An asymmetric Putnam{Fuglede theorem
for unbounded operators
Jan Stochel
IMUJ PREPRINT 1999/25
Abstract. The intertwining relations between cosubnormal and closed hyponormal (resp. cohyponormal and closed subnormal) operators are studied.
In particular, an asymmetric Putnam{Fuglede theorem for unbounded operators is proved.
Introduction
The Putnam{Fuglede theorem [12] says that if a bounded linear mapping inter-
twines two normal operators, then it intertwines their adjoints. Unfortunately, the
Putnam{Fuglede theorem is no longer true for subnormal operators. However, as
noticed by Furuta (cf. [2, 3]), an asymmetric version of it remains true for bounded
subnormals; furthermore, subnormals can be replaced by more general bounded operators (cf. [17, 18, 14, 15, 8, 21]). In particular, the following holds true: if a
bounded linear mapping intertwines a cohyponormal operator with a hyponormal
one (both are bounded), then it intertwines their adjoints. In this paper we prove a
decomposition theorem (see Theorem 4.2) for intertwining relations between cosubnormal (resp. cohyponormal) and closed hyponormal (resp. subnormal) unbounded
operators (cf. [17, 8] for the bounded case). Contrary to what has been shown
before (cf. [11, 7, 10]), here the operators in question have arbitrary domains and
their spectra may cover the whole complex plane. We also prove an asymmetric
1991 Mathematics Subject Classication. Primary 47B20; Secondary 47B15.
Key words and phrases. Normal operator, subnormal operator, hyponormal operator, intertwining relation.
This work was supported by the KBN grant # 2P03A 041 10.
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JAN STOCHEL
version of the Putnam{Fuglede theorem for couples of unbounded operators whose
rst component is cosubnormal (resp. cohyponormal) while the other one is closed
hyponormal (resp. subnormal).
1. Preliminaries
All linear spaces taken into consideration in this paper are assumed to be
complex. From now on the symbols H and K stand for Hilbert spaces. Given a
linear mapping A : H D(A) ! K, we denote by N (A), R(A), G (A) and A the
kernel, the range, the graph and the adjoint of A, respectively. As usual AjE stands
for the restriction of A to a linear subspace E of D(A); E is said to be a core for A
if G (A) G (AjE ). Denote by B (H; K) the set of all bounded linear mappings from
H into K; for simplicity we write B (H) instead of B (H; H). Set jX j = (X X )1=2
for X 2 B (H; K).
Given an operator A in H (read: a linear mapping A : H D(A) ! H), we
denote by (A) the spectrum of A. A is said to be pure if the only closed linear
subspace of H reducing A to a normal operator is f0g. If L is a closed linear
subspace of H, then AL stands for the operator in L dened by
D(AL ) = ff 2 L \ D(A) : Af 2 Lg and AL f = Af; f 2 D(AL ):
If L reduces A to the operator B , then B = A L . It may happen that A L is
densely dened in L though L is not invariant for A, i.e. A(L \ D(A)) * L.
Example 1.1. Let f1 be a normalized vector in H. Set L = ff1 g? . Let A0
be an unbounded closed densely dened operator in L. Take a normalized vector
e1 2 L n D(A0 ). Dene the operator A in H by: D(A) = D(A0 ) u C e1 u C f1
and A(h + e1 + f1 ) = A0 h + f1 for h 2 D(A0 ) and ; 2 C . It is clear that
A is closed and densely dened. Moreover, one can check that D(AL ) = D(A0 )
D(A0 ) u C e1 = L \ D(A) D(A), e1 2 L \ D(A) and Ae1 = f1 2= L .
A densely dened operator A in H is said to be hyponormal if D(A) D(A )
and kA f k kAf k for f 2 D(A). We say that a densely dened operator S in H
is subnormal if there is a Hilbert space L H and a normal operator N in L such
that S N . It is well{known that each subnormal operator is hyponormal and
that each hyponormal operator is closable (see [4, 11, 5, 6] for more details).
We begin our considerations by proving a fact which is well{known for bounded
hyponormal operators (cf. [1]; see also [5, Proposition 1] for the reverse result).
Proposition 1.2. If T is a hyponormal operator in H, then there exists a
contraction C 2 B (H) such that T T C .
Proof. Since kT f k kTf k for f 2 D(T ) D(T ), there exists a contraction
0
D 2 B (R(T ); R(T )) such that D0 T T . Let D 2 B (H) be any contraction
which extends D0 (e.g. set Df = 0 for f 2 H R(T )). Then DT T . Taking
adjoints in the last inclusion and exploiting the closability of T , we get T T (DT ) = T D . This gives us the conclusion with C = D .
The next result oers a sucient condition for a hyponormal operator to be
reduced by a closed subspace.
Proposition 1.3. Let L be a closed linear subspace of H and let T be a closed
hyponormal operator in H such that T L is densely dened in L. Then
AN ASYMMETRIC PUTNAM{FUGLEDE THEOREM
3
(i) T L is a closed hyponormal operator in L,
(ii) L reduces T to T L provided T L is normal.
Proof. (i). Since T is closed, it is easily seen that the operator S := T L is
closed as well. We now show that
(1.1)
PT jD(S ) S ;
where P is the orthogonal projection of H onto L. Indeed, by the hyponormality
of T , we have D(S ) D(T ), so hf; Sgi = hT f; gi = hPT f; gi for f; g 2 D(S ).
This, in turn, implies (1.1).
Using (1.1) and the hyponormality of T we get
(1.2)
kS f k = kPT f k kT f k kTf k = kSf k; f 2 D(S ):
Hence S is hyponormal.
(ii). It follows from the normality of S and (1.2) that kPT f k = kT f k for
f 2 D(S ). Hence T (D(S )) L and, by (1.1), we have S f = T f for f 2 D(S ).
Applying the latter equality, we obtain
hPf; S gi = hf; S gi = hf; T gi = hTf; gi = hPTf; gi; f 2 D(T ); g 2 D(S );
which implies P (D(T )) D(S ) = D(S ) D(T ) and TPf = SPf = S Pf =
PTf for f 2 D(T ). Thus PT TP which shows that L reduces T to S .
Part (ii) of Proposition 1.3 is no longer true for arbitrary operators T .
Example 1.4. Let e0 be a normalized vector in H. Set L = fe0 g? . Take a
normalized vector f0 in L. Dene pT 2 B (H) by Th = h ? hh; e0i(e0 + f0 ) for
h 2 H. One can show that kT k = 2 (use the fact that H = L u C (e0 + f0 )).
Since Th = h for h 2 L, we conclude that T L is normal. However L does not
reduce T as Te0 = ?f0 .
Our next goal is to characterize linear mappings T satisfying the condition
D(T ) R(T ). Let us begin with
Lemma 1.5. If T : H D(T ) ! K is a closed densely dened linear mapping,
then H = D(T ) + R(T ).
Proof. It is known that H K = G (T ) W G (T ) (cf. [20, Theorem 4.16]),
where W : K H ! H K is given by W (k; h) = (?h; k) for h 2 H and k 2 K. If
f 2 H, then there exist h 2 D(T ) and k 2 D(T ) such that
(f; 0) = (h; Th) W (k; T k) = (h ? T k; Th + k);
so f = h ? T k 2 D(T ) + R(T ).
It is easily seen that Lemma 1.5 is no longer true for operators which are not
closed (set Th = 0 for h 2 D, where D is a dense proper linear subspace of H).
Corollary 1.6 ([9, Theorem 3.3]). If T : H D(T ) ! K is a closed densely
dened linear mapping, then R(T ) D(T ) if and only if T 2 B (H; K).
Proof. If R(T ) D(T ), then Lemma 1.5 yields D(T ) = H, so by the closed
graph theorem T 2 B (H; K).
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JAN STOCHEL
Let T : H D(T ) ! K be a linear mapping. Dene a new linear mapping
~T : H D(T~) ! R(T ) by D(T~) = D(T ) and Th
~ = Th for h 2 D(T~). It is clear that
~
~
~
N (T ) = N (T ), R(T ) = R(T ), G (T ) = G (T ) and T~ = QT T , where QT : K ! R(T )
is the orthogonal projection of K onto R(T ). If T is densely dened, then obviously
N (T~ ) = f0g.
Theorem 1.7. If T : H D(T ) ! K is a closed densely dened linear mapping and H =
6 f0g, then the following conditions are equivalent
(i) D(T ) R(T ),
(ii) R(T ) = H,
ij
(iii) N (T ) D(T ) and supf 2D(T )nN (T ) jhkh;f
Tf k < 1 for every h 2 H,
(iv) T~?1 2 B (H; R(T )),
(v) N (T ) = f0g and T~?1 2 B (R(T ); H).
Proof. There is no loss of generality in assuming that T = T~ and R(T ) = K
because, by [16, Theorem 1], R(T ) = R(T~ ).
(i))(ii). By Lemma 1.5, we have H = D(T )+ R(T ) R(T ), so R(T ) = H.
(ii),(iii). One can show that T satises (iii) if and only if for every h 2 H there
exists ch 0 such that jhh; f ij ch kTf k for all f 2 D(T ). By [16, Theorem 1]
this is equivalent to R(T ) = H.
(ii))(iv). Since T ?1 is a closed linear mapping with D(T ?1 ) = H, the closed
graph theorem yields T ?1 2 B (H; K).
The implications (iv))(ii) and (ii))(i) are obvious.
(i),(v). According to (i))(ii), N (T ) = f0g whenever D(T ) R(T ). Thus
there is no loss of generality in assuming that N (T ) = f0g. Then T ?1 is a closed
densely dened linear mapping and (T ?1) = T ?1 . Hence D((T ?1 ) ) = R(T )
and R(T ?1) = D(T ). Applying Corollary 1.6 to T ?1 we get (i),(v).
Corollary 1.8. Let T : H D(T ) ! K be a closed densely dened linear
mapping. If R(T ) = K (resp. N (T ) = f0g), then the following conditions are
equivalent
(i) D(T ) R(T ) (resp. D(T ) R(T )),
(ii) R(T ) = H (resp. R(T ) = K),
(iii) N (T ) = f0g and R(T ) = K or equivalently R(T ) = K and R(T ) = H,
(iv) N (T ) = f0g and T ?1 2 B (K; H).
Moreover, if K = H, N (T ) = f0g and R(T ) = H, then D(T ) R(T ) if and only
if D(T ) R(T ) if and only if 0 2= (T ).
2. Intertwining normal with hyponormal operators
In this section we investigate intertwining relations between normal and hyponormal operators. We begin with the question of preserving cores by selfadjoint
intertwining operators (see [11, Theorem 2.5] for a related result).
Proposition 2.1. Let E be a core for a closed densely dened operator A in
H and let X 2 B (H) be a selfadjoint operator with N (X ) = f0g. If XA AX ,
then X (E ) is a core for A.
AN ASYMMETRIC PUTNAM{FUGLEDE THEOREM
5
Proof. Take f 2 D(A) such that f Af is orthogonal in HH to Xg AXg
for every g 2 E . Then, according to X = X and XA AX , we see that
hXf; gi = hf; Xgi = ? hAf; AXgi
= ? hAf; XAgi
= ? hXAf; Agi
= ? hAXf; Agi; g 2 E :
Since E is a core for A, we conclude that AXf 2 D(A ) and A AXf = ?Xf .
However A A is positive and selfadjoint, so Xf = 0. Since N (X ) = f0g, we get
f = 0. This completes the proof.
The next result can be extracted from [13, Theorem 1] (it is easily seen that
[13, Theorem 1] remains true for unbounded normal operators).
Theorem 2.2. Let E beTthe spectral measure of a normal operator N . If is
a Borel subset of C and f 2 z2 R(z ? N ), then E ()f = 0.
We are now in a position to prove the crucial for our further investigations
result. The idea of its proof is taken from [17, Theorem 1]).
Theorem 2.3. Assume that N is a normal operator in H, T is a closed hyponormal operator in K and X 2 B (H; K) is such that XN TX . Then
(i) jX jN N jX j,
(ii) N = T provided K = H, X 0 and N (X ) = f0g.
Proof. (i). Let E be the spectral measure of N . Take a compact subset of
C and f 2 H := R(E ()). The space H reduces ? N and H D( ? N ) for
2 C . Dene the function ' : C n ?! H by
Z
'() = ?1 z E (z.)f = ( ? N H )?1 f; 2= :
Then
(2.1)
( ? N )'() = f; 2= :
According to XN TX , we have
(2.2)
X ( ? N ) ( ? T )X; 2 C :
Combining (2.1) and (2.2), we get Xf = X ( ? N )'() = ( ? T )X'() for 2= .
Hence
(2.3)
X Xf = X ( ? T )X'(); 2= :
Since ? T is hyponormal, we conclude from Proposition 1.2 that there exists a
contraction C 2 B (K) such that
(2.4)
? T ( ? T ) C ; 2 C :
Taking adjoints in (2.2), we get
(2.5)
X ( ? T ) ( ? N ) X ; 2 C :
It follows from (2.3), (2.4) and (2.5) that
X Xf = X ( ? T )X'() = X ( ? T ) C X'()
= ( ? N ) X CX'(); 2= :
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JAN STOCHEL
T
Hence X Xf belongs to z2Cn R(z ? N ), where = fz : z 2 g. According
to Theorem 2.2, we have E (C n )X Xf = 0 (because the spectral measure F of
N is related to E via F ( ) = E ( ) for every Borel subset of C ) or equivalently
X Xf = E ()X Xf . Since f was an arbitrary element of H , we conclude that
X X (H ) H . However X X 0, so X XE () = E ()X X . This, in turn,
implies that jX jE () = E ()jX j for every compact subset of C . Since E is a
regular measure, the latter equality holds for every Borel subset of C , which
implies (i).
(ii). Since X 0, we conclude from (i) that NXf = XNf = TXf for
f 2 D(N ). Thus
(2.6)
N jX D(N ) T:
However, according to Proposition 2.1, X D(N ) is a core for N , so taking closures
in (2.6) leads to N T . Hence T N and, by the hyponormality of T , we have
D(T ) D(T ) D(N ) = D(N ). This gives us N = T .
3. More on intertwining relations
In this section we continue the study of intertwining relations. Let us rst
collect notations and denitions which play the basic role in the rest of the paper.
Given X 2 B (H; K), we set I (X ) := R(X ) = R(jX j) and F (X ) := R(X ).
Let X = X jX j be the polar decomposition of X , i.e. X 2 B (H; K) is the partial
isometry with N ( X ) = N (X ), which maps isometrically I (X ) onto F (X ). Denote by X and Xb the bounded linear mappings from I (X ) into F (X ) dened by
b := X jI (X ) . Then X is a unique unitary isomorphism in
X := X jI (X ) and X
B (I (X ); F (X )) such that
(3.1)
X jX jf = Xf; f 2 H:
Lemma 3.1. If X 2 B (H; K), then
(i) N (Xb ) = f0g,
(ii) R(X ) = R(Xb ),
(iii) Xc = Xb ,
(iv) d
jX j = jXb j,
(v) Xc = Xc = X .
Proof. The proof of (i), (ii) and (iii) is left to the reader.
(iv). Since I (X ) = I (jX j) reduces jX j and Xc = Xb , we get jXb j = (Xc Xb )1=2 =
?
(X X ) jI (X ) 1=2 = jX jjI (X ) = jd
X j, which proves (iv).
(v). It follows from (i) and (ii) that
(3.2)
I (X ) = I (Xb ) and F (X ) = F (Xb ):
jX jf = X jX jf = Xf =
On the other hand, by (iv) and (3.1), we have X jXb jf = X d
b for f 2 I (X ). This implies that X
Xf
c = X due to the uniqueness of Xc. The
equality Xc = Xc is a consequence of (3.2).
If A and B are operators in H and K, respectively, and X 2 B (H; K), then we
set AX := AI (X ) and B X := BF (X ) . To shorten notation, we write AX instead of
(AX ) . In case X intertwines A and B , AX and B X have the following properties.
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AN ASYMMETRIC PUTNAM{FUGLEDE THEOREM
7
Theorem 3.2. Let A and B be closed densely dened operators in H and K,
respectively, and let X 2 B (H; K) be such that XA BX .
(i) If I (X ) reduces A, then B X is closed densely dened in F (X ) and
b B X X:
b
XA
X
(ii) If I (X ) and F (X ) reduce A and B to normal operators, respectively, then
XA B X; jX jA AjX j; jX jB B jX j;
(3.4)
AX = X B X X :
Proof. (i). The closedness of B X follows directly from that of B . We now
show that
(3.5)
Xb (D(AX )) X (D(A )) D(B X ):
(3.3)
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Indeed, since AX = AI (X ) , we get Xb (D(AX )) = X (D(A ) \ I (X )) X (D(A )).
According to XA BX , we have X (D(A )) D(B )\F (X ) and BXf = XA f 2
F (X ) for f 2 D(A ). Hence X (D(A )) D(B X ), which proves (3.5).
Since the operator A is densely dened in H, we conclude from (3.5) that
F (X ) = X (D(A )) = X (D(A )) D(B X ), which shows that B X is densely dened
in F (X ). Applying once more the equality AX = A I (X ) and (3.5), we obtain
b for f 2 D(A ). This completes the proof of (i).
b f = XA f = BXf = B X Xf
XA
X
X
(ii). Applying the Putnam{Fuglede theorem (cf. [12, Lemma]) to (3.3), we
obtain
b X (B X ) X:
b
(3.6)
XA
jXb jAX (B X ) X jXb j, or equivalently
?
jXb jAX X (B X ) X jXb j:
Therefore, by Lemma 3.1(v), we have
(3.7)
uX
u
u
u
Since, according to Lemma 3.1(i)&(iv), N (jXb j) = N (d
jX j) = f0g, we conclude from
(3.7) via Theorem 2.3 that AX = X (B X ) X (which implies (3.4)) and
u
(3.8)
u
jXb jAX AX jXb j:
As X = Xb 0, A = AX A0 and B = B X B 0 with A0 = AN (X ) and B 0 = BN (X ) ,
we deduce from (3.6) that
b X 0 (B X ) X
b 0 = B X:
XA XA
It follows from Lemma 3.1(iv) that jX j = jXb j 0. Hence (3.8) leads to
(3.9)
jX jA jXb jAX 0 AX jXb j 0 = AjX j:
Notice that XA BX implies X B AX . Since I (X ) and F (X ) reduce B
and A to normal operators, respectively, we can replace the linear mappings X and
A in (3.9) by X and B , respectively, to get jX jB B jX j. This completes the
proof.
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JAN STOCHEL
4. Intertwining cohyponormal with hyponormal operators
In this section we describe the structure of intertwining relations between cosubnormal and hyponormal (as well as between cohyponormal and subnormal)
operators. In particular we obtain an asymmetric Putnam{Fuglede theorem for
unbounded operators. First we consider a special case of intertwining relations.
Proposition 4.1. Assume that N is a normal operator in H, T is a closed
hyponormal operator in K and X 2 B (H; K) is such that XN TX . Then I (X )
and F (X ) reduce N and T to normal operators, respectively.
Proof. It follows from Theorem 2.3(i) that jX jE () = E ()jX j for every
Borel subset of C , E being the spectral measure of N . Since I (X ) = R(jX j),
we conclude that I (X ) reduces E and, consequently, N as well. According to
Theorem 3.2(i), the operator T X is densely dened in F (X ) and
b X T X X:
b
(4.1)
XN
By Proposition 1.3(i), T X is a closed hyponormal operator in F (X ). Applying
Lemma 3.1(v) to (4.1), we get jXb jNX ( X T X X )jXb j. Since N (jXb j) = f0g and
X T X X is a closed hyponormal operator, Theorem 2.3(ii) yields NX = X T X X .
Hence T X , being unitarily equivalent to the normal operator NX , is normal as well.
By Proposition 1.3(ii), F (X ) reduces T to a normal operator. This completes the
proof.
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We are now in a position to prove the main result of the paper.
Theorem 4.2. Assume that A is a closed subnormal (resp. a closed hyponormal ) operator in H, B is a closed hyponormal (resp. a closed subnormal ) operator
in K and X 2 B (H; K) is such that XA BX . Then
(i) XA B X ,
(ii) jX jA AjX j,
(iii) jX jB B jX j,
(iv) I (X ) reduces A to the normal operator AX ,
(v) F (X ) reduces B to the normal operator B X ,
(vi) AX = X B X X .
Proof. Case 1. A is subnormal and B is hyponormal.
Let N be a normal extension of A acting in a Hilbert space L H. Dene
Y 2 B (K; L) by Y f = X f for f 2 K. We show that
(4.2)
F (X ) = F (Y ):
Indeed, since Y = JX , where J 2 B (H; L) is dened by Jh = h for h 2 H, we
get Y = XP , where P 2 B (L; H) is the orthogonal projection of L onto H. The
latter equality implies (4.2).
(v). Taking adjoints in XA BX , we obtain
(4.3)
X B AX ;
which implies Y B NY . This, in turn, forces Y N BY . Hence, by (4.2)
and Proposition 4.1, we come to the conclusion that F (X ) = F (Y ) reduces B to
the normal operator B X .
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AN ASYMMETRIC PUTNAM{FUGLEDE THEOREM
9
(iv). Applying Theorem 3.2(i) to (4.3) and Lemma 3.1(iii), we obtain AX =
A(X ) is a closed densely dened operator in I (X ) and
(4.4)
Xb (B X ) AX Xb :
Since AX A N , we see that AX is subnormal and, in consequence, hyponormal.
It follows from Proposition 4.1 applied to (4.4) that I (X ) = F (Xb ) reduces AX
to a normal operator. Hence AX is normal itself. By Proposition 1.3(ii), I (X )
reduces A to the normal operator AX .
The rest of the conclusion follows directly from Theorem 3.2(ii) applied to the
intertwining relation XA BX .
Case 2. A is hyponormal and B is subnormal.
It follows from XA BX that X B AX . According to Case 1, I (X ) =
F (X ) reduces A to the normal operator AX and F (X ) = I (X ) reduces B to
the normal operator B X . The rest of the conclusion follows once more from Theorem 3.2(ii) applied to the relation XA BX .
5. Concluding remarks and comments
We begin with a result which can be inferred from Theorem 4.2 via the implications:
a) N (X ) = f0g ) I (X ) = H,
b) N (X ) = f0g ) F (X ) = K,
c) (K = H ^ X 0 ^ N (X ) = f0g) ) X is the identity operator on H,
d) X 6= 0 ) (I (X ) 6= f0g ^ F (X ) 6= f0g).
Corollary 5.1. Assume that A is a closed subnormal (resp. a closed hyponormal ) operator in H, B is a closed hyponormal (resp. a closed subnormal )
operator in K and X 2 B (H; K) is such that XA BX . Then
(i) A is normal provided N (X ) = f0g,
(ii) B is normal provided N (X ) = f0g,
(iii) B and A are normal and unitarily equivalent provided N (X ) = f0g and
N (X ) = f0g,
(iv) B = A and A is normal provided K = H, X 0 and N (X ) = f0g,
(v) neither A nor B is pure provided X 6= 0.
Let us make some comments concerning Corollary 5.1: part (ii) of it extends [7,
Theorem 3] (the assumption (T ) 6= C can be dropped); part (iii) generalizes [11,
Theorem 4.1], [11, Proposition 4.2], [11, Corollaries 4.4 and 4.6], [7, Corollary 4]
(the assumption (N ) 6= C can be dropped), [7, Corollary 9] (the assumption
D(Sj ) = D(Sj ), j = 1; 2, can be dropped) and [10, Theorem 3.3] (the operator
C does coincides with A); part (iv) generalizes [11, Theorem 4.5], [7, Theorem 8]
(the assumptions D(Sj ) = D(Sj ), j = 1; 2, and (S1 ) 6= C can be dropped) and
[10, Proposition 3.2] (the operator B does coincides with A); nally, part (v) has
been applied in [19, Lemma 8.6] to prove that the strong symmetric commutant of
a closed hyponormal operator is equal to the orthogonal sum of strong symmetric
commutants of its normal and pure parts.
In particular part (iii) of Corollary 5.1 implies that a closed hyponormal operator A which is quasi{similar to a selfadjoint operator B is selfadjoint itself and
unitarily equivalent to B (see [11] for the denition of quasi{similarity and Corollary 4.6 therein for a related result).
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The last result of the paper is related to Remark 2.6 following [11, Theorem
2.5]. We show that the assumption imposed on the operator B in [11, Remark 2.6]
is not weaker than that in [11, Theorem 2.5].
Theorem 5.2. Let A and B be closed densely dened operators in H and K,
respectively, and let X 2 B (H; K) be such that XA BX . Assume that R(A) = H
and R(X ) = K. Then
(i) R(B ) = K,
(ii) D(B ) R(B ) if and only if 0 2= (B ),
(iii) 0 2= (A) provided 0 2= (B ) and X ?1 2 B (K; H).
Proof. Since R(X ) = K and R(A) = H, we conclude that R(XA) = K. This
and XA BX lead to R(B ) = K. Hence (ii) follows from Corollary 1.8.
Assume that 0 2= (B ) and X ?1 2 B (K; H). By XA BX , we have Af =
?
1
X XAf = X ?1 BXf for f 2 D(A), so
(5.1)
kAf k = kX ?1 BXf k kX k?1 kBXf k
(5.2)
(kX kkB ?1 k)?1 kXf k
(5.3)
(kX kkB ?1 kkX ?1 k)?1 kf k; f 2 D(A):
This means that A is bounded from below and R(A) = H. Therefore A?1 2 B (H),
which completes the proof of (iii).
We close the paper with an example which shows that the assumption X ?1 2
B (K; H) could not be removed from part (iii) of Theorem 5.2.
Example 5.3. Let H be a separable Hilbert space with an orthonormal basis
fengn2Zand let E be the linear span of fen gn2Z(Z= f: : :; ?1; 0; 1; : :: g). Dene
the sequence = fngn2Z
by
8
>
n 0;
<1
n = >1=n n 2 2 Z; n 1;
:
1=n2 n 2 2 Z+ 1; n 1:
Let B and D be the members of B (H) determined by the conditions Ben = en+1
and D en = n en (n 2 Z), and let X := B 2 D (2 B (H)). Then N (B ) = f0g,
B ?1 2 B (H) and N (X ) = f0g because N (D ) = f0g. Since Xen = n en+2 for
n 2 Z, we get E R(X ) and consequently R(X ) = H. Set A = BD , where D
is a unique normal operator in H determined by the condition D en = nen with
n = n =n+1 (n 2 Z). The operator A is closed densely dened and unbounded;
moreover N (A) = f0g. Since Aen = nen+1 for n 2 Z, we get E R(A) and
A?1 en = n??1 1en?1 for n 2 Z. Therefore R(A) = H and the operator A?1 is
unbounded (because the sequence f?n 1gn2Zis unbounded). One can check that
XAen = BXen for n 2 Z, hence that X (AjE ) BX and nally that XA BX
(because E is a core for A and B is closed). Clearly, the operator X ?1 is unbounded.
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Instytut Matematyki, Uniwersytet Jagiellonski, Krako w
E-mail address :
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