Volume 8 (2007), Issue 2, Article 54, 9 pp.
ON THE SUBCLASS OF SALAGEAN-TYPE HARMONIC UNIVALENT
FUNCTIONS
SİBEL YALÇIN, METİN ÖZTÜRK, AND MÜMİN YAMANKARADENİZ
U LUDA Ǧ Ü NIVERSITESI F EN E DEBIYAT FAKÜLTESI
M ATEMATIK B ÖLÜMÜ , 16059 B URSA , T URKEY
skarpuz@uludag.edu.tr
Received 10 October, 2006; accepted 22 April, 2007
Communicated by H.M. Srivastava
A BSTRACT. Using the Salagean derivative, we introduce and study a class of Goodman- Ronning type harmonic univalent functions. We obtain coefficient conditions, extreme points, distortion bounds, convolution conditions, and convex combination for the above class of harmonic
functions.
Key words and phrases: Harmonic, Meromorphic, Starlike, Symmetric, Conjugate, Convex functions.
2000 Mathematics Subject Classification. 30C45, 30C50, 30C55.
1. I NTRODUCTION
A continuous complex valued function f = u + iv defined in a simply connected complex
domain D is said to be harmonic in D if both u and v are real harmonic in D. In any simply
connected domain D we can write f = h + ḡ, where h and g are analytic in D. A necessary
and sufficient condition for f to be locally univalent and sense preserving in D is that |h0 (z)| >
|g 0 (z)|, z ∈ D.
Denote by SH the class of functions f = h + ḡ that are harmonic univalent and sense preserving in the unit disk U = {z : |z| < 1} for which f (0) = fz (0)−1 = 0. Then for f = h+ḡ ∈ SH
we may express the analytic functions h and g as
(1.1)
h(z) = z +
∞
X
n=2
n
an z ,
g(z) =
∞
X
bn z n ,
|b1 | < 1.
n=1
In 1984 Clunie and Sheil-Small [1] investigated the class SH as well as its geometric subclasses
and obtained some coefficient bounds. Since then, there have been several related papers on SH
and its subclasses. Jahangiri et al. [3] make use of the Alexander integral transforms of certain
analytic functions (which are starlike or convex of positive order) with a view to investigating
the construction of sense-preserving, univalent, and close to convex harmonic functions.
254-06
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S IBEL YALÇIN , M ETIN Ö ZTÜRK ,
AND
M ÜMIN YAMANKARADENIZ
Definition 1.1. Recently, Rosy et al. [4], defined the subclass GH (γ) ⊂ SH consisting of
harmonic univalent functions f (z) satisfying the following condition
0
iα zf (z)
iα
(1.2)
Re (1 + e )
−e
≥ γ,
0 ≤ γ < 1, α ∈ R.
f (z)
They proved that if f = h + ḡ is given by (1.1) and if
∞ X
2n − 1 − γ
2n + 1 + γ
(1.3)
|an | +
|bn | ≤ 2,
1−γ
1−γ
n=1
0 ≤ γ < 1,
then f is a Goodman-Ronning type harmonic univalent function in U . This condition is proved
to be also necessary if h and g are of the form
h(z) = z −
(1.4)
∞
X
n
|an |z ,
g(z) =
n=2
∞
X
|bn |z n .
n=1
Jahangiri et al. [2] has introduced the modified Salagean operator of harmonic univalent
function f as
Dk f (z) = Dk h(z) + (−1)k Dk g(z),
(1.5)
where
Dk h(z) = z +
∞
X
n k an z n
k ∈ N,
and Dk g(z) =
n=2
∞
X
n k bn z n .
n=1
We let RSH (k, γ) denote the family of harmonic functions f of the form (1.1) such that
k+1
f (z)
iα D
iα
−e
≥ γ, 0 ≤ γ < 1, α ∈ R,
(1.6)
Re (1 + e ) k
D f (z)
where Dk f is defined by (1.5).
Also, we let the subclass RS H (k, γ) consist of harmonic functions fk = h + ḡk in RSH (k, γ)
so that h and gk are of the form
(1.7)
h(z) = z −
∞
X
|an |z n ,
gk (z) = (−1)k
n=2
∞
X
|bn |z n .
n=1
In this paper, the coefficient condition given in [4] for the class GH (γ) is extended to the class
RSH (k, γ) of the form (1.6). Furthermore, we determine extreme points, a distortion theorem,
convolution conditions and convex combinations for the functions in RS H (k, γ).
2. M AIN R ESULTS
In our first theorem, we introduce a sufficient coefficient bound for harmonic functions in
RSH (k, γ).
Theorem 2.1. Let f = h + ḡ be given by (1.1). If
(2.1)
∞
X
nk [(2n − 1 − γ)|an | + (2n + 1 + γ)|bn |] ≤ 2(1 − γ),
n=1
where a1 = 1 and 0 ≤ γ < 1, then f is sense preserving, harmonic univalent in U , and
f ∈ RSH (k, γ).
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 54, 9 pp.
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O N THE S UBCLASS OF S ALAGEAN -T YPE H ARMONIC U NIVALENT F UNCTIONS
3
Proof. If the inequality (2.1) holds for the coefficients of f = h + ḡ, then by (1.3), f is sense
preserving and harmonic univalent in U. According to the condition (1.5) we only need to show
that if (2.1) holds then
Dk+1 f (z)
Re (1 + e ) k
− eiα
D f (z)
(
iα
= Re (1 + eiα )
Dk+1 h(z) − (−1)k Dk+1 g(z)
Dk h(z) + (−1)k Dk g(z)
)
− eiα
≥ γ,
where 0 ≤ γ < 1.
Using the fact that Re w ≥ γ if and only if |1 − γ + w| ≥ |1 + γ − w|, it suffices to show that
(2.2)
(1 − γ)Dk f (z) + (1 + eiα )Dk+1 f (z) − eiα Dk f (z)
− (1 + γ)Dk f (z) − (1 + eiα )Dk+1 f (z) + eiα Dk f (z) ≥ 0.
Substituting for Dk f and Dk+1 f in (2.2) yields
(1 − γ)Dk f (z) + (1 + eiα )Dk+1 f (z) − eiα Dk f (z)
− (1 + γ)Dk f (z) − (1 + eiα )Dk+1 f (z) + eiα Dk f (z)
∞
X
= (2 − γ)z +
(1 − γ − eiα + n + neiα )nk an z n
n=2
∞
X
k
iα
iα
n
−(−1)
(n + ne − 1 + γ + e )bn z n=1
∞
X
− γz −
(n + neiα − 1 − γ − eiα )nk an z n
n=2
∞
X
k
iα
iα
n
+(−1)
(n + ne + 1 + γ + e )bn z n=1
≥ (2 − γ)|z| −
∞
X
k
n
n (2n − γ)|an ||z| −
n=2
− γ|z| −
(
nk (2n + γ)|bn ||z|n
n=1
∞
X
nk (2n − γ − 2)|an ||z|n −
n=2
= 2(1 − γ)|z| − 2
∞
X
∞
X
∞
X
nk (2n + γ + 2)|bn ||z|n
n=1
∞
X
nk (2n − γ − 1)|an ||z|n − 2
n=2
∞
X
nk (2n + γ + 1)|bn ||z|n
n=1
∞
X
nk (2n − γ − 1)
nk (2n + γ + 1)
n−1
|an ||z|
−
|bn ||z|n−1
= 2(1 − γ)|z| 1 −
1
−
γ
1
−
γ
n=2
n=1
(
!)
∞
∞
k
k
X
X
n (2n − γ − 1)
n (2n + γ + 1)
> 2(1 − γ) 1 −
|an | +
|bn |
.
1
−
γ
1
−
γ
n=2
n=1
This last expression is non-negative by (2.1), and so the proof is complete.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 54, 9 pp.
)
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4
S IBEL YALÇIN , M ETIN Ö ZTÜRK ,
AND
M ÜMIN YAMANKARADENIZ
The harmonic function
(2.3)
f (z) = z +
∞
X
n=2
where
∞
X
1−γ
1−γ
n
xn z +
ȳn z̄ n ,
k
k
n (2n − γ − 1)
n (2n + γ + 1)
n=1
∞
X
|xn | +
n=2
∞
X
|yn | = 1
n=1
shows that the coefficient bound given by (2.1) is sharp.
The functions of the form (2.3) are in RSH (k, γ) because
∞ k
∞
∞
X
X
X
nk (2n + γ + 1)
n (2n − γ − 1)
|an | +
|bn | = 1 +
|xn | +
|yn | = 2.
1−γ
1−γ
n=1
n=2
n=1
In the following theorem, it is shown that the condition (2.1) is also necessary for functions
fk = h + ḡk , where h and gk are of the form (1.7).
Theorem 2.2. Let fk = h + ḡk be given by (1.7). Then fk ∈ RS H (k, γ) if and only if
∞
X
(2.4)
nk [(2n − γ − 1)|an | + (2n + γ + 1)|bn |] ≤ 2(1 − γ).
n=1
Proof. Since RS H (k, γ) ⊂ RSH (k, γ), we only need to prove the “only if” part of the theorem.
To this end, for functions fk of the form (1.7), we notice that the condition (1.6) is equivalent to
P
(1 − γ)z − ∞
nk [n − γ + (n − 1)eiα ]|an |z n
P∞ kn=2 n
P
Re
k
n
z − n=2 n |an |z + (−1)2k ∞
n=1 n |bn |z̄
P
∞
(−1)2k n=1 nk [n + γ + (n + 1)eiα ]|bn |z̄ n
P
P∞ k
−
k
n
2k
n
z− ∞
n=2 n |an |z + (−1)
n=1 n |bn |z̄
P∞ k
1 − γ − n=2 n [n − γ + (n − 1)eiα ]|an |z n−1
P
P∞ k
= Re
k
n−1 + z̄ (−1)2k
n−1
1− ∞
n=1 n |bn |z̄
n=2 n |an |z
z
P
z̄
k
iα
n−1 (−1)2k ∞
n=1 n [n + γ + (n + 1)e ]|bn |z̄
z P
P∞ k
−
k
n−1 + z̄ (−1)2k
n−1
1− ∞
n=2 n |an |z
n=1 n |bn |z̄
z
≥ 0.
The above condition must hold for all values of z, |z| = r < 1. Upon choosing the values of z
on the positive real axis, where 0 ≤ z = r < 1, we must have
P
P
k
n−1
1−γ− ∞
nk (n − γ)|an |rn−1 − ∞
n=2P
n=1 n (n + γ)|bn |r
P
Re
∞
∞
1 − n=2 nk |an |rn−1 + n=1 nk |bn |rn−1
P∞ k
P
− 1)|an |rn−1 + ∞
nk (n + 1)|bn |rn−1
iα
n=2 n (n
n=1
P
P∞ k
≥ 0.
−e
k
n−1 +
n−1
1− ∞
n=2 n |an |r
n=1 n |bn |r
Since Re(−eiα ) ≥ −|eiα | = −1, the above inequality reduces to
P
P
k
k
n−1
1−γ− ∞
− γ − 1)|an |rn−1 − ∞
n=2 n (2n
n=1 n (2n + γ + 1)|bn |r
P∞ k
P
(2.5)
≥ 0.
∞
1 − n=2 n |an |rn−1 + n=1 nk |bn |rn−1
If the condition (2.4) does not hold then the numerator in (2.5) is negative for r sufficiently
close to 1. Thus there exists a z0 = r0 in (0, 1) for which the quotient in (2.5) is negative. This
contradicts the condition for f ∈ RS H (k, γ) and hence the result.
Next we determine a representation theorem for functions in RS H (k, γ).
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 54, 9 pp.
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O N THE S UBCLASS OF S ALAGEAN -T YPE H ARMONIC U NIVALENT F UNCTIONS
5
Theorem 2.3. Let fk be given by (1.7). Then fk ∈ RS H (k, γ) if and only if
∞
X
(2.6)
fk (z) =
(Xn hn (z) + Yn gkn (z))
n=1
where h1 (z) = z,
1−γ
z n (n = 2, 3, . . . ),
− γ − 1)
1−γ
z̄ n (n = 1, 2, . . . ),
gkn (z) = z + (−1)k k
n (2n + γ + 1)
∞
X
(Xn + Yn ) = 1,
hn (z) = z −
nk (2n
n=1
Xn ≥ 0, Yn ≥ 0. In particular, the extreme points of RS H (k, γ) are {hn } and {gkn }.
Proof. For functions fk of the form (2.6) we have
∞
X
fk (z) =
(Xn hn (z) + Yn gkn (z))
=
n=1
∞
X
(Xn + Yn )z −
n=1
∞
X
n=2
+ (−1)k
∞
X
n=1
1−γ
Xn z n
− γ − 1)
nk (2n
1−γ
Yn z̄ n .
+ γ + 1)
nk (2n
Then
∞
X
nk (2n − γ − 1)
n=2
1−γ
|an | +
∞
X
nk (2n + γ + 1)
n=1
1−γ
|bn | =
∞
X
Xn +
n=2
∞
X
Yn
n=1
= 1 − X1 ≤ 1
and so fk ∈ RS H (k, γ).
Conversely, suppose that fk ∈ RS H (k, γ). Setting
nk (2n − γ − 1)
an ,
1−γ
nk (2n + γ + 1)
Yn =
bn ,
1−γ
Xn =
where
P∞
n=1 (Xn
(n = 2, 3, . . . ),
(n = 1, 2, . . . ),
+ Yn ) = 1, we obtain
fk (z) =
∞
X
(Xn hn (z) + Yn gkn (z))
n=1
as required.
The following theorem gives the distortion bounds for functions in RS H (k, γ) which yields
a covering result for this class.
Theorem 2.4. Let fk ∈ RS H (k, γ). Then for |z| = r < 1 we have
1 1−γ 3+γ
|fk (z)| ≤ (1 + |b1 |)r + k
−
|b1 | r2
2
3−γ 3−γ
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 54, 9 pp.
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6
S IBEL YALÇIN , M ETIN Ö ZTÜRK ,
AND
M ÜMIN YAMANKARADENIZ
and
1
|fk (z)| ≥ (1 − |b1 |)r − k
2
1−γ 3+γ
−
|b1 | r2 .
3−γ 3−γ
Proof. We only prove the right hand inequality. The proof for the left hand inequality is similar
and will be omitted. Let fk ∈ RS H (k, γ). Taking the absolute value of fk we obtain
|fk (z)| ≤ (1 + |b1 |)r +
∞
X
(|an | + |bn |)rn
n=2
≤ (1 + |b1 |)r +
∞
X
(|an | + |bn |)r2
n=2
∞
≤ (1 + |b1 |)r +
≤ (1 + |b1 |)r +
≤ (1 + |b1 |)r +
≤ (1 + |b1 |)r +
1 − γ X 2k (3 − γ)
(|an | + |bn |)r2
2k (3 − γ) n=2 1 − γ
∞ 1 − γ X nk (2n − γ − 1)
nk (2n + γ + 1)
|an | +
|bn | r2
k
2 (3 − γ) n=2
1−γ
1−γ
1−γ
3+γ
1−
|b1 | r2
k
2 (3 − γ)
1−γ
1 1−γ 3+γ
−
|b1 | r2 .
2k 3 − γ 3 − γ
The following covering result follows from the left hand inequality in Theorem 2.4.
Corollary 2.5. Let fk of the form (1.7) be so that fk ∈ RS H (k, γ). Then
3.2k − 1 − (2k − 1)γ 3(2k − 1) − (2k + 1)γ
w : |w| <
−
|b1 | ⊂ fk (U ).
2k (3 − γ)
2k (3 − γ)
For our next theorem, we need to define the convolution of two harmonic functions. For
harmonic functions of the form
∞
∞
X
X
n
k
fk (z) = z −
|an |z + (−1)
|bn |z̄ n
n=2
n=1
and
Fk (z) = z −
∞
X
n=2
n
k
|An |z + (−1)
∞
X
|Bn |z̄ n
n=1
we define the convolution of fk and Fk as
(2.7)
(fk ∗ Fk )(z) = fk (z) ∗ Fk (z)
∞
∞
X
X
=z−
|an ||An |z n + (−1)k
|bn ||Bn |z̄ n .
n=2
n=1
Theorem 2.6. For 0 ≤ β ≤ γ < 1, let fk ∈ RS H (k, γ) and Fk ∈ RS H (k, β). Then the
convolution
fk ∗ Fk ∈ RS H (k, γ) ⊂ RS H (k, β).
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O N THE S UBCLASS OF S ALAGEAN -T YPE H ARMONIC U NIVALENT F UNCTIONS
7
Proof. Then the convolution fk ∗ Fk is given by (2.7). We wish to show that the coefficients of
fk ∗ Fk satisfy the required condition given in Theorem 2.2. For Fk ∈ RS H (k, β) we note that
|An | ≤ 1 and |Bn | ≤ 1. Now, for the convolution function fk ∗ Fk , we obtain
∞
X
nk (2n − β − 1)
1−β
n=2
∞
X
nk (2n + β + 1)
|an ||An | +
1−β
n=1
∞
X
nk (2n − β − 1)
≤
1−β
n=2
1−γ
n=2
∞
X
nk (2n + β + 1)
1−β
n=1
∞
X
nk (2n − γ − 1)
≤
|an | +
|bn ||Bn |
|an | +
∞
X
nk (2n + γ + 1)
1−γ
n=1
|bn |
|bn | ≤ 1
since 0 ≤ β ≤ γ < 1 and fk ∈ RS H (k, γ). Therefore fk ∗ Fk ∈ RS H (k, γ) ⊂ RS H (k, β).
Next we discuss the convex combinations of the class RS H (k, γ).
Theorem 2.7. The family RS H (k, γ) is closed under convex combination.
Proof. For i = 1, 2, . . . , suppose that fki ∈ RS H (k, γ), where
∞
X
fki (z) = z −
n
k
|ain |z + (−1)
n=2
∞
X
|bin |z̄ n .
n=1
Then by Theorem 2.2,
∞
X
nk (2n − γ − 1)
(2.8)
1−γ
n=2
For
P∞
i=1 ti
|ain | +
∞
X
nk (2n + γ + 1)
n=1
1−γ
|bin | ≤ 1.
= 1, 0 ≤ ti ≤ 1, the convex combination of fki may be written as
!
!
∞
∞
∞
∞
∞
X
X
X
X
X
ti fki (z) = z −
ti |ain | z n + (−1)k
ti |bin | z̄ n .
n=2
i=1
n=1
i=1
i=1
Then by (2.8),
∞
X
nk (2n − γ − 1)
n=2
∞
X
1−γ
=
≤
!
ti |ain |
i=1
∞
X
i=1
∞
X
+
∞
X
nk (2n + γ + 1)
n=1
ti
1−γ
∞
X
nk (2n − γ − 1)
n=2
1−γ
|ain | +
∞
X
!
ti |bin |
i=1
∞
X k
n=1
!
n (2n + γ + 1)
|bin |
1−γ
ti = 1
i=1
and therefore
∞
X
ti fki (z) ∈ RS H (k, γ).
i=1
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 54, 9 pp.
http://jipam.vu.edu.au/
8
S IBEL YALÇIN , M ETIN Ö ZTÜRK ,
AND
M ÜMIN YAMANKARADENIZ
Following Ruscheweyh [5], we call the δ−neighborhood of f the set
(
∞
∞
X
X
n
k
Nδ (fk ) = Fk : Fk (z) = z −
|An |z + (−1)
|Bn |z̄ n and
n=2
n=1
∞
X
)
n(|an − An | + |bn − Bn |) + |b1 − B1 | ≤ δ
.
n=2
Theorem 2.8. Assume that
fk (z) = z −
∞
X
n
k
|an |z + (−1)
n=2
∞
X
|bn |z̄ n
n=1
belongs to RS H (k, γ). If
1
1
1
δ≤
(1 − γ) 1 − k − 3 + 2γ − k (3 + γ) |b1 | ,
3
2
2
then Nδ (fk ) ⊂ RS H (0, γ).
Proof. Let
Fk (z) = z −
∞
X
n
k
|An |z + (−1)
n=2
∞
X
|Bn |z̄ n
n=1
belong to Nδ (fk ). We have
∞
X
2n − γ − 1
n=2
≤
1−γ
|An | +
n=1
∞
X
2n − γ − 1
n=2
1−γ
+
3
1−γ
∞
X
1−γ
|an − An | +
1−γ
|Bn |
∞
X
2n + γ + 1
1−γ
n=1
∞
X
2n − γ − 1
n=2
≤
∞
X
2n + γ + 1
|an | +
∞
X
2n + γ + 1
n=1
n(|an − An | + |bn − Bn |) +
n=2
|bn − Bn |
1−γ
|bn |
3
γ
|b1 | +
|b1 |
1−γ
1−γ
∞
∞
1 X nk (2n − γ − 1)
1 X nk (2n − γ − 1)
3+γ
+ k
|an | + k
|bn | +
|b1 |
2 n=2
1−γ
2 n=2
1−γ
1−γ
3δ
γ
1
3+γ
3+γ
≤
+
|b1 | + k 1 −
|b1 | +
|b1 | ≤ 1.
1−γ 1−γ
2
1−γ
1−γ
Thus Fk ∈ RS H (0, γ) for
1
1
1
δ≤
(1 − γ) 1 − k − 3 + 2γ − k (3 + γ) |b1 | .
3
2
2
R EFERENCES
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J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 54, 9 pp.
http://jipam.vu.edu.au/
O N THE S UBCLASS OF S ALAGEAN -T YPE H ARMONIC U NIVALENT F UNCTIONS
9
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