1 Introduction

Let \({{\mathcal {A}}}\) denote the family of all functions that are analytic in the unit disk \({{\mathbb {D}}}:= \{ z:\, |z| < 1 \}\) and satisfy \(f(0)=0= f'(0)-1\). Let \({{\mathcal {B}}}\) denote the set of functions \(\omega \) that are analytic in \({{\mathbb {D}}}\) and satisfy \(|\omega (z)|\le 1 (|z|<1)\). Let S be the set of all functions \(f\in {\mathcal {A}}\) that are univalent in \({\mathbb {D}}\). Let \(S^*\) denote the subset of S consisting of all starlike functions. Let \({\mathcal {U}} \) denote the set of all \(f\in {{\mathcal {A}}}\) satisfying the condition

$$\begin{aligned} \left| \left( \frac{z}{f(z)} \right) ^{2}f'(z)-1\right| < 1,\quad z\in {{\mathbb {D}}}, \end{aligned}$$
(1)

and let \({\mathcal {P}}(2)\) be the subclass of all functions \(f\in {{\mathcal {A}}}\) such that \(f(z)\ne 0\) for \(0<|z|<1\) and

$$\begin{aligned} \left| \left( \frac{z}{f(z)}\right) ''\right| \le 2,\quad z\in {{\mathbb {D}}}. \end{aligned}$$
(2)

It is known that \({\mathcal {U}}\subset S \)(see [1]). In recent years, the class \({\mathcal {U}} \) were studied in detail (see [2,3,4,5,6]). Obradović and Ponnusamy[3] proved that

$$\begin{aligned} {\mathcal {P}}(2)\subset {\mathcal {U}}. \end{aligned}$$

For the function f defined by \( \frac{z}{f(z)}=1+\frac{1}{2}z^{3}\) , which belongs to the class \({\mathcal {U}}\), we have that

$$\begin{aligned} \left| \left( \frac{z}{f(z)}\right) ''\right| =|3z|\le 2 ~ \text{ for } \quad |z|\le \frac{2}{3}, \end{aligned}$$

i.e., \({\mathcal {P}}(2)\)-radius for the above function f is equal to \(\frac{2}{3}\). The authors considered a subclass of the class \({\mathcal {U}}\) and showed that \({\mathcal {P}}(2)\)-radius for that subclass is equal to \(\frac{2}{3}\). They conjectured that the same is valid for the class \({\mathcal {U}}\) [7]. In the second part of this paper, we shall prove that the conjecture is not true by giving the correct \({\mathcal {P}}(2)\)-radius for the class \({\mathcal {U}}\).

Let \(\Omega \) be the subset of \({\mathcal {A}}\) which consists of all functions f satisfying

$$\begin{aligned} |zf'(z)-f(z)|<\frac{1}{2}, \quad (|z|<1). \end{aligned}$$

It is known that \(\Omega \subset S^*\)[8]. In the third part of this paper, we shall give two conditions for functions to be in the class \(\Omega \).

2 \({\mathcal {P}}(2)\)-Radius for the Class \({\mathcal {U}}\)

Theorem 1

If \( f\in {\mathcal {U}}\), then

$$\begin{aligned} \left| \left( \frac{z}{f(z)}\right) ''\right| \le 2 \end{aligned}$$

for \(|z|\le r_{0}=\frac{\sqrt{5}-1}{2}=0.618...\) and the result is the best possible.

For the proof of Theorem 1, we need the next lemma given by Shaffer[9].

Lemma 1

Let \(g(z)=\sum _{n=p}^{\infty }a_{n}z^{n} \,(p\ge 1)\) be analytic in \({{\mathbb {D}}}\) and satisfy \(|g(z)|\le 1\) for \( z\in {{\mathbb {D}}}\) , then

  1. (a)

    \(|g'(z)|\le p|z|^{p-1}\) for \(|z| \le \frac{\sqrt{1+p^{2}}-1}{p}\),

  2. (b)

    \(|g'(z)|\le |z|^{p-2}\,\frac{4|z|^{2}+p^{2}(1-|z|^{2})^{2}}{4(1-|z|^{2})}\) for \(|z|> \frac{\sqrt{1+p^{2}}-1}{p}\).

These estimates are the best possible.

Proof of Theorem 1

For \(f\in {\mathcal {U}}\) let’s put

$$\begin{aligned} {\mathcal {U}}_{f}(z)=\left( \frac{z}{f(z)} \right) ^{2}f'(z)-1 . \end{aligned}$$
(3)

Then,

$$\begin{aligned} {\mathcal {U}}_{f}(z)=\frac{z}{f(z)}-z\left( \frac{z}{f(z)}\right) '-1. \end{aligned}$$

If \(f(z)=z+a_{2}z^{2}+a_{3}z^{3}+....\), then

$$\begin{aligned} {\mathcal {U}}_{f}(z)=(a_{3}-a_{2}^{2})z^{2}+... \end{aligned}$$

and

$$\begin{aligned} {\mathcal {U}}_{f}'(z)=-z\left( \frac{z}{f(z)}\right) .'' \end{aligned}$$
(4)

By using (1), previous notation and other conclusions, we can apply Lemma 1 with \(g(z)={\mathcal {U}}_{f}(z)\) and \(p=2\). By Lemma 1(a), we obtain

$$\begin{aligned} \left| {\mathcal {U}}_{f}'(z)\right| \le 2|z| ~ \text{ for } |z|\le r_{0}=\frac{\sqrt{5}-1}{2}, \end{aligned}$$

which by (4) implies

$$\begin{aligned} \left| \left( \frac{z}{f(z)}\right) ''\right| \le 2,\, |z|\le r_{0}=\frac{\sqrt{5}-1}{2}, \end{aligned}$$

i.e., f has \({\mathcal {P}}(2)\)-property in the disk \(|z|\le r_{0}=\frac{\sqrt{5}-1}{2}\), which was to be proved. \(\square \)

Similarly, by Lemma 1(b) we have

$$\begin{aligned} \left| \left( \frac{z}{f(z)}\right) ''\right| \le \frac{1-|z|^{2}+|z|^{4}}{|z|(1-|z|^{2})} =:\varphi (|z|),\, |z|> r_{0}=\frac{\sqrt{5}-1}{2}, \end{aligned}$$

where

$$\begin{aligned} \varphi (t)= \frac{1-t^{2}+t^{4}}{t(1-t^{2})},\,\,r_{0}< t <1. \end{aligned}$$

It is easy to check that \(\varphi \) is an increasing function and \(\varphi (r_{0})=2<\varphi (t)\) for \(r_{0}<t<1\). For sharpness of the theorem , let us consider the function \(f_{b}\) defined by the condition

$$\begin{aligned} \frac{z}{f_{b}(z)}=1-z\int _{0}^{z}\frac{z+b}{1+b z}\mathrm{d}z , \end{aligned}$$
(5)

where b is real and \(|b|<1.\) Since \(\omega (z)=\frac{z+b}{1+b z}:{{\mathbb {D}}}\rightarrow {{\mathbb {D}}}\), then

$$\begin{aligned} \left| z\int _{0}^{z}\frac{z+b}{1+b z}\mathrm{d}z \right| \le |z|^{2}<1,\,\,z\in {{\mathbb {D}}}, \end{aligned}$$

which by (5) implies \(\frac{z}{f_{b}(z)}\ne 0,\, z\in {{\mathbb {D}}}\), i.e., \(f_{b}\) is well defined. Also

$$\begin{aligned} \left| {\mathcal {U}}_{f_{b}}(z)\right| =\left| z ^{2}\frac{z+b}{1+b z} \right|<|z|^{2}<1, \,\,z\in {{\mathbb {D}}}, \end{aligned}$$

which gives that \(f_{b}\in {\mathcal {U}}\).

Let \(r_{1}\) be a fixed real number such that \(r_{0}<r_{1}<1\) and \(b_{1}=\frac{1-2r_{1}^2}{r^{3}_{1}}\). We claim that \(|b_{1}|<1\). In fact,

$$\begin{aligned} -1<b_{1}<1\Leftrightarrow & {} -1<\frac{1-2r_{1}^2}{r^{3}_{1}}<1 \\\Leftrightarrow & {} -r_{1}^{3}<1-2r_{1}^{2}<r_{1}^{3}\\\Leftrightarrow & {} r_{1}^{2}(1-r_{1})<1-r_{1}^{2}<r_{1}^{2}(1+r_{1}). \end{aligned}$$

The left inequality is equivalent to \(r_{1}^{2}<1+r_{1}\), which is true, and the right is equivalent to \(1-r_{1}-r_{1}^{2}<0\), which is also true since \(r_{0}<r_{1}<1\).

After simple calculations, for the function \(f_{b_{1}}\) we have

$$\begin{aligned} \left| \left( \frac{z}{f_{b_{1}}(z)}\right) ''\right| _{z=r_{1}} =\frac{1-r_{1}^{2}+r_{1}^{4}}{r_{1}(1-r_{1}^{2})} =:\varphi (r_{1})>2, \end{aligned}$$

because of the property of the function \(\varphi \) and since \(r_{0}<r_{1}<1\). It means that the function \(f_{b_{1}}\) is an extremal function for our problem, since it has \({\mathcal {P}}(2)\)-property in the disk \(|z|\le r_{0}=\frac{\sqrt{5}-1}{2}\) (because \(f_{b_{1}}\in {\mathcal {U}}\)), but not in a disk with longer radius.

3 Sufficient Conditions for Function to be in \(\Omega \)

Theorem 2

Let \( f\in {\mathcal {A}}\). If \(|f''(z)|\le 1\) then \(f\in \Omega \). The number 1 is the best possible.

Proof

Let \(g(z)=zf'(z)-f(z)\). Then, \(g'(z)=zf''(z)\). Since \(f(0)=f'(0)-1=0\) and \(|f''(z)|\le 1\) for \(z\in {\mathbb {D}}\), we have

$$\begin{aligned} g'(z)=z\omega (z) \end{aligned}$$
(6)

where \(\omega (z)\in {{\mathcal {B}}}\). It follows from (6) that

$$\begin{aligned} g(z)=\int _0^z\zeta \omega (\zeta )\mathrm{d}\zeta =z^2\int _0^1t\omega (zt)\mathrm{d}t. \end{aligned}$$

Therefore,

$$\begin{aligned} |g(z)|=|z^2\int _0^1t\omega (zt)\mathrm{d}t|<\int _0^1t\mathrm{d}t=\frac{1}{2},\quad (z\in {\mathbb {D}}). \end{aligned}$$

That is, \(|zf'(z)-f(z)|<\frac{1}{2}\) for \(z\in {\mathbb {D}}\). This implies that \(f\in \Omega \subset S^*\).

If \(|f''(z)|\le \lambda \) and \(\lambda >1\), then f may be not univalent. For example, \(f(z)=z+\frac{1}{2}\lambda z^2 \) satisfy \(|f''(z)|\le \lambda \), but \(f'(z)=1+\lambda z\) vanish at \(-\frac{1}{\lambda }\), which implies that \(f\not \in S^*\). \(\square \)

Theorem 3

Let \( f\in {\mathcal {A}}\). If

$$\begin{aligned} |z^2f''(z)+zf'(z)-f(z)|\le \frac{3}{2} \end{aligned}$$

then \(f\in \Omega \subset S^*\). The number \(\frac{3}{2}\) is the best possible.

Proof

Since \(f(0)=f'(0)-1=0\) and

$$\begin{aligned} |z^2f''(z)+zf'(z)-f(z)|\le \frac{3}{2}, \end{aligned}$$

it follows that

$$\begin{aligned} {[}z^2f'(z)-zf(z)]'=\frac{3}{2}z^2\omega (z), \end{aligned}$$

where \(\omega (z)\in {{\mathcal {B}}}\). Thus,

$$\begin{aligned} z^2f'(z)-zf(z)=\frac{3}{2}\int _0^z\omega (\zeta )\zeta ^2\mathrm{d}\zeta =\frac{3}{2}z^3\int _0^1\omega (zt)t^2\mathrm{d}t, \end{aligned}$$

and consequently,

$$\begin{aligned} |zf'(z)-f(z)|=|\frac{3}{2}z^2\int _0^1\omega (zt)t^2\mathrm{d}t| <\frac{3}{2}\int _0^1t^2\mathrm{d}t =\frac{1}{2} \end{aligned}$$

for \(z\in {\mathbb {D}}\). This implies that \(f\in \Omega \subset S^*\).

If \(|z^2f''(z)+zf'(z)-f(z)|\le \lambda \) and \(\lambda >\frac{3}{2}\), then f may be not univalent. One can see that by investigating the function \(f(z)=z+\frac{1}{2}\lambda z^2, \lambda >1 \). \(\square \)