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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

λ

s

σ

it

σ

t

=

=

=

+

(1.1)

where

a s

n

'

belong to a complex commutative Banach algebra

E

with identity element

with ||

|| 1

ω

ω =

and

λ

n

'

s

satisfy the conditions

0

<

λ

1

<

λ

2

<

λ

3

...

<

λ

n

...;

λ

n

→ ∞

as

n

→ ∞

,

lim sup

log ||

n

||

,

n

n

a

λ

→∞

= −∞

(1.2)

and

lim sup

log

.

n

n

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

of

f s

( )

is defined as

lim sup

log log

M

( )

, 0

(

)

σ

σ

ρ

ρ

σ

→∞

=

≤ ∞

________________________________________________________________________________________________

(2)

Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659

© 2011, IJMA. All Rights Reserved 2653

and for

0

<

ρ

< ∞

the type

T

of

f s

( )

is defined as (see [2])

T

lim sup

log

M

( )

, (0

T

)

e

σρ

σ

σ

→∞

=

≤ ∞

.

We shall call the entire function

f s

( )

to be of slow growth if the order

ρ

=

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. Let

Φ

denote the class of functions

h x

( )

satisfying the following conditions:

(i)

h x

( )

is defined on

[ , )

a

and is positive, strictly increasing, differentiable and tends to

as

x

→ ∞

,

(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

[ ]p

x

=

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 order

ρ α

( , )

f

of the entire function

f 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 type

T

( , )

α

f

of the entire function

f s

( )

is defined by

( , )

lim sup

(

( ))

lim sup

(log

( ))

,

[ (

)]

[ ( )]

M

M

T

T

f

e

σ ρ ρ

σ 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 integer

N

,

such that for

n

>

N

,

1/( )

n D

e

−λ

<

n

− +ε .By the definition of

M

( )

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 D

n N

O

m

σ

D

η

n

η ε

− + +

= +

+

+

+

O

(1)

+

K

( , ) (

ε η

m

σ

+

D

+

η

)

(3)

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 t

dt

,

0.

σ

σ

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 ||

||

n

n

n

M

p

a

λ

σ 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 that

α σ

( )

k

log

σ

. If

A

<

1

then there exists

ε

> 0 such that

A

+ <

ε

1

.For sufficiently large values of

σ

,

we get

M

( )

σ

<

e

σA+ε.

Hence

f s

( )

reduces to an exponential polynomial. Hence for

p

=

1, we have

A

1.

Let us suppose that

A

< ∞

.

Then for every

ε

>

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

A

k

α σ

σ

α

α σ

σ

<

=

=

,

Then

*

log ||

||

n

log ( )

A

n

a

e

λ σ

m

σ

<

c

σ

,

or

* 1

n 0

log||a ||

A

,

n

c

λ σ

σ

σ

σ

(4)

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 have

1 * * * 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 get

1/

*

(

)

(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

λ

α λ

α

− →+∞

.

When

p

=

2, 3...,

we suppose that

A

< ∞

. From the above proof, it follows that

1 *

1 * n

log ( )

[

( )]

or

log||a ||

[

( )]

n

m

A

A

σ

α

α σ

α

α σ

λ σ

<

<

.

Choose

σ

=

σ λ

(

n

)

to be the unique root of the equation

1 *

(

n

)

A

α λ

σ

α

=

,

(

σ

→ ∞ ⇔

n

→ ∞

)

1/

1/

Then

log ||

||

1

or

( log ||

||

)

(

1)

n n n n

a

a

λ λ

σ

α

α σ

>

>

.

By (1.4), when

σ

is sufficiently large, we have

α σ

(

1)

=

(1

+

o

(1)) ( )

α σ

, thus

1/

*

* 1/

(

)

( log ||

||

)

(1

(1)) ( )

(1

(1))

(

)

or

(1

(1))

( log ||

||

)

n n n n n n

a

o

o

A

o

A

a

λ λ

α λ

α

α σ

α λ

α

− −

+

=

+

+

,

Now proceeding to limits, since

ε

>

0 is arbitrary,

we obtain

1/

(

)

lim sup

(log ||

||

n

)

n n n

A

a

λ

α λ

α

− →+∞

.

The above inequality obviously holds when

A

= ∞

.

Conversely, let

1 /

(

)

lim sup

(log ||

||

n

)

n n n

B

a

λ

α 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

λ

α λ

ε

α

<

+ =

1 *

(

)

1

Hence

n

log ||

||

n n

a

B

α λ

α

λ

< −

,

(5)

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 that

λ

n

>

r

log .

n

Further, when

σ

is sufficiently large, there exists

N

>

n

0, so that

N 1

[

B

*

(

2

)]

N 1

r

λ

α

α σ

λ

+

+

. (2.3)

We have

0

0

0 1 2

1 1 1

( )

||

||

n

||

||

n

||

||

n

n N

n n n

n n n n N

M

σ

a

e

λ σ

a

e

λ σ

a

e

λ σ

A

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 is

1 *

(

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.

n

r

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 ||

||

n

n

n

M

a

ρ

ρ

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

→∞ →∞

(6)

Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659

© 2011, IJMA. All Rights Reserved 2657

We suppose

Η < ∞

. Then for a given

ε

>

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 ||

n

β

−1

[ { ( )} ]

β σ

ρ

λ σ

n

<

Η

.

We choose

σ

=

σ λ

(

n

)

to be the unique root of the equation

1/ 1

(

n

)

ρ

β λ

σ

β

=

Η

,

(

σ

→ ∞ ⇔

n

→ ∞

)

.

Then

log ||

||

1/ n

1 or

( log ||

||

1/ n

)

(

1)

n n

a

− λ

σ

β

a

− λ

β σ

>

>

.

By (2.1), when

σ

is sufficiently large, we have

β σ

(

1)

=

(1

+

o

(1)) ( )

β σ

, thus

1

(

)

[ ( log ||

||

n

)]

(1

(1))[ ( )]

(1

(1))

n

n

a

λ ρ

o

ρ

o

β λ

β

+

β σ

=

+

Η

,

or

(

1/

)

(

)

(1

(1))

log ||

||

n

n

n

o

a

ρ λ

β λ

ε

β

+

≤ Η = Η +

.

Now proceeding to limits, we obtain

(

1/

)

(

)

lim sup

log ||

||

n

n

n

n

a

λ ρ

β λ

β

→+∞ −

≤ Η

. (2.5)

The above inequality obviously holds when

Η = ∞

.

To obtain the reverse inequality, let

(

1/

)

(

)

lim sup

log ||

||

n

n

n

n

B

a

λ ρ

β λ

β

→+∞ −

=

.

We suppose that

B

< ∞

.Then for

ε

>

0 and for all

n

n

0

( ), we have

ε

(

)

* 1/

(

)

log ||

||

n

n

n

B

B

a

λ ρ

β λ

ε

β

<

+ =

or

(

)

*

[ (log ||

||

1/ n

)] .

n

B

a

n

λ ρ

β λ

α

<

Then

1/ 1

*

(

)

1

log ||

||

n

n n

a

B

ρ

β λ

β

λ

< −

.

(7)

Growth / IJMA- 2(12), Dec.-2011, Page: 2652-2659

© 2011, IJMA. All Rights Reserved 2658

1/ 1

*

(

)

||

||

exp

n

n n

a

B

ρ

β λ

λ β

<

. (2.6)

From (1.3), there exists

r

>

0

, such that

log .

n

r

n

λ

>

.

In addition, when

σ

is sufficiently large, there exists

N

>

n

0, so that

N 1

[

B

*

{ (

2

)} ]

N 1

r

ρ

λ

β

β σ

λ

+

+

. (2.7)

We have

0

0

' ' ' 0 1 2

1 1 1

( )

||

||

n

||

||

n

||

||

n

n N

n n n

n n n n N

M

σ

a

e

λ σ

a

e

λ σ

a

e

λ σ

A

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

(

)

n

exp

(

)

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 1

B

*

(

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 r

n n

n N

A

e

B

B

ρ ρ

λ

β λ

β λ

λ β

λ β

− −

= +

2 /

2

1 1

1

'

n r

n 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

→+∞

.

(8)

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.

References

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