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

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