# Weighted Anisotropic Integral Representations of Holomorphic Functions in the Unit Ball ofCn

Authors: Arman Karapetyan
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We obtain weighted integral representations for spaces of functions holomorphic in the unit ballBn and belonging to area-integrable weighted Lp-classes with “anisotropic” weight function of the type ∏i=1n(1−|w1|2−|w2|2−⋯−|wi|2)αi, w=(w1,w2,…,wn)∈Bn. The corresponding kernels of these representations are estimated, written in an integral form, and even written out in an explicit form (for n=2).


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## Body

## 1. Introduction

Denote byBn the unit ball in the complex n-dimensional space Cn:Bn={w∈Cn:|w|<1}. For 1≤p<+∞ and α>-1, denote by Hαp(Bn) the space of all functions f holomorphic in Bn and satisfying the condition
(1.1)∫Bn|f(w)|p(1-|w|2)αdm(w)<+∞,
where dm is the Lebesgue measure in Cn≡R2n. Further, for a complex number β with Reβ>-1, put
(1.2)cn(β)=Γ(n+1+β)πn⋅Γ(1+β).We have the following theorem.Theorem 1.1.
Assume that1≤p<+∞,α>-1, and that the complex number  β satisfies the condition
(1.3)Reβ≥α,p=1,Reβ>α+1p-1,1<p<∞.
Then each function  f∈Hαp(Bn) admits the following integral representations:
(1.4)f(z)=cn(β)⋅∫Bnf(w)(1-|w|2)β(1-〈z,w〉)n+1+βdm(w),z∈Bn,f(0)¯=cn(β)⋅∫Bnf(w)¯(1-|w|2)β(1-〈z,w〉)n+1+βdm(w),z∈Bn,
where 〈·,·〉 is the Hermitean  inner product in  Cn.Forn=1, that is, for the case of the unit disc D⊂C, this theorem was established in [1, 2], where the formulas (1.4) are important in the theory of factorization of meromorphic functions in the unit disc.Forn>1, the theorem was proved in [3] (when α=0) and in [4, 5] (when α>-1).In monographs [6, 7], one can find numerous applications of the formulas (1.4) in the complex analysis.In the present paper, we generalize Theorem1.1 in the following way.Assume that1≤p<+∞ and α=(α1,…,αn)∈Rn satisfies the conditions
(1.5)αn>-1,αn+αn-1>-2,αn+αn-1+αn-2>-3,⋮αn+αn-1+αn-2+⋯+α2>-(n-1),αn+αn-1+αn-2+⋯+α2+α1>-n.
Then we introduce the spaces Hαp(Bn) of functions f holomorphic in Bn and satisfying the condition
(1.6)∫Bn|f(w)|p⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)<+∞.
Section 3 contains detailed investigation of these spaces with “anisotropic” weight function.For these “anisotropic” spaces, similarities of the integral representations (1.4) are obtained, but this time a special kernels Sβ1β2⋯βn(z;w)≡Sβ(z;w) (where βi,1≤i≤n, are associated with αi,1≤i≤n, and p in a special way) appear instead of cn(β)·(1-〈z,w〉)-(n+1+β) (Theorem 4.7). Theorem 4.5 gives the description (in a multiple series form) and the main properties of these kernels. Theorem 4.8 makes it possible to represent the kernels Sβ as integrals taken over [0;1]n-1⊂Rn. Finally, in the special case n=2 we write out these kernels in an explicit form (see Theorem 4.12).


## 2. Preliminaries

In this section, we present several well-known facts which will be used in what follows.Fact 1.
Forα>-1 and k=0,1,2,…, put
(2.1)J(1)(α;k)=∫D|w|2k(1-|w|2)αdm(w),
then
(2.2)J(1)(α;k)=πΓ(k+1)Γ(α+1)Γ(k+α+2).
Moreover, if α>-1 and k,l=0,1,2,…(k≠l), then
(2.3)∫Dwkw¯l(1-|w|2)αdm(w)=0.Remark 2.1.
Assume thatρ>0 and φ is a continuous positive (i.e., φ>0) function in (0;ρ). If k,l=0,1,2,…(k≠l), then
(2.4)∫|w|<ρwkw¯lφ(|w|)dm(w)=0,
when the corresponding integral exists.Fact 2.
Forλ≠0,-1,-2,…,
(2.5)1(1-x)λ=∑k=0∞Γ(k+λ)Γ(λ)Γ(k+1)xk,|x|<1.Fact 3.
IfRep>0,Req>0, then
(2.6)∫01tp-1(1-t)q-1dt≡B(p;q)=Γ(p)Γ(q)Γ(p+q).Fact 4.
Ifg∈C[0;T](T>0), then
(2.7)lim⁡β↓0⁡1Γ(β)∫0Tg(x)(T-x)β-1dx=g(T).As a consequence of Stirling’s Formula, we have the following fact.Fact 5.
For arbitraryμ∈C and forR≥0,R→+∞(2.8)|Γ(μ+R)|≍e-R⋅RReμ+R-1/2.
In addition, if Reμ>0,Reν>0, and R≥0,R→+∞, then
(2.9)|Γ(μ+R)|≍|Γ(Reμ+R)|,|Γ(μ+R)Γ(ν+R)|≍1RReν-Reμ.Fact 6.
Assume thatp>0,ρ>0, and f∈H(|w|<ρ); then ∫02π|f(reiϑ)|pdϑ is a nondecreasing function of r, that is,
(2.10)∫02π|f(r1eiϑ)|pdϑ≤∫02π|f(r2eiϑ)|pdϑ,0≤r1<r2<ρ.Corollary 2.2.
Assume thatp>0,ρ>0,f∈H(|w|<ρ), and φ is a continuous positive (i.e., φ>0) function in (0;ρ). Then
(2.11)∫ρ1<|w|<ρ2|f(r⋅w)|p⋅φ(|w|)dm(w)≤∫ρ1<|w|<ρ2|f(w)|p⋅φ(|w|)dm(w)
if 0≤ρ1<ρ2≤ρ and 0≤r≤1. In particular,
(2.12)∫|w|<ρ|f(r⋅w)|p⋅φ(|w|)dm(w)≤∫|w|<ρ|f(w)|p⋅φ(|w|)dm(w).


## 3. Main Function Spaces

Suppose thatα=(α1,α2,…,αn)∈Rn. We put
(3.1)mα=∑αi<0-αi,lα=∑αi>0αi.
Further, we shall write α≺ (⋆) only if the following conditions are satisfied:
(⋆)αn+αn-1+⋯+αi>-(n+1-i)(1≤i≤n).
Similarly, if β=(β1,β2,…,βn)∈Cn, then we shall write β≺ (⋆) if only Reβ≡(Reβ1,Reβ2,…,Reβn)≺ (⋆).The following multidimensional analogue of Fact1 is valid.Lemma 3.1.
Forα=(α1,α2,…,αn)≺ (⋆) and for arbitrary multi-index k=(k1,k2,…,kn)∈Z+n, put
(3.2)J(n)(α;k)≡J(n)(α1,α2,…,αn;k1,k2,…,kn)=∫Bn∏i=1n|wi|2ki∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w);
then
(3.3)J(n)(α;k)=πn⋅Γ(kn+αn+αn-1+2)Γ(kn+αn+2)×Γ(kn+kn-1+αn+αn-1+αn-2+3)Γ(kn+kn-1+αn+αn-1+3)×⋯×Γ(kn+kn-1+kn-2+⋯+k3+αn+αn-1+αn-2+⋯+α3+α2+n-1)Γ(kn+kn-1+kn-2+⋯+k3+αn+αn-1+αn-2+⋯+α3+n-1)×Γ(kn+kn-1+kn-2+⋯+k3+k2+αn+αn-1+αn-2+⋯+α3+α2+α1+n)Γ(kn+kn-1+kn-2+⋯+k3+k2+αn+αn-1+αn-2+⋯+α3+α2+n)×Γ(kn+1)⋅Γ(kn-1+1)⋅Γ(kn-2+1)⋅⋯⋅Γ(k3+1)⋅Γ(k2+1)⋅Γ(k1+1)⋅Γ(αn+1)Γ(kn+kn-1+kn-2+⋯+k3+k2+k1+αn+αn-1+αn-2+⋯+α3+α2+α1+n+1).
Moreover, ifα≺ (⋆) and k=(k1,k2,…,kn),l=(l1,l2,…,ln) are arbitrary multi-indices such that k≠l, then
(3.4)∫Bn∏i=1nwiki∏i=1nwi¯li∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=0.Proof.
We intend to establish (3.3) by induction in n. For n=1 we simply arrived at (2.2). Assume the validity of (3.3) for some n and proceed to the case of n+1. Note that arbitrary w∈Bn+1 can be written as w=(w′,wn+1), where w′∈Bn,|wn+1|<1-|w′|2. Consequently, for α=(α1,α2,…,αn,αn+1)≺ (⋆) and for multi-index k=(k1,k2,…,kn,kn+1)∈Z+n+1, we have
(3.5)J(n+1)(α;k)≡J(n+1)(α1,α2,…,αn,αn+1;k1,k2,…,kn,kn+1)=∫Bn+1∏i=1n+1|wi|2ki∏i=1n+1(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=∫Bn∏i=1n|wi|2ki∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αi×∫|wn+1|<1-|w′|2|wn+1|2kn+1(1-|w′|2-|wn+1|2)αn+1dm(wn+1)dm(w′).
A simple change of variable: wn+1=1-|w′|2·ζ(ζ∈D) in the inner integral gives the following recurrent relation:
(3.6)J(n+1)(α1,α2,…,αn,αn+1;k1,k2,…,kn,kn+1)=πΓ(kn+1+1)Γ(αn+1+1)Γ(kn+1+αn+1+2)×J(n)(α1,α2,…,αn-1,αn+αn+1+kn+1+1;k1,k2,…,kn).
Note that (α1,α2,…,αn-1,αn+αn+1+kn+1+1)≺ (⋆) (for all kn+1∈Z) due to the condition α=(α1,α2,…,αn,αn+1)≺ (⋆). Consequently, in (3.6) we can apply (3.3) to J(n)(α1,α2,…,αn-1,αn+αn+1+kn+1+1;k1,k2,…,kn) due to our inductive assumption. As a result, we obtain (3.3) but this time for n+1. Thus, the inductive argument is completed and the formula (3.3) is established.
Now suppose thatk=(k1,k2,…,kn),l=(l1,l2,…,ln)∈Z+n, and k≠l⇒ there exists i0(1≤i0≤n) such that ki0≠li0. Then we can split arbitrary w=(w1,…,wn)∈Bn in wi0̂=(w1,…i0̂…,wn)∈Cn-1 and wi0∈C so that |wi0̂|2+|wi0|2<1 and dm(w)=dm(wi0)dm(wi0̂). Hence
(3.7)∫Bn∏i=1nwiki∏i=1nwi¯li∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=∫|wi0̂|2<1w1k1⋅w1¯l1⋅…i0̂…⋅wnknwn¯ln⋅∏i=1i0-1(1-|w1|2-|w2|2-⋯-|wi|2)αi×∫|wi0|2<1-|wi0̂|2wi0ki0⋅wi0¯li0⋅∏i=i0n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(wi0)dm(wi0̂).
In view of (2.4), the inner integral in (3.7) is equal to 0, so (3.4) is also proved.Corollary 3.2.
Ifα≺ (⋆), then
(3.8)∫Bn∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=πn∏i=1n(αn+αn-1+⋯+αi+n+1-i)<+∞.Remark 3.3.
In the integralsJ(n)(α;k) (see (3.2)) instead of α≺ (⋆), arbitrary β≺ (⋆) (β∈Cn) can be considered and the formulas (3.3), (3.4), and (3.8) remain true after the replacement of α by β.Definition 3.4.
Assume that0<p<+∞ and α=(α1,α2,…,αn)≺ (⋆). Denote by Lαp(Bn) the space of all complex-valued functions f in Bn with
(3.9)Mαp(f)≡∫Bn|f(w)|p⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)<+∞.
We obviously have
(3.10)Mαp(f+g)≤2p{Mαp(f)+Mαp(g)},Mαp(c⋅f)=cp⋅Mαp(f).
Correspondingly, denote by Hαp(Bn)⊂Lαp(Bn) the subspace of functions holomorphic in Bn. Note that 1≤p<q<+∞⇒Lαq(Bn)⊂Lαp(Bn),Hαq(Bn)⊂Hαp(Bn).Lemma 3.5.
Assume that0<p<+∞,α≺ (⋆), and f∈H(Bn). Then
(3.11)|f(z)|p≤22n+lαcn⋅(1-|z|)2n+lα⋅Mαp(f),∀z∈Bn,
where cn=m(Bn) is the volume of the unit ball of Cn.Proof.
Fix an arbitraryz∈Bn; put r=(1-|z|)/2 and B(z;r)={w∈Cn:|w-z|<r}⊂Bn. Since |f|p is subharmonic in Bn, we have
(3.12)|f(z)|p≤1cn⋅r2n∫B(z;r)|f(w)|pdm(w).
Note that
(3.13)w∈B(z;r)⟹|w|<r+|z|=1-|z|2+|z|=1+|z|2⟹1-|w|>1-|z|2⟹1-|w|2≥1-|w|>1-|z|2⟹1-|w1|2-|w2|2-⋯-|wi|2>1-|z|2(1≤i≤n).
Hence for w∈B(z;r)(3.14)(1-|w1|2-|w2|2-⋯-|wi|2)αi≥{1,αi≤0,(1-|z|)αi2αi,αi>0.
Combining (3.12) and (3.14), we obtain
(3.15)|f(z)|p≤1cn⋅((1-|z|)/2)2n⋅∏αi>0((1-|z|)αi/2αi)∏αi>0((1-|z|)αi/2αi)∫B(z;r)|f(w)|pdm(w)≤22n+lαcn⋅(1-|z|)2n+lα∫B(z;r)|f(w)|p⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)≤22n+lαcn⋅(1-|z|)2n+lα⋅Mαp(f).Corollary 3.6.
Hαp(Bn) is a closed subspace in Lαp(Bn).Definition 3.7.
Iff∈H(Bn) and 0≤ri≤1(1≤i≤n), then put
(3.16)fr1r2⋯rn(w)≡f(r1⋅w1,r2⋅w2,…,rn⋅wn),w=(w1,w2,…,wn)∈Bn.
Similarly, if f∈H(Bn) and 0≤r≤1, then put
(3.17)fr(w)≡f(r⋅w1,r⋅w2,…,r⋅wn),w=(w1,w2,…,wn)∈Bn.
In particular, if ri≡r∈[0;1](1≤i≤n)⇒fr1r2⋯rn(w)≡fr(w), note also that f11⋯1(w)≡f1(w)≡f(w).Lemma 3.8.
Assume that0<p<+∞,α≺ (⋆), and f∈H(Bn). Then for all ri,0≤ri≤1(1≤i≤n),
(3.18)Mαp(fr1r2⋯rn)≤Mαp(f).
In particular, f∈Hαp(Bn)⇒fr1r2⋯rn∈Hαp(Bn).Proof.
Evidently, it suffices to fixi0(1≤i0≤n) and consider the case 0≤ri0≤1,ri=1(1≤i≤n,i≠i0). In other words, it is sufficient to show that
(3.19)Mαp(f11⋯1ri01⋯11)≤Mαp(f).
To this end, we proceed as follows:
(3.20)Mαp(f11⋯1ri01⋯11)=∫Bn|f(w1,…,wi0-1,ri0⋅wi0,wi0+1,…,wn)|p×∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=∫|wi0̂|2<1∏i=1i0-1(1-|w1|2-|w2|2-⋯-|wi|2)αi×∫|wi0|2<1-|wi0̂|2|f(w1,…,wi0-1,ri0⋅wi0,wi0+1,…,wn)|p×∏i=i0n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(wi0)dm(wi0̂).
An application of Corollary 2.2 (see (2.12)) to the inner integral gives the desired inequality.Corollary 3.9.
Assume that0<p<+∞,α≺ (⋆), and f∈H(Bn). Then for all r,0≤r≤1,
(3.21)Mαp(fr)≤Mαp(f).
In particular, f∈Hαp(Bn)⇒fr∈Hαp(Bn).Lemma 3.10.
Assume that0<p<+∞,α≺ (⋆), and f∈H(Bn). Then for all ri,0≤ri≤1(1≤i≤n) and for arbitrary ε(0<ε≤1),
(3.22)∫1-ε<|w|<1|fr1r2⋯rn(w)|p⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)≤∫1-ε<|w|<1|f(w)|p⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w).Proof.
Since the caseε=1 coincides with (3.18), we can suppose that 0<ε<1. Further, similarly to the proof of Lemma 3.8, it suffices to fix i0(1≤i0≤n) and consider the case 0≤ri0≤1,ri=1(1≤i≤n,i≠i0):(3.23)∫1-ε<|w|<1|f(w1,…,wi0-1,ri0⋅wi0,wi0+1,…,wn)|p×∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)≡I1+I2,
where
(3.24)I1=∫|wi0̂|2<(1-ε)2∏i=1i0-1(1-|w1|2-|w2|2-⋯-|wi|2)αi×∫(1-ε)2-|wi0̂|2<|wi0|2<1-|wi0̂|2|f(w1,…,wi0-1,ri0⋅wi0,wi0+1,…,wn)|p×∏i=i0n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(wi0)dm(wi0̂),(3.25)I2=∫(1-ε)2<|wi0̂|2<1∏i=1i0-1(1-|w1|2-|w2|2-⋯-|wi|2)αi×∫|wi0|2<1-|wi0̂|2|f(w1,…,wi0-1,ri0wi0,wi0+1,…,wn)|p×∏i=i0n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(wi0)dm(wi0̂).
It remains to apply Corollary 2.2. More exactly, let us apply (2.11) to the inner integral of (3.24) and (2.12) to the inner integral of (3.25). By this, the proof is complete.Corollary 3.11.
Assume that0<p<+∞,α≺ (⋆), and f∈H(Bn). Then for all r,0≤r≤1 and for arbitrary ε(0<ε≤1),
(3.26)∫1-ε<|w|<1|fr(w)|p⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)≤∫1-ε<|w|<1|f(w)|p⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w).Lemma 3.12.
Assume that0<p<+∞,α≺ (⋆), and f∈Hαp(Bn). Then
(3.27)Mαp(f-fr1r2⋯rn)⟶0asri↑1(1≤i≤n).Proof.
For arbitrary fixedε∈(0;1) and for all ri,0≤ri≤1(1≤i≤n), we have
(3.28)Mαp(f-fr1r2⋯rn)=∫Bn|f(w1,w2,…,wn)-f(r1⋅w1,r2⋅w2,…,rn⋅wn)|p×∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=∫|w|<1-ε+∫1-ε<|w|<1≡J1+J2.
In view of the inequality(a+b)p≤2p(ap+bp)(a,b≥0,p>0) and due to (3.22),
(3.29)J2≤2p+1⋅∫1-ε<|w|<1|f(w1,w2,…,wn)|p×∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w).
Note that the condition f∈Hαp(Bn) implies J2→0 as ε↓0. Besides, it is evident that for a fixed ε∈(0;1):J1→0 as ri↑1(1≤i≤n). All these complete the proof.Corollary 3.13.
Assume that0<p<+∞,α≺ (⋆), and f∈Hαp(Bn). Then
(3.30)Mαp(f-fr)⟶0asr↑1.


## 4. Main Integral Representations

In fact, Lemma3.1 asserts that the system(4.1){w1k1⋅w2k2⋅⋯⋅wnkn}k1,k2,…,kn=0∞
is orthogonal in the Hilbert space (4.2)L2{Bn;∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)}.
And what about completeness of this system?Proposition 4.1.
Assume that1≤p<+∞,α≺ (⋆), and f∈Hαp(Bn). If for arbitrary multi-index k=(k1,k2,…,kn)∈Z+n(4.3)∫Bnf(w)⋅w1¯k1⋅w2¯k2⋅⋯⋅wn¯kn⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=0,
then f≡0.Proof.
Sincef∈H(Bn)⇒(4.4)f(w)=∑l1,l2,…,ln≥0al1l2⋯ln⋅w1l1w2l2⋯wnln≡∑|l|≥0al⋅wl,w∈Bn,
where |l|=l1+l2+⋯+ln for arbitrary multi-index l=(l1,l2,…,ln)∈Z+n. Moreover, for arbitrary compact K⊂Bn there exist a positive series ∑|l|≥0cl<+∞ such that |al·wl|≤cl(|l|≥0) uniformly in w∈K. Consequently, for arbitrary fixed r∈(0;1)(4.5)f(r⋅w)=∑|l|≥0al⋅r|l|wl,w∈Bn¯,
and, in addition, there exist a positive series ∑|l|≥0dl<+∞ such that |al·r|l|·wl|≤dl(|l|≥0) uniformly in w∈Bn¯.
Now let us fix arbitraryk1≥0,k2≥0,…,kn≥0. Since fr→f (as r↑1) in the space Hαp(Bn) (Corollary 3.13), we have (as r↑1)
(4.6)∫Bnf(r⋅w)⋅∏i=1nwi¯ki⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)⟶∫Bnf(w)⋅∏i=1nwi¯ki⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=0.
On the other hand,
(4.7)∫Bnf(r⋅w)⋅∏i=1nwi¯ki⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=∑|l|≥0al⋅r|l|∫Bn∏i=1nwili⋅∏i=1nwi¯ki⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=ak⋅r|k|⋅J(n)(α1,…,αn;k1,…,kn).
Combining (4.6) and (4.7), we obtain that ak·r|k|·J(n)(α1,…,αn;k1,…,kn)→0 (as r↑1) ⇒ak=0 and this takes place for any multi-index k⇒f≡0.Corollary 4.2.
Ifα≺ (⋆), then the system
(4.8){w1k1⋅w2k2⋅⋯⋅wnknJ(n)(α1,…,αn;k1,…,kn)}k1,k2,…,kn=0∞
is an orthonormal basis in the space Hα2(Bn).Corollary 4.3.
Ifα≺ (⋆), then the reproducing kernel (i.e., the kern-function) of the space Hα2(Bn) is given by the formula
(4.9)Sα(z;w)=∑k1,k2,…,kn=0∞(z1w1¯)k1⋅(z2w2¯)k2⋅⋯⋅(znwn¯)knJ(n)(α1,…,αn;k1,…,kn),z,w∈Bn.Definition 4.4.
Assume thatβ=(β1,β2,…,βn)∈Cn and β≺ (⋆) ⇔Reβ=(Reβ1,Reβ2,…,Reβn)≺ (⋆). For arbitrary z∈Bn,w∈Bn¯, let us introduce the following kernel (motivated by (4.9)):
(4.9’)Sβ(z;w)=∑k1,k2,…,kn=0∞(z1w1¯)k1⋅(z2w2¯)k2⋅⋯⋅(znwn¯)knJ(n)(β1,…,βn;k1,…,kn)
or, in a slightly different form,
(4.9’’)Sβ(z;w)=∑k=0∞∑k1+k2+⋯+kn-1+kn=k(z1w1¯)k1⋅(z2w2¯)k2⋅⋯⋅(znwn¯)knJ(n)(β1,…,βn;k1,…,kn)≡∑k=0∞Pk(z;w),
where
(4.10)Pk(z;w)=1πn⋅∑k1+k2+⋯+kn-1+kn=k(z1w1¯)k1⋅(z2w2¯)k2⋅⋯⋅(znwn¯)kn×Γ(kn+βn+2)Γ(kn+βn+βn-1+2)×Γ(kn+kn-1+βn+βn-1+3)Γ(kn+kn-1+βn+βn-1+βn-2+3)×⋯×Γ(kn+kn-1+kn-2+⋯+k3+βn+βn-1+βn-2+⋯+β3+n-1)Γ(kn+kn-1+kn-2+⋯+k3+βn+βn-1+βn-2+⋯+β3+β2+n-1)×Γ(kn+kn-1+kn-2+⋯+k3+k2+βn+βn-1+βn-2+⋯+β3+β2+n)Γ(kn+kn-1+kn-2+⋯+k3+k2+βn+βn-1+βn-2+⋯+β3+β2+β1+n)×Γ(kn+kn-1+kn-2+⋯+k3+k2+k1+βn+βn-1+βn-2+⋯+β3+β2+β1+n+1)Γ(kn+1)⋅Γ(kn-1+1)⋅Γ(kn-2+1)⋅⋯⋅Γ(k3+1)⋅Γ(k2+1)⋅Γ(k1+1)⋅Γ(βn+1).
Note that if we takeReβn>-1,βi=0(1≤i≤n-1) in (4.9’’)-(4.10),then
(4.11)Sβ(z;w)=Γ(βn+n+1)πn⋅Γ(βn+1)⋅1(1-z1w1¯-z2w2¯-z3w3¯-⋯-znwn¯)βn+n+1,
that is, we arrive at the kernel of the integral representations (1.4).
We are going to estimatePk(z;w),k=0,1,2,…⇒ to estimate the kernel Sβ(z;w). Let us put (compare with (3.1))
(4.12)mβ⋆=∑Reβi<0,1≤i≤n-1(-Reβi),lβ⋆=∑Reβi>0,1≤i≤n-1(Reβi).
Then in view of (2.9), we have (z∈Bn,w∈Bn¯,k=0,1,2,…)
(4.13)|Pk(z;w)|≤const(n;β)⋅∑k1+k2+⋯+kn-1+kn=k1(kn+1)Reβn-1×1(kn+kn-1+1)Reβn-2×⋯×1(kn+kn-1+kn-2+⋯+k3+1)Reβ2×1(kn+kn-1+kn-2+⋯+k3+k2+1)Reβ1×|Γ(kn+kn-1+⋯+k2+k1+βn+βn-1+⋯+β2+β1+n+1)||Γ(βn+1)|⋅Γ(kn+kn-1+⋯+k2+k1+1)×[Γ(kn+kn-1+⋯+k2+k1+1)⋅|z1w1¯|k1⋅|z2w2¯|k2⋅⋯⋅|znwn¯|knΓ(kn+1)⋅Γ(kn-1+1)⋅⋯⋅Γ(k2+1)⋅Γ(k1+1)]≤const(n;β)⋅∑k1+k2+⋯+kn-1+kn=k(kn+kn-1+kn-2+⋯+k3+k2+k1+1)mβ⋆(kn+1)lβ⋆×|Γ(kn+kn-1+⋯+k2+k1+Reβn+Reβn-1+⋯+Reβ2+Reβ1+n+1)||Γ(Reβn+1)|⋅Γ(kn+kn-1+⋯+k2+k1+1)×[Γ(kn+kn-1+⋯+k2+k1+1)⋅|z1w1¯|k1⋅|z2w2¯|k2⋅⋯⋅|znwn¯|knΓ(kn+1)⋅Γ(kn-1+1)⋅⋯⋅Γ(k2+1)⋅Γ(k1+1)]≤const(n;β)⋅(k+1)mβ⋆⋅Γ(k+Reβn+Reβn-1+⋯+Reβ2+Reβ1+n+1)Γ(Reβn+1)⋅Γ(k+1)×∑k1+k2+⋯+kn-1+kn=kΓ(k+1)⋅|z1w1¯|k1⋅|z2w2¯|k2⋅⋯⋅|znwn¯|knΓ(kn+1)⋅Γ(kn-1+1)⋅⋯⋅Γ(k2+1)⋅Γ(k1+1)≤const(n;β)⋅(k+1)mβ⋆⋅Γ(k+Reβn+Reβn-1+⋯+Reβ2+Reβ1+n+1)Γ(Reβn+Reβn-1+⋯+Reβ2+Reβ1+n+1)⋅Γ(k+1)×(|z1w1¯|+|z2w2¯|+⋯+|znwn¯|)k.
Further, due to (2.8),
(4.14)Γ(k+Reβn+Reβn-1+⋯+Reβ2+Reβ1+n+1)≍e-(k+1)⋅(k+1)Reβn+Reβn-1+⋯+Reβ2+Reβ1+n+k+1-1/2(k=0,1,2,…).
Consequently,(4.15)(k+1)mβ⋆⋅Γ(k+Reβn+Reβn-1+⋯+Reβ2+Reβ1+n+1)≍e-(k+1)⋅(k+1)Reβn+Reβn-1+⋯+Reβ2+Reβ1+n+mβ⋆+k+1-1/2≍Γ(k+Reβn+Reβn-1+⋯+Reβ2+Reβ1+mβ⋆+n+1).
Combining (4.13), (4.15) and taking into account the inequality
(4.16)|z1w1¯|+|z2w2¯|+⋯+|znwn¯|≤|z|⋅|w|≤|z|<1(z∈Bn,w∈Bn¯),
we obtain
(4.17)|Pk(z;w)|≤const(n;β)⋅Γ(k+Reβn+Reβn-1+⋯+Reβ2+Reβ1+mβ⋆+n+1)⋅|z|kΓ(Reβn+Reβn-1+⋯+Reβ2+Reβ1+mβ⋆+n+1)⋅Γ(k+1).
In view of the formula (2.5), the summation of (4.17) over k=0,1,2,… gives
(4.18)|Sβ(z;w)|≤const(n;β)(1-|z|)Reβn+Reβn-1+⋯+Reβ2+Reβ1+mβ⋆+n+1=const(n;β)(1-|z|)Reβn+lβ⋆+n+1.Thus, we have the following theorem.Theorem 4.5.
Ifβ≺ (⋆), then the kernel Sβ(z;w) satisfies the following properties (z∈Bn,w∈Bn¯):(a)
the series (4.9’) converges absolutely for arbitrary fixed z∈Bn,w∈Bn¯;(b)
in the expansionSβ(z;w)=∑k=0∞Pk(z;w) (see (4.9’’)), the series is majorated by a convergent positive numerical series ∑k=0∞bk uniformly in z∈K⊂⊂Bn,w∈Bn¯;(c)
Sβ(z;w) is holomorphic in z∈Bn;(d)
Sβ(z;w) is antiholomorphic in w∈Bn and continuous in w∈Bn¯;(e)
the estimate(4.19)|Sβ(z;w)|≤const(n;β)(1-|z|)n+1+Reβn+lβ⋆is valid uniformly in z∈Bn,w∈Bn¯.Definition 4.6.
Assume that1≤p<+∞ and α=(α1,α2,…,αn)≺ (⋆). For β=(β1,β2,…,βn)∈Cn, we shall write β≻(p;α) if only (compare with (1.3))(i)
whenp=1, then
(4.20)Reβi≥αi(1≤i≤n);(ii)
when1<p<+∞, then
(4.21)Reβn+Reβn-1+⋯+Reβi>αn+αn-1+⋯+αi+n+1-ip-(n+1-i).
Obviously,β≻(p;α)⇒β≺ (⋆).Theorem 4.7.
Assume that1≤p<+∞,α≺ (⋆), and β≻(p;α). Then arbitrary f∈Hαp(Bn) admits the following integral representations:
(4.22)f(z)=∫Bnf(w)Sβ(z;w)∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w),z∈Bn,(4.23)f(0)¯=∫Bnf(w)¯Sβ(z;w)∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w),z∈Bn.Proof.
Let us fix an arbitraryf∈Hαp(Bn) and an arbitrary z∈Bn. Further,
(4.24)f(w)=∑|l|≥0al⋅wl=∑m=0∞Fm(w),w∈Bn,
where
(4.25)Fm(w)=∑l1+l2+⋯+ln-1+ln=mal⋅wl(m=0,1,2,…).
Hence, for arbitrary fixed r∈(0;1),
(4.26)fr(w)=∑|l|≥0al⋅r|l|⋅wl=∑m=0∞rm⋅Fm(w),w∈Bn¯.
Obviously,
(4.27)fr(w)¯=∑|l|≥0al¯⋅r|l|⋅w¯l=∑m=0∞rm⋅Fm(w)¯,w∈Bn¯.
Note that the first series in (4.26) and (4.27) and the second series in (4.26) and (4.27) are majorated (uniformly in w∈Bn¯) by positive convergent numerical series ∑|l|≥0cl<+∞ and ∑m=0∞cm̃<+∞, respectively.
First we intend to prove both formulas (4.22), (4.23) for the function fr, that is,
(4.22’)fr(z)=∫Bnfr(w)Sβ(z;w)∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w),(4.23’)f(0)¯=∫Bnfr(w)¯Sβ(z;w)∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w).
In view of Theorem 4.5(b), (4.26)-(4.27) and due to Lebesgue’s dominated convergence theorem, we have
(4.28)∫Bnfr(w)Sβ(z;w)∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w)=∑k,m=0∞∫BnrmFm(w)⋅Pk(z;w)⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w),(4.29)∫Bnfr(w)¯Sβ(z;w)∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w)=∑k,m=0∞∫BnrmFm(w)¯⋅Pk(z;w)⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w),
and the series ∑k,m=0∞… in (4.28) and (4.29) converge absolutely.
Taking into account (3.4) and Remark 3.3, we conclude that all terms in the right-hand side of (4.29) (except of the case k=m=0) are equal to zero. Hence the left-hand side of (4.29) is equal to
(4.30)∫Bna00⋯0¯⋅1πn⋅Γ(βn+2)Γ(βn+βn-1+2)×Γ(βn+βn-1+3)Γ(βn+βn-1+βn-2+3)×⋯×Γ(βn+βn-1+βn-2+⋯+β3+n-1)Γ(βn+βn-1+βn-2+⋯+β3+β2+n-1)×Γ(βn+βn-1+βn-2+⋯+β3+β2+n)Γ(βn+βn-1+βn-2+⋯+β3+β2+β1+n)×Γ(βn+βn-1+βn-2+⋯+β3+β2+β1+n+1)Γ(βn+1)×∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w)=f(0)¯⋅∫Bn∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w)J(n)(β1,β2,…,βn;0,0,…,0)=f(0)¯.
Thus, (4.23’) has been proved.
Now let us proceed to (4.22’). If k≠m, then in view of (3.4) and Remark 3.3(4.31)∫BnrmFm(w)⋅Pk(z;w)⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w)=0.
Ifk=m, then in view of (3.4), (3.2), (3.3) and Remark 3.3(4.32)∫BnrkFk(w)⋅Pk(z;w)⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)βidm(w)=rkπn⋅∑k1+k2+⋯+kn-1+kn=kak1k2⋯kn-1knz1k1⋅z2k2⋅⋯⋅zn-1kn-1⋅znknΓ(kn+1)⋅Γ(kn-1+1)⋅⋯⋅Γ(k2+1)⋅Γ(k1+1)×Γ(kn+βn+2)Γ(kn+βn+βn-1+2)×Γ(kn+kn-1+βn+βn-1+3)Γ(kn+kn-1+βn+βn-1+βn-2+3)×⋯×Γ(kn+kn-1+kn-2+⋯+k3+βn+βn-1+βn-2+⋯+β3+n-1)Γ(kn+kn-1+kn-2+⋯+k3+βn+βn-1+βn-2+⋯+β3+β2+n-1)×Γ(kn+kn-1+kn-2+⋯+k3+k2+βn+βn-1+βn-2+⋯+β3+β2+n)Γ(kn+kn-1+kn-2+⋯+k3+k2+βn+βn-1+βn-2+⋯+β3+β2+β1+n)×Γ(kn+kn-1+kn-2+⋯+k3+k2+k1+βn+βn-1+βn-2+⋯+β3+β2+β1+n+1)Γ(βn+1)×J(n)(β1,β2,…,βn;k1,k2,…,kn)=rk⋅∑k1+k2+⋯+kn-1+kn=kak1k2⋯kn-1knz1k1⋅z2k2⋅⋯⋅zn-1kn-1⋅znkn⋅1=rk⋅Fk(z).
Consequently, (4.28), (4.31), and (4.32) together yield
(4.33)∫Bnfr(w)Sβ(z;w)∏k=1n(1-|w1|2-|w2|2-⋯-|wk|2)βkdm(w)=∑k=0∞rk⋅Fk(z)=fr(z).
Thus, (4.22’) also has been proved.
Now we intend to make passager↑1 in (4.22’) and (4.23’). Evidently, the left-hand sides tend to f(z) and f(0)¯ correspondingly. It remains to show that the right-hand sides of (4.22’) and (4.23’) tend to the right-hand sides of (4.22) and (4.23), respectively. In view of the estimate (4.19), it suffices to show that
(4.34)I(r)≡∫Bn|f(w)-fr(w)|⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)Reβidm(w)⟶r↑10.
If p=1, then due to the condition (4.20) and Corollary 3.13(4.35)I(r)=∫Bn|f(w)-fr(w)|⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)Reβi-αi+αidm(w)≤∫Bn|f(w)-fr(w)|⋅∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)αidm(w)=Mα1(f-fr)⟶r↑10.
If 1<p<∞, then an application of Hölder integral inequality to I(r) gives
(4.36)I(r)≤[Mαp(f-fr)]1/p⋅(∫Bn∏i=1n(1-|w1|2-|w2|2-⋯-|wi|2)q(Reβi-αi/p)dm(w))1/q.
Here the integral over Bn is finite due to Corollary 3.2 and the condition β≻(p;α). Consequently, I(r)→0,r↑1, in view of Corollary 3.13. Thus, the theorem is proved.Theorem 4.8.
ForReβn>-1,Reβi>0(1≤i≤n-1), and for z∈Bn,w∈Bn¯,
(4.37)Sβ(z;w)=Γ(βn+βn-1+⋯+β2+β1+n+1)πn⋅Γ(β1)Γ(β2)⋅⋯⋅Γ(βn-1)Γ(βn+1)×∫[0;1]n-1∏i=n2siβn+βn-1+⋯+βi+n+1-i⋅∏i=n2(1-si)βi-1-1dsndsn-1⋯ds3ds2(1-z1w1¯-s2z2w2¯-s2s3z3w3¯-⋯-s2s3⋯snznwn¯)βn+βn-1+⋯+β2+β1+n+1.Proof.
Fork=0,1,2,3,… according to (4.10) and (2.6), we have
(4.38)Pk(z;w)=1πn⋅∑k1+k2+⋯+kn-1+kn=k(z1w1¯)k1⋅(z2w2¯)k2⋅⋯⋅(zn-1wn-1¯)kn-1⋅(znwn¯)knΓ(k1+1)⋅Γ(k2+1)⋅⋯⋅Γ(kn-1+1)⋅Γ(kn+1)×1Γ(βn-1)∫01snkn+βn+1(1-sn)βn-1-1dsn×1Γ(βn-2)∫01sn-1kn+kn-1+βn+βn-1+2(1-sn-1)βn-2-1dsn-1×⋯×1Γ(β1)∫01s2kn+kn-1+kn-2+⋯+k2+βn+βn-1+βn-2+⋯+β2+n-1(1-s2)β1-1ds2×Γ(kn+kn-1+kn-2+⋯+k2+k1+βn+βn-1+βn-2+⋯+β2+β1+n+1)Γ(βn+1)=1πn⋅Γ(β1)Γ(β2)⋯Γ(βn-1)Γ(βn+1)×∫[0;1]n-1snβn+1⋅sn-1βn+βn-1+2⋅⋯⋅s2βn+βn-1+⋯+β2+n-1×(1-sn)βn-1-1⋅(1-sn-1)βn-2-1⋅⋯⋅(1-s2)β1-1×∑k1+k2+⋯+kn-1+kn=k(z1w1¯)k1⋅(s2z2w2¯)k2⋅(s2s3z3w3¯)k3⋅⋯⋅(s2s3⋯snznwn¯)knΓ(kn+1)⋅Γ(kn-1+1)⋅⋯⋅Γ(k2+1)⋅Γ(k1+1)×Γ(k+1)Γ(k+1)⋅Γ(k+βn+βn-1+⋯+β2+β1+n+1)dsndsn-1⋯ds3ds2=Γ(k+βn+βn-1+⋯+β2+β1+n+1)πn⋅Γ(k+1)Γ(β1)Γ(β2)⋯Γ(βn-1)Γ(βn+1)×∫[0;1]n-1snβn+1⋅sn-1βn+βn-1+2⋅⋯⋅s2βn+βn-1+⋯+β2+n-1×(1-sn)βn-1-1⋅(1-sn-1)βn-2-1⋅⋯⋅(1-s2)β1-1×(z1w1¯+s2z2w2¯+s2s3z3w3¯+⋯+s2s3⋯snznwn¯)kdsndsn-1⋯ds3ds2=Pk(z;w).
The summation of these relations over k=0,1,2,3,… yields (see (2.5))
(4.39)Sβ(z;w)=∑k=0∞Pk(z;w)=Γ(βn+βn-1+βn-2+⋯+β2+β1+n+1)πn⋅Γ(β1)Γ(β2)⋯Γ(βn-1)Γ(βn+1)×∫[0;1]n-1snβn+1⋅sn-1βn+βn-1+2⋅⋯⋅s2βn+βn-1+⋯+β2+n-1×(1-sn)βn-1-1⋅(1-sn-1)βn-2-1⋅⋯⋅(1-s2)β1-1×dsndsn-1⋯ds3ds2(1-z1w1¯-s2z2w2¯-s2s3z3w3¯-⋯-s2s3⋯snznwn¯)βn+βn-1+⋯+β2+β1+n+1.
Thus, (4.37) is proved.Remark 4.9.
Under the conditions of the theorem, the formula (4.37) easily implies the assertions (c), (d), and (e) of Theorem 4.5.Remark 4.10.
If we takeReβn>-1,βi↓0(1≤i≤n-1) in (4.37), then the formal application of (2.7) gives the following formula:
(4.40)Sβ(z;w)=Γ(βn+n+1)πn⋅Γ(βn+1)⋅1(1-z1w1¯-z2w2¯-z3w3¯-⋯-znwn¯)βn+n+1,
that is, we arrive at the kernel of the integral representations (1.4).Remark 4.11.
In fact, the kernelSβ(z;w), defined for β≺ (⋆) by (4.9’)-(4.9’’), can be considered as an analytic continuation in β of the integral Sβ(z;w), defined for Reβn>-1,Reβi>0(1≤i≤n-1) by (4.37).
Now we consider an interesting special casen=2, when (in comparison with (4.9’)-(4.9’’) or (4.37)) the kernel Sβ(z;w) can be written out in an explicit form.Theorem 4.12.
Assume thatβ=(β1,β2)≺ (⋆), that is, Reβ2>-1,Reβ2+Reβ1>-2. Then for z=(z1,z2)∈B2,w=(w1,w2)∈B2¯,
(4.41)Sβ(z;w)=1π2⋅β1(β2+1)(1-z1w1¯)β1+1(1-z1w1¯-z2w2¯)β2+2+1π2⋅(β2+1)(β2+2)(1-z1w1¯)β1(1-z1w1¯-z2w2¯)β2+3.Proof.
According to (4.9’)-(4.9’’)-(4.10) and in view of the formula (2.5),
(4.42)Sβ(z;w)=1π2⋅∑k1,k2=0∞(z1w1¯)k1⋅(z2w2¯)k2Γ(k1+1)Γ(k2+1)Γ(β2+k2+2)Γ(β2+β1+k2+2)⋅Γ(β2+β1+k2+k1+3)Γ(β2+1)=1π2⋅∑k2=0∞(z2w2¯)k2Γ(k2+1)⋅Γ(β2+k2+2)Γ(β2+β1+k2+2)⋅Γ(β2+1)×∑k1=0∞(z1w1¯)k1Γ(k1+1)⋅Γ(β2+β1+k2+k1+3)=1π2⋅∑k2=0∞(z2w2¯)k2Γ(k2+1)⋅Γ(β2+k2+2)⋅Γ(β2+β1+k2+3)Γ(β2+β1+k2+2)⋅Γ(β2+1)1(1-z1w1¯)β2+β1+k2+3=1π2⋅1(1-z1w1¯)β2+β1+3⋅∑k2=0∞(z2w2¯1-z1w1¯)k2Γ(β2+k2+2)Γ(k2+1)Γ(β2+1)⋅(β2+β1+k2+2)=1π2⋅1(1-z1w1¯)β2+β1+3⋅∑k2=0∞(z2w2¯1-z1w1¯)k2×{β1Γ(β2+k2+2)Γ(k2+1)Γ(β2+1)+Γ(β2+k2+3)Γ(k2+1)Γ(β2+1)}=1π2⋅1(1-z1w1¯)β2+β1+3{β1(β2+1)(1-z2w2¯/(1-z1w1¯))β2+2+(β2+1)(β2+2)(1-z2w2¯/(1-z1w1¯))β2+3}=1π2⋅1(1-z1w1¯)β2+β1+3{β1(β2+1)(1-z1w1¯)β2+2(1-z1w1¯-z2w2¯)β2+2+(β2+1)(β2+2)(1-z1w1¯)β2+3(1-z1w1¯-z2w2¯)β2+3}=1π2⋅{β1(β2+1)(1-z1w1¯)β1+1(1-z1w1¯-z2w2¯)β2+2+(β2+1)(β2+2)(1-z1w1¯)β1(1-z1w1¯-z2w2¯)β2+3}.
Thus, (4.41) is established.Remark 4.13.
During the proof we regroup the double series:∑k1,k2=0∞{…}=∑k2=0∞{∑k1=0∞{…}}, which is legitimate in view of Theorem 4.5(a). In fact, we apply Fubini’s theorem for double series.Remark 4.14.
The analysis of the proof shows that we have established (4.41) only for those z=(z1,z2)∈B2 and w=(w1,w2)∈B2¯, which satisfy the condition |z2w2¯/(1-z1w1¯)|<1. In fact, this is quite sufficiently since both sides of (4.41) are holomorphic in z∈Bn, antiholomorphic in w∈Bn, and continuous in w∈Bn¯ (see Theorem 4.5(c), (d)).Remark 4.15.
If one takesReβ2>-1,β1=0 in Theorem 4.12, then
(4.43)Sβ1;β2(z1,z2;w1,w2)=1π2⋅(β2+1)(β2+2)(1-z1w1¯-z2w2¯)β2+3,
which coincides with (4.11) (or (4.40)) for the case n=2.Remark 4.16.
Note that for the same casen=2 and under slightly restrictive conditions Reβ2>-1,Reβ1>0, the formula (4.37) gives (z∈B2,w∈B2¯)
(4.44)Sβ(z;w)=Γ(β2+β1+3)π2⋅Γ(β2+1)Γ(β1)⋅∫01sβ2+1⋅(1-s)β1-1(1-z1w1¯-s⋅z2w2¯)β2+β1+3ds.Remark 4.17.
It can be shown (we omit the proof) that under the conditionsReβ2>0,Reβ2+Reβ1>-2 the following interesting formula is valid (z∈B2,w∈B2¯):
(4.45)S(β1,β2)(z1,z2;w1,w2)={z2w2¯⋅∂/∂(z2w2¯)+β2+1β2}⋅S(β1+1,β2-1)(z1,z2;w1,w2).



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*Source: 354961-2011-02-10.xml*