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


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