Some general distortion results for $K(\alpha)$ and $S^{*}(\alpha)$

EDUARD ŞTEFAN GRIGORICIUC

Abstract.

In this paper we present some general distortion results for the classes $K(\alpha)$ and $S^{*}(\alpha)$ of convex, respectively starlike functions of order $\alpha$ on the unit disc. For this, we start from a classical result for the class $S$ of univalent and normalized functions on the unit disc. This inequality can be found, for example, in [2, p. 70, ex. 6]. Furthermore, since when $\alpha=0$ these classes reduce to the well-known classes of starlike and convex functions, we obtain also some general distortion results for the classes $K$ and $S^{*}$ of convex, respectively starlike functions on the unit disc.

{amssubject}

30C45, 30C50.

{keyword}

Univalent functions, starlikeness of order $\alpha$ , convexity of order $\alpha$ , coefficient estimates, distortion results.

1. Introduction

In this paper we denote $U=U(0,1)$ the open unit disc in the complex plane $\mathbb{C}$ and $S$ the family of all univalent (holomorphic and injective) normalized ( $f(0)=f^{\prime}(0)-1=0$ ) functions on the unit disc. It is well-known (see [2], [4] or [7]) or that if $f\in S$ , then

(1)

f(z)=z+\displaystyle\sum_{n=2}^{\infty}a_{n}z^{n},\quad z\in{U}.

and it is easy to prove that

(2)

f^{(k)}(z)=\sum_{p=0}^{\infty}\dfrac{(k+p)!}{p!}a_{p+k}z^{p},\quad z\in{U}.

The basic result from which we start in this paper is the following estimate which can be found, for example, in (see [2], p. 70, ex. 6):

\forall\ f\in S:\quad|f^{(k)}(z)|\leq\dfrac{k!\cdot(k+|z|)}{(1-|z|)^{k+2}},% \quad z\in{U},\quad k\in\mathbb{N}.

Taking into account the Bieberbach’s Conjecture (proved in 1984 by L. de Branges), we can prove the previous inequality using the thchnique of dominant power series. Also, there are similar results for the classes $K$ and $S^{*}$ of convex, respectively starlike functions on the unit disc (see [4], p. 117, Th. 8 and Th. 9). Recall that $f\in K$ if $f$ is univalent on $U$ and $f(U)$ is a convex domain, respectively $f\in S^{*}$ if $f$ is univalent on $U$ , $f(0)=0$ and $f(U)$ is a starlike domain with respect to the origin.

In this paper, we extend the previous inequalities for the classes $K(\alpha)$ and $S^{*}(\alpha)$ of convex, respectively starlike functions of order $\alpha$ on the unit disc, with $\alpha\in[0,1)$ . These classes was introduced by M.S. Robertson in [10]. We denote

K(\alpha)=\bigg{\{}f\in S\ :\text{Re}\bigg{[}\dfrac{zf^{\prime\prime}(z)}{f^{% \prime}(z)}+1\bigg{]}>\alpha,\quad z\in{U}\bigg{\}}

the class of convex (normalized) functions of order $\alpha$ on $U$ and

S^{*}(\alpha)=\bigg{\{}f\in S\ :\text{Re}\bigg{[}\dfrac{zf^{\prime}(z)}{f(z)}% \bigg{]}>\alpha,\quad z\in{U}\bigg{\}}

the class of starlike (normalized) functions of order $\alpha$ on $U$ .

Remark 1.1.

It is obvious that for $\alpha=0$ we obtain the well-known classes

K=\bigg{\{}f\in S\ :\text{Re}\bigg{[}\dfrac{zf^{\prime\prime}(z)}{f^{\prime}(z% )}+1\bigg{]}>0,\quad z\in{U}\bigg{\}}

and

S^{*}=\bigg{\{}f\in S\ :\text{Re}\bigg{[}\dfrac{zf^{\prime}(z)}{f(z)}\bigg{]}>% 0,\quad z\in{U}\bigg{\}}

of convex, respectively starlike functions on the unit disc $U$ in the complex plane $\mathbb{C}$ .

In the following two sections we present some general distortion results for the classes $K(\alpha)$ and $S^{*}(\alpha)$ of convex, respectively starlike functions of order $\alpha$ on the unit disc, where $\alpha\in[0,1)$ . For this, we start from a classical result for the class $S$ of univalent and normalized functions on the unit disc. This inequality can be found, for example, in [2, p. 70, ex. 6]. The classes of starlike, respectively convex functions of order $\alpha$ were first introduced by M.S. Robertson (see [10]). More results about the starlike and convex functions of order $\alpha$ can be found in [4], [5], [6], [8], [9] or [11].

2. Preliminaries

Similar to Bieberbach’s Conjecture, we have estimates of coefficients also for the classes $K(\alpha)$ and $S^{*}(\alpha)$ as we can see in the following proposition. For details and proofs, one may consult [10, p. 386], [6, Lemma 2.1] or [11, p. 65, Th. 2].

Proposition 2.1.

Let $\alpha\in[0,1)$ . The following estimates hold:

(1)

If $f\in K(\alpha)$ , then

(3) $|a_{n}|\leq\dfrac{1}{n!}\prod_{p=2}^{n}(p-2\alpha),\quad n\geq{2}.$
(2)

If $f\in S^{*}(\alpha)$ , then

(4) $|a_{n}|\leq\dfrac{1}{(n-1)!}\prod_{p=2}^{n}(p-2\alpha),\quad n\geq{2}.$

These estimates for $|a_{n}|$ are sharp. Notice also that for $\alpha=0$ , we obtain the well-known result $|a_{n}|\leq n$ .

Next, we present two distortion theorems for $K(\alpha)$ , respectively $S^{*}(\alpha)$ . These theorems are essentially due to Robertson (see [10]). For details and proofs, one may consult also [8, p. 86, Th. 4.4.5 and Th. 4.4.6] , [5, p. 56, Th. 2.3.6 and Th. 2.3.7] or [9, p. 727, Th. 3 and Th. 4].

Theorem 2.2.

(Growth and distortion theorem for $K(\alpha)$ ):
Let $\alpha\in[0,1)$ and $f\in K(\alpha)$ . Then

\dfrac{1}{(1+|z|)^{2(1-\alpha)}}\leq|f^{\prime}(z)|\leq\dfrac{1}{(1-|z|)^{2(1-% \alpha)}}.

If $\alpha=\frac{1}{2}$ , then

\log(1+|z|)\leq|f(z)|\leq-\log(1-|z|).

If $\alpha\not=\frac{1}{2}$ , then

\dfrac{(1+|z|)^{2\alpha-1}-1}{2\alpha-1}\leq|f(z)|\leq\dfrac{1-(1-|z|)^{2% \alpha-1}}{2\alpha-1},

for all $z\in{U}$ . These bounds are sharp. Equality holds in each of the above relations for

(5)

f(z)=\begin{cases}\dfrac{1-(1-z)^{2\alpha-1}}{2\alpha-1},&\alpha\not=\frac{1}{% 2}\\ \\ -\log(1-z),&\alpha=\frac{1}{2},\end{cases}

where $(1-z)^{2\alpha-1}\big{|}_{z=0}=1$ and $\log(1-z)\big{|}_{z=0}=0$ .

Theorem 2.3.

(Growth and distortion theorem for $S^{*}(\alpha)$ ):
Let $\alpha\in[0,1)$ and $f\in S^{*}(\alpha)$ . Then

\dfrac{|z|}{(1+|z|)^{2(1-\alpha)}}\leq|f(z)|\leq\dfrac{|z|}{(1-|z|)^{2(1-% \alpha)}}

and

\dfrac{1-(1-2\alpha)|z|}{(1+|z|)^{3-2\alpha}}\leq|f^{\prime}(z)|\leq\dfrac{1+(% 1-2\alpha)|z|}{(1-|z|)^{3-2\alpha}},

for all $z\in U$ . These bounds are sharp. Equality holds for

(6)

f(z)=\dfrac{z}{(1-z)^{2(1-\alpha)}},\quad z\in{U}.

Remark 2.4.

When $\alpha=0$ , we obtatin the growth and distortion results for the classes $K$ and $S^{*}$ of convex, respectively starlike functions on the unit disc.

Remark 2.5.

Let $r=|z|<1$ . Then, for every $k\in\mathbb{N^{*}}$ , the following relation hold

(7)

T_{k}=\dfrac{1}{(1-r)^{k}}=\sum_{p=0}^{\infty}\dfrac{(k+p-1)!\cdot r^{p}}{p!% \cdot(k-1)!}.

This remark will be used in the next section as part of the proofs of the main results.

Proof.

Let us consider the following Taylor series expansion

\dfrac{1}{1-r}=1+r+r^{2}+...+r^{n}+...,\quad-1<r<1.

Then

\dfrac{1}{(1-r)^{2}}=\dfrac{\partial}{\partial r}\bigg{[}\dfrac{1}{1-r}\bigg{]% }=1+2r+3r^{2}+...+nr^{n-1}+...

It is easy to prove relation (7) using mathematical induction. For this, let us consider

P(k):\dfrac{1}{(1-r)^{k}}=\sum_{p=0}^{\infty}\dfrac{(k+p-1)!\cdot r^{p}}{p!% \cdot(k-1)!},\quad k\geq{1}.

Assume that $P(k)$ is true and let us prove that $P(k+1)$ is also true, where

P(k+1):\dfrac{1}{(1-r)^{k+1}}=\sum_{p=0}^{\infty}\dfrac{(k+p)!\cdot r^{p}}{p!% \cdot k!}.

Indeed,

\dfrac{k}{(1-r)^{k+1}}=\dfrac{\partial}{\partial r}\bigg{[}\dfrac{1}{(1-r)^{k}% }\bigg{]}=\dfrac{\partial}{\partial r}\bigg{[}\sum_{p=0}^{\infty}\dfrac{(k+p-1% )!\cdot r^{p}}{p!\cdot(k-1)!}\bigg{]}

=\sum_{p=1}^{\infty}\dfrac{(k+p-1)!\cdot p\cdot r^{p-1}}{p!\cdot(k-1)!}=\sum_{% p=0}^{\infty}\dfrac{(k+p)!\cdot r^{p}}{p!\cdot(k-1)!}

and then

\dfrac{1}{(1-r)^{k+1}}=\sum_{p=0}^{\infty}\dfrac{(k+p)!\cdot r^{p}}{p!\cdot k!}

Hence, $P(k)$ is true for all $k\geq{1}$ and the relation (7) holds. ∎

3. The main results

3.1. Convexity of order $\alpha$

In this section we present a general distortion result for convex functions of order $\alpha$ .

Theorem 3.1.

Let $\alpha\in[0,1)$ and $f\in{K}(\alpha)$ . Then the following estimate hold:

|f^{(k)}(z)|\leq\dfrac{M(k,\alpha)}{(1-|z|)^{k+1-2\alpha}},\quad\forall\ z\in{% U},\quad k\geq{1},

where

M(k,\alpha)=\begin{cases}\dfrac{1}{1-2\alpha}\displaystyle\prod_{p=1}^{k}(p-2% \alpha),&\alpha\not=\dfrac{1}{2}\\ \\ (k-1)!,&\alpha=\dfrac{1}{2}\end{cases}

These bounds are sharp. Equality holds for the function given by (5).

Proof.

Let $f\in K(\alpha)$ . Then $f$ is of the form (1) and $f$ has the following form of the $k$ -th derivative:

(8)

f^{(k)}(z)=\sum_{p=0}^{\infty}\dfrac{(k+p)!}{p!}a_{p+k}z^{p},\quad z\in{U}.

Let $|z|\leq r<1$ . In view of relations (3) and (8) we obtain

|f^{(k)}(z)|=\bigg{|}\displaystyle\sum_{p=0}^{\infty}\dfrac{(k+p)!}{p!}a_{p+k}% z^{p}\bigg{|}\leq\displaystyle\sum_{p=0}^{\infty}\dfrac{(k+p)!}{p!}|a_{p+k}|% \cdot|z|^{p}

(9)

\leq\displaystyle\sum_{p=0}^{\infty}\bigg{(}\dfrac{(k+p)!}{p!}\cdot\dfrac{1}{(% p+k)!}\cdot\displaystyle\prod_{p=2}^{p+k}(p-2\alpha)r^{p}\bigg{)}=% \displaystyle\sum_{p=0}^{\infty}\dfrac{1}{p!}\displaystyle\prod_{p=2}^{p+k}(p-% 2\alpha)r^{p}

$\bullet$ Case 1: If $\alpha=\dfrac{1}{2}$ , then

|f^{(k)}(z)|\leq\displaystyle\sum_{p=0}^{\infty}\dfrac{1}{p!}\displaystyle% \prod_{p=2}^{p+k}(p-1)r^{p}=\displaystyle\sum_{p=0}^{\infty}\dfrac{(p+k-1)!r^{% p}}{p!}

=(k-1)!\cdot\displaystyle\sum_{p=0}^{\infty}\dfrac{(p+k-1)!r^{p}}{p!(k-1)!}=(k% -1)!\cdot T_{k},

where $T_{k}$ is given by (7). Hence,

|f^{(k)}(z)|\leq\dfrac{(k-1)!}{(1-|z|)^{k}},\quad z\in{U}.

$\bullet$ Case 2: If $\alpha\not=\dfrac{1}{2}$ , then

|f^{(k)}(z)|\leq\displaystyle\sum_{p=0}^{\infty}\dfrac{1}{p!}\displaystyle% \prod_{p=2}^{p+k}(p-2\alpha)r^{p}=\sum_{p=0}^{\infty}\dfrac{(p+k-2\alpha)!}{2% \alpha(1-2\alpha)p!}r^{p}

=\dfrac{(k-2\alpha)!}{2\alpha(2\alpha-1)}\cdot\sum_{p=0}^{\infty}\dfrac{(p+k-2% \alpha)!}{p!(k-2\alpha)!}r^{p}=\dfrac{(k-2\alpha)!}{2\alpha(2\alpha-1)}\cdot T% _{k+1-2\alpha}

=\dfrac{(k-2\alpha)!}{2\alpha(2\alpha-1)}\cdot\dfrac{1}{(1-r)^{k+1-2\alpha}}=% \dfrac{\displaystyle\prod_{p=2}^{k}(p-2\alpha)}{(1-r)^{k+1-2\alpha}},

where

(k-2\alpha)!=\prod_{p=0}^{k}(p-2\alpha)

is a factorial notation which depends only on $k$ , and

T_{k+1-2\alpha}=\dfrac{1}{(1-r)^{k+1-2\alpha}}=\sum_{p=0}^{\infty}\dfrac{(p+k-% 2\alpha)!}{p!(k-2\alpha)!}r^{p}.

This result holds also for $k=1$ , so we obtain that

|f^{(k)}(z)|\leq\dfrac{\displaystyle\prod_{p=2}^{k}(p-2\alpha)}{(1-r)^{k+1-2% \alpha}}=\dfrac{\dfrac{1}{1-2\alpha}\displaystyle\prod_{p=1}^{k}(p-2\alpha)}{(% 1-r)^{k+1-2\alpha}}=\dfrac{M(k,\alpha)}{(1-r)^{k+1-2\alpha}},\quad\forall\ r<1,

where

M(k,\alpha)=\dfrac{1}{1-2\alpha}\displaystyle\prod_{p=1}^{k}(p-2\alpha),\quad k% \geq{1}.

This completes the proof. ∎

Remark 3.2.

If $\alpha=0$ , then $K(0)=K$ is the class of convex functions on the unit disc and we obtain the upper bounds for the $k$ -th derivative in $K$ :

|f^{(k)}(z)|\leq\dfrac{k!}{(1-|z|)^{k+1}},\quad z\in{U},\quad k\geq{1}.

The same result can be found in [4, p. 118, Th. 9].

Remark 3.3.

Let $\alpha\in[0,1)$ and $f\in K(\alpha)$ . If $z=0$ , then we obtain

|f^{(k)}(0)|\leq\begin{cases}\dfrac{1}{1-2\alpha}\displaystyle\prod_{p=1}^{k}(% p-2\alpha),&\alpha\not=\dfrac{1}{2}\\ \\ (k-1)!,&\alpha=\dfrac{1}{2},\end{cases}

for all $k\geq{1}$ . In particular,

(10)

|f^{\prime\prime}(0)|\leq 2(1-\alpha).

3.2. Starlikeness of order $\alpha$

In this section we present a general distortion result for starlike functions of order $\alpha$ .

Theorem 3.4.

Let $\alpha\in[0,1)$ and $f\in{S}^{*}(\alpha)$ . Then the following estimate hold:

(11)

|f^{(k)}(z)|\leq\dfrac{M(k,\alpha)\big{[}k+|z|\cdot(1-2\alpha)\big{]}}{(1-|z|)% ^{k+2-2\alpha}},\quad\forall\ z\in{U},\quad k\geq{1},

where

M(k,\alpha)=\begin{cases}\dfrac{1}{1-2\alpha}\displaystyle\prod_{p=1}^{k}(p-2% \alpha),&\alpha\not=\dfrac{1}{2}\\ \\ (k-1)!,&\alpha=\dfrac{1}{2}\end{cases}

These bounds are sharp. Equality holds for the function given by (6).

Proof.

It is clear that $S^{*}(\alpha)$ is a subclass of class $S$ . From first section we know that every function $f$ from class $S$ has the following form of $k$ -th derivative

(12)

f^{(k)}(z)=\sum_{p=0}^{\infty}\dfrac{(k+p)!}{p!}a_{p+k}z^{p},\quad z\in{U}.

Let $|z|\leq r<1$ . In view of relations (4) and (12) we obtain

|f^{(k)}(z)|=\bigg{|}\displaystyle\sum_{p=0}^{\infty}\dfrac{(k+p)!}{p!}a_{p+k}% z^{p}\bigg{|}\leq\displaystyle\sum_{p=0}^{\infty}\dfrac{(k+p)!}{p!}|a_{p+k}|% \cdot|z|^{p}

\leq\sum_{p=0}^{\infty}\bigg{(}\dfrac{(p+k)!}{p!}\cdot\dfrac{1}{(p+k-1)!}\cdot% \prod_{p=2}^{p+k}(p-2\alpha)r^{p}\bigg{)}=\sum_{p=0}^{\infty}\dfrac{p+k}{p!}% \prod_{p=2}^{p+k}(p-2\alpha)r^{p}

$\bullet$ Case 1: If $\alpha=\dfrac{1}{2}$ , then

|f^{(k)}(z)|\leq\sum_{p=0}^{\infty}\dfrac{p+k}{p!}\prod_{p=2}^{p+k}(p-1)r^{p}=% \sum_{p=0}^{\infty}\dfrac{(p+k)(p+k-1)!r^{p}}{p!}

=\sum_{p=0}^{\infty}\dfrac{(p+k)!r^{p}}{p!}=k!\cdot\sum_{p=0}^{\infty}\dfrac{(% p+k)!r^{p}}{p!\cdot k!}=k!\cdot T_{k+1},

where $T_{k+1}$ is given by (7). Hence,

|f^{(k)}(z)|\leq\dfrac{k!}{(1-|z|)^{k+1}},\quad z\in{U}.

$\bullet$ Case 2: If $\alpha\not=\dfrac{1}{2}$ , then

|f^{(k)}(z)|\leq\sum_{p=0}^{\infty}\dfrac{p+k}{p!}\prod_{p=2}^{p+k}(p-2\alpha)% r^{p}=\sum_{p=0}^{\infty}\dfrac{(p+k)(p+k-2\alpha)!r^{p}}{2\alpha(2\alpha-1)p!}

=\dfrac{1}{2\alpha(2\alpha-1)}\sum_{p=0}^{\infty}\dfrac{(k+p-2\alpha)!(p+k)r^{% p}}{p!},

where

(k+p-2\alpha)!=\prod_{p=0}^{k+p}(p-2\alpha)

is a factorial notation which depends only on $k+p$ .

Next, we determine the sum of the series $\displaystyle\sum_{p=0}^{\infty}\dfrac{(k+p-2\alpha)!(p+k)r^{p}}{p!}$ . In view of the ratio test (for details see [1]) we obtain that series converges when $r<1$ .

As in the case of the class $S$ we consider $u_{p}=a_{p}+b_{p}$ , where

u_{p}=\dfrac{(k+p-2\alpha)!(p+k)r^{p}}{p!},

is the general term of the above series,

a_{p}=\dfrac{(p+k-2\alpha)!(p-2n\alpha)r^{p}}{p!(k+1-2\alpha)}

and

b_{p}=\dfrac{(p+k-2\alpha)!(k^{2}+k+pk-2\alpha k)r^{p}}{p!(k+1-2\alpha)}.

It is easy to observe that $a_{0}=0$ , so we can consider a new general term $h_{p}$ for the above series, where

h_{p}=b_{p}+a_{p+1}

h_{p}=\dfrac{(p+k+1-2\alpha)![k+r(1-2\alpha)]r^{p}}{p!(k+1-2\alpha)}.

Hence,

\sum_{p=0}^{\infty}\dfrac{(k+p-2\alpha)!(p+k)r^{p}}{p!}=\sum_{p=0}^{\infty}% \dfrac{(p+k+1-2\alpha)![k+r(1-2\alpha)]r^{p}}{p!(k+1-2\alpha)}

=(k-2\alpha)![k+r(1-2\alpha)]\cdot\sum_{p=0}^{\infty}\dfrac{(p+k+1-2\alpha)!r^% {p}}{p!(k+1-2\alpha)!}.

In view of these computations, we obtain

|f^{(k)}(z)|\leq\dfrac{(k-2\alpha)!}{2\alpha(2\alpha-1)}\cdot[k+r(1-2\alpha)]% \cdot\sum_{p=0}^{\infty}\dfrac{(p+k+1-2\alpha)!r^{p}}{p!(k+1-2\alpha)!}

=\dfrac{(k-2\alpha)!}{2\alpha(2\alpha-1)}\cdot[k+r(1-2\alpha)]\cdot T_{k+2-2\alpha}

=\dfrac{(k-2\alpha)!}{2\alpha(2\alpha-1)}\cdot\dfrac{[k+r(1-2\alpha)]}{(1-r)^{% k+2-2\alpha}}

=\prod_{p=2}^{k}(p-2\alpha)\cdot\dfrac{[k+r(1-2\alpha)]}{(1-r)^{k+2-2\alpha}}.

This result holds also for $k=1$ , so we obtain that

|f^{(k)}(z)|\leq\dfrac{1}{1-2\alpha}\prod_{p=1}^{k}(p-2\alpha)\cdot\dfrac{[k+r% (1-2\alpha)]}{(1-r)^{k+2-2\alpha}}

|f^{(k)}(z)|\leq\dfrac{M(k,\alpha)\big{[}k+r(1-2\alpha)\big{]}}{(1-r)^{k+2-2% \alpha}},\quad\forall\ r<1,

where

M(k,\alpha)=\dfrac{1}{1-2\alpha}\displaystyle\prod_{p=1}^{k}(p-2\alpha),\quad k% \in\mathbb{N}^{*}.

This completes the proof. ∎

Remark 3.5.

If $\alpha=0$ , then $S^{*}(0)=S^{*}$ is the class of starlike functions on the unit disc and we obtain the upper bounds for the $k$ -th derivative in $S^{*}$ :

|f^{(k)}(z)|\leq\dfrac{k!(k+|z|)}{(1-|z|)^{k+2}},\quad z\in{U}.

The same result can be found in [4, p. 117, Th. 8].

Remark 3.6.

The above result confirm also the inequality from [3, Th. 3.2]:

|f^{\prime\prime}(0)|\leq 2/\beta

for $\alpha\in[0,1)$ and $1/\beta=2(1-\alpha)$ .

Proof.

It is enough to take $k=2$ in (11), and we obtain that

|f^{\prime\prime}(z)|\leq\dfrac{M(2,\alpha)[2+|z|\cdot(1-2\alpha)]}{(1-|z|)^{2% +2-2\alpha}},\quad z\in{U}.

Now, for $z=0$ we obtain

|f^{\prime\prime}(0)|\leq\dfrac{(2-2\alpha)[2+0\cdot(1-2\alpha)]}{(1-0)^{2+2-2% \alpha}}=2\cdot 2(1-\alpha)=2\cdot\dfrac{1}{\beta}=\dfrac{2}{\beta},

and hence

|f^{\prime\prime}(0)|\leq 2/\beta.

∎

Remark 3.7.

Let $\alpha\in[0,1)$ and $f\in S^{*}(\alpha)$ . Then

|f^{(k)}(0)|\leq\begin{cases}\dfrac{k}{1-2\alpha}\displaystyle\prod_{p=1}^{k}(% p-2\alpha),&\alpha\not=\dfrac{1}{2}\\ \\ k!,&\alpha=\dfrac{1}{2},\end{cases}

for all $k\geq{1}$ .

Remark 3.8.

If we take $\alpha=1/2$ in Theorem 3.4, we obtain the general distortion result for the convex functions given also in [4, p. 118, Th. 9] and Remark 3.2.

Remark 3.9.

Another proof of Theorem 3.4 can be given using the duality theorem between $K(\alpha)$ and $S^{*}(\alpha)$ (see [4] or [5]).

4. Acknowledgments

The author thank the referee for his/her helpful comments and suggestions.

References

[1] Cobzaş, Ş., Differential Calculus, Cluj University Press, 1997 (in romanian).
[2] Duren, P.L., Univalent Functions, Springer-Verlag Inc., New York, 1973.
[3] Ghosh, N., Vasudevarao, Coefficient estimates for certain subclass of analytic functions defined by subordonation, Filomat, 31 (2017), no. 11, 3307–3318.
[4] Goodman, A.W., Univalent Functions, Vols. I and II, Mariner Publ. Co., Tampa, Florida, 1983.
[5] Graham, I., Kohr, G., Geometric function theory in one and higher dimensions, Marcel Deker Inc., New York, 2003.
[6] Klein, M., Functions Starlike of order $\alpha$ , Trans. Amer. Math. Soc., 131 (1968), no. 1, 99–106.
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[9] Pinchuk, B., On starlike and convex functions of order $\alpha$ , Duke Math. J., 35 (1968), 721–734.
[10] Robertson, M.S., On the theory of univalent functions, Ann. of Math. (2), 37 (1936), 374–408.
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Received June 11, 2017

Accepted September 5, 2017

“Babeş-Bolyai” University of Cluj-Napoca

Department of Mathematics

Cluj-Napoca, Romania

E-mail: eduard.grigoriciuc@ubbcluj.ro

Some general distortion results for K(α) and S*(α)

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Some general distortion results for K⁢(α) and S∗⁢(α)

Abstract.

1. Introduction

Remark 1.1.

2. Preliminaries

Proposition 2.1.

Theorem 2.2.

Theorem 2.3.

Remark 2.4.

Remark 2.5.

Proof.

3. The main results

3.1. Convexity of order α

Theorem 3.1.

Proof.

Remark 3.2.

Remark 3.3.

3.2. Starlikeness of order α

Theorem 3.4.

Proof.

Remark 3.5.

Remark 3.6.

Proof.

Remark 3.7.

Remark 3.8.

Remark 3.9.

4. Acknowledgments

References

Some general distortion results for $K(\alpha)$ and $S^{*}(\alpha)$

3.1. Convexity of order $\alpha$

3.2. Starlikeness of order $\alpha$