Palestine Journal of Mathematics
Vol. 6(Special Issue: II) (2017)
, 320326
© Palestine Polytechnic University-PPU 2017
A SUBCLASS OF HARMONIC UNIVALENT FUNCTIONS DEFINED BY A LINEAR OPERATOR Hasan Bayram and Sibel Yalçn
Communicated by Ayman Badawi MSC 2010 Classications: Primary 30C45; Secondary 30C50. Keywords and phrases: Harmonic, univalent, modied generalized Salagean operator, multiplier transformation. Abstract. In the present paper, we investigate some basic properties of a subclass of harmonic functions dened by multiplier transformations. Such as, coefcient inequalities, distortion bounds and extreme points.
1 Introduction Let H denote the family of continuous complex valued harmonic functions which are harmonic in the open unit disk U = {z : |z| < 1} and let A be the subclass of H consisting of functions which are analytic in U. A function harmonic in U may be written as f = h + g, where h and g are members of A. We call h the analytic part and g co-analytic part of f . A necessary and sufcient condition for f to be locally univalent and sense-preserving in U is that |h′ (z )| > |g ′ (z )| (see Clunie and Sheil-Small [4]). To this end, without loss of generality, we may write h(z ) = z +
∞ ∑ ak z k and k=2
g (z ) =
∞ ∑ bk z k .
(1.1)
k=1
Let SH denote the family of functions f = h + g which are harmonic, univalent, and sensepreserving in U for which f (0) = fz (0) − 1 = 0. One shows easily that the sense-preserving property implies that |b1 | < 1. The subclass SH 0 of SH consists of all functions in SH which have the additional property fz¯ (0) = 0. In 1984 Clunie and Sheil-Small [4] investigated the class SH as well as its geometric subclasses and obtained some coefcient bounds. Since then, there have been several related papers on SH and its subclasses. Also note that SH reduces to the class S of normalized analytic univalent functions in U , if the co-analytic part of f is identically zero. For f ∈ S , the differential operator Dn (n ∈ N0 = N ∪ {0}) of f was introduced by Salagean [8]. For f = h + g given by (1.1), Jahangiri et al. [7] dened the modied Salagean operator of f as Dn f (z ) = Dn h(z ) + (−1)n Dn g (z ), where Dn h(z ) = z +
∞ ∞ ∑ ∑ k n bk z k . k n ak z k and Dn g (z ) = k=2
k=1
Next, for functions f ∈ A, Cho and Srivastava [2] dened multiplier transformations. For f = h + g given by (1.1), we dene the modied multiplier transformation of f 0 Iγ,β f (z ) = D0 f (z ) = h(z ) + g (z ),
1 Iγ,β f (z ) =
γ (h(z ) + g (z )) + β (zh′ (z ) − zg ′ (z )) γD0 f (z ) + βD1 f (z ) = , γ+β γ+β ( ) n−1 n 1 Iγ,β f (z ) = Iγ,β Iγ,β f (z ) . (n ∈ N0 )
(1.2) (1.3)
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HARMONIC UNIVALENT FUNCTIONS
Where β ≥ γ ≥ 0. If f is given by (1.1), then from (1.2) and (1.3) we see that n Iγ,β f (z ) = z +
)n ∞ ( ∑ βk + γ k=2
γ+β
ak z k + (−1)n
)n ∞ ( ∑ βk − γ k=1
γ+β
bk z k .
(1.4)
Also if f is given by (1.1), then we have n Iγ,β f (z )
( ) ( ) = f ∗ ϕ1 (z ) + ϕ2 (z ) ∗ ... ∗ ϕ1 (z ) + ϕ2 (z ) | {z }
:
n times
= h ∗ ϕ1 (z ) ∗ ... ∗ ϕ1 (z ) + g ∗ ϕ2 (z ) ∗ ... ∗ ϕ2 (z ), | {z } | {z } n times
n times
where "∗" denotes the usual Hadamard product or convolution of power series and ϕ1 (z ) =
(γ + β )z − γz 2 , (γ + β )(1 − z )2
ϕ2 (z ) =
(γ − β )z − γz 2 . (γ + β )(1 − z )2
By specializing the parametres γ and n, we obtain the following operators studied by various authors: for f ∈ A, (i) I0n,1 f (z ) = Dn f (z ) ( [8]), (ii) Iλn f (z ) ( [2], [3],[5]), (iii) I1n,1 = I n f (z ) ([11]), for f ∈ H, (iv) I0n,1 f (z ) = Dn f (z ) ([7]), n n (v) Iγ, 1 f (z ) = Iγ f (z ) ([12]). Denote by SH (γ, β, n, α) the subclass of SH consisting of functions f of the form (1.1) that satisfy the condition ( ) ℜ
n+1 Iγ,β f (z ) n f (z ) Iγ,β
≥ α, 0 ≤ α < 1
(1.5)
n where Iγ,β f (z ) is dened by (1.4). We let the subclass SH (γ, β, n, α) consisting of harmonic functions fn = h + g n in SH so that h and gn are of the form
h(z ) = z −
∞ ∞ ∑ ∑ ak z k , gn (z ) = (−1)n bk z k , ak , bk ≥ 0. k=2
(1.6)
k=1
By suitably specializing the parameters, the classes SH (γ, β, n, α) reduces to the various subclasses of harmonic univalent functions. Such as, (i) SH (0, 1, 0, 0) = SH ∗ (0) ( [1], [9], [10]), (ii) SH (0, 1, 0, α) = SH ∗ (α) ( [6]), (iii) SH (0, 1, 1, 0) = KH (0) ([1], [9], [10]), (iv) SH (0, 1, 1, α) = KH (α) ([6]), (v) SH (0, 1, n, α) = H (n, α) ([7]), (vi) SH (γ, 1, n, α) = SH (γ, n, α) ([12]). Dene SH 0 (γ, β, n, α) := SH (γ, β, n, α) ∩ SH 0 and 0 SH (γ, β, n, α) := SH (γ, β, n, α) ∩ SH 0 .
2 Main results f = h + g be so that h and g are given by (1.1) with b1 = 0 . Furthermore, let )n ( ) )n ( ) ∞ ( ∑ βk + γ βk − γ βk + γ βk − γ − α |ak | + + α |bk | ≤ 1 − α, (2.1) γ+β γ+β γ+β γ+β
Theorem 2.1.
∞ ( ∑ k=2 where in
U
Let
k=2
0 ≤ γ ≤ β/2, n ∈ N0 ,
and
f∈
γ γ +β
SH 0 (γ, β, n, α).
≤α≤
β γ +β . Then
f
is sense-preserving, harmonic univalent
322
Hasan Bayram and Sibel Yalçn
If z1 ̸= z2 ,
Proof.
∞ ( ) ∑ bk z1k − z2k f (z1 ) − f (z2 ) g (z1 ) − g (z2 ) k=2 ≥ 1− =1− ∞ h(z1 ) − h(z2 ) h(z1 ) − h(z2 ) ( ) (z1 − z2 ) + ∑ ak z k − z k 1 2 k=2
∞ ∑
> 1−
k |bk |
k=2 ∞ ∑
1−
∞ βk−γ n βk−γ +α ∑ ( γ +β ) ( γ +β )
k=2
k |ak |
≥1−
1−α
k=2 ∞ ∑
1−
|bk |
n
+γ βk+γ ( βk γ +β ) ( γ +β −α)
1−α
k=2
|ak |
≥ 0,
which proves univalence. Note that f is sense preserving in U. This is because |h′ (z )| ≥
∞ ∞ ∑ ∑ 1 − k |ak | |z|k−1 > 1 − k=2
≥
∞ ∑
(
βk−γ γ +β
)n (
βk−γ γ +β
)
+α
1−α
k=2
(
)n (
βk+γ γ +β
βk+γ γ +β
−α
) |ak |
1−α
k=2
|bk | >
∞ ∑ k−1 k |bk | |z| ≥ |g ′ (z )| . k=2
Using the fact that ℜ(w) ≥ α if and only if |1 − α + w| ≥ |1 + α − w|, it sufces to show that n+1 n+1 n n f (z ) − (1 + α)Iγ,β f (z ) ≥ 0. f (z ) + Iγ,β f (z ) − Iγ,β (2.2) (1 − α)Iγ,β n+1 n Substituting for Iγ,β f (z ) and Iγ,β f (z ) in (2.2), we obtain
n n+1 n n+1 f (z ) − (1 + α)Iγ,β f (z ) + Iγ,β f (z ) f (z ) − Iγ,β (1 − α)Iγ,β )n ( ) ∞ ( ∑ βk + γ βk + γ + 1 − α |ak | |z|k ≥ 2(1 − α) |z| − γ+β γ+β k=2 ) ( ) ( ∞ n ∑ βk − γ βk − γ k − − 1 + α |bk | |z| γ+β γ+β k=2 )n ( ) ∞ ( ∑ βk + γ βk + γ k − − 1 − α |ak | |z| γ+β γ+β k=2 )n ( ) ∞ ( ∑ βk − γ βk − γ − + 1 + α |bk | |z|k γ+β γ+β k=2 { )n ( ) ∞ ( ∑ βk + γ βk + γ > 2(1 − α) |z| 1 − − α |ak | γ+β γ+β k=2 } )n ( ) ∞ ( ∑ βk − γ βk − γ − + α |bk | . γ+β γ+β k=2
This last expression is non-negative by (2.1), and so the proof is complete. Theorem 2.2.
Let
fn = h + g n
be given by (1.6) with b1
= 0.
Then
0
fn ∈ SH (γ, β, n, α) if and
only if
)n ( ∞ ( ∑ βk + γ βk + γ k=2 where
γ+β
γ+β
0 ≤ γ ≤ β/2, n ∈ N0 ,
) − α ak +
)n ( ∞ ( ∑ βk − γ βk − γ k=2
γ γ +β
≤α≤
β γ +β .
γ+β
γ+β
)
+ α bk ≤ 1 − α,
(2.3)
323
HARMONIC UNIVALENT FUNCTIONS 0
The "if" part follows from Theorem 1 upon noting that SH (γ, β, n, α) ⊂ SH 0 (γ, β, n, α). 0 For the "only if" part, we show that fn ∈/ SH (γ, β, n, α) if the condition (2.3) does not hold. Note that a necessary and sufcient condition for fn = h + g n given by (1.6), to be 0 in SH (γ, β, n, α) is that the condition (1.5) to be satised. This is equivalent to Proof.
)n ( ) )n ( ) ∞ ( ∞ ( ∑ ∑ βk+γ βk+γ βk−γ βk−γ k k ( 1 − α ) z − − α a z − + α b z k k γ +β γ +β γ +β γ +β k=2 k=2 ≥ 0. ℜ ( ( ) ) ∞ ∞ ∑ βk+γ n ∑ βk−γ n k k + z− a z b z k k γ +β γ +β k=2
k=2
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 (1 − α) −
)n ( ∞ ( ∑ βk+γ βk+γ γ +β
k=2
1−
) )n ( ) ∞ ( ∑ βk−γ βk−γ − α ak rk−1 − + α bk rk−1 γ +β γ +β
γ +β ∞ ∑
(
)n
βk+γ γ +β
k=2
k=2
ak rk−1 +
)n ∞ ( ∑ βk−γ γ +β
k=2
bk rk−1
≥ 0 (2.4)
If the condition (2.3) does not hold, then the numerator in (2.4) is negative for r sufciently close to 1. Hence there exist z0 = r0 in (0, 1) for which the quotient in (2.4) is negative. This 0 contradicts the required condition for fn ∈ SH (γ, β, n, α) and so the proof is complete. Theorem 2.3.
Let
fn
be given by (1.6). Then
fn (z ) =
∞ ∑ k=1
0
fn ∈ SH (γ, β, n, α) if and only if
(Xk hk (z ) + Yk gnk (z )) ,
where
h1 (z ) = z, hk (z ) = z −
1−α
n
+γ βk+γ ( βk γ +β ) ( γ +β −α)
z k (k = 2, 3, ...),
and gn1 (z ) = z, gnk (z ) = z + (−1)n Xk ≥ 0, Yk ≥ 0,
∞ ∑ k=1
(Xk + Yk ) = 1, 0 ≤ γ ≤ β/2, n ∈ N0 ,
In particular, the extreme points of Proof.
1−α
n
βk−γ ( βk−γ γ +β ) ( γ +β +α)
z k (k = 2, 3, ...),
γ γ +β
≤α≤
β γ +β .
0
SH (γ, β, n, α) are {hk } and {gnk }.
For functions fn of the form (1.6) we have fn (z ) =
∞ ∑
(Xk hk (z ) + Yk gnk (z ))
k=1
=
∞ ∑
(Xk + Yk ) z −
k=1
∞ ∑
(
βk+γ γ +β
)n (
∞ ∑ +
(
βk−γ γ +β
(
k=2
βk−γ γ +β
βk+γ γ +β
−α
)n (
βk−γ γ +β
1−α
k=2
=
∞ ∑
1−α
k=2
(
k=2
+(−1)n
Then
∞ ∑
βk+γ γ +β
k=2
k=2
βk+γ γ +β
1−α )n (
βk−γ γ +β
) (
+α
−α
βk+γ γ +β
+α
)n (
)
(
) Xk z k
) Yk z k .
1−α βk+γ γ +β
−α
) Xk
1−α
βk−γ γ +β
∞ ∑ Yk = 1 − X1 − Y1 ≤ 1 Xk +
∞ ∑
1−α )n (
)n (
βk−γ γ +β
) Yk
+α
324
Hasan Bayram and Sibel Yalçn 0
0
and so fn ∈ SH (γ, β, n, α). Conversely, if fn ∈ SH (γ, β, n, α), then ak ≤ (
βk+γ γ +β
and bk ≤ (
Set
( Xk = ( Yk =
βk+γ γ +β
βk−γ γ +β
βk+γ γ +β
βk−γ γ +β
βk+γ γ +β
−α
1−α )n (
βk−γ γ +β
)
).
+α
) ak , (k = 2, 3, ...) )
+α
bk , (k = 2, 3, ...)
1−α
and
−α
1−α )n (
βk−γ γ +β
)n (
1−α )n (
( X1 + Y1 = 1 −
∞ ∑ Xk + Yk
)
k=2
where Xk , Yk ≥ 0. Then, as required, we obtain ∞ ∞ ∞ ∑ ∑ ∑ Xk hk (z ) + Yk gnk (z ) = (Xk hk (z ) + Yk gnk (z )) .
fn (z ) = (X1 + Y1 )z +
k=2
Theorem 2.4.
γ γ +β
≤α≤
Let
k=2
0
fn ∈ SH (γ, β, n, α).
k=1
Then for
β γ +β we have
|fn (z )| ≤ r + (
2β +γ γ +β
and
|fn (z )| ≥ r − (
2β +γ γ +β
|z| = r < 1
(1 − α) )n (
2β +γ γ +β
(1 − α) )n (
2β +γ γ +β
−α
−α
and
0 ≤ γ ≤ β/2, n ∈ N0 ,
) r2 ,
) r2 .
We only prove the right hand inequality. The proof for the left hand inequality is similar 0 and will be omitted. Let fn ∈ SH (γ, β, n, α). Taking the absolute value of fn we have Proof.
|fn (z )|
≤ r+
∞ ∑
(ak + bk ) r2
k=2
≤ r+ ( ≤ r+ (
(1 − α) r2 )n (
2β +γ γ +β
2β +γ γ +β
2β +γ γ +β
(1 − α) )n (
2β +γ γ +β
−α −α
)
)n ( ) ( βk+γ ∞ βk+γ ∑ γ +β γ +β − α k=2
1−α
( ak +
βk−γ γ +β
)n (
βk−γ γ +β
)
1−α
) r2 .
The following covering result follows from the left hand inequality in Theorem 2.4. fn of the form β ≤ α ≤ γ+ β . Then
Corollary 2.5.
n ∈ N0 ,
γ γ +β
Let
(1.6) be so that
w : |w| < 1 − (
2β +γ γ +β
0
fn ∈ SH (γ, β, n, α),
(1 − α) )n (
2β +γ γ +β
) ⊂ fn (U ). −α
where
+α
0 ≤ γ ≤ β/2,
bk
325
HARMONIC UNIVALENT FUNCTIONS Theorem 2.6.
Proof.
The class
0
SH (γ, β, n, α) is closed under convex combinations.
0
Let fni ∈ SH (γ, β, n, α) for i = 1, 2, ..., where fni is given by fni (z ) = z −
∞ ∞ ∑ ∑ aki z k + (−1)n bki z k . k=2
Then by (2.3), ∞ ∑
(
βk+γ γ +β
∞ ∑ i=1
βk+γ γ +β
−α
1−α
k=2
For
)n (
k=2
) aki +
∞ ∑ k=2
(
βk−γ γ +β
)n (
βk−γ γ +β
)
+α
1−α
bki ≤ 1.
(2.5)
ti = 1, 0 < ti < 1, the convex combination of fni may be written as (∞ ) (∞ ) ∞ ∞ ∞ ∑ ∑ ∑ ∑ ∑ k n ti fni (z ) = z − ti aki z + (−1) ti bki z k . i=1
k=2
i=1
k=2
i=1
Then by (2.5),
)( )n ( )( ( ) ) βk−γ βk−γ ∞ ∞ ∞ ∑ ∑ ∑ γ +β γ +β + α ti aki + ti bki 1−α 1−α i=1 i=1 k=2 k=2 )n ( ) )n ( ) ( ( βk+γ βk−γ βk−γ βk+γ ∞ ∞ ∑ − α + α ∑ γ +β γ +β γ +β γ +β = ti aki + bki 1−α 1−α
∞ ∑
(
βk+γ γ +β
i=1
≤
)n (
βk+γ γ +β
−α
k=2
∞ ∑ ti = 1. i=1
This is the condition required by (2.3) and so
∞ ∑
0
ti fni (z ) ∈ SH (γ, β, n, α).
i=1
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Hasan Bayram and Sibel Yalçn
[12] E. Yasar and S. Yalcin, Certain properties of a subclasses of harmonic functions Appl. Math. Inf. Sci. 7 17491753 (2013). Author information
Hasan Bayram and Sibel Yalçn, Department of Mathematics, Faculty of Arts and Science, Uludag University, 16059, Bursa, TURKEY. E-mail:
[email protected] and
[email protected] Received:
January 28, 2016.
Accepted:
May 22, 2016.
