International Journal of Mathematical Archive-2(12), 2011,
Page: 2652-2659
Available online through
www.ijma.info
ON GENERALIZED ORDER AND GENERALIZED TYPE OF VECTOR VALUED
DIRICHLET SERIES OF SLOW GROWTH
*G. S. Srivastava and Archna Sharma
Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee- 247667, INDIA E-mail: [email protected] , [email protected]
(Received on: 10-11-11; Accepted on: 24-11-11)
________________________________________________________________________________________________ ABSTRACT
T
he concept of vector valued Dirichlet series was introduced by B. L. Srivastava [2] who characterized the growth of entire functions represented by these series. In this paper we introduce the generalized order and generalized type of entire functions slow growth.Keywords and Phrases: Vector Valued Dirichlet Series, Entire Functions, Generalized order, Generalized type.
AMS (2010) Subject Classification: 30D15, 46A 35.
________________________________________________________________________________________________
1. INTRODUCTION:
Let
1
( )
sn,
, ( , are real variables),
n n
f s
a e
λs
σ
it
σ
t
∞
=
=
=
+
(1.1)where
a s
n'
belong to a complex commutative Banach algebraE
with identity elementwith ||
|| 1
ω
ω =
andλ
n'
s
∈
satisfy the conditions0
<
λ
1<
λ
2<
λ
3...
<
λ
n...;
λ
n→ ∞
asn
→ ∞
,
lim sup
log ||
n||
,
nn
a
λ
→∞
= −∞
(1.2)and
lim sup
log
.
nn
n
D
λ
→∞
=
< ∞
(1.3)Then, the vector valued Dirichlet series in (1.1) represents an entire function
f s
( )
(see [2]). In [2], B. L. Srivastava defined the growth parameters such as order, type, lower order, lower type of the vector valued entire Dirichlet series. He also obtained the result for coefficient characterization of order and type. In this paper we obtain the generalised order and generalised type of vector valued Dirichlet series.
For the entire function
f s
( )
defined as above by (1.1) the maximum modulus, the maximum term and the index of maximum term are defined as
( )
{
}
{
}
{
}
sup || (
) ||;
,
( )
max ||
||
;
,
( )
max
; ( ) ||
||
,
.
n
n
n
n
M
f
it
t
m
a
e
n
N
N
n m
a
e
n
N
σλ
σλ
σ
σ
σ
σ
σ
+
+
=
+
∈
=
∈
=
=
∈
The order
ρ
off s
( )
is defined as
lim sup
log log
M
( )
, 0
(
)
σσ
ρ
ρ
σ
→∞
=
≤
≤ ∞
________________________________________________________________________________________________
Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659
© 2011, IJMA. All Rights Reserved 2653
and for
0
<
ρ
< ∞
the typeT
off s
( )
is defined as (see [2])T
lim sup
log
M
( )
, (0
T
)
e
σρσ
σ
→∞
=
≤
≤ ∞
.We shall call the entire function
f s
( )
to be of slow growth if the orderρ
=
0.
We obtain the characterization of growth parameters. in the context of generalised order and generalised type of vector valued Dirichlet series of slow growth. LetΦ
denote the class of functionsh x
( )
satisfying the following conditions:(i)
h x
( )
is defined on[ , )
a
∞
and is positive, strictly increasing, differentiable and tends to∞
asx
→ ∞
,(ii)
lim
( ( ))
[ ](0, ),
1, 2, 3,...
(log
p)
x
d h x
k
p
d
x
→∞
= ∈
∞
=
where
log
[0]x
=
x
, log
[1]x
=
log , log
x
[ ]px
=
log(log
[p−1]x
),
(iii) For every
c
>
0,
lim
(
)
1, lim
(
)
1
( )
( )
x x
h cx
h c
x
h x
h x
→∞ →∞
+
=
=
(1.4)that is ,
h x
( )
is slowly increasing.Let
α
( )
x
∈ Φ
,
the generalised orderρ α
( , )
f
of the entire functionf s
( )
given by (1.1) can be defined as
( , )
lim sup
(log
( ))
,
( )
M
f
σ
α
σ
ρ
ρ α
α σ
→∞
=
=
For
0
<
ρ
< ∞
the generalised typeT
( , )
α
f
of the entire functionf s
( )
is defined by
( , )
lim sup
(
( ))
lim sup
(log
( ))
,
[ (
)]
[ ( )]
M
M
T
T
f
e
σ ρ ρσ σ
α
σ
β
σ
α
α
β σ
→∞ →∞
=
=
=
where
β
(log )
x
=
α
( ).
x
2. MAIN RESULTS:
Now we prove
Lemma: 1 If the vector valued Dirichlet series given by (1.1) satisfies (1.2) and (1.3), then
lim
(log
( ))
lim
(log ( ))
,
(0, )
[ ( )]
[ ( )]
M
m
δ δ
σ σ
α
σ
α
σ
δ
α σ
α σ
→∞
=
→∞∈
∞
(2.1)Proof: From the equation (1.2), for a given
ε
>
0,
there exists an integerN
,
such that forn
>
N
,1/( )
n D
e
−λ<
n
− +ε .By the definition ofM
( )
σ
,we have( )
1 1
( ) ( ) 1
( )
||
||
||
||
,
(1)
||
||
.
n n
n n
n n
n n
D D
n n N
M
a
e
a
e
O
a
e
e
σλ σ η η λ
σ η λ η λ
σ
η
ε
∞ ∞
+ −
= =
∞
+ + − +
= +
≤
=
>
=
+
( ) /( )
1
(1)
(
)
D Dn N
O
m
σ
D
η
n
η ε∞
− + +
= +
≤
+
+
+
≤
O
(1)
+
K
( , ) (
ε η
m
σ
+
D
+
η
)
Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659
© 2011, IJMA. All Rights Reserved 2654
(log
( ))
(log
( , ) log (
)) [ (
)]
lim
lim
.
[ ( )]
[ (
)]
[ ( )]
M
K
m
D
D
D
δ
δ δ δ
σ σ
α
σ
α
ε η
σ
η
α σ
η
α σ
α σ
η
α σ
→∞ →∞
+
+
+
+
+
≤
+
+
By (1.4), it follows that
lim
(log
( ))
lim
(log ( ))
.
[ ( )]
[ ( )]
M
m
δ δ
σ σ
α
σ
α
σ
α σ
α σ
→∞
≤
→∞The reverse inequality follows from the well known relation
m
( )
σ
≤
M
( ).
σ
Hence the Lemma is proved.Lemma: 2 If the vector valued Dirichlet series given by (1.1) satisfies (1.2) and (1.3), then
0
0 ( ) 0
log ( )
m
log (
m
)
N tdt
,
0.
σσ
σ
=
σ
+
λ
σ
>
The result can be proved on the lines of [3].Hence we omit the proof.
Next we prove
Theorem: 1 Let the vector valued Dirichlet series (1.1) satisfy (1.2) and (1.3), then
(
1/)
(
)
(ln
( ))
(a) lim sup
1
lim sup
,
for
1
( )
ln ||
||
nn
n
M
p
a
λσ σ
α λ
α
σ
α σ
α
−→+∞
− =
→+∞=
,(
1/)
(
1/)
( ) (ln ( )) ( )
(b) lim sup lim sup lim sup 1, for 2, 3...
( )
ln || || n ln || || n
n n
n n
M
p
a λ a λ
σ σ σ
α λ
α
σ
α λ
α σ
α
−α
−→+∞ ≤ →+∞ ≤ →+∞ + =
Proof: We prove this theorem in two steps.
Case I: When
p
=
1,
by condition (ii), if
( )
k
ln
c
, then
1( )
exp
c
ce
/k.
k
σ
σ
α σ
σ
α
−σ
−
=
+
=
=
By Lemma 1,
lim
(log
( ))
lim
(log ( ))
( )
( )
M
m
A
σ σ
α
σ
α
σ
α σ
α σ
→+∞
=
→+∞=
.For
p
=
1,
from the given condition (ii),we can easily deduce thatα σ
( )
≅
k
log
σ
. IfA
<
1
then there existsε
> 0 such thatA
+ <
ε
1
.For sufficiently large values ofσ
,
we get
M
( )
σ
<
e
σA+ε.Hence
f s
( )
reduces to an exponential polynomial. Hence forp
=
1, we have
A
≥
1.
Let us suppose thatA
< ∞
.
Then for everyε
>
0,
∃
σ ε
0( )
>
0 such that for
σ
≥
σ
0,*
(log ( ))
( )
m
A
A
α
σ
ε
α σ
<
+ =
,or
* *
1 *
( )
log ( )
m
[
A
( )]
exp
A
c
c
Ak
α σ
σ
α
−α σ
−
σ
<
=
=
,Then
*
log ||
||
nlog ( )
An
a
e
λ σ≤
m
σ
<
c
σ
,or
* 1
n 0
log||a ||
A,
n
c
λ σ
σ
σ
σ
−
Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659
© 2011, IJMA. All Rights Reserved 2655
When
n
is large enough, setting* 1 1 * A n
A
λ
σ
=
− , we have1 * * * 1 * * 1 1 1 * * * *
log ||
||
(
)
A A A
A A
n n n
n n
a
c
A
c
A
A
A
λ
λ
λ
λ
−
− −
−
≥
−
=
−
.Now using (1.4) for the function
α
, we get1/
*
(
)
(log ||
||
)
(1)
1
n n
n
a
o
A
λ
α λ
α
−≥
+
−
.Hence we obtain
1/
(
)
lim sup
1
(log ||
||
n)
n n n
A
a
λα λ
α
− →+∞≤
−
.When
p
=
2, 3...,
we suppose thatA
< ∞
. From the above proof, it follows that1 *
1 * n
log ( )
[
( )]
or
log||a ||
[
( )]
nm
A
A
σ
α
α σ
α
α σ
λ σ
−
−
<
<
−
.Choose
σ
=
σ λ
(
n)
to be the unique root of the equation1 *
(
n)
A
α λ
σ
α
−=
,(
σ
→ ∞ ⇔
n
→ ∞
)
1/
1/
Then
log ||
||
1
or
( log ||
||
)
(
1)
n n n n
a
a
λ λσ
α
α σ
−
−
>
−
>
−
.By (1.4), when
σ
is sufficiently large, we haveα σ
(
−
1)
=
(1
+
o
(1)) ( )
α σ
, thus1/
*
* 1/
(
)
( log ||
||
)
(1
(1)) ( )
(1
(1))
(
)
or
(1
(1))
( log ||
||
)
n n n n n na
o
o
A
o
A
a
λ λα λ
α
α σ
α λ
α
− −≥
+
=
+
≤
+
,Now proceeding to limits, since
ε
>
0 is arbitrary,
we obtain1/
(
)
lim sup
(log ||
||
n)
n n n
A
a
λα λ
α
− →+∞≤
.The above inequality obviously holds when
A
= ∞
.Conversely, let
1 /
(
)
lim sup
(log ||
||
n)
n n n
B
a
λα λ
α
− →+∞=
.We suppose that
B
< ∞
.Then for a given
ε
>
0 and for all
n
≥
n
0( ), we have
ε
1 /
*
(
)
(log ||
||
n)
n n
B
B
a
λα λ
ε
α
−<
+ =
1 *
(
)
1
Hence
nlog ||
||
n n
a
B
α λ
α
λ
−< −
,Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659
© 2011, IJMA. All Rights Reserved 2656
or
||
|| exp
1(
*n)
n n
a
B
α λ
λ α
−<
−
,∀ >
n n
0 (2.2)From (1.3), there exists
r
>
0
, such thatλ
n>
r
log .
n
Further, whenσ
is sufficiently large, there existsN
>
n
0, so thatN 1
[
B
*(
2
)]
N 1r
λ
α
−α σ
λ
+
≤
+
≤
. (2.3)We have
0
0
0 1 2
1 1 1
( )
||
||
n||
||
n||
||
nn N
n n n
n n n n N
M
σ
a
e
λ σa
e
λ σa
e
λ σA
A
A
∞
= = + = +
≤
+
+
=
+
+
.Using equations (2.2) and (2.3)
0 0
1 * 1
1 *
1 1
(
)
2
||
||
exp[
{
(
)}].
exp
N
N N
n
n n
n n n n
A
e
a
B
r
B
λ σ
α λ
σα
−α σ
λ α
−= + = +
≤
≤
+
−
1 *
0
( )
1 * 1 *
1
2
2
exp[
{
(
)}].
exp[
{
(
)}]
n
N r
B
n n
B
n
C
B
r
r
α λ α
σα
α σ
σα
α σ
− −
− −
= +
=
+
≤
+
.In
A
2, we have n 1[
B
*(
2
)]
r
λ
α
−α σ
>
+
that is1 *
(
n)
2
B
r
α λ
σ
α
−<
−
.From (2.2) and the above inequality, it follows that
2
1 1
2 * *
1
(
)
(
)
exp
.exp
.
n
n n r
n n
n N
A
e
B
B
λ
α λ
α λ
λ α
λ α
∞ −
− −
= +
≤
−
−
2
2
1 1
1
where
is a constant.
nr
n N n N
e
K
K
n
λ
∞ − ∞
= + = +
=
≤
<
.Accordingly,
1 *
2
( )
(1
(1))
[
{
(
)}]
M
o
K
B
r
σ
α σα
−α σ
≤
+
+
,As in Lemma 2 (see [1]), we have
1 *
2
*( )
(1
(1)) [
{
(
)}]
(1
(1))(
1) ( )
M
o
B
o
B
r
σ
α σα
−α σ
α σ
≤
+
+
=
+
+
.Thus
(log
( ))
lim sup
1
( )
M
B
σ
α
σ
α σ
→+∞
≤
+
.Combining the inequalities obtained in case I and II, we get (a) and (b) above and the proof of Theorem 1is complete.
Theorem: 2. Let the entire function represented by the vector valued Dirichlet series (1.1) satisfiy (1.2) and (1.3), and be of generalized order
ρ
∈
(1, )
∞
. Then
(
1/)
(
)
(log
( ))
lim sup
lim sup
[ ( )]
log ||
||
nn
n
M
a
ρρ
σ σ λ
β λ
β
σ
η
β σ
β
→∞ →∞ −
=
=
. (2.4)Proof: By Lemma 1, we have
lim sup
(log
( ))
lim sup
(log ( ))
[ ( )]
[ ( )]
M
m
ρ ρ
σ σ
β
σ
β
σ
β σ
β σ
→∞ →∞
Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659
© 2011, IJMA. All Rights Reserved 2657
We suppose
Η < ∞
. Then for a givenε
>
0 ,
∃
σ ε
0( )
>
0 such that for all
σ
>
σ
0, we have
(log ( ))
[ ( )]
m
ρ
β
σ
ε
β σ
< Η + = Η
,or
log ( )
m
σ
β
−1[ { ( )} ],
β σ
ρ<
Η
or
log||a ||
nβ
−1[ { ( )} ]
β σ
ρλ σ
n<
Η
−
.We choose
σ
=
σ λ
(
n)
to be the unique root of the equation1/ 1
(
n)
ρ
β λ
σ
β
−=
Η
,(
σ
→ ∞ ⇔
n
→ ∞
)
.Then
log ||
||
1/ n1 or
( log ||
||
1/ n)
(
1)
n n
a
− λσ
β
a
− λβ σ
>
−
>
−
.By (2.1), when
σ
is sufficiently large, we haveβ σ
(
−
1)
=
(1
+
o
(1)) ( )
β σ
, thus1
(
)
[ ( log ||
||
n)]
(1
(1))[ ( )]
(1
(1))
nn
a
λ ρo
ρo
β λ
β
−≥
+
β σ
=
+
Η
,or
(
1/)
(
)
(1
(1))
log ||
||
nn
n
o
a
ρ λ
β λ
ε
β
−+
≤ Η = Η +
.Now proceeding to limits, we obtain
(
1/)
(
)
lim sup
log ||
||
nn
n
n
a
λ ρβ λ
β
→+∞ −
≤ Η
. (2.5)The above inequality obviously holds when
Η = ∞
.To obtain the reverse inequality, let
(
1/)
(
)
lim sup
log ||
||
nn
n
n
B
a
λ ρβ λ
β
→+∞ −
=
.
We suppose that
B
< ∞
.Then for
ε
>
0 and for all
n
≥
n
0( ), we have
ε
(
)
* 1/
(
)
log ||
||
nn
n
B
B
a
λ ρβ λ
ε
β
−<
+ =
or
(
)
*[ (log ||
||
1/ n)] .
n
B
a
nλ ρ
β λ
α
−<
Then
1/ 1
*
(
)
1
log ||
||
n
n n
a
B
ρ
β λ
β
λ
−
< −
.Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659
© 2011, IJMA. All Rights Reserved 2658
1/ 1
*
(
)
||
||
exp
nn n
a
B
ρ
β λ
λ β
−<
−
. (2.6)From (1.3), there exists
r
>
0
, such thatlog .
n
r
n
λ
>
.In addition, when
σ
is sufficiently large, there existsN
>
n
0, so thatN 1
[
B
*{ (
2
)} ]
N 1r
ρ
λ
β
−β σ
λ
+
≤
+
≤
. (2.7)We have
0
0
' ' ' 0 1 2
1 1 1
( )
||
||
n||
||
n||
||
nn N
n n n
n n n n N
M
σ
a
e
λ σa
e
λ σa
e
λ σA
A
A
∞
= = + = +
≤
+
+
=
+
+
.Using equations (2.6) and (2.7), we obtain
0 0
1/
' 1 * 1
1 *
1 1
(
)
2
||
||
exp
(
)
exp
N
N N
n
n n
n n n n
A
e
a
B
r
B
ρ ρ
λ σ
σβ
−β σ
λ β
−β λ
= + = +
≤
≤
+
−
1 * 1 /
0
[{ ( ) / } ]
1 * 1 *
1
2
2
exp
(
)
nexp
(
)
N
r B
n n
B
n
C
B
r
r
ρ
ρ ρ
β β λ
σβ
−β σ
− −σβ
−β σ
= +
=
+
≤
+
In
A
2', we have n 1B
*(
2
)
r
ρ
λ
β
−β σ
>
+
.Then
1/ 1
*
(
n)
2
B
r
ρ
β λ
σ
β
−<
−
.From (2.6) and the above inequality, it follows that
1/ 1/
2 /
' 1 1
2 * *
1
(
)
(
)
exp
n.exp
n n rn n
n N
A
e
B
B
ρ ρ
λ
β λ
β λ
λ β
λ β
∞
−
− −
= +
≤
−
2 /
2
1 1
1
'
n rn N n N
e
C
n
λ
∞ ∞
−
= + = +
=
≤
<
.Accordingly,
1 *
2
( )
(1
(1)) exp
(
)
M
o
C
B
r
ρ
σ
σβ
−β σ
≤
+
+
,Hence
1 *
2
(log
M
( ))
(1
o
(1))
B
(
)
r
ρ
β
σ
β σβ
−β σ
≤
+
+
.Hence, by Corollary 2 see [1], it follows that
(log
( ))
lim sup
[ ( )]
M
B
ρ σ
β
σ
β σ
→+∞
≤
.Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659
© 2011, IJMA. All Rights Reserved 2659 REFERENCES:
[1] Yinying Kong and Huilin Gan - On order and type of Dirichlet series of slow growth, Turk J Math, 34 (2010), 1-11.
[2] Srivastava B.L. -A study of spaces of certain classes of Vector Valued Dirichlet Series, Ph. D. Thesis, I.I.T Kanpur, 1983.
[3] Yu C.Y. - Sur les droites de Borel de certaines functions entireres, Ann. Sci. le’cole Norm. Sup. 68, (1951), 65-104.