On some classes of holomorphic functions whose derivatives have positive real part

Eduard Ştefan Grigoriciuc “Babeş-Bolyai” University,
Faculty of Mathematics and Computer Science
1, M. Kogălniceanu Street,
400084 Cluj-Napoca,
Romania eduard.grigoriciuc@ubbcluj.ro

Abstract.

In this paper we discuss about normalized holomorphic functions whose derivatives have positive real part. For this class of functions, denoted $R$ , we present a general distortion result (some upper bounds for the modulus of the $k$ -th derivative of a function). We present also some remarks on the functions whose derivatives have positive real part of order $\alpha$ . More details about these classes of functions can be found in [6], [8], [7, Chapter 4] and [4]. In the last part of this paper we present two new subclasses of normalized holomorphic functions whose derivatives have positive real part which generalize the classes $R$ and $R(\alpha)$ . For these classes we present some general results and examples.

Key words and phrases:

univalent function, positive real part, distortion result

1991 Mathematics Subject Classification:

30C45, 30C50

1. Introduction

In this paper we denote $U=U(0,1)$ the open unit disc in the complex plane, $\mathcal{H}(U)$ the family of all holomorphic functions on the unit disc and $S$ the family of all univalent normalized ( $f(0)=0$ and $f^{\prime}(0)=1$ ) functions on the unit disc. Also, let us denote

\mathcal{P}=\big{\{}p\in\mathcal{H}(U):p(0)=1\text{ and Re}[p(z)]>0,\quad z\in% {U}\big{\}}

the Carathéodory class and

R=\{f\in\mathcal{H}(U):f(0)=0,f^{\prime}(0)=1\text{ and Re}[f^{\prime}(z)]>0,% \quad z\in{U}\}

the class of normalized functions whose derivative has positive real part. For more details about these classes, one may consult [1], [2, Chapter 7], [3, Chapter 2] or [7, Chapter 3].

Remark 1.1.

Notice that, according to a result due to Noshiro and Warschawski (see [1, Theorem 2.16], [6] or [7, Theorem 4.5.1]), we have that each function from $R$ is also univalent on the unit disc $U$ . Hence, $R\subseteq S$ .

Remark 1.2.

Another important result (see [7, p. 87]) says that $f\in R$ if and only if $f^{\prime}\in\mathcal{P}$ .

Remark 1.3.

During this paper, we use the following notations for the series expansions of $p\in\mathcal{P}$ and $f\in S$ :

p(z)=1+p_{1}z+p_{2}z^{2}+...+p_{n}z^{n}+...

(1.1)

and

f(z)=z+a_{2}z^{2}+a_{3}z^{3}+...+a_{n}z^{n}+...

(1.2)

2. Preliminaries

First, we present some classical results regarding to the coefficient estimations and distortion results for the Carathéodory class $\mathcal{P}$ . For details and proofs, one may consult [2, Chapter 7], [3, Chapter 2], [6, Lemma 1] or [7, Chapter 3].

Proposition 2.1.

Let $p\in\mathcal{P}$ . Then

|p_{n}|\leq 2,\quad n\geq{1},

(2.1)

\dfrac{1-|z|}{1+|z|}\leq\emph{Re}[p(z)]\leq|p(z)|\leq\dfrac{1+|z|}{1-|z|}

(2.2)

and

|p^{\prime}(z)|\leq\dfrac{2}{(1-|z|)^{2}},

(2.3)

for all $z\in{U}$ . These estimates are sharp. The extremal function is $p:U\rightarrow\mathbb{C}$ given by

p(z)=\dfrac{1+z}{1-z},\quad z\in{U}.

(2.4)

The next result is another important result regarding to the coefficient estimations and distortion results for the class $R$ . For more details and proofs, one may consult [6, Theorem 1], [7, Chapter 4] or [8, Theorem A].

Proposition 2.2.

Let $f\in R$ . Then

|a_{n}|\leq\dfrac{2}{n},\quad n\geq{2},

(2.5)

\dfrac{1-|z|}{1+|z|}\leq\emph{Re}\big{[}f^{\prime}(z)\big{]}\leq|f^{\prime}(z)% |\leq\dfrac{1+|z|}{1-|z|}.

(2.6)

and

-|z|+2\log(1+|z|)\leq|f(z)|\leq-|z|-2\log(1-|z|).

(2.7)

for all $z\in{U}$ . These estimates are sharp. The extremal function is $f:U\rightarrow\mathbb{C}$ given by

f(z)=-z-\dfrac{2}{\lambda}\log(1-\lambda z),\quad|\lambda|=1,\quad z\in{U}.

(2.8)

Remark 2.3.

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

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

(2.9)

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 (2.9) 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 (2.9) holds. ∎

3. General distortion result for the class $R$

Starting from the previous proposition, we give a general distortion result (some upper bounds for the modulus of the $k$ -th derivative) for the frunction from the class $R$ .

Theorem 3.1.

If $f\in{R}$ , then the following estimate hold:

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

Proof.

It is clear that $R$ is a subclass of class $S$ . Then the $k$ -th derivative of a function $f\in{R}$ has the form

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

(3.1)

Let $|z|\leq r<1$ . In view of relations (2.5) and (3.1) we obtain that

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

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

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

Hence, we obtain that

|f^{(k)}(z)|\leq\dfrac{2(k-1)!}{(1-r)^{k}},\quad k\in\mathbb{N}^{*},\quad|z|% \leq{r}<1.

∎

Remark 3.2.

Notice that the above result is not sharp for $k=1$ (in view of relation (2.6)), but it is sharp for $k\geq{2}$ and the extremal function is $f$ given by (2.8).

4. Some remarks on the class $R(\alpha)$

Let $\alpha\in[0,1)$ . Then

R(\alpha)=\{f\in\mathcal{H}(U):f(0)=0,f^{\prime}(0)=1,\text{Re}\big{[}f^{% \prime}(z)\big{]}>\alpha,z\in{U}\}

denotes the class functions whose derivative has positive real part of order $\alpha$ . For more details about this class, one may consult [4] and [5].

Remark 4.1.

It is easy to prove that $f\in R(\alpha)$ if and only if $g\in\mathcal{P}$ , where $g:U\rightarrow\mathbb{C}$ is given by

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

(4.1)

Proposition 4.2.

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

|a_{n}|\leq\dfrac{2(1-\alpha)}{n},\quad n\geq{2},

(4.2)

and these estimates are sharp. The equality holds for the function $f:U\rightarrow\mathbb{C}$ given by

f(z)=\dfrac{(2\alpha-1)\lambda z-2(1-\alpha)\log(1-\lambda z)}{\lambda}

(4.3)

with $|\lambda|=1$ .

Proof.

Let $f\in R(\alpha)$ be of the form (1.2). Then

f^{\prime}(z)=1+\displaystyle\sum_{n=1}^{\infty}(n+1)a_{n+1}z^{n},\quad z\in{U}.

Let us consider the function $g:U\rightarrow\mathbb{C}$ given by

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

Then $g\in\mathcal{P}$ and

g(z)=\dfrac{f^{\prime}(z)-\alpha}{1-\alpha}=\dfrac{1-\alpha+\displaystyle\sum_% {n=1}^{\infty}(n+1)a_{n+1}z^{n}}{1-\alpha}=1+\displaystyle\sum_{n=1}^{\infty}% \dfrac{(n+1)}{1-\alpha}a_{n+1}z^{n}

or, equivalent

g(z)=1+\displaystyle\sum_{n=1}^{\infty}p_{n}z^{n},\quad\text{where}\quad p_{n}% =\dfrac{n+1}{1-\alpha}a_{n+1}.

(4.4)

Taking into account the relations (2.1) and (4.4) we obtain that

\bigg{|}\dfrac{n+1}{1-\alpha}a_{n+1}\bigg{|}\leq 2\Leftrightarrow|a_{n+1}|\leq% \dfrac{2(1-\alpha)}{n+1},\quad\forall\ n\geq{1}.

So we obtain that

|a_{n}|\leq\dfrac{2(1-\alpha)}{n},\quad\forall\ n\geq{2}.

The function given by relation (4.3) is obtained from the extremal function of the Carathédory class. We have the following Taylor expansion

f(z)=z+(1-\alpha)\lambda z^{2}+\dfrac{2}{3}(1-\alpha)\lambda^{2}z^{3}+...

leading to the estimates

|a_{2}|=\big{|}(1-\alpha)\lambda\big{|}=1-\alpha

|a_{3}|=\bigg{|}\dfrac{2}{3}(1-\alpha)\lambda\bigg{|}=\dfrac{2(1-\alpha)}{3}

and the equalities hold for every $n\geq{2}$ . ∎

Remark 4.3.

The previous result can be found also in [5, Theorem 3.5] with another version of the proof.

Next, we present a growth and distortion result for the class $R(\alpha)$ . Starting from this theorem we give also a general distortion result (some upper bounds for the modulus of the $k$ -th derivative) for the class $R(\alpha)$ .

Theorem 4.4.

Let $\alpha\in[0,1)$ and $f\in R(\alpha)$ . Then

|f(z)|\leq(2\alpha-1)|z|+2(\alpha-1)\log(1-|z|),

(4.5)

|f(z)|\geq-|z|-2(\alpha-1)\log(1+|z|)

(4.6)

and

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

(4.7)

for all $z\in{U}$ . These estimates are sharp. The extremal function is $f:U\rightarrow\mathbb{C}$ given by

f(z)=(2\alpha-1)z-\dfrac{2(1-\alpha)\log(1-\lambda z)}{\lambda},\quad|\lambda|% =1,\quad z\in{U}.

(4.8)

Proof.

Let $\alpha\in[0,1)$ and $f\in R(\alpha)$ . In view of Remark 4.1 and Proposition 2.1, we obtain that

\bigg{|}\dfrac{1}{1-\alpha}\big{[}f^{\prime}(z)-\alpha\big{]}\bigg{|}\leq% \dfrac{1+|z|}{1-|z|}

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

Then

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

On the other hand,

\bigg{|}\dfrac{1}{1-\alpha}\big{[}f^{\prime}(z)-\alpha\big{]}\bigg{|}\geq% \dfrac{1-|z|}{1+|z|}

\big{|}f^{\prime}(z)-\alpha\big{|}\geq\dfrac{(1-\alpha)(1-|z|)}{1+|z|}

Then

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

Hence, we obtain relations (4.7). Finally, to obtain the relations (4.5) and (4.6), it is enough to integrate the relation (4.7). ∎

Theorem 4.5.

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

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

Proof.

Let $\alpha\in[0,1)$ . It is clear that $R(\alpha)$ is a subclass of class $S$ . Then the $k$ -th derivative of a function $f\in{R(\alpha)}$ has the form

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

(4.9)

Let $|z|\leq r<1$ . According to the relations (4.2) and (4.9) we obtain that

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

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

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

Hence, we obtain that

|f^{(k)}(z)|\leq\dfrac{2(1-\alpha)(k-1)!}{(1-r)^{k}},\quad k\in\mathbb{N}^{*},% \quad|z|\leq{r}<1.

∎

Remark 4.6.

Notice that, for $k=1$ , the previous result is not sharp. The sharpness is obtained if $k\geq{2}$ for the function $f$ defined by (4.8).

Remark 4.7.

It is clear that if $\alpha=0$ , then $R(0)=R$ and we obtain the classical results from the previous section.

5. The class $R_{p}$

Let $p\in\mathbb{N}^{*}$ . Starting from the well-known class $R$ , we define

R_{p}=\{f\in\mathcal{H}(U):f(0)=0,f^{\prime}(0)=1,f^{(p)}(0)=1,\text{Re}\big{[% }f^{(p)}(z)\big{]}>0,z\in{U}\}

the class of normalized functions whose $p$ -th derivative has positive real part. This is the natural extension of the class $R$ (extension which preserves the connection with the Carathéodory class). We present for this class some important results, a few examples and structure formulas (in the particular cases $p=2$ and $p=3$ ). It is clear that if $p=1$ , then $R_{1}=R$ .

Remark 5.1.

In previous definition we have the following equivalent conditions

f^{(p)}(0)=1\Leftrightarrow a_{p}=\dfrac{1}{p!},

(5.1)

for $p\in\mathbb{N}^{*}$ arbitrary fixed. Indeed, if $f\in R_{p}$ , then

f^{(p)}(z)=\sum_{n=0}^{\infty}\dfrac{(n+p)!}{n!}a_{n+p}z^{n}=p!\cdot a_{p}+% \dfrac{(p+1)!}{1!}a_{p+1}z+\dfrac{(p+2)!}{2!}a_{p+2}z^{2}+...

For $z=0$ we obtain

f^{(p)}(0)=p!\cdot a_{p}.

Hence

f^{(p)}(0)=1\Leftrightarrow p!\cdot a_{p}=1\Leftrightarrow a_{p}=\dfrac{1}{p!}% ,\quad p\geq{1}.

Remark 5.2.

Let $p\in\mathbb{N}^{*}$ be arbitrary fixed. In view of above definition we deduce that

f\in{R_{p}}\Leftrightarrow f^{(p)}\in\mathcal{P},

so we can use the properties of Carathéodory class $\mathcal{P}$ to describe the function $f^{(p)}$ and then we can obtain some properties for $f\in R_{p}$ .

Proposition 5.3.

Let $p\in\mathbb{N}^{*}$ and $f\in R_{p}$ . Then the following relation hold:

|a_{n}|\leq\dfrac{2(n-p)!}{n!},\quad n\geq p,

(5.2)

Proof.

Let $f\in R_{p}$ . Then

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

Taking into account Remark 5.2 and Proposition 2.1 we have that

f^{(p)}\in\mathcal{P},

and

\bigg{|}\dfrac{(n+p)!}{n!}a_{n+p}\bigg{|}\leq 2,\quad\forall\ n\geq{2}.

In view of above relations we obtain

|a_{n+p}|\leq\dfrac{2\cdot n!}{(n+p)!}

or, an equivalent form

|a_{n}|\leq\dfrac{2(n-p)!}{n!},\quad\forall\ n\geq p.

∎

Theorem 5.4.

Let $p\in\mathbb{N}^{*}$ and $f\in{R_{p}}$ . Then the following estimate hold:

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

(5.3)

Proof.

Let $f\in R_{p}$ . Then

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

(5.4)

Let $|z|\leq r<1$ . Using relations (5.2) and (5.4) we obtain

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

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

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

Hence,

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

∎

Remark 5.5.

In estimates (5.3) we have the following existence condition:

\forall\ k,p\in\mathbb{N}^{*}:\quad k\geq p.

In other words, for $p\in\mathbb{N}^{*}$ arbitrary fixed we can estimate the derivatives of order $k$ with $k\geq p$ (the derivatives of order at least $p$ ). In particular, for $p=1$ (i.e. for the class $R$ ) we can estimate all derivatives of order at least 1.

Remark 5.6.

For the bounds of the modulus of the first $(p-1)$ derivatives of a function $f\in R_{p}$ we can apply the following argument

\forall\ j\in\{0,...,p-1\}:\quad|f^{(j)}(z)|\leq\underbrace{\int_{0}^{r}...% \int_{0}^{r}}_{(p-j)\text{ times}}\bigg{[}\dfrac{1+\rho}{1-\rho}\bigg{]}d\rho

(5.5)

In particular,

|f^{(p-1)}(z)|\leq-|z|-2\log(1-|z|)

and

|f^{(p-2)}(z)|\leq\dfrac{-|z|(|z|-4)}{2}-2(|z|-1)\log(1-|z|).

Hence, for $f\in R_{p}$ we obtain general upper bounds, as follows:

•

if $0\leq k<p$ , we use relation (5.3);
•

if $k\geq{p}$ , we use relation (5.5).

Remark 5.7.

If $p=1$ , then $R_{1}=R$ and we obtain the result (general result of distortion) from Theorem 3.1.

In following results we discuss about the relation between two consecutive classes of order $p$ , respectively $p+1$ , for $p\in\mathbb{N}^{*}$ arbitrary choosen.

Proposition 5.8.

Let $p\in\mathbb{N}^{*}$ . Then $R_{p}\cap R_{p+1}\not=\emptyset$ .

For $p\in\mathbb{N}^{*}$ we can find a function $f$ which belongs to both class $R_{p}$ and $R_{p+1}$ . We present two examples to illustrate this proposition (first for the case $p=1$ and second for the general case $p\geq{2}$ ).

Example 5.9.

Let $f:U\rightarrow\mathbb{C}$ be given by $f(z)=\frac{1}{2}z^{2}+z$ , $z\in{U}$ . Then $f\in{R_{1}}\cap R_{2}$ .

Proof.

Indeed, we have

f(0)=0

f^{\prime}(z)=z+1

f^{\prime\prime}(z)=1,\quad z\in{U}.

For $z=0$ we obtain

f^{\prime}(0)=f^{\prime\prime}(0)=1\quad\text{and}\quad\text{Re}f^{\prime% \prime}(z)=1>0,\quad\forall\ z\in{U}.

Then, in view of definition, $f\in{R_{2}}$ . On the other hand,

f^{\prime}(0)=1\quad\text{and}\quad\text{Re}f^{\prime}(z)=\text{Re}(z+1)=1+% \text{Re}z>0,\quad\forall\ z\in{U},

and this means that $f\in{R_{1}}$ . ∎

Example 5.10.

Let $p\geq{2}$ and let $f:U\rightarrow\mathbb{C}$ be given by

f(z)=z+\dfrac{1}{p!}z^{p}+\dfrac{1}{(p+1)!}z^{p+1},\quad z\in{U}.

Then $f\in R_{p}\cap R_{p+1}$ .

Proposition 5.11.

Let $p\in\mathbb{N}^{*}$ . In general, $R_{p}\not\subseteq R_{p+1}$ .

For $p\in\mathbb{N}^{*}$ we can find a function $f$ which belongs to the class $R_{p}$ , but does not belong to the class $R_{p+1}$ . We present two examples to illustrate this statement.

Example 5.12.

Let $f:U\rightarrow\mathbb{C}$ be given by $f(z)=z$ , $z\in{U}$ . Then $f\in{R}=R_{1}$ , but $f\not\in{R_{2}}$ .

Example 5.13.

Let $p\geq{2}$ and let $f:U\rightarrow\mathbb{C}$ be given by $f(z)=z+\frac{1}{p!}z^{p}$ , $z\in{U}$ . Then $f\in{R_{p}}$ , but $f\not\in R_{p+1}$ .

Remark 5.14.

The above example can be generalized by adding the terms between $z$ and $\frac{1}{p!}z^{p}$ . We can consider the function $f:U\rightarrow\mathbb{C}$ given by

f(z)=z+\sum_{n=2}^{p-1}a_{n}z^{n}+\dfrac{1}{p!}z^{p},\quad z\in{U}.

For $n\in\{2,3,...,p-1\}$ the coefficients $a_{n}$ can be real or complex numbers, but $a_{1}=1$ and $a_{p}=\dfrac{1}{p!}\in\mathbb{R}$ .

Proposition 5.15.

Let $p\in\mathbb{N}^{*}$ . In general, $R_{p+1}\not\subseteq{R_{p}}$ .

For $p\in\mathbb{N}^{*}$ we can find a function $f$ which belongs to the class $R_{p+1}$ , but does not belong to the class $R_{p}$ . We present also two examples to illustrate this statement.

Example 5.16.

Let $f:U\rightarrow\mathbb{C}$ be given by $f(z)=z+\frac{1}{2!}z^{2}+\frac{1}{3!}z^{3}$ , $z\in{U}$ . Then $f\in R_{2}$ , but $f\not\in R_{1}$ .

Proof.

Indeed, we have

f(0)=0,\quad f^{\prime}(z)=1+z+\frac{z^{2}}{2}\quad\text{and}\quad f^{\prime% \prime}(z)=1+z,\quad z\in{U}.

Then

f^{\prime}(0)=f^{\prime\prime}(0)=1\quad\text{and}\quad\text{Re}f^{\prime% \prime}(z)=1+\text{Re}z>0,\quad z\in{U}.

Hence, in view of definition, $f\in R_{2}$ . But,

\text{Re}f^{\prime}(z)=1+\text{Re}z+\frac{1}{2}\text{Re}z^{2}>-\frac{1}{2},% \quad z\in{U}.

Then Re $f^{\prime}(z)\not>0$ , $z\in{U}$ and hence $f\not\in R_{1}$ . ∎

Example 5.17.

Let $p\geq{2}$ and let $f:U\rightarrow\mathbb{C}$ be given by $f(z)=z+\frac{1}{(p+1)!}z^{p+1}$ , $z\in{U}$ . Then $f\in R_{p+1}$ , but $f\not\in R_{p}$ .

Remark 5.18.

Let $p\in\mathbb{N}^{*}$ . Then

(1)

$R_{p}\not\subseteq R_{p+1}$ ;
(2)

$R_{p}\not\supseteq R_{p+1}$ ;
(3)

$R_{p}\cap R_{p+1}\not=\emptyset$ .

Remark 5.19.

Let $p\geq{2}$ and consider the polynomial

q(z)=z+a_{2}z^{2}+a_{3}z^{3}+...+a_{p-1}z^{p-1}+a_{p}z^{p},\quad z\in{U}.

Then $q\in R_{p}$ if and only if $a_{p}=\dfrac{1}{p!}$ .

5.1. Structure formula for $p=2$ and $p=3$

Proposition 5.20.

Let $f:U\rightarrow\mathbb{C}$ . Then $f\in R_{2}$ if and only if there exists a function $\mu$ measurable on $[0,2\pi]$ such that

f(z)=-\dfrac{z^{2}}{2}-2\cdot\int_{0}^{2\pi}e^{it}\big{[}(z-e^{it})\log(1-ze^{% -it})-z\big{]}d\mu(t),

where $\log 1=0$ .

Proof.

According to Remark 5.2 we have that $f^{\prime\prime}\in\mathcal{P}$ . Hence, in view of Herglotz formula we obtain that

f^{\prime\prime}(z)=\int_{0}^{2\pi}\dfrac{e^{it}+z}{e^{it}-z}d\mu(t),\quad\mu% \in[0,2\pi].

Then,

f(z)=\int_{0}^{z}\bigg{(}\int_{0}^{z}\int_{0}^{2\pi}\dfrac{e^{it}+s}{e^{it}-s}% d\mu(t)ds\bigg{)}ds=\int_{0}^{z}\bigg{[}\int_{0}^{2\pi}\bigg{(}\int_{0}^{z}% \dfrac{e^{it}+s}{e^{it}-s}ds\bigg{)}d\mu(t)\bigg{]}ds.

Using [7, Theorem 3.2.2] we know that

f(z)=\int_{0}^{z}\big{[}-\zeta-2\int_{0}^{2\pi}e^{it}\log(1-\zeta e^{-it})d\mu% (t)\big{]}d\zeta,

so we obtain

f(z)=-\dfrac{z^{2}}{2}-2\cdot\int_{0}^{2\pi}e^{it}\big{[}(z-e^{it})\log(1-ze^{% -it})-z\big{]}d\mu(t).

∎

Remark 5.21.

It is possible to obtain a structure formula for the case $p=3$ :

f(z)=-\dfrac{z^{3}}{6}-2\cdot\int_{0}^{2\pi}e^{it}\bigg{[}\bigg{(}\dfrac{z^{2}% }{2}+e^{-it}-e^{it}(z-e^{it})\bigg{)}\log(1-ze^{-it})-2z-\dfrac{z^{2}}{2}\bigg% {]}d\mu(t),

where $\log 1=0$ .

6. The class $R_{p}(\alpha)$

Let $\alpha\in[0,1)$ and $p\in\mathbb{N}^{*}$ . Then we define

R_{p}(\alpha)=\{f\in\mathcal{H}(U):f(0)=0,f^{\prime}(0)=1,f^{(p)}(0)=1,\text{% Re}\big{[}f^{(p)}(z)\big{]}>\alpha,z\in{U}\}.

the class of normalized functions whose $p$ -th derivative has positive real part of order $\alpha$ .

Remark 6.1.

Let $\alpha\in[0,1)$ and $p\in\mathbb{N}^{*}$ . Then $f\in R_{p}(\alpha)$ if and only if $g\in\mathcal{P}$ , where $g:U\rightarrow\mathbb{C}$ is given by

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

Proposition 6.2.

Let $\alpha\in[0,1)$ and $p\in\mathbb{N}^{*}$ . If $f\in R_{p}(\alpha)$ , then the following relation hold:

|a_{n}|\leq\dfrac{2(1-\alpha)(n-p)!}{n!},\quad n\geq p,

(6.1)

Proof.

Similar to the proof of Proposition 4.2. ∎

Theorem 6.3.

Let $\alpha\in[0,1)$ and $p\in\mathbb{N}^{*}$ . If $f\in{R_{p}(\alpha)}$ , then the following estimate hold for all $k\in\mathbb{N}^{*}$ with $k\geq p$ :

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

(6.2)

Proof.

Similar to the proof of Theorem 4.5. ∎

Remark 6.4.

If $\alpha=0$ , then $R_{p}(0)=R_{p}$ and we obtain Proposition 5.3 and Theorem 5.4 from previous section. If, in addition, $p=1$ , then $R_{1}(0)=R$ and we obtain the coefficient estimates, respectively the growth and distortion theorem regarded to the class $R$ .

References

[1] Duren, P.L., Univalent functions, Springer-Verlag, Berlin and New York, 1983.
[2] Goodman, A.W., Univalent Functions, Vols. I and II, Mariner Publ. Co., Tampa, Florida, 1983.
[3] Graham, I., Kohr, G., Geometric function theory in one and higher dimensions, Marcel Deker Inc., New York, 2003.
[4] Krishna, D.V., RamReddy, T., Coefficient inequality for a function whose derivative has a positive real part of order $\alpha$ , Math. Bohem., 140 (2015), 43-52.
[5] Krishna, D.V., Venkateswarlu, B., RamReddy, T., Third Hankel determinant for bounded turning functions of order alpha, J. Nigerian Math. Soc., 34 (2015), 121-127.
[6] MacGregor, T.H., Functions whose derivative has a positive real part, Trans. Amer. Math. Soc., 104 (1962), 532-537.
[7] Mocanu, P.T., Bulboacă, T., Sălăgean, G.Ş., Geometric Theory of Univalent Functions, House of the Book of Science, Cluj-Napoca, 2006 (in romanian).
[8] Thomas, D.K., On functions whose derivative has positive real part, Proc. Amer. Math. Soc., 98 (1986), 68-70.

On some classes of holomorphic functions whose derivatives have positive real part

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On some classes of holomorphic functions whose derivatives have positive real part

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1. Introduction

Remark 1.1.

Remark 1.2.

Remark 1.3.

2. Preliminaries

Proposition 2.1.

Proposition 2.2.

Remark 2.3.

Proof.

3. General distortion result for the class R

Theorem 3.1.

Proof.

Remark 3.2.

4. Some remarks on the class R⁢(α)

Remark 4.1.

Proposition 4.2.

Proof.

Remark 4.3.

Theorem 4.4.

Proof.

Theorem 4.5.

Proof.

Remark 4.6.

Remark 4.7.

5. The class Rp

Remark 5.1.

Remark 5.2.

Proposition 5.3.

Proof.

Theorem 5.4.

Proof.

Remark 5.5.

Remark 5.6.

Remark 5.7.

Proposition 5.8.

Example 5.9.

Proof.

Example 5.10.

Proposition 5.11.

Example 5.12.

Example 5.13.

Remark 5.14.

Proposition 5.15.

Example 5.16.

Proof.

Example 5.17.

Remark 5.18.

Remark 5.19.

5.1. Structure formula for p=2 and p=3

Proposition 5.20.

Proof.

Remark 5.21.

6. The class Rp⁢(α)

Remark 6.1.

Proposition 6.2.

Proof.

Theorem 6.3.

Proof.

Remark 6.4.

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3. General distortion result for the class $R$

4. Some remarks on the class $R(\alpha)$

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5.1. Structure formula for $p=2$ and $p=3$

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