# Almostα-Hyponormal Operators with Weyl Spectrum of Area Zero

Authors: Vasile Lauric
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We define the class of almostα-hyponormal operators and prove that
for an operator T
in this class, (T∗T)α−(TT∗)α is trace-class and its trace is zero when
α∈(0,1]
and the area of the Weyl spectrum is zero.


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## Body

This note is dedicated to Professor Carl M. Pearcy with the occasion of his 75th birthday.Letℋ be a complex, separable, infinite-dimensional Hilbert space, and let L(ℋ) denote the algebra of all linear bounded operators on ℋ, and for 1≤p<∞, let 𝒞p(ℋ) denote the p-Schatten class on ℋ. For K∈𝒞p(ℋ), the expression ||K||p:=(∑n=1∞μn(K)p)1/p, where μ1(K)≥μ2(K)≥⋯ are the singular values of K, is a norm for p≥1, and is only a quasinorm for 0<p<1 (it does not satisfy the triangle inequality). Nevertheless, the latter case will be used in what follows.ForT∈L(ℋ),σ(T) and σw(T) will denote the spectrum and the Weyl spectrum, respectively. Recall that Weyl spectrum is the union of the essential spectrum, σe(T), and all bounded components of  ℂ∖σe(T) associated with nonzero Fredholm index. An operator T∈L(ℋ) is called  (𝒞p,α)-normal  (notation: T∈Npα(ℋ))  if   CTα:=(T*T)α-(TT*)α belongs to 𝒞p(ℋ),  and T is called  (𝒞p,α)-hyponormal  (notation: T∈Hpα(ℋ))   if  CTα is the sum of a positive definite operator and an operator in 𝒞p(ℋ), or equivalently, (CTα)- (the negative part of CTα) belongs to 𝒞p(ℋ), where α is a positive number. This note will be concerned with the particular class H1α(ℋ), which by some parallelism with some terminology used in [1], would be appropriate to be referred as almostα-hyponormal operators.Voiculescu's [1] generalization of Berger-Shaw inequality gives an estimate for the trace of CT1. The result was extended in [2]. The combination of these results will be stated after recalling some terminology and notation. The rational cyclic multiplicity of an operator T in L(ℋ), denoted by m(T), is the smallest cardinal number m with the property that there are m vectors x1,…,xm in ℋ such that (1)∨{f(T)xj∣1≤j≤m,f∈Rat(σ(T))}=H,
where Rat(σ(T)) is the algebra of complex-valued rational functions with poles off σ(T).For a Borel subsetE⊆ℂ and α>0, denote μα(E)=(α/2)∬Eρα-1dρdθ. In particular, μ2 is the planar Lebesgue measure.Theorem A (see [1, 2]).
SupposeT∈H11(ℋ). If  there exists K∈𝒞2(ℋ) such that either m(T+K)<∞ or  μ2(σ(T+K))=0, then  T∈N11(ℋ). Moreover, when  m(T+K)<∞,(2)tr⁡(CT1)≤m(T+K)π⋅μ2(σ(T+K)),
and when  μ2(σ(T+K))=0,tr(CT1)≤0,   and consequently, tr(CT1)=0.In fact, it was observed in [2] that the inequality can be improved by replacing m(T+K) with τ(T+K), where (3)τ(S):=liminf⁡[rank(I-P)SP],
and the liminf is taken over all sequences of finite-rank orthogonal projections such that P→I in the strong operator topology.Corollary B (see [2]).
LetT∈H11(ℋ) such that μ2(σw(T))=0. Then  T∈N11(ℋ) and tr(CT1)=0.On the other hand, Berger-Shaw inequality was extended to operators inH1α(ℋ) using similar circle of ideas used in [1]. This was done in [3] for the case α∈[(1/2),1] and later on in [4] for the case α∈(0,(1/2)].Theorem C (see [3, 4]).
Let0<α≤1, and let   T∈H1α(ℋ)  and  K∈𝒞2α(ℋ) with  m(T+K)<∞. Then T∈N1α(ℋ) and
(4)tr⁡(CTα)≤m(T+K)π⋅μ2α(σ(T+K)).The case in whichm(T+K)=∞ and μ2α(σ(T+K))=0 was not discussed in [4] or [3]. It is the goal of this note to make some progress towards this case. We have the following.Theorem 1.
Letα∈(0,1) and let  T∈H1α(ℋ)   and K∈𝒞α(ℋ) with  μ2α(σ(T+K))=0. Then  T∈N1α(ℋ) and  tr(CTα)=0.Remark.
It would have been desirable that Theorem1 be proved with the hypothesis that K∈𝒞2α(ℋ).Before we prove Theorem1, we extract a similar consequence to Corollary B.Corollary 2.
Letα∈(0,1] and let  T∈H1α(ℋ)   such that  μ2(σw(T))=0. Then  T∈N1α(ℋ) and  tr(CTα)=0.Proof.
Ifα=1, then conclusion holds according to Corollary B. Let α∈(0,1). First, a careful inspection of the proof of a result of Stampfli [5] leads to the following. For T∈L(ℋ) and α>0, there exists Kα∈𝒞α(ℋ) such that σ(T+Kα)∖σw(T) consists of a countable set which clusters only on σw(T). Therefore μ2(σ(T+Kα))=0 and thus Theorem 1 applies.The proof of Theorem1 makes use of the following three inequalities.Proposition D (Hansen's inequality [6]).
IfA,B∈L(ℋ), A≥0, ||B||≤1, and α∈(0,1],  then B*AαB≤(B*AB)α.Proposition E (Lowner's inequality [7]).
IfA,B∈L(ℋ), A≥B≥0, and  α∈(0,1],  then Aα≥Bα.The following is a consequence of Theorem 3.4 of [8].Proposition F (Jocic's inequality [8]).
LetA,B∈L(ℋ), A,B≥0, α∈(0,1],  and  1≤p<∞. If  A-B∈𝒞αp(ℋ), then Aα-Bα∈𝒞p(ℋ) and  ||Bα-Aα||p≤|||B-A|α||p.Proof of Theorem1.
Letα∈(0,1),  T∈H1α(ℋ), and K∈𝒞α(ℋ) with μ2α(σ(T+K))=0, and assume m(T+K)=∞, otherwise Theorem C implies T∈N1α(ℋ).
Let{en}n∈ℕ be an orthonormal basis of ℋ and let
(5)Hn=∨{r(T+K)ej∣j=1,…,n,r∈Rat(σ(T+K))}.
Assume that with respect to the decomposition ℋ=ℋn⊕ℋn⊥,  operators T and K are written as
(6)T=(T1nT2nT3nT4n),K=(K1nK2nK3nK4n).
Since ℋn is a rationally invariant subspace for T+K, we have T3n+K3n=0, and thus T3n=-K3n∈𝒞α(ℋn)⊆𝒞2α(ℋn), and σ(T1n+K1n)⊆σ(T+K), which implies μ2α(σ(T1n+K1n))=0.
LetPn be the orthogonal projection onto ℋn, and thus Pn↑I strongly. We will prove next that T1n∈H1α(ℋn) by first establishing that (7a)PnCTαPn-CT1nα=-Qn′+Kn′,
where Qn′∈L(ℋn) is positive semidefinite  and   Kn′∈𝒞1(ℋn).
Assuming that equality (7a) was already proved and writing CTα=Q+K with Q≥0 and K∈𝒞1(ℋ), then we have
(7b)CT1nα=PnQPn+PnKPn+Qn′-Kn′,that is, CT1nα is the sum of PnQPn+Qn′, which is a positive semidefinite operator, and of PnKPn-Kn′, which is a trace-class operator.
Indeed, the expressionPnCTαPn-CT1nα can be written as D1-D2, where
(8)D1=Pn(T*T)αPn-(T1n*T1n)α,D2=Pn(TT*)αPn-(T1nT1n*)α.
We can write D1=-Qn′′+Kn′′, where
(9)Qn′′=[(PnT*TPn)α-Pn(T*T)αPn],
which according to Hansen's inequality is a positive semidefinite operator, and
(10)Kn′′=[(PnT*TPn)α-(PnT*PnTPn)α],
which according to Jocic's inequality is a trace-class operator that satisfies
(11)‖Kn′′‖1≤‖|((PnT*TPn-PnT*PnTPn))|α‖1=‖(T3n*T3n)α‖1=‖T3n*T3n‖αα≤‖T3n*‖α⋅‖T3n‖αα≤‖T‖α⋅‖T3n‖αα.
Concerning operator D2, we can write D2=Qn′′′+Kn′′′, where
(12)Qn′′′=Pn(TT*)αPn-Pn(TPnT*)αPn,
which according to Lowner's inequality is a positive semidefinite operator, and
(13)Kn′′′=Pn(TPnT*)αPn-(PnTPnT*Pn)α=Pn[(TPnT*)α-(PnTPnT*Pn)α]Pn,
which is also a trace-class operator since
(14)TPnT*-PnTPnT*Pn=(TPnT*-TPnT*Pn)+(TPnT*Pn-PnTPnT*Pn)=TPnT*(I-Pn)+(I-Pn)TPnT*Pn=TT3n*+T3nT*Pn∈Cα(H),
and according to Jocic's inequality
(15)‖Kn′′′‖1≤‖(TPnT*)α-(PnTPnT*Pn)α‖1≤‖|TT3n*+T3nT*Pn|α‖1=‖TT3n*+T3nT*Pn‖αα≤C(‖TT3n*‖αα+‖T3nT*Pn‖αα)≤C‖T‖α(‖T3n*‖αα+‖T3n‖αα)=2C‖T‖α‖T3n‖αα.
Therefore,
(16)D2=Qn′′′+Kn′′′,withQn′′′≥0,Kn′′′∈C1(H),
and consequently, D1-D2=(-Qn′′+Kn′′)-(Qn′′′+Kn′′′)=-(Qn′′+Qn′′′)+(Kn′′-Kn′′′), where Qn′′+Qn′′′=:Qn′ is positive semidefinite and Kn′′-Kn′′′=:Kn′ is trace-class, which establishes equality (7a).
According to (7b), T1n∈H1α(ℋn), and since m(T1n+K1n)≤n and σ(T1n+K1n)⊆σ(T+K), Theorem C implies that tr(CT1nα)≤0,   and furthermore, by replacing T1n with T1n*,    tr(CT1nα)=0. Furthermore, equality (7a) implies
(17)PnCTαPn≤CT1nα+Kn′,
which further implies
(18)tr⁡(PnCTαPn)≤tr⁡(Kn′).
Similar utilization of Lowner's and Hansen's inequalities implies that Kn′′ and -Kn′′′ are positive semidefinite, and thus so is Kn′=Kn′′-Kn′′′. Therefore
(19)tr⁡(Kn′)≤‖(Kn′′)‖1+‖(Kn′′′)‖1≤(1+2C)‖T‖α‖T3n‖αα.
Since T3n=-K3n∈𝒞p(ℋn) and K3n→0 weakly and both |T3n| and |T3n*|≤||T||I,   we have ||T3n||α→0, and thus tr(CTα)≤0. Replacing T with T* we conclude that tr(CTα)=0.



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*Source: 801313-2011-05-22.xml*