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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2008, Article ID 768105, 12 pages doi:10.1155/2008/768105 ResearchArticleConvergenceofVectorialContinuedFractionsRelatedtotheSpectral 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 thespectral seminorm is useful to study convergence or divergence ofvectorialcontinuedfractions in Banach algebras because such convergence or divergence is relatedto 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.Fora ∈A, σa, and ρa denote, respectively, the spectrum and thespectral seminorm of a. A formal vectorialcontinued fraction is an expression ofthe form y 0 b 0 a 1 · b 1 a 2 · b 2 ··· −1 −1 , 1.1 where a n n≥1 and b n n≥1 are two sequences of elements in A. In order to discuss convergence or divergence ofthevectorialcontinued fraction 1.1, we associate a sequence s n n≥0 called sequence of nth approximants defined by: s 0 b 0 , s 1 b 0 a 1 · b −1 1 , s 2 b 0 a 1 · b 1 a 2 · b −1 2 −1 , . . . . . . s n b 0 a 1 · b 1 a 2 · b 2 ··· a n−1 · a n−1 a n · b −1 n −1 −1 −1 , . . . . . . 1.2 2 Journal of Inequalities and Applications By induction, it can be shown that s n b 0 p n · q −1 n , 1.3 where the expressions p n and q n are determined from recurrence relations p n1 p n · b n1 p n−1 · a n1 , q n1 q n · b n1 q n−1 · a n1 , 1.4 with initial conditions: p 0 0,p 1 a 1 . q 0 e, q 1 b 1 . 1.5 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 thevectorialcontinued fraction defined by e a 1 4 e a 1 9 e a ··· 1 n 2 e a ··· −1 −1 −1 −1 , 1.6 where for each positive integer n>0, we have b n 1 n 2 · e a. 1.7 So, b n a 1 n 2 e ≥ a− 1 n 2 . 1.8 Therefore, the series ∞ n1 b n diverges. By Fair 1, Theorem 2.2, we cannot ensure convergence or divergence ofthevectorialcontinued fraction 1.6. But, if we apply thespectral seminorm to 1.7,weget ρ b n ≤ 1 n 2 ρa 1 n 2 . 1.9 So, the series ∞ n1 ρb n converges. From Theorem 2.5 in Section 2 below, thevectorialcontinued fraction 1.6 diverges according tothespectral seminorm so it diverges also according tothe norm because thespectral seminorm ρ satisfies ρx ≤x, ∀x ∈A. 1.10 In Section 3, we give another example of a vectorialcontinued fraction that converges according tothespectral seminorm and diverges according tothe norm algebra. From the simple and particular example above and the example in Section 3, we see that to study convergence or divergence ofvectorialcontinuedfractions we can use thespectral seminorm ofthe algebra to include a large class ofvectorialcontinued 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. M. Hemdaoui and M. Amzil 3 2. Convergenceofvectorialcontinuedfractions In this section, we discuss some conditions upon the elements a n and b n ofthevectorialcontinued fraction 1.1with b 0 0 which are necessary to ensure the convergence. Definition 2.1. Thevectorialcontinued fraction 1.1 converges if q −1 n exists starting from a certain rank N, and the sequence of nth approximants s n converges. Otherwise, thevectorialcontinued fraction 1.1 diverges. For future use, we record the following theorem due to P. Wynn. Theorem 2.2 2. For all n ∈ N,wehave s n1 − s n p n1 · q −1 n1 − p n · q −1 n −1 n a 1 b −1 1 q 0 a 2 q −1 2 q 1 a 3 q −1 3 q 2 a 4 q −1 4 ···q n−2 a n q −1 n q n−1 a n1 q −1 n1 . 2.1 Remark 2.3. In the commutative case, Theorem 2.2 above becomes as follows. For all n ∈ N, one has s n1 − s n −1 n in1 i1 a i · q −1 n1 · q −1 n . 2.2 Since convergence or divergence ofthevectorialcontinued fraction 1.1 is not affected by the value ofthe additive term b 0 , we omit it from subsequent discussion i.e., b 0 0. Now, we give a proposition that extends a result due to Wall 3 inthecaseofscalar continued fractions. Proposition 2.4. Thevectorialcontinued 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 relation 1.5 above, we have q 1 b 1 .Soρq 1 ρb 1 0. Since coefficients of 1.1 are commuting elements in A,itiseasytoshowthatforall positive integers n and m,wehave a m · q n q n · a m ; b m · q n q n · b m . 2.3 So, ρ a m · q n ≤ ρ a m · ρ q n ; ρ b m · q n ≤ ρ b m · σ q n . 2.4 Now, suppose that for n ≥ 1, ρq 2n−1 0. From relations 1.4 and 2.4,wehave ρ q 2n1 ≤ ρ q 2n · ρ b 2n1 ρ q 2n−1 · ρ a 2n1 . 2.5 Then, ρq 2n1 0, consequently ∀n ≥ 0; ρ q 2n1 0. 2.6 So infinitely many denominators q n are not invertible. Thevectorialcontinued fraction 1.1 diverges. 4 Journal of Inequalities and Applications Theorem 2.5 below gives a necessary condition for convergence according tothespectral seminorm. This result is an extension of von Koch Theorem 4, concerning the scalar case. A similar theorem was given by Fair 1 for vectorialcontinuedfractions according tothe norm convergence. Theorem 2.5. Let a n e, for all n ≥ 1,andb n be a sequence of commuting elements in A.Ifthe vectorialcontinued fraction 1.1 converges according tospectral 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 −1 n exists, for all n ≥ N. By an induction argument, it is easy to show that for all n ∈ N,wehave ρ q 2n − e ≤ exp K 2n − 1,ρ q 2n1 ≤ exp K 2n1 , 2.7 where K 0 0andK n n k1 ρb k ; for all n ≥ 1. Since for all positive integer n, a n e,andallb n are commuting elements in A,from Remark 2.3 above, we have d n s 2n1 − s 2n q −1 2n1 · q −1 2n , ∀n ≥ E N 2 1. 2.8 So, ρ d n ≥ ρ d −1 n −1 ρ q 2n1 · q 2n −1 , ∀n ≥ E N 2 1. 2.9 Then, ρ d n ≥ ρ q 2n1 −1 · ρ q 2n −1 , ∀n ≥ E N 2 1. 2.10 From this preceding, ρ d n ≥ 1 exp K 2n1 · 1 exp K 2n − 1 ≥ 1 exp 2K 2n1 ≥ 1 exp2K > 0, 2.11 where K ∞ n1 ρb n . So, the sequence s n n≥0 is not a ρ-Cauchy sequence in A. Remark 2.6. In a Banach algebra A if ρ denotes thespectral seminorm in A it is not a multiplicative seminorm in general. Consider thevectorial subspace of A defined by Kerρ{x ∈A|ρx0}. The quotient vectorial space A/ Kerρ becomes a normed vectorial space with norm defined by ˙ρ ˙xρx,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/ Kerρ witch is a Banach space. So, ρ-Cauchy sequences in A,ρ converge in the Banach space A/ Kerρ . M. Hemdaoui and M. Amzil 5 Remark 2.7. Whenever A is commutative, thevectorialcontinued fraction 1.1 diverges, if for one character ψ, the series n≥1 |ψb n | converges. Lemma 2.8. Let u n n 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 product k 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 product k p1 eu Np is invertible as finite product of invertible elements. We have ρ k p1 e u Np ≤ k p1 1 ρ u Np ≤ ∞ p1 1 ρ u Np . 2.13 But k p1 e u Np −1 k p1 e u Np −1 k p1 ∞ n0 −1 n u n Np . 2.14 Hence, ρ k p1 e u Np −1 ≤ k p1 ∞ n0 ρ n u Np k p1 1 1 − ρ u Np 1 k p1 1 − ρ u Np ≤ 1 ∞ p1 1 − ρ u Np . 2.15 Theorem 2.9. Let in thevectorialcontinued fraction 1.1 a n e for all n ≥ 1 and b n n∈N be a sequence of commuting elements in A. If both series n≥0 ρ b 2p1 , n≥0 ρ b 2p1 · ρ 2 b 2p 2.16 converge, then, thevectorialcontinued fraction 1.1 diverges. Proof of Theorem 2.9. Since both series n≥0 ρb 2p1 and n≥0 ρb 2p1 · ρ 2 b 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. 6 Journal of Inequalities and Applications Now, consider thevectorial c ontinued fraction c 1 c 2 c 3 ··· −1 −1 −1 , 2.17 where c 2k b 2N2k1 · θ −1 k−1 · θ −1 k ,c 2k−1 −b 2N2k1 · b 2 2N2k · θ k−1 · θ k ,k 1, 2, 2.18 We will suppose that q −1 n exists for all n ≥ N otherwise, from Definition 2.1, thevectorialcontinued fraction 1.1 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 · b 2 2N2k · θ k−1 · θ k , k 1, 2, , θ 0 e . 2.19 Then, U k U k−1 · c k U k−2 , V k V k−1 · c k V k−2 , ∀k ≥ 2. 2.20 This lemma is proved by the same argument given by Wall 3, 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 ofthevectorialcontinued fraction 2.17. Since both series n≥0 ρb 2p1 , n≥0 ρb 2p1 · ρb 2p 2 converge and from Lemma 2.8 above θ k and θ −1 k are bounded, we conclude that the series k≥1 ρc k converges. Then, it follows as in the proof of Theorem 2.5, that thevectorialcontinued fraction 2.17 diverges and ρ U 2k1 · V −1 2k1 − U 2k · V −1 2k ρ p 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 n n≥1 is not a ρ-Cauchy sequence in A. M. Hemdaoui and M. Amzil 7 Now, we state Theorem 2.13 to give a sufficient condition to have convergenceofthevectorialcontinued fraction 1.1. A similar theorem was given by Peng and Hessel 5, to study convergenceofthevectorialcontinued fraction 1.1 in norm where for each positive integer n, a n e. Before stating the proof of Theorem 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,wehaveρb −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 −1 n b −1 · c n . 2.23 So, ρ b c −1 ≤ ρ b −1 · ∞ n0 ρ n b −1 · c ρ b −1 1 − ρ b −1 · c . 2.24 Lemma 2.12. Let ∈0, 1, a n n∈N and b n n∈N be two sequences of elements in A such that for each positive integer n ≥ 1, the spectra of a n · b −1 n and b −1 n lie in the open ball B0, 1/2. Then, for each positive integer n ≥ 1, q −1 n exists and ρq −1 n · q n−1 <. Where q n is the nth denominator ofthevectorialcontinued fraction 1.1. Proof. From recurrence relation 1.5 above, we have q 0 e, q 1 b 1 , 2.25 then, q −1 1 b −1 1 and ρq −1 1 · q 0 ρb −1 1 ≤ 1/2<. Now, suppose that for n ≥ 2, q −1 n−1 exists and ρq −1 n−1 · q n−2 <. Then, from recurrence relation 1.4 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 . 2.26 Put c q −1 n−1 · q n−2 · a n ,b b n . 2.27 Appling Lemma 2.11,wehave ρ b −1 · c ≤ ρ q −1 n−1 · q n−2 · ρ b −1 n · a n < 1 2 . 2.28 So b 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 1 − 1/2 <. 2.29 Therefore, q −1 n exists. So, for all n ≥ 0, q n is invertible and ρq −1 n · q n−1 <. 8 Journal of Inequalities and Applications Theorem 2.13. Let ∈0, 1, a n and b n be commuting terms ofthevectorialcontinued fraction 1.1 such that for each positive integer n ≥ 1, the spectra of a n · b −1 n and b −1 n lie in the open ball B0, 1/2. Then, thevectorialcontinued fraction 1.1 converges. Proof of Theorem 2.13. For positive integers n ≥ 1andm ≥ 1, we introduce the finite vectorialcontinued 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 s n 0 0,s 0 m s m , 2.31 where s m is the mth approximant ofthecontinued fraction 1.1. It is easily shown from 2.30 that s n m a n1 · b n1 s n1 m−1 −1 . 2.32 By the repeated use of Lemma 2.11 in each iteration in 2.30 for every n ≥ 1 and every m ≥ 1, we can show that for each n and m, b n1 s n1 m−1 −1 exists and ρ s n m <. 2.33 We have b n1 s n1 m −1 − b n1 s n1 m−1 −1 b n1 s n1 m −1 · s n1 m−1 − s n1 m · b n1 s n1 m−1 −1 . 2.34 Thus, from relations 2.32 and 2.34,wehave 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 ,form ∈ N ∗ . Then, ρ s n m1 − s n m ≤ ρ a n1 · b −1 n1 · ρ b −1 n1 · ρ K m · ρ s n1 m − s n1 m−1 · ρ K m−1 . 2.36 Since from 2.33 ρb −1 n1 · s n1 m ≤ ρb −1 n1 · ρs n1 m ≤ 1/2 2 < 1/2, then, using Lemma 2.11, ρ K m ≤ 1 1 − ρ b −1 n1 · s n1 m < 2, for m ∈ N ∗ , 2.37 we have ρa n1 · b −1 n1 ≤ 1/2 and ρb −1 n1 ≤ 1/2. M. Hemdaoui and M. Amzil 9 Then, ρ s n m1 − s n m ≤ 2 · ρ s n1 m − s n1 m−1 . 2.38 Gradually, we get ρ s n m1 − s n m ≤ 2m · ρ s nm 1 − s nm 0 . 2.39 Besides, we have s nm 0 0ands nm 1 a nm1 · b −1 nm1 . Thus, ρ s n m1 − s n m ≤ 2m · ρ s nm 1 2m · ρ a 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 1 − 2 . 2.42 Hence, the sequence s m m∈N of mth approximants ofthevectorialcontinued fraction 1.1 is a ρ-Cauchy sequence in A. Consequently, s m converges and from Lemma 2.12, q −1 n exists thus thevectorialcontinued fraction 1.1 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, thevectorialcontinued fraction a 1 e − a 2 e − a 3 e − a 4 e −··· −1 −1 −1 −1 2.43 converges. Proof. By relations 1.4 and 1.5,wehaveq 1 e, thus, σ q 1 β 1 with β 1 1 1 2 1. 2.44 And q 2 q 1 − q 0 a 2 e − a 2 , thus, σ q 2 β 2 with β 2 1 α 2 ≥ 1 − 1 4 3 4 · β 1 . 2.45 10 Journal of Inequalities and Applications By induction, we show that for all n ≥ 2 σ q n β n , 2.46 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 2 n β 0 ≥ n 1 2 n > 0; ∀n ≥ 1. 2.48 So q −1 n exists for all n ≥ 1. Since all a n are commuting elements, then by Remark 2.3 above s n s 1 n k2 s k − s k−1 s 1 n k2 d k q −1 k q −1 k−1 , 2.49 where d k −1 k−1 ik i1 − a i ik i1 a i . 2.50 We have 0 ≤ ρ d k ≤ ik i1 ρ a i ≤ 1 4 k . 2.51 Hence, ρ d k q −1 k q −1 k−1 ≤ ρ d k ρ q −1 k ρ q −1 k−1 1 β k β k−1 ρ d k ≤ 1 4 k 2 k k 1 2 k−1 k 1 2kk 1 . 2.52 Therefore, for positive integers n and m such that n>m,wehave 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 that s n n≥1 is a ρ-Cauchy sequence in A. [...]... tothespectral seminorm Consequently, thevectorialcontinued fraction 1.1 converges according tothespectral seminorm tothe value ∞ n 1 e/k 2 k 12 Journal of Inequalities and Applications References 1 W Fair, “Noncommutative continued fractions, ” SIAM Journal on Mathematical Analysis, vol 2, no 2, pp 226–232, 1971 2 P Wynn, Continuedfractions whose coefficients obey a noncommutative law of multiplication,”... Obviously, the sequence sn n≥0 is not a Cauchy sequence according tothe norm, so thevectorialcontinued fraction 1.1 does not converge in norm Now, we use thespectral seminorm, we have ρ s2n ρ s2n − 1 − s2n ∞ k 1 ·e k2 1 1 ρ u2n 2 ∞ ρ nT k n 1 1 ·e k2 1 ≤ 1 2 ρ T ≤ nρ T 1 2n2 ∞ k n 1 k2 1 1 2n2 1 2 ∞ k n −→ 0, 1 −→ 0 when n −→ ∞ k2 1 3.4 The sequence sn n≥0 ofthe nth approximants converges according to the. .. bn ⎩b e − an , ∀n ≥ 2 n Consider thevectorialcontinued fraction 1.1 formed with the sequences an n∈N and bn n∈N Using recurrence relations 1.4 and 1.5 , we can easily show that for each positive integer n ≥ 1, qn e thus qn is invertible, for all n ≥ 1 The 2n th approximant and the 2n 1 th approximant ofthevectorialcontinued fraction 1.1 are, respectively, equal to n s2n n p2n uk T k 1 n s2n 1... Example Here, we give an example of a vectorialcontinued fraction that converges according tothespectral seminorm and does not converge according tothe norm Let A be a unital complex Banach algebra and T a nonnull quasinilpotent element in A Consider the sequence in A defined for each positive integer n > 0, by un T 1 · e n2 3.1 For each positive integer n > 0, un is then invertible Let an n∈N and... 1963 3 H S Wall, Analytic Theory ofContinued Fractions, D Van Nostrand, New York, NY, USA, 1948 4 H von Koch, “Sur un th´ or` rme de Stieltjes et sur les fonctions d´ finies par des fractions continues,” e e e Bulletin de la Soci´ t´ Math´ matique de France, vol 23, pp 33–40, 1895 ee e 5 S T Peng and A Hessel, Convergenceof noncommutative continued fractions, ” SIAM Journal on Mathematical Analysis, vol . 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. Corporation Journal of Inequalities and Applications Volume 2008, Article ID 768105, 12 pages doi:10.1155/2008/768105 Research Article Convergence of Vectorial Continued Fractions Related to the Spectral. 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