# Volterra–type integral operator on analytic function spaces

R. Kargar  Young Researchers and Elite Club, Ardabil Branch, Islamic Azad
University, Ardabil, Iran<EMAIL_ADDRESS>
###### Abstract.

In this note we study the radius of convexity (hence univalency) of the
Volterra–type integral operator

$T_{g}f(z)=\int_{0}^{z}f(s)g^{\prime}(s){\rm d}s\quad(|z|<1),$

on the spaces of analytic function $f$ in the open unit disc $|z|<1$.

###### Key words and phrases:

Starlike; Convex; Radius of convexity; Volterra–type integral operator;
Linear–invariant family; Locally univalent; Boundary rotation.

###### 2010 Mathematics Subject Classification:

47B38; 30C45

##  1\. Volterra–type integral operator

Let $\Delta:=\\{z\in\mathbb{C}:|z|<1\\}$ and $H(\Delta)$ be the class of all
analytic functions in $\Delta$. In 1977 Pommerenke [16] introduced an integral
operator, called Volterra–type operator as follows:

$T_{g}:=T_{g}f(z)=\int_{0}^{z}f(s)g^{\prime}(s){\rm d}s\quad(z\in\Delta).$

He proved that $T_{g}$ is a bounded operator on the Hardy space $H^{2}$ if,
and only if, $g$ belongs to the class $BMOA$. Also, Aleman and Siskakis [1]
proved that this characterization (boundedness) is valid on each $H^{p}$ for
$1\leq p<\infty$ and that $T_{g}$ is compact on $H^{p}$ if, and only if, $g\in
VMOA$. An another natural integral operator is defined as follows:

$J_{g}f(z):=\int_{0}^{z}f^{\prime}(s)g(s){\rm d}s\quad(z\in\Delta).$

It is interesting to know this fact that

$J_{g}f(z)+T_{g}f(z)=M_{g}f-f(0)g(0),$

where $M_{g}$ is the multiplication operator and is defined by

$M_{g}f(z)=g(z)f(z)\quad(f\in H(\Delta),z\in\Delta).$

Indeed, if $f$ and $g$ are two normalized analytic functions, i.e. $f,g\in
H(\Delta),f(0)=g(0)=0$ and $f^{\prime}(0)=g^{\prime}(0)=1$, then

$J_{g}f(z)+T_{g}f(z)=g(z)f(z).$

We note that the integral operators $T_{g}$ and $J_{g}$ contain the well–known
integral operators in the analytic function theory and geometric function
theory, such as the generalized Bernardi–Libera–Livingston linear integral
operator (see [2, 7, 8]), Srivastava–Owa fractional derivative operators [11,
12] and the Cesáro integral operator, [18, 19].

Recently many researchers have been studied the integral operators $T_{g}$ and
$J_{g}$. For example, Li and Stević [6] studied the boundedness and the
compactness of $T_{g}$ and $J_{g}$ on the Zygmund space and the little Zygmund
space. We note that there exist some extensions of the integral operators
$T_{g}$, see for example [20].

The basis of this article is to study of the integral operator

(1.1) $T_{g}f(z)=\int_{0}^{z}f(s)g^{\prime}(s){\rm
d}s=\int_{0}^{1}f(tz)zg^{\prime}(tz){\rm d}t=\int_{0}^{z}f{\rm d}g.$

##  2\. Some Subclasses of Analytic Functions

In this section we recall some certain subclasses of analytic functions.
Further, we denote by $\mathcal{A}$ the class of all analytic and normalized
functions in the open unit disc $\Delta$. The subclass of $\mathcal{A}$
consisting of univalent (that is, one–to–one) functions is denoted by
$\mathcal{U}$. We say that the function $f\in\mathcal{U}$ is starlike of order
$\alpha$, $0\leq\alpha<1$ if, and only if

${\rm
Re}\left\\{z\frac{f^{\prime}(z)}{f(z)}\right\\}>\alpha\quad(z\in\Delta).$

We denote by $\mathcal{S}^{*}(\alpha)$ the class of starlike functions of
order $\alpha$. Also, $f\in\mathcal{U}$ is convex of order $\alpha$,
$0\leq\alpha<1$ iff $zf^{\prime}(z)\in\mathcal{S}^{*}(\alpha)$. The class of
convex functions of order $\alpha$ is denoted by $\mathcal{K}(\alpha)$.
Analytically, $f\in\mathcal{K}(\alpha)$ iff

${\rm
Re}\left\\{1+z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right\\}>\alpha\quad(z\in\Delta).$

Let $A$ and $B$ be two complex numbers such that $|A|>1$ and $|B|\leq 1$. We
say that a function $f\in\mathcal{A}$ belongs to the class $S^{*}(A,B)$ if it
satisfies the following subordination relation:

(2.1) $z\frac{f^{\prime}(z)}{f(z)}\prec\frac{1+Az}{1+Bz}\quad(z\in\Delta).$

We note that for real numbers $A$ and $B$, where $-1\leq B<A\leq 1$, the
function $f$ that satisfies the above subordination relation (2.1) is called
”the Janowski starlike function”. Also, we say that $f\in\mathcal{K}(A,B)$ if,
and only if, $zf^{\prime}(z)\in S^{*}(A,B)$. Indeed, if $f$ belongs to the
class $\mathcal{K}(A,B)$, then it satisfies

$1+z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\prec\frac{1+Az}{1+Bz}\quad(z\in\Delta).$

We remark that $\mathcal{K}(2,1)$ and $\mathcal{K}(2,-1)$ become to the Ozaki
conditions. Moreover, by the Lindelöf subordination principle (this principle
states that if $f(z)\prec g(z)$, then $|f^{\prime}(0)|\leq|g^{\prime}(0)|$ and
$f(\Delta_{r})\subset g(\Delta_{r})$), if $f\in\mathcal{K}(2,1)$, then we have

(2.2) ${\rm
Re}\left\\{1+z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right\\}<\frac{3}{2}\quad(z\in\Delta).$

Also, if $f\in\mathcal{K}(2,-1)$, then

(2.3) ${\rm
Re}\left\\{1+z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right\\}>-\frac{1}{2}\quad(z\in\Delta).$

Ozaki proved that if $f\in\mathcal{A}$ with $f(z)f′(z)/z\neq 0$, there, and if
either (2.2) or (2.3) holds throughout $\Delta$, then $f$ is univalent and
convex in at least one direction in $\Delta$, see [13].

Let $\mathcal{LU}$ denote the family of normalized locally univalent functions
in $\Delta$. For $\beta\in\mathbb{R}$, we consider the class
$\mathcal{G}\left(\beta\right)$ consisting of all functions $f\in\mathcal{LU}$
which satisfy the condition

(2.4) ${\rm
Re}\left\\{1+z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right\\}<1+\frac{\beta}{2}\quad(z\in\Delta).$

We note that $\mathcal{G}(1)\subset\mathcal{U}$ and
$\mathcal{G}(1)\equiv\mathcal{K}(2,1)$. Also the functions in the class
$\mathcal{G}\left(1\right)$ are starlike of order 0 in $\Delta$. The functions
class $\mathcal{G}\left(\beta\right)$ was studied extensively by Kargar et al.
[5] (see also [10]).

Let $Aut(\Delta)$ be the class of holomorphic automorphisms in $\Delta$. Any
$\phi\in Aut(\Delta)$ has the following representation:

$\phi(z)=e^{i\vartheta}\dfrac{z+a}{1+\bar{a}z}\quad(\vartheta,a\in\mathbb{R},z\in\Delta).$

The family $\mathcal{F}$ of $\mathcal{A}$ is called a linear–invariant family
(L.I.F.), if $\mathcal{F}\subset\mathcal{LU}$ and for all $f\in\mathcal{F}$
and $\phi\in Aut(\Delta)$

$F_{\phi}(f)(z):=\frac{f(\phi(z))-f(\phi(0))}{f^{\prime}(\phi(0))\phi^{\prime}(0)}\in\mathcal{F}.$

The class $\mathcal{U}$ and the class $\mathcal{K}(0)$ on $\Delta$ are a
L.I.F., but the class $\mathcal{S}^{*}(0)$ on $\Delta$ is not a L.I.F.. Also,
we define the order of the L.I.F. $\mathcal{F}$ as

${\rm
ord}f:=\sup\left\\{\left|\frac{f^{\prime\prime}(0)}{2}\right|:f\in\mathcal{F}\right\\}$

and the universal linear invariant family of order $\gamma\geq 1$ as

$\mathcal{UL}_{\gamma}:=\\{f\in\mathcal{F}:{\rm ord}f\leq\gamma\\}.$

We remark that $\mathcal{UL}_{1}\equiv\mathcal{K}(0)$ and
$\mathcal{U}\subset\mathcal{UL}_{2}$. For more details about the L.I.F. see
[4, Chapter 5].

For fixed $k\geq 2$, Let $V_{k}$ denote the class of normalized functions
$f\in\mathcal{A}$ such that

$\int_{0}^{2\pi}\left|{\rm
Re}\left\\{1+z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right\\}\right|_{z=re^{i\theta}}{\rm
d}\theta\leq k\pi.$

Indeed, a function $f\in V_{k}$ maps $\Delta$ conformally onto an image domain
$f(\Delta)$ of boundary rotation at most $k\pi$. The functions class $V_{k}$
was introduced by Loewner [9] in 1917 and was developed by Paatero [14] who
systematically developed their properties and made an exhaustive study of the
class $V_{k}$. In association with the class $V_{k}$, Robertson proved that
(see [17, Theorem 1]) the following inequlaity

(2.5)
$\left|z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-\frac{2|z|^{2}}{1-|z|^{2}}\right|\leq\frac{k|z|}{1-|z|^{2}}\quad(z\in\Delta,2\leq
k<\infty)$

holds true for each $f\in V_{k}$.

##  3\. Some Key Lemmas

In this section we recall some lemmas which help us in order to prove of main
results.

######  Lemma 3.1.

(see [3]) If $f\in\mathcal{UL}_{\gamma}$ and $\gamma\geq 1$, then

$\left|z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-\frac{2|z|^{2}}{1-|z|^{2}}\right|\leq\frac{2\gamma|z|}{1-|z|^{2}}\qquad(z\in\Delta).$

The next lemma is due to Pommerenke [15], see also [4, Lemma 5.1.3].

######  Lemma 3.2.

Let $\mathcal{F}$ be a linear–invariant family and $\delta={\rm
ord}\mathcal{F}$. Then

(3.1)
$\delta=\sup_{f\in\mathcal{F}}\sup_{|z|<1}\left|-\overline{z}+\frac{1}{2}(1-|z|^{2})\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|\quad(z\in\Delta).$

The following lemma gives a basic estimate which leads to the distortion
theorem for univalent functions.

######  Lemma 3.3.

(see [4, p. 15]) If $f\in\mathcal{U}$, then

$\left|z\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}-\frac{2r^{2}}{1-r^{2}}\right|\leq\frac{4r}{1-r^{2}}\qquad(|z|=r<1).$

The estimate is sharp for rotation of Koebe function.

######  Lemma 3.4.

(see [10]) Let $\beta\in(0,1]$ be fixed. If $f\in\mathcal{G}(\beta)$, then

$\left|\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\right|\leq\frac{\beta}{1-|z|}\quad(z\in\Delta).$

The result is sharp for the function $f^{\prime}(z)=(1-z)^{\beta}$.

In the present paper, we obtain the radius of convexity of the Volterra–type
integral operator $T_{g}$ when the functions $f$ and $g$ belonging to the some
certain subclasses of analytic functions which are defined in Section 2.

##  4\. Main Results

We begin this section with the following result.

######  Theorem 4.1.

Let $A$ and $B$ be two complex numbers, $|A|>1$ and $|B|\leq 1$ and $T_{g}$ be
given by (1.1). Also, let $0\leq\alpha<1$ be real number. If
$f\in\mathcal{S}^{*}(A,B)$ and $g\in\mathcal{K}(A,B)$, then the Volterra–type
integral operator $T_{g}$ is convex of order $\alpha$ in $|z|\leq
r_{c}(A,B,\alpha)$, where

$r_{c}(A,B,\alpha)=\left\\{\begin{array}[]{cc}\frac{2-\alpha}{2|A|}&\displaystyle{B=0}\\\
\frac{|B-A|-|(\alpha-1)B-A|}{\alpha|B|^{2}-2{Re}\\{A\overline{B}\\}}&\displaystyle{B\neq
0}.\end{array}\right.$

###### Proof.

Using the definition of convexity of order $\alpha$, where $0\leq\alpha<1$ it
is enough to find out the largest number $0<r<1$ for which

$\min_{|z|=r}{\rm
Re}\left\\{1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}\right\\}-\alpha\geq
0.$

Further, from (1.1) we have

(4.1)
$1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}-\alpha=z\frac{f^{\prime}(z)}{f(z)}-\alpha+z\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}+1.$

On the other hand, since $f\in\mathcal{S}^{*}(A,B)$, by the Lindelöf
subordination principle, we get

(4.2)
$\left|z\frac{f^{\prime}(z)}{f(z)}-\frac{1-A\overline{B}r^{2}}{1-|B|^{2}r^{2}}\right|\leq\frac{|B-A|r}{1-|B|^{2}r^{2}}\quad(|z|=r<1).$

From the above inequality (4.2), we obtain

$\displaystyle{\rm Re}\left\\{z\frac{f^{\prime}(z)}{f(z)}\right\\}-\alpha$
$\displaystyle\geq{\rm
Re}\left\\{\frac{1-A\overline{B}r^{2}}{1-|B|^{2}r^{2}}\right\\}-\frac{|B-A|r}{1-|B|^{2}r^{2}}-\alpha$
(4.3) $\displaystyle\geq\frac{1-\alpha-|B-A|r-\left({\rm
Re}\\{A\overline{B}\\}-\alpha|B|^{2}\right)r^{2}}{1-|B|^{2}r^{2}}.$

Also $g\in\mathcal{K}(A,B)$ and the above same method imply that

(4.4) ${\rm
Re}\left\\{1+z\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}\right\\}\geq\frac{1-|B-A|r-{\rm
Re}\\{A\overline{B}\\}r^{2}}{1-|B|^{2}r^{2}}.$

Now from (4.1), (4) and (4.4) we get

${\rm
Re}\left\\{1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}\right\\}-\alpha\geq\frac{2-\alpha-2|B-A|r-\left(2{\rm
Re}\\{A\overline{B}\\}-\alpha|B|^{2}\right)r^{2}}{1-|B|^{2}r^{2}}>0$

provided $\phi(r):=2-\alpha-2|B-A|r-\left(2{\rm
Re}\\{A\overline{B}\\}-\alpha|B|^{2}\right)r^{2}>0$. A simple calculation
gives that the roots of $\phi(r)$ are

$\frac{|B-A|\pm|(\alpha-1)B-A|}{\alpha|B|^{2}-2{\rm Re}\\{A\overline{B}\\}}.$

Also, it is clear that if $B=0$, then $\phi(r)$ yields that
$\phi(r)=2-\alpha-2|A|r$ and it will be positive if $r<(2-\alpha)/2|A|$. This
completes the proof. ∎

Putting $A=2$ and $B=1$ in Theorem 4.1, we get.

######  Corollary 4.1.

Let ${\rm Re}\\{zf^{\prime}(z)/f(z)\\}<3/2$ where $z\in\Delta$ and $g$
satisfies the condition (2.2). Then the radius of convexity of order $\alpha$
of the Volterra–type integral operator $T_{g}$ is $\frac{\alpha-2}{\alpha-4}$
where $(0\leq\alpha<1)$.

If we take $A=2$ and $B=-1$ in Theorem 4.1, then we have the following.

######  Corollary 4.2.

Let ${\rm Re}\\{zf^{\prime}(z)/f(z)\\}>-1/2$ where $z\in\Delta$ and $g$
satisfies the condition (2.3). Then the Volterra–type integral operator
$T_{g}$ is convex of order $\alpha$ $(0\leq\alpha<1)$ in the disc
$|z|\leq(2-\alpha)/(4+\alpha)$.

######  Theorem 4.2.

Let $0\leq\alpha<1$ and $\gamma\geq 1$. If $f\in\mathcal{S}^{*}(\alpha)$ and
$g\in\mathcal{LU}_{\gamma}$, then the radius of convexity of the Volterra–type
integral operator $T_{g}$ is

(4.5)
$r^{-}_{c}(\alpha,\gamma)=\frac{\gamma-\sqrt{\alpha^{2}+\gamma^{2}-1}}{1-\alpha}.$

###### Proof.

Let $f\in\mathcal{S}^{*}(\alpha)$ ($0\leq\alpha<1$) and $g$ belongs to the
family $\mathcal{LU}_{\gamma}$ ($\gamma\geq 1$). From (1.1) we get

(4.6)
$1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}=z\frac{f^{\prime}(z)}{f(z)}+z\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}+1.$

Since $g\in\mathcal{LU}_{\gamma}$, Lemma 3.1 implies that

(4.7) ${\rm
Re}\left\\{z\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}\right\\}\geq\frac{2r^{2}}{1-r^{2}}-\frac{2\gamma
r}{1-r^{2}}.$

Now by definition of starlikeness of order $\alpha$ and using the above
inequality (4.7), and also applying the relation (4.6), we obtain

$\displaystyle{\rm
Re}\left\\{1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}\right\\}$
$\displaystyle={\rm
Re}\left\\{z\frac{f^{\prime}(z)}{f(z)}+z\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}+1\right\\}$
$\displaystyle>\frac{1+\alpha-2\gamma r+(1-\alpha)r^{2}}{1-r^{2}}>0,$

provided $\varphi(r)=1+\alpha-2\gamma r+(1-\alpha)r^{2}>0$. It is easy to see
that the roots of $\varphi(r)$ are

$r^{\pm}_{c}(\alpha,\gamma)=\frac{\gamma\pm\sqrt{\alpha^{2}+\gamma^{2}-1}}{1-\alpha}.$

Also we see that if $0\leq\alpha<1$ and $\gamma\geq 1$, then
$0<r^{-}_{c}(\alpha,\gamma)<1$ where $r^{-}_{c}(\alpha,\gamma)$ is defined in
(4.5). This is the end of proof. ∎

Putting $\alpha=0$ and $\gamma=1$ in Theorem 4.2, we get.

######  Corollary 4.3.

Let $f$ and $g$ be starlike and convex univalent functions in the open unit
disc $\Delta$, respectively. Then the Volterra–type integral operator $T_{g}$
is convex univalent function in $\Delta$, too.

######  Theorem 4.3.

Let $0\leq\alpha<1$. If $f$ is starlike function of order $\alpha$ and
$g\in\mathcal{F}$ with ${\rm ord}\mathcal{F}=1$, then the Volterra–type
integral operator $T_{g}$ is convex univalent in the open unit disc $\Delta$.

###### Proof.

Since $f$ is starlike function of order $\alpha$, we have

(4.8) ${\rm
Re}\left\\{z\frac{f^{\prime}(z)}{f(z)}\right\\}>\alpha\quad(z\in\Delta).$

Also since $g\in\mathcal{F}$, by using the relation (3.1) and with a little
calculation we get

(4.9) ${\rm
Re}\left\\{1+z\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}\right\\}\geq\frac{1-2\delta
r+r^{2}}{1-r^{2}}\quad(|z|=r<1).$

Now, from (4.6), (4.8) and (4.9) and since ${\rm ord}\mathcal{F}=1$, we get

${\rm
Re}\left\\{1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}\right\\}\geq\frac{1+\alpha-2r+(1-\alpha)r^{2}}{1-r^{2}}$

and for $z\in\Delta$ one deduces that

${\rm
Re}\left\\{1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}\right\\}>0\quad(z\in\Delta).$

Thus by definition we conclude that $T_{g}$ is convex univalent in the open
unit disc $\Delta$. This is the end of proof. ∎

The next theorem is the following.

######  Theorem 4.4.

Let $f$ be starlike function of order $\alpha$ where $0\leq\alpha<1$ and $g$
be univalent function. Then the radius of convexity of the Volterra–type
integral operator $T_{g}$ is

$r^{-}_{c}(\alpha)=\frac{2-\sqrt{3+\alpha^{2}}}{1-\alpha}\quad(0\leq\alpha<1).$

###### Proof.

Let $f\in\mathcal{S}^{*}(\alpha)$ where $0\leq\alpha<1$ and $g\in\mathcal{U}$.
From Lemma 3.3 and by (4.6), we get

$\displaystyle{\rm
Re}\left\\{1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}\right\\}$
$\displaystyle={\rm
Re}\left\\{z\frac{f^{\prime}(z)}{f(z)}+z\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}+1\right\\}$
$\displaystyle>\frac{1+\alpha-4r+(1-\alpha)r^{2}}{1-r^{2}}>0,$

when $|z|\leq(2-\sqrt{3+\alpha^{2}})/(1-\alpha)$ and concluding the proof. ∎

######  Remark 4.1.

Taking $\alpha=0$ in Theorem 4.4, we see that if $f$ is starlike univalent
function and $g\in\mathcal{U}$, then the radius of convexity of the
Volterra–type integral operator $T_{g}$ is $2-\sqrt{3}$. Indeed, in this case
the radius of convexity the Volterra–type integral operator $T_{g}$ is equal
to familiar radius of convexity for the class $\mathcal{U}$ (see [4, Theorem
2.2.22]).

######  Theorem 4.5.

Let $f$ be starlike function of order $\alpha$ where $0\leq\alpha<1$ and $g$
be locally univalent function which satisfies the inequality (2.4) where
$0<\beta\leq 1$. Then the radius of convexity of the Volterra–type integral
operator $T_{g}$ is

$r_{c}(\alpha,\beta)=\frac{1+\alpha}{1+\alpha+\beta}.$

###### Proof.

Assume that $f\in\mathcal{S}^{*}(\alpha)$ and $g\in\mathcal{G}(\beta)$, where
$0\leq\alpha<1$ and $0<\beta\leq 1$. By Lemma 3.4 and from (4.6) we have

$\displaystyle{\rm
Re}\left\\{1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}\right\\}$
$\displaystyle>\alpha+1-\frac{\beta|z|}{1-|z|}\geq 0,$

where $|z|\leq(1+\alpha)/(1+\alpha+\beta)$. This completes the proof. ∎

Finally we have.

######  Theorem 4.6.

Let $f$ be starlike function of order $\alpha$ where $0\leq\alpha<1$ and $g\in
V_{k}$ where $k\geq 2$. Then the radius of convexity of the Volterra–type
integral operator $T_{g}$ is

$r_{c}(\alpha,k)=\frac{k-\sqrt{k^{2}-4(1-\alpha^{2})}}{2(1-\alpha)}.$

###### Proof.

From the definition of starlikeness of order $\alpha$, inequality (2.5) and
(4.6), one deduces that

$\displaystyle{\rm
Re}\left\\{1+z\frac{T_{g}^{\prime\prime}(z)}{T_{g}^{\prime}(z)}\right\\}$
$\displaystyle={\rm
Re}\left\\{z\frac{f^{\prime}(z)}{f(z)}+z\frac{g^{\prime\prime}(z)}{g^{\prime}(z)}+1\right\\}$
$\displaystyle>\frac{(1-\alpha)r^{2}-kr+1+\alpha}{1-r^{2}}=\frac{\lambda(r)}{1-r^{2}}$

where

$\lambda(r):=(1-\alpha)r^{2}-kr+1+\alpha\quad(k\geq 2,0\leq\alpha<1).$

It is easy to see that $\lambda(r)>0$ if

$r>\frac{k+\sqrt{k^{2}-4(1-\alpha^{2})}}{2(1-\alpha)}\geq 1\quad(k\geq
2,0\leq\alpha<1)$

or

$r<\frac{k-\sqrt{k^{2}-4(1-\alpha^{2})}}{2(1-\alpha)}\leq 1\quad(k\geq
2,0\leq\alpha<1).$

This completes the proof. ∎

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