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Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 570324, 18 pages
doi:10.1155/2012/570324
Research Article
A Fundamental Inequality of Algebroidal Function
Yingying Huo1, 2 and Daochun Sun1, 2
1
School of Applied Mathematics, Guangdong University of Technology, Guangdong,
Guangzhou 510520, China
2
School of Mathematics, South China Normal University, Guangdong, Guangzhou 510631, China
Correspondence should be addressed to Yingying Huo, fs hyy@yahoo.com.cn
Received 8 September 2012; Accepted 21 October 2012
Academic Editor: Ahmed El-Sayed
Copyright q 2012 Y. Huo and D. Sun. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
By using a new mapping of Ahlfors covering surfaces, a fundamental inequality in the angular
domain for the algebroidal function is obtained.
1. Introduction and Main Results
In the field of valued distribution, the fundamental inequality is an important tool. For
example, it can be used to investigate the singular direction 1. Using geometric theory,
Tsuji firstly obtained the second fundamental theory in an angular domain and proved the
existence of Borel direction 2. The value distribution theory of meromorphic functions was
extended to algebroidal functions last century 3. In 1983, Lv and Gu proved an inequality
of algebroidal function for an angular domain 4. By the inequality, some results of singular
direction are obtained; see 5, 6. In 7, the authors obtained a more accurate inequality for
angular domain. In this paper, we will use a new method to simplify and extent an inequality
of Tsuji to algebroidal functions.
First, we recall some definitions from 3.
Suppose that Av z, . . . , A0 z are analytic functions with no common zeros in the
complex plane. Ψz, W is a bivariate complex function and satisfies
Ψz, W Av zW v Av−1 zW v−1 · · · A0 z 0.
1.1
For all z in the complex plane, the equation Ψz, W 0 has v complex roots w1 z, w2 z,
. . . , wv z. Then, 1.1 defines a v-valued algebroidal function Wz; see 3, 8. If Av z 1, then Wz is called v-valued integral algebroidal function. If Ψz, W is irreducible,
2
Abstract and Applied Analysis
correspondingly Wz is called v-valued irreducible algebroidal function note that Wz is
a meromorphic function, if v 1. Now we suppose that Wz is an irreducible algebriodal
function defined by 1.1.
If Av z0 / 0, and the k-degree equation Ψz0 , W 0 and its partial derivative
∂Ψ/∂Wz0 , W 0 have no common roots i.e., z0 is not a multiple root of Ψz0 , W 0,
then z0 is said to be a regular point. The set of all regular points is called the regular set,
denoted by TW . Its complementary set SW : {z | |z| < ∞} − TW is called the critical set.
Obviously, Sw includes all branch points of Wsee 3.
The domain of a v-valued irreducible algebriodal function W is a connected Riemann
z is z and sets lying
z . A point in R
surface 8, and its single-valued domain is denoted by R
over |z| < r and {φ1 < arg z < φ2 } φ1 < φ2 are |
z| < r and Ωφ1 , φ2 . Let nr, W a and
nΩφ1 , φ2 , r, W a be the number of zeros, counted according to their multiplicities, of
W a in {|
z| < r} and {|
z| < r} Ωφ
1 , φ2 , respectively. Let nr, W a be the number of
z| < r}. Simidistinct zeros in {|
z| < r}, and let nr, Rz be the number of branch points in {|
z . Let
larly, we can define nΩφ1 , φ2 , r, W a and nΩφ1 , φ2 , r, R
1
Sr, W π
|
z|≤r
1
S Ω φ1 , φ2 , r, W π
|W z|
2
1 |Wz|2
dW
{|
z|≤r}
Ωφ1 , φ2 z reiθ ,
|W z|
1 |Wz|2
2
dW,
r
St, W
dt,
t
0
1 r S Ω φ1 , φ2 , t, W
T Ω φ1 , φ2 , r, W dt,
v 0
t
1 r nt, W a − n0, W a
n0, W a
Nr, W a dt ln r,
v 0
t
v
z
z − n 0, R
z
n 0, R
1 r n t, R
z dt ln r,
N r, R
v 0
t
v
v 2π
1 mr, W ln wk reiθ dθ.
2πv k1 0
1
T r, W v
1.2
Similarly, we can define NΩφ1 , φ2 , r, W a, NΩφ1 , φ2 , r, W a and NΩφ1 ,
z . From 3, we know that T r, w Nr, W ∞ mr, W O1 and Nr, R
z ≤
φ2 , r, R
2v − 1T r, W O1.
In this paper, we will prove the main theorem.
Theorem 1.1. Let Wz be a v-valued algebroidal function in region Ωφ1 , φ2 {|z| | φ1 <
arg z < φ2 } φ1 < φ2 . a1 , a2 , . . . , aq q 3 are q different complex numbers on the sphere with
Abstract and Applied Analysis
3
radius δ ∈ 0, 1/2. For φ, ε∗ , ε 0 < ε∗ < ε, φ1 < φ − ε < φ − ε∗ < φ ε∗ < φ ε < φ2 , and
R > R∗ > 2, we have
q − 2 S Ω φ − ε∗ , φ ε∗ , R∗ , W
q
z n Ω φ − ε, φ ε , R, W aj
n Ω φ − ε, φ ε , R, R
1.3
j1
256 vπ 24 ln R
∗
∗ 1
38
q − 2 S Ω φ − ε ,φ ε , ∗,W
.
R
δ ε − ε∗ ln R − ln R∗ By the inequality in Theorem 1.2, we will immediately have the following.
Theorem 1.2. For a meromorphic function W (a 1-valued algebroidal function with no branch points)
defined by 1.1 satisfying
lim
T r, W
R→∞
ln2 R
1.4
∞,
φ0 , such that
N Ω φ0 − ε, φ0 ε , R, W a
δ a, φ0 1 − lim lim
> 0.
ε→0 R→∞
T Ω φ0 − ε, φ0 ε , R, W
1.5
it has at least one Nevanlinna direction, that is, there exists arg z
a∈C {∞} δa, φ0 ≤ 2 holds for any finitely many deficient value a, where
2. Some Lemmas
First, it is easy to prove the following.
Lemma 2.1. Suppose that a, b > 0, then there exists p, q > 0, such that
p iq 1
,
a − bi
2.1
where
√
p
Lemma 2.2. Suppose that Az a2 b2 a
,
2a2 b2 z
0
√
a2 b2 − a
q
.
2a2 b2 2.2
dt/ t1 − t2 and Bz 1 − z/1 zi, then
1 the mapping G1 A ◦ Bz maps the unit disc U to the square Q {z x iy | 0 < x <
2h, 0 < y < 2h}, where h is constant;
2 G1 maps {|z| < r}, where 0 < r < 1, into a symmetrical convex region in Q;
4
Abstract and Applied Analysis
Q
U
G1
Figure 1
3 h in Lemma 2.2 satisfies
h
1
√
2−1
dt
t1 −
t2 √2−1
0
dt
> 1.
t1 − t2 2.3
Proof. Obviously, 1 holds. By the definition of Bz, we have B1 0, B−1 ∞, Bi 1, B−i −1.
In order to compute h, first we prove that G1 maps d0, dπ/2, dπ/4, d−π/4 to
four symmetry axes of Q, where dθ {reiθ ; −1 < r < 1}. Let z reiθ θ is fixed, r ∈ −1, 1,
G1 z A ◦ Bz B
0
− 1 i
r
1
√
dt
t1 − t2 eiθ dr
1 − r 4 ei4θ
2.4
.
Hence, when θ 0, π/4, π/2, −π/4, arg G1 reiθ π/4, π/2, −π/4, 0, respectively,
see Figure 1.
Then, we compute h. Since z 0 is the only intersection point of the lines d0 and
dπ/4. The center of the square Q, h hi, is that of the curves G1 d0 and G1 dπ/4.
Then, G1 conforms 0 onto h hi.
Hence,
h G1 e
iπ/4
√
A
2−1 √
2−1
0
1
√
2−1
1
>
√
2−1
dt
t1 − t2 dt
t1 − t2 dt
√ > 1.
t
2 At last we prove that G1 {|z| < r} is a convex region, see Figure 1.
2.5
Abstract and Applied Analysis
5
For a fixed r ∈ 0, 1, by 2.4,
G1 re
iθ
z
−1 idz
√
1 − z4
1
1
dr
1 i
√
1 − r4
r
θ
1 − idθ
.
−2
2
−
r
cos
2θ − ir −2 r 2 sin 2θ
r
0
2.6
Set G1 reiθ xθ iyθ. When θ ∈ 0, π/4, cos 2θ, sin 2θ > 0, by Lemma 2.1,
1−i
∂G1
1 − i p iq
−2
2
−2
2
∂θ
r − r cos 2θ − ir − r sin 2θ
p q − p − q i x y i,
2.7
1 − i r −2 − r 2 sin 2θ i r −2 r 2 cos 2θ
∂2 G1
∂θ2
3
r −2 − r 2 cos 2θ − ir −2 r 2 sin 2θ
1 − i p qi r −2 − r 2 sin 2θ i r −2 r 2 cos 2θ
r −2 − r 2 cos 2θ − ir −2 r 2 sin 2θ
p − q r −4 − r 4 − 2 p q sin 4θ
2
2
r −2 − r 2 cos2 2θ r −2 r 2 sin2 2θ
2 p − q sin 4θ p q r −4 − r 4
i
x y i.
2
2
2
−2
2
2
−2
2
r − r cos 2θ r r sin 2θ
2.8
where p > q > 0.
Hence,
dy y
p−q
−
,
dx x
pq
d2 y y x − y x
dx2
x 2
2.9
2 r −4 − r 4 p2 q2
> 0.
x 2 r −2 − r 2 cos2 2θ r −2 r 2 2 sin 2θ
Therefore, the image of Lr {reiθ ; 0 < θ < π/4} is a descending convex curve. By
the symmetry of square, G1 {|z| r} is a smooth curve, and G1 {|z| < r < 1} is a convex
symmetric figure in the square Q.
Then, we obtain Lemma 2.2.
6
Abstract and Applied Analysis
Lemma 2.3. Suppose that mappings
Cz h iz − h where h is defined in Lemma 2.2,
Dz, z log R − ε
yε log R ε
x ln R
i
z
z,
h
h
2h
2h
2.10
Hz e ,
z
G2 z C−1 ◦ D−1 ◦ H −1 z.
Then, the mapping G2 maps the region E {1/R < |z| < R} {| arg z| ε} into the square Q, and
∗
∗
∗
∗
∗
∗
∗
G−1
1 ◦ G2 E ⊂ {|z| < r}, where E {1/R < |z| < R } {| arg z| ε }, R ∈ 2, R, ε ∈ 0, ε,
and
ε − ε∗ 2 ε − ε∗ ln R − ln R∗ .
,
0 < 1 − r < min
4πε ln R
2πε2
2.11
Proof. The conclusion is equivalent to G2 E∗ ⊂ G1 {|z| < r}. We prove the lemma by two
cases.
Case I. When
ln R∗
ε∗
,
ε
ln R
2.12
then
0 arg G2
hε∗
h−
ε ih − h ln R∗ / ln R
ε − ε∗
ln R − ln R∗
π
i
arg h
.
ε
ln R
4
2.13
G1 {|z| < r < 1} is a convex symmetric figure if there exists a θ0 ∈ 0, π/4, such that
ε − ε∗
Re G1 reiθ0 < h
,
ε
ln R − ln R∗
,
Im G1 reiθ0 < h
ln R
2.14
then Lemma 2.3 holds, see Figure 2.
In fact, for any r ∈ 0, 1, θ ∈ 0, π/4,
G1 re
iθ
−
z
1
eiθ
1
1 idz
√
1 − z4
1 idz
√
1 − z4
reiθ
eiθ
1 idz
α β,
√
1 − z4
2.15
Abstract and Applied Analysis
7
Q
U
G1
G2
E
E∗
Figure 2
where
θ
−1 iieiθ
1 idz
dθ
α
√
√
1 − z4
1 − ei4θ
0
1
√ θ
θ
dθ
π
dθ <
√
√ πθ,
2 0 θ
sin 2θ
0
reiθ
1 idz
β
√
iθ
1 − z4
e
1
dr
1 i
r
1 − r 4 cos 2θ − i 1 r 4 sin 2θ
eiθ
1 i
1
√
r
dr
a − bi
2.16
,
where
a 1 − r 4 cos 2θ < 1,
b 1 r 4 sin 2θ,
a2 b2 1 r 8 2r 4 sin2 2θ − cos2 2θ
1−r
4
2.17
4r sin 2θ
4
2
> 4r 4 sin 2θ.
By Lemma 2.1, we have
β
1
1 i p iq dr r
ξ iη,
1
r
p − q p q i dr
2.18
8
Abstract and Applied Analysis
where
ξ
1
p − q dr r
1
r
1
p2 − q2
dr
pq
p2 q2
dr
p
r
1 √
2dr
<
√
2
a b2
r
1
dr
<
√
2
2 sin 2θ
r r
<
1−r
√
2r sin 2θ
1−r π
,
<
2r
θ
1
1
η
p q dr < 2 p dr
<
r
1
<2
2.19
r
√
4
dr
a2 b2
r
1−r π
.
<
r
2θ
Let
θ0 ε − ε∗ 2 π
< .
4
4πε2
Combing 3.1 note that by 3.1, we have r >
√
2.20
2/2,
1−r π
Re G1 reiθ0 α ξ < πθ0 2r
θ0
hε − ε∗ ,
ε
1−r π
iθ0
η<
Im G1 re
r
2θ
<
<
1 ln R − ln R∗
2
ln R
<h
ln R − ln R∗
.
ln R
2.21
Abstract and Applied Analysis
9
Q
U
G2
G1
E
E∗
Figure 3
Therefore, a vertex of G2 E∗ h−ε∗ h/ε, h−h ln R∗ / ln R ∈ G1 {|z| < r}. By Lemma 2.2,
G2 E ⊂ G1 {|z| < r}.
∗
Case II. When
ε∗ ln R∗
≤
,
ε
ln R
2.22
since G1 {|z| < r < 1} is a convex symmetric figure, we also have Lemma 2.3, see Figure 3.
For the convenience of readers, we prove the following lemma again, it can be found
in 9.
iφ
Lemma 2.4. (1) Let Gz G−1
2 ◦ G1 ze , then
|Gz | |Gz | ln R
|Gz | − |Gz |.
ε
2.23
(2) Put sx, y Re G, tx, y Im G, then
s2x s2y t2x t2y ln R sx ty − sy tx .
ε
2.24
Proof. 1 Since f H −1 ◦ G1 zeiφ and C−1 are holomorphic functions, then
−1
f z fz C
D
0,
−1
fz fz .
2.25
For
Dz yε
x ln R
i
h
h
2.26
10
Abstract and Applied Analysis
then
yh
xh
i
D−1 f, f ln R
ε
h f f
h f −f
.
2 ln R
ε
2.27
⎫
⎧
−1 −1 ⎪
⎪
⎪
D
D
⎬ |ln R ε| |ln R − ε| ln R
⎨ f
f ⎪
.
max ⎪ |ln R ε| − |ln R − ε|
⎪
ε
⎪
⎭
⎩ D−1 − D−1 ⎪
f
f
2.28
Hence,
Thus,
|Gz | |Gz | C−1
D
−1 C−1
D
f
z
f
−1
D
−1
D
f
z
−1
f
ln R −1 −1 D
−
f
D
z
f f
D−1
ε
ln R −1 −1 −1
C−1
.
−
D
f
D
f
C
z
z
f
−1
−1
f
D
D
ε
C−1
2.29
2 By
sx sz sz
2.30
sy sz − isz ,
we have
sz sx − isy
,
1−i
sz sy − isx
.
1−i
2.31
tz tx − ity
,
1−i
tz ty − itx
.
1−i
2.32
Similarly,
Then,
2 2 1 ty sx tx − sy ,
2
2 2 1 tx sy ty − sx .
|Gz |2 2
|Gz |2 2.33
Abstract and Applied Analysis
11
Therefore,
s2x s2y t2x t2y |Gz |2 |Gz |2
|Gz | |Gz |2
ln R by 2.23 |Gz |2 − |Gz |2
ε
2.34
ln R sx ty − sy tx .
ε
Lemma 2.5 see 10. Let F be a connected covering surface on F0 , F0 is bounded by q different
points with radius δ ∈ 0, 1/2, then
%
& max 0, ρF q − 2 S − 225 π 11 δ−19 L,
2.35
where L is the length of F and ρF is Euler characteristic of F, |F| is the area of F and
S
|F|
.
|F0 |
2.36
Lemma 2.6 see 10. Let V be a sphere with radius 1/2, F0 be bounded by q different points with
radius δ ∈ 0, 1/2 and Fr W ◦ GF0 then
v
1
|Fr |
S
|F0 | π k1
r 2π 2 wk sx ty − sy tx r
2 dr dθ,
0 0
1 |wk ◦ G|2
2.37
where sx, y Re G, tx, y Im G.
Proof. Suppose that wk u iv. Then
|Fr | v k1
G|
z|<r
1
1 |wk |2
2 du dv,
2.38
where
du us sx ut tx dx us sy ut ty dy,
dv vs sx vt tx dx vs sy vt ty dy.
2.39
12
Abstract and Applied Analysis
Hence, by the Jacobian determinant, we have
du dv |Fr | u2s vs2
v k1
sx ty − sy tx dx dy,
|
z|<r
u2s vs2 sx ty − sy tx
2 dx dy
1 |wk ◦ G|2
2.40
v r 2π w 2 s t − s t r
x y
y x
k
2 dr dθ.
2
k1 0 0
1 |wk ◦ G|
By |F0 | π, we have Lemma 2.6.
3. Proof of Theorem 1.1
Proof. Set Gz G−1
◦ G1 zeiφ . It conforms the unit disc U {|z| < 1} to the sector E 2
{1/R < |W| < R} {| arg z − φ| < ε} and the interior of U∗ {|z| < r} to E∗ {1/R∗ < |W| <
R∗ } {| arg z − φ| < ε∗ }, where 2 < R∗ < R, 0 < ε∗ < ε, and
ε − ε∗ 2 ε − ε∗ ln R − ln R∗ .
0 < 1 − r < min
,
4πε ln R
2πε2
3.1
Hence W ◦ G conforms {|z| < r} to the sphere V .
'r {|
Put D
z| < r}. Then by M. Hurwite Formula, we have
z − v.
'r n r, R
ρ D
3.2
'r − {z | (q w ◦ Gz − aj 0} and Fr W ◦ GF0 . Then
Put Dr D
j1
q
z − v n r, W ◦ G aj
ρFr ρDr n r, R
j1
q
z − v n 1, W ◦ G aj N.
n 1, R
3.3
j1
By Lemma 2.5, it follows that
N ρFr q − 2 Sr, W ◦ G − 227 π 11 δ−19 Lr.
3.4
Now we will prove
L2 r 24 vπr
ln R dSr, W ◦ G
.
ε
dr
3.5
Abstract and Applied Analysis
13
For any r ∈ 0, 1, k 1, 2, . . . , v and ε > 0, we have
wk ◦ G reiθ − wk ◦ Gj < ε,
3.6
where θj jπ/n j 1, 2, . . . , 2n, θ ∈ θj−1 , θj , |wk ◦ Gj | min{|wk ◦ Greiθ |, θj−1 θ θj }.
By 3.6, for any θ ∈ θj−1 , θj , we have
2
2 wk ◦ G reiθ wk ◦ Gj 2ε ε wk ◦ Gj .
3.7
2
1 wk ◦ G reiθ 2εwk ◦ Gj 2ε2
1
2
2 2
1 wk ◦ Gj 1 wk ◦ Gj 1 wk ◦ Gj √
2.
3.8
Therefore
Put wk uk s, t ivk s, t, G sr, θ itr, θ. Hence
wk ◦ G reiθj − wk ◦ G reiθj−1 2 2
n→∞
j1
1 wk ◦ G reiθj 1 wk ◦ G reiθj−1 Lk r lim
lim
2n
√
2
uk s sθ ut s tθ dθ
θj−1
2n
n→∞
)
θj
2n θj
n→∞
j1
θj−1
2
2
uk s sθ uk t tθ vk s sθ vk t tθ j 2
1 wk ◦ Gk 2π 2 2 1/2
w k sθ t θ
dθ
0
θ
j
vk s sθ vk t tθ dθ
θj−1
2
1 wk ◦ Gj j1
2 lim
2
1 |wk ◦ Gk |2
1/2
2 *1/2
3.9
dθ
.
By
sθ − sx r sin θ sy r cos θ,
tθ − tx r sin θ t − ty r sin θ,
3.10
where
x − r cos θ,
y r sin θ,
3.11
14
Abstract and Applied Analysis
we obtain
s2θ t2θ 2r 2 s2x s2y t2x t2y ,
L r 2
v
2
Lk r
k1
⎡
⎤2
1/2
v 2π w s2x s2y t2x t2y
r
k
⎢
⎥
8⎣
dθ⎦
2
1 |wk ◦ G|
k1 0
⎡
⎤2
v 2π w s t − s t 1/2 r
ln R ⎣
x
y
y
x
k
dθ⎦
by 2.24 8
2
ε
1
◦
G|
|w
0
k
k1
⎡
⎤2
1/2
2π v
wk sx ty − sy tx
r
ln R ⎣
dθ⎦
Cauchy inequality 8v
ε k1 0
1 |wk ◦ G|2
3.12
⎡
⎤
1
2
2π 2 v
wk sx ty − sy tx r ⎥ 2π
ln R ⎢
r dθ
Schwarz inequality 8v
⎣
2 dθ⎦
ε k1 0
0
1 |wk ◦ G|2
v 2π w 2 s t − s t r
ln R x y
y x
k
16vπr
2 dθ
ε k1 0
1 |wk ◦ G|2
ln R dSr, W ◦ G
.
by Lemma 2.6 16vπr
ε
dr
1 If for all r ∈ r, 1
Sr, W ◦ G N
,
q−2
3.13
then by 3.4
N 2
250 π 22
L r
S r ,W ◦ G −
2
q−2
q − 2 δ38
254 vπ 23 r ln R dSr , W ◦ G
,
by 3.5 2
dr q − 2 δ38 ε
3.14
that is,
dSr , W ◦ G
254 vπ 23 ln R
dr 2 .
2 38 q − 2 δ ε Sr , W ◦ G − N/ q − 2
3.15
Abstract and Applied Analysis
15
Hence,
1−r 1
r
254 vπ 23 ln R
dr 2
q − 2 δ38 ε
254 vπ 23 ln R
q − 2 δ38 ε
<
1
r
dSr , W ◦ G
2
Sr , W ◦ G − N/ q − 2
1
1
−
Sr, W ◦ G − N/ q − 2
SR, W ◦ G − N/ q − 2
3.16
254 vπ 23 ln R
1
.
38
δ ε
q − 2 Sr, W ◦ G − N
Therefore
254 vπ 23 ln R
q − 2 Sr, W ◦ G 38
N
δ ε1 − r
q
54
23
z n 1, W ◦ G aj 2 vπ ln R .
n 1, R
38
δ ε1 − r
j1
3.17
2 If there is a r ∈ r, 1, such that
q − 2 S r , W ◦ G − N < 0,
3.18
q − 2 Sr, W ◦ G < q − 2 S r , W ◦ G < N,
3.19
then
Equation 3.17 holds.
By 3.17 and Lemma 2.3, we have
1
q − 2 S Ω φ − ε∗ , φ ε∗ , R∗ , W − q − 2 S Ω φ − ε∗ , φ ε∗ , ∗ , W
R
q − 2 Sr, W ◦ G
q
z n Ω φ − ε, φ ε , R, W aj
by 3.1 n Ω φ − ε, φ ε , R, R
j1
256 vπ 24 ln R
.
δ38 ε − ε∗ ln R − ln R∗ 3.20
16
Abstract and Applied Analysis
4. Proof of Theorem 1.2
Proof. By the hypothesis of Theorem 1.2, there exists an increasing sequence Rn Rn → ∞,
when n → ∞, such that
lim
n→∞
T Rn , W
ln2 Rn
∞.
4.1
Then, there exist some φ0 ∈ 0, 2π, such that for arbitrary ε ∈ 0, φ0 ,
lim
T Rn , φ0 − ε, φ0 ε, W
ln2 Rn
n→∞
∞
4.2
holds. We claim that arg z φ0 is the Nevanlinna direction.
Otherwise, for a positive number ε0 , there exist some a1 , a2 , . . . , aq q ≥ 3, such that
q
δ aj , φ0 > 2 3ε0 .
4.3
j1
By the definition of δaj , φ0 , we have
q
j1
lim lim
ε → 0 R → ∞
N Ω φ0 − ε, φ0 ε , R, W aj
< q − 2 − 3ε0 .
T Ω φ0 − ε, φ0 ε , R, W
4.4
There exists ε1 > 0, such that for any ε ∈ 0, ε1 ,
q
j1
lim
R→∞
N Ω φ0 − ε, φ0 ε , R, W aj
< q − 2 − 2ε0 .
T Ω φ0 − ε, φ0 ε , R, W
4.5
Hence, for {Rn } defined earlier, when n is sufficiently large,
q
N Ω φ0 − ε, φ0 ε , Rn , W aj
j1
< q − 2 − ε0 T Ω φ0 − ε, φ0 ε , Rn , W .
4.6
Abstract and Applied Analysis
17
By Theorem 1.1, we have
ε
ε
, R, W
q − 2 S Ω φ − ,φ 2
2
ε
ε
− q − 2 S Ω φ − , φ , 1, W
2
2
q
n Ω φ − ε, φ ε , 2R, W aj
4.7
j1
252 π 24 ln 2R
.
δ38 ε ln 2
Hence,
ε
ε
, Rn , W
q − 2 T Ω φ − ,φ 2
2
q
N Ω φ − ε, φ ε , Rn , W aj
j1
4.8
54
24
2
2 π ln 2Rn
O1
δ38 ε − ε∗ ln 2
< q − 2 − ε0 O1 ln2 Rn .
Hence,
lim
n→∞
T Ω φ − ε/2, φ ε/2 , Rn , W
ln2 Rn
< O1,
4.9
which contradicts 4.2. Therefore, Theorem 1.2 holds.
Acknowledgments
The research is supported by the National Natural Science Foundation of China no.
11101096, Guangdong Natural Science Foundation no. S2012010010376, and the Startup
Foundation for Doctors of Guangdong University of Technology no. 083063.
References
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18
Abstract and Applied Analysis
5 Y. Huo and Y. Kong, “On filling discs in the strong Borel direction of algebroid function with finite
order,” Bulletin of the Korean Mathematical Society, vol. 47, no. 6, pp. 1213–1224, 2010.
6 Z.-X. Xuan and Z.-S. Gao, “The Borel direction of the largest type of algebroid functions dealing with
multiple values,” Kodai Mathematical Journal, vol. 30, no. 1, pp. 97–110, 2007.
7 Z.-X. Xuan and Z.-S. Gao, “On the Nevanlinna direction of an algebroid function dealing with
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