Normal Operators, II
C. F
OIAS, I.B. J
UNG, E. K
O,
ANDC. P
EARCYThe second, third, and fourth authors of this article wish much good health and happiness to their dear friend, Ciprian Foias, on the occasion of his birthday.
ABSTRACT. As the title indicates, this note is a sequel to [2], in which we showed that a large class of rank-one perturbations of a diagonalizable normal operator have nontrivial hyperinvari-ant subspaces. Below we establish the perhaps surprising fact that the commutants of such operators are abelian, paralleling thereby the properties of the commutants of normal operators of multiplicity one. We also show by example that this behavior does not extend to the commutants of rank-one perturbations of all normal operators of multiplicity one, and we discuss sim-ilarity and quasisimsim-ilarity questions associated with this class of operators.
1. ABELIANCOMMUTANTS
The notation and terminology we use below is consistent in every detail with that of [2], but for the reader’s convenience we begin with a short review of it. LetH
be a separable, infinite dimensional, complex Hilbert space, and denote byL(H )
the algebra of all bounded linear operators onH. ForT inL(H ), we write{T }0
for the commutant ofT (i.e., for the algebra of allS∈ L(H )such thatT S= ST )
and{T }00= ({T }0)0for the double commutant ofT. As usual in what follows,N,
Z, C, D,andT = ∂Dwill denote the sets of positive integers, integers, complex numbers, open unit disc, and unit circle inC,respectively. The ideal of compact operators inL(H )will be denoted byKand the Calkin mapL(H ) → L(H )/K
byπ. ForT inL(H )we denote byσ (T )the spectrum ofT, byσle(T )[σre(T )]
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the left essential (i.e., Calkin) [right essential] spectrum ofT, and
σe(T )= σ (π(T )) = σle(T )∪ σre(T ), σlr e(T )= σle(T )∩ σre(T ).
Moreover, we write, as usual,σp(T )for the point spectrum ofT. We next choose
an ordered orthonormal basis{en}n∈NforH, which will remain fixed throughout
the paper. IfΛ = {λn}n∈Nis any bounded sequence inC, we writeDΛ for the
normal operator inL(H )determined by the equations
(1.1) DΛ(en)= λnen, n∈ N.
This notation forΛ = {λn}n∈N andDΛwill also remain fixed throughout, as well as
the notationΛ0 for the derived set ofΛ. By definition, we shall say that an operator T inL(H )is a rank-one perturbation of a diagonalizable normal operator if there exist nonzero vectors
(1.2) u= X n∈N αnen and v= X n∈N βnen
inH and a bounded sequenceΛ = {λn}n∈NinCsuch thatT is unitarily
equiva-lent to the operatorDΛ+ u ⊗ v,where, as usual,u⊗ vis the operator of rank one defined by(u⊗ v) (x) = hx, vi u, x ∈ H .The notation{αn}n∈Nand{βn}n∈N
for the Fourier coefficients ofuandv, respectively (relative to the fixed basis{en}),
will also remain fixed throughout this note.
The main result of [2] is the following.
Theorem 1.1. LetT = DΛ+ u ⊗ vbe any rank-one perturbation of a
diagonal-izable normal operator such thatT 6∈ C1H andPn∈N(|αn|2/3+ |βn|2/3) <+∞.
ThenT has a nontrivial hyperinvariant subspace (n.h.s.).
In our proof of Theorem1.1, we first disposed of some easier cases, and then concluded the proof by establishing the following more technical result.
Theorem 1.2. With the notation as introduced above, supposeT = DΛ+ u ⊗ v
is such that
(i) the mapn→ λnofNontoΛis injective andΛ0is not a singleton;
(ii) for everyn∈ N,αnβn6=0;
(iii) Pn∈N(|αn|2/3+ |βn|2/3) <+∞(the nontrivial assumption).
Then either
(I) there exists an idempotent F with 0 6= F 6= 1H such that F ∈ {T }00, and
consequently,T has a complemented n.h.s. (i.e., there exist n.h.s. MandN of T withM ∩ N = (0)andM + N = H), or
(II) there exists an uncountable set{µ:µ∈ P}of eigenvalues ofTand an associated family{uµ}µ∈P of linearly independent eigenvectors (with T uµ = µuµ) such
We also take note of a result obtained in [4].
Proposition 1.3 ([4]). If T = DΛ+ u ⊗ v ∈ L(H )\C1H and there exists n0 ∈ N such that αn0βn0 = 0, then either λn0 ∈ σp(T ) or ¯λn0 ∈ σp(T∗).
Moreover, if there exist m0, n0 ∈ N with m0 6= n0 such that λm0 = λn0, then
λn0 ∈ σp(T ). Finally, if Λ0 is a singleton, then{T }0 contains a nonzero compact
operator. Consequently, in all casesT has a n.h.s.
Thus in what follows we restrict our attention to the class(RO)consisting of all operators T = DΛ + u ⊗ v in L(H ) for which all coefficients αn and βn are nonzero, Λ = {λn}n∈N is a one-to-one map ofNinto C, and Λ0 is not
a singleton. We remark that it follows easily that if T1 = DΛ1 + u1⊗ v1 and
T2 = DΛ2+ u2⊗ v2 belong to(RO)withT1 = T2, then the sequences Λ1and
Λ2 coincide andu1⊗ v1 = u2⊗ v2 ([4, Prop. 1.1]). It is also clear that for all T = DΛ+ u ⊗ v ∈ (RO), we haveσe(T )= σlr e(T )= σlr e(DΛ)= Λ0.
Recall also from [2, Def. 3.1] that the class(RO)1 consists, by definition,
of allT = DΛ+ u ⊗ v in (RO) such that σ (T ) = σe(T ) (= Λ0), σ (T ) is a
perfect set, andPn∈N(|αn|2/3+ |βn|2/3) <∞. Moreover, the heart of the proof
of Theorem 1.1 in [2] consisted in showing that everyT = DΛ+ u ⊗ v in(RO)1
has a n.h.s.
For operatorsT = DΛ+ u ⊗ v ∈ (RO), we now turn to a certain property of
the commutant{T }0ofT.
Theorem 1.4. SupposeT = DΛ+ u ⊗ v ∈ (RO), whereΛ = {λn}n∈N, DΛ,
anduandvare as in (1.1) and (1.2)(and the definition of the class(RO)). Then the mapϕ: {T }0 → {T }0udefined byϕ(A)= AuforA ∈ {T }0is a one-to-one,
bounded linear transformation from{T }0onto the linear manifold{T }0u.
Proof. It is obvious thatϕis linear, surjective, and bounded (bykuk). Thus it suffices to show that if A ∈ {T }0 and Au = 0, thenA = 0. The equation AT = T Ais equivalent to
(1.3) ADΛ− DΛA= (u ⊗ A∗v)− (Au ⊗ v)
and by virtue of (1.1),(ADΛ− DΛA)ek, ek
=0 fork∈ N. Thus, via (1.3) we obtain (1.4) ek, A∗v u, ek =ek, v Au, ek , k∈ N.
Since Au = 0 and hu, eki = αk 6= 0 for all k in N, we get from (1.4) that A∗v = 0. Putting this into (1.3) yields A ∈ {DΛ}0, and sinceDΛ is a normal
operator of uniform multiplicity one, we see that there exists a bounded sequence
{σk}k∈N⊂ Csuch thatAek= σkekfork∈ N. Finally, from (1.4) we get
(1.5) 0= hek, A∗vi = hAek, vi = σkhek, vi = σkβ¯k, k∈ N,
This next proposition would seem to open up several interesting possibili-ties for exploration. As usual, we abbreviate the phrase “weak [strong] operator topology onL(H )” by WOT [SOT].
Proposition 1.5. With T ∈ (RO) as in Theorem1.4, if the linear manifold {T }0uis closed(i.e., a subspace), thenT has a n.h.s.
Proof. If M = {T }0u 6= H, then M is a n.h.s. for T, and if M = H,
then u is a strictly cyclic vector for the algebra {T }0. If T has no n.h.s., then {T }0 is a transitive algebra (i.e., has no n.i.s.), and one knows (cf. [7,
Corol-lary 9.10]) that the only WOT-closed, transitive subalgebra ofL(H )that has a strictly cyclic vector isL(H ). But this implies thatT ∈ C1H, which is false since
T ∈ (RO). ❐
We now turn to a description of the commutant {T }0 of an arbitraryT in (RO)due to E. Ionascu. For completeness a proof is sketched.
Proposition 1.6 ([4]). SupposeT = DΛ+ u ⊗ v ∈ (RO), where, as always,
Λ = {λn}n∈Nandu, vare given by (1.1) and (1.2), and letA∈ {T }0. Then there
is a unique sequence{τk}k∈N= {τk(A)}k∈N⊂ Csuch that
(1.6) Au= X k∈N (τkαk)ek, and (i) X k∈N |τkαk|2<+∞, (ii) X k∈N |τkβk|2<+∞andA∗v= P k∈Nτ¯kβkek,
(iii) if (ap,q)p,q∈N is the matrix for A relative to the ordered orthonormal basis {ek}k∈N, then ap,q= hAeq, epi = αpβ¯q τp− τq λp− λq ! , p, q∈ N, p 6= q, (1.7) and ap,p= hAep, epi = τp− X q∈N q6=p αqβ¯q τp− τq λp− λq ! , p∈ N. (1.8)
Sketch of proof. Let{γk}k∈Nbe the unique square summable sequence of
com-plex numbers such that
(1.9) Au= X
k∈N γkek.
Since T ∈ (RO), αk 6= 0 for eachk ∈ N, and we defineτk = γk/αk. This
establishes (1.6) and part (i). Then, settingA∗v =Pk∈Nδkek, we obtain from
(1.4) that
(1.10) δ¯kαk= hek, A∗viαk=β¯kτkαk, k∈ N,
soδk= βkτ¯kfor k∈ N, which establishes (ii). To prove (1.7), one simply fixes p, q∈ N, begins with the equation
D A λqeq+ heq, viu , ep E = hAT eq, epi = hAeq, T∗epi (1.11) =Aeq,λ¯pep+ hep, uiv ,
and makes a short calculation. Finally, (1.8) follows from the equation (1.12) τpβ¯p= hep, A∗vi = hAep, vi = D X k∈N hAep, ekiek, X j∈N βjej E
after a short calculation. ❐
Proposition 1.6delineates the form of any element of{T }0 for T ∈ (RO),
but does not produce any specific operators in {T }0. This next result, which is
essentially a corollary of [4, Prop. 5.4], however, does.
Proposition 1.7 ([4]). SupposeT = DΛ+ u ⊗ v ∈ (RO), and letψ:σ (T )=
Λ0→ Csatisfy a Lipschitz condition of the form
(1.13) ψ(z1)− ψ(z2)
z1− z2
≤ Mψ, z1, z2∈ Λ
0
for some constantMψ (depending onψ). Define τn = ψ(λn) forn ∈ N. Then
there exists an operatorAψ∈ {T }0satisfying (1.6) whose matrix is given by (1.7) and
(1.8).
ForTin(RO)andA∈ {T }0we denote by{τ
k(A)}k∈Nthe sequence{τk}k∈N
satisfying (1.6). We are now prepared to establish the main result of this paper. Theorem 1.8. SupposeT = DΛ+u⊗v ∈ (RO), where the notation is as
estab-lished in (1.1) and (1.2). Then(the unital, WOT-closed algebra){T }0is abelian.
Proof. LetAandBbe arbitrary elements of{T }0. By virtue of Theorem1.4,
it suffices to show thatABu= BAu, and this is equivalent to (1.14) hABu, eji = hBAu, eji, j∈ N.
With the notation as in Proposition1.6, we have
(1.15) A(Bu)= A X k∈N τk(B)αkek = X k∈N τk(B)αkAek.
We next fixjarbitrary inN, and develop an expression forhABu, ejiusing (1.14), (1.15), (1.7), and (1.8): hABu, eji = X k∈N αkτk(B)hAek, eji (1.16) = X k∈{k:k6=j} αkτk(B)αjβ¯k τj(A)− τk(A) λj− λk ! + αjτj(B) τj(A)− X `∈{`:`6=j} α`β¯` τj(A)− τ`(A) λj− λ` ! = αjτj(A)τj(B)+ X k∈{k:k6=j} αjαkβ¯k λj− λk (τk(B)− τj(B))(τj(A)− τk(A)).
But the right hand side of (1.16) remains the same ifAandBare interchanged, and thus we conclude immediately that hABu, eji = hBAu, eji, j ∈ N, and
therefore thatAB= BA. ❐
This just established fact that{T }0 is abelian whenT ∈ (RO)seems quite
curious, especially in view of this next two examples, which show that ifDΛ is
replaced by nondiagonalizable normal operatorNof (uniform) multiplicity one, then{N + u ⊗ v}0is sometimes abelian and sometimes not.
Example 1.9. Let{fn}n∈Zbe an ordered orthonormal basis forH, and let U∈ L(H )be the (unweighted) bilateral shift operator characterized by the equa-tionsU fn = fn+1, n∈ Z. Then, of course, U is a (nondiagonalizable) unitary
operator of multiplicity one. Moreover, if we set u = −f1, v = f0, then, as
is also well known,U+ u ⊗ v = U − f1⊗ f0 is unitarily equivalent to a direct
sumT = S∗⊕ S ∈ L(H ⊕ H ), where, with a minor change of notation,S is an (unweighted, forward) unilateral shift inL(H ), characterized by the equations
Sgn = gn+1, n ∈ N, where {gn}n∈N is a certain ordered orthonormal basis
for H. Thus, with no loss of generality, we may identify{U + u ⊗ v}0 with {T }0, which is a well-known algebra consisting of all 2×2 (operator) matrices in L(H ⊕ H )of the form (1.17) A= Af ,g= f (S∗) Y 0 g(S) ,
where f, g are arbitrary functions inH∞(D) (the algebra of bounded analytic functions onD), andY ∈ L(H )satisfiesY S = S∗Y, (i.e., Y is a Hankel opera-tor). But calculation shows that
1H Y 0 S 1H 0 0 S = 1H 0 0 S 1H Y 0 S
if and only if(S∗−1H)Y∗=0, and since 16∈ σp(S∗), we see that{U + u ⊗ v}0
is certainly not abelian.
Next we take a closer look at this example, identifyingH ⊕ H withL2(T),
whereTis endowed with normalized arc-length measure,UwithMζ (multiplica-tion by the func(multiplica-tionζ(ζ)≡ ζ (ζ∈ T), onL2(T)), and the vectorsuandvwith the functionsu≡ −ζandv(ζ)≡ 1(ζ) =1 a.e. onT, respectively. When this is done(0)⊕ H is identified withζH2(T), whileH ⊕ (0)becomes
(1.18) nψ(· ):ψ∈ H2(T)o=nψ¯ =ψ¯
0+ψ¯1ζ¯+ψ¯2ζ¯ 2
+ · · · ∈ L2(T)o.
Clearly 0⊕ S becomesU|ζH2(T) andS∗⊕0 becomes S0 = P U|¯ P L¯ 2(T), where ¯P
is the orthogonal projection on the space defined in (1.18). Moreover, in this representation, the Hankel operator in (1.17) has the following form:
(1.19) (Y ϕ)(ζ)= (P (¯ ζh)(¯ ·)ϕ(·))(ζ) (a.e.), ϕ∈ ζH2(T),
where h(ζ) = Pn∈Zh−nζn ∈ L∞(T) and ¯P h is uniquely determined (and
uniquely determines) Y. Let A be arbitrary in {U + (−ζ) ⊗ 1}0 as in (1.17)
and define, in analogy with the definition of theτkin (1.6), the functionsτand τ∗onTby τ(ζ)= (Au)(ζ) u(ζ) = (Aζ)(ζ)ζ,¯ τ∗(ζ)= (A∗v)(ζ) v(ζ) = (A ∗1)(ζ) (a.e.).
From (1.17) recalling that we have identifiedH ⊕ H withL2(T), we obtain (Aζ)(ζ)= (Y ζ)(ζ) + (g(S)ζ)(ζ) (a.e.), where (g(S)ϕ)(ζ)= g(ζ)ϕ(ζ) (a.e.), ϕ∈ H2(T). Therefore (Y ζ)(ζ)= (P (h(¯ ·)1)(ζ) = h0+ h−1ζ¯+ h−2ζ¯2+ · · · (a.e.), (Aζ)(ζ)= g(ζ)ζ + h0+ h−1ζ¯+ h−2ζ¯2+ · · · (a.e.), and (1.20) τ(ζ)= g(ζ) + h0ζ¯+ h−1ζ¯2+ · · · (a.e.), so (1.21) τ= g + (P h)¯ ζ.¯
Moreover, from (1.17) again, we get
A∗1= (f (S0))∗1+ Y∗1=f (S˜ 0∗)1+ Y∗1 where forf =P∞n=0fnζn∈ H∞(T), we write
˜ f (ζ):= f (ζ)¯ = ∞ X n=0 ¯ fnζn (a.e.). Thus, ˜ f (S0∗)1=s- lim ρ%1 ∞ X n=0 ¯ fnρn(S0∗)n1=s- lim ρ%1 ∞ X n=0 ¯ fn(ρζ)¯ n=f .¯ Moreover, hY∗1, ϕi = h1, Y ϕi = h1,ζhϕ¯ i = ∞ X n=0 ¯ h−nϕ¯n+1 = ∞ X n=1 ¯ h−n+1ϕ¯n= DX∞ n=1 ¯ h−n+1ζn, ∞ X n=1 ϕnζn E , ϕ∈ H2(T). Therefore Y∗1(ζ)= ∞ X n=1 ¯ h−n+1ζn= ζ ∞ X n=1 h−n+1ζ¯n−1= ζ (P h)(ζ) (¯ a.e.). Thus (1.22) τ∗(ζ)=f (ζ)¯ + ζ (P h)(ζ)¯ (a.e.). and (1.23) τ∗=f¯+ζ(¯ P h).¯
Comparing (1.21) with (1.23), we obtain thatτ∗=τ¯ (a.e.) if and only iff = g
(a.e.). But in this case we have
AB= fA(S∗) YA 0 fA(S) fB(S∗) YB 0 fB(S) =
= (fAfB)(S∗) fA(S∗)YB+ YAfB(S) 0 (fAfB)(S) (fBfA)(S∗) YBfA(S)+ fB(S∗)YA 0 (fBfA)(S) = BA.
Thus, we have proved the following.
Proposition 1.10. LetA, B∈ {Mζ− ζ ⊗ 1}0with the notation as above and suppose that(τA)∗=τ¯Aand(τB)∗=τ¯B. ThenAB= BAand ¯τAB= (τBA)∗.
Corollary 1.11. The subalgebraA = {A ∈ {Mζ− ζ ⊗ 1}0: ¯τA = (τA)∗}is
a maximal abelian subalgebra of{Mζ− ζ ⊗ 1}0.
Proof. ThatAis abelian was just proved. To see thatAis maximal abelian in
{Mζ− ζ ⊗ 1}0, suppose B= fB(S∗) YB 0 gB(S) ∈ {Mζ− ζ ⊗ 1}0
and B commutes with A. Then, for every Hankel operator Y and every f ∈ H∞(D), f (S∗)fB(S∗) f (S∗)YB+ Y gB(S) 0 f (S)gB(S) = fB(S∗)f (S∗) fB(S∗)Y+ YBf (S) 0 gB(S)f (S) , which gives immediately
f (S∗)YB+ Y gB(S)= fB(S∗)Y+ YBf (S),
and sinceYBf (S)= f (S∗)YB, we getY fB(S)= fB(S∗)Y= Y gB(S), and so
(1.24) Y (gB(S)− fB(S))=0.
In (1.19) we now chooseh(ζ)=ζ¯n (a.e.). Then (1.24) becomes, withϕ(ζ)= gB− fB=
P∞
k=1ϕkζk,
0= ϕ1ζ¯n+ ϕ2ζ¯n−1+ · · · + ϕn+1 (a.e.), n∈ N.
Soϕ=0,gB= fB, andB∈ A. ❐
Example 1.12. With the notation as in Example 1.9, let us consider the commutant ofU−1
2f0⊗f1, which is an invertible bilateral weighted shift operator
(with all weights but one equal to 1 and one weight equal to 1/2). But, as is well-known (cf., e.g., [8]),U−12f0⊗ f1is similar toUand thus must have an abelian
Theorem1.8, together with Examples1.9and1.12, naturally give rise to the following question, which the authors have been unable to resolve.
Problem 1.13. For which normal operators of multiplicity oneNand which rank-one perturbationsu⊗ v is{N + (u ⊗ v)}0abelian, and why?
One obvious way to attempt to enlarge the class ofT ∈ (RO)having a n.h.s. would be to show that some operatorsT ∈ (RO)are similar or quasisimilar to operatorsTe ∈ (RO)1. In this connection, see Theorem2.5and Corollaries2.6
and2.7.
2. SIMILARITY ANDQUASISIMILARITY
The reader will recall that, by definition, a quasiaffinity inL(H )is an operatorX
satisfying kerX=kerX∗= (0), and operatorsAandBinL(H )are quasisimilar if there exist quasiaffinitiesXandY inL(H )such thatXA= BX, AY = Y B. A natural question that begs an answer is whether the operatorsT in(RO)that do satisfy the hypotheses of Theorem1.1have a n.h.s. for some elementary reason, like having σp(T ) 6= or being quasisimilar to a normal operator. (We recall
here the example given by Stampfli [9] of a T ∈ (RO) with σp(T ) = , but
that particularT does not satisfyPn∈N(|αn|2/3+ |βn|2/3) <∞. This section is
devoted to establishing some results that shed light on these problems. We begin with an easy lemma, which is certainly known.
Lemma 2.1. SupposeA, B∈ L(H )and are quasisimilar. Then:
(i) {A}0is abelian if and only if{B}0is abelian, and
(ii) A has a n.h.s. if and only if B does.
Corollary 2.2. IfT = DΛ+ u ⊗ v ∈ (RO), andT is quasisimilar to a normal
operatorN, thenNmust have(via Theorem1.8and Lemma2.1)uniform multiplic-ity one.
These results raise the question whether every operator in(RO)1is
quasisim-ilar to a normal operator and thus has a n.h.s. via Lemma2.1. The following theorem shows that this is not the case.
Theorem 2.3. There exist operatorsT ∈ (RO)1such thatT has no eigenvalues outsideσe(T )andT is not quasisimilar to(in fact, not even a quasiaffine transform
of)any normal operator.
Proof. Let{rn}n∈N\{10p:p∈N}be an enumeration of the distinct rational
num-bers in(0,1)\{1/102p:p∈ N}, and let{λ
n}n∈N,{αn}n∈N,{βn}n∈Nbe defined as follows: λn= 1 n2 ifn∈ {10 p :p∈ N}, rn ifn∈ N\{10p :p∈ N},
and αn= λn cn2, βn= − 1 n2, n∈ N,
where c = Pn∈N1/n4. Note that Pn∈N(|αn|2/3+ |βn|2/3) < +∞. We use
[4, Prop. 2.4] to show that 0 ∈ σp(T ), with T = DΛ+ u ⊗ v, where u =
P n∈Nαnen,v= P n∈Nβnen. Hence, we compute X n∈N |αn|2 |0− λn|2 = 1 c2 X n∈N 1 n4 <+∞, X n∈N αnβ¯n 0− λn = 1 c X n∈N 1 n4 =1, X n∈N |βn|2 |0− λn|2 ≥ X n∈{10p:p∈N} 1/n4 1/n4 = +∞.
Thus it follows from the aforementioned result that 0∈ σp(T )and 06∈ σp(T∗).
Moreover, it is clear that σe(T ) = Λ0 = [0,1], so to obtain that T ∈ (RO)1
it suffices to show that σp(T ) ⊂ [0,1], and this is done by showing that for µ∈ C\[0,1], X n∈N αnβ¯n µ− λn 6=1 .
Note first that ifµ6∈ R, i.e.,µ= ρ + iσ withσ 6=0, then X n∈N αnβ¯n µ− λn = X n∈N −λn/(cn4) (ρ− λn)+ iσ ! (ρ− λn)− iσ (ρ− λn)− iσ ! = X n∈N (ρ− λn)(−λn/(cn4)) (ρ− λn)2+ σ2 + i X n∈N (λn/(cn4))σ (ρ− λn)2+ σ2 6∈ R,
soρ+ iσ 6∈ σp(T )andσ (T )⊂ R. Next, supposeµ <0. Then
X n∈N αnβ¯n − |µ| − λn = 1 c X n∈N λn (|µ| + λn)n4 < 1 c X n∈N 1 n4 =1,
soµ6∈ σp(T ). On the other hand, supposeµ >1. Then
X n∈N (λn/(cn2))(−1/n2) µ− λn = − 1 c X n∈N λn n4(µ− λ n) <0,
so, once again,µ6∈ σp(T ), and we conclude thatσp(T )⊂ [0,1]= Λ0. ThusT
andT∗ belong to(RO)1. Finally, suppose thatT = DΛ+ u ⊗ vis a quasiaffine
transform of a normal operator N, which means that there exists a quasiaffinity
Xsuch thatXT = NX. Lety6= 0 be an eigenvector forT corresponding to the eigenvalue 0. Thus 0 = XT y = NXy, so 0∈ σp(N)and (sinceN is normal) N∗(Xy) = 0, i.e., 0 ∈ σp(N∗). Thus, T∗(X∗Xy) = X∗N∗Xy = 0, and
since X∗Xy 6= 0, 0 ∈ σp(T∗), which is false, as we saw above. Thus T is a
quasiaffine transform of no normal operator, and thereforeT∗ is not quasisimilar
to any normal operator. ❐
However, we still do not know the answer to the following.
Problem 2.4. Are there operatorsT in(RO)1that satisfyσp(T )∪σp(T∗)= ? (This seems to be a difficult question in the theory of Borel series.)
The following results show that the relation of quasisimilarity may sometimes be used to obtain a different version of Theorem1.1.
Theorem 2.5. SupposeT = DΛ+ u ⊗ v ∈ (RO), 0 ∈ Λ0\(Λ ∪ σp(T )∪ σp(T∗)), and DΛ1/2 is any fixed square root of the (normal) operatorDΛ. Suppose
also thatubelongs to the range ofDΛ1/2. ThenT andTe= DΛ+(DΛ−1/2u)⊗(DΛ1/2)∗v
are quasisimilar.
Proof. The hypotheses guarantee thatDΛ1/2andT = DΛ+u⊗vare quasia
ffini-ties, and thus thatDΛ−1/2is a densely defined “unbounded quasiaffinity.” Moreover the productD−Λ1/2T = D1/2
Λ + (DΛ−1/2u)⊗ v ∈ L(H )and is a quasiaffinity, and
the result now follows from the equalities
T = D1/2
Λ (DΛ1/2+ (DΛ−1/2u)⊗ v), Te= (DΛ1/2+ (DΛ−1/2u)⊗ v)DΛ1/2. ❐
As a corollary we see that some operators in(RO)1are quasisimilar to normal
operators, and Lemma2.1can be used to obtain the existence of a n.h.s. for such operators. But Theorem1.2gives more information about these operators, which seems not to be available via quasisimilarity.
Corollary 2.6. WithT as in Theorem 2.5, if Λ ⊂ [0,+∞) andDΛv = u, thenT is quasisimilar to a Hermitian (positive semi-definite) operator, and thus has a n.h.s.
Proof. TakeDΛ1/2 to be the positive semi-definite square root of DΛ and let
w = D−Λ1/2u= D1Λ/2v, soDΛ1/2w = u = DΛv andTe = DΛ+ w ⊗ w, which is
obviously positive semi-definite. ❐
Corollary 2.7. WithT as in Theorem2.5, ifPn∈N(|αn| |λn−1/2|)2/3<∞and
P
Proof. T is quasisimilar toTeandTehas a n.h.s. by Theorem1.1. ❐
Remark 2.8. The thrust of Corollary2.7is that whenT andDΛare
(nonin-vertible) quasiaffinities, the requirement in Theorem1.1thatu (=Pn∈Nαnen)
andv (=Pn∈Nβnen)satisfy
P
n∈N(|αn|2/3+ |βn|2/3) <∞can be replaced by
a stronger hypothesis onuand a weaker hypothesis onv.
In a paper to be published separately [3], we show that the operators in(RO)1
do mimic the behavior of normal operators in that they are all decomposable in the sense of [1]. This raises the question whether forT ∈ (RO)1the resolvent of T satisfies
(2.1) (T− λI)−1 ≤ MT
dist(λ, σ (T )), λ6∈ σ (T ),
for some constantMT, which is certainly true for normal operators T .Our last
example shows that some operators in(RO)1 have resolvents that do not behave
like those of normal operators.
Example 2.9. We define an operatorT = DΛ+ u ⊗ v that will turn out to
be in(RO)1 as follows. Let Λ = {λn}n∈N, where the sequence {λ2n}n∈N has
distinct entries and is dense in the setD\{(r ,0): 0≤ r <1}. Furthermore, the sequence{λ2n+1}n∈Nis given by
(2.2) λ2n+1=1− e−(2n+1), n∈ N,
(so λ2n+1 % 1 rapidly). As usual, u =
P
αnen and v =
P
βnen, where the
sequences{αn}n∈Nand{βn}n∈Nare (never zero and) defined by
(2.3) α2n= β2n= (1− |λ2n|)τ2n, n∈ N, α4n+1= 1 n2, α4n+3= (1− λ4n+3)τ4n+3, n∈ N, β4n+1= (1− λ4n+1)τ4n+1, β4n+3= 1 n2, n∈ N,
where{τn}n∈N is a sequence of positive real numbers satisfyingPn∈Nτn1/3= 1.
Note thatPn∈N(α2n/3+ βn2/3) <∞. A trivial calculation shows that, for|λ| >1,
the equation(T− λ)x = y may be written, equivalently, as
(2.4) (T− λ)−1y= x = (D
Λ− λ)−1y−hy,v¯λi
ϕλ uλ,
providedϕλ6=0, where (as in [2]) (2.5) uλ= (DΛ− λ)−1u= X n∈N αn λn− λ en, ¯ vλ= (D∗Λ−λ)¯ −1v= X n∈N βn ¯ λn−λ¯ ! en, ϕλ=1+ huλ, vi =1+ X n∈N αnβ¯n λn− λ .
Since obviouslyσe(T )= Λ0= D−, to see thatT ∈ (RO)1it suffices to show that T has no eigenvalues in the region|λ| >1. One knows from [4, Prop. 2.4] that if
|λ| >1, thenλ∈ σp(T )if and only ifϕλ=0. To see that this does not happen,
we calculate, using (2.3) and (2.5), for an arbitraryλ0satisfying|λ0| >1,
|ϕλ0−1| = X n∈N αnβ¯n λn− λ0 ≤ X n∈N |αn| |β¯n| |λ0| − |λn| (2.6) ≤ X n∈N τn(1− |λn|) |λ0| − |λn| ≤ X n∈N τn<1,
soϕλ0 6=0,σ (T )= σe(T )= D−, andT∈ (RO)1as desired.
Now suppose that there exists a constantMT >0 such that the resolvent ofT
satisfies (2.1), and putting together (2.1) and (2.4) for (λreal and)λ > 1 gives, after taking norms,
(2.7) |ϕλ|−1 hy,v¯λiuλ ≤ MT+1
λ−1 kyk, λ >1, which becomes, after settingy=v¯λand using (2.5),
(MT+1)|ϕλ| λ−1 ≥ kv¯λk kuλk (2.8) ≥ X n∈N 1 n4 1 |λ4n+3− λ|2 1/2 X n∈N 1 n4 1 |λ4n+1− λ|2 1/2 ≥ X n∈N 1 n4 1 (λ4n+1− λ)2 , λ >1,
(sinceλ4n+1< λ4n+3). Finally, note from (2.6) that|ϕλ| <2 for|λ| >1. Hence,
(2.9) X n∈N 1 n4 1 (λ4n+1− λ)2 ≤ 2(MT+1) λ−1 , λ >1.
So, multiplying each side of (2.9) by(λ−1)2, we obtain
X n∈N 1 n4 (λ−1)2 (λ4n+1− λ)2 ≤2 (MT+1)(λ−1), λ >1.
Thus, for everyn∈ N,
(2.10) 1 n4 1 1+1− λ4n+1 λ−1 2 ≤2(MT+1)(λ−1), λ >1.
Now putλ=1+ e−(4n+1)into (2.10), and calculate, yielding
1
4n4 ≤2(MT+1)e−(
4n+1), n∈ N,
which is an obvious contradiction, and therefore shows that the resolvent(T−λ)−1
ofT does not satisfy (2.1) for any constantMT.
Acknowledgment. This research was supported by the Korea Research Foun-dation Grant funded by the Korean Government (MOEHRD, Basic Research Promotion Fund) (KRF–2007–314–C00011).
REFERENCES
[1] I. COLOJOARA˘andC. FOIAS, Theory of Generalized Spectral Operators, Gordon and Breach, New York, NY, 1968.
[2] C. FOIAS,I. JUNG, E. KO, andC. PEARCY, Rank-one perturbations of normal operators, J. Funct. Anal. 253 (2007), 230–248.
[3] , Rank-one perturbations of normal operators, III (in preparation).
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39 (2001), 421–440.
[5] V. LOMONOSOV, On invariant subspaces of families of operators commuting with a completely
continuous operato, Funkcional Anal. i Prilozen 7 (1973), 55-56. (Russian)
[6] C. PEARCY, Some recent developments in operator theory, C.B.M.S. Regional Conference Series in Mathematics, Number 36, Amer. Math. Soc., Providence, Rhode Island, 1978, MR 0487495 (58 #7120).
[7] H. RADJAVIandP. ROSENTHAL, Invariant Subspaces, Springer-Verlag, New York, NY, 1973. [8] A. SHIELDS, Weighted Shift Operators and Analytic Function Theory, Math. Surveys, Number 13,
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CIPRIANFOIAS:
Department of Mathematics Texas A & M University
College Station, TX 77843, U.S.A. E-MAIL:foias@math.tamu.edu
ILBONGJUNG:
Department of Mathematics Kyungpook National University Daegu, 702-701, Korea E-MAIL:ibjung@knu.ac.kr
EUNGILKO:
Department of Mathematics Ewha Women’s University Seoul, 120-750, Korea E-MAIL:eiko@ewha.ac.kr
CARLPEARCY:
Department of Mathematics Texas A&M University
College Station, TX 77843, U.S.A. E-MAIL:pearcy@math.tamu.edu
KEY WORDS AND PHRASES: invariant subspace, hyperinvariant subspace, normal operator, rank-one perturbation, similarity, quasisimilarity.
2000MATHEMATICSSUBJECTCLASSIFICATION: 47A15; 47A55; 47B15.
Received : July 30th, 2008.
