Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 3, Article 59, 2004
CERTAIN SECOND ORDER LINEAR DIFFERENTIAL SUBORDINATIONS
V. RAVICHANDRAN
D EPARTMENT OF C OMPUTER A PPLICATIONS
S RI V ENKATESWARA C OLLEGE OF E NGINEERING
P ENNALUR , S RIPERUMBUDUR 602 105, I NDIA
vravi@svce.ac.in
URL: http://www.svce.ac.in/∼vravi
Received 29 December, 2003; accepted 03 April, 2004
Communicated by N.E. Cho
A BSTRACT. In this present investigation, we obtain some results for certain second order linear
differential subordination. We also discuss some applications of our results.
Key words and phrases: Analytic functions, Hadamard product (or convolution), differential subordination, Ruscheweyh
derivatives, univalent functions, convex functions.
2000 Mathematics Subject Classification. Primary 30C80, Secondary 30C45.
1. I NTRODUCTION
Let H denote the class of all analytic functions in ∆ := {z ∈ C : |z| < 1}. For a positive
integer n and a ∈ C, let
(
)
∞
X
H[a, n] := f ∈ H : f (z) = a +
ak z k (n ∈ N := {1, 2, 3, . . .})
k=n
and
(
A(p, n) :=
f ∈ H : f (z) = z p +
∞
X
)
ak z k
(n, p ∈ N) .
k=n+p
Set
Ap := A(p, 1), A := A1 .
For two functions f, g ∈ H, we say that the function f (z) is subordinate to g(z) in ∆ and write
f ≺g
f (z) ≺ g(z),
or
if there exists a Schwarz function w(z) ∈ H with
w(0) = 0
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
The author is thankful to the referee for his comments.
003-04
and
|w(z)| < 1
(z ∈ ∆),
2
V. R AVICHANDRAN
such that
(z ∈ ∆).
f (z) = g(w(z))
(1.1)
In particular, if the function g is univalent in ∆, the above subordination (1.1) is equivalent to
f (0) = g(0)
and
f (∆) ⊂ g(∆).
Miller and Mocanu [2] considered the second order linear differential subordination
A(z)z 2 p00 (z) + B(z)zp0 (z) + C(z)p(z) + D(z) ≺ h(z),
where A, B, C and D are complex-valued functions defined on ∆ and h(z) is any convex function and in particular h(z) = (1 + z)/(1 − z). In fact, they have proved the following:
Theorem 1.1 (Miller and Mocanu [2, Theorem 4.1a, p.188]). Let n be a positive integer and
A(z) = A ≥ 0. Suppose that the functions B(z), C(z), D(z) : ∆ → C satisfy <B(z) ≥ A
and
[=C(z)]2 ≤ n[<B(z) − A]<(nB(z) − nA − 2D(z)).
(1.2)
If p ∈ H[1, n] and if
(1.3)
<{Az 2 p00 (z) + B(z)zp0 (z) + C(z)p(z) + D(z)} > 0,
then
<p(z) > 0.
Also Miller and Mocanu [2] have proved the following:
Theorem 1.2 (Miller and Mocanu [2, Theorem 4.1e, p.195]). Let h be convex univalent in ∆
with h(0) = 0 and let A ≥ 0. Suppose that k > 4/|h0 (0)| and that B(z), C(z) and D(z) are
analytic in ∆ and satisfy
< B(z) ≥ A + |C(z) − 1| − <(C(z) − 1) + k|D(z)|.
If p ∈ H[0, 1] satisfies the differential subordination
Az 2 p00 (z) + B(z)zp0 (z) + C(z)p(z) + D(z) ≺ h(z)
then p ≺ h.
In this paper, we extend Theorem 1.1 by assuming
<{Az 2 p00 (z) + B(z)zp0 (z) + C(z)p(z) + D(z)} > α,
(0 ≤ α < 1)
and Theorem 1.2 by assuming that the function h(z) is convex of order α. Certain results of
Karunakaran and Ponnusamy [6], Juneja and Ponnusamy [7] and Owa and Srivastava [8] are
obtained as special cases. Also we give application of our results to certain functions defined
by the familiar Ruscheweyh derivatives.
For two functions f (z) and g(z) given by
p
f (z) = z +
∞
X
k
ak z ,
p
g(z) = z +
k=n+p
∞
X
bk z k
(n, p ∈ N) ,
k=n+p
the Hadamard product (or convolution) of f and g is defined by
(f ∗ g)(z) := z p +
∞
X
ak bk z k =: (g ∗ f )(z).
k=n+p
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
http://jipam.vu.edu.au/
C ERTAIN S ECOND O RDER L INEAR D IFFERENTIAL S UBORDINATIONS
3
The Ruscheweyh derivative of f (z) of order δ + p − 1 is defined by
(1.4)
Dδ+p−1 f (z) :=
zp
(1 − z)δ+p
∗ f (z)
(f ∈ A (p, n) ; δ ∈ R\ (−∞, −p])
or, equivalently, by
D
(1.5)
δ+p−1
∞ X
δ+k−1
f (z) := z +
ak z k
k
−
p
k=p+1
p
(f ∈ A (p, n) ; δ ∈ R\ (−∞, −p]) .
In particular, if δ = l (l + p ∈ N), we find from the definition (1.4) or (1.5) that
zp
dl+p−1 l−1
z f (z)
l+p−1
(l + p − 1)! dz
(f ∈ A (p, n) ; l + p ∈ N) .
Dl+p−1 f (z) =
In our present investigation of the second order linear differential subordination, we need the
following definitions and results:
Definition 1.1 (Miller and Mocanu [2, Definition 2.2b, p. 21]). Let Q be the set of functions q
that are analytic and univalent on ∆ \ E(q), where
E(q) = {ζ ∈ ∂∆ : lim q(z) = ∞}
z→ζ
and are such that q 0 (ζ) 6= 0 for ζ ∈ ∂∆ \ E(q), where ∂∆ := {z ∈ C : |z| = 1}, ∆ := ∆ ∪ ∂∆.
Theorem 1.3 (Miller and Mocanu [2, Lemma 2.2d, p. 24]). Let q ∈ Q, with q(0) = a. Let
p(z) = a + pn z n + · · · be analytic in ∆ with p(z) 6≡ a and n ≥ 1. If p(z) is not subordinate to
q(z), then there exist points z0 = r0 eθ0 ∈ ∆ and ζ0 ∈ ∂∆ − E(q), and an m ≥ n ≥ 1 for which
p(∆r0 ) ⊂ q(∆),
(i)
(ii)
(iii)
(1.6)
p(z0 ) = q(ζ0 )
z0 p0 (z0 ) = mζ0 q 0 (ζ0 ), and
<[z0 p00 (z0 )/p0 (z0 ) + 1] ≥ m<[z0 q 00 (z0 )/q 0 (z0 ) + 1],
where ∆r := {z ∈ C : |z| < r}.
Theorem 1.4 (cf. Miller and Mocanu [2, Theorem 2.3i (i), p. 35]). Let Ω be a simply connected
domain and ψ : C3 × ∆ → C satisfies the condition
ψ(iσ, ζ, µ + iη; z) 6∈ Ω
for z ∈ ∆ and for real σ, ζ, µ, η satisfying ζ ≤ −n(1 + σ 2 )/2 and ζ + µ ≤ 0. Let p(z) =
1 + pn z n + pn+1 z n+1 + · · · be analytic in ∆. If
ψ(p(z), zp0 (z), z 2 p00 (z); z) ∈ Ω,
then <p(z) > 0.
2. D IFFERENTIAL S UBORDINATION WITH C ONVEX F UNCTIONS OF O RDER α
By appealing to Theorem 1.3, we first prove the following:
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
http://jipam.vu.edu.au/
4
V. R AVICHANDRAN
Theorem 2.1. Let h be a convex univalent function of order α, 0 ≤ α < 1, in ∆ with h(0) = 0
and let A ≥ 0. Suppose that
k > 22(1−α) |h0 (0)|
and that B(z), C(z) and D(z) are analytic in ∆ and satisfy
1
(2.1)
n<B(z) ≥ n(1 − αn)A +
[|C(z) − 1| − <(C(z) − 1)] + k|D(z)|,
2β(α)
where
 α
1
4 (1 − 2α)


α
=
6

4 − 22α+1
2
(2.2)
β(α) :=

1

 (log 4)−1
α= .
2
If p ∈ H[0, n] satisfies the differential subordination
Az 2 p00 (z) + B(z)zp0 (z) + C(z)p(z) + D(z) ≺ h(z),
(2.3)
then p ≺ h.
Proof. Our proof of Theorem 2.1 is essentially similar to Theorem 1.2 of Miller and Mocanu
[2]. Let the subordination in (2.3) be satisfied so that D(0) = 0. Since
k|h0 (0)| > 22(1−α) ,
there is an r0 , 0 < r0 < 1 such that
(1 + r)2(1−α)
(1 + r0 )2(1−α)
0
2(1−α)
= k|h (0)| and 2
<
< k|h0 (0)|
r0
r
for r0 < r < 1. Since h is convex of order α in ∆, the function hr (z) = h(rz) is convex
of order α in ∆ (r0 < r < 1). By setting pr (z) = p(rz) for r0 < r < 1, we see that the
subordination (2.3) becomes
(2.4)
ur (z) := Az 2 p00r (z) + B(rz)zp0r (z) + C(rz)pr (z) + D(rz) ≺ hr (z)
(z ∈ ∆; r0 < r < 1).
Assume that pr is not subordinate to hr , for some r in (r0 , 1). Then by Theorem 1.3 there exist
points z0 ∈ ∆, w0 ∈ ∂∆ and an m ≥ n ≥ 1 such that
pr (z0 ) = hr (w0 ), z0 p0r (z0 ) = mw0 h0r (w0 ),
(2.5)
(2.6)
<
z0 p00 (z0 )
1+ 0r
pr (z0 )
≥ m<
w0 h00 (w0 )
1+ 0 r
hr (w0 )
.
Therefore we have
z02 p00r (z0 )
(2.7)
< 1+
≥ mα.
mw0 h0 (w0 )
From Equations (2.5), (2.6) and (2.7), it follows that
2 00
z0 pr (z0 )
(2.8)
<
≥ m(mα − 1).
w0 h0r (w0 )
Since hr (z) is convex of order α or equivalently
zh00r (z)
< 1+ 0
>α
hr (z)
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
(z ∈ ∆),
http://jipam.vu.edu.au/
C ERTAIN S ECOND O RDER L INEAR D IFFERENTIAL S UBORDINATIONS
5
by [2, Theorem 3.3f, p.115], we have
<
zh0r (z)
> β(α)
hr (z)
(z ∈ ∆)
where β(α) is given by Equation (2.2) and this condition is equivalent to
hr (z)
1 1
zh0 (z) − 2β(α) ≤ 2β(α) (z ∈ ∆).
r
Therefore,
1
hr (w0 )
< (C(rz0 ) − 1)
≥
{<[C(rz0 ) − 1] − |C(rz0 ) − 1|}.
w0 h0r (w0 )
2β
(2.9)
Since h is convex of order α, we have the following well-known estimate:
|h0 (z)| ≥
|h0 (0)|
(1 + r)2(1−α)
(|z| = r < 1).
By setting z = rw0 , we see that
|w0 h0r (w0 )| ≥
(2.10)
r|h0 (0)|
(1 + r)2(1−α)
(|w0 | = 1).
By setting
V :=
(2.11)
Az02 p00r (z0 ) B(rz0 )z0 p0r (z0 )
pr (z0 )
D(rz0 )
+
+ (C(rz0 ) − 1)
+
,
0
0
0
w0 hr (w0 )
w0 hr (w0 )
w0 hr (w0 ) w0 h0r (w0 )
we see that
ur (z0 ) = hr (w0 ) + V w0 h0r (w0 ).
(2.12)
From (2.8), (2.9), (2.10) and (2.11), we have
<V ≥ m(mα − 1)A + m<B(rz0 ) +
1
[<(C(rz0 ) − 1) − |C(rz0 ) − 1|]
2β(α)
(1 + r)2(1−α)
|D(rz0 )|
r|h0 (0)|
≥ m[(nα − 1)A + <B(rz0 )]
1
+
[<(C(rz0 ) − 1) − |C(rz0 ) − 1|] − k|D(rz0 )|
2β(α)
≥ n[(nα − 1)A + <B(rz0 )]
1
−
[|C(rz0 ) − 1| − <(C(rz0 ) − 1)] − k|D(rz0 )| ≥ 0,
2β(α)
−
it follows that ur (z0 ) 6∈ hr (∆), a contradiction. Therefore, pr ≺ hr for r ∈ (r0 , 1). By letting
r → 1− , we obtain the desired conclusion p ≺ h.
Remark 2.2. When α = 0, n = 1, Theorem 2.1 reduces to Theorem 1.2 of Miller and Mocanu
[2].
From the proof of Theorem 2.1, it is clear that the condition h(0) = 0 in not necessary when
C(z) = 1 and hence the following:
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
http://jipam.vu.edu.au/
6
V. R AVICHANDRAN
Corollary 2.3. Let h be a convex univalent function of order α, 0 ≤ α < 1, in ∆, h(0) = a and
let A ≥ 0. Suppose that
k > 22(1−α) /|h0 (0)|
and that B(z) and D(z) are analytic in ∆ with D(0) = 0 and
n < B(z) ≥ n(1 − αn)A + k|D(z)|
(2.13)
for all z ∈ ∆. If p ∈ H[a, n], p(0) = h(0), satisfies the differential subordination
Az 2 p00 (z) + B(z)zp0 (z) + p(z) + D(z) ≺ h(z),
(2.14)
then p ≺ h.
By taking A = 0 and D(z) = 0 in Theorem 2.1, we obtain the following:
Corollary 2.4. Let h be a convex univalent function of order α, 0 ≤ α < 1, in ∆ with h(0) = 0.
Let B(z) and C(z) be analytic functions on ∆ satisfying
<B(z) ≥
1
[|C(z) − 1| − <(C(z) − 1)],
2nβ(α)
where β(α) is as given in Theorem 2.1. If p ∈ H[0, n] satisfies the subordination
B(z)zp0 (z) + C(z)p(z) ≺ h(z),
then p(z) ≺ h(z).
By taking B(z) = 1, α = 0, n = 1, in Corollary 2.4, we have the following:
Corollary 2.5. Let h be a convex univalent function in ∆ with h(0) = 0. Let C(z) be analytic
functions on ∆ satisfying
<C(z) > |C(z) − 1|.
If the analytic function p(z) satisfies the subordination
zp0 (z) + C(z)p(z) ≺ h(z),
then p(z) ≺ h(z).
3. D IFFERENTIAL S UBORDINATION WITH C ARATHEODORY F UNCTIONS OF O RDER α
By appealing to Theorem 1.4, we now prove the following:
Theorem 3.1. Let n be a positive integer and A(z) = A ≥ 0. Suppose that the functions
B(z), C(z), D(z) : ∆ → C satisfy <B(z) ≥ A and
(3.1)
[= C(z)]2 ≤ n [<B(z) − A]
δ + 2α
2+δ
× n(<B(z) − A) −
< C(z) −
<(D(z) − α) .
1−α
1−α
If p ∈ H[1, n] and
(3.2)
<{Az 2 p00 (z) + B(z)zp0 (z) + C(z)p(z) + D(z)} > α
(α < 1),
then
<p(z) >
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
δ + 2α
.
δ+2
http://jipam.vu.edu.au/
C ERTAIN S ECOND O RDER L INEAR D IFFERENTIAL S UBORDINATIONS
7
Proof. Define the function P (z) by
P (z) :=
p(z) − γ
1−γ
where
γ :=
δ + 2α
.
δ+2
Then inequality (3.2) can be written as
<{ψ(P (z), zP 0 (z), z 2 P 00 (z); z)} > 0,
where
ψ(r, s, t; z) = At + B(z)s + C(z)r +
γC(z) + D(z) − α
.
1−γ
In view of Theorem 1.4, it is enough to show that
<ψ(iσ, ζ, µ + iη; z) ≤ 0
for all real numbers σ, ζ, µ and η with ζ ≤
−n(1+σ 2 )
,
2
ζ + µ ≤ 0 and for all z ∈ ∆. Now,
<ψ(iσ, ζ, µ + iη; z)
γC(z) + D(z) − α
= µA + ζ<B(z) − σ=C(z) + <
1−γ
γC(z) + D(z) − α
≤ ζ(<B(z) − A) − σ=C(z) + <
1−γ
1
≤ − n[<B(z) − A]σ 2 + 2=C(z)σ
2
γC(z) + D(z) − α
+n[<B(z) − A] − 2<
≤ 0,
1−γ
provided (3.1) holds. This completes the proof of our Theorem 3.1.
For α = δ = 0, Theorem 3.1 reduces to Theorem 1.1.
By taking D = 0 and C(z) = 1 in Theorem 3.1, we have the following:
Corollary 3.2. Let A ≥ 0 and <B(z) − A > δ > 0. If p ∈ H[1, n] satisfies
<{Az 2 p00 (z) + B(z)zp0 (z) + p(z)} > α
(α < 1)
then
<p(z) >
nδ + 2α
.
nδ + 2
Corollary 3.3. Let λ(z) and R(z) be functions defined on ∆ and
<λ(z) > δ +
2+δ
<R(z) ≥ 0.
(1 − α)n
If p ∈ H[1, n] satisfies
<{λ(z)zp0 (z) + p(z) + R(z)} > α
(α < 1),
then
<p(z) >
2α + δn
.
2 + δn
A special case of Corollary 3.3 is obtained by Owa and Srivastava [8, Lemma 2, p. 254].
The proof of the following theorem is similar and hence it is omitted.
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
http://jipam.vu.edu.au/
8
V. R AVICHANDRAN
Theorem 3.4. Let n be a positive integer and A(z) = A ≥ 0. Suppose that the functions
B(z), C(z), D(z) : ∆ → C satisfy <B(z) ≥ A and
δ + 2α
2+δ
2
(3.3) [=C(z)] ≤ n[<B(z) − A] n(<B(z) − A) −
< C(z) −
<(D(z) − α) .
1−α
1−α
If p ∈ H[1, n] satisfies
(3.4)
<{Az 2 p00 (z) + B(z)zp0 (z) + C(z)p(z) + D(z)} < α
then
<p(z) <
(α > 1),
δ + 2α
.
δ+2
4. A PPLICATIONS
We now give certain applications of our results obtained in Section 2 and 3.
Theorem 4.1. Let γ ∈ C with γ 6= −1, −2, −3, . . . and let φ, Φ be analytic functions on ∆ with
φ(z)Φ(z) 6= 0 for z ∈ ∆. If
<C(z) − |C(z) − 1| > 1 − 2nβ(α)<B(z),
where
Φ(z)
γΦ(z) + zΦ0 (z)
and C(z) :=
,
φ(z)
φ(z)
then the integral operator defined by
Z z
1
I(f )(z) := γ
tγ−1 f (t)φ(t)dt
z Φ(z) 0
B(z) :=
satisfies I(f )(z) ≺ h(z) for every function f (z) ≺ h(z) where h(z) is a convex function of
order α.
Proof. The result follows immediately from Corollary 2.4.
Theorem 4.2. Let h be a convex univalent function of order α in ∆, 0 ≤ α < 1 and h(0) = 1.
Let M, N, R be analytic in ∆ with R(0) = 0 and
M (z) = z n + . . . , and N (z) = z n + . . . .
Let
βN (z)
< 0
> k|R(z)|
zN (z)
22(1−α)
k> 0
|h (0)|
.
If
(4.1)
β
M 0 (z)
M (z)
+ (1 − β)
+ R(z) ≺ h(z),
0
N (z)
N (z)
then
M (z)
≺ h(z).
N (z)
Proof. Let the function p(z) be defined by
p(z) = M (z)/N (z).
Then p(0) = 1 = h(0) and it follows that
p(z) +
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
N (z) 0
M 0 (z)
zp
(z)
=
.
zN 0 (z)
N 0 (z)
http://jipam.vu.edu.au/
C ERTAIN S ECOND O RDER L INEAR D IFFERENTIAL S UBORDINATIONS
9
Also, a computation shows that the subordination in (4.1) is equivalent to
p(z) +
βN (z) 0
zp (z) + R(z) ≺ h(z).
zN 0 (z)
The result now follows by an application of Corollary 2.3
Remark 4.3. When β = 1, α = 0, Theorem 4.2 reduces to [2, Theorem 4.1h, p. 199] of Miller
and Mocanu. If α = 0 and R(z) = 0, then Theorem 4.2 reduces to a result of Juneja and
Ponnusamy [7, Corollary 1, p. 290].
More generally, we have the following:
Theorem 4.4. Let δ > −p be any real number, λ ∈ C with <λ ≥ 0. Let R(z) be a function
defined on ∆ with R(0) = 0 and h(z) a convex function of order α, 0 ≤ α < 1, h(0) = 1. Let
g ∈ Ap satisfy
Dδ+p−1 g(z)
22(1−α)
≥ µ(δ + p)|R(z)|,
k> 0
.
< λ δ+p
D g(z)
|h (0)|
If f ∈ Ap satisfies
δ+p−1
µ
µ−1
D
f (z)
Dδ+p f (z) Dδ+p−1 f (z)
(1 − λ)
+ λ δ+p
+ R(z) ≺ h(z),
Dδ+p−1 g(z)
D g(z) Dδ+p−1 g(z)
then
Dδ+p−1 f (z)
Dδ+p−1 g(z)
µ
≺ h(z).
Proof. Let the function p(z) be defined by
Dδ+p−1 f (z)
p(z) :=
Dδ+p−1 g(z)
µ
.
Then a computation shows that the following subordination holds:
B(z)zp0 (z) + p(z) + R(z) ≺ h(z),
where
λ
Dδ+p−1 g(z)
.
µ(δ + p) Dδ+p g(z)
The result follows by an application of Corollary 2.3.
B(z) :=
When R(z) = 0 and µ = 1, the Theorem 4.4 reduces to Juneja and Ponnusamy [7, Theorem
1, p. 289].
Theorem 4.5. Let α be a complex number <α > 0 and β < 1. Let M, N, R be analytic in ∆
with R(0) = 0 and
M (z) := z n + c1 z n+k + · · · ,
N (z) := z n + d1 z n+k + · · · .
Let
<
αN (z)
2 + δk
>δ+
<R(z).
0
zN (z)
(1 − β)k
If
(4.2)
<[α
M 0 (z)
M (z)
+ (1 − α)
+ R(z)] > β,
0
N (z)
N (z)
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
http://jipam.vu.edu.au/
10
V. R AVICHANDRAN
then
<
M (z)
2β + kδ
>
.
N (z)
2 + kδ
Proof. Let p(z) := M (z)/N (z). Then p(0) = 1 = h(0). It follows that
p(z) +
M 0 (z)
N (z) 0
zp
(z)
=
.
zN 0 (z)
N 0 (z)
Then
<p(z) +
M 0 (z)
M (z)
αN (z) 0
zp
(z)
+
R(z)
=
<[α
+ (1 − α)
+ R(z)]
0
0
zN (z)
N (z)
N (z)
> β.
If B(z) is defined by B(z) := αN (z)/[zN 0 (z)], then it follows that
<B(z) > δ +
2 + δk
<R(z).
(1 − β)k
The result now follows by an application of Corollary 3.3
Remark 4.6. For R(z) = 0, β = 0, Theorem 4.5 is due to Karunakaran and Ponnusamy [6,
Theorem B, p. 562].
Theorem 4.7. Let δ > −p be any real number, λ ∈ C with <λ ≥ 0. Let R(z) be a function
defined on ∆ with R(0) = 0, 0 ≤ α < 1. Let g ∈ Ap satisfies
Dδ+p−1 g(z)
µ(δ + p)(2 + δ)
< λ δ+p
> µ(δ + p)δ +
<R(z) ≥ 0.
D g(z)
1−α
If f ∈ Ap satisfies
(
)
δ+p−1
µ
µ−1
D
f (z)
Dδ+p f (z) Dδ+p−1 f (z)
< (1 − λ)
+ λ δ+p
+ R(z) > α,
Dδ+p−1 g(z)
D g(z) Dδ+p−1 g(z)
then
Dδ+p−1 f (z)
Dδ+p−1 g(z)
µ
≥
2α + δ
.
2+δ
Proof. Let
Dδ+p−1 f (z)
p(z) :=
Dδ+p−1 g(z)
µ
.
Then a computation shows that
<{B(z)zp0 (z) + p(z) + R(z)} > α,
where
B(z) :=
The result follows easily.
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
λ
Dδ+p−1 g(z)
.
µ(δ + p) Dδ+p g(z)
http://jipam.vu.edu.au/
C ERTAIN S ECOND O RDER L INEAR D IFFERENTIAL S UBORDINATIONS
11
R EFERENCES
[1] B.C. CARLSON AND S.B. SHAFFER, Starlike and prestarlike hypergeometric functions, SIAM J.
Math. Anal., 15 (1984), 737–745.
[2] S.S. MILLER AND P.T. MOCANU, Differential Subordinations: Theory and Applications, Series in
Pure and Applied Mathematics, No. 225, Marcel Dekker, New York, (2000).
[3] S.S. MILLER, P.T. MOCANU AND M.O. READE, Differential subordinations and univalent functions, Michigan Math. J., 28 (1981), 157–171.
[4] S.S. MILLER AND P.T. MOCANU, The theory and applications of second order differential subordinations, Studia Univ. Babes-Bolyai, Math., 34(4) (1989), 3–33.
[5] S.S. MILLER AND P.T. MOCANU, Classes of univalent integral operators, J. Math. Anal. Appl.,
157 (1991), 147–165.
[6] V. KARUNAKARAN AND S. PONNUSAMY, Differential subordinations and conformal mappings
I, Indian J. Pure Appl. Math., 20 (1989), 560–565.
[7] S. PONNUSAMY AND O.P. JUNEJA, Some applications of first order differential subordinations,
Glasnik Math., 25(45) (1990), 287–296.
[8] S. OWA AND H.M. SRIVASTAVA, Analytic solutions of a class of Briot-Bouqet differential equations, in Current Topics in Analytic Function Theory (H.M. Srivastava and S. Owa, Ed.), World
Scientific, Singapore, 252–259 (1992).
[9] H. SAITOH, A linear operator and its applications of first order differential subordinations, Math.
Japonica, 44 (1996), 31–38.
J. Inequal. Pure and Appl. Math., 5(3) Art. 59, 2004
http://jipam.vu.edu.au/