Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 34, 2005
ON A CONVOLUTION CONJECTURE OF BOUNDED FUNCTIONS
KRZYSZTOF PIEJKO, JANUSZ SOKÓŁ, AND JAN STANKIEWICZ
D EPARTMENT OF M ATHEMATICS
R ZESZÓW U NIVERSITY OF T ECHNOLOGY
UL . W. P OLA 2, 35-959 R ZESZÓW, P OLAND .
piejko@prz.rzeszow.pl
D EPARTMENT OF M ATHEMATICS
R ZESZÓW U NIVERSITY OF T ECHNOLOGY
UL . W. P OLA 2, 35-959 R ZESZÓW, P OLAND .
jsokol@prz.rzeszow.pl
I NSTITUTE OF M ATHEMATICS
U NIVERSITY OF R ZESZÓW
UL . R EJTANA 16A
35-959 R ZESZÓW, P OLAND .
jan.stankiewicz@prz.rzeszow.pl
Received 30 November, 2004; accepted 03 March, 2005
Communicated by K. Nikodem
A BSTRACT. We consider the convolution P (A, B) ? P (C, D) of the classes of analytic func1+Az
1+Cz
tions subordinated to the homographies 1−Bz
and 1−Dz
respectively, where A, B, C, D are
some complex numbers. In 1988 J. Stankiewicz and Z. Stankiewicz [11] showed that for certain
A, B, C, D there exist X, Y such that P (A, B) ? P (C, D) ⊂ P (X, Y ). In this paper we verify
the conjecture that P (X, Y ) ⊂ (A, B) ? P (C, D) for some A, B, C, D, X, Y.
Key words and phrases: Hadamard product, Convolution, Subordination, Bounded functions.
2000 Mathematics Subject Classification. Primary 30C45; Secondary 30C55.
1. I NTRODUCTION
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 for z ∈ ∆. 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 function ω ∈ Ω such that f (z) = g (ω(z)); z ∈ ∆. If the
function g is univalent in ∆, then f ≺ g if and only if f (0) = g(0) and f (∆) ⊂ g(∆). In this
case we have ω(z) = g −1 (f (z)).
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
229-04
2
K RZYSZTOF P IEJKO , JANUSZ S OKÓŁ ,
Let the functions f and g be of the form
∞
X
f (z) =
an z n ,
n=0
g(z) =
AND JAN
∞
X
S TANKIEWICZ
bn z n ,
z ∈ ∆.
n=0
We say that the Hadamard product of f and g is the function f ? g if
∞
X
(f ? g)(z) = f (z) ? g(z) =
an b n z n .
n=0
J. Hadamard [2] proved, that the radius of convergence of f ? g is the product of the radii of
convergence of the corresponding series f and g. The function f ?g is also called the convolution
of the functions f and g.
For the classes Q1 ⊂ H and Q2 ⊂ H the convolution Q1 ? Q2 is defined as
Q1 ? Q2 = {h ∈ H; h = f ? g, f ∈ Q1 , g ∈ Q2 }.
The problem of connections between functions f, g and their convolution f ? g or between
Q1 , Q2 and Q1 ? Q2 has 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 and G are convex mappings in ∆ and f is subordinate to F , then the convolution f ? G
is subordinate to F ? 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 and g ∈ S c , then f ? g ∈ S c , where S c is the class of univalent and
convex functions. Moreover S c ? S c = S c .
Theorem B. If F ∈ S c , G ∈ S c and f ≺ F, then f ? G ≺ F ? G.
In 1985, S. Ruscheweyh and J. Stankiewicz [10] proved some generalizations of Theorems
A and B:
Theorem C. If the functions F and G are univalent and convex in ∆, then for all functions f
and g, if f ≺ F and g ≺ G then f ? g ≺ F ? G.
For the given complex numbers A, B such that A + B 6= 0 and |B| ≤ 1 let us denote
1 + Az
.
P (A, B) = f ∈ N : f (z) ≺
1 − Bz
W. Janowski introduced the class P (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 class P (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 that A + B 6= 0, C + D 6= 0,
|B| ≤ 1, |D| ≤ 1, then
P (A, B) ? P (C, D) ⊂ P (AC + AD + BC, BD),
moreover, if |B| = 1 or |D| = 1, then
P (A, B) ? P (C, D) = P (AC + AD + BC, BD).
J. Inequal. Pure and Appl. Math., 6(2) Art. 34, 2005
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O N A C ONVOLUTION C ONJECTURE OF B OUNDED F UNCTIONS
3
The equality between the class P (A, B) ? P (C, D) and the class P (AC + AD + BC, BD)
for |B| < 1 and |D| < 1 was an open problem. In [5] K. Piejko, J. Sokół and J. Stankiewicz
proved that the above mentioned classes are different. 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 > · · · > an > 0, where n ∈ N, then
the polynomial p(z) = a0 + a1 z + a2 z 2 + · · · + an z n has no roots in ∆ = {z ∈ C : |z| ≤ 1}.
P∞
n
Theorem F (W. P
Rogosinski [8]). If thePfunction f (z)P=
n=0 αn z is subordinated to the
∞
∞
∞
n
2
2
function F (z) = n=0 βn z in ∆, then n=0 |αn | ≤ n=0 |βn | .
2. M AIN R ESULT
We prove the following theorem.
Theorem 2.1. Let A, B, C, D be some complex numbers such that B + A 6= 0, C + D 6= 0,
|B| < 1, |D| < 1, then there are not complex numbers X, Y , X + Y 6= 0, |Y | ≤ 1 such that
P (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 class P (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 functions ω ν , ν = 1, 2, 3, . . ., which
are basic in this proof. We observe that
!2
2
∞
X
1
1
= √
=
pk z k
= 1 + z + z2 + z3 + · · · ,
1−z
1−z
k=0
where
(2.1)
p0 = 1 and pk =
1 · 3 · · · · · (2k − 1)
; k = 1, 2, 3, . . .
2 · 4 · · · · · 2k
For some ν ∈ N we set
Kν (z) =
ν
X
pk z k = 1 + p1 z + p2 z 2 + p3 z 3 + · · · + pν z ν
k=0
and note that
Kν2 (z) = 1 + z + z 2 + · · · + z ν + bν+1 z ν+1 + · · · + b2ν z 2ν ,
where bν+1 , . . . , b2ν ∈ C. Let ω ν be given by
z ν+1 Kν z1
z ν + p1 z ν−1 + p2 z ν−2 + · · · + pν
ν
(2.2)
ω (z) =
=z
.
Kν (z)
1 + p1 z + p2 z 2 + · · · + pν z ν
Since
2k + 1
pk > p k
= pk+1 where k = 0, 1, 2, 3, . . . ,
2k + 2
then for ν ∈ N we have 1 > p1 > p2 > p3 > · · · > pν > 0.
Applying Theorem E to the polynomial Kν we obtain Kν (z) 6= 0 for |z| ≤ 1, hence the
function ω ν is regular in ∆. Moreover ω ν (0) = 0 and on the circle |z| = 1 we have
(ν+1)it
it
−it Kν (e )
ν it e
K
(e
)
ν
ω e = =
|Kν (eit )| = 1; t ∈ R.
Kν (eit )
J. Inequal. Pure and Appl. Math., 6(2) Art. 34, 2005
http://jipam.vu.edu.au/
4
K RZYSZTOF P IEJKO , JANUSZ S OKÓŁ ,
AND JAN
S TANKIEWICZ
In this way we conclude that for ν ∈ {1, 2, 3, . . .}
ω ν ∈ Ω.
(2.3)
Let, for a certain ν ∈ N, the function ω ν be represented by following power series expansions:
ω (z) = γ1ν z + γ2ν z 2 + γ3ν z 3 + · · · and let sνn (z) denote the partial sum
ν
sνn (z)
=
n
X
γkν z k = γ1ν z + γ2ν z 2 + γ3ν z 3 + · · · + γnν z n
k=1
for all n ∈ {1, 2, 3, . . . , ν + 1}.
Now we will estimate sνn (1) = γ1ν + γ2ν + γ3ν + · · · + γnν . If we integrate on a circle C : |z| = r
in a counterclockwise direction with 0 < r < 1, then we obtain
Z
Z
a
m
dz = 2πai,
(2.4)
az dz = 0 and
z
C
C
for all integers m 6= −1 and a ∈ C. Hence
sνn (1) = γ1ν + γ2ν + γ3ν + · · · + γnν
Z
1
1
1
1
1
ν
=
ω (z)
+
+
+ · · · + n+1 dz
2πi
z2 z3 z4
z
C
Z ν
1
ω (z)
=
1 + z + z 2 + z 3 + · · · + z n−1 dz
n+1
2πi
z
C
Z ν
1
ω (z)
=
Q(z)dz,
2πi
z n+1
C
where
Q(z) = 1 + z + z 2 + · · · + z n−1 + dn zn + dn+1 z n+1 + · · · + ds z s
is any polynomial whose first n terms are equal to 1 + z + z 2 + · · · + z n−1 .
If for n ≤ ν + 1 we set
Q(z) = Kν2 (z) = 1 + z + z 2 + · · · + z n−1 + z n + · · · + z ν + bν+1 z ν+1 + · · · + b2ν z 2ν ,
then we obtain
sνn (1)
1
=
2πi
Z
ω ν (z) 2
K (z)dz.
z n+1 ν
C
From (2.2) it follows that
sνn (1) =
1
2πi
Z
z ν+1 Kν ( z1 )
Kν (z)
Kν2 (z)dz
z n+1
C
1
z Kν
Kν (z)dz
z
C
Z
1
1
1
1
ν−n
=
z
1 + p1 + p2 2 + · · · + pν ν
1 + p1 z + p2 z 2 + · · · + pν z ν dz.
2πi
z
z
z
1
=
2πi
Z
ν−n
C
Using (2.4) we get
sνn (1) = pν−n+1 + pν−n+2 p1 + pν−n+3 p2 + · · · + pν pn−1 ,
J. Inequal. Pure and Appl. Math., 6(2) Art. 34, 2005
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O N A C ONVOLUTION C ONJECTURE OF B OUNDED F UNCTIONS
5
where (pk ) are given by (2.1), and so
sνn (1) =
(2.5)
n
X
γkν =
k=1
By the above we have
γnν
=
γ1ν
=
sνn (1)
sν1 (1)
pν−n+k pk−1 .
k=1
= pν and for all ν ∈ {1, 2, 3, . . .} and n ∈ {2, 3, 4, . . . , ν +1}
sνn−1 (1)
−
n
X
=
n
X
pν−n+k pk−1 −
k=1
n−1
X
pν−n+1+k pk−1 .
k=1
Therefore
γnν =
(2.6)
n−1
X
(pν−n+k − pν−n+1+k ) pk−1 + pν pn−1 .
k=1
The sequence (pn ) is positive and decreasing, then from (2.6) it follows that
γnν > 0 for all ν ∈ N and n ∈ {1, 2, 3, . . . , ν + 1}.
(2.7)
We conclude from (2.5) that for n = ν + 1
ν
sνν+1 (1) = γ1ν + γ2ν + · · · + γνν + γν+1
=
ν+1
X
pk−1 pk−1 = 1 +
k=1
Since
(2.8)
P∞
2
k=1 pk
ν
X
p2k .
k=1
= ∞ then
lim sνν+1 (1) = lim
ν→∞
ν→∞
1+
ν
X
!
p2k
= +∞.
k=1
ν
Now using the properties of ω we construct the function belonging to the class P (X, Y )
which is not in the class P (A, B) ? P (C, D), where |B| < 1 and |D| < 1.
Since for |x| = 1 we have P (A, B) = P (Ax, Bx), we can assume without loss of generality
that B ∈ [0; 1), D ∈ [0; 1) and Y ∈ [0; 1].
For fixed ν ∈ N let hν be given by
hν (z) =
(2.9)
1 + Xω ν (z)
,
1 − Y ω ν (z)
where wν ∈ Ω is given by (2.2). It is clearly that hν ∈ P (X, Y ). Suppose that there exist the
functions f ∈ P (A, B), g ∈ P (C, D) such that
f (z) ? g(z) = hν (z).
(2.10)
Let the functions f and g have the following form:
f (z) =
1 + Aω1 (z)
ω1 (z)
= 1 + (A + B)
1 − Bω1 (z)
1 − Bω1 (z)
and
1 + Cω2 (z)
ω2 (z)
= 1 + (C + D)
,
1 − Dω2 (z)
1 − Dω2 (z)
where ω1 , ω2 ∈ Ω. For simplicity of notation we write
f (z) = 1 + (A + B)f˜(z),
g(z) = 1 + (C + D)g̃(z),
g(z) =
where
(2.11)
f˜(z) =
ω1 (z)
1 − Bω1 (z)
J. Inequal. Pure and Appl. Math., 6(2) Art. 34, 2005
and
g̃(z) =
ω2 (z)
.
1 − Dω2 (z)
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6
K RZYSZTOF P IEJKO , JANUSZ S OKÓŁ ,
AND JAN
S TANKIEWICZ
Using these notations we can rewrite (2.10) as
h
i
1 + Xω ν (z)
˜
1 + (A + B)f (z) ? [1 + (C + D)g̃(z)] =
1 − Y ω ν (z)
and so
X +Y
h̃ν (z),
(2.12)
f˜(z) ? g̃(z) =
(A + B)(C + D)
where h̃ν (z) =
ω ν (z)
.
1−BDω ν (z)
ν ˜
Let the functions h̃ , f and g̃ have the following expansions in ∆:
∞
∞
∞
X
X
X
(2.13)
h̃ν (z) =
cνn z n , f˜(z) =
an z n , g̃(z) =
bn z n .
n=1
n=1
n=1
From (2.11) it follows, that f˜(z) ≺
and g̃(z) ≺
and hence by Theorem F (and
since 0 ≤ B < 1 and 0 ≤ D < 1) we obtain
∞
∞
X
X
1
1
2
and
|bn |2 ≤
.
|an | ≤
2
1−B
1 − D2
n=1
n=1
z
1−Bz
z
,
1−Dz
Let us note, that
∞
∞
X
X
2
2
|an | + |bn | =
(|an | − |bn |)2 + 2|an ||bn | ≤
n=1
n=1
1
1
+
.
2
1 − |B|
1 − |D|2
From (2.12) and (2.13) we obtain
an b n =
X +Y
cνn ,
(A + B)(C + D)
for
n = 1, 2, 3, . . . ,
therefore
(2.14)
∞
X
∞
X ν
1
2(X
+
Y
)
1
+
−
|cn |.
(|an | − |bn |) ≤
2
2
1
−
B
1
−
D
(A
+
B)(C
+
D)
n=1
n=1
Now we observe that
∞
X
cνn z n = h̃ν (z) =
n=1
2
ω ν (z)
1 − Y ω ν (z)
= ω ν (z) + Y [ω ν (z)]2 + Y 2 [ω ν (z)]3 + · · ·
2
3
= γ1ν z + γ2ν z 2 + · · · + Y γ1ν z + γ2ν z 2 + · · · + Y 2 γ1ν z + γ2ν z 2 + · · · + · · ·
= γ1ν z + γ2ν + Y (γ1ν )2 z 2 + γ3ν + 2Y γ1ν γ2ν + Y 2 (γ1ν )3 z 3 + · · · .
Since Y ∈ [0; 1] and since (2.7) we have
∞
X
|cνn | = |γ1ν | + γ2ν + Y (γ1ν )2 + γ3ν + 2Y γ1ν γ2ν + Y 2 (γ1ν )3 + · · ·
n=1
≥
ν+1
X
ν
= sνν+1 (1).
|cνn | ≥ γ1ν + γ2ν + γ3ν + ... + γν+1
n=1
From the above we have for all ν ∈ {1, 2, 3, . . .}
∞
X
(2.15)
|cνn | ≥ sνν+1 (1).
n=1
J. Inequal. Pure and Appl. Math., 6(2) Art. 34, 2005
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O N A C ONVOLUTION C ONJECTURE OF B OUNDED F UNCTIONS
7
Combining (2.14) and (2.15) we obtain
∞
X
ν
1
2(X + Y )
1
2
sν+1 (1).
(2.16)
(|an | − |bn |) ≤
+
−
2
2
1
−
B
1
−
D
(A
+
B)(C
+
D)
n=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 Theorem 2.1 and Theorem D we immediately have the following
Corollary 2.2. The classes P (A, B) ? P (C, D) and P (AD + AC + BC, BD) are equal if and
only if |B| = 1 or |D| = 1.
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J. Inequal. Pure and Appl. Math., 6(2) Art. 34, 2005
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