Coefficient functionals for alpha-convex functions associated with the exponential function

Abstract

Let ${\mathcal {A}}$ be the class of all normalized analytic functions f in the unit disk ${\mathbb {D}}=\{z\in {\mathbb {C}}:|z|<1 \}$, given by $f(z)=z+\sum _{n=2}^{\infty }a_n z^n$ for $z\in {\mathbb {D}}$. We give the sharp bound for the modulus of the functional $a_2 a_3-a_4$, and the second Hankel determinant $H_{2,2}(f)=a_2a_4-a_3^2$ when $f\in {\mathcal {M}}_\alpha (\exp )\subset {\mathcal {A}},$ the class of $\alpha$-convex functions ($0\le \alpha \le 1$), associated with the exponential function.

1 Introduction

Let ${\mathcal {H}}$ denote the class of all analytic functions in ${\mathbb {D}}:=\{z\in {\mathbb {C}}: |z|<1\}$ and ${\mathcal {A}}$ be the subclass of functions f of the form

$$\begin{aligned} f(z)=z+\sum _{n=2}^\infty a_nz^n,\quad z\in {\mathbb {D}}. \end{aligned}$$

(1)

Denote by ${\mathcal {S}}\subset {\mathcal {A}}$ the subclass of univalent functions.

For $\alpha \in [0,1],$ denote by ${\mathcal {M}}_\alpha \subset {\mathcal {A}}$, the so-called $\alpha$-convex functions f satisfying

$$\begin{aligned} {\text {Re}}\left\{ (1-\alpha )\frac{zf'(z)}{f(z)}+ \alpha \left( 1+\frac{zf''(z)}{f'(z)}\right) \right\} >0,\quad z\in {\mathbb {D}}. \end{aligned}$$

The class ${\mathcal {M}}_\alpha$ was introduced by Mocanu [16] (see also [8, Vol. I, pp. 142–147]), who showed that ${\mathcal {M}}_\alpha \subset {\mathcal {S}}.$

We note that when $\alpha =0$ the class ${\mathcal {M}}_0$ reduces to the class of starlike functions denoted by ${\mathcal {S}}^*$, introduced by Alexander [1] ([17], see also [8, Vol. I, Chapter 8]), and when $\alpha =1$ the class ${\mathcal {M}}_1$ reduces to the class of convex functions denoted by ${\mathcal {S}}^c$ defined by Study [24] (see also [8, Vol. I, Chapter 8]). In [15] it was shown that ${\mathcal {M}}_\alpha \subset {\mathcal {M}}_0$ for every $\alpha \in [0,1],$ and so all functions in ${\mathcal {M}}_\alpha$ are starlike, which was observed by Sakaguchi [23] before the advent of the $\alpha$-convexity concept (cf. [8, Vol. I. pp. 142-143]). Also in [15] Mocanu and Reade showed that ${\mathcal {M}}_{\alpha _1}\subset {\mathcal {M}}_{\alpha _2}$ for every $0\le \alpha _2\le \alpha _1\le 1$, and Mocanu [16], showed that functions in ${\mathcal {M}}_\alpha$ have some interesting geometrical properties.

Thus the class ${\mathcal {M}}_\alpha$ creates a “continuous passage” on $\alpha \in [0,1]$ from the family of starlike functions ${\mathcal {S}}^*={\mathcal {M}}_0$ to the family of convex functions ${\mathcal {M}}_1={\mathcal {S}}^c.$

The class ${\mathcal {M}}_\alpha$ plays an important role in geometric function theory and has been studied by many authors (e.g., [20, 19, Chapter 7] for further references).

We say that a function $f\in {\mathcal {H}}$ is subordinate to a function $g\in {\mathcal {H}}$, if there exists a function $\omega \in {\mathcal {H}}$ with $\omega (0)=0$ and $|\omega (z)|<1$ for $z\in {\mathbb {D}}$ (called a Schwarz function), such that $f(z)=g(\omega (z))$ for $z\in {\mathbb {D}}.$ We write $f\prec g.$ If g is univalent and $f(0)=g(0)$, then $f\prec g$ is equivalent to $f({\mathbb {D}})\subseteq g({\mathbb {D}})$.

Suppose that the function $\varphi$ is analytic and univalent in ${\mathbb {D}}$ and is starlike with respect to the point $\varphi (0) =1$ with $\varphi ^{\prime }(0) >0$, and is symmetric about the real axis, then Ma and Minda [13] generalized the classes of starlike and convex functions as follows:

$$\begin{aligned} {\mathcal {S}}^{*}(\varphi ):=\left\{ f\in {\mathcal {A}}:\frac{zf^{\prime }(z)}{f(z)}\prec \varphi (z),\ z\in {\mathbb {D}}\right\} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {C}}(\varphi ):=\left\{ f\in {\mathcal {A}}:1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}\prec \varphi (z),\ z\in {\mathbb {D}}\right\} . \end{aligned}$$

Clearly, $\varphi (z)=\exp (z)$, $z\in {\mathbb {D}},$ is a valid choice of the super-ordinate, which appears to have been first considered by Mendiratta et al. [14], and recently several authors have considered problems in the resulting classes of starlike and convex functions (see e.g. [25, 26], and the references therein).

Also Breaz et al. [2] have recently defined the following subclass of ${\mathcal {M}}_{\alpha }$.

Definition 1.1

A function $f\in {{\mathcal {A}}}$ is said to be in the class $\mathcal M_{\alpha }(\exp ),\ \alpha \in [0,1],$ if f satisfies the following condition:

$$\begin{aligned} (1-\alpha )\dfrac{zf'(z)}{f(z)}+\alpha \left( 1+ \dfrac{zf''(z)}{f'(z)}\right) \prec \exp (z),\quad z\in {\mathbb {D}}. \end{aligned}$$

(2)

In this paper we consider problems in the class $\mathcal M_{\alpha }(\exp )$, $\alpha \in [0,1]$, of $\alpha$-convex functions associated with the exponential function, noting that ${\mathcal {S}}^*(\exp ):={\mathcal {M}}_0(\exp )$ and ${\mathcal {C}}(\exp ):={\mathcal {M}}_1(\exp ).$

We also note that in [2], Breaz et al. gave non-sharp bounds for various coefficient functionals in ${\mathcal {M}}_{\alpha }$.

In recent years, there has been a great deal of attention given to finding bounds for the modulus of the second Hankel determinant $H_{2,2}(f)=a_2a_4-a_3^3,$ when f belongs to various subclasses of ${\mathcal {A}}$ (cf. [4] and [9] with further references).

In this paper, we find the sharp bound for $|H_{2,2}(f)|$ when $f\in {\mathcal {M}}_{\alpha }(\exp ),\ \alpha \in [0,1],$ together with the sharp bound of the functional

$$\begin{aligned} |J_{2,3}(f)|:=|a_2a_3-a_4|, \end{aligned}$$

when $f\in {\mathcal {M}}_{\alpha }(\exp ),\ \alpha \in [0,1].$

Note that $|J_{2,3}(f)|$ is a specific case of the generalized Zalcman functional $|a_na_m-a_{n+m+1}|$ investigated by Ma [12] for $f\in {\mathcal {S}}$ (cf. [21] for further references), and that sharp bounds for $|J_{2,3}(f)|$ for some specific general cases such as ${\mathcal {S}}^*(\varphi )$ and ${\mathcal {C}}(\varphi )$ have been found in [5].

2 Preliminary lemmas

Denote by ${{\mathcal {P}}}$, the class of analytic functions p in ${\mathbb {D}}$ with positive real part on ${\mathbb {D}}$ given by

$$\begin{aligned} p(z)=1+\sum _{n=1}^{\infty }c_n z^n,\quad z\in {\mathbb {D}}. \end{aligned}$$

(3)

Clearly if $\omega$ is a Schwarz function, then there exists $p\in {\mathcal {P}}$ such that

$$\begin{aligned} \omega (z)=\dfrac{p(z)-1}{p(z)+1},\quad z\in {\mathbb {D}}, \end{aligned}$$

(4)

and vice versa, if $p\in {\mathcal {P}}$, then there exists a Schwarz function $\omega \in {\mathcal {H}}$ such that

$$\begin{aligned} p(z)=\dfrac{1+\omega (z)}{1-\omega (z)},\quad z\in {\mathbb {D}}. \end{aligned}$$

In the proofs of our results, we will use the following lemma given in [6]. It contains the well known formulas (5) for $c_1$ [3] and (6) for $c_2$ (e.g., [18, p. 166]). The formula (7) for $c_3$ in the case when $\zeta _1\in [0,1]$ is due to Libera and Złotkiewicz [10] and [11]. Let $\overline{{\mathbb {D}}}:=\{z\in {\mathbb {C}}:|z|\le 1\}$ and ${\mathbb {T}}:=\{z\in {\mathbb {C}}:|z|=1\}.$

Lemma 2.1

If $p \in {{\mathcal {P}}}$ and is given by (3), then

$$\begin{aligned} c_1= & \, 2\zeta _1, \end{aligned}$$

(5)

$$\begin{aligned} c_2= & \, 2\zeta _1^2 + 2(1-|\zeta _1|^2)\zeta _2 \end{aligned}$$

(6)

and

$$\begin{aligned} c_3 = 2\zeta _1^3+2(1-|\zeta _1|^2)(2\zeta _1-\overline{\zeta _1}\zeta _2)\zeta _2 + 2(1-|\zeta _1|^2)(1-|\zeta _2|^2)\zeta _3 \end{aligned}$$

(7)

for some $\zeta _1,\zeta _2, \zeta _3 \in \overline{{\mathbb {D}}}.$

For $\zeta _1 \in {\mathbb {T}}$, there is a unique function $p \in {{\mathcal {P}}}$ with $c_1$ as in (5), namely,

$$\begin{aligned} p(z) = \frac{1+\zeta _1 z}{1-\zeta _1 z}, \quad z\in {\mathbb {D}}. \end{aligned}$$

For $\zeta _1\in {\mathbb {D}}$ and $\zeta _2 \in {\mathbb {T}}$, there is a unique function $p \in {{\mathcal {P}}}$ with $c_1$ and $c_2$ as in (6) and (7), namely,

$$\begin{aligned} p(z) = \frac{1+( \overline{\zeta }_1 \zeta _2 +\zeta _1 )z + \zeta _2 z^2}{1+( \overline{\zeta }_1 \zeta _2 -\zeta _1 )z - \zeta _2 z^2}, \quad z\in {\mathbb {D}}. \end{aligned}$$

(8)

We will also use the following lemma.

Lemma 2.2

[7] For real numbers A, B, C, let

$$\begin{aligned} Y(A,B,C) := \max \left\{ |A+Bz+Cz^2|+1-|z|^2: z\in \overline{{\mathbb {D}}}\right\} . \end{aligned}$$

If $AC\ge 0,$ then

$$\begin{aligned} Y(A,B,C)=\left\{ \begin{array}{ll} |A|+|B|+|C|, &{} |B|\ge 2(1-|C|),\\ 1+|A|+\dfrac{B^2}{4(1-|C|)}, &{} |B|<2(1-|C|). \end{array} \right. \end{aligned}$$

If $AC<0,$ then

$$\begin{aligned} Y(A,B,C)=\left\{ \begin{array}{lll} 1-|A|+\dfrac{B^2}{4(1-|C|)}, &{} -4AC(C^{-2}-1)\le B^2 \wedge |B|<2(1-|C|), \\ 1+|A|+\dfrac{B^2}{4(1+|C|)}, &{} B^2<\min \left\{ 4(1+|C|)^2,-4AC(C^{-2}-1)\right\} , \\ R(A,B,C), &{} \mathrm{otherwise} , \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} R(A,B,C):=\left\{ \begin{array}{lll} |A|+|B|-|C|, &{} |C|(|B|+4|A|)\le |AB|,\\ -|A|+|B|+|C|, &{} |AB|\le |C|(|B|-4|A|), \\ (|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}, &{} \mathrm{otherwise}. \end{array} \right. \end{aligned}$$

3 The Zalcman functional

We first consider the Zalcman functional $|a_2a_3-a_4|,$ noting that a non-sharp inequality was found in [2].

Theorem 3.1

Let $\alpha \in [0,1].$ If $f\in {\mathcal {M}}_{\alpha }(\exp )$ and is given by (1), then

$$\begin{aligned} &|a_2a_3-a_4|\\&\le {\left\{ \begin{array}{ll} \dfrac{2(\alpha +2)(4\alpha +1)J(\alpha )}{9(\alpha +1)(2\alpha +1)(3\alpha +1)(26\alpha ^3+92\alpha ^2+49\alpha +7)}, \quad &{} \alpha \in \left[ 0,\alpha '\right] ,\\ \dfrac{1}{3(3\alpha +1)}, &{}\alpha \in \left( \alpha ',1\right] , \end{array}\right. } \end{aligned}$$

(9)

where $J(\alpha ):=\sqrt{2(26\alpha ^3+92\alpha ^2+49\alpha +7)(4\alpha +1)(\alpha +2)(\alpha +1)}$ and $\alpha '\approx 0.814445$ is the unique root in [0, 1] of the equation

$$\begin{aligned} 424\alpha ^6+1728\alpha ^5+1014\alpha ^4-1134\alpha ^3-735\alpha ^2-108\alpha -1=0. \end{aligned}$$

Both inequalities are sharp.

Proof

Fix $\alpha \in [0,1]$ and let $f\in {\mathcal {M}}_{\alpha }(\exp )$ be of the form (1). Then by (2), we can write

$$\begin{aligned} (1-\alpha )\dfrac{zf'(z)}{f(z)}+\alpha \left( 1+ \dfrac{zf''(z)}{f'(z)}\right) = \exp (\omega (z)), \quad z\in {\mathbb {D}}, \end{aligned}$$

(10)

where $\omega$ is a Schwarz function. Thus there exists $p\in {\mathcal {P}}$ given by (3) such that (4) is satisfied, and so (10) can be written as

$$\begin{aligned} (1-\alpha )\dfrac{zf'(z)}{f(z)}+\alpha \left( 1+ \dfrac{zf''(z)}{f'(z)}\right) = \exp \left( \dfrac{p(z)-1}{p(z)+1}\right) , \quad z\in {\mathbb {D}}. \end{aligned}$$

(11)

Substituting (1) and (3) into (11) and equating the coefficients gives

$$\begin{aligned} \begin{aligned} a_2&= \dfrac{c_1}{2(1+\alpha )}, \quad a_3 =\dfrac{c_2}{4(1+2\alpha )}+\dfrac{c_1^2(1+4\alpha -\alpha ^2)}{16(1+2\alpha )(1+\alpha )^2},\\ a_4&= \dfrac{c_3}{6(1+3\alpha )}-\dfrac{c_1c_2(4\alpha ^2-9\alpha -1)}{24(1+3\alpha )(1+2\alpha )(1+\alpha )} \\&\quad +\dfrac{c_1^3(4\alpha ^4-31\alpha ^3+21\alpha ^2-17\alpha -1)}{288(1+\alpha )^3(1+2\alpha )(1+3\alpha )}. \end{aligned} \end{aligned}$$

(12)

Since both the class ${\mathcal {M}}_{\alpha }(\exp )$ and the functional ${\mathcal {M}}_{\alpha }(\exp )\ni f\mapsto |a_2a_3-a_4|$ are rotationally invariant, without loss of generality we may assume that $c_1 \in [0,2],$ i.e., by (5) that $\zeta _1\in [0,1].$ Using Lemma 2.1 in (12) we then obtain

$$\begin{aligned} |a_2a_3-a_4|= \frac{1}{144(3\alpha +1)(2\alpha +1)(\alpha +1)^2} |\Psi |, \end{aligned}$$

(13)

where

$$\begin{aligned} \begin{aligned} \Psi&:= (2\alpha ^3-4\alpha ^2-35\alpha -5)c_1^3-12(\alpha +1)(2\alpha ^2+1)c_1c_2\\&\quad +24(\alpha +1)^2(2\alpha +1)c_3\\&= 8\left[ (2\alpha ^3+14\alpha ^2-17\alpha -5)\zeta _1^3+6(\alpha +1)(2\alpha ^2+6\alpha +1)(1-\zeta _1^2)\zeta _1\zeta _2\right. \\&\quad -6(\alpha +1)^2(2\alpha +1)(1-\zeta _1^2)\zeta _1\zeta _2^2\\&\quad \left. +6(\alpha +1)^2(2\alpha +1)(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _3 \right] \end{aligned} \end{aligned}$$

(14)

for some $\zeta _1,\zeta _2,\zeta _3 \in \overline{{\mathbb {D}}}$.

(A) Suppose first that $\zeta _1=1.$ Note now that

$$\begin{aligned} 2\alpha ^3+14\alpha ^2-17\alpha -5<0,\quad \alpha \in [0,1], \end{aligned}$$

(15)

and that from (13) and (14) we have

$$\begin{aligned} |a_2a_3-a_4|= \frac{-2\alpha ^3-14\alpha ^2+17\alpha +5}{18(3\alpha +1)(2\alpha +1)(\alpha +1)^2}=:a. \end{aligned}$$

(16)

(B) Suppose next that $\zeta _1\in [0,1).$ Using the fact that $|\zeta _3|\le 1$, we obtain from (14) that

$$\begin{aligned} |\Psi |\le 48(1-\zeta _1^2)(2\alpha +1)(\alpha +1)^2 \Phi (A,B,C), \end{aligned}$$

where

$$\begin{aligned} \Phi (A,B,C) := \left|A +B\zeta _2 +C\zeta _2^2 \right|+ 1 -|\zeta _2|^2, \end{aligned}$$

with

$$\begin{aligned} A:=\frac{(2\alpha ^3+14\alpha ^2-17\alpha -5)\zeta _1^3}{6(2\alpha +1)(\alpha +1)^2(1-\zeta _1^2)},\quad B:=\frac{(2\alpha ^2+6\alpha +1)\zeta _1}{(2\alpha +1)(\alpha +1)},\quad C:=-\zeta _1. \end{aligned}$$

Hence and from (15) it follows that $AC>0.$

(B1) Consider first the condition $|B|\ge 2(1-|C|),$ i.e.,

$$\begin{aligned} \frac{(2\alpha ^2+6\alpha +1)\zeta _1}{(2\alpha +1)(\alpha +1)}\ge 2\left( 1-\zeta _1\right) , \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{3\zeta _1(2\alpha ^2+4\alpha +1)-2(2\alpha +1)(\alpha +1)}{(2\alpha +1)(\alpha +1)}\ge 0 \end{aligned}$$

and is true when $\zeta _1\ge \zeta ',$ where

$$\begin{aligned} \zeta ':=\frac{2(2\alpha +1)(\alpha +1)}{3(2\alpha ^2+4\alpha +1)}. \end{aligned}$$

Note that the inequality $\zeta '<1$ is equivalent to $-2\alpha ^2-6\alpha -1<0$ which is true for $\alpha \in [0,1].$

Assume now that $\zeta _1\in [\zeta ',1).$ Then applying Lemma 2.2 we have

$$\begin{aligned} |\Psi |\le 48(1-\zeta _1^2)(2\alpha +1)(\alpha +1)^2\left( |A|+|B|+|C|\right) . \end{aligned}$$

Hence, and by (13),

$$\begin{aligned} {|}a_2a_3-a_4|=\frac{1}{144(3\alpha +1)(2\alpha +1)(\alpha +1)^2}|\Psi |\le \gamma (\zeta _1), \end{aligned}$$

(17)

where

$$\begin{aligned} {\mathbb {R}}{\ni } t{\mapsto } \gamma (t):=&-\frac{t}{18(3{\alpha }+1)(2{\alpha }+1)({\alpha }+1)^2}\\&\left[ ({26\alpha }^{3}+92{\alpha }^{2}+49{\alpha }+7)t^{2} -6(4{\alpha }+1)({\alpha }+2)({\alpha }+1)\right] . \end{aligned}$$

Since $\gamma '(t)=0$ is equivalent to

$$\begin{aligned} (26\alpha ^3+92\alpha ^2+49\alpha +7)t^2-2(4\alpha +1)(\alpha +2)(\alpha +1)=0, \end{aligned}$$

it follows that $\gamma$ has the unique positive critical point

$$\begin{aligned} t':=\frac{\sqrt{2(26\alpha ^3+92\alpha ^2+49\alpha +7)(4\alpha ^3+13\alpha ^2+11\alpha +2)}}{26\alpha ^3+92\alpha ^2+49\alpha +7}, \end{aligned}$$

(18)

where the function $\gamma$ has a local maximum with

$$\begin{aligned} \gamma (t')=\frac{2(\alpha +2)(4\alpha +1)\sqrt{2(26\alpha ^3+92\alpha ^2+49\alpha +7)(4\alpha +1)(\alpha +2)(\alpha +1)}}{9(\alpha +1)(2\alpha +1)(3\alpha +1)(26\alpha ^3+92\alpha ^2+49\alpha +7)}. \end{aligned}$$

Note that $t'<1$ for all $\alpha \in [0,1]$, which is equivalent to

$$\begin{aligned} 3(6\alpha ^3+22\alpha ^2+9\alpha +1)(26\alpha ^3+92\alpha ^2+49\alpha +7)>0,\quad \alpha \in [0,1]. \end{aligned}$$

Moreover $t'\ge \zeta '$ if, and only if,

$$\begin{aligned} -64\alpha ^6 - 252\alpha ^5 - 36\alpha ^4 + 384\alpha ^3 + 258\alpha ^2 + 57\alpha + 4\ge 0 \end{aligned}$$

which is true for all $\alpha \in [0,1].$ Consequently, $\gamma (t)\le \gamma (t')$ for $t\in [\zeta ',1),$ and in particular for $t:=\zeta _1$, so we obtain $\gamma (\zeta _1)\le \gamma (t').$ Hence, and by (17) we have

$$\begin{aligned} |a_2a_3-a_4|\le \gamma (t'). \end{aligned}$$

(19)

(B2) Suppose now that $\zeta _1\in [0,\zeta '),$ then applying Lemma 2.2 we have

$$\begin{aligned} |\Psi |\le 48(1-\zeta _1^2)(2\alpha +1)(\alpha +1)^2\left( 1+|A|+\frac{B^2}{4(1-|C|)}\right) , \end{aligned}$$

and so by (13) we obtain

$$\begin{aligned} |a_2a_3-a_4|=\frac{1}{144(3\alpha +1)(2\alpha +1)(\alpha +1)^2}|\Psi |\le \varrho (\zeta _1), \end{aligned}$$

(20)

where

$$\begin{aligned} \begin{aligned} {\mathbb {R}}\ni t\mapsto \varrho (t):=&\frac{1}{36(3\alpha +1)(2\alpha +1)^2(\alpha +1)^2}\left[ (4\alpha ^4+12\alpha ^3+160\alpha ^2+90\alpha +13)t^3\right. \\&\left. -9(2\alpha ^2+1)(2\alpha ^2+4\alpha +1)t^2+12(\alpha +1)^2(2\alpha +1)^2\right] . \end{aligned} \end{aligned}$$

Since $\varrho '(t)=0$ is equivalent to

$$\begin{aligned} t\left[ t(4\alpha ^4+12\alpha ^3+160\alpha ^2+90\alpha +13)-6(2\alpha ^2+1)(2\alpha ^2+4\alpha +1)\right] =0, \end{aligned}$$

it follows that $\varrho$ has the unique positive critical point

$$\begin{aligned} t^{\prime \prime}:=\frac{6(2\alpha ^2+1)(2\alpha ^2+4\alpha +1)}{4\alpha ^4+12\alpha ^3+160\alpha ^2+90\alpha +13}, \end{aligned}$$

which is a local minimum point. Observe now that $t{\prime }{\prime }<\zeta '$ if, and only if,

$$\begin{aligned} {64\alpha }^{6} + {252\alpha }^{5} + {36\alpha }^{4} - {384\alpha }^{3} - {258\alpha }^{2} - {57\alpha } - 4<0 \end{aligned}$$

which holds for all $\alpha \in [0,1].$ Therefore,

$$\begin{aligned} \varrho (t)\le \max \{\varrho (0),\varrho (\zeta ')\},\quad 0<t<\zeta ', \end{aligned}$$

and in particular when $t=\zeta _1$ we have $\varrho (\zeta _1)\le \max \{\varrho (0),\varrho (\zeta ')\},$ and hence by (20) we obtain

$$\begin{aligned} |a_2a_3-a_4|\le \max \{\varrho (0),\varrho (\zeta ')\}. \end{aligned}$$

(21)

It is easy to check that $\gamma (\zeta ')=\varrho (\zeta '),$ so the function

$$\begin{aligned}{}[0,1] \ni t\mapsto \psi (t):={\left\{ \begin{array}{ll} \varrho (t), &{} t\in [0,\zeta '],\\ \gamma (t), &{} t\in [\zeta ',1], \end{array}\right. } \end{aligned}$$

is continuous, has a local minimum at $t=t''$ and a local maximum at $t=t'.$ Since $t''<t'$ and $\psi (1)=\gamma (1)=a,$ where a is defined by (16), it follows from (16), (19) and (21) that

$$\begin{aligned} |a_2a_3-a_4|\le \max \{\psi (t):t\in [0,1]\}=\max \{\varrho (0),\gamma (t')\}. \end{aligned}$$

A simple calculation shows that

$$\begin{aligned} \begin{aligned}&\gamma (t')-\varrho (0)\\&=\frac{2(\alpha +2)(4\alpha +1)\sqrt{2(26\alpha ^3+92\alpha ^2+49\alpha +7)(4\alpha +1)(\alpha +2)(\alpha +1)}}{9(\alpha +1)(2\alpha +1)(3\alpha +1)(26\alpha ^3+92\alpha ^2+49\alpha +7)}- \frac{1}{3(3\alpha +1)}\\&=\frac{\mu (\alpha )}{9(\alpha +1)(2\alpha +1)(3\alpha +1)(26\alpha ^3+92\alpha ^2+49\alpha +7)} \ge 0 \end{aligned} \end{aligned}$$

if, and only if,

$$\begin{aligned} \begin{aligned} \mu (\alpha ):=&-3(\alpha +1)(2\alpha +1)(26\alpha ^3+92\alpha ^2+49\alpha +7)\\&+2(\alpha +2)(4\alpha +1)\sqrt{2(26\alpha ^3+92\alpha ^2+49\alpha +7)(4\alpha +1)(\alpha +2)(\alpha +1)}\ge 0, \end{aligned} \end{aligned}$$

or equivalently, if, and only if,

$$\begin{aligned} \begin{aligned}&2(\alpha +2)(4\alpha +1)\sqrt{2(26\alpha ^3+92\alpha ^2+49\alpha +7)(4\alpha +1)(\alpha +2)(\alpha +1)}\\&\ge 3(\alpha +1)(2\alpha +1)(26\alpha ^3+92\alpha ^2+49\alpha +7). \end{aligned} \end{aligned}$$

Squaring both sides of the above inequality gives

$$\begin{aligned} (\alpha +1)(26\alpha ^3+92\alpha ^2+49\alpha +7)(424\alpha ^6+1728\alpha ^5+1014\alpha ^4-1134\alpha ^3-735\alpha ^2-108\alpha -1)\le 0 \end{aligned}$$

which is true for $\alpha \in [0,\alpha '],$ where $\alpha ' \approx 0.814445$ is the unique root in [0, 1] of the equation

$$\begin{aligned} 424\alpha ^6+1728\alpha ^5+1014\alpha ^4-1134\alpha ^3-735\alpha ^2-108\alpha -1=0. \end{aligned}$$

(C) It remains to show that both inequalities in Theorem 3.1 are sharp. If $\alpha \in (\alpha ',1],$ then the function f given by (10) with $\omega (z):=z^3,\ z\in {\mathbb {D}},$ for which $a_2=0, a_3=0$ and $a_4=1/{3(1+3\alpha )}$ is extremal for the second inequality in (9).

For the first inequality let $\alpha \in [0,\alpha '],$ and set $\tau :=t',$ where $t'$ is defined by (18). Since $\tau <1,$ the function p given by (8) with $\zeta _1=\tau$ and $\zeta _2=-1$, i.e., the function

$$\begin{aligned} p(z) := \frac{1-z^2}{1-2 \tau z+z^2} = 1 +2 \tau z +(4\tau ^2-2)z^2+\cdots ,\quad z\in {\mathbb {D}}, \end{aligned}$$

belongs to ${{\mathcal {P}}}.$ Thus the function f given by (11), with p as above and

$$\begin{aligned} \begin{aligned} a_2&= \dfrac{\tau }{1+\alpha }, \quad a_3 =\dfrac{\tau ^2(3\alpha ^2+12\alpha +5)-2(1+\alpha )^2}{4(1+2\alpha )(1+\alpha )^2},\\ a_4&= \dfrac{\tau ((52\alpha ^4+317\alpha ^3+633\alpha ^2+355\alpha +59)\tau ^2-6(8\alpha ^2+27\alpha +7)(1+\alpha )^2)}{36(1+\alpha )^3(1+2\alpha )(1+3\alpha )}. \end{aligned} \end{aligned}$$

belongs to ${\mathcal {M}}_{\alpha }(\exp )$ and is extremal for the first inequality in (9), which completes the proof of the Theorem 3.1. $\square$

For $\alpha =0$, we deduce the following ([25, Corollary 2]).

Corollary 3.1

If $f\in {\mathcal {M}}_0(\exp )$ and is given by (1), then

$$\begin{aligned} |a_2a_3-a_4|\le \dfrac{8\sqrt{7}}{63}. \end{aligned}$$

The inequality is sharp.

For $\alpha =1$, we deduce the following [25, Corollary 5].

Corollary 3.2

If $f\in {\mathcal {M}}_1(\exp )$ and is given by (1), then

$$\begin{aligned} |a_2a_3-a_4|\le \dfrac{1}{12}. \end{aligned}$$

The inequality is sharp.

4 The Hankel determinant $H_{2,2}(f)$

In this section, we find the sharp bound for the modulus of the second Hankel determinant $H_{2,2}(f)=a_2a_4-a_3^2$ when $f\in {\mathcal {M}}_{\alpha }(\exp ).$

Theorem 4.1

Let $\alpha \in [0,1].$ If $f\in {\mathcal {M}}_{\alpha }(\exp )$ and is given by (1), then

$$\begin{aligned}&|H_{2,2}(f)|= |a_2a_4-a_3^2|\\&\le {\left\{ \begin{array}{ll} \dfrac{1}{4(2\alpha +1)^2}, \quad &{} \alpha \in \left[ 0,(\sqrt{6}-1)/5\right] ,\\ \dfrac{34\alpha ^3+82\alpha ^2+27\alpha +3}{(3\alpha +1)(173\alpha ^4+546\alpha ^3+440\alpha ^2+126\alpha +11)}, &{}\alpha \in \left( (\sqrt{6}-1)/5,1\right] . \end{array}\right. } \end{aligned}$$

(22)

Both inequalities are sharp.

Proof

Fix $\alpha \in [0,1]$ and let $f\in {\mathcal {M}}_{\alpha }(\exp )$ be of the form (1). Since both the class $\mathcal M_{\alpha }(\exp )$ and the functional ${\mathcal {M}}_{\alpha }(\exp )\ni f\mapsto H_{2,2}(f)$ are rotationally invariant, without loss of generality we may assume that $c_1 \in [0,2],$ i.e., by (5) that $\zeta _1\in [0,1].$ From (12) applying Lemma 2.1 we obtain

$$\begin{aligned} |a_2a_4-a_3^2|= \frac{1}{2304(3\alpha +1)(2\alpha +1)^2(\alpha +1)^3} |\Psi |, \end{aligned}$$

(23)

where

$$\begin{aligned} \begin{aligned} \Psi&:= c_1^4(5\alpha ^4-30\alpha ^3-232\alpha ^2-162\alpha -13)-144c_2^2(1+3\alpha )(1+\alpha )^3\\&\quad -24c_1^2c_2(7\alpha ^2-2\alpha +1)(1+\alpha )^2+192c_1c_3(1+\alpha )^2(1+2\alpha )^2\\&= 16\left( (5\alpha ^4+42\alpha ^3-88\alpha ^2-90\alpha -13)\zeta _1^4\right. \\&\quad +12(7\alpha ^2+10\alpha +1)(\alpha +1)^2(1-\zeta _1^2)\zeta _1^2\zeta _2\\&\quad \left. -12(\alpha +1)^2(1-\zeta _1^2)((7\alpha ^2+4\alpha +1)\zeta _1^2+3(1+\alpha )(1+3\alpha ))\zeta _2^2\right. \\&\quad \left. +48(\alpha +1)^2(2\alpha +1)^2(1-\zeta _1^2)(1-|\zeta _2|^2)\zeta _1\zeta _3 \right) \end{aligned} \end{aligned}$$

(24)

for some $\zeta _1,\zeta _2,\zeta _3 \in \overline{{\mathbb {D}}}$.

(A) Suppose first that $\zeta _1=1.$ Since

$$\begin{aligned} \frac{-5\alpha ^4-42\alpha ^3+88\alpha ^2+90\alpha +13}{144(3\alpha +1)(2\alpha +1)^2(\alpha +1)^3}>0,\quad \alpha \in [0,1], \end{aligned}$$

(25)

from (23) and (24) we have

$$\begin{aligned} |a_2a_4-a_3^2|=\frac{-5\alpha ^4-42\alpha ^3+88\alpha ^2+90\alpha +13}{144(3\alpha +1)(2\alpha +1)^2(\alpha +1)^3}. \end{aligned}$$

(B) Now suppose that $\zeta _1\in [0,1).$ Noting from (24) that $|\zeta _3|\le 1$, we obtain

$$\begin{aligned} |\Psi |\le 768\zeta _1(1-\zeta _1^2)(2\alpha +1)^2(\alpha +1)^2 \Phi (A,B,C), \end{aligned}$$

where

$$\begin{aligned} \Phi (A,B,C) := \left|A +B\zeta _2 +C\zeta _2^2 \right|+ 1 -|\zeta _2|^2, \end{aligned}$$

with

$$\begin{aligned} \begin{aligned}&A:=\frac{(5\alpha ^4+42\alpha ^3-88\alpha ^2-90\alpha -13)\zeta _1^3}{48(2\alpha +1)^2(\alpha +1)^2(1-\zeta _1^2)},\quad B:=\frac{(7\alpha ^2+10\alpha +1)\zeta _1}{4(2\alpha +1)^2},\\&C:=-\frac{(7\alpha ^2+4\alpha +1)\zeta _1^2+9\alpha ^2+12\alpha +3}{4(2\alpha +1)^2\zeta _1}. \end{aligned} \end{aligned}$$

A simple calculation using (25) shows that $AC>0.$

(B1) Thus, we first consider the condition $|B|\ge 2(1-|C|),$ i.e.,

$$\begin{aligned} \frac{(7\alpha ^2+10\alpha +1)\zeta _1}{4(2\alpha +1)^2}>2\left( 1-\frac{(7\alpha ^2+4\alpha +1)\zeta _1^2+9\alpha ^2+12\alpha +3}{4(2\alpha +1)^2\zeta _1}\right) , \end{aligned}$$

which can be equivalently written as

$$\begin{aligned} \frac{3(7\alpha ^2+6\alpha +1)\zeta _1^2-8(2\alpha +1)^2\zeta _1+6(3\alpha +1)(\alpha +1)}{4(2\alpha +1)^2\zeta _1}>0, \end{aligned}$$

which is true for all $\alpha \in [0,1]$ and $\zeta _1\in [0,1)$. Thus, applying Lemma 2.2 we have

$$\begin{aligned} |\Psi |\le 768\zeta _1(1-\zeta _1^2)(2\alpha +1)^2(\alpha +1)^2\left( |A|+|B|+|C|\right) . \end{aligned}$$

Hence and by (23)

$$\begin{aligned} |a_2a_4-a_3^3|=\frac{1}{2304(3\alpha +1)(2\alpha +1)^2(\alpha +1)^3}|\Psi |\le \gamma (\zeta _1), \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} {\mathbb {R}}{\ni } t\mapsto \gamma (t)&:=\frac{1}{144(3\alpha +1)(2\alpha +1)^2(\alpha +1)^3}\\&\quad \times \left[ -(173\alpha ^4+546\alpha ^3+440\alpha ^2+126\alpha +11)t^4\right. \\&\quad \left. +12(5\alpha ^2+2\alpha -1)(\alpha +1)^2t^2+36(3\alpha +1)(\alpha +1)^3\right] . \end{aligned} \end{aligned}$$

Since $\gamma '(t)=0$ is equivalent to

$$\begin{aligned} \left[ (173\alpha ^4+546\alpha ^3+440\alpha ^2+126\alpha +11)t^2-6(5\alpha ^2+2\alpha -1)(\alpha +1)^2\right] t=0, \end{aligned}$$

it follows that for $(\sqrt{6}-1)/5<\alpha \le 1$ the function $\gamma$ has the unique positive critical point

$$\begin{aligned} t':=\frac{(\alpha +1)\sqrt{6(173\alpha ^4+546\alpha ^3+440\alpha ^2+126\alpha +11)(5\alpha ^2+2\alpha -1)}}{173\alpha ^4+546\alpha ^3+440\alpha ^2+126\alpha +11}, \end{aligned}$$

(26)

where the function $\gamma$ has a local maximum with

$$\begin{aligned} \gamma (t')=\frac{34\alpha ^3+82\alpha ^2+27\alpha +3}{(3\alpha +1)(173\alpha ^4+546\alpha ^3+440\alpha ^2+126\alpha +11)}. \end{aligned}$$

Note that $t'<1$, since this is equivalent to

$$\begin{aligned} (143\alpha ^4+474\alpha ^3+392\alpha ^2+126\alpha +17)(173\alpha ^4+546\alpha ^3+440\alpha ^2+126\alpha +11)>0. \end{aligned}$$

For $0\le \alpha \le (\sqrt{6}-1)/5$ we have

$$\begin{aligned} \gamma (t)\le \max \{\gamma (0),\gamma (1)\}=\gamma (0)=\frac{1}{4(2\alpha +1)^2},\quad t\in [0,1], \end{aligned}$$

since

$$\begin{aligned} \begin{aligned} \gamma (0)-\gamma (1)&=\frac{1}{4(2\alpha +1)^2}-\frac{-5\alpha ^4-42\alpha ^3+88\alpha ^2+90\alpha +13}{144(3\alpha +1)(2\alpha +1)^2(\alpha +1)^3}\\&=\frac{113\alpha ^4+402\alpha ^3+344\alpha ^2+126\alpha +23}{144(3\alpha +1)(2\alpha +1)^2(\alpha +1)^3}>0,\quad \alpha \in [0,1]. \end{aligned} \end{aligned}$$

(C) It remains to show that the inequalities in Theorem 4.1 are sharp. If $\alpha \in [0,(\sqrt{6}-1)/5],$ then the function f given by (10) with $\omega (z):=z^2,\ z\in {\mathbb {D}},$ for which $a_2=0,$ $a_3=1/(2(1+2\alpha ))$ and $a_4=0$ is extremal for the first inequality in (22).

For the second inequality, let $\alpha \in ((\sqrt{6}-1)/5,1],$ and set $\tau :=t',$ where $t'$ is given by (26). Since $\tau < 1,$ the function p given by (8) with $\zeta _1=\tau$ and $\zeta _2=-1$, i.e., the function

$$\begin{aligned} p(z) := \frac{1-z^2}{1-2 \tau z+z^2} = 1 +2 \tau z +(4\tau ^2-2)z^2+\cdots ,\quad z\in {\mathbb {D}}, \end{aligned}$$

belongs to ${{\mathcal {P}}}.$ Thus the function f given by (11) has the form (1) with

$$\begin{aligned} \begin{aligned} a_2&= \dfrac{\tau }{1+\alpha }, \quad a_3 =\dfrac{\tau ^2(3\alpha ^2+12\alpha +5)-2(1+\alpha )^2}{4(1+2\alpha )(1+\alpha )^2},\\ a_4&= \dfrac{\tau ((52\alpha ^4+317\alpha ^3+633\alpha ^2+355\alpha +59)\tau ^2-6(8\alpha ^2+27\alpha +7)(1+\alpha )^2))}{36(1+\alpha )^3(1+2\alpha )(1+3\alpha )}, \end{aligned} \end{aligned}$$

which gives equality in (22). $\square$

When $\alpha =0$, we deduce the following [25, Corollary 3].

Corollary 4.1

If $f\in {{\mathcal {S}}}^{*}(\exp ),$ then

$$\begin{aligned} |H_{2,2}(f)|\le \dfrac{1}{4}. \end{aligned}$$

The inequality is sharp.

When $\alpha =1$, we deduce the following ([25, Corollary 6]).

Corollary 4.2

If $f\in {{\mathcal {C}}}(\exp ),$ then

$$\begin{aligned} |H_{2,2}(f)|\le \dfrac{73}{2592}. \end{aligned}$$

The inequality is sharp.

Remark 4.1

We end by noting that in [22] it was recently shown that for the third Hankel determinant

$$\begin{aligned} H_{3,1}(f)=2a_{2}a_{3}a_{4}-a_{3}^{3}-a_{4}^{2}+a_5(a_{3}-a_{2}^{2}), \end{aligned}$$

when $f\in {{\mathcal {S}}}^{*}(\exp ),$ the sharp bound is $|H_{3,1}(f)|\le 1/9$, and when $f\in {{\mathcal {C}}}(\exp ),$ the sharp bound is $|H_{3,1}(f)|\le 1/144$.

Clearly finding the sharp bound for $|H_{3,1}(f)|$ when $f\in {\mathcal {M}}_{\alpha }(\exp )$ presents a significantly difficult problem.