Univalence properties of an integral operator

Abstract

In this paper, we determine conditions on $\beta , \alpha _{i}$ and $g_{i}$ such that the integral operator ${ G_{\alpha _1, \alpha _2,\ldots ,\alpha _n, \beta }}$ is univalent on the exterior of the unit disk for two subclasses of analytical functions.

1 Introduction

Let O be the class of analytical functions g defined on the exterior of the unit disk $W=\{ z \in \mathbb {C} : 1< |z|<\infty \}$.

Let $\sum$ be the subclass of O which contains the univalent functions of W.

Let $O_1$ be the subclass of O which contains the meromorphic, normalized and injective functions $g : W\longrightarrow \mathbb {C}_{\infty },$ that looks like [2]:

$$\begin{aligned} g(z)= z + \sum _{\begin{array}{c} k=3 \end{array}}^\infty \frac{b_k}{z^k}, 1< |z|<\infty \end{aligned}$$

(1.1)

With $g(\infty )= \infty , g'(\infty )=1.$

Let V be the subclass of univalent functions from O such that:

$$\begin{aligned} \Bigg | \frac{g'(z)}{z^2}+1 \Bigg | > 1, z\in W \end{aligned}$$

Let $V_2$ be the subclass of V. Let $V_{2,\mu }$ be the subclass of $V_2$ which contains the functions of the form (1.1) and satisfies the condition:

$$\begin{aligned} \Bigg | \frac{g'(z)}{z^2}+1 \Bigg | > \mu , z\in W \end{aligned}$$

for $\mu >1$ and we note $V_{2,1} = V_2.$

Let $p\in \mathbb {R}$, with $1< p < 2,$ S(p) is the subclass of O with all the functions such that:

$$\begin{aligned} \Bigg | \Bigg (\frac{g(z)}{z}\Bigg ) ^ {''} \Bigg | > p, z \in W \end{aligned}$$

Then we obtain the inequation:

$$\begin{aligned} \Bigg | \frac{g'(z)}{z^2}+1\Bigg | > \frac{p}{|z|^2} \end{aligned}$$

Let A be the class of analytic functions f(z) defined in the open unit disk $U:= \{z\in \mathbb {C}: |z|< 1 \}$ and normalized by the conditions:

$$\begin{aligned} f(0)=0=f(0)-1. \end{aligned}$$

Let S be the subclass of A consisting of univalent functions in U, of the form:

$$\begin{aligned} f(z)= z+\sum _{\begin{array}{c} k=3 \end{array}}^\infty a_k z^k . \end{aligned}$$

Let T be the univalent subclass of A which satisfies:

$$\begin{aligned} \Bigg | \frac{z^2 f'(z)}{(f(z))^2}-1 \Bigg | < 1, z\in U \end{aligned}$$

Let $T_2$ be the subclass of T for which $f''(0)=0.$

It is known that between the S class and the $\sum$ class there are the following links:

Proposition 1.1

[2]

• (i) Let f $\in$S and g($\varsigma$)=1/f(1/$\varsigma$), $\varsigma \in W$. Then g $\in \sum$ and g$(\varsigma )\ne$0, $\varsigma \in W$.

• (ii) If g $\in \sum$ and g$(\varsigma )\ne$0, $\varsigma \in W$, then f $\in$S where f(z)=1/g(1/z), $z \in U$.

Let $F_{\alpha _1, \alpha _2,\ldots ,\alpha _n, \beta }$ be the integral operator introduced by Daniel Breaz and Narayanasamy Seenivasagan [3]:

$$\begin{aligned} { F_{\alpha _1, \alpha _2,\ldots ,\alpha _n, \beta } (z) = \left\{ \beta \int _{0}^{z} t^{\beta -1} \prod _{\begin{array}{c} i=1 \end{array}}^n \left[ \frac{f_i (t)}{t}\right] ^{\frac{1}{\alpha _i}} dt \right\} ^ \frac{1}{\beta }} \in S, \end{aligned}$$

(1.2)

and we take into account that $f_i (t) \in S.$

($f_i (t) \in T_2$ which is a subclass of T, which is a subclass of A).

Let be $g_i(t) = \frac{1}{f_i \left( \frac{1}{t}\right) } \in O_1$, with $g_i(t)\ne 0; t \in O_1$

We remember that $O_1$ is the subclass of O with:

$$\begin{aligned} g(z)= z + \sum _{\begin{array}{c} k=3 \end{array}}^\infty \frac{b_k}{z^k}, 1< |z|<\infty \end{aligned}$$

(1.3)

We may say that between $T_2$ and $O_1$ there is a bijection.

We start from:

$$\begin{aligned} { F_{\alpha _1, \alpha _2,\ldots ,\alpha _n, \beta } (z) = \left\{ \beta \int _{0}^{z} t^{\beta -1} \prod _{\begin{array}{c} i=1 \end{array}}^n \left[ \frac{f_i (t)}{t}\right] ^{\frac{1}{\alpha _i}} dt \right\} ^ \frac{1}{\beta }} \end{aligned}$$

And we apply the following transformations:

$$\begin{aligned} \begin{aligned} t&\rightarrow \frac{1}{t} | ( ) ' \\&dt \rightarrow \frac{-1}{t^2}dt \end{aligned} \end{aligned}$$

(1.4)

We can form the integral operator:

$$\begin{aligned} G_{\alpha _i, \beta } (z) = \Bigg [ \beta \int _{0}^z \bigg (\frac{1}{t}\bigg )^{\beta -1}\prod _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{f_i \bigg (\frac{1}{t}\bigg )}{\frac{1}{t}} \Bigg ) ^{\frac{1}{\alpha _i}} \frac{-1}{t^2}dt \Bigg ]^\frac{1}{\beta }, \end{aligned}$$

So:

$$\begin{aligned} => G_{\alpha _i, \beta } (z) = \Bigg [- \beta \int _{0}^z t^{-1-\beta }\prod _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{t}{g_i(t)} \Bigg ) ^{\frac{1}{\alpha _i}} dt \Bigg ]^\frac{1}{\beta } \end{aligned}$$

(1.5)

Pascu proved the following theorem:

Theorem 1.1

[4, 5] Let $\beta \in \mathbb {C}, Re \beta \ge \gamma > 0$. If the function $f \in A$ satisfies the condition:

$$\begin{aligned} \frac{1-|z|^{2\gamma }}{\gamma } \Bigg | \frac{z f''(z)}{f'(z)} \Bigg | \le 1, z\in U \end{aligned}$$

(1.6)

then the integral operator:

$$\begin{aligned} F_\beta (z) = \Bigg [ \beta \int _{0}^z t^{\beta -1}f'(t)dt \Bigg ]^\frac{1}{\beta }, \; with \; f \in S. \end{aligned}$$

(1.7)

Theorem 1.2

[6] Let $\alpha , \beta \in \mathbb {C}$ and $Re \beta \ge Re \alpha \ge \frac{3}{|\alpha |}$. If the function $f \in T_2$ satisfies the condition:

$$\begin{aligned} \Bigg | \frac{z^2 f'(z)}{f^2(z)} -1 \Bigg | < 1, z\in U \; and \; | f(z)| \le 1, z \in U, \end{aligned}$$

(1.8)

then the integral operator:

$$\begin{aligned} H_{\alpha , \beta } (z) = \Bigg [ \beta \int _{0}^z t^{\beta -1} \Bigg ( \frac{f(t)}{t} \Bigg ) ^{\frac{1}{\alpha }}dt \Bigg ]^\frac{1}{\beta }, \end{aligned}$$

is in S.

Using Theorem 1.1 and Theorem 1.2, D. Breaz and N. Breaz obtained the following theorem:

Theorem 1.3

[1] Let $\alpha , \beta \in \mathbb {C}$ and $Re \beta \ge Re \alpha \ge \frac{3n}{|\alpha |},$ let $f_i \in T_2$ with:

$$\begin{aligned} f_i(z) = z + \sum _{\begin{array}{c} k=3 \end{array}}^\infty a_{k}{^i z^\mathrm {k}}, z \in U, \forall i=1, 2, \ldots , n, n\in \mathbb {N}^* \end{aligned}$$

and if $|f_i (z)|\le 1, z\in U,$

then the integral operator:

$$\begin{aligned} F_{\alpha , \beta } (z) = \Bigg [ \beta \int _{0}^z t^{\beta -1}\prod _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{f_i(t)}{t} \Bigg ) ^{\frac{1}{\alpha }} dt \Bigg ]^\frac{1}{\beta }, \end{aligned}$$

(1.9)

is in S.

2 Main results

Applying the Theorem 1.1 for the transformation $t \rightarrow \frac{1}{t} | ( ) ' ,$ we obtain:

Theorem 2.1

Let $\beta \in \mathbb {C}, Re \beta \ge \gamma > 0$. If $f \in T_2, g\in V_2$ satisfies the conditions:

$$\begin{aligned} \frac{1-|\frac{1}{z}|^{2\gamma }}{\gamma } \Bigg | \frac{\frac{1}{z} f''(\frac{1}{z})}{f'(\frac{1}{z})} \Bigg | > 1 \end{aligned}$$

and

$$\begin{aligned} \frac{ |z|^{2\gamma }-1}{ \gamma |z|^{2\gamma }} \Bigg | \frac{g''(z)}{z g'(z)} \Bigg | > 1, \end{aligned}$$

(2.1)

then the integral operator:

$$\begin{aligned} G_\beta (z) = \Bigg [ \beta \int _{0}^z \Bigg ( \frac{1}{t}\Bigg )^{\beta -1} f'\Bigg ( \frac{1}{t}\Bigg ) (-1)t^{-2})dt \Bigg ]^\frac{1}{\beta }= \Bigg [ \beta \int _{0}^z t^{-1-\beta } \frac{g'(t)}{g^2(t)} dt \Bigg ]^\frac{1}{\beta }, \end{aligned}$$

(2.2)

is in $\sum .$

Applying the Theorem 1.2 for the transformation $t \rightarrow \frac{1}{t} | ( ) ' ,$ we get :

Theorem 2.2

Let $\alpha , \beta \in \mathbb {C}$ and $Re \beta \ge Re \alpha \ge \frac{3}{|\alpha |}$. If $f\in T_2, g \in V_2$ satisfies the condition:

$$\begin{aligned} \Bigg | \frac{\frac{1}{z^2} f'(\frac{1}{z})}{f^2(\frac{1}{z})} -1 \Bigg |> 1, z\in W \nonumber \\ \Bigg | \frac{g'(z)}{z^2}+1 \Bigg | > 1 \end{aligned}$$

(2.3)

and $\bigg | f(\frac{1}{z})\bigg |=\bigg | \frac{1}{g(z)}\bigg | \ge 1, z \in W$, then the integral operator:

$$\begin{aligned} K_{\alpha , \beta } (z) = \Bigg [ \beta \int _{0}^z \bigg (\frac{1}{t}\bigg )^{\beta -1}\Bigg ( \frac{f \bigg (\frac{1}{t}\bigg )}{\frac{1}{t}} \Bigg ) ^{\frac{1}{\alpha }} \frac{-1}{t^2}dt \Bigg ]^\frac{1}{\beta } = \Bigg [ -\beta \int _{0}^z t^{-1-\beta +\frac{1}{\alpha }} g(t)^{\frac{-1}{\alpha }} dt \Bigg ]^\frac{1}{\beta }\nonumber \\ \end{aligned}$$

(2.4)

is in W.

Using Theorem 1.3, Theorem 2.1 and Theorem 2.2 for the transformation $t \rightarrow \frac{1}{t} | ( ) ',$ we get the following theorem:

Theorem 2.3

Let $\alpha , \beta \in \mathbb {C}$ and $Re \beta \ge Re \alpha \ge \frac{3n}{|\alpha |}$

Let $g_i \in V_2$, with:

$$\begin{aligned} g_i(z)= z + \sum _{\begin{array}{c} k=3 \end{array}}^\infty \frac{b^i_k}{z^k}, \forall \; i=1, 2, \ldots , n, n\in \mathbb {N}^* \end{aligned}$$

(2.5)

and if $|g_i (z)|>1, z\in W,$

then the integral operator:

$$\begin{aligned}&G_{\alpha _i, \beta } (z) = \Bigg [ \beta \int _{0}^z \bigg (\frac{1}{t}\bigg )^{\beta -1}\prod _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{f_i \bigg (\frac{1}{t}\bigg )}{\frac{1}{t}} \Bigg ) ^{\frac{1}{\alpha _i}} \frac{-1}{t^2}dt \Bigg ]^\frac{1}{\beta }\nonumber \\&\quad = \Bigg [- \beta \int _{0}^z t^{-1-\beta }\prod _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{t}{g_i(t)} \Bigg ) ^{\frac{1}{\alpha _i}} dt \Bigg ]^\frac{1}{\beta } \end{aligned}$$

(2.6)

is in $\sum .$

Theorem 2.4

Let $m > 1, g_i \in V_{2,\mu _i}$ defined by (2.5), $\alpha _i, \beta \in \mathbb {C}, Re \beta \ge \gamma$ and:

$$\begin{aligned} \gamma := \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{(1+\mu _i)m-1}{|\alpha _i|}, \mu _i > 1, i = 1, 2, \ldots , n; n\in \mathbb {N}^*. \end{aligned}$$

(2.7)

If $\Bigg | g_i(z) \Bigg | > m, z \in W, i = 1, 2, \ldots , n,$

then we obtain that the integral operator from (1.5) $G_{\alpha _i, \beta }$ is in $\sum .$

Proof

Let be the function:

$$\begin{aligned} x(z) = \int _{0}^z \prod _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{t}{g_i(t)} \Bigg ) ^{\frac{1}{\alpha _i}} dt \end{aligned}$$

We notice that:

$$\begin{aligned} x'(z)= & {} \prod _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{z}{g_i(z)} \Bigg ) ^{\frac{1}{\alpha _i}} \nonumber \\&\frac{x''(z)}{zx'(z)} =\sum _{\begin{array}{c} i=1 \end{array}}^n \frac{1}{\alpha _i} \Bigg ( \frac{-g'_i(z)}{zg_i(z)}+1 \Bigg )\nonumber \\&\Big | \frac{x"(z)}{zx'(z)}\Bigg | >\sum _{\begin{array}{c} i=1 \end{array}}^n \frac{1}{|\alpha _i|} \Bigg ( \Bigg |\frac{-g'_i(z)}{z^2}\Bigg | -1 \Bigg ) = \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{1}{|\alpha _i|} \Bigg ( \Bigg |\frac{-g'_i(z)}{z^2g^2_i(z)}\Bigg | \cdot |z \cdot g^2_i(z)| -1 \Bigg )\nonumber \\ \end{aligned}$$

(2.8)

We know that: $\Bigg | g_i(z) \Bigg | > m$

$$\begin{aligned} => \Bigg |z\cdot g^2_i(z) \Bigg |> m^2 |z| > m|z| \end{aligned}$$

(2.9)

z $\in W, i=1, 2, \ldots , n$ $\square$

Applying the relation (2.9) in the relation (2.8) we get:

$$\begin{aligned} =>&\Big | \frac{x''(z)}{zx'(z)}\Bigg | \ge \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{1}{|\alpha _i|} \Bigg ( \Bigg |\frac{-g'_i(z)}{z^2g^2_i(z)}\Bigg | \cdot m|z|-1 \Bigg )\nonumber \\&= \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{1}{|\alpha _i|} \Bigg ( \Bigg |\frac{-g'_i(z)}{z^2g^2_i(z)}-1\Bigg | \cdot |m|z|+m|z| -1 \Bigg )\nonumber \\&=\sum _{\begin{array}{c} i=1 \end{array}}^n \frac{1}{|\alpha _i|} \bigg ( (\mu +1)\cdot m|z| -1 \bigg ) \nonumber \\&=\sum _{\begin{array}{c} i=1 \end{array}}^n \frac{(\mu +1)\cdot m|z|-1}{|\alpha _i|} \end{aligned}$$

(2.10)

We know that:

$$\begin{aligned} \gamma =&\sum _{\begin{array}{c} i=1 \end{array}}^n \frac{(1+\mu _i)m-1}{|\alpha _i|} \nonumber \\&\gamma = \sum _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{(1+\mu _i)m}{|\alpha _i|} -\frac{1}{|\alpha _i|}\Bigg ) \Bigg | \cdot |z|\nonumber \\&\gamma |z|= \sum _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{(1+\mu _i)m|z|}{|\alpha _i|} -\frac{|z|}{|\alpha _i|}\Bigg ) \nonumber \\&\gamma |z| + \sum _{\begin{array}{c} i=1 \end{array}}^n\frac{|z|}{|\alpha _i|} = \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{(1+\mu _i)m|z|}{|\alpha _i|}\Bigg | - \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{1}{|\alpha _i|} \nonumber \\&\gamma |z| + \sum _{\begin{array}{c} i=1 \end{array}}^n\frac{|z|-1}{|\alpha _i|} = \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{(1+\mu _i)m|z|-1}{|\alpha _i|} > \gamma |z| \end{aligned}$$

(2.11)

Using the results from the Eqs. (2.10) and (2.11) we get:

$$\begin{aligned} \implies&\Big | \frac{x''(z)}{zx'(z)}\Bigg |> \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{(\mu +1)\cdot m|z|-1}{|\alpha _i|}>\gamma |z| \Bigg | \cdot \frac{|z|^{2 \gamma }-1}{\gamma |z|^{2 \gamma }}\nonumber \\&\Big | \frac{x''(z)}{zx'(z)}\Bigg | \cdot \frac{|z|^{2 \gamma }-1}{\gamma |z|^{2 \gamma }}> \frac{|z|^{2 \gamma }-1}{ |z|^{2 \gamma -1}}\nonumber \\&\Big | \frac{x''(z)}{zx'(z)}\Bigg | \cdot \frac{|z|^{2 \gamma }-1}{\gamma |z|^{2 \gamma }}> \frac{|z|^{2 \gamma -1 }+|z|^{2 \gamma -2 }+ ...+|z|+1}{|z|^{2\gamma -1}}>1 \end{aligned}$$

(2.12)

Given that $Re \beta \ge \gamma > 0$ result from Theorem 1.1. that:

$$\begin{aligned} G_\beta (z) = \Bigg [ \beta \int _{0}^z t^{-1-\beta } \frac{g'(t)}{g^2(t)} dt \Bigg ]^\frac{1}{\beta } \end{aligned}$$

(2.13)

is in $\sum .$

Meaning that the integral operator:

$$\begin{aligned} G_{\alpha _i, \beta } (z) = \Bigg [ - \beta \int _{0}^{z} t^{-1-\beta } \prod _{\begin{array}{c} i=1 \end{array}}^n \Bigg ( \frac{t}{g_i(t)} \Bigg ) ^{\frac{1}{\alpha _i}}dt \Bigg ]^{\frac{1}{\beta }} \end{aligned}$$

(2.14)

is in $\sum$.

Theorem 2.5

Let $m > 1, g_i \in S(p)$ and:

$$\begin{aligned} \gamma _1:= \sum _{\begin{array}{c} i=1 \end{array}}^n \frac{(1+p)m-1}{|\alpha _i|}, i = 1, 2, \ldots , n; n\in \mathbb {N}^* \end{aligned}$$

(2.15)

If

$$\begin{aligned} \bigg | g_i(z) \bigg | > m, z \in W, i = 1, 2, \ldots , n \end{aligned}$$

Then we obtain that the integral operator defined in (1.5) $G_{\alpha _i, \beta }$ is in $\sum .$

The proof of this theorem is very similar with the previous one.