The Third-Order Hermitian Toeplitz Determinant for Alpha-Convex Functions

Abstract

Sharp lower and upper bounds of the second- and third-order Hermitian Toeplitz determinants for the class of $\alpha$-convex functions were found. The symmetry properties of the arithmetic mean underlying the definition of $\alpha$-convexity and the symmetry properties of Hermitian matrices were used.

1. Introduction

Let $\mathscr{H}$ be the class of all analytic functions in $\mathbb{D}:=\{z\in \mathbb{C}:|z|<1\}$ and $\mathcal{A}$ be its subclass of functions f of the form:

$$f\left(z\right)=\sum _{n=1}^{\infty }{a}_n{z}^n,a_1=1,z\in \mathbb{D}.$$

(1)

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

For $\alpha \in [0,1],$ let ${\mathcal{M}}_{\alpha }$ denote the subclass of $\mathcal{A}$ of all functions f satisfying the condition:

$$\mathrm{Re}\left\{(1-\alpha )\frac{z{f}^{\prime }(z)}{f(z)}+\alpha \left(1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right)\right\}>0,z\in \mathbb{D}$$

(2)

The class ${\mathcal{M}}_{\alpha }$ was introduced by Mocanu [1] and its elements are called $\alpha$-convex (see also [2] [Volume I, pp. 142–147]). For $\alpha =0$, the condition (2) describes the class of starlike functions denoted by ${\mathcal{S}}^{*}$ introduced by Alexander [3] ([4], as can also be seen in [2] [Volume I, Chapter 8]). For $\alpha =1$, the condition (2) specifies the class of convex functions denoted by ${\mathcal{S}}^c$ defined by the study [5] (see also [2] [Volume I, Chapter 8]). Thus, the classes ${\mathcal{M}}_{\alpha }$ create a “continuous passage” on $\alpha \in [0,1]$ from the set of convex functions ${\mathcal{M}}_1$ to the set of starlike functions ${\mathcal{M}}_0.$ One can see that the range of $\alpha$ can be extended to the real axis as well to the complex plane. In [1], Mocanu presented a geometrical interpretation of functions in the class ${\mathcal{M}}_{\alpha }.$ In [6], it was observed that ${\mathcal{M}}_{\alpha }\subset {\mathcal{M}}_0$ for every $\alpha \in [0,1].$ This result can be found in the paper due to Sakaguchi [7] that was published before the advent of the $\alpha$-convexity concept (cf. [2] [Volume I. pp. 142–143]). Furthermore, in [6], the authors have shown that ${\mathcal{M}}_{{\alpha }_1}\subset {\mathcal{M}}_{{\alpha }_2}$ for every $0\le {\alpha }_2\le {\alpha }_1\le 1.$ The class ${\mathcal{M}}_{\alpha }$ plays an important role in the geometric function theory and was studied by various authors (e.g., [8,9] [Chapter 7] with further references).

For $q,n\in \mathbb{N},$ by $T_{q,n}\left(f\right),$ with $f\in \mathcal{A}$ of the form (1), define the matrix:

$$T_{q,n}\left(f\right):=\left[\begin{array}{cccc}{a}_n& a_{n+1}& \dots & a_{n+q-1}\\ {\overline{a}}_{n+1}& a_n& \dots & a_{n+q-2}\\ ⋮& ⋮& ⋮& ⋮\\ {\overline{a}}_{n+q-1}& {\overline{a}}_{n+q-2}& \dots & a_n\end{array}\right],$$

where ${\overline{a}}_k:=\overline{a_k}.$ In case, $a_n$ is a real number, $T_{q,n}\left(f\right)$ is the Hermitian Toeplitz matrix. Thus, the matrix $T_{q,1}\left(f\right)$ is like this. In particular:

$$det{T}_{3,1}\left(f\right)=\left|\begin{array}{ccc}1& a_2& a_3\\ {\overline{a}}_2& 1& a_2\\ {\overline{a}}_3& {\overline{a}}_2& 1\end{array}\right|=1+2\text{Re}\left(a_2^2{\overline{a}}_3\right)-2{\left|a_2\right|}^2-{\left|a_3\right|}^2.$$

(3)

In recent years, many papers have been devoted to the estimation of determinants whose entries are 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., [10,11,12], with further references), and the symmetric Toeplitz matrices (see [13]), are of particular interest.

For this reason, considering the interest of specialists, in [14,15,16], the estimation of the determinants of the Hermitian Toeplitz matrices $T_{q,1}\left(f\right)$ on the class $\mathcal{A}$ or its subclasses was started. Hermitian Toeplitz matrices play an important role in functional analysis, applied mathematics as well as in physics and technical sciences. Let us mention that only a few papers have been published that concern the estimation of Hermitian Toeplitz determinants in the basic subclasses of univalent functions. This is a new issue and it is important to find such estimates for the class of $\alpha$-convex functions, which are among the most important in geometric function theory. This paper was dedicated to finding the sharp lower and upper bounds of the second- and third-order Hermitian Toeplitz determinants for the class ${\mathcal{M}}_{\alpha }.$

Let $𝒫$ be the class of all $p\in$ $\mathscr{H}$ of the form:

$$p\left(z\right)=1+\sum _{n=1}^{\infty }{c}_n{z}^n,z\in \mathbb{D},$$

(4)

which have a positive real part.

In the proof of the main result, we will use the following lemma which contains the Carathéodory result for $c_1$ [17], and the well-known formula for $c_2$ (e.g., [18] [p. 166]).

Lemma 1. 

If $p\in$ $𝒫$ is of the form (4), then:

$$c_1=2{\zeta }_1,$$

(5)

and:

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

(6)

for some ${\zeta }_1,{\zeta }_2\in \overline{\mathbb{D}}.$

Observe that for each $\theta \in \mathbb{R},$ $det{T}_{q,1}\left(f\right)=det{T}_{q,1}\left(f_{\theta }\right),$ where $f_{\theta }\left(z\right):={\mathrm{e}}^{-\mathrm{i}\theta }f\left({\mathrm{e}}^{\mathrm{i}\theta }z\right),z\in \mathbb{D},$ i.e., $det{T}_{q,1}\left(f\right)$ is rotationally invariant.

Recall now the following observation [15].

Theorem 1. 

Let $\mathcal{F}$ be a subclass of $\mathcal{A}$ such that $\{f\in \mathcal{F}:a_2=0\}\ne \varnothing$ and $A_2\left(\mathcal{F}\right):=max\left\{\right|a_2|:f\in \mathcal{F}\}$ exists. Then:

$$1-A_2^2\left(\mathcal{F}\right)\le det{T}_{2,1}\left(f\right)\le 1.$$

Both inequalities are sharp.

2. Main Results

We now compute the sharp bounds of $det{T}_{2,1}\left(f\right)$ and $det{T}_{3,1}\left(f\right)$ in the class of $\alpha$-convex functions.

Let $\alpha \in [0,1]$. In view of (11), we see that $A_2\left({\mathcal{M}}_{\alpha }\right)=2/(\alpha +1)$ with the extremal function $f\in {\mathcal{M}}_{\alpha }$ satisfying:

$$(1-\alpha )\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+\alpha \left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)=\frac{1+z}{1-z},z\in \mathbb{D},$$

(7)

i.e., for $\alpha \in (0,1],$ with the function:

$$f\left(z\right)={\left[\frac{1}{\alpha }{\int }_0^z\frac{{\zeta }^{\frac{1}{\alpha }-1}}{{(1-\zeta )}^{\frac{2}{\alpha }}}d\zeta \right]}^{\alpha },z\in \mathbb{D},$$

called $\alpha$-convex Koebe function, and for $\alpha =0$ with the Koebe function:

$$f\left(z\right)=\frac{z}{{(1-z)}^2},z\in \mathbb{D}.$$

Hence, and by observing that the identity function belongs to the class ${\mathcal{M}}_{\alpha }$, by Theorem 1, we have:

Theorem 2. 

Let $\alpha \in [0,1]$. If $f\in {\mathcal{M}}_{\alpha }$, then:

$$-\frac{(3+\alpha )(1-\alpha )}{{(1+\alpha )}^2}\le det{T}_{2,1}\left(f\right)\le 1.$$

Both inequalities are sharp.

Particularly, for $\alpha =0$ and $\alpha =1,$ i.e., for starlike and convex functions, we have the following result [14].

Corollary 1. 

• If $f\in {\mathcal{S}}^{*}$, then:

  $$-3\le det{T}_{2,1}\left(f\right)\le 1.$$
• If $f\in {\mathcal{S}}^c$, then:

  $$0\le det{T}_{2,1}\left(f\right)\le 1.$$

All inequalities are sharp.

Now, we will compute the upper and lower bounds of $det{T}_{3,1}\left(f\right).$

Theorem 3. 

Let $\alpha \in [0,1]$. If $f\in {\mathcal{M}}_{\alpha }$, then:

$$det{T}_{3,1}\left(f\right)\le \left\{\begin{array}{cc}{\displaystyle \frac{4({\alpha }^2+3\alpha +1)({\alpha }^2+4\alpha +2){(\alpha -1)}^2}{{(\alpha +1)}^4{(2\alpha +1)}^2}},\hfill & 0\le \alpha <{\alpha }_0,\hfill \\ 1,\hfill & {\alpha }_0\le \alpha \le 1,\hfill \end{array}\right.$$

(8)

where ${\alpha }_0=0.404744\dots$ is the unique root in $(0,1)$ of the equation:

$$33{\alpha }^4+96{\alpha }^3+38{\alpha }^2-16\alpha -7=0,$$

and:

$$det{T}_{3,1}\left(f\right)\ge -\frac{{(\alpha +2)}^2{(\alpha -1)}^2}{({\alpha }^2+5\alpha +2)(-{\alpha }^2+3\alpha +2)}.$$

(9)

All inequalities are sharp.

Proof. 

Fix $\alpha \in [0,1]$ and let $f\in {\mathcal{M}}_{\alpha }$ be of the form (1). Then, by (2):

$$(1-\alpha )\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+\alpha \left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)=p\left(z\right),z\in \mathbb{D},$$

(10)

for a certain $p\in$ $𝒫$ of the form (4). Substituting the series (1) and (4) into (10), by equating the corresponding coefficients, we obtain:

$$a_2=\frac{c_1}{\alpha +1},a_3=\frac{{(\alpha +1)}^2{c}_2+(3\alpha +1)c_1^2}{2{(\alpha +1)}^2(2\alpha +1)}.$$

(11)

Since the class ${\mathcal{M}}_{\alpha }$ and the determinant $det{T}_{3,1}$ are rotationally invariant, by (5), we may assume that $c_1\in [0,2],$ i.e., that ${\zeta }_1\in [0,1]$. Furthermore, (3) with (11), (5) and (6) give:

$$\begin{array}{cc}\hfill det{T}_{3,1}\left(f\right)=& \frac{1}{{(\alpha +1)}^4{(2\alpha +1)}^2}\hfill \\ \hfill & \times \left[-({\alpha }^2+8\alpha +3)({\alpha }^2-8\alpha -5){\zeta }_1^4-8{(2\alpha +1)}^2{(\alpha +1)}^2{\zeta }_1^2\right.\hfill \\ \hfill & +{(\alpha +1)}^4{(2\alpha +1)}^2+2{(\alpha +1)}^3(1-\alpha ){\zeta }_1^2(1-{\zeta }_1^2)\mathrm{Re}({\zeta }_2)\hfill \\ \hfill & \left.-{(\alpha +1)}^4{(1-{\zeta }_1^2)}^2{\left|{\zeta }_2\right|}^2\right]\hfill \end{array}$$

(12)

for some ${\zeta }_1,{\zeta }_2\in \overline{\mathbb{D}}.$

Define:

$$\begin{array}{cc}\hfill F(x,y,t):=& -({\alpha }^2+8\alpha +3)({\alpha }^2-8\alpha -5)x^2-8{(2\alpha +1)}^2{(\alpha +1)}^{2}x\hfill \\ \hfill & +{(\alpha +1)}^4{(2\alpha +1)}^2+2{(\alpha +1)}^3(1-\alpha )(1-x)xycost\hfill \\ \hfill & -{(\alpha +1)}^4{(1-x)}^2{y}^2\hfill \end{array}$$

for $x,y\in [0,1]$ and $t\in [0,2\pi ].$

If ${\zeta }_2\ne 0,$ then ${\zeta }_2=\left|{\zeta }_2\right|{\mathrm{e}}^{\mathrm{i}\theta }$ for a unique $\theta \in [0,2\pi ).$ Thus, by (12):

$${(\alpha +1)}^4{(2\alpha +1)}^{2}det{T}_{3,1}\left(f\right)=F({\zeta }_1^2,|{\zeta }_2|,\theta ).$$

(13)

If ${\zeta }_2=0,$ then by (12):

$${(\alpha +1)}^4{(2\alpha +1)}^{2}det{T}_{3,1}\left(f\right)=F({\zeta }_1^2,0,\theta )=F({\zeta }_1^2,0,0).$$

(14)

Therefore, we will find the maximum and minimum value of $F.$

A. First, we show the inequality (8). We have:

$$\begin{array}{cc}\hfill F(x,y,t)& \le F(x,y,0)\hfill \\ \hfill & =-({\alpha }^2+8\alpha +3)({\alpha }^2-8\alpha -5)x^2-8{(2\alpha +1)}^2{(\alpha +1)}^{2}x\hfill \\ \hfill & +{(\alpha +1)}^4{(2\alpha +1)}^2+2{(\alpha +1)}^3(1-\alpha )(1-x)xy\hfill \\ \hfill & -{(\alpha +1)}^4{(1-x)}^2{y}^2=:G(x,y)\hfill \end{array}$$

(15)

for $x,y\in [0,1]$ and $t\in [0,2\pi ].$

A1. For $x=1:$

$$G(1,y)=4{(\alpha -1)}^2({\alpha }^2+3\alpha +1)({\alpha }^2+4\alpha +2),y\in [0,1].$$

(16)

A2. Let $x\in [0,1).$ Set:

$$y_w:=\frac{(1-\alpha )x}{(1+\alpha )(1-x)},x\in [0,1).$$

(a) Note that $y_w=0$ if and only if $\alpha =1.$ Then, by (15) with $\alpha =1$ we obtain:

$$F(x,y_w,0)=F(x,0,0)=144{(x-1)}^2\le 144,x\in [0,1).$$

(17)

(b) Assume now that $\alpha \in [0,1).$

Observe that $y_w>1$ if and only if $(1+\alpha )/2<x<1.$ Then:

$$\begin{array}{cc}\hfill G(x,y)& \le G(x,1)\hfill \\ \hfill & =4(5\alpha +3)(3\alpha +1)x^2-4(8{\alpha }^2+7\alpha +1){(\alpha +1)}^{2}x+4\alpha {(\alpha +1)}^5\hfill \\ \hfill & \le max\left\{G\left(\frac{\alpha +1}{2},1\right),G(1,1)\right\}\hfill \\ \hfill & =max\left\{{(\alpha -1)}^2{(2\alpha +1)}^2{(\alpha +1)}^2,4({\alpha }^2+3\alpha +1)({\alpha }^2+4\alpha +2){(\alpha -1)}^2\right\}\hfill \end{array}$$

for $x\in [0,1).$ Since the inequality:

$${(1-\alpha )}^2{(1+2\alpha )}^2{(1+\alpha )}^2\le 4{(1-\alpha )}^2({\alpha }^2+3\alpha +1)({\alpha }^2+4\alpha +2)$$

is equivalent to:

$${(1-\alpha )}^2(16{\alpha }^3+47{\alpha }^2+34\alpha +7)\ge 0,$$

which holds for $\alpha \in [0,1),$ we deduce that:

$$G(x,1)\le G(1,1)=4{(\alpha -1)}^2({\alpha }^2+3\alpha +1)({\alpha }^2+4\alpha +2),\frac{1}{2}(1+\alpha )<x<1.$$

(18)

Now, we consider the case $y_w\le 1$ which holds only for $0\le x\le (\alpha +1)/2.$ Then:

$$\begin{array}{cc}\hfill G(x,y)& \le G(x,y_w)\hfill \\ \hfill & =16{(2\alpha +1)}^2{x}^2-8{(2\alpha +1)}^2{(\alpha +1)}^{2}x+{(\alpha +1)}^4{(2\alpha +1)}^2\hfill \\ \hfill & \le max\left\{G\left(0,y_w\right),G\left(\frac{\alpha +1}{2},y_w\right)\right\}\hfill \\ \hfill & =max\left\{{(\alpha +1)}^4{(2\alpha +1)}^2,{(\alpha -1)}^2{(2\alpha +1)}^2{(\alpha +1)}^2\right\}\hfill \end{array}$$

for $0\le x\le (\alpha +1)/2.$ Since the inequality:

$${(\alpha +1)}^4{(2\alpha +1)}^2\ge {(\alpha -1)}^2{(2\alpha +1)}^2{(\alpha +1)}^2$$

is equivalent to:

$$\alpha {(2{\alpha }^2+3\alpha +1)}^2\ge 0,$$

which holds for $\alpha \in [0,1),$ we deduce that:

$$G(x,y_w)\le G(0,y_w)={(\alpha +1)}^4{(2\alpha +1)}^2,0\le x\le \frac{1}{2}(\alpha +1).$$

(19)

A3. Comparing (16), (18) and (19) with $\alpha \in [0,1),$ leads to the inequality:

$$4{(\alpha -1)}^2({\alpha }^2+3\alpha +1)({\alpha }^2+4\alpha +2)\ge {(\alpha +1)}^4{(2\alpha +1)}^2$$

equivalently written:

$$-33{\alpha }^4-96{\alpha }^3-38{\alpha }^2+16\alpha +7\ge 0,$$

which holds for $\alpha \in [0,{\alpha }_0],$ where ${\alpha }_0=0.404744\dots$ Hence, by (13)–(15), (17), we obtain (8).

B. Now, we show the inequality (9). We have:

$$\begin{array}{cc}\hfill F(x,y,t)& \ge F(x,y,\pi )\hfill \\ \hfill & =-({\alpha }^2+8\alpha +3)({\alpha }^2-8\alpha -5)x^2-8{(2\alpha +1)}^2{(\alpha +1)}^{2}x\hfill \\ \hfill & +{(\alpha +1)}^4{(2\alpha +1)}^2-2{(\alpha +1)}^3(1-\alpha )(1-x)xy\hfill \\ \hfill & -{(\alpha +1)}^4{(1-x)}^2{y}^2=:H(x,y)\hfill \end{array}$$

(20)

for $x,y\in [0,1]$ and $t\in [0,2\pi ].$

B1. For $x=1:$

$$\begin{array}{cc}\hfill H(1,y)& =4{\alpha }^6+20{\alpha }^5+8{\alpha }^4-52{\alpha }^3-12{\alpha }^2+24\alpha +8\hfill \\ \hfill & =4({\alpha }^2+3\alpha +1)({\alpha }^2+4\alpha +2){(\alpha -1)}^2\ge 0,y\in [0,1].\hfill \end{array}$$

(21)

B2. Let $x\in [0,1).$ Set:

$$y_w^{\prime }:=-\frac{(1-\alpha )x}{(\alpha +1)(1-x)}.$$

Since $y_w^{\prime }\le 0,$ we have:

$$\begin{array}{cc}\hfill H(x,y)\ge H(x,1)=& 4({\alpha }^2+5\alpha +2)(-{\alpha }^2+3\alpha +2)x^2\hfill \\ \hfill & -4(7{\alpha }^2+7\alpha +2){(\alpha +1)}^{2}x+4\alpha {(\alpha +1)}^5,x\in [0,1).\hfill \end{array}$$

For $\alpha \in [0,1],$ let:

$$x_w^{\prime }:=\frac{(7{\alpha }^2+7\alpha +2){(\alpha +1)}^2}{2({\alpha }^2+5\alpha +2)(-{\alpha }^2+3\alpha +2)}.$$

(22)

We see that $x_w^{\prime }>0,$ and $x_w^{\prime }\le 1$ if and only if:

$$(1-\alpha )({\alpha }^2+5\alpha +2)(-{\alpha }^2+3\alpha +2)(9{\alpha }^3+34{\alpha }^2+27\alpha +6)\ge 0,$$

which holds for $\alpha \in [0,1).$ Therefore:

$$\begin{array}{c}\hfill H(x,1)\ge H(x_w^{\prime },1)=-\frac{{(\alpha -1)}^2{(2\alpha +1)}^2{(\alpha +2)}^2{(\alpha +1)}^4}{({\alpha }^2+5\alpha +2)(-{\alpha }^2+3\alpha +2)}\end{array}$$

(23)

for $x\in [0,1).$

B3. Comparing (21) and (23) from (20), (13) and (14), the inequality (9) is obtained.

C. It remains to show the sharpness of all results. The first inequality in (8) is sharp for the function $f\in {\mathcal{M}}_{\alpha }$ given by (7) for which by (11),

$$a_2=\frac{2}{\alpha +1},a_3=\frac{{(\alpha +1)}^2+2(3\alpha +1)}{{(\alpha +1)}^2(2\alpha +1)}.$$

The identity function is extremal for the second inequality in (8).

Set $\tau :=\sqrt{x_w^{\prime }},$ where $x_w^{\prime }$ is given by (22). Since, as was noticed in B2, $0<x_w\le 1$ for $\alpha \in [0,1],$ the function:

$$\tilde{p}\left(z\right):=\frac{1-z^2}{1-2\tau z+z^2}=1+2\tau z+(4{\tau }^2-2)z^2+\dots ,z\in \mathbb{D},$$

(24)

belongs to $𝒫$. Thus, the function f given by (10), with $p:=\tilde{p},$ being of the form (1) with:

$$a_2=\frac{2\tau }{\alpha +1},a_3=\frac{2({\alpha }^2+5\alpha +2){\tau }^2-{(1+\alpha )}^2}{{(\alpha +1)}^2(2\alpha +1)}$$

belongs to ${\mathcal{M}}_{\alpha }$ and is extremal for the inequality (9). □

Particularly, for $\alpha =0$ and $\alpha =1$ we obtain the following result [14].

Corollary 2. 

• If $f\in {\mathcal{S}}^{*}$, then:

  $$-1\le det{T}_{3,1}\left(f\right)\le 8.$$
• If $f\in {\mathcal{S}}^c$, then:

  $$0\le det{T}_{3,1}\left(f\right)\le 1.$$

All inequalities are sharp.