BY CERTAIN STARLIKE MAPPINGS
MAMORU NUNOKAWA, NESLIHAN UYANIK and SHIGEYOSHI OWA
Communicated by the former editorial board
Let A be the class of functions f(z) which are analytic in the open unit disk Uwithf(0) = 0 andf0(0) = 1. Denote the lengthL(r) of the image curve of
|z|=r <1 byf(z)∈ A. The object of the present paper is to considerL(r) for certain starlike mappingsf(z) satisfying some subordination inU.
AMS 2010 Subject Classification: Primary 30C45.
Key words: analytic function, starlike function, subordination.
1. INTRODUCTION Let Adenote the class of functions f(z) of the form
(1.1) f(z) =z+
∞
X
n=2
anzn
which are analytic in the open unit diskU={z∈C : |z|<1}. LetS∗ be the subclass ofA consisting of functionsf(z) satisfying
(1.2) Re
zf0(z) f(z)
>0 (z∈U).
LetC(r) denote the closed curve that is the image of the circle|z|=r <1 under the mappingw=f(z)∈ A. Further, letL(r) be the length ofC(r). We denote byA(r) the area enclosed by the curve C(r). For f(z)∈ S∗, Thomas [4] has shown that
(1.3) 2p
π A(r)5L(r)52p π A(r)
1 + log
1 +r 1−r
(05r <1).
In particular,L(r)∼2p
π A(r) (r→0).
Forf(z)∈ Aandg(z)∈ A,f(z) is said to be subordinate tog(z) if there exists an analytic function w(z) in U such that f(z) =g(w(z)) (z ∈ U). We denote this subordination by
REV. ROUMAINE MATH. PURES APPL.,58(2013),1,1–7
(1.4) f(z) ≺ g(z) (z∈U).
If g(z) is univalent in U, then the subordination (1.4) is equivalent tof(0) = g(0) and f(U) ⊂ g(U) ([2]). Using the subordinations, we discuss the length L(r) of C(r) by some functionsf(z) belonging to the class S∗.
2. LENGTH OF THE CLOSED CURVE
In order to discuss our problems, we have to recall here the following lemmas.
Lemma 1([1]). If f(z)∈ Aand g(z)∈ Asatisfy f(z) ≺ g(z) (z∈U), then, forp >0 andz=reiθ (0< r <1),
(2.1)
Z 2π
0
|f(z)|pdθ 5 Z 2π
0
|g(z)|pdθ.
Lemma 2 ([3]). If F(z) is analytic in U and ReF(z) >0 (z ∈ U), and if |F(0)|51, then
(2.2) 1
2π Z 2π
0
|F(reiθ)|2dθ 5 1 + 3r2 1−r2 and
(2.3) 1
2π Z 2π
0
|F0(reiθ)|dθ 5 2 1−r2 for 05r <1.
Applying the above lemmas, we derive Theorem 1. If f(z)∈ Asatisfies (2.4) 1 +zf00(z)
f0(z) ≺ 1 + 4z+z2
1−z2 (z∈U) or
(2.5) 1 +zf00(z)
f0(z) − 1 +z2
1−z2 ≺ 4z
1−z2 (z∈U) then
(2.6) L(r) =O
A(r)log 1
1−r 12
as r→1, where O means the Landow’s symbol.
Proof. Note that (2.7) |zf0(z)|=
Z z 0
(tf0(t))0dt
=
Z z 0
(f0(t) +tf00(t)) dt
5 Z r
0
|f0(t) +tf00(t)|dρ.
This implies that, forz=reiθ and 05ρ5r <1, (2.8) L(r) =
Z 2π
0
|zf0(z)|dθ5 Z 2π
0
Z r 0
|f0(z) +zf00(z)|dρdθ
= Z 2π
0
Z r
0
|f0(z)|
1 +zf00(z) f0(z)
dρdθ
5 Z 2π
0
Z r 0
|f0(z)|2dρdθ
12 Z 2π
0
Z r 0
1 +zf00(z) f0(z)
2
dρdθ
!1
2
by Cauchy-Schwarz inequality. Let us consider r1 such that 0< r1 < r < 1.
Then
(2.9) I1 =
Z 2π
0
Z r
0
|f0(z)|2dρdθ= Z 2π
0
Z r1
0
|f0(z)|2dρdθ
+ Z 2π
0
Z r
r1
|f0(z)|2dρdθ5C1+ Z 2π
0
Z r
r1
ρ
r1|f0(z)|2dρdθ 5C1+ 1
r1 Z 2π
0
Z r 0
ρ|f0(z)|2dρdθ=C1+ 1 r1A(r), whereC1 is a bounded constant.
If f(z) satisfies the subordination (2.4), then Lemma 1 and Lemma 2 give us that
(2.10) I2= Z 2π
0
Z r 0
1 +zf00(z) f0(z)
2
dρdθ5 Z r
0
Z 2π
0
1 + 4z+z2 1−z2
2
dθdρ
<36 Z r
0
Z 2π
0
1 1−z2
2
dθdρ572π Z r
0
1 + 3ρ2 1−ρ2 dρ
<288π Z r
0
1
1−ρ2dρ= 144πlog
1 +r 1−r
.
Further, iff(z) satisfies the subordination (2.5), then we also have that I2=
Z 2π
0
Z r 0
1 +zf00(z) f0(z)
2
dθdρ5
5 Z r
0
Z 2π
0
1 +zf00(z)
f0(z) −1 +z2 1−z2
+
1 +z2 1−z2
2
dθdρ
= Z r
0
Z 2π
0
1 +zf00(z)
f0(z) −1 +z2 1−z2
2
dθdρ
+2 Z r
0
Z 2π
0
1 +zf00(z)
f0(z) −1 +z2 1−z2
1 +z2 1−z2
dθdρ+ Z r
0
Z 2π
0
1 +z2 1−z2
2
dθdρ
5 Z r
0
Z 2π
0
4z 1−z2
2
dθdρ+ Z r
0
Z 2π
0
1 +z2 1−z2
2
dθdρ
+2 (Z r
0
Z 2π
0
1 +zf00(z)
f0(z) − 1 +z2 1−z2
2
dθdρ )12(
Z r 0
Z 2π
0
1 +z2 1−z2
2
dθdρ )12
516 Z r
0
Z 2π
0
1
|1−z2|2 dθdρ+ 4 Z r
0
Z 2π
0
1
|1−z2|2dθdρ +2
(Z r 0
Z 2π
0
4z 1−z2
2
dθdρ )12
4 Z r
0
Z 2π
0
1
|1−z2|2dθdρ 12
5128π Z r
0
1
1−ρ2 dρ+ 32π Z r
0
1
1−ρ2 dρ+ 2
128π Z r
0
1 1−ρ2dρ
12
×
32π Z r
0
1 1−ρ2dρ
12
= 288π Z r
0
1
1−ρ2dρ= 144πlog
1 +r 1−r
.
Thus, we have that (2.11) L(r)5p
I1I2<
C1+ 1 r1A(r)
1
2
144πlog
1 +r 1−r
1
2
,
which means that
L(r) =O
A(r)log 1
1−r 1
2
(as r →1).
This completes the proof of the theorem.
Example1. Let us consider the Koebe function f(z) given by
(2.12) f(z) = z
(1−z)2 (z∈U).
Then, we have that
1 +zf00(z)
f0(z) = 1 + 4z+z2
1−z2 (z∈U)
and that
1 +zf00(z)
f0(z) −1 +z2
1−z2 = 4z
1−z2 (z∈U).
Therefore, we say that the Koebe function f(z) given by (1.12) satisfies the subordinations in (2.4) and (2.5) of Theorem 1.
Remark1. We note that the subordination (2.5) is equivalent to (2.13)
1 +zf00(z)
f0(z) −1 +z2 1−z2
<
4z 1−z2
(z∈U), because the function g(z) = 4z
1−z2 is starlike with respect to the origin in U, that is, univalent in U. Indeed, g(z) maps the open unit disk U onto the domain
g(U) =u+ iv=C−D,
with D = {u+ iv= ik : |k|=2}. This shows that we have the same result forf(z) whenf(z) satisfies the inequality (2.12). We give the domaing(U) as follows:
Remark 2. The subordination (2.4) in Theorem 1 shows us that zf0(z)
f(z) ≺ 1 +z
1−z (z∈U)
by the result due to Miller and Mocanu [2]. Therefore, we say that the function f(z) satisfying the subordination (2.4) is starlike in U. We also see that the functiong(z) given by
(2.14) g(z) = 1 + 4z+z2
1−z2
is starlike in U and maps the open unit disk U onto the following starlike domain.
REFERENCES
[1] J.E. Littlewood, On inequalities in the theory of functions. Proc. London Math. Soc.
23(1925), 481–519.
[2] S.S. Miller and P.T. Mocanu, Differential Subordinations. Theory and Applications, Marcel Dekker, New York, Basel, 2000.
[3] C. Pommerenke, On starlike and close-to-convex functions. Proc. London Math. Soc.
13(1963), 290–304.
[4] D.K. Thomas,A note on starlike functions. J. Lond. Math. Soc. 43(1968), 703–706.
Received 11 April 2012 University of Gunma
Hoshikuki-Cho 798-8, Chiba-Ward Chiba 260-0808
Japan
mamoru nuno@doctor.nifty.jp Atat¨urk University
Kazim Karabekir Faculty of Education Department of Mathematics
Erzurum T-25240 Turkey nesuyan@yahoo.com
Kinki University Department of Mathematics Higashi-Osaka, Osaka 577-8502
Japan
shige21@ican.zaq.ne.jp
