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

Abstract

Given α[0,1], let g α (z):=z/ ( 1 α z ) 2 , zD:={zC:|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,zD.

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.

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:={zC:|z|<1} of the form

f(z)=z+ n = 2 a n z n ,zD.
(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 fA such that

Re z f ( z ) f ( z ) >0,zD.
(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 fA such that

Re { e i δ z f ( z ) g ( z ) } >0,zD.
(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 ,zD.
(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,zD,
(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:t0}h(D) for every wh(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, zD, and then condition (1.5) is of the form

Re { e i δ f ( z ) } >0,zD.
(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 λR(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 ,zD,
(2.1)

having a positive real part in D. For each εT:={zC:|z|=1}, let

L ε (z):= 1 + ε z 1 ε z ,zD,L:= L 1 .

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

Lemma 2.1 If pP is of the form (2.1), then

| c n |2,nN,
(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 , εT. 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 θR.

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 IR, 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 λR(2/3, τ 2 (α)), as well as for α:=0 and each λR, 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,zD,
    (2.8)

where t α , λ :=α(2/(3λ)1);

  1. (ii)

    when α(0,1], for each λR( 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 ,zD,log1:=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 ,zD,log1:=0;
    (2.10)

and for each λR(0,4/3) the first equality in (2.7) is attained by the function

f 0 , 1 (z):=z2log(1z),zD,log1:=0.
(2.11)

Proof Fix α[0,1]. Observe from (1.5) that fC( g α ) if and only if

e i δ ( 1 α z ) 2 f (z)=p(z)cosδ+isinδ,zD
(2.12)

for some δ(π/2,π/2) and pP. Thus

z f (z)= e i δ g α (z) ( p ( z ) cos δ + i sin δ ) ,zD.
(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 λR. 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:=cosδ. Clearly, y(0,1] and, in view of (2.2), x[0,2]. It is convenient to use γ:=23λ instead of λ in further computation. For γR, 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 γR, 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 γR, 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+2α)|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

α00<|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 γR the function F α , γ has no critical point in (0,2)×(0,1).

Since y0 and x0, 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 4x+4α|γ|=0,x(0,2).
(2.32)

It follows at once that for α=0 and γR, as well as for α(0,1] and |γ|1/α, equation (2.32) has no root. Thus by (2.30) we consider

α0|γ|20<|γ|<1/αx>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

α0|γ|20<|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

α00<|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 γR 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=23λ, 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/(3λ)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 λR 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).

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Kowalczyk, B., Lecko, A. The Fekete-Szegö inequality for close-to-convex functions with respect to a certain starlike function dependent on a real parameter. J Inequal Appl 2014, 65 (2014). https://doi.org/10.1186/1029-242X-2014-65

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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