Fekete-Szegö inequality for a subclass of analytic functions defined by Komatu integral operator

1.   Introduction

Let $\mathcal{A}$ denote the class of functions of the form:

which are analytic in the open unit disk $\mathcal{U} = \left\{ z:\left\vert z\right\vert < 1\right\}.$ Also let $\mathcal{S}$ denote the subclass of $\mathcal{A}$ consisting of univalent functions in $\mathcal{U}$. It is well-known that for $f\in \mathcal{S},$ $\left\vert a_{3}-a_{2}^{2}\right\vert \leq 1.$ A classical theorem of Fekete-Szegö ^[8] states that for $f\in \mathcal{S}$ given by (1.1)

The latter inequality is sharp in the sense that for each $\eta$ there exists a function in $\mathcal{S}$ such that the equality holds. Later, Pfluger ^[24] has considered the complex values of $\eta$ and provided the inequality

Indeed, many authors have considered the Fekete-Szegö problem for various subclasses of $\mathcal{A},$ the upper bound for $\left\vert a_{3}-\eta a_{2}^{2}\right\vert$ is investigated by various authors ^{[1,5,6,13,16,17]}, see also recent investigations on this subject by ^{[7,11,21,22,23]}.

A function $f\in \mathcal{A}$ is said to be in the class $\mathcal{S}^{\ast }$ of starlike functions in $\mathcal{U},$ if

On the other hand, a function $f\in \mathcal{A}$ is said to be in the class of convex functions in $\mathcal{U},$ denoted by $\mathcal{C},$ if

A function $f\in \mathcal{A}$ is said to be in the class of starlike functions of complex order $b(b\in \mathbb{C} -\left\{ 0\right\}),$ denoted by $\mathcal{S}^{\ast }(b),$ provided that

Furthermore, a function $f\in \mathcal{C}(b)$ is convex functions of complex order $b(b\in \mathbb{C} -\left\{ 0\right\})$ if it satisfies the inequality

Note that $\mathcal{S}^{\ast }(1) = \mathcal{S}^{\ast }$ and $\mathcal{C}(1) = \mathcal{C}.$

The class $\mathcal{S}^{\ast }(b)$ of starlike functions of complex order $b(b\in \mathbb{C} -\left\{ 0\right\})$ was introduced by Nasr and Aouf ^[19] while the class $\mathcal{C}(b)$ of convex functions of complex order $b(b\in \mathbb{C} -\left\{ 0\right\})$ was presented earlier by Wiatrowski ^[28].

Sãlãgean ^[26] introduced the following differential operator for $f(z)\in \mathcal{A}$ which is called the Sãlãgean differential operator:

We note that,

Recently, Komatu ^[14] introduced a certain integral operator $\mathcal{L} _{a}^{\delta }$ defined by

Thus, if $f(z)\in \mathcal{A}$ is of the form (1.1), it is easily seen from (1.3) that ^[14]

We note that:

● $\mathcal{L} _{a}^{0}f(z) = f(z);$

● $\mathcal{L} _{1}^{1}f(z) = A\left[ f\right] (z)$ known as Alexander operator ^[2];

● $\mathcal{L} _{2}^{1}f(z) = \mathcal{L} \left[ f\right] (z)$ known as Libera operator ^[15];

● $\mathcal{L} _{c+1}^{1}f(z) = \mathcal{L} _{c}\left[ f\right] (z)$ called generalized Libera operator or Bernardi operator ^[3];

● For $a = 1$ and $\delta = k$ ($k$ is any integer), the multiplier transformation $\mathcal{L} _{1}^{k}f(z) = \mathcal{I}^{k}f(z)$ was studied by Flett ^[9] and Sãlãgean ^[26];

● For $a = 1$ and $\delta = -k$ $(k\in \mathbb{N} _{0} = \mathbb{N} \cup \{0\})$, the differential operator $\mathcal{L} _{1}^{-k}f(z) = \mathcal{D }^{k}f(z)$ was studied by Sãlãgean ^[26];

● For $a = 2$ and $\delta = k$ ($k$ is any integer), the operator $\mathcal{L} _{2}^{k}f(z) = \mathcal{L} ^{k}f(z)$ was studied by Uralegaddi and Somanatha ^[27];

● For $a = 2$, the multiplier transformation $\mathcal{L} _{2}^{\delta }f(z) = \mathcal{I}^{\delta }f(z)$ was studied by Jung et al. ^[10].

For $\mathcal{D}^{k}f(z)$ given by (1.2) and $\mathcal{L} _{a}^{\delta }f(z)$ is given by (1.4), we define the differential operator $\mathcal{D}^{k}\mathcal{L} _{a}^{\delta }f(z)$ as follows:

Note that, by taking $\delta = 0$ and $k = 0$ in (1.5), the differential operator $\mathcal{D}^{k}\mathcal{L} _{a}^{\delta }f(z)$ reduces to Sãl ãgean differential operator and Komatu integral operator, respectively.

Using the operator $\mathcal{D}^{k}\mathcal{L} _{a}^{\delta }f,$ we now introduce a new subclass of analytic functions as follows:

Definition 1. A function $f\in \mathcal{A}$ is said to be in the class $\mathcal{ N}_{a}^{k, \delta }(\lambda, b)$ if satisfies the inequality

Note that, $\mathcal{N}_{a}^{0, 0}(0, b) = \mathcal{S}^{\ast }(b)$ and $\mathcal{ N}_{a}^{0, 0}(1, b) = \mathcal{C}(b).$

By giving specific values to the parameters and $a, b, k, \delta$ and $\lambda,$ we obtain the following important subclasses studied by various authors in earlier works, for instance; $\mathcal{N}_{a}^{0, \delta }(0, b)$ and $\mathcal{N}_{a}^{0, \delta }(1, b)$ (Bulut ^[4]), $\mathcal{N} _{a}^{0, \delta }(\lambda, 1)$ (Mohapatra and Panigrahi ^[18]), $\mathcal{N}_{a}^{0, 0}(0, b) = \mathcal{S}^{\ast }(b)$ (Nars and Aouf ^[19]), $\mathcal{N}_{a}^{0, 0}(1, b) = \mathcal{C}(b)$ (Wiatrowski ^[28], Nars and Aouf ^[20]).

In this paper, we find an upper bound for the functional $\left\vert a_{3}-\eta a_{2}^{2}\right\vert$ for the functions $f$ belongs to the class $\mathcal{N}_{a}^{k, \delta }(\lambda, b)$.

2.   Main results

We denote by $\mathcal{P}$ a class of analytic function in $\mathcal{U}$ with $p(0) = 1$ and $\Re p(z) > 0.$ In order to derive our main results, we have to recall here the following lemma ^[25].

Lemma 1. Let $p\in \mathcal{P}$ with $p(z) = 1+c_{1}z+c_{2}z^{2}+....,$ then

If $\left\vert c_{1}\right\vert = 2$ then $p(z)\equiv p_{1}(z) = (1+\gamma _{1}z)/(1-\gamma _{1}z)$ with $\gamma _{1} = c_{1}/2.$ Conversely, if $p(z)\equiv p_{1}(z)$ for some $\left\vert \gamma _{1}\right\vert = 1,$ then $c_{1} = 2\gamma _{1}$ and $\left\vert c_{1}\right\vert = 2.$ Furthermore, we have

If $\left\vert c_{1}\right\vert < 2$ and $\left\vert c_{2}-\frac{c_{1}^{2}}{2} \right\vert \leq 2-\frac{\left\vert c_{1}\right\vert ^{2}}{2},$ then $p(z)\equiv p_{2}(z),$ where $p_{2}(z) = \frac{1+z\frac{\gamma _{2}z+\gamma _{1} }{1+\overline{\gamma }_{1}\gamma _{2}z}}{1-z\frac{\gamma _{2}z+\gamma _{1}}{ 1+\overline{\gamma }_{1}\gamma _{2}z}},$ and $\gamma _{1} = c_{1}/2,$ $\gamma _{2} = \frac{2c_{2}-c_{1}^{2}}{4-\left\vert c_{1}\right\vert ^{2}}.$ Conversely, if $p(z)\equiv p_{2}(z)$ for some $\left\vert \gamma _{1}\right\vert < 1$ and $\left\vert \gamma _{2}\right\vert = 1$ then $\gamma _{1} = c_{1}/2,$ $\gamma _{2} = \frac{2c_{2}-c_{1}^{2}}{4-\left\vert c_{1}\right\vert ^{2}}$ and $\left\vert c_{2}-\frac{c_{1}^{2}}{2}\right\vert \leq 2-\frac{\left\vert c_{1}\right\vert ^{2}}{2}.$

Now, consider the functional $\left\vert a_{3}-\eta a_{2}^{2}\right\vert$ for $b\in \mathbb{C} -\left\{ 0\right\}$ and $\eta \in \mathbb{C}.$

Theorem 1. Let $b\in \mathbb{C} -\left\{ 0\right\}$ and $0\leq \lambda \leq 1,$ $\eta \in \mathbb{C},$ $a > 0,$ $\delta \geq 0$. If $f$ of the form (1.1) is in $\mathcal{N} _{a}^{k, \delta }(\lambda, b),$ then

and

where $A_{1} = \left(\frac{a}{a+1}\right)$ and $A_{2} = \left(\frac{a}{a+2} \right).$ Consider the functions

where $p_{1},$ $p_{2}$ are given in Lemma 1. Equality in (2.1) holds if (2.4); in (2.2) if (2.4) and (2.5); for each $\eta$ in (2.3) if (2.4) and (2.5).

Proof. Denote $\mathcal{F}_{\lambda, a}^{k, \delta }(z) = (1-\lambda)\mathcal{D} ^{k}\mathcal{L} _{a}^{\delta }f(z)+\lambda z(\mathcal{D}^{k}\mathcal{L} _{a}^{\delta }f(z))^{\prime } = z+\beta _{2}z^{2}+\beta _{3}z^{3}+....,$ then

By the definition of the class $\mathcal{N}_{a}^{k, \delta }(\lambda, b),$ there exists $p\in \mathcal{P}$ such that $\frac{z\left(\mathcal{F} _{\lambda, a}^{k, \delta }(z)\right) ^{\prime }}{\mathcal{F}_{\lambda, a}^{k, \delta }(z)} = 1+b(p(z)-1),$ so that

which implies the equality

Equating the coefficients of both sides of the latter, we have

so that, on account of (2.6) and (2.7)

Taking into account (2.8) and Lemma 1, we obtain

and

Thus, we have

Then, with the aid of Lemma 1, we obtain

We now obtain sharpness of the estimates in (2.1), (2.2) and (2.3).

Firstly, in (2.1) the equality holds if $c_{1} = 2.$ Equivalently, we have $p(z)\equiv p_{1}(z) = (1+z)/(1-z).$ Therefore, the extremal functions in $\mathcal{N}_{a}^{k, \delta }(\lambda, b)$ is given by

Next, in (2.2), for first case, the equality holds if $c_{1} = c_{2} = 2.$ Therefore, the extremal functions in $\mathcal{N}_{a}^{k, \delta }(\lambda, b)$ is given by (2.11) and for the second case, the equality holds if $c_{1} = 0$, $c_{2} = 2.$ Equivalently, we have $p(z)\equiv p_{2}(z) = (1+z^{2})/(1-z^{2}).$ Therefore, the extremal functions in $\mathcal{N}_{a}^{k, \delta }(\lambda, b)$ is given by

Finally, in (2.3), the equality holds. Obtained extremal functions for (2.2) is also valid for (2.3).

Thus, the proof of Theorem 1 is completed.

Taking $k = 0$ and $\lambda = 0$ in Theorem 1, we have

Corollary 1. ^[4] Let $b\in \mathbb{C} -\left\{ 0\right\},$ $\eta \in \mathbb{C},$ $a > 0$ and $\delta \geq 0$. If $f$ of the form (1.1), is in $\mathcal{N}_{a}^{0, \delta }(0, b),$ then

and

where $A_{1} = \left(\frac{a}{a+1}\right)$ and $A_{2} = \left(\frac{a}{a+2} \right).$

If we choose $k = 0$ and $\lambda = 1$ in Theorem 1, we get

Corollary 2. ^[4] Let $b\in \mathbb{C} -\left\{ 0\right\},$ $\eta \in \mathbb{C},$ $a > 0$ and $\delta \geq 0$. If $f$ of the form (1.1), is in $\mathcal{N}_{a}^{0, \delta }(1, b),$ then

and

where $A_{1} = \left(\frac{a}{a+1}\right)$ and $A_{2} = \left(\frac{a}{a+2} \right).$

For $k = 0,$ $\delta = 0,$ $\lambda = 0$ and $b = 1$ in (2.3), we obtain

Corollary 3. ^[12] Let $\eta \in \mathbb{C}.$ If $f$ of the form (1.1), is in $\mathcal{S}^{\ast }(1),$ then

Taking $k = 0,$ $\delta = 0,$ $\lambda = 1$ and $b = 1$ in (2.3), we have

Corollary 4. ^[12] Let $\eta \in \mathbb{C}.$ If $f$ of the form (1.1), is in $\mathcal{C}(1),$ then

We next consider the case, when $\eta$ and $b$ are real. In this case, the following theorem holds.

Theorem 2. Let $b > 0$ and let $\mathcal{N}_{a}^{k, \delta }(\lambda, b).$ Then for $\eta \in \mathbb{R},$ we have

where $A_{1} = \left(\frac{a}{a+1}\right)$, $A_{2} = \left(\frac{a}{a+2} \right),$ $M_{1} = \frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{2(2\lambda +1)A_{2}^{\delta }3^{k}}$ and $M_{2} = \frac{(1+2b)(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{4b(2\lambda +1)A_{2}^{\delta }3^{k}}.$ For each $\eta,$ the equality holds for the functions given in equations (2.4) and (2.5).

Proof. First, let $\eta \leq \frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{ 2(2\lambda +1)A_{2}^{\delta }3^{k}}\leq \frac{(1+2b)(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{4b(2\lambda +1)A_{2}^{\delta }3^{k}}.$ In this case it follows from (2.8) and Lemma 1 that

Let, now $\frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{2(2\lambda +1)A_{2}^{\delta }3^{k}}\leq \eta \leq \frac{(1+2b)(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{4b(2\lambda +1)A_{2}^{\delta }3^{k}}.$ Then, using the estimations obtained above we arrived

Finally, if $\eta \geq \frac{(1+2b)(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{ 4b(2\lambda +1)A_{2}^{\delta }3^{k}},$ then

Thus, the proof of Theorem 2 is completed.

Taking $k = 0$ and $\lambda = 0$ in Theorem 2, we have

Corollary 5. ^[4] Let $b > 0$ and let $\mathcal{N}_{a}^{0, \delta }(0, b).$ Then for $\eta \in \mathbb{R},$ we have

where $A_{1} = \left(\frac{a}{a+1}\right)$ and $A_{2} = \left(\frac{a}{a+2} \right).$

Finally, considering the case of $b\in \mathbb{C} -\left\{ 0\right\}$ and $\eta \in \mathbb{R},$ we obtain

Theorem 3. Let $b\in \mathbb{C} -\left\{ 0\right\}$ and let $f\in \mathcal{N}_{a}^{k, \delta }(\lambda, b).$ For $\eta \in \mathbb{R},$ we have

where $A_{1} = \left(\frac{a}{a+1}\right)$ and $A_{2} = \left(\frac{a}{a+2} \right),$ $\left\vert b\right\vert = be^{i\theta },$ $k_{1} = \frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{2(2\lambda +1)A_{2}^{\delta }3^{k}}+\frac{ (\lambda +1)^{2}A_{1}^{2\delta }2^{2k}e^{i\theta }}{4\left\vert b\right\vert (2\lambda +1)A_{2}^{\delta }3^{k}},$ $l_{1} = \frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{4\left\vert b\right\vert (2\lambda +1)A_{2}^{\delta }3^{k}},$ $T_{1} = \Re (k_{1})-l_{1}(1-\left\vert \sin \theta \right\vert)$ and $T_{2} = \Re (k_{1})+l_{1}(1-\left\vert \sin \theta \right\vert).$ For each $\eta$ there is a function in $\mathcal{N} _{a}^{k, \delta }(\lambda, b)$ such that the equality holds.

Proof. From inequality (2.10), we may write

If we write $\left\vert b\right\vert = be^{i\theta }$ (or $b = \left\vert b\right\vert e^{^{-i\theta }}$)$,$ $k_{1} = \frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{2(2\lambda +1)A_{2}^{\delta }3^{k}}+\frac{ (\lambda +1)^{2}A_{1}^{2\delta }2^{2k}e^{i\theta }}{4\left\vert b\right\vert (2\lambda +1)A_{2}^{\delta }3^{k}}$ and $l_{1} = \frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{4\left\vert b\right\vert (2\lambda +1)A_{2}^{\delta }3^{k}}$ in the last inequality, we get

We consider the following cases for (2.13). Suppose $\eta \leq \Re (k_{1}).$ Then$\ $

Let $\eta \leq T_{1} = \Re (k_{1})-l_{1}(1-\left\vert \sin \theta \right\vert).$ On using Lemma 1 and $l_{1} = \frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{4\left\vert b\right\vert (2\lambda +1)A_{2}^{\delta }3^{k}}$ in inequality (2.14), we get

If we take $T_{1} = \Re (k_{1})-l_{1}(1-\left\vert \sin \theta \right\vert)\leq \eta \leq \Re (k_{1}),$ then (2.14) gives

Let$\ \eta \geq \Re (k_{1}).$ It then follows, from (2.13), that

Let $\ \eta \leq T_{2} = \Re (k_{1})+l_{1}(1-\left\vert \sin \theta \right\vert).$ On using (2.15) we obtain

Let $\eta \geq T_{2} = \Re (k_{1})+l_{1}(1-\left\vert \sin \theta \right\vert).$ Employing Lemma 1 together with $l_{1} = \frac{(\lambda +1)^{2}A_{1}^{2\delta }2^{2k}}{4\left\vert b\right\vert (2\lambda +1)A_{2}^{\delta }3^{k}}$ in equality (2.15), we obtain

Therefore, the proof is completed.

Corollary 6. If we take $a = 1$ in Theorems 2.1-2.3, we have the following results, respectively:

1. Let $b\in \mathbb{C} -\left\{ 0\right\}$ and $f\in \mathcal{N}_{1}^{k, \delta }(\lambda, b).$ Then, for $\eta \in \mathbb{C},$ we have

and

Equality holds for the cases $a = 1$ of (2.4) and (2.5) in Theorem 2.1.

2. Let $b > 0$ and $f\in \mathcal{N}_{1}^{k, \delta }(\lambda, b).$Then, for $\eta \in \mathbb{R},$ we have

where $Y_{1} = \frac{(\lambda +1)^{2}}{2(2\lambda +1)}(\frac{4}{3})^{^{k-\delta }}$ and $Y_{2} = \frac{(1+2b)(\lambda +1)^{2}}{4b(2\lambda +1)}(\frac{4}{3})^{^{k-\delta }}.$ For each $\eta,$ the equality holds for the cases $a = 1$ of (2.4) and (2.5).

3. Let $b\in \mathbb{C} -\left\{ 0\right\}$ and $f\in \mathcal{N}_{1}^{k, \delta }(\lambda, b).$ Then, for $\eta \in \mathbb{R},$ we have

where $\left\vert b\right\vert = be^{i\theta },$ $k_{1} = \frac{(\lambda +1)^{2} }{2(2\lambda +1)}(\frac{4}{3})^{^{k-\delta }}-(\frac{4}{3})^{^{k-\delta }} \frac{(\lambda +1)^{2}e^{i\theta }}{4\left\vert b\right\vert (2\lambda +1)},$ $l_{1} = (\frac{4}{3})^{^{k-\delta }}\frac{(\lambda +1)^{2}}{4\left\vert b\right\vert (2\lambda +1)},$ $T_{1} = \Re (k_{1})-l_{1}(1-\left\vert \sin \theta \right\vert)$ and $T_{2} = \Re (k_{1})+l_{1}(1-\left\vert \sin \theta \right\vert).$ For each $\eta$ there is a function in $\mathcal{N} _{1}^{k, \delta }(\lambda, b)$ such that the equality holds.

Acknowledgments

The authors would like to thank the anonymous referees for the useful improvements suggested.

Conflict of interest

All authors declare no conflicts of interest.