Download C. Foias, S. Hamid, C. Onica, and C. Pearcy

OPERATORS WITH COMPACT IMAGINARY PART
C. Foias, S. Hamid, C. Onica, and C. Pearcy∗†
Abstract
In this note we initiate a study of the old unsolved problem whether every
T ∈ L(H) of the form T = H + iK with K compact has a nontrivial invariant
subspace, using [6] as our main tool. In case K ≥ 0 we obtain some positive
results.
1. Let H be a separable, infinite dimensional, complex Hilbert space. We write
L(H) for the algebra bounded linear operators on H, K for the ideal of compact
operators in L(H), and π for the quotient (Calkin) map L(H) → L(H)/K. For T
in L(H) we will write, as usual, σ(T ), σe (T ), σle (T ), and kT ke , for the spectrum,
essential (Calkin) spectrum, left essential spectrum, and essential (Calkin) norm of
T , respectively, and C ∗ (S) for the unital C ∗ -algebra generated by a collection S ⊂
L(H). We remind the reader that the question whether every operator T in L(H) with
compact imaginary part (i.e., T = H + iK with H, K selfadjoint and K ∈ K) has a
nontrivial invariant subspace has been around for over fifty years, but, nevertheless, is
still open. In fact, the major results in the area, by Livsic [9], Macaev [10], Schwartz
[12] and Gohberg-Krein [7], are over forty-five years old. The purpose of this note
is to use the little known but useful main theorem from [6] to make some progress
on this old outstanding problem. For the reader’s convenience, we first state the
main result from [6], which characterizes the set of all (positive semidefinite) Q that
arise from the Aronszajn-Smith “invariant subspace procedure” applied to an arbitrary
quasitriangular operator (cf. [11, Chapter 4]), in the form this theorem from [6] was
used in [5] and [4], where it was also the main tool.
Theorem 1.1. Let A be a unital, norm-separable, norm-closed subalgebra of L(H)
and let Q ∈ L(H) be such that 0 Q 1. Then the following statements are
equivalent:
a) there exists a sequence {Qn }n∈N of projections in L(H) such that
k(1 − Qn )AQn k → 0,
A ∈ A,
and {Qn } converges to Q in the weak operator topology (WOT).
∗
†
2000 Mathematics Subject Classification. Primary 47A15.
Key words and phrases: invariant subspaces, compact imaginary part.
1
(1)
b) With M := (QH)− (6= (0)) and N := ((1−Q)QH)− (6= (0)), there exist completely
contractive algebra homomorphisms ϕ̃ : A → L(M) and θ̃ : π(A) → L(N ), with
ϕ̃(1H ) = 1M and θ̃(π(1H )) = 1N , uniquely determined by the relations
Q1/2 ϕ̃(A) = AQ1/2 |M ,
A ∈ A,
(2)
and
(1 − Q)1/2 ϕ̃(A) = θ̃(π(A))(1 − Q)1/2 |M,
A ∈ A.
(3)
Note that if for an arbitrary T in L(H), we write AT for the unital, norm-closed
algebra generated by T , then every Q that arises from Theorem 1.1 (a) (with A = AT )
gives rise to its own algebra homomorphisms ϕ
e and θe as in b) above. We also remind
the reader that the Aronszajn-Smith procedure (cf. [1]) applied to a quasitriangular
operator (cf. [8]) yielded the following (cf., e.g., [2] or [11, Ch. IV]).
Theorem 1.2. Let T be an arbitrary quasitriangular operator in L(H). If 1 ≥
Q ≥ 0 arises as in Theorem 1.1 a), and Q(1 − Q) is not a quasiaffinity (i.e., either
ker Q 6= (0) or ker(1 − Q) 6= (0)), then T has a nontrivial invariant subspace.
As a consequence of Theorem 1.2, we denote by Q(H) the set of all (quasitriangular)
T ∈ L(H) with the property that every Q satisfying a) and b) of Theorem 1.1 (with
A = AT ) is such that Q(1 − Q) is a quasiaffinity. We also denote the set of all such Q
(for a given T ) by QT , and we restrict attention in what follows to those T belonging
to Q(H) with compact imaginary part.
The following corollary of Theorem 1.1 is what will be needed below.
Corollary 1.3. Suppose T = H + iK is in Q(H), with H, K nonzero and
selfadjoint and K compact. Then T is quasitriangular, and if {Qn } is any sequence of
finite-rank projections arising from Aronszajn-Smith procedure (cf. [2] or [11, Ch. IV])
satisfying (1) (with A = AT ) and converging in the WOT to an operator Q ∈ QT ,
then, in the notation of Theorem 1.1, M = N = H, and (2) and (3) become
Q1/2 ϕ̃(A) = AQ1/2 ,
A ∈ AT ,
(4)
and
(1 − Q)1/2 ϕ̃(A) = θ̃(π(A))(1 − Q)1/2 ,
A ∈ AT .
(5)
We now introduce our first symmetrization of equations (4) and (5), with T as in
Corollary 1.3, by setting A = T n , multiplying (4) on the left by (1 − Q)1/2 , (5) on the
left by Q1/2 , and combining (4) and (5), yielding
(1 − Q)T n Q = (Q − Q2 )1/2 θ̃(π(H))n (Q − Q2 )1/2 ,
n ∈ N.
(6)
Since the right-hand-side of (6) is selfadjoint for each n ∈ N, so is the left, and writing
T n = Hn + iKn with Hn , Kn selfadjoint and Kn ∈ K, we obtain
(1 − Q)(Hn + iKn )Q = Q(Hn − iKn )(1 − Q),
2
n ∈ N,
(7)
so
QHn − Hn Q = i(Kn Q + QKn ) − 2i(QKn Q),
n ∈ N,
(8)
and
QT n − T n Q = QHn − Hn Q + i(QKn − Kn Q) = 2i(QKn − QKn Q),
n ∈ N.
(9)
Next, suppose that T = H + iK is an arbitrary operator in Q(H) with H, K
nonzero and selfadjoint, and K compact. For such T , we write RT for the unital, real,
norm-closed algebra generated by T. It follows easily by induction that Im(R) ∈ K for
every R ∈ RT .
We now introduce our second symmetrization of equations (4) and (5).
Lemma 1.5. Suppose T = H + iK belongs to Q(H), with H, K nonzero and
selfadjoint, and K compact. Then, in the notation of Corollary 1.3, for every A =
HA + iKA ∈ RT , with HA , KA selfadjoint and KA ∈ K, we have
1/2
e
e
Qθ(π(H
(1 − Q)1/2 KA Q1/2 (1 − Q)1/2 ,
A )) − θ(π(HA ))Q = 2iQ
A ∈ RT . (10)
Proof. Upon multiplying (4) on the left by Q1/2 , and on both sides by (1 − Q)1/2 ,
then multiplying (5) on the left by Q, and combining (4) and (5) we obtain
e
Qθ(π(A))(1
− Q) = (1 − Q)1/2 Q1/2 AQ1/2 (1 − Q)1/2 ,
A ∈ AT .
(11)
Thus, in particular, since KA is compact for A ∈ RT , we have
1/2 1/2
e
Qθ(π(H
Q (HA + iKA )(1 − Q)1/2 Q1/2 ,
A ))(1 − Q) = (1 − Q)
A ∈ RT
(12)
and taking the imaginary part of each side of (12) gives
1
e
{Qθ(π(H
A ))(1 − Q)
2i
e
− (1 − Q)θ(π(H
A ))Q}
1
e
e
= {Qθ(π(H
A )) − θ((π(HA ))Q},
2i
Q1/2 (1 − Q)1/2 KA (1 − Q)1/2 Q1/2 =
(13)
A ∈ RT ,
which yields (10).
Note, in particular, a special case of (10) when A = T n := Hn + iKn with Hn , Kn
selfadjoint and nonzero and Kn ∈ K:
e
e
2i(1 − Q)1/2 Q1/2 Kn (1 − Q)1/2 Q1/2 = Qθ(π(H
n )) − θ(π(Hn ))Q
n
n
e
e
= Qθ(π(H))
− θ(π(H))
Q
e
e
= Q(θ(π(T
)))n − θ((π(T
)))n Q,
Our investigation begins with the following lemma.
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(14)
n ∈ N.
Lemma 1.6. Suppose T = H + iK ∈ Q(H), with H = H ∗ , K = K ∗ 6= 0, and
K ∈ K, and there exists Q ∈ QT such that σp (Q) ∩ (0, 1) 6= ∅ – say Qe = λe with
e 6= 0 and 0 < λ < 1. Then hIm(R)e, ei = 0, for all R ∈ RT .
Proof. It suffices to show that hIm(Tn )e, ei = 0 for every n ∈ N. With Hn :=
Re(Tn ) and Kn := Im(Tn ), we have, using (9),
0 = λ hT n e, ei − hT n e, λei = h(QT n − T n Q)e, ei
= 2i h(QKn − QKn Q)e, ei = 2i(λ hKn e, ei − λ2 hKn e, ei)
= 2iλ(1 − λ) hKn e, ei ,
n ∈ N.
But since 0 < λ < 1, this gives hKn e, ei = 0 for n ∈ N, as desired.
(15)
This next result is one of our main theorems.
Theorem 1.7. Suppose T = H + iK ∈ Q(H), with H, K nonzero and selfadjoint,
and K compact and positive semidefinite. If there exists a Q ∈ QT such that σp (Q) ∩
(0, 1) 6= ∅, then T has a nontrivial reducing subspace M ⊂ ker K such that T |M is
selfadjoint. Consequently, T has a nontrivial hyperinvariant subspace (n.h.s.).
Proof. Suppose Q ∈ QT satisfies Qe = λe with e 6= 0 and 0 < λ < 1. Then
from Lemma 1.6 we obtain that Kn := Im(Tn ) satisfies hKn e, ei = 0 for all n ∈ N.
In particularW(n = 1), hKe, ei = 0, and since K ≥ 0, Ke = 0. We next show that
(0) 6= M := {T n e : n ∈ N0 } ⊂ ker K (6= H). First, since Ke = 0,
0 = hK3 e, ei = (HKH + KH 2 + (H 2 − K 2 )K)e, e
= hHKHe, ei = hKHe, Hei .
Thus KHe = 0 and KT e = KHe + iK 2 e = 0. Now suppose, by induction, that we
have shown that KHj e = 0 and Kj e = 0 for j = 1, · · · , n. We wish to show that
KHn+1 e = 0 and Kn+1 e = 0. To this end, since Kn e = 0 by the induction hypothesis,
we write
Kn+1 e = (KHn + HKn )e = 0.
(16)
Hence, we obtain via (16),
0 = hK2n+3 e, ei
= h(Kn+1 HHn+1 + Hn+1 KHn+1 + Hn+1 HKn+1 − Kn+1 KKn+1 )e, ei
= h(KHn+1 e, Hn+1 ei ,
and therefore, KHn+1 e = 0 as desired. Thus Kn e = 0 and KT n e = 0 for n ∈ N, so
M ⊂ ker K. Hence M ⊃ T M = (H + iK)M = HM, and similarly, T ∗ M = HM ⊂
M, which shows that M reduces H and T and that T |M = H|M . If T |M is not a
scalar multiple of 1M , then T |M has a nontrivial hyperinvariant subspace, and thus so
does T by [3]. On the other hand, if T |M is a (real) scalar multiple of 1M , then T has
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a nontrivial eigenspace, which cannot be all of H since K 6= 0, and this eigenspace is
hyperinvariant for T.
Corollary 1.8. Suppose T = H + iK ∈ Q(H), with H, K nonzero, K ∈ K, and
K ≥ 0. If T is purely accretive (i.e., T has no nontrivial reducing subspace on which
it is selfadjoint), then for every Q ∈ QT , σp (Q) = ∅.
Corollary 1.9. Suppose T = H +iK ∈ Q(H), with H, K nonzero and selfadjoint,
and K positive semidefinite and compact, and there exists Q ∈ QT such that σe (Q) is
a singleton {λ} with 0 < λ < 1. Then T has a nontrivial hyperinvariant subspace.
Proof. It is obvious that Q = λ1H + J where J = J ∗ is compact. If J has only
finitely many nonzero eigenvalues then ker J must be infinite dimensional, so λ is an
eigenvalue of Q, and the result follows from Theorem 1.7. On the other hand, if J
has infinitely many nonzero eigenvalues, they must converge to 0, so Q has eigenvalues
arbitrarily close to λ, and the result follows as before.
Our next result shows that under certain conditions, with T ∈ Q(H), an associated
e
Q ∈ QT , and Tb := θ(π(T
)), if Tbf = rf with r ∈ R and f 6= 0, then T f = rf also.
Theorem 1.10. Suppose T = H + iK ∈ Q(H), with H, K nonzero, K ∈ K,
and K ≥ 0. Suppose also that there exists Q ∈ QT such that the corresponding
e
Tb := θ(π(T
)) satisfies Tbf = rf for some unit vector f and r ∈ R. Then, with
M := ∨n∈N0 Qn f, T M ⊂ M and T |M = r1M . In particular, T f = rf, and the
eigenspace of T corresponding to the eigenvalue r is a n.h.s. for T .
Proof. We write R := Q1/2 (1 − Q)1/2 , note that R is a quasiaffinity, and apply (14)
with n = 1 to get
b f i − hQf, Hf
b i = r hQf, f i − hQf, rf i = 0.
2i hRKRf, f i = hQHf,
Thus, since RKR ≥ 0, f ∈ ker RKR. Next, note that by similar reasoning, we get
b − HQ
b 3 )f, f i
0 = h(Q3 H
2
b − HQ)
b + Q(QH
b − HQ)Q
b
b − HQ)Q
b
= h{Q2 (QH
+ (QH
}f, f i
= 0 + hRKRQf, Qf i + 0.
Thus Qf ∈ ker RKR also, and by an easy induction argument using the formula
b − HQ
b n=
Qn H
n
X
j=1
j−1
b − HQ)Q
b
Qn−j (QH
=
n
X
Qn−j (RKR)Qj−1 ,
n ∈ N, (17)
j=1
and the fact that
b − HQ
b n )f, f i = hQn rf, f i − hQn f, rf i = 0,
h(Qn H
n ∈ N,
we obtain by induction that
RKRQn f = 0,
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n ∈ N0 ,
(18)
b−
i.e., that M := ∨n∈N0 Qn f ⊂ ker RKR. Therefore, from (17) and (18) we get (Qn H
b n )f = 0, n ∈ N, and thus HQ
b n f = Qn Hf
b = rQn f, n ∈ N, i.e., HM
b
HQ
⊂ M and
b
b
H|M = r1M. Moreover, (11) applied to A = T yields RT R = QH(1 − Q), n ∈ N0 , and
since obviously M is reducing for Q, 1 − Q, and R, we obtain
b − Q)g = Qr(1 − Q)g = rR2 g,
RT Rg = QH(1
g ∈ M,
and since R is a quasiaffinity, T Rg = rRg, g ∈ M. Thus, since (RM)− = M,
T M ⊂M and T |M = r1M . In particular, T f = rf, and the eigenspace Er of T
associated with r is a n.h.s. for T (since K 6= 0, M ⊂ E r 6= H).
Theorems 1.7 and 1.10 can easily be seen to be equivalent (by use of the Cayley
transform) to the following, where T is the unit circle in C.
Theorem 1.11. Suppose T is a contraction in Q(H) such that 0 6= 1 − T ∗ T ∈ K
and σ(T ) ∩ T 6= T, and let Q ∈ QT . Then the following hold :
a) If σp (T ) ∩ (0, 1) 6= ∅, then T has a nontrivial reducing subspace M such that
T |M is unitary. Consequently, T has a n.h.s.
e
b) If there exists Q ∈ QT such that the corresponding Tb = θ(π(T
)) satisfies Tbf =
rf for some nonzero f in H, then T f = rf as well, and the eigenspace corresponding
to λ is a n.h.s. for T .
The above results lead to many interesting problems. Here are two:
Problem 1.12. What is the analog of Theorems 1.7 and 1.10 without the hypothesis that K ≥ 0?
Problem 1.13. If H is given to be diagonalizable or, more generally, normal,
what result can be proved?
References
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Ciprian Foias
Department of Mathematics
Texas A&M University
College Station, TX 77843, USA
Sami M. Hamid
Department of Mathematics
University of North Florida
Jacksonville, FL 32224, USA
E-mail : [email protected]
Constantin Onica
Department of Mathematics
The University of Texas - Pan American
Edinburg, TX 78539, USA
E-mail : [email protected]
Carl Pearcy
Department of Mathematics
Texas A&M University
College Station, TX 77843, USA
E-mail : [email protected]
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