**2. Main Results**

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

$$\left|\frac{zg'(z)}{g(z)}\right| \le \left|\frac{1+z}{1-z}\right|, \quad z \in \mathbb{D}.\tag{3}$$

*Then*

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

*where*

$$M(r) = \max\_{|z|=r<1} |g(z)|$$

*and* O *means Landau's symbol.*

**Proof.** Let *<sup>z</sup>* <sup>=</sup> *reiν*. We have *<sup>g</sup>* <sup>=</sup> 0 in <sup>D</sup> \ {0}. In fact, if *<sup>g</sup>* <sup>=</sup> 0 in <sup>D</sup>, it contradicts hypothesis (3). Applying [3] (Theorem 1) and the hypothesis of Theorem 1, we have

$$\begin{split} L(r) &= \int\_{0}^{2\pi} |zg'(z)| \mathbf{d}\nu = \int\_{0}^{2\pi} \left| \frac{zg'(z)}{\mathfrak{g}(z)} \right| |\mathfrak{g}(z)| \mathbf{d}\nu \\ &\leq M(r) \int\_{0}^{2\pi} \left| \frac{z\mathfrak{g}'(z)}{\mathfrak{g}(z)} \right| \mathbf{d}\nu \leq M(r) \int\_{0}^{2\pi} \left| \frac{1 + r\mathfrak{e}^{\mathrm{i}\nu}}{1 - r\mathfrak{e}^{\mathrm{i}\nu}} \right| \mathbf{d}\nu \\ &\leq M(r) \left( 2\pi + 4 \log \frac{1 + r}{1 - r} \right) \quad \text{as} \quad r \to 1. \end{split}$$

**Remark 1.** *If g satisfies the condition of Theorem 1, then g is not necessary univalent in* D*. It is well known that if g* ∈ S*, then it follows that*

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

*(for details, see [1] (Vol. 1, p. 69)). If g* ∈ A *satisfies*

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

*for some h* ∈ S<sup>∗</sup> *and some γ* ∈ (0, ∞)*, then g is said to be a Bazilevic function of type ˘ γ [13]. The class of Bazilevic functions of type ˘ γ is denoted by g* ∈ B(*γ*) *. 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* <sup>0</sup> <sup>≤</sup> *<sup>r</sup>* <sup>&</sup>lt; *R and z* <sup>=</sup> *<sup>e</sup>iν, then*

$$\frac{R-r}{R+r} \le \Re \text{e} \left\{ \frac{Re^{i\phi}+z}{Re^{i\phi}-z} \right\} = \frac{R^2-r^2}{R^2-2Rr\cos(\phi-\nu)+r^2} \le \frac{R+r}{R-r}.\tag{4}$$

*Moreover,*

$$\frac{1}{2\pi} \int\_0^{2\pi} \frac{R^2 - r^2}{R^2 - 2Rr\cos(\phi - \nu) + r^2} d\nu = 1. \tag{5}$$

**Theorem 2.** *Let g be of the form* (1) *and suppose that*

$$\left|\frac{zg'(z)}{g(z)}\right| \le \left|\frac{1+z}{1-z}\right|, \quad z \in \mathbb{D} \tag{6}$$

*and*

*M*(*r*, *β*) = max |*z*|=*r*<1 |*g*(*z*)| ≤ 1 + *z* 1 − *z β* , (7)

*where* 1 < *β. Then*

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

*where* O *means Landau's symbol.*

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

$$\begin{split} L(r) &= \int\_0^{2\pi} |zg'(z)| \mathrm{d}\nu = \int\_0^{2\pi} \left| \frac{zg'(z)}{g(z)} \right| |g(z)| \mathrm{d}\nu \\ &\leq \int\_0^{2\pi} \left| \frac{1+z}{1-z} \right| \left| \frac{1+z}{1-z} \right|^\beta \mathrm{d}\nu \leq 2^{1+\beta} \int\_0^{2\pi} \frac{1}{|1-z|^{1+\beta}} \mathrm{d}\nu \\ &= \frac{2^{1+\beta}}{(1-r)^{\beta-1}} \int\_0^{2\pi} \frac{1}{1-2r\cos\upsilon + r^2} \mathrm{d}\upsilon. \end{split}$$

From (5), we have

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

Hence, we obtain

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

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* D *and continuous on* D*. Then*

$$\int\_{-1}^{1} |h(z)|^p |\mathbf{dz}| \le \frac{1}{2} \int\_{|z|=1} |h(z)|^p |\mathbf{dz}|.$$

*where p* > 0*.*

**Theorem 3.** *Let g be of the form* (1) *and suppose that*

$$\frac{1-|z|}{1+|z|} \le \left|\frac{zg'(z)}{g(z)}\right| \le \frac{1+|z|}{1-|z|}, \quad z \in \mathbb{D}.\tag{8}$$

*Then*

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

*where*

$$m(r) = \min\_{|z|=r<1} |g(z)|\_{\prime} \quad M(r) = \max\_{|z|=r<1} |g(z)|\tag{9}$$

*and* O *means Landau's symbol.*

**Proof.** From the assumption, we have

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

because *<sup>g</sup>*(*z*) <sup>=</sup> 0 in <sup>D</sup> \ {0}. In fact, if *<sup>g</sup>*(*z*) = 0 in <sup>D</sup>, it contradicts hypothesis (8). Applying Fejér-Riesz's Lemma 2, we have

$$\begin{aligned} L(r) &\geq m(r) \int\_0^{2\pi} \left| \frac{zg'(z)}{g(z)} \right| \, \mathrm{d}\nu \geq 2m(r) \int\_{-r}^r \frac{1-\rho}{1+\rho} \mathrm{d}\rho\\ &\geq 2m(r) \log \frac{1+r}{1-r} - 2r\\ &= \mathcal{O}\left(m(r) \log \frac{1}{(1-r)}\right) \quad \text{as} \quad r \to 1. \end{aligned}$$

While, we obtain

$$\begin{split} L(r) &= \int\_{0}^{2\pi} |zg'(z)| \mathrm{d}\nu = \int\_{0}^{2\pi} \left| \frac{zg'(z)}{g(z)} \right| |g(z)| \mathrm{d}\nu \\ &= M(r) \int\_{0}^{2\pi} \frac{1+|z|}{1-|z|} \mathrm{d}\nu = 2\pi M(r) \frac{1+r}{1-r} \\ &= \mathcal{O}\left(\frac{M(r)}{1-r}\right) \quad \text{as} \quad r \to 1. \end{split}$$

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* D*. Then we have*

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

*where m*(*r*) *and M*(*r*) *are given by* (9)*, respectively.*

**Proof.** From the hypothesis, we have

$$\left|\frac{1-|z|}{1+z|}\leq \left|\frac{z\mathbf{g}'(z)}{\mathbf{g}(z)}\right|\leq \frac{1+|z|}{1-|z|},\quad z\in\mathbb{D}\_{\prime}$$

which completes the proof.

**Lemma 3** ([17] (p. 280) and [18] (p. 491))**.**

$$\int\_0^{2\pi} \frac{\mathbf{d}\nu}{|1 - r e^{i\nu}|^\beta} = \begin{cases} \mathcal{O}\left((1 - r)^{1 - \beta}\right) & \text{for the case } 1 < \beta, \\\mathcal{O}\left(\log \frac{1}{1 - r}\right) & \text{for the case } \beta = 1, \\\mathcal{O}\left(1\right) & \text{for the case } 0 \le \beta < 1, \end{cases}$$

*where* 0 < *r* < 1*,* 0 ≤ *ν* ≤ 2*π,* 0 ≤ *β and* O *means Landau's symbol.*

**Theorem 4.** *Let g be of the form* (1) *and suppose that*

$$\left|\frac{zg'(z)}{g(z)}\right| \le \frac{1}{1-|z|}, \quad z \in \mathbb{D} \tag{10}$$

*and*

$$|\lg(z)| \le \frac{1}{|1-z|^{\theta'}} \quad z \in \mathbb{D}.\tag{11}$$

*Then*

$$L(r) \le \begin{cases} \mathcal{O}\left((1-r)^{-3/2}\right) & \text{for } 1 < \beta \le 3/2, \\\mathcal{O}\left((1-r)^{-3/2}\log\frac{1}{1-r}\right) & \text{for the case } \beta = 3/2, \\\mathcal{O}\left((1-r)^{-\beta}\right) & \text{for the case } 3/2 < \beta, \end{cases}$$

*where* 0 < |*z*| = *r* < 1 *and* O *means Landau's symbol.*

**Proof.** From the hypothesis (10), it follows that *<sup>g</sup>*(*z*) <sup>=</sup> 0 in <sup>D</sup> \ {0}. Then we have

$$\begin{split} L(r) &= \int\_{0}^{2\pi} \left| r e^{i\nu} g'(re^{i\nu}) \right| \, \mathrm{d}\nu = \int\_{0}^{2\pi} \left| \frac{zg'(z)}{g(z)} \right| |g(z)| \, \mathrm{d}\nu \\ &< \int\_{0}^{2\pi} \left( \frac{1}{1-|z|} \right) \left( \frac{1}{|1-z|^{\beta}} \right) \mathrm{d}\nu \\ &= \int\_{0}^{2\pi} \left( \frac{1}{|1-z|} \right) \left( \frac{1}{|1-z|^{\beta-1}} \right) \left( \frac{1}{1-|z|} \right) \, \mathrm{d}\nu \\ &\leq \left( \int\_{0}^{2\pi} \frac{1}{|1-z|^{2}} \mathrm{d}\nu \right)^{1/2} \left( \int\_{0}^{2\pi} \left( \frac{1}{|1-z|^{2\beta-2}} \right) \frac{1}{(1-|z|)^{2}} \mathrm{d}\nu \right)^{1/2} . \end{split}$$

Applying Hayman's Lemma 3, we have

$$\begin{aligned} L(r) &\le \left(\frac{1}{1-r^2}\right)^{1/2} \left(\frac{1}{1-r}\right) \mathcal{O}(1) \\ &= \mathcal{O}\left(\frac{1}{(1-r)^{3/2}}\right) \quad \text{as} \quad r \to 1 \end{aligned}$$

for the case 1 < *β* < 3/2,

$$\begin{aligned} L(r) &\le \left(\frac{1}{1-r^2}\right)^{1/2} \left(\frac{1}{1-r}\right) \mathcal{O}\left(\log\frac{1}{1-r}\right) \\ &= \mathcal{O}\left(\frac{1}{(1-r)^{3/2}} \log \frac{1}{1-r}\right) \quad \text{as} \quad r \to 1 \end{aligned}$$

for the case *β* = 3/2 and

$$L(r) = \left(\frac{1}{1-r^2}\right)^{1/2} \left(\frac{1}{1-r}\right) \left(\frac{1}{1-r}\right)^{(2\beta-3)/2} \quad \text{as} \quad r \to 1$$

for the case 3/2 < *β*.

**Lemma 4** ([16] (p. 227))**.** *If g*(*z*) = *u*(*z*) + *iv*(*z*) *is analytic in* |*z*| ≤ *R, then*

$$\log(z) = \frac{1}{2\pi} \int\_0^{2\pi} u(Re^{i\phi}) \frac{Re^{i\phi} + z}{Re^{i\phi} - z} d\phi + iv(0). \tag{12}$$

*Moreover, if* |*z*| < *R and v*(0) = 0*, then*

$$|g(z)| = \frac{1}{2\pi} \int\_0^{2\pi} |\mu(Re^{i\phi})| \left| \frac{Re^{i\phi} + z}{Re^{i\phi} - z} \right| \,\mathrm{d}\phi\,.$$

**Theorem 5.** *Let g be of the form* (1)*. Then*

$$M(r) = \mathcal{O}\left(A(r)\log\frac{1}{1-r}\right) \quad \text{as} \ r \to 1,\tag{13}$$

*where* 0 < |*z*| = *r* < 1 *and* O *means Landau's symbol.*

**Proof.** It follows that

$$M(r) = \max\_{|z|=r<1} \left| \int\_0^z \mathbf{g}'(s) \mathbf{ds} \right| = \max\_{|z|=r<1} \left| \int\_0^r \mathbf{g}'(\rho \mathbf{e}^{i\nu}) \mathbf{d}\rho \right|.$$

Applying (12), we have

$$\begin{split} M(r) &= \max\_{|z|=r<1} \left| \frac{1}{2\pi} \int\_0^r \int\_0^{2\pi} \Re \mathbf{e} \mathbf{g}'(te^{i\nu}) \frac{te^{i\boldsymbol{\phi}} + \rho e^{i\nu}}{te^{i\boldsymbol{\phi}} - \rho e^{i\nu}} \mathbf{d}\phi \mathbf{d}\rho \right| \\ &\leq \max\_{|z|=r<1} \frac{1}{2\pi} \int\_0^r \int\_0^{2\pi} \left| \mathbf{g}'(te^{i\nu}) \right| \left| \frac{te^{i\boldsymbol{\phi}} + \rho e^{i\nu}}{te^{i\boldsymbol{\phi}} - \rho e^{i\nu}} \right| \mathbf{d}\phi \mathbf{d}\rho, \end{split}$$

where 0 ≤ *ρ* ≤ *r* < *t* < 1. Then, applying Schwarz's lemma, we have

$$\begin{split} M(r) &\leq \max\_{|z|=r<1} \left(\frac{1}{2\pi} \int\_0^r \int\_0^{2\pi} \left| \mathbf{g}'(te^{i\nu}) \right|^2 \mathbf{d}\rho \mathbf{d}\rho \right)^{1/2} \left( \int\_0^r \int\_0^{2\pi} \left| \frac{te^{i\phi} + \rho e^{i\nu}}{te^{i\phi} - \rho e^{i\nu}} \right|^2 \mathbf{d}\rho \mathbf{d}\rho \right)^{1/2} \\ &\leq \max\_{|z|=r<1} (I\_1)^{1/2} (I\_2)^{1/2} \text{ : say.} \end{split}$$

Putting 0 < *<sup>r</sup>*<sup>1</sup> < *<sup>r</sup>* and *<sup>t</sup>* = (<sup>1</sup> + *<sup>ρ</sup>*2)/2, we have

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

Then we have

$$\begin{split} I\_{1} &=& \frac{1}{2\pi} \int\_{0}^{r\_{1}} \int\_{0}^{2\pi} \left| \mathbf{g}'(te^{i\phi}) \right|^{2} \mathbf{d}\phi \mathbf{d}\rho + \frac{1}{2\pi r\_{1}^{2}} \int\_{\sqrt{(1+r\_{1}^{2})/2}}^{\sqrt{(1+r^{2})/2}} \int\_{0}^{2\pi} t \left| \mathbf{g}'(te^{i\phi}) \right|^{2} \mathbf{d}\phi \mathbf{d}t \\ &\leq& \mathcal{C} + \frac{1}{2\pi r\_{1}^{2}} A \left( \sqrt{\frac{1+r^{2}}{2}} \right) \\ &=& \mathcal{C} + \frac{1}{2\pi r\_{1}^{2}} A \left( \sqrt{\frac{1+r^{2}}{2r^{2}}} r \right) \\ &=& \mathcal{C}(A(r)) \quad \text{as } r \to 1, \end{split}$$

where *C* is a bounded positive constant. On the other hand, putting *t* → 1−, we have

$$\begin{split} I\_{2} &=& \int\_{0}^{r} \int\_{0}^{2\pi} \left| \frac{te^{i\Phi} + \rho e^{iv}}{te^{i\Phi} - \rho e^{iv}} \right|^{2} \mathrm{d}\phi \mathrm{d}\rho \\ &\leq& \int\_{0}^{r} \int\_{0}^{2\pi} \frac{4}{\left| te^{i\Phi} - \rho e^{iv} \right|^{2}} \mathrm{d}\phi \mathrm{d}\rho \\ &=& \int\_{0}^{r} \int\_{0}^{2\pi} \frac{4}{t^{2} - 2\rho t \cos(\phi - \nu) + \rho^{2}} \mathrm{d}\phi \mathrm{d}\rho. \end{split}$$

Using (5), we have

$$\begin{aligned} I\_2 &\le \ & 8\pi \int\_0^r \frac{1}{t^2 - \rho^2} \mathrm{d}\rho \\ &= & \frac{4\pi}{t} \int\_0^r \left( \frac{1}{t+\rho} + \frac{1}{t-\rho} \right) \mathrm{d}\rho \\ &= & \frac{4\pi}{t} \log \frac{t+r}{t-r} \to \mathcal{O}\left(\log \frac{1}{1-r}\right) \quad \text{as } r \to 1. \end{aligned}$$

Therefore we complete the proof of (13).

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

#### **Author Contributions:** All authors contributed equally.

**Funding:** The authors would like to express their gratitude to the referees for many valuable suggestions regarding the previous version of this paper. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2016R1D1A1A09916450).

**Conflicts of Interest:** The authors declare no conflict of interest.

## **References**


c 2018 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).
