1. Introduction
For $r \gt 0,$ let
$\mathbb{D}_r:=\{z\in\mathbb{C}: |z| \lt r \}$,
$\mathbb{D}:=\mathbb{D}_1$,
$\overline{\mathbb{D}}:=\{z\in\mathbb{C}:|z|\leqslant 1\}$ and let
$\mathbb{R} T:=\{z\in\mathbb{C}:|z|=1\}.$ Let
${\mathcal H}(\mathbb{D}_r)$ denote the class of all analytic functions f in
$\mathbb{D}_r$ and let
$\mathcal{H}:=\mathcal{H}(\mathbb{D}).$ Then
$f\in {\mathcal H}(\mathbb{D}_r)$ has the following representation
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn1.png?pub-status=live)
Let $\mathcal {A}(\mathbb{D}_r)$ be the subclass of
${\mathcal H}(\mathbb{D}_r)$ of all f normalized by
$f(0)=0=f'(0)-1$ and let
${\mathcal A}:={\mathcal A}(\mathbb{D}).$ By
${\mathcal S}$ we denote the subclass of all univalent (i.e. analytic and injective in
$\mathbb{D}$) functions in
${\mathcal A}$.
Given $\alpha\in (0,1],$ let
${\mathcal S}^*_{\alpha}$ denote class of all functions
$f\in {\mathcal A}$ such that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn2.png?pub-status=live)
and the so-called strongly starlike of order $\alpha.$ For
$\alpha:=1$ the class
$\mathcal{S}_1^*=:\mathcal{S}^*$ is the well-known class of starlike functions, i.e. functions f which map univalently
$\mathbb{D}$ onto a set which is star-shaped with respect to the origin. Then, the condition (1.2) can be written as
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU1.png?pub-status=live)
The class of strongly starlike functions was introduced by Stankiewicz [Reference Stankiewicz18] and [Reference Stankiewicz19] and independently by Brannan and Kirwan [Reference Brannan and Kirwan1] (see also [Reference Goodman6, Vol. I, pp. 137–142]). Stankiewicz [Reference Stankiewicz19] presented an external geometrical characterization of strongly starlike functions. Brannan and Kirwan found a geometrical condition called δ-visibility which is sufficient for functions to be strongly starlike. In turn, Ma and Minda [Reference Ma and Minda15] gave the internal characterization of functions in $\mathcal{S}_\alpha^*$ basing on the concept of k-starlike domains. Further results regarding the geometry of strongly starlike functions were presented in [Reference Lecko13, Chapter IV], [Reference Lecko14] and [Reference Sugawa20]. Since
$\mathcal{S}^*\subset \mathcal{S}$ (cf. [Reference Duren5, pp. 40–41]) and
${\mathcal S}^*_{\alpha}\subset \mathcal{S}^*$ for every
$\alpha\in(0,1]$, it follows that
${\mathcal S}^*_{\alpha}\subset \mathcal{S}$ for every
$\alpha\in(0,1].$
If $f\in\mathcal{S},$ then the inverse function
$F:=f^{-1}$ is well-defined and analytic in
$\mathbb{D}_{r(f)},$ where
$r(f):=\sup(\{r \gt 0:\mathbb{D}_r\subset f(\mathbb{D})\}).$ Thus
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn3.png?pub-status=live)
where $A_n:=a_n(F).$ By Koebe one-quarter theorem (e.g. [Reference Duren5, p. 31]), it follows that
$r(f) \geqslant 1/4$ for every
$f\in\mathcal{S}.$
For $f\in\mathcal{S}$ define
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU2.png?pub-status=live)
a logarithmic function associated with $f.$ The numbers
$\gamma_n:=a_n(F_f)$ are called the logarithmic coefficients of
$f.$ It is well-known that the logarithmic coefficients play a crucial role in Milin’s conjecture (see [Reference Milin16], [Reference Duren5, p. 155]).
Referring to the above idea, for $f\in\mathcal{S}$, there exists the unique function
$F_{f^{-1}}$ analytic in
$\mathbb{D}_{r(f)}$ such that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn4.png?pub-status=live)
where $\varGamma_n:=a_n\left(F_{f^{-1}}\right)$ are logarithmic coefficients of the inverse function
$f^{-1}.$
It follows from Equation (1.3) that (e.g. [Reference Goodman6, Vol. I, p. 57])
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn5.png?pub-status=live)
where $a_n:=a_n(f).$ Thus from Equation (1.4) we derive that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU3.png?pub-status=live)
and next using Equation (1.5) we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn6.png?pub-status=live)
For $q,n\in\mathbb{N},$ the Hankel matrix
$H_{q,n}(f)$ of
$f\in\mathcal{A}$ of the form (1.1) is defined as
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn7.png?pub-status=live)
In recent years, there has been a great deal of attention devoted to finding bounds for the modulus of the second and third Hankel determinants $\det H_{2,2}(f)$ and
$\det H_{3,1}(f)$, when f belongs to various subclasses of
$\mathcal{A}$ (see [Reference Cho, Kowalczyk, Kwon, Lecko and Sim2, Reference Kowalczyk, Lecko and Sim10, Reference Kowalczyk, Lecko and Thomas11] for further references).
Based on these ideas, in [Reference Kowalczyk and Lecko8] and [Reference Kowalczyk and Lecko9], the authors started the study the Hankel determinant $\det H_{q,n}(F_f)$ whose entries are logarithmic coefficients of
$f\in\mathcal{S},$ that is, an in Equation (1.7) are replaced by
$\gamma_n.$ In this paper, we continue analogous research considering the Hankel determinant
$\det H_{q,n}(F_{f^{-1}})$ whose entries are logarithmic coefficients of inverse functions, i.e. an in Equation (1.7) are now replaced by
$\varGamma_n.$ We demonstrate the sharp estimates of
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU4.png?pub-status=live)
in the classes $\mathcal{S}_\alpha^*$.
2. Preliminary lemmas
Denote by ${\mathcal P}$ the class of analytic functions
$p\in\mathcal{H}$ with positive real part given by
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn8.png?pub-status=live)
where $c_n:=a_n(p).$
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. [Reference Pommerenke17, p. 166]) and the formula for c 3 (see [Reference Cho, Kowalczyk and Lecko3, Lemma 2.4] with further remarks related to extremal functions).
Lemma 1. If $p \in {\mathcal P}$ is of the form (2.1), then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn9.png?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn10.png?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn11.png?pub-status=live)
for some $\zeta_1,\zeta_2, \zeta_3 \in \overline{\mathbb{D}}.$
For $\zeta_1 \in \mathbb{T}$, there is a unique function
$p \in {\mathcal P}$ with c 1 as in Equation (2.2), namely,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU5.png?pub-status=live)
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 Equations (2.2) and (2.3), namely,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn12.png?pub-status=live)
Lemma 2. ([Reference Choi, Kim and Sugawa4])
For real numbers A, B, C, let
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU6.png?pub-status=live)
I. If
$AC\geqslant 0,$ then
\begin{equation*} Y(A,B,C)=\left\{ \begin{array}{ll} |A|+|B|+|C|, & |B|\geqslant 2(1-|C|),\\ 1+|A|+\dfrac{B^2}{4(1-|C|)}, & |B| \lt 2(1-|C|). \end{array} \right. \end{equation*}
II. If
$AC \lt 0,$ then
\begin{equation*} Y(A,B,C)=\left\{ \begin{array}{lll} 1-|A|+\dfrac{B^2}{4(1-|C|)}, & -4AC(C^{-2}-1)\leqslant B^2 \wedge |B| \lt 2(1-|C|), \\ 1+|A|+\dfrac{B^2}{4(1+|C|)}, & B^2 \lt \min\left\{4(1+|C|)^2,-4AC(C^{-2}-1)\right\}, \\ R(A,B,C), & {\rm otherwise} , \end{array} \right. \end{equation*}
where
\begin{equation*} R(A,B,C):=\left\{ \begin{array}{lll} |A|+|B|-|C|, & |C|(|B|+4|A|)\leqslant |AB|,\\ -|A|+|B|+|C|, & |AB|\leqslant |C|(|B|-4|A|), \\ (|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}, & {\rm otherwise}. \end{array} \right. \end{equation*}
We recall now Laguerre’s rule of counting zeros of polynomials in an interval (see [Reference Jameson7], [Reference Laguerre12], [Reference Turowicz21, pp. 19–20]). We will apply Laguerre’s algorithm in the proof of the main theorem. Given a real polynomial
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU10.png?pub-status=live)
consider a finite sequence $(q_k), k = 0, 1,\dots, n,$ of polynomials of the form
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU11.png?pub-status=live)
For each $u_0\in\mathbb{R},$ let
$N(Q; u_0)$ denote the number of sign changes in the sequence
$(q_k(u_0)), k = 0, 1,\dots, n.$ Given an interval
$I \subset \mathbb{R},$ denote by
$Z(Q; I)$ the number of zeros of Q in I counted with their orders. Then the following theorem due to Laguerre holds.
Theorem 1. If a < b and $Q(a)Q(b)\neq 0,$ then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU12.png?pub-status=live)
or
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU13.png?pub-status=live)
is an even positive integer.
Note that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU14.png?pub-status=live)
Thus, when $[a, b] := [0, 1],$ Theorem 1 reduces to the following useful corollary.
Corollary 1. If $Q(0)Q(1)\neq 0,$ then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU15.png?pub-status=live)
or
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU16.png?pub-status=live)
is an even positive integer, where $N(Q; 0)$ and
$N(Q; 1)$ are the numbers of sign changes in the sequence of polynomial coefficients
$(d_k)$ and in the sequence of sums
$(\sum_{j=0}^k d_j ),$ where
$k = 0, 1,\dots , n,$ respectively.
3. Main result
The main result of this paper is the following.
Theorem 2. Let $\alpha\in(0,1].$ If
$f\in {\mathcal{S}}^*_{\alpha},$ then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn13.png?pub-status=live)
where $\alpha_0\approx 0.39059 $ is the unique root in
$(0,1]$ of the equation
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn14.png?pub-status=live)
All inequalities are sharp.
Proof. Let $f\in{\mathcal{S}}^*_{\alpha}$ be of the form (1.1). Then by Equation (1.2), there exists
$p\in\mathcal{P}$ of the form (2.1) such that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn15.png?pub-status=live)
Putting the series (1.1) and (2.1) into (3.3), by equating the coefficients we get
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn16.png?pub-status=live)
Hence and from Equation (1.6) we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU17.png?pub-status=live)
and therefore
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn17.png?pub-status=live)
Since both the class ${\mathcal{S}}^*_{\alpha}$ and
$|\det H_{2,1}\left(F_{f^{-1}}\right)|$ are rotationally invariant, without loss of generality we may assume that
$a_2\geqslant 0,$ which in view of Equation (3.4) yields
$ c_1 \geqslant 0,$ i.e. by Equation (2.2) that
$\zeta_1\in[0,1].$ Thus substituting Equations (2.2)–(2.4) into Equation (3.5), we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn18.png?pub-status=live)
for some $\zeta_1\in[0,1]$ and
$\zeta_2,\zeta_3 \in \overline{\mathbb{D}}$.
A. Suppose that $\zeta_1=0.$ Then from Equation (3.6),
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU18.png?pub-status=live)
B. Suppose that $\zeta_1=1.$ Then from Equation (3.6),
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU19.png?pub-status=live)
C. Suppose that $\zeta_1\in(0,1).$ Since
$\zeta_3 \in \overline{\mathbb{D}}$, from Equation (3.6) we get
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU20.png?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU21.png?pub-status=live)
with
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU22.png?pub-status=live)
Observe that AC < 0 and therefore we apply only the part II of Lemma 2.
C1. Let’s consider the condition $|B| \lt 2(1-|C|),$ i.e.
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU23.png?pub-status=live)
The above inequality is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn19.png?pub-status=live)
which is equivalent to $(5\alpha+1)\zeta_1^2-4\zeta_1+3 \lt 0.$ However
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU24.png?pub-status=live)
for $\zeta_1\in(0,1)$, which shows that the inequality (3.7) is false.
C2. Since
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU25.png?pub-status=live)
for $\zeta_1\in(0,1),$ we deduce that the condition
$B^2 \lt \min\{4(1+|C|)^2,-4AC(C^{-2}-1)\}$ is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn20.png?pub-status=live)
which is equivalent to $(10\alpha^2-1)\zeta_1^2+135\alpha^2+9 \lt 0$ for
$\zeta_1\in(0,1).$ However, in the case when
$10\alpha^2-1\geqslant 0$ we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU26.png?pub-status=live)
and in the case when $10\alpha^2-1 \lt 0$ we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU27.png?pub-status=live)
for all $\zeta_1\in(0,1).$ Thus the inequality (3.8) is false.
C3. The inequality $|C|(|B|+4|A|)\leqslant|AB|$ is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU28.png?pub-status=live)
which is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn21.png?pub-status=live)
where for $t\in\mathbb{R} ,$
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU29.png?pub-status=live)
Observe that the equation $175\alpha^3-70\alpha^2+35\alpha-8= 0$ has only one real root α 1 in
$(0,1],$ where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU30.png?pub-status=live)
and that the inequality (3.9) is false for $\alpha:=\alpha_1.$ Let now
$\alpha\in(0,1]\setminus\{\alpha_1\}.$ For φα, we have
$\Delta:=144(525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4) \gt 0,$ which is true for all
$\alpha\in(0,1]\setminus\{\alpha_1\}.$ Hence the square trinomial φα has two roots
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU31.png?pub-status=live)
Note that for all $\alpha\in(0,1]\setminus\{\alpha_1\}$ we have
$-6(35\alpha^2-5\alpha+4) \lt 0.$ Now for
$\alpha\in(\alpha_1,1]$ we have
$175\alpha^3-70\alpha^2+35\alpha-8 \gt 0.$ Hence
$t_2 \lt 0$ because the inequality
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU32.png?pub-status=live)
is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU33.png?pub-status=live)
which is true for all $\alpha\in(\alpha_1,1].$ On the other hand, the inequality
$t_1 \gt 1$ is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn22.png?pub-status=live)
which is evidently true for $\alpha\in(\alpha_1,\alpha_2],$ where
$\alpha_2\approx 0.82155,$ since then the right hand side of Equation (3.10) is non-positive. For
$\alpha\in(\alpha_2,1]$ by squaring both sides of Equation (3.10), we equivalently get the inequality
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU34.png?pub-status=live)
which is true for $\alpha\in(\alpha_2,1].$ Thus we conclude that for
$\alpha\in(\alpha_1,1]$ the inequality (3.9) is false.
Let $\alpha\in(0,\alpha_1).$ Then
$175\alpha^3-70\alpha^2+35\alpha-8 \lt 0$ and therefore
$t_1 \lt 0$ evidently. Moreover, the inequality
$t_2 \lt 0$ is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU35.png?pub-status=live)
which is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU36.png?pub-status=live)
which is true for all $\alpha\in(0,\alpha_1).$ Thus we conclude that for
$\alpha\in(0,\alpha_1)$ the inequality (3.9) is false.
C4. The inequality $|C|(|B|-4|A|)\geqslant|AB|$ is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU37.png?pub-status=live)
which is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn23.png?pub-status=live)
where for $t\in\mathbb{R} ,$
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU38.png?pub-status=live)
For γα we have $\Delta:=144 (525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4) \gt 0$ for all
$\alpha\in(0,1].$ Hence the square trinomial γα has two roots
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU39.png?pub-status=live)
Note that $t_4 \lt 0$ evidently. Observe now that
$t_3 \gt 0.$ Indeed, this inequality is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU40.png?pub-status=live)
which is equivalent to the evidently true inequality
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU41.png?pub-status=live)
Moreover, $t_3 \lt 1$ is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU42.png?pub-status=live)
that after squaring both sides is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU43.png?pub-status=live)
which is true for all $\alpha\in(0,1].$ Therefore the inequality (3.11) is true for
$\zeta_1\in(0,\zeta_1^0],$ where
$\zeta_1^0:=\sqrt{t_3}.$
Applying Lemma 2 for $0 \lt \zeta_1\leqslant \zeta_1^0$, we get
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU44.png?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU45.png?pub-status=live)
We have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU46.png?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU47.png?pub-status=live)
Note that for $\alpha\in(0,1/5]$ the equation
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn24.png?pub-status=live)
has no root in $(0,\zeta_1^0)$ and then evidently
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU48.png?pub-status=live)
For $\alpha\in(1/5,1]$, Equation (3.12) has a unique positive root, namely
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqn25.png?pub-status=live)
It remains to check the condition $t_5 \lt \zeta_1^0$ equivalently written as
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU49.png?pub-status=live)
which is equivalent to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU50.png?pub-status=live)
The last inequality is true for $\alpha\in(1/5,\alpha_3),$ where
$\alpha_3\approx0.812678$ is the unique root in
$(0,1)$ of the equation
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU51.png?pub-status=live)
Then ρα attains its maximum value on $(0,\zeta_1^0]$ at t 5 with
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU52.png?pub-status=live)
If $\alpha\in[\alpha_3,1],$ then evidently,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU53.png?pub-status=live)
C5. Applying Lemma 2 for $\zeta_1^0 \lt \zeta_1 \lt 1$, we get
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU54.png?pub-status=live)
where for $t\in[0,1],$
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU55.png?pub-status=live)
We have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU56.png?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU57.png?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU58.png?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU59.png?pub-status=live)
Note that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU60.png?pub-status=live)
Differentiating ψα leads to the equation
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU61.png?pub-status=live)
where for $s\in[0,1],$
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU62.png?pub-status=live)
Now we describe the number of zeros of Q in the interval $(0,1)$ by combining Descartes’ and Laguerre’s rules. To apply Descartes’ rule, we check the numbers of sign changes of coefficients of the polynomial
$Q.$ We have:
•
$ d_0(\alpha):=q_0(0)=4(35\alpha^2+1)(10\alpha^2-1) \gt 0$ iff
$\alpha\in\left(1/\sqrt{10},1\right),$
•
$ d_1(\alpha):=q_1(0)=3(175\alpha^4-315\alpha^2-4) \lt 0$ iff
$\alpha\in\left(0,1\right),$
•
$ d_2(\alpha):=q_2(0)=-18(1050\alpha^4+115\alpha^2-2) \gt 0$ iff
$\alpha\in\left(0,\alpha_4\right),$ where
\begin{equation*}\alpha_4:=\frac{1}{2}\sqrt{\frac{1}{105}(\sqrt{865}-23))}\approx 0.12355,\end{equation*}
•
$d_3(\alpha):=q_3(0)= 2295\alpha^2+108 \gt 0$ iff
$\alpha\in\left(0,1\right).$
Thus there is one change of signs in $\left(0,1/\sqrt{10}\right)$, i.e.
$N(Q,0)=1,$ and two changes of signs in
$\left[1/\sqrt{10},1\right),$ i.e.
$N(Q,0)=2.$ According to Descartes’ rule of signs, the polynomial Q has one positive real root in
$\left(0,1/\sqrt{10}\right)$ and zero or two positive real roots in
$\left[1/\sqrt{10},1\right).$
To apply Laguerres’ rule, it remains to compute the number $N(Q,1)$ of sign changes in the sequence of sums
$\sum_{j=0}^ku_j(\alpha)$, where
$k=0,1,2,3.$ We have
•
$d_0(\alpha)=4(35\alpha^2+1)(10\alpha^2-1) \gt 0$ iff
$\alpha\in\left(1/\sqrt{10},1\right),$
•
$d_0(\alpha)+d_1(\alpha)=1925 \alpha^4 - 1045\alpha^2 - 16 \gt 0$ iff
$\alpha\in\left(\alpha_5,1\right),$ where
\begin{equation*} \alpha_5:=\sqrt{(209 + 3\sqrt{5401})/770}\approx 0.74683, \end{equation*}
•
$d_0(\alpha)+d_1(\alpha)+d_2(\alpha)= -5 (3395 \alpha^4 + 623\alpha^2 - 4) \gt 0$ iff
$\alpha\in\left(0,\alpha_6\right),$ where
\begin{equation*} \alpha_6:=\sqrt{(3\sqrt{49161} - 623)/6790}\approx 0.078806, \end{equation*}
•
$d_0(\alpha)+d_1(\alpha)+d_2(\alpha)+d_3(\alpha)= -(35\alpha^2 + 4) (485\alpha^2 - 32) \gt 0$ iff
$\alpha\in\left(0,\alpha_7\right),$ where
$\alpha_7:=4\sqrt{2/485}\approx 0.25686.$
Thus there are no changes of signs in $\left(\alpha_7,1/\sqrt{10}\right)$, i.e.
$N(Q,1)=0$, and one change of sign in
$\left(0,\alpha_7\right]\cup\left[1/\sqrt{10},1\right)$ i.e.
$N(Q,1)=1$. According to Laguerre’s rule, the polynomial Q has one root in
$[0,1]$ for
$\alpha\in\left(\alpha_7,1\right)$, and no roots in
$[0,1]$ for
$\alpha\in\left(0,\alpha_7\right].$ Therefore, for
$\alpha\in\left(0,\alpha_7\right]$, the function ψα is increasing for
$\zeta_1^0 \lt t \lt 1$ and hence
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU66.png?pub-status=live)
In turn, for $\alpha\in\left(\alpha_7,1\right)$, the function ψα has a unique critical point in
$[0,1],$ where by using jointly Descartes’ and Laguerre’s rules we state that ψα attains its minimum value. Thus
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU67.png?pub-status=live)
Now we summarize results of sections C4 and C5.
(i) For
$\alpha\in\left(0,1/5\right),$ we compare
$\psi_\alpha(1)$ and
$\varrho_\alpha(0).$ Note that then
$\varrho_\alpha(0)\geqslant \psi_\alpha(1)$ since it is equivalent to
\begin{equation*}\frac14\alpha^2-\dfrac{1}{36}\alpha^2(35\alpha^2+4)=\frac{1}{36}\alpha^2(5-35\alpha^2)\geqslant 0.\end{equation*}
(ii) For
$\alpha\in\left[1/5,\alpha_3\right),$ we compare
$\psi_\alpha(1)$ and
$\varrho_\alpha(t_5).$ Note that the inequality
\begin{equation*} \frac{\alpha^2(15\alpha^2+5\alpha+2)}{35\alpha^2+30\alpha+7}\geqslant \frac{1}{36}\alpha^2(35\alpha^2+4) \end{equation*}
is equivalent to
\begin{equation*} \frac{1}{36}\alpha^2(1225\alpha^4+1050\alpha^3-155\alpha^2-60\alpha-44)\leqslant 0 \end{equation*}
which is true for
$\alpha\in[1/5,\alpha_0],$ where
$\alpha_0\approx0.390595$ is the unique root in
$(0,1]$ of Equation (3.2). Thus
$\varrho_\alpha(t_5)\geqslant \psi_\alpha(1)$ for
$\alpha\in[1/5,\alpha_0],$ and
$\varrho_\alpha(t_5) \lt \psi_\alpha(1)$ for
$\alpha\in(\alpha_0,\alpha_3).$
(iii) For
$\alpha\in\left[\alpha_3,1\right),$ we compare
$\psi_\alpha(1)$ and
$\rho_\alpha(\zeta_1^0).$ Note that the inequality
$\rho_\alpha(\zeta_1^0)\leqslant \psi_\alpha(1)$ is equivalent to
\begin{equation*} \begin{aligned} &-18375\alpha^6-16625\alpha^5-10150\alpha^4-3775\alpha^3 -1025\alpha^2-150\alpha-12\\ &+(1050\alpha^4+700\alpha^3+320\alpha^2+80\alpha+10)\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}\\ \leqslant& \frac{1}{72}(35\alpha^2+4)(175\alpha^3+70\alpha^2+35\alpha+8)^2, \end{aligned} \end{equation*}
equivalently written as
\begin{equation*} \begin{aligned} &(1050\alpha^4+700\alpha^3+320\alpha^2+80\alpha+10)\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}\\ \leqslant& \frac{1}{72}\left(1071875\alpha^8+857500\alpha^7+2045750\alpha^6 +1564500\alpha^5+881475\alpha^4+322200\alpha^3\right.\\ &\left.+85420\alpha^2+13040\alpha+1120\right), \end{aligned} \end{equation*}
which is equivalent to
\begin{equation*} \begin{aligned} \frac{25}{5184}&(35\alpha^2+4)(175\alpha^3+70\alpha^2+35\alpha+8)^2\times\\ &\times \left(42875\alpha^8+34300\alpha^7+134750\alpha^6 +110460\alpha^5+17835\alpha^4 -7344\alpha^3\right.\\ &\left.-5036\alpha^2-1120\alpha-128\right)\geqslant 0 \end{aligned} \end{equation*}
which is true for
$\alpha\in\left[\alpha_3,1\right).$
D. We now show sharpness of all inequalities by using the formula (3.5). In the first inequality in Equation (3.1), the equality is attained by the function $f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3) with
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU74.png?pub-status=live)
for which $c_1=c_3=0$ and
$c_2=-2.$
In the second inequality in Equation (3.1), the equality is attained by the function $f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3), where
$p\in\mathcal{P}$ is defined by Equation (2.5) with
$\zeta_1=t_5=:\tau$ and
$\zeta_2=1,$ i.e.
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU75.png?pub-status=live)
Here t 5 is described by Equation (3.13).
In the third inequality in Equation (3.1), the equality is attained by the function $f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3), where
$p\in\mathcal{P}$ is defined by
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU76.png?pub-status=live)
for which $c_1=c_2=c_3=2.$
This ends the proof of the theorem.
For α = 1, we have the following result:
Corollary 2. If $f\in {\mathcal{S}}^*,$ then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20241202182032052-0997:S0013091524000531:S0013091524000531_eqnU77.png?pub-status=live)
The inequality is sharp.