_{h_{a,b,\alpha}(A),h_{a,b,\alpha}(B)}\circ
h_{a,b,\alpha}.$

Combining Theorem 6 with the well known fact that $\tau(CD)\geq 0$ when $0\leq
C\in\mathcal{M}$ and $0\leq D\in\mathcal{L}_{1}$, we obtain the following
estimate

$\tau(h_{a,b,\alpha}(A)-h_{a,b,\alpha}(B))=\int_{0}^{1}\tau(h_{a,b,\alpha}^{\prime}((1-z)A+zB)(A-B))dz\geq
0.$

Furthermore, by Lemma 10 (applied to $h_{a,b,\alpha}(A)$ and
$h_{a,b,\alpha}(B)$ and $f(s)=s$) and substituting $s=h_{a,b,\alpha}(u)$, we
obtain

$\tau(h_{a,b,\alpha}(A)-h_{a,b,\alpha}(B))=\int_{0}^{c}\xi^{(3)}_{h_{a,b,\alpha}(A),h_{a,b,\alpha}(B)}(s)ds=\\\
=\int_{\mathbb{R}}\xi^{(3)}_{h_{a,b,\alpha}(A),h_{a,b,\alpha}(B)}(h_{a,b,\alpha}(u))h_{a,b,\alpha}^{\prime}(u)du=\int_{\mathbb{R}}\xi^{(4)}_{A,B}(u)h_{a,b,\alpha}^{\prime}(u)du.$

In particular, the function $\xi^{(4)}_{A,B}h^{\prime}_{a,b,\alpha}$ is
integrable. Since $h_{a,b,\alpha}^{\prime}\to\chi_{(a,b)}$ almost everywhere
as $\alpha\to 0$ and since $h_{a,b,\alpha}^{\prime}\leq h_{a,b,1}^{\prime}$
when $\alpha<1$, it follows from the Dominated Convergence Principle that

$\int_{a}^{b}\xi^{(4)}_{A,B}(u)du=\lim_{\alpha\to
0}\int_{\mathbb{R}}\xi^{(4)}_{A,B}(u)h_{a,b,\alpha}^{\prime}(u)du\geq 0.$

Since the latter inequality holds for arbitrary scalars $a,b$, it follows that
$\xi^{(4)}_{A,B}\geq 0$.

Now we show the integrability of $\xi^{(4)}_{A,B}$ on $\mathbb{R}$. Consider
the $C^{2}$-bijection $h_{\alpha}:\mathbb{R}\to(-1,1)$ given by

$h_{\alpha}(s)=\frac{s}{(\alpha^{2}+s^{2})^{\frac{1}{2}}},\quad
s\in\mathbb{R}.$

If $\xi^{(4)}_{A,B}$ is from Lemma 11, then

$\xi^{(4)}_{A,B}=\xi^{(3)}_{h_{\alpha}(A),h_{\alpha}(B)}\circ h_{\alpha}.$

By Lemma 10 (applied to the operators $h_{\alpha}(A)$ and $h_{\alpha}(B)$) we
have

$\tau(h_{\alpha}(A)-h_{\alpha}(B))=\int_{-1}^{1}\xi^{(3)}_{h_{\alpha}(A),h_{\alpha}(B)}(s)ds=\\\
=\int_{\mathbb{R}}\xi^{(3)}_{h_{\alpha}(A),h_{\alpha}(B)}(h_{\alpha}(u))h_{\alpha}^{\prime}(u)du=\int_{\mathbb{R}}\xi^{(4)}_{A,B}(u)h_{\alpha}^{\prime}(u)du.$

We have

$\left|\int_{-\infty}^{\infty}\xi^{(4)}_{A,B}(u)\cdot
h_{\alpha}^{\prime}(u)du\right|=\left|\tau(h_{\alpha}(A)-h_{\alpha}(B))\right|\leq\\\
\leq\left\|h_{\alpha}(A)-h_{\alpha}(B)\right\|_{1}\leq\frac{\mathrm{const}}{\alpha}\left\|A-B\right\|_{1},$

where the last estimate follows from Theorem 6, applied to the operators
$\alpha^{-1}A$ and $\alpha^{-1}B$ and function $h_{1}$ (note that
$h_{\alpha}(s)=h_{1}(\frac{s}{\alpha})$ for all $s$). Therefore, we have

$\alpha\int_{\mathbb{R}}h_{\alpha}^{\prime}(u)\xi_{A,B}^{(4)}(u)du=O(1).$

Since $\alpha h_{\alpha}^{\prime}\uparrow 1$ when $\alpha\to\infty$, we infer
from the Monotone Convergence Principle (which is applicable since
$\xi^{(4)}_{A,B}\geq 0$) that $\xi^{(4)}_{A,B}$ is integrable. ∎

Finally we give a simple extension of Lemma 12 to the class $W_{1}$. The
extension is based on our earlier observation in Lemma 5.

######  Lemma 13.

Suppose that the assumptions of Theorem 7 hold. The trace formula (2) holds
for every $f\in W_{1}$.

###### Proof.

Fix $f\in W_{1}$. By Lemma 5, there is a sequence $f_{n}\in
C_{b}^{1}(\mathbb{R})$ such that $f_{n}^{\prime}\in S(\mathbb{R})$ and such
that

$\lim_{n\rightarrow\infty}\left\|f-f_{n}\right\|_{W_{1}}=0.$

By Lemma 12, we have

$\tau(f_{n}(A)-f_{n}(B))=\int_{\mathbb{R}}f^{\prime}(s)\xi(s)ds.$ (8)

By Theorem 6, we have

$|\tau(f_{n}(A)-f_{n}(B))-\tau(f(A)-f(B))|\leq\|(f_{n}-f)(A)-(f_{n}-f)(B)\|_{1}\leq$
$\leq{\rm const}\cdot\|f_{n}-f\|_{W_{1}}\|A-B\|_{1}.$

Therefore,

$\lim_{n\to\infty}\tau(f_{n}(A)-f_{n}(B))=\tau(f(A)-f(B)).$ (9)

On the other hand, we have $\xi\in L_{1}$ and, therefore, $\mathcal{F}(\xi)\in
L_{\infty}$. We have $\mathcal{F}(f_{n}^{\prime})\to\mathcal{F}(f)$ in $L_{1}$
and, therefore,

$\mathcal{F}(f_{n}^{\prime})\mathcal{F}(\xi)\to\mathcal{F}(f)\mathcal{F}(\xi)$

in $L_{1}$. It follows that

$\int_{\mathbb{R}}f_{n}^{\prime}(s)\xi(s)ds=\int_{\mathbb{R}}(\mathcal{F}(f_{n}^{\prime}))(s)(\mathcal{F}(\xi))(s)ds\to\\\
\to\int_{\mathbb{R}}(\mathcal{F}(f^{\prime}))(s)(\mathcal{F}(\xi))(s)ds=\int_{\mathbb{R}}f^{\prime}(s)\xi(s)ds.$
(10)

Combining (8), (9) and (10), we conclude the proof. ∎

######  Remark 14.

We observe that Theorem 7 and the trace formula (2) can be further extended
from the class of functions $W_{1}$ to the homogeneous Besov class
$\tilde{B}^{1}_{\infty 1}$ (see e.g. [13]). In the case when
$\mathcal{M}={\mathcal{L}}(H)$, such an extension is performed in [13], in the
general case the argument is exactly the same as in [13]. We omit further
details.

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