INTRODUCTION TO DE BRANGES–ROVNYAK SPACES

Emmanuel Fricain

Description

• cours - matière potentielle : m2r

Universite LYON I cours de M2R INTRODUCTION TO DE BRANGES–ROVNYAK SPACES Emmanuel Fricain —-2008-2009—

• aa? ≤


• bb?


• bounded operators between


• classical hardy spaces


• hilbert spaces


• h1 ??


• invariant subspaces


• toeplitz operators


• a?x

Sujets

• Hilbert space

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 9
• Langue: English

Extrait

Universite LYON I
cours de M2R
INTRODUCTION TO DE
BRANGES{ROVNYAK SPACES
Emmanuel Fricain
|-2008-2009|iiContents
Preface iv
1 Hilbert space operators 1
1.1 Douglas’ factorization theorem . . . . . . . . . . . . . . . . . . . 1
1.2 The square root of a positive operator . . . . . . . . . . . . . . . 2
1.3 Partial isometries and the polar decomposition . . . . . . . . . . 5
1.4 Reproducing kernel Hilbert spaces . . . . . . . . . . . . . . . . . 9
1.5 Fredholm theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.6 The operator of multiplication by the independant variable on
2L (). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 Analytic functions on the open unit disc 17
2.1 The Poisson integral . . . . . . . . . . . . . . . . . . . . . . . . . 17
p2.2 Classical Hardy spaces H . . . . . . . . . . . . . . . . . . . . . . 18
22.3 The shift operator on H . . . . . . . . . . . . . . . . . . . . . . 22
2.4 The F. M. Riesz Theorem . . . . . . . . . . . . . . . . . . . . . . 24
22.5 Generalized Hardy spaces H () . . . . . . . . . . . . . . . . . . 26
2.6 The Cauchy integral . . . . . . . . . . . . . . . . . . . . . . . . . 27
3 Toeplitz operators 31
3.1 The multiplication operator M . . . . . . . . . . . . . . . . . . . 31’
3.2 The norm of T . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32’
3.3 The adjoint of T . . . . . . . . . . . . . . . . . . . . . . . . . . . 34’
3.4 Toeplitz operators with (anti-)analytic symbols . . . . . . . . . . 36
3.5 Composition of Toeplitz operators . . . . . . . . . . . . . . . . . 37
3.6 The compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.7 The operator K . . . . . . . . . . . . . . . . . . . . . . . . . . . 39’
23.8 Toeplitz operators on generalized Hardy spaces H () . . . . . . 41
4 The spacesM(A) andH(A) 45
4.1 The spaceM(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
4.2 Contractive inclusion ofM(A) inM(B) . . . . . . . . . . . . . . 49
4.3 Linear functionals onM(A) . . . . . . . . . . . . . . . . . . . . . 51
4.4 The complementary spaceH(A) . . . . . . . . . . . . . . . . . . 52
iii4.5 The Halmos intertwining theorem . . . . . . . . . . . . . . . . . . 54
4.6 The Lotto{Sarason theorem . . . . . . . . . . . . . . . . . . . . . 55
4.7 The overlapping space . . . . . . . . . . . . . . . . . . . . . . . . 57
4.8 Decomposition ofH(A) . . . . . . . . . . . . . . . . . . . . . . . 58
4.9 of H . . . . . . . . . . . . . . . . . . . . . . . . . 62
25 Hilbert spaces in H 65
5.1 The spaceH(b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
5.2 Model subspaces K . . . . . . . . . . . . . . . . . . . . . . . . . 66
5.3 The reproducing kernel ofH(b) . . . . . . . . . . . . . . . . . . . 68
5.4 H(b) andH(b) as invariant subspaces . . . . . . . . . . . . . . . . 69
5.5 The operator X . . . . . . . . . . . . . . . . . . . . . . . . . . . 71b
5.6 Integral representation ofH(b) . . . . . . . . . . . . . . . . . . . 72
5.7 Integral representation ofH(b) . . . . . . . . . . . . . . . . . . . 73
5.8 Multipliers ofH(b) . . . . . . . . . . . . . . . . . . . . . . . . . . 76
6 Applications ofH(b) spaces 79
6.1 Comparison of measures . . . . . . . . . . . . . . . . . . . . . . . 79
6.2 Angular derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 82
6.3 Caratheodory’s theorem . . . . . . . . . . . . . . . . . . . . . . . 85
7 The nonextreme case ofH(b) spaces 91
7.1 First properties ofH(b) . . . . . . . . . . . . . . . . . . . . . . . 91
7.2 The polynomials are dense inH(b) . . . . . . . . . . . . . . . . . 94
7.3 The shift onH(b) . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
7.4 The multipliers ofH(b) . . . . . . . . . . . . . . . . . . . . . . . 96
7.5 A result of completeness . . . . . . . . . . . . . . . . . . . . . . . 97
8 Appendix 101
8.1 Extreme points of a convex set in a Banach space . . . . . . . . . 101
1 18.2 points of the unit ball of H and H . . . . . . . . . . 104
8.3 A theorem of Helson{Szeg o . . . . . . . . . . . . . . . . . . . . . 107
Bibliography 109Chapter 1
Hilbert space operators
1.1 Douglas’ factorization theorem
Let A : H ! H and B : H ! H be bounded operators between Hilbert1 2
spaces. In certain applications, we need to know when there is a contraction
C :H ! H such that A =BC holds.1 2
Figure 1.1: The factorization A =BC
If such a contraction exists, then
BB AA =BB BCC B =B (I CC )B 0:
Douglas showed that the conditionAA BB is also su cient for the existence
of C.
Theorem1.1 (Douglas). LetH;H andH be Hilbert spaces and letA :H !1 2 1
H and B : H ! H be bounded operators. Then there is a contraction C :2
H ! H such that A =BC if and only if AA BB .1 2
Proof. The necessity was shown above. The other direction is a bit more deli-
cate. Suppose that AA BB . Hence
kA xk kB xk ; (x2H): (1.1)H H1 2
This inequality enables us to de ne an operator from the range of B to the
range of A . Let
D :R(B ) ! R (A )
B x 7 ! A x:
If an element inR(B ) has two representations, i.e. z = B x = B y, then
B (x y) = 0. Hence, by (1.1),A (x y) = 0. In other words,Dz =A x =A y
is well de ned. Moreover,
kD(B x)k kB xk ; (x2H):H H1 2
1Therefore, by continuity, D extends to a contraction from the closure ofR(B )
inH intoH . In the last step of extension, we extend D to a contraction from2 1
H to H by de ning2 1
?D(z) = 0; (z2R(B ) ):
According to our primary de nition, the contraction D satis es DB = A .
Hence A =BD . Take C =D .
Exercises
Exercise 1.1.1. Let H ;H and H be Hilbert spaces, let C :H ! H be1 2 3 1 2
a contraction, and let A :H ! H be a bounded operator. Show that2 3
A (I CC )A 0;
where I :H ! H is the identity operator.2 2
1.2 The square root of a positive operator
If B2L(H) and we de ne A = BB , then certainly A is a positive operator.
Our goal is to show that every positive operator is obtained that way. First we
need two preliminary lemmas.
Lemma1.2. LetA;B2L(H) be positive andAB =BA. ThenAB is positive.
2Proof. If TS =ST , T and S positive, it is clear that TS is positive. Without
loss of generality assume thatkBk 1. Let B =B, and let0
2
B =B B ; (n 0):n+1 n n
Then the relations
2 2B =B (I B ) +B (I B )n+1 n n nn
and
2I B = (I B ) +Bn+1 n n
imply that
0B I; (n 0):n
Thus,
nX
2B =B B B; (n 0): (1.2)n+1k
k=0Therefore, for each x2H and n 0,
n n nX X X
2 2kB xk = hB x;B xi = hB x;xik k k k
k=0 k=0 k=0
nX
2= h B x;xihBx;xi:k
k=0
P1 2Thus kB xk <1. In particular, lim B x = 0 and, by (1.2),k n!1 nk=0
nX
2lim B x =Bx:kn!1
k=0
Pn 2Since A B is positive, we conclude that AB is also positive.k=0 k
Lemma 1.3. Let A ;A ; ;B2L(H) be self adjoint. Suppose that1 2
A A =A A ; A B =BA ; (m;n 1);n m m n n n
and that
A A B; (n 1):n n+1
Then there is A2L(H), A self adjoint, such that
Ax = lim A x; (x2H):n
n!1
Proof. Let C =B A , n 1. Hencen n
0C C ; (n 1): (1.3)n+1 n
Then, by Lemma 1.2,
2 20C C C C ; (mn):n mn m
In the rst place, for each x2H,
kC xkkC xk; (mn): (1.4)n m
Since (kC xk) is a decreasing sequence of positive numbers, we concluden n1
that
lim kC xkn
n!1
exists. Secondly,
2 2 2kC x C xk kC xk k C xk ; (mn);m n m n
and thus (C x) is a Cauchy sequence. Letn n1
Cx = lim C x; (x2H):n
n!1Clearly C is linear. Moreover, by (1.4), we have
kCxk = lim kC xkkC xk; x2H;:n 1
n!+1
which proves that C2L(H) andkCkkCk. Since C is self adjoint, C is1 n
also self adjoint. Put A =B C.
Theorem 1.4. Let A 2 L(H) be positive. Then there is a unique positive
2operator B2L(H) such that A =B .
Proof. Without loss of generality assume thatkAk 1. Put B =A and0
2A BnB =B + ; (n 0):n+1 n
2
Clearly 0B I. Moreover, by Lemma 1.2, the relations0
2(I B ) + (I A)n
I B =n+1
2
and
(I B ) + (I B )n n 1
B B = (B B )n+1 n n n 1
2
imply that
0B B I; (n 0):n n+1
Hence, by Lemma 1.3, there is B2L(H), B positive, with
Bx = lim B x; (x2H):n
n!1
But
2B x AxnB x =B x + ; (n 0);n+1 n
2
2which immediately gives B x =Ax.
It remains to show that B is unique. Suppose that there is C 2L(H),
2 2 2C positive, such that C = A. Then AC = C C = CC = CA. Therefore,
p(A)C =Cp(A), wherep is any polynomial. In particular,B C =CB ,n 0,n n
and thus BC =CB. Fix x2H, and let y = (B C)x. Then
2 2h(B +C)y;yi =h(B C )x;yi = 0:
Since B and C are positive operators, we thus have
hBy;yi =hCy;yi = 0:
But these assumptions implyBy =Cy = 0. For example, to verify thatBy = 0,
1=2based on the rst paragraph, we know that B exists. Hence
1=2 2 1=2 1=2kB yk =hB y;B yi =hBy;yi = 0:
1=2Thus B y = 0 which implies By = 0. Finally,
2 2k(B C)xk =h(B C)x; (B C)xi =h(B C) x;xi =h(B C)y;xi = 0:
Therefore, B =C.The operator B whose unique existence was guaranteed by the preceding the-
1=2orem is called the square root of A and is denoted by A . The following two
results were also implicitly obtained in the proof of Theorem 1.4.
Corollary 1.5. Let A2L(H) be positive. Then there is a there is a sequence
of real polynomial (p ) with p (0) = 0 such thatn n1 n
1=2A x = lim p (A)x; (x2H):n
n!1
Corollary 1.6. Let A2L(H) be positive, and let x2H. Suppose that
hAx;xi = 0:
Then Ax =