ON A SPECIAL DIFFERENTIAL INEQUALITY
by
Gh. Oros and Georgia Irina Oros
Abstract. We find conditions on the complex-valued functions A, B, C defined in the unit disc
U and the real constants α, β, γ, δ such that the differential inequality
( ) ( ) ( ) ( ) ( ) ( )
(
( )
)
(
( )
)
( )
[
A z p4 z + B z p3 z +C z p2 z +α zp′ z 3 − β zp′ z 2 +γzp′ z +δ]
Re >0,
implies Re p
( )
z > 0, where p∈ H [1,n].AMS Classification Numbers: 30C80
Keywords: Differential subordination, dominant.
1. Introduction and preliminaries
We let H[U] denote the class of holomorfic functions in the unit disc
{
∈ <1}
= z C z
U : .
For a ∈C andn ∈N∗we let
H [a,n] = {ƒ∈ H [U], ƒ(z)=a + an z n
+ an+1 z n+1
+ …, z∈ U } and
An = {ƒ∈ H [U], ƒ(z)= z + an+1 z n+1
+ an+2 z n+2
+ …, z∈ U } with A1 = A.
In order to prove the new results we shall use the following lemma, which is a particular form of Theorema 2.3.i [1, p. 35].
Lemma A. [1, p. 35]Letψ : C2× U →C a function which satisfies Re ψ(ρi,σ, z)≤ 0,
where ρ,σ∈R,
(
1)
22 ρ
σ ≤−n + , z∈ U and n ≥1. If p∈ H [1,n] and
Reψ (p(z), zp′(z); z) > 0 then
Re p(z) > 0.
2. Main results
Theorem. Let
2 4 8
0 0 0
2
3 n n
n β γ
α δ γ β
i) Re A (z)≤ 0
ii)ReC (z)≥ 0
(1)
iii) Im2 B (z) ≤ 4
( )
( )
+ + +
− +
2 2 8 3 Re
Re 4 8
3 3 2 n3 n2 n
z C z
A n
n β α β γ
α .
If p∈ H [1,n] and
( ) ( ) ( ) ( ) ( ) ( )
(
( )
)
[
( )
(
)
( )
]
0Re
2
3 2
3 4
> + ′ + ′ −
− ′ + +
+
δ
γ
β
α
z p z z
p z
z p z z
p z C z p z B z p z A
(2)
then
Re p(z) > 0.
Proof. We let ψ : C2× U →C be defined by
( ) ( ) ( ) ( ) ( ) ( )
(
( )
)
( )
(
zp z)
zp( )
zz p z z p z C z p z B z p z A z z p z z p
′ + ′ −
− ′ + +
+ =
′
γ β
α ψ
2
3 2
3 4
) ); ( ), ( (
(3)
From (2) we have
Reψ (p(z), zp′(z); z) > 0, for z ∈U. (4)
For σ,ρ∈R satisfying
(
1)
22 ρ
σ≤−n + , and z∈ U, by using (1), we obtain: Reψ(ρi,σ, z)=
( ) ( ) ( ) ( ) ( ) ( )
(
( )
)
(
( )
)
( )
[
Az p4 z +Bz p3 z +C z p2 z +α zp′z 3−β zp′z 2+γzp′ z +δ]
Re =
=Re
[
A( )
z p4 − p3iB( )
z − p2C( )
z +ασ 3 − βσ 2 +γσ +δ]
=( )
( )
( )
(
)
(
) (
)
( )
( )
( )
( )
( )
( )
02 4 8 2 2 8 3 Re Im Re 4 8 3 8 2 4 8 2 4 8 3 Re Im Re 4 8 3 8 1 2 2 1 4 3 3 1 8 Re Im Re 2 3 2 3 2 3 2 2 6 3 2 3 2 3 2 3 2 3 4 6 3 2 4 2 2 6 4 2 3 2 3 4 ≤ + − − − + + + + + − + − − − ≤ ≤ + − − − + + + − − + + − − − ≤ ≤ + + − + + − − + + + − − − ≤
δ
γ
β
α
γ
β
α
β
α
α
δ
γ
β
α
γ
β
α
β
α
α
δ
γ
β
α
n n n n n n z C z B p z A n n p p p n n n n n n n z C p z B p z A n n p p n p n p p n p p p n z C p z B p z A pBy using Lemma A we have that Re p(z) > 0. If
2 2 8
3αn3 βn2 γn
δ = + + , then theorem can be rewritten as follows:
Corollary 1. Let α ≥ 0,β ≥ 0,γ ≥ 0, and n be a positive integer. Suppose that the function A, B, C : U →C satisfies:
i) Re A (z)≤ 0 ii)ReC (z)≥ 0
iii) Im2 B (z) ≤ 4
( )
( )
+ + + − + 2 2 8 3 Re Re 4 8
3 3 2 n3 n2 n
z C z
A n
n β α β γ
α .
If p∈ H [1,n] and
( ) ( ) ( ) ( ) ( ) ( )
(
( )
)
[
(
( )
)
( )
02 2 8 3 Re 2 3 2 3 2 3 4 > + + + + ′ + + ′ − ′ + + + n n n z p z z p z z p z z p z C z p z B z p z A γ β α δ γ β α then
Re p(z) > 0.
If
( )
z i,B( )
z z,C( )
z i,n , , , , ,Example 1. If p∈ H [1,1] and
(
) ( ) ( ) ( ) (
) ( )
(
( )
)
(
( )
)
( )
[
1 3 1 5 4 2 5 3 3]
0Re − − i p4 z + −z p3 z + − i p2 z + zp′ z 3− zp′z 2+ zp′ z + >
then
Re p(z) > 0.
If α
≡
0, then theorem can be rewritten as follows: Corollary 2. Let2 4 0
0
2
n n
,
,γ δ β γ
β ≥ ≥ ≤ + and n be a positive integer. Suppose that the function A, B, C : U →C satisfies:
i) Re A (z)≤ 0 ii)ReC (z)≥ 0
iii) Im2 B (z) ≤ 4
( )
( )
+ +
−
2 2 Re
Re 4
2
2 n n
z C z
A
n β γ
β .
If p∈ H [1,n] and
( ) ( ) ( ) ( ) ( ) ( )
[
(
( )
)
( )
]
0Re A z p4 z +B z p3 z +C z p2 z − β zp′ z 2 +γ zp′ z +δ > then
Re p(z) > 0.
If A
( )
z =−3+i,B( ) (
z =i 2 + z) ( )
,C z = 4−i,n=3,β = 2,γ = 4,δ =5, then in this case from Corollary 2 we deduce:Example 2. If p∈ H [1,3] and
(
) ( ) (
) ( ) ( ) ( )
(
( )
)
( )
[
3 2 4 2 4 5]
0Re − +i p4 z +i +z p3 z + −i p2 z − zp′z 2 + zp′ z + > then
Re p(z) > 0.
If β
≡
0, then Theorem can be rewritten as follows:Corollary 3. Let
2 8 0
0
3 n
n ,
,γ δ α γ
α ≥ ≥ ≤ + and n be a positive integer. Suppose that the function A, B, C : U →C satisfies:
i) Re A (z)≤ 0 ii)ReC (z)≥ 0
iii) Im2 B (z) ≤ 4
( )
( )
+ +
−
2 8 3 Re
Re 8
3 3 n3 n
z C z
A
n α γ
α .
( ) ( ) ( ) ( ) ( ) ( )
(
( )
)
[
( )
]
0Re A z p4 z + B z p3 z +C z p2 z +α zp′ z 3 +γ zp′ z +δ > then
Re p(z) > 0.
If
( )
( ) (
) ( )
,5 1 , 7 , 2 3 ,
4 , 2 ,
1 ,
4
5+ = + = + = = = =
−
= i B z z C z i n
β
γ
δ
z A
then in this case from Corollary 3 we deduce:
Example 3. If p∈ H [1,4] and
(
) ( ) ( ) ( ) ( ) ( )
(
( )
)
( )
03 1 7
2 3 2
1 4
5
Re 4 3 2 2 >
− + i p z + +z p z + +i p z − zp′z + zp′ z +
then
Re p(z) > 0.
If γ
≡
0, then Theorem can be rewritten as follows:Corollary 4. Let
4 8
0 0
2 3
n n
,
,β δ α β
α ≥ ≥ ≤ + and n be a positive integer.
Suppose that the function A, B, C : U →C satisfies:
i) Re A (z)≤ 0 ii)ReC (z)≥ 0
iii) Im2 B (z) ≤ 4
( )
( )
+ +
− +
2 8
3 Re
Re 4 8
3 3 2 n3 n2
z C z
A n
n β α β
α .
If p∈ H [1,n] and
( ) ( ) ( ) ( ) ( ) ( )
(
( )
)
[
(
( )
)
]
0Re A z p4 z +B z p3 z +C z p2 z +α zp′ z 3 − β zp′ z 2 +δ >
then
Re p(z) > 0.
If
( )
z = −1+ 2i,B( )
z = +4+ z,C( )
z =5−i,n =5,α = 4, β =1,δ =7A then in this
case from Corollary 4 we deduce:
(
) ( ) (
) ( ) ( ) ( )
(
( )
)
[
1 2 4 5 4 7]
0Re − + i p4 z + +z p3 z + −i p2 z + zp′ z 2 + > then
Re p(z) > 0.
References.
[1] S.S. Miller and P.T. Mocanu, Differential Subordinations. Theory and Applications, Marcel Dekker Inc. New York, Basel, 2000.
Authors:
Gh. Oros – University of Oradea, Romania
