On a Length Problem for Univalent Functions

Abstract

Let g be an analytic function with the normalization in the open unit disk. Let $L\left(r\right)$ be the length of $g\left(\right\{z:\left|z\right|=r\left\}\right)$. In this paper we present a correspondence between g and $L\left(r\right)$ for the case when g is not necessary univalent. Furthermore, some other results related to the length of analytic functions are also discussed.

1. Introduction

Let $\mathcal{A}$ be the family of functions of the form

$$g\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n$$

(1)

which are analytic in the open unit disk $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$. Let $\mathcal{S}$ denote the subfamily of $\mathcal{A}$ consisting of all univalent functions in $\mathbb{D}$.

Let $C\left(r\right)$ denote the image curve of the $\left|z\right|=r<1$ under the function $g\in \mathcal{A}$ which bound the area $A\left(r\right)$. Furthermore, let $L\left(r\right)$ be the length of $C\left(r\right)$ and $M\left(r\right)={max}_{\left|z\right|=r<1}\left|g\left(z\right)\right|$.

If $g\in \mathcal{A}$ satisfies

$$\mathfrak{Re}\left\{\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right\}>0,z\in \mathbb{D},$$

then g is said to be starlike with respect to the origin in $\mathbb{D}$ and we write $g\in {\mathcal{S}}^{*}$. It is known (for details, see [1,2]) that ${\mathcal{S}}^{*}\subset \mathcal{S}$.

The aim of the present paper is to prove, using a modified methodology, that in the following implication

$$g\in {\mathcal{S}}^{*}\Rightarrow L\left(r\right)=\mathcal{O}\left(M\left(r\right)log\frac{1}{1-r}\right)\mathrm{as}r\to 1,$$

(2)

where $\mathcal{O}$ denotes the Landau’s symbol, the assumption that g is starlike univalent can be changed by a weaker one. Result (2) was proved by Keogh [3]. Moreover, some other length problems for analytic functions are investigated. Several interesting developments related to length problems for univalent functions were considered in [4,5,6,7,8,9,10,11,12,13,14,15].

2. Main Results

Theorem 1.

Let g be of the form (1) and suppose that

$$\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\le \left|\frac{1+z}{1-z}\right|,z\in \mathbb{D}.$$

(3)

Then

$$L\left(r\right)=\mathcal{O}\left(M\left(r\right)log\frac{1}{1-r}\right)\mathrm{as}r\to 1,$$

where

$$M\left(r\right)=\underset{\left|z\right|=r<1}{max}\left|g\left(z\right)\right|$$

and $\mathcal{O}$ means Landau’s symbol.

Proof. 

Let $z=r{e}^{i\nu }$. We have $g\ne 0$ in $\mathbb{D}\backslash \left\{0\right\}$. In fact, if $g=0$ in $\mathbb{D}$, it contradicts hypothesis (3). Applying [3] (Theorem 1) and the hypothesis of Theorem 1, we have

$$\begin{array}{cc}\hfill L\left(r\right)& ={\int }_0^{2\pi }|z{g}^{\prime }\left(z\right)|\mathrm{d}\nu ={\int }_0^{2\pi }\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\left|g\left(z\right)\right|\mathrm{d}\nu \hfill \\ & \le M\left(r\right){\int }_0^{2\pi }\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\mathrm{d}\nu \le M\left(r\right){\int }_0^{2\pi }\left|\frac{1+r{e}^{i\nu }}{1-r{e}^{i\nu }}\right|\mathrm{d}\nu \hfill \\ & \le M\left(r\right)\left(2\pi +4log\frac{1+r}{1-r}\right)\mathrm{as}r\to 1.\hfill \end{array}$$

 □

Remark 1.

If g satisfies the condition of Theorem 1, then g is not necessary univalent in $\mathbb{D}$. It is well known that if $g\in \mathcal{S}$, then it follows that

$$\frac{1-\left|z\right|}{1+\left|z\right|}\le \left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\le \frac{1+\left|z\right|}{1-\left|z\right|},z\in \mathbb{D}$$

(for details, see [1] (Vol. 1, p. 69)).

If $g\in \mathcal{A}$ satisfies

$$\mathfrak{Re}\left\{\frac{z{g}^{\prime }\left(z\right)}{g^{1-\gamma }\left(z\right)h^{\gamma }\left(z\right)}\right\}>0,z\in \mathbb{D}$$

for some $h\in {\mathcal{S}}^{*}$ and some $\gamma \in (0,\infty )$, then g is said to be a Bazilevič function of type γ [13]. The class of Bazilevič functions of type γ is denoted by $g\in \mathcal{B}\left(\gamma \right)$. We note that Theorem 1 improves the implication (2) by Keogh [3] and it is also related to Theorem 3 given by Thomas [13].

We will need the following Tsuji’s result.

Lemma 1

([16] (p. 226)). (Theorem 3) If $0\le r<R$ and $z=e^{i\nu }$, then

$$\frac{R-r}{R+r}\le \mathfrak{Re}\left\{\frac{R{e}^{i\varphi }+z}{R{e}^{i\varphi }-z}\right\}=\frac{R^2-r^2}{R^2-2Rrcos(\varphi -\nu )+r^2}\le \frac{R+r}{R-r}.$$

(4)

Moreover,

$$\frac{1}{2\pi }{\int }_0^{2\pi }\frac{R^2-r^2}{R^2-2Rrcos(\varphi -\nu )+r^2}\mathrm{d}\nu =1.$$

(5)

Theorem 2.

Let g be of the form (1) and suppose that

$$\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\le \left|\frac{1+z}{1-z}\right|,z\in \mathbb{D}$$

(6)

and

$$M(r,\beta )=\underset{\left|z\right|=r<1}{max}\left|g\left(z\right)\right|\le {\left|\frac{1+z}{1-z}\right|}^{\beta },$$

(7)

where $1<\beta$. Then

$$L\left(r\right)=\mathcal{O}\left(\frac{1}{{(1-r)}^{\beta }}\right)\mathrm{as}r\to 1,$$

where $\mathcal{O}$ means Landau’s symbol.

Proof. 

From the hypotheses (6) and (7), it follows that

$$\begin{array}{cc}\hfill L\left(r\right)& ={\int }_0^{2\pi }|z{g}^{\prime }\left(z\right)|\mathrm{d}\nu ={\int }_0^{2\pi }\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\left|g\left(z\right)\right|\mathrm{d}\nu \hfill \\ & \le {\int }_0^{2\pi }\left|\frac{1+z}{1-z}\right|{\left|\frac{1+z}{1-z}\right|}^{\beta }\mathrm{d}\nu \le 2^{1+\beta }{\int }_0^{2\pi }\frac{1}{{|1-z|}^{1+\beta }}\mathrm{d}\nu \hfill \\ & =\frac{2^{1+\beta }}{{(1-r)}^{\beta -1}}{\int }_0^{2\pi }\frac{1}{1-2rcos\nu +r^2}\mathrm{d}\nu .\hfill \end{array}$$

From (5), we have

$${\int }_0^{2\pi }\frac{1}{1-2rcos\nu +r^2}\mathrm{d}\nu =\frac{2\pi }{1-r^2}.$$

Hence, we obtain

$$\begin{array}{cc}\hfill L\left(r\right)& \le \frac{2^{1+\beta }}{{(1-r)}^{\beta -1}}\frac{2\pi }{1-r^2}\hfill \\ & =\mathcal{O}\left(\frac{1}{{(1-r)}^{\beta }}\right)\mathrm{as}r\to 1.\hfill \end{array}$$

Therefore, we complete the proof of Theorem 2. □

Let us recall the following Fejér-Riesz’s result.

Lemma 2

([16]). Let h be analytic in $\mathbb{D}$ and continuous on $\overline{\mathbb{D}}$. Then

$${\int }_{-1}^1{\left|h\left(z\right)\right|}^p\left|\mathrm{d}z\right|\le \frac{1}{2}{\int }_{\left|z\right|=1}{\left|h\left(z\right)\right|}^p\left|\mathrm{d}z\right|,$$

where $p>0$.

Theorem 3.

Let g be of the form (1) and suppose that

$$\frac{1-\left|z\right|}{1+\left|z\right|}\le \left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\le \frac{1+\left|z\right|}{1-\left|z\right|},z\in \mathbb{D}.$$

(8)

Then

$$\mathcal{O}\left(m\left(r\right)log\frac{1}{1-r}\right)\le L\left(r\right)\le \mathcal{O}\left(\frac{M\left(r\right)}{1-r}\right)\mathrm{as}r\to 1,$$

where

$$m\left(r\right)=\underset{\left|z\right|=r<1}{min}\left|g\left(z\right)\right|,M\left(r\right)=\underset{\left|z\right|=r<1}{max}\left|g\left(z\right)\right|$$

(9)

and $\mathcal{O}$ means Landau’s symbol.

Proof. 

From the assumption, we have

$$\begin{array}{cc}\hfill L\left(r\right)& ={\int }_0^{2\pi }|z{g}^{\prime }\left(z\right)|\mathrm{d}\nu ={\int }_0^{2\pi }\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\left|g\left(z\right)\right|\mathrm{d}\nu \hfill \\ & \ge m\left(r\right){\int }_0^{2\pi }\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\mathrm{d}\nu \hfill \end{array}$$

because $g\left(z\right)\ne 0$ in $\mathbb{D}\backslash \left\{0\right\}$. In fact, if $g\left(z\right)=0$ in $\mathbb{D}$, it contradicts hypothesis (8). Applying Fejér-Riesz’s Lemma 2, we have

$$\begin{array}{cc}\hfill L\left(r\right)& \ge m\left(r\right){\int }_0^{2\pi }\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\mathrm{d}\nu \ge 2m\left(r\right){\int }_{-r}^r\frac{1-\rho }{1+\rho }\mathrm{d}\rho \hfill \\ & \ge 2m\left(r\right)log\frac{1+r}{1-r}-2r\hfill \\ & =\mathcal{O}\left(m\left(r\right)log\frac{1}{(1-r)}\right)\mathrm{as}r\to 1.\hfill \end{array}$$

While, we obtain

$$\begin{array}{cc}\hfill L\left(r\right)& ={\int }_0^{2\pi }|z{g}^{\prime }\left(z\right)|\mathrm{d}\nu ={\int }_0^{2\pi }\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\left|g\left(z\right)\right|\mathrm{d}\nu \hfill \\ & =M\left(r\right){\int }_0^{2\pi }\frac{1+\left|z\right|}{1-\left|z\right|}\mathrm{d}\nu =2\pi M\left(r\right)\frac{1+r}{1-r}\hfill \\ & =\mathcal{O}\left(\frac{M\left(r\right)}{1-r}\right)\mathrm{as}r\to 1.\hfill \end{array}$$

Therefore, we complete the proof of Theorem 3. □

From Theorem 3, we have the following result.

Corollary 1.

Let g be of the form (1) and suppose that g is univalent in $\mathbb{D}$. Then we have

$$\mathcal{O}\left(m\left(r\right)log\frac{1}{1-r}\right)\le L\left(r\right)\le \mathcal{O}\left(\frac{M\left(r\right)}{1-r}\right)\mathrm{as}r\to 1,$$

where $m\left(r\right)$ and $M\left(r\right)$ are given by (9), respectively.

Proof. 

From the hypothesis, we have

$$\frac{1-\left|z\right|}{1+z|}\le \left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\le \frac{1+\left|z\right|}{1-\left|z\right|},z\in \mathbb{D},$$

which completes the proof. □

Lemma 3

([17] (p. 280) and [18] (p. 491)).

$${\int }_0^{2\pi }\frac{\mathrm{d}\nu }{|1-r{e}^{i\nu }{|}^{\beta }}=\left\{\begin{array}{cc}\mathcal{O}\left({(1-r)}^{1-\beta }\right)\hfill & forthecase1<\beta ,\hfill \\ \mathcal{O}\left(log\frac{1}{1-r}\right)\hfill & forthecase\beta =1,\hfill \\ \mathcal{O}\left(1\right)\hfill & forthecase0\le \beta <1,\hfill \end{array}\right.$$

where $0<r<1$, $0\le \nu \le 2\pi$, $0\le \beta$ and $\mathcal{O}$ means Landau’s symbol.

Theorem 4.

Let g be of the form (1) and suppose that

$$\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\le \frac{1}{1-\left|z\right|},z\in \mathbb{D}$$

(10)

and

$$\left|g\left(z\right)\right|\le \frac{1}{{|1-z|}^{\beta }},z\in \mathbb{D}.$$

(11)

Then

$$L\left(r\right)\le \left\{\begin{array}{cc}\mathcal{O}\left({(1-r)}^{-3/2}\right)\hfill & for1<\beta \le 3/2,\hfill \\ \mathcal{O}\left({(1-r)}^{-3/2}log\frac{1}{1-r}\right)\hfill & forthecase\beta =3/2,\hfill \\ \mathcal{O}\left({(1-r)}^{-\beta }\right)\hfill & forthecase3/2<\beta ,\hfill \end{array}\right.$$

where $0<\left|z\right|=r<1$ and $\mathcal{O}$ means Landau’s symbol.

Proof. 

From the hypothesis (10), it follows that $g\left(z\right)\ne 0$ in $\mathbb{D}\backslash \left\{0\right\}$. Then we have

$$\begin{array}{cc}\hfill L\left(r\right)& ={\int }_0^{2\pi }\left|r{e}^{i\nu }{g}^{\prime }\left(r{e}^{i\nu }\right)\right|\mathrm{d}\nu ={\int }_0^{2\pi }\left|\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right|\left|g\left(z\right)\right|\mathrm{d}\nu \hfill \\ & <{\int }_0^{2\pi }\left(\frac{1}{1-\left|z\right|}\right)\left(\frac{1}{{|1-z|}^{\beta }}\right)\mathrm{d}\nu \hfill \\ & ={\int }_0^{2\pi }\left(\frac{1}{|1-z|}\right)\left(\frac{1}{{|1-z|}^{\beta -1}}\right)\left(\frac{1}{1-\left|z\right|}\right)\mathrm{d}\nu \hfill \\ & \le {\left({\int }_0^{2\pi }\frac{1}{{|1-z|}^2}\mathrm{d}\nu \right)}^{1/2}{\left({\int }_0^{2\pi }\left(\frac{1}{{|1-z|}^{2\beta -2}}\right)\frac{1}{{(1-|z\left|\right)}^2}\mathrm{d}\nu \right)}^{1/2}.\hfill \end{array}$$

Applying Hayman’s Lemma 3, we have

$$\begin{array}{cc}\hfill L\left(r\right)& \le {\left(\frac{1}{1-r^2}\right)}^{1/2}\left(\frac{1}{1-r}\right)\mathcal{O}\left(1\right)\hfill \\ & =\mathcal{O}\left(\frac{1}{{(1-r)}^{3/2}}\right)\mathrm{as}r\to 1\hfill \end{array}$$

for the case $1<\beta <3/2$,

$$\begin{array}{cc}\hfill L\left(r\right)& \le {\left(\frac{1}{1-r^2}\right)}^{1/2}\left(\frac{1}{1-r}\right)\mathcal{O}\left(log\frac{1}{1-r}\right)\hfill \\ & =\mathcal{O}\left(\frac{1}{{(1-r)}^{3/2}}log\frac{1}{1-r}\right)\mathrm{as}r\to 1\hfill \end{array}$$

for the case $\beta =3/2$ and

$$\begin{array}{cc}\hfill L\left(r\right)& ={\left(\frac{1}{1-r^2}\right)}^{1/2}\left(\frac{1}{1-r}\right){\left(\frac{1}{1-r}\right)}^{(2\beta -3)/2}\mathrm{as}r\to 1\hfill \end{array}$$

for the case $3/2<\beta$. □

Lemma 4

([16] (p. 227)). If $g\left(z\right)=u\left(z\right)+iv\left(z\right)$ is analytic in $\left|z\right|\le R$, then

$$g\left(z\right)=\frac{1}{2\pi }{\int }_0^{2\pi }u\left(R{e}^{i\varphi }\right)\frac{R{e}^{i\varphi }+z}{R{e}^{i\varphi }-z}\mathrm{d}\varphi +iv\left(0\right).$$

(12)

Moreover, if $\left|z\right|<R$ and $v\left(0\right)=0$, then

$$\left|g\left(z\right)\right|=\frac{1}{2\pi }{\int }_0^{2\pi }\left|u\left(R{e}^{i\varphi }\right)\right|\left|\frac{R{e}^{i\varphi }+z}{R{e}^{i\varphi }-z}\right|\mathrm{d}\varphi .$$

Theorem 5.

Let g be of the form (1). Then

$$M\left(r\right)=\mathcal{O}\left(A\left(r\right)log\frac{1}{1-r}\right)asr\to 1,$$

(13)

where $0<\left|z\right|=r<1$ and $\mathcal{O}$ means Landau’s symbol.

Proof. 

It follows that

$$\begin{array}{c}\hfill M\left(r\right)=\underset{\left|z\right|=r<1}{max}\left|{\int }_0^z{g}^{\prime }\left(s\right)\mathrm{d}s\right|=\underset{\left|z\right|=r<1}{max}\left|{\int }_0^r{g}^{\prime }\left(\rho e^{i\nu }\right)\mathrm{d}\rho \right|.\end{array}$$

Applying (12), we have

$$\begin{array}{ccc}\hfill M\left(r\right)& =& \underset{\left|z\right|=r<1}{max}\left|\frac{1}{2\pi }{\int }_0^r{\int }_0^{2\pi }\mathfrak{Re}{g}^{\prime }\left(t{e}^{i\nu }\right)\frac{t{e}^{i\varphi }+\rho e^{i\nu }}{t{e}^{i\varphi }-\rho e^{i\nu }}\mathrm{d}\varphi \mathrm{d}\rho \right|\hfill \\ & \le & \underset{\left|z\right|=r<1}{max}\frac{1}{2\pi }{\int }_0^r{\int }_0^{2\pi }\left|g^{\prime }\left(t{e}^{i\nu }\right)\right|\left|\frac{t{e}^{i\varphi }+\rho e^{i\nu }}{t{e}^{i\varphi }-\rho e^{i\nu }}\right|\mathrm{d}\varphi \mathrm{d}\rho ,\hfill \end{array}$$

where $0\le \rho \le r<t<1$. Then, applying Schwarz’s lemma, we have

$$\begin{array}{ccc}\hfill M\left(r\right)& \le & \underset{\left|z\right|=r<1}{max}{\left(\frac{1}{2\pi }{\int }_0^r{\int }_0^{2\pi }{\left|g^{\prime }\left(t{e}^{i\nu }\right)\right|}^2\mathrm{d}\varphi \mathrm{d}\rho \right)}^{1/2}{\left({\int }_0^r{\int }_0^{2\pi }{\left|\frac{t{e}^{i\varphi }+\rho e^{i\nu }}{t{e}^{i\varphi }-\rho e^{i\nu }}\right|}^2\mathrm{d}\varphi \mathrm{d}\rho \right)}^{1/2}\hfill \\ & \le & \underset{\left|z\right|=r<1}{max}{\left(I_1\right)}^{1/2}{\left(I_2\right)}^{1/2},\mathrm{say}.\hfill \end{array}$$

Putting $0<r_1<r$ and $t=\sqrt{(1+{\rho }^2)/2}$, we have

$$\rho \mathrm{d}\rho =2\sqrt{\frac{1+{\rho }^2}{2}}\mathrm{d}t<2\mathrm{d}t.$$

Then we have

$$\begin{array}{ccc}\hfill I_1& =& \frac{1}{2\pi }{\int }_0^{r_1}{\int }_0^{2\pi }{\left|g^{\prime }\left(t{e}^{i\varphi }\right)\right|}^2\mathrm{d}\varphi \mathrm{d}\rho +\frac{1}{2\pi r_1^2}{\int }_{\sqrt{(1+r_1^2)/2}}^{\sqrt{(1+r^2)/2}}{\int }_0^{2\pi }t{\left|g^{\prime }\left(t{e}^{i\varphi }\right)\right|}^2\mathrm{d}\varphi \mathrm{d}t\hfill \\ & \le & C+\frac{1}{2\pi r_1^2}A\left(\sqrt{\frac{1+r^2}{2}}\right)\hfill \\ & =& C+\frac{1}{2\pi r_1^2}A\left(\sqrt{\frac{1+r^2}{2r^2}}r\right)\hfill \\ & =& \mathcal{O}\left(A\right(r\left)\right)\mathrm{as}r\to 1,\hfill \end{array}$$

where C is a bounded positive constant. On the other hand, putting $t\to 1^{-}$, we have

$$\begin{array}{ccc}\hfill I_2& =& {\int }_0^r{\int }_0^{2\pi }{\left|\frac{t{e}^{i\varphi }+\rho e^{i\nu }}{t{e}^{i\varphi }-\rho e^{i\nu }}\right|}^2\mathrm{d}\varphi \mathrm{d}\rho \hfill \\ & \le & {\int }_0^r{\int }_0^{2\pi }\frac{4}{{\left|t{e}^{i\varphi }-\rho e^{i\nu }\right|}^2}\mathrm{d}\varphi \mathrm{d}\rho \hfill \\ & =& {\int }_0^r{\int }_0^{2\pi }\frac{4}{t^2-2\rho tcos(\varphi -\nu )+{\rho }^2}\mathrm{d}\varphi \mathrm{d}\rho .\hfill \end{array}$$

Using (5), we have

$$\begin{array}{ccc}\hfill I_2& \le & 8\pi {\int }_0^r\frac{1}{t^2-{\rho }^2}\mathrm{d}\rho \hfill \\ & =& \frac{4\pi }{t}{\int }_0^r\left(\frac{1}{t+\rho }+\frac{1}{t-\rho }\right)\mathrm{d}\rho \hfill \\ & =& \frac{4\pi }{t}log\frac{t+r}{t-r}\to \mathcal{O}\left(log\frac{1}{1-r}\right)\mathrm{as}r\to 1.\hfill \end{array}$$

Therefore we complete the proof of (13). □

Remark 2.

In Theorem 5, we do not suppose that g is univalent in $\left|z\right|<1$ and therefore, it improves the result by Pommerenke [2].