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Hadamard product of analytic functions and some special regions and curves

Abstract

In this paper we present some new applications of convolution and subordination in geometric function theory. The paper deals with several ideas and techniques used in this topic. Besides being an application to those results, it provides interesting corollaries concerning special functions, regions and curves.

MSC:30C45, 30C80.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

By we denote a class of analytic functions in the unit disc Δ={zC:|z|<1}. Let denote the subclass of consisting of functions normalized by f(0)=0, f (0)=1 and satisfying

Re ( 1 + z f ( z ) f ( z ) ) >0for zΔ.
(1.1)

Of course the functions from map the unit disc onto convex domains.

We say that the function fH is subordinate to the function gH in the unit disc Δ (written fg) if and only if there exists an analytic function wH such that w(0)=0, |w(z)|<1 and f(z)=g(w(z)) for zΔ. Therefore fg in Δ implies f(Δ)g(Δ). If, additionally, g is univalent in Δ, then

fgif and only iff(0)=g(0)andf(Δ)g(Δ).
(1.2)

Recall that the Hadamard product or convolution of two power series,

f(z)= n = 0 a n z n andg(z)= n = 0 b n z n ,

is defined as

(fg)(z)=f(z)g(z)= n = 0 a n b n z n .

The convolution has the algebraic properties of ordinary multiplication. We now look at some problems on convolution and at some of the relations between the convolution and the subordination. One can consider the following problems.

Problem 1 Find the subclasses , , of the class such that

AB=C,

where

AB= { h ( z ) = ( f g ) ( z )  for all  f A  and for all  f B } .

Problem 2 Let the sets MC and NC be given. Find a set KC, as small as possible, such that

f(Δ)Mandg(Δ)N(fg)(Δ)Kfor all f,gD,

where is some subclass of .

If M=F(Δ) and N=G(Δ), where F,GH are univalent in Δ, then, by (1.2), Problem 2 above becomes the following one:

Problem 3 Let the univalent functions F, G be given. Find a function H, H(Δ)=K, and a class DH such that

fFandgGfgHfor all f,gD.

Ruscheweyh and Sheil-Small [1] proved the Pòlya-Schoenberg conjecture that the class of convex univalent functions is preserved under convolution, namely KK=K. They proved also that the class of starlike functions and the class of close-to-convex functions are closed under convolution with the class . Another solution of Problem 1 is the following theorem.

Theorem A (see [2])

Let us denote

P x = { p H : p ( 0 ) = 1  and  R e { p ( z ) } > x } .

Suppose that α1, β1. Then

P α P β = P γ ,

where γ=12(1α)(1β).

Problem 2, when the sets M and K are discs K( s i , R i ) with center s i and radius R i , was considered in [3] and in other papers. One of results obtained there is the following theorem.

Theorem B (see [3])

Let s i C, R i R; i=1,2. If f,gH and f(0)=a, g(0)=b, then the following implication holds:

f(Δ)K( s 1 , R 1 )andg(Δ)K( s 2 , R 2 )(fg)(Δ)K(s,R)for all f,gH,

where

s=ab+m( s 1 a)( s 2 b),R=m R 1 R 2 ,m= ( R 1 2 | s 1 a | 2 ) ( R 2 2 | s 2 a | 2 ) R 1 2 R 2 2 | s 1 a | 2 | s 2 a | 2 .

In this paper we shall look for a solution of the following modified Problem 3.

Problem 4 Find a function FH and a class DH such that

gfgFfFfor all fD,gH.

The next theorem is a solution of Problem 4.

Theorem C (see [4])

Assume that fH, f (0)0, and that it satisfies

Re { 1 + z f ( z ) f ( z ) } > 1 2 for zΔ.

If gf, then

L[g]L[f],

where

L[f](z)= 2 z 0 z f(t)dt

is the Libera operator [5].

If f(z)= a 0 + a 1 z+ a 2 z 2 + a 3 z 3 + , then we have

L [ f ] ( z ) = 2 z 0 z f ( t ) d t = n = 0 2 a n n + 1 z n = f ( z ) ( 2 z log 1 1 z ) .

Therefore, Theorem C may be written as

g f g ( 2 z log 1 1 z ) f ( 2 z log 1 1 z ) for all  f K ( 1 / 2 ) , g H ,
(1.3)

where

K(α):= { f H : f ( 0 ) 0  and  Re { 1 + z f ( z ) f ( z ) } > α  for  z Δ } .
(1.4)

Theorem C improves an earlier result from [6] with the stronger assumption fK(0) and the same conclusion. Moreover, if we assume more, that fK(0) and f is univalent in Δ, then in (1.3) instead of F(z)=2/zlog(1z) we can put any convex univalent function. This is contained in the following result due to Ruscheweyh and Stankiewicz [7].

Theorem D (see [7])

Let F and G be convex univalent in Δ. Then, for all functions f,gH,

fFandgGfgFG.
(1.5)

This relationship allows us to obtain several subordination results about convolution. Note that there are no assumptions about the normalization of functions. It is one of the solutions of Problem 3. Many of the convolution properties were studied by Ruscheweyh in [8], and they have found many applications in various fields. The book [8] is also an excellent survey of the results. For the recent results on the Hadamard product in geometric function theory, see [913].

2 A family of functions

Throughout this section we consider the family of analytic functions

F s (z)= z 1 s z 2 (zΔ),
(2.1)

where s[0,1]. In Section 3 we prove that for given s[0,1) the curve F s ( e i φ ), φ[0,2π) is the Booth lemniscate. Let S denote a class of starlike functions consisting of functions f such that z f K.

Lemma 2.1 The function F s given by (2.1) is a starlike univalent function for s[0,1].

Proof We have

Re { z F s ( z ) F s ( z ) } =Re { 1 + s z 2 1 s z 2 } >0for all zΔ.

Thus, it is obvious that F s is a starlike univalent function for s[0,1]. □

Lemma 2.2 Let the function F s be of the form (2.1), and let K(1/2) be given by (1.4). If |s|6 33 0.255437, then F s K(1/2).

Proof Assume that s z 2 =r(cost+isint), 0r<1, 0t2π and set

H(t,r):= z F s ( z ) F s ( z ) +2= 6 ( 1 + s z 2 ) 4 1 ( s z 2 ) 2 = 2 + 6 r cos t + 6 i r sin t 1 r 2 cos 2 t i r 2 sin 2 t .
(2.2)

By (1.4) if Re{H(t,r)}>1/2 for all r[0, r 0 ), t[0,2π], then F s K(1/2) for all |s| r 0 . After some calculations, we obtain

Re { H ( t , r ) } = 4 r 2 x 2 + 6 r ( 1 r 2 ) x + 2 ( 1 + r 2 ) ( 1 + r 2 ) 2 4 r 2 x 2 := n ( x , r ) d ( x , r ) ,
(2.3)

where x=cost. Thus we shall find r 0 in [0,1] such that

n ( x , r ) d ( x , r ) >1/2for all r[0, r 0 ) and for all x[1,1].
(2.4)

It is easy to verify that d(x,r)>0 for all r[0,1) and for all x[1,1], and that n(1,r)=2(1 r 2 )(13r)0 for all r[1/3,1). Hence, to find r 0 in [0,1] satisfying (2.4), we may restrict our considerations to the interval [0,1/3). Then, analyzing n(x,r) as a function of one variable x, we get

min { n ( x , r ) : 1 x 1 } =n(1,r)=2 ( 1 r 2 ) (13r)>0

for all r[0,1/3). Moreover, we see that

min { d ( x , r ) : 1 x 1 } =d(1,r)= ( 1 r 2 ) 2 >0.

Therefore, if 0r<1/3, then

min { n ( x , r ) d ( x , r ) : x [ 1 , 1 ] } = 2 ( 1 3 r ) 1 r 2 .

To find r 0 in [0,1/3) satisfying (2.4), it is sufficient to solve the inequality

2 ( 1 3 r ) 1 r 2 > 1 2 ,

which is true for 0r r 0 =6 33 . □

Moreover, analogous consideration as in the above proof leads us to the fact that F s K(1/2) for s>6 33 0.255437. In Lemma 3.2 we show that F s K for 0s32 2 0.1715.

Theorem 2.3 If 0s6 33 0.255437 and fH, then

f F s L[f]L[ F s ],
(2.5)

where

L[f](z)= 2 z 0 z f(t)dt

is the Libera operator.

Proof By Lemma 2.2 we obtain

Re { 1 + z F s ( z ) F s ( z ) } > 1 2 for zΔ.

Therefore, using Theorem C, we get (2.5). □

Corollary 2.4 If 0s6 33 and

f(z)= n = 1 a n z n ,

then

n = 1 a n z n n = 1 s n 1 z 2 n 1 n = 1 a n n + 1 z n n = 1 s n 1 n + 1 z 2 n 1 .

It is known [14] that if

n = 0 α n z n n = 0 β n z n ,

then

n = 0 k | α n | 2 n = 0 k | β n | 2 .

Hence, for 0s6 33 , Corollary 2.4 gives also that

n = 1 a n z n F s ( z ) n = 1 | a n | 2 ( n + 1 ) 2 n = 1 s 2 ( n 1 ) ( n + 1 ) 2 = 1 s 2 + 1 s 2 0 t e t s 2 d t = Φ ( s 2 , 2 , 1 ) 1 s 2 ,

because for Re{a}>0, 1x<1, Re{c}>0,

Φ(x,c,a)= k = 0 x k ( k + a ) c = 1 Γ ( c ) 0 t c 1 e a t 1 x e t dt.

Theorem 2.5 If 0 s i 6 33 , i=1,2, and the functions g 1 and g 2 are in the class , then

g 1 F s 1 and g 2 F s 2 L[ g 1 ]L[ g 2 ]L [ L [ F s 1 s 2 ] ] .
(2.6)

Proof By Theorem 2.3 we have L[ g 1 ]L[ F s 1 ] and L[ g 2 ]L[ F s 2 ]. By Lemma 2.1 F s S , s[0,1], so the functions L[ F s 1 ], L[ F s 2 ] are in because L[ S ]=K. Using Theorem D, we obtain

L[ g 1 ]L[ g 2 ]L[ F s 1 ]L[ F s 2 ]

but L[ F s 1 ]L[ F s 2 ]=L[L[ F s 1 s 2 ]] because

F s 1 (z) F s 2 (z)= z 1 s 1 z 2 z 1 s 2 z 2 = z 1 s 1 s 2 z 2 = F s 1 s 2 (z).

 □

3 A family of sets

In this section we investigate the properties of a one-parameter family of the sets D(s), s[0,1], such that

D(s)= { x + i y C : ( x 2 + y 2 ) 2 x 2 ( 1 s ) 2 y 2 ( 1 + s ) 2 < 0 } ,
(3.1)

when s[0,1) and

D(1)= { x + i y C : x + i y i t  for all  t ( , 1 ) ( 1 , ) } .
(3.2)

Lemma 3.1 Let s[0,1], then D(s)= F s (Δ), where

F s (z)= z 1 s z 2 (zΔ).

Proof The function F s is analytic in the unit disc, so for the proof we need to find an image of the circle |z|=1 under the function F s . For s[0,1) and for φ[0,2π), we have

F s ( e i φ ) = e i φ 1 s e 2 i φ 1 s e 2 i φ 1 s e 2 i φ = e i φ s e i φ 1 + s 2 s ( e 2 i φ + e 2 i φ ) = ( 1 s ) cos φ + i ( 1 + s ) sin φ 1 + s 2 2 s cos 2 φ .
(3.3)

Let us define

x=Re { F s ( e i φ ) } = ( 1 s ) cos φ 1 + s 2 2 s cos 2 φ
(3.4)

and

y=Im { F s ( e i φ ) } = ( 1 + s ) sin φ 1 + s 2 2 s cos 2 φ .
(3.5)

Then, after some calculations, we can obtain from (3.4) and (3.5) that

x 2 + y 2 = 1 1 + s 2 2 s cos 2 φ and(x,y)(0,0).
(3.6)

Therefore, using the relations (3.3)-(3.5), we can find that F s ( e i φ ) is an algebraic curve of order four whose equation in orthogonal Cartesian coordinates is

( x 2 + y 2 ) 2 x 2 ( 1 s ) 2 y 2 ( 1 + s ) 2 =0and(x,y)(0,0).
(3.7)

Thus, because F s (0)=0 for s[0,1), the set F s (Δ) is bounded by the curve (3.7). Moreover, for s=1, the function F s becomes

F 1 (z)= z 1 z 2

and it is easy to see that F 1 (Δ)=D(1), see Figure 1. Then the proof is completed.  □

Figure 1
figure 1

F 1 (Δ)=D(1) .

Notice that a curve described by

( x 2 + y 2 ) 2 ( n 4 + 2 m 2 ) x 2 ( n 4 2 m 2 ) y 2 =0and(x,y)(0,0)
(3.8)

is called the Booth lemniscate, named after Booth [15, 16]. The Booth lemniscate is called elliptic if n 4 >2 m 2 , while for n 4 <2 m 2 it is termed hyperbolic. Thus it is clear that the curve (3.7) is the Booth lemniscate of elliptic type. The Booth lemniscate is a special case of a Persian curve. The Booth lemniscate of elliptic type can be described geometrically in the following equivalent ways.

  1. 1.

    Suppose that s[0,1). Let C(S,R) be a circle with the center S and the length of the radius R such that

    S= ( s 1 s 2 , 0 ) andR= 1 2 ( 1 s ) .

A ray is drawn from the point x on the circle C(S,R) through the origin o that cuts also the circle C(S,R) at the point z. A point v is also on the ray and satisfies

|ov|=|xz|.
(3.9)

If the point x goes along the circle C(S,R), then the point v describes the curve (3.7), see Figure 2.

Figure 2
figure 2

D(s) , s[0,32 2 ] .

  1. 2.

    Let s[0,1). Then the curve (3.7) consists of points M such that

    | F 1 M | 2 | F 2 M | 2 = 1 + s 2 ( 1 s 2 ) 2 | O M | 2 + s 2 ( 1 s 2 ) 4 ,

where O=(0,0) and the points F 1 , F 2 are the focuses

F 1 ( s 1 s 2 , 0 ) , F 2 ( s 1 s 2 , 0 ) ,

see Figure 3.

Figure 3
figure 3

D(s) , s(32 2 ,1) .

We say that a closed curve γ is convex when it is boundary of a convex bounded domain. Otherwise, we say that the curve γ is concave.

Lemma 3.2 Suppose that F s is given by (2.1). If 0s32 2 0.1715, then the curve F s ( e i φ ), φ[0,2π) is convex and

max { Im { F s ( e i φ ) } : φ [ 0 , 2 π ) }
(3.10)

is attained at one point only. If s(32 2 ,1), then the curve F s ( e i φ ), φ[0,2π) is concave and (3.10) is attained twice. Moreover, in both cases this curve is symmetric with respect to both axes.

Proof If s=0, then the curve F s ( e i φ ), φ[0,2π) becomes a circle and it is clear that (3.10) is attained one time only. Suppose, in the sequel, that s(0,1). From (2.2) and (2.3) we have

Re { e i φ F s ( e i φ ) F s ( e i φ ) + 1 } = 1 s 4 + 6 s ( 1 s 2 ) cos φ ( 1 + s 2 ) 2 4 s 2 cos 2 φ .

The denominator is positive, hence for the convexity of F s ( e i φ ) we need the nominator to be positive. Since s 2 6s+1>0 for 0s32 2 , then the nominator is positive because

1 s 4 + 6 s ( 1 s 2 ) cos φ 1 s 4 6 s ( 1 s 2 ) = ( 1 s 2 ) ( s 2 6 s + 1 ) .

From (3.5) we have

Im { F s ( e i φ ) } = ( 1 + s ) sin φ 1 + s 2 2 s cos 2 φ := ( 1 + s ) x 1 + s 2 2 s ( 1 2 x 2 ) :=f(x).

Then f (x)=0 if and only if x 2 = ( 1 s ) 2 /(4s). That is possible when ( 1 s ) 2 /(4s)1 because x=sinφ. Therefore, using the elementary considerations, we can find that if s(32 2 ,1), then the curve F s ( e i φ ), φ[0,2π) is concave and (3.10) is attained twice, when sinφ=(1s)/(2 s ). In this case,

max { Im { F s ( e i φ ) } : φ [ 0 , 2 π ) } = 1 + s 4 s ( 1 s ) ,

see Figure 3. Moreover, if s[0,32 2 ], then (3.10) is attained at one point only such that

max { Im { F s ( e i φ ) } : φ [ 0 , 2 π ) } =Im { F s ( e i π / 2 ) } = 1 1 + s ,

see Figures 2 and 3. □

Corollary 3.3 If 0s32 2 0.1715, then F s K.

By Lemma 2.1 the functions F s , s[0,1], are starlike univalent, hence by the geometric interpretation (1.2) of a subordination under univalent functions, we obtain from Theorem 2.3 the following corollary.

Corollary 3.4 If 0s6 33 and fH, then

f(Δ)D(s)L[f](Δ)L[ F s ](Δ).
(3.11)

Theorem 3.5 Assume that 0< s i 32 2 0.1715, i=1,2, and that the functions g 1 and g 2 are in the class with g 1 (0)= g 2 (0)=0. If

g 1 (Δ)D( s 1 )and g 2 (Δ)D( s 2 ),
(3.12)

then

( g 1 g 2 )(Δ)D( s 1 s 2 ).
(3.13)

Proof Using Lemma 3.1, we may rewrite (3.12) as

g 1 (Δ) F s 1 (Δ)and g 2 (Δ) F s 2 (Δ).
(3.14)

By Corollary 3.3, F s is convex univalent in Δ for s[0,32 2 ], so F s 1 , F s 2 are convex univalent in Δ. From (1.2) and from (3.14), we obtain

g 1 F s 1 and g 2 F s 2 .
(3.15)

By Theorem D, we get from (3.15)

g 1 g 2 F s 1 F s 2 .
(3.16)

But

F s 1 (z) F s 2 (z)= z 1 s 1 z 2 z 1 s 2 z 2 = z 1 s 1 s 2 z 2 = F s 1 s 2 (z)

and hence by Lemma 3.1 the subordination (3.16) is equivalent to (3.13). □

Theorem 3.6 Assume that 0<s32 2 0.1715 and that the functions g 1 and g 2 are in the class with g 1 (0)= g 2 (0)=0. Let K(c,R) be the disc with center cR and radius R with |c|<R. If

g 1 (Δ)D(s)and g 2 (Δ)K(c,R),
(3.17)

then

( g 1 g 2 )(Δ) R 2 c 2 R D ( c s 2 R 2 ) .
(3.18)

Proof Using Lemma 3.1, we may rewrite (3.17) as

g 1 (Δ) F s 1 (Δ)and g 2 (Δ) H c , R (Δ),
(3.19)

where

H c , R (z)= ( R 2 c 2 ) z R c z (zΔ)

maps Δ onto K(c,R). By Corollary 3.3, F s is convex univalent in Δ, also H c , R is convex univalent in Δ. Therefore, by Theorem D we obtain

( g 1 g 2 ) ( z ) z 1 s z 2 ( R 2 c 2 ) z R c z = R 2 c 2 R z 1 s c 2 z 2 / R 2 = R 2 c 2 R F s c 2 / R 2 ( z ) .

The function F s c 2 / R 2 is univalent as the convolution of a convex univalent function, so by (1.2) we obtain (3.18). □

Theorem 3.7 Assume that 0<s32 2 0.1715 and that the functions g 1 and g 2 are in the class with g 1 (0)= g 2 (0)=0. Let L(φ,d)={zC:Re{z e i φ }>d}, φR, d>0, be the half-plane. If

g 1 (Δ)D(s)and g 2 (Δ)L(φ,d),
(3.20)

then

( g 1 g 2 )(Δ)2d e i φ D(s).
(3.21)

Proof Using Lemma 3.1, we may rewrite (3.20) as

g 1 (Δ) F s 1 (Δ)and g 2 (Δ) G φ , d (Δ),
(3.22)

where

G φ , d (z)= 2 d e i φ z 1 z (zΔ)

maps Δ onto L(φ,d). By Corollary 3.3, F s is convex univalent in Δ, also G φ , d is convex univalent in Δ. Therefore, by Theorem D we obtain

( g 1 g 2 ) ( z ) z 1 s z 2 2 d e i φ z 1 z = 2 d e i φ z 1 s z 2 = 2 d e i φ F s ( z ) .

Hence by (1.2) we obtain (3.21). □

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Piejko, K., Sokół, J. Hadamard product of analytic functions and some special regions and curves. J Inequal Appl 2013, 420 (2013). https://doi.org/10.1186/1029-242X-2013-420

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Keywords

  • Hadamard product
  • univalent functions
  • convex functions
  • closed convex hull
  • convolution
  • subordination