Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 1, Article 6, 2006
SOME INEQUALITIES ASSOCIATED WITH A LINEAR OPERATOR DEFINED
FOR A CLASS OF ANALYTIC FUNCTIONS
S. R. SWAMY
P. G. D EPARTMENT OF C OMPUTER S CIENCE
R. V. C OLLEGE OF E NGINEERING
BANGALORE 560 059, K ARNATAKA , I NDIA
mailtoswamy@rediffmail.com
Received 07 March, 2005; accepted 25 July, 2005
Communicated by H. Silverman
A BSTRACT. In this paper, we give a sufficient condition on a linear operator Lp (a, c)g(z) which
can guarantee that for α a complex number with Re(α) > 0,
Lp (a + 1, c)f (z)
Lp (a, c)f (z)
+α
> ρ, ρ < 1,
Re (1 − α)
Lp (a, c)g(z)
Lp (a + 1, c)g(z)
in the unit disk E, implies
0
Lp (a, c)f (z)
Re
> ρ > ρ, z ∈ E.
Lp (a, c)g(z)
Some interesting applications of this result are also given.
Key words and phrases: Analytic functions, Differential subordination, Ruscheweyh derivatives, Linear operator.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let A(p, n) denote the class functions f normalized by
(1.1)
p
f (z) = z +
∞
X
ak z k
(p, n ∈ N = {1, 2, 3, ...}),
k=p+n
which are analytic in the open unit disk E = {z : z ∈ C, |z| < 1}.
In particular, we set A(p, 1) = Ap and A(1, 1) = A1 = A.
The Hadamard product (f ∗ g)(z) of two functions f (z) given by (1.1) and g(z) given by
p
g(z) = z +
∞
X
k=p+n
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
064-05
bk z k
(p, n ∈ N),
2
S. R. S WAMY
is defined, as usual, by
p
(f ∗ g)(z) = z +
∞
X
ak bk z k = (g ∗ f )(z).
k=p+n
The Ruscheweyh derivative of f (z) of order δ + p − 1 is defined by
zp
(1.2)
Dδ+p−1 f (z) =
∗ f (z)
(f ∈ A(p, n); δ ∈ R \ (−∞, −p])
(1 − z)δ+p
or, equivalently, by
∞ X
δ+k−1
δ+p−1
p
(1.3)
D
f (z) = z +
ak z k ,
k
−
p
k=p+n
S
where f (z) ∈ A(p, n) and δ ∈ R \ (−∞, −p]. In particular, if δ = l ∈ N {0}, we find from
(1.2) or (1.3) that
dl+k−1 l−1
zp
l+p−1
D
f (z) =
z
f
(z)
.
(l + p − 1)! dz l+p−1
The author has proved the following result in [4].
Theorem A. Let α be a complex number satisfying Re(α) > 0 and ρ < 1. Let δ > −p, f, g ∈
Ap and
Dδ+p−1 g(z)
> γ, 0 ≤ γ < Re(α), z ∈ E.
Re α δ+p
D g(z)
Then
δ+p−1
D
f (z)
2ρ(δ + p) + γ
Re
>
,
z ∈ E,
δ+p−1
D
g(z)
2(δ + p) + γ
whenever
Dδ+p−1 f (z)
Dδ+p f (z)
Re (1 − α) δ+p−1
+ α δ+p
> ρ, z ∈ E.
D
g(z)
D g(z)
The Pochhammer symbol (λ)k or the shifted factorial is given by (λ)0 = 1 and (λ)k =
λ(λ + 1)(λ + 2) · · · (λ + k − 1), k ∈ N. In terms of (λ)k , we now define the function φp (a, c; z)
by
∞
X
(a)k k+p
p
φp (a, c; z) = z +
z ,
z ∈ E,
(c)k
k=1
where a ∈ R, c ∈ R \ z0− ; z0− = {0, −1, −2, . . . }.
Saitoh [3] introduced a linear operator LP (a, c), which is defined by
(1.4)
Lp (a, c)f (z) = φp (a, c, ; z) ∗ f (z),
z ∈ E,
or, equivalently by
(1.5)
p
Lp (a, c)f (z) = z +
∞
X
(a)k
k=1
(c)k
ak+p z k+p ,
z ∈ E,
where f (z) ∈ Ap , a ∈ R, c ∈ R \ z0− .
For f (z) ∈ A(p, n) and δ ∈ R \ (−∞, −p], we obtain
(1.6)
Lp (δ + p, 1)f (z) = Dδ+p−1 f (z),
which can easily be verified by comparing the definitions (1.3) and (1.5).
The main object of this paper is to present an extension of Theorem A to hold true for a linear
operator LP (a, c) associated with the class A(p, n).
J. Inequal. Pure and Appl. Math., 7(1) Art. 6, 2006
http://jipam.vu.edu.au/
S OME I NEQUALITIES A SSOCIATED WITH A L INEAR O PERATOR D EFINED FOR
A
C LASS OF A NALYTIC F UNCTIONS
3
The basic tool in proving our result is the following lemma.
Lemma 1.1 (cf. Miller and Mocanu [2, p. 35, Theorem 2.3 i(i)]). Let Ω be a set in the complex
plane C. Suppose that the function Ψ : C 2 × E −→ C satisfies the condition Ψ(ix2 , y1 ; z) ∈
/Ω
for all z ∈ E and for all real x2 and y1 such that
1
(1.7)
y1 ≤ − n(1 + x22 ).
2
n
If p(z) = 1 + cn z + · · · is analytic in E and for z ∈ E, Ψ(p(z), zp0 (z); z) ⊂ Ω, then
Re(p(z)) > 0 in E.
2. M AIN R ESULTS
Theorem 2.1. Let α be a complex number satisfying Re(α) > 0 and ρ < 1. Let a > 0, f, g ∈
A(p, n) and
Lp (a, c)g(z)
(2.1)
Re α
> γ,
0 ≤ γ < Re(α), z ∈ E.
Lp (a + 1, c)g(z)
Then
Re
Lp (a, c)f (z)
Lp (a, c)g(z)
>
2aρ + nγ
,
2a + nγ
z ∈ E,
whenever
(2.2)
Lp (a, c)f (z)
Lp (a + 1, c)f (z)
Re (1 − α)
+α
> ρ,
Lp (a, c)g(z)
Lp (a + 1, c)g(z)
z ∈ E.
Proof. Let τ = (2aρ + nγ)/(2a + nγ) and define the function p(z) by
Lp (a, c)f (z)
−1
(2.3)
p(z) = (1 − τ )
−τ .
Lp (a, c)g(z)
Then, clearly, p(z) = 1 + cn z n + cn+1 z n+1 + · · · and is analytic in E. We set u(z) =
αLp (a, c)g(z)/Lp (a + 1, c)g(z) and observe from (2.1) that Re(u(z)) > γ, z ∈ E. Making
use of the familiar identity
z(Lp (a, c)f (z))0 = aLp (a + 1, c)f (z) − (a − p)Lp (a, c)f (z),
we find from (2.3) that
(2.4)
Lp (a, c)f (z)
Lp (a + 1, c)f (z)
u(z) 0
(1 − α)
+α
= τ + (1 − τ ) p(z) +
zp (z) .
Lp (a, c)g(z)
Lp (a + 1, c)g(z)
a
If we define Ψ(x, y; z) by
(2.5)
u(z)
Ψ(x, y; z) = τ + (1 − τ ) x +
y ,
a
then, we obtain from (2.2) and (2.4) that
{Ψ(p(z), zp0 (z); z) : |z| < 1} ⊂ Ω = {w ∈ C : Re(w) > ρ}.
Now for all z ∈ E and for all real x2 and y1 constrained by the inequality (1.7), we find from
(2.5) that
(1 − τ )
Re{Ψ(ix2 , y1 ; z)} = τ +
y1 Re(u(z))
a
(1 − τ )nγ
≤τ−
≡ ρ.
2a
J. Inequal. Pure and Appl. Math., 7(1) Art. 6, 2006
http://jipam.vu.edu.au/
4
S. R. S WAMY
Hence Ψ(ix2 , y1 ; z) ∈
/ Ω. Thus by Lemma 1.1, Re(p(z)) > 0 and hence Re
in E. This proves our theorem.
n
Lp (a,c)f (z)
Lp (a,c)g(z)
o
>τ
Remark 2.2. Theorem A is a special case of Theorem 2.1 obtained by taking a = δ + p and
c = n = 1, which reduces to Theorem 2.1 of [1], when p = 1.
Corollary 2.3. Let α be a real number with α ≥ 1 and ρ < 1. Let a > 0, f, g ∈ A(p, n) and
Lp (a, c)g(z)
Re
> γ, 0 ≤ γ < 1, z ∈ E.
Lp (a + 1, c)g(z)
Then
α(2aρ + nγ) − (1 − ρ)nγ
,
α(2a + nγ)
z ∈ E,
Lp (a + 1, c)f (z)
Lp (a, c)f (z)
Re (1 − α)
+α
> ρ,
Lp (a, c)g(z)
Lp (a + 1, c)g(z)
z ∈ E.
Re
Lp (a + 1, c)f (z)
Lp (a + 1, c)g(z)
>
whenever
Proof. Proof follows from Theorem 2.1 (Since α ≥ 1).
In its special case when α = 1, Theorem 2.1 yields:
o
n
p (a,c)g(z)
> γ, 0 ≤ γ < 1, then for
Corollary 2.4. Let a > 0, f, g ∈ A(p, n) and Re LLp (a+1,c)g(z)
ρ < 1,
Lp (a + 1, c)f (z)
Re
> ρ, z ∈ E,
Lp (a + 1, c)g(z)
implies
2aρ + nγ
Lp (a, c)f (z)
Re
>
, z ∈ E.
Lp (a, c)g(z)
2a + nγ
If we set
Lp (a + 1, c)f (z)
1
Lp (a, c)f (z)
v(z) =
−
−1
,
Lp (a + 1, c)g(z)
α
Lp (a, c)g(z)
then for a > 0, α > 0 and ρ = 0, Theorem 2.1 reduces to
Re(v(z)) > 0,
z∈E
implies
(2.6)
Re
Lp (a, c)f (z)
Lp (a, c)g(z)
>
nαγ
,
2a + nαγ
z ∈ E,
whenever Re(Lp (a, c)g(z)/Lp (a + 1, c)g(z)) > γ, 0 ≤ γ < 1. Let α → ∞.
Then (2.6) is equivalent to
Lp (a + 1, c)f (z) Lp (a, c)f (z)
Re
−
> 0 in E
Lp (a + 1, c)g(z)
Lp (a, c)g(z)
implies
Lp (a, c)f (z)
Re
> 1 in E,
Lp (a, c)g(z)
whenever Re(Lp (a, c)g(z)/Lp (a + 1, c)g(z)) > γ, 0 ≤ γ < 1.
In the following theorem we shall extend the above result, the proof of which is similar to
that of Theorem 2.1.
J. Inequal. Pure and Appl. Math., 7(1) Art. 6, 2006
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S OME I NEQUALITIES A SSOCIATED WITH A L INEAR O PERATOR D EFINED FOR
A
C LASS OF A NALYTIC F UNCTIONS
5
o
n
p (a,c)g(z)
> γ, 0 ≤ γ < 1.
Theorem 2.5. Let a > 0, ρ < 1, f, g ∈ A(p, n) and Re Lpl(a+1,c)g(z))
If
nγ(1 − ρ)
Lp (a + 1, c)f (z) Lp (a, c)f (z)
Re
−
>−
,
z ∈ E,
Lp (a + 1, c)g(z)
Lp (a, c)g(z)
2a
then
Re
Lp (a, c)f (z)
Lp (a, c)g(z)
> ρ,
z ∈ E,
and
Re
Lp (a + 1, c)f (z)
Lp (a + 1, c)g(z)
>
ρ(2a + nγ) − nγ
,
2a
z ∈ E.
Using Theorem 2.1 and Theorem 2.5, we can generalize and improve several other interesting
results available in the literature by taking g(z) = z p . We illustrate a few in the following
theorem.
Theorem 2.6. Let a > 0, ρ < 1 and f (z) ∈ A(p, n). Then
(a) for α a complex number satisfying Re(α) > 0, we have
Lp (a, c)f (z)
Lp (a + 1, c)f (z)
Re (1 − α)
+α
> ρ,
zp
zp
z ∈ E,
implies
Re
Lp (a, c)f (z)
zp
>
2aρ + n Re(α)
,
2a + n Re(α)
z ∈ E.
(b) for α real and α ≥ 1, we have
Lp (a + 1, c)f (z)
Lp (a, c)f (z)
Re (1 − α)
+α
> ρ,
zp
zp
in E
implies
Re
Lp (a + 1, c)f (z)
zp
>
(2a + n)ρ + n(α − 1)
2a + nα
in E
(c) for z ∈ E,
Re
Lp (a + 1, c)f (z) Lp (a, c)f (z)
−
zp
zp
>−
n(1 − ρ)
2a
implies
Re
Lp (a + 1, c)f (z)
zp
>
(2a + n)ρ − n
.
2a
Remark 2.7. By taking a = δ + p, c = n = 1 in Theorem 2.6 we obtain Theorem 1.6 of the
author [4], which when p = 1 reduces to Theorem 2.3 of Bhoosnurmath and Swamy [1].
In a manner similar to Theorem 2.1, we can easily prove the following, which when r = 1
reduces to part (a) of Theorem 2.6.
J. Inequal. Pure and Appl. Math., 7(1) Art. 6, 2006
http://jipam.vu.edu.au/
6
S. R. S WAMY
Theorem 2.8. Let a > 0, r > 0, ρ < 1 and f (z) ∈ A(p, n).Then for α a complex number with
Re(α) > 0, we have
r Lp (a, c)f (z)
2aρr + n Re(α)
Re
>
,
z ∈ E,
p
z
2ar + n Re(α)
whenever
(
r
r−1 )
Lp (a + 1, c)f (z)
Lp (a, c)f (z)
Lp (a, c)f (z)
> ρ,
Re (1 − α)
+α
zp
zp
zp
z ∈ E.
R EFERENCES
[1] S.S. BHOOSNARMATH AND S.R. SWAMY, Differential subordination and conformal mappings,
J. Math. Res. Expo., 14(4) (1994), 493–498.
[2] S.S. MILLER AND P.T. MOCANU, Differential Subordinations: Theory and Applications, Series
on Monographs and Text Books in Pure and Applied Mathematics (No. 225), Marcel Dekker, New
York and Besel, 2000.
[3] H. SAITOH, A linear operator and its applications of first order differential subordinations, Math.
Japon., 44 (1996), 31–38.
[4] S.R. SWAMY, Some studies in univalent functions, Ph.D thesis, Karnatak University, Dharwad,
India, 1992, unpublished.
J. Inequal. Pure and Appl. Math., 7(1) Art. 6, 2006
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