Hemdaoui, mhemdaoui@yahoo.fr Received 07 February 2008; Accepted 16 April 2008 Recommended by Charles Chidume We show that the spectral seminorm is useful to study convergence or diverge
Trang 1Volume 2008, Article ID 768105, 12 pages
doi:10.1155/2008/768105
Research Article
Convergence of Vectorial Continued Fractions
Related to the Spectral Seminorm
M Hemdaoui and M Amzil
Laboratoire G.A.F.O, D´epartement de Mathmatiques & Informatique, Facult´e des Sciences,
Universit´e Mohamed Premier, Oujda, Morocco
Correspondence should be addressed to M Hemdaoui, mhemdaoui@yahoo.fr
Received 07 February 2008; Accepted 16 April 2008
Recommended by Charles Chidume
We show that the spectral seminorm is useful to study convergence or divergence of vectorial continued fractions in Banach algebras because such convergence or divergence is related to a spectral property.
Copyright q 2008 M Hemdaoui and M Amzil This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1 Introduction
Let A be a unital complex Banach algebra We denote by e the unit element of A · is the
norm of A For a ∈ A, σa, and ρa denote, respectively, the spectrum and the spectral seminorm of a.
A formal vectorial continued fraction is an expression of the form
y0 b0 a1·b1 a2·b2 · · ·−1−1, 1.1 wherea nn≥1andb nn≥1are two sequences of elements inA
In order to discuss convergence or divergence of the vectorial continued fraction1.1,
we associate a sequences nn≥0 called sequence of nth approximants defined by:
s0 b0,
s1 b0 a1· b−1
1 ,
s2 b0 a1·b1 a2· b−1
2
−1
,
.
s n b0 a1·b1 a2·b2 · · · a n−1·a n−1 a n · b−1
n
−1−1−1
,
.
1.2
Trang 2By induction, it can be shown that
s n b0 p n · q−1
where the expressions p n and q nare determined from recurrence relations
p n1 p n · b n1 p n−1 · a n1 ,
with initial conditions:
p0 0, p1 a1.
p n and q n are respectively called nth numerator and nth denominator of 1.1
Now, consider the following example
Let a be a nonnull quasinilpotent element in A Consider the vectorial continued fraction
defined by
e a
1
4e a
1
9e a
· · ·
1
n2e a
· · ·
−1−1−1−1
where for each positive integer n > 0, we have
b n 1
So,
b n a 1
n2e
≥ a − n12
Therefore, the series ∞n1 b n diverges
By Fair1, Theorem 2.2, we cannot ensure convergence or divergence of the vectorial continued fraction1.6 But, if we apply the spectral seminorm to 1.7, we get
ρ
b n
≤ 1
n2 ρa 1
So, the series ∞n1 ρb n converges From Theorem 2.5 in Section 2 below, the vectorial continued fraction 1.6 diverges according to the spectral seminorm so it diverges also
according to the norm because the spectral seminorm ρ satisfies
ρx ≤ x, ∀x ∈ A. 1.10
In Section 3, we give another example of a vectorial continued fraction that converges according to the spectral seminorm and diverges according to the norm algebra
From the simple and particular example above and the example inSection 3, we see that
to study convergence or divergence of vectorial continued fractions we can use the spectral seminorm of the algebra to include a large class of vectorial continued fractions
First, we start by determining necessary conditions upon a n and b n to ensure the convergence
Next, we give sufficient conditions to have the convergence
Trang 32 Convergence of vectorial continued fractions
In this section, we discuss some conditions upon the elements a n and b n of the vectorial continued fraction1.1 with b0 0 which are necessary to ensure the convergence
Definition 2.1 The vectorial continued fraction 1.1 converges if q−1
n exists starting from a
certain rank N, and the sequence of nth approximants s nconverges Otherwise, the vectorial continued fraction1.1 diverges
For future use, we record the following theorem due to P Wynn
Theorem 2.2 2 For all n ∈ N, we have
s n1 − s n p n1 · q−1
n1 − p n · q−1
n
−1n a1b−11 q0a2q−12 q1a3q−13 q2a4q4−1· · · q n−2 a n q−1n q n−1 a n1 q n1−1 . 2.1
Remark 2.3 In the commutative case,Theorem 2.2above becomes as follows
For all n ∈ N, one has
s n1 − s n −1n
in1
i1
a i · q−1
n1 · q−1
Since convergence or divergence of the vectorial continued fraction1.1 is not affected
by the value of the additive term b0, we omit it from subsequent discussioni.e., b0 0 Now, we give a proposition that extends a result due to Wall3 in the case of scalar continued fractions
Proposition 2.4 The vectorial continued fraction 1.1 where its terms are commuting elements in A
diverges, if its odd partial denominators b 2n1 are all quasinilpotent elements in A.
Proof In fact, from relation1.5 above, we have q1 b1 So ρq1 ρb1 0.
Since coefficients of 1.1 are commuting elements in A, it is easy to show that for all
positive integers n and m, we have
a m · q n q n · a m; b m · q n q n · b m 2.3 So,
ρ
a m · q n
≤ ρa m
· ρq n
b m · q n
≤ ρb m
· σq n
Now, suppose that for n ≥ 1, ρq 2n−1 0
From relations1.4 and 2.4, we have
ρ
q 2n1
≤ ρq 2n
· ρb 2n1
ρq 2n−1
· ρa 2n1
Then, ρq 2n1 0, consequently
∀n ≥ 0; ρq 2n1
So infinitely many denominators q nare not invertible
The vectorial continued fraction1.1 diverges
Trang 4Theorem 2.5below gives a necessary condition for convergence according to the spectral seminorm This result is an extension of von Koch Theorem4, concerning the scalar case A similar theorem was given by Fair1 for vectorial continued fractions according to the norm convergence
Theorem 2.5 Let a n e, for all n ≥ 1, and b n be a sequence of commuting elements in A If the
vectorial continued fraction1.1 converges according to spectral seminorm, then, the series ∞
n1 ρb n
diverges.
Proof Suppose ∞n1 ρb n is a converging series, and there exists a positive integer N such that
q n−1exists, for all n ≥ N.
By an induction argument, it is easy to show that for all n ∈ N, we have
ρ
q 2n − e≤ expK 2n
q 2n1
≤ expK 2n1
where K0 0 and K n n
k1 ρb k ; for all n ≥ 1.
Since for all positive integer n, a n e, and all b n are commuting elements in A, from
Remark 2.3above, we have
d n s 2n1 − s 2n q−1
2n1 · q−1
2n , ∀n ≥ E
N
2
So,
ρ
d n
≥ρ
d−1n −1
ρ
q 2n1 · q 2n−1, ∀n ≥ E
N
2
Then,
ρ
d n
≥ρ
q 2n1
−1
·ρ
q 2n
−1
, ∀n ≥ E
N
2
From this preceding,
ρ
d n
exp
K 2n1
exp
K 2n
− 1 ≥
1 exp
2K 2n1
≥ exp2K1 > 0, 2.11
where K ∞n1 ρb n .
So, the sequences nn≥0 is not a ρ-Cauchy sequence in A.
Remark 2.6 In a Banach algebra A if ρ denotes the spectral seminorm in A it is not a
multiplicative seminorm in general
Consider the vectorial subspace of A defined by Kerρ {x ∈ A | ρx 0} The
quotient vectorial space A/Kerρ becomes a normed vectorial space with norm defined by
˙ρ ˙x ρx, x ∈ ˙x “ ˙x denotes the class of x modulo Kerρ.”
Generally, the normed vectorial space A/Kerρ is not complete Its complete normed vectorial space is A/Kerρwitch is a Banach space So, ρ-Cauchy sequences in A, ρ converge
in the Banach space A/Kerρ
Trang 5Remark 2.7 WheneverA is commutative, the vectorial continued fraction 1.1 diverges, if for
one character ψ, the series n≥1 |ψb n| converges
Lemma 2.8 Let u nn be a sequence of commuting elements in A.
If the series n≥1 ρu n converges, then, there exists a positive integer N ≥ 1 such that for every
positive integer k ≥ 1, the finite productk
p1 e u Np is invertible and ρ-bounded and its inverse is
also ρ-bounded.
Proof Since the series n≥1 ρu n converges, therefore, there exists a positive integer N ≥ 1
such that
ρ
u n
< 1; ∀n ≥ N. 2.12
Hence, for k ≥ 1 the productk
p1 eu Np is invertible as finite product of invertible elements
We have
ρ
k
p1
e u Np
≤ k
p1
1 ρu Np
≤ ∞
p1
1 ρu Np
But
k
p1
e u Np −1
k
p1
e u Np−1
k
p1
∞
n0
−1n u n Np 2.14 Hence,
ρ
k
p1
e u Np
−1
≤ k
p1
∞
n0
ρ n
u Np
k
p1
1
1− ρu Np
p1
1− ρu Np
p1
1− ρu Np
.
2.15
Theorem 2.9 Let in the vectorial continued fraction 1.1 a n e for all n ≥ 1 and b nn∈N be a sequence of commuting elements in A If both series
n≥0
ρ
b 2p1
,
n≥0
ρ
b 2p1
· ρ2
b 2p
2.16
converge, then, the vectorial continued fraction1.1 diverges.
Proof of Theorem 2.9 Since both series n≥0 ρb 2p1 and n≥0 ρb 2p1 · ρ2b 2p converge, it follows that the series n≥0 ρb 2p1 · ρb 2p converges too
Therefore, from Lemma 2.8 above, there exists a positive integer N ≥ 1 such that for
k ≥ 1, the quantity θ k k
p1 1 b 2N2p1 · b 2N2p is invertible
Trang 6Now, consider the vectorial continued fraction
c1c2c3 · · ·−1−1−1, 2.17 where
c 2k b 2N2k1 · θ−1
k−1 · θ−1
k , c 2k−1 −b 2N2k1 · b2
2N2k · θ k−1 · θ k , k 1, 2, 2.18
We will suppose that q n−1 exists for all n ≥ N otherwise, from Definition 2.1, the vectorial continued fraction1.1 diverges
Before continuing the proof, we give the following lemma that will be used later
Lemma 2.10 For all positive integers k ≥ 1, consider the quantities
U 2k p 2N2k1 · θ−1
k , V 2k q 2N2k1 · θ−1
k ,
U 2k1 p 2N2k · θ k , V 2k1 q 2N2k · θ k ,
c 2k b 2N2k1 · θ−1
k−1 · θ−1
k , c 2k−1 −b 2N2k1 · b2
2N2k · θ k−1 · θ k ,
k 1, 2, ,
θ0 e.
2.19
Then,
U k U k−1 · c k U k−2 ,
This lemma is proved by the same argument given by Wall3, Lemma 6.1 for scalar continued fractions
Lemma 2.10 shows that U n and V n are respectively the nth numerator and nth
denominator of the vectorial continued fraction2.17
Since both series n≥0 ρb 2p1 , n≥0 ρb 2p1 · ρb 2p2 converge and from Lemma 2.8
above θ k and θ−1k are bounded, we conclude that the series k≥1 ρc k converges
Then, it follows as in the proof ofTheorem 2.5, that the vectorial continued fraction2.17 diverges and
ρ
U 2k1 · V−1
2k1 − U 2k · V−1
2k
ρp 2N2k · q−1
2N2k − p 2N2k1 · q−1
2N2k1
≥ exp
2
∞
k1
ρ
c k
> 0.
2.21 So,
ρ
s 2N2k1 − s 2N2k
≥ exp
2
∞
k1
ρ
c k
> 0, ∀k ≥ 0. 2.22
This shows that the sequence of nth approximants s nn≥1 is not a ρ-Cauchy sequence in A.
Trang 7Now, we state Theorem 2.13 to give a sufficient condition to have convergence of the vectorial continued fraction1.1
A similar theorem was given by Peng and Hessel 5, to study convergence of the vectorial continued fraction1.1 in norm where for each positive integer n, a n e.
Before stating the proof ofTheorem 2.13, we give the following lemmas
Lemma 2.11 Let b and c be two commuting elements in A such that the spectrum of b−1· c is satisfied,
σb−1· c ⊂ B0, 1 Then, the element b c is invertible and its inverse satisfies ρb c−1 ≤
ρb−1/1 − ρb−1· c.
Proof Since σb−1· c ⊂ B0, 1, we have ρb−1· c < 1 So the element b c is invertible in A Its
inverse is
b c−1 b−1
e b−1· c−1 b−1·∞
n0
−1n
b−1· cn
So,
ρ
b c−1≤ ρb−1
·∞
n0
ρ n
b−1· c ρ
b−1
Lemma 2.12 Let ∈0, 1, a nn∈Nand b nn∈Nbe two sequences of elements in A such that for each
positive integer n ≥ 1, the spectra of a n · b−1
n and b−1n lie in the open ball B0, 1/2 Then, for each positive integer n ≥ 1, q−1n exists and ρq−1n · q n−1 < .
Where q n is the nth denominator of the vectorial continued fraction 1.1 .
Proof From recurrence relation1.5 above, we have
then, q−11 b−1
1 and ρq−11 · q0 ρb−1
1 ≤ 1/2 < Now, suppose that for n ≥ 2, q−1n−1 exists and ρq−1n−1 · q n−2 < .
Then, from recurrence relation1.4 above, we have
q n q n−1 · b n q n−2 · a n q n−1·b n q−1
n−1 · q n−2 · a n
Put
c q n−1−1 · q n−2 · a n , b b n 2.27 ApplingLemma 2.11, we have
ρ
b−1· c≤ ρq−1n−1 · q n−2· ρb−1n · a n< 1
Sob n q−1
n−1 · q n−2 · a n is invertible and its inverse satisfies
ρ
b n q−1
n−1 · q n−2 · a n−1< 1/2
1− 1/2 <
1/2
Therefore, q−1n exists So, for all n ≥ 0, q n is invertible and ρq−1n · q n−1 <
Trang 8Theorem 2.13 Let ∈0, 1, a n and b n be commuting terms of the vectorial continued fraction1.1
such that for each positive integer n ≥ 1, the spectra of a n · b−1
n and b n−1lie in the open ball B0, 1/2 Then, the vectorial continued fraction1.1 converges.
Proof of Theorem 2.13 For positive integers n ≥ 1 and m ≥ 1, we introduce the finite vectorial
continued fraction
s n m a n1·b n1 a n2·b n2 · · · a nm−1·b nm−1 a nm · b−1
nm
−1−1−1
2.30 with initial conditions
where s m is the mth approximant of the continued fraction 1.1
It is easily shown from2.30 that
s n m a n1·b n1 s n1 m−1 −1. 2.32
By the repeated use ofLemma 2.11in each iteration in2.30 for every n ≥ 1 and every m ≥ 1,
we can show that for each n and m, b n1 s n1 m−1 −1exists and
ρ
s n m
We have
b n1 s n1 m −1−b n1 s n1 m−1 −1b n1 s n1 m −1·s n1 m−1 − s n1 m ·b n1 s n1 m−1 −1 2.34
Thus, from relations2.32 and 2.34, we have
s n m1 − s n m a n1·b n1 s n1 m −1−b n1 s n1 m−1 −1
a n1·b n1 s n1 m −1·s n1 m−1 − s n1 m ·b n1 s n1 m−1 −1
a n1 · b−2
n1 · K m·s n1 m − s n1 m−1 · K m−1 ,
2.35
where K m e b−1
n1 · s n1 m −1, for m ∈ N∗ Then,
ρ
s n m1 − s n m ≤ ρa n1 · b−1
n1
· ρb−1n1
· ρK m
· ρs n1 m − s n1 m−1 · ρK m−1
Since from2.33 ρb−1
n1 · s n1 m ≤ ρb−1
n1 · ρs n1 m ≤ 1/22< 1/2, then, usingLemma 2.11,
ρ
K m
1− ρb−1n1 · s n1 m < 2, for m ∈ N
we have ρa n1 · b−1
n1 ≤ 1/2 and ρb−1
n1 ≤ 1/2.
Trang 9ρ
s n m1 − s n m ≤ 2· ρs n1 m − s n1 m−1 . 2.38 Gradually, we get
ρ
s n m1 − s n m ≤ 2m · ρs nm1 − s nm0 . 2.39
Besides, we have s nm0 0 and s nm
1 a nm1 · b−1
nm1
Thus,
ρ
s n m1 − s n m ≤ 2m · ρs nm1
2m · ρa nm1 · b−1
nm1
< 1
2 2m1 2.40
Now, consider m > 1, p ≥ 1, we have
ρ
s n mp − s n m ≤ip−1
i0
ρ
s n mi1 − s n mi≤ 1
2·
ip−1
i0
2m2i1
1
2· 2m1
1− 2p
1− 2 ≤ 1
2· 2m1
1− 2.
2.41
In these inequalities n is arbitrary, thus we can choose n 0.
Then,
ρ
s mp − s m≤ 1
2· 2m1
Hence, the sequences mm∈Nof mth approximants of the vectorial continued fraction 1.1 is a
ρ-Cauchy sequence in A.
Consequently, s m converges and from Lemma 2.12, q−1n exists thus the vectorial continued fraction1.1 converges
Theorem 2.14 Let a n be a sequence of commuting elements in A such that for each positive integer
n ≥ 1, σa n {α n }, where 0 ≤ α n ≤ 1/4 Then, the vectorial continued fraction
a1
e − a2
e − a3
e − a4e − · · · −1−1−1−1
2.43
converges.
Proof By relations1.4 and 1.5, we have q1 e, thus,
σ
q1
β1
with β1 1 1
And q2 q1− q0a2 e − a2, thus,
σ
q2
β2
with β2 1 α2 ≥ 1 −1
4 3
Trang 10By induction, we show that for all n ≥ 2
σ
q n
β n
such that
β n≥ n 1
2n β n−1 ,
β n β n−1 − α n β n−2
2.47 Hence,
β n≥ n 1
2n β0≥ n 1
2n > 0; ∀n ≥ 1. 2.48
So q−1n exists for all n ≥ 1.
Since all a nare commuting elements, then byRemark 2.3above
s n s1n
k2
s k − s k−1 s1n
k2
where
d k −1k−1 ik
i1
− a i ik
i1
We have
0≤ ρd k
≤ ik
i1
ρ
a i
≤ 1
Hence,
ρ
d k q−1k q−1k−1
≤ ρd k
ρ
q k−1
ρ
q k−1−1
β k β k−1 ρ
d k
≤ 1
4k
2k
k 1
2k−1
k 1
2kk 1 . 2.52
Therefore, for positive integers n and m such that n > m, we have
ρ
s n − s m≤ n
m1
ρ
d k · q−1
k · q−1
k−1
≤ n
km1
1
2kk 1 <
1
m 1 . 2.53 So,
ρ
s n − s m≤ ∞
km1
1
2kk 1 <
1
m 1 . 2.54
It follows thats nn≥1 is a ρ-Cauchy sequence in A.
... class="page_container" data-page ="7 ">Now, we state Theorem 2.13 to give a sufficient condition to have convergence of the vectorial continued fraction1.1
A similar theorem was given by... for convergence according to the spectral seminorm This result is an extension of von Koch Theorem4, concerning the scalar case A similar theorem was given by Fair1 for vectorial continued fractions. .. the open ball B0, 1/2 Then, the vectorial continued fraction1.1 converges.
Proof of Theorem 2.13 For positive integers n ≥ and m ≥ 1, we introduce the finite vectorial< /i>