Sufficient conditions for bounded turning of analytic functions
Let function $f$ be analytic in the open unit disk and be normalized such that $f(0) = f\prime (0) 1 = 0$. In this paper methods from the theory of first order differential subordinations are used for obtaining sufficient conditions for $f$ to bewith bounded turning, i.e., the read part of its first...
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| author | Bulboacă, T. Tuneski, N. Бульбоака, Т. Тунеськи, Н. |
| author_facet | Bulboacă, T. Tuneski, N. Бульбоака, Т. Тунеськи, Н. |
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| description | Let function $f$ be analytic in the open unit disk and be normalized such that $f(0) = f\prime (0) 1 = 0$. In this paper methods from the theory of first order differential subordinations are used for obtaining sufficient conditions for $f$ to bewith bounded turning, i.e., the read part of its first derivative to map the unit disk onto the right half plane. In addition, several open problems are posed. |
| first_indexed | 2026-03-24T02:09:18Z |
| format | Article |
| fulltext |
UDC 517.5
N. Tuneski (Ss. Cyril and Methodius Univ. Skopje, Macedonia),
T. Bulboacă (Babeş-Bolyai Univ., Cluj-Napoca, Romania)
SUFFICIENT CONDITIONS FOR BOUNDED TURNING
OF ANALYTIC FUNCTIONS
ДОСТАТНI УМОВИ ДЛЯ ОБМЕЖЕНОГО ПОВОРОТУ АНАЛIТИЧНИХ
ФУНКЦIЙ
Let function f be analytic in the open unit disk and be normalized such that f(0) = f \prime (0) - 1 = 0. In this paper
methods from the theory of first order differential subordinations are used for obtaining sufficient conditions for f to be
with bounded turning, i.e., the read part of its first derivative to map the unit disk onto the right half plane. In addition,
several open problems are posed.
Нехай f — функцiя, аналiтична у вiдкритому одиничному крузi, нормована так, що f(0) = f \prime (0) - 1 = 0. Методи
теорiї диференцiальних пiдпорядкувань першого порядку застосовано, щоб отримати достатнi умови того, що
функцiя f має обмежений поворот, тобто дiйсна частина її першої похiдної вiдображає одиничний круг на праву
пiвплощину. Крiм того, сформульовано кiлька вiдкритих проблем.
1. Introduction and preliminaries. Let \scrH (\BbbD ) be the class of functions that are analytic in unit
disk \BbbD = \{ z \in \BbbC : | z| < 1\} and let \scrA denote the class of functions f \in \scrH (\BbbD ) of the form
f(z) = z + a2z
2 + a3z
3 + . . . , z \in \BbbD .
The class of starlike functions, which is a subclass of the class of univalent functions, is defined
by
S\ast =
\biggl\{
f \in \scrA : \mathrm{R}\mathrm{e}
zf \prime (z)
f(z)
> 0, z \in \BbbD
\biggr\}
.
The functions f \in S\ast map the unit disk onto a starlike region, i.e., if w \in f(\BbbD ), then tw \in f(\BbbD )
for all t \in [0, 1]. More details can be found in [2].
Another subclasses of univalent functions are
R\alpha =
\bigl\{
f \in \scrA : \mathrm{R}\mathrm{e} f \prime (z) > \alpha , z \in \BbbD
\bigr\}
, 0 \leq \alpha < 1,
and
\scrR (\alpha ) =
\Bigl\{
f \in \scrA :
\bigm| \bigm| \mathrm{a}\mathrm{r}\mathrm{g} f \prime (z)
\bigm| \bigm| < \alpha \pi
2
, z \in \BbbD
\Bigr\}
, 0 < \alpha \leq 1,
which are subclasses of the class of functions with bounded turning, R = R0 = \scrR (1). The name of
the class R follows from the fact that \mathrm{R}\mathrm{e} f \prime (z) > 0 is equivalent with
\bigm| \bigm| \mathrm{a}\mathrm{r}\mathrm{g} f \prime (z)
\bigm| \bigm| < \pi
2
and \mathrm{a}\mathrm{r}\mathrm{g} f \prime (z)
is the angle of rotation of the image of a line segment starting from z under the mapping f. It is well
known that S\ast does not contain R and R does not contain S\ast [7], which brings big interest for the
class R [6 – 8].
In this paper we will study the expression
z
f \prime (z) - 1
f(z) - z
=
f \prime (z) - 1
f(z)/z - 1
, (1)
c\bigcirc N. TUNESKI, T. BULBOACĂ, 2018
1118 ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
SUFFICIENT CONDITIONS FOR BOUNDED TURNING OF ANALYTIC FUNCTIONS 1119
for receiving some results that will lead to necessary conditions for a function f \in \scrA n, n \in \BbbN ,
n \geq 2, to be with bounded turning. Here, the class \scrA n, n \in \BbbN , n \geq 2, is defined by
\scrA n =
\bigl\{
f \in \scrA : f(z) = z + anz
n + . . . , z \in \BbbD , an \not = 0
\bigr\}
.
The study will involve a method from the theory of differential subordinations, while valuable refe-
rences on this topic are [1] and [3]. Using a similar techniques as in this paper, in [8] the expression
f \prime (z) - 1
f(z)/z
is studied and results concerning the univalence and the starlikeness of f from \scrA are given.
First we introduce the notion of subordination. If f, g \in \scrA , then we say that f is subordinate
to g, and write f(z) \prec g(z), if there exists a function w, analytic in the unit disc \BbbD , such that
w(0) = 0,
\bigm| \bigm| w(z)\bigm| \bigm| < 1 and f(z) = g
\bigl(
w(z)
\bigr)
for all z \in \BbbD . Specially, if g is univalent in \BbbD , then
f(z) \prec g(z) if and only if f(0) = g(0) and f(\BbbD ) \subset g(\BbbD ).
For obtaining the main result we will use the method of differential subordinations [3]. The
general theory of differential subordinations, as well as the theory of first-order differential subordi-
nations, was introduced by Miller and Mocanu in [4] and [5]. Namely, if \varphi : \BbbC 2 \rightarrow \BbbC is analytic in
a domain D, if h is univalent in \BbbD , and if p is analytic in \BbbD with
\bigl(
p(z), zp\prime (z)
\bigr)
\in D when z \in \BbbD ,
then p is said to satisfy a first-order differential subordination if
\varphi
\bigl(
p(z), zp\prime (z)
\bigr)
\prec h(z). (2)
The univalent function q is said to be a dominant of the differential subordination (2) if p(z) \prec q(z)
for all the functions p satisfying (2). If \widetilde q is a dominant of (2) and \widetilde q(z) \prec q(z) for all dominants
of (2), then we say that \widetilde q is the best dominant of the differential subordination (2).
From the theory of first-order differential subordinations we will use the following lemma.
Lemma 1 [5]. Let q be univalent in the unit disk \BbbD , and let \theta (w) and \phi (w) be analytic in a
domain D containing q(\BbbD ), with \phi (w) \not = 0 when w \in q(\BbbD ). Set Q(z) = zq\prime (z)\phi
\bigl(
q(z)
\bigr)
, h(z) =
= \theta
\bigl(
q(z)
\bigr)
+Q(z), and suppose that:
(i) Q is starlike in the unit disk \BbbD ,
(ii) \mathrm{R}\mathrm{e}
zh\prime (z)
Q(z)
= \mathrm{R}\mathrm{e}
\Biggl[
\theta \prime
\bigl(
q(z)
\bigr)
\phi
\bigl(
q(z)
\bigr) +
zQ\prime (z)
Q(z)
\Biggr]
> 0, z \in \BbbD .
If p is analytic in \BbbD , with p(0) = q(0), p(\BbbD ) \subseteq D and
\theta
\bigl(
p(z)
\bigr)
+ zp\prime (z)\phi
\bigl(
p(z)
\bigr)
\prec \theta
\bigl(
q(z)
\bigr)
+ zq\prime (z)\phi
\bigl(
q(z)
\bigr)
= h(z), (3)
then p(z) \prec q(z), and q is the best dominant of (3).
Using Lemma 1 we will prove the following result that will be used in next section.
Lemma 2. Let f \in \scrA n, n \in \BbbN , n \geq 2, such that f(z) \not = z for all z \in \BbbD \setminus \{ 0\} . Also, let q be
univalent in the unit disk \BbbD , with q(0) = an =
f (n)(0)
n!
, and
\mathrm{R}\mathrm{e}
\biggl[
1 +
zq\prime \prime (z)
q\prime (z)
- zq\prime (z)
q(z)
\biggr]
> 0, z \in \BbbD . (4)
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
1120 N. TUNESKI, T. BULBOACĂ
If
z
f \prime (z) - 1
f(z) - z
- n \prec zq\prime (z)
q(z)
, (5)
then
f(z) - z
zn
\prec q(z),
and q is the best dominant of (5).
Proof. If we choose \theta (w) = 0 and \phi (w) =
1
w
, then \theta , \phi \in \scrH (D), where D = \BbbC \ast := \BbbC \setminus \{ 0\} .
The condition D \supset q(\BbbD ) from Lemma 1 is equivalent to q(z) \not = 0 for all z \in \BbbD , and we will prove
that this last relation holds under our assumptions. Also, let note that \phi (w) = 1/w \not = 0 for all
w \in q(\BbbD ), and let define
Q(z) := zq\prime (z)\phi
\bigl(
q(z)
\bigr)
=
zq\prime (z)
q(z)
.
Denoting
\Phi (z) := 1 +
zq\prime \prime (z)
q\prime (z)
- zq\prime (z)
q(z)
,
first we will show that the assumption (4), which is equivalent to
\mathrm{R}\mathrm{e}\Phi (z) > 0, z \in \BbbD ,
implies Q \in \scrH (\BbbD ), i.e., q(z) \not = 0 for all z \in \BbbD . In the beginning, from q(0) = an \not = 0 we receive
that Q is regular in z0 = 0. Further, let suppose that there exists z0 \in \BbbD \setminus \{ 0\} , such that q(z0) = 0.
It means that q has the form
q(z) = (z - z0)
mg(z), z \in \BbbD , m \in \BbbN \ast ,
where g \in \scrH (\BbbD ), with g(z0) \not = 0. It follows that there exists r > 0, such that g(z) \not = 0 for
all z \in U(z0; r) :=
\bigl\{
z \in \BbbC : | z - z0| < r
\bigr\}
\subset \BbbD . Now, a simple computation shows that for all
z \in U(z0; r) \setminus \{ z0\} ,
zq\prime (z)
q(z)
=
mz
z - z0
+
zg\prime (z)
g(z)
,
hence
\Phi (z) =
- mzz0
(z - z0)2
+
zg\prime (z)
g(z)
+
z2g\prime \prime (z)
g(z)
-
\biggl(
zg\prime (z)
g(z)
\biggr) 2
mz
z - z0
+
zg\prime (z)
g(z)
=
=
1
z - z0
- mzz0 + (z - z0)
2
\Biggl[
zg\prime (z)
g(z)
+
z2g\prime \prime (z)
g(z)
-
\biggl(
zg\prime (z)
g(z)
\biggr) 2
\Biggr]
mz + (z - z0)
zg\prime (z)
g(z)
.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
SUFFICIENT CONDITIONS FOR BOUNDED TURNING OF ANALYTIC FUNCTIONS 1121
Finally, having in mind that g(z0) \not = 0, from the above relation we receive that z0 \in \BbbD is a
pole of the function \Phi , which contradicts the assumption (4). Thus, we obtain that q(z) \not = 0 for all
z \in \BbbD , hence the function Q is analytic on \BbbD .
Further, q is an univalent function, implying q\prime (z) \not = 0 for all z \in \BbbD ,
Q\prime (0) =
q\prime (0)
q(0)
=
q\prime (0)
an
\not = 0
and
\mathrm{R}\mathrm{e}
zQ\prime (z)
Q(z)
= \mathrm{R}\mathrm{e}
\biggl[
1 +
zq\prime \prime (z)
q\prime (z)
- zq\prime (z)
q(z)
\biggr]
> 0, z \in \BbbD ,
meaning that Q is a starlike function. In addition, for the function h(z) = \theta
\bigl(
q(z)
\bigr)
+Q(z) = Q(z)
we have
\mathrm{R}\mathrm{e}
zh\prime (z)
Q(z)
= \mathrm{R}\mathrm{e}
zQ\prime (z)
Q(z)
> 0, z \in \BbbD .
After choosing p(z) =
f(z) - z
zn
, then p \in \scrH (\BbbD ), p(0) = an and p(z) \not = 0 for all z \in \BbbD , i.e.,
p(\BbbD ) \subset D, and all the conditions of Lemma 1 are satisfied. Concerning that the subordinations (3)
and (5) are equivalent, we receive the conclusion of Lemma 2.
2. Main results and consequences. Using Lemma 2 we will study the modulus of (1) and will
receive conclusions that will later lead to criteria for a function f to be in the class R.
Theorem 1. Let f \in \scrA n, n \in \BbbN , n \geq 2, such that f(z) \not = z for all z \in \BbbD \setminus \{ 0\} , and let
an =
f (n)(0)
n!
. If
z
f \prime (z) - 1
f(z) - z
- n \prec \lambda z
an + \lambda z
=: h1(z), (6)
where 0 < | \lambda | \leq | an| , then
f(z) - z
zn
\prec an + \lambda z (7)
and the function an + \lambda z is the best dominant of (6). Even more,\bigm| \bigm| \bigm| \bigm| f(z) - z
zn
- an
\bigm| \bigm| \bigm| \bigm| < \lambda , z \in \BbbD , (8)
and this conclusion is sharp, i.e., in the inequality (8) the parameter | \lambda | can not be replaced by a
smaller number so that the implication holds.
Proof. The function q(z) = an + \lambda z satisfies all the conditions of Lemma 2, since
\mathrm{R}\mathrm{e}
\biggl[
1 +
zq\prime \prime (z)
q\prime (z)
- zq\prime (z)
q(z)
\biggr]
= \mathrm{R}\mathrm{e}
1
1 + \lambda /anz
> 0, z \in \BbbD ,
whenever of 0 < | \lambda | \leq | an| . Further, the subordinations (5) and (6) are equivalent, and therefore (7)
follows directly from Lemma 2.
For the sharpness of our result, let assume that subordination (6) and inequality
\bigm| \bigm| \bigm| \bigm| f(z) - z
zn
- an
\bigm| \bigm| \bigm| \bigm| <
< | \lambda 1| , z \in \BbbD , holds, i.e.,
f(z) - z
zn
\prec an + \lambda 1z. But, the function an + \lambda z is the best dominant
of (6), meaning that an + \lambda z \prec an + \lambda 1z, i.e., | \lambda | \leq | \lambda 1| .
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
1122 N. TUNESKI, T. BULBOACĂ
Remark 1. It is easy to verify the following:
(i) If 0 < | \lambda | < | an| , then h1(\BbbD ) (where h1 was defined in (6)) is an open disk with the center
c =
1
2
\Bigl[
h1
\Bigl(
ei arg(an/\lambda )
\Bigr)
+ h
\Bigl(
- ei arg(an/\lambda )
\Bigr) \Bigr]
=
| \lambda | 2
| \lambda | 2 - | an| 2
,
and radius
r =
\bigm| \bigm| \bigm| h1 \Bigl( ei arg(an/\lambda )\Bigr) - c
\bigm| \bigm| \bigm| = | \lambda | | an|
| an| 2 - | \lambda | 2
.
(ii) If | \lambda | = | an| , then h1(z) =
z
z + ei arg(an/\lambda )
, and
h1(\BbbD ) =
\biggl\{
z \in \BbbC : \mathrm{R}\mathrm{e} z <
1
2
\biggr\}
.
Therefore, Theorem 1 can be written in the following equivalent form.
Theorem 1\prime . Let f \in \scrA n, n \in \BbbN , n \geq 2, such that f(z) \not = z for all z \in \BbbD \setminus \{ 0\} , and let
an =
f (n)(0)
n!
.
(i) If 0 < | \lambda | < | an| and\bigm| \bigm| \bigm| \bigm| z f \prime (z) - 1
f(z) - z
- n - | \lambda | 2
| \lambda | 2 - | an| 2
\bigm| \bigm| \bigm| \bigm| < | \lambda | | an|
| an| 2 - | \lambda | 2
, z \in \BbbD ,
then \bigm| \bigm| \bigm| \bigm| f(z) - z
zn
- an
\bigm| \bigm| \bigm| \bigm| < | \lambda | , z \in \BbbD .
(ii) If
\mathrm{R}\mathrm{e}
\biggl[
z
f \prime (z) - 1
f(z) - z
\biggr]
< n+
1
2
, z \in \BbbD ,
then \bigm| \bigm| \bigm| \bigm| f(z) - z
anzn
- 1
\bigm| \bigm| \bigm| \bigm| < 1, z \in \BbbD .
These implications are sharp, i.e., in both cases the radius of the open disk from the conclusion
is the smallest possible so that the corresponding implication holds.
Remark 2. The sharpness of Theorems 1 and 1\prime can be verified by using function f(z) =
= z + anz
n + \lambda zn+1, with an \not = 0, for which
z
f \prime (z) - 1
f(z) - z
- n =
\lambda z
an + \lambda z
.
Thus, in the case 0 < | \lambda | < | an| we have\bigm| \bigm| \bigm| \bigm| z f \prime (z) - 1
f(z) - z
- n - \lambda 2
\lambda 2 - | an| 2
\bigm| \bigm| \bigm| \bigm| = | \lambda | | an|
| an| 2 - | \lambda | 2
for z = ei arg(an/\lambda ),
while in the case | an| = | \lambda | we get
\mathrm{R}\mathrm{e}
\biggl[
z
f \prime (z) - 1
f(z) - z
\biggr]
= n+
1
2
for z = ei arg(an/\lambda ).
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
SUFFICIENT CONDITIONS FOR BOUNDED TURNING OF ANALYTIC FUNCTIONS 1123
Now we will give several corollaries and examples of Theorem 1\prime , in the case when n = 2. We
start with part (i) of Theorem 1\prime .
Corollary 1. Let f \in \scrA 2, and \lambda \in \BbbC with
4
5
| a2| \leq | \lambda | < | a2| , where a2 =
f \prime \prime (0)
2
. Also, let
denote
\mu :=
\left\{
- 2 +
| \lambda |
| a2| - | \lambda |
, if
4
5
| a2| \leq | \lambda | \leq
\sqrt{}
2
3
| a2| ,
2 +
| \lambda |
| a2| + | \lambda |
, if
\sqrt{}
2
3
| a2| \leq | \lambda | < | a2| .
If \bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < \mu
\bigm| \bigm| \bigm| \bigm| f(z)z
- 1
\bigm| \bigm| \bigm| \bigm| , z \in \BbbD \setminus \{ 0\} , (9)
then \bigm| \bigm| \bigm| \bigm| f(z) - z
z2
- a2
\bigm| \bigm| \bigm| \bigm| < \eta 1 := | \lambda | , z \in \BbbD , (10)
and \bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < \eta 2, z \in \BbbD , (11)
where
\eta 2 :=
\left\{
\bigl(
| a2| + | \lambda |
\bigr) \bigl(
2| a2| - 3| \lambda |
\bigr)
| \lambda | - | a2|
, if
4
5
| a2| \leq | \lambda | \leq
\sqrt{}
2
3
| a2| ,
2| a2| + 3| \lambda | , if
\sqrt{}
2
3
| a2| \leq | \lambda | < | a2| .
Moreover, the implication (9) \Rightarrow (10) is sharp for
\sqrt{}
2
3
| a2| \leq | \lambda | < | a2| , and the implication
(9) \Rightarrow (11) is sharp for
\sqrt{}
2
3
| a2| \leq | \lambda | < | a2| , i.e., for these ranges of | \lambda | , the values \eta 1 and \eta 2 are
the smallest ones so that the corresponding implications hold.
Also, if \eta 2 < 1, then f is univalent with bounded turning, i.e., f \in R\alpha 1 and f \in \scrR (\alpha 2), where
\alpha 1 = 1 - \eta 2 and \alpha 2 = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n} \eta 2.
Proof. First we will prove inequality (10). The assumption (9) leads to
\bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < \mu
\bigm| \bigm| \bigm| \bigm| f(z)z
- 1
\bigm| \bigm| \bigm| \bigm| = \mu
\bigm| \bigm| \bigm| \bigm| f(z) - z
z
\bigm| \bigm| \bigm| \bigm| , z \in \BbbD \setminus \{ 0\} ,
meaning that
f(z) - z
z
\not = 0 for all z \in \BbbD \setminus \{ 0\} , hence f(z) \not = z for all z \in \BbbD \setminus \{ 0\} . Also, the
inequality (9) implies \bigm| \bigm| \bigm| \bigm| z f \prime (z) - 1
f(z) - z
\bigm| \bigm| \bigm| \bigm| < \mu , z \in \BbbD \setminus \{ 0\} ,
and letting z \rightarrow 0 in the above inequality we obtain that \mu \geq 2 is a necessary condition for the above
inequality to hold in the case z = 0.
It is easy to check that
\mu =
| \lambda | | a2| -
\bigm| \bigm| 3| \lambda | 2 - 2| a2| 2
\bigm| \bigm|
| a2| 2 - | \lambda | 2
= r - | 2 + c| ,
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
1124 N. TUNESKI, T. BULBOACĂ
where r and c are defined as in the Remark 1, and that \mu \geq 2 whenever | \lambda | \geq 4
5
| a2| .
Further, we can write\bigm| \bigm| \bigm| \bigm| z f \prime (z) - 1
f(z) - z
- (2 + c) + (2 + c)
\bigm| \bigm| \bigm| \bigm| < \mu , z \in \BbbD \setminus \{ 0\} ,
and it follows that \bigm| \bigm| \bigm| \bigm| z f \prime (z) - 1
f(z) - z
- (2 + c)
\bigm| \bigm| \bigm| \bigm| < \mu + | 2 + c| = r, z \in \BbbD \setminus \{ 0\} .
The above inequality holds for z = 0, since | c| < \mu + | 2 + c| = r for 0 < | \lambda | < | a2| , and thus,
from the first part of the Theorem 1\prime (i) for the special case n = 2 we have (10).
From the assumption (9) we get
\bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < \mu
\bigm| \bigm| \bigm| \bigm| f(z)z
- 1
\bigm| \bigm| \bigm| \bigm| < \mu
\bigm| \bigm| \bigm| \bigm| f(z) - z
z2
\bigm| \bigm| \bigm| \bigm| , z \in \BbbD \setminus \{ 0\} , (12)
and the inequality (11) follows from (10) and (12), having in mind that \eta 2 = \mu
\bigl(
| a2| + \lambda
\bigr)
.
The implication (9) \Rightarrow (10) is sharp for
\sqrt{}
2
3
| a2| \leq | \lambda | < | a2| , and the implication (9) \Rightarrow (11) is
sharp for
\sqrt{}
2
3
| a2| \leq | \lambda | < | a2| , since for the function f(z) = z + a2z
2 + \lambda z3 we obtain
\bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| = | z| | 2a2 + 3\lambda z| < 2| a2| + 3| \lambda | , z \in \BbbD ,\bigm| \bigm| \bigm| \bigm| f(z)z
- 1
\bigm| \bigm| \bigm| \bigm| = | z| | a2 + \lambda z|
and \bigm| \bigm| \bigm| \bigm| f(z) - z
z2
- a2
\bigm| \bigm| \bigm| \bigm| = | \lambda | | z| < | \lambda | , z \in \BbbD .
The assertion (9) is equivalent to
\mu >
\bigm| \bigm| \bigm| \bigm| 3\lambda z + 2a2
\lambda z + a2
\bigm| \bigm| \bigm| \bigm| , z \in \BbbD \setminus \{ 0\} ,
and a simple computation shows that
\mathrm{s}\mathrm{u}\mathrm{p}
\biggl\{ \bigm| \bigm| \bigm| \bigm| 3\lambda z + 2a2
\lambda z + a2
\bigm| \bigm| \bigm| \bigm| : z \in \BbbD \setminus \{ 0\}
\biggr\}
= 2 +
| \lambda |
| a2| + | \lambda |
,
whenever | \lambda | < | a2| , hence
\mu \geq 2 +
| \lambda |
| a2| + | \lambda |
,
which holds for
\sqrt{}
2
3
| a2| \leq | \lambda | < | a2| .
Since
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
SUFFICIENT CONDITIONS FOR BOUNDED TURNING OF ANALYTIC FUNCTIONS 1125
- 2 +
| \lambda |
| a2| - | \lambda |
< 2 +
| \lambda |
| a2| + | \lambda |
, if | \lambda | <
\sqrt{}
2
3
| a2| ,
the function f(z) = z + a2z
2 + \lambda z3 shows that the implication (9) \Rightarrow (10) is not sharp for
4
5
| a2| <
< | \lambda | <
\sqrt{}
2
3
| a2| .
Finally, from (10) and the definitions of the classes R\alpha and R(\alpha ) we receive f \in R\alpha 1 and
f \in R(\alpha 2).
For \eta 2 = 1 the Corollary 1 reduces to the next example.
Example 1. Let f \in \scrA 2, with
1
5
< | a2| \leq
5
18
, where a2 =
f \prime \prime (0)
2
. Also, let
\mu \ast :=
\left\{
3
1 + | a2|
, if 0.2 =
1
5
< | a2| \leq
1
2 +
\surd
6
= 0.22474 . . . ,
1
| \lambda | \ast + | a2|
, if
1
2 +
\surd
6
\leq | a2| \leq
5
18
= 0.27 . . . ,
where
| \lambda \ast | :=
-
\bigl(
1 + | a2|
\bigr)
+
\sqrt{}
25| a2| 2 + 14| a2| + 1
6
.
If \bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < \mu \ast
\bigm| \bigm| \bigm| \bigm| f(z)z
- 1
\bigm| \bigm| \bigm| \bigm| , z \in \BbbD \setminus \{ 0\} , (13)
then \bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < 1, z \in \BbbD . (14)
This implication is sharp for
1
5
< | a2| \leq
1
2 +
\surd
6
= 0.22474 . . . . Also, the function f is univalent
with bounded turning, i.e., f \in R.
Proof. We need to prove that conditions of Corollary 1, in the case \eta 2 = 1, are equivalent to the
assumptions of this example.
For the case when
4
5
| a2| \leq | \lambda | \leq
\sqrt{}
2
3
| a2| , then \eta 2 = \mu \ast
\bigl(
| a2| + | \lambda |
\bigr)
= 1 if and only if
\mu \ast =
1
| \lambda | + | a2|
, i.e.,
- 2
\bigl(
| \lambda | + | a2|
\bigr)
+
\lambda
\bigl(
| a2| + | \lambda |
\bigr)
| a2| - | \lambda |
= 1,
or in other words
| \lambda | = | \lambda \ast | :=
-
\bigl(
1 + | a2|
\bigr)
+
\sqrt{}
25| a2| 2 + 14| a2| + 1
6
.
Here, we considered only the positive sign of the square root since the negative one leads to negative
values of | \lambda | . Further, the inequalities
4
5
| a2| \leq | \lambda | = | \lambda \ast | \leq
\sqrt{}
2
3
| a2|
are equivalent to
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
1126 N. TUNESKI, T. BULBOACĂ
0.22474 . . . =
1
2 +
\surd
6
\leq | a2| \leq
5
18
= 0.27 . . . .
In a similar way, for the case
\sqrt{}
2
3
| a2| \leq | \lambda | < | a2| we have \eta 2 = 1 if and only if 3| \lambda | +2| a2| = 1,
i.e., | \lambda | = 1 - 2| a2|
3
. A simple calculus shows that\sqrt{}
2
3
| a2| \leq | \lambda | < | a2|
is equivalent to
0.2 =
1
5
< | a2| \leq
1
2 +
\surd
6
= 0.22474 . . . ,
which completes the proof.
Remark 3. Weather implications (9) \Rightarrow (11) for
4
5
| a2| \leq | \lambda | <
\sqrt{}
2
3
| a2|
\biggl(
Corollary 1 and
(13) \Rightarrow (14) for
1
5
< | a2| \leq
1
2 +
\surd
6
= 0.22474 . . .
\biggr)
. Example 1 are sharp are still open problems.
Part (ii) from Theorem 1\prime brings the following result.
Corollary 2. Let f \in \scrA 2 and 2 \leq \mu <
5
2
. If
\bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < \mu
\bigm| \bigm| \bigm| \bigm| f(z)z
- 1
\bigm| \bigm| \bigm| \bigm| , z \in \BbbD \setminus \{ 0\} ,
then \bigm| \bigm| \bigm| \bigm| f(z) - z
a2z2
- 1
\bigm| \bigm| \bigm| \bigm| < 1, z \in \BbbD , where a2 =
f \prime \prime (0)
2
,
and \bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < 2\mu | a2| =: \eta 3, z \in \BbbD .
Even more, if \eta 3 < 1, then the function f is univalent with bounded turning, i.e., f \in R\beta 1 and
f \in \scrR (\beta 2), where \beta 1 = 1 - \eta 3 and \beta 2 = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n} \eta 3.
Proof. The assumption leads to
\bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < \mu
\bigm| \bigm| \bigm| \bigm| f(z)z
- 1
\bigm| \bigm| \bigm| \bigm| = \mu
\bigm| \bigm| \bigm| \bigm| f(z) - z
z
\bigm| \bigm| \bigm| \bigm| , z \in \BbbD \setminus \{ 0\} ,
meaning that
f(z) - z
z
\not = 0 for all z \in \BbbD \setminus \{ 0\} , hence f(z) \not = z, for all z \in \BbbD \setminus \{ 0\} . It also implies
that \bigm| \bigm| \bigm| \bigm| z f \prime (z) - 1
f(z) - z
\bigm| \bigm| \bigm| \bigm| < \mu , z \in \BbbD \setminus \{ 0\} ,
and letting z \rightarrow 0 in the above inequality we obtain that \mu \geq 2, and thus\bigm| \bigm| \bigm| \bigm| z f \prime (z) - 1
f(z) - z
\bigm| \bigm| \bigm| \bigm| \leq \mu , z \in \BbbD .
From here
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
SUFFICIENT CONDITIONS FOR BOUNDED TURNING OF ANALYTIC FUNCTIONS 1127
\mathrm{R}\mathrm{e}
\biggl[
z
f \prime (z) - 1
f(z) - z
\biggr]
\leq \mu <
5
2
, z \in \BbbD ,
and the rest of the proof follows from Theorem 1\prime (ii) for n = 2.
Similarly as in Example 1, taking \eta 3 = 1 in the previous corollary we receive:
Example 2. Let f \in \scrA 2, with
1
5
< | a2| \leq
1
4
, where a2 =
f \prime \prime (0)
2
. If
\bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < 1
2| a2|
\bigm| \bigm| \bigm| \bigm| f(z)z
- 1
\bigm| \bigm| \bigm| \bigm| , z \in \BbbD \setminus \{ 0\} , (15)
then \bigm| \bigm| f \prime (z) - 1
\bigm| \bigm| < 1, z \in \BbbD ,
and further, the function f is univalent with bounded turning, i.e., f \in R.
Remark 4. (i) It can be verified that
1
2| a2|
< \mu \ast for
1
5
< | a2| \leq
5
18
, meaning that the
condition (13) is weaker than condition (15), i.e., the result from Example 1 is better then the result
from Example 2.
(ii) It is an open problem weather the result from Corollary 2 is sharp, i.e., weather \eta 3 is the
smallest constant so the implication holds.
References
1. Bulboacă T. Differential subordinations and superordinations. New results. – Cluj-Napoca: House Sci. Book Publ.,
2005.
2. Duren P. L. Univalent functions. – Springer-Verlag,1983.
3. Miller S. S., Mocanu P. T. Differential subordinations. Theory and applications. – New York; Basel: Marcel Dekker,
2000.
4. Miller S. S., Mocanu P. T. Differential subordinations and univalent functions // Michigan Math. J. – 1981. – 28. –
P. 157 – 171.
5. Miller S. S., Mocanu P. T. On some classes of first-order differential subordinations // Michigan Math. J. – 1985. –
32. – P. 185 – 195.
6. Ibrahim R. W., Darus M. Extremal bounds for functions of bounded turning // Int. Math. Forum. – 2011. – 6, № 33. –
P. 1623 – 1630.
7. Krzyz J. A counter example concerning univalent functions // Folia Soc. Sci. Lublinensis. – 1962. – 2. – P. 57 – 58.
8. Tuneski N., Obradović M. Some properties of certain expression of analyti functions // Comput. Math. Appl. – 2011. –
62. – P. 3438 – 3445.
Received 17.06.14
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 8
|
| id | umjimathkievua-article-1622 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:09:18Z |
| publishDate | 2018 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/9a/29a400e74a4cd3b41c5bfd416280bb9a.pdf |
| spelling | umjimathkievua-article-16222019-12-05T09:21:04Z Sufficient conditions for bounded turning of analytic functions Достатнi умови для обмеженого повороту аналiтичних функцiй Bulboacă, T. Tuneski, N. Бульбоака, Т. Тунеськи, Н. Let function $f$ be analytic in the open unit disk and be normalized such that $f(0) = f\prime (0) 1 = 0$. In this paper methods from the theory of first order differential subordinations are used for obtaining sufficient conditions for $f$ to bewith bounded turning, i.e., the read part of its first derivative to map the unit disk onto the right half plane. In addition, several open problems are posed. Нехай $f$ — функцiя, аналiтична у вiдкритому одиничному крузi, нормована так, що $f(0) = f \prime (0) 1 = 0$. Методи теорiї диференцiальних пiдпорядкувань першого порядку застосовано, щоб отримати достатнi умови того, що функцiя $f$ має обмежений поворот, тобто дiйсна частина її першої похiдної вiдображає одиничний круг на праву пiвплощину. Крiм того, сформульовано кiлька вiдкритих проблем. Institute of Mathematics, NAS of Ukraine 2018-08-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1622 Ukrains’kyi Matematychnyi Zhurnal; Vol. 70 No. 8 (2018); 1118-1127 Український математичний журнал; Том 70 № 8 (2018); 1118-1127 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1622/604 Copyright (c) 2018 Bulboacă T.; Tuneski N. |
| spellingShingle | Bulboacă, T. Tuneski, N. Бульбоака, Т. Тунеськи, Н. Sufficient conditions for bounded turning of analytic functions |
| title | Sufficient conditions for bounded turning of analytic functions |
| title_alt | Достатнi умови для обмеженого повороту аналiтичних
функцiй |
| title_full | Sufficient conditions for bounded turning of analytic functions |
| title_fullStr | Sufficient conditions for bounded turning of analytic functions |
| title_full_unstemmed | Sufficient conditions for bounded turning of analytic functions |
| title_short | Sufficient conditions for bounded turning of analytic functions |
| title_sort | sufficient conditions for bounded turning of analytic functions |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1622 |
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