����������� ������ 52 (2000), 99–112

UDK 517.521

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BASES IN SEQUENCE SPACES AND EXPANSION OF A FUNCTION IN A SERIES OF POWER SERIES

Bruno de Malafosse

Abstract. In this paper we establish a relation between the existence of a basis and the solution of an infinite linear system. Then we study the expansion of a function in a series of power series.

1. Introduction

Bases in infinite dimensional sequence spaces have been studied by many authors. Concerning Schauder bases, let us cite Wilansky [15] and more recently Malkowsky [7] and [8]. New Schauder bases have been found in the spaces c_{0}(Δ^{μ}) and c(Δ^{μ}) by the last author. We also have many results concerning AK spaces, for example in the Λ-strongly null and Λ-strongly convergent sequence spaces. In this work we establish a relation between the notion of a basis and the solution of an infinite linear system. Let us recall that some results concerning linear infinite systems have been studied in Cooke [1]. We can find results concerning Pólya systems in Petersen and Baker [12] and Petersen [13]. More recently, some results concerning summability have been put together in Maddox [4], we find a study on the Walsh functions in Mursaleen [9]. R. Labas and B. de Malafosse [2, 3] gave an application of the theory of the sum of operators to the theory of infinite matrices. Note that infinite linear systems have been used for the study of the spectrum of the Cesàro operator in certain spaces, see Reade [14], Okutoyi [11] and de Malafosse [6].

The plan of this paper is organized as follows. In section 2, we recall some results concerning AK spaces and Schauder basis [7, 8]. Next, we recall the spaces s_{c} and S_{c} (see [5, 6, 9]) and define bases of α, β type. In section 3, we study how a function can have an expansion in a given series of power series. In the first subsection, we prove that this problem is equivalent to the existence of a solution of an infinite linear system and we give an application where the Cesàro operator is used. In the subsection 2 are given some properties of such expansions. In the third subsection, we study a particular expansion and we study when a function can have a unique expansion, or infinitely many expansions, or no expansion at all.

AMS Subject Classification: 46 A 15

2. Bases in certain sequence spaces

2.1. Some known results

A Banach space E of complex sequences, with the norm ∥ · ∥_{E} is called a BK space if every projection P_{n}: E → C, defined by P_{n}(X) = x_{n} is continuous. Denote by e_{n}, n ⩾ 0, the vector (0,…, 1, 0,…), where 1 is in the (n + 1)-st position and by e the vector (1, 1,…). A BK space E is said to have AK (see [15]), if for every B = (b_{n})_{n⩾0} ∈ E, B = ∑_{m=0}^{∞} b_{m}e_{m}, i.e.

∥B − ∑_{m=0}^{n} b_{m}e_{m}∥_{E} → 0 (n → ∞).

It is well-known [15], that the space l^{p}, p > 0, of the sequences X = (x_{n})_{n⩾0} such that the series ∑_{n} |x_{n}|^{p} are convergent have AK with respect to the norm ∥X∥_{l^{p}} = (∑_{n} |x_{n}|^{p})^{1/p}. It is the same for the space c of all convergent sequences, with the norm ∥X∥_{l^{∞}} = sup_{n⩾0} |x_{n}|. In this last case, if B = (b_{n})_{n⩾0} ∈ c, and (b_{n})_{n⩾0} → b, we have a unique representation B = be + ∑_{m=0}^{∞}(b_{m} − b)e_{m}, i.e.

∥B − (be + ∑_{m=0}^{N}(b_{m} − b)e_{m})∥_{l^{∞}} = sup_{m⩾N+1} |b_{m} − b| → 0, (N → ∞).

Elsewhere, (â_{m})_{m⩾0} is a Schauder basis of the normed vector space E, if for all B ∈ E there is a unique sequence X = (x_{n})_{n⩾0}, such that B = ∑_{m=0}^{∞} x_{m}â_{m}, in the space E. That is

lim_{N→∞}∥B − ∑_{m=0}^{N} x_{m}â_{m}∥_{E} = 0.

Note that l^{∞} has no Schauder basis.

Here, we shall consider sequences as column vectors and give other bases in new spaces. Then we must recall definitions of useful spaces permitting to do calculations on infinite linear systems.

2.2. Spaces S_{α} and s_{α}

In the following, we shall use infinite linear systems defined by

∑_{m=0}^{∞} a_{nm}x_{m} = b_{n}, n = 0, 1, 2,…

Such a system can be written as a matrix equation

AX = B,

where A = (a_{nm})_{n,m⩾0} and X, B are the one column matrices defined respectively by (x_{n})_{n⩾0} and (b_{n})_{n⩾0}. The following spaces have been defined, for instance, in [2] and [3]. For a sequence α = (α_{n})_{n⩾0}, where α_{n} > 0, for every n ⩾ 0, we consider the Banach algebra

S_{α} = \{ A = (a_{nm})_{n,m⩾0} | sup_{n⩾0}( ∑_{m=0}^{∞} |a_{nm}|\frac{α_{m}}{α_{n}} ) < ∞ \},  (1)

normed by

∥A∥_{S_{α}} = sup_{n⩾0}( ∑_{m=0}^{∞} |a_{nm}|\frac{α_{m}}{α_{n}} ).  (2)

We also define the Banach space s_{α} of one-column matrices by

s_{α} = \{ X = (x_{n}) | sup_{n⩾0}(\frac{|x_{n}|}{α_{n}}) < ∞ \},  (3)

normed by

∥X∥_{s_{α}} = sup_{n⩾0}(\frac{|x_{n}|}{α_{n}}).  (4)

If α = (α_{n})_{n⩾0} and β = (β_{n})_{n⩾0} are two sequences such that 0 < α_{n} ⩽ β_{n} for each n, then s_{α} ⊂ s_{β}.

A very useful special case is the one where α_{n} = r^{n}, r > 0. We denote, then, by S_{r} and s_{r}, the spaces S_{α} and s_{α}. When r = 1, we obtain the space of bounded sequences l^{∞} = s_{1}. If

∥I − A∥_{S_{α}} < 1,  (5)

then A is invertible in the space S_{α}.

S_{α} being a unit algebra, we have the useful result: if we suppose that (5) holds, then for every B ∈ s_{α}, the equation AX = B admits one and only one solution in s_{α}, given by

X = ∑_{k=0}^{∞}(I − A)^{k}B.  (6)

We have seen [3] that a matrix A, which verifies (5), for a given sequence α = (α_{n})_{n⩾0} is not necessarily of Pòlya type (see [1, 12, 13]). Recall that a matrix A satisfies the Pòlya condition if

lim inf_{m} C(n, m) = 0, n = 1, 2,…

where

C(n, m) = \frac{|a_{0m}| + |a_{1m}| + ⋯ + |a_{n−1,m}|}{|a_{nm}|}.

We see that a matrix A = (a^{|m−n|})_{n,m⩾0}, where 0 < a < 1/3, verifies (5), when we replace S_{α} by S_{1}, but, for all n ⩾ 1, ∑_{k=0}^{n−1} |\frac{a_{km}}{a_{nm}} | ⩾ a > 0, which shows that it is not of Pòlya type.

2.3. Bases of α, β type

In the following α = (α_{n})_{n⩾0} and β = (β_{n})_{n⩾0} are two given sequences that satisfy 0 < α_{n} ⩽ β_{n} for all n.

Definition 1. We shall say that (â_{m})_{m⩾0} is a basis of the α, β type if â_{m} ∈ s_{α} for all m ⩾ 0 and for every B ∈ s_{α} there is a unique sequence X = (x_{n})_{n⩾0}, such that

B = ∑_{m=0}^{∞} x_{m}â_{m}

in the space s_{β}.

If there are two reals r_{1}, r_{2} with 0 < r_{1} ⩽ r_{2} such that α = (r_{1}^{n})_{n⩾0} and β = (r_{2}^{n})_{n⩾0}, we shall say that (â_{m})_{m⩾0} is a basis of r_{1}, r_{2} type.

In order to establish a relation between the solution of an infinite linear system and the notion of basis, we need a lemma.

Lemma 2. Let â_{m} = ^{t}(a_{0m},…, a_{nm},…) ∈ s_{α} for all m ⩾ 0 and assume that the series ∑_{m=0}^{∞} x_{m}â_{m} is convergent in the space s_{β}. The following assertions are equivalent:

i) (â_{m})_{m⩾0} is a basis of α, β type;

ii) equation AX = B admits only one solution, for all B ∈ s_{α}.

Proof. Suppose that i) is satisfied. For all B ∈ s_{α} there exists X = (x_{n})_{n⩾0} such that ∥B − ∑_{m=0}^{N} x_{m}â_{m}∥_{s_{β}} tends to 0 as N tends to infinity. We have

sup_{n⩾0} (\frac{|b_{n} − ∑_{m=0}^{N} a_{nm}x_{m}|}{β_{n}}) = o(1) as N → ∞.

Then for every value of n, |b_{n} − ∑_{m=0}^{N} a_{nm}x_{m}| = β_{n}o(1) as N → ∞ and b_{n} = ∑_{m=0}^{∞} a_{nm}x_{m}, which proves that B = AX. Hence we deduce ii).

Conversely, denote by X = (x_{n})_{n⩾0} the solution of the equation AX = B, B ∈ s_{α}. We see that B = AX = (∑_{m=0}^{∞} a_{nm}x_{n})_{n⩾0}, and B ∈ s_{β} since s_{α} ⊂ s_{β}. Then for all N

∥B − ∑_{m=0}^{N} x_{m}â_{m}∥_{s_{β}} = ∥( ∑_{m=N+1}^{∞} a_{nm}x_{m})_{n⩾0}∥_{s_{β}}.

The last term of the identity is equal to ∥ ∑_{m=N+1}^{∞} x_{m}â_{m}∥_{s_{β}}, which converges to 0, as N tends to infinity, since the series ∑_{m=0}^{∞} x_{m}â_{m} is convergent in the space s_{β}. This proves that (â_{m})_{m⩾0} is a basis of α, β type.

Remark 1. Note first that we can have ∑_{m=0}^{∞} x_{m}â_{m} = 0 in a space s_{r_{2}}, for a sequence X = (x_{n})_{n⩾0} ≠ 0, although the vectors â_{0}, â_{1}, … are linearly independent (that is, if for all k ∈ N, ∑_{m=0}^{k} x_{m}â_{m} = 0, then x_{m} = 0, for all m = 0, 1,…, k. In fact, consider the family (â_{m})_{m⩾0} defined by ^{t}â_{0} = e_{0} and tâ_{n} = e_{n−1} + e_{n} for all n ⩾ 1. It is easy to see that these vectors are linearly independent. Moreover, for all m ⩾ 0, â_{m} ∈ s_{r_{2}}, for any r_{2} > 1 and we have for every integer N: ∥ ∑_{m=0}^{N}(−1)^{m}â_{m}∥_{s_{r_{2}}} = 1/r_{2}^{N+1}, which tends to 0 as N tends to infinity. Then we have ∑_{m=0}^{∞}(−1)^{m}â_{m} = 0 in the space s_{r_{2}}.

In the following we shall use the increasing sequence Λ = (λ_{n+1})_{n⩾0} of strictly positive reals tending to infinity. Denote by C = (a_{nm})_{n,m⩾0} the well-known infinite triangle matrix defined from Λ by a_{nm} = 1/λ_{n+1} for 0 ⩽ m ⩽ n, the other coefficients being equal to 0 (see [7, 8, 15]). Recall that if λ_{n} = n for all n ⩾ 1, C is called the Cesàro operator (see [6, 11, 14]). We shall assert the following reslusts in which

ĉ_{m} = ^{t}(0,…, 1/λ_{m+1}, 1/λ_{m+2},…), m = 0, 1,…

the first non-zero coefficient being in the (m + 1)-st position.

Proposition 3. i) Let (â_{m})_{m⩾0} be the sequence defined by

â_{m} = ^{t}(0,…, a_{m,m}, a_{m+1,m}, a_{m+2,m},…),

m = 0, 1, 2,…, a_{m,m} ≠ 0 being in the (m + 1)-st position. Assume that there is M > 0 such that for all n, m ⩾ 0:

|a_{nm}| ⩽ M and |\frac{a_{nm}}{a_{mm}} | ⩽ M.  (7)

For real r_{1} > M + 1 and r_{2} > r_{1}, (â_{m})_{m⩾0} is a basis of r_{1}, r_{2} type.

ii) For 2 < r_{1} < r_{2}, (ĉ_{m})_{m⩾0} is a basis of r_{1}, r_{2} type.

iii) Assume that α = (α_{n})_{n⩾0} and β = (β_{n})_{n⩾0} are two sequences for which (α_{n})_{n⩾0} is increasing and

\overline{lim}_{n→∞}(\frac{α_{n}}{β_{n}} ) = 0.  (8)

Then (ĉ_{m})_{m⩾0} is a basis of α, β type.

Proof. Assertion i). Consider for all m the sequence defined by â′_{m} = a_{mm}^{−1}â_{m} = t(0,…, 1, \frac{a_{m+1,m}}{a_{mm}} ,…). As r_{1} > 1, we deduce from (7) that

sup_{n⩾m+1}(|a_{nm}|/|a_{mm}r^{n}|) < ∞.

Define the infinite matrix A′, whose (m + 1)-st column is â′_{m}. We see that A′ is a lower triangle, whose diagonal entries are equal to 1 and the coefficient in the (n + 1)-st row and of the (m + 1)-st column, with 0 ⩽ m ⩽ n − 1 for n ⩾ 1, is equal to a_{nm}/a_{nn}. We have

∥I − A′∥_{S_{r_{1}}} = sup_{n⩾2} [\frac{1}{|a_{nn}|} (\frac{|a_{n0}|}{r_{1}^{n}} + \frac{|a_{n1}|}{r_{1}^{n−1}} + ⋯ + \frac{|a_{n,n−1}|}{r_{1}})] ,

that is, using (7),

∥I − A′∥_{S_{r_{1}}} ⩽ \frac{M}{r_{1} − 1} sup_{n⩾2} (1 − \frac{1}{r_{1}^{n}} ) .

Hence if r_{1} > M + 1 then ∥I − A′∥_{S_{r_{1}}} < 1. We deduce that for every B ∈ s_{r_{1}}, the equation A′X = B admits only one solution X = (A′)^{−1}B ∈ s_{r_{1}}. Furthermore, there exists a constant K > 0 such that

|x_{m}|∥â′_{m}∥_{s_{r_{2}}} ⩽ Kr_{1}^{m} sup_{n⩾m+1} (\frac{|a_{nm}|}{|a_{mm}r_{2}^{n}|}) ⩽ \frac{KM}{r_{2}} (\frac{r_{1}}{r_{2}} )^{m} .

Then the series with general term |x_{m}|∥â′_{m}∥_{s_{r_{2}}} is convergent, since r_{2} > r_{1}. s_{r_{2}} being a Banach space, the series ∑_{m=0}^{∞} x_{m}â′_{m} is convergent in s_{r_{2}}. Using Lemma 2 we deduce that for all B ∈ s_{r_{1}} there exists a unique sequence X = (x_{n})_{n⩾0} such that B = ∑_{m=0}^{∞} x_{m}â′_{m} = ∑_{m=0}^{∞} x_{m}a_{mm}^{−1}â_{m}. And since

sup_{n⩾m} (\frac{|a_{nm}|}{r_{1}^{n}}) ⩽ \frac{M}{r_{1}^{m}} ⩽ M,

we see that â_{m} ∈ s_{r_{1}} for all m ⩾ 0. This completes the proof of i).

Assertion ii). We can apply i) to the sequence (ĉ_{m})_{m⩾0}. Indeed, Λ being increasing, we have here a_{nm} = 1/λ_{n+1} ⩽ 1/λ_{1} and \frac{|a_{nm}|}{|a_{mm}|} = \frac{λ_{m+1}}{λ_{n+1}} ⩽ 1 for all n ⩾ 0 with m ⩽ n. It is enough to take M = sup(1, 1/λ_{1}).

Assertion iii). First we see that â_{m} ∈ s_{α} for all m ⩾ 0. In fact, lim_{n→∞} λ_{n} = ∞ and α_{n} ⩾ α_{0} imply that sup_{n⩾m}(\frac{1}{λ_{n+1}α_{n}}) < ∞ for every m. Further, the matrix whose columns are successively ĉ_{0}, ĉ_{1}, … is equal to C. The equation CX = B, where B ∈ s_{α}, admits only one solution. To calculate this solution we need the infinite lower triangle matrix Δ = (η_{nm})_{n,m⩾0} defined by η_{n,n−1} = −1 for n ⩾ 1, η_{nn} = 1 for n ⩾ 0, the other entries being equal to 0. We get then CΔD = I, where D = (λ_{n+1}δ_{nm})_{n,m⩾0}, with δ_{nn} = 1 and δ_{nm} = 0 for m ≠ n (see [6]). Hence C^{−1} = ΔD. If we denote C^{−1} = (c′_{nm})_{n,m⩾0}, then c′_{nn} = λ_{n+1} for all n ⩾ 0 and c′_{n,n−1} = −λ_{n} for n ⩾ 1. Calculating the product C^{−1}B we deduce that:

\{ \begin{matrix}x_{0} = b_{0}, \\ x_{n} = −λ_{n}b_{n−1} + λ_{n+1}b_{n}, & if n ⩾ 1. \end{matrix}  (9)

The series R_{n} = ∑_{m=N+1}^{∞} x_{m}ĉ_{m}, for a given integer N, is equal to the product

R_{N} = ( \begin{matrix}0 & & & 0 & & & 0 & & … & & & 0 \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ \frac{1}{λ_{N+2}} & & & 0 & & & 0 & & … & & & 0 \\ \frac{1}{λ_{N+3}} & & & \frac{1}{λ_{N+3}} & & & 0 & & … & & & 0 \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \end{matrix} ) ( \begin{matrix}x_{N+1} \\ x_{N+2} \\ … \\ … \\ … \end{matrix} ) .

If we set R_{N} = (y_{n})_{n⩾0}, then y_{n} = 0 for all n = 0, 1,…, N, y_{N+1} = x_{N+1}/λ_{N+2} and y_{n} = \frac{1}{λ_{n+1}} (∑_{m=N+1}^{n} x_{m}) for n ⩾ N + 2. We have

∥ ∑_{m=N+1}^{∞} x_{m}ĉ_{m}∥_{s_{β}} = sup_{n⩾N+1} (\frac{|y_{n}|}{β_{n}}) .

For n = N + 1, there is a constant K_{1} > 0 such that

\frac{|y_{N+1}|}{β_{N+1}} ⩽ \frac{K}{λ_{N+2}} \frac{α_{N+1}}{β_{N+1}} .  (10)

Using (9), we obtain for n ⩾ N + 2

\frac{|y_{n}|}{β_{n}} = \frac{|λ_{n+1}b_{n} − λ_{N+1}b_{N}|}{λ_{n+1}β_{n}}.

There is a constant K_{2} > 0 such that

\frac{|y_{n}|}{β_{n}} ⩽ K_{2} [\frac{α_{n}}{β_{n}} + \frac{λ_{N+1}}{λ_{n+1}} \frac{α_{N}}{β_{n}} ] ⩽ 2K_{2} \frac{α_{n}}{β_{n}} ,  (11)

since the sequences Λ = (λ_{n+1})_{n⩾0} and (α_{n})_{n⩾0} are increasing. Now put τ_{1}(N) = \frac{K_{1}}{λ_{N+2}} \frac{α_{N+1}}{β_{N+1}} and τ_{2}(N) = 2K_{2} sup_{n⩾N+2}(\frac{α_{n}}{β_{n}}); (10) and (11) imply

sup_{n⩾N+1} (\frac{|y_{n}|}{β_{n}}) ⩽ sup(τ_{1}(N), τ_{2}(N)).

τ_{1}(N) → 0 as N → ∞, since Λ tends to infinity and 0 < α_{n} ⩽ β_{n} for all n. Using (8), we see that τ_{2}(N) → 0 as N → ∞ and the conclusion follows from Lemma 2.

Remark 2. For any X = (x_{n})_{n⩾0} let us denote |X| = (|x_{n}|)_{n⩾0}. Consider the space of Λ-strongly null sequences w(Λ). Recall that B = (b_{n})_{n⩾0} ∈ w(Λ) if and only if C(|B|) ∈ c_{0}, that is lim_{n→∞}(\frac{1}{λ_{n}} ∑_{k=0}^{n} |b_{k}|) = 0. It is well-known (see [7]) that w(Λ) has AK with respect to the norm

∥B∥ = sup_{n⩾0}(\frac{1}{λ_{n}} ∑_{k=0}^{n} |b_{k}|),

i.e. ∥B − ∑_{m=0}^{N} b_{m}e_{m}∥ → 0 as N → ∞.

We shall consider now a new kind of basis using power series.

3. Expansion of a function in a series of power series

Until here we have studied the representation of a vector B = (b_{n})_{n⩾0} ∈ s_{α} as B = ∑_{m=0}^{∞} x_{m}(∑_{n=0}^{∞} a_{nm}e_{n}) in a space s_{β} for a given sequence (a_{nm})_{n,m⩾0}. Analogously, we are going to see how a function f can be written as f(z) = ∑_{m=0}^{∞} x_{m}(∑_{n=0}^{∞} a_{nm}z^{n}) in a given disk \overline{D}(0, R), the set of complex numbers z that satisfy |z| ⩽ R.

3.1. Definition and first properties

Let A = (a_{nm})_{n,m⩾0} be an infinite matrix and suppose that there are r, R > 0 such that

∑_{m=0}^{∞} ∑_{n=0}^{∞} |a_{nm}|R^{n}r^{m} < ∞.  (12)

Define by Â = (Â_{m})_{m⩾0} the sequence whose general term is

Â_{m}(z) = ∑_{n=0}^{∞} a_{nm}z^{n}, m = 0, 1,…

defined for |z| ⩽ R. We shall say that the infinite matrix A is associated to the sequence Â. We deduct from (12) that for every m ⩾ 0 the series ∑_{n=0}^{∞} |a_{nm}|R^{n} is convergent. Hence, if r_{m} denotes the radius of convergence of Â_{m}(z), m ⩾ 0, we must have r_{m} > 0 for all m, and we conclude that inf_{m⩾0}(r_{m}) ⩾ R.

Denote by E_{r,R}(Â) the set of functions f, defined by the power series f(z) = ∑_{n=0}^{∞} b_{n}z^{n}, such that there exists a sequence X = (x_{n})_{n⩾0} ∈ s_{r} for which

f(z) = ∑_{m=0}^{∞} x_{m}Â_{m}(z),  (13)

for all z ∈ \overline{D}(0, R).

Notice that if ρ is the radius of convergence of f, we have R < ρ, since (13) is satisfied for all z ∈ \overline{D}(0, R).

We shall give now a condition for which the vectors of the seqence Â = (Â)_{m⩾0} are linearly independent, that is: if for any integer k and z ∈ \overline{D}(0, R) the sum ∑_{m=0}^{k} x_{m}Â_{m}(z) = 0, then x_{m} = 0 for all m = 0, 1,…, k.

Proposition 4. For every integer k let (n_{i}(k))_{0⩽i⩽k} be a sequence of integers with n_{0}(k) < n_{1}(k) < ⋯ < n_{k}(k). Assume that for every k the determinant of the matrix (a_{n_{i}(k),m})_{0⩽i,m<k} is different from 0. Then the vectors Â_{m}, m = 0, 1,… are linearly independent.

Proof. From the identities ∑_{m=0}^{k} x_{m}Â_{m}(z) = ∑_{m=0}^{k} x_{m}(∑_{n=0}^{∞} a_{nm}z^{n}) = 0, where |z| ⩽ R, we obtain that ∑_{n=0}^{∞}(∑_{m=0}^{k} a_{nm}x_{m})z^{n} = 0. We obtain a linear system with infinitely many equations and a finite number of unknowns

∑_{m=0}^{k} a_{nm}x_{m} = 0,  n = 0, 1,…  (14)

Hence we have

∑_{m=0}^{k} a_{n_{i}(k),m}x_{m} = 0, i = 0, 1,…, k,

and, since the determinant of the coefficients of the variables of this last system is different from zero, we deduct that x_{0} = x_{1} = ⋯ = x_{k} = 0.

Now we can state the following results.

Proposition 5. Assume that (12) holds.

i) Let f(z) = ∑_{n=0}^{∞} b_{n}z^{n}. Then f ∈ E_{r,R}(Â) if and only if the infinite linear system

∑_{m=0}^{∞} a_{nm}x_{m} = b_{n},  n = 0, 1,…  (15)

admits a solution in the space s_{r}.

ii) If we suppose that the matrix A = (a_{nm})_{n,m⩾0} satisfies the following conditions:

sup_{n⩾0}( ∑_{m≠n}^{∞}|\frac{a_{nm}}{a_{nn}} |r^{m−n}) < 1,  \frac{b_{n}}{a_{nn}} = O(r^{n}) (n → ∞).  (16)

Then f ∈ E_{r,R}(Â) and the expansion is unique.

Proof. Assertion i). Suppose that

f(z) = ∑_{m=0}^{∞} x_{m}Â_{m}(z) = ∑_{m=0}^{∞} x_{m}( ∑_{n=0}^{∞} a_{nm}z^{n}),

for all z ∈ \overline{D}(0, R). Since we have (10), the order of symbols ∑_{n} and ∑_{m} can be interchanged. Then f(z) = ∑_{n=0}^{∞}(∑_{m=0}^{∞} a_{nm}x_{m})z^{n}, and we deduce that (15) admits a solution in s_{r}. Conversely, assume that equation (15) admits a solution in s_{r}. The double series with general term |a_{nm}||x_{m}||z|^{n} being convergent for all z ∈ \overline{D}(0, R) and for all X = (x_{m})_{n} ∈ s_{r}, we can write

f(z) = ∑_{n=0}^{∞} b_{n}z^{n} = ∑_{n=0}^{∞}( ∑_{m=0}^{∞} a_{nm}x_{m})z^{n}

for |z| ⩽ R. And (12) implies f(z) = ∑_{m=0}^{∞} x_{m}(∑_{n=0}^{∞} a_{nm}z^{n}) for all z ∈ \overline{D}(0, R), so f ∈ E_{r,R}(Â).

Assertion ii). Denote by D the infinite diagonal matrix (δ_{nm}/a_{nn})_{n,m⩾0}. From (16) we conclude ∥I − DA∥_{S_{r}} < 1 and DB ∈ s_{r}. Them (13) is equivalent to D(AX) = DB. It is easy to verify that D(AX) = (DA)X. Hence X = (DA)^{−1}DB is the unique solution of (15) in s_{r}. Finally, applying i) we conclude ii).

Definition 6. In the case when each f ∈ E_{r,R}(Â) admits a unique expansion as in (13) and satisfies Porposition 4, Â = (Â_{m})_{m⩾0} is called an r, R basis.

We are going to give some expansions in nontrivial cases. Let us consider the first example, where the infinite matrix associated with Â is of Pòlya type (see Subsection 2.2).

Example 7. Let a be a real with 0 < a < 1/3 and set

Â_{m}(z) = ∑_{n=0}^{∞} a^{|m−n|}z^{n}.

We are going to see that the function f, defined by f(z) = ∑_{n=0}^{∞} b_{n}z^{n}, where (b_{n})_{n⩾0} is bounded, belongs to E_{1,R}(Â) for all R ∈]0, 1[, is a 1, R basis. In fact, we have a_{nn} = 1 for every n and (16) is satisfied since (b_{n})_{n} ∈ s_{1} and 0 < a < 1/3 implies

sup_{n⩾0}( ∑_{m≠n}^{∞} a^{|m−n|}) ⩽ 2 ∑_{k=1}^{∞} a^{k} = \frac{2a}{1 − a} < 1.

Furthermore,

∑_{n=0}^{∞} ∑_{m=0}^{∞} a^{|m−n|}R^{n} = ∑_{n=0}^{∞}[R^{n}( ∑_{m=0}^{n} a^{m})] + ∑_{n=0}^{∞} R^{n}( ∑_{m=1}^{∞} a^{m}),

that is

∑_{n=0}^{∞} ∑_{m=0}^{∞} a^{|m−n|}R^{n} = ∑_{n=0}^{∞}[R^{n} \frac{1 − a^{n+1}}{1 − a} ] + \frac{a}{1 − a} ∑_{n=0}^{∞} R^{n},

and this last term is equal to \frac{1−a^{2}R}{(1−a)(1−R)(1−aR)}. We see that condition (12) holds

for all R < 1. Applying the previous proposition we conclude that f ∈ E_{1,R}(Â) and Â is a 1, R basis.

Now consider the sequence Λ = (λ_{n+1})_{n⩾0} defined in Subsection 2.3 and set Ĉ = (Ĉ_{m})_{m⩾0}, where Ĉ_{m} is defined by the power series Ĉ_{m} = ∑_{n=m}^{∞} \frac{1}{λ_{n+1}} z^{n}, m = 0, 1, 2,…, defined for |z| < 1. We can give the following result.

Proposition 8. Assume that Λ ∈ s_{r} for r > 1. Let f(z) = ∑_{n=0}^{∞} b_{n}z^{n} be a power series such that (b_{n})_{n⩾0} is bounded. Then f ∈ E_{r,R}(Ĉ) when R < 1/r, that is

f(z) = b_{0}Ĉ_{0}(z) + ∑_{m=1}^{∞}[λ_{m+1}b_{m} − λ_{m}b_{m−1}]Ĉ_{m}(z),  (17)

with |z| ⩽ R. Furthermore, Ĉ is an r, R basis.

Proof. We saw in Subsection 2.3 that the infinite matrix A = (a_{nm})_{n,m⩾0} associated to Ĉ is equal to C. Let us show that condition (12) is satisfied. We have

∑_{n=0}^{∞} ∑_{m=0}^{n} \frac{1}{λ_{n+1}} R^{n}r^{m} = ∑_{n=0}^{∞} \frac{1}{λ}_{n+1}R^{n} (\frac{r^{n+1} − 1}{r − 1}) ,

and since the sequence (1/λ_{n+1})_{n⩾0} is bounded, R < 1 and rR < 1, then

∑_{n=0}^{∞} ∑_{m=0}^{n} \frac{1}{λ_{n+1}} R^{n}r^{m} = \frac{1}{r − 1}(r ∑_{n=0}^{∞} \frac{1}{λ_{n+1}} (rR)^{n} − ∑_{n=0}^{∞} \frac{1}{λ_{n+1}} R^{n}).

We deduce that the series ∑_{n=0}^{∞} ∑_{m=0}^{n} \frac{1}{λ_{n+1}} R^{n}r^{m} is convergent. Then (17) is equivalent to the linear system CX = B. As in Proposition 3, X = (x_{n})_{n⩾0} is defined by (9), it can be verified that it belongs to s_{r} for r > 1. Finally, it is easy to see that Ĉ = (Ĉ_{m})_{m⩾0} is an r, R basis, since the matrix C is a lower infinite triangle matrix with non-zero entries on the main diagonal.

Remark 3. The previous proposition can be applied to the case where C is the Cesàro operator, since the sequence Λ = (n + 1)_{n⩾0} ∈ s_{r} for all r > 1.

3.2. Properties of the set E_{r,R}(Â)

In this part it is useful to associate to every power series f, defined by f(z) = ∑_{n=0}^{∞} b_{n}z^{n}, the upper triangular infinite matrix

[f] = ( \begin{matrix}b_{0} & & & b_{1} & & & b_{2} & & & … \\ 0 & & & b_{0} & & & b_{1} & & & … \\ 0 & & & 0 & & & b_{0} & & & … \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \end{matrix} ) .

That is, setting [f] = (α_{nm})_{n,m⩾0}, α(nm) = b_{m−n} when m ⩾ n, and α_{nm} = 0 when m < n (see [2]). The function

f ↦ [f]

maps the algebra of functions which can be expanded in a power series, into the algebra of the corresponding matrices. Recall that if b_{0} ≠ 0, [f] is invertible and [1/f] = [f]^{−1}.

In the following we can suppose that a_{nn} = 1 for all values of n. When a_{nn} ≠ 0 for all n, we can refer to this case by replacing Â_{m}(z) by Â′_{m}(z) = a_{mm}^{−1}Â_{m}(z). We have the next result.

Proposition 9. Assume that (12) holds for r and R > 0.

i) If f, g ∈ E_{r,R}(Â), then f + g ∈ E_{r,R}(Â).

ii) Suppose that for r ⩾ 1/R,

sup_{n⩾0}( ∑_{m≠n}^{∞} |a_{nm}|r^{m−n}) < 1.  (18)

If f, g ∈ E_{r,R}(Â), then f · g ∈ E_{r,R}(Â).

iii) Suppose that f(z) = ∑_{n=0}^{∞} b_{n}z^{n} with b_{0} ≠ 0 satisfies the condition
∑_{n=2}^{∞} |b_{n}b_{0} − b_{n−1}b_{1}|R^{n} < |b_{0}^{2}|,  (19)

with R ⩾ 1/r, and that (18) holds. Then f ∈ E_{r,R}(Â) implies 1/f ∈ E_{r,R}(Â).

Proof. If f and g belong to E_{r,R}(Â), there are two sequences (x_{n})_{n⩾0} and (y_{n})_{n⩾0} belonging to s_{r} such that (f + g)(z) = ∑_{m=0}^{∞}(x_{m} + y_{m})Â_{m}(z) for all z ∈ \overline{D}(0, R). Since s_{r} is a Banach space, (x_{n} + y_{n})_{n⩾0} ∈ s_{r}, which proves i).

Assertion ii). If f, g ∈ E_{r,R}(Â), then [f] and [g] belong to S_{R} since the expansion in (13) holds in the whole disk \overline{D}(0, R). S_{R} being an algebra, we conclude that the infinite matrix [f · g] = [f][g] ∈ S_{R}. Denote now f(z)g(z) = ∑_{n=0}^{∞} γ_{n}z^{n}. The series ∑_{n=0}^{∞} |γ_{n}|R^{n} being convergent, we conclude that the sequence γ = (γ_{n})_{n⩾0} belongs to s_{1/R}. Finally (18) is equivalent to ∥I − A∥_{S_{r}} < 1, and since γ ∈ s_{1/R} with s_{1/R} ⊂ s_{r}, the system

∑_{m=0}^{∞} a_{nm}x_{m} = γ_{n}, n = 0, 1,…

admits only one solution in s_{r}. Applying i) in Proposition 5, we conclude that f · g ∈ E_{r,R}(Â).

Assertion iii). We shall modify the expression of f(z) in order to know the behavior of ots inverse. So, let p_{1} be the function defined by p_{1}(z) = 1 − \frac{b_{1}}{b_{0}} z and consider the product

\frac{1}{b_{0}} f(z)p_{1}(z) = 1 + ∑_{n=2}^{∞} (\frac{b_{n}}{b_{0}} − \frac{b_{n−1}b_{1}}{b_{0}^{2}} ) z^{n}.

From (19) we conclude

∥I − \frac{1}{b_{0}} [f][p_{1}]∥_{S_{R}} = ∑_{n=2}^{∞} |\frac{b_{n}}{b_{0}} − \frac{b_{n−1}b_{1}}{b_{0}^{2}} | R^{n} < 1.  (20)

Consider now the equation

[f]X = Y,  (21)

for any Y ∈ s_{R}. If we set X = [p_{1}]X′, (21) is equivalent to ([f][p_{1}])X′ = Y, since [p_{1}] ∈ S_{R}. Using (20), we see that this last equation admits a unique solution X′ = ([f][p_{1}])^{−1}Y in s_{R}. Then X = [p_{1}]([f][p_{1}])^{−1}Y is the unique solution of (21) in the space s_{R}. Thus [f]^{−1} = [p_{1}]([f][p_{1}])^{−1} ∈ S_{R}, that is [f]^{−1} = [1/f] ∈ S_{R}. Now set \frac{1}{f(z)} = ∑_{n=0}^{∞} b′_{n}z^{n}, the series ∑_{n=0}^{∞} |b′_{n}|R^{n} is convergent, and the sequence B′ = (b′_{n})_{n⩾0} belongs to s_{1/R}. So B′ ∈ s_{r}, since s_{1/R} ⊂ s_{r}. From (18) we conclude that the system

∑_{m=0}^{∞} a_{nm}x_{m} = b′_{n}, n = 0, 1,…

admits only one solution in s_{r}. Using Proposition 5 we conclude that 1/f ∈ E_{r,R}(Â).

Remark 4. In the previous proposition, iii) holds when we replace the condition given by (19) by the following one: there exists R_{0} > 0 such that ∑_{n⩾1} |b_{n}|R_{0}^{n} < |b_{0}|. Indeed, we have then ∥I − \frac{1}{b_{0}} [f]∥_{S_{R_{0}}} < 1. The choice of p_{1} permits to take R = R_{1} > R_{0} as in (19) and we obtain a greater space s_{R_{1}} ⊃ s_{R_{0}} in which the equation AX = B admits only one solution (see the following application). p_{1} has been chosen such that the coefficient of z in the polynomial \frac{1}{b_{0}} f(z)p_{1}(z) is equal to 0 (see [9]). This method can be repeated, so p_{2} can be chosen so that the coefficients of z and z^{2} in \frac{1}{b_{0}} f(z)p_{1}(z)p_{2}(z) are equal to 0, and R_{2} > R_{1} satisfy ∥I − \frac{1}{b_{0}} [fp_{1}p_{2}]∥_{S_{R_{2}}} < 1, and so on.

3.2.1. An application

We shall adapt iii) in Proposition 9 in a case where we have not a_{nn} = 1 for all n. Consider the function f(z) = ∑_{n=0}^{∞} \frac{z^{n}}{(n!)^{2}} . Here we choose the sequence Λ = (λ_{n+1})_{n⩾0} of Subsection 2.3 such that the series ∑_{n=0}^{∞} 1/λ_{n+1} is convergent. We are going to verify that the following expansion

\frac{1}{f(z)} = ∑_{m=0}^{∞} x_{m}( ∑_{n=m}^{∞} \frac{1}{λ_{n+1}} z^{n})  (22)

holds for |z| ⩽ R and X ∈ s_{r}. First we see that for r = 1/R > 1 the series ∑_{n=0}^{∞} ∑_{m=0}^{∞} \frac{1}{λ_{n+1}} R^{n}r^{m} = \frac{1}{r−1}(r ∑_{n=0}^{∞} \frac{1}{λ_{n+1}} −∑_{n=0}^{∞} \frac{1}{λ_{n+1}} R^{n}) is convergent (notice

that f ∈ E_{1/R,R}(Â)). Furthermore, (19) is satisfied, since ∑_{n=2}^{∞}[\frac{1}{[(n−1)!]^{2}} − \frac{1}{(n!)^{2}} ] = 1 implies that for all R < 1, ∑_{n=2}^{∞}[\frac{1}{[(n−1)!]^{2}} − \frac{1}{(n!)^{2}} ]R^{n} < 1 (remark here that the matrix [f] does not satisfy ∥I − \frac{1}{b_{0}} [f]∥_{S_{R}} < 1 for any R > 0). Finally, recall that C is the infinite matrix associated to the sequence with general terms Ĉ_{m}(z) = ∑_{n=m}^{∞} \frac{1}{λ_{n+1}} z^{n}, m ⩾ 0. Then \frac{1}{f(z)} = ∑_{n=0}^{∞} b′_{n}z^{n} is convergent for |z| ⩽ R for all R < 1. So (22) is equivalent to CX = B′ with B′ = (b′_{n})_{n⩾0} ∈ s_{1/R}, i.e. B′ ∈ s_{r} since r = 1/R. Thus the equation CX = B′ admits a unique solution X = (x_{n})_{n⩾0} in s_{1/R}, given by (9). We see that X ∈ s_{1/R}. Using i) in Proposition 4, we conclude that (22) is satisfied for |z| ⩽ R and X ∈ s_{1/R} with R < 1. This expansion is unique and we obtain an r, R basis.

3.3. Study of a particular r, R basis

Consider now the case where Â_{0}(z) = 1 + c′_{0}z, and for m ⩾ 1, Â_{m}(z) = c_{m}z^{m−1} + z^{m} + z′_{m}z^{m+1}. Assume that rR < 1 and the series with general terms c_{n}(rR)^{n} and c′_{n}(rR)^{n} are absolutely convergent. Then (12) is satisfied and equivalent to

∑_{n=1}^{∞}(|c′_{n−1}|r^{n−1} + r^{n} + |c_{n+1}|r^{n+1})R^{n} < ∞.  (23)

One has Proposition 10. Suppose that (23) holds and R < 1/r. We have:

i) If

χ = sup\{sup_{n⩾0} (\frac{|c′_{n}|}{r} + |c_{n+2}|r) , |c_{1}|r\} < 1,  (24)

then f ∈ E_{r,R}(Â), and the expansion is unique.

ii) Assume that B ∈ s_{α}, where α = (c_{n}r^{n})_{n⩾0} with:

|c_{1}| > 1/r and sup_{n⩾1} (\frac{|c′_{n−1}| + r}{|c_{n}|} ) < r^{2}.  (25)

Then f ∈ E_{r,R}(Â) admits infinitely many expansions.

iii) Suppose that B ∈ s_{β}, where β = (c′_{n−1}r^{n})_{n⩾1} and

sup_{n⩾0} (\frac{|c_{n+2}| + 1}{|c′_{n}|}) < \frac{1}{r}.  (26)

There is a real u_{0} so that b_{0} ≠ u_{0} implies that f ∉ E_{r,R}(Â).

Proof. The infinite matrix A = (a_{nm})_{n,m⩾0} is definied here for all n ⩾ 1 by a_{n,n−1} = c′_{n−1}, a_{nn} = 1, a_{n,n+1} = c_{n+1} and a_{00} = 1, a_{01} = c_{1}, the other elements being equal to 0. Relation (24) implies ∥I − A∥_{S_{r}} < 1 (since χ = ∥I − A∥_{S_{r}}), which proves i), using Proposition 4.

Assertion ii). Denote by A(e_{0}) = (a′_{nm})_{n,m⩾0} the matrix obtained from A by addition of the supplementary row e_{0}. If D_{0} = (δ_{mn}/a′_{nn})_{n,m⩾0} is the diagonal matrix whose nonzero elements are the inverses of the diagonal entries of A(e_{0}), then (25) expresses that ∥I − D_{0}A(e_{0})∥_{S_{r}} < 1. Hence the equation AX = B with D_{0}B(u) ∈ s_{r} admits infinitely many solutions in s_{r} defined for all scalars u by X = [A(e_{0})]^{−1}B(u) (see [5]). Since D_{0}B(u) ∈ s_{r} is equivalent to b_{n} = O(c_{n}r^{n}) as n → ∞, we conclude that f(z) has infinitely many expansions which can be written as

f(z) = ∑_{m=0}^{∞}(x_{m} + uy_{m})Â_{m}(z),

where (x_{n})_{n⩾0}, (y_{n})_{n⩾0} ∈ s_{r} for all scalars u.

For iii), denote by A^{∗} the matrix obtained from A by deleting the first row, and define B^{∗} from B in the same way. D^{∗} is the diagonal matrix whose nonzero entries are the inverses of the diagonal elements of A^{∗}. (26) implies ∥I − D^{∗}A^{∗}∥_{S_{r}} < 1, so the equation A^{∗}X = B^{∗}, with D^{∗}B^{∗} = (b_{n+1}/c′_{n})_{n⩾0} ∈ s_{r}, has a unique solution X = (A^{∗})^{−1}B^{∗} in s_{r}. Hence there is a real u_{0} for which the equation AX = B does not admit any solution in s_{β}, for u_{0} ≠ b_{0}.

Example 11. Take a function f(z) = ∑_{n=0}^{∞} b_{n}z^{n}. There is a real r > 0 such that (b_{n})_{n⩾0} ∈ s_{α}, where α = (nr^{n})_{n} and there exist infinitely many expansions of the form

f(z) = x_{0} + x_{1} + ∑_{m=1}^{∞}(mx_{m−1} + m^{2}x_{m} + m^{3}x_{m+1})z^{m}.  (27)

Here we have Â_{0}(z) = 1 + z, Â_{1}(z) = 1 + z + 2z^{2} and Â_{m}(z) = (m − 1)^{3}z^{m−1} + m^{2}z^{m} + (m + 1)z^{m+1} for all m ⩾ 2. The matrix A = (a_{nm})_{n,m⩾0} is here defined by a_{00} = a_{01} = a_{10} = a_{11} = a_{12} = 1, and a_{n,n−1} = n, a_{nn} = n^{2}, a_{n,n+1} = n^{3} for all n ⩾ 2. In order to refer to the previous case, where entries on the diagonal are equal to 1, consider the product D_{1}A with D_{1} = (d_{n}δ_{nm})_{n,m⩾0} and d_{0} = 1, d_{n} = n^{−2} for n ⩾ 1. Then c′_{n} = 1/(n + 1) for all n ⩾ 0; c_{1} = 1 and c_{n} = n − 1 for

n ⩾ 2. Set τ_{1} = 1 + r and τ_{2} = sup_{n⩾2} ^{\frac{\frac{1}{n} −r}{n−1}} . Then condition ii) in the previous proposition yields

sup{τ_{1}, τ_{2}} < r^{2}.

Hence r^{2} − r − 1 > 0 that is r > (1 + \sqrt{5})/2. We conclude that for all (b_{n})_{n⩾0} ∈ s_{α} with r > (1 + \sqrt{5})/2 and for |z| ⩽ R, with R < 2/(1 + \sqrt{5}), we have the expansion in (27).

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(received 20.10.2000.)

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