Von Neumanns inequality and Andos generalization

Von Neumann's

1.

and Ando's

inequality

generalization

Summary: In this chapter, we prove (actually three times) von Neumann's inequality and its extension (due to Ando) for two mutually commuting contractions. We discuss the case of n > 2 mutually commuting contractions. We introduce the notion of semi-invariance. Finally, we show that Hilbert spaces are the only Banach spaces satisfying von Neumann's inequality. First

we

duction)

for

we

may

will prove

Neumann's

von

operator T

an

identify

T with

inequality (denoted by (vN)

in the intro-

finite dimensional Hilbert space. Equivalently n matrix with complex entries and we compute

on a

an n x

of the operators T or P(T) as operators on the n-dimensional Hilbert Cn equipped with its standard Hilbert space structure). fn (i.e. space 2 Let us first assume T unitary. Then by diagonalization, there is a unitary

the

norms

operator

v

and zj,

.

.

.

zn in o9D

,

such that 0

Z1

T

=

v*

V.

0

It follows that for any

polynomial

Zn

P

P(Zj) P(T)

=

0

v*

v

P(zn)

0

hence

IIP(T)II

=

maxlP(zj)l

<

jjPjj,,,),

so

that

we

have

(vN)

in that

case.

j
Now

assume

JIT11

we

A,

ITI

that T is

merely

a

contraction

on

By the polar decom-

fn 2.

UITI with U unitary and ITI hermitian. Note that is diagonalizable so that there is v unitary such that Moreover, ITI ITI 11. 11

position,

have T

::--

0

v* 0

An

)

v

with 0 <

Aj

< 1.

Let

us

0

ZI

T(zl,..

-,

Zn)

=

Uv*

(0

denote for all zi,

..

Zn

)

v

.

Zn in

1. Von Neumann's

so

that

T(A,,..., An)

=

inequality and Ando's generalization

T.

zn) (i,

polynomial. Then there are polynomials pij (zi, n) analytic in each variable such that

Let P be any

1, 2,..

.,

15

P(T(zi,..., Zn))

=

(Pii (ZI,

Zn))ii

This shows that the function

(Zi, is subharmonic in each

peatedly

But

.

-

-

Zn)

,

---4

11 P(T(zi,

.

.

.

Zn))

,

variable, hence by the maximum principle (applied we have in particular

re-

variable)

in each

I I P(T(A 1,

-

.

-

,

An)) I I

<

supf I I P(T(zi,

by the first part of the argument, 1, this yields zn I

.

.

since

.

zn)) 11, 1 zi IIZnI

,

T(zl,

.

.

.

,

zn)

is

=

11-

unitary when I zi I

=

IIP(T)II hence

obtain

we

(vN)

in the matrix

<

11PI1,,

case.

infinite dimensional Hilbert space H. (I am grateful to Vania Mascioni for correcting a misconception in a previous version.) For simplicity of We

now

consider

an

separable and let En I be an increasing sequence of finite H. Let Pn be the orthogonal projection dimensional subspaces such that UEn from H onto En and let Tn TnTTn Clearly JJTnJJ JIT11 hence by the first have part of the proof for any polynomial P we notation

assume

H

=

V

IIP(T-)II

n

IIPII..

:5

oo. Since sup JJTnJJ :5 1, that, for any x in H, Tnx --+ Tx when n 2 hence subsets of all over H, Tn uniformly compact Tn &X --+ T 2X when n -+ oo and more generally for all k, TnKkx --> T kX. Consequently, for any polynomial P

Now observe --+

we

T

have V

and

we

x

E

conclude that

I I P (T) I I

Theorem 1.1. Let T: H

H be

--+

FI containing H isometrically

n --4

oo,

IIP

:

El

is deduced from Sz.

Usually (vN) lowing.

when

P(T)x

P(T,,)x

H

as a

a

Nagy's

dilation theorem which is the fol-

Hilbert space unitary operator U: F1 --+ FI

contraction. Then there is

subspace

and

a

a

such that

Vn

>

0

T

n =

p

HUIIIF-

holds, U is called a dilation of T (one also says that U dilates T). The "strong dilation" is sometimes used in the literature for the same notion.

When this term

Proof: We follow the classical

argument of

and consider the Hilbertian direct

sum

[SNF]. For (1) Hn.

any

On

n

F1

in Z let we

Hn

=

H,

introduce the

nEZ

operator U: H

-+

H defined

by

the

following

matrix with operator coefficients

1. Von Neumann's

16

inequality

and Ando's

generalization

1

0

0

0

DT

-T*

T

DT*

U=

0

1

0

1

0

where T stands

U[(h,,),,Ez]

=

(0,0)-entry. Equivalently

the

as

(hn')nEZ

with h'n defined

if

nVI-1,01

T*hj

if

n=-l

DT. hi

if

hn+l h'n

H

DThO

=

Tho Let

us

mapped

into

n

=

0.

=

T n for all

coefficients of U' We claim that for

=

(I

-

T*T)

'I 2

and

TT*) ".

DT*

2

k so that we have PHUIH T and more generally (note that U has a triangular form, so the diagonal are the obvious ones). all (h,,)nEz in k and (h')nEz U[(hn)nEz] as above we have

H with

identify

PH Uj'

is

denote DT

We

-

+

(h,),,Ez

any

by

Ho n

C

=:

> 0

=

n

11h' 1112 Indeed, first

note the

(Note

that DT* P

we

=

11ho 112

=

11ho 112

+

11h, 112.

following identities

T*DT*

polynomial

+

=

f(TT*)

DTT*

and DT

(and =

TDT

f(T*T)

DT*T).

=

with

f

continuous and for any

have

T*P(TT*)

P(T*T)T*,

TP(T*T)

=

P(TT*)T,

approximating f by polynomials we obtain these identities). Then developing and the preceding identities, we obtain the above claim. As a consequence, we find that U is an isometry. Moreover U is surjective since it so

11h' 1112+ 11ho 112 using (*)

is easy to invert U. Given h' defined by hn hn-1 if n =

DT*ho. Equivalently for 2

x

(hn)nEZ in F1, we have h' Uh with DTh' 1 + T*hO and hi 10, 11, ho =

=

it is clear that U is invertible from the

2 matrices with operator entries

h =

=

(hn)nEZ

-Th'

1

+

following identity

inequality and Ando's generalization

1. Von Neumann's

1

0

DT

-T*

DT

T*

DT

T*

DT

-T*

0

1

T

DT*

-T

DT*

-T

DT*

T

DT*

Therefore

we

conclude that U is

a

surjective isometry, hence

a

17

unitary operator. 11

we

is easy to derive. polynomial P

result, (vN)

Rom this

unitary

do have for any

-

IP(U)I1 since

by spectral theory IIP(U)II

,9D. Now if T satisfies

Indeed,

first observe that if U is

(1.1)

we

=

have

IIP(T)II

:5

<

IIPII-1

supjIP(z)j I Z E Spec(U)j P(T) PHP(U)IH hence

and

Spec(U)

C

=

IIP(U)II

:

JIPII,,

and this proves (vN). Remark. When U is unitary, the spectral functional calculus shows that if P(z,. ) is a polynomial in z and, we can define f (U) P(U, U*) in the f (z) =

=

obvious way and

we

still have

If (U)II

Ilf 11.

:

=

If W1.

sup

jZ1=1 an n x n matrix, this using the diagonalization of U.

Note that if U is

can

be checked

by

the above

argument

denote by C (resp. A(D)) the space of all continuous functions on (resp. the closed linear span in C of the functions leint I n > 01). We equip C (or A(D)) with the sup norm which we denote (as above) by Note that A(D) is a subalgebra of C, it is called the disc algebra. P(z,, ) for some polynomial P in Clearly, the functions f of the form f (z)

Let

T

us

aD

=

=

two variables

are

dense in C. Therefore

f extends to the whole of C.

(fg) (U)

(1.2) whenever and

we

f and

obtain

a

are

have

f (U)g(u)

=

of this

uu: uu (f

such that uu

we

form, hence this homomorphism g

and also

C

the linear map

f (U)

-

Obviously,

by density

--

I I uU I I

remains true

on

the

completion

B (H) =

1. This shows that uU is

actually

a

*-representation. a general contraction T on H. By definition of A(D), polynomials P(z) is dense in A(D). Hence the linear (analytic) I by (vN)) extends uniquely to a map of is norm P P(T) (which mapping < 1. that such B A Moreover, again the multiplicativity of I I UT I I (H) UT: (D) UT is preserved so that we have

Let

us now

the set of all

--

--

return to

18

1. Von Neumann's

(1.3)

Vf, g

It is customary to write

inequality and Ando's generalization

A(D)

E

UT

f (T)

for

Vf

A(D)

UT(f)

(fg)

UT

:--::

(f) UT (9)

whenever

f

is in

-

A(D),

so

that

we can

write

(1.3)' In UT:

conclusion,

E

Neumann's

von

11f (T) 11 :5 11f leads to

inequality

several

complex variables. Quite surprisingly, impossible in 3 variables. We

Theorem 1.2. tions

on

H,

i.e.

(Ando's we

homomorphism

it turns out that this is

Then H

[Anl]).

theorem

start

Let

by the 2-variable

T1, T2 be

JIT211

:!- 1

and

!, 1

be embedded

Vn,k Consequently, for

any

polynomial P(zl, z2) i

Remark. Of

P. Un U2k I H

> 0

I I P(T1 T2) 11

commuting contrac-

'5-

space

in two variables

SUPf I P(Z1 Z2)1 I i

zi, z2 E

-:z:

U2 V1

::*

U2 UT

1

on

which there

we

have

9DI.

the second part follows from the first Nagy's dilation theorem. Just observe

V1 U2

F1

Tin T2k

-::::

course

follows from Sz.

in

TjT2 =T2T,.

isometrically into a Hilbert commuting unitary operators U1, U2 such that can

two

possible case.

have

JIT111

and

I

B(H)

--+

2 variables but

are

a norm

which maps the function eint to Tn for all n > 0. It is natural to wonder whether (vN) extends to the case of polynomials in

A(D)

one

exactly

as

(vN)

1

---::

U - U2

1

Ul*, hence not only U1, U2 but also U1, U2, Ul*, U2* all mutually comotherwords, all lie in a commutative unital C*-subalgebra A c B(H). By Gelfand's theorem A is isomorphic to C(K), where K is the set of all homomorphisms

U -

=

mute or, in

X: A

-+

such that

C

X(1)

=

11XII.

We have then

11 P(U1 U2)

SUP

i

I (X7 P(U1 U2))l 7

XEK

SUP xEK

hence since

IX(Ul) I

IX(U2)1

=

IP(X(Ul))X(U2))l

1

JjP(U1)V2)jj:5

SUP

JP(Zl5Z2)1-

jZ1.1=1 IZ21=1

Before proving Ando's tries. An operator T: H

Equivalently,

this

means

--4

theorem,

we

H is called

that T*T

=

first

gather more information on isomeisometry if I I Tx I I I I x I I for all x in H. In general an isometry is not surjective.

an

I.

=

1. Von Neumann's

inequality

and Ando's

19

generalization

An isometry is surjective iff it is a unitary operator. The fundamental example H (D H ED (direct of an isometry is the shift operator on the space f2 (IN, H) sum of a collection of copies of H indexed by IN), namely the operator 8 on f2(IN,H) which maps (XO,Xl,X2, ) to (0iX0iX1iX2i ). We will call a shift ...

=

...

...

operator of this kind for

any

called the

Hilbert space H. (The dimension of H is The next result is quite important. It shows

some

multiplicity of the shift.)

isometry can be decomposed as the direct sum of unitary operator, this is called the Wold decomposition.

that any a

Theorem 1.3. Let V: H

H be

--+

shift

a

above and

nVI(H)

isometry. Let H,,,,=

an

as

and let

n>O

Ho

=

He

for V

H,,.

so

(i.e. they

(i) VJH (ii) VIHO

:

:

that H

are

=

Ho ED H,, . Then H,,,) and Ho

invariant under V and

are

reducing subspaces

such that

V*)

H,,, --+ H,,o is a unitary operator from H00 onto itself Ho -4 Ho is unitarily equivalent to a shift as above with multiplicity to

equal

dim(H

E)

V(H)).

V(H(, )

Proof: The inclusion

is clear. Moreover if

H.

c

x

Vn+ly

=

then

that H,,,) and Ho

axe V* VVny V* X V'y hence V* (H,,,,) c H, This implies H,,,,, hence VIH-: H". --+ reducing subspaces for V. Moreover, clearly V(H )) H,, is a surjective isometry, hence is unitary. This proves (i). To check (ii) hote =

=

=

..

that

H.-L where

En

=

it preserves

=

Vn (H) 8 Vn+

angles

hence

(nU

p-an 1

(H)

we

Vn (H)

map U:

HJo-

f2(Eo)

-+

is

(U E,,)

pan

n

for all

have En

n

> 0.

Note that since V is

Vn (Eo) for all

=

Vn (Eo) I V' (Eo) It follows that

J-

for all

n

isometrically isomorphic to defined by U(xo,x, )

H ,L.

> 0

an

isometry

and

m.

6(Eo). Indeed, the E Vnxn is a unitary oper-

the space

=

....

:

n

n>O

Finally we clearly have (with the preceding notation) completes the proof of (ii). ator.

V

=

UsU-1. This 11

Hilbert space and let v: E --+ E be an operator defined on a subspace E C H. We will say that an operator V: H --+ H extends v if V(E) C E and V(x) v(x) for all x in E. Let H be

a

=

As

an

immediate consequence of Theorem 1.3

Lemma 1.4. a

Any isometry

V: H

i.e. there is

unitary operator, a unitary operator U: k

and

-+

--+

H

on a

we

have

Hilbert space

can

be extended to

Hilbert space F1 containing H k which extends V.

a

i ometricaljy

f2 (IN, H), we can take for U the so-called "bilateral shift" on f2 (Z, H) which is unitary and clearly extends V. On the other hand if V is unitary the above statement is trivial. By the decomposition in Theorem 1.3,

Prooh If V is

the

general

a

case

shift

s on

reduces to these two

cases.

El

20

1

Von Neumann's

-

Remark. If we wish i.e. such that the

inequality

always

we can

subspace

and Ando's

ensure

generalization

that the

span( U U'(H))

E

unitary operator U is minimal is dense in F1. Indeed, if E is

nEZ

dense, unitary.

not

simply replace

we

Lemma 1.5. Let

V1, V2 be

by T

H

and note that E is U-invariant and

commuting isometries

two

on a

Hilbert space

a Hilbert space F1 and commuting unitary operators H as an invariant subspace and such that

then there is

admitting

U11H In other

words,

two

=

commuting

and

V,

U21H

isometries

=

U1, U2

on

is

it H

V2-

be extended to two

can

UI-E

commuting

unitaries.

Remark.

Actually

reasoning)

that the

it is easy to

the argument to

modify

is true for any number

same

yield (by an inductive (and even any set) of commuting

isometries.

F1 and let Uj: F1 --+ F1 be a minimal unitary V, obtained by applying Lemma 1.4 and the remark following it,

Proof of Lemma 1.5. Let H C extension of i.e.

we assume

that the

subspace

E formed of all the finite

E Ul' h,

hn

G

of the form

sums

H

nEZ

is dense in

k.

This will, allow we

us

to define

will do this without

commute with

U, and

an

the

"spoiling" V2,

to extend

f12

Un h,,

nEZ

To

verify that

an

isometry,

this definition is

we

we

(E )

i72

of V2 which commutes with U, and good properties of V2. If we want f7, to

extension

must define

=

f72

on

E

as

follows

1: Uln (V2 hn) nEZ

unambiguous

and at the

same

time that it defines

observe that

Uln V2 hn

1: (Un V2 hn, Ulm V2 hm) n,m

(Ujn-nV2hni V2hm) n

+

E (V2hn) Um-n V2hn) 1

n<m

m

hence since U, extends V,

( V 'n-,nV2hni V2hn)

+

n>m

since

V2 and V,

(V2Vn-hn) V2hm) V2 is

an

V2hn) Vm-n V2hn),

commute

n>m

since

1: n<m

isometry

+

1: (V2hni V2Vm-n hn) n<m

1. Von Neumann's

(Vln-'hn, hm)

+

finally (applying

the

21

generalization

1: (hn7 Vm-n hm) n<m

n>m

and

and Ando's

inequality

IT)

for V2

preceding calculation

III: Ulnhn 112. This shows that V2 is

i72ho

=

V2ho for

all

ho

an

isometry and is

in H hence

f72(H)

unambigpusly

defined.

and V2 extends V2.

C H

Moreover, Finally, t he

important point is the following: we claim that if V2 already is unitary, then 72 still is unitary. Indeed, recall that an isometry is unitary iff it is surjective, or H is surjective, clearly the equivalently iff it has a dense range. But if V2: H operator f72 that we just defined has dense range (its range contains E), hence is unitary. This proves our claim. Hence if V2 hgpens to be unitary, V2 also is and we are done. To complete the proof in case V2 is not unitary, N7 e simply repeat the construction applying Lemma 1.4 this time to the isometry V2 instead of V1. By the above claim we can get a unitary extension Of f72 and a (still) unitary D extension of U1. This gives the desired-result. The following proof is essentially the original one from [An1J. Proof of Theorem 1.2. By Lemma 1.5, it suffices to show that two commuting contractions can be dilated to two commuting isometries (this is the step which is restricted to "two"!). Let T1, T2 be mutually commuting contractions on a Hilbert space H. Let H+ I;[ H G) (or H E2 (IN, H)) be the direct sum of a family of copies of H indexed by IN. Let Wi: H+ -4 H+ and W2: H+ -4 H+ be the operators defined by ---

=

V h

=

(ho, hi

=

...

....

Wi h

)

=

(Ti ho, DTi ho, 0, hl, h2 i...)

Ti* Ti) 1/2

proof of Sz. Nagy's dilation theorem. Clearly W, and W2 are isometries on H+, but in general they do not commute. We will modify them to obtain commuting isometries. Let H ED H G) H ED H and let us identify H+ and H (D H4 (D H4 H4 (D by H4 the following identification: where i

=

1,

2 and where

DT,

=

(1

-

as

above in the

...

=

h

=

(ho, f hl,..

(h,,),, o

-,

h4b f h51

On the space H4, consider a unitary operator v V: H+ -+ H+ be defined for all h (ho, k1, k2

....

h817

...

specified later) and let ho E H, k1 E H4, k2 E

(to

be

=

H4,... by the following formula Vh

Clearly

V is

=

(ho, vkl, vk2

,

...

unitary and V-1h VW1 and V2

We define

V,

key point

is that

=

v can

=

=

(ho,v-1k1,v-1k21

...

W2V-'. Clearly these

be chosen in order to

ensure

are

isometries

on

H+.

The

that V, and V2 commute.

1. Von Neumann's

22

To check this let in

H+

with ho E

and Ando's

inequality

generalization

compute V, V2 h and V2 V, h for an element h (ho, kj, k2 H, ki E H4, k2 E H4, etc. By a simple calculation we find

us

=

VIV2h

=

V2 V, h

=

(TjT2ho, vf DTT2ho, 0, DT2ho, 01, ki, k2 ...) 7

and

Hence

(since TIT2 V

By

a

ho

c

=

(T2 T, ho, f DT2T, ho, 0, DT, ho, 01, ki, k2

T2T, ) if

we

VjV2

want

=

v(f DTT2ho, 0, DT2ho, 01)

H

V2V, =

we

7...

).

must define

f DT2Tj ho, 0, DT, ho, 0 1.

simple computation, the reader will check that

jDT,Tjho,O,DT,ho,Oj have the

same norm

in

H4,

A, onto the

=

and

fDTT2ho,O,DT2ho, 01

hence this defines

v as an

isometry from the subspace

f (DT, T2ho, 0, DT2ho,O) I ho

E

HI

f (DT2Tj ho, 0, DT, ho, 0) 1 ho

E

Hj.

subspace A2

=

By density, of course v defines an isometry from the closure of Al onto the closure of A2. To extend v to an isometry of H4 onto itself (i.e. to a unitary operator,on H4) it suffices to check that H4 E) A, and H4 E) A2 have the same (Hilbertian) dimension. If H4 is finite dimensional, this is clear since A, and A2 are isometric, and if H4 is infinite dimensional it is equally clear since the dim H and the zero dimension of H4 (D A, and H4 E) A2 are at most dim H4 coordinates in the definition of A, and A2 ensures that they are at least dim H. This shows that v can be chosen so that V, and V2 commute. Thus we obtain commuting isometries V1, V2 which dilate TI, T2. Applying Lemma 1.5 to V1, V2 ==

we

obtain the conclusion.

Remark. Here is

this

a

El

third route to prove

(vN). (This

was

communicated to

me

[Dr]).

This approach algebra A(D) is the closed convex hull of the set of finite Blaschke products. Taking this for granted for the moment, it is easy to see that (vN) reduces to proving I I f (T) I I < 1 when f is a finite Blaschke product, or merely when W is a Blaschke "factor" i.e. a

by

J.

Arazy,

is based

on

proof

the fact

appears

[Fi, R2]

already

in

Drury's

survey

that the unit ball of the disc

M6bius map of the form A

Z

(P,\ (Z)

for

=

i

-

Z

But for such maps (vN) is easy: indeed B(H), JITIJ < 1 and x E H

II(T

-

A)XI12

_

11 (1

-

T)x 112

=

by

a

JAI

< 1.

simple

(JITx 112

_

calculation

IIXI12)(1

_

we

IA12)

:

find if T E

0

1. Von Neumann's

hence

(T

-

A) (1

-

AT)

-

1

11

< I

inequality and Ando's generalization

which

means

that

II

px

(T) I I

This

completes

f (T)

is

a

be

a

< 1.

our third proof of (vN). (Note that we need to know that f --> continuous to pass to the closure of the convex hull, but this

23

can

priori priori

1 in the end.) guaranteed by replacing T by rT with r < 1, and by letting r hull closed of the of the finite unit is ball that the convex us verify A(D) Blaschke products. Fortunately, there is a very elegant and simple proof of this due to Alain Bernard (cf. [BGM] for extensions), as follows. It clearly suffices to show that any polynomial f in A(D) with 11f 11 < 1 lies in the closed convex hull N of the finite Blaschke products. Let f be such a polynomial. Choosing g(z) z with N deg(f) we find an analytic polynomial g of modulus one on aD such that gf E A(D). Consider then, for any real number t, the function --*

Now let

=

=

f

ft Note that

ft

clearly analytic.

is

ft(Z) which

We claim that

if W,\ denotes the M6bius map

Indeed,

implies that I ft (z) I

is continuous

on

D,

=

=

1 when

it must be

ft has at most N zeros.) Finally, viewed as a function

a

eit9

+

1 + el t gl'

as

ft

is of modulus

above then

we can

one

on

aD.

write

W-f (Z) Wtg(z))

IzI

=

words, ft is inner. Since it product for each real t. (Note

1. In other

finite Blaschke

that

of

e t, ft

is the

boundary value of the analytic

function

f

(defined

for

E

D),

hence

we

0

+

I +

W

have

(taking

f

ft

=

0)

dt -

27r

f lies in the closed 11 products. Remark. There is a very striking analogy between the preceding argument and the known proofs of the Russo-Dye Theorem in the C*-algebra case (see e.g. the proof in [Ped, p.4]). Concerning the latter proof, it is important to observe that, when we represent a point in the open unit ball of say B(H) as the barycenter of a probability measure supported on the unitaries, the "representing" measure is actually a Jensen measure (as in the preceding remark), so that the barycenter formula is valid not only for affine functions but also for analytic ones. This leads to one more transparent proof for von Neumann's inequality. Then,

convex

a

suitable discretization of this average shows that

hull of the finite Blaschke

The problem to extend Ando's theorem to three (or more) mutually commuting contractions remained open for a while until Varopoulos [VI] found a counterexample (another example was given by Crabb-Davie [CDj). Of course

1. Von Neumann's

24

inequality

and Ando's

generalization

implies that Ando's dilation theorem (the first part of the preceding Theo1.2) is not valid for three mutually commuting contractions. For an explicit construction of three mutually commuting contractions (Ti, T2, T3) which do not dilate to three commuting unitaries, see [Parl]. Moreover, Parrott's example

this

rem

satisfies

IIP(Tli T2, T3)II

<

SUPfIP(Zli Z27 Z3)1 I (Z17 Z2, z3)

E

D31.

though it does not dilate, the triple (T1, T2, T3) does satisfy inequality. This shows that Parrott's example is quite different from the ones of Varopoulos or Crabb-Davie. See the notes on Chapter 4, for variants of Parrott's example. We will describe one of Varapoulos's counter examples in chapter 5. The CrabbDavie example in [CD] is very simple. The three commuting contractions act on an 8-dimensional space and the polynomial is homogeneous of degree 3. Let us briefly describe this example. Let H be an 8-dimensional Hilbert space with orthonormal basis e, fl, f2, f3, 91) 92) g3, h. We define the- operators on H by 1, 2, 3) Tie fi, Tifi letting (for i, j, k -gi, Tifj A (if i 4 j, with k :7: i 0. We observe that and and k 4 j), Tigj Tih 6ijh In other

the

von

words,

even

Neumann

=

=

=

=

f-h (1.4)

TiTjfk=i

ifi=j=k i,j,k are all

h

if

0

otherwise.

different

Then it is easy to check that T1, T2, T3 are mutually commuting contractions. Let P be the homogeneous polynomial of degree 3 defined by

P(Zli Z2, Z3)

=:

ZIZ2Z3

Z31

_

_

Z32

Z3. 3

_

IIPII,,,, < 4, but on the other hand 4h therefore (1.4) yields P(T,,T2,T3)e

It is any easy exercise to check that

have

TiTjTke P(TI, T2, T3) 11 >

we

TiTjfk, > I I P 11,,,.

=

4

hence This

=

yields

the announced

However, rather surprisingly the following question Let

n

> 1.

Let

us

introduce the

Cn

=

(possibly infinite) SuPf IIP(TI,

-

-

-,

example. apparently still

is

open.

constant

Tn) III

where the supremum runs over all on H (infinite dimensional Hilbert

IP(z1,...,z,,)I C2

=

I and

:5 1

we

mutually commuting contractions T1,..., T,, space) and over all polynomials P such that in for all (z1,...,zn) (o9D)n. We have just seen that C,

mentioned

(cf. [VI, CD])

that C3 > 1.

Problem. Is C3 finite? Is C,, finite for any

n

>

3?

(Note

nondecreasing.) The

following related

observation is

certainly well known:

that Cn is

clearly

1. Von Neumann's

Proposition n

--+

1.6. For all n,

and Ando's

25

generalization

CnCn. Therefore Cn

>

1, Cn+,n

>

rn

inequality

-4

00

when

oo.

Proof: We may also infinite. Fix

assume

> 0.

-

Cn and Cn are finite otherwise Cn+,, is clearly Tn and P be as in the definition of Cn with

that

Let T1,

.

.

.

HP(TI)

,

...

1

Tn)11

Cn

>

-

--

Tn+,,, and a polynomial Similarly consider commuting contractions Tn+,, such that Q (Tn+ 1, I on (0D) Tn+m) > Q (-n+ 1, Zn+m) with I Q n+ fori 1,2,...,n andTj forj C,-E. Let i j IHOTj Tjolff m

.

1,...,n

+

m.

fTj

Then

commute and if

I j

1,...,n +mj

=

.

,

=

(n +m)

are

contractions which

let

we

R(zl,..., zn+,,,) we

.

=

=

=

=

Zn+m)

zn)Q(zn+l,

P(Zi,

find

R(T,,..., Tn+m)

Tn)

P(Tl,.

=

0

Tn+m),

Q(Tn+l,

hence

JjR(Tj,...,T,,+m)jj Thus

we

subspaces

--

E c

IIP(T,,...,Tn)ll.IIQ(Tn+,,...,Tn+,.)II

>

(C,,-e)(Cm-s).

E) (Cm E). In particular, Gn C3n --+ oo. El unitary operator. It is natural to wonder what kind of H have the property that if T PEUIE we have

conclude

Let U: H

=

Cn+m ! (Cn

H be

-

-

a

=

T

Vn > 2

n =

p

,U,nI

words, when is U a (strong) dilation of T PEUIE? The operator E. U of to the called Actually this question makes compression PEUIE is usually More not we if U is sense even generally, may consider a bounded unitary. Banach --4 A defined a on algebra A and ask for which homomorphism 7r: B(H) H the E C subspaces map In other

=

(1.5) is

a

IrE: a--+

PE7r(a)JE

homomorphism. One obvious example of this V

and in that

a

E

say that E is

case we

A 7r

ir(a)E

c

situation is when

we

have

E,

invariant.

with E2 C El, it homomorphism (although El E)E2 are usually in general E is not 7r-invariant). Such spaces of the form E called semi-invariant, we will call them -x-semi-invariant when we want to specify which homomorphism is involved. When considering a single operator T in B (H) (or the algebra it generates) we will say that a subspace E is semi-invariant for T if there are invariant subspaces E2 C El (i.e. subspaces such that T(Ej) C Ej, E, eE2. The following striking discovery of Sarason j 1, 2) such that E Furthermore if

can

E2, El

be checked that for E

are

=

7r-invariant

(closed) subspaces

E, E) E2 the map

7E is

a

=

=

=

answers

the above

questions

purely algebraic fact.)

in

a

very broad context.

(It

could be stated

as a

1. Von Neumann's

26

inequality and Ando's generalization

Theorem 1.7. Let A be

homomorphism.

PE7r(a)jE. iff there

WE(a)

Note

E

7r-invariant

are

Banach

a

Let E C H be

B(E).

Then 7rE is

Proof: Consider first E Let S:

Ej/E2.

the map 7rj:

A

momorphism.

Let

A

--4

B(H) a

=

El

in

be

a

A, let

bounded 7rE

from A into

a

(a) B(E)

=

E2.

e

El E) E2. Observe that El E) E2 can be identified -+ El E) E2 be the canonical isometry. Clearly defined for j 1,2 by 7rj(a) 7r(a)IE,, is a hoEj/E2 be the quotient map and consider the map

=

Ej/E2

B(Ej)

---

7r:

of H. For

homomorphism closed subspaces E2 C El such that E

with

and let

algebra

(closed) subspace

a

Q: El

=

--

=

EI/E2. Since Ker(Q7r,(a)) D E2, there is a uniquely determined Q7r,(a): El Ej/E2 such thatTr(a)Q Q7rj(a), and the unicity and the JR(a): EI/E2 fact that 7r, is a homomorphism clearly imply that ft(a)-k(b) Fr(ab) for all a, b --4

--+

map

=

=

in A. Hence Tr is

Now it is easy to check that

homomorphism.

a

irE(a) hence 7rE itself is

a

Conversely, let homomorphism.

=

S-R(a)S-1,

homomorphism. a subspace such that

E C H be

Let El be the closed linear span of E U

(closed)

subspace and El

7r-invariant

(aEAU 7r(a)E

:) E.

Let E2

7rE

defined

by (1.5)

Then El is

is

a

obviously

a

El e E. To conclude, it

suffices to check that E2 also is 7r-invariant. For that purpose, let x E EieE E2 and let a E A. We will prove that 7r(a)x E E2. For each n, there are finite sets =

(ain)

in

A, (xin)

in E and xn in E such that

x

xn +

lim

=

7r(ani)xin

n-,oo

Note that PEx

(1.6)

=

0 since

0

=

x

El

E

PEX

E)

liM

=

n-

To check that

7r(a)x

E

PE7r(a)x

E2,

=

so

xn

that

+

lim

lim

(

PE7r(a)xn

7rE(a)xn

+

+

n-oo

hence since WE is assumed

7rE

it suffices to show that

n-oo

=

E,

a

homomorphism

(a ')x ') 7r(a)x

E

We have then

E

PE7r(a)7r(a ')x ' 2

7E(aa ')Z'

IZ

)

1. Von Neumann's

liM

=

(7rE(a)x'+E7rE(a)7rE(0)x ' ) Xn +

lim 7rE (a)

=

71

%

n-oo

27

inequality and Ando's generalization

7rE

x '

(a,,

n-oo

by (1.6) PE7r(a)x

hence

0. This shows that

==

7r(a)x

E

E' and concludes the El

proof Remark. In

particular

applies to Sz.-Nagy's dilation theorem (or Ando's). there is a unitary operator U admitting invariant subthis

For any contraction T spaces E2, El with E2 c

El such that T

This

explains why the

=

PE eE2UJE eEV

structure of the invariant

is so important to understand the structure of for much more on this theme.

Remark 1.8. Sarason's result has

algebra

an

and let V be

(where L(V)

W

w:

a

-R: A

defined

L(W)

--+

V and

E2 morphism spaces

C

El

C

V and

an

a

is

nothing

also that

vw

S-li (a)S

=

=

S:

isomorphism

--+

but the canonical

-k(a)

"compression"

It is natural to ask whether

von

spaces than Hilbert spaces. The

Theorem 1.9. Let X be

JIT11

isometric to

=

a

I

we

--+

Let W be another vector

V).

IIP(T)II

1. Then there

Ej1E2

--

are

7r to Ej/E2. We (see Chapter 4).

of

:5

is

space. Assume that for all T: X

for all

polynomials

=Z I

-

-

-

=

P. Then X is

A

V

Clearly

o,\ is

analytic

hood of V and p,\ preserves al) so that OAJ&D is in A and by the obvious extension of (1.3) if T is a contraction on X

A)(1 11x* 11

back

inequality can be valid on other negative, as shown by Foias [Foil].

11P11...

be the associated M6bius transformation.

-

come

D, let (P-X (Z)

(T

will

Hilbert space.

Proof: Consider A in

that

7r-invariant sub-

W such that the homo-

Neumann's

answer

complex Banach

a

have

v7r(a)w

L(ElIE2)

E

to this later in the Banach space framework

X with

on

W be linear maps such that the map

-+

A

E

a

homomorphism. Assume

a

purely algebraic version as follows. Let A be a homomorphism 7r: A L(V)

V

v:

[SNF]

contractions. See

by V

is

general

vector space. Consider

is the space of all linear maps

space, and let --+

a

subspaces of unitary operators

T)-. (x*, x)

Consider =

now

x, y in X with

1, and let T

=

jjxjj

=

Ilyll

=

x* (9 y, i.e. T is defined

in

a =

we

have

neighborClearly

1.

0,\(T)

=

1. Let x* be such

by T(a)

=

x* (a)y

1. Von Neumann's

28

Va,E X. Note that

T(x)

inequality and Ando's generalization

Clearly JIT11

y.

=

jjx*jj Ilyll

=

=

1 and

our

assumption

implies

jjW,\(T)jj

JI(T

=

A)(1

-

T)-111

-

< 1

hence

11 (T

-

A)xjj

<

11 (1

:

jjx

T)xjj,

-

or

Ily Exchanging

x

and y

as

well

-

Axjj

A and

as

Ily

VA c D

Consequently

we

have if t C-

=A

obtain the

Axjj

-

tX11

+

11X

that if t is real with

Itl

> I

we

obtain that if

jjxjj

=

+

an

that

yjj-

+

tyll

X

+ Y

=

11X

=

11tX

+

Y11.

I

tyll

+

=

11tX

+

Y11

underlying real Banach space has the property that for any subspace E and any two points x, y of the unit sphere of E,

that the

means

two dimensional

there is

X11

Ilyll

Vt E IR This

-

so

also have

we

11ty Finally,

jjx

inequality

0

tY + X so

--

converse

[-1, 1] 11Y

hence if t

we

yjj.

-

isometry of E which

original argument, isometric to

a

see

also

[A]

Hilbert space.

maps

for

x

more

to y. It is well known

information)

that this

(cf. [Fic]

for the

implies that

X is

Notes and Remarks

on

Chapter

1

The very simple proof of von Neumann's inequality which opens this chapter is due to Ed Nelson [Nel] (I thought it was new until Vern Paulsen directed me to John Wermer who gave For the rest of the

me

this

chapter,

reference). most references

the reader to the classical reference

After the appearance of

[SNFI,

[SNF]

for

dilation

are

in the text. We refer

given

information.

more

theory became

a

very

important

branch of operator theory, see [Pal] for more recent work in this direction. It is also closely connected with Arveson's theory of nest algebras [see Ar2 and Dad] -

After the papers [VI, CD], numerous attempts were made to prove versions of von Neumann's inequality for n-tuples of operators T1, T,, see for instance , .

.

.

[Ble, Dixl, DD, Dr, T61-2, Ho4-5-6, Gu]. Note for instance that Drury [Dr, p. 21] observed that the inequality holds for any number of commuting contractions a 2-dimensional Hilbert space. For recent results relating von Neumann's inequality to the theory of analytic functions in several variables, see [CW, CLW, Na, Lo, LoS]. The following interesting extension of Ando's theorem (Theorem 1.2), from [GR], is not so well known. A finite family f T1, T2, -, Tpj of linear operators on a Hilbert space is called cyclically commuting if TjT2 Tp- 1 Tp TpTj for in It is commuting cyclically that, T2 T3 proved [GRI every TpTj. family fT,, T2,. -, Tpj of contractions on a Hilbert space H, there exists a cyclically commuting family of unitary operators f U1, U2, Up I on a Hilbert space K D H dilating the original family. If we no longer assume that the contractions T1, T, are commuting in of of "free" the then case a n-tuple operators appears as a typical any way, example. For instance, Bo ejko [134] proved the following extension of von Neumann's inequality: let P(Xl, Xn) be an arbitrary polynomial in the non-commuting variables X1, X,,, then for any n-tuples of Hilbert space contractions (formal) on

-

-

..

...

=

-

=

...

=

...

-

.

.

.

.

(T,,..., Tn)

we

.

.

.

.

,

,

.

,

have

I IPM,

Tn) 11

:5

SuPf 11P(U1,

Un) 111

where the supremum runs over all possible n-tuples (U,,..., U,,) of unitary operators on Hilbert space. (Actually, this supremum is already achieved on the set of

n-tuples

of

unitary

matrices of

arbitrary size.) Although

this is not the

Notes and Remarks

30

on

Chapter

1

original proof, this result can be proved using the same idea as in the proof of (vN) given at the beginning of Chapter 1. In another direction, Gelu Popescu [Pol] proved an interesting extension of

(vN)

for

stated

as

Yn such that

n-tuples of operators T1, follows: let J7

=

(

C1 ED

HOm

TiTi*

< 1.

This

can

be the full Fock space with H

=

be

t2n,

M>1

be the operator and let Sj: F Let el, en be the canonical basis of tn 2 h. defined "creation of Clearly Sj are isometries by Sjh particle") (of ej 0 with orthogonal ranges, so that I I E Sj Sj* < 1. Then the main result of [Pol] =

states that for any

polynomial

11 P(TI, See also

[Po2-7, AP1-2]

-

-

P

-,

and also

as

above

11 P(Si, -Sn)

Tn) 11

:5

[Ar4]

for related results in dilation

theory.

There also has been attempts to extend von Neumann's inequality to tractions acting on Lp-spaces. In these extensions, the shift operator S: defined

tr

--4

con-

tp

by

(X1

S(XO, X1, X2,

7

X2,

crucial r6le. The sup norm of the polynomial P an zn on the right (vN) must be replaced by the norm of P(S) acting on tp. Thus, a famous conjecture due to Matsaev (1966) asserts that for any contraction T on

plays

a

hand side of

Lv (1

< p <

oo)

we

have, for

any

polynomial

JJP(T)JJLp-Lp

:!

P

as

above,

JJP(S)JJfp---+9p-

2, this is von Neumann's inequality, but for p :A 2 it is still open. However, using the dilation theory for positive contractions on L.-spaces (see [AS]), this was proved in [Pe3] and [CRW] with additional assumptions on T. For instance, it suffices that T admits a contractive majorant, i.e. there is a positive contraction T on Lp such that JTfJ :5 T(Ifl) a.e. for all f in L,,, see [Pe2] for a survey. Note however, that Matsaev's conjecture (see [HaN, p. 244-247]) remains wide open: it is not even known if T is a 2 by 2 matrix! If p

=