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Article

# The Sharp Bound of the Hankel Determinant of the Third Kind for Starlike Functions with Real Coefficients

by and
Department of Mathematics, Kyungsung University, Busan 48434, Korea
*
Author to whom correspondence should be addressed.
Mathematics 2019, 7(8), 721; https://doi.org/10.3390/math7080721
Received: 15 July 2019 / Revised: 28 July 2019 / Accepted: 3 August 2019 / Published: 8 August 2019
(This article belongs to the Special Issue Complex Analysis and Its Applications)

## Abstract

:
Let $SR *$ be the class of starlike functions with real coefficients, i.e., the class of analytic functions f which satisfy the condition $f ( 0 ) = 0 = f ′ ( 0 ) − 1$, $Re { z f ′ ( z ) / f ( z ) } > 0$, for $z ∈ D : = { z ∈ C : | z | < 1 }$ and $a n : = f ( n ) ( 0 ) / n !$ is real for all $n ∈ N$. In the present paper, it is obtained that the sharp inequalities $− 4 / 9 ≤ H 3 , 1 ( f ) ≤ 3 / 9$ hold for $f ∈ SR *$, where $H 3 , 1 ( f )$ is the third Hankel determinant of order 3 defined by $H 3 , 1 ( f ) = a 3 ( a 2 a 4 − a 3 2 ) − a 4 ( a 4 − a 2 a 3 ) + a 5 ( a 3 − a 2 2 )$.

## 1. Introduction

Let $H$ be the class of analytic functions in $D : = { z ∈ C : | z | < 1 }$ and let $A$ be the class of functions $f ∈ H$ normalized by $f ( 0 ) = 0 = f ′ ( 0 ) − 1$. That is, for $z ∈ D$, $f ∈ A$ has the following representation
$f ( z ) = z + ∑ n = 2 ∞ a n z n .$
For q, $n ∈ N$, the Hankel determinant $H q , n ( f )$ of functions $f ∈ A$ of the form (1) are defined by
$H q , n ( f ) = a n a n + 1 ⋯ a n + q − 1 a n + 1 a n + 2 ⋯ a n + q ⋮ ⋮ ⋮ ⋮ a n + q − 1 a n + q ⋯ a n + 2 q − 2 .$
Computing the upper bound of $H q , n$ over subfamilies of $A$ is an interesting problem to study. Note that $H 2 , 1 ( f ) = a 3 − a 2 2$ is the well-known functional which, for the class of univalent functions, was estimated by Bieberbach (see, e.g., [1] (Vol. I, p. 35)). Especially, the functional $H 3 , 1 ( f )$, Hankel determinant of order 3, is presented by
$H 3 , 1 ( f ) = a 1 a 2 a 3 a 2 a 3 a 4 a 3 a 4 a 5 = a 3 ( a 2 a 4 − a 3 2 ) − a 4 ( a 4 − a 2 a 3 ) + a 5 ( a 3 − a 2 2 ) .$
Let $S *$ be the class of starlike functions in $A$. That is, the class $S *$ consists of all functions $f ∈ A$ satisfying
$Re z f ′ ( z ) f ( z ) > 0 , z ∈ D .$
The leading example of a function of class $S *$ is the Koebe function k, defined by
$k ( z ) = z ( 1 − z ) − 2 = z + 2 z 2 + 3 z 3 + ⋯ , z ∈ D .$
In [2], Janteng et al. obtained the sharp inequality $| H 2 , 2 ( f ) | ≤ 1 = | H 2 , 2 ( k ) |$ for $f ∈ S *$. For the estimates on the Hankel determinant $H 3 , 1 ( f )$ over the class $S *$, Babalola [3] obtained the inequality $| H 3 , 1 ( f ) | ≤ 16$. And Zaprawa [4] improved the result by proving $| H 3 , 1 ( f ) | ≤ 1$. Next, Kwon et al. [5], recently found the inequality $| H 3 , 1 ( f ) | ≤ 8 / 9$ and we conjectured that
$| H 3 , 1 ( f ) | ≤ 4 / 9 , f ∈ S * .$
The sharp bound of $| H 3 , 1 ( f ) |$ over the class $S *$ is still open.
Let $SR *$ be the class of starlike functions in $A$ with real coefficients. Hence, if $f ∈ A$ belongs to the class $SR *$, then f has the form given by (1) with $a n ∈ R$, $n ∈ N ∖ { 1 }$ and satisfies the condition (3).
In this paper, we will prove the following.
Theorem 1.
If$f ∈ SR *$is the form (1), then the following inequalities hold:
$− 4 9 ≤ H 3 , 1 ( f ) ≤ 1 9 3 .$
The first inequality is sharp for the function$f = f 1 ∈ SR *$, where
$f 1 ( z ) : = z ( 1 − z 3 ) − 2 / 3 = z + 2 3 z 4 + 5 9 z 7 + ⋯ , z ∈ D .$
The second inequality is sharp for the function$f = f 2 ∈ SR *$, where
$f 2 ( z ) : = z exp − ∫ 0 z ( 2 / 3 ) ζ + 2 ζ 3 1 + ( 2 / 3 ) ζ 2 + ζ 4 d ζ = z − z 3 3 + 2 z 7 3 3 − 7 z 9 18 + ⋯ , z ∈ D .$

## 2. Preliminary Results

Let $P$ be the class of functions $p ∈ H$ of the form
$p ( z ) = 1 + ∑ n = 1 ∞ c n z n , z ∈ D ,$
having a positive real part in $D$, i.e., the Carathéodory class of functions. It is well known, e.g., [6] (p. 166), that for $p ∈ P$ with the form given by (6),
$2 c 2 = c 1 2 + ( 4 − c 1 2 ) ζ ,$
for some $ζ ∈ D ¯$. Moreover, the following lemma will be used for our investigation.
Lemma 1
([7]). The formula (7) with $c 1 ∈ [ 0 , 2 )$ and $ζ ∈ T$ holds only for the function $p ∈ P$ defined by
$p ( z ) = 1 + τ ( 1 + ζ ) z + ζ z 2 1 − τ ( 1 − ζ ) z − ζ z 2 , z ∈ D ,$
where $τ ∈ [ 0 , 1 )$.
Let $B 0$ be the subclass of $H$ of all self-mappings $ω$ of $D$ of the form
$ω ( z ) = ∑ n = 1 ∞ β n z n , z ∈ D ,$
i.e., the class of Schwarz functions. It is well known that $ω ∈ B 0$ if and only if $p = ( 1 + ω ) / ( 1 − ω ) ∈ P$. For coefficients of functions in $B 0$, the following properties, which can be found in [1] (Vol. I, pp. 84–85 and Vol. II, p. 78) and [8] (p. 128), will be used for our proof.
Lemma 2.
If$ω ∈ B 0$is of the form given by (8), then
(1)
$| β 1 | ≤ 1$,
(2)
$| β 2 | ≤ 1 − | β 1 | 2$,
(3)
$| β 3 ( 1 − | β 1 | 2 ) + β 1 ¯ β 2 2 | ≤ ( 1 − | β 1 | 2 ) 2 − | β 2 | 2$.
The following inequalities, which will be used, hold for the fourth coefficients for Schwarz functions with real coefficients.
Lemma 3
([9]). If $ω ∈ B 0$ is the form (8), $β n ∈ R ,$ $n ∈ N ,$ and $β 2 2 ≠ ( 1 − β 1 2 ) 2$, then
$Ψ L ≤ β 4 ≤ Ψ U ,$
where
$Ψ L : = 1 + β 1 4 + β 2 − β 2 2 − β 2 3 − 2 β 1 2 − β 1 2 β 2 + 2 β 1 β 2 β 3 − β 3 2 − 1 + β 1 2 − β 2$
and
$Ψ U : = 1 + β 1 4 − β 2 − β 2 2 + β 2 3 − 2 β 1 2 + β 1 2 β 2 − 2 β 1 β 2 β 3 − β 3 2 1 − β 1 2 − β 2 .$
For given a set A, let intA, clA and $∂$A be the sets of interior, closure and boundary, respectively, points of A. And let $R = [ 0 , 1 ] × [ − 1 , 1 ]$ be a rectangle in $R 2$. From now, we obtain several inequalities for functions, defined in subsets of R, which will be used in the proof of Theorem 1.
Proposition 1.
Define a function$F 1$by
$F 1 ( x , y ) = ∑ n = 0 4 b n ( x ) y n ,$
where
$b 4 ( x ) = ( 1 − x ) 2 ( 1 + x ) 4 , b 3 ( x ) = − x ( 1 + x ) 3 ( 10 − 11 x + x 2 ) , b 2 ( x ) = ( 1 + x ) 2 ( 7 − 16 x + 14 x 3 − 5 x 4 ) , b 1 ( x ) = x ( 10 + 9 x − 2 x 2 − 6 x 3 − 8 x 4 − 3 x 5 ) , b 0 ( x ) = − 8 + 16 x 2 + 6 x 3 − 8 x 4 − 6 x 5 .$
Then$F 1 ( x , y ) < 2 3$holds for all$( x , y ) ∈ R$.
Proof.
Let $( x , y ) ∈ R$. Since $b 4 ( x ) ≥ 0$, we have $b 4 ( x ) y 4 ≤ b 4 ( x ) y 2$ and
$F 1 ( x , y ) ≤ G ( x , y ) , ( x , y ) ∈ R ,$
where
$G ( x , y ) = b 3 ( x ) y 3 + ( b 4 ( x ) + b 2 ( x ) ) y 2 + b 1 ( x ) y + b 0 ( x ) .$
We will show that $G ( x , y ) < 2 3$ holds for $( x , y ) ∈ R$.
When $x = 0$, we have $G ( 0 , y ) = − 8 ( 1 − y 2 ) ≤ 0$, for $y ∈ [ − 1 , 1 ]$. And, when $x = 1$, we have $G ( 1 , y ) ≡ 0$.
Now, let $x ∈ ( 0 , 1 )$ be fixed and put $b i = b i ( x )$ ($i ∈ { 0 , 1 , 2 , 3 , 4 }$). Then $b 3 < 0$. Define a function $g x$ by $g x ( y ) = G ( x , y )$. Note that
$g x ( − 1 ) = 0 and g x ( 1 ) = 4 x 2 ( 1 − x 2 ) ( 5 − 2 x 2 ) ≤ 0 .$
Also,
$g x ′ ( y ) = 3 b 3 y 2 + 2 ( b 4 + b 2 ) y + b 1 = 0$
occurs at $y = ζ 1$ or $ζ 2$, where
$ζ i = − ( b 4 + b 2 ) + ( − 1 ) i + 1 ( b 4 + b 2 ) 2 − 3 b 1 b 3 3 b 3 , i ∈ { 1 , 2 } .$
It is trivial that $ζ 1 < 0 < ζ 2$. Furthermore, since $b 3 < 0$, $g x$ has the local minimum at $y = ζ 1$. Let $α = 0.322818 ⋯$ be a zero of polynomial q, where
$q ( y ) = 8 − 10 y − 42 y 2 − 14 y 3 + 7 y 4 .$
Note that $ζ 2 ≥ 1$ holds for x satisfying
$2 ( 1 − x 2 ) q ( x ) = b 1 + 2 ( b 4 + b 2 ) + 3 b 3 ≥ 0 .$
Hence we obtain
$ζ 2 ≥ 1 , when x ∈ ( 0 , α ] , ζ 2 ≤ 1 , when x ∈ [ α , 1 ) .$
(a) When $x ∈ ( 0 , α ]$, since $ζ 2 ≥ 1$, $g x$ is convex in $[ − 1 , 1 ]$. So, it holds that
$g x ( y ) ≤ max { g x ( − 1 ) , g x ( 1 ) } , y ∈ [ − 1 , 1 ] .$
Hence, by (13), we get $g x ( y ) ≤ 0 < 2 3$ for $y ∈ [ − 1 , 1 ]$.
(b) When $x ∈ [ α , 1 )$, $g x$ has its local maximum $g x ( ζ 2 )$. Using the fact that $ζ 2$ is a solution of the equation given by (14) leads us to
$g x ( ζ 2 ) = 2 3 b 1 − 2 ( b 2 + b 4 ) 2 9 b 3 ζ 2 + b 0 − b 1 ( b 2 + b 4 ) 9 b 3 .$
We claim that $g x ( ζ 2 ) − 3 < 0$ holds for all $x ∈ [ α , 1 )$. A compuation gives
$g x ( ζ 2 ) − 3 = 1 9 b 3 ( 1 − x ) ( 1 + x ) 3 [ − 2 ( 1 − x ) ( 1 + x ) κ 1 ζ 2 + x κ 2 ] ,$
where
$κ 1 = 64 − 128 x + 204 x 2 + 464 x 3 + 249 x 4 − 14 x 5 + 7 x 6$
and
$κ 2 = 910 − 11 x − 1340 x 2 − 414 x 3 + 752 x 4 + 398 x 5 − 64 x 6 + 12 x 7 .$
Since $b 3 < 0$, $g x ( ζ 2 ) − 3 < 0$ is equivalent to
$2 ( 1 − x 2 ) κ 1 ( b 4 + b 2 ) 2 − 3 b 1 b 3 < − 3 x κ 2 b 3 − 2 ( 1 − x 2 ) κ 1 ( b 4 + b 2 ) .$
We can see that the right-side of the above equation is positive for all $x ∈ [ α , 1 )$. Thus, by squaring both sides of (15), we have $g x ( ζ 2 ) < 0$ is equivalent to $Ψ > 0$, where
$Ψ = [ 3 x κ 2 b 3 + 2 ( 1 − x 2 ) κ 1 ( b 4 + b 2 ) ] 2 − 4 ( 1 − x 2 ) 2 κ 1 2 [ ( b 4 + b 2 ) 2 − 3 b 1 b 3 ] .$
By a simple calculation we have
$Ψ = − 27 x 2 ( 10 − x ) 2 ( 1 − x ) 2 ( 1 + x ) 6 Λ x ,$
where
$Λ x : = 22528 − 90112 x − 143980 x 2 + 177084 x 3 + 333021 x 4 − 21120 x 5 − 258308 x 6 − 143200 x 7 + 452 x 8 + 28728 x 9 + 37512 x 10 + 24288 x 11 + 9748 x 12 + 2720 x 13 + 968 x 14 − 48 x 15 + 36 x 16 .$
Since $Λ x < 0$ holds for all $x ∈ [ α , 1 )$, from (16), $Ψ > 0$, this implies
$g x ( ζ 2 ) < 3 .$
Finally, since
$g x ( y ) ≤ max { g x ( − 1 ) , g x ( 1 ) , g x ( ζ 2 ) } , y ∈ [ − 1 , 1 ] ,$
it follows from (13) and (17) that $g x ( y ) < 3 < 2 3$ holds for all $y ∈ [ − 1 , 1 ]$. Thus the proof of Proposition 1 is completed. □
Proposition 2.
Let
$Ω = ( x , y ) ∈ [ 0 , 1 / 2 ) × [ 0 , 1 ) : 0 ≤ x ≤ y 1 + y ⊂ R .$
Define a function$F 2 : Ω → R$by
$F 2 ( x , y ) = 1 − x 8 + y − x ( 17 + y ) H 1 ( x , y ) ,$
where$H 1 ( x , y ) = ∑ n = 0 3 d n ( y ) x n$with
$d 3 ( y ) = ( 1 + y ) 2 ( 1 − 6 y + y 2 ) , d 2 ( y ) = 17 + 24 y + 10 y 2 − 3 y 4 ,$
$d 1 ( y ) = − 8 − 26 y − y 2 + 12 y 3 + 3 y 4 and d 0 ( y ) = y ( 8 + y − 8 y 2 − y 3 ) .$
Then$F 2 ( x , y ) ≤ ( 2 / 9 ) 3$holds for all$( x , y ) ∈ Ω$.
Proof.
First of all, we note that $F 2$ is well-defined, since $8 + y − x ( 17 + y ) > 0$ holds for all $( x , y ) ∈ Ω$.
Differentiating $F 2$ with respect to x twice gives
$1 2 [ 8 + y − x ( 17 + y ) ] 3 ∂ 2 F 2 ∂ x 2 ( x , y ) = ∑ n = 0 4 d ˜ n ( y ) x n ,$
where
$d ˜ 4 ( y ) = − 3 ( 1 − 6 y + y 2 ) ( 17 + 18 y + y 2 ) 2 , d ˜ 3 ( y ) = − 4 ( 884 + 3197 y + 4605 y 2 + 2062 y 3 − 302 y 4 − 75 y 5 − 3 y 6 ) , d ˜ 2 ( y ) = 6 ( 1024 + 2344 y + 2421 y 2 + 956 y 3 − 202 y 4 − 60 y 5 − 3 y 6 ) , d ˜ 1 ( y ) = − 12 ( 8 + y ) 2 ( 4 + 7 y + 5 y 2 + y 3 − y 4 ) , d ˜ 0 ( y ) = 512 + 1088 y + 960 y 2 − 176 y 3 − 83 y 4 − 30 y 5 − 3 y 6 .$
Fix now $y ∈ [ 0 , 1 )$ and put $y 0 = y / ( 1 + y ) ∈ [ 0 , 1 / 2 )$. Let us define a function $g y : [ 0 , y 0 ] → R$ by $g y ( x ) = ∑ n = 0 4 d ˜ n ( y ) x n$. Then we have
$g y ′ ( x ) = − 12 ( 1 + y ) [ 8 + y − x ( 17 + y ) ] 2 φ ( x ) ,$
where
$φ ( x ) = 4 + 3 y + 2 y 2 − y 3 + ( 1 + y ) ( 1 − 6 y + y 2 ) x .$
Since $− 4 ≤ 1 − 6 y + y 2 ≤ 1$, we have
$φ ( x ) ≥ 4 + 3 y + 2 y 2 − y 3 − 4 x ( 1 + y ) ≥ 4 − y + 2 y 2 − y 3 > 0 , x ∈ [ 0 , y 0 ] .$
Thus, by (20), we get $g y ′ ( x ) < 0$, when $x ∈ [ 0 , y 0 ]$. So $g y$ is decreasing on the interval $[ 0 , y 0 ]$, which yields
$g y ( x ) ≥ g y ( y 0 ) = 64 ( 1 − y ) ( 8 − 7 y + 2 y 2 + 33 y 3 ) ( 1 + y ) 2 ≥ 0 , x ∈ [ 0 , y 0 ] .$
Since $8 + y − x ( 17 + y ) > 0$ holds for all $( x , y ) ∈ Ω$, by (19), $F 2 ( x , · )$ is convex on $[ 0 , y 0 ]$. This gives us that
$F 2 ( x , y ) ≤ max { F 2 ( 0 , y ) , F 2 ( y 0 , y ) } = F 2 ( 0 , y ) = y − y 3 ≤ 2 9 3 , ( x , y ) ∈ Ω ,$
as we asserted. □
Proposition 3.
Define a function$F 3$by
$F 3 ( x , y ) = 9 ( 1 − x ) ( 1 + y ) 8 − y + x ( 1 + y ) H 2 ( x , y ) ,$
where$H 2 ( x , y ) = ∑ n = 0 3 k n ( y ) x n$with
$k 3 ( y ) = ( 1 + y ) 3 , k 2 ( y ) = 1 + 7 y + 3 y 2 − 3 y 3 , k 1 ( y ) = 8 − 2 y − 15 y 2 + 3 y 3 a n d k 0 ( y ) = − y ( 8 − 9 y + y 2 ) .$
Then$F 3 ( x , y ) ≤ 2 3$holds for all$( x , y ) ∈ R$.
Proof.
First of all, by simple calculations, the equation $( ∂ F 3 / ∂ x ) ( x , y ) = 0$ gives us
$( 1 − x ) ( 8 − y + x ( 1 + y ) ) ∂ H 2 ∂ x ( x , y ) = 9 H 2 ( x , y ) .$
Also, the equation $( ∂ F 3 / ∂ y ) ( x , y ) = 0$ holds when
$− ( 1 + y ) ( 8 − y + x ( 1 + y ) ) ∂ H 2 ∂ y ( x , y ) = 9 H 2 ( x , y ) .$
Assume that the function $F 3$ has its critical point at $( x 0 , y 0 ) ∈ int R$. Since $8 − y 0 + x 0 ( 1 + y 0 ) ≠ 0$, from (22) and (23), we have
$( 1 − x 0 ) ∂ H 2 ∂ x ( x 0 , y 0 ) + ( 1 + y 0 ) ∂ H 2 ∂ y ( x 0 , y 0 ) = 0 ,$
or, equivalently, $y 0 = x 0 / ( 1 − x 0 )$. However, it holds that
$( 1 − x 0 ) ( 8 − y 0 + x 0 ( 1 + y 0 ) ) ∂ H 2 ∂ x ( x 0 , y 0 ) − 9 H 2 ( x 0 , y 0 ) = 64 ( 1 − x 0 ) ≠ 0 ,$
since $x 0 ∈ ( 0 , 1 )$. This contradicts to (22). Hence $F 3$ does not have any critical points in intR. Thus $F 3$ has its maximum on $∂ R$.
We now consider $F 3$ on $∂ R$.
(a) On the side $x = 1$, we have $F 3 ( 1 , y ) ≡ 0$.
(b) On the side $y = − 1$, we have $F 3 ( x , − 1 ) ≡ 0$.
(c) On the side $y = 1$, we have
$F 3 ( x , 1 ) = − 36 x ( 3 − 7 x + 4 x 3 ) 7 + 2 x = : φ ( x ) , x ∈ [ 0 , 1 ] .$
Since the inequality $2 ( 7 + 56 x − 126 x 2 + 72 x 4 ) > 0$ holds for all $x ∈ [ 0 , 1 ]$, it follows that $φ ( x ) < 2$ ($x ∈ [ 0 , 1 ]$). This inequality with (24) implies $F 3 ( x , 1 ) < 2 < 2 3$ holds for $x ∈ [ 0 , 1 ]$.
(d) On the side $x = 0$, we have
$F 3 ( 0 , y ) = − 9 y ( 1 − y 2 ) = : ψ ( y ) .$
And the inequality $F 3 ( 0 , y ) ≤ 2 3$ ($y ∈ [ − 1 , 1 ]$) comes directly from (25) and
$ψ ( y ) ≤ ψ ( − 1 / 3 ) = 2 3 , y ∈ [ − 1 , 1 ] .$
From (a)–(d), for all $( x , y ) ∈ ∂ R$, the inequality $F 3 ( x , y ) ≤ 2 3$ holds. Thus the proof of Proposition 3 is completed. □
Proposition 4.
For$F 1$defined by (12), the inequality
$F 1 ( x , y ) ≥ − 8$
holds for$( x , y ) ∈ [ 0 , 1 ] × [ − 1 , 0 ]$.
Proof.
Define a function $G : [ 0 , 1 ] × [ 0 , 1 ] → R$ by
$G ( x , y ) = F ( x , − y ) − b 4 ( x ) y 4 + 8 = l 3 ( x ) y 3 + l 2 ( x ) y 2 + l 1 ( x ) y + l 0 ( x ) ,$
where $l 3 ( x ) = − b 3 ( x )$, $l 2 ( x ) = b 2 ( x )$, $l 1 ( x ) = − b 1 ( x )$ and $l 0 ( x ) = b 0 ( x ) + 8$. Then we have
$F ( x , y ) + 8 ≥ G ( x , − y ) , ( x , y ) ∈ [ 0 , 1 ] × [ − 1 , 0 ] .$
We note that, when $x = 0$, $G ( 0 , y ) = 7 y 2 ≥ 0$ holds for $y ∈ [ − 1 , 1 ]$. And, when $x = 1$, $G ( 1 , y ) ≡ 8 > 0$.
Let $x ∈ ( 0 , 1 )$ be fixed and put $l i = l i ( x )$ ($i ∈ { 0 , 1 , 2 , 3 }$). Define a function $g x : [ 0 , 1 ] → R$ by $g x ( y ) = G ( x , y )$. We will show that the inequality $g x ( y ) ≥ 0$ holds for all $y ∈ [ 0 , 1 ]$.
Note that $l 3 > 0$ and $l 1 < 0$. Let
$ζ i = − l 2 + ( − 1 ) i l 2 2 − 3 l 1 l 3 l 3 , i = 1 , 2$
be the roots of the equation
$g x ′ ( y ) = 3 l 3 y 2 + 2 l 2 y + l 1 = 0 .$
Then it is easily seen that $ζ 1 < 0 < ζ 2$. Moreover $ζ 2 < 1$ holds. Indeed, $ζ 2 < 1$ is equivalent to $l 1 l 3 + 3 l 3 2 + 2 l 2 l 3 > 0$. And a computation gives
$l 1 l 3 + 3 l 3 2 + 2 l 2 l 3 = − 2 x ( 1 − x ) 2 ( 1 + x ) 4 φ ( x ) ,$
where
$φ ( x ) = − 70 − 73 x − 52 x 2 − 34 x 3 − 16 x 4 + 2 x 5 .$
Since $φ ( x ) < 0$, by (26), we get $l 1 l 3 + 3 l 3 2 + 2 l 2 l 3 > 0$ and $ζ 2 < 1$. Therefore, we have
$g x ( y ) ≥ g x ( ζ 2 ) , y ∈ [ 0 , 1 ] .$
On the other hand, simple calculations give us that
$g x ( ζ 2 ) = 1 9 l 3 [ ( 6 l 1 l 3 − 2 l 2 2 ) ζ 2 + ( 9 l 0 l 3 − l 1 l 2 ) ] = − 1 9 l 3 ( 1 − x ) ( 1 + x ) 3 [ 2 ( 1 − x 2 ) κ 1 ζ 2 + x κ 2 ] ,$
where
$κ 1 = 49 − 126 x + 255 x 2 + 472 x 3 + 204 x 4 − 24 x 5 + 16 x 6$
and
$κ 2 = − 70 + 97 x − 1352 x 2 − 429 x 3 + 746 x 4 + 401 x 5 − 56 x 6 + 15 x 7 .$
Since $l 3 > 0$, $g x ( ζ 2 ) ≥ 0$ holds, if
$2 ( 1 − x 2 ) κ 1 ζ 2 + x κ 2 ≤ 0 .$
Moreover (28) is equivalent to $Ψ ≥ 0$, where
$Ψ = [ 2 ( 1 − x 2 ) κ 1 l 2 − 3 x κ 2 l 3 ] 2 − 4 ( 1 − x 2 ) 2 κ 1 2 ( l 2 2 − 3 l 1 l 3 ) .$
We represent $Ψ$ by
$Ψ = − 27 x 4 ( 10 − x ) 2 ( 1 − x ) 2 ( 1 + x ) 6 Λ ˜ x ,$
where
$Λ ˜ x = − 17052 + 84812 x − 222415 x 2 − 10212 x 3 + 78990 x 4 − 226456 x 5 − 152793 x 6 + 198120 x 7 + 169280 x 8 − 11796 x 9 − 33413 x 10 + 1068 x 11 + 2790 x 12 − 1008 x 13 + 117 x 14 .$
Since $Λ ˜ x < 0$ holds for all $x ∈ ( 0 , 1 )$, from (29), $Ψ ≥ 0$ is true. We thus have $g x ( ζ 2 ) ≥ 0$. Finally, it follows from (27) that $g x ( y ) ≥ 0$ holds for all $y ∈ [ 0 , 1 ]$. The proof of Proposition 4 is completed. □
Proposition 5.
For a function$F 4$defined by
$F 4 ( x , y ) = F 1 ( − x , y ) ,$
where$F 1$is defined by (12), we have
$F 4 ( x , y ) ≥ − 8 , ( x , y ) ∈ [ 0 , 1 ] × [ − 1 / 3 , 1 ] .$
Proof.
It is easily checked that $F 4 ( x , y ) ≥ − 8$ holds for $y ∈ [ − 1 / 3 , 1 ]$ when $x = 0$ or $x = 1$. Let $x ∈ ( 0 , 1 )$ be fixed and put $m i = b i ( − x )$ ($i ∈ { 0 , 1 , 2 , 3 , 4 }$). Define a function $g x : [ − 1 / 3 , 1 ] → R$ by $g x ( y ) = F 4 ( x , y )$.
First, we will show that $g x ( y ) ≥ − 8$ holds for $y ∈ [ − 1 / 3 , 0 ]$. Since $m 3 > 0$ and $m 4 > 0$, we have $m 4 y 4 ≥ 0$ and $m 3 y 3 ≥ − m 3 y 2 / 3$ for $y ∈ [ − 1 / 3 , 0 ]$. Hence, we obtain
$g x ( y ) + 8 ≥ φ x ( − y ) , y ∈ [ − 1 / 3 , 0 ] ,$
where $φ x : [ 0 , 1 / 3 ] → R$ is the function defined by
$φ x ( y ) = − 1 3 m 3 + m 2 y 2 − m 1 y + m 0 + 8 , y ∈ [ 0 , 1 / 3 ] .$
Since $m 1 < 0$ and
$− 1 3 m 3 + m 2 = 1 3 ( 1 − x 2 ) 2 ( 21 − 4 x + 14 x 2 ) > 0 , x ∈ ( 0 , 1 ) ,$
we get
$φ x ′ ( y ) = 2 − 1 3 m 3 + m 2 y − m 1 > 0 , y ∈ [ 0 , 1 / 3 ] .$
Therefore $φ x$ is increasing on $[ 0 , 1 / 3 ]$ and we get
$φ x ( y ) ≥ φ x ( 0 ) = m 0 + 8 = x 2 ( 16 − 6 x − 8 x 2 + 6 x 3 ) ≥ 0 , y ∈ [ 0 , 1 / 3 ] .$
Thus, by (32), $g x ( y ) ≥ − 8$ holds for $y ∈ [ − 1 / 3 , 0 ]$.
Next, we will show that $g x ( y ) ≥ − 8$ holds for $y ∈ [ 0 , 1 ]$. For this, define a function $ψ x : [ 0 , 1 ] → R$ by
$ψ x ( y ) = g x ( y ) − m 4 y 4 + 8 = m 3 y 3 + m 2 y 2 + m 1 y + m 0 + 8 .$
It is sufficient to show that $ψ x ( y ) ≥ 0$ holds for $y ∈ [ 0 , 1 ]$, since
$g x ( y ) + 8 ≥ ψ x ( y ) , y ∈ [ 0 , 1 ] .$
Let
$ζ i = − m 2 + ( − 1 ) i m 2 2 − 3 m 1 m 3 3 m 3 , i ∈ { 1 , 2 }$
be the roots of the equation
$ψ x ′ ( y ) = 3 m 3 y 2 + 2 m 2 y + m 1 = 0 .$
Clearly, $ζ 1 < 0$. Thus we have
$ψ x ( y ) ≥ min { ψ x ( 1 ) , ψ x ( ζ 2 ) } , y ∈ [ 0 , 1 ] .$
Since
$ψ x ( 1 ) = 7 + 2 x − 19 x 2 − 4 x 3 + 29 x 4 + 2 x 5 − 9 x 6 > 0 , x ∈ ( 0 , 1 ) ,$
it is enough to show that $ψ x ( ζ 2 ) ≥ 0$ holds. A similar argument with the proof of Proposition 4, for $x ∈ ( 0 , 1 )$, $ψ x ( ζ 2 ) ≥ 0$ holds if $Λ ˜ − x < 0$, where $Λ ˜ x$ is the quantity defined by (30). It can be checked that $Λ ˜ x < 0$ holds for all $x ∈ ( − 1 , 0 )$. Consequently, $ψ x ( ζ 2 ) ≥ 0$, when $x ∈ ( 0 , 1 )$, follows. Hence, by (33), $ψ x ( y ) ≥ 0$ holds for $y ∈ [ 0 , 1 ]$. It completes the proof of Proposition 5. □

## 3. The Proof of Thereom 1

By using all lemmas and propositions in Section 2, we can prove Theorem 1 as follows.
Proof of Theorem 1.
Let $f ∈ SR *$ be of the form (1). Then by (3) there exists a $ω ∈ B 0$ of the form (8) such that
$z f ′ ( z ) f ( z ) = 1 + ω ( z ) 1 − ω ( z ) .$
Substituting the series (1) and (8) into (34), by equating the coefficients we get
$18 H 3 , 1 ( f ) = 3 β 1 4 β 2 + 6 β 1 3 β 3 + 10 β 1 β 2 β 3 − 8 β 3 2 − 11 β 1 2 β 2 2 + 9 ( β 2 − β 1 2 ) β 4 .$
Since $H 3 , 1 ( f ) = H 3 , 1 ( f ˜ )$, where $f ˜ ( z ) = − f ( − z ) ∈ SR *$, we may assume that $β 1 ∈ [ 0 , 1 ]$.
The inequality (5) will be proved case by case as in the following Table 1.
I. When $β 1 = 1$, then by Schwarz’s lemma, $β n = 0$ for all $n ≥ 2$. Thus, by (35), $H 3 , 1 ( f ) = 0$.
II. When $ω ∈ B 0$ be such that $| β 2 | = 1 − β 1 2$ and $β 1 ∈ [ 0 , 1 )$. Let $p = ( 1 + ω ) / ( 1 − ω ) ∈ P$ be of the form (6). From the relations
$c 1 = 2 β 1 and c 2 = 2 ( β 1 2 + β 2 ) ,$
it follows from that $c 1 ∈ [ 0 , 2 )$ and $2 c 2 = c 1 2 + ( 4 − c 1 2 ) ζ$, where $ζ = ± 1 ∈ T$.
II(a) Assume that $ζ = 1$. Then, by Lemma 1, $p = p 1$, where
$p 1 ( z ) = 1 + 2 τ z + z 2 1 − z 2 = 1 + 2 τ z + 2 z 2 + 2 τ z 3 + ⋯ , z ∈ D$
with $τ ∈ [ 0 , 1 )$. And, from $p = ( 1 + ω ) / ( 1 − ω )$, we have
$β 1 = τ , β 2 = 1 − τ 2 , β 3 = − τ + τ 3 and β 4 = τ 2 − τ 4 .$
Substituting (36) into (35), we get
$H 3 , 1 ( f ) = − 2 9 τ 2 ( 5 − 7 τ 2 + 2 τ 4 ) = : g ( τ 2 ) ,$
where
$g ( x ) = − 2 9 x ( 1 − x ) ( 5 − 2 x ) .$
It can be easily checked that $g ( x ) ≤ g ( 0 ) = 0$, for $x ∈ [ 0 , 1 )$. Moreover, since $g ′ ( x ) = 0$ occurs only when $x = x 1 : = ( 7 − 19 ) / 6 = 0.440184 ⋯ ∈ [ 0 , 1 )$ and $g ″ ( x 1 ) = 4 19 / 9 > 0$, it holds that
$g ( x ) ≥ g ( x 1 ) = 1 243 ( 28 − 19 19 ) ≥ − 4 9 , x ∈ [ 0 , 1 ) .$
So, from (37), the inequality (5) holds.
II(b) Now assume that $ζ = − 1$. Then, by Lemma 1 again, we get $p = p 2$, where
$p 2 ( z ) = 1 − z 2 1 − 2 τ z + z 2 = 1 + 2 τ z + ( − 2 + 4 τ 2 ) z 2 + ( − 6 τ + 8 τ 3 ) z 3 + ( 2 − 16 τ 2 + 16 τ 4 ) z 4 + ⋯ , z ∈ D$
with $τ ∈ [ 0 , 1 )$. Thus, we have
$β 1 = τ , β 2 = τ 2 − 1 , β 3 = τ 3 − τ and β 4 = τ 4 − τ 2 .$
Substituting (38) into (35), we get $H 3 , 1 ( f ) = 0$ and the inequality (5) holds.
III. Let now $| β 2 | ≠ 1 − β 1 2$ and $β 1 ≠ 1$.
At first, we will show that the second inequality in (5) holds. Since $β 1$, $β 2$ and $β 3$ are real, by Lemma 2 for $s ∈ [ 0 , 1 ]$ and t, $u ∈ [ − 1 , 1 ]$ we have
$β 1 = s , β 2 = ( 1 − s 2 ) t , β 3 = ( 1 − s 2 ) ( u ( 1 − t 2 ) − s t 2 ) .$
Substituting (39) into (10) and (11), we have
$Ψ U = ( 1 − s 2 ) [ 1 − u 2 − u ( u + 2 s ) t − ( 1 − u 2 ) t 2 + ( u + s ) 2 t 3 ]$
and
$Ψ L = ( 1 − s 2 ) [ − 1 + u 2 − u ( u + 2 s ) t + ( 1 − u 2 ) t 2 + ( u + s ) 2 t 3 ] .$
We also have $( s , t ) ∉ C$, where C is a curve defined by
$C = { ( s , t ) ∈ R : s = 1 or t = ± 1 } ⊂ ∂ R .$
III(a) Consider the case $β 2 ≥ β 1 2$, i.e., $( s , t ) ∈ Ω 1$, where $Ω 1$ is the set defined by
$Ω 1 = ( s , t ) ∈ [ 0 , 1 / 2 ) × [ 0 , 1 ) : s 2 1 − s 2 ≤ t < 1$
so that $Ω 1 ∩ C = ∅$. In this case, by (40), we have
$18 H 3 , 1 ( f ) ≤ 3 β 1 4 β 2 + 6 β 1 3 β 3 + 10 β 1 β 2 β 3 − 8 β 3 2 − 11 β 1 2 β 2 2 + 9 ( β 2 − β 1 2 ) Ψ U = − ( 1 − s 2 ) ( 1 + t ) Φ ( s , t , u ) , ( s , t , u ) ∈ Ω 1 × [ − 1 , 1 ] ,$
where
$Φ ( s , t , u ) = Φ 0 + Φ 1 u + Φ 2 u 2$
with
$Φ 0 = Φ 0 ( s , t ) : = − 9 ( 1 − t ) t − s 4 t ( 3 + 2 t − t 2 ) + s 2 ( 9 + 2 t 2 − t 3 ) , Φ 1 = Φ 1 ( s , t ) : = − 2 s ( 1 − t ) [ ( 5 − t ) t + s 2 ( 3 + 4 t + t 2 ) ] , Φ 2 = Φ 2 ( s , t ) : = ( 1 − t 2 ) [ 8 + t − s 2 ( 17 + t ) ] .$
We note that $Φ 2 > 0$, since
$8 + t − s 2 ( 17 + t ) ≥ 8 ( 1 − t ) 1 + t > 0 , ( s , t ) ∈ Ω 1 .$
Let $u 1 = − Φ 1 / ( 2 Φ 2 )$ be the root of the equation $( ∂ Φ / ∂ u ) ( s , t , u ) = 0$. Then it can be seen that $u 1 ≥ − 1$. Indeed, we note that $2 Φ 2 − Φ 1 = 2 ( 1 − t ) Υ ( s , t )$, where $Υ ( s , t ) = λ 2 ( s ) t 2 + λ 1 ( s ) t + λ 0 ( s )$, where
$λ 2 ( s ) = ( 1 − s ) 2 ( 1 + s ) , λ 1 ( s ) = 9 + 5 s − 18 s 2 + 4 s 3$
and
$λ 0 ( s ) = 8 − 17 s 2 + 3 s 3 .$
Since $λ i ( s ) ≥ 0$ when $s ∈ [ 0 , 1 / 2 )$ for $i ∈ { 1 , 2 }$, we have
$Υ ( s , t ) ≥ Υ s , s 2 1 − s 2 = 8 ( 1 + s − s 2 ) 1 + s ≥ 0 , ( s , t ) ∈ Ω 1 .$
Hence, we get $2 Φ 2 − Φ 1 ≥ 0$ and it follows from $Φ 2 > 0$ that $u 1 ≥ − 1$.
(i) Assume that $u 1 ≥ 1$. Then we have
$Φ ( s , t , u ) ≥ Φ ( s , t , 1 ) = Φ 0 + Φ 1 + Φ 2 , ( s , t , u ) ∈ Ω 1 × [ − 1 , 1 ] .$
Therefore, by (42), it holds that
$18 H 3 , 1 ( f ) ≤ − ( 1 − s 2 ) ( 1 + t ) ( Φ 0 + Φ 1 + Φ 2 ) = F 1 ( s , t ) , ( s , t ) ∈ Ω 1 ,$
where $F 1$ is the function defined by (12). From Proposition 1 and (44), we thus have $H 3 , 1 ( f ) ≤ 3 / 9$.
(ii) Assume that $− 1 ≤ u 1 ≤ 1$. Then we have
$Φ ( s , t , u ) ≥ Φ ( s , t , u 1 ) = Φ 0 − Φ 1 2 4 Φ 2 , ( s , t , u ) ∈ Ω 1 × [ − 1 , 1 ] .$
Therefore, by (42), it holds that
$18 H 3 , 1 ( f ) ≤ − ( 1 − s 2 ) ( 1 + t ) Φ 0 − Φ 1 2 4 Φ 2 = 9 F 2 ( s 2 , t ) , ( s , t ) ∈ Ω 1 ,$
where $F 2$ is the function defined by (18). Therefore, by Proposition 2, $H 3 , 1 ( f ) ≤ 3 / 9$ holds.
III(b) Consider the case $β 2 ≤ β 1 2$, i.e., $( s , t ) ∈ Ω 2$, where $Ω 2$ is the set defined by $Ω 2 = cl ( R ∖ Ω 1 ) ∖ C$. Then, from (41), we have
$18 H 3 , 1 ( f ) ≤ 3 β 1 4 β 2 + 6 β 1 3 β 3 + 10 β 1 β 2 β 3 − 8 β 3 2 − 11 β 1 2 β 2 2 + 9 ( β 2 − β 1 2 ) Ψ L = − ( 1 − s 2 ) ( 1 + t ) Φ ^ ( s , t , u ) , ( s , t , u ) ∈ Ω 2 × [ − 1 , 1 ] ,$
where
$Φ ^ ( s , t , u ) = Φ ^ 0 + Φ ^ 1 u + Φ ^ 2 u 2$
with
$Φ ^ 0 = Φ ^ 0 ( s , t ) : = 9 ( 1 − t ) t − s 4 t ( 3 + 2 t − t 2 ) − s 2 ( 9 − 20 t 2 + t 3 ) , Φ ^ 1 = Φ ^ 1 ( s , t ) : = − 2 s ( 1 − t ) [ ( 5 − t ) t + s 2 ( 3 + 4 t + t 2 ) ] , Φ ^ 2 = Φ ^ 2 ( s , t ) : = ( 1 − t ) 2 [ 8 − t + s 2 ( 1 + t ) ] .$
Using the inequality $s 2 ≥ t / ( 1 + t )$, we have $Φ ^ 2 ≥ 8 ( 1 − t ) 2 > 0$ for $( s , t ) ∈ Ω 2$. Let $u 2 = − Φ ^ 1 / ( 2 Φ ^ 2 )$ be the root of the equation $( ∂ Φ ^ / ∂ u ) ( s , t , u ) = 0$. Then, by a similar procedure with Part III(a), it can be seen that $u 2 ≥ − 1$.
(i) Assume that $u 2 ≥ 1$. Then we have
$Φ ^ ( s , t , u ) ≥ Φ ^ ( s , t , 1 ) = Φ ^ 0 + Φ ^ 1 + Φ ^ 2 , ( s , t , u ) ∈ Ω 2 × [ − 1 , 1 ] .$
Therefore, by (45), it holds that
$18 H 3 , 1 ( f ) ≤ − ( 1 − s 2 ) ( 1 + t ) ( Φ ^ 0 + Φ ^ 1 + Φ ^ 2 ) = F 1 ( s , t ) , ( s , t ) ∈ Ω 2 ,$
where $F 1$ is the function defined by (12). Thus, by Proposition 1, $H 3 , 1 ( f ) ≤ 3 / 9$ holds.
(ii) Assume that $− 1 ≤ u 2 ≤ 1$. Then we have
$Φ ^ ( s , t , u ) ≥ Φ ^ ( s , t , u 2 ) = Φ ^ 0 − Φ ^ 1 2 4 Φ ^ 2 , ( s , t , u ) ∈ Ω 2 × [ − 1 , 1 ] .$
Therefore, by (45), it holds that
$18 H 3 , 1 ( f ) ≤ − ( 1 − s 2 ) ( 1 + t ) Φ ^ 0 − Φ ^ 1 2 4 Φ ^ 2 = F 3 ( s 2 , t ) , ( s , t ) ∈ Ω 2 ,$
where $F 3$ is the function defined by (21). Therefore, by Proposition 3, we obtain $H 3 , 1 ( f ) ≤ 3 / 9$.
Next, we will show that the first inequality in (5) holds.
IV(a) Consider the case $β 2 ≥ β 1 2$. Then we have
$18 H 3 , 1 ( f ) ≥ − ( 1 − s 2 ) ( 1 + t ) Φ ^ ( s , t , u ) , ( s , t , u ) ∈ Ω 1 × [ − 1 , 1 ] ,$
where $Φ ^$ is the function defined by (46). Since $Φ ^ 1 ≤ 0$ and $Φ ^ 2 > 0$, it holds that
$Φ ^ ( s , t , u ) ≤ max { Φ ^ ( s , t , − 1 ) , Φ ^ ( s , t , 1 ) } = Φ ^ ( s , t , − 1 ) = Φ ^ 2 − Φ ^ 1 + Φ ^ 0 , ( s , t , u ) ∈ Ω 1 × [ − 1 , 1 ] .$
Hence, from (47), we obtain
$H 3 , 1 ( f ) ≥ − ( 1 − s 2 ) ( 1 + t ) ( Φ ^ 2 − Φ ^ 1 + Φ ^ 0 ) = F 4 ( s , t ) , ( s , t ) ∈ Ω 1 ,$
where $F 4$ is the function defined by (31). Thus, by Proposition 5 and (48), we get $H 3 , 1 ( f ) ≥ − 4 / 9$.
IV(b) We consider the case $β 2 ≤ β 1 2$. Then we have
$18 H 3 , 1 ( f ) ≥ − ( 1 − s 2 ) ( 1 + t ) Φ ( s , t , u ) , ( s , t , u ) ∈ Ω 2 × [ − 1 , 1 ] ,$
where $Φ$ is the function defined by (43).
For $t ∈ [ − 1 / 3 , 0 ]$, let
$s t = t 2 − 5 t t 2 + 4 t + 3$
so that $0 = s 0 ≤ s t ≤ s − 1 / 3 = 1$ holds for $t ∈ [ − 1 / 3 , 0 ]$. And let
$Ω 3 = { ( s , t ) ∈ Ω 2 : s ≤ s t } and Ω 4 = { ( s , t ) ∈ Ω 2 : s ≥ s t } .$
We note that $Ω 3 ⊂ [ 0 , 1 ] × [ − 1 , 0 ]$ and $Ω 4 ⊂ [ 0 , 1 ] × [ − 1 / 3 , 1 ]$. Then $Φ 1 ≥ 0$ when $( s , t ) ∈ Ω 3$, and $Φ 1 ≤ 0$ when $( s , t ) ∈ Ω 4$.
(i) For the case $( s , t ) ∈ Ω 3$, since $Φ 1 ≥ 0$ and $Φ 2 ≥ 0$, we have
$Φ ( s , t , u ) ≤ Φ ( s , t , 1 ) = Φ 2 + Φ 1 + Φ 0 , ( s , t , u ) ∈ Ω 3 × [ − 1 , 1 ]$
and, therefore, we get
$18 H 3 , 1 ( f ) ≥ − ( 1 − s 2 ) ( 1 + t ) ( Φ 2 + Φ 1 + Φ 0 ) = F 1 ( s , t ) , ( s , t ) ∈ Ω 3 ,$
where $F 1$ is the function defined by (12). Since $Ω 3 ⊂ [ 0 , 1 ] × [ − 1 , 0 ]$, Proposition 4 gives us that $H 3 , 1 ( f ) ≥ − 4 / 9$ holds.
(ii) For the case $( s , t ) ∈ Ω 4$, we have
$18 H 3 , 1 ( f ) ≥ − ( 1 − s 2 ) ( 1 + t ) ( Φ 2 − Φ 1 + Φ 0 ) = F 4 ( s , t ) , ( s , t ) ∈ Ω 4 ,$
where $F 4$ is the funciton defined by (31). Since $Ω 4 ⊂ [ 0 , 1 ] × [ − 1 / 3 , 1 ]$, Proposition 5 gives us that $H 3 , 1 ( f ) ≥ − 4 / 9$ holds. Thus the proof of Theorem 1 is now completed. □

## 4. Conclusions

In the present paper, we obtained that the sharp inequalities $− 4 / 9 ≤ H 3 , 1 ( f ) ≤ 3 / 9$ hold for f in the class $SR *$, i.e., starlike functions with real coefficients. Therefore, it follows that $| H 3 , 1 ( f ) | ≤ 4 / 9$ holds for $f ∈ SR *$ and this inequality is sharp with the extremal function $f 1 ∈ SR *$, where $f 1 ( z ) = z ( 1 − z 3 ) − 2 / 3$. So it can be naturally expected that the sharp inequality $| H 3 , 1 ( f ) | ≤ 4 / 9$ would hold for all $f ∈ S *$.

## Author Contributions

Writing—Original Draft Preparation, Y.J.S.; Writing—Review & Editing, O.S.K.

## Funding

This work was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIP; Ministry of Science, ICT & Future Planning) (No. NRF-2017R1C1B5076778).

## Acknowledgments

The authors would like to express their thanks to the referees for their valuable comments and suggestions.

## Conflicts of Interest

The authors declare no conflict of interest.

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Table 1. An outline of the proof.
Table 1. An outline of the proof.
CasesConditionsUsed Results for the Proof
I$β 1 = 1$Schwarz’s lemma
II(a)$β 2 = 1 − β 1 2$, $β 1 ∈ [ 0 , 1 )$Lemma 1
II(b)$β 2 = β 1 2 − 1$, $β 1 ∈ [ 0 , 1 )$Lemma 1
III(a)$| β 2 | ≠ 1 − β 1 2$, $β 1 ∈ [ 0 , 1 )$, $β 2 ≥ β 1 2$Lemma 2 and 3, Proposition 1 and 2
III(b)$| β 2 | ≠ 1 − β 1 2$, $β 1 ∈ [ 0 , 1 )$, $β 2 ≤ β 1 2$Lemma 2 and 3, Proposition 1 and 3
IV(a)$| β 2 | ≠ 1 − β 1 2$, $β 1 ∈ [ 0 , 1 )$, $β 2 ≥ β 1 2$Lemma 2 and 3, Proposition 5
IV(b)$| β 2 | ≠ 1 − β 1 2$, $β 1 ∈ [ 0 , 1 )$, $β 2 ≤ β 1 2$Lemma 2 and 3, Proposition 4 and 5

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Kwon, O.S.; Sim, Y.J. The Sharp Bound of the Hankel Determinant of the Third Kind for Starlike Functions with Real Coefficients. Mathematics 2019, 7, 721. https://doi.org/10.3390/math7080721

AMA Style

Kwon OS, Sim YJ. The Sharp Bound of the Hankel Determinant of the Third Kind for Starlike Functions with Real Coefficients. Mathematics. 2019; 7(8):721. https://doi.org/10.3390/math7080721

Chicago/Turabian Style

Kwon, Oh Sang, and Young Jae Sim. 2019. "The Sharp Bound of the Hankel Determinant of the Third Kind for Starlike Functions with Real Coefficients" Mathematics 7, no. 8: 721. https://doi.org/10.3390/math7080721

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