Coefficient Bounds for Certain Analytic Functions

Abstract

Let $\varphi$ be an analytic function with the positive real parts, $\varphi (0)= 1$ and $\varphi ^{\prime }(0) >0$. Let $f(z)= z+ a_2 z^2 +a_3 z^3 +{\cdots }$ be an analytic function satisfying the subordination $\alpha f^{\prime }(z) + (1-\alpha )zf^{\prime }(z)/f(z) \prec \varphi (z)$, $(f^{\prime }(z))^{\alpha } (zf^{\prime }(z)/f(z))^{(1-\alpha )} \prec \varphi (z)$, $(f^{\prime }(z))^{\alpha } (1+ zf^{\prime \prime }(z)/f^{\prime }(z))^{(1-\alpha )} \prec \varphi (z)$, $(f(z)/z)^{\alpha } (zf^{\prime }(z)/f(z))^{(1-\alpha )} \prec \varphi (z)$, or $(f(z)/z)^{\alpha } (1+ zf^{\prime \prime }(z)/f^{\prime }(z))^{(1-\alpha )} \prec \varphi (z)$. For these functions, the bounds for the second Hankel determinant $a_2a_4-a_3^2$ as well as the Fekete–Szegö coefficient functional are obtained. Our results include some previously known results.

1 Introduction

Let ${\mathcal {A}}$ be the class of all normalized analytic functions of the form $f(z)=z+\sum _{n=2}^\infty a_nz^n$ in the open unit disk ${\mathbb {D}}:= \{z \in {\mathbb {C}}: |z| < 1\}$, and ${\mathcal {S}}$ be the subclass of ${\mathcal {A}}$ consisting of univalent functions in ${\mathbb {D}}$. The qth Hankel determinant (denoted by $H_q(n)$) for $q=1, 2, \ldots$ and $n=1, 2, 3, \ldots$ of the function f is given by

$$\begin{aligned} H_q(n) := \left| \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_n &{} a_{n+1} &{} \cdots &{} a_{n+q-1} \\ a_{n+1} &{} a_{n+2} &{} \cdots &{} a_{n+q}\\ \vdots &{} \vdots &{} &{} \vdots \\ a_{n+q-1}&{} a_{n+q} &{} \cdots &{} a_{n+2q-2} \end{array} \right| . \end{aligned}$$

In 1996, Pommerenke [36] investigated the Hankel determinant of areally p-valent functions, univalent functions and starlike functions. Further, the Hankel determinants of areally mean p-valent functions were investigated in [29–31]. Noor studied the bounds on Hankel determinant of close-to-convex functions in [32–34]. Pommerenke [37] also established that Hankel determinant of univalent functions satisfying the following relation: $|H_q(n)|< K n^{-(\frac{1}{2}+\beta )q + \frac{3}{2}}$ $(n= 1, 2, \ldots ; q=1, 2, \ldots ),$ where $\beta > 1/40{,}000$ and K depend on q. Later, Hayman [14] proved that $|H_2(n)|< A n^{1/2} (n=1, 2, \ldots$; A is an absolute constant) for areally mean univalent functions. In 1986, Elhosh [9, 10] investigated the Hankel determinant for univalent functions with a positive Hayman index $\alpha$ and k-fold symmetric and close-to-convex functions. The second Hankel determinant $H_2(2) := a_2a_4-a_3^2$ for the class of functions derivative of which has positive real part, the classes of starlike and convex functions, close-to-starlike, and close-to-convex functions with respect to symmetric points have been studied in [13, 16], respectively. Lee et al. [26] obtained bounds on second Hankel determinant for the classes of Ma–Minda starlike and convex functions with respect to $\varphi$ and two other similar subclasses. One may refer to the survey given by Liu et al. [27] for the other works done in the research of Hankel determinant for univalent functions. They also obtained the second Hankel determinant for some other subclasses of analytic functions. Hankel determinants have been studied by several other authors for various classes of analytic functions and can be referred to in [1, 4, 23].

In 1933, Fekete and Szegö proved that

$$\begin{aligned} |a_3 -\mu a_2^2|\le \left\{ \begin{array}{l@{\quad }l} 4\mu -3, &{}(\mu \ge 1); \\ 1+\hbox {exp}\left( \frac{- 2\mu }{ 1- \mu }\right) , &{} {\left( 0\le \mu \le 1\right) ;} \\ 3-4\mu , &{} {(\mu \le 0).} \end{array} \right. \end{aligned}$$

holds for functions $f \in {\mathcal {S}}$, and the result is sharp. Keogh and Merkes [17], in 1969, obtained the sharp upper bound of the Fekete–Szegö functional $|a_3- \mu a_2^2|$ for functions in some subclasses of ${\mathcal {S}}$. The Fekete–Szegö functional problem for close-to-convex functions was investigated among others by Koepf [21], Kim et al. [6, 20] and Cho et al. [5]. The problem for starlike and convex functions were investigated in a more general settings by Ma and Minda [28]. For other general classes of p-valent functions, the Fekete–Szegö functional problem was investigated by Ali et al. [2] and Ali et al. [3]. For classes defined by quasi-subordination, see Mohd and Darus [12]. Jakubowski and Zyskowska [15] obtained the estimate for $|a_2- c a_2^2| + c|a_2|^n$ for $c \in {\mathbb {R}}$, $f \in {\mathcal {S}}$. Kiepiela et al. [19] obtained bounds for certain combinations of initial coefficients of bounded functions; these results were used later for estimating fourth coefficients of many classes [2, 18]. Other results related to this functional can be seen in [2, 3, 7, 22, 24, 25, 35].

An analytic function f is said to be subordinate to F, written $f\prec F$ or $f(z)\prec F (z),\quad (z\in {\mathbb {D}})$ if there exists an analytic function $w: {\mathbb {D}}\rightarrow {\mathbb {D}}$ satisfying $w(0)=0$ and $f(z)=F(w(z))$ in ${\mathbb {D}}$. If F is univalent in ${\mathbb {D}}$, then $f(z)\prec F (z)$ if and only if $f(0)=F(0)$ and $f({\mathbb {D}})\subseteq F({\mathbb {D}})$. Let $\varphi$ be a univalent function with positive real part, $\varphi (0)= 1$ and $\varphi ^{\prime }(0) >0$. In this paper, we determine the bounds on the second Hankel determinant $H_2(2)$ and the Fekete–Szegö functional for the functions f for which $\alpha f^{\prime }(z) + (1-\alpha ){z f^{\prime }(z)}/{f(z)}$, $(f^{\prime }(z))^{\alpha } ({z f^{\prime }(z)}/{f(z)})^{ (1-\alpha )}$, $(f^{\prime }(z))^{\alpha } (1 + {z f^{\prime \prime }(z)}/{f^{\prime }(z)})^{ (1-\alpha )}$, $({f(z)}/{z})^{\alpha } ({z f^{\prime }(z)}/{f(z)})^{(1-\alpha )}$ or $({f(z)}/{z})^{\alpha } (1 + {z f^{\prime \prime }(z)}/{f^{\prime }(z)})^{(1-\alpha )}$ is subordinate to $\varphi$. Our results include some previously known results.

2 Main Results

Let $\varphi : {\mathbb {D}}\rightarrow {\mathbb {C}}$ be a function with positive real part with

$$\begin{aligned} \varphi (z)= 1+ B_1 z +B_2 z^2 +B_3 z^3 +\cdots \quad \left( B_1>0; B_1, B_2, B_3\in {\mathbb {R}}\right) . \end{aligned}$$

(2.1)

For $0\le \alpha \le 1$, the class ${\mathcal {V}}_{\alpha }(\varphi )$ consists of functions $f\in {\mathcal {A}}$ satisfying the following subordination:

$$\begin{aligned} \alpha f^{\prime }(z) + (1-\alpha )\dfrac{z f^{\prime }(z)}{f(z)} \prec \varphi (z). \end{aligned}$$

Note that

$$\begin{aligned} {\mathcal {S}}^*(\varphi )= {\mathcal {V}}_{0}(\varphi )= \left\{ f \in {\mathcal {A}}: \dfrac{z f^{\prime }(z)}{f(z)} \prec \varphi (z)\right\} \end{aligned}$$

is the class of Ma–Minda starlike functions and

$$\begin{aligned} {\mathcal {R}}(\varphi )= {\mathcal {V}}_1(\varphi )= \left\{ f \in {\mathcal {A}}: f^{\prime }(z) \prec \varphi (z)\right\} \end{aligned}$$

is a subclass of close-to-convex function. Thus, our class provides a continuous passage from a subclass of starlike functions to the subclass of close-to-convex functions when $\alpha$ varies from 0 to 1.

For functions in the class ${\mathcal {V}}_{\alpha }(\varphi )$, we have the following result:

Theorem 2.1

Let the function $f \in {\mathcal {V}}_{\alpha }(\varphi )$ be given by $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots$.

• (1) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} {\alpha (1-\alpha )}B_1^2 + 2 (1+\alpha ) |B_2|\, \le (1+\alpha )\left( \alpha ^2 + 4\alpha + 2\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\quad \Bigl |{(\alpha -1)\left( 3\alpha ^2 + 5\alpha + 1\right) }B_1^4 - (1+\alpha )^4(3+\alpha ) {B_2^2} + (1+\alpha )^3(2 + \alpha )^2 B_1 B_3 \\&\quad +{ \alpha (1-\alpha )(1+\alpha )^2}B_1^2 B_2\Bigr | \le (1+\alpha )^4(3+\alpha )B_1^2, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|\le \dfrac{B_1^2}{(2 + \alpha )^2}. \end{aligned}$$

• (2) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} {\alpha (1-\alpha )}B_1^2 + 2{(1+\alpha )}|B_2|\ge (1+\alpha )\left( \alpha ^2 + 4\alpha + 2\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&2\Bigl |{(\alpha -1)\left( 3\alpha ^2 + 5\alpha + 1\right) }B_1^4 - (1+\alpha )^4(3+\alpha ){B_2^2} + (1+\alpha )^3(2 + \alpha )^2 B_1 B_3\\&\quad +\, { \alpha (1-\alpha )(1+\alpha )^2}B_1^2 B_2 \Bigr | \ge (1+\alpha )^2 \left[ \alpha (1-\alpha )B_1^3 + 2(1+\alpha ) |B_2|B_1\right. \\&\quad \left. +\,(1+\alpha )(2+\alpha )^2 B_1^2\right] , \end{aligned}$$

  or the conditions

  $$\begin{aligned} {\alpha (1-\alpha )}B_1^2 + 2 (1+\alpha )|B_2|\le (1+\alpha )\left( \alpha ^2 + 4\alpha + 2\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\quad \Bigl |{(\alpha -1)\left( 3\alpha ^2 + 5\alpha + 1\right) }B_1^4 - (1+\alpha )^4(3+\alpha ) {B_2^2} + (1+\alpha )^3(\alpha + 2)^2 B_1 B_3 \\&\quad \quad +{ \alpha (1-\alpha )(1+\alpha )^2}B_1^2 B_2\Bigr | \ge (1+\alpha )^4(3+\alpha )B_1^2, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|&\le \dfrac{1}{(1+\alpha )^4(2+\alpha )^2(3+\alpha )} \Bigl |{(\alpha -1)\left( 3\alpha ^2 + 5\alpha + 1\right) }B_1^4\\&\quad -\, {(1+\alpha )^4 (3+\alpha )}B_2^2+ {(1+\alpha )^3(2+\alpha )^2}B_1B_3 \\&\quad + {\alpha (1-\alpha )(1+\alpha )^2}B_1^2B_2\Bigr |. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} {\alpha (1-\alpha )}B_1^2 + 2(1+\alpha )|B_2|\, > (1+\alpha )\left( \alpha ^2 + 4\alpha + 2\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&2\Bigl |{(\alpha -1)\left( 3\alpha ^2 + 5\alpha + 1\right) }B_1^4 - (1+\alpha )^4(3+\alpha ){B_2^2}+ (1+\alpha )^3(2 + \alpha )^2 B_1 B_3 \\&\quad + {\alpha (1-\alpha )(1+\alpha )^2}B_1^2 B_2\Bigr | \le (1+\alpha )^2 \left[ \alpha (1-\alpha )B_1^3 + 2(1+\alpha ) |B_2|B_1\right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. +\,(1+\alpha )(2+\alpha )^2 B_1^2\right] , \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2| \le \dfrac{B_1^2 M}{4 (2+\alpha )^2(3+\alpha )N}, \end{aligned}$$

  where

  $$\begin{aligned} M&= 4(3+\alpha )\Big | {(\alpha -1)\left( 3\alpha ^2+5\alpha +1\right) } B_1^4 - {(1+\alpha )^4(3+\alpha )} B_2^2 \\&\qquad +\, {(1+\alpha )^3(2+\alpha )^2} B_1 B_3 + {\alpha (1-\alpha )(1+\alpha )^2}B_1^2 B_2 \Big |\\&\qquad +\Big [ 2(1+\alpha )|B_2| - (1+\alpha )\left( \alpha ^2 + 4\alpha + 2\right) B_1 + {\alpha }(1-\alpha )B_1^2 \Big ]^2 \end{aligned}$$

  and

  $$\begin{aligned} N&=\Big |{(\alpha -1)(3\alpha ^2 +5 \alpha +1)}B_1^4 - {(1+\alpha )^4(3+\alpha )}B_2^2 + {(1+\alpha )^3(2+\alpha )^2}B_1 B_3\\&\quad +\, {\alpha (1-\alpha )(1+\alpha )^2}B_1^2 B_2 \Big | - {\alpha (1-\alpha )(1+\alpha )^2}B_1^3- (1+\alpha )^3 \Big (2B_1 |B_2| +B_1^2\Big ). \end{aligned}$$

Remark 2.2

When $\varphi (z)= (1+(1-2\lambda )z)/(1-z), (0\le \lambda < 1)$, $\sqrt{1+z}$, $1 + 2/ \pi ^2 (\log ((1+\sqrt{z})/(1- \sqrt{z})))^2$ and $((1+z)/(1-z))^\beta$, $0< \beta \le 1$, the class ${\mathcal {S}}^*(\varphi )$ becomes the class ${\mathcal {S}}^*(\gamma )$ of starlike functions of order $\gamma$, the class ${\mathcal {S}}^*_{L}$ of lemniscate starlike functions, the class ${\mathcal {S}}^*_{\mathrm{p}}$ of parabolic starlike functions and the class ${\mathcal {S}}^*_{\beta }$ of strongly starlike functions of order $\beta$, respectively. When $\alpha =1$ and $\varphi = (1+z)/(1-z)$, Theorem 2.1 reduces to [16, Theorem 3.1]. When $\alpha =0$, Theorem 2.1 reduces to [26, Theorem 1].

In particular, we get the following corollary:

Corollary 2.3

[26, Theorem 1]

• (1) If $f \in {\mathcal {S}}^*(\gamma )$, then $|a_2 a_4 - a_3^2|\le (1-\gamma )^2$.

• (2) If $f \in {\mathcal {S}}^*_{L}$, then $|a_2 a_4 - a_3^2|\le 1/16 = 0.0625$.

• (3) If $f \in {\mathcal {S}}^*_{p}$, then $|a_2 a_4 - a_3^2|\le 16/\pi ^4 \approx 0.164255$.

• (4) If $f \in {\mathcal {S}}^*_{\beta }$, then $|a_2 a_4 - a_3^2|\le \beta ^2$.

Theorem 2.1 is proved by expressing the coefficient of the function f in terms of the coefficient of a function with positive real part. Recall that the class ${\mathcal {P}}$ of functions with positive real part consists of all analytic functions $p(z)= 1 + \sum _{n=0}^{\infty }c_n z^n$ with ${\text {Re}} p(z) >0$ for $z \in {\mathbb {D}}$. Let $\varOmega$ be the class of all analytic functions $w:{\mathbb {D}}\rightarrow {\mathbb {D}}$ of the form $w(z)= w_1 z + w_2 z^2 + \cdots$. To prove our results, we need the following lemmas:

Lemma 2.4

[8] If the function

$$\begin{aligned} p(z)= 1+c_1 z +c_2 z^2 +c_3 z^3 +\cdots \end{aligned}$$

(2.2)

is in ${\mathcal {P}}$, then the following sharp estimate holds: $|c_n|\le 2 \quad (n=1, 2, 3, \ldots )$.

Lemma 2.5

[11] If the function given by (2.2) is in ${\mathcal {P}}$, then,

$$\begin{aligned}&\displaystyle 2c_2 =c_1^2 + x\left( 4-c_1^2\right) , \end{aligned}$$

(2.3)

$$\begin{aligned}&\displaystyle 4c_3 = c_1^3 + 2\left( 4-c_1^2\right) c_1 x - c_1 \left( 4- c_1^2\right) x^2 +2 \left( 4-c_1^2\right) \left( 1-|x|^2\right) y, \end{aligned}$$

(2.4)

for some x, y with $|x|\le 1$ and $|y|\le 1$.

Lemma 2.6

[28] Let $p \in {\mathcal {P}}$ be given by (2.2). Then,

$$\begin{aligned} |c_2-\nu c_1^2| \le \left\{ \begin{array}{l@{\quad }l} -4\nu + 2, &{} \text {if }\nu \le 0; \\ 2, &{}\text {if }0 \le \nu \le 1; \\ 4 \nu -2, &{}\text {if }\nu \ge 1. \end{array} \right. \end{aligned}$$

Proof of Theorem 2.1

Since $f \in {\mathcal {V}}_{\alpha }(\varphi ),$ there is an analytic function $w(z)= w_1 z + w_2 z^2 + \cdots \in \varOmega$, such that

$$\begin{aligned} \alpha f^{\prime }(z)+(1-\alpha ) \dfrac{z f^{\prime }(z)}{f(z)}=\varphi (w(z)). \end{aligned}$$

(2.5)

Define $p_1(z)$ by

$$\begin{aligned} p_1(z)=\dfrac{1+ w(z)}{1-w(z)}=1 + c_1 z + c_2 z^2 + \cdots . \end{aligned}$$

Then

$$\begin{aligned} w(z)= \dfrac{p_1(z)-1}{p_1(z)+1}= \dfrac{1}{2}\left( c_1 z +\left( c_2-\dfrac{c_1^2}{2}\right) z^2+\cdots \right) . \end{aligned}$$

Clearly $p_1$ is analytic in ${\mathbb {D}}$ with $p_1(0)=1$ and $p_1\in {\mathcal {P}}$. Since $\varphi (z)= 1 + B_1 z + B_2 z^2 + B_3 z^3+ \cdots$, we get

$$\begin{aligned} \varphi \left( \dfrac{p_1(z)-1}{p_1(z)+1}\right) =1 + \dfrac{1}{2}B_1 c_1z + \left( \dfrac{1}{2}B_1\left( c_2-\dfrac{c_1^2}{2}\right) +\dfrac{1}{4}B_2 c_1^2\right) z^2 +\cdots . \end{aligned}$$

(2.6)

Also, the Taylor series expansion of f gives

$$\begin{aligned} \begin{aligned} \alpha f^{\prime }(z)+(1-\alpha )z \dfrac{f^{\prime }(z)}{f(z)}&= 1+ a_2(1+\alpha )z + \Big ((2+\alpha )a_3 - (1-\alpha )a_2^2 \Big )z^2 \\&\quad +\,\Big ( (3+\alpha )a_4 - (1-\alpha )\left( 3a_2a_3 - a_2^3\right) \Big )z^3 +\cdots . \end{aligned} \end{aligned}$$

(2.7)

Then from (2.5), (2.6), and (2.7), we get

$$\begin{aligned} a_2&= \dfrac{B_1 c_1}{2(1+\alpha )}. \end{aligned}$$

(2.8)

$$\begin{aligned} a_3&=\dfrac{1}{4(2+\alpha )} \left[ 2B_1 c_2+\left( \dfrac{(1-\alpha )}{(1+\alpha )^2}B_1^2 +(B_2 - B_1)\right) c_1^2 \right] . \end{aligned}$$

(2.9)

$$\begin{aligned} a_4&=\dfrac{1}{8(3+\alpha )} \Bigg [4B_1 c_3+ \left( B_1-\dfrac{3(1-\alpha )}{(1+\alpha )(2+\alpha )}B_1^2 +\dfrac{(1-\alpha )(1-4\alpha )}{(1+\alpha )^3 (2+\alpha )}B_1^3 - 2B_2\right. \nonumber \\&\left. \quad +\, \dfrac{3(1-\alpha )}{(1+\alpha )(2+\alpha )}B_1B_2 + B_3\right) c_1^3 +\left( -4B_1 {+} \dfrac{6(1-\alpha )}{(1+\alpha )(2+\alpha )}B_1^2 + 4B_2\right) c_1 c_2 \Bigg ] \end{aligned}$$

(2.10)

Therefore,

$$\begin{aligned} a_2 a_4 - a_3^2&=\dfrac{1}{16}B_1\left[ \left\{ \dfrac{(\alpha -1)\left( 3\alpha ^2+5\alpha +1\right) }{(1+\alpha )^4(2+\alpha )^2(3+\alpha )}B_1^3\right. \right. \\&\quad +\,\dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )}B_1(B_2 - B_1) \\&\quad \left. +\, \dfrac{1}{(2+\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{1}{(1+\alpha )(3+\alpha )}B_3 + \dfrac{1}{(1+\alpha )(2+\alpha )^2(3+\alpha )}(B_1 - 2B_2) \right\} c_1^4 \\&\quad +\, \left\{ \dfrac{4}{(1+\alpha )(2+\alpha )^2(3+\alpha )}(B_2 - B_1) +\dfrac{2\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )} B_1^2 \right\} c_2 c_1^2\\&\quad \left. -\, \dfrac{4}{(2+\alpha )^2}B_1 c_2^2 +\dfrac{4}{(1+\alpha )(3+\alpha )}B_1 c_1 c_3\right] . \end{aligned}$$

Since the function $p(\hbox {e}^{i \theta }z)$ $(\theta \in {\mathbb {R}})$ is in the class ${\mathcal {P}}$ for any $p \in {\mathcal {P}}$, without loss of generality, we can assume that $c_1=c>0$. Substituting the values of $c_2$ and $c_3$ from (2.3) and (2.4) in the above expression, we get

$$\begin{aligned} |a_2 a_4 - a_3^2|&= \dfrac{1}{16}B_1\bigg |\bigg \{\dfrac{(\alpha -1)\left( 3\alpha ^2+5\alpha +1\right) }{(1+\alpha )^4(2+\alpha )^2(3+\alpha )}B_1^3 + \dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )}B_1B_2\\&\quad -\, \dfrac{1}{(2+\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{1}{(1+\alpha )(3+\alpha )}B_3 \bigg \}c^4 + \bigg \{ \dfrac{2}{(1+\alpha )(2+\alpha )^2(3+\alpha )}B_2 \\&\quad +\, \dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )}B_1^2 \bigg \} c^2 \left( 4-c^2\right) x - \bigg \{ \dfrac{1}{(1+\alpha )(2+\alpha )^2(3+\alpha )}c^2 \\&\quad + \,\dfrac{4}{(2+\alpha )^2} \bigg \} B_1 \left( 4-c^2\right) x^2 + \dfrac{2}{(1+\alpha )(3+\alpha )}B_1 c\left( 4-c^2\right) \left( 1-|x|^2\right) y \bigg |. \end{aligned}$$

Replacing |x| by $\mu$ and by making use of the triangle inequality and the fact that $|y|\le 1$ in the above expression, we get

$$\begin{aligned} |a_2 a_4 - a_3^2|&\le \dfrac{1}{16}B_1 \bigg [\bigg |\bigg \{\dfrac{(\alpha -1)\left( 3\alpha ^2+5\alpha +1\right) }{(1+\alpha )^4(2+\alpha )^2(3+\alpha )}B_1^3 - \dfrac{1}{(2+\alpha )^2}\dfrac{B_2^2}{B_1}\nonumber \\&\quad +\, \dfrac{1}{(1+\alpha )(3+\alpha )}B_3 +\dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )}B_1B_2 \bigg \}\bigg |c^4 \nonumber \\&\quad +\, \dfrac{2c}{(1+\alpha )(3+\alpha )}B_1 \left( 4-c^2\right) +\bigg \{ \dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )}B_1^2 \nonumber \\&\quad +\,\dfrac{2}{(1+\alpha )(2+\alpha )^2(3+\alpha )}|B_2|\bigg \}c^2\left( 4-c^2\right) \,\mu \nonumber \\&\quad +\bigg \{\dfrac{c^2}{(1+\alpha )(2+\alpha )^2(3+\alpha )} -\dfrac{2c }{(1+\alpha )(3+\alpha )} + \dfrac{4}{(2+\alpha )^2}\bigg \}\nonumber \\&\quad \times B_1\left( 4-c^2\right) \mu ^2 \bigg ] = F(c, \mu ). \end{aligned}$$

(2.11)

We shall now maximize $F(c, \mu )$ for $(c, \mu )\in [0, 2]\times [0, 1]$. Differentiating $F(c, \mu )$ in (2.11) partially with respect to $\mu$, we get

$$\begin{aligned} \dfrac{\partial {F}}{\partial {\mu }}&= \dfrac{1}{16}B_1\left[ \dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )} c^2\left( 4-c^2\right) B_1^2\right. \nonumber \\&\quad +\,\dfrac{2}{(1+\alpha )(2+\alpha )^2(3+\alpha )} c^2\left( 4-c^2\right) |B_2|\nonumber \\&\quad +\, 2\mu B_1 \left( 4-c^2\right) \left( \dfrac{1}{(1+\alpha )(2+\alpha )^2(3+\alpha )}c^2 \right. \nonumber \\&\left. \left. \quad -\, \dfrac{2}{(1+\alpha )(3+\alpha )}c + \dfrac{4}{(2+\alpha )^2}\right) \right] . \end{aligned}$$

(2.12)

For $0<\mu <1$, and for any fixed $c\in [0, 2]$, we observe that $\partial {F}/\partial {\mu }>0$. Thus, $F(c, \mu )$ is an increasing function of $\mu$, and for $c\in [0, 2]$, $F(c, \mu )$ has a maximum value at $\mu =1$. Thus, we have

$$\begin{aligned} \text {max} F(c, \mu )&= F(c, 1)= G(c). \end{aligned}$$

(2.13)

The Eqs. (2.11) and (2.13), upon a little simplification, yield

$$\begin{aligned} G(c)&= \dfrac{1}{16}B_1\bigg [\bigg \{\bigg | \dfrac{(\alpha -1)\left( 3\alpha ^2+5\alpha +1\right) }{(1+\alpha )^4(2+\alpha )^2(3+\alpha )}B_1^3 - \dfrac{1}{(2+\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{1}{(1+\alpha )(3+\alpha )}B_3\nonumber \\&\quad +\, \dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )}B_1B_2\bigg | - \dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )} B_1^2 \nonumber \\&\quad -\, \dfrac{1}{(1+\alpha )(2+\alpha )^2(3+\alpha )}\big ( 2|B_2|+ B_1\big ) \bigg \} c^4 + 4 \bigg \{\dfrac{\alpha (1-\alpha )}{(1+\alpha )^2(2+\alpha )^2(3+\alpha )}B_1^2 \nonumber \\&\quad +\, \dfrac{2}{(1+\alpha )(2+\alpha )^2(3+\alpha )} |B_2|- \dfrac{\alpha ^2 +4\alpha +2}{(1+\alpha )(2+\alpha )^2(3+\alpha )}B_1\bigg \}c^2 \nonumber \\&\quad +\, \dfrac{16}{(2+\alpha )^2}B_1\bigg ]\nonumber \\&= \dfrac{B_1}{16} \left( P c^4 + Q c^2 + R\right) , \end{aligned}$$

(2.14)

where

$$\begin{aligned} P&= {(1+\alpha )^{-4}(2+\alpha )^{-2}(3+\alpha )^{-1}}\Big (\Big |{(\alpha -1)\left( 3\alpha ^2+5\alpha +1\right) }B_1^3 \nonumber \\&\quad -\, {(1+\alpha )^4(3+\alpha )}\dfrac{B_2^2}{B_1} + {(1+\alpha )^3(2+\alpha )^2}B_3 + {\alpha (1-\alpha )(1+\alpha )^2}B_1B_2\Big |\nonumber \\&\quad -\, {\alpha (1-\alpha )(1+\alpha )^2} B_1^2 - {(1+\alpha )^3} (2|B_2|+ B_1)\Big ), \end{aligned}$$

(2.15)

$$\begin{aligned} Q&= {4(1+\alpha )^{-2}(2+\alpha )^{-2}(3+\alpha )^{-1}} \Big ({\alpha (1-\alpha )}B_1^2 + {2(1+\alpha )} |B_2| \nonumber \\&\quad -\, {(1+\alpha )(\alpha ^2+4\alpha +2 )}B_1\Big ), \end{aligned}$$

(2.16)

and

$$\begin{aligned} R&= 16(2+\alpha )^{-2}B_1. \end{aligned}$$

(2.17)

We know that

$$\begin{aligned} \mathop {\max }\limits _{0 \le t \le 4} \left( Pt^2 + Qt + R\right) = \left\{ \begin{array}{ll} R, &{} \quad Q\le 0, P\le -\frac{Q}{4}; \\ 16P + 4Q + R, &{}\quad Q\ge 0, P\ge -\frac{Q}{8}\quad \text {or}\quad Q\le 0, P\ge -\frac{Q}{4}; \\ \frac{4PR -Q^2}{4P}, &{}\quad Q > 0, P\le -\frac{Q}{8}. \end{array} \right. \end{aligned}$$

(2.18)

Thus, we have, from (2.14),

$$\begin{aligned} |a_2 a_4 - a_3^2| \le \dfrac{B_1}{16}\left\{ \begin{array}{l@{\quad }l} R, &{} Q\le 0, P\le -\frac{Q}{4}; \\ 16P + 4Q + R, &{} Q\ge 0, P\ge -\frac{Q}{8}\quad \text {or}\quad Q\le 0, P\ge -\frac{Q}{4}; \\ \frac{4PR -Q^2}{4P}, &{} Q > 0, P\le -\frac{Q}{8}. \end{array} \right. \end{aligned}$$

where P, Q,  and R are given by (2.15), (2.16), and (2.17), respectively. A simple computation will give the results stated in the theorem. $\square$

Our next theorem gives the Fekete–Szegö inequality for functions in the class ${\mathcal {V}}_{\alpha }(\varphi )$.

Theorem 2.7

Let $\varphi$ be defined as in (2.1) and let the function $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots \in {\mathcal {V}}_{\alpha }(\varphi )$. Then we have the following:

• (1) If $B_1$, $B_2$ and $\lambda$ satisfy the condition

  $$\begin{aligned} (2+\alpha )B_1^2 \lambda \le (1-\alpha )B_1^2 + (B_2-B_1)(1+\alpha )^2, \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \lambda a_2^2|\le \dfrac{1}{(2+\alpha )}\left[ B_2 +\dfrac{(1-\alpha )}{(1+\alpha )^2}B_1^2 -\dfrac{(2+\alpha )\lambda }{(1+\alpha )^2}B_1^2 \right] . \end{aligned}$$

• (2) If $B_1$, $B_2$ and $\lambda$ satisfy the condition

  $$\begin{aligned}&(1-\alpha )B_1^2 + (B_2-B_1)(1+\alpha )^2\le (2+\alpha ) B_1^2 \lambda \\&\quad \le (1-\alpha )B_1^2 + (B_2+B_1)(1+\alpha )^2, \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \lambda a_2^2|\le \dfrac{B_1}{2+\alpha }. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $\lambda$ satisfy the condition

  $$\begin{aligned} (1-\alpha )B_1^2 + (B_2 + B_1)(1+\alpha )^2 \le (2+\alpha ) B_1^2 \lambda , \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \lambda a_2^2| \le \dfrac{1}{2+\alpha }\left[ -B_2 - \dfrac{1-\alpha }{(1+\alpha )^2}B_1^2 + \dfrac{(2+\alpha )\lambda }{(1+\alpha )^2}B_1^2\right] . \end{aligned}$$

Proof

By using (2.8) and (2.9), we get

$$\begin{aligned} a_3- \lambda a_2^2&=\dfrac{1}{2+\alpha }\left[ \dfrac{c_2 B_1}{2}-\dfrac{c_1^2 B_1}{4} + \dfrac{c_1^2 B_2}{4} - \dfrac{(1-\alpha )c_1^2 B_1^2}{4(1+\alpha )^2} - \dfrac{\lambda B_1^2 c_1^2}{4(1+\alpha )^2}\right] \\&= \dfrac{B_1}{2(2+\alpha )}\big (c_2-\nu c_1^2\big ), \end{aligned}$$

where

$$\begin{aligned} \nu = \dfrac{1}{2}\left( 1 - \dfrac{B_2}{B_1} - \dfrac{(1-\alpha ) B_1}{(1+\alpha )^2}+\dfrac{(2+\alpha )B_1}{(1+\alpha )^2}\lambda \right) . \end{aligned}$$

Now, by using Lemma 2.6, we get the desired result. $\square$

Remark 2.8

Bounds for the second and the third coefficients for f in ${\mathcal {V}}_{\alpha }(\varphi )$ can be directly obtained from Eq. (2.8) and Theorem 2.7 as follows:

$$\begin{aligned} |a_2|&\le \dfrac{B_1}{1+ \alpha }, \end{aligned}$$

and

$$\begin{aligned} |a_3|&\le \left\{ \begin{array}{l@{\quad }l} \dfrac{B_2 + \frac{1-\alpha }{(1+\alpha )^2}B_1^2 }{2+\alpha }, &{} (1-\alpha )B_1^2 + (B_2 - B_1)(1+\alpha )^2 \ge 0; \\ \dfrac{B_1}{2+\alpha }, &{} (1-\alpha )B_1^2 + (B_2 - B_1)(1+\alpha )^2 \le 0 \quad \text {or} \\ &{} (1-\alpha )B_1^2 + (B_2 + B_1)(1+\alpha )^2 \ge 0;\\ \dfrac{-B_2 - \frac{1-\alpha }{(1+\alpha )^2}B_1^2}{2+\alpha }, &{}(1-\alpha )B_1^2 + (B_2 + B_1)(1+\alpha )^2 \le 0. \end{array} \right. \end{aligned}$$

Let $\varphi : {\mathbb {D}}\rightarrow {\mathbb {C}}$ be an analytic function given by (2.1). For $0\le \alpha \le 1$, the class ${\mathcal {M}}_{\alpha }(\varphi )$ consists of functions $f\in {\mathcal {A}}$ satisfying the following subordination

$$\begin{aligned} \left( f^{\prime }(z)\right) ^\alpha \Big (\dfrac{z f^{\prime }(z)}{f(z)}\Big )^{1-\alpha } \prec \varphi (z). \end{aligned}$$

We see that ${\mathcal {M}}_{0}(\varphi )={\mathcal {S}}^*(\varphi )$ is the class of Ma–Minda starlike functions and

$$\begin{aligned} {\mathcal {M}}_1(\varphi )={\mathcal {R}}(\varphi )= \left\{ f \in {\mathcal {A}}: f^{\prime }(z) \prec \varphi (z)\right\} \end{aligned}$$

is a subclass of close-to-convex function. Thus, this class also provides a passage from a subclass of starlike functions to the subclass of close-to-convex functions when $\alpha$ varies from 0 to 1. Also, for different functions of $\varphi$, we get different subclasses of starlike functions as stated earlier.

Theorem 2.9

Let the function $f \in {\mathcal {M}}_{\alpha }(\varphi )$ be given by $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots$. Then,

• (1) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 |B_2| \le ((1+\alpha )(3 + \alpha )-1)B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\left| - (2 + \alpha )^2 (3 + \alpha ) B_1^4 - 12(1+\alpha )^3(3+\alpha ) {B_2^2} + 12(1+\alpha )^2(2 + \alpha )^2 B_1 B_3\right| \nonumber \\&\quad -\, 12(1+\alpha )^3(3+ \alpha )B_1^2 \le 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|\le \dfrac{B_1^2}{(2 + \alpha )^2}. \end{aligned}$$

• (2) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 |B_2| \ge ((1+\alpha )(3 + \alpha )-1)B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |{-(1-\alpha )(2+\alpha )^2(3+\alpha ) }B_1^4 - 12(1+\alpha )^3(3+\alpha ){B_2^2}\\&\quad +\, 12 (1+\alpha )^2(2 + \alpha )^2 B_1 B_3\Big | - 12(1+\alpha )^2 |B_2|B_1 \\&\quad -\,6(1+\alpha )^2((1+\alpha )(3 + \alpha )+1) B_1^2 \ge 0, \end{aligned}$$

  or the conditions

  $$\begin{aligned} 2 |B_2| \le ((1+\alpha )(3 + \alpha )-1)B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |- (2 + \alpha )^2 (3 + \alpha ) B_1^4 - 12(1+\alpha )^3(3+\alpha ) {B_2^2} + 12(1+\alpha )^2(2 + \alpha )^2 B_1 B_3\Big |\\&- 12(1+\alpha )^3(3 + \alpha )B_1^2 \ge 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|&\le \dfrac{1}{12(1+\alpha )^3(2+\alpha )^2(3+\alpha )} \Big |{-12(1+\alpha )^3(3+\alpha )}B_2^2 \\&\quad -\, {(2+\alpha )^2(3+\alpha )}B_1^4+ {12(1+\alpha )^2(2+\alpha )^2}B_1B_3 \Big |. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 |B_2| > \left( (1+\alpha )(3 + \alpha )-1\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |{-(1-\alpha )(2+\alpha )^2(3+\alpha )}B_1^4 - 12(1+\alpha )^3(3+\alpha ){B_2^2}\\&\quad +\, 12 (1+\alpha )^2(2 + \alpha )^2 B_1 B_3\Big |- 12(1+\alpha )^2 |B_2|B_1 \\&\quad -\,6(1+\alpha )^2((1+\alpha )(3 + \alpha )+1) B_1^2 \le 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2| \le \dfrac{B_1^2 M}{( (2+\alpha )^2(3+\alpha ))N}, \end{aligned}$$

  where

  $$\begin{aligned} M&= \Big | {-(1-\alpha )(2+\alpha )^2(3+\alpha )^2} B_1^4 - {12(1+\alpha )^3(3+\alpha )^2} B_2^2 \\&\qquad +\, {12(1+\alpha )^2(2+\alpha )^2(3+\alpha )} B_1 B_3 \Big | - 12{(1+\alpha )(2+\alpha )^2}B_1 |B_2| \\&\qquad -\, {12(1+\alpha )}B_2^2 - {3(1+\alpha )(2+ \alpha )^4}B_1^2, \end{aligned}$$

  and

  $$\begin{aligned} N&=\Big |{-(1-\alpha )(2+\alpha )^2(3+\alpha )}B_1^4 - {12(1+\alpha )^3(3+\alpha )}B_2^2 \\&\qquad +\, {12(1+\alpha )^2(2+\alpha )^2}B_1 B_3\Big | - {12(1+\alpha )^2}B_1^2 - {24(1+\alpha )^2}B_1 |B_2|. \end{aligned}$$

Proof

Since $f \in {\mathcal {M}}_{\alpha }(\varphi ),$ there is an analytic function $w(z)= w_1 z + w_2 z^2 + \cdots \in \varOmega$, such that

$$\begin{aligned} (f^{\prime }(z))^{\alpha } \Big (\dfrac{z f^{\prime }(z)}{f(z)}\Big )^{1-\alpha }= \varphi (w(z)). \end{aligned}$$

(2.19)

Define $p_1(z)$ by

$$\begin{aligned} p_1(z)=\dfrac{1+ w(z)}{1-w(z)}=1 + c_1 z + c_2 z^2 + \cdots , \end{aligned}$$

then this implies

$$\begin{aligned} w(z)= \dfrac{p_1(z)-1}{p_1(z)+1}= \dfrac{1}{2}\left( c_1 z +\left( c_2-\dfrac{c_1^2}{2}\right) z^2+\cdots \right) . \end{aligned}$$

Clearly $p_1$ is analytic in ${\mathbb {D}}$ with $p_1(0)=1$ and $p_1\in {\mathcal {P}}$. Then, since $\varphi (z)= 1 + B_1 z + B_2 z^2 + B_3 z^3+ \cdots$, we get

$$\begin{aligned} \varphi \left( \dfrac{p_1(z)-1}{p_1(z)+1}\right) =1 + \dfrac{1}{2}B_1 c_1z + \left( \dfrac{1}{2}B_1\left( c_2-\dfrac{c_1^2}{2}\right) +\dfrac{1}{4}B_2 c_1^2\right) z^2 +\cdots . \end{aligned}$$

(2.20)

Also, the Taylor series expansion of f gives

$$\begin{aligned}&(f^{\prime }(z))^\alpha \left( \dfrac{z f^{\prime }(z)}{f(z)}\right) ^{1-\alpha }= 1+ a_2(1+\alpha )z + \frac{1}{2}\Bigg ((2 + \alpha )(2 a_3 -(1-\alpha )a_2^2\Bigg )z^2\nonumber \\&\quad +\,\frac{1}{6}(3 + \alpha )(6 a_4 - 6(1-\alpha ) a_2 a_3 + (1-\alpha )(2-\alpha ) a_2^3))z^3 +\cdots . \end{aligned}$$

(2.21)

Then from (2.19), (2.20) and (2.21), we get

$$\begin{aligned} a_2&= \dfrac{B_1 c_1}{2(1+\alpha )}. \end{aligned}$$

(2.22)

$$\begin{aligned} a_3&=\dfrac{1}{2(2+\alpha )}\left[ B_1 c_2 - \left( \dfrac{2(1+\alpha )^2(B_1 - B_2)-(1-\alpha )(2+\alpha )B_1^2}{4(1+ \alpha )^2}\right) c_1^2\right] , \end{aligned}$$

(2.23)

$$\begin{aligned} a_4&=\dfrac{1}{2(3+\alpha )}\left[ B_1 c_3+ \left( \dfrac{B_1}{4}-\dfrac{(1-\alpha )(3+\alpha )}{4(1+\alpha )(2+\alpha )}B_1^2 +\dfrac{(1-\alpha )(1-2\alpha )(3+\alpha )}{24(1+\alpha )^3 }B_1^3 \right. \right. \nonumber \\&\left. \quad +\, \dfrac{B_2}{2} + \dfrac{(1-\alpha )(3+\alpha )}{4(1+\alpha )(2+\alpha )}B_1B_2 + \dfrac{B_3}{4}\right) c_1^3 +\left( -B_1 + \dfrac{(1-\alpha )(3+\alpha )}{2(1+\alpha )(2+\alpha )}B_1^2 \right. \nonumber \\&\left. \left. \quad +\, B_2\right) c_1 c_2\right] . \end{aligned}$$

(2.24)

Thus,

$$\begin{aligned} a_2 a_4 - a_3^2&=\dfrac{1}{16}B_1\left[ \left\{ \dfrac{- (1-\alpha )}{12(1+\alpha )^3}B_1^3 - \dfrac{1}{(2+\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{1}{(1+\alpha )^3(3+\alpha )}B_3\right. \right. \\&\left. \left. \quad +\, \dfrac{1}{(1+\alpha )(2+\alpha )^2(3+\alpha )}(B_1-2B_2) \right\} c_1^4 \right. \\&\left. \quad +\, \dfrac{4}{(1+\alpha )(2+\alpha )^2(3+\alpha )}( B_2 - B_1)c_2 c_1^2\right. \\&\left. \quad -\, \dfrac{4}{(2+\alpha )^2}B_1 c_2^2 +\dfrac{4}{(1+\alpha )(3+\alpha )}B_1 c_1 c_3\right] . \end{aligned}$$

Since the function $p(\hbox {e}^{i \theta }z)$ $(\theta \in {\mathbb {R}})$ is in the class ${\mathcal {P}}$ for any $p \in {\mathcal {P}}$, without loss of generality, we can assume that $c_1=c>0$. Substituting the values of $c_2$ and $c_3$ from (2.3) and (2.4) in the above expression, we get

$$\begin{aligned} |a_2 a_4 - a_3^2|&= \dfrac{1}{16}B_1\bigg |\bigg \{\dfrac{\alpha -1}{12(1+\alpha )^3}B_1^3 - \dfrac{1}{(2+\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{1}{(1+\alpha )(3+\alpha )}B_3 \bigg \}c^4 \\&\quad +\, \bigg \{ \dfrac{2}{(1+\alpha )(2+\alpha )^2(3+\alpha )}B_2 \bigg \} c^2 \left( 4-c^2\right) x - \bigg \{ \dfrac{1}{(1+\alpha )(2+\alpha )^2(3+\alpha )}c^2 \\&\quad +\, \dfrac{4}{(2+\alpha )^2} \bigg \} B_1 \left( 4-c^2\right) x^2 + \dfrac{2}{(1+\alpha )(3+\alpha )}B_1 c\left( 4-c^2\right) \left( 1-|x|^2\right) y \bigg |. \end{aligned}$$

Replacing |x| by $\mu$ and by making use of the triangle inequality and the fact that $|y|\le 1$ in the above expression, we get

$$\begin{aligned} |a_2 a_4 - a_3^2|&\le \dfrac{1}{16}B_1 \bigg [\bigg |\bigg \{\dfrac{\alpha -1}{12(1+\alpha )^3}B_1^3 - \dfrac{1}{(2+\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{1}{(1+\alpha )(3+\alpha )}B_3 \bigg \}\bigg |c^4 \nonumber \\&\quad +\, \dfrac{2c}{(1+\alpha )(3+\alpha )}B_1 \left( 4-c^2\right) + \dfrac{2}{(1+\alpha )(2+\alpha )^2(3+\alpha )}|B_2|c^2\left( 4-c^2\right) \,\mu \nonumber \\&\quad +\, \bigg \{\dfrac{c^2}{(1+\alpha )(2+\alpha )^2(3+\alpha )} -\dfrac{2c }{(1+\alpha )(3+\alpha )} + \dfrac{4}{(2+\alpha )^2}\bigg \}B_1\left( 4-c^2\right) \mu ^2 \bigg ]\nonumber \\&= F(c, \mu ). \end{aligned}$$

(2.25)

We shall now maximize $F(c, \mu )$ for $(c, \mu )\in [0, 2]\times [0, 1]$. Differentiating $F(c, \mu )$ in (2.11) partially with respect to $\mu$, we get

$$\begin{aligned} \dfrac{\partial {F}}{\partial {\mu }}&= \dfrac{1}{16}B_1\left[ \dfrac{2}{(1+\alpha )(2+\alpha )^2(3+\alpha )} c^2\left( 4-c^2\right) |B_2| \right. \nonumber \\&\quad +\, 2\mu B_1 \left( 4-c^2\right) \left\{ \dfrac{1}{(1+\alpha )(2+\alpha )^2(3+\alpha )}c^2 \right. \nonumber \\&\left. \left. \quad -\, \dfrac{2}{(1+\alpha )(3+\alpha )}c + \dfrac{4}{(2+\alpha )^2}\right\} \right] . \end{aligned}$$

(2.26)

For $0<\mu <1$, and for any fixed $c\in [0, 2]$, we observe that $\partial {F}/\partial {\mu }>0$. Thus $F(c, \mu )$ is an increasing function of $\mu$, and for $c\in [0, 2]$, $F(c, \mu )$ has a maximum value at $\mu =1$. Thus, we have

$$\begin{aligned} \text {max} F(c, \mu )&= F(c, 1)= G(c). \end{aligned}$$

(2.27)

The Eqs. (2.25) and (2.27), upon a little simplification, yield

$$\begin{aligned} G(c)&= \dfrac{B_1}{16}\left[ c^4\left\{ \left| - \dfrac{1}{ (2+\alpha )^2} \dfrac{B_2^2}{B_1 } - \dfrac{ 1-\alpha }{12 (1+\alpha )^3}B_1^3 + \dfrac{1}{(1+\alpha )(3+\alpha )}B_3\right| \right. \right. \nonumber \\&\left. \quad +\, \dfrac{1}{(1+\alpha )(2+\alpha )^2(3+\alpha )}(2|B_2|+ B_1)\right\} \nonumber \\&\left. \quad +\, c^2 \left\{ \dfrac{1 }{(1+\alpha )(2+\alpha )^2(3+\alpha )}4(B_1 +2 |B_2|) - \dfrac{4 }{(2+\alpha )^2}B_1 \right\} + \dfrac{16 }{(2+\alpha )^2}B_1\right] \nonumber \\&=\dfrac{B_1}{16} \left( P c^4 + Q c^2 + R\right) , \end{aligned}$$

(2.28)

where

$$\begin{aligned}&P = \frac{1}{12}(1+\alpha )^{-3}(2+\alpha )^{-2}(3+\alpha )^{-1} \Big (\Big | 12(1+\alpha )^3(3+\alpha )\dfrac{-B_2^2}{B_1 }\nonumber \\&\qquad -\,(1-\alpha ) (2+\alpha )^2 (3+\alpha )B_1^3 + 12(1+\alpha )^2 (2+\alpha )^2 B_3 \Big |\nonumber \\&\qquad - 12(1+\alpha )^2 (B_1+ 2|B_2|)\Big ), \end{aligned}$$

(2.29)

$$\begin{aligned}&Q = 4(1+\alpha )^{-1}(2+\alpha )^{-2}(3+\alpha )^{-1}\Big ( (B_1+ 2|B_2|)- (1+\alpha )(3+\alpha ) B_1 \Big ),\end{aligned}$$

(2.30)

$$\begin{aligned}&R = 16(2+\alpha )^{-2} B_1. \end{aligned}$$

(2.31)

Thus, using (2.18) and (2.28) we get,

$$\begin{aligned} |a_2 a_4 - a_3^2| \le \dfrac{B_1}{16}\left\{ \begin{array}{l@{\quad }l} R, &{} Q\le 0, P\le -\frac{Q}{4}; \\ 16P + 4Q + R, &{} Q\ge 0, P\ge -\frac{Q}{8}\quad \text {or}\quad Q\le 0, P\ge -\frac{Q}{4}; \\ \frac{4PR -Q^2}{4P}, &{} Q > 0, P\le -\frac{Q}{8}. \end{array}\right. \end{aligned}$$

where P, Q, andR are given by (2.29), (2.30), and (2.31), respectively. $\square$

Remark 2.10

When $\alpha =1$ and $\varphi = (1+z)/(1-z)$, Theorem 2.9 reduces to [16, Theorem 3.1]. When $\alpha =0$, Theorem 2.9 reduces to [26, Theorem 1]. Therefore, Corollary 2.3 follows as a particular case.

Our next theorem gives the Fekete–Szegö inequality for functions in the class ${\mathcal {M}}_{\alpha }(\varphi )$.

Theorem 2.11

Let $\varphi$ be defined as in (2.1) and let the function $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots \in {\mathcal {M}}_{\alpha }(\varphi )$. Then we have the following:

• (1) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} 2(2+\alpha )B_1^2 \mu \le (2+\alpha )(1-\alpha )B_1^2 + 2(B_2-B_1)(1+\alpha )^2, \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{1}{(2+\alpha )}\Big (B_2 +\dfrac{(2+\alpha )(1-\alpha )}{2(1+\alpha )^2}B_1^2 -\dfrac{(2+\alpha )\mu }{(1+\alpha )^2}B_1^2 \Big ). \end{aligned}$$

• (2) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned}&(1-\alpha )(2+\alpha )B_1^2 + 2(B_2-B_1)(1+\alpha )^2\le 2 (2+\alpha )B_1^2 \mu \\&\quad \le (1-\alpha )(2+\alpha )B_1^2 + 2(B_2+B_1)(1+\alpha )^2, \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{B_1}{2+\alpha }. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} (1-\alpha )(2+\alpha )B_1^2 + 2(B_2 + B_1)(1+\alpha )^2 \le 2(2+\alpha )B_1^2 \mu , \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{1}{2+\alpha }\Big (-B_2 - \dfrac{(1-\alpha )(2+\alpha )}{2(1+\alpha )^2}B_1^2 + \dfrac{(2+\alpha )\mu }{(1+\alpha )^2}B_1^2\Big ). \end{aligned}$$

Proof

Using (2.22) and (2.23), we get,

$$\begin{aligned} a_3- \mu a_2^2&=\dfrac{1}{(2+\alpha )}\left( \dfrac{c_2 B_1}{2} - \dfrac{c_1^2 B_1}{4}+\dfrac{c_1^2 B_2}{4}+\dfrac{(1-\alpha )(2+\alpha )}{4(1+\alpha )^2}{c_1^2 B_1^2}\right. \\&\left. \quad -\, \dfrac{\mu (2+\alpha ) B_1^2 c_1^2}{4(1+\alpha )^2}\right) \\&= \dfrac{B_1}{2(2+\alpha )}\left( c_2 - c_1^2 \Big ( \dfrac{1}{2}\Big (1-\dfrac{B_2}{B_1}\Big ) - \dfrac{(1-\alpha )(2+\alpha )}{4(1+\alpha )^2}B_1 + \dfrac{(2+\alpha )}{2(1+\alpha )^2}\mu B_1\Big )\right) \\&= \dfrac{B_1}{2(2+\alpha )}(c_2-\nu c_1^2), \end{aligned}$$

where $\nu = \dfrac{1}{2}\left( 1-\dfrac{B_2}{B_1} - \dfrac{(1-\alpha ) (2+\alpha )}{2(1+\alpha )^2}B_1+\dfrac{\mu (2+\alpha )}{(1+\alpha )^2}B_1 \right)$. Using Lemma 2.6, we get the desired result. $\square$

Remark 2.12

Bounds for the second and the third coefficients for f can be directly obtained from Theorem 2.11 as follows:

$$\begin{aligned} |a_2|\le \dfrac{B_1}{1+ \alpha }, \end{aligned}$$

and

$$\begin{aligned} |a_3|\le \left\{ \begin{array}{ll} \dfrac{B_2 + \frac{(1-\alpha )(2+\alpha )}{2(1+\alpha )^2}B_1^2 }{2+\alpha }, &{} (1-\alpha )(2+\alpha )B_1^2 + 2(B_2 - B_1)(1+\alpha )^2 \ge 0; \\ \dfrac{B_1}{2+\alpha }, &{} (1-\alpha )(2+\alpha )B_1^2 + 2(B_2 - B_1)(1+\alpha )^2 \le 0 \quad \text {or} \\ &{}(1-\alpha )(2+\alpha )B_1^2 + 2(B_2 + B_1)(1+\alpha )^2 \ge 0;\\ \dfrac{-B_2 - \frac{(1-\alpha )(2+\alpha )}{2(1+\alpha )^2} B_1^2}{2+\alpha }, &{}(1-\alpha )(2+\alpha )B_1^2 + 2(B_2 + B_1)(1+\alpha )^2 \le 0. \end{array} \right. \end{aligned}$$

Let $\varphi : {\mathbb {D}}\rightarrow {\mathbb {C}}$ be an analytic function given by (2.1). For $0\le \alpha \le 1$, the class ${\mathcal {L}}_{\alpha }(\varphi )$ consists of functions $f\in {\mathcal {A}}$ satisfying the following subordination:

$$\begin{aligned} (f^{\prime }(z))^\alpha \Big (1+\dfrac{z f^{\prime \prime }(z)}{f^{\prime }(z)}\Big )^{1-\alpha } \prec \varphi (z). \end{aligned}$$

We see that ${\mathcal {L}}_{0}(\varphi )={\mathcal {K}}(\varphi )$ is the class of Ma–Minda convex functions and

$$\begin{aligned} {\mathcal {L}}_1(\varphi )={\mathbb {R}}(\varphi )= \{f \in {\mathcal {A}}: f^{\prime }(z) \prec \varphi (z)\} \end{aligned}$$

is a subclass of close-to-convex function. Thus, this class also provides a continuous passage from a subclass of convex functions to the subclass of close-to-convex functions when $\alpha$ varies from 0 to 1.

Theorem 2.13

Let the function $f \in {\mathcal {L}}_{\alpha }(\varphi )$ be given by $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots$. Then,

• (1) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 |B_2| (9 \alpha ^2 -20 \alpha +12)+ B_1^2(1-\alpha )(6+5\alpha )\le \left( -9\alpha ^2 + 4\alpha + 12\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\quad \Big |(1- \alpha )(2 \alpha ^2 - 5\alpha -6) B_1^4 - 8(3-2\alpha ) {B_2^2} + 9\left( 2 - \alpha \right) ^2 B_1 B_3 \\&\quad + (1-\alpha )(6+5\alpha )B_1^2 B_2\Big | - 8(3 -2 \alpha )B_1^2 \le 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|\le \dfrac{B_1^2}{9(2 - \alpha )^2}. \end{aligned}$$

• (2) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 |B_2| (9 \alpha ^2 -20 \alpha +12)+ B_1^2(1-\alpha )(6+5\alpha )\ge \left( -9\alpha ^2 + 4\alpha + 12\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&2\Big |{(1-\alpha )\left( 2\alpha ^2 - 5\alpha -6\right) }B_1^4 - 8(3-2\alpha ){B_2^2} + 9 \left( 2 - \alpha \right) ^2 B_1 B_3 \\&\quad + (1-\alpha )(6+5\alpha )B_1^2 B_2\Big | - 2(9\alpha ^2-20\alpha +12) |B_2|B_1 \\&\quad -\,(1-\alpha )(6+5\alpha )B_1^3 - 9(2- \alpha )^2 B_1^2 \ge 0, \end{aligned}$$

  or the conditions

  $$\begin{aligned} 2 |B_2| (9 \alpha ^2 -20 \alpha +12)+ B_1^2(1-\alpha )(6+5\alpha )\le \left( -9\alpha ^2 + 4\alpha + 12\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |(1- \alpha )(2 \alpha ^2 - 5\alpha -6) B_1^4 - 8(3-2\alpha ) {B_2^2} + 9\left( 2 - \alpha \right) ^2 B_1 B_3 \\&\quad + (1-\alpha )(6+5\alpha )B_1^2 B_2\Big | - 8(3 -2 \alpha )B_1^2 \ge 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|&\le \dfrac{1}{72(2-\alpha )^2(3-2\alpha )} \Big |{-8(3-2\alpha )}B_2^2 + {(1-\alpha )\left( 2\alpha ^2-5\alpha -6\right) }B_1^4 \\&\quad + {9(2-\alpha )^2}B_1B_3 + (1-\alpha )(6+5\alpha )B_1^2 B_2 \Big |. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 |B_2| (9 \alpha ^2 -20 \alpha +12)+ B_1^2(1-\alpha )(6+5\alpha )> \left( -9\alpha ^2 + 4\alpha + 12\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&2\Big |{(1-\alpha )\left( 2\alpha ^2 - 5\alpha -6\right) }B_1^4 - 8(3-2\alpha ){B_2^2} + 9 \left( 2 - \alpha \right) ^2 B_1 B_3 \\&\quad +\, (1-\alpha )(6+5\alpha )B_1^2 B_2\Big | - 2(9\alpha ^2-20\alpha +12) |B_2|B_1\\&\quad -(1-\alpha )(6+5\alpha )B_1^3 - 9(2- \alpha )^2 B_1^2 \le 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2| \le \dfrac{B_1^2 M}{1152 (2-\alpha )^2(3-2\alpha )N}, \end{aligned}$$

  where

  $$\begin{aligned} M&= 4\left( \Big | {32(1-\alpha )(3-2\alpha )\left( 2\alpha ^2-5\alpha -6\right) } B_1^4 - {256(3-2\alpha )^2} B_2^2 \right. \\&\quad +\, {288(2-\alpha )^2(3-2\alpha )} B_1 B_3 + {32(1-\alpha )(6+5\alpha )(3-2\alpha )}B_1^2 B_2 \Big |\\&\quad -\, 18 (1-\alpha )(2-\alpha )^2(6+5\alpha )B_1^3 - 36{(2 -\alpha )^2\left( 12 -20\alpha +9\alpha ^2\right) }B_1 |B_2| \\&\quad -\, {81(2-\alpha )^4}B_1^2 -{(1-\alpha )^2(6+5\alpha )^2}B_1^4 - {4\left( 12-20\alpha +9\alpha ^2\right) ^2}B_2^2 \\&\quad \left. -\, {4(1-\alpha )(6+5\alpha )\left( 12-20\alpha +9\alpha ^2\right) }B_1^2 |B_2|\right) \end{aligned}$$

  and

  $$\begin{aligned} N&=\Big |{(1-\alpha )(2\alpha ^2 -5 \alpha -6)}B_1^4 - {8(3-2\alpha )}B_2^2 + {9(2-\alpha )^2}B_1 B_3\\&\quad +\, {(1-\alpha )(6+5\alpha )}B_1^2 B_2\Big | - {(1-\alpha )(6+5\alpha )}B_1^3 - {2(9\alpha ^2-20\alpha +12)}B_1 |B_2|\\&\quad -\, {(9\alpha ^2-20\alpha +12)}B_1^2. \end{aligned}$$

Proof

Since $f \in {\mathcal {L}}_{\alpha }(\varphi ),$ there is an analytic function $w(z)= w_1 z + w_2 z^2 + \cdots \in \varOmega$, such that

$$\begin{aligned} (f^{\prime }(z))^{\alpha } \Big (1+\dfrac{z f^{\prime \prime }(z)}{f^{\prime }(z)}\Big )^{1-\alpha }= \varphi (w(z)). \end{aligned}$$

(2.32)

Define $p_1(z)$ by

$$\begin{aligned} p_1(z)=\dfrac{1+ w(z)}{1-w(z)}=1 + c_1 z + c_2 z^2 + \cdots , \end{aligned}$$

which then implies

$$\begin{aligned} w(z)= \dfrac{p_1(z)-1}{p_1(z)+1}= \dfrac{1}{2}\left( c_1 z +\left( c_2-\dfrac{c_1^2}{2}\right) z^2+\cdots \right) . \end{aligned}$$

Clearly $p_1$ is analytic in ${\mathbb {D}}$ with $p_1(0)=1$ and $p_1\in {\mathcal {P}}$. Then, since $\varphi (z)= 1 + B_1 z + B_2 z^2 + B_3 z^3+ \cdots$, we get

$$\begin{aligned} \varphi \left( \dfrac{p_1(z)-1}{p_1(z)+1}\right) =1 + \dfrac{1}{2}B_1 c_1z + \left( \dfrac{1}{2}B_1\left( c_2-\dfrac{c_1^2}{2}\right) +\dfrac{1}{4}B_2 c_1^2\right) z^2 +\cdots . \end{aligned}$$

(2.33)

Also, the Taylor series expansion of f gives

$$\begin{aligned}&(f^{\prime }(z))^\alpha \Big (1+\dfrac{z f^{\prime \prime }(z)}{f^{\prime }(z)}\Big )^{1-\alpha }= 1+ 2 a_2 z + (3(2-\alpha ) a_3 -4(1-\alpha )a_2^2)z^2 \nonumber \\&\quad +\,\Bigg (4(3-2\alpha ) a_4 - 18(1-\alpha ) a_2 a_3 + 8(1-\alpha ) a_2^3\Bigg )z^3 +\cdots . \end{aligned}$$

(2.34)

Then from (2.32), (2.33), and (2.34), we get

$$\begin{aligned}&\displaystyle a_2= \dfrac{B_1 c_1}{4}, \end{aligned}$$

(2.35)

$$\begin{aligned}&\displaystyle a_3=\dfrac{1}{6(2+\alpha )}\left[ B_1 c_2 - \left( \dfrac{(B_1 - B_2)-(1-\alpha )B_1^2}{2}\right) c_1^2\right] , \end{aligned}$$

(2.36)

$$\begin{aligned}&\displaystyle a_4=\dfrac{1}{8(3-2\alpha )}\left[ B_1 c_3+ \left( \dfrac{B_1}{4}-\dfrac{3(1-\alpha )}{4(2-\alpha )}B_1^2 +\dfrac{(1-\alpha )(1-2\alpha )}{4(2-\alpha ) }B_1^3 - \dfrac{B_2}{2} \right. \right. \nonumber \\&\displaystyle \left. \left. \qquad +\, \dfrac{3(1-\alpha )}{4(2-\alpha )}B_1B_2 + \dfrac{B_3}{4}\right) c_1^3 +\left( -B_1 + \dfrac{3(1-\alpha )}{2(2-\alpha )}B_1^2 + B_2\right) c_1 c_2\right] . \end{aligned}$$

(2.37)

Thus,

$$\begin{aligned} a_2 a_4 - a_3^2&=\dfrac{1}{1152}B_1\left[ \left\{ \dfrac{ (1-\alpha )(2\alpha ^2 -5\alpha -6)}{(2-\alpha )^2 (3-2\alpha )}B_1^3 - \dfrac{8}{(2-\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{9}{3-2\alpha }B_3\right. \right. \\&\quad \left. -\, \dfrac{9\alpha ^2 -20\alpha +12}{(2-\alpha )^2(3-2\alpha )}(B_1-2B_2) +\dfrac{(1-\alpha )(6+5\alpha )}{(2-\alpha )^2(3-2\alpha )}B_1( B_2- B_1) \right\} c_1^4 \\&\quad -\, \left\{ \dfrac{4(9\alpha ^2 -20\alpha +12)}{(2-\alpha )^2(3-2\alpha )}(B_2- B_1) +\dfrac{2(1-\alpha )(6+5\alpha )}{(2-\alpha )^2(3-2\alpha )} B_1^2 \right\} c_2 c_1^2 \\&\left. \quad -\, \dfrac{32}{(2-\alpha )^2}B_1 c_2^2 +\dfrac{36}{3-2\alpha }B_1 c_1 c_3\right] . \end{aligned}$$

Since the function $p(\hbox {e}^{i \theta }z)$ $(\theta \in {\mathbb {R}})$ is in the class ${\mathcal {P}}$ for any $p \in {\mathcal {P}}$, without loss of generality, we can assume that $c_1=c>0$. Substituting the values of $c_2$ and $c_3$ from (2.3) and (2.4) in the above expression, we get

$$\begin{aligned} |a_2 a_4 - a_3^2|&= \dfrac{1}{1152}B_1\bigg |\bigg \{\dfrac{(1-\alpha )\left( 2\alpha ^2-5\alpha -6\right) }{(2-\alpha )^2(3-2\alpha )}B_1^3 + \dfrac{(1-\alpha )(5\alpha + 6)}{(2-\alpha )^2(3-2\alpha )}B_1B_2\\&\quad -\, \dfrac{8}{(2-\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{8}{(3-2\alpha )}B_3 \bigg \}c^4 + \bigg \{ \dfrac{2(9\alpha ^2-20\alpha +12)}{(2-\alpha )^2(3-2\alpha )}B_2 \\&\quad +\, \dfrac{(1-\alpha )(6+5\alpha )}{(2-\alpha )^2(3-2\alpha )}B_1^2 \bigg \} c^2 \left( 4-c^2\right) x - \bigg \{ \dfrac{9\alpha ^2-20\alpha +12}{(2-\alpha )^2(3-2\alpha )}c^2 \\&\quad {} + \dfrac{32}{(2-\alpha )^2} \bigg \} B_1 \left( 4-c^2\right) x^2 + \dfrac{18}{3-2\alpha }B_1 c\left( 4-c^2\right) \left( 1-|x|^2\right) y \bigg |. \end{aligned}$$

Replacing |x| by $\mu$ and by making use of the triangle inequality and the fact that $|y|\le 1$ in the above expression, we get

$$\begin{aligned} |a_2 a_4 - a_3^2|&\le \dfrac{1}{1152}B_1 \bigg [\bigg |\bigg \{\dfrac{(1-\alpha )\left( 2\alpha ^2-5\alpha -6\right) }{(2-\alpha )^2(3-2\alpha )}B_1^3 - \dfrac{8}{(2-\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{8}{3-2\alpha }B_3 \nonumber \\&\quad +\,\dfrac{(1-\alpha )(5\alpha +6)}{(2-\alpha )^2(3-2\alpha )}B_1B_2 \bigg \}\bigg |c^4 + \dfrac{18c}{3-2\alpha }B_1 \left( 4-c^2\right) \nonumber \\&\quad +\,\bigg \{ \dfrac{(1-\alpha )(6+5\alpha )}{(2-\alpha )^2(3-2\alpha )}B_1^2 + \dfrac{2(9\alpha ^2 -20\alpha +12) }{(2-\alpha )^2(3-2\alpha )}|B_2|\bigg \}c^2\left( 4-c^2\right) \,\mu \nonumber \\&\quad +\,\bigg \{\dfrac{(9\alpha ^2 -20\alpha +12) c^2}{(2-\alpha )^2(3-2\alpha )} -\dfrac{18c }{3-2\alpha } + \dfrac{32}{(2-\alpha )^2}\bigg \}B_1\left( 4-c^2\right) \mu ^2 \bigg ]\nonumber \\&= F(c, \mu ). \end{aligned}$$

(2.38)

We shall now maximize $F(c, \mu )$ for $(c, \mu )\in [0, 2]\times [0, 1]$. Differentiating $F(c, \mu )$ in (2.11) partially with respect to $\mu$, we get

$$\begin{aligned} \dfrac{\partial {F}}{\partial {\mu }}=&\dfrac{1}{1152}B_1\left[ \dfrac{(1-\alpha )(6+5\alpha )}{(2-\alpha )^2(3-2\alpha )} c^2\left( 4-c^2\right) B_1^2 \right. \nonumber \\&\left. +\,\dfrac{2(9\alpha ^2 -20\alpha +12)}{(2-\alpha )^2(3-2\alpha )} c^2\left( 4-c^2\right) |B_2| -\, 2\mu B_1 \left( 4-c^2\right) \right. \nonumber \\&\left. \times \, \left( \dfrac{9\alpha ^2 -20\alpha +12}{(2-\alpha )^2(3-2\alpha )}c^2 - \dfrac{18}{3-2\alpha }c + \dfrac{32}{(2-\alpha )^2}\right) \right] . \end{aligned}$$

(2.39)

For $0<\mu <1$, and for any fixed $c\in [0, 2]$, we observe that $\partial {F}/\partial {\mu }>0$. Thus $F(c, \mu )$ is an increasing function of $\mu$, and for $c\in [0, 2]$, $F(c, \mu )$ has a maximum value at $\mu =1$. Thus, we have

$$\begin{aligned} \text {max} F(c, \mu )&= F(c, 1)= G(c). \end{aligned}$$

(2.40)

The Eqs. (2.38) and (2.40), upon a little simplification, yield

$$\begin{aligned} G(c)= & {} \dfrac{B_1}{1152}\left[ c^4\left\{ \left| - \dfrac{8 }{ (2-\alpha )^2}\dfrac{B_2}{B_1} + \dfrac{ (1-\alpha )\left( 2\alpha ^2-5\alpha -6\right) }{(2-\alpha )^2 (3-2\alpha )}B_1^3 + \dfrac{9}{3-2\alpha } B_3\right. \right. \right. \nonumber \\&\left. \left. +\,\dfrac{(1-\alpha )(6+5\alpha )}{(2-\alpha )^2(3-2\alpha )}B_1 B_2\right| - \dfrac{(1-\alpha )(6+5\alpha )}{(2-\alpha )^2(3-2\alpha )}B_1^2- \dfrac{9\alpha ^2 -20\alpha +12}{(2-\alpha )^2(3-2\alpha )}(B_1 + 2 |B_2|)\right\} \nonumber \\&+\, 4c^2 \left\{ \dfrac{2(9\alpha ^2 -20\alpha +12)}{(2-\alpha )^2(3-2\alpha )}|B_2|+\dfrac{(1-\alpha )(6+5\alpha ) }{(2-\alpha )^2(3-2\alpha )}B_1^2+ \dfrac{9\alpha ^2 -4\alpha -12}{(2-\alpha )^2(3-2\alpha )} B_1 \right\} \nonumber \\&\left. +\, \dfrac{128 }{(2-\alpha )^2}B_1\right] \nonumber \\= & {} \dfrac{B_1}{1152}( P c^4 + Q c^2 + R), \end{aligned}$$

(2.41)

where

$$\begin{aligned} P&= (2-\alpha )^{-2} (3-2\alpha )^{-1}\Big (\Big | -8(3-2\alpha )\dfrac{ B_2^2}{B_1 } -(1-\alpha )\left( 2\alpha ^2-5\alpha -6\right) B_1^3 \nonumber \\&\quad + 9(2-\alpha )^2 B_3 + (1-\alpha )(6+5\alpha ) B_1 B_2\Big |- 2 (9\alpha ^2 -20\alpha +12) |B_2|\nonumber \\&\quad -\, (1-\alpha )(6+5\alpha )B_1^2- (9\alpha ^2 -20\alpha +12) B_1\Big ), \end{aligned}$$

(2.42)

$$\begin{aligned} Q&= (2-\alpha )^{-2}(3-2\alpha )^{-1}\Big (4(1-\alpha )(6+5\alpha ) B_1^2+ 4(9\alpha ^2 -4\alpha -12)(B_1+2 |B_2|) \Big ), \end{aligned}$$

(2.43)

$$\begin{aligned} R&= 128 (2-\alpha )^{-2} B_1. \end{aligned}$$

(2.44)

Thus using (2.18) and (2.41) we get,

$$\begin{aligned}|a_2 a_4 - a_3^2| \le \dfrac{B_1}{1152}\left\{ \begin{array}{l@{\quad }l} R, &{} Q\le 0, P\le -\frac{Q}{4}; \\ 16P + 4Q + R, &{} Q\ge 0, P\ge -\frac{Q}{8}\quad \text {or}\quad Q\le 0, P\ge -\frac{Q}{4}; \\ \frac{4PR -Q^2}{4P}, &{} Q > 0, P\le -\frac{Q}{8}. \end{array}\right. \end{aligned}$$

where P, Q, R are given by (2.42), (2.43) and (2.44), respectively. $\square$

Remark 2.14

When $\alpha =1$ and $\varphi = (1+z)/(1-z)$, Theorem 2.13 reduces to [16, Theorem 3.1]. When $\alpha =0$, Theorem 2.13 reduces to [26, Theorem 2].

Our next theorem gives the Fekete–Szegö inequality for functions in the class ${\mathcal {L}}_{\alpha }(\varphi )$.

Theorem 2.15

Let $\varphi$ be defined as in (2.1) and let the function $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots \in {\mathcal {L}}_{\alpha }(\varphi )$. Then we have the following:

• (1) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} 3(2-\alpha )B_1^2 \mu \le 4(1-\alpha )B_1^2 + 4(B_2-B_1), \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{1}{3(2-\alpha )}\Big (B_2 +{(1-\alpha )}B_1^2 - \dfrac{3(2-\alpha )\mu }{4}B_1^2 \Big ). \end{aligned}$$

• (2) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} 4(1-\alpha )B_1^2 + 4(B_2-B_1)\le 3 (2-\alpha )B_1^2 \mu \le 4(1-\alpha )B_1^2 + 4(B_2+B_1), \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{B_1}{3(2+\alpha )}. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} 4(1-\alpha )B_1^2 + 4(B_2 + B_1) \le 3 (2-\alpha )B_1^2 \mu , \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{1}{3(2+\alpha )}\Big (-B_2 - {(1-\alpha )}B_1^2 + \dfrac{3(2-\alpha )\mu }{4}B_1^2\Big ). \end{aligned}$$

Proof

Using (2.35) and (2.36) we get,

$$\begin{aligned} a_3- \mu a_2^2&=\dfrac{1}{(2-\alpha )} \left( \dfrac{c_2 B_1}{6} - \dfrac{c_1^2 B_1}{12}+\dfrac{c_1^2 B_2}{12}+\dfrac{1-\alpha }{12}{c_1^2 B_1^2} - \dfrac{\mu (2-\alpha ) B_1^2 c_1^2}{16}\right) \\&= \dfrac{B_1}{6(2-\alpha )}\left( c_2 - \dfrac{c_1^2}{2} \Big ( \Big (1-\dfrac{B_2}{B_1}\Big ) - (1-\alpha )B_1 + \dfrac{3(2-\alpha )}{4}\mu B_1\Big )\right) \\&= \dfrac{B_1}{6(2+\alpha )}(c_2-\nu c_1^2), \end{aligned}$$

where $\nu = \dfrac{1}{2}\left( 1-\dfrac{B_2}{B_1} - {(1-\alpha )}B_1+\dfrac{3 \mu (2-\alpha )}{4}B_1 \right)$. Using Lemma 2.6 we get the desired result. $\square$

Remark 2.16

Bounds for the second and the third coefficient for f can be directly obtained from Theorem 2.13 as follows:

$$\begin{aligned} |a_2|\le \dfrac{B_1}{2}, \end{aligned}$$

and

$$\begin{aligned}|a_3|\le \left\{ \begin{array}{l@{\quad }l} \dfrac{B_2+{(1-\alpha )}B_1^2 }{3(2-\alpha )}, &{} (1-\alpha )B_1^2 + (B_2 - B_1) \ge 0; \\ \dfrac{B_1}{3(2-\alpha )}, &{} (1-\alpha )B_1^2 + (B_2 - B_1) \le 0 \quad \text {or} \\ &{}(1-\alpha )B_1^2 + (B_2 + B_1) \ge 0;\\ \dfrac{-B_2 - (1-\alpha )B_1^2}{3(2-\alpha )}, &{}(1-\alpha )B_1^2 + (B_2 + B_1) \le 0. \end{array} \right. \end{aligned}$$

Let $\varphi : {\mathbb {D}}\rightarrow {\mathbb {C}}$ be an analytic function given by (2.1). For $0\le \alpha \le 1$, the class ${\mathcal {K}}_{\alpha }(\varphi )$ consists of functions $f\in {\mathcal {A}}$ satisfying the following subordination:

$$\begin{aligned} \Big (\dfrac{f(z)}{z}\Big )^\alpha \Big (\dfrac{z f^{\prime }(z)}{f(z)}\Big )^{1-\alpha } \prec \varphi (z). \end{aligned}$$

We see that ${\mathcal {K}}_{0}(\varphi )={\mathcal {S}}^*(\varphi )$ is the class of Ma–Minda starlike functions.

Theorem 2.17

Let the function $f \in {\mathcal {K}}_{\alpha }(\varphi )$ be given by $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots$. Then,

• (1) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 (1-\alpha )^2|B_2| +(1-\alpha )\alpha B_1^2\le \left( 2-\alpha ^2\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |-(1- \alpha ) B_1^4 - (3-2\alpha ) {B_2^2} + \left( 2 - \alpha \right) ^2 B_1 B_3 + \alpha (1-\alpha )B_1^2 B_2\Big | \\&\quad -\, (3 -2 \alpha )B_1^2 \le 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|\le \dfrac{B_1^2}{(2 - \alpha )^2}. \end{aligned}$$

• (2) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 (1-\alpha )^2|B_2| +(1-\alpha )\alpha B_1^2\ge \left( 2-\alpha ^2\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\quad \Big |{-2(1-\alpha )}B_1^4 - 2(3-2\alpha ){B_2^2} + 2 \left( 2 - \alpha \right) ^2 B_1 B_3 +2\alpha (1-\alpha )B_1^2 B_2\Big | \\&\quad -\, 2(1-\alpha )^2 |B_2|B_1 -(1-\alpha )\alpha B_1^3 - (2- \alpha )^2 B_1^2 \ge 0, \end{aligned}$$

  or the conditions

  $$\begin{aligned} 2 (1-\alpha )^2|B_2| +(1-\alpha )\alpha B_1^2\le \left( 2-\alpha ^2\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |-(1- \alpha ) B_1^4 - (3-2\alpha ) {B_2^2} + \left( 2 - \alpha \right) ^2 B_1 B_3 + \alpha (1-\alpha )B_1^2 B_2\Big |\\&\quad -\, (3 -2 \alpha )B_1^2 \ge 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|&\le \dfrac{1}{(2-\alpha )^2(3-2\alpha )} \Big |{-(3-2\alpha )}B_2^2 - {(1-\alpha )}B_1^4 \\&\quad + {(2-\alpha )^2}B_1B_3 + \alpha (1-\alpha )B_1^2 B_2 \Big |. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned} 2 (1-\alpha )^2|B_2| +(1-\alpha )\alpha B_1^2 > \left( 2-\alpha ^2\right) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\quad \Big |{-2(1-\alpha )}B_1^4 - 2(3-2\alpha ){B_2^2} + 2 \left( 2 - \alpha \right) ^2 B_1 B_3 +2\alpha (1-\alpha )B_1^2 B_2\Big | \\&\quad -\, 2(1-\alpha )^2 |B_2|B_1 -(1-\alpha )\alpha B_1^3 - (2- \alpha )^2 B_1^2 \le 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2| \le \dfrac{B_1^2 M}{(2-\alpha )^2(3-2\alpha )N}, \end{aligned}$$

  where

  $$\begin{aligned} M&= \Big | {-(1-\alpha )(3-2\alpha )} B_1^4 - {(3-2\alpha )^2} B_2^2 + {(2-\alpha )^2(3-2\alpha )} B_1 B_3 \\&\qquad +\, {\alpha (1-\alpha )(3-2\alpha )}B_1^2 B_2 \Big | - \frac{\alpha }{2}(1-\alpha )(2-\alpha )^2 B_1^3 - {(1-\alpha )^2(2-\alpha )^2}B_1 |B_2| \\&\qquad -\, \frac{(2-\alpha )^4}{4}B_1^2 -{\frac{\alpha ^2}{4}(1-\alpha )^2}B_1^4 - {(1-\alpha )^4}B_2^2 - {\alpha (1-\alpha )^3}B_1^2 |B_2| \end{aligned}$$

  and

  $$\begin{aligned} N&=\Big |{-(1-\alpha )}B_1^4 - {(3-2\alpha )}B_2^2 + {(2-\alpha )^2}B_1 B_3 + {\alpha (1-\alpha )}B_1^2 B_2\Big |\\&\qquad -\, {\alpha (1-\alpha )}B_1^3 - {2(1-\alpha )^2}B_1 |B_2| - {(1-\alpha )^2}B_1^2. \end{aligned}$$

Proof

Since $f \in {\mathcal {K}}_{\alpha }(\varphi ),$ there is an analytic function $w(z)= w_1 z + w_2 z^2 + \cdots \in \varOmega$, such that

$$\begin{aligned} \Big (\dfrac{f(z)}{z}\Big )^{\alpha } \Big (\dfrac{z f^{\prime }(z)}{f(z)}\Big )^{1-\alpha }= \varphi (w(z)). \end{aligned}$$

(2.45)

Define $p_1(z)$ by

$$\begin{aligned} p_1(z)=\dfrac{1+ w(z)}{1-w(z)}=1 + c_1 z + c_2 z^2 + \cdots , \end{aligned}$$

which implies

$$\begin{aligned} w(z)= \dfrac{p_1(z)-1}{p_1(z)+1}= \dfrac{1}{2}\left( c_1 z +\left( c_2-\dfrac{c_1^2}{2}\right) z^2+\cdots \right) . \end{aligned}$$

Clearly $p_1$ is analytic in ${\mathbb {D}}$ with $p_1(0)=1$ and $p_1\in {\mathcal {P}}$. Then, since $\varphi (z)= 1 + B_1 z + B_2 z^2 + B_3 z^3+ \cdots$, we get

$$\begin{aligned} \varphi \left( \dfrac{p_1(z)-1}{p_1(z)+1}\right) =1 + \dfrac{1}{2}B_1 c_1z + \left( \dfrac{1}{2}B_1\left( c_2-\dfrac{c_1^2}{2}\right) +\dfrac{1}{4}B_2 c_1^2\right) z^2 +\cdots . \end{aligned}$$

(2.46)

Also, the Taylor series expansion of f gives

$$\begin{aligned} \Big (\dfrac{f(z)}{z}\Big )^\alpha \Big (\dfrac{z f^{\prime }(z)}{f(z)}\Big )^{1-\alpha }&= 1+ a_2 z + \Big ((2-\alpha ) a_3 -(1-\alpha )a_2^2\Big )z^2 \nonumber \\&\quad +\, \left( (3-2\alpha ) a_4 - 3(1-\alpha ) a_2 a_3 + (1-\alpha ) a_2^3\right) z^3 +\cdots . \end{aligned}$$

(2.47)

Then from (2.45), (2.46) and (2.47), we get

$$\begin{aligned}&a_2= \dfrac{B_1 c_1}{2}, \end{aligned}$$

(2.48)

$$\begin{aligned}&a_3=\dfrac{1}{4(2-\alpha )}\left[ 2B_1 c_2 - \Big ({(B_1 - B_2)-(1-\alpha )B_1^2}\Big )c_1^2\right] , \end{aligned}$$

(2.49)

$$\begin{aligned}&a_4=\dfrac{1}{8(3-2\alpha )}\Big [4B_1 c_3+ \Big ({B_1}-\dfrac{3(1-\alpha )}{2-\alpha }B_1^2 +\dfrac{(1-\alpha )(1-2\alpha )}{2-\alpha }B_1^3 - {2B_2} \nonumber \\&\qquad +\, \dfrac{3(1-\alpha )}{2-\alpha }B_1B_2 + {B_3}\Big )c_1^3 +\Big (-4 B_1 + \dfrac{6(1-\alpha )}{2-\alpha }B_1^2 + 4 B_2\Big )c_1 c_2\Big ]. \end{aligned}$$

(2.50)

Thus,

$$\begin{aligned} a_2 a_4 - a_3^2&=\dfrac{1}{16}B_1\left[ \left\{ -\dfrac{1-\alpha }{(2-\alpha )^2 (3-2\alpha )}B_1^3 - \dfrac{1}{(2-\alpha )^2}\dfrac{B_2^2}{B_1} + \dfrac{1}{3-2\alpha }B_3\right. \right. \\&\left. \quad +\, \dfrac{(1-\alpha )^2}{(2-\alpha )^2(3-2\alpha )}(B_1 - 2 B_2 ) + \dfrac{\alpha (1-\alpha )}{(2-\alpha )^2(3-2\alpha )}B_1 (B_2 - B_1)\right\} c_1^4 \\&\quad +\, \left\{ \dfrac{4(1-\alpha )^2}{(2-\alpha )^2(3-2\alpha )}(B_2-B_1) +\dfrac{2 \alpha (1-\alpha )}{(2-\alpha )^2(3-2\alpha )} B_1^2 \right\} c_2 c_1^2 \\&\left. \quad -\, \dfrac{4}{(2-\alpha )^2}B_1 c_2^2 -\dfrac{4}{3-2\alpha }B_1 c_1 c_3\right] . \end{aligned}$$

Proceeding similarly as in the proof of Theorem 2.1, we would see that $|a_2 a_4 - a_3^2|$ will be bounded by

$$\begin{aligned} G(c)&= \dfrac{B_1}{16}\left[ c^4\left\{ \Bigg | - \dfrac{1 }{ (2-\alpha )^2}\dfrac{B_2^2}{B_1} - \dfrac{1-\alpha }{(2-\alpha )^2 (3-2\alpha )}B_1^3 + \dfrac{1}{3-2\alpha } B_3\right. \right. \nonumber \\&\quad +\,\dfrac{\alpha (1-\alpha )}{(2-\alpha )^2(3-2\alpha )}B_1 B_2\Bigg | - \dfrac{\alpha (1-\alpha )}{(2-\alpha )^2(3-2\alpha )}B_1^2\nonumber \\&\left. \quad -\, \dfrac{(1-\alpha )^2}{(2-\alpha )^2(3-2\alpha )}(B_1+2|B_2|)\right\} + 4c^2 \left\{ \dfrac{2(1-\alpha )^2}{(2-\alpha )^2(3-2\alpha )}|B_2|\right. \nonumber \\&\left. \quad +\,\dfrac{\alpha (1-\alpha ) }{(2-\alpha )^2(3-2\alpha )}B_1^2- \dfrac{ \left( 2-\alpha ^2\right) }{(2-\alpha )^2(3-2\alpha )}B_1 \right\} \left. +\, \dfrac{16}{(2-\alpha )^2}B_1\right] \nonumber \\&=\dfrac{B_1}{16}( P c^4 + Q c^2 + R), \end{aligned}$$

(2.51)

where

$$\begin{aligned}&P = (2-\alpha )^{-2}(3-2\alpha )^{-1}\Big (\Big | -(3-2\alpha )\dfrac{ B_2^2}{B_1 } - (1-\alpha ) B_1^3 + (2-\alpha )^2 B_3 \nonumber \\&\qquad +\, \alpha (1-\alpha ) B_1 B_2\Big | - 2 (1-\alpha )^2 |B_2|- \alpha (1-\alpha )B_1^2 + (1-\alpha )^2 B_1\Big ), \end{aligned}$$

(2.52)

$$\begin{aligned}&Q = 4(2-\alpha )^{-2}(3-2\alpha )^{-1} \Big (2 (1-\alpha )^2|B_2|+\alpha (1-\alpha ) B_1^2- \left( 2-\alpha ^2\right) B_1 \Big ) \end{aligned}$$

(2.53)

$$\begin{aligned}&R = 16 (2-\alpha )^{-2} B_1. \end{aligned}$$

(2.54)

Thus, using (2.18) and (2.51), we get,

$$\begin{aligned} |a_2 a_4 - a_3^2| \le \dfrac{B_1}{16}\left\{ \begin{array}{ll} R, &{} Q\le 0, P\le -\frac{Q}{4}; \\ 16P + 4Q + R, &{} Q\ge 0, P\ge -\frac{Q}{8}\quad \text {or}\quad Q\le 0 , P\ge -\frac{Q}{4}; \\ \frac{4PR -Q^2}{4P}, &{} Q > 0, P\le -\frac{Q}{8}. \end{array}\right. \end{aligned}$$

where P, Q, areR are given by (2.52), (2.53), and (2.54), respectively. $\square$

Remark 2.18

When $\alpha =0$, Theorem 2.17 reduces to [26, Theorem 2]. Then Corollary 2.3 comes as a particular case.

Our next theorem gives the Fekete–Szegö inequality for functions in the class ${\mathcal {K}}_{\alpha }(\varphi )$.

Theorem 2.19

Let $\varphi$ be defined as in (2.1) and let the function $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots \in {\mathcal {K}}_{\alpha }(\varphi )$. Then we have the following:

• (1) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} (2-\alpha )B_1^2 \mu \le (1-\alpha )B_1^2 + (B_2-B_1), \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{1}{2-\alpha }\Big (B_2 +{(1-\alpha )}B_1^2 - (2-\alpha )\mu B_1^2 \Big ). \end{aligned}$$

• (2) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} (1-\alpha )B_1^2 + (B_2-B_1)\le (2-\alpha )B_1^2 \mu \le (1-\alpha )B_1^2 + (B_2+B_1), \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{B_1}{2-\alpha }. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} (1-\alpha )B_1^2 + (B_2 + B_1) \le (2-\alpha )B_1^2 \mu , \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{1}{2-\alpha }\Big ({-}B_2 - {(1-\alpha )}B_1^2 + (2-\alpha )\mu B_1^2\Big ). \end{aligned}$$

Proof

Using (2.48) and (2.49) we get,

$$\begin{aligned} a_3- \mu a_2^2&=\dfrac{1}{4(2-\alpha )}\Big ( 2 c_2 B_1 - {c_1^2 B_1}+ {c_1^2 B_2}+(1-\alpha ){c_1^2 B_1^2} - {\mu (2-\alpha ) B_1^2 c_1^2}\Big )\\&= \dfrac{B_1}{2(2-\alpha )}\left( c_2 - \dfrac{c_1^2}{2} \Big ( \Big (1-\dfrac{B_2}{B_1}\Big ) - (1-\alpha )B_1 + {(2-\alpha )}\mu B_1\Big )\right) \\&= \dfrac{B_1}{2(2-\alpha )}(c_2-\nu c_1^2), \end{aligned}$$

where $\nu = \dfrac{1}{2}\left( \dfrac{-B_2}{B_1}+1 - {(1-\alpha ) }B_1+ \mu (2-\alpha ) B_1 \right)$. Using Lemma 2.6, we get the desired result. $\square$

Remark 2.20

Bounds for the second and the third coefficients for f can be directly obtained from Theorem 2.15 as follows:

$$\begin{aligned} |a_2|\le {B_1}, \quad \end{aligned}$$

and

$$\begin{aligned} |a_3|\le \left\{ \begin{array}{ll} \dfrac{ B_2+(1-\alpha )B_1^2 }{2-\alpha }, &{} (1-\alpha )B_1^2 + (B_2 - B_1) \ge 0; \\ \dfrac{B_1}{2-\alpha }, &{} (1-\alpha )B_1^2 + (B_2 - B_1) \le 0 \quad \text {or} \\ &{} (1-\alpha )B_1^2 + (B_2 + B_1)\ge 0;\\ \dfrac{-B_2 - (1-\alpha )B_1^2}{2-\alpha }, &{}(1-\alpha )B_1^2 + (B_2 + B_1) \le 0. \end{array} \right. \end{aligned}$$

Let $\varphi : {\mathbb {D}}\rightarrow {\mathbb {C}}$ be an analytic function given by (2.1). For $0\le \alpha \le 1$, the class ${\mathcal {T}}_{\alpha }(\varphi )$ consists of functions $f\in {\mathcal {A}}$ satisfying the following subordination:

$$\begin{aligned} \Big (\dfrac{f(z)}{z}\Big )^\alpha \Big (1 + \dfrac{z f^{\prime \prime }(z)}{f^{\prime }(z)}\Big )^{1-\alpha } \prec \varphi (z). \end{aligned}$$

We see that

$$\begin{aligned} {\mathcal {T}}_{0}(\varphi )={\mathcal {K}}(\varphi )= \left\{ f \in {\mathcal {A}}:1+ \dfrac{z f^{\prime }(z)}{f(z)} \prec \varphi (z)\right\} \end{aligned}$$

is the class of Ma–Minda convex functions.

Theorem 2.21

Let the function $f \in {\mathcal {T}}_{\alpha }(\varphi )$ be given by $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots$. Then,

• (1) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned}&4 (1-\alpha )(2-\alpha )|6-7\alpha ||B_2| +2 (1-\alpha )|7\alpha ^2-8\alpha -6| B_1^2 \\&\quad \le ((2-\alpha )^2(12-11\alpha ) -2(1-\alpha )\left( 2 - \alpha \right) |6-7\alpha |) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |(1- \alpha )(67\alpha ^4 - 329 \alpha ^3 + 516 \alpha ^2 + 168 \alpha -576) B_1^4 - 12(2-\alpha )^4 (12-11\alpha ) {B_2^2}\\&\quad + 12 (6- 5\alpha )^2\left( 2 - \alpha \right) ^3 B_1 B_3 -24 (1-\alpha ) (2-\alpha )^2 (7\alpha ^2 -8 \alpha -6 )B_1^2 B_2\Big |\\&\quad -\, 12(2-\alpha )^4 (12-11\alpha ) B_1^2 \le 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2|\le \dfrac{B_1^2}{(6 - 5\alpha )^2}. \end{aligned}$$

• (2) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned}&4 (1-\alpha )(2-\alpha )|6-7\alpha ||B_2| +2 (1-\alpha )|7\alpha ^2-8\alpha -6| B_1^2\\&\quad \ge ((2-\alpha )^2(12-11\alpha )-2(1-\alpha )\left( 2 - \alpha \right) |6-7\alpha |) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |{(1-\alpha )(67\alpha ^4 - 329 \alpha ^3 + 516 \alpha ^2 + 168 \alpha -576)}B_1^4 - 12(2-\alpha )^4 (12-11\alpha ){B_2^2} \\&\quad + 12 (6- 5\alpha )^2\left( 2 - \alpha \right) ^3 B_1 B_3 -24 (1-\alpha ) (2-\alpha )^2 (7\alpha ^2 -8 \alpha -6 )B_1^2 B_2\Big |\\&\quad -\, 24(1-\alpha )(2-\alpha )^3 |6-7\alpha | |B_2|B_1 -12(1-\alpha )(2-\alpha )^2|7\alpha ^2-8\alpha -6| B_1^3\\&\quad -\, 6 (2- \alpha )^3 (2(1-\alpha ) |6- 7 \alpha | + (2-\alpha ) (12-11\alpha ))B_1^2 \ge 0, \end{aligned}$$

  or the conditions

  $$\begin{aligned}&4 (1-\alpha )(2-\alpha )|6-7\alpha ||B_2| +2 (1-\alpha )|7\alpha ^2-8\alpha -6| B_1^2 \\&\quad \le \Big ((2-\alpha )^2(12-11\alpha ) -2(1-\alpha )\left( 2 - \alpha \right) |6-7\alpha |\Big ) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&\Big |(1- \alpha )(67\alpha ^4 - 329 \alpha ^3 + 516 \alpha ^2 + 168 \alpha -576) B_1^4 - 12(2-\alpha )^4 (12-11\alpha ) {B_2^2} \\&\quad +12 (6- 5\alpha )^2\left( 2 - \alpha \right) ^3 B_1 B_3 -24 (1-\alpha ) (2-\alpha )^2 (7\alpha ^2 -8 \alpha -6 )B_1^2 B_2\Big | \\&\quad -\, 12(2-\alpha )^4(12-11\alpha ) B_1^2 \ge 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned}&\Big |a_2a_4 - a_3^2| \le \dfrac{1}{12(2-\alpha )^4(6-5\alpha )^2 (12-11\alpha )} \Big |{-12(2-\alpha )^4(12-11\alpha )}B_2^2 \\&\quad + {(1-\alpha )(67\alpha ^4 - 329 \alpha ^3 + 516 \alpha ^2 + 168 \alpha -576)}B_1^4 \\&\quad +12 {(2-\alpha )^3 (6-5\alpha )^2}B_1B_3 -24 (1-\alpha )(2-\alpha )^2 (7\alpha ^2 - 8\alpha -6)B_1^2 B_2 \Big |. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $B_3$ satisfy the conditions

  $$\begin{aligned}&4 (1-\alpha )(2-\alpha )|6-7\alpha ||B_2| +2 (1-\alpha )|7\alpha ^2-8\alpha -6| B_1^2 \\&\quad > \Big ((2-\alpha )^2(12-11\alpha ) -2(1-\alpha )\left( 2 - \alpha \right) |6-7\alpha |\Big ) B_1, \end{aligned}$$

  and

  $$\begin{aligned}&|{(1-\alpha )(67\alpha ^4 - 329 \alpha ^3 + 516 \alpha ^2 + 168 \alpha -576)}B_1^4 - 12(2-\alpha )^4 (12-11\alpha ){B_2^2} \\&\quad + 12 (6- 5\alpha )^2\left( 2 - \alpha \right) ^3 B_1 B_3 -24 (1-\alpha ) (2-\alpha )^2 (7\alpha ^2 -8 \alpha -6 )B_1^2 B_2|\\&\quad -\, 24(1-\alpha )(2-\alpha )^3 |6-7\alpha | |B_2|B_1 -12(1-\alpha )(2-\alpha )^2|7\alpha ^2-8\alpha -6| B_1^3\\&\quad -\, 6 (2- \alpha )^3 \Big (2(1-\alpha ) |6- 7 \alpha | + (2-\alpha ) (12-11\alpha )\Big )B_1^2 \le 0, \end{aligned}$$

  then the second Hankel determinant satisfies

  $$\begin{aligned} |a_2a_4 - a_3^2| \le \dfrac{B_1^2 M}{(2 (6-5\alpha )^2(12-11\alpha ))N}, \end{aligned}$$

  where

  $$\begin{aligned} M&= \Big | {2(1-\alpha )(12-11\alpha )(67\alpha ^4-329\alpha ^3 + 516\alpha ^2 + 168\alpha -576)} B_1^4 \\&\quad -\, {24(2-\alpha )^4(12-11\alpha )^2} B_2^2 + {24(2-\alpha )^3(12-11\alpha )(6-5\alpha )^2} B_1 B_3 \\&\quad - 48(1-\alpha )(2-\alpha )^2(12-11\alpha )(7\alpha ^2 -\,8\alpha -6)B_1^2 B_2 \Big |\\&\quad -\, 24(1-\alpha )(2-\alpha )|7\alpha ^2 -8\alpha -6|\Big ((2-\alpha )(12-11\alpha )+2(1-\alpha )|6-7\alpha |\Big )B_1^3 \\&\quad -\, {48(1-\alpha )(2-\alpha )^2|6-7\alpha |\Big (2(1-\alpha )|6-7\alpha | +(2-\alpha )(12-11\alpha )\Big )}B_1 |B_2| \\&\quad -\, 6(2-\alpha )^2\Big (4(1-\alpha )|6-7\alpha |(2-\alpha )(12-11\alpha ) +(2-\alpha )^2(12-11\alpha )^2 \\&\quad +\, 4(1-\alpha )^2(6-7\alpha )^2\Big )B_1^2 -{24(1-\alpha )^2(7\alpha ^2-8\alpha -6)^2}B_1^4\\&\quad - {96(1-\alpha )^2(6-7\alpha )^2(2-\alpha )^2}B_2^2 -\, 96(1-\alpha )^2(2-\alpha )|6-7\alpha ||7\alpha ^2\\&\quad -8\alpha -6|B_1^2 |B_2| \end{aligned}$$

  and

  $$\begin{aligned} N&=\Big |{(1-\alpha )(67\alpha ^4 -329 \alpha ^3 +516 \alpha ^2 +168 \alpha -576)}B_1^4 \\&\quad -\, {12(2-\alpha )^4(12-11\alpha )}B_2^2 + {12(2-\alpha )^3(6-5\alpha )^2}B_1 B_3 \\&\quad -24 {(1-\alpha )(2-\alpha )^2(7\alpha ^2-8\alpha -6)}B_1^2 B_2\Big | \\&\quad - {24 (1-\alpha )(2-\alpha )^2|7\alpha ^2 -8\alpha -6|}B_1^3 - 48(1-\alpha )(2-\alpha )^3|6\\&\quad -\, 7\alpha |B_1 |B_2|-\, {24(1-\alpha )(2-\alpha )^3|6-7\alpha |}B_1^2. \end{aligned}$$

Proof

Since $f \in {\mathcal {T}}_{\alpha }(\varphi ),$ there is an analytic function $w(z)= w_1 z + w_2 z^2 + \cdots \in \varOmega$, such that

$$\begin{aligned} \Big (\dfrac{f(z)}{z}\Big )^{\alpha } \Big (1+\dfrac{z f^{\prime \prime }(z)}{f^{\prime }(z)}\Big )^{1-\alpha }= \varphi (w(z)). \end{aligned}$$

(2.55)

Define $p_1(z)$ by

$$\begin{aligned} p_1(z)=\dfrac{1+ w(z)}{1-w(z)}=1 + c_1 z + c_2 z^2 + \cdots , \end{aligned}$$

then this implies

$$\begin{aligned} w(z)= \dfrac{p_1(z)-1}{p_1(z)+1}= \dfrac{1}{2}\left( c_1 z +\left( c_2-\dfrac{c_1^2}{2}\right) z^2+\cdots \right) . \end{aligned}$$

Clearly $p_1$ is analytic in ${\mathbb {D}}$ with $p_1(0)=1$ and $p_1\in {\mathcal {P}}$. Then, since $\varphi (z)= 1 + B_1 z + B_2 z^2 + B_3 z^3+ \cdots$, we get

$$\begin{aligned} \varphi \left( \dfrac{p_1(z)-1}{p_1(z)+1}\right) =1 + \dfrac{1}{2}B_1 c_1z + \left( \dfrac{1}{2}B_1\left( c_2-\dfrac{c_1^2}{2}\right) +\dfrac{1}{4}B_2 c_1^2\right) z^2 +\cdots . \end{aligned}$$

(2.56)

Also, the Taylor series expansion of f gives

$$\begin{aligned} \Big (\dfrac{f(z)}{z}\Big )^\alpha \Big (1+\dfrac{z f^{\prime \prime }(z)}{f^{\prime }(z)}\Big )^{1-\alpha }&= 1+ (2-\alpha )a_2 z + \Big ((6-5\alpha ) a_3-\,\frac{1}{2}(1-\alpha )(8+\alpha )a_2^2\Big )z^2 \nonumber \\&\quad +\, \Big ((12-11\alpha ) a_4 - (1-\alpha )(18+5\alpha ) a_2 a_3 \nonumber \\&\quad +\,\frac{1}{6} (1-\alpha )(\alpha ^2 +28 \alpha + 48) a_2^3\Big )z^3 +\cdots . \end{aligned}$$

(2.57)

Then from (2.55), (2.56) and (2.57), we get

$$\begin{aligned}&a_2= \dfrac{B_1 c_1}{2(2-\alpha )}. \end{aligned}$$

(2.58)

$$\begin{aligned}&a_3=\dfrac{1}{8(6- 5\alpha )}\left[ 4B_1 c_2 - \dfrac{2(B_1 - B_2)(2-\alpha )^2-(1-\alpha )(8+\alpha )B_1^2}{(2-\alpha )^2} c_1^2\right] ,\nonumber \\ \end{aligned}$$

(2.59)

$$\begin{aligned}&a_4=\dfrac{1}{8(12-11\alpha )}\left[ 4 B_1 c_3+ \left( {B_1}-\dfrac{(1-\alpha )(18+ 5\alpha )}{(2-\alpha )(6- 5\alpha )}B_1^2 - {2B_2} + {B_3} \right. \right. \nonumber \\&\left. \qquad -\,\dfrac{(1-\alpha )(10 \alpha ^3 + 25 \alpha ^2+ 186 \alpha -144)}{6(2-\alpha )^3 (6-5\alpha ) }B_1^3 + \dfrac{(1-\alpha )(18+ 5\alpha )}{(2-\alpha )(6-5\alpha ) }B_1B_2 \right) c_1^3 \nonumber \\&\left. \qquad +\,\left( -{4} B_1 + \dfrac{2(1-\alpha )(18+ 5\alpha )}{(2-\alpha )(6-5\alpha )}B_1^2 + 4 B_2\right) c_1 c_2\right] . \end{aligned}$$

(2.60)

Thus,

$$\begin{aligned} a_2 a_4 - a_3^2&=\dfrac{1}{8}B_1\left[ \left\{ \dfrac{ (1-\alpha )(67 \alpha ^4 -329\alpha ^3 + 516 \alpha ^2 +168 \alpha -576)}{24(2-\alpha )^4 (6-5\alpha )^2 (12-11\alpha )}B_1^3 \right. \right. \\&\quad -\, \dfrac{1}{2(6-5\alpha )^2}\dfrac{B_2^2}{B_1}+ \dfrac{1}{2(2-\alpha )(12- 11\alpha )}B_3\\&\quad +\, \dfrac{(1-\alpha )(6- 7\alpha )}{(2-\alpha )(6-5\alpha )^2 (12-11\alpha )}(B_1 - 4B_2) \\&\left. \quad +\, \dfrac{(1-\alpha )(7\alpha ^2 -8\alpha -6)}{(2-\alpha )^2(6-5\alpha )^2 (12-11\alpha )}B_1(B_1 - B_2) \right\} c_1^4 \\&\quad +\, \left\{ \dfrac{4(1-\alpha )(6- 7\alpha )}{(2-\alpha )(6-5\alpha )^2 (12-11\alpha )}(B_2-B_1) \right. \\&\left. \quad -\,\dfrac{2 (1-\alpha )(7\alpha ^2 -8\alpha -6)}{(2-\alpha )^2(6-5\alpha )^2 (12-11\alpha )} B_1^2 \right\} c_2 c_1^2 - \dfrac{2}{(6-5\alpha )^2}B_1 c_2^2\\&\left. \quad +\,\dfrac{2}{(2-\alpha )(12- 11 \alpha )}B_1 c_1 c_3\right] . \end{aligned}$$

Again, proceeding as in the proof of Theorem 2.1, we see that $|a_2 a_4 - a_3^2|$ is bounded by

$$\begin{aligned} G(c)&= \dfrac{B_1}{8}\left[ c^4\left\{ \Bigg | - \dfrac{1}{2(6-5\alpha )^2}\dfrac{B_2^2}{B_1}\right. \right. \nonumber \\&\quad +\, \dfrac{ (1-\alpha )(67 \alpha ^4 -329\alpha ^3 + 516 \alpha ^2 +168 \alpha -576)}{24(2-\alpha )^4 (6-5\alpha )^2 (12-11\alpha )}B_1^3 \nonumber \\&\quad +\, \dfrac{1}{2(2-\alpha )(12- 11\alpha )}B_3 - \dfrac{(1-\alpha )(7\alpha ^2 -8 \alpha -6)}{(2-\alpha )^2(6-5\alpha )^2 (12-11\alpha )}B_1 B_2\Bigg | \nonumber \\&\quad -\, \dfrac{(1-\alpha )|7\alpha ^2 -8 \alpha -6|}{(2-\alpha )^2(6-5\alpha )^2 (12-11\alpha )}B_1^2\nonumber \\&\left. \quad -\, \dfrac{(1-\alpha )|6-7\alpha |}{(2-\alpha )(6-5\alpha )^2 (12-11\alpha )}(B_1 + 2 |B_2|)\right\} \nonumber \\&\quad +\,4c^2 \left\{ \dfrac{(1-\alpha )|6-7\alpha |}{(2-\alpha )(6-5\alpha )^2 (12-11\alpha )}( B_1 + 2|B_2| )\right. \nonumber \\&\quad +\,\dfrac{(1-\alpha )|7\alpha ^2 -8 \alpha -6|}{(2-\alpha )^2(6-5\alpha )^2 (12-11\alpha )} B_1^2 \nonumber \\&\left. \left. \quad -\, \dfrac{1}{2(6-5\alpha )^2}B_1\right\} +\dfrac{8B_1}{(6-5\alpha )^2}\right] \nonumber \\ =&\, \dfrac{B_1}{8}\left( P c^4 + Q c^2 + R\right) , \end{aligned}$$

(2.61)

where

$$\begin{aligned} P&= \frac{1}{24}(2-\alpha )^{-4} (6-5\alpha )^{-2} (12-11\alpha )^{-1}\Big (\Big | - 12(2-\alpha )^4(12-11\alpha )\dfrac{B_2^2}{B_1} \nonumber \\&\quad +\,(1-\alpha )(67 \alpha ^4 -329\alpha ^3 + 516 \alpha ^2 +168 \alpha -576)B_1^3\nonumber \\&\quad +\, 12(2-\alpha )^3(6-5\alpha )^2 B_3 - 24(1-\alpha )(2-a)^2(7\alpha ^2 -8 \alpha -6) B_1 B_2\Big |\nonumber \\&\quad - 48(1-\alpha )(2-\alpha )^3|6-7\alpha | |B_2| -\,24 (1-\alpha )(2-\alpha )^2|7\alpha ^2 -8 \alpha - 6|B_1^2\nonumber \\&\quad - 24(1-\alpha )(2-\alpha )^3|6-7\alpha |B_1\Big ), \end{aligned}$$

(2.62)

$$\begin{aligned} Q&= 2(2-\alpha )^{-2}(6-5\alpha )^{-2} (12-11\alpha )^{-1} \Big (2(1-\alpha )(2-\alpha )|6-7\alpha |(2|B_2| + B_1)\nonumber \\&\quad +2(1-\alpha )|7\alpha ^2 -8 \alpha -6| B_1^2- (2-\alpha )^2(12-11\alpha )B_1\Big ), \end{aligned}$$

(2.63)

$$\begin{aligned} R&= 8(6-5\alpha )^{-2} B_1. \end{aligned}$$

(2.64)

Thus using (2.18) and (2.61) we get,

$$\begin{aligned} |a_2 a_4 - a_3^2| \le \dfrac{B_1}{8}\left\{ \begin{array}{l@{\quad }l} R, &{} Q\le 0, P\le -\frac{Q}{4}; \\ 16P + 4Q + R, &{} Q\ge 0, P\ge -\frac{Q}{8} or Q\le 0, P\ge -\frac{Q}{4}; \\ \frac{4PR -Q^2}{4P}, &{} Q > 0, P\le -\frac{Q}{8}. \end{array} \right. \end{aligned}$$

where P, Q, R are given by (2.62), (2.63) and (2.64), respectively. $\square$

Remark 2.22

When $\alpha =0$, Theorem 2.21 reduces to [26, Theorem 2].

Our next theorem gives the Fekete–Szegö inequality for functions in the class ${\mathcal {T}}_{\alpha }(\varphi )$.

Theorem 2.23

Let $\varphi$ be defined as in (2.1) and let the function $f(z)=z + a_2 z^2 + a_3 z^3 +\cdots \in {\mathcal {T}}_{\alpha }(\varphi )$. Then we have the following:

• (1) If $B_1$, $B_2$, and $\mu$ satisfy the condition

  $$\begin{aligned} 2(6-5\alpha )B_1^2 \mu \le (1-\alpha )(8+\alpha )B_1^2 + 2(2-\alpha )^2(B_2-B_1), \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{1}{2(6-5\alpha )}\Big (2B_2 +\dfrac{(1-\alpha )(8+\alpha )}{(2-\alpha )^2}B_1^2 - \dfrac{2(6-5\alpha )}{(2-\alpha )^2}\mu B_1^2 \Big ). \end{aligned}$$

• (2) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned}&(1-\alpha )(8+\alpha )B_1^2 + 2(2-\alpha )^2(B_2-B_1)\\&\quad \le 2(6-5\alpha )B_1^2 \mu \le (1-\alpha )(8+\alpha )B_1^2 + 2(2-\alpha )^2(B_2+B_1), \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{B_1}{6-5\alpha }. \end{aligned}$$

• (3) If $B_1$, $B_2$ and $\mu$ satisfy the condition

  $$\begin{aligned} (1-\alpha )(8+\alpha )B_1^2 + 2(2-\alpha )^2(B_2 + B_1) \le 2(6-5\alpha )B_1^2 \mu , \end{aligned}$$

  then

  $$\begin{aligned} |a_3 - \mu a_2^2|\le \dfrac{1}{2(6-5\alpha )}\Big ({-}2B_2 - \dfrac{(1-\alpha )(8+\alpha )}{(2-\alpha )^2}B_1^2 + \dfrac{2(6-5\alpha )}{(2-\alpha )^2}\mu B_1^2\Big ). \end{aligned}$$

Proof

Using (2.58) and (2.59) we get,

$$\begin{aligned} a_3- \mu a_2^2&=\dfrac{1}{8(6-5\alpha )}\Big ( 4 c_2 B_1 - {2 c_1^2 B_1}+ {2 c_1^2 B_2}+\dfrac{(1-\alpha )(8+\alpha )}{(2-\alpha )^2}{c_1^2 B_1^2} \\&\qquad -\, \mu \dfrac{2(6-5\alpha )}{(2-\alpha )^2} B_1^2 c_1^2 \Big )\\&= \dfrac{B_1}{2(6-5\alpha )}\left( c_2 - \dfrac{c_1^2}{4} \Big ( 2\Big (1-\dfrac{B_2}{B_1}\Big ) - \dfrac{(1-\alpha )(8+\alpha )}{(2-\alpha )^2}B_1\right. \\&\qquad \left. +\, \dfrac{2(6-5\alpha )}{(2-\alpha )^2}\mu B_1\Big )\right) \\&= \dfrac{B_1}{2(6-5\alpha )}\left( c_2-\nu c_1^2\right) , \end{aligned}$$

where $\nu = \dfrac{1}{4}\left[ 2\Big (1-\dfrac{B_2}{B_1}\Big ) - \dfrac{(1-\alpha )(8+\alpha )}{(2-\alpha )^2}B_1 + \dfrac{2(6-5\alpha )}{(2-\alpha )^2}\mu B_1 \right]$. Using Lemma 2.6 we get the desired result. $\square$

Remark 2.24

Bounds for the second and the third coefficients for f can be directly obtained from Theorem 2.17 as follows:

$$\begin{aligned}|a_2|\le \dfrac{B_1}{2-\alpha },\end{aligned}$$

and

$$\begin{aligned}|a_3|\le \left\{ \begin{array}{l@{\quad }l} \dfrac{ B_2+\frac{(1-\alpha )(8+\alpha )}{2(2-\alpha )^2}B_1^2 }{6-5\alpha }, &{} (1-\alpha )(8+\alpha )B_1^2 + 2(2-\alpha )^2(B_2-B_1) \ge 0; \\ \dfrac{B_1}{6-5\alpha }, &{} (1-\alpha )(8+\alpha )B_1^2 + 2(2-\alpha )^2(B_2-B_1) \le 0 \quad \text {or} \\ &{} (1-\alpha )(8+\alpha )B_1^2 + 2(2-\alpha )^2(B_2+B_1)\ge 0;\\ \dfrac{-B_2 - \frac{(1-\alpha )(8+\alpha )}{2(2-\alpha )^2}B_1^2}{6-5\alpha }, &{}(1-\alpha )(8+\alpha )B_1^2 + 2(2-\alpha )^2(B_2 + B_1)\le 0. \end{array} \right. \end{aligned}$$