AJMS

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ISSN 2581-3463
RESEARCH ARTICLE
On a Class of Functions Analytic in a Half Plane
Kirti Chauhan
Department of Mathematics, P.P.N. College, Kanpur, Uttar Pradesh, India
Received: 15-07-2021; Revised: 20-08-2021; Accepted: 15-09-2021
ABSTRACT
In this paper, we have defined a class of functions represented by generalized Dirichlet series whose
coefficients satisfy certain given conditions. Region of convergence is obtained depending on the fixed
Dirichlet series. This class is complete and becomes an Hilbert space with respect to the inner product
defined. Schauder basis is also obtained.
Key words: Abscissa of convergence, abscissa of absolute convergence, Dirichlet series, inner
product, norm, Schauder basis
A.M.S. Subject Classification: 46C05
Address for correspondence:
Kirti Chauhan
E-mail: kirtichauhan98@gmail.com
INTRODUCTION
In the present work, we shall throughout consider
the functions `f’ represented by generalized
Dirichlet series f s a e
k
s
k
k
( ) = =
∞
∑ 
1
with positive
exponents rather than negative ones. It is just to
have an analogy with the power series a z
k
k
k=
∞
∑ 1
.
The only difference in the characteristics of the
Dirichlet series with positive exponents, as
compared to the same, with negative exponents
lies in the region of their convergence that is what
we call the half plane of convergence. The
Dirichlet series with positive exponents converges
in the left half plane whereas the one with negative
exponents converges in the right half plane. We
shall also take the liberty of interpreting the results
of all those workers who have considered the
Dirichlet series with negative exponents in our
terminology.[1-5]
It is well known that the Dirichlet series[2,5]
with
positive exponents (frequencies), in its most
generalized form, is given by
f s a e
k
s
k
k
( ) = =
∞
∑ 
1
(1.1)
Where, s=σ+it (σ,t real variables),{ak
} in general,
is a complex sequence, {λk
} is a strictly increasing
sequence of positive real numbers, and λk
→∞ as
k→∞. It has been proved in[2]
that the exponents
{λk
} satisfy
limsupk
k
logk
D
→ ∞ = ∞

(1.2)
The abscissa of ordinary convergence (abs.
convergence) σc
(σa
) of the Dirichlet series (1.1)
is defined as the least upper bound of all those `σ’
for which the series converges (abs.). These are
given as follows
c
k
m
k
m
k
k k
log a
provided
or a diverg
=
−
≤
→ ∞
=
=
∞
∑
∑
limsup
1
1
0 e
es
log a
provided
or a conver
k
m k m
k
k
k
−
→ ∞
= +
∞
=
∞
∑
∑
limsup
1
1
0
g
ges













(1.3)
σ
σ
a
k
m
k
m
k
k
k
log a
or a diverges abs
=
− ≤
−
→ ∞
=
=
∞
∑
∑
limsup
l
1
1
0

:
( .)
i
imsupk
m k m
k
k
k
log a
or a converges abs
→ ∞
= +
∞
=
∞
∑
∑



1
1
0

:
( .)
σ











(1.4)

2. Chauhan: On a class of functions in a half plane
AJMS/Jul-Sep-2021/Vol 5/Issue 3 21
Also
0 ≤ − ≤
 
c a D (1.5)
If D given in Equation (1.2) is zero, then the
abscissa of ordinary convergence, that is, σc
and
the abscissa of absolute convergence, that is, σa
have the same value which is given by,
σ σ
α
λ
c a k
k
k
log
= = − → ∞
limsup (1.6)
Let
u s
( ) = =
∞
∑ α λ
k
s
k
e k
1
(1.7)
be a fixed Dirichlet serieswith given exponents or
frequencies {λk
} satisfying,
limsupk
k
logk
D
→ ∞ = ∞

(1.8)
and the coefficients {αk
} being a fixed sequence
of non-zero complex numbers satisfying the
condition,
−limsupk
k
k
log
A
→ ∞ =
α
λ
(1.9)
Where, A is any arbitrary but fixed real number. If
D is zero, then the Dirichlet series u(s) has
coincident abscissa of absolute convergence ( )
a
u
and abscissa of ordinary convergence ( )
c
u
, which
is given by the formula,
σ σ
α
λ
c
u
a
u k
k
k
log
A
= = − =
→ ∞
limsup (1.10)
or equivalently by − =
→ ∞
−
limsupk k
a e
k
A
1
 .
The left plane σA is denoted by Ru
and is called
the region of convergence of the fixed Dirichlet
series u(s). Furthermore, the series u(s) converges
uniformly in each left half plane σ=A-ϵ, ϵ0. The
sum of the series u(s) is an analytic function in the
left half plane σA and an entire function if A=+∞.
The coefficients of the fixed Dirichlet series can
be expressed in terms of its sum function by the
Hadamard formula,
k
T
T
T
' ' A it
T
u it e dt
k
'
= +
( )
→
−
− +
( )
∫
lim

1
2
Let the class Ω(u,p) is defined as follows,
Ω(u,p)={ :
f f s a e
k
s
k
k
( ) = =
∞
∑ 
1
and
a
±
1
k
k
p
k
∞ ≤
=
∑ for p ∞
∞
}
1
It can easily be seen that the function `u’ given by
the fixed Dirichlet series u(s) does not belong to
Ω(u,p) for p≥1.
Now, for f f s a e
k
k
s k
, ( ) = =
∞
∑ 1

and
g g s u,p
, ( ) = ∈ ( )
=
∞
∑ b e
k
k
s k
1

We define the following pointwise linear
operations and norm in Ω(u,p) as in the space Ωu,c
)
in the following way.
a. ( ) ( )
f a b e
k k
s
k
k
+ ( ) = +
=
∞
∑
g s 
1
b. µ µ λ
f s a e
k
s
k
k
( )( ) = =
∞
∑ 1
where μ is a
scalar
c. f p
p
k
= =
∞
∑
a
±
k
k
p
1
Clearly, f p
p
exists. It can further be seen that
Ω(u,p) is a normed linear space also.
Ω(u,p) As a Banach space[2]
Here, we have proved that Ω(u,p) is a Banach
space for 1≤p∞ which cannot become a Hilbert
space unless p=2 and possess the property of
uniform convergence[3]
over every compact subset
in the region of convergence Ru
of u.
Theorem [1]:
Ω(u,p) is a Banach space for p≥1.
Proof:
Let {fl
} be a Cauchy sequence in Ω(u,p) such that
f s a e
l lk
s
k
k
( ) = =
∞
∑ 
1
then
a
±
lk
k
p
k
∞
=
∑ k
1

Since, {fl
} is a Cauchy sequence in Ω(u,p) so, for
a given ∈0, there exists a positive integer N0
(∈)
such that
f f
l m p
− ∈ for l,m N
∈
( )
0
or
a a
lk mk
k
k
p p
−
=
∞
∑





 ∈

1
1
⇒
a a
lk mk
k
p
p
k
−
=
∞
∈
∑ 
1
(2.1)
⇒
a a
lk mk
k
p
p
−
∈

for each k

3. Chauhan: On a class of functions in a half plane
AJMS/Jul-Sep-2021/Vol 5/Issue 3 22
⇒
a a
lk mk
k
−
∈

� for each k
This shows that
alk
k







is a Cauchy sequence in C
for each k and so converges to
a k
k
0

(say) as l→∞.
i.e., liml
lk
k
k
k
a a
→ ∞ =
 
0
Define a f s a e
k
s
k
k
0 0
1
( ) = =
∞
∑ 
, we show that fl
→f0
as l→∞ and f0
∈Ω(u,p).
From Equation (2.1), we see that for any positive
integer m,
a a
lk mk
k
p
p
k
m −
=
∈
∑ 
1
. Let m→∞ then
a a
lk k
k
p
p
k
−
=
∞
∈ ∈
∑ 0
1

for lN0
(∈). This shows
that (fl
-f0
) belongs to Ω(u,p) such that f f
l p
− ∈
0
i.e. fl
→f0
as →∞. Now, it follows then f0
=fl
+(f0
-fl
)
∈Ω(u,p).
Theorem [2]:
fl
→f in Ω(u,p)⇒ fl
(s)→f(s) uniformly over every
compact subset of the region of convergence Ru
of u.
Proof:
Let S be a compact subset in the region of
convergence Ru
, then we get a rectangle T in Ru
containing S where
T={(σ,t):σ1
≤σ≤σ2
,t1
≤t≤t2
}
Let
σ3
satisfies σ2
σ3
σ so that θ
σ
σ
=
e
e
2
3
1. Then, for
any given ϵ0 choose a `η’ such that
η θλ
.M k
k=
∞
∑
1
 where ± e M
k
k
3
≤ ∞
and
f f
l p
−  for l l
≥ ( )
0 
⇒
a a
lk k
k
p
k
−
≤
=
∞
∑ 
· p
1
for l l
≥ ( )
0 
a a
lk k
k
p
p
−
α
η for each k=1,2,3,…. l≥l0
(η)
Then for s ∈S and l ≥l0
, we have
f s f s a a e
l lk k
k
k
( )− ( ) ≤ −
=
∞
∑ σλ
1
≤
−
=
∞
∑
a a
lk k
k
k

± e
k
k
1
3 k k
k
k 1 k 1
e M
∞ ∞
σ λ λ
= =
≤ η α ≤ η θ ε
∑ ∑
which proves uniform convergence[3]
over every
compact subset of the region of convergence
Ru
of u.
Theorem [3]:
Ω(u,p) cannot become a Hilbert space unless p=2.
Proof:
Let f s es
( ) = +
 
1 2
1 2
λ λ
es
,
g s es
( ) = + −
α α
λ λ
1 2
1 2
( )es
. Clearly, f,g ∈Ω(u,p).
For p≠2, we see that
fp
p
p p
k
= +
=
=
∑
a
±
k
k
p




1
1
2
2
1
2
⇒ =
fp
p
2
1
,
similarly
1
2p
p
g =
Also f g p
p
p
+ = =
2
2
1
1


p
⇒ f g p
+ = 2
and f g p
p
p
− = =
2
2
2
2


p
⇒ f g p
− = 2
Clearly f g f g f g
+ − ≠ +
+
p p p p
2 2 2 2
2 2
. . as
8 ≠ +






2 2 2
2 2
p p
for p≠2. Thus, Parallelogram
law[1,6]
does not hold showing that it is not a Hilbert
space.
Next, we find Schauder basis[4]
for Ω(u,p) (1≤p∞)
in the following Lemma.
Theorem [4]:
The set ±k k
es k
λ ∞
{ } =1
is the Schauder basis for
Ω(u,p).
Proof:
Let f∈ Ω(u,p), where f f s a e
k
s
k
k
: ( ) = =
∞
∑ 
1
=
k
k 1
a k

=
∑ ′
δ where δ λ
k
s
s e k
( ) = ±k and a
a
±
k
' k
k
=
fork=1,2,3. Then, for a given ε0, there exists a
positive integer `n’ such that
a
±
k
k
p
k n
= +
∞
∑
1

But then
n ' '
k k
k 1 k n 1
lim a lim a
p
p
k p k
n n p
f

 = = +
→ →∞
=
− ∑ ∑
δ δ
p
k
1
k
a
lim 0
á
k n
n
∞
= +
→∞
= =
∑

4. Chauhan: On a class of functions in a half plane
AJMS/Jul-Sep-2021/Vol 5/Issue 3 23
Thus '
k
k 1
a k
f


=
= ∑ ⇒ ( ) ( )
'
k
k 1
a s
k
f s


=
= ∑
(2.2)
Now, we will show the uniqueness of the above
representation (2.2). If possible, let us assume
'
k
k 1
b k
f


=
= ∑ then
n ' '
k k
k 1
n n
' '
k k
k 1 k 1
(a b )
a b
p
k p
p p
k k p
p
f f

 
=
=
−
− + −
≤
∑
∑ ∑
→0 as n→∞.
Therefore,
p
n ' '
k k
k 1
a b 0 as n .

=
− → →
∑
Consequently, ' '
k k
a b
= for each k. Hence, ak
=bk
for each k.
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Duke. Math J 1997;86:1-37.
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New York: Springer Verlag; 1980.
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