The reflexivity of an extension of unilateral weighted shift operator by a nilpotent operator

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The reflexivity of an extension of unilateral weighted shift operator by a nilpotent operator

Khalid Elhachimi

To cite this version:

Khalid Elhachimi. The reflexivity of an extension of unilateral weighted shift operator by a nilpotent operator. 2015. �hal-01238199�

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The reflexivity of an extension of unilateral weighted shift operator by a nilpotent operator

Khalid Elhachimi

Department of Mathematics, Faculty of Sciences, PO. Box 1014, Mohammed V University, Rabat, Morocco

Abstract. A bounded linear operatorTon a Hilbert space is said to be reflexive if the operators which leave invariant the invariant subspaces of T are wot-limits of polynomials in T. In this paper we give a necessary and sufficient condition for an extension of a weighted shift operator by a nilpotent one to be reflexive.

Résumé.Un opérateur linéaire bornéTsur un espace de Hilbert est dit réflexif si les opérateurs qui laissent invariant les sous-espaces invariants pourTsont limites, pour la topologie faible des opérateurs, de polynômes enT. Dans ce papier nous donnons une condition nécessaire et suffisante pour que l’extension d’un shift à poids unilateral opérateur par un opéra- teur nilpotent soit réflexive.

1 Introduction

LetB(H)be the algebra of bounded linear operators on a separable Hilbert space H. For T ∈ _B(H) we denote by LatT the lattice of linear closed subspaces of H that are invariant by T, AlgLat(T)will denote de set of operatorsSinB(H)such thatLatT ⊂ LatS. We say that the operatorTis reflexive ifAlgLat(T)is a weakly subalgebraW(T)ofB(H)generated by T.

The concept of reflexivity has been studied extensively and starting with Sarason [9] who established the reflexivity of the normal operator. In [6] the author show that every isometry is reflexive. Deddens and Fillmore presented necessary and sufficient conditions for reflexivity of operators

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acting on finite-dimensional Hilbert spaces in [7]. In [1] the authors char- acterize the reflexivity of nilpotent operator definite on separable Hilbert space of infinite dimension. The reflexivity of weight shifts operators has given in [10].

The study of reflexivity of the extensions of operator is more difficult to that of only one operator. Indeed, while it follows from [7] that direct sums of reflexive operators acting on finite-dimensional spaces remain reflexive, in [8], Larson and Wogen constructed a reflexive operator acting on

`² whose direct sum with the zero operator fails to be reflexive. In [2] the authors have characterized the extension of normal operators by a nilpotent operator which are reflexive and in [3] the authors have characterized the extension of subnormal operators by an algebraic operator which are reflexive.

Since the weighted shift operators are interesting in providing examples and counter-examples to illustrate many properties, we will continue in this paper the study of reflexivity of the extensions of operators, such as the reflexivity of an extension of injective unilateral weight shift operator by a nilpotent operator. Precisely, let A ∈ _B(H)be an injective unilateral weighted shift operator,N ∈ _B(K)be a nilpotent operator on a separable Hilbert spaceK_{of index}n ≥ _1, e ∈ K_{such that} Nⁿ⁻¹e 6= _{0 and}T be the extension ofAbyNdefined onH ⊕ Kby

T=

A X

0 N

, whereX ∈_B(K)_.

Denoter₁(A) := lim[m(Aⁿ)]^1/n wherem(A) := inf{||A f|| : ||f|| = 1}is the lower bound ofA.

In this paper we show that ifr₁(A) > 0, thenTis reflexive if and only if one of the following condition is satisfied:

• The adjoint operator ofAhas a nonzero eigenvalue.

• Nis reflexive.

• Tⁿe∈/ Im(A).

2 Preliminary

Let {β_k}_k be a sequence of positive numbers with β0 = 1 and consider the space of the sequences

φb(k) _k such that the series∑k|φb(k)|²|β_k|²con- verges. We will write φ(z) = _∑_kφb(k)z^k whether or not the series con- verges for any value ofz. The spaceH²(β)is the set of formal power series

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φ(z) =_∑_kφb(k)z^k. This space is a Hilbert space with the inner product defined byhφ,ψi=_∑_kφb(k)ψb(k)β²_k.H^∞(β)denotes the set of formal power se- riesφ(z) =_∑_kφb(k)z^k such thatφH²(β)⊂ H²(β). Recall that the setH^∞(β) is a commutative algebra (see [10]). We denote byM_zthe operator of multiplication byzon H²(β)defined by

M_z(φ) =

∑

k

φb(k)f_k₊₁,

whereφ= _∑_kφb(k)f_kand{f_k}_k is an orthogonal basis ofH²(β)defined by f_k(z) = z^k. For given φ ∈ H^∞(β)the operator of multiplication by φ is denoted byM_φ. Now, according to [10, proposition 7], we can suppose that A= MzandH = H²(β)whenAis an injective weighted shift operator on Hilbert spaceH. In this case, we see that Ais bounded if and only if the sequence {^β^k⁺¹/β_k}_k is bounded, and in this case ||Aⁿ|| = sup_k{^β^k⁺¹/β_k}_k, for all integern.

For w ∈ _C, λ_w denoted the evaluation at w defined on power polynomials p by λ_w(p) = p(w). We say that w is a bounded point evaluation forH²(β)if the functionalλwextended a bounded linear functional on H²(β). Hence in this case there exists a unique functionk_win H²(β)such thatλ_w(f) =hf,k_wifor all f inH²(β). Recall that the set

∆:=

z∈_C:|z|< lim

n→+_∞infβ^1/n_n

is a largest disc in which all power series inH²(β)converge and the span of{kw:w∈ _∆}is dense inH²(β)(see [10, Theorem 10]).

Also, we shall denote by Vect_A(x), A ∈ _B(H) and x ∈ H, the linear subspace ofH invariant by Aand generated byx and byVect_A(x)_its closure inH. The spaceCm[y]denotes the linear space of polynomials in y of degree less or equal to m, while C[y] will denote the algebra of all polynomials iny. Finally, fornintegerW_A(Aⁿ)denoted the ideal ofW(A) generated byAⁿ.

Let mention here that if

T=

A X

0 N

.

is an extension of an operatorA∈_B(H)by a nilpotent operatorN∈_B(K) of indexn≥1 andkan integer, then we can show easily that

T^k =

A^k X_k 0 N^k

, where X_k =

k−1 i

∑

=₀

A^k⁻ⁱ⁻¹XNⁱ

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and

(1) A^kX_n−AⁿX_k = X_nN^k

We start with some lemma which we will be useful in the characterization of the reflexivity of an extension of a weighted shift operator by a nilpotent operator. Indeed

Lemma 2.1. Let A ∈ _B(H)be an injective unilateral weighted shift operator such that r₁(A) > 0, N ∈ _B(K)be a nilpotent operator on a separable Hilbert spaceKof index n≥1and T be the extension of A by N defined onH ⊕ Kby

T=

A X

0 N

, where X∈ _B(K). We have W(T) =_C_n₋₁[T]⊕W_T(Tⁿ)and

W_T(Tⁿ) =

B Y 0 0

:B∈W_A(Aⁿ)and BX_n= AⁿY

Proof. The first statement is immediate. We have to prove the second one.

Let

S= B Y

0 0

withB∈W_A(Aⁿ)andBX_n = AⁿY.

SinceB ∈W_A(Aⁿ), then there exists a functionφin H^∞(β)and a sequence of polynomial{p_α}such that B = M_φ and Aⁿp_α(A) −→ M_φ (WOT) see [10, Theorem 3]. Put pα(z) = _∑_ja_j,αz^j, so for 0 ≤ k ≤ n−1 we have hAⁿp_α(A)f₀,f_ki = a_k₋_n,αβ²_k and hB f₀,f_ki = φb(k)β²_k. Therefore,φb(k) = 0 sincea_j,α =_{0 for all}_j<_{0 and}β_k 6=0. Consequently, there exists a function ψinH^∞(β)such thatφ=zⁿψbecauser₁(A)> 0 . Sinceψ∈ H^∞(β), there exists, from [10, Theorem 12], a sequence of polynomials{q_i}_i such that q_i(A)−→M_ψ(SOT). Therefore

q_i(T)Tⁿ−→

B M_ψX_n

0 0

(SOT)

Finally, using the fact thatBXn = AⁿYand the injectivity ofAwe getY = MψX_n. Thusq_i(T)Tⁿ−→S(WOT), this impliesS∈W_T(Tⁿ).

The second lemma can be stated as follow

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(2) T=

A X

0 N

, where X∈ _B(K). If A^∗has a nonzero eigenvalue, then AlgLat(T)⊂ {Tⁿ}⁰.

Proof. Since A^∗ has a nonzero eigenvalue, then the operator Ais reflexive and soAlgLat(A)⊂ {A}⁰see [10]. It is easy to show that, ifSis an operator inAlgLat(T), then

S=

B Y

0 D

whereY ∈ _B(K,H), B ∈ {A}⁰ andD ∈ AlgLat(A). Thus to conclude it suffices to show thatBX_n = AⁿY+X_nD. Indeed, let x ∈ H andy ∈ K. SinceS(x+y)is inVect_T(x+y)and

Vect_T(x+y) =_C_n₋₁[T](x+y) +Vect_A(Aⁿx+X_ny)

there exists a polynomialq(z) = _∑_jb_jz^j in Cn−1[z] depending onx andy such thatS(x+y)−q(x+y)∈ Vect_A(Aⁿx+X_ny). Thusq(N)y= Dyand

Bx+Yy−q(A)x−

∑

j

b_jX_jy ∈Vect_A(Aⁿx+X_ny).

Moreover, by identity (1) we getAⁿ∑jb_jX_jy=q(A)X_ny−X_nDy. Therefore BAⁿx+AⁿYy+XnDy ∈ Vect_A(Aⁿx+Xny). Now, since B ∈ AlgLat(A), then

BX_ny−AⁿYy−X_nDy ∈ ^\

x∈H

Vect_A(Aⁿx+X_ny). Next, letw∈_∆such thatw6=0 and defineψ∈ H²(β)by

ψ=−^λ^w(X_ny) wⁿ k_w.

Thus, it is easy to show that λ_w(Aⁿψ+X_ny) = 0 for all w ∈ _∆− {0}. Moreover, we haveλw(Aⁿψ+Xny) = 0 for allw ∈ _∆ since∆ is an open set in C. This implies that Aⁿψ+X_ny = 0 on span{k_w,w∈_∆}. Con- sequently Aⁿψ+X_ny = 0 because the set span{k_w,w∈_∆} is dense in H²(β). Whence, we get Vect_A(Aⁿψ+Xny) = {0}and thereforeBXny = AⁿYy+X_nDy. This completes the proof.

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(3) T=

A X

0 N

, where X∈_B(K),

e ∈ K, S =

B Y

0 D

such that Nⁿ⁻¹e 6=₀and De = 0. If S ∈ AlgLat(T)and A^∗has a nonzero eigenvalue, then B∈W_A(Aⁿ).

Proof. Suppose thatS∈ AlgLat(T). Since

Se∈_C_n₋₁[T]e+Vect_A(X_ne)

there exists a polynomialq(z) =_∑_jb_jz^jinCn−1[z]depending onesuch that q(N)e = _{0 and}Ye−_∑_jb_jX_je ∈ Vect_A(X_ne). Thus, from the factNⁿ⁻¹e 6=₀ we see thatq=0 and consequentlyYe∈Vect_A(Xne). SinceS∈ AlgLat(T), the operatorB∈ {A}⁰and therefore there exists a functionφinH^∞(β)such thatB= M_φby [10, Theorem 3]. Thus, by applying Lemma 2.2, we get

(4) φg =zⁿh

with g = X_ne and h = Ye. Now, we have to show that φb(k) = 0 for all integer 0≤ k ≤ n−1. Indeed, let 0≤ k ≤ n−1 the smallest integer such thatφb(k) 6= 0, thus there existsϕ∈ H^∞(β)such thatϕb(0) =φb(k)6=0 and φ=z^kϕ. By Equation (4), we getbg(0) =0 and the fact thatgis analytic on the disc D(0,r₁(A))there exists by [10, Proposition 18] an integerm > 0 and a functionξ ∈ H²(β)such that ξ(0) 6= 0 and g = z^mξ. More, since h ∈ Vect_A(z^mξ), the point z = 0 is a root of hof index great or equal m.

Hence there existsψ∈ H²(β)such thath= z^mψand soϕξ =zⁿ⁻^kψby (4).

Hence ϕ(₀)ξ(₀) = 0, a contradiction. Thus, necessarily k ≥ n. Therefore there exists a function f in H^∞(β)such that φ = zⁿf. From [10, theorem 12], we haveσ_i(A)−→M_f (SOT) whereσ_iis the Cesaro means for f. Since the sequence{Aⁿσ_i(A)}_i is bounded, thenAⁿσ_i(A) −→B(SOT) and soB is inW_A(Aⁿ).

Lemma 2.4. Let A be an injective unilateral weighed shift operator on H²(β), n≥1an integer and Q∈_C_n₋₁[z]. If there exists f ∈ H²(β)such that Q(A)f ∈ AⁿVect_A(f)and bf(0)6=0, then Q=0.

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Proof. If f ∈ H²(β) such that Q(A)f ∈ AⁿVect_A(f)and f(0) 6= 0, then there exists a sequence of polynomial {p_i}_i such that M_zⁿ_p_i_f −→ M_{Q f} (SOT). Hence, for any integer q ≥ 0, we have pc_if(q−n) −→ Q fc(q). Since p_if is in H²(β), then pc_if(q−n) = 0 for all 0 ≤ q ≤ n−1 and so Q fc(q) = 0 for every 0 ≤ q ≤ n−1. Moreover, forQ(z) := _∑ⁿ_k₌⁻₀¹b_kz^k we haveQ fc(q) =_∑_kⁿ₌⁻₀¹b_kbf(q−k)and bf(q−k) =0 for allk>q. Therefore

∑

q k=0

b_kbf(q−k) =₀ ∀₀≤ q≤n−_1.

For q = 0, we get b0f(0) = 0 and so b0 = 0. And for q = 1 we obtains b₁f(0) = 0 and sob₁ = 0, and so forth. Thusb_k = 0 for all 0 ≤ k ≤ n−1 and thereforeQ=0.

Remark 2.5. LetA ∈ _B(H)be an injective unilateral weighted shift operator such thatr₁(A)>_0,N∈_B(K)be a nilpotent operator on a separable Hilbert spaceK of indexn ≥ 1 andTbe the extension of AbyN defined onH ⊕ Kby

T=

A X

0 N

, whereX∈_B(K),

e ∈ Ksuch that Nⁿ⁻¹e 6= 0. By [7, Theorem 3], there existF ∈ LatNsuch that K = VectN(e)⊕F. Now proceeding as in the proof of [7, Lemma 9]

and using Lemmas 2.1, 2.2 and 2.3 with the condition A^∗ has a nonzero eigenvalue, we can show that any operatorSin AlgLat(T)can be written in the unique form

S= S₀+ 0 Y

0 D

whereS₀ ∈ W(T),Y ∈ _B(K,H), D ∈ AlgLat(N), De = 0, D|_F = 0 and AⁿY+XnD=0.

3 The main result

Theorem 3.1. Let A ∈ _B(H)be an injective unilateral weighted shift operator such that r₁(A)>0, N ∈_B(K)a nilpotent operator of index n≥1, e ∈ Ksuch that Nⁿ⁻¹e 6=0and

T =

A X

0 N

where X∈_B(K,H)

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an extension of A by N. Then T is reflexive if and only if A^∗ has a nonzero eigenvalue or Tⁿe ∈/ Im(A)or N is reflexive.

Proof. SupposeA^∗has a nonzero eigenvalue and show thatTis reflexive, Indeed, letS ∈ AlgLat(T)ande ∈ K such thatNⁿ⁻¹e 6= 0. According to the previous Remark 2.5 we can suppose

S= 0 Y

0 D

,

whereY ∈ _B(K,H)with D ∈ AlgLat(N), De = 0, D|_F = 0 and AⁿY = X_nD. In the other hand, since N^je ∈ Vect_N(e), there exists for each ja polynomial P ∈ _C[z] depending on jsuch that DN^je = P(N)N^je. Now since S(x+N^je) ∈ Vect_T(x+N^je)for all x ∈ H²(β), then there exists a polynomial Q(z) = _∑_kb_kz^k ∈ _C_n₋₁[z] depending onx,e and jsuch that DN^je=P(N)N^je= Q(N)N^jeand

YN^je−Q(A)x−

∑

k

b_kX_kN^je ∈Vect_A(Aⁿx+X_nN^je). According toAⁿY=XnDand Identity (1), it follows

(5) Q(A)(Aⁿx+X_nN^je)∈ AⁿVect_A(Aⁿx+X_nN^je).

We have to show thatQ = 0. Indeed, this follows easily from Lemma 2.4 whenX_nN^je(0) 6= 0. If X_nN^je(0) = 0, then by [10, Proposition 18], there exist an integermandψin H²(β)such that ψ(0) 6= 0 and XnN^je = z^mψ.

Here, we have to distinguish tow cases. Ifm< n, then from the injectivity ofAand (5) we see

Q(A)(zⁿ⁻^mx+ψ)∈ AⁿVect_A(zⁿ⁻^mx+ψ).

Since(zⁿ⁻^mx+ψ)(0) 6= 0, then Lemma 2.4 impliesQ = 0. Ifm ≥ n, then by (5) we get

Q(A)(x+z^m⁻ⁿψ)∈ AⁿVect_A(x+z^m⁻ⁿψ).

Since P is independent on x, then the polynomial Q is also independent onx. Sincexis arbitrary in H²(β), then we can choose it such thatx(0)∈/ {0,−ψ(0)}. HenceQ= 0 since(x+z^m⁻ⁿψ)(0)6=0. ThereforeD =0 and soY = 0 by the injectivity ofAand the equationAⁿY = XnD. This yields S=0 and soTis reflexive according to Remark 2.5.

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Suppose thatNis reflexive orTⁿe∈/ Im(A)and show thatTis reflexive.

ifS∈ AlgLat(T), then, by the remark 2.5, we can suppose that S=

0 Y 0 D

,

whereY ∈ _B(K,H)with D ∈ AlgLat(N), De = 0, D|_F = 0 and AⁿY = XnD.

IfNis reflexive, there existppolynomial such thatD= p(N)and there- foreD=0. From whereS=_0.

Suppose Tⁿe ∈/ Im(A). Since D ∈ AlgLat(N), there existe for all integer 0 ≤ k ≤ n, p_k polynomial such that DN^ke = p_k(N)e. The fact that p_k(T)Tⁿ = Tⁿp_k(T)_{we get} p_k(A)X_ne−X_np_k(N)e ∈ ImAⁿ and since AⁿY+XnD=0, we obtainXnp_k(N)e= XnDN^ke= AⁿYN^ke, this is imply, for all integer 0≤ k≤ n, p_k(A)X_ne∈ Im(Aⁿ)and thereforezⁿDivided p_k, thusDN^ke=0, thereforeD=0 and consequentlyS=_0.

Conversely, suppose A^∗ has no nonzero eigenvalue andTⁿe ∈ Im(A) andNis not reflexive. The fact that Tⁿe ∈ Im(A), there is a vectory₀ ∈ H such that Tⁿe = Ay₀ and by [Theorem 3 [1]] the order of nilpotency pof N|_Fless than or equal ton−2. Consider the operatorD ∈ _B(K)defined by

D|_F=0;De=0;DN^ke=0 ifk≥n−pandDN^ke = Nⁿ⁻¹eif 1≤k ≤n−1−p.

Note that D ∈/ W(N)and D ∈ AlgLat(N) by [Theorem 3 [1]]. LetY ∈ B(K,H)defined by

Y|_F =0;Ye=0;YN^ke=0 ifk≥n−pandYN^ke= Xn−1e−y0if 1≤k ≤n−1−p.

Thus obtainedAⁿY+X_nD=0. LetS∈ B(H ⊕ K)defined by S=

0 Y

0 D

It is clear thatS ∈/ W(T)sinceD ∈/ W(N). Show thatS ∈ AlgLat(T). Let x ∈ H _andy ∈ K_{. Let} ppolynomial depends onysuch thatDy = p(N)y.

SinceAⁿY+XnD=0 we haveTⁿS=0 and

Tⁿ(S−P(T))(x+y) =−Tⁿp(T)(x+y)∈Vect_T(x+y).

In the fact thatr₁(A) > 0, I ∈ W_A(A)by [[10], proposition 19 ] and therefore

(S−P(T))(x+y)∈Vect_T(x+y),

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thusS ∈ AlgLat(T)and therefore T is not reflexive. This completes the proof of the theorem

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