volume 5, issue 2, article 30, 2004.
Received 03 March, 2004;
accepted 30 March, 2004.
Communicated by:H.M. Srivastava
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Journal of Inequalities in Pure and Applied Mathematics
PARTIAL SUMS OF CERTAIN MEROMORPHIC FUNCTIONS
NAK EUN CHO AND SHIGEYOSHI OWA
Department of Applied Mathematics Pukyong National University Pusan 608-737, KOREA.
EMail:necho@pknu.ac.kr Department of Mathematics Kinki University
Higashi-Osaka, Osaka 577-8502 JAPAN.
EMail:owa@math.kindai.ac.jp
c
2000Victoria University ISSN (electronic): 1443-5756 048-04
Partial Sums Of Certain Meromorphic Functions Nak Eun Cho and Shigeyoshi Owa
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Abstract
The purpose of the present paper is to establish some results concerning the partial sums of meromorphic starlike and meromorphic convex functions anal- ogous to the results due to H. Silverman [J. Math. Anal. Appl. 209 (1997), 221-227]. Furthermore, we consider the partial sums of certain integral opera- tor.
2000 Mathematics Subject Classification:Primary 30C45.
Key words: Partial sums, Meromorphic starlike functions, Meromorphic convex func- tions, Meromorphic close-to-convex functions, Integral operators.
This work was supported by theKorea Research Foundation Grant (KRF-2003-015- C00024).
Contents
1 Introduction. . . 3 2 Main Results . . . 6
References
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1. Introduction
LetΣbe the class consisting of functions of the form
(1.1) f(z) = 1
z +
∞
X
k=1
akzk
which are analytic in the punctured open unit diskD= {z : 0< |z| <1}. Let Σ∗(α)andΣk(α)be the subclasses ofΣconsisting of all functions which are, respectively, meromorphic starlike and meromorphic convex of order α (0 ≤ α <1)inD. We also denote byΣc(α)the subclass ofΣwhich satisfies
−Re{z2f0(z)} > α (0≤α <1; z ∈ U =D ∪ {0}).
We note that every function belonging to the classΣc(α)is meromorphic close- to-convex of orderαinD(see [2]). Iff(z) = P∞
k=0akzkandg(z) =P∞ k=0bkzk are analytic in U, then their Hadamard product (or convolution), denoted by f ∗g, is the function defined by the power series
(f ∗g)(z) =
∞
X
k=0
akbkzk (z ∈ U).
A sufficient condition for a functionf of the form (1.1) to be inΣ∗(α)is that (1.2)
∞
X
k=1
(k+α)|ak| ≤1−α
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and to be inΣk(α)is that (1.3)
∞
X
k=1
k(k+α)|ak| ≤1−α.
Further, we note that these sufficient conditions are also necessary for functions of the form (1.1) with positive or negative coefficients ([6, 13], also see [7]).
Recently, Silverman [10] determined sharp lower bounds on the real part of the quotients between the normalized starlike or convex functions and their se- quences of partial sums. Also, Li and Owa [4] obtained the sharp radius which for the normalized univalent functions inU, the partial sums of the well-known Libera integral operator [5] imply starlikeness. Further, for various other in- teresting developments concerning partial sums of analytic univalent functions, the reader may be (for examples) refered to the works of Brickman et al. [1], Sheil-Small [9], Silvia [11], Singh and Singh [12] and Yang and Owa [14].
Since to a certain extent the work in the meromorphic univalent case has paralleled that of the analytic univalent case, one is tempted to search results analogous to those of Silverman [10] for meromorphic univalent functions in D. In the present paper, motivated essentially by the work of Silverman [10], we will investigate the ratio of a function of the form (1.1) to its sequence of par- tial sumsfn(z) = 1z+Pn
k=1akzkwhen the coefficients are sufficiently small to satisfy either condition (1.2) or (1.3). More precisely, we will determine sharp lower bounds for Re{f(z)/fn(z)}, Re{fn(z)/f(z)}, Re{f0(z)/fn0(z)}, and Re{fn0(z)/f0(z)}. Further, we give a property for the partial sums of certain in- tegral operators in connection with meromorphic close-to-convex functions. In the sequel, we will make use of the well-known result thatRe{(1 +w(z))/(1−
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w(z))} > 0 (z ∈ U)if and only ifw(z) = P∞
k=1ckzk satisfies the inequality
|w(z)|<|z|. Unless otherwise stated, we will assume thatf is of the form (1.1) and its sequence of partial sums is denoted byfn(z) = 1z +Pn
k=1akzk.
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2. Main Results
Theorem 2.1. Iff of the form (1.1) satisfies condition (1.2), then
Re
f(z) fn(z)
≥ n+ 2α
n+ 1 +α (z ∈ U).
The result is sharp for everyn, with extremal function
(2.1) f(z) = 1
z + 1−α
n+ 1 +αzn+1 (n ≥0).
Proof. We may write n+ 1 +α
1−α
f(z)
fn(z) − n+ 2α n+ 1 +α
= 1 +Pn
k=1akzk+1+n+1+α1−α P∞
k=n+1akzk+1 1 +Pn
k=1akzk+1 := 1 +A(z)
1 +B(z).
Set(1 +A(z))/(1 +B(z)) = (1 +w(z))/(1−w(z)), so thatw(z) = (A(z)− B(z))/(2 +A(z) +B(z)). Then
w(z) =
n+1+α 1−α
P∞
k=n+1akzk+1 2 + 2Pn
k=1akzk+1+n+1+α1−α P∞
k=n+1akzk+1 and
|w(z)| ≤
n+1+α 1−α
P∞
k=n+1|ak| 2−2Pn
k=1|ak| − n+1+α1−α P∞
k=n+1|ak|.
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Now|w(z)| ≤1if and only if 2
n+ 1 +α 1−α
∞ X
k=n+1
|ak| ≤2−2
n
X
k=1
|ak|,
which is equivalent to (2.2)
n
X
k=1
|ak|+n+ 1 +α 1−α
∞
X
k=n+1
|ak| ≤1.
It suffices to show that the left hand side of (2.2) is bounded above byP∞ k=1((k+
α)/(1−α))|ak|, which is equivalent to
n
X
k=1
k+ 2α−1 1−α
|ak|+
∞
X
k=n+1
k−n−1 1−α
|ak| ≥0.
To see that the functionf given by (2.1) gives the sharp result, we observe for z =reπi/(n+2)that
f(z)
fn(z) = 1 + 1−α
n+ 1 +αzn+2 −→1− 1−α
n+ 1 +α = n+ 2α
n+ 1 +α whenr→1−. Therefore we complete the proof of Theorem2.1.
Theorem 2.2. Iff of the form (1.1) satisfies condition (1.3), then
Re
f(z) fn(z)
≥ (n+ 2)(n+α)
(n+ 1)(n+ 1 +α) (z ∈ U).
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The result is sharp for everyn, with extremal function
(2.3) f(z) = 1
z + 1−α
(n+ 1)(n+ 1 +α)zn+1 (n ≥0).
Proof. We write (n+ 1)(n+ 1 +α)
1−α
f(z)
fn(z)− (n+ 2)(n+α) (n+ 1)(n+ 1 +α)
= 1 +Pn
k=1akzk+1+(n+1)(n+1+α) 1−α
P∞
k=n+1akzk+1 1 +Pn
k=1akzk+1 := 1 +w(z)
1−w(z),
where
w(z) =
(n+1)(n+1+α) 1−α
P∞
k=n+1akzk+1 2 + 2Pn
k=1akzk+1+ (n+1)(n+1+α) 1−α
P∞
k=n+1akzk+1. Now
|w(z)| ≤
(n+1)(n+1+α) 1−α
P∞
k=n+1|ak| 2−2Pn
k=1|ak| − (n+1)(n+1+α) 1−α
P∞
k=n+1|ak| ≤1, if
(2.4)
n
X
k=1
|ak|+(n+ 1)(n+ 1 +α) 1−α
∞
X
k=n+1
|ak| ≤1.
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The left hand side of (2.4) is bounded above byP∞
k=1(k(k+α)/(1−α))|ak|if 1
1−α ( n
X
k=1
(k(k+α)−(1−α)|ak|
+
∞
X
k=n+1
(k(k+α)−(n+ 1)(n+ 1 +α))|ak| )
≥0,
and the proof is completed.
We next determine bounds forRe{fn(z)/f(z)}.
Theorem 2.3. (a) Iff of the form (1.1) satisfies condition (1.2), then
Re
fn(z) f(z)
≥ n+ 1 +α
n+ 2 (z ∈ U).
(b) Iff of the form (1.1) satisfies condition (1.3), then
Re
fn(z) f(z)
≥ (n+ 1)(n+ 1 +α)
(n+ 1)(n+ 2)−n(1−α) (z ∈ U).
Equalities hold in (a) and (b) for the functions given by (2.1) and (2.3), respec- tively.
Proof. We prove (a). The proof of (b) is similar to (a) and will be omitted. We
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write
n+ 2 1−α
fn(z)
f(z) − n+ 1 +α n+ 2
= 1 +Pn
k=1akzk+1+n+1+α1−α P∞
k=n+1akzk+1 1 +Pn
k=1akzk+1 := 1 +w(z)
1−w(z), where
|w(z)| ≤
n+2 1−α
P∞
k=n+1|ak| 2−2Pn
k=1|ak| − n+2α1−α P∞
k=n+1|ak| ≤1.
This last inequality is equivalent to (2.5)
n
X
k=1
|ak|+n+ 1 +α 1−α
∞
X
k=n+1
|ak| ≤1.
Since the left hand side of (2.5) is bounded above byP∞
k=1((k+α)/(1−α))|ak|, the proof is completed.
We turn to ratios involving derivatives. The proof of Theorem 2.4 below follows the pattern of those in Theorem 2.1and (a) of Theorem2.3and so the details may be omitted.
Theorem 2.4. Iff of the form (1.1) satisfies condition (1.2) withα= 0, then
(a) Re
f0(z) fn0(z)
≥ 0 (z∈ U),
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(b) Re
fn0(z) f0(z)
≥ 1
2 (z ∈ U).
In both cases, the extremal function is given by (2.1) withα = 0.
Theorem 2.5. Iff of the form (1.1) satisfies condition (1.3), then
(a) Re
f0(z) fn0(z)
≥ n+ 2α
n+ 1 +α (z ∈ U),
(b) Re
fn0(z) f0(z)
≥ n+ 1 +α
n+ 2 (z ∈ U).
In both cases, the extremal function is given by (2.3).
Proof. It is well known that f ∈ Σk(α) ⇔ −zf0 ∈ Σ∗(α). In particular, f satisfies condition (1.3) if and only if−zf0 satisfies condition (1.2). Thus, (a) is an immediate consequence of Theorem2.1and (b) follows directly from (a) of Theorem2.3.
For a functionf ∈Σ, we define the integral operatorF as follows:
F(z) = 1 z2
Z z
0
tf(t)dt= 1 z +
∞
X
k=1
1
k+ 2akzk (z ∈ D).
Then-th partial sumFnof the integral operatorF is given by Fn(z) = 1
z +
n
X
k=1
1
k+ 2akzk (z ∈ D).
The following lemmas will be required for the proof of Theorem2.8below.
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Lemma 2.6. For0≤θ ≤π,
1 2 +
m
X
k=1
cos(kθ) k+ 1 ≥ 0.
Lemma 2.7. LetP be analytic in U withP(0) = 1andRe{P(z)} > 12 inU. For any functionQanalytic inU, the functionP ∗Qtakes values in the convex hull of the image onU underQ.
Lemma 2.6 is due to Rogosinski and Szegö [8] and Lemma 2.7 is a well- known result (c.f. [3,12]) that can be derived from the Herglotz’ representation forP.
Finally, we derive
Theorem 2.8. Iff ∈Σc(α), thenFn ∈Σc(α)
Proof. Letf be of the form (1.1) and belong to the classΣc(α)for0≤ α <1.
Since−Re{z2f0(z)}> α, we have
(2.6) Re
(
1− 1
2(1−α)
∞
X
k=1
kakzk+1 )
> 1
2 (z∈ U).
Applying the convolution properties of power series toFn0, we may write
−z2Fn0(z) = 1−
n
X
k=1
k
k+ 2akzk+1 (2.7)
= 1− 1
2(1−α)
∞
X
k=1
kakzk+1
!
∗ 1 + 2(1−α)
n+1
X
k=1
1 k+ 1zk
! .
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Putting z = reiθ(0 ≤ r < 1, 0 ≤ |θ| ≤ π), and making use of the minimum principle for harmonic functions along with Lemma2.6, we obtain
Re (
1 + 2(1−α)
n+1
X
k=1
1 k+ 1zk
)
= 1 + 2(1−α)
n+1
X
k=1
rkcoskθ k+ 1 (2.8)
>1 + 2(1−α)
n+1
X
k=1
coskθ k+ 1 ≥α.
In view of (2.6), (2.7), (2.8) and Lemma2.7, we deduce that
−Re{z2Fn0(z)} > α (0≤α <1; z ∈ U).
Therefore we complete the proof of Theorem2.8.
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