Radius properties for analytic and p-valently starlike functions

Abstract

Let ${\mathcal{A}}_p$ be the class of functions f(z) which are analytic in the open unit disk $U$ and satisfy $\frac{z^p}{f\left(z\right)}\ne 0\left(z\in U\right)$. Also, let ${\mathcal{S}}_p^{*}\left(\alpha \right)$ denotes the subclass of ${\mathcal{A}}_p$ consisting of f(z) which are p-valently starlike of order α(0 ≦ α < p). A new subclass ${\mathcal{U}}_p\left(\lambda \right)$ of ${\mathcal{A}}_p$ is introduced by

$$\left|z^2{\left(\frac{z^{p-1}}{f\left(z\right)}-\frac{1}{z}\right)}^{\prime }\right|\leqq \lambda \left(z\in U\right)$$

for some real λ > 0. The object of the present paper is to consider some radius properties for $f\left(z\right)\in {\mathcal{S}}_p^{*}\left(\alpha \right)$ such that ${\delta }^{-p}f\left(\delta z\right)\in {\mathcal{U}}_p\left(\lambda \right)$.

2010 Mathematics Subject Classification: Primary 30C45.

1 Introduction

Let ${\mathcal{A}}_p$ be the class of functions f(z) of the form

$$f\left(z\right)=z^p+\sum _{n=p+1}^{\infty }{a}_n{z}^n\left(p=1,2,3,\dots \right)$$

(1.1)

which are analytic in the open unit disk $U=\left\{z\in ℂ:\left|z\right|<1\right\}$ and satisfy

$$\frac{z^p}{f\left(z\right)}=1+\sum _{n=p+1}^{\infty }{b}_n{z}^{n-p}\ne 0\left(z\in U\right).$$

(1.2)

For $f\left(z\right)\in {\mathcal{A}}_p$, we say that f(z) belongs to the class ${\mathcal{U}}_p\left(\lambda \right)$ if it satisfies

$$\left|z^2{\left(\frac{z^{p-1}}{f\left(z\right)}-\frac{1}{z}\right)}^{\prime }\right|\leqq \lambda \left(z\in U\right)$$

(1.3)

for some real number λ > 0.

Let us consider a function f _δ (z) given by

$$f_{\delta }\left(z\right)=\frac{z^p}{{\left(1-z\right)}^{\delta }}\left(\delta \in ℝ\right).$$

(1.4)

Then, we can write that

$$f_{\delta }\left(z\right)=\frac{z^p}{1+{\sum }_{n=1}^{\infty }{a}_n{z}^n}$$

with

$$a_n={\left(-1\right)}^n\left(\begin{array}{c}\hfill \delta \hfill \\ \hfill n\hfill \end{array}\right)$$

and

$$\begin{array}{ll}\hfill \left|z^2{\left(\frac{z^{p-1}}{f_{\delta }\left(z\right)}-\frac{1}{z}\right)}^{\prime }\right|& =\left|\sum _{n=1}^{\infty }\left(n-1\right)a_n{z}^n\right|\\ <\sum _{n=1}^{\infty }\left(n-1\right)\left|a_n\right|.\end{array}$$

Thus, if δ = 2, then

$$\left|z^2{\left(\frac{z^{p-1}}{f_2\left(z\right)}-\frac{1}{z}\right)}^{\prime }\right|<1.$$

This shows that $f_2\left(z\right)\in {\mathcal{U}}_p\left(\lambda \right)$ for λ ≧ 1.

If δ = 3, then we have that

$$\left|z^2{\left(\frac{z^{p-1}}{f_3\left(z\right)}-\frac{1}{z}\right)}^{\prime }\right|<5$$

Which shows that $f\left(z\right)\in {\mathcal{U}}_p\left(\lambda \right)$ for λ ≧ 5.

Further, if δ = 4, then

$$\left|z^2{\left(\frac{z^{p-1}}{f_4\left(z\right)}-\frac{1}{z}\right)}^{\prime }\right|<11$$

which shows that $f\left(z\right)\in {\mathcal{U}}_p\left(\lambda \right)$ for λ ≧ 11.

If p = 1, then $f\left(z\right)\in {\mathcal{U}}_1\left(\lambda \right)$ is defined by

$$\left|z^2{\left(\frac{1}{f\left(z\right)}-\frac{1}{z}\right)}^{\prime }\right|\leqq \lambda \left(z\in U\right)$$

(1.5)

for some real number λ > 0. Note that (1.5) is equivalent to

$$\left|f^{\prime }\left(z\right){\left(\frac{z}{f\left(z\right)}\right)}^2-1\right|\leqq \lambda \left(z\in U\right).$$

Therefore, this class ${\mathcal{U}}_1\left(\lambda \right)$ was considered by Obradović and Ponnusamy [1]. Further-more, this class was extended as the class $\mathcal{U}\left({\beta }_1,{\beta }_2;\lambda \right)$ by Shimoda et al. [2].

Let ${\mathcal{S}}_p^{*}\left(\alpha \right)$ denotes the subclass of ${\mathcal{A}}_p$ consisting of f(z) which satisfy

$$Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)>\alpha \left(z\in U\right)$$

(1.6)

for some real α (0 ≦ α < p).

A function $f\left(z\right)\in {\mathcal{S}}_p^{*}\left(\alpha \right)$ is said to be p-valently starlike of order α in $U$ (cf. Robertson [3]).

2 Coefficient inequalities

For $f\left(z\right)\in {\mathcal{A}}_p$, we consider the sufficient condition for f(z) to be in the class ${\mathcal{U}}_p\left(\lambda \right)$.

Lemma 1 If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\sum _{n=p+2}^{\infty }\left(n-p-1\right)\left|b_n\right|\leqq \lambda ,$$

(2.1)

then $f\left(z\right)\in {\mathcal{U}}_1\left(\lambda \right)$.

Proof We note that

$$\begin{array}{ll}\hfill \left|z^2{\left(\frac{z^{p-1}}{f\left(z\right)}-\frac{1}{z}\right)}^{\prime }\right|& =\left|\sum _{n=p+1}^{\infty }\left(n-p-1\right)b_n{z}^{n-p}\right|\\ <\sum _{n=p+1}^{\infty }\left(n-p-1\right)\left|b_n\right|.\end{array}$$

Therefore, if

$$\sum _{n=p+1}^{\infty }\left(n-p-1\right)\left|b_n\right|=\sum _{n=p+2}^{\infty }\left(n-p-1\right)\left|b_n\right|\leqq \lambda ,$$

then $f\left(z\right)\in {\mathcal{U}}_p\left(\lambda \right)$.

Example 1 If we consider a function $f\left(z\right)\in {\mathcal{A}}_p$ given by

$$\frac{z^p}{f\left(z\right)}=1+b_{p+1}z+\sum _{n=p+2}^{\infty }\frac{\lambda {\mathsf{\text{e}}}^{i\phi }}{\left(n-p\right){\left(n-p-1\right)}^2}{z}^{n-p}\ne 0\left(z\in U\right)$$

with

$$b_n=\frac{\lambda {\mathsf{\text{e}}}^{i\phi }}{\left(n-p\right){\left(n-p-1\right)}^2}\left(\lambda >0,\phi \in ℝ\right)$$

for n ≧ p + 2, then we see that

$$\begin{array}{c}\sum _{n=p+2}^{\infty }\left(n-p-1\right)\left|b_n\right|=\sum _{n=p+2}^{\infty }\frac{\lambda {\mathsf{\text{e}}}^{i\phi }}{\left(n-p\right)\left(n-p-1\right)}\\ <\lambda \sum _{n=p+2}^{\infty }\left(\frac{1}{n-p-1}-\frac{1}{n-p}\right)=\lambda .\end{array}$$

Thus, this function f(z) satisfies the inequality (2.1). Also, we see that

$$\begin{array}{l}\left|z^2{\left(\frac{z^{p-1}}{f(x)}-\frac{1}{z}\right)}^{\prime }\right|=\left|{\displaystyle \sum _{n=p+2}^{\infty }\frac{\lambda {\text{e}}^{i\phi }}{n-p-1)(n-p)}{z}^{n-p}}\right|\\ <\lambda {\displaystyle \sum _{n=p+2}^{\infty }\left(\frac{1}{n-p-1}-\frac{1}{n-p}\right)}=\lambda .\end{array}$$

Therefore, we say that $f\left(z\right)\in {\mathcal{U}}_p\left(\lambda \right)$.

Next, we discuss the necessary condition for the class ${\mathcal{S}}_p^{*}\left(\alpha \right)$.

Lemma 2 If $f\left(z\right)\in {\mathcal{S}}_p^{*}\left(\alpha \right)$ satisfies

$$\frac{z^p}{f\left(z\right)}=1+\sum _{n=p+1}^{\infty }{b}_n{z}^{n-p}\ne 0\left(z\in U\right)$$

with b _n = |b _n | e^i(n-p)θ(n = p + 1, p + 2, p + 3,...), then

$$\sum _{n=p+1}^{\infty }\left(n+\alpha -2p\right)\left|b_n\right|\leqq p-\alpha .$$

Proof Let us define the function F(z) by

$$F\left(z\right)=\frac{z^p}{f\left(z\right)}=1+\sum _{n=p+1}^{\infty }{b}_n{z}^{n-p}.$$

It follows that

$$\begin{array}{c}Re\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)=Re\left(p-\frac{z{F}^{\prime }\left(z\right)}{F\left(z\right)}\right)\\ =Re\left(\frac{p-{\sum }_{n=p+1}^{\infty }\left(n-2p\right)b_n{z}^{n-p}}{1+{\sum }_{n=p+1}^{\infty }{b}_n{z}^{n-p}}\right)\\ =Re\left(\frac{p-{\sum }_{n=p+1}^{\infty }\left(n-2p\right)\left|b_n\right|{\mathsf{\text{e}}}^{i\left(n-p\right)\theta }{z}^{n-p}}{1+{\sum }_{n=p+1}^{\infty }\left|b_n\right|{\mathsf{\text{e}}}^{i\left(n-p\right)\theta }{z}^{n-p}}\right)>\alpha \end{array}$$

for $z\in U$. Letting z = |z| e^-iθ, we have that

$$\frac{p-{\sum }_{n=p+1}^{\infty }\left(n-2p\right)\left|b_n\right||z{|}^{n-p}}{1+{\sum }_{n=p+1}^{\infty }\left|b_n\right||z{|}^{n-p}}>\alpha \left(z\in U\right).$$

If we take |z| → 1^-, we obtain that

$$\frac{p-{\sum }_{n=p+1}^{\infty }\left(n-2p\right)\left|b_n\right|}{1+{\sum }_{n=p+1}^{\infty }\left|b_n\right|}\geqq \alpha$$

which implies that

$$\sum _{n=p+1}^{\infty }\left(n+\alpha -2p\right)\left|b_n\right|\leqq p-\alpha .$$

Remark 1 If we take p = 1 in Lemmas 1 and 2, then we have that

(i) $f\left(z\right)\in {\mathcal{A}}_1,\sum _{n=2}^{\infty }\left(n-2\right)\left|b_n\right|\leqq \lambda \Rightarrow f\left(z\right)\in {\mathcal{U}}_1\left(\lambda \right)$

and

(ii) $f\left(z\right)\in {\mathcal{S}}^{*}\left(\alpha \right),\left|b_n\right|=\left|b_n\right|{\mathsf{\text{e}}}^{i\left(n-1\right)\theta }\Rightarrow \sum _{n=2}^{\infty }\left(n+\alpha -2\right)\left|b_n\right|\leqq 1-\alpha .$

3 Radius problems

Our main result for the radius problem is contained in

Theorem 1 Let $f\left(z\right)\in {\mathcal{S}}_p^{*}\left(\alpha \right)$ (p - 1 ≦ α < p) with

$$\frac{z^p}{f\left(z\right)}=1+\sum _{n=p+1}^{\infty }{b}_n{z}^{n-p}\ne 0\left(z\in U\right).$$

and b _n = | b _n | e^i(n-p)θ(n = p + 1, p + 2, p + 3, ...). If $\delta \in ℂ\left(\left|\delta \right|<1\right)$, then $\frac{1}{{\delta }^p}f\left(\delta z\right)$ belongs to the class ${\mathcal{U}}_p\left(\lambda \right)$ for $0<\left|\delta \right|\leqq \left|{\delta }_0\left(\lambda \right)\right|$, where |δ₀(λ)| is the smallest positive root of the equation

$$|\delta {|}^2\sqrt{1-\alpha }-\left(1-|\delta {|}^2\right)\lambda =0,$$

(3.1)

that is,

$$\left|{\delta }_0\left(\lambda \right)\right|=\sqrt{\frac{\lambda }{\lambda +\sqrt{1-\alpha }}.}$$

(3.2)

Proof Since

$$f\left(\delta z\right)={\delta }^p{z}^p+\sum _{n=p+1}^{\infty }{a}_n{\delta }^n{z}^n,$$

we have that

$$\frac{z^p}{\frac{1}{{\delta }^p}f\left(\delta z\right)}=1+\sum _{n=p+1}^{\infty }{b}_n{\delta }^{n-p}{z}^{n-p}.$$

In view of Lemma 1, we have to show that

$$\sum _{n=p+2}^{\infty }\left(n-p-1\right)\left|b_n\right||\delta {|}^{n-p}\leqq \lambda .$$

Note that $f\left(z\right)\in {\mathcal{S}}_p^{*}\left(\alpha \right)$ satisfies

$$\left|b_n\right|\leqq \frac{p-\alpha }{n+\alpha -2p}<1\left(p-1\leqq \alpha <p\right).$$

Applying Cauchy-Schwarz inequality, we obtain that

$$\begin{array}{ll}\hfill \sum _{n=p+2}^{\infty }\left(n-p-1\right)\left|b_n\right||\delta {|}^{n-p}& \leqq {\left(\sum _{n=p+2}^{\infty }\left(n-p-1\right)|b_n{|}^2\right)}^{\frac{1}{2}}{\left(\sum _{n=p+2}^{\infty }\left(n-p-1\right)|\delta {|}^{2\left(n-p\right)}\right)}^{\frac{1}{2}}\\ \leqq {\left(\sum _{n=p+2}^{\infty }\left(n-p-1\right)|\delta {|}^{2\left(n-p\right)}\right)}^{\frac{1}{2}}\sqrt{p-\alpha }.\end{array}$$

Let |δ|² = x. Then, we have that

$$\begin{array}{ll}\hfill \sum _{n=p+2}^{\infty }\left(n-p-1\right)x^{n-p}& =x^2\left(\sum _{n=p+2}^{\infty }\left(n-p-1\right)x^{n-p-2}\right)\\ =x^2{\left(\sum _{n=p+2}^{\infty }{x}^{n-p-1}\right)}^{\prime }\\ =x^2\left(\sum _{n=1}^{\infty }{x}^{n-1}\right)\\ =\frac{x^2}{{\left(1-x\right)}^2}.\end{array}$$

This gives us that

$$\sum _{n=p+2}^{\infty }\left(n-p-1\right)\left|b_n\right||\delta {|}^{n-p}\leqq \frac{|\delta {|}^2\sqrt{p-\alpha }}{1-|\delta {|}^2}.$$

Let us define the function h(|δ|) by

$$h\left(\left|\delta \right|\right)=|\delta {|}^2\sqrt{p-\alpha }-\left(1-|\delta {|}^2\right)\lambda .$$

Then, h (|δ|) satisfies h (0) = -λ < 0 and $h\left(1\right)=\sqrt{p-\alpha }>0$. Indeed, we have that h (|δ₀(λ) |) = 0 for

$$0<\left|{\delta }_0\left(\lambda \right)\right|=\sqrt{\frac{\lambda }{\lambda +\sqrt{p-\alpha }}}<1.$$

This completes the proof of the theorem.

Corollary 1 Let $f\left(z\right)\in {\mathcal{S}}_1^{*}\left(\alpha \right)\left(0\leqq \alpha <1\right)$ with

$$\frac{z}{f\left(z\right)}=1+\sum _{n=2}^{\infty }{b}_n{z}^{n-1}\ne 0\left(z\in U\right)$$

and b _n = |b _n | e^i(n-1)θ(n = 2, 3, 4,...). If δ ∈ ℂ (|δ| < 1), then $\frac{1}{\delta }f\left(\delta z\right)$ belongs to the class ${\mathcal{U}}_1\left(\lambda \right)$ for $0<\left|\delta \right|\leqq \left|{\delta }_0\left(\lambda \right)\right|$, where |δ₀(λ)| is the smallest positive root of the equation

$$|\delta {|}^2\sqrt{1-\alpha }-\left(1-|\delta {|}^2\right)\lambda =0,$$

that is,

$$\left|{\delta }_0\left(\lambda \right)\right|=\sqrt{\frac{\lambda }{\lambda +\sqrt{1-\alpha }}.}$$

Remark 2 In view of (3.2), we define the function g(λ) by

$$g\left(\lambda \right)=\left|{\delta }_0\left(\lambda \right)\right|=\sqrt{\frac{\lambda }{\lambda +\sqrt{p-\alpha }}}.$$

Then, we have that

$$g^{\prime }\left(\lambda \right)=\frac{1}{2}\sqrt{\frac{p-\alpha }{\lambda {\left(\lambda +\sqrt{p-\alpha }\right)}^3}}>0$$

for λ > 0. Therefore, |δ₀(λ)| given by (3.2) is increasing for λ > 0.

Remark 3 If we put $\alpha =p-\frac{1}{2}$ in Theorem 1, then

$$\left|{\delta }_0\left(\lambda \right)\right|=\sqrt{\frac{2\lambda }{2\lambda +\sqrt{2}}.}$$

Therefore, if we consider $\lambda =\frac{1}{2}$, then we see that

$$\left|{\delta }_0\left(\frac{1}{2}\right)\right|=\sqrt{\frac{1}{1+\sqrt{2}}}=0.64359\dots$$

and if we make λ = 5, then we have that

$$\left|{\delta }_0\left(5\right)\right|=\sqrt{\frac{10}{10+\sqrt{2}}}=0.93600\dots$$