Volume 8 (2007), Issue 4, Article 104, 15 pp.
SECOND ORDER DIFFERENTIAL SUBORDINATIONS OF HOLOMORPHIC
MAPPINGS ON BOUNDED CONVEX BALANCED DOMAINS IN Cn
YU-CAN ZHU AND MING-SHENG LIU
D EPARTMENT OF M ATHEMATICS
F UZHOU U NIVERSITY,
F UZHOU , 350002 F UJIAN , P. R. C HINA
zhuyucan@fzu.edu.cn
D EPARTMENT OF M ATHEMATICS
S OUTH C HINA N ORMAL U NIVERSITY,
G UANGZHOU , 510631 G UANGDONG , P. R. C HINA
liumsh@scnu.edu.cn
Received 09 November, 2006; accepted 04 November, 2007
Communicated by G. Kohr
A BSTRACT. In this paper, we obtain some second order differential subordinations of holomorphic mappings on a bounded convex balanced domain Ω in Cn . These results imply some first
order differential subordinations of holomorphic mappings on a bounded convex balanced domain Ω in Cn . When Ω is the unit disc in the complex plane C, these results are just ones of
Miller and Mocanu et al. about differential subordinations of analytic functions on the unit disc
in the complex plane C.
Key words and phrases: Differential subordination, biholomorphic convex mapping, convex balanced domain, Minkowski
functional.
2000 Mathematics Subject Classification. 32H02, 30C45.
1. I NTRODUCTION
Let Cn be the space of n complex variables z =p(z1 , z2 , . . . , zn ) with the Euclidian inner
P
product hz, wi = nj=1 zj wj and the norm kzk = hz, zi. A domain Ω is called a balanced
domain in Cn if λz ∈ Ω for all z ∈ Ω and λ ∈ C with |λ| ≤ 1. The Minkowski functional of
the balanced domain Ω is
n
o
z
ρ(z) = inf t > 0, ∈ Ω , z ∈ Cn .
t
The authors thank the referee for his helpful comments and suggestions to improve our manuscript.
This research is partly supported by the National Natural Science Foundation of China(No.10471048), the Doctoral Foundation of the
Education Committee of China(No.20050574002), the Natural Science Foundation of Fujian Province, China (No.Z0511013) and the Education
Commission Foundation of Fujian Province, China (No.JB04038).
288-06
2
Y U -C AN Z HU AND M ING -S HENG L IU
Suppose that Ω is a bounded convex balanced domain in Cn , and ρ(z) is the Minkowski
functional of Ω. Then ρ(·) is a norm of Cn such that
Ω = {z ∈ Cn : ρ(z) < 1},
ρ(λz) = |λ|ρ(z)
for λ ∈ C, z ∈ Cn (see [20]).
Let pj > 1 (j = 1, 2, . . . , n). Then
(
Dp =
(z1 , z2 , . . . , zn ) ∈ Cn :
n
X
)
|zj |pj < 1
j=1
is a bounded convex balanced domain, and the Minkowski functional ρ(z) of Dp satisfies
n X
zj pj
(1.1)
ρ(z) = 1.
j=1
1/p
n
o
P
Pn
n
p
n
p
ρ(z) =
|z
|
is
the
Minkowski
functional
of
domain
B
=
z
∈
C
:
|z
|
<
1
,
p
j=1 j
j=1 j
where p > 1.
Let Df (z) and D2 f (z)(·, ·) denote the first Fréchet derivative and the second Fréchet derivative for a holomorphic mapping f : Ω → Cn respectively. Then they have the matrix
representation
!
n
2
X
∂fj (z)
∂
f
(z)
j
Df (z) =
, D2 f (z)(b, ·) =
bl
,
∂zk
∂z
k ∂zl
1≤j, k≤n
l=1
1≤j, k≤n
where b = (b1 , b2 , . . . , bn ) ∈ Cn . The mapping f : Ω → Cn is called locally biholomorphic if
the matrix Df (z) is nonsingular at each point z in Ω.
The class of all holomorphic mappings f : Ω → Cn is denoted by H(Ω, Cn ). Assume
f, g ∈ H(Ω, Cn ). Then we say that the mapping f is subordinate to g, written f ≺ g or
f (z) ≺ g(z), if there exists a holomorphic mapping w : Ω → Ω with w(0) = 0 such that
f (z) ≡ g(w(z)) for all z ∈ Ω. If g is a biholomorphic mapping, then f (z) ≺ g(z) if and only
if f (Ω) ⊂ g(Ω) and f (0) = g(0).
In classical results of geometric function theory, differential subordinations provide some
simple proofs. They play a key role in the study of some integral operators, differential equations, and properties of subclasses of univalent functions, etc. S.S. Miller and P.T. Mocanu et al.
have obtained some deep results for differential subordinations [10, 11, 12, 13, 16, 14]. There
is a excellent text Differential Subordinations Theory and Applications, by S.S. Miller and P.T.
Mocanu [15].
The geometric function theory of several complex variables has been studied by many authors. Many important results for biholomorphic convex or starlike mappings in Cn have been
obtained (see [2, 3]). Some differential subordinations of analytic functions in the complex
plane are also extended to Cn [4, 6, 8, 15, 22]. But there are very few results on second order
differential subordinations of holomorphic mappings in Cn .
In this paper, we obtain some second order differential subordinations of holomorphic mappings on a bounded convex balanced domain Ω in Cn . These results imply some first order
differential subordinations of holomorphic mappings on a bounded convex balanced domain Ω
in Cn . When Ω is the unit disc in the complex plane C, these results are just those of Miller
and Mocanu et al. about differential subordinations of analytic functions on the unit disc in the
complex plane C.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 104, 15 pp.
http://jipam.vu.edu.au/
S ECOND O RDER D IFFERENTIAL S UBORDINATIONS OF H OLOMORPHIC M APPINGS
3
2. M AIN R ESULTS AND T HEIR P ROOFS
In the following, we always assume that the domain Ω is a bounded convex balanced domain
in Cn and ρ(z) is the Minkowski functional of Ω. Then ρ(·) is a norm of Cn such that
Ω = {z ∈ Cn : ρ(z) < 1},
ρ(λz) = |λ|ρ(z)
for λ ∈ C, z ∈ Cn .
In order to derive our main results, we need the following lemmas.
Lemma 2.1. Suppose that ρ(z) is twice differentiable in Ω − {0}, and let w ∈ H(Ω, Cn ) with
w(z) 6≡ 0 and w(0) = 0. If z0 ∈ Ω − {0} satisfies
ρ(w(z0 )) =
max ρ(w(z)),
ρ(z)≤ρ(z0 )
then there exists a real number t ≥ 1/2 such that
∂ρ
(2.1)
Dw(z0 )(z0 ), (w0 ) = tρ(w0 ),
∂z
and
(2.2)
∂ρ
Re D w(z0 )(z0 , z0 ), (w0 )
∂z
)
( n
n
X ∂2ρ
X
∂2ρ
(w0 )bj bl − tρ(w0 ),
≥ Re
(w0 )bj bl −
∂z
∂z
j ∂zl
j ∂z l
j,l=1
j,l=1
2
where w0 = w(z0 ), Dw(z0 )(z0 ) = (b1 , b2 , . . . , bn ).
Proof. Since ρ(w0 ) =
max ρ(w(z)), then we have w0 6= 0. Otherwise, there is w(z) ≡ 0,
ρ(z)≤ρ(z0 )
which contradicts the hypothesis of Lemma 2.1.
Let w(z) = (w1 (z), w2 (z), . . . , wn (z)), γ(t) = w(eit z0 ) = (γ1 (t), γ2 (t), . . . , γn (t)). Then
we have γj (t) = wj (eit z0 ) (j = 1, 2, . . . , n), γ(0) = w(z0 ) = w0 and
!
n
n
X
X
dγj (t)
∂wj (eit z0 ) 0
dγj (t)
∂wj (eit z0 ) 0
it
−it
= ie
zk ,
= −ie
zk ,
dt
∂zk
dt
∂zk
k=1
k=1
where z0 = (z10 , z20 , . . . , zn0 ). Set L(t) = ρ(γ(t)) (−π ≤ t ≤ π). Some straightforward
calculations yield
n
n
X
∂ρ
dγj (t) X ∂ρ
dγj (t)
(γ(t)) ·
+
(γ(t)) ·
∂zj
dt
∂z j
dt
j=1
j=1
"
#
n
it
X
∂ρ
∂w
(e
z
)
j
0
= −2 Im eit
(γ(t)) ·
zk0
∂z
∂z
j
k
j,k=1
∂ρ
it
it
= −2 Im Dw(e z0 )(e z0 ), (γ(t)) ,
∂z
L0 (t) =
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 104, 15 pp.
http://jipam.vu.edu.au/
4
Y U -C AN Z HU AND M ING -S HENG L IU
"
#
n
n
n
2
X
X
X
∂ρ
∂w
∂ρ
∂
w
j
j
L00 (t) = −2 Im ieit
(γ(t)) ·
+ ie2it
(γ(t))
zk0 zl0
∂z
∂z
∂z
∂z
j
k
j
k ∂zl
j,k=1
j,k=1
l=1
#
" n
n
X X
∂2ρ
∂wl
∂w
j
0
− 2 Im i
(γ(t)) ·
· (eit zm
)
· (eit zk0 )
∂z
∂z
∂z
∂z
j
l
m
k
j,k=1 l,m=1
#
" n
n
X X
∂w
∂2ρ
∂wl
j
0 )
+ 2 Im i
(γ(t)) ·
· (eit zm
· (eit zk0 ) .
∂z
∂z
∂z
j ∂z l
m
k
j,k=1 l,m=1
Noting L(0) = max L(t) , we have L0 (0) = 0 and L00 (0) ≤ 0. It follows that
−π≤t≤π
∂ρ
Im Dw(z0 )(z0 ), (w0 ) = 0,
∂z
(2.3)
and
(2.4)
∂ρ
∂ρ
Re D w(z0 )(z0 , z0 ), (w0 ) + Re Dw(z0 )(z0 ), (w0 )
∂z
∂z
" n
#
X ∂2ρ
∂2ρ
+ Re
(w0 )bj bl −
(w0 )bj bl ≥ 0.
∂zj ∂zl
∂zj ∂zl
j,l=1
2
On the other hand, by Schwarz’s Lemma in Cn [19], we have
ρ(w0 )
ρ(w(z))
≤
ρ(z)
ρ(z0 )
for 0 < ρ(z) ≤ ρ(z0 ).
Let
ϕ(r) =
ρ(w(rz0 ))
ρ(w(rz0 ))
=
.
ρ(rz0 )
rρ(z0 )
Then ϕ(1) = max ϕ(r) . It follows that
0<r≤1
ϕ0 (1) = lim−
r→1
ϕ(r) − ϕ(1)
≥ 0.
r−1
By a simple calculation, we obtain
ρ(w0 )
2
∂ρ
ϕ (1) = −
+
Re Dw(z0 )(z0 ), (w0 ) ≥ 0.
ρ(z0 )
ρ(z0 )
∂z
0
If we let
1
∂ρ
t=
Re Dw(z0 )(z0 ), (w0 ) ,
ρ(w0 )
∂z
then we have t ≥ 1/2, therefore (2.1) of Lemma 2.1 holds, and (2.2) follows from (2.3) and
(2.4). This completes the proof.
Remark 2.2. Since ρ(tz) = tρ(z) for t > 0, then for z ∈ Cn − {0}, we have
n
n
X
X
∂ρ
dρ(tz) ∂ρ
∂ρ
(2.5)
ρ(z) =
=
zj +
z j = 2 Re z, (z) .
∂z
dt t=1 j=1 ∂zj
∂z j
j=1
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 104, 15 pp.
http://jipam.vu.edu.au/
S ECOND O RDER D IFFERENTIAL S UBORDINATIONS OF H OLOMORPHIC M APPINGS
5
z
For any z ∈ Cn − {0}, we have ρ( ρ(z)
) = 1. Letting w(z) ≡ z in (2.1), we obtain that there
1
exists a real number t ≥ 2 such that
∂ρ
z, (z) = tρ(z) ≥ 0, z ∈ Cn − {0}.
∂z
Hence it follows from (2.5) that
∂ρ
ρ(z) = 2 z, (z) , z ∈ Cn − {0}.
∂z
Lemma 2.3 ([10]). Let g(ξ) = a + b1 ξ + b2 ξ 2 + · · · be analytic in |ξ| < 1 with g(ξ) 6≡ 0. If
ξ0 = r0 eiθ0 (0 < r0 < 1) and Re g(ξ0 ) = min Re g(ξ), then
|ξ|≤r0
(2.6)
ξ0 g 0 (ξ0 ) ≤ −
| a − g(ξ0 ) |2
,
2 Re(a − g(ξ0 ))
and
(2.7)
Re{ξ02 g 00 (ξ0 ) + ξ0 g 0 (ξ0 )} ≤ 0.
Lemma 2.4. Suppose that ρ(z) is differentiable in Ω−{0}. Let h : Ω → Cn be a biholomorphic
convex mapping with h(0) = 0. Then for every z ∈ Ω − {0}, we have
∂ρ
2 Dh(z)−1 h(z), (z) − ρ(z) ≤ ρ(z).
∂z
(z)i for |ξ| ≤ 1. Then
Proof. For each z ∈ Ω − {0}, we let g(ξ) = hDh(z)−1 (h(z) − h(ξz)), ∂ρ
∂z
g(ξ) is analytic in |ξ| ≤ 1 and
∂ρ
−1
g(ξ) = Dh(z) h(z), (z) + b1 ξ + · · · .
∂z
From the result in [3, 7], we have Re g(ξ) > 0 for all |ξ| < 1. Hence we obtain
0 = Re g(1) = min Re g(ξ).
|ξ|≤1
By a simple calculation, we may obtain
∂ρ
∂ρ
ρ(z)
0
−1
.
g (1) = − Dh(z) Dh(z)(z), (z) = − z, (z) = −
∂z
∂z
2
By (2.6), we have
−ρ(z) Re a + |a|2 ≤ 0,
E
D
(z) . It follows that
where a = (Dh(z)−1 h(z), ∂ρ
∂z
∂ρ
−1
2 Dh(z) h(z), (z) − ρ(z) ≤ ρ(z).
∂z
Lemma 2.5 ([23]). Suppose that ρ(z) is twice differentiable in Ω − {0}. If f : Ω → Cn is a
biholomorphic convex mapping, then we have
( n
)
n
X ∂2ρ
X
∂2ρ
∂ρ
−1 2
(2.8) Re
bl bm +
bl bm − Df (z) D f (z)(b, b),
≥0
∂zl ∂zm
∂zl ∂z m
∂z
l,m=1
l,m=1
for every z = (z1 , z2 , . . . , zn ) ∈ Ω − {0}, b = (b1 , b2 , . . . , bn ) ∈ Cn with Re b, ∂ρ
= 0.
∂z
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 104, 15 pp.
http://jipam.vu.edu.au/
6
Y U -C AN Z HU AND M ING -S HENG L IU
Lemma 2.6. Assume that ρ(z) is differentiable in Ω − {0}. Then
∂ρ
(2.9)
ρ(z) = 2 z, (z) , z ∈ Cn − {0},
∂z
and
(2.10)
∂ρ
2 w, (z) ≤ ρ(w),
∂z
z ∈ Cn − {0}, w ∈ Cn .
Proof. From Remark 2.2, we only need to prove (2.10). Let z ∈ Cn − {0} and
Ωz = {w ∈ Cn : ρ(w) < ρ(z)}.
Then Ωz is a convex domain in Cn , and
∂ρ
(z)
∂z
Dis the normal Evector of ∂Ωz at z. For every
(z) ≥ 0. It follows that
z, w ∈ Cn with ρ(z) = 1, ρ(w) = 1, we have Re z − w, ∂ρ
∂z
∂ρ
∂ρ
(2.11)
2 Re w, (z) ≤ 2 Re z, (z) = ρ(z) = 1.
∂z
∂z
E
D
When w, ∂ρ
(z) = 0, it is obvious that (2.10) holds.
D ∂z
E
∂ρ
z
w
to substitute for z and ρ(w)
e−iθ to substitute
When w, ∂z (z) 6= 0, then ρ(w) 6= 0. Using ρ(z)
for w in (2.11), we obtain
2 Re w, ∂ρ (z) ≤ ρ(w), z ∈ Cn − {0}, w ∈ Cn ,
∂z
D
E
where θ = arg w, ∂ρ
(z)
and ∂ρ
(λz) = ∂ρ
(z) for all λ ∈ (0, +∞) and z ∈ Ω − {0}. This
∂z
∂z
∂z
completes the proof.
Lemma 2.7. Suppose that ρ(z) is differentiable in Ω − {0}, and let h : Ω → Cn be a biholomorphic convex mapping with h(0) = 0. Then for every z ∈ Ω − {0} and vector ξ ∈ Cn , the
inequality
2 Dh(z)−1 (ξ), ∂ρ (z) ≤ (1 + ρ(z))2 ρ(Dh(0)−1 (ξ))
∂z
holds.
Proof. Without loss of generality, we may assume that h is a biholomorphic convex mapping
on Ω. If not, then we can replace h(z) by hr (z) = h(rz), where 0 < r < 1.
For any fixed z ∈ Ω − {0}, from the proof of Theorem 2.1 in [5, 9], there exist ze ∈ ∂Ω and
µ ∈ (0, 1) such that h(z) = µh(e
z ) and
1−µ≥
1 − ρ(z)
.
1 + ρ(z)
Let g(w) = h−1 [(1 − µ)h(w) + µh(e
z )]. Since h is a biholomorphic convex mapping on Ω, then
g ∈ H(Ω, Cn ) with g(Ω) ⊂ Ω and g(0) = z. For every ξ ∈ Cn − {0}, we set
ξ
∂ρ
ψ(λ) = 2 g λ
, (z) ,
ρ(ξ)
∂z
then ψ(λ) is an analytic function in |λ| < 1. By Lemma 2.6, we obtain
ξ
|ψ(λ)| ≤ ρ g λ
<1
ρ(ξ)
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 104, 15 pp.
http://jipam.vu.edu.au/
S ECOND O RDER D IFFERENTIAL S UBORDINATIONS OF H OLOMORPHIC M APPINGS
7
for all |λ| < 1, and
ψ(λ) = ρ(z) + 2 Dg(0)
ξ
ρ(ξ)
∂ρ
, (z) λ + · · · .
∂z
From the classical result in [1], we have |ψ 0 (0)| ≤ 1 − |ψ(0)|2 . It follows that
∂ρ
2 Dg(0)(ξ), (z) ≤ (1 − ρ(z)2 )ρ(ξ).
∂z
Since Dh(z)Dg(0) = (1 − µ)Dh(0), then
2 Dh(z)−1 Dh(0)(ξ), ∂ρ (z) ≤ 1 (1 − ρ(z)2 )ρ(ξ) ≤ (1 + ρ(z))2 ρ(ξ)
1−µ
∂z
for all ξ ∈ Cn , z ∈ Ω − {0}.
Set ζ = Dh(0)(ξ), then ξ = Dh(0)−1 ζ and
2 Dh(z)−1 (ζ), ∂ρ (z) ≤ (1 + ρ(z))2 ρ(Dh(0)−1 (ζ)),
∂z
which completes the proof.
Theorem 2.8. Let f, g ∈ H(Ω, Cn ) with f (0) = g(0), and let g be biholomorphic convex on
Ω. Suppose that ρ(z) is twice differentiable in Ω − {0}. If f is not subordinate to g, then there
exist points z0 ∈ Ω − {0}, w0 ∈ ∂Ω with 0 < ρ(z0 ) < 1, ρ(w0 ) = 1 and there is a real number
t ≥ 1/2 such that
(1) fD(z0 ) = g(w0 ),
E
(w
)
= t, and
(2) Dg(w0 )−1 Df (z0 )(z0 ), ∂ρ
0
∂z
D
E
∂ρ
−1 2
(3) Re Dg(w0 ) D f (z0 )(z0 , z0 ), ∂z (w0 ) ≥ −t.
Proof. If f is not subordinate to g, then there exist points z0 ∈ Ω − {0}, w0 ∈ ∂Ω with 0 <
ρ(z0 ) < 1, ρ(w0 ) = 1 such that f (z0 ) = g(w0 ) and f (Dr ) ⊂ g(Ω), where Dr = {z ∈ Cn :
ρ(z) < r} and r = ρ(z0 ).
Let w(z) = g −1 (f (z)). Then w : Dr → Ω is a holomorphic mapping with w(z) 6≡ 0 and
w(0) = 0 satisfying f (z) = g(w(z)) for z ∈ Dr . Hence
1 = ρ(w0 ) =
max ρ(w(z)).
ρ(z)≤ρ(z0 )
By a simple calculation, we have
Dw(z0 )(z0 ) = Dg(w0 )−1 Df (z0 )(z0 ),
Dg(w0 )−1 D2 f (z0 )(z0 , z0 ) = Dg(w0 )−1 D2 g(w0 )(Dw(z0 )(z0 ), Dw(z0 )(z0 ))
+ D2 w(z0 )(z0 , z0 ).
From (2.1), there is a real number t ≥ 1/2 such that
∂ρ
Dw(z0 )(z0 ), (w0 ) = tρ(w0 ) = t.
∂z
So we obtain
∂ρ
Dg(w0 ) Df (z0 )(z0 ), (w0 ) = t,
∂z
−1
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 104, 15 pp.
http://jipam.vu.edu.au/
8
Y U -C AN Z HU AND M ING -S HENG L IU
and
∂ρ
Re Dg(w0 ) D f (z0 )(z0 , z0 ), (w0 )
∂z
∂ρ
−1 2
= Re Dg(w0 ) D g(w0 )(Dw(z0 )(z0 ), Dw(z0 )(z0 )), (w0 )
∂z
∂ρ
+ Re D2 w(z0 )(z0 , z0 ), (w0 )
∂z
∂ρ
−1 2
≥ Re Dg(w0 ) D g(w0 )(a, a), (w0 )
∂z
( n
)
n
2
X ∂2ρ
X
∂ ρ
+ Re
(w0 )aj al −
(w0 )aj al − t,
∂zj ∂z l
∂zj ∂zl
j,l=1
j,l=1
−1
2
where a = Dw(z0 )(z0 ) = (a1 , a2 , . . . , an ). If we let b = (b1 , b2 , . . . , bn ) with bj = iaj , then we
have
∂ρ
∂ρ
Re b, (w0 ) = Re i Dw(z0 )(z0 ), (w0 )
= Re{it} = 0.
∂z
∂z
From Lemma 2.5, we obtain
∂ρ
−1 2
Re Dg(w0 ) D f (z0 )(z0 , z0 ), (w0 )
∂z
∂ρ
−1 2
≥ − Re Dg(w0 ) D g(w0 )(b, b), (w0 )
∂z
)
( n
n
2
X
X ∂2ρ
∂ ρ
(w0 )bj bl − t
+ Re
(w0 )bj bl +
∂zj ∂zl
∂zj ∂z l
j,l=1
j,l=1
≥ −t.
This completes the proof.
Remark 2.9. When n = 1, Ω is the unit disc in the complex plane C and ρ(z) = |z|(z ∈ C), we
may obtain Lemma 1 in [14] from Theorem 2.8. Theorem 2.8 will play a key role in studying
some second order differential subordinations of holomorphic mappings on a bounded convex
balanced domain Ω in Cn .
Let Ω1 be a set of Cn , and let h be a biholomorphic convex mapping on Ω. Suppose that ρ(z)
is twice differentiable in Ω − {0}. We define Ψ(Ω1 , h) to be the class of maps ψ : Cn × Cn ×
Cn × Ω → Cn that satisfy the following conditions:
(1) ψ(h(0), 0, 0, 0) ∈ Ω1 , and
D
E
D
E
∂ρ
−1
(2) ψ(α, β, γ, z) ∈
/ Ω1 for α = h(w), Dh(w)−1 (β), ∂ρ
(w)
=
t,
Re
Dh(w)
(γ),
(w)
∂z
∂z
≥ −t, and z ∈ Ω, where ρ(w) = 1 and t ≥ 1/2.
Theorem 2.10. Let ψ ∈ Ψ(Ω1 , h). If f ∈ H(Ω, Cn ) with f (0) = h(0) satisfies
(2.12)
ψ(f (z), Df (z)(z), D2 f (z)(z, z), z) ∈ Ω1
for all z ∈ Ω, then f (z) ≺ h(z).
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S ECOND O RDER D IFFERENTIAL S UBORDINATIONS OF H OLOMORPHIC M APPINGS
9
Proof. If f is not subordinate to h, then by Theorem 2.8, there exist points z0 ∈ Ω − {0},
w0 ∈ ∂Ω with 0 < ρ(z0 ) < 1, ρ(w0 ) = 1 and there is a real number t ≥ 1/2 such that
∂ρ
−1
f (z0 ) = h(w0 ),
Dh(w0 ) Df (z0 )(z0 ), (w0 ) = t,
∂z
and
∂ρ
−1 2
Re Dh(w0 ) D f (z0 )(z0 , z0 ), (w0 ) ≥ −t.
∂z
2
Set α = f (z0 ), β = Df (z0 )(z0 ), γ = D f (z0 )(z0 , z0 ), then according to the definition of
Ψ(Ω1 , h), we have
ψ(f (z0 ), Df (z0 )(z0 ), D2 f (z0 )(z0 , z0 ), z0 ) ∈
/ Ω1
which contradicts (2.12). Hence f (z) ≺ h(z), and the proof of Theorem 2.10 is complete. Theorem 2.11. Let A ≥ 0, h ∈ H(Ω, Cn ) be biholomorphic convex with h(0) = 0, and
let ψ(z) ∈ H(Ω, Cn ) with ψ(0) = 0. Suppose that ρ(z) is twice differentiable in Ω − {0},
k > 4kDh(0)−1 k, and ϕ, φ : Ω1 × Ω → C are holomorphic such that
Re φ(α, z) ≥ A + |ϕ(α, z) − 1| − Re[ϕ(α, z) − 1] + kρ(ψ(z))
for all (α, z) ∈ h(Ω) × Ω, where Ω1 is a domain of Cn with h(Ω) ⊂ Ω1 and kDh(0)−1 k =
sup ρ(Dh(0)−1 (ξ)). If f ∈ H(Ω, Cn ) with f (0) = 0 satisfies
ρ(ξ)≤1
AD2 f (z)(z, z) + φ(f (z), z)Df (z)(z) + ϕ(f (z), z)f (z) + ψ(z) ≺ h(z),
then f (z) ≺ h(z).
Proof. Without loss of generality, we may assume that f and g satisfy the conditions of Theorem
2.11 on Ω. If not, then we can replace f (z) by fr (z) = f (rz), ψ(z) by ψr (z) = ψ(rz), and h(z)
by hr (z) = h(rz), where 0 < r < 1. We would then prove fr (z) ≺ hr (z) for all 0 < r < 1. By
letting r → 1− , we obtain f (z) ≺ h(z).
Let
ψ(α, β, γ, z) = Aγ + φ(α, z)β + ϕ(α, z)α + ψ(z),
E
D
E
D
∂ρ
−1
(w)
=
t,
Re
Dh(w)
(γ),
(w)
≥ −t, where
and let α = h(w), Dh(w)−1 (β), ∂ρ
∂z
∂z
ρ(w) = 1, t ≥ 1/2. If we set
ψ(α, β, γ, z) = h(w) + λDh(w)(w),
then we have
λw = ADh(w)−1 (γ) + φ(α, z)Dh(w)−1 (β)
+ [ϕ(α, z) − 1]Dh(w)−1 h(w) + Dh(w)−1 (ψ(z)).
D
E
Since 2 w, ∂ρ
(w)
= ρ(w) = 1 from Remark 2.2, we obtain
∂z
∂ρ
∂ρ
−1
−1
(2.13) λ = 2A Dh(w) (γ), (w) + 2φ(α, z) Dh(w) (β), (w)
∂z
∂z
∂ρ
∂ρ
−1
−1
+ 2[ϕ(α, z) − 1] Dh(w) h(w), (w) + 2 Dh(w) (ψ(z)), (w) .
∂z
∂z
By Lemma 2.4, we have
2 Dh(w)−1 h(w), ∂ρ (w) − 1 ≤ 1.
∂z
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10
Y U -C AN Z HU AND M ING -S HENG L IU
By Lemma 2.7, we obtain
Re λ ≥ −2At + 2t Re φ(α, z) + Re[ϕ(α, z) − 1]
− |ϕ(α, z) − 1| − 4kDh(0)−1 kρ(ψ(z))
≥ (2t − 1){|ϕ(α, z) − 1| − Re[ϕ(α, z) − 1]}
+ (k − 4kDh(0)−1 k)ρ(ψ(z)) ≥ 0.
(2.14)
Now we verify that ψ(α, β, γ, z) ∈
/ h(Ω). Suppose not, then there exists w1 ∈ Ω such that
ψ(α, β, γ, z) = h(w1 ). From the result in [3, 7, 18, 19], we have
∂ρ
−1
− Re λ = 2 Re Dh(w) (h(w) − h(w1 )), (w) > 0,
∂z
which contradicts (2.14), hence ψ(α, β, γ, z) ∈
/ h(Ω). By Theorem 2.10, we obtain f (z) ≺
h(z), and the proof is complete.
Corollary 2.12. Let A ≥ 0, h ∈ H(Ω, Cn ) be biholomorphic convex with h(0) = 0, and let
ψ(z) ∈ H(Ω, Cn ) with ψ(0) = 0. Suppose that k > 4kDh(0)−1 k, ρ(z) is twice differentiable
in Ω − {0}, and B(z), C(z) ∈ H(Ω, C) satisfy
Re B(z) ≥ A + |C(z) − 1| − Re[C(z) − 1] + kρ(ψ(z))
for all z ∈ Ω, where kDh(0)−1 k = sup ρ(Dh(0)−1 (ξ)). If f ∈ H(Ω, Cn ) with f (0) = 0
ρ(ξ)≤1
satisfies
AD2 f (z)(z, z) + B(z)Df (z)(z) + C(z)f (z) + ψ(z) ≺ h(z),
then f (z) ≺ h(z).
Corollary 2.13. Let A ≥ 0, h ∈ H(Ω, Cn ) be biholomorphic convex. Suppose that ρ(z) is
twice differentiable in Ω − {0}, and B(z) ∈ H(Ω, C) with Re B(z) ≥ A for all z ∈ Ω. If
f ∈ H(Ω, Cn ) with f (0) = h(0) satisfies
AD2 f (z)(z, z) + B(z)Df (z)(z) + f (z) ≺ h(z),
then f (z) ≺ h(z).
Corollary 2.14. Let h ∈ H(Ω, Cn ) be biholomorphic convex with h(0) = 0. Suppose that ρ(z)
is twice differentiable in Ω − {0} and φ : Ω1 → C is holomorphic such that Re φ(h(z)) ≥ 0 for
all z ∈ Ω. If f ∈ H(Ω, Cn ) with f (0) = 0 satisfies
f (z) + φ(f (z))Df (z)(z) ≺ h(z),
then f (z) ≺ h(z).
Remark 2.15. When n = 1, we have Df (z)(z) = zf 0 (z) and D2 f (z)(z, z) = z 2 f 00 (z). From
Corollary 2.12, we may obtain Theorem 2 in [13], Theorem 3.1a in [15], Theorem 1 for case 1
in [12] and Theorem 1 in [14]. From Corollary 2.13, we may obtain Corollary 2.1 in [13].
Example 2.1. Let β > 0 and γ ∈ C with 2 Re γ ≥ β. The unit ball in Cn is denoted by
z
B = {z ∈ Cn : kzk < 1}. If u ∈ Cn with kuk = 1, then h(z) = 1−hz,ui
is a biholomorphic
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S ECOND O RDER D IFFERENTIAL S UBORDINATIONS OF H OLOMORPHIC M APPINGS
11
convex mapping on B (see [17]). By a simple calculation, we have
z
Re γ|1 − hz, ui|2 + β[Rehz, ui − |hz, ui|2 ]
(2.15)
Re β
,u + γ =
1 − hz, ui
|1 − hz, ui|2
β|1 − hz, ui|2 + 2β[Rehz, ui − |hz, ui|2 ]
≥
2|1 − hz, ui|2
β(1 − |hz, ui|2 )
=
>0
2|1 − hz, ui|2
for all z ∈ B. If f ∈ H(B, Cn ) with f (0) = 0, then by Corollary 2.14, we have
f (z) +
Df (z)(z)
z
z
≺
=⇒ f (z) ≺
.
βhf (z), ui + γ
1 − hz, ui
1 − hz, ui
Example 2.2. Let A ≥ 0, β ≥ 0 and γ ∈ C with Re γ ≥ β/2 + A, u ∈ Cn with kuk = 1. If
f ∈ H(B, Cn ) with f (0) = 0, then by Theorem 2.11, Corollary 2.13 and (2.15), we have
hz, ui
z
z
2
AD f (z)(z, z) + β
+ γ Df (z)(z) + f (z) ≺
=⇒ f (z) ≺
,
1 − hz, ui
1 − hz, ui
1 − hz, ui
and
AD2 f (z)(z, z) + [βhf (z), ui + γ]Df (z)(z) + f (z) ≺
z
z
=⇒ f (z) ≺
.
1 − hz, ui
1 − hz, ui
Let ρ(z) be differentiable in Ω − {0}. For M > 0, we define Ψ(M ) to be the class of maps
ψ : Cn × Cn × Cn × Ω → Cn that satisfy the following conditions:
(1) ρ(ψ(0, 0, 0, 0)) < M and
D
E
D
E
∂ρ
(2) ρ(ψ(α, β, γ, z)) ≥ M for all ρ(α) = M, 2 β, ∂ρ
(α)
=
tM,
2
Re
γ,
(α)
≥
∂z
∂z
(t2 − t)M, and t ≥ 1.
Theorem 2.16. Let ψ ∈ Ψ(M ). If w ∈ H(Ω, Cn ) with w(0) = 0 satisfies
ρ(ψ(w(z), Dw(z)(z), D2 w(z)(z, z), z)) < M
(2.16)
for all z ∈ Ω − {0}, then ρ(w(z)) < M for z ∈ Ω.
Proof. Suppose that the conclusion of Theorem 2.16 is false. Then there exists a point z0 ∈ Ω −
{0} such that ρ(w(z0 )) = M and ρ(w(z)) ≤ M for ρ(z) ≤ ρ(z0 ). It implies w0 = w(z0 ) 6= 0.
Let
z0
∂ρ
ϕ(ξ) = 2 w
ξ , (w0 ) ,
ξ ∈ C.
ρ(z0 )
∂z
Then ϕ(ξ) is an analytic function in |ξ| < 1. By Lemma 2.6, we have
z0
∂ρ
z0
|ϕ(ξ)| ≤ 2 w
ξ , (w0 ) ≤ ρ w
ξ
≤ ρ(w(z0 ))
ρ(z0 )
∂z
ρ(z0 )
for all |ξ| ≤ ρ(z0 ), and
∂ρ
ϕ(ρ(z0 )) = 2 w(z0 ), (w0 ) = ρ(w(z0 )) = max |ϕ(ξ)|.
|ξ|≤ρ(z0 )
∂z
By a simple calculation, we have
∂ρ
ρ(z0 )ϕ (ρ(z0 )) = 2 Dw(z0 )(z0 ), (w0 ) ,
∂z
0
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12
Y U -C AN Z HU AND M ING -S HENG L IU
∂ρ
ρ(z0 ) ϕ (ρ(z0 )) = 2 D w(z0 )(z0 , z0 ), (w0 ) .
∂z
Using Lemma A in [10] (also see [15, p. 19]), there exists a real number t ≥ 1 such that
∂ρ
2 Dw(z0 )(z0 ), (w0 ) = tM,
∂z
∂ρ
2
2 Re D w(z0 )(z0 , z0 ), (w0 ) ≥ (t2 − t)M.
∂z
2
00
2
By the definition of ψ, we have
ρ(ψ(w(z0 ), Dw(z0 )(z0 ), D2 w(z0 )(z0 , z0 ), z0 )) ≥ M,
which contradicts (2.16). Hence ρ(w(z)) < M for z ∈ Ω, and the proof is complete.
Theorem 2.17. Let ρ(z) be differentiable in Ω−{0}. Suppose that A(z), B(z), C(z) ∈ H(Ω, C)
with A(z) 6= 0 for all z ∈ Ω satisfy
B(z)
1
C(z) ρ(ϕ(z))
(2.17)
Re
≥ max −1,
− Re
+
,
A(z)
|A(z)|
A(z)
|A(z)|
or
C(z)
1
ρ(ϕ(z))
(2.18)
Re
≥
+
+1
A(z)
|A(z)|
|A(z)|
s
C(z)
1
ρ(ϕ(z))
B(z)
and 1 − 2 Re
−
−
≤ Re
≤ −1
A(z) |A(z)|
|A(z)|
A(z)
for all z ∈ Ω. If w(z) ∈ H(Ω, Cn ) with w(0) = 0 satisfies
ρ(A(z)D2 w(z)(z, z) + B(z)Dw(z)(z) + C(z)w(z) + ϕ(z)) < 1
for all z ∈ Ω, then ρ(w(z)) < 1 for z ∈ Ω.
Proof. Let
ψ(α, β, γ, z) = A(z)γ + B(z)β + C(z)α + ϕ(z),
E
E
D
∂ρ
(α)
≥ (t2 − t) and t ≥ 1. From (2.9) and
(α)
=
t,
2
Re
γ,
where ρ(α) = 1, 2 β, ∂ρ
∂z
∂z
(2.10), we have
−iθ ∂ρ
ρ(ψ(α, β, γ, z)) ≥ 2 ψ(α, β, γ, z)e , (α) ∂z
∂ρ
∂ρ
−iθ
= |A(z)|2 γ, (α) + B(z)e 2 β, (α)
∂z
∂z
∂ρ
∂ρ
−iθ
−iθ
ϕ(z), (α) + C(z)e 2 α, (α) + 2e
∂z
∂z
B(z)
C(z) ρ(ϕ(z))
2
≥ |A(z)| t + t Re
− 1 + Re
−
,
A(z)
A(z)
|A(z)|
D
where θ = arg A(z). Let
B(z)
C(z) ρ(ϕ(z))
L(t) = t + t Re
− 1 + Re
−
A(z)
A(z)
|A(z)|
2
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S ECOND O RDER D IFFERENTIAL S UBORDINATIONS OF H OLOMORPHIC M APPINGS
13
for t ≥ 1. Then we have
L0 (t) = 2t + Re
B(z)
− 1.
A(z)
If Re B(z)
≥ −1 for z ∈ Ω, then L0 (t) ≥ Re B(z)
+ 1 ≥ 0. Hence we obtain
A(z)
A(z)
min L(t) = L(1) = Re
t≥1
C(z) ρ(ϕ(z))
1
B(z)
+ Re
−
≥
.
A(z)
A(z)
|A(z)|
|A(z)|
It follows that ρ(ψ(α, β, γ, z)) ≥ 1.
If Re B(z)
≤ −1 for z ∈ Ω, then
A(z)
1
B(z)
min L(t) = L
1 − Re
t≥1
2
A(z)
2
1
B(z)
C(z) ρ(ϕ(z))
=−
Re
− 1 + Re
−
4
A(z)
A(z)
|A(z)|
1
≥
.
|A(z)|
It also follows that ρ(ψ(α, β, γ, z)) ≥ 1.
Hence we have ψ ∈ Ψ(1). From Theorem 2.16, we obtain ρ(w(z)) < 1 for z ∈ Ω.
Remark 2.18. Setting n = 1, ϕ(z) ≡ 0, A(z) ≡ A and C(z) = 1 − B(z) in Theorem 2.17, we
get Theorem 4 in [13].
Corollary 2.19. Suppose that B(z) ∈ H(B, C) and A ≥ 0 satisfy Re B(z) ≥ 0 for all z ∈ B.
If w(z) ∈ H(B, Cm ) with w(0) = 0 satisfy
kAD2 w(z)(z, z) + B(z)Dw(z)(z) + w(z)k < 1
for all z ∈ B, then kw(z)k < 1 for z ∈ B.
Example 2.3. Let k and n1 be positive integers and let α = (α1 , α2 , . . . , αm ) ∈ Cm . We define
k
αk = (α1k , α2k , . . . , αm
). Suppose A1 , A2 , . . . , An1 ∈ H(B, C) (n1 ≥ 2) satisfy
Re A1 (z) ≥
n1
X
|Ak (z)|
k=2
for z ∈ B, where B is the unit ball in Cn . If w(z) = (w1 (z), w2 (z), . . . , wm (z)) ∈ H(B, Cm )
with w(0) = 0 satisfies
2
n1
m X
n
n
2
X
X
X
∂ wυ (z)
∂wυ (z)
k
zj zl +
zj +
Ak (z)[wυ (z)] < 1
∂zj ∂zl
∂zj
υ=1 j,l=1
j=1
k=1
for all z ∈ B, then
In fact, if we let
Pm
υ=1
|wυ (z)|2 < 1 for all z ∈ B.
ψ(α, β, γ, z) = γ + β +
n1
X
Ak (z)αk
k=1
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14
Y U -C AN Z HU AND M ING -S HENG L IU
for kαk = 1, hβ, αi = t, Rehγ, αi ≥ t2 − t and t ≥ 1, then we have
n1
X
k
kψ(α, β, γ, z)k ≥ hγ, αi + hβ, αi + A1 (z) +
Ak (z)hα , αi
k=2
≥ Re{hγ, αi + hβ, αi + A1 (z)} −
n1
X
k=2
≥ t2 + Re A1 (z) −
n1
X
|Ak (z)|
m
X
!
|αj |k+1
j=1
|Ak (z)| ≥ 1.
k=2
Hence ψ ∈ Ψ(1) for ρ(z) =
1 for all z ∈ B.
qP
n
j=1
|zj |2 . According to Theorem 2.16, we have
Pm
υ=1
|wυ (z)|2 <
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