General Mathematics Vol. 13, No. 3 (2005), 37–46
Differential Subordination Defined by
Sălăgean Operator
Georgia Irina Oros
Dedicated to Professor Dumitru Acu on his 60th anniversary
Abstract
By using the Sălăgean operator Dn f (z), z ∈ U , we introduce
a class of holomorphic functions, denoted by Sαn (β), and we obtain
some inclusion relations and differential subordination.
2000 Mathematics Subject Classification: Primary 30C80, Secondary
30C45, 30A20, 34A40.
Keywords and phrases: Holomorphic function, convex function,
α-convex function (Mocanu function), dominant.
1
Introduction and preliminaries
Denote by U the unit disc of the complex plane:
U = {z ∈ C : |z| < 1}.
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Georgia Irina Oros
Let H(U ) be the space of holomorphic functions in U . 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 + . . . ,
z ∈ U.
We let
An = {f ∈ H(U ), f (z) = z + an+1 z n+1 + an+2 z n+2 + . . . , z ∈ U }
with A1 = A.
For α real, let
zf 0 (z)
zf 00 (z)
J(α, f ; z) = (1 − α)
+α 1+ 0
.
f (z)
f (z)
(1)
The class of α-convex functions (Mocanu functions) in the unit disc, are
defined by
Mα = {f ∈ A; Re J(α, f ; z) > 0, z ∈ U }.
If f and g are analytic in U , then we say that f is subordinate to g,
written f ≺ g, or f (z) ≺ g(z), if there is a function w analytic in U with
w(0) = 0, |w(z)| < 1, for all z ∈ U such that f (z) = g[w(z)] for z ∈ U . If g
is univalent, then t ≺ g if and only if f (0) = g(0) and f (U ) ⊂ g(U ).
We use the following subordination results.
Lemma A. (Hallenbeck and Ruscheweyh [2, p.71]) Let h be a convex function with h(0) = a and let γ ∈ C∗ be a complex with Re γ ≥ 0. If p ∈ H(U ),
with p(0) = a and
1
p(z) + zp0 (z) ≺ h(z)
γ
then
p(z) ≺ q(z) ≺ h(z)
Differential subordination defined by Sălăgean operator
39
where
Z z
γ
γ
q(z) =
h(t)t n −1 dt.
γ
nz n 0
The function q is convex and is the best dominant.
(The definition of best dominant is given in [1]).
Lemma B. (Miller and Mocanu [1]) Let g be a convex function in U and
let
h(z) = g(z) + nαzg 0 (z),
where α > 0 and n is a positive integer. If
p(z) = g(0) + pn z n + . . .
is holomorphic in U and
p(z) + αzp0 (z) ≺ h(z)
then
p(z) ≺ g(z),
z ∈ U,
and this results is sharp.
Definition 1. (see [3]) For f ∈ A and n ∈ N∗ ∪ {0} the operator Dn f is
defined by
D0 f (z) = f (z)
Dn+1 f (z) = z[Dn f (z)]0 ,
z ∈ U.
Remark 1. We have
Dn f (z) = (K ∗ K ∗ K ∗ ... ∗ K ∗ f )(z)
where ∗ stands for convolution, K(z) =
n
D f (z) = z +
∞
X
j=2
z
is the Koebe function and
(1 − z)2
j n aj z j ,
z ∈ U.
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Georgia Irina Oros
2
Main results
Definition 2. We denote
J1 (α, f ; z) = zJ(α, f ; z),
z ∈ U,
where J(α, f ; z) is given by (1).
Definition 3. If 0 ≤ β < 1, α ∈ R and n ∈ N∗ ∪ {0}, let Sαn (β) denote the
class of functions f ∈ A which satisfy the inequality
Re [Dn J1 (α, f ; z)]0 > β,
z ∈ U.
Theorem 1. If 0 ≤ β < 1, α ∈ R and n ∈ N∗ ∪ {0}, then
Sαn+1 (β) ⊂ Sαn (δ),
where
δ = δ(β) = 2β − 1 + 2(1 − β) ln 2.
Proof. Let f ∈ Sαn+1 (β). By using the properties of the operator Dn f we
have
(2)
Dn+1 J1 (α, f ; z) = z[Dn J1 (α, f ; z)]0 ,
z ∈ U.
Differentiating (2), we obtain
(3)
[Dn+1 J1 (α, f ; z)]0 = [Dn J1 (α, f ; z)]0 + z[Dn J1 (α, f ; z)]00 ,
If we let p(z) = [Dn J1 (α, f ; z)]0 , then (3) becomes
(4)
[Dn+1 J1 (α, f ; z)]0 = p(z) + zp0 (z),
z ∈ U.
z ∈ U.
Differential subordination defined by Sălăgean operator
41
Since f ∈ Sαn+1 (β), by using Definition 3 we have
Re [p(z) + zp0 (z)] > β,
(5)
z ∈ U,
which is equivalent to
p(z) + zp0 (z) ≺
1 + (2β − 1)z
≡ h(z),
1+z
z ∈ U.
By using Lemma A, we have
p(z) ≺ q(z) ≺ h(z)
i.e.
[Dn J1 (α, f ; z)]0 ≺ q(z),
where
1
q(z) =
z
Z
z
0
1 + (2β − 1)t
ln(1 + z)
dt = 2β − 1 + 2(1 − β)
,
1+t
z
z ∈ U.
The function q is convex and is the best dominant.
Hence p(z) ≺ q(z), z ∈ U , it results that
Re p(z) > Re g(1) = 2β − 1 + 2(1 − β) ln 2
from which we deduce that Sαn+1 (α) ⊂ Sαn (δ).
Theorem 2. Let g be a convex function, g(0) = 1, and let h be a function
such that
h(z) = g(z) + zg 0 (z),
z ∈ U.
If f ∈ Sαn (β) and verifies the differential subordination
(6)
[Dn+1 J1 (α, f ; z)]0 ≺ h(z),
z ∈ U,
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Georgia Irina Oros
then
[Dn J1 (α, f ; z)]0 ≺ g(z),
z ∈ U,
and this result is sharp.
Proof. By using the properties of the operator Dn f we have
Dn+1 J1 (α, f ; z) = z[Dn J1 (α, f ; z)]0 ,
z ∈ U,
we obtain
[Dn+1 J1 (α, f ; z)]0 = [Dn J1 (α, f ; z)]0 + z[Dn J1 (α, f ; z)]00 .
If we let
p(z) = [Dn J1 (α, f ; z)]0 ,
then we obtain
[Dn+1 J1 (α, f ; z)]0 = p(z) + zp0 (z),
z ∈ U,
and (6), becomes
p(z) + zp0 (z) ≺ g(z) + zg 0 (z),
z ∈ U.
By using Lemma B, we have
p(z) ≺ g(z),
z∈U
i.e.
[Dn J1 (α, f ; z)]0 ≺ g(z)
and this result is sharp.
Theorem 3. Let g be a convex function, g(0) = 1 and
h(z) = g(z) + zp0 (z),
z ∈ U.
Differential subordination defined by Sălăgean operator
43
If f ∈ Sαn (β) and verifies the differential subordination
[Dn J1 (α, f ; z)]0 ≺ h(z),
(7)
z ∈ U,
then
Dn J1 (α, f ; z)
≺ g(z),
z
and this result is sharp.
z ∈ U,
Proof. We let
Dn J1 (α, f ; z)
,
z
z ∈ U,
Dn J1 (α, f ; z) = zp(z),
z ∈ U.
(8)
p(z) =
and we obtain
(9)
By differentiating (9), we obtain
[Dn J1 (α, f ; z)]0 = p(z) + zp0 (z),
z ∈ U.
Then (7), becomes
p(z) + zp0 (z) ≺ g(z) + zg 0 (z) ≡ h(z),
z ∈ U.
By using Lemma B, we have
p(z) ≺ g(z),
z ∈ U,
i.e.
Dn J1 (α, f ; z)
≺ g(z),
z
and this result is sharp.
z ∈ U,
Theorem 4. Let h ∈ H(U ), with h(0) = 1, h0 (0) 6= 0, which verifies the
inequality
Re
zh00 (z)
1
1+ 0
>− ,
h (z)
2
z ∈ U.
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Georgia Irina Oros
If f ∈ Sαn (β) and verifies the differential subordination
[Dn+1 J1 (α, f ; z)]0 ≺ h(z),
(10)
z ∈ U,
where
Z
1 z
g(z) =
h(t)dt, z ∈ U.
z 0
The function g is convex and is the best dominant.
Proof. A simple application of the differential subordination technique [2,
Corollary 2.6.g.2, p.66] shows that the function g is convex. From
Dn+1 J1 (α, f ; z) = z[Dn J1 (α, f ; z)]0 ,
z ∈ U,
we obtain
[Dn+1 J1 (α, f ; z)]0 = [Dn J1 (α, f ; z)]0 + z[Dn J1 (α, f ; z)]00 ,
z ∈ U.
If we let
p(z) =]Dn J1 (α, f ; z)]0 ,
then we obtain
[Dn+1 J1 (α, f ; z)]0 = p(z) + zp0 (z),
z ∈ U.
By using Lemma A, we have
1
p(z) ≺ g(z) =
z
Z
z
h(t)dt,
0
i.e.
Z
1 z
h(t)dt, z ∈ U.
[D J1 (α, f ; z)] ≺
z 0
Theorem 5. Let h ∈ H(U ), with h(0) = 1, h0 (0) 6= 0, which verifies the
n
inequality
0
Re
zh00 (z)
1
1+ 0
>− ,
h (z)
2
z ∈ U.
Differential subordination defined by Sălăgean operator
45
If f ∈ Sαn (β) and verifies the differential subordination
(11)
then
[Dn J1 (α, f ; z)]0 ≺ h(z),
Dn J1 (α, f ; z)
≺ g(z),
z
z ∈ U,
z ∈ U,
where
Z
1 z
g(z) =
h(t)dt, z ∈ U.
z 0
The function g is convex and is the best dominant.
Proof. A simple application of the differential subordination technique [2,
Corollary 2.6.g.2, p.66] shows that the function g is convex.
We let
Dn J1 (α, f ; z)
p(z) =
,
z
z ∈ U,
Dn J1 (α, f ; z) = zp(z),
z ∈ U.
and we obtain
By differentiating, we obtain
[Dn J1 (α, f ; z)]0 = p(z) + zp0 (z),
z ∈ U.
Then (11) becomes
p(z) + zp0 (z) ≺ h(z),
z ∈ U.
By using Lemma A, we have
p(z) ≺ g(z),
i.e.
z ∈ U,
Dn J1 (α, f ; z)
≺ g(z),
z
z ∈ U,
where
Z
1 z
h(t)dt,
g(z) =
z 0
is convex and is the best dominant.
z ∈ U,
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Georgia Irina Oros
References
[1] S. S. Miller and P. T. Mocanu, On some classes of first-order differential
subordinations, Michigan Math. J., 32(1985), 185-195.
[2] S. S. Miller and P. T. Mocanu, Differential Subordinations. Theory and
Applications, Marcel Dekker Inc., New York, Basel, 2000.
[3] Grigore Şt. Sălăgean, Subclasses of univalent functions, Lecture Notes
in Math., Springer Verlag, Berlin, 1013(1983), 362-372.
Department of Mathematics
University of Oradea
Str. Armatei Române, No. 5
410087 Oradea, Romania
E-mail: georgia oros ro@yahoo.co.uk