Abstract
The sharp lower and upper bounds of the Hermitian Toeplitz determinants of the second and third order for certain close-to-star functions are computed.
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1 Introduction
Let \(\mathbb D:=\{ z \in \mathbb {C} : \vert z\vert <1 \}\) and \(\overline{\mathbb D}:=\{z\in \mathbb C: \vert z\vert \le 1\}.\) Given \(r>0,\) let \(\mathbb T_r:=\{ z \in \mathbb {C} : \vert z\vert =r\}\) and \(\mathbb T:=\mathbb T_1.\) Let \(\mathcal {H}\) denote the class of all analytic functions in \(\mathbb {D}\) and \(\mathcal {A}\) be its subclass of all functions f normalized by \(f(0)=0\) and \(f'(0)=1,\) i.e., of the form
Let \(\mathcal {S}\) be the subclass of \(\mathcal {A}\) of univalent functions.
Given \(q,n\in \mathbb {N},\) define the matrix \(T_{q,n}(f)\) of \(f\in \mathcal {A}\) of the form (1.1) by
where \(\overline{a}_k:=\overline{a_k}.\) When \(a_n\) is a real number, \(T_{q,n}(f)\) is the Hermitian Toeplitz matrix. In particular, any matrix \(T_{q,1}(f)\) so is.
In recent years, many authors studied 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 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 (see, e.g., [5,6,7, 12, 14, 16,17,18,19, 23]. The determinants of symmetric Toeplitz matrices, the study of which was initiated in [1], are another example of such interest.
In [8, 10, 13], the study to estimate the determinants \(T_{q,n}(f)\) whose entries are coefficients of functions in subclasses of \(\mathcal {A}\) was initiated. As it is well known, Hermitian Toeplitz matrices play an important role in functional analysis, applied mathematics and in technical sciences. Further results in this direction were obtained in [15].
In this paper, we continue this research by computing the sharp upper and lower bounds of determinants \(T_{2,1}(f)\) and \(T_{3,1}(f)\) over subclasses of close-to-star functions.
It can be observed that for each \(\theta \in \mathbb {R},\) \(\det T_{q,1}(f)=\det T_{q,1}(f_\theta ),\) where \(f_\theta (z):=\mathrm {e}^{-\mathrm {i}\theta }f(\mathrm {e}^{\mathrm {i}\theta }z),\ z\in \mathbb {D},\) i.e., \(\det T_{q,1}(f)\) is rotation invariant.
Theorem 1.1
[8] Let \(\mathcal {F}\) be a subclass of \(\mathcal {A}\) such that \(\{f\in \mathcal {F}: a_2=0\}\not =\emptyset \) and \(A_2(\mathcal {F}):=\max \{\vert a_2\vert : f\in \mathcal {F}\}\) exists. Then,
Both inequalities are sharp.
Let \(\mathcal {S}^*\) denote the subclass of \(\mathcal {S}\) of starlike functions, i.e., \(f\in \mathcal {S}^*\) if \(f\in \mathcal {A}\) and
A function \(f\in \mathcal {A}\) is called close-to-star if there exist \(g\in \mathcal {S}^*\) and \(\beta \in \mathbb {R}\) such that
Recall that the class \(\mathcal {CST}\) of all close-to-star functions introduced by Reade [24] bear the same relation to the class of close-to-convex functions as the class of starlike functions bear to the class of convex functions ([24, p. 61])). This relationship is called Alexander type. Namely, \(f\in \mathcal {CST}\) if and only if a function
is close-to-convex ([11, 9, Vol. II, p. 3]). The class of close-to-convex functions was introduced by Kaplan [11], where the following geometrical interpretation was shown: \(f\in \mathcal {A}\) is close-to-convex if and only if there are no sections of the curve \(f(\mathbb {T}_r),\) for every \(r\in (0,1),\) in which the tangent vector turns backward through an angle not less then \(\pi \) (cf. [9, Vol. II, p. 4]). An analogous geometrical interpretation for close-to-star functions was shown by Reade: \(f\in \mathcal {A}\) is close-to-star if and only if there are no sections of the curve \(f(\mathbb {T}_r),\) for every \(r\in (0,1),\) in which the radius vector to the curve \(f(\mathbb {T}_r)\) turns backward through an angle not less than \(\pi \) ( [24, p. 61]). Recall also that Lewandowski [20] proved that the class of close-to-convex functions is identical to the class of linearly accessible functions introduced by Biernacki [2].
The class of close-to-star functions and their subclasses were studied by various authors (e.g., MacGregor [21], Sakaguchi [26], Causey and Merkes [4]; for further references, see [9, Vol. II, pp. 97–104]).
Given \(g\in \mathcal {S}^*\) and \(\beta \in \mathbb R,\) let \(\mathcal {CST}_\beta (g)\) are the subclass of \(\mathcal {CST}\) of all f satisfying (1.2). The four classes \(\mathcal {CST}_0(g_i),\ i=1,\dots ,4\) where
and
are particularly interesting and were separately studied by various authors. In [10], the sharp bounds of the second- and third-order Hermitian Toeplitz determinants were computed for the classes \(\mathcal {CST}_0(g_1)\) and \(\mathcal {CST}_0(g_2).\) In this paper, we will estimate the second- and third-order Hermitian Toeplitz determinants for the other two classes, i.e., for \(\mathcal {CST}_0(g_3)=:\mathcal F_1\) and \(\mathcal {CST}_0(g_4)=:\mathcal F_2,\) in which elements f in view of (1.2) satisfy the condition
and
respectively. Let us add that every function f in \(\mathcal F_1\) analytic in the closed disk \(\overline{\mathbb {D}}\) is starlike in one direction as defined by Robertson [25] (see also [9, Vol. I, pp. 207–208] and [9, Vol. I, p. 210, Theorem 27] with \(\mu =0\) and \(\nu =2\pi /3\))), i.e., intersection of \(f(\mathbb {D})\) with the real axis is a line segment covered by f univalently. Note that
Let \(\mathcal {P}\) be the class of all \(p\in \mathcal {H}\) of the form
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 \(c_2\) ([3, 22, p. 166]) and further remarks in [7]). In fact, the formulas below follow also from the Carathéodory–Toeplitz theorem.
Lemma 1.2
If \(p \in {\mathcal P}\) is of the form (1.6), then
Moreover,
and
for some \(\zeta _i \in \overline{\mathbb {D}}\), \(i \in \{ 1,2 \}\).
For \(\zeta _1 \in \mathbb {T}\), there is a unique function \(p \in {\mathcal P}\) with \(c_1\) as in (1.8), namely,
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 (1.8) and (1.9), namely,
2 The Class \({\mathcal {F}}_1\)
Let \(f \in {\mathcal {F}}_1\) be the form (1.1). Then by (1.3), there exists \(p\in {\mathcal P}\) of the form (1.6) such that
Substituting the series (1.1) and (1.6) into (2.1) by equating the coefficients, we get
By (1.7), it follows that \(A_2({\mathcal {F}}_1)=3\) for the function \(f\in \mathcal {A}\) defined by
which is extremal. Observe also that \(a_2=0\) for the function \(f\in {\mathcal {F}}_1\) defined by
Therefore by Theorem 1.1, we have
Theorem 2.1
If \(f\in \mathcal {F}_1\), then
Both inequalities are sharp.
Now, we will compute the bounds of \(\det T_{3,1}(f)\) in the class \(\mathcal {F}_1.\)
Theorem 2.2
If \(f \in \mathcal {F}_1,\) then
The inequality is sharp.
Proof
By (2.2) and (1.7), we see that \(\vert a_2\vert \le 3\) and \(\vert a_3\vert \le 4.\) Since \({{\,\mathrm{Re}\,}}(a_2^2 \overline{a}_3) \le \vert a_2\vert ^2\vert a_3\vert \) from (1.5) we get
where
Observe that the point (1, 1) is the unique solution \((0,3)\times (0,4)\) of the system of equations
However,
so (1, 1) is a saddle point of F.
We now consider F on the boundary of \([0,3]\times [0,4].\)
-
(1)
On the side \(x=0,\)
$$\begin{aligned} F(0,y)=1-y^2 \le 1,\quad 0\le y\le 4. \end{aligned}$$ -
(2)
On the side \(x=3,\)
$$\begin{aligned} F\left( 3,y\right) =-17+18y-y^2 \le F\left( 3,4\right) =39,\quad 0\le y\le 4. \end{aligned}$$ -
(3)
On the side \(y=0,\)
$$\begin{aligned} F(x,0)=1-2x^2 \le 1,\quad 0\le x\le 3. \end{aligned}$$ -
(4)
On the side \(y=4\)
$$\begin{aligned} F\left( x,4\right) =-15+6x^2 \le 39,\quad 0\le x\le 3. \end{aligned}$$
Therefore, the inequality \(F(x,y) \le 39\) holds for all \((x,y)\in [0,3]\times [0,4],\) which in view of (2.5) shows (2.4).
For the function (2.3), \(a_2=-3\) and \(a_3=4,\) which makes equality in (2.4). \(\square \)
Theorem 2.3
If \(f \in \mathcal {F}_1,\) then
The inequality is sharp.
Proof
Substituting (1.8) and (1.9) into (2.2), we get
with \(\zeta _i \in \overline{\mathbb {D}},\) \(i=1,2.\) Therefore from (1.5), we get
where
and
A. Let \(\zeta _1\zeta _2\not =0.\) Thus \(\zeta _1 =r \mathrm {e}^{\mathrm {i}\theta }\) and \(\zeta _2 = s \mathrm {e}^{\mathrm {i}\psi }\) with \(r,s\in (0,1]\) and \(\theta ,\psi \in [0,2\pi ).\) Then,
where
and
where \(\alpha \in \mathbb {R}\) is the quantity satisfying
with
Since \(\sin (\psi +\alpha ) \ge -1\) and \(s\le 1\), we have
Therefore, from (2.9) to (2.11), we get
Taking into account that \(\vert \zeta _2\vert \le 1\) from (2.8), we have
Thus from (2.7), (2.14) and (2.15), it follows that
where
with
and
for \(t\in [0,1]\) and \(x\in [-1,1].\)
Let \(\varOmega :=[0,1]\times [-1,1]\) and
Now, we will show that
A1. We now deal with the critical points of G in the interior of \(\varOmega \), i.e., in \((0,1)\times (-1,1).\) Note that \(g_2(t,x)\ge 0.\) Moreover, \(g_2(t,x)=0\) holds for \(t=\sqrt{2}/2\) and \(x=\sqrt{2}/2.\) A1.1. When \(t=\sqrt{2}/2\) and \(x=\sqrt{2}/2,\) from (2.8), (2.10) and (2.11), we have
which yields
A1.2. Suppose now that \(g_2(t,x)>0.\) Differentiating G with respect to x yields
A1.2.a. Assume first that \(\partial g_1/\partial x=0.\) Then from (2.18), it follows that \(\partial g_2/\partial x=0,\) which is possible only for \(t=\sqrt{2}/2\) and \(x=\sqrt{2}/2.\) Further argumentation is as in A1.1. A1.2.b Assume now that \(\partial g_1/\partial x\ne 0.\) Then we can write the equation (2.18) as
or equivalently by substituting (2.17) as
Furthermore, note that by (2.19),
and this inequality holds for
or
Differentiating G with respect to t and using (2.19) yield
where for \((t,x)\in (0,1)\times (-1,1),\)
Therefore, each critical point of G satisfies
or
I. Assume that (2.22) holds. Then, \( x=x(t)= t. \) Thus by (2.20), we see that
occurs only when \(t=\hat{t}_i,\ i=1,2,\) where \( \hat{t}_1 := \sqrt{3}/2\) and \(\hat{t}_2 = \sqrt{2}/2.\) Thus,
However, it can be seen that
which means that the inequality (2.21) is not satisfied for \(t=\hat{t}_1\) and \(x=\hat{x}_1\).
Note that the case \(\hat{t}_2=\sqrt{2}/2\) and \(\hat{x}_2=\sqrt{2}/2\) was considered in A1.1.
Therefore, G does not have critical point in the interior of \(\varOmega \) in the case of (2.22).
II. Suppose that (2.23) is satisfied. The equation (2.23) as a quadratic one of x with \(\Delta :=17-44t^2+36t^4>0,\ t\in (0,1),\) has two roots, namely,
a. Let \(x=x_1.\) Then, \(\varPhi (t,x_1(t))=0\) is equivalent to the equation
Squaring both sides of (2.25) leads to
where for \(t\in (0,1),\)
We see that there is a unique root \(t=\sqrt{2}/2\) of the equation (2.26), which does not satisfy (2.25). b. Let \(x=x_2.\) Then \(\Phi (t,x_2(t))=0\) is equivalent to the equation
Squaring the both sides of (2.27) yields again the equation (2.26) having a unique root \(t=\sqrt{2}/2,\) which satisfies (2.27) also. Since by (2.24), \(x_2(t_2)=\sqrt{2}/2,\) we see that this case was considered in A1.1.
A2. It remains to consider G in the boundary of \(\Omega \).
-
(1)
On the side \(t=0\),
$$\begin{aligned} G(0,x) \equiv -9 > \Theta ,\quad x\in [-1,1]. \end{aligned}$$ -
(2)
On the side \(t=1\),
$$\begin{aligned} G(1,x)=11-20x+8x^2\ge G(1,1) = -1 > \Theta ,\quad x\in [-1,1]. \end{aligned}$$ -
(3)
On the side \(x=-1\),
$$\begin{aligned} G(t,-1)=-9-12t+12t^2+32t^3+16t^4=:\varrho _1(t),\quad t\in [0,1]. \end{aligned}$$Since \(\varrho _1'(t) = 0\) occurs only when
$$\begin{aligned} t=t':=\frac{1}{4}\root 3 \of {4+2\sqrt{2}}+\frac{1}{2\root 3 \of {4+2\sqrt{2}}}-\frac{1}{2}=0.23784\dots \in [0,1] \end{aligned}$$(2.28)and \(\varrho _1''(t')>0,\) we have
$$\begin{aligned} \varrho _1(t) \ge \varrho _1(t')= \Theta , \quad t\in [0,1]. \end{aligned}$$ -
(4)
On the side \(x=1\), we have
$$\begin{aligned} G(t,1)=-9+12t+12t^2-32t^3+16t^4:=\varrho _2(t),\quad t\in [0,1]. \end{aligned}$$Since \(\varrho _2\) is increasing,
$$\begin{aligned} \varrho _2(t) \ge \varrho _2(0) = -9 > \Theta , \quad t\in [0,1]. \end{aligned}$$
B. Suppose that \(\zeta _1=0.\) Then,
and, therefore,
C. Suppose that \(\zeta _2=0\) and \(\zeta _1=r \mathrm {e}^{\mathrm {i}\theta },\) where \(r\in (0,1]\) and \(\theta \in [0,2\pi ).\) Then,
and
Thus,
where
for \(t\in (0,1]\) and \(x\in [-1,1].\) Set
Note that \(-1<x_w\) holds for \(t\in \left( (-2+\sqrt{10})/6, 1\right) ,\) and \(x_w<1\) holds for \(t\in \left( 0,(2+\sqrt{10})/6\right) .\) Hence for \(t\in (\left( -2+\sqrt{10})/6,(2+\sqrt{10})/6\right) ,\) we have
When \( t<(-2+\sqrt{10})/6,\) then
Since \(\phi _1'(t)=0\) occurs only when \(t=\left( -3+\sqrt{15}\right) /6= 0.145497\cdots \) and
it follows that
When \(t>(\sqrt{10}+ 2)/6,\) then
Since \(\phi _2\) is decreasing, we have
Summarizing, from Parts A–C, it follows that the inequality (2.6) holds.
It remains to show that the inequality (2.6) is sharp. It is observed from (2.7), (2.14), (2.15) and (2.16) that \(\det T_{3,1}(f) = \Theta \) holds when the following conditions are satisfied:
where \(t'\) is given by (2.28), and where \(\alpha \) is determined by the condition (2.12) with \(\kappa _1\) and \(\kappa _2\) given by (2.13). Then \(\theta =\pi ,\) \(\kappa _1 = 0\) and \(\kappa _2 = 1/8((4+2\sqrt{2})^{(2/3)}+(2-\sqrt{2})(4+2\sqrt{2})^{(1/3)}+8) = 1.588825\dots \) Thus, (2.12) is satisfied if we take \( \alpha = \pi /2. \) Thus if we put \( \psi = \pi , \) then \(\psi \) satisfies the fourth condition in (2.29). Now, let us consider a function \(\tilde{p}\) which has the form (1.10) with \(\zeta _1 = -t'\) and \(\zeta _2 = -1\), i.e.,
Since \(\zeta _1\in \mathbb {D}\) and \(\zeta _2 \in \mathbb {T}\), from Lemma 1.2 it follows that \(\tilde{p}\) belongs to \({\mathcal P}\). Therefore, the extremal function f in the class \(\mathcal F_1\) for which equality in (2.6) holds satisfies (2.1) with \(p=\tilde{p}.\) \(\square \)
3 The Class \(\mathcal {F}_2\)
Let \(f \in \mathcal {F}_2\) be the form (1.1). Then by (1.4), there exists \(p\in {\mathcal P}\) of the form (1.6) such that
Putting the series (1.1) and (1.6) into (3.1) by equating the coefficients, we get
By (1.7), it follows that \(A_2(\mathcal {F}_2)=3\) with the extremal function
Note that \(a_2=0\) for the function \(f\in \mathcal {F}_2\) given by
Thus by Theorem 1.1, we have
Theorem 3.1
If \(f\in \mathcal {F}_2\), then
Both inequalities are sharp.
Now, we estimate \(\det T_{3,1}(f)\) in the class \(\mathcal {F}_2.\)
Theorem 3.2
If \(f \in \mathcal {F}_2,\) then
The inequality is sharp.
Proof
By (3.2) and (1.7), we see that \(\vert a_2\vert \le 3\) and \(\vert a_3\vert \le 5.\) As in the proof of Theorem 2.2, the inequality (2.5) holds with the function
Repeating argumentation in the in the proof of Theorem 2.2, we see that the function F does not have any relative maxima in \((0,3)\times (0,5).\)
We consider F on the boundary of \([0,3]\times [0,5].\)
-
(1)
On the side \(x=0\),
$$\begin{aligned} F(0,y)=1-y^2 \le 1,\quad y\in [0,5]. \end{aligned}$$ -
(2)
On the side \(x=3\),
$$\begin{aligned} F(3,y)= -17+18y-y^2 \le F(3,5) = 48,\quad y\in [0,5]. \end{aligned}$$ -
(3)
On the side \(y=0\),
$$\begin{aligned} F(x,0)=1-2x^2 \le 1,\quad x\in [0,3]. \end{aligned}$$ -
(4)
On the side \(y=5\),
$$\begin{aligned} F(x,5)= -24 + 8x^2 \le F(3,5) = 48,\quad x\in [0,3]. \end{aligned}$$
Therefore, the inequality \(F(x,y) \le 48\) holds for all \((x,y)\in [0,3]\times [0,5],\) which in view of (2.5) shows (3.4).
For the function (3.3), \(a_2=3\) and \(a_3=5,\) which makes the equality in (3.4). \(\square \)
Theorem 3.3
If \(f \in \mathcal {F}_2,\) then
Proof
Substituting (1.8) and (1.9) into (3.2) yields
for some \(\zeta _i \in \overline{\mathbb {D}}\) (\(i=1,2\)). Therefore from (1.5), we get
where
and
A. Suppose that \(\zeta _1\zeta _2\not =0.\) Thus, \(\zeta _1 =r \mathrm {e}^{\mathrm {i}\theta }\) and \(\zeta _2 = s \mathrm {e}^{\mathrm {i}\psi }\) with \(r,s\in (0,1]\) and \(\theta ,\psi \in [0,2\pi ).\) Then
and
where \(\alpha \) is the quantity satisfying (2.12) with
Since \(\sin (\psi +\alpha )\ge -1\) and \(s\le 1,\) from (3.8) we have
Thus, from (3.6), (3.7) and (3.10), it follows that
where, for \(t\in [0,1]\) and \(x\in [-1,1],\)
with
Let \(\varOmega :=[0,1]\times [-1,1]\) and
Now, we will show that
A1. For this, we first we find the critical points of G in the interior of \(\varOmega \), i.e., in \((0,1)\times (-1,1).\) Note first that \(g(t,x)>0\) in \((0,1)\times (-1,1).\) From the equation
it follows that
Hence,
Under the condition (3.14), the equation (3.13) by substituting (3.12) can be equivalently written as
Differentiating G with respect to t leads to the equation
which by using (3.13) is equivalent to
The above equation is a quadratic one of x with \(\Delta =\Delta (t):=12-36t^2+25t^4,\) and \(\Delta (t)\ge 0\) if and only if \(t\in (0,t_1]\cup [t_2,1),\) where
For \(\Delta (t_1)= 0\), the equation (3.16) has a unique root
Analogously, for \(\Delta (t_2) = 0,\) the equation (3.16) has a unique root
It can be verified that the equation in (3.15) does not hold for \(t = t_1\) and \(x =x_0',\) and for \(t = t_2\) and \(x = x_0''.\)
Assume now that \(t\in (0, t_1)\cup (t_2, 1).\) Thus, there are two roots \(x_1\) and \(x_2\) of (3.16), namely
I. Consider the case \(x=x_1.\) Note that \(x_1>-1\) is equivalent to
Observe that \(-2+8t-5t^2 > 0\) if and only if \(t\in (t_3, 1),\) where
Thus for \(t\in (t_3, t_1)\cup (t_2, 1)\) by squaring the both sides of (3.17), we get the inequality
which holds if and only if \(t\in (t_4, t_1)\cup (t_2, 1)=:J_1,\) where \(t_4:=\left( 5+\sqrt{65}\right) /20=0.653112\cdots .\) Moreover, \(x_1 < 1\) is equivalent to the inequality
which is true for \(t\in J_1.\)
Substituting \(x = x_1\) into the equation (3.15), we get
where for \(t\in J_1,\)
Since \(Q_1(t) < 0\) for \(t\in J_1,\) and \( Q_2(t) < 0\) for \(t\in (t_4,t_5)\), where \(t_5:=\sqrt{10-2\sqrt{17}}/2=0.662153\dots ,\) by squaring both sides of (3.19) we equivalently get the equation
which has no solution for \(t\in (t_4,t_5).\)
II. Consider now the case \(x = x_2.\) Note that \(x_2 >-1\) is equivalent to
Since \(2-8t+5t^2 < 0\) for \(t\in (t_3, 1),\) where \(t_3\) is given by (3.18), consider \(t\in (0, t_3].\) By squaring both sides of (3.21), we get the inequality
which is true for \(t\in (0, t_3].\) Thus, \(x_2 >-1\) holds for all \(t\in (0, t_1)\cup (t_2, 1).\) Moreover, \(x_2 < 1\) is equivalent to the inequality
By squaring both sides, we equivalently get the inequality
which is true for \(t\in (t_6, t_1)\cup (t_2, 1)=:J_2,\) where \(t_6:=(-5+\sqrt{65})/20=0.153113\cdots .\) Thus, we restrict our consideration for \(t\in J_2.\)
Substituting \(x = x_2\) into the equation (3.15) yields the equation (3.20), which has three roots in \(J_2,\) namely,
Since \(x_2(\tilde{t}_1)=\tilde{x}_1=-\sqrt{2}/2,\) it follows that \(4\tilde{x}_1 + 3\tilde{t}_1 < 0\) which contradicts (3.14). Therefore, \((\tilde{t}_1,\tilde{x}_1)\) is not a critical point of G.
Since
it follows that
which contradicts (3.14). Therefore, \((\tilde{t}_2,\tilde{x}_2)\) is not a critical point of G.
Further,
Since
then \((\tilde{t}_3,\tilde{x}_3)\) satisfies (3.14), and therefore it is a unique critical point of G. Denote
Since \(\lambda _1<0\) and
the function G has a local maximum at \((\tilde{t}_3,\tilde{x}_3).\)
A2. It remains to consider G in the boundary of \(\varOmega \).
-
(1)
On the side \(t=0\),
$$\begin{aligned} G(0,x) \equiv 1 < \Theta ,\quad x\in [-1,1]. \end{aligned}$$ -
(2)
On the side \(t=1\),
$$\begin{aligned} G(1,x)=-2-6x-4x^2\le G\left( 1,-\frac{3}{4}\right) = \frac{1}{4} < \Theta ,\quad x\in [-1,1]. \end{aligned}$$ -
(3)
On the side \(x=-1\),
$$\begin{aligned} G(t,-1)=1+2t-7t^2+4t^3=:\varrho _1(t),\quad t\in [0,1]. \end{aligned}$$Since \(\varrho _1'(t)=0\) if and only if \(t=1/6\) or \( t=1,\) we see that
$$\begin{aligned} \varrho _1(t)\le \varrho _1\left( \frac{1}{6}\right) =\frac{125}{108}<\Theta ,\quad t\in [0,1]. \end{aligned}$$ -
(4)
On the side \(x=1\), we have
$$\begin{aligned} G(t,1)=1+2t-3t^2-8t^3-4t^4=:\varrho _2(t),\quad t\in [0,1]. \end{aligned}$$
Since \(\varrho _2'(t) = 0\) occurs only when \(t=(\sqrt{3}-1)/4\in (0,1)\) and \(\varrho _2''((\sqrt{3}-1)/4)=-6(1+\sqrt{3})<0,\) it follows that
B. Suppose that \(\zeta _1=0.\) Then,
C. Suppose that \(\zeta _2=0\) and \(\zeta _1=r \mathrm {e}^{\mathrm {i}\theta },\) where \(r\in (0,1]\) and \(\theta \in [0,2\pi ).\) Then,
and
Thus,
where
for \(t\in (0,1]\) and \(x\in [-1,1].\) Set
Since \(-1<x_w<0,\)
By the fact that \(\phi '(t) = 0\) occurs when \(t=\sqrt{6}/3=0.816496\cdots \) and \(\phi ''(\sqrt{6}/3)=16>0,\) we see that
Summarizing, from Parts A–C, it follows that the inequality (3.5) holds.
Now, we discuss the sharpness of (3.5). It is observed from (3.7), (3.10) and (3.11) that \(\det T_{3,1}(f) = -4\Theta \) holds when the following conditions are satisfied:
where \(\tilde{t}_3\) and \(\tilde{x}_3\) are given by (3.22) and (3.23), respectively, and \(\alpha \) is determined by the condition (2.12) with \(\kappa _1\) and \(\kappa _2\) given by (3.9). Set \(\theta =\mathrm{{Arccos}}(\tilde{x}_3)\) so that it satisfies the second condition in (3.24). Then, \(\kappa _1 = 0.884712\cdots >0\) and \(\kappa _2 = -0.542022\cdots <0\). Thus, (2.12) is satisfied if we take
Thus if we put
then \(\psi \) satisfies the fourth condition in (3.24). Now, let us consider a function \(\tilde{p}\) which has the form (1.10) with \(\zeta _1 = \tilde{t}_3\mathrm {e}^{\mathrm {i}\theta }\) and \(\zeta _2 = \mathrm {e}^{\mathrm {i}\psi }\). Since \(\zeta _1\in \mathbb {D}\) and \(\zeta _2 \in \mathbb {T}\), in view of Lemma 1.2, we see that \(\tilde{p}\) belongs to the class \({\mathcal P}\). Taking now a function
we see that \(\tilde{f} \in \mathcal {F}_2\) and \(\det T_{3,1}(\tilde{f})=-4\Theta ,\) which completes the proof. \(\square \)
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Lecko, A., Śmiarowska, B. Sharp Bounds of the Hermitian Toeplitz Determinants for Certain Close-to-Star Functions. Bull. Iran. Math. Soc. 48, 3077–3098 (2022). https://doi.org/10.1007/s41980-022-00687-y
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DOI: https://doi.org/10.1007/s41980-022-00687-y
Keywords
- Hermitian Toeplitz determinant
- Univalent function
- Close-to-star function
- Close-to-convex functgion
- Carathéodory class