A bounded operator T on a Hilbert space H is said to be decomposable (cf. [LN, Definition 1.1.1]), if for every open cover C=U∪V of the complex plane, there are T-invariant closed subspacesH andH of H, such that the spectra of the restrictions of T satisfyσ (T|H)⊆U andσ (T|H )⊆V, and such thatH=H +H .
Given T∈B(H), a spectral capacity for T is a mapping E from the set of closed subsets ofC into the set of all closed, T-invariant subspaces ofH, such that
(i) E(∅)= {0}and E(C)=H,
(ii) E(U1)+ · · · +E(Un)=Hfor every (finite) open cover{U1, . . . ,Un}ofC, (iii) E(∞
n=1Fn)=∞
n=1E(Fn)for every countable family(Fn)∞n=1 of closed subsets ofC,
(iv) σ (T|E(F)) ⊆ F for every closed subset F of C (with the convention that σ (T|{0})= ∅).
Finally, given T∈B(H)andξ ∈H, the local resolvent set,ρT(ξ ), of T atξ is the union of all open subsets U ofC, for which there exist holomorphic vector-valued functions fU: U→H, such that (T−λ1)fU(λ)=ξ for allλ∈U. Note that according to Neumann’s lemma,
{z∈C| |z|>T} ⊆ρT(ξ ),
and therefore,σT(ξ ):=C\ρT(ξ ), the local spectrum of T atξ, is compact. For any subset A ofC, the corresponding local spectral subspace of T is
HT(A)= {ξ ∈H|σT(ξ )⊆A}. (9.1)
It is not hard to see thatHT(A)is T-hyperinvariant.
Now, the three definitions given above, i.e. that of decomposability, that of a spec-tral capacity and that of a local specspec-tral subspace, are closely related, as the following theorem indicates:
Theorem9.1 [LN, Proposition 1.2.23]. — Let T be a bounded operator on a Hilbert space H. Then the following are equivalent:
(i) T is decomposable, (ii) T has a spectral capacity,
(iii) for every closed subset F ofC,HT(F)is closed and σ ((1−pT(F))T|HT(F)⊥)⊆σ (T)\F, where pT(F)denotes the projection ontoHT(F).
Moreover, if T is decomposable, then the map F→HT(F)is the unique spectral capacity for T.
Now, if the operator T appearing in our generalized version of Theorem1.1is de-composable, how is the local spectral subspaceHT(B)related to the T-invariant subspace KT(B)for B∈B(C)?
Proposition9.2. — If T is a decomposable operator in the II1-factorM, then for every B∈ B(C),KT(B)=HT(B).
Proof. — SinceσT(ξ )is compact for everyξ ∈H, we have that HT(B)=
K⊆B,K compact
HT(K).
Moreover, by definition
KT(B)=
K⊆B,K compact
KT(K).
Hence, it suffices to prove that for every compact set K⊆C,KT(K)=HT(K). Since T is decomposable, Theorem9.1implies that for every closed subset F ofC,
σ (T|HT(F))⊆F, and
σ ((1−QT(F))T|HT(F)⊥)⊆σ (T)\F,
where QT(F)∈W∗(T)denotes the projection ontoHT(F). In particular, supp(μT|HT(F))⊆F,
and
supp(μT|(1−QT(F))T|HT(F)⊥)⊆σ (T)\F.
It follows that QT(F)≤PT(F), where PT(F)∈W∗(T)denotes the projection ontoKT(F).
Now,
τ (PT(F))=μT(F)
=τ (QT(F))μT|HT(F)(F)+τ (QT(F)⊥)μ(1−QT(F))T|
HT(F)⊥(F), and since F∩σ (T)\F⊆∂F, we get that
τ (PT(F))≤τ (QT(F))+τ (QT(F)⊥)μ(1−QT(F))T|
HT(F)⊥(∂F).
Hence, ifμT(∂F)=0, thenτ (PT(F))≤τ (QT(F)), and it follows that PT(F)=QT(F).
For a general closed subset F ofC, define Ft= z∈Cdist(z,F)≤ 1
t
, (t>0).
Then FtF as t ∞. Take 0<t1≤t2≤ · · ·,such that tn→ ∞ and such that for all n∈N,μT(∂Ftn)=0. Then, since F→QT(F)is a spectral capacity for T, we have:
QT(F)= ∞ n=1
QT(Ftn)
= ∞ n=1
PT(Ftn)
=PT(F).
Corollary9.3. — If T∈Mis decomposable, then supp(μT)=σ (T).
Proof. — We know from [Br] that supp(μT)⊆σ (T)always holds. Now, let T∈M be decomposable, and let F=supp(μT). Then KT(F)=H. Hence, by Proposition 9.2 and Theorem9.1,
HT(F)=HT(F)=KT(F)=H.
Therefore, by condition (iv) in the definition of a spectral capacity,
σ (T)=σ (T|HT(F))⊆F=supp(μT).
Corollary9.4. — Every II1-factorMcontains a non–decomposable operator.
Proof. — According to [DH1, Example 6.6],N =∞
k=2B(Ck)contains an opera-tor T for whichσ (T)=D andμT=δ0. Since N embeds into the hyperfinite II1-factor R, and since every II1-factor containsR as a von Neumann subalgebra,Mcontains a copy of T. According to Corollary9.3, T is not decomposable.
Remark 9.5. — By [DH2], Voiculescu’s circular operator is decomposable. More generally, every DT-operator is decomposable.
Appendix A: Proof of Theorem6.2
Let 0<p<1 and consider a fixed map f : [a,b] →Lp(M, τ )which is Hölder continuous with exponentα > 1−pp. That is, f satisfies (6.2) for some positive constant C. For every closed subinterval[c,d]of[a,b]we letd
c f(x)dx be given by Definition6.1.
LemmaA.1. — Let a=x0<x1<· · ·<xm−1<xm=b and a=y0<y1<· · ·<yn−1<
yn=b be partitions of the interval[a,b], and define Tx=
m i=1
f(xi)(xi−xi−1),
Ty= n
j=1
f(yj)(yj−yj−1).
Then withδx=max1≤i≤m(xi−xi−1)andδy=max1≤j≤n(yj−yj−1)one has that Tx−Typp≤Cp(m+n)max{δx, δy}p+αp.
(A.1)
Proof. — Let a=z0 <z1<· · · <zr−1<zr =b be the partition of the interval [a,b] containing all of the points x1, . . . ,xm−1 and y1, . . . ,yn−1. For 1≤k≤r, let i(k)∈ {1, . . . ,m}and j(k)∈ {1, . . . ,n}be those indices for which
[zk−1,zk] ⊆ [xi(k)−1,xi(k)] and
[zk−1,zk] ⊆ [yj(k)−1,yj(k)]. Then
Tx= r
k=1
f(xi(k))(zk−zk−1), (A.2)
and
Ty= r
k=1
f(yj(k))(zk−zk−1).
(A.3)
Since zk ∈ [xi(k)−1,xi(k)] ∩ [yj(k)−1,yj(k)], both of the points xi(k)and yj(k) must belong to the interval[zk,zk+max{δx, δy}], and it follows that
f(xi(k))−f(yj(k))p≤C max{δx, δy}α. (A.4)
Combining (A.2), (A.3) and (A.4) we find that Tx−Typp≤
r k=1
(zk−zk−1)pCpmax{δx, δy}αp
≤(m+n)Cpmax{δx, δy}αp+p. LemmaA.2. — For every c∈(a,b),
b a
f(x)dx= c a
f(x)dx+ b c
f(x)dx.
(A.5)
Proof. — Let (Sn)∞n=1, (S(1)n )∞n=1 and (S(2)n )∞n=1 be the sequences defining b
a f(x)dx, c
af(x)dx andb
c f(x)dx, respectively (cf. Definition6.1). That is, with xi=a+ib−a get from LemmaA.1that
Sn−(S(1)n +S(2)n )pp≤3·2nCp
Hence
Proof. — At first we consider the case c=b. Taking (6.4) into account we obtain:
(b−a)f(b)−
and then by LemmaA.3, It follows now that for arbitrary c∈(a,b),
(c−a)f(c)−
Proof Theorem6.2. — According to LemmaA.2, M−
and by LemmaA.4,
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U. H.
Department of Mathematics and Computer Science, University of Southern Denmark,
Campusvej 55,
5230 Odense M, Denmark haagerup@imada.sdu.dk H. S.
Danske Bank,
Holmens Kanal 2–12,
1092 København K, Denmark
Manuscrit reçu le 9 septembre 2007 publié en ligne le 26 mars 2009.