Abstract
We consider the Reinhardt domain We express the explicit closed form of the Bergman kernel for using the exponential generating function for the Stirling number of the second kind. As an application, we show that the Bergman kernel for has zeros if and only if . The study of the zeros of is reduced to some real polynomial with coefficients which are related to Bernoulli numbers. This result is a complete characterization of the existence of zeros of the Bergman kernel for for all positive integers .
1. Introduction
Let be a bounded domain in . The space is denoted by the set of all holomorphic functions on satisfyingwhere is the volume measure on . For any , defined by is a bounded linear functional on . By Riesz representation theorem, there exists the unique element such that ; namely, for all . Define the Bergman kernel function for . It is defined for arbitrary bounded domains, but it is difficult to obtain the explicit form of the Bergman kernel for general bounded domains. Recently, the Bergman kernels for various domains have been computed explicitly in [1–6].
Over the last decade the Hartogs domainwas investigated, where is a suitably chosen continuous function on a bounded domain . The Bergman kernel for was obtained explicitly in [7] when is an irreducible bounded symmetric domain. This result was generalized to the cases when is the product of bounded symmetric domains in [8] and when is a bounded homogeneous domain in [9].
Also the problem of determining whether the Bergman kernels are zero-free has been a well-known open problem in several complex variables ever since Lu Qi-Keng raised the question related to the existence of Bergman representative coordinates. If the Bergman kernel for a bounded domain is zero-free for all , then is called the Lu Qi-Keng domain. One can see many examples of Lu Qi-Keng domains and non-Lu Qi-Keng domains in [8–13].
If is symmetric [8] or homogeneous [9], then the main part of the Bergman kernel for is the polynomial whose coefficients are written as the forms containing the Stirling number of the second kind. The Routh-Hurwitz theorem (see Lemma 11) gives the condition that a real polynomial has no zeros in the closed right half-plane, and using this criterion we have the algorithmic method of determining whether the Hartogs domain is a Lu Qi-Keng domain or not. The existence of zeros of all Hartogs domains is classified in [8, 9] only when the dimension of the base domain is low (less than 4). However it looks hard to study Lu Qi-Keng problem for all dimensions, since Routh-Hurwitz theorem involves too many terms when the order of the polynomial is large.
In this paper we consider the Reinhardt domain defined by
From Theorem 2.5 in [8], the Bergman kernel for can be obtained explicitly as the following.
Theorem 1. The Bergman kernel for is written as where where is the Stirling number of the second kind.
In Section 2, we prove Theorem 1 using the result in [8] and express in terms of the coefficients of a certain generating function (see Theorem 8). We use the well-known formal series for exponential generating function, where is the Stirling number of the second kind.
For the study of the existence of zeros of , we need to define and byThe zero set of the Bergman kernel with reduces to the zero set of the polynomial with . Now we write
In Section 3, we introduce the Routh-Hurwitz theorem that is efficient on checking whether the real polynomial has zeros in the right half plane. Using the generating form of coefficients of (see Proposition 12), we will show that if , then does not satisfy Routh-Hurwitz conditions, so we obtain the following main result of this paper.
Theorem 2. The Bergman kernel for is zero-free if and only if .
For the proof of Theorem 2, we will show that if , then at least one of , , or is negative (see Theorem 13). In Section 4, we discuss the properties of and prove Theorem 13 using properties of the Bernoulli numbers and Genocchi numbers.
Remark 3. In Theorem 5.2(ii) and Theorem 5.3 of [8], one can see that has no zeros and has zeros. The main contribution, Theorem 2 in this paper, is the complete classification of the answer to the Lu Qi-Keng problem for for all dimensions .
2. Explicit Form of the Bergman Kernel for
In [8], we know the explicit form of the Bergman kernel for Cartan-Hartogs domain in the case whenwhere is the generic norm with respect to the bounded symmetric domain . Using the numerical invariants , , with respect to the bounded symmetric domain, we define the Hua polynomial where for and . Then for any , we define bywhere . We also define satisfying
Theorem 4 (see [8]). Let , where and are defined as in (10). Then the Bergman kernel for is given by where
For any , the number of the partitions of into blocks is denoted by and called the Stirling number of the second kind.
Lemma 5 (see [14]). For any positive integer , it holds that
Proof of Theorem 1. For any , let be the unit disk and . Then and Since the Hua polynomial is , by (12) we have so that and .
Thus the function in Theorem 4 is reduced toNow we claim that is equal to . If , then, by (13), we haveBy Lemma 5, we have since . Thus we obtainIf we compare (20) and (22), then for all . Then from (19), it completes the proof of Theorem 1.
Then the polynomial can be written aswhere Note that Thus we have Note that for and for ; we have
Lemma 6. For any nonnegative integer , one has
Proof. Note that for any nonnegative integer , Note that Thus which completes the proof.
By Lemma 6, we have
Definition 7. Define
Let be the operator which gives the th coefficient in the series expansion of a generating function. It is well-known that the exponential generating function of is the formal power series
Using the above generating function, we prove the following.
Theorem 8. Let be the polynomial defined as in (8) with Then the coefficient is written as where .
Proof. By (27) and (32), the coefficient of can be expressed asNote that, by (34), we have which follows that It completes the proof.
3. Lu Qi-Keng Domains
In this section we investigate the explicit form of and prove that the Bergman kernel for has zeros for any positive integer .
Note that if , then Thus by Theorem 1 and (8), we obtain the following.
Lemma 9. The zero set is equal to the zero set where .
Since the holomorphic function maps the unit disk onto the right half plane, we obtain the following consequence of Theorem 1 and Lemma 9.
Lemma 10. For any positive integers , the domain is a Lu Qi-Keng domain if and only if all zeros of the polynomial lie in the closed left half plane .
The Routh-Hurwitz criterion is the most efficient method for determining whether the polynomial has zeros in the open left half plane. Let with real coefficients and , and define for by where if or .
Lemma 11 (Routh-Hurwitz/Liénard-Chipart [15]). All zeros of given polynomial lie in the open left half plane if and only if This condition is also equivalent to any one of the following four forms: (i), , ; , ,(ii), , ; , ,(iii); , ; , ,(iv); , ; ,
Proposition 12. Let be the operator which gives the th coefficient in the series expansion of a generating function
Proof. Note that By Theorem 8,
By Proposition 12, we have
Theorem 13. For each , at least one of , , or is negative. More precisely, one has (i) if ,(ii) if ,(iii) if .
Proof. We will prove it in Section 4.
Remark 14. (i) Note thathas two negative real zeros. Thus, is a Lu Qi-Keng domain.
(ii) In fact, we see that has one positive real zero . Thus, we conclude that the Bergman kernel for has zeros, so is not a Lu Qi-Keng domain.
(iii) By using a computer program (Maple or Mathematica), we computed the explicit values of , , and for . One can check Theorem 13 holds for in Tables 1 and 2.
Now we can prove the main theorem of this paper using Theorem 13.
Theorem 2 (again). The Bergman kernel for is zero-free if and only if .
Proof. (i) If or for , then by Theorem 13. So, the polynomial does not satisfy any condition in Lemma 11. It follows that is not a Lu Qi-Keng domain.
(ii) If for , then by Theorem 13. So, the polynomial does not satisfy conditions (iii) and (iv) in Lemma 11. It follows that is not a Lu Qi-Keng domain.
(iii) If for , then by Theorem 13. So, the polynomial does not satisfy conditions (i) and (ii) in Lemma 11. It follows that is not a Lu Qi-Keng domain.
By (i), (ii), and (iii), does not satisfy any condition of Routh-Hurwitz theorem, so the Bergman kernel for has zeros for all .
4. Proof of Theorem 13
In this section, we investigate the properties of , , and and prove Theorem 13. For convenience, we denote the functions , , , by
It is interesting that the numbers ’s are related to the following Bernoulli numbers. Bernoulli [16] introduced Bernoulli numbers for satisfying the identity where , , and The Genocchi numbers for are defined [17] by
Lemma 15 (see [18]). Let be any positive integer. Then (i)’s are positive for all positive integers ;(ii).
Lemma 16. For any , one has Moreover .
Proof. By Lemma 15(ii), we have It follows that Note that for all by Lemma 15(i). Thus and for all .
Lemma 17. For any , one has (i),(ii),(iii),(iv),(v),(vi).
Proof. From the identity we obtain It follows that Thus we obtain (i). From the identity we obtain It follows that Thus we obtain (ii). From the identity we have . Since for in Lemma 16, we obtain (iii). From the identity we obtain (iv). Similarly as the previous proofs, we can easily see that (v) comes from the identity and (vi) comes from the identity
Proposition 18. For any , it holds that (i), , ,(ii), , , ,(iii), ,(iv) for .
Proof. (ii) Note that (iii) Note that . Since , we have Here by Proposition 18(i) and (ii), (iv) Similarly as the proof of (iii), we obtain for .
Now we prove Theorem 13 using the above proposition.
Theorem 13 (again). For any positive integers , one has (i) if or ,(ii) if ,(iii) if .
Proof. (i) Note that By Proposition 18(ii), it follows that (ii) Note that since By Proposition 18(ii) and (iii), it follows that (iii) Note that since By Lemma 17(iii), we have Moreover .
Appendix
We add the explicit forms of Taylor expansions for some functions which have been discussed in Section 4.
Competing Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
This work was supported by NRF-2015R1D1A1A01060295 from National Research Foundation of Korea.