# Spectrum and Analytic Functional Calculus in Real and Quaternionic
Frameworks

Florian-Horia Vasilescu
Department of Mathematics, University of Lille,
59655 Villeneuve d’Ascq, France
<EMAIL_ADDRESS>
###### Abstract

We present an approach to the spectrum and analytic functional calculus for
quaternionic linear operators, following the corresponding results concerning
the real linear operators. In fact, the construction of the analytic
functional calculus for real linear operators can be refined to get a similar
construction for quaternionic linear ones, in a classical manner, using a
Riesz-Dunford-Gelfand type kernel, and considering spectra in the complex
plane. A quaternionic joint spectrum for pairs of operators is also discussed,
and an analytic functional calculus is constructed, via a Martinelli type
kernel in two variables.

Keywords: spectrum in real and quaternionic contexts; holomorphic stem
functions; analytic functional calculus for real and quaternionic operators

AMS Subject Classification: 47A10; 30G35; 47A60

Keywords: real and quaternionic operators; spectra; analytic
functionalcalculus.

Mathematics Subject Classification 2010: 47A10; 47A60; 30G35

##  1 Introduction

In this text we consider ${\mathbb{R}}$-, ${\mathbb{C}}$-, and
${\mathbb{H}}$-linear operators, that is, real, complex and quaternionic
linear operators, respectively.

While the spectrum of a linear operator is traditionally defined for complex
linear operators, it is sometimes useful to have it also for real linear
operators, as well as for quaternionic linear ones. The definition of the
spectrum for a real linear operator goes seemingly back to Kaplansky (see
[10]), and it can be stated as follows. If $T$ is a real linear operator on
the real vector space ${\mathcal{V}}$, a point $u+iv$ ($u,v\in{\mathbb{R}}$)
is in the spectrum of $T$ if the operator $(u-T)^{2}+v^{2}$ is not invertible
on ${\mathcal{V}}$, where the scalars are identified with multiples of the
identity on ${\mathcal{V}}$. Although this definition involves only operators
acting in ${\mathcal{V}}$, the spectrum is, nevertheless, a subset of the
complex plane. As a matter of fact, a motivation of this choice can be
illustrated via the complexification of the space ${\mathcal{V}}$ (see Section
2).

The spectral theory for quaternionic linear operators is largely discussed in
numerous work, in particular in the monographs [5] and [4], and in many of
their references as well. In these works, the construction of an analytic
functional calculus (called $S$-analytic functional calculus) means to
associate to each function from the class of the so-called slice
hyperholomorphic or slice regular functions a quaternionic linear operator,
using a specific noncommutative kernel.

The idea of the present work is to replace the class of slice regular
functions by a class holomorphic functions, using a commutative kernel of type
Riesz- Dunford-Gelfand. These two classes are isomorphic via a Cauchy type
transform (see [21]), and the image of the analytic functional calculus is the
same, as one might expect (see Remark 8).

As in the case of real operators, the verbatim extension of the classical
definition of the spectrum for quaternionic operators is not appropriate, and
so a different definition using the squares of operators and real numbers was
given, which can be found in [5] (see also [4]). We discuss this definition in
our framework (see Definition 1), showing later that its ”complex border“
contains the most significant information, leading to the construction of an
analytic functional calculus, equivalent to that obtained via the slice
hyperholomorphic functions.

In fact, we first consider the spectrum for real operators on real Banach
spaces, and sketch the construction of an analytic functional calculus for
them, using some classical ideas (see Theorem 2). Then we extend this
framework to a quaternionic one, showing that the approach from the real case
can be easily adapted to the new situation.

As already mentioned, and unlike in [5] or [4], our functional calculus is
obtained via a Riesz-Dunford-Gelfand formula, defined in a partially
commutatative context, rather than the non-commutative Cauchy type formula
used by previous authors. Our analytic functional calculus holds for a class
of analytic operator valued functions, whose definition extends that of stem
functions, and it applies, in particular, to a large family of quaternionic
linear operators. Moreover, we can show that the analytic functional calculus
obtained in this way is equivalent to the analytic functional calculus
obtained in [5] or [4], in the sense that the images of these functional
calculi coincide (see Remark 8).

We finally discuss the case of pairs of commuting real operators, in the
spirit of [20], showing some connections with the quaternionic case.
Specifically, we define a quaternionic spectrum for them and construct an
analytic functional calculus using a Martinelli type formula, showing that for
such a construction only a sort of ”complex border“ of the quaternionic
spectrum should be used.

This work is just an introductory one. Hopefully, more contributions on this
line will be presented in the future.

##  2 Spectrum and Conjugation

Let ${\mathcal{A}}$ be a unital real Banach algebra, not necessarily
commutative. As mentioned in the Introduction, the (complex) spectrum of an
element $a\in{\mathcal{A}}$ may be defined by the equality

$\sigma_{\mathbb{C}}(a)=\\{u+iv;(u-a)^{2}+v^{2}\,\,{\rm
is\,\,not\,\,invertible},u,v\in{\mathbb{R}}\\},$ (1)

This set is conjugate symmetric, that is $u+iv\in\sigma_{\mathbb{C}}(a)$ if
and only if $u-iv\in\sigma_{\mathbb{C}}(a)$. A known motivation of this
definition comes from the following remark.

Fixing a unital real Banach algebra ${\mathcal{A}}$, we denote by
${\mathcal{A}}_{\mathbb{C}}$ the complexification of ${\mathcal{A}}$, which is
given by $A_{\mathbb{C}}={\mathbb{C}}\otimes_{\mathbb{R}}{\mathcal{A}}$,
written simply as ${\mathcal{A}}+i{\mathcal{A}}$, where the sum is direct,
identifying the element $1\otimes a+i\otimes b$ with the element $a+ib$, for
all $a,b\in{\mathcal{A}}$.

Then ${\mathcal{A}}_{\mathbb{C}}$ is a unital complex algebra, which can be
organized as a Banach algebra, with a (not necessarily unique) convenient
norm. To fix the ideas, we recall that the product of two elements is given by
$(a+ib)(c+id)=ac-bd+i(ad+bc)$ for all $a,b,c,d\in{\mathcal{A}}$, and the norm
may be defind by $\|a+ib\|=\|a\|+\|b\|$, where $\|*\|$ is the norm of
${\mathcal{A}}$.

In the algebra ${\mathcal{A}}_{\mathbb{C}}$, the complex numbers commute with
all elements of ${\mathcal{A}}$. Moreover, we have a conjugation given by

${\mathcal{A}}_{\mathbb{C}}\ni a+ib\mapsto
a-ib\in{\mathcal{A}}_{\mathbb{C}},\,a,b\in{\mathcal{A}},$

which is a unital conjugate-linear automorphism, whose square is the identity.
In particular, an arbitrary element $a+ib$ is invertible if and only if $a-ib$
is invertible.

The usual spectrum, defined for each element $a\in{\mathcal{A}}_{\mathbb{C}}$,
will be denoted by $\sigma(a)$. Regarding the algebra ${\mathcal{A}}$ as a
real subalgebra of ${\mathcal{A}}_{\mathbb{C}}$, one has the following.

###### Lemma 1

For every $a\in{\mathcal{A}}$ we have the equality
$\sigma_{\mathbb{C}}(a)=\sigma(a)$.

Proof. The result is well known but we give a short proof, because a similar
idea will be later used.

Let $\lambda=u+iv$ with $u,v\in{\mathbb{R}}$ arbitrary. Assuming $\lambda-a$
invertible, we also have $\bar{\lambda}-a$ invertible. From the obvious
identity

$(u-a)^{2}+v^{2}=(u+iv-a)(u-iv-a),$

we deduce that the element $(u-a)^{2}+v^{2}$ is invertible, implying the
inclusion $\sigma_{\mathbb{C}}(a)\subset\sigma(a)$.

Conversely, if $(u-a)^{2}+v^{2}$ is invertible, then both $u+iv-a,u-iv-a$ are
invertible via the decomposition from above, showing that we also have
$\sigma_{\mathbb{C}}(a)\supset\sigma(a)$.

###### Remark 1

The spectrum $\sigma(a)$ with $a\in{\mathcal{A}}$ is always a conjugate
symmetric set.

We are particularly interested to apply the discussion from above to the
context of linear operators. The spectral theory for real linear operators is
well known, and it is developed actually in the framework of linear relations
(see [1]). Nevertheless, we present here a different approach, which can be
applied, with minor changes, to the case of some quaternionic operators.

For a real or complex Banach space $\mathcal{V}$, we denote by
$\mathcal{B(V)}$ the algebra of all bounded ${\mathbb{R}}$-( respectively
${\mathbb{C}}$-)linear operators on $\mathcal{V}$. As before, the multiples of
the identity will be identified with the corresponding scalars.

Let ${\mathcal{V}}$ be a real Banach space, and let
${\mathcal{V}}_{\mathbb{C}}$ be its complexification, which, as above, is
identified with the direct sum ${\mathcal{V}}+i{\mathcal{V}}$. Each operator
$T\in\mathcal{B(V)}$ has a natural extension to an operator
$T_{\mathbb{C}}\in\mathcal{B}({\mathcal{V}}_{\mathbb{C}})$, given by
$T_{\mathbb{C}}(x+iy)=Tx+iTy,\,x,y\in{\mathcal{V}}$. Moreover, the map
$\mathcal{B(V)}\ni T\mapsto
T_{\mathbb{C}}\in\mathcal{B}({\mathcal{V}}_{\mathbb{C}})$ is unital,
${\mathbb{R}}$-linear and multiplicative. In particular, $T\in\mathcal{B(V)}$
is invertible if and only if
$T_{\mathbb{C}}\in\mathcal{B}({\mathcal{V}}_{\mathbb{C}})$ is invertible.

Fixing an operator $S\in\mathcal{B}(\mathcal{V}_{\mathbb{C}})$, we define the
operator $S^{\flat}\in\mathcal{B}(\mathcal{V}_{\mathbb{C}})$ to be equal to
$CSC$, where $C:{\mathcal{V}}_{\mathbb{C}}\mapsto{\mathcal{V}}_{\mathbb{C}}$
is the conjugation $x+iy\mapsto x-iy,\,x,y\in{\mathcal{V}}$. It is easily seen
that the map $\mathcal{B}(\mathcal{V}_{\mathbb{C}})\ni S\mapsto
S^{\flat}\in\mathcal{B}(\mathcal{V}_{\mathbb{C}})$ is a unital conjugate-
linear automorphism, whose square is the identity on
$\mathcal{B}(\mathcal{V}_{\mathbb{C}})$. Because
$\mathcal{V}=\\{u\in\mathcal{V}_{\mathbb{C}};Cu=u\\}$, we have $S^{\flat}=S$
if and only if $S(\mathcal{V})\subset\mathcal{V}$. In particular, we have
$T_{\mathbb{C}}^{\flat}=T_{\mathbb{C}}$. In fact, because of the
representation

$S=\frac{1}{2}(S+S^{\flat})+i\frac{1}{2i}(S-S^{\flat}),\,\,S\in\mathcal{B}(\mathcal{V}_{\mathbb{C}}),$

where
$(S+S^{\flat})({\mathcal{V}})\subset{\mathcal{V}},i(S-S^{\flat})({\mathcal{V}})\subset{\mathcal{V}}$,
the algebras $\mathcal{B}(\mathcal{V}_{\mathbb{C}})$ and
$\mathcal{B(V)}_{\mathbb{C}}$ are isomorphic and they will be often
identified, and $\mathcal{B(V)}$ will be regarded as a (real) subalgebra of
$\mathcal{B}(\mathcal{V})_{\mathbb{C}}$. In particular, if $S=U+iV$, with
$U,V\in\mathcal{B(V)}$, we have $S^{\flat}=U-iV$, so the map $S\mapsto
S^{\flat}$ is the conjugation of the complex algebra
$\mathcal{B}(\mathcal{V})_{\mathbb{C}}$ induced by the conjugation $C$ of
${\mathcal{V}}_{\mathbb{C}}$.

For every operator $S\in\mathcal{B}(\mathcal{V}_{\mathbb{C}})$, we denote, as
before, by $\sigma(S)$ its usual spectrum. As $\mathcal{B(V)}$ is a real
algebra, the (complex) spectrum of an operator $T\in\mathcal{B(V)}$ is given
by the equality (1):

$\sigma_{\mathbb{C}}(T)=\\{u+iv;(u-T)^{2}+v^{2}\,\,{\rm
is\,\,not\,\,invertible},u,v\in{\mathbb{R}}\\}.$

###### Corollary 1

For every $T\in\mathcal{B}({\mathcal{V}})$ we have the equality
$\sigma_{\mathbb{C}}(T)=\sigma(T_{\mathbb{C}})$.

##  3 Analytic Functional Calculus for Real
Operators

Having a concept of spectrum for real operators, an important step for further
development is the construction of an analytic functional calculus. Such a
construction has been done actually in the context of real linear relations in
[1]. In what follows we shall present a similar construction for real linear
operators. Although the case of linear relations looks more general, unlike in
[1], we perform our construction using a class of operator valued analytic
functions insted of scalar valued analytic functions. Moreover, our arguments
look simpler, and the construction is a model for a more general one, to get
an analytic functional calculus for quaternionic linear operators.

If ${\mathcal{V}}$ is a real Banach space, and so each operator
$T\in\mathcal{B}({\mathcal{V}})$ has a complex spectrum
$\sigma_{\mathbb{C}}(T)$, which is compact and nonempty, one can use the
classical Riesz-Dunford functional calculus, in a slightly generalized form
(that is, replacing the scalar-valued analytic functions by operator-valued
analytic ones, which is a well known idea).

The use of vector versions of the Cauchy formula is simplified by adopting the
following definition. Let $U\subset{\mathbb{C}}$ be open. An open subset
$\Delta\subset U$ will be called a Cauchy domain (in $U$) if
$\Delta\subset\bar{\Delta}\subset U$ and the boundary of $\Delta$ consists of
a finite family of closed curves, piecewise smooth, positively oriented. Note
that a Cauchy domain is bounded but not necessarily connected.

###### Remark 2

If $\mathcal{V}$ is a real Banach space, and $T\in\mathcal{B(V})$, we have the
usual analytic functional calculus for the operator
$T_{\mathbb{C}}\in\mathcal{B}(\mathcal{V}_{\mathbb{C}})$ (see [6]). That is,
in a slightly generalized form, and for later use, if
$U\supset\sigma(T_{\mathbb{C}})$ is an open set in ${\mathbb{C}}$ and
$F:U\mapsto\mathcal{B}(\mathcal{V}_{\mathbb{C}})$ is analytic, we put

$F(T_{\mathbb{C}})=\frac{1}{2\pi i}\int_{\Gamma}F(\zeta)(\zeta-
T_{\mathbb{C}})^{-1}d\zeta,$

where $\Gamma$ is the boundary of a Cauchy domain $\Delta$ containing
$\sigma(T_{\mathbb{C}})$ in $U$. In fact, because $\sigma(T_{\mathbb{C}})$ is
conjugate symmetric, we may and shall assume that both $U$ and $\Gamma$ are
conjugate symmetric. Because the function $\zeta\mapsto F(\zeta)(\zeta-
T_{\mathbb{C}})^{-1}$ is analytic in $U\setminus\sigma(T_{\mathbb{C}})$, the
integral does not depend on the particular choice of the Cauchy domain
$\Delta$ containing $\sigma(T_{C})$.

A natural question is to find an appropriate condition to we have
$F(T_{\mathbb{C}})^{\flat}=F(T_{\mathbb{C}})$, which would imply the
invariance of $\mathcal{V}$ under $F(T_{\mathbb{C}})$.

With the notation of Remark 2, we have the following.

###### Theorem 1

Let $U\subset{\mathbb{C}}$ be open and conjugate symmetric. If
$F:U\mapsto\mathcal{B}(\mathcal{V}_{\mathbb{C}})$ is analytic and
$F(\zeta)^{\flat}=F(\bar{\zeta})$ for all $\zeta\in U$, then
$F(T_{\mathbb{C}})^{\flat}=F(T_{\mathbb{C}})$ for all
$T\in\mathcal{B}(\mathcal{V})$ with $\sigma_{\mathbb{C}}(T)\subset U$.

Proof. We use the notation from Remark 2, assuming, in addition, that $\Gamma$
is conjugate symmetric as well. We put
$\Gamma_{\pm}:=\Gamma\cap{\mathbb{C}}_{\pm}$, where ${\mathbb{C}}_{+}$ (resp.
${\mathbb{C}}_{-}$) equals to $\\{\lambda\in{\mathbb{C}};\Im\lambda\geq 0\\}$
(resp. $\\{\lambda\in{\mathbb{C}};\Im\lambda\leq 0\\}$). We write
$\Gamma_{+}=\cup_{j=1}^{m}\Gamma_{j+}$, where $\Gamma_{j+}$ are the connected
components of $\Gamma_{+}$. Similarly, we write
$\Gamma_{-}=\cup_{j=1}^{m}\Gamma_{j-}$, where $\Gamma_{j-}$ are the connected
components of $\Gamma_{-}$, and $\Gamma_{j-}$ is the reflexion of
$\Gamma_{j+}$ with respect of the real axis.

As $\Gamma$ is a finite union of Jordan piecewise smooth closed curves, for
each index $j$ we have a parametrization $\phi_{j}:[0,1]\mapsto{\mathbb{C}}$,
positively oriented, such that $\phi_{j}([0,1])=\Gamma_{j+}$. Taking into
account that the function $t\mapsto\overline{\phi_{j}(t)}$ is a
parametrization of $\Gamma_{j-}$ negatively oriented, and setting
$\Gamma_{j}=\Gamma_{j+}\cup\Gamma_{j-}$, we can write

$F_{j}(T_{\mathbb{C}}):=\frac{1}{2\pi i}\int_{\Gamma_{j}}F(\zeta)(\zeta-
T_{\mathbb{C}})^{-1}d\zeta=$ $\frac{1}{2\pi
i}\int_{0}^{1}F(\phi_{j}(t))(\phi_{j}(t)-T_{\mathbb{C}})^{-1}\phi_{j}^{\prime}(t)dt$
$-\frac{1}{2\pi
i}\int_{0}^{1}F(\overline{\phi_{j}(t)})(\overline{\phi_{j}(t)}-T_{\mathbb{C}})^{-1}\overline{\phi_{j}^{\prime}(t)}dt.$

Therefore,

$F_{j}(T_{\mathbb{C}})^{\flat}=-\frac{1}{2\pi
i}\int_{0}^{1}F(\phi_{j}(t))^{\flat}(\overline{\phi_{j}(t)}-T_{\mathbb{C}})^{-1}\overline{\phi_{j}^{\prime}(t)}dt$
$+\frac{1}{2\pi
i}\int_{0}^{1}F(\overline{\phi_{j}(t)})^{\flat}(\phi_{j}(t)-T_{\mathbb{C}})^{-1}\phi_{j}^{\prime}(t)dt.$

According to our assumption on the function $F$, we obtain
$F_{j}(T_{\mathbb{C}})=F_{j}(T_{\mathbb{C}})^{\flat}$ for all $j$, and
therefore

$F(T_{\mathbb{C}})^{\flat}=\sum_{j=1}^{m}F_{j}(T_{\mathbb{C}})^{\flat}=\sum_{j=1}^{m}F_{j}(T_{\mathbb{C}})=F(T_{\mathbb{C}}),$

which concludes the proof.

###### Remark 3

If ${\mathcal{A}}$ is a unital real Banach algebra,
${\mathcal{A}}_{\mathbb{C}}$ its complexification, and $U\subset{\mathbb{C}}$
is open, we denote by $\mathcal{O}(U,{\mathcal{A}}_{\mathbb{C}})$ the algebra
of all analytic ${\mathcal{A}}_{\mathbb{C}}$-valued functions. If $U$ is
conjugate symmetric, and ${\mathcal{A}}_{\mathbb{C}}\ni
a\mapsto\bar{a}\in{\mathcal{A}}_{\mathbb{C}}$ is its natural conjugation, we
denote by $\mathcal{O}_{s}(U,{\mathcal{A}}_{\mathbb{C}})$ the real subalgebra
of $\mathcal{O}(U,{\mathcal{A}}_{\mathbb{C}})$ consisting of those functions
$F$ with the property $F(\bar{\zeta})=\overline{F(\zeta)}$ for all $\zeta\in
U$. Adapting a well known terminology, such functions will be called
(${\mathcal{A}}_{\mathbb{C}}$-valued $)$ stem functions.

When ${\mathcal{A}}={\mathbb{R}}$, so
${\mathcal{A}}_{\mathbb{C}}={\mathbb{C}}$, the space
$\mathcal{O}_{s}(U,{\mathbb{C}})$ will be denoted by $\mathcal{O}_{s}(U)$,
which is a real algebra. Note that
$\mathcal{O}_{s}(U,{\mathcal{A}}_{\mathbb{C}})$ is also a bilateral
$\mathcal{O}_{s}(U)$-module.

In the next result, we identify the algebra $\mathcal{B}(\mathcal{V})$ with a
subalgebra of $\mathcal{B}(\mathcal{V})_{\mathbb{C}}$. In ths case, when
$F\in\mathcal{O}_{s}(U,\mathcal{B}(\mathcal{V})_{\mathbb{C}})$, we shall write

$F(T)=\frac{1}{2\pi i}\int_{\Gamma}F(\zeta)(\zeta-T)^{-1}d\zeta,$

noting that the right hand side of this formula belongs to
$\mathcal{B}(\mathcal{V})$, by Theorem 1.

The properties of the map $F\mapsto F(T)$, which can be called the (left)
analytic functional calculus of $T$, are summarized by the following.

###### Theorem 2

Let ${\mathcal{V}}$ be a real Banach space, let $U\subset{\mathbb{C}}$ be a
conjugate symmetric open set, and let $T\in\mathcal{B}(\mathcal{V})$, with
$\sigma_{\mathbb{C}}(T)\subset U$. Then the assignment

${\mathcal{O}}_{s}(U,\mathcal{B}(\mathcal{V})_{\mathbb{C}})\ni F\mapsto
F(T)\in\mathcal{B}(\mathcal{V})$

is an ${\mathbb{R}}$-linear map, and the map

${\mathcal{O}}_{s}(U)\ni f\mapsto f(T)\in\mathcal{B}(\mathcal{V})$

is a unital real algebra morphism.

Moreover, the following properties are true:

(1) For all
$F\in\mathcal{O}_{s}(U,\mathcal{B}(\mathcal{V})_{\mathbb{C}}),\,f\in{\mathcal{O}}_{s}(U)$,
we have $(Ff)(T)=F(T)f(T)$.

(2) For every polynomial
$P(\zeta)=\sum_{n=0}^{m}A_{n}\zeta^{n},\,\zeta\in{\mathbb{C}}$, with
$A_{n}\in\mathcal{B}(\mathcal{V})$ for all $n=0,1,\ldots,m$, we have
$P(T)=\sum_{n=0}^{m}A_{n}T^{n}\in\mathcal{B}(\mathcal{V})$.

Proof.  The arguments are more or less standard (see [6]). The
${\mathbb{R}}$-linearity of the maps

${\mathcal{O}}_{s}(U,\mathcal{B}(\mathcal{V})_{\mathbb{C}})\ni F\mapsto
F(T)\in\mathcal{B}(\mathcal{V}),\,{\mathcal{O}}_{s}(U)\ni f\mapsto
f(T)\in\mathcal{B}(\mathcal{V}),$

is clear. The second one is actually multiplicative, which follows from the
multiplicativiry of the usual analytic functional calculus of $T$.

In fact, we have a more general property, specifically $(Ff)(T)=F(T)f(T)$ for
all
$F\in\mathcal{O}_{s}(U,\mathcal{B}(\mathcal{V})_{\mathbb{C}}),\,f\in{\mathcal{O}}_{s}(U)$.
This follows from the equalities,

$(Ff)(T)=\frac{1}{2\pi
i}\int_{\Gamma_{0}}F(\zeta)f(\zeta)(\zeta-T)^{-1}d\zeta=$ $\left(\frac{1}{2\pi
i}\int_{\Gamma_{0}}F(\zeta)(\zeta-T)^{-1}d\zeta\right)\left(\frac{1}{2\pi
i}\int_{\Gamma}f(\eta)(\eta-T)^{-1}d\eta\right)=F(T)f(T),$

obtained as in the classical case (see [6], Section VII.3), which holds
because $f$ is ${\mathbb{C}}$-valued and commutes with the operators in
$\mathcal{B}(\mathcal{V})$. Here $\Gamma,\,\Gamma_{0}$ are the boundaries of
two Cauchy domains $\Delta,\,\Delta_{0}$ respectively, such that
$\Delta\supset\bar{\Delta}_{0}$, and $\Delta_{0}$ contains $\sigma(T)$.

Note that, in particular, for every polynomial
$P(\zeta)=\sum_{n=0}^{m}A_{n}\zeta^{n}$ with
$A_{n}\in\mathcal{B}(\mathcal{V})$ for all $n=0,1,\ldots,m$, we have
$P(T)=\sum_{n=0}^{m}A_{n}q^{n}\in\mathcal{B}(\mathcal{V})$ for all
$T\in\mathcal{B}(\mathcal{V})$.

###### Example 1

Let $\mathcal{V}={\mathbb{R}}^{2}$, so
$\mathcal{V}_{\mathbb{C}}={\mathbb{C}}^{2}$, endowed with its natural Hilbert
space structure. Let us first observe that we have

$S=\left(\begin{array}[]{cc}a_{1}&a_{2}\\\
a_{3}&a_{4}\end{array}\right)\,\,\Longleftrightarrow
S^{\flat}=\left(\begin{array}[]{cc}\bar{a}_{1}&\bar{a}_{2}\\\
\bar{a}_{3}&\bar{a}_{4}\end{array}\right),$

for all $a_{1},a_{2},a_{3},a_{4}\in{\mathbb{C}}$.

Next we consider the operator $T\in\mathcal{B}({\mathbb{R}}^{2})$ given by the
matrix

$T=\left(\begin{array}[]{cc}u&v\\\ -v&u\end{array}\right),$

where $u,v\in{\mathbb{R}},v\neq 0$. The extension $T_{\mathbb{C}}$ of the
operator $T$ to ${\mathbb{C}}^{2}$, which is a normal operator, is given by
the same formula. Note that

$\sigma_{\mathbb{C}}(T)=\\{\lambda\in{\mathbb{C}};(\lambda-u)^{2}+v^{2}=0\\}=\\{u\pm
iv\\}=\sigma(T_{\mathbb{C}}).$

Note also that the vectors $\nu_{\pm}=(\sqrt{2})^{-1}(1,\pm i)$ are normalized
eigenvectors for $T_{\mathbb{C}}$ corresponding to the eigenvalues $u\pm iv$,
respectively. The spectral projections of $T_{\mathbb{C}}$ corresponding to
these eigenvalues are given by

$E_{\pm}(T_{\mathbb{C}}){\bf w}=\langle{\bf
w},\nu_{\pm}\rangle\nu_{\pm}=\frac{1}{2}\left(\begin{array}[]{cc}1&\mp i\\\
\pm i&1\end{array}\right)\left(\begin{array}[]{c}w_{1}\\\
w_{2}\end{array}\right),$

for all ${\bf w}=(w_{1},w_{2})\in{\mathbb{C}}^{2}$.

Let $U\subset{\mathbb{C}}$ be an open set with $U\supset\\{u\pm iv\\}$, and
let $F:U\mapsto\mathcal{B}({\mathbb{C}}^{2})$ be analytic. We shall compute
explicitly $F(T_{\mathbb{C}})$. Let $\Delta$ be a Cauchy domain contained in
$U$ with its boundary $\Gamma$, and containing the points $u\pm iv$. Assuming
$v>0$, we have

$F(T_{\mathbb{C}})=\frac{1}{2\pi i}\int_{\Gamma}F(\zeta)(\zeta-
T_{\mathbb{C}})^{-1}d\zeta=$
$F(u+iv)E_{+}(T_{\mathbb{C}})+F(u-iv)E_{-}(T_{\mathbb{C}})=$
$\frac{1}{2}F(u+iv)\left(\begin{array}[]{cc}1&-i\\\
i&1\end{array}\right)+\frac{1}{2}F(u-iv)\left(\begin{array}[]{cc}1&i\\\
-i&1\end{array}\right).$

Assume now that $F(T_{\mathbb{C}})^{\flat}=F(T_{\mathbb{C}})$. Then we must
have

$(F(u+iv)-F(u-iv)^{\flat})\left(\begin{array}[]{cc}1&-i\\\
i&1\end{array}\right)=(F(u+iv)^{\flat}-F(u-iv))\left(\begin{array}[]{cc}1&i\\\
-i&1\end{array}\right).$

We also have the equalities

$\left(\begin{array}[]{cc}1&-i\\\
i&1\end{array}\right)\left(\begin{array}[]{c}1\\\
i\end{array}\right)=2\left(\begin{array}[]{c}1\\\
i\end{array}\right),\,\,\left(\begin{array}[]{cc}1&-i\\\
i&1\end{array}\right)\left(\begin{array}[]{c}1\\\ -i\end{array}\right)=0,$
$\left(\begin{array}[]{cc}1&i\\\
-i&1\end{array}\right)\left(\begin{array}[]{c}1\\\
-i\end{array}\right)=2\left(\begin{array}[]{c}1\\\
-i\end{array}\right),\,\,\left(\begin{array}[]{cc}1&i\\\
-i&1\end{array}\right)\left(\begin{array}[]{c}1\\\ i\end{array}\right)=0,$

Using these equalities, we finally deduce that

$(F(u+iv)-F(u-iv)^{\flat})\left(\begin{array}[]{c}1\\\ i\end{array}\right)=0,$

and

$(F(u-iv)-F(u+iv)^{\flat})\left(\begin{array}[]{c}1\\\
-i\end{array}\right)=0,$

which are necessary conditions for the equality
$F(T_{\mathbb{C}})^{\flat}=F(T_{\mathbb{C}})$. As a matter of fact, this
example shows, in particular, that the condition
$F(\zeta)^{\flat}=F(\bar{\zeta})$ for all $\zeta\in U$, used in Theorem 1, is
sufficient but it might not be always necessary.

##  4 Analytic Functional Calculus for Quaternionic
Operators

###  4.1 Quaternionic Spectrum

We now recall some known definitions and elementary facts (see, for instance,
[5], Section 4.6, and/or [21]).

Let ${\mathbb{H}}$ be the abstract algebra of quaternions, which is the four-
dimensional ${\mathbb{R}}$-algebra with unit $1$, generated by the ”imaginary
units“ $\\{\bf{j,k,l}\\}$, which satisfy

${\bf jk=-kj=l,\,kl=-lk=j,\,lj=-jl=k,\,jj=kk=ll}=-1.$

We may assume that ${\mathbb{H}}\supset{\mathbb{R}}$ identifying every number
$x\in{\mathbb{R}}$ with the element $x1\in{\mathbb{H}}$.

The algebra ${\mathbb{H}}$ has a natural multiplicative norm given by

$\|{\bf x}\|=\sqrt{x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{0}^{2}},\,\,{\bf
x}=x_{0}+x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l},\,\,x_{0},x_{1},x_{2},x_{3}\in{\mathbb{R}},$

and a natural involution

${\mathbb{H}}\ni{\bf x}=x_{0}+x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l}\mapsto{\bf x}^{*}=x_{0}-x_{1}{\bf j}-x_{2}{\bf k}-x_{3}{\bf
l}\in{\mathbb{H}}.$

Note that ${\bf x}{\bf x}^{*}={\bf x}^{*}{\bf x}=\|{\bf x}\|^{2}$, implying,
in particular, that every element ${\bf x}\in{\mathbb{H}}\setminus\\{0\\}$ is
invertible, and ${\bf x}^{-1}=\|{\bf x}\|^{-2}{\bf x}^{*}$.

For an arbitrary quaternion ${\bf x}=x_{0}+x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l},\,\,x_{0},x_{1},x_{2},x_{3}\in{\mathbb{R}}$, we set $\Re{\bf x}=x_{0}=({\bf
x}+{\bf x}^{*})/2$, and $\Im{\bf x}=x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l}=({\bf x}-{\bf x}^{*})/2$, that is, the real and imaginary part of ${\bf
x}$, respectively.

We consider the complexification
${\mathbb{C}}\otimes_{\mathbb{R}}{\mathbb{H}}$ of the ${\mathbb{R}}$-algebra
${\mathbb{H}}$ (see also [8]), which will be identified with the direct sum
${\mathbb{M}}={\mathbb{H}}+i{\mathbb{H}}$. Of course, the algebra
${\mathbb{M}}$ contains the complex field ${\mathbb{C}}$. Moreover, in the
algebra ${\mathbb{M}}$, the elements of ${\mathbb{H}}$ commute with all
complex numbers. In particular, the ”imaginary units“ $\bf j,k,l$ of the
algebra ${\mathbb{H}}$ are independent of and commute with the imaginary unit
$i$ of the complex plane ${\mathbb{C}}$.

In the algebra ${\mathbb{M}}$, there also exists a natural conjugation given
by $\bar{\bf a}={\bf b}-i{\bf c}$, where ${\bf a}={\bf b}+i{\bf c}$ is
arbitrary in ${\mathbb{M}}$, with ${\bf b},{\bf c}\in{\mathbb{H}}$ (see also
[8]). Note that $\overline{\bf a+b}=\bar{\bf a}+\bar{\bf b}$, and
$\overline{\bf ab}=\bar{\bf a}\bar{\bf b}$, in particular $\overline{r\bf
a}=r\bar{\bf a}$ for all ${\bf a},{\bf b}\in{\mathbb{M}}$, and
$r\in{\mathbb{R}}$. Moreover, $\bar{{\bf a}}={\bf a}$ if and only if ${\bf
a}\in{\mathbb{H}}$, which is a useful characterization of the elements of
${\mathbb{H}}$ among those of ${\mathbb{M}}$.

###### Remark 4

In the algebra ${\mathbb{M}}$ we have the identities

$(\lambda-{\bf x}^{*})(\lambda-{\bf x})=(\lambda-{\bf x})(\lambda-{\bf
x}^{*})=\lambda^{2}-\lambda({\bf x}+{\bf x}^{*})+\|{\bf
x}\|^{2}\in{\mathbb{C}},$

for all $\lambda\in{\mathbb{C}}$ and ${\bf x}\in{\mathbb{H}}$. If the complex
number $\lambda^{2}-2\lambda\Re{\bf x}+\|{\bf x}\|^{2}$ is nonnull, then both
element $\lambda-{\bf x}^{*},\,\lambda-{\bf x}$ are invertible. Conversely, if
$\lambda-{\bf x}$ is invertible, we must have $\lambda^{2}-2\lambda\Re{\bf
x}+\|{\bf x}\|^{2}$ nonnull; otherwise we would have $\lambda={\bf
x}^{*}\in{\mathbb{R}}$, so $\lambda={\bf x}\in{\mathbb{R}}$, which is not
possible. Therefore, the element $\lambda-{\bf x}\in{\mathbb{M}}$ is
invertible if and only if the complex number $\lambda^{2}-2\lambda\Re{\bf
x}+\|{\bf x}\|^{2}$ is nonnull. Hence, the element $\lambda-{\bf
x}\in{\mathbb{M}}$ is not invertible if and only if $\lambda=\Re{\bf x}\pm
i\|\Im{\bf x}\|$. In this way, the spectrum of a quaternion ${\bf
x}\in{\mathbb{H}}$ is given by the equality $\sigma({\bf x})=\\{s_{\pm}(\bf
x)\\}$, where $s_{\pm}(\bf x)=\Re{\bf x}\pm i\|\Im{\bf x}\|$ are the
eigenvalues of $\bf x$ (see also [20, 21]).

The polynomial $P_{\bf x}(\lambda)=\lambda^{2}-2\lambda\Re{\bf x}+\|{\bf
x}\|^{2}$ is the minimal polynomial of $\bf x$. In fact, the equality
$\sigma({\bf y})=\sigma({\bf x})$ for some ${\bf x,y}\in{\mathbb{H}}$ is an
equivalence relation in the algebra ${\mathbb{H}}$, which holds if and only if
$P_{\bf x}=P_{\bf y}$. In fact, setting
$\mathbb{S}=\\{\mathfrak{\kappa}\in{\mathbb{H}};\Re\mathfrak{\kappa}=0,\|\mathfrak{\kappa}\|=1\\}$
(that is the unit sphere of purely imaginary quaternions), representig an
arbitrary quaternion $\bf x$ under the form
$x_{0}+y_{0}\mathfrak{\kappa}_{0}$, with $x_{0},y_{0}\in{\mathbb{R}}$ and
$\mathfrak{\kappa}_{0}\in\mathbb{S}$, a quaternion $\bf y$ is equivalent to
$\bf x$ if anf only if it is of the form $x_{0}+y_{0}\mathfrak{\kappa}$ for
some $\mathfrak{\kappa}\in\mathbb{S}$ (see [3] or [21] for some details).

###### Remark 5

Following [5], a right ${\mathbb{H}}$-vector space $\mathcal{V}$ is a real
vector space having a right multiplication with the elements of
${\mathbb{H}}$, such that $(x+y){\bf q}=x{\bf q}+y{\bf q},\,x({\bf q}+{\bf
s})=x{\bf q}+x{\bf s},\,x({\bf q}{\bf s})=(x{\bf q}){\bf s}$ for all
$x,y\in\mathcal{V}$ and ${\bf q},{\bf s}\in{\mathbb{H}}$.

If $\mathcal{V}$ is also a Banach space the operator $T\in\mathcal{B(V)}$ is
right ${\mathbb{H}}$-linear if $T(x{\bf q})=T(x){\bf q}$ for all
$x\in\mathcal{V}$ and ${\bf q}\in{\mathbb{H}}$. The set of right
${\mathbb{H}}$ linear operators will be denoted by $\mathcal{B^{\rm r}(V)}$,
which is, in particular, a unital real algebra.

In a similar way, one defines the concept of a left ${\mathbb{H}}$-vector
space. A real vector space $\mathcal{V}$ will be said to be an
${\mathbb{H}}$-vector space if it is simultaneously a right ${\mathbb{H}}$\-
and a left ${\mathbb{H}}$-vector space. As noticed in [5], it is the framework
of ${\mathbb{H}}$-vector spaces an appropriate one for the study of right
${\mathbb{H}}$-linear operators.

If ${\mathcal{V}}$ is ${\mathbb{H}}$-vector space which is also a Banach
space, then ${\mathcal{V}}$ is said to be a Banach ${\mathbb{H}}$-space. In
this case, we also assume that $R_{\bf q}\in\mathcal{B}({\mathcal{V}})$, and
the map ${\mathbb{H}}\ni{\bf q}\mapsto R_{\bf q}\in\mathcal{B}({\mathcal{V}})$
is norm continuous, where $R_{\bf q}$ is the right multiplication of the
elements of $\mathcal{V}$ by a given quaternion ${\bf q}\in{\mathbb{H}}$.
Similarly, if $L_{\bf q}$ is the left multiplication of the elements of
$\mathcal{V}$ by the quaternion ${\bf q}\in{\mathbb{H}}$, we assume that
$L_{\bf q}\in\mathcal{B}({\mathcal{V}})$ for all ${\bf q}\in{\mathbb{H}}$, and
that the map ${\mathbb{H}}\ni{\bf q}\mapsto L_{\bf
q}\in\mathcal{B}({\mathcal{V}})$ is norm continuous. Note also that

$\mathcal{B^{\rm r}(V)}=\\{T\in\mathcal{B(V)};TR_{\bf q}=R_{\bf q}T,\,{\bf
q}\in{\mathbb{H}}\\}.$

To adapt the discussion regarding the real algebras to this case, we first
consider the complexification ${\mathcal{V}}_{\mathbb{C}}$ of ${\mathcal{V}}$.
Because ${\mathcal{V}}$ is an ${\mathbb{H}}$-bimodule, the space
${\mathcal{V}}_{\mathbb{C}}$ is actually an ${\mathbb{M}}$-bimodule, via the
multiplications

$({\bf q}+i{\bf s})(x+iy)={\bf q}x-{\bf s}y+i({\bf q}y+{\bf s}x),(x+iy)({\bf
q}+i{\bf s})=x{\bf q}-y{\bf s}+i(y{\bf q}+x{\bf s}),$

for all ${\bf q}+i{\bf s}\in{\mathbb{M}},\,{\bf q},{\bf
s}\in{\mathbb{H}},\,x+iy\in{\mathcal{V}}_{\mathbb{C}},\,x,y\in{\mathcal{V}}$.
Moreover, the operator $T_{\mathbb{C}}$ is right ${\mathbb{M}}$-linear, that
is $T_{\mathbb{C}}((x+iy)({\bf q}+i{\bf s}))=T_{\mathbb{C}}(x+iy)({\bf
q}+i{\bf s})$ for all ${\bf q}+i{\bf
s}\in{\mathbb{M}},\,x+iy\in{\mathcal{V}}_{\mathbb{C}}$, via a direct
computation.

Let $C$ be the conjugation of ${\mathcal{V}}_{\mathbb{C}}$. As in the real
case, for every $S\in\mathcal{B}({\mathcal{V}}_{\mathbb{C}})$, we put
$S^{\flat}=CSC$. The left and right multiplication with the quaternion ${\bf
q}$ on ${\mathcal{V}}_{\mathbb{C}}$ will be also denoted by $L_{\bf q},R_{\bf
q}$, respectively, as elements of $\mathcal{B}({\mathcal{V}}_{\mathbb{C}})$.
We set

$\mathcal{B}^{\rm
r}({\mathcal{V}}_{\mathbb{C}})=\\{S\in\mathcal{B}({\mathcal{V}}_{\mathbb{C}});SR_{\bf
q}=R_{\bf q}S,\,{\bf q}\in{\mathbb{H}}\\},$

which is a unital complex algebra containing all operators $L_{\bf q},{\bf
q}\in{\mathbb{H}}$. Note that if $S\in\mathcal{B}^{\rm
r}({\mathcal{V}}_{\mathbb{C}})$, then $S^{\flat}\in\mathcal{B}^{\rm
r}({\mathcal{V}}_{\mathbb{C}})$. Indeed, because $CR_{\bf q}=R_{\bf q}C$, we
also have $S^{\flat}R_{\bf q}=R_{\bf q}S^{\flat}$. In fact, as we have
$(S+S^{\flat})({\mathcal{V}})\subset{\mathcal{V}}$ and
$i(S-S^{\flat})({\mathcal{V}})\subset{\mathcal{V}}$, it folows that the
algebras $\mathcal{B}^{\rm r}({\mathcal{V}}_{\mathbb{C}}),\,\mathcal{B}^{\rm
r}({\mathcal{V}})_{\mathbb{C}}$ are isomorphic, and they will be often
identified, where $\mathcal{B^{\rm r}(V)}_{\mathbb{C}}=\mathcal{B^{\rm
r}(V)}+i\mathcal{B^{\rm r}(V)}$ is the complexification of $\mathcal{B^{\rm
r}(V)}$, which is also a unital complex Banach algebra.

Looking at the Definition 4.8.1 from [5] (see also [4]), we give the folowing.

###### Definition 1

For a given operator $T\in\mathcal{B^{\rm r}(V)}$, the set

$\sigma_{\mathbb{H}}(T):=\\{{\bf q}\in{\mathbb{H}};T^{2}-2(\Re{\bf q})T+\|{\bf
q}\|^{2})\,\,{\rm not}\,\,{\rm invertible}\\}$

is called the quaternionic spectrum (or simply the $Q$-spectrum) of $T$.

The complement
$\rho_{\mathbb{H}}(T)={\mathbb{H}}\setminus\sigma_{\mathbb{H}}(T)$ is called
the quaternionic resolvent (or simply the $Q$-resolvent) of $T$.

Note that, if ${\bf q}\in\sigma_{\mathbb{H}}(T)$), then $\\{{\bf
s}\in{\mathbb{H}};\sigma({\bf s})=\sigma({\bf
q})\\}\subset\sigma_{\mathbb{H}}(T)$.

Assuming that ${\mathcal{V}}$ is a Banach ${\mathbb{H}}$-space, then
$\mathcal{B^{\rm r}(V)}$ is a unital real Banach ${\mathbb{H}}$-algebra (that
is, a Banach algebra which also a Banach ${\mathbb{H}}$-space), via the
algebraic operations $({\bf q}T)(x)={\bf q}T(x)$, and $(T{\bf q})(x)=T({\bf
q}x)$ for all ${\bf q}\in{\mathbb{H}}$ and $x\in{\mathcal{V}}$. Hence the
complexification $\mathcal{B^{\rm r}(V)}_{\mathbb{C}}$ is, in particular, a
unital complex Banach algebra. Also note that the complex numbers, regarded as
elements of $\mathcal{B^{\rm r}(V)}_{\mathbb{C}}$, commute with the elements
of $\mathcal{B^{\rm r}(V)}$. For this reason, for each $T\in\mathcal{B^{\rm
r}(V)}$ we have the resolvent set

$\rho_{\mathbb{C}}(T)=\\{\lambda\in{\mathbb{C}};(T^{2}-2(\Re\lambda)T+|\lambda|^{2})^{-1}\in\mathcal{B^{\rm
r}(V)}\\}=$ $\\{\lambda\in{\mathbb{C}};(\lambda-
T_{\mathbb{C}})^{-1}\in\mathcal{B^{\rm
r}(V}_{\mathbb{C}})\\}=\rho(T_{\mathbb{C}}),$

and the associated spectrum $\sigma_{\mathbb{C}}(T)=\sigma(T_{\mathbb{C}})$.

Clearly, there exists a strong connexion between $\sigma_{\mathbb{H}}(T)$ and
$\sigma_{\mathbb{C}}(T)$. In fact, the set $\sigma_{\mathbb{C}}(T)$ looks like
a ”complex border“ of the set $\sigma_{\mathbb{H}}(T)$. Specifically, we can
prove the following.

###### Lemma 2

For every $T\in\mathcal{B^{\rm r}(V)}$ we have the equalities

$\sigma_{\mathbb{H}}(T)=\\{{\bf
q}\in{\mathbb{H}};\sigma_{\mathbb{C}}(T)\cap\sigma({\bf q})\neq\emptyset\\}.$
(2)

and

$\sigma_{\mathbb{C}}(T)=\\{\lambda\in\sigma({\bf q});{\bf
q}\in\sigma_{\mathbb{H}}(T)\\}.$ (3)

Proof. Let us prove (2). If ${\bf q}\in\sigma_{\mathbb{H}}(T)$, and so the
$T^{2}-2(\Re{\bf q})T+\|{\bf q}\|^{2}$ is not invertible, choosing
$\lambda\in\\{\Re{\bf q}\pm i\|\Im{\bf q}\|\\}=\sigma({\bf q})$, we clearly
have $T^{2}-2(\Re\lambda)T+|\lambda|^{2}$ not invertible, implying
$\lambda\in\sigma_{\mathbb{C}}(T)\cap\sigma({\bf q})\neq\emptyset$.

Conversely, if for some ${\bf q}\in{\mathbb{H}}$ there exists
$\lambda\in\sigma_{\mathbb{C}}(T)\cap\sigma({\bf q})$, and so
$T^{2}-2(\Re\lambda)T+|\lambda|^{2}=T^{2}-2(\Re{\bf q})T+\|{\bf q}\|^{2}$ is
not invertible, implying ${\bf q}\in\sigma_{\mathbb{H}}(T)$.

We now prove (3). Let $\lambda\in\sigma_{\mathbb{C}}(T)$, so the operator
$T^{2}-2(\Re\lambda)T+|\lambda|^{2}$ is not invertible. Setting ${\bf
q}=\Re(\lambda)+\|\Im\lambda\|\kappa$, with $\kappa\in\mathbb{S}$, we have
$\lambda\in\sigma({\bf q})$. Moreover, $T^{2}+2(\Re{\bf q})T+\|{\bf q}\|^{2}$
is not invertible, and so ${\bf q}\in\sigma_{\mathbb{H}}(T)$.

Conversely, if $\lambda\in\sigma({\bf q})$ for some ${\bf
q}\in\sigma_{\mathbb{H}}(T)$, then $\lambda\in\\{\Re{\bf q}\pm i\|\Im({\bf
q})\|\\}$, showing that $T^{2}-2\Re(\lambda)T+|\lambda|^{2}=T^{2}+2(\Re{\bf
q})T+\|{\bf q}\|^{2}$ is not invertible.

Remark As expected, the set $\sigma_{\mathbb{H}}(T)$ is nonempty and bounded,
which follows easily from Lemma 2. It is also compact, as a consequence of
Definition 1, because the set of invertible elements in $\mathcal{B^{\rm
r}(V)}$ is open.

We recall that a subset $\Omega\subset{\mathbb{H}}$ is said to be spectrally
saturated (see [20],[21]) if whenever $\sigma({\bf h})=\sigma({\bf q})$ for
some ${\bf h}\in{\mathbb{H}}$ and ${\bf q}\in\Omega$, we also have ${\bf
h}\in\Omega$. As noticed in [20] and [21], this concept coincides with that of
axially symmetric set, introduced in [5].

Note that the subset $\sigma_{\mathbb{H}}(T)$ spectrally saturated.

###  4.2 Analytic Functional Calculus

If ${\mathcal{V}}$ is a Banach ${\mathbb{H}}$-space, because $\mathcal{B^{\rm
r}({\mathcal{V}})}$ is real Banach space, each operator $T\in\mathcal{B^{\rm
r}({\mathcal{V}})}$ has a complex spectrum $\sigma_{\mathbb{C}}(T)$.
Therefore, applying the corresponding result for real operators, we may
construct an analytic functional calculus using the classical Riesz-Dunford
functional calculus, in a slightly generalized form. In this case, our basic
complex algebra is $\mathcal{B}^{\rm r}(\mathcal{V})_{\mathbb{C}}$, endowed
with the conjugation $\mathcal{B}^{\rm r}(\mathcal{V})_{\mathbb{C}}\ni
S\mapsto S^{\flat}\in\mathcal{B}^{\rm r}(\mathcal{V})_{\mathbb{C}}$.

###### Theorem 3

Let $U\subset{\mathbb{C}}$ be open and conjugate symmetric. If
$F:U\mapsto\mathcal{B}^{\rm r}(\mathcal{V}_{\mathbb{C}})$ is analytic and
$F(\zeta)^{\flat}=F(\bar{\zeta})$ for all $\zeta\in U$, then
$F(T_{\mathbb{C}})^{\flat}=F(T_{\mathbb{C}})$ for all $T\in\mathcal{B}^{\rm
r}(\mathcal{V})$ with $\sigma_{\mathbb{C}}(T)\subset U$.

Both the statement and the proof of Theorem 3 are similar to those of Theorem
1, and will be omitted.

As in the real case, we may identify the algebra $\mathcal{B}^{\rm
r}(\mathcal{V})$ with a subalgebra of $\mathcal{B}^{\rm
r}(\mathcal{V})_{\mathbb{C}}$. In ths case, when
$F\in\mathcal{O}_{s}(U,\mathcal{B}^{\rm
r}(\mathcal{V})_{\mathbb{C}})=\\{F\in{\mathcal{O}}(U,\mathcal{B}^{\rm
r}({\mathcal{V}})_{\mathbb{C}});F(\bar{\zeta})=F(\zeta)^{\flat}\,\,\forall\zeta\in
U\\}$ (see also Remark 3), we can write, via the previous Theorem,

$F(T)=\frac{1}{2\pi
i}\int_{\Gamma}F(\zeta)(\zeta-T)^{-1}d\zeta\in\mathcal{B}^{\rm
r}(\mathcal{V}),$

for a suitable choice of $\Gamma$.

The next result provides an analytic functional calculus for operators from
the real algebra $\mathcal{B}^{\rm r}(\mathcal{V})$.

###### Theorem 4

Let ${\mathcal{V}}$ be a Banach ${\mathbb{H}}$-space, let
$U\subset{\mathbb{C}}$ be a conjugate symmetric open set, and let
$T\in\mathcal{B}^{\rm r}(\mathcal{V})$, with $\sigma_{\mathbb{C}}(T)\subset
U$. Then the map

${\mathcal{O}}_{s}(U,\mathcal{B}^{\rm r}(\mathcal{V})_{\mathbb{C}})\ni
F\mapsto F(T)\in\mathcal{B}^{\rm r}(\mathcal{V})$

is ${\mathbb{R}}$-linear, and the map

${\mathcal{O}}_{s}(U)\ni f\mapsto f(T)\in\mathcal{B}^{\rm r}(\mathcal{V})$

is a unital real algebra morphism.

Moreover, the following properties are true:

$(1)$ For all $F\in\mathcal{O}_{s}(U,\mathcal{B}^{\rm
r}(\mathcal{V})_{\mathbb{C}}),\,f\in{\mathcal{O}}_{s}(U)$, we have
$(Ff)(T)=F(T)f(T)$.

$(2)$ For every polynomial
$P(\zeta)=\sum_{n=0}^{m}A_{n}\zeta^{n},\,\zeta\in{\mathbb{C}}$, with
$A_{n}\in\mathcal{B}^{\rm r}(\mathcal{V})$ for all $n=0,1,\ldots,m$, we have
$P(T)=\sum_{n=0}^{m}A_{n}T^{n}\in\mathcal{B}^{\rm r}(\mathcal{V})$.

The proof of this result is similar to that of Theorem 2 and will be omitted.

###### Remark 6

The algebra ${\mathbb{H}}$ is, in particular, a Banach ${\mathbb{H}}$-space.
As already noticed, the left multiplications $L_{\bf q},\,{\bf
q}\in{\mathbb{H}},$ are elements of $\mathcal{B}^{\rm r}({\mathbb{H}})$. In
fact, the map ${\mathbb{H}}\ni{\bf q}\mapsto L_{\bf q}\in\mathcal{B}^{\rm
r}({\mathbb{H}})$ is a injective morphism of real algebras allowing the
identification of ${\mathbb{H}}$ with a subalgebra of $\mathcal{B}^{\rm
r}({\mathbb{H}})$.

Let $\Omega\subset{\mathbb{H}}$ be a spectrally saturated open set, and let
$U=\mathfrak{S}(\Omega):=\\{\lambda\in{\mathbb{C}},\exists{\bf
q}\in\Omega,\lambda\in\sigma({\bf q})\\}$, which is open and conjugate
symmetric (see [21]). Denotig by $f_{\mathbb{H}}$ the function $\Omega\ni{\bf
q}\mapsto f({\bf q}),{\bf q}\in\Omega$, for every $f\in\mathcal{O}_{s}(U)$, we
set

$\mathcal{R}(\Omega):=\\{f_{\mathbb{H}};f\in\mathcal{O}_{s}(U)\\},$

which is a commutative real algebra. Defining the function $F_{\mathbb{H}}$ in
a similar way for each $F\in\mathcal{O}_{s}(U,{\mathbb{M}})$, we set

$\mathcal{R}(\Omega,{\mathbb{H}}):=\\{F_{\mathbb{H}};F\in\mathcal{O}_{s}(U,{\mathbb{M}})\\},$

which, according to the next theorem, is a right $\mathcal{R}(\Omega)$-module.

The next result is an analytic functional calculus for quaternions (see [21],
Theorem 5), obtained as a particular case of Theorem 4 (see also its
predecessor in [5]).

###### Theorem 5

Let $\Omega\subset{\mathbb{H}}$ be a spectrally saturated open set, and let
$U=\mathfrak{S}(\Omega)$. The space $\mathcal{R}(\Omega)$ is a unital
commutative ${\mathbb{R}}$-algebra, the space
$\mathcal{R}(\Omega,{\mathbb{H}})$ is a right $\mathcal{R}(\Omega)$-module,
the map

${\mathcal{O}}_{s}(U,{\mathbb{M}})\ni F\mapsto
F_{\mathbb{H}}\in\mathcal{R}(\Omega,{\mathbb{H}})$

is a right module isomorphism, and its restriction

${\mathcal{O}}_{s}(U)\ni f\mapsto f_{\mathbb{H}}\in\mathcal{R}(\Omega)$

is an ${\mathbb{R}}$-algebra isomorphism.

Moreover, for every polynomial
$P(\zeta)=\sum_{n=0}^{m}a_{n}\zeta^{n},\,\zeta\in{\mathbb{C}}$, with
$a_{n}\in{\mathbb{H}}$ for all $n=0,1,\ldots,m$, we have
$P_{\mathbb{H}}(q)=\sum_{n=0}^{m}a_{n}q^{n}\in{\mathbb{H}}$ for all
$q\in{\mathbb{H}}$.

Most of the assertions of Theorem 5 can be obtained directly from Theorem 4.
The injectivity of the map ${\mathcal{O}}_{s}(U)\ni f\mapsto
f_{\mathbb{H}}\in\mathcal{R}(\Omega)$, as well as an alternative complete
proof, can be obtained as in the proof of Theorem 5 from [21].

###### Remark 7

That Theorems 3 and 4 have practically the same proof as Theorems 1 and 2
(respectively) is due to the fact that all of them can be obtained as
particular cases of more general results. Indeed, considering a unital real
Banach algebra ${\mathcal{A}}$, and its complexification
${\mathcal{A}}_{\mathbb{C}}$, identifying ${\mathcal{A}}$ with a real
subalgebra of ${\mathcal{A}}_{\mathbb{C}}$, for a function
$F\in\mathcal{O}_{s}(U,A_{\mathbb{C}})$, where $U\subset{\mathbb{C}}$ is open
and conjugate symmetric, the element $F(b)\in{\mathcal{A}}$ for each
$b\in{\mathcal{A}}$ with $\sigma_{\mathbb{C}}(b)\subset U$. The assertion
follows as in the proof of Theorem 1. The other results also have their
counterparts. We omit the details.

###### Remark 8

The space $\mathcal{R}(\Omega,{\mathbb{H}})$ can be independently defined, and
it consists of the set of all ${\mathbb{H}}$-valued functions, which are slice
regular in the sense of [5], Definition 4.1.1. They are used in [5] to define
a quaternionic functional calculus for quaternionic linear operators (see also
[4]). Roughly speaking, given a quaternionic linear operator, each regular
quaternionic-valued function defined in a neighborhood $\Omega$ of its
quaternionic spectrum is associated with another quaternionic linear operator,
replacing formally the quaternionic variable with that operator. This
constraction is largely explained in the fourth chapter of [5].

Our Theorem 4 constructs an analytic functional calculus with functions from
${\mathcal{O}}_{s}(U,\mathcal{B}^{\rm r}(\mathcal{V})_{\mathbb{C}})$, where
$U$ is a a neighborhood of the complex spectrum of a given quaternionic linear
operator, leading to another quaternionic linear operator, replacing formally
the complex variable with that operator. We can show that those functional
calculi are equivalent. This is a consequence of the fact that the class of
regular quaternionic-valued function used by the construction in [5] is
isomorphic to the class of analytic functions used in our Theorem 5. The
advantage of our approach is its simplicity and a stronger connection with the
classical approach, using spectra defined in the complex plane, and Cauchy
type kernels partially commutative.

Let us give a direct argument concerning the equivalence of those analytic
functional calculi. For an operator $T\in\mathcal{B}^{\rm r}(\mathcal{V})$,
the so-called right $S$-resolvent is defined via the formula

$S_{R}^{-1}({\bf s},T)=-(T-{\bf s}^{*})(T^{2}-2\Re({\bf s})T+\|{\bf
s}\|)^{-1},\,\,{\bf s}\in\rho_{\mathbb{H}}(T)$ (4)

(see [5], formula (4.27)). Fixing an element $\kappa\in\mathbb{S}$, and a
spectrally saturated open set $\Omega\subset{\mathbb{H}}$, for
$\Phi\in\mathcal{R}(\Omega,{\mathbb{H}})$ one sets

$\Phi(T)=\frac{1}{2\pi}\int_{\partial(\Sigma_{\kappa})}\Phi({\bf s})d{\bf
s}_{\kappa}S_{R}^{-1}({\bf s},T),$ (5)

where $\Sigma\subset\Omega$ is a spectrally saturated open set containing
$\sigma_{\mathbb{H}}(T)$, such that
$\Sigma_{\kappa}=\\{u+v\kappa\in\Sigma;u,v\in{\mathbb{R}}\\}$ is a subset
whose boundary $\partial(\Sigma_{\kappa})$ consists of a finite family of
closed curves, piecewise smooth, positively oriented, and $d{\bf
s}_{\kappa}=-\kappa du\wedge dv$. Formula (5) is a (right) quaternionic
functional calculus, as defined in [5], Section 4.10.

Because the space $\mathcal{V}_{\mathbb{C}}$ is also an ${\mathbb{H}}$-space,
we may extend these formulas to the operator
$T_{\mathbb{C}}\in\mathcal{B}^{\rm r}(\mathcal{V}_{\mathbb{C}})$, extending
the operator $T$ to $T_{\mathbb{C}}$, and replacing $T$ by $T_{\mathbb{C}}$ in
formulas (4) and (5). For the function
$\Phi\in\mathcal{R}(\Omega,{\mathbb{H}})$ there exists a function
$F\in{\mathcal{O}}_{s}(U,\mathcal{B}^{\rm r}(\mathcal{V}_{\mathbb{C}}))$ such
that $F_{\mathbb{H}}=\Phi$. Denoting by $\Gamma_{\kappa}$ the boundary of a
Cauchy domain in ${\mathbb{C}}$ containing the compact set $\cup\\{\sigma({\bf
s});{\bf s}\in\overline{\Sigma_{\kappa}}\\}$, we can write

$\Phi(T_{\mathbb{C}})=\frac{1}{2\pi}\int_{\partial(\Sigma_{\kappa})}\left(\frac{1}{2\pi
i}\int_{\Gamma_{\kappa}}F(\zeta)(\zeta-{\bf s})^{-1}d\zeta\right)d{\bf
s}_{\kappa}S_{R}^{-1}({\bf s},T_{\mathbb{C}})=$ $\frac{1}{2\pi
i}\int_{\Gamma_{\kappa}}F(\zeta)\left(\frac{1}{2\pi}\int_{\partial(\Sigma_{\kappa})}(\zeta-{\bf
s})^{-1}d{\bf s}_{\kappa}S_{R}^{-1}({\bf s},T_{\mathbb{C}})\right)d\zeta.$

It follows from the complex linearity of $S_{R}^{-1}({\bf s},T_{\mathbb{C}})$,
and from formula (4.49) in [5], that

$(\zeta-{\bf s})S_{R}^{-1}({\bf s},T_{\mathbb{C}})=S_{R}^{-1}({\bf
s},T_{\mathbb{C}})(\zeta-T_{\mathbb{C}})-1,$

whence

$(\zeta-{\bf s})^{-1}S_{R}^{-1}({\bf s},T_{\mathbb{C}})=S_{R}^{-1}({\bf
s},T_{\mathbb{C}})(\zeta-T_{\mathbb{C}})^{-1}+(\zeta-{\bf s})^{-1}(\zeta-
T_{\mathbb{C}})^{-1},$

and therefore,

$\frac{1}{2\pi}\int_{\partial(\Sigma_{\kappa})}(\zeta-{\bf s})^{-1}d{\bf
s}_{\kappa}S_{R}^{-1}({\bf
s},T_{\mathbb{C}})=\frac{1}{2\pi}\int_{\partial(\Sigma_{\kappa})}d{\bf
s}_{\kappa}S_{R}^{-1}({\bf s},T_{\mathbb{C}})(\zeta-T_{\mathbb{C}})^{-1}+$
$\frac{1}{2\pi}\int_{\partial(\Sigma_{\kappa})}(\zeta-{\bf s})^{-1}d{\bf
s}_{\kappa}(\zeta-T_{\mathbb{C}})^{-1}=(\zeta-T_{\mathbb{C}})^{-1},$

because

$\frac{1}{2\pi}\int_{\partial(\Sigma_{\kappa})}d{\bf
s}_{\kappa}S_{R}^{-1}({\bf s},T_{\mathbb{C}})=1\,\,\,{\rm
and}\,\,\,\frac{1}{2\pi}\int_{\partial(\Sigma_{\kappa})}(\zeta-{\bf
s})^{-1}d{\bf s}_{\kappa}=0,$

as in Theorem 4.8.11 from [5], since the ${\mathbb{M}}$-valued function ${\bf
s}\mapsto(\zeta-{\bf s})^{-1}$ is analytic in a neighborhood of the set
$\overline{\Sigma_{\kappa}}\subset{\mathbb{C}}_{\kappa}$ for each
$\zeta\in\Gamma_{\kappa}$, respectively. Therefore
$\Phi(T_{\mathbb{C}})=\Phi(T)_{\mathbb{C}}=F(T_{\mathbb{C}})=F(T)_{\mathbb{C}}$,
implying $\Phi(T)=F(T)$.

##  5 Some Examples

###### Example 2

One of the simplest Banach ${\mathbb{H}}$-space is the space ${\mathbb{H}}$
itself. As already noticed (see Remark 6), taking
${\mathcal{V}}={\mathbb{H}}$, so ${\mathcal{V}}_{\mathbb{C}}={\mathbb{M}}$,
and fixing an element ${\bf q}\in{\mathbb{H}}$, we may consider the operator
$L_{\bf q}\in\mathcal{B}^{\rm r}({\mathbb{H}})$, whose complex spectrum is
given by $\sigma_{\mathbb{C}}(L_{\bf q})=\sigma({\bf q})=\\{\Re{\bf q}\pm
i\|\Im{\bf q}\|\\}$. If $U\subset{\mathbb{C}}$ is conjugate symmetric open set
containing $\sigma_{\mathbb{C}}(L_{\bf q})$, and
$F\in\mathcal{O}_{s}(U,{\mathbb{M}})$, then we have

$F({L_{\bf q}})=F(s_{+}({\bf q}))\iota_{+}(\mathfrak{s}_{\tilde{\bf
q}})+F(s_{-}({\bf q}))\iota_{-}(\mathfrak{s}_{\tilde{\bf q}})\in{\mathbb{M}},$
(6)

where $s_{\pm}({\bf q})=\Re{\bf q}\pm i\|\Im{\bf q}\|$, $\tilde{\bf q}=\Im\bf
q,\,\mathfrak{s}_{\tilde{\bf q}}=\tilde{\bf q}\|\tilde{\bf q}\|^{-1}$, and
$\iota_{\pm}(\mathfrak{s}_{\tilde{\bf q}})=2^{-1}(1\mp
i\mathfrak{s}_{\tilde{\bf q}})$ (see [21], Remark 3).

###### Example 3

Let ${\mathfrak{X}}$ be a topological compact space, and let
$C({\mathfrak{X}},{\mathbb{M}})$ be the space of ${\mathbb{M}}$-valued
continuous functions on ${\mathfrak{X}}$. Then
$C({\mathfrak{X}},{\mathbb{H}})$ is the real subspace of
$C({\mathfrak{X}},{\mathbb{M}})$ consisting of ${\mathbb{H}}$-valued
functions, which is also a Banach ${\mathbb{H}}$-space with respect to the
operations $({\bf q}F)(x)={\bf q}F(x)$ and $(F{\bf q})(x)=F(x){\bf q}$ for all
$F\in C({\mathfrak{X}},{\mathbb{H}})$ and $x\in{\mathfrak{X}}$. Moreover,
$C({\mathfrak{X}},{\mathbb{H}})_{\mathbb{C}}=C({\mathfrak{X}},{\mathbb{H}}_{\mathbb{C}})=C({\mathfrak{X}},{\mathbb{M}})$.

We fix a function $\Theta\in C({\mathfrak{X}},{\mathbb{H}})$ and define the
operator $T\in\mathcal{B}(C({\mathfrak{X}},{\mathbb{H}}))$ by the relation
$(TF)(x)=\Theta(x)F(x)$ for all $F\in C({\mathfrak{X}},{\mathbb{H}})$ and
$x\in{\mathfrak{X}}$. Note that $(T(F{\bf q}))(x)=\Theta(x)F(x){\bf
q}=((TF){\bf q})(x)$ for all $F\in C({\mathfrak{X}},{\mathbb{H}}),{\bf
q}\in{\mathbb{H}}$, and $x\in{\mathfrak{X}}$. In othe words,
$T\in\mathcal{B}^{\rm r}(C({\mathfrak{X}},{\mathbb{H}}))$. Note also that the
operator $T$ is invertible if and only if the function $\Theta$ has no zero in
${\mathfrak{X}}$.

Let us compute the $Q$-spectrum of $T$. According to Definition 1, we have

$\rho_{\mathbb{H}}(T)=\\{{\bf q}\in{\mathbb{H}};(T^{2}-2\Re{\bf q}\,T+\|{\bf
q}\|^{2})^{-1}\in\mathcal{B}^{\rm r}(C({\mathfrak{X}},{\mathbb{H}}))\\}.$

Consequently, ${\bf q}\in\sigma_{\mathbb{H}}(T)$ if and only if zero is in the
range of the function

$\tau({\bf q},x):=\Theta(x)^{2}-2\Re{\bf q}\,\Theta(x)+\|{\bf
q}\|^{2},\,x\in\mathfrak{X}.$

Similarly,

$\rho_{\mathbb{C}}(T)=\\{\lambda\in{\mathbb{C}};(T^{2}-2\Re\lambda\,T+\|\lambda\|^{2})^{-1}\in\mathcal{B}^{\rm
r}(C({\mathfrak{X}},{\mathbb{H}}))\\},$

and so $\lambda\in\sigma_{\mathbb{C}}(T)$ if and only if zero is in the range
of the function

$\tau(\lambda,x):=\Theta(x)^{2}-2\Re\lambda\,\Theta(x)+|\lambda|^{2},\,x\in\mathfrak{X}.$

Looking for solutions $u+iv,u,v\in{\mathbb{R}}$, of the equation
$(u-\Theta(x))^{2}+v^{2}=0$, a direct calculation shows that $u=\Re\Theta(x)$
and $v=\pm\|\Im\Theta(x)\|$. Hence

$\sigma_{\mathbb{C}}(T)=\\{\Re\Theta(x)\pm
i\|\Im\Theta(x)\|;x\in\mathfrak{X}\\}=\cup_{x\in\mathfrak{X}}\sigma(\Theta(x)).$

Of course, for every open conjugate symmetric subset $U\subset{\mathbb{C}}$
containing $\sigma_{\mathbb{C}}(T)$, and for every function
$\Phi\in\mathcal{O}_{c}(U,\mathcal{B}(C({\mathfrak{X}},{\mathbb{M}})))$, we
may construct the operator $\Phi(T)\in\mathcal{B}^{\rm
r}(C({\mathfrak{X}},{\mathbb{H}}))$, using Theorem 4.

##  6 Quaternionic Joint Spectrum of Paires

In many applications, it is more convenient to work with matrix quaternions
rather than with abstract quaternions. Specifically, one considers the
injective unital algebra morphism

${\mathbb{H}}\ni x_{1}+y_{1}{\bf j}+x_{2}{\bf k}+y_{2}{\bf
l}\mapsto\left(\begin{array}[]{cc}x_{1}+iy_{1}&x_{2}+iy_{2}\\\
-x_{2}+iy_{2}&x_{1}-iy_{1}\end{array}\right)\in{\mathbb{M}}_{2},$

with $x_{1},y_{1},x_{2},y_{2}\in{\mathbb{R}},$ where ${\mathbb{M}}_{2}$ is the
complex algebra of $2\times 2$-matrix, whose image, denoted by
${\mathbb{H}}_{2}$ is the real algebra of matrix quaternions. The elements of
${\mathbb{H}}_{2}$ can be also written as matrices of the form

$Q({\bf z})=\left(\begin{array}[]{cc}z_{1}&z_{2}\\\
-\bar{z}_{2}&\bar{z_{1}}\end{array}\right),\,\,{\bf
z}=(z_{1},z_{2})\in{\mathbb{C}}^{2}.$

A strong connection between the spectral theory of pairs of commuting
operators in a complex Hilbert space and the algebra of quaternions has been
firstly noticed in [17]. Another connection will be presented in this section.

If ${\mathcal{V}}$ is an arbitrary vector space, we denote by
${\mathcal{V}}^{2}$ the Cartesian product ${\mathcal{V}}\times{\mathcal{V}}$.

Let $\mathcal{V}$ be a real Banach space, and let ${\bf
T}=(T_{1},T_{2})\in\mathcal{B(V)}^{2}$ be a pair of commuting operators. The
extended pair ${\bf
T}_{\mathbb{C}}=(T_{1{\mathbb{C}}},T_{2{\mathbb{C}}})\in\mathcal{B(V_{\mathbb{C}})}^{2}$
also consists of commuting operators. For simplicity, we set

$Q({\bf
T}_{\mathbb{C}}):=\left(\begin{array}[]{cc}T_{1{\mathbb{C}}}&T_{2{\mathbb{C}}}\\\
-T_{2{\mathbb{C}}}&T_{1{\mathbb{C}}}\end{array}\right)$

which acts on the complex Banach space $\mathcal{V}_{\mathbb{C}}^{2}$.

We now define the quaternionic resolvent set and spectrum for the case of a
pair of operators, inspired by the previous discussion concerning a single
operator.

###### Definition 2

Let $\mathcal{V}$ be a real Banach space. For a given pair ${\bf
T}=(T_{1},T_{2})\in\mathcal{B(V)}^{2}$ of commuting operators, the set of
those $Q({\bf z})\in{\mathbb{H}}_{2},\,{\bf
z}=(z_{1},z_{2})\in{\mathbb{C}}^{2}$, such that the operator

$T_{1}^{2}+T_{2}^{2}-2\Re{z_{1}}T_{1}-2\Re{z_{2}}T_{2}+|z_{1}|^{2}+|z_{2}|^{2}$

is invertible in $\mathcal{B(V)}$ is said to be the quaternionic joint
resolvent (or simply the $Q$-joint resolvent) of ${\bf T}$, and is denoted by
$\rho_{\mathbb{H}}({\bf T})$.

The complement $\sigma_{\mathbb{H}}({\bf
T})={\mathbb{H}}_{2}\setminus\rho_{\mathbb{H}}({\bf T})$ is called the
quaternionic joint spectrum (or simply the $Q$-joint spectrum) of ${\bf T}$.

For every pair ${\bf
T}_{\mathbb{C}}=(T_{1{\mathbb{C}}},T_{2{\mathbb{C}}})\in\mathcal{B(V_{\mathbb{C}})}^{2}$
we put ${\bf
T}_{\mathbb{C}}^{c}=(T_{1{\mathbb{C}}},-T_{2{\mathbb{C}}})\in\mathcal{B(V_{\mathbb{C}})}^{2}$,
and for every pair ${\bf z}=(z_{1},z_{2})\in{\mathbb{C}}^{2}$ we put ${\bf
z}^{c}=(\bar{z}_{1},-z_{2})\in{\mathbb{C}}^{2}$

###### Lemma 3

A matrix quaternion $Q({\bf z})$ $({\bf z}\in{\mathbb{C}}^{2})$ is in the set
$\rho_{\mathbb{H}}({\bf T})$ if and only if the operators $Q({\bf
T}_{\mathbb{C}})-Q({\bf z}),\,Q({\bf T}_{\mathbb{C}}^{c})-Q({\bf z}^{c})$ are
invertible in $\mathcal{B}(\mathcal{V}_{\mathbb{C}}^{2})$.

Proof The assertion follows from the equalities

$\left(\begin{array}[]{cc}T_{1{\mathbb{C}}}-z_{1}&T_{2{\mathbb{C}}}-z_{2}\\\
-T_{2{\mathbb{C}}}+\bar{z}_{2}&T_{1{\mathbb{C}}}-\bar{z}_{1}\end{array}\right)\left(\begin{array}[]{cc}T_{1{\mathbb{C}}}-\bar{z}_{1}&-T_{2{\mathbb{C}}}+z_{2}\\\
T_{2{\mathbb{C}}}-\bar{z}_{2}&T_{1{\mathbb{C}}}-z_{1}\end{array}\right)=$
$\left(\begin{array}[]{cc}T_{1{\mathbb{C}}}-\bar{z}_{1}&-T_{2{\mathbb{C}}}+z_{2}\\\
T_{2{\mathbb{C}}}-\bar{z}_{2}&T_{1{\mathbb{C}}}-z_{1}\end{array}\right)\left(\begin{array}[]{cc}T_{1{\mathbb{C}}}-z_{1}&T_{2{\mathbb{C}}}-z_{2}\\\
-T_{2{\mathbb{C}}}+\bar{z}_{2}&T_{1{\mathbb{C}}}-\bar{z}_{1}\end{array}\right)=$
$[(T_{1{\mathbb{C}}}-z_{1})(T_{1{\mathbb{C}}}-\bar{z}_{1})+(T_{2{\mathbb{C}}}-z_{2})(T_{2{\mathbb{C}}}-\bar{z}_{2})]{\bf
I}.$

for all ${\bf z}=(z_{1},z_{2})\in{\mathbb{C}}^{2}$, where $\bf I$ is the
identity. Consequently, the operators $Q({\bf T}_{\mathbb{C}})-Q({\bf
z}),\,Q({\bf T}_{\mathbb{C}}^{c})-Q({\bf z}^{c})$ are invertible in
$\mathcal{B}({\mathcal{V}}_{\mathbb{C}}^{2})$ if and only if the operator
$(T_{1{\mathbb{C}}}-z_{1})(T_{1{\mathbb{C}}}-\bar{z}_{1})+(T_{2{\mathbb{C}}}-z_{2})(T_{2{\mathbb{C}}}-\bar{z}_{2})$
is invertible in $\mathcal{B}(\mathcal{V}_{\mathbb{C}})$. Because we have

$T_{1{\mathbb{C}}}^{2}+T_{2{\mathbb{C}}}^{2}-2\Re{z_{1}}T_{1{\mathbb{C}}}-2\Re{z_{2}}T_{2{\mathbb{C}}}+|z_{1}|^{2}+|z_{2}|^{2}=$
$[T_{1}^{2}+T_{1}^{2}-2\Re{z_{1}}T_{1}-2\Re{z_{2}}T_{2}+|z_{1}|^{2}+|z_{2}|^{2}]_{\mathbb{C}},$

the operators $Q({\bf T}_{\mathbb{C}})-Q({\bf z}),\,Q({\bf
T}_{\mathbb{C}}^{c})-Q({\bf z}^{c})$ are invertible in
$\mathcal{B}({\mathcal{V}}_{\mathbb{C}}^{2})$ if and only if the operator
$T_{1}^{2}+T_{1}^{2}-2\Re{z_{1}}T_{1}-2\Re{z_{2}}T_{2}+|z_{1}|^{2}+|z_{2}|^{2}$
is invertible in $\mathcal{B(V)}$.

Lemma 3 shows that we have the property $Q({\bf z})\in\sigma_{\mathbb{H}}({\bf
T})$ if and only if $Q(z^{c})\in\sigma_{\mathbb{H}}({\bf T}^{c})$. Putting

$\sigma_{{\mathbb{C}}^{2}}({\bf T}):=\\{{\bf z}\in{\mathbb{C}}^{2};Q({\bf
z})\in\sigma_{\mathbb{H}}({\bf T})\\},$

the set $\sigma_{{\mathbb{C}}^{2}}({\bf T})$ has a similar property,
specifically $\bf z\in\sigma_{{\mathbb{C}}^{2}}({\bf T})$ if and only if $\bf
z^{c}\in\sigma_{{\mathbb{C}}^{2}}({\bf T}^{c})$. As in the quaternionic case,
the set $\sigma_{{\mathbb{C}}^{2}}({\bf T})$ looks like a ”complex border“ of
the set $\sigma_{\mathbb{H}}({\bf T})$.

###### Remark 9

For the extended pair ${\bf
T}_{\mathbb{C}}=(T_{1{\mathbb{C}}},T_{2{\mathbb{C}}})\in{B(V_{\mathbb{C}})}^{2}$
of the commuting pair ${\bf T}=(T_{1},T_{2})\in\mathcal{B(V)}$ there is an
interesting connexion with the joint spectral theory of J. L. Taylor (see [15,
16]; see also [19]). Namely, if the operator
$T_{1{\mathbb{C}}}^{2}+T_{2{\mathbb{C}}}^{2}-2\Re{z_{1}}T_{1{\mathbb{C}}}-2\Re{z_{2}}T_{2{\mathbb{C}}}+|z_{1}|^{2}+|z_{2}|^{2}$
is invertible, then the point ${\bf z}=(z_{1},z_{2})$ belongs to the joint
resolvent of ${\bf T}_{\mathbb{C}}$. Indeed, setting

$R_{j}({\bf T}_{\mathbb{C}},{\bf
z})=(T_{j{\mathbb{C}}}-\bar{z}_{j})(T_{1{\mathbb{C}}}^{2}+T_{2{\mathbb{C}}}^{2}-2\Re{z_{1}}T_{1{\mathbb{C}}}-2\Re{z_{2}}T_{2{\mathbb{C}}}+|z_{1}|^{2}+|z_{2}|^{2})^{-1},$

$q=Q({\bf z})$ for $j=1,2$, we clearly have

$(T_{1{\mathbb{C}}}-z_{1})R_{1}({\bf T}_{\mathbb{C}},{\bf
z})+(T_{2{\mathbb{C}}}-z_{2})R_{2}({\bf T}_{\mathbb{C}},{\bf z})={\bf I},$

which, according to [15], implies that ${\bf z}$ is in the joint resolvent of
${\bf T}_{\mathbb{C}}$. A similar argument shows that, in this case the point
${\bf z}^{c}$ belongs to the joint resolvent of ${\bf T}_{\mathbb{C}}^{c}$. In
addition, if $\sigma(T_{\mathbb{C}})$ designates the Taylor spectrum of
$T_{\mathbb{C}}$, we have the inclusion
$\sigma(T_{\mathbb{C}})\subset\sigma_{{\mathbb{C}}^{2}}({\bf T})$. In
particular, for every complex-valued function $f$ analytic in a neighborhood
of $\sigma_{{\mathbb{C}}^{2}}({\bf T})$, the operator $f(\bf T_{\mathbb{C}})$
can be computed via Taylor’s analytic functional calculus. In fact, we have a
Martinelli type formula for the analytic functional calculus:

###### Theorem 6

Let $\mathcal{V}$ be a real Banach space, let ${\bf
T}=(T_{1},T_{2})\in\mathcal{B(V)}^{2}$ be a pair of commuting operators, let
$U\subset{\mathbb{C}}^{2}$ be an open set, let $D\subset U$ be a bounded
domain containing $\sigma_{{\mathbb{C}}^{2}}({\bf T})$, with piecewise-smooth
boundary $\Sigma$, and let $f\in\mathcal{O}(U)$. Then we have

$f({\bf T}_{\mathbb{C}})=\frac{1}{(2\pi i)^{2}}\int_{\Sigma}f({\bf z}))L({\bf
z,T_{\mathbb{C}}})^{-2}(\bar{z}_{1}-T_{1{\mathbb{C}}})d\bar{z}_{2}-(\bar{z}_{2}-T_{2{\mathbb{C}}})d\bar{z}_{1}]dz_{1}dz_{2},$

where

$L({\bf
z,T_{\mathbb{C}}})=T_{1{\mathbb{C}}}^{2}+T_{2{\mathbb{C}}}^{2}-2\Re{z_{1}}T_{1{\mathbb{C}}}-2\Re{z_{2}}T_{2{\mathbb{C}}}+|z_{1}|^{2}+|z_{2}|^{2}.$

Proof. Theorem III.9.9 from [19] implies that the map $\mathcal{O}(U)\ni
f\mapsto f({\bf T}_{\mathbb{C}})\in\mathcal{B(V_{\mathbb{C}})}$, defined in
terms of Taylor’s analytic functional calculus, is unital, linear,
multiplicative, and ordinary complex polynomials in ${\bf z}$ are transformed
into polynomials in ${\bf T}_{\mathbb{C}}$ by simple substitution, where
$\mathcal{O}(U)$ is the algebra of all analytic functions in the open set
$U\subset{\mathbb{C}}^{2}$, provided $U\supset\sigma({\bf T}_{\mathbb{C}})$.

The only thing to prove is that, when $U\supset\sigma_{{\mathbb{C}}^{2}}({\bf
T})$, Taylor’s functional calculus is given by the stated (canonical) formula.
In order to do that, we use an argument from the proof of Theorem III.8.1 in
[19], to make explicit the integral III(9.2) from [19] (see also [12]).

We consider the exterior algebra

$\Lambda[e_{1},e_{2},\bar{\xi_{1}},\bar{\xi_{2}},\mathcal{O}(U)\otimes\mathcal{V}_{\mathbb{C}}]=\Lambda[e_{1},e_{2},\bar{\xi_{1}},\bar{\xi_{2}}]\otimes\mathcal{O}(U)\otimes\mathcal{V}_{\mathbb{C}},$

where the indeterminates $e_{1},e_{2}$ are to be associated with the pair
${\bf T}_{\mathbb{C}}$, we put $\bar{\xi_{j}}=d\bar{z}_{j},\,j=1,2$, and
consider the operators $\delta=(z_{1}-T_{1{\mathbb{C}}})\otimes
e_{1}+(z_{2}-T_{2{\mathbb{C}}})\otimes
e_{2},\,\bar{\partial}=(\partial/\partial\bar{z_{1}})\otimes\bar{\xi_{1}}+(\partial/\partial\bar{z_{2}})\otimes\bar{\xi_{2}}$,
acting naturally on this exterior algebra, via the calculus with exterior
forms.

To simplify the computation, we omit the symbol $\otimes$, and the exterior
product will be denoted simply par juxtaposition.

We fix the exterior form $\eta=\eta_{2}=fye_{1}e_{2}$ for some
$f\in\mathcal{O}(U)$ and $y\in\mathcal{X}_{\mathbb{C}}$, which clearly satisfy
the equation $(\delta+\bar{\partial})\eta=0$, and look for a solution $\theta$
of the equation $(\delta+\bar{\partial})\theta=\eta$. We write
$\theta=\theta_{0}+\theta_{1}$, where $\theta_{0},\theta_{1}$ are of degree
$0$ and $1$ in $e_{1},e_{2}$, respectively. Then the equation
$(\delta+\bar{\partial})\theta=\eta$ can be written under the form
$\delta\theta_{1}=\eta,\,\delta\theta_{0}=-\bar{\partial}\theta_{1}$, and
$\bar{\partial}\theta_{0}=0$. Note that

$\theta_{1}=fL({\bf
z,T_{\mathbb{C}}})^{-1}[(\bar{z}_{1}-T_{1{\mathbb{C}}})ye_{2}-(\bar{z}_{2}-T_{2{\mathbb{C}}})]ye_{1}$

is visibly a solution of the equation $\delta\theta_{1}=\eta$. Further, we
have

$\bar{\partial}\theta_{1}=fL({\bf
z,T_{\mathbb{C}}})^{-2}[(z_{1}-T_{1{\mathbb{C}}})(\bar{z}_{2}-T_{2{\mathbb{C}}})y\bar{\xi}_{1}e_{1}-(z_{1}-T_{1{\mathbb{C}}})(\bar{z}_{1}-T_{1{\mathbb{C}}})y\bar{\xi}_{2}e_{1}+$
$(z_{2}-T_{2{\mathbb{C}}})(\bar{z}_{2}-T_{2{\mathbb{C}}})y\bar{\xi}_{1}e_{2}-(z_{2}-T_{2{\mathbb{C}}})(\bar{z}_{1}-T_{1{\mathbb{C}}})y\bar{\xi}_{2}e_{2}]=$
$\delta[fL({\bf
z,T_{\mathbb{C}}})^{-2}(\bar{z}_{1}-T_{1{\mathbb{C}}})y\bar{\xi}_{2}-fL({\bf
z,T_{\mathbb{C}}})^{-2}(\bar{z}_{2}-T_{2{\mathbb{C}}})y\bar{\xi}_{1}],$

so we may define

$\theta_{0}=-fL({\bf
z,T_{\mathbb{C}}})^{-2}(\bar{z}_{1}-T_{1{\mathbb{C}}})y\bar{\xi}_{2}+fL({\bf
z,T_{\mathbb{C}}})^{-2}(\bar{z}_{2}-T_{2{\mathbb{C}}})y\bar{\xi}_{1}.$

Formula III(8.5) from [19] shows that

$f({\bf T}_{\mathbb{C}})y=-\frac{1}{(2\pi
i)^{2}}\int_{U}\bar{\partial}(\phi\theta_{0})dz_{1}dz_{2}=$ $\frac{1}{(2\pi
i)^{2}}\int_{\Sigma}f({\bf z}))L({\bf
z,T_{\mathbb{C}}})^{-2}[(\bar{z}_{1}-T_{1{\mathbb{C}}})yd\bar{z}_{2}-(\bar{z}_{2}-T_{2{\mathbb{C}}})yd\bar{z}_{1}]dz_{1}dz_{2},$

for all $y\in\mathcal{X}_{\mathbb{C}}$, via Stokes’s formula, where $\phi$ is
a smooth function such that $\phi=0$ in a neighborhood of
$\sigma_{{\mathbb{C}}^{2}}({\bf T})$, $\phi=1$ on $\Sigma$ and the support of
$1-\phi$ is compact.

###### Remark 10

(1) We may extend the previous functional calculus to
$\mathcal{B(V}_{\mathbb{C}})$-valued analytic functions, setting, for such a
function $F$ and with the notation from above,

$F({\bf T}_{\mathbb{C}})=\frac{1}{(2\pi i)^{2}}\int_{\Sigma}F({\bf z}))L({\bf
z,T_{\mathbb{C}}})^{-2}(\bar{z}_{1}-T_{1{\mathbb{C}}})d\bar{z}_{2}-(\bar{z}_{2}-T_{2{\mathbb{C}}})d\bar{z}_{1}]dz_{1}dz_{2}.$

In particular, if $F({\bf z})=\sum_{j,k\geq
0}A_{jk{\mathbb{C}}}z_{1}^{j}z_{2}^{k}$, with $A_{j,k}\in\mathcal{B(V)}$,
where the series is convergent in neighborhood of
$\sigma_{{\mathbb{C}}^{2}}({\bf T})$, we obtain

$F({\bf T}):=F({\bf T}_{\mathbb{C}})|\mathcal{V}=\sum_{j,k\geq
0}A_{jk}T_{1}^{j}T_{2}^{k}\in\mathcal{B(V)}.$

(2) The connexion of the spectral theory of pairs with the algebra of
quaternions is even stronger in the case of complex Hilbert spaces.
Specifically, if $\mathcal{H}$ is a complex Hilbert space and ${\bf
V}=(V_{1},V_{2})$ is a commuting pair of bounded linear operators on
$\mathcal{H}$, a point ${\bf z}=(z_{1},z_{2})\in{\mathbb{C}}^{2}$ is in the
joint resolvent of ${\bf V}$ if and only if the operator $Q({\bf V})-Q({\bf
z})$ is invertible in $\mathcal{H}^{2}$, where

$Q({\bf V})=\left(\begin{array}[]{cc}V_{1}&V_{2}\\\
-V_{2}^{*}&V_{1}^{*}\end{array}\right).$

(see [17] for details). In this case, there is also a Martinelli type formula
which can be used to construct the associated analytic functional calculus
(see [18],[19]). An approach to such a construction in Banach spaces, by using
a so-called splitting joint spectrum, can be found in [14].

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