New Applications of the Bernardi Integral Operator

Abstract

Let $A(p,n)$ be the class of $f\left(z\right)$ which are analytic p-valent functions in the closed unit disk $\overline{\mathbb{U}}=\left\{z\in \mathbb{C}:\left|z\right|\le 1\right\}$. The expression $B_{-m-\lambda }f\left(z\right)$ is defined by using fractional integrals of order $\lambda$ for $f\left(z\right)\in A(p,n).$ When $m=1$ and $\lambda =0,$ $B_{-1}f\left(z\right)$ becomes Bernardi integral operator. Using the fractional integral $B_{-m-\lambda }f\left(z\right),$ the subclass $T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right)$ of $A(p,n)$ is introduced. In the present paper, we discuss some interesting properties for $f\left(z\right)$ concerning with the class $T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right).$ Also, some interesting examples for our results will be considered.

1. Introduction

Let $A(p,n)$ be the class of functions $f\left(z\right)$ of the form

$$f\left(z\right)=z^p+\sum _{k=p+n}^{\infty }{a}_k{z}^k,n\in \mathbb{N}=\{1,2,3,\dots \}$$

(1)

that are analytic p-valent functions in the closed unit disk $\overline{\mathbb{U}}=\left\{z\in \mathbb{C}:\left|z\right|\le 1\right\}$. For functions $f\left(z\right)\in A(p,n),$ we consider

$$B_{-1}f\left(z\right)=\frac{p+\gamma }{z^{\gamma }}{\int }_0^z{t}^{\gamma -1}f\left(t\right)dt=z^p+\sum _{k=p+n}^{\infty }\frac{p+\gamma }{k+\gamma }{a}_k{z}^k,\gamma \in \mathbb{N}.$$

(2)

If $p=1,$ for $f\left(z\right)\in A(1,n)$

$$B_{-1}f\left(z\right)=\frac{1+\gamma }{z^{\gamma }}{\int }_0^z{t}^{\gamma -1}f\left(t\right)dt=z+\sum _{k=n+1}^{\infty }\frac{1+\gamma }{k+\gamma }{a}_k{z}^k,\gamma \in \mathbb{N}$$

(3)

is considered by Bernardi [1]. Therefore, $B_{-1}f\left(z\right)$ in (3) is said to be the Bernardi integral operator. Further, if $p=1$ and $\gamma =1,$ for $f\left(z\right)\in A(1,n)$

$$L_{-1}f\left(z\right)=\frac{2}{z}{\int }_0^{z}f\left(t\right)dt=z+\sum _{k=n+1}^{\infty }\frac{2}{k+1}{a}_k{z}^k$$

(4)

is defined by Libera [2]. Therefore, $L_{-1}f\left(z\right)$ in (4) is called the Libera integral operator.

For $B_{-1}f\left(z\right)$ in (2), we consider

$$B_{-2}f\left(z\right)=B_{-1}\left(B_{-1}f\left(z\right)\right)=z^p+\sum _{k=p+n}^{\infty }{\left(\frac{p+\gamma }{k+\gamma }\right)}^2{a}_k{z}^k$$

(5)

and

$$B_{-m}f\left(z\right)=B_{-1}\left(B_{-m+1}f\left(z\right)\right)=z^p+\sum _{k=p+n}^{\infty }{\left(\frac{p+\gamma }{k+\gamma }\right)}^m{a}_k{z}^k$$

(6)

with $m\in \mathbb{N}$ and $B_{0}f\left(z\right)=f\left(z\right).$

From the various definitions of fractional calculus of $f\left(z\right)\in A(p,n)$ (that is, fractional integrals and fractional derivatives) given in the literature, we would like to recall here the following definitions for fractional calculus which were used by Owa [3] and Owa and Srivastava [4].

Definition 1.

The fractional integral of order λ for $f\left(z\right)\in A(p,n)$ is defined by

$$D_z^{-\lambda }f\left(z\right)=\frac{1}{\mathsf{\Gamma }\left(\lambda \right)}{\int }_0^z\frac{f\left(t\right)}{{(z-t)}^{1-\lambda }}dt,\lambda >0$$

(7)

where the multiplicity of ${(z-t)}^{\lambda -1}$ is removed by requiring $log(z-t)$ to be real when $z-t>0$ and Γ is the Gamma function.

With the above definitions, we know that

$$D_z^{-\lambda }f\left(z\right)=\frac{\mathsf{\Gamma }(p+1)}{\mathsf{\Gamma }(p+1+\lambda )}{z}^{p+\lambda }+\sum _{k=p+n}^{\infty }\frac{\mathsf{\Gamma }(k+1)}{\mathsf{\Gamma }(k+1+\lambda )}{a}_k{z}^{k+\lambda }$$

(8)

for $\lambda >0$ and $f\left(z\right)\in A(p,n).$ Using the fractional integral operator over $A(p,n),$ we consider

$$B_{-\lambda }f\left(z\right)=\frac{\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )}{z}^{1-\gamma -\lambda }{D}_z^{-\lambda }\left(z^{\gamma -1}f\left(z\right)\right)=z^p+\sum _{k=p+n}^{\infty }\frac{\mathsf{\Gamma }(k+\gamma )\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )\mathsf{\Gamma }(k+\gamma +\lambda )}{a}_k{z}^k,$$

(9)

where $0\le \lambda \le 1.$ If $\lambda =0$ in (9), then $B_{0}f\left(z\right)=f\left(z\right)$ and if $\lambda =1$ in (9), then we see that

$$B_{-1}f\left(z\right)=\frac{p+\gamma }{z^{\gamma }}{\int }_0^z{t}^{\gamma -1}f\left(t\right)dt.$$

(10)

With the operator $B_{-\lambda }f\left(z\right)$ given by (9), we know

$$B_{-m-\lambda }f\left(z\right)=B_{-m}\left(B_{-\lambda }f\left(z\right)\right)=z^p+\sum _{k=p+n}^{\infty }{\left(\frac{p+\gamma }{k+\gamma }\right)}^m\frac{\mathsf{\Gamma }(k+\gamma )\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )\mathsf{\Gamma }(k+\gamma +\lambda )}{a}_k{z}^k,$$

(11)

where $0\le \lambda \le 1$ and $m\in \mathbb{N}.$ The operator $B_{-m-\lambda }f\left(z\right)$ is a generalization of the Bernardi integral operator $B_{-1}f\left(z\right)$. From the definition of $B_{-m-\lambda }f\left(z\right),$ we know that

$$B_{-m-\lambda }f\left(z\right)=B_{-m}\left(B_{-\lambda }f\left(z\right)\right)=B_{-\lambda }\left(B_{-m}f\left(z\right)\right).$$

(12)

From s different boundary points $z_l(l=1,2,3,\dots ,s)$ with $|z_l|=1,$ we consider

$${\alpha }_s=\frac{1}{s}\sum _{l=1}^s\frac{B_{-m-\lambda }f\left(z_l\right)}{z_l^p},$$

(13)

where ${\alpha }_s\in e^{i\beta }{B}_{-m-\lambda }f\left(\mathbb{U}\right),$${\alpha }_s\ne 1,$ $-\frac{\pi }{2}\le \beta \le \frac{\pi }{2}$ and $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}$ is the open unit disk. For such ${\alpha }_s,$ if $f\left(z\right)\in A(p,n)$ satisfies

$$\left|\frac{e^{i\beta }\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-{\alpha }_s}{e^{i\beta }-{\alpha }_s}-1\right|<\rho ,z\in \mathbb{U}$$

(14)

for some real $\rho >0,$ we say that the function $f\left(z\right)$ belongs to the class $T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right).$

It is clear that a function $f\left(z\right)\in A(p,n)$ belongs to the class $T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right)$ provided that the condition

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|,z\in \mathbb{U},$$

(15)

is satisfied. If we consider the function $f\left(z\right)\in A(p,n)$ given by

$$f\left(z\right)=z^p+{\left(\frac{p+n+\gamma }{p+\gamma }\right)}^m\frac{\mathsf{\Gamma }\left(p+\gamma \right)\mathsf{\Gamma }\left(p+n+\gamma +\lambda \right)}{\mathsf{\Gamma }\left(p+n+\gamma \right)\mathsf{\Gamma }\left(p+\gamma +\lambda \right)}\rho (e^{i\beta }-{\alpha }_s)z^{p+n}$$

(16)

then $f\left(z\right)$ satisfies

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|=\rho \left|e^{i\beta }-{\alpha }_s\right|{\left|z\right|}^m<\rho \left|e^{i\beta }-{\alpha }_s\right|,z\in \mathbb{U}.$$

(17)

Therefore, $f\left(z\right)$ given by (16) is in the class $T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right).$

Discussing our problems for $f\left(z\right)\in T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right),$ we have to recall here the following lemma due to Miller and Mocanu [5,6] (refining the old one in Jack [7].)

Lemma 1.

Let the function $w\left(z\right)$ given by

$$w\left(z\right)=a_n{z}^n+a_{n+1}{z}^{n+1}+a_{n+2}{z}^{n+2}+\dots ,n\in \mathbb{N}$$

(18)

be analytic in $\mathbb{U}$ with $w\left(0\right)=0.$ If $\left|w\right(z\left)\right|$ attains its maximum value on the circle $\left|z\right|=r$ at a point $z_0,(0<|z_0|<1)$ then there exists a real number $k\ge n$ such that

$$\frac{z_0{w}^{\prime }\left(z_0\right)}{w\left(z_0\right)}=k$$

(19)

and

$$Re\left(1+\frac{z_0{w}^{″}\left(z_0\right)}{w^{\prime }\left(z_0\right)}\right)\ge k.$$

(20)

2. Properties of Functions Concerning with the Class T_p,n (α_s, β, ρ; m, λ)

We begin with a sufficient condition on a function $f\left(z\right)\in A(p,n)$ which makes it a member of $T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right).$

Theorem 1.

If $f\left(z\right)\in A(p,n)$ satisfies

$$\left|\frac{B_{-m-\lambda +1}f\left(z\right)}{B_{-m-\lambda }f\left(z\right)}-1\right|<\frac{\left|e^{i\beta }-{\alpha }_s\right|n\rho }{(p+\gamma )(1+\left|e^{i\beta }-{\alpha }_s\right|\rho )},z\in \mathbb{U}$$

(21)

for some ${\alpha }_s$ given by (13) with ${\alpha }_s\ne 1$ such that $z_g\in \partial \mathbb{U}(g=1,2,3,\dots ,s),$ and for some real $\rho >1,$ then

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|,z\in \mathbb{U}$$

(22)

that is, $f\left(z\right)\in T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right).$

Proof. 

We introduce the function $w\left(z\right)$ defined by

$$w\left(z\right)=\frac{e^{i\beta }\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-{\alpha }_s}{e^{i\beta }-{\alpha }_s}-1=\frac{e^{i\beta }}{e^{i\beta }-{\alpha }_s}\left\{\sum _{k=p+n}^{\infty }{\left(\frac{p+\gamma }{k+\gamma }\right)}^m\frac{\mathsf{\Gamma }(k+\gamma )\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )\mathsf{\Gamma }(k+\gamma +\lambda )}{a}_k{z}^{k-p}\right\}.$$

(23)

Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ with $w\left(0\right)=0$ and

$$\frac{B_{-m-\lambda }f\left(z\right)}{z^p}=1+(1-e^{-i\beta }{\alpha }_s)w\left(z\right).$$

(24)

Noting that

$$B_{-m-\lambda +1}f\left(z\right)=\frac{\gamma }{p+\gamma }{B}_{-m-\lambda }f\left(z\right)+\frac{1}{p+\gamma }z{\left(B_{-m-\lambda }f\left(z\right)\right)}^{\prime },$$

(25)

we obtain that

$$\frac{B_{-m-\lambda +1}f\left(z\right)}{B_{-m-\lambda }f\left(z\right)}-1=\frac{(1-e^{-i\beta }{\alpha }_s)z{w}^{\prime }\left(z\right)}{(p+\gamma )(1+(1-e^{-i\beta }{\alpha }_s)w\left(z\right))}$$

(26)

and that

$$\left|\frac{B_{-m-\lambda +1}f\left(z\right)}{B_{-m-\lambda }f\left(z\right)}-1\right|=\frac{1}{p+\gamma }\left|\frac{(1-e^{-i\beta }{\alpha }_s)z{w}^{\prime }\left(z\right)}{1+(1-e^{-i\beta }{\alpha }_s)w\left(z\right)}\right|<\frac{|e^{i\beta }-{\alpha }_s|n\rho }{(p+\gamma )(1+|e^{i\beta }-{\alpha }_s\left|\rho \right)}.$$

(27)

by employing (21). Assume, to arrive at a contradiction, that there exists a point $z_0,(0<|z_0|<1)$ such that

$$\max \left\{\right|w\left(z\right)|;|z|\le |z_0\left|\right\}=|w\left(z_0\right)|=\rho >1.$$

(28)

Then, we can write that $w\left(z_0\right)=\rho e^{i\theta },(0\le \theta \le 2\pi )$ and $z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right),(k\ge n)$ by Lemma 1. For such a point $z_0\in \mathbb{U},$ $f\left(z\right)$ satisfies

$$\left|\frac{B_{-m-\lambda +1}f\left(z_0\right)}{B_{-m-\lambda }f\left(z_0\right)}-1\right|=\frac{1}{p+\gamma }\left|\frac{(1-e^{-i\beta }{\alpha }_s)z_0{w}^{\prime }\left(z_0\right)}{1+(1-e^{-i\beta }{\alpha }_s)w\left(z_0\right)}\right|$$

$$=\frac{1}{p+\gamma }\left|\frac{(1-e^{-i\beta }{\alpha }_s)k\rho }{1+(1-e^{-i\beta }{\alpha }_s)\rho e^{i\theta }}\right|$$

$$\ge \frac{|1-e^{-i\beta }{\alpha }_s|n\rho }{(p+\gamma )(1+|1-e^{-i\beta }{\alpha }_s\left|\rho \right)}$$

$$=\frac{|e^{i\beta }-{\alpha }_s|n\rho }{(p+\gamma )(1+|e^{i\beta }-{\alpha }_s\left|\rho \right)}.$$

(29)

Since this contradicts our condition (21), we see that there is no $z_0,(0<|z_0|<1)$ such that $\left|w\right(z_0\left)\right|=\rho >1.$ This shows us that

$$\left|w\left(z\right)\right|=\left|\frac{e^{i\beta }\left(\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right)}{e^{i\beta }-{\alpha }_s}\right|<\rho ,z\in \mathbb{U},$$

(30)

that is, that

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|,z\in \mathbb{U}.$$

(31)

This completes the proof of the theorem.  □

Example 1.

We consider a function $f\left(z\right)\in A(p,n)$ given by

$$f\left(z\right)=z^p+a_{p+n}{z}^{p+n},z\in \mathbb{U}$$

(32)

with $0<|a_{p+n}|<\frac{1}{2Q}$, where

$$Q={\left(\frac{p+\gamma }{p+n+\gamma }\right)}^m\frac{\mathsf{\Gamma }(p+n+\gamma )\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )\mathsf{\Gamma }(p+n+\gamma +\lambda )}.$$

(33)

For such $f\left(z\right)$, we have

$$\frac{B_{-m-\lambda +1}f\left(z\right)}{B_{-m-\lambda }f\left(z\right)}=\frac{z^p+\left(\frac{p+n+\gamma }{p+\gamma }\right)Q{a}_{p+n}{z}^{p+n}}{z^p+Q{a}_{p+n}{z}^{p+n}}=\frac{1+\left(\frac{p+n+\gamma }{p+\gamma }\right)Q{a}_{p+n}{z}^n}{1+Q{a}_{p+n}{z}^n}$$

(34)

that is, that

$$\left|\frac{B_{-m-\lambda +1}f\left(z\right)}{B_{-m-\lambda }f\left(z\right)}-1\right|=\left|\frac{\left(\frac{n}{p+\gamma }\right)Q{a}_{p+n}{z}^n}{1+Q{a}_{p+n}{z}^n}\right|<\frac{\left(\frac{n}{p+\gamma }\right)Q\left|a_{p+n}\right|}{1-Q|a_{p+n}|},z\in \mathbb{U}.$$

(35)

Now, we consider five boundary points such that

$$z_1=e^{-i\frac{arg\left(a_{p+n}\right)}{n}}$$

(36)

$$z_2=e^{i\frac{\pi -6arg\left(a_{p+n}\right)}{6n}}$$

(37)

$$z_3=e^{i\frac{\pi -4arg\left(a_{p+n}\right)}{4n}}$$

(38)

$$z_4=e^{i\frac{\pi -3arg\left(a_{p+n}\right)}{3n}}$$

(39)

and

$$z_5=e^{i\frac{\pi -2arg\left(a_{p+n}\right)}{2n}}.$$

(40)

For these five boundary points, we know that

$$\frac{B_{-m-\lambda }f\left(z_1\right)}{z_1^p}=1+Q{a}_{p+n}{e}^{-iarg\left(a_{p+n}\right)}=1+Q\left|a_{p+n}\right|,$$

(41)

$$\frac{B_{-m-\lambda }f\left(z_2\right)}{z_2^p}=1+Q{a}_{p+n}{e}^{i\left(\frac{\pi }{6}-arg\left(a_{p+n}\right)\right)}=1+\frac{\sqrt{3}+i}{2}Q\left|a_{p+n}\right|,$$

(42)

$$\frac{B_{-m-\lambda }f\left(z_3\right)}{z_3^p}=1+Q{a}_{p+n}{e}^{i\left(\frac{\pi }{4}-arg\left(a_{p+n}\right)\right)}=1+\frac{\sqrt{2}(1+i)}{2}Q\left|a_{p+n}\right|,$$

(43)

$$\frac{B_{-m-\lambda }f\left(z_4\right)}{z_4^p}=1+Q{a}_{p+n}{e}^{i\left(\frac{\pi }{3}-arg\left(a_{p+n}\right)\right)}=1+\frac{1+\sqrt{3}i}{2}Q\left|a_{p+n}\right|,$$

(44)

and

$$\frac{B_{-m-\lambda }f\left(z_5\right)}{z_5^p}=1+Q{a}_{p+n}{e}^{i\left(\frac{\pi }{2}-arg\left(a_{p+n}\right)\right)}=1+iQ\left|a_{p+n}\right|.$$

(45)

Thus ${\alpha }_5$ is given by

$${\alpha }_5=\frac{1}{5}\sum _{l=1}^5\frac{B_{-m-\lambda }f\left(z_l\right)}{z_l^p}=1+\frac{(3+\sqrt{2}+\sqrt{3})(1+i)}{10}Q\left|a_{p+n}\right|.$$

(46)

This gives us that

$$\left|1-e^{-i\beta }{\alpha }_5\right|=\frac{\sqrt{2}(3+\sqrt{2}+\sqrt{3})}{10}Q\left|a_{p+n}\right|$$

(47)

with $\beta =0.$ For such ${\alpha }_5$ and $\beta ,$ we take $\rho >1$ with

$$\frac{\left(\frac{n}{p+\gamma }\right)Q\left|a_{p+n}\right|}{1-Q|a_{p+n}|}\le \frac{|e^{i\beta }-{\alpha }_5|n\rho }{(p+\gamma )(1+|e^{i\beta }-{\alpha }_5\left|\rho \right)}.$$

(48)

It follows from the above that

$$\rho \ge \frac{10}{\sqrt{2}(3+\sqrt{2}+\sqrt{3})(1-2Q|a_{p+n}\left|\right)}>\frac{10}{\sqrt{2}(3+\sqrt{2}+\sqrt{3})}>1.$$

(49)

For such ${\alpha }_5$ and $\rho >1$, $f\left(z\right)$ satisfies

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<Q|a_{p+n}|\le \rho |e^{i\beta }-{\alpha }_5|,z\in \mathbb{U}.$$

(50)

Our next result reads as follows.

Theorem 2.

If $f\left(z\right)\in A(p,n)$ satisfies

$$\left|\left(\frac{B_{-m-\lambda +1}f\left(z\right)}{B_{-m-\lambda }f\left(z\right)}-1\right)\left(\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right)\right|<\frac{{\left|e^{i\beta }-{\alpha }_s\right|}^{2}n{\rho }^2}{(p+\gamma )(1+\left|e^{i\beta }-{\alpha }_s\right|\rho )},z\in \mathbb{U}$$

(51)

for some ${\alpha }_s$ defined by (13) with ${\alpha }_s\ne 1$ such that $z_g\in \partial \mathbb{U}(g=1,2,3,\dots s),$ and for some real $\rho >1,$ then

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|,z\in \mathbb{U}$$

(52)

that is, $f\left(z\right)\in T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right).$

Proof. 

Define a function $w\left(z\right)$ by (23). Using (24) and (26), we have

$$\left|\left(\frac{B_{-m-\lambda +1}f\left(z\right)}{B_{-m-\lambda }f\left(z\right)}-1\right)\left(\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right)\right|=\left|\frac{{\left(1-e^{-i\beta }{\alpha }_s\right)}^{2}zw\left(z\right)w^{\prime }\left(z\right)}{(p+\gamma )(1+\left(1-e^{-i\beta }{\alpha }_s\right)w\left(z\right))}\right|.$$

(53)

We suppose that there exists a point $z_0,(0<|z_0|<1)$ such that

$$\max \left\{\right|w\left(z\right)|;|z|\le |z_0\left|\right\}=|w\left(z_0\right)|=\rho >1.$$

(54)

Then, Lemma 1 leads us that $w\left(z_0\right)=\rho e^{i\theta },(0\le \theta \le 2\pi )$ and $z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right),(k\ge n).$ It follows from the above that

$$\left|\left(\frac{B_{-m-\lambda +1}f\left(z_0\right)}{B_{-m-\lambda }f\left(z_0\right)}-1\right)\left(\frac{B_{-m-\lambda }f\left(z_0\right)}{z_0^p}-1\right)\right|=\left|\frac{{\left(1-e^{-i\beta }{\alpha }_s\right)}^2{z}_{0}w\left(z_0\right)w^{\prime }\left(z_0\right)}{(p+\gamma )(1+\left(1-e^{-i\beta }{\alpha }_s\right)w\left(z_0\right))}\right|$$

$$=\frac{{\left|e^{i\beta }-{\alpha }_s\right|}^2{\rho }^{2}k}{(p+\gamma )\left|1+(1-e^{-i\beta }{\alpha }_s)\rho e^{i\theta }\right|}$$

$$\ge \frac{{\left|e^{i\beta }-{\alpha }_s\right|}^{2}n{\rho }^2}{(p+\gamma )(1+\left|e^{i\beta }-{\alpha }_s\right|\rho )}.$$

(55)

This contradicts our condition (51) for $f\left(z\right)$. Therefore, there is no $z_0,(0<|z_0|<1)$ such that $\left|w\right(z_0\left)\right|=\rho >1.$ This means that

$$\left|\left(\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right)\right|<\rho |e^{i\beta }-{\alpha }_s|,z\in \mathbb{U}.$$

(56)

 □

Example 2.

Consider a function $f\left(z\right)$ given by (32) with $0<|a_{p+n}|<\frac{1}{Q}$, where Q is given by (33). For this function $f\left(z\right),$ we have

$$\left|\left(\frac{B_{-m-\lambda +1}f\left(z\right)}{B_{-m-\lambda }f\left(z\right)}-1\right)\left(\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right)\right|=\left|\frac{n{Q}^2{a}_{p+n}^2{z}^{2n}}{(p+\gamma )(1+Q{a}_{p+n}{z}^n)}\right|$$

$$<\frac{n{Q}^2{\left|a_{p+n}\right|}^2}{(p+\gamma )(1-Q|a_{p+n}\left|\right)},z\in \mathbb{U}.$$

(57)

Consider five boundary points $z_1,z_2,z_3,z_4$ and $z_5$ in Example 1. Then, we have

$$\left|1-e^{-i\beta }{\alpha }_5\right|=\frac{\sqrt{2}(3+\sqrt{2}+\sqrt{3})}{10}Q\left|a_{p+n}\right|$$

(58)

with $\beta =0.$ With such ${\alpha }_5$ and $\beta ,$ we take $\rho >1$ by

$$\frac{n{Q}^2{\left|a_{p+n}\right|}^2}{(p+\gamma )(1-Q|a_{p+n}\left|\right)}\le \frac{{\left|e^{i\beta }-{\alpha }_5\right|}^{2}n{\rho }^2}{(p+\gamma )(1+\left|e^{i\beta }-{\alpha }_5\right|\rho )}.$$

(59)

Then, this ρ satisfies

$$\rho \ge \frac{10}{\sqrt{2}(3+\sqrt{2}+\sqrt{3})Q\left|a_{p+n}\right|}>1.$$

(60)

For such ${\alpha }_5$ and ρ, we know that

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<\rho |e^{i\beta }-{\alpha }_5|,z\in \mathbb{U}.$$

(61)

Next, we derive the following result.

Theorem 3.

If $f\left(z\right)\in A(p,n)$ satisfies

$$\left|\frac{B_{-m-\lambda +g}f\left(z\right)}{z^p}-1\right|<\rho |e^{i\beta }-{\alpha }_s|\left(\frac{p+n+\gamma }{p+\gamma }\right),z\in \mathbb{U}$$

(62)

for some ${\alpha }_s$ defined by (13) with ${\alpha }_s\ne 1,g=1,2,3,\dots m,$ and for some real $\rho >1,$ then

$$\left|\frac{B_{-m-\lambda +g-1}f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|,z\in \mathbb{U}.$$

(63)

Proof. 

Define the function $w\left(z\right)$ by

$$w\left(z\right)=\frac{e^{i\beta }\frac{B_{-m-\lambda +g-1}f\left(z\right)}{z^p}-{\alpha }_s}{e^{i\beta }-{\alpha }_s}-1$$

$$=\frac{e^{i\beta }}{e^{i\beta }-{\alpha }_s}\left\{\sum _{k=p+n}^{\infty }{\left(\frac{p+\gamma }{k+\gamma }\right)}^{m-g}\frac{\mathsf{\Gamma }(k+\gamma )\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )\mathsf{\Gamma }(k+\gamma +\lambda )}{a}_k{z}^{k-p}\right\}.$$

(64)

It follows from the above that

$$B_{-m-\lambda +g-1}f\left(z\right)=z^p+(1-e^{-i\beta }{\alpha }_s)z^{p}w\left(z\right).$$

(65)

By the definition of $B_{-m-\lambda +g}f\left(z\right),$ we know that

$$B_{-m-\lambda +g}f\left(z\right)=\frac{z^{1-\gamma }}{p+\gamma }{\left(z^{\gamma }{B}_{-m-\lambda +g-1}f\left(z\right)\right)}^{\prime }=z^p\left\{1+(1-e^{-i\beta }{\alpha }_s)w\left(z\right)\left(1+\frac{z{w}^{\prime }\left(z\right)}{(p+\gamma )w\left(z\right)}\right)\right\}.$$

(66)

Our condition implies that

$$\left|\frac{B_{-m-\lambda +g}f\left(z\right)}{z^p}-1\right|=\left|(1-e^{-i\beta }{\alpha }_s)w\left(z\right)\left(1+\frac{z{w}^{\prime }\left(z\right)}{(p+\gamma )w\left(z\right)}\right)\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|\left(\frac{p+n+\gamma }{p+\gamma }\right)$$

(67)

for all $z\in \mathbb{U}.$ Suppose that there exists a point $z_0,(0<|z_0|<1)$ such that

$$\max \left\{\right|w\left(z\right)|;|z|\le |z_0\left|\right\}=|w\left(z_0\right)|=\rho >1.$$

(68)

Then, Lemma 1 says that $w\left(z_0\right)=\rho e^{i\theta },(0\le \theta \le 2\pi )$ and $z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right),(k\ge n)$. Therefore, we have

$$\left|\frac{B_{-m-\lambda +g}f\left(z_0\right)}{z_0^p}-1\right|=\left|(1-e^{-i\beta }{\alpha }_s)w\left(z_0\right)\left(1+\frac{z_0{w}^{\prime }\left(z_0\right)}{(p+\gamma )w\left(z_0\right)}\right)\right|$$

$$=\rho \left|e^{i\beta }-{\alpha }_s\right|\left(1+\frac{k}{p+\gamma }\right)\ge \rho \left|e^{i\beta }-{\alpha }_s\right|\left(\frac{p+n+\gamma }{p+\gamma }\right)$$

(69)

which contradicts the inequality (67). This means that there is no $z_0\in \mathbb{U}$ such that $\left|w\right(z_0|=\rho >1.$ Thus we know that

$$\left|w\left(z\right)\right|=\left|\frac{e^{i\beta }\frac{B_{-m-\lambda +g-1}f\left(z\right)}{z^p}-{\alpha }_s}{e^{i\beta }-{\alpha }_s}-1\right|=\left|\frac{e^{i\beta }}{e^{i\beta }-{\alpha }_s}\left(\frac{B_{-m-\lambda +g-1}f\left(z\right)}{z^p}-1\right)\right|<\rho .$$

(70)

This completes the proof of the theorem.  □

Theorem 3 implies the following one.

Theorem 4.

If $f\left(z\right)\in A(p,n)$ satisfies

$$\left|\frac{B_{-m-\lambda +g}f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|{\left(\frac{p+n+\gamma }{p+\gamma }\right)}^g,z\in \mathbb{U}$$

(71)

for some ${\alpha }_s$ given by (13) with ${\alpha }_s\ne 1,g=1,2,3,\dots ,m,$ and for some real $\rho >1,$ then

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|,z\in \mathbb{U}$$

(72)

or, equivalently, $f\left(z\right)\in T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right).$

Proof. 

By means of Theorem 3, we see that if $f\left(z\right)$ satisfies the inequality (71), then

$$\left|\frac{B_{-m-\lambda +g-1}f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|{\left(\frac{p+n+\gamma }{p+\gamma }\right)}^{g-1},z\in \mathbb{U}$$

(73)

Similarly, we have

$$\left|\frac{B_{-m-\lambda +g-2}f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|{\left(\frac{p+n+\gamma }{p+\gamma }\right)}^{g-2},z\in \mathbb{U}.$$

(74)

Continuing this consideration, we obtain that

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<\rho \left|e^{i\beta }-{\alpha }_s\right|,z\in \mathbb{U}.$$

(75)

 □

Example 3.

Consider the function

$$f\left(z\right)=z^p+a_{p+n}{z}^{p+n},z\in \mathbb{U}$$

(76)

which satisfies

$$B_{-m-\lambda +g}f\left(z\right)=z^p+\frac{\mathsf{\Gamma }(p+n+\gamma )\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )\mathsf{\Gamma }(p+n+\gamma +\lambda )}{\left(\frac{p+\gamma }{p+n+\gamma }\right)}^{m-g}{a}_{p+n}{z}^{p+n}.$$

(77)

It follows from (77) that

$$\left|\frac{B_{-m-\lambda +g}f\left(z\right)}{z^p}-1\right|=\frac{\mathsf{\Gamma }(p+n+\gamma )\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )\mathsf{\Gamma }(p+n+\gamma +\lambda )}{\left(\frac{p+\gamma }{p+n+\gamma }\right)}^{m-g}|a_{p+n}{\left|\right|z|}^n$$

$$<Q{\left(\frac{p+n+\gamma }{p+\gamma }\right)}^g\left|a_{p+n}\right|,z\in \mathbb{U},$$

(78)

where Q is given by (33). Now, we consider the five boundary points $z_1,z_2,z_3,z_4$ and $z_5$ as in Example 1. Then we see

$$\left|e^{i\beta }-{\alpha }_5\right|=\frac{\sqrt{2}(3+\sqrt{2}+\sqrt{3})}{10}Q\left|a_{p+n}\right|$$

(79)

where $\beta =0.$ With the above relation (79), we consider $\rho >1$ such that

$$Q{\left(\frac{p+n+\gamma }{p+\gamma }\right)}^g|a_{p+n}|\le \rho |e^{i\beta }-{\alpha }_5|,$$

(80)

that is, ρ satisfies

$$\rho \ge \frac{10}{\sqrt{2}(3+\sqrt{2}+\sqrt{3})}{\left(\frac{p+n+\gamma }{p+\gamma }\right)}^g>1.$$

Thus, we have that

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|\le Q\left|a_{p+n}\right|$$

(81)

$$\le \rho |e^{i\beta }-{\alpha }_5|{\left(\frac{p+\gamma }{p+n+\gamma }\right)}^g$$

$$<\rho |e^{i\beta }-{\alpha }_5|,z\in \mathbb{U}.$$

(82)

Remark 1.

If we take $\gamma =1$ in the results of this section, then these results correspond to applications of the Libera integral operator as introduced by Libera [2].

Let us write that

$$B_{-m-\lambda }f\left(z\right)=L_{-m-\lambda }f\left(z\right)=z^p+\sum _{k=p+n}^{\infty }{\left(\frac{p+1}{k+1}\right)}^m\frac{\mathsf{\Gamma }(p+1+\lambda )k!}{\mathsf{\Gamma }(k+1+\lambda )p!}{a}_k{z}^k$$

(83)

for $\gamma =1$ in (11). Then Theorem 1 says that if $f\left(z\right)\in A(p,n)$ satisfies

$$\left|\frac{L_{-m-\lambda +1}f\left(z\right)}{L_{-m-\lambda }f\left(z\right)}-1\right|<\frac{|e^{i\beta }-{\alpha }_s|n\rho }{(p+1)(1+|e^{i\beta }-{\alpha }_s\left|\rho \right)},z\in \mathbb{U},$$

(84)

for some ${\alpha }_s$ given by

$${\alpha }_s=\frac{1}{s}\sum _{l=1}^s\frac{L_{-m-\lambda }f\left(z_l\right)}{z_l^p},z_l\in \overline{\mathbb{U}},$$

(85)

where $-\frac{\pi }{2}\le \beta \le \frac{\pi }{2},$ and for some real $\rho >1,$ then

$$\left|\frac{L_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<|e^{i\beta }-{\alpha }_s|\rho ,z\in \mathbb{U}.$$

(86)

For another result, we consider again the Libera integral operator with $\gamma =1.$

3. Application of Carathéodory Lemma

In this section, we will apply Carathéodory Lemma for coefficients of functions $f\left(z\right)\in A(p,n).$

In 1907, Carathéodory [8] gave the following result.

Lemma 2.

Let a function $g\left(z\right)$ given by

$$g\left(z\right)=1+\sum _{k=1}^{\infty }{c}_k{z}^k$$

(87)

be analytic in $\mathbb{U}$ and $Reg\left(z\right)>0,z\in \mathbb{U}.$ Then $g\left(z\right)$ satisfies

$$|c_k|\le 2,(k=1,2,3,\dots ).$$

(88)

The inequality (88) is sharp for each $k.$

Applying the above lemma, we derive the following thorem.

Theorem 5.

If $f\left(z\right)\in A(p,n)$ is in the class $T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right),$ then

$$\left|a_k\right|\le \frac{2|e^{i\beta }-{\alpha }_s|\rho }{R},(k=p+n,p+n+1,\dots ),$$

(89)

where

$$R={\left(\frac{p+\gamma }{k+\gamma }\right)}^m\frac{\mathsf{\Gamma }(k+\gamma )\mathsf{\Gamma }(p+\gamma +\lambda )}{\mathsf{\Gamma }(p+\gamma )\mathsf{\Gamma }(k+\gamma +\lambda )}.$$

(90)

The result is sharp for $f\left(z\right)$ given by

$$B_{-m-\lambda }f\left(z\right)=z^p\frac{e^{i\theta }-(1+2|e^{i\beta }-{\alpha }_s\left|\rho \right)z}{e^{i\theta }-z}.$$

(91)

Proof. 

For $f\left(z\right)\in T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right),$ we see that

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<|e^{i\beta }-{\alpha }_s|\rho ,z\in \mathbb{U}.$$

(92)

If we define a function $g\left(z\right)$ with $f\left(z\right)\in T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right)$ by

$$g\left(z\right)=\frac{\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-(1-|e^{i\beta }-{\alpha }_s\left|\rho \right)}{|e^{i\beta }-{\alpha }_s|\rho },z\in \mathbb{U},$$

(93)

then $g\left(z\right)$ is analytic in $\mathbb{U}$ with $g\left(0\right)=1$ and $Reg\left(z\right)>0,z\in \mathbb{U}.$ Also, $g\left(z\right)$ has the following power series expansion:

$$g\left(z\right)=1+\sum _{k=p+n}^{\infty }\frac{R}{|e^{i\beta }-{\alpha }_s|\rho }{a}_k{z}^{k-p}.$$

Therefore, by applying Lemma 2 to $g\left(z\right),$ we obtain

$$\frac{R}{|e^{i\beta }-{\alpha }_s|\rho }\left|a_k\right|\le 2,(k=p+n,p+n+1,\dots ).$$

(94)

This shows the coefficient inequalities (89). Note that

$$g\left(z\right)=\frac{e^{i\theta }+z}{e^{i\theta }-z}=1+\sum _{k=1}^{\infty }2e^{i\theta }{z}^k=1+\sum _{k=1}^{\infty }{c}_k{z}^k$$

(95)

is analytic in $\mathbb{U},$ $g\left(0\right)=1,$ $Reg\left(z\right)>0,(z\in \mathbb{U})$ and $|c_k|=2,(k=1,2,3,\dots ).$ Therefore, considering $f\left(z\right)$ such that

$$g\left(z\right)=\frac{\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-(1-|e^{i\beta }-{\alpha }_s\left|\rho \right)}{|e^{i\beta }-{\alpha }_s|\rho }=\frac{e^{i\theta }+z}{e^{i\theta }-z},$$

(96)

we have

$$B_{-m-\lambda }f\left(z\right)=z^p\frac{e^{i\theta }-(1+2|e^{i\beta }-{\alpha }_s\left|\rho \right)z}{e^{i\theta }-z}.$$

(97)

This completes the proof of the theorem.  □

Remark 2.

If we take $\gamma =1$ in Theorem 5, then we get the following result for the Libera integral operator.

If $f\left(z\right)\in A(p,n)$ satisfies

$$\left|\frac{L_{-m-\lambda }f\left(z\right)}{z^p}-1\right|<|e^{i\beta }-{\alpha }_s|\rho ,z\in \mathbb{U},$$

(98)

then

$$\left|a_k\right|\le \frac{2|e^{i\beta }-{\alpha }_s|\rho }{R_0},(k=p+n,p+n+1,\dots ),$$

(99)

where

$$R_0={\left(\frac{p+1}{k+1}\right)}^m\frac{\mathsf{\Gamma }(p+1+\lambda )k!}{\mathsf{\Gamma }(k+1+\lambda )p!}.$$

(100)

The result is sharp for $f\left(z\right)$ given by

$$L_{-m-\lambda }f\left(z\right)=z^p\frac{e^{i\theta }-(1+2|e^{i\beta }-{\alpha }_s\left|\rho \right)z}{e^{i\theta }-z}.$$

(101)

Finally, we derive

Theorem 6.

If $f\left(z\right)\in A(p,n)$ satisfies

$$\sum _{k=p+n}^{\infty }R|a_k|\le |e^{i\beta }-{\alpha }_s|\rho ,$$

(102)

then $f\left(z\right)\in T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right),$ where R is given by (90).

Proof. 

For $f\left(z\right)\in A(p,n),$ we consider

$$\left|\frac{B_{-m-\lambda }f\left(z\right)}{z^p}-1\right|=\left|\sum _{k=p+n}^{\infty }R{a}_k{z}^k\right|$$

$$<\sum _{k=p+n}^{\infty }R|a_k|\le |e^{i\beta }-{\alpha }_s|\rho ,z\in \mathbb{U}.$$

(103)

Therefore, if $f\left(z\right)\in A(p,n)$ satisfies (102), then we know $f\left(z\right)\in T_{p,n}\left({\alpha }_s,\beta ,\rho ;m,\lambda \right).$ □

Remark 3.

Letting $\gamma =1$ in Theorem 6, we have the result concerning with the Libera integral operator $L_{-m-\lambda }f\left(z\right).$