Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 1, Article 9, 2002
ON HARMONIC FUNCTIONS CONSTRUCTED BY THE HADAMARD PRODUCT
METIN ÖZTÜRK, SIBEL YALÇIN, AND MÜMIN YAMANKARADENİZ
ometin@uludag.edu.tr
U LUDAG U NIVERSITY
FACULTY OF S CIENCE
D EPARTMENT OF M ATHEMATICS
16059 B URSA /T URKEY.
Received 11 May, 2001; accepted 11 October, 2001.
Communicated by S.S. Dragomir
A BSTRACT. 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.
Key words and phrases: Harmonic functions, Hadamard product and extremal problems.
2000 Mathematics Subject Classification. 30C45, 31A05.
1. I NTRODUCTION
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 sense-preserving functions
on U and its subclasses [2, 5].
Here we discuss two classes obtained by the Hadamard product.
2. T HE C LASS 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
n
(2.1)
h(z) = z +
an z
and
g(z) =
bn z n ,
n=2
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
040-01
n=2
2
M ETIN Ö ZTÜRK , S IBEL YALÇIN , AND M ÜMIN YAMANKARADENIZ
then the class of functions f = h + g is denoted by PeH0 [5].
The function
1
1
z2 + · · · +
zn + · · ·
(2.2)
tα (z) = z +
1+α
1 + (n − 1)α
is analytic on U when α is a complex number different from −1, − 12 , − 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
F (z) = H(z) + G(z)
∞
∞
X
X
bn
an
n
= z+
z +
zn
1
+
(n
−
1)α
1
+
(n
−
1)α
n=2
n=2
(2.4)
= z+
∞
X
∞
X
An z n +
n=2
Bn z n ,
z∈U
n=2
PeH0 (α).
is in
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 ∈ Pe0 (α) then there exists f ∈ Pe0 so that
H
H
α[zFz (z) + zFz (z)] + (1 − α)F (z) = f (z).
Conversely, for any function f ∈ Pe0 , there exists F ∈ Pe0 (α) satisfying (2.5).
(2.5)
H
H
Proof. Let F ∈ PeH0 (α). If f ∈ PeH0 , then since
0
αztα (z) + (1 − α)tα (z) = t0 (z),
as F = f ∗ (tα + tα ) we obtain that
0
f (z) = α[f (z) ∗ (ztα (z) + zt0α (z))] + (1 − α)[f (z) ∗ (tα (z) + tα (z))].
Therefore,
f (z) = α[zFz (z) + zFz (z)] + (1 − α)F (z).
Conversely, for f ∈ PeH0 , from (2.1), (2.2) and (2.5),
∞
∞
∞
∞
X
X
X
X
n
n
n
bn z = z +
[1 + (n − 1)α]An z +
[1 + (n − 1)α]Bn z n .
z+
an z +
n=2
n=2
n=2
n=2
From these one obtains
(2.6)
An =
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)].
J. Inequal. Pure and Appl. Math., 3(1) Art. 9, 2002
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O N H ARMONIC F UNCTIONS C ONSTRUCTED BY THE H ADAMARD P RODUCT
3
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)}.
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) + zG0n (z)] + (1 − α)[Hn (z) + Gn (z)] ∈ PeH0 .
Since PeH0 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.
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
α[zH (z) + zG0 (z)] + (1 − α)[H(z) + G(z)] = f (z).
Since by [5, Proposition 2.2]
−r + 2 ln(1 + r) ≤ Re f (z) ≤ −r − 2 ln(1 − r),
J. Inequal. Pure and Appl. Math., 3(1) Art. 9, 2002
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4
M ETIN Ö ZTÜRK , S IBEL YALÇIN , AND M ÜMIN YAMANKARADENIZ
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
Z
1
1
ζ α −1
0
z
dζ,
1 − zζ
|ζ| ≤ 1,
Re α > 0,
and for f = h + g ∈ PeH0
h(z) ∗
z
h(zζ)
=
,
1 − zζ
ζ
g(z) ∗
z
g(zζ)
=
,
1 − zζ
ζ
we have
1
H(z) = h(z) ∗ tα (z) =
α
Z
1
G(z) = g(z) ∗ tα (z) =
α
Z
1
1
ζ α −2 h(zζ) dζ
0
and
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 ).
Remark 2.7. For some values of α, PeH0 (α) ⊂ PeH0 is not true. It is known [5, Corollary 2.5] that
the sharp inequalities
2n − 1
n
are true. Hence, for example, the function
|an | ≤
(2.9)
f (z) = z +
and |bn | ≤
∞
X
2n − 1
n=2
n
n
z +
2n − 3
n
∞
X
2n − 3
n=2
n
zn
belongs to PeH0 . In this case
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O N H ARMONIC F UNCTIONS C ONSTRUCTED BY THE H ADAMARD P RODUCT
F (z) = z +
∞
X
n=2
5
∞
X
2n − 1
2n − 3
n
z +
zn
n[1 + (n − 1)α]
n[1
+
(n
−
1)α]
n=2
2
belongs to the class PeH0 (α) for α ∈ C, α 6= −1/n, n ∈ N. However, for Re α ∈ − |α|3 , 0 , α 6=
of PeH0 given in (2.9) are not satisfied, F ∈
/ PeH0 . Hence
−1, − 12 , . . . as the coefficientconditions
2
for each α ∈ C with Re α ∈ − |α|3 , 0 , α 6= −1, − 12 , . . . , PeH0 (α) − PeH0 6= φ.
Theorem 2.8. Let F = H + G ∈ PeH0 (α). Then
2
(i) ||An | − |Bn || ≤
, n≥1
n|1 + (n − 1)α|
(ii) If F is sense-preserving, then
2n − 1
1
|An | ≤
,
n |1 + (n − 1)α|
n = 1, 2, . . .
and
2n − 3
1
,
n |1 + (n − 1)α|
Equality occurs for the functions of type (2.3) where
|Bn | ≤
n = 2, 3, . . .
2z
3z̄ − z̄ 2
f (z) =
+ ln(1 − z) −
− 3 ln(1 − z̄),
1−z
1 − z̄
z ∈ U.
Proof. By (2.6),
||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).
3. T HE C LASS PH (β, α)
Let f = h + g for analytic functions
∞
X
h(z) = 1 +
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
1
1
1
(3.1)
kα (z) = 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)
F = f ∗ (kα + kα ) = (h ∗ kα ) + (g ∗ kα ) = H + G,
by PH (β, α). If α = 0, then since F = f , PH (β, 0) = PH (β).
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6
M ETIN Ö ZTÜRK , S IBEL YALÇIN , AND M ÜMIN YAMANKARADENIZ
Therefore,
(3.3)
F (z) = H(z) + G(z)
∞
∞
X
X
bn
an
n
zn
= 1+
z +
1
+
nα
1
+
nα
n=1
n=1
= 1+
∞
X
n
An z +
n=1
∞
X
Bn z n ,
z∈U
n=1
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)
f (z) = α[zFz (z) + zFz (z)] + F (z).
Conversely, let f = h + g ∈ PH (β) be given by (3.4). Hence, we can write
(3.5)
h(z) = αzH 0 (z) + H(z),
g(z) = αzG0 (z) + G(z).
From the system (3.5) the analytic functions H and G are in the form
∞
X
an
H(z) = 1 +
z n = h(z) ∗ kα (z),
1
+
nα
n=1
G(z) =
∞
X
n=1
bn
z n = g(z) ∗ kα (z).
1 + nα
Hence the function F = H + G belongs to the class PH (β, α).
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.
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O N H ARMONIC F UNCTIONS C ONSTRUCTED BY THE H ADAMARD P RODUCT
7
Proof. Since
Z
1 1 1 −1 1
kα (z) =
tα
dt, Re α > 0,
α 0
1 − zt
and for f = h + g ∈ PH (β),
1
1
h(z) ∗
= h(zt) and g(z) ∗
= g(zt),
1 − zt
1 − zt
we obtain
Z
1 1 1 −1
t α h(zt)dt
H(z) = h(z) ∗ kα (z) =
α 0
and
Z
1 1 1 −1
G(z) = g(z) ∗ kα (z) =
t α g(zt)dt.
α 0
Therefore, F = H + G is of type (3.7).
Theorem 3.4. Let F ∈ PH (β, α). Then
2(1 − β)
, n≥1
(i) ||An | − |Bn || ≤
|1 + nα|
(ii) If F is sense- preserving, then for n = 1, 2, . . .
(1 − β)(n + 1)
(1 − β)(n − 1)
|An | ≤
and |Bn | ≤
.
|1 + nα|
|1 + nα|
Equality is valid for the functions of type (3.2) where
1+z
1 + (1 − 2β)z
(3.8)
f (z) = Re
+ i Im
.
1−z
1−z
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)
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)
kα (ηz) dµ(η).
|η|=1
Here µ is the probability measure defined on the Borel sets on X.
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
Taking the Hadamard product of both sides by kα (z), we get
Z
1
z
H(z) + G(z) =
kα (z) ∗
+ (1 − 2β)η kα (z) ∗
dµ(η)
1 − zη
1 − zη
|η |=1
Z
kα (ηz)
=
kα (ηz) + (1 − 2β)η
dµ(η).
η
|η|=1
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M ETIN Ö ZTÜRK , S IBEL YALÇIN , AND M ÜMIN YAMANKARADENIZ
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)}.
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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