A-Weyl’s定理和（WE）性质的摄动.pdf

本文研究使有界线性算子的 aWeyl 定理和(WE)性质都成立的充分必要条件.同时，研究 aWeyl定理和(WE)性质在拟幂零或紧摄动下的稳定性.作为应用，研究特殊算子的相关稳定性.关键词AWeyl 定理(WE)性质紧摄动拟幂零摄动doi:10.3969/j.issn.10068074.2023.03.0041IntroductionIn 1909,Weyl 1 examined the compact perturbations of some selfadjoint operators and found thatthe intersection of their spectrums consisted precisely of those points of the spectrums which were notisolated eigenvalues of finite multiplicity.Later,this observation was abstracted into“Weyls theorem”.In recent decades,combining with the relationships between different spectrum subsets,mathematiciansput forward a series of variants of Weyls theorem.In 2 and 3,Harte and Lee and Rakoevicintroducedthe definition of aWeyls theorem Berkani and Kachad 4 gave another variantproperty(WE).In thispaper,we investigate the necessary and sufficient conditions such that both aWeyls theorem and theThis work is supported by the Natural Science Foundation of Shannxi Province(No.2021JM519),2021 Talent Project of WeinanNormal University(No.2021RC16)and 2022 Horizontal Project of Weinan Normal University(No.2022HX126)Corresponding author:Che Yuhong(1982 ),Lecturer,MSc Email:hongmeigui_ 收稿日期:2023 年 6 月 19 日82数学理论与应用property(WE)hold for a bounded linear operator,and study the stability of aWeyls theorem and theproperty(WE)under quasinilpotent or compact perturbations.To begin with,we introduce some terminologies and notations.Throughout this paper,C(resp.N)denotes the set of all complex numbers(resp.nonnegative integers).H denotes an infinitedimensionalcomplex separable Hilbert space and B(H)(resp.K(H)denotes the set of all bounded linear(pact)operators on H.In addition,F(H)denotes the set of all finite rank operators in B(H).For T B(H),we use N(T),R(T)and(T)to denote the kernel,the range and the spectrum of T,respectively.Let(T)=C(T).If R(T)is closed and n(T)(resp.d(T),then T is said to be an upper(resp.lower)semiFredholm operator,where n(T)=dimN(T)and d(T)=codimR(T).Especially,ifT isanuppersemiFredholmoperatorwithn(T)=0,thenT iscalledaboundedbelowoperator.Now,theindex of T is defined by ind(T)=n(T)d(T).If ind(T)is finite,then T is called a Fredholm operator.T B(H)is called a Weyl operator if T is a Fredholm operator with ind(T)=0.The ascent and descentof T are closely related with the kernel and range of the power of T,which are respectively defined byasc(T)=infn N:N(Tn)=N(Tn+1)and des(T)=infn N:R(Tn)=R(Tn+1).If theinfimum does not exist,then we write asc(T)=(resp.des(T)=).It is known from Proposition38.3 in 5 that asc(T)=des(T)if they are finite simultaneously.T is called a Browder operator if T isa Fredholm operator with both finite ascent and finite descent.The Browder spectrum b(T),the Weylspectrum w(T),the approximate point spectrum a(T)are respectively defined by:b(T)=C:T I is not a Browder operator,w(T)=C:T I is not a Weyl operatoranda(T)=C:T I is not a bounded below operator.Let ea(T)=C:T I is an upper semiFredholmoperator with ind(T I)0 and ab(T)=C:T I is an upper semiFredholm opreator withasc(T I).We call ea(T)=Cea(T)and ab(T)=Cab(T)the essential approximatepoint spectrum and Browder essential approximate point spectrum respectively.Besides,for E C,weuse isoE and accE to denote the set of all isolated points and all accumulation points of E respectively.Let T B(H).T is said to satisfy aWeyls theorem,ifa(T)ea(T)=a00(T),where a00(T)=isoa(T):0 n(T I)(3).T is said to satisfy the property(WE)and denoted by T (WE),if(T)w(T)=E(T),where E(T)=iso(T):0 0 such that ea(T)and N(T I)n=1R(T I)n)if 0|0.Since 0/1(T),it followsthat ea(T)and N(T I)n=1R(T I)n)if 0|0|is small enough.Based on thefact that aWeyls theorem holds for T implies asc(T I),we have N(T I)=N(T I)n=1R(T I)n)=0(Lemma 3.4,9).So 0 isoa(T).We will take the following two cases intoconsideration.Case 10/acc(T).It follows that 0 E(T).Since T (WE),we know T 0I is a Browder opreator,which means0/ab(T).Case 20/C:n(T I)=.It follows that 0 a00(T).Since aWeyls theorem holds for T,we obtain 0/ab(T).Now we get ab(T)1(T)a(T):n(T I)=0 acc(T)C:n(T I)=.The converse inclusion is obvious.Then ab(T)=1(T)a(T):n(T I)=0 acc(T)C:n(T I)=.Remark 2.1(1)In Theorem 2.1,if T (WE)and aWeyls theorem holds for T,the three partsof the decomposition form of ab(T)are essential.We explain it with the following examples.(i)Let T B(2)be defined byT(x1,x2,)=(x2,x3,).ThenbothaWeylstheoremandtheproperty(WE)holdforT.Butab(T)=a(T):n(TI)=0 acc(T)C:n(T I)=.This means 1(T)cannot be removed.(ii)Let T B(2)be defined byT(x1,x2,)=(0,x1,x22,x33,).Then both aWeyls theorem and the property(WE)hold for T.However ab(T)=1(T)acc(T)C:n(T I)=.(iii)Let A,B B(2)be defined byA(x1,x2,)=(0,x1,x2,),B(x1,x2,)=(0,x2,0,x4,0,x6,).Put T=(A00B).Then both aWeyls theorem and the property(WE)hold for T.However,ab(T)=1(T)a(T):n(T I)=0.(2)If T (WE)and aWeyls theorem holds for T,then ab(T)1(T)a(T):n(T I)=0acc(T)and ab(T)=1(T)a(T):n(T I)=0 C:n(T I)=.But the converse is not true.For example,let A,B B(2)be defined byA(x1,x2,x3,)=(0,x1,x2,x3,),B(x1,x2,x3,)=(x2,x3,x4,).AWeyls 定理和(WE)性质的摄动85Put T=(A00B).Then ab(T)=1(T)a(T):n(T I)=0 acc(T)=C:|1.But since(T)=a(T)=C:|1,w(T)=ea(T)=C:|=1,andE(T)=a00(T)=,it follows neither aWeyls theorem nor the property(WE)holds for T.If ab(T)=1(T)a(T):n(T I)=0 C:n(T I)=,we claim thataWeyls theorem holds for T.In fact,since a(T)ea(T)1(T)a(T):n(T I)=0 C:n(T I)=and a00(T)1(T)a(T):n(T I)=0 C:n(T I)=,it follows that a(T)ea(T)ab(T)and a00(T)ab(T).This impliesa(T)ea(T)=a00(T),which means aWeyls theorem holds for T.Buttheproperty(WE)doesnotholdforT B(H)ifab(T)=1(T)a(T):n(TI)=0 C:n(T I)=.Consider the following example.Let T B(2)be defined byT(x1,x2,x3,)=(0,x2,0,x4,).Then ab(T)=1(T)a(T):n(T I)=0 C:n(T I)=0,1.Butsince(T)=w(T)=0,1 and E(T)=0,1,we see the property(WE)does not hold for T.If T (WE),we can prove acc(T)=1(T)acc C:n(T I)d(T I).ByTheorem 2.1,the following fact holds.Corollary 2.1Let T B(H).Then both aWeyls theorem and the property(WE)hold for Tif and only if ab(T)=1(T)a(T):n(T I)=0 acc C:n(T I)d(T I)C:n(T I)=.In Theorem 2.1 and Corollary 2.1,the set a(T):n(T I)=0 cannot be replaced by theset (T):n(T I)=0.Since (T):n(T I)=0 b(T),we have the following corollary.Corollary 2.2Let T B(H).Then both aWeyls theorem and the property(WE)hold for T ifand only if b(T)=1(T)(T):n(T I)=0 acc(T)C:n(T I)=isoa(T)SF+(T),where SF+(T)=C:T I is an upper semiFredhom operatorwith ind(T I)0.Proof Necessity.From Theorem 2.1,we know ab(T)=1(T)a(T):n(T I)=0 acc(T)C:n(T I)=.Since b(T)=ab(T)ab(T)b(T)and ab(T)b(T)=a(T)(T)a(T):T I is an upper semiFredholm operator with asc(T I)b(T)(T):n(T I)=0 isoa(T)SF+(T),we get b(T)1(T)(T):n(T I)=0acc(T)C:n(T I)=isoa(T)SF+(T).The converseinclusion is obvious.Sufficiency.Since ab(T)(T):n(T I)=0=a(T):n(T I)=0,ab(T)isoa(T)SF+(T)=and 1(T)acc(T)C:n(T I)=ab(T)86数学理论与应用and ab(T)b(T),it follows that ab(T)=ab(T)b(T)=1(T)a(T):n(T I)=0 acc(T)C:n(T I)=.Then T (WE)and aWeyls theorem holds for Taccording to Theorem 2.1.Let s(T)=C:T I is not a surjective operator.We have the following corollary.Corollary 2.3Let T B(H).Then the following statements are equivalent:(1)T (WE)and aWeyls theorem holds for T(2)ab(T)=1(T)a(T):n(T I)=0 acc C:des(T I)=C:n(T I)=;(3)ab(T)=1(T)a(T):n(T I)=0 accs(T)C:n(T I)=;(4)b(T)=1(T)(T):n(T I)=0 acc C:des(T I)=C:n(T I)=isoa(T)SF+(T);(5)b(T)=1(T)(T):n(T I)=0 accs(T)C:n(T I)=isoa(T)SF+(T).From Theorem 2.1,if ab(T)=1(T),then both aWeyls theorem and the property(WE)hold forT.But the converse is not true.For example,let T B(2)be defined byT(x1,x2,x3,)=(0,x1,x22,x33,).Then(T)=w(T)=a(T)=ea(T)=0,and E(T)=a00(T)=,which means both aWeylstheorem and the property(WE)hold for T.But since ab(T)=0,1(T)=,we can see ab(T)=1(T).We call T B(H)finiteaisoloid if isoa(T)C:0 n(T I),and callT B(H)aisoloid if isoa(T)C:0 n(T I).Clearly,finiteaisoloid implies aisoloid.If ab(T)=1(T),then T is finiteaisoloid.In fact,let 0 isoa(T),then 0/1(T).Byab(T)=1(T),we have 0/ab(T).Then 0 ab(T)a(T).This implies 0 C:0 n(T I)0 such that ea(T)andN(T I)n=1R(T I)n)if 0|0|.Similar to the proof of Theorem 2.1,we can get 0 isoa(T)a(T),where a(T)=Ca(T).Since T is finiteaisoloid,we know0 a00(T)a(T),and since aWeyls theorem holds for T,we have 0/ab(T).In Corollary 2.4,if we change the condition“T is finiteaisoloid”to“T is aisoloid”,the result willnot be true.For example,let A,B B(2)be defined byA(x1,x2,x3,)=(0,x1,x2,x3,),B(x1,x2,x3,)=(0,x1,0,x33,).AWeyls 定理和(WE)性质的摄动87Put T=(A00B).Then(1)(T)=w(T)=C:|1,a(T)=ea(T)=0 C:|=1,andE(T)=a00(T)=,which means both aWeyls theorem and the property(WE)hold for T.(2)isoa(T)=0 C:0 n(T I),so T is aisoloid.But since ab(T)=0 C:|=1 and 1(T)=C:|=1,we get ab(T)=1(T).Corollary 2.5Let T B(H).Then T is aisoloid and both aWeyls theorem and the property(WE)hold for T if and only if ab(T)=1(T)acc(T)C:n(T I)=.Proof From Theorem 2.1,if ab(T)=1(T)acc(T)C:n(T I)=,then bothaWeyls theorem and the property(WE)hold for T.Since isoa(T):n(T I)=01(T)acc(T)C:n(T I)=,it follows isoa(T):n(T I)=0 ab(T).Then isoa(T):n(T I)=0 a(T),thus isoa(T):n(T I)=0=,whichmeans T is aisoloid.For the converse,we suppose T is aisoloid and both aWeyls theorem and the property(WE)holdfor T.By Theorem 2.1,ab(T)=1(T)a(T):n(T I)=0 acc(T)C:n(T I)=.Since both aWeyls theorem and the property(WE)hold for T,similar to theproof of Theorem 2.1,we can get acca(T)1(T).Since T is aisoloid,we know isoa(T):n(T I)=0=,then a(T):n(T I)=0=isoa(T):n(T I)=0 acca(T):n(T I)=0 1(T).Soab(T)1(T)acc(T)C:n(T I)=.The converse inclusion is clear.Then ab(T)=1(T)acc(T)C:n(T I)=.From Remark 2.1,we see the condition“T is isoloid”is essential in Corollary 2.5.For T B(H),let Hol(T)be the class of all complexvalued functions which are analytic in aneighborhood of(T)and are not constant on any component of(T).In the following,we will exploreaWeyls theorem and the property(WE)for the functions of operator T.Remark 2.2For T B(H),T satisfies both aWeyls theorem and the property(WE)cannotimply f(T)satisfies both aWeyls theorem and the property(WE)for f Hol(T).For instance,let A,B B(2)be defined byA(x1,x2,x3,)=(0,x1,x2,),B(x1,x2,x3,)=(x2,x3,x4,).Put T=(A+I00B I),then(T)=w(T)=C:|1|1 C:|+1|1,a(T)=ea(T)=C:|1|=1 C:|+1|1 and E(T)=a00(T)=.It meansthat T satisfies both aWeyls theorem and the property(WE).Let f0(T)=T2.Since 1 (T2)w(T2)and 1 a(T2)ea(T2)but 1/E(T2)a00(T2),it follows that f0(T)satisfies neither aWeyls theorem nor the property(WE).88数学理论与应用Next,weconsideraWeylstheoremandtheproperty(WE)underfunctionalcalculus.LetPab(T)=a(T)ab(T)and SF+(T)=C:T I is an upper semiFredholm operator.Theorem 2.2Let T B(H).Then both aWeyls theorem and the property(WE)hold for f(T)for any f Hol(T)if and only if the following conditions hold:(1)T satisfies aWeyls theorem and has the property(WE)(2)ind(T I)ind(T I)0 for each pair,SF+(T)(3)if Pab(T)=,then ab(T)=1(T)acc(T)C:n(T I)=.Proof Necessity.(1)is obvious.In the following,we will prove(2)and(3).If there exists 0,0 SF+(T)such thatind(T 0I)=m 0 ind(T 0I)=n,where n is finite and m is finite or infinite.If m is finite,we put f(T)=(T 0I)n(T 0I)m ifm is infinite,we put f(T)=(T 0I)(T 0I).Then 0 a(f(T)ea(f(T).Since aWeylstheorem holds for f(T),it follows that f(T)is an upper semiFredholm operator with finite ascent.Thenasc(T0I)0.Then(2)is true.If Pab(T)=,take 0 Pab(T).From Corollary 2.5,we need to prove T is aisoloid.Supposethere exists 0 isoa(T):n(T I)=0,and let f(T)=(T 0I)(T 0I).By0,0 isoa(T),we get 0 isoa(f(T)(10).Since n(f(T)=n(T 0I)+n(T 0I),we know 0 n(f(T).Then 0 a00(f(T).Since aWeyls theorem holds for f(T),f(T)is an upper semiFredholm operator,which implies T 0I is an upper semiFredholm operator.ThenT 0I is bounded below since 0 isoa(T):n(T I)=0.It is in contradiction to the fact0 isoa(T),therefore T is aisoloid,and(3)is true.Sufficiency.We will take the following two cases into consideration.Case 1Pab(T)=.It follows that(T)=w(T),a(T)=ea(T),E(T)=a00(T)=,and it is not difficult to showE(f(T)f(E(T)=and a00(f(T)f(a00(T)=for any f Hol(T).By(2),w()andea()satisfy the spectral mapping theorem.Since()and a()satisfy the spectral mapping theorem,itfollows that(f(T)=f(T)=f(w(T)=w(f(T)and a(f(T)=f(a(T)=f(ea(T)=ea(f(T)for any f Hol(T).Then both aWeyls theorem and the property(WE)hold for f(T)for any f Hol(T).Case 2ab(T)=1(T)acc(T)C:n(T I)=.By Corollary 2.5,we see T is aisoloid.For any f Hol(T),take 0(f(T)w(f(T).Supposef(T)0I=(T 1I)n1(T 2I)n2(T tI)ntg(T),where i=jif i=j and g(T)is invertible.Then for each i,T iI is a Fredholm operator andAWeyls 定理和(WE)性质的摄动89ti=1niind(T iI)=0.By(2),we know T iI is a Weyl opreator.Since the property(WE)holds forT,T iI is a Browder opreator.Then f(T)0I is a Browder opeartor,which means 0 E(f(T).Take 0 a(f(T)ea(f(T)and assume f(T)0I has the same decomposition as above.Since ea()satisfies the spectral mapping theorem(by(2),it follows i ea(T)(1 i t).ThenT iI is bounded below or i a(T)ea(T).Since aWeyls theorem holds for T,T iI has finiteascent.Then f(T)0I has finite ascent.This shows 0 a00(f(T).For the converse,let 0 a00(f(T)and assume f(T)0I has the same decomposition as above.Then n(T iI).Without loss of generality,we suppose i a(T).Thus i isoa(T).Since Tis aisoloid,it follows i a00(T).Then T iI is an upper operator with ind(T iI)0.So f(T)0I is an upper semiFredholm operator with ind(f(T)0I)0,that is,0 a(f(T)ea(f(T).If 0 E(f(T)and f(T)0I has the same decomposition as above,without loss of generality,we suppose i(T),thus i iso(T).Since T is aisoloid,it follows i E(T).Then T iI is aWeyl operator,and so f(T)0I is also a Weyl operator,that is,0(f(T)w(f(T).As showed above,we get both aWeyls theorem and the property(WE)hold for f(T)for anyf Hol(T).3AWeyls theorem and the property(WE)under perturbationsAn operator R B(H)is called a Riesz operator if T I is a Fredholm operator for all nonzero C.Evidently,quasinilpotent operators and compact operators are Riesz operators.It is known ifT B(H),then b(T+R)=b(T),w(T+R)=w(T)and ea(T+R)=ea(T)for every Rieszoperator R commuting with T.Remark 3.1(1)Let T B(H)and K K(H)(or K F(H)with KT=TK.Even if bothaWeyls theorem and the property(WE)hold for T,we can not induce T+K satisfies aWeyls theoremor T+K (WE).For example,let A,B,C,P B(2)be defined byA(x1,x2,x3,)=(0,x1,0,x22,0,x33,),B(x1,x2,x3,)=(x1,0,0,),C(x1,x2,x3,)=(0,0,x2,0,x3,0,x4,),P(x1,x2,)=(x1,0,0,).Put T=(AC0B)and K=(000P).Then we have(i)(T)=a(T)=0,1,w(T)=ea(T)=0 and E(T)=a00(T)=1,which meansboth aWeyls theorem and the property(W