Volume 10 (2009), Issue 2, Article 40, 8 pp.
ON A CERTAIN SUBCLASS OF STARLIKE FUNCTIONS WITH NEGATIVE
COEFFICIENTS
A. Y. LASHIN
D EPARTMENT OF M ATHEMATICS
FACULTY OF S CIENCE
M ANSOURA U NIVERSITY
M ANSOURA , 35516, EGYPT.
aylashin@yahoo.com
Received 11 December, 2007; accepted 29 May, 2009
Communicated by S.S. Dragomir
A BSTRACT. We introduce the class H(α, β) of analytic functions with negative coefficients. In
this paper we give some properties of functions in the class H(α, β) and we obtain coefficient
estimates, neighborhood and integral means inequalities for the function f (z) belonging to the
class H(α, β). We also establish some results concerning the partial sums for the function f (z)
belonging to the class H(α, β).
Key words and phrases: Univalent functions, Starlike functions, Integral means, Neighborhoods, Partial sums.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let A denote the class of functions of the form
∞
X
(1.1)
f (z) = z +
ak z k ,
k=2
which are analytic in the unit disc U = {z : |z| < 1}. And let S denote the subclass of A
consisting of univalent functions f (z) in U.
A function f (z) in S is said to be starlike of order α if and only if
0 zf (z)
Re
>α
(z ∈ U ),
f (z)
for some α (0 ≤ α < 1). We denote by S ∗ (α) the class of all functions in S which are starlike
of order α. It is well-known that
S ∗ (α) ⊆ S ∗ (0) ≡ S ∗ .
The author would like to thank the referee of the paper for his helpful suggestions.
368-07
2
A. Y. L ASHIN
Further, a function f (z) in S is said to be convex of order α in U if and only if
00
zf (z)
Re 1 + 0
>α
(z ∈ U ),
f (z)
for some α (0 ≤ α < 1). We denote by K(α) the class of all functions in S which are convex
of order α.
The classes S ∗ (α), and K(α) were first introduced by Robertson [8], and later were studied
by Schild [10], MacGregor [4], and Pinchuk [7].
Let T denote the subclass of S whose elements can be expressed in the form:
f (z) = z −
(1.2)
∞
X
ak z k
(ak ≥ 0).
k=2
∗
We denote by T (α) and C(α), respectively, the classes obtained by taking the intersections of
S ∗ (α) and K(α) with T,
T ∗ (α) = S ∗ (α) ∩ T
and
C(α) = K(α) ∩ T.
The classes T ∗ (α) and C(α) were introduced by Silverman [11].
Let H(α, β) denote the class of functions f (z) ∈ A which satisfy the condition
2 00
αz f (z) zf 0 (z)
Re
+
>β
f (z)
f (z)
for some α ≥ 0, 0 ≤ β < 1, f (z)
6= 0 and z ∈ U.
z
The classes H(α, β) and H(α, 0) were introduced and studied by Obraddovic and Joshi [5],
Padmanabhan [6], Li and Owa [2], Xu and Yang [14], Singh and Gupta [13], and others.
Further, we denote by H(α, β) the class obtained by taking intersections of the class H(α, β)
with T , that is
H(α, β) = H(α, β) ∩ T.
We note that
H(0, β) = T ∗ (β)
(Silverman [11]).
2. C OEFFICIENT E STIMATES
Theorem 2.1. A function f (z) ∈ T is in the class H(α, β) if and only if
(2.1)
∞
X
[(k − 1)(αk + 1) + (1 − β)]ak ≤ 1 − β.
k=2
The result is sharp.
Proof. Assume that the inequality (2.1) holds and let |z| < 1. Then we have
2 00
P∞
αz f (z) zf 0 (z)
− k=2 (k − 1)(αk + 1)ak z k−1 P
f (z) + f (z) − 1 = k−1
1− ∞
k=2 ak z
P∞
(k − 1)(αk + 1)ak
P
≤ k=2
≤ 1 − β.
1− ∞
k=2 ak
2 00
0
(z)
lie in the circle centered at w = 1 whose radius
This shows that the values of αzff(z)(z) + zff (z)
is 1 − β. Hence f (z) is in the class H(α, β).
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O N A C ERTAIN S UBCLASS OF S TARLIKE F UNCTIONS
3
To prove the converse, assume that f (z) defined by (1.2) is in the class H(α, β). Then
P
2 00
1− ∞
− 1) + k)]ak z k−1
αz f (z) zf 0 (z)
k=2 [αk(k
P
(2.2)
Re
+
= Re
>β
k−1
f (z)
f (z)
1− ∞
k=2 ak z
2 00
0
(z)
for z ∈ U. Choose values of z on the real axis so that αzff(z)(z) + zff (z)
is real. Upon clearing the
−
denominator in (2.2) and letting z → 1 through real values, we have
!
∞
∞
X
X
[αk(k − 1) + k]ak ,
β 1−
ak ≤ 1 −
k=2
k=2
which obviously is the required result (2.1).
Finally, we note that the assertion (2.1) of Theorem 2.1 is sharp, with the extremal function
being
1−β
(2.3)
f (z) = z −
zk
(k ≥ 2).
[(k − 1)(αk + 1) + (1 − β)]
Corollary 2.2. Let f (z) ∈ T be in the class H(α, β). Then we have
1−β
(2.4)
ak ≤
(k ≥ 2).
[(k − 1)(αk + 1) + (1 − β)]
Equality in (2.4) holds true for the function f (z) given by (2.3).
3. S OME P ROPERTIES OF THE C LASS H(α, β)
Theorem 3.1. Let 0 ≤ α1 < α2 and 0 ≤ β < 1. Then H(α2 , β) ⊂ H(α1 , β).
Proof. It follows from Theorem 2.1. That
∞
∞
X
X
[(k − 1)(α1 k + 1) + (1 − β)]ak <
[(k − 1)(α2 k + 1) + (1 − β)]ak ≤ 1 − β
k=2
k=2
for f (z) ∈ H(α2 , β). Hence f (z) ∈ H(α1 , β).
Corollary 3.2. H(α, β) ⊆ T ∗ (β).
The proof is now immediate because α ≥ 0.
4. N EIGHBORHOOD R ESULTS
Following the earlier investigations of Goodman [1] and Ruscheweyh [9], we define the δ−
neighborhood of function f (z) ∈ T by:
(
)
∞
∞
X
X
Nδ (f ) = g ∈ T : g(z) = z −
bk z k ,
k |ak − bk | ≤ δ .
k=2
k=2
In particular, for the identity function
e(z) = z,
we immediately have
(
(4.1)
Nδ (e) =
g ∈ T : g(z) = z −
∞
X
bk z k ,
k=2
Theorem 4.1. H(α, β) ⊆ Nδ (e), where δ =
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 40, 8 pp.
∞
X
)
k |bk | ≤ δ
.
k=2
2(1−β)
.
(2α+2−β)
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4
A. Y. L ASHIN
Proof. Let f (z) ∈ H(α, β). Then, in view of Theorem 2.1, since [(k − 1)(αk + 1) + (1 − β)]
is an increasing function of k (k ≥ 2), we have
(2α + 2 − β)
∞
X
∞
X
ak ≤
k=2
[(k − 1)(αk + 1) + (1 − β)]ak ≤ 1 − β,
k=2
which immediately yields
∞
X
(4.2)
ak ≤
k=2
1−β
.
(2α + 2 − β)
On the other hand, we also find from (2.1)
∞
∞
∞
X
X
X
(α + 1)
kak − β
ak ≤
[(α(k − 1) + 1)k − β)]ak
k=2
k=2
k=2
=
(4.3)
∞
X
[(k − 1)(αk + 1) + (1 − β)]ak ≤ 1 − β.
k=2
From (4.3) and (4.2), we have
(α + 1)
∞
X
kak ≤ (1 − β) + β
k=2
∞
X
ak
k=2
1−β
(2α + 2 − β)
2(α + 1)(1 − β)
≤
,
(2α + 2 − β)
≤ (1 − β) + β
that is,
∞
X
(4.4)
kak ≤
k=2
2 (1 − β)
= δ,
(2α + 2 − β)
which in view of the definition (4.1), prove Theorem 4.1.
Letting α = 0, in the above theorem, we have:
Corollary 4.2. T ∗ (β) ⊆ Nδ (e), where δ =
2(1−β)
.
(2−β)
5. I NTEGRAL M EANS I NEQUALITIES
We need the following lemma.
Lemma 5.1 ([3]). If f and g are analytic in U with f ≺ g, then
Z 2π
Z 2π
g(reiθ )δ dθ ≤
f (reiθ )δ dθ,
0
0
iθ
where δ > 0, z = re and 0 < r < 1.
Applying Lemma 5.1, and (2.1), we prove the following theorem.
Theorem 5.2. Let δ > 0. If f (z) ∈ H(α, β), then for z = reiθ , 0 < r < 1, we have
Z 2π
Z 2π
f (reiθ )δ dθ ≤
f2 (reiθ )δ dθ,
0
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0
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O N A C ERTAIN S UBCLASS OF S TARLIKE F UNCTIONS
5
where
f2 (z) = z −
(5.1)
(1 − β)
z2.
(2α + 2 − β)
Proof. Let f (z) defined by (1.2) and f2 (z) be given by (5.1). We must show that
δ
δ
Z 2π Z 2π ∞
X
(1
−
β)
k−1 1 −
dθ.
ak z dθ ≤
z
1 −
(2α
+
2
−
β)
0
0
k=2
By Lemma 5.1, it suffices to show that
∞
X
1−
ak z k−1 ≺ 1 −
k=2
(1 − β)
z.
(2α + 2 − β)
Setting
(5.2)
1−
∞
X
ak z k−1 = 1 −
k=2
(1 − β)
w(z).
(2α + 2 − β)
From (5.2) and (2.1), we obtain
∞
X (2α + 2 − β)
|w(z)| = ak z k−1 (1
−
β)
k=2
≤ |z|
∞
X
[(k − 1)(αk + 1) + (1 − β)]
1−β
k=2
ak ≤ |z| .
This completes the proof of the theorem.
Letting α = 0 in the above theorem, we have:
Corollary 5.3. Let δ > 0. If f (z) ∈ T ∗ (β), then for z = reiθ , 0 < r < 1, we have
Z 2π
Z 2π
δ
iθ
f2 (reiθ )δ dθ,
f (re ) dθ ≤
0
0
where
f2 (z) = z −
(1 − β) 2
z .
(2 − β)
6. PARTIAL S UMS
In this section we will examine the ratio of a function
Pnof the kform (1.2) to its sequence of
partial sums defined by f1 (z) = z and fn (z) = z − k=2 ak z when the coefficients of f
are sufficiently
small
to
satisfy
0thecondition(2.1).
We will determine sharp lower bounds for
f (z)
fn (z)
f (z)
fn0 (z)
Re fn (z) , Re f (z) , Re f 0 (z) and Re f 0 (z) .
n
In what follows, we will use the well known result that
1 − w(z)
Re
> 0,
z ∈ U,
1 + w(z)
if and only if
w(z) =
∞
X
ck z k
k=1
satisfies the inequality |w(z)| ≤ |z| .
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 40, 8 pp.
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6
A. Y. L ASHIN
Theorem 6.1. If f (z) ∈ H(α, β), then
(6.1)
Re
f (z)
1
≥1−
fn (z)
cn+1
(z ∈ U, n ∈ N )
and
(6.2)
where ck =:
Re
fn (z)
f (z)
[(k−1)(αk+1)+(1−β)]
1−β
≥
cn+1
1 + cn+1
(z ∈ U, n ∈ N ) ,
. The estimates in (6.1) and (6.2) are sharp.
Proof. We employ
P∞ the same technique used by Silverman [12]. The function f (z) ∈ H(α, β),
if and only if k=2 ck ak ≤ 1. It is easy to verify that ck+1 > ck > 1. Thus,
n
X
(6.3)
ak + cn+1
k=2
∞
X
ak ≤
k=n+1
∞
X
ck ak ≤ 1.
k=2
We may write
P
P
k−1
1 − nk=2 ak z k−1 − cn+1 ∞
f (z)
1
1 + D(z)
k=n+1 ak z
P
− 1−
=
=
.
cn+1
n
k−1
fn (z)
cn+1
1 − k=2 ak z
1 + E(z)
Set
1 + D(z)
1 − w(z)
=
,
1 + E(z)
1 + w(z)
so that
w(z) =
Then
and
E(z) − D(z)
.
2 + D(z) + E(z)
P
k−1
cn+1 ∞
k=n+1 ak z
Pn
P
w(z) =
k−1
2 − 2 k=2 ak z k−1 − cn+1 ∞
k=n+1 ak z
P
cn+1 ∞
k=n+1 ak
Pn
P
|w(z)| ≤
.
2 − 2 k=2 ak − cn+1 ∞
k=n+1 ak
Now |w(z)| ≤ 1 if and only if
n
X
ak + cn+1
k=2
∞
X
ak ≤ 1,
k=n+1
which is true by (6.3). This readily yields the assertion (6.1) of Theorem 6.1.
To see that
z n+1
(6.4)
f (z) = z −
cn+1
gives sharp results, we observe that
f (z)
zn
=1−
.
fn (z)
cn+1
Letting z → 1− , we have
f (z)
1
=1−
,
fn (z)
cn+1
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O N A C ERTAIN S UBCLASS OF S TARLIKE F UNCTIONS
7
which shows that the bounds in (6.1) are the best possible for each n ∈ N. Similarly, we take
P
P
k−1
1 − nk=2 ak z k−1 + cn+1 ∞
fn (z)
1 − w(z)
cn+1
k=n+1 ak z
P∞
(1 + cn+1 )
−
=
:=
,
k−1
f (z)
1 + cn+1
1 − k=2 ak z
1 + w(z)
where
P
(1 + cn+1 ) ∞
k=n+1 ak
Pn
P
|w(z)| ≤
.
2 − 2 k=2 ak + (1 − cn+1 ) ∞
k=n+1 ak
Now |w(z)| ≤ 1 if and only if
n
X
∞
X
ak + cn+1
k=2
ak ≤ 1,
k=n+1
which is true by (6.3). This immediately leads to the assertion (6.2) of Theorem 6.1.
The estimate in (6.2) is sharp with the extremal function f (z) given by (6.4). This completes
the proof of Theorem 6.1.
Letting α = 0 in the above theorem, we have:
Corollary 6.2. If f (z) ∈ T ∗ (β), then
Re
f (z)
n
≥
,
fn (z)
(n + 1 − β)
(z ∈ U )
and
fn (z)
n+1−β
≥
,
(z ∈ U ) .
f (z)
(n + 2 − 2β)
The result is sharp for every n, with the extremal function
Re
f (z) = z −
(6.5)
1−β
z n+1 .
(n + 1 − β)
We now turn to the ratios involving derivatives. The proof of Theorem 6.3 below follows the
pattern of that in Theorem 6.1, and so the details may be omitted.
Theorem 6.3. If f (z) ∈ H(α, β), then
Re
(6.6)
f 0 (z)
n+1
≥
1
−
fn0 (z)
cn+1
(z ∈ U ) ,
and
(6.7)
Re
fn0 (z)
f 0 (z)
≥
cn+1
n + 1 + cn+1
(z ∈ U, n ∈ N ) .
The estimates in (6.6) and (6.7) are sharp with the extremal function given by (6.4).
Letting α = 0 in the above theorem, we have:
Corollary 6.4. If f (z) ∈ T ∗ (β), then
f 0 (z)
βn
Re 0
≥
,
fn (z)
(n + 1 − β)
(z ∈ U ) ,
and
fn0 (z)
n+1−β
≥
,
(z ∈ U ) .
0
f (z)
n + (1 − β)(n + 2)
The result is sharp for every n, with the extremal function given by (6.5).
Re
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 40, 8 pp.
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8
A. Y. L ASHIN
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