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volume 6, issue 2, article 34, 2005.

Received 30 November, 2004;

accepted 03 March, 2005.

Communicated by:K. Nikodem

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Journal of Inequalities in Pure and Applied Mathematics

ON A CONVOLUTION CONJECTURE OF BOUNDED FUNCTIONS

1KRZYSZTOF PIEJKO, ¹JANUSZ SOKÓŁ AND²JAN STANKIEWICZ

1Department of Mathematics Rzeszów University of Technology ul. W. Pola 2, 35-959 Rzeszów, Poland.

EMail:piejko@prz.rzeszow.pl EMail:jsokol@prz.rzeszow.pl

2Institute of Mathematics University of Rzeszów ul. Rejtana 16A

35-959 Rzeszów, Poland.

EMail:jan.stankiewicz@prz.rzeszow.pl

2000c Victoria University ISSN (electronic): 1443-5756 229-04

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On a Convolution Conjecture of Bounded Functions

Krzysztof Piejko, Janusz Sokół and Jan Stankiewicz

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J. Ineq. Pure and Appl. Math. 6(2) Art. 34, 2005

http://jipam.vu.edu.au

Abstract

We consider the convolutionP(A, B)? P(C, D)of the classes of analytic functions subordinated to the homographies ^1+Az_1−Bz and ^1+Cz_1−Dz respectively, where A, B, C, Dare some complex numbers. In 1988 J. Stankiewicz and Z. Stankiewicz [11] showed that for certain A, B, C, D there exist X, Y such thatP(A, B)? P(C, D) ⊂ P(X, Y). In this paper we verify the conjecture thatP(X, Y) ⊂ (A, B)? P(C, D)for someA, B, C, D, X, Y.

2000 Mathematics Subject Classification:Primary 30C45; Secondary 30C55.

Key words: Hadamard product, Convolution, Subordination, Bounded functions.

1. Introduction

Let ∆ = {z ∈ C : |z| < 1} denote the open unit disc and let H be the class of functions regular in ∆. We will denote by N the class of functions f ∈ H normalized by f(0) = 1. The class of Schwarz functions Ω is the class of functions ω ∈ H,such that ω(0) = 0 and|ω(z)| < 1 forz ∈∆. We say that a function f is subordinate to a function g in∆ (and write f ≺ g or f(z)≺g(z)) if there exists a functionω∈Ωsuch thatf(z) =g(ω(z));z ∈∆.

If the function g is univalent in ∆, thenf ≺ g if and only iff(0) = g(0) and f(∆)⊂g(∆). In this case we haveω(z) = g⁻¹(f(z)).

Let the functionsf andg be of the form f(z) =

∞

X

n=0

a_nzⁿ, g(z) =

∞

X

n=0

b_nzⁿ, z ∈∆.

We say that the Hadamard product off andgis the functionf ? gif (f ? g)(z) =f(z)? g(z) =

∞

X

n=0

a_nb_nzⁿ.

J. Hadamard [2] proved, that the radius of convergence off ? gis the product of the radii of convergence of the corresponding seriesfandg.The functionf ? g is also called the convolution of the functionsf andg.

For the classesQ₁ ⊂ HandQ₂ ⊂ Hthe convolutionQ₁? Q₂is defined as Q₁? Q₂ ={h∈ H; h=f ? g, f ∈Q₁, g ∈Q₂}.

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The problem of connections between functionsf, gand their convolutionf ? gor betweenQ1, Q2andQ1? Q2has often been investigated. Many conjectures have been given, however, many of them have still not been verified.

In 1958, G. Pólya and I.J. Schoenberg [7] conjectured that the Hadamard product of two convex mappings is a convex mapping. In 1961 H.S. Wilf [12]

gave the more general supposition that if F andG are convex mappings in∆ andf is subordinate toF, then the convolutionf ? Gis subordinate toF ? G.

In 1973, S. Rusheweyh and T. Sheil-Small [9] proved both conjectures and more results of this type. Their very important results we may write as:

Theorem A. If f ∈ S^c andg ∈ S^c, thenf ? g ∈ S^c,where S^c is the class of univalent and convex functions. MoreoverS^c? S^c =S^c.

Theorem B. IfF ∈S^c, G ∈S^c andf ≺F,thenf ? G≺F ? G.

In 1985, S. Ruscheweyh and J. Stankiewicz [10] proved some generaliza- tions of TheoremsAandB:

Theorem C. If the functionsF andGare univalent and convex in∆,then for all functionsf andg, iff ≺F andg ≺Gthenf ? g≺F ? G.

For the given complex numbersA, B such thatA+B 6= 0and|B| ≤1let us denote

P(A, B) =

f ∈ N : f(z)≺ 1 +Az 1−Bz

.

W. Janowski introduced the classP(A, B)in [3] and considered it for some real A and B. If A = B = 1 then the class P(1,1) is the class of functions with positive real part (Carathéodory functions). Note, that for |B| < 1 the

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classP(A, B)is the class of bounded functions. In 1988 J. Stankiewicz and Z.

Stankiewicz [11] investigated the convolution properties of the class P(A, B) and proved the following theorem:

Theorem D. If A, B, C, D are some complex numbers such thatA+B 6= 0, C+D6= 0,|B| ≤1,|D| ≤1,then

P(A, B)? P(C, D)⊂P(AC+AD+BC, BD),

moreover, if|B|= 1or|D|= 1,then

P(A, B)? P(C, D) =P(AC+AD+BC, BD).

The equality between the class P(A, B)? P(C, D)and the classP(AC + AD+BC, BD)for|B|<1and|D|<1was an open problem. In [5] K. Piejko, J. Sokół and J. Stankiewicz proved that the above mentioned classes are dif- ferent. In this paper we give an extension of this result. First we need two theorems.

Theorem E (G. Eneström [1], S. Kakeya [4]). If a₀ > a₁ > · · · > a_n > 0, wheren∈N,then the polynomialp(z) =a₀+a₁z+a₂z²+· · ·+a_nzⁿhas no roots in∆ = {z ∈C: |z| ≤1}.

Theorem F (W. Rogosinski [8]). If the functionf(z) =P∞

n=0α_nzⁿis subordi- nated to the functionF(z) = P∞

n=0β_nzⁿin∆,thenP∞

n=0|α_n|² ≤P∞

n=0|β_n|².

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2. Main Result

We prove the following theorem.

Theorem 2.1. LetA, B, C, Dbe some complex numbers such thatB+A6= 0, C+D6= 0, |B| < 1, |D| < 1, then there are not complex numbers X, Y, X+Y 6= 0,|Y| ≤1such thatP(X, Y)⊂P(A, B)? P(C, D).

Proof. As in [5], the proof will be divided into three steps. First we give a family of bounded functions ω^ν; ν = 1,2,3. . . having special properties of coefficients. Afterwards we construct, using ω^ν, a function belonging to the classP(X, Y)and finally we will show that such a function is not in the class P(A, B)? P(C, D).

Now we use a method of E. Landau [6] and find some functionsω^ν, ν = 1,2, 3, . . ., which are basic in this proof. We observe that

1 1−z =

1

√1−z 2

=

∞

X

k=0

p_kz^k

!2

= 1 +z+z² +z³+· · · ,

where

(2.1) p₀ = 1 and p_k = 1·3· · · · ·(2k−1)

2·4· · · · ·2k ; k = 1,2,3, . . . For someν ∈Nwe set

Kν(z) =

ν

X

k=0

pkz^k = 1 +p1z+p2z²+p3z³+· · ·+pνz^ν

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and note that

K_ν²(z) = 1 +z+z²+· · ·+z^ν +b_ν+1z^ν+1+· · ·+b_2νz^2ν,

whereb_ν+1, . . . , b_2ν ∈C.Letω^ν be given by (2.2) ω^ν(z) = z^ν+1K_ν ¹_z

Kν(z) =zz^ν +p₁z^ν−1+p₂z^ν−2 +· · ·+p_ν 1 +p1z+p2z²+· · ·+pνz^ν . Since

p_k > p_k2k+ 1

2k+ 2 =p_k+1 where k = 0,1,2,3, . . . , then forν ∈Nwe have1> p₁ > p₂ > p₃ >· · ·> p_ν >0.

Applying TheoremEto the polynomialK_νwe obtainK_ν(z)6= 0for|z| ≤1, hence the function ω^ν is regular in∆. Moreoverω^ν(0) = 0 and on the circle

|z|= 1we have

ω^ν e^it =

e^(ν+1)itK_ν(e^−it) K_ν(e^it)

=

K_ν(e^it)

|K_ν(e^it)| = 1; t∈R. In this way we conclude that forν ∈ {1,2,3, . . .}

(2.3) ω^ν ∈Ω.

Let, for a certainν ∈N,the functionω^ν be represented by following power series expansions: ω^ν(z) = γ₁^νz+γ₂^νz² +γ₃^νz³ +· · · and lets^ν_n(z)denote the

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partial sum

s^ν_n(z) =

n

X

k=1

γ_k^νz^k =γ₁^νz+γ₂^νz² +γ₃^νz³+· · ·+γ_n^νzⁿ

for alln ∈ {1,2,3, . . . , ν+ 1}.

Now we will estimates^ν_n(1) =γ₁^ν +γ₂^ν +γ₃^ν +· · ·+γ_n^ν.If we integrate on a circle C : |z| = rin a counterclockwise direction with 0 < r < 1,then we obtain

(2.4)

Z

C

az^mdz = 0 and Z

C

a

zdz = 2πai,

for all integersm 6=−1anda∈C.Hence s^ν_n(1) =γ₁^ν+γ₂^ν +γ₃^ν+· · ·+γ_n^ν

= 1 2πi

Z

C

ω^ν(z) 1

z² + 1 z³ + 1

z⁴ +· · ·+ 1 zⁿ⁺¹

dz

= 1 2πi

Z

C

ω^ν(z)

zⁿ⁺¹ 1 +z+z²+z³+· · ·+zⁿ⁻¹ dz

= 1 2πi

Z

C

ω^ν(z)

zⁿ⁺¹ Q(z)dz, where

Q(z) = 1 +z+z²+· · ·+zⁿ⁻¹+dnzn+dn+1zⁿ⁺¹+· · ·+dsz^s

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is any polynomial whose firstnterms are equal to1 +z+z²+· · ·+zⁿ⁻¹. If forn≤ν+ 1we set

Q(z) = K_ν²(z) = 1+z+z²+· · ·+zⁿ⁻¹+zⁿ+· · ·+z^ν+b_ν+1z^ν+1+· · ·+b_2νz^2ν, then we obtain

s^ν_n(1) = 1 2πi

Z

C

ω^ν(z)

zⁿ⁺¹ K_ν²(z)dz.

From (2.2) it follows that s^ν_n(1) = 1

2πi Z

C

z^ν+1Kν(¹_z)

Kν(z)

zⁿ⁺¹ K_ν²(z)dz

= 1 2πi

Z

C

z^ν−nK_ν 1

z

K_ν(z)dz

= 1 2πi

Z

C

z^ν−n

1 +p₁1

z +p₂ 1

z² +· · ·+p_ν 1 z^ν

× 1 +p₁z+p₂z²+· · ·+p_νz^ν dz.

Using (2.4) we get

s^ν_n(1) =pν−n+1+pν−n+2p₁+pν−n+3p₂+· · ·+p_νpn−1, where(p_k)are given by (2.1), and so

(2.5) s^ν_n(1) =

n

X

k=1

γ_k^ν =

n

X

k=1

pν−n+kpk−1.

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By the above we have γ₁^ν = s^ν₁(1) = p_ν and for all ν ∈ {1,2,3, . . .} and n ∈ {2,3,4, . . . , ν+ 1}

γ_n^ν =s^ν_n(1)−s^ν_n−1(1) =

n

X

k=1

p_ν−n+kp_k−1−

n−1

X

k=1

p_ν−n+1+kp_k−1.

Therefore

(2.6) γ_n^ν =

n−1

X

k=1

(pν−n+k−pν−n+1+k)pk−1 +pνpn−1.

The sequence(p_n)is positive and decreasing, then from (2.6) it follows that (2.7) γ_n^ν >0 for all ν ∈N and n∈ {1,2,3, . . . , ν+ 1}.

We conclude from (2.5) that forn=ν+ 1 s^ν_ν+1(1) =γ₁^ν +γ₂^ν +· · ·+γ^ν_ν +γ_ν+1^ν =

ν+1

X

k=1

pk−1pk−1 = 1 +

ν

X

k=1

p²_k.

SinceP∞

k=1p²_k=∞then

(2.8) lim

ν→∞s^ν_ν+1(1) = lim

ν→∞ 1 +

ν

X

k=1

p²_k

!

= +∞.

Now using the properties of ω^ν we construct the function belonging to the classP(X, Y)which is not in the classP(A, B)? P(C, D),where|B|<1and

|D|<1.

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Since for|x| = 1we have P(A, B) =P(Ax, Bx),we can assume without loss of generality thatB ∈[0; 1), D∈[0; 1)andY ∈[0; 1].

For fixedν ∈Nleth^ν be given by

(2.9) h^ν(z) = 1 +Xω^ν(z)

1−Y ω^ν(z),

wherew^ν ∈ Ωis given by (2.2). It is clearly thath^ν ∈ P(X, Y).Suppose that there exist the functionsf ∈P(A, B), g∈P(C, D)such that

(2.10) f(z)? g(z) = h^ν(z).

Let the functionsf andg have the following form:

f(z) = 1 +Aω₁(z)

1−Bω₁(z) = 1 + (A+B) ω₁(z) 1−Bω₁(z) and

g(z) = 1 +Cω₂(z)

1−Dω₂(z) = 1 + (C+D) ω₂(z) 1−Dω₂(z), whereω₁, ω₂ ∈Ω.For simplicity of notation we write

f(z) = 1 + (A+B) ˜f(z), g(z) = 1 + (C+D)˜g(z),

where

(2.11) f˜(z) = ω1(z)

1−Bω₁(z) and ˜g(z) = ω2(z) 1−Dω₂(z).

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Using these notations we can rewrite (2.10) as h

1 + (A+B) ˜f(z)i

?[1 + (C+D)˜g(z)] = 1 +Xω^ν(z) 1−Y ω^ν(z) and so

(2.12) f˜(z)?g(z) =˜ X+Y (A+B)(C+D)

˜h^ν(z),

where˜h^ν(z) = _1−BDω^ω^ν^(z)ν(z).

Let the functions˜h^ν,f˜andg˜have the following expansions in∆:

(2.13) ˜h^ν(z) =

∞

X

n=1

c^ν_nzⁿ, f˜(z) =

∞

X

n=1

a_nzⁿ, ˜g(z) =

∞

X

n=1

b_nzⁿ.

From (2.11) it follows, that f(z)˜ ≺ _1−Bz^z and g(z)˜ ≺ _1−Dz^z , and hence by TheoremF(and since0≤B <1and0≤D <1) we obtain

∞

X

n=1

|a_n|² ≤ 1

1−B² and

∞

X

n=1

|b_n|² ≤ 1 1−D². Let us note, that

∞

X

n=1

|a_n|²+|b_n|²

=

∞

X

n=1

(|a_n| − |b_n|)²+ 2|a_n||b_n|

≤ 1

1− |B|² + 1 1− |D|².

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From (2.12) and (2.13) we obtain a_nb_n= X+Y

(A+B)(C+D)c^ν_n, for n= 1,2,3, . . . , therefore

(2.14)

∞

X

n=1

(|a_n| − |b_n|)² ≤ 1

1−B² + 1 1−D² −

2(X+Y) (A+B)(C+D)

∞

X

n=1

|c^ν_n|.

Now we observe that

∞

X

n=1

c^ν_nzⁿ= ˜h^ν(z) = ω^ν(z) 1−Y ω^ν(z)

=ω^ν(z) +Y [ω^ν(z)]² +Y²[ω^ν(z)]³+· · ·

= γ₁^νz+γ₂^νz²+· · ·

+Y γ₁^νz+γ₂^νz²+· · ·2

+Y² γ₁^νz+γ₂^νz² +· · ·3

+· · ·

=γ₁^νz+ γ₂^ν +Y(γ₁^ν)²

z²+ γ₃^ν + 2Y γ₁^νγ₂^ν +Y²(γ₁^ν)³

z³+· · · . SinceY ∈[0; 1]and since (2.7) we have

∞

X

n=1

|c^ν_n|=|γ₁^ν|+

γ₂^ν +Y(γ₁^ν)² +

γ₃^ν + 2Y γ₁^νγ₂^ν +Y²(γ₁^ν)³ +· · ·

≥

ν+1

X

n=1

|c^ν_n| ≥γ₁^ν+γ₂^ν +γ₃^ν+...+γ_ν+1^ν =s^ν_ν+1(1).

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From the above we have for allν ∈ {1,2,3, . . .}

(2.15)

∞

X

n=1

|c^ν_n| ≥s^ν_ν+1(1).

Combining (2.14) and (2.15) we obtain (2.16)

∞

X

n=1

(|a_n| − |b_n|)² ≤ 1

1−B² + 1 1−D²−

2(X+Y) (A+B)(C+D)

s^ν_ν+1(1).

From (2.8) it follows that we are able to choose a suitable ν such that the right side of (2.16) is negative. In this way (2.16) follows the contradiction and the proof is complete.

From Theorem2.1and TheoremDwe immediately have the following Corollary 2.2. The classesP(A, B)? P(C, D)andP(AD+AC+BC, BD) are equal if and only if|B|= 1or|D|= 1.

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[9] S. RUSCHEWEYH AND T. SHEIL-SMALL, Hadamard product of schlicht functions and the Pólya-Schoenberg conjecture, Comtnt. Math.

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