©Hindawi Publishing Corp.
HYPERINVARIANT SUBSPACES FOR SOME OPERATORS
ON LOCALLY CONVEX SPACES
EDVARD KRAMAR
(Received 27 January 1999)
Abstract.Some results concerning hyperinvariant subspaces of some operators on locally convex spaces are considered. Denseness of a class of operators which have a hyperinvari-ant subspace in the algebra of locally bounded operators is proved.
Keywords and phrases. Locally convex space, hyperinvariant subspace, operator. 2000 Mathematics Subject Classification. Primary 47A15, 47B99, 46A32.
1. Introduction. Let X be a locally convex space over the complex field C. Each system of seminormsPinducing its topology will be called acalibration. We denote by
ᏼ(X)the collection of all calibrations onXand byΛ(X)all continuous seminorms with respect to the given topology. Let us denote byᏸ(X)the set of all linear continuous operators onXand by(X)the set of compact operators onX, i.e.,T∈(X)if there exists a neighborhoodUsuch thatT (U)is a relatively compact set. We shall denote by im(T )the range ofTand by kerTthe null space ofT. For a givenP∈ᏼ(X), letP= {pα: α∈∆}, where∆is some index set. Choose anyα∈∆and letXα:=X/ker(pα)denote the quotient space which is a normed space with respect to the normxαα=pα(x) wherexα=x+ker(pα). The completeness ofXαwe denote by ˜Xα. It is well known that for the dual spaces the following relation holdsX={(Xα)˜ , α∈∆}(Floret and Wloka [2]). LetBbe an absolutely convex and bounded set, thenXB:={nB, n∈N}
is a normed subspace inXwith respect to the normxB:=inf{λ >0 :x∈λB}. It is easy to see that for anypα∈P there is someλα≥0 such that
pα(x)≤λαxB, x∈XB. (1.1)
A linear operatorT onXis callednuclearif it can be written in the form
T x= ∞
j=1
λjcj(x)aj, x∈X, (1.2)
where(cj)is an equicontinuous sequence inX,(aj)is a sequence contained in an absolutely convex bounded setB inX, such thatXB is complete and(λj)∈l1(cf.
For a givenP∈ᏼ(X)we denote byBP(X)the algebra ofuniversally bounded oper-ators onX, i.e., the collection of all linear operatorsT onXfor which
pα(T x)≤Cpα(x), x∈X, pα∈P, (1.3)
whereC≥0 is independent ofpα∈P. The algebraBP(X)is a unital normed algebra with respect to the norm
TP=suppα(T x):pα(x)≤1, x∈X, pα∈P. (1.4)
Lemma1.1. LetXbe a Hausdorff locally convex space andT ∈ᏸB(X), then there exists a calibrationP∈ᏼ(X)such thatT∈BP(X).
Proof. Choose any calibrationP ∈ᏼ(X). Then there exists a neighborhoodUγ such thatT (Uγ)is bounded. Without loss of generality we may assume thatUγis the open semiball corresponding to the seminormpγ∈P. For anypα∈Pthere is some Cα≥0 such that sup{pα(T x):x∈Uγ} ≤Cα. Hence forx∈Xfor whichpγ(x) <1 it followspα(T x)≤Cα. Let us prove that
pα(T x)≤Cαpγ(x), pα∈P, x∈X. (1.5)
Fix anypα∈P and assume first thatCα>0. If for somez∈X there is pα(T z) > Cαpγ(z)then there existsµ >0 such thatpα(T z) > µ > Cαpγ(z)and then for the vector(Cα/µ)z we have a contradiction with the above implication. If Cα=0 then (1.5) holds with zero on both sides. Especially, we havepγ(T x)≤Cγpγ(x), x∈X. Let us defineP= {p
α:pα∈P}, wherepα(x)=max{pα(x), Cαpγ(x)}, x∈X. We readily verify thatPis again a calibration. Now, we can for anyp
α∈Pestimate
p
α(T x)=max
pα(T x), Cαpγ(T x)≤CαC0pγ(x)≤C0pα(x), (1.6) whereC0=max{1,Cγ}. HenceT∈BP(X).
Byinvariant subspace of T ∈ᏸ(X)we mean a closed subspace M ⊂X with the propertyT (M)⊂M. Likewise, M is hyperinvariantforT if it is invariant for every operatorB∈ᏸ(X)that commutes withT.
2. Main results. We consider some instances of the existence of a hyperinvariant subspace for some operators on locally convex spaces. We start with the following result obtained by Ma in [9] as a generalization from normed spaces (Kim et al. [4]) to locally convex spaces.
Theorem2.1. LetX be a Hausdorff locally convex space,T ∈ᏸ(X)and let K∈ (X)be such that the commutator T K−KT has rank at most one. ThenT has a nontrivial hyperinvariant subspace.
We consider two cases in which the operator has the above property. DenoteN0:= N∪{0}.
(i) {Tna:n∈N0} ⊂B, for some absolutely convex bounded setBinXfor whichXB is complete,
(ii) there is somepγ∈P, such that(T)−nϕ∈X
γfor alln∈N0, (iii) ∞n=0(T)−nϕγTnaB<∞.
Then there exists a nontrivial hyperinvariant subspace forT.
Proof. SinceT is invertible, the adjoint operatorTis invertible, too. Define an operator sequence as follows
Kn=
n
k=0
Tka⊗T−kϕ, n∈N. (2.1)
LetqM
α be a member of a calibration inducing the topologyτb onᏸ(X)of the uni-form convergence on bounded subsets ofX, whereMis a bounded set onXandqM α corresponds toMand somepα∈P. Letm > n, m,n∈N, then we can estimate
qM
αKm−Kn=sup x∈Mpα
Kmx−Knx≤
m
k=n+1
sup x∈Mpα
T−kϕ(x)Tka
≤
m
n+1
T−kϕ γsup
x∈Mpγ(x)pα
Tka
≤λα·sup x∈Mpγ(x)
m
n+1
T−kϕ
γTkaB (2.2)
and by (iii),{Kn}is aτb-Cauchy sequence onᏸ(X). SinceXis barreled and complete,
ᏸ(X)is quasicomplete with respect toτb(Köthe [6]) and hence also sequentially com-plete. Thus, the sequence{Kn}converges to some operatorK∈ᏸ(X). Denote byJB the injection map ofXBintoXand byπγthe natural mapXXγ. ThenK=JBKγ πγ, whereKγ can be written in the form
Kγxγ= ∞
k=0
ψγk
xγak, (2.3)
whereψkγ(xγ)=((T)−kϕ)(x)forxγ=x+ker(pγ)andak=Tka, k∈N0. Taking into account the above assumptions, the operatorKγ acting between two normed spaces is nuclear, and sinceJBandπγare continuous,Kis nuclear and then also compact [2]. It is not hard to verify thatT K−KT= −a⊗Tϕ, thus, it has rank one and applying Theorem 2.1, the proof is complete.
Corollary2.3. LetXbe a complete barreled Hausdorff locally convex space,T∈ ᏸ(X)andC∈ᏸ(X)an invertible operator such thatC∈ {T}(in the second commu-tant)and suppose that there exista∈X,ϕ∈XandP∈ᏼ(X)such that:
(i) {Cna:n∈N0} ⊂Bfor some absolutely convex bounded setBinXfor whichXB is complete,
(ii) there is somepγ∈P, such that(C)−nϕ∈X
γ,for alln∈N0, (iii) ∞n=0(C)−nϕγCnaB<∞.
Proof. By the above theorem,Chas a nontrivial hyperinvariant subspace and since
{T}⊂ {C}, the operatorT has a nontrivial hyperinvariant subspace, too.
As in Kim et al. [5] we can prove for locally convex spaces the following theorem.
Theorem2.4. LetXbe a Hausdorff locally convex space andT∈ᏸ(X)a nonscalar operator. IfTorThas an eigenvector, thenThas a nontrivial hyperinvariant subspace. Proof. LetT z=λz, z=0, and chooseϕ∈Xsuch that (T−λI)ϕ=0. Then forK:=z⊗ϕwe haveT K−KT=z⊗(λI−T )ϕand by Theorem 2.1 the conclusion follows. IfThas an eigenvector the proof is similar.
For a givenT ∈ᏸ(X), the numberλ∈Cis in theresolvent setofT, orλ∈&(T ), if and only if there exists(T−λI)−1∈ᏸ(X). Thespectrumis the setσ (T )=C\&(T ). As
in a normed space we can define the following main subsets of the spectrum:σp(T ), σc(T ), andσr(T )—the point spectrum, the continuous, andthe residual spectrum, re-spectively, as follows (see Uss [10]):λ∈σp(T )if and only ifT−λIis not one-to-one, λ∈σr(T )if and only ifλ∉σp(T )and im(T−λI)=X and λ∈σc(T )if and only ifλ∉(σp(T )∪σr(T ))and(T−λI)−1is not continuous. WhenX is complete, then
σ (T )=σp(T )∪σr(T )∪σc(T )(see [10]).
Theorem2.5. LetX be a complete Hausdorff locally convex space. Then the set of all continuous linear operators which have a nontrivial hyperinvariant subspace is τb-dense inᏸB(X).
Proof. We may assume thatXis not normable since for normed spaces such a result holds (see Kim et al. [5]). Choose any nonzeroT∈ᏸB(X). Clearly,T=µI, since I∉ᏸB(X). LetP= {pα:α∈∆}be a calibration onX. As in the proof of Lemma 1.1 there is somepγ∈P such that for anypα∈P there is someCα>0 such that (1.5) holds. By Uss [10],σ (T )= ∅, so, we can choose someλ∈σ (T ). Ifλ∈σp(T ), thenT has, by Theorem 2.4, a hyperinvariant subspace. Forλ∉σp(T )we still have two cases. Ifλ∈σr(T ), then im(T−λI)=Xand, by the Hahn-Banach theorem, there exists a nontrivialϕ∈Xsuch thatϕ((T−λI)x)=0 for allx∈X. Consequently,λ∈σp(T) and by Theorem 2.4,T has a nontrivial hyperinvariant subspace. Letλ∈σc(T ). Then there exists a net(aδ)δ∈Γ such that(T−λI)aδ→0 andaδ0. First, suppose that λ=0 and prove thatpγ(aδ)0. For anypα∈Pwe have
|λ|pαaδ≤pα(λI−T )aδ+pαT aδ≤pα(λI−T )aδ+Cαpγaδ. (2.4)
Ifpγ(aδ)→0, thenpα(aδ)→0 for allα∈∆, henceaδ→0 which is a contradiction. So, we may assume that
pγaδ≥1, δ∈Γ. (2.5)
Now, letλ=0. If for someδ∈Γ :pγ(aδ)=0, then by (1.5), pα(T aδ)=0 for all α∈∆, and consequentlyT aδ =0, which is a contradiction with λ∉σp(T ). Thus,
pγ(aδ)=0 for allδ∈Γ and we may again assume that (2.5) holds. Now, for eachδ∈Γ
by the Hahn-Banach theorem, there exists someϕδ∈Xsuch that
Anyx∈Xcan be written in the form
x=pγϕδ(x)
aδaδ+zδ, δ∈Γ, (2.7)
wherezδ∈ker(ϕδ). Define an operator net(Tδ)δ∈Γ as follows
Tδx:= ϕδ(x)
pγaδλaδ+T zδ, x∈X, δ∈Γ. (2.8)
Let us take anypα∈P and any bounded setMonXwhich generate one of the semi-normsqM
α of a calibration defining the topologyτb. Then we can estimate
qM
αTδ−T=sup x∈Mpα
Tδ−Tx≤sup x∈M
ϕδ(x)
pγ(aδ) pα
T aδ−λaδ
≤sup
x∈Mpγ(x)pα
T aδ−λaδ, (2.9)
where the right-hand side tends to zero. So, the differenceTδ−T isτb-convergent to zero and sinceλ∈σp(Tδ), by Theorem 2.4 all operatorsTδhave a nontrivial hyperin-variant subspace.
In the sequel, we shall generalize another result from Kim et al. [3] to locally convex spaces. In the proof, one needs some lemmas which we shall write down for such spaces.
A linear operatorT∈ᏸ(X)is said to bebounded belowif for any seminormp∈Λ(X) there is someq∈Λ(X)such that
q(T x)≥p(x), x∈X. (2.10)
Lemma2.6. LetXbe a complete Hausdorff locally convex space andT∈ᏸ(X). Then T is invertible inᏸ(X)if and only if it is bounded below andim(T )=X.
Proof. IfT is invertible then im(T )=XandT−1is continuous, thusTis bounded
below. Conversely, letT be bounded below with dense range. Let us prove that im(T ) is closed. Take any convergent net(yδ)γ∈Γ⊂im(T ), yδ→yand letyδ=T xδ, xδ∈ X, δ∈Γ. Then by (2.10) it follows that(xδ)is a Cauchy net and thus it is convergent xδ→x. By continuity ofT it follows thatT xδ→T xhencey∈im(T ). The range is closed and dense, hence im(T )=X.
In the vector space ˜X:=X×Xwe define the topology ˜τby the following system of seminorms ˜P= {p˜}, where ˜p(˜x)=p1(x1)+p2(x2), p1,p2∈Pfor someP∈ᏼ(X)and
˜
x=(x1,x2)∈X. Let us denote by˜ Λ(X)˜ all continuous seminorms on ˜X. It is easy to
see that ˜Xis complete wheneverXis complete. Let us define(A⊕B)(x,y):=(Ax,By).
Lemma2.7. LetXbe a complete Hausdorff locally convex space andA1,A2∈ᏸ(X). Then the operator A˜:= A1⊕A2 is invertible in ᏸ(X)˜ if and only if A1 and A2 are invertible inᏸ(X).
Proof. Let us suppose thatAi, i=1,2, are invertible. Then it is easy to see that for ˜Athe inverse operator isA−1
For any ˜p∈Λ(X)˜ there is some ˜q∈Λ(X)˜ such that ˜q(A˜˜x)≥p(˜˜x), x˜∈X. Writing˜ ˜
p=p1+p2and ˜q=q1+q2, withp1,p2,q1,q2∈Λ(X)we have
q1A1x1+q2A2x2≥p1x1+p2x2, x˜=x1,x2∈X.˜ (2.11)
Replacing ˜xonce with(x,0)and then by(0,x)we obtain
q1A1x≥p1(x), q2A2x≥p2(x), x∈X. (2.12)
Since ˜pis arbitrary so arep1andp2which means thatA1andA2are bounded below.
By invertibility of ˜Ainᏸ(X)˜ there exists the operator(A˜)−1which is equal to(A˜−1) and it is continuous in the strong topology in ᏸ(X˜) (see [2]), where ˜A=A
1⊕A2.
Thus, ˜Ais bounded below. In the same manner as above,A
1andA2are then bounded
below and hence injective. By the relations im(Ai)=ker(A
i)⊥, i=1,2 (Köthe [6]) the operatorsA1andA2have dense range. By Lemma 2.6 both are invertible.
Lemma2.8. Let X be a Hausdorff locally convex space, T ∈ ᏸB(X) and {Kn} a sequence of operators in BP(X)for some P ∈ᏼ(X)and such that KnP → 0. Let K0∈(X)and
KnT=T Kn+1 or Kn+1T=T Kn, n=0,1,2,... . (2.13) ThenT has a nontrivial hyperinvariant subspace.
The proof with the first assumption in (2.13) is made by Kramar [7], the proof with the second one is similar. In the next lemma one needs the Riesz functional calculus which can be generalized to locally convex spaces for operators of the form T =
αI+T1, whereα∈CandT1is a locally bounded operator (see Uss [10]).
Lemma2.9. LetXbe a complete barreled Hausdorff locally convex space,T∈ᏸ(X) a nonscalar operator,K∈(X)a nonzero operator andfan analytic function on an open neighborhoodᐁofσ (K)and such thatKT=T f (K). Letf (0)=0andσp(T )=
σp(T)= ∅. Thenfmapsσ (T )onto itself and for every integerk, the functionfkhas no fixed point onσ (K)except zero and the following relations hold
fn(K)T=T fn+1(K), KTn=Tnfn(K), n∈N. (2.14) In the proof one may restrict to nonnormable case, then for K∈(X) we have
{0} ⊂σ (K)(see [10]). The proof is then similar to that one in normed space using Lemma 2.7 and the following facts in locally convex setting:
(i) ifλ=0 andK∈(X)thenλ∈σp(K)if and only ifλ∈σp(K)(see Edwards [1]), (ii) ifK1andK2are locally bounded thenK1⊕K2is locally bounded too,
(iii) the Riesz functional calculus and spectral mapping theorem for locally bounded operators (see Uss [10]).
Lemma2.10. LetXbe a complete barreled Hausdorff locally convex space,K, T , f, andᐁas in the previous lemma. Thenσ (K)= {0}if either|f(0)|<1or|f(0)|>1 andker(K)= {0}.
Theorem2.11. Let X be a complete barreled Hausdorff locally convex space, T∈ᏸ(X)a nonscalar andK∈(X)a nonzero operator, f an analytic function on an open neighborhoodᐁof the spectrumσ (K)such thatKT=T f (K). ThenT has a nontrivial hyperinvariant subspace whether one of the following conditions holds
(i) |f(0)|<1,
(ii) |f(0)|>1andσ (K)= {0}, (iii) |f(0)|>1andker(K)= {0}.
Proof. As in normed case, we may assume thatf (0)=0 and thatσp(T )=σp(T)= ∅(Theorem 2.4) and in all three cases by Lemma 2.10 it follows thatσ (K)= {0}. Let us treat the case|f(0)|<1, then there are 0< c <1 andδ >0 such that|f (ζ)|< c|ζ| for 0<|ζ|< δ. Define a sequence of operators:K0=KandKn=fn(K0),n∈N. Let
us prove that for this sequence the assumptions of Lemma 2.8 are satisfied. Choose ε∈(0,δ)and denoteSε= {λ, λ∈C,|λ| =ε}then we may use the functional calculus (see Uss [10]) in the following manner
Kn=2πi1
Sεf
n(λ)(λI−K)−1dλ
=2πi1
Sε fn(λ)I
λ dλ+
1 2πi
Sε
fn(λ)K(λI−K)−1
λ dλ
=fn(0)I+ 1 2πi
Sε
fn(λ)K(λI−K)−1
λ dλ,
(2.15)
wherefn(0)=0. For a givenP= {pα} ∈ᏼ(X)sinceK∈(X)there is some semiball Uγsuch thatK(Uγ)is bounded and the set(λI−K)−1K(Uγ)is forλ∈Sεbounded too.
Fix anypα∈P, letCλ
γ,α=sup{pα((λI−K)−1Kx), x∈Uγ}andCγ,α=sup{Cγ,αλ , λ∈ Sε}. Then as in the proof of Lemma 1.1 we havepα((λI−K)−1Kx)≤Cγ,αpα(x), x∈X.
If we defineP= {p
α}, wherepα =max{pα, Cγ,α·pα}all operators(λI−K)−1Kare inBP(X)and
p α
(λI−K)−1Kx≤Mγp
α(x), x∈X, λ∈Sε (2.16)
withMγ=max{1,Cγ,γ}. So, we can for eachp
α∈Pandx∈Xestimate
p
αKnx≤sup λ∈Sε
fn(λ)
|λ| ·supλ∈Sε
pαK(λI−K)−1x·ε≤εcnMγp
α(x). (2.17)
HenceKnP≤cn·Mγεand we haveKnP→0. By Lemma 2.8 the operatorT has a
nontrivial hyperinvariant subspace. In the case|f(0)|>1 the proof is similar. First, one can find a function h analytic on a neighborhood ᐁ1 of the origin such that h(f (ζ))=ζ, ζ∈ᐁ1. Then as above one can verify that all assumptions of Lemma 2.8 are satisfied forhn(K), n∈Nand the conclusion follows.
References
[1] R. E. Edwards,Functional Analysis. Theory and Applications, Holt, Rinehart and Winston, New York, 1965. MR 36#4308. Zbl 182.16101.
[3] H. W. Kim, R. Moore, and C. M. Pearcy,A variation of Lomonosov’s theorem, J. Operator Theory2(1979), no. 1, 131–140. MR 81b:47007. Zbl 433.47010.
[4] H. W. Kim, C. Pearcy, and A. L. Shields, Rank-one commutators and hyperinvariant subspaces, Michigan Math. J. 22 (1975), no. 3, 193–194 (1976). MR 52#11626. Zbl 319.47027.
[5] ,Sufficient conditions for rank-one commutators and hyperinvariant subspaces,
Michigan Math. J.23(1976), no. 3, 235–243 (1977). MR 57#3881. Zbl 346.47026. [6] G. Köthe,Topological Vector Spaces. II, Springer-Verlag, Berlin, Heidelberg, New York,
1979. MR 81g:46001. Zbl 417.46001.
[7] E. Kramar,Invariant subspaces for some operators on locally convex spaces, Comment. Math. Univ. Carolin.38(1997), no. 4, 635–644. MR 99a:47013.
[8] G. L. Litvinov,On the traces of linear operators in locally convex spaces, Selecta Math. Soviet.8(1989), no. 3, 203–212. Zbl 686.47021.
[9] T. W. Ma,On rank one commutators and triangular representations, Canad. Math. Bull. 29(1986), no. 3, 268–273. MR 87g:47028. Zbl 555.47004.
[10] P. Uss,Sur les opérateurs bornés dans les espaces localement convexes, Studia Math.37 (1970/71), 139–158. MR 46#2466. Zbl 212.15901.
Kramar: University of Ljubljana, Department of Mathematics, Jadranska19, 1000 Ljubljana, Slovenia
