Reproducing kernels of some weighted Bergman spaces
aa r X i v : . [ m a t h . C V ] S e p Reproducing kernels of some weighted Bergman spaces
Guan-Tie Deng ∗ , Yun Huang † , Tao Qian ‡ Abstract
Herein, the theory of Bergman kernel is developed to the weighted case. A general formof weighted Bergman reproducing kernel is obtained, by which we can calculate concreteBergman kernel functions for specific weights and domains.
Keywords reproducing kernel, reproducing kernel Hilbert space, weigted Bergman spaces
The theory of Bergman spaces has, in the past several decades, become important in complexanalysis of both one and several complex variables, see [5] and [8]. Recall that, for anarbitrary domain Ω ⊂ C n , the Bergman space A p (Ω) is defined as the collection of analyticfunctions F that satisfy k F k A p = (cid:26)Z Ω | F ( z ) | p dA ( z ) (cid:27) p < ∞ , where dA ( z ) = dxdy is the Lebesgue measure on C n . When p = 2, the reproducing ker-nel plays an important role in the Hilbert space. For the classical Bergman spaces thereproducing properties and biholomorphic invariance are investigated in [1] and [9]. ∗ School of Mathematical Sciences, Beijing Normal University, Beijing, 100875. Email:[email protected] work was partially supported by NSFC (Grant 11971042) and by SRFDP (Grant 20100003110004) † Corresponding author. Department of Mathematics, Faculty of Science and Technology, University ofMacau, Macao (Via Hong Kong). Email: [email protected]. ‡ Institute of Systems Engineering, Macau University of Science and Technology, Avenida Wai Long,Taipa, Macau). Email: [email protected]. The work is supported by the Macau Science and Technol-ogy foundation No.FDCT079/2016/A2 , FDCT0123/2018/A3, and the Multi-Year Research Grants of theUniversity of Macau No. MYRG2018-00168-FST. K ( z, w ) = P ∞ n =1 φ n ( z ) φ n ( w ), where { φ n } is any complete orthonormal basis obtainedthrough orthonormalization of polynomials. However, for unbounded regions, density ofpolynomials can not be guaranteed. In such case appropriate weight functions are intro-duced to make the weighted polynomials dense in certain Bergman spaces of unboundeddomains. Then the kernel representations can be similarly obtained. In this paper, we applythe Laplace transform to the case of Bergman spaces on tube domains, which will be aneffective and new method to calculate reproducing kernels.Herein, we develop the theory of the weighted Bergman spaces and obtain a generalrepresentation formula of the kernel function for the spaces on tubular domains. As acomplementary party to the general study, we calculate the concrete forms of the Bergmankernels for some special weights on the tube domains.In some previous studies the concerned reproducing kernels can be computed by usingour general representation formula, since the given weight functions satisfy the set condi-tions in our theorems. For example, taking the weight function ρ ( iy ) = y q − π Γ(2 q − , we canderive that the Bergman–Selberg reproducing kernel on the upper half plane is in the formof K q ( z, ω ) = Γ(2 q ) (cid:0) iz − ¯ ω (cid:1) q with q > , which is introduced in [3]. For the weight func-tion ρ ( iy ) = y v − , a direct computation gives that the corresponding kernel is of the form K ( z, w ) = v − vπ (cid:0) z − ¯ wi (cid:1) − v − , which is studied in [7]. For weighted Bergman spaces associatedwith Lorentz cones, referring to [6], the kernel functions can be also obtained by our formula.The related calculation process is given as an example in the final section. Especially, in theun-weighted case, i.e., letting ρ ( iy ) = 1, it follows that the classical Bergman kernel on theupper half plane is K ( z, w ) = π (cid:0) iz − ¯ w (cid:1) . 2 Preliminaries
Let Ω be an arbitrary domain (an open connected set) in the complex n -dimensional Eu-clidean space C n = { z = x + iy : x, y ∈ R n } . Suppose that ρ ( z ) is a positive continuousfunction on Ω that takes the value 0 for z / ∈ Ω. We consider the weighted volume measure dA ρ ( z ) = ρ ( z ) dA ( z ) , where dA ( z ) = dxdy is the Lebesgue measure on C n . For p >
0, we denote by L pρ the spaceof measurable functions on Ω such that k F k L pρ = (cid:18)Z Ω | F ( z ) | p dA ρ ( z ) (cid:19) p < ∞ . (1)The space of such functions is called the weighted Lebesgue space with weight ρ . Thequantity k F k L pρ is called the norm of the function F , which is a true norm if p ≥ A pρ the collection of functions F that are holomorphic on Ω and satisfy thecondition (1). Such a class is called the weighted Bergman space with weight ρ . It is obviousthat A pρ ⊂ L pρ .We first assert that functions in the weighted Bergman space cannot grow too rapidlynear the boundary. Lemma . Point-evaluation is a bounded linear functional in each weighted Bergman space A pρ . More specifically, each function F ∈ A pρ has the property | F ( z ) | ≤ ( ω n ε z ) − p ( δ z ) − np k F k A pρ . (2) Here, ω n = π n n ! is the volume of unit ball B n (0 , , δ z = min { , − dist ( z, ∂ Ω) } wheredist ( z, ∂ Ω) } is the distance from z to the boundary of Ω , and ε z = min { ρ ( ζ ) : | ζ − z | ≤ δ z } .Proof. For fixed point z ∈ Ω, the bounded closed ball B n ( z, δ z ) lies in Ω. Since ρ ( ζ ) is apositive continuous function on Ω, then for any ζ ∈ B n ( z, δ z ), we have ε z = min ρ ( ζ ) > | F ( z ) | p ≤ ω n δ nz Z B n ( z,δ z ) | F ( ζ ) | p dA ( ζ ) ≤ ω n δ nz Z B n ( z,δ z ) | F ( ζ ) | p ρ ( ζ ) dA ρ ( ζ ) ≤ ε z ω n δ nz Z B n ( z,δ z ) | F ( ζ ) | p dA ρ ( ζ ) ≤ ε z ω n δ nz Z Ω | F ( ζ ) | p dA ρ ( ζ )= 1 ε z ω n δ nz k F k pA pρ , A pρ is a Banach space when 1 ≤ p < ∞ and a complete metric space when 0 < p < Lemma . Suppose that ρ ( z ) is a positive continuous function on Ω that takes the value for z Ω . For < p < ∞ , the weighted Bergman space A pρ is closed in L pρ .Proof. Let { F n } be a sequence in A pρ and assume F n → F in L pρ . In particular, { F n } is a Cauchy sequence in L pρ . Applying the previous Lemma, we see that { F n } convergesuniformly on every compact subset of Ω. Combining with the assumption that F n → F in L pρ , we conclude that F n → F uniformly on every compact subset of Ω. Therefore, F isanalytic in Ω and belongs to A pρ .Now let p = 2, A ρ is a Hilbert space with inner product h F, G i ρ = Z Ω F ( z ) G ( z ) dA ρ ( z )for F, G ∈ A ρ .Since each point evaluation functional T [ F ] = F ( z ) of A ρ is bounded, the Riesz rep-resentation theorem for Hilbert space guarantees existence of a unique function K ( ζ , z ) = K z ( ζ ) ∈ A ρ such that F ( z ) = h F, K z i ρ for every F ∈ A ρ . The function K ( ζ , z ) is known asthe reproducing kernel with weight ρ , or the weighted Bergman kernel function. It has thereproducing property F ( z ) = Z Ω F ( ζ ) K z ( ζ ) dA ρ ( ζ ) (3)for each function F ∈ A ρ . Taking F ( z ) = K ( z, ζ ) for some ζ ∈ Ω, we see that K ( z, ζ ) = Z Ω K ( η, ζ ) K ( η, z ) dA ρ ( ζ ) = K ( ζ , z ) . (4)Thus the kernel function has the symmetry property K ( z, ζ ) = K ( ζ , z ), which also showsthat K ( z, ζ ) is analytic in z and anti-analytic in ζ . Another consequence is the formula K ( z, z ) = Z Ω | K ( z, ζ ) | dA ρ ( ζ ) = k K ( z, · ) k L ρ > . (5)In view of (5), applying the Schwarz inequality to (3), there holds | F ( z ) | ≤ p K ( z, z ) k F k A ρ . ζ ∈ Ω, 1 p K ( ζ , ζ ) ≤ k F k A ρ for all F ∈ A (Ω , ρ ) with F ( ζ ) = 1. In fact, the lower bound is sharp and uniquely attainedby the function F ( z ) = K ( z,ζ ) K ( ζ,ζ ) .The theory of reproducing kernel Hilbert spaces guarantees that the existence of thereproducing kernel K ( · , · ) is unique.Recall that a holomorphic mapping w = Φ( z ) from a domain Ω to a domain Ω is saidto be biholomorphic if it is one-to-one, onto, and its holomorphic inverse exists.In fact, the kernel function with weight is biholomorphic invariant in the sense of thefollowing Lemma. Lemma . Suppose that w = Φ( z ) is a biholomorphic mapping of a domain Ω onto a domain Ω , ρ ( z ) and ρ ( w ) are two positive continuous functions on domains Ω and Ω respectively, ρ ( z ) = 0 for z Ω and ρ ( w ) = 0 for w Ω , K ρ ( z, ζ ) and K ρ ( w, ς ) are reproducingkernels of two weighted Bergman spaces A ρ and A ρ respectively. If ρ ( z ) = ρ (Φ( z )) for all z ∈ Ω , then K ρ ( z, ζ ) = ( D Φ)( z ) K ρ (Φ( z ) , Φ( ζ ))( D Φ)( ζ ) , (6) where ( D Φ)( z ) is the determinant of the holomorphic Jacobian matrix of w = Φ( z ) .Proof. Let F ∈ A ρ , after a change of variables ς = Φ( ζ ) in the following integral, Z Ω ( D Φ)( z ) K ρ (Φ( z ) , Φ( ζ ))( D Φ)( ζ ) F ( ζ ) ρ ( ζ ) dA ( ζ )= Z Ω ( D Φ)( z ) K ρ (Φ( z ) , ς )( D Φ)(Φ − ( ς )) F (Φ − ( ς )) ρ ( ς )( D R Φ − )( ς ) dA ( ς ) , where D R Φ − is the determinant of the real Jacobian matrix of Φ − . Based on the rela-tionship between the determinant of the real Jacobian matrix and that of the holomorphicJacobian matrix, i.e., D R Φ − = | D Φ − | (see [5] Proposition 1.4.10), the above formulasimplifies to Z Ω ( D Φ)( z ) K ρ (Φ( z ) , ς ) n(cid:0) ( D Φ)(Φ − ( ς )) (cid:1) − F (Φ − ( ς )) o ρ ( ς ) dA ( ς ) . On the other hand, by hypothesis, the expression in braces is an element of A ρ . So the lastline equals ( D Φ)( z ) (cid:0) ( D Φ)(Φ − (Φ( z ))) (cid:1) − F (Φ − (Φ( z ))) = F ( z ) . By the uniqueness of the reproducing kernel of the weighted Bergman space A ρ , we see that(6) holds. This completes the proof of Lemma 3.5 Main results
In order to obtain the explicit reproducing kernel of the weighted Bergman kernel corre-sponding to a specific weights ρ in a concrete domain, we suppose that ρ ( z ) is a positivecontinuous function on a tube domainΩ = T B = { z = x + iy : y ∈ B } over an open connected subset B of the real n -dimensional Euclidean space R n . In addition,we assume that ρ ( x + iy ) = ρ ( iy ) for all x ∈ R n , y ∈ B and ρ ( z ) = 0 for z T B . Inthis case, the computation of the weighted Bergman kernels on those tube domain is greatlybenefited from the homogeneity of T B in the real directions. An important tool is the Laplacetransforms F = L ( f ) of a function f , that is, F = ( L f )( z ) = Z R n f ( t ) e πiz · t dt, (7)where z · t = x · t + iy · t = n P k =1 z k · t k , and x · t, y · t are the Euclidean scalar products for x, y, t ∈ R n . The definition will be further justified together with the specific spaces that thetest f belongs to. It is obvious that F is well defined only when f decays sufficiently fast at ∞ . Let I ( t ) = Z B ρ ( iy ) e − πy · t dy, (8)then the set U I = { t : I ( t ) < ∞} is a convex set, and log I ( t ) is a convex function on U I .The weighted L pI space is the set of the measurable function defined on R n such that k f k L pI = (cid:18)Z R n | f ( t ) | p I ( t ) dt (cid:19) p < ∞ . Notice that if f ∈ L pI , then f ( t ) = 0 almost everywhere for all t U I , so we can assume thatthe support of f is contained in the closure of U I . We also see that A ρ and L I are Hilbertspaces with the inner product h F, G i ρ = Z B Z R n F ( x + iy ) G ( x + iy ) ρ ( iy ) dxdy for F, G ∈ A ρ and h f, g i I = Z R n f ( t ) g ( t ) I ( t ) dt for f, g ∈ L I respectively.The main result herein is established as follows.6 heorem . The weighted Bergman kernel K ( z, w ) of A ρ is given by K ( z, w ) = Z R n e πit ( z − w ) I − ( t ) dt, (9) where ρ ( z ) is a positive continuous function on the tube domain T B and satisfies that ρ ( x + iy ) = ρ ( iy ) for all x ∈ R n , y ∈ B . To prove Theorem 1, we need the following lemma, which is also an important result byitself.
Lemma . The Laplace transform L is an isometry from L I to A ρ preserving the Hilbertspace norms, i.e., k L f k A ρ = k f k L I . Proof.
First, we prove that if F ( z ) ∈ A ρ , there exists f ∈ L I such that F ( z ) = ( L f )( z ),which means that the Laplace transform L is surjective.Let B ⊆ B be a bounded connected open set, so there exists a positive constant R suchthat B ⊆ D (0 , R ). Assume that l ε ( z ) = (1 + ε ( z + · · · + z n )) N , where N is a integer and N > n . Then for ε ≤ R and z = x + iy with | y | ≤ R , | l ε ( z ) | = | ((1 + ε ( z + · · · + z n )) ) N | = (cid:16)(cid:0) ε ( | x | − | y | ) (cid:1) + 4 ε ( x · y ) (cid:17) N ≥ (cid:0) ε ( | x | − | y | ) (cid:1) N ≥ (cid:18)
12 + ε | x | (cid:19) N , i.e., | l − ε ( z ) | ≤ ( + ε | x | ) N . Set F ε ( z ) = F ( z ) l − ε ( z ), then base on H¨older’s inequality, Z R n | F ε,y ( x ) | dx ≤ (cid:18)Z R n | F y ( x ) | dx (cid:19) (cid:18)Z R n (cid:12)(cid:12) l − ε ( x + iy ) (cid:12)(cid:12) dx (cid:19) < ∞ , which implies that F ε,y ( x ) = F ε ( x + iy ) ∈ L ( R n ) andlog Z R n | F ε ( x + iy ) | dx is a convex function on B . Therefore, for any compact K ⊆ B , we havesup (cid:26)Z R n | F ε ( x + iy ) | dx : y ∈ K (cid:27) < ∞ . (10)7or any a, b, y ∈ B , t ∈ R n , let G ε ( z ) = F ε ( z ) e πiz · t ; J ( x ′ , x n , τ ) = F ε ( x + i ( a + τ ( b − a ))) , x = ( x ′ , x n ) , ≤ τ ≤ I ε ( y, t ) = ˇ F ε,y ( t ) e − πy · t . In order to show that, for all a, b ∈ B , I ε ( a, t ) = I ε ( b, t ) and I ( a, t ) = I ( b, t ) almosteverywhere for all t ∈ R n , we first assume that a = ( a ′ , a n ), b = ( a ′ , b n ) and the closedinterval [ a, b ] = { a + t ( b − a ) : 0 ≤ t ≤ } is contained in B . Then (10) implies that theintegral Z ∞ Z Z R n − ( | J ( x ′ , x n , τ ) | + | J ( x ′ , − x n , τ ) | ) dx ′ dτ dx n is finite. This meanslim R →∞ Z Z R n − ( | J ( x ′ , R, τ ) | + | J ( x ′ , − R, τ ) | ) dx ′ dτ = 0 . Therefore, | I ε ( a, t ) − I ε ( b, t ) | = (cid:12)(cid:12)(cid:12)(cid:12)Z R n ( G ε ( x + ib ) − G ε ( x + ia )) dx (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z R n Z ∂∂τ G ε ( x + i ( a + τ ( b − a )) dτ dx (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z R n Z ∂∂y n ( G ε ( x + i ( a ′ , y n ))) (cid:12)(cid:12) y n = a n + τ ( b n − a n ) ( b n − a n ) dτ dx (cid:12)(cid:12)(cid:12)(cid:12) = | b n − a n | (cid:12)(cid:12)(cid:12)(cid:12)Z R n Z i ∂∂x n ( G ε ( x + i ( a + τ ( b − a )))) dτ dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ( t ) lim R →∞ Z Z R n − ( | J ( x ′ , R, τ ) | + | J ( x ′ , − R, τ ) | ) dx ′ dτ = 0 , where C ( t ) = | b n − a n | e − π | t | ( | a | + | b − a | ) . Remark that B is connected and by an iterationargument, we can show that g ε,y ( t ) = ˇ F ε,y ( t ) e − πy · t is a function independent of y ∈ B .Hence g ε ( t ) = g ε,y ( t ) is independent of y ∈ B and g ε ( t ) e πy · t = ˇ F ε,y ( t ) ∈ L ( R n ) for all y ∈ B .On the other hand, it is obvious that F ε,y ( x ) → F y ( x ) as ε →
0. Let Iy ( t ) = ˇ F y ( t ) e − πy · t ,we can also prove that ˇ F y ( t ) e − πy · t is independent of y ∈ B and g ( t ) = ˇ F y ( t ) e − πy · t almosteverywhere. Indeed, for a, b ∈ B and any compact subset K ⊂ R n , let R = max {| t | : t ∈ K } . Then Plancherel’s theorem implies that k I b − I a k L ( K ) ≤ k I b − g ε k L ( K ) + k g ε − I a k L ( K ) ≤ e πR R (cid:0) k ˇ F a − ˇ F ε,a k L ( K ) + k ˇ F ε,b − ˇ F b k L ( K ) (cid:1) = e πR R (cid:0) k F a − F ε,a k L ( R n ) + k F b − F ε,b k L ( R n ) (cid:1) → ε →
0. Hence ˇ F a ( t ) e − πa · t = ˇ F b ( t ) e − πb · t almost everywhere on R n .Next, we show that g ( t ) e πy · t ∈ L ( R n ). In order to prove this affirmance, decompose R n into the union of finite non-overlapping cones { Γ k } ∞ k =1 with common vertex at the origin,i.e., R n = ∪ Nk =1 Γ k and let B δ = D ( y , δ ) ⊂ B . Then for any y ∈ D ( y , δ ) and y k ∈ ( y + Γ k )satisfying δ ≤ | y k − y | < δ , we have ( y k − y ) · t ≥ | y k − y |√ | t | − | y − y || t | ≥ ( √ − ) δ | t | ≥ δ | t | for y k − y , t ∈ Γ k . Thus, it follows from H¨older’s inequality and Plancherel’s theorem that Z Γ k | g ( t ) e πy · t | dt ≤ Z Γ k | ˇ F y k ( t ) e − π δ | t | | dt ≤ (cid:18)Z Γ k | ˇ F y k ( t ) | dt (cid:19) (cid:18)Z Γ k | e − π δ | t | | dt (cid:19) = (cid:18)Z Γ k | F y k ( t ) | dt (cid:19) (cid:18)Z Γ k | e − π δ | t | | dt (cid:19) < ∞ , which shows that g ( t ) e πy · t ∈ L (Γ k ). Then g ( t ) e πy · t ∈ L ( R n ). Therefore, we can seethat the function G ( z ) = R R n g ( t ) e − πi ( x + iy ) · t dt is well-defined and holomorphic on the tubedomain T B .Now we can prove thatlim ε → Z R n g ε ( t ) e − πi ( x + iy ) · t dt = Z R n g ( t ) e − πi ( x + iy ) · t dt. For y ∈ B , (cid:12)(cid:12)(cid:12)(cid:12)Z R n ( g ε ( t ) − g ( t )) e − πi ( x + iy ) · t dt (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z R n (cid:12)(cid:12)(cid:0) ˇ F ε,y ( t ) e − πy · t − ˇ F y ( t ) e − πy · t (cid:1) e πiz · t (cid:12)(cid:12) dt = n X k =1 Z Γ k (cid:12)(cid:12)(cid:0) ˇ F ε,y k ( x ) − ˇ F y k ( x ) (cid:1) e − πi ( y k − y ) · t (cid:12)(cid:12) dt ≤ n X k =1 (cid:18)Z Γ k | ˇ F ε,y k ( x ) − ˇ F y k ( x ) | dt (cid:19) (cid:18)Z Γ k | e − π δ | t | | dt (cid:19) ≤ " max ≤ k ≤ n (cid:18)Z Γ k | e − π δ | t | | dt (cid:19) n X k =1 (cid:18)Z Γ k | F ε,y k ( x ) − F y k ( x ) | dt (cid:19) → ε →
0. It follows that lim ε → F ε ( z ) = G ( z ). Combining with lim ε → F ε ( z ) = F ( z ), we get G ( z ) = F ( z ) for z ∈ T B . Therefore, there exists a measurable function g ( t ) such that g ( t ) e πy · t ∈ L ( R n ) for all y ∈ B , then F ( z ) = Z R n g ( t ) e − πiz · t dt holds for z ∈ T B . Hence g ( t ) = ˇ F y ( t ) e − πy · t for all y ∈ B . Since B is connected, wecan choose a sequence of bounded domains { B k } such that B ⊂ B ⊂ B ⊂ . . . and B = ∪ ∞ k =0 B δ k . Then ˇ F y k ( t ) e − πy k · t = ˇ F y ( t ) e − πy · t for y ∈ B and y k ∈ B k , where k >
0. These9mply that g ( t ) = ˇ F ( t ) e − πy · t holds for all y ∈ B . In other words, F ( z ) = R R n g ( t ) e − πiz · t dt holds for all z ∈ T B . By letting f ( t ) = g ( − t ), we see that f ∈ L I and F ( z ) = ( L f )( t ) forany given F ( z ) ∈ A ρ , which means that the Laplace transform L is surjective.Now we prove that (7) is well-defined, injective and preserves norm, i.e., k L f k A ρ = k f k L I . Let F ( x + iy ) = F y ( x ) = ( L f )( x + iy ) for every fixed y ∈ B . Based on Plancherel’stheorem, there holds Z R n | F y ( x ) | dx = Z R n | f ( t ) | e − πy · t dt. Multiplying by ρ ( iy ) and performing integral over B on both sides of the above equation,we have k F k A ρ = Z B (cid:18)Z R n | F ( x + iy ) | dx (cid:19) ρ ( iy ) dy = Z B (cid:18)Z R n | e − πy · t f ( t ) | dt (cid:19) ρ ( iy ) dy = Z R n | f ( t ) | I ( t ) dt = k f k L I . It then follows that f ( t ) = 0 almost everywhere for all t U I , where U I = { t : I ( t ) < ∞} .Therefore, f ( t ) is supported in U I . This completes the proof of Lemma 4.Let us now prove Theorem 1. Proof.
For F ( z ) ∈ A ρ ( T B ), there exists f ∈ L I such that F ( z ) = ( L f )( z ) = R R n f ( t ) e πiz · t dt .And for the kernel K z ( z ) = K ( z, z ) ∈ A ρ , there also exists f z ∈ L I such that K z ( z ) =( L f z )( z ) = R R n f z ( t ) e πiz · t dt for z , z ∈ T B . On the other hand, Lemma 1 claims that L is an isometry from L I to A ρ preserving the Hilbert space norm. Using the polarizationidentity of k F k A ρ = k f k L I , it then follows that the inner product is also preserved. Hencewe have F ( z ) = h F, K z i ρ = h f, f z i I = Z R n f ( t ) f z ( t ) I ( t ) dt. Hence Z R n f ( t ) e πiz · t dt = Z R n f ( t ) f z ( t ) I ( t ) dt holds for every f ∈ L I , which implies that e πiz · t = f z ( t ) I ( t ) almost everywhere on U I = { t ∈ R n : I ( t ) < ∞} . Then, f z ( t ) = e − πi ¯ z · t I − ( t ). Here, I ( t ) − takes 0 when I ( t ) = ∞ bythe definition of I ( t ). Hence, K z ( z ) = K ( z, z ) = Z R n f z ( t ) e πiz · t dt = Z R n e πi ( z − ¯ z ) · t I − ( t ) dt. K ( z, z ) = K ( z , z ) for all z, z ∈ T B ;(ii) K ( z, z ) reproduces every element in A ρ in the following sense F ( z ) = Z B Z R n K ( z, u + iv ) F ( u + iv ) ρ ( iv ) dudv for every F ∈ A ρ ;(iii) K z ∈ A ρ for all z ∈ T B , where K z ( z ) = K ( z, z ).We shall show that (9) admits these properties. We first prove the symmetric propertyas follows, K ( z , z ) = Z R n e πi ( z − ¯ z ) · t I − ( t ) dt = Z R n e − πi ( ¯ z − z ) · t I − ( t ) dt = Z R n e πi ( z − ¯ z ) · t I − ( t ) dt = K ( z, z ) , which means (i) holds for the Bergman kernel in the form of (9). We then show K ( z, z )reproduces every element in A ρ . Indeed, for F ( z ) , K z ( z ) ∈ A ρ , F ( z ) = Z U I f ( t ) e πiz · t dt and K z ( z ) = Z R n (cid:18) e − πi ¯ z · t I ( t ) (cid:19) e πit · z dt. Then the polarization identity implies that h F, K z i A ρ = Z B Z R n K ( z , z ) F ( z ) dA ρ ( z )= (cid:28) f ( t ) , e − πi ¯ z · t I ( t ) (cid:29) L I = Z U I f ( t ) e − πi ¯ z · t I ( t ) I ( t ) dt = Z U I f ( t ) e πiz · t dt = F ( z ) . Hence, the second property is proved.Finally, we prove that K z ( z ) ∈ A ρ . For fixed z = u + iv ∈ T B , there exists δ > v + P δ ⊂ B , where P δ = [ − δ, δ ] n ⊂ R n . Let ε = min { ρ ( iy ) : y ∈ v + P δ } >
0, then I ( t ) e πv · t = Z B e − π ( y − v ) · t ρ ( y ) dy ≥ ε Z P δ e − πη · t dη = ε n Y k =1 sinh(4 πδt k )2 πδt k . h K z , K z i A ρ = Z T B | K z ( z ) | dA ρ ( z ) = h K z , K z i A ρ = (cid:28) e − πi ¯ z · t I ( t ) , e − πi ¯ z · t I ( t ) (cid:29) L I = Z R n (cid:12)(cid:12)(cid:12)(cid:12) e − πi ¯ z · t I ( t ) (cid:12)(cid:12)(cid:12)(cid:12) I ( t ) dt = Z R n e − πv · t I ( t ) dt ≤ Z R n ε n Y k =1 sinh(4 πδt k )2 πδt k ! − dt < ∞ . Therefore, K z ( z ) ∈ A ρ for z ∈ T B . The weighted Bergman kernel have never been computed explicitly. However, the theorywould be a bit hollow if we do not compute at least one weighted Bergman kernel. In thissection, we calculate some weighted Bergman kernels as examples. The first example is theweighted Bergman kernel for tube over the following cone.
Example 1.
Suppose that B = { y = ( y ′ , y n ) : y n > | y ′ | } . Denote by A α the space ofanalytic functions on tube domain Ω = { z = x + iy : x ∈ R n , y ∈ B } such that k F k A α = (cid:18)Z Ω | F ( x + iy ) | ( y n − | y ′ | ) α dxdy (cid:19) < ∞ . (11)Then the reproducing kernel for the Hilbert space A α is K α ( z, w ) = C ,α (( z ′ − ¯ w ′ ) − i ( z n − ¯ w n )) − n − α − , (12)where z ′ = ( z , . . . , z n − ), w ′ = ( w , . . . , w n − ) and C ,α = 2 n +1+2 α Γ( n + α + 1)Γ( α + 1) π n . (13)Here, ℜ{ ( z ′ − ¯ w ′ ) − i ( z n − ¯ w n ) } = 2( y n + v n ) + ( x ′ − u ′ ) · ( x ′ − u ′ ) − ( y ′ + v ′ ) · ( y ′ + v ′ ) > | arg(( z ′ − ¯ w ′ ) − i ( z n − ¯ w n )) | < π , in which z = ( z ′ , z n ) = ( x ′ , x n ) + i ( y ′ , y n ), w =( w ′ , w n ) = ( u ′ , u n ) + i ( v ′ , v n ) ∈ T B and z ′ = z ′ · z ′ = z + z + . . . + z n − . Proof.
We fist compute I ( t ) as follows, I ( t ) = Z B e − πy · t ( y n − | y ′ | ) α dy = Z ∞ Z | y ′ | 0, For t n > I ( t ) = Z ∞ x αn e − πx n t n n − Y k =1 Z R e − πx k t n − πx k t k dx k ! dx n = Z ∞ x αn e − πx n t n n − Y k =1 Z R e − πt n (cid:16) x k − xktktn + ( tk tn ) − ( tk tn ) (cid:17) dx k ! dx n = Z ∞ x αn e − πx n t n n − Y k =1 Z R e − πt n ( x k − tk tn ) e πt ktn dx k ! dx n . Let s = x k − t k t n and 4 πt n s = η , then I ( t ) = Z ∞ x αn e − πx n t n n − Y k =1 e πt ktn Z R e − πt n s ds ! dx n = Z ∞ x αn e − πx n t n n − Y k =1 e πt ktn √ πt n Z R e − η dη dx n = Z ∞ x αn e − πx n t n e π | t ′| tn (cid:18) √ π √ πt n (cid:19) n − dx n = e π | t ′| tn (2 √ t n ) − n Z ∞ x αn e − πx n t n dx n . Putting u = − πx n t n , we have I ( t ) = e π | t ′| tn (2 √ t n ) − n Z ∞ πt n ) α +1 u α e u du = 2 − n Γ( α + 1)(4 π ) α +1 e π | t ′| tn t (1 − n ) − α − n . Now we can compute the formula of weighted Bergman kernel. According to the repre-sentation form of reproducing kernel, K α ( z, z ) = Z R n e πi ( z − ¯ z ) · t I − ( t ) dt = (4 π ) α +1 n − Γ( α + 1) Z R n e πi ( z − ¯ z ) · t − π | t ′| tn t ( n − α +1 n dt. w = z − ¯ z and C α = (4 π ) α +1 n − Γ( α +1) , then K α ( z, z ) = C α Z R n e πiw · t − π | t ′| tn t ( n − α +1 n dt = C α Z ∞ Z R n − e πi ( w n t n + w ′ · t ′ ) − π | t ′| tn t ( n − α +1 n dt ′ dt n = C α Z ∞ e πiw n t n t ( n − α +1 n (cid:18)Z R n − e πiw ′ · t ′ − π | t ′| tn dt ′ (cid:19) dt n = C α Z ∞ e πiw n t n t ( n − α +1 n n − Y k =1 Z R e πiw k t k − πt ktn dt k ! dt n , in which Z R e πiw k t k − πt ktn dt k = Z R e − πtn ( t k − iw k t n t k +( iw k t n ) + w k t n ) dt k = e − πw kt ntn Z R e − πtn ( t k − iw k t n ) dt k . Let s = t k − iw k t n and η = πt n s , then Z R e πiw k t k − πt ktn dt k = e − πw kt ntn Z R e − πtn s ds = e − πw kt ntn Z R (cid:18) t n π (cid:19) e − η dη = e − πw kt ntn t n . Therefore, K α ( z, z ) = C α Z ∞ e πiw n t n t ( n − α +1 n n − Y k =1 (cid:18) e − πw kt ntn t n (cid:19) dt n = C α Z ∞ e πiw n t n t ( n − α +1 n e − π P n − k =1 w kt ntn t ( n − n dt n = C α Z ∞ e πt n (2 iw n − w ′ · w ′ ) t n + αn dt n . In order to compute K α ( z, z ), it suffices to show that ℜ (2 iw n − w ′ · w ′ ) < 0, where w = z − ¯ z z, z ∈ T B . In fact, since ℜ ( iw n ) = −ℑ w n , ℜ ( w ′ · w ′ ) = ( ℜ w ′ ) − ( ℑ w ′ ) , for w n = z n − ¯ z ,n , ℜ (2 iw n − w ′ · w ′ ) = − ℑ w n − n − X k =1 (cid:0) ( ℜ w k ) − ( ℑ w k ) (cid:1) = − y n + y ,n ) + n − X k =1 ( y k + y ,k ) − n − X k =1 ( x k − x ,k ) ≤ − y n + y ,n ) + n − X k =1 ( y k + y ,k + 2 y k y ,k ) ≤ − y n + y ,n ) + 2 n − X k =1 ( y k + y ,k )= 2 − y n + n − X k =1 y k ! + − y ,n + n − X k =1 y ,k !! < . Now, we can continue to calculate K α ( z, z ). Let u = πt n ( w ′ · w ′ − iw n ), K α ( z, z ) = C α Z ∞ e − πt n ( w ′ · w ′ − iw n ) t n + αn dt n = C α Z ∞ π ( w ′ · w ′ − iw n )) n + α +1 e − u u n + α du = C α Γ( n + α + 1)( π ( w ′ · w ′ − iw n )) n + α +1 = C α, ( w ′ · w ′ − iw n ) − n − α − = C α, ( z ′ − ¯ z ′ ) − i ( z n − ¯ z ,n )) − n − α − , which shows that (12) holds. Example 2. Denote by Ω n the Siegel domain in C n , and define it asΩ n = { z = x + iy : y n > | z ′ | } , where z = ( z ′ , z n ) , z n = x n + iy n . Let α ∈ R and denote by A α (Ω n ) the space of analyticfunctions F ( z ) in the Siegel domain Ω n satisfying k F k A α (Ω n ) = (cid:18)Z Ω n | F ( x + iy ) | ( y n − | z ′ | ) α dxdy (cid:19) < ∞ . (15)Then the reproducing kernel for the Hilbert space A α (Ω n ) is K α, Ω n ( z, w ) = C ,α ( i ( ¯ w n − z n ) − z ′ · ¯ w ′ ) − n − α − , (16)where C ,α = 2 − − α C ,α , C ,α is defined by (13).15 roof. Define a transform Φ : Ω n → T B as w = ( w ′ , w n ) = Φ( z ) = (2 z ′ , z n − iz ′ · z ′ ). Thenwe observe that the Siegel domain Ω n is biholomorphically equivalent to the tube domain T B over B = { v ∈ R n : v n > | v ′ | } via Φ. The inverse of Φ is z = Φ − ( w ) = (2 − w ′ , w n + i w ′ · w ′ ),and the determinants of the holomorphic Jacobian matrixes of w = Φ( z ) and z = Φ − ( w ) are( D Φ)( z ) = 2 ( n − and ( D Φ − )( w ) = 2 − ( n − respectively. For w = u + iv , z = x + iy ∈ T B , y n − | z ′ | = y n − n − X k =1 ( x k + y k )= ℑ ( w n + i w ′ · w ′ ) − | w ′ | = v n + 12 ( u ′ · u ′ − v ′ · v ′ ) − | w ′ | = v n + 12 n − X k =1 ( u k − v k ) − n − X k =1 ( u k + v k )= v n − | v ′ | . By Lemma 3, the reproducing kernel K α, Ω n of the Hibert space A α (Ω n ) is K α, Ω n ( z, w ) = 2 n − K α (Φ( z ) , Φ( w )) = C ,α − − α ( i ( ¯ w n − z n ) − z ′ · ¯ w ′ ) − n − α − , (17)where C ,α is defined by (13). Example 3. For α ∈ R , denote by B n the unit ball { z : | z | < } and A α ( B n ) the spaceof analytic functions F ( z ) on B n satisfying k F k A α ( B n ) = (cid:18)Z B n | F ( z ) | (1 − | z | ) α | z n | α dxdy (cid:19) < ∞ . (18)Then the reproducing kernel for the Hilbert space A α ( B n ) is K α,B n ( z, w ) = C ,α (1 + ¯ w n ) α (1 + z n ) α (1 − ¯ w · z ) − n − α − , (19)with C ,α = 2 − α − n C ,α = 2 − α − n − C ,α , where C ,α is defined by (13). Proof. Note that the Siegel domain Ω n is an unbounded realization of B n , i.e., Ω n is biholo-morphically equivalent to B n . The corresponding biholomorphic automorphism (so-calledthe Cayley transform) w = Φ( z ) : B n Ω n can be written in the following explicit form, w = Φ( z ) = (cid:18) z ′ z n + 1 , i (1 − z n )1 + z n (cid:19) for z ∈ B n . And its inverse form is z = Φ − ( w ) = (cid:18) w ′ − i w n , i w n − i w n (cid:19) for w ∈ Ω n . w = Φ( z ) and z = Φ − ( w ) are( D Φ)( z ) = − i n +2 (1 + z n ) n +1 and ( D Φ − )( w ) = i − i ) n +1 respectively. By Lemma 3, the reproducing kernel K α,B n of the Hibert space A α ( B n ) is K α,B n ( z, w ) = − i n +2 (1 + z n ) n +1 K α, Ω n (Φ( z ) , Φ( w )) i n +2 (1 + ¯ w n ) n +1 = C ,α (1 + ¯ w n ) α (1 + z n ) α (1 − ¯ w · z ) − n − α − (20)with C ,α = 2 − α − n C ,α = 2 − α − n − C ,α , where C ,α is defined by (13). Example 4. Suppose that Λ n is the Lorentz cone (or a forward light cone) defined by { y = ( y ′ , y n ) : y n > | y ′ |} . The quadratic function ∆( y ) = y n − | y ′ | is call the Lorentz form.Let α ∈ R and denote by A α ( T Λ n ) the space of analytic functions F ( z ) in the tube domain T Λ n over the forward cone Λ n such that k F k A α (Λ n ) = (cid:18)Z Λ n Z R n | F ( x + iy ) | (∆( y )) α dxdy (cid:19) < ∞ . (21)Then the reproducing kernel for the Hilbert space A α ( T Λ n ) is K α, Λ n ( z, w ) = C ,α P ( z − ¯ w ) − α − n , (22)where C ,α is defined by C ,α = 4 α Γ( α + n + 1)Γ(2 α + 2 n )Γ( α + n + ) π n Γ( α + 1)Γ(2 α + n )Γ( α + n + ) (23)and P ( z ) = z + · · · + z n − − z n satisfying P ( z ) ⊂ C \ ( −∞ , 0] for z ∈ T Λ n . Proof. For t = ( t ′ , t n ) ∈ R n and Let a = 4 πt = ( a ′ , a n ), then based on the form of kernel inTheorem 1, we have I ( t ) = Z B e − πy · t (∆( y )) α dy = Z ∞ Z | y ′ | 0] for z ∈ T Λ n .Indeed, P ( z ) = z + · · · + z n − − z n = ( x + iy ) + · · · + ( x n − + iy n − ) − z n = ( x − y ) + · · · + ( x n − − y n − ) − ( x n − y n ) + 2 i ( x y + · · · + x n − y n − − x n y n )= n − X k =1 x k − x n ! + y n − n − X k =1 y k ! + 2 i n − X k =1 x k y k − x n y n ! . It follows from P n − k =1 x k y k = x n y n that P n − k =1 x k y k y n = x n , then x n = ( P n − k =1 x k y k y n ) . H¨older’sinequality implies that x n ≤ n − X k =1 x k ! n − X k =1 y k y n ! < n − X k =1 x k . Hence, P ( z ) ∈ C \ ( −∞ , 0] for z ∈ T B . Therefore, ( P ( z )) − α − n is well defined for z ∈ T Λ n and P ( iy ) = − y − y − · · · − y n − + y n = ∆( y ) for y ∈ Λ n and P ( iy ) α = ρ ( iy ) is a weightfunction on T Λ n . Then K ( iy ) = 4 α Γ( α + n + 1)Γ(2 α + 2 n )Γ( α + n + ) π n Γ( α + 1)Γ(2 α + n )Γ( α + n + ) P ( iy ) − α − n . For z = x + iy ∈ T Λ n , K ( x + iy ) admits Taylor expansion formula K ( x + iy ) = ∞ X k =0 · · · ∞ X k n =0 K ( k + ··· + k n ) ( iy ) k ! · · · k n ! x k · · · x k n n , K ( z ) = 4 α Γ( α + n + 1)Γ(2 α + 2 n )Γ( α + n + ) π n Γ( α + 1)Γ(2 α + n )Γ( α + n + ) P ( z ) − α − n . As a result, for z, w ∈ T Λ n , K α, Λ n ( z, w ) = K ( z − ¯ w ) = C ,α P ( z − ¯ w ) − α − n . where C ,α is defined by (23). This proves (22). References [1] S. 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