commuting toeplitz operators with pluriharmonic symbols...

COMMUTING TOEPLITZ OPERATORS WITH PLURIHARMONICSYMBOLS ON THE FOCK SPACE

WOLFRAM BAUER, BOO RIM CHOE, AND HYUNGWOON KOO

Abstract. In the setting of the Bergman space over the disk or the ball, it has been knownthat two Toeplitz operators with bounded pluriharmonic symbols can (semi-)commute onlyin the trivial cases. In this paper we study the analogues on the Fock space over the multi-dimensional complex space. As is the case in various other settings, we are naturally ledto the problem of characterizing certain type of fixed points of the Berezin transform. Forsuch fixed points, we obtain a complete characterization by means of eigenfunctions of theLaplacian. We also obtain other characterizations. In particular, it turns out that thereare many nontrivial cases on the Fock space for (semi-)commuting Toeplitz operators withpluriharmonic symbols. All in all our results reveal that the situation on the Fock spaceappears to be much more complicated than that on the classical Bergman space setting,which partly is caused by the unboundedness of the operator symbols. Some of our resultsare restricted to the one-variable case and the corresponding several-variable case is leftopen.

1. Introduction

In 1964 Brown and Halmos [9] showed that Toeplitz operators with bounded symbols,acting on the Hardy space over the unit disk, can commute only in the trivial cases. Initiatedby such a seminal work of Brown and Halmos, the problem of characterizing when twoToeplitz operators commute has been one of the topics of constant interest in the studyof Toeplitz operators on classical function spaces over various domains. The problem witharbitrary bounded symbols being still far from its solution, all the results known so far(except for [9]) are restricted to certain subclasses of symbols; see, for example, [5, 6, 7,10, 12, 13, 17, 19, 20, 24, 27] and references therein. In particular, for Toeplitz operators

with bounded harmonic symbols over the unit disk, Axler and Cucukovic [2] obtained thecomplete Bergman space analogue of the aforementioned result of Brown and Halmos. LaterZheng [27] extended these results to Toeplitz operators with bounded pluriharmonic symbolsacting on the Hardy/Bergman space over the ball. The purpose of the current paper is toexplore commuting Toeplitz operators with pluriharmonic symbols acting on the Fock spacedescribed below.

We first recall the function space to work on. Let n be a fixed positive integer, reservedfor the dimension of the underlying complex n-space Cn. For z = (z1, . . . , zn) and w =(w1, . . . , wn) in Cn, we write z · w =

∑nj=1 zjwj so that z · w is the Hermitian inner product

Date: July 25, 2014; (Revised) February 26, 2015.2010 Mathematics Subject Classification. Primary 47B35; Secondary 30H20, 32A37.Key words and phrases. Commuting Toeplitz operators; Fock space; Berezin transform.W. Bauer was supported by an Emmy-Noether grant of Deutsche Forschungsgemeinschaft (DFG),

B. R. Choe was supported by NRF(2013R1A1A2004736) of Korea, and H. Koo was supported byNRF(2012R1A1A2000705) of Korea.

1

2 W. BAUER, B. CHOE, AND H. KOO

of z, w ∈ Cn and |z| = (z · z)1/2. We denote by dµ the normalized Gaussian measure on Cn

given by

dµ(z) :=1

πne−|z|

2

dv(z)

where dv = dvn denotes the Lebesgue measure on Cn. The Fock space H2(Cn) is then thespace of all Gaussian square integrable entire functions on Cn, i.e.,

H2(Cn) := L2(Cn, dµ) ∩Hol(Cn).

Here, and elsewhere, Hol( · ) stands for the class of functions holomorphic over the domainspecified in the parenthesis. As is well known, H2(Cn) is a closed subspace of L2(Cn, dµ)and thus is a Hilbert space. In the literature it frequently is denoted as Bargmann spaceor Segal-Bargmann space. The above naming is due to the fact that H2(Cn) canonically isisomorphic to the usual Fock space over Cn. We refer the reader to [29] for a systematictreatment of both function theoretic and operator theoretic aspects of the one-dimensionalFock spaces.

We now recall the operators to work with. Given a complex valued measurable function uon Cn satisfying suitable growth condition at ∞, the Toeplitz operator Tu with symbol u isdefined by

(1.1) Tu := PMu

where P : L2(Cn, dµ) → H2(Cn) denotes the Hilbert space orthogonal projection and Mu

denotes the operator of pointwise multiplication by u. In general the symbol function of theToeplitz operator in (1.1) is unbounded and therefore may lead to an unbounded operatoron H2(Cn). We will specify later (see Section 2.2) the class of symbols, denoted by Sym(Cn),which is not small in the sense that it contains all the functions with at most linear exponentialgrowth at ∞. Our symbol class is an algebra and has the property that products of finitelymany Toeplitz operators with symbols therein are always densely defined on H2(Cn). Inparticular, given two Toeplitz operators with symbols in Sym(Cn), their (semi-)commutatorsare densely defined on H2(Cn).

Recall that a complex valued function on Cn is said to be pluriharmonic if its restrictionto an arbitrary complex line is harmonic as a function of one complex variable. As is wellknown, a pluriharmonic function on Cn is precisely a function which can be expressed assum of a holomorphic function and a co-holomorphic function on Cn; see, for example, [18,Proposition 2.2.3]. Our main aim in this paper is to explore the problem of when two Toeplitzoperators with pluriharmonic symbols (semi-)commute. The start line for our approach is thesame as the one for earlier studies on the Bergman space over the classical bounded domainssuch as the disk ([2]), the ball ([27]) and the polydisk ([10]). That is, we convert the originaloperator theoretic problem to a function theoretic one via the Berezin transform B; seeSection 2.3 for precise definition. In fact, the Berezin transform converts our problem to thatof characterizing B-fixed points of the form fg+hk with entire functions f, g, h, k ∈ Sym(Cn);see Section 2.5. However, we immediately encounter the main difficulty for our Fock spacesetting: B-fixed points might not have any harmonicity in general, unlike the aforementionedBergman space setting.

In the next section we collect some known material and prove some basic facts that willbe needed subsequently. In Section 3, in connection with the remarks in the preceding

COMMUTING TOEPLITZ OPERATORS 3

paragraph, we study B-fixed points of the form

N∑j=1

fjgj(1.2)

where all functions fj, gj ∈ Sym(Cn) are entire. We first handle the simplest case whereN = 1 and functions involved are both polynomials; see Theorem 3.2. We then characterizefunctions in the algebra generated by the holomorphic polynomials and the reproducingkernels; see Theorem 3.5. This result reveals an intimate relation between eigenfunctions ofthe Laplacian and the B-fixed points of the form under consideration. We are thus eventuallyled to the “eigenfunction-series” characterization of B-fixed points of the form (1.2); seeTheorem 3.11. In the course of the proofs we obtain a characterization (Corollary 3.12) byharmonicity when each pair fj, gj contains a polynomial.

In Section 4, we apply our results obtained in earlier sections to study the one-variable caseof (semi-)commuting Toeplitz operators with harmonic functions whose (co-)holomorphicparts have at most linear exponential growth at∞. This amounts to studying B-fixed pointsof the form (1.2) with N ≤ 2 where functions involved all have the specified growth rate at∞.We obtain quite explicit characterizations other than the eigenfunction-series characterizationmentioned above; see Theorems 4.1 and 4.14. These characterizations show that there areextra cases for the Fock space, which have no analogue on the Bergman space over the ballor the polydisk. The several-variable case appears to be more subtle and is left open. As anapplication of our results, we obtain a “zero-product property” (Theorem 4.2) for Toeplitzoperators under consideration. In addition, we provide a counter example (Example 4.3) tothe analogue of a question posed by Louhichi and Rao.

2. Preliminaries

In this section we recall some known results and prove some basic facts related to thesymbol class and the Berezin transform.

2.1. Auxiliary Fock spaces. Consider a family of normalized Gaussian measures on Cn

given by

dµt(z) :=1

(tπ)ne−|z|2t dv(z)

for t > 0. Associated with the measure dµt is the t-scaled Fock space

H2t (Cn) := L2(Cn, dµt) ∩Hol(Cn),

which is regarded as a closed subspace of L2(Cn, dµt). We write 〈·, ·〉t and ‖ · ‖t for the innerproduct and norm of L2(Cn, dµt), respectively. As is well known, H2

t (Cn) is a reproducingkernel Hilbert space whose reproducing kernel Kt

w at w ∈ Cn is given by

Ktw(z) = e

z·wt ;

see, for example, [3, 29]. Note ‖Ktw‖2

t = Ktw(w) = e

|w|2t by the reproducing property. We

denote by

ktw(z) :=Ktw(z)

‖Ktw‖t

= exp

z · wt− |w|

2

2t


the normalized kernel at w. If we omit the index t in these notation, we always mean t = 1.We note a simple but useful integral identity∫

Cne2Re (z·w) dµt(w) = et|z|

2

,(2.1)

which can be easily verified via the relation Kw = Kttw.

2.2. Symbol space. Given c > 0, consider the space Dc of all complex measurable functionsu on Cn such that u(z)e−c|z|

2is essentially bounded on Cn. Clearly, each Dc is a Banach space

equipped with the norm ‖u‖Dc := ‖ue−c|·|2‖L∞(Cn,dv). Using these spaces, we now define oursymbol space by

Sym(Cn) :=⋂c>0

Dc.

Note that Sym(Cn) is a ∗-algebra under pointwise multiplication and complex conjugation.Given 0 < s < t, note that there is a constant Cs,t > 0 such that

Cs,t‖f‖s ≤ ‖f‖D1/2t≤ ‖f‖t(2.2)

for f ∈ H2t (Cn); the first inequality is clear and the second one can be easily verified via the

reproducing property. It follows that

Sym(Cn) ∩Hol(Cn) =⋂t>0

H2t (Cn).

We equip Sym(Cn) with the Frechet topology which is induced by the system of normspj := ‖ · ‖D 1

j

j∈N; the symbol N denotes the set of all positive integers. While we refer to

[23] for the precise notion of Frechet topology and related facts, we would like to mentionhere that Sym(Cn) now has the structure of a complete metric space with the property: asequence in Sym(Cn) converges if and only if it converges with respect to each of the abovenorms pj, j ∈ N.

Consider the vector space of entire function

(2.3) H(Cn) :=

⋃0<c< 1

2

Dc

∩Hol(Cn).

Clearly, H(Cn) is contained in H2(Cn) and contains all the reproducing kernels of H2(Cn).We thus see that H(Cn) is a dense subset of H2(Cn). It has been shown in [4] that H(Cn) isinvariant under the action of each Toeplitz operator with symbol in Sym(Cn). So, H(Cn) isinvariant under products of finitely many such operators. In particular, products of finitelymany Toeplitz operators with symbols in Sym(Cn) are always densely defined on H2(Cn).

In what follows we use the standard multi-index notation. That is, for α = (α1, . . . , αn) ∈Nn

0 where N0 := N ∪ 0, we use the notation

|α| : = α1 + · · ·+ αn,

α! : = α1! · · ·αn!,

xα : = xα11 · · · xαnn

for x = (x1, . . . , xn). Setting

∂ := (∂1, . . . , ∂n)


where ∂j denotes the differentiation with respect to the j-th component of given variable, wealso use the notation

f (α) := ∂αf

for f ∈ Hol(Cn). Finally, we will use the scaled Laplacian

∆ := ∂ · ∂ =n∑j=1

∂j∂j

which differs from the standard one by a factor of 4.

Lemma 2.1. If f ∈ Sym(Cn) ∩Hol(Cn), then f (α) ∈ Sym(Cn) for all α ∈ Nn0 .

Proof. Let f ∈ Sym(Cn) ∩Hol(Cn) and α ∈ Nn0 . Let ε > 0. Consider t > 0 to be specified

later. Pick t < (2ε)−1 to begin with. Given z ∈ Cn, we have by the reproducing property

f(z) = 〈f,Ktz〉t =

∫Cnf(w)e

z·wt dµt(w)

and thus differentiation under the integral sign yields

(2.4) f (α)(z) =1

t|α|

∫Cnwαf(w)e

z·wt dµt(w).

Note that |f(z)| ≤ ‖f‖Dεeε|z|2. Also, note that there is a constant C = C(|α|, ε) > 0 and

|wα| ≤ |w||α| ≤ Ceε|w|2

for all w ∈ Cn. It follows from these observations and (2.4) that∣∣f (α)(z)∣∣ ≤ C‖f‖Dε

πnt|α|+n

∫Cn

exp

Re (z · w)

t−(

1

t− 2ε

)|w|2

dv(w)

=C‖f‖Dε

t|α|(1− 2tε)nexp

|z|2

4t(1− 2εt)

;

the last equality comes from (2.1). In particular, choosing t := 14ε

, we have∣∣f (α)(z)∣∣ ≤ 2n(4ε)|α|C‖f‖Dεe2ε|z|2

so that

‖f (α)‖D2ε ≤ 2n(4ε)|α|C‖f‖Dε .Since ε > 0 is arbitrary, this yields f (α) ∈ Sym(Cn), as required.

2.3. Heat transform. Given t > 0 and u ∈ Sym(Cn), define

(2.5)

Bt[u](z) : =

∫Cnu(z ± w) dµt(w)

=1

(πt)n

∫Cnu(w) exp

−|z − w|

2

t

dv(w)

for z ∈ Cn. Put

Λu(t, z) := Bt[u](z)(2.6)

for t > 0 and z ∈ Cn.


As is well known, the function Λu satisfies the heat equation

(∆z − ∂t) Λu(t, z) = 0(2.7)

where ∂t = ∂∂t

and ∆z denotes the Laplacian with respect to z-variable; see [8] or, for theone-variable case, [29, Theorem 3.14]. For this reason Bt[u] is often called the heat transformof u (at time t). Using the mean value property one may easily check that harmonic functionsin the symbol space are fixed by the heat transform. Also is well known that the Laplaciancommutes with the heat transform when applied to functions which together with theirderivatives have a suitable growth rate at ∞. For example, that is the case when they acton the second-order Sobolev type subspace

(2.8) Sym2(Cn) :=u ∈ C2(Cn) : u, ∂ju, ∂j∂ju ∈ Sym(Cn) for all j = 1, . . . , n

of the symbol space. In fact, for u ∈ Sym2(Cn), using the identity

∆z

[exp

−|z − w|

2

t

]= ∆w

[exp

−|z − w|

2

t

],

one may check via two times partial integration in (2.5) (the boundary terms vanish due tothe growth assumption on the derivatives of u) that

∆Bt[u] = Bt[∆u]

for t > 0. The next lemma shows that the symbol space is invariant under the heat transform.

Lemma 2.2. Given ε, t > 0 with 0 < ε ≤ 12t

, the inequality

‖Bt[u]‖D2ε ≤ 2n‖u‖Dεholds for u ∈ Sym(Cn). In particular, for all α, β ∈ Nn

0 and t > 0, the operators

∂α∂βBt : Sym(Cn)→ Sym(Cn)

are well-defined and continuous in the Frechet topology of Sym(Cn).

Proof. Let 0 < ε ≤ 12t

, u ∈ Sym(Cn) and fix z ∈ Cn. Note |u(z)| ≤ ‖u‖Dεeε|z|2. Thus (2.5)

yields ∣∣Bt[u](z)∣∣ ≤ ‖u‖Dε

(πt)n

∫Cn

exp

ε|z + ξ|2 − |ξ|

2

t

dv(ξ)

=‖u‖Dε(πt)n

eε|z|2

∫Cn

exp

2εRe (z · ξ)−

(1

t− ε)|ξ|2dv(ξ)

=‖u‖Dε

(1− tε)nexp

ε

(1 +

tε

1− tε

)|z|2

;

we used (2.1) for the last equality. Since 0 < tε ≤ 1/2, this implies the asserted inequality.

In the case where α = β = 0 the second part of the lemma is immediate from the firstpart. The statement for general α, β ∈ Nn

0 follows from this special case, differentiation underthe integral sign and the fact that multiplication by polynomials in z and z is a continuousoperation on Sym(Cn).

Remark 2.3. With our previous notation, it follows from Lemma 2.2 that each Bt maps thesymbol space Sym(Cn) into Sym2(Cn).


Now, it is legitimate by Lemma 2.2 to consider Bt Bs for any s, t > 0 on the symbol space.In conjunction with this remark, we note the well-known semi-group property

(2.9) Bt Bs = Bs+t;see, for example, [8] or, for the one-variable case, [29, Theorem 3.13]. When t = 1, we writeB := B1. Note

|kz(w)|2 = exp

2Re (z · w)− |z|2

= e|w|2−|z−w|2

and thus

B[u](z) =⟨ukz, kz

⟩=⟨P (ukz), kz

⟩=⟨Tukz, kz

⟩for u ∈ Sym(Cn). This relation allows us to extend the definition of B to operators as follows.Let T be the algebra consisting of all finite sums of finite products of Toeplitz operatorswith symbols in Sym(Cn). Given T ∈ T we define

B[T ](z) :=⟨Tkz, kz

⟩for z ∈ Cn. We call B[T ] the Berezin transform of T . Injectivity of the Berezin transform iswell known.

Lemma 2.4. The Berezin transform B : T −→ Cω(Cn) is one-to-one. Here, Cω(Cn)denotes the class of all real analytic functions on Cn.

We refer to [4, Lemma 12] for a proof of the above lemma.

2.4. Periodic entire functions. Given u ∈ Sym(Cn), consider the function Λu on (0,∞)×Cn as in (2.6). From the growth condition of u and (2.5) it is easy to see that the functionΛu(·, z) with z ∈ Cn fixed holomorphically extends to the right half-plane λ ∈ C : Reλ > 0.Assume now that u is fixed under the Berezin transform, i.e., B[u] = u. It then follows fromthe semigroup property (2.9) that

(2.10) Λu(t+ 1, z) = Bt+1[u](z) = (Bt B)[u](z) = Bt[u](z) = Λu(t, z)

for all t > 0 and z ∈ Cn. By the identity theorem we find that Λu(·, z) defines a 1-periodicfunction (i.e. a periodic function with period 1) on the right half-plane and therefore extendsto a periodic entire function on the complex plane.

In conjunction with the observation in the preceding paragraph, we recall the characteri-zation of periodic entire functions on C of at most linear exponential growth at ∞. We saythat a measurable function h on Cn is of exponential type and write h ∈ E(Cn), if there issome s > 0 such that |h(z)|e−s|z| is essentially bounded on Cn. Clearly, we have

E(Cn) ⊂ Sym(Cn).

A short proof of the next (well-known) lemma can be found in [4].

Lemma 2.5. Let h ∈ Hol(C) ∩ E(C) be 1-periodic. Then h is a trigonometric polynomial,i.e., of the type

h(λ) =N∑

m=−N

cme2πimλ (i =

√−1)

for some N ∈ N0 and coefficients cm.

If the “exponential type” hypothesis is relaxed in the hypothesis of Lemma 2.5, we get atrigonometric series rather than a trigonometric polynomial.


Proposition 2.6. Let h ∈ Hol(C) be 1-periodic. Then h is a trigonometric series, i.e., ofthe type

h(λ) =∞∑

m=−∞

cme2πimλ

for some coefficients cm which makes the series above converge absolutely and uniformly oneach horizontal strip in C.

Proof. Define

h(λ) := h

(Logλ

2πi

),

where Logλ denotes the principal branch of the logarithmic function. Since h is 1-periodic,

h extends to a holomorphic function on C \ 0. For λ = x+ iy with x and y real, note

h(e2πiλ

)= h

(e−2πye2πix

)= h

(−2πy + 2πix

2πi

)= h(λ);

the second equality holds for −12< x < 1

2by definition of Log and thus for general x by

periodicity and continuity. Thus, using the Laurent expansion of h near the origin

(2.11) h(λ) =∞∑

m=−∞

cmλm, λ 6= 0

the original function h can be represented in the form

(2.12) h(λ) = h(e2πiλ

)=

∞∑m=−∞

cme2πimλ

for λ ∈ C. The series above converges absolutely and uniformly on each horizontal strip,because the series in (2.11) converges absolutely and uniformly on each compact annularregion in C \ 0. This completes the proof.

2.5. (Semi-)commutators of Toeplitz operators. Given Tu and Tv with u, v ∈ Sym(Cn),we denote by

(Tu, Tv] : = Tuv − TuTv

and

[Tu, Tv] : = TuTv − TvTuthe semi-commutator and the commutator of Tu and Tv, respectively. Note that these oper-ators are densely defined on H2(Cn).

In connection with Lemma 2.4, we compute the Berezin transform of (semi-)commutatorsof two Toeplitz operators with pluriharmonic symbols. First, let’s compute the Berezintransform of the (semi-)commutator (Tf , Tg] = [Tg, Tf ] for f, g ∈ Sym(Cn) ∩ Hol(Cn). Forz ∈ Cn, we have by the reproducing property[

TgKz

](ξ) =

⟨gKz, Kξ

⟩= 〈gKξ, Kz〉 = g(z)Kz(ξ)

for all ξ ∈ Cn. In other words,[TgKz

]= g(z)Kz. It follows that

B[TfTg](z) =

⟨fTgkz, kz

⟩= g(z)

⟨fkz, kz〉 = g(z)f(z);


the last equality holds, because the Berezin transform fixes holomorphic functions. Combin-ing these observations, we obtain

(2.13) B[(Tf , Tg]

]= B[fg]− fg.

Next, consider a pair of symbol functions u, v ∈ Sym(Cn) of the form

u = f + k and v = h+ g(2.14)

where f, g, h, k ∈ Hol(Cn). In fact one may check f, g, h, k ∈ Sym(Cn) under the assumptionu, v ∈ Sym(Cn), because the growth rate of a holomorphic function is controlled by that ofits real part.

A direct calculation yields (Tv, Tu] = (Th, Tk] and thus

[Tu, Tv] = (Tv, Tu]− (Tu, Tv] = (Th, Tk]− (Tf , Tg].

We thus have by (2.13)

(2.15) B[[Tu, Tv]

]= B[hk − fg] + fg − hk.

Now, the next proposition is immediate from (2.13), (2.15) and Lemma 2.4.

Proposition 2.7. Let f, g, h, k ∈ Sym(Cn) ∩Hol(Cn). Then the following statements hold:

(a) (Tf , Tg] = 0 ⇐⇒ B[fg] = fg.

(b) [Tf+k, Th+g] = 0 ⇐⇒ B[fg − hk] = fg − hk.

2.6. Miscellany. By Proposition 2.7 we are naturally led to the study on B-fixed points. Wenote here a known result in this direction. Let S(Cn) be the Schwartz space over Cn ∼= R2n.As is well known, the Berezin transform B can be regarded as a continuous convolutionoperator

B : S(Cn) −→ S(Cn) : B[u] = u ∗ h = (2π)−n∫Cnu(w)h(· − w) dv(w),

where h = 2n exp−| · |2. Therefore it naturally admits an extension to the dual spaceS ′(Cn) of tempered distributions. The B-fixed points in S ′(Cn) are completely characterizedas in the next lemma; see [14].

Lemma 2.8. Let u ∈ S ′(Cn). If B[u] = u, then u is a harmonic polynomial.

We also recall the following well-known “complexification” lemma; see, for example [16,Proposition 1.69].

Lemma 2.9. Let Ω be a domain in Cn and assume that Q is holomorphic on Ω×Ω∗ whereΩ∗ = z : z ∈ Ω. If Q(z, z) = 0 for all z ∈ Ω, then Q = 0 on Ω× Ω∗.

3. Fixed points of the Berezin transform

In this section, in view of Proposition 2.7, we explore solutions of the equation

B

[N∑`=1

f`g`

]=

N∑`=1

f`g`(3.1)

where f`, g` ∈ Sym(Cn) ∩Hol(Cn) for each `. To begin with, we first consider the simplestcase

B[fg] = fg,(3.2)


where f, g ∈ Sym(Cn) ∩Hol(Cn).When f and g are polynomials, we see from Lemma 2.8 that (3.2) holds if and only if fg is

harmonic. Such type of harmonic functions are characterized as in the next lemma. In whatfollows, a function u ∈ C2(Cn) is said to be n-harmonic if u is harmonic in each variableseparately. Thus, fg with f, g ∈ Hol(Cn) is n-harmonic if and only if ∂jf = 0 or ∂jg = 0 foreach j = 1, . . . , n.

Lemma 3.1. Given f, g ∈ Hol(Cn), the following statements are equivalent:

(a) fg is harmonic.(b) (fg) U is n-harmonic for some unitary operator U on Cn.

Proof. Note that n-harmonicity implies harmonicity and that harmonicity is invariant undercomposition with unitary operators. Thus the implication (b) =⇒ (a) holds. Conversely,

assume (a). Then we have ∂f(z) · ∂g(z) = 0 and thus by Lemma 2.9

∂f(z) · ∂g(w) = 0

for all z, w ∈ Cn. So, setting

Sf := span ∂f(z) : z ∈ Cn and Sg := span ∂g(z) : z ∈ Cn,we see that Sf ⊥ Sg. Now, put d := dimSf , pick an orthonormal basis B1, . . . , Bd of Sf ,and extend it to an orthonormal basis B1, . . . , Bn of Cn. Let U be the n×n matrix whosej-th column vector is Bj. We then have

∂j(f U)(z) = [(∂jf)(Uz)] ·Bj = 0 for j > d,

because Bjj>d ⊂ Cn Sf . Similarly, ∂j(g U) = 0 for j ≤ d. Accordingly, (fg) U isn-harmonic, as asserted. This completes the proof.

As an immediate consequences of Lemmas 2.8 and 3.1, we obtain the following character-ization.

Theorem 3.2. Given polynomials p, q ∈ Hol(Cn), the following statements are equivalent:

(a) B[pq] = pq.(b) pq is harmonic.(c) (pq) U is n-harmonic for some unitary operator U on Cn.

When n = 1, the statements above reduce to

(d) either p or q is constant.

However, if the function class is broadened to a larger class, say the class of exponentialtype, the characterization by means of harmonicity is no longer true. To see an example onC, put

ha,b(z) = eaRe z+bIm z

for a, b ∈ C. One may compute

B[ha,b] = ea2+b2

4 ha,b;

see [14] or [29, pp. 114–115]. Also, note ∆ha,b = a2+b2

4ha,b. Thus, if 0 6= a2 + b2 ∈ 8πiZ,

then ha,b is non-harmonic but fixed by the Berezin transform. Here, and in what follows, thesymbol Z stands for the set of all integers.

We now proceed to the investigation into the solutions, suggested by the examples in thepreceding paragraph, of (3.2). To be more explicit, we introduce some notation. Let Pn be


the algebra of holomorphic polynomials on Cn and let Kn be the algebra generated by allreproducing kernels for H2(Cn). We denote by An the algebra generated by Pn and Kn.More explicitly, we have

(3.3) An :=

N∑j=1

pjKaj : N ∈ N and pj ∈Pn, aj ∈ Cn for j = 1, . . . , N

;

recall that Ka denotes the reproducing kernel at a ∈ Cn for the space H2(Cn). We willcharacterize solutions of (3.2) in An. In what follows we use the notation

Ka,b := KaKb

for a, b ∈ Cn. Also, we putf ∗(z) := f(z)

for f ∈ Hol(Cn).

Lemma 3.3. The equality

B[pqKa,b

](z) = eb·aKa,b(z) q∗

(∂z + z + a

)p(z + b), z ∈ Cn,

holds for a, b ∈ Cn and p, q ∈Pn.

Proof. Fix a, b ∈ Cn and p, q ∈Pn. Using the representation q(w) =:∑

α cαwα, we have

B[pqKa,b

](z) = e−|z|

2∑α

cαIα(z)

where

Iα :=

∫Cnp(w)wα exp

w · (a+ z) + (b+ z) · w

dµ(w).

For each α, note from the reproducing property that

Iα(z) =

∫Cnp(w)∂αz

[exp

w · (a+ z) + (z + b) · w

]dµ(w)

= ∂α∫Cp(w)ew·(z+a)Kz+b(w) dµ(w)

= ∂α[p(z + b)e(z+b)·(z+a)

]and, in addition, that the relation

∂α[p(z + b)e(z+b)·(z+a)

]= e(z+b)·(z+a)

(∂z + z + a

)αp(z + b)

= eb·a+|z|2Ka,b(z)(∂z + z + a

)αp(z + b)

holds. Combining these observations, we conclude the lemma.

We remark that the Berezin transform of a polynomial is again a polynomial by the casea = b = 0 of Lemma 3.3. In fact the Berezin transform is an isomorphism on the space of allpolynomials in z and z preserving the degree; see [4].

Lemma 3.4. For a, b ∈ Cn and p, q ∈Pn \ 0, assume

B[pqKa,b

]= pqKa,b.

Then a · b ∈ 2πiZ and the following statements hold:

(a) If a 6= 0, then q is constant.


(b) If b 6= 0, then p is constant.

Proof. By Lemmas 3.3 and 2.9 we have

(3.4) p(z)q∗(w) = eb·aq∗(∂z + w + a

)p(b+ z)

for all z, w ∈ Cn. This, with z fixed, can be regarded as a w-polynomial identity. Comparingthe coefficients of the highest w-degree terms, we obtain a z-polynomial identity

p(z) = eb·ap(z + b).

Since p is not the zero polynomial, it follows that eb·a = 1, or equivalently, a · b ∈ 2πiZ.Moreover, if b 6= 0, then p is constant by b-periodicity. This shows (b). Note that (a) is thesame statement as (b) via the complex conjugation.

Note that a function f ∈ An admits a canonical representation

(3.5) f =J∑j=0

pjKaj

where p0 ∈Pn, pj ∈Pn \ 0 for each j ≥ 1 and a0 = 0, . . . , aJ is a collection of distinctpoints in Cn. Consider another g ∈ An with canonical representation

(3.6) g =L∑`=0

q`Kb` .

The following characterization shows that the action of the Berezin transform on An is quiterigid.

Theorem 3.5. For functions f, g ∈ An with canonical representations as in (3.5) and (3.6),the following statements are equivalent:

(a) B[fg] = fg.(b) B[pjq`Kaj ,b` ] = pjq`Kaj ,b` for each j and `.

If, in addition, neither f nor g is a polynomial, then either condition of the above is alsoequivalent to

(c) Both pj, q` are constants and ajb` ∈ 2πiZ for each j and `.

Proof. Using the notation

Ra,b,p,q = Ra,b,p,q(z, z) := eb·aq∗(∂z + z + a

)p(b+ z),(3.7)

we have by Lemma 3.3

B[fg]

=∑j,`

B[pjq`Kaj ,b`

]=∑j,`

Raj ,b`,pj ,q`Kaj ,b` .

So, assuming (a), we have ∑j,`

[Raj ,b`,pj ,q` − pjq`

]Kaj ,b` = 0.

Recall that each Raj ,b`,pj ,q` is a polynomial in z and z. Note that the functions Kaj ,b` forma linearly independent set over the polynomials, because (aj, b`)’s are all distinct. It followsfrom the above that Raj ,b`,pj ,q` = pjq` for each j and `. So, (b) holds. The implication(b) =⇒ (a) is clear. Finally, when neither f nor g is a polynomial, the equivalence (b)⇐⇒ (c)holds by Lemmas 3.3 and 3.4.


For any a, b ∈ Cn, note that Ka,b is a ∆-eigenfunction with eigenvalue a · b. Thus Theorem3.5 suggests a special role played by ∆-eigenfunctions with eigenvalues in 2πiZ. Motivatedby this observation, we now proceed to the investigation into the B-fixed points in question,i.e., the solutions of (3.1), by means of such eigenfunctions.

Proposition 3.6. Given f, g ∈ Sym(Cn) ∩Hol(Cn), the series

S(λ, z) :=∑α∈Nn0

λ|α|

α!f (α)(z)g(α)(z), (λ, z) ∈ C× Cn,

converges absolutely. Moreover, S(λ, ·) ∈ Sym(Cn) for each λ ∈ C.

Proof. Let f, g ∈ Sym(Cn) ∩Hol(Cn). Given λ ∈ C with λ 6= 0, put t := |λ| > 0. Since f isfixed by the heat transform, we have

f(z) =1

(πt)n

∫Cnf(w) exp

−|z − w|

2

t

dv(w)

for z ∈ Cn. Using this, we now estimate the growth rate of derivatives of f . Given α ∈ Nn0 ,

differentiating under the integral sign, we obtain

f (α)(z) =1

πntn+|α|

∫Cnf(w)(w − z)α exp

−|z − w|

2

t

dv(w)

=1

πntn+|α|

∫Cnf(ξ + z)ξαe−

|ξ|2t dv(ξ).

(3.8)

Thus, setting

If (α, z) :=

∫Cn

∣∣f(ξ + z)||ξα|e−|ξ|2t dv(ξ),

we have |f (α)(z)| ≤ (πntn+|α|)−1If (α, z). The same inequality holds with g in place of f . Wethus obtain

t|α||f (α)(z)g(α)(z)|α!

≤ 1

(tπ)2n

If (α, z)Ig(α, z)

α!t|α|(3.9)

for all z ∈ Cn.We now estimate the right hand side of (3.9). Let 0 < ε < 1

2t. Since

|f(ξ + z)| ≤ ‖f‖Dεeε|ξ+z|2 ≤ ‖f‖Dεe2ε(|ξ|2+|z|2),

it follows that

|If (α, z)| ≤ ‖f‖Dεe2ε|z|2n∏j=1

∫C|ζ|αj exp

(2ε− t−1)|ζ|2

dv1(ζ)

= (2π)n‖f‖Dεe2ε|z|2n∏j=1

∫ ∞0

rαj+1e−(t−1−2ε)r2

dr

=: (∗).

Applying the identity ∫ ∞0

r2b−1e−ar2

dr =Γ (b)

2ab


valid for positive numbers a and b, we have

(∗) =πntn+

|α|2 ‖f‖Dεe2ε|z|2

(1− 2εt)n+|α|2

n∏j=1

Γ(αj

2+ 1).

The same estimate holds with g in place of f . It follows that the right hand side of (3.9) isless than or equal to

(3.10)‖f‖Dε‖g‖Dεe4ε|z|2

(1− 2εt)2n· 1

(1− 2εt)|α|

n∏j=1

Γ2(αj

2+ 1)

αj!.

In conjunction with the product of quotients of Gamma functions above, we note via Stirling’sformula the estimate

Γ2(m2

+ 1)

m!=

Γ2(m2

+ 1)

Γ(m+ 1)≤ C

√m+ 1

2m, m = 0, 1, 2, . . . ,

with an absolute constant C > 0. Applying this to each factor of the product in (3.10), weobtain

(3.11)1

(tπ)2n

If (α, z)Ig(α, z)

α!t|α|≤ Cε(z)

(1 + |α|)n/2

2|α|(1− 2tε)|α|

where Cε(z) := Cn‖f‖Dε‖g‖Dεe4ε|z|2(1−2tε)−2n. Note that ε can be chosen arbitrarily small.

Also, given an integer m ≥ 0, the number of multi-indices α with |α| = m is (n+m−1)!(n−1)!m!

which

is comparable to mn−1. So, choosing sufficiently small ε > 0, say ε ≤ 15t

, the sequence on theright hand side of (3.11) is summable over α ∈ Nn

0 and the assertion follows from (3.9).

The formal expression of the heat transform

Bt =∞∑j=0

(t∆)j

j!= et∆

is well known; see, for example, [15]. The next lemma shows that this formal expression isactually the case for functions under consideration. Note that the series in the next lemmaconverges according to Proposition 3.6.

Lemma 3.7. Given f, g ∈ Sym(Cn) ∩Hol(Cn), the series expansion

(3.12) Bt[fg](z) =∑α∈Nn0

t|α|

α!f (α)(z)g(α)(z)

is valid for all t > 0 and z ∈ Cn.

Proof. Let f, g ∈ Sym(Cn) ∩Hol(Cn) and z ∈ Cn. Note from (3.8)

f (α)(z)g(α)(z) =1

t2|α|

∫Cn

∫Cnf(ξ + z)g(w + z)wαξ

αdµt(ξ) dµt(w)

for each α ∈ Nn0 and t > 0. Multiplying by t|α|

α!both sides of the above equality and then

taking the sum over all α ∈ Nn0 under the integral sign (justified by the growth estimate


(3.11)), we obtain∑α∈Nn0

t|α|

α!f (α)(z)g(α)(z) =

∫Cn

∫Cnf(ξ + z)g(w + z)

∑α∈Nn0

wαξα

α!t|α|

dµt(ξ) dµt(w)

=

∫Cn

∫Cnf(ξ + z)g(w + z)e

w·ξt dµt(ξ) dµt(w).

Recall that ew·ξt is the reproducing kernel for H2

t (Cn). So, the double integral of the abovereduces to ∫

Cnf(w + z)g(w + z) dµt(w) = Bt[fg](z),

which completes the proof.

Given a B-fixed point u ∈ Sym(Cn), recall that Λu(·, z) (with z ∈ Cn fixed) extends to a1-periodic entire function on C; see (2.10). In the next lemma Λu denotes such extension onCn+1. Recall that E(Cn) denotes the class of all functions over Cn of exponential type.

Lemma 3.8. If u ∈ E(Cn) and B[u] = u, then Λu ∈ E(Cn+1).

Proof. Let u ∈ E(Cn) and assume B[u] = u. Let λ ∈ C be given. We may assume Reλ > 0by periodicity. Since Reλ > 0, we have

Λu(λ, z) =1

(πλ)n

∫Cnu(z + w)e−

|w|2λ dv(w).

Thus, setting t := Reλ > 0 and choosing s > 0 such that

‖u‖Es := supz∈Cn|u(z)|e−s|z| <∞,

we have

|λ|ne−s|z||Λu(λ, z)| ≤‖u‖Esπn

∫Cn

exp

s|w| − t

|λ|2|w|2

dv(w)

= cn‖u‖Es∫ ∞

0

r2n−1 exp

sr − tr2

|λ|2

dr,

where cn is a constant depending only on the dimension n. Meanwhile, the integral above isdominated by some constant, depending on a, times∫ ∞

−∞exp

2sr − tr2

|λ|2

dr = |λ|

√π

texp

s2|λ|2

t

.

Now, further assuming by 1-periodicity that t ≥ max(|Imλ|, 1), we obtain

|Λu(λ, z)| ≤C‖u‖Es√t|λ|n−1

exp

s|z|+ s2|λ|2

t

≤ C‖u‖Eses|z|+2s2|λ|

with a constant C = C(n, s) > 0. This completes the proof.

Proposition 3.9. Given a positive integer N , let f`, g` ∈ Sym(Cn) ∩ Hol(Cn) for ` =1, . . . , N . Put

(3.13) u :=N∑`=1

f`g`


and assume B[u] = u. Then there is a sequence of functions ϕmm∈Z ⊂ Sym2(Cn) with thefollowing properties:

(a) For z ∈ Cn and λ ∈ C

Λu(λ, z) =∑α∈Nn0

λ|α|

α!

[N∑`=1

f(α)` (z)g

(α)` (z)

]=

∞∑m=−∞

ϕm(z)e2πimλ,

where both series converge absolutely, and moreover, for each fixed z, the series onthe right hand side converges uniformly on each horizontal strip in C.

(b) Each ϕm admits representations

ϕm(z) =∑α∈Nn0

γ(|α|,m)

α!

[N∑`=1

f(α)` (z)g

(α)` (z)

]

=

∫ 1

0

Bt

[N∑`=1

f`g`

](z)e−2πimt dt

where coefficients γ(|α|,m) are given by

γ(|α|,m) :=

∫ 1

0

t|α|e−2πimt dt.

(c) Given 0 < ε ≤ 12

and j ∈ N,

supm∈Z|m|j‖ϕm‖D2ε <∞.

(d) If, in addition, u ∈ E(Cn), then ϕm 6≡ 0 only for finitely many m’s.

Proof. Since u is a B-fixed point by assumption, the function Λu(·, z) with z fixed extends toa 1-periodic entire function on C, still denoted by Λu(·, z). Thus, by Proposition 2.6, Λu(·, z)can be expressed in the form

(3.14) Λu(λ, z) =∞∑

m=−∞

ϕm(z)e2πimλ

with coefficients ϕm(z) depending on z. Now, (a) is a consequence of Lemma 3.7. Whenλ = t ∈ [0, 1] (with z fixed), note that the right hand side of (3.14) is the Fourier series ofΛu(t, z). Thus (b) holds by (a). Note that due to Lemma 2.1 functions of the form (3.13)define elements in Sym2(Cn). According to Lemma 2.2 it thus follows from the integralexpression in (b) that functions ϕm are elements in Sym2(Cn) for all m ∈ Z.

Now, we prove (c). Fix j ∈ N. We have by periodicity of Λu(·, z) for any z ∈ Cn and m:

ϕm(z) =

∫ 1

0

Λu(t, z)e−2πimt dt

=1

(−2πim)j

∫ 1

0

Λu(t, z)∂jt [e−2πimt] dt

=1

(2πim)j

∫ 1

0

∂jtΛu(t, z)e−2πimt dt.(3.15)


In conjunction with this, recall that Λu(t, z) is a solution of the heat equation (2.7) and thatthe Laplacian ∆ commutes with Bt on Sym2(Cn). Therefore, setting

hj := ∆ju =∑|α|=j

j!

α!

N∑`=1

f(α)` g

(α)` ,

we obtain by an inductive argument

(3.16) ∂jtΛu(t, z) = ∆jzΛu(t, z) = Bt

[hj](z);

note u,∆u, . . . ,∆j−1u ∈ Sym2(Cn) by Lemma 2.1. Let 0 < ε ≤ 12. For any t ∈ (0, 1], we

have by Lemma 2.2

‖∂jtΛu(t, ·)‖D2ε = ‖Bt[hj]‖D2ε ≤ 2n‖hj‖Dε .This, together with the identity (3.15), yields

|m|j‖ϕm‖D2ε ≤1

(2π)j

∫ 1

0

‖∂jtΛu(t, ·)‖D2ε dt ≤‖hj‖Dε2n(2π)j

and thus we conclude (c), as asserted.Finally, we prove (d). Assume u ∈ E(Cn). We then have Λu ∈ E(Cn+1) by Lemma 3.8.

So, there is some s > 0 such that |Λu(λ, z)|e−s(|λ|+|z|) is bounded on Cn+1. It follows that ϕmdisappears for all m with |m| > s by Lemma 2.5. The proof is complete.

Let ϕ be a ∆-eigenfunction with eigenvalue λ. Then the function Φ(t, z) := eλtϕ(z) clearlysolves the initial value problem for the heat equation:

(∆z − ∂t)Φ(t, z) = 0, Φ(0, z) = ϕ(z).

If, in addition, ϕ ∈ Sym2(Cn), then the next lemma shows that the heat transform Bt[ϕ](z)turns out to coincide with Φ(t, z).

Lemma 3.10. Let ϕ ∈ Sym2(Cn) be a ∆-eigenfunction with eigenvalue λ ∈ C. Then

Bt[ϕ] = eλtϕ

for all t > 0. In particular, B[ϕ] = ϕ whenever λ ∈ 2πiZ.

Proof. Let z ∈ Cn. As was noticed before, the function Φ(t, z) := Bt[ϕ](z) with z fixedextends to a holomorphic function on the right half-plane. Since ∆ϕ = λϕ, we have as in(3.16)

∂jtBt[ϕ](z) = Bt[∆jϕ

](z) = λjBt[ϕ](z)

for t > 0 and each nonnegative integer j. This shows that the j-th Taylor coefficient at s > 0of the function Φ(·, z) is λj

j!Bs[ϕ](z). Accordingly, we have

(3.17) Bt[ϕ](z) =∞∑j=0

(t− s)jλj

j!Bs[ϕ](z) = eλ(t−s)Bs[ϕ](z)

for any t, s > 0.Meanwhile, by an elementary change of variables in the definition of heat transform (2.5),

we have

Bs[ϕ](z) =

∫Cnϕ(z +

√sw) dµ(w)


for s > 0. In the integral on the right hand side of the above, note that the integrand tendsto ϕ(z) as s ↓ 0. In addition, since ϕ ∈ Sym(Cn), we have

|ϕ(z +√sw)| ≤ ‖ϕ‖D1/4

exp

|z +√sw|2

4

≤ ‖ϕ‖D1/4

exp

|z|2 + |w|2

2

for 0 < s ≤ 1. Since the function above is µ-integrable as a function of w, we deduce by theLebesgue dominated convergence theorem that lims↓0 Bs[ϕ](z) = ϕ(z). Consequently, takingthe limit s ↓ 0 in (3.17), we conclude the lemma.

We are now ready to prove the following eigenfunction-series characterization for solutionsof (3.1) in the symbol space. In what follows, we denote by Xm(Cn) the ∆-eigenspace inSym2(Cn) with eigenvalue 2πmi, i.e.,

Xm(Cn) := u ∈ Sym2(Cn) : ∆u = 2πmiufor each integer m.

Theorem 3.11. Given a positive integer N , let f`, g` ∈ Sym(Cn)∩Hol(Cn) for ` = 1, . . . , Nand put

u :=N∑`=1

f`g`.

Then the following statements are equivalent:

(a) B[u] = u.(b) There are functions ϕm ∈ Xm(Cn) for m ∈ Z such that

u =∞∑

m=−∞

ϕm,

where the series converges in the Frechet topology of Sym(Cn).

If, in addition, u ∈ E(Cn), then the series in (b) reduces to a finite sum.

Proof. First, assume (b). Recall from Lemma 2.2 that the Berezin transform B acts contin-uously on Sym(Cn). Therefore we obtain by Lemma 3.10

B[u] =∞∑

m=−∞

B[ϕm] =∞∑

m=−∞

ϕm = u

so that (a) holds.Conversely, assume (a). By Proposition 3.9(a) there are functions ϕm ∈ Sym2(Cn) such

that

(3.18)∑α∈Nn0

t|α|

α!

[N∑`=1

f(α)` g

(α)`

]=

∞∑m=−∞

ϕme2πimt

for t ≥ 0; recall that both series converge absolutely. Given 0 < ε ≤ 12, we have by Proposition

3.9(c)

‖ϕm‖D2ε ≤Cε|m|2

, m = ±1,±2, . . . ,


for some constant Cε > 0 independent of m. This implies that the series on the right handside of (3.18) actually converges in the Frechet topology of Sym(Cn) (uniformly in t real).In particular, taking t = 0, we have

(3.19) u =∞∑

m=−∞

ϕm

with the series convergent in the Frechet topology of Sym(Cn).Recall that the function Λu(t, z) satisfies the heat equation (2.7). In addition, since u is

fixed by the Berezin transform by assumption, Λu(t, z) is 1-periodic as a function of t. Also,by Proposition 3.9(b), we have the integral representation

ϕm(z) =

∫ 1

0

Λu(t, z)e−2πimt dt

for each m. This shows that each ϕm is real analytic. In addition, applying the Laplacianunder the integral sign, we have

∆ϕm(z) =

∫ 1

0

∆zΛu(t, z)e−2πimt dt

=

∫ 1

0

∂tΛu(t, z)e−2πimt dt

= −∫ 1

0

Λu(t, z)∂t[e−2πimt] dt

= 2πimϕm(z)

for each m. So, we conclude that (b) holds.When u ∈ E(Cn), note that the series on the right hand side of (3.18) and thus that of

(3.19) is actually a finite sum by Proposition 3.9(d). This completes the proof.

We now observe a consequence of Proposition 3.9.

Corollary 3.12. Given a positive integer N , let f`, g` ∈ Sym(Cn)∩Hol(Cn) for ` = 1, . . . , Nand assume that f` or g` is a polynomial for each `. Then the following statements areequivalent:

(a) B

[N∑`=1

f`g`

]=

N∑`=1

f`g`.

(b)N∑`=1

f`g` is harmonic.

Proof. The implication (b) =⇒ (a) is clear, because harmonic functions in the symbol space

are fixed by the Berezin transform. Conversely, assume (a) and put u :=∑N

`=1 f` g`. Sincef` or g` is a polynomial for each ` by assumption, we see from Proposition 3.9(a) thatΛu(·, z) with fixed z ∈ Cn is a 1-periodic polynomial and thus is constant. It follows that∑N

`=1 f(α)` g

(α)` = 0 for any multi-index α 6= 0. Now, the case |α| = 1 implies (b).

In conjunction with harmonic functions of the type in the above corollary, we remark thatfurther characterizations are available for n = 1:


Remark 3.13. When n = 1, condition (b) of Corollary 3.12 can be described more explicitlyby the condition

N∑`=1

[f` − f(0)][g` − g(0)] = 0;

see [11, Theorem 3.3]. In particular, when both pairs f1, f2 and g1, g2 contain a noncon-stant function, it is elementary to check that f1g1 + f2g2 is harmonic if and only if

cf1 + df2 = c1 and dg1 − cg2 = c2(3.20)

for some constants c, d, c1, c2 with (c, d) 6= (0, 0).

We also observe a consequence of Theorem 3.11. To see sequences satisfying property (b)of the next corollary, one may take aj := 2πimjζ and b` := n`η where ζ, η ∈ Cn with ζ ·η = 1and mj, n` are sequences of distinct integers. Also, when the series in (3.22) are finitesums, note that the next corollary is contained in Theorem 3.5.

Corollary 3.14. Let aj, bj be sequences of points in Cn \ 0 such that (aj, b`)’s are alldistinct and let cj, dj be sequences of complex numbers such that

(3.21)∞∑j=1

(|cj|et|aj |

2

+ |dj|et|bj |2)<∞

for any t > 0. Put

(3.22) f :=∞∑j=1

cjKaj and g :=∞∑j=1

djKbj .

Then the following statements are equivalent:

(a) B[fg] = fg.(b) For each j and `, either cjd` = 0 or aj · b` ∈ 2πiZ.

Proof. Note ‖Ka‖t = ‖Ktta‖t = e

t|a|22 . Thus we have by (3.21)

∞∑j=1

|cj|‖Kaj‖t <∞ and∞∑j=1

|dj|‖Kbj‖t <∞

for any t > 0. Thus we see from (2.2) that the defining series in (3.22) for f and g convergein the Frechet topology of the symbol space. Note

‖KaKb‖t = ‖Ka+b‖t = exp

t|a+ b|2

2

≤ exp

t(|a|2 + |b|2)

for a, b ∈ Cn and t > 0. Using this inequality, we also see from (3.21) and (2.2) that theseries

fg =∑j,`

cjd`KajKb` =∑j,`

cjd`Kaj ,b`

converges in the Frechet topology of the symbol space. Note that each Kaj ,b` is a ∆-eigenfunction with eigenvalue aj · b`. Also, note that any finite sub-collection of Kaj ,b`j,`∈Zis linearly independent, because (aj, b`)’s are all distinct by assumption. The assertion nowfollows from Theorem 3.11.

We now close this section with a remark on the range of the Berezin transform.


Remark 3.15. The Berezin transform BD associated with the Bergman space over the unitdisk D of the complex plane is defined in a similar way by means of the normalized reproducingkernels; we refer to [28] for the precise definition and related facts. In [1, 22] solutions to anequation of the following form are completely determined:

BD[u] =N∑`=1

f`g`

where f`, g` ∈ Hol(D) and u ∈ L1(D). It turns out that this type of equation can be solvedonly when the right hand side is of very restricted form. Here, we want to point out thatthe analogous equation for the Berezin transform B associated with the Fock space can besolved rather easily for a large class of entire functions in the symbol space.

Given f`, g` ∈ Sym(Cn) for ` = 1, . . . , N , consider the equation

N∑`=1

f`g` = B[u] with u ∈ L2(Cn, dµ),(3.23)

or more explicitly,

N∑`=1

f`(z)g`(z) =e−|z|

2

πn

∫Cnu(ξ) exp

z · ξ + z · ξ − |ξ|2

dv(ξ).

By Lemma 2.9 we have

Q(z, w) : =N∑`=1

f`(z)g∗` (w)

=e−z·w

πn

∫Cnu(ξ) exp

z · ξ + w · ξ − |ξ|2

dv(ξ)

for z, w ∈ Cn. By Lemma 2.9 again, this is equivalent to

e−|z|2

4 Q

(iz

2,iz

2

)=

1

πn

∫Cnu(ξ)e−|ξ|

2

eiRe (z·ξ) dv(ξ)

=: F(ue−|·|

2)(z),

where F stands for the Fourier transform. Note that the function z 7→ Q(iz2, iz

2

)belongs to

the symbol space. Applying the inverse Fourier transform F−1 on L2(Cn, dv), we obtain

u(ξ) = e|ξ|2F−1

[e−|z|2

4 Q

(iz

2,iz

2

)](ξ), ξ ∈ Cn.

So, the function u above is the unique solution of (3.23).

4. (Semi-)Commuting Toeplitz operators: One-dimensional case

In this section our discussion is restricted to the one-dimensional case n = 1; the multi-dimensional case is left open.

We aim to characterize (semi-)commuting Toeplitz operators with symbols of exponentialtype as in (2.14). We note that the holomorphic part and co-holomorphic part of a plurihar-monic symbol of exponential type must be individually of the same type, as in the case of


general pluriharmonic symbols (see Section 2.5). So, according to the equivalence in Propo-sition 2.7, we make the following assumption throughout the discussion unless otherwisespecified:

B[fg + hk] = fg + hk(FB)

for f, g, h, k ∈ Hol(C) ∩ E(C).

Put u := fg+hk. Since functions under consideration are all exponential type, Proposition3.9 gives a representation of the form

∞∑j=0

(λ∆)j

j!u(z) =

L∑m=−L

ϕm(z)e2πimλ, λ, z ∈ C

for some integer L ≥ 0. Given an integer j ≥ 1, comparing coefficients of λj of both sides,we obtain a system of equations

∆ju(z) =L∑

m=−L

ϕm(z)(2πim)j, j = 0, 1, . . . , 2L+ 1,

or more explicitly in a matrix form,u(z)

∆u(z)...

∆2L+1u(z)

=

1 · · · 1

−2πiL · · · 2πiL...

......

(−2πiL)2L+1 · · · (2πiL)2L+1

ϕ−L(z)ϕ−L+1(z)

...ϕL(z)

.

Since the rows of the matrix on the right hand side are linearly dependent, one concludesthat there is a non-trivial relation of the form

2L+1∑`=0

γ`∆`u(z) = 0

for some coefficients γ`, not all 0. This can be rephrased as a more explicit equation of theform

J∑j=1

γnj

[f (nj)(z)g(nj)(z) + h(nj)(z)k(nj)(z)

]= 0

where coefficients γnj are all nonzero. Now, using Lemma 2.9, we obtain an equation

(4.1)J∑j=1

γnj[f (nj)(z)(g∗)(nj)(w) + h(nj)(z)(k∗)(nj)(w)

]= 0

in the independent variables z, w ∈ C. We may assume (when J ≥ 2) that nj < nj+1 for allj. To simplify notation we also express (4.1) in a vector form. Put

V :=

∂n1

...

...∂nJ

J×1

and W :=

γn1∂

n1

...

...γnJ∂

nJ

J×1

.


Also, put Vf := V f and Wf := Wf , etc. Then, (4.1) becomes

(4.2) W tg∗(w)Vf (z) +W t

k∗(w)Vh(z) = 0,

where the super-script t means the transpose.We introduce further notation. Recall that A1 denotes the algebra generated by the

polynomial algebra P1 and the reproducing kernel algebra K1; see (3.3). Put

D1 := p(∂) : 0 6= p ∈P1for the collection of all linear differential operators with constant coefficients. From elemen-tary theory of the ordinary differential equations with constant coefficients, note

A1 =⋃D∈D1

kerD

where kernels are taken in Hol(C).

We now pause to characterize semi-commuting Toeplitz operators with symbols underconsideration. In the next theorem the possibility of the second case of (c) is an extra casefor the Fock space, which has no analogue on the Bergman space over the ball or the polydisk;see [10, 26].

Theorem 4.1. Given f, g ∈ Hol(C) ∩ E(C), the following statements are equivalent:

(a) (Tf , Tg] = 0.(b) B[fg] = fg.(c) Either (i) or (ii) holds;

(i) f or g is constant.(ii) There are finite collections ajNj=1 and b`M`=1 of distinct complex numbers such

that

f ∈ span Ka1 , . . . , KaN and g ∈ span Kb1 , . . . , KbM

with ajb` ∈ 2πiZ for each j and `.

Proof. By Proposition 2.7 and Theorem 3.5 we only need to prove the implication (b) =⇒ (c).So, assume (b). If f or g is a polynomial, then Corollary 3.12 implies that (i) holds. So,assume that neither f nor g is a polynomial. By (4.2) (with h = 0 or k = 0) we haveW tg∗(w)Vf (z) = 0, which is the same as[

W tg∗(w)V

]f(z) = 0 or

[W tf (z)V

]g∗(w) = 0.

So, when w is fixed this is a linear differential equation in z with constant coefficients, andvice versa. Also, note J ≥ 2, because f and g∗ are not polynomials. Thus, fixing w suchthat Wg∗(w) 6= 0, we see that f must be of the form

f =N∑j=1

pjKaj ∈ A1 \P1,

where pj ∈P1 for each j and aj’s are distinct complex numbers. Exchanging the roles of zand w, we see that g also must be of the same form, i.e.,

g =M∑`=1

q`Kb` ∈ A1 \P1,


where q` ∈ P1 for each ` and b`’s are distinct complex numbers. We see from Theorem 3.5that (ii) holds. Overall, we conclude that (b) implies (c). The proof is complete.

As an application we obtain a zero-product property as in the next theorem. We do notknow whether the hypothesis TuTv = 0 = TvTu can be relaxed to TuTv = 0.

Theorem 4.2. For f, g, h, k ∈ Hol(C)∩ E(C) put u := f + k and v := h+ g. If TuTv = 0 =TvTu, then either u = 0 or v = 0.

Proof. Assume TuTv = 0 = TvTu. We claim that at least one of f, g, h, k is constant. Withthis claim granted, it is easily seen that either u = 0 or v = 0. For example, assume that gis constant. Then, since v is holomorphic, we have 0 = TuTv = Tuv, which implies uv = 0.So, we conclude by real-analyticity that either u = 0 or v = 0. Other cases can be treatedsimilarly.

We now proceed to prove that at least one of f, g, h, k is constant. To derive a contradiction,assume that none of them is constant. By a straightforward calculation using Lemma 2.4,the assumption TuTv = 0 = TvTu is equivalent to

B[hk] = −fh− gk − fg(4.3)

and

B[fg] = −fh− gk − hk.(4.4)

Applying the Berezin transform to both sides of (4.4), we have B2[fg] = fg by (4.3), whichcan be rephrased via an elementary change of variables as B[(fg)√2] = (fg)√2 where (fg)√2

denotes the dilated function z 7→ (fg)(√

2z). It follows from Theorem 4.1 that there arefinite collections of ajN1

j=1 and bjM1j=1 of distinct complex numbers (with ajb` ∈ πiZ for all

j, `) such that

f =

N1∑j=1

αjKaj and g =

M1∑j=1

βjKbj(4.5)

for some nonzero coefficients αj and βj. Similarly, there are finite collections of cjN2j=1

and djM2j=1 of distinct complex numbers (with cjd` ∈ πiZ for all j, `) such that

h =

N2∑j=1

γjKcj and k =

M2∑j=1

δjKdj(4.6)

for some nonzero coefficients γj and δj. We may assume without loss of generality thatthe numbers bj and dj are all nonzero.

Using the representations in (4.5), we have fg =∑

j,` αjβ`Kaj ,b` and no term in this

expansion is holomorphic, because bj 6= 0 for all j. Also, using the representations in (4.6),we have by Lemma 3.3

B[hk] =∑j,`

γjδ`B[Kcj ,d` ] =∑j,`

γjδ`ecjd`Kcj ,d` .(4.7)

Note again that no term in the above sum is holomorphic, because dj 6= 0 for all j. Thus,comparing the holomorphic parts of both sides of (4.3), we see that fh = (constant) and this


constant must be nonzero, because f and h are not identically zero. Now, since f, h ∈ Hol(C)are nonvanishing and of exponential type, we have

f(z) = f0eaz and h(z) = h0e

−az

for some nonzero numbers a, f0 and h0. A similar argument yields

g(z) = g0ebz and k(z) = k0e

−bz

for some nonzero numbers b, g0 and k0. Now, we have by (4.7) and (4.3)

h0k0eabK−a,−b + f0g0Ka,b + (f0h0 + g0k0) = 0,

which is a contradiction to the fact that Ka,b, K−a,−b, 1 is linearly independent. Thiscompletes the proof.

Example 4.3. In the setting of Toeplitz operators Tu1 , Tu2 , Tu3 with bounded symbols actingon the Bergman space over the unit disc D, none of them being a constant multiple of theidentity, Louhichi and Rao conjectured the following in [21]:

(C) If [Tu1 , Tu2 ] = 0 = [Tu1 , Tu3 ], then [Tu2 , Tu3 ] = 0.

To our knowledge this conjecture is still open. One may also consider the analogues onvarious other settings. In this context, Vasilevski [25] gave a counterexample in the caseof the Bergman space over the multi-dimensional unit ball. Also, Bauer and Issa [5] founda counterexample, involving at least one unbounded symbol, to the case of the Fock spaceover C. Here we wish to point out the failure of (C) in the case of the Fock space overC, even when one additionally assumes harmonicity of the symbols. For example, considernonconstant (co-)holomorphic symbols

u1(z) := ez, u2(z) := eπz and u3(z) := e2πiz.

Clearly, one has [Tu1 , Tu2 ] = 0, since u1 and u2 are holomorphic. Moreover, Theorem 4.1implies that [Tu1 , Tu3 ] = −(Tu1 , Tu3 ] = 0. On the other hand, Theorem 4.1 again implies that[Tu2 , Tu3 ] = −(Tu2 , Tu3 ] 6= 0, since 2π2i /∈ 2πiZ.

Remark 4.4. Note that the implication (b) =⇒ (c) of Theorem 4.1 does not generalize tohigher dimensions n > 1. In fact, harmonic functions of the form fg with f, g ∈ Hol(Cn) ∩E(Cn) are all fixed by the Berezin transform.

We now continue to investigate properties of solutions of (FB). We consider the followingthree cases separately:

(A) All of f, g, h, k belong to A1.(S) Some, but not all, of f, g, h, k belong to A1.(N) None of f, g, h, k belongs to A1.

First, we consider solutions as in (A) of (FB). For harmonic functions as in property (ii) ofthe next proposition, recall that there are more explicit characterizations; see Remark 3.13.

Proposition 4.5. Assume (A). Then (FB) holds if and only if one of the following twocases is fulfilled:

(i) B[fg] = fg and B[hk] = hk.(ii) There are functions A,B ∈ A1 such that B

[AB

]= AB and fg+hk−AB is harmonic.


Proof. Before proceeding, we first introduce some temporary notation. We denote by Ef thenon-polynomial part of f so that

Pf := f − Ef ∈P1

is the polynomial part of f . We define Eg, Pg, etc, in a similar way.Since the Berezin transform fixes harmonic functions of exponential type, the sufficiency

is clear. We now prove the necessity. Assume (FB). Using the polynomial and the non-polynomial parts, decompose

fg + hk =(PfPg + PhPk

)+(PfEg + PhEk

)+(EfPg + EhPk

)+(EfEg + EhEk

).

By Theorem 3.5 the Berezin transform of each group is invariant. Namely, we have

(PP) B[PfPg + PhPk

]= PfPg + PhPk.

(PE) B[PfEg + PhEk

]= PfEg + PhEk.

(EP) B[EfPg + EhPk

]= EfPg + EhPk.

(EE) B[EfEg + EhEk

]= EfEg + EhEk.

We split the proof into four cases.

(Case 1): Suppose that both pairs f, g and h, k contain a polynomial. In this case wesee from Corollary 3.12 that (ii) holds.

(Case 2): Suppose that both pairs Ef , Eh and Eg, Ek are linearly independent. Then,from (EE) together with Theorem 3.5, we have

B[EfEg

]= EfEg and B

[EhEk

]= EhEk.

Also, from (PE) and (EP), we see that Pf , Ph, Pg, Pk are all constants again by Theorem 3.5.Thus (i) holds.

(Case 3): Suppose that only one of the pairs Ef , Eh and Eg, Ek is linearly independent.Without loss of generality assume that Ef , Eh is linearly independent but Eg, Ek islinearly dependent. Note from (Case 1) that we may assume either Eg 6= 0 or Ek 6= 0. Wemay thus further assume Eg 6= 0 and Ek = cEg for some constant c. Since Ef , Eh is linearlyindependent, we see from (EP) that Pg and Pk are constants, as in the proof of (Case 2).Thus k − cg = c2 where c2 = Pk − cPg. Also, it follows from (PE) that Pf + cPh = c1 forsome constant c1 and thus f + ch = E + c1 where E := Ef + cEh. Accordingly, we have

fg + hk = Eg + c1g + c2h.

Note B[EEg] = EEg by (EE) and hence B[Eg]

= Eg. So, (ii) holds in this case.

(Case 4): Suppose that both pairs Ef , Eh and Eg, Ek are linearly dependent. We mayassume by (Case 1) that both Ef and Eg are nontrivial. Thus, by the linear dependence,we have Eh = cEf and Ek = dEg for some constants c and d. It follows from (PE) and (EP)that

(4.8) Pf + dPh = c1 and Pg + cPk = c2

for some constants c1 and c2. This, together with (PP), yields

(4.9) B[(1 + cd)PhPk

]= (1 + cd)PhPk.


First, assume cd+ 1 = 0. We then have

Ef =1

cEh = −dEh and Eg =

1

dEk = −cEk

and thus by (4.8)

f + dh = c1 and g − 1

dk = g + ck = c2.

Accordingly, fg + hk is harmonic by (3.20) and thus (ii) holds. Next, assume cd + 1 6= 0.Then we have B

[PhPk

]= PhPk by (4.9) and hence, according to Theorem 3.2, either Ph

or Pk is constant. Assume that Ph is constant. Then Pf is also constant by (PE) so thath = cf + c3 for some constant c3. It follows that

fg + hk = f(g + ck) + c3k

and thus (ii) is fulfilled. If Pk is constant, we can argue in a similar way. This completes theproof.

Next, we consider solutions as in (S) of (FB). We need some preliminary lemmas.

Lemma 4.6. Assume (FB). If f ∈ A1, then h ∈ A1 or k ∈P1.

Proof. Assuming f ∈ A1, we can find some D ∈ D1 for which Df = 0. We apply D = Dz toboth sides of (4.2) to get

0 = Dz[Wtk∗(w)Vh(z)] = [W t

k∗(w)V ]Dh(z).

Now, either k is a polynomial or Wk∗ 6≡ 0. In the latter case, fixing a suitable w, we have

Dh = 0 where D := [W tk∗(w)V ]D ∈ D1. Thus h ∈ A1, as required.

Lemma 4.7. Assume (FB) and (S). If f, h ∈ A1 \P1, then g, k /∈ A1 and cf + dh ∈ P1

for some constant c and d, not both 0.

Proof. Pick D ∈ D1 of minimal order such that f, h ∈ kerD. By factorization (up to aconstant factor) we can rewrite D in the form

D =M∏j=1

(∂ − λj)mj ,

where λ1, . . . , λM are distinct complex numbers and mj ∈ N for each j. Since f and h arenot polynomials by assumption, there is some j0 with λj0 6= 0. Without any restriction weassume j0 = 1 and consider the differential operator

D := (∂ − λ1)m1−1

M∏j=2

(∂ − λj)mj .

Also, regarding D as an operator acting on A1, fix a basis pj,` ∈ A1 : j = 1, . . . ,M and ` =1, . . . ,mj of kerD such that

span pj,` = ker(∂ − λj

)` ker(∂ − λj

)`−1

for each j = 1, . . . ,M and ` = 1, . . . ,mj. For short we rewrite such basis as q1, . . . , qNwith q1 = p1,m1 . Since f, g ∈ kerD, we can form the expansions

f =N∑`=1

c`q` and h =N∑`=1

d`q`


for some coefficients c` and d`. Note either c1 6= 0 or d1 6= 0 by minimality of D. Insertingthese into (4.2), we obtain

0 =N∑`=1

c`Wtg∗(w)Vq`(z) +

N∑`=1

d`Wtk∗(w)Vq`(z)

=N∑`=1

W tc`g∗+d`k∗

(w)Vq`(z).

Note from the construction that q2, . . . , qN are all annihilated by D. Now, an application of

D = Dz to both sides yields

0 = W tc1g∗+d1k∗(w)VDq1(z) =

[W tDq1

(z)V]

(c1g∗ + d1k

∗)(w).(4.10)

Note WDq1(z) is never zero, because Dq1 ∈ ker(∂ − λ1) is a constant multiple of eλ1z with

λ1 6= 0. Thus (4.10) shows that c1g∗ + d1k

∗ ∈ A1. Since either c1 6= 0 or d1 6= 0, we mayassume c1 = 1 for simplicity so that g∗ + d1k

∗ ∈ A1, or said differently, g + d1k ∈ A1. Thusg, k /∈ A1 by (S). Note

fg + hk = (g + d1k)f + k(h− d1f).

Thus the right hand side is also fixed by the Berezin transform. Since g + d1k ∈ A1 andk /∈ A1, we deduce from Lemma 4.6 that h− d1f ∈P1. This concludes the proof.

Lemma 4.8. Assume (FB) and (S) with f, g, h, k all nonconstant. Assume that at least oneof f, g, h, k is a polynomial. Then fg + hk is harmonic.

Proof. Assume that f, g, h, k are all nonconstant and, without loss of generality, assumeh ∈ P1. We will show that either f or g is a polynomial and thus conclude the lemma byCorollary 3.12.

Suppose that neither f nor g is a polynomial. We will show that f, g, h, k ∈ A1, which isimpossible by (S). An application of ∂N+1

z with N := deg h to both sides of (4.2) yields[W tg∗(w)V

]∂N+1f(z) = 0

for z, w ∈ C. Note that Wg∗(w) is not identically zero, because g is not a polynomial. Thusthe differential equation above implies f ∈ A1. Similarly, we have g ∈ A1. We now proceedto prove k ∈ A1. Since h ∈P1, a manipulation as in the proof of Lemma 3.3 yields

B[kh](z) = h∗(∂z + z

)k(z).

Put

Q(z) := B[kh](z)− k(z)h(z) =[h∗(∂z + z

)− h∗(z)

]k(z)

and define

(4.11) Q(z, w) :=[h∗(∂z + w

)− h∗(w)

]k(z)

so that Q is an entire function on C2 with the property Q(z, z) = Q(z). Meanwhile, notefrom (FB)

Q = gf − B[gf ].


Recall f, g ∈ A1. Thus, using the notation in (3.7), we see from Lemma 3.3 that Q is of theform

Q(z) =∑j,`

finite

[pj(z)q`(z)−Raj ,b`,pj ,q`(z, z)

]Kaj ,b`(z)

with holomorphic polynomials pj, q` and distinct (aj, b`) ∈ C2. This representation of Q,together with Lemma 2.9, yields

Q(z, w) =∑j,`

finite

[pj(z)q∗` (w)−Raj ,b`,pj ,q`(z, w)

]Kaj(z)K∗b`(w).

Accordingly, we see that Q(·, w) ∈ A1 for each fixed w ∈ C. Thus, for each fixed w ∈ C,

we can choose a differential operator D(w) ∈ D1 which annihilates Q(·, w). Accordingly, weobtain by (4.11)

D(w)[h∗(∂z + w)− h∗(w)

]k = D(w)Q(·, w) = 0.

Recall that h is nonconstant by assumption. So, choosing a suitable w for which h∗(∂z+w

)6=

h∗(w), we conclude k ∈ A1, as asserted. This completes the proof.

Assuming (S) and combining the results in Lemmas 4.6-4.8, we obtain the following prop-erties of solutions of (FB).

Proposition 4.9. Assume (S) with f, g, h, k all nonconstant. Then (FB) holds if and onlyif there are functions A,B ∈ Hol(C) ∩ E(C) such that B

[AB

]= AB and fg + hk − AB is

harmonic.

Proof. The sufficiency is clear as in the proof of Proposition 4.5. We now prove the necessity.Assume (FB). By Lemma 4.8 we may assume none of f, g, h, k is a polynomial. We may alsoassume f ∈ A1 without loss of generality. Since f ∈ A1 and k /∈ P1, we have h ∈ A1 byLemma 4.6. According to Lemma 4.7, we have

g, k /∈ A1 and p := cf + dh ∈P1

for some constants c and d, not both 0. We may assume that d = 1 so that

fg + hk = f(g − ck) + pk.

So, if p or g− ck is constant, then the conclusion clearly holds. Otherwise, applying Lemma4.8 to the right hand side of the above, we deduce that fg+ hk is harmonic. This completesthe proof.

Finally, we consider solutions as in (N) of (FB).

Proposition 4.10. Assume (N). Then (FB) holds if and only if fg + hk is harmonic.

Proof. The sufficiency being again clear, we only need to prove the necessity. Assume (FB).We claim

p := h− cf ∈P1 and q := g − dk ∈P1(4.12)

for some (nonzero) constant c and d. With this granted, we can write

fg + hk = f(dk + q + ck) + pk.

If p is constant, then we see from (FB) that f(dk + q + ck) is fixed by the Berezin transformand thus f ∈ A1 by Theorem 4.1. So, p cannot be constant by (N). Thus we can applyLemma 4.8 to the right hand side of the above to conclude the proposition.


It remains to prove (4.12). Note that f (n1), . . . , f (nJ ) is linearly independent as a set offunctions, because f /∈ A1 by (N). This equivalently means that CJ Yf = 0 where

Yf := span Vf (z) ∈ CJ : z ∈ C.In other words, dimYf = J . Thus, choosing λ1, . . . , λJ ∈ C such that Vf (λ1), . . . , Vf (λJ)form a basis of Yf , we see that the matrix

(4.13) Af := [Vf (λ1), . . . , Vf (λJ)]J×J

is invertible. Define (possibly non-invertible) Ah by replacing f with h in (4.13). Then putΦ := (Af )

−1Vf ∈ Hol(CJ) so that

(4.14) Vf (z) = AfΦ(z).

Applying (4.2) to each z = λj and then rewriting all J equations as a single matrix equation,we obtain

W tg∗(w)Af +W t

k∗(w)Ah = 0.

From (4.14) and (4.2) we thus have

0 = W tg∗(w)Vf (z)−W t

g∗(w)AfΦ(z)

= −W tk∗(w)Vh(z) +W t

k∗(w)AhΦ(z)

= W tk∗(w)

(AhΦ(z)− Vh(z)

).

This shows that

AhΦ(z)− Vh(z) ∈ CJ span Wk∗(w) ∈ CJ : w ∈ C = 0;the last equality holds as above, because k 6∈ A1 by (N). Since z is arbitrary, this yields

Vh = AhΦ = Ah(Af )−1Vf .

Accordingly,

h(nj) ∈ span f (n1), . . . , f (nJ ) =: S1(4.15)

for each j = 1, . . . , J . Applying ∂nJ−n1 to the case j = 1, we see that

h(nJ ) ∈ span f (nJ ), . . . , f (2nJ−n1) =: S2.(4.16)

Note that f (n1), . . . , f (nJ ), . . . , f (2nJ−n1) is also linearly independent by (N) as before. So,we deduce from (4.15) with j = J and (4.16)

h(nJ ) ∈ S1 ∩ S2 = span f (nJ ),which implies the first half of (4.12). Since g, h /∈ A1 by (N), one shows the second half of(4.12) in the same way. This completes the proof.

Putting together what we have proved so far, we obtain the following characterization forthe solutions of (FB).

Corollary 4.11. Let f, g, h, k ∈ Hol(C) ∩ E(C). Then (FB) holds if and only if one of thefollowing two cases is fulfilled:

(i) B[fg]

= fg and B[hk]

= hk.

(ii) There are functions A,B ∈ Hol(C)∩E(C) such that B[AB

]= AB and fg+hk−AB

is harmonic.


Proof. The sufficiency is clear as before. For the necessity, assume (FB). If one of f, g, h, kis constant, then (i) holds. Otherwise, either (i) or (ii) holds by Propositions 4.5, 4.9 and4.10.

In fact Condition (ii) above can be made more explicit via the next lemma.

Lemma 4.12. Let f, g, h, k ∈ Hol(C) ∩ E(C) and assume that none of them is constant.Assume that there are functions A,B ∈ Hol(C) ∩ E(C) such that B[AB] = AB and fg +hk − AB is harmonic. Then one of the following three cases is fulfilled:

(i) fg + hk is harmonic.(ii) There are constants a, b and c with ab 6= 0 such that

af − bh = c and B[f(bg + ak)] = f(bg + ak).

(iii) There are constants a, b and c with ab 6= 0 such that

ag − bk = c and B[(bf + ah)g] = (bf + ah)g.

Proof. If A or B is constant, then (i) holds. So, assume that both A and B are nonconstant.By harmonicity and Lemma 2.9 we have

f ′(z)g′(w) + h′(z)k′(w) = A′(z)B′(w)(4.17)

for any z, w ∈ C. Put F := f ′

A′, G := g′

B′, H := h′

A′and K := k′

B′for simplicity. Since A′ and

B′ are not identically zero, we see from the above

F (z)G(w) +H(z)K(w) = 1.(4.18)

From this one can see that functions F and H are both constant or both nonconstant. Thesame is true for functions G and K.

First, consider the case where F and H are nonconstant. Taking z-derivative of both sidesof (4.18) and then rearranging the resulting equation, we obtain

K(w)

G(w)+F ′(z)

H ′(z)= 0,

which implies K = c1G for some constant c1. Inserting this into (4.18), we obtain

[F (z) + c1H(z)]G(w) = 1

and hence deduce that G and K must be constant. Now, writing

g = bB + b′ and k = aB + a′

where a, b, a′ and b′ are constants with ab 6= 0, we see that the first part of (iii) holds.Meanwhile, since A = bf + ah+ c′ for some constant c′ by (4.17), we have

bAB − (bf + ah)g = −b′(bf + ah) + c′(g − b′).In particular, the function on the left hand side of the above is harmonic and thus is fixedby the Berezin transform. Now, since B[AB] = AB by assumption, we see that the secondpart of (iii) also holds.

Next, when G and K are nonconstant, the same argument shows that (ii) holds. Finally,when functions F , G, H and K are all constant, the argument above shows that both (ii)and (iii) hold. The proof is complete.

In conjunction with Lemma 4.12, we note the following.


Lemma 4.13. Let f, g, h, k ∈ Hol(C) ∩ E(C). If there are constants a, b and c with ab 6= 0such that

af − bh = c and B[f(bg + ak)] = f(bg + ak),(4.19)

then

B[fg + hk] = fg + hk.

Proof. Suppose (4.19). Note

b(fg + hk) = f(bg + ak)− ck.

Since f(bg+ ak) is fixed by the Berezin transform by assumption and ck is harmonic, we seethat the function on the left hand side of the above is fixed by the Berezin transform. Sinceb 6= 0, this completes the proof.

We finally obtain the following characterization for the solutions of (FB).

Theorem 4.14. Let f, g, h, k ∈ Hol(C) ∩ E(C). Then (FB) holds if and only if one of thefollowing four cases is fulfilled:

(i) B[fg] = fg and B[hk] = hk.(ii) fg + hk is harmonic.

(iii) There are constants a, b and c with ab 6= 0 such that

af − bh = c and B[f(bg + ak)] = f(bg + ak).

(iv) There are constants a, b and c with ab 6= 0 such that

ag − bk = c and B[(bf + ah)g] = (bf + ah)g.

Proof. When none of f , g, h and k is constant, the theorem is immediate from Corollary4.11, Lemma 4.12 and Lemma 4.13. Otherwise, (FB) is clearly equivalent to (i).

As a consequence, we have the following characterization of commuting Toeplitz operatorswith symbols under consideration.

Theorem 4.15. Given f, g, h, k ∈ Hol(C) ∩ E(C), the following three statements are equiv-alent:

(a) [Tf+k, Th+g] = 0.

(b) B[fg − hk] = fg − hk.(c) One of the following four conditions is fulfilled;

(i) (Tf , Tg] = (Th, Tk] = 0.

(ii) f + k, h+ g, 1 is linearly dependent.(iii) There are constants a, b and c with ab 6= 0 such that

af + bh = c and B[f(bg + ak)] = f(bg + ak).

(iv) There are constants a, b and c with ab 6= 0 such that

ag + bk = c and B[(bf + ah)g] = (bf + ah)g.

Proof. The equivalence (a) ⇐⇒ (b) holds by Proposition 2.7. The equivalence (b) ⇐⇒ (c)holds by Theorem 4.14 together with Theorem 4.1 and (3.20).


As was mentioned earlier before Theorem 4.1, compared to the corresponding result in [2]for the Bergman space over the unit disk, the first condition of (c) above contains extra casesfor the Fock space. Note that the last two conditions of (c) above also contain extra cases forthe Fock space. For example, using Theorem 4.1, pick any functions f, ϕ, ψ ∈ Hol(C)∩E(C)such that B[fϕ] 6= fϕ and B[fψ] = fψ. One may then check [Tf+ϕ+ψ, Tf+ϕ−ψ] = 0 byTheorem 4.15.

Acknowledgement: The first-named author acknowledges financial support through thefunds NRF(2013R1A1A2004736) and NRF(2012R1A1A2000705) of Korea which enabled himto visit the Korea University, Seoul where the discussion on commuting Toeplitz operatorswith pluriharmonic symbols on the Fock space has been started.

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Institut fur Analysis, Leibniz UniversitatWelfengarten 1, 30167 Hannover, GermanyE-mail address: [email protected]

Department of Mathematics, Korea UniversitySeoul 136-713, KoreaE-mail address: [email protected]

Department of Mathematics, Korea UniversitySeoul 136-713, KoreaE-mail address: [email protected]

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