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Article

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

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.

References

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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

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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

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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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