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STUDIA UNIV. “BABEŞ–BOLYAI”, MATHEMATICA, Volume LII, Number 3, September 2007
A COUNTER-EXAMPLE CONCERNING STARLIKE FUNCTIONS
RÓBERT SZÁSZ
Dedicated to Professor Petru Blaga at his 60th anniversary
Abstract. Let A denote the Alexander integral operator, let C and S ∗
denote the class of close-to-convex functions and the class of starlike functions, respectively. In the paper it is proved that the inclusion A(C) ⊂ S ∗
does not hold.
1. Introduction
Let A be the class of analytic functions defined in the unit disk
U = {z ∈ C : |z| < 1}
and having the form f (z) = z + a2 z 2 + a3 z 3 + . . .
The analytic description of the class S ∗ and C are given by
zf 0 (z)
> 0, z ∈ U
S ∗ = f ∈ A | Re
f (z)
and
zf 0 (z)
∗
> 0, z ∈ U .
C = f ∈ A | ∃ g ∈ S : Re
g(z)
In this article we discuss a relation between these two classes which involve the integral
operator of Alexander defined by
Z
A(f )(z) =
0
z
f (t)
dt.
t
In [2] pp. 310 and [3] pp. 361 the authors have proved the following theorem:
Received by the editors: 09.11.2006.
2000 Mathematics Subject Classification. 30C45.
Key words and phrases. the operator of Alexander, starlike functions, close-to-convex functions.
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RÓBERT SZÁSZ
Theorem 1. Let g ∈ A be a function which has the property:
z(zg 0 (z))0 zg 0 (z) Re
≥ Im
, z ∈ U.
g(z)
g(z)
If f ∈ A and
Re
zf 0 (z)
> 0, z ∈ U
g(z)
Re
zf 0 (z)
> 0, z ∈ U
g 0 (z)
or
then F = A(f ) ∈ S ∗ .
This Theorem rises the question, if the following better results can be valid:
A(C) ⊂ S ∗
2. Preliminaries
To prove the main result, we need the Lemma
Lemma 1. [1] pp. 18 (2.8)
The functions
f (z) =
z − 21 (x + y)z 2
, |x| = |y| = 1
(1 − yz)2
belong to class C.
These functions are the extreme points of class C if x 6= y.
3. The Main Result
Theorem 2.
A(C) 6⊂ S ∗ .
Proof. We shall prove that there are two complex numbers x, y ∈ ∂U, x 6= y so that
A(f ) 6∈ S ∗
where
f (z) =
168
z − 21 (x + y)z 2
(1 − yz)2
A COUNTER-EXAMPLE CONCERNING STARLIKE FUNCTIONS
and
Z
A(f )(z)
z
=
0
1 − 12 (x + y)t
dt =
(1 − yt)2
0
1
x
z
1
x log(1 − yz)
1−
−
1+
.
2
y 1 − yz
2
y
y
Z
A(f )(z)
=
=
f (t)
dt.
t
z
The branch of log(1 − yz) is chosen so that Im (log(1 − y(z)) ∈ [−π, π].
Let F be the function defined by the equality:
x
2
−
1
+
0
y yz
z(A(f )) (z)
F (z) =
=
A(f )(z)
1 − x (1 − yz) − 1 + x (1 − yz)2 log(1−yz)
y
y
yz
If A(f ) ∈ S ∗ then Re F (z) > 0, z ∈ U and from the continuity it follows that
Re F (z) ≥ 0 for every z ∈ ∂U for which F (z) is defined.
(1)
We will prove that the assertion (1) is not valid.
If we introduce the notations
x
= cos u + i sin u
y
yz = cos α + i sin α
we get that
F (z)
=
=
1 − i tg u2 − cos α − i sin α
=
−i tg u2 (1 − cos α − i sin α) + 4 sin2 α2 log(1 − cos α − i sin α)
1 − cos α − i tg u2 + sin α
4 sin2 α2 ln 2 sin α2 − tg u2 sin α − 2i sin2 α2 π − α + tg u2
Re F (z) =
(1 − cos α) − tg u2 sin α + 4 sin2 α2 ln 2 sin α2 + tg u2 + sin α 2 sin2 α2 tg
=
2
2
− tg u2 sin α + 4 sin2 α2 ln 2 sin α2
+ 4 sin4 α2 tg u2 + π − α
u
2
+π−α
The numerator of Re F (z) is a polinomial of degree two with respect to tg u2 .
The discriminant of the polinomial is
αh
α α i
∆(α) = 4 sin4
(π − α)2 − 4(π − α) sin α − 16 sin2 ln 2 sin
.
2
2
2
169
RÓBERT SZÁSZ
Because
∆(α)
= 4π 2
α→0 sin4 α
2
lim
there are α ∈ (−π, π), for which ∆(α) > 0.
The inequality ∆(α) > 0 for some α ∈ (−π, π), means that there are two
points x, y ∈ ∂U, x 6= y and z ∈ ∂U so that Re F (z) < 0, which contradicts (1).
References
[1] Hallenbeck, D.J., MAc Gregor, T.H., Linear problems and convexity techniques in geometric function theory, Pitman Advanced Publiching Program (1984).
[2] Miller, S.S., Mocanu, P.T., Differential Subordinations, Marcel Decker Inc. New York.
Basel, 2000.
[3] Mocanu, P.T., Bulboacă, T., Sălăgean, G.S., Teoria geometrică a funcţiilor univalente,
Casa Cărţii de Ştiinţă, Cluj-Napoca, 1991.
Str. Rovinari, Bl. 32/b/14, Tg. Mureş, Romania
E-mail address: szasz [email protected]
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