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TOEPLITZ OPERATORS IN SEGAL-BARGMANN SPACES OF VECTOR-VALUED FUNCTIONS

DARIUSZ CICHO ´N and HAROLD S. SHAPIRO

Abstract

We discuss new results concerning unbounded Toeplitz operators defined in Segal-Bargmann spaces of (vector-valued) functions, i.e. the space of all entire functions which are square summable with respect to the Gaussian measure inCn. The problem of finding adjoints of analytic Toeplitz operators is solved in some cases. Closedness of the range of analytic Toeplitz operators is studied.

We indicate an example of an entire function inducing a Toeplitz operator, for which the space of polynomials is not a core though it is contained in its domain.

1. Introduction

We begin by introducing basic definitions and notations. LetH be a complex separable Hilbert space. ConsiderL2(µ)⊗H, the Hilbert space of all complex Borel functions taking values inH which are square-integrable on Cn with respect to the measureµgiven bydµ(z) = π−ne−z·zdV (z), whereV is the Lebesgue measure inCnandz·z= |z1|2+· · ·+|zn|2forz=(z1, . . . , zn)∈Cn. The inner product inL2(µ)H is given by

f, g(H) =

Cn f (ζ ), g(ζ )H dµ(ζ), f, gL2(µ)H, where ·,·H stands for the inner product inH. The norm induced by the above defined inner product is denoted by·(H)and in caseH =CN by·(N). The Segal-Bargmann spaceBnH (abbreviatedBH) is a closed subspace ofL2(µ)H composed of all entire functions belonging toL2(µ)H. By PIH we mean the orthogonal projection ofL2(µ)H ontoBH. We will use the following identifications: L2(µ) = L2(µ)⊗C, B = B ⊗C, ·,· = ·,·(1),· = ·(1) andP = PIC. Observe thatB⊗CN can be identified withB. . .⊕B(N-times), and following this we haveB⊗CN = {(f1, . . . , fN):f1, . . . , fNB}and(f1, . . . , fN)2(N)=

Cn(|f1|2+ · · · +

|fN|2) dµ.

The first named author is supported by KBN under grant no. 2 P03A 004 17.

Received October 1, 2001; in revised form June 4, 2002.

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Givenf : Cn → CandhH we define(fh)(z) := f (z)h,z ∈Cn. Put ea(z) := ez·a, where z· a := n

k=1zkak and a := (a1, . . . , an) for a = (a1, . . . , an) ∈ Cn and z = (z1, . . . , zn) ∈ Cn. Denote by E (=En) the linear span of the set of functions{ea : a ∈ Cn}. It can be checked that f (z), hH = f, ezh(H) forfBH,hH andz ∈ Cn (cf. [4]), which is referred to as the reproducing property forBH. ByP (=Pn) we denote the space of all analytic polynomials inCn. BothP andE are dense subsets of B. The sequence fk(z) := zk/

k!, k ∈ Nn, z ∈ Cn, forms an orthonormal basis forB, where according to the standard multiindex notation zk := zk11 ·. . .·zknn, k! := k1. . .·kn! and |k| := k1+ . . .+kn for all z=(z1, . . . , zn)∈Cnandk=(k1, . . . , kn)∈Nn(N= {0,1,2, . . .}).

Ais called a linear operator in a complex Hilbert space H if its domain D(A)is a linear subspace ofH andA: D(A)H is a linear mapping. As usual,A, A, R(A)stand for the adjoint (provided it exists), the closure of Aand the range ofA, respectively. We say that a linear subspaceDD(A) is a core for a closable operatorAif(A|D) = A. All these notions are also meaningful for linear operators acting between different Hilbert spaces. By B(H,K)(H,K-Hilbert spaces) we denote the space of all bounded linear operators A : HK, andB(H) := B(H,H). The graph norm·A is defined byf2A = f2H + Af2H,fD(A).

WritePH for the space of polynomials (inCn) taking values inH, i.e.

the totality of functions of typeK

j=0pjhj, wherepjPn,hjH and K ∈ N(in the case whereH = CN the elements of this space are identified withN-tuples of polynomials). SpacesEH andPB(H,K)are defined analogously. Observe thatpPB(CM,CN)can be written as a matrix [pkl]l=k=11,...,M,...,N with some polynomials pklP. Given pPB(H,K), p(z)=

|j|<KAjzj,AjB(H,K),z ∈ Cn, definep#PB(K,H) via p#(z) = p(z), z ∈ Cn, and a differential operator (p(D)F)(z) =

|j|<KAjDjF(z), z ∈ Cn, where F : CnH is an analytic function (hereDj = ∂zj1|j|

1 ...∂zjnn forj = (j1, . . . , jn)). We will regardp(D)as an op- eratorin BH and its domain is defined as D(p(D))= {F ∈ BH : p(D)FBK}.

Letϕ : CnB(H,K)be an analytic function. Defineϕ#(z) := ϕ(z), which amounts to taking adjoints of all coefficients in the series expansion of ϕ, andϕ(z) := ϕ(z) forz ∈ Cn. A Toeplitz operator1 with symbolϕ is defined byD(Tϕ) = {f ∈ BH : ϕfBK}and Tϕf = ϕf for fD(Tϕ), where(ϕf )(z) =ϕ(z)f (z),z∈Cn. We need one more operator

1What we have defined here is, in fact, ananalyticToeplitz operator, cf. [4] for the general case.

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inBH denoted byΠϕ, which is defined via Πϕf (z)=

Cnϕ(ζ)f (ζ )ez·ζdµ(ζ), z∈Cn,

and its domain consists of allf :CnK such that the above integral exists2 for allz∈Cnand the function inzdefined by it lies inBH. SinceΠp = p#(D)forpPB(H,K)(cf. [4]), we may regardΠϕ as a generalized differential operator. It is known thatΠp =(Tp|E⊗H)=(Tp|P⊗H)(cf. [4, Corollary 7.3]). The following abstract lemma reflects the connection between TpandΠp.

Lemma1.1. LetAbe a closed operator inH andVD(A)be a dense subspace ofH. Then forB :=(A|V)the following conditions are equivalent:

(i) A=B;

(ii) ker(I+AB)= {0}; (iii) ABis symmetric;

(iv) ABis non-negative;

(v) ABis selfadjoint;

(vi) V is a core forA.

InterchangingAandBin(i)–(v)gives another condition equivalent to(vi).

Proof. Observe that AB and B is closed. Both these facts are in frequent use in the following proof. We begin with (i)⇔(ii). Let%(A)denote the graph of A, i.e. the set {(x, y) : xD(A), Ax = y}. Then one can easily show that%(B)%(A)consists of all (x, y)such thatxD(B), y = BxD(A)andx +Ay = 0, which immediately implies the desired equivalence. Equivalence (i)⇔(vi) is implied by taking adjoints in (i). By standard operator theory (i) implies (iii), (iv) and (v). Since each of (iv) and (v) implies (iii) by definition, it suffices to prove implication (iii)⇒(i). Observe thatAAis a selfadjoint operator contained inAB. Hence by (iii)AA=AB, which implies (ii) and, consequently, (i). So the equivalence of conditions (i)–(vi) is established.

Condition (i) may be rewritten asA = B. Thus the last statement of the assertion follows.

Theadjointness hypothesisconsists in equalityTp = Πp forpPB(H), which was proved in some cases, e.g.pPn, cf. [10], orpP1B(H)and its leading coefficient is surjective, cf. [4]. In the present paper we present new results concerning the hypothesis.

2i.e. it is weakly convergent, cf. [4]

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Toeplitz operators of type Tp as above, and their adjoints, play an im- portant role in extending known results on partial differential operators and convolution operators in the space of entire functions to the context of Segal- Bargmann spaces. Suppose for simplicity we consider one-variable Segal- Bargmann spaceB. In the classical theory of linear differential equations of infinite order, with constant coefficients (one variable), one starts with a “sym- bol function” ϕ, which is entire and of exponential type, and the thematic problem is to show that every entire functionuwhich satisfies the equation ϕ(D)u=0 is, in the sense of uniform approximation on compact sets, the limit of a sequence each term of which is a finite linear combination of “monomial exponential” solutions of this same equation, that is solutions of the form m(z)=zrexp(wz), wherew∈Candr ∈N. There are various ways, all equi- valent, to define the operatorϕ(D), for example as an infinite order differential operator, or as convolution with a certain compactly supported measure onC, whose Fourier-Laplace transform isϕ. This theory originated in the study of difference and differential-difference equations, and there is a solid treatment of this e.g. [6], for more modern versions see e.g. [5].

A more refined question (arising from Delsarte’s notion of mean periodic functions, originally in the context of functions of a real variable but extended to the holomorphic category by Laurent Schwartz and others) is this: Givenu entire such that its translates (equivalently its derivatives) fail to span Hol(C) (the entire functions), which is equivalent to the existence of some nontrivialϕ of exponential type such thatϕ(D)u=0, do those translates none the less span enough monomial exponentials to, in turn, span (“synthesize”)u? This is the spectral synthesis question for Hol(C). It is known that this question too has an affirmative answer; in higher dimensions however there are counterexamples to the spectral synthesis version, whereas the positive solution to the versions in the preceding paragraph holds in all dimensions.

Using this as a guide, one can formulate analogous questions in other topo- logical vector spaces of entire functions. For example, inBlet us ask whether the solutions toϕ#(D)u = 0 are in the closure of the monomial exponential solutions. Here we may take as our definition ofϕ#(D)the operatorΠϕ. It is no longer necessary to supposeϕof exponential type, we get a sensible problem whenever the product ofϕby each exponential is inB; however, to fix ideas think first of the case whereϕis of exponential type; even here the question to be posed is still unsolved. The monomial exponentials in question consist precisely of, for each zerowofϕ, the functionszkexp(wz),k =0, . . . , r−1 whereris the multiplicity of the zero.

The question whether these span all solutions toϕ#(D)u= 0 in thenorm topologyofB, then translates, by duality, to this: supposehinBis orthogonal to the monomial exponentials belonging to kerϕ#(D); this is equivalent to it

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vanishing on all zeros of ϕ, with the appropriate multiplicities, i.e. to the assertion: h/ϕ is an entire function. We ask if h must then be orthogonal to each u that is, in turn, orthogonal to allϕ(z)exp(tz) with t ∈ C. Since h = ϕF for some entire F, the question is whetherϕF is approximable in B-norm by exponential multiples of ϕ. And this is just the question: do the finite linear combinations of exponentials span every entire function, in the metric ofL2(|ϕ|2dµ)?

In [10] an affirmative solution was given forϕwhich are exponential poly- nomials, and indeed in any number of variables. But, for other classes of entire functionsϕthis is unsolved.

The counterexample of Borichev [1] (and our version of it below, in Sec- tion 4) are a borderline case: they can be said to disprove spectral synthesis for some “symbol function”ϕof very large growth – so large, indeed, that the equationϕ#(D)u= 0 has to be interpreted in a generalized sense, namelyu (inB) is orthogonal to all thepolynomialmultiples ofϕ(this is all we have to work with, since theexponentialmultiples, needed for the natural definition, do not lie inB). So, it is of interest to know whether a corresponding counter- example exists whereϕis of smaller growth, at least multiplying exponentials intoB.

2. Adjointness

In what follows-(w, R) ⊆ Cn stands for the open polydisk with radiusR centered atw. LetPk : Cn → C, k = 1, . . . , n, denote projection onto the k-axis, i.e.Pk(z):=zk for allz=(z1, . . . , zn)∈Cn.

Lemma2.1. LetXbe a Borel measurable set such thatPk(X)is bounded for somek ∈ {1, . . . , n}andR:=inf{r >0 :Pk(X)is contained in-(w, r) with somew}. Then there exists a constantCdepending only onRsuch that (2.1)

Cn|f|2C

Cn\X|f|2 for all entire functionsf :Cn→C.

Proof. First consider the one-dimensional case. FixR > 0, w ∈ Cand an entire functionf defined onC. PutI (r):=2π

0 |f (reit +w)|2dt,r ≥0.

SinceI (r)is increasing with respect tor we have

re−r2I (r)(R+r)e3R2e−(R+r)2I (R+r), r ∈[0, R].

Both sides of this inequality are functions inr, so integrating over the interval

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[0, R] (with respect to the Lebesgue measure) yields (2.2)

-(w,R)|f|2e3R2

-(w,2R)\-(w,R)|f|2dµ.

Adding

C\-(w,R)|f|2to both sides of (2.2) implies the assertion withX= -(w, R)andC:=e3R2+1.

We now proceed with the multi-dimensional case. Assume thatXsatisfies the assumptions of the lemma with somek. LetRbe the constant defined in the statement of Lemma, thenPk(X)-(w, R+1)with somew∈ C. Let f :Cn→Cbe an entire function. SetYk := {(z1, . . . , zn)∈Cn:|zkw|>

R+1}. An easy application of the one-dimensional case leads to

Cn|f|2C

Yk

|f|2dµ,

withC :=e3(R+1)2+1. SinceYk ⊆Cn\Xthe result follows.

Lemma 2.2. Given fB1, pP1\ {0} and 4 > 0 there exists a polynomialqP1such that f −q ≤ 4 andf = q (modp), i.e. there existsh∈Hol(C)satisfyingfq=ph.

Proof. We first state the following auxiliary fact:

Lemma2.3. LetpP1\ {0}and letρ >0be such that-(0, ρ)contains all zeroes ofp. Then there exists a constantC >0depending only onpandρ such that for everyF ∈Hol(-(0, ρ))one can find a uniqueQP1such that

(i) degQ≤degp−1, (ii) F−Qp ∈Hol(-(0, ρ)),

(iii) supz∈-(0,ρ)|Q(z)| ≤Csupz∈-(0,ρ)|F(z)|.

To prove this suppose thatz1, . . . , zr are distinct zeroes ofpwith multi- plicitiesm1, . . . , mr (resp.). Observe thatm:= m1+. . . mr =degp. By the Cauchy inequality there existsC1>0 (depending onρ) such that

|g(l)(zj)| ≤C1 sup

z∈-(0,ρ)|g(z)|, l =1, . . . , mj,

for everyj ∈ {1, . . . , r}andg ∈Hol(-(0, ρ)). We can find another constant C2>0, which depends only onpandρ, such that

sup

z∈-(0,ρ)|h(z)| ≤C2max{|h(lj)(zj)|:lj =1, . . . , mj andj =1, . . . , r},

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for every hP1 with deghm−1, because by the Lagrange-Hermite interpolation the right-hand side of the above inequality defines a norm in the space of polynomials of degree at mostm−1.

Choose arbitraryF ∈Hol(-(0, ρ)). Applying the Lagrange-Hermite inter- polation we infer that there exists a uniqueQP1of degree at mostm−1 such that

Q(l)(zj)=F(l)(zj), l =1, . . . , mj.

for everyj = 1, . . . , r. HenceQsatisfies (i) and (ii). By the choice of C1

andC2we see that (iii) holds withC := C1C2. This completes the proof of Lemma 2.3.

We now turn to the proof of Lemma 2.2. FixfB1,pP1andδ >0.

Choose a polynomialg such thatf −g ≤ δandρ > 0 such that-(0, ρ) contains all zeroes ofp. SetF :=fg. Then, by the reproducing property forB,|F(z)| ≤δe|z|2/2,z∈C, and, consequently,

|F(z)| ≤δeρ2/2, z-(0, ρ).

We can now apply Lemma 2.3 to obtain the polynomial Q, degQm−1 (herem:=degp), such thatF−Qp is analytic in-(0, ρ)(hence entire) and

|Q(z)| ≤Cδeρ2/2, z-(0, ρ), withC >0 depending only onpandρ. This implies that

|Q(z)| ≤Cδeρ2/2max |z|

ρ m−1

,1

, z∈C.

This inequality leads toQ ≤ Cδ with a new constantC > 0 depending only onpandρ. Settingq :=g+Qwe obtain a polynomial, for whichfq is divisible byp. Moreover,

f −q = fgQ ≤ fg + Q ≤δ(1+C).

Sinceδcan be made arbitrarily small, the result follows.

Theorem2.4. LetpP1⊗B(CM,CN)withM > N ≥1andfD(Tp). Assume thatp(z)is of maximal rank for at least one point z ∈ C. Suppose that Tpf = 0. Then for any 4 > 0 there exists qP1⊗CM such that f −q(M)4andTpq=0.

Proof. We can represent p by means of the matrix [pkl]l=k=11,...,M,...,N, where pklP1. Putp˜ :=[pkl]l=k=11,...,N,...,N. By assumption on the rank ofp, rearranging columns if necessary, we can assume thatd(z):= detp(z)˜ is not identically

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zero; moreover, it can be done so that degdis maximal among degrees of all possible minors of dimensionN, which arise from the matrix representingp. Given 40 > 0 by Lemma 2.2 we can find polynomials qN+1, . . . , qMP1 such that fj = qj (modd) and fjqj < 40, j = N +1, . . . , M. By the Cramer formulas there exist rational functions q1, . . . , qN such that Tp(q1, . . . , qM) = 0. Thus every qj, j = 1, . . . , N, can be expressed by qN+1, . . . , qMvia

(2.3)

qj =d1det



p11 . . . p1j−1M

l=N+1p1lql p1j+1 . . . p1N

... ... ... ... ...

pN1 . . . pN j1M

l=N+1pNlql pN j+1 . . . pNN



forj =1, . . . , N.

Observe that allqj’s are polynomials! Indeed, one can verify that the differ- encefj−qjis an entire function. To see this it suffices to check that the formula (2.3) holds ifqj is replaced byfj,j = 1, . . . , M, in both members of (2.3).

Then it turns out that for eachj ∈ {1, . . . , N}functionfjqjis a linear com- bination with polynomial coefficients offlql,l =N+1, . . . , M, divided by d. By the choice ofqjthe differencefl−qlis divisible byd,l=N+1, . . . , M. Thusfjqjis entire and, consequently,qjis entire. Since every entire rational function has to be a polynomial, we deduce that everyqj is a polynomial.

It remains to estimate the norm offjqj forj = 1, . . . , N. Note that by (2.3) for fixedj ∈ {1, . . . , N}we get

fjqj = M l=N+1

αlMl

d (flql),

where Ml is a properly chosen minor of dimensionN in p (obviously, Ml

depends onj, which has been omitted in the notation) andαlis equal to 1 or

−1. Choose R > 0 such that all zeros ofd lie in-R (= -(0, R)). By the assumption on degd we infer that every quotient Mdl is bounded outside-R. Setc˜l =supMd(z)l(z):z∈C\-R

. Then

|fj(z)qj(z)| ≤ M l=N+1

˜

cl|fl(z)ql(z)|, z∈C\-R.

Taking squares, integrating overC\-Rand applying Lemma 2.1 yield fjqj2CR402(MN) M

l=N+1

˜ cl2,

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whereCR is some positive constant. If40 is small enough, then polynomial q:=(q1, . . . , qM)satisfies all the required conditions.

Note. The authors thank John McCarthy for a valuable suggestion in connection with the proof of Theorem 2.4.

Theorem2.5. LetpP1B(CM,CN). ThenTp=Πp.

Proof. It suffices to check the density ofP1⊗CM inD(Tp)with respect to the graph norm ·Tp. Choose fD(Tp). We are going to show that there existsqP1⊗CM such that f − qTp is arbitrarily small. Define

˜

pP1B(CM+N,CN) byp(z)˜ = [p(z),−IN], whereIN stands for the identity matrix of dimensionN. It is easily seen thatp˜ satisfies the assump- tions of Theorem 2.4. SinceTp˜(f, Tpf ) = 0, given4 > 0 we can findq˜ = (q1, . . . , qM+N)P1⊗CM+N such thatTp˜q˜ =0 and(f, Tpf )− ˜q(M+N)4. Settingq :=(q1, . . . , qN)we see thatTpq =(qN+1, . . . , qN+M), as a con- sequence of the equalityTp˜q˜ =0, andf−q2Tp = (f, Tpf )− ˜q2(M+N)42. It seems that the case of homogeneous polynomials is much easier to deal with, even in the case of several complex variables and values being operators on (possibly) infinite-dimensional Hilbert spaceH.

Proposition2.6.IfpPB(H)is a homogeneous polynomial then Tp =Πp.

Proof. The main idea of the proof is to writeB⊗H as an orthogonal sum of subspaces reducingp#(D)Tpto a non-negative operator. LetFkPH denote the space of homogeneous polynomials of degree k (with the zero polynomial included). Obviously,BH =

k=0Fk. It is easily seen that p#(D)Tp(Fk)Fk for allk ≥ 0. Hence, eachFk reducesp#(D)Tp. Pick arbitraryfFk and compute

p#(D)Tpf, f(H)=

Cn p#(D)(pf )(z), f (z)Hdµ(z)

=

Cn

Cn p(ζ )p(ζ )f (ζ), f (z)Hez·¯ζdµ(ζ ) dµ(z).

The last equality follows from [4, Lemma 2.3], but the reader may obtain it applying the theorem on differentiating under the integral sign. We may now change the order of integration, which is allowed because the function under the integral sign is summable with respect toµµ. Then it suffices to apply the reproducing property forBH to see that

p#(D)Tpf, f(H)= ppf, f(H)≥0.

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Since3p#(D)Tp=

k=0p#(D)Tp|Fkwe infer thatp#(D)Tpis non-negative, so by Lemma 1.1 the assertion follows.

Remark2.7. The above prove works also for the wider class oft-homo- geneous polynomials(cf. [9]).

Proposition2.8.Assume thatp = [pkl]k,l=1,...,NPB(CN)is such that allpkl are homogeneous. Ifdegpkldepends only onl, thenTp =Πp.

Proof. (Based on an idea of J. Janas.) We are going to prove thatTpp#(D) is non-negative. LetFk,jP ⊗Cj denote the space of homogeneous poly- nomials of degreekfork≥0. Observe thatB⊗CN =

k=0Fk,N. We claim that eachFk,N is invariant underTpp#(D)(hence reducing for this operator).

OperatorTpp#(D)|Fk,N : Fk,NB ⊗CN may be written as an operator matrix [Akl]k,l=1,...,N, where

Akl = N j=0

Tpkjp#lj(D), D(Akl)=Fk,1.

By assumption onp we see that iff is homogeneous of degreek then so is Aklf.

It now remains to show thatTpp#(D)|Fk,N is non-negative. Note first that PICN(pf )(z)=

Cnp(ζ )f (ζ )ez·¯ζdµ(ζ)

=p#(D)f (z)

, z∈Cn, fFk,N,

for allk ≥0 (cf. [4, Lemma 2.3]). Thus

Tpp#(D)f, f(H) = pp#(D)f, f(H) = p#(D)f, PICN(pf )(H) ≥0 for allfFk,N andk≥0. Applying Lemma 1.1 completes the proof.

One of the natural questions which arise when studying the adjointness hypothesis is whether it is possible to find the solution to this problem only by means of coefficients ofp. This idea is presented in the following theorem.

The set of operatorsAB(H)is calledjointly subnormalif there exists a setAB(K)of commuting normal operators defined on a larger Hilbert space containing H as a closed subspace such that for everyAA there existsAAsatisfyingA=A|H.

3 Note thatp#(D)Tpis closed.

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Theorem2.9. Suppose thatp=N

k=0pk⊗AkPB(H)withpkP andAkB(H). If the sequence{Ak}Nk=0is jointly subnormal, thenTp=Πp.

The proof of this theorem requires the following lemma

Lemma2.10. Suppose thatqP⊗B(H)takes values only among normal operators,fD(Tq)andgBH. Then the function

Cn×Cn (z, w)→ g(w), q(w)f (z)ez·w ∈C is summable with respect to the measureµµinCn×Cn.

Proof. We begin with a change of variablesw=z+ζ, which yields

Cn

g(w), q(w)f (z)ez·wdµ(w)=

Cn

g(z+ζ), q(z+ζ)f (z)e−z·ζdµ(ζ ) Thus

J :=

Cn

Cn

g(w), q(w)f (z)ez·wdµ(w) dµ(z)

Cng(· +ζ)e−ζ(H)q(· +ζ)f(H)dµ(ζ).

Butg(· +ζ )e−ζ(H) = e12ζ2g(H), which together with the normality of q(z)for allz∈Cnimplies that

J ≤ g(H)

Cnq(· +ζ)f(H)e12ζ2dµ(ζ).

Applying the isometry theorem (cf. [10], [4]), which states that4 ph2(H)=

j≥0

1

j!(Djp#)(D)h2

(H), pPB(H), hBH,

we infer thatq(· +ζ )f(H)is of polynomial growth, soJ <∞. Proof of Theorem2.9. Letq(z) := N

j=0pjBj, where{Bk}Nk=0 is the sequence of commuting normal operators defined on a larger Hilbert space K such thatAk = Bk|H,k = 0, . . . , N. Thusq takes its values only among normal operators and, consequently,Tq = q#(D), according to [4, Theorem

4The symbol “

j≥0” should be read “sum over all multiindicesj”. In the following equality we putF(H)= ∞wheneverF /L2(µ)H

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7.8]. We will show that p#(D)Tpf, f(H) ≥ 0 for everyfD(p#(D)Tp). Observe that

p#(D)Tpf, f(H)=

Cn p#(D)(pf )(z), f (z)dµ(z)

=

Cn q#(D)(pf )(z), f (z)dµ(z)

=

Cn

Cn q(w)p(w)f (w), f (z)ez·wdµ(w) dµ(z) (2.4).

We know that pfBH, which implies that qfBK, because p(z)= q(z)|H for allz∈Cn. By Lemma 2.10 we are allowed to change the order of integration in (2.4). So we arrive at

p#(D)Tpf, f(H) =

Cn

Cn q(w)p(w)f (w), f (z)ez·wdµ(z) dµ(w)

=

Cn q(w)p(w)f (w), f (w)dµ(w)

=

Cnp(w)f (w)2dµ(w)≥0.

Thus ker(IH +p#(D)Tp)= {0}, which implies thatTp=p#(D).

Remark 2.11. If the propertyTp = Πp is proved for all polynomials, whose coefficients commute, then it can easily be shown for arbitrary poly- nomialp. To see this pick arbitrarypPB(H)and define polynomial qPB(HH)via

q(z)=

0 0 p(z) 0

, z∈Cn.

Note thatq(z)can be written as

|j|≤NBjzj,z∈Cn, whereBjB(HH) andBjBk =0 for all admissiblej andk. Suppose thatTq =q#(D). Then for anygD(Tq)one can find a sequence{hj}j=0inP(HH)such that hjgandqhjqg. One can readily check thatq(z)(f⊕g) = p(z)f for allz∈Cnandf, gH.

Pick arbitraryfD(Tp). Thenf ⊕0 belongs to the domain ofTq. Let {hj}j=0P(HH)be the sequence chosen so thathjf ⊕0 and qhjq(f⊕0). LetPH0be the orthogonal projection ofH⊕H ontoH⊕0.

By the choice ofqit is apparent that the sequence{PH0hj}j=0P⊗Htends

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tof in the graph norm ofTp. ThusPH is a core forTpand, consequently, Tp =p#(D).

3. Closedness of range

LetpPB(CN). In what follows, detpstands for the polynomial defined in the natural way as(detp)(z) = det(p(z)), andpklB(CN−1)originates frompby removing thek-th row and thel-th column.

Theorem3.1. Assume thatpP1B(CN),p=0. ThenTpis bounded below if and only ifdeg detp≥maxk,l=1,...,Ndeg detpkl≥0.

Proof. The “only if” part of assertion was proved in [4, Proposition 8.6].

It remains to consider the “if” part. Suppose that p = [pkl]k,l=1,...,N with pklP1 and Tp(f1, . . . , fN) = (g1, . . . , gN) for (f1, . . . , fN)D(Tp), which means that

p11 . . . p1N

... ...

pN1 . . . pNN

f1

f...N

=

g1

g...N

.

Note that by assumption detpis not identically 0, so the above equation may be solved forf1, . . . , fN. ChooseR > 0 such that all zeroes of detp lie in -(0, R). Hence, by the Cramer formulas, we infer that

|fj(z)| ≤sup

detpkl(ζ ) detp(ζ )

:k, l =1, . . . , N,|ζ| ≥R

(|g1(z)|+. . .|gN(z)|)

for allj = 1, . . . N andz∈C\-(0, R). LetM denote the supremum in the above inequality. Then it follows that

C\-(0,R)

N j=1

|fj|2MNN

j=1

gj2.

Applying Lemma 2.1 yields

Tp(f1, . . . , fN)(N)= (g1, . . . , gN)(N)C(f1, . . . , fN)(N)

with an appropriate constantC >0 depending only onpandR. This means thatTpis bounded below.

Theorem 3.1 is the one variable refinement of [4, Proposition 8.6], which gives the necessary condition for boundedness below ofTpin the multivariable case. Although it does not seem likely one can obtain the multivariable version of the above theorem we have not been able to give any counterexample.

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The following proposition deals with non-injective operators with closed range.

Proposition3.2.LetpP1B(CN,C). ThenR(Tp)is closed.

Proof. Assume thatp = [p1, . . . , pN] andq is the greatest common di- visor of allpj’s,j = 1, . . . , N. We may choosepsuch thatpN =0, which involves no loss of generality. We are going to establish the following descrip- tion of the range:

R(Tp)=

gB1: g

q extends to an entire function

.

SinceTp(f1, . . . , fN)=N

j=1pjfjthe inclusion “⊆” is obvious. To prove the reverse inclusion pickgB1such thatgq is entire. By easy algebra there exist polynomialsq1, . . . , qj such thatN

j=1qjpj = qr, wherer is a polynomial chosen so thatg−qrp

N is entire (use the interpolation property). Put fj :=qj for j =1, . . . , N−1, and fN := gqr

pN +qN. It is easily seen thatfNB1andTp(f1, . . . , fN)=g, thusgR(Tp).

Observe thatR(Tp)=R(Tq), whereTqis Toeplitz operator defined inB1. By the Newman-Shapiro Isometry Theorem [10], [4] operatorTq is bounded below, henceR(Tq)is closed.

We now turn to an example showing that the above proposition is not true in the multivariable case.

Example 3.3. R(Tp) need not be closed even in case of homogeneous polynomials. Define p = [p1 p2], wherep1(z, w, ζ ) = z3w3ζ6 and p2(z, w, ζ )= w6,(z, w, ζ)∈ C3. Observe thatR(Tp)contains all functions of typew3ζ6q, whereqis an arbitrary polynomial depending only onz. This follows fromTp(−w3q, z3q) = w3ζ6q. Suppose R(Tp)is closed. Then it follows that the closure of{w3ζ6q :q is a polynomial depending only onz}is contained inR(Tp). But this closure is equal to{w3ζ6f :fB3, f depends only onz}, which is an immediate consequence ofw3ζ6q2=3!6!q2, with q depending only onz. Putf0(z) = (ez2/2−1)(zz1)1(zz2)1, where z1, z2 are two different zeroes of function ez2/2−1. Observe thatf0B1. Hence, there aref, gB3such thatTp(f, g)(z, w, ζ )=w3ζ6f0(z), or more explicitly

(z3w3ζ6)f (z, w, ζ )+w6g(z, w, ζ)=w3ζ6 ez2/2−1

(zz1)(zz2), z, w, ζ ∈C.

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Fix arbitraryx >0. Substitutingz:=x,w:= 1x andζ :=1 we get g

x,1

x,1

=x3 ex2/2−1

(xz1)(xz2), x >0.

This leads to a contradiction, because the left hand side of the above equality can be estimated from above bygexpx2

2 + 21x2 + 12

,x > 0, whereas this is impossible for the right hand side. So we have proved that the range ofTp

is not closed.

By a similar argument one can show that the range of operator Tq with q(z, w, ζ )=[q1q2],q1(z, w, ζ)=z3w3ζ6,q2(z, w, ζ )=w4,(z, w, ζ )∈ C3, is not closed. Observe that degq1 =degq2in opposition to the previous example. Hence, if we drop the assumption on equality of degrees of coordinate polynomials, then it does not follow that the range ofTp is closed in case of several complex variables.

4. The non-density example

Below we show an example of an analytic functionϕ for which polynomials do notform a core for the operatorTϕthough they are contained in its domain.

In other words: we will indicate an analyticϕfor which the space of all entire functions square-integrable with respect to the measure(1+|ϕ|2) dµcontains polynomials as a non-dense subset. Before we go into details we need the following theorem, which is of independent interest.

Theorem4.1. Let{zk}k=1⊆ Cbe a sequence of non-zero numbers satis- fyingzk+1

zkλfor allk ∈Nwith someλ >1. Then the formula ψ(z):=

k=1

1− z

zk

, z∈C,

defines an entire function for which there existsc >0such thatTψfcf for allfD(Tψ).

Proof. Since we want to impose some additional conditions on the given sequence{zk}k=1⊆C, we will show that it suffices to deal only with a modi- fication of this sequence obtained by removing a few initial terms. Indeed, if the theorem is proved forψ1(z):=

k=1

1− zk+jz

with some integerj ≥0, then applying the boundedness below ofTpwith the polynomialp:= ψψ1 we get Tψf = Tpψ1fc1ψ1f, fD(Tψ),

wherec1is a positive constant. Thus ifTψ1 is bounded below, then so isTψ.

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From now on we are going to assume that|z1| ≥1,

(4.5) λ≥ 3

|zk|+1, k =1,2, . . . , and

(4.6) |zkzj|>4, k =j,

which can be easily obtained by omitting a suitable number of initial terms in the sequence{zk}k=1.

Denote by-(w, r)the open disk with radiusrcentered atw∈C. We claim that there exists a positive constantasuch that

(4.7) |ψ(z)| ≥a for z∈C\ k=1

-(zk,1).

To prove this we first show that there existsa >0 such that

(4.8) |ψ(zk+ζ)| ≥a whenever |ζ| =1 and k=1,2, . . . . By assumption (4.6) the unit disks centered at thezk’s are disjoint so

min{|ψ(zk+ζ)|:k =1,2,|ζ| =1}>0. Fixk ≥3 and observe that

ψ(zk+ζ )=

1− zk+ζ z1

k−1 j=2

1− zk+ζ

zjζ

zk

j=k+1

1− zk+ζ zj

We are going to find uniform (ink) estimates for terms appearing in the above equality. We have

1− zk+ζ

z1ζ

zk

≥ |zk| − |z1| −1

|z1zk|

≥ 1

|z1|− 1

|z2|− 1

|z1z2|

>0.

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The next term is estimated as follows k−1

j=2

1− zk+ζ zj

= k−2

j=1

1− zk

zk−jζ zk−j

k−

2 j=1

zk

zk−j

− 1

|zk−j|−1

k−

2 j=1

λj− 1

|z1|−1

.

The last term tends to infinity ask → ∞. By (4.5) we see that it is always positive, so there exists a uniform positive lower bound for allk’s. We now turn to the remaining part:

j=k+1

1− zk+ζ zj

= j=1

1− zk+ζ zk+j

j=1

1− |zk| +1 λj|zk|

=

j0

j=1

1− |zk| +1 λj|zk|

j=j0+1

1− |zk| +1 λj|zk|

,

wherej0is the minimal integer for which 1− λ2j >0 forjj0. Hence

j=j0+1

1− |zk| +1 λj|zk|

j=j0+1

1− 2

λj

>0.

To complete the proof of (4.8) we proceed to show that (4.9) 1− |zk| +1

λj|zk| >c,˜ j =1, . . . , j0, with somec >˜ 0. Indeed,

1− |zk| +1

λj|zk| ≥1− |zk| +1

λ|zk| =1− 1 λ− 1

λ|zk| ≥1− 1 λ− 1

λ|z2|, which implies (4.9). So we have proved (4.8).

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