Working Papers of the University of Vaasa,
Department of Mathematics and Statistics
1
On Kre˘ın’s extension theory of
nonnegative operators
Seppo Hassi, Mark Malamud,
and Henk de Snoo
Preprint, August 2002
University of Vaasa
Department of Mathematics and Statistics
P.O. Box 700, FIN-65101 Vaasa, Finland
Preprints are available at: http://www.uwasa.fi/julkaisu/sis.html
ON KREI˘N’S EXTENSION THEORY OF NONNEGATIVE OPERATORS
SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
Abstract. The famous M.G. Kre˘ın’s extension theory of nonnegative operators is being
presented in elementary terms, building on the completion of nonnegative operator blocks.
The present treatment includes the refinements to the theory due to T. Ando and K. Nishio.
Moreover, the extension theory not only includes the case of operator extensions but also
the case of relation (multivalued-operator) extensions.
1. Introduction
Let S be a densely defined nonnegative operator in a Hilbert space H. A description of
all nonnegative selfadjoint extensions of S is due to M.G. Kre˘ın, see [26]. In particular, it
was shown that there are two extremal extensions of S, the Friedrichs (hard) extension and
the Kre˘ın-von Neumann (soft) extension. Further results were obtained by T. Ando and
K. Nishio [7], who considered the case that S need not be densely defined.
In order to obtain such results several approaches have been used by various authors.
Kre˘ın [26] employed the Cayley transform to reduce the extension of nonnegative operators
to the extension of symmetric contractive operators. He also used semibounded sesquilinear
forms, cf. [23]; a technique which can be extended to the case of linear relations along the
lines of [15]. Extensions of symmetric contractive operators has been the subject of many
papers, see [9], [13], and the references in those papers. An approach in terms of “boundary
conditions” to the extensions of a positive operator S was proposed by M.I. Vishik [35] and
M.S. Birman [10]. (See also the later exposition of this theory based on the investigation
of quadratic forms in [2].) This approach was subsequently formalized in the concept of
“boundary triplet” and was further developed by many authors (see for instance [17] and
the references therein).
In this paper the main results of Kre˘ın’s work and later refinements on extensions of
nonnegative operators are presented in an elementary manner starting from the solution to
a simple completion problem. The systematic use of block representations of the extreme
contractive extensions of a not everywhere defined symmetric contraction makes it possible
to simplify and unify the proofs and to make the exposition quite brief.
The contents of this paper are as follows. In Section 2 certain incomplete block operators
are completed to nonnegative operators. There is a close connection to shorted operators,
which were introduced originally by M.G. Kre˘ın [26]. Later shorted operators, or Schur
complements were used in system theory, see for instance [3], [4], [5], [33]. The results
of Section 2 are translated in Section 3 to the case of selfadjoint contractive extensions of
symmetric contractions. The nonemptyness of the set of such extensions follows from the
1991 Mathematics Subject Classification. Primary 47A57, 47B25; Secondary 47A55, 47B65.
Key words and phrases. Completion, shorted operator, generalized Schur complement, selfadjoint con-
tractive extension, nonnegative selfadjoint extension, Friedrichs and Kre˘ın-von Neumann extension.
The first and the second author were supported by the Academy of Finland (projects 40362, 49349, 79774).
1
2 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
existence of solutions to associated completion problems. All the main results are then
presented with quite elementary arguments, for instance, Kre˘ın’s uniqueness criterion about
the equality of the extreme extensions. The actual extension theory of nonnegative operators
is given in Section 4. The connections between the completion problems treated in this
paper and the extension theory are briefly explained in an Appendix. There somewhat more
sophisticated tools than elsewhere in this paper will be used; namely the presentation is
based on the notions of abstract boundary conditions and Weyl functions.
The following notations will be used throughout the paper. For Hilbert spaces H and K
the bounded linear operators from H to K will be denoted by [H,K], or by [H] when K = H. A
(closed) linear relation A in a Hilbert space H is just a (closed) linear subspace of H⊕H. For
the convenience of the reader some facts concerning linear relations will be briefly reviewed.
Recall the following notions
domA = { f ∈ H : {f, g} ∈ A for some g ∈ H }, kerA = { f ∈ H : {f, 0} ∈ A },
ranA = { g ∈ H : {f, g} ∈ A for some f ∈ H }, mulA = { g ∈ H : {0, g} ∈ A }.
With A−1 = {{g, f} : {f, g} ∈ A} it is clear that domA−1 = ranA, ranA−1 = domA,
kerA−1 = mulA, and mulA−1 = kerA. A linear relation A is (the graph of) an operator
precisely when mulA = {0}. The (Moore-Penrose) pseudo-inverse of a relation A is the
operator A(−1) defined by
A(−1) = { {g, f} ∈ H2 : {f, g} ∈ A, f ⊥ kerA },
which coincides with the usual notion when A is an operator. For a closed linear relation A
define As = { {f, g} ∈ A : g ⊥ mulA }. Clearly As is a closed linear operator; it is called
the (orthogonal) operator part of A. Note that domAs = domA and domA = (mulA)
⊥.
The adjoint A∗ of a linear relation A is defined by
A∗ = { {f, g} ∈ H2 : (g, h) = (f, k) for all {h, k} ∈ A }.
The adjoint is always linear and closed. A linear relation A is symmetric if A ⊂ A∗, and
selfadjoint if A = A∗. A linear relation A is nonnegative if (f I, f ) ≥ 0 for all {f, f I} ∈ A,
or equivalently, if (Asf, f) ≥ 0 for all f ∈ domA. A nonnegative relation is symmetric.
Let A be an operator which is semibounded from below A ≥ β. Then dom [A] stands for
the closure of domA with respect to the norm ,f,2A = (1 − β),f,2 + (Af, f), f ∈ domA.
The closure of the form (Af, f ) is denoted by t[f ] or by A[f ] = A[f, f ]. It is well-known
[1], [23] that dom [A] = dom (A − β)1/2 when A is selfadjoint. If A is a semibounded linear
relation then define dom [A] = dom [As]. The notation ran [A] stands for dom [A
−1]. Let
A be a symmetric relation in a Hilbert space. If A has a nontrivial multivalued part, then
then every selfadjoint extension of A has a nontrivial multivalued part. If A is a densely
defined operator, then mulA∗ = (domA)⊥ is trivial, and A and all its selfadjoint extensions
are operators. If A is an operator which is nondensely defined, then mulA∗ = (domA)⊥
is nontrivial, and there exist selfadjoint extensions with a nontrivial multivalued part. The
componentwise sum or equivalently the linear span of the graphs of linear relations A and
B is denoted by A +B.
2. Nonnegative operator blocks and shorted operators
In this section incomplete operator blocks of a special form are completed to nonnegative
operators. The solutions to the completion problem are described. The connection to the
EXTENSION THEORY 3
notion of shorted operators [26] and the relation between completions and extensions are
explained.
2.1. Completion to nonnegative operator blocks. Let H = H1 ⊕ H2 be an orthogonal
decomposition of the Hilbert space H and let A0 be an incomplete block operator of the form
(2.1) A0 =
w
A11 A12
A21 ∗
W
,
where Aij ∈ [Hj,Hi], i, j = 1, 2. If the set
(2.2) {A22 ∈ [H2] : A = (Aij)2i,j=1 ≥ 0 }
is nonempty, then A11 ≥ 0 and A21 = A∗12. The converse question will now be addressed.
Proposition 2.1. Let A0 be the incomplete block operator in the Hilbert space H = H1⊕H2,
given by (2.1). Assume that A11 ≥ 0 and A21 = A∗12. Then:
(i) There exists a completion A ∈ [H] of A0 with some operator A22 if and only if
ranA12 ⊂ ranA1/211 .
(ii) In this case the operator S = A
(−1/2)
11 A12 is well defined and S ∈ [H2,H1]. Moreover,
S∗S is the smallest operator in the set (2.2).
Proof. (i) Assume that there exists a completion A22 belonging to the set (2.2). Then
0 ∈ ρ(A11 + ε) for all ε > 0, and therefore
(2.3)
w
I 0
−A21(A11 + ε)−1 I
Ww
A11 + ε A12
A21 A22 + ε
Ww
I −(A11 + ε)−1A12
0 I
W
=
w
A11 + ε 0
0 A22 + ε− A21(A11 + ε)−1A12
W
.
The operator in the right side of (2.3) is nonnegative if and only if
(2.4) A21(A11 + ε)
−1A12 ≤ A22 + ε,
or equivalently
(2.5)
8 ,A11,
0
(t+ ε)−1 d,EtA12f,2 ≤ ε,f,2 + ,A
1
2
22f,2, ε > 0,
where Et denotes the spectral family of A11. Letting ε ↓ 0 in (2.5), the monotone convergence
theorem implies that A12f ∈ ranA1/211 for all f ∈ H2. Thus, ranA12 ⊂ ranA
1/2
11 .
Conversely, if ranA12 ⊂ ranA1/211 , then the operator S := A
(−1/2)
11 A12 is well defined and
S ∈ [H2, H1]. Since A12 = A1/211 S, it follows from A21 = S∗A
1/2
11 and
(2.6) A =
w
A1/211
S∗
Wp
A1/211 S
Q
≥ 0,
that the operator A22 = S
∗S gives a completion for A0.
(ii) According to (i) A21 = S
∗A1/211 , and S
∗S ∈ [H2] gives a solution to the completion
problem (2.2). Now
s− lim
ε↓0
A21(A11 + ε)−1A12 = s− lim
ε↓0
S∗A1/211 (A11 + ε)
−1A1/211 S = S
∗S,
4 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
and if A22 is an arbitrary operator in the set (2.2), then by letting ε ↓ 0 in (2.4) one concludes
that S∗S ≤ A22. Therefore, S∗S satisfies the desired minimality property. -
The above proposition establishes the existence of a minimal solution A = Amin to the
completion problem (2.1) in block form via (2.6):
(2.7) Amin =
w
A
1/2
11
S∗
Wp
A
1/2
11 S
Q
.
Now every other completion A ∈ [H] can be represented in a similar block form as follows:
(2.8) A =
w
A11 A12
A21 A22
W
=
w
A
1/2
11
S∗
Wp
A
1/2
11 S
Q
+
w
0 0
0 A22 − S∗S
W
.
Corollary 2.2. Let A = (Aij)
2
i,j=1 be a block representation of A ∈ [H] with respect to the
decomposition H = H1 ⊕ H2. Then A is nonnegative precisely when:
(i) A11 ≥ 0, A21 = A∗12;
(ii) ranA12 ⊂ ranA1/211 ;
(iii) A22 ≥ S∗S, S = A(−1/2)11 A12.
Moreover, if ranA12 ⊂ ranA11 then the condition (iii) takes the form A22 ≥ A21A(−1)11 A12.
Proof. The first assertion follows immediately from Proposition 2.1 and the above block
representations. As to the last statement it is enough to notice that if ranA12 ⊂ ranA11,
then S∗S can be rewritten as S∗S = A21A
(−1)
11 A12. -
Observe, that the inclusion ranA12 ⊂ ranA11 holds if ranA11 is closed, which is so in
particular if 0 ∈ ρ(A11), i.e., if A11 is invertible. In this last case the difference A22 − S∗S =
A22 −A21A−111 A12 is called a Schur complement of A = (Aij)2i,j=1. In this sense Corollary 2.2
contains Sylvester’s criterion: the operator A = (Aij)
2
i,j=1 with 0 ∈ ρ(A11) is nonnegative if
and only if A11 ≥ 0, A21 = A∗12, and A22 − A21A−111 A12 ≥ 0.
For each solution A ∈ [H] of the completion problem (2.1) the identities (2.7) and (2.8)
give the following result concerning the kernels kerA and kerAmin:
(2.9) kerA = kerAmin ∩ ker
w
0 0
0 A22 − S∗S
W
,
as, in general, nonnegative bounded linear operators C1 and C2 satisfy the following identity
ker (C1 + C2) = kerC1 ∩ kerC2. Hence, among all solutions A ∈ [H] of the completion
problem (2.1) the minimal solution Amin has the largest kernel: kerA ⊂ kerAmin .
Proposition 2.3. Let Amin be the minimal solution to the completion problem (2.1). Then
(2.10) kerAmin =
Fw
−A−1/211 (S ϕ)
ϕ
W
: Sϕ ∈ ranA1/211 , ϕ ∈ H2
k
,
where A−1/211 stands for the preimage. Moreover, if A ∈ [H] is a solution to the completion
problem (2.1) given in the block form (2.8) and if R := A22 − S∗S, then
(i) kerA = kerAmin if and only if S
−1(ranA1/211 ) ⊂ kerR.
(ii) If ranS ∩ ranA1/211 = {0}, then kerA = kerAmin if and only if kerS ⊂ kerR.
EXTENSION THEORY 5
(iii) The minimal solution to the completion problem (2.1) is the only solution A ∈ [H] for
which kerA = kerAmin if and only if ranA12 ⊂ ranA11. This condition is satisfied
in particular if ranA11 is closed.
Proof. It follows from (2.7) that
kerAmin =
Fw
ψ
ϕ
W
: A
1/2
11 ψ + Sϕ = 0
k
,
which leads to the identity (2.10). In order to prove (i), (ii), and (iii), observe that kerA =
kerAmin if and only if
(2.11) kerAmin ⊂ ker
w
0 0
0 A22 − S∗S
W
,
which follows from (2.9).
(i) According to (2.10) and (2.11) the identity kerA = kerAmin holds if and only if Rϕ = 0
for all ϕ ∈ S−1(ranA1/211 ).
(ii) If ranS ∩ ranA1/211 = {0}, then S−1(ranA1/211 ) = kerS, and the assertion follows from
the result in (i).
(iii) According to (i) Amin is the only solution A to the completion problem (2.1) for
which equality holds in (2.10) if and only if S−1(ranA1/211 ) = H2. But this is equivalent to
ranS ⊂ ranA1/211 and hence also to ranA12 ⊂ ranA11. The last statement in (iii) follows
from ranA12 ⊂ ranA1/211 ⊂ ranA11 = ranA11. -
Remark 2.4. Let A = (Aij)
2
i,j=1 be a block representation of A ≥ 0, A ∈ [H], with respect
to the decomposition H = H1 ⊕ H2. Observe, that
(2.12) ker
D
A11 A12
i
= ker
p
A1/211 S
Q
.
It follows from (2.12) that the identity (2.10) can be rewritten as
(2.13) kerAmin =
Fw
−A−111 (A12 ϕ)
ϕ
W
: A12ϕ ∈ ranA11, ϕ ∈ H2
k
,
where A−111 stands for the preimage.
Since A11 ≥ 0, Proposition 2.1 shows that kerA11 = kerA1/211 ⊂ kerA21. In particular,
kerA11 ⊕ {0} ⊂ kerA. Therefore the completion problem (2.1) can be reduced initially.
With respect to the decomposition H1 = kerA11⊕ ranA11 the operators A11 and A21 can be
decomposed as A11 = 0⊕AI11 with kerAI11 = {0} and A21 = 0⊕AI21. Hence, the completion
problem (2.1) can be reduced to a completion problem in H8 kerA11 via
(2.14) A0 =


0 0 0
0 AI11 A
I
12
0 AI21 ∗

 ,
where the decomposition is with respect to H = kerA11 ⊕ ranA11 ⊕ H2. Therefore, without
loss of generality, it may be assumed in the sequel that the incomplete block operator A0 in
(2.1) satisfies kerA11 = {0}. In this case the preimages in (2.10) and (2.13) coincide with
the usual inverses.
6 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
2.2. Shorted operators. Let H be a Hilbert space, let N be a closed linear subspace of H,
and let A ∈ [H] be a nonnegative operator. Then the set
(2.15) {D ∈ [H] : 0 ≤ D ≤ A, ranD ⊂ N }
contains a unique maximal element, which is called the shortening AN of A to the subspace
N. The operator AN was introduced by M.G. Kre˘ın [26]; the mapping A→ AN is called the
Kre˘ın transformation. It will be shown that the existence and uniqueness of the operator AN
are consequences of Proposition 2.1 and the corresponding decomposition (2.8) with H2 = N.
Proposition 2.5. (Cf. [26], [34], [27]) Let A = (Aij)
2
i,j=1 be a block representation of A ≥ 0,
A ∈ [H], with respect to the decomposition H = H1 ⊕N. The set (2.15) contains a unique
maximal element AN given by
(2.16) AN =
w
0 0
0 A22 − S∗S
W
, S = A
(−1/2)
11 A12.
Moreover, each element D ∈ [H] in (2.15) satisfies the inequality
(2.17) (Df, f) ≤ inf
g∈H1
(A(f − g), f − g), f ∈ H,
and AN is the only element in (2.15) for which equality holds:
(2.18) (ANf, f) = inf
g∈H1
(A(f − g), f − g), f ∈ H.
Proof. According to Corollary 2.2 AN given by (2.16) belongs to the set in (2.15). Now let D
be any operator in the set in (2.15). Since D ≥ 0 and ranD ⊂ N, it has the representation
(2.19) D =
w
0 0
0 D22
W
, D22 ≥ 0.
Moreover, it follows from A−D ≥ 0 and Proposition 2.1 that A22−D22 ≥ S∗S, or in other
words, that D ≤ AN. Therefore, AN is the unique maximal element in (2.15).
Now let f = f1 ⊕ f2 ∈ H1 ⊕N and let g ∈ H1. Then it follows from (2.8) and (2.16) that
(A(f − g), f − g) = ,A1/211 (f1 − g) + Sf2,2 + ((A22 − S∗S)f2, f2)
= ,A1/211 (f1 − g) + Sf2,2 + (ANf2, f2)
= ,A1/211 (f1 − g) + Sf2,2 + (ANf, f).
Recall that ranS ⊂ (kerA1/211 )⊥ = ranA
1/2
11 , which leads to (2.18).
Since AN is the unique maximal element in (2.15), (2.18) implies that every operator D in
the set in (2.15) satisfies (2.17). Assume that for some D ∈ [H] with 0 ≤ D ≤ A, ranD ⊂ N:
(2.20) (Df, f) = inf
g∈H1
(A(f − g), f − g), f ∈ H.
Then (Df, f) = (ANf, f), f ∈ H, and consequently D = AN. -
Example 2.6. Let H = H1 ⊕N, R ∈ [N], R ≥ 0, let H3 be a Hilbert space, X ∈ [H1,H3],
Y ∈ [N,H3], and let ranY ⊂ ranX . Let the operator A ∈ [H] be given by
(2.21) A =
w
X∗
Y ∗
W
(X, Y ) +
w
0 0
0 R
W
≥ 0.
EXTENSION THEORY 7
It follows from the definition (2.21) of A that
(A(f − g), f − g) = ,X(f1 − g1) + Y f2,2 + (Rf2, f2), f1, g1 ∈ H1, f2 ∈ N,
where f = f1 ⊕ f2 and g = g1 ⊕ 0. Hence, (2.18) and ran Y ⊂ ranX imply that
(ANf, f) =
ww
0 0
0 R
W
f, f
W
, f ∈ H.
The operator in the right side is nonnegative, is majorized by A, and has its range in N, so
that the shortening of A to the subspace N is given by:
(2.22) AN =
w
0 0
0 R
W
.
In particular, AN = 0 if and only if R = 0.
The identity (2.18) provides a characterization of shorted operators from which various
well-known monotonicity and continuity properties follow, cf. [3], [4], [33].
Proposition 2.7. Let A,B ∈ [H] be nonnegative operators, let M,N be closed linear sub-
spaces of H, and let λ ≥ 0. Then:
(i) (λA)N = λAN;
(ii) AN +BN ≤ (A+B)N, in particular A ≤ B ⇒ AN ≤ BN;
(iii) AM∩N = (AM)N, in particular M ⊂ N⇒ AM ≤ AN.
Moreover, if An ∈ [H] and An ≥ A, then
(2.23) A = w − lim
n→∞
An ⇒ AN = s− lim
n→∞
(An)N.
Proof. The statements (i)—(iii) follow easily from (2.18). As to the last statement, let f ∈ H.
According to (2.18) for each ε > 0 there exists gε ∈ H1, such that
(ANf, f) ≤ (A(f − gε), f − gε) ≤ (ANf, f ) + ε/2.
Since w − limn→∞An = A and An ≥ A, there exists some n0 = n0(gε), such that
(ANf, f ) ≤ (An(f − gε), f − gε) ≤ (ANf, f) + ε, n ≥ n0,
and, therefore, also
(ANf, f) ≤ ((An)Nf, f) ≤ (ANf, f ) + ε, n ≥ n0.
As ε > 0 is arbitrary, it follows that limn→∞((An)Nf, f ) = (ANf, f), f ∈ H; and hence
AN = w − limn→∞(An)N. Now An ≥ A implies that (An)N ≥ AN. For the nonnegative
operators Bn := (An)N − AN the weak convergence to zero implies the strong convergence:
w − limBn = 0⇒ s− limBn = 0. -
Remark 2.8. Items (i) and (ii) express the sublinear property and item (iii) expresses the
multiplicative property of shortening. The implication (2.23) need not be true without the
condition An ≥ A, even if the strong limit is replaced with the weak limit AN = w −
limn→∞(An)N. For instance, let X ∈ [H1,H3] be such that ranX = H3, let Y ∈ [N,H3] be
nontrivial, and let εn ↓ 0 as n→∞. Then
An =
w
εnX∗
Y ∗
W D
εnX Y
i
→
w
0 0
0 Y ∗Y
W
= A,
8 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
in the strong sense as n →∞. However, Example 2.6 shows that (An)N = 0 for all n ∈ N,
and AN = Y ∗Y W= 0. In addition, notice that strong convergence of nonnegative operators
does not follow from their weak convergence. For instance, when Cn = (I + Un∗)(I + Un),
where U denotes the unilaterial shift, then Cn → 2I in the weak sense, whereas the limit in
the strong sense does not exist.
Clearly, with N = H2, a solution A to (2.1) is minimal precisely when AN is trivial.
Corollary 2.9. Let A ∈ [H] be a solution to the completion problem (2.1) with a block
representation A = (Aij)2i,j=1. Then A is equal to the minimal solution if and only if
inf
g∈H1
(A(f − g), f − g) = 0, f ∈ H.
This result can be stated in a different, but an equivalent, manner. Let A ∈ [H] be a
solution to the completion problem (2.1). Provide the Hilbert space H with the scalar product
(Af, g), f, g ∈ H, and denote by HA the semi-Hilbert space obtained by the completion of H
with respect to (Af, g). Then A is minimal if and only if H1 is dense in HA, or equivalently,
if and only if H1/kerA is dense in the quotient space HA/kerA.
2.3. Completions and extensions. Let A ∈ [H] be a solution to the completion problem
(2.1). The restriction of this bounded linear operator to the subspace H1⊕ {0} is a bounded
symmetric operator A1:
A1 = AH1 =
w
A11
A21
W
.
The completion problem is equivalent to the description of all bounded selfadjoint extensions
in [H] of A1, see the Appendix. As the operator A1 is not defined everywhere, its adjoint A
∗
1
in H is a linear relation (multivalued operator). In order to determine this adjoint observe
that, in the language of relations,
(2.24) A1 = A ∩ Z∗, Z = {0}⊕H2.
Lemma 2.10. Let A ∈ [H] be a solution to the completion problem (2.1). Then
(2.25) A∗1 = A +({0}⊕ H2).
Proof. The representation (2.24) implies that A∗1 = clos (A +Z). In fact A +Z is closed,
which implies (2.25). To see that A +Z is indeed closed, assume that {fn, gn} → {f, g}
as n → ∞ for a sequence {fn, gn} ∈ A +({0} ⊕ H2), where gn = Afn + hn and hn ∈ H2.
Then Afn → Af , since A is bounded, and hn = gn − Afn → g − Af =: h ∈ H2 as n →∞.
Therefore, {f, g} = {f, Af} + {0, h} ∈ A + ({0}⊕ H2). -
Lemma 2.11. Let A ∈ [H] be a solution to the completion problem (2.1). Then
(2.26) ranA ⊂ ranA∗1 ∩ ranA1/2.
Moreover, equality holds in (2.26) if A = Amin:
ranAmin = ranA
∗
1 ∩ ranA
1/2
min.
Proof. Since A is a selfadjoint extension of A1, one has A ⊂ A∗1 and therefore (2.26) holds.
EXTENSION THEORY 9
Now let A = Amin. To prove the reverse inclusion, let g ∈ ranA∗1 ∩ ranA
1/2
min. Since
g ∈ ranA∗1, it follows from Lemma 2.24 with A = Amin, that g = Aminf + h for some f ∈ H
and h ∈ H2. Moreover, since g ∈ ranA1/2min one obtains
(2.27) h ∈ ranA1/2min = ran
X
A1/211
A21A
(−1/2)
11
~
,
cf. (2.7). Since h ∈ H2, (2.27) implies that h = 0 and therefore g ∈ ranAmin. -
It follows from Lemma 2.10 that the defect subspace of A1 at z ∈ ρ(A11) is given by
(2.28) Nz(A
∗
1) = ker (A
∗
1 − z) =
Fw
−(A11 − z)−1A∗21f2
f2
W
: f2 ∈ H2
k
.
Let N z(A∗1) = { {f, zf} : f ∈ Nz(A∗1) } and define a special extension Az of A1 by
(2.29) Az = A1 + N z(A∗1).
Lemma 2.12. For z ∈ ρ(A11) the extension Az in (2.29) has the block representation:
(2.30) Az =
w
A11 A∗21
A21 A22(z)
W
,
where
(2.31) A22(z) = zI +A21(A11 − z)−1A∗21.
Proof. Assume that z ∈ ρ(A11), so that (2.28) holds. LetFw
0
f2
W
,
w
g1
g2
Wk
∈ Az.
Then f1 = (A11 − z)−1A∗21f2 and consequently g1 = A11f1 − z(A11 − z)−1A∗21f2 = A∗21f2,
g2 = A21f1+zf2 = (zI+A21(A11−z)−1A∗21)f2. Therefore, Az has the form (2.30), (2.31). -
If z ∈ ρ(A11), then the block representation (2.30) shows that (Az)∗ = Az¯. In particular,
if z ∈ ρ(A11) ∩R, then Az is selfadjoint. Since A11 ≥ 0, each z < 0 belongs to ρ(A11).
Lemma 2.13. Let Az, z < 0, be defined by (2.29). Then
s− lim
z↑0
Az = Amin.
Proof. The statement follows from (2.31) and s− limz↑0 (zI +A21(A11 − z)−1A∗21) = S∗S; cf.
the proof of Proposition 2.1. -
3. Selfadjoint contractive extensions of symmetric contractions
Let H1 be a closed linear subspace of the Hilbert space H and let T1 ∈ [H1,H] be a
symmetric contraction. The set of all selfadjoint contractive extensions of T1 is denoted by
Ext T1(−1, 1):
Ext T1(−1, 1) = { T ∈ [H] : T1 ⊂ T = T ∗, ,T, ≤ 1 }.
It was shown by Kre˘ın [26], cf. also [1], that the set Ext T1(−1, 1) is nonempty. Here Kre˘ın’s
result and a description of Ext T1(−1, 1) is obtained via the preceding completion problem.
An important tool in the description is provided by the notion of defect operator.
10 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
3.1. Defect operators. Let H1 and H2 be Hilbert spaces and let T ∈ [H1,H2] be a contrac-
tion. The corresponding defect operators DT and DT ∗ are defined by
DT = (I − T ∗T )1/2, DT∗ = (I − TT ∗)1/2.
They satisfy the well-known commutation properties
(3.1) TDT = DT ∗T, T
∗DT∗ = DTT
∗,
which follow from the spectral theorem.
Corollary 3.1. Let T ∈ [H1,H2] be a contraction. Then
(3.2) T (kerDT ) = kerDT∗, T
∗(kerDT∗) = kerDT .
In particular,
(3.3) kerDT = {0} if and only if kerDT∗ = {0}.
Proof. It suffices to show the first identity in (3.2). Let ϕ ∈ kerDT , then according to (3.1)
Tϕ ∈ kerDT ∗. Hence, T (kerDT ) ⊂ kerDT ∗. Conversely, let ϕ ∈ kerDT∗. Then ϕ = TT ∗ϕ,
and T ∗ϕ ∈ kerDT according to (3.1), which shows the reverse inclusion. -
The ranges ranT and ranDT∗ are complementary subspaces in the sense of [11]: H =
ranT + ranDT ∗ with overlapping space ran T ∩ ranDT ∗, cf. also [6]. The intersection
ranT ∩ ranDT∗ is in general nontrivial; it can be described as follows.
Lemma 3.2. Let T ∈ [H1,H2] be a contraction. Then
ranT ∩ ranDT∗ = ranTDT = ranDT ∗T.
Proof. The commutation relations (3.1) show that ran TDT = ranDT ∗T ⊂ ranT ∩ ranDT∗.
Hence, it suffices to prove the inclusion
ranT ∩ ranDT ∗ ⊂ ran TDT .
Suppose that ϕ ∈ ranT ∩ ranDT∗. Then it follows from (3.1) that DT∗ϕ ∈ ranT and
T ∗ϕ ∈ ranDT . Hence, DT∗ϕ = Tf and T ∗ϕ = DTg for some f, g ∈ H. Therefore, ϕ =
(I − TT ∗)ϕ+ TT ∗ϕ = DT ∗Tf + TDTg = TDT (f + g). -
3.2. Reduction to a completion problem. Let H1 be a closed linear subspace of the
Hilbert space H and let T1 ∈ [H1,H] be a symmetric contraction. Write T1 in the block form
T1 =
D
T11
T21
i
with respect to the orthogonal decomposition H = H1 ⊕H2. In the next theorem
it is shown that the completion problems associated with
(3.4) A0± =
w
I ± T11 ±T12
±T21 ∗
W
have solutions. Moreover, the corresponding minimal solutions A+ and A− are shown to be
connected with two extreme selfadjoint contractive extensions of T1.
Theorem 3.3. Let T1 =
D
T11
T21
i
∈ [H1, H] be a symmetric contraction from the Hilbert space
H1 into the Hilbert space H = H1 ⊕ H2. Then:
(i) The completion problems for A0± in (3.4) have minimal solutions A±.
(ii) The operators Tm := A+ − I and TM := I − A− belong to Ext T1(−1, 1).
EXTENSION THEORY 11
(iii) The operators Tm and TM have the block form
(3.5) Tm =
w
T11 DT11V
∗
V DT11 −I + V (I − T11)V ∗
W
, TM =
w
T11 DT11V
∗
V DT11 I − V (I + T11)V ∗
W
,
where V is given by V := T21D
(−1)
T11
.
(iv) The operators Tm and TM are extremal extensions of T1:
(3.6) T ∈ Ext T1(−1, 1) if and only if T = T ∗ ∈ [H], Tm ≤ T ≤ TM .
(v) The operators Tm and TM are connected via
(3.7) (−T )m = −TM , (−T )M = −Tm.
Proof. (i) The condition ,T1, ≤ 1 is equivalent to T ∗21T21 ≤ I − T 211 = D2T11, where DT11 is
the (selfadjoint) defect operator of T11, so that
(3.8) ,T21f, ≤ ,DT11f,, f ∈ H1.
This implies the existence of a contraction V ∈ [H1,H2] such that V DT11f = T21f , f ∈ H1,
cf. [19], [20]. Moreover, V is uniquely determined if the extra condition kerV ⊃ kerDT11 is
assumed. Since
(3.9) T ∗21 = DT11V
∗ = (I + T11)
1/2(I − T11)1/2V ∗,
the range inclusions
(3.10) ran T ∗21 ⊂ ran (I ± T11)1/2
hold. By Proposition 2.1 the following operators
(3.11) S+ = (I + T11)
(−1/2)T ∗21, S− = (I − T11)(−1/2)T ∗21.
are well defined. According to Proposition 2.1 there are minimal solutions A± to the com-
pletion problems for A0± in (3.4).
(ii)&(iii) The operators Tm = A+ − I and TM = I − A− can be rewritten as follows:
(3.12) Tm = A+ − I =
w
T11 T
∗
21
T21 −I + S∗+S+
W
, TM = I − A− =
w
T11 T
∗
21
T21 I − S∗−S−
W
.
Observe that
(3.13) S± = (I ± T11)(−1/2)DT11V ∗ = P±(I ∓ T11)1/2V ∗ = (I ∓ T11)1/2P±V ∗,
where P± are the orthogonal projections onto
(ker (I ± T11)1/2)⊥ = (ker (I ± T11))⊥ = ran (I ± T11) = ran (I ± T11)1/2.
Since kerV ⊃ kerDT11 implies ranV ∗ ⊂ ranDT11 ⊂ ran (I ± T11)1/2, it follows that
(3.14) S+ = (I − T11)1/2V ∗, S− = (I + T11)1/2V ∗.
Consequently,
(3.15) S∗+S+ = V (I − T11)V ∗, S∗−S− = V (I + T11)V ∗,
12 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
which implies the representations for Tm and TM in (iii). Clearly, Tm and TM are selfadjoint
extensions of T1, which satisfy the inequalities I + Tm ≥ 0 and I − TM ≥ 0. Moreover, it
follows from (3.5) that
(3.16) TM − Tm =
w
0 0
0 2(I − V V ∗)
W
.
Hence, TM ≥ Tm and consequently I − Tm ≥ I − TM ≥ 0 and I + TM ≥ I + Tm ≥ 0.
Therefore, Tm and TM are contractions, which proves (ii).
(iv) Observe, that T ∈ Ext T1(−1, 1) if and only if T = T ∗ ⊃ T1 and I ± T ≥ 0. By
Proposition 2.1 this is equivalent to
(3.17) S∗+S+ − I ≤ T22 ≤ I − S∗−S−.
The inequalities (3.17) are equivalent to (3.6).
(v) The relations (3.7) follow from (3.11) and (3.12). -
The formulas for Tm and TM in (3.5) were obtained by V. Kolmanovich and M.M. Mala-
mud, see [24, p.5], [25]. Apparently they are a part of mathematical folklore.
3.3. Selfadjoint contractive extensions of symmetric contractions. The first result
in this subsection is the well-known result of M.G. Kre˘ın concerning the nonemptyness of
Ext T1(−1, 1), which is now just an immediate consequence of the previous theorem. Notice,
that Theorem 3.3 also gives an explicit structure for the set Ext T1(−1, 1).
Theorem 3.4. ([26], cf. also [1]) Let the Hilbert space H have the orthogonal decomposition
H = H1⊕H2 and let T1 =
D
T11
T21
i
∈ [H1, H] be a symmetric contraction. Then Ext T1(−1, 1) is
a nonempty operator interval with endpoints Tm and TM ; i.e. the equivalence (3.6) holds.
The result in Theorem 3.4 can also be stated in terms of operator balls. The defect
operator DV ∗ ∈ [H2] gives rise to an orthogonal decomposition of the Hilbert space H2:
H2 = ranDV ∗ ⊕ kerDV ∗.
Denote the set of all selfadjoint contractions on a Hilbert space H by
CH = {K ∈ [H] : K = K∗, ,K, ≤ 1 }.
Corollary 3.5. [26, 27] Let the Hilbert space H have the orthogonal decomposition H =
H1 ⊕ H2, let T1 =
D
T11
T21
i
∈ [H1, H], and let T0 = (Tm + TM)/2, where Tm and TM are as in
Theorem 3.3. Then the formula
(3.18) T = T0 + 2
−1(TM − Tm)1/2
w
0 0
0 K
W
(TM − Tm)1/2,
or, equivalently, the formula
(3.19) T =
w
T11 DT11V
∗
V DT11 −V T ∗11V ∗
W
+
w
0 0
0 DV ∗KDV ∗
W
,
establishes a one-to-one correspondence between the set Ext T1(−1, 1) and the operator ball
CH, where H = ranDV ∗.
EXTENSION THEORY 13
Proof. The inequality in (3.6) can be rewritten in the form
−(TM − Tm)/2 = Tm − T0 ≤ T − T0 ≤ TM − T0 = (TM − Tm)/2.
These inequalities are equivalent to the existence of a selfadjoint contraction 4K in H with
ran 4K ⊂ H = ranDV ∗, such that 2(T − T0) = (TM − Tm)1/2 4K(TM − Tm)1/2, cf. (3.16).
This implies the representation (3.18) for T . The one-to-one correspondence between T ∈
Ext T1(−1, 1) and K ∈ CH via (3.18) is obvious from the given identities. To see the equiv-
alence of (3.18) and (3.19) observe that (3.5) implies
T0 =
w
T11 DT11V
∗
V DT11 −V T11V ∗
W
.
The form of the second term in the right side of (3.19) is clear from (3.16) and (3.18). -
The extensions Tm and TM can be characterized in a different manner, cf. Corollary 2.9.
Corollary 3.6. ([26]) Let T ∈ Ext T1(−1, 1). Then
(i) T = Tm if and only if infg∈H1((I + T )(f − g), f − g) = 0 for every f ∈ H;
(ii) T = TM if and only if infg∈H1((I − T )(f − g), f − g) = 0 for every f ∈ H.
Let T ∈ [H] be a contraction, in general, with ker (I ± T ) W= {0}. Denote by H± the
semi-Hilbert spaces obtained by the completion of H with respect to the seminorms , · ,I±T ,
(3.20) (f, g)I±T = ((I ± T )f, g), ,f,I±T = ((I ± T )f, f )1/2.
Since (3.20) define only seminorms, quotient spaces H/ker (I ± T ) have to be considered to
obtain Hilbert spaces. This yields another result due to M.G. Kre˘ın.
Corollary 3.7. ([26], cf. also [1]) Assume T ∈ Ext T1(−1, 1). Then T = Tm if and only if
H1 is dense in H+, and T = TM if and only if H1 is dense in H−.
Remark 3.8. Let T1 =
D
T11
T21
i
∈ [H1, H] be a symmetric contraction presented in the block
form with respect to H = H1⊕H2. If T ∈ Ext T1(−1, 1), then f1 ∈ ker (I ± T11) implies that
f1 ⊕ 0 ∈ ker (I ± T ), cf. Remark 2.4. This indicates that the extension problem for T1 =D
T11
T21
i
∈ [H1, H] could have been reduced to an extension problem with ker (I ± T11) = {0}.
In fact, let K1 and K2 be defined by
(3.21) K1 := ker (I + T11), K2 := ker (I − T11).
It follows from (3.8) that T1 and T , when written in the block form with respect to the
decomposition H = K1 ⊕ K2 ⊕ HI1 ⊕ H2, where HI1 = H1 8 (K1 ⊕ K2), have the form
T11 = (−IK1)⊕ IK2 ⊕ T I11, T21 = (0, 0, T I21), T12 = T ∗21,
with T I11 ∈ [HI1], T I21 ∈ [HI1,H2]. Hence, the description of the selfadjoint contractive exten-
sions T of T1 in H is reduced to the description of the selfadjoint contractive extensions T
I
of T I1 =
D
T I11
T I21
i
in HI = HI1 ⊕ H2.
Proposition 2.3 gives rise to a full analog for the contractions Tm and TM . Here only a
useful description for the kernels ker (I + Tm) and ker (I − TM) will be given.
Lemma 3.9. If ker (I + T11) = {0}, then
ker (I + Tm) =
Fw
(I + T11)−1DT11V
∗ f2
−f2
W
: V ∗f2 ∈ ran (I + T11)1/2, f2 ∈ H2
k
.
14 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
If ker (I − T11) = {0}, then
ker (I − TM ) =
Fw
(I − T11)−1DT11V ∗ f2
f2
W
: V ∗f2 ∈ ran (I − T11)1/2, f2 ∈ H2
k
.
Proof. It follows from (3.11), (3.12), and Proposition 2.3 (or directly from (3.5)) that
ker (I + Tm) =
Fw
(I + T11)
−1/2S+ f2
−f2
W
: S+f2 ∈ ran (I + T11)1/2, f2 ∈ H2
k
.
Here S+ = (I − T11)1/2V ∗, so that (I + T11)−1/2S+ f2 = (I + T11)−1DT11V ∗f2. It is easy to
check that
(I − T11)1/2V ∗f2 ∈ ran (I + T11)1/2 ⇐⇒ V ∗f2 ∈ ran (I + T11)1/2.
This proves the first statement. The proof of the second statement is similar. -
As a consequence one also obtains a description for the kernels ker (I+TM ) and ker (I−Tm).
Lemma 3.10. Let ker (I ± T11) = {0} and let f = f1 ⊕ f2 ∈ H1 ⊕ H2. Then the following
statements hold:
(i) f ∈ ker (I + TM ) if and only if f ∈ ker (I + Tm) and DV ∗f2 = 0;
(ii) f ∈ ker (I − Tm) if and only if f ∈ ker (I − TM ) and DV ∗f2 = 0.
Moreover
(iii) ker (I + TM) = {0} if and only if kerDV ∩ ran (I + T11)1/2 = {0};
(iv) ker (I − Tm) = {0} if and only if kerDV ∩ ran (I − T11)1/2 = {0}.
Proof. (i) According to (3.16)
I + TM = I + Tm +
w
0 0
0 2(I − V V ∗)
W
,
where the bounded linear operators I + TM and I + Tm are nonnegative, and V ∗ is a con-
traction. Hence the statement follows from ker (I − V V ∗) = kerDV ∗.
(iii) It follows from Lemma 3.9 and part (i) that ker (I + TM) = {0} if and only if
(3.22) { f2 ∈ kerDV ∗ : V ∗f2 ∈ ran (I + T11)1/2 } = {0}.
The assertion now follows from the identity V ∗(kerDV ∗) = kerDV , cf. Corollary 3.1.
The proofs of (ii) and (iv) are similar to the proofs of (i) and (iii). -
3.4. Completions and extensions. Assume that T1 =
D
T11
T21
i
∈ [H1, H] is a symmetric
contraction presented in the block form with respect to H = H1 ⊕ H2. Therefore its adjoint
T ∗1 is a closed linear relation. Let T ∈ Ext T1(−1, 1), so that T is a contractive selfadjoint
extension of T1 ∈ [H1,H]. The following result is straightforward, cf. Lemma 2.10.
Lemma 3.11. Let T ∈ Ext T1(−1, 1), then
T ∗1 = T +({0}⊕ H2).
The next lemma contains some useful facts, which follow directly from Lemma 2.11.
Lemma 3.12. Let T1, Tm and TM be as in Theorem 3.3. Then:
(i) ran (I + Tm) = ran (I + T1)
∗ ∩ ran (I + Tm)1/2;
(ii) ran (I − TM ) = ran (I − T1)∗ ∩ ran (I − TM )1/2.
EXTENSION THEORY 15
It follows from Lemma 3.11 that the defect subspace Nz(T
∗
1 ) at z ∈ ρ(T11) is given by
Nz(T
∗
1 ) = ker (T
∗
1 − z) =
Fw
−(T11 − z)−1T ∗21f2
f2
W
: f2 ∈ N = (domT1)⊥
k
.
Let N z(T ∗1 ) = { {f, zf} : f ∈ Nz(T ∗1 ) } and define the special extension Tz of T1 by
(3.23) Tz = T1 + N z(T ∗1 ).
Lemma 3.13. For z ∈ ρ(T11) the extension Tz in (3.23) has the block representation
(3.24) Tz =
w
T11 T
∗
21
T21 T22(z)
W
,
where
(3.25) T22(z) = zI + T21(T11 − z)−1T ∗21.
If z ∈ ρ(T11), then (Tz)∗ = Tz¯, and if z ∈ ρ(T11) ∩ R, then Tz is selfadjoint. Since T11
is a contraction each z < −1 or z > 1 belongs to ρ(T11). A translation for Lemma 2.13 is
obtained by using the following limits, cf. (3.15):
(i) s− limr↑1 T21(I + rT11)−1T ∗21 = V (I − T11)V ∗ = S∗+S+;
(ii) s− limr↑1 T21(I − rT11)−1T ∗21 = V (I + T11)V ∗ = S∗−S−.
Lemma 3.14. Let Tz, z ∈ ρ(T11), be given by (3.24), (3.25). Then
(3.26) s− lim
z↑−1
Tz = Tm, s− lim
z↓1
Tz = TM .
If T1 ∈ [H1,H] is a contraction, then rT1 with 0 ≤ r < 1 is also a contraction. The
selfadjoint contractive extensions of rT1 form a nonempty operator interval whose endpoints
are denoted by (rT )m and (rT )M . The next result concerns the limiting behaviour of these
extensions as r→ 1.
Proposition 3.15. Let H1 be a closed linear subspace of H and let T1 ∈ [H1,H] be a sym-
metric contraction. Then
(i) s− limr↑1(rT )m = Tm;
(ii) s− limr↑1(rT )M = TM .
Proof. According to (3.5) the extension (rT )m of rT1 has the block representation:
(rT )m =
w
rT11 DrT11V
∗(r)
V (r)DrT11 −I + V (r)(I − rT11)V ∗(r)
W
,
where the operator V (r) is given by
V (r) = rT21D
−1
rT11
= rV DT11D
−1
rT11
,
cf. Theorem 3.3. An application of the Lebesgue dominated convergence theorem shows
that for all f ∈ H1,
lim
r↑1
,rDT11D−1rT11f − f,
2 = lim
r↑1
8 1
−1
eeeeer
5
1− t2
1− r2t2 − 1
eeeee
2
d(Etf, f) = 0,
where Et stands for the spectral family of T11. Consequently,
s− lim
r↑1
V (r) = V, s− lim
r↑1
V ∗(r) = V ∗.
16 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
In view of (3.5) these equalities imply (i).
Part (ii) can be proved similarly. -
According to B.N. Parlett [32, p.252], W.M. Kahan has obtained the following formula
T0 = lim
z↓1
w
T11 T
∗
21
T21 −T21(zI − T 211)−1T11T ∗21
W
for one of the elements of the class Ext T1(−1, 1) in the case that H is finite-dimensional. It
follows from (3.5) that T0 = (Tm + TM )/2.
3.5. Kre˘ın’s uniqueness criterion. Although the set Ext T1(−1, 1) is a nonempty operator
interval it may degenerate when the endpoints Tm and TM coincide. This situation has been
characterized by M.G. Kre˘ın [26]. Here a geometric approach to obtain Kre˘ın’s criterion is
presented.
The next result characterizes isometric operators in the class of all contractions.
Proposition 3.16. For a contraction T ∈ [H1,H2] the following statements are equivalent:
(i) T is isometric;
(ii) kerT = {0} and ran T ∩ ranDT∗ = {0};
(iii) for some, and equivalently for every, subspace L with ran T ⊂ L (= closL) one has
(3.27) sup
f∈L
|(f, Tϕ)|
,DT∗f,
=∞ for every ϕ ∈ H1\{0}.
Proof. (i)⇒ (iii) Let L be an arbitrary subspace with ranT ⊂ L. Assume that the supremum
in (3.27) is finite for some ϕ ∈ H1. Then there exists C > 0, such that
|(f, Tϕ)| ≤ C,DT∗f, for every f ∈ L.
Since ranT ⊂ L, also the following inequality holds:
(3.28) ,ϕ,2 = ,Tϕ,2 ≤ C,DT ∗Tϕ,.
Here D2T∗Tϕ = TD
2
Tϕ = 0, since T is isometric. Therefore, (3.28) implies ϕ = 0. Conse-
quently (3.27) holds for every ϕ W= 0.
(iii)⇒ (ii) Assume that (3.27) is satisfied with some subspace L. If (ii) does not hold, then
either kerT W= {0}, in which case (3.27) does not hold for 0 W= ϕ ∈ kerT , or ranT ∩ranDT∗ W=
{0}. However, then with 0 W= Tϕ = DT ∗h the supremum in (3.27) is finite even if f varies
over the whole space H2. Thus, if (ii) does not hold then (3.27) fails to be true.
(ii) ⇒ (i) Let ranT ∩ ranDT ∗ = {0}. Then by Lemma 3.2 TDT = 0 and it follows from
ker T = {0} that DT = 0, i.e., T is isometric. This completes the proof. -
As a direct consequence one obtains Kre˘ın’s uniqueness criterion.
Proposition 3.17. [26] Let H1 be a closed linear subspace of the Hilbert space H and let
T1 ∈ [H1,H] be a symmetric contraction. Then Tm = TM if and only if
(3.29) sup
f∈H1
|(T1f,ϕ)|2
,f,2 − ,T1f,2
=∞ for every ϕ ∈ H2 \ {0}.
EXTENSION THEORY 17
Proof. The definition of V in the proof of Theorem 3.3 (see (3.8), (3.9)) implies that
(T1f,ϕ) = (T21f,ϕ) = (DT11f, V
∗ϕ), ,f,2 − ,T1f,2 = ,DVDT11f,2.
In view of (3.16) Tm = TM if and only if V
∗ is an isometry. Since ran V ∗ ⊂ ranDT11,
the assertion follows from the equivalence of (i) and (iii) in Proposition 3.16 with L =
ranDT11 . -
3.6. Extremal selfadjoint contractive extensions. The next result contains a character-
ization for the extensions T = Tm and T = TM among the class of all selfadjoint contractive
extensions of T1, which goes back to [26] and [27].
Proposition 3.18. [26, 27] Let H = H1 ⊕N and let T ∈ Ext T1(−1, 1). Then:
(i) T = Tm + (I + T )N, T = TM − (I − T )N;
(ii) TM − Tm = (I + TM)N = (I − Tm)N;
(iii) (D2Tm)N = 0, (D
2
TM
)N = 0.
Proof. (i) Let T ∈ Ext T1(−1, 1), so that I ± T ≥ 0. It follows from (2.8) and Proposition
2.5 that I + T = I + Tm + (I + T )N and I − T = I − TM + (I − T )N.
(ii) Apply the first equality in (i) to T = TM and the second equality in (i) to T = Tm.
(iii) Observe, that D2Tm ≤ 2(I + Tm) and D2TM ≤ 2(I − TM ). Hence, by Proposition 2.7,
(3.30) 0 ≤ (D2Tm)N ≤ 2(I + Tm)N, 0 ≤ (D
2
TM
)N ≤ 2(I − TM )N.
By part (i) (I + Tm)N = (I − TM)N = 0. Hence, (3.30) gives (D2Tm)N = 0, (D2TM )N = 0. -
Part (i) of Proposition 3.18 gives the following characterization for Tm and TM : if T ∈
Ext T1(−1, 1), then T = Tm if and only if (I+T )N = 0, and T = TM if and only if (I−T )N = 0.
Part (iii) deserves some further attention.
Lemma 3.19. Let H = H1⊕N and let T1 =
D
T11
T21
i
∈ [H1, H] be a symmetric contraction with
ker (I ± T11) = {0}. Let T ∈ Ext T1(−1, 1) and K ∈ CH with H = ranDV ∗ be connected via
(3.18) or (3.19). Then
(D2T )N =
w
0 0
0 DV ∗D
2
KDV ∗
W
.
Proof. A simple calculation using the correspondence (3.19), the commutation relations (3.1),
and the selfadjointness of T11 and K leads to
(3.31) D2T = I − T ∗T =
w
X∗
Y ∗
W
(X Y ) +
w
0 0
0 DV ∗D2KDV ∗
W
,
where X = DVDT11 and Y = −(DV T11V ∗ + V ∗KDV ∗). Since ker (I ± T11) = {0}, one
has ranDT11 = H1 and consequently ranX = ranDV . Therefore, ranDV T11V
∗ ⊂ ranX.
Furthermore, with K ∈ [H] and H = ranDV ∗, it follows that
ranV ∗KDV ∗ ⊂ V ∗(H) ⊂ ranV ∗DV ∗ = ranDV V ∗ ⊂ ranDV .
Hence, ranY ⊂ ranX. The result is now obtained from Example 2.6. -
The nonempty operator interval Ext T1(−1, 1) of all selfadjoint contractive extensions of
T1 is a closed convex set; the set of extreme points of Ext T1(−1, 1) will be denoted by
Ext ET1(−1, 1). According to (3.18) and (3.19) T ∈ Ext
E
T1
(−1, 1) if and only ifK = K∗ = K−1,
i.e. K is selfadjoint and unitary in H. Clearly, K = −IH and K = IH correspond to the
18 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
extensions Tm and TM , respectively. In fact, one obtains from Lemma 3.19 the following
characterization for the class of extremal selfadjoint contractive extensions of T1.
Proposition 3.20. Let H = H1⊕N and let T ∈ Ext T1(−1, 1). Assume that ker (I ±T11) =
{0}. Then the following statements are equivalent:
(i) T ∈ Ext ET1(−1, 1);
(ii) (D2T )N = 0;
(iii) infg∈H1 ,DT (f − g), = 0 for every f ∈ H.
Proof. It follows from (3.18) and (3.19) that T ∈ Ext ET1(−1, 1) if and only if DK = 0. Hence,
the equivalence of (i) and (ii) is obtained from Lemma 3.19.
The equivalence of (ii) and (iii) is a consequence of Proposition 2.5. -
4. Extensions of nonnegative operators and relations
In this section the main results of M.G. Kre˘ın’s work [26] on the extension theory of
nonnegative densely defined operators in a Hilbert space are presented. In fact, the general
case of nonnegative nondensely defined operators and multivalued extensions is considered,
cf. [15]. Also some complements due to T. Ando and K. Nishio [7] are treated.
4.1. Linear fractional transformations. The extension theory of nonnegative relations
can be connected to the extension theory of symmetric contractive operators. Define the
linear fractional transformation X, taking a linear relation A into a linear relation X(A), by
(4.1) X(A) = { {f + f I, f − f I} : f = {f, f I} ∈ A } = −I + 2(I +A)−1.
Clearly, X maps the (closed) linear relations one-to-one onto themselves, X2 = I, and
(4.2) X(A)−1 = X(−A),
for every linear relation A. Moreover,
domX(A) = ran (I +A), ranX(A) = ran (I − A),
ker (X(A) − I) = kerA, ker (X(A) + I) = mulA.(4.3)
In addition, X preserves closures, adjoints, componentwise sums, orthogonal sums, inter-
sections, and inclusions. The relation X(A) is symmetric if and only if A is symmetric. It
follows from (4.1) and
,f + f I,2 − ,f − f I,2 = 4Re (f I, f)
that X gives a one-to-one correspondence between nonnegative linear relations and symmet-
ric contractions. Moreover, X provides a one-to-one correspondence between nonnegative
selfadjoint relations and selfadjoint contractions.
4.2. Nonnegative selfadjoint extensions of nonnegative relations. Let A be a closed
nonnegative relation in the Hilbert space H, and let the set of all nonnegative selfadjoint
extensions 4A = 4A∗ of A be denoted by Ext A(0,∞). The linear fractional transformation
T1 = X(A) of A defined by (4.1) is a symmetric contraction with domT1 = ran (I + A).
Moreover, the formula (4.1) gives a bijective correspondence between the contractive self-
adjoint extensions T ∈ Ext T1(−1, 1) of T1 and the nonnegative selfadjoint extensions 4A =
EXTENSION THEORY 19
4A∗ ∈ Ext (0,∞) of A ≥ 0. Let Tm and TM be the selfadjoint contractive extensions of T1 as
in Theorem 3.3, and define the selfadjoint relations AF and AK by
(4.4) AF = X
−1(Tm) = −I + 2(I + Tm)−1, AK = X−1(TM ) = −I + 2(I + TM)−1.
The following theorem is a translation of the corresponding facts for selfadjoint contractive
extensions of a contractive symmetric operator.
Theorem 4.1. Let A be a closed nonnegative relation in H. Then Ext A(0,∞) is nonempty;
in fact AF and AK belong to Ext A(0,∞). If 4A ∈ Ext A(0,∞), then
(4.5) (AF + a)
−1 ≤ ( 4A+ a)−1 ≤ (AK + a)−1, a > 0.
Moreover, the selfadjoint extensions AF and AK of A are connected via
(4.6) (A−1)F = (AK)
−1, (A−1)K = (AF )
−1.
Proof. The fact that AF and AN are selfadjoint extensions of A follows from the preservation
of adjoints and inclusions of the linear fractional transformation X . Moreover, since Tm and
TM are contractive, the extensions AF and AK are nonnegative, so that Ext A(0,∞) is
nonempty. Let 4A ∈ Ext A(0,∞) and let T := X( 4A). Then T ∈ Ext T1(−1, 1) and
( 4A+ a)−1 = 1
a− 1 I −
2
(a− 1)2
w
T +
a+ 1
a− 1
W−1
, a > 0,
where |(a− 1)−1(a+ 1)| > 1 for a > 0. Hence, (4.5) follows from (3.6). The relations (4.6)
follow from (3.7), (4.2), and (4.4). -
The selfadjoint extensions AF and AK of A are called the Friedrichs (hard) and the Kre˘ın-
von Neumann (soft) extension, respectively. The extremal properties (4.5) of the Friedrichs
and Kre˘ın-von Neumann extensions have been discovered by Kre˘ın [26] in the case when A
is a densely defined operator. The case when A is not densely defined was considered by
T. Ando and K. Nishio [7], and E.A. Coddington and H.S.V. de Snoo [15] (who allowed
relation extensions). The formulas (4.6) can be found in [7] and [15].
4.3. The Friedrichs and the Kre˘ın-von Neumann extension. The Friedrichs and the
Kre˘ın-von Neumann extension can be characterized via semibounded sesquilinear forms. The
form domain generated by A ≥ 0 is denoted by dom [A]. It coincides with the completion
of domA with respect to the inner product (f, g)A = (f, g) + (Af, g), f, g ∈ domA, when
A is an operator, and with respect to the inner product ( f , g )A = (f, g) + (f I, g), where
f = {f, f I}, g = {g, gI} ∈ A, when A is a linear relation. The corresponding form is denoted
by 4A[·, ·] =: 4A[·]. Observe, that if T = X( 4A) and 4A = 4A∗ ≥ 0 then
(4.7) ran (I + T )1/2 = dom 4A1/2, ran (I − T )1/2 = ran 4A1/2.
For the Friedrichs extension AF of A one has the following result.
Theorem 4.2. Let A be a closed nonnegative relation in H. Then:
(i) dom [A] = dom [AF ];
(ii) 4A[·] ≤ AF [·] and in particular dom [AF ] ⊂ dom [ 4A] for every 4A ∈ Ext A(0,∞);
(iii) AF = { {f, f I} ∈ A∗ : f ∈ dom [A] }, in particular mulAF = mulA∗.
Moreover, AF is the only extension 4A ∈ Ext A(0,∞) for which dom 4A ⊂ dom [A].
20 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
Proof. Let T = X( 4A) = −I+2(I+A)−1. Then f = {f, f I} ∈ 4A is equivalent to {f+f I, 2f} ∈
I + T . Therefore,
(4.8) 2, f ,2A = (f + f
I, 2f) = (g, (I + T )g) = ,g,2I+T ,
where g = f + f I ∈ domT . Now (i) and the last statement of the theorem follow from
Corollary 3.7 and the identities T1 = X(A), Tm = X(AF ).
(ii) As it is well known, cf. [26], [23] (see also [15]) this assertion is equivalent to the first
inequality in (4.5).
(iii) Part (i) shows that AF ⊂ L := { {f, f I} ∈ A∗ : f ∈ dom [A]}. To prove the
reverse inclusion assume that {f, f I} ∈ L. Then f ∈ domA∗ = ran (I + T ∗1 ) and in addition
f ∈ domA1/2F = ran (I+Tm)1/2 by (4.7). Now Lemma 3.12 gives f ∈ ran (I+Tm) = domAF .
Therefore, {f, k} ∈ AF ⊂ A∗ for some k ∈ H. This implies {0, f I − k} ∈ A∗, i.e. f I − k ∈
mulA∗. It remains to prove that mulA∗ = mulAF , which then finally gives {f, f I} ∈ AF .
By construction ker (I + Tm) = (ran (I + T1))
⊥ and moreover mulAF = ker (I + Tm) and
domA = ran (I + T1), cf. (4.3). Therefore, mulAF = (domA)
⊥ = mulA∗. -
The Kre˘ın-von Neumann extension AK of A can be characterized analogously, by inter-
changing the roles of ranges and domains, cf. [26], [7], [15].
Theorem 4.3. Let A be a closed nonnegative relation in H. Then:
(i) ran [A] = ran [AK ];
(ii) AK [·] ≤ 4A[·] and in particular dom [ 4A] ⊆ dom [AK ] for every 4A ∈ Ext A(0,∞);
(iii) AK = { {f, f I} ∈ A∗ : f I ∈ ran [A] }, in particular kerAK = kerA∗.
Moreover, AK is the only extension 4A ∈ Ext A(0,∞) for which ran 4A ⊂ ran [A].
Proof. Part (i) follows from ran [A] = dom [A−1] = dom ((A−1)F )1/2 = ranA
1/2
K .
Part (ii) follows from the second inequality in (4.5), or from (ii) in Theorem 4.2 and the
fact that 0 ≤ H1 ≤ H2 is equivalent to 0 ≤ H−12 ≤ H−11 , cf. [26], [23], and [15, Theorem 1].
Similarly (iii) and the last statement are obtained from the correspondings results in
Theorem 4.2 by means of (4.6). -
The proof of the above theorem was based on Theorem 4.2 and the relations in (4.6). When
one applies the same method as was used in the proof of Theorem 4.2 and the definitions of
AF and AK in (4.4), one arrives at the following characterizations for AF and AK .
Corollary 4.4. Let f = {f, f I} ∈ A∗ and let f A = {fA, f IA} ∈ A. Then:
(i) f ∈ AF if and only if
(4.9) inf{ ,f − fA,2 + (f I − f IA, f − fA) : f A = {fA, f IA} ∈ A } = 0;
(ii) f ∈ AK if and only if
(4.10) inf{ ,f I − f IA,2 + (f I − f IA, f − fA) : f A = {fA, f IA} ∈ A } = 0.
Proof. (i) This is just a reformulation from Theorem 4.2. In fact, (4.8) implies that
2,f − fA,2 + 2(f I − f IA, f − fA) = ,g − h,2I+T ,
where g = f + f I and h = fA + f
I
A ∈ domT1 = H1. Now (i) follows from Corollary 3.7.
EXTENSION THEORY 21
(ii) Observe, that f = {f, f I} ∈ 4A if and only if {f + f I, 2f I} ∈ I − T . Then with
g = f + f I one has 2[(f I, f) + ,f I,2] = ((I − T )g, g) and this gives
2[(f I − f IA, f − fA) + ,f I − f IA,2] = ((I − T )(g − h), g − h) = ,g − h,2I−T ,
where h = fA + f
I
A ∈ H1. Hence, the statement is again obtained from Corollary 3.7. -
Corollary 4.4 gives immediately the following descriptions for AF and AK .
Corollary 4.5. Let A be a closed nonnegative relation in H. Then:
(i) The Friedrichs extension AF admits the decomposition
AF = (PA)F ⊕ ({0}⊕mulA∗),
where P is the orthogonal projection from H onto domA. In particular, if A is an
operator then the orthogonal operator part of AF is (PA)F .
(ii) The Kre˘ın-von Neumann extension AK admits the decomposition
AK = AK ⊕ (kerA∗ ⊕ {0}), A = { { P f, f I} : {f, f I} ∈ A },
where P is the orthogonal projection from H onto ranA. In particular, if kerA = {0}
then the orthogonal injective part of AK is AK.
Proof. (i) Let f I ∈ mulA∗. Then {0, f I} ∈ AF since
,0− fA,2 + (f I − f IA, 0− fA) = ,fA,2 + (f IA, fA)
and the infimum in (4.9) is achieved when {fA, f IA} = {0, 0}.
(ii) Similarly for f ∈ kerA∗ one has
,0− f IA,2 + (0− f IA, f − fA) = ,f IA,2 + (f IA, fA)
and the infimum in (4.10) is achieved when {fA, f IA} = {0, 0}. -
Remark 4.6. (a) The construction of the extension AF for a densely defined operator A
contained in Theorem 4.2 goes back to K.O. Friedrichs [21]. It was extended to the setting of
relations in [14]. The other assertions of Theorem 4.2 have been obtained by M.G. Krein [26]
for a densely defined operator A.
(b) Assertions (ii) and (iii) of Theorem 4.3 have also been discovered by M.G. Krein [26]
for a densely defined operator A.
(c) The description (4.9) of AF has been discovered by K.O. Friedrichs [21], while the
description (4.10) of AK has been obtained by T. Ando and K. Nishio [7]. These descriptions
are related to the following characterizations of dom [A] and ran [A], respectively:
f ∈ dom [A] if and only if fn → f, (A(fn − fm), fn − fm)→ 0, m, n→∞,
g ∈ ran [A] if and only if Afn → g, (A(fn − fm), fn − fm)→ 0, m,n→∞,
for some sequence fn ∈ domA.
(d) Assertion (i) in Corollary 4.5 has been obtained in [14]. In the case of a positive
definite operator A ≥ cI > 0 the equality in part (ii) of Corollary 4.5 as well as the equality
dom [AK ] = dom [A] + N0, Na := ker (A∗ − a), has been discovered in [26]. Both of these
identities have been generalized in [29, Corollary 5] to the case of a nonnegative operator
A ≥ 0 with compact inverse A−1.
Note also the following equivalence (see [29]): dom [AK ] ⊃ domA∗ if and only if the
extensions AF and AK are transversal, i.e., A∗ = AF + AK .
22 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
4.4. Limiting characterizations of AF and AK. The Friedrichs and the Kre˘ın-von Neu-
mann extensions AF and AK can be obtained as strong resolvent limits of a sequence of
extensions of A, which is useful for instance for spectral considerations. Let
(4.11) N x = { {fx, xfx} : fx ∈ Nx } ⊂ A∗, x < 0,
be a defect subspace of A∗ and define a selfadjoint extension of A by
(4.12) 4Ax = A + N x.
Proposition 4.7. Let A be a closed nonnegative relation in H and let 4Ax be defined by (4.12).
Then
(4.13) AK = s− R − lim
x↑0
4Ax, AF = s− R − lim
x↓−∞
4Ax.
Moreover, if 0 ∈ ρ (A) then AK = A + N 0, and if A is a bounded operator then AF =
A +({0}⊕mulA∗).
Proof. Let T1 = X(A) be a symmetric contraction defined by (4.1) and for x < 0 let
β = (1 − x)(1 + x)−1 ∈ R \ [−1, 1]. Then Tβ := T1 + N β(T1) = X( 4Ax) is a selfadjoint
extension of T1. Since I+Tβ = 2(I+ 4Ax)−1, the identities in (4.13) follow from Lemma 3.14:
s − lim
x↑0
2(I + 4Ax)−1 = s − lim
β↓1
(I + Tβ) = I + TM = 2(I +AK)
−1,
s − lim
x↓−∞
2(I + 4Ax)−1 = s− lim
β↑−1
(I + Tβ) = I + Tm = 2(I +AF )
−1.
(4.14)
As to the second part, observe that if 0 ∈ ρ (AK) then s − limx↑0(I + Ax)−1 = (I + A0)−1.
Similarly, if A ∈ [H] then s− limx↓−∞(I +Ax)−1 = (I +A + ({0}⊕mulA∗))−1. -
Remark 4.8. The limiting results (4.13) in Proposition 4.7 for operator extensions are
contained in [7] and the representation of AK in case 0 ∈ ρ (A) has been discovered by
M.G. Kre˘ın [26] when A is a densely defined operator.
Observe that, when A is a bounded operator the representation of AF can be also obtained
from Lemma 2.10 by using part (iii) of Theorem 4.2. Moreover, Corollary 4.5 gives also the
following orthogonal representations:
AF = PA⊕ ({0}⊕mulA∗), AK = A ⊕ N 0,
if, respectively, A is a bounded operator and 0 ∈ ρ (A). Namely, then PA, respectively
A −1 = P A−1, is automatically selfadjoint.
In [15] it is shown that the linear relation A +({0} ⊕mulA∗) is selfadjoint if and only if
domA = domA ∩ domA∗, in which case this linear relation coincides with the Friedrichs
extension AF of A, and similarly that the linear relation A +(kerA∗ ⊕ {0}) is selfadjoint if
and only if ranA = ranA ∩ ranA∗, in which case it coincides with the Kre˘ın-von Neumann
extension AN of A.
4.5. Positively closable operators. Let A be a nonnegative relation in the Hilbert space
H. Since mulA ⊂ mulA∗ = mulAF , AF is an operator if and only if A is densely defined. In
this case, each 4A ∈ Ext A(0,∞) is an operator. Moreover, if A is a nonnegative relation and
some 4A ∈ Ext A(0,∞) is an operator (so that also A is an operator), then AK is an operator.
For, if not then ker (AK + a)
−1 = mulAK , a > 0, is not trivial, but then according to (4.5)
EXTENSION THEORY 23
also ker ( 4A+ a)−1 = mul 4A is not trivial. Hence, if A is a nonnegative operator, then there
exist selfadjoint operator extensions if and only if the Kre˘ın-von Neumann extension AK is
an operator. According to [7] a nonnegative operator A is called positively closable if
(4.15) lim
n→∞
(Afn, fn) = 0 and lim
n→∞
Afn = g imply g = 0.
Note that (Afn, fn) → 0, n → ∞, implies (A(fn − fm), fn − fm) → 0, as follows from the
Cauchy-Schwarz inequality for (A·, ·), i.e., |(Afn, fm)|2 ≤ (Afn, fn)(Afm, fm). Recall that
the usual characterization associated with AK involves the limiting conditions Afn → g and
(A(fn − fm), fn − fm)→ 0, cf. Remark 4.6 (c). Clearly, if A is a nonnegative operator and
AI is the operator induced by A in HI = H8 kerA, then A is positively closable if and only
if AI is positively closable. The following interpretation for positive closability in (4.16) does
not seem to have been noted earlier in the literature.
Proposition 4.9. Let A be a nonnegative operator in H. Then for a sequence fn ∈ domA,
(4.16) lim
n→∞
(Afn, fn) = 0 and lim
n→∞
Afn = g if and only if g ∈ mulAK ,
and the following statements are equivalent:
(i) A is positively closable;
(ii) the Kre˘ın-von Neumann extension AK of A is an operator.
Moreover, if ranA is closed, then (i) and (ii) are equivalent to
(iii) (Aϕ,ϕ) = 0 =⇒ Aϕ = 0.
Proof. Assume that limn→∞(Afn, fn) = 0 and limn→∞Afn = g. Then the Cauchy-Schwarz
inequality for (A·, ·) implies that for all h ∈ domA,
|(g, h)|2 = lim
n→∞
|(Afn, h)|2 ≤ lim
n→∞
(Afn, fn)(Ah, h) = 0.
Hence, g ⊥ domA so that {0, g} ∈ A∗. Now part (ii) of Corollary 4.4 shows that {0, g} ∈ AK .
Conversely, if {0, g} ∈ AK then the existence of a sequence fn ∈ domA with the desired
limiting properties follows from (4.10). This proves (4.16).
The equivalence of (i) and (ii) is now clear, since in view of (4.16) the definition (4.15) of
positive closability of A ≥ 0 can be restated as: {0, g} ∈ AK implies g = 0.
Next assume that ranA is closed. Then limn→∞Afn = g = Aϕ for some ϕ ∈ domA and
this implies that
(Aϕ,ϕ)2 = lim
n→∞
(Afn,ϕ)2 ≤ lim
n→∞
(Afn, fn)(Aϕ,ϕ) = 0.
Therefore, the implication (4.15) reduces to (iii). -
Similarly one obtains an interpretation of the positive closability of A−1.
Proposition 4.10. Let A be a closed nonnegative relation in H with kerA = {0}. Then for
a sequence fn ∈ domA,
(4.17) lim
n→∞
(Afn, fn) = 0 and lim
n→∞
fn = f if and only if f ∈ kerAF ,
and the following statements are equivalent:
(i) A−1 is positively closable;
(ii) the Friedrichs extension AF of A satisfies kerAF = {0}.
Moreover, if domA is closed, then (i) and (ii) are equivalent to
24 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
(iii) (Aϕ,ϕ) = 0 =⇒ ϕ = 0.
Positive closability can be translated for contractions via the transformation (4.1). This
yields some further equivalent conditions given for positive closability in Proposition 4.9.
Proposition 4.11. Let A be a nonnegative operator in H and let T1 = X(A) =
D
T11
V DT11
i
be
decomposed as in Theorem 3.3. Then the positive closability of A is equivalent to each of the
following statements:
(iv) limn→∞ ,DT1ψn, = 0, limn→∞(I − T1)ψn = g =⇒ g = 0;
(v) kerDV ∩ ran (I + T11)1/2 = {0}.
Moreover, if ran (I − T1) is closed, then the positive closability of A is equivalent to
(vi) DT1ψ = 0 =⇒ (I − T1)ψ = 0.
In particular, AK is an operator if DV ∗ in (3.19) satisfies kerDV ∗ = {0}.
Proof. To see that (iv) is equivalent to (4.15), observe that with ϕn = (I + T1)ψn, ψn ∈ H1,
(Aϕn,ϕn) = ((−I + 2(I + T1)−1)ϕn,ϕn) = ((I + T1)ψn, (I − T1)ψn) = ,DT1ψn,2,
Aϕn = −ϕn + 2(I + T1)−1ϕn = (I − T1)ψn.
(4.18)
As to (v) observe, that mulAK = ker (I + TM); cf. (4.3) and (4.4). Moreover, since ker (I +
T11) = ker (I + T1) = mulA = {0}, the equivalence of the positive closability of A to (v)
follows from part (iii) of Lemma 3.10.
To see (vi) notice that, if ran (I − T1) = ranA is closed, then it follows from part (iii) of
Proposition 4.9 that the condition (iv) simplifies to (vi).
The last statement follows from (v) by taking into account (3.3) in Corollary 3.1. -
Similarly one obtains further equivalent conditions given for the positive closability of A−1
in Proposition 4.10.
Proposition 4.12. Let A be a closed nonnegative relation in H with kerA = {0} and let
T1 = X(A) =
D
T11
V DT11
i
be decomposed as in Theorem 3.3. Then the positive closability of A−1
is equivalent to each of the following statements:
(iv) limn→∞ ,DT1ψn, = 0, limn→∞(I + T1)ψn = g =⇒ g = 0;
(v) kerDV ∩ ran (I − T11)1/2 = {0}.
Moreover, if ran (I + T1) is closed, then the positive closability of A
−1 is equivalent to
(vi) DT1ψ = 0 =⇒ (I + T1)ψ = 0.
In particular, kerAF = {0} if DV ∗ in (3.19) satisfies kerDV ∗ = {0}.
Remark 4.13. The translation of positive closability for contractions yields another ap-
proach to prove Propositions 4.9, 4.10, 4.11, and 4.12. For instance, the equivalence of (ii)
in Proposition 4.9 and (iv) in Proposition 4.11 is seen as follows. Let
lim
n→∞
(I − T1)ψn = lim
n→∞
(I − TM )ψn = g.
Then for all f ∈ H,
(4.19) lim
n→∞
((I − T 2M )ψn, f) = lim
n→∞
((I − TM )ψn, (I + TM)f ) = (g, (I + TM )f),
while lim
n→∞
DT1ψn = 0 implies
(4.20) lim
n→∞
|(I − T 2M)ψn, f )| ≤ lim
n→∞
,DT1ψn, ,DTMf, = 0.
EXTENSION THEORY 25
Together (4.19) and (4.20) show that g ⊥ ran (I + TM ), i.e., g ∈ ker (I + TM ). In particular,
ker (I + TM) (= mulAK) = {0} guarantees g = 0.
Conversely, for each g ∈ ker (I + TM) it is easy to construct a sequence ψn ∈ H1 satisfying
the limiting properties in part (iv) of Proposition 4.11. In fact, in view of Lemmas 3.9, 3.10,
g =
w
(I − T11)1/2ψ
−V h
W
,
where h = (I + T11)
1/2ψ ∈ kerDV ∩ ran (I + T11)1/2. Now ran (I − T11)1/2 = H1, since
ker (I − T11) (= ker (I − T1) = kerA) = {0}. Therefore,
ψ = lim
n→∞
(I − T11)1/2ψn
for some sequence ψn ∈ H1. Then limn→∞DT11ψn = (I + T11)1/2ψ = h and consequently
lim
n→∞
,DT1ψn,2 = lim
n→∞
,DVDT11ψn,2 = ,DV h,2 = 0,
lim
n→∞
(I − T1)ψn = lim
n→∞
w
(I − T11)ψn
−V DT11ψn
W
= g.
4.6. Kre˘ın’s uniqueness result. The following uniqueness criterion results from Proposi-
tion 3.17.
Theorem 4.14. ([26]) Let A be a closed nonnegative relation in H. Then AF = AK (i.e.
Ext A(0,∞) consists of one element) if and only if for some (and hence for all) a > 0
(4.21) sup
{f,f I}∈A
|(f,ϕ)|2
(f I, f)
=∞ for every ϕ ∈ N−a \ {0}.
Proof. The result can be obtained from Proposition 4.14 by using the equivalence {f, f I} ∈ A
if and only if {f + f I, 2f} ∈ I + T1, and the identities
(T1g,ϕ) = ((T1 + I)g,ϕ) = 2(f,ϕ), ,g,2 − ,T1g,2 = 4(f I, f),
where g = f + f I and ϕ ∈ (domT1)⊥ = ker (I +A∗). -
Example 4.15. Let A be a symmetric operator in H, such that domA2 = H. Then the set
Ext A2(0,∞) consists of more than one element, and hence it is infinite. Actually, if ϕ ∈
Ni(A), i.e. {ϕ, iϕ} ∈ A∗, then clearly {ϕ,−ϕ} ∈ (A∗)2. Since the inclusion (A∗)2 ⊂ (A2)∗ is
always satisfied, it follows that {ϕ,−ϕ} ∈ (A2)∗. Observe that
|(f,ϕ)|2
(A2f, f)
=
|(Af,ϕ)|2
,Af,2 ≤ ,ϕ,
2 for every f ∈ domA2.
Thus the condition (4.21) is not satisfied and hence (A2)F W= (A2)K .
It is easily seen that A∗A, AA∗ ∈ Ext A2(0,∞), and moreover A∗A = (A2)F , but in
general AK W= AA∗. In particular, if A ≥ 0 then (AF )2 W= (A2)F = A∗A and (A2)K W= (AK)2.
Moreover, the inequality m(A2) ≥ m(A)2 holds true for the lower bounds m(A) and m(A2)
of the operators A and A2, and here the equality may also occur. For more information
concerning this and similar examples, see also [8].
It should be pointed out that even in the case domA = H it is possible that domA2 = {0}.
In [12] it is shown that each symmetric operator A1 has a restriction A, such that domA = H
and domA2 = {0}.
26 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
Remark 4.16. The conditions domA = H and n+(A) = n (A) < ∞ imply (A2)∗ = (A∗)2.
However, if n±(A) = ∞, then the inclusion dom (A∗)2 ⊂ dom (A2)∗ is in general strict.
For example, if A = −∆min is the minimal Laplace operator on L2(Ω), Ω being a bounded
domain in Rn with a regular boundary, then A∗ = −∆max is the maximal operator with
domA∗ = { f ∈ L2(Ω) : ∆f ∈ L2(Ω) } and the inclusion (A∗)2 ⊂ (A2)∗ is strict. Actually,
it is clear that dom (A∗)2 ⊂ domA∗ but dom (A2)∗ W⊂ domA∗. Indeed, by the closed graph
theorem the inclusion dom (A2)∗ ⊂ domA∗ is equivalent to the validity of the estimate
(4.22) ,∆f,L2(Ω) ≤ c
J
,∆2f,L2(Ω) + ,f,L2(Ω)
o
for all f ∈ C∞(Ω).
However, according to Ho¨rmander’s Theorem [22] the last estimate is impossible. Hence,
dom (A2)∗ W⊂ domA∗ and consequently (A∗)2 W= (A2)∗.
In fact, even the weaker inclusion ∩Nj=2dom (Aj)∗ ⊂ domA∗ does not hold, i.e., for an
arbitrary N , ∩Nj=2dom (Aj)∗ W⊂ domA∗. Otherwise one would arrive at the following estimate
(4.23) ,∆f,L2(Ω) ≤ c
^
N3
j=2
,∆jf,L2(Ω) + ,f,L2(Ω)
„
for all f ∈ C∞(Ω),
which, as it is shown in [30], is not valid.
Note also that similar conclusions are valid for an arbitrary minimal symmetric differential
operator A = P (D) in L2(Ω) with constant coefficients, which is not of the form P (D) =
P (ξ1D1+. . .+ξnDn) where (ξ1, . . . , ξn) ∈ Rn and Dj := d/dxj . Namely, this result is implied
by [30, Theorem 2] which guarantees the absence of the estimate (4.23) with ∆ replaced by
P (D), P (D) not being of the form mentioned above.
4.7. Extremal selfadjoint extensions. Denote by Ext EA(0,∞)(⊂ Ext A(0,∞)) the set of
images of the operators T ∈ Ext ET1(−1, 1) under the transformation (4.1). Since Tm and TM
belong to Ext ET1(−1, 1), see Propositions 3.18, 3.20, the extensions AF and AK belong to
Ext EA(0,∞). The aim is to describe all extensions 4A in the class Ext EA(0,∞).
The next result is well known, cf. e.g. [6], [7].
Lemma 4.17. For a closed linear operator T in H with kerT = {0} the following equivalence
holds:
Cg := sup
f∈dom T
|(f, g)|
,Tf, <∞ ⇐⇒ g ∈ ran T
∗.
In this case Cg = ,h, with h ∈ (kerT ∗)⊥ and g = T ∗h.
Proposition 4.18. [7] Let A be a closed nonnegative positively closable operator. Then
domA1/2K = dom [AK ] =
F
h ∈ H : C2h := sup
f∈domA
|(Af, h)|2
(Af, f)
<∞
k
.
Moreover, the equality ,A1/2K h, = Ch is valid for each h ∈ dom [AK ].
Proof. Without loss of generality assume that kerA = {0}. Let B = A−1 and ϕ = Af . Then
(4.24) C2h = sup
f∈domA
|(Af,h)|2
(Af, f)
= sup
ϕ∈domB
|(ϕ, h)|2
(Bϕ,ϕ)
= sup
ϕ∈dom [B]
|(ϕ, h)|2
,B1/2F ϕ,2
.
By Lemma 4.17 the conditions h ∈ ranB1/2F and Ch <∞ are equivalent and, moreover, for
each h ∈ ranB1/2F = domA
1/2
K the equality Ch = ,B
−1/2
F h, = ,A
1/2
K h, is satisfied. -
EXTENSION THEORY 27
It follows from Proposition 4.18 that
(4.25) inf{ ,A1/2K (f − h), : h ∈ domA } = 0,
for every f ∈ domA1/2K . In fact, if AK is not an operator then (4.25) still holds when
,A1/2K (f−h),2 is replaced by the corresponding quadratic form AK [f−h] := AK [f−h, f−h],
cf. also (4.10) in Corollary 4.4. Actually, one has the following result.
Proposition 4.19. For A ≥ 0 the following statements are equivalent:
(i) 4A ∈ Ext EA(0,∞);
(ii) for every {f, f I} ∈ 4A,
inf{ (f I − f IA, f − fA) : {fA, f IA} ∈ A } = 0;
(iii) the form associated with 4A satisfies 4A[f, g] = AK [f, g], f, g ∈ dom 4A1/2.
Proof. Let T = X( 4A) and let {f, f I} ∈ 4A. Then equivalently {f + f I, 2f} ∈ I + T , i.e.,
(I + T )g = 2f for g = f + f I, and moreover,
,DTg,2 = (D2T g, g) = 4(f I, f).
Thus, with {fA, f IA} ∈ A and h = fA + f IA one can write ,DT (g− h),2 = 4(f I − f IA, f − fA).
Therefore, the equivalence of (i) and (ii) is obtained from Proposition 3.20.
To see that (ii) implies (iii) assume that {f, f I} ∈ 4A. Then it follows from AK ≤ 4A that
for every ε > 0 there exists {fA, f IA} ∈ A, such that
0 ≤ AK [f − h] ≤ (f I − f IA, f − fA) < ε.
This implies that 0 ≤ (f I, f )− AK [f ] < ε. Therefore,
(4.26) 4A[f ] = (f I, f ) = AK [f ]
and (iii) follows from the polarization identity.
Conversely, assume that (iii) is satisfied. Then (4.26) holds for every {f, f I} ∈ 4A and
consequently (f I − f IA, f − fA) = AK [f − fA] with {fA, f IA} ∈ A. Now (ii) follows from the
fact that the form AK [·] associated with AK satisfies the condition in (ii). -
In the case where A ≥ 0 is a densely defined operator the results in Proposition 4.19 can
be found in the literature (cf. e.g. [8, Theorem 4.4] and the references therein).
4.8. The completion problem revisited. The extension theory presented above was built
on the (minimal) solution to the completion problem (2.1) for A0. The completion problem
for A0 can be viewed as an extension problem for the bounded nonnegative operator A1 =D
A11
A21
i
, domA1 = H1, in H = H1 ⊕ H2 and can be restated as follows: are there bounded
nonnegative selfadjoint extensions for A1? According to Proposition 2.1 the criterion is
ranA∗21 ⊂ ranA
1/2
11 . In this case, the solution Amin in (2.7) is not only minimal in the class
of all (bounded nonnegative) completions of A1, but also minimal in the wider class of all
nonnegative selfadjoint extensions ofA1, that is, Amin coincides with the Kre˘ın-von Neumann
extension AK of A1. By allowing arbitrary nonnegative selfadjoint (operator) completions,
one arrives at a completion problem for A0 which in terms of A1 can be restated as follows:
when is A1 positively closable? The next result gives a solution to this problem by means
of A11 and A21; hence it is a generalization of Proposition 2.1.
28 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
Proposition 4.20. Let A1 =
D
A11
A21
i
be a bounded nonnegative operator in H1 ⊕ H2 with
domA1 = H1 and assume that kerA11 = {0}. Then the following conditions are equivalent:
(i) A1 is positively closable;
(ii) T := A21A
−1/2
11 is a densely defined closable operator in H1;
(iii) S := A−1/211 A
∗
21 is a densely defined operator in H2.
In this case, A0 in (2.1) has a minimal completion Amin given by
Amin = AK =
w
A11 A12
A21 S∗S
W
,
and moreover the closed form corresponding to AK is tK [f, f ] = ,A1/211 f1 + Sf2,2.
Proof. (a) First the equivalence (i) ⇐⇒ (ii) will be proved. Assume that (i) holds. Let
{hn, gn} ∈ A21A−1/211 , such that {hn, gn} → {0, g} as n → ∞. Let fn = A−1/211 hn. Then
limn→∞A
1/2
11 fn = 0 and limn→∞A21fn = limn→∞ gn = g. Consequently the sequence fn
satisfies the limiting conditions in (4.15). Therefore, g = 0 by positive closability of A1.
This shows that A21A
−1/2
11 is closable.
Conversely, assume that (ii) holds and let fn ∈ H1 be a sequence with the limiting prop-
erties in (4.15). Let gn = A21fn and hn = A
1/2
11 fn. Then {hn, gn} ∈ A21A−1/211 . Moreover,
limn→∞ hn = limn→∞A
1/2
11 fn = 0 and g = limn→∞A21fn. Hence, {0, g} belongs to the clo-
sure of A21A
−1/2
11 and therefore g = 0 by the closability of A21A
−1/2
11 . This proves that A1 is
positively closable.
(b) To prove the equivalence (ii) ⇐⇒ (iii) observe that T is densely defined and T ∗ =
(A21A
−1/2
11 )
∗ = A
−1/2
11 A
∗
21 = S since A21 is bounded. Thus, if T is closable then S = T
∗ is
densely defined.
Conversely, if S is densely defined then T as a densely defined operator is closable and
clos (T ) = clos (A21A
−1/2
11 ) = S
∗.
(c) Clearly,
4A :=
w
A
1/2
11
S∗
Wp
A
1/2
11 S
Q
=
w
A11 A12
A21 S
∗S
W
is a nonnegative selfadjoint extension of A1. Moreover, it is easy to check that
ran [A1] = ran
w
A
1/2
11
S∗
W
= ran 4A1/2.
Hence, ran 4A ⊂ ran [A1]. Now Theorem 4.3 shows that 4A = AK .
The operator
p
A
1/2
11 S
Q
is closed, since A11 is bounded and S is closed. Therefore, the
closed form associated with AK is tK [f, f ] = ,A1/211 f1 + Sf2,2, f = f ⊕ f2 ∈ H. -
Example 4.21. Let the bounded nonnegative operator A1 in H1⊕H2 be given by A1 =
D
A11
I
i
with domA1 = H1 = H2. Then ranA1 is closed, since it follows from g = limn→∞A1fn,
fn ∈ domA, that
g2 := lim
n→∞
A21fn = lim
n→∞
fn, g1 := limA11fn = A11g2,
EXTENSION THEORY 29
i.e., g = A1g2. By part (iii) of Proposition 4.9 A1 is positively closable if and only if
(A1ϕ,ϕ) = 0 implies A1ϕ = 0. Clearly, this condition is now equivalent to kerA11 = {0}. In
this case S = A
−1/2
11 = S
∗ is a closed densely defined operator and AK is given by
Amin = AK =
w
A1/211
S∗
Wp
A1/211 S
Q
=
w
A11 I
I A−111
W
.
It is bounded if and only if A11 is boundedly invertible.
Example 4.22. Let A1 =
D
A11
(·,g)g
i
be a bounded nonnegative operator in H1 ⊕ H2 with
g ∈ H1 = H2 = domA1 and assume that kerA11 = {0}. Then (A1ϕ,ϕ) = 0 implies A1ϕ = 0,
but A1 need not be positively closable since ranA1 is not closed in general (compare with
Proposition 4.9).
In fact, Proposition 4.20 shows that A1 is positively closable if and only if g ∈ ranA1/211 .
In this case S = (·, g)A−1/211 g ∈ [H1], S∗ = (·, A−1/211 g)g, and
Amin = AK =
w
A11 (·, g)g
(·, g)g ,A−1/211 g,2(·, g)g
W
.
If g W∈ ranA1/211 then S = 0 {g}⊥ is not densely defined and AK is a selfadjoint relation.
5. Appendix
The aim of this appendix is to further clarify the connection between the completions of
A0 in (2.1) and the extensions of A1 =
D
A11
A21
i
. For this purpose boundary triplets and Weyl
functions will be used. For these notions the reader is referred to [17], [16], and especially
to [18, Section 3], where closely related results can be found.
Proposition 5.1. Let A1 =
D
A11
A21
i
be a bounded nonnegative operator in H = H1 ⊕ H2 with
domA1 = H1 and let
4A0 =
w
A11 A∗21
A21 A22
W
be a bounded selfadjoint (not necessarily nonnegative) operator extension of A1. Then:
(i) A1 has equal defect numbers (d, d), d = dimH2 ≤ ∞.
(ii) The adjoint linear relation A∗1 of A1 is given by
A∗1 = { f = {f, 4A0f + h} : f ∈ H, h ∈ H2 }.
(iii) A boundary triplet for A∗1 is given by Π = {H2,Γ0,Γ1}, where
Γ0 f = −h, Γ1 f = f2; f = f1 ⊕ f2 ∈ H, f = {f, 4A0f + h} ∈ A∗1.
(iv) The corresponding γ-field is γ(λ) = ( 4A− λ)−1H2 and the Weyl function is given by
M(λ) = P2( 4A0 − λ)−1H2 = DA22 − z − A21(A11 − z)−1A∗21
i−1
,
where P2 stands for the orthogonal projection onto H2.
(v) The selfadjoint extensions 4Aτ = ker (Γ0 + τΓ1) of A1 in H are in one to one corre-
spondence with the selfadjoint relations τ in H2 via
4Aτ =
w
A11 A12
A21 A22 + τ
W
,
30 SEPPO HASSI, MARK MALAMUD, AND HENK DE SNOO
and their resolvents are connected by
( 4Aτ − λ)−1 = ( 4A0 − λ)−1 − γ(λ)(M(λ) + τ−1)−1γ(λ¯)∗, λ ∈ ρ( 4Aτ ) ∩ ρ( 4A0).
Proof. (i) It is easy to check that Nλ(A∗1) = ker (A
∗
1−λ) is equal to ( 4A0−λ)−1H2, λ ∈ ρ( 4A0).
Hence, d+ = d− = dimNλ(A∗1) = dimH2, λ ∈ ρ (A1).
(ii) The form of the adjoint A∗1 is obtained from Lemma 2.10.
(iii) Let f = {f, f I} = {f, 4A0f + h} ∈ A∗1 and g = {g, gI} = {g, 4A0g + k} ∈ A∗1. Then
(f I, g)− (f, gI) = ( 4A0f + h, g) − (f, 4A0g + k) = (h, g)− (f, k) = (Γ1 f ,Γ0 g )− (Γ0 f ,Γ1 g ),
so that the abstract Green’s identity holds. The surjectivity of the mapping Γ := (Γ0,Γ1) :
A∗1 → H2 ⊕H2 is clear.
(iv) Each f λ ∈ N λ(A∗1) = { {fλ,λfλ} : fλ ∈ Nλ(A∗1) } can be represented as
{fλ,λfλ} = {( 4A0 − λ)−1h,λ( 4A0 − λ)−1h} = {( 4A0 − λ)−1h, 4A0( 4A0 − λ)−1h− h}.
Hence,
Γ0 f λ = h, Γ1 f λ = P2( 4A0 − λ)−1h,
which gives the indicated form for γ(λ) and M(λ).
(v) Since τ is a linear relation the set ker (Γ0 + τΓ1) is equal to
{ f ∈ A∗1 : {Γ1 f ,−Γ0 f } ∈ τ } = { f ∈ A∗1 : {f2, h} ∈ τ },
which implies the desired form for 4Aτ = ker (Γ0 + τΓ1). The last statement is just Kre˘ın’s
resolvent formula. -
Passing to the boundary triplet 4Π = {H2,Γ1,−Γ0} one obtains another Weyl function for
A1 which is of the form
(5.1) MF (λ) = −M(λ)−1 = −A22 + z +A21(A11 − z)−1A∗21.
In fact, MF (λ) is associated with the Friedrichs extension of A1, i.e., kerΓ1 = AF .
Proposition 5.2. Let the assumptions and notations be as in Proposition 5.1, let MF be
given by (5.1), let kerA11 = {0}, and let S = A−1/211 A∗21. Then:
(i) there exists the strong resolvent limit
MF (0) := s−R − lim
x↑0
MF (x) = S
∗S − A22;
(ii) the Friedrichs and the Kre˘ın-von Neumann extension of A1 are given by
AF = kerΓ1 =
w
A11 A12
A21 {0}⊕H2
W
, AK = ker (Γ0 +MF (0)Γ1) =
w
A11 A12
A21 MF (0)
W
.
Proof. (i) It is easy to see that
lim
x↑0
(A21(A11 + x)
−1A∗21h, h) = lim
x↑0
((A11 + x)
−1A∗21h,A
∗
21h) = (Sh, Sh).
This implies that the linear relation S∗S associated with the closed nonnegative form (Sh, Sh)
is the strong resolvent limit s−R− limx↑0A21(A11+x)−1A∗21 = S∗S, see [18, Proposition 4.1].
Now (5.1) shows that s− R − limx↑0MF (x) = S∗S − A22.
(ii) The statement for AF follows directly from Proposition 4.7, since AF = ker (Γ0+ τΓ1)
must correspond to the selfadjoint relation τ = {0} ⊕ H2. Moreover, for x < 0 the proof
EXTENSION THEORY 31
of Proposition 5.1 shows that f ∈ 4Ax = A + N x if and only if {Γ0 f ,Γ1 f } ∈ M(x).
Equivalently, {Γ1 f ,−Γ0 f } ∈MF (x), i.e., f ∈ ker (Γ0 +MF (x)Γ1). Now (4.13) implies
AK = s −R − lim
x↑0
4Ax = ker (Γ0 +MF (0)Γ1),
where the second identity can be shown with the aid of Kre˘ın’s formula, cf. [18, p. 172]. -
Corollary 5.3. Let A1 be as in Proposition 5.1 with kerA11 = {0} and let S = A−1/211 A∗21.
Then the Friedrichs and the Kre˘ın-von Neumann extension AF and AK of A1 satisfy the
following statements:
(i) AK = AF if and only if MF (0) = {0}⊕ H2, or equivalently, domS = {0};
(ii) AK is an operator if and only if MF (0) is an operator in H2, or equivalently, S is a
densely defined operator in H2;
(iii) AK is a bounded operator if and only if MF (0) is a bounded operator in H2, or
equivalently, S ∈ [H2].
Observe, that if A22 = 0 then MF (0) is associated with the closed nonnegative form
t[h, h] = (Sh, Sh) = lim
x↑0
(MF (x)h, h),
where
dom t = domS = {h ∈ H2 : lim
x↑0
(MF (x)h, h) <∞ }, (dom t)⊥ = H28domS = mulMF (0).
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Department of Mathematics and Statistics, University of Vaasa, P.O. Box 700, 65101
Vaasa, Finland
E-mail address : sha@uwasa.fi
Department of Mathematics, Donetsk National University, Universitetskaya str. 24,
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E-mail address : mmm@univ.donetsk.ua
Department of Mathematics, University of Groningen, Postbus 800, 9700 AV Groningen,
Nederland
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