Volume 10 (2009), Issue 1, Article 27, 9 pp.
ON A SUBCLASS OF HARMONIC UNIVALENT FUNCTIONS
K. K. DIXIT AND SAURABH PORWAL
D EPARTMENT OF M ATHEMATICS
JANTA C OLLEGE , BAKEWAR , E TAWAH
(U. P.), I NDIA – 206124
kk.dixit@rediffmail.com
Received 04 June, 2008; accepted 27 September, 2008
Communicated by H.M. Srivastava
A BSTRACT. The class of univalent harmonic functions on the unit disc satisfying the condition
P
∞
m
n
k=2 (k − αk )(|ak | + |bk |) ≤ (1 − α)(1 − |b1 |) is given. Sharp coefficient relations and distortion theorems are given for these functions. In this paper we find that many results of Özturk
and Yalcin [5] are incorrect. Some of the results of this paper correct the theorems and examples
of [5]. Further, sharp coefficient relations and distortion theorems are given. Results concerning the convolutions of functions satisfying the above inequalities with univalent, harmonic and
convex functions in the unit disc and harmonic functions having positive real part are obtained.
Key words and phrases: Convex harmonic functions, Starlike harmonic functions, extremal problems.
2000 Mathematics Subject Classification. 30C45, 31A05.
1. I NTRODUCTION
Let U denote the open unit disc and SH denote the class of all complex valued, harmonic,
orientation-preserving, univalent functions f in U normalized by f (0) = fz (0) − 1 = 0. Each
f ∈ SH can be expressed as f = h + ḡ where h and g belong to the linear space H(U ) of all
analytic functions on U .
Firstly, Clunie and Sheil-Small [3] studied SH together with some geometric subclasses of
SH . They proved that although SH is not compact, it is normal with respect to the topology of
0
uniform convergence on compact subsets of U . Meanwhile, the subclass SH
of SH consisting
of the functions having the property that fz̄ (0) = 0 is compact.
In this article we concentrate on a subclass of univalent harmonic mappings defined in Section
2. The technique employed by us is entirely different to that of Özturk and Yalcin [5].
2. T HE C LASS HS(m, n; α)
Let Ur = {z : |z| < r, 0 < r ≤ 1} and U1 = U . A harmonic, complex-valued, orientationpreserving, univalent mapping f defined on U can be written as:
(2.1)
265-08
f (z) = h(z) + g(z),
2
K. K. D IXIT AND S AURABH P ORWAL
where
h(z) = z +
(2.2)
∞
X
k
ak z ,
g(z) =
∞
X
bk z k
k=1
k=2
are analytic in U .
Denote by HS(m, n, α) the class of all functions of the form (2.1) that satisfy the condition:
∞
X
(2.3)
(k m − αk n )(|ak | + |bk |) ≤ (1 − α)(1 − |b1 |),
k=2
where m ∈ N, n ∈ N0 , m > n, 0 ≤ α < 1 and 0 ≤ |b1 | < 1.
The class HS(m, n, α) with b1 = 0 will be denoted by HS o (m, n, α).
We note that by specializing the parameter we obtain the following subclasses which have
been studied by various authors.
(1) The classes HS(1, 0, α) ≡ HS(α) and HS(2, 1, α) ≡ HC(α) were studied by Özturk
and Yalcin [5].
(2) The classes HS(1, 0, 0) ≡ HS and HS(2, 1, 0) ≡ HC were studied by Avci and
Zlotkiewicz [2]. If h, g, H, G, are of the form (2.2) and if f (z) = h(z) + g(z) and
F (z) = H(z) + G(z), then the convolution of f and F is defined to be the function:
(f ∗ F )(z) = z +
∞
X
k
ak A k z +
k=2
∞
X
bk Bk z k ,
k=1
while the integral convolution is defined by:
(f ♦ F )(z) = z +
∞
X
a k Ak
k=2
k
k
z +
∞
X
bk Bk
k=1
k
zk .
The δ – neighborhood of f is the set:
(
)
∞
X
Nδ (f ) = F :
k(|ak − Ak | + |bk − Bk |) + |b1 − B1 | ≤ δ
k=2
(see [1], [6]). In this case, let us define the generalized δ-neighborhood of f to be the set:
(
)
∞
X
N (f ) = F :
(k − α)(|ak − Ak | + |bk − Bk |) + (1 − α)|b1 − B1 | ≤ (1 − α)δ .
k=2
In the present paper we find that many results of Özturk and Yalcin [5, Theorem 3.6, 3.8] are
incorrect, and we correct them. It should be noted that the examples supporting the sharpness
of [5, Theorem 3.6, 3.8] are not correct and we remedy this problem. Finally, we improve
Theorem 3.15 of Özturk and Yalcin [5].
3. M AIN R ESULTS
First, let us give the interrelation between the classes HS(m, n, α1 ) and HS(m, n, α2 ) where
0 ≤ α1 ≤ α2 < 1.
Theorem 3.1. HS(m, n, α2 ) ⊆ HS(m, n, α1 ) where 0 ≤ α1 ≤ α2 < 1. Consequently
HS o (m, n, α2 ) ⊆ HS o (m, n, α1 ). In particular HS(m, n, α) ⊆ HS(m, n, 0) andHS o (m, n, α) ⊆
HS o (m, n, 0).
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 27, 9 pp.
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H ARMONIC U NIVALENT F UNCTIONS
3
Proof. Let f ∈ HS(m, n, α2 ). Thus we have:
∞
X
k m − α2 k n
(3.1)
k=2
1 − α2
(|ak | + |bk |) ≤ (1 − |b1 |).
Now, using (3.1),
∞
X
k m − α1 k n
1 − α1
k=2
(|ak | + |bk |) ≤
∞
X
k m − α2 k n
1 − α2
k=2
(|ak | + |bk |)
≤ (1 − |b1 |).
Thus f ∈ HS(m, n, α1 ).
This completes the proof of Theorem 3.1.
Theorem 3.2. HS(m, n, α) ⊆ HS(α), ∀m ∈ N, ∀n ∈ N0 , HS(m, n, α) ⊆ HC(α), ∀m ∈
N − {1}, ∀n ∈ N0 ,where 0 ≤ α < 1.
Proof. Let f ∈ HS(m, n, α). Then
∞
X
k m − αk n
(3.2)
k=2
1−α
(|ak | + |bk |) ≤ (1 − |b1 |).
Now using (3.2),
∞
X
k−α
k=2
1−α
(|ak | + |bk |) ≤
∞
X
k n (k − α)
k=2
=
(|ak | + |bk |)
∞
X
k n+1 − αk n
k=2
≤
1−α
1−α
∞
X
k m − αk n
k=2
1−α
(|ak | + |bk |)
(|ak | + |bk |)
(since m > n)
≤ (1 − |b1 |).
Thus f ∈ HS(α) and we have HS(m, n, α) ⊆ HS(α).
We have to show that HS(m, n, α) ⊆ HC(α). By (3.2),
∞
X
k(k − α)
k=2
1−α
(|ak | + |bk |) ≤
∞
X
k m − αk n
k=2
1−α
(|ak | + |bk |)
(since, m ≥ 2)
≤ (1 − |b1 |).
Thus f ∈ HC(α). So we have HS(m, n, α) ⊆ HC(α).
Theorem 3.3. The class HS(m, n, α) consists of univalent sense preserving harmonic mappings.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 27, 9 pp.
http://jipam.vu.edu.au/
4
K. K. D IXIT AND S AURABH P ORWAL
Proof. If z1 6= z2 then:
f (z1 ) − f (z2 ) ≥ 1 − g(z1 ) − g(z2 ) h(z1 ) − h(z2 ) h(z1 ) − h(z2 ) P∞
k
k
b
(z
−
z
)
k
1
2
k=1
P∞
= 1 − z1 − z2 + k=2 ak (z1k − z2k ) P∞
k=1 k|bk |
P
>1−
1− ∞
k|ak |
P∞ k=2km −αkn
|bk |
k=1
P∞ 1−α
≥ 0,
≥1−
km −αkn
1 − k=2 1−α |ak |
which proves univalence.
Note that f is sense preserving in U because
∞
X
1 h (z) ≥ 1 −
k|ak ||z|k−1
>1−
k=2
∞
X m
k=2
≥
k − αk n
|ak |
1−α
∞
X
k m − αk n
>
k=1
∞
X
k=1
≥
∞
X
1−α
|bk |
k m − αk n
|bk ||z|k−1
1−α
k|bk ||z|k−1 ≥ |g 1 (z)|.
k=1
Theorem 3.4. If f ∈ HS(m, n, α) then
|f (z)| ≤ |z|(1 + |b1 |) +
1−α
(1 − |b1 |)|z|2
− α2n
2m
and
1−α
2
|f (z)| ≥ (1 − |b1 |) |z| − m
|z| .
2 − α2n
Equalities are attained by the functions:
(3.3)
fθ (z) = z + |b1 |eiθ z̄ +
1−α
(1 − |b1 |)z 2
− α2n
2m
and
(3.4)
fθ (z) = z + |b1 |eiθ z̄ +
1−α
(1 − |b1 |)z̄ 2
n
− α2
2m
for properly chosen real θ.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 27, 9 pp.
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H ARMONIC U NIVALENT F UNCTIONS
5
Proof. We have
2
|f (z)| ≤ |z|(1 + |b1 |) + |z|
∞
X
(|ak | + |bk |)
k=2
∞
1 − α X k m − αk n
≤ |z|(1 + |b1 |)|z| m
(|ak | + |bk |)
2 − α2n k=2 1 − α
2
≤ |z|(1 + |b1 |) + |z|2
1−α
(1 − |b1 |)
− α2n
2m
and
|f (z)| ≥ (1 − |b1 |)|z| −
∞
X
(|ak | + |bk |)|z|k ≥ (1 − |b1 |)|z| − |z|2
k=2
≥ (1 − |b1 |)|z| − |z|2
∞
X
(|ak | + |bk |)
k=2
1−α
− α2n
2m
∞
X
k m − αk n
k=2
1−α
(|ak | + |bk |)
1−α
≥ (1 − |b1 |)|z| − |z|2 m
(1 − |b1 |)
n
2
−
α2
2 (1 − α)
= (1 − |b1 |) |z| − |z| m
.
2 − α2n
It can be easily seen that the function f (z) defined by (3.3) and (3.4) is extremal for Theorem
3.4.
Thus the class HS(m, n, α) is uniformly bounded, and hence it is normal by Montel’s Theorem.
Remark 1.
(i) For m = 1, n = 0, HS(1, 0, α) = HS(α). The above theorem reduces to:
1−α
(3.5)
|f (z)| ≤ |z|(1 + |b1 |) +
(1 − |b1 |)|z|2 ,
2−α
(3.6)
1−α 2
|f (z)| ≥ (1 − |b1 |) |z| −
|z| .
2−α
This result is different from that of Özturk and Yalcin [5, Theorem 3.6]. Also, our result
gives a better estimate than that of [5] because
1−α
|f (z)| ≤ |z|(1 + |b1 |) +
(1 − |b1 |)|z|2
2−α
(1 − α2 )
≤ |z|(1 + |b1 |) +
(1 − |b1 |)|z|2
2
and
1−α 2
(1 − α2 ) 2
|f (z)| ≥ (1 − |b1 | |z| −
|z| ≥ (1 − |b1 |) |z| −
|z| .
2−α
2
Although, Özturk and Yalcin [5] state that the result is sharp for the function
(1 − |b1 |)
(1 − α2 )z̄ 2 ,
2
it can be easily seen that the function fθ (z) does not satisfy the coefficient condition
for the class HS(α) defined by them. Hence, the function fθ (z) does not belong in
fθ (z) = z + |b1 |eiθ z̄ +
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 27, 9 pp.
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6
K. K. D IXIT AND S AURABH P ORWAL
HS(α). Therefore the results of Özturk and Yalcin are incorrect. The correct results are
mentioned in (3.5) and (3.6) and these results are sharp for functions in (3.3) and (3.4)
with m = 1, n = 0.
(ii) For m = 2, n = 1, HS(2, 1, α) = HC(α). Theorem 3.4 reduces to
1−α
(1 − |b1 |)|z|2 ,
|f (z)| ≤ |z|(1 + |b1 |) +
4 − 2α
and
1−α 2
|f (z)| ≥ (1 − |b1 |) |z| −
|z| .
4 − 2α
This result is different from the result of Özturk and Yalcin [5, Theorem 3.8], and it can
be easily seen that our result gives a better estimate. Also, it can be easily verified that
the sharp result for [5, Theorem 3.8] given by the function
3 − α − 2α2 2
z̄
2α
does not belong to HC(α). Hence the results of Özturk and Yalcin [5] are incorrect.
The correct result is obtained by Theorem 3.4 by putting m = 2, n = 1.
fθ (z) = z + |b1 |eiθ z̄ +
Theorem 3.5. The extreme points of HS o (m, n, α) are functions of the form z +ak z k or z +bl z l
with
1−α
1−α
|ak | = m
, |bl | = m
,
0 ≤ α < 1.
n
k − αk
l − αln
Proof. Suppose that
∞ X
k
k
f (z) = z +
ak z + b k z
k=2
is such that
∞
X
k m − αk n
k=2
1−α
(|ak | + |bk |) < 1,
ak > 0.
Then, if λ > 0 is small enough we can replace ak by ak −λ, ak +λ and we obtain two functions
that satisfy the same condition, for which one obtains f (z) = 21 [f1 (z) + f2 (z)]. Hence f is not
a possible extreme point of HS o (m, n, α).
Now let f ∈ HS o (m, n, α) be such that
∞
X
k m − αk n
(3.7)
(|ak | + |bk |) = 1,
ak 6= 0, bl 6= 0.
1
−
α
k=2
If λ > 0 is small enough and if µ, τ with |µ| = |τ | = 1 are properly chosen complex numbers,
then leaving all but ak , bl coefficients of f (z) unchanged and replacing ak , bl by
1−α
1−α
ak + λ m
µ,
bl − λ m
τ,
n
k − αk
l − αln
1−α
1−α
ak − λ m
µ,
b
+
λ
τ,
l
k − αk n
lm − αln
we obtain functions f1 (z), f2 (z) that satisfy (3.2) such that f (z) = 12 [f1 (z) + f2 (z)].
In this case f cannot be an extreme point. Thus for |ak | = km1−α
, |bl | = lm1−α
, f (z) =
−αkn
−αln
k
o
l
z + ak z or f (z) = z + bl z are extreme points of HS (m, n, α).
Remark 2.
(1) If m = 1, n = 0 the extreme points of the class HS o (α) are obtained.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 27, 9 pp.
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H ARMONIC U NIVALENT F UNCTIONS
7
(2) If m = 2, n = 1 the extreme points of the class HC o (α) are obtained.
o
Let KH
denote the class of harmonic univalent functions of the form (2.1) with b1 = 0
that map U onto convex domains. It is known [3, Theorem 5.10] that the sharp inequalities
|Ak | ≤ k+1
, |Bk | ≤ k−1
are true. These results will be used in the next theorem.
2
2
Theorem 3.6. Suppose that
F (z) = z +
∞ X
Ak z k + Bk z k
k=2
o
KH
.
o
o
belongs to
If f ∈ HS (m, n, α) then f ∗ F ∈ HS o (m − 1, n − 1; α) if n ≥ 1 and
f ♦ F ∈ HS (m, n; α).
Proof. Since f ∈ HS o (m, n; α), then
∞
X
(3.8)
(k m − αk n )(|ak | + |bk |) ≤ 1 − α.
k=2
Using (3.8), we have
∞
∞
X
X
Ak Bk m−1
n−1
m
n
(k
− αk )(|ak Ak | + |bk Bk |) =
(k − αk ) |ak | + |bk | k
k
k=2
k=2
≤
∞
X
(k m − αk n )(|ak | + |bk |) ≤ 1 − α.
k=2
o
It follows that f ∗ F ∈ HS (m − 1, n − 1; α). Next, again using (3.8),
∞
X
ak Ak bk Bk m
n
(k − αk ) + k k
k=2
∞
X
Ak Bk m
n
≤
(k − αk ) |ak | + |bk | k
k
k=2
≤
∞
X
(k m − αk n )(|ak | + |bk |)
k=2
≤ 1 − α.
Thus we have f ♦F ∈ HS o (m, n; α).
P∞
Let S denote the class of analytic univalent functions of the form F (z) = z + k=2 Ak z k . It
is well known that the sharp inequality |Ak | ≤ k is true. It is needed in next theorem.
Theorem 3.7. If f ∈ HS o (m, n; α) and F ∈ S then for | ∈ | ≤ 1, f ∗ (F + ∈ F̄ ) ∈
HS o (m − 1, n − 1; α) if n ≥ 1.
Proof. Since f ∈ HS o (m, n; α), we have
∞
X
(3.9)
(k m − αk n )(|ak | + |bk |) ≤ 1 − α.
k=2
Now, using (3.9)
∞
∞
X
X
m−1
n−1
(k
− αk )(|ak Ak | + |bk Bk |) ≤
(k m − αk n )(|ak | + |bk |)
k=2
k=2
≤ 1 − α.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 27, 9 pp.
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8
K. K. D IXIT AND S AURABH P ORWAL
It follows that f ∗ (F + ∈ F̄ ) ∈ HS o (m − 1, n − 1; α) if n ≥ 1.
Let PHo denote the class of functions F complex and harmonic in U, f = h + ḡ such that
Re f (z) > 0, z ∈ U and
H(z) = 1 +
∞
X
k
Ak z ,
G(z) =
k=1
∞
X
Bk z k .
k=2
It is known [4, Theorem 3] that the sharp inequalities |Ak | ≤ k + 1, |Bk | ≤ k − 1 are true.
Theorem 3.8. Suppose that
F (z) = 1 +
∞ X
k
Ak z + Bk
zk
k=1
belong to PHo . Then f ∈ HS o (m, n; α) and for
1, α) if n ≥ 1 and A11 f ♦F ∈ HS o (m, n; α)
3
2
1
f
A1
≤ |A1 | ≤ 2,
∗ F ∈ HS o (m − 1, n −
Proof. Since f ∈ HS o (m, n; α), then we have
∞
X
(3.10)
(k m − αk n )(|ak | + |bk |) ≤ 1 − α.
k=2
Now, using (3.10),
∞
X
(k
k=2
m−1
− αk
n−1
ak Ak bk Bk ) +
A1 A1 ∞
X
|ak | k + 1
|bk | k − 1
m
n
≤
(k − αk )
+
|A
k
|A1 | k
1|
k=2
≤
∞
X
(k m − αk n )(|ak | + |bk |)
k=2
≤ 1 − α.
Thus A11 f ∗ F ∈ HS o (m − 1, n − 1, α) if n ≥ 1. Similarly, we can show that
HS o (m, n; α).
1
f
A1
♦F ∈
Theorem 3.9. Let
f (z) = z + b1 z +
∞ X
ak z k + b k z k
k=2
be a member of HS(m, n, α). If δ ≤ (
n ≥ 1.
2n −1
2n
)(1 − |b1 |), then N (f ) ⊂ HS(α), provided that
Proof. Let f ∈ HS(m, n; α) and
F (z) = z + B1 z +
∞ X
Ak z k + Bk z k
k=2
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H ARMONIC U NIVALENT F UNCTIONS
9
belong to N (f ). We have
∞
X
(k − α)(|Ak | + |Bk |)
(1 − α)|B1 | +
k=2
≤ (1 − α)|B1 − b1 | + (1 − α)|b1 |
∞
∞
X
X
+
(k − α)|(|Ak − ak | + |Bk − bk |) +
(k − α)(|ak | + |bk |)
k=2
k=2
≤ (1 − α)δ + (1 − α)|b1 | +
1
2n
∞
X
(k n+1 − αk n )(|ak | + |bk |)
k=2
1
≤ (1 − α)δ + (1 − α)|b1 | + n (1 − α)(1 − |b1 |)
2
≤ (1 − α),
if δ ≤
2n −1
2n
(1 − |b1 |). Thus F (z) ∈ HS(α).
Remark 3. For f ∈ HS(2, 1, α) ≡ HC(α), our result is different from the result given by
Özturk and Yalcin [5, Theorem 3.15]. It can be easily seen that our result improves it.
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J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 27, 9 pp.
http://jipam.vu.edu.au/