Open Access

The Fekete-Szegö inequality for close-to-convex functions with respect to a certain starlike function dependent on a real parameter

Journal of Inequalities and Applications20142014:65

https://doi.org/10.1186/1029-242X-2014-65

Received: 9 December 2013

Accepted: 31 January 2014

Published: 13 February 2014

Abstract

Given α [ 0 , 1 ] , let g α ( z ) : = z / ( 1 α z ) 2 , z D : = { z C : | z | < 1 } . An analytic standardly normalized function f in D is called close-to-convex with respect to g α if there exists δ ( π / 2 , π / 2 ) such that

Re { e i δ z f ( z ) g α ( z ) } > 0 , z D .

For the class C ( g α ) of all close-to-convex functions with respect to g α , the Fekete-Szegö problem is studied.

MSC:30C45.

Keywords

Fekete-Szegö problem close-to-convex functions close-to-convex functions with respect to the Koebe function close-to-convex functions with argument δ functions convex in the positive direction of the imaginary axis

1 Introduction

The classical problem settled by Fekete and Szegö [1] is to find for each λ [ 0 , 1 ] the maximum value of the coefficient functional
Φ λ ( f ) : = | a 3 λ a 2 2 |
over the class S of univalent functions f in the unit disk D : = { z C : | z | < 1 } of the form
f ( z ) = z + n = 2 a n z n , z D .
(1.1)
By applying the Loewner method they proved that
max f S Φ λ ( f ) = { 1 + 2 exp ( 2 λ / ( 1 λ ) ) , λ [ 0 , 1 ) , 1 , λ = 1 .

The problem of calculating max f F Φ λ ( f ) for various compact subclasses of the class A of all analytic functions f in D of the form (1.1), as well as for λ being an arbitrary real or complex number, was considered by many authors (see, e.g., [210]).

Let S denote the class of starlike functions, i.e., the class of all functions f A such that
Re z f ( z ) f ( z ) > 0 , z D .
(1.2)
Given δ ( π / 2 , π / 2 ) and g S , let C δ ( g ) denote the class of functions called close-to-convex with argument δ with respect to g, i.e., the class of all functions f A such that
Re { e i δ z f ( z ) g ( z ) } > 0 , z D .
(1.3)
Let
C ( g ) : = δ ( π / 2 , π / 2 ) C δ ( g ) , C δ : = g S C δ ( g )
denote the classes of functions called close-to-convex with respect to g and close-to-convex with argument δ, respectively, and let
C : = δ ( π / 2 , π / 2 ) C δ = δ ( π / 2 , π / 2 ) g S C δ ( g )

denote the class of close-to-convex functions (see [[11], pp.184-185], [12, 13]). It is well known that S and C are the subclasses of S .

By using a specific starlike function g, inequality (1.3) defines the related class C δ ( g ) . Given α [ 0 , 1 ] , let
g α ( z ) : = z ( 1 α z ) 2 = n = 1 n α n 1 z n , z D .
(1.4)
It is easy to check that each g α satisfies (1.2), i.e., g α S for every α [ 0 , 1 ] . Then (1.3) is of the form
Re { e i δ ( 1 α z ) 2 f ( z ) } > 0 , z D ,
(1.5)

and defines the class C δ ( g α ) , and further the class C ( g α ) . Such classes of functions were studied in [14, 15] and [16], where some generalization of the Robertson condition for convexity in one direction [17] was discussed.

Note that for α : = 1 we get the Koebe function g 1 = : k . Then condition (1.5) defines the class C δ ( k ) and further the class C ( k ) of functions close-to-convex with respect to the Koebe function. Such functions have a well-known geometrical meaning, namely condition (1.5) geometrically says that the function h : = e i δ f has the boundary normalization
lim t h 1 ( h ( z ) + t ) = 1

and h ( D ) is a domain such that { w + t : t 0 } h ( D ) for every w h ( D ) . Such functions h, clearly univalent as close-to-convex, and domains h ( D ) are called convex in the positive (negative) direction of the real axis and are related to functions convex in the direction of the imaginary axis (see, e.g., [1720], [[21], Chapter VI], [22]).

For α : = 0 we have the identity function g 0 ( z ) = z , z D , and then condition (1.5) is of the form
Re { e i δ f ( z ) } > 0 , z D .
(1.6)

Functions f having such a property are called of bounded turning with argument δ and form the class C δ ( g 0 ) denoted usually as P ( δ ) . Functions in the class P : = C ( g 0 ) are usually called of bounded turning (cf. [[23], Vol. I, p.101]). On the other hand, condition (1.6) is known as the famous criterium of univalence due to Noshiro [24] and Warschawski [25] (cf. [[23], Vol. I, p.88]). In this way condition (1.5) creates a simple parametric passage from the class P ( δ ) to the class C δ ( k ) .

The main goal of this paper is to study the Fekete-Szegö problem for the classes C ( g α ) , α [ 0 , 1 ] . For the class C ( k ) , i.e., for α = 1 , the Fekete-Szegö problem was examined in [26], where it was shown that
max f C ( k ) Φ λ ( f ) { | 3 4 λ | , λ ( , 1 / 3 ] [ 1 , + ) , 1 3 ( 2 3 λ ) 2 2 | 2 3 λ | + | 1 λ | + 2 3 , λ [ 1 / 3 , 1 ] ,
with sharpness of the result for λ R ( 2 / 3 , 1 ) . Recall here that in [3] Keogh and Merkes proved that
max f C 0 Φ λ ( f ) = { | 3 4 λ | , λ ( , 1 / 3 ] [ 1 , + ) , 1 / 3 + 4 / ( 9 λ ) , λ [ 1 / 3 , 2 / 3 ] , 1 , λ [ 2 / 3 , 1 ] .
For λ [ 0 , 1 ] Koepf in [5] extended the above result for the class of close-to-convex functions showing that
max f C Φ λ ( f ) = max f C 0 Φ λ ( f ) .

For other results on the Fekete-Szegö problem for various subclasses of close-to-convex functions, particularly for strongly close-to-convex functions, see [2729] and [30].

For the class P and λ [ 0 , 1 ] , we get the following sharp result published, among other results, in [[6], Theorem 2.3], namely
max f P Φ λ ( f ) = 2 3 .

2 Main result

By P we denote the class of all analytic functions p in D of the form
p ( z ) = 1 + n = 1 c n z n , z D ,
(2.1)
having a positive real part in D . For each ε T : = { z C : | z | = 1 } , let
L ε ( z ) : = 1 + ε z 1 ε z , z D , L : = L 1 .

Inequalities (2.2) and (2.3) below are well known (see, e.g., [[31], pp.41, 166]).

Lemma 2.1 If p P is of the form (2.1), then
| c n | 2 , n N ,
(2.2)
and
| c 2 c 1 2 2 | 2 | c 1 | 2 2 .
(2.3)
Both inequalities are sharp. The equality in (2.2) holds for every function L ε , ε T . The equality in (2.3) holds for every function
p t , θ ( z ) : = t L ( e i θ z ) + ( 1 t ) L ( e 2 i θ z 2 ) = 1 + 2 t e i θ z + 2 e 2 i θ z 2 + , z D ,
(2.4)

where t [ 0 , 1 ] and θ R .

Now we prove the main theorem of this paper. The source of the method of proof is in Koepf’s paper [5], where the upper bound of Φ λ for close-to-convex functions with λ restricted to the interval ( 1 / 3 , 2 / 3 ) was calculated. However, we modify this technique and use it homogeneously for the class C ( g α ) for all real λ, partially analogously as in [26] for the class C ( k ) , and in [32]. We apply also the powerful Laguerre’s rule of counting zeros of polynomials in an interval. We propose Laguerre’s algorithm for such a computation by its simplicity, usefulness and efficiency.

We shortly recall Laguerre’s rule of counting zeros of polynomials in an interval (see [33, 34], [[35], pp.19-20]). Given a real polynomial
Q ( u ) = a 0 u n + a 1 u n 1 + + a n 1 u + a n ,
(2.5)
consider a finite sequence ( q k ) , k = 0 , 1 , , n , of polynomials of the form
q k ( u ) = j = 0 k a j u k j .
(2.6)

For each u 0 R , let N ( Q ; u 0 ) denote the number of sign changes in the sequence ( q k ( u 0 ) ) , k = 0 , 1 , , n . Given an interval I 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 2.2 If a < b , Q ( a ) 0 and Q ( b ) 0 , then Z ( Q ; ( a , b ) ) = N ( Q ; a ) N ( Q ; b ) or N ( Q ; a ) N ( Q ; b ) Z ( Q ; ( a , b ) ) is an even positive integer.

Note that q k ( 0 ) = a k and q k ( 1 ) = j = 0 k a j . Thus, when ( a , b ) : = ( 0 , 1 ) , Theorem 2.2 reduces to the following useful corollary.

Corollary 2.3 If Q ( 0 ) 0 and Q ( 1 ) 0 , then Z ( Q ; ( 0 , 1 ) ) = N ( Q ; 0 ) N ( Q ; 1 ) or N ( Q ; 0 ) N ( Q ; 1 ) Z ( Q ; ( 0 , 1 ) ) 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 ( a k ) and in the sequence of sums ( j = 0 k a j ) , with k = 0 , 1 , , n , respectively.

The main theorem of the paper is as follows.

Theorem 2.4 Let α [ 0 , 1 ] . Then
max f C ( g α ) Φ λ ( f ) { | 2 3 + 4 3 α + α 2 ( 1 + α ) 2 λ | , λ R ( τ 1 ( α ) , τ 2 ( α ) ) , 2 3 + α 2 ( 1 3 ( 2 3 λ ) 2 2 | 2 3 λ | + | 1 λ | ) , λ [ τ 1 ( α ) , τ 2 ( α ) ] ,
(2.7)
where
τ 1 ( α ) : = 2 α 3 ( 1 + α ) , τ 2 ( α ) : = 2 ( 2 + α ) 3 ( 1 + α ) .
For each α ( 0 , 1 ] and each λ R ( 2 / 3 , τ 2 ( α ) ) , as well as for α : = 0 and each λ R , the inequality is sharp and the equality is attained by a function in C 0 ( g α ) . In particular,
  1. (i)
    when α ( 0 , 1 ] , for each λ [ τ 1 ( α ) , 2 / 3 ] the second equality in (2.7) is attained by the function f α , t α , λ given by the differential equation
    f α , t α , λ ( z ) = p t α , λ , 0 ( z ) ( 1 α z ) 2 , f α , t α , λ ( 0 ) : = 0 , z D ,
    (2.8)
     
where t α , λ : = α ( 2 / ( 3 λ ) 1 ) ;
  1. (ii)
    when α ( 0 , 1 ] , for each λ R ( τ 1 ( α ) , τ 2 ( α ) ) the first equality in (2.7) is attained by the function f α , 1 , given by (2.8) with t α , λ 1 , i.e., when α ( 0 , 1 ) , by the function
    f α , 1 ( z ) = 2 ( 1 α ) 2 log 1 α z 1 z 1 + α 1 α z 1 α z , z D , log 1 : = 0 ,
    (2.9)
     
and when α = 1 , by the Koebe function f 1 , 1 : = k ;
  1. (iii)
    when α = 0 , for each λ [ 0 , 4 / 3 ] the second equality in (2.7) is attained by the function
    f 0 , 0 ( z ) : = z + log 1 + z 1 z , z D , log 1 : = 0 ;
    (2.10)
     
and for each λ R ( 0 , 4 / 3 ) the first equality in (2.7) is attained by the function
f 0 , 1 ( z ) : = z 2 log ( 1 z ) , z D , log 1 : = 0 .
(2.11)
Proof Fix α [ 0 , 1 ] . Observe from (1.5) that f C ( g α ) if and only if
e i δ ( 1 α z ) 2 f ( z ) = p ( z ) cos δ + i sin δ , z D
(2.12)
for some δ ( π / 2 , π / 2 ) and p P . Thus
z f ( z ) = e i δ g α ( z ) ( p ( z ) cos δ + i sin δ ) , z D .
(2.13)
Setting the series (1.1), (1.4) and (2.1) into (2.13), by comparing coefficients, we get
a 2 = 1 2 ( c 1 e i δ cos δ + 2 α )
(2.14)
and
a 3 = 1 3 ( c 2 e i δ cos δ + 2 α c 1 e i δ cos δ + 3 α 2 ) .
(2.15)
Let λ R . Using (2.3), from (2.14) and (2.15), we have
Φ λ ( f ) = | a 3 λ a 2 2 | = | 1 3 c 2 e i δ cos δ + 2 3 α c 1 e i δ cos δ + α 2 1 4 λ ( c 1 2 e 2 i δ cos 2 δ + 4 α c 1 e i δ cos δ + 4 α 2 ) | = | α 2 ( 1 λ ) + 1 3 ( c 2 c 1 2 2 ) e i δ cos δ + c 1 2 6 ( 1 3 2 λ e i δ cos δ ) e i δ cos δ + α ( 2 3 λ ) c 1 e i δ cos δ | α 2 | 1 λ | + 1 3 ( 2 | c 1 | 2 2 ) cos δ + | c 1 | 2 6 | 1 3 2 λ e i δ cos δ | cos δ + α | 2 3 λ | | c 1 | cos δ = α 2 | 1 λ | + ( 2 3 + | c 1 | 2 6 ( 1 ( 3 λ 9 4 λ 2 ) cos 2 δ 1 ) + α | 2 3 λ | | c 1 | ) cos δ .
(2.16)
Set x : = | c 1 | and y : = cos δ . Clearly, y ( 0 , 1 ] and, in view of (2.2), x [ 0 , 2 ] . It is convenient to use γ : = 2 3 λ instead of λ in further computation. For γ R , let
s γ ( y ) : = 1 ( 1 1 4 γ 2 ) y 2 , y [ 0 , 1 ] .
Set R : = [ 0 , 2 ] × [ 0 , 1 ] . For α [ 0 , 1 ] and γ R , define
F α , γ ( x , y ) : = 1 3 α 2 | 1 + γ | + 1 3 ( 2 + x 2 2 ( s γ ( y ) 1 ) + α | γ | x ) y , ( x , y ) R .
Consequently, in view of (2.16) we have
max f C ( g α ) Φ λ ( f ) max ( x , y ) R F α , γ ( x , y ) .
Now, for each α [ 0 , 1 ] and γ R , we find the maximum value of F α , γ on the rectangle R.
  1. 1.
    In the corners of R we have
    F α , γ ( 0 , 0 ) = F α , γ ( 2 , 0 ) = 1 3 α 2 | 1 + γ | ,
    (2.17)
    F α , γ ( 0 , 1 ) = 1 3 α 2 | 1 + γ | + 2 3 ,
    (2.18)
    F α , γ ( 2 , 1 ) = 1 3 α 2 | 1 + γ | + 1 3 ( 1 + 2 α ) | γ | .
    (2.19)
     
  2. 2.

    For x = 0 and y ( 0 , 1 ) we have a linear function and for x ( 0 , 2 ) and y = 0 we have a constant function.

     
  3. 3.
    For x ( 0 , 2 ) and y = 1 , let
    G α , γ ( x ) : = F α , γ ( x , 1 ) = 1 3 ( 1 4 ( | γ | 2 ) x 2 + α | γ | x + α 2 | 1 + γ | + 2 ) .
     
For | γ | = 2 we get the linear functions, so let | γ | 2 . Then G α , γ ( x ) = 0 if and only if
x = 2 α | γ | 2 | γ | = : x α , γ .
Thus x α , γ ( 0 , 2 ) if and only if
α 0 0 < | γ | < 2 1 + α .
(2.20)
Moreover, we have
F α , γ ( x α , γ , 1 ) = 1 3 ( α 2 γ 2 2 | γ | + α 2 | 1 + γ | + 2 ) .
(2.21)
  1. 4.
    For x = 2 and y ( 0 , 1 ) , let
    H α , γ ( y ) : = F α , γ ( 2 , y ) = 1 3 ( α 2 | 1 + γ | + 2 y s γ ( y ) + 2 α | γ | y ) .
     
For | γ | = 2 we have the linear functions, evidently, so let | γ | 2 . Note first that
s γ ( y ) > 0 , y ( 0 , 1 ) .
(2.22)
Taking into account (2.22), we have
y s γ ( y ) = ( 1 1 4 γ 2 ) y 2 1 ( 1 1 4 γ 2 ) y 2 = s γ 2 ( y ) 1 s γ ( y ) , y ( 0 , 1 ) .
(2.23)
Using (2.23) we get
H α , γ ( y ) = 2 3 ( s γ ( y ) + s γ 2 ( y ) 1 s γ ( y ) + α | γ | ) = 0 , y ( 0 , 1 )
(2.24)
if and only if
2 s γ 2 ( y ) + α | γ | s γ ( y ) 1 = 0 , y ( 0 , 1 ) ,
i.e., in view of (2.22) if and only if
s γ ( y ) = α | γ | + 8 + α 2 γ 2 4 = : s α , γ , y ( 0 , 1 ) .
Since | γ | 2 , so from the above we get the equation
y 2 = 4 α 2 γ 2 + α | γ | 8 + α 2 γ 2 2 ( 4 γ 2 ) , y ( 0 , 1 ) .
(2.25)
Thus the solution of equation (2.25), and hence of (2.24), exists if and only if
0 < 4 α 2 γ 2 + α | γ | 8 + α 2 γ 2 2 ( 4 γ 2 ) < 1 .
(2.26)
Elementary computing shows that (2.26) holds if and only if
| γ | < 2 1 + α .
(2.27)
Thus the function H α , γ has a critical point in ( 0 , 1 ) , namely
y = 4 α 2 γ 2 + α | γ | α 2 γ 2 + 8 2 ( 4 γ 2 ) = : y α , γ ,
as the unique solution of (2.25), if and only if (2.27) holds. Moreover,
F α , γ ( 2 , y α , γ ) = 1 3 α 2 | 1 + γ | + 1 6 4 α 2 γ 2 + α | γ | 8 + α 2 γ 2 2 ( 4 γ 2 ) ( 8 + α 2 γ 2 + 3 α | γ | ) .
(2.28)
  1. 5.

    We will prove that for each α [ 0 , 1 ] and γ R the function F α , γ has no critical point in ( 0 , 2 ) × ( 0 , 1 ) .

     
Since y 0 and x 0 , we have
F α , γ x = 0
if and only if
s γ ( y ) = 1 α | γ | x , y ( 0 , 1 ) .
(2.29)
Since x > 0 , by comparing (2.29) and (2.22), we see that x > α | γ | . By a simple observation, we deduce that the solution of (2.29) can exist only when
( α = 0 | γ | = 2 ) ( α 0 γ 0 | γ | 2 x > α | γ | ) .
(2.30)
Squaring then (2.29), we obtain
s γ 2 ( y ) 1 = 2 α | γ | x + α 2 γ 2 x 2 .
(2.31)
Since by (2.22), s γ ( y ) 0 for y ( 0 , 1 ) , taking into account (2.23), we have
F α , γ y = 2 3 + x 2 6 ( s γ ( y ) 1 ) + 1 3 α | γ | x + x 2 6 s γ 2 ( y ) 1 s γ ( y ) .
Thus, by using (2.29) and (2.31), after simplifying we have
F α , γ y = 0
if and only if
α | γ | x 2 4 x + 4 α | γ | = 0 , x ( 0 , 2 ) .
(2.32)
It follows at once that for α = 0 and γ R , as well as for α ( 0 , 1 ] and | γ | 1 / α , equation (2.32) has no root. Thus by (2.30) we consider
α 0 | γ | 2 0 < | γ | < 1 / α x > α | γ | .
(2.33)
Solving now (2.32), we have
x = 2 2 1 α 2 γ 2 α | γ | = : x 1 ; α , γ , x = 2 + 2 1 α 2 γ 2 α | γ | = : x 2 ; α , γ .
Clearly, x 2 ; α , γ ( 0 , 2 ) and it remains to consider x 1 ; α , γ . It is easy to check that x 1 ; α , γ > α | γ | . Thus setting x : = x 1 ; α , γ into (2.31) and computing, we have
y 2 = 2 α | γ | x 1 ; α , γ α 2 γ 2 x 1 ; α , γ 2 1 1 4 γ 2 = ( 4 α 2 γ 2 4 1 α 2 γ 2 ) α 2 γ 2 ( 1 1 α 2 γ 2 ) 2 ( 4 γ 2 ) .
(2.34)
A solution in ( 0 , 1 ) of (2.34) exists if and only if
0 < ( 4 α 2 γ 2 4 1 α 2 γ 2 ) α 2 γ 2 ( 1 1 α 2 γ 2 ) 2 ( 4 γ 2 ) < 1 .
(2.35)
By (2.33) consider
α 0 | γ | 2 0 < | γ | < 1 α .
(2.36)
We prove that then condition (2.35) is false. When 2 < | γ | < 1 / α , then an easy computation shows that the left-hand inequality in (2.35) is false. Thus by (2.36) it remains to consider
α 0 0 < | γ | < 1 α 2 .
(2.37)
By an easy computation we check that then the left-hand inequality in (2.35) holds. Since 4 γ 2 > 0 , write the right-hand inequality in (2.35) as
( 8 2 ( 1 + 2 α 2 ) γ 2 ) 1 α 2 γ 2 < ( α 4 + α 2 ) γ 4 2 ( 1 + 4 α 2 ) γ 2 + 8 .

The last step is to show, which does not cause difficulties, that under the assumption (2.37) the above inequality is false. We omit the details.

Summarizing, we proved that condition (2.35) is false, so equation (2.34) has no solution in ( 0 , 1 ) .

Thus the proof that for α [ 0 , 1 ] and γ R the function F α , γ has no critical point in ( 0 , 2 ) × ( 0 , 1 ) is finished.
  1. 6.

    Now we calculate the maximum value of F α , γ in R, which, as was shown, is attained on the boundary of R. Let α [ 0 , 1 ] . Taking into account Part 3 with (2.20) and Part 4 with (2.27), we consider the following cases.

     
  2. (A)
    | γ | 2 / ( 1 + α ) . Then the maximum value of F α , γ is attained in a corner of R. Thus by (2.17)-(2.19) an easy computation shows that
    max ( x , y ) R F α , γ ( x , y ) = F α , γ ( 2 , 1 ) = 1 3 | α 2 + ( 1 + α ) 2 γ | .
    (2.38)
     
  3. (B)
    2 / ( 1 + α ) | γ | < 2 / ( α + 1 ) . Then the maximum value of F α , γ is attained in a corner of R or in ( x α , γ , 1 ) . Thus, by (2.17)-(2.19) and (2.21), we calculate that
    max ( x , y ) R F α , γ ( x , y ) = F α , γ ( x α , γ , 1 ) = 1 3 α 2 γ 2 2 | γ | + 1 3 α 2 | 1 + γ | + 2 3 .
    (2.39)
     
  4. (C)
    γ = 0 . Then the maximum value of F α , 0 is attained in a corner of R or in the point ( 2 , y α , 0 ) = ( 2 , 1 / 2 ) . Thus, by (2.17)-(2.19) and (2.28) with γ : = 0 , we see that
    max ( x , y ) R F α , 0 ( x , y ) = F α , 0 ( 0 , 1 ) = 1 3 α 2 + 2 3 .
    (2.40)
     
  5. (D)
    0 < | γ | < 2 / ( α + 1 ) . Then we compare all values (2.17)-(2.19), and by (2.21) and (2.28), F α , γ ( x α , γ , 1 ) and F α , γ ( 2 , y α , γ ) . We will show that the value F α , γ ( x α , γ , 1 ) is the largest one. As it is easy to observe, it is enough to prove that
    F α , γ ( x α , γ , 1 ) F α , γ ( 2 , y α , γ )
    (2.41)
     
i.e., in view of (2.21) and (2.28), after a simple computation, we have
2 ( 2 + α 2 γ 2 2 | γ | ) 4 α 2 γ 2 + α | γ | 8 + α 2 γ 2 2 ( 4 γ 2 ) ( 8 + α 2 γ 2 + 3 α | γ | ) .
(2.42)
As | γ | < 2 , so squaring (2.42) and computing, we equivalently have
α 4 | γ | 5 + ( 6 α 4 8 α 2 ) γ 4 + ( 20 α 2 + 8 ) | γ | 3 ( 8 α 2 + 16 ) γ 2 24 | γ | + 48 α | γ | ( 2 | γ | ) ( α 2 γ 2 + 8 ) 3 / 2 .
(2.43)
To verify that (2.43) holds, setting u α : = 2 / ( α + 1 ) , we will show that for every α [ 0 , 1 ] we have
Q α ( u ) : = α 4 u 5 + ( 6 α 4 8 α 2 ) u 4 + ( 20 α 2 + 8 ) u 3 ( 8 α 2 + 16 ) u 2 24 u + 48 α u ( 2 u ) ( α 2 u 2 + 8 ) 3 / 2 = : S α ( u ) , u [ 0 , u α ] .
(2.44)

( 1 o ) For α = 0 , inequality (2.43) is evidently true.

( 2 o ) For α = 1 , we have u 1 = 1 and inequality (2.44) after computing is equivalent to the evidently true inequality
( u 1 ) 2 ( 2 u 6 + 32 u 4 + 40 u 3 92 u 2 + 144 u + 144 ) 0 , u [ 0 , 1 ] .
( 3 o ) Let α ( 0 , 1 ) . We will show that for u [ 0 , u α ] ,
V α ( u ) : = Q α 2 ( u ) S α 2 ( u ) = ( Q α ( u ) S α ( u ) ) ( Q α ( u ) + S α ( u ) ) > 0 .
(2.45)
Further, taking into account that Q α and S α are continuous functions with
Q α ( 0 ) S α ( 0 ) = 48 > 0 ,
from (2.45) we deduce that
Q α ( u ) S α ( u ) > 0 , u [ 0 , u α ] ,

which confirms (2.44) and further (2.43).

Now we prove that (2.45) holds, i.e., after computation we have
V α ( u ) = ( α 8 α 6 ) u 9 + ( 2 α 8 5 α 6 + 5 α 4 ) u 8 + ( 20 α 6 16 α 4 8 α 2 ) u 7 + ( 12 α 6 + 6 α 4 + 36 α 2 + 4 ) u 6 + ( 16 α 4 24 α 2 16 ) u 5 ( 8 α 4 + 124 α 2 + 8 ) u 4 + ( 272 α 2 + 96 ) u 3 ( 176 α 2 + 60 ) u 2 144 u + 144 > 0 , u [ 0 , u α ] .
As in (2.6), let ( q k ) , k = 0 , 1 , , 9 , be a sequence of polynomials of the form
q k ( u ) = j = 0 k a j u k j , u [ 0 , u α ] ,
corresponding to the polynomial Q : = V α in (2.5) for Laguerre’s rule in [ 0 , u α ] .
  1. (a)
    First we check the signs of the elements of the sequence ( q k ( 0 ) ) , i.e., of the sequence ( a k ) for k = 0 , 1 , , 9 . A simple computation shows that for α ( 0 , 1 ) we have q 0 ( 0 ) < 0 , q 1 ( 0 ) > 0 , q 2 ( 0 ) < 0 , q 3 ( 0 ) > 0 , q 4 ( 0 ) < 0 , q 5 ( 0 ) < 0 , q 6 ( 0 ) > 0 , q 7 ( 0 ) < 0 , q 8 ( 0 ) < 0 and q 9 ( 0 ) > 0 . Hence
    N ( V α ; 0 ) = 7 , α ( 0 , 1 ) .
    (2.46)
     
  2. (b)
    Now we check the signs of the elements of the sequence ( q k ( u α ) ) for k = 0 , 1 , , 9 . After the detailed computation and arguments based on Laguerre’s rule, we show that q 0 ( u α ) < 0 , q 1 ( u α ) > 0 , q 2 ( u α ) < 0 , q 3 ( u α ) > 0 , q 5 ( u α ) < 0 , q 6 ( u α ) > 0 , q 7 ( u α ) > 0 , q 8 ( u α ) < 0 and q 9 ( u α ) > 0 . Moreover, we show that there exists a unique α 0 ( 0 , 1 ) such that q 4 ( u α 0 ) = 0 and q 4 ( u α ) < 0 for α ( 0 , α 0 ) and q 4 ( u α ) > 0 for α ( α 0 , 1 ) . Thus for three cases, namely for α ( 0 , α 0 ) , α : = α 0 and α ( α 0 , 1 ) , we have
    N ( V α ; u α ) = 7 .
     

Hence, by (2.46) and by Corollary 2.3, we conclude that for each α ( 0 , 1 ) the polynomial V α has no zero in ( 0 , u α ) , and since V α ( 0 ) = 144 > 0 , so (2.45) holds.

Now we shortly explain the method of describing the signs of q k ( u α ) , k = 1 , , 9 . Note that the case k = 0 is evident since
q 0 ( u α ) = α 6 ( α 2 1 ) < 0 , α ( 0 , 1 ) .
For other cases, i.e., for k = 1 , , 9 , we use Laguerre’s rule in each case in the same manner. We clarify this for the case k = 2 . We have
q 2 ( u ) = ( α 8 α 6 ) u 2 + ( 2 α 8 5 α 6 + 5 α 4 ) u + 20 α 6 16 α 4 8 α 2 , u [ 0 , u α ] .
We will show that
q 2 ( u α ) < 0 , α ( 0 , 1 ) ,
(2.47)
i.e., after computing we get
2 α 2 ( 2 α 4 5 α 2 + 5 ) < ( 2 α 5 18 α 4 + 16 α 2 + 8 ) α + 1 , α ( 0 , 1 ) .
(2.48)
To verify that (2.48) holds, we will show that
r ( t ) : = 2 t 2 ( 2 t 4 5 t 2 + 5 ) < ( 2 t 5 18 t 4 + 16 t 2 + 8 ) t + 1 = : s ( t ) , t [ 0 , 1 ] .
(2.49)
Applying Corollary 2.3, we see that
w ( t ) : = s 2 ( t ) r 2 ( t ) = 8 t 12 + 4 t 11 + 116 t 10 + 396 t 9 + 170 t 8 640 t 7 508 t 6 64 t 5 82 t 4 + 256 t 3 + 256 t 2 + 64 t + 64 = : j = 0 12 b j t 12 j > 0 , t [ 0 , 1 ] .
(2.50)

Indeed, the numbers of sign changes in the sequence of polynomial coefficients ( b k ) and in the sequence of sums ( j = 0 k b j ) , where k = 0 , 1 , , 12 , equal 3, i.e., N ( w ; 0 ) = N ( w ; 1 ) = 3 . Thus we conclude that the polynomial w has no zero in the interval ( 0 , 1 ) and, since w ( 0 ) = 64 > 0 , so (2.50) holds. Hence, and by the fact that s ( 0 ) > r ( 0 ) , we deduce that (2.49) holds, which confirms (2.47).

Note here that for the case k = 4 we show that the equation
q 4 ( u α ) = 0
has a unique solution α = : α 0 ( 0 , 1 ) . In this case, we show by using Laguerre’s rule that the corresponding polynomial w as in (2.50) has a unique zero in ( 0 , 1 ) and further we deduce that
q 4 ( u α ) < 0 , α ( 0 , α 0 )
and
q 4 ( u α ) > 0 , α ( α 0 , 1 ) .
Summarizing, taking into account (2.38)-(2.41), we have
max ( x , y ) R F α , γ ( x , y ) = { 1 3 | α 2 + ( 1 + α ) 2 γ | , | γ | 2 1 + α , 1 3 α 2 γ 2 2 | γ | + 1 3 α 2 | 1 + γ | + 2 3 , | γ | 2 1 + α .

Finally, substituting γ = 2 3 λ , the above yields (2.7).

Now we deal with the sharpness of the result. Let α ( 0 , 1 ] . We prove that for each λ ( , 2 / 3 ] [ τ 2 ( α ) , + ) inequality (2.7) is sharp. Let λ [ τ 1 ( α ) , 2 / 3 ] . Since then
α 2 ( 1 3 ( 2 3 λ ) 2 2 | 2 3 λ | + | 1 λ | ) + 2 3 = α 2 ( 4 9 λ 1 3 ) + 2 3 ,
inequality (2.7) is of the form
max f C ( g α ) Φ λ ( f ) α 2 ( 4 9 λ 1 3 ) + 2 3 , λ [ τ 1 ( α ) , 2 / 3 ] .
(2.51)
Let t α , λ : = α ( 2 / ( 3 λ ) 1 ) . Then t α , λ [ 0 , 1 ] and, in view of (2.4), p t α , λ , 0 P , with c 1 = 2 t α , λ and c 2 = 2 . Setting δ : = 0 and p : = p t α , λ , 0 into (2.12), we get the function f α , t α , λ given by equation (2.8) for which, by (2.14) and (2.15),
a 2 = t α , λ + α = 2 α 3 λ , a 3 = 1 3 ( 2 + 4 α t α , λ + 3 α 2 ) = 2 3 + α 2 ( 8 9 λ 1 3 ) .
Hence
Φ λ ( f α , t α , λ ) = 2 3 + α 2 ( 4 9 λ 1 3 ) ,

which makes the equality in (2.51), so in (2.7). Clearly, f α , t α , λ C 0 ( g α ) because (2.12) is satisfied for δ = 0 .

Let λ ( , τ 1 ( α ) ] [ τ 2 ( α ) , + ) . Then inequality (2.7) is of the form
max f C ( g α ) Φ λ ( f ) | 2 3 + 4 3 α + α 2 ( 1 + α ) 2 λ | .
(2.52)
Set δ : = 0 and p : = L into (2.12). Then for α ( 0 , 1 ) we get the function f α , 1 given by (2.9) and for α = 1 we get the Koebe function f 1 , 1 = k , with
a 2 = 1 + α , a 3 = 1 3 ( 2 + 4 α + 3 α 2 ) , α ( 0 , 1 ] .
Hence
Φ λ ( f α , 1 ) = | 2 3 + 4 3 α + α 2 ( 1 + α ) 2 λ | ,

which makes the equality in (2.52), so in (2.7). Clearly, f α , 1 C 0 ( g α ) for α ( 0 , 1 ] .

Let α : = 0 . We prove that for each λ R inequality (2.7) is sharp. For λ [ τ 1 ( 0 ) , τ 2 ( 0 ) ] = [ 0 , 4 / 3 ] inequality (2.7) is of the form
max f P Φ λ ( f ) 2 3 .
(2.53)
Setting δ : = 0 and, by (2.4), p : = p 0 , 0 into (2.12), we get the function f 0 , 0 given by (2.10) with a 2 = 0 and a 3 = 2 / 3 . Hence
Φ λ ( f 0 , 0 ) = 2 3 ,

which makes the equality in (2.53), so in (2.7).

For λ ( , 0 ] [ 4 / 3 , + ) inequality (2.7) is of the form
max f P Φ λ ( f ) | 2 3 λ | .
(2.54)
Setting δ : = 0 and p : = L into (2.12), we get the function f 0 , 1 given by (2.11) with a 2 = 1 and a 3 = 2 / 3 . Hence
Φ λ ( f 0 , 1 ) = | 2 3 λ | ,

which makes the equality in (2.54), so in (2.7). □

Remark 2.5 Particularly, from (2.7) for α [ 0 , 1 ] we have
max f C ( g α ) Φ λ ( f ) = { | 2 3 + 4 3 α + α 2 ( 1 + α ) 2 λ | , λ R ( τ 1 ( α ) , τ 2 ( α ) ) , 2 3 + α 2 ( 4 9 λ 1 3 ) , λ [ τ 1 ( α ) , 2 / 3 ] .

For α : = 1 we have τ 1 ( 1 ) = 1 / 3 and τ 2 ( 1 ) = 1 , g 1 = k , and then Theorem 2.4 reduces to the result of [26] as follows.

Corollary 2.6
max f C ( k ) Φ λ ( f ) { | 3 4 λ | , λ ( , 1 / 3 ] [ 1 , + ) , 2 3 + 1 3 ( 2 3 λ ) 2 2 | 2 3 λ | + | 1 λ | , λ [ 1 / 3 , 1 ] .
(2.55)

For each λ ( , 2 / 3 ] [ 1 , + ) , the inequality is sharp and the equality is attained by a function in C 0 ( k ) . In particular, for each λ [ 1 / 3 , 2 / 3 ] the second equality in (2.55) is attained by the function f t λ : = f 1 , t 1 , λ given by differential equation (2.8), where t λ : = t 1 , λ . For each λ ( , 1 / 3 ] [ 1 , + ) , the first equality in (2.55) is attained by the Koebe function f 1 : = k .

For α : = 0 we have τ 1 ( 0 ) = 0 and τ 2 ( 0 ) = 4 / 3 , and Theorem 2.4 yields the following.

Corollary 2.7
max f P Φ λ ( f ) = { | 2 3 λ | , λ ( , 0 ] [ 4 / 3 , + ) , 2 3 , λ [ 0 , 4 / 3 ] .
(2.56)

For each λ [ 0 , 4 / 3 ] the second equality in (2.56) is attained by the function given by (2.10). For each λ ( , 0 ] [ 4 / 3 , + ) , the first equality in (2.56) is attained by function (2.11).

Declarations

Authors’ Affiliations

(1)
Department of Applied Mathematics, University of Warmia and Mazury
(2)
Department of Analysis and Differential Equations, University of Warmia and Mazury

References

  1. Fekete M, Szegö G: Eine Bemerkung über ungerade schlichte Funktionen. J. Lond. Math. Soc. 1933, 8: 85–89.View ArticleMATHGoogle Scholar
  2. Jakubowski ZJ:Sur le maximum de la fonctionnelle | A 3 − α A 2 2 | ( 0 ≤ α < 1 ) dans la famille de fonctions F M . Bull. Soc. Sci. Lett. Lódź 1962.,13(1): Article ID 19Google Scholar
  3. Keogh FR, Merkes EP: A coefficient inequality for certain classes of analytic functions. Proc. Am. Math. Soc. 1969, 20: 8–12.MathSciNetView ArticleMATHGoogle Scholar
  4. Pfluger A: The Fekete-Szegö inequality for complex parameter. Complex Var. Theory Appl. 1986, 7: 149–160. 10.1080/17476938608814195MathSciNetView ArticleMATHGoogle Scholar
  5. Koepf W: On the Fekete-Szegö problem for close-to-convex functions. Proc. Am. Math. Soc. 1987, 101: 89–95.MathSciNetMATHGoogle Scholar
  6. Kanas S, Lecko A: On the Fekete-Szegö problem and the domain of convexity for a certain class of univalent functions. Folia Sci. Univ. Tech. Resov. 1990, 73: 49–57.MathSciNetMATHGoogle Scholar
  7. London RR: Fekete-Szegö inequalities for close-to-convex functions. Proc. Am. Math. Soc. 1993,117(4):947–950.MathSciNetMATHGoogle Scholar
  8. Kim YC, Choi JH, Sugawa T: Coefficient bounds and convolution properties for certain classes of close-to-convex functions. Proc. Jpn. Acad. 2000, 76: 95–98. 10.3792/pjaa.76.95MathSciNetView ArticleMATHGoogle Scholar
  9. Bhowmik B, Ponnusamy S, Wirths KJ: On the Fekete-Szegö problem for concave univalent functions. J. Math. Anal. Appl. 2011, 373: 432–438. 10.1016/j.jmaa.2010.07.054MathSciNetView ArticleMATHGoogle Scholar
  10. Kanas S: An unified approach to the Fekete-Szegö problem. Appl. Math. Comput. 2012, 218: 8453–8461. 10.1016/j.amc.2012.01.070MathSciNetView ArticleMATHGoogle Scholar
  11. Ozaki S: On the theory of multivalent functions. Sci. Rep. Tokyo Bunrika Daigaku, Sect. A. Math. Phys. Chem. 1935, 2: 167–188.MATHGoogle Scholar
  12. Kaplan W: Close to convex schlicht functions. Mich. Math. J. 1952, 1: 169–185.View ArticleMATHMathSciNetGoogle Scholar
  13. Goodman AW, Saff EB: On the definition of close-to-convex function. Int. J. Math. Math. Sci. 1978, 1: 125–132. 10.1155/S0161171278000150MathSciNetView ArticleMATHGoogle Scholar
  14. Lecko A: Some generalization of analytic condition for class of functions convex in a given direction. Folia Sci. Univ. Tech. Resov. 1993,121(14):23–34.MathSciNetMATHGoogle Scholar
  15. Lecko A: A generalization of analytic condition for convexity in one direction. Demonstr. Math. 1997,XXX(1):155–170.MathSciNetMATHGoogle Scholar
  16. Lecko A, Yaguchi T: A generalization of the condition due to Robertson. Math. Jpn. 1998,47(1):133–141.MathSciNetMATHGoogle Scholar
  17. Robertson MS: Analytic functions star-like in one direction. Am. J. Math. 1936, 58: 465–472. 10.2307/2370963View ArticleMATHMathSciNetGoogle Scholar
  18. Hengartner W, Schober G: On schlicht mappings to domains convex in one direction. Comment. Math. Helv. 1970, 45: 303–314. 10.1007/BF02567334MathSciNetView ArticleMATHGoogle Scholar
  19. Ciozda W: Sur la classe des fonctions convexes vers l’axe réel négatif. Bull. Acad. Pol. Sci., Sér. Sci. Math. 1979,27(3–4):255–261.MathSciNetMATHGoogle Scholar
  20. Lecko A: On the class of functions convex in the negative direction of the imaginary axis. J. Aust. Math. Soc. 2002, 73: 1–10. 10.1017/S1446788700008430MathSciNetView ArticleMATHGoogle Scholar
  21. Lecko A: Some Methods in the Theory of Univalent Functions. Oficyna Wydawnicza Politechniki Rzeszowskiej, Rzeszów; 2005.MATHGoogle Scholar
  22. Elin M, Khavinson D, Reich S, Shoikhet D: Linearization models for parabolic dynamical systems via Abel’s functional equation. Ann. Acad. Sci. Fenn., Math. 2010, 35: 439–472.MathSciNetView ArticleMATHGoogle Scholar
  23. Goodman AW: Univalent Functions. Mariner, Tampa; 1983.MATHGoogle Scholar
  24. Noshiro K: On the theory of schlicht functions. J. Fac. Sci., Hokkaido Univ., Ser. 1 1934–1935, 2: 129–155.MATHGoogle Scholar
  25. Warschawski SE: On the higher derivatives at the boundary in conformal mapping. Trans. Am. Math. Soc. 1935,38(2):310–340. 10.1090/S0002-9947-1935-1501813-XMathSciNetView ArticleMATHGoogle Scholar
  26. Kowalczyk, B, Lecko, A: Fekete-Szegö problem for close-to-convex functions with respect to the Koebe function (submitted)Google Scholar
  27. Koepf W: On the Fekete-Szegö problem for close-to-convex functions II. Arch. Math. 1987, 49: 420–433. 10.1007/BF01194100MathSciNetView ArticleMATHGoogle Scholar
  28. Abdel-Gawad HR, Thomas DK: The Fekete-Szegö problem for strongly close-to-convex functions. Proc. Am. Math. Soc. 1992,114(2):345–349.MathSciNetMATHGoogle Scholar
  29. Darus M, Thomas DK: The Fekete-Szegö theorem for strongly close-to-convex functions. Sci. Math. 2000,3(2):201–212.MathSciNetMATHGoogle Scholar
  30. Choi JH, Kim YC, Sugawa T: A general approach to the Fekete-Szegö problem. J. Math. Soc. Jpn. 2007,59(3):707–727. 10.2969/jmsj/05930707MathSciNetView ArticleMATHGoogle Scholar
  31. Pommerenke C: Univalent Functions. Vandenhoeck & Ruprecht, Göttingen; 1975.MATHGoogle Scholar
  32. Kowalczyk, B, Lecko, A: Fekete-Szegö problem for a certain subclass of close-to-convex functions. Bull. Malays. Math. Soc. (accepted)Google Scholar
  33. Laguerre EN: Sur la théorie des équations numériques. J. Math. Pures Appl. 1883, 9: 99–146. In: Oeuvres de Laguerre, vol. 1, pp. 3–47, Paris (1898)MATHGoogle Scholar
  34. Jameson GJO: Counting zeros of generalized polynomials: Descartes’ rule of signs and Leguerre’s extensions. Math. Gaz. 2006,90(518):223–234.Google Scholar
  35. Turowicz A: Geometria Zer Wielomianów. PWN, Warszawa; 1967. (Geometry of Zeros of Polynomials)Google Scholar

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© Kowalczyk and Lecko; licensee Springer. 2014

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