# Uniform Extendibility of the Bergman Kernel for Generalized Minimal Balls

Authors: Jong-Do Park
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We consider a class of convex domains which contains non-Reinhardt domains with nonsmooth boundary. We show that the domains of this class satisfy the conditionQ.


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## Body

## 1. Introduction

LetΩ be a bounded domain in Cn. The Bergman space A2(Ω) is the space of holomorphic functions on Ω which are square integrable. The Bergman space has the Bergman kernel K(z,w), which is uniquely determined by the properties that it is an element of A2(Ω) in z, is conjugate symmetric, and reproduces A2(Ω). The Bergman projection P of L2(Ω) onto A2(Ω) is given by
(1)Pfw=∫Ω‍fzKz,w¯dνz,
where f∈L2(Ω) and w∈Ω.A bounded domainΩ is said to satisfy the condition Q if the Bergman projection maps C0∞(Ω) into the space O(Ω¯) of all holomorphic functions on Ω that can be extended holomorphically to a neighborhood of Ω¯.It was also known that the conditionQ holds on a domain Ω whenever the ∂¯-Neumann problem is globally real analytic hypoelliptic on Ω [1–3]. When a domain is a smoothly bounded pseudoconvex domain with real analytic boundary, there are many well-known examples on which the globally real analytic hypoellipticity of the ∂¯-Neumann problem holds:(i)
strictly pseudoconvex domains [4–6],(ii)
smooth bounded pseudoconvex circular domains with∑j=1n‍zj(∂r/∂zj)≠0 near the boundary, where r(z) is the defining function for domains [7],(iii)
Reinhardt domains [8].In this paper, we consider a class of convex circular domains which neither are Reinhardt nor have smooth boundaries and prove that the conditionQ holds on each domain in the class defined above. Recently, there are many results of computing the Bergman kernels for various domains explicitly. See [9–14].Let us restrict our attention to circular domains which are not Reinhardt. Forz=(z1,…,zn)∈Cn, consider the norm
(2)N*z=z2+z•z2,
where z•w=∑j=1n‍zjwj and |z|2=z•z¯. The norm N* is the smallest norm in Cn that coincides with the Euclidean norm in Rn, under certain restriction [15]. The minimal ball B*∶={z∈Cn:N*(z)<1} is the first known bounded domain in Cn which is neither Reinhardt nor homogeneous and for which the Bergman kernel can be computed explicitly as the closed form [16].For any positive integersp and q, we denote by Mp,q(C) the pq-dimensional complex vector space of all (p×q)-matrices with complex coefficients. If Z=(zjk)1≤j≤p;1≤k≤q is an element of Mp,q(C), we set
(3)Z*∶=∑j=1p‍∑k=1q‍zjk2+∑k=1q‍zjk21/2.
Let d be a positive integer and let m=(m1,…,md) and n=(n1,…,nd) be multi-indices with mj≥1 and nj≥1 for all j=1,…,d. We denote by j0 the number of those nj’s that are equal to 1. If j0≥1, we may assume without loss of generality that n1=⋯=nj0=1 and nj≥2 for j=j0+1,…,d. Let a=(a1,…,ad)∈Rd with aj≥1, and consider the function ρa,m,n(Z) defined for Z=(Z(1),…,Z(d))∈Mm1,n1(C)×⋯×Mmd,nd(C) by
(4)ρa,m,nZ∶=∑j=1d‍Zj*2aj,
and define thegeneralized minimal ballsΩa,m,n∶={Z:ρa,m,n(Z)<1}, where
(5)Zj∶=Z1j⋮Zmjj,Zlj∶=Zl,1j,…,Zl,njj∈Cnj.
Note that
(6)Zj*=∑l=1mj‍∑r=1nj‍Zl,rj2+∑r=1nj‍Zl,rj2=∑l=1mj‍NZlj2,
where N(·)∶=2N*(·).Remark 1.
(i) Note that, forn=(1,…,1), the domain Ωa,m,n is a complex ellipsoid, which is a smoothly bounded Reinhardt domain and clearly satisfies the condition Q [8].
(ii) Note also that, ford=1 and a=m=(1), the domain Ωa,m,n is the minimal ball with radius 1/2, which is not Reinhardt but satisfies the condition Q [16]. If at least one of nj’s is greater than 1, then Ωa,m,n is neither smooth nor Reinhardt.In 2002, Youssfi [17] computed the Bergman kernel for the generalized minimal ball Ωa,m,n using the transformation formula of the Bergman kernel under proper holomorphic mappings [18–20]. See the Bergman kernel for Ωa,m,n in Theorem 4. The Bergman kernel for Ωa,m,n contains an infinite series Ha,m,n(x) defined by
(7)∑k1=0∞‍⋯∑kd=0∞×∏j=j0+1d‍Γkj+mjnj-mjaj-1Γ1+∑j=1j0‍kj+mjaj+∑j=j0+1d‍kj+mjnj-mjaj·∏j=1j0‍Γkj+mjaj·∏j=j0+1d‍Γkj+mjnj-mjaj-1·x1k1⋯xdkd.
It is the crucial part to investigate the infinite series Ha,m,n(x), when we study the Bergman kernel for Ωa,m,n.In [16], it was proved that the minimal ball satisfies the condition Q. In this paper, we generalize this result to the generalized minimal balls Ωa,m,n as follows.Theorem 2.
The conditionQ holds on each domain
(8)Ωa,m,n=Z:ρa,m,nZ<1,
where ρa,m,n(Z) is defined as in (4) and all aj’s are positive integers.Similarly as in [16], we obtain the properties of the proper holomorphic mappings on Ωa,m,n.Corollary 3.
LetD⊂Cm1n1+⋯+mdnd be any bounded circular domain which contains the origin.(i)
Iff:Ωa,m,n→D is a proper holomorphic mapping, then f extends holomorphically to a neighborhood of Ω¯a,m,n.(ii)
IfD is smooth then there is no proper holomorphic mapping from Ωa,m,n into D.In Section2, we review the notation and the Bergman kernel for Ωa,m,n following [17]. In Section 3, we prove the theorem using the characterization theorem (Lemma 6). In the final section, the explicit formulas of Ha,m,n(x) have been obtained in some cases and we show that the minimal ball satisfies the condition Q using Theorem 2.


## 2. Bergman Kernel Function forΩa,m,n

In this section we review the Bergman kernel forΩa,m,n following the notations which were introduced in [17]. For u=(uj,l)l=1,…,mj;j=1,…,d with uj,l∈C, we define
(9)tu∶=tj,luj,ll=1,…,mj;j=1,…,d,sett2=tt.Consider the differential operatorD=D(a,m,n) acting on functions f(t) for t=(tj,l) by
(10)Dft∶=a1⋯adD1,1⋯D1,m1⋯Dd,1⋯Dd,mdft,
where
(11)Dj,lft∶=4ftifj≤j0,2nj-1!2tj,ldfdtj,l+nj-1ftifj>j0.We also consider the differential operatorD~=D~(a,m,n) acting on functions g(x) for x=(x1,…,xd) defined by
(12)D~gx∶=D~1⋯D~d∏j=1j0‍xjmj-1∏j=j0+1d‍xjmjnj-mj-1gx,
where
(13)D~j∶=dmj-1dxjmj-1ifj≤j0,dmjnj-mj-1dxjmjnj-mj-1ifj>j0.Note thatj0 depends on n. Since nj≥2 for j>j0, then we have mjnj-mj-1≥mj-1≥0.We also consider the product group(14)Λ∶=-1,1mj0+1×⋯×-1,1md
and extend each element ε=(εj,l), l=1,…,mj, j=j0+1,…,d, of Λ to an element ε~ of {-1,1}m by
(15)ε~j,l∶=1ifj≤j0,εj,lifj>j0.Iff(t) is a smooth function defined on the region of t=(tj,l) with ∑j=1d‍∑l=1mj‍tj,l<1, we consider the function f~ on the region of (t,u) with ∑j=1d‍∑l=1mj‍(tj,l+uj,l)<1 in Cm×Cm by
(16)f~t,u∶=∑ε∈Λ‍ft+ε~u∏j=j0+1d‍∏l=1mj‍εj,luj,l,
where Cm=Cm1×⋯×Cmd. Then define Lf(t,u)∶=f~(t,u). For example, we consider d=2 and j0=1. Then,
(17)f~t,u=∑ε∈Λ‍ft1,1+u1,1,t2,1+ε2,1u2,1ε2,1u2,1=ft1,1+u1,1,t2,1+u2,1u2,1-ft1,1+u1,1,t2,1-u2,1u2,1.Consider the series defined forx=(x1,…,xd)∈Cd by
(18)Gx∶=D~Ha,m,nx,
where Ha,m,n(x) is defined in (7). The series G(x) converges for x<1. Finally, we set
(19)Rt∶=G∑l=1m1‍t1,l,…,∑l=1md‍td,l,Ht,u∶=LDRt,u.We extend the operation• to a Cm-valued bilinear mapping defined for Z,W∈Mm1,n1(C)×⋯×Mmd,nd(C) by
(20)Z•W∶=Zlj•Wljl=1,…,mj;j=1,…,d
if Z=(Z(1),…,Z(d)), W=(W(1),…,W(d)), and
(21)Zj=Z1j⋮Zmjj,Wj=W1j⋮Wmjj.In 2002, Youssfi [17] computed the Bergman kernel for Ωa,m,n as follows.Theorem 4 ([17], Theorem A).
The Bergman kernel function for the domainΩ=Ωa,m,n is given by the formula
(22)KΩZ,W=1vΩ·Ht,uH0,0,forZ,W∈Ω,
where t=Z•W¯ and u=(Z•Z)(W•W¯). The latter product is understood in the sense of (9). More precisely, one has t=(tj,l)l=1,…,mj;j=1,…,d and u=(uj,l)l=1,…,mj;j=1,…,d, where
(23)tj,l=Zlj•Wlj¯=∑r=1nj‍Zl,rjWl,rj¯,uj,l=Zlj•ZljWlj•Wlj¯=∑r=1nj‍Zl,rj2∑r=1nj‍Wl,rj2¯.Remark 5.
In fact, the weighted Bergman kernel and Szegö kernel have been obtained in [17].


## 3. Proof of Theorem2

For the study ofHa,m,n(x), we need to introduce Appell’s hypergeometric functions. Recall that Appell’s hypergeometric function [21] is defined by
(24)FAnα;β1,…,βn;γ1,…,γn;ζ1,…,ζn=∑k1=0∞‍⋯∑kn=0∞‍αk1+⋯+knβ1k1⋯βnknk1!⋯kn!γ1k1⋯γnknζ1k1⋯ζnkn,
where (a)m=Γ(a+m)/Γ(a).Now we prove the main theorem of this paper. At first we need to representHa,m,n(x) in terms of Appell hypergeometric functions FA(d) when aj’s are positive integers. If we write
(25)hj=mj,j≤j0;mjnj-1,j>j0,
then we have
(26)Ha,m,nx=∑k1=0∞‍⋯∑kd=0∞‍Γ1+∑j=1d‍kj+hj/aj∏j=1d‍Γkj+hj/ajxk=∏j=1d‍xj1-hj∑k1=h1-1∞‍⋯∑kd=hd-1∞‍Γ1+∑j=1d‍kj+1/aj∏j=1d‍Γkj+1/ajxk=∏j=1d‍xj1-hjIa;x+P0x,
where P0(x)=P0(x1,…,xd) is a polynomial such that there exists a j with 1≤j≤d satisfying that the degree in xj is strictly less than hj-1 and
(27)Ia;x∶=∑k1=0∞‍⋯∑kd=0∞‍Γ1+∑j=1d‍kj+1/aj∏j=1d‍Γkj+1/ajxk.Thus we have(28)Gx=D~1⋯D~d∏j=1d‍xjhj-1Ha,m,nx=∂h1-1∂x1h1-1⋯∂hd-1∂xdhd-1Ia;x.Ifaj’s are positive integers, then there are positive integers pj and qj such that kj=ajpj+qj and 0≤qj≤aj-1. It follows that
(29)I(a;x)=∑q1=0a1-1‍⋯∑qd=0ad-1‍∑p1=0∞⋯∑pd=0∞‍∏j=1d‍Γpj+qj+1aj-1Γ1+∑j=1d‍pj+∑j=1d‍qj+1aj·∏j=1d‍Γpj+qj+1aj-1·∏j=1d‍xjajpj+qj=∑q1=0a1-1‍⋯∑qd=0ad-1‍∏j=1d‍cpj,qj,ajxjqj×FA(d)1+∑j=1d‍pj+∑j=1d‍qj+1aj;∑j=1d‍qj+1aj1;q+1a;xa,
where
(30)cpj,qj,aj∶=Γ1+∑j=1d‍pj+∑j=1d‍qj+1/aj∏j=1d‍Γqj+1/aj.
Here we used the notation 1=(1,…,1︸d) and (q+1)/a=(q1+1)/a1,…,(qd+1)/ad.Note that Appell’s hypergeometric functionsFA(n)(α;β;γ;x) converge for x<1 and diverge for x>1. Thus there exist a nonnegative constant CJ and an analytic function P1(x) such that
(31)Ia;x=P1x∏J⊂{1,…,d}‍1-∑j∈J‍xjajCJ.From (28), there exist a nonnegative constant CJ* and an analytic function P2(x) such that
(32)Gx=P2x∏J⊂{1,…,d}‍1-∑j∈J‍xjajCJ*.
From (19), there exists an analytic function P3(x) such that
(33)DRt=∏j=1d‍aj∏l=1mj‍Dj,lRt=P3tP4t,
where
(34)P4t∶=∏J⊂{1,…,d}‍1-∑j∈J‍∑l=1mj‍tj,lajCJ*.The following lemma is useful when we show that a bounded domain satisfies the conditionQ.Lemma 6 ([22, 23]).
For a bounded domainΩ, the condition Q holds on Ω if and only if, for each compact subset L of Ω, there exists an open set UL containing Ω¯ such that(i)
for each fixedw0∈LK(z,w0) is holomorphic in z on UL,(ii)
K(z,w) is continuous on UL×L.The above property for smoothly bounded domains was proved by Chen [22] and duplicated for general bounded domains by Thomas [23].At first we need to study the singular points off~(t,u) defined as in (16).Lemma 7.
Letf(t) be a smooth function on a region of t=(tj,l) with ∑j=1d‍∑l=1mj‍tj,l<1. Denote
(35)Pf∶=t=tj,l∈Cm:fthasapoleatt.
If f~(t,u) is defined as in (16), then there exists a function F(t,u) such that(i)
PF∶={(t,u)∈Cm×Cm:Ft,uhasapoleat(t,u)} equal to {(t+εu)∈Pf:ε∈Λ},(ii)
f~(t,u)=F(t,u)×2∑j=j0+1d‍mj for (t,u)∉PF.Proof.
If we defineaj,l+=tj,l+uj,l and aj,l-=tj,l-uj,l, then t+ε~u is equal to one of
(36)a1,1+⋮a1,m1+,…,aj0,1+⋮aj0,mj0+,aj0+1,1±⋮aj0+1,mj0+1+,…,ad,1±⋮ad,md±,
where aj,l± is one of either aj,l+ or aj,l-.
LetN0 be the number of aj,l-’s. For fixed j and l with j0+1≤j≤d, 1≤l≤mj, we have
(37)f~t,u=1∏j=j0+1d‍∏l=1mj‍uj,l∑ε∈Λ‍-1N0ft+ε~u=1∏j=j0+1d‍∏l=1mj‍uj,l·∑aj,l+‍-1N0ft+ε~u+∑aj,l-‍-1N0+1ft+ε~u=1∏j=j0+1d‍∏l=1mj‍uj,l·∑aj,l+‍-1N0ft+ε~u-∑aj,l-‍-1N0ft+ε~u,
where ∑aj,l+‍(∑aj,l-‍, resp.) means summation running over ε∈Λ which contains aj,l+(∑aj,l-‍, resp.).
From the above formula, we see that a numerator has a factoraj,l+-aj,l-. Repeating this process for all j0+1≤j≤d and 1≤l≤mj, we obtain that f~(t,u) is equal to certain function F(t,u) times
(38)∏j=j0+1d‍∏l=1mj‍aj,l+-aj,l-=∏j=j0+1d‍∏l=1mj‍2uj,l=2∑j=j0+1d‍mj∏j=j0+1d‍∏l=1mj‍uj,l,
which means that F satisfies condition (ii).Lettj,l and uj,l be defined as in (23). By Lemma 7 the Bergman kernel of Ωa,m,n has poles at (t,u)∈Cm×Cm with
(39)∑j∈J‍∑l=1mj‍tj,l±uj,l1/2aj=1
for some J⊂{1,…,d}. Write tj,l and uj,l as
(40)tj,l=∑r=1nj‍arbr¯,uj,l=∑r=1nj‍ar2∑r=1nj‍br2¯,
where ar=Zl,r(j) and br=Wl,r(j). Then it follows that
(41)NZlj=∑r=1nj‍ar2+∑r=1nj‍ar2,NWlj=∑r=1nj‍br2+∑r=1nj‍br2.The following lemma is needed when we estimate the functionP4(x).Lemma 8.
Fora=(a1,…,an) and b=(b1,…,bn) with ar,br∈C, let
(42)t=∑r=1n‍arbr¯,u=∑r=1n‍ar2·∑r=1n‍br2¯.
Then one has
(43)t±u≤∑‍ar2+∑‍ar2∑‍br2+∑‍br2.Proof.
By the Schwartz inequality, we have(44)t±u≤∑‍arbr+∑‍ar2·∑‍br2¯≤∑‍ar2∑‍br2+∑‍ar2·∑‍br2¯≤∑‍ar2+∑‍ar2∑‍br2+∑‍br2.Then, by using Lemma8, the identity (41), and the inequality (xy)1/2a≤(1/2)(xa+ya) for x,y>0, we have
(45)1-∑j∈J‍∑l=1mj‍tj,l±uj,l1/2aj≥1-∑j∈J‍∑l=1mj‍tj,l±uj,l1/2aj≥1-∑j∈J‍∑l=1mj‍NZlj2·NWlj2(1/2)aj≥1-12∑j∈J‍∑l=1mj‍NZlj2aj+NWlj2aj=1-12∑j∈J‍∑l=1mj‍NZlj2aj+∑l=1mj‍NWlj2aj≥1-12∑j∈J‍∑l=1mj‍NZlj2aj+∑l=1mj‍NWlj2aj≥1-12ρZ+ρW.
Thus by Lemma 6, each domain Ωa,m,n satisfies the condition Q.


## 4. Examples

In fact, it is difficult to writeHa,m,n(x) explicitly. In the final section, we give some explicit formulas for Ha,m,n(x).Lemma 9.
Letα=(α1,…,αn) with αj∈N∪{0} and t=(t1,…,tn)∈Cn. For any s>0 and t∈Cn with ∑j=1n‍tj<1, one has
(46)∑α‍Γ∑j=1n‍αj+sΓsΓαj+1∏j=1n‍tjαj=11-∑j=1n‍tjs.Proof.
Consider the Taylor expansion of the function in the right hand side of (46). See the details in [11] or [13].Lemma 10.
Leta(k) be a polynomial of degree r with variable k. Then there exist unique constants cl for 0≤l≤r such that
(47)∑k=0∞‍akxk=∑l=0r‍cl1-xl+1.Proof.
For a given polynomiala(k), there exist unique constants bl for 1≤l≤r+1 such that
(48)ak=b0+b1k+1+b2k+1k+2+⋯+brk+1⋯k+r.
Note that, for l≥1,
(49)11-xl+1=1l!dldxl11-x=1l!dldxl∑k=0∞‍xk=1l!∑k=0∞‍k+1⋯k+lxk.
If we write cl=bl/l!, then we obtain (47).Example 11.
The complexd-dimensional ball with radius 1/2: if d≥1 and a=m=n=(1,…,1), then using Lemma 9, we have
(50)Ha,m,nx=∑h∈N0d‍1+∑j=1d‍hj+1∏j=1d‍Γhj+1xh=Γd+11-∑j=1d‍xjd+1.Example 12.
The Thullen domain{(z1,z2)∈C2:|z1|2+|z2|2p<1} with a positive integer p: if d=2, a=(1,p), and n=m=(1,1), then
(51)Ωa,m,n=z1,z2∈C2:2z12+2pz22p<1
is linearly biholomorphic to the Thullen domain {(z1,z2)∈C2:|z1|2+|z2|2p<1}. In this case, we have
(52)Ha,m,nx=∑h1=0∞‍∑h2=0∞‍Γ2+h1+h2+1/pΓh1+1Γh2+1/px1h1x2h2.
If we apply Lemma 9 to a variable x1, then we obtain
(53)Ha,m,n(x)=∑h2=0∞‍Γ2+h2+1/pΓh2+1/px2h21-x12+(h2+1)/p=∑h2=0∞‍h2+1p1+h2+1px21-x11/ph2·11-x12+1/p=1p2∑h2=0∞‍h2+1h2+1+px21-x11/ph2·11-x12+1/p.
By Lemma 10, there exist constants bl such that
(54)Ha,m,nx=1p2∑h2=0∞‍∑l=02‍cl1-x11/p-x2l+11-x12-l/p=1p2∑h2=0∞‍∑l=02‍cl·gp1-x11/p,x21-x1-x2pl+11-x12-l/p,
where gp(x1,x2)=(x1p-x2p)/(x1-x2) is a polynomial for a positive integer p. Note that 2-l/p≥2-2/p>0.Example 13.
A complex ellipsoid{(z1,z2)∈C2:|z1|4+|z2|4<1}: if d=2, a=(2,2), and m=n=(1,1), then the following formula was proved in [13]:
(55)Ha,m,nx=xπ+2arcsinxfx2,y24π1-x23/21-x2-y23+yπ+2arcsinyfy2,x24π1-y23/21-x2-y23+2xy1-x2-y23+gx2,y22π1-x21-y21-x2-y22,
where f(a,b)=3-6a+6b+3a2-6ab-b2 and g(a,b)=2-a-b-a2+2ab-b2.Example 14.
The minimal ball with radius1/2: if d=1, a=m=(1), and n≥2, the infinite series Ha,m,n(x) is equal to
(56)∑h=0∞‍Γh+nΓh+n-1xh=∑h=0∞‍h+n-1xh=∑h=0∞‍h+1xh+n-2∑h=0∞‍xh=n-1-n-2x1-x2=x2-nn-1xn-2-n-2xn-11-x2.
Since n≥2, we have (n-1)xn-2-(n-2)xn-1 which is analytic in x and this shows that the minimal ball satisfies the condition Q.Example 15.
Letd=2, m=(m1,m2), and n=(1,n2) with n2≥2 and a=(1,1). Then the infinite series Ha,m,n(x) is equal to
(57)∑h‍Γ1+h1+m1+h2+m2n2-m2Γh1+m1Γh2+m2n2-m2xh=∑h1=0∞‍x1h1Γh1+m1·∑h2=0∞‍Γh2+1+h1+m1+m2n2-m2Γh2+m2n2-m2x2h2=∑h1=0∞‍x1h1Γh1+m1x21-m2n2+m2·∑k2=m2n2-m2-1∞‍Γk2+2+h1+m1Γk2+1x2k2=x21-m2n2+m2∑k2=m2n2-m2-1∞‍x2k2Γk2+1·∑h1=0∞‍Γh1+k2+2+m1Γh1+m1x1h1=x11-m1x21-m2n2+m2·∑k1=m1-1∞‍∑k2=m2n2-m2-1∞‍Γk1+k2+3Γk1+1Γk2+1x1k1x2k2.
Next, for nonnegative integers l1 and l2, we need to compute the infinite series
(58)∑k1=l1∞‍∑k2=l2∞‍Γk1+k2+3Γk1+1Γk2+1x1k1x2k2.
It is interesting to compare this series (58) with the left hand side in (46). We can express theinfinite series (58) in terms offinite series using Lemma 9. Precisely, the series (58) is equal to
(59)21-x1-x23-∑k1=0l1-1‍x1k1Γk1+1Γk1+31-x2k1+3-∑k2=0l2-1‍x2k2Γk2+1Γk2+31-x1k2+3+∑k1=0l1-1‍∑k2=0l2-1‍Γk1+k2+3Γk1+1Γk2+1x1k1x2k2.



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*Source: 919465-2015-01-14.xml*