Geometric Properties of Certain Analytic Functions Associated with the Dziok-Srivastava Operator

Abstract

The objective of the present paper is to derive certain geometric properties of analytic functions associated with the Dziok–Srivastava operator.

1. Introduction

Throughout this paper, we assume that:

$$n,p\in \mathbb{N},-1\le B<A\le 1,\alpha >0\mathrm{and}\beta <1.$$

(1)

Let $A_n\left(p\right)$ denote the class of functions of the form:

$$f\left(z\right)=z^p+\sum _{k=n+p}^{\infty }{a}_k{z}^k$$

(2)

which are analytic in the open unit disk $U=\{z:|z|<1\}$. If $f\left(z\right)=z^p+{\sum }_{k=n+p}^{\infty }{a}_k{z}^k\in A_n\left(p\right)$ and $g\left(z\right)=z^p+{\sum }_{k=n+p}^{\infty }{b}_k{z}^k\in A_n\left(p\right)$, then the Hadamard product (or convolution) of f and g is defined by:

$$(f*g)\left(z\right)=z^p+\sum _{k=n+p}^{\infty }{a}_k{b}_k{z}^k$$

For:

$${\alpha }_j\in \mathbb{C}(j=1,2,\cdots ,l)\mathrm{and}{\beta }_j\in \mathbb{C}\backslash \{0,-1,-2,\cdots \}(j=1,2,\cdots ,m)$$

the generalized hypergeometric function ${}_l{F}_m({\alpha }_1,\cdots ,{\alpha }_l;{\beta }_1,\cdots ,{\beta }_m;z)$ is defined by:

$${}_l{F}_m({\alpha }_1,\cdots ,{\alpha }_l;{\beta }_1,\cdots ,{\beta }_m;z)=\sum _{k=0}^{\infty }\frac{{\left({\alpha }_1\right)}_k\cdots {\left({\alpha }_l\right)}_k}{{\left({\beta }_1\right)}_k\cdots {\left({\beta }_m\right)}_k}\frac{z^k}{k!}$$

$$(l\le m+1;l,m\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\};z\in U)$$

where ${\left(x\right)}_k$ is the Pochhammer symbol given by ${\left(x\right)}_k=x(x+1)\cdots (x+k-1)$ for $k\in \mathbb{N}$ and ${\left(x\right)}_0=1$. Corresponding to the function $z^p{}_l{F}_m({\alpha }_1,\cdots ,{\alpha }_l;{\beta }_1,\cdots ,{\beta }_m;z)$, the well-known Dziok–Srivastava operator [1] $H({\alpha }_1,\cdots ,{\alpha }_l;{\beta }_1,\cdots ,{\beta }_m):A_n\left(p\right)\to A_n\left(p\right)$ is defined by:

$$H({\alpha }_1,\cdots ,{\alpha }_l;{\beta }_1,\cdots ,{\beta }_m)f\left(z\right)=\left({z^p}_l{F}_m({\alpha }_1,\cdots ,{\alpha }_l;{\beta }_1,\cdots ,{\beta }_m;z\right)*f\left(z\right)$$

$$(l\le m+1;l,m\in {\mathbb{N}}_0;z\in U)$$

If $f\in A_n\left(p\right)$ is given by (2), then we have:

$$H({\alpha }_1,\cdots ,{\alpha }_l;{\beta }_1,\cdots ,{\beta }_m)f\left(z\right)=z^p+\sum _{k=n+p}^{\infty }\frac{{\left({\alpha }_1\right)}_k\cdots {\left({\alpha }_l\right)}_k}{{\left({\beta }_1\right)}_k\cdots {\left({\beta }_m\right)}_k}\frac{a_k}{k!}{z}^k$$

For convenience, we write:

$$H_m^l\left({\alpha }_1\right)=H({\alpha }_1,\cdots ,{\alpha }_l;{\beta }_1,\cdots ,{\beta }_m)(l\le m+1;l,m\in {\mathbb{N}}_0)$$

It is noteworthy to mention that the Dziok–Srivastava operator is a generalization of certain linear operators considered in earlier investigations.

Next, we consider the function $h(A,B;z)=(1+Az)/(1+Bz)$ for $z\in U$. It is known that the function $h(A,B;z)$ is the conformal map of U onto a disk, symmetrical with respect to the real axis, which is centered at the point $(1-AB)/(1-B^2)$$(B\ne \pm 1)$ and with its radius equal to $(A-B)/$$(1-B^2)$$(B\ne \pm 1)$. Furthermore, the boundary circle of this disk intersects the real axis at the points $(1-A)/(1-B)$ and $(1+A)/(1+B)$ with $B\ne \pm 1$.

Let $P[A,B]$ denote the class of functions of the form $p\left(z\right)=1+p_{1}z+\cdots$, which are analytic in U and satisfy the subordination $p\left(z\right)\prec h(A,B;z)$. It is clear that $p\in P[A,B]$ if and only if:

$$\left|p\left(z\right)-\frac{1-AB}{1-B^2}\right|<\frac{A-B}{1-B^2}(-1<B<A\le 1;z\in U)$$

and:

$$\mathrm{Re}p\left(z\right)>\frac{1-A}{2}(B=-1;z\in U)$$

For two functions f and g analytic in U, f is said to be subordinate to g, written by $f\left(z\right)\prec g\left(z\right)(z\in U)$, if there exists a Schwarz function w in U such that:

$$\left|w\right(z\left)\right|\le \left|z\right|\mathrm{and}f\left(z\right)=g\left(w\right(z\left)\right)(z\in U)$$

Furthermore, if the function g is univalent in U, then:

$$f\left(z\right)\prec g\left(z\right)(z\in U)⟺f\left(0\right)=g\left(0\right)\mathrm{and}f\left(U\right)\subset g\left(U\right)$$

Many properties of analytic functions have been investigated by several authors(see [1,2,3,4,5,6,7,8,9,10,11]). In this paper, we derive certain geometric properties of analytic functions associated with the well-known Dziok–Srivastava operator.

2. Main Results

Theorem 1.

Let f belong to the class $A_n\left(p\right)$. Furthermore, let:

$$\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}\in P[A,B].$$

(3)

Then:

$$\begin{array}{c}\mathrm{Re}\left\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\right\}\hfill \\ \le \left\{\begin{array}{}\mathrm{(4)}& \frac{1+(A+B+n\alpha (A-B))r^n+AB{r}^{2n}}{{(1+B{r}^n)}^2}\hfill & if M_n(A,B,\alpha ,r)\le 0,\hfill \mathrm{(5)}& \frac{L_n^2-4{\alpha }^2{K}_A{K}_B}{4\alpha (A-B)r^{n-1}(1-r^2)K_B}\hfill & if M_n(A,B,\alpha ,r)\ge 0,\hfill \end{array}\right.\hfill \end{array}$$

where:

$$\left\{\begin{array}{c}{K}_A=1-A^2{r}^{2n}+nA{r}^{n-1}(1-r^2),\hfill \\ K_B=1-B^2{r}^{2n}+nB{r}^{n-1}(1-r^2),\hfill \\ L_n=2\alpha (1-AB{r}^{2n})+n\alpha (A+B)r^{n-1}(1-r^2)+(A-B)r^{n-1}(1-r^2),\hfill \\ M_n(A,B,\alpha ,r)=2\alpha K_B(1+A{r}^n)-L_n(1+B{r}^n).\hfill \end{array}\right.$$

(6)

The result is sharp.

Proof. 

For $z=0$, the equality in (4) holds true. Thus, we assume that $0<\left|z\right|=r<1$. From (3), we can write:

$$\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}=\frac{1+A{z}^n\phi \left(z\right)}{1+B{z}^n\phi \left(z\right)}(z\in U),$$

(7)

where $\phi \left(z\right)$ is analytic and $\left|\phi \right(z\left)\right|\le 1$ in U. From (7), we have:

$$\begin{array}{c}\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\hfill \\ =\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\frac{\alpha (A-B)z^n(n\phi \left(z\right)+z{\phi }^{\prime }\left(z\right))}{{(1+B{z}^n\phi \left(z\right))}^2}\hfill \\ =\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\frac{n\alpha }{A-B}(A-B{H}_m^l\left({\alpha }_1\right)f\left(z\right)/z^p)(H_m^l\left({\alpha }_1\right)f\left(z\right)/z^p-1)\hfill \\ +\frac{\alpha (A-B)z^{n+1}{\phi }^{\prime }\left(z\right)}{{(1+B{z}^n\phi \left(z\right))}^2}\hfill \end{array}$$

(8)

By using the Carathéodory inequality:

$$|{\phi }^{\prime }\left(z\right)|\le \frac{1-{\left|\phi \left(z\right)\right|}^2}{1-r^2},$$

we get:

$$\begin{array}{cc}\mathrm{Re}\left\{\frac{z^{n+1}{\phi }^{\prime }\left(z\right)}{{(1+B{z}^n\phi \left(z\right))}^2}\right\}\hfill & \le \frac{r^{n+1}{(1-|\phi \left(z\right)|}^2)}{(1-r^2){|1+B{z}^n\phi \left(z\right)|}^2}\hfill \\ & =\frac{r^{2n}|A-B{H}_m^l\left({\alpha }_1\right)f\left(z\right)/z^p{|}^2-{|H_m^l\left({\alpha }_1\right)f\left(z\right)/z^p-1|}^2}{{(A-B)}^2{r}^{n-1}(1-r^2)}\hfill \end{array}$$

(9)

Set $\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}=u+iv(u,v\in \mathbb{R})$. Then, (8) and (9) give:

$$\begin{array}{c}\mathrm{Re}\left\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\right\}\hfill \\ \le \left(1+n\alpha \frac{A+B}{A-B}\right)u-\frac{n\alpha A}{A-B}-\frac{n\alpha B}{A-B}(u^2-v^2)\hfill \\ +\alpha \frac{r^{2n}({(A-Bu)}^2+{\left(Bv\right)}^2)-({(u-1)}^2+v^2)}{(A-B)r^{n-1}(1-r^2)}\hfill \\ =\left(1+n\alpha \frac{A+B}{A-B}\right)u-\frac{n\alpha }{A-B}(A+B{u}^2)\hfill \\ +\alpha \frac{r^{2n}{(A-Bu)}^2-{(u-1)}^2}{(A-B)r^{n-1}(1-r^2)}+\frac{\alpha }{A-B}\left(nB-\frac{1-B^2{r}^{2n}}{r^{n-1}(1-r^2)}\right)v^2\hfill \end{array}$$

(10)

Note that:

$$\begin{array}{cc}\frac{1-B^2{r}^{2n}}{r^{n-1}(1-r^2)}\hfill & \ge \frac{1-r^{2n}}{r^{n-1}(1-r^2)}=\frac{1}{r^{n-1}}(1+r^2+r^4+\cdots +r^{2(n-2)}+r^{2(n-1)})\hfill \\ & =\frac{1}{2r^{n-1}}[(1+r^{2(n-1)})+(r^2+r^{2(n-2)})+\cdots +(r^{2(n-1)}+1)]\hfill \\ & \ge n\ge nB\hfill \end{array}$$

(11)

Using (10) and (11), we obtain:

$$\begin{array}{cc}\mathrm{Re}\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\}\hfill & \le \left(1+n\alpha \frac{A+B}{A-B}\right)u-\frac{n\alpha }{A-B}(A+B{u}^2)\hfill \\ & +\alpha \frac{r^{2n}{(A-Bu)}^2-{(u-1)}^2}{(A-B)r^{n-1}(1-r^2)}={\psi }_n\left(u\right)\hfill \end{array}$$

(12)

It is known that for $\left|\xi \right|\le \sigma (\sigma <1)$,

$$\left|\frac{1+A\xi }{1+B\xi }-\frac{1-AB{\sigma }^2}{1-B^2{\sigma }^2}\right|\le \frac{(A-B)\sigma }{1-B^2{\sigma }^2}$$

(13)

and:

$$\frac{1-A\sigma }{1-B\sigma }\le \mathrm{Re}\left\{\frac{1+A\xi }{1+B\xi }\right\}\le \frac{1+A\sigma }{1+B\sigma }$$

(14)

Furthermore, (7) and (14) show that:

$$\frac{1-A{r}^n}{1-B{r}^n}\le \mathrm{Re}\left\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}\right\}\le \frac{1+A{r}^n}{1+B{r}^n}.$$

Now, we calculate the maximum value of ${\psi }_n\left(u\right)$ on the segment $\left[\frac{1-A{r}^n}{1-B{r}^n},\frac{1+A{r}^n}{1+B{r}^n}\right]$. Obviously,

$${\psi }_n^{\prime }\left(u\right)=1+n\alpha \frac{A+B}{A-B}-\frac{2n\alpha B}{A-B}u+2\alpha \frac{(1-AB{r}^{2n})-(1-B^2{r}^{2n})u}{(A-B)r^{n-1}(1-r^2)}$$

$${\psi }_n^{″}\left(u\right)=-\frac{2\alpha }{A-B}\left(nB+\frac{1-B^2{r}^{2n}}{r^{n-1}(1-r^2)}\right)<0\left(\mathrm{see} \left(11\right)\right)$$

(15)

and ${\psi }_n^{\prime }\left(u\right)=0$ if and only if:

$$\begin{array}{cc}u=u_n=\hfill & \frac{2\alpha (1-AB{r}^{2n})+n\alpha (A+B)r^{n-1}(1-r^2)+(A-B)r^{n-1}(1-r^2)}{2\alpha [1-B^2{r}^{2n}+nB{r}^{n-1}(1-r^2)]}\hfill \\ & =\frac{L_n}{2\alpha K_B}\left(\mathrm{see} \left(6\right)\right)\hfill \end{array}$$

(16)

Since:

$$\begin{array}{cc}2\alpha \hfill & K_B(1-A{r}^n)-L_n(1-B{r}^n)\hfill \\ & =2\alpha [(1-A{r}^n)(1-B^2{r}^{2n})-(1-B{r}^n)(1-AB{r}^{2n})]\hfill \\ & -n\alpha r^{n-1}(1-r^2)[(A+B)(1-B{r}^n)-2B(1-A{r}^n)]-(A-B)r^{n-1}(1-r^2)(1-B{r}^n)\hfill \\ & =-2\alpha (A-B)r^n(1-B{r}^n)-n\alpha (A-B)r^{n-1}(1-r^2)(1+B{r}^n)\hfill \\ & -(A-B)r^{n-1}(1-r^2)(1-B{r}^n)<0\hfill \end{array}$$

we see that:

$$u_n>\frac{1-A{r}^n}{1-B{r}^n}$$

(17)

However, $u_n$ is not always less than $\frac{1+A{r}^n}{1+B{r}^n}$. The following two cases arise.

Case (I). $u_n\ge \frac{1+A{r}^n}{1+B{r}^n}$, that is $M_n(A,B,\alpha ,r)$$\le 0$. In view of ${\psi }_n^{\prime }\left(u_n\right)=0$ and (15), the function ${\psi }_n\left(u\right)$ is increasing on the segment $\left[\frac{1-A{r}^n}{1-B{r}^n},\frac{1+A{r}^n}{1+B{r}^n}\right]$. Thus, we deduce from (12) that, if $M_n(A,B,\alpha ,r)\le 0$, then:

$$\begin{array}{cc}\mathrm{Re}\left\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\right\}\hfill & \le {\psi }_n\left(\frac{1+A{r}^n}{1+B{r}^n}\right)\hfill \\ & =\left(1+n\alpha \frac{A+B}{A-B}\right)\left(\frac{1+A{r}^n}{1+B{r}^n}\right)\hfill \\ & -\frac{n\alpha }{A-B}\left(A+B{\left(\frac{1+A{r}^n}{1+B{r}^n}\right)}^2\right)\hfill \\ & =\frac{1+A{r}^n}{1+B{r}^n}-\frac{n\alpha }{A-B}\left(1-\frac{1+A{r}^n}{1+B{r}^n}\right)\left(A-B\frac{1+A{r}^n}{1+B{r}^n}\right)\hfill \\ & =\frac{1+(A+B+n\alpha (A-B))r^n+AB{r}^{2n}}{{(1+B{r}^n)}^2}\hfill \end{array}$$

This gives (4).

Next, we consider the function f defined by:

$$\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}=\frac{1+A{z}^n}{1+B{z}^n}$$

which satisfies the condition (3). It is easy to check that:

$$\frac{H_m^l\left({\alpha }_1\right)f\left(r\right)}{r^p}+\alpha r{(\frac{H_m^l\left({\alpha }_1\right)f\left(r\right)}{r^p})}^{\prime }=\frac{1+(A+B+n\alpha (A-B))r^n+AB{r}^{2n}}{{(1+B{r}^n)}^2}$$

which implies that the inequality (4) is sharp.

Case (II). $u_n\le \frac{1+A{r}^n}{1+B{r}^n}$, that is $M_n(A,B,\alpha ,r)\ge 0$. In this case, we easily have:

$$\mathrm{Re}\left\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\right\}\le {\psi }_n\left(u_n\right)$$

(18)

In view of (6), ${\psi }_n\left(u\right)$ in (12) can be written as:

$${\psi }_n\left(u\right)=\frac{-\alpha K_B{u}^2+L_{n}u-\alpha K_A}{(A-B)r^{n-1}(1-r^2)}$$

(19)

Therefore, if $M_n(A,B,\alpha ,r)\ge 0$, then it follows from (16), (18), and (19) that:

$$\begin{array}{cc}\mathrm{Re}\left\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\right\}\hfill & \le \frac{-\alpha K_B{u}_n^2+L_n{u}_n-\alpha K_A}{(A-B)r^{n-1}(1-r^2)}\hfill \\ & =\frac{L_n^2-4{\alpha }^2{K}_A{K}_B}{4\alpha (A-B)r^{n-1}(1-r^2)K_B}\hfill \end{array}$$

To show the sharpness, we take:

$$\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}=\frac{1+A{z}^n\phi \left(z\right)}{1+B{z}^n\phi \left(z\right)}\mathrm{and}\phi \left(z\right)=\frac{z-c_n}{1-c_{n}z}$$

where $c_n\in \mathbb{R}$ is determined by:

$$\frac{H_m^l\left({\alpha }_1\right)f\left(r\right)}{r^p}=\frac{1+A{r}^n\phi \left(r\right)}{1+B{r}^n\phi \left(r\right)}=u_n\in \left(\frac{1-A{r}^n}{1-B{r}^n},\frac{1+A{r}^n}{1+B{r}^n}\right]$$

Clearly, $-1<\phi \left(r\right)\le 1,-1\le c_n<1,\left|\phi \left(z\right)\right|\le 1(z\in U)$, and so, f satisfies the condition (3). Since:

$${\phi }^{\prime }\left(r\right)=\frac{1-c_n^2}{{(1-c_{n}r)}^2}=\frac{1-{\left|\phi \left(r\right)\right|}^2}{1-r^2}$$

from the above argument, we find that:

$$\frac{H_m^l\left({\alpha }_1\right)f\left(r\right)}{r^p}+\alpha r{(\frac{H_m^l\left({\alpha }_1\right)f\left(r\right)}{z^p})}^{\prime }={\psi }_n\left(u_n\right)$$

The proof of the theorem is now completed. □

Corollary 1.

Let $f\in A_1\left(p\right)$, and satisfy $\mathrm{Re}\{H_m^l\left({\alpha }_1\right)f\left(z\right)/z^p\}>\beta$$(\beta <1;z\in U)$. Then, for $\left|z\right|=r<1$,

$$\mathrm{Re}\left\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\right\}\le \beta +(1-\beta )\frac{1+2\alpha r-r^2}{{(1-r)}^2}$$

The result is sharp.

Proof. 

By considering $\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)/z^p-\beta }{1-\beta }$ instead of $H_m^l\left({\alpha }_1\right)f\left(z\right)/z^p$, we only need to prove the corollary for $\beta =0$. Putting $n=A=1$ and $B=-1$ in (6), we get:

$$K_1=2(1-r^2),K_{-1}=0,L_1=2\alpha (1+r^2)+2(1-r^2)$$

and:

$$M_1(1,-1,\alpha ,r)=-2(1-r)[1+\alpha -(1-\alpha )r^2]\le 0$$

Consequently, an application of (4) in Theorem 2.1 yields:

$$\mathrm{Re}\left\{\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p})}^{\prime }\right\}\le \frac{1+2\alpha r-r^2}{{(1-r)}^2}$$

The sharpness follows immediately from that of Theorem 1. □

Theorem 2.

Let ${\alpha }_j$$(j=1,2,\cdots ,l)$ and ${\beta }_s$$(s=1,2,\cdots ,m)$ be positive real numbers. Furthermore, let $f\left(z\right)=z^p+{\sum }_{k=n+p}^{\infty }{a}_k{z}^k\in A_n\left(p\right)$, and satisfy:

$$\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{\left(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}\right)}^{\prime }\in P[A,B]$$

(20)

Then:

$$|a_k|\le \frac{k!(A-B){\left({\beta }_1\right)}_k\cdots {\left({\beta }_m\right)}_k}{(1+\alpha (k-p)){\left({\alpha }_1\right)}_k\cdots {\left({\alpha }_l\right)}_k}(k\ge n+p)$$

(21)

The result is sharp for each $k\ge n+p$.

Proof. 

It is well known that if:

$$g\left(z\right)=\sum _{k=1}^{\infty }{b}_k{z}^k\prec \phi \left(z\right)(z\in U)$$

where $g\left(z\right)$ is analytic in U and $\phi \left(z\right)=z+\cdots$ is convex univalent in U, then $|b_k|\le 1$$(k=1,2,3,\cdots )$.

From (20), we have:

$$\begin{array}{c}\frac{1}{A-B}\left(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}+\alpha z{\left(\frac{H_m^l\left({\alpha }_1\right)f\left(z\right)}{z^p}\right)}^{\prime }-1\right)\hfill \\ =\frac{1}{A-B}\sum _{k=n+p}^{\infty }\frac{(1+\alpha (k-p)){\left({\alpha }_1\right)}_k\cdots {\left({\alpha }_l\right)}_k·a_k}{k!{\left({\beta }_1\right)}_k\cdots {\left({\beta }_m\right)}_k}{z}^{k-p}\hfill \\ \prec \frac{z}{1+Bz}(z\in U)\hfill \end{array}$$

(22)

In view of the function $\frac{z}{1+Bz}$ being convex univalent in U, it follows from (22) that:

$$\frac{(1+\alpha (k-p)){\left({\alpha }_1\right)}_k\cdots {\left({\alpha }_l\right)}_k}{k!(A-B){\left({\beta }_1\right)}_k\cdots {\left({\beta }_m\right)}_k}\left|a_k\right|\le 1(k\ge n+p)$$

which gives (21).

Next, we consider the function $f_{k-p}\left(z\right)$ defined by:

$$f_{k-p}\left(z\right)=z^p+(A-B)\sum _{q=1}^{\infty }\frac{{(-B)}^{q-1}{\left({\beta }_1\right)}_{qk}\cdots {\left({\beta }_m\right)}_{qk}\left(qk\right)!}{(1+\alpha q(k-p)){\left({\alpha }_1\right)}_{qk}\cdots {\left({\alpha }_l\right)}_{qk}}{z}^{q(k-p)+p}(z\in U;k\ge n+p)$$

Since:

$$\frac{H_m^l\left({\alpha }_1\right)f_{k-p}\left(z\right)}{z^p}+\alpha z{\left(\frac{H_m^l\left({\alpha }_1\right)f_{k-p}\left(z\right)}{z^p}\right)}^{\prime }=\frac{1+A{z}^{k-p}}{1+B{z}^{k-p}}\prec \frac{1+Az}{1+Bz}(z\in U)$$

and:

$$f_{k-p}\left(z\right)=z^p+\frac{k!(A-B){\left({\beta }_1\right)}_k\cdots {\left({\beta }_m\right)}_k}{(1+\alpha (k-p)){\left({\alpha }_1\right)}_k\cdots {\left({\alpha }_l\right)}_k}{z}^k+\cdots$$

for each $k\ge n+p$, the proof of Theorem 2 is completed. □