1 Introduction
Let \(\mathcal{A}_{n}\) denote the class of functions of the form
which are analytic in the open unit disk \(\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}\), and let \(\mathcal{A}_{1}=\mathcal{A}\).
$$ f(z)=z+\sum^{\infty}_{k=n+1}a_{k}z^{k} \quad \bigl(n\in\mathbb{N}=\{1,2,3,\ldots\}\bigr), $$
(1.1)
A function \(f(z) \in\mathcal{A}\) is said to be in the class \(\mathcal{S}^{*}(\alpha)\) in \(\mathbb{U}\) if it satisfies
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
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 \(\mathcal{C}(0)=\mathcal{C}\).
$$ \operatorname{Re}\frac{zf'(z)}{f(z)}>\alpha \quad (z\in\mathbb{U}) $$
(1.2)
$$ \operatorname{Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)} \biggr\} >\alpha\quad (z\in \mathbb{U}) $$
(1.3)
Anzeige
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
for some real β (\(0<\beta\leqq1\)). We denote this class by \(\widetilde{\mathcal{S}}^{*}(\beta)\). Note that \(\widetilde{\mathcal {S}}^{*}(1)=\mathcal{S}^{*}\).
$$ \frac{zf'(z)}{f(z)}\prec \biggl( \frac{1+z}{1-z} \biggr)^{\beta} $$
(1.4)
Define
Mocanu [2] considered the problem of finding λ such that
Mocanu [2] has shown that:
$$ \mathcal{P}_{n}(\lambda)=\bigl\{ f(z) \in \mathcal{A}_{n}: \bigl\vert f''(z)\bigr\vert \leqq \lambda\ (\lambda> 0; z\in\mathbb{U})\bigr\} . $$
(1.5)
$$f(z) \in\mathcal{P}_{n}(\lambda)\quad \text{implies}\quad f(z) \in \mathcal{S}^{*}. $$
Theorem A
([2])
Anzeige
If
then
\(\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}\).
$$\lambda=\frac{n(n+1)}{2n+1}\quad (n\in\mathbb{N}), $$
Ponnusamy and Singh [3] proved the following results.
Theorem B
Let
If
\(0<\lambda\leqq\lambda_{n}\), then
\(\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}(\beta)\), where
$$\lambda_{n}= \frac{n(n+1)}{\sqrt{(n+1)^{2}+1}}\quad (n\in\mathbb{N}). $$
$$\beta=\beta_{n}(\lambda)=\left \{ \begin{array}{l@{\quad}l} \frac{(n+1)(n-\lambda)}{n(n+1)+\lambda},& \textit{if }0<\lambda\leqq \frac {n(n+1)}{n+2}, \\ \frac{n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})},& \textit{if }\frac{n(n+1)}{n+2}\leq \lambda\leq\lambda_{n}. \end{array} \right . $$
Theorem C
Let
\(0<\beta\leqq 1\)
and
If
\(0<\lambda\leqq\lambda'_{n}\), then
\(\mathcal{P}_{n}(\lambda)\subset \widetilde{\mathcal{S}}^{*}(\beta)\).
$$\lambda'_{n}= \frac{n(n+1)\sin \frac{\pi\beta}{2}}{ \sqrt {1+(n+1)^{2}+2(n+1)\cos \frac{\pi\beta}{2}}} \quad (n\in\mathbb{N}). $$
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
where
\(\operatorname{Re} (c)\geqq0\)
and
\(c\neq0\).
$$z^{-c}\int_{0}^{z}t^{c-1}f(t)\, dt\prec\frac{1}{n}z^{- \frac{c}{n}}\int_{0}^{z}t^{ \frac{c}{n}-1}g(t) \, dt, $$
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
The bound
\(\frac{n\lambda}{n(n+1)-\lambda}\)
in (2.1) is sharp.
$$ \biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert < \frac{n\lambda}{n(n+1)-\lambda} \quad (z\in \mathbb{U}). $$
(2.1)
Proof
Let
Then we have
Applying the lemma with \(c=1\), it follows from (2.2) that
which yields
and hence
By (2.3) we can write
where \(w(z)\) is analytic in \(\mathbb{U}\) with \(w(0)=0\) and \(|w(z)|<1\) (\(z\in\mathbb{U}\)). Since
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
In view of (2.6), we deduce that
and so
$$f(z)=z+\sum^{\infty}_{k=n+1}a_{k}z^{k} \in \mathcal{P}_{n}(\lambda)\quad \text{and} \quad 0<\lambda<n(n+1) \quad (n\in \mathbb{N}). $$
$$ zf''(z)=n(n+1)a_{n+1}z^{n}+\cdots \prec\lambda z. $$
(2.2)
$$\frac{1}{z}\int_{0}^{z}tf''(t) \, dt\prec \frac{\lambda}{n}z^{-\frac{1}{n}}\int_{0}^{z}t^{\frac{1}{n}} \, dt, $$
$$ f'(z)-\frac{f(z)}{z}\prec\frac{\lambda z}{n+1}, $$
(2.3)
$$ \biggl\vert f'(z)-\frac{f(z)}{z}\biggr\vert < \frac{\lambda}{n+1}\quad (z\in\mathbb{U}). $$
(2.4)
$$ f'(z)-\frac{f(z)}{z}=\frac{\lambda w(z)}{n+1}, $$
(2.5)
$$f'(z)-\frac{f(z)}{z}=na_{n+1}z^{n}+\cdots, $$
$$ \int_{0}^{z} \biggl( \frac{f'(t)}{t}- \frac{f(t)}{t^{2}} \biggr)\, dt= \frac{\lambda}{n+1}\int_{0}^{z} \frac{w(t)}{t}\, dt. $$
(2.6)
$$\begin{aligned} \biggl\vert \frac{f(z)}{z}-1\biggr\vert &=\frac{\lambda}{n+1}\biggl\vert \int_{0}^{1} \frac{w(uz)}{u}\, du\biggr\vert \leqq\frac{\lambda}{n+1}\int_{0}^{1} \frac{|w(uz)|}{u}\, du \\ &\leqq\frac{\lambda |z|^{n}}{n+1}\int_{0}^{1}u^{n-1} \, du<\frac{\lambda}{n(n+1)} \end{aligned}$$
$$ \biggl\vert \frac{f(z)}{z}\biggr\vert >1- \frac{\lambda}{n(n+1)}>0\quad (z \in\mathbb{U}). $$
(2.7)
Now, by using (2.4) and (2.7), we find that
for \(z\in\mathbb{U}\), which shows (2.1).
$$\begin{aligned} \biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert &=\biggl\vert \frac{z}{f(z)}\biggr\vert \biggl\vert f'(z)- \frac{f(z)}{z}\biggr\vert \\ &< \frac{ \frac{\lambda}{n+1}}{1- \frac{\lambda}{n(n+1)}}= \frac{n\lambda}{n(n+1)-\lambda} \end{aligned}$$
For sharpness, we consider the function
for \(0<\lambda<n(n+1)\). Obviously \(f(z)\in \mathcal{P}_{n}(\lambda)\). Furthermore we have
as \(z\rightarrow e^{ \frac{\pi i}{n}}\). This completes the proof of Theorem 1. □
$$ f(z)=z+ \frac{\lambda}{n(n+1)}z^{n+1}\quad (z\in\mathbb{U}) $$
(2.8)
$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert =\biggl\vert \frac{ \frac{\lambda}{n+1}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\vert \rightarrow\frac{n\lambda }{n(n+1)-\lambda} $$
Next, we prove the following.
Theorem 2
Let
\(0<\lambda<n(n+1)\) (\(n\in \mathbb{N}\)). Then
where
The result is sharp, that is, the order
α
is best possible.
$$\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}(\alpha), $$
$$ \alpha=\alpha_{n}(\lambda)= \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}. $$
(2.9)
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
Hence \(f(z)\in\mathcal{S}^{*}(\alpha)\) where \(\alpha=\alpha_{n}(\lambda)\) is given by (2.9).
$$1-\operatorname{Re} \frac{zf'(z)}{f(z)}<\frac{n\lambda}{n(n+1)-\lambda}\quad (z\in\mathbb{U}). $$
For the function \(f(z)\in\mathcal{P}_{n}(\lambda)\) defined by (2.8), we have
as \(z\rightarrow e^{ \frac{\pi i}{n}}\). Therefore the order α cannot be increased. □
$$\operatorname{Re} \frac{zf'(z)}{f(z)}=\operatorname{Re} \biggl\{ \frac{1+ \frac{\lambda }{n}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\} \rightarrow \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}=\alpha $$
Remark 1
Let us compare Theorem 2 with Theorem B. Clearly
Also, for \(\frac{n(n+1)}{n+2}\leqq\lambda\leqq\lambda_{n}\), we have
Thus we conclude that Theorem 2 extends and improves Theorem B by Ponnusamy and Singh [3].
$$n(n+1)>\lambda_{n}\quad \text{and} \quad \alpha_{n}( \lambda)>\beta_{n}(\lambda)\quad \biggl(0<\lambda\leqq \frac{n(n+1)}{n+2} \biggr). $$
$$\begin{aligned} \alpha_{n}(\lambda)-\beta_{n}(\lambda)&=\frac{(n+1)(n-\lambda )}{n(n+1)-\lambda} -\frac{n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})} \\ &=\frac{2(n+1)(n-\lambda)(n(n+1)+\lambda) -(n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2})}{ 2(n^{2}(n+1)^{2}-\lambda^{2})} \\ &=\frac{n^{2}(n+1-\lambda)^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})}>0. \end{aligned}$$
Taking
Theorem 2 reduces to the following.
$$\lambda= \frac{n(n+1)}{2n+1}\quad \text{and}\quad \lambda=n, $$
Corollary 1
For
\(n\in\mathbb{N}\)
we have
The results are sharp.
$$ \mathcal{P}_{n} \biggl( \frac{n(n+1)}{2n+1} \biggr)\subset\mathcal {S}^{*} \biggl( \frac{1}{2} \biggr) \quad \textit{and}\quad \mathcal{P}_{n}(n)\subset \mathcal{S}^{*}. $$
(2.10)
Further, applying Theorem 1, we derive the following.
Theorem 3
Let
\(0<\beta\leqq1\)
and
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.
$$ \widetilde{\lambda}_{n}= \frac{n(n+1)\sin \frac{\pi\beta}{2}}{n+\sin \frac{\pi\beta}{2}}\quad (n\in \mathbb{N}). $$
(2.11)
Proof
Let
where \(\widetilde{\lambda}_{n}\) is given by (2.11). Then \(\widetilde{\lambda}_{n}\leqq n\) and it follows from Theorem 1 that
This implies that
Hence \(f(z)\in \widetilde{\mathcal{S}}^{*}(\beta)\).
$$0<\beta\leqq1,\qquad f(z)\in \mathcal{P}_{n}(\lambda) \quad \text{and} \quad 0<\lambda\leqq \widetilde{\lambda}_{n}, $$
$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert < \frac{n \widetilde{\lambda}_{n}}{n(n+1)- \widetilde{\lambda}_{n}} =\sin \frac{\pi\beta}{2}\quad (z\in\mathbb{U}). $$
$$\biggl\vert \arg\frac{zf'(z)}{f(z)}\biggr\vert < \frac{\pi\beta}{2}\quad (z\in \mathbb{U}). $$
Furthermore, for the function \(f\in\mathcal{P}_{n}(\lambda)\) defined by (2.8) and \(\widetilde{\lambda}_{n}<\lambda<n(n+1)\), we have
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. □
$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert \rightarrow\frac{n\lambda }{n(n+1)-\lambda} > \frac{n \widetilde{\lambda}_{n}}{n(n+1)- \widetilde{\lambda}_{n}} =\sin\frac{\pi\beta}{2} $$
Remark 2
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
where
The result is sharp, that is, the bound
\(\alpha_{n}(\sigma,\lambda)\)
cannot be increased.
$$ \operatorname{Re} \biggl\{ \sigma \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)} \biggr\} >\alpha \quad (z\in\mathbb{U}), $$
(2.12)
$$ \alpha=\alpha_{n}(\sigma,\lambda)=\sigma \frac{n-(n+1)\lambda}{n-\lambda}+(1- \sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}. $$
(2.13)
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
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
which leads to
Also, by Theorem 2, we have
Let us define the function \(g(z)\) by
where \(0<\sigma\leqq1\) and α is given by (2.13). Then \(g(z)\) is analytic in \(\mathbb{U}\) and
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
Thus, in view of (2.15)-(2.18) and (2.13), we find that
This contradicts the expression (2.14). Hence, we say that \(\operatorname{Re} g(z)>0\) (\(z\in\mathbb{U}\)) and (2.12) is proved.
$$ zf''(z)=\lambda w(z)\quad (z\in\mathbb{U}), $$
(2.14)
$$f'(z)-1=\int_{0}^{z}f''(t) \, dt=\lambda\int_{0}^{z} \frac{w(t)}{t}\, dt= \lambda\int_{0}^{1} \frac{w(uz)}{u}\, du, $$
$$\begin{aligned} \bigl\vert f'(z)\bigr\vert &\geqq1-\lambda\int _{0}^{1} \frac{|w(uz)|}{u}\, du \\ &>1-\lambda|z|^{n}\int_{0}^{1}u^{n-1} \, du \\ &>1-\frac{\lambda}{n}>0 \quad (z\in\mathbb{U}). \end{aligned}$$
(2.15)
$$ \operatorname{Re} \frac{zf'(z)}{f(z)}>\frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}\quad (z\in \mathbb{U}). $$
(2.16)
$$ g(z)=\sigma \biggl(1+ \frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)}- \alpha, $$
(2.17)
$$\begin{aligned} g(0)&=1-\alpha=1-\sigma\frac{n-(n+1)\lambda}{n-\lambda}-(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda} \\ &=\sigma\frac{n\lambda}{n-\lambda}+(1-\sigma) \frac{n\lambda}{n(n+1)-\lambda}>0. \end{aligned}$$
$$ \operatorname{Re} g(z)>0 \quad \bigl(\vert z\vert <|z_{0}|\bigr) \quad \text{and}\quad \operatorname{Re} g(z_{0})=0. $$
(2.18)
$$\begin{aligned} \sigma\bigl\vert z_{0}f''(z_{0}) \bigr\vert &=\bigl\vert f'(z_{0})\bigr\vert \biggl\vert g(z_{0})+ \alpha-\sigma -(1-\sigma) \frac{z_{0}f'(z_{0})}{f(z_{0})}\biggr\vert \\ &\geqq\bigl\vert f'(z_{0})\bigr\vert \biggl\vert \operatorname{Re} g(z_{0})+ \alpha-\sigma-(1-\sigma)\operatorname{Re} \frac{z_{0}f'(z_{0})}{f(z_{0})}\biggr\vert \\ & > \biggl(1-\frac{\lambda}{n} \biggr) \biggl(\sigma-\alpha+(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda} \biggr) \\ &=\sigma\lambda>0. \end{aligned}$$
For the function \(f(z)\in \mathcal{P}_{n}(\lambda)\) (\(0<\lambda<n\)) defined by (2.8), we get
as \(z\rightarrow e^{ \frac{\pi i}{n}}\). Therefore the bound α is best possible. □
$$\begin{aligned}& \operatorname{Re} \biggl\{ \sigma \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)} \biggr\} \\& \quad =\sigma \biggl(1+\operatorname{Re} \biggl\{ \frac{\lambda z^{n}}{1+ \frac{\lambda}{n}z^{n}} \biggr\} \biggr)+(1-\sigma) \operatorname{Re} \biggl\{ \frac{1+ \frac{\lambda}{n}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\} \\& \quad \rightarrow\sigma\frac{n-(n+1)\lambda}{n-\lambda}+(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}=\alpha \end{aligned}$$
Making \(\sigma=1\) in Theorem 4, we have the following.
Corollary 2
Let
\(0<\lambda<n\) (\(n\in\mathbb{N}\)). Then
The result is sharp. In particular, for
\(n\in\mathbb{N}\), we have
and the results are sharp.
$$ \mathcal{P}_{n}(\lambda)\subset\mathcal{C} \biggl( \frac{n-(n+1)\lambda}{n-\lambda} \biggr). $$
(2.19)
$$ \mathcal{P}_{n} \biggl( \frac{n}{2n+1} \biggr)\subset\mathcal{C} \biggl( \frac{1}{2} \biggr),\qquad \mathcal{P}_{n} \biggl( \frac{n}{n+1} \biggr)\subset \mathcal{C}, $$
(2.20)
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
The result is sharp.
$$ \operatorname{ Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)}+\frac{zf'(z)}{f(z)} \biggr\} > \frac{n-(n+1)\lambda}{n-\lambda}+\frac{(n+1)(n-\lambda)}{n(n+1)-\lambda } \quad (z\in \mathbb{U}). $$
(2.21)
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.
Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Competing interests
The authors declare that there is no conflict of interests regarding the publication of this article.
Authors’ contributions
The main idea was proposed by NX and D-GY participated in the research. All authors read and approved the final manuscript.