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# 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.

## References

1. Duren, PL: Univalent Functions. Springer, New York (1983)

2. Mocanu, PT: Two simple sufficient conditions for starlikeness. Mathematica 34, 175-181 (1992)

3. Ponnusamy, S, Singh, V: Criteria for strongly starlike functions. Complex Var. Theory Appl. 34, 267-291 (1997)

4. Hallenbeck, DJ, Ruscheweyh, S: Subordination by convex functions. Proc. Am. Math. Soc. 52, 191-195 (1975)

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## Acknowledgements

This work was partially supported by the National Natural Science Foundation of China (Grant Nos. 11171045; 11471163). The authors would like to express deep appreciation to Professor Shigeyoshi Owa for enlightening discussions and help.

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Correspondence to Neng Xu.

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The main idea was proposed by NX and D-GY participated in the research. All authors read and approved the final manuscript.

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Xu, N., Yang, DG. Criteria for starlike and convex functions of order α . J Inequal Appl 2015, 28 (2015). https://doi.org/10.1186/s13660-015-0548-0

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### Keywords

• starlike function
• convex function
• strongly starlike function
• subordination