Abstract
The sharp bounds for the fourth-order Hermitian Toeplitz determinant over the class of convex functions are computed.
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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
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
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
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
and the fourth Toeplitz determinant \(|T_{4,1}(f)|\) is given by
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
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
and
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
Both inequalities are sharp with equalities attained by
and by the identity, respectively.
Proof
Let \(f\in \mathcal S^c\) be the form (1). Then by (2),
for some \(p\in \mathcal P\) of the form (5). Substituting the series (1) and (5) into (10) by equating the coefficients we get
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
Now by using (6) and (7) we get
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
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
Hence by using (6) and (7) after tedious computation we get
A. Suppose that \(\zeta =0.\) Then
Suppose that \(\eta =0.\) Then
We have
Hence and from (15) it follows that \(|T_{4,1}(f)|\ge 0.\)
We have
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
where
For \(c\in [0,2]\) and \(x,y\in (0,1]\) we have
where for \(x,y\in [0,1],\)
and
C. Let us observe that
Hence and from (16) with (17) it follows that
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 \(y\in (0,1],\)
Let now \(x\in (0,1).\) Then
Therefore we consider two cases.
D1. Assume that \(y_w< 1\), i.e., equivalently that \(x\in (0,x_1(c)),\) where
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
where
-
(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 \(c\in (0,2),\)
We will show that \(\gamma \) increasis. Note that
is equivalent to
where for \(t\in {\mathbb {R}},\)
and
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
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
Indeed, the above inequality is equivalent to
where
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
which confirms (21), (22), and ends the proof of (19) so of (18).
Summarizing,
-
(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
has solutions \((c_i,x_i), i=1,2,3,\) only when
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
where for \((c,x)\in \varDelta _2,\)
-
(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
has the solution \(c=x=0\) evidently. Let \(x\not =0\) and \(x\not =3.\) From the first equation we get
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
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
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).
\(\square \)
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Lecko, A., Sim, Y.J. & Śmiarowska, B. The fourth-order Hermitian Toeplitz determinant for convex functions. Anal.Math.Phys. 10, 39 (2020). https://doi.org/10.1007/s13324-020-00382-3
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DOI: https://doi.org/10.1007/s13324-020-00382-3