The second Hankel determinant for strongly convex and Ozaki close-to-convex functions

Abstract

Let f be analytic in the unit disk ${\mathbb {D}}=\{z\in {\mathbb {C}}:|z|<1 \}$, and ${{\mathcal {S}}}$ be the subclass of normalized univalent functions given by $f(z)=z+\sum _{n=2}^{\infty }a_n z^n$ for $z\in {\mathbb {D}}$. We give sharp bounds for the modulus of the second Hankel determinant $H_2(2)(f)=a_2a_4-a_3^2$ for the subclass ${\mathcal F_{O}}(\lambda ,\beta )$ of strongly Ozaki close-to-convex functions, where $1/2\le \lambda \le 1$, and $0<\beta \le 1$. Sharp bounds are also given for $|H_2(2)(f^{-1})|$, where $f^{-1}$ is the inverse function of f. The results settle an invariance property of $|H_2(2)(f)|$ and $|H_2(2)(f^{-1})|$ for strongly convex functions.

1 Introduction

Let ${{\mathcal {A}}}$ denote the class of analytic functions f in the unit disk ${\mathbb {D}}:=\{ z\in {\mathbb {C}}: |z|<1 \}$ normalized by $f(0)=0=f'(0)-1$. Then, for $z\in {\mathbb {D}}$, $f\in {{\mathcal {A}}}$ has the following representation

$$\begin{aligned} f(z) = z+ \sum _{n=2}^{\infty }a_n z^n. \end{aligned}$$

(1)

Let ${{\mathcal {S}}}$ denote the subclass of all univalent (i.e., one-to-one) functions in ${{\mathcal {A}}}$.

Denote by ${{\mathcal {S}}}^*$ the subclass of ${{\mathcal {S}}}$ consisting of starlike functions, i.e., functions f which map ${\mathbb {D}}$ onto a set which is star-shaped with respect to the origin. Then, it is well known that a function $f \in {{\mathcal {S}}}^*$ if, and only if, for $z\in {\mathbb {D}},$

$$\begin{aligned} {{\,\mathrm{Re}\,}}\left\{ \frac{zf'(z)}{f(z)} \right\} >0, \end{aligned}$$

and that ${\mathcal {S}}^*\subset {\mathcal {S}}.$

Next denote by ${{\mathcal {K}}}$ the subclass of ${{\mathcal {S}}}$ consisting of functions which are close-to-convex, i.e., functions f which map ${\mathbb {D}}$ onto a close-to-convex domain, if, and only if, there exist $0\le \delta \le 2\pi$ and $g\in {\mathcal {S}}^*$ such that for $z\in {\mathbb {D}},$

$$\begin{aligned} {{\,\mathrm{Re}\,}}\left\{ \mathrm {e}^{\mathrm {i}\delta } \frac{zf'(z)}{g(z)} \right\} >0. \end{aligned}$$

(2)

Again it is well known that ${\mathcal {K}}\subset {\mathcal {S}}$.

The class ${\mathcal {C}}(\alpha )$ for $0\le \alpha <1$ of convex functions of order $\alpha$ consisting of functions $f\in {\mathcal {A}}$ satisfying

$$\begin{aligned} {{\,\mathrm{Re}\,}}\left\{ 1+ \frac{zf''(z)}{f'(z)} \right\} >\alpha \end{aligned}$$

for $z\in {\mathbb {D}},$ is also well known and has been widely studied. When $\alpha =0$, we obtain the class ${\mathcal {C}}$ of convex functions.

Little attention has been given to the case when $\alpha <0$, but several authors have considered the class ${\mathcal {C}}(-1/2)$, consisting of functions $f\in {\mathcal {A}}$ satisfying

$$\begin{aligned} {{\,\mathrm{Re}\,}}\left\{ 1+ \frac{zf''(z)}{f'(z)} \right\} >-\dfrac{1}{2} \end{aligned}$$

for $z\in {\mathbb {D}}$, whose members are known to be close-to-convex, and so univalent.

The class ${\mathcal {F}}_O(\lambda )$, defined for $1/2\le \lambda \le 1$ by

$$\begin{aligned} {{\,\mathrm{Re}\,}}\left\{ 1+ \frac{zf''(z)}{f'(z)} \right\} >\dfrac{1}{2}-\lambda \end{aligned}$$

(3)

for $z\in {\mathbb {D}}$, was formally introduced in [1] and is also known to be a subclass of the close-convex functions. More information concerning the coefficients of functions in ${\mathcal {F}}_O(\lambda )$ (called Ozaki close-to-convex functions), has also be found by Allu, Lecko and Thomas (to appear). We note that clearly ${\mathcal {F}}_O(1/2)={\mathcal {C}}$, and ${\mathcal {F}}_O(1)={\mathcal {C}}(-1/2)$.

Also in [1], the following notion of strongly Ozaki functions was introduced, and some basic properties were obtained.

Definition 1

Let $0< \beta \le 1$ and $1/2\le \lambda \le 1$. Then, $f\in {\mathcal {A}}$ is called a strongly Ozaki close-to-convex if, and only if, for $z\in {\mathbb {D}},$

$$\begin{aligned} \left| \arg \left( \dfrac{2\lambda -1}{2\lambda +1}+\dfrac{2}{2\lambda +1} \left( 1+\dfrac{zf''(z)}{f'(z)}\right) \right) \right| <\dfrac{\beta \pi }{2}. \end{aligned}$$

(4)

We denote this class of functions by ${\mathcal {F}}_{O}(\lambda ,\beta )$, noting that when $\beta =1$ this reduces to (3), and when $\lambda =1/2$ we obtain the class ${\mathcal {C}}^{\beta }$ of strongly convex functions considered in [9].

For $q,n \in {\mathbb {N}},$ the Hankel determinant $H_{q,n}(f)$ of function $f \in {{\mathcal {A}}}$ of the form (1) is defined as

$$\begin{aligned} H_{q,n}(f) := \begin{vmatrix} a_{n}&a_{n+1}&\cdots&a_{n+q-1} \\ a_{n+1}&a_{n+2}&\cdots&a_{n+q} \\ \vdots&\vdots&\vdots&\vdots \\ a_{n+q-1}&a_{n+q}&\cdots&a_{n+2(q-1)} \end{vmatrix}. \end{aligned}$$

Hankel matrices are used in the theory of Markov processes, of non-stationary signals, in Hamburger moment problem, and many other issues in both pure mathematics and technical applications.

To find the upper bound of $H_{q,n}(f)$ for the whole class ${\mathcal {S}}$ of univalent functions as well as for its subclasses is an interesting problem to study. For the class ${\mathcal {S}}$, a basic result was shown by Pommerenke [7], who found the growth of $H_{q,n}(f)$ dependent on q and n. In recent years there has been a great deal of attention devoted 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 ${\mathcal {A}}$. Most authors have used the method introduced in [4], where the sharp bound $|H_2(2)(f)|\le 1$ was found for $f\in {\mathcal {S}}^*$. However, finding the sharp bound for $|H_2(2)(f)|$ when $f\in {\mathcal {K}}$ and even when $\delta =0$ remains an open problem.

In [2], a sharp bound was obtained for $|H_2(2)(f)|$ when $f\in {\mathcal {F}}_O(\lambda )$. The main purpose of this paper is to give the sharp bounds for $|H_2(2)(f)|$, when $f\in {\mathcal {F}}_O(\lambda ,\beta )$, together with sharp bounds for $|H_2(2)(f^{-1})|$, where $f^{-1}$ is the inverse function of f.

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

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

(5)

We will use the following properties for the coefficients of functions ${{\mathcal {P}}}$, given by (5).

Lemma 1

[5, 6] If $p\in {\mathcal {P}}$ and is given by (5) with $c_1\ge 0$, then for some complex valued $\zeta$ with $|\zeta |\le 1$, and some complex valued $\eta$ with $|\eta | \le 1$,

$$\begin{aligned} 2c_{2}&=c_{1}^{2}+\zeta (4-c_{1}^{2}),\\ 4c_{3}&=c_{1}^{3}+2(4-c_{1}^{2})c_{1}\zeta -c_{1}(4-c_{1}^{2})\zeta ^{2}+2(4-c_{1}^{2})(1-|\zeta |^{2})\eta .\\ \end{aligned} $$

(6)

Lemma 2

[3] Let $\overline{{\mathbb {D}}} := \{z\in {\mathbb {C}}:|z|\le 1\}$, and for real numbers A, B, C, let

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

(7)

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

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

(8)

where

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

(9)

We also note that from (4), we can write, for $z\in {\mathbb {D}},$

$$\begin{aligned} 1+ \dfrac{zf''(z)}{f'(z)} = \left( \frac{1}{2}+\lambda \right) p(z)^{\beta } + \frac{1}{2}-\lambda \end{aligned}$$

(10)

for some $p\in {\mathcal {P}}$ and so equating coefficients we have

$$\begin{aligned} a_2&= \dfrac{\beta }{4} \Big (1 + 2 \lambda \Big ) c_1,\\ a_3&= \dfrac{\beta }{12} \Big (1 + 2 \lambda \Big ) \Big (c_2 - \frac{1}{2}(1-2\beta -2\beta \lambda )c_1^2 \Big ),\\ a_4&= \dfrac{\beta }{24}\Big (1 + 2 \lambda \Big )\Big (c_3-\dfrac{1}{4}(4-7\beta -6\beta \lambda ) c_1c_2\\ &\quad +\dfrac{1}{24}(8-21\beta +16\beta ^2-18\beta \lambda +30\beta ^2\lambda +12\beta ^2\lambda ^2)c_1^3\Big ). \end{aligned} $$

(11)

3 $H_2(2)(f)$ for strongly Ozaki functions

We prove the following.

Theorem 1

Let $\beta \in (0,1]$ and $\lambda \in [1/2,1]$. If $f\in {{\mathcal {F}}}_{O}(\lambda ,\beta )$, then

$$\begin{aligned}&|H_{2}(2)(f)|\\ &\quad\le {\left\{ \begin{array}{ll} \dfrac{1}{36}\beta ^2(1+2\lambda )^2, &{} (5+2\lambda )\beta \le 2, \\ \dfrac{ \beta ^2(1+2\lambda )^2 ( 68+60\beta +25\beta ^2 +24\beta \lambda -12\beta ^2\lambda -60\beta ^2\lambda ^2 ) }{ 576( 4 +5\beta +2\beta \lambda -2\beta ^2\lambda -4\beta ^2\lambda ^2 )}, &{}\text {otherwise}. \end{array}\right. } \end{aligned} $$

The inequalities are sharp.

Proof

From (11) we have

$$\begin{aligned} a_2a_4-a_3^2 = -\frac{1}{2304}\beta ^2(1+2\lambda )^2 \varPsi , \end{aligned}$$

(12)

where

$$\begin{aligned} \varPsi:=& 16c_2^2 -24c_1c_3 -2c_1^2c_2\left[ -4+\beta (5+2\lambda )\right] \\& +c_1^4\left[ -4+2\beta ^2\lambda (1+2\lambda )+\beta (5+2\lambda )\right] . \end{aligned} $$

Since both the class ${{\mathcal {F}}}_{O}(\lambda ,\beta )$ and the functional are rotationally invariant, without loss of generality we may assume that $c = c_1 \in [0,2]$. By Lemma 1, we obtain

$$\begin{aligned} \varPsi&= 2c^4\left[ -1+\beta ^2\lambda (1+2\lambda )\right] -c^2(4-c^2)\beta (5+2\lambda )\zeta \\&\quad + 2(32-4c^2-c^4)\zeta ^2 - 12c(4-c^2)(1-|\zeta |^2)\eta , \end{aligned} $$

(13)

for some $\zeta$, $\eta \in \overline{{\mathbb {D}}}$.

Since $|\eta |\le 1$ and $|\zeta |\le 1,$ by applying the triangle inequality to (13), we have

$$\begin{aligned} |\varPsi |&\le 2\left| -1+\beta ^2\lambda (1+2\lambda )\right| c^4 +\beta (5+2\lambda )c^2(4-c^2)|\zeta |\\&\quad + 2(4-c^2)(4-c)(2-c)|\zeta |^2 + 12c(4-c^2)\\&\le 2\left| -1+\beta ^2\lambda (1+2\lambda )\right| c^4 +\beta (5+2\lambda )c^2(4-c^2)\\&\quad + 2(4-c^2)(4-c)(2-c) + 12c(4-c^2)=G_1(c^2), \end{aligned} $$

(14)

where for $x\in [0,4],$

$$\begin{aligned} G_1(x) := 64+2b_1 x+b_2x^2 \end{aligned}$$

with

$$\begin{aligned} b_1 := 2\left[ \beta (5+2\lambda )-2 \right] \quad \text {and}\quad b_2 := 2\left| \beta ^2\lambda (1+2\lambda ) -1\right| -\beta (5+2\lambda ) -2. \end{aligned}$$

I. Consider first the case $(5+2\lambda )\beta \le 2$.

Since $b_1 \le 0$, and since $\beta \le 2/(5+2\lambda )$ and $1/2 \le \lambda \le 1$, the inequalities

$$\begin{aligned} \beta ^2\lambda (1+2\lambda ) \le \frac{ 4\lambda (1+2\lambda ) }{ (5+2\lambda )^2 } < 1 \end{aligned}$$

hold. Thus, we get

$$\begin{aligned} b_2 = -2\left[ \beta ^2\lambda (1+2\lambda )-1\right] -\beta (5+2\lambda ) -2 = -\beta \left[ 2\beta \lambda (1+2\lambda ) +5+2\lambda \right] <0. \end{aligned}$$

Hence, $G_1'(x) = 2b_1 +2b_2x \le 0$ holds for all $x\in [0,4]$, and so $G_1$ is decreasing on [0, 4]. Since $c\in [0,2]$, it follows from (14) that

$$\begin{aligned} |\varPsi | \le G_1(c^2) \le G_1(0) = 64, \end{aligned}$$

which establishes the first inequality in Theorem 1.

We next divide the case $(5+2\lambda )\beta >2$ into two cases.

II(a). First suppose that $\beta ^2\lambda (1+2\lambda ) \ge 1$. Then, we see that $b_1>0$ and $b_2<0$. Indeed,

$$\begin{aligned} b_1+4 =2\beta (5+2\lambda ) \ge \frac{2(5+2\lambda )}{ \sqrt{\lambda (1+2\lambda )} } \ge \frac{ 14}{\sqrt{3}} >4 \end{aligned}$$

and

$$\begin{aligned} b_2 = 2\beta ^2\lambda (1+2\lambda ) -\beta (5+2\lambda ) -4 \le 2-\beta (5+2\lambda ) <0, \end{aligned}$$

since $\beta ^2 \lambda (1+2\lambda ) \le 3$ and $(5+2\lambda )\beta >2$. Moreover, by putting

$$\begin{aligned} \tau :=-\frac{b_1}{b_2}, \end{aligned}$$

(15)

we see that

$$\begin{aligned} 0<\tau <4 \end{aligned}$$

(16)

holds, since

$$\begin{aligned} b_1+4b_2 = 2\beta \left[ 4\beta \lambda (1+2\lambda ) -(5+2\lambda ) \right] -20 \le 2\beta ( 8\lambda ^2 +2\lambda -5) -20 \le -10<0. \end{aligned}$$

Therefore, $G_1'(\tau )=0$ and $G_1''(\tau )<0$ imply that for $x\in [0,4],$

$$\begin{aligned} G_1(x) \le G_1(\tau ) = \frac{ 4\left[ -68 -12(5+2\lambda )\beta + (-25 +12\lambda +60\lambda ^2)\beta ^2\right] }{ -4 -(5+2\lambda )\beta +2\lambda (1+2\lambda )\beta ^2}, \end{aligned}$$

and so from (12) and (14), we obtain the second inequality in Theorem 1 in this case.

II(b). Now assume that $\beta ^2\lambda (1+2\lambda ) < 1$ and $(5+2\lambda )\beta >2$. By (13), when $c=2$, we have

$$\begin{aligned} |\varPsi | \le 32 | -1 +\beta ^2 \lambda (1+2\lambda ) | \le 64, \end{aligned}$$

and, when $c=0$, since $\zeta \in \overline{{\mathbb {D}}}$, we also have

$$\begin{aligned} |\varPsi | = 64|\zeta |^2 \le 64. \end{aligned}$$

Now let $c \in (0,2)$. Then, since $\eta \in \overline{{\mathbb {D}}}$, we obtain

$$\begin{aligned} |\varPsi | \le 12c(4-c^2) \varPhi (A,B,C), \end{aligned}$$

where

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

with

$$\begin{aligned} A := \frac{c^3\left[ -1+\beta ^2\lambda (1+2\lambda )\right] }{6(4-c^2)}, \quad B := -\frac{1}{12}\beta (5+2\lambda ) c, \quad C := \frac{8+c^2}{6c}. \end{aligned}$$

Noting that $AC<0$ in this case, we now apply Lemma 2.

Simple calculations show that the first two alternatives when $AC<0$ are not satisfied, and so we must consider R(A, B, C).

We first show that $|C|(|B|+4|A|) > |AB|$ holds for all $c\in (0,2)$. A calculation shows that

$$\begin{aligned} |AB|-|C|(|B|+4|A|) = \frac{-1}{72(4-c^2)}H_1(c^2), \end{aligned}$$

where for $x\in [0,4],$

$$\begin{aligned} H_1(x) := k_0 +2k_1x+k_2x^2 \end{aligned}$$

with

$$\begin{aligned} k_0 := 32\beta (5+2\lambda ), \quad k_1 := -2\left[ -16 +16\beta ^2\lambda (1+2\lambda ) +\beta (5+2\lambda )\right] \end{aligned}$$

and

$$\begin{aligned} k_2 := 8 -8\beta ^2\lambda (1+2\lambda ) -2\beta (5+2\lambda ) +\beta ^3\lambda (5+12\lambda +4\lambda ^2). \end{aligned}$$

Note that $k_2 \ge 0$ implies $k_1>0$. Indeed, since $\beta ^2 \lambda (1+2\lambda ) <1$, we have $3-2\beta ^2\lambda (1+2\lambda )>0$. Therefore, if $k_2 \ge 0$, then

$$\begin{aligned} k_1 = 2\beta (5+2\lambda )\left[ 3-2\beta ^2\lambda (1+2\lambda ) \right] +4k_2>0. \end{aligned}$$

Thus, when $k_2 \ge 0$, from $k_0>0$ and $k_1>0$, we have $H_1(x)>0$, for $x\in [0,4]$.

When $k_2<0$, since $H_1''\equiv 2k_2<0$, we get for $x\in [0,4],$

$$\begin{aligned} H_1(x) \ge \min \{ H_1(0), H_1(4) \}. \end{aligned}$$

(17)

Next fix $\lambda \in [1/2,1]$. Let $I:=\left( 2/(5+2\lambda ),1/\sqrt{\lambda (1+2\lambda )}\right)$ and $\varphi :I\rightarrow {\mathbb {R}}$ be defined by

$$\begin{aligned} \varphi (x) := 24 -(5+2\lambda )x - 24\lambda (1+2\lambda )x^2 +\lambda (1+2\lambda )(5+2\lambda )x^3. \end{aligned}$$

Since

$$\begin{aligned} \varphi '(x)&= -(5+2\lambda ) -48\lambda (1+2\lambda )x +3\lambda (1+2\lambda )(5+2\lambda )x^2 \\&< -48\lambda (1+2\lambda )x +2(5+2\lambda ) \\&< \frac{ 2(25-28\lambda -92\lambda ^2) }{ 5+2\lambda } <0, \end{aligned} $$

so $\varphi$ is decreasing on I and for $x\in I,$

$$\begin{aligned} \varphi (x) > \varphi \left( \frac{1}{\sqrt{\lambda (1+2\lambda )}}\right) = 0. \end{aligned}$$

Thus, $H_1(4) = k_0+8k_1+16k_2 = 16 \varphi (\beta ) >0$. Also, since $H_1(0)=k_0>0$, by (17), $H_1(x)>0$ for all $x\in [0,4]$, which implies that $|C|(|B|+4|A|) > |AB|$.

Next a calculation gives

$$\begin{aligned} |AB|-|C|(|B|-4|A|) = \frac{-1}{72(4-c^2)} H_2(c^2), \end{aligned}$$

where for $x\in [0,4],$

$$\begin{aligned} H_2(x) := l_0 +2l_1x +l_2x^2 \end{aligned}$$

with

$$\begin{aligned} l_0 := 32\beta (5+2\lambda ), \quad l_1 := 2\left[ -16 +16\beta ^2\lambda (1+2\lambda ) -\beta (5+2\lambda )\right] \end{aligned}$$

and

$$\begin{aligned} l_2 := -8 +8\beta ^2\lambda (1+2\lambda ) -2\beta (5+2\lambda ) +\beta ^3\lambda (5+12\lambda +4\lambda ^2). \end{aligned}$$

Note that $l_2<0$, and let

$$\begin{aligned} \xi _i := \frac{-l_1 +(-1)^{i+1} \sqrt{l_1^2 - l_0l_2} }{l_2}, \quad i=1,2, \end{aligned}$$

be the zeroes of $H_2$. Since $l_2<0$, so $\xi _1<\xi _2$. Moreover, since $l_0>0$, we have $-l_1+\sqrt{l_1^2-l_0l_2} >0$ and $\xi _1<0$. Also $\xi _2 < 4$, since this is equivalent to

$$\begin{aligned} l_0l_2+8l_1l_2 +16l_2^2 >0, \end{aligned}$$

(18)

which is true. In fact, a computation shows that

$$\begin{aligned} l_0l_2+8l_1l_2 +16l_2^2 = 16(8+t)\left[ t(2-u) +8(1-u) \right] (1-u), \end{aligned}$$

where $t =\beta (5+2\lambda )$ and $u=\beta ^2\lambda (1+2\lambda )$. From the inequalities $t>0$ and $0<u<1$, we obtain the inequality (18). Therefore, since $l_2<0$, $H_2(x) \ge 0$ holds for $x\in (0,\xi _2]$, and $H_2(x) \le 0$ holds for $x\in [\xi _2,4)$. Thus, $|AB| \le |C|(|B|-4|A|)$ holds when $c\in (0,\sqrt{\xi _2}]$.

Thus, when $c\in (0,\sqrt{\xi _2}]$, Lemma 2 gives

$$\begin{aligned} |\varPsi | \le 12c(4-c^2)(-|A|+|B|+|C|) = G_2(c^2), \end{aligned}$$

(19)

where for $x\in [0,4],$

$$\begin{aligned} G_2(x) := 64+2d_1 x+d_2x^2 \end{aligned}$$

(20)

with

$$\begin{aligned} d_1:=2\left[ -2+(5+2\lambda )\beta \right] \quad \text {and} \quad d_2:=-4-(5+2\lambda )\beta +2\lambda (1+2\lambda )\beta ^2. \end{aligned}$$

Note that $d_1>0$ and $d_2<0$. We also note that $G_2'(x)=0$ holds only for

$$\begin{aligned} x=-\frac{d_1}{d_2}=:\tau . \end{aligned}$$

(21)

Clearly, $\tau >0$. Now we claim $\tau < \xi _2$. Since

$$\begin{aligned} H_2(\tau ) = \frac{1}{d_2^2} ( l_0 d_2^2 - 2l_1d_1d_2 +l_2d_1^2 ), \end{aligned}$$

the inequality $\tau < \xi _2$ is equivalent to

$$\begin{aligned} l_0 d_2^2 - 2l_1d_1d_2 +l_2d_1^2 >0. \end{aligned}$$

(22)

Moreover, by putting $t =\beta (5+2\lambda )$ and $u=\beta ^2\lambda (1+2\lambda )$ again, we get

$$\begin{aligned}&l_0 d_2^2 - 2l_1d_1d_2 +l_2d_1^2 \\&\quad = 4\left[ t^3(4+u) +4t^2(7+2u) +32(7-11u+4u^2) + 4t(26-9u-8u^2) \right] . \end{aligned} $$

Since $t>0$ and $0<u<1$, all terms in the right side in the above are positive, and the inequality (22) follows. So, $\tau < \xi _2$ and, since $d_2<0$, the function $G_2$ has its maximum at $\tau$ in $(0,\xi _2]$. Therefore, by (19), we have

$$\begin{aligned} |\varPsi | \le G_2(\tau ) = \frac{ 4\left[ -68 -12(5+2\lambda )\beta + (-25 +12\lambda +60\lambda ^2)\beta ^2\right] }{ -4 -(5+2\lambda )\beta +2\lambda (1+2\lambda )\beta ^2}. \end{aligned}$$

Thus, using (12), the second inequality in Theorem 1 holds.

We are therefore left to consider the interval $c\in [\sqrt{\xi _2},2)$, where Lemma 2 gives

$$\begin{aligned} |\varPsi | \le 12c(4-c^2)(|C|+|A|)\sqrt{1-\frac{B^2}{4AC}} = 2g_1(c^2)\sqrt{g_2(c^2)}, \end{aligned}$$

(23)

and where for $x\in [0,4],$

$$\begin{aligned} g_1(x) := 32-4x-\lambda (1+2\lambda )\beta ^2x^2, \end{aligned}$$

and

$$\begin{aligned} g_2(x) := 1 + \frac{ (5+2\lambda )^2\beta ^2 (4-x) }{ 16[ 1 - \lambda \beta ^2(1+2\lambda ) ](8+x) }. \end{aligned}$$

It is easily checked that $g_1$ and $g_2$ are both decreasing on $[\xi _2,2)$. Hence, from (23), we have

$$\begin{aligned} |\varPsi | \le 2g_1(\xi _2)\sqrt{g_2(\xi _2)} = G_2(\xi _2), \end{aligned}$$

(24)

where $G_2$ is defined by (20). Since $G_2(\xi _2) \le G_2(\tau )$, it follows from (12) and (24) that the second inequality in Theorem 1 holds once more.

In order to show that the inequalities are sharp, first let $f_1$ be defined for $z\in {\mathbb {D}}$ by

$$\begin{aligned} 1+\frac{zf_1''(z)}{f_1'(z)} = \left( \frac{1}{2}+\lambda \right) p_1^{\beta }(z) + \frac{1}{2} - \lambda , \end{aligned}$$

where $p_1(z) := (1+z^2)/(1-z^2)$ for $z\in {\mathbb {D}}.$ Then, $f_1 \in {{\mathcal {F}}}_{O}(\lambda ,\beta )$ with

$$\begin{aligned} f_1(z) = z+ \frac{1}{6}(1+2\lambda )\beta z^3 + \cdots \end{aligned}$$

for $z\in {\mathbb {D}}.$ Thus, the first bound in Theorem 1 is sharp when $(5+2\lambda )\beta \le 2$.

Next let $f_2$ be defined for $z\in {\mathbb {D}}$ by

$$\begin{aligned} 1+\frac{zf_2''(z)}{f_2'(z)} = \left( \frac{1}{2}+\lambda \right) p_2^{\beta }(z) + \frac{1}{2} - \lambda , \end{aligned}$$

where

$$\begin{aligned} p_2(z) := \frac{1-z^2}{1-bz+z^2},\quad z\in {\mathbb {D}}, \end{aligned}$$

with

$$\begin{aligned} b := \sqrt{\tau }=\sqrt{ \frac{ 2[2-\beta (5+2\lambda )] }{ 2\beta ^2\lambda (1+2\lambda ) -\beta (5+2\lambda ) -4 } }. \end{aligned}$$

(25)

Here $\tau$ is defined by (15) or by (21), and in the first case $0<\tau <4$ in view of (16), or in the second case, $0<\tau<\xi _2<4$ as was shown in II(b). Therefore, $b\in (0,2)$ and $p_2\in {\mathcal {P}}.$ Thus, $f_2 \in {{\mathcal {F}}}_{O}(\lambda ,\beta )$ with

$$\begin{aligned} a_2 = \frac{1}{4}(1+2\lambda )\beta b,\quad a_3 = \frac{1}{24}(1+2\lambda )\beta \left[ -4 +(1+2(1+\lambda )\beta ) b^2\right] , \end{aligned}$$

and

$$\begin{aligned} a_4 = \frac{1}{576}(1+2\lambda )\beta b \left[ 2(8+15\lambda +6\lambda ^2)\beta ^2b^2 -3\beta (7+6\lambda )(4-\tau ^2) -8(3-b^2) \right] . \end{aligned}$$

Hence,

$$\begin{aligned} H_2(2)(f)&= - \frac{\beta ^2(1+2\lambda )^2}{2304}\left[ 64+ 4(-2+(5+2\lambda )\beta )b^2\right. \\&\quad \left. + (-4-(5+2\lambda )\beta +2\lambda (1+2\lambda )\beta ^2)b^4 \right] . \end{aligned} $$

(26)

Finally, substituting b from (25) into (26), shows that equality is attained in the second inequality in Theorem 1 when $(5+2\lambda )\beta >2$. $\square$

4 $H_2(2)(f^{-1})$ for strongly Ozaki functions

Since each class ${{\mathcal {F}}}_{O}(\lambda ,\beta )$ is compact and every $f\in {{\mathcal {F}}}_{O}(\lambda ,\beta )$ is invertible, there exists $r_0\in (0,1)$ such that ${\mathbb {D}}_{r_0}\subset f({\mathbb {D}})$ for every $f\in {{\mathcal {F}}}_{O}(\lambda ,\beta ),$ where ${\mathbb {D}}_{r_0}:=\{z\in {\mathbb {C}}:|z|<r_0\}.$ Therefore, for $w\in {\mathbb {D}}_{r_0}$ the inverse function $f^{-1}$ can be written as

$$\begin{aligned} f^{-1}(w) = w+\sum _{n=2}^{\infty }\delta _n w^n. \end{aligned}$$

(27)

Moreover, when $f\in {{\mathcal {F}}}_{O}(\lambda ,\beta )$ is of the form (1), then the following relations hold (see e.g., [8])

$$\begin{aligned} \delta _2=-a_2, \quad \delta _3=2a_2^2-a_3,\quad \delta _4=-5a_2^3 +5a_2a_3 -a_4. \end{aligned}$$

(28)

The following proposition will be used in our proof.

Proposition 1

Let $t\in (2,11]$ and $u\in (1,3].$ Define $H:[0,4]\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} H(x) := h_1(x) \sqrt{ h_2(x) }, \end{aligned}$$

(29)

where

$$\begin{aligned} h_1(x):= 64 -8x +ux^2, \quad \text {and} \quad h_2(x):= 1+ \frac{ t^2(4-x) }{ 8(2+u)(8+x) }. \end{aligned}$$

Then, H is convex on [0, 4].

Proof

By differentiating H twice, we obtain

$$\begin{aligned}&(h_2(x))^{3/2} H''(x) \\&\quad = h_1''(x) (h_2(x))^2 + h_1'(x) h_2(x) h_2'(x) -\frac{1}{4} h_1(x) (h_2'(x))^2 +\frac{1}{2} h_1(x) h_2(x) h_2''(x) \\&\quad = \frac{ G(x) }{ 32(2+u)^2(8+x)^4 }, \end{aligned} $$

where

$$\begin{aligned} G(x):=& 64u(2+u)^2(8+x)^4 +t^4\left[ 1920 -624x +u(1024 -640x +6x^2 +14x^3 +x^4) \right] \\& -16t^2(8+x)\left[ -768 +2u(-448 +48x +15x^2 +x^3)\right. \\& \left. +u^2(-256 +48x +15x^2 +x^3) \right] . \end{aligned} $$

We show that our assertion is true by proving that $G(x) \ge 0$ for $x\in [0,4]$.

Let $x\in [0,4]$, $u\in (1,3]$ be fixed, and

$$\begin{aligned} F(s) := A_0 + A_1s +A_2s^2, \quad s\in [4,121], \end{aligned}$$

where

$$\begin{aligned}A_0 :=& 64u(2+u)^2(8+x)^4, \\A_1 :=& -16(8+x)\left[ -768 +2u(-448 +48x +15x^2 +x^3) \right. \\& \left. +u^2(-256 +48x +15x^2 +x^3)\right] \end{aligned}$$

and

$$\begin{aligned} A_2 = 1920 -624x +u(1024 -640x +6x^2 +14x^3 +x^4). \end{aligned}$$

I. Consider first the case $A_2 \le 0$. Then, we note that

$$\begin{aligned} F(4)&= 16\left[ 4u^3(8+x)^4 +144(184+17x)\right. \\&\quad +u(95232 +32640x +4806x^2 +342x^3 + 9x^4) \\&\quad \left. +12u^2(6144 +2688x +456x^2 +35x^3 +x^4) \right] , \end{aligned} $$

and so clearly $F(4)>0$.

Also when $s=121$,

$$\begin{aligned} F(121) = k_0 +k_1 u +k_2u^2 +k_3u^3, \end{aligned}$$

where

$$\begin{aligned}&k_0 := 17424 (2296 -439x)>0, \\&k_1 := 3 \left( 9972736 - 2866048x - 154782x^2 + 41370x^3 + 3675x^4\right) , \\&k_2 := 48 \left( 104448 +5760x -4728x^2 -757x^3 -35x^4\right) ,\quad k_3 := 64(8+x)^4 . \end{aligned}$$

Since $k_3>0$ and $1<u\le 3$, we have

$$\begin{aligned} F(121) \ge k_0 +k_1u +(k_2+k_3)u^2, \end{aligned}$$

and

$$\begin{aligned} k_2+k_3 = 16 \left( 329728 +25472x -12648x^2 -2143x^3 -101x^4\right) >0. \end{aligned}$$

(30)

Thus, when $k_1 \ge 0$, it is clear that $F(121)>0$.

If $k_1<0$, then since $3k_1 < k_1u$ we have

$$\begin{aligned} F(121) \ge k_0 +3k_1 +(k_2+k_3)u^2, \end{aligned}$$

(31)

Now that

$$\begin{aligned} k_0 +3k_1 = 9 \left( 14417792 -3715952x -154782x^2 +41370x^3 +3675x^4\right) >0, \end{aligned}$$

(32)

which, from (30), (31) and (32), implies that $F(121)>0$. Thus, since $A_2 \le 0$,

$$\begin{aligned} F(s) \ge \min \{ F(4), F(121) \} >0, \quad s\in [4,121]. \end{aligned}$$

II. Next we consider the case $A_2>0$. In this case

$$\begin{aligned} F(s) \ge A_0 +(A_1+4A_2)s =: \tilde{F}(s), \quad s\in [4,121]. \end{aligned}$$

(33)

First note that

$$\begin{aligned} \tilde{F}(4)=&16\left[ 4u^3(8+x)^4 +144(184+17x)\right. \\&+u( 95232 +32640x +4806x^2 +342x^3 +9x^4 ) \\&\left. + 12u^2(6144 +2688x +456x^2 +35x^3 +x^4)\right] >0. \end{aligned} $$

Also when $s=121$, we have

$$\begin{aligned} \tilde{F}(121) = q_0 +q_1 u +q_2 u^2 +q_3 u^3, \end{aligned}$$

where

$$\begin{aligned}&q_0 := 69696 (184 +17x),\\&q_1 := 12 \left( 1285120 +38528x -45774x^2 -6174x^3 -261x^4\right) ,\\&q_2 := 48 \left( 104448 +5760x -4728x^2 -757x^3 -35x^4\right) ,\quad q_3 := 64(8+x)^4. \end{aligned}$$

Since $q_3 >0$ and $u>1$, we obtain

$$\begin{aligned} \tilde{F}(121) > q_0 +q_1u +(q_2+q_3)u^2, \end{aligned}$$

and further since $q_0>0$, $q_1>0$, and $q_2+q_3>0$, it follows that $\tilde{F}(121) > 0$. Hence, since the function $\tilde{F}$ is linear with respect to s, $\tilde{F}(4)>0$ and $\tilde{F}(121)>0$, we obtain

$$\begin{aligned} \tilde{F}(s) \ge \min \{ \tilde{F}(4), \tilde{F}(121) \} >0, \quad s\in [4,121], \end{aligned}$$

and so from (33) it follows that $F(s)>0$ for $s\in [4,121]$.

Finally, note that I and II implied that $F(s)>0$ for $s\in (4,121]$, $x\in [0,4]$, and $u\in (1,3]$, which therefore shows that $G(x) \ge 0$, and the proof of Proposition 1 is complete. $\square$

Theorem 2

Let $\beta \in (0,1]$ and $\lambda \in [1/2,1]$. If $f\in {{\mathcal {F}}}_{O}(\lambda ,\beta )$, then

$$\begin{aligned}&|H_{2}(2)(f^{-1})|\\&\quad\le {\left\{ \begin{array}{ll} \dfrac{\beta ^2(1+2\lambda )^2}{36}, \quad &{} (1+10\lambda )\beta \le 2, \\ \dfrac{\beta ^2(1+2\lambda )^2}{576} \left( 16 + \dfrac{((1+10\lambda )\beta -2)^2 }{ 4 + (1+10\lambda )\beta + (8\lambda ^2-3 -2\lambda )\beta ^2 } \right) , &{} (1+10\lambda )\beta >2. \end{array}\right. } \end{aligned} $$

(34)

Both inequalities are sharp.

Proof

Note first that since $H_{2}(2)$ and the class ${{\mathcal {F}}}_{O}(\lambda ,\beta )$ are rotationally invariant, by (28) together with (11) and (6) we can write

$$\begin{aligned} H_{2}(2)(f^{-1}) = \delta _2\delta _4-\delta _3^2=\frac{1}{2304}\beta ^2(1+2\lambda )^2 \varPhi , \end{aligned}$$

(35)

where

$$\begin{aligned} \varPhi:=& \left[ 2 -(3+2\lambda -8\lambda ^2)\beta ^2 \right] c^4 -(1+10\lambda ) \beta c^2(4-c^2) \zeta \\& -2(c^2+8)(4-c^2)\zeta ^2 + 12c(4-c^2)(1-|\zeta |^2) \eta , \end{aligned} $$

(36)

for some $c\in [0,2]$ and $\zeta$, $\eta \in \overline{{\mathbb {D}}}$.

I. Assume first that $(1+10\lambda )\beta \le 2$. Then, applying the triangle inequality to (36), and since $|\zeta | \le 1$, we obtain

$$\begin{aligned} |\varPhi |&\le [ 2-(3+2\lambda -8\lambda ^2)\beta ^2 ] c^4 + 12c(4-c^2)\\&\quad + (1+10\lambda ) \beta c^2 (4-c^2)|\zeta | +2(4-c^2)(2-c)(4-c)|\zeta |^2 \\&\le 64 + 4\left[ -2 + (1+10\lambda )\beta \right] c^2 + \beta \left[ -1 -3\beta +8\lambda ^2\beta -2\lambda (5+\beta )\right] c^4. \end{aligned} $$

(37)

Since $-1 -3\beta +8\lambda ^2 \beta -2\lambda (5+\beta ) < 0,$ and $-2 + (1+10\lambda )\beta \le 0$, by (37), it follows that

$$\begin{aligned} |\varPhi | \le 64, \end{aligned}$$

and so from (35), the first inequality in (34) is proved.

II. Next assume that $(1+10\lambda )\beta > 2$. We consider two cases.

II(A) When $\beta =1/2$ and $\lambda =1$, from (36), we obtain

$$\begin{aligned} \varPhi = -2(4-c^2)\zeta \left[ 8\zeta +c^2(3+\zeta )\right] . \end{aligned}$$

Therefore, since $|\zeta | \le 1$ and $c \in [0,2]$, we have

$$\begin{aligned} |\varPhi | \le 2(4-c^2)|\zeta |\left[ 8|\zeta |+c^2(3+|\zeta |) \right] \le 8(4-c^2)(2+c^2) \le 72, \end{aligned}$$

and so from (35), we obtain inequality (34).

II(B) Now assume that $\beta \not =1/2$ or $\lambda \not =1,$ so that $-2 < (-3-2\lambda +8\lambda ^2)\beta ^2 \le 3$.

When $c=0$, (36), gives

$$\begin{aligned} |\varPhi | = 64|\zeta |^2 \le 64, \end{aligned}$$

(38)

and when $c=2$, we have

$$\begin{aligned} |\varPhi | = 16[ 2-(3+2\lambda -8\lambda ^2)\beta ^2 ]. \end{aligned}$$

(39)

We suppose therefore that $c\in (0,2)$ and use Lemma 2. Applying the triangle inequality to (36), we obtain

$$\begin{aligned} |\varPhi | \le 12c(4-c^2) \varPsi (A,B,C), \end{aligned}$$

where for $\zeta \in \overline{{\mathbb {D}}},$

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

with

$$\begin{aligned} A := \frac{ -\left[ 2-(3+2\lambda -8\lambda ^2)\beta ^2 \right] c^3 }{ 12(4-c^2) }, \quad B := \frac{1}{12}(1+10\lambda )\beta c, \quad C := \frac{c^2+8}{6c}. \end{aligned}$$

Clearly, $A<0$, $B>0$ and $C>0$, and so $AC<0$. Furthermore, the following conditions hold for all $\lambda$ and $\beta$ such that $(1+10\lambda )\beta >2$:

• (i) $B^2 > -4AC( C^{-2} -1)$,

• (ii) $2(1-|C|) < |B|$,

• (iii) $|C|( |B|+4|A| ) > |AB|$.

Indeed (i) follows from

$$\begin{aligned} C^{-2}-1 = \frac{ -(16-c^2)(4-c^2) }{ (c^2+8)^2 }<0 \quad \text {and} \quad AC<0, \end{aligned}$$

and (ii) written as $B-2(1-C)>0$ is true, since

$$\begin{aligned} 12c[ B-2(1-C) ] = [ (1+10\lambda )\beta +4] c^2 -24c +32> 6c^2 -24c +32 >0. \end{aligned}$$

Finally, note that since $B-4C<0$,

$$\begin{aligned} 12c(B-4C) = [ (1+10\lambda )\beta -8 ]c^2 -64 \le 3c^2-64 <0 \end{aligned}$$

and $BC>0$, so condition (iii) written as $BC +A(B-4C) >0$ is true.

Next note that the condition $|AB| \le |C|(|B|-4|A|)$ is equivalent to $BC+4AC+AB \ge 0$, and a computation gives

$$\begin{aligned} 144(4-c^2)(BC+4AC+AB) = \varphi (c^2), \end{aligned}$$

where for $x\in [0,4],$

$$\begin{aligned} \varphi (x) := b_0 +2b_1x +b_2x^2 \end{aligned}$$

(40)

with

$$\begin{aligned}&b_0 := 64\beta (1+10\lambda ),\quad b_1 := -4 \left[ 16 +(1+10\lambda )\beta +8(-3-2\lambda +8\lambda ^2)\beta ^2 \right] ,\\&b_2 := -16 -4(1+10\lambda )\beta -8(-3-2\lambda +8\lambda ^2)\beta ^2 +(3+32\lambda +12\lambda ^2-80\lambda ^3) \beta ^3. \end{aligned}$$

We note that $b_0>0$, $b_1<0$ and $b_2<0$. Furthermore, $\varphi '(x)=2b_1+2b_2x<0$ for $x\in (0,4)$. Clearly, $\varphi (0)=b_0>0$, and $\varphi (4)=16b_2 +8b_1 +b_0 <0$. Indeed, by putting $t=(1+10\lambda )\beta$ and $u=(-3-2\lambda +8\lambda ^2)\beta ^2$, since $t>2$ and $u>-2$, we have

$$\begin{aligned} 16b_2 +8b_1 +b_0 = -16(u+2)(24+t) < 0. \end{aligned}$$

Therefore, the function $\varphi$ has a unique zero $x_0\in (0,4)$, where

$$\begin{aligned} x_0 := \frac{ -b_1 -\sqrt{b_1^2 - b_0b_2} }{ b_2 }. \end{aligned}$$

Thus,

$$\begin{aligned} {\left\{ \begin{array}{ll} \varphi (x) \ge 0, \quad &{}\text {when} \; x\in (0,x_0], \\ \varphi (x) \le 0, &{}\text {when} \; x\in [x_0,4), \end{array}\right. } \end{aligned}$$

(41)

and it follows that

$$\begin{aligned} {\left\{ \begin{array}{ll} |AB| \le |C|(|B|-4|A|), \quad &{}\text {when} \; c\in (0,\sqrt{x_0}], \\ |AB| \ge |C|(|B|-4|A|), &{}\text {when} \; c\in [\sqrt{x_0},2). \end{array}\right. } \end{aligned}$$

II(B)-a. Let $c\in (0,\sqrt{x_0}]$. Then, by Lemma 2,

$$\begin{aligned} |\varPhi | \le 12c(4-c^2)(-|A|+|B|+|C|) = g(c^2), \end{aligned}$$

(42)

where for $x\in [0,4],$

$$\begin{aligned} g(x) := 64 +2d_1x +d_2x^2, \end{aligned}$$

(43)

with

$$\begin{aligned} d_1:=2\left[ -2+(1+10\lambda )\beta \right] ,\quad d_2 := -\left[ 4 +(1+10\lambda )\beta +(-3-2\lambda +8\lambda ^2)\beta ^2 \right] . \end{aligned}$$

We note that $g'(x)=0$ only when $x=\tilde{\tau }:=-d_1/d_2$, and clearly, $\tilde{\tau }>0$.

We now show that $\tilde{\tau } < x_0$. Since

$$\begin{aligned} \varphi (\tilde{\tau }) = \frac{1}{d_2^2} \left( b_0d_2^2 - 2b_1d_1d_2 +b_2d_1^2\right) , \end{aligned}$$

by (41), the inequality $\tilde{\tau } < x_0$ is equivalent to

$$\begin{aligned} b_0d_2^2 - 2b_1d_1d_2 +b_2d_1^2 >0. \end{aligned}$$

(44)

Moreover, by putting $t=(1+10\lambda )\beta$, and $u=(-3-2\lambda +8\lambda ^2)\beta ^2$, we obtain

$$\begin{aligned}&b_0d_2^2 - 2b_1d_1d_2 +b_2d_1^2 \\&\quad = 128(14+11u+2u^2) +16(52+9u-4u^2)t +32(7-u)t^2\\&\quad \quad +4(8-u)t^3=: L(u,t). \end{aligned} $$

(45)

Since $t>2$ and $-2<u\le 3$, the quantities $16(52+9u-4u^2)$, $32(7-u)$ and $4(8-u)$ are all positive. Thus, we obtain

$$\begin{aligned} L(u,t)> L(u,2) = 128(6+u)^2 >0, \end{aligned}$$

which, by (45), implies that (44) holds. So $\tilde{\tau }<x_0,$ and since $d_2<0$, the function g has its maximum at $\tilde{\tau }$ in $(0,x_0]$. Thus, by (42), we have

$$\begin{aligned} |\varPhi | \le g(\tilde{\tau }) = 64 +\frac{ 4((1+10\lambda )\beta -2)^2 }{ 4 +(1+10\lambda )\beta +(8\lambda ^2-3-2\lambda )\beta ^2 }. \end{aligned}$$

(46)

II(B)-b. Let $c\in [\sqrt{x_0},2)$. Then, by Lemma 2, we obtain

$$\begin{aligned} |\varPhi | \le 12c(4-c^2)( |C|+|A| )\sqrt{ 1-\frac{B^2}{4AC} } = h_1(c^2) \sqrt{ h_2(c^2) }, \end{aligned}$$

(47)

where for $x\in [0,4],$

$$\begin{aligned} h_1(x) := 64 -8x +(8\lambda ^2-3-2\lambda )\beta ^2 x^2, \end{aligned}$$

and

$$\begin{aligned} h_2(x) := 1 + \frac{ \beta ^2(1+10\lambda )^2 (4-x) }{ 8\left[ 2+(-3-2\lambda +8\lambda ^2)\beta ^2\right] (8+x) }. \end{aligned}$$

It is easy to see that $h_2$ is decreasing in $[x_0,2)$.

(i) Moreover, when $(8\lambda ^2-3-2\lambda )\beta ^2 \le 1$, $h_1$ is also decreasing in $[x_0,2)$. Therefore, from (47), we obtain

$$\begin{aligned} |\varPhi | \le h_1(x_0) \sqrt{ h_2(x_0) }= g(x_0) \le g(\tilde{\tau }), \end{aligned}$$

(48)

where g is the function defined by (43).

(ii) When $(8\lambda ^2-3-2\lambda )\beta ^2 > 1$, by putting $t=(1+10\lambda )\beta$ and $u=(8\lambda ^2-3-2\lambda )\beta ^2$ so that $2<t\le 11$ and $1<u\le 3$, we obtain

$$\begin{aligned} h_1(c^2) \sqrt{h_2(c^2)} = H(c^2), \end{aligned}$$

(49)

where H is the function defined by (29). By Proposition 1, we deduce that

$$\begin{aligned} H(c^2)&\le \max \{ H(x_0), H(4) \}\\&= \max \{ g(x_0), 32 +16(8\lambda ^2-3-2\lambda )\beta ^2 \} \le g(\tilde{\tau }), \end{aligned} $$

(50)

and so by (47), (49) and (50), we obtain the inequality (48) once more.

Thus, from (38), (39), (46) and (48), inequality (46) holds for all $c\in [0,2]$ and $\zeta$, $\eta \in \overline{{\mathbb {D}}}$, and so from (35), the second inequality in (34) is proved.

In order to show that the inequalities are sharp, first let $f_1$ be defined for $z\in {\mathbb {D}}$ by

$$\begin{aligned} 1+\frac{zf_1''(z)}{f_1'(z)} = \left( \frac{1}{2}+\lambda \right) p_1^{\beta }(z) + \frac{1}{2} - \lambda , \end{aligned}$$

where $p_1(z) := (1+z^2)/(1-z^2)$ for $z\in {\mathbb {D}}.$ Then, $f_1 \in {{\mathcal {F}}}_{O}(\lambda ,\beta )$ with

$$\begin{aligned} f_1(z) = z+ \frac{1}{6}(1+2\lambda )\beta z^3 + \cdots \end{aligned}$$

for $z\in {\mathbb {D}}.$ Hence, $f_1^{-1}$ is given by

$$\begin{aligned} f_1^{-1}(w) = w -\frac{1}{6}(1+2\lambda )\beta w^3 + \cdots \end{aligned}$$

for $w\in {\mathbb {D}}_{r_0}.$ Thus, $H_{2,2}(f_1^{-1}) = -\beta ^2(1+2\lambda )^2/36$, which shows that the first bound in (34) is sharp when $(1+10\lambda )\beta \le 2$.

Next let $f_2$ be defined, for $z\in {\mathbb {D}}$ by

$$\begin{aligned} 1+\frac{zf_2''(z)}{f_2'(z)} = \left( \frac{1}{2}+\lambda \right) p_2^{\beta }(z) + \frac{1}{2} - \lambda , \end{aligned}$$

where

$$\begin{aligned} p_2(z) := \frac{1+bz+z^2}{1-z^2},\quad z\in {\mathbb {D}}, \end{aligned}$$

with

$$\begin{aligned} b := \sqrt{ \frac{ 2[ (1+10\lambda )\beta -2] }{ 4 + (1+10\lambda )\beta +(8\lambda ^2-3-2\lambda )\beta ^2 } }. \end{aligned}$$

(51)

Since $b \in (0,2)$, it follows that $p_2 \in {{\mathcal {P}}}$. Therefore, $f_2 \in {{\mathcal {F}}}_{O}(\lambda ,\beta )$ with

$$\begin{aligned} a_2 = \frac{1}{4}(1+2\lambda )\beta b, \quad a_3 = \frac{1}{24}(1+2\lambda )\beta \left[ 4+ (-1+2(1+\lambda )\beta ) b^2 \right] \end{aligned}$$

and

$$\begin{aligned} a_4&= \frac{1}{576}(1+2\lambda )\beta b\\&\quad \times \left[ 2(8+15\lambda +6\lambda ^2)\beta ^2b^2 -3(7+6\lambda )\beta (-4+b^2) +8(-3+b^2) \right] . \end{aligned} $$

A simple calculation gives

$$\begin{aligned}H_{2}(2)(f_2^{-1}) &= a_2^4 - a_2^2a_3 +a_2a_4 -a_3^2\\& = -\frac{\beta ^2(1+2\lambda )^2}{2304}\left[ 64 +4(-2+(1+10\lambda )\beta )b^2\right. \\&\quad\left. -(4+(1+10\lambda )\beta +(-3-2\lambda +8\lambda ^2)\beta ^2)b^4 \right] , \end{aligned} $$

(52)

and substituting for b in (51) into (52) we obtain

$$\begin{aligned} H_{2}{(2)}(f_2^{-1}) = -\frac{\beta ^2(1+2\lambda )^2}{2304} \left( 64 + \frac{ 4((1+10\lambda )\beta -2)^2 }{ 4 + (1+10\lambda )\beta + (8\lambda ^2-3 -2\lambda )\beta ^2 } \right) . \end{aligned}$$

This shows that the bound in (34) is sharp for the case $(1+10\lambda )\beta > 2$. $\square$

5 Strongly convex functions $(\lambda =1/2)$

When $\lambda =1/2,$ i.e. when $f\in {\mathcal {C}}^{\beta },$ the class of convex functions of order $\beta$, it was shown in [9] that the bounds for $|H_2(2)(f)|$ and $|H_2(2)(f^{-1})|$ were the same, reflecting other invariant properties concerning the coefficients of f and $f^{-1}$. It was shown that when $f\in {\mathcal {C}}^{\beta },$

$$\begin{aligned} |H_{2}(2)(f)|\, \left( |H_{2}(2)(f^{-1})|\right) \le {\left\{ \begin{array}{ll} \dfrac{\beta ^2}{9},&{} 0<\beta \le \dfrac{1}{3},\\ \dfrac{\beta (1+\beta )(1+17\beta )}{72(3+\beta )}, &{} \dfrac{1}{3}< \beta \le 1.\\ \end{array}\right. } \end{aligned}$$

Although these bounds are correct, they are not (as claimed), best possible, since $\lambda =1/2$ in Theorems 1 and 2 give the following sharp bounds.

Theorem 3

If $f\in {\mathcal {C}}^\beta$, then for $0<\beta \le 1,$

$$\begin{aligned} |H_{2}(2)(f)|\, \left( |H_2(2)(f^{-1})|\right) \le {\left\{ \begin{array}{ll} \dfrac{\beta ^2}{9},&{} 0<\beta \le \dfrac{1}{3},\\ \dfrac{\beta ^2(1+\beta )(17+\beta )}{72(2+3\beta -\beta ^2)}, &{} \dfrac{1}{3}< \beta \le 1.\\ \end{array}\right. } \end{aligned}$$

When $\beta =1$, we deduce the following [4].

Corollary 1

If $f\in {\mathcal {C}}$, then

$$\begin{aligned} |H_{2}(2)(f)|\,\left( |H_2(2)(f^{-1})|\right) \le \dfrac{1}{8}. \end{aligned}$$

We note that the proof of the weaker result above used Lemma 1, which alone was not strong enough to give the sharp estimate, whereas the additional use of Lemma 2 produces the correct estimates given in Theorem 3.

Remark 1

We finally note that when $\lambda =1$, Theorems 1 and 2 show that there is no invariance between $|H_{2}(2)(f)|$ and $|H_2(2)(f^{-1})|$; however, a simple calculation gives the following strange invariance property when $\beta =1$.

Corollary 2

If $f\in {\mathcal {F}}_{O}(\lambda )$ with $\lambda =\dfrac{1}{18}(8 + \sqrt{73})$, then

$$\begin{aligned} |H_{2}(2)(f)|\,\left( |H_2(2)(f^{-1})|\right) \le \dfrac{2395 + 233 \sqrt{73}}{15552}. \end{aligned}$$

Both inequalities are sharp.