# Construction of Biholomorphic Convex Mappings of Orderα on Bpn

Authors: Ming-Sheng Liu; Fengying Yang; Kit Ian Kou; Changqin Huang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

Some sufficient conditions for biholomorphic convex
mappings of orderα on the Reinhardt domain Bpn in Cn are given; from that,
criteria for biholomorphic convex mappings of order α with particular form
become direct. As applications of these sufficient conditions, some concrete
biholomorphic convex mappings of order α   on Bpn are provided.


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## Body

## 1. Introduction and Preliminaries

The analytic functions of one complex variable, which map the unit diskU={z∈C:|z|<1} onto starlike domains or convex domains, have been extensively studied. These functions are easily characterized by simple analytic or geometric conditions. In the case of one complex variable, the following notions are well known.LetH(U) = {f :U↦ℂ be analytic in U with f(0)=f′(0)-1=0}. A function f∈H(U) is said to be convex if f(U) is convex, that is, given w1,w2∈f(U), tw1+(1-t)w2∈f(U) for all t∈[0,1]. We let K denote the class of univalent convex functions in U. Suppose α∈[0,1). If f∈H(U) satisfies f′(z)≠0 for all z∈U and the following inequality:
(1)Re{zf′′(z)f′(z)+1}>α,∀z∈U,
then we call f(z) a convex function of order α in U. We let K(α) denote the class of convex functions of order α in U. It is evident that K≡K(0).In higher dimensions, demanding that a mapping takes the unit ball to a convex domain turned out to be a very restrictive condition. It is rather hard to construct concrete biholomorphic convex mappings on some domains inℂn, even on the Euclidean unit ball.Supposen is a fixed positive integer, p>1. Let Cn be the space of n complex variables z=(z1,z2,…,zn) with the usual inner product 〈z,w〉=∑j=1nzjw¯j, where w=(w1,w2,…,wn)∈Cn. We introduce the p-norm of Cn:  ∥z∥p=(∑j=1n|zj|p)1/p, and let Bpn={z∈Cn:∥z∥p<1}; it is evident that Bpn is a Reinhardt domain. For simplicity, let ∥z∥=∥z∥2=〈z,z〉.LetH(Bpn) be the class of holomorphic mappings f(z)=(f1(z),…,fn(z)) in the Reinhardt domain Bpn, where z=(z1,…,zn)∈Cn. A mapping f∈H(Bpn) is said to be locally biholomorphic in Bpn if f has a local inverse at each point z∈Bpn or, equivalently, if the first Fréchet derivative Df(z)=(∂fj(z)/∂zk)1≤j,k≤n is nonsingular at each point in Bpn.The second Fréchet derivative of a mappingf∈H(Bpn) is a symmetric bilinear operator D2f(z)(·,·) on Cn×Cn, and D2f(z)(z,·) is the linear operator obtained by restricting D2f(z) to {z}×Cn. The matrix representation of D2f(z)(b,·) is
(2)D2f(z)(b,·)=(∑l=1n∂2fj(z)∂zk∂zlbl‍)1≤j,k≤n,
where f(z)=(f1(z),…,fn(z)),b=(b1,…,bn)∈Cn.LetN(Bpn) denote the class of all locally biholomorphic mappings f:Bpn→Cn such that f(0)=0,Df(0)=I, where I is the unit matrix of n×n. If f∈N(Bpn) is a biholomorphic mapping on Bpn and f(Bpn) is a convex domain in Cn, then we call f a biholomorphic convex mapping on Bpn. The class of all biholomorphic convex mappings on Bpn is denoted by K(Bpn). Obviously, K=K(Bp1). The biholomorphic convex mapping of order α on Bpn was introduced and investigated in [1–5]; the ε starlike and ε quasi-convex mappings were investigated in [4, 6].Definition 1 (see [1–3, 5]).
Suppose0≤α<1,p≥2,u(z)=∑j=1n|zj|p, and f∈N(Bpn). Assume that for any z=(z1,…,zn)∈Bpn and b=(b1,…,bn)∈Cn with Re〈b,∂u/∂z¯〉=0, we have
(3)Jf(z,b)=Re{p24∑k=1n|zk|p-2|bk|2‍+p2(p2-1)×∑k=1n|zk|pzk2bk2-〈Df(z)-1D2f(z)(b,b),∂u∂z¯〉}≥α·p2∑k=1n|zk|p-2|bk|2,
where ∂u/∂z¯=(∂u/∂z1¯,…,∂u/∂zn¯). Then, f(z) is called a biholomorphic convex mapping of order α on Bpn. The class of all biholomorphic convex mappings of order α on Bpn is denoted by K(Bpn,α). It is evident that K(Bpn)≡K(Bpn,0) and K(Bp1,α)≡K(α).In 1995, Roper and Suffridge [7] proved that if f∈K and F(f)(z)=(f(z1),f′(z1)z0), where z=(z1,z0)∈Bn,z1∈U,z0=(z2,…,zn)∈Cn-1, then F(f)(z)∈K(B2n). F(f) is popularly referred to as the Roper-Suffridge operator. Using this operator, we may construct a lot of concrete biholomorphic convex mappings on B2n. Roper and suffridge [8] also obtained some sufficient conditions for biholomorphic convex mappings on the Euclidean unit ball. Liu and Zhu [9] had given some sufficient conditions and concrete examples of biholomorphic convex mappings on the Reinhardt domain Bpn. Liu [3] also gave some sufficient conditions for biholomorphic convex mappings of order α on Bpn. A problem is naturally posed: can we give several direct criteria for biholomorphic convex mapping of order α on Bpn? For example, can we get some sufficient conditions such that the mapping of the form
(4)f(z)=(p1(z1,z2,…,zn),p2(z2,zn),…,pn-1(zn-1,zn),pn(zn))
is a biholomorphic convex mapping of order α on Bpn?The aim of this paper is to give an answer to the above problem. From these, we may construct some concrete biholomorphic convex mappings of orderα on Bpn.


## 2. Main Results

Theorem 2.
Suppose thatn≥2,p≥2, 0≤α<1, q=p/(p-1). Let
(5)f(z)=(f1(z1,zn),f2(z2,zn)…,fn-1(zn-1,zn),pn(zn)),
where z=(z1,z2,…,zn)∈Bpn,pn(ζ)∈H(U) and fj(zj,zn):Bp2→C is holomorphic with fj(0,0)=0,(∂fj/∂zj)(0,0)=1,(∂fj/∂zn)(0,0)=0(j=1,2,…,n-1). If f satisfies the following conditions:
(6)(1)∏j=1n-1∂fj∂zj·pn′(zn)‍≠0,|znpn′′(zn)|≤(1-α)|pn′(zn)|,(2)|zj∂2fj∂zj2|+|zj∂2fj∂zj∂zn|≤(1-α)|∂fj∂zj|(j=1,2,…,n-1),(3)(1-|zn|p)1/q(∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj|p‍)1/p+(1-|zn|p)1/q(∑j=1n-1|∂2fj/∂zj2∂fj/∂zj|p‍)1/p+(1-|zn|p)1/q×(∑j=1n-1|∂fj/∂zn∂fj/∂zj||pn′′(zn)pn′(zn)|p)1/p≤(1-α-|znpn′′(zn)pn′(zn)|)|zn|p-2
for all z=(z1,…,zn)∈Bpn, then f∈K(Bpn,α).Proof.
By direct computation of the Fréchet derivatives off(z), we obtain(7)Df(z)=(∂f1∂z10⋯0∂f1∂zn0∂f2∂z2⋯0∂f2∂zn⋯⋯⋯⋯⋯00⋯∂fn-1∂zn-1∂fn-1∂zn00⋯0pn′(zn)),Df(z)-1=(1∂f1/∂z10⋯0-∂f1/∂zn(∂f1/∂z1)pn′(zn)01∂f2/∂z2⋯0-∂f2/∂zn(∂f2/∂z2)pn′(zn)⋯⋯⋯⋯⋯00⋯1∂fn-1/∂zn-1-∂fn-1/∂zn(∂fn-1/∂zn-1)pn′(zn)00⋯01pn′(zn)),D2f(z)(b,b)=(B10⋯0A10B2⋯0A2⋯⋯⋯⋯⋯00⋯Bn-1An-100⋯0An)(b1b2⋯bn-1bn)=(∂2f1∂z12b12+2∂2f1∂z1∂znb1bn+∂2f1∂zn2bn2∂2f2∂z22b22+2∂2f2∂z2∂znb2bn+∂2f2∂zn2bn2⋮∂2fn-1∂zn-12bn-12+2∂2fn-1∂zn-1∂znbn-1bn+∂2fn-1∂zn2bn2pn′′(zn)bn2), where
(8)Aj=∂2fj∂zn∂zjbj+∂2fj∂zn2bn(j=1,2,…,n-1),An=pn′′(zn)bn,Bj=∂2fj∂zj2bj+∂2fj∂zj∂znbn(j=1,2,…,n-1).
Taking z=(z1,…,zn)∈Bpn,b=(b1,…,bn)∈Cn such that Re〈b,(∂u/∂z¯)〉=0, by the hypothesis of Theorem 2, we have
(9)2pJf(z,b)-α∑j=1n|bj|2|zj|p-2≥(1-α)∑j=1n|bj|2|zj|p-2-2pRe〈Df(z)-1D2f(z)(b,b),∂u∂z¯〉=(1-α)∑j=1n|bj|2|zj|p-2-Re{∑j=1n-11∂fj/∂zj[∂2fj∂zj2bj2+2∂2fj∂zj∂znbjbn+∂2fj∂znbn2-∂fj∂znpn′′(zn)pn′(zn)bn2]×|zj|pzj+pn′′(zn)pn′(zn)bn2|zn|pzn{∑j=1n-11∂fj/∂zj[∂2fj∂zj2bj2+2∂2fj∂zj∂znbjbn}≥(1-α)∑j=1n-1|bj|2|zj|p-2-∑j=1n|1∂fj/∂zj|[|∂2fj∂zj2||bj|2+|∂2fj∂zj∂zn||bj|2+|∂2fj∂zj∂zn||bn|2+|∂2fj∂zn2||bn|2+|∂fj∂zn||pn′′(zn)pn′(zn)||bn|2[|∂2fj∂zj2|]|zj|p-1-|pn′′(zn)pn′(zn)||bn|2|zn|p-1=∑j=1n-1‍|bj|2|zj|p-2×[1-α-|∂2fj/∂zj2|+|∂2fj/∂zj∂zn||∂fj/∂zj||zj|]+|bn|2[-∑j=1n-1|∂fj/∂zn∂fj/∂zj||pn′′(zn)pn′(zn)||zj|p-1](1-α-|znpn′′(zn)pn′(zn)|)|zn|p-2-∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj||zj|p-1-∑j=1n-1|∂2fj/∂zn2∂fj/∂zj||zj|p-1-∑j=1n-1|∂fj/∂zn∂fj/∂zj||pn′′(zn)pn′(zn)||zj|p-1].
By Hölder’s inequality, we have
(10)∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj||zj|p-1≤(∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj|p‍)1/p(∑j=1n-1|zj|(p-1)q‍)1/q≤(∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj|p‍)1/p(1-|zn|p)1/q,∑j=1n-1|∂2fj/∂zn2∂fj/∂zj||zj|p-1≤(∑j=1n-1|∂2fj/∂zn2∂fj/∂zj|p‍)1/p(∑j=1n-1|zj|(p-1)q‍)1/q≤(∑j=1n-1|∂2fj/∂zn2∂fj/∂zj|p‍)1/p(1-|zn|p)1/q,∑j=1n-1|∂fj/∂zn∂fj/∂zj|‍|pn′′(zn)pn′(zn)||zj|p-1≤(∑j=1n-1|∂fj/∂zn∂fj/∂zj|p|pn′′(zn)pn′(zn)|p)1/p(∑j=1n-1|zj|(p-1)q‍)1/q≤(∑j=1n-1|∂fj/∂zn∂fj/∂zj|p‍|pn′′(zn)pn′(zn)|p)1/p(1-|zn|p)1/q.
Hence, we conclude from the above inequalities and the hypothesis of Theorem 2 that
(11)2pJf(z,b)-α∑j=1n|bj|2|zj|p-2≥|bn|2[×0.4cm-(1-|zn|p)1/q(∑j=1n-1|∂fj/∂zn∂fj/∂zj|p‍·|pn′′(zn)pn′(zn)|p)1/p](1-α-|znpn′′(zn)pn′(zn)|)|zn|p-2-(1-|zn|p)1/q×(∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj|p‍)1/p-(1-|zn|p)1/q(∑j=1n-1|∂2fj/∂zj2∂fj/∂zj|p‍)1/p-(1-|zn|p)1/q×(∑j=1n-1|∂fj/∂zn∂fj/∂zj|p‍·|pn′′(zn)pn′(zn)|p)1/p]≥0,
for all z=(z1,…,zn)∈Bpn,b=(b1,…,bn) such that Re〈b,∂u/∂z¯〉=0. Thus, it follows from Definition 1 that f∈K(Bpn,α). The proof is complete.Remark 3.
Settingα=0, fj(zj,zn)=pj(zj)+fj(zn),(j=1,2,…,n-1) in Theorem 2, we get Theorem 1 of [9].Let us give two examples to illustrate the application of Theorem2 in the following.Example 4.
Suppose thatp≥2,0≤α<1,0<|λ|≤1-α and k is a positive integer such that k<p≤k+1. Let (12)f(z)=(eλzn-1λz1+a1z1znk+1,z2+a2z2znk+1,…,zn-1+an-1zn-1znk+1,eλzn-1λ),M(p)={(1-α-|λ|)p(1+|λ|)p(k+1)p,p=k+1,kk(1-α-|λ|)p(1+|λ|)p(k+1)p(p-1)p-1(k-p+1)k-p+1,k<p<k+1.
Ifmax⁡{|aj|:j=1,2,…,n-1}≤(1-α)/(k+2-α) and
(13)∑j=1n-1|aj|p(1-|aj|)p‍≤M(p),
then f(z)∈K(Bpn,α).Proof.
Let(14)fj(zj,zn)=zj+ajzjznk+1(j=1,2,…,n-1),pn(zn)=eλzn-1λ.
Then,
(15)∂fj∂zj=1+ajznk+1,∂fj∂zn=(k+1)ajzjznk,∂2fj∂zj∂zn=(k+1)ajznk,∂2fj∂zj2=0,pn′(zn)=eλzn,pn′′(zn)=λpn′(zn).
So it follows from |aj|≤(1-α)/(k+2-α)<1 that
(16)|zj∂2fj∂zj2|+|zj∂2fj∂zj∂zn|=|zj(k+1)ajznk|≤(k+1)|aj|≤(1-α)(1-|aj|)≤(1-α)(1-|ajznk+1|)≤(1-α)|1+ajznk+1|=(1-α)|∂fj∂zj|,(j=1,2,…,n-1),|znpn′′(zn)pn′(zn)|=|λ||zn|≤|λ|≤1-α⟹|pn′′(zn)pn′(zn)|≤|λ|.
Sincemax⁡{|aj|:j=1,2,…,n-1}≤(1-α)/(k+2-α), we have
(17)|∂fj∂zj|=|1+ajznk+1|≥1-|aj|>0.
By straightforward calculations, we obtain
(18)(∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj|p‍)1/p≤{∑j=1n-1|(k+1)ajznk1+ajznk+1|p‍}1/p≤(k+1){∑j=1n-1|aj|p(1-|aj|)p‍}1/p×|zn|k-1,(∑j=1n-1|∂2fj/∂zj2∂fj/∂zj|p‍)1/p=0,(∑j=1n-1|∂fj/∂zn∂fj/∂zj|p|pn′′(zn)pn′(zn)|p‍)1/p≤|λ|{∑j=1n-1|(k+1)ajzjznk1+ajznk+1|p‍}1/p≤(k+1)|λ|×{∑j=1n-1|aj|p(1-|aj|)p‍}1/p×|zn|k-1.
Set q=p/(p-1). Then,
(19)(1-|zn|p)1/q(∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj|p‍)1/p≤(k+1){∑j=1n-1|aj|p(1-|aj|)p‍}1/pφ(|zn|)|zn|p-2,(1-|zn|p)1/q(∑j=1n-1|∂fj/∂zn∂fj/∂zj|p|pn′′(zn)pn′(zn)|p‍)1/p≤(k+1)|λ|{∑j=1n-1|aj|p(1-|aj|)p‍}1/pφ(|zn|)|zn|p-2,
where φ(x)=(1-xp)1/qxk-p+1,x∈[0,1].
Whenp=k+1, we have max⁡0≤x≤1⁡φ(x)=1.
Whenk<p<k+1, we have 0<k-p+1<1 and
(20)φ′(x)=(1-xp)(1/q)-1xk-p[(k-p+1)-kxp],
so
(21)max⁡0≤x≤1⁡φ(x)=φ(k-p+1kp)=(p-1k)1/q(k-p+1k)(k-p+1)/p.
Hence, when
(22)∑j=1n-1|aj|p(1-|aj|)p‍≤M(p),
we have
(23)(1-|zn|p)1/q(∑j=1n-1|∂2fj/∂zj∂zn∂fj/∂zj|p‍)1/p+(1-|zn|p)1/q(∑j=1n-1|∂2fj/∂zj2∂fj/∂zj|p‍)1/p+(1-|zn|p)1/q(∑j=1n-1|∂fj/∂zn∂fj/∂zj|p‍|pn′′(zn)pn′(zn)|p)1/p≤(1-α-|znpn′′(zn)pn′(zn)|)|zn|p-2,zj∈U,j=1,2,…,n.
By Theorem 2, we obtain that f∈K(Bpn,α). The proof is complete.Example 5.
Suppose thatp≥2,0≤α<1,|an|≤(1-α)/(4-2α) and k is a positive integer such that k<p≤k+1. Let(24)f(z)=(z1+a1z1znk+1,z2+a2z2znk+1,…,zn-1+an-1zn-1znk+1,zn+anzn2),M′(p)={[1-α-(4-2α)|an|]p(k+1)p,p=k+1,kk[1-α-(4-2α)|an|]p(k+1)p(p-1)p-1(k-p+1)k-p+1,k<p<k+1. If max⁡{|aj|:j=1,2,…,n-1}≤(1-α)/(k+2-α)<1 and
(25)∑j=1n-1|aj|p(1-|aj|)p≤M′(p),
then f(z)∈K(Bpn,α).By applying the same method of the proof for Theorem2, we may prove the following result.Theorem 6.
Suppose thatn≥2,0≤α<1,p≥2 and fj:U→C are analytic on U, fj(0)=fj′(0)=0(j=1,2,…,n-1), pj∈H(U),Pj′(ξ)≠0,|ξpj′′(ξ)|≤(1-α)|pj′(ξ)|(ξ∈U,j=1,2,…n). Let
(26)f(z)=(pk(zk),…,pn-1(zn-1)+fn-1(zn),pn(zn))p1(z1)+f1(zk),p2(z2)+f2(zk)…,pk(zk),…,pn-1(zn-1)+fn-1(zn),pn(zn)),(1≤k≤n, when k=j,fj(zk)=0).
If for anyz=(z1,z2,…,zn)∈Bpn, we have
(27)(1-|zk|p)1/q(∑j=1k-1|fj′′(zk)pj′(zj)|p‍)1/p+(1-|zk|p)1/q×(∑j=1k-1|fj′(zk)pj′(zj)|p|pk′′(zk)pk′(zk)|p‍)1/p≤(1-α-|zkpk′′(zk)pk′(zk)|)|zk|p-2,(1-|zn|p)1/q(∑j=k+1n-1|fj′′(zn)pj′(zj)|p‍)1/p+(1-|zn|p)1/q(∑j=k+1n-1|fj′(zn)pj′(zj)|p‍|pn′′(zn)pn′(zn)|p)1/p≤(1-α-|znpn′′(zn)pn′(zn)|)|zn|p-2,
then f(z)∈K(Bpn,α).Remark 7.
Settingk=n,α=0 in Theorem 6, we get Theorem 1 in [9].Example 8.
Suppose thatp≥2,0≤α<1,0<|λ|≤1-α and k is a positive integer such that k<p≤k+1. Let
(28)f(z)=(z1+a1′z12+a1z2k+1+b1z2k+2,eλz2-1λ,z3+a3′z32+a3znk+1+b3znk+2,…,zn-1+an-1′zn-12+an-1znk+1+bn-1znk+2,eλzn-1λ),M′′(p)={(1-2c)p(1-α-|λ|)p(k+1+|λ|)p,p=k+1,(1-2c)pkk(1-α-|λ|)p(k+1+|λ|)p(p-1)p-1(k-p+1)k-p+1,k<p<k+1,
where c=max⁡{|aj′|:j=1,2,…,n-1}. If c≤(1-α)/4<1 and
(29)[(k+1)|a1|+(k+2)|b1|]p≤M′′(p),∑j=3n-1‍[(k+1)|aj|+(k+2)|bj|]p≤M′′(p),
then f(z)∈K(Bpn,α).
Now, we give another sufficient condition forK(Bpn,α), which gives an answer to the problem mentioned in the introduction.Theorem 9.
Suppose thatn≥2,0≤α<1,p≥2 and k is a positive integer such that k<p≤k+1. Let
(30)f(z)=(p2(z2,zn),…,pn-1(zn-1,zn),pn(zn))p1(z1,z2,…,zn),p2(z2,zn),…,pn-1(zn-1,zn),pn(zn)),
where z=(z1,z2,…,zn)∈Bpn,pn∈H(U),pj(zj,zn):Bp2→C is holomorphic with pj(0,0)=0,(∂pj/∂zj)(0,0)=1,(∂pj/∂zn)(0,0)=0 and p1(z1,…,zn):Bpn→C is holomorphic with p1(0,0,…,0)=0,(∂p1/∂z1)(0,0,…,0)=1,(∂p1/∂zl)(0,0,…,0)=0 for 2≤l≤n. If f(z) satisfies the following conditions:
(31)(1)∏j=1n-1∂pj∂zjpn′(zn)‍≠0,|znpn′′(zn)|≤(1-α)|pn′(zn)|;(2)∑l=1n|z1∂2p1∂z1∂zl|≤(1-α)|∂p1∂z1|,|zj∂2pj∂zj2|+|zj∂2pj∂zj∂zn|≤(1-α)|∂pj∂zj|ffffff(j=2,3,…,n-1);(3)|z1|p-1|∂p1/∂z1|(|∂p1/∂zj|(|∂2pj/∂zj2|+|∂2pj/∂zj∂zn|)|∂pj/∂zj|+∑l=1n|∂2p1∂zj∂zl|‍)≤|zj|p-2(1-α-|zj∂2pj/∂zj2|+|zj∂2pj/∂zj∂zn||∂pj/∂zj|)hhhhhhh(j=2,3,…,n-1);(4)∑j=2n-1|∂2pj/∂zj∂zn∂pj/∂zj|‍|zj|p-1+∑j=2n-1|∂2pj/∂zn2∂pj/∂zj||zj|p-1+∑j=2n-1|∂pj/∂zn∂pj/∂zj|‍|pn′′(zn)pn′(zn)||zj|p-1+|z1|p-1|∂p1/∂z1|×(∑l=1n|∂2p1∂zn∂zl|+∑j=2n-1(×|∂pj∂zj|-1)|∂p1∂zj|(|∂2pj∂zj∂zn|+|∂2pj∂zn2|)×|∂pj∂zj|-1)+|∂p1∂zn||pn′′(zn)pn′(zn)|+∑j=2n-1|∂p1∂zj||(∂pj/∂zn)(∂pj/∂zj)||pn′′(zn)pn′(zn)|‍)≤|zn|p-2(1-α-|znpn′′(zn)pn′(zn)|);
then f∈K(Bpn,α).Proof.
By calculating the Fréchet derivatives off(z) straightforwardly, we obtain(32)Df(z)=(∂p1∂z1∂p1∂z2⋯∂p1∂zn-1∂p1∂zn0∂p2∂z2⋯0∂p2∂zn⋯⋯⋯⋯⋯00⋯∂pn-1∂zn-1∂pn-1∂zn00⋯0pn′(zn)),Df(z)-1=(1∂p1/∂z1-∂p1/∂z2(∂p1/∂z1)(∂p2/∂z2)⋯-∂p1/∂zn-1(∂p1/∂z1)(∂pn-1/∂zn-1)-∂p1/∂zn(∂p1/∂z1)pn′(zn)+∑j=2n-1∂pj/∂zn∂pj/∂zj∂p1/∂zj(∂p1/∂z1)pn′(zn)‍01∂p2/∂z2⋯0-∂p2/∂zn(∂p2/∂z2)pn′(zn)⋯⋯⋯⋯⋯00⋯1∂pn-1/∂zn-1-∂pn-1/∂zn(∂pn-1/∂zn-1)pn′(zn)00⋯01pn′(zn))D2f(z)(b,b)=(∑l=1n∂2p1∂z1∂zlbl‍∑l=1n∂2p1∂z2∂zlbl‍⋯∑l=1n∂2p1∂zn-1∂zlbl‍C10D2⋯0C2⋯⋯⋯⋯⋯00⋯Dn-1Cn-100⋯0Cn)(b1b2⋯bn-1bn)=(∑j=1n∑l=1n∂2p1∂zj∂zlblbj‍‍∂2p2∂z22b22+2∂2p2∂z2∂znb2bn+∂2p2∂zn2bn2⋮∂2pn-1∂zn-12bn-12+2∂2pn-1∂zn-1∂znbn-1bn+∂2pn-1∂zn2bn2pn′′(zn)bn2), where
(33)C1=∑l=1n∂2p1∂zn∂zlbl,Cj=∂2pj∂zn∂zjbj+∂2pj∂zn2bn(j=2,3,…,n-1),Cn=pn′′(zn)bn,Dj=∂2pj∂zj2bj+∂2pj∂zj∂znbn(j=2,3,…,n-1).
Takingz=(z1,…,zn)∈Bpn,b=(b1,…,bn)∈Cn such that Re〈b,∂u/∂z¯〉=0, by Definition 1 and the hypothesis of Theorem 9, we have
(34)2pJf(z,b)-α∑j=1n|bj|2|zj|p-2≥(1-α)∑j=1n|bj|2|zj|p-2-2pRe〈Df(z)-1D2f(z)(b,b),∂u∂z¯〉=(1-α)∑j=1n|bj|2|zj|p-2-Re{|zj|pzj1∂p1/∂z1×[+∑j=2n-1(∂pj/∂zn)(∂p1/∂zj)∂pj∂zjpn′′(zn)pn′(zn)bn2∑j=1n∑l=1n∂2p1∂zj∂zlblbj‍‍-∑j=2n-1∂p1/∂zj∂pj/∂zj‍×(∂2pj∂zj2bj2+2∂2pj∂zj∂znbjbn+∂2pj∂zn2bn2)-∂p1∂znpn′′(zn)pn′(zn)bn2+∑j=2n-1(∂pj/∂zn)(∂p1/∂zj)∂pj/∂zjpn′′(zn)pn′(zn)bn2[+∑j=2n-1(∂pj/∂zn)(∂p1/∂zj)∂pj∂zjpn′′(zn)pn′(zn)bn2∑j=1n∑l=1n∂2p1∂zj∂zlblbj‍‍‍]|z1p|z1+∑j=2n-11∂pj/∂zj(∂2pj∂zj2bj2+2∂2pj∂zj∂znbjbn+∂2pj∂zn2bn2-∂pj∂znpn′′(zn)pn′(zn)bn2)×|zj|pzj+pn′′(zn)pn′(zn)bn2|zn|pzn}≥(1-α)∑j=1n|bj|2|zj|p-2-1|∂p1/∂z1|×[∑j=1n∑l=1n|∂2p1∂zj∂zl||bj|2+∑j=2n-1|∂p1/∂zj∂pj/∂zj|×(|∂2pj∂zj2||bj|2+|∂2pj∂zj∂zn||bj|2+|∂2pj∂zj∂zn||bn|2+|∂2pj∂zn||bn2|)+|∂p1∂zn|×|pn′′(zn)pn′(zn)||bn|2+∑j=2n-1|(∂pj/∂zn)(∂p1/∂zj)∂pj/∂zj||pn′′(zn)pn′(zn)||bn|2]|z1|p-1-∑j=2n-11|∂pj/∂zj|(|∂2pj∂zj2||bj|2+|∂2pj∂zj∂zn||bj|2+|∂pj∂zj∂zn||bn|2+|∂2pj∂zn2||bn|2+|∂pj∂zn||pn′′(zn)pn′(zn)||bn|2)|zj|p-1-|pn′′(zn)pn′(zn)||bn|2|zn|p-1=|b1||z1|p-2×[1-α-∑l=1n|z1∂2p1/∂z1∂zl||∂p1/∂z1|]+∑j=2n-1|bj|2[|∂p1∂zj||∂2pj/∂zj2|+|∂2pj/∂zj∂zn||∂pj/∂zj|)]|zj|p-2×(-|zj∂2pj/∂zj2|+|zj∂2pj/∂zj∂zn||∂pj/∂zj|)1-α-|zj∂2pj/∂zj2|+|zj∂2pj/∂zj∂zn||∂pj/∂zj|)-|z1|p-1|∂p1/∂z1|×(∑l=1n|∂2p1∂zj∂zl|+|∂p1∂zj|×|∂2pj/∂zj2|+|∂2pj/∂zj∂zn||∂pj/∂zj|)]+|bn|2[-∑j=2n-1|∂pj/∂zn∂pj/∂zj||pn′′(zn)pn′(zn)||zj|p-1]×(∑l=1n|∂2p1∂zn∂zl|∑l=1n|∂2p1∂zn∂zl||zn|p-2(1-α-|znpn′′(zn)pn′(zn)|)-|z1|p-1|∂p1/∂z1|×(∑l=1n|∂2p1∂zn∂zl|+∑j=2n-1|∂p1∂zj|×|∂2pj/∂zn2|+|∂2pj/∂zj∂zn||∂pj/∂zj|+|∂p1∂zn||pn′′(zn)pn′(zn)|+∑j=2n-1|∂p1∂zj|×|∂pj/∂zn∂pj/∂zj||pn′′(zn)pn′(zn)|(∑l=1n|∂2p1∂zn∂zl|)-∑j=2n-1|∂2pj/∂zj∂zn∂pj/∂zj||zj|p-1-∑j=2n-1|∂2pj/∂zn2∂pj/∂zj||zj|p-1-∑j=2n-1|∂pj/∂zn∂pj/∂zj||pn′′(zn)pn′(zn)||zj|p-1]≥0
for all z=(z1,…,zn)∈Bpn,b=(b1,…,bn) such that Re〈b,∂u/∂z¯〉=0. Thus, it follows from Definition 1 that f∈K(Bpn,α). The proof is complete.Corollary 10.
Suppose that0≤α<1, n≥2,p≥2 and k is a positive integer such that k<p≤k+1. Let
(35)f(z)=(pn-1(zn-1)+fn-1(zn),pn(zn))p1(z1,z2,…,zn),p2(z2)+f2(zn),…,pn-1(zn-1)+fn-1(zn),pn(zn)),
where z=(z1,z2,…,zn)∈Bpn,fj:U→C is holomorphic with fj(0)=0,fj′(0)=0(j=2,3,…,n-1),pj∈H(U)(j=2,3,…,n) and p1(z1,…,zn):Bpn→C is holomorphic with p1(0,0,…,0)=0,(∂p1/∂z1)(0,0,…,0)=1,(∂p1/∂zl)(0,0,…,0)=0(l=2,3,…,n). If f satisfies the following conditions:
(36)(1)∂p1∂z1·∏j=2npj′(zj)‍≠0,|zjpj′′(zj)|≤(1-α)|pj′(zj)|,j=2,…,n;(2)∑l=1n|z1∂2p1∂z1∂zl|‍≤(1-α)|∂p1∂z1|;(3)|z1|p-1|(∂p1/∂zj)·(pj′′(zj)/pj′(zj))∂p1/∂z1|+|z1|p-1×∑l=1n|∂2p1/∂zj∂zl∂p1/∂z1|≤(1-α-|zjpj′′(zj)pj′(zj)|)|zj|p-2,j=2,…,n-1;(4)∑j=2n-1|fj′′(zn)pj′(zj)||zj|p-1+∑j=2n-1|fj′(zn)pj′(zj)||pn′′(zn)pn′(zn)||zj|p-1+|z1|p-1∑j=2n-1|(fj′′(zn)/pj′(zj))(∂p1/∂zj)∂p1/∂z1|+|z1|p-1×∑j=2n-1|fj′(zn)pj′(zj)pn′′(zn)pn′(zn)∂p1∂zj(∂p1∂z1)-1|+|z1|p-1∑l=1n∂2p1/∂zl∂zn∂p1/∂z1+|z1|p-1|(pn′′(zn)/pn′(zn))(∂p1/∂zn)∂p1/∂z1|≤(1-α-|znpn′′(zn)pn′(zn)|)|zn|p-2,
for all z=(z1,…,zn)∈Bpn, then f∈K(Bpn,α).Remark 11.
Settingpj(zj,zn)=fj(zj),2≤j≤n-1,α=0 in Theorem 9 or fj(zn)=0,2≤j≤n,α=0 in Corollary 10, we get Theorem 2 in [9].Example 12.
Suppose thatp≥2,0≤α<1,0<|λ|≤1-α and k is a positive integer such that k<p≤k+1. Let(37)f(z)=(z1+∑j=2n-1ajzjk+1‍+anz1znk+1,z2+b2z2znk+1,…,zn-1+bn-1zn-1znk+1,eλzn-1λ),N(p)={(∑j=2n-1|bj|p1-|bj|p‍)1/p+a1-a(1+(k+1)∑j=2n-1|bj|1-|bj|),p=k+1,(∑j=2n-1|bj|p1-|bj|p)1/p(p-1k)1/q(k-p+1k)(k-p+1)/p+a1-a(1+(k+1)∑j=2n-1|bj|1-|bj|),k<p<k+1,
wherea=max⁡{|aj|:j=2,…,n},b=max⁡{|bj|:j=2,…,n-1}. If
(38)a≤1-|λ|-α(k+1)2(k+1+|λ|)+1-|λ|-α<1,b≤(k+|λ|)(1-α)+|λ|(k+2-α)(k+1+|λ|)<1,N(p)≤1-α-|λ|(k+1)(k+1+|λ|),
for all z=(z1,…,zn)∈Bpn, then f(z)∈K(Bpn,α).Proof.
Put(39)p1(z1,z2,…,zn)=z1+∑j=2n-1ajzjk+1‍+anz1znk+1,pn(zn)=eλzn-1λ,pj(zj,zn)=zj+bjzjznk+1(j=2,3,…,n-1).
Then,
(40)∂p1∂z1=1+anznk+1,∂p1∂zj=(k+1)ajzjk(j=2,3,…,n-1),∂p1∂zn=(k+1)anz1znk,∂2p1∂zn2=k(k+1)anz1znk-1,∂2p1∂zj2=k(k+1)ajzjk-1,∂2p1∂z1∂zn=∂2p1∂zn∂z1=(k+1)anznk,∂2p1∂z1∂zj=0(j=2,3,…,n-1),∂pj∂zj=1+bjznk+1,∂pj∂zn=(k+1)bjzjznk,∂2pj∂zj∂zn=∂2pj∂zn∂zj=(k+1)bjznk,∂2pj∂zn2=k(k+1)bjzjznk-1,pn′(zn)=eλzn,pn′′(zn)=λpn′(zn),
so it follows from a=max⁡{|aj|:j=2,…,n}≤(1-|λ|-α)/((k+1)2(k+1+|λ|)+1-|λ|-α)<(1-α)/(k+2-α)<1 and b=max⁡{|bj|:j=2,…,n-1}≤((k+|λ|)(1-α)+|λ|)/(k+2-α)(k+1+|λ|)≤(1-α)/(k+2-α)<1 that
(41)|∂p1∂z1|=|1+anznk+1|≥1-|an|≥1-a>0,|∂pj∂zj|=|1+bjznk+1|≥1-|bj|≥1-b>0(j=2,3,…,n-1),pn′(zn)=eλzn≠0.
By calculating straightforwardly, we obtain
(42)(1-α)|∂p1∂z1|-∑l=1n|z1∂2p1∂z1∂zl|‍=(1-α)|1+anznk+1|-|(k+1)anz1znk|≥(1-α)(1-|an|)-(k+1)|an|=1-α-(k+2-α)|an|>1-α-(k+2-α)1-αk+2-α=0,(1-α)|∂pj∂zj|-|zj∂2pj∂zj2|-|zj∂2pj∂zj∂zn|=(1-α)|1+bjznk+1|-(k+1)|bjzjznk|≥(1-α)(1-|bj|)-(k+1)|bj|=1-α-(k+2-α)|bj|≥1-α-(k+2-α)1-αk+2-α=0,(j=2,…,n-1),|znpn′′(zn)pn′(zn)|=|λ||zn|≤|λ|≤1-α.
By calculating straightforwardly, we also obtain
(43)|zj|p-2(|zj(∂2pj/∂zj2)|+|zj(∂2pj/∂zj∂zn)||(∂pj/∂zj)|1-α-|zj∂2pj/∂zj2|+|(∂2pj/∂zj∂zn)|(∂pj/∂zj))-|z1|p-1|∂p1/∂z1|×(|∂p1/∂zj|(|∂2pj/∂zj2|+|∂2pj/∂zj∂zn|)|∂pj/∂zj|+∑l=1n|∂2p1∂zj∂zl|)=|zj|p-2(1-α-|(k+1)zjbjznk1+bjznk+1|)-|z1|p-1|1+anznk+1|×(|(k+1)ajzjk(k+1)bjznkbjznk+1|+|k(k+1)ajzjk-1|(|(k+1)ajzjk(k+1)bjznkbjznk+1|)≥|zj|p-2(1-α-(k+1)|bj|1-|bj|)-11-|an|((k+1)2|aj||bj|1-|bj|+k(k+1)|aj|)×|zj|k-1≥|zj|p-2(1-α-(k+1)b1-b)-11-a((k+1)2ab1-b+(k+1)2a)|zj|p-2≥|zj|p-2(1-α-(k+1)b1-b-(k+1)2a1-a11-b)=|zj|p-21-b((1-α)(1-b)-(k+1)b-(k+1)21-aa)=|zj|p-21-b(1-α-(k+2-α)b-(k+1)2a1-a)≥|zj|p-21-b((k+1)2(k+1+|λ|)+1-|λ|-α(k+1)2(k+1+|λ|))1-α-(k+2-α)×(k+|λ|)(1-α)+|λ|(k+2-α)(k+1+|λ|)-(k+1)2(1-|λ|-α)(k+1)2(k+1+|λ|)+1-|λ|-α×(k+1)2(k+1+|λ|)+1-|λ|-α(k+1)2(k+1+|λ|))=|zj|p-21-b(1-α-(k+|λ|)(1-α)+|λ|k+1+|λ|-1-|λ|-αk+1+|λ|)=0.
Set q=p/(p-1), then by Hölder’s inequality, we have
(44)∑j=2n-1|∂2pj/∂zj∂zn∂pj/∂zj||zj|p-1=∑j=2n-1|(k+1)bjznk1+bjznk+1||zj|p-1≤(k+1)∑j=2n-1|bj||zn|k-11-|bj||zj|p-1≤(k+1)[∑j=2n-1|bj|p(1-|bj|)p]1/p×[∑j=2n-1|zj|(p-1)q]1/q|zn|k-1≤(k+1)[∑j=2n-1|bj|p(1-|bj|)p]1/p×(1-|z1|p-|zn|p)1/q|zn|k-1≤(k+1)[∑j=2n-1|bj|p(1-|bj|)p]1/pφ(|zn|,|z1|)|zn|p-2,∑j=2n-1|∂2pj/∂zn2∂pj/∂zj||zj|p-1=∑j=2n-1|k(k+1)bjzjznk+11+bjznk+1||zj|p-1≤k(k+1)∑j=2n-1|bj||zn|k-11-|bj||zj|p-1≤k(k+1)[∑j=2n-1|bj|p(1-|bj|)p]1/p×[∑j=2n-1|zj|(p-1)q]1/q|zn|k-1,≤k(k+1)[∑j=2n-1|bj|p(1-|bj|)p]1/p×(1-|z1|p-|zn|p)1/q|zn|k-1≤k(k+1)[∑j=2n-1|bj|p(1-|bj|)p]1/p×φ(|zn|,|z1|)|zn|p-2,∑j=2n-1|∂pj/∂zn∂pj/∂zj||pn′′(zn)pn′(zn)||zj|p-1=∑j=2n-1|(k+1)bjzjznk1+bjznk+1||λ||zj|p-1≤(k+1)|λ|∑j=2n-1|bj||zn|k-11-|bj||zj|p-1≤(k+1)|λ|[∑j=2n-1|bj|p(1-|bj|)p]1/p×[∑j=2n-1|zj|(p-1)q]1/q|zn|k-1≤(k+1)|λ|[∑j=2n-1|bj|p(1-|bj|)p]1/p×(1-|z1|p-|zn|p)1/q|zn|k-1≤(k+1)|λ|[∑j=2n-1|bj|p(1-|bj|)p]1/p×φ(|zn|,|z1|)|zn|p-2,
where ψ(x)=(1-xp-yp)1/qxk-p+1,x,y∈[0,1].
Whenp=k+1, we have max⁡0≤x,y≤1ψ(x,y)=1.
Whenk<p<k+1, we have 0<k-p+1<1 and
(45)ψx(x,y)=(1-xp-yp)(1/q)-1xk-p×[(k-p+1)(1-yp)-kxp],ψy(x,y)=(1-p)yp-1(1-xp-yp)(1/q)-1xk-p+1,
so
(46)max⁡0≤x,y≤1⁡ψ(x,y)=ψ(k-p+1kp,0)=(p-1k)1/q(k-p+1k)(k-p+1)/p,|z1|p-1|∂p1/∂z1|×(∑l=1n|∂2p1∂zn∂zl|+∑j=2n-1((|∂pj/∂zj|)-1|∂p1∂zj|×(|∂2pj∂zj∂zn|+|∂2pj∂zn2|)×(|∂pj∂zj|)-1)+|∂p1∂zn||pn′′(zn)pn′(zn)|+∑j=2n-1|∂p1∂zj|‍|∂pj/∂zn∂pj/∂zj||pn′′(zn)pn′(zn)|)≤11-|an|[+(k+1)|an||λ|+∑j=2n-1(k+1)2|λ∥aj∥bj|1-|bj|](k+1)|an|+k(k+1)|an|+∑j=2n-1(k+1)|aj|(k+1)2|bj|1-|bj|+(k+1)|an||λ|+∑j=2n-1(k+1)2|λ||aj||bj|1-|bj|]×|zn|p-2≤(k+1)(k+1+λ)a1-a×[1+(k+1)∑j=2n-1|bj|1-|bj|‍]|zn|p-2.
Hence, we have
(47)∑j=2n-1‍|∂2pj/∂zj∂zn∂pj/∂zj||zj|p-1+∑j=2n-1|∂2pj/∂zn2∂pj/∂zj||zj|p-1+∑j=2n-1‍|∂pj/∂zn∂pj/∂zj||pn′′(zn)pn′(zn)||zj|p-1+|z1|p-1|∂p1/∂z1|×(∑j=2n-1|∂2p1∂zn∂zj|+∑j=2n-1‍|∂p1/∂zj|(|∂2pj/∂zj∂zn|+|∂2pj/∂zn2|)|∂pj/∂zj|+|∂p1∂zn||pn′′(zn)pn′(zn)|+∑j=2n-1|∂p1∂zj|‍|∂pj/∂zn∂pj/∂zj||pn′′(zn)pn′(zn)|)≤(k+1)(k+1+λ)N(p)|zn|p-2≤(k+1)(k+1+λ)1-α-|λ|(k+1)(k+1+|λ|)|zn|p-2≤(1-α-|λ|)|zn|p-2≤(1-α-|λzn|)|zn|p-2=|zn|p-2(1-α-|znpn′′(zn)pn′(zn)|).
By Theorem 6, we obtain that f∈K(Bpn,α). The proof is complete.By applying the same method of the proof for Example12, we only need to let 2|bn|/(1-2|bn|) instead of |λ|, we may prove the following result.Example 13.
Suppose thatp≥2,0≤α<1,0≤|bn|≤(1-α)/(4-2α) and k is a positive integer such that k<p≤k+1. Let
(48)f(z)=(z1+∑j=2n-1ajzjk+1‍+anz1znk+1,z2+b2z2znk+1,…,zn-1+bn-1zn-1znk+1,zn+bnzn2∑j=2n-1ajzjk+1‍+anz1znk+1,z2),
where a=max⁡{|aj|:j=2,…,n} and b=max⁡{|bj|:j=2,…,n-1}. If (49)a≤1-(2|bn|/(1-2|bn|))-α(k+1)2(k+1+2|bn|/(1-2|bn|))+1-(2|bn|/(1-2|bn|))-α<1,b≤((k+(2|bn|/(1-2|bn|)))(1-α)+(2|bn|/(1-2|bn|)))/(k+2-α)(k+1+(2|bn|/(1-2|bn|)))<1, and
(50)N(p)≤1-α-(2|bn|/(1-2|bn|))(k+1)(k+1+(2|bn|/(1-2|bn|)))=1-(4-2α)|bn|-α(k+1)(k+1-2k|bn|),
where N(p) is defined in Example 12, then f(z)∈K(Bpn,α).By applying the same method of the proof for Theorem2, we may get the following result.Theorem 14.
Suppose that0≤α<1,n≥2,p≥2 and l is a positive integer such that l<p≤l+1. Let
(51)f(z)=(pk(zk),…pn(zn)+fn(zk))p1(z1,z2,…,zn),p2(z2)+f2(zk),…,pk(zk),…pn(zn)+fn(zk))(2≤k≤n),
where z=(z1,z2,…,zn)∈Bpn, fk(zk)=0, fj:U→C is holomorphic with fj(0)=0,fj′(0)=0(j=2,3,…,n-1),pj∈H(U)(j=2,3,…,n) and p1(z1,…,zn):Bpn→C is holomorphic with p1(0,0,…,0)=0,(∂p1/∂z1)(0,0,…,0)=1,(∂p1/∂zl)(0,0,…,0)=0(l=2,3,…,n). If f satisfies the following conditions:
(52)(1)∂p1∂z1·‍∏j=2npj′(zj)≠0,|zjpj′′(zj)|≤(1-α)|pj′(zj)|,j=2,…,n;(2)∑l=1n|z1∂2p1∂z1∂zl|‍≤(1-α)|∂p1∂z1|;(3)|z1|p-1|(∂p1/∂zj)·(pj′′(zj)/pj′(zj))∂p1/∂z1|+|z1|p-1∑l=1n|∂2p1/∂zj∂zl∂p1/∂z1|≤(1-α-|zjpj′′(zj)pj′(zj)|)|zj|p-2,(j=2,…,k-1,k+1,…,n-1);(4)∑j=2,j≠kn|fj′′(zk)pj′(zj)||zj|p-1‍+∑j=2,j≠kn|fj′(zk)pj′(zj)||pk′′(zk)pk′(zk)||zj|p-1‍+∑j=2,j≠kn|fj′′(zk)pj′(zj)∂p1∂zj(∂p1∂z1)-1||z1|p-1‍+∑j=2,j≠kn|fj′(zk)pj′(zj)pk′′(zk)pk′(zk)×∂p1∂zj(∂p1∂z1)-1|×|z1|p-1+∑l=1n|(∂2p1/∂zl∂zk)(∂p1/∂z1)||z1|p-1‍+|(pk′′(zk)/pk′(zk))(∂p1/∂zk)(∂p1/∂z1)||z1|p-1≤(1-α-|zkpk′′(zk)pk′(zk)|)|zk|p-2,
for all z=(z1,…,zn)∈Bpn, then f∈K(Bpn,α).



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*Source: 719652-2014-04-08.xml*