On the difference of inverse coefficients of 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 normalised univalent functions given by $f(z)=z+\sum _{n=2}^{\infty }a_n z^n$ for $z\in {\mathbb {D}}$. Let F be the inverse function of f defined in some set $|\omega |\le r_{0}(f)$, and be given by $F(\omega )=\omega +\sum _{n=2}^{\infty }A_n \omega ^n$. We prove the sharp inequalities $-1/3 \le |A_4|-|A_3| \le 1/4$ for the class ${\mathcal {K}}\subset {\mathcal {S}}$ of convex functions, thus providing an analogue to the known sharp inequalities $-1/3 \le |a_4|-|a_3| \le 1/4$, and giving another example of an invariance property amongst coefficient functionals of convex functions.

1 Introduction and definitions

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

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

(1.1)

In 1985, de Branges [1] solved the famous Bieberbach conjecture by showing that if $f\in {{\mathcal {S}}}$, then $|a_n| \le n$ for $n \ge 2$, with equality when $f(z)=k(z):=z/(1-z)^2$, or a rotation. It was therefore natural to ask if for $f\in {\mathcal S}$, the inequality $||a_{n+1}|-|a_{n}|| \le 1$ is true when $n \ge 2$. This was shown not to be the case even when $n=2$ [2], and that the following sharp bounds hold.

$$\begin{aligned} -1 \le |a_3| - |a_2| \le \frac{3}{4} + e^{-\lambda _0}(2e^{-\lambda _0}-1) = 1.029\cdots , \end{aligned}$$

where $\lambda _0$ is the unique value of $\lambda$ in $0< \lambda <1$, satisfying the equation $4\lambda = e^{\lambda }$.

Hayman [3] showed that if $f \in {{\mathcal {S}}}$, then $| |a_{n+1}| - |a_n| | \le C$, where C is an absolute constant. The exact value of C is unknown, the best estimate to date being $C=3.61\cdots$ [4], which because of the sharp estimate above when $n=2$, cannot be reduced to 1.

Although the expected inequality $| |a_{n+1}| - |a_n| | \le 1$ has been verified for $n\ge 2$ by Leung [5] in the case of starlike functions $\mathcal {S^*}$, sharp upper and lower bounds for $|a_{n+1}| - |a_n|$ are only known when $n=2$ for some subclasses of ${\mathcal {S}}$, such as the classes ${\mathcal {K}}$ of convex and close-to-convex functions [6, 7]. An exception to this was provided by Ming and Sugawa [8], who showed that if $f\in {\mathcal {K}}$, then $-1/3\le |a_{4}| - |a_3| \le 1/4,$ and that both of these inequalities are sharp. It turns out that finding sharp bounds for $| |a_{n+1}| - |a_n| |$ in the case of convex functions presents a significantly difficult problem.

If $f\in {\mathcal {S}}$, then since f is univalent, it possesses an inverse function F, given by $F(\omega )=\omega +\sum _{n=2}^{\infty }A_n \omega ^n$ defined in some set $|\omega |\le r_{0}(f)$.

Little information is known about the the difference of coefficients of the inverse functions for $f\in {\mathcal {S}}$, and even finding the order of growth of $||A_{n+1}| - |A_n||$ appears to be an open problem. On the other hand sharp upper and lower bounds for $|A_{n+1}| - |A_n|$ have recently been found when $n=2$ for a number of subclasses of ${\mathcal {S}}$ [9].

In this paper we will show that if $f\in {\mathcal {K}}$, then $-1/3\le |A_{4}| - |A_3| \le 1/4,$ providing another example of an invariance property amongst coefficient functionals for convex functions, noticed in [9,10,11].

We note first that equating coefficients easily gives

$$\begin{aligned} A_{2}=-a_{2}, \quad A_{3}=2a_{2}^2-a_{3} \quad \text {}\quad A_4 = -5a_2^3 +5a_2a_3 -a_4. \end{aligned}$$

(1.2)

2 Preliminary Lemmas

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

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

(2.1)

We will use the following lemmas for the coefficients of functions ${{\mathcal {P}}}$, given by (2.1).

Lemma 2.1

[2, p. 41] For $p \in {{\mathcal {P}}}$, $|c_n|\le 2$ for $n\ge 1.$ The inequalities are sharp.

Lemma 2.2

[10] If $p \in {{\mathcal {P}}}$ is of the form (2.1) with $c_1\ge 0$, then

$$\begin{aligned} 2c_2 = c_1^2 + (4-c_1^2)\zeta \end{aligned}$$

(2.2)

and

$$\begin{aligned} 4c_3 = c_1^3 +(4-c_1^2)c_1\zeta (2-\zeta ) + 2(4-c_1^2)(1-|\zeta |^2) \eta \end{aligned}$$

(2.3)

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

The next lemma is a special case of more general results due to Choi et al. [12] (see also [13]). Let $\overline{\mathbb {D}}:=\{z\in \mathbb {C}: |z|\le 1\},$ and define

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

Lemma 2.3

[12] 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

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

where

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

(2.4)

We recall that the discriminant $\Delta _P$ of the real quadratic polynomial $P(t)=\alpha +2\beta t+\gamma t^2$ is defined as $\beta ^2 -\alpha \gamma$. We will allow a degenerate case such as $\gamma =0$. The next lemma will also feature in our proof.

Lemma 2.4

Let P(t) and Q(t) be (possibly degenerate) real quadratic polynomials. Suppose that $P(t)>0$ and $Q(t)>0$ on an interval $I \subset {\mathbb {R}}$ and that $\Delta _P >0$. If there exists a positive constant T such that

• (i) $\Delta _Q \ge T^{3/2} \Delta _P$, and

• (ii) $T P(t) \ge Q(t)$ for $t\in I$,

then the function $G(t) = \sqrt{P(t)} - \sqrt{Q(t)}$ is convex on I.

3 Preliminary results

In order to prove the main result in this paper we will need the following two theorems.

Theorem 3.1

Let $f \in {{\mathcal {K}}}$ and be given by (1.1). Then

$$\begin{aligned} |A_4| \le {\left\{ \begin{array}{ll} \dfrac{1}{24} (4+ 21s^2 +s^3), \quad &{}\text {if} \; 0 \le s \le \dfrac{2}{7}, \\ s(1-2s^2), &{}\text {if} \; \dfrac{2}{7} \le s \le \sqrt{ \dfrac{5}{59} }, \\ \dfrac{ (1+5s^2)\sqrt{25-s^2} }{ 12\sqrt{6} }, &{}\text {if} \; \sqrt{ \dfrac{5}{59} } \le s \le \sqrt{ \dfrac{5}{11} }, \\ \dfrac{1}{3}s(2+s^2), &{}\text {if} \; \sqrt{ \dfrac{5}{11} } \le s \le 1, \end{array}\right. } \end{aligned}$$

(3.1)

where $s = |a_2| \in [0,1]$. All the inequality are sharp.

Proof

We note at this point that if $f \in {{\mathcal {K}}}$, then $|a_2|\le 1$, and we can write $1+zf''(z)/f'(z)=p(z)$, for some $p \in {{\mathcal {P}}}$ with the form (2.1). So equating coefficients gives

$$\begin{aligned} a_2 = \frac{1}{2}c_1, \quad a_3 = \frac{1}{6}(c_1^2+c_2) \quad \text {and}\quad a_4 = \frac{1}{24}(c_1^3+3c_1c_2+2c_3). \end{aligned}$$

(3.2)

Hence from (1.2) we obtain

$$\begin{aligned} A_4 = \frac{1}{24}(-6c_1^3 +7c_1c_2 -2c_3). \end{aligned}$$

(3.3)

Since ${{\mathcal {P}}}$ and $|A_4|$ are rotationally invariant, using Lemma 2.1, we may assume that $c_1=c$ with $0 \le c \le 2$. Now using Lemma 2.2, we obtain

$$\begin{aligned} 48A_4 = -6c^3 +5c(4-c^2)\zeta +c(4-c^2)\zeta ^2 -2(4-c^2)(1-|\zeta |^2)\eta , \end{aligned}$$

(3.4)

where $\eta \in \overline{{\mathbb {D}}}$. If $a_2=c/2=1$, then $A_4 = -1$, and so the inequality (3.1) is true when $s=|a_2|=1$. Also if $a_2=c/2=0$, then $|A_4| = (1-|\zeta |^2)|\eta |/6 \le 1/6$, and so the inequality (3.1) is true when $s=|a_2|=0$.

We now assume that $0<a_2<1$, or equivalently, $0<c<2$. Then, since $|\eta | \le 1$, (3.4) implies that

$$\begin{aligned} 48|A_4| \le 2(4-c^2)[ |A+B\zeta +C\zeta ^2| +(1-|\zeta |^2) ], \end{aligned}$$

(3.5)

where

$$\begin{aligned} A=\frac{-3c^3}{4-c^2}, \quad B=\frac{5}{2}c, \quad C=\frac{1}{2}c. \end{aligned}$$

We note that $AC<0$.

• (a) We first consider the case $0 < a_2 \le 2/7$. Since $a_2=c/2$, we have $0 < c \le 4/7$. and the condition $-4AC(C^{-2}-1) \le B^2$ and $|B|<2(1-|C|)$ in Lemma 2.3 is satisfied. Therefore from Lemma 2.3 we obtain

  $$\begin{aligned} 48|A_4| \le 2(4-c^2)\left[ 1-|A|+\frac{B^2}{4(1-|C|)} \right] = \frac{1}{4}(32+42c^2+c^3). \end{aligned}$$

  (3.6)

  We note that for $c \ge 4/7$ (i.e. the case $a_2 \ge 2/7$), the conditions $|B| \ge 2(1-|C|)$ and $-4AC(C^{-2}-1) \ge B^2$ are valid. Hence from Lemma 2.3 we have

  $$\begin{aligned} \max _{\zeta \in \overline{{\mathbb {D}}}} \left\{ |A+B\zeta +C\zeta ^2| +(1-|\zeta |^2) \right\} = R(A,B,C), \end{aligned}$$

  where R is given by (2.4).

• (b) For the case $2/7 \le a_2 \le \sqrt{5/59}$, the conditions $4/7 \le c \le 2\sqrt{5/59}$ and $|AB| \le |C|(|B|-4|A|)$ are valid, and so Lemma 2.3 gives

  $$\begin{aligned} 48|A_4| \le 2(4-c^2)( -|A|+|B|+|C| ) = 12c(2-c^2). \end{aligned}$$

  (3.7)

• (c) For the case $\sqrt{5/11} \le a_2 < 1$ we have $2\sqrt{5/11} \le c < 2$, and the condition $|C|(|B|+4|A|) \le |AB|$ is satified. Therefore Lemma 2.3 gives

  $$\begin{aligned} 48|A_4| \le 2(4-c^2)(|A|+|B|-|C|) = 2c(8+c^2). \end{aligned}$$

  (3.8)

• (d) For the case $\sqrt{5/59} \le a_2 \le \sqrt{5/11}$, we have $2\sqrt{5/59} \le c \le 2\sqrt{5/11}$, and the conditions $|C|(|B|+4|A|) \ge |AB|$ and $|AB| \ge |C|(|B|-4|A|)$ are satified. Therefore Lemma 2.3 gives

  $$\begin{aligned} 48|A_4| \le 2(4-c^2)( |A|+|C| ) \sqrt{ 1-\frac{B^2}{4AC} } = \frac{1}{2}c(4+5c^2) \sqrt{ \frac{50}{3c^2} - \frac{1}{6} }. \end{aligned}$$

  (3.9)

Finally replacing c in (3.6), (3.7), (3.8) and (3.9) by 2s, we obtain the inequalities in (3.1).

We now show that the inequalities in (3.1) are sharp by constructing extreme functions for each case.

Given $p \in {{\mathcal {P}}}$, let $f_p \in {{\mathcal {K}}}$ be defined by

$$\begin{aligned} 1+ \frac{zf_p''(z)}{f_p'(z)} = p(z). \end{aligned}$$

(3.10)

For the case $2/7 \le s \le \sqrt{5/59}$, we consider a function $p_1 \in {{\mathcal {P}}}$ defined by $p_1(z) = (1+2sz+z^2)/(1-z^2)$. Then it is easy to see that the coefficients of $p_1$ are given by $c_1=2s$, $c_2=2$ and $c_3=2s$. From (3.3), we obtain $A_4 = s(1-2s^2)$.,and so the inequality (3.1) in this case is sharp for $f_{p_1} \in {{\mathcal {K}}}$.

In a similar way we can see that the inequality (3.1) in the case $\sqrt{5/11} \le s \le 1$ is sharp for $f_{p_2} \in {\mathcal K}$, where

$$\begin{aligned} p_2(z) = \frac{1-z^2}{1-2sz+z^2} = 1 +2sz +(4s^2-2)z^2+ (8s^3-6s)z^3+ \cdots . , \end{aligned}$$

and the inequality (3.1) in the case $0 \le s \le 2/7$ is sharp for $f_{p_3} \in {{\mathcal {K}}}$, where

$$\begin{aligned} \begin{aligned} p_3(z)&= \frac{2-3sz+3sz^2-2z^3}{2-7sz-7sz^2+2z^3} \\&= 1 +2sz +s(5+7s)z^2+ \frac{1}{2}(-4+49s^2+49s^3)z^3+ \cdots . \end{aligned} \end{aligned}$$

Finally we consider the case $\sqrt{5/59} \le s \le \sqrt{5/11}$. Let $p_4 \in {{\mathcal {P}}}$ be defined by

$$\begin{aligned} p_4(z) = \frac{ 1+s(\xi +1)z +\xi z^2 }{ 1+s(\xi -1)z -\xi z^2 }, \end{aligned}$$

where $\xi = e^{i\theta }$ with

$$\begin{aligned} \theta = \arccos \left( \frac{ 5(1-7s^2) }{ 24s^2 } \right) , \end{aligned}$$

(3.11)

then it is easy to see that the coefficients of $p_4$ are given by

$$\begin{aligned} c_1 = 2s, \quad c_2 = 2(s^2(1-\xi ) + \xi ) \quad \text {and}\quad c_3 = 2s[ s^2(1-\xi )^2 +(2-\xi )\xi ], \end{aligned}$$

which by (3.3) implies that

$$\begin{aligned} A_4 = \frac{1}{6}s[ \xi (5+\xi ) -s^2(6+5\xi +\xi ^2) ]. \end{aligned}$$

(3.12)

Since $\xi = e^{i\theta }$, simple computations using (3.11) show that

$$\begin{aligned} \begin{aligned}&| \xi (5+\xi ) -s^2(6+5\xi +\xi ^2) |^2 \\&= | (1-7s^2)\cos \theta +5(1-s^2) +i (1+5s^2)\sin \theta |^2 \\&= -24s^2(1-s^2)\cos ^2\theta + 10(1-s^2)(1-7s^2)\cos \theta + 25(1-s^2)^2 +(1+5s^2)^2 \\&= \frac{ (25-s^2)(1+5s^2)^2 }{ 24s^2 }. \end{aligned} \end{aligned}$$

Hence from (3.12) we obtain

$$\begin{aligned} |A_4| = \frac{1}{6}s | \xi (5+\xi ) -s^2(6+5\xi +\xi ^2) | = \frac{ (1+5s^2)\sqrt{25-s^2} }{ 12\sqrt{6} }, \end{aligned}$$

and so (3.1) in the case $\sqrt{5/59} \le s \le \sqrt{5/11}$ is sharp for $f_{p_4} \in {{\mathcal {K}}}$, which completes the proof of Theorem 3.1. $\square$

We next denote by ${\mathcal {K}}_+$ the class of functions $f \in {\mathcal {A}}$ with non-negative second coefficient, and prove the following.

Theorem 3.2

Let $f \in {\mathcal {K}}_{+}$ and be given by (1.1). Then

$$\begin{aligned} |A_3+A_4| \le {\left\{ \begin{array}{ll} \dfrac{1}{3}(1-s)(-1+2s+6s^2), \quad &{}\dfrac{4}{7} \le s \le 1, \\ \dfrac{1}{24}(8-16s+25s^2+s^3), &{}0 \le s \le \dfrac{4}{7}, \end{array}\right. } \end{aligned}$$

(3.13)

where $s=a_2 \in [0,1]$. The inequalities are sharp.

Proof

Since $f \in {\mathcal {K}}_{+}$ we can write $1+zf''(z)/f'(z) = p(z)$, for some $p \in {\mathcal {P}}$ given by (2.1). From (3.2) we obtain (3.3) and

$$\begin{aligned} A_3 = 2a_2^2 - a_3 = \frac{1}{6}(2c_1^2-c_2). \end{aligned}$$

We also note that $c_1 \in [0,2]$ since $a_2 \ge 0$, and so

$$\begin{aligned} A_3+A_4 = \frac{1}{24}( 8c_1^2 -6c_1^3 +7c_1c_2 -4c_2 -2c_3). \end{aligned}$$

Thus using Lemma 2.2 we have

$$\begin{aligned} 48(A_3+A_4) = 6c^2(2-c) +(5c-4)(4-c^2)\zeta +c(4-c^2)\zeta ^2 -2(4-c^2)(1-|\zeta |^2)\eta , \end{aligned}$$

(3.14)

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

If $c=2s=2$, then $A_3+A_4=0$, hence (3.13) is true when $s=1$.

Next assume that $0 \le s < 1$ (i.e. $0 \le c < 2$). Then since $|\eta | \le 1$, (3.14) implies that

$$\begin{aligned} 48|A_3+A_4| \le 2(4-c^2)[ |A +B\zeta +C\zeta ^2| - (1-|\zeta |^2) ], \end{aligned}$$

where

$$\begin{aligned} A=\frac{3c^2}{2+c}, \quad B=\frac{5c-4}{2}, \quad C=\frac{1}{2}c. \end{aligned}$$

We note that $AC \ge 0$.

• (a) First we consider the case $8/7 \le c < 2$., then it is easy to see that $|B| \ge 2(1-|C|)$ is valid, and so by Lemma 2.3 we obtain

  $$\begin{aligned} \begin{aligned} 48|A_3+A_4|&\le 2(4-c^2)( |A|+|B|+|C| ) \\&= 4(2-c)(-2+2c+3c^2). \end{aligned} \end{aligned}$$

  (3.15)

• (b) Next consider the case $0 \le c < 8/7$, then the condition $|B| <2(1-|C|)$ is satisfied, and so by Lemma 2.3, we obtain

  $$\begin{aligned} \begin{aligned} 48|A_3+A_4|&\le 2(4-c^2)( 1+|A|+\frac{B^2}{4(1-|C|)} ) \\&= \frac{1}{4}(64-64c+50c^2+c^3). \end{aligned} \end{aligned}$$

  (3.16)

Replacing c in (3.15) and (3.16) by 2s, we therefore obtain (3.13) when $s \in [0,1)$.

Now we shall show the inequality (3.13) is sharp. For given $p \in {\mathcal {P}}$, we recall the function $f_p$ defined by (3.10). For the case $4/7 \le s \le 1$, we can easily check that the equality in (3.13) holds for $f_{p_1} \in {\mathcal {K}}$, where $p_1(z) = (1+2sz+z^2)/(1-z^2)$. For the case $0 \le s \le 4/7$, we can check the inequality (3.13) is sharp for $f_{p_5} \in {\mathcal {K}}$, where $p_5 \in {\mathcal {P}}$ is defined by

$$\begin{aligned} \begin{aligned} p_5(z)&= \frac{(1-z)(2+(4-3s)z+2z^2)}{(1+z)(2-7sz+2z^2)} \\&= 1+ 2sz + (-2+3s+7s^2)z^2 + \frac{1}{2}s(-24+21s+49s^2)z^3 + \cdots . \end{aligned} \end{aligned}$$

Now the proof of Theorem 3.2 is finished. $\square$

Proposition 3.3

For a fixed constant $c \in [1,2]$, define $F_c:[0,1]\times [-1,1] \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \begin{aligned} F_c(r,t)&= 16(1+6r^2+r^4) -4c^2( 1+31r^2+2r^4+24rt+10r^3t ) \\&\quad +c^4[ 36+r^4+60rt+10r^3t+r^2(13+24t^2) ]. \end{aligned} \end{aligned}$$

(3.17)

Then $F_c(r,t) \ge 0$ for all $(r,t) \in [0,1]\times [-1,1]$.

Proof

We first show that $F_c$ does not have any critical points in $(0,1)\times (-1,1)$. Since

$$\begin{aligned} \frac{ \partial F_c }{ \partial t }(r,t) = 2c^2r[ -4(12+5r^2)+c^2(30+5r^2+24rt) ], \end{aligned}$$

$(\partial F_c)/(\partial t)(r,t) = 0$ when $t=t_0$, where

$$\begin{aligned} t_0 = \frac{ 48 +20r^2 -5c^2(6+r^2) }{ 24c^2r}. \end{aligned}$$

Furthermore a simple computation gives

$$\begin{aligned} \frac{ \partial F_c }{ \partial r }(r,t) \Big |_{t=t_0} = -\frac{1}{6}r \varphi _c(r), \end{aligned}$$

where $\varphi _c(r) = (4-c^2)^2 r^2 -6(32-108c^2 +c^4)$. It is now easy to see that $\varphi _c(r) >0$ for $r \in (0,1)$. Thus the system of equations $(\partial F_c)/(\partial t) = 0 = (\partial F_c)/(\partial r) = 0$ has no roots in $(0,1)\times (-1,1)$.

Next we show that $F_c \ge 0$ on the boundary of $[0,1]\times [-1,1]$.

• (i) We first note that $F_c(0,t) = 4(4-c^2+9c^4) >0$.

• (ii) Let $g_1(t) = F_c(1,t)$, $t\in [-1,1]$. If $c \le 2 \sqrt{17/59}$, then $g_1$ is decreasing on $[-1,1]$, since

  $$\begin{aligned} g_1'(t) = -136c^2 +70c^4 +48c^4 t \le c^2(-136+118c^2) \le 0, \quad t\in [-1,1]. \end{aligned}$$

  Thus $g_1(t) \ge g_1(1) = 16(8-17c^2+9c^4) \ge 0$, $t\in [-1,1]$. If $c > 2 \sqrt{17/59}$, we have $g_1'(t_0) =0$, where $t_0 = (68-35c^2)/(24c^2) \in (-1,1)$. Since $g_1''(t_0)>0$ we have

  $$\begin{aligned} g_1(t) \ge g_1(t_0) = \frac{1}{24} (-1552 +1496c^2 -25c^4) \ge 0, \quad t\in [-1,1]. \end{aligned}$$

• (iii) For a fixed $r \in [0,1]$, define $G_1:[1,4]\rightarrow {\mathbb {R}}$ by $G_1(x) = b_2 x^2 +b_1x +b_0$, where

  $$\begin{aligned} b_2 = (6-5r+r^2)^2, \quad b_1 = -4(1-24r+31r^2-10r^3+2r^4) \quad \text {and}\quad b_0 = 16(1+6r^2+r^4). \end{aligned}$$

  Then since $b_2 \ge 0,$

  $$\begin{aligned} G_1'(x) = 2b_2 x +b_1 \ge 2b_2+b_1 = 68 -24r -50r^2 +20r^3 -6r^4 >0, \quad x\in [1,4], \end{aligned}$$

  and so $G_1(x) \ge G_1(1) = 48+36r+9r^2+30r^3+9r^4 > 0$. Thus $G_1(c^2) > 0$ holds for all $c \in [1,4]$, and $0 \le r \le 1$, and we obtain the inequality $F_c(r,-1)>0$ for $r\in [0,1]$ since $F_c(r,-1) = G_1(c^2)$.

• (iv) Now let $G_2$ be defined by $G_2(x) = b_2 x^2 +b_1x +b_0$, where

  $$\begin{aligned} b_2 = (6+5r+r^2)^2, \quad b_1 = -4(1+24r+31r^2+10r^3+2r^4) \quad \text {and}\quad b_0 = 16(1+6r^2+r^4). \end{aligned}$$

  By the same methods used in (iii) we have

  $$\begin{aligned} G_2(x) \ge G_2(1) = 3(1-r)(16+4r+7r^2-3r^3) \ge 0, \quad x\in [1,4], \end{aligned}$$

  and so the identity $F_c(r,1) = G_2(c^2)$ shows that $F_c(r,1) \ge 0$, $r\in [0,1]$, which completes the proof of Proposition 3.3.

$\square$

Since all the proofs of the inequalities in the following proposition are similar to those used in the proof of Proposition 3.3 above, we omit the details.

Proposition 3.4

Define $G_1:[0,2]\times [0,1]\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} G_1(c,r) = -6c^2-5r(4-c^2)+(4-c^2)r^2. \end{aligned}$$

Then $G_1(c,r) \le 0$ for all $(c,r) \in [0,2]\times [0,1]$.

We are able to now state and prove our main result.

4 Main result

Theorem 4.1

Let $f \in {{\mathcal {K}}}$ and be given by (1.1). Then

$$\begin{aligned} -\frac{1}{3} \le |A_4|-|A_3| \le \frac{1}{4}. \end{aligned}$$

(4.1)

Both inequalities are sharp.

Proof

Since $f \in {{\mathcal {K}}}$ we again write $1+zf''(z)/f'(z)=p(z)$ for some $p \in {{\mathcal {P}}}$ with the form (2.1), then equating coefficients gives

$$\begin{aligned} a_2 = \frac{1}{2}c_1, \quad a_3 = \frac{1}{6}(c_1^2+c_2) \quad \text {and}\quad a_4 = \frac{1}{24}(c_1^3+3c_1c_2+2c_3). \end{aligned}$$

Thus from (1.2) we have

$$\begin{aligned} |A_4|-|A_3| = \frac{1}{24}|-6c_1^3 +7c_1c_2 -2c_3| - \frac{1}{6}|2c_1^2 -c_2|. \end{aligned}$$

Since ${{\mathcal {P}}}$ and $|A_4|-|A_3|$ are rotationally invariant, using Lemma 2.1, we may assume that $c_1=c$ with $0 \le c \le 2$. Moreover by Theorem 3.1, $|A_4|-|A_3| \le |A_4| \le 1/4$ when $0 \le c \le 2\sqrt{5/59}$, and so it is enough to consider c satisfying $2\sqrt{5/59} \le c \le 2$.

We now use Lemma 2.2, and the fact that $|\eta | \le 1$ to obtain

$$\begin{aligned} \begin{aligned}&48(|A_4|-|A_3|) \\&\le | -6c^3 +5c(4-c^2)\zeta +c(4-c^2)\zeta ^2 | - |12c^2-4(4-c^2)\zeta | +2(4-c^2)(1-|\zeta |^2) \\&=: \Psi _U(c,\zeta ), \end{aligned} \end{aligned}$$

(4.2)

where $c \in [2\sqrt{5/59},2]$ and $\zeta \in \overline{{\mathbb {D}}}$. We now prove that

$$\begin{aligned} \Psi _U(c,\zeta ) \le 12 \end{aligned}$$

for $c \in [2\sqrt{5/59},2]$ and $\zeta \in \overline{{\mathbb {D}}}$. Note that since $\Psi _U(c,0) = 8-c^2(14-6c) \le 8$, we may assume that $\zeta \in \overline{{\mathbb {D}}}$ with $\zeta \not =0$.

I(a) We first assume that $2\sqrt{5/59} \le c \le 1$, then

$$\begin{aligned} \begin{aligned}&\Psi _U(c,\zeta ) \\&\le |-6c^2 +5(4-c^2)\zeta +(4-c^2)\zeta ^2| - |12c^2-4(4-c^2)\zeta | +2(4-c^2)(1-|\zeta |^2). \end{aligned} \end{aligned}$$

(4.3)

Putting $\zeta =re^{i\theta }$, $r\in (0,1]$, $\theta \in {\mathbb {R}}$, and $t=\cos \theta \in [-1,1]$, we have

$$\begin{aligned} \begin{aligned}&|-6c^2 +5(4-c^2)\zeta +(4-c^2)\zeta ^2| - |12c^2-4(4-c^2)\zeta | \\&= \sqrt{ b_0 +b_1t +b_2t^2 } - \sqrt{ d_0 +d_1t} \\&=: \Lambda _1(c,r,t), \end{aligned} \end{aligned}$$

(4.4)

where

$$\begin{aligned} \begin{aligned} b_0 = b_0(c,r)&:= 36c^4 +25(4-c^2)^2r^2 +(4-c^2)^2r^4 +12c^2(4-c^2)r^2, \\ b_1 = b_1(c,r)&:= -60c^2(4-c^2)r +10(4-c^2)^2r^3, \\ b_2 = b_2(c,r)&:= -24c^2(4-c^2)r^2, \\ d_0 = d_0(c,r)&:= 16[ 9c^4 +r^2(4-c^2)^2 ], \\ d_1 = d_1(c,r)&:= -96rc^2(4-c^2). \end{aligned} \end{aligned}$$

(4.5)

• (i) Now assume that $\Lambda _1$ is increasing with respect to $t \in [-1,1]$. Then by (4.3) and (4.4), we have

  $$\begin{aligned} \Psi _U(c,\zeta ) \le \Lambda _1(c,r,1) + 2(4-c^2)(1-r^2) =: H_1(c,r), \end{aligned}$$

  where

  $$\begin{aligned} H_1(c,r) = |-6c^2 +5(4-c^2)r +(4-c^2)r^2| - |12c^2-4(4-c^2)r| +2(4-c^2)(1-r^2). \end{aligned}$$

  Let

  $$\begin{aligned} k_1= \sqrt{ \frac{4r}{r+3} } \quad \text {and}\quad k_2 = \sqrt{ \frac{4r(5+r)}{6+5r+r^2} }. \end{aligned}$$

  (4.6)

  We note that $0 < k_1 \le k_2 \le \sqrt{2}$ for $r \in (0,1]$ and consider the following cases.

  • (a) If $c \le k_1$, then

    $$\begin{aligned} \begin{aligned} H_1(c,r)&= 8 +4r -4r^2 +c^2(4-r+r^2) \\&\le \frac{12(2+3r-r^2)}{3+r} \le 12, \quad r\in (0,1]. \end{aligned} \end{aligned}$$

  • (b) If $k_1 \le c \le k_2$, then

    $$\begin{aligned} \begin{aligned} H_1(c,r)&= 8 +36r -4r^2 +c^2(-20-9r+r^2) \\&\le \frac{ 12(2+3r-r^2) }{ 3+r } \le 12, \quad r\in (0,1]. \end{aligned} \end{aligned}$$

  • (c) If $c \ge k_2$, then

    $$\begin{aligned} \begin{aligned} H_1(c,r)&= c^2(-8+r+3r^2) +4(2-r-3r^2) \\&\le \frac{48(1-3r-2r^2)}{ 6+5r+r^2 } \le 8 < 12, \quad r\in (0,1]. \end{aligned} \end{aligned}$$

    Thus by (a), (b) and (c), $H_1(c,r) \le 12$ follows, and $\Psi _U(c,\zeta ) \le 12$ holds.

  • (ii) Next assume that $\Lambda _1$ is decreasing with respect to $t \in [-1,1]$. Then by Proposition 3.4 and the inequality $12c^2+4(4-c^2)r \ge 0$, we obtain

    $$\begin{aligned} \begin{aligned} \Psi _U(c,\zeta )&\le \Lambda _1(c,r,-1) +2(4-c^2)(1-r^2) \\&= |G_1(c,r)| - |12c^2+4(4-c^2)r| +2(4-c^2)(1-r^2) \\&= 4(2+r-3r^2) + c^2(-8-r+3r^2) \\&\le 4(2+r-3r^2) < 12, \end{aligned} \end{aligned}$$

    since $-8-r+3r^2<0$, and $G_1$ is the function defined by (3.5). Thus we obtain $\Psi _U(c,\zeta ) < 12$.

  • (iii) Now we assume that $\Lambda _1$ is neither increasing nor decreasing with respect to $t \in [-1,1]$, and let $\lambda _1(t) = \Lambda _1(\cdot ,\cdot ,t)$. Then there exists $t_1 \in (-1,1)$ such that $\lambda _1'(t_1)=0$, which implies that

    $$\begin{aligned} \Lambda _1(\cdot ,\cdot ,t_1) = \lambda _1(t_1) = \left( \frac{ b_1 +2b_2t_1 -d_1 }{d_1} \right) \sqrt{d_0+d_1t_1}. \end{aligned}$$

    Since $b_2<0$,

    $$\begin{aligned} b_1+2b_2t_1-d_1> b_1+2b_2-d_1 = 2r(4-c^2)[ 20r^2 +c^2(18-24r-5r^2) ] > 0. \end{aligned}$$

    (4.7)

    and so using $d_1<0$ and (4.7), it follows that $\lambda _1(t_1) \le 0$. Hence it follows from (4.3), (4.4), (i) and (ii) that

    $$\begin{aligned} \Psi _U(c,\zeta ) \le \max _{t \in \Omega _1} \Lambda _1(c,r,t) +2(4-c^2)(1-r^2) \le 12, \end{aligned}$$

    where

    $$\begin{aligned} \Omega _1 = \{-1,1\} \cup \{t_1 \in (-1,1): \lambda _1'(t_1)=0 \}. \end{aligned}$$

    I(b) Now assume that $1 \le c \le 2$, then by (4.2) we have

    $$\begin{aligned} \Psi _U(c,\zeta ) \le (4-c^2) \left[ {\tilde{\Psi }}(c,\zeta ) + 2(1-|\zeta |^2) \right] \le 3 \left[ {\tilde{\Psi }}(c,\zeta ) + 2(1-|\zeta |^2) \right] , \end{aligned}$$

    where ${\tilde{\Psi }}(c,\zeta )$ is defined by

    $$\begin{aligned} {\tilde{\Psi }}(c,\zeta ) = \left| \frac{-6c^3}{4-c^2} + 5c\zeta +c\zeta ^2 \right| - \left| \frac{12c^2}{4-c^2} - 4\zeta \right| . \end{aligned}$$

    We now show that

    $$\begin{aligned} {\tilde{\Psi }}(c,\zeta ) + 2(1-|\zeta |^2) \le 4. \end{aligned}$$

    (4.8)

    A computation putting $\zeta =re^{i\theta }$, $r\in (0,1]$ and $\theta \in {\mathbb {R}}$, gives

    $$\begin{aligned} {\tilde{\Psi }}(c,\zeta ) = \frac{c}{4-c^2} \sqrt{b_0+b_1t+b_2t^2} - \frac{1}{4-c^2} \sqrt{d_0+d_1t}, \end{aligned}$$

    where $t=\cos \theta \in [-1,1]$ and $b_i$, $i\in \{0,1,2\}$, and $d_j$, $j\in \{0,1\}$, are given by (4.5). So the inequality (4.8) is equivalent to

    $$\begin{aligned} c^2(b_0+b_1t+b_2t^2) - (d_0+d_1t) -4(1+r^2)^2(4-c^2)^2 \le 4(1+r^2)(4-c^2) \sqrt{ d_0 +d_1 t}. \end{aligned}$$

    (4.9)

    Next write

    $$\begin{aligned} c^2(b_0+b_1t+b_2t^2) - (d_0+d_1t) -4(1+r^2)^2(4-c^2)^2 = -(4-c^2)F_c(r,t), \end{aligned}$$

    where $F_c$ is defined by (3.17). Since $c\in [1,2]$, Proposition 3.3, shows that $F_c(r,t) \ge 0$, which gives

    $$\begin{aligned} c^2(b_0+b_1t+b_2t^2) - (d_0+d_1t) -4(1+r^2)^2(4-c^2)^2 \le 0. \end{aligned}$$

    Therefore since $4(1+r^2)(4-c^2) \sqrt{ d_0 +d_1 t} \ge 0$, (4.9) is true. Thus (4.8) follows, and $\Psi _U(c,\zeta ) \le 12$ holds.

Thus the proof of the upper bound in (4.1) is complete.

II. We next prove the lower bound in (4.1). Since ${{\mathcal {P}}}$ and $|A_3|-|A_4|$ are rotationally invariant, using Lemma 2.1 we may assume that $c_1=c$, with $0 \le c \le 2$. Moreover by Theorem 3.2, $|A_3|-|A_4| \le |A_3+A_4| \le 1/3$ hold when $0 \le c \le 4/3$ or $\sqrt{2} \le c \le 2$. Hence it is enough to consider c satisfying $4/3 \le c \le \sqrt{2}$.

Using Lemma 2.2 and the fact that $|\eta | \le 1$, we obtain

$$\begin{aligned} \begin{aligned}&48(|A_3|-|A_4|) \\&\le |12c^2-4(4-c^2)\zeta | - | -6c^3 +5c(4-c^2)\zeta +c(4-c^2)\zeta ^2 | + 2(4-c^2)(1-|\zeta |^2) \\&=: \Psi _L(c,\zeta ), \end{aligned} \end{aligned}$$

where $\zeta \in \overline{{\mathbb {D}}}$, and so we must show that

$$\begin{aligned} \Psi _L(c,\zeta ) \le 16 \end{aligned}$$

(4.10)

holds for all $c\in [4/3,\sqrt{2}]$ and $\zeta \in \overline{{\mathbb {D}}}$.

Note that we can assume that $\zeta \not =0$, since

$$\begin{aligned} \Psi _L(c,0) = 8 +2c^2(5-3c) < 16. \end{aligned}$$

Also since $c \ge 4/3$, we have

$$\begin{aligned} \begin{aligned}&|12c^2-4(4-c^2)\zeta | - | -6c^3 +5c(4-c^2)\zeta +c(4-c^2)\zeta ^2 | \\&< |12c^2-4(4-c^2)\zeta | - \frac{4}{3} | -6c^2 +5(4-c^2)\zeta +(4-c^2)\zeta ^2 | \\&= 4(4-c^2) [ |A^*-\zeta | - \frac{1}{3} | -2A +5\zeta +\zeta ^2 | ], \end{aligned} \end{aligned}$$

(4.11)

where $A^* = 3c^2/(4-c^2)$. Moreover putting $\zeta =re^{i\theta }$, $r\in (0,1]$, $\theta \in {\mathbb {R}}$, and $t=\cos \theta \in [-1,1]$, we have

$$\begin{aligned} |A^*-\zeta | - \frac{1}{3} | -2A^* +5\zeta +\zeta ^2 | = \sqrt{d_0+d_1t} - \sqrt{ (b_0+b_1t+b_2t^2) / 9 } =: \Lambda _{2}(c,r,t), \end{aligned}$$

(4.12)

where

$$\begin{aligned} d_0=A^2+r^2, \quad d_1=-2Ar, \\ b_0 = 25r^2+(2A^*+r^2)^2, \quad b_1=10r(-2A^*+r^2),\quad b_2=-8A^*r^2. \end{aligned}$$

In order to apply Lemma 2.4, we put

$$\begin{aligned} \Delta _1 = d_1^2, \quad \Delta _2 = \frac{1}{81}(b_1^2-4b_0b_2) \quad \text {and}\quad T=\frac{5}{8}c^2. \end{aligned}$$

We now show that

$$\begin{aligned} \Delta _2 \ge T^{3/2}\Delta _1, \end{aligned}$$

(4.13)

and

$$\begin{aligned} T(d_0+d_1t) \ge \frac{1}{9}(b_0+b_1t+b_2t^2), \quad t \in [-1,1]. \end{aligned}$$

(4.14)

First let

$$\begin{aligned} G(c,r) = \frac{ \Delta _2 }{ \Delta _1 } = \frac{ (100-c^2)[ -4r^2+c^2(-6+r^2) ]^2 }{ 729c^4(4-c^2) }. \end{aligned}$$

(4.15)

Then it is easy to see that

$$\begin{aligned} G(c,r) \ge G(c,0) = \frac{4(100-c^2)}{81(4-c^2)} \ge G \left( \frac{4}{3},0 \right) = \frac{884}{405} > 2. \end{aligned}$$

(4.16)

Since $c \le \sqrt{2}$, we obtain

$$\begin{aligned} T^{3/2} = \left( \frac{5}{8}c^2 \right) ^{3/2} \le \left( \frac{5}{4} \right) ^{3/2} = \frac{5\sqrt{5}}{8} < 2. \end{aligned}$$

(4.17)

Thus by (4.15), (4.16) and (4.17), we obtain (4.13).

Next we will show that (4.14) holds. To do this, we let

$$\begin{aligned} -\frac{1}{9} b_2 t^2 +\left( - \frac{1}{9}b_1 +T d_1\right) t + T d_0 - \frac{1}{9}b_0 = \frac{1}{ 72(4-c^2)^2 } G_2(t), \end{aligned}$$

where

$$G_2(t)= 45c^6(9+r^6rt) -128r^2(25+r^2+10rt) -\, 8c^4[36+r^4+195rt+10r^3t +r^2(58+24t^2)] + 16c^2r[4r^3+120t+40r^2t +r(121+48t^2)]. $$

We consider the following five cases.

• (a) Write

  $$\begin{aligned} G_2(-1) = B_1 +B_2 r +B_3r^2 +B_4r^3 +B_5r^4, \end{aligned}$$

  where

  $$\begin{aligned} B_1= & {} -288c^4+405c^6, \quad B_2 = -1920c^2 +1560c^4 -270c^6, \\ B_3= & {} -3200 +2704c^2 -656c^4 +45c^6, \quad B_4 = 1280 -640c^2 +80c^4, \\ B_5= & {} -128 +64c^2 -8c^4. \end{aligned}$$

  Then it is easy to check that

  $$\begin{aligned} B_2>0, \quad B_4>0, \quad B_3<0, \quad B_5<0. \end{aligned}$$

  Therefore we obtain

  $$\begin{aligned} \begin{aligned} B_1 +B_2 r +B_3r^2 +B_4r^3 +B_5r^4&\ge B_1 +B_3 +B_5 \\&= -3328 +2768c^2 -952c^4 +450c^6 \\&>0. \end{aligned} \end{aligned}$$

  Hence $G_2(-1) > 0$.

• (b) We next show that $G_2(1)>0$.

  First note that

  $$\begin{aligned} G_2(1) = B_6 + B_7s +B_8s^2 +B_9s^3 =: H_1(s), \end{aligned}$$

  where

  $$\begin{aligned} B_6=\, & {} -128r^2(5+r^2), \\ B_7=\, & {} 16r(120+169r+40r^2+4r^3), \\ B_8=\, & {} - 8(36+195r+82r^2+10r^3+r^4), \\ B_9 =\,&{} 45(3+r)^2, \end{aligned}$$

  and $s=c^2 \in [16/9,2]$, and since $B_8<0$ and $B_9>0$, we have

  $$\begin{aligned} \begin{aligned} H_1'(s)&= B_7 +2B_8s +3B_9s^2 \\&\ge B_7 +4B_8 + \frac{256}{27}B_9 \\&= \frac{16}{3} (504 - 330r +95r^2 +60r^3 +6r^4) \\&>0. \end{aligned} \end{aligned}$$

  Thus $H_1$ is increasing with respect to $s \in [16/9,2]$, and so

  $$\begin{aligned} \begin{aligned} H_1(s)&\ge H_1\left( \frac{16}{9} \right) \\&= \frac{128}{81} (864 -135r^2 -250r^3 -25r^4) >0. \end{aligned} \end{aligned}$$

  Thus $G_2(1)>0$ as claimed.

• (c) We note that $G_2$ has its (unique) critical value at $t=t_0$, where

  $$\begin{aligned} t_0 = \frac{5(-48c^2 +27c^4 +32r^2 -8c^2r^2)}{ 192c^2r }. \end{aligned}$$

  We now show that $G_2(t_0)>0$, in which case $G_2(t)>0$ for all $t\in [-1,1]$, and (4.14) follows. Write

  $$\begin{aligned} G_2(t_0) = \frac{1}{192c^2} G_3(c^2,r^2), \end{aligned}$$

  (4.18)

  where

  $$\begin{aligned} G_3(x,y)\,=\, & {} B_{10} + B_{11}y + B_{12}y^2, \\ B_{10}\,=\, & {} B_{10}(x)\, =\, 9x^2(-25600 +29056x -6660x^2 +2025x^3), \\ B_{11}\,= \,& {} B_{11}(x) =\, -48x (6400 - 944 x - 344 x^2 + 45 x^3), \\ B_{12}\,=\, & {} B_{12}(x) = -64 (100 - x) (4 - x)^2. \end{aligned}$$

  It is easy to see that $B_{11}<0$ and $B_{12}<0$, so

  $$\begin{aligned} \begin{aligned} G_3(x,y)&\ge G_3(x,1) = B_{10} +B_{11} +B_{12} \\&= -102400 - 254976 x - 192000 x^2 + 278080 x^3 - 62100 x^4 + 18225 x^5 > 0. \end{aligned} \end{aligned}$$

  holds for $x\in [16/9,2]$ and $y\in [0,1]$. Thus we obtain $G_2(t_0)>0$, by (4.18).

  Hence by Lemma 2.4, the function $\Lambda _2$ is convex on $[-1,1]$, and the convexity of $\Lambda _2$ (on $[-1,1]$), (4.11) and (4.12) implies that

  $$\begin{aligned} \Psi _L(c,\zeta ) \le \max \{ \Psi _L(c,r) ; \Psi _L(c,-r) \}, \end{aligned}$$

  (4.19)

  where $c\in [4/3,\sqrt{2}]$ and $r\in (0,1]$. Thus we will obtain (4.10) if we show that $\Psi _L(c,r) \le 4$ and $\Psi _L(c,-r) \le 4$ for $c \in [4/3,\sqrt{2}]$ and $r\in (0,1]$, where

  $$\begin{aligned} \Psi _L(c,-r) = \left| 3c^2+r(4-c^2) \right| - \frac{1}{3} \left| -6c^2 -5r(4-c^2) +r^2(4-c^2) \right| + \frac{1}{2}(4-c^2)(1-r^2) \end{aligned}$$

  and

  $$\begin{aligned} \Psi _L(c,r) = \left| 3c^2-r(4-c^2) \right| - \frac{1}{3} \left| -6c^2 +5r(4-c^2) +r^2(4-c^2) \right| + \frac{1}{2}(4-c^2)(1-r^2). \end{aligned}$$

• (d) Clearly $3c^2+r(4-c^2) \ge 0$ holds. Also

  $$\begin{aligned} -6c^2 -5r(4-c^2) +r^2(4-c^2) = -(6-5r+r^2)c^2 -20r+4r^2 \le -20r+4r^2 <0. \end{aligned}$$

  Hence

  $$\begin{aligned} \begin{aligned} \Psi _L(c,-r)&= 3c^2+r(4-c^2) + \frac{1}{3} [ -6c^2 -5r(4-c^2) +r^2(4-c^2) ] + \frac{1}{2}(4-c^2)(1-r^2) \\&= \frac{1}{6}[ 4(3-4r-r^2) +c^2(3+4r+r^2) ] \\&\le \frac{1}{3}(9-4r-r^2) < 4, \end{aligned} \end{aligned}$$

  which gives $\Psi _L(c,-r) \le 4$.

• (e) Next clearly $3c^2 -r(4-c^2)> 4(c^2-1) >0$. If $-6c^2 +5r(4-c^2) +r^2(4-c^2) \le 0$, then, since $c \le \sqrt{2}$, we have

  $$\begin{aligned} \begin{aligned} \Psi _L(c,r)&= \frac{1}{6} [ 4(3+4r-r^2) +c^2(3-4r+r^2) ] \\&\le \frac{1}{3}(9+4r-r^2) \le 4. \end{aligned} \end{aligned}$$

  If $-6c^2 +5r(4-c^2) +r^2(4-c^2) \ge 0$, then

  $$\begin{aligned} c^2 \le \frac{2(5r+r^2)}{6+5r+r^2}. \end{aligned}$$

  Using this inequality, we see that

  $$\begin{aligned} \begin{aligned} \Psi _L(c,r)&= \frac{1}{6} [ 4(3-16r-5r^2) +c^2(27+16r+5r^2) ] \\&\le \frac{12(1+3r)}{6+5r+r^2} \le 4. \end{aligned} \end{aligned}$$

  Hence we obtain the desired inequality $\Psi _L(c,r) \le 4$,and by (4.19), the inequality (4.10) is established. Thus the proof of the lower bound in (4.1) is complete.

We end the proof of Theorem 4.1 by giving extreme functions for the inequalities in (4.1).

First consider $f_1 \in {{\mathcal {K}}}$ defined by $1+zf_1''(z)/f_1'(z) = (1+z^2)/(1-z^2)$. Comparing coefficients, we obtain $a_2=0$, $a_3=1/3$, $a_4=0$, and so from (1.2) we have $A_4=0$, $A_3=-1/3$, and $|A_4|-|A_3| = -1/3$, so that the lower bound in (4.1) is sharp for $f_1$.

Next consider $f_2 \in {{\mathcal {K}}}$ defined by $1+zf_2''(z)/f_2'(z) = (1+z+z^2)/(1-z^2)$. Then the coefficients of $f_2$ are given by $a_2=1/2$, $a_3=1/2$, $a_4=3/8$, and so from (1.2) we have $A_4=1/4$, $A_3=0$ and $|A_4|-|A_3| = 1/4$. Thus the upper bound in (4.1) is sharp for $f_2$. $\square$