$\displaystyle\;2^{n-1}\sum_{k=1}^{p}|f_{k}\,g_{k}||\\{|E_{x,y}|(M_{k})+|E_{Jx,y}|(M_{k})\\}$
$\displaystyle\leq$
$\displaystyle\;2^{n-1}\sum_{k=1}^{p}|f_{k}\,g_{k}|\left(\sqrt{{\rm
Re}\,E_{x}(M_{k})}\,\sqrt{{\rm Re}\,E_{y}(M_{k})}+\sqrt{{\rm
Re}\,E_{Jx}(M_{k})}\,\sqrt{{\rm Re}\,E_{y}(M_{k})}\right)$ $\displaystyle=$
$\displaystyle\;2^{n-1}\sum_{k=1}^{p}|f_{k}\,g_{k}|\left(\sqrt{{\rm
Re}\,E_{x}(M_{k})}\,\sqrt{{\rm Re}\,E_{y}(M_{k})}+\sqrt{{\rm
Re}\,E_{x}(M_{k})}\,\sqrt{{\rm Re}\,E_{y}(M_{k})}\right)$ $\displaystyle\leq$
$\displaystyle\;2^{n}\left(\sum_{k=1}^{p}|f_{k}|^{2}{\rm
Re}\,E_{x}(M_{k})\right)^{1/2}$ $\displaystyle\times$
$\displaystyle\;\left(\sum_{k=1}^{p}|g_{k}|^{2}{\rm
Re}\,E_{y}(M_{k})\right)^{1/2}.$

Thus, (5.15) holds for simple functions. ∎

###  5.3. Spectral integrals of unbounded measurable functions

Fix $\mathfrak{I}\in\mathbb{S}$. 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., $E(\\{\lambda\in\Omega:f(\lambda)=\infty\\})=0$.

######  Definition 5.3.1 (bounding sequence).

Let $(M_{j})_{j=1}^{\infty}$ be a sequence of sets, where
$M_{j}\in\mathscr{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$ and
$M_{j}\subseteq M_{j+1}$ for $j=1,2,\ldots$. We will call
$(M_{j})_{j=1}^{\infty}$ a bounding sequence for a subset $\mathfrak{G}$ of
$\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$ if every
$f\in\mathfrak{G}$ is bounded on $M_{j}$ and $E(\cup_{j=1}^{\infty}M_{j})=I$.

######  Remark 5.3.2.

We note that if $(M_{j})_{j=1}^{\infty}$ is a bounding sequence, then for the
results appearing in Subsection 5.1, we have $E(M_{j})\preceq E(M_{j+1})$ for
$j=1,2,\ldots$,

$\lim_{j\to\infty}E(M_{j})x=x\quad\quad{\rm for}\quad x\in\mathcal{H}_{n}$

and $\cup_{j=1}^{\infty}E(M_{j})\mathcal{H}_{n}$ is dense in
$\mathcal{H}_{n}$.

######  Remark 5.3.3.

Every finite subset
$\\{f_{1},\ldots,f_{k}\\}\subseteq\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$
has a bounding sequence. To see this, let

$M_{j}:=\\{\lambda\in\Omega:|f_{j}(\lambda)|\leq n\quad{\rm for}\quad
j=1,\ldots,k\\}$

and $M=\cup_{j=1}^{\infty}M_{j}$. Thus,
$\Omega\,\setminus\,M\subseteq\cup_{j=1}^{k}\\{\lambda\in\Omega:f_{j}(\lambda)=\infty\\}$,
in which case $E(\Omega\,\setminus\,M)=0$. Consequently,
$(M_{j})_{j=1}^{\infty}$ is a bounding sequence for
$\\{f_{1},\ldots,f_{k}\\}$.

######  Theorem 5.3.4.

Let $f\in\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$ and

(5.16)
$\mathcal{D}(\mathbb{I}(f)):=\\{x\in\mathcal{H}_{n}:\int_{\Omega}|f(\lambda)|^{2}d({\rm
Re}\langle E(\lambda)x,x\rangle)<\infty\\}.$

For any bounded sequence $(M_{j})_{j=1}^{\infty}$ for $f$, we have the
following:

  1. (i)

$\mathcal{D}(\mathbb{I}(f))=\\{x\in\mathcal{H}_{n}:\text{$(\mathbb{I}(f\chi_{M_{j}})x)_{j=1}^{\infty}$
converges in
$\mathcal{H}_{n}$}\\}=\\{x\in\mathcal{H}_{n}:\sup_{j\in\mathbb{N}}\|\mathbb{I}(f\chi_{M_{j}})x\|<\infty\\}$.

  2. (ii)

For any $x\in\mathcal{D}(\mathbb{I}(f))$, the limit of the sequence
$(\mathbb{I}(f\chi_{M_{j}})_{k=1}^{\infty}$ does not depend on the choice of
the bounding sequence $(M_{j})_{j=1}^{\infty}$. Moreover, there is a linear
operator $\mathbb{I}(f)\in\mathcal{L}(\mathcal{H}_{n})$ with domain
$\mathcal{D}(\mathbb{I}(f))$ given by

(5.17)
$\mathbb{I}(f)x=\lim_{j\to\infty}\mathbb{I}(f\chi_{M_{j}})x\quad\quad{\rm
for}\quad x\in\mathcal{D}(\mathbb{I}(f)).$

  3. (iii)

The right submodule
$\cup_{j=1}^{\infty}E(M_{j})\mathcal{H}_{n}\subseteq\mathcal{D}(\mathbb{I}(f))\subseteq\mathcal{H}_{n}$
and is a core for $\mathbb{I}(f)$. Moreover,

(5.18)
$E(M_{j})\mathbb{I}(f)\subseteq\mathbb{I}(f)E(M_{j})=\mathbb{I}(f\chi_{M_{j}})\quad\quad{\rm
for}\quad j=1,2,\ldots.$

###### Proof.

The proof is broken into steps.

Step 1: Prove (i).

Let $x\in\mathcal{D}(\mathbb{I}(f))$. Thus, by definition, $f\in
L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x})$ and we may use
Lebesgue’s dominated convergence theorem to obtain $f\chi_{M_{i}}\to f$ in
$L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{x})$. Since
$(M_{j})_{j=1}^{\infty}$ is a bounding sequence for $f$, we have that $f$ is
bounded on $M_{j}$. Thus,
$f\chi_{M_{j}}\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}})$
and $\mathbb{I}(f\chi_{M_{j}})\in\mathcal{B}(\mathcal{H}_{n})$ given by
Definition 5.2.3. Using Theorem 5.2.4(v), we obtain

$\displaystyle\|\mathbb{I}(f\chi_{M_{i}})x-\mathbb{I}(f\chi_{M_{j}})x\|^{2}=$
$\displaystyle\;\|\mathbb{I}(f\chi_{M_{i}}-f\chi_{M_{j}})x\|^{2}$
$\displaystyle=$
$\displaystyle\;\int_{\Omega}|f(\lambda)\chi_{M_{i}}(\lambda)-f(\lambda)\chi_{M_{j}}(\lambda)|^{2}d({\rm
Re}\langle E(\lambda)x,x\rangle)$ $\displaystyle=$
$\displaystyle\;\|f\chi_{M_{i}}-f\chi_{M_{j}}\|^{2}_{L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{x})}\quad{\rm
for}\quad i,j=1,2,\ldots.$

Thus, putting all of the above observations together, we have that
$(\mathbb{I}(f\chi_{M_{i}})x)_{i=1}^{\infty}$ is a Cauchy sequence in
$\mathcal{H}_{n}$ and hence $\mathbb{I}(f\chi_{M_{i}})x)_{i=1}^{\infty}$
converges in $\mathcal{H}_{n}$.

Next, if $\\{\mathbb{I}(f\chi_{M_{i}})x:i=1,2,\ldots\\}$ converges in
$\mathcal{H}_{n}$, then $\\{\|\mathbb{I}(f\chi_{M_{i}})x\|:i=1,2,\ldots\\}$ is
bounded.

Finally, suppose that $\\{\|\mathbb{I}(f\chi_{M_{i}})x\|:i=1,2,\ldots\\}$ is
bounded. Then

$\sup_{i\in\mathbb{N}}\|\mathbb{I}(f\chi_{M_{i}})x\|<\infty.$

Since $|f(\lambda)\chi_{M_{i}}(\lambda)|^{2}$ converges monotonically to
$|f(\lambda)|^{2}$ $E_{x}$-a.e. on $\Omega$, we may use Lebesgue’s monotone
convergence theorem to obtain

$\displaystyle\int_{\Omega}|f(\lambda)|^{2}d({\rm Re}\langle
E(\lambda)x,x\rangle)=$
$\displaystyle\;\lim_{i\to\infty}\int_{\Omega}f(\lambda)\chi_{M_{i}}(\lambda)|^{2}d({\rm
Re}\langle E(\lambda)x,x\rangle)$ $\displaystyle=$
$\displaystyle\;\lim_{i\to\infty}\|\mathbb{I}(f\chi_{M_{i}})x\|^{2}<\infty.$

Thus, $f\in L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x})$ and
hence $x\in\mathcal{D}(\mathbb{I}(f))$. We have managed to show

$\displaystyle\mathcal{D}(\mathbb{I}(f))\subseteq$
$\displaystyle\;\\{x\in\mathcal{H}_{n}:\text{$(\mathbb{I}(f\chi_{M_{j}})x)_{j=1}^{\infty}$
converges in $\mathcal{H}_{n}$}\\}$ $\displaystyle\subseteq$
$\displaystyle\;\\{x\in\mathcal{H}_{n}:\sup_{j\in\mathbb{N}}\|\mathbb{I}(f\chi_{M_{j}})x\|<\infty\\}$
$\displaystyle\subseteq$ $\displaystyle\;\mathcal{D}(\mathbb{I}(f)),$

in which case we have (i).

Step 2: Prove (ii).

Suppose $(N_{i})_{i=1}^{\infty}$ is also a bounding sequence for $f$ and
$x\in\mathcal{D}(\mathbb{I}(f)).$ Using Theorem 5.2.4(i) and (v), we have

$\displaystyle\|\mathbb{I}(f\chi_{M_{i}}x-\mathbb{I}(f\chi_{N_{j}})x\|=$
$\displaystyle\;\|\mathbb{I}(f\chi_{M_{i}}-f\chi_{N_{j}})x\|_{L_{2}(\Omega,\mathbb{C}_{\mathfrak{I}},E_{x})}$
$\displaystyle\leq$
$\displaystyle\;\|f\chi_{M_{i}}-f\|_{L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x})}+\|f\chi_{N_{j}}-f\|_{L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x})}$
$\displaystyle\to$ $\displaystyle\;0\quad\quad{\rm as}\quad i,j\to\infty.$

Thus,

$\lim_{i\to\infty}\mathbb{I}(f\chi_{M_{i}})x=\lim_{j\to\infty}\mathbb{I}(f\chi_{N_{j}})x\quad\quad{\rm
for}\quad x\in\mathcal{D}(\mathbb{I}(f)).$

By (i), we have that $\mathcal{D}(\mathbb{I}(f))$ is a right linear subspace
of $\mathcal{H}_{n}$ and (5.17) gives rise to a right linear operator
$\mathbb{I}(f)\in\mathcal{L}(\mathcal{H}_{n})$.

Step 3: Prove (iii).

Suppose $y\in\mathcal{H}_{n}$. Recall that $E(M_{i})=\mathbb{I}(\chi_{M_{i}})$
and hence we may use Theorem 5.2.4 to obtain

$\displaystyle\mathbb{I}(f\chi_{M_{i}})y=$
$\displaystyle\;\mathbb{I}(f\chi_{M_{i}}\chi_{M_{j}})y=\mathbb{I}(f\chi_{M_{i}})E(M_{j})y$
(5.19) $\displaystyle=$
$\displaystyle\;E(M_{j})\mathbb{I}(f\chi_{M_{i}})y\quad\quad{\rm
whenever}\quad i\geq j.$

Consequently, $\sup\\{\|\mathbb{I}(f\chi_{M_{i}})y\|:i=1,2,\ldots\\}<\infty$
and hence $E(M_{i})y\in\mathcal{D}(\mathbb{I}(f))$, i.e., the union
$\cup_{i=1}^{\infty}E(M_{i})\mathcal{H}_{n}\subseteq\mathcal{D}(\mathbb{I}(f))$.

Letting $i\to\infty$ in (5.19), we obtain

$\mathbb{I}(f)E(M_{i})y=\mathbb{I}(f\chi_{M_{i}})y\quad\quad{\rm for}\quad
y\in\mathcal{H}_{n}.$

Suppose $x\in\mathcal{D}(\mathbb{I}(f))$ and letting $i\to\infty$ in (5.19),
we obtain the equality in (5.18).

Finally, using the fact that $E(M_{i})y\to y$ as $i\to\infty$ for all
$y\in\mathcal{H}_{n}$ and
$\mathbb{I}(f)E(M_{i})x=E(M_{i})\mathbb{I}(y)\to\mathbb{I}(f)x$ for all
$x\in\mathcal{D}(\mathbb{I}(f))$, we have that
$\cup_{i=1}^{\infty}E(M_{i})\mathcal{H}_{n}$ is a core for $\mathbb{I}(f)$. ∎

######  Remark 5.3.5.

If $f\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$, then
$(M_{i})_{i=1}^{\infty}$, where $M_{i}=\Omega$ for $i=1,2,\ldots$, is a
bounding sequence for $f$. Consequently, the operator
$\mathbb{I}(f)\in\mathcal{L}(\mathcal{H}_{n})$ given by (5.17) coincides with
$\mathbb{I}(f)\in\mathcal{B}(\mathcal{H}_{n})$ in Definition 5.2.3.

######  Theorem 5.3.6.

Suppose $f,g\in\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$.
Then

$\displaystyle\langle\mathbb{I}(f))x,\mathbb{I}(g)y\rangle=$
$\displaystyle\;\int_{\Omega}{\rm Re}(f(\lambda)\overline{g(\lambda)})d\langle
E(\lambda)x,y\rangle$ (5.20) $\displaystyle+$
$\displaystyle\;\int_{\Omega}{\rm Im}(f(\lambda)\overline{g(\lambda)})d\langle
JE(\lambda)x,y\rangle$

for $x\in\mathcal{D}(\mathbb{I}(f))$ and $y\in\mathcal{D}(\mathbb{I}(g))$ and

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

###### Proof.

We will first show (5.20). Let $(M_{i})_{i=1}^{\infty}$ be a bounding sequence
for $\\{f,g\\}$ and hence
$fg\chi_{M_{i}}\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$.
We may make use of Theorem 5.2.4(i) and (ii) to obtain

$\displaystyle\;\int_{\Omega}{\rm Re}(f(\lambda)\overline{g(\lambda)})d\langle
E(\lambda)x,y\rangle+\int_{\Omega}{\rm
Im}(f(\lambda)\overline{g(\lambda)})d\langle JE(\lambda)x,y\rangle$
$\displaystyle=$
$\displaystyle\;\langle\mathbb{I}(f\bar{g}\chi_{M_{i}})x,y\rangle$ (5.22)
$\displaystyle=$
$\displaystyle\;\langle\mathbb{I}(f\chi_{M_{i}})x,\mathbb{I}(g\chi_{M_{i}})y\rangle.$

Since $x\in\mathcal{D}(\mathbb{I}(f)$, we have $f\in
L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x})$ by (5.16).
Similarly, since $y\in\mathcal{D}(\mathbb{I}(g)$, we have $g\in
L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{y})$. One can use Lemma
5.2.5(ii) to see that the integrals

$\int_{\Omega}{\rm Re}(f(\lambda)\overline{g(\lambda)}d\langle
E(\lambda)x,y\rangle\;{\rm and}\;\int_{\Omega}{\rm
Im}(f(\lambda)\overline{g(\lambda)})d\langle E(\lambda)Jx,y\rangle$

are both convergent. We may also use Lemma 5.2.5(ii) and the fact that $f\to
f\chi_{M_{i}}$ in $L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x})$
to see that

$\displaystyle\;|\int_{\Omega}{\rm
Re}(f(\lambda)\overline{g(\lambda)})\chi_{M_{i}}(\lambda)d\langle
E(\lambda)x,y\rangle-\int_{\Omega}{\rm
Re}(f(\lambda)\overline{g(\lambda)})d\langle E(\lambda)x,y\rangle$
$\displaystyle+$ $\displaystyle\;\int_{\Omega}{\rm
Im}(f(\lambda)\overline{g(\lambda)})\chi_{M_{i}}(\lambda)d\langle
E(\lambda)Jx,y\rangle-\int_{\Omega}{\rm
Im}(f(\lambda)\overline{g(\lambda)})d\langle E(\lambda)Jx,y\rangle|$
$\displaystyle=$ $\displaystyle\;|\int_{\Omega}{\rm
Re}(\\{f(\lambda)\chi_{M_{i}}(\lambda)-f(\lambda)\\}\overline{g(\lambda)})d\langle
E(\lambda)x,y\rangle$ $\displaystyle+$ $\displaystyle\;\int_{\Omega}{\rm
Im}(\\{f(\lambda)\chi_{M_{i}}(\lambda)-f(\lambda)\\}\overline{g(\lambda)})d\langle
E(\lambda)Jx,y\rangle|$ $\displaystyle\leq$
$\displaystyle\;2^{n}\|f\chi_{M_{i}}-f\|_{L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{x})}\|g\|_{L_{2}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E_{y})}$
$\displaystyle\to$ $\displaystyle\;0\quad{\rm as}\quad i\to\infty.$

Thus, if we let $i\to\infty$ in (5.22), we obtain (5.20).

Formula (5.21) is easily obtained from (5.20) by putting $y=x$ and $g=f$. ∎

######  Theorem 5.3.7.

Let $J$ be an imaginary operator associated with the spectral measure $E$. For
any $f,g\in\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$ and
$c\in\mathbb{C}_{\mathfrak{I}}$, where $\mathfrak{I}\in\mathbb{S}$, we have
the following:

  1. (i)

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

  2. (ii)

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

  3. (iii)

$\mathbb{I}(fc+g)=\overline{\mathbb{I}(f)({\rm Re}(c)I+{\rm
Im}(c)J)+\mathbb{I}(g)}$ for all $c\in\mathbb{C}_{\mathfrak{I}}$.

  4. (iv)

$\mathbb{I}(f)$ is a closed normal operator belonging to
$\mathcal{L}(\mathcal{H}_{n})$.

  5. (v)

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

###### Proof.

Let $(M_{i})_{i=1}^{\infty}$ be a bounding sequence
$f,g\in\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$ (see
Remark 5.3.3). We note that it is easy to check that $(M_{i})_{I=1}^{\infty}$
is a bounded sequence for the functions $f+g,fg$ and $\overline{f}$.
Consequently, Theorem 5.3.4(iii) can be used to see that

(5.23)
$\mathcal{E}:=\\{\bigcup_{i=1}^{\infty}E(M_{i})x:x\in\mathcal{H}_{n}\\}$

is a core for $\mathbb{I}(f+g)$ and $\mathbb{I}(fg)$. The remainder of the
proof is broken into steps.

Step 1: Prove (i).

Suppose $x\in\mathcal{D}(\mathbb{I}(f))$ and $y\in\mathcal{D}(\mathbb{I}(g))$.
Using (5.18) and Theorem 5.2.4(i) we obtain

$\displaystyle\langle E(M_{i})\mathbb{I}(f)x,y\rangle=$
$\displaystyle\;\langle\mathbb{I}(f\chi_{M_{i}})x,y\rangle=\langle
x,\mathbb{I}(\overline{f\chi_{M_{i}}y})\rangle=\langle
x,\mathbb{I}(\overline{f}\chi_{M_{i}})y\rangle$ $\displaystyle=$
$\displaystyle\;\langle x,E(M_{i})\mathbb{I}(\overline{f})y\rangle.$

Letting $i\to\infty$ above results in $\langle\mathbb{I}(f)x,y\rangle=\langle
x,\mathbb{I}(\overline{f})y\rangle$, i.e.,
$\mathbb{I}(\overline{f})\subseteq\mathbb{I}(f)^{*}$. Next, suppose
$x\in\mathcal{H}_{n}$ and $y\in\mathcal{D}(\mathbb{I}(f)^{*})$. Then we can
again make use of (5.18) and Theorem 5.2.4(i) to obtain

$\langle
x,E(M_{i})\mathbb{I}(f)^{*}y\rangle=\langle\mathbb{I}(f)E(M_{i})x,y\rangle=\langle\mathbb{I}(f\chi_{M_{i}})\,x,y\rangle=\langle
x,\mathbb{I}(\overline{f\chi_{M_{i}}})y\rangle.$

Thus, $E(M_{i})\mathbb{I}(f)^{*}y=\mathbb{I}(\overline{f\chi_{M_{i}}})y$ and
hence

$\displaystyle\|\mathbb{I}(\overline{f\chi_{M_{i}}})y\|^{2}=$
$\displaystyle\;{\rm
Re}\langle\mathbb{I}(\overline{f\chi_{M_{i}}})y,\mathbb{I}(\overline{f\chi_{M_{i}}})y\rangle$
$\displaystyle=$ $\displaystyle\;{\rm Re}\langle
E(M_{i})\mathbb{I}(f)^{*}y,E(M_{i})\mathbb{I}(f)^{*}y\rangle$ $\displaystyle=$
$\displaystyle\;\|E(M_{i})\mathbb{I}(f)^{*}y\|^{2}.$

But then

$\sup_{i=1,2,\ldots}\|\mathbb{I}(f\chi_{M_{i}})\|\leq\|\mathbb{I}(f)^{*}y\|,$

in which case, Theorem 5.3.4(i) ensures that
$y\in\mathcal{D}(\mathbb{I}(\overline{f}))$. Thus,
$\mathcal{D}(\mathbb{I}(f)^{*})\subseteq\mathcal{D}(\mathbb{I}(\overline{f}))$
and hence $\mathbb{I}(\overline{f})\subseteq\mathbb{I}(f)^{*}$ implies that
$\mathbb{I}(f)^{*}=\mathbb{I}(\overline{f})$ as required.

Step 2: Prove (ii).

We begin by noting that (5.18) asserts that $\mathcal{E}$ is a dense right
submodule such that
$\mathcal{E}\subseteq\mathcal{D}(\mathbb{I}(\overline{g})\mathbb{I}(\overline{f}))$.
Consequently, we may use (i) and Theorem 3.0.24 to see that
$\mathbb{I}(f)\mathbb{I}(g)$ is closable. One can use (5.18) and Theorem
5.2.4(i) to check that

(5.24) $\mathbb{I}(fg)E(M_{i})=\mathbb{I}(f)\mathbb{I}(g)E(M_{i})$

and hence

(5.25)
$E(M_{i})\mathbb{I}(f)\mathbb{I}(g)\subseteq\mathbb{I}(f)\mathbb{I}(g)E(M_{i})=\mathbb{I}(fg)E(M_{i}).$

If we let $i\to\infty$ in (5.24) and we make use of the fact that
$\mathcal{E}$ is a core for $\mathbb{I}(fg)$, then we obtain
$\mathbb{I}(fg)\subseteq\overline{\mathbb{I}(f)\mathbb{I}(g)}$. On the other
hand, as $\mathbb{I}(fg)E(M_{i})\in\mathcal{B}(\mathcal{H}_{n})$, we may use
(5.25) to conclude
$E(M_{i})\overline{\mathbb{I}(f)\mathbb{I}(g)}\subseteq\mathbb{I}(fg)E(M_{i})$.
Letting $i\to\infty$ in
$E(M_{i})\overline{\mathbb{I}(f)\mathbb{I}(g)}\subseteq\mathbb{I}(fg)E(M_{i})$
and using (5.17), we obtain
$\overline{\mathbb{I}(f)\mathbb{I}(g)}\subseteq\mathbb{I}(fg)$. But then we
have (ii).

Step 3: Prove (iii).

The fact that $\mathbb{I}(fc)=\mathbb{I}(f)({\rm Re}(c)I+{\rm Im}(c)J)$, where
$I$ is the identity operator and $J$ is the imaginary operator, is an
immediate consequence of Theorem 5.2.4(iii) and (ii) in the present theorem.
Thus, in order to prove (iii), we need only consider $\mathbb{I}(f+g)$. Let
$\mathcal{E}$ be as in (5.23). Then, in view of (i), the set containments

$\mathcal{E}\subseteq\mathcal{D}(\mathbb{I}(\overline{f})+\mathbb{I}(\overline{g}))=\mathcal{D}(\mathbb{I}(f)^{*}+\mathbb{I}(g)^{*})\subseteq\mathcal{D}(\\{\mathbb{I}(f)+\mathbb{I}(g)\\}^{*})$

demonstrate that $\mathbb{I}(f)+\mathbb{I}(g)$ is closable. Moreover,
$\mathcal{E}$ is a dense right submodule of $\mathcal{H}_{n}$. Using (5.18)
and Theorem 5.2.4(i), we obtain

$\displaystyle(\mathbb{I}(f)+\mathbb{I}(g))E(M_{i})=$
$\displaystyle\;\mathbb{I}(f\chi_{M_{i}})+\mathbb{I}(g\chi_{M_{i}})$
$\displaystyle=$ $\displaystyle\;\mathbb{I}(f\chi_{M_{i}}+g\chi_{M_{i}})$
(5.26) $\displaystyle=$ $\displaystyle\;\mathbb{I}(f+g)E(M_{i})$

and

$\displaystyle E(M_{i})(\mathbb{I}(f)+\mathbb{I}(g))=$
$\displaystyle\;E(M_{i})\mathbb{I}(f)+E(M_{i})\mathbb{I}(g)$
$\displaystyle\subseteq$
$\displaystyle\;\\{\mathbb{I}(f)+\mathbb{I}(g)\\}E(M_{i})$ (5.27)
$\displaystyle=$ $\displaystyle\;\mathbb{I}(f+g)E(M_{i}).$

If we let $i\to\infty$ in (5.26) and make use of the fact that

$\\{\bigcup_{i=1}^{\infty}E(M_{i})x:x\in\mathcal{H}_{n}\\}$

is a core for $\mathbb{I}(f+g)$, then we obtain
$\mathbb{I}(f+g)\subseteq\overline{\mathbb{I}(f)+\mathbb{I}(g)}$. Finally,
(5.27) implies that
$E(M_{i})\overline{\mathbb{I}(f)+\mathbb{I}(g)}\subseteq\mathbb{I}(f+g)E(M_{i})$
and letting $i\to\infty$ we have
$\overline{\mathbb{I}(f)+\mathbb{I}(g)}\subseteq\mathbb{I}(f+g)$.

Step 4: Prove (iv).

It follows from (i) and (ii) that
$\mathbb{I}(f)^{*}\mathbb{I}(f)=\mathbb{I}(\overline{f})\mathbb{I}(f)\subset\mathbb{I}(\overline{f}f)$.
Note that $\mathbb{I}(f)^{*}\mathbb{I}(f)\in\mathcal{L}(\mathcal{H}_{n})$ is
closed. Thus, we may use Theorem 3.2.3(ii) to conclude that
$\mathbb{I}(f)^{*}\mathbb{I}(f)$ is self-adjoint. On the other hand, (i)
implies that $\mathbb{I}(\overline{f}f)$ is self-adjoint, in which case, we
have $\mathbb{I}(f)^{*}\mathbb{I}(f)=\mathbb{I}(\overline{f}f)$. We can
replicated the above argument to obtain
$\mathbb{I}(f)\mathbb{I}(f)^{*}=\mathbb{I}(f\overline{f})$ and hence we have
$\mathbb{I}(f)^{*}\mathbb{I}(f)=\mathbb{I}(f)\mathbb{I}(f)^{*}$.

Step 5: Prove (v).

We begin by noting that (ii) implies that
$\mathcal{D}(\mathbb{I}(f)\mathbb{I}(g))\subseteq\mathcal{D}(\mathbb{I}(fg))\cap\mathcal{D}(g)$.
Thus, in order to prove (v), we need only show the opposite set containment.
Suppose $x\in\mathcal{D}(\mathbb{I}(fg))\cap\mathcal{D}(\mathbb{I}(g))$. Then
we may use (5.24) to obtain
$\mathbb{I}(fg)E(M_{i})x=\mathbb{I}(f)\mathbb{I}(g)E(M_{i})x$ and letting
$i\to\infty$ and using the fact that $\mathbb{I}(f)$ is closed, we have
$\mathbb{I}(g)x\in\mathcal{D}(\mathbb{I}(f))$. Thus, we have
$x\in\mathcal{D}(\mathbb{I}(f)\mathbb{I}(f))$ as required. ∎

######  Theorem 5.3.8.

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

  1. (i)

If $f=g$ $E$-a.e. on $\Omega$, then $\mathbb{I}(f)=\mathbb{I}(g)$.

  2. (ii)

If $f$ is real-valued $E$-a.e. on $\Omega$, then
$\mathbb{I}(f)\in\mathcal{L}(\mathcal{H}_{n})$ is a self-adjoint operator.

  3. (iii)

If $f(\lambda)\geq 0$ $E$-a.e. on $\Omega$, then
$\mathbb{I}(f)\in\mathcal{L}(\mathcal{H}_{n})$ is a positive operator.
Moreover, $(\mathbb{I}(\sqrt{f}))^{2}=\mathbb{I}(f)$ and hence $\mathbb{I}(f)$
has a positive square root.

###### Proof.

Suppose $f=g$ $E$-a.e.. Then $E(\\{\lambda\in\Omega:f(\lambda)\neq
g(\lambda)\\})=0$ and then it follows from (5.16) that
$\mathcal{D}(\mathbb{I}(f))=\mathcal{D}(\mathbb{I}(g))$. We also have that any
bounding sequence $(M_{i})_{i=1}^{\infty}$ for $f$ is also a bounding sequence
for $g$. Consequently, we may use (5.17) to deduce that
$\mathbb{I}(f)=\mathbb{I}(g)$. Thus, we have proved (i).

Next, suppose $f$ is real-valued $E$-a.e.. Then from (5.20) with
$g(\lambda)=1$ and $y=x$ we obtain

(5.28) $\langle\mathbb{I}(f)x,x\rangle=\int_{\Omega}{\rm
Re}(f(\lambda))d\langle E(\lambda)x,x\rangle=\int_{\Omega}f(\lambda)d\langle
E(\lambda)x,x\rangle\in\mathfrak{S}(\mathbb{R}_{n})$

for all $x\in\mathcal{D}(\mathbb{I}(f))$. Thus, $\mathbb{I}(f)$ is self-
adjoint and we have proved (ii).

Next, suppose $f\geq 0$ $E$-a.e.. Then (5.28) can be used to show that
$\mathbb{I}(f)$ is a positive operator. Finally, in view of the first part of
(iii), it is obvious that $\mathbb{I}(\sqrt{f})$ is a positive operator. The
fact that $(\mathbb{I}(\sqrt{f}))^{2}=\mathbb{I}(f)$ is a direct consequence
of Theorem 5.3.7(ii) and (v). ∎

######  Definition 5.3.9
($L_{\infty}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$).

Let $f\in\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$. We
shall let

$L_{\infty}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E):=\\{\text{$E$-measurable
$f:\Omega\to\mathbb{C}_{\mathfrak{I}}$: $f$ is bounded on $\Omega$}\\}$

be endowed with the norm

$\|f\|_{\infty}:=\inf_{{\stackrel{{\scriptstyle M\in\mathscr{A}}}{{{\rm
with}\;E(M)=0}}}}\sup\\{|f(\lambda)|:\lambda\in\Omega\,\setminus\,M\\}.$

######  Lemma 5.3.10.

Let $f\in\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$. The
operator $\mathbb{I}(f)\in\mathcal{B}(\mathcal{H}_{n})$ if and only if $f\in
L_{\infty}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$. In this case,
$\|\mathbb{I}(f)\|=\|f\|_{\infty}$.

###### Proof.

Suppose $f\in L_{\infty}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$.
Then $|f(\lambda)|\leq\kappa$ $E$-a.e. for some $\kappa>0$ and

$\|\mathbb{I}(f)x\|^{2}=\int_{\Omega}|f(\lambda)|^{2}d({\rm Re}\langle
E(\lambda)x,x\rangle)\leq\|f\|_{\infty}\|x\|^{2}.$

Thus, $\mathbb{I}(f)\in\mathcal{B}(\mathcal{H}_{n})$.

Next, suppose $\mathbb{I}(f)\in\mathcal{B}(\mathcal{H}_{n})$ and put

$M_{i}:=\\{\lambda\in\Omega:|f(\lambda)|\geq\|\mathbb{I}(f)\|+2^{-i}\\}\quad\quad{\rm
for}\quad i=1,2,\ldots.$

and $M:=\cup_{i=1}^{\infty}M_{i}$. Note that
$M=\\{\lambda\in\Omega:|f(\lambda)|>\|\mathbb{I}(f)\|\\}$. Using Theorem
5.3.7(ii) and (5.21), we obtain

$\displaystyle\|\mathbb{I}(f)\|^{2}\cdot\|E(M_{i})x\|^{2}\geq$
$\displaystyle\;\|\mathbb{I}(f)E(M_{i})x\|^{2}$ $\displaystyle=$
$\displaystyle\;\|\mathbb{I}(f\chi_{M_{i}})x\|^{2}$ $\displaystyle=$
$\displaystyle\;\int_{\Omega}|f(\lambda)\chi_{M_{i}}(\lambda)|^{2}d({\rm
Re}\langle E(\lambda)x,x\rangle)$ $\displaystyle=$
$\displaystyle\;\int_{\Omega}|f(\lambda)|^{2}d({\rm Re}\langle
E(\lambda)x,x\rangle)$ $\displaystyle\geq$
$\displaystyle\;(\|\mathbb{I}(f)\|+2^{-i})^{2}\|E(M_{i})x\|^{2}\quad{\rm
for}\quad i\in\mathbb{N}\;{\rm and}\;x\in\mathcal{H}_{n}.$

But then $E(M_{i})=0$ for all $i=1,2,\ldots$, in which case $E(M)=0$, i.e.,
$\|f\|_{\infty}\leq\|\mathbb{I}(f)\|$. Thus, we have the characterisation of
elements in $L_{\infty}(\Omega,E)$ and $\|\mathbb{I}(f)\|=\|f\|_{\infty}$. ∎

######  Lemma 5.3.11.

Let $f\in\mathfrak{F}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$. The
right linear operator $\mathbb{I}(f)\in\mathcal{L}(\mathcal{H}_{n})$ is
invertible if and only if $f(\lambda)\neq 0$ $E$-a.e. on $\Omega$. In this
case, $\mathbb{I}(f)^{-1}=\mathbb{I}(1/f)$, with the understanding that
$\frac{1}{\infty}=0$ and $\frac{1}{0}=\infty$.

###### Proof.

Let $N:=\\{\lambda\in\Omega:f(\lambda)=0\\}$. We may use Theorem 5.3.7(iii)
and (v) to obtain

$\mathbb{I}(f)E(N)=\mathbb{I}(f\chi_{N})=\mathbb{I}(0)=0.$

Consequently, $\mathbb{I}(f)$ is not invertible whenever $E(N)\neq 0$.

Next, suppose $E(N)=0$ and hence $1/f\in\mathfrak{F}(\Omega,E)$. We may use
the fact that $f(1/f)=1$ $E$-a.e. on $\Omega$ to see that
$\mathcal{D}(\mathbb{I}(f(1/f)))=\mathcal{D}(\mathbb{I}(g))=\mathcal{H}_{n}$,
where $g(\lambda)=1$. Consequently, Theorem 5.3.7(v) implies that

$\mathcal{D}(\mathbb{I}(1/f)\mathbb{I}(f))=\mathcal{D}(\mathbb{I}(f)).$

Moreover, Theorem 5.3.7(iii) implies that

$\mathbb{I}(1/f)\mathbb{I}(f)\subseteq\mathbb{I}((1/f)f)=\mathbb{I}(1)=I.$

Thus, putting these observations together, we have that $\mathbb{I}(f)$ is
invertible and $\mathbb{I}(f)^{-1}\subseteq\mathbb{I}(1/f)$. To see that
$\mathbb{I}(1/f)\subseteq\mathbb{I}(f)$, we may simply replace $f$ by $1/f$ in
the above proof. ∎

######  Definition 5.3.12 (intrinsic function).

Let $\mathfrak{I}\in\mathbb{S}$. Let $\Omega\subseteq\mathbb{R}^{n+1}$ be an
axially symmetric open set and let
$\mathcal{U}=\\{(u,v)\in\mathbb{R}^{2}:u+\mathbb{S}v\subseteq\Omega\\}$. A
function $f:\Omega\to\mathbb{R}_{n}$ is called a left slice function, if it is
of the form

$f(\lambda)=f_{0}(u,v)+\mathfrak{I}f_{1}(u,v)\qquad\text{for
}\lambda=u+\mathfrak{I}v\in\Omega$

with two functions $f_{0},f_{1}:\mathcal{U}\to\mathbb{R}_{n}$ that satisfy the
compatibility conditions

(5.29) $f_{0}(u,-v)=f_{0}(u,v),$ (5.30) $f_{1}(u,-v)=-f_{1}(u,v).$

We say that $f$ is an intrinsic function if in addition $f_{0}$, and $f_{1}$
are real valued. We denote by $\mathcal{S}(\Omega)$ the set of intrinsic
functions defined on $\Omega$.

######  Remark 5.3.13.

For intrinsic functions, when we restrict $f$ to
$\Omega_{\mathfrak{I}}\subseteq\mathbb{C}_{\mathfrak{I}}$, we have that
$f(\Omega_{\mathfrak{I}})$ belongs to $\mathbb{C}_{\mathfrak{I}}$, since
$f_{0}$, and $f_{1}$ are real-valued. To stress this fact we will often use
the notation
$\mathfrak{F}(\Omega_{\mathfrak{I}},\mathscr{B}(\Omega_{\mathfrak{I}}),\mathbb{C}_{\mathfrak{I}})$
instead of $\mathfrak{F}(\Omega,\mathscr{B}(\Omega))$ when we work on a
complex plane $\mathbb{C}_{\mathfrak{I}}$, and when we consider the half plane
$\mathbb{C}_{\mathfrak{I}}^{+}$ we write
$\mathfrak{F}(\Omega_{\mathfrak{I}}^{+},\mathscr{B}(\Omega_{\mathfrak{I}})^{+},\mathbb{C}_{\mathfrak{I}}^{+})$.
Observe that condition (5.30) forces $f_{1}(u,v)=0$ for every
$u+\mathfrak{I}v\in\Omega_{\mathfrak{I}}^{+}\cap\mathbb{R}$ whenever $v=0$.

######  Theorem 5.3.14.

Let $E$ be a spectral measure on $\mathscr{B}(\Omega_{\mathfrak{I}}^{+})$,
where $\Omega_{\mathfrak{I}}^{+}\subseteq\mathbb{C}_{\mathfrak{I}}^{+}$ for
$\mathfrak{I}\in\mathbb{S}$, and
$f\in\mathfrak{F}(\Omega_{\mathfrak{I}}^{+},\mathscr{B}(\Omega_{\mathfrak{I}}^{+}),\mathbb{C}_{\mathfrak{I}},E)$
be an intrinsic function. Then

$\displaystyle\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}^{+}_{\mathfrak{I}}=$
$\displaystyle\;\\{s\in\mathbb{C}^{+}_{\mathfrak{I}}:E(\\{\lambda\in\Omega:|f(\lambda)^{2}-2{\rm
Re}(s)f(\lambda)+|s|^{2}|<\varepsilon\\})$ $\displaystyle\;\;\neq 0\text{ for
all $\varepsilon>0$}\\}.$

######  Remark 5.3.15.

The assumption that
$f\in\mathfrak{F}(\Omega,\mathscr{B}(\Omega),\mathbb{C}_{\mathfrak{I}},E)$ is
intrinsic (see Definition 5.3.12) is necessary for preserving the fact that
$\sigma_{S}(f(T))$ is axially symmetric (see Remark 3.1.7).

###### Proof of Theorem 5.3.14.

Notice that $s\in\rho_{S}(\mathbb{I}(f))$ if and only if
$(\mathbb{I}(f)^{2}-2{\rm
Re}(s)\mathbb{I}(f)+|s|^{2}I)^{-1}\in\mathcal{B}(\mathcal{H}_{n})$. Thus,
Lemma 5.3.10 and Lemma 5.3.11 imply that
$s\in\rho_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}^{+}$ if and only if
$f^{2}-2{\rm Re}(s)f+|s|^{2}\neq 0$ $E$-a.e. on $\Omega$ and $1/(f^{2}-2{\rm
Re}(s)f+|s|^{2})\in
L_{\infty}(\Omega,\mathscr{B}(\Omega),\mathbb{C}_{\mathfrak{I}},E)$, or,
equivalently, there exists $\kappa>0$ such that

$E(\\{\lambda\in\Omega:|\ f(\lambda)-2{\rm
Re}(s)f(\lambda)-|s|^{2}|\geq\kappa\\})=0.$

Thus, $s\in\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}^{+}$ if and
only if

$E(\\{\lambda\in\Omega:|f(\lambda)^{2}-2{\rm
Re}(s)f(\lambda)+|s|^{2}|<\varepsilon\\})\neq 0$

for all $\varepsilon>0$. ∎

######  Remark 5.3.16.

We wish to highlight that Theorem 5.3.14 will be very useful for showing that
an operator-valued integral (9.3) appearing in the spectral theorem for a
bounded normal operator is on the $S$-spectrum of $T$.

######  Theorem 5.3.17.

Let $E$ be a spectral measure on $\mathscr{B}(\Omega)$ and
$W\in\mathcal{B}(\mathcal{H}_{n})$. Then $E(M)W=WE(M)$ for all
$M\in\mathscr{B}(\Omega)$ if and only if
$W\mathbb{I}(f)\subseteq\mathbb{I}(f)W$ for every
$f\in\mathfrak{F}(\Omega,\mathscr{B}(\Omega),\mathbb{C}_{\mathfrak{I}},E)$.

###### Proof.

Suppose $E(M)W=WE(M)$ for all $M\in\mathscr{B}(\Omega)$. Then, as
$\mathbb{I}(\chi_{M})=E(M)$, we have $W\mathbb{I}(f)\subseteq\mathbb{I}(f)W$
for every
$f\in\mathfrak{F}(\Omega,\mathscr{B}(\Omega),\mathbb{C}_{\mathfrak{I}},E)$.
Conversely, suppose $W\mathbb{I}(f)\subseteq\mathbb{I}(f)W$ for every
$f\in\mathfrak{F}(\Omega,\mathscr{B}(\Omega),\mathbb{C}_{\mathfrak{I}},E)$.
Then

$\displaystyle{\rm Re}\,\langle E(M)Wx,Wx\rangle=$
$\displaystyle\;\|E(M)Wx\|^{2}$ $\displaystyle=$
$\displaystyle\;\|WE(M)x\|^{2}$ $\displaystyle\leq$
$\displaystyle\;\|W\|^{2}\|E(M)x\|^{2}$ $\displaystyle=$
$\displaystyle\;\|W\|^{2}{\rm Re}\langle E(M)x,x\rangle$

for all $M\in\mathscr{B}(\Omega)$ and $x\in\mathcal{H}_{n}$. Thus,
$Wx\in\mathcal{D}(\mathbb{I}(f))$ whenever $x\in\mathcal{D}(\mathbb{I}(f))$.
It is immediate to see that $W\mathbb{I}(f)=\mathbb{I}(f)W$ whenever
$f\in\mathfrak{B}_{s}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$ and
that this observation can be extended to all
$f\in\mathfrak{B}(\Omega,\mathscr{A},\mathbb{C}_{\mathfrak{I}},E)$.
Consequently, we have
$W\mathcal{D}(\mathbb{I}(f))\subseteq\mathcal{D}(\mathbb{I}(f))$ and hence
$W\mathbb{I}(f)\subseteq\mathbb{I}(f)W$. ∎

######  Theorem 5.3.18.

Suppose $S\in\mathcal{B}(\mathcal{H}_{n})$ and $E$ is a spectral measure on
$(\Omega,\mathscr{A})$. Then $SE(M)=E(M)S$ for every $M\in\mathscr{A}$ if and
only if $S\mathbb{I}(f)\subseteq\mathbb{I}(f)S$ for every
$f\in\mathfrak{F}(\Omega,\mathscr{B}(\Omega),\mathbb{C}_{\mathfrak{I}},E)$.

###### Proof.

If $S\mathbb{I}(f)\subseteq\mathbb{I}(f)S$ for every
$f\in\mathfrak{F}(\Omega,\mathscr{A},E)$, then $SE(M)=E(M)S$ for every
$M\in\mathscr{A}$ is obvious. Conversely, suppose $SE(M)=E(M)S$ for every
$M\in\mathscr{A}$. Since

$\displaystyle{\rm Re}(E_{Sx}(M))=$ $\displaystyle\;{\rm Re}(\langle
E(M)Sx,E(M)Sx\rangle)={\rm Re}(\langle SE(M)x,SE(M)x\rangle)$ $\displaystyle=$
$\displaystyle\;\|SE(M)x\|^{2}$ $\displaystyle\leq$
$\displaystyle\;\|S\|^{2}\|E(M)x\|^{2}$ $\displaystyle=$
$\displaystyle\;\|S\|^{2}{\rm Re}(\langle E(M)x,E(M)x\rangle,$

we may use (5.16) to conclude that $Sx\in\mathcal{D}(\mathbb{I}(f))$ whenever
$x\in\mathcal{D}(\mathbb{I}(f))$. Since $SE(M)=E(M)S$ for every
$M\in\mathscr{A}$, we have that $S\mathbb{I}(f_{0})=S\mathbb{I}(f_{0})$ for
every simple function $f_{0}\in\mathfrak{B}_{s}$, which can easily be
extended, by taking limits, to $S\mathbb{I}(f)=\mathbb{I}(f)S$ for every
$f\in\mathfrak{B}$. The fact that $S\mathbb{I}(f)=\mathbb{I}(f)S$ for every
$f\in\mathfrak{F}(\Omega,\mathscr{A},E)$ follows immediately from (5.17). ∎

######  Definition 5.3.19 (transformation of a spectral measure).

Let $E$ be a spectral measure on a Borel $\sigma$-algebra $\mathscr{A}$
generated by a set $\Omega$, $J$ be an associated imaginary operator with $E$
and $\psi$ be a function such that $\psi:\Omega\to\Omega^{\prime}$ and
$\mathscr{A}^{\prime}$ be the $\sigma$-algebra of all subsets
$M^{\prime}\subseteq\Omega^{\prime}$ such that
$\psi^{-1}(M^{\prime})\in\mathscr{A}$. We shall let

(5.31) $E^{\prime}(M^{\prime}):=E(\psi^{-1}(M^{\prime}))\quad\quad{\rm
for}\quad M^{\prime}\in\mathscr{A}^{\prime}.$

######  Remark 5.3.20.

It is very easy to check that
$E^{\prime}:\mathscr{A}^{\prime}\to\mathcal{B}(\mathcal{H}_{n})$ is a spectral
measure on $\mathscr{A}^{\prime}$. Moreover, in view of (5.31) if $J$ is an
imaginary operator associated with $E$, then $J$ is also an imaginary operator
associated with the $E^{\prime}$.

######  Theorem 5.3.21.

Let $\Omega,\Omega^{\prime},\psi,J,E$ and $E^{\prime}$ be as in Definition
5.3.19. Suppose $h\in\mathfrak{F}(\Omega^{\prime},\mathscr{A}^{\prime},F)$.
Then $h\circ\psi\in\mathfrak{F}(\Omega,\mathscr{A},E)$ and

$\displaystyle\;\int_{\Omega^{\prime}}{\rm
Re}(h(\lambda^{\prime}))\,dE^{\prime}(\lambda^{\prime})+\int_{\Omega^{\prime}}{\rm
Im}(h(\lambda^{\prime}))\,dE^{\prime}(\lambda^{\prime})J$ (5.32)
$\displaystyle=$ $\displaystyle\;\int_{\Omega}{\rm
Re}(h(\psi(\lambda)))\,dE(\lambda)+\int_{\Omega}{\rm
Im}(h(\psi(\lambda)))\,dE(\lambda)J$

###### Proof.

The fact that $h\circ\psi\in\mathfrak{F}(\Omega,\mathscr{A},E)$ follows
immediately from Definition 5.3.19. Since $\langle E(\cdot)x,x\rangle$ and
$\langle F(\cdot)x,x\rangle$ are positive $\mathbb{R}_{n}$-valued measures on
$\mathscr{A}$ and $\mathscr{A}^{\prime}$ for any $x\in\mathcal{H}_{n}$,
respectively, we have invoke Theorem 4.0.10 to obtain

$\int_{\Omega^{\prime}}|h(\lambda^{\prime})|^{2}d\langle
E^{\prime}(\lambda^{\prime})x,x\rangle=\int_{\Omega}|h(\psi(\lambda))|^{2}d\langle
E(\lambda)x,x\rangle\quad\quad{\rm for}\quad x\in\mathcal{H}_{n}$

and

$\displaystyle\;\int_{\Omega^{\prime}}{\rm Re}(h(\lambda^{\prime}))d\langle
E^{\prime}(\lambda^{\prime})y,y\rangle+\int_{\Omega^{\prime}}{\rm
Im}(h(\lambda^{\prime}))d\langle E^{\prime}(\lambda^{\prime})Jy,y\rangle$
(5.33) $\displaystyle=$ $\displaystyle\;\int_{\Omega}{\rm
Re}(h(\psi(\lambda)))d\langle E(\lambda)y,y\rangle+\int_{\Omega}{\rm
Im}(h(\psi(\lambda)))d\langle E(\lambda)Jy,y\rangle$

for all $y\in\mathcal{H}_{n}$ such that $h$ is $\langle
E^{\prime}(\cdot)y,y\rangle$-integrable on $\Omega^{\prime}$. We may use
(5.16) and (5.33) to realise that
$\mathcal{D}(\mathbb{I}_{E^{\prime}}(h))=\mathcal{D}(\mathbb{I}_{E}(h\circ\psi))$,
where $\mathbb{I}_{E^{\prime}}$ and $\mathbb{I}_{E}$ denote the spectral
integrals with respect to $E^{\prime}$ and $E$, respectively. Notice that
(5.20) with $g(\lambda)=1$ and (5.33) can be used to obtain
$\langle\mathbb{I}_{E^{\prime}}(h)y,y\rangle=\langle\mathbb{I}_{E}(h\circ\psi)y,y\rangle$
for all $y\in\mathcal{D}(\mathbb{I}_{E}(h\circ\psi))$ (which coincides with
$\mathbb{I}_{E^{\prime}}(h)$). Finally, the polarisation formula (2.20), to
deduce (5.32).

∎

##  6\. Spectral theorem for a bounded self-adjoint operator

The main goal of this section is to formulate and prove the spectral theorem
for bounded self-adjoint operators on a Clifford module (see Theorem 6.0.1).

######  Theorem 6.0.1 (spectral theorem for bounded self-adjoint operators).

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be self-adjoint. Then there exists a
spectral measure $E$ on the Borel $\sigma$-algebra
$\mathscr{B}(\sigma_{S}(T))$ such that

(6.1) $T=\int_{\sigma_{S}(T)}t\,dE(t).$

The spectral measure $E$ is unique in the sense that if $F$ is a spectral
measure on $\mathscr{B}(\mathbb{R})$ such that $T=\int_{\mathbb{R}}t\,dF(t)$,
then $E(M\cap\sigma_{S}(T))=F(M)$ for all $M\in\mathscr{B}(\mathbb{R})$.
Moreover, $W\in\mathcal{B}(\mathcal{H}_{n})$ commutes with $T$ if and only if
$WE(M)=E(M)W$ for every $M\in\mathscr{B}(\sigma_{S}(T))$.

Before we can prove Theorem 6.0.1, we need a number of lemmas. In the
following one, we wish to stress that the polynomial $p$ appearing in the
spectral mapping identity (6.2) is a polynomial with real coefficients.

######  Lemma 6.0.2.

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

(6.2) $\sigma_{S}(p(T))=p(\sigma_{S}(T))\quad\quad{\rm for}\quad
p\in\mathbb{R}[t].$

###### Proof.

Suppose $p(t)=\sum_{j=0}^{k}p_{j}t^{j}\in\mathbb{R}[t]$, with $p_{k}\neq 0$.
We begin by noting that
$p(T):=\sum_{j=0}^{n}p_{j}T^{j}\in\mathcal{B}(\mathcal{H}_{n})$ is self-
adjoint. Thus, Lemma 3.2.4 asserts that $\sigma_{S}(p(T))\subseteq\mathbb{R}$.
If $k=0$, then (6.2) holds trivially.

Suppose $k>0$. We will first show that
$p(\sigma_{S}(T))\subseteq\sigma_{S}(p(T))$. We will only show that
$\sigma_{S}(p(T))\subseteq p(\sigma_{S}(T))$ (the proof of the other
containment is very standard and does not divert at all from the classical
case). Choose $t_{0}\in\sigma_{S}(p(T))$. Consider the polynomial
$\varphi(t):=p(t)-t_{0}\in\mathbb{R}[t]$. If $\varphi$ has all real zeros, say
$t_{1},\ldots t_{k}$ (not necessarily distinct), then
$\varphi(t)=c\prod_{j=1}^{k}(t-t_{j})$ for some constant $c\in\mathbb{R}$.
Therefore,

$\varphi(T)=p(T)-t_{0}\,I=c\prod_{j=1}^{k}(T-t_{j}\,I).$

Since $\varphi(T)$ is not invertible, $T-t_{j^{\prime}}\,I$ must not be
invertible for some $j^{\prime}\in\\{1,\ldots,k\\}$. Thus,
$t_{j^{\prime}}\in\sigma_{S}(T)$, in which case $p(t_{j^{\prime}})=t_{0}$.

Next, if $\varphi$ has non-real zeros, say
$\zeta_{1},\bar{\zeta}_{1},\ldots,\zeta_{\ell},\bar{\zeta}_{\ell}$. Then

$\varphi(t)=\psi(t)\prod_{j=1}^{\ell}(t-\zeta_{k})(t-\bar{\zeta}_{m})$

for some polynomial $\psi\in\mathbb{R}[t]$ which is constant or has all real
zeros. Write $\zeta_{j}=u_{j}+v_{j}\,\mathfrak{I}$, where $v_{j}\neq 0$ for
$j=1,\ldots,\ell$. Since

$(t-\zeta_{j})(t-\bar{\zeta}_{j})=(t-u_{j})^{2}+v_{j}^{2}\quad\quad{\rm
for}\quad j=1,\ldots,\ell,$

we have

$\varphi(T)=\psi(T)\prod_{j=1}^{\ell}\\{(T-u_{j}\,I)^{2}+v_{j}^{2}\,I\\}.$

Since $T-u_{j}\,I\in\mathcal{B}(\mathcal{H}_{n})$ is self-adjoint, we have
that $(T-u_{j}\,I)^{2}$ is a positive operator, in which case
$(T-u_{j}\,I)^{2}+v_{j}^{2}\,I$ is invertible for $j=1,\ldots,\ell$.
Consequently, $\varphi(T)$ not being invertible implies that $\psi(T)$ is not
invertible. Hence, $\psi$ cannot be a constant polynomial. We may now proceed
as in the case when $\varphi$ has all real zeros to obtain the desired
conclusion. ∎

######  Lemma 6.0.3.

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

(6.3) $\|p(T)\|=\max_{t\in\sigma_{S}(T)}|p(t)|\quad\quad{\rm for}\quad
p\in\mathbb{R}[t].$

###### Proof.

In view of (3.10) and (3.23), we have

$\displaystyle\|p(T)\|^{2}=$
$\displaystyle\;\|p(T)^{*}\,p(T)\|=\|p(T)^{2}\|=r_{S}(p(T)^{2})$
$\displaystyle=$ $\displaystyle\;\max\\{|\tau|:\tau\in\sigma_{S}(p(T)^{2})\\}$
$\displaystyle=$
$\displaystyle\;\max\\{p(t_{0})^{2}:t_{0}\in\sigma_{S}(T)\\},$

i.e., (6.3) holds. ∎

######  Theorem 6.0.4 (functional calculus for bounded self-adjoint
operators).

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be self-adjoint. Then corresponding to
any $f,g\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$, there exist self-adjoint
operators $f(T),g(T)\in\mathcal{B}(\mathcal{H}_{n})$ which exhibit the
following properties:

  1. (i)

$(fg)(T)=f(T)g(T)$.

  2. (ii)

$(f+g)(T)=f(T)+g(T)$.

  3. (iii)

$\|f(T)\|=\|f\|_{\infty}$.

  4. (iv)

If $f|_{\sigma_{S}(T)}\geq 0$, then $f(T)$ is a positive operator.

###### Proof.

Fix $f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$. By the Weierstraß
approximation theorem, $\mathbb{R}[t]$ is uniformly dense in
$\mathscr{C}(\sigma_{S},\mathbb{R})$ with supremum norm

$\|f\|_{\infty}:=\sup_{t\in\sigma_{S}(T)}\,|f(t)|\quad\quad{\rm for}\quad
f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}).$

Thus, there exists a sequence of real-valued polynomials
$(f_{i})_{i=1}^{\infty}$ such that $f$ is the uniform limit of
$(f_{i})_{i=1}^{\infty}$ on $\sigma_{S}(T)$, i.e.,

$\lim_{n\to\infty}\,\max_{t_{0}\in\sigma_{S}(T)}|f(t_{0})-f_{n}(t_{0})|.$

Therefore, $(f_{i})_{i=1}^{\infty}$ is a Cauchy sequence and (6.3) applied to
$f_{i}$ implies that

$\lim_{i,j\to\infty}\|f_{i}(T)-f_{j}(T)\|=0.$

Since $\mathcal{B}(\mathcal{H}_{n})$ is complete, as must have that
$(f_{i}(T))_{i=1}^{\infty}$ has a limit in the uniform operator topology,
which we will denote by $f(T)$. One can easily check that $f(T)$ does not
depend on the choice of the Cauchy sequence $(f_{i})_{i=1}^{\infty}$.

Assertions (i)-(iii) are a direct consequence of the definition of $f(T)$ for
$f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$, given above, just as in the
classical complex Hilbert space setting. To prove (iv), observe that if
$f|_{\sigma_{S}(T)}\geq 0$, then there exists $g\in\mathbb{R}[t]$ such that
$f_{\sigma_{S}(T)}=g^{2}|_{\sigma_{S}(T)}$. Thus,

$\displaystyle\langle f(T)x,x\rangle=$ $\displaystyle\;\langle
g(T)^{2}x,x\rangle$ $\displaystyle=$ $\displaystyle\;\langle
g(T)x,g(T)x\rangle$ $\displaystyle\succeq$ $\displaystyle\;0\quad\quad{\rm
for}\quad x\in\mathcal{H}_{n}.$

Thus, $f(T)$ is a positive operator. ∎

We are now ready to prove Theorem 6.0.1.

###### Proof of Theorem 6.0.1.

Fix $T\in\mathcal{B}(\mathcal{H}_{n})$ and $x\in\mathcal{H}_{n}$. We shall
utilise the functional calculus given in Theorem 6.0.4 to make sense of $f(T)$
for $f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$. Let
$\ell_{x}:\mathscr{C}(\sigma_{S}(T),\mathbb{R})\to\mathfrak{S}(\mathbb{R}_{n})$,
where $\mathfrak{S}(\mathbb{R}_{n})$ denotes the real vector space of self-
adjoint Clifford number in $\mathbb{R}_{n}$, be the linear functional given by

$\ell_{x}(f)=\langle f(T)x,x\rangle\quad\quad{\rm for}\quad
f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}).$

The fact that $\ell_{x}(f)\in\mathfrak{S}(\mathbb{R}_{n})$ follows at once
from the fact that $f(T)=f(T)^{*}$ (see Theorem 6.0.4 and $\ell_{x}(f)=\langle
f(T)x,x\rangle$.

We claim that $\ell_{x}$ is a positive linear functional. If
$f|_{\sigma_{S}(T)}\geq 0$, then $f|_{\sigma_{S}(T)}=g^{2}|_{\sigma_{S}(T)}$
for some $g\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$. Thus,

$\displaystyle\ell_{x}(f)=$ $\displaystyle\;\langle g(T)^{2}x,x\rangle$
$\displaystyle=$ $\displaystyle\;\,\langle g(T)x,g(T)x\rangle$
$\displaystyle\geq$ $\displaystyle\;0\quad\quad{\rm for}\quad
x\in\mathcal{H}_{n},$

in which case,
$\ell_{x}:\mathscr{C}(\sigma_{S}(T),\mathbb{R})\to\mathfrak{S}(\mathbb{R}_{n})$
is a positive linear functional. Consequently, it follows from Theorem 4.0.11
that there exists a unique positive $\mathbb{R}_{n}$-valued Borel measure
$\mu_{x}$ on $\mathscr{B}(\sigma_{S}(T))$ such that

(6.4) $\ell_{x}(f)=\int_{\sigma_{S}(T)}f(t)\,d\mu_{x}(t)\quad\quad{\rm
for}\quad f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}).$

Fix $x,y\in\mathcal{H}_{n}$. Utilising the polarisation formula (2.20), one
can check that

(6.5) $\langle
f(T)x,y\rangle=\int_{\sigma_{S}(T)}f(t)\,d\mu_{x,y}(t)\quad\quad{\rm for}\quad
f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}),$

where $\mu_{x,y}$ denotes the $\mathbb{R}_{n}$-valued measure on
$\mathscr{B}(\sigma_{S}(T))$ given by

(6.6)
$\mu_{x,y}:=\frac{\sum_{\alpha}(\mu_{x+ye_{\alpha}}-\mu_{x-ye_{\alpha}})e_{\alpha}}{4\,\dim\mathfrak{S}(\mathbb{R}_{n})}\quad\quad{\rm
for}\quad x,y\in\mathcal{H}_{n}.$

Moreover, one can use (6.6) and the uniqueness of the positive Borel measure
in (6.4) to show that for any $x,y\in\mathcal{H}_{n}$, there is one and only
one $\mathbb{R}_{n}$-valued measure on $\mathscr{B}(\sigma_{S}(T))$ such that
(6.5) holds.

An immediate consequence of the uniqueness of $\mu_{x,y}$ in (6.5) is for any
$x,y,z\in\mathcal{H}_{n}$ and $a,b\in\mathbb{R}_{n}$, we have

$\mu_{xa+yb,z}(M)=\mu_{x,z}(M)\,a+\mu_{y,z}(M)b$

and

$\mu_{x,ya+zb}(M)=\bar{a}\mu_{x,y}(M)+\bar{b}\mu_{x,z}(M)$

for all $M\in\mathscr{B}(\sigma_{S}(T))$. We can also use (6.5) to show that
for any $x,y\in\mathcal{H}_{n}$, we have

$\displaystyle|\mu_{x,y}(M)|\leq$ $\displaystyle\;|\mu_{x,y}(\sigma_{S}(T))|$
$\displaystyle=$ $\displaystyle\;|\langle x,y\rangle|$ $\displaystyle\leq$
$\displaystyle\;\|x\|\|y\|\quad\quad{\rm for}\quad
M\in\mathscr{B}(\sigma_{S}(T)).$

Thus, for any $M\in\mathscr{B}(\sigma_{S}(T))$ we may use Theorem 3.0.26 to
obtain an operator $E(M)\in\mathcal{B}(\mathcal{H}_{n})$ such that

(6.7) $\langle E(M)x,y\rangle:=\mu_{x,y}(M).$

We now wish to show that
$E:\mathscr{B}(\sigma_{S}(T))\to\mathcal{B}(\mathcal{H}_{n})$ is a spectral
measure (see Definition 5.1.2). To see that $E$ is a spectral measure, we
shall make use of Lemma 5.1.6. Since $\langle E(\cdot)x,x\rangle=\mu_{x}$ is a
positive Borel measure (and hence $\mu_{x}$ is countably additive) for any
$x\in\mathcal{H}_{n}$, it suffices to show that $E(\sigma_{S}(T))=I$,
$E(M)=E(M)^{*}$ and $E(M)^{2}=E(M)$ for $M\in\mathscr{B}(\sigma_{S}(T))$.

Putting $f(t)=1$ into (6.5), we obtain $\langle
x,y\rangle=\mu_{x,y}(\sigma_{S}(T))=\langle E(\sigma_{S}(T)x,y\rangle$ for all
$x,y\in\mathcal{H}_{n}$ and hence $E(\sigma_{S}(T))=I$. Next, since

$\overline{\langle
f(T)x,y\rangle}=\int_{\sigma_{S}(T)}f(t)d\bar{\mu}_{x,y}(t)\quad\quad{\rm
for}\quad f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}).$

On the other hand,

$\displaystyle\overline{\langle f(T)x,y\rangle}=$ $\displaystyle\;\langle
y,f(T)x\rangle$ $\displaystyle=$ $\displaystyle\;\langle f(T)y,x\rangle$
$\displaystyle=$
$\displaystyle\;\int_{\sigma_{S}(T)}f(t)d\mu_{y,x}(t)\quad\quad{\rm for}\quad
f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}).$

Thus, for any $x,y\in\mathcal{H}_{n}$, we have

$\int_{\sigma_{S}(T)}f(t)d\bar{\mu}_{x,y}(t)=\int_{\sigma_{S}(T)}f(t)d\mu_{y,x}(t)\quad\quad{\rm
for}\quad f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$

and the aforementioned uniqueness of the $\mathbb{R}_{n}$-valued measure
$\mu_{x,y}$ such that (6.5) forces $\bar{\mu}_{x,y}=\mu_{y,x}$. Consequently,
(6.7) implies that $E(M)=E(M)^{*}$ for all $M\in\mathscr{B}(\sigma_{S}(T))$.

We will now prove that $E(M)^{2}=E(M)$ for all
$M\in\mathscr{B}(\sigma_{S}(T))$. Using the definition of $f(T)$ for
$f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$, one can extend the identity in
Theorem 6.0.4(i) to the case when
$f,g\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$. Thus, for any
$x,y\in\mathcal{H}_{n}$, we have

$\langle
f(T)g(T)x,y\rangle=\langle(fg)(T)x,y\rangle=\int_{\sigma_{S}(T)}f(t)g(t)d\mu_{x,y}(t)$

for all $f,g\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}).$ On the other hand,

$\langle
f(T)g(T)x,y\rangle=\int_{\sigma_{S}(T)}f(t)d\mu_{g(T)x,y}\quad\quad{\rm
for}\quad f,g\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}).$

Thus, using the uniqueness of the $\mathbb{R}_{n}$-valued measure $\mu_{x,y}$
in (6.5), we have $d\mu_{g(T)x,y}=g\,d\mu_{x,y}$ for all
$x,y\in\mathcal{H}_{n}$. Consequently, for any $x,y\in\mathcal{H}_{n}$, we
have $\langle E(M)g(T)x,y\rangle=\int_{M}g(t)d\mu_{x,y}$ for all
$M\in\mathscr{B}(\sigma_{S}(T))$. Putting these observations together, for any
$x,y\in\mathcal{H}_{n}$, we have

$\int_{\sigma_{S}(T)}g(t)d\mu_{x,E(M)y}=\int_{\sigma_{S}(T)}g(t)\chi_{M}(t)d\mu_{x,y}(t)$

for all $g\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$. Thus, using the
uniqueness of the $\mathbb{R}_{n}$-valued measure in (6.5), we have
$d\mu_{x,E(M)y}=\chi_{M}d\mu_{x,y}$, i.e.,

$\mu_{x,E(M)y}(N)=\int_{N}\chi_{M}(t)d\mu_{x,y}(t)=\mu_{x,y}(M\cap
N)\quad\quad{\rm for}\quad N\in\mathscr{B}(\sigma_{S}(T)),$

i.e., $\langle E(N)x,E(M)y\rangle=\langle E(M\cap N)x,y\rangle$. Using the
fact that $E(M)=E(M)^{*}$, we arrive at $\langle E(M)E(N)x,y\rangle=\langle
E(M\cap N)x,y\rangle$ for all $M,N\in\mathscr{B}(\sigma_{S}(T))$. Setting
$N=M$, we get $E(M)^{2}=E(M)$ for all $M\in\mathscr{B}(\sigma_{S}(T))$. Thus,
$E$ is a spectral measure on $\sigma_{S}(T)$.

Formula (6.1) follows at once from (6.5) with $f(t)=t$ together with (6.7).

Let $F$ be another spectral measure on $\mathscr{B}(\mathbb{R})$ such that
$\eqref{eq:SPECSA}$ holds with $F$ in place of $E$. We shall let
$\mathbb{I}_{F}(f)$ denote the spectral integral for
$f\in\mathfrak{F}(\mathscr{B}(\mathbb{R}),F)$ (see Theorem 5.3.7). Let
$g:\sigma_{S}(T)\to\mathbb{R}$ given by $g(t)=t$. Then

$T=\mathbb{I}_{F}(g)=\int_{\mathscr{B}(\mathbb{R})}t\,dF(t).$

Theorem 5.3.14 asserts that

$\displaystyle\sigma_{S}(\mathbb{I}(g))=$
$\displaystyle\;\\{s\in\mathbb{R}:F(\\{t\in\mathscr{B}(\mathbb{R}):|g(t)^{2}-2{\rm
Re}(s)g(t)+|s|^{2}|<\varepsilon\\})\neq 0$ $\displaystyle\;\;\;\text{ for all
$\varepsilon>0$}\\}$ $\displaystyle=$
$\displaystyle\;\\{s\in\mathbb{R}:F(\\{t\in\mathscr{B}(\mathbb{R}):|t^{2}-2st+s^{2}|<\varepsilon\\})\neq
0$ $\displaystyle\;\;\;\text{ for all $\varepsilon>0$}\\}$ $\displaystyle=$
$\displaystyle\;\\{s\in\mathbb{R}:F(\\{t\in\mathscr{B}(\mathbb{R}):|t-s|^{2}<\varepsilon\\})\neq
0$ $\displaystyle\;\;\;\text{ for all $\varepsilon>0$}\\}$ $\displaystyle=$
$\displaystyle\;{\rm supp}\,F,$

where we used (5.4) to obtain the last equality. Thus, as
$\sigma_{S}(T)=\sigma_{S}(\mathbb{I}_{F}(g))$, we have $\sigma_{S}(T)={\rm
supp}\,F$. On the other hand, let $\mathbb{I}(g)$ denote the spectral integral
of $g\in\mathfrak{F}(\sigma_{S}(T),E)$. Then, using the argument above, with
$F$ replaced by $E$, we obtain $\sigma_{S}(T)={\rm supp}\,E$ and hence
$\sigma_{S}(T)={\rm supp}\,E={\rm supp}\,F$. Using Theorem 5.2.4(v), we obtain
for any $x\in\mathcal{H}_{n}$,

$\int_{\sigma_{S}(T)}|f(t)|^{2}d({\rm Re}\langle
E(t)x,x\rangle)=\int_{\sigma_{S}(T)}|f(t)|^{2}d({\rm Re}\langle
F(t)x,x\rangle)$

for every $f\in\mathfrak{B}$ and, in particular, for every
$f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R})$. Thus, Theorem 4.0.7 forces
$\langle E(M)x,x\rangle=\langle F(M)x,x\rangle$ for all
$M\in\mathscr{B}(\sigma_{S}(T))$. Consequently, the polarisation formula
(2.20) forces $E=F|_{\sigma_{S}(T)}$.

Finally, suppose $WT=WT$ for $W\in\mathcal{B}(\mathcal{H}_{n})$. Using Theorem
5.2.4(iv), we have for any $x,y\in\mathcal{H}_{n}$,

$\langle f(T)Wx,y\rangle=\int_{\sigma_{S}(T)}{\rm Re}(f(t))d\langle
E(t)Wx,y\rangle\quad\quad{\rm for}\quad f\in\mathfrak{B}.$

In view of Definition 5.2.3, we have that $Wf(T)=f(T)W$ for every
$f\in\mathfrak{B}$ and hence, for any $x,y\in\mathcal{H}_{n}$,

$\langle f(T)Wx,y\rangle=\langle f(T)x,W^{*}y\rangle=\int_{\sigma_{S}(T)}{\rm
Re}(f(t))d\langle E(t)x,W^{*}y\rangle\quad{\rm for}\;f\in\mathfrak{B}.$

Putting these observations together, we have, in particular,

$\int_{\sigma_{S}(T)}f(t)d\langle
E(t)Wx,y\rangle=\int_{\sigma_{S}(T)}f(t)d\langle E(t)x,W^{*}y\rangle\quad{\rm
for}\;f\in\mathscr{C}(\sigma_{S}(T),\mathbb{R}).$

Thus, using Corollary 4.0.8, we have $\langle E(M)Wx,y\rangle=\langle
E(M)x,W^{*}y$, i.e., $\langle E(M)Wx,y\rangle=\langle WE(M)x,y\rangle$ for
every $x,y\in\mathcal{H}_{n}$ and $M\in\mathscr{B}(\sigma_{S}(T))$, i.e.,
$E(M)W=WE(M)$ for every $M\in\mathscr{B}(\sigma_{S}(T))$.

∎

######  Remark 6.0.5.

Let $T$ and $E$ be as in Theorem 6.0.1. We wish to record that in the proof of
Theorem 6.0.1 we showed that the support of the spectral measure of $A$ is
precisely the $S$-spectrum of $T$, i.e.,

${\rm supp}\,E=\sigma_{S}(T).$

######  Corollary 6.0.6.

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be a positive operator. Then there
exists a unique positive operator $A\in\mathcal{B}(\mathcal{H}_{n})$ such that
$A^{2}=T$. Moreover, if $(f_{j})_{j=1}^{\infty}$ is any sequence of
polynomials such that $f_{j}(t)\to t^{1/2}$ uniformly on $[0,d]$, where
$d=\max\sigma_{S}(T)$, then

(6.8) $\lim_{j\to\infty}\|f_{j}(T)-T^{1/2}\|=0.$

###### Proof.

The spectral theorem for a bounded self-adjoint operator (see Theorem 6.0.1
asserts that there exists a uniquely determined spectral measure $E$ on
$\sigma_{S}(T)$ such that $T=\int_{\sigma_{S}(T)}t\,dE(t)$. In view of Lemma
3.2.5, we have $\sigma_{S}(T)\subseteq[0,d]$, where $d\geq 0$. Notice that
have that
$f(t)=t^{1/2}\in\mathfrak{B}(\sigma_{S}(T),\mathscr{B}(\sigma_{S}(T)),\mathbb{C}_{\mathfrak{I}})$
(see Definition 5.2.3), where $\mathfrak{I}\in\mathbb{S}$, and

$A:=f(T)=\int_{\sigma_{S}(T)}t^{1/2}\,dE(t).$

Thus,

$\langle Ax,x\rangle=\int_{[0,d]}t^{1/2}d\langle E(t)x,x\rangle\succeq
0\quad\quad{\rm for}\quad x\in\mathcal{H}_{n}$

and hence $A$ is a positive operator and we may use Theorem 5.2.4(i) to check
that $A^{2}=T$. The limit (6.8) is an immediate consequence of Theorem
5.2.4(vii). We will now prove that there is only one positive operator
$W\in\mathcal{B}(\mathcal{H}_{n})$ such that $W^{2}=T$. Suppose
$\widetilde{A}\in\mathcal{B}(\mathcal{H}_{n})$ is a positive operator such
that $\widetilde{A}^{2}=T$. Then it is easy to see that
$\widetilde{A}T=T\widetilde{A}$ and hence
$\widetilde{A}g(T)=g(T)\widetilde{A}$ for any $g\in\mathbb{R}[t]$.
Consequently, we have

$\widetilde{A}\left(\lim_{j\to\infty}f_{j}(T)\right)=\left(\lim_{j\to\infty}f_{j}(T)\right)\widetilde{A},$

where $(f_{j})_{j=0}^{\infty}$ is any sequence of polynomials in
$\mathbb{R}[t]$ such that $f_{j}(t)\to t^{1/2}$ uniformly on $[0,d]$, i.e.,

(6.9) $\widetilde{A}A=A\tilde{A}.$

For any $x\in\mathcal{H}_{n}$, let $y:=(A-\widetilde{A})x$. Thus, using (6.9),
we obtain

(6.10) $\langle
Ay,y\rangle+\langle\widetilde{A}y,y\rangle=\langle(A+\widetilde{A})(A-\widetilde{A})x,y\rangle=\langle(A^{2}-\widetilde{A}^{2})x,y\rangle=0.$

Therefore, as $\langle Ay,y\rangle\succeq 0$ and
$\langle\widetilde{A}y,y\rangle\succeq 0$, we must have

(6.11) $\langle Ay,y\rangle=\langle\widetilde{A}y,y\rangle=0.$

But then, as $A=D^{*}D$ for some $D\in\mathcal{B}(\mathcal{H}_{n})$ and
$\widetilde{A}=\widetilde{D}^{*}\widetilde{D}$ for some
$\widetilde{D}\in\mathcal{B}(\mathcal{H}_{n})$, we have that (6.11) implies
that $\langle
Dy,Dy\rangle=\langle\widetilde{D}y,\widetilde{D}y\rangle=0\in\mathbb{R}_{n}$,
i.e.,

${\rm Re}\,\langle Dy,Dy\rangle={\rm
Re}\,\langle\widetilde{D}y,\widetilde{D}y\rangle=0.$

Thus, $\|Dy\|=\|Dy\|=0$, i.e., $Dy=\widetilde{D}y=0$. But then we have

$Ay=D^{*}Dy=\widetilde{D}^{*}D=\widetilde{A}y=0.$

Therefore, for any $x\in\mathcal{H}_{n}$, we have

$\displaystyle\|(A-\widetilde{A})x\|^{2}=$ $\displaystyle\;{\rm
Re}\,\langle(A-\widetilde{A})x,(A-\widetilde{A})x\rangle$ $\displaystyle=$
$\displaystyle\;{\rm Re}\,\langle(A-\widetilde{A})^{2}x,x\rangle$
$\displaystyle=$ $\displaystyle\;{\rm Re}\,\langle(A-\widetilde{A})y,x\rangle$
$\displaystyle=$ $\displaystyle\;0$

and hence $(A-\widetilde{A})x=0$ for every $x\in\mathcal{H}_{n}$, i.e.,
$A=\widetilde{A}$. ∎

######  Corollary 6.0.7.

Suppose $A\in\mathcal{B}(\mathcal{H}_{n})$ and
$B\in\mathcal{B}(\mathcal{H}_{n})$ are self-adjoint operators with spectral
measures $E_{A}$ and $E_{B}$, respectively. Then $AB=BA$ if and only if
$E_{A}(M)E_{B}(M)=E_{B}(M)E_{A}(M)$ for all $M\in\mathscr{B}(\mathbb{R})$.

###### Proof.

The last assertion in Theorem 6.0.1 with $T=A$ and $W=B$ ensures that $AB=BA$
if and only if $BE_{A}(M)=E_{A}(M)B$. Applying the last assertion in Theorem
6.0.1 with $T=B$ and $W=E_{B}(M)$ we see that

$AB=BA\Longleftrightarrow BE_{A}(M)=E_{A}(M)B\Longleftrightarrow
E_{A}(M)E_{B}(M)=E_{B}(M)E_{A}(M)$

for any $M\in\mathscr{B}(\mathbb{R})$. ∎

##  7\. Polar decomposition for bounded operators

Let $T\in\mathcal{B}(\mathcal{H}_{n})$. The aim of this section it show that
there exists a uniquely determined positive operator
$P\in\mathcal{B}(\mathcal{H}_{n})$ and a partial isometry $Q$ such that
$T=UQ$. This will be important for the spectral theorem for a normal operator
in Section 9.

######  Theorem 7.0.1 (polar decomposition for bounded operators).

Every $T\in\mathcal{B}(\mathcal{H}_{n})$ admits a factorisation

(7.1) $T=UQ,$

where $Q:=|T|$ is uniquely determined, where $|T|:=(T^{*}T)^{1/2}$, and
$U:\overline{{\rm Ran}(Q)}\to\overline{{\rm Ran}(T)}$ is a partial isometry.
In the particular case that $T\in\mathcal{B}(\mathcal{H}_{n})$ is normal, we
can choose $U$ such that $U$ is unitary and we have the following:

(7.2) $WT=TW\;{\rm and}\;WT^{*}=T^{*}W\Longrightarrow WQ=QW\;{\rm
for}\;W\in\mathcal{B}(\mathcal{H}_{n})$

and

(7.3) $WT=TW\;{\rm and}\;WT^{*}=T^{*}W\Longrightarrow WU=UW\;{\rm
for}\;W\in\mathcal{B}(\mathcal{H}_{n}).$

Moreover, $T$ is normal if and only if $QU=UQ$.

###### Proof.

Note that $T^{*}T\in\mathcal{B}(\mathcal{H}_{n})$ is a positive operator and
has a unique positive square root
$Q:=(T^{*}T)^{1/2}\in\mathcal{B}(\mathcal{H}_{n})$ (see Corollary 6.0.6). For
any $x\in\mathcal{H}_{n}$, we have

$\displaystyle\|Tx\|^{2}=$ $\displaystyle\;{\rm Re}\,\langle Tx,Tx\rangle$
$\displaystyle=$ $\displaystyle\;{\rm Re}\,\langle T^{*}Tx,x\rangle$
$\displaystyle=$ $\displaystyle\;{\rm Re}\,\langle Q^{2}x,x\rangle$
$\displaystyle=$ $\displaystyle\;{\rm Re}\,\langle Qx,Qx\rangle$ (7.4)
$\displaystyle=$ $\displaystyle\;\|Qx\|^{2}.$

Thus, if we let $x=y-z$, with $y,z\in\mathcal{H}_{n}$, then (7.4) implies that
$Ty=Tz$ whenever $Qy=Qz$ for any $y,z\in\mathcal{H}_{n}$. Consequently, we may
let $U:{\rm Ran}\,Q\to{\rm Ran}\,T$ denote the operator belonging to
$\mathcal{B}(\mathcal{H}_{n})$ given by $U(Qx)=Tx$ for $x\in\mathcal{H}_{n}$.
Next, extend $U$ to all of $\mathcal{H}_{n}$ via

(7.5) $Uy=\begin{cases}Tx&\quad{\rm if}\quad y=Qx,\\\ 0&\quad{\rm if}\quad
y\in\overline{{\rm Ran}\,\,Q}^{\perp}.\end{cases}$

Since $\langle Uy,z\rangle=\langle y,U^{*}z\rangle=0$ for all $y\in{\rm
Ran}\,\,Q$ and $z\in\mathcal{H}_{n}$, we have
$U^{*}:\mathcal{H}_{n}\to\overline{{\rm Ran}\,\,Q}^{\perp}$ and hence ${\rm
Ran}\,\,U\subseteq\overline{{\rm Ran}\,\,Q}$.

We now wish to show that

(7.6) $U^{*}Ux=x\quad\quad{\rm for}\quad x\in\overline{{\rm Ran}\,\,Q}.$

Suppose $y,z\in{\rm Ran}\,Q$. Then

$\langle y,z\rangle=\langle Uy,Uz\rangle=\langle y,U^{*}Uz\rangle$

and hence $\langle U^{*}Uy-y,w\rangle=0$ for all $w\in{\rm Ran}\,Q$. We have
already noted that $U^{*}:\mathcal{H}_{n}\to\overline{{\rm Ran}\,Q}$. Hence
for all $y\in\overline{{\rm Ran}\,Q}$, we have $U^{*}Uy-y\in\overline{{\rm
Ran}\,Q}$. Consequently, we have (7.6). Putting all of these observations
together, we arrive at the factorisation (7.1). The asserted uniqueness can be
justified precisely the same way as the complex Hilbert space case.

Let us now suppose that $T$ is normal. We first note that $\langle
TT^{*}x,x\rangle=\langle T^{*}Tx,x\rangle$ implies that $\langle
Tx,Tx\rangle=\langle T^{*}x,T*x\rangle$ and hence ${\rm Re}\,\langle
Tx,Tx\rangle={\rm Re}\,\langle T^{*}x,T*x\rangle$, i.e., $\|Tx\|=\|T^{*}x\|$
for all $x\in\mathcal{H}_{n}$. Therefore, ${\rm Ker}\,\,T={\rm Ker}\,\,T^{*}$
and hence

${\rm Ran}\,\,Q^{\perp}={\rm Ker}\,\,Q={\rm Ker}\,\,T={\rm
Ker}\,\,T^{*}=\overline{{\rm Ran}\,\,T}^{\perp}.$

Thus, we can extend $U:\overline{{\rm Ran}\,Q}\to\overline{{\rm Ran}\,Q}$ from
(7.5) to $U:\mathcal{H}_{n}\to\mathcal{H}_{n}$ (with a slight abuse of
notation we shall denote the extension of $U$ by $U$ as well) via

$Uy=y\quad{\rm if}\quad y\in\overline{{\rm Ran}\,Q}^{\perp}.$

As $\mathcal{H}_{n}=\overline{{\rm Ran}\,Q}\oplus\overline{{\rm
Ran}\,Q}^{\perp}$, we have that $U:\mathcal{H}_{n}\to\mathcal{H}_{n}$ is
unitary.

The implication 7.2 is an immediate consequence of the final assertion of
Theorem 6.0.1 and Corollary 6.0.6. We will now show (7.3). In view of (7.2),
we have

(7.7) $(WU-UW)Q=0.$

Thus, as $T\in\mathcal{B}(\mathcal{H}_{n})$ is normal, we have
$\mathcal{H}_{n}={\rm Ker}T\oplus\overline{{\rm Ran}\,T}$. Thus we may write
any $x\in\mathcal{H}_{n}$ as $x=y+z$, where $y\in{\rm Ker}\,T={\rm Ker}\,Q$
and $z\in\overline{{\rm Ran}\,T}$, we may use the fact that ${\rm Ker}\,Q={\rm
Ker}\,T$ and (7.7) to obtain

$(WU-UW)x=(WU-UW)(y+z)=0,$

i.e., $WU=UW$.

The final assertion can be justified in the same way as the complex Hilbert
space case (bearing in mind that one must use Corollary 6.0.6 which guarantees
that bounded positive operators have a unique positive square root). ∎

##  8\. An additive decomposition for bounded operators and imaginary
operators

In this section, we will show that all bounded operators on a Clifford module
admit an additive decomposition which is analogous to the well-known additive
decomposition

$T=\left(\frac{T+T^{*}}{2}\right)+i\left(\frac{T-T^{*}}{2i}\right),$

which holds in the complex Hilbert space case. In the particular case that the
bounded operator is normal, this additive decomposition will be useful for
proving the spectral theorem for a bounded normal operators in Section 9.

######  Lemma 8.0.1.

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

(8.1) $\sigma_{S}(T)\subseteq\\{s\in\mathbb{R}^{n+1}:|s|=1\\}.$

###### Proof.

We claim that
$\rho_{S}(T)\subseteq\mathbb{R}^{n+1}\setminus\\{s\in\mathbb{R}^{n+1}:|s|=1\\}$.
Since $TT^{*}=I$, it is obvious that $\|T\|=1$. Thus, Theorem 3.1.8 ensures
that

$\\{s\in\mathbb{R}^{n+1}:|s|>1\\}\subseteq\rho_{S}(T).$

Since $\mathcal{Q}_{0}(T)=T^{2}$ is a unitary operator, we have that
$\mathcal{Q}_{0}(T)$ is invertible, i.e., $0\in\rho_{S}(T)$. For any
$s=s_{0}+s_{1}\in\mathbb{R}^{n+1}$, with $0<|s|<1$, we have

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

Thus, as $T^{*}$ is unitary and $|s^{-1}|>1$, we have that $s^{-1}$ belongs to
$\rho_{S}(T^{*})$, in which case $s\in\mathbb{R}^{n+1}$, with $0<|s|<1$,
belongs to $\rho_{S}(T)$. Putting everything together we have (8.1). ∎

######  Definition 8.0.2 (imaginary operator).

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 $J^{*}=-J$.

######  Theorem 8.0.3.

Corresponding to any operator $T\in\mathcal{B}(\mathcal{H}_{n})$, there exist
a self-adjoint operator $A\in\mathcal{B}(\mathcal{H}_{n})$, a partial
imaginary operator $J_{0}\in\mathcal{B}(\mathcal{H}_{n})$ and a positive
operator $B\in\mathcal{B}(\mathcal{H}_{n})$ such that

(8.2) $T=A+J_{0}B,$

where $A:=(T+T^{*})/2$ and $B:=|T-T^{*}|/2$ are uniquely determined by $T$.
Moreover, we may choose $J_{0}$ in (8.2) to be an imaginary operator.

###### Proof.

If we let $A:=(T+T^{*})/2$, then $A\in\mathcal{B}(\mathcal{H}_{n})$ is
obviously self-adjoint. Consider the anti self-adjoint operator
$Y:=T-A=(T-T^{*})/2\in\mathcal{B}(\mathcal{H}_{n})$. Since $Y$ is normal, we
may use Theorem 7.0.1 to find a positive operator
$B\in\mathcal{B}(\mathcal{H}_{n})$ and a partial isometry $J_{0}$ such that
$Y=J_{0}B$. Moreover, by Theorem 7.0.1, we have $BJ_{0}=J_{0}B$. Since $Y$ is
anti self-adjoint, we must have $J_{0}^{*}=-J_{0}$. Thus, $J_{0}$ is a partial
imaginary operator.

The uniqueness of $A$ is obvious. The uniqueness of $B$ follows from Theorem
7.0.1. Finally, the fact that we may choose $J_{0}$ to be unitary follows from
Theorem 7.0.1 applied to the bounded normal operator $(T-T^{*})/2$. ∎

In the case that $T\in\mathcal{B}(\mathcal{H}_{n})$ is normal, we have the
following refinement of Theorem 8.0.3.

######  Theorem 8.0.4.

Corresponding to any normal operator $T\in\mathcal{B}(\mathcal{H}_{n})$, there
exist a self-adjoint operator $A\in\mathcal{B}(\mathcal{H}_{n})$, a partial
imaginary operator $J_{0}\in\mathcal{B}(\mathcal{H}_{n})$ and a positive
operator $B\in\mathcal{B}(\mathcal{H}_{n})$ such that $A$, $J_{0}$ and $B$
mutually commute and satisfy

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

In this case, $A$ and $B$ are as in Theorem 8.0.3 and

(8.4) $TJ_{0}=J_{0}T.$

Moreover, we may choose $J_{0}$ to be an imaginary operator (in this case we
shall write $J$ in place of $J_{0}$).

###### Proof.

In view of Theorem 8.0.4, we only have to show that
$A:=(T+T^{*})/2,B:=|T-T^{*}|/2$ and $J_{0}$ mutually commute and that (8.4)
holds. The fact that $B$ and $J_{0}$ commute follows from the fact that
$J_{0}B$ is a polar factorisation for the bounded normal operator
$Y:=(T-T^{*})/2$. Since $T$ is normal, we have $A(T-T^{*})=(T-T^{*})A$ and
$A(T-T^{*})^{*}=(T-T^{*})^{*}A$. Thus, the fact that $A$, $B$ and $J_{0}$
mutually commute follows at once from (7.2) and (7.3). The final assertion
(8.4) is an immediate consequence of (8.3) and the fact that $A$, $J$ and $B$
mutually commute. ∎

######  Theorem 8.0.5.

Let $J$ be an imaginary operator on a Clifford module $\mathcal{H}_{n}$, fix
$\mathfrak{I}\in\mathbb{S}$ and define

(8.5)
$\mathcal{H}_{\pm}(J,\mathfrak{I}):=\\{x\in\mathcal{H}_{n}:J\,x=x(\pm\mathfrak{I})\\}.$

Then the following statements hold:

  1. (i)

$\mathcal{H}_{\pm}(J,\mathfrak{I})$ are nontrivial, i.e.,
$\mathcal{H}_{\pm}(J,\mathfrak{I})\neq\\{0\\}$ and

(8.6)
$\mathcal{H}_{+}(J,\mathfrak{I})\cap\mathcal{H}_{-}(J,\mathfrak{I})=\\{0\\}.$

  2. (ii)

$\mathcal{H}_{\pm}(J,\mathfrak{I})$ are $\mathbb{C}_{\mathfrak{I}}$ closed
right linear subspaces of $\mathcal{H}_{n}$, with respect to
$\mathbb{C}_{\mathfrak{I}}$, i.e., if
$x,y\in\mathcal{H}_{\pm}(J,\mathfrak{I})$ and
$\lambda\in\mathbb{C}_{\mathfrak{I}}$, then
$x\,\lambda+y\in\mathcal{H}_{\pm}(J,\mathfrak{I})$. Consequently,
$\mathcal{H}_{\pm}(J,\mathfrak{I})$ may be both be viewed as a complex Hilbert
space with respect to the $\mathbb{C}_{\mathfrak{I}}$-valued inner product
given by

(8.7) $\langle x,y\rangle_{\mathbb{C}_{\mathfrak{I}}}:={\rm Re}(\langle
x,y\rangle)-{\rm Re}(\langle
x,y\rangle\mathfrak{I})\mathfrak{I}\in\mathbb{C}_{\mathfrak{I}}\quad\quad{\rm
for}\quad x,y\in\mathcal{H}_{\pm}(J,\mathfrak{I}).$

  3. (iii)

$\mathcal{H}_{n}=\mathcal{H}_{+}(J,\mathfrak{I})\oplus\mathcal{H}_{-}(J,\mathfrak{I})$.

  4. (iv)

For any orthonormal basis $(\eta_{i})_{i\in\mathcal{I}}$ of the complex
Hilbert space $\mathcal{H}_{+}(J,\mathfrak{I})$, we have that, for any choice
of $\mathfrak{J}\in\mathbb{S}$ with
$\mathfrak{I}\,\mathfrak{J}=-\mathfrak{J}\mathfrak{I}$,
$(\eta_{i}\,\mathfrak{J})_{i\in\mathcal{I}}$ is an orthonormal basis of
$\mathcal{H}_{-}(J,\mathfrak{I})$.

Moreover, for any orthonormal basis $(\tilde{\eta})_{i\in\mathcal{I}}$ of the
complex Hilbert space $\mathcal{H}_{-}(J,\mathfrak{I})$, we have that, for any
choice of $\mathfrak{J}\in\mathbb{S}$ with
$\mathfrak{I}\,\mathfrak{J}=-\mathfrak{J}\mathfrak{I}$,
$(\eta_{i}\,\mathfrak{J})_{i\in\mathcal{I}}$ is an orthonormal basis of
$\mathcal{H}_{+}(J,\mathfrak{I})$.

  5. (v)

For any orthonormal basis $(\eta)_{i\in\mathcal{I}}$ of the complex Hilbert
space $\mathcal{H}_{+}(J,\mathfrak{I})$, $(\eta_{i})_{i\in\mathcal{I}}$ is an
orthonormal basis of the Clifford module $\mathcal{H}_{n}$.

  6. (vi)

$\sigma_{S}(J)\cap\mathbb{C}_{\mathfrak{I}}^{+}=\\{\mathfrak{I}\\}$.

  7. (vii)

For any orthonormal basis $(\eta_{i})_{i\in\mathcal{I}}$ of the complex
Hilbert space $\mathcal{H}_{+}(J,\mathfrak{I})$ and
$\mathfrak{I}\in\mathbb{S}$, we have

(8.8) $Jx=\sum_{i\in\mathcal{I}}\eta_{i}\,\mathfrak{I}\,\langle
x,\eta_{i}\rangle\quad\quad{\rm for}\quad x\in\mathcal{H}_{n}.$

###### Proof.

We will first prove (i). Suppose there exists
$x\in\mathcal{H}_{n}\setminus\\{0\\}$ such that $x-Jx\mathfrak{I}\neq 0$. Then

$\displaystyle J(x-xJx\mathfrak{I})=$ $\displaystyle\;Jx+x\mathfrak{I}$
$\displaystyle=$ $\displaystyle\;(x-Jx\mathfrak{I})\mathfrak{I}.$

Thus, $y:=x-Jx\mathfrak{I}\in\mathcal{H}_{n}\setminus\\{0\\}$ and
$Jy=y\mathfrak{I}$ and hence $\mathcal{H}_{+}(J,\mathfrak{I})\neq\\{0\\}$.

On the other hand, if there exists $x\in\mathcal{H}_{n}\setminus\\{0\\}$ such
that $x=Jx\mathfrak{I}$, then choose $\mathfrak{J}\in\mathbb{S}$ such that
$\mathfrak{I}\mathfrak{J}=-\mathfrak{J}\mathfrak{I}$. Thus,
$Jx\mathfrak{I}\mathfrak{J}=x\mathfrak{J}$ implies that
$-J(x\mathfrak{J})\mathfrak{I}=x\mathfrak{J}$ and hence

$J(x\mathfrak{J})=(x\mathfrak{J})\mathfrak{I}.$

Thus, we again arrive at the conclusion
$\mathcal{H}_{+}(J,\mathfrak{I})\neq\\{0\\}$. The justification that
$\mathcal{H}_{-}(J,\mathfrak{I})\neq\\{0\\}$ can be carried out in a similar
fashion.

We will now show (8.6). Suppose to the contrary that there exists a nonzero
vector
$x\in\mathcal{H}_{+}(J,\mathfrak{I})\cap\mathcal{H}_{-}(J,\mathfrak{I})$. Then
we have $x\mathfrak{I}=x(-\mathfrak{I})$, i.e., $x\mathfrak{I}=0$ which forces
$x=0$, a contradiction.

We will now prove (ii). Suppose $x,y\in\mathcal{H}_{+}(J,\mathfrak{I})$ and
$\lambda\in\mathbb{C}_{\mathfrak{I}}$. Then

$J(x\lambda+y)=x\mathfrak{I}\lambda+y\mathfrak{I}=(x\lambda+y)\mathfrak{I}.$

Thus, $\mathcal{H}_{+}(J,\mathfrak{I})$ is a $\mathbb{C}_{\mathfrak{I}}$ right
linear subspace of $\mathcal{H}_{n}$. The fact that
$\mathcal{H}_{+}(J,\mathfrak{I})$ is closed follows at once via the continuity
of multiplication. The justification that $\mathcal{H}_{-}(J,\mathfrak{I})$ is
a closed right-linear subspace, with respect to $\mathbb{C}_{\mathfrak{I}}$,
can be carried out in a similar fashion. Finally, the fact that
$\mathcal{H}_{+}(J,\mathfrak{I})$ and $\mathcal{H}_{-}(J,\mathfrak{I})$ may be
viewed as a complex Hilbert spaces with respect to the
$\mathbb{C}_{\mathfrak{I}}$-valued inner product given by (8.7) follows
immediately from $\mathcal{H}_{\pm}(J,\mathfrak{I})$ being closed right
subspaces of $\mathcal{H}_{n}$ and checking that (8.7) is an
$\mathbb{C}_{\mathfrak{I}}$-valued inner product (to this end, it will be
helpful to use the easily verified fact that $\langle
x,y\rangle\lambda=\pm\lambda\langle x,y\rangle$ for all
$\lambda\in\mathbb{C}_{\mathfrak{I}}$ and
$x,y\in\mathcal{H}_{\pm}(J,\mathfrak{I})$).

We will now prove (iii). Let $y=(x-Jx\mathfrak{I})/2$ and
$z=(x+Jx\mathfrak{I})/2$. Then $y\in\mathcal{H}_{+}(J,\mathfrak{I})$ since

$Jy=\frac{1}{2}(Jx+x\mathfrak{I})=\frac{1}{2}(x-Jx\mathfrak{I})\mathfrak{I}.$

Similarly, $z\in\mathcal{H}_{-}(J,\mathfrak{I})$ since

$Jz=\frac{1}{2}(Jx-x\mathfrak{I})=\frac{1}{2}(x+Jx\mathfrak{I})(-\mathfrak{I}).$

Thus, as $x=y+z$ and (8.6) holds we have
$\mathcal{H}_{n}=\mathcal{H}_{+}(J,\mathfrak{I})\oplus\mathcal{H}_{-}(J,\mathfrak{I})$.

We will now prove (iv). Let $\mathfrak{J}\in\mathbb{S}$ be such that
$\mathfrak{I}\mathfrak{J}=-\mathfrak{J}\mathfrak{I}$. Then for any
$y\in\mathcal{H}_{+}(J,\mathfrak{I})$, we have

$Jy\,\mathfrak{J}=(y\mathfrak{I})\mathfrak{J}=(\eta_{i}\mathfrak{J})(-\mathfrak{I}).$

Thus, $y\mathfrak{J}\in\mathcal{H}_{-}(J,\mathfrak{I})$. One can easily push
this observation further and establish that
$\varphi:\mathcal{H}_{+}(J,\mathfrak{I})\to\mathcal{H}_{-}(J,\mathfrak{I})$
given by $\varphi(y)=y\mathfrak{J}$ is an isomorphism. Consequently, if
$(\eta_{i})_{i\in\mathcal{I}}$ is an orthonormal basis of
$\mathcal{H}_{+}(J,\mathfrak{I})$, then
$(\eta_{i}\,\mathfrak{J})_{i\in\mathcal{I}}$ is an orthonormal basis of
$\mathcal{H}_{-}(J,\mathfrak{I})$.

We will now prove (v). Let $(\eta_{i})_{i\in\mathcal{I}}$ be any orthonormal
basis of $\mathcal{H}_{+}(J,\mathcal{I})$. Then we may apply (iv) and (iii) to
see that $(\eta_{i})_{i\in\mathcal{I}}$ is also an orthonormal basis of
$\mathcal{H}_{n}$.

We will now prove (vi). Recall that $J^{*}=-J$ and $JJ^{*}=I$. Thus, combining
(3.28) and (8.1), we have

$\sigma_{S}(J)\cap\mathbb{C}_{\mathfrak{I}}^{+}=\\{\lambda\in\mathbb{C}_{\mathfrak{I}}^{+}:{\rm
Re}(\lambda)=0\\}\cap\\{\lambda\in\mathbb{C}_{\mathfrak{I}}^{+}:|\lambda|=1\\}=\\{\mathfrak{I}\\}.$

We will now prove (vii). Let $(\eta_{i})_{i\in\mathcal{I}}$ be any orthonormal
basis of $\mathcal{H}_{+}(J,\mathcal{I})$. In view of (v), we have that
$(\eta_{i})_{i\in\mathcal{I}}$ is an orthonormal basis of $\mathcal{H}_{n}$.
Consequently, in view of Lemma 2.2.6(iv), we have

$x=\sum_{i\in\mathcal{I}}\eta_{i}\langle x,\eta_{i}\rangle\quad\quad{\rm
for}\quad x\in\mathcal{H}_{n}$

and hence

$Jx=\sum_{i\in\mathcal{I}}J\,\eta_{i}\langle
x,\eta_{i}\rangle=\sum_{i\in\mathcal{I}}\eta_{i}\,\mathfrak{I}\langle
x,\eta_{i}\rangle\quad\quad{\rm for}\quad x\in\mathcal{H}_{n}.$

∎

##  9\. Spectral theorem for a bounded normal operator and some consequences

In this section we will be prove one of the main results of this manuscript,
namely the spectral theorem for a bounded normal operator on a Clifford
module.

Let $\mathfrak{I}\in\mathbb{S}$ be arbitrary. In the following lemma, we shall
identify $\mathbb{R}\times[0,\infty)$ with
$\mathbb{C}^{+}_{\mathfrak{I}}:=\\{\lambda\in\mathbb{C}^{+}_{\mathfrak{I}}:{\rm
Im}(\lambda)\geq 0\\}$ in the natural way. Consequently, a spectral measure on
$\mathscr{B}(\mathbb{R}\times[0,\infty))$ may be viewed as a spectral measure
on $\mathscr{B}(\mathbb{C}^{+}_{\mathfrak{I}})$.

######  Lemma 9.0.1.

Let $E_{1}$ be a spectral measure on $\mathscr{B}(\mathbb{R})$, $E_{2}$ be a
spectral measure on $\mathscr{B}([0,\infty))$ and $\mathfrak{I}\in\mathbb{S}$.
Suppose $E_{1}(M_{1})E_{2}(M_{2})=E_{2}(M_{2})E_{1}(M_{1})$ for every
$M_{1}\in\mathscr{B}(\mathbb{R})$ and $M_{2}\in\mathscr{B}([0,\infty))$ and
let $E(M_{1}\times M_{2}):=E(M_{1})E(M_{2})$ be the uniquely determined
spectral measure on $\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+})$ (see Theorem
5.1.15). For any real-valued function
$f\in\mathfrak{F}(\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+},E)$, we have

(9.1)
$\int_{\mathbb{R}}f(t)\,dE_{1}(t)=\int_{\mathbb{C}_{\mathfrak{I}}^{+}}f({\rm
Re}(\lambda))dE(\lambda)$

and

(9.2)
$\int_{[0,\infty)}g(u)\,dE_{2}(u)=\int_{\mathbb{C}_{\mathfrak{I}}^{+}}g({\rm
Im}(\lambda))dE(\lambda)$

###### Proof.

We will verify (9.1) for simple functions. For any
$M\in\mathscr{B}(\mathbb{R})$, we have

$\displaystyle\int_{\mathbb{C}_{\mathfrak{I}}^{+}}\chi_{M}({\rm
Re}(\lambda))\,dE(\lambda)=$ $\displaystyle\;E(M\times[0,\infty))$
$\displaystyle=$ $\displaystyle\;E_{1}(M)E_{2}([0,\infty))=E_{1}(M)$
$\displaystyle=$ $\displaystyle\;\int_{\mathbb{R}}\chi_{M}(t)dE_{1}(t)$

Thus, by linearity, (9.1) holds for simple functions. Passing to a limit, we
obtain (9.1) for any
$f\in\mathfrak{F}(\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+}),E)$.

Formula (9.2) can be justified in much the same way as (9.1).

∎

We are now ready to state and proof the spectral theorem for a bounded normal
operator on a Clifford module.

######  Theorem 9.0.2 (spectral theorem for bounded normal operators).

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be normal. Then corresponding to any
choice of $\mathfrak{I}\in\mathbb{S}$, there exists a spectral measure $E$ on
the Borel $\sigma$-algebra
$\mathscr{B}(\sigma_{S}(T)\cap\mathbb{C}^{+}_{\mathfrak{I}})$ such that

(9.3) $T=\int_{\sigma_{S}(T)\,\cap\,\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda)\,dE(\lambda)+\int_{\sigma_{S}(T)\,\cap\,\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda)\,dE(\lambda)\,J,$

where $J$ an anti self-adjoint and unitary operator obeying (8.3). $E$ is
unique in the sense that if $F$ is a spectral measure on
$\mathscr{B}(\mathbb{C}^{+}_{\mathfrak{I}})$ such that

$T=\int_{\mathbb{C}^{+}_{\mathfrak{I}}}{\rm
Re}(\lambda)\,dF(\lambda)+\int_{\mathbb{C}^{+}_{\mathfrak{I}}}{\rm
Im}(\lambda)dF(\lambda)J,$

then $E(M\cap\sigma_{S}(T)\cap\mathbb{C}^{+}_{\mathfrak{I}})=F(M)$ for all
$M\in\mathscr{B}(\mathbb{C}^{+}_{\mathfrak{I}})$. Moreover,
$W\in\mathcal{B}(\mathcal{H}_{n})$ commutes with $T$ if and only if
$WE(M)=E(M)W$ for every
$M\in\mathscr{B}(\sigma_{S}(T)\cap\mathbb{C}^{+}_{\mathfrak{I}})$.

###### Proof.

We begin by using (8.3) to write $T=A+JB$, where
$A\in\mathcal{B}(\mathcal{H}_{n})$ is self-adjoint,
$B\in\mathcal{B}(\mathcal{H}_{n})$ is positive and $J$ is anti self-adjoint
and unitary and $A,B$ and $J$ all mutually commute. Since $A$ is self-adjoint
and $B$ is positive, we may use Theorem 6.0.1 to obtain spectral measures
$E_{1}$ (resp., $E_{2}$) on $\mathscr{B}(\sigma_{S}(A))$ (resp.,
$\mathscr{B}(\sigma_{S}(B))$) such that

$A=\int_{\sigma_{S}(A)}t\,dE_{1}(t)\quad{\rm and}\quad
B=\int_{\sigma_{S}(B)}t\,dE_{2}(t).$

Note that since $A\in\mathcal{B}(\mathcal{H}_{n})$ is self-adjoint, we have
that $\sigma_{S}(T)$ is a non-empty compact subset of $\mathbb{R}$ and since
$B\in\mathcal{B}(\mathcal{H}_{n})$ is positive, we have that $\sigma_{S}(B)$
is a non-empty compact subset of $[0,\infty)$ (see Lemma 3.2.4). Moreover, in
view of Remark 6.0.5, we have

${\rm supp}\,E_{1}=\sigma_{S}(A)\quad{\rm and}\quad{\rm
supp}\,E_{2}=\sigma_{S}(B).$

Since $A$ and $B$ commute, we have that
$E_{1}(M_{1})E_{2}(M_{2})=E_{1}(M_{1})E_{2}(M_{2})$ for every
$M\in\mathscr{B}(\sigma)$. Thus, we invoke Theorem 5.1.15 with
$\Omega_{1}=\mathscr{B}(\sigma_{S}(A))$ and
$\Omega_{2}=\mathscr{B}(\sigma_{S}(B))$ to obtain a uniquely determined
spectral measure $E$ on $\mathscr{B}(\sigma_{S}(A)\times\sigma_{S}(B))$ given
by $E(M_{1}\times M_{2})=E(M_{1})E(M_{2})$ for
$M_{1}\in\mathscr{B}(\sigma_{S}(A))$ and $M_{2}\in\mathscr{B}(\sigma_{S}(B))$.

Let $\mathfrak{I}\in\mathbb{S}$ be arbitrary and identify
$\mathbb{R}\times[0,\infty)$ with
$\mathbb{C}^{+}_{\mathfrak{I}}:=\\{\lambda\in\mathbb{C}^{+}_{\mathfrak{I}}:{\rm
Im}(\lambda)\geq 0\\}$ in the natural way. Consequently, we will view $E$ as a
spectral measure on the non-empty compact subset
$\sigma_{S}(A)\times\sigma_{S}(B)\subseteq\mathscr{B}(\mathbb{C}^{+}_{\mathfrak{I}})$.
Lemma 9.0.1 with $f(\lambda)={\rm Re}(\lambda)$ and $g(\lambda)={\rm
Im}(\lambda)$ imply that

(9.4)
$A=\int_{\mathbb{R}}t\,dE_{1}(t)=\int_{\mathbb{C}^{+}_{\mathfrak{I}}}{\rm
Re}(\lambda)dE(\lambda)$

and

(9.5)
$B=\int_{[0,\infty)}u\,dE_{2}(u)=\int_{\mathbb{C}^{+}_{\mathfrak{I}}}{\rm
Im}(\lambda)dE(\lambda),$

respectively. Since ${\rm supp}\,E_{1}=\sigma_{S}(A)$, ${\rm
supp}\,E_{2}=\sigma_{B}(A)$ and ${\rm
supp}\,E\subseteq\sigma_{S}(A)\times\sigma_{B}(A)$, we may rewrite (9.4) and
(9.5) as

(9.6)
$A=\int_{\sigma_{S}(A)}t\,dE_{1}(t)=\int_{\sigma_{S}(A)\times\sigma_{S}(B)}{\rm
Re}(\lambda)dE(\lambda)$

and

(9.7)
$B=\int_{\sigma_{S}(B)}u\,dE_{2}(u)=\int_{\sigma_{S}(A)\times\sigma_{S}(B)}{\rm
Im}(\lambda)dE(\lambda),$

respectively. Thus , as $T=A+BJ$, we can use (9.4) and (9.5) to obtain

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

Theorem 5.3.14 with $f(\lambda)=\lambda$ implies that

$\displaystyle\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}=$
$\displaystyle\;\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}^{+}$
$\displaystyle=$
$\displaystyle\;\\{s\in\mathbb{C}^{+}_{\mathfrak{I}}:E(\\{\lambda\in\Omega:|f(\lambda)^{2}-2{\rm
Re}(s)f(\lambda)+|s|^{2}|<\varepsilon\\})\neq 0$ $\displaystyle\;\;\text{for
all $\varepsilon>0$}\\}$ $\displaystyle=$
$\displaystyle\;\\{s\in\mathbb{C}^{+}_{\mathfrak{I}}:E(\\{\lambda\in\Omega:|\lambda^{2}-2{\rm
Re}(s)\lambda+|s|^{2}|<\varepsilon\\})\neq 0$ (9.9)
$\displaystyle\;\;\text{for all $\varepsilon>0$}\\}$ $\displaystyle=$
$\displaystyle\;\\{s\in\mathbb{C}^{+}_{\mathfrak{I}}:E(\\{\lambda\in\Omega:|\lambda-s|\cdot|\lambda-\bar{s}|<\varepsilon\\})\neq
0$ $\displaystyle\;\;\text{for all $\varepsilon>0$}\\}$ (9.10)
$\displaystyle=$
$\displaystyle\;\\{s\in\mathbb{C}^{+}_{\mathfrak{I}}:E(\\{\lambda\in\Omega:|\lambda-s|^{2}<\varepsilon\\})\neq
0$ $\displaystyle\;\;\text{for all $\varepsilon>0$}\\},$

since, if we write
$\lambda=\lambda_{0}+\lambda_{1}\mathfrak{I}\in\mathbb{C}_{\mathfrak{I}}^{+}$,
with $\lambda_{0}\in\mathbb{R}$ and $\lambda_{1}\geq 0$, and
$s=s_{0}+s_{1}\mathfrak{I}\in\mathbb{C}_{\mathfrak{I}}^{+}$, with
$s_{0}\in\mathbb{R}$ and $s_{1}\geq 0$, then

$\displaystyle|\lambda-\bar{s}|^{2}=$
$\displaystyle\;(\lambda_{0}-s_{0})^{2}+\lambda_{1}^{2}+2\lambda_{1}s_{1}+s_{1}^{2}$
$\displaystyle\geq$
$\displaystyle\;(\lambda_{0}-s_{0})^{2}+\lambda_{1}^{2}-2\lambda_{1}s_{1}+s_{1}^{2}$
$\displaystyle=$ $\displaystyle\;|\lambda-s|^{2}.$

Thus, in view of (5.4), (9.10) implies that
$\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}={\rm supp}\,E$. Thus, we may
rewrite (9.8) to obtain (9.3).

We will now show that $E$ is unique. Suppose $E^{\prime}$ on
$\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+})$ is another spectral measure such
that (9.3) holds. One can argue as in the proof of Theorem 6.0.1 to see that
${\rm supp}\,E^{\prime}=\sigma_{S}(T)$ which is non-empty and compact by
Theorem 3.1.8. Consequently, we may use Theorem 5.2.4(iv) with $y=x$ and
$f\in\mathscr{C}(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+},\mathbb{R})$
to obtain

$\int_{\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}}f(\lambda)d\langle
E(\lambda)x,x\rangle=\int_{\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}}f(\lambda)d\langle
E^{\prime}(\lambda)x,x\rangle.$

Thus, one can show that $E=E^{\prime}$ as in the proof of Theorem 6.0.1.

The proof of the final assertion can be completed in the same way as the proof
of the final assertion in Theorem 6.0.1. ∎

######  Remark 9.0.3.

Let $\mathfrak{I},\mathfrak{J}\in\mathbb{S}$ and
$\gamma:\mathbb{C}^{+}_{\mathfrak{I}}\to\mathbb{C}^{+}_{\mathfrak{J}}$ denote
the bijective map given by

(9.11)
$\gamma(\lambda_{0}+\lambda_{1}\mathfrak{I})=\lambda_{0}+\lambda_{1}\mathfrak{J},$

where $\lambda_{0}\in\mathbb{R}$ and $\lambda_{1}\geq 0$. An immediate
consequence of the proof of Theorem 9.0.2, is that if $E_{\mathfrak{I}}$,
where $\mathfrak{I}\in\mathbb{S}$, is a spectral measure for a normal operator
$T\in\mathcal{B}(\mathcal{H}_{n})$ and $E_{\mathfrak{J}}$, where
$\mathfrak{J}\in\mathbb{S}$, then since
$\gamma(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+})=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$
and ${\rm
supp}\,E_{\mathfrak{I}}=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$, we
have

(9.12) $E_{\mathfrak{I}}(M)=E_{\mathfrak{J}}(\gamma(M))\quad\quad{\rm
for}\quad M\in\mathscr{B}(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}).$

In view of the above observations, we are justified in calling a spectral
measure $E$ on $\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$ the spectral
measure of $T$.

######  Corollary 9.0.4 (Borel functional calculus in the bounded case).

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be normal, $J$ be an imaginary operator
satisfying (8.3) and $E$ be the spectral measure of $T$. Fix
$\mathfrak{I}\in\mathbb{S}$ and put
$\Omega_{\mathfrak{I}}^{+}:=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$.
For any
$f,g\in\mathfrak{B}(\Omega^{+}_{\mathfrak{I}},\mathscr{B}(\Omega_{\mathfrak{I}}^{+}),\mathbb{C}_{\mathfrak{I}},E)$
(see Subsection 5.2), we have the spectral integrals
$\mathbb{I}(f),\mathbb{I}(g)\in\mathcal{B}(\mathcal{H}_{n})$ are normal
operators with the following properties:

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

  4. (iv)

For all $x,y\in\mathcal{H}_{n}$, we have

$\langle\mathbb{I}(f)x,y\rangle=\int_{\Omega_{\mathfrak{I}}^{+}}{\rm
Re}(f(\lambda))d\langle
E(\lambda)x,y\rangle+\int_{\Omega_{\mathfrak{I}}^{+}}{\rm
Im}(f(\lambda))d\langle JE(\lambda)x,y\rangle$

  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)

$\|\mathbb{I}(f)\|\leq\|f\|_{\infty}$. Moreover,
$\|\mathbb{I}(f)\|=\|f\|_{\infty}$ if and only if $f\in
L_{\infty}(\Omega_{\mathfrak{I}}^{+},\mathscr{B}(\Omega_{\mathfrak{I}}^{+},\mathbb{C}_{\mathfrak{I}},E)$.

  7. (viii)

For any sequence of functions $(f_{j})_{j=1}^{\infty}$, where
$f_{j}\in\mathfrak{B}$ 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

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

  8. (viii)

If we let $\Pi_{\pm}(J,\mathfrak{I})$ denote the orthogonal projection onto
the right complex subspace $\mathcal{H}_{\pm}(J,\mathfrak{I})$, respectively
(see Theorem 8.0.5(ii)), then

$\displaystyle\langle\mathbb{I}(f)x,y\rangle=$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}d\langle
E(\lambda)\Pi_{+}(J,\mathfrak{I})x,y\rangle\,f(\lambda)$ $\displaystyle+$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}d\langle
E(\lambda)\Pi_{-}(J,\mathfrak{I})x,y\rangle\overline{f(\lambda)}\quad\quad{\rm
for}\quad x,y\in\mathcal{H}_{n},$

where both integrals above are meant in the sense of (4.5).

  9. (ix)

If
$f\in\mathfrak{B}(\Omega_{\mathfrak{I}}^{+},\mathscr{B}(\Omega_{\mathfrak{I}}^{+}),\mathbb{C}_{\mathfrak{I}},E)$
is nonnegative $E$-a.e., then $\mathbb{I}(f)\succeq 0$.

  10. (x)

$\mathbb{I}(f)^{-1}\in\mathcal{B}(\mathcal{H}_{n})$ if and only if
$f(\lambda)\neq 0$ $E$-a.e. and

$f^{-1}\in
L_{\infty}(\Omega_{\mathfrak{I}}^{+},\mathscr{B}(\Omega_{\mathfrak{I}}^{+}),\mathbb{C}_{\mathfrak{I}},E).$

In this case,

$\mathbb{I}(f)^{-1}=\mathbb{I}(1/f).$

  11. (xi)

The spectral measure $E_{\mathbb{I}(f)}$ of $\mathbb{I}(f)$ satisfies the
identity

(9.13) $E_{\mathbb{I}(f)}(M)=E(f^{-1}(M\cap f(\Omega_{\mathfrak{I}}^{+})).$

###### Proof.

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be normal and fix
$\mathfrak{I}\in\mathbb{S}$. Then by Theorem 9.0.2, we can find a spectral
measure $E$ on
$\Omega_{\mathfrak{I}}^{+}:=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$
such that (9.3) holds. Thus, we may invoke Theorem 5.2.4 with
$\Omega=\Omega_{\mathfrak{I}}^{+}$ and
$\mathscr{A}=\mathscr{B}(\Omega_{\mathfrak{I}}^{+})$ to obtain (i)-(v), the
first assertion in (vi) and (vii). The fact that
$\mathbb{I}(f)\in\mathcal{B}(\mathcal{H}_{n})$ is normal is an easy
consequence of (i) and (ii). Indeed, as

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

we have that $\mathbb{I}(f)$ is normal.

The second assertion in (vi) follows immediately from Lemma 5.3.10. We will
now prove (vii). We may use Theorem 8.0.5(iii) to write $x=x_{+}+x_{-}$ and
$y=y_{+}+y_{-}$, where $x_{+},y_{+}\in\mathcal{H}_{+}(J,\mathfrak{I})$ and
$x_{-},y_{-}\in\mathcal{H}_{-}(J,\mathfrak{I})$. We note that
$\Pi_{\pm}(J,\mathfrak{I})x=x_{\pm}$. Then we may use (iv) and (8.5) to obtain

$\displaystyle\langle\mathbb{I}(f)x,y\rangle=$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}{\rm Re}(f(\lambda))\,d\langle
E(\lambda)x,y\rangle$ $\displaystyle+$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}{\rm Im}(f(\lambda))\,d\langle
E(\lambda)J(x_{+}+x_{-}),y\rangle$ $\displaystyle=$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}{\rm Re}(f(\lambda))\,d\langle
E(\lambda)(x_{+}+x_{-}),y\rangle$ $\displaystyle+$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}{\rm Im}(f(\lambda))\,d\langle
E(\lambda)(x_{+}\mathfrak{I}-x_{-}\mathfrak{I}),y\rangle$ $\displaystyle=$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}\,d\langle
E(\lambda)x_{+},y\rangle f(\lambda)$ $\displaystyle+$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}\,d\langle
E(\lambda)x_{-},y\rangle\overline{f(\lambda)},$

in which case we have the desired identity.

We will now prove (ix). If we let $y=x$ in (iv) with $f(\lambda)\geq 0$
$E$-a.e., then we may use the fact that $E(M)\succeq 0$ for all
$M\in\mathscr{B}(\Omega_{\mathfrak{I}}^{+}$ to obtain

$\langle\mathbb{I}(f)x,x\rangle=\int_{\Omega_{\mathfrak{I}}^{+}}f(\lambda)d\langle
E(\lambda)x,x\rangle\succeq 0.$

Assertion (x) is an immediate consequence of Lemma 5.3.10 and Lemma 5.3.11.
Finally, we will prove (xi). Using Theorem 5.3.21 with
$\Omega=\Omega_{\mathfrak{I}}^{+}$,
$\Omega^{\prime}=f(\Omega_{\mathfrak{I}}^{+})\setminus\\{\infty\\}$,
$\psi(\lambda)=f(\lambda)$, $h(\lambda^{\prime})=\lambda^{\prime}$ and $F$ be
the spectral measure given by (9.13), we obtain

$\displaystyle\;\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda^{\prime})dF(\lambda^{\prime})+\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda^{\prime})dF(\lambda^{\prime})J$ $\displaystyle=$
$\displaystyle\;\int_{f(\Omega_{\mathfrak{I}}^{+})\setminus\\{\infty\\}}{\rm
Re}(\lambda^{\prime})dF(\lambda^{\prime})+\int_{f(\Omega_{\mathfrak{I}}^{+})\setminus\\{\infty\\}}{\rm
Im}(\lambda^{\prime})dF(\lambda^{\prime})J$ $\displaystyle=$
$\displaystyle\;\int_{f(\Omega_{\mathfrak{I}}^{+})\setminus\\{\infty\\}}{\rm
Re}(f(\lambda))dE(\lambda)+\int_{f(\Omega_{\mathfrak{I}}^{+})\setminus\\{\infty\\}}{\rm
Im}(f(\lambda))dE(\lambda)J$ $\displaystyle=$ $\displaystyle\;\mathbb{I}(f).$

Thus, the asserted uniqueness in Theorem 9.0.2 ensures that the spectral
measure for $\mathbb{I}(f)$ is given by (9.13). ∎

######  Corollary 9.0.5 (subclasses of bounded normal operators).

Let $T\in\mathcal{B}(\mathcal{H}_{n})$ be normal with spectral measure $E$ and
$J$ be an imaginary operator associated with $T$ (see Theorem 8.0.4). We have
the following:

  1. (i)

$T$ is self-adjoint if and only if $\sigma_{S}(T)\subseteq\mathbb{R}$. In this
case,

$T=\int_{\mathbb{R}}t\,dE(t).$

  2. (ii)

$T$ is positive if and only if $\sigma_{S}(T)\subseteq[0,\infty)$. In this
case,

$T=\int_{[0,\infty)}t\,dE(t).$

  3. (iii)

$T$ is anti self-adjoint if and only if
$\sigma_{S}(T)\subseteq\\{s\in\mathbb{R}^{n+1}:{\rm Re}(s)=0\\}$. In this
case,

$T=\int_{\mathbb{R}}t\,dE(t)J.$

  4. (v)

$T$ is unitary if and only if
$\sigma_{S}(T)\subseteq\\{s\in\mathbb{R}^{n+1}:|s|=1\\}$. In this case,

$T=\int_{|\lambda|=1}{\rm Re}(\lambda)\,dE(\lambda)+\int_{|\lambda|=1}{\rm
Im}(\lambda)\,dE(\lambda).$

  5. (vi)

$T$ is imaginary if and only if
$\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}=\\{\mathfrak{I}\\}$. In this
case,

$Tx=\sum_{i\in\mathcal{I}}\eta_{i}\,\mathfrak{I}\langle
x,\eta_{i}\rangle\quad\quad{\rm for}\quad x\in\mathcal{H}_{n},$

where $(\eta_{i})_{i\in\mathcal{I}}$ is any orthonormal basis of
$\mathcal{H}_{+}(T,\mathfrak{I})$ (see (8.5)).

###### Proof.

Let us prove the claims. We prove (i). If $T$ is self-adjoint, then Lemma
3.2.4 asserts that $\sigma_{S}(T)\subseteq\mathbb{R}$. The integral
representation $T=\int_{\mathbb{R}}t\,dE(t)$ appeared in Theorem 6.0.1.
Conservely, suppose $T\in\mathcal{B}(\mathcal{H}_{n})$ is normal and
$\sigma_{S}(T)\subseteq\mathbb{R}$. Then the fact that $T=T^{*}$ follows at
once from Corollary 9.0.4 with $f(\lambda)=\lambda$. Assertions (ii) (with the
help of Lemma 3.2.5), (iii) (with the help of Lemma 8.0.1) and (iv) (with the
help of Lemma 3.2.6) can be proved in much the same manner using Theorem 9.0.2
and Corollary 9.0.4.

Finally, Assertion (vi) is just (8.8). ∎

##  10\. Spectral theorem for an unbounded normal operator and some
consequences

Before we can formulate and prove the spectral theorem for an unbounded normal
operator on a Clifford module, we will need to define the bounded transform of
a densely defined closed operator and also a lemma which highlights various
properties of the aforementioned bounded transform.

######  Definition 10.0.1.

Suppose $T\in\mathcal{L}(\mathcal{H}_{n})$ is a densely defined closed
operator. Let $C_{T}$ be as in the statement of Theorem 3.2.3, i.e.,
$C_{T}:=(I+T^{*}T)^{-1}$. In view of Theorem 3.2.3(ii),
$C_{T}\in\mathcal{B}(\mathcal{H}_{n})$ and $C_{T}$ is positive. In view of
Corollary 6.0.6, $C_{T}$ has a unique positive square root
$C_{T}^{1/2}\in\mathcal{B}(\mathcal{H}_{n})$. If we let

(10.1) $Z_{T}:=TC_{T}^{1/2},$

then $Z_{T}$ will be called the bounded transform of $T$ (we will justify this
nomenclature in Lemma 10.0.2).

######  Lemma 10.0.2.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be densely defined. Then the following
statements hold:

  1. (i)

$Z_{T}\in\mathcal{B}(\mathcal{H}_{n})$. Moreover,

(10.2) $\|Z_{T}\|\leq 1$

and

(10.3) $C_{T}=(I+T^{*}T)^{-1}=I-Z_{T}^{*}Z_{T}.$

  2. (ii)

$(Z_{T})^{*}=Z_{T^{*}}$ and hence $Z_{T}$ is self-adjoint whenever $T$ is
self-adjoint.

  3. (iii)

$Z_{T}$ is normal whenever $T$ is normal.

###### Proof.

The proof in the classical complex Hilbert space (see, e.g., Lemma 5.8 in
[66]) can be carried over into the Clifford module setting. For completeness,
we will provide the proof. The proof is broken into steps.

Step 1: Prove (i).

First note that

(10.4) $\\{\text{$Cx$:
$x\in\mathcal{H}$}\\}=\mathcal{D}(I+T^{*}T)=\mathcal{D}(T^{*}T)$

and hence if $x\in\mathcal{H}_{n}$, then

$\displaystyle\|TC_{T}^{1/2}C_{T}^{1/2}x\|^{2}=$ $\displaystyle\;\langle
T^{*}TC_{T}x,C_{T}x\rangle$ $\displaystyle\leq$
$\displaystyle\;\langle(I+T^{*}T)C_{T}x,C_{T}x\rangle$ $\displaystyle=$
$\displaystyle\;\langle C_{T}^{-1}C_{T}x,C_{T}x\rangle$ $\displaystyle=$
$\displaystyle\;\langle x,C_{T}x\rangle$ $\displaystyle=$
$\displaystyle\;\|C_{T}^{1/2}x\|^{2}.$

Thus, if $y\in\\{\text{$C_{T}^{1/2}x$: $x\in\mathcal{H}_{n}$}\\}$, then

(10.5) $\|Z_{T}y\|=\|TC_{T}^{1/2}y\|\leq\|y\|.$

Since ${\rm Ker}(C_{T})=\\{0\\}$, we have that ${\rm
Ker}(C_{T}^{1/2})=\\{0\\}$ and hence $\\{\text{$C_{T}^{1/2}x$:
$x\in\mathcal{H}$}\\}$ is a dense subset of $\mathcal{H}_{n}$. By assumption,
$T$ is a closed operator and since
$C_{T}^{1/2}\in\mathcal{B}(\mathcal{H}_{n})$, we have that $Z_{T}$ is closed
as well. Thus, we have

$\\{\text{$C_{T}^{1/2}x$: $x\in\mathcal{H}_{n}$}\\}\subseteq\mathcal{D}(T),\ \
\ \ \mathcal{D}(Z_{T})=\mathcal{H}_{n}$

and, in view of (10.5), we have $\|Z_{T}\|\leq 1$.

Next, it follows from (10.5) and $C^{1/2}T^{*}\subseteq Z_{T}^{*}$ that

$\displaystyle(I-C_{T})C_{T}^{1/2}=$
$\displaystyle\;C_{T}^{1/2}(I+T^{*}T)C_{T}-C_{T}^{1/2}C_{T}$ $\displaystyle=$
$\displaystyle\;C_{T}^{1/2}T^{*}TC_{T}^{1/2}C_{T}^{1/2}$
$\displaystyle\subseteq$ $\displaystyle\;Z_{T}^{*}Z_{T}C_{T}^{1/2}.$

Thus, $Z_{T}^{*}Z_{T}C_{T}^{1/2}=(I-C_{T})C_{T}^{1/2}$ and, as
$\\{\text{$C_{T}^{1/2}x$: $x\in\mathcal{H}$}\\}$ is a dense subset of
$\mathcal{H}_{n}$, we get (10.3).

Step 2: Prove (ii).

Using (10.3) we get that $C_{T^{*}}=(I+TT^{*})^{-1}$. If
$x\in\mathcal{D}(T^{*})$, then let $y=C_{T^{*}}x$. Therefore,

$x=(I+TT^{*})y$

and

$T^{*}x=T^{*}(I+TT^{*})y=(I+T^{*}T)T^{*}y.$

Thus, $C_{T^{*}}x\in\mathcal{D}(T^{*})$ and hence

(10.6) $C_{T}T^{*}x=T^{*}y=T^{*}C_{T^{*}}x.$

It follows easily from (10.6) and (10.3) that
$p(C_{T^{*}})x\in\mathcal{D}(T^{*})$ and

$p(C_{T})T^{*}x=T^{*}p(C_{T^{*}})x$

for any real polynomial $p\in\mathbb{R}[t]$. By the Weierstraß approximation
theorem, there exists a sequence of real polynomials $(p_{n})_{n=0}^{\infty}$
which converge uniformly in supremum norm to the function $t\mapsto t^{1/2}$
on $[0,1]$. We may use Corollary 6.0.6 to obtain

$\lim_{j\to\infty}\|p_{j}(C_{T})-C_{T}^{1/2}\|=\lim_{j\to\infty}\|p_{j}(C_{T})-C_{T}^{1/2}\|=0.$

Since $T$ is a closed operator, $T^{*}$ is also a closed operator. Thus, we
have

$\displaystyle C_{T}^{1/2}T^{*}x=$
$\displaystyle\;\lim_{j\to\infty}p_{j}(C_{T})T^{*}x=\lim_{j\to\infty}T^{*}p_{n}(C_{T^{*}})x$
$\displaystyle=$ $\displaystyle\;T^{*}(C_{T^{*}})^{1/2}x\quad{\rm for}\quad
x\in\mathcal{D}(T^{*}).$

As $C_{T}^{1/2}T^{*}\subseteq(TC_{T}^{1/2})^{*}=Z_{T^{*}}$, we get that

$Z_{T^{*}}x=C_{T}^{1/2}T^{*}x=T^{*}(C_{T^{*}})^{1/2}x=(Z_{T})^{*}x$

for $x\in\mathcal{D}(T^{*})$. Finally, since $\mathcal{D}(T^{*})$ is dense in
$\mathcal{H}$ and $Z_{T}\in\mathcal{B}(\mathcal{H}_{n})$, we have that
$Z_{T^{*}}x=(Z_{T})^{*}x$ for $x\in\mathcal{H}_{n}$, i.e.,
$Z_{T^{*}}=(Z_{T})^{*}$.

Step 3: Prove (iii).

Using the first of (ii) on $T$ and $T^{*}$ and the assumption $TT^{*}=T^{*}T$,
we have

$\displaystyle I-(Z_{T})^{*}Z_{T}=$
$\displaystyle\;(I+T^{*}T)^{-1}=(I+TT^{*})^{-1}=I-(Z_{T^{*}})^{*}Z_{T^{*}}$
$\displaystyle=$ $\displaystyle\;I-Z_{T}(Z_{T})^{*}$

in which case it is clear that $Z_{T}$ is normal. ∎

Suppose $T\in\mathcal{L}(\mathcal{H}_{n})$ is an unbounded normal operator and
consider the bounded transform $Z_{T}\in\mathcal{B}(\mathcal{H}_{n})$ of $T$
given by (10.1). In view of Lemma 10.0.2, we have $Z_{T}$ is a bounded normal
operator. Thus, we may use Theorem 8.0.4 to find a self-adjoint operator
$A_{Z_{T}}\in\mathcal{B}(\mathcal{H}_{n})$, a positive operator
$B_{Z_{T}}\in\mathcal{B}(\mathcal{H}_{n})$ and a imaginary operator $J$ such
that $A,B$ and $J$ mutually commute and obey

(10.7) $Z_{T}=A_{Z_{T}}+JB_{Z_{T}}.$

######  Definition 10.0.3 (strongly commuting operators).

Suppose $T,\widetilde{T}\in\mathcal{L}(\mathcal{H}_{n})$ are unbounded normal
operators . We will say that $T$ and $\widetilde{T}$ strongly commute if the
bounded transforms of $T$ and $\widetilde{T}$, i.e., $Z_{T}$ and
$Z_{\widetilde{T}}$, respectively, commute.

We are now ready to formulate and prove the spectral theorem for an unbounded
normal operator.

######  Theorem 10.0.4.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be an unbounded normal operator. Then
corresponding to any choice of $\mathfrak{I}\in\mathbb{S}$ and an imaginary
operator $J$ in (10.7), there exists a spectral measure $E$ on the Borel
$\sigma$-algebra $\mathscr{B}(\sigma_{S}(T)\cap\mathbb{C}^{+}_{\mathfrak{I}})$
such that

(10.8) $T=\int_{\sigma_{S}(T)\,\cap\,\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda)\,dE(\lambda)+\int_{\sigma_{S}(T)\,\cap\,\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda)\,dE(\lambda)\,J.$

$E$ is unique in the sense that if $F$ is a spectral measure on
$\mathscr{B}(\mathbb{C}^{+}_{\mathfrak{I}})$ such that
$T=\int_{\mathbb{C}^{+}_{\mathfrak{I}}}{\rm
Re}(\lambda)\,dF(\lambda)+\int_{\mathbb{C}^{+}_{\mathfrak{I}}}{\rm
Im}(\lambda)dF(\lambda)J$, then
$E(M\cap\sigma_{S}(T)\cap\mathbb{C}^{+}_{\mathfrak{I}})=F(M)$ for all
$M\in\mathscr{B}(\mathbb{C}^{+}_{\mathfrak{I}})$.

######  Remark 10.0.5.

Before proceeding to the proof of Theorem 10.0.4, we wish to point out that a
consequence of Theorem 10.0.4 is that for any unbounded normal operator
$T\in\mathcal{L}(\mathcal{H}_{n})$, we have

(10.9) $\sigma_{S}(T)\neq\emptyset.$

Indeed, for any choice of $\mathfrak{I}\in\mathbb{S}$, Theorem 5.3.14 with
$f(\lambda)=\lambda$ implies that ${\rm
supp}\,E=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$. Since ${\rm
supp}\,E\neq\emptyset$, we have (10.9).

###### Proof of Theorem 10.0.4.

The proof is broken into steps.

Step 1: Show that there exists a spectral measure $E$ on
$\sigma_{S}(T)\cap\mathbb{C}^{+}_{\mathfrak{I}}$ such that (9.3) holds.

In view of Lemma 10.0.2, we have $Z_{T}$ is a bounded normal operator.
Moreover, $\|Z_{T}\|\leq 1$. Let $J$ be the imaginary operator appearing in
(10.7). Since $Z_{T}\in\mathcal{B}(\mathcal{H}_{n})$ is normal, we can use
Theorem 6.0.1 to obtain a uniquely determined spectral measure $E_{Z_{T}}$ on
$\mathscr{B}(\sigma_{S}(Z_{T}))\subseteq\overline{\mathbb{D}^{+}_{\mathfrak{I}}}$,where

$\mathbb{D}_{\mathfrak{I}}:=\\{\lambda\in\mathbb{C}_{\mathfrak{I}}:|\lambda|<1\\}\quad\quad{\rm
for}\quad\mathfrak{I}\in\mathbb{S},$

such that

(10.10) $Z_{T}=\int_{\mathbb{D}_{\mathfrak{I}}}{\rm
Re}(\lambda)dE_{Z_{T}}(\lambda)+\int_{\mathbb{D}_{\mathfrak{I}}}{\rm
Im}(\lambda)dE_{Z_{T}}(\lambda)J.$

We may use (10.10) and (5.20) to see that, for any $x\in\mathcal{H}_{n}$ and
$M\in\mathscr{B}(\mathbb{C}^{+}_{\mathfrak{I}})$, we have

(10.11)
$\langle(I-(Z_{T})^{*}Z_{T})E_{Z_{T}}(M),E_{Z_{T}}(M)x\rangle=\int_{M}(1-|\lambda|^{2})d\langle
E_{Z_{T}}(\lambda)x,x\rangle.$

A simple consequence of Theorem 3.2.3 is $C_{T}^{-1}=I-(Z_{T})^{*}Z_{T}$ is
invertible and positive. Thus, ${\rm Ker}\,(I-(Z_{T})^{*}(Z_{T}))=\\{0\\}$ and
$I-(Z_{T})^{*}Z_{T}$ is positive. Thus, (10.11) implies that
$\sigma_{S}(Z_{T})\cap\mathbb{C}_{\mathfrak{I}}^{+}={\rm
supp}\,E_{Z_{T}}\subseteq\overline{\mathbb{D}^{+}_{\mathfrak{I}}}$ and
$E_{Z_{T}}(\mathbb{T}_{\mathfrak{I}}^{+})=0$. Consequently,

$E_{Z_{T}}(\mathbb{D}^{+}_{\mathfrak{I}})=E_{Z_{T}}(\overline{\mathbb{D}^{+}_{\mathfrak{I}}}\,\setminus\,\mathbb{T}_{\mathfrak{I}}^{+})=I,$

where
$\mathbb{T}_{\mathfrak{I}}:=\\{\lambda\in\mathbb{C}_{\mathfrak{I}}:|\lambda|=1\\}$.

Let $\psi(\lambda)=\lambda(1-|\lambda|)^{-1/2}$. Notice that $\psi$ is an
intrinsic function (see Defintion 5.3.12). We claim that
$\mathbb{I}(\psi)=Z_{T}(C_{T}^{1/2})^{-1}$, where $\mathbb{I}$ denotes the
spectral integral with respect to the spectral measure $E_{Z_{T}}$. Indeed,
since $E_{Z_{T}}(\mathbb{D}_{\mathfrak{I}}^{+})=I$ and $\psi$ is clearly
finite $E_{Z_{T}}$-a.e. on $\mathbb{C}_{\mathfrak{I}}^{+}$, (5.16) asserts
that
$\mathcal{D}(\mathbb{I}(\psi))=\mathcal{D}(\mathbb{I}((1-\lambda)^{-1/2}))$.
Thus, we may make us of Theorem 5.3.7(ii) and (iv) to obtain

$\mathbb{I}(\psi)=\mathbb{I}(\lambda)\mathbb{I}((1-|\lambda|^{2})^{-1/2}).$

Next, in view of Lemma 3.0.29, we have that

$\mathbb{I}((1-|\lambda|^{2})^{-1/2})=\mathbb{I}((1-|\lambda|^{2})^{1/2})^{-1}.$

Finally, since $Z_{T}=\mathbb{I}(\lambda)$ and the square root of a positive
bounded operator is unique (see Corollary 6.0.6), we have that

$C_{T}^{1/2}=(I-(Z_{T})^{*}Z_{T})^{1/2}=\mathbb{I}((1-|\lambda|^{2})^{1/2}).$

Therefore, we have proved the claim
$\mathbb{I}(\psi)=Z_{T}(C_{T}^{1/2})^{-1/2}$.

In what follows, with a slight abuse of notation, we will view $\psi$ as
$\psi|_{\mathbb{D}_{\mathfrak{I}}^{+}}$. Let $E(M):=E_{Z_{T}}(\psi^{-1}(M))$
for $M\in\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+})$. In view of Remark
5.3.20, $E$ is a spectral measure on
$\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+})$ and $J$ is an imaginary
associated operator with the spectral measure $E$. Using the fact that
$T=Z_{T}\\{(C_{T})^{1/2}\\}^{-1}$,
$\mathbb{I}(\psi)=Z_{T}\\{(C_{T})^{1/2}\\}^{-1}$ and Theorem 5.3.21, we obtain

$\displaystyle T=$
$\displaystyle\;\mathbb{I}(\psi)=\int_{\mathbb{D}_{\mathfrak{I}}^{+}}{\rm
Re}(\psi(\lambda))dE_{Z_{T}}(\lambda)+\int_{\mathbb{D}_{\mathfrak{I}}^{+}}{\rm
Im}(\psi(\lambda))dE_{Z_{T}}(\lambda)J$ $\displaystyle=$
$\displaystyle\;\int_{\mathbb{D}_{\mathfrak{I}}^{+}}\frac{1}{\sqrt{1-|\lambda|^{2}}}\,dE_{Z_{T}}(\lambda)$
(10.12) $\displaystyle=$
$\displaystyle\;\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda)dE(\lambda)+\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda)dE(\lambda)J.$

Finally, using Theorem 5.3.14, we have that ${\rm
supp}\,E=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$, i.e., (10.12) is
(10.8).

Step 2: Show that $E$ is unique.

Suppose $F$ is another spectral measure on
$\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+})$ such that the imaginary operator
$J$ is associated with $F$ and

$T=\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda)\,dF(\lambda)+\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda)\,dF(\lambda)\,J.$

Let $F^{\prime}(M):=F(\psi(M))$ for
$M\in\mathscr{B}(\mathbb{D}_{\mathfrak{I}}^{+})$. In view of Remark 5.3.20,
$F^{\prime}$ is a spectral measure on
$\mathscr{B}(\mathbb{D}_{\mathfrak{I}}^{+})$ and the imaginary operator $J$ is
associated with $F^{\prime}$. Moreover, since
$E_{Z_{T}}(\mathbb{D}_{\mathfrak{I}}^{+})$, we have

$\displaystyle Z_{T}=$
$\displaystyle\;\int_{\mathbb{D}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda)dF^{\prime}(\lambda)+\int_{\mathbb{D}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda)dF^{\prime}(\lambda)\,J$ $\displaystyle=$
$\displaystyle\;\int_{\mathbb{D}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda)dE_{Z_{T}}(\lambda)+\int_{\mathbb{D}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda)dE_{Z_{T}}(\lambda)\,J.$

But then the Borel functional calculus applied to
$Z_{T}\in\mathcal{B}(\mathcal{H}_{n})$ (see Corollary 9.0.4) asserts that for
any real-valued polynomial $p\in\mathbb{R}[\lambda]$ and
$x\in\mathcal{H}_{n}$, we have

$\displaystyle\langle p(\lambda)x,x\rangle=$
$\displaystyle\;\int_{\mathbb{D}_{\mathfrak{I}}^{+}}p(\lambda)dF^{\prime}(\lambda)x,x\rangle$
$\displaystyle=$
$\displaystyle\;\int_{\mathbb{D}_{\mathfrak{I}}^{+}}p(\lambda)dE_{Z_{T}}(\lambda)x,x\rangle.$

Since $\mathbb{R}[\lambda]$ is dense in the Banach space of real-valued
continuous functions on the compact set
$\overline{\mathbb{D}_{\mathfrak{I}}^{+}}$ (denoted by
$\mathscr{C}(\overline{\mathbb{D}_{\mathfrak{I}}^{+}},\mathbb{R})$), we can
use Corollary 4.0.8 to see that $\langle F^{\prime}(M)x,x\rangle=\langle
E_{Z_{T}}(M)x,x\rangle$ for any
$M\in\mathscr{B}(\mathbb{D}_{\mathfrak{I}}^{+})$, But then the polarisation
formula (2.20) enables us to deduce that $F^{\prime}=E_{Z_{T}}$. Finally since
$F(M)=F^{\prime}(\psi^{-1}(M))$ for
$M\in\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+})$, we have $F=E$ on
$\mathscr{B}(\mathbb{C}_{\mathfrak{I}}^{+})$ as required. ∎

######  Remark 10.0.6.

Let $\mathfrak{I},\mathfrak{J}\in\mathbb{S}$ and
$\gamma:\mathbb{C}^{+}_{\mathfrak{I}}\to\mathbb{C}^{+}_{\mathfrak{J}}$ denote
the bijective map given by

(10.13)
$\gamma(\lambda_{0}+\lambda_{1}\mathfrak{I})=\lambda_{0}+\lambda_{1}\mathfrak{J},$

where $\lambda_{0}\in\mathbb{R}$ and $\lambda_{1}\geq 0$. An immediate
consequence of the proof of Theorem 10.0.4, is that if $E_{\mathfrak{I}}$,
where $\mathfrak{I}\in\mathbb{S}$, is a spectral measure for an unbounded
normal operator $T\in\mathcal{L}(\mathcal{H}_{n})$ and $E_{\mathfrak{J}}$,
where $\mathfrak{J}\in\mathbb{S}$, then since
$\gamma(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+})=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$
and ${\rm
supp}\,E_{\mathfrak{I}}=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$, we
have

(10.14) $E_{\mathfrak{I}}(M)=E_{\mathfrak{J}}(\gamma(M))\quad\quad{\rm
for}\quad M\in\mathscr{B}(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}).$

In view of the above observations, we are justified in calling a spectral
measure $E$ on $\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$ the spectral
measure of $T$.

######  Corollary 10.0.7 (Borel functional calculus in the unbounded case).

Let $T,E$ and $J$ be as in Theorem 10.0.4 and
$\Omega_{\mathfrak{I}}^{+}:=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$
for $\mathfrak{I}\in\mathbb{S}$. For any
$f,g\in\mathfrak{F}(\Omega_{\mathfrak{I}}^{+},\mathscr{B}(\Omega_{\mathfrak{I}}^{+}),E)$
and $c\in\mathbb{C}_{\mathfrak{I}}$, we have the spectral integrals
$\mathbb{I}(f),\mathbb{I}(g)\in\mathcal{L}(\mathcal{H}_{n})$ have the
following properties:

  1. (i)

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

  2. (ii)

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

  3. (iii)

$\mathbb{I}(fc+g)=\overline{\mathbb{I}(f)({\rm Re}(c)I+{\rm
Im}(c)J)+\mathbb{I}(g)}$ for all $c\in\mathbb{C}_{\mathfrak{I}}$.

  4. (iv)

$\mathbb{I}(f)$ is a closed normal operator belonging to
$\mathcal{L}(\mathcal{H}_{n})$.

  5. (v)

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

  6. (vi)

If we let $\Pi_{\pm}(J,\mathfrak{I})$ denote the orthogonal projection onto
the right complex subspace $\mathcal{H}_{\pm}(J,\mathfrak{I})$, respectively
(see Theorem 8.0.5(ii)), then

$\displaystyle\langle\mathbb{I}(f)x,y\rangle=$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}d\langle
E(\lambda)\Pi_{+}(J,\mathfrak{I})x,y\rangle\,f(\lambda)$ $\displaystyle+$
$\displaystyle\;\int_{\Omega_{\mathfrak{I}}^{+}}d\langle
E(\lambda)\Pi_{-}(J,\mathfrak{I})x,y\rangle\overline{f(\lambda)}\quad\quad{\rm
for}\quad x\in\mathcal{D}(T),$

where both integrals above are meant in the sense of (4.5).

  7. (vii)

If
$f\in\mathfrak{F}(\Omega_{\mathfrak{I}}^{+},\mathscr{B}(\Omega_{\mathfrak{I}}^{+}),\mathbb{C}_{\mathfrak{I}},E)$
is nonnegative $E$-a.e., then $\mathbb{I}(f)\succeq 0$.

  8. (vii)

$\mathbb{I}(f)^{-1}\in\mathcal{L}(\mathcal{H}_{n})$ if and only if
$f(\lambda)\neq 0$ $E$-a.e. In this case,
$\mathbb{I}(f)^{-1}=\mathbb{I}(1/f)$.

  9. (viii)

The spectral measure $E_{\mathbb{I}(f)}$ of $\mathbb{I}(f)$ satisfies the
identity

(10.15) $E_{\mathbb{I}(f)}(M)=E(f^{-1}(M\cap f(\Omega_{\mathfrak{I}}^{+})).$

###### Proof.

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be normal and fix
$\mathfrak{I}\in\mathbb{S}$. Then by Theorem 10.0.4, we can find a spectral
measure $E$ on $\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$ such that
(10.8) holds. Thus, we may invoke Theorem 5.3.7 with
$\Omega=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$ and
$\mathscr{A}=\mathscr{B}(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+})$ to
obtain (i)-(v).

Assertions (vi)-(viii) are straight forward adaptations of
Theorem 5.2.4(viii)-(xi) to the unbounded setting. The corresponding proofs
can be completed in much the same way as the bounded setting. ∎

######  Corollary 10.0.8 (subclasses of bounded normal operators).

Let $T,E$ and $J$ be as in Theorem 10.0.4 and
$\Omega_{\mathfrak{I}}^{+}:=\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$
for $\mathfrak{I}\in\mathbb{S}$. We have the following:

  1. (i)

$T$ is self-adjoint if and only if $\sigma_{S}(T)\subseteq\mathbb{R}$. In this
case,

$T=\int_{\mathbb{R}}t\,dE(t).$

  2. (ii)

$T$ is positive if and only if $\sigma_{S}(T)\subseteq[0,\infty)$. In this
case,

$T=\int_{[0,\infty)}t\,dE(t).$

  3. (iii)

$T$ is anti self-adjoint if and only if
$\sigma_{S}(T)\subseteq\\{s\in\mathbb{R}^{n+1}:{\rm Re}(s)=0\\}$. In this
case,

$T=\int_{\mathbb{R}}t\,dE(t)J.$

###### Proof.

The proofs of (i)-(iii) can be completed in much the same way of the proofs of
Corollary 9.0.5(i)-(iii) with Theorem 10.0.4 in place of Theorem 9.0.2. ∎

The following corollary is an unbounded analogue of the Teichmüller
decomposition.

######  Corollary 10.0.9.

Corresponding to any normal operator $T\in\mathcal{L}(\mathcal{H}_{n})$, there
exist a self-adjoint operator $A\in\mathcal{L}(\mathcal{H}_{n})$, an imaginary
operator $J\in\mathcal{B}(\mathcal{H}_{n})$ and a positive operator
$B\in\mathcal{L}(\mathcal{H}_{n})$ such that $A$, $J$ and $B$ strongly commute
and satisfy

(10.16) $T=A+JB.$

In this case, $A=\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Re}(\lambda)\,dE(\lambda)$ and $B=\int_{\mathbb{C}_{\mathfrak{I}}^{+}}{\rm
Im}(\lambda)\,dE(\lambda)$, where $E$ is the spectral measure for $T$, and

(10.17) $T$ and $J$ strongly commute.

###### Proof.

The additive decomposition (10.16) follows at once from (10.8). The fact that
$A$, $B$ and $J$ strongly commute follows from the fact that
$Z_{A}=A_{Z_{T}}$, $Z_{B}=B_{Z_{T}}$, where $A_{Z_{T}}$ and $B_{Z_{T}}$ are as
in (10.7), and the fact that $A_{Z_{T}},B_{Z_{T}}$ and $J$ mutually commute.
Finally, (10.17) is an immediate consequence of (10.16) and the fact that
$A,B$ and $J$ strongly commute ∎

######  Theorem 10.0.10 (spectral mapping theorem for an unbounded normal
operator).

Let $T\in\mathcal{L}(\mathcal{H}_{n})$ be normal with spectral measure $E$ and
$\Omega_{\mathfrak{I}}^{+}:=\sigma_{S}(T)\cap\mathbb{C}^{+}_{\mathfrak{I}}$,
where $\mathfrak{I}\in\mathbb{S}$. For any intrinsic continuous function
$f\in\mathfrak{B}(\Omega_{\mathfrak{I}}^{+},\mathscr{B}(\Omega_{\mathfrak{I}}^{+}),\mathbb{C}_{\mathfrak{I}},E)$,
we have

(10.18)
$\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}=\overline{f(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}})},$

If $T\in\mathcal{B}(\mathcal{H}_{n})$ is normal, then (10.18) becomes

(10.19)
$\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}=f(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}).$

###### Proof.

We begin by noting that image of $f$ is symmetric about the real axis in
$\mathbb{C}_{\mathfrak{I}}$ (see (5.29) and (5.30)). We will first show that

$\overline{f(\sigma_{S}(T)\cap\mathbb{C}_{{\mathfrak{I}}})}\subseteq\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}.$

Suppose $w_{0}\in\overline{f(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}})}$ and
$w_{0}\in\mathbb{C}_{\mathfrak{I}}^{+}$. For any $\varepsilon>0$, we can find
$\lambda_{0}\in\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}$ such that
$f(\lambda_{0})\in\mathbb{C}_{\mathfrak{I}}^{+}$ and

(10.20) $|f(\lambda_{0})-w_{0}|<\frac{\varepsilon}{2}.$

Since $f$ is continuous, we have the existence of $\delta>0$ such that

$\displaystyle U_{\delta}:=$
$\displaystyle\;\\{\lambda\in\mathbb{C}_{\mathfrak{I}}:|\lambda-\lambda_{0}|<\delta\\}$
$\displaystyle\subseteq$
$\displaystyle\;\\{\lambda\in\mathbb{C}_{\mathfrak{I}}:|f(\lambda)-f(\lambda_{0})|<\frac{\varepsilon}{2}\\}$
$\displaystyle\subseteq$
$\displaystyle\;\\{\lambda\in\mathbb{C}_{\mathfrak{I}}:|f(\lambda)-w_{0}|<\varepsilon\\}$
$\displaystyle=:$ $\displaystyle\;V_{\varepsilon}.$

Since $w_{0}$ and $f(\lambda_{0})$ both belong to
$\mathbb{C}_{\mathfrak{I}}^{+}$, we may proceed as in the verification of
(9.10) and show that

$V_{\varepsilon}\cap\mathbb{C}_{\mathfrak{I}}^{+}=\\{\lambda\in\mathbb{C}^{+}_{\mathfrak{I}}:|f(\lambda)^{2}-2{\rm
Re}(w_{0})f(\lambda)+|w_{0}|^{2}|<\varepsilon^{2}\\}.$

Thus, as $\lambda_{0}\in\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$ and
$\lambda_{0}\in U_{\delta}$, we have that $E(U_{\delta})\neq\emptyset$, in
which case $E(V_{\varepsilon}\cap\mathbb{C}_{\mathfrak{I}}^{+})\neq\emptyset$.
But then Theorem 5.3.14 ensures that
$w_{0}\in\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}^{+}$.

If $w_{0}$ and $f(\lambda_{0})$ both belong to
$\mathbb{C}_{\mathfrak{I}}^{-}$, then we can use the fact that the image of
$f$ is symmetric about the real axis and the fact that the $S$-spectrum is
axially symmetric (see Remark 3.1.7) and repeat the argument above with
$\bar{w}_{0}$ and $\bar{\lambda}_{0}$ in place of $w_{0}$ and $\lambda_{0}$,
respectively, to arrive at the same conclusion.

We will now show that
$\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}\subseteq\overline{f(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}})}$.
The inclusion
$\sigma_{S}(\mathbb{I}(f))\cap\mathbb{C}_{\mathfrak{I}}\subseteq\overline{f(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}})}$
follows immediately from comparing the formula for
$\sigma_{S}(\mathbb{I}(f)\cap\mathbb{C}_{\mathfrak{I}}$ coming from Theorem
5.3.14 and the formula for
$\overline{f(\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}})}$ derived from (9.9).

Finally, (10.19) drops out easily from the fact that if
$T\in\mathcal{B}(\mathcal{H}_{n})$, then
$\sigma_{S}(T)\cap\mathbb{C}_{\mathfrak{I}}^{+}$ is compact. ∎

##  Appendix A Background on the $S$-spectrum and connections with functional
calculi and function theory

In 1936, Birkhoff and von Neumann, in their paper [9] on the logic of quantum
mechanics, showed that an set-theoretic abstraction of quantum mechanics can
be formulated on Hilbert spaces over the reals, complex numbers and
quaternions. Consequently, there was a strong motivation to prove the spectral
theorem in the quaternionic setting (i.e., a Clifford module over
$\mathbb{R}_{2}$) and since that time several attempts have appeared in the
literature. The main contributions are due to O. Teichmüller [67] in 1936 and
to K. Viswanath [69] in 1971. However, both authors do not make clear the
notion of spectrum that is in use for quaternionic linear operators.
Nevertheless, there are useful results on quaternionic operator theory in [67]
and [69].

The major breakthrough came in 2006 when I. Sabadini and the first author
discovered the $S$-spectrum $\sigma_{S}(T)$ and the $S$-functional calculus
for a quaternionic linear operator $T$. A prime motivation for this
investigation was to give quaternionic quantum mechanics a rigorous
mathematical foundation. The strategy for the identification of the
$S$-spectrum was purely based on hyperholomorphic analysis methods and not on
physical arguments (see the introduction of the book [16] for a detailed
explanation). The definition of $S$-spectrum $\sigma_{S}(T)$ for an linear
operator $T$ on a quaternionic Banach space $V$ is somewhat counter intuitive
because it involves the second order operator

$Q_{s}(T):=T^{2}-2s_{0}T+|s|^{2}\,I$

and is given by

$\sigma_{S}(T)=\\{s\in\mathbb{H}\ :\ Q_{s}(T)\ \ {\rm is\ not\ invertible}\ \
\mathcal{B}(V)\\},$

where $\mathbb{H}$ is the algebra of quaternions, $s_{0}$ the real part of
$s\in\mathbb{H}$, $|s|^{2}$ is the modulus squared and $\mathcal{B}(V)$ is the
space of all bounded linear operators.

Before 2006 in the literature there were two different notions of spectrum in
the quaternionic setting (as well as in the Clifford setting): the left
spectrum $\sigma_{L}(T)$ and the right spectrum $\sigma_{R}(T)$ and both
definitions mimic the eigenvalue problem for complex operators. We point out
that just in the finite dimensional case the quaternionic spectral theorem was
precisely proved using the notion of right spectrum by D. R. Farenick, B. A.
F. Pidkowich in [30] that was published in 2003. We also want to point out
that in the literature on quaternionic quantum mechanics physicists used the
right spectrum to describe the bounded states where there are just the
eigenvalues, see the book of S. Adler [1] and for more recent advanced see the
paper of J. Gantner [34] on the equivalence of complex and quaternionic
quantum mechanics. Only in 2015 the authors with D. Alpay proved the spectral
theorem for quaternionic normal operators based on the $S$-spectrum
$\sigma_{S}(T)$ in [4] which was published in 2016. Fairly recently, there has
been a renewed quaternionic quantum mechanics, which utilises the notion of
$S$-spectrum (see, e.g., [57, 58, 59, 68]).

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