Journal of Inequalities in Pure and
Applied Mathematics
ON HARMONIC FUNCTIONS BY THE HADAMARD PRODUCT
M. ÖZTÜRK, S. YALÇIN AND M. YAMANKARADENIZ
Uludag University,
Faculty of Science,
Department of Mathematics
16059 Bursa/Turkey.
EMail: ometin@uludag.edu.tr
volume 3, issue 1, article 9,
2002.
Received 11 May, 2001;
accepted 11 October, 2001.
Communicated by: S.S. Dragomir
Abstract
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2000
School of Communications and Informatics,Victoria University of Technology
ISSN (electronic): 1443-5756
040-01
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Abstract
A function f = u + iv defined in the domain D ⊂ C is harmonic in D if u, v are
real harmonic. Such functions can be represented as f = h + ḡ where h, g are
analytic in D. In this paper the class of harmonic functions constructed by the
Hadamard product in the unit disk, and properties of some of its subclasses are
examined.
2000 Mathematics Subject Classification: 30C45, 31A05.
Key words: Harmonic functions, Hadamard product and extremal problems.
On Harmonic Functions
Constructed by the Hadamard
Product
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Class PeH0 (α) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3
The Class PH (β, α) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
References
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1.
Introduction
Let U denote the open unit disk in C and let f = u + iv be a complex valued
harmonic function on U . Since u and v are real parts of analytic functions, f
admits a representation f = h + g for two functions h and g, analytic on U .
The Jacobian of f is given by Jf (z) = |h0 (z)|2 − |g 0 (z)|2 . The necessary
and sufficient conditions for f to be local univalent and sense-preserving is
Jf (z) > 0, z ∈ U [1].
Many mathematicians studied the class of harmonic univalent and sensepreserving functions on U and its subclasses [2, 5].
Here we discuss two classes obtained by the Hadamard product.
On Harmonic Functions
Constructed by the Hadamard
Product
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
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2.
The Class PeH0 (α)
Let PH denote the class of all functions f = h+ḡ so that Re f > 0 and f (0) = 1
where h and g are analytic on U.
If the function fz +fz = h0 +g 0 belongs to PH for the analytic and normalized
functions
∞
∞
X
X
(2.1)
h(z) = z +
an z n
and
g(z) =
bn z n ,
n=2
n=2
then the class of functions f = h + g is denoted by PeH0 [5].
The function
1
1
(2.2)
tα (z) = z +
z2 + · · · +
zn + · · ·
1+α
1 + (n − 1)α
is analytic on U when α is a complex number different from −1, − 21 , − 13 , . . . .
For f ∈ PeH0 , we denote, by PeH0 (α), the class of functions defined by
F = f ∗ (tα + tα ).
(2.3)
Here f ∗ (tα + tα ) is the Hadamard product of the functions f and tα + tα .
Therefore
(2.4)
F (z) = H(z) + G(z)
∞
∞
X
X
an
bn
n
= z+
z +
zn
1
+
(n
−
1)α
1
+
(n
−
1)α
n=2
n=2
= z+
∞
X
n=2
An z n +
∞
X
n=2
Bn z n ,
z∈U
On Harmonic Functions
Constructed by the Hadamard
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Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
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is in PeH0 (α).
Conversely, if F is in the form (2.4), with an , bn being the coefficients of
f ∈ PeH0 , then F ∈ PeH0 (α).
Furthermore, if α = 0, then as F = f , we have PeH0 (0) = PeH0 . Moreover
PeH0 (∞) = {I : I(z) ≡ z, z ∈ U } and since I ∈ PeH0 , PeH0 ∩ PeH0 (α) 6= φ.
Theorem 2.1. If F ∈ PeH0 (α) then there exists f ∈ PeH0 so that
On Harmonic Functions
Constructed by the Hadamard
Product
α[zFz (z) + zFz (z)] + (1 − α)F (z) = f (z).
(2.5)
Conversely, for any function f ∈ PeH0 , there exists F ∈ PeH0 (α) satisfying (2.5).
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
Proof. Let F ∈ PeH0 (α). If f ∈ PeH0 , then since
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0
αztα (z) + (1 − α)tα (z) = t0 (z),
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as F = f ∗ (tα + tα ) we obtain that
JJ
J
0
f (z) = α[f (z) ∗ (ztα (z) + zt0α (z))] + (1 − α)[f (z) ∗ (tα (z) + tα (z))].
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Therefore,
f (z) = α[zFz (z) + zFz (z)] + (1 − α)F (z).
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Conversely, for f ∈ PeH0 , from (2.1), (2.2) and (2.5),
z+
∞
X
n=2
II
I
n
an z +
∞
X
n=2
bn
zn
= z+
∞
X
n=2
n
[1+(n−1)α]An z +
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∞
X
n=2
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[1 + (n − 1)α]Bn
zn.
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From these one obtains
An =
(2.6)
an
1 + (n − 1)α
and
Bn =
bn
.
1 + (n − 1)α
Therefore,
F (z) = z +
∞
X
n=2
∞
X
an
bn
n
z +
zn
1 + (n − 1)α
1
+
(n
−
1)α
n=2
= f (z) ∗ [tα (z) + tα (z)].
On Harmonic Functions
Constructed by the Hadamard
Product
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
Corollary 2.2. A function F = H + G of the form (2.4) belongs to PeH0 (α), if
and only if
(2.7)
Re{z(αH 00 (z) + αG00 (z)) + H 0 (z) + G0 (z)} > 0,
z ∈ U.
Proof. If F = H + G ∈ PeH0 (α), then from Theorem 2.1
α[zH 0 (z) + zG0 (z)] + (1 − α)[H(z) + G(z)] = h(z) + g(z) ∈ PeH0
and h0 + g 0 ∈ PH . Hence
0 < Re{h0 (z) + g 0 (z)}
= Re{αzH 00 (z) + αH 0 (z) + (1 − α)H 0 (z)
+αzG00 (z) + αG0 (z) + (1 − α)G0 (z)}
= Re{z(αH 00 (z) + αG00 (z)) + H 0 (z) + G0 (z)}.
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Conversely, if the function F = H + G of the form (2.4) satisfies (2.7), then by
Theorem 2.1, h0 + g 0 ∈ PH and the function
f (z) = h(z) + g(z) = α[zH 0 (z) + zG0 (z)] + (1 − α)(H(z) + G(z))
is from the class PeH0 . Hence by Theorem 2.1, F = H + G ∈ PeH0 (α).
Proposition 2.3. PeH0 (α) is convex and compact.
Proof. Let F1 = H1 + G1 , F2 = H2 + G2 ∈ PeH0 (α) and let λ ∈ [0, 1]. Then
Re{z[α(λH100 (z) + (1 − λ)H200 (z)ᾱ(λG001 (z) + (1 − λ)G002 (z))]
+ λ[H10 (z) + G01 (z)] + (1 − λ)[H20 (z) + G02 (z)]}
= λ Re{z[αH100 (z) + ᾱG001 (z)] + H10 (z) + G01 (z)}
+ (1 − λ) Re{z[αH200 (z) + ᾱG002 (z)] + H20 (z) + G02 (z)}
> 0.
Hence, from Corollary 2.2, λ F1 + (1 − λ)F2 ∈ PeH0 (α). Therefore, PeH0 (α) is
convex.
On the other hand, let Fn = Hn + Gn ∈ PeH0 (α) and let Fn → F = H + G.
By Corollary 2.2,
0
α[zHn (z) + zGn (z)] + (1 − α)[Hn (z) + Gn (z)] ∈ PeH0 .
Since
PeH0
On Harmonic Functions
Constructed by the Hadamard
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Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
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0
is compact, [5],
0
α[zH (z) + zG0 (z)] + (1 − α)[H(z) + G(z)] ∈ PeH0 .
Hence, by Theorem 2.1, F = H + Ḡ ∈ PeH0 (α). Therefore, PeH0 (α) is compact.
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Proposition 2.4. If F = H + G ∈ PeH0 (α) and |z| = r < 1 then
−r + 2 ln(1 + r) ≤ Re{α[zH 0 (z) + zG0 (z)] + (1 − α)[H(z) + G(z)]}
≤ −r − 2 ln(1 − r).
Equality is obtained for the function (2.3) where
f (z) = 2z + ln(1 − z) − 3z − 3 ln(1 − z),
z ∈ U.
Proof. From Theorem 2.1, if F = H + G ∈ PeH0 (α), then there exists f =
h + ḡ ∈ PeH0 so that
0
0
α[zH (z) + zG (z)] + (1 − α)[H(z) + G(z)] = f (z).
Since by [5, Proposition 2.2]
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
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−r + 2 ln(1 + r) ≤ Re f (z) ≤ −r − 2 ln(1 − r),
the proof is complete.
Proposition 2.5. If F = H + G ∈ PeH0 (α) and Re α > 0, then there exists an
f ∈ PeH0 so that
Z
1 1 1 −2
(2.8)
F (z) =
ζ α f (zζ) dζ, z ∈ U.
α 0
Proof. Since
1
tα (z) =
α
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Z
0
1
1
ζ α −1
z
dζ,
1 − zζ
|ζ| ≤ 1,
Re α > 0,
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and for f = h + g ∈ PeH0
h(z) ∗
z
h(zζ)
=
,
1 − zζ
ζ
g(z) ∗
we have
1
H(z) = h(z) ∗ tα (z) =
α
Z
1
α
Z
1
ζ α −2 h(zζ) dζ
0
and
G(z) = g(z) ∗ tα (z) =
1
z
g(zζ)
=
,
1 − zζ
ζ
1
1
ζ α −2 g(zζ) dζ.
0
Hence F is type (2.8).
Theorem 2.6. If Re α > 0, then PeH0 (α) ⊂ PeH0 . Further, for any 0 < Re α1 ≤
Re α2 , PeH0 (α2 ) ⊂ PeH0 (α1 ).
Proof. Let F ∈ PeH0 (α) and Re α > 0. Then there exists f ∈ PeH0 so that
F = H + G = f ∗ (tα + tα ) = (h ∗ tα ) + (g ∗ tα ).
Hence, 0 < Re{h0 + g 0 } = Re{h0 + g 0 } and since Re α > 0, Re{H 0 + G0 } > 0,
and H(0) = 0, H 0 (0) = 1, G(0) = G0 (0) = 0 and hence F = H + G ∈ PeH0 .
For 0 < Re α1 ≤ Re α2 , if F ∈ PeH0 (α2 ), from Corollary 2.2
0 < Re{z(α2 H 00 (z) + α2 G00 (z)) + H 0 (z) + G0 (z)}
≤ Re{z(α1 H 00 (z) + α1 G00 (z)) + H 0 (z) + G0 (z)}
we get F ∈ PeH0 (α1 ).
On Harmonic Functions
Constructed by the Hadamard
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Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
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Remark 2.1. For some values of α, PeH0 (α) ⊂ PeH0 is not true. It is known [5,
Corollary 2.5] that the sharp inequalities
|an | ≤
(2.9)
2n − 1
n
|bn | ≤
and
2n − 3
n
are true. Hence, for example, the function
f (z) = z +
∞
X
2n − 1
n=2
n
n
z +
∞
X
2n − 3
n=2
n
zn
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
belongs to PeH0 . In this case
F (z) = z +
∞
X
n=2
∞
X
2n − 1
2n − 3
n
z +
zn
n[1 + (n − 1)α]
n[1
+
(n
−
1)α]
n=2
belongs tothe class PeH0 (α) for α ∈ C, α 6= −1/n, n ∈ N. However, for Re α ∈
2
− |α|3 , 0 , α 6= −1, − 12 , . . . as the coefficient conditions of PeH0 given in (2.9)
2
are not satisfied, F ∈
/ PeH0 . Hence for each α ∈ C with Re α ∈ − |α|3 , 0 ,
α 6= −1, − 1 , . . . , Pe0 (α) − Pe0 6= φ.
2
H
H
Theorem 2.7. Let F = H + G ∈ PeH0 (α). Then
(i) ||An | − |Bn || ≤
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2
,
n|1 + (n − 1)α|
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n≥1
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(ii) If F is sense-preserving, then
|An | ≤
1
2n − 1
,
n |1 + (n − 1)α|
n = 1, 2, . . .
|Bn | ≤
2n − 3
1
,
n |1 + (n − 1)α|
n = 2, 3, . . .
and
On Harmonic Functions
Constructed by the Hadamard
Product
Equality occurs for the functions of type (2.3) where
f (z) =
2z
3z̄ − z̄ 2
+ ln(1 − z) −
− 3 ln(1 − z̄),
1−z
1 − z̄
z ∈ U.
Proof. By (2.6),
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
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||An | − |Bn || =
1
||an | − |bn || .
|1 + (n − 1)α|
Also by [5, Theorem 2.3], we have
||an | − |bn || ≤
2
n
the required results are obtained.
On the other hand, from (2.6) and from the coefficient relations in PeH0 given
in (2.9), we obtain the coefficient inequalities for PeH0 (α).
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3.
The Class PH (β, α)
Let f = h + g for analytic functions
h(z) = 1 +
∞
X
an z
n
and
g(z) =
n=1
∞
X
bn z n
n=1
on U. The class PH (β) of all functions with Re f (z) > β, 0 ≤ β < 1 and
f (0) = 1 is studied in [5].
Let us consider the function
(3.1) kα (z) = 1 +
1
1
1
z +···+
z n + · · · , α ∈ C, α 6= −1, − , . . .
1+α
1 + nα
2
which is analytic on U.
For f ∈ PH (β), let us denote the class of functions
(3.2)
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F = f ∗ (kα + kα ) = (h ∗ kα ) + (g ∗ kα ) = H + G,
F (z) = H(z) + G(z)
∞
∞
X
X
an
bn
n
= 1+
z +
zn
1
+
nα
1
+
nα
n=1
n=1
= 1+
∞
X
n=1
An z n +
∞
X
n=1
Bn z n ,
Metin Öztürk, Sibel Yalçin and
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by PH (β, α). If α = 0, then since F = f , PH (β, 0) = PH (β).
Therefore,
(3.3)
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Theorem 3.1. If F ∈ PH (β, α) then there exists an f ∈ PH (β), so that
(3.4)
α[zFz (z) + zFz (z)] + F (z) = f (z).
Conversely, for f ∈ PH (β), there is a solution of (3.4) belonging to PH (β, α).
Proof. Since k0 (z) = αzkα0 (z) + kα (z), for f ∈ PH (β), using the fact that,
f = f ∗ (k0 + k0 ),
f (z) = α[f (z) ∗
(zkα0 (z)
+
zkα0 (z))]
+ [f (z) ∗ (kα (z) + kα (z))]
is obtained. Hence, for F ∈ PH (β, α)
On Harmonic Functions
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Metin Öztürk, Sibel Yalçin and
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f (z) = α[zFz (z) + zFz (z)] + F (z).
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Conversely, let f = h + g ∈ PH (β) be given by (3.4). Hence, we can write
(3.5)
0
h(z) = αzH (z) + H(z),
0
g(z) = αzG (z) + G(z).
From the system (3.5) the analytic functions H and G are in the form
H(z) = 1 +
∞
X
n=1
G(z) =
∞
X
n=1
an
z n = h(z) ∗ kα (z),
1 + nα
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bn
z n = g(z) ∗ kα (z).
1 + nα
Hence the function F = H + G belongs to the class PH (β, α).
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Corollary 3.2. The necessary and sufficient conditions for a function F of form
(3.3) to belong to PH (β, α) are
(3.6)
Re{z(αH 0 (z) + αG0 (z)) + H(z) + G(z)} > β,
z ∈ U.
Proof. If F ∈ PH (β, α) then by Theorem 3.1,
β < Re{f (z)}
= Re{α[zFz (z) + zFz (z)] + F (z)}
= Re{z(αH 0 (z) + αG0 (z)) + H(z) + G(z)}, z ∈ U.
Conversely, if a function F = H + G of form (3.3) satisfies (3.6), then
zαH 0 (z) + H(z) + αzG0 (z) + G(z) ∈ PH (β).
Hence, from Theorem 3.1, we have F = H + Ḡ ∈ PH (β, α).
Proposition 3.3. If F ∈ PH (β, α), Re α > 0 then there exists an f ∈ PH (β) so
that
Z
1 1 1 −1
(3.7)
F (z) =
t α f (zt)dt, z ∈ U.
α 0
The converse is also true.
On Harmonic Functions
Constructed by the Hadamard
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Mümin Yamankaradeniz
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Proof. Since
kα (z) =
1
α
Z
0
1
1
t α −1
1
dt,
1 − zt
Re α > 0,
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and for f = h + g ∈ PH (β),
h(z) ∗
1
1
= h(zt) and g(z) ∗
= g(zt),
1 − zt
1 − zt
we obtain
1
H(z) = h(z) ∗ kα (z) =
α
Z
1
α
Z
1
t α −1 h(zt)dt
0
and
G(z) = g(z) ∗ kα (z) =
1
1
1
t α −1 g(zt)dt.
0
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
Therefore, F = H + G is of type (3.7).
Theorem 3.4. Let F ∈ PH (β, α). Then
(i) ||An | − |Bn || ≤
2(1 − β)
,
|1 + nα|
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(ii) If F is sense- preserving, then for n = 1, 2, . . .
|An | ≤
(1 − β)(n + 1)
|1 + nα|
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and
|Bn | ≤
(1 − β)(n − 1)
.
|1 + nα|
Equality is valid for the functions of type (3.2) where
1+z
1 + (1 − 2β)z
+ i Im
.
(3.8)
f (z) = Re
1−z
1−z
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Proof. Let F ∈ PH (β, α). Then from (3.3), as the coefficient relation for PH (β)
is
||an | − |bn || ≤ 2(1 − β)
[5, Proposition 3.4], the required inequalities are obtained.
On the other hand, from (3.3), as the coefficient relations for PH (β) are
|an | ≤ (1 − β)(n + 1)
and
|bn | ≤ (1 − β)(n − 1)
On Harmonic Functions
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the required inequalities are obtained.
Proposition 3.5. If F = H + G ∈ PH (β, α), then for X = {η : |η| = 1} and
z ∈ U,
Z
H(z) + G(z) = 2(1 − β)
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
kα (ηz) dµ(η).
|η|=1
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Here µ is the probability measure defined on the Borel sets on X.
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Proof. From [5, Corollary 3.3] there exists a probability measure µ defined on
the Borel sets on X so that
Z
1 + (1 − 2β) zη
h(z) + g(z) =
dµ(η).
1 − zη
|η|=1
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Taking the Hadamard product of both sides by kα (z), we get
H(z) + G(z)
Z
=
kα (z) ∗
|η |=1
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1
1 − zη
+ (1 − 2β)η kα (z) ∗
z
1 − zη
dµ(η)
J. Ineq. Pure and Appl. Math. 3(1) Art. 9, 2002
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Z
=
|η|=1
kα (ηz)
kα (ηz) + (1 − 2β)η
η
dµ(η).
Theorem 3.6. If Re α ≥ 0, then PH (β, α) ⊂ PH (β). Further if 0 ≤ Re α1 ≤
Re α2 , then PH (β, α2 ) ⊂ PH (β, α1 ).
Proof. Let F ∈ PH (β, α) and Re α ≥ 0. Then as Re{h0 + g 0 } > β, we have
Re{H 0 + G0 } > β and F (0) = 1. Hence F ∈ PH (β). Further as 0 ≤ Re α1 ≤
Re α2 , for F ∈ PH (β, α2 )
β < Re{z(α2 H 0 (z) + α2 G0 (z)) + H(z) + G(z)}
< Re{z(α1 H 0 (z) + α1 G0 (z)) + H(z) + G(z)}.
On Harmonic Functions
Constructed by the Hadamard
Product
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
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Therefore, by Corollary 3.2, F ∈ PH (β, α1 ).
For f ∈ PH , the class BH (α) consisting of the functions F = f ∗ (kα + kα )
is studied in [2]. The relation between the classes PH (β, α) and BH (α) is given
as follows.
Proposition 3.7. For Re α ≥ 0, PH (β, α) ⊂ BH (α).
Proof. If F ∈ PH (β, α) then there exists an f ∈ PH (β) so that F = f ∗ (kα +
kα ). Since Re f (z) > β, f (0) = 1 and 0 ≤ β < 1, Re f (z) > 0. Hence,
f ∈ PH . By the definition of BH (α), F ∈ BH (α).
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References
[1] J. CLUNIE AND T. SHEIL-SMALL, Harmonic univalent functions, Ann.
Acad. Sci. Fenn. Series A.I, Math., 9 (1984), 3–25.
[2] Z.J. JAKUBOWSKI, W. MAJCHRZAK AND K. SKALSKA, Harmonic
mappings with a positive real part, Materialy XIV Konferencjiz Teorii Zagadnien Exstremalnych, Lodz, 17–24, 1993.
[3] H. LEWY, On the non vanishing of the Jacobian in certain one to one mappings, Bull. Amer. Math. Soc., 42 (1936), 689–692.
[4] H. SILVERMAN, Harmonic univalent functions with negative coefficients,
J. Math. Anal. Appl., 220 (1998), 283–289.
[5] S. YALÇIN, M. ÖZTÜRK AND M. YAMANKARADENIZ, On some subclasses of harmonic functions, in Mathematics and Its Applications, Kluwer
Acad. Publ.; Functional Equations and Inequalities, 518 (2000), 325–331.
On Harmonic Functions
Constructed by the Hadamard
Product
Metin Öztürk, Sibel Yalçin and
Mümin Yamankaradeniz
Title Page
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