The Third-Order Hermitian Toeplitz Determinant for Classes of Functions Convex in One Direction

Abstract

In this paper, the sharp bounds for the third Hermitian Toeplitz determinant over classes of functions convex in the direction of the imaginary axis and convex in the direction of the positive real axis are computed.

1 Introduction

Let $\mathcal {H}$ be the class of analytic functions in $\mathbb {D}:=\{z\in \mathbb {C}: |z|<1\}$, and let $\mathcal {A}$ be the 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)

Let $\mathcal {S}$ be the subclass of $\mathcal {A}$ of univalent functions.

In this paper, we estimate the Hermitian Toeplitz determinants for functions convex in the direction of the imaginary axis and convex in the direction of the positive real axis. Hermitian Toeplitz matrices play an important role in the applied mathematics as well as in technical sciences, e.g., in the Szegö theory the stochastic filtering, the signal processing, the biological information processing and other engineering problems.

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

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

where ${\overline{a}}_k:=\overline{a_k}.$ Let $|T_{q,n}(f)|$ denote the determinant of $T_{q,n}(f).$

Recently, Ali et al. [1] introduced the concept of the symmetric Toeplitz determinant $T_q(n)$ for $f\in \mathcal {A}$ in the following way:

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

They found a number of estimates for $T_2(n),$$T_3(1),$$T_3(2)$ and $T_2(3)$ over selected subclasses of $\mathcal {A}.$

In recent years, a lot of papers have been devoted to the estimation of determinants built on 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 and the generalized Zalcman functional $J_{m,n}(f):=a_{m+n-1}-a_ma_n,\ m,n\in \mathbb {N},$ are of particular interest. From the large number of papers in this direction, we recall [2, 3, 5,6,7, 15, 16, 19,20,21, 26,27,28, 30, 33], where the second- and third-order Hankel determinants over selected subclasses of $\mathcal {A}$ have been studied. Some of these papers and many others concern also the generalized Zalcman functional, particularly the functional $J_{2,3}(f).$

Being in interest in this research topic in [11], the study of the Hermitian Toeplitz determinants on classes of analytic normalized functions has been initiated. In this paper, we compute the second and third Toeplitz determinants over class of functions convex in the imaginary axis and the class of functions convex in the positive direction of the real axis.

Let us recall some properties of the Toeplitz determinant $|T_{q,1}(f)|$ (see [11]).

• For each $\theta \in \mathbb {R},$$|T_{q,1}(f)|= |T_{q,1}(f_\theta )|,$ where $f_\theta (z):=\mathrm {e}^{-\mathrm {i}\theta }f(\mathrm {e}^{\mathrm {i}\theta }z),\ z\in \mathbb {D},$ i.e., $|T_{q,1}(f)|$ is rotation invariant.

• Since $a_1=1$ is a real number, $T_{q,1}(f)$ is a Hermitian matrix, i.e., $T_{q,1}(f)=\overline{T_{q,1}^{\mathrm{T}}(f)}=:T^*,$ where $\overline{T_{q,1}^{\mathrm{T}}(f)}$ is the conjugate transpose matrix of $T_{q,1}(f).$

• Since $|T_{q,1}(f)|$ for $f\in \mathcal {A}$ is a determinant of Hermitian matrix, it is a real number.

Given a subclass $\mathcal {F}$ of $\mathcal {A},$ let $A_2(\mathcal {F}):=\max \{|a_2|: f\in \mathcal {F}\}$ if exists. Since for $f\in \mathcal {A},$

$$\begin{aligned} |T_{2,1}(f)|=1-|a_2|^2, \end{aligned}$$

we get the result below. The equality for the lower bound is attained by a function in $\mathcal {F}$ which is extremal for $A_2(\mathcal {F}).$ The identity makes equality for the upper bound.

Theorem 1

Let $\mathcal {F}$ be a subclass of $\mathcal {A}$ and $A_2(\mathcal {F})$ exists. If the identity is an element of $\mathcal {F},$ then

$$\begin{aligned} 1-A_2^2(\mathcal {F})\le |T_{2,1}(f)| \le 1. \end{aligned}$$

Both inequalities are sharp.

By $\mathcal {CV}(\mathrm {i})$ and $\mathcal {CV}(1)$, we denote the subclasses of ${\mathcal A}$ of functions f which satisfy

$$\begin{aligned} {{\,\mathrm{Re}\,}}\{ (1-z^2)f'(z)\} > 0,\quad z\in {\mathbb {D}}, \end{aligned}$$

(2)

and

$$\begin{aligned} {{\,\mathrm{Re}\,}}\{ (1-z)^2f'(z)\} > 0,\quad z\in {\mathbb {D}}, \end{aligned}$$

(3)

respectively. Both classes play an important role in the geometric function theory in view of their geometrical properties. Each function $f\in \mathcal {CV}(\mathrm {i})$ maps univalently ${\mathbb {D}}$ onto a domain $f({\mathbb {D}})$ convex in the direction of the imaginary axis, i.e., for $w_1,w_2\in f({\mathbb {D}})$ such that ${{\,\mathrm{Re}\,}}w_1={{\,\mathrm{Re}\,}}w_2$ the line segment $[w_1,w_2]$ lies in $f({\mathbb {D}}),$ with the additional property that there exist two points $\omega _1,\omega _2$ on the boundary of $f({\mathbb {D}})$ for which $\{\omega _1+\mathrm {i}t: t> 0\}\subset {\mathbb {C}}\setminus f({\mathbb {D}})$ and $\{\omega _2-\mathrm {i}t: t> 0\}\subset {\mathbb {C}}\setminus f({\mathbb {D}})$ (see, e.g., [13, p. 199]). In fact, the class $\mathcal {CV}(\mathrm {i})$ is the subclass of the class $\mathcal {CV}$ of functions convex in the direction of the imaginary axis which was introduced by Robertson [31] in 1936. Robertson’s analytic condition for the class $\mathcal {CV}$ was shown by him under some regularity of functions in $\mathcal {CV}$ on the unit circle. The proof of Robertson’s conjecture for the whole class $\mathcal {CV}$ was finally completed by Hengartner and Schober [14] who divided the class $\mathcal {CV}$ into three subclasses with the class $\mathcal {CV}(\mathrm {i})$ as one of them (see also [13, pp. 193–206]).

Each function in the class $\mathcal {CV}(1)$ maps univalently ${\mathbb {D}}$ onto a domain $f({\mathbb {D}})$ called convex in the positive direction of the real axis, i.e., $\{w+it:t\ge 0\}\subset f({\mathbb {D}})$ for every $w\in f({\mathbb {D}})$ [4, 8,9,10, 12, 24, 25].

The condition (3) was generalized by replacing the polynomial $1-z^2$ by quadratic polynomials [22, 23] and by any polynomials having their roots in ${\mathbb {C}}\setminus {\mathbb {D}}$ [17, 18].

In this paper, we compute sharp lower and upper bounds for

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

(4)

over the classes $\mathcal {CV}(\mathrm {i})$ and $\mathcal {CV}(1).$

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 $\mathbb {D}.$

In the proof of the main result, we will use the following lemma which contains the well-known formula for $c_2$ (see, e.g., [29, p. 166]) and further remarks in [7]).

Lemma 1

If $p \in {{\mathcal {P}}}$ is of the form (5), then

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

(6)

and

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

(7)

for some $\zeta _i \in \overline{{\mathbb {D}}}$, $i=1,2.$

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

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

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

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

(8)

2 Functions Convex in the Direction of the Imaginary Axis

Since $A_2({\mathcal {C}}{\mathcal {V}}(\mathrm {i}))=1$ ([14], see also [13, Vol. I, pp. 200–201]) with the extremal function

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

and since the identity belongs to ${\mathcal {C}}{\mathcal {V}}(\mathrm {i}),$ by Theorem 1, we have

Theorem 2

Let $\alpha \in [0,1)$. If $f\in {{\mathcal {C}}{\mathcal {V}}(\mathrm {i})}$, then

$$\begin{aligned} 0\le |T_{2,1}(f)| \le 1. \end{aligned}$$

Both inequalities are sharp.

Now, we will compute the bounds of $|T_{3,1}(f)|.$

Theorem 3

If $f \in {{\mathcal {C}}{\mathcal {V}}(\mathrm {i}),}$ then

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

(9)

The inequality is sharp.

Proof

Let $f \in {{\mathcal {C}}{\mathcal {V}}}(\mathrm {i})$ be the form (1). Since $|a_2| \le 1,$$|a_3| \le 1$ ([14], see also [13, Vol. I, pp. 200–201]) and ${{\,\mathrm{Re}\,}}\left( a_2^2 {\overline{a}}_3\right) \le |a_2^2 a_3|,$ from (4), we get

$$\begin{aligned} |T_{3,1}(f)|\le F(|a_2|,|a_3|), \end{aligned}$$

(10)

where

$$\begin{aligned} F(x,y):=2x^2 y -2x^2 - y^2 +1,\quad (x,y)\in [0,1]\times [0,1]. \end{aligned}$$

We have

$$\begin{aligned} \frac{\partial F}{\partial x}= -4x(1-y) \le 0, \quad (x,y)\in [0,1]\times [0,1]. \end{aligned}$$

Thus,

$$\begin{aligned} F(x,y) \le F(0,y) = 1-y^2 \le 1, \quad (x,y)\in [0,1]\times [0,1], \end{aligned}$$

which in view of (10) shows (9).

Clearly, the identity makes the inequality (9) sharp. $\square$

Theorem 4

If $f \in {{\mathcal {C}}{\mathcal {V}}(\mathrm {i})},$ then

$$\begin{aligned} |T_{3,1}(f)| \ge -\frac{1}{2}. \end{aligned}$$

(11)

The inequality is sharp.

Proof

By (2), there exists $p\in {{\mathcal {P}}}$ of the form (5) such that

$$\begin{aligned} (1-z^2)f'(z)=p(z),\quad z\in \mathbb {D}. \end{aligned}$$

(12)

Putting the series (1) and (5) into (12) by equating the coefficients, we get

$$\begin{aligned} a_2 = \frac{1}{2}c_1, \qquad a_3 = \frac{1}{3}(c_2+1). \end{aligned}$$

(13)

By (13), (6) and (7), we have

$$\begin{aligned} a_2 = \zeta _1,\qquad a_3 = \frac{1}{3}\left( 1+2\zeta _1^2 +2(1-|\zeta _1|^2)\zeta _2\right) \end{aligned}$$

with $\zeta _i \in \overline{{\mathbb {D}}}$, $i=1,2.$ Therefore, from (4), we get

$$\begin{aligned} |T_{3,1}(f)| = \frac{1}{9}( \varPsi _1 + \varPsi _2), \end{aligned}$$

(14)

where

$$\begin{aligned} \varPsi _1 := 8 -18|\zeta _1|^2 +8|\zeta _1|^4 -4(1-|\zeta _1|^2)^2|\zeta _2|^2 \end{aligned}$$

and

$$\begin{aligned} \varPsi _2 := 2{{\,\mathrm{Re}\,}}\zeta _1^2 -4(1-|\zeta _1|^2){{\,\mathrm{Re}\,}}\zeta _2+ 4(1-|\zeta _1|^2){{\,\mathrm{Re}\,}}( \zeta _1^2 {\overline{\zeta }}_2). \end{aligned}$$

A. When $\zeta _1=0,$ then

$$\begin{aligned} |T_{3,1}(f)|=\frac{1}{9}\left( 8-4|\zeta _2|^2-4{{\,\mathrm{Re}\,}}(\zeta _2)\right) \ge \frac{1}{9}\left( 8-4|\zeta _2|^2-4|\zeta _2|\right) \ge 0. \end{aligned}$$

When $\zeta _2=0,$ then

$$\begin{aligned} \begin{aligned} |T_{3,1}(f)|&=\frac{1}{9}\left( 8-18|\zeta _1|^2+8|\zeta _1|^4+2{{\,\mathrm{Re}\,}}(\zeta _1^2)\right) \\&\ge \frac{4}{9}\left( 2-5|\zeta _1|^2+2|\zeta _2|^4\right) \ge -\frac{4}{9}. \end{aligned} \end{aligned}$$

B. Suppose that $\zeta _1,\zeta _2\in \overline{{\mathbb {D}}}\setminus \{0\}.$ Then, there exist unique $\theta$ and $\psi$ in $[0,2\pi )$ such that $\zeta _1 =r \mathrm {e}^{\mathrm {i}\theta }$ and $\zeta _2 = s \mathrm {e}^{\mathrm {i}\psi },$ where $r:=|\zeta _1|\in (0,1]$ and $s:=|\zeta _2|\in (0,1].$ Thus,

$$\begin{aligned} \begin{aligned} \varPsi _2&= 2r^2 \cos 2\theta - 4s(1-r^2)\cos \psi + 4r^2 s(1-r^2)\cos (2\theta -\psi ) \\&= 2r^2 \sin (2\theta +\alpha ) \sqrt{ 1+4s^2(1-r^2)^2 +4s(1-r^2)\cos \psi } - 4s(1-r^2)\cos \psi , \end{aligned} \end{aligned}$$

(15)

where $\alpha \in [0,2\pi )$ is a unique quantity satisfying

$$\begin{aligned} \cos \alpha = \frac{\kappa _1}{ \sqrt{\kappa _1^2 +\kappa _2^2} },\qquad \sin \alpha = \frac{\kappa _2}{ \sqrt{\kappa _1^2 +\kappa _2^2} } \end{aligned}$$

(16)

with

$$\begin{aligned} \kappa _1 := 2s(1-r^2)\sin \psi ,\qquad \kappa _2 := 1 + 2s(1-r^2)\cos \psi . \end{aligned}$$

From (15), we have

$$\begin{aligned} \begin{aligned} -\varPsi _2&\le 2r^2 \sqrt{1+4s^2(1-r^2)^2+4s(1-r^2)\cos \psi } + 4s(1-r^2)\cos \psi \\&\le 2r^2-4sr^4+4s. \end{aligned} \end{aligned}$$

(17)

Therefore, by (14) and (17), we obtain

$$\begin{aligned} 9|T_{3,1}(f)|= \varPsi _1 + \varPsi _2\ge F(r,s), \end{aligned}$$

(18)

where

$$\begin{aligned} F(x,y):=4(2-5x^2+2x^4) -4(1-x^4)y -4(1-x^2)^2 y^2,\quad x,y\in (0,1]. \end{aligned}$$

Since

$$\begin{aligned} \frac{\partial F}{\partial y} = -4(1-x^2)\left( 1+x^2+2(1-x^2)y\right) \le 0, \quad x,y\in (0,1], \end{aligned}$$

we see that

$$\begin{aligned} F(x,y) \ge F(x,1) = 4x^2(-3+2x^2) \ge -\frac{9}{2}, \quad x,y\in (0,1]. \end{aligned}$$

Hence, by (18) and part A, it follows that the inequality (11) is true.

The inequality is sharp with the equality attained by the function

$$\begin{aligned} f(z)=\int _{0}^{z} \frac{ 1+t^2 }{ (1-t^2)(1- \mathrm {i}\sqrt{3} t - t^2) } \mathrm {d}t,\quad z\in \mathbb {D}, \end{aligned}$$

which belongs to ${{\mathcal {C}}{\mathcal {V}}}(\mathrm {i})$ and for which $a_2=\mathrm {i}\sqrt{3}/2$ and $a_3=0.$$\square$

3 Functions Convex in the Positive Direction of the Real Axis

Since $A_2({\mathcal {C}}{\mathcal {V}}(1))=2$ [10] with the Koebe function

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

(19)

as the extremal, and the identity belongs to ${\mathcal {C}}{\mathcal {V}}(1),$ by Theorem 1, we have

Theorem 5

If $f\in {{\mathcal {C}}{\mathcal {V}}}(1)$, then

$$\begin{aligned} -3\le |T_{2,1}(f)| \le 1. \end{aligned}$$

Both inequalities are sharp.

Now, we will compute the bounds of $|T_{3,1}(f)|.$

Theorem 6

If $f \in {{\mathcal {C}}{\mathcal {V}}}(1),$ then

$$\begin{aligned} |T_{3,1}(f)| \le 8. \end{aligned}$$

(20)

The inequality is sharp.

Proof

Let $f \in {{\mathcal {C}}{\mathcal {V}}}(1)$ be the form (1). Since $|a_2| \le 2,$$|a_3| \le 3$ [10] and ${{\,\mathrm{Re}\,}}\left( a_2^2 {\overline{a}}_3\right) \le |a_2^2 a_3|,$ from (4), we get

$$\begin{aligned} T_{3,1}(f) \le F(|a_2|,|a_3|), \end{aligned}$$

(21)

where

$$\begin{aligned} F(x,y):=2x^2 y -2x^2 - y^2 +1,\quad (x,y)\in [0,2]\times [0,3]. \end{aligned}$$

Solving the system of equations $\partial F/\partial x=0=\partial F/\partial y,$ we see that (1, 1) is the unique critical point in $(0,2)\times (0,3).$ Since

$$\begin{aligned} \left( \frac{\partial ^2F}{\partial x^2}\cdot \frac{\partial ^2F}{\partial y^2}-\frac{\partial ^2F}{\partial x\partial y}\right) (1,1)=-4<0, \end{aligned}$$

F has a saddle point at (1, 1). On the boundary of $[0,2]\times [0,3]$, we have

• (1) $F(0,y)=1-y^2 \le 1,\quad y\in [0,3]$;

• (2) $F(2,y)=-7+8y-y^2 \le 8,\quad y\in [0,3]$;

• (3) $F(x,0)=1-2x^2 \le 1,\quad x\in [0,2]$;

• (4) $F(x,3)=-8+4x^2 \le 8,\quad x\in [0,2]$.

Hence and from (21), the inequality (20) follows.

Equality in (20) holds for the Koebe function k given by (19) for which $a_2=2$ and $a_3=3.$$\square$

Theorem 7

If $f \in {{\mathcal {C}}{\mathcal {V}}}(1),$ then

$$\begin{aligned} \begin{aligned} |T_{3,1}(f)|&\ge -\frac{1}{225} \left( -558+286\sqrt{5} + 19\sqrt{54+14\sqrt{5}} -5\sqrt{270+70\sqrt{5}} \right) \\&\approx -0.68328. \end{aligned} \end{aligned}$$

(22)

The inequality is sharp.

Proof

By (3), there exists $p\in {{\mathcal {P}}}$ of the form (5) such that

$$\begin{aligned} (1-z)^2f'(z)=p(z),\quad z\in \mathbb {D}. \end{aligned}$$

(23)

Putting the series (1) and (5) into (23) by equating the coefficients, we get

$$\begin{aligned} a_2 = \frac{1}{2}(2+c_1), \qquad a_3 = \frac{1}{3}(3+2c_1+c_2). \end{aligned}$$

(24)

Substituting (6) and (7) into the equalities (24), we get

$$\begin{aligned} a_2 = 1+\zeta _1,\qquad a_3 = \frac{1}{3}( 3 +4\zeta _1 +2\zeta _1^2 +2(1-|\zeta _1|^2)\zeta _2 ) \end{aligned}$$

for some $\zeta _i \in \overline{{\mathbb {D}}}$, $i=1,2$. Furthermore, from (4), we obtain

$$\begin{aligned} |T_{3,1}(f)| = \frac{1}{9} (\varPsi _1 + \varPsi _2), \end{aligned}$$

(25)

where

$$\begin{aligned} \varPsi _1 := 14|\zeta _1|^2+ 8|\zeta _1|^4 -4(1-|\zeta _1|^2)^2 |\zeta _2|^2 \end{aligned}$$

(26)

and

$$\begin{aligned} \begin{aligned} \varPsi _2&:= 18{{\,\mathrm{Re}\,}}\zeta _1^2+32|\zeta _1|^2{{\,\mathrm{Re}\,}}\zeta _1 +8(1-|\zeta _1|^2){{\,\mathrm{Re}\,}}(\zeta _1 {\overline{\zeta }}_2)\\&\quad +\, 4(1-|\zeta _1|^2){{\,\mathrm{Re}\,}}( \zeta _1^2 {\overline{\zeta }}_2). \end{aligned} \end{aligned}$$

(27)

A. Suppose that $\zeta _1,\zeta _2\in \overline{{\mathbb {D}}}\setminus \{0\}.$ Then, there exist unique $\theta$ and $\psi$ in $[0,2\pi )$ such that $\zeta _1 =r \mathrm {e}^{\mathrm {i}\theta }$ and $\zeta _2 = s \mathrm {e}^{\mathrm {i}\psi },$ where $r:=|\zeta _1|\in (0,1]$ and $s:=|\zeta _2|\in (0,1].$ From (26) and (27), we, respectively, have

$$\begin{aligned} \varPsi _1 \ge 14r^2 +8r^4 -4(1-r^2)^2=4r^4+22r^2-4 \end{aligned}$$

(28)

and

$$\begin{aligned} \varPsi _2 = 18r^2 \cos 2\theta + 32r^3 \cos \theta +4(1-r^2)rs \sqrt{\kappa _1^2 +\kappa _2^2} \sin (\psi +\alpha ), \end{aligned}$$

where $\alpha \in [0,2\pi )$ is a unique quantity satisfying (16) with

$$\begin{aligned} \kappa _1 := 2\sin \theta +r\sin 2\theta ,\qquad \kappa _2 := 2\cos \theta +r\cos 2\theta . \end{aligned}$$

(29)

Hence,

$$\begin{aligned} \varPsi _2\ge 18r^2\cos 2\theta +32r^3 \cos \theta -4(1-r^2)r \sqrt{4+4r\cos \theta +r^2}. \end{aligned}$$

(30)

Let $\varOmega :=(0,1]\times [-1,1].$ From (25), (28) and (30), it follows that

$$\begin{aligned} |T_{3,1}(f)| \ge -\frac{4}{9} F(r,\cos \theta ), \end{aligned}$$

(31)

where

$$\begin{aligned} F(x,y) := 1-x^2-x^4-9x^2 y^2 -8x^3 y +x(1-x^2)\sqrt{g(x,y)},\quad (x,y)\in \varOmega , \end{aligned}$$

with

$$\begin{aligned} g(x,y):=4+4xy+x^2,\quad (x,y)\in \varOmega . \end{aligned}$$

Let

$$\begin{aligned} \varTheta := \frac{1}{100} \left( -558 +286\sqrt{5} + 19\sqrt{54+14\sqrt{5}} -5\sqrt{270+70\sqrt{5}} \right) = 1.53738\cdots . \end{aligned}$$

Now, we will show that

$$\begin{aligned} \max \{ F(x,y) : (x,y) \in \varOmega \} = \varTheta . \end{aligned}$$

(32)

(A1) For this, we first find the critical points of F in the interior of $\varOmega$, i.e., in $(0,1)\times (-1,1).$ Note that in ${{\,\mathrm{Int}\,}}\varOmega$, the equation

$$\begin{aligned} \frac{\partial F}{\partial y}= -18x^2 y -8x^3 +2x^2(1-x^2) g(x,y)^{-1/2}=0, \end{aligned}$$

is equivalent to

$$\begin{aligned} g(x,y)^{-1/2} = \frac{9y+4x}{1-x^2}. \end{aligned}$$

(33)

Furthermore, note that

$$\begin{aligned} 9y+4x \ge 0 \end{aligned}$$

(34)

holds, since $g(x,y)^{1/2} \ge 0$ and $1-x^2 >0$. Under the condition (34), Eq. (33) can be written as

$$\begin{aligned} 324xy^3+(369x^2+324)y^2+(136x^3+288x)y+15x^4+66x^2-1=0. \end{aligned}$$

(35)

The equation

$$\begin{aligned} \begin{aligned} \frac{\partial F}{\partial x}=&-2x -4x^3 -18xy^2 -24x^2y + (1-3x^2)g(x,y)^{1/2}\\&+x(1-x^2)(2y+x)g(x,y)^{-1/2}=0, \end{aligned} \end{aligned}$$

is equivalent to

$$\begin{aligned} 63x^2 y^2 +2(9x+14x^3)y -(1-12x^2+3x^4) = 0. \end{aligned}$$

(36)

Note that $\varDelta =\varDelta (x):=144-504x^2 +385x^4 \ge 0$ iff $x\in (0,x_1]\cup [x_2,1),$ where

$$\begin{aligned} x_1=\frac{2}{\sqrt{55}}\sqrt{9-6\sqrt{\frac{2}{7}}}\approx 0.64908,\qquad x_2=\frac{2}{\sqrt{55}}\sqrt{9+6\sqrt{\frac{2}{7}}}\approx 0.94223. \end{aligned}$$

Since $\varDelta (x_1)=0,$ Eq. (36) has a unique root $y_0'=-(9+14x_1^2)/(63x_1)\approx -\,0.36433.$ Analogously, $\varDelta (x_2)=0,$ so Eq. (36) has a unique root $y_0''=-(9+14x_2^2)/(63x_2)\approx -\,0.36100$. As easy to check, the polynomial in (35) does not vanish for $x=x_1,\ y=y_0'$ and for $x=x_2,\ y=y_0''.$

Let now $x\in (0,x_1)\cup (x_2,1).$ Thus, there are two roots $y_1$ and $y_2$ of (36), namely

$$\begin{aligned} y_j = \frac{ -\,(9+14x^2) + (-1)^j \sqrt{ 144-504x^2 +385x^4 }}{ 63x }, \quad j=1,2. \end{aligned}$$

(37)

(1) Consider the case $y=y_1$. Note that $y_1>-1$ is equivalent to

$$\begin{aligned} -\,9+63x-14x^2>\sqrt{ 144-504x^2 +385x^4 }. \end{aligned}$$

(38)

We have $-9+63x-14x^2>0$ iff $x\in (x_3,1),$ where $x_3=(63-\sqrt{3465})/28\approx 0.14771$. Thus, for $x\in (x_3,x_1)\cup (x_2,1)$ by squaring the both sides of (38), we get the inequality

$$\begin{aligned} 189x^4+1764x^3-4725x^2+1134x+63<0 \end{aligned}$$

which is true for $x\in (x_4,x_1)\cup (x_2,1),$ where $x_4\approx 0.32137$. Moreover, $y_1<1$ is equivalent to the inequality

$$\begin{aligned} -\,9-63x-14x^2<\sqrt{ 144-504x^2 +385x^4 }, \end{aligned}$$

which is clearly true for $x\in (x_4,x_1)\cup (x_2,1)$.

Substituting $y=y_1$ into Eq. (35), we get

$$\begin{aligned} Q_1(x)\sqrt{144-504x^2+385x^4}=Q_2(x), \end{aligned}$$

(39)

where

$$\begin{aligned} \begin{aligned}&Q_1(x):=242028-234738x^2-204120x^4,\\&Q_2(x):=-\,3096792+9103752x^2-3419010x^4-3572100x^6. \end{aligned} \end{aligned}$$

(40)

Since $Q_1(x)> 0$ for $x\in (x_4,x_1),$$Q_1(x)< 0$ for $x\in (x_2,1),$$Q_2(x)<0$ for $x\in (x_4,x_1)$ and $Q_2(x)>0$ for $x\in (x_2,1),$ Eq. (39) has no solution.

(2) Consider now the case $y=y_2$. Note that $y_2>-1$ is equivalent to

$$\begin{aligned} \sqrt{ 144-504x^2 +385x^4 }>9-63x+14x^2. \end{aligned}$$

(41)

Since $9-63x+14x^2<0$ for $x\in (x_5,1),$ where

$$\begin{aligned} x_5=\frac{3}{4}\left( 3-\sqrt{\frac{55}{7}}\right) \approx 0.14771, \end{aligned}$$

let us consider $x\in (0,x_5]$. By squaring both sides of (41) and grouping, we get the inequality

$$\begin{aligned} 189x^4+1764x^3-4725x^2+1134x+63\ge 0, \end{aligned}$$

which is true for $x\in (0,x_5]$. Thus, $y_2>-1$ holds for all $x\in (0,x_1)\cup (x_2,1)$. Moreover, $y_2<1$ is equivalent to the inequality

$$\begin{aligned} \sqrt{ 144-504x^2 +385x^4 }<9+63x+14x^2. \end{aligned}$$

Since the right hand of the above inequality is positive, by squaring both sides and grouping, we get the inequality

$$\begin{aligned} 189x^4-1764x^3-4725x^2-1134x+63>0, \end{aligned}$$

which is true for $x\in (x_6,x_1)\cup (x_2,1),$ where $x_6\approx 0.04624$. Thus, we now consider $x\in (x_6,x_1)\cup (x_2,1)$.

Substituting $y=y_2$ into Eq. (35), we get

$$\begin{aligned} Q_1(x)\sqrt{144-504x^2+385x^4}=-Q_2(x), \end{aligned}$$

(42)

where $Q_1$ and $Q_2$ are given by (40). Since $Q_1(x)Q_2(x)>0$ for $x\in (x_6,x_1)\cup (x_2,1),$ by squaring both sides of (42) and grouping, we equivalently get the equation

$$\begin{aligned} 26248933872(x^2-2)(x^2-1)(5x^2-2)(25x^4+45x^2-11)=0 \end{aligned}$$

which has two roots

$$\begin{aligned} {\tilde{x}}_1= \sqrt{\frac{2}{5}}\approx 0.63246,\qquad {\tilde{x}}_2=\sqrt{ - \frac{9}{10} + \frac{ \sqrt{5} }{2} }\approx 0.46694. \end{aligned}$$

(43)

Substituting $x={\tilde{x}}_1$ into $y_1$ given by (37), we get ${\tilde{y}}_1=-1/\sqrt{10}\approx -0.31623$. But $9{\tilde{y}}_1 + 4{\tilde{x}}_1<0$ which contradicts (34). Therefore, $({\tilde{x}}_1,{\tilde{y}}_1)$ is not a critical point of F in ${{\,\mathrm{Int}\,}}\varOmega$.

Substituting $x={\tilde{x}}_2$ into $y_1$ given by (37), we get

$$\begin{aligned} {\tilde{y}}_2= - \frac{1}{220} \left( 15\sqrt{2(-9+5\sqrt{5})} + \sqrt{10(-9+5\sqrt{5})} \right) =-\,0.163604\cdots \end{aligned}$$

(44)

Since

$$\begin{aligned} 9{\tilde{y}}_2+4\tilde{x_2} = \frac{1}{220}(-135\sqrt{2}+79\sqrt{10}) \sqrt{-9+5\sqrt{5}} = 0.395333\cdots >0, \end{aligned}$$

$({\tilde{x}}_2,{\tilde{y}}_2)$ satisfies (34), and thus, it is a unique critical point of F.

Denote

$$\begin{aligned} \lambda _1 := F_{xx}({\tilde{x}}_2,{\tilde{y}}_2),\qquad \lambda _2 := F_{xy}({\tilde{x}}_2,{\tilde{y}}_2),\qquad \lambda _3 := F_{yy}({\tilde{x}}_2,{\tilde{y}}_2). \end{aligned}$$

Numerical calculations yield

$$\begin{aligned} \lambda _1 = -\,6.74042\cdots <0,\qquad \lambda _1 \lambda _3 - \lambda _2^2 = 22.9037\cdots >0. \end{aligned}$$

Thus, F has a local maximum at $(\tilde{x_2},{\tilde{y}}_2)$ with

$$\begin{aligned} F({\tilde{x}}_2,{\tilde{y}}_2) = \varTheta . \end{aligned}$$

(A2) It remains to consider F in the boundary of $\varOmega$.

• (1) On the side $x=0,$ we have $F(0,y) \equiv 1<\varTheta ,\ y\in [-1,1]$.

• (2) On the side $x=1$, we have

  $$\begin{aligned} F(1,y)=-1-8y-9y^2 \le F\left( 1,-\frac{4}{9}\right) = \frac{7}{9}<\varTheta ,\quad y\in [-1,1]. \end{aligned}$$

• (3) On the side $y=-1$, we have

  $$\begin{aligned} \begin{aligned} F(x,-1)&\le F\left( \frac{1}{18}(11-\sqrt{85}),-1\right) \\&= \frac{1}{486}(-251+85\sqrt{85}) = 1.09601\cdots <\varTheta , \quad x\in [0,1]. \end{aligned} \end{aligned}$$

• (4) On the side $y=1$, we have

  $$\begin{aligned} F(x,1)= 1+2x-9x^2-10x^3-2x^4,\quad x\in [0,1]. \end{aligned}$$

  It is easy to see that

  $$\begin{aligned} F(x,1) \le \gamma (x)\le \gamma (x_7) = 1+\frac{3}{2}x_7 = 1.43153\cdots <\varTheta ,\quad x\in [0,1], \end{aligned}$$

  where $x_7=42^{-1/3}\approx 0.28769$ and

  $$\begin{aligned} \gamma (x):=1+2x-21x^4,\quad x\in [0,1]. \end{aligned}$$

B. When $\zeta _1=0,$ then

$$\begin{aligned} T_{3,1}(f)=-\frac{4}{9}|\zeta _2|^2\ge -\frac{4}{9}. \end{aligned}$$

C. Let $\zeta _2=0$ and $\zeta _1\not =0$. Then,

$$\begin{aligned} \varPsi _1 := 14|\zeta _1|^2+ 8|\zeta _1|^4,\quad \varPsi _2=18{{\,\mathrm{Re}\,}}\zeta _1^2 + 32|\zeta _1|^2{{\,\mathrm{Re}\,}}\zeta _1. \end{aligned}$$

Thus, taking $\zeta _1 =r \mathrm {e}^{\mathrm {i}\theta },$ where $r\in [0,1]$ and $\theta \in [0,2\pi ),$ we have

$$\begin{aligned} \varPsi _1=14r^2 +8r^4,\quad \varPsi _2 = 18r^2 \cos 2\theta + 32r^3 \cos \theta . \end{aligned}$$

Since the inequalities (28) and (30) hold, further argumentation of part A remains valid.

Summarizing from parts A–C, it follows that $F(x,y) \le \varTheta$ holds for all $(x,y)\in [0,1]\times [-1,1]$. This together with (31) proves (22).

Now, we discuss the sharpness of (22). From (25), (28) and (30), that $|T_{3,1}(f)| = -\,(4/9)\varTheta$ holds when the following conditions are satisfied:

$$\begin{aligned} x={\tilde{x}}_2, \qquad \cos \theta = {\tilde{y}}_2, \qquad s=1,\quad \sin (\psi +\alpha )=-\,1, \end{aligned}$$

(45)

where ${\tilde{x}}_2$ and ${\tilde{y}}_2$ are given by (43) and (44), and where $\alpha$ is determined by the condition (16) with $\kappa _1$ and $\kappa _2$ given in (29). Set $\theta ={\mathrm{Arccos}}({\tilde{y}}_2)$ so that it satisfies the second condition in (45). Then, we have $\kappa _1 = 1.82232\cdots >0$ and $\kappa _2 = -\,0.76915\cdots <0$. Thus, (16) is satisfied if we take $\alpha$ by

$$\begin{aligned} \alpha = -\, {\mathrm{Arccos}}\left( \frac{ \kappa _1 }{ \sqrt{ \kappa _1^2 + \kappa _2^2 } } \right) = -\,{\mathrm{Arccos}}\left( \frac{2\sin \theta +x_2\sin 2\theta }{ \sqrt{ 4+4x_2\cos \theta +x_2^2 } } \right) . \end{aligned}$$

Thus, if we put

$$\begin{aligned} \psi = \frac{3\pi }{2} - \alpha = \frac{3\pi }{2} + {\mathrm{Arccos}}\left( \frac{2\sin \theta +{\tilde{x}}_2\sin 2\theta }{ \sqrt{ 4+4{\tilde{x}}_2\cos \theta +{\tilde{x}}_2^2 } } \right) = 5.11178\cdots , \end{aligned}$$

then $\psi$ satisfies the fourth condition in (45).

Now, let us consider a function ${\tilde{p}}$ which has the form (8) with $\zeta _1 = {\tilde{x}}_2\mathrm {e}^{\mathrm {i}\theta }$ and $\zeta _2 = \mathrm {e}^{\mathrm {i}\psi }$. Since $\zeta _1\in {\mathbb {D}}$ and $\zeta _2 \in {\mathbb {T}}$, in view of Lemma 1, we see that ${\tilde{p}}$ belongs to the class ${{\mathcal {P}}}$. Finally, let

$$\begin{aligned} {\tilde{f}}(z) := \int _{0}^{z} \frac{{\tilde{p}}(\zeta )}{(1-\zeta )^2}\mathrm {d}\zeta ,\quad z\in {\mathbb {D}}. \end{aligned}$$

Clearly, ${\tilde{f}} \in {{\mathcal {C}}{\mathcal {V}}}(1)$ and $|T_{3,1}({\tilde{f}})|=-(4/9)\varTheta$. Thus, the proof of the theorem is completed. $\square$