Xianfeng ZHAO

1 Introduction

Let dμ be the Gaussian measure on the complex plane C.It is well-known that,in terms of the standard area measure dA(z)=dxdy=drdθ on C,we have

Recall that the Fock space F2is defined to be the subspace

It is easy to check that the functions zn(n≥0)are orthogonal in F2and their linear span is dense in F2.Using polar coordinates we get that.Thus

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forms an orthonormal basis of the Fock space F2.

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Given ? in L∞(C),the Fock Toeplitz operator with symbol ? is defined by

which shows that

For more information on the topics of the Fock space and Fock Toeplitz operators,we refer to[2–3,9].

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As usual,let kzdenote the normalized reproducing kernel for F2.That is,

For a bounded operator A on the Fock space F2,the Berezin transform of A is the function A on the complex plane defined by

For ? ∈L∞(C),is called the Berezin transform of ? given by

The Berezin transform is a very usefultool in studying Toeplitz operators on the Bergman space and the Fock space.For instance,the compactness,boundedness,positivity,invertibility and Fredholmness of the Toeplitz operators are partially or completely characterized by their Berezin transforms(please see[1,4–5,7–8]).

Recently,the author and Zheng[6]studied the positive Toeplitz operators on the Bergman space via their Berezin transforms.They showed that the positivity of a Toeplitz operator on the Bergman space is not completely determined by the positivity of the Berezin transform of its symbol.Indeed,they constructed a quadratic polynomial of|z|on the unit disk and showed that even if the minimal value of the Berezin transform of the polynomial is positive,the Toeplitz operator with the function as the symbol may not be positive.

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Motivated by this result,we try to study the positivity of the Toeplitz operators on the Fock space in the present paper.Observe that from the definition of the Berezin transform we see that the function ? is nonnegative on C provided that T?≥0,and T?≥0 if the function ?(z)≥0 for all z∈C.Thus,it is natural to ask the following question.

Question 1.1Is the positivity of a Fock Toeplitz operator completely determined by the positivity of the Berezin transform of its symbol?If not,then how to construct the “simplest”function ? which satisfies that ? is positive on the complex plane but the Toeplitz operator T?is not positive?

As we mentioned above,one can find an example from the set

such that T?is not positive on the Bergman space but ? is strictly positive on the unit disk.So,it is natural to try to use the radial functions(i.e.,?(z)= ?(|z|)for all z∈ C)of the formto construct a suitable example and give a negative answer to the above question.However,these simple radial functions are not bounded on the complex plane C.Our main idea in this paper is to simplify our calculations and estimations by the bounded radial functions e?λ|z|2(λ >0).More precisely,we will consider the bounded radial functions of the following form:

Question 1.2Is there a function ? in(1.1)such thatbut T?is not positive on F2for all 2α=β>0?

Based on the answer to Question 1.1,we will give a negative answer to Question 1.2 by taking 2α= β=and showing that T?is positive if and only if its Berezin transform is a nonnegative function in this case.That is,for this type of α and β,there do not exist a,b,c such that ? is positive but T?is not.The proofs will be given in the last section.

2 Preliminaries

It is difficult to study the positivity of the Toeplitz operators on function spaces in the general case even if the symbols are continuous functions.However,if ? is a radial function on C,we can find the relationship between the positivity of T?and the Berezin transform ? by its matrix,since the matrix representation of this type of Toeplitz operator is a diagonal matrix under the orthonormal basis.

Lemma 2.1Suppose that?is a bounded radial function onC.Then the matrix represen-tation of the Toeplitz operatorT?under the basisis

In particular,if

withλ >0,then the matrix representation ofT?is given by

ProofFor each n≥0,we have

If ?(z)=e?λ|z|2(λ >0)and n≥0,then we have

Thus,

This completes the proof of Lemma 2.1.

Note that for the Bergman space case,the Berezin transforms of the functions|z|l(l≥0)are so complicated even if these functions are very simple(see[6,Lemma 3.3]).However,the Berezin transform of the function e?λ|z|2(λ >0)on C has a good expression,that is the following lemma,which is very usefulfor proving our main results.

Lemma 2.2Suppose that?is a bounded radial function onC.Then the Berezin transform of?is given by

for some N＞2.However,this is impossible.The contradiction shows that

In particular,if?(z)=e?λ|z|2withλ >0,then

for allz∈C.

For the special case,we have the following simple calculations:

ProofBy the definition of the Berezin transform,we have

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where the last“=”comes from the proof of Lemma 2.1.Therefore,

This gives that,so(3)? (2).This completes the proof.

as desired.

Before studying the answer to Question 1.1,we first consider the function ? in(1.1)with c=0.Combining the above two lemmas,we get the following result.

Proposition 2.1Let?(z)=a ? e?λ|z|2,wherea ∈ Randλ >0.Then the following conditions are equivalent:

(1)The Fock Toeplitz operatorT?is positive;

(2)a

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(3)The Berezin transformfor allz∈C.

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ProofBy Lemma 2.1,we get that the matrix representation of T?is

Thus,T?is positive if and only iffor all n≥0,which gives thatThis proves(1)?(2).Now Lemma 2.2 implies

So we obtain thatis nonnegative on C if and only if

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Remarks 2.1The above proposition tells us that there exists a function ? ∈ L∞(C)such that the Fock Toeplitz operatobut ? is not a nonnegative function on C.To see this,we only need to takewith λ >0 in Proposition 2.1.Furthermore,this result also implies that to give a negative answer to Question 1.1,we only need to consider the case ofandin(1.1).

3 A Negative Answer to Question 1.1

In this section,we construct a function ? such that the Berezin transform of ? is positive on the complex plane but the corresponding Fock Toeplitz operator is not positive.To do so,we first consider the case 2α=β=1.Without loss of generality,we may assume c=1.Then we have the following theorem.

Theorem 3.1Suppose that,wherea,b∈ R.Then

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(1)ifb?orb≤0,thenT?is positive if and only ifis a nonnegative function on the complex plane;

(2)for eachb∈(?,?],there exists a real numberasuch thatfor allz∈C,butT?is not a positive Toeplitz operator.

ProofUse Lemmas 2.1–2.2 and let λ =,1,respectively.Then we obtain that the matrix representation of T?is given by

and the Berezin transform of ? is

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ProofBy Lemma 2.1,we see that the matrix representation of T?is

for all,andif and only if

for all z∈C.Therefore,is equivalent to

Letting∈[0,1],thenis equivalent to

First,we prove part(1)of the above theorem.Now we are going to determine the minimal value of the functionand the minimal term of the sequenceTo do this,we consider the following two cases.

Case ISuppose that b≥0.It is easy to see that f(t)is increasing on[0,1]andis decreasing for n＞1.So we have

and

Case IISupposeA simple calculation gives that f(t)is decreasing ifand f(t)is increasing ifNote that,so

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To find the minimal term of the sequence,observe thatand we claim thatis the minimal term.Indeed,if there exists some N>1 such that

then we obtain

where P:L2(C,dμ)→ F2is the orthogonal projection.Using the reproducing kernel Kz(w)=,we express the Toeplitz operator as an integral operator:

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Combining the above two cases gives the conclusion in part(1).

For the second part,we turn to consider b ∈ (?,?].First,observe that T?is not positive if and only if

To finish the proof,it suffices to show that for each b ∈ (?,?],there exists a real constant a such that

and T?is not positive.As shown above,we need the following inequality:

Indeed,we will show that

for each b∈ (?,?].To do this,we first find the minimal value of f(t).Note thatifUsing the monotonicity of the function f(see the argument in Case II),we get that

On the other hand,we have the following inequality:

for eachIndeed,if we consider the function

then the first inequality follows immediately from the mean value theorem.This finishes the proof of Theorem 3.1.

One may ask that if we remove the condition 2α = β,can we easily construct an example from(1.1)to give a negative answer to Question 1.1?Actually,we will show that the answer is yes by the following corollary,which can be proved by the same method as the one in the above theorem,so we omit its proof here.

Corollary 3.1Letwherea,b∈ R.Then for eachthere exists a real numberasuch thatfor allz∈C,butT?is not a positive Toeplitz operator onF2.

4 A Negative Answer to Question 1.2

In Section 3,we study the radial function ?(z)=a+be?α|z|2+ce?β|z|2(a,b,c∈ R,α,β >0)and choose real numbers a,b and c such thatis nonnegative on C and T?is not positive under the assumption 2α= β=1.In the final section,we will show that this is not true for all 2α=β>0.More precisely,we have the following theorem,which gives a negative answer to Question 1.2 with the condition 2α=β=

Theorem 4.1Suppose thatwherea,b∈ R.ThenT?is positiveif and only if?(z)is a nonnegative function onC.

These imply thatif and only if

So,T??0 if and only if

On the other hand,using Lemma 2.2 we obtain thatfor all z∈C if and only if for all z∈C.

Letting[0,1]andwe see that the above condition on the positivity of ? is equivalent to

Using the same idea as the one in the proof of Theorem 3.1,we divide the proof into the following four cases.

Case IWe first consider the case.It is easy to see that

which shows that T?is positive if and only if ? is a nonnegative function on C.

Case IISuppose thatIt is easy to check that g(t)is increasing ifand decreasing ifObserve thatso we have

On the other hand,we have

for allprovided thatThis implies that the sequenceis decreasing,so the minimal term isas desired.

Case IIIIn this case,we considerNote thatand

It follows from the argument in Case II that

Also,it is easy to see thatfor allThis gives that

Case IVFinally,we deal with the caseThenandIt follows that

Now using the same techniques as the one in Case II of Theorem 3.1,we see that the minimal term of the above sequence is the first term.We conclude that

for all b∈R.This completes the whole proof.

AcknowledgementThe author thanks the referees for their many suggestions.

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