aa r X i v : . [ m a t h . C V ] S e p ZEROS OF NEW BERGMAN KERNELS
NOUREDDINE GHILOUFI AND SAFA SNOUN
Abstract.
In this paper we determine explicitly the kernels K α,β associated with new Bergman spaces A α,β ( D ) considered recentlyby the first author and M. Zaway. Then we study the distributionof the zeros of these kernels essentially when α ∈ N where the zerosare given by the zeros of a real polynomial Q α,β . Some numericalresults are given throughout the paper. Introduction
The notion of Bergman kernels has several applications and repre-sents an essential tool in complex analysis and geometry. Sometimesit is necessary to determine explicitly these kernels, however this isnot simple in general. In fact if an orthonormal basis of a Hilbertspace is given then the Bergman kernel of this space can be obtainedas a series using the basis elements. For example, the Bergman ker-nel of the space A α ( D ) of holomorphic functions on the unit disk D of C that are square integrable with respect to the positive measure dµ α ( z ) = ( α + 1)(1 − | z | ) α dA ( z ) is given by K α ( z, w ) = − zw ) α +2 .Hence this kernel has no zero in D . For more details about Bergmanspaces one can see [2]. In order to obtain kernels with zeros in D ,Krantz consider in his book [3] some subspaces of A α ( D ). In our state-ment, instead of considering subspaces, we modify slightly the measure dµ α to obtain a Bergman kernel that is comparable in some sense withthe previous one with some zeros in D . These spaces are consideredrecently by the first author and Zaway in [1]. We recall the main back-ground of this paper:Throughout the paper, D will be the unit disk of the complex plane C as it was mentioned before and D ∗ = Dr { } . We set N := { , , , . . . } the set of positive integers and R the set of real numbers. We claimthat a real number x is said to be positive (resp. negative) if x ≥ x ≤ Mathematics Subject Classification.
Key words and phrases.
Bergman spaces, Bergman Kernels, zeros of holomorphicfunctions, algebraic sets.
For every − < α, β < + ∞ , we consider the positive measure µ α,β on D defined by dµ α,β ( z ) := 1 B ( α + 1 , β + 1) | z | β (1 − | z | ) α dA ( z )where B is the beta function defined by B ( s, t ) = Z x s − (1 − x ) t − dx = Γ( s )Γ( t )Γ( s + t ) , ∀ s, t > dA ( z ) = 1 π dxdy = 1 π rdrdθ, z = x + iy = re iθ the normalized area measure on D .We denote by A α,β ( D ) the set of holomorphic functions on D ∗ thatbelongs to the space: L ( D , dµ α,β ) = { f : D −→ C ; measurable function such that k f k α,β, < + ∞} where k f k α,β, := Z D | f ( z ) | dµ α,β ( z ) . The set A α,β ( D ) is a Hilbert space and A α,β ( D ) = A α,m ( D ) if β = β + m with m ∈ N and − < β ≤ A α,β ( D ) = A α ( D ) is the classical Bergman space equipped withthe new norm k . k α,β , . Moreover, for any α, β > −
1, if we set(1.1) e n ( z ) = s B ( α + 1 , β + 1) B ( α + 1 , n + β + 1) z n for every n ≥ − m , then the sequence ( e n ) n ≥− m is an orthonormal basisof A α,β ( D ). Furthermore, if f, g ∈ A α,β ( D ) with f ( z ) = + ∞ X n = − m a n z n , g ( z ) = + ∞ X n = − m b n z n then h f, g i α,β = + ∞ X n = − m a n b n B ( α + 1 , n + β + 1) B ( α + 1 , β + 1)where h ., . i α,β is the inner product in A α,β ( D ) inherited from L ( D , dµ α,β ).The following main result determine the reproducing kernel of A α,β ( D ). EROS OF NEW BERGMAN KERNELS 3
Theorem 1.
Let − < α, β < + ∞ and K α,β be the reproducingBergman kernel of A α,β ( D ) . Then K α,β ( w, z ) = Q α,β ( wz )( wz ) m (1 − wz ) α where Q α,β ( ξ ) = ( α + 1) B ( α + 1 , β + 1) if β ∈ N β B ( α + 1 , β + 1) B ( α + 1 , β + 1) + ∞ X n =0 ( − ξ ) n n + β (cid:18) α + 1 n (cid:19) if β N with β = β − ⌊ β ⌋ − β − m . As a consequence of this main result, the study can be reduced tothe case β = β ∈ ] − , K α,β ( w, z ) = K α,β ( wz ) and M : A α,β ( D ) −→ A α,β ( D ) f B ( α + 1 , β + 1) B ( α + 1 , β + 1) z m f then the linear operator M is invertible and bi-continuous and K α,β = M − ◦ K α,β . Thus we can assume that m = 0 i.e. β = β .The proof of the main result is the aim of the following section. Thenas a consequence, we will prove that for α ∈ N and β ∈ ] − , K α,β is a totally real submanifold of D ∗ × D ∗ with real dimensionone formed by at most ( α + 1) connected components. This set isreduced to one connected component for β closed to − β → ( − + )and it is empty for β near 0 ( β → − ). These zeros are related tothe zeros set Z Q α,β of Q α,β in C . Hence we will concentrate essentiallyon the distribution of Z Q α,β . This will be the aim of the third sectionof the paper where we start by a general study and we conclude that Z Q α,β is formed by exactly ( α + 1) connected regular curves when β varies in the interval ] − , Proof of the main result
The proof of the first case is simple (it was done in [1]) however,the proof of the second one is more delicate and it will be done bysteps. Using the sequence ( e n ) n ≥− m given by (1.1), we deduce that the N. GHILOUFI AND S. SNOUN reproducing kernel of A α,β ( D ) can be written as follows K α,β ( w, z ) = + ∞ X n = − m e n ( w ) e n ( z ) = + ∞ X n = − m B ( α + 1 , β + 1) B ( α + 1 , n + β + 1) w n z n = B ( α + 1 , β + 1)( wz ) m + ∞ X n =0 B ( α + 1 , n + β − m + 1) ( wz ) n = R α,β ( wz )( wz ) m =: K α,β ( wz )where R α,β ( ξ ) = B ( α + 1 , β + 1) + ∞ X n =0 ξ n B ( α + 1 , n + β − m + 1) . If β = m ∈ N , then R α,m ( ξ ) = B ( α + 1 , m + 1) + ∞ X n =0 ξ n B ( α + 1 , n + 1)= ( α + 1) B ( α + 1 , m + 1)(1 − ξ ) α . We consider now the case β ∈ ] m − , m [ with m ∈ N and we prove theresult in two steps: • First step: the case α ∈ N . We start by proving the followingpreliminary lemma. Lemma 1.
We have R α,β ( ξ ) = Q α,β ( ξ )(1 − ξ ) α where Q α,β is a polynomial of degree α +1 with Q α,β (1) = 0 that satisfiesthe recurrence formula: Q α +1 ,β ( ξ ) = 1 α + β + 2 (cid:2) ξ (1 − ξ ) Q ′ α,β ( ξ ) + ( α + β − m + 2 + ( m − β ) ξ ) Q α,β ( ξ ) (cid:3) . Proof. If α = 0, then we have R ,β ( ξ ) = B (1 , β + 1) + ∞ X n =0 ξ n B (1 , n + β − m + 1)= 1 β + 1 + ∞ X n =0 ( n + β − m + 1) ξ n = Q ,β ( ξ )(1 − ξ ) with Q ,β ( ξ ) = 1 β + 1 (( m − β ) ξ + β − m + 1) . EROS OF NEW BERGMAN KERNELS 5
Assume that the result is proved for α ∈ N i.e. R α,β ( ξ ) = Q α,β ( ξ )(1 − ξ ) α where Q α,β is a polynomial of degree α + 1 with Q α,β (1) = 0. R α +1 ,β ( ξ ) = B ( α + 2 , β + 1) + ∞ X n =0 ξ n B ( α + 2 , n + β − m + 1)= B ( α + 2 , β + 1) + ∞ X n =0 Γ( α + 3 + n + β − m )Γ( α + 2)Γ( n + β − m + 1) ξ n = B ( α + 1 , β + 1) α + β + 2 + ∞ X n =0 ( α + 2 + n + β − m ) B ( α + 1 , n + β − m + 1) ξ n = 1 α + β + 2 (cid:0) ξ R ′ α,β ( ξ ) + ( α + β − m + 2) R α,β ( ξ ) (cid:1) = 1 α + β + 2 (cid:18) ξ Q ′ α,β ( ξ )(1 − ξ ) α + ξ (2 + α ) Q α,β ( ξ )(1 − ξ ) α +( α + β − m + 2) Q α,β ( ξ )(1 − ξ ) α (cid:19) = Q α +1 ,β ( ξ )(1 − ξ ) α , with Q α +1 ,β ( ξ ) = ξ (1 − ξ ) Q ′ α,β ( ξ ) + ( α + β − m + 2 + ( m − β ) ξ ) Q α,β ( ξ ) α + β + 2 . Thus Q α +1 ,β is a polynomial of degree α + 2 and Q α +1 ,β (1) = α + 2 α + β + 2 Q α,β (1) = 0 . (cid:3) Now, we can deduce the proof of Theorem 1 in the case α ∈ N . Thiswill be done by induction on α . The result is true for α = 0. Indeed,we have Q ,β ( ξ ) = 1 β + 1 (1 + β − β ξ ) = β B (1 , β + 1) B (1 , β + 1) (cid:18) β − ξ β (cid:19) . N. GHILOUFI AND S. SNOUN
Assume that the result is true until the value α . Thanks to Lemma 1,we have Q α +1 ,β ( ξ ) = 1 α + β + 2 (cid:0) ξ (1 − ξ ) Q ′ α,β ( ξ ) + ( α + 2 + β − β ξ ) Q α,β ( ξ ) (cid:1) = β B ( α + 1 , β + 1)( α + β + 2) B ( α + 1 , β + 1) " α +1 X j =1 j ( − j j + β (cid:18) α + 1 j (cid:19) ξ j + α +2 X j =1 ( j −
1) ( − j +1 j − β (cid:18) α + 1 j − (cid:19) ξ j + ( α + 2 + β ) α +1 X j =0 ( − j j + β (cid:18) α + 1 j (cid:19) ξ j + β α +2 X j =1 ( − j j − β (cid:18) α + 1 j − (cid:19) ξ j = β B ( α + 2 , β + 1) B ( α + 2 , β + 1) α +2 X j =0 ( − ξ ) j j + β (cid:18) α + 2 j (cid:19) . This achieves the first step. • Second step: The general case ( α > − ). We set S α,β ( ξ ) := B ( α + 1 , β + 1) β B ( α + 1 , β + 1) Q α,β ( ξ )= (1 − ξ ) α +2 β ∞ X n =0 B ( α + 1 , β + 1) B ( α + 1 , n + β + 1) ξ n = Q α,β ( ξ ) β and G α,β ( ξ ) := + ∞ X n =0 ( − n n + β (cid:18) α + 1 n (cid:19) ξ n . To prove the result it suffices to attest that S α,β = G α,β on D . Toshow this equality we will prove that both functions S α,β and G α,β satisfy the following differential equation:(2.1) ξF ′ ( ξ ) = − β F ( ξ ) + (1 − ξ ) α +1 , ∀ ξ ∈ D . It follows that S α,β − G α,β satisfies on D ∗ the homogenous differentialequation: ξF ′ ( ξ ) = − β F ( ξ ). In particular it satisfies the same ho-mogenous differential equation on ]0 , σ ∈ R such that for every t ∈ ]0 ,
1[ we have S α,β ( t ) − G α,β ( t ) = σt − β .Since S α,β − G α,β is differentiable at 0, we get σ = 0 i.e S α,β = G α,β on ]0 ,
1[ and by the analytic extension principle we conclude that S α,β = G α,β on D .To finish the proof we will show that both functions S α,β and G α,β EROS OF NEW BERGMAN KERNELS 7 satisfy the differential equation (2.1). For G α,β the result is obvious.Indeed ξG ′ α,β ( ξ ) = + ∞ X n =0 nn + β (cid:18) α + 1 n (cid:19) ( − ξ ) n = + ∞ X n =0 (cid:18) − β n + β (cid:19) (cid:18) α + 1 n (cid:19) ( − ξ ) n = (1 − ξ ) α +1 − β G α,β ( ξ ) . Now for S α,β , it is not hard to prove that ξS ′ α,β ( ξ ) = − (1 − ξ ) α +1 + ∞ X n =0 ( α + 1) B ( α + 1 , β + 1)( α + 1 + n + β ) B ( α + 1 , n + β + 1) ξ n = (1 − ξ ) α +1 − β S α,β ( ξ ) . Thus the proof of Theorem 1 is finished.As a first consequence of Theorem 1, we obtain the following identity:
Corollary 1.
Let − < α < + ∞ and − < β < . For every n ∈ N , n X k =0 (cid:18) α + 2 k (cid:19) ( − k B ( α + 1 , n − k + β + 1) = β B ( α + 1 , β + 1) (cid:18) α + 1 n (cid:19) ( − n n + β . Proof.
Thanks to Theorem 1, we have S α,β ( ξ ) = + ∞ X n =0 ( − n n + β (cid:18) α + 1 n (cid:19) ξ n = B ( α + 1 , β + 1) β (1 − ξ ) α +2 + ∞ X n =0 ξ n B ( α + 1 , n + β + 1)= B ( α + 1 , β + 1) β " + ∞ X n =0 (cid:18) α + 2 n (cid:19) ( − ξ ) n + ∞ X n =0 ξ n B ( α + 1 , n + β + 1) = B ( α + 1 , β + 1) β + ∞ X n =0 d n ξ n where d n = n X k =0 (cid:18) α + 2 k (cid:19) ( − k B ( α + 1 , n − k + β + 1) . So the result follows. (cid:3)
Using the proof of Theorem 1, one can conclude the following corol-lary:
N. GHILOUFI AND S. SNOUN
Corollary 2.
For every − < α and − < β < , the function G α,β ( ξ ) = β − Q α,β ( ξ ) = + ∞ X n =0 ( − n n + β (cid:18) α + 1 n (cid:19) ξ n satisfies: ξG ′ α,β ( ξ ) = (1 − ξ ) α +1 − βG α,β ( ξ ) and G α +1 ,β ( ξ ) = 1 α + β + 2 (cid:0) ξ (1 − ξ ) G ′ α,β ( ξ ) + ( α + β + 2 − βξ ) G α,β ( ξ ) (cid:1) = 1 α + β + 2 (cid:0) ( α + 2) G α,β ( ξ ) + (1 − ξ ) α +2 (cid:1) . Remarks 1. (1) Using the Stirling formula, one can prove that G α,β is boundedon the closed unit disk D . This fact will be used frequently inthe hole of the paper.(2) Thanks to Lemma 1, for α ∈ N , one has G α,β (1) = 0. Forthe general case, if G α ,β (1) = 0 for some − < α ≤ G α + n,β (1) = 0 for every n ∈ N .In the rest of the paper, we assume that G α,β (1) = 0. This may betrue for any − < α and − < β < Zeros of Bergman kernels
Using Theorem 1, the function K α, has no zero in the unit disk D .However if − < β < K α,β may have some zeros in D . We claimthat if ξ ∈ D ∗ is a zero of K α,β then the sets { ( z, w ) ∈ D ; wz = ξ } and { ( z, w ) ∈ D ; zw = ξ } define two totally real algebraic surfaces(of real dimension equal to 2) of C that are contained in the zeros setof the Bergman kernel K α,β . Thus it suffices to study the zeros set of K α,β .Due to an algebraic problem, we focus sometimes on the case α ∈ N ,because in this case the zeros of K α,β are given by the zeros of thepolynomial G α,β contained in D . Thus for α ∈ N , we will study thezeros set of G α,β in the hole complex plane C . It is interesting to discussthe variations of these sets in terms of the parameter β . All results on G α,β can be viewed as particular cases of those of the following lineartransformation. EROS OF NEW BERGMAN KERNELS 9
The Linear transformation T β . If O ( D (0 , R )) is the space ofholomorphic function on the disk D (0 , R ) and − < β <
0, then wedefine T β on O ( D (0 , R )) by T β f ( z ) = + ∞ X n =0 a n n + β z n for any f ( z ) = P + ∞ n =0 a n z n . The transformation T β is linear and bi-jective from O ( D (0 , R )) onto itself. It transforms any polynomial to apolynomial with the same degree. We start by the study of zeros of T β f in general then we specialize the study to the case f ( z ) = P α ( z ) =(1 − z ) α +1 where T β P α is exactly G α,β . Theorem 2.
Let < R ≤ + ∞ and f be a holomorphic function on D (0 , R ) such that ( f (0) , f ′ (0)) = (0 , . Then for every < r < R ,there exist − < β ( f, r ) ≤ − | f (0) || f (0) | + r | f ′ (0) | ≤ β ( f, r ) < that depend on f and r such that the function T β f has no zero in D (0 , r ) for every β ( f, r ) < β < and has exactly one simple zero in D (0 , r ) for every − < β < β ( f, r ) . When f (0) = 0 and f ′ (0) = 0 the result is reduced to ” 0 is the uniquezero (simple) of the function T β f in D (0 , r ) for every − < β < f ′ (0) = 0 and f (0) = 0 then ” the function T β f has nozero in D (0 , r ) for every − < β <
0. ”
Proof. If f ( z ) = P + ∞ n =0 a n z n for every z ∈ D (0 , R ) with ( a , a ) = (0 , F β,f ( z ) = a β + a β z. If | z | = r we have | T β f ( z ) − F β,f ( z ) | ≤ + ∞ X n =2 | a n | n + β r n . Moreover, if we set ψ ( β ) = (cid:12)(cid:12)(cid:12)(cid:12) | a | β + | a | r β (cid:12)(cid:12)(cid:12)(cid:12) − + ∞ X n =2 | a n | n + β r n then ψ (cid:18) − | a || a | + r | a | (cid:19) < and lim β → − ψ ( β ) = + ∞ ( resp. lim β → ( − + ψ ( β ) = + ∞ ) when a = 0 (resp. a = 0). It follows that there exist − < β ≤ − | a || a | + r | a | ≤ β < f and r such that for every β ∈ ] − , β [ ∪ ] β ,
0[ onehas ψ ( β ) >
0. Hence, for every β ∈ ] − , β [ ∪ ] β ,
0[ and | z | = r , wehave | T β f ( z ) − F β,f ( z ) | < | F β,f ( z ) | . Thus by Rouch´e Theorem, T β f and F β,f have the same number of zeros counted with their multiplicities inthe disk D (0 , r ). (cid:3) In the following lemma we collect some useful properties of T β f thatwill be used frequently in the sequel. Lemma 2. If f is a holomorphic function on D (0 , R ) and − < β < then the following assertions hold (1) The number is a zero of f if and only if it is a zero of T β f (with the same multiplicity). (2) The derivative of T β f satisfies z ( T β f ) ′ ( z ) = f ( z ) − βT β f ( z ) , ∀ z ∈ D (0 , R ) . (3) The functions f and T β f have a commun zero in D ∗ (0 , R ) ifand only if f ≡ . (4) The function T β f has a zero in D ∗ (0 , R ) with multiplicity greateror equal to if and only if f ≡ . Now we consider a fixed holomorphic function f on D (0 , R ) with f (0) = 0. We set H f ( β, z ) := T β f ( z ) = + ∞ X n =0 a n n + β z n for ( β, z ) ∈ ] − , × D (0 , R ) and D f := { ( β, z ) ∈ ] − , × D (0 , R ); H f ( β, z ) = 0 } . We assume that the set D f is not empty. Indeed if f ≡ c is a constantfunction then T β f ≡ cβ , thus D c = ∅ if c = 0 and D c = C if c = 0.Moreover it is easy to find some examples of non constant holomorphicfunctions g where T β g has no zero for some value of β . But we don’tknow if there exists a non constant function g such that D g is empty. Proposition 1.
The set D f is a submanifold of (real) dimension onein R formed by at most countable connected components ( Y f,k ) k .If Y is a connected component of D f then there exist − ≤ a Y < b Y ≤ and a C ∞ − function X :] a Y , b Y [ −→ D (0 , R ) such that Y = Graph ( X ) := { ( β, X ( β )); β ∈ ] a Y , b Y [ } . EROS OF NEW BERGMAN KERNELS 11
Moreover for every β ∈ ] a Y , b Y [ , one has (3.1) X ′ ( β ) = X ( β ) f ( X ( β )) + ∞ X n =0 a n ( n + β ) ( X ( β )) n . Proof.
For every ( β, z ) ∈ D α we have ∂H α ∂z ( β, z ) = ( T β f ) ′ ( z ) = 1 z f ( z ) = 0 . The result follows using the implicit functions theorem. (cid:3)
It is easy to see that if 0 < R < + ∞ then a Y > − Y of D f except the unique component Y f, given by The-orem 2 where a Y f, = −
1. However, b Y = 0 if and only if R = + ∞ i.e. f is an entire function. In this case, of entire functions, all functions X f,k are defined on ] − , Remark 2.
Using the same proof, the previous result can be improvedto the complex case as follows:If we set Ω := { β ∈ C ; − < ℜ e ( β ) < } and D f := { ( β, z ) ∈ Ω × D (0 , R ); H f ( β, z ) = 0 } then D f is a submanifold of (complex) dimension one in Ω × D (0 , R )formed by connected components. Thus, the Lelong number of thecurrent [ D f ] of integration over D f is equal to one at every point of D f .(This is due to the fact that all zeros of H f are simple).The following theorem gives the asymptotic behaviors of functions X f near − f is a polynomial. We claim that if f ( z ) = a + a z then the solution is explicitly determined by X f ( β ) = − a a β + 1 β . Hence we will consider the case when deg ( f ) ≥ Theorem 3.
Let f ( z ) = P pn =0 a n z n be a polynomial of degree p ≥ with f (0) = 0 . We set a n = | a n | e iθ n for any ≤ n ≤ p . The set D f is formed exactly by p connected components ( Y f,k ) ≤ k ≤ p − withthe corresponding functions X f,k :] − , −→ C . Again we keep X f, toindicate the function related to the unique component given by Theorem2. (1) For every ≤ k ≤ p − , we have lim β → − |X f,k ( β ) | = + ∞ and (3.2) X f,k ( β ) ∼ β → − (cid:18) − p | a | β | a p | (cid:19) p e i θ − θp +2 jkπp where j k ∈ Z that depends on k . (2) If f ′ (0) = 0 then for every ≤ k ≤ p − β → ( − + |X f,k ( β ) | = + ∞ and lim β → ( − + X f, ( β ) = 0 . Moreover, we have (3.3) X f, ( β ) ∼ β → ( − + a a (1 + β ) . and (3.4) X f,k ( β ) ∼ β → ( − + (cid:18) ( p − | a | (1 + β ) | a p | (cid:19) p − e i ( θ − θp +(2 sk +1) π ) p − , ∀ ≤ k ≤ p − for some s k ∈ Z that depends on k . If f (0) = 0 and f ′ (0) = 0 then all functions X f,k are bounded near − Proof.
Let 0 ≤ k ≤ p −
1. As a = 0 then using the equality a β + p X n =1 a n n + β ( X f,k ( β )) n = 0we obtain lim β → − X f,k ( β ) = ∞ and − a β ∼ β → − a p p + β ( X f,k ( β )) p . That means ( X f,k ( β )) p ∼ β → − − pa a p β so we get Equation (3.2).With the same way if a = 0 then for every 0 ≤ k ≤ p − β → ( − + X f,k ( β ) ∈ { , ∞} . Thanks to Theorem 2,lim β → ( − + X f, ( β ) = 0 and lim β → ( − + X f,k ( β ) = ∞ , ∀ ≤ k ≤ p − . Thus, we obtain X f, ( β ) ∼ β → ( − + − a a ββ . EROS OF NEW BERGMAN KERNELS 13
Therefore, Equation (3.3) follows.For 1 ≤ k ≤ p − a β ∼ β → ( − + − a p p + β ( X f,k ( β )) p − thus, ( X f,k ( β )) p − ∼ β → ( − + − ( p − a a p (1 + β )and Equation (3.4) follows. (cid:3) Application on the Bergman kernels.
As mentioned before,for any α ∈ N , T β P α = G α,β where P α ( z ) = (1 − z ) α +1 . Hence allprevious results are valid and more precisions are needed to accomplishthe study of X α,k := X P α ,k , ≤ k ≤ α . We start by claiming that if x < β x ∈ ] − ,
0[ such that G α,β x ( x ) = 0. It followsthat ( β x , x ) is in a component (says Y α, ) of D α := D P α . Hence, thecorresponding function X α, maps ] − ,
0[ onto ] − ∞ , X ′ α, ( β ) < β ∈ ] − ,
0[ thus it is a decreasing functionand lim β → − X α, ( β ) = −∞ , lim β → ( − + X α, ( r ) = 0 . Using Corollary 2, we can deduce that X α, ( β ) ≥ X α +1 , ( β ) for every β ∈ ] − ,
0[ (See Figure 1).
Remark 3.
For every α ∈ N , we set − < s α < X α, ( β ) = −
1. The polynomial G α,β has no zero in ] − ,
0[ forevery s α < β < − ,
0[ for every − < β < s α .We claim that ( s α ) α is an increasing sequence (See again Figure 1).The following lemma explain differently the conclusion of Theorem2 in the current statement (See Table 1 for numerical values of β and β given by Theorem 2 for this example). Lemma 3.
For every α > − , the family of functions ( β (1+ β ) G α,β ) β ∈ ] − , converges uniformly on D to the constant (resp. to the polynomial ( α + 1) ξ ) as β → − (resp. as β → ( − + ).In particular, for every m ∈ N (resp. m ∈ N ∗ ) the family of kernels ( K α,β ) β ∈ ] m − ,m [ converges uniformly on every compact subset of D ∗ to K α,m (resp. to K α,m − ) as β → m − (resp. as β → ( m − + ).Proof. The lemma is a simple consequence of the following equality: β (1 + β ) G α,β ( ξ ) = (1 + β ) − β (1 + α ) ξ + β (1 + β ) + ∞ X n =2 ( − ξ ) n n + β (cid:18) α + 1 n (cid:19) Figure 1.
Graphs of X α, for 0 ≤ α ≤ D (obtained usingthe Stirling formula). (cid:3) Table 1.
Numerical values of β ( P α ,
1) and β ( P α , α β ( P α , β ( P α , − . − . − . − . − . − . − . − . − . − . − . − . − . − . − . − . K α,β in terms of the parameter β . Essentially the factthat the Bergman kernel K α,β converges uniformly on every compactsubset of D to the classical Bergman kernel K α, when β → − .Now we will focus on the other components of D α . We use X α,k , ≤ k ≤ α to indicate the corresponding functions such that ℑ m ( X α,k ( β )) ≤ EROS OF NEW BERGMAN KERNELS 15 ≤ k ≤ ⌊ α +12 ⌋ and X α,α +1 − k ( β ) = X α,k ( β ) for every 1 ≤ k ≤ α . Theorem 3 can be written as follows: Proposition 2.
For every α ∈ N , we have X α,k ( β ) ∼ β → − (cid:18) − α + 1 β (cid:19) α +1 e i (2 k − α − πα +1 , ∀ ≤ k ≤ α X α,k ( β ) ∼ β → ( − + (cid:18) α ( α + 1)1 + β (cid:19) α e i (2 k − α − πα , ∀ ≤ k ≤ α X α, ( β ) ∼ β → ( − + − βα + 1 . The following figures (figures 2 and 3) explain numerically the resultof Proposition 2.
Figure 2.
Graphs of X , • (in red) with asymptoticcurves ( to C X , in blue and to C X , in green)3.3. Even and odd Bergman kernels.
Following the idea of Krantzdeveloped in [3], we consider the subspaces E α,β ( D ) and L α,β ( D ) of A α,β ( D ) generated respectively by the even ( e n ) n ≥ and the odd ( e n +1 ) n ≥ sequences. Hence E α,β ( D ) and L α,β ( D ) are Hilbert subspaces of A α,β ( D )formed respectively by even and odd functions. The reproducing Bergmankernels of these spaces are given by E α,β ( z, w ) = E α,β ( zw ) and L α,β ( z, w ) = L α,β ( zw ) where E α,β ( ξ ) = ( K α,β ( ξ ) + K α,β ( − ξ ))= 12(1 − ξ ) α +2 (cid:0) (1 + ξ ) α +2 Q α,β ( ξ ) + (1 − ξ ) α +2 Q α,β ( − ξ ) (cid:1) =: I α,β ( ξ )2(1 − ξ ) α +2 Figure 3.
Graphs of X , • (in red) with asymptotic curveto C X , (in green)and L α,β ( ξ ) = ( K α,β ( ξ ) − K α,β ( − ξ ))= 12(1 − ξ ) α +2 (cid:0) (1 + ξ ) α +2 Q α,β ( ξ ) − (1 − ξ ) α +2 Q α,β ( − ξ ) (cid:1) =: J α,β ( ξ )2(1 − ξ ) α +2 Again, to study the zeros of even and odd Bergman kernels, it sufficesto study the zeros of the corresponding functions I α,β and J α,β . Let ε α,β (resp. Θ α,β ) be the number of zeros of the function I α,β (resp. J α,β ) inthe unit disk D counted with their multiplicities. To determine ε α,β andΘ α,β in the case when α ∈ N , we start by the case β = 0. In this case itis easy to check that the zeros of I α, are given by z k := − i tan (cid:16) (2 k +1) π α +2) (cid:17) where 0 ≤ k ≤ α + 1 (we omit the value k for which cos( (2 k +1) π α +2) ) = 0whenever α is odd). Similarly to the even case, the zeros of J α, aregiven by w k := − i tan (cid:0) kπα +2 (cid:1) , ≤ k ≤ α +1. It follows that if α = 4 τ + r with τ ∈ N and 0 ≤ r ≤ ε α, = (cid:26) τ if r = 02 τ + 2 if ≤ r ≤ α, = (cid:26) τ + 1 if ≤ r ≤ τ + 3 if r = 3 Proposition 3.
Let α = 4 τ + r ∈ N with τ ∈ N and ≤ r ≤ . (1) If r = 0 then: (a) There exists − < β < such that for every β < β ≤ we have ε α,β = ε α, . (b) There exists − < β < such that for every − < β < β we have Θ α,β = ε α, + 1 . EROS OF NEW BERGMAN KERNELS 17 (2) If r = 2 then: (a) There exists − < β < such that for every − < β < β we have ε α,β = Θ α, + 1 . (b) There exists − < β < such that for every β < β ≤ we have Θ α,β = Θ α, .Proof. We claim that ± i are zeros of I α, (resp. J α, ) when α = 4 τ (resp. α = 4 τ + 2). For this reason we omit the corresponding valuesof α in the proposition in order to use the Rouch´e theorem. Hence itsuffices to study the convergence in terms of the parameter β .Thanks to Lemma 3, the family of polynomials ((1 + β ) Q α,β ( ξ )) − <β< converges to 1 on D when β → − and to the polynomial ( α + 1) ξ when β → ( − + . It follows that (1 + β ) I α,β ( ξ ) converges to I α, ( ξ ) on D as β → − and to ( α + 1) ξ J α, ( ξ ) on D as β → ( − + .For the odd case, the family (1 + β ) J α,β ( ξ ) converges to J α, ( ξ ) when β → − and to ( α + 1) ξ I α, ( ξ ) when β → ( − + . Using the Rouch´etheorem the result follows. (cid:3) To improve the previous result, we consider the number of zeros b ε α, (resp. b Θ α, ) of the function I α, (resp. J α, ) in the closed unit disk D given by b ε α, = 2 τ + 2 and b Θ α, = (cid:26) τ + 1 if ≤ r ≤ τ + 3 if ≤ r ≤ α = 4 τ + r . Using the same idea, one can prove the followingcorollary: Corollary 3.
Let α ∈ N and η = tan (cid:0) π + πα +2 (cid:1) . (1) There exist − < β < β < that depend on α such thatfor every < η < η , the polynomial I α,β ( ηξ ) has exactly b ε α, zeros in D for every β ∈ ] β , and b Θ α, + 1 zeros in D for every β ∈ ] − , β [ . (2) There exist − < β < β < that depend on α such thatfor every < η < η , the polynomial J α,β ( ηξ ) has exactly b Θ α, zeros in D for every β ∈ ] β , and b ε α, + 1 zeros in D for every β ∈ ] − , β [ . If the conditions of the previous proposition are satisfied then onecan take η = 1 in the corollary to obtain the same result given by theproposition. 4. Open problems
It is interesting to study the asymptotic distribution of zeros of G α,β when α ∈ N and goes to infinity. In other words, can we find a positive measure µ such that the sequence of measures µ α,β := 1 α + 1 α +1 X j =0 δ X α,j ( β ) converges weakly to the measure µ as α → + ∞ ? Geometrically, thedistribution of the set {X α,j ( β ) , ≤ j ≤ α + 1 } may depend on α insome non trivial way. For example, the distribution of sets X α, • ( − − )for α = 31 , , , ,
36 are similar (see figure 4) however these are dif-ferent to the one that correspond to α = 35 (see figure 5).Can we find explicitly the equation of the parametric curve that de-scribe the set X α, • ( β )? (this curve may be a circle in figure 4 for α = 31 , , , , ≤ k ≤ α , there exists t α,k ∈ ] − ,
0[ such that |X α,k ( t α,k ) | = min − <β< |X α,k ( β ) | and satisfies α +1 X j,k =0 (cid:18) α + 1 j (cid:19)(cid:18) α + 1 k (cid:19) ( − j + k ( j + t α,k ) R j + kα,k cos( θ α,k ( j − k )) = 0with X α,k ( t α,k ) = R α,k e iθ α,k . One of the most important question is to see if the critical value t α,k of β that realizes the minimum of |X α,k ( β ) | doesn’t depend on k .It means that all functions attempt their minimums at the same ”time”.For even and odd kernels, can we prove Corollary 3 with η = 1?Indeed, if we show that the zeros of I α,β ( ξ ) and J α,β ( ξ ) that convergesto ± i are in D then we conclude the result. We note that this fact isconfirmed numerically for some values of α . References [1]
N. Ghiloufi and M. Zaway , Meromorphic Bergman spaces, preprint(2020) 1-16.[2]
H. Hedenmalm, B. Korenblum and K. Zhu , Theory of Bergmanspaces, Graduate texts in Mathematics, 199 (2000).[3]
S.G. Krantz , Geometric analysis of the Bergman kernel and metric,Graduate text in Mathematics 268, (2013).
EROS OF NEW BERGMAN KERNELS 19
Annex: Numerical results
All figures of this paper were produced using Python software. Wegive here the used code.****************************************** from scipy import specialfrom sympy.abc import x, y, zdef A(beta, alpha):s=0for j in range(0, alpha+2):s=s+ ((1)/(j+beta)*special.binom(alpha+1,j)*(-x)**j)return simport numpyfrom cmath import *from sympy.solvers import solveimport csvDATA_PATH = ’/content/drive/My Drive/graphes data/alpha7_7.csv’i=1with open(DATA_PATH, mode=’w’, newline=’’) as points_file:points_writer = csv.writer(points_file, delimiter=’,’)for beta in numpy.arange(10**(-6), 1, 10**(-2)):row = []for s in solve(A(beta, 6), x):sol = complex(s)row.append(-beta)row.append(sol.real)row.append(sol.imag)points_writer.writerow(row)print(i)i=i+1i=1for beta in numpy.arange(1-10**(-2), 1, 10**(-3)):row = []for s in solve(A(beta, 6), x):sol = complex(s)row.append(-beta)row.append(sol.real)row.append(sol.imag)points_writer.writerow(row)print(i)i=i+1 **********************************************In the following, we present some zeros sets of G α,β for β = − − and α ∈ { , , , , , , , , } . The values of α and β are chosenarbitrary just to see that there is no geometric stability of these zeros.It is possible that the geometric distribution of zeros of G α,β dependson both α and β in a complicate manner. It is also possible that if wesee numerically the geometric distribution of zeros of G α,β for α largeenough, then some new limit curve appear. However as a materialproblem, it was not possible for us to exceed the value α = 101. Figure 4.
The Sets X α, • ( − − ) for α ∈ { , , , , } E-mail address : [email protected], [email protected] Department of mathematics, College of science, P.O. box 400 KingFaisal University, Al-Ahsa, 31982, Kingdom of Saudi Arabia.University of Gabes, Faculty of Sciences of Gabes, LR17ES11 Math-ematics and Applications, 6072, Gabes, Tunisia.
E-mail address : [email protected] University of Gabes, Faculty of Sciences of Gabes, LR17ES11 Math-ematics and Applications, 6072, Gabes, Tunisia.
EROS OF NEW BERGMAN KERNELS 21
Figure 5.
The Sets X α, • ( − − ) for α = 35 (in violet)at left and α = 49 (in red) and α = 51 (in blue) at right Figure 6.
The Set X , • ( − −4