# The spectral theorem for normal operators on a Clifford module

Fabrizio Colombo  (FC) Dipartmento di Matematica
Politecnico di Milano
Via E. Bonardi, 9
I-20133 Milano, Italy<EMAIL_ADDRESS>and  David P. Kimsey  (DPK)
School of Mathematics, Statistics and Physics
Newcastle University
Newcastle upon Tyne NE1 7RU UK<EMAIL_ADDRESS>
###### Abstract.

In this paper, using the recently discovered notion of the $S$-spectrum, we
prove the spectral theorem for a bounded or unbounded normal operator on a
Clifford module (i.e., a two-sided Hilbert module over a Clifford algebra
based on units that all square to be $-1$). Moreover, we establish the
existence of a Borel functional calculus for bounded or unbounded normal
operators on a Clifford module. Towards this end, we have developed many
results on functional analysis, operator theory, integration theory and
measure theory in a Clifford setting which may be of an independent interest.
Our spectral theory is the natural spectral theory for the Dirac operator on
manifolds in the non-self adjoint case. Moreover, our results provide a new
notion of spectral theory and a Borel functional calculus for a class of
$n$-tuples of commuting or non-commuting operators on a real or complex
Hilbert space. Moreover, for a special class of $n$-tuples of operators on a
Hilbert space our results provide a complementary functional calculus to the
functional calculus of J. L. Taylor.

AMS Classification 47A10, 47A60.

Keywords: Clifford spectral theorem, $S$-spectrum, Clifford measure theory,
Dirac operator on manifolds, continuous and Borel functional calculi, spectral
integrals.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Clifford algebras
    2. 2.2 Clifford modules
  3. 3 Linear operators on Clifford modules
    1. 3.1 $S$-resolvent set, $S$-spectrum for a linear operator and the spectral radius formula
    2. 3.2 Basic facts for normal operators
  4. 4 Measure theory and integration theory for $\mathbb{R}_{n}$-valued measures
  5. 5 Spectral integrals
    1. 5.1 Basic properties
    2. 5.2 Spectral integrals of bounded measurable functions
    3. 5.3 Spectral integrals of unbounded measurable functions
  6. 6 Spectral theorem for a bounded self-adjoint operator
  7. 7 Polar decomposition for bounded operators
  8. 8 An additive decomposition for bounded operators and imaginary operators
  9. 9 Spectral theorem for a bounded normal operator and some consequences
  10. 10 Spectral theorem for an unbounded normal operator and some consequences
  11. A Background on the $S$-spectrum and connections with functional calculi and function theory

##  1\. Introduction

The spectral theorem for a normal operator on a complex Hilbert space is an
incredibly elegant result which lies at the heart of operator theory, harmonic
analysis and mathematical physics. In this paper we endeavour to generalise
the spectral theorem to a noncommutative setting where a complex Hilbert space
is replaced by a Hilbert module over a Clifford algebra
$\mathcal{H}_{n}:=\mathcal{H}\otimes\mathbb{R}_{n}$, where $\mathcal{H}$ is a
real Hilbert space and $\mathbb{R}_{n}:=\mathbb{R}_{0,n}$ is a Clifford
algebra which is generated by units $e_{0}=1$ and $e_{1},\ldots,e_{n}$, where
$e_{i}^{2}=-1$ and $e_{i}e_{j}=-e_{j}e_{i}$ for $i,j=1,\ldots,n$ with $n>0$
(we will favour the term Clifford module for brevity) and the notion of
spectrum is replaced by the recently discovered notion of $S$-spectrum. We
wish to stress that given the well-known classification of Clifford algebras
$\mathbb{R}_{n}$ for $n=1,2,\ldots$ (see, e.g., [50]), our results can easily
be translated to handle spectral theory for a linear operator on any Hilbert
module over a finitely generated unital ring.

The $S$-spectrum can be characterised by the invertibility of a second order
operator and it is defined to be a subset of the set of paravectors in
$\mathbb{R}_{n}$ (where one can instil a natural complex structure
corresponding to any paravector). More precisely, corresponding to a right
linear operator $T$ on a Clifford module $\mathcal{H}_{n}$ over
$\mathbb{R}_{n}$, we define

(1.1)
$\sigma_{S}(T):=\\{s=\sum_{j=0}^{n}s_{j}e_{j}:(T^{2}-2s_{0}T+(s_{0}^{2}+\ldots+s_{n}^{2})I)^{-1}\notin\mathcal{B}(\mathcal{H}_{n})\\},$

where $\mathcal{B}(\mathcal{H}_{n})$ denotes the set of bounded linear
operators on $\mathcal{H}_{n}$. The $S$-spectrum was discovered by the first
author and I. Sabadini in the context of arbitrary operators on a quaternionic
Banach space (i.e., a Clifford module over $\mathbb{R}_{2}$) and paravector
operators $T$ on a Banach module over a Clifford algebra
$\mathscr{X}_{n}=\mathscr{X}\otimes\mathbb{R}_{n}$, where $\mathscr{X}$ is a
real Banach space and $\mathbb{R}_{n}$ is as above with $n>2$, i.e., right
linear operators of the form

$T=T_{0}+\sum_{j=1}^{n}e_{j}T_{j},$

where $T_{0},\ldots,T_{n}$ are linear operators on the real Banach space
$\mathscr{X}$ (see Appendix A for more details).

Recently we realised that we can dispense with the restriction that $T$ is a
paravector operator, i.e., the $S$-spectrum can be defined for an arbitrary
right linear operator on $\mathscr{X}_{n}$. This observation will turn out to
be critical for this manuscript. Moreover, the consequences of this
observation on the $S$-functional calculus and the slice hyperholomorphic
function theory have been investigated in [17] and [18], respectively.

It turns out that Clifford numbers of the form $s=\sum_{j=0}^{n}s_{j}e_{j}$
embed naturally into a complex plane and hence

$\sigma_{S}(T)\subseteq\bigcup_{\mathfrak{I}\,\in\,\mathbb{S}}\mathbb{C}_{\mathfrak{I}},$

where
$\mathbb{C}_{\mathfrak{I}}=\\{\lambda_{1}+\lambda_{2}\mathfrak{I}:\lambda_{1},\lambda_{2}\in\mathbb{R}\\}$
and $\mathbb{S}:=\\{s=\sum_{j=1}^{n}s_{j}e_{j}:s^{2}=-1\\}$. A moment’s
consideration of (1.1) will reveal that the upper half plane
$\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$ can identified in a natural
way with $\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{J}}^{+}$ for any choice of
$\mathfrak{I},\mathfrak{J}\in\mathbb{S}$. Thus, one may think of the
$S$-spectrum as a complex notion of spectrum with an elegant symmetry. For a
concise background of the history of the $S$-spectrum, related function theory
and known results in the quaternionic case see Appendix A.

We shall see that corresponding to every densely defined normal operator $T$
on a Clifford module $\mathcal{H}_{n}$, there exists an imaginary operator $J$
(i.e., $J^{*}J=I$ and $J^{*}=-J$) and a uniquely determined spectral measure
$E:=E_{\mathfrak{I}}$ such that

(1.2) $T=\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda)dE(\lambda)+\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda)dE(\lambda)\,J\quad\quad{\rm for}\quad\mathfrak{I}\in\mathbb{S}.$

We shall also see that

(1.3) ${\rm supp}\,\,E=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}.$

Moreover, we establish a full analogue of spectral integrals corresponding to
a spectral measure $E$, the Borel functional calculus associated with $T$ and
the spectral mapping theorem in the bounded and unbounded case. The fact that
$J$ is an imaginary operator on $\mathcal{H}_{n}$ allows one to think of $J$
as an operator-valued analogue of an imaginary unit. It is worth pointing out
that in the present setting the spectral measure $E$ gives rise to a family of
$\mathbb{R}_{n}$-valued measures which are positive as elements of the
Clifford algebra $\mathbb{R}_{n}$. This creates a significant technical
difficulty when building the requisite machinery to prove (1.2) and the Borel
functional calculus.

Motivation

We mention some of the main motivations to study the spectral theorem on a
Clifford module.

(I) Spectral theory for vector operators.

The spectral theory on the $S$-spectrum is a very natural tool for studying
vector operators that come from vector analysis such as the gradient operators
with nonconstant coefficients

(1.4) $T=\sum_{j=1}^{n}e_{j}a_{j}(x)\partial_{x_{j}},\ \ \
x=(x_{1},...,x_{n})\in\mathbb{R}^{n}$

where $e_{j}$ is an orthonormal basis in $\mathbb{R}^{n}$ and
$a_{j}:\Omega\to\mathbb{R}$, $j=1,...,n$ are given functions with suitable
regularity and $\Omega\subseteq\mathbb{R}^{n}$ is an open set. With our
spectral theory we can define, for example, the fractional powers of operators
of the form (1.4) which can be used to represent fractional Fourier’s law for
the propagation of the heat in nonhomogeneous materials contained in $\Omega$.
In the quaternionic setting, fractional Fourier’s law, the has been considered
in various papers, see for example [15, 19], and the references therein.

(II) Dirac operator on manifolds.

In a great preponderance of the papers on spectral theory for the Dirac
operators the self-adjoint case is considered. This is most likely the cause
due to the associated difficulties of defining in an appropriate way the
spectrum of the non self-adjoint Dirac operator. However, the Dirac operator
is just a particular case of a Clifford operator, and on manifolds, it has in
general non-constant coefficients, so its natural notion of spectrum is the
$S$-spectrum. The $S$-spectrum of a self-adjoint Clifford operator (in
particular the Dirac operator) is real. Let us explain with more details the
above motivation. Assume that $g:U\to\mathbb{R}^{n\times n}$, given by

$g(x)=(g_{ij}(x))_{i,j=1}^{n},$

is a smooth matrix-valued function defined on the open set $U$ in
$\mathbb{R}^{n}$ where $g(x)$ will always be taken to be positive-definite and
symmetric. Then

(1.5) $g_{x}(\xi,\eta)=\sum_{i,j=1}^{n}g_{ij}(x)\xi_{i}\eta_{j},\ \ \xi,\
\eta\in\mathbb{R}^{n}$

is a positive-definite inner product on $\mathbb{R}^{n}$ and

(1.6) $g_{x}(X,Y)=\sum_{i,j=1}^{n}g_{ij}(x)a_{i}b_{j},\ \ {\rm where}\ \
X=\sum_{i=1}^{n}a_{i}(x)\partial_{x_{i}},\ \ \
Y=\sum_{j=1}^{n}b_{j}(x)\partial_{x_{j}},$

defines a positive-definite inner product space on the tangent space
$\mathcal{T}_{x}(U)$ to $U$ at $x$. So $U$ can be seen as a coordinate
neighbourhood for the Riemannian manifold $M$ taking $x=(x_{1},...,x_{n})$ as
coordinates and (1.6) as inner product, making it possible to study the Dirac
operator on $M$, by introducing it as a nonconstant coefficients
nonhomogeneous first-order systems of differential operators on
$\mathcal{C}^{\infty}(U,\mathcal{H})$. To give a precise definition of the
spectrum of the Dirac operator on $M$, (i..e, the $S$-spectrum), we need to
perform the following steps.

(a) We need to give the definition of the Dirac operator $D$ on $M$.

(b) Next, we need to write $D^{2}$ explicitly in terms of a second-order
Laplacian and a curvature operator, via the Bochner-Weitzenböck theorem.

The precise expression of the operator $D^{2}$ is of crucial importance in
order to define the $S$-spectrum because it is associated with the operator

$\mathcal{Q}_{s}(D):=D^{2}-2s_{0}D+|s|^{2}\,I,$

here $s$ is a paravector in the Clifford algebra $\mathbb{R}_{n}$. Before we
can define the Dirac operator, we require some additional notions. Let
$g:\mathbb{R}^{n\times n}\to U$ be an invertible function on
$\mathbb{R}^{n\times n}$ and write

(1.7) $g^{-1}:U\to\mathbb{R}^{n\times n},\ \
g^{-1}(x)=(g^{ij}(x))_{i,j=1}^{n}.$

Next, let

$\gamma(x)=(\gamma_{ij}(x))_{i,j=1}^{n},\ \
\gamma^{-1}(x)=(\gamma^{ij}(x))_{i,j=1}^{n}:U\to\mathbb{R}^{n\times n},$

be the unique square roots of $g$ and $g^{-1}$, respectively. Let
$e_{1},...,e_{n}$ be the skew-adjoint operators satisfying the Clifford
relations $e_{j}e_{k}+e_{k}e_{j}=-2\delta_{jk}$, set

$e_{i}(x)=\sum_{j=1}^{n}\gamma^{ij}(x)e_{j},\ \ \ x\in U,$

for every $i=1,...,n.$ By definition we have that

(1.8) $e_{j}(x)e_{k}(x)+e_{k}(x)e_{j}(x)=-2g^{jk}(x),\ \ \ x\in U.$

Let $d\tau:{\rm spin}(n)\to\mathcal{U}(\mathcal{H})$ be the representation of
the Lie algebra of ${\rm spin}(n)$ derived from the representation $\tau:{\rm
spin}(n)\to\mathcal{U}(\mathcal{H})$. For more details see [39].

As a differential operator on $\mathcal{C}^{\infty}(U,\mathcal{H})$ the
standard Dirac operator $D$ on $(U,g)$ is defined by

$D=\sum_{i=1}^{n}e_{i}(x)\Big{(}\partial_{x_{i}}+d\tau(\omega_{i}(x))\Big{)}$

where $\omega_{1},...,\omega_{n}:U\to{\rm spin}(n)$ are smooth functions
uniquely determined by $g$.

Some more work is necessary to define $D^{2}$, but clearly the spectral theory
on the $S$-spectrum allows one to consider non-self adjoint Dirac operators.
We recall that the scalar curvature of $(U,g)$ is defined by

(1.9) $\kappa(x)=-\sum_{i,j=1}^{n}R_{ijij}(x)$

where $R_{ijk\ell}(x)$ is the Riemann curvature tensor. As an example we
recall Spinor Laplacian, that is we assume $\mathcal{H}=\Theta_{n}$ and $\tau$
is the Spin representation of $S$ of ${\rm spin}(n)$ on $\Theta_{n}$, $n=2m$,
(see [39]) the second order operator $D^{2}$ is often called spinor Laplacian
is given by the Lichnerowicz’s formula

$D^{2}=-\Delta_{S}+\frac{1}{4}\kappa(x).$

So in similar cases the $S$-spectrum can be written explicitly. Several
results, such as [11, 32, 47, 48, 52] and also [31, 43, 49, 56, 60, 61] in
spectral theory for the Dirac operators can now be seen in a new light using
the spectral theory on the $S$-spectrum for the Dirac operator, also using the
$S$-functional calculus [3, 7, 24, 35] and the function theory [2, 21, 22, 23]
on which this calculus is based on for $n$-tuples of operators.

(III) Complementarity to the Taylor spectrum for a class of tuples of
commuting operators or non-commuting operators and harmonic analysis.

Given an $n$-tuple of bounded or unbounded operators $(T_{1},…,T_{n})$ on a
real Hilbert space $\mathcal{H}$, we can form the right linear operator

$T=\sum_{j=1}^{n}e_{\alpha_{j}}T_{j},$

where
$\alpha_{1},\ldots,\alpha_{n}\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})$.
Now if there exists a configuration of units such that $T$ is normal, then one
may define the spectrum of $(T_{1},\ldots,T_{n})$ to be the $S$-spectrum of
$T$, i.e., $\sigma_{S}(T)$. Moreover, one has a Borel functional calculus at
hand for a reasonably large class of functions of $T$, which is helpful in
problems in harmonic analysis and partial differential equations. It is worth
mentioning that one can find the relations between the monogenic functional
calculus, Taylor functional calculus and the Weyl functional calculus (see,
[45, 46, 53, 54, 63]). In harmonic analysis in higher dimensions, singular
integrals and in the study of the Fourier transform one can find various
connections with Clifford analysis in the recent book [64]. Boundary value
problems treated with quaternionic techniques can be found in [41]. Clifford
wavelets, singular integrals, and Hardy spaces are studied in [55].

(IV) Spectral theory for linear operators on a Hilbert module over a finitely
generated unital ring.

We first note that we may use the classification of Clifford modules (see,
e.g., [50]) and the fact that every finitely generated unital ring is
isomorphic to $\mathbb{R}^{m\times m}$ to see that there is a natural
isomorphism between any Hilbert module over a finitely generated unital ring
$R$ and a Clifford module $\mathcal{H}_{n}$ for an appropriate choice of $n$.
Thus, given a normal right linear operator $T$ on a Hilbert module over a
finitely generated unital ring $R$, we may view $T$ as a being a right linear
operator on a Clifford module $\mathcal{H}_{n}$ for a suitable choice of $n$.
We can then utilise the spectral theory and functional calculi in the Clifford
setting and translate back to the Hilbert module setting in a straight forward
manner.

For the function theory of slice hyperholomorphic functions the main books are
[5, 25, 26, 27, 33, 36], while for the spectral theory on the $S$-spectrum we
mention [6, 12, 14, 16, 27]. For the Fueter and monogenic function theory and
related topics see the books [10, 20, 29, 39, 40, 44, 65].

Strategy

We wish to summarise our strategy for proving the spectral theorem for a
normal operator on a Clifford module.

(I) It turns out that the $S$-spectrum and $S$-resolvent set can be defined
for all bounded or unbounded operators (not necessarily paravector operators).
This is the first crucial intuition for the decomposition of normal operators
and we show that the $S$-spectrum of a bounded operator is a non-empty compact
subset of $\\{s\in\mathbb{R}_{n}:s=s_{0}+\sum_{j=1}^{n}s_{j}e_{j}\\}$.

(II) We define a spectral measure on a Clifford module and spectral integrals
$\mathbb{I}(f)$, where $f$ belongs to a suitable class of functions (see
Section 4).

Moreover, we point out that an absolutely key result is Theorem 5.3.14 which
connects $\sigma_{S}(\mathbb{I}(f))$ to ${\rm supp}\,E$. Dealing with the
$S$-spectrum requires to overcome substantially different difficulties with
respect to the classical complex or quaternionic Hilbert space case (where the
spectral measure $E$ gives rise to a family of positive measures in the usual
sense).

(III) We prove a spectral theorem for a bounded self-adjoint operator (see
Theorem 6.0.1) where the important result given in Theorem 5.3.14 is being
used to show that the spectral measure $E$ has the property that ${\rm
supp}\,E=\sigma_{S}(T)$.

(IV) We prove a polar decomposition for a bounded Clifford operator $T$ and
specialise the result to the case when $T$ is normal (see Theorem 7.0.1). This
then enables one to prove that every bounded normal operator can be written as
$T=A+BJ$, where $A=A^{*}\in\mathcal{B}(\mathcal{H}_{n})$,
$B\in\mathcal{B}(\mathcal{H}_{n})$ is positive, $JJ^{*}=I$ and $J^{*}=-J$ and
$A$, $B$ and $J$ all mutually commute (see Theorem 8.0.4).

(V) To prove the spectral theorem for bounded normal operators (see Theorem
9.0.2), one needs to apply a technical result (see Theorem 5.3.17) to
manufacture a uniquely determined spectral measure $E$ which lives on
$\sigma_{S}(A)\times\sigma_{S}(B)$ which will be identified with the complex
plane $\mathbb{C}_{\mathfrak{I}}$, where $\mathfrak{I}\in\mathbb{S}$.

One applies the key result Theorem 5.3.14 with the identity function
$f(\lambda)=\lambda$ to see that
$\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}={\rm
supp}\,E\cap\mathbb{C}_{\mathfrak{I}}$. One can get a suitable integral
representation for $T$ which resembles the quaternionic case.

(VI) We use the bounded case of the spectral theorem (see Theorem 9.0.2) to
prove the unbounded case (see Theorem 10.0.4).

For the convenience of the reader we have compiled a list of commonly used
notation that appears throughout this manuscript.

Notation

$\mathbb{R}_{n}:=\mathbb{R}_{0,n}$ will denote the Cifford algebra generated
by the units $e_{0}:=1$ and $e_{1},\ldots,e_{n}$, where
$e_{i}e_{j}=-e_{j}e_{i}$ for $i\neq j$. An element $a\in\mathbb{R}_{n}$ can be
written as

$a=\sum_{\alpha}a_{\alpha}e_{\alpha}:=\sum_{\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})}a_{\alpha}e_{\alpha},$

where $\emptyset$ can be identified with $0$ and $\alpha\neq\emptyset$ can be
interpreted as a $n$-tuple $(i_{1},\ldots,i_{n})$ with $i_{1}<\ldots<i_{n})$.

$\mathbb{S}:=\\{\mathfrak{I}=(0,\mathfrak{I}_{1},\ldots,\mathfrak{I}_{n})\in\mathbb{R}^{n+1}:\sum_{i=1}^{n}\mathfrak{I}_{i}=1\\}$.
A typical element of $\mathbb{S}$ will be denoted by $\mathfrak{I}$ or
$\mathfrak{J}$.

For any $\mathfrak{I}\in\mathbb{S}$, we shall let
$\mathbb{C}_{\mathfrak{I}}=\\{\lambda_{0}+\lambda_{1}\,\mathfrak{I}:\lambda_{0},\lambda_{1}\in\mathbb{R}\\}$
and
$\mathbb{C}^{+}_{\mathfrak{I}}=\\{\lambda_{0}+\lambda_{1}\,\mathfrak{I}:\lambda_{0}\in\mathbb{R},\lambda_{1}\geq
0\\}$.

$\mathfrak{S}(\mathbb{R}_{n})$ will denote the set of self-adjoint Clifford
numbers in $\mathbb{R}_{n}$, i.e., all $a\in\mathbb{R}_{n}$ such that
$\bar{a}=a$.

$\mathfrak{P}(\mathbb{R}_{n})$ will denote the set of positive semidefinite
Clifford numbers in $\mathbb{R}_{n}$, i.e., all $a\in\mathbb{R}_{n}$ such that
$a=b\bar{b}$ for some $b\in\mathbb{R}_{n}$. In this case, we will write
$a\succeq 0$.

$\chi:\mathbb{R}_{n}\to\mathbb{R}^{2^{n}\times 2^{n}}$ will denote the
injective $*$-homomorphism from the Clifford algebra $\mathbb{R}_{n}$ to the
set of real matrices of size $2^{n}$ (see Definition 2.1.7).

$\mathcal{H}_{n}=\mathcal{H}\otimes\mathbb{R}_{n}$ will denote a two-sided
Clifford module over $\mathbb{R}_{n}$, where $\mathcal{H}$ is a real Hilbert
space (see Definition 2.1.6).

$\langle\cdot,\cdot\rangle:\mathcal{H}_{n}\times\mathcal{H}_{n}\to\mathbb{R}_{n}$
given by (2.9) and $\|x\|:=\sqrt{{\rm Re}\,\langle x,x\rangle}$ for
$x\in\mathcal{H}_{n}$.

For any closed submodule $Y\subseteq\mathcal{H}_{n}$, we shall let

$Y^{\perp}:=\\{y\in\mathcal{H}_{n}:\text{$\langle x,y\rangle=0$ for all
$x\in\mathcal{H}_{n}$}\\}.$

$\mathcal{L}(\mathcal{H}_{n})$ will denote the set of right linear operators
on $\mathcal{H}_{n}$. The domain of an operator
$T\in\mathcal{L}(\mathcal{H}_{n})$ will be denoted by $\mathcal{D}(T)$ (see
Definition 3.0.1).

${\rm Ran}\,T$ and ${\rm Ker}\,T$ will denote the range and kernel of
$T\in\mathcal{L}(\mathcal{H}_{n})$, respectively (see Definition 3.0.3).

$\mathcal{B}(\mathcal{H}_{n})$ will denote the set of bounded linear operators
on $\mathcal{H}_{n}$ (see Definition 3.0.4. We shall let

$\|T\|:=\sup_{\|x\|\leq 1}\|Tx\|=\lim_{\|x\|=1}\|Tx\|.$

$\mathcal{G}(T)$ will denote the graph of a linear operator (see Definition
3.0.6).

For $T\in\mathcal{L}(\mathcal{H}_{n})$, we shall let $\langle
x,y\rangle_{T}:=\langle x,y\rangle+\langle Tx,Ty\rangle$ and
$\|x\|_{T}:=(\|x\|+\|Tx\|^{2})^{1/2}$ for $x\in\mathcal{D}(T)$.

For $S,T\in\mathcal{L}(\mathcal{H}_{n})$, we will write $S\subseteq T$ if
$\mathcal{D}(S)\subseteq\mathcal{D}(T)$ and $Sx=Tx$ for $x\in\mathcal{D}(S)$.

$|T|:=(T^{*}T)^{1/2}$ (see Theorem 7.0.1).

For any closable operator $T\in\mathcal{L}(\mathcal{H}_{n})$, we shall let
$\overline{T}$ denote the closure of $T$ (see Definition 3.0.14).

We shall let $\mathcal{Q}_{s}(T):=T^{2}-2{\rm Re}(s)T+|s|^{2}I$.

$\rho_{S}(T)$ denotes the $S$-resolvent set of $T$ and $\sigma_{S}(T)$ denotes
the $S$-spectrum of $T$ (see Definition 3.1.1).

We shall let $S_{L}^{-1}(s,T)$ and $S_{R}^{-1}(s,T)$ denote the left
$S$-resolvent operator and right $S$-resolvent operator of $T$ at $s$,
respectively (see Definition 3.1.4).

For $T\in\mathcal{B}(\mathcal{H}_{n})$, we shall let $r_{S}(T)$ denote the
spectral radius of $T$ with respect to the $S$-spectrum (see (3.22)).

For $T\in\mathcal{B}(\mathcal{H}_{n})$, we shall let $\sigma_{PS}(T)$,
$\sigma_{RS}(T)$ and $\sigma_{CS}(T)$ denote the point $S$-spectrum of $T$,
residual $S$-spectrum of $T$ and continuous $S$-spectrum of $T$, respectively
(see Definition 3.1.11).

For $T\in\mathcal{B}(\mathcal{H}_{n})$, we shall let $\Pi_{S}(T)$ and
$\Gamma_{S}(T)$ denote the approximate point $S$-spectrum of $T$ and
compression $S$-spectrum of $T$, respectively (see Definition 3.1.11).

$T\in\mathcal{L}(\mathcal{H}_{n})$ will be called normal if $T$ is densely
defined, closed, $\mathcal{D}(T)=\mathcal{D}(T^{*})$ and $TT^{*}=T^{*}T$.

$T\in\mathcal{L}(\mathcal{H}_{n})$ will be called self-adjoint if
$\mathcal{D}(T)=\mathcal{D}(T^{*})$ and $Tx=T^{*}x$ for all
$x\in\mathcal{D}(T)$.

$T\in\mathcal{L}(\mathcal{H}_{n})$ will be called positive if $T$ is
$\mathcal{D}(T)=\mathcal{D}(T^{*})$ and $\langle Tx,x\rangle\succeq 0$ for all
$x\in\mathcal{D}(T)$, i.e., $\langle
Tx,x\rangle\in\mathfrak{P}(\mathbb{R}_{n})$ for all $x\in\mathcal{D}(T)$.

$T\in\mathcal{L}(\mathcal{H}_{n})$ will called anti self-adjoint if
$\mathcal{D}(T)=\mathcal{D}(T^{*})$ and $T=-T^{*}$ for $x\in\mathcal{D}(T)$.

$T\in\mathcal{B}(\mathcal{H}_{n})$ will be called unitary if $TT^{*}=I$.

$J\in\mathcal{B}(\mathcal{H}_{n})$ will be called imaginary if $J$ is anti
self-adjoint and $J$ is unitary.

$\mathcal{P}(\mathcal{H}_{n})$ will denote the set of orthogonal projections
on $\mathcal{H}_{n}$.

$s-\lim_{i\to\infty}T_{i}$ will denote the limit of a sequence of operators
$(T_{i})_{i=1}^{\infty}$, where $T_{i}\in\mathcal{B}(\mathcal{H}_{n})$ for
$i=1,2,\ldots$, in the strong operator topology.

$E$ will denote a spectral measure on $(\Omega,\mathscr{A})$, where
$\mathscr{A}$ is a $\sigma$-algebra of sets generated by $\Omega$.

The set of Borel sets generated by $\Omega$ will be denoted by
$\mathscr{B}(\Omega)$.

Given a spectral measure $E$ on $(\Omega,\mathscr{B}(\Omega))$, ${\rm
supp}\,E$ will denote the support of $E$ (see Definition 5.1.9).

Given an $\mathbb{R}_{n}$-valued measure $\nu$ the total variation of $\nu$
will be denoted by $|\nu|$ (see (5.5)).

Given an imaginary operator $J\in\mathcal{B}(\mathcal{H}_{n})$, we will say
that $J$ is associated with a spectral measure $E$ on $(\Omega,\mathscr{A})$
if $E(M)J=JE(M)$ for all $M\in\mathscr{A}$ (see Definition 5.2.1).

$\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ will denote the
Banach space of all bounded $\mathscr{A}$-measurable functions
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}$ equipped with the norm

$\|f\|_{\infty}=\sup_{\lambda\in\Omega}|f(\lambda)|.$

$\mathfrak{B}_{s}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ will denote
the subspace of $\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$
of all simple functions.

Given $f\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ and a
spectral measure $E$ on $(\Omega,\mathscr{A})$, we shall let $\mathbb{I}(f)$
denote the spectral measure of $f$ with respect to $E$ (see Definition 5.2.3).

Given a spectral measure $E$ on $(\Omega,\mathscr{A})$, we shall let
$\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$ denote the set
of all $\mathscr{A}$-measurable functions
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}\cup\\{\infty\\}$ which are $E$-a.e.
finite, i.e., $E(\\{\lambda\in\Omega:f(\lambda)=\infty\\})=0$.

Given $f\in\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$ and a
spectral measure $E$ on $(\Omega,\mathscr{A})$, we shall let $\mathbb{I}(f)$
denote the spectral measure of $f$ with respect to $E$ (see Theorem
5.3.4(ii)).

Given an imaginary operator $J\in\mathcal{B}(\mathcal{H}_{n})$ and
$\mathfrak{I}\in\mathbb{S}$,
$\mathcal{H}_{\pm}(J,\mathfrak{I})=\\{x\in\mathcal{H}_{n}:Jx=x(\pm\mathfrak{I})\\}$
is a complex subspace of $\mathcal{H}_{n}$ (see (8.5)).

Given a densely defined operator $T\in\mathcal{L}(\mathcal{H}_{n})$, we shall
let $C_{T}:=(I+T^{*}T)^{-1}$ and $Z_{T}:=TC_{T}^{1/2}$ (see Definition
10.0.1).

##  2\. Preliminaries

In this section we will formulate a number of definitions and results on
Clifford algebras, Clifford modules, linear operators on Clifford modules, the
$S$-spectrum and measure theory and integration theory with respect to a
Clifford algebra-valued measure. While the topic of Clifford algebras is very
classical and well-known, the theory of linear operators on Clifford modules
and measure theory and integration theory with respect to a Clifford algebra-
valued measure are not so well developed. We have furnished proofs for
Clifford algebra/module analogues of results whenever the proof differs from
the classical case with the aim of making the present manuscript as self
contained as possible.

###  2.1. Clifford algebras

######  Definition 2.1.1.

A collection of $n$ elements $e_{1},\ldots,e_{n}$, with $n=p+q$ and
$p,q\in\mathbb{N}_{0}$ will be called imaginary units if

$\displaystyle e_{i}^{2}=$ $\displaystyle\;+1\quad\quad{\rm for}\quad
i=1,\ldots,p$ $\displaystyle e_{j}^{2}=$ $\displaystyle\;-1\quad\quad{\rm
for}\quad j=p+1,\ldots,n$ $\displaystyle e_{j}e_{k}+e_{k}e_{j}=$
$\displaystyle\;0\quad\quad{\rm for}\quad j\neq k$ (2.1) $\displaystyle
e_{1}\ldots e_{n}\neq$ $\displaystyle\;\pm 1.$

We shall denote the real algebra generated by the imaginary units
$e_{1},\ldots,e_{n}$ the universal Clifford algebra and denote it by
$\mathbb{R}_{p,q}$. An element of $\mathbb{R}_{p,q}$ is called a Clifford
number.

######  Remark 2.1.2.

We note that it is only necessary to assume (2.1) if $p-q\equiv 1({\rm
mod}\;4)$. It is easy to check that
$\\{e_{\alpha}\\}_{\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})}$
is linearly independent and hence $\mathbb{R}_{p,q}$ has dimension $2^{n}$ (as
a real vector space).

Let $\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})$ and if
$\alpha\neq\emptyset$, then we may write $\alpha=\\{i_{1},\ldots,i_{r}\\}$,
with $i_{1}<\ldots<i_{r}$. Then we may let

$\displaystyle e_{\alpha}:=$
$\displaystyle\;e_{i_{1},\ldots,i_{r}}:=e_{i_{1}}\ldots
e_{i_{r}}\quad\quad{\rm if}\quad\alpha=\\{i_{1},\ldots,i_{r}\\},$
$\displaystyle e_{0}:=$ $\displaystyle\;e_{\emptyset}=1\quad\quad{\rm
if}\quad\alpha=\emptyset$

and write an arbitrary Clifford number $a\in\mathbb{R}_{p,q}$ as

$a=\sum_{\alpha}a_{\alpha}e_{\alpha}:=\sum_{\alpha\,\in\,\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})}a_{\alpha}e_{\alpha},$

where the sum is taken over $\emptyset$ and subsets $\\{i_{1},\ldots,i_{r}\\}$
with $i_{1}<\ldots<i_{r}$.

######  Remark 2.1.3.

The only Clifford algebra considered in the remainder of this paper will be
$\mathbb{R}_{n}:=\mathbb{R}_{0,n}$.

We will let

$\bar{a}:=\sum_{\alpha}a_{\alpha}\bar{e}_{\alpha}=\sum_{\alpha}a_{\alpha}(-e_{\alpha})$

and

$|a|:=\left(\sum_{\alpha}a_{\alpha}^{2}\right)^{1/2}\quad\quad{\rm for}\quad
a=\sum_{\alpha}a_{\alpha}e_{\alpha}\in\mathbb{R}_{n}.$

Hence one can easily check that for all $a,b\in\mathbb{R}_{n}$, we have

$\displaystyle\overline{ab}=$ $\displaystyle\;\bar{b}\,\bar{a}$
$\displaystyle\overline{a+b}=$ $\displaystyle\;\bar{a}+\bar{b}$ (2.2)
$\displaystyle|a+b|\leq$ $\displaystyle\;|a|+|b|$ (2.3) $\displaystyle|ab|=$
$\displaystyle\;|a|\,|b|\quad\quad{\rm whenever}\quad b\bar{b}=|b|^{2}$ (2.4)
$\displaystyle|ab|\leq$ $\displaystyle\;2^{n-1}|a|\,|b|.$

######  Definition 2.1.4.

Given $s=(s_{0},s_{1},\ldots,s_{n})\in\mathbb{R}^{n+1}$, we may identify the
vector $s\in\mathbb{R}^{n+1}$ with the paravector
$\mathfrak{s}=\sum_{i=0}^{n}s_{i}e_{i}\in\mathbb{R}_{n}$. With a slight abuse
of notation, for the remainder of this paper, we will use $s$ in place of
$\mathfrak{s}$.

Let $\mathbb{S}$ denote the unit sphere of vectors in $\mathbb{R}^{n+1}$,
i.e.,

$\mathbb{S}:=\\{\mathfrak{I}=(0,\mathfrak{I}_{1},\ldots,\mathfrak{I}_{n})\in\mathbb{R}^{n+1}:\sum_{i=1}^{n}\mathfrak{I}^{2}_{i}=1\\}.$

It is easy to see that $\mathbb{S}$ is an $(n-1)$-sphere in $\mathbb{R}^{n}$
and $\mathfrak{I}\in\mathbb{S}$ implies that $\mathfrak{I}^{2}=-1$. Note that
the real two-dimensional subspace of $\mathbb{R}^{n+1}$ generated by $1$ and
$\mathfrak{I}$ is complex plane
$\mathbb{C}_{\mathfrak{I}}:=\mathbb{R}+\mathbb{R}\,\mathfrak{I}$. It is not
hard to see that $\mathbb{C}_{\mathfrak{I}}$ is isomorphic to the usual
complex plane $\mathbb{C}$.

Given a paravector $s=\sum_{i=0}^{n}s_{i}e_{i}\in\mathbb{R}^{n+1}$, it is
possible to find $\mathfrak{I}_{s}\in\mathbb{S}$ such that
$s\in\mathbb{C}_{\mathfrak{I}_{s}}$. Indeed, if $s\neq 0$, then we may write

$s=s_{0}+\left(\frac{\sum_{i=1}^{n}s_{i}e_{i}}{|s|}\right)|s|.$

Thus, if we let $\mathfrak{I}_{s}:=\frac{\sum_{i=1}^{n}s_{i}e_{i}}{|s|}$, then
one can check that $\mathfrak{I}_{s}\in\mathbb{S}$ and hence
$s\in\mathbb{C}_{\mathfrak{I}_{s}}$.

######  Definition 2.1.5 (self-adjoint Clifford numbers).

We will call $a\in\mathbb{R}_{n}$ self-adjoint if $a=\bar{a}$. The set of all
self-adjoint Clifford numbers in $\mathbb{R}_{n}$ shall be denoted by
$\mathfrak{S}(\mathbb{R}_{n})$. We note that $\mathfrak{S}(\mathbb{R}_{n})$ is
a real vector space.

######  Definition 2.1.6 (positive semidefinite Clifford numbers).

We will call $a\in\mathbb{R}_{n}$ positive semidefinite if there exists
$b\in\mathbb{R}_{n}$ such that $a=b\bar{b}$. In this case, we shall write
$a\succeq 0$. The set of all positive semidefinite Clifford numbers in
$\mathbb{R}_{n}$ shall be denoted by $\mathfrak{P}(\mathbb{R}_{n})$.

We will need to recall a well-known injective $*$-homomorphism
$\chi:\mathbb{R}_{n}\to\mathbb{R}^{2^{n}\times 2^{n}}$ which can be found,
e.g., in [55].

######  Definition 2.1.7.

Let $\chi:\mathbb{R}_{n}\to\mathbb{R}^{2^{n}\times 2^{n}}$ be the injective
$*$-homomorphism given by the following inductive construction. We will first
give meaning to $\chi(e_{j})$ for $j=0,1,\ldots,n$. Let
$\chi(e_{0})=I_{2^{n}}$ and $\chi(e_{j}):=E_{j}^{n}$ for $j=1,\ldots,n$, where
$\\{E_{j}^{k}\\}_{j=1}^{k}$ are inductively defined via

$E_{1}^{1}:=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix},\quad
E_{j}^{k+1}:=\begin{pmatrix}E_{j}^{k}&0\\\ 0&-E_{j}^{k}\end{pmatrix}\quad{\rm
and}\quad E_{k+1}^{k+1}=\begin{pmatrix}0&-I_{2^{k}}\\\
I_{2^{k}}&0\end{pmatrix}$

for $j=1,\ldots,k$ and $k=1,\ldots,n-1$. Next, we let

$\chi(e_{\alpha}):=\chi(e_{i_{1}})\cdots\chi(e_{i_{k}})\quad{\rm
for}\quad\alpha=\\{i_{1},\ldots,i_{k}\\},$

where $i_{1}<\ldots<i_{k}$. Finally for
$a=\sum_{\alpha}a_{\alpha}e_{\alpha}\in\mathbb{R}_{n}$, we let

$\chi(a):=\sum_{\alpha}a_{\alpha}\chi(e_{\alpha}).$

######  Remark 2.1.8.

Let $a\in\mathbb{R}_{n}$. It is easy to check that $\chi(a)$ is a positive
semidefinite matrix in $\mathbb{R}^{2^{n}\times 2^{n}}$ if and only if $a$ is
a positive semidefinite Clifford number.

###  2.2. Clifford modules

######  Definition 2.2.1 (Clifford modules over $\mathbb{R}_{n}$).

Let $\mathcal{H}$ be a real Hilbert space with an inner product
$\langle\cdot,\cdot\rangle_{\mathcal{H}}$ and a natural norm
$\|x\|_{\mathcal{H}}:=\langle x,x\rangle_{\mathcal{H}}^{1/2}$ for
$x\in\mathcal{H}$. Then by $\mathcal{H}_{n}$ we mean the two-sided Clifford
module generated by $\mathcal{H}$ and $\mathbb{R}_{n}$ with $n>0$, i.e.,
$\mathcal{H}_{n}$ consists of all vectors of the form

(2.5) $x=\sum_{\alpha}x_{\alpha}\otimes e_{\alpha},$

with

(2.6) $\displaystyle x+y:=$
$\displaystyle\;\sum_{\alpha}(x_{\alpha}+y_{\alpha})\otimes e_{\alpha},$ (2.7)
$\displaystyle xa:=$
$\displaystyle\;\sum_{\alpha}x_{\alpha}\otimes(e_{\alpha}a)=\sum_{\alpha,\beta}(x_{\alpha}a_{\beta})\otimes(e_{\alpha}e_{\beta})$
(2.8) $\displaystyle ax:=$
$\displaystyle\;\sum_{\alpha}x_{\alpha}\otimes(ae_{\alpha})=\sum_{\alpha,\beta}(x_{\alpha}a_{\beta})\otimes(e_{\beta}e_{\alpha})$

for $y=\sum_{\alpha}y_{\alpha}\otimes e_{\alpha}$ and
$a=\sum_{\alpha}a_{\alpha}e_{\alpha}$. We will employ the short-hand
$x=\sum_{\alpha}x_{\alpha}e_{\alpha}$ in place of
$x=\sum_{\alpha}x_{\alpha}\otimes e_{\alpha}$. We may then endow a Clifford
module $\mathcal{H}_{n}$ with the $\mathbb{R}_{n}$-valued “inner product”
$\langle\cdot,\cdot\rangle$ given by

(2.9) $\langle x,y\rangle:=\sum_{\alpha,\beta}e_{\alpha}\bar{e}_{\beta}\langle
x_{\alpha},y_{\beta}\rangle_{\mathcal{H}}\quad\quad{\rm for}\quad
x,y\in\mathcal{H}_{n}.$

######  Remark 2.2.2.

For the great preponderance of this work, we will not need the fact that
$\mathcal{H}_{n}$ is a two-sided module. Thus, whenever possible we will make
use of the fact that $\mathcal{H}_{n}$ is a right module. However, for Theorem
3.1.10 we will need to use the two-sided which will be utilised when proving
the spectral theorem for a bounded self-adjoint operator in Section 6.

######  Remark 2.2.3.

It is easy to check that the following facts hold:

(2.10) $\displaystyle\langle x,x\rangle\succeq$ $\displaystyle\;0\quad{\rm
and}\quad\langle x,x\rangle=0\Longleftrightarrow x=0$ (2.11)
$\displaystyle\langle x+y,z\rangle=$ $\displaystyle\;\langle
x,y\rangle+\langle y,z\rangle,$ (2.12) $\displaystyle\overline{\langle
x,y\rangle}=$ $\displaystyle\;\langle y,x\rangle$ (2.13) $\displaystyle\langle
xa,y\rangle=$ $\displaystyle\;\langle x,y\rangle a\quad\quad{\rm for}\quad
x,y,z\in\mathcal{H}_{n}\quad{\rm and}\quad a\in\mathbb{R}_{n}.$

Furthermore, if we define

(2.14)
$\|x\|:=\left(\sum_{\alpha}\|x_{\alpha}\|^{2}_{\mathcal{H}}\right)^{1/2}\quad\quad{\rm
for}\quad x=\sum_{\alpha}x_{\alpha}e_{\alpha},$

then $\|\cdot\|$ is a “norm” on $\mathcal{H}_{n}$, i.e., for all
$x,y\in\mathcal{H}_{n}$ and $a\in\mathbb{R}_{n}$, we have

(2.15) $\displaystyle\|x\|>$ $\displaystyle\;0\quad\quad\text{ whenever $x$ is
nonzero,}$ (2.16) $\displaystyle\|xa\|=$
$\displaystyle\;\|ax\|=|a|\cdot\|x\|\quad{\rm for}\quad a\in\mathbb{R}^{n+1},$
(2.17) $\displaystyle\|xa\|=\|ax\|\leq$
$\displaystyle\;2^{n-1}|a|\|x\|\quad{\rm for}\quad a\in\mathbb{R}_{n}$ (2.18)
$\displaystyle\|x+y\|\leq$ $\displaystyle\;\|x\|+\|y\|.$

######  Remark 2.2.4.

In view of (2.15), (2.16) and (2.18), the fact that $\mathcal{H}$ is complete
and the definition of a Clifford module (see Definition 2.1.6), it is very
easy to see that $\mathcal{H}_{n}$ together with $\|\cdot\|$ can be viewed as
a complex Banach space over $\mathbb{C}_{\mathfrak{I}}$ for any choice of
$\mathfrak{I}\in\mathbb{S}$.

######  Definition 2.2.5 (orthonormal basis in $\mathcal{H}_{n}$).

We will call $(\xi_{i})_{i\in\mathcal{I}}\subseteq\mathcal{H}_{n}$ linearly
independent if for any finite subset
$\widetilde{\mathcal{I}}\subseteq\mathcal{I}$, we have that the equation

$\sum_{i\in\widetilde{\mathcal{I}}}\xi_{i}a_{i}=0,$

where $(a_{i})_{i\in\widetilde{\mathcal{I}}}\subseteq\mathbb{R}_{n}$, only has
the trivial solution, i.e., $a_{i}=0$ for all $i\in\widetilde{\mathcal{I}}$.
Next, we will say that
$\mathcal{B}:=(\xi_{i})_{i\in\mathcal{I}}\subseteq\mathcal{H}_{n}$ is an
orthonormal basis for $\mathcal{H}_{n}$ if $\mathcal{B}$ is linearly
independent, the closed right linear span of $\mathcal{B}$ is
$\mathcal{H}_{n}$, i.e., every $x\in\mathcal{H}_{n}$ belongs to the closure
(with respect to the norm topology on $\mathcal{H}_{n}$) of the set

$\\{\sum_{i\in\widetilde{\mathcal{I}}}\xi_{i}a_{i}:\text{$a_{i}\in\mathbb{R}_{n}$
for all $i\in\widetilde{\mathcal{I}}$ with $\widetilde{\mathcal{I}}$
finite}\\}$

(in this case, we will utilise the short-hand
$x=\sum_{i\in\mathcal{I}}\xi_{i}a_{i}$) and

$\langle\xi_{i},\xi_{j}\rangle=\begin{cases}1&{\rm if}\quad i=j,\\\ 0&{\rm
if}\quad i\neq j.\end{cases}$

While, it is not true that

$\|x\|=\langle x,x\rangle^{1/2}\quad\quad{\rm for}\quad x\in\mathcal{H}_{n},$

we do have the following facts which are reminiscent of the classical Hilbert
space setting.

######  Lemma 2.2.6.

Let $\mathcal{H}_{n}$ be a Clifford module over $\mathbb{R}_{n}$. Then the
following facts hold:

  1. (i)

$\|x\|^{2}={\rm Re}\,\langle x,x\rangle$ for $x\in\mathcal{H}_{n}$.

  2. (ii)

$|\langle x,y\rangle|\leq\|x\|\,\|y\|$ for $x,y\in\mathcal{H}_{n}$.

  3. (iii)

Every Clifford module $\mathcal{H}_{n}$ has an orthonormal basis. Moreover, if
$\mathcal{B}:=(\xi_{i})_{i\in\mathcal{I}}$ is a basis for $\mathcal{H}$, then
$\mathcal{B}$ is an orthonormal basis for $\mathcal{H}_{n}$.

  4. (iv)

For any orthonormal basis $(\xi_{i})_{i\in\mathcal{I}}$, we have

(2.19) $x=\sum_{i\in\mathcal{I}}\xi_{i}\langle x,\xi_{i}\rangle.$

###### Proof.

See, e.g., Proposition 1.9 in [55] for (i) and (ii). We will now prove (iii).
Let $\mathcal{B}:=(\xi_{i})_{i\in\mathcal{I}}$ be an orthonormal basis for the
real Hilbert space $\mathcal{H}$. Then, in view of (2.5), $\mathcal{B}$ is a
basis for $\mathcal{H}_{n}$. In view of (2.9), we have

$\langle\xi_{i},\xi_{j}\rangle=\langle\xi_{i},\xi_{j}\rangle_{\mathcal{H}}=\begin{cases}1&{\rm
if}\quad i=j\\\ 0&{\rm if}\quad i\neq j.\end{cases}$

Thus, $\mathcal{B}$ is an orthonormal basis of $\mathcal{H}_{n}$.

Let $\mathcal{B}$ be as above. Since $\mathcal{B}$ is a basis for the real
Hilbert basis $\mathcal{H}$, we may use the classical orthonormal expansion
for a real Hilbert space to obtain

$\displaystyle x=$ $\displaystyle\;\sum_{\alpha}x_{\alpha}e_{\alpha}$
$\displaystyle=$
$\displaystyle\;\sum_{\alpha}\left(\sum_{i\in\mathcal{I}}\langle
x_{\alpha},\xi_{i}\rangle\right)e_{\alpha}$ $\displaystyle=$
$\displaystyle\;\sum_{i\in\mathcal{I}}\sum_{\alpha}\xi_{i}\langle
x_{\alpha},\xi_{i}\rangle e_{\alpha}$ $\displaystyle=$
$\displaystyle\;\sum_{i\in\mathcal{I}}\sum_{\alpha}\xi_{i}\langle
x_{\alpha}e_{\alpha},\xi_{i}\rangle$ $\displaystyle=$
$\displaystyle\;\sum_{i\in\mathcal{I}}\xi_{i}\langle x,\xi_{i}\rangle.$

∎

In the following lemma, by $\dim\mathfrak{S}(\mathbb{R}_{n})$, we mean the
dimension of the real vector space of self-adjoint Clifford numbers in
$\mathbb{R}_{n}$ (see Definition 2.1.5). For example,
$\dim\mathfrak{S}(\mathbb{R}_{0})=\dim\mathfrak{S}(\mathbb{R}_{1})=\dim\mathfrak{S}(\mathbb{R}_{2})=1$
and $\dim\mathfrak{S}(\mathbb{R}_{3})=4$.

######  Lemma 2.2.7 ([38], Polarisation formula for a Clifford module).

Let $\mathcal{H}_{n}$ be a Clifford module over $\mathbb{R}_{n}$. Then

(2.20) $4\dim\mathfrak{S}(\mathbb{R}_{n})\langle
x,y\rangle=\sum_{\alpha}(\langle x+ye_{\alpha},x+ye_{\alpha}\rangle-\langle
x-ye_{\alpha},x-ye_{\alpha}\rangle)e_{\alpha}$

for all $x,y\in\mathcal{H}_{n}$.

######  Lemma 2.2.8 (Parallelogram law).

Let $\mathcal{H}_{n}$ be a Clifford module. Then for any
$x,y\in\mathcal{H}_{n}$, we have

(2.21) $\|x\|^{2}+\|y\|^{2}=2\|x\|^{2}+2\|y\|^{2},$

where $\|\cdot\|$ is defined by (2.14).

###### Proof.

Since

$\displaystyle\|x+y\|^{2}+\|x+y\|^{2}=$ $\displaystyle\;{\rm Re}(\langle
x,x\rangle+\langle x,y\rangle+\langle y,x\rangle+\langle y,y\rangle)$
$\displaystyle+$ $\displaystyle\;{\rm Re}(\langle x,x\rangle-\langle
x,y\rangle-\langle y,x\rangle+\langle y,y\rangle)$ $\displaystyle=$
$\displaystyle\;2{\rm Re}\,\langle x,x\rangle+2{\rm Re}\,\langle y,y\rangle$
$\displaystyle=$ $\displaystyle\;2\|x\|^{2}+2\|y\|^{2},$

we have (2.21). ∎

######  Definition 2.2.9.

Let $K\subseteq\mathcal{H}_{n}$. We will call $K$ convex if for any choice of
$x,y\in K$, we have

$c\,x+(1-c)\,y\in K\quad\quad{\rm for}\quad 0\leq c\leq 1.$

The next lemma provides a natural generalisation of the closed point in a
closed convex subset property that holds for any complex Hilbert space.

######  Lemma 2.2.10.

Let $K$ be a non-empty, closed and convex subset of a Clifford module
$\mathcal{H}_{n}$. For any $x\in K$, there exists a unique vector $y\in K$
such that $\|x-y\|$ is as small as possible.

###### Proof.

We begin by letting

$m:=\inf_{z\in K}\|x-z\|.$

Let $(y_{i})_{i=1}^{\infty}$ be any minimising sequence for $x$, i.e.,

(2.22) $\lim_{i\to\infty}m_{i}=m\quad\quad{\rm for}\quad m_{i}:=\|x-y_{i}\|.$

Next, if we apply (2.21) to the vectors $x=(x-y_{i})/2$ and $y=(x-y_{j})/2$,
we obtain

(2.23)
$\left\|x-\frac{y_{i}+y_{j}}{2}\right\|^{2}+\left\|\frac{y_{i}-y_{j}}{2}\right\|^{2}=\frac{d_{i}^{2}+d_{j}^{2}}{2}.$

Since $K$ is a convex set, we must have that $(y_{i}+y_{j})/2\in K$.
Consequently, we have $\|x-(y_{i}+y_{j})/2\|\geq m$. Thus, (2.22) and (2.23)
imply that $(y_{i})_{i=0}^{\infty}$ is a Cauchy sequence in $\mathcal{H}_{n}$.
Thus, as $K$ is closed, we have that

$y:=\lim_{i\to\infty}y_{i}\in K.$

Notice that $y$ has the property that $\|x-y\|$ is as small as possible, since

$\|x-y\|=\lim_{i\to\infty}\|x-y_{i}\|=m.$

Finally, suppose $y^{\prime}\in K$ such that $\|x-y^{\prime}\|=m$. Then simply
use (2.21) with $x-y$ and $x-y^{\prime}$ to deduce that $y=y^{\prime}$. ∎

######  Theorem 2.2.11.

Let $Y$ be a closed right submodule of a Clifford module $\mathcal{H}_{n}$ and

$Y^{\perp}:=\\{x\in\mathcal{H}_{n}:\text{$\langle x,y\rangle=0$ for all $y\in
Y$}\\}.$

Then the following statement hold:

  1. (i)

$Y^{\perp}$ is a closed right submodule of $\mathcal{H}_{n}$.

  2. (ii)

$\mathcal{H}_{n}=Y\oplus Y^{\perp}$.

  3. (iii)

$(Y^{\perp})^{\perp}$.

###### Proof.

We will first show (i). Fix any $y\in Y$ and suppose $x,z\in Y^{\perp}$ and
$a\in\mathbb{R}_{n}$. Then

$\displaystyle\langle xa+z,y\rangle=$ $\displaystyle\;\langle x,y\rangle
a+\langle z,y\rangle$ $\displaystyle=$ $\displaystyle\;0+0=0.$

Thus, $Y^{\perp}$ is a right submodule of $\mathcal{H}_{n}$. Next, suppose
$(x_{i})_{i=0}^{\infty}$ is a convergent sequence, where $x_{i}\in Y^{\perp}$
for $i=0,1,\ldots$ and let

$x:=\lim_{i\to\infty}x_{i}.$

Then

$\langle x,y\rangle=\langle x-x_{i},y\rangle+\langle x_{i},y\rangle=\langle
x-x_{i},y\rangle$

and hence for any $\epsilon>0$, we have

$\displaystyle|\langle x,y\rangle|=$ $\displaystyle\;|\langle
x-x_{i},y\rangle|\leq\|x-x_{i}\|\cdot\|y\|$ $\displaystyle<$
$\displaystyle\;\varepsilon\quad\quad{\rm for}\quad\text{$i$ sufficiently
large.}$

But then we have $|\langle x,y\rangle|=0$, in which case $x\in Y^{\perp}$.
Thus, (i) holds.

We will now prove (ii). Suppose $x\in\mathcal{H}_{n}$ is arbitrary. Then by
Lemma 2.2.10, we can find a unique vector $y\in Y$ such that $\|x-y\|$ is as
small as possible. Consequently, if we put $z:=x-y$, then for any $v\in Y$ and
$t\in\mathbb{R}$, we have

$\|z\|^{2}\leq\|z+t\,v\|^{2}.$

Thus,

$\displaystyle\|z\|^{2}\leq$ $\displaystyle\;{\rm Re}\,\langle
z+t\,v,z+t\,v\rangle$ $\displaystyle=$ $\displaystyle\;{\rm Re}\,\langle
z,z\rangle+t({\rm Re}\langle z,v\rangle+{\rm Re}\,\langle
v,z\rangle)+t^{2}{\rm Re}\,\langle v,v\rangle$ $\displaystyle=$
$\displaystyle\;\|z\|^{2}+2t\,{\rm Re}\langle z,v\rangle+t^{2}\|v\|^{2},$

in which case we have

(2.24) ${\rm Re}\,\langle z,v\rangle=0\quad\quad{\rm for}\quad v\in Y.$

Since $Y$ is a submodule, we can replace $y$ by $ye_{\alpha}$ in (2.24) and
realise that $\langle z,v\rangle=0$ for all $y\in Y$. Thus, $z\in Y^{\perp}$
and we have the decomposition $\mathcal{H}_{n}=Y\oplus Y^{\perp}$.

To prove that the decomposition $\mathcal{H}_{n}=Y\oplus Y^{\perp}$ is unique,
suppose $x\in\mathcal{H}_{n}$ can be written as $x=y+z$ and
$x=y^{\prime}+z^{\prime}$, where $y,y^{\prime}\in Y$ and $z,z^{\prime}\in
Y^{\perp}$. But then $y-y^{\prime}=z-z^{\prime}$ will simultaneously belong to
$Y$ and $Y^{\perp}$. Consequently, $\langle
y-y^{\prime},y-y^{\prime}\rangle=0$ forces $y=y^{\prime}$ (see Definition
2.2.1) and hence $z=z^{\prime}$.

Finally, we note that (iii) is a direct consequence of (ii). ∎

##  3\. Linear operators on Clifford modules

######  Definition 3.0.1 (linear operator on a Clifford module).

Let $\mathcal{L}(\mathcal{H}_{n})$ denote the set of linear operators
$\widetilde{T}:\mathcal{D}(\widetilde{T})\to\mathcal{H}$, where
$\mathcal{D}(T)\subseteq\mathcal{H}_{n}$. The subspace
$\mathcal{D}(\widetilde{T})\subseteq\mathcal{H}_{n}$ will be called the domain
of $\widetilde{T}\in\mathcal{L}(\mathcal{H}_{n})$. We shall let
$\mathcal{L}(\mathcal{H}_{n})$ denote the set of all operators
$T:\mathcal{D}(T)\to\mathcal{H}_{n}$ of the form
$T=\sum_{\alpha}e_{\alpha}T_{\alpha}$, where
$T_{\alpha}\in\mathcal{L}(\mathcal{H}_{n})$ which acts on
$\mathcal{D}(T)=\bigcap_{\alpha}\mathcal{D}(T_{\alpha})\subseteq\mathcal{H}_{n}$
via

(3.1)
$Tx=\sum_{\alpha,\beta}T_{\alpha}(x_{\beta})e_{\alpha}e_{\beta}\quad\quad{\rm
for}\quad x=\sum_{\alpha}x_{\alpha}e_{\alpha}\in\mathcal{D}(T).$

######  Remark 3.0.2.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$. One can use (3.1) to check that for
all $x,y\in\mathcal{D}(T)$ and $a\in\mathbb{R}_{n}$, we have

(3.2) $\displaystyle T(xa+y)=$ $\displaystyle\;(Tx)a+Ty$

Thus, in view of (3.2), $\mathcal{L}(\mathcal{H}_{n})$ consists of right
linear operators.

######  Definition 3.0.3 (kernel and range).

Given $T\in\mathcal{L}(\mathcal{H}_{n})$, with domain of $T$ denoted by
$\mathcal{D}(T)$. Then the range and kernel of $T$ will be given by

${\rm Ran}\,\,T=\\{\text{$y\in\mathcal{H}_{n}${\rm:} $Tx=y$ for
$x\in\mathcal{D}(T)$}\\}$

and

${\rm Ker}\,\,T=\\{\text{$x\in\mathcal{D}(T)${\rm:} $Tx=0$}\\},$

respectively.

######  Definition 3.0.4 (bounded linear operator on a Clifford module).

We will call an operator $T\in\mathcal{L}(\mathcal{H}_{n})$ bounded if

(3.3) $\mathcal{D}(T)=\mathcal{H}_{n}$

and

(3.4) $\|T\|:=\sup_{\|x\|\leq 1}\|Tx\|<\infty\quad\quad{\rm for}\quad
x\in\mathcal{H}_{n}.$

We will let $\mathcal{B}(\mathcal{H}_{n})$ denote the set of all operators
$T\in\mathcal{L}(\mathcal{H}_{n})$ such that (3.3) and (3.4) hold.

######  Remark 3.0.5.

One can use (2.15), (2.16) and (2.18) and the definition of the operator norm
on $\mathcal{B}(\mathcal{H}_{n})$ (see (3.4)) to see that
$\delta(T,W)=\|T-W\|$ is a metric on $\mathcal{B}(\mathcal{H}_{n})$. One can
proceed as in the classical case when $\mathcal{H}_{n}$ is a complex Hilbert
space to see that $\mathcal{B}(\mathcal{H}_{n})$ is complete with respect to
the metric $\delta$.

######  Definition 3.0.6 (graph of a linear operator).

Suppose $T\in\mathcal{L}(\mathcal{H}_{n})$. The graph of $T$ is the set

$\mathcal{G}(T):=\\{(x,Tx):x\in\mathcal{D}(T)\\}.$

######  Lemma 3.0.7.

Let $\mathcal{H}_{n}$ be a Clifford module over $\mathbb{R}_{n}$. A right
submodule $\mathcal{K}$ of $\mathcal{H}_{n}\oplus\mathcal{H}_{n}$ satisfies

(3.5) $\mathcal{K}=\\{(x,Tx):x\in\mathcal{D}(T)\\},$

for some $T\in\mathcal{L}(\mathcal{H}_{n})$ if and only if

(3.6) $(0,y)\in\mathcal{K}\Longrightarrow y=0.$

###### Proof.

If $\mathcal{K}$ is as in (3.5), then (3.6) obviously holds. Conversely, if
(3.6) holds, then $(x,y)$ and $(x,z)$ belonging to $\mathcal{K}$ implies that
$y=z$, i.e., there exists a function $T:\mathcal{D}(T)\to\mathcal{H}_{n}$. The
fact that $T\in\mathcal{L}(\mathcal{H}_{n})$ follows easily from the right
linearity of $\mathcal{K}$. Thus, (3.5) holds. ∎

A very simple consequence of the polarisation formula (2.20) is the following
lemma.

######  Lemma 3.0.8.

Suppose $T\in\mathcal{L}(\mathcal{H}_{n})$ is a densely defined operator. If
$\langle Tx,x\rangle=0$ for all $x\in\mathcal{D}(T)$, then $Tx=0$ for all
$x\in\mathcal{D}(T)$.

######  Lemma 3.0.9.

Suppose $S,T\in\mathcal{L}(\mathcal{H}_{n})$ such that $S\subseteq T$, $S$ is
surjective and $T$ is injective. Then $S=T$.

###### Proof.

Fix $x\in\mathcal{D}(T)$. Thus, as $S$ is surjective, we can find
$y\in\mathcal{D}(S)$ such that $Sy=Tx$. Since $S\subseteq T$, we have $Ty=Tx$.
By the injectivity of $T$, we have $x=y$. Thus,
$\mathcal{D}(T)\subseteq\mathcal{D}(S)$ and hence
$\mathcal{D}(T)=\mathcal{D}(S)$, in which case $S=T$. ∎

######  Definition 3.0.10 (graph norm).

Suppose $T\in\mathcal{L}(\mathcal{H}_{n})$. It is easy to check that
$\mathcal{D}(T)$ is a right submodule of $\mathcal{H}_{n}$ which can be
endowed with
$\langle\cdot,\cdot\rangle_{T}:\mathcal{D}(T)\times\mathcal{D}(T)\to\mathbb{R}_{n}$
given by

(3.7) $\langle x,y\rangle_{T}:=\langle x,y\rangle+\langle
Tx,Ty\rangle\quad\quad{\rm for}\quad x,y\in\mathcal{D}(T),$

where $\langle\cdot,\cdot\rangle_{T}$ obeys (2.10)-(2.13) and the
corresponding norm

(3.8) $\|x\|_{T}:=(\|x\|^{2}+\|Tx\|^{2})^{1/2}\quad\quad{\rm for}\quad
x\in\mathcal{D}(T).$

######  Definition 3.0.11 (closed operator).

An operator $T\in\mathcal{L}(\mathcal{H}_{n})$ is called closed if the set
$\\{(x,Tx):x\in\mathcal{H}_{n}\\}$ is a closed subset of
$\mathcal{H}_{n}\times\mathcal{H}_{n}$ (endowed with the product topology).
Let $S$ and $T$ both belong $\mathcal{L}(\mathcal{H}_{n})$. We will write
$S=T$ if $\mathcal{D}(S)=\mathcal{D}(T)$ and $Sx=Tx$ for all
$x\in\mathcal{D}(S)=\mathcal{D}(T)$. We will write $S\subseteq T$ if
$\mathcal{D}(S)\subseteq\mathcal{D}(T)$ and $Sx=Tx$ for all
$x\in\mathcal{D}(S)$. Clearly, $S=T$ if and only if $S\subseteq T$ and
$T\subseteq S$. An operator $T\in\mathcal{L}(\mathcal{H}_{n})$ is called
closable if there exists a closed operator $X\in\mathcal{L}(\mathcal{H}_{n})$
so that $T\subseteq X$.

######  Theorem 3.0.12.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$. The follow statements are equivalent:

  1. (i)

$T$ is closed.

  2. (ii)

For any sequence $(x_{i})_{i=1}^{\infty}$, with $x_{i}\in\mathcal{D}(T)$ for
$i=1,2,\ldots$, such that

$\lim_{i\to\infty}x_{i}=x,$

where $x\in\mathcal{D}(T)$, and

$\lim_{i\to\infty}T(x_{n})=y,$

where $y\in\mathcal{H}_{n}$, we have $Tx=y$.

  3. (iii)

$\mathcal{D}(T)$ together with $\|\cdot,\cdot\|_{T}$ (see (3.11)) is a
complete normed right module over $\mathbb{R}_{n}$.

###### Proof.

In view of Definition 3.0.11, ${\rm(i)}\Longleftrightarrow{\rm(ii)}$ is
immediate. We will now show ${\rm(i)}\Longleftrightarrow{\rm(iii)}$. In view
of (3.8), $\mathcal{D}(T)$ together with $\|\cdot\|_{T}$ is a complete normed
right module over $\mathbb{R}_{n}$ if and only if $\mathcal{G}(T)$ is
complete, i.e., $\mathcal{G}(T)$ is a closed. ∎

######  Theorem 3.0.13.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$. The following statements are
equivalent:

  1. (i)

$T$ is closable.

  2. (ii)

$\overline{\\{(x,Tx):x\in\mathcal{D}(T)\\}}=\\{(x,Wx):\text{for some operator
$W\in\mathcal{L}(\mathcal{H}_{n})$}\\}$.

  3. (iii)

For any sequence $(x_{i})_{i=1}^{\infty}$, where $x_{i}\in\mathcal{D}(T)$ for
$i=1,2,\ldots$, such that

$\lim_{i\to\infty}x_{i}=0$

and

$\lim_{i\to\infty}T(x_{i})=y,$

where $y\in\mathcal{H}_{n}$, then $y=0$.

###### Proof.

We will first show ${\rm(i)}\Longrightarrow{\rm(ii)}$. If
$S\in\mathcal{L}(\mathcal{H}_{n})$ is any closed operator such that
$T\subseteq S$, then

$\\{(x,Tx):x\in\mathcal{D}(T)\\}\subseteq\\{(x,Sx):x\in\mathcal{D}(S)\\}.$

Hence, as $S$ is closed,

$\overline{\\{(x,Tx):x\in\mathcal{D}(T)\\}}\subseteq\\{(x,Sx):x\in\mathcal{D}(S)\\}.$

Therefore, in view of Lemma 3.0.7, (ii) holds.

We will now show ${\rm(ii)}\Longrightarrow{\rm(i)}$. If (ii) holds, then
$T\subseteq W$ and hence $W$ is closed since

$\\{(x,Wx):x\in\mathcal{D}(W)\\}$

is closed. Thus, $T$ is closable.

The proof of ${\rm(ii}\Longrightarrow{\rm(iii)}$ follows immediately from the
Lemma 3.0.7 and the fact that $(0,y)\in\overline{\mathcal{G}(T)}$ implying
that $y=0$.

∎

######  Definition 3.0.14.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be closable. We let

$\overline{T}x:=\lim_{i\to\infty}T(x_{i})$

denote the operator in $\mathcal{L}(\mathcal{H}_{n})$ with domain

$\displaystyle\mathcal{D}(T)=$
$\displaystyle\;\\{x\in\mathcal{H}_{n}:x=\lim_{i\to\infty}i_{n}\ \text{for
$(x_{i})_{n=0}^{\infty}\subseteq\mathcal{D}(T)$ and
$\\{T(x_{i})\\}_{i=0}^{\infty}$}$ $\displaystyle\;\text{converges in
$\mathcal{H}_{n}$}\\}.$

In view of Theorem 3.0.13, the definition of $\overline{T}$ is independent of
the choice of sequence $(x_{i})_{i=0}^{\infty}$. Note that for any closed
operator $W\in\mathcal{L}(\mathcal{H}_{n})$ such that $T\subseteq W$,

$\overline{T}\subseteq W.$

######  Definition 3.0.15 (continuous operator).

Let $T\in\mathcal{L}(\mathcal{H}_{n})$. $T$ will be called continuous if for
any sequence $(x_{i})_{i=1}^{\infty}$, where $x_{i}\in\mathcal{D}(T)$ for
$i=1,2,\ldots$, such that

$\lim_{i\to\infty}x_{i}=x,$

where $x\in\mathcal{D}(T)$, we have

$\lim_{i\to\infty}T\,x_{i}=Tx.$

######  Theorem 3.0.16.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$. Then $T$ is continuous if and only if
$T\in\mathcal{B}(\mathcal{H}_{n})$.

###### Proof.

The proof can be carried out as in the classical complex Hilbert space case,
see, e.g., [51]. ∎

######  Definition 3.0.17.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$. A subset $\mathcal{E}$ of
$\mathcal{D}(T)$ will be called core of $\mathcal{D}(T)$ if $\mathcal{E}$ is a
right submodule of $\mathcal{D}(T)$ and $\mathcal{E}$ is dense in
$(\mathcal{D}(T),\|\cdot\|_{T})$.

######  Definition 3.0.18 (adjoint operator).

Given $T\in\mathcal{L}(\mathcal{H}_{n})$ which is densely defined, we let
$T^{*}\in\mathcal{L}(\mathcal{H}_{n})$ denote the unique operator so that

$\langle Tx,y\rangle=\langle x,T^{*}y\rangle,\quad
x\in\mathcal{D}(T),y\in\mathcal{D}(T^{*}),$

where the domain of $T^{*}$ is given by

$\displaystyle\mathcal{D}(T^{*})=$
$\displaystyle\;\\{\text{$y\in\mathcal{H}_{n}$ : there exists
$z\in\mathcal{H}_{n}$ with $\langle Tx,y\rangle=\langle x,z\rangle$}$
$\displaystyle\text{for every $x\in\mathcal{D}(T)$}\\}.$

######  Remark 3.0.19.

Let $T\in\mathcal{B}(\mathcal{H}_{n})$. Then, in view of Definition 3.0.18, it
is easy to check that $(T^{*})^{*}=T$.

######  Lemma 3.0.20.

Let $T,W\in\mathcal{B}(\mathcal{H}_{n})$. Then
$TW\in\mathcal{B}(\mathcal{H}_{n})$,

(3.9) $\|TW\|\leq\|T\|\cdot\|W\|$

and

(3.10) $\|T^{*}T\|=\|T^{*}\|^{2}=\|T\|^{2}.$

###### Proof.

The fact that $TW\in\mathcal{B}(\mathcal{H}_{n})$, (3.9) and (3.10) can be
proved exactly as in the classical complex Banach algebra case, see, e.g.,
Theorem 8 on page 168 of [51] for a proof of the classical complex Hilbert
space case of (3.9) and [28] for a classical complex Hilbert space case of
(3.10). ∎

######  Definition 3.0.21 (self-adjoint, anti self-adjoint and unitary).

Let $T\in\mathcal{L}(\mathcal{H}_{n})$. We will call $T$ self-adjoint, anti
self-adjoint and unitary if $T=T^{*}$ with
$\mathcal{D}(T)=\mathcal{D}(T^{*})$, $T=-T^{*}$ with
$\mathcal{D}(T)=\mathcal{D}(T^{*})$ and $TT^{*}=T^{*}T=I$, respectively.

######  Remark 3.0.22.

In view of Theorem 3.0.23(i), we have that if
$T\in\mathcal{L}(\mathcal{H}_{n})$ is self-adjoint or anti self-adjoint
operator, then $T$ is closed.

######  Theorem 3.0.23.

If $T\in\mathcal{L}(\mathcal{H}_{n})$ is densely defined and
$W\in\mathcal{L}(\mathcal{H}_{n})$, then:

  1. (i)

$T^{*}\in\mathcal{L}(\mathcal{H}_{n})$ is closed.

  2. (ii)

${\rm{\rm Ran}\,}(T)^{\perp}={\rm Ker}(T^{*})$.

  3. (iii)

If $T\subseteq W$, then $W^{*}\subseteq T^{*}$.

###### Proof.

The proofs can completed in much the same way as the case when
$\mathcal{H}_{n}$ is a complex Hilbert space (see, e.g., Proposition 1.6 in
[66]). ∎

######  Theorem 3.0.24.

If $T\in\mathcal{L}(\mathcal{H}_{n})$ is densely defined, then:

  1. (i)

$T$ is closable if and only if $\mathcal{D}(T^{*})$ is dense in
$\mathcal{H}_{n}$.

  2. (ii)

If $T$ is closable, then $\overline{T}=T^{**}$, where $T^{**}:=(T^{*})^{*}$.

  3. (iii)

$T$ is closed if and only if $T=T^{**}$.

  4. (iv)

If $T$ is closable and ${\rm Ker}(T)=\\{0\\}$ and ${\rm Ran}\,(T)$ is dense in
$\mathcal{H}_{n}$, then $T^{*}$ is invertible and $(T^{*})^{-1}=(T^{-1})^{*}$.

  5. (v)

If $T$ is closable and ${\rm Ker}\,(T)=\\{0\\}$, then $T^{-1}$ is closable if
and only if ${\rm Ker}\,(\overline{T})=\\{0\\}$. In this case, we have
$(\overline{T})^{-1}=\overline{T^{-1}}$.

  6. (vi)

Suppose $T$ is invertible. Then $T$ is closed if and only if $T^{-1}$ is
closed.

###### Proof.

The proofs can completed in much the same way as the case when
$\mathcal{H}_{n}$ is a complex Hilbert space (see, e.g., Theorem 1.8 in [66]).
∎

######  Lemma 3.0.25.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be a densely defined operator. Then the
graph of $T^{*}$ satisfies

(3.11)
$\mathcal{G}(T^{*})=V(\mathcal{G}(T))^{\perp}=V(\mathcal{G}(T)^{\perp}),$

where
$V:\mathcal{H}_{n}\oplus\mathcal{H}_{n}\to\mathcal{H}_{n}\oplus\mathcal{H}_{n}$
denotes the unitary operator given by

$V(x,y)=(-y,x).$

###### Proof.

Suppose $x\in\mathcal{D}(T)$ and $y\in\mathcal{D}(T^{*})$. Using Definition
3.0.18, we have

$\displaystyle\langle\langle V(x,Tx),(y,T^{*}y)\rangle=$
$\displaystyle\;\langle(-Tx,x),(y,T^{*}y)\rangle$ $\displaystyle=$
$\displaystyle\;\langle-Tx,y\rangle+\langle x,T^{*}y\rangle$ $\displaystyle=$
$\displaystyle\;0,$

in which case we have $\mathcal{G}(T^{*})\subseteq V(\mathcal{G}(T))^{\perp}$.

Conversely, suppose $(y,z)\in V(\mathcal{G}(T))^{\perp}$. Then for any
$x\in\mathcal{D}(T)$, we have

$\langle V(x,Tx),(y,z)\rangle=\langle-Tx,y\rangle+\langle x,z\rangle=0,$

and hence $\langle Tx,y\rangle=\langle x,z\rangle$. But then we have again use
Definition 3.0.18 to obtain $y\in\mathcal{D}(T^{*})$ and $z=T^{*}y$, i.e.,
$(y,z)\in\mathcal{G}(T^{*})$. Therefore, we have
$V(\mathcal{G}(T))^{\perp}\subseteq\mathcal{G}(T^{*})$.

The second equality is an immediate consequence of $V$ being unitary on
$\mathcal{H}_{n}\oplus\mathcal{H}_{n}$. ∎

######  Theorem 3.0.26 (Riesz representation theorem for Clifford modules).

Let $\mathcal{H}_{n}$ be a Clifford module over $\mathbb{R}_{n}$. Suppose
$B:\mathcal{H}_{n}\times\mathcal{H}_{n}\to\mathbb{R}_{n}$ is bounded, i.e.,
there exists $M\geq 0$ such that

$|B(x,y)|\leq M\|x\|\|y\|\quad\quad{\rm for}\quad x,y\in\mathcal{H}_{n},$

and satisfies the following:

  1. (i)

$B(x+y,z)=B(x,z)+B(y,z)$ and $B(x,y+z)=B(x,y)+B(x,z)$ for
$x,y,z\in\mathcal{H}_{n}$.

  2. (ii)

$B(x\,a,y)=B(x,y)a$ and $B(x,y\,a)=\bar{a}\,B(x,y)$ for
$x,y\in\mathcal{H}_{n}$ and $a\in\mathbb{R}_{n}$.

Then there exists a unique $T\in\mathcal{B}(\mathcal{H}_{n})$ such that

(3.12) $B(x,y)=\langle Tx,y\rangle\quad\quad{\rm for}\quad
x,y\in\mathcal{H}_{n}.$

###### Proof.

The proof of Corollary 1.11 in [55] can easily be adjusted to our present
right Clifford module setting. ∎

The following theorem appears in [55].

######  Theorem 3.0.27 (Hahn-Banach theorem for a Clifford module).

Let $\mathcal{H}_{n}$ be a Clifford module over $\mathbb{R}_{n}$ and
$\mathcal{S}_{n}$ be a right submodule of $\mathcal{H}_{n}$. Suppose
$\mathscr{L}:\mathcal{S}_{n}\to\mathbb{R}_{n}$ is continuous (i.e., bounded)
and satisfies

(3.13) $\mathscr{L}(xa+y)=\mathscr{L}(x)a+\mathscr{L}(y)\quad\quad{\rm
for}\quad x,y\in\mathcal{S}_{n}\quad{\rm and}\quad a\in\mathbb{R}_{n}.$

Then $\mathscr{L}$ has a continuous extension
$\mathscr{L}:\mathcal{H}_{n}\to\mathbb{R}_{n}$ (with a slight abuse of
notation, we shall use $\mathscr{L}$ to denote the extension) such that (3.13)
holds for all $x,y\in\mathcal{H}_{n}$ and $a\in\mathbb{R}_{n}$.

######  Theorem 3.0.28 (closed graph theorem for a Clifford module).

Suppose $T\in\mathcal{L}(\mathcal{H}_{n})$ is a closed operator with
$\mathcal{D}(T)=\mathcal{H}_{n}$. Then $T\in\mathcal{B}(\mathcal{H}_{n})$.

###### Proof.

The proof given in Theorem 2.2.7 in [62] can easily be adjusted to the present
Clifford module setting. ∎

######  Lemma 3.0.29.

Suppose $C\in\mathcal{B}(\mathcal{H}_{n})$ is invertible in
$\mathcal{B}(\mathcal{H}_{n})$. Then the following statements hold:

  1. (i)

$C-D$ is invertible in $\mathcal{B}(\mathcal{H}_{n})$ whenever

$\|D\|<\frac{1}{\|C^{-1}\|}.$

  2. (ii)

Suppose $C$ is invertible in $\mathcal{B}(\mathcal{H}_{n})$ and
$\|C-D\|<\|C^{-1}\|^{-1}$. Then $D$ is invertible in
$\mathcal{B}(\mathcal{H}_{n})$.

###### Proof.

The proof of (i) and (ii) is exactly the same as in the classical case, see,
e.g., Theorem 2 in Chapter 17 of [51] for the proof of (i). ∎

######  Definition 3.0.30 (normal operator on a Clifford module).

Suppose $T\in\mathcal{L}(\mathcal{H}_{n})$. We will call $T$ normal if $T$ is
densely defined, $T$ is closed, $\mathcal{D}(T)=\mathcal{D}(T^{*})$ and
$TT^{*}=T^{*}T$.

######  Lemma 3.0.31.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be normal. If
$S\in\mathcal{L}(\mathcal{H}_{n})$ so that $T\subseteq S$ and
$\mathcal{D}(S)\subseteq D(S^{*})$, then $S=T$.

###### Proof.

If $T\subseteq S$, then $S^{*}\subseteq T^{*}$ and hence

$\mathcal{D}(T)\subseteq\mathcal{D}(S)\subseteq\mathcal{D}(S^{*})\subseteq\mathcal{D}(T^{*})=\mathcal{D}(T),$

i.e., $\mathcal{D}(S)=\mathcal{D}(T)$. Therefore, $S=T$. ∎

###  3.1. $S$-resolvent set, $S$-spectrum for a linear operator and the
spectral radius formula

######  Definition 3.1.1 ($S$-resolvent set and $S$-spectrum for
$T\in\mathcal{L}(\mathcal{H}_{n})$).

Let
$T\in\mathcal{L}(\mathcal{H})$ be densely defined and
$\mathcal{Q}_{s}(T):\mathcal{D}(T^{2})\to\mathcal{H}$ be given by

(3.14) $\mathcal{Q}_{s}(T)x=(T^{2}-2{\rm Re}(s)T+|s|^{2}I)x\quad{\rm for}\quad
x\in\mathcal{D}(T^{2}).$

The $S$-resolvent set of $T$ is defined as follows

$\displaystyle\rho_{S}(T)=$ $\displaystyle\;\\{\text{$s\in\mathbb{R}^{n+1}$
{\rm:} ${\rm Ker}\,\,\mathcal{Q}_{s}(T)=\\{0\\}$, ${\rm
Ran}\,\,\mathcal{Q}_{s}(T)$ is dense in $\mathcal{H}_{n}$ and}$
$\displaystyle\;\text{$\mathcal{Q}_{s}(T)^{-1}:{\rm
Ran}\,\mathcal{Q}_{s}(T)\to\mathcal{D}(T^{2}$) is bounded}\\}.$

The $S$-spectrum is defined as

$\sigma_{S}(T)=\mathbb{R}^{n+1}\,\setminus\,\rho_{S}(T).$

######  Remark 3.1.2.

In [27], the $S$-spectrum for $T\in\mathcal{L}(\mathcal{H}_{n})$ was
considered for paravector operators, i.e., operators of the form

$T=\sum_{i=0}^{n}e_{i}T_{i},$

because the purpose of this theory was to define a functional calculus for
$(n+1)$\- tuples of noncommuting operators. Quite early on in our
investigation of the Clifford spectral theorem, we directed our attention to
fully Clifford operators because of the crucial decomposition

(3.15) $T=A+J_{0}B.$

of normal bounded operators given in Theorem 8.0.4. This decomposition implies
that, even if $T$ is a paravector operator, the operators $J_{0}$ and $B$ will
not be paravector operators in general.

A further observation is that in the case $T$ is a paravector operator with
noncommuting components the $S$-spectrum is defined by the operator
$T^{2}-2{\rm Re}(s)T+|s|^{2}I$ that is not a paravector.

For an explanation of why this theory is so flexible see the introduction of
the paper [18], where some considerations are made on various Cauchy formulas
that define various holomorphic functional calculi. The properties of the
$S$-spectrum and the $S$-resolvent operators for fully Clifford operators
remain the same with the same proofs valid for paravector operators, see [17,
18, 37].

######  Remark 3.1.3.

For any $T\in\mathcal{L}(\mathcal{H}_{n})$, it is easy to show that the
$S$-resolvent set is equal to

$\displaystyle\rho_{S}(T)=\\{s\in\mathbb{R}^{n+1}:\mathcal{Q}_{s}(T)^{-1}:$
$\displaystyle\;{\rm Ran}\,\,\mathcal{Q}_{s}(T)\to\mathcal{D}(T^{2})$ (3.16)
$\displaystyle\;\text{is bijective and bounded}\\}$

######  Definition 3.1.4 (left and right $S$-resolvent operator).

Suppose $T\in\mathcal{B}(\mathcal{H}_{n})$. For any choice of
$s\in\rho_{S}(T)$, we shall define the left $S$-resolvent operator via

(3.17)
$S_{L}^{-1}(s,T):=-\mathcal{Q}_{s}(T)^{-1}(T-\bar{s}\,I)\in\mathcal{B}(\mathcal{H}_{n})$

and the right $S$-resolvent operator via

(3.18)
$S_{R}^{-1}(s,T):=-(T-\bar{s}\,I)\mathcal{Q}_{s}(T)^{-1}\in\mathcal{B}(\mathcal{H}_{n}).$

######  Remark 3.1.5.

One can check that
$S_{L}^{-1}(s,T):\rho_{S}(T)\to\mathcal{B}(\mathcal{H}_{n})$ is a right slice
hyperholomorphic function (see Lemma 3.10 in [17] for details). Analogously,
one can check that
$S_{R}^{-1}(s,T):\rho_{S}(T)\to\mathcal{B}(\mathcal{H}_{n})$ is a left slice
hyperholomorphic function.

######  Definition 3.1.6 (axially symmetric).

We will call a set $\Omega\subseteq\mathbb{R}^{n+1}$ axially symmetric, if
whenever $s=s_{0}+s_{1}\mathfrak{I}\in\Omega$, where
$s_{0},s_{1}\in\mathbb{R}$ and $\mathfrak{I}\in\mathbb{S}$, then
$s_{0}+s_{1}\mathfrak{J}\in\Omega$ for all $\mathfrak{J}\in\mathbb{S}$.

######  Remark 3.1.7 (the $S$-spectrum is axially symmetric).

Let $T\in\mathscr{L}(\mathcal{H}_{n})$. Then $\sigma_{S}(T)$ is axially
symmetric. If $\sigma_{S}(T)=\emptyset$, then there is nothing to prove. If
$\sigma_{S}(T)\not=\emptyset$, then one need only notice that
$s\in\rho_{S}(T)$ depends only on ${\rm Re}(s)$ and $|s|$.

######  Theorem 3.1.8 (properties of the $S$-spectrum for
$T\in\mathcal{B}(\mathcal{H}_{n})$).

Suppose
$T\in\mathcal{B}(\mathcal{H}_{n})$. Then $\sigma_{S}(T)$ is a non-empty
compact subset of

$\\{s\in\mathbb{R}^{n+1}:0\leq|s|\leq\|T\|\\}.$

###### Proof.

We will first show that $\sigma_{S}(T)$ is non-empty. For any choice of
$\varepsilon>0$, the series $\sum_{i=0}^{\infty}T^{i}s^{-i-1}$ converges
uniformly in norm on
$s\in\Omega_{\varepsilon}:=\\{a\in\mathbb{R}^{n+1}:|a|=\|T\|+\varepsilon\\}$
to $S_{L}^{-1}(s,T)$. Thus, for any $\mathfrak{I}\in\mathbb{S}$, we may use
the fact that

$\int_{\Omega_{\varepsilon}\cap\mathbb{C}_{\mathfrak{I}}}s^{-i-1}\,ds(-\mathfrak{I})=\begin{cases}2\pi&{\rm
if}\quad i=0,\\\ 0&{\rm if}\quad i\neq 0,\end{cases}$

to obtain

(3.19)
$\int_{\Omega_{\varepsilon}}S_{L}^{-1}(s,T)ds(-\mathfrak{I})=\sum_{i=0}^{\infty}T^{i}\left(\int_{\Omega_{\varepsilon}}s^{-i-1}ds(-\mathfrak{I})\right)=2\pi\,I.$

If
$M_{\varepsilon}:=\overline{\\{a\in\mathbb{R}^{n+1}:|a|\leq\|T\|+\varepsilon\\}}\subseteq\rho_{S}(T)$,
then Remark 3.1.5 asserts that $S_{L}^{-1}(s,T)|_{M_{\varepsilon}}$ is a right
slice hyperholomorphic $\mathcal{B}(\mathcal{H}_{n})$-valued function.
However, an analogue of Cauchy’s integral formula for
$\mathcal{B}(\mathcal{H}_{n})$-valued functions asserts that (3.19) must be
$0$, which is clearly not the case. Thus, $M_{\varepsilon}$ is not a subset of
$\rho_{S}(T)$. But then $\sigma_{S}(T)$ cannot be empty.

We will now show that $\sigma_{S}(T)$ is a closed subset of
$\mathbb{R}^{n+1}$. Notice that
$\varphi:\mathbb{R}^{n+1}\to\mathcal{B}(\mathcal{H}_{n})$ given by
$\varphi(s)=\mathcal{Q}_{s}(T)$ is a continuous function. Lemma 3.0.29 can be
used to show that the set of invertible operators in
$\mathcal{B}(\mathcal{H}_{n})$ is open. Thus,
$\rho_{S}(T)=\varphi^{-1}(\\{W\in\mathcal{B}(\mathcal{H}_{n}):W^{-1}\in\mathcal{B}(\mathcal{H}_{n})\\})$
is open. Consequently, $\sigma_{S}(T)=\mathbb{R}^{n+1}\setminus\rho_{S}(T)$ is
closed.

Next, we will show that
$\sigma_{S}(T)\subseteq\\{s\in\mathbb{R}^{n+1}:|s|\leq\|T\|\\}$. In view of
(3.9), we have $\|T^{i}\|\leq\|T\|^{i}$ for $i=0,1,\ldots$. Thus, as
$s\in\mathbb{R}^{n+1}$, we have the estimate

$\|T^{i}s^{-i-1}\|\leq
2^{n-1}|s^{-i-1}|\cdot\|T^{i}\|=\frac{\|T\|^{i}}{|s|^{i+1}}\quad\quad{\rm
for}\quad i=0,1,\ldots$

and hence

(3.20) $\text{$\sum_{i=0}^{\infty}\|T^{i}s^{-i-1}\|$ converges if and only if
$|s|<\|T\|$}.$

We now claim that

(3.21) $(T^{2}-2{\rm
Re}(s)T+|s|^{2}\,I)\left(\sum_{i=0}^{\infty}T^{i}s^{-i-1}\right)=\bar{s}\,I-T.$

Indeed, notice that

$\displaystyle\;(T^{2}-2{\rm
Re}(s)T+|s|^{2}\,I)\left(\sum_{i=0}^{\infty}T^{i}s^{-i-1}\right)$
$\displaystyle=$
$\displaystyle\;\sum_{i=0}^{\infty}\left(T^{i+1}s^{-i-1}-T^{i+1}s^{-i-1}(s+\bar{s})+T^{i}s^{-i-1}s\bar{s}\right)$
$\displaystyle=$
$\displaystyle\;\sum_{i=0}^{\infty}\left(T^{i+1}s^{-i}-T^{i+1}s^{-i}-T^{i+1}s^{-i-1}\bar{s}+T^{i}s^{-i}\bar{s}\right)$
$\displaystyle=$ $\displaystyle\;\bar{s}\,I-T.$

Thus, we have (3.21). Putting together (3.20) and (3.21), we have that
$\sigma_{S}(T)\subseteq\\{s\in\mathbb{R}^{n+1}:|s|\leq\|T\|\\}$. As
$\sigma_{S}(T)$ is closed, we have that is a closed subset of a compact set,
i.e., $\sigma_{S}(T)$ is compact.

Finally, we will show that $\sigma_{S}(T)$ is closed. Suppose
$s\in\rho_{S}(T)$ and $\tilde{s}\in\mathbb{R}^{n+1}$ be such that
$|s-\tilde{s}|$ is sufficiently small. Then

$\displaystyle\mathcal{Q}_{s}(T)-\mathcal{Q}_{\tilde{s}}(T)=$
$\displaystyle\;2{\rm Re}(\tilde{s}-s)T+(|s|^{2}-|\tilde{s}|^{2})I$
$\displaystyle=$
$\displaystyle\;\frac{1}{|s|^{2}-|\tilde{s}|^{2}}\left(\frac{2{\rm
Re}(\tilde{s}-s)}{|s|^{2}-|\tilde{s}|^{2}}\,T+I\right),$

one can use Lemma 5.3.11 to see that $\mathcal{Q}_{\tilde{s}}$ is invertible
in $\mathcal{B}(\mathcal{H}_{n})$. Thus, $\tilde{s}\in\rho_{S}(T)$ and
$\rho_{S}(T)$ is open, i.e., $\sigma_{S}(T)$ is closed. ∎

######  Definition 3.1.9 (spectral radius).

Let $T\in\mathcal{B}(\mathcal{H}_{n})$. Then we shall denote the spectral
radius of $T$ by $r_{S}(T)$ which is given by

(3.22) $r_{S}(T):=\sup_{s\in\sigma_{S}(T)}|s|=\max_{s\in\sigma_{S}(T)}|s|.$

######  Theorem 3.1.10 (spectral radius formula).

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be normal. Then

(3.23) $\|T\|=r_{S}(T).$

###### Proof.

Theorem 6.7 in [[17] asserts, in particular, that for any
$T\in\mathcal{B}(\mathcal{H}_{n})$, we have

$\lim_{i\to\infty}(\|T^{i}\|_{1})^{\frac{1}{i}}=r_{S}(T),$

where

$\|T\|_{1}:=\sum_{\alpha}\|T_{\alpha}\|_{\mathcal{H}}\quad\quad{\rm for}\quad
T=\sum_{\alpha}T_{\alpha}e_{\alpha}.$

Notice that

$\displaystyle\|T\|:=$
$\displaystyle\;\sup_{\|x\|=1}\|Tx\|=\sup_{\|x\|=1}\|\sum_{\alpha,\beta}T_{\alpha}x_{\beta}e_{\alpha}e_{\beta}\|$
$\displaystyle\leq$
$\displaystyle\;\sup_{\|x\|=1}\sum_{\alpha,\beta}\|T_{\alpha}x_{\beta}e_{\alpha}e_{\beta}\|$
$\displaystyle\leq$
$\displaystyle\;2^{n-1}\sup_{\|x\|=1}\sum_{\alpha,beta}\|T_{\alpha}x_{\beta}\|_{\mathcal{H}}\cdot|e_{\alpha}e_{\beta}|$
$\displaystyle\leq$
$\displaystyle\;2^{n-1}2^{n-1}\sum_{\alpha,\beta}\left(\sup_{\|x\|=1}\|T_{\alpha}x_{\beta}\|_{\mathcal{H}}\right)$
$\displaystyle\leq$
$\displaystyle\;2^{2(n-1)}\sum_{\alpha,\beta}\|T_{\alpha}\|_{\mathcal{H}}$
$\displaystyle=$ $\displaystyle\;2^{3n-2}\|T\|_{1}.$

In view of (3.10), one can use induction on $i$ to show that
$\|T^{2^{i}}\|^{\frac{1}{2^{i}}}=\|T\|$. In view of Theorem 3.1.8, we have
$r_{S}(T)\leq\|T\|$. Thus,

$r_{S}(T)\leq\|T\|=\|T^{2^{i}}\|^{\frac{1}{2i}}\quad\quad{\rm for}\quad
i=1,2,\ldots$

and hence

$\displaystyle r_{S}(T)\leq$
$\displaystyle\;\lim_{i\to\infty}\|T^{2^{i}}\|^{\frac{1}{2i}}$
$\displaystyle=$
$\displaystyle\;\lim_{i\to\infty}\left(2^{\frac{3n-2}{i}}(\|T^{2^{i}}\|_{1})^{\frac{1}{2^{i}}}\right)$
$\displaystyle=$ $\displaystyle\;r_{S}(T).$

and hence we have (3.23). ∎

Following the possible splittings of the classical spectrum for operators on
complex Banach space we can give the same spitting also for fully linear
operators on Clifford modules. We recall that the splitting of the spectrum is
defined according to where an operators is not invertible. We will mention two
possible splittings. The first one is the point, residual, continuous
$S$-spectrum of a Clifford operator.

######  Definition 3.1.11 (point, residual, continuous $S$-spectrum).

Let $T:\mathcal{D}(T)\to\mathcal{H}_{n}$. We split the $S$-spectrum into the
three disjoint sets:

  * (P)

The point $S$-spectrum of $T$:

$\sigma_{PS}(T)=\\{s\in\mathbb{R}^{n+1}\ :\ \ {\rm
Ker}(\mathcal{Q}_{s}(T))\not=\\{0\\}\\}.$

  * (R)

The residual $S$-spectrum of $T$:

$\sigma_{RS}(T)=\left\\{s\in\mathbb{R}^{n+1}\ :\ \ {\rm
Ker}(\mathcal{Q}_{s}(T))=\\{0\\},\ \overline{{\rm
Ran}(\mathcal{Q}_{s}(T))}\not=\mathcal{H}_{n}\right\\}.$

  * (C)

The continuous $S$-spectrum of $T$:

$\displaystyle\sigma_{CS}(T)=$ $\displaystyle\;\\{s\in\mathbb{R}^{n+1}\ :\
{\rm Ker}(\mathcal{Q}_{s}(T))=\\{0\\},\ \overline{{\rm
Ran}(\mathcal{Q}_{s}(T))}=\mathcal{H}_{n}$ $\displaystyle\;\;{\rm
and}\;\mathcal{Q}_{s}(T)^{-1}\not\in\mathcal{B}(\mathcal{H}_{n})\\}.$

######  Remark 3.1.12.

Notice that if $A\in\mathcal{B}(\mathcal{H}_{n})$ that satisfies the two
conditions:

  1. (i)

There exists $K>0$ such that $\|Av\|\geq K\|v\|$ for $v\in\mathcal{D}(A)$
(bounded from below)

  2. (ii)

the range of $A$ is dense in $\mathcal{H}_{n}$,

then $A$ is invertible.

So in analogy to the classical case for the $S$-spectrum we have:

######  Definition 3.1.13 (approximate point and compression $S$-spectrum).

Let $T$ be a Clifford bounded linear operator. The approximate point
$S$-spectrum of $T$, denoted by $\Pi_{S}(T)$, is defined as

$\Pi_{S}(T)=\\{s\in\mathbb{R}^{n+1}\ :\ T^{2}-2{\rm
Re}(s)T+|s|^{2}\mathcal{I}\text{ is not bounded from below}\\}.$

The compression $S$-spectrum of $T$, denoted by $\Gamma_{S}(T)$, is defined as

$\Gamma_{S}(T)=\\{s\in\mathbb{R}^{n+1}\ :\ \text{ the range of }\ T^{2}-2{\rm
Re}T+|s|^{2}\mathcal{I}\text{ is not dense}\\}.$

The set $\Pi_{S}(T)$ contains the $S$-eigenvalues.

###  3.2. Basic facts for normal operators

In this section we are defining self-adjoint, anti self-adjoint operators and
positive operators on a Clifford module. We are also formulating and proving a
number of facts which will be useful when proving the spectral theorem for a
bounded self-adjoint operator (see Section 6) and also the spectral theorem
for an unbounded normal operator (see 10).

For the quaternionic setting similar properties hold, see the book [16] and
the references therein.

######  Lemma 3.2.1.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be given. The following statements are
equivalent:

  1. (i)

$T$ is self-adjoint.

  2. (ii)

$\overline{\langle Tx,x\rangle}=\langle Tx,x\rangle$ for all
$x\in\mathcal{D}(T)$.

###### Proof.

If $T$ is self-adjoint, then

$\langle Tx,x\rangle=\langle x,Tx\rangle\quad\quad{\rm for}\quad
x\in\mathcal{D}(T)$

and hence

(3.24) $\overline{\langle x,Tx\rangle}=\langle Tx,x\rangle\quad\quad{\rm
for}\quad x\in\mathcal{D}(T).$

On the other hand, if (ii) is in force, then (2.20) can be used to show that
$\langle Tx,y\rangle=\langle x,Ty\rangle$ for all $x,y\in\mathcal{D}(T)$.
Thus, (i) holds. ∎

######  Definition 3.2.2 (positive operator on a Clifford module).

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be given. The operator $T$ is called
positive if $\langle Tx,x\rangle\succeq 0$ for all $x\in\mathcal{D}(T)$.

The following theorem will be useful when considering the unbounded case of
the spectral theorem for normal operators.

######  Theorem 3.2.3.

Suppose $T\in\mathcal{L}(\mathcal{H}_{n})$ is a densely defined closed
operator. Then the following statement hold:

  1. (i)

$I+T^{*}T$ is a bijective mapping on $\mathcal{H}_{n}$. If
$C_{T}:=(I+T^{*}T)^{-1}\in\mathcal{B}(\mathcal{H}_{n})$, then
$C_{T}\in\mathcal{B}(\mathcal{H}_{n})$, $C_{T}$ is positive and $I-C_{T}$ is
positive.

  2. (ii)

The operator $T^{*}T\in\mathcal{L}(\mathcal{H}_{n})$ is positive and
$\mathcal{D}(T^{*}T)$ is a core for $T$. In particular, if $T$ is self-
adjoint, then $\mathcal{D}(T^{2})$ is a core for $T$.

###### Proof.

We will first prove (i). In view of Lemma 3.0.25, we have
$\mathcal{G}(T^{*})=V(\mathcal{G}(T))^{\perp}$, where $V$ denotes the unitary
operator in the statement of Lemma 3.0.25. Notice that
$\mathcal{H}_{n}\oplus\mathcal{H}_{n}=\mathcal{G}(T^{*})\oplus
V(\mathcal{G}(T))$. Consequently, corresponding to every
$z\in\mathcal{H}_{n}$, there exist $x\in\mathcal{D}(T)$ and
$y\in\mathcal{D}(T^{*})$ such that

$(0,z)=(y,T^{*}y)=V(x,Tx)=(y-Tx,T^{*}y+x).$

Thus, $y=Tx$ and $z=x+T^{*}y=(I+T^{*}T)x$, in which case the operator
$I+T^{*}T\in\mathcal{L}(\mathcal{H}_{n})$ is surjective. To see that
$I+T^{*}T$ is injective, notice that for any $x,y\in\mathcal{D}(T^{*}T)$, we
have

$\displaystyle\|(I+T^{*}T)(x-y)\|^{2}=$ $\displaystyle\;{\rm
Re}\langle(x-y)+T^{*}T(x-y),(x-y)+T^{*}T(x-y)\rangle$ (3.25) $\displaystyle=$
$\displaystyle\;{\rm Re}\,\langle x-y,x-y\rangle+2{\rm Re}\,\langle
T(x-y),T(x-y)\rangle$ $\displaystyle+$ $\displaystyle\;{\rm Re}\,\langle
T^{*}T(x-y),T^{*}T(x-y)\rangle$ (3.26) $\displaystyle=$
$\displaystyle\;\|x-y\|^{2}+2\|T(x-y)\|^{2}+\|T^{*}T(x-y)\|^{2}.$

Thus, as $I+T^{*}T$ is injective and surjective, we have that $I+T^{*}T$ is
bijective and let $C_{T}:=(I+T^{*}T)^{-1}$.

Suppose $z=(I+T^{*}T)x$ for some $x\in\mathcal{D}(T^{*}T)$. Then $C_{T}z=x$
and one can use (3.26) with $y=0$ to obtain

$\|C_{T}z\|=\|x\|\leq\|(I+T^{*}T)x\|=\|z\|$

and hence $C_{T}\in\mathcal{B}(\mathcal{H}_{n})$. The fact that $C_{T}$ is
self-adjoint is an immediate consequence of

$\langle C_{T}z,z\rangle=\langle x,z\rangle=\langle
x,(I+T^{*}T)x\rangle=\langle(I+T^{*}T)x,x\rangle.$

The remaining conclusions in (i) can be easily justified via (3.26).

We will now prove (ii). Since $C_{T}\in\mathcal{B}(\mathcal{H}_{n})$ is self-
adjoint, we have that $C_{T}^{-1}$ is also self-adjoint by Theorem 3.0.24(iv),
i.e., $I+T^{*}T$ is self-adjoint. The fact that $T^{*}T$ is positive follows
from

$\langle T^{*}Tx,x\rangle=\langle Tx,Tx\rangle\succeq 0\quad{\rm for}\quad
x\in\mathcal{D}(T^{*}T).$

To check that $\mathcal{D}(T^{*}T)$ is a core for $T$, we must check that
$\mathcal{D}(T^{*}T)$ is dense in the Clifford module
$(\mathcal{D}(T),\|\cdot\|_{T})$. If $y\in\mathcal{D}(T)$ satisfies $\langle
y,x\rangle_{T}=0$ for all $x\in\mathcal{D}(T^{*}T)$, then realise that

$0=\langle y,x\rangle+\langle Ty,Tx\rangle=\langle y,(I+T^{*}T)x\rangle$

for every $x\in\mathcal{D}(T^{*}T)$. Consequently, $y=0$ since ${\rm
Ran}\,(I+T^{*}T)=\mathcal{H}_{n}$, in which case we have that
$\mathcal{D}(T^{*}T)$ is a core for $T$.

The second assertion is very obvious. ∎

######  Lemma 3.2.4.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be self-adjoint. Then
$\sigma_{S}(T)\subseteq\mathbb{R}$.

###### Proof.

In order to show that $\sigma_{S}(T)\subseteq\mathbb{R}$, it suffices to show
that $\rho_{S}(T)\subseteq\\{s\in\mathbb{R}^{n+1}:{\rm Re}(s)=0\\}$. Let
$s=s_{0}+s_{1}\in\mathbb{R}^{n+1}$, where $s_{0}\in\mathbb{R}$ and
$s_{1}\in{\rm Im}(\mathbb{R}^{n+1})\setminus\\{0\\}$. Notice that

$\mathcal{Q}_{s}(T):=T^{2}-2{\rm
Re}(s)T+|s|^{2}I=(T-s_{0}\,I)^{2}+|s_{1}|^{2}I$

and hence $\mathcal{D}(T^{2})=\mathcal{D}((T-s_{0}\,I)^{2})$ and as $T$ is
self-adjoint, we have that $(T-s_{0}\,I)^{2}+|s|^{2}I$ is self-adjoint.
Moreover, in view of the second assertion in Theorem 3.2.3(ii), we have that
$\mathcal{D}(T^{2})$ is dense in $\mathcal{H}_{n}$. We claim that ${\rm
Re}(\langle(T-s_{0}\,I)^{2}\,x,x\rangle)\geq 0$ for all
$x\in\mathcal{D}(T^{2})$. Indeed,

$\displaystyle{\rm Re}(\langle(T-s_{0}\,I)^{2}x,x\rangle=$
$\displaystyle\;{\rm Re}(\langle(T-s_{0}\,I)x,(T-s_{0}\,I)x\rangle)$
$\displaystyle=$ $\displaystyle\;\|(T-s_{0}\,I)x\|^{2}\geq 0\quad\quad{\rm
for}\quad x\in\mathcal{H}_{n}.$

Notice that

$\displaystyle\|\mathcal{Q}_{s}(T)x\|^{2}=$ $\displaystyle\;{\rm
Re}\langle\\{(T-s_{0}\,I)^{2}+|s_{1}|^{2}I\\}x,\\{(T-s_{0}\,I)^{2}+|s_{1}|^{2}\,I\\}x\rangle$
$\displaystyle=$ $\displaystyle\;\|(T-s_{0}\,I)^{2}x\|^{2}+2{\rm
Re}\langle(T-s_{0}\,I)^{2}x,|s_{1}|^{2}x\rangle+|s_{1}|^{4}\|x\|^{2}$
$\displaystyle=$
$\displaystyle\;\|(T-s_{0}\,I)^{2}x\|^{2}+2|s_{1}|^{2}\|(T-s_{0}\,I)x\|^{2}+|s_{1}|^{4}\|x\|^{2}$
(3.27) $\displaystyle\geq$ $\displaystyle\;|s_{1}|^{4}\|x\|^{2}\quad\quad{\rm
for}\quad x\in\mathcal{D}(T^{2})$

and hence $\mathcal{Q}_{s}(T)^{-1}:{\rm
Ran}\,\mathcal{Q}_{s}(T)\to\mathcal{D}(T^{2})$ is bounded (just take
$x=\mathcal{Q}_{s}(T)^{-1}y$ for $y\in{\rm Ran}\,\mathcal{Q}_{s}(T)$ in
(3.27)) and ${\rm Ker}\,\mathcal{Q}_{s}(T)=\\{0\\}.$ As $\mathcal{Q}_{s}(T)$
is self-adjoint, we have that

$\displaystyle\overline{{\rm Ran}\,\mathcal{Q}_{s}(T)}=$ $\displaystyle\;{\rm
Ran}\,(\mathcal{Q}_{s}(T)^{\perp})^{\perp}$ $\displaystyle=$
$\displaystyle\;{\rm Ker}\,(\mathcal{Q}_{s}(T)^{*})^{\perp}$ $\displaystyle=$
$\displaystyle\;{\rm Ker}\,\mathcal{Q}_{s}(T))^{\perp}$ $\displaystyle=$
$\displaystyle\;\\{0\\}^{\perp}=\mathcal{H}_{n}.$

Thus, for all $s=s_{0}+s_{1}$, with $s_{1}\in{\rm
Im}(\mathcal{H}_{n})\setminus\\{0\\}$, we have $s\in\rho_{S}(T)$, i.e.,
$\rho_{S}(T)\subseteq\mathbb{R}^{n+1}\setminus\mathbb{R}$, i.e.,
$\sigma_{S}(T)\subseteq\mathbb{R}$.

∎

######  Lemma 3.2.5.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be a positive operator. Then
$\sigma_{S}(T)\subseteq[0,\infty)$.

###### Proof.

A careful inspection of the proof of Lemma 3.2.4, bearing in mind the
additional hypothesis that $T$ is a positive operator, will reveal that for
all $s<0$, we have that
$\mathcal{Q}_{s}(T)^{-1}\in\mathcal{B}(\mathcal{H}_{n})$ and hence
$(-\infty,0)\subseteq\rho_{S}(T)$, i.e., $\sigma_{S}(T)\subseteq[0,\infty)$. ∎

######  Lemma 3.2.6.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be anti self-adjoint. Then

(3.28) $\displaystyle\sigma_{S}(T)\subseteq{\rm
Im}(\mathbb{R}^{n+1}):=\\{a\in\mathbb{R}^{n+1}:{\rm Re}(a)=0\\}.$

###### Proof.

Since $T=-T^{*}$, we have

$\displaystyle\|\mathcal{Q}_{s}(T)x\|^{2}=$ $\displaystyle\;{\rm Re}\langle
T^{2}x,T^{2}x\rangle+2(s_{0}^{2}-|s_{1}|^{2}){\rm Re}\langle
Tx,Tx\rangle+(s_{0}^{2}+|s_{1}|^{2}){\rm Re}\langle x,x\rangle$ (3.29)
$\displaystyle=$
$\displaystyle\;\|T^{2}x\|+2(s_{0}^{2}-|s_{1}|^{2})\|Tx\|^{2}+(s_{0}^{2}+|s_{1}|^{2})\|x\|^{2}\quad{\rm
for}\;x\in\mathcal{D}(T^{2}).$

Therefore, if $|s_{0}|\geq|s_{1}|$, then we have

(3.30) $\|\mathcal{Q}_{s}(T)x\|^{2}\geq s_{0}^{4}\|x\|^{2}\quad\quad{\rm
for}\quad x\in\mathcal{D}(T^{2})$

and we may proceed as in Lemma 3.2.4 to show that
$s=s_{0}+s_{1}\in\rho_{S}(T)$, i.e., (3.28) holds.

If $|s_{0}|<|s_{1}|$, then
$(s_{0}^{2}-|s_{1}|^{2})\|T^{2}x\|\|x\|\leq\|(s_{0}^{2}-|s_{1}|^{2})\|Tx\|^{2}$
for $x\in\mathcal{D}(T)$ together with (3.29) can be used to show that (3.30)
holds. But then we may proceed as above to obtain (3.28). ∎

######  Theorem 3.2.7.

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be a normal operator. Then we have

$\sigma_{PS}(T)=\sigma_{PS}(T^{*}),\ \ \ \ \ \
\sigma_{RS}(T)=\sigma_{RS}(T^{*})=0,\ \ \ \ \
\sigma_{CS}(T)=\sigma_{CS}(T^{*}).$

###### Proof.

Since $T$ is normal and $\mathcal{Q}_{s}(T)^{*}=\mathcal{Q}_{s}(T^{*})$ it is
clear that $\mathcal{Q}_{s}(T)^{*}$ is normal. For bounded linear operators
the kernel $T$ and the kernel of its adjoint are equal so

${\rm Ker}\,(\mathcal{Q}_{s}(T))={\rm Ker}\,(\mathcal{Q}_{s}(T^{*}))$

so by the definition of point $S$-spectrum se have

$\sigma_{PS}(T)=\sigma_{PS}(T^{*}).$

The fact that $\sigma_{RS}(T)=\sigma_{RS}(T^{*})=0$ follows by contradiction,
in fact if $0\not=s\in\sigma_{RS}(T)$ we get

$\\{0\\}={\rm Ker}\,(\mathcal{Q}_{s}(T))={\rm
Ker}\,(\mathcal{Q}_{s}(T^{*}))=({\rm
Ran}\,(\mathcal{Q}_{s}(T))^{\perp}\not=\\{0\\}.$

In the same way we can prove that $\sigma_{RS}(T^{*})=0$. Since $T$ and
$T^{*}$ have the same $S$-spectrum and the three components of the
$S$-spectrum, by definition, are pairwise disjoint it follows that
$\sigma_{CS}(T)=\sigma_{CS}(T^{*})$.

∎

##  4\. Measure theory and integration theory for $\mathbb{R}_{n}$-valued
measures

######  Definition 4.0.1 (positive $\mathbb{R}_{n}$-valued measure).

Let $\Omega$ be a non-empty set and $\mathscr{A}$ denote a $\sigma$-algebra on
$\Omega$. We will call $\mu:\mathscr{A}\to\mathbb{R}_{n}\cup\\{\infty\\}$
positive if $\mu(M)\succeq 0$ (see Definition 2.1.6) for every
$M\in\mathscr{A}$ such that $\mu(M)\neq\infty$ and $\mu$ is $\sigma$-additive,
i.e.,

(4.1)
$\mu\left(\bigcup_{i=1}^{\infty}M_{n}\right)=\sum_{i=1}^{\infty}\mu(M_{n})$

for every sequence $M:=(M_{i})_{i=1}^{\infty}$, where $M_{i}\cap
M_{j}=\emptyset$ for $i\neq j$ and $M_{i}\in\mathscr{A}$ for $i=1,2,\ldots$.
In this case, we shall write $\mu$ is $\mathfrak{P}(\mathbb{R}_{n})$-valued.

######  Definition 4.0.2 (finite, semi-finite and $\sigma$-finite
$\mathfrak{P}(\mathbb{R}_{n})$-valued measure).

Let $\Omega$ be a non-empty set and $\mathscr{A}$ denote a $\sigma$-algebra on
$\Omega$. We will call a $\mathfrak{P}(\mathbb{R}_{n})$-valued measure $\mu$
finite if $\mu(\Omega)<\infty$, semi-finite if for every $M\in\mathscr{A}$
such that

$\mu(M)=\infty$, there exists $N\in\mathscr{A}$ such that $\mu(N)<\infty$ and
$N\subseteq M$

and $\sigma$-finite if

$\Omega=\bigcup_{i=1}^{\infty}M_{i},$

where $\mu(M_{i})<\infty$ for $i=1,2,\ldots$. Note that if $\mu$ is finite,
then the finite additivity of $\mu$ together with $\mu(\Omega)<\infty$ implies
that $\mu(M)<\infty$ for every $M\in\mathscr{A}$.

######  Definition 4.0.3 (Borel measure).

Let $X$ be a Hausdorff space. We will call a
$\mathfrak{P}(\mathbb{R}_{n})$-valued measure $\mu$ on the Borel
$\sigma$-algebra generated by $X$ a positive $\mathbb{R}_{n}$-valued Borel
measure.

######  Definition 4.0.4 ($\mathbb{R}_{n}$-valued measure).

Let $\Omega$ be a non-empty set and $\mathscr{A}$ denote a $\sigma$-algebra on
$\Omega$. We will call $\mu:\mathscr{A}\to\mathbb{R}_{n}$ a
$\mathbb{R}_{n}$-valued measure if $\mu$ is $\sigma$-additive. In this case,
$\mu$ has the Jordan decomposition

(4.2)
$\mu=\sum_{\alpha}\\{\mu^{(\alpha)}_{+}-\mu^{(\alpha)}_{-}\\}e_{\alpha},$

where $\mu_{\pm}^{(\alpha)}$ are positive measures (in the usual sense) for
every
$\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})$. The support of
a $\mathbb{R}_{n}$-valued measure $\mu$ shall be denoted by ${\rm supp}\,\mu$
and is defined as the set $N$ which satisfies

$\mu(M)=\mu(M\cap N)\quad\quad{\rm for}\quad M\in\mathscr{A}.$

We will call $\mu$ finite if $\mu(\Omega)<\infty$.

######  Definition 4.0.5 (Integral with respect to a $\mathbb{R}_{n}$-valued
measure).

Let $\mu$ be a finite $\mathbb{R}_{n}$-valued measure on a non-empty set
$\Omega$, $\mathscr{A}$ be a $\sigma$-algebra generated by $\Omega$,
$\mathfrak{I}\in\mathbb{S}$ and
$\mu=\sum_{\alpha}\\{\mu_{+}^{(\alpha)}-\mu_{-}^{(\alpha)}\\}e_{\alpha}$ be
the Jordan decomposition for $\mu$. Then for any $\mathscr{A}$-measurable
function $f:\Omega\to\mathbb{R}$, we shall define

(4.3)
$\int_{\Omega}f(\lambda)\,d\mu(\lambda):=\sum_{\alpha}\left(\int_{\Omega}f(\lambda)\,d\mu_{+}^{(\alpha)}-\int_{\Omega}f(\lambda)\,d\mu_{-}^{(\alpha)}\right)e_{\alpha}$

and for any $\mathscr{A}$-measurable function
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}$, we shall define

$\displaystyle\int_{\Omega}f(\lambda)\,d\mu(\lambda):=$
$\displaystyle\;\sum_{\alpha}\left(\int_{\Omega}{\rm
Re}(f(\lambda))\,d\mu_{+}^{(\alpha)}-\int_{\Omega}{\rm
Re}(f(\lambda))\,d\mu_{-}^{(\alpha)}\right)e_{\alpha}$ (4.4) $\displaystyle+$
$\displaystyle\;\sum_{\alpha}\left(\int_{\Omega}{\rm
Im}(f(\lambda))\,d\mu_{+}^{(\alpha)}-\int_{\Omega}{\rm
Im}(f(\lambda))d\mu_{-}^{(\alpha)}\right)\,\mathfrak{I}\,e_{\alpha},$

provided that all of the four integrals on the right-hand side exist and we do
not end up with the indeterminate expression $\infty-\infty$. Similarly, for
any $\mathscr{A}$-measurable function $f:\Omega\to\mathbb{C}_{\mathfrak{I}}$,
we can define an integral with $f(\lambda)$ on the right via

$\displaystyle\int_{\Omega}d\mu(\lambda)\,f(\lambda):=$
$\displaystyle\;\sum_{\alpha}\left(\int_{\Omega}{\rm
Re}(f(\lambda))\,d\mu_{+}^{(\alpha)}-\int_{\Omega}{\rm
Re}(f(\lambda))\,d\mu_{-}^{(\alpha)}\right)e_{\alpha}$ (4.5) $\displaystyle+$
$\displaystyle\;\sum_{\alpha}\left(\int_{\Omega}{\rm
Im}(f(\lambda))\,d\mu_{+}^{(\alpha)}-\int_{\Omega}{\rm
Im}(f(\lambda))\,d\mu_{-}^{(\alpha)}\right)\,\mathfrak{I}\,e_{\alpha}.$

######  Definition 4.0.6 (measure space and $\mu$-integrability).

Let $\Omega$ be a non-empty set and $\mathscr{A}$ denote a $\sigma$-algebra on
$\Omega$. Suppose $\mu=\sum_{\alpha}\mu^{(\alpha)}e_{\alpha}$ is a
$\mathfrak{P}(\mathbb{R}_{n})$-valued measure on $\mathscr{A}$. We shall call
the triple $(\Omega,\mathscr{A},\mu)$ a measure space. We will write that a
property holds $\mu$-a.e. on $\Omega$ whenever the desired property holds
except on a set $M\in\mathscr{A}$, where $\mu(M)=0$. A function
$f:\Omega\to\mathbb{R}\cup\\{\infty\\}$ is called measurable if
$\\{\lambda\in\Omega:f(\lambda)\leq t\\}\in\mathscr{A}$ for every
$t\in\mathbb{R}$. We will call a function
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}\cup\\{\infty\\}$, where
$\mathfrak{I}\in\mathbb{S}$, $\mu$-integrable if
$\int_{\Omega}f(\lambda)d\mu^{(\alpha)}(\lambda)$ converge for all
$\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\}).$

######  Theorem 4.0.7.

Let $X$ be a compact Hausdorff space and $\mathscr{C}(X,\mathbb{R})$ denote
the normed space of real-valued continuous functions on $X$ together with the
supremum norm $\|\cdot\|_{\infty}$. Corresponding to any bounded positive
linear functional $\mathscr{L}:\mathscr{C}(X,\mathbb{R})\to\mathbb{R}$, there
exists a unique positive Borel measure $\mu$ on $X$ such that

$\mathscr{L}(p)=\int_{X}p(t)\,d\mu(t)\quad\quad{\rm for}\quad
p\in\mathscr{C}(X,\mathbb{R}).$

In this case, $\mu(M)\leq\|\mathscr{L}\|$ for every set $M$ that belongs to
the Borel $\sigma$-algebra generated by $X$, i.e., $\mathscr{B}(X)$.

###### Proof.

The existence and uniqueness of $\mu$ is a special case of Theorem D in
Section 56 of [42]. The last assertion follows immediately from the fact that

$\|\mathscr{L}\|\geq\mathscr{L}(1)=\int_{X}d\mu(t)=\mu(X)\geq\mu(M)\quad\quad{\rm
for}\quad M\in\mathscr{B}(X).$

∎

######  Corollary 4.0.8.

Let $X$ be a compact Hausdorff space. Corresponding to any bounded linear
functional $\mathscr{L}:\mathscr{C}(X,\mathbb{R})\to\mathbb{R}_{n}$, there
exists a unique $\mathbb{R}_{n}$-valued Borel measure $\mu$ on $X$ such that

(4.6) $\mathscr{L}(p)=\int_{X}p(t)\,d\mu(t)\quad\quad{\rm for}\quad
p\in\mathscr{C}(X,\mathbb{R}).$

In this case, $|\mu(M)|\leq\|\mathscr{L}\|$ for every set
$M\in\mathscr{B}(X)$.

###### Proof.

Write
$\mathscr{L}(f)=\sum_{\alpha}\\{\mathscr{L}_{+}^{(\alpha)}(f)-\mathscr{L}_{-}^{(\alpha)}(f)\\}e_{\alpha}$,
where $\mathscr{L}_{\pm}^{(\alpha)}:\mathscr{C}(X,\mathbb{R})\to\mathbb{R}$
are positive linear functionals for
$\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})$. It is easy to
see that $\mathscr{L}_{\pm}^{(\alpha)}$ are bounded linear functionals and we
may apply Theorem 4.0.7 to obtain the unique positive Borel measures
$\mu^{(\alpha)}_{\pm}$ so that

$\mathscr{L}_{\pm}^{(\alpha)}(f)=\int_{X}f(t)d\mu_{\pm}^{(\alpha)}(t)\quad\quad{\rm
for}\quad f\in\mathscr{C}(X,\mathbb{R}).$

Thus, if we let
$\mu=\sum_{\alpha}\\{\mu_{+}^{(\alpha)}-\mu_{-}^{(\alpha)}\\}e_{\alpha}$, then
we obtain (4.6).

To see that $\mu$ is unique, suppose there is another $\mathbb{R}_{n}$-valued
measure $\nu$ on $X$ so that (4.6) holds. Write the Jordan decomposition
$\nu=\sum_{\alpha}\\{\nu^{(\alpha)}_{+}-\nu^{(\alpha)}_{-}\\}e_{\alpha}$,
where $\nu^{(\alpha)}_{\pm}$ are positive Borel measures on $X$. Then, for any
$\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})$, we must have
$\mu_{\pm}^{(\alpha)}=\nu_{\pm}^{(\alpha)}$. Thus, $\mu=\nu$.

The final assertion can be shown in much the same way as the final assertion
of Theorem 4.0.7. ∎

######  Definition 4.0.9 (transformation of a $\mathbb{R}_{n}$-valued
measure).

Suppose $\mu$ is a positive $\mathbb{R}_{n}$-valued measure on a
$\sigma$-algebra of sets $\mathscr{A}$ generated by a set $\Omega$ and
$\psi:\Omega\to\Omega^{\prime}$. Let $\mathscr{A}^{\prime}$ be the family of
sets given by $M^{\prime}\in\mathscr{A}^{\prime}$ if
$\psi^{-1}(M^{\prime})\in\mathscr{A}$. Notice that $\mathscr{A}^{\prime}$ is a
$\sigma$-algebra on $\Omega^{\prime}$ and
$\mu^{\prime}(M^{\prime}):=\mu(\psi^{-1}(M^{\prime}))$ is a positive
$\mathbb{R}_{n}$-valued measure on $\mathscr{A}^{\prime}$.

######  Theorem 4.0.10.

Let $\Omega,\Omega^{\prime},\mathscr{A},\mathscr{A}^{\prime},\mu$ and
$\mu^{\prime}$ be as in Definition 4.0.9. Suppose
$f:\Omega^{\prime}\to\mathbb{C}_{\mathfrak{I}}$, where
$\mathfrak{I}\in\mathbb{S}$, is a $\mu^{\prime}$-a.e. finite
$\Omega^{\prime}$-measurable function. Then $f\circ\psi$ is a $\mu$-a.e.
finite $\mathscr{A}$-measurable function on $\Omega$ and

(4.7)
$\int_{\Omega^{\prime}}f(\lambda^{\prime})d\mu^{\prime}(\lambda^{\prime})=\int_{\Omega}f(\psi(\lambda))d\mu(\lambda).$

###### Proof.

Let
$\mu^{\prime}=\sum_{\alpha}\\{\nu_{+}^{(\alpha)}-\nu_{-}^{(\alpha)}\\}e_{\alpha}$
and $\mu=\sum_{\alpha}\\{\mu_{+}^{(\alpha)}-\mu_{-}^{(\alpha)}\\}e_{\alpha}$
be Jordan decompositions (see 4.2) for $\mu^{\prime}$ and $\mu$, respectively.
Applying the classical result to $\nu_{\pm}^{(\alpha)}$ and
$\mu_{\pm}^{(\alpha)}$ (which are just positive measures in the usual sense),
see, e.g., Theorem C in Section 39 of [42], we obtain both assertions. ∎

######  Theorem 4.0.11.

Let $X$ be a compact Hausdorff space. Corresponding to any bounded positive
linear functional
$\mathscr{L}:\mathscr{C}(X,\mathbb{R})\to\mathfrak{S}(\mathbb{R}_{n})$, there
exists a unique positive $\mathbb{R}_{n}$-valued Borel measure $\mu$ on $X$
such that

(4.8) $\mathscr{L}(p)=\int_{X}p(t)\,d\mu(t)\quad\quad{\rm for}\quad
p\in\mathscr{C}(X,\mathbb{R}).$

In this case, $|\mu(M)|\leq\|\mathscr{L}\|$ for every set
$M\in\mathscr{B}(X)$.

###### Proof.

First, let $K\geq 0$ be such that

$|\mathscr{L}(p)|\leq K\|p\|_{\infty}\quad\quad{\rm for}\;{\rm
every}\;p\in\mathscr{C}(X,\mathbb{R}).$

Next, note that for every $p\in\mathscr{C}(X,\mathbb{R})$, we may uniquely
decompose $\mathscr{L}(p)$ as
$\mathscr{L}(p)=\sum_{\alpha}\mathscr{L}_{\alpha}(p)e_{\alpha}$, where
$\mathscr{L}_{\alpha}(p)\in\mathbb{R}$. Let
$L(p):\mathscr{C}(X,\mathbb{R})\to\chi(\mathbb{R}_{n})$ given by
$L(p)=\chi(\mathscr{L}(p))$ for $p\in\mathscr{C}(X,\mathbb{R})$. Since $\chi$
is a $*$-homomorphism and
$\mathscr{L}:\mathscr{C}(X,RR)\to\mathfrak{S}(\mathbb{R}_{n})$, we have that
$L(p)$ is a real Hermitian matrix for every $p\in\mathscr{C}(X,\mathbb{R})$.
Moreover, in view of the first assertion in Remark 2.1.8 and the estimate

$\displaystyle\|L(p)\|_{\infty}=$
$\displaystyle\;\|\chi(\mathscr{L}(p))\|=\sum_{\alpha}|\mathscr{L}_{\alpha}(p)|$
$\displaystyle\leq$
$\displaystyle\;2^{n/2}(\sum_{\alpha}\mathscr{L}_{\alpha}(p)^{2})^{1/2}$
$\displaystyle=$ $\displaystyle\;2^{n/2}|\mathscr{L}(p)|$ $\displaystyle\leq$
$\displaystyle\;2^{n/2}K\|p\|_{\infty}$

where $\|\cdot\|_{\infty}$ denotes the maximum row sum norm of a matrix, we
have that $L(p)$ is a bounded positive $\mathbb{R}^{2^{n}\times 2^{n}}$-valued
linear functional. Consequently, a finite dimensional version of the operator-
valued version of the Riesz representation theorem (see, e.g., Theorem 19 in
[8]) gives us the existence of a uniquely determined positive
$\mathbb{R}^{2^{n}\times 2^{n}}$-valued measure
$\nu:=(\nu_{ik})_{i,j=1}^{2^{n}}$ such that

$L(p)=\int_{X}p(t)\,d\nu(t):=\left(\int_{X}p(t)d\nu_{ij}(t)\right)_{i,j=1}^{2^{n}}\quad\quad{\rm
for}\quad p\in\mathscr{C}(X,\mathbb{R}).$

If we write $L(p)=(L_{ij}(p))_{i,j=1}^{2^{n}}$, then
$L_{ij}:\mathscr{C}(X,\mathbb{R})\to\mathbb{R}$ is a bounded linear functional
such that

(4.9) $L_{ij}(p)=\int_{X}p(t)d\nu_{ij}(t)\quad\quad{\rm for}\quad
p\in\mathscr{C}(X,\mathbb{R}).$

Consequently, Corollary 4.0.8, with $n=0$, asserts that $\nu_{ij}$ is the only
$\mathbb{R}$-valued Borel measure such that (4.9) holds. If we use the fact
that $L(p)=\chi(\mathscr{L}(p))$ and the aforementioned uniqueness of the
$\mathbb{R}$-valued measure in (4.9), then for any $M\in\mathscr{B}(X)$, we
have that $\nu(M)=\chi(a_{M})$ for some $a_{M}\in\mathbb{R}_{n}$. In view of
Remark 2.1.8, the fact that $\nu(M)$ is a positive semidefinite matrix implies
and $\nu(M)=\chi(a_{M})$ implies that $a_{M}\in\mathfrak{P}(\mathbb{R}_{n})$
for all $M\in\mathscr{B}(X)$. It is easy to check that
$\mu(M):=a_{M}=\chi^{-1}(\nu(M))$ is a positive $\mathbb{R}_{n}$-valued
measure with the property that

$\mathscr{L}(p)=\chi^{-1}(L(p))=\int_{X}p(t)\,d\mu(t)\quad\quad{\rm for}\quad
p\in\mathscr{C}(X,\mathbb{R}),$

i.e., (4.8) holds.

The uniqueness of $\mu$ such that (4.8) holds follows immediately from the
uniqueness of $\nu$ and the injectivity of $\chi$. The final assertion can be
proved in much the same way as the proof of the final assertion in Theorem
4.0.7. ∎

##  5\. Spectral integrals

Throughout this section, $\mathscr{A}$ will denote an algebra of subsets of
$\Omega$, $\mathcal{H}_{n}$ will denote a Clifford module over
$\mathbb{R}_{n}$ and all infinite sums of operators in
$\mathcal{B}(\mathcal{H}_{n})$ will be meant in the strong operator topology,
i.e., $\sum_{k=1}^{\infty}T_{k}=T$ if and only if

$Tx=\lim_{k\to\infty}\sum_{j=1}^{k}T_{j}x\quad\quad{\rm for}\quad
x\in\mathcal{H}_{n}.$

###  5.1. Basic properties

######  Definition 5.1.1 (orthogonal projection).

An operator $T\in\mathcal{B}(\mathcal{H}_{n})$ is called an orthogonal
projection if $T$ is self-adjoint and $T^{2}=T$. The set of orthogonal
projections on $\mathcal{H}_{n}$ will be denoted by
$\mathcal{P}(\mathcal{H}_{n})$.

######  Definition 5.1.2 (spectral premeasure and spectral measure).

Let $\mathscr{A}$ be an algebra of subsets in $\Omega$. We will call
$E:\mathscr{A}\to\mathcal{P}(\mathcal{H}_{n})$ a spectral premeasure if the
following conditions hold:

  1. (i)

$E$ is countably additive, i.e.,

$E\left(\bigcup_{i=1}^{\infty}M_{i}\right)=\sum_{i=1}^{\infty}E(M_{i})$

for every sequence of mutually disjoint sets $(M_{i})_{i=1}^{\infty}$ such
that $M_{i}\in\Omega$ for $i=1,2,\ldots$ and
$\cup_{i=1}^{\infty}M_{i}\in\Omega$.

  2. (ii)

$E(\Omega)=I$.

If $\mathscr{A}$ is a $\sigma$-algebra, then $E$ will be called a spectral
measure. In this case, we will write that $E$ is a spectral premeasure (resp.,
measure) on $(\Omega,\mathscr{A})$.

######  Lemma 5.1.3.

Let $\mathscr{A}$ be an algebra of sets in $\Omega$. Suppose
$E:\mathscr{A}\to\mathcal{P}(\mathcal{H}_{n})$ is a finitely additive map,
i.e.,

$E\left(\bigcup_{i=1}^{k}M_{i}\right)=\sum_{i=1}^{k}E(M_{i})$

for every collection of mutually disjoint sets $(M_{n})_{i=1}^{k}$ such that
$M_{i}\in\Omega$ for $i=1,\ldots,k$ and $\bigcup_{i=1}^{k}M_{i}\in\Omega$.
Then

(5.1) $E(M)E(N)=E(M\cap N)\quad\quad{\rm for}\quad M,N\in\mathscr{A}.$

In particular, we have $E(M)E(N)=0$ if $M\cap N=\emptyset$.

###### Proof.

The proof can be carried out in the same way as the complex Hilbert space case
(see, e.g., Lemma 4.3 in [66]). ∎

######  Lemma 5.1.4.

Suppose $\mathscr{A}$ is an algebra of sets in $\Omega$. The mapping
$E:\mathscr{A}\to\mathcal{P}(\mathcal{H}_{n})$ is a spectral premeasure if and
only if the following conditions hold:

  1. (i)

For any $x\in\mathcal{H}_{n}$, the set function $E_{x}(\cdot)$ given by

$E_{x}(M):=\langle E(M)x,x\rangle,$

where $M\in\mathscr{A}$, is a countably additive
$\mathfrak{P}(\mathbb{R}_{n})$-valued set function.

  2. (ii)

$E(\Omega)=I$.

###### Proof.

Suppose $E$ is a spectral premeasure. Then, by Definition 5.1.2, (ii) holds.
It follows from Lemma 5.1.3 with $M=N$ that $E(M)^{2}x=E(M)x$ for
$x\in\mathcal{H}_{n}$ and $M\in\mathscr{A}$, we have

$E_{x}(M)=\langle
E(M)x,E(M)x\rangle\in\mathfrak{P}(\mathbb{R}_{n})\quad\text{for
$M\in\mathscr{A}$ and $x\in\mathcal{H}_{n}$.}$

Thus, (i) holds.

Conversely, suppose (i) and (ii) hold. Suppose $(M_{i})_{i=1}^{\infty}$ be a
sequence of mutually disjoint sets such that $M_{i}\in\mathscr{A}$ for
$i=1,2,\ldots$ and hence $M:=\cup_{i=1}^{\infty}M_{n}\in\mathscr{A}$, since
$\mathscr{A}$ is a $\sigma$-algebra. For any $x\in\mathcal{H}_{n}$, we note
that $E_{x}$ is finitely additive and hence $E$ is finitely additive as well.
Thus, we may use Lemma 5.1.3 to deduce that $(E(M_{i}))_{i=1}^{\infty}$ is a
sequence of orthogonal projections such that $E(M_{j})E(M_{i})=0$ whenever
$j\neq n$. Consequently, $\sum_{i=1}^{\infty}E(M_{i})$ has a limit in the
strong operator topology of $\mathcal{B}(\mathcal{H}_{n})$. Since

$\displaystyle E_{x}(M)=$ $\displaystyle\;\sum_{i=1}^{\infty}E_{x}(M_{i})$
$\displaystyle=$ $\displaystyle\;\sum_{i=1}^{\infty}\langle
E(M_{i})x,x\rangle$ $\displaystyle=$
$\displaystyle\;\left\langle\sum_{i=1}^{\infty}E(M_{i})x,x\right\rangle\quad\quad{\rm
for}\quad x\in\mathcal{H}_{n},$

we may use the polarisation formula (2.20) to obtain

$\langle
E(M)x,y\rangle=\langle\sum_{i=1}^{\infty}E(M_{i})x,y\rangle\quad\quad{\rm
for}\quad x,y\in\mathcal{H}_{n},$

i.e., $E(M)=\sum_{i=1}^{\infty}E(M_{i})$. Thus, $E$ is a spectral premeasure.
∎

######  Remark 5.1.5.

In the event that $\mathscr{A}$ is a $\sigma$-algebra, a careful inspection of
the proof of Lemma 5.1.4 shows that we can easily adapt the proof to obtain
the following result where spectral premeasure and countably additive
$\mathfrak{P}(\mathbb{R}_{n})$-valued set function are replaced by spectral
measure and $\mathfrak{P}(\mathbb{R}_{n})$-valued measure, respectively.

######  Lemma 5.1.6.

Suppose $\mathscr{A}$ is a $\sigma$-algebra of sets in $\Omega$. The mapping
$E:\mathscr{A}\to\mathcal{P}(\mathcal{H}_{n})$ is a spectral measure if and
only if the following conditions hold:

  1. (i)

For any $x\in\mathcal{H}_{n}$, the set function $E_{x}(\cdot)$ given by
$E_{x}(M):=\langle E(M)x,x\rangle$, where $M\in\mathscr{A}$, is a
$\mathfrak{P}(\mathbb{R}_{n})$-valued measure.

  2. (ii)

$E(\Omega)=I$.

Moreover, $E_{x}(\cdot)$ is a finite $\mathfrak{P}(\mathbb{R}_{n})$-valued
measure.

###### Proof.

See Remark 5.1.5 for the characterisation of spectral measures. The second
assertion is a direct consequence of $E(\Omega)=I$. Indeed,
$E_{x}(\Omega)=\langle E(\Omega)x,x\rangle=\langle x,x\rangle<\infty$. ∎

######  Remark 5.1.7.

Fix $x,y\in\mathcal{H}_{n}$ and let $E$ be a spectral measure on
$\mathscr{A}$. Then consider the $\mathbb{R}_{n}$-valued measure
$E_{x,y}(M):=\langle E(M)x,y\rangle$ for $M\in\mathscr{A}$. One can use the
polarisation formula (2.20) to verify

(5.2)
$E_{x,y}=\frac{\sum_{\alpha}(E_{x+ye_{\alpha}}-E_{x-ye_{\alpha}})e_{\alpha}}{4\dim\mathfrak{S}(\mathbb{R}_{n})}.$

It turns out that the $\mathbb{R}_{n}$-valued measure $E_{x,y}$ has the
properties detailed in the following lemma.

######  Lemma 5.1.8.

Let $\mathscr{A}$ be an algebra of sets in $\Omega$. Suppose $E$ is a spectral
premeasure on $\mathscr{A}$. Then for any sequences of sets
$(M_{i})_{i=1}^{\infty}$ and $(N_{i})_{i=1}^{\infty}$ in $\mathscr{A}$ such
that $M_{i+1}\subseteq M_{i}$, $N_{i}\subseteq N_{i+1}$ for $i=1,2,\ldots$ and
$M:=\cap_{i=1}^{\infty}M_{i}\in\mathscr{A}$, we have

(5.3) $E(M)=s-\lim_{i\to\infty}E(M_{i})\quad{\rm and}\quad
E(N)=s-\lim_{i\to\infty}E(N_{i}),$

where $N:=\cup_{i=1}^{\infty}N_{i}$.

###### Proof.

The proof is very straight forward. ∎

We will now define the support of a spectral measure. We shall suppose
$\Omega$ is a Hausdorff topological space which has a countable base of open
sets. In what follows, $\mathscr{B}(\Omega)$ will denote the Borel
$\sigma$-algebra of sets generated by $\Omega$, i.e., the smallest
$\sigma$-algebra which contains all open sets in $\Omega$. For the sake of
brevity, we shall sometimes write that $E$ is a spectral measure on
$\mathscr{B}(\Omega)$ in place of $E$ is a spectral measure on
$(\Omega,\mathscr{B}(\Omega))$.

######  Definition 5.1.9 (support of a spectral measure).

Let $E$ be a spectral measure on $\mathscr{B}(\Omega)$. We shall define the
support of $E$ by

${\rm
supp}\,E:=\Omega\setminus\bigcup_{\stackrel{{\scriptstyle\text{$M\in\mathscr{B}(\Omega)$
open}}}{{E(M)=0}}}M.$

The following lemma will be important when proving the bounded case for the
spectral theorem for normal operators on a Clifford module.

######  Lemma 5.1.10.

Let $E$ be a spectral measure on $(\Omega,\mathscr{B}(\Omega))$. Then

(5.4) ${\rm supp}\,E=\\{\lambda\in\Omega:\text{$E(M)\neq 0$ for every open set
$M\in\mathscr{B}(\Omega)$ with $\lambda\in M$}\\}.$

###### Proof.

The characterisation (5.4) is a straight forward consequence of Definition
5.1.9. ∎

Let $\mathscr{A}$ be an algebra of sets in $\Omega$. For any
$\mathbb{R}_{n}$-valued measure $\nu$ on $\Omega$, we shall let the total
variation of $\nu$ be given by

(5.5) $|\nu|:=\sup_{(M_{i})_{i=1}^{k}}|\nu(M_{i})|,$

where $(M_{i})_{i=1}^{k}$ is any sequence of mutually disjoint sets with
$M_{i}\in\mathscr{A}$ for $i=1,\ldots,k$ and $k\in\mathbb{N}$ is arbitrary.

######  Definition 5.1.11 (An $L_{2}$-space with respect to
$\mathbb{R}_{n}$-valued measure).

Let $\mathscr{A}$ be an algebra of sets in $\Omega$,
$\mathfrak{I}\in\mathbb{S}$ and $\nu$ be a $\mathbb{R}_{n}$-valued measure on
$\mathscr{A}$. Then we shall let

$L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},\nu):=\\{\text{$\mathscr{A}$-measurable
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}$}:\int_{\Omega}|f(\lambda)|^{2}d|\nu|(\lambda)<\infty\\}.$

For $x\in\mathcal{H}_{n}$ and a spectral measure $E$ on $\mathscr{B}(\Omega)$
we let $\mu_{x}(M)=\langle E(M)x,x\rangle$ for $M\in\mathscr{B}(\Omega)$ and

$L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x}):=\\{\text{$\mathscr{A}$-measurable
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}$}:\int_{\Omega}|f(\lambda)|^{2}d|\mu_{x}|(\lambda)<\infty\\}.$

######  Remark 5.1.12.

Let $\Omega,\mathscr{A},\mathfrak{I}$, $E$, $x$, $E_{x}$ and $\nu$ be as in
Definition 5.1.11. Write the Jordan decomposition
$\nu=\sum_{\alpha}=\\{\nu_{+}^{(\alpha)}-\nu_{-}^{(\alpha)}\\}e_{\alpha}$ for
$\nu$. Then it is easy to see that

(5.6)
$L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},\nu)=\bigcap_{\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})}\\{L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},\nu_{+}^{(\alpha)})\cap
L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},\nu_{-}^{(\alpha)})\\},$

and

$\displaystyle
L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x})=\\{\text{$\mathscr{A}$-measurable
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}$}:$
$\displaystyle\;\int_{\Omega}|f(\lambda)|^{2}d({\rm Re}\mu_{x}(\lambda))$
(5.7) $\displaystyle<$ $\displaystyle\;\infty\\},$

where $\mu_{x}(M):=\langle E(M)x,x\rangle$ for $M\in\mathscr{A}$.

######  Lemma 5.1.13.

Let $E^{\prime}$ be a spectral premeasure on $(\Omega,\mathscr{A}^{\prime})$.
Then there exists a spectral measure $E$ on $(\Omega,\mathscr{A})$, where
$\mathscr{A}$ is the $\sigma$-algebra generated by $\mathscr{A}^{\prime}$,
such that $E(M)=E^{\prime}(M)$ for all $M\in\mathscr{A}^{\prime}$.

###### Proof.

For any $x\in\mathcal{H}_{n}$, we may use Lemma 5.1.4 to see that
$\mu^{\prime}_{x}(\cdot)$ given by $\mu^{\prime}_{x}(M)=\langle
E(M)x,x\rangle$ is a premeasure on $\mathscr{A}^{\prime}$. Moreover, using
Lemma 2.2.6, we obtain

$\displaystyle|\mu^{\prime}_{x}(\Omega)|=$
$\displaystyle\;|\langle\mu^{\prime}(\Omega)x,x\rangle|=|\langle x,x\rangle|$
$\displaystyle\leq$ $\displaystyle\;\|x\|^{2}.$

Thus, $\mu^{\prime}_{x}$ is a finite $\mathbb{R}_{n}$-valued premeasure on
$\mathscr{A}^{\prime}$ and hence all of the positive measure components in the
Jordan decomposition of

$\mu^{\prime}_{x}=\sum_{\alpha}(\mu^{\prime}_{x,\alpha,+}-\mu^{\prime}_{x,\alpha,-})e_{\alpha},$

namely
$(\mu_{x,\alpha,\pm})_{\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})}$,
are also finite positive measures. Thus, we may appeal to Theorem A in §13 of
[42] to produce uniquely determined measures
$(\mu_{x,\alpha,+})_{\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})}$
on the $\sigma$-algebra $\mathscr{A}$. Consequently,
$\mu_{x}:=\sum_{\alpha}(\mu_{x,\alpha,+}-\mu_{x,\alpha,-})e_{\alpha}$ is the
unique $\mathfrak{S}(\mathbb{R}_{n})$-valued measure on $\mathscr{A}$ such
that $\mu_{x}(M)=\mu^{\prime}_{x}(M)$ for $M\in\mathscr{A}^{\prime}$.

Let $x,y\in\mathcal{H}_{n}$ and $\mu_{x,y}$ be the $\mathbb{R}_{n}$-valued
measure given by

(5.8)
$\mu_{x,y}(M):=\frac{\sum_{\alpha}(\mu_{x+ye_{\alpha}}-\mu_{x-ye_{\alpha}})e_{\alpha}}{4\dim\mathfrak{S}(\mathbb{R}_{n})}\quad\quad{\rm
for}\quad M\in\mathscr{A}.$

Next, let

$\tilde{\mathscr{A}}:=\\{M\in\mathscr{A}:\text{the map $x\mapsto\mu_{x,y}(M)$
is right linear for all $y\in\mathcal{H}_{n}$}\\}.$

One can use (5.8) and Lemma 5.1.8 to show that $\tilde{\mathscr{A}}$ has the
following property. The union $\cup_{m=1}^{\infty}M_{i}\in\tilde{\mathscr{A}}$
for every sequence of sets $(M_{i})_{m=1}^{\infty}$ such that
$M_{i}\in\tilde{\mathscr{A}}$ for $i=1,2,\ldots$ with $M_{i}\subseteq
M_{i+1}$. Thus, as $\mathscr{A}^{\prime}\subseteq\tilde{\mathscr{A}}$, we may
use Theorem B in §6 of [42] to deduce that
$\mathscr{A}\subseteq\tilde{\mathscr{A}}$ and hence
$\tilde{\mathscr{A}}=\mathscr{A}$.

In much the same way as above, one can show that the map
$y\mapsto\mu_{x,y}(M)$ is anti-right linear, i.e.,
$\mu_{x,ya+z}(M)=\bar{a}\mu_{x,y}(M)+\mu_{x,z}(M)$ for all $M\in\mathscr{A}$
and $x,z\in\mathcal{H}_{n}$. It is easy to check that

$\mu^{\prime}_{x+xe_{\alpha}}(M)-\mu^{\prime}_{x-xe_{\alpha}}(M)=2(\mu^{\prime}_{x,x}(M)e_{\alpha}-e_{\alpha}\mu^{\prime}_{x,x}(M))$

and hence by the uniqueness of $\mu$ we have

$\mu_{x+xe_{\alpha}}(M)-\mu_{x-xe_{\alpha}}(M)=2(\mu_{x}(M)e_{\alpha}-e_{\alpha}\mu_{x}(M)).$

Consequently,

$\mu_{x,x}(M)=2\sum_{\alpha}(\mu_{x}(M)-e_{\alpha}\mu_{x}(M)e_{\alpha}),$

in which case $\mu_{x,x}(M)\in\mathfrak{P}(\mathbb{R}_{n})$ and ${\rm
Re}\,\mu_{x,x}(M)\geq 0$ for all $M\in\mathscr{A}$ and $x\in\mathcal{H}_{N}$.
For any $M\in\mathscr{A}$, we may use the polarisation formula (2.20) to check
that $\langle E^{\prime}(M)x,y\rangle=\mu^{\prime}_{x,y}(M)$ and Lemma
2.2.6(ii) to deduce

$\displaystyle|\mu_{x,y}(M)|\leq$ $\displaystyle\;|\mu_{x,y}(\Omega)|$
$\displaystyle=$ $\displaystyle\;|\mu^{\prime}_{x,y}(\Omega)|$
$\displaystyle=$ $\displaystyle\;|\langle
E^{\prime}(\Omega)x,y\rangle|=|\langle x,y\rangle$ $\displaystyle=$
$\displaystyle\;\|x\|^{2}\|y\|^{2}\quad\quad{\rm for}\quad
x,y\in\mathcal{H}_{n}.$

Thus, for any $M\in\mathscr{A}$, we may invoke Theorem 3.0.26 to obtain a
positive operator $E(M)\in\mathcal{B}(\mathcal{H}_{n})$ such that $\langle
E(M)x,y\rangle=\mu_{x,y}(M)$ for $x,y\in\mathcal{H}_{n}$. For
$M\in\mathscr{A}^{\prime}$, we have

$E^{\prime}(M)x,x\rangle=\langle E(M)x,x\rangle\quad\quad{\rm for}\quad
x\in\mathcal{H}_{n}$

and, hence, one can use the polarisation formula (2.20) to check that
$E^{\prime}(M)=E(M)$ for $M\in\mathscr{A}^{\prime}$.

Finally, let
$\mathscr{P}:=\\{M\in\mathscr{A}:E(M)\in\mathcal{P}(\mathcal{H}_{n})\\}$. It
is very easy to see that $\mathscr{P}$ has the monotone property and hence
$\mathscr{A}^{\prime}\subseteq\mathscr{P}$. But then
$\mathscr{A}\subseteq\mathscr{P}$. Thus, we may use Lemma 5.1.6 to deduce that
$E$ is a spectral measure. ∎

######  Lemma 5.1.14.

Suppose $\Omega$ is a locally compact Hausdorff space which has a countable
base of open sets and $E$ is a spectral measure on $\mathscr{B}(\Omega)$. For
any $x\in\mathcal{H}_{n}$, let $\mu_{x}(M):=\langle E(M)x,x\rangle$ for
$M\in\mathscr{B}(\Omega)$ and let
$\mu_{x}=\sum_{\alpha}\\{\mu_{x,+}^{(\alpha)}-\mu_{x,-}^{(\alpha)}\\}e_{\alpha}$
be the Jordan decomposition for $\mu_{x}$. Then

(5.9) $\mu_{x,+}^{(\alpha)}$ is a finite positive Borel measure for every
$\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})$.

###### Proof.

The last assertion of Lemma 5.1.6 ensures that $\mu_{x}$ is a finite
$\mathfrak{P}(\mathbb{R}_{n})$-valued Borel measure. Consequently,
$\mu_{x,+}^{(\alpha)}$ is a finite positive Borel measure for all
$\alpha\in\raisebox{1.79993pt}{\Large$\wp$}(\\{1,\ldots,n\\})$. But then we
may use the well-known fact that every positive Borel measure on a locally
compact Hausdorff space which has a countable base of open sets is regular
(see, e.g., Proposition 7.2.3 in [13]) to obtain (5.9). ∎

######  Theorem 5.1.15.

Suppose $\Omega_{i}$ is a locally compact Hausdorff space which has a
countable base of open sets and
$E_{i}:\mathscr{B}(\Omega_{i})\to\mathcal{P}(\mathcal{H}_{n})$ is a spectral
measure on $\mathscr{B}(\Omega_{i})$ for $i=1,\ldots,d$. If
$E_{i}E_{j}=E_{j}E_{i}$ for $i,j=1,\ldots,d$, then there exists a unique
spectral measure $E:\mathscr{B}(\Omega)\to\mathcal{P}(\mathcal{H}_{n})$, where
$\Omega:=\Omega_{1}\times\cdots\times\Omega_{d}$, such that

(5.10) $E(M_{1}\times\cdots\times M_{d})=E(M_{1})\cdots E(M_{d})\quad{\rm
for}\quad M_{i}\in\mathscr{B}(\Omega_{i})\;{\rm and}\;i=1,\ldots,d.$

###### Proof.

Let us consider the case $d=2$ (the more general case follows in much the same
way). Let $\mathscr{A}^{\prime}$ denote the algebra of sets generated by sets
of the form $M_{1}\times M_{2}$, where $M_{i}\in\mathscr{B}(\Omega_{i})$ for
$i=1,2$. Thus, every $N\in\mathscr{A}^{\prime}$ can be written as
$N=\cup_{j=1}^{k}N_{j}$, where $N_{1},\ldots N_{k}$ are mutually disjoint and
are of the form

$N_{j}=M_{1j}\times M_{2j}\in\mathscr{A}^{\prime}\quad\quad{\rm for}\quad
j=1,\ldots,k.$

We shall now define $E(N):=\sum_{j=1}^{k}E_{1}(M_{1j})E_{2}(M_{2j})$.

We claim that $E(N)\in\mathcal{P}(\mathcal{H}_{n})$. To this end, note that if
$i\neq j$, then $N_{i}\cap N_{j}=\emptyset$ forces $M_{1i}\cap
M_{1j}=\emptyset$ or $M_{2i}\cap M_{2j}=\emptyset$. In either case, we can
make use of Lemma 5.1.3 to deduce

$\displaystyle 0=$ $\displaystyle\;E_{1}(M_{1i}\cap M_{1j})E_{2}(M_{2i}\cap
M_{2j})$ $\displaystyle=$
$\displaystyle\;E_{1}(M_{1i})E_{1}(M_{1j})E_{2}(M_{2i})E_{2}(M_{2j})$
$\displaystyle=$
$\displaystyle\;E_{1}(M_{1i})E_{2}(M_{2i})E_{1}(M_{1j})E_{2}(M_{2j}).$

Consequently, $E(N)\in\mathcal{P}(\mathcal{H}_{n})$. One can easily modify the
proof of Theorem E in §8 of [42] to show that $E$ is independent of the
representation that we choose for $N\in\mathcal{B}(\Omega)$.

What remains is to show that $E$ is countably additive, i.e., for any monotone
sequence of sets $(N_{i})_{i=1}^{\infty}$, where
$N_{i}\in\mathscr{B}(\Omega^{\prime})$, with $N:=\cup_{i=1}^{\infty}N_{i}$, we
must show that $E(N)=\sum_{i=1}^{\infty}E(N_{i})$. In view of the fact that
${\rm Re}\langle E(N^{\prime})x,x\rangle$ is a positive measure for any choice
of $N^{\prime}\in\mathscr{B}(\Omega^{\prime})$ and $x\in\mathcal{H}_{n}$, we
can prove

${\rm Re}\langle E(N)x,x\rangle\leq\sum_{i=1}^{\infty}{\rm Re}\langle
E(N_{i}x,x\rangle$

in much the same way as the proof of Theorem 4.10 in [66] with the caveat that
terms of the form $\langle E(N^{\prime})x,x\rangle$ must be replaced by ${\rm
Re}\langle E(N^{\prime})x,x\rangle$. Since $N_{i}\subseteq N$, we immediately
have $\sum_{i=1}^{k}{\rm Re}\langle E(N_{i})x,x\rangle\leq{\rm Re}\langle
E(N)x,x\rangle$ and hence we arrive at

(5.11) ${\rm Re}\,\langle E(N)x,x\rangle=\sum_{i=1}^{\infty}{\rm Re}\,\langle
E(N_{i})x,x\rangle.$

Since $E\in\mathcal{P}(\mathcal{H}_{n})$, we have

$\|E(N^{\prime})x\|^{2}={\rm Re}\,\langle
E(N^{\prime})x,E(N^{\prime})x\rangle={\rm Re}\,\langle
E(N^{\prime})x,x\rangle\quad\quad{\rm for}\quad
N^{\prime}\in\mathscr{B}(\Omega)$

and thus we may use (5.11) and the fact that all sums are taken with respect
to the strong operator topology to obtain $E(N)=\sum_{i=1}^{\infty}E(N_{i})$,
i.e., $E$ is countably additive.

Finally, since the $\sigma$-algebra generated by $\mathscr{A}^{\prime}$ is
$\mathscr{B}(\Omega)$, we may use Lemma 5.1.13 to deduce that the spectral
premeasure $E$ has an extension to a spectral measure. With a slight abuse of
notation, the aforementioned spectral measure on $\mathscr{B}(\Omega)$ will be
denoted by $E$. One can use the definition of $E$ to verify (5.10). The
uniqueness of a spectral $E$ which obeys (5.10) drops out immediately from the
fact that the $\sigma$-algebra generated by the algebra of sets of the form
$M_{1}\times M_{2}$, where $M_{i}\in\mathscr{B}(\Omega_{i})$ for $i=1,2$, is
$\mathscr{B}(\Omega)$.

∎

###  5.2. Spectral integrals of bounded measurable functions

We anticipate the definition of imaginary operator given in Definition 8.0.2.
We will call an operator $J_{0}\in\mathcal{B}(\mathcal{H}_{n})$ a partial
imaginary operator if $J_{0}$ is a partial isometry and $J_{0}^{*}=-J_{0}$. We
will call $J\in\mathcal{B}(\mathcal{H}_{n})$ an imaginary operator if $J$ is
unitary and $J^{*}=-J$.

######  Definition 5.2.1.

Given a spectral measure $E$ on $(\Omega,\mathscr{A})$ and an imaginary
operator $J\in\mathcal{B}(\mathcal{H}_{n})$, we will say that $J$ is
associated with the spectral measure $E$ if

(5.12) $JE(M)=E(M)J\quad\quad{\rm for}\quad M\in\mathscr{A}.$

Given a spectral measure $E$ on $(\Omega,\mathscr{A})$ as above and an
imaginary $J$ associated with $E$, we wish to give meaning to the integral of
a measurable function $f:\Omega\to\mathbb{C}_{\mathfrak{I}}$, where
$\mathfrak{I}\in\mathbb{S}$, against a spectral measure $E$. Let
$\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ denote the Banach
space of all bounded $\mathscr{A}$-measurable functions
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}$, where $\mathfrak{I}\in\mathbb{S}$,
equipped with the norm

$\|f\|_{\infty}=\sup\\{|f(\lambda)|:\lambda\in\Omega\\}.$

We let $\chi_{M}$ denote the characteristic function with respect to
$M\in\Omega$, i.e.,

$\chi_{M}(\lambda)=\begin{cases}1&{\rm if}\quad\lambda\in M\\\ 0&{\rm
if}\quad\lambda\notin M.\end{cases}$

Let $\mathfrak{B}_{s}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ denote
the subspace of simple functions in
$\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$, i.e., the
subspace of functions
$f\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ which are of
the form

(5.13) $f(\lambda)=\sum_{j=1}^{k}c_{j}\,\chi_{M_{j}}(\lambda),$

where $c_{1},\ldots,c_{k}\in\mathbb{C}_{\mathfrak{I}}$ and
$M_{1},\ldots,M_{k}$ are pairwise disjoint sets belonging to
$\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$.

Given $f\in\mathfrak{B}_{s}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$, we
shall let

(5.14) $\mathbb{I}(f):=\sum_{j=1}^{k}\\{{\rm Re}(c_{j})E(M_{i})+{\rm
Im}(c_{j})E(M_{j})J\\}\in\mathcal{B}(\mathcal{H}_{n}).$

It can be easily checked that the finite additivity of $E$ implies that the
definition of $\mathbb{I}(f)$ in (5.14) is independent of the (5.13).

######  Lemma 5.2.2.

Let $f\in\mathfrak{B}_{s}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$. Then
$\|\mathbb{I}(f)\|\leq\|f\|_{\infty}$.

###### Proof.

Since $M_{1}\ldots,M_{k}$ are mutually disjoint sets belonging to
$\mathscr{B}(\Omega)$, Lemma 5.1.3 asserts that $E(M_{i})E(M_{j})=0$ whenever
$i\neq j$. Thus, for any $x\in\mathcal{H}_{n}$, we have

$\displaystyle\|\mathbb{I}(f)x\|^{2}=$ $\displaystyle\;\sum_{i,j=1}^{k}{\rm
Re}\left\langle\\{{\rm Re}(c_{i})I+{\rm Im}(c_{i})J\\}E(M_{i})x,\\{{\rm
Re}(c_{j})I+{\rm Im}(c_{j})J\\}E(M_{j})x\right\rangle$ $\displaystyle=$
$\displaystyle\;\sum_{i=1}^{k}{\rm Re}\left\langle\\{{\rm Re}(c_{i})I+{\rm
Im}(c_{i})J\\}E(M_{i})x,\\{{\rm Re}(c_{i})+{\rm Im}(c_{i})J\\}x\right\rangle$
$\displaystyle=$ $\displaystyle\;\sum_{i=1}^{k}{\rm Re}\left\langle\\{{\rm
Re}(c_{i})I+{\rm Im}(c_{i})J\\}E(M_{i})x,\\{{\rm Re}(c_{i})I+{\rm
Im}(c_{i})J\\}E(M_{i})x\right\rangle$ $\displaystyle=$
$\displaystyle\;\sum_{i=1}^{k}\|({\rm Re}(c_{i})I+{\rm
Im}(c_{i})J)E(M_{i})x\|^{2}$ $\displaystyle\leq$
$\displaystyle\;\sum_{i=1}^{k}|c_{i}|^{2}\|E(M_{i})x\|^{2}$
$\displaystyle\leq$
$\displaystyle\;\sum_{i=1}^{k}\|f\|^{2}_{\infty}\|E(M_{i})x\|^{2}=\|f\|_{\infty}^{2}\left\|\sum_{i=1}^{k}E(M_{i})x\right\|^{2}$
$\displaystyle\leq$ $\displaystyle\;\|f\|_{\infty}^{2}\|x\|^{2}.$

Consequently, $\|\mathbb{I}(f)\|\leq\|f\|_{\infty}$ holds. ∎

######  Definition 5.2.3 ($\mathbb{I}(f)$ for
$f\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$).

We will now give a definition for $\mathbb{I}(f)$ when
$f\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$. Since
$\mathfrak{B}_{s}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ is a dense
subset of the Banach space
$\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ with respect to
the supremum norm $\|\cdot\|_{\infty}$, given any
$f\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$, we have the
existence of a Cauchy sequence $(f_{j})_{j=1}^{\infty}$, where
$f_{j}\in\mathfrak{B}_{s}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ for
$j=1,2\ldots$, such that

$\lim_{j\to\infty}\|f-f_{j}\|_{\infty}=0.$

Lemma 5.2.2 can be used to show that $(\mathbb{I}(f_{j}))_{j=1}^{\infty}$ is a
Cauchy sequence of operators belonging to $\mathcal{B}(\mathcal{H}_{n})$. As
$\mathcal{B}(\mathcal{H}_{n})$ is a complete metric space (see Remark 3.0.5),
we have the existence of an operator
$\mathbb{I}(f)\in\mathcal{B}(\mathcal{H}_{n})$ such that

$\lim_{j\to\infty}\|\mathbb{I}(f)-\mathbb{I}(f_{j})\|=0.$

Moreover, Lemma 5.2.2 can be used to show that the limit $\mathbb{I}(f)$ does
not depend on the choice of Cauchy sequence.

######  Theorem 5.2.4.

For any $f,g\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$, we
have the following:

  1. (i)

$\mathbb{I}(\bar{f})=\mathbb{I}(f)^{*}$.

  2. (ii)

$\mathbb{I}(fg)=\mathbb{I}(f)\mathbb{I}(g)$.

  3. (iii)

$\mathbb{I}(\mathfrak{I})=J$ and
$\mathbb{I}(cf+g)=c\,\mathbb{I}(f)+\mathbb{I}(g)$ for all
$c\in\mathbb{C}_{\mathfrak{I}}$ and $\mathfrak{I}\in\mathbb{S}$.

  4. (iv)

$\langle\mathbb{I}(f)x,y\rangle=\int_{\Omega}{\rm Re}(f(\lambda))d\langle
E(\lambda)x,y\rangle+\int_{\Omega}{\rm Im}(f(\lambda))d\langle
JE(\lambda)x,y\rangle$ for all $x,y\in\mathcal{H}_{n}$.

  5. (v)

$\|\mathbb{I}(f)x\|^{2}=\int_{\Omega}|f(\lambda)|^{2}d({\rm Re}\,\langle
E(\lambda)x,x\rangle)$ for all $x\in\mathcal{H}_{n}$.

  6. (vi)

$\|\mathbb{I}(f)\|\leq\|f\|_{\infty}$.

  7. (vii)

For any sequence of functions $(f_{j})_{j=1}^{\infty}$, where
$f_{j}\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ for
$j=1,2,\ldots$, which converges pointwise $E$-a.e. on $\Omega$ to $f$ and
there exists $\kappa>0$ such that $|f_{j}(\lambda)|\leq\kappa$ for all
$\lambda\in\Omega$ and $j=1,2,\ldots$, we have

$s-\lim_{j\to\infty}\mathbb{I}(f_{n})=\mathbb{I}(f).$

###### Proof.

In view of Definition 5.2.3, it suffices to prove (i)-(vi) for
$f,g\in\mathfrak{B}_{s}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$. Let
$f(\lambda)=\sum_{j=1}^{k}c_{j}\chi_{M_{j}}(\lambda)$ and
$g(\lambda)=\sum_{j=1}^{k}d_{j}\chi_{N_{j}}(\lambda)$. Using the fact that
$JE(M)=E(M)J$ for all $M\in\mathscr{A}$ and $J^{*}=-J$, we have

$\displaystyle\mathbb{I}(f)^{*}=$
$\displaystyle\;\sum_{j=1}^{k}E(M_{j})^{*}\\{{\rm Re}(c_{j})I+{\rm
Im}(c_{j})J^{*}\\}E(M_{j})$ $\displaystyle=$
$\displaystyle\;\sum_{j=1}^{k}E(M_{j})\\{{\rm Re}(c_{j})I-{\rm
Im}(c_{j})J\\}E(M_{j})$ $\displaystyle=$ $\displaystyle\;\sum_{j=1}^{k}\\{{\rm
Re}(c_{j})I-{\rm Im}(c_{j})J\\}E(M_{j})$ $\displaystyle=$
$\displaystyle\;\mathbb{I}(\bar{f}).$

Thus, we have proved (i).

Next, since

$\displaystyle\mathbb{I}(f)\mathbb{I}(g)=$
$\displaystyle\;\left(\sum_{i=1}^{k}\\{{\rm Re}(c_{i})I+{\rm
Im}(c_{i})J\\}E(M_{i})\right)\left(\sum_{j=1}^{k}\\{{\rm Re}(d_{j})I+{\rm
Im}(d_{j})J\\}E(N_{j})\right)$ $\displaystyle=$
$\displaystyle\;\sum_{i,j=1}^{k}\\{{\rm Re}(c_{i}){\rm Re}(d_{j})I-{\rm
Im}(c_{i}){\rm Im}(d_{j})I$ $\displaystyle\;\;\;\;+$ $\displaystyle({\rm
Re}(c_{i}){\rm Im}(d_{j})+{\rm Re}(d_{j}){\rm Im}(c_{i}))J\\}E(M_{i}\cap
N_{j})$ $\displaystyle=$ $\displaystyle\;\mathbb{I}(f)\mathbb{I}(g),$

we have proved (ii).

To prove the first statement in (iii), simply observe that
$f(\lambda)=\mathfrak{I}\,\chi_{\Omega}(\lambda)=\mathfrak{I}$ and hence
$\mathbb{I}(\mathfrak{I})=J$. The second statement in (iii) is an easy
consequence of (ii) and the fact that $\mathbb{I}(\mathfrak{I})=J$.

The proof of (iv) is very straight forward. To prove (v), we may use (i), (ii)
and (iv) to verify that

$\displaystyle\|\mathbb{I}(f)x\|^{2}=$ $\displaystyle\;{\rm
Re}\langle\mathbb{I}(f)x,\mathbb{I}(f)x\rangle$ $\displaystyle=$
$\displaystyle\;{\rm Re}\,\langle\mathbb{I}(\bar{f})\mathbb{I}(f)x,x\rangle$
$\displaystyle=$ $\displaystyle\;{\rm
Re}\,\langle\mathbb{I}(|f|^{2})x,x\rangle$ $\displaystyle=$
$\displaystyle\;{\rm Re}\left(\int_{\Omega}|f(\lambda)|^{2}d\langle
E(\lambda)x,x\rangle\right)$ $\displaystyle=$
$\displaystyle\;\int_{\Omega}|f(\lambda)|^{2}d({\rm Re}\,\langle
E(\lambda)x,x\rangle)\quad\quad{\rm for}\quad x\in\mathcal{H}_{n}.$

Statement (v) is a direct consequence of (iv). Indeed, using (iv), we have

$\|\mathbb{I}(f)x\|^{2}\leq\|f\|_{\infty}\,{\rm Re}\langle
x,x\rangle=\|f\|_{\infty}\|x\|^{2}\quad\quad{\rm for}\quad
x\in\mathcal{H}_{n},$

in which case we have (v).

Finally, to prove (vii), we may use (vi) to see that

$\|\\{\mathbb{I}(f)-\mathbb{I}(f_{n})\\}x\|^{2}=\int_{\Omega}|f(\lambda)-f_{n}(\lambda)|^{2}d({\rm
Re}\langle E(\lambda)x,x\rangle)\quad\quad{\rm for}\quad x\in\mathcal{H}_{n}$

and hence (vii) follows immediately from the Lebesgue dominated convergence
theorem. ∎

######  Lemma 5.2.5.

Let $\mathscr{A}$ be an algebra of sets in $\Omega$ and $E$ be a spectral
measure on $\mathscr{A}$. Then the following facts hold:

  1. (i)

$|E_{x,y}|(M)\leq\sqrt{{\rm Re}\,E_{x}(M)}\,\sqrt{{\rm Re}\,E_{y}(M)}$ for all
$x,y\in\mathcal{H}_{n}$ and $M\in\mathscr{A}$.

  2. (ii)

Let $f\in L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{x})$ and $g\in
L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{y})$. Then

$\displaystyle\;\left|\int_{\Omega}{\rm Re}(f(\lambda)g(\lambda))\,d\langle
E(\lambda)x,y\rangle+\int_{\Omega}{\rm Im}(f(\lambda)g(\lambda))\,d\langle
E(\lambda)Jx,y\rangle\right|$ $\displaystyle\leq$
$\displaystyle\;\int_{\Omega}|f(\lambda)g(\lambda)|d|E_{x,y}|(\lambda)$ (5.15)
$\displaystyle\leq$
$\displaystyle\;2^{n}\|f\|_{L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{x})}\|g\|_{L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{y})},$

where $E_{x,y}(M)=\langle E(M)x,y\rangle$ for $M\in\mathscr{B}(\Omega)$ and
for all $x,y\in\mathcal{H}_{n}$.

###### Proof.

We will first prove (i). Suppose $M=\cup_{k=1}^{p}M_{k}$, where
$M_{k}\in\mathscr{A}$ and $M_{k}\cap M_{j}=\emptyset$ whenever $k\neq j$.
Using Lemma 2.2.6, we have

$\displaystyle|E_{x,y}(M_{k})|=$ $\displaystyle\;|\langle
E(M_{k})x,y\rangle|=|\langle E(M_{k})x,E(M_{k})y\rangle|$ $\displaystyle\leq$
$\displaystyle\;\|E(M_{k})x\|\|E(M_{k})y\|$ $\displaystyle=$
$\displaystyle\;\sqrt{{\rm Re}\,E_{x}(M_{k})}\,\sqrt{{\rm Re}\,E_{y}(M_{k})}.$

Consequently, we have

$\displaystyle\sum_{k=1}^{p}|E_{x,y}(M_{k})|\leq$
$\displaystyle\;\sum_{k=1}^{p}\sqrt{{\rm Re}\,E_{x}(M_{k})}\,\sqrt{{\rm
Re}\,E_{y}(M_{k})}$ $\displaystyle\leq$
$\displaystyle\;\left(\sum_{k=1}^{p}{\rm
Re}\,E_{x}(M_{k})\right)^{1/2}\,\left(\sum_{k=1}^{p}{\rm
Re}\,E_{y}(M_{k})\right)^{1/2}$ $\displaystyle=$ $\displaystyle\;\sqrt{{\rm
Re}\,E_{x}(M)}\,\sqrt{{\rm Re}\,E_{y}(M)}.$

Finally, if we take the supremum over all possible disjoint unions of $M$ in
$\mathscr{A}$, then we obtain (i).

We will now prove (ii). We will verify (5.15) for simple functions, in which
case (5.15) will hold by the density of simple functions in both
$L_{2}$-spaces. Suppose $f(\lambda)=\sum_{k=1}^{p}f_{k}\chi_{M_{k}}(\lambda)$
and $g(\lambda)=\sum_{k=1}^{p}g_{k}\chi_{M_{k}}(\lambda)$ are simple functions
in $L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{x})$ and
$L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{y})$, respectively. Using (2.4),
Assertion (i) and (2.4), we have

$\displaystyle\;\left|\int_{\Omega}{\rm Re}(f(\lambda)g(\lambda))d\langle
E(\lambda)x,y\rangle+\int_{\Omega}{\rm Im}(f(\lambda)g(\lambda))d\langle
E(\lambda)Jx,y\rangle\right|$ $\displaystyle=$
$\displaystyle\;\left|\sum_{k=1}^{p}\\{{\rm Re}(f_{k}g_{k})\langle
E(M_{k})x,y\rangle+{\rm Im}(f_{k}g_{k})\langle E(M_{k})Jx,y\rangle\\}\right|$
$\displaystyle\leq$