Properties for certain subclasses of analytic functions with nonzero coefficients

Abstract

In the present paper, we obtain some mapping and inclusion properties for subclasses of analytic functions by using a linear operator defined by the Gaussian hypergeometric function.

MSC:30C45.

1 Introduction

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

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_n{z}^n(a_n\ne 0),$$

(1.1)

which are analytic in the open unit disk $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}$. We also denote by $\mathcal{S}$ the class of all functions in $\mathcal{A}$ which are univalent in $\mathbb{U}$.

A function $f\in \mathcal{A}$ is said to be in the class ${\mathrm{\Re }}^t(A,B,\alpha )$ if

$$\left|\frac{f^{\prime }(z)-1}{t(A-B)-B(f^{\prime }(z)-1)}\right|<\alpha (z\in \mathbb{U}),$$

(1.2)

where A and B are complex numbers with $A\ne B$, $t\in \mathbb{C}\setminus \{0\}$, and α is a positive real number.

In particular, for some real numbers A and B with $-1\le B<A\le 1$ and $\alpha =1$ without any restriction of the coefficients $a_n$ ($n\in \mathbb{N}=\{1,2,\dots \}$) the class ${\mathrm{\Re }}^t(A,B,\alpha )$ was introduced by Dixit and Pal [1]. Moreover, by giving specific values t, A, B, and α in (1.2), we obtain subclasses studied by various researchers in earlier works (see [2–6]).

A function $f\in \mathcal{A}$ is said to be in the class $\mathcal{UST}(\alpha )$ if

$$\mathrm{\Re }\left\{\frac{f(z)-f(\xi )}{(z-\xi )f^{\prime }(z)}\right\}>\alpha (z\times \xi \in \mathbb{U}\times \mathbb{U};0\le \alpha <1).$$

Furthermore, a function $f\in \mathcal{A}$ is said to be in the class $\mathcal{UCV}(\alpha )$ if

$$\mathrm{\Re }\{1+\frac{(z-\xi )f^{″}(z)}{f^{\prime }(z)}\}>\alpha (z\times \xi \in \mathbb{U}\times \mathbb{U};0\le \alpha <1).$$

The classes $\mathcal{UST}(0)\equiv \mathcal{UST}$ and $\mathcal{UCV}(0)\equiv \mathcal{UCV}$ are introduced by Goodman [7, 8] (they are called the classes of uniformly starlike and uniformly convex functions, respectively). The classes of uniformly starlike and uniformly convex functions have been extensively studied by Ma and Minda [9] and Rønning [10].

Let us consider the Gaussian hypergeometric function $F(a,b;c;z)$ defined by

$$F(a,b;c;z)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(a)}_n{(b)}_n}{{(c)}_n}{z}^n(a,b,c\in \mathbb{C};c\ne 0,-1,-2,\dots ;z\in \mathbb{U}),$$

where ${(\nu )}_n$ is the Pochhammer symbol (or the shifted factorial) defined (in terms of the gamma function) by

$$\begin{array}{rl}{(\nu )}_n:& =\frac{\mathrm{\Gamma }(\nu +n)}{\mathrm{\Gamma }(\nu )}\\ =\{\begin{array}{ll}1& \text{if }n=0\text{ and }\nu \in \mathbb{C}\setminus \{0\},\\ \nu (\nu +1)\cdots (\nu +n-1)& \text{if }n\in \mathbb{N}\text{ and }\nu \in \mathbb{C}.\end{array}\end{array}$$

We note that $F(a,b;c;z)=F(b,a;c;z)$ and

$$F(a,b;c;1)=\frac{\mathrm{\Gamma }(c-a-b)\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(c-a)\mathrm{\Gamma }(c-b)}(\mathrm{\Re }\{c-a-b\}>0).$$

We also recall (see [11, 12]) that the function $F(a,b;c;z)$ is bounded if $\mathrm{\Re }\{c-a-b\}>0$, and it has a pole at $z=1$ if $\mathrm{\Re }\{c-a-b\}\le 0$. Moreover, univalence, starlikeness and convexity properties of $zF(a,b;c;z)$ have been extensively studied by Ponnusamy and Vuorinen [13] and Ruscheweyh and Singh [14].

For $f\in \mathcal{A}$, we define the operator $I_{a,b;c}f$ by

$$I_{a,b;c}f(z)=zF(a,b;c;z)\ast f(z),$$

(1.3)

where ∗ denotes the usual Hadamard product (or convolution) of power series. For a special case of the operator $I_{a,b;c}f$, we can refer to the paper by Swaminathan [15] and the references cited therein.

In this paper, we obtain a necessary condition for the class ${\mathrm{\Re }}^t(A,B,\alpha )$ and sufficient conditions for the classes ${\mathrm{\Re }}^t(A,B,\alpha )$, $\mathcal{UST}(\alpha )$, and $\mathcal{UCV}(\alpha )$, respectively. Moreover, we find a condition for univalency of the operator $I_{a,b;c}f$ defined by (1.3).

2 Main result

Theorem 1 Let a function f of the form (1.1) be in the class ${\mathrm{\Re }}^t(A,B,\alpha )$ with $a_n=|a_n|e^{i\frac{(3n+1)\pi }{2}}$ ($n\in \mathbb{N}\setminus \{1\}$), then

$$\sum _{n=2}^{\mathrm{\infty }}n(1-\alpha |B|)|a_n|\le \alpha |t||A-B|.$$

(2.1)

Proof From the definition of ${\mathrm{\Re }}^t(A,B,\alpha )$, we have

$$|f^{\prime }(z)-1|<\alpha |t(A-B)-B(f^{\prime }(z)-1)|(z\in \mathbb{U}),$$

and so

$$\left|\sum _{n=2}^{\mathrm{\infty }}n{a}_n{z}^{n-1}\right|<\alpha |t(A-B)-B\sum _{n=2}^{\mathrm{\infty }}n{a}_n{z}^{n-1}|.$$

(2.2)

If we take $z=r{e}^{i\frac{\pi }{2}}$, then we see that

$$a_n{z}^{n-1}=|a_n|r^{n-1}(0\le r<1).$$

(2.3)

Then, by using (2.3) to (2.2), we have

$$\begin{array}{rl}\sum _{n=2}^{\mathrm{\infty }}n|a_n|r^{n-1}& <\alpha |t(A-B)-B\sum _{n=2}^{\mathrm{\infty }}n|a_n|r^{n-1}|\\ <\alpha |t(A-B)|+\alpha |B|\sum _{n=2}^{\mathrm{\infty }}n|a_n|r^{n-1},\end{array}$$

or, equivalently,

$$\sum _{n=2}^{\mathrm{\infty }}n(1-\alpha |B|)|a_n|r^{n-1}<\alpha |t||A-B|.$$

(2.4)

Letting $r\to 1^{-}$ in (2.4), we have the inequality (2.1). Thus we complete the proof of Theorem 1. □

Theorem 2 Let a function f of the form (1.1) be in the class $\mathcal{A}$. If

$$\sum _{n=2}^{\mathrm{\infty }}n(1+\alpha |B|)|a_n|\le \alpha |t||A-B|,$$

(2.5)

where A and B are complex numbers with $A\ne B$, $t\in \mathbb{C}\setminus \{0\}$, and α is a positive real number, then $f\in {\mathrm{\Re }}^t(A,B,\alpha )$. The result is sharp for the function defined by

$$\begin{array}{c}f(z)=z+\sum _{n=2}^{\mathrm{\infty }}\frac{\alpha t(A-B)ϵ}{n^2(n-1)(1+\alpha |B|)}{z}^n\hfill \\ (A,B\in \mathbb{C};A\ne B;t\in \mathbb{C}\setminus \{0\};|ϵ|=1;z\in \mathbb{U}).\hfill \end{array}$$

Proof In view of the definition of ${\mathrm{\Re }}^t(A,B,\alpha )$, it suffices to prove that

$$|f^{\prime }(z)-1|<\alpha |t(A-B)-B(f^{\prime }(z)-1)|(z\in \mathbb{U}).$$

(2.6)

From (2.6), we have

$$\left|\sum _{n=2}^{\mathrm{\infty }}n{a}_n{z}^{n-1}\right|<\alpha |t(A-B)-B\sum _{n=2}^{\mathrm{\infty }}n{a}_n{z}^{n-1}|.$$

Hence it is sufficient to show that

$$\sum _{n=2}^{\mathrm{\infty }}n|a_n|r^{n-1}<\alpha (|t||A-B|-|B|\sum _{n=2}^{\mathrm{\infty }}n|a_n|r^{n-1}),$$

which is equivalent to the relation

$$\sum _{n=2}^{\mathrm{\infty }}n(1+\alpha |B|)|a_n|r^{n-1}<\alpha |t||A-B|.$$

(2.7)

Letting $r\to 1^{-}$ in (2.7), we have

$$\sum _{n=2}^{\mathrm{\infty }}n(1+\alpha |B|)|a_n|\le \alpha |t||A-B|.$$

Therefore, by the assumption (2.5), we prove that $f\in {\mathrm{\Re }}^t(A,B,\alpha )$. □

Theorem 3 Let a function f of the form (1.1) be in the class $\mathcal{A}$. If

$$\sum _{n=2}^{\mathrm{\infty }}((3-\alpha )n-2)|a_n|\le 1-\alpha (0\le \alpha <1),$$

(2.8)

then $f\in \mathcal{UST}(\alpha )$. The result is sharp for the function

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}\frac{(1-\alpha )ϵ}{n(n-1)((3-\alpha )n-2)}{z}^n(0\le \alpha <1;|ϵ|=1).$$

Proof It suffices to show that

$$|\frac{f(z)-f(\xi )}{(z-\xi )f^{\prime }(z)}-1|\le 1-\alpha (0\le \alpha <1;(z,\xi )\in \mathbb{U}\times \mathbb{U}).$$

We have

$$\begin{array}{c}|\frac{f(z)-f(\xi )}{(z-\xi )f^{\prime }(z)}-1|\hfill \\ =\left|\frac{{\sum }_{n=2}^{\mathrm{\infty }}{a}_n({\xi }^{n-1}+z{\xi }^{n-2}+\cdots +z^{n-1})-{\sum }_{n=2}^{\mathrm{\infty }}n{a}_n{z}^{n-1}}{1+{\sum }_{n=2}^{\mathrm{\infty }}n{a}_n{z}^{n-1}}\right|\hfill \\ \le \frac{{\sum }_{n=2}^{\mathrm{\infty }}2(n-1)|a_n|}{1-{\sum }_{n=2}^{\mathrm{\infty }}n|a_n|},\hfill \end{array}$$

which is bounded by $1-\alpha$ if the assumption (2.8) is satisfied. Thus we complete the proof of Theorem 3. □

Theorem 4 Let a function f of the form (1.1) be in the class $\mathcal{A}$. If

$$\sum _{n=2}^{\mathrm{\infty }}n(2n-1-\alpha )|a_n|\le 1-\alpha (0\le \alpha <1),$$

(2.9)

then $f\in \mathcal{UCV}(\alpha )$. The result is sharp for the function

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}\frac{(1-\alpha )ϵ}{n^2(n-1)(2n-1-\alpha )}{z}^n(0\le \alpha <1;|ϵ|=1;z\in \mathbb{U}).$$

Proof It suffices to show that

$$\left|\frac{(z-\phi )f^{″}(z)}{f^{\prime }(z)}\right|<1-\alpha (0\le \alpha <1;(z,\phi )\in \mathbb{U}\times \mathbb{U}).$$

We have

$$\left|\frac{(z-\phi )f^{″}(z)}{f^{\prime }(z)}\right|=\left|\frac{(z-\phi ){\sum }_{n=2}^{\mathrm{\infty }}n(n-1)a_n{z}^{n-2}}{1-{\sum }_{n=2}^{\mathrm{\infty }}n{a}_n{z}^{n-1}}\right|\le \frac{2{\sum }_{n=1}^{\mathrm{\infty }}n(n-1)|a_n|}{1-{\sum }_{n=2}^{\mathrm{\infty }}n|a_n|},$$

which is bounded by $1-\alpha$ if the assumption (2.9) is satisfied. Thus we complete the proof of Theorem 4. □

Theorem 5 Let $a,b\in \mathbb{C}\setminus \{0\}$ and $c>|a|+|b|$. If $f\in {\mathrm{\Re }}^t(A,B,\alpha )$ with $a_n=|a_n|e^{i\frac{(3n+1)\pi }{2}}$, $0<|B|<1$, and

$$\frac{\mathrm{\Gamma }(c-|a|-|b|)\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(c-|a|)\mathrm{\Gamma }(c-|b|)}\le \frac{1-\alpha |B|}{1+\alpha |B|}+1,$$

then $I_{a,b;c}f\in {\mathrm{\Re }}^t(A,B,\alpha )$, where the operator $I_{a,b;c}f$ is defined by (1.3).

Proof We want to prove from Theorem 2 that

$$T_1:=\sum _{n=2}^{\mathrm{\infty }}n(1+\alpha |B|)|A_n|\le \alpha |t||A-B|,$$

where

$$A_n=\frac{{(a)}_{n-1}{(b)}_{n-1}}{{(c)}_{n-1}{(1)}_{n-1}}{a}_n.$$

Since, from Theorem 1,

$$\begin{array}{c}|a_n|\le \frac{\alpha |t||A-B|}{n(1-\alpha |B|)},\hfill \\ T_1\le \frac{\alpha |t||A-B|(1+\alpha |B|)}{1-\alpha |B|}\sum _{n=2}^{\mathrm{\infty }}\frac{{(|a|)}_{n-1}{(|b|)}_{n-1}}{{(c)}_{n-1}{(1)}_{n-1}}\hfill \\ \phantom{T_1}=\frac{\alpha |t||A-B|(1+\alpha |B|)}{1-\alpha |B|}(\sum _{n=0}^{\mathrm{\infty }}\frac{{(|a|)}_n{(|b|)}_n}{{(c)}_n{(1)}_n}-1)\hfill \\ \phantom{T_1}=\frac{\alpha |t||A-B|(1+\alpha |B|)}{1-\alpha |B|}(\frac{\mathrm{\Gamma }(c-|a|-|b|)\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(c-|a|)\mathrm{\Gamma }(c-|b|)}-1)\hfill \\ \phantom{T_1}\le \alpha |t||A-B|,\hfill \end{array}$$

which completes the proof of Theorem 5. □

We introduce the following lemma which is needed for the proof of the next theorem.

Lemma 1 [16]

Let ω be regular in the unit disk $\mathbb{U}$ with $\omega (0)=0$. Then, if $|\omega (z)|$ attains a maximum value on the circle $|z|=r$ ($0\le r<1$) at a point $z_0$, we can write

$$z_0{\omega }^{\prime }(z_0)=k\omega (z_0)(k\ge 1).$$

Theorem 6 Let a function f of the form (1.1) be in the class $\mathcal{A}$. Assume

$$\left|\frac{{(I_{a,b;c}f(z))}^{\prime }-1}{1-\alpha }{|}^{\beta }\right|\frac{z{(I_{a,b;c}f(z))}^{″}}{{(I_{a,b;c}f(z))}^{\prime }-\alpha }{|}^{\gamma }<\frac{1}{2^{\gamma }}(z\in \mathbb{U})$$

(2.10)

for some real α ($0\le \alpha <1$), $\beta >0$, and $\gamma >0$. Then

$$|{(I_{a,b;c}f(z))}^{\prime }-1|<1-\alpha (z\in \mathbb{U}).$$

(2.11)

Proof Let us define ω by

$$\omega (z)=\frac{{(I_{a,b;c}f(z))}^{\prime }-1}{1-\alpha }(z\in \mathbb{U}).$$

Then it follows that ω is analytic in $\mathbb{U}$ with $\omega (0)=0$. By (2.10), we have

$$\begin{array}{rl}|\omega (z){|}^{\beta }|\frac{z{\omega }^{\prime }(z)}{\omega (z)+1}{|}^{\gamma }& =|\omega (z){|}^{\beta +\gamma }|\frac{z{\omega }^{\prime }(z)}{\omega (z)}\frac{1}{\omega (z)+1}{|}^{\gamma }\\ <\frac{1}{2^{\gamma }}(z\in \mathbb{U}).\end{array}$$

(2.12)

Suppose that there exists a point $z_0\in \mathbb{U}$ such that

$$\underset{|z|\le |z_0|}{max}|\omega (z)|=|\omega (z_0)|=1.$$

Then, by Lemma 1, we can put

$$\frac{z_0{\omega }^{\prime }(z_0)}{\omega (z_0)}=k\ge 1.$$

Hence, we obtain

$$\begin{array}{rl}|\omega (z_0){|}^{\beta }|\frac{z_0{\omega }^{\prime }(z_0)}{\omega (z_0)+1}{|}^{\gamma }& =|\frac{z_0{\omega }^{\prime }(z_0)}{\omega (z_0)+1}{|}^{\gamma }\\ \ge {\left(\frac{k}{2}\right)}^{\gamma }\ge \frac{1}{2^{\gamma }},\end{array}$$

which contradicts the condition (2.12). This shows that

$$|\omega (z)|=\left|\frac{{(I_{a,b;c}f(z))}^{\prime }-1}{1-\alpha }\right|<1(z\in \mathbb{U}).$$

Thus we complete the proof of Theorem 6. □

Remark 1 The condition (2.11) in Theorem 6 implies that

$$\mathrm{\Re }\left\{{(I_{a,b;c}f(z))}^{\prime }\right\}>0(z\in \mathbb{U}).$$

Therefore, by the Noshiro-Warschawski theorem [17], the operator $I_{a,b;c}f$ is univalent in $\mathbb{U}$ under the restrictions of Theorem 6.