M. ABSOLUTE CONTINUITY AND OPERATORS HYPONORMAL

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I. J. Math. M. Sei.
Vol. 4 No. 2 (1981) 321-335
321
ABSOLUTE CONTINUITY AND HYPONORMAL OPERATORS
C.R. PUTNAM
Department of Mathematics
Purdue University
West Lafayette, Indiana 47907
(Received July 18, 1980)
Let
T
representation
T
ABSTRACT.
eigenvalue of
unitary.
T*
be a completely hyponormal operator, with the rectangular
A + iB, on a separable Hilbert space.
then
T
It is known that
also has a polar factorization
A, B
and
U
If
T
0
is not an
UP,
with
U
are all absolutely continuous
operators.
Conversely, given an arbitrary absolutely continuous selfadjoint
or unitary
U, it is shown that there exists a corresponding completely
hyponormal operator as above.
A
It is then shown that these ideas can be used to
establish certain known absolute continuity properties of various unitary
operators by an appeal to a lemma in which, in one interpretation, a given
unitary operator is regarded as a polar factor of some completely hyponormal
The unitary operators in question are chosen from a number of sources:
operator.
the F. and M. Riesz theorem, dissipative and certain mixing transformations in
ergodic theory, unitary dilation theory, and minimal normal extensions of subnormal contractions."
KEY WORDS AND PHRASES. Szlfadjoint operators, uLtary op6rors, hyponormal
operators, zrgodi theory, uitary dilations, subnormal opors.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES:
26A46, 28AI0, 47A35, 47A55, 47B47.
Primary:
47B15, 47B20;
Seaonday:
i.
INTRODUCTION.
Let
T
be a bounded operator, with the rectangular decomposition
322
C.R. PUTNAM
A + iB,
T
on a complex, separable Hilbert space
u (T)
and point spectrum by
u (T)
P
and
hyponormal if
T*T
TT*
and completely hyponormal if, in addition,
there is no nontrivial reducing space of
T
T
-D
-iC (C
BA
u(A)
u(B)
and
completely hyponormal then
then
llEtxll
2
(1.1)
Re (T),
A
(For a
is hyponormal
T
If
onto the
u(T)
are just the real projections of
absolutely continuous, that is, if
A
0),
on which it is normal.
see Putnam [2], p. 46.
real and imaginary axes;
is said to be
has no normal part, that is, if
recent survey of hyponormal operators, see Clancey [i].
then the sets
T
Then
respectively.
D -> 0, equivalently, AB
_
H, and denote its spectrum
It is known that if
as well as
Im (T)
B
T
is
Re (-iT),
is
has the spectral resolution
A
I tdEt
(1.2)
is an absolutely continuous function of
t
for every
x
in
H;
see [2], p. 42.
Since (i.i) holds if
T- zI, it is clear
is replaced by its translate
T
(T) C U (T*),
that if
T
is hyponormal then
that if
T
is completely hyponormal then
U
P
P
and it is not hard to show
up (T)
is empty.
is
T
Also, if
completely hyponormal then it has a (unique) polar factorization
UP,
T
if and only if
T
U
unitary
% (T*);
0
ence
P
(T’T) 1/2),
(1.3)
see the discussion in Putnam [3], pp. 419-420.
is completely hyponormal with the factorization (1.3) then
see [2], p. 21 and Lemma 4 of [4].
absolutely continuous;
U
If
is also
Thus, if
U
has
the spectral resolution
itdGt
U
0
then
IIGtxll
2
(1.4)
0
0
is an absolutely continuous function of
t
for every
x
in
H.
CONTINUITY AND HYPONORMAL OPERATORS
323
A converse of the above assertions concerning absolute continuity also holds,
namely:
(*)
Every bounded absolutely continuous
operator
selfadoint
A
i__s th__e
real part of some completely hyponormal operator.
(**)
Every absolutely continuous unitary operator
U
is the
unitary
factor in the polar factorization (1.3) of some completely hyponormal operator.
That there exists some nonnormal hyponormal operator
Re (T)
A
was shown by Kato [5], and that
T
for which
can even be constructed so that
T
it is completely hyponormal can be deduced from Kato’s results.
A direct proof
of (*) will be given below however.
Let
T
A
O
+ iB0
0
(B0x) (t)
n
x
where, for
=-(i)
[_
-I
2
(s-t)-ix(s)ds
(A0x)
A0B0
-iC
B0A0
2
and
T
O
-1
CoX
where
0
(x,1)1
{A nl },
Since the vectors
L (E).
2
0,1,2,..., span L (E), it is clear that
2
L (E)
tx(t)
(t)
the integral being a Cauchy principal
denotes the identity function on
1
having positive measure and let
(-=,=)
L (E),
in
It is easily verified that
value.
and
be any bounded subset of
E
0
is completely hyponormal on
see, e.g., Clancey and Putnam [6], p. 452.
It is now easy to prove (*).
According to spectral multiplicity theory
applied to absolutely continuous operators,
is unitarily equivalent to, and
A
here will simply be assumed to be equal to, the following canonical representation.
There exist disjoint Borel sets
a nonempty subset of
and
H
E
n
nee
Z
and
H
n
(n
E u)
(AnXn) (t)
H
n
then
where each
tXn(t)
T
n
A
0 },
of
for
n
Xn
A
is the direct sum of
in
Hn.
T
T
Since each
A
O
T
n
0
E u
n
+ iB0
on
H
An
u
and
is
has positive measure
n
Z
A
is a direct sum
H
n
copies of
Re(T), where
En
and where each
The operator
is a bQunded set.
the direct sum,
space
{1,2,...,
{En} on the real line, when
A
n
on
L2(En
copies of
is the real part of
on the corresponding
H
n
is completely hyponormal is a consequence of the properties of
(n
T
O
E
u).
on
That
2
L (En)
T
324
C.R. PUTNAM
as described earlier, and so (*) is proved.
T,T
Although the selfcommutator
of the above operator
D
TT*
not be of trace class, it is clear that such a
T
need
T
could be constructed. This
n
could as well be replaced by
is apparent from the fact that T
n
k=l
(n 6
where the
k
l,...,n) are any positive constants whose total sum
[
(A0+itnkB0),
,
tnk
is finite.
In order to prove (**), first choose
operator on the Hilbert space
U
0
H
S
{l }, n
0, i,
n
2,...,
sequence of numbers satisfying, say,
2
S
0
0
O
U0P 0
Izl
where
i.
lim
An
1
lim A
n
n+
and
2.
If
H
(bij) be
doubly infinite matrices defined by aij
ijlj and bij ijAj_l. Then
WBW* where W is the unitary bilateral shift x
(...,X_l (x0),Xl,...) +
(...,x0, (Xl),X2,...), the parentheses referring to the 0 position. If
is the
x
O
be any bounded, strictly increasing
n+-
A
S
with a polar factorization
is unitary with simple multiplicity on the entire unit circle
For example, let
the
to be any completely hyponormal
O
bilateral sequence Hilbert space,’let
UoA1/2
Op(D),
(S 0).
of [3].
U
where
S
O
W*
0
then
Also,
is completely hyponormal.
(S
In fact,
In addition,
U
0
{z: 1
0)
A
S0*S 0 SoS0*
Izl
S
D
B
any Borel set
O
G (8)U
0
U
0
has the desired properties.
have the spectral resolution
8 c [0,2]
(See also section 3 below.)
O
G (8)H.
S0(B)
[7], pp. 326-327.
e
0
O
G (S)
(Note that
see also (3.2) below.)
then a simple calculation shows that
addition
U
it
dG
0
t
and, for
0
of positive measure, consider the unitary operator
on the Hilbert space
has positive measure;
and, since
see, e.g., Th. 1 and Th. 9
2
Next, let
0
B
is nonsingular, that is,
O
/ };
and
(aij)
A
S0(8)
is completely hyponormal on
If S (8)
0
M
0
S
=G0(B)U0[G 0(8) S0*S0G0(8)] %
is hyponormal on
O
G (8)H;
if and only if
o
G (8)H.
In
see Lemma 8 of Putnam
From here on, the proof of (**) is essentially identical to
that of (*) but where unitary operators now play the role formerly played by
CONTINUITY AND HYPONORMAL OPERATORS
selfadjoint ones.
325
Also, corresponding to a similar argument above, the
completely hyponormal operator
of (**) can be chosen so that
T
T*T
TT*
is
of trace class.
Absolutely continuous selfadjoint and unitary operators occur in various
connections.
The results of (*) and (**) indicate that the study of such
operators can be reduced, at least in principle, to the study of the real part,
or of the unitary factor in the polar factorization, of an appropriate
completely hyponormal operator.
In the sections below this (somewhat tendentious)
point of view will be exploited to investigate certain unitary operators arising
bilateral shifts and their role in the
in the following situations:
F. ad
M. Riesz theorem and in dissipative and certain mixing transformations in
ergodic theory;
unitary dilations of contractions; minimal normal extensions of
subnormal contractions.
2.
A LEMMA.
The following lemma (I) asserts some absolute continuity properties of
operators effecting the unitary equivalence of two selfadjoint operators having
a positive difference.
As will be noted below, it concerns the unitary operators
of polar factorizations of certain corresponding completely hyponormal operators.
It will play a key role in later applications.
LEMMA (I)
space
H
and
Le__t
J
and
D
b_e bounded, selfado.int operators on
D ->_ 0
suppose that
an__d
0.
Le__t
U
b_e unitary
a Hilbert
with the
spectral resolution (1.4) and suppose that
J
(i)
UJU*
If
H
is the least space reducing both
containing the range of
then
U
(2.1)
D
is absolutely continuous.
D,
J
and
U
and
(2.2)
326
C.R. PUTNAM
(ii)
l__f
8
is any Borel subset
o__f [0,2]
G (8)
for which
(and
I
whether or not (2.2) is satisfied) then
2IIDII
lSl
<- d(J)
Here, max (J)
(J)
181
and
% (D),
0
that
IIDII <-d (J)
IIDII > 0
(I)
denote the maximum and minimum points of
J
=>
k
const > 0.
so that
T
(The proof of a slightly
is given in Putnam [8] under the added hypothesis
or even if there exists a sequence
D
D
1
8
Note
for which
and a
J2
Jl
is any Borel subset of
J
J
limllDnl
d(Jn)
for which
2
In order to prove (i), note that
real constant
8.
and that, in (ii), if there exists a
whenever
0
(2.3)
rain s(J)
in which case, condition (2.2) is automatically assured.)
corresponding sequence
181
max (J)
denotes linear Lebesgue measure of
that
then
(J)
rain
and
different formulation of
d(J)
d(J)
where
[0,2]
> 0
for which
J
can be replaced by
kI
(finite),
G(8)
0.
for any
and hence there is no loss of generality in supposing that
Then, in particular,
is hyponormal.
T
That
T
UJ
satisfies (I.i) by virtue of (2.1),
is completely hyponormal is a consequence
of (2.2) and, as noted in section 1 above,
U
is absolutely continuous.
The proof of (ii) essentially has been given in [8].
An alternate proof can
be obtained from an inequality concerning hyponormal operators derived in [4].
In fact, if
T
and if, for
0
et
is any completely hyponormal operator with factorization (1.3)
t < 2
(T) N {z: arg z
then, for any Borel set
8
of
t}
and
M(t)
I
et
(2.4)
rdr
[0,2),
IIG(8)DG(8)II
-<_
[
(2.5)
M(t)dt
See Theorem 1 of [4], where the additional hypothesis that
hyponormal is imposed.
T
is normal,
D
0
T
be completely
This restriction is not necessary, however.
and (2.5) becomes trivial.
If
T
In fact, if
is not normal, then one
327
CONTINUITY AND HYPONORMAL OPERATORS
need only split off its normal part and then apply the result in [4] to the
remaining completely hyponormal part of
_
to a reducing subspace is a subset of the spectrum of
T
restriction of
(Note that the spectrum of the
T.
the entire space.)
UJ
T
If
1/2 as
defined above, then clearly
nax (J))1/2.
M(t)
Consequently,
1/2ax (J)
on
is a
1/2
in (J))
subset of the annulus centered at the origin with inner radius
and outer radius
(T)
T
min (J))
and
(2.3) follows from (2.5).
3.
BILATERAL SHIFTS
Let
U
be unitary on
K I
Then
space
U
UnK
for
and let
H
n
be a subspace of
0
of (constant) multiplicity
H
(A
dim K.
has the spectral resolution (1.4) then it is known that
U
for any Borel subset
8 of
[0,2],
181
0.
This fact can be derived from ) as follows.
Let
{l }, n
n
be any bounded, nondecreasing sequence of real numbers.
For any
G(8)
if and only if
0
J
on
be defined by
H
and so .i) holds with
Z(A
Dx
strictly increasing, then
An
J
and
D
satisfy
0
and
d (J)
AnXn.
An-i )xn
x
(3.2)
UnK.
n 6
Clearly,
D
Next, if the
l
n
IIDII
1
for
1
{A
n
n
0, +/-I, +/-2
x
in
one
H
Let the selfadjoint
It is seen that
and hence (2.2) holds.
D),
0
for n
0
E
Jx
n
(3.2) now follows from (i) of (I).
that, say,
E xn where
x
has the (unique) representation
operators
(3.1)
satisfying the first part of (3.1) is called a wandering subspace of U.)
If a bilateral shift
operator
for which
H
UnK
I
H
and
+2
+i
is a bilateral shift on
K
K
0.
UJU*x
If now
Z
An_iXn
{A }
n
The first half of
sequence is chosen so
i, then the corresponding
and so the second part of (3.2)
follows from (ii) of (I).
The F. and M. Riesz theorem ] furnishes an interesting instance of how
is
C.R. PUTNAM
328
bilateral shifts can arise.
Let
A be normalized linear Lebesgue measure and
any nonzero, finite, complex Borel measure on the unit circle
C:
Izl
i.
The F. and M. Riesz theorem is the assertion that
znd
for
0
(3.3)
1,2,...,
n
C
implies
(3.4)
I[
that is,
and
2
operator on
L2(II)
L
A
(II)
spanned by
are equivalent measures on
z
of multiplication by
C.
is the subspace of
M
and if
is the unitary
U
If
{z,z 2 ,...} then
[ UnK
M
(3.5)
n=0
where
is the one-dimensional wandering subspace
K
I
UnK.
M
UM
and
L2 (I I)
See Sarason [10], pp. 14-18, Sarason [11], and Putnk [8].
It turns
out ([81) that (3.4) is simply an interpretation of (3.2).
It may be noted that the representation (3.5) is a form of what is usually
called the Wold decomposition for isometries.
(See Sz. Nagy and Foias [12], pp.
for a statement of the theorem and some of its history.
3, 51
Perhaps it
should be referred to as the von Neumann-Wold-Halmos decomposition.
_
Another instance where the bilateral shift occurs comes from ergodic theory.
Here, one has a system
x
(X,S,m,#)
and a measure, m, defined on some -algebra, S, of subsets of
0 < re(X)
(R).
In addiion,
preserving transformation of
Let
consisting of a measure space, X, of points
U
X, where
is here assumed to be an invertible measure
onto itself.
X
(See e.g., Brown [13], Halmos [14].)
be the unitary operator defined by
U: f(x)
In case there exists a set
/
f((x))
A c S
on
2
L (X,dm).
for which
m (A) > 0
(3.6)
and the images
CONTINUITY AND HYPONORMAL OPERATORS
n(A)
A
n
(n
0, _+i, +_2,...)
generating set of
X
p. 11, Hopf [15], p. 46.
In case
In fact, if
m(X) <
In case
2
L (A,dm)
K
in this case it may happen that
2
L (X,dm)
eigenvalue
I.
e
R,
where
U
X
called a K-automorphism.
see Halmos [14],
K
m(X)
U
of (3.6) has
U
1
in its point
being an eigenvector.
However, even
is a bilateral shift on the orthogonal
R
denotes the eigenspace corresponding to the
(X,S,m,)
is a Kolmogorov
is a special type of mixing transformation
For the definitions of these concepts, see Arnold and
2
L (X,dm)
and the space
2
L (X,dm)
R
Then
is the
admits a
R
decomposition of the form (3.1) for some wandering subspace
in this case,
U
and
is a wandering
Arez [16], pp. 32-35, 153-157, Brown [13], p. 39, problem 25.
space of constants in
is a
2
This happens, for instance, if
(or K-) system, in which case
then
A
then
L (X,dm).
H
the unitary operatory
spectrum, the characteristic function of
complement
A
n
is dissipative, then
U, so that (3.1) holds with
subspace of
-0
X
is said to be dissipative;
and
is a bilateral shift.
are disjoint and
329
Incidentally,
K.
may be a bilateral shift of infinite multiplicity;
cf. [16]
pp. 156-157.
4.
UNITARY DILATIONS
Let
T
be a contraction
(IITII =<
i)
on the Hilbert space
known result of B. Sz.-Nagy that there exists a Hilbert space
unitary operator
U
on
K
n
T
where
P
n
pUn lH
T.
and a
n
(4.1)
0,i,2,...,
P:K + H.
Such a unitary
T
(It is easily verified that also
*n
U
is called
PU
*n
H
for
The minimal unitary dilation, which is known to be unique, is
1,2,
then obtained by restricting
and contains
K D H
with the property that
is the orthogonal projection
a unitary dilation of
It is a well-
H.
H.
U
to the least subspace of
K
which reduces
U
(A brief discussion of unitary dilation theory, its history,
applications, and references, can be found in Halmos
I17].
For an extensive
C.R. PUTNAM
330
treatment of the subject see Sz.-Nagy and Foias [12] and the more recent survey
article of Douglas [18]
Suppose now that the contraction
T
is completely nonunitary (c.n.u.), so
that there is no nontrivial reducing space of
Thus, either
unitary.
TT* # I.
or
T*T # I
T
on which its restriction is
If the minimal unitary dilation,
U, of such a contraction has the spectral resolution (1.4) then it is an
important property of
U
that it satisfies (3.2), a result first established by
B. Sz.-Nagy and C. Foias [19].
U
In general, however,
is itself not a bilateral
shift, although (3.2) does imply that it always contains such a shift.
precisely, there is a subspace of
bilateral shift.
K
U
which reduces
More
and on which it is a
This is clear if one notes that (3.2) implies that
spectral multiplicity of at least 1 at every point of the unit circle
U
has a
zl
i.
It will be shown below how the assertion (3.2) for the minimal unitary
dilation
of a c.n.u, contraction can be deduced from the lemma (I) of section 2.
First, a few preliminaries will be discussed.
For the given contraction,
the so-called defect operators are defined by
D
DT,
TT*)
(I
1/2,
TD
T
(I
T’T)
T,
1/2 and
and it is fairly easy to show that
DT,T
T
see, e.g., [18], p. 165.
T*DT,
and
DTT*
(4.2)
Let
(4.3)
where the parentheses refer to the
0
position.
If one defines the operator
matrix
"I
0
D
U
1
T
(T)
0
-T*
(4.4)
DT,
I
CONTINUITY AND HYPONORMAL OPERATORS
(0,0)
refers to the
(T)
where
on the space
T
unitary dilation of
U
element, then
is readily verified to be a
1
of (4.3).
K
1
331
This explicit construction
In order to obtain the
of a unitary dilation is due to J. J. Schaffer [20].
U
minimal unitary dilation one need only restrict
That even
{G
U
which reduces
K
1
t}
U
1
and contains
to the least subspace of
1
H.
is absolutely continuous (that is, the "if" part of (3.2) if
1
U
refers now to the spectral family of
application of (i) of (I) of section 2.
Let
will be obtained as an
I)
{l }, n
0, +/-i, +/-2,...,
n
be a
real sequence satisfying
10 i 2
-i
-2
and
const <
n
and define the selfadjoint diagonal operator matrix
(4.5)
of (4.3) by
on
J
(4.6)
0 Ii
A straightforward calculation using (4.2) and the (crucial) equality
of (4.5) shows that
UJU*
J
(l_l- 10)I
D
Also, in view of (4.5), D
n
clear from (4.7) that if
and
J
is c.n.u., then
T
and since
with the
M
I0 i
H
D
M
then
M
that is,
0
occurring there now replaced by
absolutely continuous.
In_ 1
and
In
>
+/-
c H.
M
n
for
M
i, it is
which reduces both
is the least subspace of
and contains the range of
U
(4.7)
(I 12)I
0.
I’s so that
By choosing the
(0)
Since
K
I.
and so
M
Thus,
U
1
+/-
reduces
U
(2.2) holds
of (4.4) is
This implies, of course, that its restriction to any
nontrivial reducing subspace, and, in particular, the minimal unitary dilation,
must also be absolutely continuous.
Moreover, corresponding to a similar argument in section 3, one may choose
n
1
if
k
1
and
1
n
0
if
n
2.
Then (4.5) holds and if the
332
C.R. PUTNAM
IIDII.
d(J)
1
G(8)
is defined by (4.6), with
J
corresponding
0,
8
family of
U
given by (4.7), then clearly
D
II
It follows from (ii) of the lamina (I) that
[0,27]
being a Borel subset of
whenever
being the spectral
However, the corresponding assertion for the minimal
of (4.4).
1
{G t}
and
0
unitary dilation does not immediately follow.
of
U
1
{Gt}
(3.2) has been established when
Thus, so far,
is the spectral family
{G t}
but only the "if" part of (3.2) has been proved when
(T
U
spectral family of the minimal unitary dilation
being a c.n.u, contractior.
x
0
of
is in
H
DT,H
for
and
n ->_ I.
Then
U
[12], pp. 17-18.)
for definiteness, that
Note that
similar.
-I
and
T
D
is c.n.u., either
0.
for
0
n
(If
D
M
T
0.
+
where
1
DT,
0, then clearly
[0,27]
1
n
I.
Suppose,
for
Hence, by (ii)
and if the minimal
has the spectral resolution (1.4), then
This establishes
(See
is nc restricted
J
IIDII
d(J)
0.
T.
and
0, the treatment is
invariant and
K
U
reducing
K
or
0
be
and in the closure
-i
DT,
and
0
T
is any Borel subset of
UIIK
U
unitary dilation
G(8)
8
n
I
is the minimal unitary dilation of
K
1
of (4.6) leaves
J
of the lemma (I), if
whenever
D
n
for
+ x
0)
If one chooses the sequence of (4.5) so that
to this space.
n
T
Since
(x
is the least subspace of
K
U
X_l +
DTH
is in the closure of
Xn
H, so that
containing
+
x
of those vectors
K
In fact, let
It is not hard to establish the "only if" part too however.
the subspace of
is the
181
0
(3.2) where the minimal unitary dilation
has the spectral resolution (1.4).
Incidentally, it can be shown that if
dilation of a c.n.u. ontraction
and
L*
closure of (U*- T*)H
U*2L *
K
See [12], p. 57.
restriction
U
U’L*
1
K
is the minimal unitary
K
closure of CO-T)H
L
then
U
are wandering subspaces of
L*
H
L
UL
The argument applied above to
U
on
H (c K)
on
T
U
and
U2L
U
1
can now be applied directly to
(4.8)
and then to its
on
U
on
K
of (4.8) in
CONTINUITY AND HYPONORMAL OPERATORS
order to establish, again, the relation (3.2) for
5.
U.
SUBNORMAL CONTRACTIONS
Let
T
be subnormal on
and a normal operator
N
a normal extension of
T.
extension, and
T
333
N
K’
on
K’
so that there exists a Hilbert space
H
NH c H
such that
NIH.
T
and
D H
Then
N
is
It is known that there is a unique minimal normal
will henceforth be assumed to be of this type.
is said to be completely subnormal if
A subnormal
It may be noted,
has no normal part.
T
incidentally, that the subnormal operators constitute a proper subset of the set
of hyponormal operators.
(Concerning the basic properties of subnormal operators,
A more elaborate treatment of subnormal operator theory can be found
see [17].
in the recent tract of Conway and Olin [21].
Suppose now that
T
minimal normal extension
of
(The space
K.
H
is a completely subnormal contraction on
N
K’
on
and with the minimal unitary dilation
is regarded as a subspace of
and
P
denote the orthogonal projections
P’ :K’
and
U
have the spectral resolutions
I
U
zdG
C
has
zl
Tnx
< 1
(C
z
{z: z
Nnx punx
1}).
and
and continuous on
zl
N
i!
P:K
/
H, and let
zdEz)
Let
P
|
hdG x
z
in
n
and
H
Hence, if
x
H,
in
x
in
H
8 be any closed subset of the unit circle
continuous function on
C
such that
(Clearly, such functions exist.
#
1
N
and
r
one
1,2,...,
f
is analytic on
f(z)dE x
z
f(z)dG x and P’
f(z)dE x
f(z)dG x. Consequently, if h
P
z
z
z
C
C
then
z <- 1
z < 1 and continuous on
(real) function harmonic in
hdEz x
P’
Let
is any
P
P
U
z -I
then, for
1
K’.
and of
and
H
pu*nx.
P’N*nx
<=
/
K
ZdEz(=
x
Then, for
T*nx
with the
H
on
C
8 and
_
(5.1)
and let
0
< I
# be a real
on
C
8.
Then it is well-known that there exists an
as described above (which can be given explicitly by the Poisson integral) for
h
C.R. PUTNAM
334
0 on
h
which
In view of (5.1), one has
C.
"Iz
I I I
P’
hdE
+
z
0dEz )x
fr
x
(5.2)
H
in
C
C
<i
0dG zx
P
From basic max-rain properties of harmonic functions it is clear that
1
h
0
zl
in
and so, on forming inner products of the vectors in
< 1
dllEzXll 2 -<_
x, one obtains
(5.2) with
C
of functions
8
On
0
(n
1,2,...)
0dllGzXll
C
2
On choosing a sequence
converging to the characteristic function of
one obtains
IIE(8)xII
IIG(8)xll
-<_
x
in
(5.3)
H.
Since (5.3) holds for an arbitrary closed set
also for any Borel set
on
C.
on
8
C.
181
In particular, if
Let
181
C
8 on
it clearly holds
denote Lebesgue arc length measure
0, then, in view of the absolute continuity of
the minimal unitary dilation of a c.n.u, contraction (cf. section 4 above),
G(8)
for
0 (on K), and so
n
Since
0,1,2,
spanned by the vectors
Thus, if
E(8)
O.
E(8)P’
8
N
{N*Nx: x
0 (on
K’).
Hence
0
is the minimal normal extension of
H, n
is any Borel subset of
0,i,2,...
C
},
for which
E(8)N*np
N*nE(8)P
E(8)
and so
181
K’
T
is
0.
then also
0
In other words, the min:kl normal extension of a completely subnoal
contraction is absolutely continuous on the unit circle
Izl
1.
Other proofs
of this last assertion can be found in Conway and Olin [21], p. 35, Olin [22],
and Putnam [23].
ACknOWLEDGMENT. This work was supported by a National Science Foundation research
grant.
REFERENCES
74__2,
I.
CLANCEY, K. Seminormal operators, Lecture notes in mathematics,
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2.
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22.
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