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ACTA UNIVERSITATIS APULENSIS

No 13/2007

NOTE ON CERTAIN ANALYTIC FUNCTIONS

Shigeyoshi Owa, Toshio Hayami and Kazuo Kuroki
Abstract.Let A be the class of all analytic functions f (z) in the open
unit disk U. For f (z) ∈ A, a subclass Bk (α, β, γ) of A is introduced. The
object of the present paper is to discuss some properties of functions f (z)
belonging to the class Bk (α, β, γ).
2000 Mathematics Subject Classification : 30C45.
Kewwords and Phrases : Analytic function, Carath´eodory function.
1.Introduction
Let A be the class of functions f (z) of form
f (z) = z +

∞
X

an z n

(1)

n=2

which are analytic in the open unit disk U = {z ∈ C : |z| < 1}. A function
f (z) ∈ A is said to be a member of the subclass Bk (α, β, γ) of A if it satisfies
Re{αf (k) (z) + βzf (k+1) (z)} > γ

(k ∈ N = {1, 2, 3, . . . } ; z ∈ U)

(2)

for some aj ∈ R (j = 2, 3, 4, . . . , k), α ∈ R, β ∈ R (β 6= 0), and γ ∈ R (0 ≤
γ < k!αak ; a1 = 1). We consider some properties for functions f (z) belonging to the class Bk (α, β, γ).
Remark 1. Bk (α, β, γ) is convex.
Because, for f (z) ∈ Bk (α, β, γ) and g(z) ∈ Bk (α, β, γ), we define

97

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions

F (z) = (1 − t)f (z) + tg(z)

(0 ≤ t ≤ 1).

Then
 (k)

Re αF (z) + βF (k+1) (z)


= Re α(1 − t)f (k) (z) + αtg (k) (z) + β(1 − t)zf (k+1) (z) + βtzg (k+1) (z)




= (1 − t)Re αf (k) (z) + βzf (k+1) (z) + tRe αg (k) (z) + βzg (k+1) (z)
> (1 − t)γ + tγ = γ.

Therefore F (z) ∈ Bk (α, β, γ), that is, Bk (α, β, γ) is convex.
In the present paper, we consider some properties of functions f (z) belonging to the class Bk (α, β, γ).
2.Properties of the class B1 (α, β, γ) and B2 (α, β, γ)
We begin with the statement and the proof of the following result.
For cases k = 1, we obtain
Theorem 1. A function f (z) ∈ A is in the class of B1 (α, β, γ) if and
only if
!
Z
∞
X
1
xn−1 z n dµ(x)
(3)
f (z) = z + 2(α − γ)
n((n
−
1)β
+
α)

|x|=1
n=2
where µ(x) is the probability measure on X = {x ∈ C : |x| = 1}.
Proof. For f (z) ∈ A, we define
p(z) =

αf ′ (z) + βzf ′′ (z) − γ
.
α−γ
98

(4)

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions
Then p(z) is Carath´eodory function . Therefore we can write
Z
αf ′ (z) + βzf ′′ (z) − γ
1 + xz
=
dµ(x)

(see[1]).
α−γ
|x|=1 1 − xz

(5)

It follows from (5) that



Z
1 + xz
1 αβ −1
α ′
′′
z
f (z) + zf (z) = z
γ + (α − γ)
dµ(x)
β

β
|x|=1 1 − xz


Z
1 αβ −1
2 2
= z
γ + (α − γ)
(1 + xz)(1 + xz + x z + . . . )dµ(x) .
β
|x|=1
α
−1
β



(6)

Integrating the both sides of (6), we know that


Z z
α
α ′
−1
′′
β
f (ζ) + ζf (ζ) dζ =
ζ
β
0
1
=
β

Z

|x|=1

(Z

z

αζ

α
−1
β

+ 2(α − γ)

0

∞
X

xn ζ

n=1

−1
n+ α
β

!

dζ

)

dµ(x),

that is, that
α
β

z f ′ (z) =

1
β

Z

|x|=1

(

α
β

= z + 2(α − γ)z

Z

α
β

|x|=1

!)

β
xn z
dµ(x)
nβ + α
n=1
!
∞
X
1
xn z n dµ(x).
nβ + α
n=1

βz + 2(α − γ)

α
β

∞
X

n+ α
β

Thus, we have
′

f (z) = 1 + 2(α − γ)

Z

|x|=1

∞
X
n=1

!
1
n n
x z dµ(x).
nβ + α

An integration of both sides in (7) gives us that
!
)
Z z(
Z z
Z
∞
X
1
f ′ (ζ)dζ =
1 + 2(α − γ)
xn ζ n dµ(x) dζ,
nβ
+
α
0
|x|=1
0
n=1
99

(7)

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions
or
f (z) = z + 2(α − γ)

Z

|x|=1

= z + 2(α − γ)

Z

|x|=1

!
1
xn z n+1 dµ(x)
(n
+
1)(nβ
+
α)
n=1
!
∞
X
1
xn−1 z n dµ(x).
n
((n
−
1)β
+
α)
n=2
∞
X

This completes the proof of Theorem 1.
Corollary 1. The extreme points of B1 (α, β, γ) are
fx (z) = z + 2(α − γ)

∞
X
n=2

xn−1
zn
n ((n − 1)β + α)

(|x| = 1).

In view of Theorem 1, we have the following corollary for an .
Corollary 2. If f (z) ∈ A is in the class B1 (α, β, γ), then
|an | ≤

2(α − γ)
n ((n − 1)β + α)

(n = 2, 3, 4, . . . ).

Equality holds for the function f (z) given by
f (z) = z + 2(α − γ)

∞
X
n=2

xn−1
zn
n ((n − 1)β + α)

(|x| = 1).

Further, the following distortion inequality follows from Theorem 1.
Corollary 3. If f (z) ∈ A is in the class B1 (α, β, γ), then
!
∞
X
|z|n
|f (z)| ≤ |z| + 2(α − γ)
(z ∈ U).
n ((n − 1)β + α)
n=2

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S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions

Remark 2. If β > 0 and
see that

∞
X
n=2

α
= j (j = 2, 3, 4, . . . ) in Corollary 3, then we
β

∞
|z|n
|z|2 X
1
≤
n ((n − 1)β + α)
β n=2 n(n + j − 1)

∞ 
X
1
1
|z|2
=
−
β(j − 1) n=2 n n + j − 1
j
X
1
log(j)
|z|2
<
|z|2 .
=
β(j − 1) n=2 n
β(j − 1)

Therefore, we have that

|f (z)| < |z| +
< 1+

2(α − γ)log(j) 2
|z|
β(j − 1)

2(α − γ)log(j)
.
β(j − 1)

Next, for cases k = 2 we show
Theorem 2. A function f (z) ∈ A is in the class B2 (α, β, γ) if and only
if
f (z) = z + a2 z 2 + 2(2αa2 − γ)

Z

|x|=1

∞
X
n=3

!
xn−2
z n dµ(x)
n(n − 1) ((n − 2)β + α)

where µ(x) is the probability measure on X = {x ∈ C : |x| = 1}.
Proof.For f (z) ∈ A, we define
p(z) =

αf ′′ (z) + βzf ′′′ (z) − γ
.
2αa2 − γ

Then p(z) is Carath´eodory function. Hence, we can write
Z
1 + xz
αf ′′ (z) + βzf ′′′ (z) − γ
=
dµ(x).
2αa2 − γ
|x|=1 1 − xz
In view of (8), we have that
101

(8)

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions




α ′′
′′′
z
f (z) + zf (z) =
β
(
!
)
Z
∞
X
1 αβ −1
z
γ + (2αa2 − γ)
1+2
xn z n dµ(x)
β
|x|=1
n=1
!
Z
∞
X
α
α
1
=
2αa2 z β −1 + 2(2αa2 − γ)
xn z n+ β −1 dµ(x).
β |x|=1
n=1
α
−1
β

(9)

Integrating the both sides of (9), we have that
Z

z

ζ

α
−1
β

0

1
=
β

Z

|x|=1




α ′′
′′′
f (ζ) + ζf (ζ) dζ
β

(Z

z

2αa2 ζ

α
−1
β

∞
X

+ 2(2αa2 − γ)

0

xn ζ

n+ α
−1
β

n=1

!!

dζ

)

dµ(x),

that is, that
1
z f ′′ (z) =
β
α
β

Z

(

∞
X

α
β

2βa2 z + 2(2αa2 − γ)

|x|=1

n=1

α
β
xn z n+ β
nβ + α

!)

dµ(x).

This implies that

f ′′ (z) =

Z

|x|=1

(

2a2 + 2(2αa2 − γ)

∞
X
n=1

xn
zn
nβ + α

!)

dµ(x).

(10)

An integration of both sides in (10) gives us that
!
)
Z
Z z(
Z z
∞
X
xn
n
′′
ζ
dµ(x) dζ
2a2 + 2(2αa2 − γ)
f (ζ)dζ =
nβ + α
0
|x|=1
0
n=1

or

f ′ (z) − 1 = 2a2 z + 2(2αa2 − γ)

Z

|x|=1

102

∞
X
n=1

xn
z n+1
(n + 1)(nβ + α)

!

dµ(x).

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions
Therefore, we know that

f ′ (z) = 1 + 2a2 z + 2(2αa2 − γ)

Z

|x|=1

∞
X
n=2

!
xn−1
z n dµ(x). (11)
n ((n − 1)β + α)

Applying the same method for (11), we see that
Z z
f ′ (ζ)dζ =
0

=

Z

0

z

(

1 + 2a2 ζ + 2(2αa2 − γ)

Z

|x|=1

∞
X
n=2

!
)
xn−1
ζ n dµ(x) dζ.
n ((n − 1)β + α)

Thus, we obtain that

f (z) = z + a2 z 2 + 2(2αa2 − γ)

∞
X

Z

|x|=1

= z + a2 z 2 + 2(2αa2 − γ)

Z

|x|=1

xn−1
z n+1
(n + 1)n ((n − 1)β + α)

n=2
∞
X
n=3

!

dµ(x)

!
xn−2
z n dµ(x)
n(n − 1) ((n − 2)β + α)

This completes the proof of Theorem 2.
Corollary 4. The extreme points of B2 (α, β, γ) are
fx (z) = z + a2 z 2 + 2(2αa2 − γ)

∞
X
n=3

xn−2
zn
n(n − 1) ((n − 2)β + α)

!

In view of Theorem 2, we have the following corollary for an .
Corollary 5. If f (z) ∈ A is in the class B2 (α, β, γ), then
|an | ≤

2(2αa2 − γ)
n(n − 1) ((n − 2)β + α)

103

(n = 3, 4, 5, . . . ).

(|x| = 1).

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions
Equality holds for the function f (z) given by
f (z) = z+a2 z 2 +2(2αa2 −γ)

∞
X
n=3

xn−2
zn
n(n − 1) ((n − 2)β + α)

!

(|x| = 1).

Further, the following distortion inequality follows from Theorem 2.
Corollary 6. If f (z) ∈ A is in the class B2 (α, β, γ), then
!
∞
n
X
|z|
|f (z)| ≤ |z|+|a2 ||z|2 +2(2αa2 −γ)
n(n − 1) ((n − 2)β + α)
n=3

(z ∈ U).

3.Properties of the class Bk (α, β, γ)
For cases k is any natural number, we have
Theorem 3.A function f (z) ∈ A belongs to the class Bk (α, β, γ) if and
only if
f (z) = z + a2 z 2 + . . .
k

+ak z +2(k!αak −γ)

Z

|x|=1

∞
X

xn−k z n
n(n − 1) . . . (n − k + 1) ((n − k)β + α)
n=k+1

!

dµ(x)

for k = 1, 2, 3, . . . , where µ(x) is the probability measure on X = {x ∈ C :
|x| = 1}.
Proof.For f (z) ∈ A, we define
p(z) =

αf (k) (z) + βzf (k+1) (z) − γ
.
k!αak − γ

Since p(z) is Carath´eodory function, we can write that
Z
1 + xz
αf (k) (z) + βzf (k+1) (z) − γ
=
dµ(x).
k!αak − γ
|x|=1 1 − xz
104

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S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions
This means that


α
α
−1
(k)
(k+1)
zβ
f (z) + zf
(z) =
β
(

1 αβ −1
z
β

1
β

Z

γ + (k!αak − γ)

Z

1+2

|x|=1

k!αak z

α
−1
β

+ 2(k!αak − γ)

|x|=1

∞
X

!

xn z n dµ(x)

n=1

∞
X

)

−1
n n+ α
β

x z

n=1

!

=

(13)

dµ(x).

Integrating the both sides of (13), we obtain that


Z z
α
α
−1
(k)
(k+1)
ζβ
f (ζ) + ζf
(ζ) dζ
β
0
!! )
(Z
Z
∞
z
X
α
α
1
xn ζ n+ β −1
dζ dµ(x),
k!αak ζ β −1 + 2(k!αak − γ)
=
β |x|=1
0
n=1
that is, that
α
β

z f (k) (z) =

1
β

Z

(

α
β

k!βak z + 2(k!αak − γ)

|x|=1

∞
X
n=1

β
xn z
nβ + α

α
−1
β

!)

dµ(x).

This is equivalent to

f

(k)

(z) =

Z

(

k!ak + 2(k!αak − γ)

|x|=1

∞
X
n=1

xn
zn
nβ + α

!)

dµ(x).

(14)

Now, since f (0) = 0, f ′ (0) = 1, and f (m) (0) = m!am (m = 2, 3, 4, . . . ),
we see that
Z z
f (m) (ζ)dζ = f (m−1) (z) − f (m−1) (0)
0

= f (m−1) (z) − (m − 1)!am−1 .
105

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions
Furthermore, we know that
Z
Z z Z ζm
...
0

0

and

∞
X

ζ2

m!am dζ1 dζ2 . . . dζm = am z m ,

0

xn z n+k
=
(n + k)(n + k − 1) . . . (n + 1)(nβ + α)

n=1

∞
X

xn−k z n
.
n(n − 1) . . . (n − k + 1) ((n − k)β + α)
n=k+1
Therefore, integrating k times the both sides in (14), we obtain that
z

Z

0

=

Z

z

0

Z

ζk

...

0

Z

ζ2

Z

f (k) (ζ1 )dζ1 dζ2 . . . dζk

0

ζk

...

0

ζ2

Z

(

Z

k!ak + 2(k!αak − γ)

0

|x|=1

∞
X
xn ζ1n
nβ + α
n=1

!

)

dµ(x) dζ1 dζ2 . . . dζk ,

that is, that
z

Z
f (z) = f (0)+

0

0

+

Z

0

z

Z

0

ζk

...

Z

Z
f (0)dζ1 +
′

ζ2

z

Z

0

ζ2

Z
f (0)dζ1 dζ2 +
′′

0

(

k!ak + 2(k!αak − γ)

0

Z

|x|=1

Thus, we conclude that

z

Z

ζ3

0

Z

ζ2

f ′′′ (0)dζ1 dζ2 dζ3 +. . .

0

∞
X
xn ζ1n
nβ + α
n=1

!

)

dµ(x) dζ1 dζ2 . . . dζk .

f (z) = z + a2 z 2 + a3 z 3 + · · · + ak z k
+2(k!αak − γ)

Z

|x|=1

∞
X

xn−k z n
n(n − 1) . . . (n − k + 1) ((n − k)β + α)
n=k+1

The proof of Theorem 3 is complete.
Corollary 7.The extreme points of Bk (α, β, γ) are

106

!

dµ(x).

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions
fx (z) = z + a2 z 2 + a3 z 3 + · · · + ak z k
∞
X

xn−k z n
+2(k!αak − γ)
n(n − 1) . . . (n − k + 1) ((n − k)β + α)
n=k+1

!

(|x| = 1).

In view of Theorem 3, we see that
Corollary 8.If f (z) belongs to the class Bk (α, β, γ), then

|an | ≤

2(k!αak − γ)
n(n − 1) . . . (n − k + 1) ((n − k)β + α)

(n = k+1, k+2, k+3, . . . ).

Equality holds for the function f (z) given by
f (z) = z + a2 z 2 + a3 z 3 + · · · + ak z k
∞
X

xn−k z n
+2(k!αak −γ)
n(n − 1) . . . (n − k + 1) ((n − k)β + α)
n=k+1

!

(|x| = 1).

Further, the following distortion inequality follows from Theorem 3.
Corollary 9.If f (z) belongs to the class Bk (α, β, γ), then

|f (z)| ≤ |z| + |a2 ||z|2 + |a3 ||z|3 + · · · + |ak ||z|k

+2(k!αak −γ)

∞
X

n

|z|
n(n − 1) . . . (n − k + 1) ((n − k)β + α)
n=k+1

107

!

(z ∈ U).

S. Owa, T. Hayami, K. Kuroki - Note on certain analytic functions
References
[1]. D. J. Hallenbeck and T. H. MacGregor, Linear Problems and Convexity Techniques in Geometric Function Theory, Monographs and Studies
in Mathematics 22, Pitman (1984).
Authors:
Shigeyoshi Owa, Toshio Hayami and Kazuo Kuroki
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502
Japan
E-mail : owa@math.kindai.ac.jp

108