Application of a subordination theorem associated with certain new generalized subclasses of analytic and univalent functions

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:

$$ f(z)=z+a_{2}z^{2}+a_{3}z^{3}+... $$

(1)

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:

$$ \Re \left(\frac{zf^{\prime }(z)}{f(z)}\right)>\beta,\;\;z\in U $$

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:

$$\Re \left(1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)>\beta,\;\;z\in U $$

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:

$$ g(z)=z+z^{3}+z^{5}+...\;\;\;z\in U $$

(2)

be in the class S^∗ while the function g(z) of the form:

$$ g(z)=z+z^{2}+z^{3}+...\;\;\;z\in U $$

(3)

be in the class K. With reference to (2) and (3), we can write that:

$$ g_{\alpha }(z)=\frac{z}{1-z^{\alpha }}=z+\sum\limits_{k=1}^{\infty }z^{1+k\alpha }\;\;\;\;z\in U, $$

(4)

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 f_α(z) of the form (4). Here, we present a more generalized form of (4) such that:

$$ g_{\alpha,n}(z)=\frac{A^{n}z}{\left(A+Bz^{\alpha }\right)^{n}} =z+\sum\limits_{k=1}^{\infty }(-1)^{k}\frac{B^{k}}{A^{k}}n_{k}z^{1+k\alpha }\;\;\;\;z\in U $$

(5)

for some real α(0<α≤2),−1≤B<A≤1,n≥0 and n_k 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 H_α,n of analytic function f_α,n(z) which is a convolution (or Hadamard product) of f(z) and g_α,n (f(z)∗g_α,n(z)) such that:

$$ f_{\alpha,n}(z)=z+\sum\limits_{k=1}^{\infty }(-1)^{k}\frac{B^{k}}{A^{k}} n_{k}a_{k+1}z^{1+k\alpha }\;\;\;\;z\in U $$

(6)

In addition, if f_α,n(z)∈H_α,n satisfies the following condition:

$$ \Re \left(\frac{zf_{\alpha,n}^{\prime }(z)}{f_{\alpha,n}(z)}\right)>\gamma \;\;\;z\in U $$

(7)

for some real α(0<α≤2),n>0, and γ(0≤γ<1), then f_α,n belong to the starlike class $S_{\alpha,n}^{\ast }(A,B,\gamma)$ (of order γ). Also, if f_α,n(z)∈H_α,n satisfies the following condition:

$$ \Re \left(1+\frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right)>\gamma \;\;\;z\in U $$

(8)

for some real α(0<α≤2),n>0, and γ(0≤γ<1), then f_α,n belong to the convex class $K_{\alpha,n}^{\ast }(A,B,\gamma)$ (of order γ). Here, it is noted that f_α,n(z)∈H_α,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:

$$f(z)\prec g(z)\,\,\Rightarrow f(0)=g(0)\ {and}\ f(U)\subset g(U). $$

Furthermore, if the function g is univalent in U:

$$ f(z)\prec g(z)\,\,\,\Leftrightarrow f(0)=g(0)\ { and }\ f(U)\subset g(U). $$

(9)

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:

$$ \sum\limits_{k=1}^{\infty }a_{k}b_{k}z^{k}\prec f(z)\;\;\;\left(z\in U,\;\;a_{1}=1\right) \ (see {[4-6]}). $$

(10)

Lemma 1

The sequence $\left \{ b_{k}\right \}_{k=1}^{\infty }$ is a subordinating factor sequence if and only if:

$$ \Re \left\{ 1+2\sum\limits_{k=1}^{\infty }b_{k}z^{k}\right\} >0\;\;\;(z\in U). $$

(11)

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:

$$ \Re \left\{ 1+z\frac{s^{\prime \prime }(z)}{s^{\prime }(z)}-z\frac{s^{\prime }(z)}{s(z)}\right\} >{\text{max}}\left\{ 0,\Re \Big(\frac{\mu -1}{\mu }s(z)\Big) \right\}. $$

(12)

Suppose that r (r(z)≠0) satisfies the differential equation:

$$ (1-\mu)(r(z)-1)+\mu \frac{zr^{\prime }(z)}{r(z)}\prec (1-\mu)(s(z)-1)+\mu \frac{zs\prime (z)}{s(z)},\;z\in U $$

(13)

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:

$$ z_{0}\omega^{\prime }(z_{0})=\sigma \omega (z_{0}), $$

(14)

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 f_α,n(z) given by (6) belonging to both classes $S_{\alpha,n}^{\ast }(A,B,\gamma)$ and K_α,n(A,B,γ) in the unit disk U.

Theorem 1

Let the function f_α,n(z) of the form (6) satisfy the inequality:

$$ \sum_{k=1}^{\infty }\left(k\alpha -\gamma +1\right)n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}|\leq 1-\gamma. $$

(15)

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 f_α,n(z)given by:

$$f_{\alpha,n}(z)=z+\frac{\left(1-\gamma \right) e^{i\pi }}{\big(k\alpha -\gamma +1\big)n_{k}\frac{|B|^{k}}{A^{k}}}z^{1+k\alpha }\ \ \ (k\geq 1). $$

Proof

Suppose that the function f_α,n(z) given by (6) satisfies (15), then:

$$\left|\frac{zf_{\alpha,n}^{\prime }(z)}{f_{\alpha,n}(z)}-1\right|=\left| \frac{\sum_{k=1}^{\infty }(-1)^{k}k\alpha n_{k}\frac{B^{k}}{A^{k}} a_{k+1}z^{k\alpha }}{1+\sum_{k=1}^{\infty }(-1)^{k}n_{k}\frac{B^{k}}{A^{k}} a_{k+1}z^{k\alpha }}\right| $$

$$\leq \frac{\sum_{k=1}^{\infty }k\alpha n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}||z|^{k\alpha }}{1-\sum_{k=1}^{\infty }n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}||z|^{k\alpha }}|<\frac{\sum_{k=1}^{\infty }k\alpha n_{k}\frac{ |B|^{k}}{A^{k}}|a_{k+1}|}{1-\sum_{k=1}^{\infty }n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}|}\leq 1-\gamma. $$

This shows that $f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)$, and this ends the proof. □

Corollary 1

Let the function f_α,n(z)of the form (6) satisfy the inequality:

$$\sum_{k=1}^{\infty }\big(k\alpha +1\big)n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}|\leq 1. $$

Then, $f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,0)$.

Theorem 2

Let the function f_α,n(z) of the form (6) satisfy the inequality:

$$ \sum\limits_{k=1}^{\infty }\left(k\alpha +1\right)\left(k\alpha -\gamma +1\right)n_{k} \frac{|B|^{k}}{A^{k}}|a_{k+1}|\leq 1-\gamma. $$

(16)

Then, f_α,n(z)∈K_α,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 f_α,n(z)given by:

$$f_{\alpha,n}(z)=z+\frac{\left(1-\gamma \right) e^{i\pi }}{\left(k\alpha +1 \right)\left(k\alpha -\gamma +1\right)n_{k}\frac{|B|^{k}}{A^{k}}}z^{1+k\alpha }\ \ \ \ \ (k\geq 1). $$

Proof

The proof is similar to that of Theorem 1. □

Corollary 2

Let the function f_α,n(z) of the form (6) satisfy the inequality:

$$\sum\limits_{k=1}^{\infty }\left(k\alpha +1\right)^{2}n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}|\leq 1. $$

Then, f_α,n(z)∈K_α,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 f_α,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:

$$\Re \left\{ 1+\frac{z[1-\sigma (1-2z)]}{(1-z)(1+\sigma z)}\right\} >{\text{max}}\left\{ 0,\Re \left(\frac{\mu -1}{\mu }\right)\left(\frac{1+\sigma z}{1-z} \right)\right\}. $$

If

$${}(1-\mu)\left(f_{\alpha,n}^{\prime }(z)-1\right)+\mu \left(\frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right)\prec (1-\mu)\left(\left(\frac{1+\sigma z}{1-z}\right)-1\right)+\mu \left(\frac{(1+\sigma)z}{ (1+\sigma z)(1-z)}\right) $$

holds true, then $f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)$.

Proof

Suppose that we let:

$$ r(z)=f_{\alpha,n}^{\prime }(z)\;\;\;{\text{and}}\;\;\;s(z)=\frac{1+\sigma z}{1-z}. $$

(17)

Then,

$${}\Re \left\{ 1+\frac{zs^{\prime \prime }(z)}{s^{\prime }(z)}-\frac{zs^{\prime }(z)}{s(z)}\right\} >{\text{max}}\left\{ 0,\Re \left(\frac{\mu -1}{\mu }\right)\left(\frac{1+\sigma z}{1-z}\right)\right\} ={\text{max}}\left\{ 0,\Re \left(\frac{\mu -1}{\mu }s(z)\right)\right\} $$

and

$$(1-\mu)(r(z)-1)+\mu \frac{zr^{\prime }(z)}{r(z)}=(1-\mu)\left(f_{\alpha,n}^{\prime }(z)-1\right)+\mu \left(\frac{zf_{\alpha,n}^{\prime \prime }(z)}{ f_{\alpha,n}^{\prime }(z)}\right) $$

$$ \prec (1-\mu)\left(\left(\frac{1+\sigma z}{1-z}\right)-1\right)+\mu \left(\frac{ (1+\sigma)z}{(1+\sigma z)(1-z)}\right)=(1-\mu)(s(z)-1)+\mu \frac{zs^{\prime }(z)}{s(z)}. $$

(18)

Using Lemma 2 in (18), then we obtain the desired result. □

Theorem 4

Let the analytic function f_α,n(z)be defined as in (6). Suppose that f_α,n(z) satisfies the condition that:

$$ \Re \left\{ \frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right\} <-\frac{1+\sigma }{2(1-\sigma)}. $$

(19)

Then, for 0<α≤2,n>0 and $\sigma >1, f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)$.

Proof

Setting:

$$f_{\alpha,n}^{\prime }(z)=\left(\frac{1+\sigma \omega (z)}{1-\omega (z)} \right),\;\;\;\omega (z)\neq 1. $$

Then, ω is regular in U, and since σ≠−1, then ω(0)=0. Also, it follows that:

$$\Re \left\{ \frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right\} =\Re \left\{ \frac{(1+\sigma)z\omega^{\prime }(z)}{(1-\omega (z))(1+\sigma \omega (z))}\right\} <\frac{\sigma +1}{2(\sigma -1)},\;\;\sigma \neq 1. $$

Next, we show that |ω(z)|<1. So, let there exists a point z₀∈U such that for |z|≤|z₀|:

$${\text{max}}|\omega (z)|=|\omega (z)|=1. $$

Then, appealing to Lemma 3 and setting ω(z₀)=e^iθ,z₀ω^′(z₀)=δe^iθ and for δ≥1,σ>1, we have that:

$$\begin{aligned} \Re \left\{\frac{zf_{\alpha,n}^{\prime \prime}(z)}{f_{\alpha,n}^{\prime}(z)}\right\} &=\Re \left\{\frac{(1+\sigma)z_{0}\omega^{\prime}(z_{0})}{(1-\omega (z_{0}))(1+\sigma \omega (z_{0}))}\right\} =\Re \left\{\frac{\delta e^{i\theta}(1+\sigma)}{(1-e^{i\theta})(1+\sigma e^{i\theta})} \right\} \\&=\frac{\delta(\sigma +1)}{2(\sigma -1)}\geq \frac{(\sigma +1)}{2(\sigma -1)}. \end{aligned} $$

Therefore,

$$\Re \left\{ \frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right\} \geq -\frac{1+\sigma }{2(1-\sigma)}\;\;\;z\in U $$

which negates the hypothesis (19).

Hence, we conclude that |ω(z)|<1 for all z∈U and:

$$f_{\alpha,n}^{\prime }(z)\prec \left(\frac{1+\sigma z}{1-z}\right),\;\;\sigma \neq 1,\;z\in U $$

and this obviously ends the proof. □

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 f_a,n∈H_a,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:

$$ \frac{n(\alpha -\gamma +1)|B|}{2\big[n\alpha |B|+(1-\gamma)(A+n|B|)\big]} \left(f_{\alpha,n}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z) $$

(20)

for every function g_α in K_α and:

$$ \Re \left(f_{\alpha,n}(z)\right)>-\frac{\left[n\alpha |B|+(1-\gamma)(A+n|B|) \right]}{n(\alpha -\gamma +1)|B|}. $$

(21)

The constant factor:

$$\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} $$

in the subordination result (20) is sharp.

Proof

Let $f_{\alpha,n}\in \overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)$ and let g_α be any function in K_α. Then:

$$\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} \left(f_{\alpha,n}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z) $$

$$=\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} \left(z+\sum\limits_{k=1}^{\infty }a_{k+1}b_{k+1}z^{k\alpha +1}\right). $$

Thus, by Definition 2, the subordination result (20) will hold true if:

$$\left\{ \frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|) \right]}a_{k}\right\}_{k=1}^{\infty } $$

is a subordinating factor sequence, with a₁=1, appealing to Lemma 1, this is equivalent to:

$$ \Re \left\{ 1+\sum\limits_{k=1}^{\infty }\frac{n(\alpha -\gamma +1)|B|}{\left[ n\alpha |B|+(1-\gamma)(A+n|B|)\right]}a_{k}z^{(k-1)\alpha +1}\right\} >0\;\;\;(z\in U). $$

(22)

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:

$$\Re \left\{ 1+\sum\limits_{k=1}^{\infty }\frac{n(\alpha -\gamma +1)|B|}{\left[ n\alpha |B|+(1-\gamma)(A+n|B|)\right]}a_{k}z^{(k-1)\alpha +1}\right\} $$

$$=\Re \left\{ 1+\frac{n(\alpha -\gamma +1)|B|}{M}z+\frac{A}{M} \sum\limits_{k=2}^{\infty }n(\alpha -\gamma +1)\frac{|B|}{A}a_{k}z^{(k-1)\alpha +1}\right\} $$

$$\geq 1-\frac{n(\alpha -\gamma +1)|B|}{M}r-\frac{A}{M}\sum\limits_{k=2}^{\infty }n_{k-1}\Big((k-1)\alpha -\gamma +1\Big)\frac{|B|^{k-1}}{A^{k-1}} a_{k}r^{(k-1)\alpha +1} $$

$$>1-\frac{n(\alpha -\gamma +1)|B|}{\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} r-\frac{A\left(1-\gamma \right) }{\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} r^{\alpha +1} $$

$$ >1-\frac{n(\alpha -\gamma +1)|B|}{\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} r-\frac{A\left(1-\gamma \right) }{\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} r=1-r>0 $$

(23)

$$\left(|z|=r<1\right), $$

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:

$$g_{\alpha }(z)=\frac{z}{1-z^{\alpha }}\in K_{\alpha } $$

in (20). Now, suppose that we consider the function q_α,n(z) of the form:

$$q_{\alpha,n}(z)=z-\frac{1-\gamma }{n(\alpha -\gamma +1)\frac{|B|}{A}} z^{\alpha +1} $$

which belongs to the class $\overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)$. Then, using (20), we have that:

$$\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]}.q_{\alpha,n}(z)\prec \frac{z}{1-z^{\alpha }}\;\;\;(z\in U) $$

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:

$${\text{min}}\left\{ \Re \left(\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]}.q_{\alpha,n}(z)\right)\right\} =-\frac{1}{2} \;\;(z\in U) $$

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. □

Corollary 3

Let $f_{\alpha,n}(z)\in \overline {S}_{\alpha,n}^{\ast }(1,-1,\gamma)$. Then:

$$\frac{n(\alpha -\gamma +1)}{2\left[n\alpha +(1-\gamma)(1+n)\right]}\left(f_{\alpha,n}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z) $$

for every function g_α in K_α and:

$$\Re \left(f_{\alpha,n}(z)\right)>-\frac{\left[n\alpha +(1-\gamma)(1+n)\right]}{ n(\alpha -\gamma +1)}. $$

The constant factor:

$$\frac{n(\alpha -\gamma +1)}{2\left[n\alpha +(1-\gamma)(1+n)\right]} $$

is sharp.

Corollary 4

Let $f_{\alpha,1}(z)\in \overline {S}_{\alpha,1}^{\ast }(1,-1,\gamma)$. Then:

$$\frac{(\alpha -\gamma +1)}{2(\alpha -2\gamma +2)}\left(f_{\alpha,1}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z) $$

for every function g_α in K_α and:

$$\Re \left(f_{\alpha,1}(z)\right)>-\frac{\left(\alpha -2\gamma +2\right)}{(\alpha -\gamma +1)}. $$

The constant factor:

$$\frac{(\alpha -\gamma +1)}{2\left(\alpha -2\gamma +2\right)} $$

is sharp.

Corollary 5

[9,10] Let $f_{1,1}(z)\in \overline {S} _{1,1}^{\ast }(1,-1,\gamma)$. Then:

$$\frac{(2-\gamma)}{2\left(3-2\gamma \right)}\left(f_{1,1}\ast g_{1}\right)(z)\prec g_{1}(z) $$

for every function g₁ in K₁ and:

$$\Re \left(f_{1,1}(z)\right)>-\frac{\left(3-2\gamma \right)}{(2-\gamma)}. $$

The constant factor:

$$\frac{(2-\gamma)}{2\left(3-2\gamma \right)} $$

is sharp.

Corollary 6

[9–11] Let $f_{1,1}(z)\in \overline {S} _{1,1}^{\ast }(1,-1,0)$. Then:

$$\frac{1}{3}\left(f_{1,1}\ast g_{1}\right)(z)\prec g_{1}(z) $$

for every function g₁ in K₁ and:

$$\Re \left(f_{1,1}(z)\right)>-\frac{3}{2}. $$

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:

$$ \frac{n(\alpha +1)(\alpha -\gamma +1)|B|}{2\left[n\alpha (\alpha +1)|B|+(1-\gamma)(A+n(\alpha +1)|B|)\right]}\left(f_{\alpha,n}\ast g_{\alpha } \right)(z)\prec g_{\alpha }(z) $$

(24)

for every function g_α in K_α and:

$$ \Re \left(f_{\alpha,n}(z)\right)>-\frac{\left[n\alpha (\alpha +1)|B|+(1-\gamma)(A+n(\alpha +1)|B|)\right]}{n(\alpha +1)(\alpha -\gamma +1)|B|}. $$

(25)

The constant factor:

$$\frac{n(\alpha +1)(\alpha -\gamma +1)|B|}{2\left[n\alpha (\alpha +1)|B|+(1-\gamma)(A+n(\alpha +1)|B|)\right]} $$

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:

$$ \frac{n(\alpha +1)(\alpha -\gamma +1)}{2\left[n\alpha (\alpha +1)+(1-\gamma)(1+n(\alpha +1))\right]}\left(f_{\alpha,n}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z) $$

(26)

for every function g_α in K_α and:

$$ \Re \left(f_{\alpha,n}(z)\right)>-\frac{\left[n\alpha (\alpha +1)+(1-\gamma)(1+n(\alpha +1))\right]}{n(\alpha +1)(\alpha -\gamma +1)}. $$

(27)

The constant factor:

$$\frac{n(\alpha +1)(\alpha -\gamma +1)}{2\left[n\alpha (\alpha +1)+(1-\gamma)(1+n(\alpha +1))\right]} $$

cannot be replaced by a larger one.

Corollary 8

Let $f_{\alpha,1}(z)\in \overline {K}_{\alpha,1}(1,-1,\gamma)$. Then:

$$ \frac{(\alpha +1)(\alpha -\gamma +1)}{2\left[\alpha (\alpha +1)+(1-\gamma)(\alpha +2)\right]}\left(f_{\alpha,1}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z) $$

(28)

for every function g_α in K_α and:

$$ \Re \left(f_{\alpha,1}(z)\right)>-\frac{\left[\alpha (\alpha +1)+(1-\gamma)(\alpha +2)\right]}{(\alpha +1)(\alpha -\gamma +1)}. $$

(29)

The constant factor:

$$\frac{(\alpha +1)(\alpha -\gamma +1)}{2\left[\alpha (\alpha +1)+(1-\gamma)(\alpha +2)\right]} $$

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:

$$ \frac{2-\gamma }{5-3\gamma }\left(f_{1,1}\ast g_{1}\right)(z)\prec g_{1}(z) $$

(30)

for every function g₁ in K₁ and:

$$ \Re \left(f_{1,1}(z)\right)>-\frac{5-3\gamma }{2(2-\gamma)}. $$

(31)

The constant factor:

$$\frac{2-\gamma }{5-3\gamma } $$

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:

$$ \frac{2}{5}\left(f_{1,1}\ast g_{1}\right)(z)\prec g_{1}(z) $$

(32)

for every function g₁ in K₁ and:

$$ \Re \left(f_{1,1}(z)\right)>-\frac{5}{4}. $$

(33)

The constant factor:

$$\frac{2}{5} $$

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].