* Uji
L IB RAR
THE PERTURBATION THEORY OF SOME VOLTERRA OPERATORS
by
John Markham Freeman
B.A.,
University of Florida
(June, 1957)
SUBMITTED IN PARTIAL FULFILLMENT
OF THE REQUIREMENT FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
June, 1963
Signature of Author,
.........
De'lartment of Mathema tics, May 15, 1963
Certified by.
Thesis Supervisor
Accepted by........
Chairman,
f
Departmental Comrittee on
Graduate Students
The Perturbation Theory of Volterra Operators
John Markham Freeman
Submitted to the Department of Mathematics on May 17, 1963 in
partial fulfillment of the requirements for the degree of
Doctor of Philosophy.
ABST RACT
A general procedure is derived for obtaining sufficient
conditions for the similarity of operators T and T + P.
This
is applied to obtain sharp conditipns for the similarity of
the Volterra operators J: f(x) -+ / f(y)dy and J + P where
P: f(x)
/+/fp(x,y)f(y)dy. By the same methods perturbations
of the one sided shift operator S on LP(Ofo) by certain trace
class operators P are shown to be similar to S.
In the last chapter solvability conditions are obtained
for the operator equation
TX - XS = A
where T and S are normal operators.
Thesis Supervis.or:
Gian-Carlo Rota
Title: Associate Professor of
Mathematics
BIOGRAPHY
John Markham Freeman was born in Sarasota,
and attended high school there.
Florida
He graduated OBK from the
University of Florida and attended graduate school in mathematics at the Massachusetts Institute of Technology.
He was,
for one year, at the University of Heidelberg
on a Fulbright Scholarship.
ACKNOWLEDGEMENT
The author wishes to thank Professor Gian-Carlo Rota
for his assistance and encouragement.
TABLE OF CONTENTS
Page
Introduction
0.1
Chapter I.
Spaces
Chapter II.
The Operator
1.
2.
3.
J: f(x)
i.1
-+./f(y)dy
Preliminaries
Solution of the Commutator Equation
Solution of the Operator Equation
A+Pr(A) =
P
4.
The Similarity of
5.
Applications
Chapter III.
Chapter IV.
of Regular Perturbations
II.1
11.6
1.10
J
and
J+P
II.13
11.16
The Shift Operator
1.
2.
Preliminaries
Spaces of Regular Perturbations
3.
The Similarity of
The Operator Equation
S
and
III.'
of S
S+P
111.2
III.2
SX - XT = A
Introduction
IV.1
1.
2.
Preliminaries on Rectangles
The Tensor Product E 0 F(-)
on 13(H)
IV. 3
3.
F(*)
Complete Additivity of E
on Schmidt Class
OX = A when A is
Solvability of
of Schmidt Class
The Convolution E*F(') and
Applications
4.
5.
Iv.6
IV.9
IV.13
IV.16
0.1
INTRODUCTION
In (3 ] Friedrichs studies perturbations of the selfadjoint operator
f(s) -+ sf(s) on
T:
integral operators
P
L2 (a,b)
with regular kernels.
termine conditions on the perturbation
P
by Fredholm
In order to de-
sufficient to en-
sure the similarity of T + P and the unperturbed operator
T,
Friedrichs used a method which has since been abstractly formulated by Schwartz [21] and applied to the perturbation theory
of a number of self-adjoint operators.
In Chapters II and III of this thesis the perturbation
theory of certain non self-adjoint operators will be approached
in a manner similar in broad outline to these methods of
Friedrichs-Schwartz.
Chapter II will be concerned with the
quasi-nilpotent Volterra operator "indefinite integration" on
LP(0,1), and Chapter III with the discrete Volterra operator
"shift right" on tP(Q,
io)
.
In the paper of Friedrichs mentioned above it is assumed
that the kernel
p(s, t)
P:
(1)
of the perturbing Fredholmoperator
b
f(s) -+/a p(s, t) f(t)dt
be regular in the sense that H6lder conditions of order
a (0 < a < 1)
be satisfied;
fp(sl
t) - p(s 2 ,t)t
<
K js,- s2'
(2)
jp(s,t 1 ) - p(s,t
2
)|
<
Kit 1 - t2 1a
II
0.2
It Is then proved that
perturbed operator
T:
T + P
f(s)
-+
sf(S)
provided that
is similar to the
INt
is small enough,
where
iP|
(3)
|p(s,t )-p(st2)1
=
+_u
|p(s 1 ,t)-p(s 2 9t)
_upp__t)_+_p
it 1 t 2 1 C
"l'21a
The crux of the method used in proving this result lies
in the observation that for a regular Fredholm integral operaA, the commutator equation
tor
TF~(A)
(4)
F~(A)T = A
-
is solved by the singular integral operator,
b
F~(A) f(s)
(5)
=
(c)
/a
askt) f(t)dt
(where (c) denotes the Cauchy principal value).
Chapter IV will deal with the solvability of (4) when
T
is any normal operator--without restrictions as to type
and multiplicity of spectrum.
to (5)
A singular integral analogous
will be defined which solves (4) for operators which
are "regular" with respect to
T.
I.1
CHAPTER I
SPACES OF REGULAR PERTURBATIONS
Let
space.
T
and
P
be fixed bounded operators on a Banach
The operators
T
and
T + P
are said to be similar
provided that there exists a bounded invertible operator S
such that
T = S
(T + P)S
In terms of the notion of a "regular perturbation of
T" to
be formulated in this chapter, it will be possible to state
T + P
sufficient conditions for the similarity of
unperturbed operator
and the
T.
The basic observation leading to the abstract notion
of regularity with respect to an operator
If
X
is the following.
simultaneously solves the two operator equations
(1)
TX - XT = A
(2)
A + PX
then
T
= -P,
(I + X)T = (T + P)(I + X).
(This is seen by multiplying
out both sides and collecting terms according to (1)
Hence
T + P
is similar to
vertible (e.g. if |IXII
T
provided that
I + X
and (2).)
is in-
lim ,,jijll/n <
n-+oo
In order to apply this observation to the perturbation
<
1
or merely
theory of
T, one first determines a class
operators
A for which the commutator equation (1)
CZ
of "regular"
is explicitly
I-
mt=t=t=
- i"Mma
-
A
-,
' -
- -
1.2
solvable by a bounded operator
X = F~(A).
In the following
chapters it will be seen that the operator
rule,
f(A)
"Isingular", i.e. does not belong to
a2
Now, having determined
and a map
is,
as a
.
FT
from
(,into
the bounded operators such that
(3)
T[~(A)
the equations
(1)
PF~(A) = -P ;
A+
A e
a
r~(A)T = A,
then reduce to
and (2)
(A)
any solution
-
of this equation also satisfies
(5)
[I + f~(A)]T = (T + P)[I +
and hence
T
(I+
r~(A)]O-
and
T + P
r(A)I
are similar provided that
exists.
In terms of the map
(6)
A
r~ :
-
Pf(A)
equation (4) becomes
(I +
p)A = -P
which is solved formally by the Neumann series
A =
Z
(
1
)n r-n(.p)
n=O
However, in order to make even the individual terms of
the series meaningful one must assume first that
P e
,
I.3
(so that
F. (P)
operator
r(A)
P
a
,
i.e.
is defined) and also that the "singular"
be "smoothed" by left multiplication by
if
P and
A e
(2
,
Jp(A) = Pr(A) e
then
These considerations suggest the definition (below) of a space
T.
of regular perturbations of an operator
Let
space
T
X
be a fixed (bounded) linear operator on a Banach
, and denote by
B
bounded linear operators on X
note the norm on
Definition 1.1.
6 (
( X
)
the Banach space of
A linear set
te|
on
will de-
).
a
c
8
(
) is called
a space of regular perturbations (s.r.p.) of
exists a norm
11*|
Throughout
.
a
T
if there
Q,:
and a linear map
-+
6
(X
such that
(a)
L,
(b)
Tfl(A) -
(c)
I|lr(A)H| < KjAf
(d)
if
[(A)T = A
P, A C (
, then
|P[-(A)|I
In what follows
and P
Ca.
.|
is a Banach space under |
The map
operator on
2 .
denoted by
|n.
Pr(A) C
K1|P| JAI
2
1
,
and
.
is assumed to be an s.r.p. of
given by (5)
T
is then a bounded
Its norm and those of its iterates will be
n = 12,...
)
I1
Proposition 1.2.
A sufficient condition for the (unique)
solvability of'
(I + [~p)(A) =
(7)
is that
A C'(
for
P
FE-n I1/n
urnli tPh/
< 1
n-+oo
Proof:
If this condition is satisfied, then the series
fpn
(-1) n
converges (absolutely) in the operator norm.
n=O
Its sum is
(I +
F-P)"'
.
The following lemma (cf. [2],
several times
Lemm
k
.
in the next chapters.
Let
(S, Z, V)
be a positive measure space and
a measurable function on
ess-sup /
s
page 518) will be needed
SxS with
|k(s,t)J '(dt)
<
M < o
and
S
ess-sup / Ik(s,t)II(ds) < M.
t
S
Then
tor on
Kf(s) = / k(s,t) f(t)dt
S
LP(S, Z,
)(1 <p :
oo)
defines a bounded linear operaand
||K| P
M.
II.]
CHAPTER II
THE OPERATOR
J:
J: f(x) +
0 f(y)d y
In this chapter perturbations of the Volterra operatar
x
f(x)
/0 f(y)dy on LP(O,l) will be treated.
Sufficient
conditions which are in a precise sense sharp will be obtained
for the similarity of J and J + P, where P is also a Volterra
X
operator P: f(x)
/'J+ p(x,y) f(y)dy.
Preliminaries
1.
x
Given two Volterra operators K:
f(x)
-*
/0 k(x,y) f(y)dy arnd
x
f(x)
L:
//
4(x,y) f(y)dy
then (under restrictions to be
st.ated below on the kernels k and 4) KL is the Volterra operator
x
KL:
f(x)
-+
f(y)dy
V/
where
k*4(x,y)
(1)
= /
X
y
k(x,n) ~1,y)dn
To begin with we prove several facts concerning the c omposition k*4'.
By 'kernel'
we will mean simply a (measurable )
real or complex valued function k(xy) on 0 < y < x < 1.
a > 0,
(2)
let
| Jk| I a =
19
Lemma 2.1.
lk(x,y)(x-y) 1-a
sup
03<<1
If IIk I
,
< Q
then K:
is a bounded operator on LP(0,1) (1 < p < 00) and
K
<
I|| I atoo
tk
x
For
.
Proof:
We have ,
2.
immediately from (2),
x
ess-sup
O<x<l
sup
O<x< 1
Ik(x,y)Jdy < IIk!I
/e
dy
(x-y )
and
ess-sup 1' 1
<y_< 1
(x,y)fdx < IkI
a
dx
00
sup
0 Y.1l
I
( x-y )1-a
1
and hence by 1.4 we get
Lemma 2.2.
11"
P0 0 <
dx
x
and
then
00,
(where B(a,@)
-:1 B(Ia
E+ctOu
)u|k
a
I
I|
P 00
is the beta function).
Since lk(xsn)
4
IJkJ Ica
(r1,y)J
(x-n) 1
it
II k I!
are kernels for which
If k and t
IIk*0i II
Proof:
IlK! I p: C I kf L,00 with C -/o
00111
4,00
( i-y )
follows that
|k*(-(x,y ) I :< I k IIa9uII IIP O J
x
dn
(-y
(x-'i)
=
IIk I(
VI)a+p
-I
Since this last integral is B(ap),
1
./0
dt
)l"P
t11a~lt(
1
this is equivalent to the
asserted inequality.
The following (known)
out explicit
ention.
facts will be used freely and with-
a
11.3
If k, 4,
(A)
ImJ
!y,
and m are kernels with 1k| L.
and
p, Y > 0, then
finite for some a,
all
,,
i1
l|
( k*J-)-,m = k*(44m)
(B)
If ||kja and
are finite, and K and L are the
lV4I1
Volterra operators defined by k and 4 respectively,
kernel is ki4-.
is a Volterra operator and its
I jkfj a9
For a kernel k with
< 0 we define
= k*k*. .*k
(3)
For example,
then KL
(n factors).
f (x)
the iterates of J:
/ 0 f(y)dy are
x
(L[)
Jn:
x)
f
(n)(x,y ) f(y)dy
/
-+
where
.
(n)
-
xy=)
If I|k|Jl,0
IIk(n) IIna,
( n-1
<
-<
then
al
Proof:
This holds for n = 1.
holds for n, we have by 2.2
< B(nasca)
In
aS"0
denotes the gamma function).,
r
)k
|kI
r n -
(where
1(
n-1
)n-
(x-
I I(n+1l)a,-
r (a_)n
r na )
k|k(n+l) II(n+l)a,
< B(na.*a)
I . n+1
alkoo,
Assuming inductively that
=
Ik
IkJ
S(a )n+l
({n1)
|
I Ik|
=
|ltW
n+1
.a
it
114.
Lemma 2. .
If Ik|l
, then the norms of the operators
x
k(n) (x,y)
: f( x) -+ /0
satisfy
p-IAL-r a) n
K
Itk I
.nO0
lKnl1/n = 0, i.e., K is a quasi-nilpotent operator
Thus lim
p
n-oo
on
f(y) dy
LP(0,l).
By 2.1,
Proof:
|IKnI
p
<
a
.
1
lemma, this in turn is majorized by
r (a) n
I k, I
a r (na-)
= 0 now follows since lim
I|Kn I/ln
p
That lim
n-+o
n-+o
when
By the preceding
.
r(na) l/n S00
a > 0.
Lemma 2
If kernels k and 4 are continuous on 0 < y < x < 1
and I|k|
,
ItIP
0 < y < x
1.
If a + 0 > 1, then k*4 is continuous on
0 < y
<
< 0,o
then k*t is continuous on
x < 1 with k*L(x,x) = 0.
k(x,y)
By the assumptions,
Proof:
=M(X* .
and
(x-y)1a
(x,y) =
n
0 < y < x < 1.
where m and n are continuous and bounded on
When y < x the variable change ni = y + t(x-y)
gives
k*4-(x,y)
=(x-y)a+,-1
Thus,
if
0 < y0 < X
n
1 m[xy+t(x)
(x-y)a+p-l
[Y+t (x-1),yd
(1-t)l-a
)-
f XY)(t)
f 1 and (xy)
dt
converges to (x Oy0),
then
l1-11.
laii ilmself
lisailmIIl |
Ti -1
na -
-
II.5
the number
k*((x,y)
converges to
k*L(x 0 ,yo), by the dominated
convergence theorem.
For we have
(t)
(xy)-(x(xy
and
()M
If
That
<
when
(yf ) (t)
constant /(l-t)l-a
0 < t < 1
1-P
k*-t converges to 0 as (x,y) converges to (x0 ,xO) follows
also from the above expression for k*i
Lemma 2.6.
If
k (x,y) = a
k(x,y)
providing a +
is continuous on
>
1.
0 < x < y < 1,
k(x,y) and C(x,y) are continuous on 0 < y < x < 1,
and ||k,
|I
t o
< W , then
k* 4 (x,y) = kl* 4 (x,y) + k(x,x)4(x,y).
Proof:
For 0
y < x < 1,
k*'dx+hhy) - k*4(xy)
=
h
h
h
y
x+h k(x+h)
As h
-+
x+h
- k(x Tl),(11y)dn +
X k(x+he1n)
x
k( xtn) t(1qqy)d n
- k(xn)
0, the first integral converges to k *C(xy) by dominated
convergence,
the second to k(x,x)-t(x,y) by continuity of the
11.6
integrand (recalling that y < x), and the third to 0 since the
integrand is integrable, uniformly in h, in an interval about x.
Solution of the Commutator Equation.
§2.
Let U,
a
(a > 0) be the class of kernels a satisfying
(i) a and a1 are continuous on 0 < y
(ii)
(iii)
<
x< 1
a(x,x) = a1 (x,x) = 0
a1 1
exists and is continuous on 0< y < x < 1 and
||all Iat
(The subscript 1 continues to denote differentiation with
respect to x.)
By (ii) it follows that a1 = 1*a
1 1
and a = 1
* a
and
hence, by 2.2,
|IaIay ||a o
||a||a+2,
(5)
| lal|l
From this it
+
I all+1I
a s
I
is clear that
jaj =
(and not just a pseudo-norm) on R2
|
*
a11 !||00" is a norm
and that I*I is equivalent
on G2 to the norm
a
(6)
(aIa = IIa Ia+2,o + IIaIIa+1,+
I ato
*
11.7
Proposition 2..
is a Banach space under 1-.1
I*-Cauchy
-
am| =
1 (xy)-
0 < y < x
Ian
a
-
a
|ax -+ 0
Hence
1.
and bounded there.
an
= c
CL
as
m, n -+ 0
converges uniformly to 0 on
a n(x,y)(xa
1-a converges uniformly
c(xy)(x-y)1-,
to a function
Now setting
a
=',
continuous
we have a c
and
a| =
an
11
c 1 a ,0 -+ 0
as
n-
oo
We now solve the commutator equation
Jr(A) -
(7)
when
A
r(A)J = A
is a Volterra operator with kernel
By the general remarks made earlier (7)
a e
becomes
l1*r(a) - r(a)*l = a
(8)
if one assumes a solution to (7)
(9)
be a
this means that
(x,y)](x Y1-a
on 0 < y < x < 1
a1
an e
So let
s equenc e;
By the definition of 11.*1
[a
*J.
is complete in the norm
,an
.
By the remark made above, it suffices to show that
Proof:
a
2
1-(A):
x
f (x) -+ j 0
of the form
(a)(x,y) f(y)dy.
a
,
11.8
If
Proposition 2.8.
satisfies (8),
I
a
r (a)(x,y)
(10)
(a) t
2
Since a
defined by
and a
<
r(a)
Thus
y
<
x < 1, and
represents a bounded quasi-
LP(0,1) with t|r(A) I
on
r-(A)
0< y < x < 1
a(+x-y,)d
is continuous on 0
Ea .
r~(a)
then the kernel
y
/
axay
nilpot ent operat or
Proof:
a as
a e
are continuous on 0 < y + C
|1a|
< X <
.
1
0) the Leibniz rule for differentiating an integral with
y
parameter can be applied twice to /0 a(F,+x-y, )dt. This gives
(c
>
(applying either
a2
aX8Y
a
y
or
r(a)(xy)
2
)
+ a(x,y).
0 a 11(F+x-y.,5)dF
From this follows the continuity of
r-(a)
on
0 <y < x <l
and
I-(a) (x,y)|I
_<f y
a , <"
I|all
0(x-y)1a
+ IjalIa+1 ,O(Xmy)
IalIIC
I1~
0 + Ia ||a+
,
1
(x-y) 1-a
<
< Ia|
and hence IIr(a) I
Since
le[~(a) (x,y)
x
2
=/
dn[ a
y
=-f
y
(+-y,)d
]
+
a, (9n-,)
dt + a~nay) 7
y
= a(x,y) .- al (+x-y,
)d+/al(
y
jal
. a
( X-Y) 1-a
11.9
and
71
2
x
Tj = X
71
1
a1 ((+x-iT,()dCj
y1=
=/
/0
(,)d
al(9+x-y,9)dt
we have
x
1*r~(a) - fl(a)*1 = a(x,y) -/
a
But the last two terms vanish since a e a
satisfied by
fl(a).
so that (4t)
is
The last assertion now follows directly
from 24.1.
Remark.
For a kernel
k
of the form
k(x,y) = m(y)/m(x),
it can be shown that the commutator equation
k*r(a)
-
r(a)*k = a
is formally solved by
r(a)(x,y) = m(
1
a(t+x-y,t) M
provided a(x,x) = a 1 (xx)= 0.
,2
d]
By using this,
results analogous to those of the present chapter can be
obtained for Volterra operators
above type.
K
with kernels
k
of the
A
II.10
§3 .
Solution of the Operator Equation
P
For Volterra operators
and
cz
a e
a + p*r (a)
=
-p
=
0P
with kernels
A
and
A + PF-(A) = -P
the equation
(11)
A + PF (A)
p C Q,
is equivalent to
,
i.e. to the integro-differential equation
x
a(x,y) +
sy1 p(x, i)
Lemma 2.9.
p
r* (b)
Ca
a2
zC
+,
y
aIay /
If p
|p *
Proof:
C
a(g+n-y,)didri = -p(x,y) .
and
b e
x.
, then
and
r(b)I
I| a |bI
: B(G,)
Using 2.6 we differentiate
respect to
a
p *
.
r(b)
twice with
This yields first
ax p*r(b)(xy) =
p 1 *[~(b)(xy) + p(x,x) F(b)(x,y)
= pl-*[~(b) (x,y)
(since
p(x,x) = 0),
and then
2
ap
F(b) (x,y)
-
p
1 1*F(b)(x,y)
=pli 41- ~( b) ( x,y)
Thus by 2.5 and 2.8, p*[f~(b)
+ pl(x,x) F~(b)(x,y)
(since pl(x,x) = 0).
and (p*r~(b))
are
1 =b)
lpl*'F(b)
continuous on
0 < y < x < 1 and (p*[(b)) 1 1
continuous on
0 < y < x < 1. Three applications of 2.2 now
yield
=
is
II.11
I Ip*F(b)I
II(p*F(b))
1
B(C+2,9)
-
1U+9+2Soo
11 C+ +, 00 :
|I IP|Y+2,ol IF (b)I |
||p, I a+ 1ol ||1 (b)II
B(a+ljp)
II(p*r(b)) ,I IO+Poo .: B(C0,9)
I1p1 1 Icy
Finally, using the fact that
|
B(Y,9)
<
B(aY2)
Y > c,
when
b) I
"
0
r(b) I
bI and
we get
B(U,
Ip*~F(b)jI
I [~(
I
) lpla Ibt
by adding the three inequalities.
The next lemma will give bounds for the norms of the
iterates of the operator
(12)
a -+ p *
Lemma 2.10.
If
p
C0.0
r-(a)
and
.
(2
a c
,t then fl(a)
p
eanc+a
and
r (a(n)r
I1n(a)
Proof:
Taking
and
b = a
ric
Y ra)
p(a)
(since
P = a
B(G,a) =
IpI
r'(a) r'(a)/
I P|,
jal
in 2.9 yields
|al
r' ((C+a)).
Now assume inductively that the lemma holds for
P= n
+ a
and
b = F.(a) in 2.9.
p
n
and take
11.12
Then
p
r (b)
eaC
(n+1) c+a
B(c,na+a)
lp I
l11~(a)I
+1 a)
Pn+1(a)|I
p
ja)
* B(c,na+a) |p.1
nf
jp1n+1
C+a]
(n+)
r
and
the last inequality following by induction assumption.
Proposition 2.11.
bounded operator on
11
n-o
Proof:
If
p C
Z
az
then
, CL
r:p
a -+ pir- (a)
and
1 Fn I1/n = 0
p
a
By (2) and (6) it is clear that the norms I*!a inThus for
crease with a.
(
1F~(a) I
a c aa
(n+l)
r
n+1
Ip In
#ri JTIF)
al
the last inequality being a special case of 2.10.
r'(a)n+l1
[ -n|
lim
n-+w
is a
'(na)
/n=
, [ (n+l) a I
Hence
from which 2.11 follows since
II.13
The Similarity of
4.
J + P
and
J.
Having now established the axioms 1.1 for G.,
we pass
to the question of similarity of the perturbed and unperturbed
operators.
Theorem A.
If
p e Cl
C,
then the operators
J
and
J + P,
where
J:
f(x)
0 f(y)dy
and
P: f(x) -/0
LP (O,1)
are similar on
Proof:
A
Y
for any
p(x,y) f(y)dy
p
with
By 2.7, 2.8 and 2.9, the class of Volterra operators
a and
a C
with kernels
= |aa
|AI
[(A):
f(x) --,/o
r~(a)(x,y) f(y)dy
is a space of regular perturbations of
A + P F(A) = -P is solvable for
with
1 < p < oo.
p Ca
.
FE(A)
Since
(I + |~(A)]
exists.
Chapter I, J
and
A
J.
with
By 2.11 and 1.2,
a
C
a
given
P
is quasi-nilpotent,
Hence by the general considerations of
J + P
are similar.
The preceding theorem can be strengthened by a procedure
used by Volterra-Peres
[22) and Kalisch (
].
I.14
x
Let
G:
f(x) -+ /, g(x,y) f(y)dy
be a Volterra operator
whose kernel satisfies
(1)
g(x,y)
and
gl(x,y) are continuous on 0 < y 5 x < 1
1
g(x,x) > 0
(ii)
(iii)
Z t g(t)
/0
gi(t)
and d
= 0
g(x,x)dx
and
g1t
Corollary A':
where
,O < o
G
0
is similar to
a
<
= gl(tt).
0 < y < x < 1
g1 1 (xy) is continuous on
11g1
0 < t < 1.
are continuous on
g~(t) = g(t,t)
where
(iv)
and
and
< 1.
cJ.
This will follow easily from the next lemmas.
Lemma 2.12.
Let
be as above with
G
c = 1, and set
x
r(x) =0 g(t,t)dt.
Then
non-singular operator on
f(x) -+ f(r(x))
Sr:
LP(0,l).
a Volterra operator whose kernel
is a bounded
Moreover
h
H = S 1 lGSr is
satisfies h(x,x) = 1
and the conditions (i) to (iv) above.
Proof:
Since
1
and
/0 g(tt)dt
g(t,t)
is continuous and > 0 on 0 < t < 1,
-1
= 1,
m = r"w
exists and both
r
and
m
are continuously differentiable:
and
dr= g(x,x)
dx
Thus
Sr and
rd
(bounds
<
dm =1
dx
Sr1
d~j/
= Sm
d
and
g~iT7~ii"xTT3
are bounded operators on
IIjX1c1AP
LP(0,1)
respectively).
Moreover,
since
m(x)
1G
S4*
r 6GS r f(X)
X
/0
H = S ;GSr
h(x,y) =
f(m(y) )dy
g(m(x),y)
0
-.g
x).m(v.)
f(y)dy
m iy Om.
,
is a Volterra operator with kernel
M
h(x,x) = 1.
s atis fying
Now
h (xy)
M(y))
g(m(y))g~(m(x))
81(m1
1
h 1 1 (x,y)
and,
-
In view of the above expression for
h.,
tions (i)
-
is continuous on
(iv) on
d (
)-Ft(M(
x)
the continuity of
dih 1/dt follows from the continuity of
Similarly, h 1 1
m(Y)
( X) )
g(m(y))
and
g 1 (m(x)
g.
0 < y < x < 1
To see that
growth condition at the diagonal,
h
1
dg~ /dt.
by the assump-
satisfies the proper
h
(x,y)
notice that in the above expression for
containing gll(m(x),m(y))
and
g
=0 (
1 -- I
(X-y)
h i,
only the term
can be unbounded near
x = y.
But
by the assumption (iv) on
g 1 1 (m(x),m(y))
which in turn
(m(x)-m(y))l,
m(x)
is 0(
1
(x-y)
since x-y = r(m(x))-r(m(y))
S/(y)
g( tqt)dt.e
II.16
Lemma 2.12.
Let
H
be a Volterra operator whose kernel
h(x,x) = 1
satisfies
k(x) = exp /
and (i) to (iv) above and set
h (tt)dt.
Proof:
q(x,x) = 1,
is
is a
Moreover,
q
satisfies
ql(x,x) = 0.
Since
f(x) -+ /
MkIMk:
Q
LP(0,1).
is a Volterra operator whose kernel
(iv) and
(i),
f(x) -+ k(x)f(x)
M
Then
bounded non-singular operator on
M= HMk
h
h(x,y) f(y)dy
a Volterra operator with kernel
x
q(x,y) = h(x,y) exp[-/
so that
h1 (tt)dt]
q(x,x) = h(x,x) = 1 and
ql(xy) = [h 1 (xy)-h1 (x,x)h(x,y)] exp(-/Y hl(t,t)dt]
qll(xy) = [h 1 1 (x,y)-h(x,y)t- (x) + f1 (x)2 h(x,y)]exp[-
Thus
q1 (x,x) = h 1 (x,x) - h 1 (x,x)h(x,x)
ties
(i) and (iv) hold for
sions for
on
q, q1
(tt)dt
That the proper-
follows from the above expres-
and the assumptions (i) to (iv)
h.
Proof of A':
that
g
and
q
= 0.
1
Multiplying by 1/c,
/0 g(tt)dt = 1.
G
can be normalized so
Then by the lemmas, G
a Volterra operator
Q. whose kernel satisfies
ql(x,x) = 0,
and (iv).
has kernel
and (1)
p = q-1 e a
is similar to
J.
is similar to
q(x,x) = 1,
But then the operator
P = q - J
and hence by Theorem A, Q = J + P
1 ..17
§5.
Applications.
x
The Volterra operator
J
similar to
if,
G:
f(x) -+/
g(xy)
f(y)dy
is
say,
g(x,y)
= eX(Xmy) (where
g(x,y)
=
X
is any complex number)
or if
4-1
1
This latter example shows that
when
p > 2
J":
where
P
f(x)
J
P < 2.
J
is similar to
is the fractional
1
By a result of Kalisch [iI],
when
> 2.
where
+
J + JP
integral operator
x
/
(x-y)
J
"
f(y)dy
is not similar to
J + JP
Thus Theorem A is sharp with respect to the
allowable algebraic singularity of
p1 1
at the diagonal.
III.1
CHAPTER III
THE SHIFT OPERATOR
In the present chapter perturbations of the isometric
operator
S:
on
4P(Oo)
(X
l ,x2
*'''
by certain trace class operators will be shown
to be similar to the unperturbed operator
§1.
Preliminaries
With respect to the basis [On:
00 =
S.
01 = (0,1,0,...),
(1,0,0,...),
n =0,1,2,...] where
etc.,
by the matrix
S =
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
0*~
a
0.-i
0
The matrix of the operator Itshif t lef t",9
S:
(X
,x
,Xlx2pe***-
l'KP2'''
S
is represented
111.2
is
0
1
0
0
0
o
1
0
0
*
o
o
1
Q
S..
S
S
s
0
0
It
S
is well-known tat
LP( 0,oo)
S*
and
S S = I
and satisfy
both have norm 1 on
E
SS"
and
where
is
E1
the projection
E1 :
(x 0 ,x1 x2 ,...)
-+
(0,xlx
2
,''')
0
More generally,
Sn
= En
where
En:
The projections
(x
0
) -+ (0,0...PXn'Xn+'0'*
.xll''''xn,...
)
*
are represented by the matrices
En
n-zeros
en
diag(0,0, ... ,0,
For an infinite matrix
, 1, 1,..
a =
[a,
)
we define
00
(1)
Ia I
and will denote by
ms n0O
M
the class of matrices
a
with
Ja I < ".*
111.3
e M
a
The matrices
A:
represent bounded operators on
(x00,x ,x2 '.)-' (yo-yl'y2 ... )
0(0,o):
,
m0
As an operator
on
is of trace class (see
A
42(0,o),
[13J).
Its trace is given by
tr(A)
E0 ann *
n=0
M, then the series
If a
Lemma 3.
=
E SkAS
k=O
converges (conditionally)
Y(A)
satisfies
in the norm of
IIY(A)Il
<
6
(t);
its sum
and is represented by the
1a
matrix
Y(A)
Proof:
We first
observe
matrix left
one column,
a + S*kask
shifts
the partial
sums
= [tr(S~nASMl
that the operation
and
a matrix
N
YN(A) = E
a -+ s*a
-+
as
shifts
up one row.
a
Hence
units diagonally upwards.
k
S
shifts
a
*
AS
Thus
have matrices
k=0
N
N
Y N (ak)
0 S
=
k=0
as
= [ r an+k,m+k
k=O
Now to establish 3-1, we first
a bounded operator
Y(A)
on
show that
Y(a)
0(0,00) with IIY(A)IfI
represents
j al.
111.14
We ha ve
sup
m>O
< sup
rItr(?AS"'
n=0
m>O
n=0 k=0
ian+km+kI < IaI
a nd
00
sup
n>O
E |tr(S*nASm)J
M=O
which,
<
r
sup
using 1-3, establishes
Finally we show that
E
m=O k=0
n>O
the assertion.
YN(A)
To do this we observe that
lan+km+kl < I a I
Y(A)
in the norm of
-+
Y(A)
-
YN-l(A)
'6
(4 p).
is represented by
the ma trix
Y(a ) -
YN-1(a ) =
tr (S""
'ASN~
a nd
00
r- Itr(S*N+nASN+m)
m=0
sup
n>O
=
sup
n>O
m=0 k=0 I N+n+k,N+m+kI
s NasN
sup
n>N
00
m=N k=0
and (similarly)
sup
m>O
n=0
ltr(S*N+n ASN+m )I
Hence (again by 1-3)
*N
.: aI
N
we have
>..kT
IY(A)
-YN-(A )Ip :
But the latter converges to 0 as N
wr
Is *N as14
-+
10,
0
which proves thbe lemma .
III .
*2.
Spaces of Regular Perturbations of S
r
(2)
F(a)
we define
a c M
For a matrix
(a) = s"Y(a).
is thus the matrix of
F(A)
=
S"Y(A),
i.e. of the
operator
(A)
Since
||S'1|1
= 1,
=
Z S
k=0
we have by 3.1,
IJ (A)|I
0..
Proposition
I
and only if
Proof:
!
tr(ASn)
F(A)S = A
- 0
for
S S = I
Recalling that
jai .
|IIY(A )||I I
satisfies the commutator equation
F(A)
Sfl(A) if
AS
SF(A) - T- (A)S
and
= B1 Y(A)
= Y(A)
=
n
0,1,2,...
SS" =E
- S*Y(A)S
- S Y(A)S -
= A - (I-E )Y(A).
But (I-E1 )Y(A)
(I-e 1 )Y(a)
=
we have
has as matrix
~tr(A)
tr(AS)
tr(AS2 )
0
0
0
9
0
from which the result follow immediately.
(I-E )Y(A)
111.6
Lemma 3.3.
If
p
and
a e M,
pF-(a)
then
c
M
and
|pr~~(a)l < jpj|aj
Proof:
By definition (1),
00
=
|pF(a)I
2
n=0 m=0
00
00
<
Z
:
n=0 m=0
00
| 2I pnk tr(S
ASm
k=0
00
00
2k P
Z
|ak+1+Jm+j
nk = j=
00
00
z0 1Pnk'I
nz
<
00
00
00
Z0
Z~ Iak+l+jm+jI]
jp|ja| .
We are now in a position to determine some spaces of regular perturbations of
a c M
S.
Let
a
denote the space of matrices
whose entries vanish on and above the main diagonal.
Q Cta
More generally,
(a > 0)
will be the space of
whose entries vanish on and above the
First, it is clear that the
the norm
'*1
of
M.
ath
a
a e M
sub-diagonal.
are Banach spaces under
Moreover, since
tr(asn), n = 0,1,2,...,
are the diagonal and super-diagonal sums of the entries of
it follows from 3.2 that for
sr(a)
Lemma 3.-4
If
p
,
a C Q
-
f-(a)s
and
b c
r(a)
q
then
and
IpF(b)t
Ife ,p
surely solves
a.
=
I
bI.
a,
pr(b) e
III .7
, then the entries of
b e a
If
Proof:
and above the
(0-1)th sub-diagonal.
For
has only O's on and above the
Y(b)
b c q P
and above the
In particular
has only Os on and above the (a+P)th row.
pr(b) = e
r(b) = s"Y(b) and
sub-diagonal when
Pth
sub-diagonal.
(V+P)th
vanish on
has entries vanishing on
pF(b)
Thus
(see 3.1).
r(b)
pr_(b)
Hence
so the result now follows by 3.3.
pr(b)
We now investigate the bounds of the iterates of the
operator
:a
*
on a
Lemma 3. .
If
pe (2,
and
and
the Fn(a)
ae(q,
n
t (a)| <
p
Proof:
for
-+ p I'(a)
n
I -r
k=l
Ieka+LP|]raj
By 3.4 this is true for
b =
and taking
rn(a)
n = 1.
and
Assuming validity
= na+a in
p
F
ana
~(b)a
=
(n+)
(b) e
+
and
n+1
(
k=-1
IIeko+ajp|]
tat
As an immediate consequence of 3.5 we have
3.4 gives
111.8
p ea
Let
Proposition 3.6.
fl
norms of the powers of
aa.
p
Then
I
denote the
pnja
as operators on the Banach space
n
p
a+pF(a)
Hence
and
a
=
T
k--l te(k+l)p
a
-Mp
a C
is (uniquely) solvable for
2
provided that
n
TT le(k+l)aP
lrn
*3 .
1/n <1
k--1
n-+x*
Similarity of S + P
and
0
S
We can now easily deduce some sufficient conditions for
the similarity of
If
Theorem B.
S + P
p e
0
and
S.
and
jpI
<
then
-1
S
P
+
and
S
are similar.
Proof:.P f |IPt <1 2
Hence a+pr (a)
we have by 3.6 that IrnI
p 0
= -p
is solvable for
jal < jpj+jpF-(a)j
so that
|al
11r(A)I|P
s + P
and
< 1.
Proposition Bt.
[I +
:
(1+jaj)-
T(A)]
a
jp3
.
<
n
0
Moreover,
1
exists, since
Hence by the considerations of Chapter I,
< ta.
S
a c Q
< (1+lal)lpl
But then
<
-
are similar.
If
p e 62 1
low a certain row, then
and
p
has only 0 entries be-
S + P and
S
are similar.
-
m
r
P
III.9
Proof:
enp = 0
By assumption
Fn(p) =
by 3.5,
p
for large
0
p
n
large enough.
Thus
n.
vanishes on and above the first
Hence,
Z (- 1 )F
_P)
n=0
Moreover,
sub-diagonal and
the same is true of
below a certain row,
a =
a + pfl(a) = -p.
is a finite sum and satisfies
since
for
Thus
a.
|(a)
vanishes on and above the main diagonal and below some row.
Such a matrix is nilpotent and hence
Thus by Chapter I,
and
S + P
0
0
0
0
...
0
0
0
0.
1
0
0
0
...
pi
0
0
0
0
1
0
0
...
P2 1
0
0
0
0
1
0
...
+
20
exists.
are similar.
Theorem B refers to perturbations of
Remark.
s+p
S
[I + fl(A)]"
of the form
s
.
p3 0 p3 2 p3 2 0
Stronger results can be obtained from 3.6 when the first subdiagonal of
a > 1.
s
is not perturbed, i.e. when
For then,
p C
a
with
instead of the estimate
IEn 1
given by 3.6 when
p e
pn
00
(since
e=
I)
we have
n
But, since left
the first
le
pI -. 0
multiplication by the projections em replace
m rows of p by rows containing only O's, we have
as
n -+ao
when
a >0.
IV.l
CHAPTER IV
THE OPERATOR EQUATION SX-XT = A
Introduction:
In this chapter we will obtain solvability
conditions for the comrtutator equation
TX - XT = A
(1)
when T is a normal operator on a Hilbert space H.
The results
will apply equally well to
SX - XT = A
(2)
when S and T are both normal.
For two bounded (not necessarily normal) operators S and
T on H we define
JX = SX - XT
(3)
for
X
8B(H).
B(H)and,
space
Then
r
is a bounded operator on the Banach
by a result of Kleinecke (see [17]),
has
bs spectrum
(4)
For
a(0)
C
=(S)
- S(T).
one has the Dunford operational calculus
f
-
f()
defined by
(5)
/
f(0) = -
for functions
f
f(z)(E-mz)"*dz
aD
holomorphic on a neighborhood
A more useful representation of
f(0)
f(O)A =
1+r
/
f(S-z)A(z-T)
5G
of
a(EJ).
is obtained by Rosenblua
(J'7];
(6)
D
1 dz
IV.2
where
G
0
when
is a certain neighborhood of
'
a(T).
In particular,
gives the explicit inversion formula for
a(O), (6)
DX= A,
1
(A) =
(7)
(S-z)" A(z-T)
In [8] Heinz shows that if
T + T
-1 exists as a bounded operator on
010
O"I(A)
=
(8)
-
dz,
c-G
< b < a < S + S", then
S(H) and is given by
-tT
etS Ae
dt
/
0
where the integral is absolutely convergent:
/ 00|e1 tsAe-.tT||dt < 1(a-b)'
(9)
||A||
.
We, on the other hand, are principally interested in
solving (1),
()4),
OX = A
i.e.
when
S = T.
0 then belongs to the spectrum of
not exist as a bounded operator on
By Kleinecke's result
Q, so that
b(H).
01will
Thus the above
formulas do not apply to the problem of inverting the commutator equation (1).
It is the object of this chapter to construct an operational calculus for
0
when
S = T
is normal and, from this,
deduce sufficient conditions for the solvability of
OX = A.
The explicit solution will have the form of a singular integral
operator
|(A)
analogous to (5 ) of page 0.2.
By carrying through the analysis to include the case
S
# T,
we will also be able to formulate the question of ex-
istence of the wave operators
IV.3
e
U+ = w -lim
(10)
e
These are the unitary (or partial-isometric]
in a new way.
operators which implement the unitary equivalence of
S and
[or of their absolutely continous parts) in the Kato-Cook-
T
Rosenblum treatment of self-adjoint perturbations
of a self-adjoint operator
In what follows
denoted by
S+
O
-
T
and
S+
*
T.
X -+ SX
operators
T.
S = T + P
and
X -+ XT will be
Using this notation we have
From (41),IV. 10
below it
f ollows that,
as oper-
ators on the Hilbert space of Schmidt class operators, S+ and
T
W
(S*)+ and (T-)* = (T )
have as their adjoints (S+)
and hence
El
= (S*)+ -
This implies that if
then
El
S
and
(T )
T
are normal operators on
H,
is normal as an operator on Schmidt class.
The goal of the next few sections is to calculate explicitly the spectral resolution of
and
T.
§1.
Preliminaries on Rectangles
Given two sets
are of the form
rectangles.
S
and
0
in terms of those of
S2, the subsets of
6xY with 8 C S1 and Y c S2
S
S, x S2 which
will be called
The symbol 10' will denote a union whose summands
are pairwise disjoint.
IV.4
Lemma
If
rectangle
1 x Y
6xY
and 62 X2
are non-void, then a third
is their disjoint union,
6 XY=(6 y
1 1
)rj( 62 Y2 ) if
and only if
either
6= 61 0 62
and
Y = Y
= Y2
or
6= 61 = 82
and
Y = Y
0 Y2
( bxy )-(6 ix 1
[ (6)
Lemmal..
Proofs:
(See Halmos [5 1.)
Lemm
.
If
If
6 C 68
and only if either
8 C 81, then (8 xY)-(6 1 xY1 ) =
Conversely, suppose (8xY)-(6 1 yi)
and Y-Y
6
1
O,
or
The
is a rectangle.
If
then 4.2 expresses this rectangle as
we must then have either Y = Y-Y
n Y,
is impossible since
Y
the rectangles
and
void intersection).
Y C y1 ,
x(Y-Y1 ) by 4.2.
the disjoint union of two non-void rectangles.
6xY
Y1 )
1
Y c Y .
argument is the same if
61 J
6-1 )X
(6xy)fl( 8xY1 ) is non-void, then (6xy)-(
is a rectangle if
Proof:
1)XY-Y 1) IOU(_
1
1
or
6-.
=
0 and 6 n
Y,
Hence by 4.1
n 6
81
But this
0 (recall that
were assumed to have a non-
Thus either 6-8
= $or
Y-Y
=
$,
and
this proves the lemma.
Lemm
The smallest rectangle containing a union U(
of non-void rectangles is the rectangle (U6 I)x(UY ).
i
i
Proof:
(Straightforward.)
i xyi)
Iv.
6
if
.
Lemma
i = 1,2,3, are non-void pairwise dis-
1 xY1 ,
joint rectangles whose union is a rectangle
6xY, then
(6 xY)-(
1
Proof:
By 4.3 it suffices to show that either
xy1 ) =
I or Y = Y
tj 6 xY
c 6,0
6 = 61 for some
for some 1.
if
We first show:
6
is a rectangle for some I = 1,2, or 3.
For example:
or 62 C 81.
0, then
86
61
0,
if 61 n 62
on the contrary,
Suppose,
either 6 c 8
or
then either 61 C2
that
81
62 and
62 ' 61 . Since
61 and take x 1 f1 - 62 and x2
62
61 n 2 0, we must have Y1 f y2 = 0 (by the assumption that
61
1and 62 xY2 are disjoint).
Now by 4.4,
3
8xY =
U
W)x(
i=1
Take y1 e Y, and Y2
2*
U Y ) and hence (x1 ,y2 ) and (x2 ,yl)
3
i=1
belong to 6xY.
But then, since x1
must belong to
3 xY3
81, (x1 ,y2 ) and (x2 ,yl)
But this is also impossible -since,
by
assumption
0
= (653 xy 3 )n(6
1
and hence either 6 fl
3
(x
1
,y 2)
1
) = (8 3
n )(Y
=
Y
'
3 xY3 or (x2 ,yl)
or
y3 fy
663 xy 3 *
3
=
ly1)t
so that either
This contradiction es"
tablishes the assertion made at the beginning of the paragraph.
The analogous assertion holds for the Y1 .
Case I:
y C Y-Yy.
6 n6
=
$
Then (x,y) c
In particular (x,y)
i
for
6x 8
c 6xY
J.
7 Then
for
Y1 implies that (x,y)
/
1
x e 81.
/
Y1.
=
For suppose
3
61 (see 4.).
1=1
But this is impossible
for all x e 6
since x c 61 implies that (x,y)
y
Y =Yi
2 xY 2 and (x,y)
/
3 Y3 , and
IV.6
C 6
65
Case II:
for some
i
j with i
and
or Y =
3=6
or
6=
Then either
2 C 6i.
the above we know that if 6
63
c 61.
or 6
Since
62 C6
that 6=6
and 6
c
assume that
61 n
62 C
y e
1
implies that
=
3
Then
6
2
n
Y = Y .
This implies that
3
From
then either 6
6=
and
/ j, say
C 63
C3
implies
6 c
1=3
6=61. Hence we can
U 6i,
=
0'
also, since
For otherwise take
Then (x,y) e 6xY for all x e 6 and, in particular,
Y0%3.
But this is a contradiction,
for x e 6 *
since x e 63 implies
that x p (61 Y1 )U(62 y 2 ), and x X Y3 implies that (x,y)
X 63 xy
It follows from the first part of the proof that Cases I
and II are exhaustive, so the lemma is proved.
§2.
The Tensor Product
(.
denote the 0-ring of Borel subsets of the complex
Let F
C
plane
E
A resolution of the identity on a Hilbert space
.
H is a function
whose values are (orthogonal)
E(.) .on (3
projections on H and which satisfies
(1) E(O) = 0,
E(6f6t) = E(
(ii)
= I
E(C)
6
)E(
6
1)
for all 6, 6' e
i.e. if
cc
6
e6
and
6
6n then
n=1
z E(n )f for every
E(6)f=
f e H.
n=1
be the ring generated by the rectangles
Let
6, Y
15
,
i.e.
Borel rectangles.
k
f
6xY with
is the set of finite disjoint unions of
If
E( *) and F(*) ars resolutions of the
IV.7
identity we define the (bounded) operator
E ® F(6xY)
on
by
4(H)
E & F(6xY)A = E(6)AF(Y).
(I)
n
Lemma k.6.
If
6xY
=3
(%xYi) then
i=l
EOF(6 xY) =
6 xY
If
Proof:
n
Y E ® F ( 1 xY)
i=1
= (61 xy)0(2xY 2 )
6 = 56 2 and
that, say,
then by 14.1 we can assume
Y = Y1 = Y2.
We then calculate
directly
E(6 1 )AF(Y)+E(62 )AF(Y) = [E(%1)+E(6 2 )]AF(Y) = E( 6 )AF(Y)
If
5xY =
( 6 2 xY2)(
3
0 (6 xY i), then by 1 . we can assume that, say,
i=1
Thus the case n = 3 follows
3 xY3 ) is a rectangle.
by applying twice the case n = 2.
Now assume the result for k < n - 1.
Since
n
6xY
= (
xY1 )n
i=2
where
(6ixY ) = (61 Y
81 (Y-Y 1 )
A=
1
)J
1
and A2 = (bm1)xY,
2
we have by an appli-
cation of the case of three disjcint summands
E ® F(byxY) = E ® F( 5 1 xY1)+E ® F(
n
Now
=
and A=
2
[1 xY )Aa]
1
i=2
1
)+E ® F(A2)*
n
0 [((
i=2
Yi)n
2 ].
Thus, since
the intersection of two (Borel) rectangles is again a (Borel)
rectangle, we can apply the inductive assumption to get
IV.8
n
n
Z E ® F[(6 xY )lA]+ Z E ® F[(61 Y
1=2
1=2
n
and this is equal to Z E ® F( 6 xy ) since for i > 2,
1=2
E ® F(A1)+EOF(A
6 xY
2
)
= [(6 xy )Ao[(6
)na 2]
xY )n62
n
Hence
=2
E 9 F(6xY)
E
F(i
xy )
1=1
Lemm
..
If
A
(I = 1,...,n) and A
(j = 1,2,...,m) are
two families of pairwise disjoint Borel rectangles and
m
n
m
n
S=
A
A
,then
Z E
F(A )
ZE
F(A').
Proof:
j=l
=1
j=1 3
1=1
A,
n
a
ii
A
1,...,n; j = 1,2,...,m) is a
(i
1
family of disjoint rectangles and A
Moreover that
E
) =
2 E ( F(®
j
Z E ® F(A)
follows from 4.!5.
Z E ® F(A ) are equal to
a
ii,j
=
Z
i
UP a I
) and E
=
A.
F(
=
Hence both Z E ® F(A1 ) and
E ® F(A
.
In virtue of this lemma we can define the operator
E ® F(A)
on
(2)
E ® F(A)A =
B(H) for any
A
'\
by
n
ZE(6 )AF(Yi)
i=1
(i = 1,...,n) is any finite family of disjoint
n
Borel rectangles with A =
t 61 xY.
Then as a consequence
where 61 xY
1=1
of 1.+.7 we have
IV.9
Lemma 4...
E ® F(A) on
The operator
additive function of
Lemma 14.9.
(1) E 9 F(O) = 0,
(i)
A e
E ® F(c x .) = I.
= E 9 F(A)E ( F(A') for all
is clear from (1 ).
If A
m
i b
=
XY
4te p.
then for
13(H)
E ® F(A)E ® F(AI)A
=
Z E(8 i n
Complete Additivity of
Let
H
)AE(YifY
3
)
E ® F(4fA
x(y fYP,); = 3 (6 xy )(6 xy' )
i
ii
i
since 3 (6 nf6)
j
911,
§3.
is a finitely
(H)
A e
(ii) E ® F(A nl At)
Proof:
6
=A
1 )A,
2
E ( F( *) on Schmidt Class
now be a separable Hilbert space and [n] a comH.
plete orthonormal set in
An operator
A
on
H
is said
to be of Schmidt class if
00
JJAJ 12
l s| =
Z I |Anj 12<
n=l
The Schmidt class of operators forms a Hilbert space with
(AB)
8
Zz (An, Bn)
n=.
This Hilbert space is independent of [0 ] and is (unitarily
equivalent to) the tensor product
linear functional (.,
the form
on
H.
g) by
H ® H*
If we denote the
', then the elements of H ® H* of
are identified with the operators h -+ (hg)f
n
More generally, Z ci f ® i is the operator on H
f ® g
i=l
IV.10
of finite rank given by
n
n
(Z
z
g )h =
1f
iJ~l
c1 (h,g )f.
is the closure of the set of operators of finite rank
H 9 H"
||
in the norm
For these remarks and the facts which we
||s.
list next the reference is Schatten [gO].
(i)
(f e 1)X = f
X*g.
If
X e B
(ii)
For
A c H
For each
projection on
then
Proof:
If a =
H ® H*
®
and
(AX,B)5 = (ABX )8 .
and
(A, f
H,
lar, (0 ® ', f ® 1g) s
Lemma I+.10.
) = (Xf)
(H) and A,B e H ® H , then
(XA,B)s = (AX B)
(iii)
X(f ®
In particular
XA C H 9 H*.
A cH®H, the AX and
and
Xe b3(H)
If
®g)s
= (Ag,
f).
In partiou-
*(0,f)(T.
A e Pt
,
is an orthogonal
E ® F(A)
Moreover,
if
4, &1
e it
and A c At.
F(6) < E 0 F(').
E
From 4.9 it follows that E ® F(A) 2 = E 0 F(A).
n
6 xY
and A,B e H ® H", then, by (ii)
above,
(E ® F(A)AB)5 =
n
2 (E(8 i)AF(Yi),B), =
i=1
n
i=l (A,E(
i=1
and hence
E 0 F(A)
gonal projection.
6 i)BF(Yi)),
= E ® F(A).
= (A,E ® F(A)B)8
Thus
E ® F(A) is an ortho-
IV.ll
If
A a A I, then 6 ' = A U (At-A) and hence,
finite additivity of
E ® F( * ) on )
E ® F(A t) = E 0 F(A)
by the
,
+ E ® F(A ..A)
from which the last assertion of the lemma follows.
From now on the
A e
E ® F(A) with
preted exclusively as operators on
H
will be interH* (and not on
8B(H)).
0..E ® F(*) is strongly completely additive on V
Lemma
Proof:
If A n C TL
an d An O
, then by 4.10,
E ® F(An) is an
increasing sequence of projections on the Hilbert space H
Thus
.
E ® F(A n ) converges strongly to a projection
P
H*.
on
U An £ E , then E ( F(A) = P.
n=l
, this will
Since E ® F( *) is, by 4.8, finitely additive on
H ® H".
We now show that if A
prove the lemma.
From (i)nabove we have E ® F( 6 xY)f ®g = [E(6)f]
Thus, if A
1E
=
xY,
then
F(A)f
Z
|E(6 )f
i=l
F(Y)g|I
J
I|E(6 )fII 2 |F(Yg)gl|2
=
i=1
-
(p.
x
v
()
the cartesian product of the two measures
P(O)= |IE( *)f||
Hence, if An,
A e
2
and
and
v(.) = I|F( R)g1
An o A, then
2
.
IV.12
F(A-.nA)f'
I) A 1 12,1E
(AE ® F(A-A )f
I IIA 12( xV)(vis a finite measure and
since PxV
2
) -+ 0 as n
%.
A-a
Thus
4i so that E ( F(A)
lim E ® F(An)f ® g = E ® F(A)f
= P at all
n-ow
elements of
H 0 H
f ® '.
of the form
But then, since H
0H'
is the closed linear span of elements of this form, we have
E
= P throughout
F()
Proposition h.12.
E 9 F( *) has a unique extension to a reso-
lution of the identity on
(i)
H
H', i.e.
E ® F( *) is defined and strongly completely additive
on the a-ring
8 x13 of Borel subsets o
E ® F(AfA') = E ® F(A)E ® F(A 1)
(ii)
(iii)
Proof:
H".
H
(Cx
and
E 9 F(A)" = E ® F(A)
for all A,A' 9 15
E ® F(O) = 0 and E
F(
x C) = I.
The set functions 4A,B( *) = (E
pletely additive on the ring
(
Schwartz's inequality fIA,B(A)|
xl.
F( *)A,B)s are com-
of Borel rectangles and by
.
|1A1111B||, for all A ev1.
Hence by standard theorems (see e.g. 12t , p. 136) on the extension of measures IAB(.) can be uniquely extended to a
measure on
11A||
6
3 |1B|1 8.
x'
.
That
tAB(A)
The extended measure also has the bound
in B for fixed A e )5 x6
is linear in A and conjugate linear
follows from the uniqueness
extended measure and the fact that
for A c
.
Hence for each A e
AB (A)
of the
has this property
1B x)3 there exists a unique
IV.13
on H ® H* such that "AB(6) =
bounded operator E ® F(A)
That the E ® F( *) thus extended satisfies (ii)
(E ® F(A)A,B).
follows again from the uniqueness of the measures PAB(*) and
satisfies (ii) on R
the fact that E ® F( P)
Thus the ex-
.
tended E 9 F(O) is a weak resolution of the identity.
How-
ever, the strong complete additivity now follows, since (ii)
I |E
implies that
A
e
e
® F( *)At I2= (E 9 F( *)4,A)5 and hence for
with An *
x13
n
||E ® F(A.-on)A|
we have
A
n -+ O
as
n -+ c.
In the case of Schmidt class operators of rank
Remark 4.13.
and B = f ® g , we have
one, A =
(E ® F(A)AB)
and v(
(E( *),f)
where P(e)
immediately for A e TI
(E ® F( 6 xY)0 e Y,
= (Px )(A)
) = (
This follows
TF(TT,).
using that
f ®g) = (E(6)g,f) (F(Y) 7,)
uniqueness the two measures are equal on )
.
x1
Then by
.
Solvability of 7X = A when A is of Schmidt Class
14..
with resolutions
be normal operators on
H
of the identity E( o)
and F( e)
Then E ® F( #) has
support included in
(S)xa(T).
Let
S
and
T
and measurable on G(S)xa(T),
on
(1)
H
respectively.
For a function f(X,§),
we define the operator f(S+,T_)
H 4by
(f(S+,T_)AB)
bounded
/
f d(E ® F(*)A,B)s
-I(S)xG(T)
IV.l1
and g(X,), bounded and measurable
For two functions f(X,)
on a(S)xa(T), we have
= f (S~qT.)g(S~tT.)
(f 9) (S+IT.)
(2)
This follows from (ii) above by approximating f and g uniA similar argument shows that if
formly by simple functions.
h(X) is bounded on a(S) and k( ) is bounded on 0(T) then
h(S+)
(3)
where h(S)
=/hdE
Theorem C.
Let
and
T
= k(T)
=/kdF.
k(T)
and
S
k(T.)
and
= h(S)+
be bounded normal operators on
with spectral resolutions E( P)
and F( .)
Schmidt class operator on
If
(*)
F(A) = (c)/
.
H.
and let
A
H
be a
dE ® F( *)A exists in the weak operator
topology of H, and
(**) E 0 F( 6 )A = 0 where 6 is the diagonal of 0(S)xO(T),
then
Proof:
- r(A)T = A.
SF(A)
Let
X
be the characteristic function of
Ae= [(X,t):IX-fI > el and set
Then f. is a bounded function on
fe(X,9)
a(S)xa(T) and the assumption
(*) means that the Schmidt class operators f (S+.,T)A converge
in the weak operator topology of H to F~(A).
We have by (2)
and (3) above
3[fr(S,
T.)A)
[fe(S+,T.)AJT
=
=
(S gT )f(S+,T)A
= E ( F(A )A
IV.15
to A - E 9 F( 8 )A.
which converges in 1|*11,
But then since
F(A) = w - lim f,(S+,T)A we have
F~(A)T = A - E ® F( 8 )A,
-
Sr~(A)
from which the theorem follows.
Examples.
H = L 2 (-l,1) and S = T be the operator
Let
f(s) -+ sf(s) and
f (s)
A:
+
For a subset
+1
a(s,t)f(t)dt
.1
A
of the square,
ator with kernel
tion f(s,t)
nel
where
'(a a.
E '0 E(A)A is the
More generally,
on the square,
f(s,t)a(s,t).
7/a(s,t)I 2 dsdt
f(T+,T_)A
Thus the
integral oper-
f or a bounded func-
is the
operator
"0
<
operator with ker-
r-(A),
if
it
exists,
is
the weak operator limit of the Schmidt class operators
/
1
dE ® F(*)A:
f(s)
/(
Y
) f(t)dt,
t
Is-t|ze
is-tl>Ct
+1
i.e.
fl(A)f(s) = (c) 4
Here
s-t) f(t)dt.
E OE(5)A = 0, since its
the condition (**)
kernel is
X
6
a = 0,
so that
is vacuously fulfilled.
The situation is reversed if H is finite-dimensional.
In this case, E ® F(6)
=
0 is necessary and sufficient for
the solvability of SX-XT = A.
a finite
A= [(,
For a(S)xa(T)
number of points so that f(X,)
X):
X
/
=
exists and Sr(A)
E 9 F(A)A
= A - E ® F(6)A,
-
just
where
is bounded on a(S)xa(T).
f(S+,T.)A
consists of
Thus
r~(A)
A
r~(A)T = (S+-T.)f(S+,T_)A =
so that E ® F(
6
)A = 0 is sufficient.
IV.16
It
is necessary since E @ F(6)[SX-XT] = g(S+,T. )X where
,=(A-C)
Remark.
2(. 6 (x
-)0.
The conditions
(*) and (**) are easily shown to be
also necessary for the solvability of SX-XT = A for X in
Schmidt class where the operators f(S+,T.) and E ®F(A ) are
defined.
The difficulty in showing the necessity of these
conditions for solvability in
b(H) stems from the fact that
we may well have solvability in
class.
'6(H) but not in Schmidt
It can be shown that in the first
example considered
above TX-XT = A possesses a Schmidt class solution if
and only
if
411 ,
t) 12dsdt < GO,
a property not enjoyed by the regular
studied by Friedrichs,
for which,
Fredholm kernels
on the other hand,
commutator equation is solvable in
the
a(H)
These difficulties will be partially overcome in the
last section of the chapter.
E 9 F(
*
§5.
6
By other devices,
the condition
)A= 0 will be shown to be necessary for solvability in
(H).
The Convolution E * F(*)
and Applications
We now assume that S and T are self-adjoint operators on
H with resolutions of the identity E(*)
and F(*).
E
F()
is then defined on the Borel subsets of RxR where R is the real
line.
For a Borel subset 8 of R we define
IV.17
= E ® F(A)
E * F(6)
where A = [(X, ): X-9 e 6].
Then
E * F(*)
H ® H" defined on R.
of the identity on
is a resolution
For functions f(x)
bounded and measurable on C(S)-a(T) we have (recalling that
0 = S+
f(!)
-
T)
=
f(X-
/
/
)d E ® F( *) =
f(x)d E * F( )
o(S)-a(T)
O(S)xa(T)
Thus, in particular, we have the expression
*it]
for
/40
itx d
E *F(
*)
itC] (A) = its -itT%
e
Ae-o
, interpreted as an operator on H ® H".
e.
Thus for
A e H ® H" and f, g e H
its Ae--itT
(a
we have
+0itX
gf) =
d (E * F( *)A,
f
g)s
and hence
Theorem D.
If
A
is of Schmidt class then (e itsAeitT gf)
is the Fourier transform of the finite Borel measure
(E *
Examples *
If
F(*)A, f ®j)
A = 0 ®
, then
(E * F( *)A,
where
r
® g)
P(.) = (E(*)$,f) and v(.)
= (puv)( *)
(F( *)Yg),
since by 4.13,
(E 9 F( *)A,
f
s
PXV)
IV.18
If
S
(E( *)V,f)
have absolutely continuous spectrum i.e.
T
and
and (F(
are absolutely continuous measures
e)Y,g)
on
R, then
f[(E(*)$,f)
d±*(8)
=
(F(*)Y,g)]dx
so that
itS
(e
it x(
x
-itT
Ae.
e
e'
g9f)
If, finally,
S
L2 (,+o)
)
-E
T is the operator
F(*)y
f(s)
-+
sf(s)
J]d
on
then
itx[(Of)*(Tg)]dx
(eitT Ae-itT
Solvability of DX = A
Since de
il
=
ito
we have
/eiso (A)ds = A - e
(A)
0
this suggests, as a solution of OX = A, the integral
r~(A) = -i /
e
(A)dt
0
If this integral exists in the sense of the weak operator topology of H, then 01~(A) = A.
Thus sufficient conditions for
solvability can be expressed as integrability conditions on the
functions (a
Ag,f)
which,
if
A is of Schmidt class,
Fourier transforms of the finite Borel measures (E
are the
F(*)Af 0 g) .
IV.19
Existence of the Wave Operators U
Then
Define
U
dt t =
eitS Pe-itT
.
= eitS e-itT and set P = S - T.
it (P)
U= I + i
/
so that
te is(P)ds
0
Thus the existence of U+ = w - lim U
of
/
depends on the existence
00itO]
e
(P)dt as a weak operator integral, for which, if P
0
is of Schmidt class, sufficient conditions are again expressible
in terms of the integrability of the Fourier transforms of
finite Borel measures
Remark.
(E 4:F( *)P,
f 9 g) .
These considerations suggest how to formulate ab-
stractly the notion of "regularity" or "smoothness" of a
Schmidt class operator A with respect to a self-adjoint operator T (or self-adjoint operators T and S) in such a way that
regularity of A => solvability of OX
=
A (or the existence of Uk).
Namely, the Fourier transforms of certain finite Borel measures
on R should be integrable.
The H6lder-regularity of the kernels
a(s,t) assumed in the Friedrichs example T: f(s) -+ sf(s) is
clearly expressible in the above terms.
--
-,
-'
I~ I
~~II
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