Volume 8 (2007), Issue 4, Article 95, 10 pp.
COEFFICIENT INEQUALITIES FOR CERTAIN CLASSES OF ANALYTIC AND
UNIVALENT FUNCTIONS
TOSHIO HAYAMI, SHIGEYOSHI OWA, AND H.M. SRIVASTAVA
D EPARTMENT OF M ATHEMATICS
K INKI U NIVERSITY
H IGASHI -O SAKA , O SAKA 577-8502
JAPAN
ha_ya_to112@hotmail.com
D EPARTMENT OF M ATHEMATICS
K INKI U NIVERSITY
H IGASHI -O SAKA , O SAKA 577-8502
JAPAN
owa@math.kindai.ac.jp
D EPARTMENT OF M ATHEMATICS AND S TATISTICS
U NIVERSITY OF V ICTORIA
V ICTORIA , B RITISH C OLUMBIA V8W 3P4
C ANADA
harimsri@math.uvic.ca
Received 09 August, 2007; accepted 12 September, 2007
Communicated by Th.M. Rassias
A BSTRACT. For functions f (z) which are starlike of order α, convex of order α, and λ-spirallike of order α in the open unit disk U, some interesting sufficient conditions involving coefficient
inequalities for f (z) are discussed. Several (known or new) special cases and consequences of
these coefficient inequalities are also considered.
Key words and phrases: Coefficient inequalities, Analytic functions, Univalent functions, Spiral-like functions, Starlike
functions, Convex functions.
2000 Mathematics Subject Classification. Primary 30A10, 30C45; Secondary 26D07.
1. I NTRODUCTION , D EFINITIONS AND P RELIMINARIES
Let A0 be the class of functions f (z) of the form:
(1.1)
f (z) = a0 + a1 z +
∞
X
an z n ,
n=2
The present investigation was supported, in part, by the Natural Sciences and Engineering Research Council of Canada under Grant
OGP0007353.
260-07
2
T OSHIO H AYAMI , S HIGEYOSHI OWA , AND H.M. S RIVASTAVA
which are analytic in the open unit disk
U = {z : z ∈ C
and |z| < 1} .
If f (z) ∈ A0 is given by (1.1), together with the following normalization:
a0 = 0 and
a1 = 1,
then we say that f (z) ∈ A.
If f (z) ∈ A satisfies the following inequality:
0 zf (z)
>α
(z ∈ U; 0 5 α < 1),
(1.2)
R
f (z)
then f (z) is said to be starlike of order α in U. We denote by S ∗ (α) the subclass of A consisting
of functions f (z) which are starlike of order α in U. Similarly, we say that f (z) is in the class
K(α) of convex functions of order α in U if f (z) ∈ A satisfies the following inequality:
zf 00 (z)
(1.3)
R 1+ 0
>α
(z ∈ U; 0 5 α < 1).
f (z)
It is easily observed from (1.2) and (1.3) that (see, for details, [3])
f (z) ∈ K(α) ⇐⇒ zf 0 (z) ∈ S ∗ (α)
(0 5 α < 1).
As usual, in our present investigation, we write
S ∗ := S ∗ (0)
and K := K(0).
Furthermore, we let B denote the class of functions p(z) of the form:
∞
X
p(z) = 1 +
pn z n ,
n=1
which are analytic in U.
Each of the following lemmas will be needed in our present investigation.
Lemma 1. A function p(z) ∈ B satisfies the following condition:
R[p(z)] > 0
(z ∈ U)
if and only if
p(z) 6=
ζ −1
ζ +1
(z ∈ U; ζ ∈ C; |ζ| = 1).
Proof. For the sake of completeness, we choose to give a proof of Lemma 1, even though it is
fairly obvious that the following bilinear (or Möbius) transformation:
z−1
w=
z+1
maps the unit circle ∂U onto the imaginary axis R(w) = 0. Indeed, for all ζ such that
|ζ| = 1 (ζ ∈ C), we set
ζ −1
w=
(ζ ∈ C; |ζ| = 1).
ζ +1
Then
1 + w
= 1,
|ζ| = 1 − w
which shows that
ζ −1
R(w) = R
=0
(ζ ∈ C; |ζ| = 1).
ζ +1
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 95, 10 pp.
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C OEFFICIENT I NEQUALITIES FOR C ERTAIN C LASSES OF A NALYTIC AND U NIVALENT F UNCTIONS
Moreover, by noting that p(0) = 1 for p(z) ∈ B, we know that
ζ −1
p(z) 6=
(z ∈ U; ζ ∈ C; |ζ| = 1).
ζ +1
This evidently completes the proof of Lemma 1.
3
∗
Lemma 2. A function f (z) ∈ A is in the class S (α) if and only if
∞
X
(1.4)
1+
An z n−1 6= 0,
n=2
where
n + 1 − 2α + (n − 1)ζ
an .
2 − 2α
An =
Proof. Upon setting
zf 0 (z)
f (z)
p(z) =
−α
f (z) ∈ S ∗ (α) ,
1−α
we find that
p(z) ∈ B
Using Lemma 1, we have
zf 0 (z)
f (z)
(1.5)
−α
1−α
(z ∈ U).
and R[p(z)] > 0
6=
ζ −1
ζ +1
(z ∈ U; ζ ∈ C; |ζ| = 1),
which readily yields
(ζ + 1)zf 0 (z) + (1 − 2α − ζ)f (z) 6= 0
f (z) ∈ S ∗ (α); z ∈ U; ζ ∈ C; |ζ| = 1 .
Thus we find that
(ζ + 1)z + (ζ + 1)
∞
X
!
nan z n
+ (1 − 2α − ζ) z +
n=2
∞
X
!
an z n
6= 0
n=2
(z ∈ U; ζ ∈ C; |ζ| = 1),
that is, that
(1.6)
2(1 − α)z
1+
∞
X
n + 1 − 2α + (n − 1)ζ
n=2
2(1 − α)
!
an z n−1
6= 0
(z ∈ U; ζ ∈ C; |ζ| = 1).
Now, dividing both sides of (1.6) by 2(1 − α)z (z 6= 0), we obtain
∞
X
n + 1 − 2α + (n − 1)ζ
1+
an z n−1 6= 0
2(1
−
α)
n=2
(z ∈ U; ζ ∈ C; |ζ| = 1),
which completes the proof of Lemma 2 (see also Remark 2 below).
Remark 1. It follows from the normalization conditions:
a0 = 0 and a1 = 1
that
A0 =
1 − 2α − x
a0 = 0 and
2 − 2α
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 95, 10 pp.
A1 =
2 − 2α
a1 = 1.
2 − 2α
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T OSHIO H AYAMI , S HIGEYOSHI OWA , AND H.M. S RIVASTAVA
Remark 2. The assertion (1.4) of Lemma 2 is equivalent to
!
2
z + ζ+2α−1
z
1
2−2α
f (z) ∗
6= 0
(z ∈ U),
z
(1 − z)2
which was given earlier by Silverman et al. [2]. Furthermore, in its special case when α = 0,
Lemma 2 yields a recent result of Nezhmetdinov and Ponnusamy [1] for the sufficient conditions
involving the coefficients of f (z) to be in the class S ∗ .
The object of the present paper is to give some generalizations of the aforementioned result
due to Nezhmetdinov and Ponnusamy [1]. We also briefly discuss several interesting corollaries
and consequences of our main results.
2. C OEFFICIENT C ONDITIONS FOR F UNCTIONS IN THE C LASS S ∗ (α)
Our first result for functions f (z) to be in the class S ∗ (α) is contained in Theorem 1 below.
Theorem 1. If f (z) ∈ A satisfies the following condition:
n " k
#
∞
X X
X
β
γ
(−1)k−j (j + 1 − 2α)
aj
k−j
n−k n=2
k=1 j=1
" k
#
!
∞
X
X
β
γ
+
(−1)k−j (j − 1)
aj
(2.1)
5 2(1 − α)
k
−
j
n
−
k
k=1 j=1
(0 5 α < 1; β ∈ R; γ ∈ R),
then f (z) ∈ S ∗ (α).
Proof. First of all, we note that
(1 − z)β 6= 0 and (1 + z)γ 6= 0
(z ∈ U; β ∈ R; γ ∈ R).
Hence, if the following inequality:
!
∞
X
(2.2)
1+
An z n−1 (1 − z)β (1 + z)γ 6= 0
(z ∈ U; β ∈ R; γ ∈ R)
n=2
holds true, then we have
1+
∞
X
An z n−1 6= 0,
n=2
which is the relation (1.4) of Lemma 2. It is easily seen that (2.1) is equivalent to
! ∞
! ∞
!
∞
X
X
X
n−1
n
n
n
(2.3)
1+
6= 0,
An z
(−1) bn z
cn z
n=2
n=0
n=0
where, for convenience,
β
γ
bn :=
and cn :=
.
n
n
Considering the Cauchy product of the first two factors, (2.3) can be rewritten as follows:
! ∞
!
∞
X
X
(2.4)
1+
Bn z n−1
cn z n 6= 0,
n=2
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 95, 10 pp.
n=0
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C OEFFICIENT I NEQUALITIES FOR C ERTAIN C LASSES OF A NALYTIC AND U NIVALENT F UNCTIONS
where
Bn :=
n
X
5
(−1)n−j Aj bn−j .
j=1
Furthermore, by applying the same method for the Cauchy product in (2.4), we find that
!
∞
n
X
X
Bk cn−k z n−1 6= 0
1+
(z ∈ U)
n=2
k=1
or, equivalently, that
1+
" n
∞
k
X
X X
n=2
k=1
!
(−1)k−j Aj bk−j
#
cn−k z n−1 6= 0
(z ∈ U).
j=1
Thus, if f (z) ∈ A satisfies the following inequality:
!
∞ X
n
k
X
X
(−1)k−j Aj bk−j cn−k 5 1,
n=2
k=1
j=1
that is, if
!
∞ X
n
k
X
X
1
(−1)k−j [(j + 1 − 2α) + (j − 1)ζ]aj bk−j cn−k 2(1 − α) n=2 k=1 j=1
" k
#
∞
n
X
X
X
1
k−j
5
(−1)
(j + 1 − 2α)aj bk−j cn−k 2(1 − α) n=2 k=1 j=1
n " k
!
#
X X
+|ζ| (−1)k−j (j − 1)bk−j aj cn−k k=1
51
j=1
(0 5 α < 1; ζ ∈ C; |ζ| = 1),
then f (z) ∈ S ∗ (α). This completes the proof of Theorem 1.
Setting α = 0 in Theorem 1, we deduce the following corollary.
Corollary 1. If f (z) ∈ A satisfies the following condition:
n " k
#
∞
X X
X
β
γ
(−1)k−j (j + 1)
aj
k−j
n−k n=2
k=1 j=1
" k
#
!
∞
X
X
β
γ
(2.5)
+
(−1)k−j (j − 1)
aj
52
k−j
n−k k=1 j=1
(β ∈ R; γ ∈ R),
then f (z) ∈ S ∗ .
Remark 3. If, in the hypothesis (2.5) of Corollary 1, we set
β − 1 = γ = 0 or
β = γ = 1 or β − 2 = γ = 0,
we arrive at the result given by Nezhmetdinov and Ponnusamy [1]. Moreover, for β = γ = 0 in
Theorem 1, we obtain Corollary 2 below.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 95, 10 pp.
http://jipam.vu.edu.au/
6
T OSHIO H AYAMI , S HIGEYOSHI OWA , AND H.M. S RIVASTAVA
Corollary 2. If f (z) ∈ A satisfies the following coefficient inequality:
∞
X
(2.6)
(n − α)|an | 5 1 − α
(0 5 α < 1),
n=2
then f (z) ∈ S ∗ (α).
In particular, by putting α = 0 in (2.6), we get the following well-known coefficient condition
for the familiar class S ∗ of starlike functions in U.
Corollary 3. If f (z) ∈ A satisfies the following coefficient inequality:
∞
X
(2.7)
n|an | 5 1,
n=2
∗
then f (z) ∈ S .
We next derive the coefficient condition for functions f (z) to be in the class K(α).
Theorem 2. If f (z) ∈ A satisfies the following condition:
" k
#
∞
n
X
X
X
β
γ
(−1)k−j j(j + 1 − 2α)
aj
k−j
n−k n=2
k=1 j=1
∞ " k
#
!
X X
β
γ
(2.8)
+
(−1)k−j j(j − 1)
aj
5 2(1 − α)
k−j
n−k k=1 j=1
(0 5 α < 1; β ∈ R; γ ∈ R),
then f (z) ∈ K(α).
Proof. Since zf 0 (z) belongs to the class S ∗ (α) if and only if f (z) is in the class K(α), and since
f (z) = z +
(2.9)
∞
X
an z n
n=2
and
zf 0 (z) = z +
(2.10)
∞
X
nan z n ,
n=2
upon replacing aj in Theorem 1 by jaj , we readily prove Theorem 2.
By considering some special values for the parameters α, β and γ, we can deduce the
following corollaries.
Corollary 4. If f (z) ∈ A satisfies the following condition:
" k
#
∞
n
X
X
X
β
γ
(2.11)
(−1)k−j j(j + 1)(−1)k−j
aj
k
−
j
n
−
k
n=2
k=1 j=1
∞ " k
#
!
X X
β
γ
+
(−1)k−j j(j − 1)
aj
5 2 (β ∈ R; γ ∈ R),
k−j
n−k k=1
j=1
then f (z) ∈ K.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 95, 10 pp.
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C OEFFICIENT I NEQUALITIES FOR C ERTAIN C LASSES OF A NALYTIC AND U NIVALENT F UNCTIONS
7
Corollary 5. If f (z) ∈ A satisfies the following coefficient inequality:
∞
X
(2.12)
n(n − α)|an | 5 1 − α
(0 5 α < 1),
n=2
then f (z) ∈ K(α).
Corollary 6. If f (z) ∈ A satisfies the following coefficient inequality:
∞
X
(2.13)
n2 |an | 5 1,
n=2
then f (z) ∈ K.
3. C OEFFICIENT C ONDITIONS FOR F UNCTIONS IN THE C LASS SP(λ, α)
In this section, we consider the subclass SP(λ, α) of A, which consists of functions f (z) ∈ A
if and only if the following inequality holds true:
0
zf (z)
π
π
iλ
(3.1)
R e
−α
>0
z ∈ U; 0 5 α < 1; − < λ <
.
f (z)
2
2
For f (z) ∈ SP(λ, α), we first derive Lemma 3 below.
Lemma 3. A function f (z) ∈ A is in the class SP(λ, α) if and only if
1+
(3.2)
∞
X
Cn z n−1 6= 0,
n=2
where
Cn :=
n − 1 + 2(1 − α)e−iλ cos λ + (n − 1)ζ
an .
2(1 − α)e−iλ cos λ
Proof. Letting
eiλ
p(z) =
zf 0 (z)
f (z)
− α − i(1 − α) sin λ
,
(1 − α) cos λ
we see that
p(z) ∈ B
and R[p(z)] > 0
It follows from Lemma 1 that
0
zf (z)
iλ
− α − i(1 − α) sin λ
e
ζ −1
f (z)
6=
(3.3)
(1 − α) cos λ
ζ +1
(z ∈ U).
(z ∈ U; ζ ∈ C; |ζ| = 1).
We need not consider Lemma 1 for the case when z = 0, because (3.3) implies that
p(0) 6=
ζ −1
ζ +1
(ζ ∈ C; |ζ| = 1).
It also follows from (3.3) that
eiλ [zf 0 (z) − αf (z)] − i(1 − α)f (z) sin λ
6=
(1 − α) cos λ
(z ∈ U; ζ ∈ C; |ζ| = 1) ,
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 95, 10 pp.
ζ −1
ζ +1
f (z)
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8
T OSHIO H AYAMI , S HIGEYOSHI OWA , AND H.M. S RIVASTAVA
which readily yields
(ζ + 1) eiλ [zf 0 (z) − αf (z)] − i(1 − α)f (z) sin λ 6= (ζ − 1)(1 − α)f (z) cos λ
(z ∈ U; ζ ∈ C; |ζ| = 1)
or, equivalently,
(3.4) (ζ + 1)eiλ zf 0 (z) − αeiλ f (z) − ζαeiλ f (z) − i(1 − α)f (z) sin λ − iζ(1 − α)f (z) sin λ
6= ζ(1 − α)f (z) cos λ − (1 − α)f (z) cos λ
(z ∈ U; ζ ∈ C; |ζ| = 1) .
We find from (3.4) that
(ζ + 1)eiλ zf 0 (z) − αeiλ f (z) − ζαeiλ f (z) − ζ(1 − α)eiλ f (z) + (1 − α)e−iλ f (z) 6= 0
(z ∈ U; ζ ∈ C; |ζ| = 1) ,
that is, that
(1 + ζ)eiλ zf 0 (z) + (e−iλ − 2α cos λ − ζeiλ )f (z) 6= 0
(z ∈ U; ζ ∈ C; |ζ| = 1) ,
which, in light of (1.1) with a0 = a1 − 1 = 0, assumes the following form:
!
!
∞
∞
X
X
(ζ + 1)eiλ z +
nan z n + (e−iλ − ζeiλ − 2α cos λ) z +
an z n 6= 0
n=2
n=2
(z ∈ U; ζ ∈ C; |ζ| = 1)
or, equivalently,
(3.5)
2(1 − α)z cos λ
1+
∞
X
n + e−2iλ − 2αe−iλ cos λ + (n − 1)ζ
2(1 − α)e−iλ cos λ
n=2
!
an z n−1
6= 0
(z ∈ U; ζ ∈ C; |ζ| = 1) .
Finally, upon dividing both sides of (3.5) by
2(1 − α)z cos λ 6= 0
and noting that
e−2iλ = −1 + 2e−iλ cos λ,
we obtain
1+
∞
X
n − 1 + 2(1 − α)e−iλ cos λ + (n − 1)ζ
an 6= 0
2(1 − α)e−iλ cos λ
π
π
0 5 α < 1; − < λ < ; ζ ∈ C; |ζ| = 1 ,
2
2
n=2
which completes the proof of Lemma 3 (see also the proof of a known result [1, Theorem
3.1]).
By applying Lemma 3, we now prove Theorem 3 below.
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 95, 10 pp.
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C OEFFICIENT I NEQUALITIES FOR C ERTAIN C LASSES OF A NALYTIC AND U NIVALENT F UNCTIONS
9
Theorem 3. If f (z) ∈ A satisfies the following condition:
" k
#
∞
n
X
X
X
β
γ
(−1)k−j [j − α + (1 − α)e−2iλ ]
aj
k−j
n−k n=2
k=1 j=1
" k
#
!
∞
X
X
β
γ
(3.6)
+
(−1)k−j (j − 1)
aj
5 2(1 − α) cos λ
k
−
j
n
−
k
k=1 j=1
π
π
0 5 α < 1; − < λ < ; β ∈ R; γ ∈ R ,
2
2
then f (z) ∈ SP(λ, α).
Proof. Applying the same method as in the proof of Theorem 1, we see that f (z) is in the class
SP(λ, α) if
n
!
∞ X
k
X
X
(3.7)
(−1)k−j Cj bk−j cn−k 5 1
n=2 k=1
j=1
where, as before,
β
γ
bn :=
and
cn :=
,
n
n
the coefficients Cn being given as in Lemma 3. It follows from the inequality (3.7) that
1
|2(1 − α)e−iλ cos λ|
" k
#
∞ X
n
X
X
k−j
−iλ
·
(−1) (j − 1 + 2(1 − α)e cos λ) + ζ(j − 1) aj bk−j cn−k n=2 k=1
j=1
1
2(1 − α) cos λ
" k
#
n
∞
X
X
X
k−j
−iλ
·
(−1)
j
−
α
+
(1
−
α)(−1
+
2e
cos
λ)
b
a
c
k−j j
n−k n=2
k=1 j=1
!
" k
#
n
X
X
+ |ζ| (−1)k−j (j − 1)bk−j aj cn−k k=1 j=1
π
π
(3.8)
51
0 5 α < 1; − < λ < ; ζ ∈ C; |ζ| = 1 ,
2
2
which implies that, if f (z) satisfies the hypothesis (3.6) of Theorem 3, then f (z) ∈ SP(λ, α).
This completes the proof of Theorem 3.
5
In its special case when
β−1=γ =0
β=γ=1
or
or
β − 2 = γ = 0,
Theorem 3 would immediately yield the following corollary.
Corollary 7 (cf. [1]). If f (z) ∈ A satisfies any one of the following conditions:
(3.9)
∞ X
[n − α + (1 − α)e−2iλ ](an − an−1 ) + an−1 + |(n − 1)(an − an−1 ) + an−1 |
n=2
5 2(1 − α) cos λ
J. Inequal. Pure and Appl. Math., 8(4) (2007), Art. 95, 10 pp.
π
π
0 5 α < 1; − < λ <
2
2
http://jipam.vu.edu.au/
10
T OSHIO H AYAMI , S HIGEYOSHI OWA ,
AND
H.M. S RIVASTAVA
or
(3.10)
∞ X
[n − α + (1 − α)e−2iλ ](an − an−2 ) + 2an−2 + |(n − 1)(an − an−2 ) + 2an−2 |
n=2
5 2(1 − α) cos λ
π
π
0 5 α < 1; − < λ <
2
2
or
(3.11)
∞ X
[n − 1 − α + (1 − α)e−2iλ ](an − 2an−1 + an−2 ) + an − an−2 n=2
+ |(n − 2)(an − 2an−1 + an−2 ) + an − an−2 |
π
π
,
5 2(1 − α) cos λ
0 5 α < 1; − < λ <
2
2
then f (z) ∈ SP(λ, α).
Remark 4. For λ = 0, Theorem 3 implies Theorem 1. Furthermore, by setting α = 0 in
Theorem 3, we arrive at the following sufficient condition for functions f (z) ∈ A to be in the
class SP(λ).
Corollary 8. If f (z) ∈ A satisfies the following condition:
" k
#
∞
n
X
X
X
β
γ
(3.12)
(−1)k−j (j + e−2iλ )
aj
k
−
j
n
−
k
n=2
k=1 j=1
" k
#
!
∞
X
X
β
γ
+
(−1)k−j (j − 1)
aj
k−j
n−k k=1 j=1
π
π
,
5 2 cos λ
0 5 α < 1; β ∈ R; γ ∈ R; − < λ <
2
2
then
f (z) ∈ SP(λ) := SP(λ, 0).
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http://jipam.vu.edu.au/