paper 11-17-2009. 116KB Jun 04 2011 12:03:06 AM
ACTA UNIVERSITATIS APULENSIS
No 17/2009
HARMONIC UNIVALENT FUNCTIONS DEFINED BY AN
INTEGRAL OPERATOR
˘
Luminit
¸ a Ioana Cotˆırla
Abstract.We define and investigate a new class of harmonic univalent
functions defined by an integral operator. We obtain coefficient inequalities,
extreme points and distortion bounds for the functions in our classes.
2000 Mathematics Subject Classification: 30C45, 30C50, 31A05.
Keywords and phrases: Harmonic univalent functions,integral operator.
1. Introduction
A continuous complex-valued function f = u + iv defined in a 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 we can write f = h + g, 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 |h′ (z)| > |g ′ (z)|, z ∈ D. (See Clunie and
Sheil-Small [2].)
Denote by H the class of functions f = h + g that are harmonic univalent
and sense preserving in the unit disc U = {z : |z| < 1} so that f = h + g is
normalized by f (0) = h(0) = fz′ (0) − 1 = 0.
Let H(U ) be the space of holomorphic functions in U . We let:
An = {f ∈ H(U ), f (z) = z + an+1 z n+1 + . . . , z ∈ U },
with A1 = A.
We let H[a, n] denote the class of analytic functions in U of the form
f (z) = a + an z n + an+1 z n+1 + . . . ,
95
z ∈ U.
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
The integral operator I n is defined in [5] by:
(i) I 0 f (z) = f (z);
Z
z
f (t)t−1 dt;
1
(ii) I f (z) = If (z) =
0
(iii) I n f (z) = I(I n−1 f (z)), n ∈ N and f ∈ A.
Ahuja and Jahongiri [3] defined the class H(n) (n ∈ N) consisting of all
univalent harmonic functions f = h + g that are sense preserving in U and h
and g are of the form
h(z) = z +
∞
X
ak z k ,
g(z) =
k=2
∞
X
bk z k ,
|b1 | < 1.
(1)
k=1
For f = h + g given by (1) the integral operator I n of f is defined as
I n f (z) = I n h(z) + (−1)n I n g(z),
where
n
I h(z) = z +
∞
X
−n
(k)
k
n
ak z and I g(z) =
k=2
∞
X
(2)
(k)−n bk z k .
k=1
For 0 ≤ α < 1, n ∈ N, z ∈ U , let H(n, α) the family of harmonic functions
f of the form (1) such that
n
I f (z)
> α.
(3)
Re
I n+1 f (z)
The families H(n + 1, n, α) and H − (n + 1, n, α) include a variety of wellknown classes of harmonic functions as well as many new ones. For example
HS(α) = H(1, 0, α) is the class of sense-preserving,harmonic univalent functions f which are starlike of order α ∈ U , and HK(α) = H(2, 1, α) is the class
of sense-preserving,harmonic univalent functions f which are convex of order
α in U , and H(n + 1, n, α) = H(n, α) is the class of S˘al˘agean-type harmonic
univalent functions.
Let we denote the subclass H − (n, α) consist of harmonic functions fn =
h + g n in H − (n, α) so that h and gn are of the form
h(z) = z −
∞
X
k
ak z ,
n−1
gn (z) = (−1)
k=2
∞
X
k=1
96
bk z k ,
(4)
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
where ak , bk ≥ 0, |b1 | < 1.
For the harmonic functions f of the form (1) with b1 = 0 Avci and
Zlotkiewich [1] show that if
∞
X
k(|ak | + |bk |) ≤ 1,
k=2
then f ∈ HS(0), where HS(0) = H(1, 0, 0), and if
∞
X
k 2 (|ak | + |bk |) ≤ 1,
k=2
then f ∈ HK(0), where HK(0) = H(2, 1, 0).
For the harmonic functions f of the form (4) with n = 0, Jahangiri in [4]
showed that f ∈ HS(α) if and only if
∞
X
(k − α)|ak | +
k=2
∞
X
(k + α)|bk | ≤ 1 − α
k=1
and f ∈ H(2, 1, α) if and only if
∞
X
k(k − α)|ak | +
k=2
∞
X
k(k + α)|bk | ≤ 1 − α.
k=1
2.Main results
In our first theorem, we deduce a sufficient coefficient bound for harmonic
functions in H(n, α).
Theorem 2.1. Let f = h + g be given by (1). If
∞
X
{ψ(n, k, α)|ak | + θ(n, k, α)|bk |} ≤ 2,
k=1
where
ψ(n, k, α) =
(k)−n − α(k)−(n+1)
(k)−n + α(k)−(n+1)
and θ(n, k, α) =
,
1−α
1−α
97
(5)
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
a1 = 1, 0 ≤ α < 1, n ∈ N. Then f is sense-preserving in U and f ∈ H(n, α).
Proof. According to (2) and (3) we only need to show that
n
I f (z) − αI n+1 f (z)
Re
≥ 0.
I n+1 f (z)
The case r = 0 is obvious. For 0 < r < 1, it follows that
n
I f (z) − αI n+1 f (z)
Re
I n+1 f (z)
"
n+1 #
∞
n
X
1
1
z(1 − α) +
ak z k
−α
k
k
k=2
= Re
∞ n+1
n n
X
X
1
1
k
n+1
bk z k
ak z + (−1)
z +
k
k
k=2
k=1
"
#
n
n+1
∞
X
1
1
(−1)n
+α
bk z k
k
k
k=1
+
∞
∞
n+1
n+1
X
X
1
1
k
n+1
z+
bk z k
ak z + (−1)
k
k
k=2
k=1
"
n+1 #
∞
n
X
1
1
1−α+
ak z k−1
−α
k
k
k=2
= Re
∞ n+1
∞ n+1
X
X
1
1
k−1
n+1
ak z
+ (−1)
bk z k z −1
1 +
k
k
k=1
k=2
"
#
n
n+1
∞
X
1
1
bk z k z −1
(−1)n
+α
k
k
k=1
+
∞
∞
n+1
n+1
X
X
1
1
k−1
n+1
ak z
+ (−1)
1+
bk z k z −1
k
k
k=1
k=2
1 − α + A(z)
.
= Re
1 + B(z)
For z = reiθ we have
"
n+1 #
X 1 n
1
A(reiθ ) =
−α
ak rk−1 e(k−1)θi
k
k
k=2
98
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
Setting
"
n+1 #
∞
n
X
1
1
+α
+(−1)n
bk rk−1 e−(k+1)θi
k
k
k=1
∞ n+1
X
1
iθ
ak rk−1 e(k−1)θi
B(re ) =
k
k=2
∞ n+1
X
1
n+1
bk rk−1 e−(k+1)θi
+(−1)
k
k=1
1 − α + A(z)
1 + w(z)
= (1 − α)
,
1 + B(z)
1 − w(z)
the proof will be complete if we can show that |w(z)| ≤ 1. This is the case
since, by the condition (5), we can write
A(z) − (1 − α)B(z)
|w(z)| =
A(z) + (1 − α)B(z) + 2(1 − α)
"
n+1 #
∞ n
X
1
1
ak rk−1 e(k−1)θi
−
k
k
k=2
=
∞
X
P∞
(k−1)θi
k−1
n
2(1−α) + k=2 C(n, k, α)ak r e
+ (−1)
D(n, k, α)bk rk−1 e−(k+1)θi
k=1
"
#
∞
X
1 n+1
1 n
k−1 −(k+1)θi
n
+
bk r e
(−1)
k
k
k=1
+
∞
∞
X
X
2(1 − α) +
C(n, k, α)ak rk−1 e(k−1)θi + (−1)n
D(n, k, α)bk rk−1 e−(k+1)θi
k=2
k=1
"
n+1 #
∞
n
X
1
1
|ak |rk−1
−
k
k
k=2
≤
2(1 − α) −
∞
X
C(n, k, α)|ak |r
k=2
k−1
−
∞
X
k=1
99
D(n, k, α)|bk |rk−1
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
"
n+1 #
∞
n
X
1
1
+
|bk |rk−1
k
k
k=1
+
2(1 − α) −
∞
X
C(n, k, α)|ak |r
k=2
k−1
−
∞
X
D(n, k, α)|bk |rk−1
k=1
"
n+1 #
n
1
1
−
|ak |rk−1
k
k
k=1
∞
X
=
4(1 − α) −
∞
X
{C(n, k, α)|ak | + D(n, k, α)|bk |}rk−1
k=1
"
n+1 #
n
1
1
+
|bk |rk−1
k
k
k=1
∞
X
+
4(1 − α) −
∞
X
{C(n, k, α)|ak | + D(n, k, α)|bk |}rk−1
k=1
"
n+1 #
∞
n
X
1
1
|ak |
−
k
k
k=1
<
4(1 − α) −
∞
X
{C(n, k, α)|ak | + D(n, k, α)|bk |}
k=1
+
"
n+1 #
∞
n
X
1
1
|bk |
+
k
k
k=1
4(1 − α) −
∞
X
≤ 1,
{C(n, k, α)|ak | + D(n, k, α)|bk |}
k=1
where
and
n
n+1
1
1
C(n, k, α) =
+ (1 − 2α)
k
k
n+1
n
1
1
− (1 − 2α)
.
D(n, k, α) =
k
k
The harmonic univalent functions
∞
∞
X
X
1
1
k
f (z) = z +
yk z k ,
xk z +
ψ(n,
k,
α)
θ(n,
k,
α)
k=2
k=1
100
(6)
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
where n ∈ N and
∞
X
|xk | +
k=2
∞
X
|yk | = 1, show that the coefficient bound given
k=1
by (5) is sharp. The functions of the form (6) are in H(n, α) because
∞
X
{ψ(n, k, α)|ak | + θ(n, k, α)|bk |} = 1 +
k=1
∞
X
k=2
|xk | +
∞
X
|yk | = 2.
k=1
In the following theorem it is shown that the condition (5) is also necessary
for functions fn = h + g n where h and gn are of the form (4).
Theorem 2.2. Let fn = h + g n be given by (4). Then fn ∈ H − (n, α) if
and only if
∞
X
{ψ(n, k, α)ak + θ(n, k, α)bk } ≤ 2
(7)
k=1
where a1 = 1, 0 ≤ α < 1, n ∈ N.
Proof. Since H − (n, α) ⊂ H(n, α) we only need to prove the ”only if” part
of the theorem. For functions fn of the form (4), we note that the condition
n
I fn (z)
>α
Re
I n+1 fn (z)
is equivalent to
"
n+1 #
∞
n
X
1
1
(1 − α)z −
ak z k
−α
k
k
k=2
Re
∞ n+1
∞ n+1
X
X
1
1
k
2n
ak z + (−1)
bk z k
z −
k
k
k=1
k=2
"
n+1 #
∞
n
X
1
1
(−1)2n−1
+α
bk z k
k
k
k=1
≥ 0.
+
∞
∞
n+1
n
X
X
1
1
k
2n
ak z + (−1)
bk z k
z−
k
k
k=1
k=2
(8)
The above required condition (8) must hold for all values of z in U . Upon
choosing the values of z on the positive real axis where 0 ≤ z = r < 1, we
101
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
must have
"
n+1 #
∞
n
X
1
1
1−α−
−α
ak rk−1
k
k
k=2
∞ n+1
∞ n+1
X
X
1
1
k−1
ak r
+
bk rk−1
1−
k
k
k=1
k=2
#
"
∞
n
n+1
X
1
1
bk rk−1
+α
k
k
k=1
−
≥ 0.
∞ n+1
∞
X
X 1 n+1
1
k−1
k−1
ak r
+
bk r
1−
k
k
k=1
k=2
(9)
If the condition (7) does not hold, then the expression in (9) is negative for r
sufficiently close to 1. Hence there exist z0 = r0 in (0, 1) for which the quotient
in (9) is negative. This contradicts the required condition for fn ∈ H − (n, α).
So the proof is complete.
Next we determine the extreme points of the closed convex hull of H − (n, α)
denoted by clcoH − (n, α).
Theorem 2.3. Let fn be given by (4). Then fn ∈ H − (n, α) if and only if
fn (z) =
∞
X
[xk hk (z) + yk gnk (z)],
k=1
where
h(z) = z,
hk (z) = z −
and
gnk (z) = z + (−1)n−1
xk ≥ 0,
yk ≥ 0,
1
zk ,
ψ(n, k, α)
1
zk
θ(n, k, α)
xp = 1 −
(k = 2, 3, . . . )
(k = 1, 2, 3, . . . )
∞
X
k=2
xk −
∞
X
yk .
k=1
In particular, the extreme points of H − (n, α) are {hk } and {gnk } .
Proof. For functions fn of the form (5)
fn (z) =
∞
X
k=1
[xk hk (z) + yk gnk (z)] =
∞
X
k=1
102
(xk + yk )z −
∞
X
k=2
1
xk z k
ψ(n, k, α)
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
n−1
+(−1)
∞
X
k=1
Then
∞
X
ψ(n, k, α)
k=2
1
xk
ψ(n, k, α)
=
∞
X
xk +
k=2
∞
X
1
yk z k .
θ(n, k, α)
+
∞
X
θ(n, k, α)
k=1
1
yk
θ(n, k, α)
yk = 1 − x1 ≤ 1
k=1
and so fn (z) ∈ clcoH − (n, α).
Conversely, suppose fn (z) ∈ clcoH − (n, α, β). Letting
x1 = 1 −
∞
X
xk −
k=2
xk = ψ(n, k, α)ak ,
∞
X
yk ,
k=1
(k = 2, 3, . . . ) and yk = θ(n, k, α)bk ,
(k = 1, 2, 3, . . . )
we obtain the required representation, since
fn (z) = z −
∞
X
ak z k + (−1)n−1
k=2
=z−
∞
X
k=2
k=1
∞
X
[z − hk (z)]xk −
k=2
= 1−
bk z k
∞
X
1
1
k
n−1
xk z + (−1)
yk z k
ψ(n, k, α)
θ(n,
k,
α)
k=1
=z−
"
∞
X
∞
X
xk −
k=2
[z − gnk (z)]yk
k=1
∞
X
k=1
=
∞
X
∞
X
#
yk z +
∞
X
xk hk (z) +
k=2
∞
X
yk gnk (z)
k=1
[xk hk (z) + yk gnk (z)].
k=1
The following theorem gives the distortion bounds for functions in H − (n, α)
which yields a covering results for this class.
103
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
Theorem 2.4. Let fn ∈ H − (n, α). Then for |z| = r < 1 we have
|fn (z)| ≤ (1 + b1 )r + {φ(n, k, α) − Ω(n, k, α)b1 }rn+1
and
|fn (z)| ≥ (1 − b1 )r − {φ(n, k, α) − Ω(n, k, α)b1 }rn+1 ,
where
1−α
φ(n, k, α) = n
n+1
1
1
−α
2
2
and
1+α
Ω(n, k, α) = n
n+1 .
1
1
−α
2
2
Proof. We prove the right hand side inequality for |fn |. The proof for the
left hand inequality can be done using similar arguments. Let fn ∈ H − (n, α).
Taking the absolute value of fn then by Theorem 2.2, we obtain:
∞
∞
X
X
|fn (z)| = z −
ak z k + (−1)n−1
bk z k
k=2
≤r+
∞
X
k=2
≤ r + b1 r +
∞
X
k
ak r +
∞
X
k=1
k
bk r = r + b1 r +
k=1
∞
X
(ak + bk )rk
k=2
∞
X
(ak + bk )r2 = (1 + b1 )r + φ(n, k, α)
k=2
k=2
≤ (1 + b1 )r + φ(n, k, α)rn+2
"
∞
X
k=2
1
(ak + bk )r2
φ(n, k, α)
#
ψ(n, k, α)ak + θ(n, k, α)bk
≤ (1 + b1 )r + {φ(n, k, α) − Ω(n, k, α)b1 }rn+2 .
The following covering result follows from the left hand inequality in Theorem 2.4.
Corollary 2.1. Let fn ∈ H − (n, α), then for |z| = r < 1 we have
{w : |w| < 1 − b1 − [φ(n, k, α) − Ω(n, k, α)b1 ] ⊂ fn (U )}.
104
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
References
[1] Y. Avci, E. Zlotkiewicz, On harmonic univalent mappings, Ann. Univ.
Marie Crie-Sklodowska, Sect. A., 44(1991).
[2] J. Clunie, T. Sheil-Small, Harmonic univalent functions, Ann. Acad.
Sci. Fenn. Ser. A.I. Math., 9(1984), 3-25.
[3]O.P. Ahuja, J.M. Jahangiri, Multivalent harmonic starlike functions,
Ann. Univ. Marie Curie-Sklodowska Sect. A, LV 1(2001), 1-13.
[4] J.M. Jahangiri, Harmonic functions starlike in the unit disc, J. Math.
Anal. Appl., 235(1999).
[5] G.S. S˘al˘agean, Subclass of univalent functions, Lecture Notes in Math.
Springer-Verlag, 1013(1983), 362-372.
Author:
Luminit¸a-Ioana Cotˆırl˘a
Department of Mathematics and Computer Science
University of Babe¸s-Bolyai
400084 Cluj-Napoca, Romania
email: uluminita@math.ubbcluj.ro, luminitacotirla@yahoo.com
105
No 17/2009
HARMONIC UNIVALENT FUNCTIONS DEFINED BY AN
INTEGRAL OPERATOR
˘
Luminit
¸ a Ioana Cotˆırla
Abstract.We define and investigate a new class of harmonic univalent
functions defined by an integral operator. We obtain coefficient inequalities,
extreme points and distortion bounds for the functions in our classes.
2000 Mathematics Subject Classification: 30C45, 30C50, 31A05.
Keywords and phrases: Harmonic univalent functions,integral operator.
1. Introduction
A continuous complex-valued function f = u + iv defined in a 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 we can write f = h + g, 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 |h′ (z)| > |g ′ (z)|, z ∈ D. (See Clunie and
Sheil-Small [2].)
Denote by H the class of functions f = h + g that are harmonic univalent
and sense preserving in the unit disc U = {z : |z| < 1} so that f = h + g is
normalized by f (0) = h(0) = fz′ (0) − 1 = 0.
Let H(U ) be the space of holomorphic functions in U . We let:
An = {f ∈ H(U ), f (z) = z + an+1 z n+1 + . . . , z ∈ U },
with A1 = A.
We let H[a, n] denote the class of analytic functions in U of the form
f (z) = a + an z n + an+1 z n+1 + . . . ,
95
z ∈ U.
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
The integral operator I n is defined in [5] by:
(i) I 0 f (z) = f (z);
Z
z
f (t)t−1 dt;
1
(ii) I f (z) = If (z) =
0
(iii) I n f (z) = I(I n−1 f (z)), n ∈ N and f ∈ A.
Ahuja and Jahongiri [3] defined the class H(n) (n ∈ N) consisting of all
univalent harmonic functions f = h + g that are sense preserving in U and h
and g are of the form
h(z) = z +
∞
X
ak z k ,
g(z) =
k=2
∞
X
bk z k ,
|b1 | < 1.
(1)
k=1
For f = h + g given by (1) the integral operator I n of f is defined as
I n f (z) = I n h(z) + (−1)n I n g(z),
where
n
I h(z) = z +
∞
X
−n
(k)
k
n
ak z and I g(z) =
k=2
∞
X
(2)
(k)−n bk z k .
k=1
For 0 ≤ α < 1, n ∈ N, z ∈ U , let H(n, α) the family of harmonic functions
f of the form (1) such that
n
I f (z)
> α.
(3)
Re
I n+1 f (z)
The families H(n + 1, n, α) and H − (n + 1, n, α) include a variety of wellknown classes of harmonic functions as well as many new ones. For example
HS(α) = H(1, 0, α) is the class of sense-preserving,harmonic univalent functions f which are starlike of order α ∈ U , and HK(α) = H(2, 1, α) is the class
of sense-preserving,harmonic univalent functions f which are convex of order
α in U , and H(n + 1, n, α) = H(n, α) is the class of S˘al˘agean-type harmonic
univalent functions.
Let we denote the subclass H − (n, α) consist of harmonic functions fn =
h + g n in H − (n, α) so that h and gn are of the form
h(z) = z −
∞
X
k
ak z ,
n−1
gn (z) = (−1)
k=2
∞
X
k=1
96
bk z k ,
(4)
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
where ak , bk ≥ 0, |b1 | < 1.
For the harmonic functions f of the form (1) with b1 = 0 Avci and
Zlotkiewich [1] show that if
∞
X
k(|ak | + |bk |) ≤ 1,
k=2
then f ∈ HS(0), where HS(0) = H(1, 0, 0), and if
∞
X
k 2 (|ak | + |bk |) ≤ 1,
k=2
then f ∈ HK(0), where HK(0) = H(2, 1, 0).
For the harmonic functions f of the form (4) with n = 0, Jahangiri in [4]
showed that f ∈ HS(α) if and only if
∞
X
(k − α)|ak | +
k=2
∞
X
(k + α)|bk | ≤ 1 − α
k=1
and f ∈ H(2, 1, α) if and only if
∞
X
k(k − α)|ak | +
k=2
∞
X
k(k + α)|bk | ≤ 1 − α.
k=1
2.Main results
In our first theorem, we deduce a sufficient coefficient bound for harmonic
functions in H(n, α).
Theorem 2.1. Let f = h + g be given by (1). If
∞
X
{ψ(n, k, α)|ak | + θ(n, k, α)|bk |} ≤ 2,
k=1
where
ψ(n, k, α) =
(k)−n − α(k)−(n+1)
(k)−n + α(k)−(n+1)
and θ(n, k, α) =
,
1−α
1−α
97
(5)
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
a1 = 1, 0 ≤ α < 1, n ∈ N. Then f is sense-preserving in U and f ∈ H(n, α).
Proof. According to (2) and (3) we only need to show that
n
I f (z) − αI n+1 f (z)
Re
≥ 0.
I n+1 f (z)
The case r = 0 is obvious. For 0 < r < 1, it follows that
n
I f (z) − αI n+1 f (z)
Re
I n+1 f (z)
"
n+1 #
∞
n
X
1
1
z(1 − α) +
ak z k
−α
k
k
k=2
= Re
∞ n+1
n n
X
X
1
1
k
n+1
bk z k
ak z + (−1)
z +
k
k
k=2
k=1
"
#
n
n+1
∞
X
1
1
(−1)n
+α
bk z k
k
k
k=1
+
∞
∞
n+1
n+1
X
X
1
1
k
n+1
z+
bk z k
ak z + (−1)
k
k
k=2
k=1
"
n+1 #
∞
n
X
1
1
1−α+
ak z k−1
−α
k
k
k=2
= Re
∞ n+1
∞ n+1
X
X
1
1
k−1
n+1
ak z
+ (−1)
bk z k z −1
1 +
k
k
k=1
k=2
"
#
n
n+1
∞
X
1
1
bk z k z −1
(−1)n
+α
k
k
k=1
+
∞
∞
n+1
n+1
X
X
1
1
k−1
n+1
ak z
+ (−1)
1+
bk z k z −1
k
k
k=1
k=2
1 − α + A(z)
.
= Re
1 + B(z)
For z = reiθ we have
"
n+1 #
X 1 n
1
A(reiθ ) =
−α
ak rk−1 e(k−1)θi
k
k
k=2
98
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
Setting
"
n+1 #
∞
n
X
1
1
+α
+(−1)n
bk rk−1 e−(k+1)θi
k
k
k=1
∞ n+1
X
1
iθ
ak rk−1 e(k−1)θi
B(re ) =
k
k=2
∞ n+1
X
1
n+1
bk rk−1 e−(k+1)θi
+(−1)
k
k=1
1 − α + A(z)
1 + w(z)
= (1 − α)
,
1 + B(z)
1 − w(z)
the proof will be complete if we can show that |w(z)| ≤ 1. This is the case
since, by the condition (5), we can write
A(z) − (1 − α)B(z)
|w(z)| =
A(z) + (1 − α)B(z) + 2(1 − α)
"
n+1 #
∞ n
X
1
1
ak rk−1 e(k−1)θi
−
k
k
k=2
=
∞
X
P∞
(k−1)θi
k−1
n
2(1−α) + k=2 C(n, k, α)ak r e
+ (−1)
D(n, k, α)bk rk−1 e−(k+1)θi
k=1
"
#
∞
X
1 n+1
1 n
k−1 −(k+1)θi
n
+
bk r e
(−1)
k
k
k=1
+
∞
∞
X
X
2(1 − α) +
C(n, k, α)ak rk−1 e(k−1)θi + (−1)n
D(n, k, α)bk rk−1 e−(k+1)θi
k=2
k=1
"
n+1 #
∞
n
X
1
1
|ak |rk−1
−
k
k
k=2
≤
2(1 − α) −
∞
X
C(n, k, α)|ak |r
k=2
k−1
−
∞
X
k=1
99
D(n, k, α)|bk |rk−1
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
"
n+1 #
∞
n
X
1
1
+
|bk |rk−1
k
k
k=1
+
2(1 − α) −
∞
X
C(n, k, α)|ak |r
k=2
k−1
−
∞
X
D(n, k, α)|bk |rk−1
k=1
"
n+1 #
n
1
1
−
|ak |rk−1
k
k
k=1
∞
X
=
4(1 − α) −
∞
X
{C(n, k, α)|ak | + D(n, k, α)|bk |}rk−1
k=1
"
n+1 #
n
1
1
+
|bk |rk−1
k
k
k=1
∞
X
+
4(1 − α) −
∞
X
{C(n, k, α)|ak | + D(n, k, α)|bk |}rk−1
k=1
"
n+1 #
∞
n
X
1
1
|ak |
−
k
k
k=1
<
4(1 − α) −
∞
X
{C(n, k, α)|ak | + D(n, k, α)|bk |}
k=1
+
"
n+1 #
∞
n
X
1
1
|bk |
+
k
k
k=1
4(1 − α) −
∞
X
≤ 1,
{C(n, k, α)|ak | + D(n, k, α)|bk |}
k=1
where
and
n
n+1
1
1
C(n, k, α) =
+ (1 − 2α)
k
k
n+1
n
1
1
− (1 − 2α)
.
D(n, k, α) =
k
k
The harmonic univalent functions
∞
∞
X
X
1
1
k
f (z) = z +
yk z k ,
xk z +
ψ(n,
k,
α)
θ(n,
k,
α)
k=2
k=1
100
(6)
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
where n ∈ N and
∞
X
|xk | +
k=2
∞
X
|yk | = 1, show that the coefficient bound given
k=1
by (5) is sharp. The functions of the form (6) are in H(n, α) because
∞
X
{ψ(n, k, α)|ak | + θ(n, k, α)|bk |} = 1 +
k=1
∞
X
k=2
|xk | +
∞
X
|yk | = 2.
k=1
In the following theorem it is shown that the condition (5) is also necessary
for functions fn = h + g n where h and gn are of the form (4).
Theorem 2.2. Let fn = h + g n be given by (4). Then fn ∈ H − (n, α) if
and only if
∞
X
{ψ(n, k, α)ak + θ(n, k, α)bk } ≤ 2
(7)
k=1
where a1 = 1, 0 ≤ α < 1, n ∈ N.
Proof. Since H − (n, α) ⊂ H(n, α) we only need to prove the ”only if” part
of the theorem. For functions fn of the form (4), we note that the condition
n
I fn (z)
>α
Re
I n+1 fn (z)
is equivalent to
"
n+1 #
∞
n
X
1
1
(1 − α)z −
ak z k
−α
k
k
k=2
Re
∞ n+1
∞ n+1
X
X
1
1
k
2n
ak z + (−1)
bk z k
z −
k
k
k=1
k=2
"
n+1 #
∞
n
X
1
1
(−1)2n−1
+α
bk z k
k
k
k=1
≥ 0.
+
∞
∞
n+1
n
X
X
1
1
k
2n
ak z + (−1)
bk z k
z−
k
k
k=1
k=2
(8)
The above required condition (8) must hold for all values of z in U . Upon
choosing the values of z on the positive real axis where 0 ≤ z = r < 1, we
101
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
must have
"
n+1 #
∞
n
X
1
1
1−α−
−α
ak rk−1
k
k
k=2
∞ n+1
∞ n+1
X
X
1
1
k−1
ak r
+
bk rk−1
1−
k
k
k=1
k=2
#
"
∞
n
n+1
X
1
1
bk rk−1
+α
k
k
k=1
−
≥ 0.
∞ n+1
∞
X
X 1 n+1
1
k−1
k−1
ak r
+
bk r
1−
k
k
k=1
k=2
(9)
If the condition (7) does not hold, then the expression in (9) is negative for r
sufficiently close to 1. Hence there exist z0 = r0 in (0, 1) for which the quotient
in (9) is negative. This contradicts the required condition for fn ∈ H − (n, α).
So the proof is complete.
Next we determine the extreme points of the closed convex hull of H − (n, α)
denoted by clcoH − (n, α).
Theorem 2.3. Let fn be given by (4). Then fn ∈ H − (n, α) if and only if
fn (z) =
∞
X
[xk hk (z) + yk gnk (z)],
k=1
where
h(z) = z,
hk (z) = z −
and
gnk (z) = z + (−1)n−1
xk ≥ 0,
yk ≥ 0,
1
zk ,
ψ(n, k, α)
1
zk
θ(n, k, α)
xp = 1 −
(k = 2, 3, . . . )
(k = 1, 2, 3, . . . )
∞
X
k=2
xk −
∞
X
yk .
k=1
In particular, the extreme points of H − (n, α) are {hk } and {gnk } .
Proof. For functions fn of the form (5)
fn (z) =
∞
X
k=1
[xk hk (z) + yk gnk (z)] =
∞
X
k=1
102
(xk + yk )z −
∞
X
k=2
1
xk z k
ψ(n, k, α)
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
n−1
+(−1)
∞
X
k=1
Then
∞
X
ψ(n, k, α)
k=2
1
xk
ψ(n, k, α)
=
∞
X
xk +
k=2
∞
X
1
yk z k .
θ(n, k, α)
+
∞
X
θ(n, k, α)
k=1
1
yk
θ(n, k, α)
yk = 1 − x1 ≤ 1
k=1
and so fn (z) ∈ clcoH − (n, α).
Conversely, suppose fn (z) ∈ clcoH − (n, α, β). Letting
x1 = 1 −
∞
X
xk −
k=2
xk = ψ(n, k, α)ak ,
∞
X
yk ,
k=1
(k = 2, 3, . . . ) and yk = θ(n, k, α)bk ,
(k = 1, 2, 3, . . . )
we obtain the required representation, since
fn (z) = z −
∞
X
ak z k + (−1)n−1
k=2
=z−
∞
X
k=2
k=1
∞
X
[z − hk (z)]xk −
k=2
= 1−
bk z k
∞
X
1
1
k
n−1
xk z + (−1)
yk z k
ψ(n, k, α)
θ(n,
k,
α)
k=1
=z−
"
∞
X
∞
X
xk −
k=2
[z − gnk (z)]yk
k=1
∞
X
k=1
=
∞
X
∞
X
#
yk z +
∞
X
xk hk (z) +
k=2
∞
X
yk gnk (z)
k=1
[xk hk (z) + yk gnk (z)].
k=1
The following theorem gives the distortion bounds for functions in H − (n, α)
which yields a covering results for this class.
103
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
Theorem 2.4. Let fn ∈ H − (n, α). Then for |z| = r < 1 we have
|fn (z)| ≤ (1 + b1 )r + {φ(n, k, α) − Ω(n, k, α)b1 }rn+1
and
|fn (z)| ≥ (1 − b1 )r − {φ(n, k, α) − Ω(n, k, α)b1 }rn+1 ,
where
1−α
φ(n, k, α) = n
n+1
1
1
−α
2
2
and
1+α
Ω(n, k, α) = n
n+1 .
1
1
−α
2
2
Proof. We prove the right hand side inequality for |fn |. The proof for the
left hand inequality can be done using similar arguments. Let fn ∈ H − (n, α).
Taking the absolute value of fn then by Theorem 2.2, we obtain:
∞
∞
X
X
|fn (z)| = z −
ak z k + (−1)n−1
bk z k
k=2
≤r+
∞
X
k=2
≤ r + b1 r +
∞
X
k
ak r +
∞
X
k=1
k
bk r = r + b1 r +
k=1
∞
X
(ak + bk )rk
k=2
∞
X
(ak + bk )r2 = (1 + b1 )r + φ(n, k, α)
k=2
k=2
≤ (1 + b1 )r + φ(n, k, α)rn+2
"
∞
X
k=2
1
(ak + bk )r2
φ(n, k, α)
#
ψ(n, k, α)ak + θ(n, k, α)bk
≤ (1 + b1 )r + {φ(n, k, α) − Ω(n, k, α)b1 }rn+2 .
The following covering result follows from the left hand inequality in Theorem 2.4.
Corollary 2.1. Let fn ∈ H − (n, α), then for |z| = r < 1 we have
{w : |w| < 1 − b1 − [φ(n, k, α) − Ω(n, k, α)b1 ] ⊂ fn (U )}.
104
L.I. Cotˆırl˘a-Harmonic univalent functions defined by an integral operator
References
[1] Y. Avci, E. Zlotkiewicz, On harmonic univalent mappings, Ann. Univ.
Marie Crie-Sklodowska, Sect. A., 44(1991).
[2] J. Clunie, T. Sheil-Small, Harmonic univalent functions, Ann. Acad.
Sci. Fenn. Ser. A.I. Math., 9(1984), 3-25.
[3]O.P. Ahuja, J.M. Jahangiri, Multivalent harmonic starlike functions,
Ann. Univ. Marie Curie-Sklodowska Sect. A, LV 1(2001), 1-13.
[4] J.M. Jahangiri, Harmonic functions starlike in the unit disc, J. Math.
Anal. Appl., 235(1999).
[5] G.S. S˘al˘agean, Subclass of univalent functions, Lecture Notes in Math.
Springer-Verlag, 1013(1983), 362-372.
Author:
Luminit¸a-Ioana Cotˆırl˘a
Department of Mathematics and Computer Science
University of Babe¸s-Bolyai
400084 Cluj-Napoca, Romania
email: uluminita@math.ubbcluj.ro, luminitacotirla@yahoo.com
105