94

Appendix l: Halmos' lemma

Halmos' lemma: Let the map

A : Xl --> X2

denote a linear contraction between two Hilbert spaces. Then there

exist an orthogonal projection P, a unitary map U and an isometry

J, such that

A p·U·J

Proof: We explicitly define the maps.

J : Xl Xl X2

J (xl) = xl 0 .

U Xl X2 X2 Xl

U = [_: : * ]1

*;;;

with S (I-A A) Xl Xl1

*;;;

and T (I-AA ) }{2 X2

The projection P is defined by identifying the spaces X2 and

X2 0 ,

P : X2 Xl : X2

P(x2 xl) = x 2 .

It is an easy exercise to show that the maps have the required

specifications. But to prove that the map U is unitary, we need to

know that Sand T intertwine with A

AS = TA

This we show as follows. Consider the identity

2 * * 2A'S = A'(I-A A) = (I-AA )'A = T 'A .

95

By a simple induction in nErn this extends to

A.S 2n = A.S 2(n-' ).s2 = T2(n-' ).A.S 2 = T2(n-' ).T2.A

We have shown that

where pol denotes a polynomial.

We now choose polynomials Pn(·) on [0,'] in such a way that

converges to v't uniformly on [0, , ] By the functional

calculus we get

Pn(S2) ------> S in ?1i (X, )n

Pn(T2)

------> T in ?1i(X 2)n

Hence

AS TA .

For x, EX, 'Ne get

p·U·J(X, ) = p·U(x,l& 0) P(Ax,l& (-Sx, )) A(X, )

•

96

Appendix 2: Gaussian measures

The results in this appendix come from [1 8]. We provide the

construction of the gaussian measure on infinite dimensional spaces,

uniqueness of the measure and some properties that are both useful and

have illustrative purposes.

The numbering is independent of the numbering in other parts of

these notes.

A: The gaussian measure ona>

IR

We define the space moo as the set of

\Roo = ... ,an"") I aiE\R

We define a topology in moo by the metric

all real sequences,

for all iEIN } .

Le.

00

= = l 2-nn=l

!Xn-Yn!

l+lx -y In nwhere

The well known real Hilbert space is a subset of \Ra>

n=l

Writing \Roo-n for the subset of \Roo consisting of zeros in the

first n positions, and identifying the space \Rn as a subset of \Roo

by adding zeros after the first n positions, we get

\Roo \Rn ffi 1R 00 - n .

00

In \R we define the cylinder sets

B ffi 1R00

-n

where n are natural numbers and B runs over Borel sets in IRn

.

These cylinder sets form a base for the Borel sets of \Roo, I- I

We define the gaussian content on the cylinder sets in !Roo

by setting, for nEIN and a Borel set B C !Rn ,

ffi \Roo-n) = ! .. .. dxnwhere dx

1dx2... dx

ndenotes the Lebesgue integration over \Rn

97

We have that

'Y(moo) = 1

and for every fixed nEm 'Y becomes a measure on mn

We shall use the Kolmogorov extension theorem saying that if 'Y

is a content on the cylinder sets in

'Y(moo) =

00

m such that

and if 'Y is a measure on mn for every nEm , then 'Y extends

uniquely to a Borel measure on moo, I- IWe denote the measure by the same symbol -r .

For a > 0 we define

'Ya(B e moo - n ) = (2va)-n/2 f .. .. dxnB

It is known that for different a's the corresponding measures are

pairwise singular. For simplicity we develop the theory for a =

In particular we shall need the measures for the numbers

a = and

These measures will be denoted by 'Y

a =

and

Theorem lA: Consider a sequence

numbers. We define the set

of positive real

00

If the series 2an is summable then -r(E)=ln=l

otherwise 'Y(E)=O .

Proof: We construct a net of functions converging to the

indicator function of E, lE' Given A>O, we define the function00

fA : m m

by00

exp(-A 1n=l

98

It is easy to prove that

{: if x(Ef?\ = positive real number if xEEand that f?\ l pointwise for ?\----->O

The functions

N

= exp(-?\ 2 for NEllin=l

are obviously integrable over ffioo and for fixed ?\ form a decreasing

sequence. By the Lebesgue theorem of monotone convergence we get that

J = J lim f?\ = lim J f?\a" [Roo N- ' N- [Roo '

in case the limit exists.

Notice that in the case where f depends only on a finite

number of variables xl ,x2' .. o,xn the integration reduces to

N

2 -N/2 1\ 2 d d= exp(-2L xn) x x 2o .. dxNn=l

t )1+2?\a·x we computenN N

J Ne xp(-?\ 2 2 .. dxN[R n=l n=l

Using the substitution

Nnn=l

[ J[R

N/2 N [ J 1 2 dtn exp(-2 t ))1+2?\an=l ffi n

nN -1/2 J 2(1+2?\a ) since )dtn=l n

[R

../2; .

00

200

If an 00 then n (1 +2?\a ) = 00 , thusn=l

n=l n

J f?\ 0ffioo

Otherwise we have that

00

l an < 00 ,n=l

and by the inequality for non-negative x

99

o In ( 1+x) xwe get that

oc

TT (l+2Aan) < '" •n=lThis implies that

{ oco

TTn=l

(l+2Aa )-1/2n

if

if

cc

2 a conn=l

'"2 a < '"n

n=l

By using the Lebesgue theorem

= JIR'"

co

If 1an < "', we have thatn=l

cc

of dominated convergence we get

lim JA->O IR

lim TT (1+2Aan)A->O n=l

which concludes the proof.

•

The difficulty in developing the theory of gaussian measures in

IR'" is due to the fact that contrary to the finite dimensional case

where = we have, as shown in the above, that2(L ) = 0 .

To shorten the notation we define the measures

nEIN, by

on

(d) (2 1 2 2))d dw ,+xn xl'" xn

If we identify the spaces IR'" = IR n ffi IRoo- n and IR n x IR

oo- n , and denote

by oo-n the gaussian measure onoo-n

IR , constructed in the same way

as the measure on IRoo

the measure can be identified with

the product measure ® '

= ®n oo-n

Definition 2A: Let f

100

denote a function defined on 1R00 and

integrable with respect to For nErn we define

= J ,1R00-n

where xElRn and sElRoo-n and IR n is treated as a subset of

By the Fubini theorem Enf is well defined, and Enf

integrable function over withn

J = JIR n 1R00

Lemma 3A: For fELl and nErn we have that

00IR

is an

Proof: Assume that f is a non-negative function, i.e. assume

that f20 Then we have that Enf 2 0 and

II En fill = J(En f ) • n = Lf • II fillIR n IR

An arbitrary f can be written uniquely as

f = f+ - f- ,

where f+ and f

Then we have

are non-negative integrable functions such that

f+·f- 0

I f I = f+ + f

IIEn(f+ - f-1II 1 IIEn(f+1II 1 + IIEn(f-1II 1

= If I = .

•

Definition 4A: A function f defined on 1R00 is called tame if

there exists an nErn such that

where

1R00 ------> IRn

101

denotes the natural projection.

Notice that if is a tame function, depending only on

the first n variables, then

Lemma SA: The tame functions are dense in,

L ('Y) •

Proof: Since the cylinder sets form a base for the Borel sets in

moo every Borel function can be approximated by finite linear

combination9 of indicator functions of cylinder sets.

functions of cylinder sets are tame.

Theorem 6A: (Jessen) For fELl ('Y) we have that

-----+l f in L' ('Y) •n-

Indicator

•

Proof: Consider E>O. Find a tame function gEL' ('Y) such that

Ilf - gill < E/2 .

As g is tame, there exists an such that g depends only on the

first N variables. For nLN we have that

and

IIEnf - fill IIEn(f-g)11 1 + Ilf-Engll,

liEn (f-g) II, + II f-gll, 2·llf-gll, < E .

•

Corollary 7A: Let f denote a function defined on which is

integrable with respect to If f is constant with respect to

every finite number of variables (depending only on the tail), then f

is constant a.e. in

102

coProof: Considered as a tame function on ffi Enf depends only

on the first n variables. Hence E f is constant a.e. Since f isn

constant with respect to every finite number of variables, we have for

all m,nEIN that

fffico- n

fffico-m

Emf (x, , x 2 ' ... , xm)

Thus

is obviously constant a.e.

•

Remark: Since the arctan of a measurable function is integrable,

corollary 7A holds for every measurable f .

The following corollary is often called the Kolmogorov zero-one

law.

Corollary SA: Let B denote a Borel set inco

ffi If the

indicator function of B is constant with respect to every

finite number of variables, then B is either of measure zero or one.

Proof: From corollary 7A the function 'B is constant a.e., by

which it is obvious that 1B = 0 a.e. or 'B = ,a.e.

•

B: The linear measurable functionals onco

IR

Definition 1B:

ffico if A fulfills

is called a linear measurable functional on

is linear on E and measurable with respect to .

1) A

2) A

is defined on a linear setco

E C lR of full measure

103

In what follows we shall provide a representation for these

linear measurable functionals defined on rn°O

Proposition 2B: (The Kolmogorov inequality) For00

we define

n

= l akxk·k=l

For arbitrary E>O we have that

nEIN and

sup

n

I > E }) l / E 2 .k=l

Proof: We start by noticing that

1 J 1 2 d 0rn t·exp(-2 t ) tand

1 f 2 1 2 drn t exp(-2 t ) t = 1For arbitrary, pairwise disjoint, measurable sets

meIN we have

Q crn°Om

n

lk=l

n

l 1 frn 2 (' 2 dxk'exp - 2xk) xkk=ln n

ld f 2 (' 2 d l L 2 2xk'exp - 2Xk) xk a k xkk=l rn k=l rn

n

foo ( lrn k=l

and since

estimate

the relation

Z + 2sm(sn-Sm)

s2 + 2s (s -s ) + (Sn-Sm)2m m n m

we can continue

implies the

n

Z l f + .m=l Qm

We construct the sets Qm' mEIN, as the inverse images by the

function

104

the smallest kErn such that I > E

inf { kErn I > E }Hence

Notice

-1 {Qm=f (m)=

that the function

00xEIR f = m } •

depends only on the first m

coordinates

We return to the integration. Since the functions and

depends only on the variablesfunction

andvariablesmonly on the firstdepend

we get by the Fubini theorem

o ,

J (lQ1R00 m

J 1Q JIR 00 m IR 00

sn-sm being linear in the variables xm+ 1 ,xm+2'· .,xn .

J s (s -sm n m -Qm

We estimate the second expression using the fact that if

then and hence

JQm

> E •

L JQm

Then we get

2= E .

We return to the first estimate

n n n

2 L 2 J L 2 2 n Qm)= Ek=l m=1 Qm m=1

• Ii 00 therexEIR exist such thatxEIR

00there such that= E exist

( xEIR00

I I > E })sup ,

})

})

which concludes the proof.

•

Theorem 3B: (The Kolmogorov large number theorem) Consider a

sequence of posi tive numbers If

00

2 < 00n=l

then the

105

series

'"= 2anXn

n=l

converges a.e. in rn'" with respect to .

Proof: We shall prove that the measure of

'"{ cc l diverges }xErn a xn n

n=l

is zero.

n

We define = 2akxkk=l

lim supn

lim infn

For mEIN we have

= +

S S sup Is - s (x) Iq p q -

+ sup -

lim sup - lim infn n

S sup s (x) - infp -

sup {s (x) -p-

S sup Is (x) -pLm p-

2 sup Is - Ip

For arbitrary E>O we define

M - I > 2E } .Hence for all mE IN

:= {co

xErn sup - > E }PLm

and thus

S for all mEIN .

Using the Kolmogorov inequality, we compute

106

'Y(M) s lim sup 'Y(Mm)mlim sup 'Y({

00

I - s (x) I > })x IR supm m -00

s lim sup 22 1 2 = 0a km k=m

•

We have seen that series of the form

converges a.e. in

i\ = 2n=l

00

with

n=l

and thus defines a linear measurable

functional. The converse is true as well.

We introduce some notation. Let us define

r 0 r 1 r 0 r

where the number takes the n-th position rand

00on IR

Theorem 4B: Consider a linear measurable functional i\ defined

Then we have00

1) 2!i\(en ) 12

107

subspace of lRoo

= {00

xEIR there exists an NEIN such that x = 0n for } .

Theorem 5B: Consider a measurable set

following statements are equivalent,

B C lRoo

Then the

(' )

(2 )

,(B) > 0

,(y + B) > 0 for all

Proof: It is easy to see that the theorem holds in the case

where , is the gaussian measure ,n

in the finite dimensional space

Let us first prove the restricted form of theorem 5B, where B

are cylinder sets. Consider the set

C = B Ell lRoo-n

for an nEIN and a Borel set B C IR n

arbi trary there exists an meIN

consider two cases.

Assume that ,(C) > O. For

such that yElRm. We have to

m n Then we regard y as being in the space IRn, and the

result follows by the finite dimensional case.

m > n: We rewrite the set C as

C = B Ell lR oo - n = B Ell IRm- n Ell !Roo-m ,

where B Ell !Rm-n is a Borel set in !Rm. Then we refer to the case m

n .

To prove the full version of the theorem, take a non-negative

integrable function00

f:IR--->IR.00

Consider yEIRO ' then there exists

an mE IN such that yElRm If n m we have

TyEn(f) EnTy(f),

where (Enf)(zl ,z2" "zn) JlRoo-n

and = (Z1 ,z2'" ,zn)

By the Fubini theorem the function

for all nEIN .

108

Enf : ([;

is integrable over IR n with respect to and

J = JIRn IR

Notice that f i 0 implies that Enf i 0 .

Assume that

J > 0 .IR

Hence

J > 0IR n

Using the theorem for the finite dimensional case and the relation

T E (f) = E T (f) ,Y n n Y

we get that

for all nEIN .

Hence

Setting f

o < J = JIRn IR

'B' the theorem follows.

•

Corollary 6B: For arbitrary linear measurable functional A we

have that

C ,

where denotes the domain of A •

Proof:

will prove that

contains a linear subset E of full measure. We

C E C •

Assume that there exists an E \ E. For positive t we

define the sets

+ E

Since E

109

is a linear set, the sets are pairwise disjoint for

different indices. Using theorem 5B, we get

Hence

-r (Et ) > °is a family of pairwise disjoint sets with positive

measures. This contradicts the fact that00

-r(IR)=l

•

Proposition 7B: Denote by A a linear measurable functional on

IRoo

Then we have that00

l IA ( en) I2 < 00 •n=l

Proof: For x = (Xl ,x2'. "xn" .)EIRoo

we define

= (0,0, .. ,O,xn+ l ,xn+2'" )EIRoo

Thus we have that

n

l xkek + .k=l

For we get that

n

= l XkA(ek) + =k=l

J =00IR

where = ) is measurable for all n and has the same

domain of definition as A Hence exp ( i- A( ) and )

integrable00

are over IR

For arbitrary u > ° we getnJ ooeXP[i'u l

IR k=l

11 JeXP[i,u'A(ek)t dt • Jk= 1 IR .,J2; IR oo

n 2 2 Jn IA(ek)1 ).k=l IR

Elementary computation ascertains that

110

Using this we get that

nII I exp(_&u 2 2 IA(ek) 1 2 ) for all n .ffi k=l

00

Assume that

for

limn_

2IA(en) 1 2 = 00. This yieldsn=l

J aIR

all u > a. By the dominated convergence theorem

J J limIRoo !Roo n-

J00 1 • ( 1,IR

we get

which is a contradiction.

•

Proposition BB: Consider a linear measurable functional A. If

then

A a a.e.

Proof: A(en) a for all implies that

and define

L a }

s a }E+ = -E and since the measure

E {xEE

AI 00 = aIRa

Let E denote the domain of definition for A

E+ xEE

Since A is linear, we have that

is symmetric, we get

+(E )

Then it is obvious that

for every and likewise for E

111

are constant with respect to every finite numberboth 1 + and 1E E

of variables. From the Kolmogorov zero-one law we get that

either 0 or and likewise with E Then

'"Y(E+) = '"Y (E- ) =

and hence

'"Y({ xEE I o }) '"Y(E+ n E-}

+'"Y (E ) is

•

We are now ready to prove theorem 4B.

Proof (Theorem 4B): Since 1) amounts to proposition 7B, it

remains to prove 2}, i.e.00

I a.e.n=l

Since by proposition 7B00

I IA (en) I2 < 00 rn=l

theorem 3B ascertains that00

In=l

converges a.e. in moo and defines a linear measurable functional A .

We must prove that A = ". For we have00

A(ek) = I A(en)n=l

By proposition 8B we conclude00

I a.e.n=l

•

For arbitrary functionals "1 """n' we wish to calculate the

measure of sets of the form

112

where

{ co I k=l, .. ,n }xE:lR a k< bk , ,ak,bk are real numbers. We shall often use the

shorter notation

= {co

I }[i\>c] xE:1R

Lemma 9B: Consider a sequence {fn}n=l of measurable functions

in !R'" with values in lR. If {fn}n=l converges almost everywhere

to a function f, i.e.

f n f pointwise for a.e.co

xE:lR

then to every cE:lR there exists a sequence {Ck}kE:m fulfilling

kck- - - ...., c

and

-r[f>c]

Proof: Assume that [f=c] o and denote by M the

zero-measure set on which {fn}n=lget

does not converge to f Then we

pointwise on the set

IR'"

l[fniC] n 'l[f>c]

lR'" \ (M U [f=c]) i.e. almost everywhere on

By using the dominated convergence theorem we get

-r[fniC] ----.... -r[f>c] .n

The case when the set [f=c] has positive measure now follows.

We find a sequence {Ck}kE:m of real numbers fulfilling

ck c and -r[f=ck] = 0 for every kEmk

This is possible, since there would otherwise exist an

uncountable family of disjoint sets with positive measure,

contradicting the fact that the measure -r is finite.

113

By applying the above established result to the zero-measure

sets [f>ck] and using the dominated convergence theorem, we get,

= limk--

The measure of the sets

lim lim .k--n--

•

where

f 2>c 2 , .. , fm>c m] ,

f 1 ,f 2' .. ,fm are measurable functions, can be calculated in a

similar way.

We shall apply the lemma with a linear measurable functional in

ffioo , denoted by A. It has been proved earlier that00 00

n

xErnOO IJ

2anXn with 2 < 00n=l n=l

We denote by AN the sum of the terms with indices from

Then by lemma 9B we get

= lim lim ,k--N--

where the sequence {ck}kErn converges to cErn .

We now calculate

1: akxk i cm }k=l

... .. dxnffin

where

to N.

n

M = { I 2akxk i Cm }k=l

By choosing an orthogonal transformation in

spanned by

rn n sending the line

n

lI!!n112 = l into the line spanned by the first natural basisk=l

vector in rnn, we get by using the transformation theorem

114

J

112}

00

J

cm/ 112

We choose A wi th By first lettingi.e. 2a;n=l

go to infinity the above expression reducesand afterwards m,n ,

to

with A

00

,[A>C] = J

c

being a linear measurable functional in00

IR with00

2 1A ( en) I 2 = 1n=l

The expression can easily be extended to00 00

cm

= J ... J

c,where denote linear measurable functionals in

00

IR all

with00

and the vectors

2 1Ai (en) I 2 = 1n=l

{ain

for i=l, .. ,m

orthogonal for different indices

i . One hereby obtains an orthogonal transformation in rnN with

A simple set theoretical argument together with the additive

property of the measure give us the expression

b1

bm

.. = ... J ..

a 1 am

where A1, .. ,Am are measurable functionals in00

IR withco

'\ IA.(e )12 =1 for i=l, .. ,m,L n

and {a in

n=l

Ai(en)}nErn orthogonal for different indices i

115

c: Linear transformations in 00IR

We shall extend unitary transformations of

maps in 1R00 that preserve the gaussian measure.

onto to

Definition 1C: A weak measurable linear transformation in 1R00

is a map

A y 00Ax E IR

where

1) The domain of definition EA is a linear set of full measure.

2) The map A is linear.

3) Every coordinate function Ym IR is a linear measurable

functional in00

IR

There are some comments in connection with this definition.

1) According to the earlier paragraph concerning functionals in00

IR we have that00

2amnXnn=l

with

matrix

00IR wein

(infinite)

L 2 .

thebygiventhenisAtransformation

{amn}m,nEm where the rows {amn}nEm are elements in

2) By the maximal domain for a linear functional

The

understand the set

If we let

00

{ I 2A(en)Xn converges}n=l

Em denote the maximal domain for the functional and00

set E = n Em' we get thatm=1

EA C E and = 1

If EA # E we extend in the obvious manner the transformation A to

the whole E, thus getting A maximally defined.

A

116

3) It follows from the earlier results that

is uniquely determined by its values on

[2 C EA and that

It is indeed

determined by its values on the set

where

en (0 I 0 I •• ,0,1 1 0 . . . )

icoordinate number n

4) The converse holds as well; if elements of [2 are taken as

the rows of a matrix A = {amn}m/nEm I

linear transformation in moo

then A is a weak measurable

Let A denote a linear bounded transformation

A : [2 [2 .

Then we extend A uniquely to a weak measurable linear transformation

in by extending the domain of definition for the matrix

given by

a rnn = .

Since the rows are elements of [2 I by the Parseval identity00

'\ a 2 =L mnn=l

00

l 1< Aen/em >1 2n=l

00

ll< en/A*em >12 = IIA* emI1

2< 00

n=l

the extension is well defined.

Theorem 2C: Let u denote a unitary

transformation. We extend u00

to a weak measurable linear map in m

by the matrix {u =mn Then the gaussian measure

is invariant under the transformation U

Proof: Consider nEm and ak/bkEm I k=l / .. ,n We define

B { xEm00 I ak

and that

-r(B)

b,

(21T)-ft n J ..a,

117

bn

J .. .. dx nan

U(B) = { I ak < bk for k=' , .. ,n } ,*where U denote the extension of the adjoint of U Since the rows

*in the matrix of U are the columns in the matrix of U

that00

U(B) = { I a k< 2UikX i bk for k=' , .. ,n } .i='

We define the functionals00

we get

k=', .. ,n ,

U are pairwise orthogonal,

fork=" .. ,n}

and observe that since the rows in

-r(U(B)) = -r{ I ak< bkb, b n

(21T)-ft n J.. J .. .. dx na, an

D: The gaussian measure on

Consider the linear space00

= { I for all iEIN }lC = ( z, ,z 2 ' ... , zn ' .. ) z.ElCl.and the Hilbert space

00

£2 (lC) = {co 2 I z 12 < 00 }Z E lC n

n=1and introduce in lCn , nEIN , the gaussian measure

where

•

118

and

and dx,dy, ... dxndYn indicates the Lebesgue integration

in 1R 2n .

Then all former results concerning the gaussian measure are easilyIX)

extended to this new gaussian measure on

E: Hilbert-Schmidt enlargements

Let }{, denote a real or complex Hilbert space. A

Hilbert-Schmidt enlargement X, of }(, is itself a new Hilbert

space containing as a linear dense subset and such that the

inclusion mapping of }{ into X is a Hilbert-Schmidt operator, i.e.IX)

for every orthonormal basis where 11,11 and

denote the norms corresponding to the inner products

respectively.

and

119

Lemma 2E: Let A : HxH [ denote a map which is conjugate

linear in the first term and linear in the second. If there exists a

constant K fulfilling

IA(X,y) I s K·llxll- flyII for all x,yEH,then there exists a linear bounded operator

A : H H

fulfilling

A(X,y) for all x,yEH.

We have the following fundamental result.

Theorem 3E: There exists an orthonormal basis

which is a complete orthogonal system in X .

Proof: We define

in H

A(X,y) HxH ---..., [.

By lemma lE there exists a constant K fulfilling

!A(X,y)! K-llxll·fIyll for all x,yEH

by which there exists an operator J : H H with

for all x,yEH

It is easily seen that J is a strictly positive (hence self-adjoint)

operator and that if

then00 00

denotes an orthonormal basis in

00

H ,

tr(J)

00

l IIen 11 2 < 00 ,n=l

i.e. J is a trace class operator.

Now choose an orthonormal basis

eigenvectors for J with eigenvalues

{bn}nEIN in H consisting of

120

We then have00

1 ) The series {An}nEm is convergent, i.e. l An < 00 sincen=l

J is of trace class.

2) The orthogonality follows easily for n,mEm

= = An = AnOOn,m

3) It is obvious that {b } is total in H,n nEm

•

Corollary 4E: There exists an orthonormal basis {bn}nEm in

H, and a convergent series of strictly positive numbers {An}nEm

such that

{00

'\ x bL n nn=l

00

l An0 I xn 1 2n=l

Proof: We choose {A } as constructed in then nEm

proof of theorem 3E and the corollary follows.

•

F: The gaussian measure on Hilbert-Schmidt enlargements (cf. [23])

Using the orthonormal basis in H and the positive

eigenvalues {An}nEm from the above paragraph we get

= = A for xEHn n n n

Let us define00

[2 { lC00 l IXnl 2 < 00 }x En=l

00

I 2 { c00 l Anlxnl2 < 00 }x En=l

121

we can define the gaussian

By identifying

!t - L2

)( - 12

via the orthonormal basis {bn}nEIN in !t ,measure a = on the Borel sets of )( as the measure

a'Y

on the corresponding Borel sets in L , and then get

= 'Ya (Z2)

Since unitary transformations of L 2 onto extend to orthogonal

weakly measurable linear transformations in and the measure a'Y

is invariant under these transformations, the measure is invariant

under the corresponding extensions of the unitary maps between Hilbert

spaces. In particular, we have that the measure does not depend on

the chosen orthonormal basis in !t used for identifying !t with L2 .

We would like to show that the gaussian measure does not depend

on the selected Hilbert-Schmidt enlargement. This can be done by

showing that if !t, and denote two different Hilbert-Schmidt

enlargements then there exist a Hilbert-Schmidt enlargement Kfulfilling

i=' ,2 .

If f denotes a continuous function defined on !t with values

and f 2 are equal on )(,

in and f, and

respectively, then

are continuous extensions of f to !t, and

hence equal

almost everywhere.

Definition , F: Let and denote

Hilbert-Schmidt enlargements of a Hilbert space !t, •

!t, is finer than !t2 if the identity

I : !t ---> !t

extends to a continuous and one-to-one map

We say that

122

It is obvious that if *1 is finer than *2' then there exist

a constant C > 0 such that

for all xEl{. The smallest C fulfilling the above is the operator

Lenuna 2F: Consider seminorms {II· Iln}nEIN and define 11'11 * in l{

setting

Ilxll; = 2n=l

If a sequence fulfills

and for every nEIN

------> 0p,q

then

a ,

a .

Proof: The proof amounts to an adjustment of the well known

method of verification that the countable direct sum of Hilbert spaces

is complete.

•

Lenuna 3F: Let l{1 ' : l{ ------>

is a continuous linear functional, then there exists a Hilbert-Schmidt

enlargement l{2'2 •

123

Proof: Use theorem 3E to choose an orthonormal basis

in H, such that {en}nEm are pairwise orthogonal with respect to

Then

with

and

n ' 1

Expanding aEH we get

a

II e 11 21

•

124

* is well defined on

We define the inner product

125

00

2 Ilepll; < 00 •p=l

Then the completion *2 of *' 0p

126

and thereby

o .

•

Lemma 4F: If denotes a

Hilbert-Schmidt enlargements of a Hilbert space X,

sequence of

all finer than

then there exist aa fixed Hilbert-Schmidt enlargements

Hilbert-Schmidt enlargement

nEIN .

XO'

and conclude that

127

is well defined.

Let denote the completion of and let {ep}pEm

be an orthonormal basis in *,. Then we have00 00

2 2 2p=l p=l n=l

00 00 00

2a 2 = 2a - c2n n nn=l p=l n=l

We have to prove that H is a

00

22-n < 00 •n s l

Hilbert-Schmidt enlargement of

* . Take nEm We must show that H is finer than *n It isobvious that the inclusion mapping

is continuous.

It remains to prove that the extension of I over is

one-to-one. Assume that {Xp}pEm C * is a Cauchy-sequence in the II- II

Then for every nEm----> 0- norm and that IIxp 11 0

We know that for every n

----->, 0 .p,q

the enlargement is finer than *0 and

hence

Since

----> 0p

implies

o .

we get by lemma 2F that

00

'\ a -llx 11 2L n p nn=l

II xp II ----> 0pwhich proves that the inclusion mapping of H into *0 is one-to-one.

Since the inclusion mapping of into *0 is one-to-one, the

injection of H into *n must be one-to-one as well.

•

We are ready to prove the announced principal result.

128

Theorem SF: To every pair of Hilbert-Schmidt enlargements of a

Hilbert space there always exists a Hilbert-Schmidt enlargement that is

finer than any enlargements from the pair.

and H2'

129

the functionals {An}nEm are continuous in both the - norm and

the II· 11 3 - norm and that these functionals are dense in Ut, 1)' and (K3'

References

[1] V. Bargmann, On a Hilbert Space of Analytic Functions and an

Associated Integral Transform Part I,

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[9] R.L. Hudson and R.F. Streater, Ito's formula is the chain

rule with Wick ordering, Physics Letters, vol 86A, number 5,

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butions in infinitely many dimensions 1,11,111

I Commun. Math .Phys. 1 (1965) page 175-214.

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page 99-124.

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real wave representation 2,3,53,72,77,90,91

second quantization 30Stone 29,30tame 1°°,101 , 102total 16,53,66,67,92,118,120ultracoherent 57vacuum 1,4,10,20,23,65,83,85,92value of 66,90weak measurable linear trans-

formation 115,116,121Weyl 1,45,52,91Wick 22,45,59,79,81,83,88Wiener 16,20Wigner 3,86

Subject index

annihilation operator 1,3,6,8,10,21,25,61,64,79,81,83,85,92

anti-commutator 1anti-normal 92base space 1,2,4,10,23,65,83,85,

87Bose albebra 1,2,3,4,6,10,11,20,

22,23,65,79,83,84,85,88- extende 36,65,83- Fock space 1,2,3Campbell-Baker-Hausdorff 50,51,

59,91,92coherent 2,33,34,43,44,63,64,66,

83,89,92commutation 9,10,11,12,45,55,59,

60,65,82,85,93complex wave representation 2,53,

66,69,70,71,78,80,87,90conjugation 1,2,3,53,54,55,57,69,

73,74,76,77,80,84,88,89,90creation operator 1,3,6,8,10,21,

25,64,79,81,83,85,92cylinder set 96,97,101,107derivation 1,6,10,11,30Fourier transformation 25,31,32,71free commutative algebra 1,4- product 1,4,18gaussian content 96- measure 3,68,76,78,80,86,96,99,

107,115,116,117,118,120,121,122Halmos 27,94Heisenberg 33,43Hermite 2,31,32Hilbert-Schmidt enlargement 68,71,76,

80,86,87,90,91,118,120,121,122,125,126,127,128

Kolmogorov extension theorem 97inequality 103,105

- large number theorem 104- zero one law 102,111Leibniz rule 6,30,31,36,37,39,61,82linear measurable functional 102,103,

106,108,109,110,111 ,113,114,115,117,122,123,128,129

Nelson 27one-parameter group 29t- picture 65,84,88

product 59,61,62,63,64,65,78,88value 72,90