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Application of a subordination theorem associated with certain new generalized subclasses of analytic and univalent functions
Journal of the Egyptian Mathematical Society volume 28, Article number: 33 (2020)
Abstract
The prime focus of the present work is to investigate some fascinating relations of some analytic and univalent functions using a subordination theorem.
Introduction
Let H denote the class of normalized analytic functions f(z) having the form:
in the unit disk \(U=\left \{ z\in \mathbb {C}:|z|<1\right \} \). Also, let S denote the subclass of H univalent in U. Suppose that S∗ denote the subclass of S consisting of the functions f(z) which are starlike in U. A function f(z)∈K is said to be convex in U if f(z)∈S satisfies the condition that zf′(z)∈S∗. If f(z)∈H satisfies the geometric condition:
for some real β(0≤β<1), then we say that f(z) belongs to the class S∗(β) starlike of order β, and if f(z)∈H satisfies the geometric condition:
for some real β(0≤β<1), then we say that f(z) belongs to the class K(β) convex of order β (see [1, 2]). Let the function g(z) of the form:
be in the class S∗ while the function g(z) of the form:
be in the class K. With reference to (2) and (3), we can write that:
where we consider the principal value of z kα for some real α(0<α≤2). See Darus and Owa [3] for some properties of functions fα(z) of the form (4). Here, we present a more generalized form of (4) such that:
for some real α(0<α≤2),−1≤B<A≤1,n≥0 and nk is given by \(n_{k}=\Pi _{j=1}^{k}\left (\frac {n+j-1}{j!}\right)\).
In view of (1) and (5), we introduce a class Hα,n of analytic function fα,n(z) which is a convolution (or Hadamard product) of f(z) and gα,n (f(z)∗gα,n(z)) such that:
In addition, if fα,n(z)∈Hα,n satisfies the following condition:
for some real α(0<α≤2),n>0, and γ(0≤γ<1), then fα,n belong to the starlike class \(S_{\alpha,n}^{\ast }(A,B,\gamma)\) (of order γ). Also, if fα,n(z)∈Hα,n satisfies the following condition:
for some real α(0<α≤2),n>0, and γ(0≤γ<1), then fα,n belong to the convex class \(K_{\alpha,n}^{\ast }(A,B,\gamma)\) (of order γ). Here, it is noted that fα,n(z)∈Hα,n(z) belong to the convex class \(K_{\alpha,n}(A,B,\gamma)\Leftrightarrow zf_{\alpha,n}^{\prime }(z)\) belong to the starlike class \(S_{\alpha,n}^{\ast }(A,B,\gamma)\). For the purpose of the present investigation, we shall call to mind the following definitions and lemmas.
Definition 1
(Subordination principle) For two functions f and g analytic in U, we say that f is subordinate to g, and write f≺g in U or f(z)≺g(z), if there exists a Schwarz function w(z), which is analytic in U with w(0)=0 and |w(z)|<1 (z∈U), such that f(z)=g(w(z)). It is known that:
Furthermore, if the function g is univalent in U:
Also, we say that g(z)is superordinate to f(z) in U (see [4–6]).
Definition 2
(Subordinating factor sequence) A sequence \(\left \{ b_{k}\right \}_{k=1}^{\infty }\) of complex numbers is called subordinating factor sequence if for every univalent function f(z) in K, we have the subordination given by:
Lemma 1
The sequence \(\left \{ b_{k}\right \}_{k=1}^{\infty }\) is a subordinating factor sequence if and only if:
The lemma above is due to Wilf [7]. Interested reader can also refer to [4–6].
Lemma 2
Let s(z)(s(z)≠0)be a univalent function in U. Also, let μ≠0 be a complex number, then we have that:
Suppose that r (r(z)≠0) satisfies the differential equation:
then r≺s and s is the best dominant (see [8] among others).
Lemma 3
Let ωbe regular in H with ω(0)=0. Also, suppose that |ω(z)| attains its maximum value on the circle |z|<1 at a point z0, then:
where σ is any real number and σ≥1 (see [8] among others).
Coefficient inequality
In this section, we consider the coefficient inequalities for function fα,n(z) given by (6) belonging to both classes \(S_{\alpha,n}^{\ast }(A,B,\gamma)\) and Kα,n(A,B,γ) in the unit disk U.
Theorem 1
Let the function fα,n(z) of the form (6) satisfy the inequality:
Then, \(f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)\) for 0≤γ<1s,0<α≤2,−1≤B<A≤1,0<A≤1 and n>0. The equality holds true for fα,n(z)given by:
Proof
Suppose that the function fα,n(z) given by (6) satisfies (15), then:
This shows that \(f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)\), and this ends the proof. â–¡
Corollary 1
Let the function fα,n(z)of the form (6) satisfy the inequality:
Then, \(f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,0)\).
Theorem 2
Let the function fα,n(z) of the form (6) satisfy the inequality:
Then, fα,n(z)∈Kα,n(A,B,γ) for 0≤γ<1, 0<α≤2,−1≤B<A≤1,0<A≤1 and n>0. The equality holds true for fα,n(z)given by:
Proof
The proof is similar to that of Theorem 1. â–¡
Corollary 2
Let the function fα,n(z) of the form (6) satisfy the inequality:
Then, fα,n(z)∈Kα,n(A,B,0).
Remark 1
Putting A=n=1 and B=−1 in Theorems 1 and 2, we obtain the results obtained by Darus and Owa [[3], Theorems 3 and 4].
Next, we present some subordination results.
Some subordination results
Our prime objective here is to establish sufficient conditions for functions belonging to the analytic class \(S_{\alpha,n}^{\ast }(A,B,\gamma)\).
Theorem 3
Suppose that the function fα,n(z) is as defined in (6). Let 0<α≤2,n>0,σ≠−1 and μbe a non-zero complex number in U such that:
If
holds true, then \(f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)\).
Proof
Suppose that we let:
Then,
and
Using Lemma 2 in (18), then we obtain the desired result. â–¡
Theorem 4
Let the analytic function fα,n(z)be defined as in (6). Suppose that fα,n(z) satisfies the condition that:
Then, for 0<α≤2,n>0 and \(\sigma >1, f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)\).
Proof
Setting:
Then, ω is regular in U, and since σ≠−1, then ω(0)=0. Also, it follows that:
Next, we show that |ω(z)|<1. So, let there exists a point z0∈U such that for |z|≤|z0|:
Then, appealing to Lemma 3 and setting ω(z0)=eiθ,z0ω′(z0)=δeiθ and for δ≥1,σ>1, we have that:
Therefore,
which negates the hypothesis (19).
Hence, we conclude that |ω(z)|<1 for all z∈U and:
and this obviously ends the proof. â–¡
Application of a subordination theorem
Let \(\overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)\) and \(\overline {K} _{\alpha,n}(A,B,\gamma)\) denote the classes of functions fa,n∈Ha,n whose coefficients satisfy conditions (15) and (16), respectively. We note that \(\overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)\subseteq S_{\alpha,n}^{\ast }(A,B,\gamma)\ \)and \(\overline {K}_{\alpha,n}(A,B,\gamma)\subseteq K_{\alpha,n}(A,B,\gamma).\) Here, we consider an application of the subordination result given in Lemma 1 to both classes \(\overline {S} _{\alpha,n}^{\ast }(A,B,\gamma)\) and \(\overline {K}_{\alpha,n}(A,B,\gamma) \).
Theorem 5
Let \(f_{\alpha,n}(z)\in \overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)\). If 0≤γ<1,0<α≤2,−1≤B<A≤1,0<A≤1 and n>0, then:
for every function gα in Kα and:
The constant factor:
in the subordination result (20) is sharp.
Proof
Let \(f_{\alpha,n}\in \overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)\) and let gα be any function in Kα. Then:
Thus, by Definition 2, the subordination result (20) will hold true if:
is a subordinating factor sequence, with a1=1, appealing to Lemma 1, this is equivalent to:
Since \(n_{k}\left (k\alpha -\gamma +1\right)\frac {|B|^{k}}{A^{k}}\) is an increasing function of k (k≥1), we have that:
where M=[nα|B|+(1−γ)(A+n|B|)]. Therefore, (22) holds true in U and this obviously proves the inequality (20) while (21) follows by taking:
in (20). Now, suppose that we consider the function qα,n(z) of the form:
which belongs to the class \(\overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)\). Then, using (20), we have that:
which can easily be verified that for 0≤γ<1,0<α≤2,−1≤B<A≤1,0<A≤1,n≥0 and |z|≤r:
and this evidently completes the proof of Theorem 5. For various choices of the parameters involved, several interesting results are obtained. Given below are few instances. â–¡
Corollary 3
Let \(f_{\alpha,n}(z)\in \overline {S}_{\alpha,n}^{\ast }(1,-1,\gamma)\). Then:
for every function gα in Kα and:
The constant factor:
is sharp.
Corollary 4
Let \(f_{\alpha,1}(z)\in \overline {S}_{\alpha,1}^{\ast }(1,-1,\gamma)\). Then:
for every function gα in Kα and:
The constant factor:
is sharp.
Corollary 5
[9,10] Let \(f_{1,1}(z)\in \overline {S} _{1,1}^{\ast }(1,-1,\gamma)\). Then:
for every function g1 in K1 and:
The constant factor:
is sharp.
Corollary 6
[9–11] Let \(f_{1,1}(z)\in \overline {S} _{1,1}^{\ast }(1,-1,0)\). Then:
for every function g1 in K1 and:
Theorem 6
Let \(f_{\alpha,n}(z)\in \overline {K}_{\alpha,n}(A,B,\gamma)\). If 0≤γ<1,0<α≤2,−1≤B<A≤1 and n>0, then:
for every function gα in Kα and:
The constant factor:
in the subordination result (24) cannot be replaced by a larger one, and the proof of which is similar to that of Theorem 3.
Corollary 7
Let \(f_{\alpha,n}(z)\in \overline {K}_{\alpha,n}(1,-1,\gamma)\). Then:
for every function gα in Kα and:
The constant factor:
cannot be replaced by a larger one.
Corollary 8
Let \(f_{\alpha,1}(z)\in \overline {K}_{\alpha,1}(1,-1,\gamma)\). Then:
for every function gα in Kα and:
The constant factor:
cannot be replaced by a larger one.
Corollary 9
[9,10] Let \(f_{1,1}(z)\in \overline {K} _{1,1}(1,-1,\gamma)\). Then:
for every function g1 in K1 and:
The constant factor:
cannot be replaced by a larger one.
Corollary 10
[9,10] Let \(f_{1,1}(z)\in \overline {K} _{1,1}(1,-1,0)\). Then:
for every function g1 in K1 and:
The constant factor:
cannot be replaced by a larger one.
For further illustrations on the applications of the subordination result stated in Lemma 1, interested reader can see [4,6,8–11].
Availability of data and materials
All data generated or analyzed during this study are included in this published article.
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The authors would like to express their sincerest thanks to the referees for a careful reading and various suggestions made for the improvement of the paper.
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Hamzat, J.O., El-Ashwah, R.M. Application of a subordination theorem associated with certain new generalized subclasses of analytic and univalent functions. J Egypt Math Soc 28, 33 (2020). https://doi.org/10.1186/s42787-020-00094-4
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DOI: https://doi.org/10.1186/s42787-020-00094-4