Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 4, Article 140, 2006
PARTIAL SUMS OF SOME MEROMORPHIC FUNCTIONS
S. LATHA AND L. SHIVARUDRAPPA
D EPARTMENT OF M ATHEMATICS
M AHARAJA’ S C OLLEGE
U NIVERSITY OF M YSORE
M YSORE - 570005, INDIA.
drlatha@gmail.com
D EPARTMENT OF M ATHEMATICS , S.J.C.E
V ISWESHWARAYAH T ECHNOLOGICAL U NIVERSITY
B ELGAM , INDIA.
shivarudrappa@lycos.com
Received 05 July, 2005; accepted 18 October, 2006
Communicated by G. Kohr
A BSTRACT. In the present paper we give some results concerning partial sums of certain meromorphic functions.We also consider the partial sums of certain integral operator.
Key words and phrases: Partial sums, Meromorphic functions, Integral operators, Meromorphic starlike functions, Meromorphic convex functions, Meromorphic close to convex functions.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let Σ be the class consisting of functions of the form
∞
(1.1)
f (z) =
1 X
+
ak z k
z k=1
which are regular in the punctured disc E = {z : 0 < |z| < 1} with a simple pole at the origin
and residue 1 there. P
Let fn (z) = z1 + nk=1 ak z k be the nth partial sum of the series expansion for f (z) ∈ Σ.
Let Σ∗ (A, B), ΣK (A, B), Σc (A, B), −1 ≤ A < B ≤ 1 be the subclasses of functions in Σ
satisfying
0 zf (z)
1 + Az
(1.2)
−
≺
,
z ∈ U = E ∪ {0}
f (z)
1 + Bz
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
203-05
2
S. L ATHA AND L. S HIVARUDRAPPA
−
(1.3)
zf 00 (z)
1 + Az
+1 ≺
,
0
f (z)
1 + Bz
−z 2 f 0 (z) ≺
(1.4)
1 + Az
,
1 + Bz
z ∈ U.
z∈U
respectively [5]. The classes Σ∗ (2α − 1, 1) and ΣK (2α − 1, 1) are respectively the well known
subclasses of Σ consisting of functions meromorphic starlike of order α and meromorphic convex of order α and meromorphically close to convex of order α denoted by Σ∗ (α), ΣK (α) and
Σc (α) respectively.P
P∞
1
k
k
If f (z) = z1 + ∞
k=1 ak z and g(z) = z +
k=1 bk z , then their Hadamard product (or
convolution), denoted by f (z) ∗ g(z) is the function defined by the power series
∞
1 X
f (z) ∗ g(z) = +
ak b k z k .
z k=1
In the present paper, we give sufficient conditions for f (z) to be in Σ∗ (A, B), ΣK (A, B) and
further investigate the ratio of a function of the form (1.1) to its sequence of partial sums when
the coefficients are sufficiently small to satisfy conditions
∞
X
k{k(1 + B) + (1 + A)}|ak | ≤ B − A,
k=1
∞
X
{k(1 + B) + (1 + A)}|ak | ≤ B − A.
k=1
n
o
n
o
n 0 o
fn (z)
f (z)
More precisely, we will determine sharp lower bounds for < ffn(z)
,
<
,
<
(z)
f (z)
fn0 (z)
n 0 o
(z)
. Further, we give a property for the partial sums of certain integral operators in
and < ffn0 (z)
connection with functions belonging to the class Σc (A, B).
Theorem 2.1. Let f (z) =
(2.1)
∞
X
1
z
2. S OME P RELIMINARY R ESULTS
P
k
+ ∞
z ∈ E. If
k=1 ak z ,
k{k(1 + B) + (1 + A)}|ak | ≤ B − A,
then
f (z) ∈ ΣK (A, B).
k=1
Proof. It suffices to show that
1 + zf 00 (z) + 1 f (z)
< 1,
zf 00 (z)
B 1 + f 0 (z) + A that is,
zf 00 (z) + 2f 0 (z)
Bzf 00 (z) + (A + B)f 0 (z) < 1.
J. Inequal. Pure and Appl. Math., 7(4) Art. 140, 2006
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PARTIAL S UMS
3
Consider
(2.2)
00
0
zf
(z)
+
2f
(z)
Bzf 00 (z) + (A + B)f 0 (z) P∞
k(k + 1)ak z k+1
k=1
P∞
P
=
∞
(B − A) + B k=1 k(k + 1)ak z k+1 − (2B − A) k=1 kak z k+1 Σk(k + 1)|ak |
P
≤
.
(B − A) − ∞
k=1 k(kB + A)|ak |
(2.2) is bounded by 1 if
∞
X
k(k + 1)|ak | ≤ (B − A)
k=1
∞
X
k(kB + A)|ak |
k=1
which reduces to (2.1)
Similarly we can prove the following theorem.
P
k
Theorem 2.2. Let f (z) = z1 + ∞
z ∈ E. If
k=1 ak z ,
(2.3)
∞
X
{k(1 + B) + (1 + A)}|ak | ≤ B − A,
then
f (z) ∈ Σ∗ (A, B).
k=1
3. M AIN R ESULTS
Theorem 3.1. If f (z) of the form (1.1) satisfies (2.3), then
2(n + 1 + A)
f (z)
<
≥
,
fn (z)
2n + 2 + A + B
z ∈ U.
The result is sharp for every n, with extremal function
f (z) =
(3.1)
1
B−A
+
z n+1 ,
z 2n + 2 + A + B
n ≥ 0.
Proof. Consider
f (z)
2n + A + B
−
fn (z) 2n + 2 + A + B
P∞
P
k+1
1 + nk=1 ak z k+1 + 2n+2+A+B
k=n+1 ak z
B−A
P
=
1 + nk=1 ak z k+1
1 + w(z)
=
,
1 − w(z)
2n + 2 + A + B
B−A
where
w(z) =
2+
P∞
2n+2+A+B
k+1
k=n+1 ak z
B−A
P∞
P
2 nk=1 ak z k+1 − 2n+2+A+B
k=n+1
B−A
and
∞
2n+2+A+B
k=1 |ak |
B−A
P∞
Pn
2n+2+A+B
2 k=1 |ak | − B−A
k=n+1
ak z k+1
P
|w(z)| ≤
2−
|ak |
.
Now
|w(z)| ≤ 1
J. Inequal. Pure and Appl. Math., 7(4) Art. 140, 2006
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4
S. L ATHA AND L. S HIVARUDRAPPA
if and only if
2
2n + 2 + A + B
B−A
X
∞
|ak | ≤ 2 − 2
k=n+1
n
X
|ak |,
k=1
which is equivalent to
n
X
(3.2)
|ak | +
k=1
∞
2n + 2 + A + B X
|ak | ≤ 1.
B−A
k=n+1
It suffices to show that the left hand side of (3.2) bounded above by
∞
X
2k + A + B
k=1
(B − A)
|ak |,
which is equivalent to
n X
2(k + A)
k=1
B−A
∞ X
2(k − n − 1)
|ak | +
|ak | ≥ 0.
B−A
k=n+1
To see that the function f (z) given by (3.1) gives the sharp result, we observe for
πi
z = re n+2
that
f (z)
B−A
B−A
2(n + 1 + A)
=1+
z n+2 → 1 −
=
fn (z)
2n + 2 + A + B
2n + 2 + A + B
2(n + 1) + A + B
when r → 1− .
Therefore we complete the proof of Theorem 3.1.
Corollary 3.2. For A = 2α − 1, B = 1, we get Theorem 2.1 in [3] which states as follows:
If f (z) of the form (1.1) satisfies condition
∞
X
(k + α)|ak | ≤ 1 − α,
1
then
<
f (z)
fn (z)
≥
n + 2α
,
n+1+α
z ∈ U.
The result is sharp for every n, with extremal function
f (z) =
1
1 − α n+1
+
z ,
z n+1+α
n ≥ 0.
Theorem 3.3. If f (z) of the form (1.1) satisfies (2.1), then
f (z)
(n + 2)(2n + A + B)
<
≥
,
fn (z)
(n + 1)(2n + 2 + A + B)
z ∈ U.
The result is sharp for every n, with extremal function
(3.3)
f (z) =
1
B−A
+
z n+1 ,
z (n + 1)(2n + 2 + A + B)
J. Inequal. Pure and Appl. Math., 7(4) Art. 140, 2006
n ≥ 0.
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PARTIAL S UMS
5
Proof. Consider
f (z)
(n + 2)(2n + A + B)
−
fn (z) (n + 1)(2n + 2 + A + B)
P
P∞
k+1
1 + nk=1 ak z k+1 + (n+1)(2n+2+A+B)
1 + w(z)
k=n+1 ak z
B−A
P
:=
,
=
n
k+1
1 + k=1 ak z
1 − w(z)
(n + 1)(2n + 2 + B + A)
B−A
where
w(z) =
2+
(n+1)(2n+2+A+B) P∞
k+1
k=n+1 ak z
B−A
P
P∞
2 nk=1 ak z k+1 + (n+1)(2n+2+B+A)
k=n+1
B−A
Now
|w(z)| ≤
2−
(n+1)(2n+2+A+B) P∞
k=n+1 |ak |
B−A
Pn
(n+1)(2n+2+A+B) P∞
2 k=1 |ak | −
k=n+1
B−A
ak z k+1
|ak |
.
≤1
if
∞
(n + 1)(2n + 2 + A + B) X
|ak | +
|ak | ≤ 1.
B
−
A
k=1
k=n+1
n
X
(3.4)
The left hand side of (3.4) is bounded above by
∞
X
k(2k + A + B)
k=1
B−A
|ak |
if
1
B−A
( n
X
(k(2k + A + B) − (B − A)) |ak |
k=1
∞
X
+
)
(k(2k + A + B) − (n + 1)(2n + 2 + A + B)) |ak |
≥ 0,
k=n+1
and the proof is completed.
Corollary 3.4. For A = 2α − 1, B = 1, we get Theorem 2.2 in [3] which reads:
If f (z) of the form (1.1) satisfies condition
∞
X
k(k + α)|ak | ≤ 1 − α,
1
then
f (z)
(n + 2)(n + α)
<
≥
, z ∈ U.
fn (z)
(n + 1)(n + 1 + α)
The result is sharp for every n, with extremal function
1
1−α
f (z) = +
z n+1 , n ≥ 0.
z (n + 1)(n + 1 + α)
o
n
(z)
.
We next determine bounds for < ffn(z)
Theorem 3.5.
(a) If f (z) of the form (1.1) satisfies the condition (2.3), then
fn (z)
2n + 2 + A + B
<
≥
, z ∈ U.
f (z)
n+2
J. Inequal. Pure and Appl. Math., 7(4) Art. 140, 2006
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6
S. L ATHA AND L. S HIVARUDRAPPA
(b) If f (z) of the form (1.1) satisfies condition (2.1), then
fn (z)
2(n + 1)(2n + 2 + A + B)
<
≥
,
f (z)
2(n + 1)(n + 2) − n(B − A)
z ∈ U.
Equalities hold in (a) and (b) for the functions given by (3.1) and (3.3) respectively.
Proof. We prove (a). The proof of (b) is similar to (a) and will be omitted.
Consider
2(n + 2) fn (z) 2n + 2 + A + B
−
B−A
f (z)
2(n + 2)
P
P∞
1 + nk=1 ak z k+1 + 2n+2+A+B
ak z k+1
1 + w(z)
Pn B−A k+1 k=n+1
:=
,
=
1 + k=1 ak z
1 − w(z)
where
n+2
B−A
|w(z)| ≤
Pn
P∞
k=n+1 |ak |
P∞
n+A+B
k=n+1
B−A
1 − k=1 |ak | −
|ak |
This last inequality is equivalent to
n
∞
X
2n + 2 + A + B X
(3.5)
|ak | +
|ak | ≤ 1.
B−A
k=1
k=n+1
Since the left hand side of (3.5) is bounded above by
∞
X
2k + A + B
k=1
B−A
≤ 1.
|ak |,
the proof is completed.
Corollary 3.6. For A = 2α − 1, B = 1, we get Theorem 2.3 in [3] which reads:
(a) If f (z) of the form (1.1) satisfies condition
∞
X
(k + α)|ak | ≤ 1 − α,
1
then
fn (z)
(n + 1 + α)
≥
, z ∈ U.
<
f (z)
(n + 2)
(b) If f (z) of the form (1.1) satisfies condition
∞
X
k(k + α)|ak | ≤ 1 − α,
1
then
fn (z)
(n + 1)(n + 1 + α)
<
≥
, z ∈ U.
f (z)
(n + 1)(n + 2) − n(1 − α)
Equalities hold in (a) and (b) for the functions given by
1
1−α
f (z) = +
z n+1 , n ≥ 0,
z (n + 1 + α)
1
1−α
f (z) = +
z n+1 , n ≥ 0
z (n + 1)(n + 1 + α)
respectively.
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PARTIAL S UMS
7
We turn to ratios involving derivatives. The proof of Theorem 3.7 is similar to that in Theorem 3.1 and (a) of Theorem 3.5 and so the details may be omitted.
Theorem 3.7. If f (z) of form (1.1) satisfies condition (2.3) with A = −B, then
0 f (z)
<
(a)
≥ 0, z ∈ U,
fn0 (z)
0
fn (z)
1
(b)
<
≥ , z ∈ U.
0
f (z)
2
In both the cases, the extremal function is given by (3.1) with A = −B.
Theorem 3.8. If f (z) of form (1.1) satisfies condition (2.1) then,
0 f (z)
2(n + A + B)
(a)
<
≥
, z ∈ U,
0
fn (z)
2n + 2 + A + B
0
fn (z)
2n + 2 + A + B
(b)
<
≥
, z ∈ U.
0
f (z)
2(n + 2)
In both the cases, the extremal function is given by (3.3)
Proof. It is well known that f (z) ∈ ΣK (A, B) if and only if −zf 0 (z) ∈ Σ∗ (A, B). In particular,
f (z) satisfies condition (2.1) if and only if −zf 0 (z) satisfies condition (2.3). Thus (a) is an
immediate consequence of Theorem 3.1 and (b) follows directly from (a) of Theorem 3.5. For a function f (z) ∈ Σ, we define the integral operator F (z) as follows
Z
∞
1 X 1
1 z
F (z) = 2
tf (t)dt = +
ak z k , z ∈ E.
z 0
z k=1 k + 2
The nth partial sum Fn (z) of the integral operator F (z) is given by
n
Fn (z) =
1 X 1
+
ak z k ,
z k=1 k + 2
z ∈ E.
The following lemmas will be required for the proof of Theorem 3.11 below.
P
cos(kθ)
Lemma 3.9. For 0 ≤ θ ≤ π, 12 + m
k=1 k+1 ≥ 0
Lemma 3.10. Let P be analytic in U with P (0) = 1 and <{P (z)} > 12 in U. For any function
Q analytic in U the function P ∗ Q takes values in the convex hull of the image on U under Q.
Lemma 3.9 is due to Rogosinski and Szego [4] and Lemma 3.10 is a well known result ([2]
and [6]) that can be derived from the Herglotz representation for P. Finally we derive
Theorem 3.11. If f (z) ∈ Σc (A, B), then Fn (z) ∈ Σc (A, B).
Proof. Let f (z) be the form (1.1) and belong to the class Σc (A, B).
We have,
(
)
∞
1 X
1
(3.6)
< 1−
kak z k+1 > , z ∈ U.
B − A k=1
2
J. Inequal. Pure and Appl. Math., 7(4) Art. 140, 2006
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8
S. L ATHA AND L. S HIVARUDRAPPA
Applying the convolution properties of power series to Fn0 (z) we may write
n
X
k
2 0
−z Fn (z) = 1 −
(3.7)
ak z k+1
k
+
2
k=1
!
!
∞
∞
X
1
1 X
kak z k+1 ∗ 1 + (B − A)
= 1−
zk .
B − A k=1
k
+
1
k=n+1
Putting z = reiθ , 0 ≤ r < 1, 0 ≤ |θ| ≤ π, and making use of the minimum principle for
harmonic functions along with Lemma 3.9, we obtain
(
)
n+1 k
n+1
X
X
r cos(kθ)
1 k
(3.8)
< 1 + (B − A)
z
= 1 + (B − A)
k+1
k+1
k=1
k=1
> 1 + (B − A)
n+1
X
cos kθ
k+1
k=1
B−A
≥ 1−
.
2
In view of (3.6), (3.7), (3.8) and Lemma 3.10 we deduce that
B−A
2 0
−<{z Fn (z)} > 1 −
, 0 ≤ A + B < 2, z ∈ U,
2
which completes the proof of Theorem 3.11
Corollary 3.12. For A = 2α − 1, B = 1, we obtain Theorem 2.8 in [3] which reads:
If f (z) ∈ Σc (α), then Fn (z) ∈ Σc (α).
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Montreal, 1982.
[2] A.W. GOODMAN, Univalent Functions, Vol. I, Mariner Publ. Co., Tampa, Fl., (1983).
[3] NAK EUN CHO AND S. OWA, On partial sums of certain meromorphic functions, J. Inequal.
in Pure and Appl. Math., 5(2) (2004), Art. 30. [ONLINE: http://jipam.vu.edu.au/
article.php?sid=377].
[4] W. ROGOSINSKI AND G. SZEGO, Uber die abschimlte von potenzreihen die in ernein kreise be
schranket bleiben, Math. Z., 28 (1928), 73–94.
[5] S. LATHA, Some studies in the theory of univalent and multivalent functions, PhD Thesis, (1994).
[6] R. SINGH AND S. SINGH, Convolution properties of a class of starlike fuctions, Proc. Amer. Math.
Soc., 106 (1989), 145–152.
J. Inequal. Pure and Appl. Math., 7(4) Art. 140, 2006
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