On Some Integral Operators

Abstract

Let P(n, β), 0 ≤ β < 1, be the class of functions $p: p(z) = 1 + c_{n}z^{n} + c_{n+1}z^{n+1}+\ldots$ analytic in the unit disc E such that Re{p(z)} > β. The class P _k (n, β), k ≥ 2 is defined as follows: An analytic function p ∈ P _k (n, β), k ≥ 2, 0 ≤ β < 1 if and only if there exist $p_{1},p_{2} \in P(n,\beta )$ such that

$$\displaystyle{p(z) = \left (\frac{k} {4} + \frac{1} {2}\right )p_{1}(z) -\left (\frac{k} {4} -\frac{1} {2}\right )p_{2}(z).}$$

In this paper, we discuss some integral operators for certain classes of analytic functions defined in E and related with the class P _k (n, β).

1 Introduction

Let $\mathcal{A}(n)$ denote the class of functions f of the form

$$\displaystyle\begin{array}{rcl} f(z) = z +\sum _{ k=n+1}^{\infty }a_{ k}z^{k},\quad \left (n = N =\{ 1,2,3,\ldots,\}\right ),& &{}\end{array}$$

(1)

analytic in the unit disc E = { z:  | z |  < 1}. Let P(n, β) be the class of functions h(z) of the form

$$\displaystyle\begin{array}{rcl} h(z) = 1 + c_{n}z^{n} + c_{ n+1}z^{n+1}+\ldots,& &{}\end{array}$$

(2)

which are analytic in E and satisfy $Re\{h(z)\} >\beta,0 \leq \beta < 1,z \in$ We note that $P(1,0) \equiv P$ is the class of functions with positive real part.

Let P _k (n, β), k ≥ 2, 0 ≤ β < 1, be the class of functions p, analytic in E, such that

$$\displaystyle\begin{array}{rcl} p(z) = \left (\frac{k} {4} + \frac{1} {2}\right )p_{1}(z) -\left (\frac{k} {4} -\frac{1} {2}\right )p_{2}(z)\end{array}$$

if and only if $p_{1},p_{2} \in P(n,\beta )$ for z ∈ E. The class $P_{k}(1,0) \equiv P_{k}$ was introduced in [6]. We note that p ∈ P _k (n, β) if and only if there exists h ∈ P _k (n, 0) such that

$$\displaystyle\begin{array}{rcl} p(z) = (1-\beta )h(z)+\beta,\end{array}$$

Let f and g be analytic in E with f(z) given by (1) and $\quad g(z) = z +\sum _{ k=n+1}^{\infty }b_{k}z^{k}.$ Then the convolution (or Hadamard product ) of f and g is defined by

$$\displaystyle\begin{array}{rcl} (f \star g)(z) = z +\sum _{ k=n+1}^{\infty }a_{ k}b_{k}z^{k}.\end{array}$$

A function $f \in \mathcal{A}(n)$ is said to belong to the class R _k (n, β), k ≥ 2, 0 ≤ β < 1, if and only if $\frac{zf^{\prime}} {f} \in P_{k}(n,\beta )$ for z ∈ E. 

We note that $R_{k}(1,0) \equiv R_{k}$ is the class of functions with bounded radius rotation, first discussed by Tammi, see [1] and R ₂(1, 0) consists of starlike univalent functions.

Similarly $f \in \mathcal{A}(n)$ belongs to V _k (n, β) for z ∈ E if and only if $\frac{(f^{\prime})^{\prime}} {f^{\prime}} \in P_{k}(n,\beta ).$ It is obvious that

$$\displaystyle\begin{array}{rcl} f \in V _{k}(n,\beta )\quad \mbox{ if and only if }\quad zf^{\prime}\in R_{ k}(n,\beta ).& &{}\end{array}$$

(3)

It may be observed that $V _{2}(1,0) \equiv C,$ the class of convex univalent functions and $V _{k}(1,0) \equiv V _{k}$ is the class of functions with bounded boundary rotation first discussed by Paatero, see [1].

2 Preliminary Results

We need the following results in our investigation.

Lemma 2.1 ([5]).

Let $u = u_{1} + iu_{2},\quad v = v_{1} + iv_{2}$ and Ψ(u,v) be a complex-valued function satisfying the following conditions:

• (i). Ψ(u,v) is continuous in a domain D ⊂ C ²

• (ii). (1,0) ∈ D and Ψ{(1,0)} > 0.

• (iii). $Re\varPsi (iu_{2},v_{1}) \leq 0$ whenever $(iu_{2},v_{1}) \in D$ and $v_{1} \leq \frac{-1} {2} (1 + u_{2}^{2}).$

Let h(z), given by (2) , be analytic in E such that $\left (h(z),zh^{\prime}(z)\right ) \in D$ and $Re\varPsi \left (h(z),zh^{\prime}(z)\right ) > 0$ for all z ∈ E, then Re{h(z)} > 0 in E.

We shall need the following result which is a modified version of Theorem 3.3e in [4, p113].

Lemma 2.2.

Let β > 0,β + δ > 0 and α ∈ [α ₀ ,1), where

$$\displaystyle\begin{array}{rcl} \alpha _{0} =\max \left \{\frac{\beta -\delta - 1} {2\beta }, \frac{-\delta } {\beta } \right \}.\end{array}$$

If $\left \{h(z) + \frac{zh^{\prime}(z)} {\beta h(z)+\delta } \right \} \in P(1,\alpha )$ for z ∈ E, then h ∈ P(1,σ) in E, where

$$\displaystyle\begin{array}{rcl} \sigma (\alpha,\beta,\delta ) = \left [ \frac{(\beta +\delta )} {\beta \{{}_{2}F_{1}(2\beta (1-\alpha ),1,\beta +\delta + 1; \frac{r} {1+r}\}} -\frac{\delta } {\beta }\right ],& &{}\end{array}$$

(4)

where $_{2}F_{1}$ denotes hypergeometric function. This result is sharp and external function is given as

$$\displaystyle\begin{array}{rcl} p_{0}(z) = \frac{1} {\beta g(z)} -\frac{\delta } {\beta },& &{}\end{array}$$

(5)

with

$$\displaystyle\begin{array}{rcl} g(z)& =& \int _{0}^{1}\left ( \frac{1 - z} {1 - tz}\right )^{2\beta (1-\alpha )}t^{(\beta +\delta -1)}dt {}\\ & =& _{2}F_{1}\left (2\beta (1-\alpha ),1,\beta +\delta + 1; \frac{z} {z - 1}\right ).\left (\beta +\delta \right )^{-1}. \end{array}$$

3 Main Results

Theorem 3.1.

Let $f \in R_{k}(n,\beta ),\quad g \in R_{k}(n,\beta ),\alpha,c,\delta$ and ν be positively real and $\delta =\nu =\alpha.$ Then the function F defined by

$$\displaystyle\begin{array}{rcl} \left [F(z)\right ]^{\alpha } = cz^{\alpha -c}\int _{ 0}^{z}t^{(c-\delta -\nu )-1}\left (f(t)\right )^{\delta }\left (g(t)\right )^{\nu }dt& &{}\end{array}$$

(6)

belongs to R _k (n,σ), where

$$\displaystyle\begin{array}{rcl} \sigma = \frac{2(2\beta c_{1} + n\alpha _{1})} {(n\alpha _{1} - 2\beta + 2c_{1}) + \sqrt{(n\alpha _{1 } - 2\beta + 2c_{1 } )^{2 } + 8(2\beta c_{1 } + n\alpha _{1 } )}},& &{}\end{array}$$

(7)

with

$$\displaystyle\begin{array}{rcl} c_{1} = \frac{c-\alpha } {\alpha },\quad \alpha _{1} = \frac{1} {\alpha }.\end{array}$$

Proof.

First we show that there exists a function $F \in \mathcal{A}(n)$ satisfying (6). Let

$$\displaystyle\begin{array}{rcl} G(z) = z^{-(\nu +\delta )}\left (f(z)\right )^{\delta }\left (g(z)\right )^{\nu } = 1 +\alpha _{ n}z^{n} +\alpha _{ n+1}z^{n+1}+\ldots,\end{array}$$

and choose the branches which equal 1 when z = 0. For

$$\displaystyle\begin{array}{rcl} K(z) = z^{(c-\nu -\delta )-1}\left (f(z)\right )^{\delta }\left (g(z)\right )^{\nu } = z^{c-1}G(z),\end{array}$$

we have

$$\displaystyle\begin{array}{rcl} L(z) = \frac{c} {z^{c}}\int _{0}^{z}K(t)dt = 1 + \frac{c} {n + 1}\alpha _{n}z^{n}+\ldots,\end{array}$$

where L is well defined and analytic in E. Now let

$$\displaystyle\begin{array}{rcl} F(z) = \left [z^{\alpha }L(z)\right ]^{\frac{1} {\alpha } } = z\left [L(z)\right ]^{\frac{1} {\alpha } },\end{array}$$

where we choose the branch of $\left [L(z)\right ]^{\frac{1} {\alpha } }$ which equals 1 when z = 0. Thus $F \in \mathcal{A}(n)$ and satisfies (6).

Now, from (6), we have

$$\displaystyle\begin{array}{rcl} z^{(c-\alpha -1)}\left [F(z)\right ]^{\alpha }\left [(c-\alpha ) +\alpha \frac{zF^{\prime}(z)} {F(z)} \right ] = c\left [z^{(c-\delta -\nu )-1}\left (f(z)\right )^{\delta }\left (g(z)\right )^{\nu }\right ].& &{}\end{array}$$

(8)

We write

$$\displaystyle\begin{array}{rcl} \frac{zF^{\prime}(z)} {F(z)} = p(z) = \left (\frac{k} {4} + \frac{1} {2}\right )p_{1}(z) -\left (\frac{k} {4} -\frac{1} {2}\right )p_{2}(z).& &{}\end{array}$$

(9)

Then $p(z) = 1 + c_{n}z^{n} + c_{n+1}z^{n+1}+\ldots,$ is analytic in E. 

Logarithmic differentiation of (8) and use of (9) yields

$$\displaystyle\begin{array}{rcl} (c -\alpha -1) +\alpha p(z) + \frac{\alpha zp^{\prime}(z)} {(c-\alpha ) +\alpha p(z)} = (c -\delta -\nu - 1) + \frac{\delta zf^{\prime}(z)} {f(z)} + \frac{\nu zg^{\prime}(z)} {g(z)}.\end{array}$$

Since $\nu +\delta =\alpha: f,g \in P_{k}(n,\beta )$ and it is known [2] that P _k (n, β) is a convex set, it follows that

$$\displaystyle\begin{array}{rcl} \left \{p + \frac{\frac{1} {\alpha } zp^{\prime}} {p + \left (\frac{c-\alpha } {\alpha } \right )}\right \} \in P_{k}(n,\beta ),\quad z \in E.\end{array}$$

Define

$$\displaystyle\begin{array}{rcl} \varPhi _{\alpha,c}(z) = \frac{1} {1 + c_{1}} \frac{z} {(1 - z)^{\alpha _{1}+1}} + \frac{c_{1}} {1 + c_{1}} \frac{z} {(1 - z)^{\alpha _{1}+2}},\end{array}$$

with $\quad \alpha _{1} = \frac{1} {\alpha },\quad c_{1} = \frac{c-\alpha } {\alpha }.$

Then, using (9), we have

$$\displaystyle\begin{array}{rcl} \left (p \star \frac{\varPhi _{\alpha,c}} {z}\right )& =& p(z) + \frac{\alpha _{1}zp^{\prime}(z)} {p(z) + c_{1}} {}\\ & =& \left (\frac{k} {4} + \frac{1} {2}\right )\left [p_{1}(z) + \frac{\alpha _{1}zp_{1}^{\prime}(z)} {p_{1}(z) + c_{1}}\right ] {}\\ & & -\left (\frac{k} {4} -\frac{1} {2}\right )\left [p_{2}(z) + \frac{\alpha _{1}zp_{2}^{\prime}(z)} {p_{2}(z) + c_{1}}\right ]. \end{array}$$

Since $\left \{p + \frac{\alpha _{1}zp^{\prime}} {p+c_{1}} \right \} \in P_{k}(n,\beta ),$ it follows that

$$\displaystyle\begin{array}{rcl} \left \{p_{i} + \frac{\alpha _{1}zp_{i}^{\prime}} {p_{i} + c_{1}}\right \} \in P_{k}(n,\beta ),\quad \mbox{ for}\quad i = 1,2,\quad z \in E.\end{array}$$

Writing $p_{i}(z) = (1-\sigma )H_{i}(z)+\sigma,\quad i = 1,2,$ we have, for z ∈ E, 

$$\displaystyle\begin{array}{rcl} \left [(1-\sigma )H_{i} +\sigma + \frac{\alpha _{1}(1-\sigma )H_{i}^{\prime}} {(1-\sigma )H_{i} +\sigma +c_{1}}-\beta \right ] \in P(n,0).\end{array}$$

We now form the functional Ψ(u, v) by taking u = H _i and $v = zH_{i}^{\prime}$ and so

$$\displaystyle\begin{array}{rcl} \varPsi (u,v) = (\sigma -\beta ) + (1-\sigma )u + \frac{\alpha _{1}(1-\sigma )v} {(1-\sigma )u +\sigma +c_{1}}.\end{array}$$

It can easily be seen that:

• (i) Ψ(u, v) is continuous in $\mathcal{D} = \left (\mathcal{C}-\left \{\frac{\sigma +c_{1}} {1-\sigma }\right \}\right ) \times \mathcal{C}.$

• (ii) $(i,0) \in \mathcal{D}$ and $Re\{\varPsi (i,0) = 1-\beta > 0.$

To verify the condition (iii) of Lemma 2.1, we proceed as follows:

For all $(iu_{2},v_{1}) \in \mathcal{D}$ such that $\quad v_{1} \leq \frac{-n(1+u_{2}^{2})} {2},$ and

$$\displaystyle\begin{array}{rcl} \mathfrak{R}\left \{\varPsi (iu_{2},v_{1})\right \}& =& (\sigma -\beta ) + \frac{\alpha _{1}(1-\sigma )(\sigma +c_{1})v_{1}} {(\sigma +c_{1})^{2} + (1-\sigma )^{2}u_{2}^{2}} \\ & \leq & \frac{2(\sigma -\beta )\left \{(\sigma +c_{1})^{2}+(1-\sigma )^{2}u_{2}^{2}\right \}-n\alpha _{1}(1-\sigma )(\sigma +c_{1})(1+u_{2}^{2})} {2(\sigma +c_{1})^{2}+(1-\sigma )^{2}u_{2}^{2}} \\ & =& \frac{A + Bu_{2}^{2}} {2C} \\ & \leq & 0,\quad \mbox{ if}\quad A \leq 0\quad \mbox{ and}\quad B \leq 0, {}\end{array}$$

(10)

where

$$\displaystyle\begin{array}{rcl} A& =& 2(\sigma -\beta )(\sigma +c_{1})^{2} - n\alpha _{ 1}(1-\sigma )(\sigma +c_{1}), {}\\ B& =& 2(\sigma -\beta )(1-\sigma )^{2} - n\alpha _{ 1}(1-\sigma )(\sigma +c_{1}) {}\\ C& =& (\sigma +c_{1})^{2} + (1-\sigma )^{2}u_{ 2}^{2} > 0. \end{array}$$

From A = 0, we obtain σ as given by (7) and B ≤ 0 ensures that 0 ≤ σ < 1. Thus using Lemma 2.1, it follows that H _i  ∈ P(n, 0) and therefore $p_{i} \in P(n,\sigma ),\quad i = 1,2.$ Consequently p ∈ P _k (n, σ) and this completes the proof. □ 

Corollary 3.1.

For $0 = c = n = 1,\beta = 0$ and f = g,F ∈ V _k implies that $F \in R_{k}(\frac{1} {2})$ and this, with k = 2, gives us a well-known result that every convex function is starlike of order $\frac{1} {2}$ in E.

Corollary 3.2.

For n = 1, let f ∈ R _k (1,σ) in Theorem 3.1 . Then $F \in R_{k}(1,\sigma _{0}),$ where σ ₀ is given by (2.1) with $\beta =\alpha,\delta = (1-\alpha ).$ This result is sharp.

Corollary 3.3.

In (2) , we take $\nu +\delta = 1,c = 2,f = g$ and obtain Libera’s integral operator [3, 6]as:

$$\displaystyle\begin{array}{rcl} F(z) = \frac{2} {z}\int _{0}^{z}f(t)dt,& &{}\end{array}$$

(11)

where f ∈ R _k (n,β). Then, by Theorem 3.1 , it follows that $F \in R_{k}(n,\sigma _{1}),$ where

$$\displaystyle\begin{array}{rcl} \sigma _{1} = \frac{2(2\beta + n)} {\left [(n - 2\beta + 2) + \sqrt{(n - 2\beta + 2)^{2 } + 8(2\beta + n)}\right ]}.& &{}\end{array}$$

(12)

For β = 0 and n = 1, we have Libera’s operator for the class R _k of bounded radius rotation. That is, if f ∈ R _k and F is given by (3.6), then

$$\displaystyle\begin{array}{rcl} F \in R_{k}(1,\sigma _{2}),\quad \mbox{ with}\quad \sigma _{2} = \frac{2} {3 + \sqrt{17}}.\end{array}$$

Using Theorem 3.1 and relation (3), we can prove the following.

Theorem 3.2.

Let f and g belong to V _k (n,β), and let F be defined by (6) with α,c,δ,ν positively real, $\delta +\nu =\alpha.$ Then F ∈ V _k (n,σ), where σ is given by (7) .

By taking $\alpha = 1,c + \frac{1} {\lambda },\nu +\delta =\alpha = 1$ and f = g in (6), we obtain the integral operator I _λ (f) = F, defined as:

$$\displaystyle\begin{array}{rcl} F(z) = \frac{1} {\lambda } \int _{0}^{z}t^{\frac{1} {\lambda } -2}f(t)dt,\quad (\lambda > 0).& &{}\end{array}$$

(13)

With the similar techniques, we can easily prove the following result which is stronger version than the one proved in Theorem 3.1.

Theorem 3.3.

Let f ∈ R _k (n,γ) and let, for 0 < λ ≤ 1,F be defined by (13) . Then $F \in R_{k}(n,\delta ^{{\ast}}),$ where δ ^∗ satisfies the conditions given below:

• (i) If $0 <\lambda \leq \frac{1} {2}$ and $\frac{n\lambda } {2(\lambda -1)} \leq \gamma < 1,$ then

  $$\displaystyle\begin{array}{rcl} \delta ^{{\ast}} =\delta _{ 1} = \frac{1} {4\lambda }\left [A_{1} + \sqrt{A_{1 }^{2 } + 8B_{1}}\right ] \geq 0,\end{array}$$

  where

  $$\displaystyle\begin{array}{rcl} A_{1}& =& 2\gamma \lambda + 2\lambda - n\lambda {}\\ B_{1}& =& \lambda \{2\gamma (1-\lambda ) + n\lambda \}. \end{array}$$

• (ii) If $\frac{1} {2} <\lambda \leq 1,$ $\frac{n(\lambda -1)} {2\lambda } \leq \frac{n(3\lambda -\sqrt{8\lambda })} {2\lambda } \leq \gamma,$ then

  $$\displaystyle\begin{array}{rcl} \delta ^{{\ast}} =\delta _{ 2} = \frac{1} {4\lambda }\left [A_{2} + \sqrt{A_{2 }^{2 } + 8B_{2}}\right ] \geq 0,\end{array}$$

  where

  $$\displaystyle\begin{array}{rcl} A_{2}& =& 2\lambda + 2\lambda \gamma - n\lambda {}\\ B_{2}& =& \lambda (2\lambda \gamma + n - n\lambda ). \end{array}$$

• (iii) If $\frac{1} {2} <\lambda \leq 1,\quad \frac{n(\lambda -1)} {2\lambda } < \frac{n(3\lambda -\sqrt{8\lambda })} {2\lambda } <\gamma < 1,$ then $\delta _{3} =\delta _{1}.$

Special Cases

• (1). Let $\lambda = \frac{1} {2}$ in (13). Then we have Libera’s operator and (i) gives us

  $$\displaystyle\begin{array}{rcl} \delta ^{{\ast}} =\delta _{ 1} = \frac{2(2\gamma + n)} {(n - 2\gamma + 2) + \sqrt{(n - 2\gamma + 2)^{2 } + 8(2\gamma + n)}}.\end{array}$$

• (2). When $\gamma = 0,\lambda = \frac{1} {2},n = 1,$ and f ∈ R _k , then $F \in R_{k}(1,\delta _{1}),$ where

  $$\displaystyle\begin{array}{rcl} \delta ^{{\ast}} =\delta _{ 1} = \frac{2} {3 + \sqrt{17}}.\end{array}$$

• (3). Let $\lambda = 1,\gamma = 0,n = 1$ and f ∈ R _k . Then, from (3.8), it follows that

  $$\displaystyle\begin{array}{rcl} F(z) =\int _{ 0}^{z}\frac{f(t)} {t} dt\end{array}$$

  and, by Theorem 3.3, $F \in R_{k}(\frac{1} {2}).$ By using relation (3) and k = 2, we obtain a well-known result that every convex function is starlike of order $\frac{1} {2}.$

Theorem 3.4.

Let $f \in R_{k}(n,0),g \in R_{k}(n,\alpha ),0 \leq \alpha \leq 1.$ Let the function F, for b ≥ 0, be defined as

$$\displaystyle\begin{array}{rcl} F(z) = \frac{1 + b} {z^{b}} \int _{0}^{z}f^{\alpha }(t)t^{b-\alpha -1}g(t)dt.& &{}\end{array}$$

(14)

Then F ∈ R _k (n,η), z ∈ E, where

$$\displaystyle\begin{array}{rcl} \eta = \frac{2n} {(2b + n) + \sqrt{(2b + n)^{2 } + 8n}}.& &{}\end{array}$$

(15)

Proof.

Set

$$\displaystyle\begin{array}{rcl} \frac{zF^{\prime}(z)} {F(z)} = p(z) = \left (\frac{k} {4} + \frac{1} {2}\right )p_{1}(z) -\left (\frac{k} {4} -\frac{1} {2}\right )p_{2}(z).\end{array}$$

Then p(z) is analytic in E and p(0) = 1. From (14), we have

$$\displaystyle\begin{array}{rcl} p(z) + \frac{zp^{\prime}(z)} {p(z) + b}& =& \left [\alpha \frac{zf^{\prime}(z)} {f(z)} + (1-\alpha )\right ] + \frac{zg^{\prime}(z)} {g(z)} - 1 {}\\ & =& \left [\alpha h_{1} + (1-\alpha )\right ] + \left [(1-\alpha )h_{2}(z)+\alpha \right ] - 1 {}\\ & =& \alpha h_{1}(z) + (1-\alpha )h_{2}(z) = h(z),\quad h \in P_{k}(n,0). \end{array}$$

Since $g \in P_{k}(n,\alpha ),\quad f \in R_{k}(n,0),$ it follows that $h_{1},h_{2} \in P_{k}(n,0)$ and P _k (n, 0) is a convex set. Now following the similar technique of Theorem 3.1 and using Lemma 2.1, we obtain the required result that $\frac{zF^{\prime}(z)} {F(z)} = p(z) \in P_{k}(n,\eta ),$ where η is given by (15). □ 

Remark 3.1.

When n = 1, we obtain best possible value of η = σ given by (2.1) with $\alpha = 0,\beta = 1,\delta = b.$

Conclusion. In this paper, we have introduced and considered a new class P _k (n, β) of analytic function. We have discussed several special cases of this new class. We have discussed some integral operators for certain classes of analytic functions in the unit disc E and related with the new class P _k (n, β). Results obtained in this paper can be viewed as an refinement and improvement of the previously known results in this field.