55
Internat. J. Math. & Math. Sci.
(1988) 55-70
VOL. ii NO.
NEW CLASSIFICATION OF ANALYTIC FUNCTIONS
WITH NEGATIVE COEFFICIENTS
SHIGEYOSHI OWA
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577
Japan
MILUTIN
OBRADOVI(
Department of Mathematics
Faculty of Technology and Metallurgy
4 Karnegieva Street
[I000 Belgrade
Yugoslavi a
(Received April 8, 1986)
New classification of analytic functions with negative coefficients is
A(p,n,Bk) of
the coefficients inequality, that is, new subclass
ABSTRACT.
given
by
using
The object of the present
analytic functions with negative coefficient is defined.
A(p,n,Bk)
paper is to prove various distortion theorems for functions in
fractional calculus of functions belonging to
A(p,n,Bk).
and for
Further, some properties of
A(p,n,Bk) are shown.
the class
KEYWORDS AND PHRASES.
theorem,
distortion
Analytic function,
fractional integral,
fractional derivative, extreme point.
1980 AMS SUBJECT CLASSIFICATION CODE. 30C45, 26A24.
I.
INTRODUCTION.
Let
A
p,n
.
denote the class of functions of the form
z p-
f(z)
akz
k=p+n
k
(ak
A function f(z) belonging to
Re
for
A
p,n
some
(0 <
zf’(z)
f(z)
<
p)
Izl
{z:
which are analytic in the unit disk U
<
{1,2,3
I}, where N
is said to be in the class
A
p,n
>
N; n
O; p
S
p,n
is said to be in the class
}.
() if and only if
([.2)
(,
and
(I.I)
N)
for
K
p,n
all
z
D
Also,
() if and only if
function
f(z)
belonging
to
OBRADOVI6
S. OWA and M.
56
z f"(z)
for some
<
(0 < e
a
>
+----;---c
t z)
Re
p), and for
(1.3)
z e 0
al[
S
K
We note that
(,) and
(e) are the
p,n
p,n
functions and p--va]ent convx fuuct[ons of ordec
note tlat S
f(z) e K
p ,n
p,n
(a)
c
S
p ,n
(0), K
() if and only
.
In view of the
reqults
(k
k=p+n
and that
f(z) c K
p,n
zf’(z)/p
i[
by
K
c
a ill,
ak
p ,n
g S
L
() for
we know that
p-valeqt
a
>
0
starlike
Furthermore,
we
p, and that
<
a
f(z) e S
p.
p ,n
(x)
[ and oaly if
,
p
k(k- a)
ak
p-
.
A(p,n,Bk) denote the subclass of
th following
(0) for 0
p,n
of
respectively.
(a) if and only if
k=p+n
Let
)
p ,n
(a)
subclasses
A
p,n
consisting of functions which satisfy
|nequa[ity
\i
k=p+n
(1.4)
0).
(Bk>
Bkak
It follows from (1.4) that
A(p,,n,Bk)
A(p,n,Ck)
(,3
< Ck Bk).
Therefore we can classify the analytic functions belonging to
above inequality (1.4).
A
p,n
according to the
,
REMARK 1. A(l,l,(k-e)/(1-=))= T () (Silverman [2]),
,
C(e) (Silverman [2]), A(l,l,k(l+8)/28(l-e))
P (e,8)
A(l,l,k(k-)/(l-e))
,
S (e,B) (Gupta and Jain [4]),
(Gupta and Jain [3]), A(I,I {(k-l)+B(k+l-2a)}/28(l-a))
,
C (e,8) (Gupta and Jain [4]),
A(l,l,k{k-l)+B(k+l-2a)}/2B(l-e))
R(a) (Sarangi and Uralegaddi [5]), A(l,l,k/(l-a))
A(I,I,I/(I-))
Q(a) (Sarangi and
Uralegaddi [5]), A(l,l,(k+m-l)!(2k+m-l)/(k-l)!(m+l)!)
K (Owa [6]),
m
A(l,l,(m-k+k)r(m+e+k)/(k-l)!r(m++2))
R(m+) (Owa [7]),
A(l,l,(k-B)C(a,k)/(l-B))
R[a,3] (Silverman and Silvla [8]),
A(I,I,k(I+Y)C(a,k)/2Y(I-8))
Pe[8,] (Owa and Ahuja [9]), and
K(m+a) (Owa [I0]),
A(l,l,(m+a+2k-l)F(m++k)/(k-l)!F(m+2+a))
k
where C(a,k)
(j-2a)/(k-l)!.
j=2
REMARK 2. A(l,n,(k-cO/(l-e))
(n) (Chatterjea [II]),
C (n) (Chatterjea [II]) and A(I n,k/(l-e))
A(l,n k(k-)/(l-e))
(Sekine and Owa [12]).
REMARK 3. A(p,l,(l+b)k/2b(l-a)p)
A(p,l,(l+b)k2/2b(l-a) p2 )--C,P (a,b)
(a,b) (Owa [13]),
P
(Owa [13]),
A(p,l,.(l-ap+bk)/(b-a)p)--T (a,b) (Goel and Sohi [14]),
P
C (a,b) (Goel and Sohi [14]),
A(p,l,k(l-ap+bk)/
P
(b_a)p2
C(a n)
NEW CLASSIFICATION OF ANALYTIC FUNCTIONS WITH NEGATIVE COEFFICIENTS
57
,
(Owa [15]),
A(p,l,(k+m-l)!(2kn-p)/(k-p).’(m+p)’), Km+p_l
T (p,a,B)(Shukla and Dashrath [16]),
C(p,.,B) (Owa and Srivastava [17]),
A(p,l,k(l+B)/(B-a)p)
A(p,l,k2(l+B)/(-x)p 2)
T (p,) (Owa [I]), and A(p,l,k(k-)/p(p-a))
A(p,I,(k-a)/(p-))
(,"A,a [1]).
2.
DISTORTION THEOREMS.
We begin with the statement and the proof of the following result.
THEOREM I.
Bk Bk+l
max
for
0
Let the fnction f(z) defined by (i.I) be in the class
A(p,n,Bk) with
Then
p+n
I.1 p
B
p+n
(2.1)
The equalities in (2.1) are attained for the function f(z) given by
z e U
z p-
f (z)
PROOF.
p+n
Izl p+n
Bp+n
or
K= +n
V
k;p+n
.
ak
k=p+n
p+n
O,
max
O,
B
Bk
k+l,
we have
(2.3)
Bkak
(2.4)
ak
Hence, it follows from (2.4) that
max
(2.2)
A(p,n,B k) and
f(z)
Since
-1-- z p+n
Fp+n
Izl p
p+n
and
-
I a k}
I=1 I=l p+n k=p+n
p+n
(2.5)
k=p+n
B
(2.6)
p+n
Furthermore, it is clear that the equalities in (2.1) are attained the function f(z)
given by (2.2).
REMARK 4.
-
Note that if
max
{0,
I,I
B
p+n
> I, then
I’lP+n:}-- I1- i,i1_._ p*n
+n
From [I], f(z) is p-valent starlike in U if and only if
Therefore, we have
Izl"- Bp+n Izl
I (z>l,
Bp+n)
Izlp+n
Izl’ +--u
Bp+n
for p-valent starlike functions of the form (I.I).
(z CU).
(p+n)/p.
58
OBRADOVI6
S. OWA and M.
THEOREM
2.
A(p,n,kB k) with
{0,
max
Let
the
Bk 8K+
p]zl p-I
p+a
f(z)
funceion
defined
by
(I. I)
be
ia
tIe
cl ass
Then
Izl
’ I’<-)I pI’-I p-+
+n-I
p+n-I
p+n
U. The equalities in (2.7) are attained for the function f(z) given by
for z
zp
f(z)
PROOF.
(p+n)B
Note that, for
8p+n
zP+n
f(z)
+nkak
k=Lp
(2.8)
p+n
A(p,n,kBk) and
kB
k=p+n
ka k’
Bka Bk+
1,
(2.9)
tha is, that
k=p+n
kak Bp+n
(2. lO)
lis ,gives that
k=p+n
’
max 0, plI p-I
p+n-I
p+n
and
k--’p+n
p+n
Further, the equaItttes tn (2.7) are attained for the function f() given by (2.8).
REMARK 5.
If
Bp+n)
I/p, then
p+n
for
p+n
z
Thus, from [17], we know that, for p-valent starlike functions of the form
(I.I), Theorem 2 gives
Plzlp-IIBp+n Izl p+n-I
p+n
Next, we derive the following lemma.
LEMMA I.
Let
n
(k-
+ i)
i=l
for
i=l
Aik
(2.13)
2.. Then we have
Ai(P
i=l
PROOF.
+ n) i-.
J
n (p +
n-
+ i).
i=2
In case of j=2, it is clear from (2.13) that
(2.14)
NEW CLASSIFICATION OF ANALYTIC FUNCTIONS WITH NEGATIVE COEFFICIENTS
2
;[
or, that
AI=
and
+ i)
(k-
k2+
(2.15)
k,
I. Thus we have
A2=
2
+ n) i-l=
i At(P
i=l
which proves
k(k + I)=
59
+ (p + a)
p + n +
(2,16)
(2.14) for j=2.
Assume that (2.14) holds true for j=j.
Then
j+l
+ i)
H (k
(k + j)
H (k
i=l
Aiki)
(k + j)
i=l
where
BI=
+ i)
i=l
jAI, Bj+I= Aj,
I Biki
jA i (i
Bi= Ai_l+
(2.17)
i=l
2,3
j).
Hence we obtain
i=l Bi(P n)i-l= JA1+i !2(Ai-l+ jAi)
i i(p+n) i-I+
i i(p+n)
i Ai(P
(P+n) i-I+
+
(p+n)
A
A
i=l
EJ(P+n)J
i-I
i=l
(p + n + j)
+ n)
i-I
i=I
(p + n + j)
II (p + n
+ i)
1=2
j+l
l[
(p + n-
+ i).
(2.19)
i=2
Consequently, by the mathematical induction, we complete the proof of Lemma I.
Applying Lemma I, we prove
THEOREM
A(p,n,kmBk)
Let
3.
the
f(z)
function
k’ Bk+land ,
with B
2
defined
by
(I. I)
be
in
the
class
p. Then we have
m
II (p+n-l+i)
If(J)(z)J
{0,
max
(i=2) izlp+n-J}
(p+n)m-IB p+n
II (p+l-i))Iz] p-ji=l
and
(2.20)
j
(p+n- l+i)
If(j)(z)l
)(
J (p+l_i))Izlp-j+ (i=2
(p+n)
i=l
for
z
e
PROOF.
and 2
(
j
(
Since f(z)
(p +
m-
IBp+n
p+n-j
,)Iz
(2.21)
m.
e A(p,n,
kmBk)
t
and
n)m-tBp+n k=p+n k ak
)
Bk 8k+l,
I
k=p+n
m
k
Bka k
we note that
’
(2.22)
OBRADOVI6
S. OWA and M.
60
that is, that
ak<
k
k=p+n
for
2
(2.23)
m-t
B
(p + n)
p+n
For f(z) defined by (I.I),
m.
f(J) (z)
we have
n (p+l-[))z p-ji=]
for
2
p. Hence, by using Lemma
m
i=l
and (2.23), we obtain
i=l
i=l
J
IzlP-J+
<p+,.--,>]
i=l
(p+l-i))IzlP-J+
max
_I-.I p+n-L
(p+n)
m-
1B p+n
i Ai(P+n) t=1)
i=t
I=2
(p+n)
i=l
IfCJ)(z)l
(p+n)m-iB p+n
R (p+n-l+i)
j
l[
which shows (2.21).
k
i=l
(p+l-i))IzlP-J+ Izl p+n-j Ai(
kiakl
i=1
k=p+n
It
i=l
-(
i Aikl )a
(p+l-i))IzlP-J+ Izl p+n-j
k=p+n
R
(2.24)
jN (k-l+i))a k
(p+1-i))IzlP-J+ lzl p+n-j
=I
k=p+n i
If(J)(z)l
i=l
-i))ak zk-j
(k +
k=p+n i=l
m-
(2.25)
IBp+n
Similarly,
{o,(
J
r[
<:t:,+t t:>)
i=l
X (i n
lzl p-...+ I,.I p:’’--+ k=p+n
(k-l+i))a k}
J
m+,x
(p+n-l+i)
+;
{0,1
’-::’
,:,,+,.-:,.:,)I=I
(p+n) m--
i=l
which gives (2.20).
REMARK 6. If
p+’’-j
1B p+n Izl
(2.26)
Thus we have the theorem.
j
J
n (p+n-l+i))/(p+n) m-I n (p+l-i),
Bp+n)
i=2
then
i=l
J
(p+n- l+i)
=ax
I,.1+’-+-
{o,(
i=1
j
1I
i=l
THEOREM
A(p,n,
kmBk
4.
with
i=2
(p+n)
m-
B
n
(p+l-i))
Let
Izl
Izl p+n-.-i}
p+n
(p+n-l+i)
p-j- i=2
the
Bk+land
(p+n)m-lB p+n
lZl p+n-j
function
f(z)
p+l
p+n. Then
m
defined
by
we have
(1.1)
be
in
the
class
NEW CLASSIFICATION OF ANALYTIC FUNCTIONS WITH NEGATIVE COEFFICIENTS
61
H (p+n- [+i)
(2.27)
(p+n)m- IB p+n
for
z
g
U and p+l
PROOF.
m.
Note that
i kta k
(p+n)
k=p+n
p+l
for
m-
Since
m.
j
f(J)(z)
(k+l-[)akz k-j
|I
k=p+n i
for
(2.29)
and (2.28), we have
m, by using Lemma
p+l
(2.28)
tBp+n
k;p+n i--1
(k-l+i))a k
(p+n- l+ i)
tI
(p+n)m-IB p+n Iz IP+n-J
i =2
<
(2.30)
which completes the proof of Theorem 4.
3.
FRACTIONAL CALCULUS.
Many essentially equivalent definitions of the fractional calculus, that is, the
fractional derivatives and the fractional integrals, have been in the literature (cf.,
[18], [19], [20],
and [21]).
We find it to be convenient to recall here the following
definitions which were used recently by Owa
DEFINITION I.
D
where
%
>
containing
-X
z
O, f(z)
z
f(z)
()- d
(z-
(3 I)
)I-
the multiplicity of
(z- )
The fractional derivative of order
I, f(z)
z
d
order
is
removed by
requiring
is defined by
f(-) d
f
zo
r(-x)
(3 "))
(z-)
is an analytic function in a simply connected region of the z-
to be real when
DEFINITION 3.
-I
(z- ) & O.
plane containing the origin and the multiplicity of
log(z -)
is defined by
is an analytic function in a simply connected region of the z-plane
DXf(z)
z
<
f
o
r(),)
the origin and
DEFINITION 2.
0
%
The fractional Integral or order
log(z -) to be real when
where
([221, [23]).
(z- )
>
(z
)
is removed by requiring
0.
Under the hypotheses of Definition 2, the fractional derivative of
(n + ) is defined by
D
n+
z
f(z)
n
d
dz
n
D f(z),
z
(3.3)
S. OWA and M.
62
4
0
ahere
<
and n
-
OBRADOVI6
N U {0}.
,N
0
With the above definitions of the fractional calculus, we prove
THEOREM
Let
5.
A(p,n,B k) with
II):Xf(z)l
f,mcion
the
Bk Bk+ I.
r(p+l)_Izx I
max {0,
$-+Xf
f(z)
defined
by
(I. I)
be
i
class
the
Then
r (_+n+)
-F+n+--F-p$f)B p+n
Izln)}
(.4)
and
for
4
>
p+
F(_+_ I)
ID-f(z)
z
(I
F(p+I+X)
0 and z
U.
g
F (+n+1) F (_+I+X)
F(p+n+l+X)
n)
F(p+l)Bp+n Iz[
+
le equalities in (3.4) and (3.5) are attained for the function
f(z) given by (2.2).
PROOF.
We define the function F(z) by
r_. +x)
F(z)
>
0. Then the function
f(z)
k
(3.6)
(k) defined by
r(k+
r.(_p+l +4)
r(k+l+,) r(p+l)
@(k)
(k p+n)
(3.7)
Hence we have
is decreasing in k.
0
Dz
r(---k+l-+X---- r(p+l---- akz
k=p+n
X
-4
r(k+1) r!p.+l+),)
zPfor
z
r(p+F)--
<
(k)
(p+n)
F(p+n+ I) r(p+l+4)
(3.8)
r(p/d41$X)-r(p/i
Therefore, it follows from (2.4) and (].8) that
IF(z)1
max
which implies
Izl p-
max {0
{0,
,(p+n)
Y- a
l.l p+n k=p+n
k
r(p+n+l) r(p+l+X)
Izl p-
rCp----+n+l+X) FOp+l) B p+n
Izl p+n
(3.9)
(].4), and
IF(z)
z
which gives (3.5).
IzlP+
,(p+n)
p+ __r(p_n+
Izt p+nk;p+n k
a
r.(p+ +X)
F(p+n+l+;k) r(p+l)
Bp+ n
z
p+n
(3.1o)
NEW CLASSIFICATION OF ANALYTIC FUNCTIONS WITH NEGATIVE COEFFICIENTS
63
Furthermore, since the equalities in (3.9) and (3.10) are attained the function
f(z) defined by
D
F_!p(+1)
-k
f(z)
z
z
p+x
{1
r(p+l+),)
[’(p+n+l)
z n}
F(p+n+l+),) r(p+l) B
p+n
(3.11)
(3.4) and (3.5) ace attained for the function f(z)
we can show that the equalities in
given by (2.2).
REMARK 7.
{F(p+n+l) F(p+I+X))/[F(p+n+I+X) F(p+l)) for A
B
If
p+n
>
O,
then
Next, we derive
THEOREM 6. Let the function f(z) defined by (I.I) be in the class
A(p,n, kB k) with
I. Then we have
Bk Bk+
IDXf(z)l
z
max
r(p+1)_[zlP-X (I
{0
and
r(p+l)
[DXf(Z)lz
for
0
[z[p-)"
r(ik)
X
<
r (p+n) r(p+1-x)
F(p+n+l-X) F(p+l)
(pi
(I +
r(p+n) t’(p+l-),)
r(p+l) B
p+n
r(---n+l-X)
The equalities in
and z
Bp+n [z[n)}
[z[n}
(3.12)
(3.13)
(3.12) and (3.13) are attained for the
function f(z) given by (2.8).
Define the function G(z) by
PROOF.
G(z)
r(p+l)
z
z
f(z)
r(k+l) r (p+l-x)
r(k+l-),) r(p+l)
for
0
h
<
I.
k
(3.14)
Setting
r(k) r (p+l-.)
r(k+l-X) r(p+l)
,(k)
we can see that
akz
(k
p + n),
(3.15)
(k) is a decreasing function of k, that is, that
o >
(k)
(p+n)
r (p+n) r(p+l-x)
r(p+n+l-X) r(p+l)
(3.16)
Consequently, it follows from (2.10) and (3.16) that
max
{0,
k=p+n
max
{0,
{ziP
r(p+n) r(p+l-X)
r(p+n+l-X) r(p+l)
Bp+ n
(3.17)
S. OWA and M.
64
which proves
OBRADOVI6
(3.12), and
IG(z)
IzlP+
Izl p+nk=p+n kak
(p+n)
(3.)
F(p+n+l-k) F(p+l) B
p+n
which shows (3.13).
Finally, we note that the equalities in (3.17) and (3.18) are attained for the
function f(z) def[ned by
-r-ip-+l)zP-)"
D)’f(z)
z
This
(_p_+_n_r__(_.[+l-),)
{I
I’(p+l-k)
zn}.
l(p+n+l-),) l(p+l) B
p+n
(3.19)
implies that the equalities in (3.12) and (3.13) are attained for the function
f(z) given by (2.8).
REMARK 8. If
F(P/I)-[zLX
F(p+I-A)
{0
max
THEOREM
Let
7.
the
Bk Bk+land
function
f(z)
m
p+l
j+l
p+n. Then
p+n-
(p+n)
0
<
<
I, p+l < j
U
PROOF.
p+l
n
z
n}.
1. I)
be
in
the
class
p+n-j
p+n
where
U
(p+l <
U- {0}
(j
p+n-l)
0
No e
.
ha t
m and 0
O(k)
we know that
U0,
by
F(p+n-j
+i
I, then
Bp+n
Bp+n
defined
r(p+n+l-j-%) S
m, and z
Dkf(j)(z
z
for
F(p+n+l-;) r(p+l)
F(p+n) F(p+l-A)
r(p+n+l-),) F(p+l)
i2
for
r(p+n) r(p+l-)
(I
F(p4[ii)
A(p,n, kmB k) with
<
{F(p+n) F(p+l-l)}/{F(p+n+l-k) F(p+l)} for 0
Bp+n)
<
p+n).
J
II (k+
k=p+n i=1
<
i
r(k+l-J)
f(-k+ ILj-- ak
zk-j
(3.21)
I. Denoting
r(k-j)
(k) p+n),
r(k+l-j-).)
O(k) is a decreasing function of k, so that
0
<
O(k) < O(p+n)
Consequently, with the aid of Lemma
r (p+n,j)
#(p+n-l-j--)"
(3.22)
and (2.28), we have
r (p+n-j)
p+n-j -,
Ok+l-1)
k=p+n i
)ak
NEW CLASSIFICATION OF ANALYTIC FUNCTIONS WITH NEGATIVE COEFFICIENTS
F__p_+_n._)
p+n-j
j+l
I!
--2
_[ (_+__ )__
F(p+n+l-j-)
j+l
A.
i=l
(p+n)m-ig p+n
65
p+n- l--i
p+n-j-k
z
(p+n)m-IB p+n
(3.2 3)
which shows the iqequality (3.20).
4.
SOME PROPERTIES OF THE CLASS
A(p,n,Bk).
We shall give some properties of the class
A(p,n,Bk) conslting of functions of
the form (I.I) satisfying the inequality (1.4).
A(p ,n, Bk)
THEOREM 8.
PROOF.
is convex
se.
We need only to prove that the function h(z) defined by
fl(z)
h(z)
+ (I
6)f2(z)
A(p,n,Bk) for functions
is in the class
f.(z)
p
z
A(p,n,
h(z)
.
Bk).
z
p
z
(a
k,jz
{ak,l+
(I
p
Ck zk,
k;p+n
Ck=
ao,l+
(j=l,2) belonging
k,j
O;j
I)
(4.1)
to
A(p,n, Bk). Let
1,2)
(4.2)
Then we have
k=p+n
where
f.(z)j
a
k=p+n
be in the class
(0
(I
)a
k=p+n
BkCk=k
From this,
k ,2
+n
Bk{ak I+
k=p+n
Bkak
+ (1
)ak, 2 }z
k
it is easy to see that
(1
+ (I
)a
.
k 2
6)
k=p+n
Bkak,2
8)
(4.4)
which implies that h(z)
THEOREM 9.
A(p,n, Bk).
Let
fl(z)
z
fk(z)
z
p
(4.5)
and
Then f(z) is in the class
f(z)
el
z
k
(k
A(p,n,Bk) if and only if
Ifi(z)
+
k=p+n
kfk(z),
p+n).
(4.6)
it can be expressed in the form
(4.7)
66
S. OWA and M.
where
i7
0, 6 k ) 0 (k
PROOF.
p+n), and
E
k=P +n
OBRADOVI6
6
6k=
I.
We assume that tile function f(z) can be expressed in tle form (4.7).
Sinc e
(61+’K
f(z)
+n
6k)zP-k=p+n k
6k
Z
.
k
k
k p +n
z
p
k=p+n
(4.8)
dkZ
we observe that
k=p+n
Conversely,
61,
+n k
k=
I,
(4.9)
A(p,n, Bk).
that i,s, that f(z)
that
assne
A(p,n,Bk). Then,
6
Bkdk
the
function
f(z)
defined
by
(I.I)
is
in
the
class
[t follows that
(k
k’ Bk
a
(4.10)
p+n).
Therefore, we may put
6k= Bkak
(k
p+n)
and
61=
6
k=p+n
Thus we prove that
f(z)
z
p
k.
.
.
k=p+n
61fl(z)
+
61f1(z
+
akz
6k
Bk
k=p+n
k--p+n
z
k
6kfk(z)
(4.11)
This completes the assertion of Theorem 9.
By virtue of Theorem 8 and Theorem 9, we have
COROLLARY I.
A(p,n,Bk) are
The extreme points of
fl(Z)
and
fk(z)
(k
p+n) defined
in Theorem 9.
Next, we prove
THEOREM I0.
Let
f.(z)j
(j=l,2) defined by (4.2) be in the class
A(p,n,B,_,jm
NEW CLASSIFICATION OF ANALYTIC FUNCTIONS WITH NEGATIVE COEFFICIENTS
67
Then the function h(z) defined by
h(z)
z
p
’ ak,
a
L
,2
k=p+n
E(p,n,
is in the class
PROOF.
Bk,3),
wheme
(4.12)
z
Bk,3 Bk,lBk, 2.
We need to prove that
(4.13)
k=p+n
for
Bk,3 Bk,IBk,2"
Since
1,2),
(j
(4.14)
k=p+n
by using the Cuchy-Schwarz inequality, we have
(4.1)
Hence, if
,lak,2
k,l k,2
k=p+n
Bk’3ak lak’2
or
SkBk,3Bk_2
I
,lak,2
(k
Bk,l Bk,2 ak,l ak,2
(k
then the inequality (4.13) is satisfied.
(4.16)
p+n),
(4.17)
p+n),
Since
(k
(4.18)
p+n)
by means of (4.15), we can show that if
BklBk=__,______m2==_
Bk,3
that is, if
(4.19)
p+n),
then (4.13) is satisfied. Thus we have Theorem
p+n)
Bk,3 Bk,iBk,2(k
(k
I0.
Finally, we derive
THEOREM
Let
11.
fj(z)
(J=|
,2,... ,m) defined
by
(4.2)
be
in
the
class
A(p,n,Bk). Then the function
h(z)
is in the class
PROOF.
z
m
p
I (I
k=p+n
A(p,n, Ck)., where C
k
2
ak, j iz
k
(4.20)
B/m.
It is sufficient to show that
m
Ck(
k:p+n
j
2
ak, j
(4.21)
S. OWA and M.
68
for
?
Ck B/m.
Note that, for
fj(z)
g
A(p,
OBRADOVI
B
kn
(j=l,2. ,...,m),
2
(4.22)
k=p+n
k=p+n
It follows from (4.22) that
m
(4.23)
k=p+n
j=l
Consequently, we have
2
k=p+n
for
Ck
B
k2/m
ak,j
wh[ch completes the
m
2
B
I
k
k=p+n
2
=I
(4.24)
ak,
proof of Theorem II.
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I.
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2.
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3.
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OWA, S.
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Integral.
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ith
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20.
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