# Stable Functions of Janowski Type

Koneri Chandrasekran  Koneri Chandrasekran
Department of Mathematics
Jeppiaar SRR Engineering College, Affiliated to Anna University
Chennai 603 103, India<EMAIL_ADDRESS>,  Devasir John Prabhakaran
Devasir John Prabhakaran
Department of Mathematics
MIT Campus, Anna University
Chennai 600 044, India<EMAIL_ADDRESS>and  Priyanka Sangal  Priyanka
Sangal
Department of Mathematics
R.C.U. Govt. P.G. College Uttarkashi
Uttarakhand 249193, India<EMAIL_ADDRESS>
###### Abstract.

A function $f\in\mathcal{A}_{1}$ is said to be stable with respect to
$g\in\mathcal{A}_{1}$ if

$\displaystyle\frac{s_{n}(f(z))}{f(z)}\prec\frac{1}{g(z)},\qquad
z\in\mathbb{D},$

holds for all $n\in\mathbb{N}$ where $\mathcal{A}_{1}$ denote the class of
analytic functions $f$ in the unit disk
$\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$ normalized by $f(0)=1$. Here
$s_{n}(f(z))$, the $n^{th}$ partial sum of
$f(z)=\displaystyle\sum_{k=0}^{\infty}a_{k}z^{k}$ is given by
$s_{n}(f(z))=\displaystyle\sum_{k=0}^{n}a_{k}z^{k},\
n\in\mathbb{N}\cup\\{0\\}$. In this work, we consider the following function

$\displaystyle v_{\lambda}(A,B,z)=\left(\frac{1+Az}{1+Bz}\right)^{\lambda}$

for $-1\leq B<A\leq 1$ and $0\leq\lambda\leq 1$ for our investigation. The
main purpose of this paper is to prove that $v_{\lambda}(A,B,z)$ is stable
with respect to $\displaystyle v_{\lambda}(0,B,z)=\frac{1}{(1+Bz)^{\lambda}}$
for $0<\lambda\leq 1$ and $-1\leq B<A\leq 0$. Further, we prove that
$v_{\lambda}(A,B,z)$ is not stable with respect to itself, when $0<\lambda\leq
1$ and $-1\leq B<A<0$.

###### Key words and phrases:

Analytic functions, Starlike Functions, Subordination and Stable Functions

###### 1991 Mathematics Subject Classification:

Primary: 30C45

## Introduction & Main Results

Let $\mathcal{A}$ denote the family of functions $f$ that are analytic in the
unit disk $\mathbb{D}:=\\{z:\,|z|<1\\}$. Let $\mathcal{A}_{1}$ is the subset
of $\mathcal{A}$ with the normalization $f(0)=1$. A single valued function
$f\in\mathcal{A}_{1}$ is said to be univalent in a domain
$\Delta\subseteq\mathbb{C}$ if $f$ is one-to-one in $\Delta$. The class of all
univalent functions with the normalization $f(0)=0=f^{\prime}(0)-1$ is denoted
by $\mathcal{S}$. Let $\Omega$ be the family of functions $\omega$, regular in
$\mathbb{D}$ and satisfying the conditions $\omega(0)=0$ and $|\omega(z)|<1$
for all $z\in\mathbb{D}$. For $f,g\in\mathcal{A}$, the function $f$ is said to
be subordinate to $g$, denoted by $f\prec g$ if and only if there exists an
analytic function $\omega\in\Omega$ such that $f=g\circ\omega$. In particular,
if $g$ is univalent in $\mathbb{D}$ then $f(0)=g(0)$ and
$f(\mathbb{D})\subseteq g(\mathbb{D})$ hold.

The function $zf(z)\in\mathcal{A}_{1}$ is starlike of order $\lambda$ if ${\rm
Re}\left(\dfrac{zf^{\prime}(z)}{f(z)}\right)>\lambda$ for all $z\in\mathbb{D}$
and $0\leq\lambda<1$. The class of all starlike functions, denoted by
$\mathcal{S}^{\ast}(\lambda)$ is a subclass of $\mathcal{S}$. The $n^{th}$
partial sum $s_{n}(f(z))$ of $f(z)=\displaystyle\sum_{k=0}^{\infty}a_{k}z^{k}$
is given by
$s_{n}(f(z))=\displaystyle\sum_{k=0}^{n}a_{k}z^{k},n=0,1,2,\ldots$. For more
details about the univalent functions, its subclasses and subordination
properties, we refer [2, 3, 5].

The concept of stable functions was first introduced by Ruscheweyh and Salinas
[6], while discussing the class of starlike functions of order $\lambda$,
where $1/2\leq\lambda<1$. However, the class of starlike functions of order
$\lambda\in[1/2,1)$ is comparatively a much narrow class but it has many
interesting properties too. Ruscheweyh and Salinas [6] proved the following
result.

######  Theorem 1.

[6] Let $\lambda\in[1/2,1)$ and $zf\in\mathcal{S}^{\ast}(\lambda)$, then

$\displaystyle\frac{s_{n}(f,z)}{f(z)}\prec(1-z)^{\lambda},\qquad
n\in\mathbb{N},z\in\mathbb{D}.$

Theorem 1 has several applications in Gegenbauer polynomial sums and motivated
by Theorem 1, Ruscheweyh and Salinas [6] introduced the concept of Stable
functions which is stated as follows. For some $n\in\mathbb{N}$, a function
$F$ is said to be $n$-stable function with respect to $G$ if

$\displaystyle\frac{s_{n}(F(z))}{F(z)}\prec\frac{1}{G(z)},\quad\mbox{for
$F,G\in\mathcal{A}_{1}$ and $z\in\mathbb{D}$}.$

Moreover, the function $F$ is said to be stable with respect to $G$, if $F$ is
$n$-stable with respect to $G$ for every $n\in\mathbb{N}$. Particularly, if
the function $F$ is $n$-stable with respect to itself. Then for every
$n\in\mathbb{N}$, $F$ is stable. In the present context, for $-1\leq B<A\leq
1$, we define a function

$\displaystyle
v_{\lambda}(A,B,z):=\left(\frac{1+Az}{1+Bz}\right)^{\lambda}\quad\mbox{ for
$z\in\mathbb{D}$ and $\lambda\in(0,1]$}.$

For $\lambda=1/2$, Ruscheweyh and Salinas [7] proved that $v_{1/2}(1,-1,z)$ is
stable function with respect to itself. The stability of $v_{1/2}(1,-1,z)$ is
equivalent to the simultaneous non-negativity of general class of sine and
cosine sums given by Vietoris [11], the most celebrated theorem of positivity
of trigonometric sums. Ruscheweyh and Salinas [7] conjectured that
$v_{\lambda}(1,-1,z)$ is stable for $0<\lambda<1/2$. Using computer algebra,
for $\lambda=1/4$ it was shown in [7] that $v_{1/4}(1,-1,z)$ is $n$-stable for
$n=1,2,3,\cdots,5000$. In the limiting case, the validation of stability of
$v_{\lambda}(1,-1,z)$ for $0<\lambda<1/2$ interpreted in terms of positivity
of trigonometric polynomials.

Further extensions of Vietoris Theorem and stable functions to Cesàro stable
functions and Generalized Cesàro stable functions have been studied in [4] and
[9] respectively. In this direction, conjectures are also proposed in [9] that
linked Generalized Cesàro stable functions with the positivity of
trigonometric sums. Chakraborty and Vasudevarao [1] considered $A=1-2\alpha$,
$B=-1$ and proved the following result.

######  Theorem 2.

[1] For $0<\lambda\leq 1$ and $1/2\leq\alpha<1$,
$v_{\lambda}(1-2\alpha,-1,z)=\left(\dfrac{1+(1-2\alpha)z}{1-z}\right)^{\lambda}$
is stable with respect to $v_{\lambda}(0,-1,z)=\dfrac{1}{(1-z)^{\lambda}}$.

Chakraborty and Vasudevarao [1] also proved that $v_{\lambda}(1-2\alpha,-1,z)$
is not stable with repsect to itself when $1/2<\alpha<1$ and $0<\lambda\leq
1$. For $\lambda=1$, the function $v_{1}(A,B,z)=\dfrac{1+Az}{1+Bz}$ have been
studied widely by many researchers. The analytic functions of
$\mathcal{A}_{1}$ subordinate to $\dfrac{1+Az}{1+Bz}$ have been studied by
Janowski [3] and the class of such functions is denoted by $\mathcal{P}(A,B)$.
The functions of $\mathcal{P}(A,B)$ are called Janowski functions. Moreover,
the set of functions $zf\in\mathcal{A}_{1}$, for which
$\dfrac{zf^{\prime}(z)}{f(z)}\prec\dfrac{1+Az}{1+Bz}$ holds, called Janowski
starlike functions and the class of such functions is denoted by
$\mathcal{S}^{\ast}(A,B)$. It can be easily seen that
$\mathcal{S}^{\ast}(1,-1)\equiv\mathcal{S}^{\ast}$.

In this paper, we show that $v_{\lambda}(A,B,z)$ is stable with respect to
$v_{\lambda}(0,B,z)=1/(1+Bz)^{\lambda}$ for $0<\lambda\leq 1$ and $-1\leq
B<A\leq 0$. Further, $v_{\lambda}(A,B,z)$ is not stable with respect to
itself, when $0<\lambda\leq 1$ and $-1\leq B<A<0$. We can write
$v_{\lambda}(A,B,z)$ as,

$\displaystyle v_{\lambda}(A,B,z)$
$\displaystyle=\left(\frac{1+Az}{1+Bz}\right)^{\lambda}$
$\displaystyle=\left({1+Az}\right)^{\lambda}\left({1+Bz}\right)^{-\lambda}$
$\displaystyle=\left(1+\sum_{k=1}^{\infty}\frac{[\lambda]_{k}}{k!}A^{k}z^{k}\right)\left(1+\sum_{k=1}^{\infty}\frac{(\lambda)_{k}}{k!}(-B)^{k}z^{k}\right)$
$\displaystyle=1+\sum_{n=1}^{\infty}\left(\sum_{k=0}^{n}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k}}{(n-k)!}A^{k}(-B)^{n-k}\right)z^{n},$
(1)

where $[\lambda]_{k}$ and $(\lambda)_{k}$ denote the factorial polynomials
given as

$\displaystyle\left\\{\begin{array}[]{ll}[\lambda]_{k}=&\lambda(\lambda-1)(\lambda-2)(\lambda-3)\cdots(\lambda-k+1),\quad\mbox{
and }\\\
(\lambda)_{k}=&\frac{\Gamma(\lambda+k)}{\Gamma(\lambda)}=\lambda(\lambda+1)\cdots(\lambda+k-1),\quad\mbox{
Where $\Gamma$ is well-known gamma function, }\end{array}\right.$

for $k=1,2,\cdots$ respectively with $[\lambda]_{0}=1=(\lambda)_{0}$. So
$v_{\lambda}(A,B,z)$ can be written as

$\displaystyle
v_{\lambda}(A,B,z)=1+\sum_{n=1}^{\infty}a_{n}(A,B,\lambda)z^{n},$

where

$\displaystyle
a_{n}:=a_{n}(A,B,\lambda)=\sum_{k=0}^{n}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k}}{(n-k)!}A^{k}(-B)^{n-k}.$

Now, we state two lemmas which will helpful to prove our main results.

######  Lemma 1.

For $0<\lambda\leq 1$ and $-1\leq B<A\leq 0$, we have

$\displaystyle\sum_{k=0}^{n}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k}}{(n-k)!}A^{k}(-B)^{n-k}>0.$

######  Lemma 2.

Let $v_{\lambda}(A,B,z)$ be defined by (1). Then for $\lambda\in(0,1]$ and
$-1\leq B<A\leq 0$,

$\displaystyle(m+1)(n+1)\left(\sum_{k=0}^{n+1}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n+1-k}}{(n+1-k)!}A^{k}B^{n+1-k}\right)$
$\displaystyle-
mn\left(\sum_{k=0}^{n}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k}}{(n-k)!}A^{k}B^{n-k}\right)\geq
0$ (2)

holds for all $m,\ n\in\mathbb{N}$.

Now, we state main results of this paper which are about the stability of
$v_{\lambda}(A,B,z)$ with repsect to $v_{\lambda}(0,B,z)$ and
$v_{\lambda}(A,B,z)$ itself.

######  Theorem 3.

For $\lambda\in(0,1]$ and $-1\leq B<A\leq 0$, $v_{\lambda}(A,B,z)$ given in
(1) is stable with respect to
$v_{\lambda}(0,B,z)=\dfrac{1}{(1+Bz)^{\lambda}}$.

If we substitute $A=0$ in Theorem 3, we get the following corollary which is
also a generalization of the result given by Ruscheweyh and Salinas [6].

######  Corollary 1.

For $\lambda\in(0,1]$ and $-1\leq B<0$,
$v_{\lambda}(0,B,z)=\dfrac{1}{(1+Bz)^{\lambda}}$ is stable function.

Now for $0<\mu\leq\lambda\leq 1$, we have the following corollary of Theorem
3.

######  Corollary 2.

For $0<\mu\leq\lambda\leq 1$ and for $-1\leq B<0$ we have

$\displaystyle\frac{s_{n}(v_{\mu}(0,B,z))}{v_{\lambda}(0,B,z)}\prec\frac{1}{v_{\lambda}(0,B,z)},\qquad\mbox{
for $z\in\mathbb{D}$}.$

Theorem 3 also generalizes result of Chakraborty and Vasudevarao [1] as if we
substitute $A=1-2\alpha$ and $B=-1$ in Theorem 3, reduces to Theorem 2. In
other words, Theorem 2 is a particular case of Theorem 3.

######  Theorem 4.

For $\lambda\in(0,1]$ and $-1\leq B<A<0$,
$v_{\lambda}(A,B,z)=\left(\dfrac{1+Az}{1+Bz}\right)^{\lambda}$ is not stable
with respect to itself.

## Proof of Main Results

###### Proof of Lemma 1.

Consider,

$\displaystyle 1$ $\displaystyle=(1-z)^{\lambda}(1-z)^{-\lambda}$
$\displaystyle=1+\sum_{n=1}^{\infty}\left(\sum_{k=0}^{n}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k}}{(n-k)!}(-1)^{k}\right)z^{n}$

Comparing the coefficients of $z^{n}$ on both the sides we have

$\displaystyle\sum_{k=0}^{n}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k}}{(n-k)!}(-1)^{k}=0,$

which can be expanded as

$\displaystyle\frac{(\lambda)(\lambda+1)\cdots(\lambda+n-1)}{(n)!}+\frac{(\lambda)(\lambda+1)\cdots(\lambda+n-2)}{(n-1)!}\left(\frac{\lambda}{1!}\right)(-1)$
$\displaystyle\qquad+\frac{(\lambda)(\lambda+1)\cdots(\lambda+n-3)}{(n-2)!}\frac{\lambda(\lambda-1)}{2!}(-1)^{2}+\cdots+\frac{\lambda}{1!}\frac{(\lambda)(\lambda-1)\cdots(\lambda-n+2)}{(n-1)!}(-1)^{n-1}$
$\displaystyle\qquad\qquad+\frac{(\lambda)(\lambda-1)\cdots(\lambda-n+1)}{n!}(-1)^{n}=0.$

Since $0\leq\lambda<1$, so only first term in the above equation is positive.
By multiplying $2^{nd},3^{rd},\cdots$, $(n+1)^{th}$ terms by
$\dfrac{\alpha}{\beta}$, $\dfrac{\alpha^{2}}{\beta^{2}}$, $\cdots$,
$\dfrac{\alpha^{n}}{\beta^{n}}$ respectively, we obtain for
$0\leq\alpha<\beta$,

$\displaystyle\frac{(\lambda)(\lambda+1)\cdots(\lambda+n-1)}{(n)!}+\frac{(\lambda)(\lambda+1)\cdots(\lambda+n-2)}{(n-1)!}\left(\frac{\lambda}{1!}\right)(-1)\frac{\alpha}{\beta}$
$\displaystyle\quad+\frac{(\lambda)(\lambda+1)\cdots(\lambda+n-3)}{(n-2)!}\frac{\lambda(\lambda-1)}{2!}\left(\frac{-\alpha}{\beta}\right)^{2}+\cdots+\frac{\lambda}{1!}\frac{(\lambda)(\lambda-1)\cdots(\lambda-n+2)}{(n-1)!}\left(\frac{-\alpha}{\beta}\right)^{n-1}$
$\displaystyle\qquad\qquad+\frac{(\lambda)(\lambda-1)\cdots(\lambda-n+1)}{n!}\left(\frac{-\alpha}{\beta}\right)^{n}\geq
0.$

After multiplying by $\beta^{n}$ we obtain

$\displaystyle\beta^{n}\sum_{k=0}^{n}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k}}{(n-k)!}(-1)^{k}\left(\frac{\alpha}{\beta}\right)^{k}\geq
0$ (3)

By substituting $\alpha=-A$, $\beta=-B$ in (3) so that for $-1\leq B<A\leq 0$,
the lemma is proved. ∎

###### Proof of Lemma 2.

Let $v_{\lambda}(A,B,z)$ be defined by (1). Then,

$\displaystyle v_{\lambda}(A,B,z)$
$\displaystyle=\left(\frac{1+Az}{1+Bz}\right)^{\lambda}=1+a_{1}z+a_{2}z^{2}+a_{3}z^{3}+\cdots$
$\displaystyle v_{\lambda}^{\prime}(A,B,z)$
$\displaystyle=\lambda\left(\frac{1+Az}{1+Bz}\right)^{\lambda-1}\left(\frac{(1+Bz)A-(1+Az)B}{(1+Bz)^{2}}\right)$
$\displaystyle=\lambda(A-B)\frac{(1+Az)^{\lambda-1}}{(1+Bz)^{\lambda+1}}$
$\displaystyle(1+Bz)v_{\lambda}^{\prime}(A,B,z)$
$\displaystyle=\lambda(A-B)(1+Az)^{\lambda-1}(1+Bz)^{-\lambda}$ (4)

Since $0>A>B$, $0<\lambda\leq 1$,
$(1+Az)^{\lambda-1}=1+(\lambda-1)Az+\frac{(\lambda-1)(\lambda-2)}{2!}A^{2}z^{2}+\cdots$
and $(1+Bz)^{-\lambda}=1-\lambda
Bz+\frac{\lambda(\lambda+1)}{2!}B^{2}z^{2}+\cdots$ have positive Taylor series
coefficients. A simple computation yields that

$\displaystyle(1+Bz)v_{\lambda}^{\prime}(A,B,z)$
$\displaystyle=(a_{1}+2a_{2}z+3a_{3}z^{2}+\cdots)(1+Bz)$
$\displaystyle=a_{1}+\sum_{n=1}^{\infty}((n+1)a_{n+1}+Bna_{n})z^{n}.$ (5)

Since right hand side of (4) has positive Taylor coefficients, from (4) and
(5) we conclude that

$\displaystyle(n+1)a_{n+1}+Bna_{n}>0,\quad n\in\mathbb{N}.$ (6)

The left hand side of the expression given in (2) can be rewritten as

$\displaystyle(m+1)(n+1)a_{n+1}+mnBa_{n}.$ (7)

Equivalently, (7) can be written as

$\displaystyle m((n+1)a_{n+1}+Bna_{n})+(n+1)a_{n+1}.$

Using (6) and the fact that $a_{n}\geq 0$ for $m,n\in\mathbb{N}$, the lemma is
proved for $\lambda\in(0,1]$ and $-1\leq B<A\leq 0$. ∎

Before going to proceed further for the proof of Theorem 3, it is easy to
verify the following relations.

$\begin{split}s^{\prime}_{n}(v_{\lambda}(A,B,z),z)&=s_{n-1}(v^{\prime}_{\lambda}(A,B,z),z),\\\
zs^{\prime}_{n}(v_{\lambda}(A,B,z),z)&=s_{n}(zv^{\prime}_{\lambda}(A,B,z),z),\\\
z^{2}s^{\prime}_{n}(v_{\lambda}(A,B,z),z)&=s_{n}(z^{2}v^{\prime}_{\lambda}(A,B,z),z).\end{split}$
(8)

Now, we are ready to give the proof of Theorem 3.

###### Proof of Theorem 3.

To show that $v_{\lambda}(A,B,z)$ is stable with respect to
$v_{\lambda}(0,B,z)$, it is enough to show that

$\displaystyle\frac{s_{n}(v_{\lambda}(A,B,z),z)}{v_{\lambda}(A,B,z),z}\prec\frac{1}{v_{\lambda}(0,B,z)},\qquad
z\in\mathbb{D}$

for all $n\in\mathbb{N}$, i.e., to prove that

$\displaystyle\frac{(1+Bz)^{\lambda}s_{n}(v_{\lambda}(A,B,z),z)}{(1+Az)^{\lambda}}\prec(1+Bz)^{\lambda},\qquad
z\in\mathbb{D},$

which can be equivalently written as

$\displaystyle\frac{(1+Bz)s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}}}{(1+Az)}\prec(1+Bz).$

To show that , it is enough to prove that

$\displaystyle\left|\displaystyle\frac{(1+Bz)s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}}}{(1+Az)}-1\right|\leq|B|\leq
1,\quad z\in\mathbb{D}.$

For fixed $n$ and $\lambda$, we consider the following function

$\displaystyle
h(z)=1-\displaystyle\frac{(1+Bz)s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}}}{(1+Az)},\quad
z\in\mathbb{D}.$

It is easy to see that

$\displaystyle
v^{\prime}_{\lambda}(A,B,z)=\lambda(A-B)\frac{(1+Az)^{\lambda-1}}{(1+Bz)^{\lambda+1}}=\lambda(A-B)\frac{v_{\lambda}(A,B,z)}{(1+Bz)(1+Az)},$

which can be rewritten in the following form

$\displaystyle
v_{\lambda}(A,B,z)-\frac{(1+(A+B)z+ABz^{2})}{\lambda(A-B)}v^{\prime}_{\lambda}(A,B,z)=0\quad\mbox{for
$\ z\in\mathbb{D}$}.$ (9)

A simple calculations gives that

$\displaystyle h^{\prime}(z)$
$\displaystyle=\frac{A-B}{(1+Az)^{2}}s_{n}(v_{\lambda}(A,B,z),z)^{1/\lambda}-\frac{(1+Bz)}{(1+Az)\lambda}s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}-1}s^{\prime}_{n}(v_{\lambda}(A,B,z),z)$
$\displaystyle=\frac{(A-B)s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}}}{(1+Az)^{2}}\left(s_{n}(v_{\lambda}(A,B,z),z)-\frac{(1+Az)(1+Bz)}{(A-B)\lambda}s^{\prime}_{n}(v_{\lambda}(A,B,z),z)\right)$
(10)

Using relations (8) in (10), we get

$\begin{split}h^{\prime}(z)=&\frac{(A-B)s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}-1}}{(1+Az)^{2}}\bigg{[}s_{n}\left(v_{\lambda}(A,B,z)-\frac{(1+Az)(1+Bz)}{(A-B)\lambda}v^{\prime}_{\lambda}(A,B,z),z\right)\\\
&\qquad\qquad\qquad+\frac{(n+1)}{\lambda(A-B)}\sum_{k=0}^{n+1}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k+1}}{(n-k+1)!}A^{k}(-B)^{n-k+1}z^{n}\\\
&\qquad\qquad\qquad\qquad-\frac{nAB}{\lambda(A-B)}\sum_{k=0}^{n}\frac{[\lambda]_{k}}{k!}\frac{(\lambda)_{n-k}}{(n-k)!}A^{k}(-B)^{n-k}z^{n+1}\bigg{]}.\end{split}$
(11)

Substituting (9) in (11) and using definition of $a_{n}$, the following form
of $h^{\prime}(z)$ can be obtained.

$\displaystyle h^{\prime}(z)=$
$\displaystyle\frac{z^{n}s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}-1}}{\lambda(1+Az)^{2}}\left((n+1)a_{n+1}-ABzna_{n}\right)$
$\displaystyle=$
$\displaystyle\frac{z^{n}s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}-1}}{\lambda}\left((n+1)a_{n+1}-ABzna_{n}\right)(1+Az)^{-2}$
$\displaystyle=$
$\displaystyle\frac{z^{n}s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}-1}}{\lambda}\left((n+1)a_{n+1}-ABzna_{n}\right)(1-2Az+3A^{2}z^{2}-4A^{3}z^{3}+\cdots)$
$\displaystyle=$
$\displaystyle\frac{z^{n}s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}-1}}{\lambda}\left((n+1)a_{n+1}+\sum_{m=1}^{\infty}(m+1)(n+1)a_{n+1}+mnBa_{n}(-A)^{m}z^{m}\right).$

Since $A\in(-1,0]$, we have $-A\geq 0$. Therefore in view of Lemma 1, we
obtain $a_{n}>0$ for all $n\in\mathbb{N}$. Further, from Lemma 2, we obtain
$(m+1)(n+1)a_{n+1}+Bmna_{n}>0$ for all $n,m\in\mathbb{N}$. Thus

$\displaystyle(n+1)a_{n+1}+\sum_{m=0}^{\infty}[(m+1)(n+1)a_{n+1}+Bmna_{n}](-A)^{m}z^{m}$

represents a series of positive Taylor’s coefficients. Since $a_{n}>0$ for all
$n\in\mathbb{N}$, the function $v_{\lambda}(A,B,z)$ has a series
representation with positive Taylor coefficients. Hence,

$|s_{n}(v_{\lambda}(A,B,z),z)|\leq s_{n}(v_{\lambda}(A,B,z),|z|)$

holds and consequently

$\displaystyle|h^{\prime}(z)|\leq h^{\prime}(|z|),\quad\mbox{ for all
$z\in\mathbb{D}$ holds.}$ (12)

Since $h(0)=0$ and $h(-B)=1$, using (12) we obtain

$\displaystyle|h(z)|=\left|\int_{0}^{z}h^{\prime}(t)dt\right|\leq\int_{0}^{-B}h^{\prime}\left(\frac{-tz}{B}\right)dt\leq\int_{0}^{-B}h^{\prime}(t)dt=1,\quad\
z\in\mathbb{D}.$

Therefore,

$\displaystyle\left|\frac{(1+Bz)(s_{n}(v_{(}A,B,z),z))^{\frac{1}{\lambda}}}{(1+Az)}-1\right|<1,\quad\
z\in\mathbb{D}.$

which implies that

$\displaystyle\frac{s_{n}(v_{\lambda}(A,B,z),z)}{v_{\lambda}(A,B,z)}\prec\frac{1}{v_{\lambda}(0,B,z)}.$

Therefore, $v_{\lambda}(A,B,z)$ is $n$-stable with respect to
$v_{\lambda}(0,B,z)$ for all $n\in\mathbb{N}$. Hence $v_{\lambda}(A,B,z)$ is
stable with respect to $v_{\lambda}(0,B,z)$ for all $0<\lambda\leq 1$ and
$-1\leq B<A\leq 0$. ∎

For the proof of Corollary 2, we need the following proposition which follows
the same procedure as given in [8].

######  Proposition 1.

Let $\alpha,\beta>0$ and $B\in[-1,0)$. If $F\prec(1+Bz)^{\alpha}$ and
$G\prec(1+Bz)^{\beta}$ then $FG\prec(1+Bz)^{\alpha+\beta}$ for
$z\in\mathbb{D}$.

###### Proof.

The function $\log(1+Bz)$ is convex univalent for $z\in\mathbb{D}$ and
$B\in[-1,0)$. Our claim follows from

$\displaystyle\frac{1}{\alpha+\beta}\log(F(z)G(z))$
$\displaystyle=\frac{\alpha}{\alpha+\beta}\log(1+Bu(z))+\frac{\beta}{\alpha+\beta}\log(1+Bv(z))$
$\displaystyle\prec\log(1+Bz),$

where $u,v$ are analytic functions such that $|u(z)|\leq|z|$ and
$|v(z)|\leq|z|$ for $z\in\mathbb{D}$. ∎

Now we are ready to give proof of Corollary 2.

###### Proof of Corollory 2.

For $0<\mu\leq\lambda\leq 1$ we have,

$\displaystyle\frac{1}{\lambda}\log\left((1+Bz)^{\lambda}s_{n}(v_{\mu}(0,B,z),z)\right)$
$\displaystyle=\frac{1}{\lambda}\log\left[(1+Bz)^{\lambda-\mu}(1+Bz)^{\mu}s_{n}(v_{\mu}(0,B,z),z)\right]$
$\displaystyle=\frac{1}{\lambda}\log(1+Bz)^{\lambda-\mu}+\frac{1}{\lambda}\log\left[(1+Bz)^{\mu}s_{n}(v_{\mu}(0,B,z),z)\right]$
$\displaystyle=\frac{1}{\lambda}\log(1+Bu(z))^{\lambda-\mu}+\frac{1}{\lambda}\log(1+Bw(z))^{\mu}$
$\displaystyle\prec(1+Bz)^{\lambda}$

for $|u(z)|\leq|z|$ and $|w(z)|\leq|z|$. Therefore,
$(1+Bz)^{\lambda}s_{n}(v_{\mu}(0,B,z),z)\prec(1+Bz)^{\lambda}$ holds for all
$z\in\mathbb{D}$ and $0<\mu\leq\lambda\leq 1$. ∎

Now we prove that $v_{\lambda}(A,B,z)$ is not stable with respect to itself
for $\lambda\in(0,1]$ and $-1\leq B<A\leq 0$.

###### Proof of Theorem 4.

For $-1\leq B<A\leq 0$, to prove that $v_{\lambda}(A,B,z)$ is stable with
respect to itself, we need to show that

$\displaystyle\frac{s_{n}(v_{\lambda}(A,B,z),z)}{v_{\lambda}(A,B,z)}\prec\frac{1}{v_{\lambda}(A,B,z)},\quad
z\in\mathbb{D}.$ (13)

Equivalently $G(z)\prec H(z)$ where

$\displaystyle
G(z):=\frac{(1+Bz)s_{n}(v_{\lambda}(A,B,z),z)^{\frac{1}{\lambda}}}{1+Az}\quad\mbox{and}\quad
H(z):=\frac{1+Bz}{1+Az}$

Since $G(z)$ and $H(z)$ are analytic in $\mathbb{D}$ for $-1\leq B<A\leq 0$
and $H(z)$ is univalent in $\mathbb{D}$. In the point of view of the
subordination, we have $G(z)\prec H(z)$ if and only if $G(0)=H(0)$ and
$G(\mathbb{D})\subseteq H(\mathbb{D})$ and $G=H\circ\omega_{1}$, where
$\omega_{1}\in\Omega$ analytic in $\mathbb{D}$ satisfying $\omega_{1}(0)=0$
and $|\omega_{1}(z)|<1$.

In view of the Schwartz Lemma, we have $|\omega_{1}(z)|\leq|z|$ for
$z\in\mathbb{D}$ and $|\omega_{1}^{\prime}(0)|\leq 1$. If $G\prec H$, it
follows that $|G^{\prime}(0)|\leq|H^{\prime}(0)|$ and $G(|z|\leq r)\subseteq
H(|z|\leq r),0\leq r<1$.

Let $\displaystyle\omega=H(z)=\frac{1+Bz}{1+Az}$, then $\displaystyle
z=\frac{1+B\omega}{1+A\omega}$. Therefore, the image of $|z|\leq r$ under
$H(z)$ is $\left|\dfrac{1+B\omega}{1+A\omega}\right|\leq r$ which after
simplification is equivalent to $|w-C(r,A,B)|\leq R(r,A,B)$ where

$\displaystyle C(r,A,B):=\frac{r^{2}A-B}{B^{2}-r^{2}A^{2}}\quad\mbox{and}\quad
R(r,A,B):=\frac{r(A-B)}{B^{2}-r^{2}A^{2}}$

To show that $G\nprec H$, it is enough to show that $G(|z|\leq r)\nsubseteq
H(|z|\leq r)$. To prove that $G(|z|\leq r)\nsubseteq H(|z|\leq r)$, it is
enough to choose a point $z_{0}$ with $|z_{0}|\leq r_{0}$ such that $G(z_{0})$
does not lie in the disk $|\omega-C(r,A,B)|\leq R(r,A,B)$ for some $-1\leq
B<A\leq 0$.

Figure 1. $G(z_{0},A,B)\nprec H(z_{0},A,B)$ for $z_{0}=0.915282-0.357037i$,
$A=-0.679,B=-0.97,$ and $\lambda=0.3$.

Choose $z_{0}=0.915282-0.357037i$, $r_{0}=0.98$, $A=-0.679,B=-0.97,$
$\lambda=0.3$ and $n=1$. Then $G(z_{0})=0.8697+0.5845i$,
$C(r_{0},A,B)=0.634444$ and $R(r_{0},A,B)=0.576521$. Clearly $G(z_{0})$ does
not lie in the disk $|\omega-C(r_{0},A,B)|\leq R(r_{0},A,B)$. Therefore
$G\nprec H$ i.e., (13) does not hold. The graphical illustration of these
values is also given here in Figure 1. Hence $v_{\lambda}(A,B,z)$ is not
stable with respect to itself. ∎

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