arXiv:1511.06946v1 [math.CV] 22 Nov 2015
Criteria of Biholomorphic Convex Mappings on the bounded convex balanced domain Dpn Ni Li and Ming-Sheng Liu∗ School of Mathematical Sciences, South China Normal University, Guangzhou 510631, Guangdong, People’s Republic of China
Abstract. In this paper, we first establish several general sufficient conditions for the biholomorphic convex mappings on the bounded convex balanced domain Dpn (pj ≥ 2, j = 1, · · · , n) in C n , which extend some related results of earlier authors. From these, some concrete examples of biholomorphic convex mappings on Dpn are also provided. Keywords: Locally biholomorphic mapping, biholomorphic convex mapping, bounded convex balanced domain 2010 Mathematics Subject Classification: 32H02
1
Introduction and Preliminaries
Suppose that C n is the vector space of n complex variables z = (z1 , z2 , · · · , zn ) with n P the Euclidean inner product hz, wi = zj w ¯j , where w = (w1 , w2 , · · · , wn ) ∈ C n . If Ω is j=1
a domain in C n , and for every z ∈ Ω, |λ| ≤ 1, we have λz ∈ Ω, then we call Ω a balanced domain. Its Minkowski functional is defined by z ρ(z) = inf{t > 0, ∈ Ω}, z ∈ C n . t Assume Ω is a bounded convex balanced domain in C n , let ρ(z) be its Minkowski functional, then ρ(·) is a norm of C n , ρ(λz) = |λ|ρ(z) for every λ ∈ C, z ∈ C n , and Ω = {z ∈ C n : ρ(z) < 1}, and ρ(z) = 0 if and only if z = 0. Let p = (p1 , · · · , pn ) with pj > 1 (j = 1, 2, · · · , n), and Dpn
n
= {(z1 , z2 , · · · , zn ) ∈ C :
n X
|zj |pj < 1},
(1)
j=1
∗
Correspondence should be addressed to Ming-Sheng Liu;
[email protected] This research is supported by Guangdong Natural Science Foundation (Grant No.2014A030307016, 2014A030313422).
1
then Dpn is a bounded convex balanced domain in C n , and its Minkowski functional ρ(z) satisfies the following equality n X zj pj ρ(z) = 1.
(2)
j=1
p p
When p1 = · · · = pn = p, we denote Dpn by Bpn , at this time, we have ρ(z) = |z1 |p + · · · + |zn |p . In particular, let U = Bp1 be the unit disk in the complex plane
C.
Let Ω be a domain in C n , a mapping f : Ω → C n is said to be locally biholomorphic in Ω if f has a locally inverse at each point z ∈ Ω or, equivalently, if the first Fr´echet derivative ∂fj (z) Df (z) = ∂zk 1≤j,k≤n is nonsingular at each point in Ω. The second Fr´echet derivative of a mapping f : Ω → C n is a symmetric bilinear operator D 2 f (z)(·, ·) on C n × C n , and D 2 f (z)(z, ·) is the linear operator obtained by restricting D 2 f (z) to {z} × C n . The matrix representation of D 2 f (z)(b, ·) is X n ∂ 2 fj (z) 2 bl D f (z)(b, ·) = ∂zk ∂zl 1≤j,k≤n l=1
where f (z) = (f1 (z), · · · , fn (z)), b = (b1 , · · · , bn ) ∈ C n . Let N (Dpn ) be the class of all locally biholomorphic mappings f (z) = (f1 (z), · · · , fn (z)) : Dpn → C n such that f (0) = 0, Df (0) = I, where z = (z1 , · · · , zn ) ∈ C n and I is the unit matrix of n × n. If f ∈ N (Dpn ) is a biholomorphic mapping on Dpn and f (Dpn ) is a convex domain in C n , then we say that f is a biholomorphic convex mappings on Dpn . Let K(Dpn ) denote the class of all biholomorphic convex mappings on Dpn with f (0) = 0, Df (0) = I. It is not easy to construct concrete biholomorphic convex mappings on some domains in C n , even on the unit ball B2n . In 1995, Roper and Suffridge[7] proved that: If f ∈ K p and F (z) = (f (z1 ), f ′ (z1 )z0 ), where z = (z1 , z0 ) ∈ B2n , z1 ∈ U, z0 = (z2 , · · · , zn ) ∈ C n−1 , then F ∈ K(B2n ). Which is popularly referred to as the Roper-Suffridge operator. Using this operator, we may construct a lot of concrete biholomorphic convex mappings on Bn2 . Gong and Liu [2] generalized the Roper-Suffridge operator to the Reinhardt domain n P Dp = {z = (z1 , z2 , · · · , zn ) ∈ C n : |zj |pj < 1}, where p1 = 2, p2 ≥ 1, p3 ≥ 1, · · · , pn ≥ j=1
1. However, according to the result in [1, 9], none of these concrete examples belongs to K(Dpn )(pj > 2, j = 1, 2, · · · , n). In 1999, Roper and Suffridge[8] gave two concrete examples of biholomorphic convex mappings on Bp2 in C n , Liu and Zhu[4] gave some sufficient conditions for biholomorphic convex mappings on Bpn . Hamada and Kohr[3], Zhu[10] gave a necessary and sufficient condition for biholomorphic convex mappings on bounded convex balanced domain Dpn as follows. Theorem A[3, 10] Suppose that pj ≥ 2(j = 1, 2, ..., n), ρ(z) is the Minkowski functional of Dpn , and f ∈ N (Dpn ). Then f ∈ K(Dpn ) if and only if for any z = (z1 , z2 , · · · , zn ) ∈ 2
Dpn \{0}, and b = (b1 , b2 , · · · , bn ) ∈ C n \{0} such that Re
X n j=1
we have
zj pj bj pj = 0, ρ(z) zj
X n n z pj b X p2j |zj |pj −2 pj j j 2 p ( |b | + − 1) Jf (z, b) = Re ( )2 j j 2 ρ(z)pj 2 ρ(z) zj j=1 j=1 n X pj zj pj ∂ρ −1 2 −2 ≥ 0. Df (z) D f (z)(b, b), ρ(z) ρ(z) ∂z j=1
Liu and Zhu[5] had established some sufficient conditions of biholomorphic convex mappings on Dpn for mappings of the following forms f (z) = (f1 (z1 ) + g1 (zn ), f2 (z2 ) + g2 (zn ), · · · , fn−1 (zn−1 ) + gn−1 (zn ), fn (zn )), and f (z) = (f1 (z1 , z2 , · · · , zn ), f2 (z2 ), · · · , fn (zn )). Liu and Li [6] did the further promotion on Dpn , they had established a sufficient condition of biholomorphic convex mappings on Dpn for mappings of the following form f (z) = (f1 (z1 , z2 , · · · , zn ), f2 (z2 ) + g2 (zn ), · · · , fn−1 (zn−1 ) + gn−1 (zn ), fn (zn )). Now we pose two problems as follows: Problem I: Can we get some sufficient conditions such that mapping of the form : f (z) = (f1 (z1 , z2 , · · · , zn ), f2 (z2 , · · · , zn ), · · · , fn−1 (zn−1 , zn ), fn (zn )) is a biholomorphic convex mapping on Dpn ? Problem II: Can we get some sufficient conditions such that the mapping f (z) = (f1 (z1 , z2 ), f2 (z1 , z2 )),
z = (z1 , z2 )
is a biholomorphic convex mapping on Dp2 in C 2 ? The object of this paper is to give partial answers to the above problems. From these, we may construct some concrete biholomorphic convex mappings on Dpn .
2
Main results
In order to give a partial answer to Problem I, we first establish the following theorem. Theorem 1 Suppose that n ≥ 2, pj ≥ 2, j = 1, 2, · · · , n, and fj : U → C is holomorphic with fj (0) = 0, fj′ (0) = 1(j = 3, · · · , n−1, n), f1 (z1 , · · · , zn ) : Dpn → C is holomorphic ∂f1 1 with f1 (0, 0, · · · , 0) = 0, ∂f ∂z1 (0, 0, · · · , 0) = 1, ∂zl (0, 0, · · · , 0) = 0(l = 2, 3, · · · , n). Let f (z) = (f1 (z1 , z2 , · · · , zn ), f2 (z2 , z3 , · · · , zn ), f3 (z3 ), · · · , fn−1 (zn−1 ), fn (zn )). 3
If f (z) satisfies the following conditions (1)
n ∂f1 ∂f2 Y ′ fj (zj ) 6= 0, ∂z1 ∂z2
|zj fj′′ (zj )| ≤ |fj′ (zj )|(j = 3, 4, · · · , n);
j=3
n X
n X ∂ 2 f1 ∂ 2 f2 ∂f1 ∂f2 |z2 |≤| |, |≤| |; ∂z1 ∂zl ∂z1 ∂z2 ∂zl ∂z2 l=1 l=2 X n ∂f1 ∂ 2 f2 n z2 ∂ 2 f2 n ∂ 2 f1 X ∂z2 ∂z2 ∂zl ∂z2 ∂zl ∂z2 ∂zl X p2 −2 + (3) p1 ≤ 1 − ; ∂f2 ∂f1 ∂f2 p2 |z2 | ∂f1
(2)
|z1
∂z1
l=1
∂z2
l=2
∂z1
∂z2
l=2
2 2 ∂f1 ′′ 1 ∂f2 ′′ X n ∂ f1 n ∂f1 ∂ f2 ∂zj ∂zl X ∂zj fj (zj ) ∂z2 ∂zj ∂zl ∂f ∂z2 ∂zj fj (zj ) (4) p1 ∂f1 + ∂f2 ∂f1 + ∂f1 ∂f2 f ′ (zj ) + ∂f1 f ′ (zj )
∂z1
l=1
∂z2
l=2
∂z1
∂z1
∂z2
j
j
∂z1
2 ∂f X n ∂ f2 ∂zj ∂zl ∂z2j fj′′ (zj ) fj′′ (zj ) pj −2 (j = 3, 4, · · · , n) +p2 1 − zj ′ ∂f2 + ∂f2 f ′ (zj ) ≤ pj |zj | f (zj )
∂z2
l=2
j
∂z2
j
K(Dpn ).
Dpn
\ {0}, then f ∈ for all z = (z1 , · · · , zn ) ∈ Proof By direct computation of the Fr´echet derivatives of f (z) , we get that ∂f1 ∂f1 ∂f1 ∂f1 · · · ∂z2 ∂zn−1 ∂zn ∂z1 ∂f ∂f2 ∂f2 2 0 · · · ∂z ∂z ∂z n 2 n−1 Df (z) = · · · · · · · · · ··· ··· ′ 0 · · · fn−1 (zn−1 ) 0 0 0 0 ··· 0 fn′ (zn )
Df (z)−1
1
∂f1 ∂z1 0 = · · · 0 0
n P
−
∂f2 ∂z2
∂f ∂f ∂f1 ∂f2 − ∂z1 ∂z2 ∂z2 ∂z3 3 2 ∂f1 ∂f2 ′ f (z ) ∂z1 ∂z2 3 3 ∂f2 ∂z3 ∂f2 ′ f (z ) ∂z2 3 3
··· 0
··· ···
··· ···
0
···
0
∂f1 ∂z2 ∂f1 ∂f2 ∂z1 ∂z2
1
−
n P
∂ 2 f1 ∂z1 ∂zl bl
l=1 0 D 2 f (z)(b, b) = ··· 0 0
From (2) and |zj |pj = zj
···
l=1 n P
l=2
pj 2
zj
∂ 2 f1 ∂z2 ∂zl bl ∂2f
···
···
∂z2 ∂zl bl
···
··· 0 0
··· ··· ···
pj 2
∂ρ = ∂zl
2
n P
∂f1 ∂f2 ∂f ∂f − ∂z 1 ∂z2 ∂z2 ∂zn−1 n−1 2 ∂f1 ∂f2 ′ f (z ) n−1 ∂z1 ∂z2 n−1 ∂f2 ∂zn−1 ∂f2 ′ f (z ) ∂z2 n−1 n−1
∂f ∂f ∂f1 ∂f2 − ∂z 1 ∂z2 ∂z2 ∂zn n 2 ∂f1 ∂f2 ′ f (z n) n ∂z1 ∂z2 ∂f2 ∂zn ∂f2 ′ f (z ) ∂z2 n n
···
··· 0
−
−
1 ′ fn−1 (zn−1 )
1 fn′ (zn )
0
∂ 2 f1 ∂zn ∂zl bl
n n P P
∂ 2 f1 ∂zj ∂zl bl bj
j=1 l=1 b1 n n P P ∂ 2 f2 b 2 2 ∂zn ∂zl bl ∂zj ∂zl bl bj l=2 . · · · = j=2 l=2 ··· f3′′ (z3 )b23 bn−1 0 ··· bn fn′′ (zn )bn fn′′ (zn )b2n l=1 n P
∂2f
, direct computation yields pl |zl |pl n zj pj , P 2zl ρ(z)pl −1 pj ρ(z) j=1
4
(3)
zj and ρ(z) ≤ 1(j = 1, 2, · · · , n) for all z = (z1 , z2 , · · · , zn ) ∈ Dpn . Taking z = (z1 , · · · , zn ) ∈ Dpn \ {0}, b = (b1 , · · · , bn ) ∈ C n \ {0} such that X n zj pj bj pj Re = 0, ρ(z) zj j=1
by applying the fact 0 < ρ(z) < 1 and the hypothesis of Theorem 1, we have Jf (z, b) ≥
n X pj |zj |pj −2 j=1
=
n X j=1
ρ(z)pj
|bj |2 − 2
j=1
∂f1 ∂z2 ∂f2 j=2 l=2 ∂z2
n n n n X X pj |zj |pj −2 1 X X ∂ 2 f1 2 |bj | − Re ∂f1 bl bj − ρ(z)pj ∂zj ∂zl ∂z1
n ∂f1 X ∂z2 + ∂f2 ∂z2 j=3
j=1 l=1
∂ 2 f2 bj bl ∂zj ∂zl
∂f2 ∂f1 fj′′ (zj ) 2 p1 |z1 |p1 − b ∂zj ∂zj fj′ (zj ) j z1 ρ(z)p1
∂ 2 f2 ∂zj ∂zl bl bj ∂f2 j=2 l=2 ∂z2
∂f2 ∂zj ∂f2 j=3 ∂z2
n X fj′′ (zj ) 2 pj |zj |pj fj′′ (zj ) 2 p2 |z2 |p2 b + b − + fj′ (zj ) j z2 ρ(z)p2 f ′ (z ) j zj ρ(z)pj j=3 j j X n n X n n ∂f1 n X X ∂ 2 f1 2 X ∂z2 ∂ 2 f2 2 pj |zj |pj −2 1 2 |bj | − ∂f1 ∂zj ∂z |bj | + ∂f2 ∂zj ∂z |bj | ρ(z)pj | | l l n X
X n n X
≥
n X pj zj pj ∂ρ i RehDf (z)−1 D 2 f (z)(b, b), ρ(z) ρ(z) ∂z
∂z1
j=1
+
n | ∂f1 || ∂f2 | X j=3
∂z2 ∂zj 2 | ∂f ∂z2 | ∂2f
j=1 l=1
∂z2
j=2 l=2
n ′′ fj (zj ) 2 pj |zj |pj −1 ∂f1 fj′′ (zj ) 2 p1 |z1 |p1 −1 X |bj | − + f ′ (zj ) |bj | ρ(z)pj ∂zj fj′ (zj ) ρ(z)p1 j j=3
X n ∂f2 ′′ n X n ∂zj fj (zj ) 2 p2 |z2 |p2 −1 ∂zj ∂z2 l 2 X − ∂f2 f ′ (zj ) |bj | ∂f2 |bj | + ρ(z)p2 j=2 l=2
j=3
∂z2
∂z2
j
∂2f
=
n z1 n z 2 ∂ 2 f2 1 p1 −2 2 X X ∂z2 ∂zl p2 |z2 |p2 −2 p |z | |b | 1 1 2 ∂z ∂z 1 l + |b1 |2 1 − 1 − ∂f1 ∂f2 ρ(z)p2 −1 ρ(z)p1 ρ(z) ∂z1 ∂z2 l=1 l=2 2f 2f ∂ ∂ ∂f n n n 1 2 1 |z | p1 −1 X X ∂z ∂z ∂z X zj fj′′ (zj ) |bj |2 pj |zj |pj −2 ∂z2 ∂zl 1 2 2 l −p1 1− ′ ∂f1 + ∂f2 ∂f1 + ρ(z) ρ(z) ρ(z)pj −1 fj (zj ) l=1
∂z1
l=2
∂z2
∂z1
j=3
2 2 ∂f1 ∂f2 ′′ ∂f n ∂f1 ∂ f2 n ∂ f1 |z | p1 −1 X X ∂z2 ∂zj fj (zj ) ∂z1j fj′′ (zj ) ∂z ∂z ∂z 1 j ∂z2 j ∂zl l −p1 ∂f2 ∂f1 + ∂f1 ∂f2 f ′ (zj ) + ∂f1 f ′ (zj ) ∂f1 + ρ(z)
l=1
l=2
≥
∂z1
l=2
∂z2
∂z1
2 ∂f n ∂ f2 |z | p2 −1 X ∂zj ∂zl ∂z2j fj′′ (zj ) 2 −p2 ∂f2 + ∂f2 f ′ (zj ) ρ(z)
∂z2
∂z2
∂z1
∂z2
j
∂z1
j
2 2 n n p1 −2 X X z2 ∂z∂ 2f∂z2 l z1 ∂z∂ 1f∂z1 l 2 p1 |z1 | 2 p2 −2 1− |b1 | 1− ∂f2 ∂f1 + |b2 | p2 |z2 | ρ(z)p1 ∂z1 ∂z2 l=1 l=2 ∂f1 ∂ 2 f2 X n ∂ 2 f1 n n X ∂z2 ∂zl X ∂z2 ∂z2 ∂zl zj fj′′ (zj ) 2 pj −2 −p1 |bj | pj |zj | 1− ′ ∂f1 + ∂f2 ∂f1 + fj (zj )
l=1
∂z1
l=2
∂z2
∂z1
5
j=3
j
2 2 ∂f 1 ∂f2 ′′ X n ∂f1 ∂ f2 n ∂ f1 ∂z2 ∂zj ∂zl ∂f ∂zj fj (zj ) ∂z1j fj′′ (zj ) ∂zj ∂zl X ∂z 2 −p1 ∂f2 ∂f1 + ∂f1 ∂f2 f ′ (zj ) + ∂f1 f ′ (zj ) ∂f1 +
l=1
∂z1
∂2f
−p2
X n ∂zj ∂z2 l ∂f2 + l=2
∂z2
∂z1 l=2 ∂z2 ∂f2 ′′ ∂zj fj (zj ) ∂f2 f ′ (zj ) j ∂z2
∂z1
j
∂z2
∂z1
j
≥ 0.
✷ Thus it follows from Theorem A that f ∈ K(Dpn ). Remark 1 Setting f2 (z2 , · · · , zn ) = f2 (z2 ) in Theorem 1, we get Theorem 2 of [5]. Setting n = 3 in Theorem 1, we get the following corollary, which gives an answer to Problem I for the case n = 3. Corollary 1 Suppose that pj ≥ 2, j = 1, 2, 3. Let f3 : U → C is holomorphic with f3 (0) = 0, f3′ (0) = 1, f1 (z1 , z2 , z3 ), f2 (z2 , z3 ) : Dp3 → C are holomorphic with f1 (0, 0, 0) = ∂f2 ∂f1 ∂f2 1 0, ∂f ∂z1 (0, 0, 0) = 1, ∂zl (0, 0, 0) = 0(l = 2, 3) and f2 (0, 0) = 0, ∂z2 (0, 0) = 1, ∂z3 (0, 0) = 0. Set f (z) = (f1 (z1 , z2 , z3 ), f2 (z2 , z3 ), f3 (z3 )). If f (z) satisfies the following conditions ∂f1 ∂f2 ′ f (z3 ) 6= 0, |z3 f3′′ (z3 )| ≤ |f3′ (z3 )|; ∂z1 ∂z2 3 3 3 X ∂ 2 f1 ∂f1 X ∂ 2 f2 ∂f2 (2) z1 , ; ≤ ≤ z2 ∂z1 ∂zl ∂z1 ∂z2 ∂zl ∂z2 l=1 l=2 X 3 ∂f1 ∂ 2 f2 3 z2 ∂ 2 f2 3 ∂ 2 f1 X ∂z2 ∂zl X ∂z2 ∂z2 ∂zl ∂z2 ∂zl p2 −2 ; (3) p1 ∂f + ∂f ∂f ≤ 1 − ∂f p2 |z2 | (1)
1
l=1
∂z1
l=2
2
1
∂z2
∂z1
2
∂z2
l=2
1 ∂f2 ′′ ∂f1 ′′ X 3 ∂f1 ∂ 2 f2 3 ∂ 2 f1 ∂z2 ∂z3 ∂zl ∂f ∂z3 ∂zl X ∂z3 f3 (z3 ) ∂z2 ∂z3 f3 (z3 ) (4) p1 ∂f2 ∂f1 + ∂f1 ∂f2 f ′ (z3 ) + ∂f1 f ′ (z3 ) ∂f1 + l=1
∂z1
l=2
∂z2
∂z1
∂z1
3
∂z2
∂z1
j
∂2f
2 ′′ X n ∂z3 ∂z2 l ∂f f3′′ (z3 ) ∂z3 f3 (z3 ) p3 −2 +p2 1 − z3 ′ ∂f2 + ∂f2 f ′ (z3 ) ≤ p3 |z3 | f (z3 ) l=2
∂z2
∂z2
3
3
for all z = (z1 , z2 , z3 ) ∈ Dp3 \ {0}, then f ∈ K(Dp3 ). Now let us give two examples to illustrate the application of Theorem 1 in the following. Example 1. Suppose that pj ≥ p1 ≥ 2(j = 2, · · · , n), 0 < |λ| ≤ 1, and k is a positive integer such that k < max{pj : j = 1, · · · , n} ≤ k + 1, let f (z) = (z1 + a1 z12 +
n X
aj zjk+1 , z2 + a2 z22 +
n X
aj zjk+1 ,
j=3
j=2
where a = max{|aj | : j = 1, 2, · · · , n}. If a ≤
1−|λ| (k+1)2 +4
0, ∂z1 ∂z2 ∂ 2 f2 ∂ 2 f1 z2 ∂z ∂z z1 ∂z1 ∂zl 2a 2a |2a1 z1 | |2a2 z2 | = ≤ ≤ 1, ∂f22 l = ≤ ≤ 1, ∂f1 |1 + 2a1 z1 | 1 − 2a |1 + 2a2 z2 | 1 − 2a ∂z1 ∂z2 zj fj′′ (zj ) f ′ (zj ) = |λ||zj | ≤ 1, j = 3, · · · , n. j |
By calculating straightforwardly, we also obtain X n ∂f1 ∂ 2 f2 n z 2 ∂ 2 f2 n ∂ 2 f1 X ∂z2 ∂z2 ∂zl ∂z2 ∂zl ∂z2 ∂zl X p2 −2 p2 |z2 | 1− ∂f2 ∂f1 ∂f2 − p1 ∂f1 + l=2
∂z2
l=1
∂z1
l=2 k−1 1)a2 z2 |
∂z2
∂z1
|k(k + |(k + 1)a2 z2k | |2a2 | |2a2 z2 | + − p1 = p2 |z2 | 1− |1 + 2a2 z2 | |1 + 2a1 z1 | |1 + 2a2 z2 | |1 + 2a1 z2 | 2|a2 | (k + 1)|a2 | 2|a2 | k(k + 1)|a2 | p2 −2 p2 −2 p2 −2 ≥ p2 |z2 | 1− |z2 | + |z2 | − p1 1 − 2|a2 | 1 − 2|a1 | 1 − 2|a2 | 1 − 2|a1 | k(k + 1)a (k + 1)a 2a − − ≥ p1 |z2 |p2 −2 1 − 1 − 2a 1 − 2a 1 − 2a p −2 2 p1 |z2 | = {1 − [(k + 1)2 + 4]a} > 0. 1 − 2a Since for j = 3, · · · , n, we have 2 n ∂ f1 X ∂zj ∂zl |k(k + 1)aj zjk−1 | k(k + 1)|aj | k(k + 1)|aj | = ≤ |zj |k−1 ≤ |zj |pj −2 , ∂f1 |1 + 2a1 z1 | 1 − 2|a1 | 1 − 2|a1 | p2 −2
l=1
∂z1
2 n ∂f1 ∂ f2 X ∂z2 ∂zj ∂zl ∂f2 ∂f1 =
l=2
∂z2
∂z1
∂f1 ∂f2 ′′ ∂z2 ∂zj fj (zj ) ∂f1 ∂f2 f ′ (zj ) = ∂z1
∂z2
j
∂f1 ′′ ∂zj fj (zj ) ∂f1 f ′ (zj ) = j ∂z1
2 n ∂ f2 X ∂zj ∂zl ∂f1 =
∂z1 l=2 ∂f2 ′′ ∂zj fj (zj ) ∂f2 f ′ (zj ) j ∂z2
=
k−1 k(k + 1)2 |a2 ||aj | |(k + 1)a2 z2k | |k(k + 1)aj zj | ≤ |zj |pj −2 , |1 + 2a2 z2 | |1 + 2a1 z1 | (1 − 2|a2 |)(1 − 2|a1 |)
|(k + 1)a2 z2k | |(k + 1)aj zjk | (k + 1)2 |λ||a2 ||aj | |λ| ≤ |zj |pj −2 , |1 + 2a1 z1 | |1 + 2a2 z2 | (1 − 2|a1 |)(1 − 2|a2 |) |(k + 1)aj zjk | (k + 1)|λ||aj | |λ| ≤ |zj |pj −2 , |1 + 2a1 z1 | 1 − 2|a1 | |k(k + 1)aj zjk−1 | |1 + 2a1 z1 |
≤
k(k + 1)|aj | |zj |pj −2 , 1 − 2|a1 |
|(k + 1)aj zjk | (k + 1)|aj ||λ| |λ| ≤ |zj |pj −2 . |1 + 2a2 z2 | 1 − 2|a2 | 7
Therefore 2 2 ∂f1 ′′ 1 ∂f2 ′′ X n ∂ f1 n ∂f1 ∂ f2 ∂zj ∂zl X ∂zj fj (zj ) ∂z2 ∂zj ∂zl ∂f ∂z2 ∂zj fj (zj ) p1 ∂f1 + ∂f2 ∂f1 + ∂f1 ∂f2 f ′ (zj ) + ∂f1 f ′ (zj )
l=1
≤
≤
= ≤ ≤
∂z1
l=2
∂z2
∂z1
∂z1
∂z2
j
∂z1
j
2 ∂f X n ∂ f2 ∂zj ∂zl ∂z2j fj′′ (zj ) +p2 ∂f1 + ∂f2 f ′ (zj ) j ∂z1 ∂z2 l=2 k(k + 1)|aj | k(k + 1)2 |a2 ||aj | (k + 1)2 |λ||a2 ||aj | p1 |zj |pj −2 + + 1 − 2|a1 | (1 − 2|a1 |)(1 − 2|a2 |) (1 − 2|a1 |)(1 − 2|a2 |) (k + 1)|aj ||λ| k(k + 1)|aj | (k + 1)|λ||aj | + + + p2 |zj |pj −2 1 − 2|a1 | 1 − 2|a1 | 1 − 2|a2 | 2 k(k + 1) a|aj | (k + 1)2 |λ|a|aj | (k + 1)|λ||aj | pj −2 k(k + 1)|aj | p1 |zj | + + + 1 − 2a (1 − 2a)2 (1 − 2a)2 1 − 2a k(k + 1)|aj | (k + 1)|λ||aj | + +p2 |zj |pj −2 1 − 2a 1 − 2a p |a | p1 (k + 1)a|aj | p2 |aj | 1 j pj −2 |zj | (k + 1)(k + |λ|) + + 1 − 2a (1 − 2a)2 1 − 2a pj (1 − |λ|) = pj |zj |pj −2 (1 − |λ|) |zj |pj −2 (k + 1)(k + |λ|) (k + 1)(k + λ) zj fj′′ (zj ) pj −2 . pj |zj | 1− ′ f (zj )
j
✷ Hence it follows from Theorem 1 that f ∈ K(Dpn ). Example 2 Suppose that pj ≥ p1 ≥ 2(j = 2, · · · , n), and k is a positive integer such that k < max{pj : j = 1, · · · , n} ≤ k + 1, let f (z) = (z1 + a1 z12 +
n X
aj zjk+1 , z2 + a2 z22 +
aj z32 , z3 + a3 z32 , · · · , zn + an zn2 )
j=3
j=2
where a = max{|aj | : j = 1, 2, · · · , n}. If a ≤
n X
1 (k+1)2 +4
< 41 , and
2|aj | ) pj (1 − 1−2|a p1 + p2 p1 (k + 1)a j| , + |a | ≤ j 2|aj | 1 − 2a (1 − 2a)2 ) (k + 1)(k + 1−2|a j|
j = 3, · · · , n.
Then f (z) ∈ K(Dpn ). Next, we establish a sufficient condition for the biholomorphic convex mapping on Dpn , which extend the main result of [6]. Theorem 2 Suppose that n ≥ 2, pj ≥ 2, j = 1, 2, · · · , n. Let f (z) = (p1 (z1 , z2 , · · · , zn ), p2 (z2 , zn ), · · · , pn−1 (zn−1 , zn ), pn (zn )), where z = (z1 , z2 , · · · , zn ) ∈ Dpn , pn (zn ) ∈ N (U ), pj (zj , zn ) : Dp2 → C is holomorphic with ∂p
∂p
pj (0, 0) = 0, ∂zjj (0, 0) = 1, ∂znj (0, 0) = 0, and p1 (z1 , · · · , zn ) : Dpn → C is holomorphic with 8
∂p1 1 p1 (0, 0, · · · , 0) = 0, ∂p ∂z1 (0, 0, · · · , 0) = 1, ∂zl (0, 0, · · · , 0) = 0 for 2 ≤ l ≤ n. If f (z) satisfies the following conditions n−1 Y
∂pj ′ p (zn ) 6= 0, |zn p′′n (zn )| ≤ |p′n (zn )|; ∂zj n j=1 n 2p X ∂ 2 p1 ∂p1 ∂ 2 pj ∂pj ∂ j z1 (2) ∂z1 ∂z ≤ ∂z1 , zj ∂z 2 + zj ∂zj ∂zn ≤ ∂zj (j = 2, 3, · · · , n − 1); l j l=1 2 2 2 ∂ pj ∂ 2 pj | ∂p1 |(| ∂ p2j | + | ∂ pj |) X n + z z 2 j j ∂ p1 ∂zj ∂zj ∂zn ∂zj ∂zn ∂zj ∂zj2 p1 ≤ pj |zj |pj −2 1 − (3) ∂p1 + ∂zj ∂zl ∂p ∂pj | ∂z1 | | ∂zjj | ∂zj l=1 (1)
(j = 2, 3, · · · , n − 1); X n ∂ 2 p1 pj ∂ 2 pj ∂ 2 pj ∂pj p′′n (zn ) p1 + + + (4) ∂zn ∂z ∂zj ∂zn ∂z 2 ∂zn p′ (zn ) ∂pj ∂p1 l | | n n j=2 ∂zj l=1 ∂z1 2 2 ∂p ∂ pj ∂ pj n−1 ∂pj n−1 X ∂zj1 ( ∂zj ∂zn + ∂zn2 ) ∂p1 p′′ (zn ) X ∂p1 ∂zn p′′n (zn ) n + + + ∂pj ∂zn p′n (zn ) ∂zj ∂pj p′n (zn ) ∂zj j=2 j=2 ∂zj ′′ zn p (zn ) , ≤ pn |zn |pn −2 1 − ′ n pn (zn ) n−1 X
for all z = (z1 , · · · , zn ) ∈ Dpn \ {0}, then f ∈ K(Dpn ). Proof By calculating the Fr´ echet derivatives of f (z) straightforwardly, we obtain ∂p1 ∂p1 ∂p1 ∂p1 ∂z1 ∂z2 · · · ∂zn−1 ∂zn ∂p2 2 0 ∂p 0 ∂z2 · · · ∂zn Df (z) = · · · · · · · · · ··· ··· , ∂pn−1 ∂pn−1 0 0 · · · ∂zn−1 ∂zn ′ 0 0 ··· 0 pn (zn )
Df (z)−1
1
∂p1 ∂z1 0 = ··· 0 0
−
∂p1 ∂z2 ∂p1 ∂p2 ∂z1 ∂z2
1 ∂p2 ∂z2
···
···
−
∂p1 ∂zn−1 ∂p1 ∂pn−1 ∂z1 ∂zn−1
−
∂p1 ∂zn ∂p1 ′ p (z ) ∂z1 n n
+
n−1 P j=2
∂p2
···
0
− ∂p2 ∂z′n ∂z2
···
0
···
0
···
··· 1 ∂pn−1 ∂zn−1
0
9
∂p1 ∂pj ∂zj ∂zn ∂pj ∂p1 ′ p (zn ) ∂zj ∂z1 n
pn (zn )
··· −
∂pn−1 ∂zn ∂pn−1 ′ p (z ) ∂zn−1 n n
1 p′n (zn )
,
D 2 f (z)(b, b) =
=
where
n P
l=1
n P
∂ 2 p1 ∂z1 ∂zl bl
l=1
0 ··· 0 0
∂ 2 p1 ∂z2 ∂zl bl
D2 ··· 0 0 n n P P
···
n P
l=1
··· ··· ··· ···
∂ 2 p1 ∂zn−1 ∂zl bl
j=1 l=1 2p 2 + 2 ∂z∂2 ∂z b2 bn n
C2 ··· Cn−1 Cn C1
0 ··· Dn−1 0
∂ 2 p1 ∂zj ∂zl bl bj
b1 b2 ··· bn−1 bn
+ , ··· ∂ 2 pn−1 2 ∂ 2 pn−1 2 ∂ 2 pn−1 b b + + 2 b b n−1 n 2 2 n n−1 ∂z ∂z ∂zn ∂z n n−1 ∂ 2 p2 2 b ∂z22 2
∂ 2 p2 2 2 bn ∂zn
n−1
p′′n (zn )b2n
n X ∂ 2 p1 bl , = ∂zn ∂zl
C1 Cj
=
Dj
=
l=1 ∂2p
∂ 2 pj bn (j = 2, 3, · · · , n − 1), ∂zn ∂zj ∂zn2 ∂ 2 pj ∂ 2 pj b + bn (j = 2, 3, · · · , n − 1). j ∂zj ∂zn ∂zj2 j
bj +
Cn = p′′n (zn )bn ,
Taking z = (z1 , · · · , zn ) ∈ Dpn \ {0}, b = (b1 , · · · , bn ) ∈ C n such that X n zj pj bj pj Re = 0, ρ(z) zj j=1
by the hypothesis of Theorem 2, we have Jf (z, b) ≥
n X pj |zj |pj −2 j=1
=
n X j=1
ρ(z)pj
|bj |2 − 2
n X pj zj pj ∂ρ i RehDf (z)−1 D 2 f (z)(b, b), ρ(z) ρ(z) ∂z j=1
∂p1 ∂zj ∂pj j=2 ∂zj
n n n−1 X pj |zj |pj −2 1 X X ∂ 2 p1 2 |b | − Re b b − j l j ∂p1 ρ(z)pj ∂zj ∂zl ∂z1
∂ 2 pj + 2 b2n ∂zn
j=1 l=1
n−1 ∂pj ∂p1 ∂zn ∂zj ∂pj ∂p1 j=2 ∂zj ∂z1
∂p1 p′′n (zn ) 2 X b + − ∂zn p′n (zn ) n
∂ 2 pj ∂ 2 pj 2 bj bn bj + 2 2 ∂zj ∂zn ∂zj
p′′n (zn ) 2 p1 |z1 |p1 b p′n (zn ) n z1 ρ(z)p1
n−1 X ∂ 2 pj ∂ 2 pj 2 ∂pj p′′n (zn ) 2 pj |zj |pj p′′n (zn ) 2 pn |zn |pn 1 ∂ 2 pj 2 b +2 b − bj bn + + ′ b b + ∂pj ∂zj ∂zn ∂zn2 n ∂zn p′n (zn ) n zj ρ(z)pj pn (zn ) n zn ρ(z)pn ∂zj2 j j=2 ∂zj
≥
n X pj |zj |pj −2 j=1
ρ(z)pj
X 1 2 n n X X ∂p ∂ 2 p1 2 n−1 ∂ pj 2 ∂ 2 pj 2 ∂z j |bj | − ∂p1 ∂pj ∂z 2 |bj | + ∂zj ∂zn |bj | ∂zj ∂zl |bj | + | | 2
1
∂z1
j=2
j=1 l=1
j
∂zj
2 2 n−1 ∂pj ∂p1 ′′ X ∂ pj ∂zn ∂zj pn (zn ) 2 p1 |z1 |p1 −1 ∂ pj 2 ∂p1 p′′n (zn ) 2 2 + |bn | + 2 |bn | + |bn | + ∂pj p′ (zn ) |bn | ∂zj ∂zn ∂z ∂zn p′ (zn ) ρ(z)p1 n
n
10
j=2
∂zj
n
n−1 X
2 ∂ pj 2 ∂ 2 pj 2 ∂ 2 pj 2 − ∂z 2 |bj | + ∂zj ∂zn |bj | + ∂zj ∂zn |bn | ∂pj | | j j=2 ∂zj 2 ′′ pn −1 pj −1 ∂ pj 2 ∂pj p′′n (zn ) pn (zn ) 2 pj |zj | |bn |2 pn |zn | + 2 |bn | + |b | − n p′ (zn ) ∂zn ∂zn p′n (zn ) ρ(z)pj ρ(z)pn n n P ∂ 2 p1 ∂2p ∂2p |z | n−1 1 |zj ∂z 2j | + |zj ∂zj ∂zj n | ∂z pj −2 p −2 1 ∂zl X 1 p |z | p |z | j j 1 1 j 1 − l=1 ∂p1 1− |bj |2 + = |b1 |2 pj ∂p ρ(z)p1 ρ(z) | | | j| 1
∂z1
p1 |z1 |p1 −1 − 1 p ρ(z) 1 ∂p ∂z1
X n l=1
j=2
∂zj
∂ 2 pj ∂p1 | ∂zj2 |
∂ 2 p1 | + | ∂zj ∂zl ∂zj
∂2p + | ∂zj ∂zj n |
2
+ |bn |
∂p
| ∂zjj |
pn |zn |pn −2 zn p′′n (zn ) (1 − | |) ρ(z)pn p′n (zn )
∂ 2 pj ∂ 2 pj ∂pj ′′ n−1 X ∂p1 ∂z pn (zn ) X ∂p1 | ∂zn2 | + | ∂zj ∂zn | ∂p1 p′′n (zn ) n−1 n + + + ∂zj ∂zj ∂pj p′ (zn ) ∂pj ∂zn p′n (zn ) l | | n j=2 j=2 ∂zj ∂zj
X n ∂ 2 p1 − ∂zn ∂z ρ(z)p1 | ∂p1 | p1 |z1 |p1 −1 ∂z1
l=1
2 ∂2p ∂pj ′′ n−1 ∂ pj X ∂z 2j pj |zj |pj −1 n−1 X ∂z X pn (zn ) pj |zj |pj −1 ∂zj ∂zn pj |zj |pj −1 n−1 n n − ∂pj ρ(z)pj − ∂pj p′ (zn ) ρ(z)pj ∂pj ρ(z)pj −
j=2
j=2
∂zj
|p1 −2
p1 |z1 ≥ |b1 |2 ρ(z)p1 −
−
p1 1 | ∂p ∂z1 |
p1 ∂p1 | ∂z1 |
X n l=1
n P
l=1
1−
j=2
∂zj
∂zj
n
2
|z1 ∂z∂ 1p∂z1 l | 1 | ∂p ∂z1 |
+
n−1 X
|bj |2 pj |zj |pj −2 1 −
j=2
|zj
∂ 2 pj |+ ∂zj2
∂2p
|zj ∂zj ∂zj n |
∂p
| ∂zjj |
∂ 2 pj ∂ 2 pj ∂p1 | ∂zj2 | + | ∂zj ∂zn | ∂ 2 p1 zn p′′n (zn ) 2 p−2 |+ |) | + |b | p |z | (1 − | n n n ∂p ∂zj ∂zl ∂zj p′n (zn ) | j| ∂zj
∂ 2 pj ∂ 2 pj ∂p n−1 n−1 ∂p1 ∂znj p′′n (zn ) X ∂p1 | ∂zn2 | + | ∂zj ∂zn | ∂p1 p′′n (zn ) X + + + ∂zj ∂zj ∂pj p′ (zn ) ∂pj ∂zn p′n (zn ) l | | n j=2 j=2 ∂zj ∂zj
X n ∂ 2 p1 ∂zn ∂z l=1
2 ∂2p ∂pj ′′ n−1 n−1 n−1 ∂ pj X ∂z 2j X ∂z X p (zn ) ∂zj ∂zn n n n − ∂pj pj − ∂pj p′ (zn ) pj ∂pj pj −
j=2
∂zj
j=2
∂zj
j=2
∂zj
n
≥ 0.
Thus it follows from Theorem A that f ∈ K(Dpn ). The proof is complete. ✷ Remark 2 Setting pj (zj , zn ) = fj (zj ) + pj (zn ) j = 2, · · · , n in Theorem 2, we get Theorem 2.1 of [6]. Setting n = 3 in Theorem 2, we get Corollary 1. Example 3 Suppose that pj ≥ p1 ≥ 2(j = 2, · · · , n), 0 < |λ| ≤ 1, and k is a positive integer such that k < max{pj : j = 1, · · · , n} ≤ k + 1. Let p(z) = (z1 +
n−1 X
aj zjk+1 + an z1 znk+1 , z2 + a2 z2 znk+1 , · · · , zn−1 + an−1 zn−1 znk+1 ,
j=2
11
eλzn −1 ), λ
where a = max |aj | : j = 2, · · · , n . If a ≤
1−|λ| 2(k+1)2 (k+1+|λ|)+1+|λ|
< 1, and
n−1 n−1 X X |aj | pj |aj | a p1 pn (1 − |λ|) + , 1 + (k + 1) ≤ 1 − |aj | 1 − a 1 − |aj | (k + 1)(k + 1 + |λ|) j=2
j=2
then p(z) ∈ K(Dpn ). Proof Put
p1 (z1 , z2 , · · · , zn ) = z1 +
n−1 X
aj zjk+1 + an z1 znk+1 ,
j=2
eλzn − 1 . pj (zj , zn ) = zj + aj zj znk+1 (j = 2, 3, · · · , n − 1), pn (zn ) = λ 1−|λ| 1 < k+2 Then it follows from a = max |aj | : j = 2, · · · , n ≤ 2(k+1)2 (k+1+|λ|)+1−|λ| 0, ∂pj k+1 ∂zj = |1 + aj zn | ≥ 1 − |aj | ≥ 1 − a > 0,
p′n (zn ) = eλzn 6= 0, n ∂p1 X ∂ 2 p1 k+1 k − z1 ∂z1 ∂z1 ∂z = |1 + an zn | − |(k + 1)an z1 zn | ≥ 1 − (k + 2)|an | > 0, l l=1 2 2 ∂pj ∂ pj zj ∂ pj k+1 k ∂zj − zj ∂z 2 − ∂zj ∂zn = |1 + aj zn | − (k + 1)|aj zj zn | ≥ 1 − (k + 2)|aj | > 0, j ′′ and znp′pn(z(znn) ) = |λ||z| ≤ |λ| < 1. By calculating straightforwardly, we also obtain n
= ≥ ≥ = ≥
∂ 2 pj ∂ 2 pj ∂p1 ∂ 2 pj ∂ 2 pj zj 2 + zj ∂z ∂z 2 + ∂z ∂z n ∂zj ∂zj ∂zn X n j j ∂ 2 p1 j p1 pj −2 + pj |zj | − ∂p1 1− ∂zj ∂zl ∂pj ∂pj | | ∂z1 l=1 ∂zj ∂zj |(k + 1)aj zjk ||(k + 1)aj znk | |(k + 1)zj aj znk | p1 k−1 pj −2 − + |k(k + 1)aj zj | pj |zj | 1− |1 + aj znk+1 | |1 + an znk+1 | |1 + aj znk+1 | (k + 1)|aj | (k + 1)2 |aj |2 pj pj −2 pj |zj | + k(k + 1)|aj | |zj |k−1 1− − 1 − |aj | 1 − |an | 1 − |aj | (k + 1)a (k + 1)2 a2 pj pj −2 2 pj |zj | 1− + (k + 1) a |zj |pj −2 − 1−a 1−a 1−a pj |zj |pj −2 (k + 1)2 a 1 − (k + 2)a − 1−a 1−a p −2 pj |zj | j 1 − |λ| (k + 2)(1 − |λ|) − 1− 1−a 2(k + 1)2 (k + 1 + |λ|) + 1 − |λ| 2(k + 1 + |λ|) 12
pj |zj |pj −2 [(k + 2) + (k + 1)2 ](1 − |λ|) 1− > 0, 1−a 2(k + 1)2 (k + 1 + |λ|)
≥
j = 2, · · · , n − 1.
Since ∂ 2 pj n−1 n−1 X pj |aj | ∂zj ∂zn X (k + 1)aj znk ≤ (k + 1) pj ∂p = pj |zn |pn −2 , j 1 − |aj | 1 + a z k+1
n−1 X j=2
∂zj
j n
j=2
j=2
∂2p
j n−1 n−1 X pj |aj | X k(k + 1)aj zj z k−1 2 ∂zn n = |zn |pn −2 , ≤ k(k + 1) p j 1 + a z k+1 ∂pj 1 − |aj | n j j=2 j=2 j=2 ∂zj ∂p n−1 n−1 X pj |aj | X ∂z j p′′ (zn ) n−1 X (k + 1)aj zj z k n = |λ| ≤ |λ|(k + 1) pj ∂pn n′ p |zn |pn −2 , j 1 + a z k+1 j pn (zn ) 1 − |a | j j n j=2 j=2 j=2 ∂zj
n−1 X
pj
we have
n−1 X
X 2 n ∂ 2 p1 ∂ pj ∂ 2 pj ∂pj p′′n (zn ) + + p1 + ∂zn ∂zl ∂zj ∂zn ∂z 2 ∂zn p′ (zn ) ∂p1 ∂pj | | | | n n j=2 ∂zj ∂z1 l=1 ∂ 2 pj ∂ 2 pj ∂p1 ∂pj ′′ n−1 | ∂z | ∂z ∂zn + ∂z 2 X X ∂p1 ∂z j j pn (zn ) ∂p1 p′′n (zn ) n−1 n n + + + ∂zj ∂pj p′ (zn ) ∂pj ∂zn p′n (zn ) | | n j=2 j=2 ∂zj ∂zj n−1 n−1 n−1 X pj |aj | X pj |aj | X pj |aj | + k(k + 1) + |λ|(k + 1) ≤ |zn |pn −2 (k + 1) 1 − |aj | 1 − |aj | 1 − |aj | pj
j=2
j=2
+
p1 (k + 1)|an | + k(k + 1)|an | + 1 − |an |
+(k + 1)|an ||λ| +
n−1 X j=2
pn −2
≤ |zn |
(k + 1)(k + 1 + |λ|)
n−1 X
n−1 X (k + 1)|aj |(k + 1)2 |aj | j=2
j=2
j=2
(k + 1)|aj |[(k + 1)|aj | + k(k + 1)|aj |] 1 − |aj |
(k + 1)|aj ||λ| (k + 1)|aj | |zn |k−1 1 − |aj |
j=2
+
n−1 X
1 − |aj |
≤ |zn |pn −2 (k + 1)(k + 1 + |λ|)
p1 pj |aj | + (k + 1)a + k(k + 1)a 1 − |aj | 1 − a
+ (k + 1)a|λ| +
n−1 X j=2
n−1 X (k + 1)|aj |(k + 1)|aj ||λ| 1 − |aj |
j=2
|zn |pn −2
pj |aj | 1 − |aj |
n−1 X |aj | p1 pn −2 (k + 1)(k + 1 + |λ|)a|zn | + 1 + (k + 1) 1−a 1 − |aj | j=2
X n−1 n−1 X |aj | pj |aj | a p1 pn −2 + ≤ |zn | (k + 1)(k + 1 + |λ|) 1 + (k + 1) 1 − |aj | 1 − a 1 − |aj | j=2 j=2 zn p′′ (zn ) , ≤ pn |zn |pn −2 (1 − |λ|) ≤ pn |zn |pn −2 1 − ′ n pn (zn ) 13
hence by Theorem 2, we obtain that f ∈ K(Dpn ).
✷ 2|an | 1−2|an |
By applying the same method of the proof for example 3, we only need to let instead of |λ|, we may prove the following result. Example 4 Suppose that pj ≥ p1 ≥ 2(j = 2, · · · , n), 0 < |an | ≤ 14 , and k is a positive integer such that k < max{pj : j = 1, · · · , n} ≤ k + 1. Let p(z) = (z1 +
n−1 X
aj zjk+1 + an z1 znk+1 , z2 + a2 zn znk+1 , · · · , zn−1 + an−1 zn−1 znk+1 , zn + an zn2 ),
j=2
where a = max |aj | : j = 2, · · · , n . If a≤
1− 2(k + 1)2 (k + 1 +
2|an | 1−2|an | 2|an | 1−2|an | ) +
1−
2|an | 1−2|an |
< 1,
and n−1 X j=2
2|an | n−1 X |aj | ) pn (1 − 1−2|a pj |aj | p1 a n| , + 1 + (k + 1) ≤ 2|an | 1 − |aj | 1 − a 1 − |aj | ) (k + 1)(k + 1 + 1−2|a j=2 n|
then p(z) ∈ K(Dpn ). Applying the same method as our proof of Theorem 1, we may show the following theorem. Theorem 3 Suppose that n ≥ 2, pj ≥ 2, j = 1, 2, · · · , n, k is a positive integer such that 2 ≤ k ≤ n, and fj , pj : U → C is holomorphic with fj (0) = 0, fj′ (0) = 0(j = 2, 3, · · · , n), fk (zk ) = 0, p1 (z1 , z2 , · · · , zn ) : Dp → C is holomorphic with p1 (0, 0, · · · , 0) = ∂p1 1 0, ∂p ∂z1 (0, 0, · · · , 0) = 1, ∂zl (0, 0, · · · , 0) = 0(l = 2, 3, · · · , n). Let f (z) = (p1 (z1 , z2 , · · · , zn ), p2 (z2 ) + f2 (zk ), · · · , pk−1 (zk−1 ) + fk−1 (zk ), pk (zk ), pk+1 (zk+1 ) + fk+1 (zk ), · · · , pn (zn ) + fn (zk )). If for any z = (z1 , z2 , · · · , zn ) ∈ Dpn \{0}, we have n ∂p1 Y ′ (1) pj (zj ) 6= 0, |zj p′′j (zj )| ≤ |p′j (zj )|, j = 2, · · · , n; · ∂z1 j=2 n X ∂ 2 p1 ∂p1 z1 (2) ∂z1 ∂zl ≤ ∂z1 ; l=1
′′
2 ∂p1 · pj′ (zj ) ′′ n ∂ p1 X ∂zj pj (zj ) ∂zj ∂zl zj pj (zj ) pj −2 , (3) p1 1− ′ + p1 ∂p1 ≤ pj |zj | ∂p1 p (zj )
∂z1
l=1
(j = 2, · · · , k − 1, k + 1, · · · ′′ n n X X fj (zk ) (4) p′ (zj ) pj + j j=2,j6=k
j
∂z1
j=2,j6=k
, n − 1); ′′ ′ fj′ (zk ) ∂p1 n X fj (zk ) p′′k (zk ) pj (zj ) ∂zj p1 p′ (zj ) p′ (zk ) pj + ∂p1 j k j=2,j6=k
14
∂z1
′
fj′ (zk ) p′′k′ (zk ) ∂p1 p′′k (zk ) ∂p1 n n ∂ 2 p1 X X pj (zj ) pk (zk ) ∂zj ∂zl ∂zk p′k (zk ) ∂zk p1 + + ∂p1 p1 + ∂p1 p1 ∂p1 ∂z ∂z ∂z j=2,j6=k l=1 1 1 1 zk p′′k (zk ) pk |zk |pk −2 , ≤ 1 − ′ pk (zk )
then f ∈ K(Dpn ). Finally, we give a partial answer to Problem II by verifying the following theorem. Theorem 4 Suppose that p ≥ 2. Let f (z) = (z1 + a1 z12 + a′1 z22 , a2 z12 + z2 + a′2 z22 ), where z = (z1 , z2 ). If (a1 , a2 , a′1 , a′2 ) satisfies the following conditions: 4|a1 | + 2|a′2 | + 8|a1 ||a′2 | + 2|a2 | + 4|a1 ||a2 | + 8|a′1 ||a2 | + 4|a1 ||a2 | ≤ 1, 2|a1 | +
2|a′1 | +
4|a′1 ||a′2 |
+
4|a′2 | +
8|a1 ||a′2 |
+
4|a′1 ||a′2 | +
8|a′1 ||a2 |
≤ 1.
(4) (5)
Then f ∈ K(Bp2 ). Proof By calculating the Fr´echet derivatives of f (z) straightforwardly, we obtain ! ! 1+2a′2 z2 −2a′1 z2 1 + 2a1 z1 2a′1 z2 A , Df (z)−1 = −2aA2 z1 1+2a Df (z) = , 1 z1 2a2 z1 1 + 2a′2 z2 A A ! ! ! 2 + 2a′ b2 ′b 2a b b 2a b 2a 1 1 1 1 1 1 2 1 2 , = D 2 f (z)(b, b) = 2a2 b21 + 2a′2 b22 b2 2a2 b1 2a′2 b2 ! 2a′1 z2 (2a2 b21 +2a′2 b22 ) (1+2a′2 z2 )(2a1 b21 +2a′1 b22 ) − −1 2 A A Df (z) D f (z)(b, b) = 2a z (2a b2 +2a′ b2 ) (1+2a1 z1 )(2a2 b21 +2a′2 b22 ) , − 2 1 1A1 1 2 + A where A = 4a1 a′2 z1 z2 − 4a′1 a2 z1 z2 + 1 + 2a′2 z2 + 2a1 z1 .
(6)
Taking z = (z1 , z2 ) ∈ Bp2 \ {0}, b = (b1 , b2 ) ∈ C 2 such that Rehb, ∂u ∂z i = 0, by the hypothesis of Theorem 2, we have 2
X 2 ∂u 2 |bj |2 |zj |p−2 − RehDf (z)−1 D 2 f (z)(b, b), Jf (z, b) ≥ i p p ∂z j=1
2 X
(1 + 2a′2 z2 )(2a1 b21 + 2a′1 b22 ) − 2a′1 z2 (2a2 b21 + 2a′2 b22 ) |z1 |p |bj | |zj | − Re = A z1 j=1 (1 + 2a1 z1 )(2a2 b21 + 2a′2 b22 ) − 2a2 z1 (2a1 b21 + 2a′1 b22 ) |z2 |p + A z2 2 X 2a1 + 4a1 a′2 z2 2 2a′1 + 4a′1 a′2 z2 2 4a′1 a2 z2 2 2 p−2 |b1 | + |b2 | + |bj | |zj | − ≥ A |b1 | A A j=1 ′ ′ 4a1 a2 z1 2 4a′1 a2 z1 2 4a1 a2 2 p−1 |b2 | |z1 | |b1 | + + + A |b2 | A A 2
p−2
15
2a2 + 4a1 a2 z1 2 2a′2 + 4a1 a′2 z1 2 p−1 + |b1 | + |b2 | |z2 | A A 2|a1 | + 4|a1 ||a′2 | + 2|a2 | + 4|a1 ||a2 | + 4|a′1 ||a2 | + 4|a1 ||a2 | ≥ |b1 |2 |z1 |p−2 1 − 1 − 4|a1 ||a′2 | − 4|a′1 ||a2 | − 2|a′2 | − 2|a1 | 2|a′1 | + 4|a′1 ||a′2 | + 2|a′2 | + 4|a1 ||a′2 | + 4|a′1 ||a′2 | + 4|a′1 ||a2 | 2 p−2 +|b2 | |z2 | 1− 1 − 4|a1 ||a′2 | − 4|a′1 ||a2 | − 2|a′2 | − 2|a1 | ≥ 0, hence it follows from Theorem A that f (z) ∈ K(Bp2 ).
✷
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