Bases of reproducing kernels in de Branges spaces E Fricain
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Bases of reproducing kernels in de Branges spaces E Fricain

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33 Pages
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Bases of reproducing kernels in de-Branges spaces E. Fricain? August 18, 2004 Abstract This paper deals with geometric properties of sequences of reproducing kernels related to de-Branges spaces. If b is a nonconstant function in the unit ball of H∞, and Tb is the Toeplitz operator, with symbol b, then the de- Branges space, H(b), associated to b, is defined byH(b) = (Id?TbTb)1/2H2, where H2 is the Hardy space of the unit disk. It is equiped with the inner product such that (Id? TbTb)1/2 is a partial isometry from H2 onto H(b). First, following a work of Ahern-Clark, we study the problem of orthogonal basis of reproducing kernels in H(b). Then we give a criterion for sequences of reproducing kernels which form an unconditionnal basis in their closed linear span. As far as concerns the problem of complete unconditionnal basis in H(b), we show that there is a dichotomy between the case where b is an extreme point of the unit ball of H∞ and the opposite case. Keywords: de-Branges spaces, Riesz bases, reproducing kernels. 2000 ams subject classification: 46E22, 47B32, 47B38, 30H05, 46J15, 46B15. Acknowledgements.

  • inner product

  • nontrivial invariant

  • invariant subspace

  • geometric properties

  • branges space

  • facts concerning

  • problems such

  • positive borel

  • subspace bh2


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Bases of reproducing kernels in de-Branges spaces
∗E. Fricain
August 18, 2004
Abstract
This paper deals with geometric properties of sequences of reproducing
kernels related to de-Branges spaces. If b is a nonconstant function in the
∞unitballofH ,andT istheToeplitzoperator,withsymbolb,thenthede-b
1/2 2Brangesspace,H(b),associatedtob,isdefinedbyH(b) = (Id−T T ) H ,b b
2where H is the Hardy space of the unit disk. It is equiped with the inner
1/2 2product such that (Id−T T ) is a partial isometry from H ontoH(b).b b
First,followingaworkofAhern-Clark,westudytheproblemoforthogonal
basisofreproducingkernelsinH(b). Thenwegiveacriterionforsequences
of reproducing kernels which form an unconditionnal basis in their closed
linear span. As far as concerns the problem of complete unconditionnal
basis in H(b), we show that there is a dichotomy between the case where
∞b is an extreme point of the unit ball of H and the opposite case.
Keywords: de-Branges spaces, Riesz bases, reproducing kernels.
2000 ams subject classification: 46E22, 47B32, 47B38, 30H05, 46J15, 46B15.
Acknowledgements. This work was done while the author was visiting the
Department of Mathematics and Statistics of the University of Laval (Quebec).
He is grateful to the Department members, specially to T. Ransford, for their
warm hospitality.
∗Institut Girard Desargues, UFR de Math´ematiques, Universit´e Claude Bernard Lyon 1,
69622 Villeurbanne Cedex, France. fricain@igd.univ-lyon1.fr.
11 Introduction
This paperis devoted to geometric properties of sequences of reproducing kernels
inde-Brangesspaces. Thesespaces,firststudiedbyL.deBrangesandJ.Rovnyak
2[6], are (not necessarily closed) subspaces of the Hardy space H of the unit disk,
. Recall first that
Z
2 2H := f : → analytic : sup |f(rζ)| dm(ζ)<∞ ,
06r<1
where is the unit circle and dm is the normalized Lebesgue measure on .
2 2 2As usual, H will be identified (via radial limits) with the space of L = L ( )
functions whose negatively indexed Fourier coefficients vanish. Norm and inner
2 2product in L or H will be denoted byk·k andh·,·i , respectively.2 2
2 2 ∞Let P denote the orthogonal projection of L onto H . For ϕ∈ L , let T+ ϕ
2denote the Toeplitz operator with symbol ϕ defined on H by T f = P (ϕf).ϕ +
2Thede-Brangesspace,H(ϕ),associatedtoϕconsistsofthoseH functionswhich
1/2belong to the range of the operator (Id−T T ) . It is a Hilbert space whenϕ ϕ
equipped with the inner product
hf,gi :=hP ⊥f ,P ⊥g i ,ϕ Ker(Id−T T ) 1 Ker(Id−T T ) 1 2ϕ ϕϕ ϕ
1/2 1/2where f = (Id−T T ) f , g = (Id−T T ) g and P ⊥ denotes theϕ ϕ 1 ϕ ϕ 1 Ker(Id−T T )ϕ ϕ
2 ⊥orthogonal projection of H onto Ker(Id−T T ) . Note thatH(ϕ) is containedϕ ϕ
2contractively in H and the inner product is defined in order to make (Id−
1/2 2T T ) a partial isometry of H ontoH(ϕ). The norm ofH(ϕ) will be denotedϕ ϕ
byk·k .ϕ
2For λ∈ , we let k denote the kernel function for the functionnal on H ofλ
−1evaluation at λ; it is given by k (z) = (1−λz) (z ∈ ) and satisfies f(λ) =λ
2 2hf,k i (f ∈ H ). Since H(ϕ) is contained contractively in H , the restrictionλ 2
to H(ϕ) of evaluation at λ is a bounded linear functionnal on H(ϕ). It is thus
ϕinduced, relative to the inner product inH(ϕ), by a vector k inH(ϕ). It is easy
λ
ϕ
to see ([19], (II-3)) that k = (Id−T T )k andϕ ϕ λλ
ϕ
f(λ)=hf,k i ,ϕλ
for all f ∈H(ϕ). From now on, b will be a nonconstant function in the unit ball
∞of H , that is an holomorphic and bounded function in , withkbk 6 1. Then∞
since T k =b(λ)k , we haveλ λb
1−b(λ)bbk =(Id−T T )k = .b λλ b
1−λz
2It is easy to see that H(b) is a closed subspace of H if and only if T is ab
partial isometry. That happens if and only if b is an inner function, that is a
2




∞function in H whose radial limits are of modulus one almost everywhere. Then
2H(b) is the orthogonal complement of the Beurling invariant subspace bH , the
typical nontrivial invariant subspace of the shift operator S. Hence, the space
H(b), with b inner, are the nontrivial invariant subspaces of the backward shift
∗S . In this case, starting with the work of S.V. Hruscev, N.K. Nikolski and
B.S. Pavlov, a whole direction of research has investigated geometric properties
of reproducting kernels inH(b) (see [4], [9], [10], [11]). One of the motivation to
study geometric properties of reproducting kernels inH(b) is the link being with
nontrigonometric exponentials systems. Recall that in the special case where
z+1 bb(z) = exp(a ), a > 0, the reproducing kernels k , with λ ∈ , arise as
λz−1
1+λthe range of the exponential functions exp(−iμw)χ , with μ = i , under a(0,a) 1−λ
2natural unitary map going fromL (0,a) toH(b). Geometric properties of family
of exponentials arise in many problems such as scaterring theory, controllability
and analysis of convolution equations (see [3] and [11] for details). We intend
to provide a comprehensive treatment of geometric properties of reproducing
kernels of H(b), emphasing the parallel with the particular case where b is an
inner function.
We now recall some basic definitions concerning geometric properties of se-
quences in an Hilbert space. For most of the definitions and facts below, one can
use [14] as a main reference.
Let H be a complex Hilbert space. If (x ) ⊂H, we denote by Span(x :n n>1 n
n> 1) the closure of the linear hull generated by (x ) . The sequence (x )n n>1 n n>1
is called:
(Co) complete if Span(x :n> 1)=H;n
(M) minimal if for all n> 1, x 6∈ Span(x :m =n);n m

xn
(UM) uniformly minimal if inf dist ,Span(x :m =n) >0;m
n>1 kx kn
(UBS) an unconditionnal basis in its closed linear span if every element x ∈
Span(x : n > 1) can be uniquely decomposed in an uncondionnal con-n X
vergent series x = a x ;n n
n>1
(RS) a Riesz basis in its closed linear span if there are positive constants c,C
such that X X X
22 2 c |a | 6 a x 6 C |a | , (1)n n n n
n>1 n>1 n>1
finite complex sequences (a ) ;n n≥1
(UB) an unconditionnal basis of H if it is complete and an unconditionnal basis
in its closed linear span.
3
66Obviously we have
(UB) =⇒ (RS)=⇒ (USB) =⇒ (UM) =⇒ (M).
In general, all the converse implications are false but K¨othe-Topelitz theorem
asserts that ifkx k 1, then (USB)⇐⇒ (RS).n
The Gram matrix of the sequence (x ) is Γ = (hx ,x i) . Uncondi-n n≥1 n m n,m≥1
tionnal basis are characterized by the fact that Γ defines an invertible operator
2on ` .
2We recall some well-known facts concerning reproducing kernels in H . Let
kλnΛ = (λ ) be a sequence of distinct points in and denote by x =n n>1 n
kk kλ 2n
the normalized reproducing kernel. Then we have
– (k ) isminimalifandonlyif(λ ) isBlaschkesequence(whichmeansλ n>1 n n>1n P Q
that (1−|λ |) < ∞). As usual, we denote by B = B = b ,n Λ λnn>1 n>1
|λ |n λ −znwhere b (z) = .λn λn 1−λ zn
– If (λ ) is a Blaschke sequence, then (k ) is complete inH(B).n n>1 λ n>1n
– (x ) is a Riesz basis ofH(B)if andonly if itis uniformly minimal whichn n>1
is equivalent to (λ ) satisfies the Carleson conditionn n>1
inf|B (λ )|>0,n n
n>1
where B =B/b ; we will write in this case (λ ) ∈ (C).n λ n n>1n
In this paper, we intend to study the property of unconditionnal basis for se-
quences of reproducing kernels inH(b). The study of the spacesH(b) frequently
∞bifurcates into two cases depending b is an extreme point of the unit ball of H
or not. We will show that for the property of unconditionnal basis inH(b), there
exists a dichotomy between the two cases. Recall that K. de Leeuw and W.
∞Rudin [7] proved that b is an extreme point of the unit ball of H (abbreviated
by b is extreme) if and only if
Z
2log(1−|b| )dm=−∞.
We now precise some notations that will be used in this paper. For a positive
2finite Borel measure ν on and q a function in L (ν), we let
Z
iθq(e ) iθ(K q)(z) := dν(e ), z∈ \ ,ν −iθ1−e z
2and we think ofK as a linear transformation ofL (ν) into the space of holomor-ν
2 nphic functions in \ . Moreover, we let H (ν) be the closed linear span of z ,
4






2n> 0, (for the norm of L (ν)) and we denote by Z the operator of multiplica-ν
2tion by the independant variable on H (ν). If ν is absolutely continuous and ρ is
its Radon-Nikodym derivative with respect to nomalized Lebesgue measure, we
2 2write K in place of K , H (ρ) in place of H (ν) and Z in place of Z . Noticeρ ν ρ ν
2 2that if q∈L (ρ) then qρ∈L and
K q =P (qρ).ρ +
The planof the paperis the following: the next section deals with the problem of
orthogonal basis of reproducing kernels inH(b). As for the classical case where b
is inner, this problem depends on the spectral study of a rank one perturbation
∗ ∗of X , where X =S . In particular, we prove (Corollary 2.2) that if b is not|H(b)
an inner function, then H(b) does not possess orthogonal basis of reproducing
kernels. In section 3, we give a criterion for the property of unconditionnal basis
in its closed linear span (Theorem 3.1 and Theorem 3.2). Then we give some
applications of this criterion, which are generalizations of results concerning the
classical case. In section 4, we study the case where b is extreme and prove that
Id−T T isaninvertible operatorfromH(u)ontoH(b), withuaninnerfunction,b b
∞ ∞if and only if dist(ub,H ) < 1 and dist(zub,H ) = 1 (Theorem 4.1). Then we
buse this result to caracterize sequences (k ) which form an unconditionnaln>1λn
basis of H(b) (Theorem 4.2). In section 5, we study the case where b is not
an extreme point. Contrary to the extreme case, we show that H(b) cannot
possess unconditionnal basis of reproducing kernels (Corollary 5.1). Then, we
getacaracterizationofcompleteness (Proposition5.2)andfinallymaking further
assumptiononthemultiplierofH(b),wegivearesultconcerningsummationbasis
of reproducing kernels (Theorem 5.1).
2 Orthogonal bases of reproducing kernel
bIt is clear that if (λ ) ⊂ , then the family (k ) cannot be orthogonal. Inn n>1 n>1λn
some cases, it is possible, however, to consider reproducing kernels with poles on
the unit circle. Let
ZY|a | a −z ζ +zn nNb(z) =z exp − dμ(ζ) ,
a 1−a z ζ−zn n
n
P
be the canonical factorisation of b, where (1−|a |) < ∞ and where μ is ann
positive Borel measure on and set
( )Z
2X1−|a | dμ(t)n
E := ζ ∈ : + < +∞ .b 2 2|ζ−a | |t−ζ|nn
Recall that we say that b has an angular derivative in the sense of Carath´eodory
0at the point λ of if b and b have a nontangential limit at λ and |b(λ)| = 1.
5




1−b(λ)b
bThen we have the following criterion for the inclusion k := ∈ H(b),λ
1−λz
λ∈ .
∞Theorem A (Ahern-Clark-Sarason, see [2] and [19]) Letb∈H ,kbk 6 1∞
and λ∈ . Then the following assertions are equivalent:
1−cb(z)
(i) thereisacomplexnumbercofunit modulussuchthatthe function
1−λz
is in H(b);
(ii) λ∈ E ;b
1−|b(z)|
(iii) liminf < +∞;
z→λ 1−|z|
(iv) b has an angular derivative in the sense of Carath´eodory at λ;
(v) every function in H(b) has a nontangential limit at the point λ.
Moreover, in this case, the number c is unique and is given by c = b(λ) :=
1−b(λ)b
blim b(rλ). If k := , then for all f ∈H(b), we haver→1 λ
1−λz
bf(λ)=hf,k i .bλ
0 0 0Let now λ,λ ∈E , λ =λ and assume that b(λ) =b(λ) =α,|α|= 1. Thenb
01−b(λ)b(λ)b b b 0hk ,k i =k (λ) = =0.0 bλ λ λ 01−λλ
bSo if we want to get an orthogonal sequence of reproducing kernel (k ) , wen>1λn
have to choose sequence (λ ) such that (λ ) ⊂ E and b(λ ) = α, n> 1,n n>1 n n>1 b n
|α| =1. FollowingtheworkofAhern-Clark[1]concerningtheclassicalcasewhere
b is an inner function, we proceed first to a study of rank one perturbations of
∗ ∗ ∞X which are isometry, where X = S . Recall that if ϕ∈ H , thenH(b) is|H(b)
invariant under T and the norm of T as an operator in H(b) does not exceedϕ ϕ
∗kϕk . Hence S =T acts as a contraction inH(b) (see [19], (II-7)). Recall also∞ z
that we have (see [19], (II-9))
∗ ∗X h =Sh−hh,S bi b (h∈H(b)). (2)b
6

6∗2.1 Spectral properties of rank one perturbation of X
In this subsection, we proceed to an investigation of spectral properties of rank
∗one perturbations of X which are isometry. Actually, our study goes beyond
what is necessary for our treatment of the existence of orthogonal basis. First
∗we give results concerning spectral properties for X . We will see that these
properties depend whether b is an extreme point or not (for the analogue results
for X, see [19], (IV-5), (V-7) and (V-8)).
∞Lemma 2.1 Let b∈ H , kbk 6 1 and h∈H(b). Then∞
∗ ∗kX hk =khk ⇐⇒hh,S bi =0b b b
Proof: using the relation (2), we get
∗ ∗ ∗ ∗ ∗XX h =S (Sh−hh,S bi b) =h−hh,S bi S b.b b
Hence
∗ 2 ∗ 2 ∗ 2kX hk =hXX h,hi =khk −|hh,S bi | , (3)b bb b
which gives the lemma.
∞ ∗Lemma 2.2 Let b∈ H , kbk 6 1. Then σ (X )⊂ .∞ p
∗ ∗Proof: theinclusionσ (X )⊂ followsfromthefactthatX isacontraction.p
∗Assume that there exist λ ∈ ∩ σ (X ) and let h ∈ H(b), h 6≡ 0 such thatp
∗ ∗ ∗X h = λh. Then kX hk = khk and Lemma 2.1 implies that hh,S bi = 0.b b b
∗Hence X h = Sh = λh, which gives that λ∈ σ (S), which is absurd and provesp
the lemma.
Proposition 2.1 (a) If b is extreme then
∗ ∗ ∗σ (X ) ={λ∈ :b(λ) =0} and σ(X )=σ (X )∪σ(b),p p
where σ(b) := \ρ(b) and ρ(b) denotes the set of points ζ ∈ such that
there exist an open arc I, ζ ∈ I and b can be continued analytically across
I with|b| =1 on I. Moreover if b(λ) = 0, then

b
∗Ker(X −λId) = . (4)
z−λ
∗(b) If b is nonextreme then σ(X )= .
7



∗Proof: recall that X is completely nonunitary and if Θ ∗ denotes the char-X
∗acteristic operator function of X , in the theory of Sz-Nagy and Foias, we have
b
∗ ∗(see [17]) Θ = b (in the extreme case) and Θ = (in the nonextremeX X
a
∗case). Spectral properties ofX follow now froma theorem of Sz-Nagy and Foias
(see [20], Theorem 4.1. p. 247). It just remains to check equality (4). Let λ∈ ,
∗b(λ) =0 and f ∈ Ker(X −λId), f 6≡ 0. Then using (2), we have
∗(z−λ)f =hf,S bi b,b

b
∗which implies that f ∈ (b/(z−λ)). Thus Ker(X −λId)⊂ and an
z−λ
argument of dimension shows that there is equality.
∗Rank one perturbations of X that we will interest in are defined as follows.
Definition 2.1 If λ is a complex number of modulus 1, define the operator Uλ
of H(b) by
∗ −1 b ∗U :=X +λ(1−λb(0)) k ⊗S b.λ 0
Proposition 2.2 The operator U is an isometry of H(b). Moreover, it is aλ
unitary operator of H(b) if and only if b is extreme and in this case, the U areλ
∗the only one-dimensional perturbations of X which are unitary.
Proof: denote by μ the measure on whose Poisson integral is the real partλ
1+λb 2of , denote by V the transformation defined on L (μ ) by V q(z) =λλb λb
1−λb
(1−λb(z))K q(z), and finally denote by Z the operator of multiplication byμ μλ λ
2the independant variable on H (μ ). We know (see [19], (III-8)) that we haveλ
−1U =V Z V (5)λ μλb λ λb
2and moreover V is an isometry of H (μ ) onto H(b). Hence U is cleary anλ λλb
isometry of H(b). We see also that this isometry is onto if and only if Z isμλ
2 2onto, which is equivalent to H (μ )=L (μ ). But a theorem of Szeg¨o says thatλ λ
2 2H (μ ) = L (μ ) if and only if the Radon-Nikodym derivative of the absolutelyλ λ
continuous part of μ with respect to normalized Lebesgue measure is not log-λ
integrable. Now a theorem of Fatou shows that this Radon-Nikodym derivative
21−|b|
2equals to . Since log|1−λb| is always integrable (being the logarithm
2|1−λb|
∞ 2 2of the modulus of the H function 1−λb), we see that H (μ ) =L (μ ) if andλ λ
2only if log(1−|b| ) is not integrable, which is exactely the condition that b is
extreme.
∗Now, assume that b is extreme and that U := X +h⊗k, h,k ∈H(b), is a
∗ ∗unitaryoperator. Iff ⊥ k, thenwe haveUf =X f, which giveskX fk =kfk .b b
8


∗Lemma 2.1 implies that f ⊥ S b. It follows that there exist c ∈ such that
∗ ∗ ∗k = cS b, which gives U = X +h ⊗S b, with h = ch. Taking the adjoint of1 1
this relation, we see that if f ⊥ h , then kXfk = kfk . Now recall (see [19],1 b b
2 2 2 b(VIII-4)) thatkXfk =kfk −|f(0)| , which gives f(0) = 0, that is f ⊥ k . Itb b 0
b ∗ b ∗follows that there exist c ∈ such thath =c k and thus U =X +c k ⊗S b.1 1 1 10 0
−1It remains to show that there exist λ∈ such that c = λ(1−λb(0)) . Notice1
that for all f ∈H(b), we have

2 2 ∗ 2 2 ∗ 2 b 2 ∗ b ∗kfk =kUfk =kX fk +|c | |hf,S bi | kk k +2Re c hf,S bihk ,X fi .1 b 1 b bb b b 0 b 0
∗In particular for f =S b, using relation (3), we get
∗ 2 2 ∗ 2 2 ∗ ∗0 =−kS bk +|c | kS bk (1−|b(0)| )+2Re(c (X S b)(0)).1 1b b
∗ ∗ ∗ ∗ 2 ∗ ∗ ∗ 2Since X S b =SS b−kS bk b, it follows that (X S b)(0) =−kS bk b(0), whichb b
implies that
2 20 =−1+|c | (1−|b(0)| )−2Re(c b(0)).1 1
−1Nowdefineλ :=c +b(0). Usingtheprevious equality, easycomputationsshow1
−1that λ∈ and c =λ(1−λb(0)) , which ends the proof of the proposition. 1
The following lemma is a generalization of a result of Ahern-Clark [1] for the
case where b is an inner function.
Lemma 2.3 Let ζ ∈ . The following assertions are equivalent:
(i) b has an angular derivative in the sense of Carath´eodory at ζ;
b ∗(ii) k ∈ Im(Id−ζX ).0
∗ b bMoreover, in that case, we have (Id−ζX )k =k .ζ 0
Proof: (i) =⇒ (ii): sinceb hasangularderivative inthe sense of Carath´eodory
b bat ζ, we know (see [19], (VI-4)) that k tends to k in norm as z tends nontan-z ζ
gentially to ζ. Notice we have
∗ b ∗ b ∗ b b ∗ ∗ bk(Id−zX )k −(Id−ζX )k k =k(Id−zX )(k −k )+((Id−zX )−(Id−ζX ))k kz ζ z ζ ζ
b b ∗ b62kk −k k+|z−ζ|kX kkk k.z ζ ζ
∗ b ∗ bHence (Id− zX )k tends to (Id− ζX )k as z tends nontangentially to ζ.z ζ
Moreover we have
∗ b b(Id−zX )k =k ,z 0
b ∗ b(see [19], (V-8)) which implies that k = (Id−ζX )k .0 ζ
b ∗(ii) =⇒ (i): assume that there exists g ∈H(b) such that k = (Id−ζX )g.0
We have
b ∗ −1 bk =(Id−zX ) kz 0
∗ −1 ∗=(Id−zX ) (Id−ζX )g
∗ ∗=g+(z−ζ)(Id−zX )X g,
9




which gives that

b ∗ −1 ∗kk k6kgk 1+|z−ζ|k(Id−zX ) kkX k .z
∗ −1 −1Using the fact thatk(Id−zX ) k6 (1−|z|) , we deduce that

|z−ζ|b ∗kk k6kgk 1+ kX k .z 1−|z|
As |z−ζ|/(1−|z|) stays bounded as z tends nontangentially to ζ, we get that
bkk kstaysboundedasz tendsnontangentiallytoζ,whichbyTheoremA,impliesz
that b has an angular derivative in the sense of Carath´eodory at ζ.
Since U is an isometry, its point spectrum is located on the unit circle. Theλ
notionofangularderivativewillleadustocaracterizeit. Thisresultwasobtained
by Ahern-Clark [1] for the case where b is inner.
Theorem 2.1 Let λ ∈ . Then a complex number ζ is an eigenvalue of U ifλ
and only if b has an angular derivative in the sense of Caratheodory at ζ and
bb(ζ)=λ. Moreover we have Ker(U −ζId)= k .λ ζ
Proof: assume that b has an angular derivative in the sense of Caratheodory
b ∗ bat ζ and b(ζ)=λ. Using Lemma 2.3, we have k =(Id−ζX )k . Hence0 ζ
b ∗ b −1 b ∗ b(U −ζId)k =(X −ζId)k +λ(1−λb(0)) hk ,S bi kλ bζ ζ ζ 0
b −1 b ∗ b=−ζk +λ(1−λb(0)) hk ,S bi k .b0 ζ 0
Take now a sequence (z ) which tends nontangentially to ζ and notice thatn n
b(z )−b(0) λ−b(0)n∗ b ∗ bhS b,k i = lim hS b,k i = lim = .b bζ zn
n→+∞ n→+∞ z ζn
That implies
b b −1 b(U −ζId)k =−ζk +λ(1−λb(0)) ζ(λ−b(0))k = 0,λ ζ 0 0
bwhich proves that ζ ∈σ (U ) and k ⊂ Ker(U −ζId).p λ λζ
Reciprocally, let ζ ∈ σ (U ) and f ∈ H(b), f 6≡ 0 such that (U −ζId)f =p λ λ
∗ −1 ∗ b0. Then, we have (X − ζId)f = −λ(1− λb(0)) hf,S bi k . Notice that ifb 0
∗ ∗hf,S bi = 0, then ζ ∈ σ (X ), which is absurd thanks to Lemma 2.2. Henceb p
∗ b ∗hf,S bi = 0, and there exists c∈ , c = 0, such that k = (Id− ζX )(cf).b 0
Lemma 2.3 implies that b has an angular derivative in the sense of Carath´edory
b ∗ b b ∗at ζ and k = (Id− ζX )k . We deduce that k − cf ∈ Ker(X − ζId) and0 ζ ζ
b bLemma 2.2 implies that k = cf. Hence k ∈ Ker(U − ζId). But previousλζ ζ
computations show that
!
b(ζ)−b(0)b −1 b(U −ζId)k = −ζ +λ(1−λb(0)) k ,λ ζ 0
ζ
10
66