Chapter XI The H∞ Functional Calculus

Chapter XI

The H" Functional Calculus

1. Construction of the Calculus.

The construction of the minimal dilation, which we explained in the previous Chapter, will now allow us to construct an H w functional calculus for contractions on a Hilbert space. So, as before, let H be a Hilbert space, T a contraction on H. Throughout this Chapter, we assume T to be completely non-unitary, which, as we recall, means that there is no subspace F with T F = F and the restriction TIF being unitary. Let now # in H w ( I I ) . We will give a meaning t o the operator # ( T ). For this, we first write the Fourier series of #

and, for 0 < r

< 1, we set

:

So we may define :

#r(T)

=

xakrkTk, k/O

and this series converges in L(H).

Proposition 1.1.

~

For every z in H , the limit lim #,(T)z

r- 1-

exists in H. The operator #(T)thus defined satisfies :

ll#(T)II I Il4lt".

(4)

Chapter X I

234

Proof. - Let (A', U) be the rn.u.d. of (H,T). We first show that the limit limr+l- &(U)z exists in U . Since q5 E H - , it is bounded on the unit circle, and 4 ( U ) can be defined by means of the integral (see Chapter VII, Theorem 4.2) :

d(U) =

1

#JdE,

4U)

where E is the spectral measure of U . By Theorem 4.3, Chapter VII, it satisfies :

since the measures EZ,+are absolutely continuous with respect to Lebesgue measure (see Chapter X, Theorem 2.5). Of course, the function fz is in L1. Furthermore, we know that #r ---t 4 a.e. (Chapter VIII, Section 2), and that SUP r.8

Idr(eie)I

i II4tIoo

(Chapter VIII, Section 1). From Lebesgue's Theorem, we deduce that, in I / , dr(U)Z - 4 ( U ) .

4

0 ,

r

-+

1-

,

We observe that the convergence obtained is a strong convergence, not a convergence in operator norm. Since T = P U , we get :

and this implies that &(T)z converges, when r 4 1-, to a limit in H . We denote this limit by $(T)z,and from ( 5 ) and (6) we get :

and

ll#(T)ll I11411~, as we announced.

H"

235

Functional Calculus

We now determine the adjoint of the operator 4 ( T ) . If 4 E H w , we denote by $t the function defined by (see Chapter 11, Proposition 2.2) :

&A) This function is in H" is :

, and

=

=

D.

Ck,,, - a k e i k e , the Taylor series of @

if d ( e i e ) =

@(e")

xE

+(X) ,

C

akeik'

,

k10

Proposition 1.2. - The adjoint of 4 ( T ) is given b y .-

Proof. - First, quite obviously, if T is c.n.u., so is T' immediately that :

47(T)*

. For r < 1 , one checks

4!(T*).

When r + I - , the right-hand side converges, at every point 2 , to @ ( T ' ) z , and the left-hand side, to 4 ( T ) * z .This proves our Proposition. The mapping phism :

4

-

4 ( T ) ,from H m

into L ( H ) , is an algebra homomor-

41$

(4?lJ)(T)= d ( T ) 4 ( T )

1

It defines a functional calculus. Obviously, if with the usual definition. If f$ =

xk,O

akeike,

with

~

I q

4

E

H"

.

is a polynomial,

I < 00 , then

4 ( T ) = x k >-O

#(T)coincides akTk.

Finally, if T is normal, we also get :

where E is the spectral measure of T. In short, the H m functional calculus coincides with the previously known ones when both make sense. We now study the continuity properties of the

H m functional calculus.

Theorem 1.3. - Let T be a c.n.u. contraction. Then : a) I f

#,

+

4 in H", then 4,(T)+ d(T)in f ( H ) . I14,,llm < 00 and 4,, + I$ a.e. on C , then 4,,(T) -, 4 ( T ) ,

b) I f sup,, strongly and ultrast rongly.

Chapter XI

296

c) If #,, -+ 4 for a ( L , , L l ) (which is the same as o ( H w , L 1 / H , ' ) , see B.B. [l],p . 41) then &(T)-+ 4(T) weakly.

Proof. - a) follows from Proposition 1.1, formula (4). To prove b), c), we may of course assume q5 = 0. We write, for z E H : when n by Lebesgue Theorem. This proves b). For c), we have the same way, for every

( M W , ~ ) = (M+, Y)

=

1:

2,

-+ 00,

y in H ,

d8

~n(ei~)jz,v(e)

-+

o,

when n -+

00,

and our Proposition is proved, Let AT be the subalgebra of L ( H ) generated by T and I : this is the space of polynomials p ( T ), p E P+ .

Corollary 1.4. - For every qi in H w , the operator 4 ( T ) is in the closure of the algebra AT for the weak topology. Indeed, polynomials are dense in H w for o ( H m , L 1 / H , ' ) , and we just apply the previous Proposition. This Corollary will be improved in Section 3. We now study the composition of functions.

Proposition 1.5. Let u in H w , with u(T), and, for every u E H w , ~

It11

< 1 in D . I f T

is c.n.u., so is

u(v(T)) = uou(T).

Proof. We consider the following transformation of the complex plane (Moebius transformation) : ~

2

-+

z-a

B,(z) = 1-iiz

'

where a is a fixed point in D. We already met this transformation in Chapter VIII. Then IBa(z)l = 1 if and only if operator on H , the operator V :

v

121

= 1. Therefore, if U is a unitary

= (U-a)(l-nU)-'

is unitary, and conversely. V is called the Cayley Transform of U .

H"

Functional Calculus

Here, we take w E H w with IuI W(2)

=

< 1 on D .

Thus lw(O)(

237

< 1, and

weset :

44 - 4 0 )

1 - v(o)u(z) '

which is a function in H", satisfying lw(z)I < 1 on D . Therefore w ( T ) is a contraction. Set T' = w(T). Let Ho be the space where T' ia unitary. From what we said follows that HO is also the space where w ( T ) is unitary, since w ( T ) is the Cayley transform of T'. We now show that T itself has to be unitary on H o . By Schwarz Lemma (see Cartan [I]), since w ( 0 ) = 0 and 1w(z)1 < 1 on D , we can find a function a ,analytic in D , such that w ( z ) = Z L Z ( Z ) , and 5 1 on D . So a ( T ) is a contraction and :

w(T) = T a p ) . For z E HO we get :

which proves that T is an isometry on Ho. Formula (1) implies that T is surjective on H o , since w ( T ) is. So TIM, is unitary, which implies Ho = {0}, since we assumed T to be c.n.u. We have proved that v ( T ) was c.n.u. So w(w(T))makes sense. For every polynomial p , we get : p(T') = p o u ( T ) . For w E H w , let pn be a sequence of polynomials converging t o w for u ( L w ,L1) (for example the C e s k o means of the Fourier series of w ,see Chapter VIII, Section 2). Then, by Theorem 1.3, c), p,(T') converges to w(T') weakly. But also pnow converges to wow in o ( L m , L1), so p , o w ( T ) + w o v ( T ) weakiy. Therefore, w(T') = w o u(T). The outer functions play a special role in this functional calculus :

Proposition 1.6. Let T be a c.n.u. contraction, and F be ar~outer function, in H m . The operator F ( T ) is injective and has dense range.

Proof. - Let

5

be such that F ( T ) z = 0 . For n 2 0 , we have :

Chapter X I

238

This implies that the function G = F.f, is in H,', the subspace of H' consisting of functions which vanish at 0. We set :

so H is an outer function, and IH(eie)I = If+(e)l, a.e. (see Chapter VIII, Proposition 5.1). Set G = Gi.G,, where G , is inner and Go outer. Then [HI = lGl/lFl on C. The two functions F H and Go are outer, and, on the unit circle :

they are equal in the unit disk (up to a constant of modulus l ) , by Chapter VIII, Proposition 5.2. So FH = G o . This implies that :

so

is in H' . So we see that fz is in H ' . But fz is real valued, so, by Chapter VIII, Proposition 3.2, must be constant, that is f+ = C , a.e. We deduce from the fact that G was in H i that C . F is in Hd , meaning that C F ( 0 ) = 0 . Since an outer function never vanishes inside the unit disk, C = 0 , and fz = 0 a.e. But :

-

and so z = 0 . Let's show finally that F ( T ) has dense range. The operator T' is c.n.u. and Fa is also outer (check on the definition) and in H". So Fu(T') = F(T)' is injective, and F ( T ) has dense range. Conversely, we have : P r o p o s i t i o n 1.6. - For every function u which is not outer, there exists a completely non-unitary contraction T on a Hilbert space, such that u(T)= 0 .

Proof. - We decompose u = m.F, with m inner, non constant, and we build T such that m ( T ) = 0 . We know (cf. Chapter VIII) that m.H2 is strictly contained in H ' . Set H = H 2 8 m.H2 : this will be the required Hilbert space. Let M be the multiplication by z , on H 2 . One checks easily that the adjoint M' is given by :

M'f

=

1

;(fk)

-

m)),

H"

239

Functional Calculus

which implies that MLnf-3 0 when n + 00, for every function f in H2.The space m.H2 is invariant by M , so H is invariant by M' . We put :

T

=

(M*IH)*.

We obtain a contraction T on H , with T*" = M * " ) H This implies that M and T are c.n.u.

-+

0 ,strongly, when

n + 00.

By definition, if P is the projection of H 2 onto H :

T" = P ( M " I H ) , for n 2 1. Therefore, for every w in H m ,

But we know that w ( M ) is the operator of multiplication by w on H2.In particular, for f in H , we have : m(T)f = Pm(M)f = P(m.1).

But m./ is in m . H 2 , so P(m.1) = 0, and this implies m ( T ) = 0 2. The Soectral Theorem for the H"

functional calculus.

For the analytic functional calculus, we met (Chapter 11, Theorem 2.3), the theorem :

f ( 4 T ) ) = a(f(T)),

(1)

where f is analytic on a ( T ) .

n.

We are now dealing with contractions, so u ( T ) c But f is not analytic on a neighborhood of 0, only inside D . So formula (1) does not make perfect sense : if u ( T ) n C is not empty, how should f ( u ( T )n C) be understood ? The function f is only defined a.e. on C, by means of a radial limit. So the left-hand side of (1) is not well defined, though the right-hand side is. For points inside the open disk, there is no difficulty :

Chapter X I

24 0

Proposition 2.1. - Let f be Then :

a

function in H - , and T a c.n.u. contraction.

f ( m n D) c o(f(T)).

(2)

Proof. - Set, for X E D :

Then g is in

H”. Moreover,

So, if X E o ( T ) , X - T is not invertible, and f(T) - f(A) either. This proves our claim.

is not invertible

If f is in A ( D ) ,formula (2) can be extended to :

More precisely, we have :

Theorem 2 . 2 . - Let T be a c.n.u. contraction with spectral radius I , and let X E a(T),with 1x1 = 1. Assume that f E H m is continuous at A , that is :

exists. Then o(f(T))contains !(A). Proof. - Replacing T by T/X,we m a y assume X h ( t ) = f(z) - f ( 1 ) . Then h(1) = 0, and : lim

z+A,lzl
We have subset of

for simplicity. Put

h ( z ) = 0.

We consider the following sequence of functions

with, for instance,

= 1,

:

..

un(l) = 0 , IlunIlm 5 2, and u,(z) + 1, uniformly on every compact

o\l

(2).

H m Functional Calculus Lemma 2.3.

~

If the function h in Hm satisfies (I), then : Ilunh - hllm

Proof. - Let

E

-+

when n -+ 00.

0 ,

> 0 be given. By (l),we can find

of 1 :

V,

a such that the neighborhood

= { z ; z E D , Iz - 1) < a}

has the property : if t E V,, then lh(z)I < E . Since radial limits exist a.e., we also get that lh(z)I < E

Put

C,

= C n { I z - 11

, if

It- 11 <

a , and IzI

5 1.

(3)

< a}, and CL = C\C,. Then, on Ci,

sup lun(z)h(z) - h(z)l

-+

ZEtb

0

,

when n

-+ 00

,

by property (2) of the sequence ( U n ) n > o . So, we can find no such that, for n 2. n o , I u n ( z ) h ( z )- h ( t ) I

<

E

n 2 no

,

,

z E Cb

.

On C,, we have, by (3) : SUP

ZEC,

therefore llunh - hll

<4

Iun(z)h(z) - h(z)l

< 3.5

9

~ for , n 2 no, and this proves the Lemma.

The properties of the H m functional calculus (Proposition 1.1) imply :

h(T)u,,(T)- h ( T )

+

0

,

in f (H).

(4)

We observe that un(T) - Z cannot tend t o 0 in f (H) . Otherwise, for n large enough, un(T)would be invertible and so would be un(T)" = Z - T. But we have assumed 1 E a ( T ) . Therefore, h ( T ) cannot be invertible, which means that f ( T ) - f(1) is not invertible, so f ( 1 ) E u(f(T)). The converse formula :

f(o(T))

' o(f(T))

is false in general : there are examples with u ( f ( T ) )= I'j, and a(T) reduced to a point. See Foiq, Nagy, Pearcy [l],and J. Esterle f l ] . To finish this Section, we wish to point out that if the spectral radius r(T) is strictly smaller than one, the analytic functional calculus should be preferred to the H w functional calculus. Indeed, it allows us to consider functions which are anaIytic on a neighborhood of {IzI 5 r(T)}, a class which is much larger than H m .

3. The HDDfunctional calculus on an akebra eenerated by a sinde operator.

Let T be an operator on a Banach space E. We have already met AT , the algebra generated by T in t(H): this is just the set of finite sums z k >-O akTk. We consider now ZT,cIosure of AT in L ( H ) for the ultraweak topology. Then we know (Chapter V, Proposition 2.12) that IT is a dual space, namely the dual of the quotient of the nuclear operators, N ( H ) , by ' A T , the pre-orthogonal of Z T , that is : 'AT

=

{N E U(H); t r ( N p ( T ) )= o , for every polynomial p)

Since the trace is linear, we have also

To simplify our notation, we just put :

and

ZT

is the dual of NT. There is a weak topology on Z T , which is and, by Proposition 2.12, Chapter V, this topology is the trace

u ( ~ TN ,T ) ,

on

AT of

o ( t ( ~ N) (,H ) ) .

We observe that N T , being a quotient of a separable space, is itself separable. So the balls of ZT are metrizable for the weak topology (B.B. [l],p. 57).

The algebra 3~ will be called ultraweakly closed algebra generated by T. It contains the closure of AT in operator norm, and is contained in the closure of AT for the weak topology of L ( H ) . The behavior of the H m functional calculus with respect to this algebra is given by the following theorem :

Theorem 3.1. - Let T be a c.n.u. contraction on H and let ultraweakly closed algebra generated by T . Then :

AT

be the

a) If ( h , ) , , ? ~ is a bounded sequence in H-, converging to 0 a t every point of the open disk D ,h,(T) converges to 0 in AT for U ( ~ TUT). , U the convergence is uniform in D , then h,(T) converges to O in operator norm.

b) If (hn),,>O - is a bounded sequence in H a , converging to 0 a.e. on C, then h , ( T ) converges to 0 for the ultrastrong topology.

H" c) The mapping \Ir :

h

Functional Calculus +

24 3

h(T) is continuous from Ha U ( H " , L I / H ; )

into X T , u(XjiT,UT). d ) If 0 is an isometry, \Ir is also an homeomorphism for the above weak topologies.

Proof. - a) Since the sequence (hn),,?o is bounded in H-, (h ,(T )) is bounded in L(H).Since weak and ultraweak topologies coincide on bounded subsets of L(H) (Proposition 2.9, Chapter V), we just need to show that hn(T) 4 0 weakly, that is :

,

(h,(T)z,y)

---*

(h,(U)z,y)

+ 0 ,

0

for all z,y E H .

This follows from : for all z , y H ~

But since T is c.n.u. :

where fi,v is in L1 (see Chapter IX). Since h , ( z ) -+ 0,for all J E D , this convergence is uniform in any disk of radius r < 1. We deduce that the derivatives h', h", . .., satisfy : h(k)(0)= 0 ,

for all k E

IN.

From this follows obviously by induction that for every k 2 0 , the k-th Fourier coefficient of h, , i n ( k ) , satisfies :

Consequently, for any polynomial p ,

But polynomials are dense in L1, and the sequence ( h , ) , ? ~ is bounded in L,. Therefore, for every f in Ll ,

244

Chapter X I

and this implies our claim. The second part of a) is just Theorem 1.3, a). b) was proved in Theorem 1.3,b). c ) We need to show that if h , + 0 for u ( H , , L l I H , ' ) , then h, (T ) -+ 0 weakly. But the assumption implies that for every k 2 0 , k , ( k ) + 0 , when n + 00, and we proceed as in the proof of a). If 4 is in HOD, the operator @) is in A T . Indeed, let (pn)*?o be a sequence of polynomials, bounded in L, and converging to 4 for u(L,, L I ) (for example the CesAro means of the Fourier series of 4, see Chapter VIII, Section 2). Then each p , is in A T , and the sequence p,(T) converges for the weak topology, by what we just proved. But it is bounded in L(H), and so, by Chapter V, Proposition 2.9, it converges ultraweakly. So d ( T ) is in AT.

d) We put L = Irn Q . Since Q is an isometry,

L is a closed subspace of

f ( H ) , and Q is surjective from H m onto L. Moreover, L c A T . The weak topology on L is the topology induced by o ( ~ TAT). , We show that L is closed for this topology. We use the following lemma, which is due to S. Banach [I]:

L e m m a 3.2. - A subspace L of a dual E' is a ( E * ,E) closed if and only if L n BE. is a ( E ' , E ) closed. Proof. - The reader may consult S. Banach [ 11 or Bourbaki [ 11 for more general versions of this result. However, we include a proof, which follows rather closely Banach's original ideas ([l], pp. 121-125).

It's enough to show that for any fo @ L,there exists x E E with fo(z) = 1 and f ( z ) = 0 for all f E L ( a subspace L with this property is called regularly closed in Banach's terminology). Indeed, if L is not a ( E ' , E ) closed, take fo 6 F , fo $ F. With the above point 2, the neighborhood of fo :

should intersect L : a contradiction. So, let fo 4 L , with l/foll = 1. Of course, fo 4 ~ B L for , any IZ 2 0. Since ~ B is L compact, we can strictly separate fo and ~ B :L there is a point z, in E, with llznll = 1, and reals a, > 0 , E , > 0, such that :

Ha Functional Calculue

24 5

Next, we observe that : f:i

distw(/o,nBL)

>

(3)

0.

Indeed, assume not. Then, there is a sequence (fn)n20, with fn E nBL for all n, such that /n + fo in E* . But the fn's are bounded in norm, so all of them belong to a given no B L . We have :

Ilfn -foil 2

l(/n - fo)(znoI

2

Eno

3

a contradiction. Let E = inf, d i s t ~ . ( f O , n B ~Let ). :

Kn =

&

- B E - + ~ B L ,

2

n2O.

This is a compact set, €or a ( E * , E ) ,which does not contain fo, by (3). So there is a point yn in E , lJynll= 1, and a real > 0 , such that :

on

fo(Yn) > &

Isg(Yn)

+ nJ(yn)I

<

Pn

,

(4)

Ptt

BE- , f E B L , n 2 0.

E

(5)

From ( 5 ) , taking f = 0, g such that g(yn) = 1, we deduce : &

Pn 2 -2 , and from (4),

L

(6)

1. From ( 5 ) , we also deduce :

Extracting a subsequence from the yn 's if necessary (we keep the same notation), we can assume that there is a I such that : fo(Yn)

+

1 s

and 1 _> ~ / *2

(8)

We consider :

G = {(Xf(yn))n ;

E

a

f E BL}-

This is a vector subspace of co ,by (7). In the space (c) of convergent sequences, condition ( 8 ) implies that the distance of the sequence (fo ( Y ~ ) to ) ~ G is 2 ~ / 2 . The dual of ( c ) is 1 1 , so there is a sequence of complex numbers (C,) such that :

n

n

Take z = EnCnyn : this is the required point, and our Lemma is proved.

24 6

Chapter XI

Here, since \li is an isometry, L n Bc(H) is the image of the ball of H", which is a ( L , , L 1 ) compact. Therefore, B L ( H ) is a(f(H),.M(H)) compact, and therefore closed. By Lemma 3.2, L itself is closed for a ( f ( H ) ,. M ( H ) ) . This proves that L = AT,and that \Ir is surjective. But then, \k' is an injective and continuous map from the compact equipped with a ( L m ,L 1 ) , onto &T , equipped with @T, NT). Therefore, Q in an homeomorphism between these two spaces, equipped with their weak topologies, and our Theorem is proved.

H-

Functional Calculus

Exercises on Chapter XI.

Exercise 1. - Let T be a c.n.u. contraction on H , and let h in H" be a non-zero function. Show that for every z in H such that h ( T ) z = 0 , we have T " z + 0 when n + 00. Exercise 2. - Let B, be the inner function :

B,(z) =

z-a -

1-iiz'

la1 < 1.

Construct an operator T such that B,(T) = 0.

Exercise 3. - (Mean ergodic Theorem). Let T be a contraction on a Hilbert T k converge strongly in H , space. Prove that the Cesaiiro averages when n + 0 0 .

c,"-'

Let T be an invertible isometry on a Banach space E . Let Exercise 4. A(n) be the algebra of absolutely convergent Fourier series : f is in A(n) if f = x r m c j e i j e , with CrmIcj( < 00. Show that f(T)can be defined by means of a normally convergent series, and that the usual properties of a functional calculus hold. ~

Exercise 5. - Let S be the right shift on l z ( Z ) and (p,,),,?~ be a sequence of polynomials. a) Show that p,(S)

+

0 in

L(1,) if and only if pn + 0 in L b o ( n ) .

b) Show that p,(S) + 0 strongly if and only if (p,,),,zo is bounded in &,(TI) and pn + 0 in L Z ( n ) . c) Show that p,(S) --t 0 weakly if and only if (p,,),,?~ is bounded in Lm(II) and, for every j 2 0 ,the j t h coefficient of p,, --t 0 when n + 00.

Exercise 6. - Let S be the right shift on I z ( Z ) . To every element of the algebra generated by S , that is to every finite sum Ck>O - a k S k , we associate the polynomial x k >-O a k z k , a) Show that the closure of As for the operator norm can be identified to the algebra A (D) .

b) Show that the closure of As for the strong topology can be identified

to H".

c) Show that the closure of As for the weak topology can also be identified to H".

Chapter X I

948

Exercise 7. - Let S be the right shift on l l ( Z ) . To every element of the algebra generated by S and S-' , that is to every finite sum If='= akSk, we associate the trigonometric polynomial Ck=-al N akt k . a) Show that the closure of the algebra AS,S-I generated by S and S-', for the operator norm, can be identified to C(rI).

b) Show that the closure of As,s-l to

for the strong topology can be identified

L,.

c) Show that the closure of A s , s - ~ for the weak topology can also be identified to L,.

Notes and C o m m e n t s : Section 1 comes from Nagy-Foiq 111. In Section 2, Theorem 2.2 is due to Foiq-Mlak [I] ; our proof follows J. Esterle 121.

Complements on Chapter XI :

On the algebra generated by the right shift S, the weak and ultraweak topologies coincide. This property holds for a large class of operators, as described by Bercovici-Foiq-Pearcy 111. A first example of an operator for which this property does not hold was built by D. Westwood [I]. Another was built by G. Cassier 111. Cassier's example has some interesting properties. Let T be the operator. Then the algebra AT is equal to the commutant {T}',and also to the bicommutant {T}".Moreover, there is a nuclear operator A with the property that no finite rank operator B satisfies :

tr ( A U ) = tr (BU),

for aII B E

IT .