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(ABC).

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 \((1+10\lambda )\beta >2\):

  1. (i)

    \(B^2 > -4AC( C^{-2} -1)\),

  2. (ii)

    \(2(1-|C|) < |B|\),

  3. (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.