The fourth-order Hermitian Toeplitz determinant for convex functions

Abstract

The sharp bounds for the fourth-order Hermitian Toeplitz determinant over the class of convex functions are computed.

1 Introduction

Let $\mathcal H$ be the class of analytic functions in $\mathbb D:=\{z\in \mathbb C: |z|<1\}$, $\mathcal A$ be its subclass normalized by $f(0):=0,f'(0):=1,$ that is, functions of the form

$$\begin{aligned} f(z)=\sum _{n=1}^\infty a_nz^n,\quad a_1:=1,\ z\in \mathbb D, \end{aligned}$$

(1)

and $\mathcal S$ be the subclass of $\mathcal A$ of univalent functions. Let $\mathcal S^c$ denote the subclass of $\mathcal S$ of convex functions, that is, univalent functions $f\in \mathcal A$ such that $f(\mathbb D)$ is a convex domain in ${\mathbb {C}}.$ By the well-known result of Study [11] (see also [5, p. 42]), a function f is in $\mathcal S^c$ if and only if

$$\begin{aligned} {{\,\mathrm{Re}\,}}\left\{ 1+\dfrac{zf^{''}(z)}{f'(z)}\right\} >0,\quad z\in \mathbb D. \end{aligned}$$

(2)

Given $q,n\in {\mathbb {N}},$ the Hermitian Toeplitz matrix $T_{q,n}(f)$ of $f\in \mathcal A$ of the form (1) is defined by

$$\begin{aligned} T_{q,n}(f):=\left[ \begin{matrix} a_n&{}\quad a_{n+1}&{}\quad \ldots &{}\quad a_{n+q-1}\\ {\overline{a}}_{n+1}&{}\quad a_n&{}\quad \ldots &{}\quad a_{n+q-2}\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ {\overline{a}}_{n+q-1}&{}\quad {\overline{a}}_{n+q-2}&{}\quad \ldots &{}\quad a_n\end{matrix}\right] , \end{aligned}$$

where ${\overline{a}}_k:=\overline{a_k}.$ Let $|T_{q,n}(f)|$ denote the determinant of $T_{q,n}(f).$ In particular, the third Toeplitz determinant $|T_{3,1}(f)|$ is given by

$$\begin{aligned} |T_{3,1}(f)|=\begin{vmatrix} 1&\quad a_2&\quad a_3\\ {\overline{a}}_2&\quad 1&\quad a_2\\ {\overline{a}}_3&\quad {\overline{a}}_2&\quad 1\end{vmatrix}=1+2{{\,\mathrm{Re}\,}}\left( a_2^2 {\overline{a}}_3\right) -2 |a_2|^2-|a_3|^2 \end{aligned}$$

(3)

and the fourth Toeplitz determinant $|T_{4,1}(f)|$ is given by

$$\begin{aligned} \begin{aligned} |T_{4,1}(f)|=&\begin{vmatrix} a_1&\quad a_2&\quad a_3&\quad a_4\\ {\overline{a}}_2&\quad a_1&\quad a_2&\quad a_3\\ {\overline{a}}_3&\quad {\overline{a}}_2&\quad a_1&\quad a_2\\ {\overline{a}}_4&\quad {\overline{a}}_3&\quad {\overline{a}}_2&\quad a_1\end{vmatrix}\\ =&1 -2{{\,\mathrm{Re}\,}}\left( a_2^3 {\overline{a}}_4\right) + 4{{\,\mathrm{Re}\,}}\left( a_2^2 {\overline{a}}_3\right) - 2{{\,\mathrm{Re}\,}}\left( a_2{\overline{a}}_3^2 a_4\right) \\&+4{{\,\mathrm{Re}\,}}\left( a_2 a_3 {\overline{a}}_4\right) + |a_2|^4- 3|a_2|^2+ |a_3|^4- 2|a_3|^2 \\&+ |a_2|^2 |a_4|^2 - 2 |a_2|^2 |a_3|^2 - |a_4|^2. \end{aligned} \end{aligned}$$

(4)

In recent years a lot of papers has been devoted to the estimation of determinants built with using coefficients of functions in the class $\mathcal A$ or its subclasses. Hankel matrices i.e., square matrices which have constant entries along the reverse diagonal (see e.g., [3] with further references), and the symmetric Toeplitz determinant (see [1]) are of particular interest.

For this reason looking on the interest of specialists in [4] and [7] the study of the Hermitian Toeplitz determinants on the class $\mathcal A$ or its subclasses has begun. Hermitian Toeplitz matrices play an important role in functional analysis, applied mathematics as well as in physics and technical sciences.

In [4] the conjecture that the sharp inequalities $0\le |T_{q,1}(f)|\le 1$ for all $q\ge 2,$ holds over the class $\mathcal S^c$ was proposed and was confirmed for $q=2$ and $q=3.$ The purpose of this paper is to prove this conjecture for $q=4.$

Let $\mathcal P$ be the class of all $p\in \mathcal H$ of the form

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

(5)

having a positive real part in $\mathbb D.$

The key to the proof of the main result is the following lemma. It contains the well-known formula for $c_2$ (see e.g., [10, p. 166]) and the formula for $c_3$ due to Libera and Zlotkiewicz [8, 9].

Lemma 1

If $p\in \mathcal P$ is of the form (5) with $c_1\ge 0,$ then

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

(6)

and

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

(7)

for some $\zeta ,\eta \in {\overline{\mathbb D}}:=\{z\in {\mathbb {C}}:|z|\le 1\}$

2 Main result

In [4] the sharp bounds for the Hermitian-Toeplitz determinants of the second and third-order for the class of convex functions of order $\alpha$ were computed. In particular for $\alpha =0,$ the results obtained are reduced for the class of convex functions, namely, $0\le |T_{2,1}(f)|\le 1$ and $0\le |T_{3,1}(f)|\le 1$ when $f\in \mathcal S^c.$ Both results suggested the conjecture that $0\le |T_{q,1}(f)|\le 1$ for every $q\ge 2.$ In this paper, this conjecture was confirmed for $q = 4.$ For the sake of consistency, we also provide a short proof of the case $q = 3.$

Theorem 1

If $f\in \mathcal S^c,$ then

$$\begin{aligned} 0\le |T_{3,1}(f)|\le 1. \end{aligned}$$

(8)

Both inequalities are sharp with equalities attained by

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

(9)

and by the identity, respectively.

Proof

Let $f\in \mathcal S^c$ be the form (1). Then by (2),

$$\begin{aligned} f'(z)+zf''(z)=p(z)f'(z), \quad z\in \mathbb D, \end{aligned}$$

(10)

for some $p\in \mathcal P$ of the form (5). Substituting the series (1) and (5) into (10) by equating the coefficients we get

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

(11)

Noting that the class $\mathcal S^c$ and $|T_{3,1}(f)|$ are rotationally invariant, we may assume that $c:=c_1$ with $c\in [0,2]$ ([2], see also [6, Vol. I, page 80, Theorem 3]), i.e., by (11), $a_2\in [0,1].$ Then by (3) and (11) we have

$$\begin{aligned} \begin{aligned} |T_{3,1}(f)|&=1+2a_2^2{{\,\mathrm{Re}\,}}a_3-2a_2^2-|a_3|^2\\&=\dfrac{1}{72}\left( 72-36c^2+4c^4+2c^2{{\,\mathrm{Re}\,}}c_2-2|c_2|^2\right) . \end{aligned} \end{aligned}$$

Now by using (6) and (7) we get

$$\begin{aligned} 0\le |T_{3,1}(f)|=\dfrac{1}{144}(4-c^2)^2(9-|\zeta |^2)\le 1, \end{aligned}$$

which shows (8).

It is clear that the function (9) and the identity make the results sharp. $\square$

We will now estimate the fourth-order Toeplitz determinant $|T_{4,1}(f)|$ for $f\in \mathcal S^c.$

Theorem 2

If $f\in \mathcal S^c,$ then

$$\begin{aligned} 0\le |T_{4,1}(f)|\le 1. \end{aligned}$$

(12)

Both inequalities are sharp with equalities attained by the function (9) and the identity, respectively.

Proof

Assuming as in the proof of Theorem 1 that $c:=c_1$ with $c\in [0,2],$ so $a_2\in [0,1],$ from (4) and (11) we obtain

$$\begin{aligned} \begin{aligned} |T_{4,1}(f)| =&1+ a_2^4 - 3 a_2^2+ 4 a_2^2{{\,\mathrm{Re}\,}}a_3- 2 a_2^2 |a_3|^2- 2a_2^3{{\,\mathrm{Re}\,}}a_4 + a_2^2 | a_4|^2\\&+ 4 a_2{{\,\mathrm{Re}\,}}(a_3 {\overline{a}}_4) - 2a_2{{\,\mathrm{Re}\,}}(a_3^2 {\overline{a}}_4) - 2 |a_3|^2+ |a_3|^4 - |a_4|^2\\ =&\frac{1}{20736}\left( 20736- 15552 c^2+ 3600 c^4- 252 c^6+c^8+ 1152 c^2{{\,\mathrm{Re}\,}}c_2\right. \\&\left. - 288 c^4{{\,\mathrm{Re}\,}}c_2 + 64 c^4({{\,\mathrm{Re}\,}}c_2)^2 - 2c^6{{\,\mathrm{Re}\,}}c_2 + 252 c^2 |c_2|^2\right. \\&\left. -31c^4 |c_2|^2+ 16 |c_2|^4 - 1152|c_2|^2-24 c^4{{\,\mathrm{Re}\,}}c_2^2\right. \\&\left. - 8 c^2|c_2|^2{{\,\mathrm{Re}\,}}c_2 - 12 c^5{{\,\mathrm{Re}\,}}c_3 + 144 c{{\,\mathrm{Re}\,}}(c_2 {\overline{c}}_3) \right. \\&\left. + 12 c^3{{\,\mathrm{Re}\,}}( c_2{\overline{c}}_3)- 48 c{{\,\mathrm{Re}\,}}( c_2^2{\overline{c}}_3) - 144|c_3|^2+36 c^2 |c_3|^2 \right) . \end{aligned} \end{aligned}$$

(13)

Hence by using (6) and (7) after tedious computation we get

$$\begin{aligned} \begin{aligned} |T_{4,1}(f)|&=\frac{1}{82944}(4-c^2)^3\left[ 1296-9(c^2+32)|\zeta |^2\right. \\&\quad +\,6c^2|\zeta |^2{{\,\mathrm{Re}\,}}\zeta +(16-c^2)|\zeta |^4-36c(1-|\zeta |^2){{\,\mathrm{Re}\,}}(\zeta {\overline{\eta }})\\&\quad \left. +12c(1-|\zeta |^2){{\,\mathrm{Re}\,}}(\zeta ^2{\overline{\eta }})-36(1-|\zeta |^2)^2|\eta |^2\right] . \end{aligned} \end{aligned}$$

(14)

A. Suppose that $\zeta =0.$ Then

$$\begin{aligned}0\le |T_{4,1}(f)|=\frac{1}{64}(4-c^2)^3\le 1. \end{aligned}$$

Suppose that $\eta =0.$ Then

$$\begin{aligned} \begin{aligned} |T_{4,1}(f)|=&\frac{1}{82944}(4-c^2)^3\left[ 1296-9(c^2+32)|\zeta |^2\right. \\&\left. +6c^2|\zeta |^2{{\,\mathrm{Re}\,}}\zeta +(16-c^2)|\zeta |^4\right] . \end{aligned} \end{aligned}$$

(15)

We have

$$\begin{aligned} \begin{aligned}&1296-9(c^2+32)|\zeta |^2+6c^2|\zeta |^2{{\,\mathrm{Re}\,}}\zeta +(16-c^2)|\zeta |^4\\&\ge 1296-9(c^2+32)|\zeta |^2-6c^2|\zeta |^3+(16-c^2)|\zeta |^4\\&=1296-288|\zeta |^2+16|\zeta |^4-|\zeta |^2(3+|\zeta |)^2c^2\\&\ge 1296-288|\zeta |^2+16|\zeta |^4-4|\zeta |^2(3+|\zeta |)^2\\&=1296-324|\zeta |^2-24|\zeta |^3+12|\zeta |^4\ge 960,\quad 0\le |\zeta |\le 1. \end{aligned} \end{aligned}$$

Hence and from (15) it follows that $|T_{4,1}(f)|\ge 0.$

We have

$$\begin{aligned} \begin{aligned}&1296-9(c^2+32)|\zeta |^2+6c^2|\zeta |^2{{\,\mathrm{Re}\,}}\zeta +(16-c^2)|\zeta |^4\\&\le 1296-9(c^2+32)|\zeta |^2+6c^2|\zeta |^3+(16-c^2)|\zeta |^4\\&=1296-288|\zeta |^2+16|\zeta |^4-|\zeta |^2(3-|\zeta |)^2c^2\\&\le 1296-288|\zeta |^2+16|\zeta |^4\le 1296,\quad 0\le |\zeta |\le 1. \end{aligned} \end{aligned}$$

Hence and from (15) it follows that $|T_{4,1}(f)|\le 1.$

B. Let now $\zeta ,\eta \in {\overline{\mathbb D}}\setminus \{0\}.$ By setting $\zeta :=x\mathrm {e}^{\mathrm {i}\theta },\ \eta :=y\mathrm {e}^{\mathrm {i}\psi },\ x,y\in (0,1],\ \theta ,\psi \in [0,2\pi ),$ from (14) we get

$$\begin{aligned} |T_{4,1}(f)|=\frac{1}{82944}(4-c^2)^3 F(c,x,y,\theta ,\psi ), \end{aligned}$$

(16)

where

$$\begin{aligned} \begin{aligned} F(c,x,y,\theta ,\psi ):=&1296-9(c^2+32)x^2+6c^2x^3\cos \theta +(16-c^2)x^4\\&-36c(1-x^2)xy\cos (\theta -\psi )+12c(1-x^2)x^2y\cos (2\theta -\psi )\\&-36(1-x^2)^2y^2. \end{aligned} \end{aligned}$$

For $c\in [0,2]$ and $x,y\in (0,1]$ we have

$$\begin{aligned} G(c,x,y)\le F(c,x,y,\theta ,\psi )\le H(c,x,y), \end{aligned}$$

(17)

where for $x,y\in [0,1],$

$$\begin{aligned} \begin{aligned}&G(c,x,y):=F(c,x,y,\pi ,\pi )\\&=1296-9(c^2+32)x^2-6c^2x^3+(16-c^2)x^4\\&\quad -12c(1-x^2)(3+x)xy-36(1-x^2)^2y^2 \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} H(c,x,y):=&1296-9(c^2+32)x^2+6c^2x^3+(16-c^2)x^4\\&+12c(1-x^2)(3+x)xy-36(1-x^2)^2y^2. \end{aligned} \end{aligned}$$

C. Let us observe that

$$\begin{aligned} \begin{aligned} G(c,x,y)=&1296-9(c^2+32)x^2-6c^2x^3+(16-c^2)x^4\\&-12c(1-x^2)(3+x)xy-36(1-x^2)^2y^2\\ \ge&1296-324x^2-24x^3-24(1-x^2)(3+x)x-36(1-x^2)^2\\ \ge&1296-480=816,\quad c\in [0,2],\ x,y\in (0,1]. \end{aligned} \end{aligned}$$

Hence and from (16) with (17) it follows that

$$\begin{aligned} \begin{aligned} |T_{4,1}(f)|&=\frac{1}{82944}(4-c^2)^3 F(c,x,y,\theta ,\psi )\\&\ge \frac{1}{82944}(4-c^2)^3G(c,x,y)\ge \frac{17}{1728}(4-c^2)^3\ge 0,\quad c\in [0,2], \end{aligned} \end{aligned}$$

which together with part A shows the lower bound in (12).

D. Now will discuss the upper bound of $|T_{4,1}(f)|.$

Let first $x=1.$ Then for $c\in [0,2]$ and $y\in (0,1],$

$$\begin{aligned} H(c,1,1)=H(c,1,y)=1024-4c^2\le 1024,\quad c\in [0,2]. \end{aligned}$$

Let now $x\in (0,1).$ Then

$$\begin{aligned}y_w:=\frac{cx(x+3)}{6(1-x^2)}\ge 0,\quad -36(1-x^2)^2<0. \end{aligned}$$

Therefore we consider two cases.

D1. Assume that $y_w< 1$, i.e., equivalently that $x\in (0,x_1(c)),$ where

$$\begin{aligned} x_1(c):=\frac{-3c+\sqrt{9c^2+24c+144}}{2(c+6)},\quad c\in [0,2]. \end{aligned}$$

Note that $x_1(c)\le 1$ for all $c\in [0,2]$ and $x_1(0)=1.$ Let $\varDelta _1:=\{(c,x): 0\le c\le 2,\ 0\le x\le x_1(c)\}.$ We have

$$\begin{aligned}H(c,x,y)\le H(c,x,y_w)=h(c,x),\quad (c,x)\in \varDelta _1,\ y\in (0,1], \end{aligned}$$

where

$$\begin{aligned}h(c,x):=1296-288x^2+12c^2x^3+16x^4, \quad (c,x)\in \varDelta _1. \end{aligned}$$

• (i) On the vertices of $\varDelta _1,$

  $$\begin{aligned} \begin{aligned}&h(0,0)=1296, \quad h(0,x_1(0))= h(0,1)=1296,\quad h(2,0)=1296,\\&h(2,x_1(2))=h\left( 2,\frac{1}{8}\left( \sqrt{57}-3\right) \right) =\frac{9}{32}\left( 3461+113\sqrt{57}\right) \approx 1213.349. \end{aligned} \end{aligned}$$

• (ii) On the side $x = 0,$

  $$\begin{aligned}h(c,0)=1296, \quad c\in (0,2).\end{aligned}$$

• (iii) On the side $x = x_1(c)$ for $c\in (0,2),$

$$\begin{aligned} \begin{aligned}&h(c,x_1(c))\\ =&\frac{18}{(c+6)^4}\left( -9c^6-72c^5-180c^4+216c^3+11840c^2+52224c+73728\right. \\&\left. +(3c^5+20c^4+36c^3+344c^2+768c)\sqrt{9c^2+24c+144}\right) =:\gamma (c). \end{aligned} \end{aligned}$$

We will show that $\gamma$ increasis. Note that

$$\begin{aligned} \begin{aligned} \gamma '(c)=&\frac{-36\left( \varrho _1(c)+\varrho _2(c)\sqrt{9c^2+24c+144}\right) }{(c+6)^5\sqrt{9c^2+24c+144}}>0, \quad c\in (0,2), \end{aligned} \end{aligned}$$

(18)

is equivalent to

$$\begin{aligned} \varrho _1(c)+\varrho _2(c)\sqrt{9c^2+24c+144}<0, \quad c\in (0,2), \end{aligned}$$

(19)

where for $t\in {\mathbb {R}},$

$$\begin{aligned}\varrho _1(t):=-27t^7-630t^6-4224t^5-15084t^4-55800t^3-77472t^2-214272t-331776\end{aligned}$$

and

$$\begin{aligned}\varrho _2(t):=9t^6+198t^5+1080t^4+2268t^3+9896t^2+7296t-9216.\end{aligned}$$

Since $\varrho _1'$ has only two real zeros at $c\approx -12.8436$ and $c\approx -5.43834,$ and $\varrho _1'(0)<0,$ so $\varrho _1$ decreases on (0, 2). This with $\varrho _1(0)<0$ yields

$$\begin{aligned} \varrho _1(c)<0,\quad c\in (0,2). \end{aligned}$$

(20)

Since $\varrho _2'>0$ for $c\in [0,2],$ so $\varrho _2$ increases from $\varrho _2(0)=-9216$ to $\varrho _2(2)=87296.$ Thus $\varrho _2(c)\le 0$ for $c\in (0,c']$ and $\varrho _2(c)> 0$ for $c\in (c',2),$ where $c'\approx 0.627225$ is the unique zero of $\gamma _2$ in (0, 2).

Hence and from (20) it follows that the inequality (19) occurs for $c\in (0,c'].$

Note now that for $c\in (c',2)$ we have

$$\begin{aligned} \varrho _1^2(c)>(9c^2+24c+144)\varrho _2^2(c). \end{aligned}$$

(21)

Indeed, the above inequality is equivalent to

$$\begin{aligned} 48(c+6)^5\varrho _3(c)<0,\quad c\in (c',2), \end{aligned}$$

(22)

where

$$\begin{aligned}\varrho _3(t):=9t^6-30t^5-4248t^4-31392t^3-70272t^2-208896t-262144,\quad t\in \mathbb R.\end{aligned}$$

As easy to see $\varrho _3'(c)<0$ for $c\in (0.6,2),$ so $\varrho _3$ decreases on (0.6, 2),  and therefore on $(c',2).$ This and $\varrho _3(0.6)<0$ yields

$$\begin{aligned}\varrho _3(c)<0,\quad c\in (c',2),\end{aligned}$$

which confirms (21), (22), and ends the proof of (19) so of (18).

Summarizing,

$$\begin{aligned}h(c,x_1)=\gamma (c)\le \gamma (2)<1296,\quad c\in (0,2).\end{aligned}$$

• (iv) When $c=0,$ then $x_1(0)=1$ and

  $$\begin{aligned}h(0,x)=1296-288x^2+16x^4\le 1296,\quad x\in [0,1).\end{aligned}$$

• (v) When $c = 2,$ then

  $$\begin{aligned}h(2,x)=1296-288x^2+48x^3+16x^4\le 1296,\quad x\in (0,x_1(2)).\end{aligned}$$

  Indeed, as easy to see the function $(0,x_1(2))\ni x\mapsto H_2(2,x)$ decreases.

• (vi) It remains to consider the interior of $\varDelta _1.$ Since the system of equations

$$\begin{aligned} {\left\{ \begin{array}{ll} 24cx^3=0\\ -576x+36c^2x^2+64x^3=0 \end{array}\right. } \end{aligned}$$

has solutions $(c_i,x_i), i=1,2,3,$ only when

$$\begin{aligned} {\left\{ \begin{array}{ll} c_1=0\\ x_1=0 \end{array}\right. }\quad {\left\{ \begin{array}{ll} c_2=0\\ x_2=-3 \end{array}\right. }\quad {\left\{ \begin{array}{ll} c_3=0\\ x_3=3 \end{array}\right. } \end{aligned}$$

so h has no critical points in the interior of $\varDelta _1.$

D2. Assume that $y_w\ge 1$, i.e., equivalently that $x\in [x_1(c),1].$ Let $\varDelta _2:=\{(c,x): 0\le c\le 2,\ x_1(c)\le x\le 1\}.$ Then

$$\begin{aligned}H(c,x,y)\le H(c,x,1)=g(c,x),\quad (c,x)\in \varDelta _2,\ y\in (0,1],\end{aligned}$$

where for $(c,x)\in \varDelta _2,$

$$\begin{aligned}g(c,x):=1260+36cx-3(3c^2-4c+72)x^2+6(c^2-6c)x^3-(c^2+12c+20)x^4.\end{aligned}$$

• (i) On the vertices of $\varDelta _2,$

  $$\begin{aligned} \begin{aligned}&g(0,x_1(0))=g(0,1)=1024,\quad g(2,1)=1008,\\&g(2,x_1(2))=g\left( 2,(\sqrt{57}-3)/8\right) =\frac{9}{32}\left( 3461+113\sqrt{57}\right) \approx 1213.349. \end{aligned} \end{aligned}$$

• (ii) On the side $x = x_1(c)$ we have the case B1(iii).

• (iii) On the side $x = 1,$

  $$\begin{aligned}g(c,1)=-4c^2+1024\le 1024,\quad c\in (0,2).\end{aligned}$$

• (iv) On the side $c = 2,$

  $$\begin{aligned}g(2,x)=-48x^4-48x^3-228x^2+72x+1260, \quad x\in \left[ (\sqrt{57}-3)/8,1\right) .\end{aligned}$$

  Since $-192x^3-144x^2-456x+72=0$ if and only if $x\approx 0.14944,$ so the function

  $$\begin{aligned}\left[ (\sqrt{57}-3)/8,1\right) \ni x\mapsto H_3(2,x)\end{aligned}$$

  is decreasing. Therefore

  $$\begin{aligned}g(2,x)\le g\left( 2,(\sqrt{57}-3)/8\right) \approx 1213.349,\quad x\in \left[ (\sqrt{57}-3)/8,1\right) .\end{aligned}$$

• (v) It remains to consider the interior of $\varDelta _2.$ The system of equations

$$\begin{aligned} {\left\{ \begin{array}{ll} 36x-6(3c-2)x^2+12(c-3)x^3-2(c+6)x^4&{}=0\\ 36c-6(3c^2-4c+72)x+18(c^2-6c)x^2-4(c^2+12c+20)x^3&{}=0 \end{array}\right. } \end{aligned}$$

(23)

has the solution $c=x=0$ evidently. Let $x\not =0$ and $x\not =3.$ From the first equation we get

$$\begin{aligned} c=\frac{6(1-x^2)(x+3)}{x(x-3)^2} \end{aligned}$$

(24)

which satisfies the inequality $0\le c\le 2$ only when $x\in [x',1),$ where $x'\approx 0.81244.$ Now substituting (24) into the second equation of (23) we obtain the equation

$$\begin{aligned}4x^6+45x^5-333x^4+162x^3-486x^2+945x-81=0,\end{aligned}$$

which has the unique solution in (0, 1),  namely $x''\approx 0.089756<x'.$ Thus g has no critical point in the interior of $\varDelta _2.$

Summarizing, from par D it follows that

$$\begin{aligned} |T_{4,1}(f)|=\frac{1}{82944}(4-c^2)^3 F(c,x,y,\theta ,\psi )\le \frac{1}{82944}\cdot 4^3\cdot 1296=1. \end{aligned}$$

(25)

It is clear that equality for the upper bound in (12) holds for the identity function, and for the lower bound for the function (9).

$\square$