Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2008, Article ID 561638, 12 pages
doi:10.1155/2008/561638
Research Article
Subordination and Superordination Results for
a Class of Analytic Multivalent Functions
S. P. Goyal,1 Pranay Goswami,1 and H. Silverman2
1
2
Department of Mathematics, University of Rajasthan, Jaipur 302004, India
Department of Mathematics, College of Charleston, Charleston, SC 29424, USA
Correspondence should be addressed to H. Silverman, silvermanh@cofc.edu
Received 25 September 2007; Revised 19 February 2008; Accepted 21 May 2008
Recommended by Dorothy Wallace
We derive subordination and superordination results for a family of normalized analytic functions
in the open unit disk defined by integral operators. We apply this to obtain sandwich results and
generalizations of some known results.
Copyright q 2008 S. P. Goyal et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
1. Introduction
Let H denote the class of analytic functions in the unit disk Δ : {z : |z| < 1}, and let Ha, p
be the subclass of H of the form
fz a ap zp ap1 zp1 · · · ,
p ∈ N {1, 2, 3, . . .}.
1.1
Let Ap be the subclass of H of the form
fz zp ∞
ak zk ,
p ∈ N.
1.2
kp1
If f and g are analytic and there exists a Schwarz function wz, analytic in Δ with
w0 0,
|wz| < 1,
z ∈ Δ,
1.3
such that fz gwz, then the function f is called subordinate to g and is denoted by
f ≺g
or fz ≺ gz,
z ∈ Δ.
1.4
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In particular, if the function g is univalent in Δ, the above subordination is equivalent to
f0 g0,
fΔ ⊂ gΔ.
1.5
Suppose h and k are analytic functions in Δ and φr, s, t; z : C 3 × Δ→C. If h and
φhz, zh z, z2 h z; z are univalent and if h satisfies the second-order superordination
kz ≺ φhz, zh z, z2 h z; z,
1.6
then h is a solution of the differential superordination 1.6. Note that if f is subordinate to
g, then g is superordinate to f. An analytic function q is called subordinant if q ≺ h forall h
satisfying 1.6. A univalent subordinant q that satisfies q ≺ q for all subordinants q of 1.6 is
said to be the best subordinant. Miller and Mocanu 1 have obtained conditions on k, q, and
φ for which the following implication holds:
kz ≺ φhz, zh z, z2 h z; z ⇒ qz ≺ hz.
1.7
Ali et al. 2 have obtained sufficient conditions for certain normalized analytic functions
fz to satisfy
q1 z ≺
zf z
≺ q2 z,
fz
1.8
where q1 and q2 are given univalent functions in Δ with q1 0 1 and q2 0 1.
Recently, Shanmugam et al. 3, 4 have also obtained sandwich results for certain classes
of analytic functions. Further subordination results can be found in 5–8.
2. Definitions and Preliminaries
Definition 2.1. For fz ∈ Ap, Shams et al. 9 defined the following integral operator:
z
σ−1
z
ftdt
t
0
∞ p1 σ
zp an zn , σ > 0.
n
1
np1
Iσ fz p 1σ
zΓσ
log
2.1
2.2
For the operator, one easily gets
zIσ fz p 1Iσ−1 fz − Iσ fz.
2.3
Also for −1 ≤ B < A ≤ 1 and λ ≥ 0, Shams et al. 9 defined a class Ωσp A, B; λ of functions
fz ∈ Ap, so that
p − λ Iσ fz
λ Iσ−1 fz
1 Az
≺
.
p
zp
p
zp
1 Bz
2.4
The family Ωσp A; B; λ is a general family containing various new and known classes of
analytic functions see, e.g., 10, 11.
S. P. Goyal et al.
3
Definition 2.2 see 1. Denote by Q the set of all functions fz that are analytic and injective
on Δ − Ef, where
Ef ζ ∈ ∂Δ : limfz ∞ ,
2.5
z→ζ
0 for ζ ∈ ∂Δ − Ef.
and are such that f ζ /
We will require certain results due to Miller and Mocanu 1, 12, Bulboacă 13, and
Shanmugam et al. 4 contained in the following lemmas.
Lemma 2.3 see 12. Let qz be univalent in the unit disk Δ, and let θ and φ be analytic in the
domain D containing qΔ with φw /
0 when w ∈ qΔ. Set Qz zq zφqz, hz θqz Qz. Suppose that
i Qz is starlike univalent in Δ;
ii Rezh z/Qz > 0 for z ∈ Δ.
If pz is analytic in Δ, with p0 q0, pΔ ∈ D, and
θpz zp zφpz ≺ θqz zq zφqz,
2.6
then pz ≺ qz and qz is the best dominant.
Lemma 2.4 see 4. Let qz be a convex univalent function in Δ and ψ, γ ∈ C with Re1 zq z/q z > max{0, −Reψ/γ}. If pz is analytic in Δ and
ψpz γzp z ≺ ψqz γzq z,
2.7
then pz ≺ qz and qz is the best dominant.
Lemma 2.5 see 12. Let qz be univalent in Δ, and let φz be analytic in a domain containing
qΔ. If zq z/ϕqz is starlike and
zp zϕpz ≺ zq zϕqz,
2.8
then pz ≺ qz and qz is the best dominant.
Lemma 2.6 see 13. Let qz be convex univalent in the unit disk Δ, and let ϑ and ϕ be analytic in
a domain D containing qΔ. Suppose that
i Reϑ qz/ϕqz > 0 for z ∈ Δ;
ii zq zϕqz is starlike univalent in z ∈ Δ.
If pz ∈ Hq0, 1 ∩ Q, with pΔ ⊆ D, and if ϑpz zp zφpz is univalent in Δ and
ϑqz zq zϕqz ≺ ϑpz zp zϕpz,
2.9
then qz ≺ pz and qz is the best subordinant.
Lemma 2.7 see 1. Let qz be convex univalent in Δ and γ ∈ C. Further assume that Reγ > 0.
If pz ∈ Hq0, 1 ∩ Q and pz γzp z is univalent in Δ, then
qz γzq z ≺ pz γzp z,
which implies that qz ≺ pz and qz is the best subordinant.
2.10
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The main object of this paper is to apply a method based on the differential subordination
in order to derive several subordination results.
3. Subordination for analytic functions
Theorem 3.1. Let qz be univalent in the unit disk Δ, λ ∈ C, and
zq z
pp 1
Re 1 > max 0, − Re
,
q z
λ
λ/
0 p ∈ N.
3.1
If fz ∈ Ap satisfies the subordination
p − λ Iσ fz
λzq z
λ Iσ−1 fz
≺
qz
,
p
zp
p
zp
pp 1
3.2
where Iσ fz is defined by 2.1, then
Iσ fz
zp
≺ qz
3.3
Iσ fz
.
zp
3.4
and qz is the best dominant.
Proof. Consider
hz :
Differentiating 3.4 with respect to z logarithmically, we get
zh z zIσ fz
− p.
σ
hz
I fz
3.5
Now, in view of 2.3, we obtain from 3.5 the following subordination:
hz λzq z
λzh z
≺ qz .
pp 1
pp 1
3.6
An application of Lemma 2.4, with γ λ/pp 1 and ψ 1, leads to 3.3.
Taking qz 1 Az/1 Bz in Theorem 3.1, we arrive at the following.
Corollary 3.2. Let −1 ≤ B < A ≤ 1 and Re1 − Bz/1 Bz > max{0, −Repp 1/λ}λ / 0, p ∈ N. If f ∈ Ap and
p − λ Iσ fz
λ
A − Bz
1 Az
λ Iσ−1 fz
≺
,
3.7
p
p
p
z
p
z
1 Bz pp 1 1 Bz2
then
Iσ fz 1 Az
≺
zp
1 Bz
and 1 Az/1 Bz is the best dominant.
3.8
S. P. Goyal et al.
5
Putting p 1 and qz 1 z/1 − z in Theorem 3.1, we get the following corollary.
Corollary 3.3. Let Re1 z/1 − z > max{0, − Re2/λ} and λ / 0. If f ∈ A1 and
λIσ−1 fz 1 − λIσ fz 1 z
λz
≺
,
z
z
1 − z 1 − z2
3.9
Iσ fz 1 z
≺
z
1−z
3.10
then
and 1 z/1 − z is the best dominant.
Theorem 3.4. Let qz be univalent in Δ and 0 /
γ, μ ∈ C, and α, β ∈ C such that α β /
0. Let
f ∈ Ap and suppose that q satisfies
zq z zq z
Re 1 −
> 0.
3.11
q z
qz
If
1 γμ
αzIσ−1 fz βzIσ fz
zq z
−
p
≺1γ
,
σ−1
σ
qz
αI fz βI fz
3.12
then
αIσ−1 fz βIσ fz
α βzp
μ
≺ qz,
3.13
μ/
0, α β /
0.
3.14
and qz is the best dominant.
Proof. Let us consider a function hz defined by
hz :
αIσ−1 fz βIσ fz
α βzp
μ
,
Now, differentiating 3.14 logarithmically, we get
αzIσ−1 fz βzIσ fz
zh z
μ
−p .
hz
αIσ−1 fz βIσ fz
3.15
By setting
θw 1,
φw γ
,
w
3.16
it can be easily observed that θw is analytic in C and that φw /
0 is analytic in C/{0}. Also,
we let
Qz zq zφqz γ
zq z
,
qz
pz θqz Qz 1 γ
zq z
.
qz
3.17
3.18
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International Journal of Mathematics and Mathematical Sciences
From 3.11 we see that Qz is starlike univalent in the unit disk Δ, and from 3.18 we get
zq z zq z
zp z
Re 1 > 0.
3.19
Re
−
Qz
q z
qz
An application of Lemma 2.3 to 3.12 yields the result.
Putting α 0, β 1, γ 1, and qz 1 Az/1 Bz in Theorem 3.4, we obtain the
following corollary.
Corollary 3.5. If fz ∈ Ap and for −1 ≤ A < B ≤ 1, μ / 0,
zIσ fz
A − Bz
1μ
−
p
≺1
,
Iσ fz
1 Az1 Bz
then
Iσ fz
zp
μ
≺
1 Az
1 Bz
3.20
3.21
and 1 Az/1 Bz is the best dominant.
By setting α 0, β 1, γ 1, σ 0, p 1, and qz 1 BzμA−B/B inTheorem 3.4, we
get the following corollary.
Corollary 3.6. Suppose fz ∈ A1 and let −1 ≤ B < A ≤ 1 and B /
0. If
zf z
1 Az
1μ
−1 ≺
,
fz
1 Bz
then
fz
z
3.22
μ
≺ 1 BzμA−B/B
3.23
and 1 BzμA−B/B is the best dominant.
Remark 3.7. qz 1 BzμA−B/B is univalent if and only if |μA − B/B − 1| ≤ 1 or |μA −
B/B 1| ≤ 1 see 5.
Again by setting β 1, μ 1, α 0, γ 1/b, p 1, and σ 0, and by qz 1/1 −
z2b b ∈ C \ {0} in Theorm 3.4, we get the following corollary.
Corollary 3.8. Suppose fz ∈ A1 and b is a nonzero complex number for which
1z
1 zf z
−1 ≺
.
1
b fz
1−z
3.24
Then,
fz
1
≺
z
1 − z2b
and 1/1 − z2b is the best dominant.
3.25
S. P. Goyal et al.
7
The result contained in Corollary 3.8 was earlier given by Srivastava and Lashin 7.
Theorem 3.9. Let q be univalent in the unit disk Δ, and let μ, γ / 0, η, δ, α, β ∈ C, and fz ∈ Ap.
Suppose that q satisfies
zq z
η
Re 1 > max 0, −Re
.
3.26
q z
γ
Let
ψz αIσ−1 fz βIσ fz
α βzp
μ αzIσ−1 fz βzIσ fz
η γμ
−
p
δ.
αIσ−1 fz βIσ fz
3.27
ψz ≺ ηqz δ γzq z,
3.28
If
then
αIσ−1 fz βIσ fz
α βzp
μ
≺ qz,
αβ
/ 0,
3.29
and qz is the best dominant.
Proof. Define a function hz by
hz :
αIσ−1 fz βIσ fz
α βzp
μ
.
3.30
Then, a computation shows that
and hence
αzIσ−1 fz βzIσ fz
zh z
μ
−
p
hz
αIσ−1 fz βIσ fz
3.31
zαIσ−1 fz zβIσ fz
−
p
.
zh z μhz
αIσ−1 fz βIσ fz
3.32
Set
θw ηw δ,
φw γ,
3.33
and let
Qz zq zφqz γzq z,
pz θqz Qz ηqz δ γzq z.
From 3.26, we see that Qz is starlike in Δ and that
zp z
η
zq z
Re
Re
1 > 0,
Qz
γ
q z
3.34
3.35
by the hypothesis 3.26 of Theorem 3.9. Thus, applying Lemma 2.3, the proof of Theorem 3.9
is completed.
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International Journal of Mathematics and Mathematical Sciences
By setting β 1, γ 1, α 0, and qz 1 Az/1 Bz, we obtain the following
corollary.
Corollary 3.10. Let fz ∈ Ap and Reη > 0. Suppose that
1 − Bz
Re
> max{0, −Reη}.
1 Bz
3.36
If
Iσ fz
zp
μ zIσ fz
1 Az
A − B
−
p
δ ≺η
δz
ημ
,
Iσ fz
1 Bz
1 Bz2
then
Iσ fz
zp
μ
≺
1 Az
1 Bz
3.37
3.38
and 1 Az/1 Bz is the best dominant.
Again by setting β 1, γ 1, α 0, p 1, and σ 0, and by qz 1 z/1 − z, we
get the following corollary.
Corollary 3.11. Let fz ∈ A1 and
zf z
fz μ
1z
2z
ημ
,
−1
δ ≺η
δ
z
fz
1−z
1 − z2
3.39
then
fz μ 1 z
≺
z
1−z
3.40
and 1 z/1 − z is the best dominant.
4. Superordination for analytic functions
Theorem 4.1. Let q be convex univalent in the unit disk Δ, and λ ∈ C. Suppose λ satisfies Re{λ} > 0
and Iσ fz/zp ∈ Hq0, 1 ∩ Q. Suppose that
p − λ Iσ fz
λ Iσ−1 fz
p
zp
p
zp
4.1
is univalent in the unit disk Δ. If
qz λzq z
p − λ Iσ fz
λ Iσ−1 fz
,
≺
pp 1 p
zp
p
zp
4.2
Iσ fz
zp
4.3
then
qz ≺
and qz is the best subordinant.
S. P. Goyal et al.
9
Proof. Let
pz Iσ fz
,
zp
z
/ 0.
4.4
Differentiating logarithmically, we get
zp z zIσ fz
−p.
pz
Iσ fz
4.5
After some computation, we get
pz p − λ Iσ fz
λzp z λ Iσ−1 fz
.
pp 1 p
zp
p
zp
4.6
Now, using Lemma 2.7, we get the desired result 4.3.
Corollary 4.2. Let q be convex univalent in Δ, and λ ∈ C. Suppose λ satisfies R{λ} > 0 and
Iσ fz/zp ∈ Hq0, 1 ∩ Q. Let
p − λ Iσ fz
λ Iσ−1 fz
4.7
p
zp
p
zp
be univalent in the unit disk Δ. If
p − λ Iσ fz
1 Az λ Iσ−1 fz
,
≺
zp
p
zp
pp 11 Bz2 1 Bz p
4.8
1 Az Iσ fz
≺
1 Bz
zp
4.9
λA − Bz
then
and 1 Az/1 Bz is the best subordinant.
Since the proofs of Theorems 4.3 and 4.4 are similar to the proofs of the previous
theorems, we only give statements of these theorems without proofs.
Theorem 4.3. Let qz be convex univalent in Δ, and 0 /
γ, μ ∈ C, and α, β ∈ C such that α β /
0.
μ
Let fz ∈ Ap. Suppose that αIσ−1 fz βIσ fz/α βzp ∈ Hq0, 1 ∩ Q, and
1 γμ
αzIσ−1 fz βzIσ fz
−p
αIσ−1 fz βIσ fz
4.10
is univalent in Δ. If
1γ
αzIσ−1 fz βIσ fz
zq z
−
p
,
≺ 1 γμ
qz
αIσ−1 fz βIσ fz
then
qz ≺
and qz is the best subordinant.
αIσ−1 fz βIσ fz
α βzp
4.11
μ
4.12
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International Journal of Mathematics and Mathematical Sciences
Theorem 4.4. Let q be convex univalent in the unit disk Δ, and let γ /
0 ∈ C, η, δ, α, β ∈ C with
αβ
/ 0, and fz ∈ Ap. Suppose that Iσ fz/zp ∈ Hq0, 1 ∩ Q, and
ηq z
Re
> 0.
4.13
γ
If
αIσ−1 fz βIσ fz
ηqz δ γzq z ≺
α βzp
then
qz ≺
μ zαIσ−1 fz zβIσ fz
ηγμ
−p
δ,
αIσ−1 fz βIσ fz
4.14
αIσ−1 fz βIσ fz
α βzp
μ
4.15
and qz is the best subordinant.
5. Sandwich results
Combining results of differential subordinations and superordinations, we arrive at the
following “sandwich results.”
Theorem 5.1. Let q1 z be convex univalent, and let q2 z be univalent in Δ, and λ ∈ C. Suppose q1
satisfies Re{λ} > 0 and q2 satisfies 3.1. If Iσ fz/zp ∈ Hq0, 1 ∩ Q and
σ−1
μ
αI fz βIσ fz
, αβ/
0,
5.1
α βzp
is univalent in Δ, and if
q1 z λzq1 z λIσ−1 fz p − λIσ fz
λzq2 z
≺
q
z
≺
,
2
pp 1
zp
zp
pp 1
then
q1 z ≺
Iσ fz
zp
5.2
≺ q2 z
5.3
and q1 z and q2 z are, respectively, the best subordinant and the best dominant.
Theorem 5.2. Let q1 z be convex univalent, and let q2 z be univalent in Δ, and λ ∈ C. Suppose that
q2 satisfies 3.11. Further suppose that αIσ−1 fz βIσ fz/α βzp μ ∈ Hq0, 1 ∩ Q and
1 γμαzIσ−1 fz βzIσ fz /αIσ−1 fz βIσ fz − p is univalent in Δ.
If
zq z
zq2 z
αzIσ−1 fz βzIσ fz
1γ 1
−
p
≺
1
γ
≺ 1 γμ
,
5.4
q1 z
q2 z
αIσ−1 fz βIσ fz
then
q1 z ≺
αIσ−1 fz βIσ fz
α βzp
μ
≺ q2 z,
αβ
/ 0,
and q1 z and q2 z are, respectively, the best subordinant and the best dominant.
5.5
S. P. Goyal et al.
11
Theorem 5.3. Let q1 z be convex univalent, and let q2 z be univalent in Δ, and λ ∈ C.
Suppose that q1 z satisfies 4.13 and q2 z satisfies 3.28. Further suppose that αIσ−1 fz βIσ fz/α βzp μ ∈ Hq0, 1 ∩ Q with α β /
0, and that
αIσ−1 fz βIσ fz
α βzp
μ η γμ
zαIσ−1 fz zβIσ fz
−p
αIσ−1 fz βIσ fz
δ
5.6
is univalent in Δ. If
ηq1 z δ γzq1 z ≺
αIσ−1 fz βIσ fz
α βzp
μ η γμ
zαIσ−1 fz zβIσ fz
−
p
δ
αIσ−1 fz βIσ fz
≺ ηq2 z δ γzq2 z,
5.7
then
q1 z ≺
αIσ−1 fz βIσ fz
α βzp
μ
≺ q2 z
5.8
and q1 z and q2 z are, respectively, the best subordinant and the best dominant.
Acknowledgments
The authors are thankful to the referee for his useful suggestions. P. Goswami is also thankful
to CSIR, India, for providing the Junior Research Fellowship under Research Scheme no.
09/1350434/2006-EMR-1.
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