Criteria for starlike and convex functions of order α

Abstract

Let $\mathcal{A}_{n}$ ($n\in\mathbb{N}$) be the class of certain analytic functions $f(z)$ in the open unit disk $\mathbb{U}$ and $\mathcal{P}_{n}(\lambda)$ be the subclass of $\mathcal{A}_{n}$ consisting of $f(z)$ which satisfy $|f''(z)| \leqq \lambda$ ($\lambda> 0$) in $\mathbb{U}$. Some properties for the class $\mathcal{P}_{n}(\lambda)$, which are the improvements of the previous results due to Ponnusamy and Singh (Complex Var. Theory Appl. 34:276-291, 1997), are discussed.

1 Introduction

Let $\mathcal{A}_{n}$ denote the class of functions of the form

$$ f(z)=z+\sum^{\infty}_{k=n+1}a_{k}z^{k} \quad \bigl(n\in\mathbb{N}=\{1,2,3,\ldots\}\bigr), $$

(1.1)

which are analytic in the open unit disk $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}$, and let $\mathcal{A}_{1}=\mathcal{A}$.

A function $f(z) \in\mathcal{A}$ is said to be in the class $\mathcal{S}^{*}(\alpha)$ in $\mathbb{U}$ if it satisfies

$$ \operatorname{Re}\frac{zf'(z)}{f(z)}>\alpha \quad (z\in\mathbb{U}) $$

(1.2)

for some real α ($\alpha<1$). If $f(z) \in \mathcal{S}^{*}(\alpha)$ with $0 \leqq\alpha< 1$, then $f(z)$ is said to be univalent and starlike of order α in $\mathbb{U}$. We denote $\mathcal{S}^{*}(0) = \mathcal{S}^{*}$. A function $f(z) \in\mathcal{A}$ is said to be in the class $\mathcal{C}(\alpha)$ if it satisfies

$$ \operatorname{Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)} \biggr\} >\alpha\quad (z\in \mathbb{U}) $$

(1.3)

for some real α ($\alpha<1$). If $f(z) \in \mathcal{C}(\alpha)$ with $0 \leqq\alpha< 1$, then $f(z)$ is said to be univalent and convex of order α in $\mathbb{U}$. We write $\mathcal{C}(0)=\mathcal{C}$.

Let $f(z)$ and $g(z)$ be analytic in $\mathbb{U}$. Then we say that $f(z)$ is subordinate to $g(z)$ in $\mathbb{U}$, written $f(z)\prec g(z)$, if there exists a function $w(z)$ analytic in $\mathbb{U}$ which satisfies $w(0)=0$, $|w(z)|<1$ ($z \in \mathbb{U}$) and $f(z)=g(w(z))$ for $z\in\mathbb{U}$. If $g(z)$ is univalent in $\mathbb{U}$, then the subordination $f(z)\prec g(z)$ is equivalent to $f(0)=g(0)$ and $f(\mathbb{U})\subset g(\mathbb{U})$ (cf. Duren [1]).

A function $f(z) \in\mathcal{A}$ is said to be strongly starlike of order β in $\mathbb{U}$ if it satisfies

$$ \frac{zf'(z)}{f(z)}\prec \biggl( \frac{1+z}{1-z} \biggr)^{\beta} $$

(1.4)

for some real β ($0<\beta\leqq1$). We denote this class by $\widetilde{\mathcal{S}}^{*}(\beta)$. Note that $\widetilde{\mathcal {S}}^{*}(1)=\mathcal{S}^{*}$.

Define

$$ \mathcal{P}_{n}(\lambda)=\bigl\{ f(z) \in \mathcal{A}_{n}: \bigl\vert f''(z)\bigr\vert \leqq \lambda\ (\lambda> 0; z\in\mathbb{U})\bigr\} . $$

(1.5)

Mocanu [2] considered the problem of finding λ such that

$$f(z) \in\mathcal{P}_{n}(\lambda)\quad \text{implies}\quad f(z) \in \mathcal{S}^{*}. $$

Mocanu [2] has shown that:

Theorem A

([2])

If

$$\lambda=\frac{n(n+1)}{2n+1}\quad (n\in\mathbb{N}), $$

then $\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}$.

Ponnusamy and Singh [3] proved the following results.

Theorem B

Let

$$\lambda_{n}= \frac{n(n+1)}{\sqrt{(n+1)^{2}+1}}\quad (n\in\mathbb{N}). $$

If $0<\lambda\leqq\lambda_{n}$, then $\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}(\beta)$, where

$$\beta=\beta_{n}(\lambda)=\left \{ \begin{array}{l@{\quad}l} \frac{(n+1)(n-\lambda)}{n(n+1)+\lambda},& \textit{if }0<\lambda\leqq \frac {n(n+1)}{n+2}, \\ \frac{n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})},& \textit{if }\frac{n(n+1)}{n+2}\leq \lambda\leq\lambda_{n}. \end{array} \right . $$

Theorem C

Let $0<\beta\leqq 1$ and

$$\lambda'_{n}= \frac{n(n+1)\sin \frac{\pi\beta}{2}}{ \sqrt {1+(n+1)^{2}+2(n+1)\cos \frac{\pi\beta}{2}}} \quad (n\in\mathbb{N}). $$

If $0<\lambda\leqq\lambda'_{n}$, then $\mathcal{P}_{n}(\lambda)\subset \widetilde{\mathcal{S}}^{*}(\beta)$.

It is easy to verify that Theorem B and Theorem C are better than Theorem A in two different ways.

In this paper we generalize and refine the above theorems. Furthermore we find λ such that $f(z)\in \mathcal{P}_{n}(\lambda)$ implies $f(z)\in \mathcal{C}(\alpha)$ ($\alpha<1$). These results are sharp.

2 Main results

To derive our first result, we need the following lemma due to Hallenbeck and Ruscheweyh [4].

Lemma

Let $g(z)$ be analytic and convex univalent in $\mathbb{U}$ and $f(z)=g(0)+\sum^{\infty}_{k=n}a_{k}z^{k}$ ($n\in\mathbb{N}$) be analytic in $\mathbb{U}$. If $f(z)\prec g(z)$, then

$$z^{-c}\int_{0}^{z}t^{c-1}f(t)\, dt\prec\frac{1}{n}z^{- \frac{c}{n}}\int_{0}^{z}t^{ \frac{c}{n}-1}g(t) \, dt, $$

where $\operatorname{Re} (c)\geqq0$ and $c\neq0$.

Now, we derive the following.

Theorem 1

Let $0<\lambda<n(n+1)$ ($n\in \mathbb{N}$). If $f(z)\in \mathcal{P}_{n}(\lambda)$, then

$$ \biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert < \frac{n\lambda}{n(n+1)-\lambda} \quad (z\in \mathbb{U}). $$

(2.1)

The bound $\frac{n\lambda}{n(n+1)-\lambda}$ in (2.1) is sharp.

Proof

Let

$$f(z)=z+\sum^{\infty}_{k=n+1}a_{k}z^{k} \in \mathcal{P}_{n}(\lambda)\quad \text{and} \quad 0<\lambda<n(n+1) \quad (n\in \mathbb{N}). $$

Then we have

$$ zf''(z)=n(n+1)a_{n+1}z^{n}+\cdots \prec\lambda z. $$

(2.2)

Applying the lemma with $c=1$, it follows from (2.2) that

$$\frac{1}{z}\int_{0}^{z}tf''(t) \, dt\prec \frac{\lambda}{n}z^{-\frac{1}{n}}\int_{0}^{z}t^{\frac{1}{n}} \, dt, $$

which yields

$$ f'(z)-\frac{f(z)}{z}\prec\frac{\lambda z}{n+1}, $$

(2.3)

and hence

$$ \biggl\vert f'(z)-\frac{f(z)}{z}\biggr\vert < \frac{\lambda}{n+1}\quad (z\in\mathbb{U}). $$

(2.4)

By (2.3) we can write

$$ f'(z)-\frac{f(z)}{z}=\frac{\lambda w(z)}{n+1}, $$

(2.5)

where $w(z)$ is analytic in $\mathbb{U}$ with $w(0)=0$ and $|w(z)|<1$ ($z\in\mathbb{U}$). Since

$$f'(z)-\frac{f(z)}{z}=na_{n+1}z^{n}+\cdots, $$

the function $w(z)$ in (2.5) satisfies $|w(z)|\leq|z|^{n}$ ($z\in\mathbb{U}$) by the Schwarz lemma. Also (2.5) leads to

$$ \int_{0}^{z} \biggl( \frac{f'(t)}{t}- \frac{f(t)}{t^{2}} \biggr)\, dt= \frac{\lambda}{n+1}\int_{0}^{z} \frac{w(t)}{t}\, dt. $$

(2.6)

In view of (2.6), we deduce that

$$\begin{aligned} \biggl\vert \frac{f(z)}{z}-1\biggr\vert &=\frac{\lambda}{n+1}\biggl\vert \int_{0}^{1} \frac{w(uz)}{u}\, du\biggr\vert \leqq\frac{\lambda}{n+1}\int_{0}^{1} \frac{|w(uz)|}{u}\, du \\ &\leqq\frac{\lambda |z|^{n}}{n+1}\int_{0}^{1}u^{n-1} \, du<\frac{\lambda}{n(n+1)} \end{aligned}$$

and so

$$ \biggl\vert \frac{f(z)}{z}\biggr\vert >1- \frac{\lambda}{n(n+1)}>0\quad (z \in\mathbb{U}). $$

(2.7)

Now, by using (2.4) and (2.7), we find that

$$\begin{aligned} \biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert &=\biggl\vert \frac{z}{f(z)}\biggr\vert \biggl\vert f'(z)- \frac{f(z)}{z}\biggr\vert \\ &< \frac{ \frac{\lambda}{n+1}}{1- \frac{\lambda}{n(n+1)}}= \frac{n\lambda}{n(n+1)-\lambda} \end{aligned}$$

for $z\in\mathbb{U}$, which shows (2.1).

For sharpness, we consider the function

$$ f(z)=z+ \frac{\lambda}{n(n+1)}z^{n+1}\quad (z\in\mathbb{U}) $$

(2.8)

for $0<\lambda<n(n+1)$. Obviously $f(z)\in \mathcal{P}_{n}(\lambda)$. Furthermore we have

$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert =\biggl\vert \frac{ \frac{\lambda}{n+1}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\vert \rightarrow\frac{n\lambda }{n(n+1)-\lambda} $$

as $z\rightarrow e^{ \frac{\pi i}{n}}$. This completes the proof of Theorem 1. □

Next, we prove the following.

Theorem 2

Let $0<\lambda<n(n+1)$ ($n\in \mathbb{N}$). Then

$$\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}(\alpha), $$

where

$$ \alpha=\alpha_{n}(\lambda)= \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}. $$

(2.9)

The result is sharp, that is, the order α is best possible.

Proof

If $f(z)\in\mathcal{P}_{n}(\lambda)$ and $0<\lambda<n(n+1)$ ($n\in\mathbb{N}$), then an application of Theorem 1 yields

$$1-\operatorname{Re} \frac{zf'(z)}{f(z)}<\frac{n\lambda}{n(n+1)-\lambda}\quad (z\in\mathbb{U}). $$

Hence $f(z)\in\mathcal{S}^{*}(\alpha)$ where $\alpha=\alpha_{n}(\lambda)$ is given by (2.9).

For the function $f(z)\in\mathcal{P}_{n}(\lambda)$ defined by (2.8), we have

$$\operatorname{Re} \frac{zf'(z)}{f(z)}=\operatorname{Re} \biggl\{ \frac{1+ \frac{\lambda }{n}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\} \rightarrow \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}=\alpha $$

as $z\rightarrow e^{ \frac{\pi i}{n}}$. Therefore the order α cannot be increased. □

Remark 1

Let us compare Theorem 2 with Theorem B. Clearly

$$n(n+1)>\lambda_{n}\quad \text{and} \quad \alpha_{n}( \lambda)>\beta_{n}(\lambda)\quad \biggl(0<\lambda\leqq \frac{n(n+1)}{n+2} \biggr). $$

Also, for $\frac{n(n+1)}{n+2}\leqq\lambda\leqq\lambda_{n}$, we have

$$\begin{aligned} \alpha_{n}(\lambda)-\beta_{n}(\lambda)&=\frac{(n+1)(n-\lambda )}{n(n+1)-\lambda} -\frac{n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})} \\ &=\frac{2(n+1)(n-\lambda)(n(n+1)+\lambda) -(n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2})}{ 2(n^{2}(n+1)^{2}-\lambda^{2})} \\ &=\frac{n^{2}(n+1-\lambda)^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})}>0. \end{aligned}$$

Thus we conclude that Theorem 2 extends and improves Theorem B by Ponnusamy and Singh [3].

Taking

$$\lambda= \frac{n(n+1)}{2n+1}\quad \text{and}\quad \lambda=n, $$

Theorem 2 reduces to the following.

Corollary 1

For $n\in\mathbb{N}$ we have

$$ \mathcal{P}_{n} \biggl( \frac{n(n+1)}{2n+1} \biggr)\subset\mathcal {S}^{*} \biggl( \frac{1}{2} \biggr) \quad \textit{and}\quad \mathcal{P}_{n}(n)\subset \mathcal{S}^{*}. $$

(2.10)

The results are sharp.

Further, applying Theorem 1, we derive the following.

Theorem 3

Let $0<\beta\leqq1$ and

$$ \widetilde{\lambda}_{n}= \frac{n(n+1)\sin \frac{\pi\beta}{2}}{n+\sin \frac{\pi\beta}{2}}\quad (n\in \mathbb{N}). $$

(2.11)

If $0<\lambda\leqq\widetilde{\lambda}_{n}$, then $\mathcal{P}_{n}(\lambda)\subset \widetilde{\mathcal{S}}^{*}(\beta)$ and the bound $\widetilde{\lambda}_{n}$ cannot be increased.

Proof

Let

$$0<\beta\leqq1,\qquad f(z)\in \mathcal{P}_{n}(\lambda) \quad \text{and} \quad 0<\lambda\leqq \widetilde{\lambda}_{n}, $$

where $\widetilde{\lambda}_{n}$ is given by (2.11). Then $\widetilde{\lambda}_{n}\leqq n$ and it follows from Theorem 1 that

$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert < \frac{n \widetilde{\lambda}_{n}}{n(n+1)- \widetilde{\lambda}_{n}} =\sin \frac{\pi\beta}{2}\quad (z\in\mathbb{U}). $$

This implies that

$$\biggl\vert \arg\frac{zf'(z)}{f(z)}\biggr\vert < \frac{\pi\beta}{2}\quad (z\in \mathbb{U}). $$

Hence $f(z)\in \widetilde{\mathcal{S}}^{*}(\beta)$.

Furthermore, for the function $f\in\mathcal{P}_{n}(\lambda)$ defined by (2.8) and $\widetilde{\lambda}_{n}<\lambda<n(n+1)$, we have

$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert \rightarrow\frac{n\lambda }{n(n+1)-\lambda} > \frac{n \widetilde{\lambda}_{n}}{n(n+1)- \widetilde{\lambda}_{n}} =\sin\frac{\pi\beta}{2} $$

as $z \rightarrow e^{ \frac{\pi i}{n}}$. This shows that $f \notin \widetilde{S}^{*}(\beta)$ and so the proof of Theorem 3 is completed. □

Remark 2

Since $\widetilde{\lambda}_{n}>\lambda_{n}'$ (cf. Theorem C) we see that Theorem 3 is better than Theorem C by Ponnusamy and Singh [3].

Finally we discuss the following.

Theorem 4

Let $0<\lambda<n$ ($n\in \mathbb{N}$) and $0<\sigma\leqq1$. If $f(z)\in \mathcal{P}_{n}(\lambda)$, then

$$ \operatorname{Re} \biggl\{ \sigma \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)} \biggr\} >\alpha \quad (z\in\mathbb{U}), $$

(2.12)

where

$$ \alpha=\alpha_{n}(\sigma,\lambda)=\sigma \frac{n-(n+1)\lambda}{n-\lambda}+(1- \sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}. $$

(2.13)

The result is sharp, that is, the bound $\alpha_{n}(\sigma,\lambda)$ cannot be increased.

Proof

Let $f(z)\in\mathcal{P}_{n}(\lambda)$ and $0<\lambda<n$. Then, by (2.2) (used in the proof of Theorem 1) and the Schwarz lemma, we can write

$$ zf''(z)=\lambda w(z)\quad (z\in\mathbb{U}), $$

(2.14)

where $w(z)$ is analytic in $\mathbb{U}$ and $|w(z)|\leq |z|^{n}$ ($z\in\mathbb{U}$). Further, we deduce from (2.14) that

$$f'(z)-1=\int_{0}^{z}f''(t) \, dt=\lambda\int_{0}^{z} \frac{w(t)}{t}\, dt= \lambda\int_{0}^{1} \frac{w(uz)}{u}\, du, $$

which leads to

$$\begin{aligned} \bigl\vert f'(z)\bigr\vert &\geqq1-\lambda\int _{0}^{1} \frac{|w(uz)|}{u}\, du \\ &>1-\lambda|z|^{n}\int_{0}^{1}u^{n-1} \, du \\ &>1-\frac{\lambda}{n}>0 \quad (z\in\mathbb{U}). \end{aligned}$$

(2.15)

Also, by Theorem 2, we have

$$ \operatorname{Re} \frac{zf'(z)}{f(z)}>\frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}\quad (z\in \mathbb{U}). $$

(2.16)

Let us define the function $g(z)$ by

$$ g(z)=\sigma \biggl(1+ \frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)}- \alpha, $$

(2.17)

where $0<\sigma\leqq1$ and α is given by (2.13). Then $g(z)$ is analytic in $\mathbb{U}$ and

$$\begin{aligned} g(0)&=1-\alpha=1-\sigma\frac{n-(n+1)\lambda}{n-\lambda}-(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda} \\ &=\sigma\frac{n\lambda}{n-\lambda}+(1-\sigma) \frac{n\lambda}{n(n+1)-\lambda}>0. \end{aligned}$$

We claim that $\operatorname{Re} g(z)>0$ for $z\in\mathbb{U}$. Otherwise there exists a point $z_{0}\in\mathbb{U}$ such that

$$ \operatorname{Re} g(z)>0 \quad \bigl(\vert z\vert <|z_{0}|\bigr) \quad \text{and}\quad \operatorname{Re} g(z_{0})=0. $$

(2.18)

Thus, in view of (2.15)-(2.18) and (2.13), we find that

$$\begin{aligned} \sigma\bigl\vert z_{0}f''(z_{0}) \bigr\vert &=\bigl\vert f'(z_{0})\bigr\vert \biggl\vert g(z_{0})+ \alpha-\sigma -(1-\sigma) \frac{z_{0}f'(z_{0})}{f(z_{0})}\biggr\vert \\ &\geqq\bigl\vert f'(z_{0})\bigr\vert \biggl\vert \operatorname{Re} g(z_{0})+ \alpha-\sigma-(1-\sigma)\operatorname{Re} \frac{z_{0}f'(z_{0})}{f(z_{0})}\biggr\vert \\ & > \biggl(1-\frac{\lambda}{n} \biggr) \biggl(\sigma-\alpha+(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda} \biggr) \\ &=\sigma\lambda>0. \end{aligned}$$

This contradicts the expression (2.14). Hence, we say that $\operatorname{Re} g(z)>0$ ($z\in\mathbb{U}$) and (2.12) is proved.

For the function $f(z)\in \mathcal{P}_{n}(\lambda)$ ($0<\lambda<n$) defined by (2.8), we get

$$\begin{aligned}& \operatorname{Re} \biggl\{ \sigma \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)} \biggr\} \\& \quad =\sigma \biggl(1+\operatorname{Re} \biggl\{ \frac{\lambda z^{n}}{1+ \frac{\lambda}{n}z^{n}} \biggr\} \biggr)+(1-\sigma) \operatorname{Re} \biggl\{ \frac{1+ \frac{\lambda}{n}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\} \\& \quad \rightarrow\sigma\frac{n-(n+1)\lambda}{n-\lambda}+(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}=\alpha \end{aligned}$$

as $z\rightarrow e^{ \frac{\pi i}{n}}$. Therefore the bound α is best possible. □

Making $\sigma=1$ in Theorem 4, we have the following.

Corollary 2

Let $0<\lambda<n$ ($n\in\mathbb{N}$). Then

$$ \mathcal{P}_{n}(\lambda)\subset\mathcal{C} \biggl( \frac{n-(n+1)\lambda}{n-\lambda} \biggr). $$

(2.19)

The result is sharp. In particular, for $n\in\mathbb{N}$, we have

$$ \mathcal{P}_{n} \biggl( \frac{n}{2n+1} \biggr)\subset\mathcal{C} \biggl( \frac{1}{2} \biggr),\qquad \mathcal{P}_{n} \biggl( \frac{n}{n+1} \biggr)\subset \mathcal{C}, $$

(2.20)

and the results are sharp.

Taking $\sigma=\frac{1}{2}$ in Theorem 4, we obtain the following.

Corollary 3

Let $0<\lambda<n$ ($n\in\mathbb {N}$). If $f(z)\in \mathcal{P}_{n}(\lambda)$, then

$$ \operatorname{ Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)}+\frac{zf'(z)}{f(z)} \biggr\} > \frac{n-(n+1)\lambda}{n-\lambda}+\frac{(n+1)(n-\lambda)}{n(n+1)-\lambda } \quad (z\in \mathbb{U}). $$

(2.21)

The result is sharp.