Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 2, Article 51, 2006
APPROXIMATION OF ENTIRE FUNCTIONS OF TWO COMPLEX VARIABLES
IN BANACH SPACES
RAMESH GANTI AND G.S. SRIVASTAVA
D EPARTMENT OF M ATHEMATICS
I NDIAN I NSTITUTE OF T ECHNOLOGY ROORKEE
ROORKEE - 247 667, I NDIA .
girssfma@iitr.ernet.in
Received 05 January, 2006; accepted 27 January, 2006
Communicated by S.P. Singh
A BSTRACT. In the present paper, we study the polynomial approximation of entire functions
of two complex variables in Banach spaces. The characterizations of order and type of entire
functions of two complex variables have been obtained in terms of the approximation errors.
Key words and phrases: Entire function, Order, type, Approximation, Error.
2000 Mathematics Subject Classification. 30B10, 30D15.
1. I NTRODUCTION
Let f (z1 , z2 ) =
am1 m2 z1m1 z2m2 be a function of the complex variables z1 and z2 , regular
for | zt | ≤ rt , t = 1, 2. If r1 and r2 can be taken arbitrarily large, then f (z1 , z2 ) represents an
entire function of the complex variables z1 and z2 . Following Bose and Sharma [1], we define
the maximum modulus of f (z1 , z2 ) as
P
M (r1 , r2 ) = max |f (z1 , z2 )| ,
|zt |≤rt
t = 1, 2.
The order ρ of the entire function f (z1 , z2 ) is defined as [1, p. 219]:
log log M (r1 , r2 )
= ρ.
log(r1 r2 )
r1 ,r2 →∞
lim sup
For 0 < ρ < ∞, the type τ of an entire function f (z1 , z2 ) is defined as [1, p. 223]:
log M (r1 , r2 )
= τ.
r1ρ + r2ρ
r1 ,r2 →∞
lim sup
Bose and Sharma [1], obtained the following characterizations for order and type of entire
functions of two complex variables.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
008-06
2
R AMESH G ANTI AND G.S. S RIVASTAVA
Theorem 1.1. The entire function f (z1 , z2 ) =
only if
P∞
m1 ,m2 =0
am1 m2 z1m1 z2m2 is of finite order if and
m2
1
log(mm
1 m2 )
−1 )
m1 ,m2 →∞ log (|am1 m2 |
is finite and then the order ρ of f (z1 , z2 ) is equal to µ.
µ = lim sup
Define
α = lim sup
q
ρ
m2
1
mm
1 m2 |am1 m2 | .
(m1 +m2 )
m1 ,m2 →∞
We have
Theorem 1.2. If 0 < α < ∞, the function f (z1 , z2 ) =
function of order ρ and type τ if and only if α = eρτ .
P∞
m1 ,m2 =0
am1 m2 z1m1 z2m2 is an entire
Let Hq , q > 0 denote the space of functions f (z1 , z2 ) analytic in the unit bi-disc U =
{z1 , z2 ∈ C : |z1 | < 1, |z2 | < 1} such that
kf kHq =
lim
r1 ,r2 →1−0
Mq (f ; r1 , r2 ) < ∞,
where
1q
Z πZ π
1
f (r1 eit1 , r2 eit2 ) q dt1 dt2
Mq (f ; r1 , r2 ) =
,
4π 2 −π −π
0
and let Hq , q > 0 denote the space of functions f (z1 , z2 ) analytic in U and satisfying the
condition
Z
1q
Z
1
q
|f (z1 , z2 )| dx1 dy1 dx2 dy2
< ∞.
kf kHq0 =
π 2 |z1 |<1 |z2 |<1
Set
0 = kf kH
kf kH∞
∞ = sup {|f (z1 , z2 )| : z1 , z2 ∈ U }.
0
Hq and Hq are Banach spaces for q ≥ 1. In analogy with spaces of functions of one variable,
0
we call Hq and Hq the Hardy and Bergman spaces respectively.
The function f (z1 , z2 ) analytic in U belongs to the space B(p, q, κ), where 0 < p < q ≤ ∞,
and 0 < κ ≤ ∞, if
κ1
Z 1 Z 1
κ(1/p−1/q)−1
κ
< ∞,
kf kp,q,κ =
{(1 − r1 )(1 − r2 )}
Mq (f, r1 , r2 )dr1 dr2
0
0
0 < κ < ∞,
kf kp,q,∞ = sup {[(1 − r1 )(1 − r2 )}(1/p−1/q)−1 Mq (f, r1 , r2 ) : 0 < r1 , r2 < 1} < ∞.
The space B(p, q, κ) is a Banach space for p > 0 and q, κ ≥ 1, otherwise it is a Fréchet space.
Further, we have
q
0
(1.1)
Hq ⊂ Hq = B , q, q , 1 ≤ q < ∞.
2
Let X be a Banach space and let Em,n (f, X) be the best approximation of a function f (z1 , z2 ) ∈
X by elements of the space P that consists of algebraic polynomials of degree ≤ m + n in two
complex variables:
(1.2)
Em,n (f, X) = inf {kf − pkx ; p ∈ P }.
To the best of our knowledge, characterizations for the order and type of entire functions of
two complex variables in Banach spaces have not been obtained so far. In this paper, we have
made an attempt to bridge this gap.
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
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E NTIRE F UNCTIONS OF T WO C OMPLEX VARIABLES
3
Notation: For reducing the length of expressions we use the following notations in the main
results.
1 1
1/κ
B
(n + 1)κ + 1; κ
−
= B[n, p, 2, κ]
p 2
1 1
1/κ
B
−
= B[m, p, 2, κ]
(m + 1)κ + 1; κ
p 2
1 1
1/κ
B
(n + 1)κ + 1; κ
−
= B[n, p, q, κ]
p q
1 1
1/κ
B
(m + 1)κ + 1; κ
−
= B[m, p, q, κ].
p q
2. M AIN R ESULTS
P∞
Theorem 2.1. Let f (z1 , z2 ) =
is of finite order ρ, if and only if
m,n=0
amn z1m z2n , then the entire function f (z1 , z2 ) ∈ B(p, q, κ)
ρ = lim sup
(2.1)
m,n→∞
ln (mm nn )
.
− ln Em,n (f, B(p, q, κ))
Proof. We prove the above result in two steps. First we consider the space B(p, q, κ), q = 2,
0 < p < 2 and κ ≥ 1. Let f (z1 , z2 ) ∈ B(p, q, κ) be of order ρ. From Theorem 1.1, for any
> 0, there exists a natural number n0 = n0 () such that
|amn | ≤ m−m/ρ+ n−n/ρ+
(2.2)
m, n > n0 .
We denote the partial sum of the Taylor series of a function f (z1 , z2 ) by
Tm,n (f, z1 , z2 ) =
m X
n
X
aj1 j2 z1j1 z2j2 .
j1 =0 j2 =0
We write
(2.3)
Em,n (f, B(p, 2, κ))
= kf − Tm,n (f )kp,2,κ

1
! κ2
Z 1 Z 1
κ
X X 2j 2j
κ(1/p−1/2)−1
2
2
1
=
{(1 − r1 )(1 − r2 )}
r1 r2 |aj1 j2 |
dr1 dr2
,
 0 0

j1
where
XX
j1
S1 =
r12j1 r22j2 |aj1 j2 |2
= S1 + S 2 +
j2
m
∞
X
X
∞
X
j2
∞
X
r12j1 r22j2 |aj1 j2 |2 ,
j1 =m+1 j2 =n+1
r12j1 r22j2 |aj1 j2 |2
j1 =0 j2 =n+1
and S2 =
∞
n
X
X
r12j1 r22j2 |aj1 j2 |2 .
j1 =m+1 j2 =0
Since S1 , S2 are bounded and r1 , r2 < 1, therefore the above expression (2.3) becomes
) 21
Z 1
( X
∞
∞
X
,
Em,n (f, B(p, 2, κ)) ≤ C
{(1 − r)κ(1/p−1/2)−1 }r(s+1)κ dr
|aj1 j2 |2
0
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
j1 =m+1 j2 =n+1
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4
R AMESH G ANTI AND G.S. S RIVASTAVA
where
Z
1
κ(1/p−1/2)−1
{(1 − r)
0
(s+1)κ
}r
Z
dr
1
κ(1/p−1/2)−1
{(1 − r1 )
=
(m+1)κ
}r1
dr1
0
Z
1
×
{(1 −
(n+1)κ
dr2
r2 )}κ(1/p−1/2)−1 r2
.
0
Therefore
(
Em,n (f, B(p, 2, κ)) ≤ CB[m, p, 2, κ]B[n, p, 2, κ]
(2.4)
∞
X
∞
X
) 21
|aj1 j2 |2
,
j1 =m+1 j2 =n+1
where C is a constant and B(a, b) (a, b > 0) denotes the beta function.
By using (2.2), we have
∞
X
∞
X
∞
X
2
|aj1 j2 | ≤
j1 =m+1 j2 =n+1
≤
∞
X
2j
2j
1 − 2
− ρ+
j2 ρ+
j1
j1 =m+1 j2 =n+1
∞
∞
2j
2j1
X
X
− 2
− ρ+
j2 ρ+
j1
j2 =n+1
j1 =m+1
≤ O(1)(m + 1)−2(m+1)/ρ+ (n + 1)−2(n+1)/ρ+ .
Using the above inequality in (2.4), we have
Em,n (f, B(p, 2, κ)) ≤ CB[m, p, 2, κ]B[n, p, 2, κ](m + 1)−(m+1)/ρ+ (n + 1)−(n+1)/ρ+ .
⇒ρ+≥
ln [(m + 1)(m+1) (n + 1)(n+1) ]
.
− ln {Em,n (f, B(p, 2, κ))} + ln {B[m, p, 2, κ]} + ln {B[n, p, 2, κ]}
Now
Γ((n + 1)κ + 1)Γ κ 1 − 1
p
2
1 1
B (n + 1)κ + 1; κ
−
=
.
p 2
Γ n + 12 + p1 κ + 1
Hence
(n+1)κ+3/2
1
1
−[(n+1)κ+1]
e
[(n + 1)κ + 1]
Γ p−2
1 1
B (n + 1)κ + 1; κ
−
' [(n+1/2+1/p)κ+1]
.
p 2
e
[(n+ 12 + p1 )κ + 1](n+1/2+1/p)κ+3/2
Thus
(2.5)
1
(n+1)
1 1
∼
B (n + 1)κ + 1; κ
−
= 1.
p 2
Now proceeding to limits, we obtain
(2.6)
ρ ≥ lim sup
m,n→∞
ln (mm nn )
.
− ln {Em,n (f, B(p, 2, κ))}
For the reverse inequality, since from the right hand side of the inequality (2.4), we have
(2.7)
|am+1n+1 |B[m, p, 2, κ]B[n, p, 2, κ] ≤ Em,n (f, B(p, 2, κ)),
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
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E NTIRE F UNCTIONS OF T WO C OMPLEX VARIABLES
5
we have
ln (mm nn )
ln (mm nn )
≥
.
− ln Em,n (f, B(p, 2, κ))
− ln |am+1n+1 | + ln {B[m, p, 2, κ]} + ln {B[n, p, 2, κ]}
Now proceeding to limits, we obtain
lim sup
(2.8)
m,n→∞
ln (mm nn )
≥ ρ.
− ln Em,n (f, B(p, 2, κ))
From (2.6) and (2.8), we get the required result.
In the second step, for the general case B(p, q, κ), q 6= 2, we have
Em,n (f, B(p, q, κ))
(2.9)
≤ kf − Tm,n (f )kp,q,κ
1

! κq
κ
Z 1 Z 1
X X qj qj
κ(1/p−1/q)−1
q
1
2
=
{(1 − r1 )(1 − r2 )}
r1 r2 |aj1 j2 |
dr1 dr2
,

 0 0
j1
where
XX
j1
r12j1 r22j2 |aj1 j2 |2
j2
∞
X
∞
X
r12j1 r22j2 |aj1 j2 |2 ,
j1 =m+1 j2 =n+1
m
∞
X
X
S1 =
= S1 + S 2 +
j2
r12j1 r22j2 |aj1 j2 |2
∞
n
X
X
and S2 =
j1 =0 j2 =n+1
r12j1 r22j2 |aj1 j2 |2 .
j1 =m+1 j2 =0
Since S1 , S2 are bounded and r1 , r2 < 1, therefore the above expression (2.9) becomes
) 1q
Z 1
( X
∞
∞
X
0
Em,n (f, B(p, q, κ)) ≤ C
{(1 − r)κ(1/p−1/q)−1 }r(s+1)κ dr
,
|aj1 j2 |q
0
where
Z
j1 =m+1 j2 =n+1
1
κ(1/p−1/q)−1
{(1 − r)
0
(s+1)κ
}r
Z
dr
1
κ(1/p−1/q)−1
{(1 − r1 )
=
(m+1)κ
}r1
dr1
0
1
Z
×
(n+1)κ
r2 )}κ(1/p−1/q)−1 r2
dr2
{(1 −
.
0
Therefore
(
0
Em,n (f, B(p, q, κ)) ≤ C B[m, p, q, κ]B[n, p, q, κ]
(2.10)
∞
X
) 1q
∞
X
|aj1 j2 |q
,
j1 =m+1 j2 =n+1
0
where C is constant and B[m, p, q, κ] is Euler’s integral of the first kind. By using (2.2), we
get
∞
X
∞
X
q
|aj1 j2 | ≤
j1 =m+1 j2 =n+1
∞
X
j1 =m+1
∞
X
−qj1
(ρ+)
j1
j2
j2 =n+1
≤ O(1)(m + 1)
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
−qj2
(ρ+)
−q(m+1)
(ρ+)
(n + 1)
−q(n+1)
(ρ+)
.
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6
R AMESH G ANTI AND G.S. S RIVASTAVA
Using above inequality in (2.10), we get
0
Em,n (f, B(p, q, κ)) ≤ C B[m, p, q, κ]B[n, p, q, κ](m + 1)−(m+1)/ρ+ (n + 1)−(n+1)/ρ+ .
⇒ρ+≥
ln [(m + 1)m+1 (n + 1)n+1 ]
.
− ln Em,n (f, B(p, q, κ)) + ln {B[m, p, q, κ]} + ln {B[n, p, q, κ]}
Now proceeding to limits, we obtain
ρ ≥ lim sup
(2.11)
m,n→∞
ln (mm nn )
.
− ln Em,n (f, B(p, q, κ))
Let 0 < p < q < 2, and κ, q ≥ 1. Since
κ1 − κ1
1
1
1
Em,n (f, B(p, q, κ)),
−
Em,n (f, B(p1 , q1 , κ1 )) ≤ 21/q−1/q1 κ
p q
where p1 = p, q1 = 2 and κ1 = κ, and the condition (2.1) is already proved for the space
B(p, 2, κ), we get
(2.12)
lim sup
m,n→∞
ln (mm nn )
ln (mm nn )
≥ lim sup
= ρ.
− ln Em,n (f, B(p, q, κ))
m,n→∞ − ln Em,n (f, B(p, 2, κ))
Now let 0 < p ≤ 2 < q. Since
M2 (f, r1 , r2 ) ≤ Mq (f, r1 , r2 ),
0 < r1 , r2 < 1,
therefore
Z
(2.13)
1
1
Z
{(1 − r1 )(1 − r2 )}κ(1/p−1/q)−1 Qdr1 dr2
Em,n (f, B(p, q, κ)) ≥
0
κ1
0
≥ |am+1n+1 |B[m, p, q, κ]B[n, p, q, κ],
where Q = inf [M2κ (f − p; r1 , r2 ) : p ∈ P ]. Hence we have
ln (mm nn )
ln (mm nn )
≥
.
− ln Em,n (f, B(p, q, κ))
− ln |am+1n+1 | + ln {B[m, p, q, κ]} + ln {B[n, p, q, κ]}
Now proceeding to limits, we obtain
lim sup
(2.14)
m,n→∞
ln (mm nn )
≥ ρ.
− ln Em,n (f, B(p, q, κ))
From (2.11) and (2.14), we get the required result.
Now we prove
Theorem 2.2. Let f (z1 , z2 ) =
finite order ρ, if and only if
P∞
(2.15)
ρ = lim sup
m,n=0
amn z1m z2n , then the entire function f (z1 , z2 ) ∈ Hq is of
m,n→∞
Proof. Let f (z1 , z2 ) =
entire, we have
P∞
m,n=0
ln (mm nn )
.
− ln Em,n (f, Hq )
amn z1m z2n ∈ Hq be an entire transcendental function. Since f is
(2.16)
lim
m,n→∞
p
(m+n)
|amn | = 0,
and f ∈ Hq , therefore
Mq (f ; r1 , r2 ) < ∞,
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
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E NTIRE F UNCTIONS OF T WO C OMPLEX VARIABLES
7
and f (z1 , z2 ) ∈ B(p, q, κ), 0 < p < q ≤ ∞; q, κ ≥ 1. By (1.1) we obtain
(2.17)
Em,n (f, B(q/2, q, q)) ≤ ςq Em,n (f, Hq ),
1 ≤ q < ∞,
where ςq is a constant independent of m, n and f . In the case of space H∞ ,
(2.18)
Em,n (f, B(p, ∞, ∞)) ≤ Em,n (f, H∞ ),
0 < p < ∞.
From (2.17), we have
ln (mm nn )
ξ(f ) = lim sup
m,n→∞ − ln Em,n (f, Hq )
ln (mm nn )
≥ lim sup
m,n→∞ − ln Em,n (f, B(q/2, q, q))
≥ ρ,
1 ≤ q < ∞,
(2.19)
and using estimate (2.18) we prove inequality (2.19) for the case q = ∞.
For the reverse inequality
ξ(f ) ≤ ρ,
(2.20)
since
Em,n (f, Hq ) ≤ O(1)
∞
X
∞
X
|aj1 j2 (f )|,
j1 =m+1 j2 =n+1
using (2.2), we have
Em,n (f, Hq ) ≤ O(1)
≤
∞
X
∞
X
j
j
1
− ρ+
− 2
j2 ρ+
j1
j1 =m+1 j2 =n+1
∞
∞
j1
j
X
X
− ρ+
− 2
O(1)
j1
j2 ρ+
j1 =m+1
j2 =n+1
≤ O(1)(m + 1)−(m+1)/ρ+ (n + 1)−(n+1)/ρ+ .
ln [(m + 1)(m+1) (n + 1)(n+1) ]
.
− ln [Em,n (f, Hq )]
Now proceeding to limits and since is arbitrary, then we will get (2.20). From (2.19) and
(2.20) we will obtain the required result.
Now we prove sufficiency. Assume that the condition (2.15) is satisfied. Then it follows that
ln [1/Em,n (f, Hq )]1/(m+n) → ∞ as m, n → ∞.
This yields
q
lim (m+n) Em,n (f, Hq ) = 0.
⇒ρ+≥
m,n→∞
This relation and the estimate |am+1n+1 (f )| ≤ Em,n (f, Hq ) yield the relation (2.16). This
means that f (z1 , z2 ) ∈ Hq is an entire transcendental function.
Now we prove
P
m n
Theorem 2.3. Let f (z1 , z2 ) = ∞
m,n=0 amn z1 z2 , then the entire function f (z1 , z2 ) ∈ B(p, q, κ)
of finite order ρ, is of type τ if and only if
1
1
ρ
(2.21)
τ=
lim sup {mm nn Em,n
(f, B(p, q, κ))} m+n .
eρ m,n→∞
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
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8
R AMESH G ANTI AND G.S. S RIVASTAVA
Proof. We prove the above result in two steps.
First we consider the space B(p, q, κ), q = 2, 0 < p < 2 and κ ≥ 1. Let f (z) ∈ B(p, q, κ)
be of order ρ. From Theorem 1.2, for any > 0, there exists a natural number n0 = n0 () such
that
|amn | ≤ m−m/ρ n−n/ρ [eρ(τ + )]
(2.22)
m+n
ρ
.
We denote the partial sum of the Taylor series of a function f (z1 , z2 ) by
m X
n
X
Tm,n (f, z1 , z2 ) =
aj1 j2 z1j1 z2j2 ,
j1 =0 j2 =0
we write
(2.23)
Em,n (f, B(p, 2, κ))
= kf − Tm,n (f )kp,2,κ

1
! κ2
Z 1 Z 1
κ
X
X
κ(1/p−1/2)−1
2j1 2j2
2
{(1 − r1 )(1 − r2 )}
r1 r2 |aj1 j2 |
=
dr1 dr2
,
 0 0

j1
where
XX
j1
S1 =
j2
m
X
r12j1 r22j2 |aj1 j2 |2
∞
X
= S1 + S 2 +
r12j1 r22j2 |aj1 j2 |2
∞
X
j2
∞
X
r12j1 r22j2 |aj1 j2 |2 ,
j1 =m+1 j2 =n+1
∞
n
X
X
and S2 =
j1 =0 j2 =n+1
r12j1 r22j2 |aj1 j2 |2 .
j1 =m+1 j2 =0
Since S1 , S2 are bounded, and r1 , r2 < 1 therefore the above expression (2.23) becomes
) 21
( ∞
∞
X X
,
(2.24)
Em,n (f, B(p, 2, κ)) ≤ DB[m, p, 2, κ]B[n, p, 2, κ]
|aj1 j2 |2
j1 =m+1 j2 =n+1
where D is a constant and B(a, b) (a, b > 0) denotes the beta function. By using (2.22), we
have
∞
∞
∞
∞
2j
2j
X
X
X
X
2(j1 +j2 )
− 1 − 2
2
|aj1 j2 | ≤
j1 ρ j2 ρ [eρ(τ + )] ρ
j1 =m+1 j2 =n+1
≤
j1 =m+1 j2 =n+1
∞
2j
X
− 1
j1 ρ [eρ(τ
j1 =m+1
+ )]
2j1
ρ
∞
X
2j
2
− ρ+
j2
[eρ(τ + )]
2j2
ρ
j2 =n+1
≤ O(1)(m + 1)−2(m+1)/ρ (n + 1)−2(n+1)/ρ [eρ(τ + )]
2(m+n+2)
ρ
.
Using the above inequality in (2.24), we get
ρ
Em,n
(f, B(p, 2, κ)) ≤ Dρ B ρ [m, p, 2, κ]B ρ [n, p, 2, κ]Y [eρ(τ + )](m+n+2) ,
where Y = (m + 1)−(m+1) (n + 1)−(n+1) .
Now proceeding to limits and since is arbitrary, we have
1
1
ρ
(2.25)
lim sup {mm nn Em,n
(f, B(p, 2, κ))} m+n ≤ τ.
eρ m,n→∞
For the reverse inequality, since from the right hand side of (2.24),
|am+1n+1 |B[m, p, 2, κ]B[n, p, 2, κ] ≤ Em,n (f, B(p, 2, κ))
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
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E NTIRE F UNCTIONS OF T WO C OMPLEX VARIABLES
9
we have
ρ
ρ
mm/(m+n) nn/(m+n) |am+1n+1 |ρ/(m+n) B (m+n) [m, p, 2, κ]B (m+n) [n, p, 2, κ]
ρ
≤ {Em,n
mm nn }1/(m+n) .
Now proceeding to limits, we obtain
1
1
ρ
lim sup {mm nn Em,n
(f, B(p, 2, κ))} m+n .
eρ m,n→∞
τ≤
(2.26)
From (2.25) and (2.26), we get the required result.
In the second step, for the general case B(p, q, κ), q 6= 2, we have
Em,n (f, B(p, q, κ))
(2.27)
≤ kf − Tm,n (f )kp,q,κ

1
! κq
Z 1 Z 1
κ
X
X
κ(1/p−1/q)−1
qj1 qj2
q
=
{(1 − r1 )(1 − r2 )}
r1 r2 |aj1 j2 |
dr1 dr2
,
 0 0

j1
j2
where
XX
j1
r12j1 r22j2 |aj1 j2 |2 = S1 + S2 +
j2
∞
X
r12j1 r22j2 |aj1 j2 |2 ,
j1 =m+1 j2 =n+1
m
∞
X
X
S1 =
∞
X
r12j1 r22j2 |aj1 j2 |2
and S2 =
∞
n
X
X
r12j1 r22j2 |aj1 j2 |2 .
j1 =m+1 j2 =0
j1 =0 j2 =n+1
Since S1 , S2 are bounded, therefore the above expression (2.27) becomes
1
Z
{(1 − r)κ(1/p−1/q)−1 }r(s+1)κ dr
Em,n (f, B(p, q, κ)) ≤ G
( X
∞
0
∞
X
) 1q
|aj1 j2 |q
,
j1 =m+1 j2 =n+1
where
Z
1
κ(1/p−1/q)−1
{(1 − r)
0
(s+1)κ
}r
Z
dr
1
κ(1/p−1/q)−1
{(1 − r1 )
=
(m+1)κ
}r1
dr1
0
Z
×
1
(n+1)κ
r2 )}κ(1/p−1/q)−1 r2
dr2
{(1 −
.
0
Since r1 , r2 < 1, therefore we have
(
(2.28)
Em,n (f, B(p, q, κ)) ≤ GB[m, p, q, κ]B[n, p, q, κ]
∞
X
∞
X
) 1q
|aj1 j2 |q
,
j1 =m+1 j2 =n+1
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
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10
R AMESH G ANTI AND G.S. S RIVASTAVA
where G is constant and B[m, p, q, κ] is Euler’s integral of the first kind. Using (2.22), we get
∞
X
∞
X
∞
X
|aj1 j2 |q ≤
j1 =m+1 j2 =n+1
≤
∞
X
−
j1
qj1
ρ
j1 =m+1 j2 =n+1
∞
qj
X
− 1
j1 ρ [eρ(τ
j1 =m+1
−
j2
qj2
ρ
+ )]
[eρ(τ + )]
qj1
ρ
∞
X
q(j1 +j2 )
ρ
qj
2
− ρ+
j2
[eρ(τ + )]
qj2
ρ
j2 =n+1
≤ O(1)(m + 1)−q(m+1)/ρ (n + 1)−q(n+1)/ρ [eρ(τ + )]
q(m+n+2)
ρ
.
Using the above inequality in (2.28), we get
ρ
Em,n
(f, B(p, q, κ)) ≤ Gρ B ρ [m, p, q, κ]B ρ [n, p, q, κ]Y [eρ(τ + )](m+n+2) ,
where Y = (m + 1)−(m+1) (n + 1)−(n+1) . Now proceeding to limits, since is arbitrary, we have
1
1
ρ
lim sup {mm nn Emn
(f, B(p, q, κ))} m+n ≤ τ.
eρ m,n→∞
(2.29)
Let 0 < p < q < 2, and κ, q ≥ 1. Since
Em,n (f, B(p1 , q1 , κ1 )) ≤ 21/q−1/q1 [κ(1/p − 1/q)]1/κ−1/κ1 Em,n (f, B(p, q, κ)),
where p1 = p, q1 = 2 and κ1 = κ, and the condition (2.21) has already been proved for the
space B(p, 2, κ), we get
1
ρ
lim sup {mm nn Em,n
(f, B(p, q, κ))} m+n
m,n→∞
1
ρ
≥ lim sup {mm nn Em,n
(f, B(p, 2, κ))} m+n = τ.
m,n→∞
Now let 0 < p ≤ 2 < q. Since, in this case we have
M2 (f, r1 , r2 ) ≤ Mq (f, r1 , r2 ),
0 < r1 , r2 < 1,
therefore
(2.30)
1
1
ρ
lim sup {mm nn Em,n
(f, B(p, q, κ))} m+n ≥ lim sup {mm nn |amn |ρ } m+n
m,n→∞
m,n→∞
= eρτ.
From (2.29) and (2.30), we get the required result.
Lastly we prove
P
m n
Theorem 2.4. Let f (z1 , z2 ) = ∞
m,n=0 amn z1 z2 , then the entire function f (z1 , z2 ) ∈ Hq having
finite order ρ is of type τ if and only if
(2.31)
Proof. Since f (z1 , z2 ) =
(2.32)
τ=
P∞
1
1
ρ
lim sup {mm nn Em,n
(f, Hq )} m+n .
eρ m,n→∞
m,n=0
amn z1m z2n is an entire transcendental function, we have
p
lim m+n |amn | = 0.
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
m,n→∞
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E NTIRE F UNCTIONS OF T WO C OMPLEX VARIABLES
11
Therefore f (z1 , z2 ) ∈ B(p, q, κ), 0 < p < q ≤ ∞; q, κ ≥ 1. We have
1
1
ρ
ξ(f ) =
lim sup {mm nn Em,n
(f, Hq )} m+n
(2.33)
eρ m,n→∞
n
q
o 1
1
ρ
m+n = τ
≥
lim sup mm nn Em,n
f, B , q, q
eρ m,n→∞
2
for 1 ≤ q < ∞. Using the estimate (2.18) we prove inequality (2.33) in the case q = ∞. For
the reverse inequality
ξ(f ) ≤ τ,
(2.34)
we have
Em,n (f, Hq ) ≤
∞
X
∞
X
|aj1 j2 (f )|.
j1 =m+1 j2 =n+1
Using (2.22), we get
ρ
(f, Hq ) ≤ O(1)(m + 1)−(m+1) (n + 1)−(n+1) [eρ(τ + )](m+n+2)
Em,n
1
1
ρ
(f, Hq )} (m+n+2) .
⇒ τ + ≥ {(m + 1)(m+1) (n + 1)(n+1) Em,n
eρ
Now proceeding to limits, since is arbitrary, we get
1
1
ρ
(2.35)
τ≥
lim sup {mm nn Em,n
(f, Hq )} m+n .
eρ m,n→∞
From (2.33) and (2.35), we obtain the required result.
Now we prove sufficiency. Assume that the condition (2.31) is satisfied. Then it follows that
ρ
{Em,n
(f, Hq )}1/(m+n) → 0 as m, n → ∞. This yields
q
(m+n)
Em,n (f, Hq ) = 0.
lim
m,n→∞
This relation and the estimate |am+1n+1 (f )| ≤ Em,n (f, Hq ) yield the inequality (2.32). This
implies that f (z1 , z2 ) ∈ Hq is an entire transcendental function.
R EFERENCES
[1] S.K. BOSE, AND D. SHARMA, Integral functions of two complex variables, Compositio Math., 15
(1963), 210–226.
[2] S.B. VAKARCHUK AND S.I. ZHIR, On some problems of polynomial approximation of entire
transcendental functions, Ukrainian Mathem. J., 54(9) (2002), 1393–1401.
J. Inequal. Pure and Appl. Math., 7(2) Art. 51, 2006
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