# Lebesgue type decompositions and Radon-Nikodym derivatives for pairs of
bounded linear operators

S. Hassi  and  H.S.V. de Snoo  Department of Mathematics and Statistics
University of Vaasa
P.O. Box 700, 65101 Vaasa
Finland<EMAIL_ADDRESS>Bernoulli Institute for Mathematics, Computer Science
and Artificial Intelligence
University of Groningen
P.O. Box 407, 9700 AK Groningen
Nederland<EMAIL_ADDRESS>
###### Abstract.

For a pair of bounded linear Hilbert space operators $A$ and $B$ one considers
the Lebesgue type decompositions of $B$ with respect to $A$ into an almost
dominated part and a singular part, analogous to the Lebesgue decomposition
for a pair of measures (in which case one speaks of an absolutely continuous
and a singular part). A complete parametrization of all Lebesgue type
decompositions will be given, and the uniqueness of such decompositions will
be characterized. In addition, it will be shown that the almost dominated part
of $B$ in a Lebesgue type decomposition has an abstract Radon-Nikodym
derivative with respect to the operator $A$.

###### Key words and phrases:

Lebesgue type decompositions, operator range, pair of bounded operators,
almost dominated part, singular part, Radon-Nikodym derivative

###### 2010 Mathematics Subject Classification:

47A05, 47A06, 47A65

##  1\. Introduction

Let ${\mathfrak{E}}$, ${\mathfrak{H}}$, and ${\mathfrak{K}}$ be Hilbert spaces
and $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ be bounded linear operators.
In the present paper it will be shown that there are so-called Lebesgue type
decompositions of the operator
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ relative to the operator
$A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ of the form

(1.1) $B=B_{1}+B_{2},\quad
B_{1},B_{2}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}}),$

where ${\rm ran\,}B_{1}\perp{\rm ran\,}B_{2}$, $B_{1}$ is almost dominated by
$A$, and $B_{2}$ is singular with respect to $A$; the terminology will be
explained below. The collection of all Lebesgue type decompositions will be
parametrized and a criterion for the uniqueness of such decompositions will be
established. Furthermore, it will be shown that if $B$ has the above Lebesgue
type decomposition (1.1) with respect to $A$, then there exists a uniquely
determined closed linear operator $C$ from ${\mathfrak{H}}$ to
${\mathfrak{K}}$ satisfying a certain minimality condition, that, in general,
is unbounded, such that

(1.2) $B_{1}=CA;$

for details, see Theorem 7.4 below. This operator $C$ will be called the
Radon-Nikodym derivative of $B_{1}$ with respect to $A$. The above results are
the abstract analogs of the usual Lebesgue decomposition of a finite measure
into an absolutely continuous part and a singular part, and of the
corresponding Radon-Nikodym derivative for the absolutely continuous part.
There are, indeed, situations in measure theory where actually there is more
than one Lebesgue type decomposition.

The above results can be interpreted as special cases of corresponding results
for linear relations. For any linear relation $T$ from a Hilbert space
${\mathfrak{H}}$ to a Hilbert space ${\mathfrak{K}}$ there exist the so-called
Lebesgue type decompositions of the form

$T=T_{1}+T_{2}$

such that ${\rm ran\,}T_{1}\perp{\rm ran\,}T_{2}$, $T_{1}$ is a regular
operator (an operator whose closure in ${\mathfrak{H}}\times{\mathfrak{K}}$ is
an operator, i.e., a closable operator), and $T_{2}$ is a singular relation (a
relation whose closure is the product of closed linear subspaces in
${\mathfrak{H}}$ and ${\mathfrak{K}}$, respectively). A general treatment of
Lebesgue type decompositions of a linear relation $T$ was carried out in the
recent paper [7]. For instance, in that paper an explicit parametrization of
all Lebesgue type decompositions of $T$ was established and the case where the
Lebesgue type decomposition of $T$ is unique has been characterized therein.
In the setting of a pair of positive operators this type of uniqueness result
goes back to Ando [1].

In the present paper Lebesgue type decompositions for linear relations which
are simultaneously operator ranges are studied (see [4]); in what follows such
linear relations are called shortly operator range relations. Operator range
relations coincide with the linear relations of the form

(1.3) $L(A,B)=\\{\,\\{Af,Bf\\}:\,f\in{\mathfrak{E}}\,\\},$

where $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. The Lebesgue type
decompositions of the relation $L(A,B)$ correspond to the Lebesgue type
decompositions of the operator $B$ with respect to the operator $A$ in (1.1).
In particular, $B$ is almost dominated by $A$ precisely if the corresponding
relation $L(A,B)$ is regular, i.e., $L(A,B)$ is a closable operator; moreover,
$B$ is singular with respect to $A$ precisely if $L(A,B)$ is singular. If
$L(A,B)$ is regular, then its closure is the Radon-Nikodym derivative of $B$
with respect to $A$ mentioned in (1.2); the precise meaning of this statement
will be explained later. The approach in the present paper makes it possible
to further develop the results established in [7], including uniqueness
results, to the setting of operator range relations; see also [8, 6, 10],

The topic of the present paper was inspired by the work of Ando about Lebesgue
type decompositions for pairs of bounded nonnegative operators [1], and the
work of Simon about Lebesgue decompositions for nonnegative forms [29], see
also [6] and [8], and Kosaki’s work on the Radon-Nikodym derivative in the
setting of $C^{*}$-algebras [21]. The context of pairs of bounded linear
operators which are not necessarily nonnegative, is connected with the work of
Mac Nerney, Kaufman, and others; see [16, 17, 18, 19, 20]. Moreover, there are
strong connections with the work of Izumino [11, 12, 13] and of Izumino and
Hirasawa [14]. They treated the case where $L(A,B)$ in (1.3) is a densely
defined operator. The parametrization of the Lebesgue type decompositions and
the notion of Radon-Nikodym seem to be new even in the case where $L(A,B)$ is
densely defined. Moreover, at this point, it should be mentioned that,
although the paper is inspired by [1], [29], and [21], the decompositions
(1.1) here are concerned with pairs $A$ and $B$, where $B_{1}$ is almost
dominated by $A$ and $B_{2}$ is singular with respect to $A$. However, the
decompositions of Ando and Simon, and Kosaki’s Radon-Nikodym derivatives
belong to a slightly different setting; this setting will be considered in
further work.

Here is a brief description of the contents of the paper. The concept of
operator ranges, including their natural topologies, will be reviewed in
Section 2. Operator range relations and a special normalized class of them are
treated in Section 3. As an application, a construction of the operator range
representation of a closed relation is derived in Section 4. This construction
resembles a measure theoretic treatment of Radon-Nikodym derivatives for a
pair of positive measures, and, as a bonus, leads to a natural introduction of
Radon-Nikodym derivatives for pairs of bounded linear operators in Section 7.
As a preparation for Lebesgue type decompositions for pairs of bounded
operators, some characterizations of regular and singular operator range
relations along the lines of [7] are given in Section 5. The corresponding
classification for pairs of bounded linear operators can be found in Section
6. This involves the notions of domination and almost domination of a bounded
operator with respect to another bounded operator, which correspond to the
concept of absolute continuity of a positive measure with respect to another
positive measure. Similarly the notion of singularity of a pair of bounded
operators is defined as an operator analog for the concept of singularity of a
pair of positive measures. In Section 7 the abstract Radon-Nikodym derivative
is introduced and investigated. The definition of Radon-Nikodym derivative
given here involves an optimality property, see Lemma 7.1 and Theorem 7.4 and,
in fact, this notion is uniquely determined in the general setting of operator
range relations. Furthermore, it is shown that the Radon-Nikodym derivatives
for operator range relations admit similar properties known to hold for pairs
of positive measures; see Theorem 7.7. Finally, all Lebesgue type
decompositions for pairs of bounded linear operators are described in Section
8 with a uniqueness result analogous to that of Ando [1].

Further work will be concerned with the situation that the operators $A$ and
$B$ are nonnegative; such cases have been considered by Ando, Simon, Kosaki,
and by later authors. In particular, there will be an explicit connection to
the recent papers by Z. Sebestyén, Zs. Tarcsay, and T. Titkos; see for
instance [27, 28, 30]. Moreover, further work will also be connected with the
situation where at least one of the operators $A$ and $B$ is not bounded.

##  2\. Linear relations admitting an operator range representation

In this section the special class of linear relations which, in addition, are
operator ranges, will be introduced. In what follows, such linear relations
are briefly called operator range relations. This class extends the class of
closed linear relations and the operator range relations have a number of
useful properties. The introduction will be facilitated by a brief treatment
of operator ranges in Hilbert spaces. The notions of operator ranges and
operator range relations in the present sense go back to [4], [17], [22],
[23], [24], [25]. A brief survey is given in this section. For the convenience
of the reader, some proofs are included.

######  Definition 2.1.

Let ${\mathfrak{X}}$ be a Hilbert space with inner product $(\cdot,\cdot)$. A
subspace ${\mathfrak{M}}$ of ${\mathfrak{X}}$, together with an inner-product
$(\cdot,\cdot)_{+}$ on ${\mathfrak{M}}$, is said to be an operator range in
${\mathfrak{X}}$, if

  1. (a)

${\mathfrak{M}}$ is a Hilbert space when equipped with $(\cdot,\cdot)_{+}$;

  2. (b)

$\|u\|_{+}\geq c\|u\|_{\mathfrak{H}}$, $u\in{\mathfrak{M}}$, for some $c>0$.

In particular, a closed linear subspace is an operator range.

The terminology operator range is motivated the following lemma. If
${\mathfrak{M}}\subset{\mathfrak{X}}$ is the range of a bounded operator, then
there is a natural inner-product on ${\mathfrak{M}}$, that makes
${\mathfrak{M}}$ an operator range.

######  Lemma 2.2.

Let ${\mathfrak{X}}$ and ${\mathfrak{Y}}$ be Hilbert spaces and let
$Z\in\mathbf{B}({\mathfrak{Y}},{\mathfrak{X}})$. Then the linear space
${\mathfrak{M}}={\rm ran\,}Z$, equipped with the inner-product

(2.1) $(Zx,Zy)_{+}=(x,y),\quad x,y\in{\mathfrak{Y}}\ominus{{\rm ker\,}\,}Z,$

is an operator range.

###### Proof.

Assume that $Z$ is not the zero operator. Note that ${\mathfrak{M}}={\rm
ran\,}Z$ with $(\cdot,\cdot)_{+}$ in (2.1) is indeed an inner-product space.
To see that it is complete, let $(Zx_{n})$ be a Cauchy sequence in
$({\mathfrak{M}},(\cdot,\cdot)_{+})$ with $x_{n}\in{\mathfrak{Y}}\ominus{{\rm
ker\,}\,}Z$. Then $(x_{n})$ is a Cauchy sequence in
${\mathfrak{Y}}\ominus{{\rm ker\,}\,}Z$. Thus $x_{n}\to x$ for some
$x\in{\mathfrak{Y}}\ominus{{\rm ker\,}\,}Z$ and $Zx_{n}\to Zx$, since
$Z\in\mathbf{B}({\mathfrak{Y}},{\mathfrak{X}})$. This shows (a) in Definition
2.1. By definition $\|Zx\|_{+}=\|x\|$, $x\in{\mathfrak{Y}}\ominus{{\rm
ker\,}\,}Z$, and $\|Zx\|\leq\|Z\|\|x\|$, $x\in{\mathfrak{Y}}$, lead to the
inequality

$\|Zx\|_{+}=\|x\|\geq\frac{1}{\|Z\|}\|Zx\|_{\mathfrak{X}},\quad
x\in{\mathfrak{Y}}\ominus{{\rm ker\,}\,}Z,$

which gives (b) in Definition 2.1. ∎

There is also a converse to Lemma 2.2.

######  Lemma 2.3.

Let $({\mathfrak{M}},(\cdot,\cdot)_{+})$ be an operator range in
${\mathfrak{X}}$. Then there exists an operator
$Z\in\mathbf{B}({\mathfrak{X}})$ such that ${\mathfrak{M}}={\rm ran\,}Z$ and

(2.2) $(Zx,Zy)_{+}=(x,y),\quad x,y\in{\mathfrak{X}}\ominus{{\rm ker\,}\,}Z.$

The operator $Z$ may be chosen to be nonnegative.

###### Proof.

Let $\imath:{\mathfrak{M}}\to{\mathfrak{X}}$ be the identification map, where
each space has its own topology. Then it follows from (b) that
$c\,\|\imath\,x\|\leq\|x\|_{+}$, so that $\imath$ is bounded. Its adjoint
$\imath^{\times}$ from ${\mathfrak{X}}$ to ${\mathfrak{M}}$ is a bounded
mapping and one has the polar decomposition (cf. [15, Section VI 2.7])

$\imath^{\times}=Z\,|\,\imath^{\times}|,$

where $Z:{\mathfrak{X}}\to{\mathfrak{M}}$ is the unique partial isometry with
initial space ${\rm\overline{ran}\,}|\,\imath^{\times}|$ and final space
${\rm\overline{ran}\,}\imath^{\times}$. Since the last space is the orthogonal
complement in ${\mathfrak{M}}$ of ${{\rm ker\,}\,}\imath$, one sees that
${\rm\overline{ran}\,}\imath^{\times}={\mathfrak{M}}$. Consequently, ${\rm
ran\,}Z={\mathfrak{M}}$, and (2.2) holds as $Z$ is a partial isometry from
${\mathfrak{X}}$ to ${\mathfrak{M}}$. Finally, it remains to observe that

$c\|Zx\|\leq\|Zx\|_{+}\leq\|x\|,\quad x\in{\mathfrak{X}},$

thanks to (b) and the fact that $Z$ is a partial isometry. Thus
$Z\in\mathbf{B}({\mathfrak{X}})$.

For the proof of the last statement, consider
$Z\in{\mathbf{B}}({\mathfrak{X}})$ and its polar decomposition $Z=|Z^{*}|C$,
where $C\in{\mathbf{B}}({\mathfrak{X}})$ is the unique partial isometry with
initial space ${\rm\overline{ran}\,}Z^{*}$ and final space
${\rm\overline{ran}\,}|Z^{*}|$. Observe that, by the Douglas Lemma, ${\rm
ran\,}|Z^{*}|={\rm ran\,}Z$, so that ${\rm ran\,}|Z^{*}|={\mathfrak{M}}$ and,
clearly,

$(|Z^{*}|Cx,|Z^{*}|Cy)_{+}=(x,y)=(Cx,Cy),\quad x,y\in({{\rm
ker\,}\,}Z)^{\perp}.$

This implies that

$(|Z^{*}|u,|Z^{*}|v)_{+}=(u,v),\quad u,v\in({{\rm ker\,}\,}|Z^{*}|)^{\perp},$

where $|Z^{*}|\in\mathbf{B}({\mathfrak{X}})$ is nonnegative. ∎

According to Lemma 2.2 and Lemma 2.3, an operator range ${\mathfrak{M}}$ is
naturally parametrized by a bounded operator
$Z\in\mathbf{B}({\mathfrak{Y}},{\mathfrak{X}})$. The subspace ${{\rm
ker\,}\,}Z\subset{\mathfrak{Y}}$ is called the redundant part of this
parametrization, as it does not contribute to ${\rm ran\,}Z$. The restriction
$Z_{0}$ of $Z$ to ${\mathfrak{Y}}\ominus{{\rm ker\,}\,}Z$ is called the
reduced part or reduction of $Z$:

(2.3) $Z=(Z_{0}\,;\,O_{{{\rm ker\,}\,}Z}).$

The topology of ${\mathfrak{M}}$ is uniquely determined by the reduced part of
the representing operator in a sense to b explained below.

######  Lemma 2.4.

Let ${\mathfrak{M}}$ be an operator range in ${\mathfrak{X}}$. Assume that
there exist Hilbert spaces ${\mathfrak{Y}}$ and ${\mathfrak{Y}}_{1}$ and
operators $Z\in\mathbf{B}({\mathfrak{Y}},{\mathfrak{X}})$ and
$Z_{1}\in\mathbf{B}({\mathfrak{Y}}_{1},{\mathfrak{X}})$, which each satisfy
the conditions in Lemma 2.2. Then there is a bounded and boundedly invertible
operator $W\in\mathbf{B}({\mathfrak{Y}}_{1},{\mathfrak{Y}})$ such that

(2.4) $Z_{1}=ZW,\quad{{\rm ker\,}\,}W={{\rm ker\,}\,}Z_{1},\quad{\rm
ran\,}W={\rm\overline{ran}\,}Z^{*}.$

Consequently, the topologies induced on ${\mathfrak{M}}$ by $Z$ and by $Z_{1}$
are equivalent.

###### Proof.

If ${\mathfrak{M}}={\rm ran\,}Z_{1}$ with
$Z_{1}\in\mathbf{B}({\mathfrak{Y}}_{1},{\mathfrak{X}})$ with a Hilbert space
${\mathfrak{Y}}_{1}$, then it is clear that the operators $Z_{1}$ and $Z$ have
the same range. Hence, by the Douglas lemma, there is an operator
$W\in\mathbf{B}({\mathfrak{Y}}_{1},{\mathfrak{Y}})$ such that (2.4) holds.
Note that the operator $W$ in Lemma 2.4 is a bounded bijective mapping from
$({{\rm ker\,}\,}Z_{1})^{\perp}$ onto $({{\rm ker\,}\,}Z)^{\perp}$, as follows
from the closed graph theorem. ∎

######  Lemma 2.5.

Let ${\mathfrak{M}}={\rm ran\,}Z$ be an operator range as in Lemma 2.1 and let
$Z_{0}$ be the reduced part of $Z$. Then the following statements are
equivalent:

  1. (i)

${\mathfrak{M}}$ is an operator range in ${\mathfrak{X}}$ that is closed;

  2. (ii)

${\rm ran\,}Z$ or, equivalently, ${\rm ran\,}Z_{0}$ is closed;

  3. (iii)

$Z_{0}$ is bounded and boundedly invertible.

Operator ranges in a Hilbert space ${\mathfrak{X}}$ extend the notion of
closed linear subspaces of ${\mathfrak{X}}$. In particular, operator ranges
form a lattice; cf. [4].

The previous notion of operator range will now be extended to subspaces of
product spaces; see for instance [23].

######  Definition 2.6.

A linear relation $T$ from ${\mathfrak{H}}$ to ${\mathfrak{K}}$ is said to be
an operator range relation if its graph is an operator range in
${\mathfrak{H}}\times{\mathfrak{K}}$, thus $T={\rm ran\,}\Phi$ for some
$\Phi\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}}\times{\mathfrak{K}})$,
where ${\mathfrak{E}}$ is some Hilbert space.

######  Lemma 2.7.

Let $T$ be a closed linear relation from the Hilbert space ${\mathfrak{H}}$ to
the Hilbert space ${\mathfrak{K}}$. Then $T$ is an operator range relation.

###### Proof.

Consider the orthogonal decomposition
${\mathfrak{E}}={\mathfrak{H}}\times{\mathfrak{K}}=T\operatorname{\,\widehat{\oplus}\,}T^{\perp}$
and let $\Phi=P_{T}$ be the orthogonal projection from
${\mathfrak{E}}={\mathfrak{H}}\times{\mathfrak{K}}$ to $T$, so that $T={\rm
ran\,}\Phi$. ∎

To characterize operator range relations the following notions and notations
are needed. For a pair of operators
$A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ the linear relation $L(A,B)$
from ${\mathfrak{H}}$ to ${\mathfrak{K}}$ is defined by

$L(A,B)=\\{\,\\{Af,Bf\\}:\,f\in{\mathfrak{E}}\,\\}.$

Let $P_{1}$ and $P_{2}$ be the orthogonal projections from
${\mathfrak{H}}\times{\mathfrak{K}}$ onto the component subspaces
${\mathfrak{H}}\times\\{0\\}$ and $\\{0\\}\times{\mathfrak{K}}$, respectively.
The mappings $\iota_{1}\\{\varphi,0\\}=\varphi$ and
$\iota_{2}\\{0,\psi\\}=\psi$ identify ${\mathfrak{H}}\times\\{0\\}$ with
${\mathfrak{H}}$ and $\\{0\\}\times{\mathfrak{K}}$ with ${\mathfrak{K}}$,
respectively. In the following theorem the operator range relations are
characterized; see [2, Theorem 1.10.1].

######  Theorem 2.8.

Let ${\mathfrak{H}}$ and ${\mathfrak{K}}$ be Hilbert spaces and let $T$ be a
linear relation from ${\mathfrak{H}}$ to ${\mathfrak{K}}$. Then there are the
following statements:

  1. (a)

If $T={\rm ran\,}\Phi$, where
$\Phi\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}}\times{\mathfrak{K}})$, then
$T=L(A,B)$ with

$A=\iota_{1}P_{1}\Phi\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})\quad\mbox{and}\quad
B=\iota_{2}P_{2}\Phi\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}}).$

  2. (b)

If $T=L(A,B)$, where $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, then $T={\rm ran\,}\Phi$
with

$\Phi\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}}\times{\mathfrak{K}})\quad\mbox{and}\quad\Phi
f=\\{Af,Bf\\},\quad f\in{\mathfrak{E}}.$

######  Corollary 2.9.

If $T$ is an operator range relation from ${\mathfrak{H}}$ to
${\mathfrak{K}}$, then ${\rm dom\,}T$ and ${{\rm ker\,}\,}T$ are operator
ranges in ${\mathfrak{H}}$, while ${\rm ran\,}T$ and ${\rm mul\,}T$ are
operator ranges in ${\mathfrak{K}}$.

The concept of an operator range relation extends in a certain way the notion
of a closed linear relation. For instance, sums, and intersections, as well as
Cartesian products of operator range relations are again operator range
relations. In particular, operator range relations form a lattice; cf. [4].
Notice also that if the relation $T$ is itself a range of some closed linear
relation $H$ from a Hilbert space ${\mathfrak{E}}$ to the Hilbert space
${\mathfrak{H}}\times{\mathfrak{K}}$, then $T$ is still an operator range
relation.

##  3\. Operator range relations and normalized pairs

Let ${\mathfrak{E}}$, ${\mathfrak{H}}$, and ${\mathfrak{K}}$ be Hilbert spaces
and consider the pair of operators
$A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. It will be convenient to
recall from [5, 26], see also [4], the following auxiliary column and row
operators. The column operator $c(A,B)$ from ${\mathfrak{E}}$ to
${\mathfrak{H}}\times{\mathfrak{K}}$ is defined by

(3.1) $c(A,B)=\begin{pmatrix}A\\\ B\end{pmatrix},\quad\mbox{i.e.,}\quad
c(A,B)\varphi=\begin{pmatrix}A\varphi\\\
B\varphi\end{pmatrix},\quad\varphi\in{\mathfrak{E}}.$

Then $c(A,B)$ belongs to
${\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}}\times{\mathfrak{K}})$. Likewise,
the row operator from ${\mathfrak{H}}\times{\mathfrak{K}}$ to ${\mathfrak{E}}$
is defined by

(3.2) $r(A,B)=\begin{pmatrix}A&B\end{pmatrix}\quad\mbox{i.e.,}\quad
r(A,B)\begin{pmatrix}h\\\ k\end{pmatrix}=Ah+Bk,\quad
h\in{\mathfrak{H}},\,\,k\in{\mathfrak{K}}.$

Then $r(A,B)$ belongs to
${\mathbf{B}}({\mathfrak{H}}\times{\mathfrak{K}},{\mathfrak{E}})$ and

(3.3) ${\rm ran\,}r(A,B)={\rm ran\,}A+{\rm ran\,}B.$

Moreover, it is clear from (3.1) and (3.2) that

(3.4) $c(A,B)^{*}=r(A^{*},B^{*}).$

Therefore, it follows from (3.1) and (3.2) that

(3.5) $c(A,B)^{*}c(A,B)=A^{*}A+B^{*}B,$

and that

(3.6) $c(A,B)c(A,B)^{*}=\begin{pmatrix}AA^{*}&AB^{*}\\\
BA^{*}&BB^{*}\end{pmatrix}.$

For a pair of operators $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ the linear relation $L(A,B)$
from ${\mathfrak{H}}$ to ${\mathfrak{K}}$ is defined by

(3.7) $L(A,B)=\\{\,\\{Af,Bf\\}:\,f\in{\mathfrak{E}}\,\\},$

so that $L(A,B)$ is an operator range relation as in Theorem 2.8, since in the
present notation

(3.8) $L(A,B)={\rm ran\,}c(A,B).$

The operator range relation $L(A,B)$ is sometimes called a quotient, as,
indeed, in the sense of the product of linear relations one can write $L(A,B)$
as $BA^{-1}$; cf. [16]. Note that the domain and the range of $L(A,B)$ in
(3.7) are given by

${\rm dom\,}L(A,B)={\rm ran\,}A,\quad{\rm ran\,}L(A,B)={\rm ran\,}B,$

while the kernel and the multivalued part of $L(A,B)$ are given by

(3.9) ${{\rm ker\,}\,}L(A,B)=A({{\rm ker\,}\,}B),\quad{\rm
mul\,}L(A,B)=B({{\rm ker\,}\,}A).$

Recall that for a linear relation $T$ from ${\mathfrak{H}}$ to
${\mathfrak{K}}$ the adjoint $T^{*}$ is an automatically closed linear
relation from ${\mathfrak{K}}$ to ${\mathfrak{H}}$ given by

(3.10) $T^{*}=JT^{\perp}=(JT)^{\perp},$

where $J$ stands the flip-flop operator $\\{f,g\\}\mapsto\\{g,-f\\}$; in other
words

(3.11)
$T^{*}=\\{\,\\{h,k\\}\in{\mathfrak{K}}\times{\mathfrak{H}}:\,(g,h)=(f,k)\mbox{
for all }\\{f,g\\}\in T\,\\}.$

With $J$ the definition of the adjoint can be expressed in geometric terms as
follows $T^{*}=(JT)^{\perp}$. In particular, $T^{**}=T^{\perp\perp}$.

Now apply the definition (3.10) to the linear relation $L(A,B)$ in (3.7).
Then, by (3.8), the adjoint of $L(A,B)$ is given by $JL(A,B)^{\perp}=J({\rm
ran\,}c(A,B))^{\perp}$, i.e.

$L(A,B)^{*}=J\,{{\rm ker\,}\,}c(A,B)^{*}=J\,{{\rm ker\,}\,}r(A^{*},B^{*}).$

In other words, the adjoint of $L(A,B)$ is given by

(3.12)
$L(A,B)^{*}=\\{\,\\{k,h\\}\in{\mathfrak{K}}\times{\mathfrak{H}}:\,B^{*}k=A^{*}h\,\\},$

cf. (3.11). At this point, it is useful to introduce the following linear
subspaces ${\mathfrak{D}}(A,B)$ and ${\mathfrak{R}}(A,B)$ of ${\mathfrak{K}}$
and ${\mathfrak{H}}$ by

(3.13) ${\mathfrak{D}}(A,B)=\\{k\in{\mathfrak{K}}:\,B^{*}k\in{\rm
ran\,}A^{*}\\},\quad{\mathfrak{R}}(A,B)=\\{h\in{\mathfrak{H}}:\,A^{*}h\in{\rm
ran\,}B^{*}\\},$

respectively. In other words, ${\mathfrak{D}}(A,B)$ is the pre-image
$(B^{*})^{-1}({\rm ran\,}A^{*})$ and ${\mathfrak{R}}(A,B)$ is the pre-image
$(A^{*})^{-1}({\rm ran\,}B^{*})$. Note that

${\mathfrak{D}}(A,B)={\mathfrak{K}}\,\,\Leftrightarrow\,\,{\rm
ran\,}B^{*}\subset{\rm
ran\,}A^{*},\quad{\mathfrak{R}}(A,B)={\mathfrak{H}}\,\,\Leftrightarrow\,\,{\rm
ran\,}A^{*}\subset{\rm ran\,}B^{*}.$

From the expression (3.12) it is seen that the domain and the range of
$L(A,B)^{*}$ are given by the sets in (3.13):

(3.14) ${\rm dom\,}L(A,B)^{*}={\mathfrak{D}}(A,B),\quad{\rm
ran\,}L(A,B)^{*}={\mathfrak{R}}(A,B).$

Moreover, the kernel and multivalued part of $L(A,B)^{*}$ are given by

(3.15) ${{\rm ker\,}\,}L(A,B)^{*}={{\rm ker\,}\,}B^{*},\quad{\rm
mul\,}L(A,B)^{*}={{\rm ker\,}\,}A^{*},$

respectively. Observe that (3.14) implies that

(3.16) ${\rm mul\,}L(A,B)^{**}={\mathfrak{D}}(A,B)^{\perp},\quad{{\rm
ker\,}\,}L(A,B)^{**}={\mathfrak{R}}(A,B)^{\perp}.$

It is helpful to state some general properties for the operators
$A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, in which case
$c(A,B)\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}}\times{\mathfrak{K}})$. By
means of the Douglas lemma, one sees, for instance, from (3.5) and (3.4), that

(3.17) $\begin{split}{\rm ran\,}(A^{*}A+B^{*}B)^{\frac{1}{2}}&={\rm
ran\,}(c(A,B)^{*}c(A,B))^{\frac{1}{2}}={\rm ran\,}c(A,B)^{*}\\\ &={\rm
ran\,}r(A^{*},B^{*})={\rm ran\,}A^{*}+{\rm ran\,}B^{*},\end{split}$

cf. [4], and from (3.6) that

(3.18) ${\rm ran\,}\begin{pmatrix}AA^{*}&AB^{*}\\\
BA^{*}&BB^{*}\end{pmatrix}^{\frac{1}{2}}={\rm
ran\,}(c(A,B)c(A,B)^{*})^{\frac{1}{2}}={\rm ran\,}c(A,B).$

Moreover, it is clear that ${\rm ran\,}c(A,B)$ and ${\rm ran\,}c(A,B)^{*}$ are
simultaneously closed, and therefore the following spaces

(3.19) ${\rm ran\,}c(A,B),\quad{\rm ran\,}c(A,B)^{*}c(A,B),\quad{\rm
ran\,}c(A,B)^{*},\quad{\rm ran\,}c(A,B)^{*}c(A,B),$

are closed simultaneously; see e.g. [2, Theorem 1.3.5]. The following
characterization of closedness is a direct consequence of these
considerations.

######  Lemma 3.1.

Let the linear relation $L(A,B)$ be given by (3.7). Then the following
statements are equivalent:

  1. (i)

$L(A,B)$ is closed;

  2. (ii)

${\rm ran\,}(A^{*}A+B^{*}B)$ is closed in ${\mathfrak{E}}$;

  3. (iii)

${\rm ran\,}A^{*}+{\rm ran\,}B^{*}$ is closed in ${\mathfrak{E}}$.

###### Proof.

Thanks to the equivalences in (3.19), the assertions in (i), (ii), and (iii)
follow from (3.8), (3.5), and (3.4) together with (3.3), respectively. ∎

The next corollary shows that range space relations admit some specific
properties which are well known in the case of closed operators; cf. the
closed graph theorem. The following result goes back to Foias [4], see also
[23]; the present proof seems to be new.

######  Corollary 3.2.

Let $T$ be a range space relation from ${\mathfrak{H}}$ to ${\mathfrak{K}}$.
Then the following implications hold:

  1. (i)

If $\,{\rm mul\,}T=\\{0\\}$, then ${\rm dom\,}T$ is closed implies that $T$ is
a bounded operator;

  2. (ii)

If $\,{{\rm ker\,}\,}T=\\{0\\}$, then ${\rm ran\,}T$ is closed implies that
$T^{-1}$ is a bounded operator.

###### Proof.

(i) By assumption one can write $T=L(A,B)$ with some operators $A$ and $B$ as
in (3.7). Now the assumption ${\rm mul\,}T=\\{0\\}$ means that

${{\rm ker\,}\,}A\subset{{\rm ker\,}\,}B\quad\mbox{or,
equivalently,}\quad{\rm\overline{ran}\,}B^{*}\subset{\rm\overline{ran}\,}A^{*}.$

If ${\rm dom\,}T={\rm ran\,}A$ is closed, then also ${\rm ran\,}A^{*}$ closed.
Hence, in fact, ${\rm ran\,}B^{*}\subset{\rm ran\,}A^{*}$ and

${\rm ran\,}A^{*}+{\rm ran\,}B^{*}={\rm ran\,}A^{*}\quad\text{is closed}.$

By Lemma 3.1 $L(A,B)$ a closed operator and the statement follows from the
closed graph theorem.

(ii) This is obtained by applying (i) to the inverse $T^{-1}$. ∎

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$,
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ and let the linear relation
$L(A,B)$ be given by (3.7). In general, there will be some redundance in the
representation (3.7); cf. (2.3). In fact, the redundant part is given by the
closed linear subspace

${{\rm ker\,}\,}c(A,B)={{\rm ker\,}\,}A\cap{{\rm ker\,}\,}B.$

Observe that it follows from the identities

$({\rm ran\,}(A^{*}A+B^{*}B))^{\perp}={{\rm ker\,}\,}(A^{*}A+B^{*}B)={{\rm
ker\,}\,}A\cap{{\rm ker\,}\,}B,$

that the space ${\mathfrak{E}}$ has the following orthogonal decomposition:

(3.20) ${\mathfrak{E}}={\rm\overline{ran}\,}(A^{*}A+B^{*}B)\oplus({{\rm
ker\,}\,}A\cap{{\rm ker\,}\,}B).$

Hence, one may reduce the representation in (3.7) by introducing the
restrictions $A_{0}$ and $B_{0}$ of $A$ and $B$ to the closed linear subspace
${\mathfrak{E}}_{0}={\rm\overline{ran}\,}(A^{*}A+B^{*}B)$ of ${\mathfrak{E}}$,
cf. (2.3), so that

(3.21) $A=r(A_{0},O_{{{\rm ker\,}\,}A\,\cap\,{{\rm ker\,}\,}B}),\quad
B=r(B_{0},O_{{{\rm ker\,}\,}A\,\cap\,{{\rm ker\,}\,}B}),$

with respect to the decomposition (3.20). It is clear that ${{\rm
ker\,}\,}A_{0}\cap{{\rm ker\,}\,}B_{0}=\\{0\\}$, and thus

(3.22) $L(A,B)=\\{\,\\{A_{0}f,B_{0}\,f\\}:\,f\in{\mathfrak{E}}_{0}\,\\}$

is a representation in terms of the reduced part $c(A_{0},B_{0})$ of $c(A,B)$.
Recall that reduced representations as in (3.22) are uniquely defined, up to
everywhere defined operators which are bounded and boundedly invertible; cf.
Lemma 2.4. Furthermore, via (3.6), one has from (3.21) that, with respect to
the decomposition (3.20),

(3.23) $A^{*}A+B^{*}B=\begin{pmatrix}(A_{0})^{*}A_{0}+(B_{0})^{*}B_{0}&0\\\
0&0\end{pmatrix}.$

It is clear that
${\rm\overline{ran}\,}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})={\mathfrak{E}}_{0}$;
hence one obtains the following lemma.

######  Lemma 3.3.

Let the linear relation $L(A,B)$ be given by (3.7) and let $A_{0}$ and $B_{0}$
as in (3.21) Then the following statements are equivalent:

  1. (i)

$L(A,B)$ is closed;

  2. (ii)

${\rm ran\,}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})={\mathfrak{E}}_{0}$.

In the rest of this section attention is paid to the case where the linear
relation $L(A,B)$ is closed. In this case the operators $A$ and $B$
representing $L(A,B)$ can be replaced by a normalized pair $A^{*}A+B^{*}B=I$,
in which case certain formulas simplify. In fact, it is instructive to first
consider the case where $A^{*}A+B^{*}B$ is an orthogonal projection.

######  Lemma 3.4.

Assume that there exists an orthogonal projection $Q$ in ${\mathfrak{E}}$ such
that

(3.24) $A^{*}A+B^{*}B=Q,$

in which case the redundant part is given by ${{\rm ker\,}\,}A\cap{{\rm
ker\,}\,}B={{\rm ker\,}\,}Q$. Then the relation $L(A,B)$ is closed and the
orthogonal projection from ${\mathfrak{H}}\times{\mathfrak{K}}$ onto $L(A,B)$
is given by

(3.25) $P_{L(A,B)}=\begin{pmatrix}AA^{*}&AB^{*}\\\
BA^{*}&BB^{*}\end{pmatrix}.$

Moreover, the orthogonal projection from ${\mathfrak{H}}\times{\mathfrak{K}}$
onto $L(A,B)^{*}$ is given by

(3.26) $P_{L(A,B)^{*}}=\begin{pmatrix}I-BB^{*}&BA^{*}\\\
AB^{*}&I-AA^{*}\end{pmatrix}.$

###### Proof.

Since ${\rm ran\,}Q$ is closed, it follows from Lemma 2.8 that $L(A,B)$ is
closed. Let $\Phi=c(A,B)$, so that
$\Phi\in\mathbf{B}({\mathfrak{E}},{\mathfrak{H}}\times{\mathfrak{K}})$. Then
(3.24) means that $\Phi^{*}\Phi=Q$ and ${\rm ran\,}\Phi^{*}={\rm ran\,}Q$.
Observe that

$(\Phi\Phi^{*})(\Phi\Phi^{*})=\Phi(\Phi^{*}\Phi)\Phi^{*}=\Phi
Q\Phi^{*}=\Phi\Phi^{*},$

so that $\Phi\Phi^{*}$ is an orthogonal projection, mapping onto ${\rm
ran\,}\Phi\Phi^{*}={\rm ran\,}\Phi$. Thus the statement follows from (3.6);
cf. [2, Appendix D, p. 703].

Note that the condition (3.24) implies that ${{\rm ker\,}\,}A\cap{{\rm
ker\,}\,}B={{\rm ker\,}\,}\Phi={{\rm ker\,}\,}Q$. Hence it follows that ${\rm
ran\,}r(A^{*},B^{*})$ is dense in ${\mathfrak{E}}_{0}$ and thus ${\rm
ran\,}r(A^{*},B^{*})={\mathfrak{E}}_{0}={\rm ran\,}Q$ by Lemma 3.1. Note that
therefore also ${\rm ran\,}r(B^{*},-A^{*})={\mathfrak{E}}_{0}$, so that the
selfadjoint mapping

$\begin{pmatrix}B\\\ -A\end{pmatrix}\begin{pmatrix}B\\\
-A\end{pmatrix}^{*}=\begin{pmatrix}B\\\ -A\end{pmatrix}(B^{*}\,\,-A^{*})$

takes ${\mathfrak{K}}\times{\mathfrak{H}}$ onto $JL(A,B)$. Moreover, this
mapping is, due to (3.24), idempotent. Thus the orthogonal projection from
${\mathfrak{K}}\times{\mathfrak{H}}$ onto $JL(A,B)$ is given by

$P_{JL(A,B)}=\begin{pmatrix}B\\\ -A\end{pmatrix}\begin{pmatrix}B\\\
-A\end{pmatrix}^{*}=\begin{pmatrix}BB^{*}&-BA^{*}\\\
-AB^{*}&AA^{*}\end{pmatrix},$

which is equivalento to (3.25).

Since $L(A,B)^{*}=(JL(A,B))^{\bot}$, the orthogonal projection onto
$L(A,B)^{*}$ is given by

$P_{L(A,B)^{*}}=I-P_{JL(A,B)}=\begin{pmatrix}I-BB^{*}&BA^{*}\\\
AB^{*}&I-AA^{*}\end{pmatrix},$

which gives (3.26). ∎

Recall that the adjoint relation $L(A,B)^{*}$ has the representation (3.12).
As a closed linear relation from ${\mathfrak{K}}$ to ${\mathfrak{H}}$ it is an
operator range relation. Under the condition (3.24) such a parametrization can
be made explicit via Lemma 3.4.

######  Corollary 3.5.

Let the linear relation $L(A,B)$ be given by (3.7) and assume that the
condition (3.24) holds. Then the adjoint relation $L(A,B)^{*}$ is given by

$L(A,B)^{*}=\bigl{\\{}\\{(I-BB^{*})\varphi+BA^{*}\psi,\,AB^{*}\varphi+(I-AA^{*})\psi\\}:\,\varphi\in{\mathfrak{K}},\,\psi\in{\mathfrak{H}}\bigr{\\}}.$

The pair of bounded linear operators
$A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ in (3.7) is said to be
normalized if

(3.27) $A^{*}A+B^{*}B=I.$

In this case, the relation $L(A,B)$ is closed by Lemma 3.4 and the
representation of $L(A,B)$ in (3.7) is reduced.

######  Lemma 3.6.

Let the linear relation $L(A,B)$ be given by (3.7) and assume that $L(A,B)$ is
closed. Then $L(A,B)$ can be represented by a normalized pair.

###### Proof.

Since $L(A,B)$ is closed, it is known by Lemma 3.1 that ${\rm
ran\,}(A^{*}A+B^{*}B)$ is closed. Thus for the reduced representation in
(3.22) one now has

${\rm ran\,}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})={\mathfrak{E}}_{0}$

by Lemma 3.3. Replacing $A_{0}$ and $B_{0}$ by the equivalent pair

$A_{0}^{\prime}=A_{0}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})^{-{\frac{1}{2}}}\quad\mbox{and}\quad
B_{0}^{\prime}=B_{0}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})^{-{\frac{1}{2}}}$

leads to $L(A,B)=L(A_{0}^{\prime},B_{0}^{\prime})$, where
$A_{0}^{\prime}\in{\mathbf{B}}({\mathfrak{E}}_{0},{\mathfrak{H}})$ and
$B_{0}^{\prime}\in{\mathbf{B}}({\mathfrak{E}}_{0},{\mathfrak{K}})$ form a
normalized pair. ∎

If $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ are normalized as in (3.27),
then the orthogonal operator part of $L(A,B)=BA^{-1}$ can be rewritten in
terms of the Moore-Penrose inverse $A^{(-1)}$ of $A$, which is an operator
from ${\rm ran\,}A\subset{\mathfrak{H}}$ to $({{\rm
ker\,}\,}A)^{\perp}\subset{\mathfrak{E}}$. In fact, it is defined as follows:
for $\psi\in{\rm ran\,}A$, there exists a unique $\varphi\in({\rm
dom\,}A)^{\perp}$ such that $A\varphi=\psi$, and $A^{(-1)}\psi:=\varphi$; cf.
[2]. Note that in the literature one often also extends $A^{(-1)}$ by
$A^{(-1)}\psi=0$ for $\psi\in({\rm ran\,}A)^{\perp}$.

######  Lemma 3.7.

Let the linear relation $L(A,B)$ be given by (3.7) and assume that the
condition (3.27) holds. Then the orthogonal operator part $L(A,B)_{\rm s}$ is
given by

(3.28) $L(A,B)_{\rm s}=\\{\,\\{A\varphi,B\varphi\\}:\,\varphi\in({{\rm
ker\,}\,}A)^{\perp}\\},$

and, consequently,

(3.29) $L(A,B)_{\rm s}=BA^{(-1)}.$

###### Proof.

Consider the representation (3.7). Since
$A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$, the orthogonal
decomposition

${\mathfrak{E}}=({{\rm ker\,}\,}A)^{\perp}\oplus{{\rm ker\,}\,}A$

leads to the alternative representation

(3.30) $L(A,B)=\\{\,\\{A\varphi,B\varphi+B\psi\\}:\,\varphi\in({{\rm
ker\,}\,}A)^{\perp},\,\psi\in{{\rm ker\,}\,}A\,\\},$

cf. [7]. Note that for all $\varphi\in({{\rm ker\,}\,}A)^{\perp}$ and
$\psi\in{{\rm ker\,}\,}A$, one has by (4.10)

$\begin{split}(B\varphi,B\psi)&=(B^{*}B\varphi,\psi)=((I-A^{*}A)\varphi,\psi)\\\
&=(\varphi,\psi)-(A\varphi,A\psi)=0.\end{split}$

Consequently, the range decomposition in the representation (3.30) is
orthogonal. Recall from (3.9) that ${\rm mul\,}L(A,B)=B({{\rm ker\,}\,}A)$.
Thus the orthogonal operator part of $L(A,B)$ has the representation (3.28).
The representation (3.29) follows from the definition of $A^{(-1)}$. ∎

If $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ satisfy (3.24), then the
statement in Corollary 3.5 is a direct consequence of Lemma 3.4. Under the
stronger condition (3.27), a slightly different looking result can be obtained
by invoking the polar decomposition of the operator $A$, i.e.,

$A=V_{A}(A^{*}A)^{1/2},$

where $V_{A}$ is the unique partial isometry from ${\mathfrak{E}}$ to
${\mathfrak{H}}$ with initial space ${\rm\overline{ran}\,}A^{*}$ and final
space ${\rm\overline{ran}\,}A$. Recall that $I-V_{A}(V_{A})^{*}=P_{\,{{\rm
ker\,}\,}A^{*}}$. By means of the polar decomposition of $A$ one sees that the
normalization (3.27) leads to

$\begin{split}I-AA^{*}&=I-V_{A}A^{*}A(V_{A})^{*}\\\
&=I-V_{A}(V_{A})^{*}+V_{A}B^{*}B(V_{A})^{*}=P_{\,{{\rm
ker\,}\,}A^{*}}+V_{A}B^{*}B(V_{A})^{*}.\end{split}$

Moreover, by the well-known commutation relations

$B(I-B^{*}B)^{1/2}=(I-BB^{*})^{1/2}B,\quad(I-B^{*}B)^{1/2}B^{*}=B^{*}(I-BB^{*})^{1/2},$

one obtains from the normalization (3.27) that

$AB^{*}=V_{A}(A^{*}A)^{1/2}B^{*}=V_{A}(I-B^{*}B)^{{1/2}}B^{*}=V_{A}B^{*}(I-BB^{*})^{{1/2}}.$

Therefore the orthogonal projection from ${\mathfrak{H}}\times{\mathfrak{K}}$
onto $L(A,B)^{*}$ in Lemma 3.4 is given by

$\begin{split}P_{L(A,B)^{*}}&=\begin{pmatrix}I-BB^{*}&(I-BB^{*})^{1/2}B(V_{A})^{*}\\\
V_{A}B^{*}(I-BB^{*})^{1/2}&P_{\,{{\rm
ker\,}\,}A^{*}}+V_{A}B^{*}B(V_{A})^{*}\end{pmatrix}\\\
&=\begin{pmatrix}(I-BB^{*})^{1/2}\\\
V_{A}B^{*}\end{pmatrix}\begin{pmatrix}(I-BB^{*})^{1/2}\\\
V_{A}B^{*}\end{pmatrix}^{*}+\begin{pmatrix}0&0\\\ 0&P_{\,{{\rm
ker\,}\,}A^{*}}\end{pmatrix}.\end{split}$

From this, the following result is now clear.

######  Corollary 3.8.

Let the linear relation $L(A,B)$ be given by (3.7) and assume that the
condition (3.27) holds. Then

$L(A,B)^{*}=\\{\,\\{(I-BB^{*})^{1/2}k,V_{A}B^{*}k+P_{\,{{\rm
ker\,}\,}A^{*}}h\\}:\,h\in{\mathfrak{H}},\,k\in{\mathfrak{K}}\,\\},$

where the range decomposition is orthogonal.

Note that the results in Corollary 3.5 and Corollary 3.8 are of practical
importance in the description of boundary value problems (when one speaks of
boundary values in parametrized from); cf. [2].

##  4\. Construction of the closure of operator range relations

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Then the linear relation
$L(A,B)$ from ${\mathfrak{H}}$ to ${\mathfrak{K}}$ in (3.7) is an operator
range relation. By Lemma 2.7 the closure $L(A,B)^{**}$ of $L(A,B)$ is an
operator range relation and it will be shown how the closure $L(A,B)^{**}$ can
be represented in terms of the original operators $A$ and $B$.

First return to the treatment involving the pair of bounded operators $A$ and
$B$. The following construction is inspired by arguments appearing in measure
theory. Due to the obvious inequalities

(4.1) $A^{*}A\leq A^{*}A+B^{*}B,\quad B^{*}B\leq A^{*}A+B^{*}B,$

an application of the Douglas lemma [3] shows that there exists a pair of
contractions $C_{A}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$C_{B}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, such that

(4.2) $A=C_{A}\,(A^{*}A+B^{*}B)^{1/2},\quad B=C_{B}\,(A^{*}A+B^{*}B)^{1/2},$

or, equivalently

(4.3) $A^{*}=(A^{*}A+B^{*}B)^{1/2}(C_{A})^{*},\quad
B^{*}=(A^{*}A+B^{*}B)^{1/2}(C_{B})^{*}.$

The contractions $C_{A}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$C_{B}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ are uniquely determined
by the conditions that $C_{A}$ and $C_{B}$ vanish on $({\rm
ran\,}(A^{*}A+B^{*}B)^{\frac{1}{2}})^{\perp}={{\rm ker\,}\,}A\cap{{\rm
ker\,}\,}B$ or, equivalently,

(4.4) ${\rm
ran\,}(C_{A})^{*}\subset{\rm\overline{ran}\,}(A^{*}A+B^{*}B),\quad{\rm
ran\,}(C_{B})^{*}\subset{\rm\overline{ran}\,}(A^{*}A+B^{*}B),$

and with these conditions one also has as a consequence

${{\rm ker\,}\,}(C_{A})^{*}={{\rm ker\,}\,}A^{*},\quad{{\rm
ker\,}\,}(C_{B})^{*}={{\rm ker\,}\,}B^{*}.$

It follows from (4.2) that $L(A,B)$ can be written as

(4.5)
$L(A,B)=\\{\,\\{C_{A}\,(A^{*}A+B^{*}B)^{1/2}f,C_{B}\,(A^{*}A+B^{*}B)^{1/2}f\\}:\,f\in{\mathfrak{E}}\,\\}.$

This representation will be called the canonical representation of $L(A,B)$.
Of course the pair $C_{A}$ and $C_{B}$ itself induces a linear relation
$L(C_{A},C_{B})$ in ${\mathfrak{H}}\times{\mathfrak{K}}$, in other words,

(4.6) $L(C_{A},C_{B})=\\{\,\\{C_{A}f,C_{B}\,f\\}:\,f\in{\mathfrak{E}}\,\\}.$

Analogous to (3.1) one may now introduce the column operator $c(C_{A},C_{B})$
from ${\mathfrak{E}}$ to ${\mathfrak{H}}\times{\mathfrak{K}}$ by

(4.7) $c(C_{A},C_{B})=\begin{pmatrix}C_{A}\\\
C_{B}\end{pmatrix},\quad\mbox{i.e.,}\quad
c(C_{A},C_{B})\varphi=\begin{pmatrix}C_{A}\varphi\\\
C_{B}\varphi\end{pmatrix},\quad\varphi\in{\mathfrak{E}}.$

Then $c(C_{A},C_{B})$ belongs to
${\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}}\times{\mathfrak{K}})$ and

$L(C_{A},C_{B})={\rm ran\,}c(C_{A},C_{B});$

cf. (3.8). It follows from (4.5) that

(4.8) $L(A,B)={\rm ran\,}c(A,B)\subset{\rm
ran\,}c(C_{A},C_{B})=L(C_{A},C_{B}).$

Note that if $L(A,B)$ is closed, then $L(C_{A},C_{B})=L(A,B)$ by (4.5),
(3.17), and Lemma 3.1.

The next lemma contains a first step in establishing the connection between
the pairs $A,B$ and $C_{A},C_{B}$.

######  Lemma 4.1.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$,
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, and let $C_{A}$, $C_{B}$ be
uniquely defined by (4.2) and (4.4). Then

(4.9) ${{\rm ker\,}\,}C_{A}\cap{{\rm ker\,}\,}C_{B}={{\rm ker\,}\,}A\cap{{\rm
ker\,}\,}B,$

and

(4.10)
$[(C_{A})^{*}C_{A}+(C_{B})^{*}C_{B}]h=\left\\{\begin{array}[]{ll}h,&h\in{\rm\overline{ran}\,}(A^{*}A+B^{*}B),\\\
0,&h\in{{\rm ker\,}\,}A\cap{{\rm ker\,}\,}B.\end{array}\right.$

###### Proof.

It follows from (4.2), (4.3), and (4.4) that

$(A^{*}A+B^{*}B)^{1/2}[(C_{A})^{*}C_{A}+(C_{B})^{*}C_{B}](A^{*}A+B^{*}B)^{1/2}=A^{*}A+B^{*}B.$

Consequently, thanks to (4.4),

$[(C_{A})^{*}C_{A}+(C_{B})^{*}C_{B}](A^{*}A+B^{*}B)^{1/2}=(A^{*}A+B^{*}B)^{1/2},$

which by continuity of $(C_{A})^{*}C_{A}+(C_{B})^{*}C_{B}$ implies that

(4.11) $[(C_{A})^{*}C_{A}+(C_{B})^{*}C_{B}]h=h,\quad
h\in{\rm\overline{ran}\,}(A^{*}A+B^{*}B).$

Hence the first part of (4.10) holds.

Keeping (3.20) in mind, it is clear that in the present context one also has

${\mathfrak{E}}={\rm\overline{ran}\,}((C_{A})^{*}C_{A}+(C_{B})^{*}C_{B})\oplus({{\rm
ker\,}\,}C_{A}\cap{{\rm ker\,}\,}C_{B}).$

Observe that it follows from (4.4) that

${\rm\overline{ran}\,}((C_{A})^{*}C_{A}+(C_{B})^{*}C_{B})\subset{\rm\overline{ran}\,}(A^{*}A+B^{*}B).$

or, equivalently,

${{\rm ker\,}\,}A\cap{{\rm ker\,}\,}B={{\rm
ker\,}\,}(A^{*}A+B^{*}B)\subset{{\rm ker\,}\,}C_{A}\cap{{\rm ker\,}\,}C_{B}.$

Assume that these inclusions are strict. Then there exists a nontrivial
element $h\in{\rm\overline{ran}\,}(A^{*}A+B^{*}B)$ with $h\in{{\rm
ker\,}\,}C_{A}\cap{{\rm ker\,}\,}C_{B}$, which contradicts (4.11). Thus

${\rm\overline{ran}\,}((C_{A})^{*}C_{A}+(C_{B})^{*}C_{B})={\rm\overline{ran}\,}(A^{*}A+B^{*}B),$

and (4.9) follows. Thus also the second part of (4.10) holds. ∎

Note that the pair $C_{A}$ and $C_{B}$ is normalized precisely when it is
reduced or, equivalently, when the pair $A$ and $B$ is reduced.

The precise connection between the pairs $A,B$ and $C_{A},C_{B}$ is now
established in terms of the _polar decomposition_ (cf. [15, Section VI 2.7])
of the column operator $c(A,B)$ in (3.1); cf. (4.3).

######  Lemma 4.2.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$,
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, and let $C_{A}$, $C_{B}$ be
uniquely defined by (4.2) and (4.4). Let $c(A,B)$ and $c(C_{A},C_{B})$ be as
defined in (3.1) and (4.7), respectively. Then the identity

$\begin{pmatrix}A\\\ B\end{pmatrix}=\begin{pmatrix}C_{A}\\\
C_{B}\end{pmatrix}(A^{*}A+B^{*}B)^{1/2},$

is the polar decomposition of the column operator $c(A,B)$, where the column
operator $c(C_{A},C_{B})$ is the unique partial isometry with initial space
${\rm\overline{ran}\,}(A^{*}A+B^{*}B)$ and final space
${\rm\overline{ran}\,}c(A,B)$.

###### Proof.

By Lemma 4.1 $(C_{A})^{*}C_{A}+(C_{B})^{*}C_{B}$ is the orthogonal projection
from ${\mathfrak{E}}$ onto

${\rm
ran\,}((C_{A})^{*}C_{A}+(C_{B})^{*}C_{B})={\rm\overline{ran}\,}(A^{*}A+B^{*}B).$

In particular, ${\rm ran\,}[(C_{A})^{*}C_{A}+(C_{B})^{*}C_{B}]$ is closed and,
equivalently, ${\rm ran\,}c(C_{A},C_{B})=L(C_{A},C_{B})$ is closed; cf. Lemma
3.1. Then the column operator $c(C_{A},C_{B})$ is a partial isometry with
initial space ${\rm\overline{ran}\,}(A^{*}A+B^{*}B)$ and final space ${\rm
ran\,}c(C_{A},C_{B})$ (cf. [2, Appendix D]). Since ${\rm
ran\,}(A^{*}A+B^{*}B)^{1/2}$ is dense in
${\rm\overline{ran}\,}(A^{*}A+B^{*}B)$, also its image under $c(C_{A},C_{B})$
is dense in ${\rm ran\,}c(C_{A},C_{B})$. By construction $c(C_{A},C_{B})$ maps
${\rm ran\,}(A^{*}A+B^{*}B)^{1/2}$ onto ${\rm ran\,}c(A,B)$, cf. (4.5), and
thus the final space of $c(C_{A},C_{B})$ is equal to
${\rm\overline{ran}\,}c(A,B)$. ∎

Recall that the reduction of $L(A,B)$ in (3.21) is with respect to the
orthogonal decomposition (3.20) of the space ${\mathfrak{E}}$:

${\mathfrak{E}}={\mathfrak{E}}_{0}\oplus({{\rm ker\,}\,}A\cap{{\rm
ker\,}\,}B)\quad\mbox{with}\quad{\mathfrak{E}}_{0}={\rm\overline{ran}\,}(A^{*}A+B^{*}B)^{1/2}.$

Thus $A$ and $B$ are row operators:

(4.12) $A=r(A_{0},O_{{{\rm ker\,}\,}A\,\cap\,{{\rm
ker\,}\,}B})\quad\mbox{and}\quad B=r(B_{0},O_{{{\rm ker\,}\,}A\,\cap\,{{\rm
ker\,}\,}B}),$

such that $A_{0}\in\mathbf{B}({\mathfrak{E}}_{0},{\mathfrak{H}})$ and
$B_{0}\in\mathbf{B}({\mathfrak{E}}_{0},{\mathfrak{K}})$. As in (4.2), there
exists a pair of operators
$C_{A_{0}}\in\mathbf{B}({\mathfrak{E}}_{0},{\mathfrak{H}})$ and
$C_{B_{0}}\in\mathbf{B}({\mathfrak{E}}_{0},{\mathfrak{K}})$, such that

(4.13) $A_{0}=C_{A_{0}}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})^{\frac{1}{2}},\quad
B_{0}=C_{B_{0}}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})^{\frac{1}{2}},$

and these operators are unique. It will be shown that, in fact, the operators
$C_{A_{0}}$ and $C_{B_{0}}$ are the reductions of the pair $C_{A}$ and $C_{B}$
in (4.2).

######  Lemma 4.3.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$,
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, let $C_{A}$, $C_{B}$ be
uniquely defined by (4.2) and (4.4), and let $A_{0}$ and $B_{0}$ be the
restrictions of $A$ and $B$ as in (3.21). Then the operators $C_{A_{0}}$ and
$C_{B_{0}}$ in (4.13) are the restrictions of $C_{A}$ and $C_{B}$. In other
words:

$L(C_{A},C_{B})=\\{\,\\{C_{A_{0}}f,C_{B_{0}}\,f\\}:\,f\in{\mathfrak{E}}_{0}\,\\}$

is a reduced representation of $L(C_{A},C_{B})$.

###### Proof.

With the restrictions $A_{0}$ and $B_{0}$, it follows from (3.23) and (4.2)
that

(4.14)
$\begin{split}A&=C_{A}(A^{*}A+B^{*}B)^{\frac{1}{2}}=C_{A}\begin{pmatrix}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})^{\frac{1}{2}}&0\\\
0&0\end{pmatrix},\\\
B&=C_{B}(A^{*}A+B^{*}B)^{\frac{1}{2}}=C_{B}\begin{pmatrix}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})^{\frac{1}{2}}&0\\\
0&0\end{pmatrix}.\end{split}$

It has been shown in (4.9) that ${\mathfrak{E}}_{0}={{\rm ker\,}\,}A\cap{{\rm
ker\,}\,}B={{\rm ker\,}\,}C_{A}\cap C_{B}$. Now let
${{\widetilde{C}}}_{A}\in\mathbf{B}({\mathfrak{E}}_{0},{\mathfrak{H}})$ and
${{\widetilde{C}}}_{B}\in\mathbf{B}({\mathfrak{E}}_{0},{\mathfrak{K}})$ be the
restrictions of $C_{A}$ and $C_{B}$, so that $C_{A}=({{\widetilde{C}}}_{A},0)$
and $C_{B}=({{\widetilde{C}}}_{B},0)$. Then with $A=r(A_{0},0)$ and
$B=r(B_{0},0)$, it follows from (4.14), that

$A_{0}={{\widetilde{C}}}_{A}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})^{\frac{1}{2}},\quad
B_{0}={{\widetilde{C}}}_{B}((A_{0})^{*}A_{0}+(B_{0})^{*}B_{0})^{\frac{1}{2}}.$

A comparison with (4.13) gives that ${{\widetilde{C}}}_{A}=C_{A_{0}}$ and
${{\widetilde{C}}}_{B}=C_{B_{0}}$. ∎

The description of $L(A,B)^{**}$ and its orthogonal operator part follows
easily from Lemma 4.2 and Lemma 3.7. The Moore-Penrose inverse
$(C_{A})^{(-1)}$ of $C_{A}$ is an operator from ${\rm
ran\,}C_{A}\subset{\mathfrak{H}}$ to $({{\rm
ker\,}\,}C_{A})^{\perp}\subset{\mathfrak{E}}$, which is defined as follows:
for $\psi\in{\rm ran\,}C_{A}$, there exists a unique $\varphi\in({\rm
dom\,}C_{A})^{\perp}$ such that $C_{A}\varphi=\psi$, and
$(C_{A})^{(-1)}\psi:=\varphi$; cf. [2]. Note that in the literature one often
also extends $A^{(-1)}$ by $A^{(-1)}\psi=0$ for $\psi\in({\rm
ran\,}A)^{\perp}$.

######  Theorem 4.4.

Let the linear relation $L(A,B)$ be given by (3.7) and let $C_{A}$ and $C_{B}$
be the uniquely defined operators satisfying (4.2) and (4.4). Then the linear
relation $L(C_{A},C_{B})$ is closed and

(4.15)
$L(A,B)^{**}=L(C_{A},C_{B})=\\{\,\\{C_{A}\varphi,C_{B}\varphi\\}:\,\varphi\in{\mathfrak{E}}\\}=C_{B}(C_{A})^{-1}.$

Consequently, the orthogonal operator part of $L(A,B)^{**}$ is given by

(4.16) $(L(A,B)^{**})_{\rm
s}=\\{\,\\{C_{A}\varphi,C_{B}\varphi\\}:\,\varphi\in({{\rm
ker\,}\,}C_{A})^{\perp}\\}=C_{B}(C_{A})^{(-1)}.$

###### Proof.

As shown in Lemma 4.2 one has ${\rm
ran\,}c(C_{A},C_{B})={\rm\overline{ran}\,}c(A,B)$. Hence, it follows from
(4.8), after taking closures, that

$L(A,B)^{**}={\rm\overline{ran}\,}c(A,B)={\rm
ran\,}c(C_{A},C_{B})=L(C_{A},C_{B}),$

and it is clear that $L(C_{A},C_{B})=C_{B}(C_{A})^{-1}$. Thus (4.15) has been
shown.

In order to show (4.16), observe that
$C_{A}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ gives the orthogonal
decomposition

${\mathfrak{E}}=({{\rm ker\,}\,}C_{A})^{\perp}\oplus{{\rm ker\,}\,}C_{A}.$

Hence, (4.15) leads to the alternative representation

(4.17)
$L(C_{A},C_{B})=\\{\,\\{C_{A}\varphi,C_{B}\varphi+C_{B}\psi\\}:\,\varphi\in({{\rm
ker\,}\,}C_{A})^{\perp},\,\psi\in{{\rm ker\,}\,}C_{A}\,\\}.$

It follows from (4.4) that

$({{\rm
ker\,}\,}C_{A})^{\perp}={\rm\overline{ran}\,}(C_{A})^{*}\subset{\rm\overline{ran}\,}(A^{*}A+B^{*}B).$

Therefore, for all $\varphi\in({{\rm ker\,}\,}C_{A})^{\perp}$ and
$\psi\in{{\rm ker\,}\,}C_{A}$ one has by (4.10)

$\begin{split}(C_{B}\varphi,C_{B}\psi)&=((C_{B})^{*}C_{B}\varphi,\psi)=((I-(C_{A})^{*}C_{A})\varphi,\psi)\\\
&=(\varphi,\psi)-(C_{A}\varphi,C_{A}\psi)=0.\end{split}$

Consequently, the range decomposition in the representation (4.17) is
orthogonal. Furthermore, recall that

${\rm mul\,}L(C_{A},C_{B})=C_{B}({{\rm ker\,}\,}C_{A}),$

cf. (3.9), where the right-hand side is closed as the left-hand side is
closed. Thus the orthogonal operator part of $L(C_{A},C_{B})$ has the
representation

$L(C_{A},C_{B})_{\rm
s}=\\{\,\\{C_{A}\varphi,C_{B}\varphi\\}:\,\varphi\in({{\rm
ker\,}\,}C_{A})^{\perp}\\},$

which proves (4.16). ∎

Finally, the representation of the adjoint relation $L(C_{A},C_{B})^{*}$ will
be considered. Recall that the identity (3.12) gives the following
representation:

$L(C_{A},C_{B})^{*}=\\{\,\\{k,h\\}\in{\mathfrak{K}}\times{\mathfrak{H}}:\,(C_{B})^{*}k=(C_{A})^{*}h\,\\}.$

An application of Lemma 3.4 and Lemma 4.1 gives the following result; cf.
Corollary 3.5.

######  Corollary 4.5.

Let the linear relation $L(A,B)$ be given by (3.7) and let $C_{A}$ and $C_{B}$
be the uniquely defined operators satisfying (4.2) and (4.4). Then the adjoint
relation $L(C_{A},C_{B})^{*}$ is given by

$\bigl{\\{}\\{(I-C_{B}(C_{B})^{*})\varphi+C_{B}(C_{A})^{*}\psi,\,C_{A}(C_{B})^{*}\varphi+(I-C_{A}(C_{A})^{*})\psi\\}:\,\varphi\in{\mathfrak{K}},\,\psi\in{\mathfrak{H}}\bigr{\\}}.$

In order to rewrite the result in Corollary 4.5, consider the polar
decomposition of the contraction
$C_{A}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$:

$C_{A}=W_{A}\big{(}(C_{A})^{*}C_{A}\big{)}^{1/2},$

where $W_{A}$ is the unique partial isometry from ${\mathfrak{H}}_{A,B}$ to
${\mathfrak{H}}$ with initial space and final space given by

$({{\rm
ker\,}\,}W_{A})^{\perp}={\rm\overline{ran}\,}\big{(}(C_{A})^{*}C_{A}\big{)}^{1/2}\quad\mbox{and}\quad{\rm
ran\,}W_{A}={\rm\overline{ran}\,}C_{A},$

respectively. Under the assumption that the pair $A$ and $B$ is reduced it
follows that the pair $C_{A}$ and $C_{B}$ is normalized, see Lemma 4.1. Hence
the argument preceding Corollary 3.8 now gives the following result.

######  Corollary 4.6.

Let the linear relation $L(A,B)$ be given by (3.7) and let $C_{A}$ and $C_{B}$
be the uniquely defined operators satisfying (4.2) and (4.4). Assume that the
pair $A$ and $B$ is reduced. Then the adjoint relation $L(C_{A},C_{B})^{*}$ is
given by

$L(C_{A},C_{B})^{*}=\\{\\{\big{(}I-C_{B}(C_{B})^{*}\big{)}^{1/2}h,W_{A}(C_{B})^{*}h+k\\}:\,h\in{\mathfrak{K}},\,k\in{{\rm
ker\,}\,}(C_{A})^{*}\\},$

where the range decomposition is orthogonal.

##  5\. Regular and singular operator range relations

Let ${\mathfrak{E}}$, ${\mathfrak{H}}$, and ${\mathfrak{K}}$ be Hilbert spaces
and consider the pair of bounded linear operators
$A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Let the operator range
relation $L(A,B)$ be given by (3.7). Criteria will be given for the regularity
and singularity of $L(A,B)$ in terms of $A~{}$ and $B$, and in terms of
$C_{A}$ and $C_{B}$ in (4.2).

First the regularity and the singularity of $L(A,B)$ will be expressed in
terms of the pair $A$ and $B$; see [7] for some further equivalent statements
which hold for general linear relations. The regularity of $L(A,B)$ is
described by the following lemma.

######  Lemma 5.1.

Let the linear relation $L(A,B)$ and ${\mathfrak{D}}(A,B)$ be given by (3.7)
and (3.13), respectively. Then the following statements are equivalent:

  1. (i)

$L(A,B)$ is regular;

  2. (ii)

the set ${\mathfrak{D}}(A,B)$ is dense in ${\mathfrak{K}}$.

Moreover, the following statements are equivalent:

  1. (iii)

$L(A,B)$ is a bounded operator;

  2. (iv)

${\mathfrak{D}}(A,B)={\mathfrak{K}}$, i.e., ${\rm ran\,}B^{*}\subset{\rm
ran\,}A^{*}$.

Finally, the following statements are equivalent:

  1. (v)

$L(A,B)\in{\mathbf{B}}({\mathfrak{H}},{\mathfrak{K}})$.

  2. (vi)

${\rm ran\,}B^{*}\subset{\rm ran\,}A^{*}$ and ${\rm ran\,}A={\mathfrak{H}}$.

###### Proof.

(i) $\Leftrightarrow$ (ii) Recall that $L(A,B)$ is regular if and only if
${\rm mul\,}L(A,B)^{**}=\\{0\\}$. It follows from (3.16) that this is
equivalent to ${\mathfrak{D}}(A,B)$ being dense in ${\mathfrak{K}}$; see
(3.14). The other description for ${\mathfrak{D}}(A,B)$ is obtained from
(3.13).

(iii) $\Leftrightarrow$ (iv) The relation $L(A,B)$ is a bounded operator
precisely if there exists $c\geq 0$ such that

(5.1) $\|Bf\|\leq c\|Af\|,\quad f\in{\mathfrak{E}},$

or, equivalently, if $B^{*}B\leq c^{2}A^{*}A$. By the Douglas lemma [3] this
is equivalent to ${\rm ran\,}B^{*}\subset{\rm ran\,}A^{*}$. This last
inclusion is the same as saying ${\rm
dom\,}L(A,B)^{*}={\mathfrak{D}}(A,B)={\mathfrak{K}}$; see (3.13) and (3.14).

(v) $\Leftrightarrow$ (vi) This is now clear. ∎

Likewise, the singularity of $L(A,B)$ can be expressed in various equivalent
useful ways.

######  Lemma 5.2.

Let the linear relation $L(A,B)$ be given by (3.7). Then the following
statements are equivalent:

  1. (i)

$L(A,B)$ is singular;

  2. (ii)

$L(A,B)^{**}={\rm\overline{ran}\,}A\times{\rm\overline{ran}\,}B$;

  3. (iii)

${\mathfrak{D}}(A,B)\subset{{\rm ker\,}\,}B^{*}$;

  4. (iv)

${\mathfrak{R}}(A,B)\subset{{\rm ker\,}\,}A^{*}$;

  5. (v)

${\rm ran\,}A^{*}\cap{\rm ran\,}B^{*}=\\{0\\}$;

  6. (vi)

$L(A,B)^{*}={{\rm ker\,}\,}B^{*}\times{{\rm ker\,}\,}A^{*}$.

###### Proof.

This is a straightforward application of [7, Proposition 2.8], together with
(3.14) and (3.15). It suffices to prove (iii) $\Leftrightarrow$ (v).

(iii) $\Rightarrow$ (v) Let $\ell\in{\rm ran\,}A^{*}\cap{\rm ran\,}B^{*}$.
Then $\ell\in{\rm ran\,}A^{*}$ and $\ell=B^{*}h$ for some
$h\in{\mathfrak{H}}$. Thus $h\in{\mathfrak{D}}(A,B)$ and by assumption one
sees that $h\in{{\rm ker\,}\,}B^{*}$. Thus it follows that $\ell=0$, and hence
(v) holds.

(v) $\Rightarrow$ (iii) This implication is trivial. ∎

Next the regularity and the singularity of $L(A,B)$ will be expressed in terms
of the pair $C_{A}$ and $C_{B}$ in (4.2). Recall from Theorem 4.4 that
$L(A,B)^{**}=L(C_{A},C_{B})$. Thus the following characterization of the
regularity of $L(A,B)$ is clear.

######  Lemma 5.3.

Let the linear relation $L(A,B)$ be given by (3.7) and let $C_{A}$ and $C_{B}$
be the uniquely defined operators satisfying (4.2) and (4.4). Then the
following statements are equivalent:

  1. (i)

$L(A,B)$ is regular;

  2. (ii)

$L(C_{A},C_{B})$ is an operator;

  3. (iii)

${{\rm ker\,}\,}C_{A}\subset{{\rm ker\,}\,}C_{B}$.

Likewise, the singularity of $L(A,B)$ can be expressed as follows.

######  Corollary 5.4.

Let the linear relation $L(A,B)$ be given by (3.7) and let $C_{A}$ and $C_{B}$
be the uniquely defined operators satisfying (4.2) and (4.4). Then the
following statements are equivalent:

  1. (i)

$L(A,B)$ is singular;

  2. (ii)

${{\rm ker\,}\,}C_{A}+{{\rm ker\,}\,}C_{B}={\mathfrak{E}}$.

###### Proof.

Due to $L(A,B)^{**}=L(C_{A},C_{B})$ (see Theorem 4.4), it follows from Lemma
5.2 that $L(A,B)$ is singular if and only if $L(C_{A},C_{B})={\rm
ran\,}C_{A}\times{\rm ran\,}C_{B}$, or, equivalently, precisely when

(5.2) ${\rm ran\,}c(C_{A},C_{B})={\rm ran\,}C_{A}\times{\rm ran\,}C_{B}.$

(i) $\Rightarrow$ (ii) Assume that $L(A,B)$ is singular. Then, thanks to
(5.2), for every $e\in{\mathfrak{E}}$ there exist
$e_{1},e_{2}\in{\mathfrak{E}}$ such that

$c(C_{A},C_{B})e=\begin{pmatrix}h_{1}\\\ h_{2}\end{pmatrix},\quad
c(C_{A},C_{B})e_{1}=\begin{pmatrix}h_{1}\\\ 0\end{pmatrix},\quad
c(C_{A},C_{B})e_{2}=\begin{pmatrix}0\\\ h_{2}\end{pmatrix}.$

Then, clearly, $e_{1}\in{{\rm ker\,}\,}C_{B}$, $e_{2}\in{{\rm ker\,}\,}C_{A}$,
and

$e_{0}=e-e_{1}-e_{2}\in{{\rm ker\,}\,}c(C_{A},C_{B})={{\rm
ker\,}\,}C_{A}\cap{{\rm ker\,}\,}C_{B}.$

This shows that $e\in{{\rm ker\,}\,}C_{A}+{{\rm ker\,}\,}C_{B}$. Therefore,
${\mathfrak{E}}={{\rm ker\,}\,}C_{A}+{{\rm ker\,}\,}C_{B}$.

(ii) $\Rightarrow$ (i) Assume that ${\mathfrak{E}}={{\rm ker\,}\,}C_{A}+{{\rm
ker\,}\,}C_{B}$. Let $a=a_{1}+a_{2}$ and $b=b_{1}+b_{2}$ with
$a_{1},b_{1}\in{{\rm ker\,}\,}C_{B}$ and $a_{2},b_{2}\in{{\rm ker\,}\,}C_{A}$.
Then the equality

$\begin{pmatrix}C_{A}a\\\ C_{B}b\end{pmatrix}=\begin{pmatrix}C_{A}a_{1}\\\
C_{B}b_{2}\end{pmatrix}=\begin{pmatrix}C_{A}(a_{1}+b_{2})\\\
C_{B}(a_{1}+b_{2})\end{pmatrix},$

shows that ${\rm ran\,}C_{A}\times{\rm ran\,}C_{B}\subset{\rm
ran\,}c(C_{A},C_{B})$. Hence, (5.2) is satisfied. ∎

##  6\. Classification of pairs of bounded linear operators

Let ${\mathfrak{E}}$, ${\mathfrak{H}}$, and ${\mathfrak{K}}$ be Hilbert spaces
and let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ be a pair of bounded linear
operators. The characterizations of the operator range relation $L(A,B)$ from
(3.7) will now be augmented by further characterizations in terms of $A$ and
$B$, that are influenced by similar observations in measure theory.

######  Definition 6.1.

Let ${\mathfrak{E}}$, ${\mathfrak{H}}$, and ${\mathfrak{K}}$ be Hilbert spaces
and let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Then the operator $B$ is
said to be dominated by $A$, denoted by $B\prec A$, if there exists some $c>0$
such that

$\|Bf\|\leq c\|Af\|\quad\mbox{for all}\quad f\in{\mathfrak{E}}.$

By the Douglas lemma this definition is equivalent to the factorization $B=CA$
where $C\in{\mathbf{B}}({\mathfrak{H}},{\mathfrak{K}})$; the operator $C$ is
uniquely determined when ${\rm ran\,}C^{*}\subset{\rm\overline{ran}\,}A$, in
which case ${{\rm ker\,}\,}C^{*}={{\rm ker\,}\,}B^{*}$. Note that in the
present context Definition 6.1 agrees with the definition of domination in
[9]. For the following, it is useful to recall from the Douglas lemma that $B$
is dominated by $A$ if and only if ${\rm ran\,}B^{*}\subset{\rm ran\,}A^{*}$,
i.e., ${\mathfrak{D}}(A,B)={\mathfrak{K}}$.

The following simple result is immediate from Lemma 5.1.

######  Lemma 6.2.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ and let the relation
$L(A,B)$ be defined by (3.7). Then the following statements are equivalent:

  1. (i)

$B$ is dominated by $A$;

  2. (ii)

${\mathfrak{D}}(A,B)={\mathfrak{K}}$;

  3. (iii)

$L(A,B)$ is a bounded operator.

The notion of domination in Definition 6.1 is now extended.

######  Definition 6.3.

Let ${\mathfrak{E}}$, ${\mathfrak{H}}$, and ${\mathfrak{K}}$ be Hilbert spaces
and let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Then the operator $B$ is
said to be almost dominated by $A$, if there exists a sequence of bounded
operators $B_{n}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}}_{n})$, where
${\mathfrak{K}}_{n}$ are Hilbert spaces, and a sequence $c_{n}\geq 0$, such
that for all $f\in{\mathfrak{E}}$:

  1. (a)

$\|B_{n}f\|\leq c_{n}\|Af\|$;

  2. (b)

$\|B_{n}f\|\leq\|B_{n+1}f\|$;

  3. (c)

$\|B_{n}f\|\nearrow\|Bf\|$.

It is clear that if $B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ is
dominated by $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$, then $B$ is
automatically almost dominated by $A$ by taking $B_{n}=B$ and $c_{n}=c$.

The analog of Lemma 6.2 for the almost dominated case is contained in the
following theorem.

######  Theorem 6.4.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ and let the relation
$L(A,B)$ be defined by (3.7). Then the following statements are equivalent:

  1. (i)

$B$ is almost dominated by $A$;

  2. (ii)

${\mathfrak{D}}(A,B)$ is dense in ${\mathfrak{K}}$;

  3. (iii)

$L(A,B)$ is regular.

###### Proof.

(i) $\Rightarrow$ (iii) Assume that $B$ is almost dominated by $A$. Then there
exists a sequence of operators
$B_{n}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}}_{n})$ as in Definition
6.3. Note that if $f\in{{\rm ker\,}\,}A$ then $B_{n}f=0$ due to (a), and hence
$\|Bf\|=\sup\|B_{n}f\|=0$ due to (c). One concludes that ${{\rm
ker\,}\,}A\subset{{\rm ker\,}\,}B$, so that $L(A,B)$ is an operator, see
(3.9). Define the sequence of linear relations $T_{n}$ from
${\rm\overline{ran}\,}A$ to ${\mathfrak{K}}_{n}$ by

$T_{n}={\rm clos\,}\\{\,\\{Af,B_{n}f\\}:\,f\in{\mathfrak{E}}\,\\}.$

Due to (a) it follows that each $T_{n}$ is a closed bounded operator from
${\rm\overline{ran}\,}A$ to ${\mathfrak{K}}_{n}$. Furthermore, by (b) one sees
that for $m\leq n$

$\|T_{m}Af\|=\|B_{m}f\|\leq\|B_{n}f\|=\|T_{n}Af\|,\quad f\in{\mathfrak{E}},$

which implies that

(6.1) $\|T_{m}h\|\leq\|T_{n}h\|,\quad h\in{\rm\overline{ran}\,}A.$

Moreover, if $h\in{\rm dom\,}L(A,B)$ so that $h=Af$ for some
$f\in{\mathfrak{E}}$, then it follows from (c) that

(6.2) $\|T_{n}h\|=\|T_{n}Af\|=\|B_{n}f\|\nearrow\|Bf\|=\|T_{A,B}h\|.$

Hence the sequence
$T_{n}\in{\mathbf{B}}({\rm\overline{ran}\,}A,{\mathfrak{K}}_{n})$ satisfies
(6.1) and (6.2). Thus [7, Theorem 8.8] implies that the operator $L(A,B)$ is
closable.

(ii) $\Rightarrow$ (i) Assume that $L(A,B)$ is regular, so that $L(A,B)$ is a
closable operator from ${\mathfrak{H}}$ to ${\mathfrak{K}}$. Then by [7,
Theorem 8.9] there exists a sequence
$T_{n}\in{\mathbf{B}}({\rm\overline{ran}\,}A,{\mathfrak{H}})$ of bounded
operators with the property (6.1), such that

(6.3) $\|T_{n}h\|\nearrow\|T_{A,B}h\|,\quad h\in{\rm dom\,}L(A,B).$

Define the operators
$B_{n}=T_{n}A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$. It will be shown
that the conditions of Definition 6.3 are satisfied with
${\mathfrak{K}}_{n}={\mathfrak{H}}$. First note that

$\|B_{n}f\|\leq\|T_{n}\|\,\|Af\|,\quad f\in{\mathfrak{E}},$

so that (a) is satisfied. Secondly, observe that for all $m\leq n$ it follows
from (6.1) that

$\|B_{m}f\|=\|T_{m}Af\|\leq\|T_{n}Af\|=\|B_{n}f\|,\quad f\in{\mathfrak{E}},$

so that (b) is satisfied. Finally note that (6.3) implies that

$\|B_{n}f\|=\|T_{n}Af\|\nearrow\|L(A,B)Af\|=\|Bf\|,\quad f\in{\mathfrak{E}},$

so that (c) is satisfied. Thus, $B$ is almost dominated by $A$.

(ii) $\Leftrightarrow$ (iii) See Lemma 5.1. ∎

The following definition finds its inspiration in a similar notion which is
current in measure theory.

######  Definition 6.5.

Let ${\mathfrak{E}}$, ${\mathfrak{H}}$, and ${\mathfrak{K}}$ be Hilbert spaces
and let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Then the operator $B$ is
said to be singular with respect to $A$ or, equivalently, the operator $A$ is
said to be singular with respect to $B$, if for every
$D\in{\mathbf{B}}({\mathfrak{E}})$

$D\prec A\quad\text{and}\quad D\prec B\quad\Rightarrow\quad D=0.$

Note that an equivalent statement is that ${\rm ran\,}D^{*}\subset{\rm
ran\,}A^{*}$ and ${\rm ran\,}D^{*}\subset{\rm ran\,}B^{*}$ imply that $D=0$.
It is straightforward to characterize the property ”$B$ is singular with
respect to $A$” in terms of the operators $A$ and $B$.

######  Theorem 6.6.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ and let the relation
$L(A,B)$ be defined by (3.7). Then the following statements are equivalent:

  1. (i)

$B$ is singular with respect to $A$;

  2. (ii)

${\rm ran\,}A^{*}\cap{\rm ran\,}B^{*}=\\{0\\}$;

  3. (iii)

$L(A,B)$ is singular.

###### Proof.

(i) $\Rightarrow$ (ii) Assume that $B$ is singular with respect to $A$. To
prove (ii), suppose that ${\rm ran\,}A^{*}\cap{\rm ran\,}B^{*}\neq\\{0\\}$.
Then there exists a proper orthogonal projection $D$ in ${\mathfrak{E}}$ with
${\rm ran\,}D\subset{\rm ran\,}A^{*}\cap{\rm ran\,}B^{*}$, or

${\rm ran\,}D\subset{\rm ran\,}A^{*}\quad\mbox{and}\qquad{\rm
ran\,}D\subset{\rm ran\,}B^{*}.$

Since $D$ is an orthogonal projection, it is selfadjoint and one concludes
that $D\prec A$ and $D\prec B$. Hence $D=0$. This contradiction implies that
${\rm ran\,}A^{*}\cap{\rm ran\,}B^{*}=\\{0\\}$.

(ii) $\Rightarrow$ (i) Assume that ${\rm ran\,}A^{*}\cap{\rm
ran\,}B^{*}=\\{0\\}$. To prove (i), suppose that
$D\in{\mathbf{B}}({\mathfrak{E}})$ satisfies $D\prec A$ and $D\prec B$ or,
equivalently, ${\rm ran\,}D^{*}\subset{\rm ran\,}A^{*}$ and ${\rm
ran\,}D^{*}\subset{\rm ran\,}B^{*}$. This leads to ${\rm
ran\,}D^{*}\subset{\rm ran\,}A^{*}\cap\,{\rm ran\,}B^{*}$. Hence $D^{*}=0$ and
thus $D=0$. Therefore, $B$ is singular with respect to $A$.

(ii) $\Leftrightarrow$ (iii) See Lemma 5.2. ∎

##  7\. Almost domination and the Radon-Nikodym derivative

Let ${\mathfrak{E}}$, ${\mathfrak{H}}$, and ${\mathfrak{K}}$ be Hilbert spaces
and let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Let $L(A,B)$ be the
corresponding operator range relation defined in (3.7). If $B$ is dominated or
almost dominated by $A$, then there is a factorization of $B$ with respect to
$A$, which gives the notion of the Radon-Nikodym derivative in the abstract
setting of the operator range relation $L(A,B)$.

Observe the following straightforward remarks. Let $L(A,B)$ be an operator
range relation as in (3.7) and recall that $L(A,B)$ is equal to the quotient
$BA^{-1}$. In the case that $L(A,B)$ is an operator, one may write

(7.1) $Bf=L(A,B)Af\quad\mbox{for all}\quad f\in{\mathfrak{E}}.$

If also $L(A,B)^{**}$ is an operator, then it follows from

$L(A,B)=\\{\\{Af,Bf\\}:\,f\in{\mathfrak{E}}\\}\subset L(A,B)^{**},$

that one may write

(7.2) $Bf=L(A,B)^{**}Af\quad\mbox{for all}\quad f\in{\mathfrak{E}}.$

Note that in this identity only the reduced part of the pair $A$ and $B$ plays
a role.

First the case of domination will be considered.

######  Lemma 7.1.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Then the following
statements are equivalent:

  1. (i)

$B$ is dominated by $A$;

  2. (ii)

$B=CA$ holds for some bounded linear operator $C$ from ${\mathfrak{H}}$ to
${\mathfrak{K}}$;

  3. (iii)

$B=CA$ holds for some closed bounded linear operator $C$ from ${\mathfrak{H}}$
to ${\mathfrak{K}}$.

If one of these conditions is satisfied, then $L(A,B)$ is a bounded linear
operator and $B=L(A,B)A$. Moreover, if $B=CA$ holds for some bounded linear
operator $C$ from ${\mathfrak{H}}$ to ${\mathfrak{K}}$, then $L(A,B)\subset
C$; and if $B=CA$ holds for some closed bounded linear operator $C$ from
${\mathfrak{H}}$ to ${\mathfrak{K}}$, then $L(A,B)^{**}\subset C$.

###### Proof.

(i) $\Rightarrow$ (ii) It follows from Lemma 6.2 that $L(A,B)$ is a bounded
linear operator. Hence it follows from (7.1) that (ii) holds.

(ii) $\Rightarrow$ (iii) To see this replace $C$ in $B=CA$ by its closure
$C^{**}$.

(iii $\Rightarrow$ (i) This is clear.

These equivalent statements imply that $L(A,B)$ is a bounded linear operator
and it follows from (3.7) that $B=L(A,B)A$. Therefore, if $B=CA$ holds for
some bounded linear operator $C$ from ${\mathfrak{H}}$ to ${\mathfrak{K}}$,
then $L(A,B)\subset C$ and, if in addition $C$ is closed, it follows that
$L(A,B)^{**}\subset C$. ∎

Notice that the Radon-Nikodym derivative in the following definition satisfies
the minimality property expressed in Lemma 7.1.

######  Definition 7.2.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ and assume that $B$ is
dominated by $A$. Then the Radon-Nikodym derivative $R(A,B)$ of $B$ with
respect to $A$ is the bounded closed operator $L(A,B)^{**}$ from
${\mathfrak{H}}$ to ${\mathfrak{K}}$.

For an illustration of such a Radon-Nikodym derivative, return to the
inequalities $A^{*}A\leq A^{*}A+B^{*}B$ and $B^{*}B\leq A^{*}A+B^{*}B$. These
inequalities imply that there are
$C_{A}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$C_{B}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, such that the
identities in (4.2) hold, and they are unique when (4.4) is assumed. It is
clear from Definition 6.1 that $A$ and $B$ are dominated by the operator
$(A^{*}A+B^{*}B)^{\frac{1}{2}}$, so that each of the following relations from
${\mathfrak{E}}$ to ${\mathfrak{H}}$ and from ${\mathfrak{E}}$ to
${\mathfrak{K}}$, respectively,

$L((A^{*}A+B^{*}B)^{{\frac{1}{2}}},A)\quad\mbox{and}\quad
L((A^{*}A+B^{*}B)^{{\frac{1}{2}}},B)$

is not only regular, but (the graph of) a bounded operator. Recall that
$A_{0}$ and $B_{0}$ are the reduction of the pair $A$ and $B$; see (4.12) and
Lemma 4.3.

######  Lemma 7.3.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, and let $C_{A}$ and $C_{B}$
be the uniquely defined operators satisfying (4.2) and (4.4). Then the Radon-
Nikodym derivatives of $A$ and $B$ with respect to
$(A^{*}A+B^{*}B)^{{\frac{1}{2}}}$ are given by

(7.3) $\begin{array}[]{c}R((A^{*}A+B^{*}B)^{{\frac{1}{2}}},A)=C_{A_{0}},\quad
R((A^{*}A+B^{*}B)^{{\frac{1}{2}}},B)=C_{B_{0}},\end{array}$

###### Proof.

Since the operators $C_{A}$ and $C_{B}$ satisfy the identities (4.2), if
follows from Lemma 7.1 that

$R((A^{*}A+B^{*}B)^{{\frac{1}{2}}},A)^{**}\subset C_{A},\quad
R((A^{*}A+B^{*}B)^{{\frac{1}{2}}},B)^{**}\subset C_{B}.$

Since all these operators are closed and bounded,

${\rm dom\,}R((A^{*}A+B^{*}B)^{{\frac{1}{2}}},A)^{**}={\rm
dom\,}R((A^{*}A+B^{*}B)^{{\frac{1}{2}}},B)^{**}={\rm\overline{ran}\,}(A^{*}A+B^{*}B)$

and the identities (7.3) follow. ∎

Next, the notion of almost domination in the general case will be taken up
again.

######  Theorem 7.4.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Then the following
statements are equivalent:

  1. (i)

$B$ is almost dominated by $A$;

  2. (ii)

$B=CA$ holds for some closable linear operator $C$ from ${\mathfrak{H}}$ to
${\mathfrak{K}}$;

  3. (iii)

$B=CA$ holds for some closed linear operator $C$ from ${\mathfrak{H}}$ to
${\mathfrak{K}}$.

If one of these conditions is satisfied, then $L(A,B)$ is a closable linear
operator such that $B=L(A,B)A$. Moreover, if $B=CA$ holds for some closable
linear operator $C$ from ${\mathfrak{H}}$ to ${\mathfrak{K}}$, then
$L(A,B)\subset C$ and if $B=CA$ holds for some closed linear operator $C$ from
${\mathfrak{H}}$ to ${\mathfrak{K}}$, then $L(A,B)^{**}\subset C$.

###### Proof.

(i) $\Rightarrow$ (ii) Assume that $B$ is almost dominated by $A$. By Theorem
6.4 this implies that $L(A,B)$ is a closable operator. In particular, it
follows from (7.2) that (ii) holds.

(ii) $\Rightarrow$ (iii) As in the previous lemma this is seen by replacing
$C$ in $B=CA$ by its closure.

(iii) $\Rightarrow$ (i) Assume that $B=CA$ with a closed operator $C$. It
follows from $B=CA$ that $A^{*}C^{*}\subset B^{*}$, in other words

$\\{k,k^{\prime}\\}\in C^{*}\quad\Rightarrow\quad\\{k,A^{*}k^{\prime}\\}\in
B^{*}\quad\Rightarrow\quad B^{*}k=A^{*}k^{\prime}.$

Hence ${\rm dom\,}C^{*}\subset{\mathfrak{D}}(A,B)$. Since $C$ is a closed
operator, it follows that ${\rm dom\,}C^{*}$ is dense and thus that
${\mathfrak{D}}(A,B)$ is dense. By Theorem 6.4 this means that $B$ is almost
dominated by $A$.

It remains to prove the last statements. Notice that if $C$ satisfies (ii),
then ${\rm ran\,}A\subset{\rm dom\,}C$ and

$L(A,B)=\\{\\{Ah,Bh\\}:\,h\in{\mathfrak{E}}\\}=\\{\\{Ah,CAh\\}:\,h\in{\mathfrak{E}}\\}\subset
C.$

If, in addition, $C$ is closed, then one sees that $L(A,B)^{**}\subset C$. ∎

Similar to what happens in the dominated case, the Radon-Nikodym derivative in
the following definition satisfies the minimality property expressed in
Theorem 7.4.

######  Definition 7.5.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ and assume that $B$ is
almost dominated by $A$. Then the Radon-Nikodym derivative $R(A,B)$ of $B$
with respect to $A$ is the closed operator $L(A,B)^{**}$ from ${\mathfrak{H}}$
to ${\mathfrak{K}}$.

######  Corollary 7.6.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ and assume that $B$ is
almost dominated by $A$. Then the Radon-Nikodym derivative $R(A,B)$ of $B$
with respect to $A$ is bounded if and only if $B$ is dominated by $A$.

Now recall that $L(A,B)^{**}=L(C_{A},C_{B})$ and, moreover, that
$L(C_{A},C_{B})$ is an operator precisely when ${{\rm
ker\,}\,}C_{A}\subset{{\rm ker\,}\,}C_{B}$. The Radon-Nikodym derivative
$R(A,B)$ can be expressed in terms of the Radon-Nikodym derivatives in Lemma
7.3.

######  Theorem 7.7.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ and assume that $B$ is
dominated or almost dominated by $A$. Moreover, let $C_{A}$ and $C_{B}$ be as
defined in (4.2) and (4.4) and let $C_{A_{0}}$ and $C_{B_{0}}$ be the Radon-
Nikodym derivatives in (7.3). Then the Radon-Nikodym derivative $R(A,B)$ of
$B$ with respect to $A$ is given by the quotient

(7.4) $R(A,B)=C_{B_{0}}(C_{A_{0}})^{-1},$

where ${{\rm ker\,}\,}C_{A_{0}}\subset{{\rm ker\,}\,}C_{B_{0}}$.

###### Proof.

The Radon-Nikodym derivative $R(A,B)$ is given by the closed operator
$L(C_{A},C_{B})$. Now consider the reduction $A_{0}$ and $B_{0}$ of the pair
$A$ and $B$. By Lemma 4.3 one has

$R(A,B)=(L(A,B)^{**}=L(C_{A},C_{B})=L(C_{A_{0}},C_{B_{0}}),$

with ${{\rm ker\,}\,}C_{A_{0}}\subset{{\rm ker\,}\,}C_{B_{0}}$. The identity
(7.4) by rewriting the above result as a quotient of the operators $C_{A_{0}}$
and $C_{B_{0}}$. ∎

It will be helpful to compare the results in Lemma 7.3 and Theorem 7.7,
together with Theorem 4.4, with the following construction known from measure
theory. Let $(\mu,\nu)$ be a pair of finite positive measures. Then $\mu$ and
$\nu$ are absolutely continuous with respect to the sum measure
$\rho=\mu+\nu$: $\mu\ll\rho$, $\nu\ll\rho$. This gives rise to the existence
of the corresponding Radon-Nikodym derivatives $f=\frac{d\mu}{d\rho}$ and
$g=\frac{d\nu}{d\rho}$ and in this case

(7.5) $f+g=1\quad\rho\text{-a.e.}$

If, in addition, $\nu\ll\mu$ with the Radon-Nikodym derivative
$h=\frac{d\nu}{d\mu}$, then $\nu\ll\mu\ll\rho$ implies that

$g=\frac{d\nu}{d\rho}=\frac{d\nu}{d\mu}\,\frac{d\mu}{d\rho}=hf\quad\rho\text{-a.e.}$

Since $\rho=\mu+\nu$ and $\nu\ll\mu$, one has also $\rho\ll\mu$. Consequently,
one has $f>0$ $\rho$-a.e. ($\Leftrightarrow$ $\mu$-a.e.) and thus, in fact,
the Radon-Nikodym derivative $h$ is given by

(7.6) $\frac{d\nu}{d\mu}=\frac{g}{f}\quad\mu\text{-a.e.}$

######  Remark 7.8.

The concept of Radon-Nikodym derivative given in Definitions 7.2 and 7.5 is
applicable and uniquely determined for general operator range relations, which
are regular (i.e. closable operators). Indeed, by Theorem 2.8 the operator $T$
has the representation $T=L(A,B)$ and if $T={\rm ran\,}Z$ for some other
operator
$Z\in{\mathbf{B}}({\mathfrak{Y}},{\mathfrak{H}}\times{\mathfrak{K}})$, then
Lemma 2.4 shows that there exists a bounded and boundedly invertible operator
$W\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{Y}})$ such that $c(A,B)=ZW$. Then

$L(A,B)={\rm ran\,}c(A,B)={\rm ran\,}ZW={\rm ran\,}Z$

and taking closures leads to ${\rm\overline{ran}\,}Z=R(A,B)$.

##  8\. Lebesgue type decompositions for pairs of bounded linear operators

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ be bounded linear operators.
In this section it will be shown that there exist Lebesgue type decompositions
$B=B_{1}+B_{2}$ such that $B_{1}$ is almost dominated by $A$ and $B_{2}$ is
singular with respect to $A$. The main idea is to go back to the corresponding
operator range relation $L(A,B)$ and to use the Lebesgue type decompositions
of $L(A,B)$; cf. [7].

######  Definition 8.1.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ be bounded linear operators
and let $P$ be the orthogonal projection onto ${\mathfrak{D}}(A,B)^{\perp}$.
The regular part $B_{\rm{reg}}$ and the singular part $B_{\rm{sing}}$ are
defined by

(8.1) $B_{\rm{reg}}=(I-P)B,\quad B_{\rm{sing}}=PB.$

The corresponding decomposition

(8.2) $B=B_{\rm{reg}}+B_{\rm{sing}}.$

is called the Lebesgue decomposition of $B$ with respect to $A$.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ be as in Definition 8.1. Let
$L(A,B)$ from ${\mathfrak{H}}$ to ${\mathfrak{K}}$ be defined as in (3.7) and
recall that

${\rm dom\,}L(A,B)^{*}={\mathfrak{D}}(A,B)\quad\mbox{and}\quad{\rm
mul\,}L(A,B)^{**}={\mathfrak{D}}(A,B)^{\perp};$

cf. (3.16). Then the relation $L(A,B)$ has the Lebesgue decomposition

(8.3) $L(A,B)=L(A,B)_{\rm{{reg}}}+L(A,B)_{\rm{{sing}}},$

where the regular and singular components are given by

(8.4) $L(A,B)_{\rm{{reg}}}=(I-P)L(A,B),\quad L(A,B)_{\rm{{sing}}}=PL(A,B);$

here $P$ stands for the orthogonal projection from ${\mathfrak{K}}$ onto ${\rm
mul\,}L(A,B)^{**}={\mathfrak{D}}(A,B)^{\perp}$; cf. [7]. Via the decomposition
(8.3) one may now obtain the Lebesgue decomposition of $B$ with respect to $A$
as in Definition 8.1.

######  Theorem 8.2 (Lebesgue decomposition).

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ be bounded linear operators.
Then $B_{\rm{reg}}$ is almost dominated by $A$, $B_{\rm{sing}}$ is singular
with respect to $A$, and $B$ has the Lebesgue decomposition (8.2) with respect
to $A$. The regular part $B_{\rm reg}$ can be written as

(8.5) $B_{\rm reg}=R(A,B_{\rm reg})A,$

where $R(A,B_{\rm reg})$ is the Radon-Nikodym derivative of $B_{\rm{reg}}$
with respect to $A$:

(8.6) $R(A,B_{\rm reg})=L(A,B_{\rm reg})^{**}.$

In fact, if $L(A,B)$ is given by (8.3) and (8.4), then

(8.7) $L(A,B)_{\rm{reg}}=L(A,B_{\rm reg}),\quad L(A,B)_{\rm{sing}}=L(A,B_{\rm
sing}).$

###### Proof.

The decomposition (8.2) follows from (8.1). It follows from (8.4) that
$L(A,B)_{\rm{{reg}}}$ and $L(A,B)_{\rm{{sing}}}$ have the representations

$\begin{split}&L(A,B)_{\rm{reg}}=\\{\,\\{Af,(I-P)Bf\\}:\,f\in{\mathfrak{E}}\,\\}=\\{\,\\{Af,B_{\rm{reg}}f\\}:\,f\in{\mathfrak{E}}\,\\},\\\
&L(A,B)_{\rm{sing}}=\\{\,\\{Af,PBf\\}:\,f\in{\mathfrak{E}}\,\\}=\\{\,\\{Af,B_{\rm{sing}}f\\}:\,f\in{\mathfrak{E}}\,\\},\end{split}$

which give (8.7). Since the relation $L(A,B)_{\rm{reg}}$ is regular, it
follows from Theorem 6.4 that $B_{\rm{reg}}$ is almost dominated by $A$.
Likewise, since the relation $L(A,B)_{\rm{sing}}$ is singular, it follows from
Theorem 6.6 that $B_{\rm{sing}}$ is singular with respect to $A$.

The statements about the Radon-Nikodym derivative in (8.5) and (8.6) follow
from Theorem 7.4. ∎

The Lebesgue decomposition in (8.2) is an example of a so-called Lebesgue type
decomposition.

######  Definition 8.3.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ be bounded operators. The
operator $B$ is said to have a Lebesgue type decomposition with respect to
$A$, if $B=B_{1}+B_{2}$ where
$B_{1},B_{2}\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ have the
properties:

  1. (a)

${\rm ran\,}B_{1}\perp{\rm ran\,}B_{2}$;

  2. (b)

$B_{1}$ is almost dominated by $A$;

  3. (c)

$B_{2}$ is singular with respect to $A$.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ as in Definition 8.3. Let
the linear relation $L(A,B)$ from ${\mathfrak{H}}$ to ${\mathfrak{K}}$ be
defined by (3.7). According to [7] the Lebesgue type decompositions of
$L(A,B)$ are in one-to-one correspondence with the closed linear subspaces
${\mathfrak{L}}\subset{\mathfrak{K}}$ such that

(8.8) ${\mathfrak{L}}\subset{\rm\overline{dom}\,}L(A,B)^{*}\setminus{\rm
dom\,}L(A,B)^{*},$

which satisfy the condition

(8.9) ${\rm
clos\,}({\mathfrak{L}}^{\perp}\cap{\mathfrak{D}}(A,B))={\mathfrak{L}}^{\perp}\cap{\rm
clos\,}{\mathfrak{D}}(A,B).$

Define the closed linear subspace ${\mathfrak{M}}$ by

(8.10) ${\mathfrak{M}}={\mathfrak{D}}(A,B)^{\perp}\oplus{\mathfrak{L}},$

and let $P_{\mathfrak{M}}$ be the orthogonal projection from ${\mathfrak{K}}$
onto ${\mathfrak{M}}$. Then the corresponding Lebesgue type decomposition of
$L(A,B)$ is given by

(8.11) $L(A,B)=L(A,B)_{1}+L(A,B)_{2},$

where the regular and singular components are given by

(8.12) $L(A,B)_{1}=(I-P_{\mathfrak{M}})L(A,B),\quad
L(A,B)_{2}=P_{\mathfrak{M}}L(A,B),$

respectively. Via the decomposition (8.11) one may now obtain the Lebesgue
type decompositions of $B$ with respect to $A$ as in Definition 8.3.

######  Theorem 8.4.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$ be bounded linear operators.
Then the Lebesgue type decompositions of $B$ with respect to $A$ are in one-
to-one correspondence with the closed linear subspaces
${\mathfrak{L}}\subset{\mathfrak{K}}$ in (8.8) which satisfy the condition
(8.9). In particular, the corresponding Lebesgue type decomposition is given
by

$B=B_{1}+B_{2},\quad B_{1}=(I-P_{\mathfrak{M}})B,\quad
B_{2}=P_{\mathfrak{M}}B,$

where ${\mathfrak{M}}$ is as in (8.10) and $P_{\mathfrak{M}}$ is the
orthogonal projection from ${\mathfrak{K}}$ onto ${\mathfrak{M}}$, while
$B_{1}$ is almost dominated by $A$ and $B_{2}$ is singular with respect to
$A$. The regular part $B_{1}$ can be written as

$B_{1}=R(A,B_{1})A,$

where $R(A,B_{1})$ is the Radon-Nikodym derivative of $B_{1}$ with respect to
$A$:

$R(A,B_{1})=L(A,B_{1})^{**}.$

In fact, $L(A,B)$ is given by (8.11) and (8.12) precisely, when

$L(A,B)_{\rm{1}}=L(A,B_{1}),\quad L(A,B)_{2}=L(A,B_{2}).$

###### Proof.

Consider the linear relation $L(A,B)$ from ${\mathfrak{H}}$ to
${\mathfrak{K}}$ defined by (3.7).

First it is shown that every Lebesgue type decomposition of the linear
relation $L(A,B)$ (in the sense of [7]) generates a Lebesgue type
decomposition of $B$ with respect to $A$ as in Definition 8.3. To see this let
${\mathfrak{L}}\subset{\rm\overline{dom}\,}L(A,B)^{*}\setminus{\rm
dom\,}L(A,B)$ be a linear subspace which satisfies (8.9) and let
${\mathfrak{M}}$ be as defined in (8.10) with the corresponding orthogonal
projection $P_{\mathfrak{M}}$ onto ${\mathfrak{M}}$. According to [7, Theorem
5.4] the formula

$L(A,B)=(I-P_{\mathfrak{M}})L(A,B)+P_{\mathfrak{M}}L(A,B)$

determines a Lebesgue type decomposition of $L(A,B)$, where
$(I-P_{\mathfrak{M}})L(A,B)$ is the regular part and $P_{\mathfrak{M}}L(A,B)$
is the singular part generated uniquely by the subspace ${\mathfrak{L}}$. From
the representation of the regular part

$(I-P_{\mathfrak{M}})L(A,B)=\\{\,\\{Af,(I-P_{\mathfrak{M}})Bf\\}:\,f\in{\mathfrak{E}}\,\\}$

and Theorem 6.4 it follows that $B_{1}=(I-P_{\mathfrak{M}})B$ is almost
dominated by $A$. Likewise from the representation of the singular part

$P_{\mathfrak{M}}L(A,B)=\\{\,\\{Af,P_{\mathfrak{M}}Bf\\}:\,f\in{\mathfrak{E}}\,\\}$

and Theorem 6.6 it follows that $B_{2}=P_{\mathfrak{M}}B$ is singular with
respect to $A$. Hence, the identity $B=B_{1}+B_{2}$ is a Lebesgue type
decomposition of $B$ with respect to $A$ in the sense of Definition 8.3.

Conversely, assume that $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$,
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$, and that $B$ has a Lebesgue
type decomposition as in Definition 8.3. Then it is clear that the
corresponding relations satisfy

(8.13) $L(A,B)=L(A,B_{1})+L(A,B_{2}),$

where, due to Theorem 6.4 and Theorem 6.6, the relation $L(A,B_{1})$ is
regular and the relation $L(A,B_{2})$ is singular. Hence (8.13) is a Lebesgue
type decomposition for $L(A,B)$. Again by [7, Theorem 5.4] there exists a
linear subspace ${\mathfrak{L}}$, such that (8.8) and (8.9) are satisfied, and
a subspace ${\mathfrak{M}}$ given by (8.10) such that

(8.14)
$\begin{split}L(A,B_{1})&=(I-P_{{\mathfrak{M}}})L(A,B)=L(A,(I-P_{{\mathfrak{M}}})B),\\\
L(A,B_{2})&=P_{{\mathfrak{M}}}L(A,B)=L(A,P_{{\mathfrak{M}}}B).\end{split}$

Thanks to the first identities in (8.14), for every $h\in{\mathfrak{E}}$ there
exists $f\in{\mathfrak{E}}$, such that

$Ah=Af,\quad(I-P_{{\mathfrak{M}}})Bh=B_{1}f.$

Thus $f=h+\varphi$ for some $\varphi\in{{\rm ker\,}\,}A$. Since $L(A,B_{1})$
is regular, it is an operator and hence ${\rm mul\,}L(A,B_{1})=B_{1}({{\rm
ker\,}\,}A)=0$; cf. (3.9). This shows that $B_{1}\varphi=0$, and thus
$(I-P_{{\mathfrak{M}}})Bh=B_{1}h$. Therefore, $(I-P_{{\mathfrak{M}}})B=B_{1}$
and, consequently, $P_{{\mathfrak{M}}}B=B_{2}$. This proves the one-to-one
correspondence between the Lebesgue type decompositions of $B=B_{1}+B_{2}$ in
Definition (8.3) and of $L(A,B)$ in (8.11) and (8.12). The one-to-one
correspondence between the closed subspaces ${\mathfrak{L}}$ satisfying the
conditions (8.8) and (8.9) is obtained from [7, Theorem 5.4].

The statement about the Radon-Nikodym derivative of $B_{1}$ with respect to
$A$ follows from Theorem 6.4. ∎

For any Lebesgue type decomposition of $B$ with respect to $A$, the part
$B_{1}$, which is almost dominated by $A$, is, in fact, dominated by the
regular part $B_{\rm reg}$.

######  Corollary 8.5.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Let $B=B_{1}+B_{2}$ be a
Lebesgue type decomposition of $B$, then

$\|B_{1}h\|\leq\|B_{\rm reg}h\|,\quad h\in{\mathfrak{E}}.$

###### Proof.

Let $B=B_{1}+B_{2}$ be a Lebesgue type decomposition of $B$ with respect to
$A$. Then as in the proof of Theorem 8.4 one finds that

$B_{1}=(I-P_{{\mathfrak{M}}})B=(I-P_{\mathfrak{M}})(I-P)B=(I-P_{\mathfrak{M}})B_{\rm
reg},$

where it was used that ${\rm ran\,}P\subset{\rm ran\,}P_{{\mathfrak{M}}}$; cf.
(8.10). ∎

The uniqueness of Lebesgue type decompositions of $B$ with respect to $A$ in
Definition 8.3 can be characterized as follows.

######  Corollary 8.6.

Let $A\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{H}})$ and
$B\in{\mathbf{B}}({\mathfrak{E}},{\mathfrak{K}})$. Then the following
statements are equivalent:

  1. (i)

$B$ admits a unique Lebesgue type decomposition with respect to $A$;

  2. (ii)

$L(A,B)$ admits a unique Lebesgue type decomposition;

  3. (iii)

${\mathfrak{D}}(A,B)$ is closed;

  4. (iv)

$B_{\rm reg}$ is dominated by $A$;

  5. (v)

the Radon-Nikodym derivative $R(A,B_{\rm reg})$ is a bounded operator.

In this case, all Lebesgue type decompositions of $B$ with respect to $A$
coincide with the Lebesgue decomposition of $B=B_{\rm reg}+B_{\rm sing}$ in
Theorem 8.2.

###### Proof.

(i) $\Leftrightarrow$ (ii) The Lebesgue type decompositions of $B=B_{1}+B_{2}$
with respect to $A$ correspond to the Lebesgue type decompositions of
$L(A,B)=T_{1}+T_{2}$ via Theorem 8.4. Hence $B$ has a unique Lebesgue type
decomposition with respect to $A$ if and only $L(A,B)$ has a unique Lebesgue
type decomposition.

(ii) $\Leftrightarrow$ (iii) By [7, Theorem 6.1] $L(A,B)$ has a unique
Lebesgue type decomposition if and only if ${\rm dom\,}L(A,B)^{*}$ is closed,
i.e., ${\mathfrak{D}}(A,B)$ is closed; cf. (3.14).

(ii) $\Leftrightarrow$ (iv) Again by [7, Theorem 6.1] $L(A,B)$ has a unique
Lebesgue type decomposition if and only if $L(A,B)_{\rm{reg}}=L(A,B_{\rm
reg})$ is bounded; cf. Theorem 8.2. Now $L(A,B_{\rm reg})$ is bounded if and
only if $B_{\rm reg}$ is dominated by $A$; cf Lemma 6.2.

(iv) $\Leftrightarrow$ (v) This follows from Corollary 7.6.

The last statement is clear from (8.8), since if ${\mathfrak{D}}(A,B)={\rm
dom\,}L(A,B)^{*}$ is closed, then ${\mathfrak{L}}=\\{0\\}$ and
${\mathfrak{M}}$ in (8.10) coincides with ${\mathfrak{D}}(A,B)^{\perp}$. ∎

## References

  * [1] T. Ando, “Lebesgue-type decomposition of positive operators”, Acta Sci. Math. (Szeged), 38 (1976), 253–260.
  * [2] J. Behrndt, S. Hassi, and H.S.V. de Snoo, Boundary value problems, Weyl functions, and differential operators, Monographs in Mathematics, Vol. 108, Birkhäuser, 2020.
  * [3] R.G. Douglas, “On majorization, factorization and range inclusion of operators in Hilbert space”, Proc. Amer. Math. Soc., 17 (1966), 413–416.
  * [4] P.A. Fillmore and J.P. Williams, “On operator ranges”, Adv. Math., 7 (1971), 254–281.
  * [5] S. Hassi, J.-Ph. Labrousse, and H.S.V. de Snoo, ”Operational calculus for rows, columns, and blocks of linear relations”, Advances in Operator Theory, 5 (2020), 1297–1313.
  * [6] S. Hassi, Z. Sebestyén, and H.S.V. de Snoo, “Lebesgue type decompositions for nonnegative forms”, J. Functional Analysis, 257 (2009), 3858–3894.
  * [7] S. Hassi, Z. Sebestyén, and H.S.V. de Snoo, “Lebesgue type decompositions for linear relations and Ando’s uniqueness criterion”, Acta Sci. Math. (Szeged), 84 (2018), 465–507.
  * [8] S. Hassi, Z. Sebestyén, H.S.V. de Snoo, and F.H. Szafraniec, “A canonical decomposition for linear operators and linear relations”, Acta Math. Hungarica, 115 (2007), 281–307.
  * [9] S. Hassi and H.S.V. de Snoo, “Factorization, majorization, and domination for linear relations”, Annales Univ. Sci. Budapest, 58 (2015), 53–70.
  * [10] S. Hassi, H.S.V. de Snoo, and F.H. Szafraniec, “Componentwise and canonical decompositions of linear relations”, Dissertationes Mathematicae, 465, 2009 (59 pages).
  * [11] S. Izumino, “Decomposition of quotients of bounded operators with respect to closability and Lebesgue-type decomposition of positive operators”, Hokkaido Math. J., 18 (1989), 199–209.
  * [12] S. Izumino, “Quotients of bounded operators”, Proc. Amer. Math. Soc., 106 (1989), 427–435.
  * [13] S. Izumino, “Quotients of bounded operators and their weak adjoints”, J. Operator Theory, 29 (1993), 83–96.
  * [14] S. Izumino and G. Hirasawa, “Positive symmetric quotients and their selfadjoint extensions”, Proc. Amer. Math. Soc., 129 (2001), 2987–2995.
  * [15] T. Kato, Perturbation theory for linear operators, Springer-Verlag, Berlin, 1980.
  * [16] W.E. Kaufman, “Representing a closed operator as a quotient of continuous operators”, Proc. Amer. Math. Soc., 72 (1978), 531–534.
  * [17] W.E. Kaufman, “Semiclosed operators in Hilbert spaces”, Proc. Amer. Math. Soc., 76 (1979), 67–73.
  * [18] W.E. Kaufman, “Closed operators and pure contractions in Hilbert space”, Proc. Amer. Math. Soc., 87 (1983), 83–87.
  * [19] W.E. Kaufman, “A stronger metric for closed operators in Hilbert space”, Proc. Amer. Math. Soc., 90 (1984), 83–87.
  * [20] J.J. Koliha, “On Kaufman’s theorem”, J. Math. Anal. Appl., 411 (2014), 688–692.
  * [21] H. Kosaki, ”Linear Radon-Nikodym theorems for states on a von Neumann algebra”, Pub!. RIMS Kyoto Univ., 18 (1982). 379–386.
  * [22] J.-Ph. Labrousse, ”Les opérateurs quasi Fredholm: une généralisation des opérateurs semi Fredholm”, Rend. Circ. Mat. Palermo (2), 29 (1980), 161–258.
  * [23] J.Ph. Labrousse, A. Sandovici, H.S.V. de Snoo, and H. Winkler, ”The Kato decomposition of quasi-Fredholm relations”, Operators Matrices, 4 (2010), 1–51.
  * [24] J.S. Mac Nerney, “Investigations concerning positive definite continued fractions, Duke Math. J., 26 (1959), 663–678.
  * [25] J.S. Mac Nerney, “Continuous imbeddings in Hilbert spaces”, Rend. Circ. Mat. Palermo, (2) 19 (1970), 109–112.
  * [26] M. Möller and F.H. Szafraniec, ”Adjoints and formal adjoints of matrices of unbounded operators”, Proc. Amer. Math. Soc., 136 (2008) 2165–2176.
  * [27] Z. Sebestyén, Zs. Tarcsay, and T. Titkos, “Lebesgue decomposition theorems”, Acta Sci. Math. (Szeged), 79(1-2) (2013), 219–233.
  * [28] Z. Sebestyén, T. Titkos, ”A Radon-Nikodym type theorem for forms”, Positivity 17 (2013) 863–873.
  * [29] B. Simon, “A canonical decomposition for quadratic forms with applications to monotone convergence theorems”, J. Functional Analysis, 28 (1978), 377–385.
  * [30] Zs. Tarcsay, ”Radon-Nikodym theorems for nonnegative forms, measures and representable functionals”, Complex Analysis and Operator Theory 10 (2016) 479–494.