Invariant subspaces for some operators on locally convex spaces
Edvard Kramar
Abstract. The invariant subspace problem for some operators and some operator alge- bras acting on a locally convex space is studied.
Keywords: invariant subspace, locally convex space, locally bounded operator, univer- sally bounded operator, compact operator
Classification: 47A15, 46A32, 46A99
1. Introduction
Let X be a locally convex Hausdorff space over the complex field C. Each system of seminormsP inducing its topology will be called acalibration([11]). We denote byP(X) the collection of all calibrations onX. GivenP ∈ P(X), we call itbasic calibration if the corresponding “semiballs”U(ε, p) ={x∈X :p(x)< ε}, ε >0, p∈ P, form a neighborhood base at 0. As it is easily seen, P is basic if and only if for each p1, p2 ∈ P there is some p0 ∈ P such that pi(x) ≤ p0(x), i= 1,2. For any P ∈ P(X) we can generate a basic calibration P′ ∈ P(X) by taking maxima of finite seminorms fromP. For a givenP ∈ P(X) we denote by QP(X) the algebra ofquotient bounded operators on X, i.e. the collection of all linear operatorsT onX for which
p(T x)≤cpp(x), x∈X, p∈P,
and byBP(X) the algebra ofuniversally bounded operators onX, i.e. the set of all T ∈ QP(X) for which c = cp is independent of p∈ P ([11]). The algebra QP(X) is a unital locally m-convex algebra with respect to seminormsPb ={p}b (see eg. [6]) where
p(Tb ) = sup{p(T x) : x∈X, p(x)≤1}, p∈P, andBP(X) is a unital normed algebra with respect to the norm
kTkP = sup{p(Tb ) : p∈P}.
Let us define still some other families of linear operators. A linear operatorT on X islocally bounded, orT ∈ LB(X), if there exists a neighborhoodU such that
T(U) is bounded, andT iscompact, orT ∈ K(X), if there exists a neighborhood U such thatT(U) is a relatively compact set. Let us denote
B0(X) =∪{BP(X), P ∈ P(X)},
and byL(X) the set of all linear continuous operators onX (similarly L(X, Y) for two spaces X and Y). The following inclusions hold: K(X) ⊂ LB(X) ⊂ B0(X)⊂ L(X) (the second inclusion which is not so obvious will be verified later, or see [11]).
Given any linear operatorT onX, we define the spectrum and the resolvent set ofT with respect to various algebras. ForT∈ L(X): λ∈ρ(T) iff (λI−T)−1 exists inL(X), for T ∈ QP(X): λ ∈ρ(QP, T) iff (λI −T)−1 exists in QP(X) and similarly ρ(BP, T) for T ∈ BP(X). The corresponding complements in C will be denoted byσ(T),σ(QP, T) andσ(BP, T). Obviously,σ(T)⊂σ(QP, T)⊂ σ(BP, T) for T ∈BP(X). It is known thatσ(BP, T) is bounded and closed for T ∈BP(X) ([2]), but in general the above spectra can be unbounded. In the case whenσ(T) is bounded we denote
r(T) = sup{|λ|:λ∈σ(T)}.
ByR(T) we shall denote the range of an operatorT. LetS be a map onX which may be nonlinear. If there existP∈ P(X) andc >0 such that
p(Sx)≤cp(x), x∈X, p∈P, S will be called, as in [5], aP-bounded map.
2. Main results
Let us first prove two useful lemmas.
Lemma 1. Letp, qbe two seminorms onX such that: q(x)≤1 for eachx∈X for which p(x)<1. Then
q(x)≤p(x), x∈X.
Proof: Let 0≤p(z)< q(z) for somez∈X. Then there is someλ >0 such that p(z)< λ < q(z), hencep(z/λ)<1 andq(z/λ)>1 which is a contradiction.
Lemma 2. LetXbe a Hausdorff locally convex space andT1, T2∈ LB(X), then there exists a common calibrationP′∈ P(X)such thatT1, T2∈BP′(X).
Proof: We may take a basic calibration P ∈ P(X). Then there exist neigh- borhoods U1, U2 such thatTi(Ui), i= 1,2 are bounded. Without loss of gene- rality we may assume that Ui is the open semiball corresponding to the semi- norm pi ∈ P, i = 1,2. For every p ∈ P there are λ(p)1 , λ(p)2 ≥ 0 such that
sup{p(Tix) : x∈Ui} ≤λ(p)i ,i= 1,2. We assume firstly thatλ(p)i >0,i= 1,2.
For x ∈ X for which pi(x) < 1 it follows p(Tix/λ(p)i ) ≤ 1, i = 1,2, and by Lemma 1 we obtain
p(Tix)≤λ(p)i pi(x), x∈X, i= 1,2.
Since P is a basic calibration there is some p0 ∈ P such that pi(x) ≤ p0(x), i= 1,2. Hence for λp = max{λ(p)1 , λ(p)2 }we have
p(Tix)≤λpp0(x), p∈P, x∈X, i= 1,2.
If one ofλ(p)i is zero, then p(Tix) = 0 for eachx∈X and the above inequality trivially holds. Especially, we have p0(Tix)≤λ0p0(x), x∈X,i = 1,2. Let us defineP′={p′, p∈P}, where
p′(x) = max{p(x), λpp0(x)}, x∈X.
We readily verify that P′ is again a calibration. Now, we can estimate for any p′ ∈P′ and i= 1,2
p′(Tix) = max{p(Tix), λpp0(Tix)} ≤λpc0p0(x)≤c0p′(x), i= 1,2, wherec0 = max{1, λ0}. HenceTi∈BP′(X),i= 1,2.
TakingT1 =T2 we obtain
Corollary. EachT ∈ LB(X)is inB0(X).
If we take T ∈ LB(X), then T ∈ BP(X) for some P ∈ P(X) and hence σ(BP, T) is bounded and then σ(T) is bounded, too. We shall first prove some generalizations of some results from [5].
Lemma 3. Let X, Y be Hausdorff locally convex spaces, T ∈ L(X, Y) and K∈ LB(Y). LetS be a map onX such that for someP′∈ P(X)and someε >0 (1) p′(Sx)≤(r(K) +ε)−1p′(x), p′ ∈P′, x∈X.
If T =KT S, thenT = 0.
Proof: Let us choose anyP ∈ P(Y). Then there exists a neighborhood of zero U0onY such thatK(U0) is bounded. We may assume thatU0is an open semiball corresponding top0∈P. Let us denoteB =cobK(U0) the absolute convex closed hull of K(U0) and YB =span(B) the linear span ofB. This is a normed space with respect to the norm k.kB, the Minkowski’s functional ofB. It is not hard to see that the topology induced by this norm is finer than the relative topology
induced byP. Clearly,K(Y)⊂YB sinceU0 is absorbent andK(U0)⊂B and it followskKxkB≤1 for eachx∈Y such thatp0(x)<1. By Lemma 1 we obtain
(2) kKxkB≤p0(x), x∈Y,
hence the map K : Y → YB is continuous. Let us prove that KB := K|YB is continuous onYB. SinceB is bounded there is someλ >0 such thatB ⊂λU0, henceK(B)⊂λK(U0)⊂λB. Consequently, for allx∈YB such thatkxkB<1 it follows thatλ−1kKxkB≤1 and by Lemma 1 we have
kKxkB≤λkxkB, x∈YB.
Denote by J : YB → Y the inclusion map, then clearly KB = KJ. Since the norm topology onYB is finer than the relative one, we obtain ([3])σ(K)− {0}= σ(KB)− {0}. Thus,r(K) =r(KB). Without loss of generality we may assume that P′ is a basic calibration and (1) again holds. By the supposed equality it follows that T x∈YB for eachx∈ X and T =KnT Sn for alln ∈N. Fix any x∈X and n∈ N, then by the continuity of KB andT and by the inequalities (1) and (2) we can estimate
kT xkB=kKn+1T Sn+1xkB=kKBnKT Sn+1xkB≤ kKBnkB.kKT Sn+1xkB
≤ kKBnkB.p0(T Sn+1x)≤ kKBnkB.C.p′1(Sn+1x)
≤C.kKBnkB.(r(K) +ε)−(n+1)p′1(x),
where p′1 ∈ P′. For the above ε > 0 take any δ ∈ (0, ε) andn ∈ Nsufficiently large to yieldkKBnkB<(r(KB) +δ)n. Then
kT xkB≤C.(r(K) +δ)n.(r(K) +ε)−(n+1).p′1(x).
Sendingn→ ∞ we obtainT x= 0 and sincex∈X is arbitrary we haveT = 0.
As in [5] we call K ∈ LB(X) decomposable at 0 if for each ε > 0 we have a decompositionX =M ⊕N, whereM andN are nontrivial invariant subspaces ofK andr(K|M)< ε.
Let us prove the following result for locally convex spaces.
Theorem 4. LetX be a Hausdorff locally convex space andY a complete Haus- dorff locally convex space,T ∈ L(X, Y),K∈ LB(Y)andS aP-bounded map on X for some P ∈ P(X)and such that T=KT S. Then
(i) if r(K) = 0, thenT = 0;
(ii) if K∈ K(Y), then T has finite rank;
(iii) if K is decomposable at 0, thenR(T)is not dense inY.
Proof: (i) SinceS isP-bounded we havep(Sx)≤cp(x),x∈X,p∈P, for some c >0. Let us choose ε >0 such that ε < 1/c. Then p(Sx) ≤ε−1p(x), p∈P, x∈X, and by Lemma 3,T = 0.
(ii) Now, letε >0 be such thatε <(2c)−1. SinceK is compact, its spectrum σ(K) is a compact set, it has no limit point other than 0 and each λ∈ σ(K), λ6= 0, is an eigenvalue ([3]). For a locally bounded operator one can generalize the Riesz functional calculus to locally convex spaces (see [10]). Denoteσε={λ∈ σ(K) : |λ| < ε} and by Pε the corresponding projector for which PεK =KPε
and σ(K|R(Pε)) = σε. By the same calibration P as in (i) we have: p(Sx) ≤ (2ε)−1p(x) ≤ (r(PεK) +ε)−1p(x), p ∈ P, x ∈ X, since PεT = Pε2KT S = PεKPεT S, by Lemma 3,PεT = 0, henceT = (I−Pε)T. Thus,R(T) is contained in the finite-dimensional subspaceR(I−Pε).
(iii) Again chooseε >0 as in (ii) and use the decompositionX =M⊕N where r(K|M)< ε. Denote byPM :Y →M the corresponding projector. As in (ii) we obtainPMT = 0, and sinceR(T)⊂ R(I−PM), the range ofT is not dense.
As it is shown in [5], for two given operatorsA,B withR(A)⊂ R(B) acting between Banach spaces there exists a bounded mapS(which need not be linear) such that A=BS. This result can be generalized to the case in which the final space is locally convex.
Lemma 5. LetX,Z be Banach spaces andY a Hausdorff locally convex space.
Let A ∈ L(X, Y), B ∈ L(Z, Y) such that R(A)⊂ R(B). Then there exists a mapS(not linear in general)fromX intoZ such thatA=BS and such that for someC >0
kSxk ≤Ckxk, x∈X.
The proof is the same as in [5] and we omit it.
Theorem 6. Let Y be a complete Hausdorff locally convex space, K∈ LB(Y) and M := R(T)⊂ Y for some continuous operator T from a Banach space X intoY and letM ⊂K(M). Then the following statements hold:
(i) if r(K) = 0, thenM ={0};
(ii) if K∈ K(Y), then M is finite-dimensional;
(iii) if K is decomposable at0, thenM is not dense inY.
Proof: SinceR(T)⊂ R(KT), by Lemma 5 there is somek.k-bounded mapS : X →Xsuch thatT =KT Sand by Theorem 4 all statements follow immediately.
We shall now consider some invariant subspace problems on locally convex spaces. Let us denote byLb(X) the spaceL(X) endowed with the topologyτb of uniform convergence on bounded sets.
Theorem 7. Let X be a complete Hausdorff locally convex space and A an operator algebra in L(X), such that A=R(S) for some continuous operatorS
from a Banach spaceY into Lb(X). Let there exist an operatorK1∈ K(X)and an operatorK2∈ LB(X)which is decomposable at0, such that
AK1 ⊂K2A.
ThenAhas a nontrivial invariant subspace.
Proof: IfAhad no invariant subspace then by a generalized Lomonosov’s theo- rem (see [7]) there exists anA0 ∈ Asuch thatA0K1z=z,z6= 0,z∈X. Define T y := (Sy)(z), y ∈ Y, and let us prove that T ∈ L(Y, X). Let us choose any P ∈ P(X), any p∈P and any bounded set M which containsz ∈X. Then by the continuity ofS there is some CpM >0 such that qMp (Sy) := sup{p((Sy)x) : x∈M} ≤CpMkykand hence for anyy∈Y
p(T y) =p((Sy)z)≤CpMkyk.
Obviously, R(T) = Az = {Az, A ∈ A}. If Az = {0}, then V = span{z}
is an invariant subspace for A. If Az 6= {0} then Az is a range of a nonzero continuous operatorT and clearly,Azis invariant forA. For anyA∈ Awe have Az =AA0K1z =K2A2z for some A2 ∈ A and henceAz ⊂K2(Az). By part (iii) of Theorem 6,Azis not dense inX, henceAzis a proper invariant subspace
forA.
Corollary 8. LetXbe a complete Hausdorff locally convex space andA 6=C.Ia Banach algebra inL(X)with a norm topology finer then the topologyτbinherited from L(X)and let there be some K1 ∈ K(X)and K2 ∈ LB(X), decomposable at0, such that
AK1 ⊂K2A.
ThenAhas a nontrivial invariant subspace.
The algebra of universally bounded operators is a normed algebra with respect to the normk.kP for eachP ∈ P(X) and it is complete wheneverX is complete (see [11]). Thus, we have
Corollary 9. LetXbe a complete Hausdorff locally convex space andP ∈ P(X) such thatBP(X)6=C.Iand let existK1 ∈ K(X)andK2∈ LB(X), decomposable at0, such that
BP(X)K1 ⊂K2BP(X).
ThenBP(X)has a nontrivial invariant subspace.
Theorem 10. LetX be a complete Hausdorff locally convex space andA 6=C.I an operator algebra inL(X). Let there be some continuous operator T from a Banach spaceY intoLb(X)such thatA=R(T)and let there be someK1, K2∈ K(X)such that
AK1 ⊂K2A.
Then the commutant ofAhas a nontrivial invariant subspace.
Proof: If the commutant A′ had no invariant subspace then by Lomonosov’s theorem [7] there exist an operator B ∈ A′ and a nonzero z ∈ X such that BK1z =z. For anyA ∈ Ait follows: Az =ABK1z =BAK1z =BK2A1z for someA1∈ A. Hence the linear manifoldAzsatisfies the inclusionAz⊂(BK2)Az and as in the above proof we see that Az = R(T), where T is a continuous operator from a Banach space. By part (ii) of Theorem 6 it follows that Az is finite-dimensional. Let us choose A0 ∈ A such that A0 6= λI. If Az = {0}
then A0 has a nontrivial nullspace M ⊃ span{z}. If Az 6= {0} then it is a finite-dimensional invariant subspace forA0. ThusA0has a nontrivial eigenspace which is invariant for all operators commuting withA0, andA′ has a nontrivial
invariant subspace.
Corollary 11. Let X be a complete infra-barrelled locally convex space and A∈ L(X),A6=λI and such that for someP ∈ P(X)it satisfies the condition:
p(Anx)≤Cpp(x), x∈X, p∈P, Cp≥0, n∈N. Let there be somek∈NandK∈ K(X)such that
AK=KAk. ThenAhas a nontrivial hyperinvariant subspace.
Proof: Let us choose any sequence{an} ∈l1 and define Snx=
Xn j=0
ajAjx, x∈X, n∈N.
Givenε >0, we can find for arbitraryp∈P and any bounded setM, sufficiently largem, n∈N,m > n, such that the following estimations hold
qpM(Sm−Sn) = sup
x∈M
p(
Xm
j=n+1
ajAjx)≤Cp sup
x∈M
p(x).
Xm
j=n+1
|aj|< ε.
Thus, {Sn} is a Cauchy sequence in Lb(X), since it is quasicomplete ([9]) it is also sequentially complete and we have for each sequence {an} ∈ l1 an operator S = P
ajAj ∈ L(X). Denote A = {S := P
ajAj : {aj} ∈ l1}. Then by an estimation similar to the one given above we can prove that the map {aj} →S is a continuous map ofl1 intoLb(X). So, Ais a range of a continuous operator from a Banach space and clearlyA is an algebra. In the same manner as in [5]
we haveSK=KS1 whereS, S1 ∈ Aand the conclusion follows by Theorem 10.
Let us now generalize a result from [8].
Theorem 12. Let X be a Hausdorff locally convex space, A ∈ LB(X) and {Kn}∞n=0 a sequence of operators from BP(X) for some P ∈ P(X) such that kKnkP →0andK0∈ K(X). Let the following relations hold
KnA=AKn+1, n= 0,1, . . . . ThenAhas a nontrivial hyperinvariant subspace.
Proof: By the above relations it immediately follows thatK0An = AnKn for n= 0,1,2, . . . and clearly K0A is compact, too. DenoteA={A}′. IfAhad no invariant subspace, then by [7] there existsA1∈ Asuch that 1∈σp(A1K0A) (the point spectrum). SinceA1K0Ais also compact, then 1∈σp((A1K0A)∗), too ([3]).
Thus, there is somef ∈X′,f 6= 0 such that (A1K0A)∗f =f. Consequently, for eachn∈N:
(3) Kn∗A∗1A∗(A∗)n−1f = (A∗)nK0∗A∗1f = (A∗)n−1f.
If (A∗)nf = 0 for somen∈N, thenker(A∗)6={0} and then R(A)⊥ 6={0} ([9]) (where forM ⊂X :M⊥={f ∈X′:f(x) = 0, x∈M}). So,R(A)6=X. In this case this set is a proper hyperinvariant subspace ofA. Ifgn:= (A∗)n−1f 6= 0 for eachn∈N, then (3) implies
Kn∗A∗1A∗gn=gn, n∈N.
Let us prove that there exists someP′∈ P(X) such that allKn andA1A are in BP′(X). Clearly,AA1is also locally bounded, hence there is some neighborhood U0 for which AA1(U0) is bounded. We may assume that U0 is the semiball corresponding to some p0 ∈ P. Thus, we have sup{p(AA1x) : x ∈ U0} ≤ λp, p∈P. Without loss of generality we may also assume thatλp>0 for eachp∈P. By Lemma 1 we obtain
p(AA1x)≤λpp0(x), x∈X, p∈P,
and especially also p0(AA1x)≤λ0p0(x),x∈X. At the same time we have p(Knx)≤ kKnkP.p(x), x∈X, p∈P.
Let us defineP′={p′}, where
p′(x) = max{p(x), λpp0(x)}, x∈X, p∈P.
It is easy to see that P′ is again a calibration on X and for each x ∈ X and p′ ∈P′ we can estimate
p′(AA1x) = max{p(AA1x), λpp0(AA1x)} ≤λpc0p0(x)≤c0p′(x),
wherec0= max{1, λ0}, and by a simple verification we also have p′(Knx)≤ kKnkP.p′(x), x∈X, p′∈P′.
Thus, all Kn and AA1 are in BP′(X) and kKnkP′ ≤ kKnkP for each n ∈ N. Let us take an arbitrary n ∈ N. Since gn ∈ X′, there is some p′n ∈ P′ with the corresponding quotient spaceXn:=X/kerp′n(which is a normed space with respect to the normkxˆnkn=p′n(x), where ˆxn=x+kerp′n) such thatgn∈(Xn)′ (see [4]). For anyx∈X we can now estimate
|gn(x)|=|gn(AA1Knx)| ≤ kgnknp′n(AA1Knx)≤ kgnknkAA1kP′kKnkP.p′n(x).
Taking supremum over allx∈X for whichp′n(x) =kxˆnkn≤1 we obtain kgnkn≤ kgnknkAA1kP′kKnkP,
hence
1≤ kAA1kP′kKnkP.
Sincen∈Nis arbitrary andkKnkP →0, we have a contradiction.
Finally, we give some generalization of some results from [1].
Theorem 13. LetX be a Hausdorff locally convex space andA∈ L(X),A6=λI.
Let
AK=µKA
for some nonzeroK∈ K(X)and µ∈C. ThenAhas a nontrivial hyperinvariant subspace.
The proof of this theorem and of the following one is for a locally convex space the same as for a normed space and we omit it.
Theorem 14. LetX be a Hausdorff locally convex space andA∈ L(X),A6=λI, andM a subspace of L(X)of finite dimension such that AM=MA and such thatM ∩ K(X)6={0}. ThenAhas a nontrivial hyperinvariant subspace.
We shall give the following variant of generalization of a result from [1].
Theorem 15. Let X be a Hausdorff locally convex space and A ∈ LB(X), B ∈ L(X) and K ∈ K(X) nontrivial operators such that there exist λ, θ ∈ C,
|λ|<1and|θ| ≤1with the properties
BA=λAB and BK =θKB.
ThenAhas a nontrivial invariant subspace.
Proof: Since alsoK∈ LB(X), by Lemma 2 there exists a calibrationP ∈ P(X) such that A, K ∈ BP(X). If A had no nontrivial invariant subspace then the
same would be true for the algebra A generated by Ak, k ∈ N. By [7] then there exist S ∈ A and x 6= 0 such that SKx = x. Since S = Pn
j=1λjAj for some {λj} ⊂ C, we have (P
λjAj)Kx = x and for each m = 0,1,2. . . also Bm(Pn
j=1λjAj)Kx=Bmx. Taking into account the supposed relations we have BmAjK=λmjθmAjKBm, m= 0,1,2, . . . , j= 1,2, . . .
and we obtain
(4) [(λ1λmθmA+λ2λ2mθmA2+· · ·+λnλmnθmAn)K]Bmx=Bmx.
Denote by Tm the operator in the square brackets. Then for each p ∈ P and y∈X we can estimate p(Tmy)≤Mm,np(y), where
Mm,n =|λ|m|θ|mkAkP[|λ1|+|λ2||λ|mkAkP+· · ·+|λn||λ|(n−1)mkAknP−1].kKkP. Thus,Tm ∈BP(X) andkTmkP →0 form→ ∞. In virtue of (4) we obtain for anyp∈P andx∈X
p(Bmx) =p(TmBmx)≤ kTmkP.p(Bmx)
and if we choosek∈Nsuch that kTkkP <1, we havep(Bkx) = 0 for all p∈P. Consequently, Bkx = 0. So, B has a nontrivial kernel which is an invariant
subspace forA.
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Department of Mathematics, University of Ljubljana, Jadranska 19, 1000 Ljubljana, Slovenia
(Received November 26, 1996)
