����������� ������ 53 (2001), 91–102

UDK 517.521

���������� ������ ��� research paper

SOME NEW PROPERTIES OF SEQUENCE SPACES AND APPLICATION TO THE CONTINUED FRACTIONS

Bruno de Malafosse

Abstract. We give two methods of approximation of a solution of an infinite linear system. First, we will construct a sequence of finite matrices which approaches a solution, this one being defined by an infinite sequence. Then, we will apply these results to the continued fractions.

1. Introduction

Infinite linear systems have been studied by many authors, let us cite for instance Cooke [1], Defranza, Zeller [2], Maddox [6], Pòlya [1], Reade [13]. Here, we construct a natural sequence which converges to a solution of the system, by the means of a sequence of finite matrices deduced from an infinite matrix A. This principle has been developed by Pòlya, but the space used by this author contains infinitely many solutions, and the matrices are very particular. We propose another class of matrices, a space in which the system has one and only one solution, and the possibility to do a calculus of error.

This paper is organized as follows; in section 2 we recall [1], [10], [11] the Pòlya’s method, which illustrate the construction of a solution of an infinite linear system. This method consists in defining a sequence of finite matrices converging to a solution of a system. We give, further, in other spaces denoted s_{c} or s_{r}, see [4], [7] another sequence of finite matrices which converges to a solution of such a space. Under other conditions, we have a second approximation of this solution, with a calculus of the error. In Section 3, we recall (see E. Hellinger and H.S. Wall [3], [14], [15]) some results concerning the continued fractions and the bounded matrices. Finally, we apply the results of Section 2 to those. In fact, to obtain the formal expansion into a continued fraction, we have to give an approximation of the coefficient a′_{11}, in the first row, and in the first column of a particular right reciprocal matrix. a′_{11} is called the leading coefficient, we have in fact a′_{11} = a′_{11}(z), and it has been shown [14] that a′_{11}(z) = \frac{1}{z} + \frac{O(1)}{z Im(z)}, where O(1) represents a function of z which is numerically less than a constant independent of z for all z with Im(z) > 0.

AMS Subject Classification: 46 A 15, 40 A 15

2. Some definitions and properties of infinite linear systems

2.1. Linear infinite systems and Pòlya matrices In this work, we shall study linear infinite systems

∑_{m=1}^{+∞} a_{nm}x_{m} = b_{n},  n = 1, 2,…,  (1)

where the sequences (a_{nm}) and (b_{n}) are given, (x_{n}) being the unknown sequence. This system is equivalent to the single matrix equation

AX = B,  (2)

where A = (a_{nm}), n being the index of the n-th row, m the one of the m-th column, n and m being integers greater than 1; X = (x_{n}) and B = (b_{n}) are one-column matrices. Define, now, for every p, by [A]_{p} the matrix, whose elements of the p first rows, and of the p first columns are those of A. In this section the goal is to give a method permitting to calculate an approximation of a solution of an infinite linear system, by the means of a sequence of finite matrices {[A]_{p}}. This principle has been developed by Pòlya, see [1], but, as we shall see, the matrices used and the results obtained are totally different.

Let us recall that a Pòlya matrix A satisfies the following conditions a_{1m} ≠ 0 for infinitely many values of m, and

lim inf_{m→∞} ∑_{k=1}^{n−1} |\frac{a_{km}}{a_{nm}} | = 0, n = 2, 3,…

Pòlya’s theorem, [1], [10], is formulated as follows:

Theorem 1. If A is a Pòlya matrix, then the equation AX = B, admits for any B, a solution such that the series ∑_{m} a_{nm}x_{m} are absolutely convergent, for each value of n.

Recall briefly the well-known construction of a solution of such a system, (see [1] for a detailed study). We consider, here, the particular case where the finite matrices deducted from A, are successively

Ã_{1} = (a_{11}), Ã_{2} = ( \begin{matrix}a_{12} & a_{13} \\ a_{22} & a_{23} \end{matrix} ) , …,

Ã_{n} = ( \begin{matrix}a_{1_{,}α_{n}} & . & . & a_{1_{,}α_{n}+n−1} \\ . & . & . & . \\ . & . & . & . \\ a_{n_{,}α_{n}} & . & . & a_{n,α_{n}+n−1} \end{matrix} ) , …

with α_{n} = (n^{2} − n + 2)/2, for all n ≥ 2, and are invertible; then a solution of (1) can be determined by the following method:

Ã_{1}X_{1} = B̃_{1}, with X_{1} = (x_{1}), B̃_{1} = (b_{1}) admits x_{1} = b_{1}/a_{11} as solution; in the same way, with x_{1} defined by the preceding equation, we let B̃_{2}, such that tB̃_{2} = (0, b_{2} − a_{21}x_{1}), and consider the equation:

Ã_{2}X_{2} = B̃_{2},

with ^{t}X_{2} = (x_{2}, x_{3}). This one has X_{2} = (Ã_{2})^{−1}B̃_{2}, as its unique solution. Step by step, setting, for i ≥ 3

tB̃_{i} = (0, 0,…, b_{i} − ∑_{m=1}^{α_{i}−1} a_{i,m}x_{m}),

where x_{1},…, x_{α_{i}−1}, are determined by X_{j} = (Ã_{j})^{−1}B̃_{j}, 1 ≤ j ≤ i − 1, it has been proved that the vector Z, defined by Z = (X_{n})_{n≥1} is a solution of (1) satisfying Theorem 1.

Remark 1. In the case where the matrices Ã_{n} are not all invertible, it is necessary to consider the first integer p, such that Ã_{p} is not invertible; it has been proved by Pòlya that there exists an integer k_{1} > p, such that the matrix

Ã′_{1} = ( \begin{matrix}a_{1,α_{p}} & . & . & a_{1,α_{p}+k_{1}−1} \\ . & . & . & . \\ . & . & . & . \\ a_{k_{1},α_{p}} & . & . & a_{k_{1},α_{p}+k_{1}−1} \end{matrix} )

is invertible. So, we obtain by induction a strictly increasing sequence of integers k_{1} < k_{2} < ⋯ < k_{n} < ⋯, and a sequence of corresponding matrices. Using a reasoning analogous to the preceding one, we can construct a solution satisfying Theorem 1.

A Pólya’s system admits, indeed, infinitely many solutions, see [1]. In this work we introduce spaces in which the system contains a unique solution, and we impose other hypotheses on A. Let us recall [4], [7] the spaces that we shall use.

2.2. The spaces S_{c} and s_{c}

For a sequence c = (c_{n}), with c_{n} > 0 for every n, we define the Banach algebra S_{c} by

S_{c} = \{ A = (a_{nm}) | sup_{n}(∑_{m} |a_{nm}| \frac{c_{m}}{c_{n}} ) < ∞ \},  (3)

normed by ∥A∥_{S_{c}} = sup_{n}(∑_{m} |a_{nm}| \frac{c_{m}}{c_{n}} ). We also define the Banach space s_{c} of one-row matrices, by

s_{c} = \{ (x_{n}) | sup_{n}(\frac{|x_{n}|}{c_{n}}) < ∞ \},  (4)

normed by

∥X∥_{s_{c}} = sup_{n} (\frac{|x_{n}|}{c_{n}}) .  (5)

If c = (c_{n}), and c′ = (c′_{n}) are two sequences, such that 0 < c_{n} < c′_{n} ∀n, then:

s_{c} ⊂ s_{c′}.

A special, very useful case is the one where c_{n} = r^{n}, r > 0. Then we denote by S_{r} and s_{r}, the spaces S_{c}, and s_{c}. When r = 1, we obtain the space of the bounded sequences l^{∞} = s_{1}. Finally, φ is the set of all sequences that have only a finite number of nonvanishing terms.

If ∥I − A∥_{S_{c}} < 1, we shall say that A satisfies the condition Γ_{c}. If c = (r^{n}), Γ_{c} is replaced by Γ_{r}.

S_{c} being a unit algebra, we have the useful result: if A satisfies the condition Γ_{c}, A is invertible in the space S_{c}, and for every B ∈ s_{c}, the equation AX = B admits one and only one solution in s_{c}, given by

X = ∑_{n=0}^{∞}(I − A)^{n}B.  (6)

We have seen [4] that a matrix A, which verifies the condition Γ_{c}, for a given sequence c = (c_{n}) is not necessarily of Pòlya type. In fact, the matrix A = (ζ^{|m−n|}), with 0 < ζ < 1/3 satisfies the condition Γ_{1}; but: ∑_{k=1}^{n−1} \frac{|a_{km}|}{|a_{nm}|} ≥ ζ > 0, which shows that this matrix is not of Pòlya type.

2.3. Construction of a sequence converging to a solution of an infinite linear system In the following, A is a matrix in S_{c} with a_{nn} ≠ 0 for all n. For any positive

integer q let A′_{q} denote the matrix with entries a′_{nm} = a_{nm} for 1 ≤ n, m ≤ q and a′_{nm} = 0 otherwise. We denote by B_{q} the matrix deduced from B by the same way. As we have seen in 2-1 we associate to the matrix A′_{q}, the finite matrix [A]_{q} of order q, whose entries in the first q rows and columns are equal to those in A. In the same way we define [B]_{q} from B. When [A]_{q} is invertible, we denote by Â′_{q} the matrix

( \begin{matrix}[A] & {−1 \atop q} \\ & O \end{matrix} ) .

When A = I, A′_{q} is denoted by I′_{q}. Then we have

A′_{q}Â′_{q} = Â′_{q}A′_{q} = I′_{q}.

Proposition 2. There exists a non-invertible infinite matrix A such that all the finite matrices [A]_{q} , q ≥ 1, are invertible.

Proof. Consider, indeed, the matrix A defined for α > 1 by a_{11} = 1, a_{1m} = 0 if m ≥ 2; a_{2m} = 1 for all values of m; a_{n,n−2} = −1 if n ≥ 3, a_{nn} = \frac{1}{α^{2}} if n ≥ 3; the other elements being equal to 0. That is,

A = ( \begin{matrix}1 & 0 & 0 & . & . & . \\ 1 & 1 & 1 & . & . & . \\ −1 & 0 & \frac{1}{α^{2}} & 0 & O \\ 0 & −1 & 0 & \frac{1}{α^{2}} & . \\ & O & . & . & . & . \end{matrix} ) .  (7)

Then A is an element of S_{r}, with 0 < r < 1. The matrices [A]_{1}, [A]_{2} and [A]_{3} are obviously invertible; and for q ≥ 4, the determinant of [A]_{q} is equal to

\frac{1}{(α^{2})^{q−3}} (1 + \frac{1}{α^{2}} ),

and is different from 0. If A were invertible, the equation AX = B, where B ∈ φ, would have a solution. Denote then by C the matrix obtained from A, by deleting the two first rows and by D the matrix obtained from B in the same way. Let us describe the set of the solutions of CX = D. First we remark that C is invertible, being upper triangular. Considering the matrix obtained from C by adding of the rows e′_{1} = (1, 0,…) and e′_{2} = (0, 1, 0,…), we see that it is lower triangular with non-zero elements on the main diagonal, and then invertible. We conclude [4], [7], that the solutions of CX = D form a linear space of dimension 2. The function −1 + \frac{1}{α^{2}} z^{2} admitting z = ±α as roots, the solutions of CX = 0 are given by X = (λα^{n} + μ(−α)^{n})_{n≥1}. Then the solutions of CX = D are

X = (λα^{n} + μ(−α)^{n})_{n≥1} + C^{−1}D,

where λ and μ are arbitrary scalars. Finally the evaluation of the sum of the series ∑_{m} a_{2m}x_{m} shows then, that the product AX cannot converge, which is contradictory.

We see that we must give additional conditions for A, so that the sequence defined by Â′_{q}B_{q} converges to a limit, as q tends to infinity, in a given space. So, we impose the following hypotheses on A.

Definition 3. Let c = (c_{n}) be a decreasing sequence, such that for all n: 0 < c_{n} ≤ 1; A ∈ S_{c} (3), is called c-invertible, if the following conditions are satisfied:

1. For any q, [A]_{q} is invertible, (we shall denote the elements of this inverse by a′_{nm}(q)).

2. Letting, for all q ≥ 2:

k_{q} = sup_{n≥q}( ∑_{m=1}^{q−1} \frac{|a_{nm}|}{c_{n}}),

the series ∑_{q} k_{q} is convergent.

3. The sequence of general term τ_{q} = sup_{n,m≤q}{|a′_{nm}(q)|} is bounded.

When c_{n} = r^{n}, with r ∈ ]0, 1] we shall say that A is r-invertible. Denote, now, for every q, by X_{q} = (x_{n}(q))_{n≥1} the vector Â′_{q}B_{q}, we have the following result.

Theorem 4. When A is c-invertible, for every B ∈ φ the sequence (X_{q}) converges in s_{c} to a limit Z, a solution of AX = B.

Proof. Let B = (b_{n}) be an element of φ, there exists an integer N such that b_{n} = 0, for n > N. First let us show that (X_{q}) is a Cauchy sequence in s_{c}. For q > N, we have

X_{q} − X_{q−1} = Â′_{q}[A′_{q−1} − A′_{q}]X_{q−1}.

In fact, this last expression is equal to

Â′_{q}[I′_{q−1}B_{q} − A′_{q}Â′_{q−1}B_{q}],

where I′_{q−1}B_{q} = B_{q} and Â′_{q}A′_{q} = I′_{q}. We know that for n ≥ q + 1, the terms δ_{n}, of X_{q} − X_{q−1}, are zero; and if n ≤ q:

δ_{n} = −a′_{nq}(q) ∑_{m=1}^{q−1} a_{qm}x_{m}(q − 1).

Elsewhere x_{n}(q − 1) = ∑_{m=1}^{N} a′_{nm}(q − 1)b_{m} for n ≤ q. Then we have

\frac{|δ_{n}|}{c_{n}} ≤ \frac{|a′_{nq}(q)|}{c_{n}} ∑_{m=1}^{q−1} ∑_{k=1}^{N} |a_{qm}| |a′_{mk}(q − 1)| |b_{k}| .

If |τ_{q}| ≤ M for every q, then

\frac{|δ_{n}|}{c_{n}} ≤ \frac{M^{2}}{c_{n}} c_{q} ∑_{m=1}^{q−1} \frac{|a_{qm}|}{c_{q}} ∑_{1}^{N} |b_{k}|,

(c_{n}) being decreasing, there exists a constant C > 0, such that \frac{|δ_{n}|}{c_{n}} ≤ CM^{2}k_{q}, then

∥X_{q} − X_{q−1}∥_{s_{c}} ≤ CM^{2}k_{q}.

We conclude easily that (X_{q}) is a Cauchy sequence in s_{c}. In fact ∀k, l, with k ≥ N:

∥X_{k+l} − X_{k}∥_{s_{c}} ≤ ∑_{q=k+1}^{k+l} ∥X_{q} − X_{q−1}∥_{s_{c}} ≤ CM^{2} ∑_{q=k+1}^{k+l} k_{q}.

Since s_{c} is a Banach space, the Cauchy sequence (X_{q}) has a limit Z in s_{c}. Now we have to verify that AZ = B. For this, let Δ_{q} = (A − A′_{q})Z_{q}. The coordinates of Δ_{q}, whose indices n are less than q, are equal to 0, and for n ≥ q + 1 they are equal to: ∑_{m=1}^{q} a_{nm}(∑_{l=1}^{N} a′_{ml}b_{l}). We have

∥(A − A′_{q})Z_{q}∥_{s_{c}} ≤ M[ sup_{n≥q+1}( ∑_{m=1}^{q} |a_{nm}| \frac{1}{c_{n}})](∑_{l=1}^{N} |b_{l}|).

Hence ∥(A − A′_{q})Z_{q}∥_{s_{c}} ≤ KMk_{q+1}, with K = ∑_{l=1}^{N} |b_{l}|; then Δ_{q} converges to 0, in s_{c}. Furthermore, since the map X ↦ AX from s_{c} into itself is continuous, we conclude that the sequence (AZ_{q}) converges to AZ, as q → ∞, which concludes the proof.

This theorem shows not only the existence of a solution for the system, but mainly gives a natural method to approximate it. Here, we see, once more, that a Pòlya matrix does not necessarily satisfy the preceding hypotheses; consider the lower triangular matrix A, defined by a_{nm} = 0 if m ≠ n, n − 1, a_{nn} = 1 and a_{nn−1} = ζ^{n−1}, where 0 < ζ < 1. We have k_{q} = ζ^{q−1}, then the series ∑_{q} k_{q} is convergent; A verifies the condition Γ_{1}, which proves that the sequence (τ_{q}) is convergent, and we see that all the matrices [A]_{q} are invertible. Hence A is 1-invertible, but is not of Pòlya type, since it is lower triangular. When we used the condition Γ_{r}, (see 2-2), we also had the existence (and uniqueness) of the solution, but this one was written in a more complicated form of a sum of a series of infinite matrices (6). When A is r-invertible, and satisfies the condition Γ_{r}, which is possible, as we shall see, the unique solution in s_{r} can be written as

Z = lim_{q→∞} Â′_{q}B_{q} = ∑_{n=0}^{∞}(I − A)^{n}B.

Denote by E_{r} the set of r-invertible matrices, and by K_{u}, u > 0, the set of the matrices, whose entries on the main diagonal are equal to 1, (case which can be referred considering the product DA, where D = (δ_{nm}/a_{nm})), and which satisfy the condition Γ_{u}. For every pair (r, u) ∈]0, 1] × R^{+∗}, the set E_{r} ∩ K_{u} is not empty, and contains the unit matrix I. If T denotes the set of all upper triangular infinite matrices, whose diagonal elements are all equal to 1, then we have the following result

Proposition 5. For every pair (r, u) of reals, with 0 < r ≤ 1 and u ≥ 1, one has K_{u} ∩ T ⊂ E_{r}.

Proof. Let A be a matrix of K_{u} ∩ T, we have here

sup_{n}( ∑_{m=n+1}^{∞} |a_{nm}| u^{m−n}) < 1.

Since r ≤ u, A belongs to S_{r}. The matrix A′_{q} being upper triangular with non zero entries on the main diagonal, is invertible for all values of q. Furthermore, k_{q} = 0 for all q; we have now to verify that (τ_{q}) is bounded. For this, let us consider the matrix A_{q}^{∗}, whose entries are those of A′_{q}, except those of the main diagonal, whose indices are larger than q which are equal to 1, that is

A_{q}^{∗} = ( \begin{matrix}[A] & q \\ & 1 & & O \\ O & & 1 \\ & & & . \end{matrix} ) .  (8)

Then A_{q}^{∗} is invertible, and (A_{q}^{∗})^{−1} can be written under the form

( \begin{matrix}[A] & {−1 \atop q} \\ & 1 & & O \\ O & & 1 \\ & & & . \end{matrix} ) .  (9)

The sequence of general term (A_{q}^{∗})^{−1} is then bounded in S_{1}. Indeed, for n less than q − 1

∑_{m=n+1}^{q} |a_{nm}| ≤ ∥I − A∥_{S_{u}} ,

since u ≥ 1. Hence ∥I − A_{q}^{∗}∥_{S_{1}} ≤ ∥I − A∥_{S_{u}} < 1 and

∥(A_{q}^{∗})^{−1}∥_{S_{1}} ≤ \frac{1}{1 − ∥I − A∥_{S_{u}}} .

Finally, ∥Â′_{q}∥_{S_{1}} being less than ∥(A_{q}^{∗})^{−1}∥_{S_{1}}, we conclude that for every n, m, q,

|a′_{nm}(q)| ≤ sup_{i}( ∑_{j=1}^{q} |a′_{ij}(q)|) ≤ \frac{1}{1 − ∥I − A∥_{S_{u}}} .

Then the sequence (τ_{q}) is bounded; this completes the proof.

In the same way, considering the case where c_{n} = 1, for every n, k_{q} is, then defined by sup_{n≥q}(∑_{m=1}^{q−1} |a_{nm}|), we have the following result.

Corollary 6. If A is a matrix belonging to K_{1}, such that the series ∑_{q} k_{q} is convergent, then A belongs to E_{1}.

Proof. In fact [A]_{q} is invertible for every q, and |τ_{q}| ≤ \frac{1}{1−∥I−A∥_{Su}}, for any u strictly larger than 1.

Example 7. If A = (σ^{|m−n|m}), with σ ∈ ]0, 1[ and ∑_{m≥1,m≠n} σ^{|m−n|m} < 1, then we see that A satisfies the condition Γ_{1} and

k_{q} ≤ σ^{q−1} + σ^{2(q−2)} + ⋯ + σ^{q−1} ≤ (q − 1)σ^{q−1}.

The second condition of Definition 3 is then satisfied, and A is 1-invertible. We see that this matrix is of Pòlya, which proves that this system has infinitely many solutions; and here we have determined a space in which we have one and only one solution approximated by the sequence X_{q}.

Let us come back to the matrix A, defined by (7). Consider a real 0 < θ < 1 and the matrix M_{θ} = AP′_{θ}, with P′_{θ} = (θ^{n−1}δ_{nm}). Denote, as in Definition 3, [M_{θ}]_{q}^{−1} = (α_{nm}(q))_{n,m≤q}, we have

Corollary 8. ∀r ≤ 1 M_{θ} ∉ E_{r}, and the sequence defined by:

τ′_{q} = sup_{n,m≤q}{|α_{nm}(q)|}

is not bounded.

Proof. First we show that there exists no real r ≤ 1, such that M_{θ} is rinvertible. Let B be any matrix of φ; if Y were an element of s_{r}, such that M_{θ}Y = B, we would conclude that B = A(P′_{θ}Y), where P′_{θ}Y ∈ s_{r} (since θ ≤ 1). A would be surjective from s_{r} into φ, which is contradictory, as we have seen in Proposition 2. By application of Theorem 4, we see that, at least one condition of Definition 3 is not verified. It is easy to prove that it is the third, where τ_{q} is replaced by τ′_{q} (notice that this property does not depend on r). In fact, the first hypothesis of Definition 3 is obviously satisfied, and for the second hypothesis, we have ∀q ≥ 4

k_{q} = (\frac{θ}{r} )^{q} \frac{1}{r^{3}} ;

and if r ∈ ]θ, 1[, the series ∑_{q} k_{q} is convergent.

2.4. A second method of approximation of a solution of an infinite system We suppose that all the diagonal elements are equal to 1, we can give

an analogous result, whose the advantage is to approximate the solution Z = ∑_{n=0}^{∞}(I − A)^{n}B, of the equation AX = B, by the means of the sequence

Y_{q} = (A_{q}^{∗})^{−1}B = ∑_{n=0}^{∞}(I − A_{q}^{∗})^{n}B,

A_{q}^{∗} being defined in (8). When B ∈ φ, Y_{q} is equal to the sequence X_{q} = (x_{n}(q))_{n}, given in Definition 3. Given a real r > 0, let us define from the infinite matrix A two sequences:

γ_{q} = sup_{n≤q}( ∑_{m=q+1}^{∞} |a_{nm}| r^{m−n}), γ′_{q} = sup_{n≥q+1}( ∑_{m=1,m≠n}^{∞} |a_{nm}| r^{m−n}),

so that to assert the following result

Proposition 9. Assume that:

i) The sequences (γ_{q}) and (γ′_{q}) converge to 0, as q tends to infinity;

ii) A satisfies the condition Γ_{r}.

Then for any B ∈ s_{r}, (Y_{q}) converges to Z in s_{r}, and

∥Y_{q} − Z∥_{s_{r}} ≤ sup(γ_{q}, γ′_{q})\frac{∥B∥_{s_{r}}}{(1 − ρ)^{2}} ,

where ρ = ∥I − A∥_{S_{r}}.

Proof. If we put ρ_{q} = ∥A − A_{q}^{∗}∥_{S_{r}}, (then ρ_{q} ≤ ρ, for every q), it can be easily proved that

∥Y_{q} − Z∥_{s_{r}} ≤ ∥A − A_{q}^{∗}∥_{S_{r}} ∑_{n=1}^{∞} ∑_{i=1}^{n} ρ_{q}^{n−i}ρ^{i−1} ∥B∥_{s_{r}} .

We can evaluate the double series in the second member of the inequality, since

∑_{n=1}^{∞} ∑_{i=1}^{n} ρ_{q}^{n−i}ρ^{i−1} = ∑_{n=1}^{∞} \frac{ρ_{q}^{n}}{ρ_{q} − ρ} − ∑_{n=1}^{∞} \frac{ρ^{n}}{ρ_{q} − ρ} = \frac{1}{(1 − ρ_{q})(1 − ρ)},

this last term being less than 1/(1 − ρ)^{2}. Finally ρ_{q} being equal to sup(γ_{q}, γ′_{q}), tends to 0 as q → ∞, this complete the proof.

Example 10. This result can be applied to the matrix A = (σ^{|m−n|m}), with 0 < σ < 1/3. So A satisfies the condition Γ_{1}. Considering the sequence defined by χ_{q} = \frac{σ^{q}}{1−σ^{q}}, we have:

γ_{q} ≤ sup_{n≤q}{ ∑_{k=1}^{∞} σ^{(q+1)(q+k−n)}} ≤ χ_{q+1},
γ′_{q} ≤ sup_{n≥q+1}\{(n − 1)σ^{n−1} + \frac{σ^{n+1}}{1−σ^{n+1}} \} ≤ qσ^{q} + χ_{q+2}.

Since the sequences (χ_{q}) and (qσ^{q}) converge to 0, as q tends to infinity, it is the same for ∥A − A_{q}^{∗}∥_{S_{1}}. Furthermore there exists an integer N, such that

qσ^{q} + χ_{q+2} − χ_{q+1} = qσ^{q} [1 + \frac{σ^{2}}{q (1 − σ^{q+2})} − \frac{σ}{q (1 − σ^{q+1})}] > 0

for all q ≥ N. Hence applying the proposition we get

∥Y_{q} − Z∥_{s_{1}} ≤ (qσ^{q} + χ_{q+2}) \frac{∥B∥_{s_{1}}}{(1 − ρ)^{2}} for all q ≥ N.

Remark 2. We note that a matrix r-invertible does not necessarily satisfy the previous proposition. Indeed, take A ∈ K_{1} ∩ T, defined for a given real ρ, 0 < ρ < 1, by

A = ( \begin{matrix}1 & ρ \\ & 1 & ρ & O \\ O & & . & . \\ & & & . \end{matrix} );

we deduct from the Proposition 5, that it belongs to E_{1}; but γ_{q} = ρ does not converge to 0, as q → ∞. This shows that the first condition of the preceding proposition cannot be satisfied.

3. An application to the continued fractions

Let us consider the system of linear equations

\{ \begin{matrix}(β_{1} + z)x_{1} & − a_{1}x_{2} & & = b_{1}, \\ & − a_{1}x_{1} & + (β_{2} + z)x_{2} − a_{2}x_{3} & = b_{2}, \\ & & − a_{2}x_{2} + (β_{3} + z)x_{3} − a_{3}x_{4} & = b_{3}, \\ … & … & … & … & … & … & … \end{matrix}  (10)

If ^{t}B = e′_{1} = (1, 0, 0,…), it is well known that we may write the linear equations in the form

x_{1} = \frac{1}{β_{1} + z − \frac{a_{1}x_{2}}{x_{1}}} , \frac{a_{1}x_{2}}{x_{1}} = \frac{a_{1}^{2}}{β_{2} + z − \frac{a_{2}x_{3}}{x_{2}}} , \frac{a_{2}x_{3}}{x_{2}} = \frac{a_{2}^{2}}{β_{3} + z − \frac{a_{3}x_{4}}{x_{3}}} , …

If we substitute in succession from each into the preceding, we obtain the formal expansion of x_{1} into a continued fraction, also called the A-fraction, that is

x_{1} = \frac{1}{β_{1} + z − _{\frac{a_{1}^{2}}{β_{2}+z− \frac{a^{2}_{2}}{β_{3}+z−.._{.}}}}} .

The system defined by (10), is equivalent to the matrix equation AX = B. The infinite tridiagonal matrix A admitting infinitely many right inverses it is necessary to recall some results on continued fractions and bounded matrices. l^{2} is the normed space of the sequences X = (x_{n}) such that ∑_{n} |x_{n}|^{2} < ∞, with ∥X∥_{l^{2}} = (∑_{n} |x_{n}|^{2})^{1/2}. A is called bounded, if there exists a constant M > 0, such that for all X, Y ∈ l^{2}, X = (x_{n}), Y = (y_{n})

|^{t}XAY | = | ∑_{n,m} a_{nm}x_{n}y_{m}| ≤ M ∥X∥_{l^{2}} ∥Y ∥_{l^{2}} .

We define by ∥A∥_{2} the smallest number M > 0, such that this inequality is verified; it is easy to see that ∥·∥_{2}is a norm defined in the space of the bounded matrices. It is known that there exists one bounded inverse, such that x_{1} can be written, as above, in a continued fraction. If we denote by a′_{nm} the elements of this bounded inverse, B being equal to ^{t}e′_{1}, we have x_{1} = a′_{11}. This number is called the leading coefficient, see [14], it is said too that “a′_{11} is formally equal to the A-fraction”. We shall write more precisely a′_{11} = a′_{11}(z).

In order to give explicitly this bounded inverse we give the following result [14].

Proposition 11. If we have ∥I − A∥_{2} < 1, A admits a bounded reciprocal, and for every B ∈ l^{2} the equation AX = B admits only one solution in l^{2}.

Writing

A = zI + J_{0} = z (I + \frac{1}{z}J_{0}) ,

we deduce that for |z| > ∥J_{0}∥_{2}, A admits the bounded inverse we have been searching for. We can, now, give an application of Proposition 9 where the sequence Y_{q} = X_{q} = Â_{q}^{′ t}e′_{1}, given in 2-3 is used. Let us recall that x_{1}(q) is the first component of this vector.

Proposition 12. Assume that

i) β_{n} = O(1), as n → ∞;

ii) (a_{n}) is decreasing and converges to 0 as n tends to infinity. Putting K = sup_{n}(a_{n}, |β_{n}|), we deduce that if |z| > 3K, then x_{1}(q) → a′_{11}(z), as q → ∞, and

|x_{1}(q) − a′_{11}(z)| ≤ 2a_{q} \frac{1}{(|z| − 3K)^{2}} .

Proof. In order to reduce to the case where all the elements of the main diagonal are equal to 1, we consider the product DA, where D = (δ_{nm}/β_{n} + z). We have ∥I − DA∥_{S_{1}} = sup(τ_{1}, τ_{2}), with

τ_{1} = \frac{a_{1}}{|β_{1} + z|}, τ_{2} = sup_{n} (\frac{a_{n} + a_{n+1}}{|β_{n} + z|} ) .

Putting ρ′ = ∥I − DA∥_{S_{1}}, we have for |z| > 3K

ρ′ ≤ sup_{n} (\frac{2a_{n}}{|z| − K} ) ≤ \frac{2K}{|z| − K} < 1.

Since D ∈ S_{1}, and A^{−1} = (DA)^{−1} D, we deduce that A is invertible in S_{1}. We have

γ_{q} = \frac{a_{q}}{|β_{q} + z|}, γ′_{q} = sup_{n≥q} (\frac{a_{n} + a_{n+1}}{|β_{n} + z|} ) ,

both less than \frac{2a_{q}}{|z|−K}, as |z| > 3K. Let us show, now, that ∥J_{0}∥_{2} ≤ 3K. We have

tXJ_{0}Y = − ∑_{n=2}^{∞} a_{n−1}x_{n}y_{n−1} + ∑_{n=1}^{∞} β_{n}x_{n}y_{n} − ∑_{n=1}^{∞} a_{n}x_{n}y_{n+1}.

Hence |^{t}XJ_{0}Y | ≤ 3K ∥X∥_{l^{2}} ∥Y ∥_{l^{2}}, and ∥J_{0}∥_{2} ≤ 3K. If we take, |z| > 3K, then A admits a bounded reciprocal, as above, which is equal to the preceding inverse A^{−1}. Denote now, by x′_{1}(q) the first component of Â′_{q}D^{−1 t}e_{1}, and by α′_{nm} the elements of the matrix (DA)^{−1}, then we have

x′_{1}(q) = (β_{1} + z)x_{1}(q), α′_{11} = (β_{1} + z)a′_{11}.

Hence, applying Proposition 9, we obtain

|x_{1}(q) − a′_{11}(z)| = |\frac{x′_{1}(q) − α′_{11}}{β_{1} + z} | ≤ \frac{1}{|β_{1} + z|} \frac{2a_{q}}{|z| − K} \frac{1}{(1 − ρ′)^{2}}

and since \frac{1}{1−ρ′} ≤ \frac{|z|−K}{|z|−3K}, we deduce the result.

We remark that the map A: X → AX defined in the previous proposition is a bijection from s_{1} into itself, and its restriction to l^{2} is, also unrecognized, bijective from this space into l^{2}. A^{−1} is a bounded inverse, and belongs to S_{1}.

Acknowledgements. The author thanks the referee for his aid concerning the presentation of this paper.

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(received 03.02.2000, in revised form 15.05.2002)

Universite du Havre, I.U.T., B.P. 4006, Le Havre, France