Internat. J. Math. & Math. Sci.
Vol. 22, No. 3 (1999) 445–458
S 0161-17129922445-3
© Electronic Publishing House
ON THE RITT ORDER AND TYPE OF A CERTAIN CLASS
OF FUNCTIONS DEFINED BY BE -DIRICHLETIAN
ELEMENTS
MARCEL BERLAND
(Received 2 September 1997)
Abstract. We introduce the notions of Ritt order and type to functions defined by the
series
∞
(∗)
fn σ + iτ0 exp − sλn , s = σ + iτ, (σ , τ) ∈ R × R
n=1
∞
indexed by τ0 on R, where λn 1 is a D-sequence and (fn )∞
1 is a sequence of entire
functions of bounded index with at most a finite number of zeros. By definition, the series
are BE -Dirichletian elements. The notions of order and type of functions, defined by BDirichletian elements, are considered in [3, 4]. In this paper, using a technique similar to
that used by M. Blambert and M. Berland [6], we prove the same properties of Ritt order
and type for these functions.
Keywords and phrases. Ritt order and type, entire functions of bounded index, BE Dirichletian elements, Dirichletian elements.
1991 Mathematics Subject Classification. 30A16, 30A64.
1. Preliminary lemmas
Definition 1.1 (B. Lepson [10]). An entire function f is said to be of bounded
index if there exists a nonnegative integer ν such that
f (k) (s) f (j) (s)
(0)
max
k
∈
{0,
1,
.
.
.
,
ν}
≥
,
f (s) = f (s)
(1.1)
k!
j!
for all j and for all s. The least such integer ν is called the index of f .
Theorem A (F. Gross [8]). An entire function with at most a finite number of zeros
is of bounded index if and only if it is of the form P (s) exp(αs), where P (s) is polynomial
and α is a complex constant.
Theorem B (S. M. Shah [16]). Let f (s) = P (s) exp(αs), where α is any complex number and P (s) is a polynomial of degree less than n. Then f is of bounded index and the
index ν ≤ p, where p is any integer such that p ≥ n − 1 and
n|α|
n(n − 1)
|α|n
1
+
|α|2
+···+
≤ 1.
(1.2)
p +1
2!
p(p + 1)
(p − n + 2) · · · (p + 1)
∞
Let λn 1 be a D-sequence (that is a positive strictly increasing unbounded sequence) and (fn )∞
1 be a sequence of entire functions fn of bounded index νn with
446
MARCEL BERLAND
at most a finite number of zeros from Theorem A. As a result of the two theorems,
we have ∀s ∈ C, ∀n ∈ N\{0}
fn (s) = Pn (s) exp αn s ,
(1.3)
where Pn (s) is a polynomial of degree mn and αn is a complex constant, that is,
s → Pn (s) =
m
n
an,j s j
with an,mn = 0 and s ∈ C.
(1.4)
j=0
Let us suppose that ∃k ∈]0, λ1 [, ∀n ∈ N\{0}
αn ∈ d(0,k) ,
(1.5)
where d(0,k) is the closed disc centered at 0 and of radius k.
Consider the space of elements
∞
f τ0 :
fn σ + iτ0 exp − sλn ,
s = σ + iτ, (σ , τ) ∈ R × R
(1.6)
n=1
indexed by τ0 on R. By definition, {fτ0 } is the BE -Dirichletian element. Let
mn
,
β = lim sup
λn
n →∞
∀n ∈ N\{0}.
An = max anj | j ∈ 0, 1, . . . , mn
(1.7)
(1.8)
Consider the associated Dirichletian element
∞
An exp − sλn ,
fA :
(1.9)
n=1
f
whose coefficients are strictly positive and denote, by σc A , the abscissa of convergence
of {fA }.
Let us state three lemmas due to M. Blambert and M. Berland [6] which we use later.
These demonstrations are obvious because this sequence (αn )∞
1 is bounded.
f
Lemma 1.1. If σc A = −∞, β < ∞, ∀n ∈ N\{0}, αn ∈ d(0,k) , then fτ0 converges
absolutely on C for any arbitrary τ0 in R.
f
Lemma 1.2. If σc A = −∞, β < ∞, ∀n ∈ N\{0}, αn ∈ d(0,k) , we have ∀τ0 ∈ R, ∀τ ∈ R
lim fτ0 (σ + iτ) = 0.
σ →∞
f
(1.10)
Lemma 1.3. If σc A = −∞, ∀n ∈ N\{0}, αn ∈ d(0,k) , we have ∀τ0 ∈ R, ∀σ ∈ R
Pn σ + iτ0 exp αn σ + iτ0 − σ λn
(1.11)
1 τ2
= lim
fτ0 (σ + iτ) exp iτλn dτ ,
τ2 →∞ τ2 τ
1
where τ1 is any arbitrary real number.
ON THE RITT ORDER AND TYPE OF A CERTAIN CLASS OF FUNCTIONS . . .
447
2. Main theorems. Let us define the following quantities. For each σ on C,
M σ ; fτ0 = sup fτ0 σ + iτ | σ ≥ σ , τ ∈ R ,
Mn σ ; fτ0 = sup fτ0 ,n σ + iτ | σ ≥ σ , τ ∈ R ,
µ σ ; fτ0 = sup fn σ + iτ0 exp − σ λn | n ∈ N\{0} ,
µn σ ; fτ0 = sup fn σ + iτ0 exp − σ λn | n ≥ n ;
(2.1)
(2.2)
(2.3)
(2.4)
where
fτ0 ,n (s) =
∞
fn σ + iτ0 exp − sλn .
(2.5)
n=n
The quantities defined above are finite.
Remark. The function σ M(σ ; fτ0 ) is decreasing onto R.
f
Theorem 2.1. If σc A = −∞, β < ∞, ∀n ∈ N\{0}, αn ∈ d(0,k) , we have
lim M σ ; fτ0 = 0
σ →∞
and
lim M σ ; fτ0 = ∞.
σ →−∞
(2.6)
Proof. We have ∀ε ∈]0, 1[, ∃n = nε ∈ N\{0}, ∀n ≥ n , mn /λn < β + ε and
∃n (= n
ε ) ∈ N\{0}, ∀n ≥ n , k/λn < ε, ∀σ > 0 such that
∞
fn σ + iτ0 exp − σ λn
n=n1 (=max{n , n })
≤
∞
n=n1
β + ε log 1 + |σ | + |τ0 |
|τ0 |
+ε 1+
,
σ
σ
An exp − σ λn 1 −
(2.7)
∀ε ∈ ]0, 1 − ε[, ∃σ = σε > 0, ∀σ > σ ,
β + ε log 1 + |σ | + |τ0 | + ε|τ0 |
< ε
σ
(2.8)
and
(β + ε) log 1 + |σ | + |τ0 | + ε|τ0 |
+ε
σ 1−
σ
> σ (1 − ε) − ε > σ (1 − ε) − ε (> 0).
(2.9)
Therefore, ∃n1 ∈ N\{0} such that
∞
An exp − σ (1 − ε) − ε λn ,
Mn1 σ ; fτ0 < fA,n1 σ (1 − ε) − ε =
(2.10)
n=n1
where
lim Mn1 σ ; fτ0 = 0.
σ →∞
(2.11)
448
MARCEL BERLAND
On the other hand, we have, ∀n ∈ {1, 2, . . . , n1 − 1}
lim Pn (σ + iτ0 ) exp αn (σ + iτ0 ) − σ λn = 0 with λ1 > k and αn ∈ d(0,k) . (2.12)
σ →∞
We have
lim M σ ; fτ0 = 0.
(2.13)
σ →∞
On the other hand, ∀σ < 0, if M(σ ; fτ0 ) is bounded onto R implies that (from Lemma
1.3)
∀n ∈ N\{0} : j ∈ {0, 1, . . . , mn } ⇒ an,j = 0 .
(2.14)
Or, thus, we get the contradiction that
an,mn = 0
and
∀n ∈ N\{0}
(2.15)
lim M σ ; fτ0 = ∞.
(2.16)
σ →−∞
Thus, (2.13) and (2.16) prove the theorem.
Furthermore, let
fτ 0
ρR
fτ
λR 0
fτ
= lim sup
σ →−∞
= lim inf
σ →−∞





 log+ log+ M σ ; fτ0




−σ





 log+ log+ M σ ; fτ0




−σ
,
(2.17)
.
(2.18)
fτ
By definition, ρR 0 and λR 0 are the Ritt-order and the lower Ritt-order of function
fτ0 defined by BE -Dirichletian element {fτ0 }. Also, M(σ ; fA ) is defined in a similar
manner with fA in the place of fτ0 . It is trivial that




 log+ log+ fA (σ ) 
f
,
(2.19)
ρRA = lim sup


−σ

σ →−∞ 
f
λRA
= lim inf
σ →−∞




 log+ log+ fA (σ ) 




−σ
.
(2.20)
f
Theorem 2.2. If σc A = −∞, β < ∞, ∀n ∈ N\{0}, αn ∈ d(0,k) , and L(= lim sup{log n/
n→∞
λn }) < ∞, we have ∀τ0 ∈ R,
fτ 0
ρR
f
= ρRA
and
fτ
f
f
fτ
λR 0 = λRA .
(2.21)
Proof. (1) We get the inequalities, ∀τ0 ∈ R,
f
fτ 0
ρRA ≤ ρR
and
λRA ≤ λR 0 .
(2.22)
τ0 is any arbitrary real number. Consider the closed interval I(s, λ) = {s ∈ C | |σ −
σ | ≤ λ > 0, τ = τ0 }, where σ = Re(s ), τ = Im(s ), and s = σ + iτ0 . Let ∀n ∈ N\{0},
ON THE RITT ORDER AND TYPE OF A CERTAIN CLASS OF FUNCTIONS . . .
449
pn (s, λ) = sup Pn s | s ∈ d(s,λ) ,
(2.23)
∗
(s, λ) = sup Pn s | s ∈ I(s, λ) .
pn
(2.24)
Using Lemma 1.3, we have ∀σ ∈ [σ − λ, σ + λ], ∀n ∈ N\{0},
Pn σ + iτ0 exp Re αn − λn σ − Im αn τ0 ≤ M σ − λ; fτ ,
0
(2.25)
and then (M. Blambert and M. Berland [6])
∗
(s, λ) exp − (σ + λ) λn − Re αn − Im αn τ0 ≤ M σ − λ; fτ0 ,
pn
−mn
An
∗
pn
(s, λ),
6
pn (s, λ) ≤
−mn
≤ pn (s, λ) ∀λ ≥ 1,
1 + |s|
(2.26)
(2.27)
(M. Berland [1]).
(2.28)
Therefore, we have ∀λ ≥ 1, ∀σ ∈ R, ∀n ∈ N\{0}
An ≤ M σ − λ; fτ0 exp
σ +λ+
Im αn
mn
+
τ0 λn , (2.29)
log 6 1 + |σ | + |τ0 |
λn
λn
where λn = λn − Re(αn ).
We have ∀ε ∈]0, 1[, ∃n1 ∈ N\{0}, ∀n ≥ n1
An ≤ M σ − λ; fτ0 exp σ + λ + β + ε log 6 1 + |σ | + |τ0 | + ε λn ,
(2.30)
and ∀ε1 ∈]0, 1[, ∃σ1 = σε1 > 0, ∀σ < −σ1
λ + β + ε log 6 1 + |σ | + |τ0 | + ε
< ε1 ,
−σ
(2.31)
where β = lim sup{mn /λn }(< ∞) which implies that, ∀ε1 ∈]0, 1[, ∀n ≥ n1 , ∀σ < σ1
n→∞
An ≤ M σ − λ; fτ0 exp σ 1 − ε1 λn .
(2.32)
Now, ∀n ∈ N\{0}, Re(αn ) ≤ k (because αn ∈ d(0,k) )
λn = λn 1 −
Re αn
k
.
≥ λn 1 −
λn
λn
(2.33)
We have, ∀ε2 ∈]0, 1[, ∃n ∈ N\{0}, ∀n ≥ n ,
k
< ε2
λn
and
λn ≥ λn 1 − ε2
( ⇒ β = β)
(2.34)
which implies that, ∀λ ≥ 1, ∀n ≥ max{n1 , n }(= n2 ), ∀σ < −σ1
An exp − σ 1 − ε1 1 − ε2 λn ≤ M σ − λ; fτ0 .
(2.35)
Put, ∀n ∈ N\{0}
µn(1,2) = 1 − ε1 1 − ε2 λn
(> 0),
(2.36)
450
MARCEL BERLAND
(µn(1,2) ) is a D-sequence. Consider the Dirichletian element
∞
fA(1,2) :
An exp − sµn(1,2)
(2.37)
n=1
fA
indexed by the couple (1, 2) and denote, by σc
{fA(1,2) }. We have,
Ritt-order is
fA
(1,2)
σc
fA
ρR
= −∞ (since
(1,2)
= lim sup
σ →−∞
and its lower Ritt-order is
fA
λR
(1,2)
= lim inf
σ →−∞
Now, ∀σ ∈ R,
f
σc A
(1,2)
, the abscissa of convergence of
= −∞), fA(1,2) is an entire function and, its


 log+ log+ fA(1,2) (σ ) 



 log+ log+ fA(1,2) (σ ) 

,
(2.38)
.
(2.39)

−σ

−σ
fA(1,2) (σ ) = fA σ 1 − ε1 1 − ε2
(2.40)
and
fA
ρR
fA
λR
(1,2)
(1,2)
f
= 1 − ε1 1 − ε2 ρRA ,
= 1 − ε1
f
1 − ε2 λRA .
Put (Q. S. Liu [11]) ∀σ ∈ R,
µn2 σ ; fA(1,2) = sup An exp −σ µn(1,2) n ≥ n2 .
(2.41)
(2.42)
(2.43)
We have ∀ε > 0, ∀σ ∈ R
fA(1,2),n (σ ) ≤
2
!
"
!
"
An exp −(σ − L − ε)µn(1,2) exp −(L + ε)µn(1,2)
∞ n=n2
≤ µn2 σ − L − ε; fA(1,2) Kn2 (ε)
with Kn2 (ε) = n=n2 exp − (L + ε)µn(1,2) .
Hence, ∀ε > 0, ∀σ ∈ R
fA(1,2),n (σ ) ≤ µn2 σ − L − ε; fA(1,2) Kn2 (ε).
(2.44)
#∞
2
(2.45)
Then we have, ∀λ ≥ 1, ∃n2 = max{n1 , n }, ∀n ≥ n2 , ∀σ < −σ1
An exp −σ µn(1,2) ≤ M σ − λ; fτ0 .
This implies that
µn2 σ ; fA(1,2) ≤ M σ − λ; fτ0 .
(2.46)
(2.47)
From (2.45) and (2.47), we have
fA(1,2),n (σ + L + ε) ≤ M σ − λ; fτ0 Kn2 (ε),
2
fA
fτ
fτ
σ −λ
(1,2),n2
ρR
≤ ρR 0 lim
= ρR 0 .
σ →−∞ σ + L + ε
(2.48)
(2.49)
ON THE RITT ORDER AND TYPE OF A CERTAIN CLASS OF FUNCTIONS . . .
451
Or
fA
ρR
(1,2),n2
fA
= ρR
(1,2)
(M. Blambert [5]).
(2.50)
We have, ∀ε1 ∈ ]0, 1[, ∀ε2 ∈]0, 1[
fA
(1,2)
f
fτ
= 1 − ε1 1 − ε2 ρRA ≤ ρR 0 ,
(2.51)
fA
(1,2)
f
fτ
= 1 − ε1 1 − ε2 λRA ≤ λR 0 .
(2.52)
ρR
λR
As ε1 and ε2 are arbitrary, we have
fτ
f
ρRA ≤ ρR 0 ,
f
fτ
and then λRA ≤ λR 0 .
(2.53)
(2) We get the inequalities, ∀τ0 ∈ R,
fτ 0
ρR
f
≤ ρRA
and
fτ
f
λR 0 ≤ λRA .
From Theorem 2.1, we have ∀ε ∈]0, 1[, ∀ε ∈]0, 1 − ε[, ∃σ > 0, ∀σ < −σ log (1 + |σ | + |τ0 |)
|τ0 |
+ε 1+
θσ
σ 1− β+ε
|σ |
|σ |
log(1 + |σ | + |τ0 |)
|τ0 |
+ε 1+
= σ 1+ β+ε
|σ |
|σ |
"
!
> σ 1 + β + 2ε ε + ε = σ 1 + ε1 ,
where ε1 = β + 2ε ε + ε, θσ = 1 if σ > 0 and θσ = −1 if σ < 0.
Hence, ∀σ < −σ , ∃n1 ∈ N\{0},
Mn1 σ ; fτ0 ≤ fA,n1 σ 1 + ε1 ,
(2.54)
(2.55)
(2.56)
which implies that
fτ0 ,n1
ρR
fA,n1 ≤ ρR
f
1 + ε1 = 1 + ε1 ρRA ,
(2.57)
and where
fτ0 ,n1
≤ ρRA
fτ ,n1
λR 0
f
(2.58)
M σ ; fτ0 ≤ Mn0 1 σ ; fτ0 + Mn1 σ ; fτ0 ,
(2.59)
Mn0 1 σ ; fτ0 = sup fτ00 ,n1 σ + iτ | σ ≥ σ , τ ∈ R
(2.60)
Now, σ ∈ R,
where
f
≤ λRA .
ρR
and
and
1 −1
n
fn σ + iτ0 exp − sλn .
fτ00 ,n1 :
(2.61)
n=1
Then ∀τ0 ∈ R,
fτ 0
ρR
fτ00 ,n1
since ρR
= 0.
fτ0 ,n
fτ ,n
fτ ,n
≤ max ρR 0 1 , ρR 0 1 = ρR 0 1
(2.62)
452
MARCEL BERLAND
Finally, we have
fτ 0
ρR
f
(2.63)
f
(2.64)
≤ ρRA ,
and, similarly, we can show that
fτ
λR 0 ≤ λRA .
Hence, (1) and (2) implies (2.21) which proves this theorem.
fτ 0
If ρR
> 0, we put
fτ 0
By definition, τR
fτ
τR 0




 log M σ ; f 
τ0
,
= lim sup

 exp − σ ρ fτ0 
σ →−∞ 
R
(2.65)
fτ
νR 0




 log M σ ; f 
τ0
.
= lim inf
fτ


σ →−∞ 
exp − σ ρR 0 
(2.66)
fτ 0
and νR
It is trivial that if
f
ρRA
are the Ritt-type and the lower Ritt-type of order of fτ0 .
> 0,
f
τRA


 log f (σ ) 
A
,
= lim sup
f
σ →−∞  exp −σ ρ A 
(2.67)


 log f (σ ) 
A
.
= lim inf
f
σ →−∞ 
exp −σ ρ A 
(2.68)
R
f
νRA
R
f
Theorem 2.3. If σc A = −∞, β < ∞, L(= limn→∞ (log n/λn )) = 0, ∀n ∈ N\{0}, αn ∈
fτ
d(0,k) , ρR 0
> 0, we have ∀τ0 ∈ R,
fτ 0
τR
f
= τRA .
(2.69)
Proof. (1) We have the inequality, ∀τ0 ∈ R,
f
fτ
τRA ≤ τR 0 ,
(2.70)
τ0 is any arbitrary real number. From Theorem 2.2, we have, ∀ε ∈]0, 1[, ∃n1 ∈ N\{0},
∀n ≥ n1 , ∀σ ∈ R; ∀λ ≥ 1,
(2.71)
An ≤ M σ − λ; fτ0 exp σ + λ + β + ε log 6 1 + |σ | + |τ0 | + ε λn ,
where
λn = λn − Re αn
Now, ∀σ < 0,
An ≤ M σ − λ; fτ0 exp
and
β = lim sup
n→∞
mn
λn
β = β < ∞.
(2.72)
σ 2λ + β + ε log 6 1 + |σ | + |τ0 | + ε
λn .
σ −λ−
−σ
(2.73)
453
ON THE RITT ORDER AND TYPE OF A CERTAIN CLASS OF FUNCTIONS . . .
Also, we have ∀ε1 ∈]0, 1[, ∃σ1 = σε1 > 0, ∀σ < −σ1 ,
2λ(β + ε) log 6 1 + |σ | + |τ0 | + ε
< ε1 ,
−σ
An ≤ M σ − λ; fτ0 exp σ 1 − ε1 − λ λn
λ
µn(1,2) ,
≤ M σ − λ; fτ0 exp σ −
1 − ε1
where ∀n ≥ n2 = max n1 , n ,
µn(1,2) = 1 − ε1 1 − ε2 λn ,
λn ≥ λn 1 − ε2
ε2 ∈]0, 1[
(2.74)
(2.75)
(2.76)
which implies that, ∀λ ≥ 1, ∀n ≥ n2 , ∀σ < −σ1
An ≤ M σ − λ; fτ0 exp σ 1 −
fτ 0
τR
λ
µn(1,2) .
1 − ε1
(2.77)
is the Ritt-type of order of fτ0 , we have, ∀ε > 0, ∃σ = σε > 0, ∀σ < −σ !
"
fτ
fτ
log M σ − λ; fτ0 ≤ τR 0 + ε exp − (σ − λ)ρR 0 .
(2.78)
Hence, ∀n ≥ n2 , ∀ε > 0,
fτ
!
" fτ
log An ≤ τR 0 + ε exp − (σ − λ)ρR 0 + σ −
λ
µn(1,2) .
(1 − ε1 )
(2.79)
Let us consider fn , the function defined by
fn (σ ) = a exp − (σ − λ)b + µn(1,2) (σ + c),
(2.80)
and indexed by n > n2 . Choosing
fτ 0
a = τR
+ ε > 0,
fτ0
b = ρR
> 0,
c=
λ
,
ε1 − 1
(2.81)
we get ∀n ≥ n2 , ∀σ ∈ R\{σn },
fn (σ ) > fn σn
(2.82)
with
σn − λ =
ab
1
log
b
µn(1,2)
(2.83)
and
lim σn = −∞ ⇒ ∃n3 ∈ N\{0},
n→∞
∀n ≥ max n2 , n3 ,
σn < − max σ1 , σ ,
(2.84)
(2.85)
454
MARCEL BERLAND
where
log An ≤ fn σn =
b/µn(1,2)
µn(1,2) An
µn(1,2)
b
$
%
&
ε1 λ
ab
1
+
+ log
1 − ε1 b
µn(1,2)
µn(1,2)
(2.86)
≤ eab exp
ε1 λb
.
1 − ε1
Or, if L(= limn→∞ (log n/λn )) = 0, we have

 fA



(1,2)


fA
fA
ρ
/µ
n
R
(1,2)
(1,2)
(1,2) 
eρR
= lim sup µn(1,2) An
,
τR



n→∞ 
(2.87)
(M. Berland [3], following the theorem of Lindelöf-Blambert-Yu) and
fτ 0
ρR
f
= ρRA = 1 − ε1
1
fA
1 − ε2
ρR
fA
f
τRA = τR
(1,2)
(1,2)
,
(Theorem 2.2),
,
(2.89)
from which
1 − ε1
1 − ε2 λn
f
ρ A /[(1−ε1 )(1−ε2 )λn ]
AnR
≤e
(2.88)
fτ
τR 0
+ε

f
ρRA
fτ 0

ε1 λρR 
exp 
1 − ε1
(2.90)
and
f
ρ A /λn
AnR
f
ρ A /[(1−ε1 )(1−ε2 )λn ]
≤ AnR
.
(2.91)
f
Then, ∀ε1 ∈]0, 1[, ∀ε2 ∈]0, 1[, ∀ε > 0, ρRA > 0,
f
τRA


fτ
fτ 0
λρ
ε
τR 0 + ε
1
R
,
exp 
≤
1 − ε1
1 − ε1 1 − ε 2
(2.92)
as ε1 , ε2 , and ε are arbitrary, we deduce immediately that
f
fτ
∀τ0 ∈ R : τRA ≤ τR 0 .
(2.93)
(2) We get, when τ0 is a fixed real number, ∀ε > 0, ∃σε > 0, ∀σ < −σε ,
fτ (σ + iτ) ≤ fA (σ 1 + ε) .
0
(2.94)
In particular, ∀ε > 0, ∃σε > 0, ∀ε ∈ ]0, ε /σε [, ∃σε > 0, ∀σ < − max{σε , σε }
fA σ (1 + ε) ≤ fA σ − ε ,
(M. Berland [3])
(2.95)
and, hence, ∀ε > 0,
∀σ < − max{σε , σε } : M σ ; fτ0 ≤ fA σ − ε .
(2.96)
ON THE RITT ORDER AND TYPE OF A CERTAIN CLASS OF FUNCTIONS . . .
fτ 0
From ρR
455
f
= ρRA > 0, we get the inequality, ∀ε > 0,
fτ 0
τR
f
f
≤ τRA exp ε ρRA .
(2.97)
As ε is arbitrary, we have
fτ 0
∀τ0 ∈ R : τR
f
≤ τRA .
(2.98)
fτ 0
As a result of this theorem, we have an expression for τR
f
in terms of λn and An .
If σc A = −∞, β < ∞, ∀n ∈ N\{0}, αn ∈ d(0,k) , ∀τ0 ∈ R,
log n
= 0,
L = lim
n→∞
λn
fτ 0
ρR
> 0,
(2.99)
we have
fτ
τR 0
fτ
e ρR 0
= lim sup λn
n→∞
fτ
ρ 0 /λn
AnR
.
(2.100)
Remark. The notions of Ritt-type of order of functions, defined by B−Dirichletian
elements, are considered in [3] with the same result of this theorem.
f
Theorem 2.4. If σc A = −∞, β < ∞, ∀n ∈ N\{0}, αn ∈ d(0,k) , L = 0, λn ∼ λn+1 , ϕ
defined by
log An /An+1
,
(2.101)
ϕ(n) =
λn+1 − λn
f
is a nondecreasing function of n ≥ n1 , and ρRA > 0, we have, ∀τ0 ∈ R,
fτ 0
νR
f
= νRA .
(2.102)
Proof. (1) We have the inequality, ∀τ0 ∈ R,
fτ
f
νRA ≤ νR 0 .
(2.103)
Suppose that the inequality is false. Then
fτ 0
∃τ0 ∈ R : νR
f
fτ
f
< νRA .
fA
f
(2.104)
fτ
f
Let ε ∈]0, νRA −νR 0 [, ε ∈]0, ε/νB A [ and ν = νR −ε /(1−ε ); then νR 0 < ν < νRA . Under
the conditions stated in Theorem 2.4, R. K. Srivastava [17] proved that
ρfA /λ f
f
n
,
νRA e ρRA = lim inf λn AnR
n→∞
(2.105)
which implies that ∃n ∈ N\{0}, ∀n ≥ n ,
ρfA /λ f
n
νeρRA < λn AnR
.
(2.106)
456
MARCEL BERLAND
Now, ∀ε1 ∈]0, 1[, ∀ε2 ∈]0, 1[, ∃σ1 = σε1 > 0, ∀σ < −σ1 , ∀n ≥ n2 (= max{n1 , n }),
λ
µn(1,2) ,
1 − ε1
(2.107)
(see Theorem 2.3),
(2.108)
An ≤ M σ − λ; fτ0 exp σ −
where λ is a constant lying in [1, ∞[ and
µn(1,2) = 1 − ε1 1 − ε2 λn
which gives
log An − σ −
log An +
µn(1,2) ≤ log M σ − λ; fτ0 ,
(2.109)
λ
µn(1,2) − σ µn(1,2) ≤ log M σ − λ; fτ0 .
1 − ε1
(2.110)
λ
1 − ε1
Let us consider ϕn , the function defined by
ϕn (σ ) =
αn − βn σ
,
f exp − (σ − λ)ρRA
(2.111)
and indexed by n ≥ n2 . Choose
αn = log An +
λ
µn(1,2) ,
1 − ε1
This takes the maximum value at
βn = µn(1,2)
(> 0).

= log An + λ − 1  ,
µn(1,2)
1 − ε1 ρ fA
R
µn(1,2) ρfA /µn
ε1 λ f A
(1,2) exp
max ϕn (σ ) | σ ∈ R = f
ρR
A R
1 − ε1
ρRA e
log M σn − λ; fτ0
fτ0
fA
≤
! fτ0 " (for ρR = ρR ).
exp − σn − λ ρR
αn
1
σn =
− f
βn ρ A
R
(2.112)

(2.113)
(2.114)
As ∀n ∈ N\{0},
f
ρRA /µn(1,2)
µn(1,2) < λn ⇐⇒ A
f
ρ A /λn
> AnR
(2.115)
which gives
1 − ε 1 1 − ε2
f
ρRA e
log M σn − λ; fτ0
ε1 λ fA
ρ fA /λn
ρ
λn An
≤
exp
!
fτ " .
1 − ε1 R
exp − σn − λ ρR 0
(2.116)
∞
Finally, we have, ∀ε3 > 0, ∃ σn 1 , lim σn = −∞,
n→∞
log M σn − λ; fτ0
fτ 0
!
f τ 0 " ≤ ν R + ε3 .
exp − σn − λ ρR
(2.117)
ON THE RITT ORDER AND TYPE OF A CERTAIN CLASS OF FUNCTIONS . . .
457
Hence, we get, ∀ε3 > 0, ∀ε1 ∈]0, 1[, ∀ε2 ∈]0, 1[,
1 − ε1
1 − ε2 exp
ε1 λ fA
ρ
1 − ε1 R
fτ 0
ν ≤ νR
+ ε3 .
(2.118)
Choosing ε3 = ε1 = ε2 of ]0, 1[, we get
fτ
ν ≤ νR 0 .
(2.119)
Thus, we get the contradiction that
fτ 0
νR
fτ0
< (ν ≤)νR
(2.120)
which proves, under the stated conditions, that it is impossible to find a τ0 of R such
fτ
f
that νR 0 < νRA .
(2) We have the inequality, ∀τ0 ∈ R,
fτ 0
νR
f
≤ νRA
(see Theorem 2.3, 2).
(2.121)
fτ 0
As a result of this theorem, we have an expression for νR
∀τ0 ∈ R,
fτ
νR 0
fτ
e ρR 0
= lim inf λn
n→∞
fτ
ρ 0 /λn
AnR
in terms of λn and An ,
.
(2.122)
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Berland: 4, rue Dupin, 75006 Paris, France
E-mail address: berland@email.enst.fr