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
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).
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1 Introduction
Let \(\mathcal{A}(n)\) denote the class of functions f of the form
analytic in the unit disc E = { z: | z | < 1}. Let P(n, β) be the class of functions h(z) of the form
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
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
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
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
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
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
where \(_{2}F_{1}\) denotes hypergeometric function. This result is sharp and external function is given as
with
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
belongs to R k (n,σ), where
with
Proof.
First we show that there exists a function \(F \in \mathcal{A}(n)\) satisfying (6). Let
and choose the branches which equal 1 when z = 0. For
we have
where L is well defined and analytic in E. Now let
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
We write
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
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
Define
with \(\quad \alpha _{1} = \frac{1} {\alpha },\quad c_{1} = \frac{c-\alpha } {\alpha }.\)
Then, using (9), we have
Since \(\left \{p + \frac{\alpha _{1}zp^{\prime}} {p+c_{1}} \right \} \in P_{k}(n,\beta ),\) it follows that
Writing \(p_{i}(z) = (1-\sigma )H_{i}(z)+\sigma,\quad i = 1,2,\) we have, for z ∈ E,
We now form the functional Ψ(u, v) by taking u = H i and \(v = zH_{i}^{\prime}\) and so
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
where
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:
where f ∈ R k (n,β). Then, by Theorem 3.1 , it follows that \(F \in R_{k}(n,\sigma _{1}),\) where
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
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:
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
Then F ∈ R k (n,η), z ∈ E, where
Proof.
Set
Then p(z) is analytic in E and p(0) = 1. From (14), we have
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.
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Acknowledgements
The author would like to thank Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Pakistan, for providing excellent research and academic environment. This research work is supported by the HEC project NRPU No: 20-1966/R&D/11-2553, titled: Research Unit of Academic Excellence in Geometric Function Theory and Applications.
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Noor, K.I. (2014). On Some Integral Operators. In: Rassias, T., Tóth, L. (eds) Topics in Mathematical Analysis and Applications. Springer Optimization and Its Applications, vol 94. Springer, Cham. https://doi.org/10.1007/978-3-319-06554-0_25
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