Bull. Soc. math. France,
127, 1999, p. 25–41.
GENERALIZATION OF VON NEUMANN’S
SPECTRAL SETS AND
INTEGRAL REPRESENTATION OF OPERATORS
BY
Bernard DELYON and François DELYON (*)
ABSTRACT. — We extend von Neumann’s theory of spectral sets, in order to
deal with the numerical range of operators. An integral representation for arbitrary
operators is given, allowing to extend functional calculus to non-normal operators.
We apply our results to the proof of the Burkholder conjecture: let T be an operator
consisting in a finite product of conditional expectation, then for any square integrable
function f , the iterates T n f converge almost surely to some limit.
RÉSUMÉ. — GÉNÉRALISATION DES ENSEMBLES SPECTRAUX ET REPRÉSENTATION
— Nous modifions la théorie des ensembles spectraux
de von Neumann pour l’appliquer à l’image numérique des opérateurs. Nous donnons
une représentation intégrale pour des opérateurs bornés quelconques; ceci étend le
calcul fonctionnel aux opérateurs non normaux. Comme application, nous démontrons
la conjecture de Burkholder: soit T un opérateur produit d’un nombre fini d’espérances
conditionnelles, alors pour toute fonction de carré sommable f , les itérées T n f
convergent presque sûrement.
INTÉGRALE DES OPÉRATEURS.
1. Introduction
The spectral sets theory was introduced by von Neumann [3] in order to
extend functional calculus to non-normal operators on an Hilbert space.
Let us recall basic principles of von Neumann’s theory as described in [4].
(*) Texte reçu le 9 octobre 1997, révisé le 4 avril 1998, accepté le 3 juillet 1998.
B. DELYON, IRISA-INRIA, Campus de Beaulieu, 35042 Rennes CEDEX (France).
Email: [email protected].
F. DELYON, Centre de Physique Théorique, École Polytechnique, 91128 Palaiseau,
CEDEX (France); UMR-C7644 du Centre National de la Recherche Scientifique.
Email: [email protected]
Keywords: numerical range, field of values, spectral sets, spectral measures
AMS classification: 47 A 12, 47 A 25, 47 A 60, 60 F 15.
BULLETIN DE LA SOCIÉTÉ MATHÉMATIQUE DE FRANCE
c Société mathématique de France
0037–9484/1999/25/$ 5.00
26
B. DELYON AND F. DELYON
DEFINITION 1. — A set σ ⊂ C is a spectral set of the operator T on an
Hilbert space if it is closed and if for any rational function u(z) one has
u(T ) ≤ supu(z).
(1)
z∈σ
This formula is valid even if u(z) has poles in σ. On the other hand,
rational functions are dense, in the sense of Runge Theorem [6], in the set
of analytic functions. This usual formulation in terms of rational functions
is nothing but a way to avoid technicalities about boundary values and
singularities of analytic functions.
The following theorem, due to von Neumann [3], gives necessary and
sufficient conditions under which a ball, its complement, or a half-plane
is a spectral set of T :
THEOREM 1 (von Neumann). — A necessary and sufficient condition
for one of the domains
|z − α| ≤ r,
|z − α| ≥ r,
Re(αz) ≥ β
to be a spectral set of T is that 1
T − α ≤ r, (T − αI)−1 ≤ r−1 ,
Re(αT ) ≥ β.
This implies obviously that for any T ,
z : |z| ≤ T is a spectral set of T .
Notice that Theorem 1 implies that for any point α out of the spectrum
of T , there exists an r such that {z : |z − α| ≥ r} is a spectral set and as
a consequence, the intersection of the spectral sets of T is its spectrum.
This implies that the intersection of two spectral sets is not necessarily a
spectral set, what is apparently a serious drawback of the theory.
The concept of numerical range if closely related:
DEFINITION 2. — The numerical range (or field of values) of the
operator T is
Θ(T ) = z = (f ,T f ): f = 1 .
It is well-known that this set is convex [5, p. 267], and that it is a real
interval if and only if T is symetric [5, p. 269].
1
With the convention that Re(T ) =
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THEOREM 2. — The intersection of the spectral half-planes is the closure
of the numerical range.
Denote by Σ(T ) the intersection of the spectral half-planes. Since both
sets are convex and closed, it is enough to prove that for any closed halfplane H, H contains Σ(T ) if and only if it contains Θ(T ). If H has the
form
H = z : Re(az + b) ≥ 0
we have
Σ(T ) ⊂ H ⇐⇒ Re(aT + b) ≥ 0
⇐⇒ (f, Re(aT + b)f ) ≥ 0,
f = 1
⇐⇒ Re(a(f, T f ) + b) ≥ 0, f = 1
⇐⇒ Θ(T ) ⊂ H.
Our main contribution is the following theorem:
THEOREM 3. — Let T be an operator on an Hilbert space and σ be a
bounded convex subset of C containing 2 Θ(T ). There exists a constant Cσ
such that for any rational function u(z) one has
u(T ) ≤ Cσ supu(z).
(2)
z∈σ
More generally, for any finite sequence of rational functions u1 , . . . ,un ,
the following holds
(3)
ui (z)2 .
ui (T )∗ ui (T ) ≤ Cσ2 sup
z∈σ
i
i
The constant Cσ may be chosen as
Cσ =
2πD2
S
3
+3
where S is the area of σ and D its diameter.
A natural choice for σ is the closure of Θ(T ); however if S is small, one
may prefer to choose σ larger.
2
By Theorem 2 this condition means that Re(ᾱ(T − z)) ≥ 0 for any point z of ∂σ
and α the (inwards) normal vector at z.
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B. DELYON AND F. DELYON
This theorem is a consequence of the following representation theorem.
THEOREM 4. — Let σ be as in Theorem 3 and assume that is has a
piecewise C 1 boundary. Denote by C(∂σ) the space of continuous functions on ∂σ endowed with the uniform norm. Under the assumptions of
Theorem 3, there exist a continuous linear operator Sσ on C(∂σ) and a
measure µT (dz) on ∂σ with values on the set of non-negative self-adjoint
operators such that
(4)
u(T ) =
(Sσ u)(z) µT (dz)
∂σ
for any rational function with poles outside of σ, and
µT (dz) = I , Sσ ≤ Cσ .
∂σ
The operator Sσ and the measure µT are detailed in section 3; Sσ is
the inverse of the operator R defined in Lemma 6. Let us shortly mention
here, that:
dz
µT (dz) = Re
2iπ(z − T )
properly defined when (z − T ) is not invertible, thus µT depends on σ
and T , but Sσ only depends on σ. We assume that ∂σ is piecewise C 1
for avoiding technicalities in the definition of this measure, however
some extra effort would probably lead to the same result without this
assumption.
In Section 5, we provide two useful results in order to control spectral
sets of product of operators.
EXAMPLE: the unit circle. — Let T an operator such that the set σ
of Theorem 3 is the unit circle. Let us show the differences between our
theorems and the von Neumann result. In this case, the assumption of
Theorem 3 is:
(f, T f ) ≤ f 2
(5)
which is weaker than von Neumann’s condition, since T can be 2.
Furthermore, if σ is the unit circle, one can compute explicitly the
operator Sσ (as the inverse of the operator R of Lemma 6, Section 3):
1
Sσ (u) = 2u −
2π
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GENERALIZATION OF VON NEUMANN’S SPECTRAL SETS
Thus, we have:
29
u(T ) ≤ 3 sup u(z).
z≤1
Let us give an explicit example. Consider the nilpotent matrix:
0 2
T =
.
0 0
This matrix satisfies equation (5) and we can evaluate µT on the unit
circle:
dθ 1
dz
e−iθ
.
=
(6)
Re
iθ
1
2iπ(z − T )
2π e
Thus µT is positive on the unit circle and T = 2.
The Burkholder Conjecture. — Consider T = P1 P2 · · · Pn a finite
product of conditional expectations w.r.t σ-fields F1 · · · Fn .
It is well-known (see [8], [9]) that for any f in L2 , T k f converges in L2
to the conditional expectations w.r.t F1 ∩ F2 ∩ . . . ∩ Fn (i.e., T k converges
strongly to the orthogonal projection on the largest invariant space of T );
this convergence holds even in Lp for 1 < p < +∞ (see [9]).
The conjecture of Burkholder is that the convergence holds almost
surely for any function of L2 :
lim T k f = E f |F1 ∩ F2 ∩ . . . ∩ Fn , a.s.
k→∞
Since T is a positive contraction on L1 and T 1 = 1, the almost sure
convergence of the averages
An f =
n−1
1 k
T f
n
k=0
is a consequence Chacon’s Theorem (th 3.7 in [2]). By using a technique
borrowed to Stein [7] one proves that almost sure convergence of T n f
holds if
∞
(n + 1)(I − T )∗ T ∗n T n (I − T ) < ∞.
n=0
We shall prove this result in Section 6. We first use the results on products
of operators for proving that an Θ(T ) is included in some sectorial domain
(this result has been proved independently in a unpublished paper of
A. Brunel [1, prop. 9]); we then use equation (3).
We would like to thank Jean-Luc Sauvageot for fruitful discussions.
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B. DELYON AND F. DELYON
2. The von Neumann theory revisited
The following Theorem [3], which is the basis of the proof of Theorem 1,
is actually the cornerstone of the theory:
THEOREM (von Neumann). — If T ≤ 1, the unit disk U is a spectral
set for T .
We give here a new proof of it which introduces in a simplified context
the essential ingredients of the proof of Theorem 3.
Let us assume first that the spectrum of T is contained in the open
unit disk U . For any rational function u, analytic on the disk, we have
1
T ∗ dz
dz
1
= u(z)
+
u(z)
u(T ) =
2iπ ∂U
z−T
z−T
1 − zT ∗ 2iπ
since the added term is analytic in the disk; thus
dz
u(T ) = u(z)(z −1 − T ∗ )−1 (1 − T ∗ T )(z − T )−1
2iπz
what rewrites
(7)
u(T ) =
∂U
u(z)µT (dz)
for some measure µT with values on the set of non-negative self-adjoint
operators. Applying this identity to u(z) = 1, we get that
µT (∂U ) = 1.
Since µT is self-adjoint, we have
∗
u(T ) =
(8)
∂U
u(z)µT (dz).
The equation
∗
u(z) − u(T ) µT (dz) u(z) − u(T ) ≥ 0
rewrites, thanks to (7) and (8)
∗
u(T ) u(T ) ≤
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u(z)2 µT (dz).
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This completes the proof in the case where the spectrum of T is in the
open unit disk. If not, we can consider the operators (1 − ε)T , apply the
result to them and
u((1 − ε)T ) ≤ sup u(z)
|z|≤1
implying
v(T ) ≤ sup v
|z|≤1
z 1−ε
for any v rational in the neighborhood of U . We can let ε tend to zero
and the result is proved.
3. Proof of Theorem 4
The proof of Theorem 4 relies on the following lemma:
LEMMA 6. — Let σ be a bounded convex subset of C and C(∂σ) be the
set of complex functions continuous on σ and harmonic on its interior,
with the supremum norm. The following equation
dz
)
ϕ(z) Re(
(9)
(Rϕ)(z0 ) =
2iπ(z
− z0 )
∂σ
defines a continuous operator on C(∂σ) (the value for z0 ∈ ∂σ is defined
by continuity). This operator has a bounded inverse and
−1 2πD2 3
R ≤
+3
S
Proof of Lemma 6. — For a real continuous function ϕ, equation (9)
rewrites:
1
(10)
(Rϕ)(z0 ) =
ϕ(z) d arg(z − z0 )
2π ∂σ
well defined for z0 ∈ σ \ ∂σ. First let us show that this equation defines
a continuous operator on C(∂σ). Let z0 ∈ ∂σ and {zn } be a sequence of
points of σ \ ∂σ converging to z0 . Let E be a small interval of ∂σ containing z0 , with boundary points z1 and z2 (z0 = z1 and z0 = z2 ). On the one
hand:
d arg(z − zn ) = (z1 − zn , z2 − zn ) −→ (z1 − z0 , z2 − z0 )
(11)
E
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B. DELYON AND F. DELYON
by continuity of the angle (z1 − z, z2 − z) in the neighbourhood of z0 .
Notice that, by convexity of σ, this limit is at least π (= π at a regular
point).
On the other hand, it is clear that
ϕ(z) d arg(z − zn ) −→
ϕ(z) d arg(z − z0 ).
(12)
∂σ\E
∂σ\E
Let us denote:
d arg(z − z0 )
·
2π
Equations (11) and (12) prove that for any continuous function ϕ on ∂σ,
constant in the neighbourhood of z0 , µzn (ϕ) converges to some limit
µz0 (ϕ) if zn → z0 and this limit does not depend on the particular
sequence zn ∈ σ \ ∂σ. This clearly extends to any continuous function ϕ,
and the measures µzn (dz) converges weakly to µz0 (dz). Moreover the
function z → µz (ϕ) is continuous (since its value on the boundary is
defined as an existing limit).
Consequently R is well-defined and continuous on C(∂σ).
Let us set
R = 12 (I + P ).
µz0 (dz) =
Since the limit (11) is at least π, we have µz0 ({z0 }) ≥
This operator rewrites
(P ϕ)(z0 ) =
ϕ(z) νz0 (dz)
1
2
and P ≤ 1.
∂σ
where νz0 is a positive measure of total mass 1 and such that νz0 ({z0 }) = 0
at any regular point.
In order to complete the proof of Lemma 6, we have to prove that R
has a continuous inverse. This is a direct consequence of the following:
PROPOSITION 7. — There exist a positive measure ν ∗ on ∂σ such that
for ϕ ≥ 0 and any z0 ∈ ∂σ:
3 S 3
.
(P 3 ϕ)(z0 ) ≥ ν ∗ (ϕ) and ν ∗ (∂σ) =
4 πD2
Proof of Proposition 7. — Because of the translation invariance of the
result we can assume that 0 is the center of gravity of σ:
(13)
z dx dy = 0.
∂σ
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The proof is based on a lower bound of P by a non-negative operator of
rank 2; we have indeed
(z̄ − z¯ ) dz dz
0
≥ Re
(14)
νz0 (dz) = Re
2iπ(z − z0 )
2iπD2
where D is the diameter of σ; this lower bound is non-negative. Now, if
we define the operator Q by
1
ϕ(z) Im((z̄ − z¯0 ) dz)
(15)
(Qϕ)(z0 ) =
2S ∂σ
where S is the total area of σ, equation (14) rewrites, for any positive
function ϕ,
S
(Qϕ)(z0 )
P ϕ(z0 ) ≥
πD2
We obtain the effect of Q on linear functions by use of the Green formula:
for any analytical function f (z)
1
(Qf )(z0 ) =
f (z) Im((z̄ − z¯0 ) dz)
2S
1
=
f (z) (x − x0 ) dy − (y − y0 ) dx
2S
1
=
∂x (f (z)(x − x0 )) + ∂y (f (z)(y − y0 ) dx dy
2S
1
f (z)(z − z0 ) + 2f (z) dx dy.
=
2S
Hence using (13):
(Q1)(z0 ) = 1, Q(z)(z0 ) = − 12 z0 , Q Im(āz) + b = − 12 Im(āz0 ) + b.
By (15), the function Qϕ is a real linear function of z0 :
(Qϕ)(z0 ) = Im(āz0 ) + b
with
1
a=
2S
ϕ(z) dz,
1
b=
2S
ϕ(z) Im(z̄ dz).
Since the effect of the kernel Q on a such a function is to multiply a by − 12
and to leave b unchanged, we obtain
πD2 3
(P 3 ϕ)(z0 ) ≥ (Q3 ϕ)(z0 )
S
= Q2 Im(āz) + b = b + 14 Im(āz0 )
3b
≥
4
since Im(āz0 ) ≥ −b (i.e. Qϕ is positive). This states the proposition.
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B. DELYON AND F. DELYON
Proof of Lemma 6 (continued). — Proposition 7 rewrites with ε =
ν ∗ (∂σ)
ν ∗ (dz)
+ (1 − ε)
ε
∂σ
= εL + (1 − ε)K
P3 = ε
ϕ(z)
∂σ
ϕ(z) νz
0 (dz)
where νz
0 ( · ) is for any z0 a positive measure, K = 1, L = 1,
and P L = L. Hence
(I + P 3 ) I − 12 εL = I + (1 − ε)K.
Since I + (1 − ε)K is invertible, this implies that
(I + P 3 )−1 = (I + (1 − ε)K)−1 I − 12 εL ,
(I + P 3 )−1 ≤ 2 + ε
2ε
and finally,
2πD2 3
3(2 + ε)
=
+ 3.
R−1 = 2(I + P 3 )−1 (I − P + P 2 ) ≤
ε
S
Proof of Theorem 4. — Set
Sσ = R−1 .
Lemma 6 implies that for any analytic function u on σ, with real part ϕ,
one has
dz
dz
= Re
ϕ(z0 ) =
Sσ ϕ(z)Re
Sσ ϕ(z)
2iπ(z − z0 )
2iπ(z − z0 )
∂σ
∂σ
hence, the function
u(z0 ) −
∂σ
Sσ ϕ(z)
dz
2iπ(z − z0 )
is analytic inside σ with zero real part; consequently there exists a real
number u0 such that
dz
·
Sσ ϕ(z)
u(z0 ) = iu0 +
2iπ(z
− z0 )
∂σ
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This implies, if the spectrum of T lies in the interior of σ
dz
,
Sσ ϕ(z)
u(T ) = iu0 +
2iπ(z
− T)
∂σ
∗
dz
u(T )∗ = −iu0 +
Sσ ϕ(z)
.
2iπ(z − T )
∂σ
Thus
(16)
Re u(T ) =
∂σ
Sσ ϕ(z)Re
dz
2iπ(z − T )
where Re(U ) for an operator U is defined as 12 (U + U ∗ ). Denoting by ψ
the imaginary part of u and applying identity (16) to the function −iu
we obtain
dz
(17)
Im u(T ) =
Sσ ψ(z)Re
2iπ(z − T )
∂σ
1
where Im(U ) for an operator U is defined as 2i
(U − U ∗ ); putting together
equations (16) and (17), we deduce
dz
=
Sσ u(z)Re
Sσ u(z)µT (dz).
u(T ) =
2iπ(z − T )
∂σ
∂σ
Let us check that the assumptions imply that µT is a measure on ∂σ with
values on the set of non-negative self-adjoint operators. The assumption
on σ rewrites, thanks to Theorem 1
αT + ᾱT ∗ ≥ αz + ᾱz̄
for any z ∈ ∂σ regular and ᾱ parallel to i dz. This implies
−ᾱ ∗
2Re
= −ᾱ(z − T )−1 − α(z − T )−1
z−T
∗
= −(z − T )−1 (ᾱ(z − T )∗ + α(z − T ))(z − T )−1
≥ 0.
Thus, if z is a regular point,
Re
dz
≥ 0;
2iπ(z − T )
since the spectrum of T is inside σ, the mass of the non-regular points
is zero and we conclude that µT is a positive measure. Note that equation (16) and R1 = 1 imply that µT (1) = I. This completes the proof in
the case where the spectrum of T lies in the interior of σ.
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If not, we have to define the measure µT as a suitable limit.
Pick a point z ∗ in the interior of σ and set
Tε = (1 − ε)T + εz ∗ .
The convexity of σ implies that the spectrum of Tε is in the interior of σ,
and the half-planes containing σ are spectral sets of Tε (use Theorem 1).
For any analytic function u and ε
> ε:
∂σ
Sσ (u)(z)(dµTε − dµTε ) = u(Tε ) − u(Tε ) = u(Tε ) − v(Tε )
where
v(z) = u
1 − ε
1−ε
(z − εz ∗ ) + ε
z ∗ .
At this point, Theorem 3 applies to {Tε }ε>0 :
(18)
∂σ
Sσ (u)(z)(dµTε − dµTε ) ≤ Cσ supu(z) − v(z).
z∈σ
As ε
goes to ε, left hand side of (18) goes to 0 (u is uniformly continuous
on σ). The same inequality holds for ū, and thus for any continuous
function on ∂σ. Thus, since Sσ is one-to-one, for any continuous function u
lim
ε,ε →0
∂σ
u(z)(dµTε − dµTε ) = 0.
Thus for any vector f , the positive measures (f, µTε f ) are weakly continuous on ε and converges to some limit measure µT,f . For any u, this limit µT,f (u) is bilinear on f , thus defining an operator valued measure µT .
The theory for these operator valued measures µT can be carried out
like the projector valued measures of the spectral theory of self-adjoint
operators. One easily checks that the following properties, satisfied by Tε ,
are also satisfied by T :
µT (A ∪ B) = µT (A) + µT (B),
A ∩ B = ∅,
µT (∅) = 0, µT (∂σ) = I,
∞
µT
Ai = s-lim µT (Ai ), Ai ⊂ Ai+1
i=1
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and3 that for any bounded Borel function u on ∂σ, there exists a unique
operator u(T ) such that, for any f ∈ L2 :
Sσ (u)(f, dµT f ).
f, u(T )f =
∂σ
4. Proof of Theorem 3
Let us assume first that ∂σ is piecewise C 1 . For any rational function u,
one has
∗
Sσ u(z) − u(T ) µT (dz) Sσ u(z) − u(T ) ≥ 0
what rewrites, thanks to (4)
u(T )∗ u(T ) ≤
(19)
Sσ u(z)2 µT (dz).
Since for any z the map u → (Su)(z) is continuous on C(∂σ), there exist
measures νz (ds) such that
Sσ u(z) = u(s)νz (ds) and Sσ = sup νz .
z
Hence, by the Cauchy-Schwarz inequality
2 2 2
(20) Sσ u(z) ≤ νz u(s) · νz (ds) ≤ Sσ u(s) · νz (ds).
Finally equations (19) and (20) imply
∗
(Sσ ui )(z)2 µT (dz)
ui (T ) ui (T ) ≤
i
i
≤ Sσ ui (s)2 · νz (ds) µT (dz)
i
ui (s)2 .
≤ Sσ sup
2
s∈σ
i
1
If now ∂σ is not piecewise C , we consider the sets
σε = z : d(z, σ) ≤ ε .
These sets are closed and convex and we can take the limit ε → 0 in the
equation
ui (s)2 .
ui (T )∗ ui (T ) ≤ Cσε 2 sup
i
s∈σε
i
3
s-lim denotes the strong limit of operators: A = s-lim An if and only if An f − f converges to 0 for any vector f .
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5. Spectral sets of a product of operators
Let us consider a contraction T , i.e. T ≤ 1. The unit circle is a
spectral set for T . Obviously the unit circle is a spectral set for products
of contractions. In this part we give two examples of properties which
remain true through product of operators. These properties will be useful
for the proof of Burkholder’s conjecture.
PROPOSITION 8. — For any 0 ≤ α ≤ 1, denote by Cα the circle of radius
1 − α centered at (α,0). Let Ti (i = 1,2), be operators such that Cαi is a
spectral set of Ti . Then Cα1 α2 is spectral set for T1 T2 .
Proof. — From Theorem 1:
Ti − αi I ≤ 1 − αi ,
then,
1 > αi > 0
T1 T2 − α1 α2 I = (T1 − α1 )T2 + α1 (T2 − α2 )
≤ (1 − α1 ) + α1 (1 − α2 ) = 1 − α1 α2 .
For half-planes, the stability through product is more complicated.
PROPOSITION 9. — For any real ε, denote Hε the half-plane, containing
(0,0), and having (1,0) on its boundary, defined by:
Hε = z : Re((1 + iε)(1 − z)) ≥ 0 .
Let Ti (i = 1,2), be contractions with Hεi as spectral set. Assume in
addition that Cα is spectral set of T2 for some α > 0. If ε1 ε2 > 0, there
exist ε = 0 with the same sign as the εi ’s, such that Hε is a spectral set
of T1 T2 .
Proof. — Choose γ < 1 and set z1 = 1 + iγε1 . From the assumptions,
setting Si = 1 − Ti , {z : Re((1 + iεi )z) ≥ 0} is a spectral set of Si and
one has
2 Re(z1 S1 ) = 2γ Re (1 + iε1 )S1 + (1 − γ)(S1 + S1∗ )
≥ (1 − γ)(S1 + S1∗ )
= (1 − γ) I + S1∗ S1 − (I − S1 )∗ (I − S1 )
≥ (1 − γ)S1∗ S1 .
By the way, this proves that Hγε is spectral set of a contraction T as soon
as Hε is spectral set of T . On the other hand:
I − T1 T2 = (I − T1 )T2 + (I − T2 ) = T2∗ S1 T2 + S2∗ S1 T2 + S2 ,
Re z1 (I − T1 T2 ) = Re T2∗ (z1 S1 )T2 + Re(z1 S2∗ S1 T2 ) + Re(z1 S2 ).
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For the second term, using the inequality
2 Re(AB) ≥ −AA∗ − B ∗ B
√
√
with A = z1 ( 1 − γ )−1 S2∗ and B = 1 − γ S1 T2 , we obtain:
2 Re z1 (I − T1 T2 ) ≥ 2T2∗ Re(z1 S1 )T2 − (1 − γ)T2∗ S1∗ S1 T2
−
≥−
|z1 |2 ∗
S S2 + 2 Re(z1 S2 )
1−γ 2
|z1 |2 ∗
S S2 + 2 Re(z1 S2 ).
1−γ 2
By the assumption on T2 and setting β = 1 − α:
S2∗ S2 = (S2 − β)∗ (S2 − β) − β 2 + 2β Re(S2 ) ≤ 2β Re(S2 ).
Consequently,
|z1 |2
Re(βS2 ) + Re(z1 S2 )
Re z1 (I − T1 T2 ) ≥ −
1−γ
ε1 γ S2
= 2C Re 1 + i
C
with
C =1−
(1 + γ 2 ε21 )β
·
1−γ
Thus, for γ small enough C is positive and |ε1 |γ/C ≤ |ε2 |.
Finally, Re(z1 (I − T1 T2 )) ≥ 0 and the proof is complete.
6. Proof of Burkholder’s conjecture
Consider T = P1 P2 · · · Pn a finite product of projectors. The set {0, 1}
is the spectral set of any self-adjoint projector P . A fortiori, the circle C1/2 ,
as well as any half-plane Hε , are spectral sets of P . From Propositions 8
and 9, there exists α, ε > 0 such that Cα , Hε and H−ε are spectral sets
of T . The set σ = Cα ∩ Hε ∩ H−ε satisfies the assumptions of Theorem 3.
Notice that:
|1 − z|
< ∞.
sup
z∈σ 1 − |z|
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B. DELYON AND F. DELYON
Theorem 3 yields:
p
(n + 1)(I − T )∗ T ∗n T n (I − T )
n=0
∞
(n + 1)|z|2n
≤ Cσ sup |1 − z|2
(21)
z∈σ
n=0
|1 − z|2
= Cσ sup
2 2
z∈σ (1 − |z| )
|1 − z| 2
≤ Cσ sup
< ∞.
z∈σ 1 − |z|
Once this is obtained we conclude by using classical arguments. Define
An =
n−1
1 k
T
n
k=0
and let us recall that Chacon’s Theorem implies the almost sure convergence of An f if f ∈ L2 . First we have
An − T n =
1
(I − T )(I + 2T + · · · + nT n−1 )
n
and setting g = (I − T )f ,
1
(I + 2T + · · · + nT n−1 )g,
n
1
sup(An − T n )f ≤ sup g + 2T g + · · · + nT n−1 g n
n n
n−1
1/2
1
k(T k g)2
≤ sup (1 + 2 + · · · + n)1/2
n n
k=0
∞
1/2
k 2
≤
k(T g)
.
(An − T n )f =
k=0
Equation (21) implies now, for any f ∈ L2 , the following bound:
sup |(An − T n )f | ≤ Cf 2 .
(22)
2
n
This allows us to conclude by a density argument.
Indeed, since T is a contraction, {f : f = T ∗ f } coincides with {f : f =
T f }. On the other hand, the range of I − T is dense in the orthogonal
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of {f : f = T ∗ f }. Thus, for any ε > 0 and any f ∈ L2 , there exists
g, h, f1 ∈ L2 such that:
f = (I − T )g + h + f1 ,
h2 < ε, T f1 = f1
We can assume that f1 is 0. Notice that equation (21) implies that
T n (I − T )g converges a.s to zero. Furthermore, Chacon’s Theorem implies
the almost sure convergence of An h, thus:
lim sup |T n f | = lim sup |T n h − An h|
n
n
Equation (22) applied to h, provides:
lim sup |T n f | ≤ Cε.
2
n
Since ε is arbitrary, T n f converges a.s, and the conjecture is proved.
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