International Journal of Pure and Applied Mathematics ————————————————————————– Volume 60 No. 1 2010, 5-14
SUBCLASSES OF HARMONIC UNIVALENT FUNCTIONS ASSOCIATED WITH HYPERGEOMETRIC FUNCTIONS Sayali S. Joshi Department of Mathematics Walchand College of Engineering Sangli, Maharastra, 416 415, INDIA e-mail: sayali
[email protected]
Abstract: In the present paper two subclasses of SH are defined using convolution of harmonic univalent functions with hypergeometric functions and several interesting properties like coefficient bound, distortion theorem and extreme points are obtained. AMS Subject Classification: 30C45, 30C55, 50E20 Key Words: harmonic functions, univalent, sense-preserving, analytic, hypergeometric functions
1. Introduction A continuous function f = u + iv in any simply connected domain D ⊂ C is a complex-valued, harmonic function if both u and v are real harmonic in D. In [4], Clunie and Sheil-Small developed the basic theory of harmonic functions which are univalent in U = {z : |z| < 1} having normalization f (0) = 0 and fz (0) = 1. Such functions admit representation f = h + g¯,
(1.1) |h′ (z)|
where h and g are analytic. In this case f is sense-preserving if > |g′ (z)| ′ in U or equivalently, if dilation function ω = hg ′ satisfies |ω| < 1 for z ∈ U . To this end, without loss of generality we may write Received:
December 3, 2009
c 2010 Academic Publications
6
S.S. Joshi
h(z) = z +
∞ X
n=2
n
An z ,
g(z) =
∞ X
Bn z n ,
z ∈ U,
|B1 | < 1.
(1.2)
n=1
Let SH denote the family of functions of the form (1.1) which are harmonic, univalent and sense-preserving in U . For harmonic functions f = h + g¯ we call h as the analytic part of f and g as the co-analytic part of f . Note that the familiar class S of analytic functions is contained in SH . SH reduces to S, if the co-analytic part of f is identically zero. ∗ and K denote the subclasses of S consisting of harmonic univalent Let SH H H functions which map U onto starlike and convex domains respectively. In [7] ∗ (α) and K (α)(0 ≤ α < 1), which and [8], Jahangiri studied the subclasses SH H are harmonic starlike functions of order α and harmonic convex functions of order α in U respectively. Since then, there have been several related papers on SH and its subclasses. For more information see [3, 5, 6, 11].
SH
Recently Ahuja [2], studied the connection between various subclasses of and hypergeometric functions.
In the present paper we define two new subclasses of SH which are generated by taking convolution of harmonic univalent functions with hypergeometric functions. Consider the Gaussian hypergeometric function defined by ∞ X (ai )n (bi )n n z , ai , bi , ci ∈ C and ci 6= 0, −1, −2, . . . . F (ai , bi ; ci ; z) = (ci )n (1)n n=0
It is well-known that hypergeometric functions play an important role in the theory of univalent functions. The extensive use of such functions in geometric function theory has been shown by a number of authors which include Srivastava and Manocha [12], Owa and Srivastava [9], and Ruscheweyh and Singh [10]. We shall consider the functions Φ1 (z) and Φ2 (z) defined by ∞ X (a1 )n−1 (b1 )n−1 n Φ1 (z) = zF (a1 , b1 ; c1 ; z) = z + z (c1 )n−1 (1)n−1
(1.3)
n=2
and
Φ2 (z) = zF (a2 , b2 ; c2 ; z) =
∞ X (a2 )n−1 (b2 )n−1
n=1
(c2 )n−1 (1)n−1
zn .
(1.4)
∗ (Φ , Φ , α) denote the subclass of S Definition 1. Let SH 1 2 H consisting of the functions f = h + g¯ that satisfies the condition ) ( ′ z(h(z) ∗ Φ1 (z))′ − z(g(z) ∗ Φ2 (z)) ≥ α, (1.5) Re (h(z) ∗ Φ1 (z)) + (g(z) ∗ Φ2 (z))
where 0 ≤ α < 1, Φ1 (z) and Φ2 (z) are given by (1.3) and (1.4) respectively.
SUBCLASSES OF HARMONIC UNIVALENT FUNCTIONS...
7
Definition 2. Let KH (Φ1 , Φ2 , α) denote subclass of SH consisting of the functions f = h + g¯ that satisfies the condition 2 z (h(z) ∗ Φ1 (z))′′ + z(h(z) ∗ Φ1 (z))′ 2 ′′ ′ +z (g(z) ∗ Φ2 (z)) + z(g(z) ∗ Φ2 (z)) ≥ α, (1.6) Re z(h(z) ∗ Φ1 (z))′ − z(g(z) ∗ Φ2 (z))′
where 0 ≤ α < 1, Φ1 (z) and Φ2 (z) are given by (1.3) and (1.4) respectively.
We denote by TH the subclass of SH consisting functions of the form f = h + g¯ where h and g are given by ∞ ∞ X X n |Bn |z n , z ∈ U, |B1 | < 1. (1.7) |An |z , g(z) = h(z) = z − n=1
n=2
Also let
∗ ∗ T SH (Φ1 , Φ2 , α) = SH (Φ1 , Φ2 , α)∩TH and T KH (Φ1 , Φ2 , α) = KH (Φ1 , Φ2 , α)∩TH . ∗ (Φ , Φ , α) and T K (Φ , Φ , α) are non-empty as Note that the subclasses T SH 1 2 H 1 2 for the particular values of Φ1 and Φ2 that is by taking Φ1 (z) = zF (1, 1; 1; z) ∗ (Φ , Φ , α) and T K (Φ , Φ , α) and Φ2 (z) = zF (1, 1; 1; z) we observe that T SH 1 2 H 1 2 reduces to the classes studied by Jahangiri in [7, 8].
2. Results and Proofs Theorem 2.1. Let f = h + g¯, be such that h and g are given by (1.2). Furthermore let ∞ X n − α (a1 )n−1 (b1 )n−1 n + α (a2 )n−1 (b2 )n−1 |An | + |Bn | ≤ 2 (2.1) 1 − α (c1 )n−1 (1)n−1 1 − α (c2 )n−1 (1)n−1 n=1 (a1 )n−1 (b1 )n−1 n+α (a2 )n−1 (b2 )n−1 with A1 = 1, 0 ≤ α < 1 and n ≤ n−α 1−α (c1 )n−1 (1)n−1 and n ≤ 1−α (c2 )n−1 (1)n−1 . ∗ Then f is harmonic, univalent and sense-preserving in U and f ∈ SH (Φ1 , Φ2 , α). Proof. First we have to show that f = h + g¯ is locally univalent and sensepreserving in U , for which it is sufficient to show that |h′ (z)| > |g′ (z)|. Consider |h′ (z)| ≥ 1 −
∞ X
n=2
n|An |r n−1 > 1 −
∞ X
n=2
n|An |
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S.S. Joshi
≥1−
∞ X n − α (a1 )n−1 (b1 )n−1
n=2
1−α
(c1 )n−1 (1)n−1
|An | ≥
∞ X n + α (a2 )n−1 (b2 )n−1 1−α
n=1
≥
∞ X
n|Bn | ≥
∞ X
(c2 )n−1 (1)n−1
|Bn |
n|Bn |r n−1 > |g′ (z)|.
n=1
n=1
Now to show f is univalent in U , suppose z1 , z2 ∈ U such that z1 6= z2 then f (z1 ) − f (z2 ) ≥ 1 − g(z1 ) − g(z2 ) h(z1 ) − h(z2 ) h(z1 ) − h(z2 ) P∞ P∞ Bn (z1n − z2n ) nBn n=1P n=1 P∞ =1− n n > 1 − 1 − (z1 − z2 ) + ∞ n=2 An (z1 − z2 ) n=2 nAn P∞ n+α (a2 )n1 (b2 )n−1 n=1 1−α (c2 )n1 (1)n1 |Bn | ≥1− ≥ 0. P n−α (a1 )n−1 (b1 )n−1 1− ∞ |A | n n=2 1−α (c1 )n−1 (1)n−1
∗ (Φ , Φ , α). Now, we show that f ∈ SH 1 2
Using the fact that Re ω ≥ α if and only if |1 − α + ω| ≥ |1 + α − ω|, it suffices to show that |A(z) + (1 − α)B(z)| − |A(z) − (1 + α)B(z)| ≥ 0,
(2.2)
′
where A(z) = z(h(z) ∗ Φ1 (z))′ − z(g(z) ∗ Φ2 (z)) and B(z) = (h(z) ∗ Φ1 (z)) + (g(z)∗Φ2 (z)) substituting values of A(z) and B(z) in equation (2.2) and making use of (2.1) we obtain |A(z)+(1−α)B(z)|−|A(z)−(1+α)B(z)| = |z(h(z)∗Φ1 (z))′ −z(g(z) ∗ Φ2 (z)) + (1 − α)(h(z) ∗ Φ1 (z)) + (g(z) ∗ Φ2 (z))| − |z(h(z) ∗ Φ1 (z))′ − z(g(z) ∗ Φ2 (z)) − (1 + α)(h(z) ∗ Φ1 (z)) + (g(z) ∗ Φ2 (z))| ∞ X (a1 )n−1 (b1 )n−1 An z n (n + 1 − α) = (2 − α)z + (c1 )n−1 (1)n−1 n=2 ∞ X (a2 )n−1 (b2 )n−1 − Bn z n (n − 1 + α) (c2 )n−1 (1)n−1 n=1 ∞ X (a1 )n−1 (b1 )n−1 (n − 1 − α) − −αz + An z n (c1 )n−1 (1)n−1 n=2 ∞ X (a2 )n−1 (b2 )n−1 (n + 1 + α) Bn z n − (c2 )n−1 (1)n−1 n=1
′
′
SUBCLASSES OF HARMONIC UNIVALENT FUNCTIONS... ∞ X
≥ 2(1 − α)|z| −
(n + 1 − α)
n=2
−
∞ X
(n − 1 + α)
n=1
− α|z| −
∞ X
(a1 )n−1 (b1 )n−1 |An ||z|n (c1 )n−1 (1)n−1
(a2 )n−1 (b2 )n−1 |Bn ||z|n (c2 )n−1 (1)n−1
(n − 1 − α)
n=2 ∞ X
9
(a1 )n−1 (b1 )n−1 |An ||z|n (c1 )n−1 (1)n−1
(a2 )n−1 (b2 )n−1 |Bn ||z|n (c2 )n−1 (1)n−1 n=1 ( ∞ X n − α (a1 )n−1 (b1 )n−1 = 2(1 − α)|z| 1 − |An ||z|n−1 1 − α (c ) (1) 1 n−1 n−1 n=2 ) ∞ X n + α (a2 )n−1 (b2 )n−1 n−1 |Bn ||z| − 1 − α (c2 )n−1 (1)n−1 n=1 ( ∞ X n − α (a1 )n−1 (b1 )n−1 |An | ≥ 2(1 − α) 1 − 1 − α (c1 )n−1 (1)n−1 n=2 ) ∞ X n + α (a2 )n−1 (b2 )n−1 |Bn | ≥ 0, − 1 − α (c2 )n−1 (1)n−1 n=1 −
(n + 1 + α)
∗ (Φ , Φ , α). by the condition (2.1). Hence f ∈ SH 1 2
The coefficient bound is sharp for the function ∞ X 1 − α (c1 )n−1 (1)n−1 f (z) = z + xn z n n − α (a1 )n−1 (b1 )n−1 n=2 ∞ X 1 − α (c2 )n−1 (1)n−1 y¯n z¯n , + n + α (a ) (b ) 2 n−1 2 n−1 n=1 P∞ P∞ where n=2 |xn | + n=1 |yn | = 1.
(2.3)
Theorem 2.2. Let the function f = h + g¯ be so that h and g are given ∗ (Φ , Φ , α) if and only if by (1.7). Then f ∈ T SH 1 2 ∞ X n − α (a1 )n−1 (b1 )n−1 n + α (a2 )n−1 (b2 )n−1 |An | + |Bn | ≤ 2, (2.4) 1 − α (c1 )n−1 (1)n−1 1 − α (c2 )n−1 (1)n−1 n=1 (a1 )n−1 (b1 )n−1 n+α where A1 = 1, 0 ≤ α < 1 and n ≤ n−α 1−α (c1 )n−1 (1)n−1 and n ≤ 1−α (a2 )n−1 (b2 )n−1 (c2 )n−1 (1)n−1 .
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S.S. Joshi
∗ Proof. In view of Theorem 2.1 we need only to prove that f ∈ 6 T SH (Φ1 , Φ2 , α) if the coefficient inequality (2.4) does not hold. To this end we ∗ (Φ , Φ , α) then have, if f is in T SH 1 2 ) ( ′ z(h(z) ∗ Φ1 (z))′ − z(g(z) ∗ Φ2 (z)) −α ≥ 0 Re (h(z) ∗ Φ1 (z)) + (g(z) ∗ Φ2 (z))
which is equivalent to P (a1 )n1 (b1 )n1 (1 − α)|z| − ∞ (n − α) (c |An ||z|n n=2 ) (1) 1 n−1 n1 P∞ (a ) (b ) 2 n−1 2 n−1 n − n=1 (n + α) (c2 )n−1 (1)n−1 |Bn ||¯ z| ≥ 0. Re P∞ (a1 )n−1 (b1 )n−1 P∞ (a2 )n−1 (b2 )n−1 n− n |z|− |A ||z| |B ||¯ z | n n n=2 (c1 )n−1 (1)n−1 n=1 (c2 )n−1 (1)n−1
Upon choosing the values of z on the positive real axis where 0 ≤ z = r < 1, the above inequality reduces to P (a1 )n−1 (b1 )n−1 n−1 (1 − α) − ∞ n=2 (n − α) (c1 )n−1 (1)n−1 |An |r (a2 )n−1 (b2 )n−1 n−1 n=1 (n + α) (c2 )n−1 (1)n−1 |Bn |r P P∞ (a2 )n−1 (b2 )n−1 (a1 )n−1 (b1 )n−1 n−1 − n−1 − ∞ n=2 (c1 )n−1 (1)n−1 |An |r n=1 (c2 )n−1 (1)n−1 |Bn |r
−
1
P∞
≥ 0.
(2.5)
If condition (2.4) does not hold then the numerator in (2.5) is negative for r sufficiently close to 1. Thus there exists z0 = r0 in (0, 1) for which the quotient of (2.5) is negative. This contradicts the required condition for f ∈ ∗ (Φ , Φ , α) and so the proof is completed. T SH 1 2 Theorem 2.3. If f = h + g¯ be such that h and g are given by (1.2). Also let ∞ X n − α (a1 )n−1 (b1 )n−1 n + α (a2 )n−1 (b2 )n−1 n |An | + n |Bn | 1 − α (c1 )n−1 (1)n−1 1 − α (c2 )n−1 (1)n−1
n=1
≤2
n−α (a1 )n−1 (b1 )n−1
(2.6) (a ) (b 2 n−1 2 )n−1 n+α
with A1 = 1, 0 ≤ α < 1 and n ≤ 1−α (c1 )n−1 (1)n−1 and n ≤ 1−α (c2 )n−1 (1)n−1 . Then f is harmonic, univalent and sense-preserving in U and f ∈ KH (Φ1 , Φ2 , α). Theorem 2.4. If f = h + g¯ be such that h and g are given by (1.7). Then f ∈ T KH (Φ1 , Φ2 , α) if and only if (2.6) is satisfied. We omit the proofs of Theorems 2.3 and 2.4 since they are similar to that of Theorem 2.2.
SUBCLASSES OF HARMONIC UNIVALENT FUNCTIONS...
11
∗ (Φ , Φ , α) and Theorem 2.5. Let f ∈ T SH 1 2 (n − α) (a1 )n−1 (b1 )n−1 (2 − α)a1 b1 ≤ (1 − α)c1 (1 − α) (c1 )n−1 (1)n−1
and (2 − α)a1 b1 (n + α) (a2 )n−1 (b2 )n−1 ≤ (1 − α)c1 (1 − α) (c2 )n−1 (1)n−1 for n ≥ 2. Then we have (1 − α) (1 + α) c1 − |B1 | r 2 , |f (z)| ≤ (1 + |B1 |)r + a1 b1 (2 − α) (2 − α) and c1 (1 − α) (1 + α) |f (z)| ≥ (1 − |B1 |)r − − |B1 | r 2 , a1 b1 (2 − α) (2 − α)
|z| = r < 1 (2.7)
|z| = r < 1. (2.8)
∗ (Φ , Φ , α), then we have, Proof. Let f ∈ T SH 1 2 ∞ X (|An | + |Bn |)r n |f (z)| ≤ (1 + |B1 |)r +
≤ (1 + |B1 |)r +
n=2 ∞ X
(|An | + |Bn |)r 2
n=2
1−α 2−α
∞
c1 X a1 b1 n=2 2 − α a1 b1 2 − α a1 b1 |An | + |Bn | r 2 × 1−α c1 1−α c1 ∞ 1−α c1 X n − α (a1 )n−1 (b1 )n−1 ≤ (1 + |B1 |)r + |An | 2 − α a1 b1 1 − α (c1 )n−1 (1)n−1 n=2 n + α (a2 )n−1 (b2 )n−1 + |Bn | r 2 1 − α (c2 )n−1 (1)n−1 1−α c1 1+α ≤ (1 + |B1 |)r + |B1 | r 2 , by (2.4) 1− 2 − α a1 b1 1−α 1−α 1+α c1 − |B1 | r 2 ≤ (1 + |B1 |)r + a1 b1 2 − α 2 − α and similarly we can prove the other inequality (2.8). ≤ (1 + |B1 |)r +
Theorem 2.6. Let f ∈ T KH (Φ1 , Φ2 , α) with 2(2 − α)a1 b1 (n − α) (a1 )n−1 (b1 )n−1 ≤ (1 − α)c1 (1 − α) (c1 )n−1 (1)n−1
12
S.S. Joshi
and 2(2 − α)a1 b1 (n + α) (a2 )n−1 (b2 )n−1 ≤ . (1 − α)c1 (1 − α) (c2 )n−1 (1)n−1 Then (1 − α) (1 + α) − |B1 | r 2 (2 − α) (2 − α)
|z| = r < 1
(1 − α) (1 + α) − |B1 | r 2 (2 − α) (2 − α)
|z| = r < 1.
c1 2a1 b1
c1 |f (z)| ≥ (1 − |B1 |)r − 2a1 b1
|f (z)| ≤ (1 + |B1 |)r + and
The details of the proof of Theorem 2.6 are omitted. ∗ (Φ , Φ , α) and T K (Φ , Next, we will derive extreme points for classes T SH 1 2 H 1 Φ2 , α). 1−α (c1 )n−1 (1)n−1 n n−α (a1 )n−1 (b1 )n−1 z , (n ≥ 2) ∗ (Φ , Φ , α) if and Then f ∈ T SH 1 2
Let h1 (z) = z, hn (z) = z −
Theorem 2.7.
1−α (c2 )n−1 (1)n−1 n and gn (z) = z + n+α ¯ , (n ≥ 1). (a2 )n−1 (b2 )n−1 z P only if it can be expressed as f (z) = ∞ n=1 (xn hn + yn gn ), where xn ≥ 0, yn ≥ P∞ 0, n=1 (xn + yn ) = 1. ∗ (Φ , Φ , α) are given by {h } and In particular the extreme points of T SH 1 2 n {gn }.
Proof. Let f (z) = =
∞ X
(xn hn + yn gn )
n=1 ∞ X
∞ X 1 − α (c1 )n−1 (1)n−1 (xn + yn )z − xn z n n − α (a1 )n−1 (b1 )n−1 n=2
n=1 ∞ X
+
Then
1 − α (c2 )n−1 (1)n−1 yn z¯n . n + α (a ) (b ) 2 n−1 2 n−1 n=1
∞ X n − α (a1 )n−1 (b1 )n−1 1 − α (c1 )n−1 (1)n−1 xn 1 − α (c1 )n−1 (1)n−1 n − α (a1 )n−1 (b1 )n−1 n=2 ∞ X n + α(a2 )n−1 (b2 )n−1 1 − α (c2 )n−1 (1)n−1 + yn 1 − α(c2 )n−1 (1)n−1 n + α (a2 )n−1 (b2 )n−1 n=1 =
∞ X
n=2
xn +
∞ X
n=2
yn = 1 − x1 ≤ 1,
SUBCLASSES OF HARMONIC UNIVALENT FUNCTIONS...
13
∗ (Φ , Φ , α). Conversely, if f ∈ T S ∗ (Φ , Φ , α), then and hence f ∈ T SH 1 2 1 2 H 1 − α (c2 )n−1 (1)n−1 1 − α (c1 )n−1 (1)n−1 and |Bn | ≤ . |An | ≤ n − α (a1 )n−1 (b1 )n−1 n + α (a2 )n−1 (b2 )n−1 Set n − α (a1 )n−1 (b1 )n−1 xn = for n ≥ 2 1 − α (c1 )n−1 (1)n−1 and n + α (a2 )n−1 (b2 )n−1 for n ≥ 1. yn = 1 − α (c2 )n−1 (1)n−1 Then note that by Theorem 2.2, 0 ≤ ≥ 2) and 0 ≤ yn ≤ 1(n ≥ 1) and Px∞n ≤ 1(n P in view of this we define x1 = 1 − n=2 xn − ∞ ≥ 0. Consequently, we n=1 yn P can see that f (z) can be expressed in the form f (z) = ∞ n=1 (xn hn + yn gn ) as required.
Similarly, we have Theorem 2.8.
Let h1 (z) = z, hn (z) = z −
1−α (c1 )n−1 (1)n−1 n n(n−α) (a1 )n−1 (b1 )n−1 z ,
1−α (c2 )n−1 (1)n−1 n (n ≥ 2) and gn (z) = z + n(n+α) ¯ , (n ≥ 1). Then f ∈ T KH (Φ1 , Φ2 , (a2 )n−1 (b2 )n−1 z P∞ α) if and only P if it can be expressed as f (z) = n=1 (xn hn + yn gn ), where xn ≥ 0, yn ≥ 0, ∞ (x + y ) = 1. n n n=1
In particular the extreme points of T KH (Φ1 , Φ2 , α) are given by {hn } and {gn }.
Acknowledgments This work is supported by Department of Science and Technology, SERC Division, New Delhi under the Young Scientist Project (SR/FTP/MS-17/2007). The author is grateful to Dr. R.N. Mohapatra for reading the manuscript and making suggestions for improvement. References [1] O.P. Ahuja, J.M. Jahangiri, H. Silverman, Convolutions for special classes of harmonic univalent functions, App. Math. Lett., 16 (2003), 905–909. [2] O.P. Ahuja, Planar harmonic convolution operators generated by hypergeometric functions, Integral trans. and Spec. Funct., 18, No. 3 (2007), 165–177.
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