General Mathematics Vol. 10, No. 1–2 (2002), 33–36
On a differential inequality II
Georgia Irina Oros
Abstract
We find conditions on the complex-valued function A defined in
the unit disc U and the real constants α, β, γ, such that the differential inequality
Re [A(z)p2 (z) − α(zp0 (z))2 + βzp0 (z) + γ] > 0
implies Re p(z) > 0, where p ∈ H[1, n].
2000 Mathematical Subject Classification: 30C80
1
Introduction and preliminaries
We let H[U ] denote the class of holomorphic functions in the unit disc
U = {z ∈ C : |z| < 1}.
For a ∈ C and n ∈ N∗ we let
H[a, n] = {f ∈ H[U ], f (z) = a + an z n + an+1 z n+1 + . . . , z ∈ U }
and
An = {f ∈ H[U ], f (z) = z + an+1 z n+1 + an+2 z n+2 + . . . , z ∈ U }
33
34
Georgia Irina Oros
with A1 = A.
In order to prove the new results we shall use the following lemma, which
is a particular form of Theorem 2.3.i [1, p. 35].
Lemma A. [1, p. 35] Let ψ : C2 × U → C a function which satisfies
Re ψ(ρi, σ; z) ≤ 0,
n
where ρ, σ ∈ R, σ ≤ − (1 + ρ2 ), z ∈ U and n ≥ 1.
2
If p ∈ H[1, n] and
Re ψ(p(z), zp0 (z); z) > 0
then
Re p(z) > 0.
2
Main results
αn2 βn
+
and let n be a positive integer.
Theorem. Let α ≥ 0, β ≥ 0, γ ≤
4
2
Suppose that the function A : U → C satisfies
Re A(z) ≥ −
(1)
αn2 βn
−
.
2
2
If p ∈ H[1, n] and
(2)
Re [A(z)p2 (z) − α(zp0 (z))2 + βzp0 (z) + γ] > 0,
then
Re p(z) > 0.
On a differential inequality II
35
Proof. We let ψ : C2 × U → C be defined by
(3)
ψ(p(z), zp0 (z); z) = A(z)p2 (z) − α(zp0 (z))2 + βzp0 (z) + γ.
From (2) we have
(4)
Re ψ(p(z), zp0 (z); z) > 0, for z ∈ U.
n
n2
For σ, ρ ∈ R satisfying σ ≤ − (1 + ρ2 ), hence −σ 2 ≤ − (1 + ρ2 )2 , and
2
4
z ∈ U , by using (1) we obtain:
Re ψ(ρi, σ; z) = Re [A(z)(ρi)2 − ασ 2 + βσ + γ] =
= −ρ2 Re A(z) − ασ 2 + βσ + γ ≤
= −ρ2 Re A(z) −
αn2
βn
(1 + ρ2 )2 −
(1 + ρ2 ) + γ ≤
4
2
αn2 αn2 2 αn2 4 βn βn 2
−
ρ −
ρ −
−
ρ ≤
4
2
4
2
2
·
¸
αn2 4
αn2 βn
αn2 βn
2
≤−
ρ − ρ Re A(z) +
+
−
−
+ γ ≤ 0.
4
2
2
4
2
≤ −ρ2 Re A(z) −
By using Lemma A we have that Re p(z) > 0.
If p = 0 then we obtain the Theorem from [2].
αn2 βn
If γ =
+
, Theorem can be rewritten as follows:
4
2
Corollary. Let α ≥ 0, β ≥ 0 and let n be a positive integer. Suppose that
the function A : U → C satisfies
Re A(z) ≥ −
αn2 βn
−
.
2
2
If p ∈ H[1, n] and
·
¸
αn2 βn
2
0
2
0
Re A(z)p (z) − α(zp (z)) + βzp (z) +
+
>0
4
2
36
Georgia Irina Oros
then
Re p(z) > 0.
If β = 0 then we obtain the Corollary from [2].
1
If α = , n = 1, β = 3, A(z) = 1 + z. In this case from Corollary we
2
deduce
Example 1. If p ∈ H[1, 1] then
¸
·
1 0
13
2
2
0
>0
Re (1 + 2z)p (z) − (zp (z)) + 3zp (z) +
2
8
implies
Re p(z) > 0.
1
If α = , n = 2, A(z) = 1 + 2z. In this case from Corollary 1 we deduce
2
Example 2. If p ∈ H[1, 2] then
·
¸
1 0
9
2
0
2
Re (1 + 3z)p (z) − 2(zp (z)) + zp (z) +
>0
4
4
implies 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.
[2] Georgia Irina Oros, On a differential inequality I, (submitted)
Faculty of Mathematics and Computer Sciences
Babeş-Bolyai University
3400 Cluj-Napoca, Romania