����������� ������ 57 (2005), 41–60

UDK 517.521

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THE BANACH ALGEBRA B(X), WHERE X IS A BK SPACE AND APPLICATIONS

Bruno de Malafosse

Abstract. In this paper we give some properties of Banach algebras of bounded operators B(X), when X is a BK space. We then study the solvability of the equation Ax = b for b ∈ {s_{α}, sα^{∘}, s_{α}^{(c)}, l_{p}(α)} with α ∈ U^{+} and 1 ⩽ p < ∞. We then deal with the equation T_{a}x = b, where b ∈ χ(Δ^{k}) for k ⩾ 1 integer, χ ∈ {s_{α}, sα^{∘}, s_{α}^{(c)}, l_{p}(α)}, 1 ⩽ p < ∞ and T_{a} is a Toeplitz triangle matrix. Finally we apply the previous results to infinite tridiagonal matrices and explicitly calculate the inverse of an infinite tridiagonal matrix. These results generalize those given in [4, 9].

1. Preliminary results

Let A = (a_{nm})_{n,m⩾1} be an infinite matrix and consider the sequence x = (x_{n})_{n⩾1}. We will define the product Ax = (A_{n}(x))_{n⩾1} with A_{n}(x) = ∑_{m=1}^{∞} a_{nm}x_{m} whenever the series are convergent for all n ⩾ 1. Throughout this paper we use the convention that any term with subscript less than 1 is equal to naught. Let s denote the set of all complex sequences. We write φ, c_{0}, c and l_{∞} for the sets of finite, null, convergent and bounded sequences respectively. For any given subsets X, Y of s, we shall say that the operator represented by the infinite matrix A = (a_{nm})_{n,m⩾1} maps X into Y, that is A ∈ (X, Y), see [5], if

i) the series defined by A_{n}(x) = ∑_{m=1}^{∞} a_{nm}x_{m} are convergent for all n ⩾ 1 and for all x ∈ X;

ii) Ax ∈ Y for all x ∈ X.

For any subset X of s, we shall write

AX = { y ∈ s : there is x ∈ X, y = Ax }.

If Y is a subset of s, we shall denote the so-called matrix domain by

Y (A) = { x ∈ s : y = Ax ∈ Y }.

AMS Subject Classification: 40C05, 46A45

Keywords and phrases: Infinite linear system, sequence space, BK space, Banach algebra, bounded operator.

Let X ⊂ s be a Banach space, with norm ∥ · ∥_{X}. By B(X) we will denote the set of all bounded linear operators, mapping X into itself. We shall say that A ∈ B(X) if and only if A: X → X is a linear operator and

∥A∥_{B(X)}^{∗} = sup_{x≠0}(∥Ax∥_{X}/∥x∥_{X}) < ∞.

It is well known that B(X) is a Banach algebra with the norm ∥A∥_{B(X)}^{∗}, see [1]. A Banach space X ⊂ s is a BK space if the projection P_{n}: x ↦ x_{n} from X into ℂ is continuous for all n. A BK space X ⊃ φ is said to have AK if for every x ∈ X, x = lim_{p→∞} ∑_{k=1}^{p} x_{k}e_{k}, where e_{k} = (0, …, 1, …), 1 being in the k-th position. It is well known that if X has AK then B(X) = (X, X), see [19]. In the following we shall explicitly give some new properties of particular algebras.

2. The set B(l_{p}(α)) for 1 ⩽ p ⩽ ∞

2.1. The set B(l_{p}(α)) for p ⩾ 1 real

Put now

U^{+} = { x = (x_{n})_{n⩾1} ∈ s : x_{n} > 0 for all n }.

Recall that l_{p}, for p > 0 is the set of sequences x = (x_{n})_{n⩾1} such that ∑_{n=1}^{∞} |x_{n}|^{p} < ∞. Using Wilansky’s notations [21], for any given α = (α_{n})_{n⩾1} ∈ U^{+} and p ⩾ 1 real we have

l_{p}(α) = (\frac{1}{α})^{−1} ∗ l_{p} = \{ x ∈ s : ∑_{n=1}^{∞} (\frac{|x_{n}|}{α_{n}})^{p} < ∞ \}.

Define the diagonal matrix D_{ξ} = (ξ_{n}δ_{nm})_{n,m⩾1}, (where δ_{nm} = 0 for all n ≠ m and δ_{nm} = 1 otherwise), we then have D_{α}l_{p} = l_{p}(α). It is easy to see that l_{p}(α) is a Banach space with the norm

∥x∥_{l_{p}(α)} = ∥D_{\frac{1}{α}} x∥_{l_{p}} = [ ∑_{n=1}^{∞} (\frac{|x_{n}|}{α_{n}})^{p}]^{\frac{1}{p}} .

We have the following lemma.

Lemma 1. Let α = (α_{n})_{n⩾1}, β = (β_{n})_{n⩾1} ∈ U^{+} and p ⩾ 1 a real. The condition α/β ∈ l_{∞} implies that l_{p}(α) ⊂ l_{p}(β).

Proof. Since p ⩾ 1, we get α/β ∈ l_{∞} if and only if (α/β)^{p} ∈ l_{∞} and for all x ∈ l_{p}(α)

∥x∥_{l_{p}(β)}^{p} = ∑_{n=1}^{∞} (|x_{n}|\frac{1}{α_{n}} \frac{α_{n}}{β_{n}} )^{p} ⩽ sup_{n⩾1} [(\frac{α_{n}}{β_{n}} )^{p}] ∥x∥_{l_{p}(α)}^{p}.

This gives the conclusion.

We also have l_{p}(α) ⊂ l_{p′}(α) for 1 ⩽ p ⩽ p′.

As we will see later, l_{p}(α) has AK, so B(l_{p}(α)) = (l_{p}(α), l_{p}(α)) and B(l_{p}(α)) is a Banach algebra with identity. So we get

∥Ax∥_{l_{p}(α)} ⩽ ∥A∥_{B(l_{p}(α))}^{∗}∥x∥_{l_{p}(α)} for all x ∈ l_{p}(α).

The Banach algebra B(X) and applications 43

We have l_{p} = l_{p}(e), where e = (1, …, 1, …) and

∥D_{\frac{1}{α}} AD_{α}∥_{B(l_{p})}^{∗} = ∥A∥_{B(l_{p}(α))}^{∗} for all A ∈ B(l_{p}(α)).

Indeed, writing D_{α}x = y, we get

sup_{x≠0} (\frac{∥(D_{\frac{1}{α}} AD_{α})x∥_{l_{p}}}{∥x∥_{l_{p}}} ) = sup_{y≠0} (\frac{∥D_{\frac{1}{α}} Ay∥_{l_{p}}}{∥D_{\frac{1}{α}} y∥_{l_{p}}} ) = ∥A∥_{B(l_{p}(α))}^{∗}.

So we can say that A ∈ B(l_{p}(α)) if and only if D_{1/α}AD_{α} ∈ B(l_{p}). When α = (r^{n})_{n⩾1}, for a given real r > 0, l_{p}(α) is denoted l_{p}(r). When p = ∞, we obtain the next results.

2.2. The case p = ∞ and Banach algebra S_{α}

Let α = (α_{n})_{n⩾1} ∈ U^{+}. Using Wilansky’s notation [20], we have l_{∞}(α) = (1/α)^{−1} ∗ l_{∞} = D_{α}l_{∞}. We will write

s_{α} = l_{∞}(α) = { x ∈ s : x_{n}/α_{n} = O(1) (n → ∞) },

see [6–15]. The set s_{α} is a Banach space with the norm ∥x∥_{s_{α}} = sup_{n⩾1}(|x_{n}|/α_{n}). The set

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

is a Banach algebra with identity normed by

∥A∥_{S_{α}} = sup_{n⩾1}( ∑_{m=1}^{∞} |a_{nm}|\frac{α_{m}}{α_{n}} ).  (2)

Recall that if A ∈ (s_{α}, s_{α}), then ∥Ax∥_{s_{α}} ⩽ ∥A∥_{S_{α}}∥x∥_{s_{α}} for all x ∈ s_{α}. Thus we obtain the following result where we put B(s_{α}) = B(s_{α}) ∩ (s_{α}, s_{α}).

Lemma 2. For any given α ∈ U^{+} we have B(s_{α}) = S_{α} = (s_{α}, s_{α}).

As we have seen above when α = (r^{n})_{n⩾1}, r > 0, S_{α} and s_{α} are denoted S_{r} and s_{r}. When r = 1, s_{1} = l_{∞} is the set of all bounded sequences.

Recall [14] that s_{α} = s_{β} if and only if there are K_{1}, K_{2} > 0 such that K_{1} ⩽ α_{n}/β_{n} ⩽ K_{2} for all n.

In the same way we will define the sets

sα^{∘} = \{ x ∈ s : \frac{x_{n}}{α_{n}} = o(1) (n → ∞) \}

and

s_{α}^{(c)} = \{ x ∈ s : \frac{x_{n}}{α_{n}} → l (n → ∞) for some l \} .

The sets sα^{∘} and s_{α}^{(c)} are Banach spaces with the norm ∥ · ∥_{s_{α}}.

3. New Banach algebras

3.1. The Banach algebra B(χ) for χ ∈ {s_{α}, sα^{∘}, s_{α}^{(c)}, l_{p}(α)} with 1 ⩽ p < ∞

In this section we will give an explicit expression for the norm ∥A∥_{B(l_{p}(α))}^{∗} for 1 ⩽ p ⩽ ∞ and give some properties of the equation Ax = b for A ∈ B(χ), b ∈ χ

with χ ∈ {s_{α}, sα^{∘}, s_{α}^{(c)}, l_{p}(α)} and 1 ⩽ p < ∞.

We have the next result where U is the set of all sequences u = (u_{n})_{n⩾1} with u_{n} ≠ 0 for all n and L is the set of lower triangular infinite matrices, (that is A ∈ L if a_{nm} = 0 for m > n).

Lemma 3. i) Let T ∈ L be a triangle, that is t_{nn} ≠ 0 for all n, and X a BK space. Then X(T) is a BK space with the norm

∥x∥_{X(T)} = ∥T x∥_{X}.  (3)

ii) Let T = D_{a} = (a_{n}δ_{nm})_{n,m⩾1} be a diagonal matrix with a ∈ U and X a BK space with AK. Then X(T) has AK with the norm given by (3).

Proof. i) was shown in [17] and ii) comes from [20, Theorem 4.3.6 pp. 52]. We then have the following result.

Lemma 4. i) Let α ∈ U^{+}. Then s_{α}, sα^{∘} and s_{α}^{(c)} are BK spaces with the norm ∥ · ∥_{s_{α}} and sα^{∘} has AK. The set l_{p}(α) for 1 ⩽ p < ∞ is a BK space and has AK with the norm ∥ · ∥_{l_{p}(α)}.

ii) Let χ be any of the spaces s_{α}, sα^{∘}, or s_{α}^{(c)}. Then
|P_{n}(x)| = |x_{n}| ⩽ α_{n}∥x∥_{s_{α}} for all n ⩾ 1 and for all x ∈ χ.

Proof. i) First we have s_{α} = s_{1}(D_{1/α}), since x ∈ s_{α} if and only if D_{1/α}x = x/α ∈ s_{1}. It is well known that s_{1} is a BK space with respect to the norm ∥ · ∥_{s_{1}}, so by Lemma 3, the set s_{1}(D_{1/α}) is also a BK space with

∥x∥_{s_{1}(D_{1/α})} = ∥D_{1/α}x∥_{s_{1}} = ∥x∥_{s_{α}}.

We also have sα^{∘} = c_{0}(D_{1/α}) and s_{α}^{(c)} = c(D_{1/α}). We conclude since c_{0} and c are BK spaces with respect to the norm ∥∥_{s_{1}}, and c_{0} has AK. Finally l_{p} for 1 ⩽ p < ∞ being a BK space with AK, it follows that l_{p}(α) = l_{p}(D_{1/α}) also has AK.

ii) is a direct consequence of the definition of the sets s_{α}, sα^{∘}, and s_{α}^{(c)}. Remark 1. Note that if X is a BK space with the norm ∥ · ∥_{X}, then

(X, X) ⊂ B(X).

Indeed, by [17 Theorem 4.2.8 p. 57], since X is a BK space, the matrix map A ∈ (X, X) is continuous and there is M > 0 such that

∥Ax∥_{X} ⩽ M∥x∥_{X} for all x ∈ X.

To obtain other results we require some definitions and lemmas. We will write B_{X}(0, 1) = { x ∈ X : ∥x∥_{X} ⩽ 1 } for the unit ball, where X is any given BK space. Thus we get

∥A∥_{B(X)}^{∗} = sup_{x≠0}(∥Ax∥_{X}/∥x∥_{X}) = sup_{x∈B_{X}(0,1)}(∥Ax∥_{X}) for all A ∈ B(X).

Recall that for all a = (a_{n})_{n⩾1} ∈ X such that the series ∑_{n=1}^{∞} a_{n}x_{n} is convergent for all x ∈ X the identity

∥a∥_{X}^{∗} = sup_{x∈B_{X}(0,1)}(| ∑_{n=1}^{∞} a_{n}x_{n}|)

is defined and finite. In Malkowsky [17], we have

Lemma 5. Let X be a BK space. Then A ∈ (X, l_{∞}) if and only if

sup_{n⩾1}(∥A_{n}∥_{X}^{∗}) = sup_{n⩾1}( sup_{x∈B_{X}(0,1)}| ∑_{n=1}^{∞} a_{nk}x_{k}|) < ∞.

We also have, (cf. [17, Theorem 1.2.3, p. 155]),

Lemma 6. For every a ∈ l_{1}, ∥a∥_{c_{0}}^{∗} = ∥a∥_{c}^{∗} = ∥a∥_{l_{∞}}^{∗} = ∥a∥_{l_{1}} = ∑_{n=1}^{∞} |a_{n}|.

It can be easily deduced that if X is a BK space, then A ∈ (X, s_{β}) if and only if

sup_{n⩾1} ∥\frac{1}{β_{n}} A_{n}∥_{X}^{∗} < ∞.

Since there is no characterization of the set (l_{p}(α), l_{p}(α)) for 1 < p < ∞ and p ≠ 2, we need to define a subset B̂_{p}(α) of (l_{p}(α), l_{p}(α)) permitting us to obtain the inverse of some well chosen matrix map A ∈ (l_{p}(α), l_{p}(α)). In this way we are led to define the number

N_{p,α}(A) = [ ∑_{n=1}^{∞}( ∑_{m=1}^{∞} (|a_{nm}|\frac{α_{m}}{α_{n}} )^{q})^{p−1}]^{\frac{1}{p}} ,

for 1 < p < ∞ and q = p/(p − 1). Thus we can state the following

Proposition 7. Let α ∈ U^{+}. Then

i) for every A ∈ S_{α}

∥A∥_{B(s_{α})}^{∗} = ∥A∥_{S_{α}} = sup_{n⩾1}( ∑_{m=1}^{∞} |a_{nm}|\frac{α_{m}}{α_{n}} );

ii) a) B(l_{1}(α)) = (l_{1}(α), l_{1}(α)) and A ∈ B(l_{1}(α)) if and only if A^{t} ∈ S_{1/α},
b) ∥A∥_{B(l_{1}(α))}^{∗} = ∥A^{t}∥_{S_{1/α}} for all A ∈ B(l_{1}(α)).
iii) For 1 < p < ∞ we have B̂_{p}(α) ⊂ B(l_{p}(α)), where

B̂_{p}(α) = { A = (a_{nm})_{n,m⩾1} : N_{p,α}(A) < ∞ },

and for every A ∈ B̂_{p}(α), ∥A∥_{B(l_{p}(α))}^{∗} ⩽ N_{p,α}(A).

Proof. i) First we have

∥Ax∥_{s_{α}} = sup_{n⩾1}(\frac{1}{α_{n}} | ∑_{k=1}^{∞} a_{nk}x_{k}|) = sup_{n⩾1} (\frac{1}{α_{n}} |A_{n}(x)|) for all x ∈ s_{α}  (4)

then

∥A∥_{B(s_{α})}^{∗} = sup_{x∈B_{sα} (0,1)} (sup_{n⩾1} (\frac{1}{α_{n}} |A_{n}(x)|)) = sup_{n⩾1} (\frac{1}{α_{n}} sup_{x∈B_{sα} (0,1)}(|A_{n}(x)|)) . (5)

Writing x = αy in (5) we obtain

∥A∥_{B(s_{α})}^{∗} = sup_{n⩾1} (\frac{1}{α_{n}} sup_{y∈B_{s_{1}} (0,1)}(|A_{n}(αy)|)) = sup_{n⩾1} (\frac{1}{α_{n}} ∥A_{n}D_{α}∥_{l_{∞}}^{∗}) .

Now by Lemma 6 we have

∥A_{n}D_{α}∥_{l_{∞}}^{∗} = ∥(a_{nk}α_{k})_{k⩾1}∥_{l_{1}} = ∑_{k=1}^{∞} |a_{nk}|α_{k} for all n.

We conclude that ∥A∥_{B(s_{α})}^{∗} = sup_{n⩾1}(∑_{m=1}^{∞} |a_{nm}|α_{m}/α_{n}).

ii) By [17, Theorem 2.27, p. 175] we have B(l_{1}) = (l_{1}, l_{1}) and the condition A ∈ (l_{1}(α), l_{1}(α)) is equivalent to D_{1/α}AD_{α} ∈ (l_{1}, l_{1}). It is well known that D_{1/α}AD_{α} ∈ (l_{1}, l_{1}) if and only if (D_{1/α}AD_{α})^{t} ∈ S_{1}. This means that D_{α}A^{t}D_{1/α} ∈ S_{1} and A^{t} ∈ S_{1/α}. Furthermore, since ∥A∥_{B(l_{1})}^{∗} = ∥A^{t}∥_{S_{1}}, so ∥A∥_{B(l_{1}(α))}^{∗} = ∥A^{t}D_{1/α}∥_{S_{1}} = ∥A^{t}∥_{S_{α}}. This permits us to conclude for ii).

iii) Let A ∈ B̂_{p}(α) be a given infinite matrix and take any x ∈ l_{p}. We have

∥Ax∥_{l_{p}}^{p} = ∥( ∑_{m=1}^{∞} a_{nm}x_{m})_{n⩾1}∥_{l_{p}}^{p} = ∑_{n=1}^{∞}| ∑_{m=1}^{∞} a_{nm}x_{m}|^{p} ⩽ ∑_{n=1}^{∞}( ∑_{m=1}^{∞} |a_{nm}x_{m}|)^{p},

and from the Hölder’s inequality, we get for every n

∑_{m=1}^{∞} |a_{nm}x_{m}| ⩽ ( ∑_{m=1}^{∞} |a_{nm}|^{q})^{\frac{1}{q}} ( ∑_{m=1}^{∞} |x_{m}|^{p})^{\frac{1}{p}} = ( ∑_{m=1}^{∞} |a_{nm}|^{q})^{\frac{1}{q}} ∥x∥_{l_{p}},

with q = p/(1 − p). We deduce that

∥Ax∥_{l_{p}}^{p} ⩽ ∑_{n=1}^{∞}[( ∑_{m=1}^{∞} |a_{nm}|^{q})^{\frac{1}{q}} ∥x∥_{l_{p}}]^{p} = ∑_{n=1}^{∞}( ∑_{m=1}^{∞} |a_{nm}|^{q})^{\frac{p}{q}} ∥x∥_{l_{p}}^{p};

and since p/q = p − 1, we have

∥Ax∥_{l_{p}} ⩽ [ ∑_{n=1}^{∞}( ∑_{m=1}^{∞} |a_{nm}|^{q})^{p−1}]^{\frac{1}{p}} ∥x∥_{l_{p}}

and

∥A∥_{B(l_{p})}^{∗} = sup_{x≠0}(\frac{∥Ax∥_{l_{p}}}{∥x∥_{l_{p}}} ) ⩽ [ ∑_{n=1}^{∞}( ∑_{m=1}^{∞} |a_{nm}|^{q})^{p−1}]^{\frac{1}{p}} .  (6)

We have proved that if A ∈ B̂_{p}(e), then A ∈ B(l_{p}). So if A ∈ B̂_{p}(α) and D_{1/α}AD_{α} ∈ B̂_{p}(e), then D_{1/α}AD_{α} ∈ B(l_{p}) and A belongs to B(l_{p}(α)). This concludes the proof.

3.2. Application to the solvability of the equation Ax = b

For a = (a_{nn})_{n⩾1} ∈ U we get the following elementary but very usefull result

Proposition 8. Let X ⊂ s be a BK space. Assume that D_{1/a}A ∈ (X, X) and

∥I − D_{1/a}A∥_{B(X)}^{∗} < 1.  (7)

Then the equation Ax = b with D_{1/a}b = (b_{n}/a_{nn})_{n⩾1} ∈ X admits a unique solution in X given by x = (D_{1/a}A)^{−1}D_{1/a}b.

Proof. First we see that since D_{1/a}A ∈ (X, X) and condition (7) holds, then D_{1/a}A is invertible in B(X). Since B(X) is a Banach algebra of operators with identity, then

(D_{1/a}A)^{−1}(D_{1/a}Ax) = [(D_{1/a}A)^{−1}o(D_{1/a}A)](x) = I(x) = x for all x ∈ X.

Thus the equation Ax = b with D_{1/a}b ∈ X is equivalent to D_{1/a}Ax = D_{1/a}b which in turn is x = (D_{1/a}A)^{−1}(D_{1/a}b) ∈ X, this concludes the proof.

We can express a similar result in a more general case.

In the following we will write

Γ_{α} = { A = (a_{nm})_{n,m⩾1} ∈ S_{α} : ∥I − A∥_{S_{α}} < 1 },

and for 1 < p < ∞,

Γ′_{p,α} = { A = (a_{nm})_{n,m⩾1} ∈ (l_{p}(α), l_{p}(α)) : N_{p,α}(I − A) < 1 }, for 1 < p < ∞.

Note that since S_{α} is a Banach algebra, the condition A ∈ Γ_{α} means that A is invertible and A^{−1} ∈ S_{α}.

In the following we will put |a| = (|a_{nn}|)_{n⩾1} for any given a = (a_{nn})_{n⩾1}.

Corollary 9. Let α ∈ U^{+} and A be an infinite matrix with a ∈ U. Assume that D_{1/a}A ∈ Γ_{α}. Then

i) a) for any given b ∈ s_{|a|α}, the equation Ax = b admits in s_{α} a unique solution given by

x^{0} = (D_{1/a}A)^{−1}(D_{1/a}b) = A^{−1}b  (8)

with A^{−1} ∈ (s_{|a|α}, s_{α});

b) if lim_{n→∞}(a_{nm}/a_{nn}α_{n}) = 0 for all m ⩾ 1, for any given b ∈ s|a|α^{∘}, the equation Ax = b admits a unique solution in sα^{∘} given by (8) and A^{−1} ∈ (s|a|α^{∘}, sα^{∘});

c) if lim_{n→∞}(a_{nm}/a_{nn}α_{n}) = l_{m} for some l_{m}, m ⩾ 1 and
lim_{n→∞}(\frac{1}{α_{n}a_{nn}} ∑_{m=1}^{∞} a_{nm}α_{m}) = l for some l  (9)

then for any given b ∈ s_{|a|α}^{(c)}, the equation Ax = b admits a unique solution in s_{α}^{(c)}

given by (8) with A^{−1} ∈ (s_{|a|α}^{(c)}, s_{α}^{(c)}).

ii) a) If A^{t}D_{1/a} ∈ Γ_{1/α} then for given b ∈ l_{1}(|a|α) the equation Ax = b admits a unique solution in l_{1}(α) given by (8) with A^{−1} ∈ (l_{1}(|a|α), l_{1}(α)).

b) Let p > 1 real. If D_{1/a}A ∈ Γ′_{p,α}, then for any given b ∈ l_{p}(|a|α) the equation Ax = b admits a unique solution in l_{p}(α) given by (8) with A^{−1} ∈ (l_{p}(|a|α), l_{p}(α)).

Proof. i) By Proposition 7 i), if D_{1/a}A ∈ Γ_{α}, then

∥I − D_{1/a}A∥_{S_{α}} = ∥I − D_{1/a}A∥_{B(s_{α})}^{∗} < 1.

Thus D_{1/a}A is invertible in B(s_{α}) ∩ S_{α}. Then (D_{1/a}A)^{−1} ∈ S_{α}, that is A^{−1} ∈ (s_{|a|α}, s_{α}) and we conclude by Proposition 8.

b) The set sα^{∘} being a BK space with AK, we have B(sα^{∘}) = (sα^{∘}, sα^{∘}). Since D_{1/a}A ∈ Γ_{α} and lim_{n→∞}(a_{nm}/a_{nn}α_{n}) = 0 for all m ⩾ 1, we deduce that

D_{1/α}(D_{1/a}A)D_{α} ∈ (c_{0}, c_{0}). So D_{1/a}A ∈ (sα^{∘}, sα^{∘}). Now by Lemma 6 we get

∥I − D_{1/a}A∥_{B(s_{α}^{∘})}^{∗} = sup_{x∈B_{sα}}(∥(I − D_{1/a}A)x∥_{s_{α}}) = ∥I − D_{1/a}A∥_{S_{α}} < 1

and we conclude by Proposition 8. c) Here we have

∥I − D_{1/a}A∥_{B(s_{α}^{(c)})}^{∗} = ∥I − D_{1/a}A∥_{S_{α}} < 1.

Then (D_{1/a}A)^{−1} ∈ S_{α} ∩ B(s_{α}^{(c)}), so (D_{1/a}A)^{−1} is an operator represented by an infinite matrix, since (D_{1/a}A)^{−1} ∈ S_{α} and the condition (D_{1/a}A)^{−1} ∈ B(s_{α}^{(c)}) implies (D_{1/a}A)^{−1} ∈ (s_{α}^{(c)}, s_{α}^{(c)}). We conclude again by Proposition 8.

ii) a) By Proposition 7 ii) b) the condition A^{t}D_{1/a} ∈ Γ_{1/α} implies

∥I − D_{1/a}A∥_{B(l_{1}(α))}^{∗} = ∥I − (D_{1/a}A)^{t}∥_{S_{1/α}} < 1

and (D_{1/a}A)^{−1} ∈ B(l_{1}(α)) = (l_{1}(α), l_{1}(α)). Thus A^{−1} ∈ (l_{1}(|a|α), l_{1}(α)) and we conclude by Proposition 8. ii) b) By Proposition 7 iii) we have (D_{1/a}A)^{−1} ∈ B(l_{p}(α)) = (l_{p}(α), l_{p}(α)), so A^{−1} ∈ (l_{p}(|a|α), l_{p}(α)) and again we conclude by Proposition 8.

4. Matrix transformations mapping in the set χ(Δ^{k})

for k ⩾ 1 integer and χ ∈ {s_{α}, sα^{∘}, s_{α}^{(c)}, l_{p}(α)} with 1 ⩽ p < ∞.

In this section we will give some properties of the set χ(Δ) for χ = s_{α}, sα^{∘}, s_{α}^{(c)}, or l_{p}(α). The characterization of the set (χ(Δ^{k}), χ(Δ^{k})) given in [15] being complicated, we will deal with the subset (χ(Δ^{k}), χ(Δ^{k}))_{L′} of infinite Toeplitz triangles that map χ(Δ^{k}) into itself.

4.1. The sets X = χ(Δ^{k}) for χ = s_{α}, sα^{∘}, s_{α}^{(c)}, or l_{p}(α), with 1 ⩽ p < ∞ Recall that the operator of first difference [1, 3, 6, 7, 8], [10–13] and [14–18],

is defined by Δ = (η_{nm})_{n,m⩾1} ∈ L, with η_{nn} = 1 for all n ⩾ 1, η_{n,n−1} = −1 for all n ⩾ 2 and η_{nm} = 0 otherwise. It is well known that Σ = Δ^{−1} = (η′_{nm})_{n,m⩾1}, with η′_{nm} = 1 for m ⩽ n and η′_{nm} = 0 otherwise. We need to use the following sets

Ĉ_{1} = { α ∈ U^{+} : (α_{1} + ⋯ + α_{n})/α_{n} = O(1) (n → ∞) },
Ĉ = { α ∈ U^{+} : lim_{n→∞}(α_{1} + ⋯ + α_{n})/α_{n} = l for some l ∈ ℂ },
Γ = \{ α ∈ U^{+} : \overline{lim}_{n→∞} (\frac{α_{n−1}}{α_{n}} ) < 1 \},
Γ̂ = \{ α ∈ U^{+} : lim_{n→∞} (\frac{α_{n−1}}{α_{n}} ) < 1 \}.

It can be easily seen that Γ̂ ⊂ Γ and it was shown in [12] that Γ ⊂ Ĉ_{1} and by [15] we have Ĉ = Γ̂. So we have

Lemma 10. Ĉ = Γ̂ ⊂ Γ ⊂ Ĉ_{1}.

We also have

Lemma 11. Let α ∈ U^{+} and k ⩾ 1 be an integer. Then

i) s_{α}^{(c)}(Δ^{k}) = s_{α}^{(c)} if and only if α ∈ Γ̂.

ii) If α ∈ Γ, then s_{α}(Δ^{k}) = s_{α}, sα^{∘}(Δ^{k}) = sα^{∘} and l_{p}(α)(Δ^{k}) = l_{p}(α) for 1 ⩽ p < ∞.

Proof. i) comes from [12, Theorem 2.6, p. 1789]. ii) By [12, Proposition 2.1, p. 1786], the condition α ∈ Γ implies α ∈ Ĉ_{1}, so Δ and Δ^{k} are bijective from s_{α} to itself and from sα^{∘} to itself then s_{α}(Δ^{k}) = s_{α} and sα^{∘}(Δ^{k}) = sα^{∘}. It remains to show that if α ∈ Γ, then l_{p}(α)(Δ) = l_{p}(α) for 1 ⩽ p < ∞. If we put l = \overline{lim}_{narrow∞}(α_{n−1}/α_{n}) < 1, for given ε_{0}, such that 0 < ε_{0} < 1 − l, there exists N_{0} such that sup_{n⩾N_{0}+1}(α_{n−1}/α_{n}) ⩽ l + ε_{0} < 1. Consider now the infinite matrix

Σ_{α}^{(N_{0})} = ( \begin{matrix}[Δ_{α}^{(N_{0})}]^{−1} & & & O \\ & 1 \\ O & & 1 \\ & & & . \end{matrix} ) ,

Δ_{α}^{(N_{0})} being the finite matrix whose entries are those of Δ_{α} = D_{1/α}ΔD_{α} for all n, m ⩽ N_{0}. We get

Q = Σ_{α}^{(N_{0})}Δ_{α} = (q_{nm})_{n,m⩾1},

with

q_{nm} = \{ \begin{matrix}1, & for m = n, \\ −\frac{α_{m}}{α_{m+1}} , & for m = n − 1 ⩾ N_{0}, \\ 0, & otherwise. \end{matrix}

For every x ∈ l_{p} we get (I − Q)x = (0, …, 0, (α_{N_{0}}/α_{N_{0}+1})x_{N_{0}}, …, (α_{n−1}/α_{n})x_{n−1}, …)^{t}, where α_{N_{0}}x_{N_{0}}/α_{N_{0}+1} is in the (N_{0} + 1) position. So we get

∥(I − Q)x∥_{l_{p}}^{p} = ∑_{n=N_{0}+1}^{∞} (\frac{α_{n−1}}{α_{n}} )^{p} |x_{n−1}|^{p} ⩽ sup_{n⩾N_{0}+1} [(\frac{α_{n−1}}{α_{n}} )^{p}] ( ∑_{n=N_{0}}^{∞} |x_{n}|^{p}),

and

∥I − Q∥_{B(l_{p})}^{∗} = sup_{x≠0} (\frac{∥(I − Q)x∥_{l_{p}}}{∥x∥_{l_{p}}} ) ⩽ [ sup_{n⩾N_{0}+1} (\frac{α_{n−1}}{α_{n}} )^{p}]^{\frac{1}{p}} .

Since α_{n−1}/α_{n} ⩽ l + ε_{0} < 1 for all n ⩾ N_{0} + 1, we deduce

sup_{n⩾N_{0}+1} (\frac{α_{n−1}}{α_{n}} )^{p} < 1.

Hence

∥I − Q∥_{B(l_{p})} ⩽ [ sup_{n⩾N_{0}+1} (\frac{α_{n−1}}{α_{n}} )^{p}]^{\frac{1}{p}} < 1.

We have shown that Q is invertible in B(l_{p}). Now let b ∈ l_{p}. The equations

Δ_{α}x = b and Qx = Σ^{(N_{0})}b

are equivalent in l_{p}. Since Q^{−1} ∈ B(l_{p}), reasoning as in Proposition 8 that Q^{−1}(Qx) = (Q^{−1}Q)x = x = (Δ_{α})^{−1}b for all x ∈ l_{p}. This shows the map Δ_{α} is bijective from l_{p} to l_{p} and Δ is bijective from l_{p}(α) to l_{p}(α).

4.2. The set (χ(Δ), χ(Δ)) for χ ∈ {s_{α}, sα^{∘}, s_{α}^{(c)}}

First recall the characterizations of the set (χ(Δ), χ(Δ)) for χ = s_{α}, sα^{∘}, or s_{α}^{(c)}. For this we will consider the following properties

lim_{l→∞}( ∑_{j=1}^{l} α_{j}|∑_{k=l}^{∞} (\frac{a_{nk} − a_{n−1,k}}{α_{n}} )|) = 0 for all n;  (10)
sup_{n⩾1}| ∑_{j=1}^{∞} α_{j}( ∑_{k=j}^{∞} (\frac{a_{nk} − a_{n−1,k}}{α_{n}} ))| < ∞;  (11)
lim_{n→∞}[α_{j}( ∑_{k=j}^{∞} (\frac{a_{nk} − a_{n−1,k}}{α_{n}} ))] = 0 for all j;  (12)

sup_{l⩾1}( ∑_{j=1}^{l} α_{j}|∑_{k=l}^{∞} (\frac{a_{nk} − a_{n−1,k}}{α_{n}} )|) < ∞ for all n;  (13)
lim_{n→∞}(α_{j}( ∑_{k=j}^{∞} (\frac{a_{nk} − a_{n−1,k}}{α_{n}} ))) = l_{j} for all j;  (14)

lim_{n→∞} ∑_{k=1}^{∞}[(\frac{a_{nk} − a_{n−1,k}}{α_{n}} ) ( ∑_{j=1}^{k} α_{j})] = l exists.  (15)

The following Proposition was proved in [15] Proposition 12. Let α ∈ U^{+}. Then

i) A ∈ (s_{α}(Δ), s_{α}(Δ)) if and only if (10) and (11) hold;

ii) A ∈ (sα^{∘}(Δ), sα^{∘}(Δ)) if and only if (11), (12) and (13) hold;

iii) A ∈ (s_{α}^{(c)}(Δ), s_{α}^{(c)}(Δ)) if and only if (11), (13), (14) and (15) hold.

Note that A ∈ (χ(Δ^{k}), χ(Δ^{k})) if and only if

D_{1/α}Δ^{k}A ∈ (χ(Δ^{k}), Y) with Y ∈ {l_{∞}, c_{0}, c},

see [16]. Now we will give necessary conditions for a matrix map to be bijective from χ(Δ^{k}) into itself when χ ∈ {s_{α}, sα^{∘}, s_{α}^{(c)}, l_{p}(α)} and p ⩾ 1 real. We have

Proposition 13. Let α ∈ U^{+} and k ⩾ 1 an integer.

i) Let b ∈ χ(Δ^{k}), χ ∈ {s_{α}, s_{α}^{∘}, s_{α}^{(c)}} and assume that there is τ ∈]0, 1[ such that

∥Δ^{k}[(I − A)x]∥_{s_{α}} ⩽ τ∥Δ^{k}x∥_{s_{α}} for all x ∈ χ(Δ^{k}).  (16)

Then a) the map A: x ↦ Ax is bijective from χ(Δ^{k}) into itself;

b) if (16) holds and α ∈ Γ̂ then the matrix map A: x ↦ Ax is bijective from χ into itself.

ii) a) If α ∈ Γ and A^{t} ∈ Γ_{1/α}, then A: x ↦ Ax is bijective from l_{1}(α)(Δ^{k}) into itself.

b) Let 1 < p < ∞. If α ∈ Γ and A ∈ Γ′_{p,α}, then the map A: x ↦ Ax is bijective from l_{p}(α)(Δ^{k}) into itself.

Proof. i) a) It is enough to show the proposition in the case when X = s_{α}. Using Lemma 3, condition (16) means that

∥I − A∥_{B(s_{α}(Δ^{k}))}^{∗} = sup_{x≠0} (\frac{∥Δ^{k}[(I − A)x]∥_{s_{α}}}{∥Δ^{k}x∥_{s_{α}}} ) ⩽ τ < 1,

so A is invertible in B(s_{α}(Δ^{k})) and by Proposition 8 the equation Ax = b for b ∈ s_{α}(Δ^{k}) admits a unique solution in s_{α}(Δ^{k}). i) b) Since Γ̂ ⊂ Γ, by Lemma 10 and Lemma 11 ii), we easily deduce that α ∈ Γ̂ implies χ(Δ^{k}) = χ, ∥x∥_{χ} = ∥x∥_{s_{α}} for χ ∈ {s_{α}, sα^{∘}, s_{α}^{(c)}}. We conclude since condition (16) means that

∥I − A∥_{B(s_{α})} = ∥I − A∥_{S_{α}} ⩽ τ < 1. ii) By Lemma 11, the condition α ∈ Γ implies Δ is bijective from l_{p}(α) to itself

for any given p with 1 ⩽ p < ∞. Then Δ^{k} is also bijective and l_{p}(α)(Δ^{k}) = l_{p}(α). We conclude applying Corollary 9 ii) with a = e.

Example 14. Let γ, η ∈ ℂ and put

M(γ, η) = ( \begin{matrix}1 & η \\ γ & 1 & η & 0 \\ & γ & 1 & η \\ & 0 & . & . & . \end{matrix} ) .  (17)

As a direct consequence of Proposition 13 we can state the following. For α ∈ U^{+} and

ξ_{α} = |γ| sup_{n⩾2} (\frac{α_{n−1}}{α_{n}} ) + |η| sup_{n⩾1} (\frac{α_{n+1}}{α_{n}} ) < 1,  (18)

the equation M(γ, η)x = b with b ∈ s_{α}(Δ) admits a unique solution x^{∘} = [M(γ, η)]^{−1}b ∈ s_{α}(Δ). Furthermore if α ∈ Γ the equation M(γ, η)x = b with b ∈ s_{α}admits a unique solution x^{∘} ∈ s_{α}.

Proof. First we have [M(γ, η)−I]x = (γx_{n−1} +ηx_{n+1})_{n⩾1}, with x_{0} = 0. Then we easily get Δ[(M(γ, η) − I)x] = (y_{n})_{n⩾1}with

y_{n} = η(x_{n+1} − x_{n}) + γ(x_{n−1} − x_{n−2}) = 0 and x_{−1} = x_{0} = 0.

Hence

\frac{1}{α_{n}} |y_{n}| ⩽ |η||x_{n+1} − x_{n}|\frac{1}{α_{n+1}} \frac{α_{n+1}}{α_{n}} + |γ||x_{n−1} − x_{n−2}|\frac{1}{α_{n−1}} \frac{α_{n−1}}{α_{n}}

⩽ |η|\frac{α_{n+1}}{α_{n}} ∥Δx∥_{s_{α}} + |γ|\frac{α_{n−1}}{α_{n}} ∥Δx∥_{s_{α}}.

Thus

sup_{n⩾1} (\frac{1}{α_{n}} |y_{n}|) = ∥Δ[(I − M(γ, η))x]∥_{s_{α}} ⩽ ξ_{α}∥Δx∥_{s_{α}} < ∥Δx∥_{s_{α}}

and

∥I − M(γ, η)∥_{B(s_{α}(Δ))}^{∗} < 1.

As above we conclude that if (18) holds the equation M(γ, η)x = b with b ∈ s_{α}(Δ) admits in s_{α}(Δ) the unique solution x^{∘} = [M(γ, η)]^{−1}b. If α ∈ Γ, then s_{α}(Δ) = s_{α} by Lemma 11 and we conclude from the preceding that the equation M(γ, η)x = b with b ∈ s_{α} admits a unique solution in s_{α}.

4.3. The equation T_{a}x = b, where T_{a} is a Toeplitz triangle matrix We will denote by L′ ⊂ L the set of all infinite Toeplitz triangles. We will

say that T = (t_{nm})_{n,m⩾1} ∈ L′ if there is a sequence a = (a_{n})_{n⩾1} ∈ s, such that t_{nm} = a_{n−m+1} for m ⩽ n and t_{nm} = 0 otherwise; then we will write T = T_{a}; L′ is a subset of the set of Toeplitz matrices, see [1]. It can easily be seen that T_{a}T_{a′} = T_{a′}T_{a} for all a, a′ ∈ s. We then have the next result.

Proposition 15. Let k ⩾ 1 be an integer and let T_{a} ∈ L′ for a ∈ U. Then i) T_{a} ∈ B(s_{α}(Δ^{k})) if and only if T_{a} ∈ S_{α}; ii) T_{a} ∈ B(sα^{∘}(Δ^{k})) if and only if T_{a} ∈ S_{α} and

lim_{n→∞} \frac{a_{n−m+1}}{α_{n}} = 0 for all m;  (19)

iii) T_{a} ∈ B(s_{α}^{(c)}(Δ^{k})) if and only if T_{a} ∈ S_{α},

lim_{n→∞} \frac{a_{n−m+1}}{α_{n}} = l_{m} for all m  (20)

and

lim_{n→∞}( ∑_{m=1}^{n} a_{n−m+1} \frac{α_{m}}{α_{n}} ) = l.  (21)

iv) ∥T_{a}∥_{B(s_{α}^{∘}(Δ^{k}))}^{∗} = ∥T_{a}∥_{B(s_{α}(Δ^{k}))}^{∗} ⩽ ∥T_{a}∥_{S_{α}}and ∥T_{a}∥_{B(s_{α}(Δ^{k}))}^{∗} = ∥T_{a}∥_{S_{α}} for α ∈ Γ.

v) Let 1 ⩽ p < ∞. Then T_{a} ∈ B(l_{p}(α)(Δ^{k})) if and only if T_{a} ∈ (l_{p}(α), l_{p}(α)). Proof. i) The set s_{α}(Δ^{k}) is a BK space by Lemma 3 in which T = Δ^{k}. So

by Remark 1, T_{a} ∈ B(s_{α}(Δ^{k})) if and only if T_{a} ∈ (s_{α}(Δ^{k}), s_{α}(Δ^{k})). Thus the condition T_{a} ∈ B(s_{α}(Δ^{k})) is equivalent to T_{a}(Σ^{k}x) ∈ s_{α}(Δ^{k}) for all x ∈ s_{α}. So T_{a} ∈ B(s_{α}(Δ^{k})) if and only if

Δ^{k}[T_{a}(Σ^{k}x)] ∈ s_{α} for all x ∈ s_{α}.

Since Δ^{k}, Σ^{k} ∈ L′, we have

Δ^{k}[T_{a}(Σ^{k}x)] = (Δ^{k}T_{a}Σ^{k})x = (T_{a}(ΔΣ)^{k})x = T_{a}x for all x ∈ s_{α}.

We conclude that T_{a} ∈ B(s_{α}(Δ^{k})) if and only if T_{a} ∈ (s_{α}, s_{α}), that is T_{a} ∈ S_{α}. ii) As above sα^{∘}(Δ^{k}) is a BK space and T_{a} ∈ B(sα^{∘}(Δ^{k})) if and only if T_{a} ∈ (sα^{∘}, sα^{∘}). Now we have D_{1/α}T_{a}D_{α} = (ξ_{nm})_{n,m⩾1}, with ξ_{nm} = a_{n−m+1}α_{m}/α_{n} for m ⩽ n and ξ_{nm} = 0 otherwise. Thus we get D_{1/α}T_{a}D_{α} ∈ (c_{0}, c_{0}) if and only if T_{a} ∈ S_{α} and (19) holds. We get iii) by a similar argument. So T_{a} ∈ B(sα^{∘}(Δ^{k})) if and only if

T_{a} ∈ (sα^{∘}, sα^{∘}) and T_{a} ∈ B(s_{α}^{(c)}(Δ^{k})) if and only if T_{a} ∈ (s_{α}^{(c)}, s_{α}^{(c)}).

iv) Let us show that ∥T_{a}∥_{B(s_{α}(Δ^{k}))}^{∗} ⩽ ∥T_{a}∥_{S_{α}}. We have

∥T_{a}∥_{B(s_{α}(Δ^{k}))}^{∗} = sup_{x≠0} (\frac{∥T_{a}x∥_{s_{α}(Δ^{k})}}{∥x∥_{s_{α}(Δ^{k})}} ) < ∞.

Since Δ^{k} ∈ L′, we have Δ^{k}T_{a} = T_{a}Δ^{k} and for every x ∈ s_{α}(Δ^{k})

∥T_{a}x∥_{s_{α}(Δ^{k})} = ∥Δ^{k}(T_{a}x)∥_{s_{α}} = ∥T_{a}(Δ^{k}x)∥_{s_{α}},

so

∥T_{a}x∥_{s_{α}(Δ^{k})} ⩽ ∥T_{a}∥_{S_{α}}∥Δ^{k}x∥_{s_{α}} = ∥T_{a}∥_{S_{α}}∥x∥_{s_{α}(Δ^{k})}

and

∥T_{a}∥_{B(s_{α}(Δ^{k}))}^{∗} ⩽ ∥T_{a}∥_{S_{α}}.

Now it can easily be seen that ∥T_{a}∥_{B(s_{α}(Δ^{k}))}^{∗} = ∥T_{a}∥_{B(s_{α}^{∘}(Δ^{k}))}^{∗}. In the case when α ∈ Γ we have s_{α}(Δ^{k}) = s_{α} and by Proposition 7 i) we get

∥T_{a}∥_{B(s_{α}(Δ^{k}))}^{∗} = ∥T_{a}∥_{S_{α}}.

v) By Lemma 3 where T = Δ^{k}, we see that l_{p}(α)(Δ^{k}) is a BK space. We have T_{a} ∈ B(l_{p}(α)(Δ^{k})) if and only if T_{a} ∈ (l_{p}(α)(Δ^{k}), l_{p}(α)(Δ^{k})). Now T_{a} ∈ (l_{p}(α)(Δ^{k}), l_{p}(α)(Δ^{k})) if and only if Δ^{k}(T_{a}Σ^{k}x) = T_{a}x ∈ l_{p}(α) for all x ∈ l_{p}(α). This means that T_{a} ∈ (l_{p}(α), l_{p}(α)).

Proposition 16. Let T_{a} ∈ L′ with a_{1} ≠ 0. Then

i) If (1/a_{1})T_{a} ∈ Γ_{α} then for any given b ∈ s_{α}(Δ^{k}) the equation T_{a}x = b admits in s_{α}(Δ^{k}) a unique solution

x = T_{a}^{−1}b.  (22)

ii) If (1/a_{1})T_{a} ∈ Γ_{α} and (19) holds, then for any given b ∈ sα^{∘}(Δ^{k}) the equation T_{a}x = b admits in sα^{∘}(Δ^{k}) a unique solution given by (22).

iii) If (1/a_{1})T_{a} ∈ Γ_{α}, (20) and (21) hold, then for any given b ∈ s_{α}^{(c)}(Δ^{k}) the equation T_{a}x = b admits in s_{α}^{(c)}(Δ^{k}) a unique solution given by (22).

iv) a) If (1/a_{1})T_{a}^{t} ∈ Γ_{1/α}, then for any given b ∈ l_{1}(α)(Δ^{k}) the equation T_{a}x = b admits in l_{1}(α)(Δ^{k}) a unique solution given by (22).

b) Let 1 < p < ∞ and assume that (1/a_{1})T_{a} ∈ Γ′_{p,α}. Then for any given b ∈ l_{p}(α)(Δ^{k}) the equation T_{a}x = b admits in l_{p}(α)(Δ^{k}) a unique solution given by (22).

Proof. i) Since T_{a} ∈ L the equation T_{a}x = b is equivalent to x = T_{a}^{−1}b. Now the condition (1/a_{1})T_{a} ∈ Γ_{α} implies T_{a}^{−1} ∈ S_{α} and by Proposition 15 i), this means that T_{a}^{−1} ∈ B(s_{α}(Δ^{k})). We conclude T_{a}^{−1}b ∈ s_{α}(Δ^{k}).

ii) By Proposition 15 ii) we have T_{a} ∈ B(sα^{∘}(Δ^{k})) since T_{a} ∈ (sα^{∘}, sα^{∘}); and the condition (1/a_{1})T_{a} ∈ Γ_{α} and Proposition 15 iv) imply

∥I − \frac{1}{t_{11}} T_{a}∥_{B(s_{α}(Δ^{k}))}^{∗} = ∥I − \frac{1}{t_{11}} T_{a}∥_{B(s_{α}^{∘}(Δ^{k}))}^{∗} ⩽ ∥I − \frac{1}{t_{11}} T_{a}∥_{S_{α}} < 1.

Then T_{a}^{−1} ∈ B(sα^{∘}(Δ^{k})) and the unique solution of the equation T_{a}x = b for b ∈ sα^{∘}(Δ^{k}) is x = T_{a}^{−1}b ∈ sα^{∘}(Δ^{k}). We get iii) from the characterization of (s_{α}^{(c)}, s_{α}^{(c)}) and reasoning as above. iv) a) is a direct consequence of Corollary 9 ii) a) and Proposition 15 v). iv) b) Here the condition (1/a_{1})T_{a} ∈ Γ′_{p,α} implies T_{a}^{−1} ∈ (l_{p}(α), l_{p}(α)), since B(l_{p}(α)) = (l_{p}(α), l_{p}(α)). Then from Proposition 15 v) T_{a}^{−1} ∈ (l_{p}(α)(Δ^{k}), l_{p}(α)(Δ^{k})) and for all b ∈ l_{p}(α)(Δ^{k}) the unique solution x = T_{a}^{−1}b of the equation T_{a}x = b belongs to l_{p}(α)(Δ^{k}).

Remark 2. The conditions (1/a_{1})T_{a} ∈ Γ_{α} and (1/a_{1})T_{a}^{t} ∈ Γ_{1/α} are equivalent to

sup_{n⩾2}( ∑_{m=1}^{n−1} |a_{n−m+1}|\frac{α_{m}}{α_{n}} ) < |a_{1}| and sup_{n⩾1}( ∑_{m=n+1}^{∞} |a_{m−n+1}|\frac{α_{n}}{α_{m}} ) < |a_{1}|;

respectively. Note that the condition (1/a_{1})T_{a} ∈ Γ′_{p,α} means

∑_{n=2}^{∞}( ∑_{m=1}^{n−1} |a_{n−m+1}|^{p/(p−1)})^{p−1} < |a_{1}|^{p/(p−1)}.

Furthermore, it can be verified from Proposition 15 that if T_{a} ∈ L′, then i) T_{a} ∈ B(c_{0}(Δ^{k})) if and only if T_{a} ∈ S_{1} and a ∈ c_{0};

ii) T_{a} ∈ B(c(Δ^{k})) if and only if T_{a} ∈ S_{1} and a ∈ c_{s}.

5. Application to infinite tridiagonal matrices

In this section we will consider infinite tridiagonal matrices. These matrices are used in many applications, let us cite for instance the case of continued fractions [9], or the finite differences method, see [20]. We deal with some properties of the matrix map M(γ, a, η) between particular sequence spaces. Then we will explicitly calculate the inverse of M(γ, a, η).

5.1. Properties of infinite matrices considered as operators in certain BK spaces

Let γ = (γ_{n})_{n⩾1}, η = (η_{n})_{n⩾1}, a′ = (a_{n})_{n⩾1} be sequences with a′ ∈ U. Consider the infinite tridiagonal matrix

M(γ, a, η) = ( \begin{matrix}a_{1} & η_{1} \\ γ_{2} & a_{2} & η_{2} & & 0 \\ & . & . & . \\ 0 & & γ_{n} & a_{n} & η_{n} \\ & & & . & . \end{matrix} ) .

We then have

Proposition 17. Assume that D_{1/a}M(γ, a, η) ∈ Γ_{α}, that is

sup_{n⩾1} [\frac{1}{a_{n}} (|γ_{n}|\frac{α_{n−1}}{α_{n}} + |η_{n}|\frac{α_{n+1}}{α_{n}} )] < 1.

Then i) M(γ, a, η) is bijective from s_{α}into s_{|a|α} and M(γ, a, η)^{−1} ∈ (s_{|a|α}, s_{α});

ii) M(γ, a, η) is bijective from sα^{∘} into s|a|α^{∘} and M(γ, a, η)^{−1} ∈ (s|a|α^{∘}, sα^{∘});

iii) if

lim_{n→∞} [\frac{1}{a_{n}} (γ_{n} \frac{α_{n−1}}{α_{n}} + η_{n} \frac{α_{n+1}}{α_{n}} )] = l ≠ 0,

then M(γ, a, η) is bijective from s_{α}^{(c)} into s_{|a|α}^{(c)} and M(γ, a, η)^{−1} ∈ (s_{|a|α}^{(c)}, s_{α}^{(c)}).

iv) Let p ⩾ 1 be a real. If K̃_{p,α} = K_{1} + K_{2} < 1 with

K_{1} = sup_{n⩾1} (|\frac{γ_{n}}{a_{n}} | \frac{α_{n−1}}{α_{n}} ) and K_{2} = sup_{n⩾1} (|\frac{η_{n}}{a_{n}} | \frac{α_{n+1}}{α_{n}} ) ,

then M(γ, a, η) is bijective from l_{p}(α) into l_{p}(|a|α) and M(γ, a, η)^{−1} ∈ (l_{p}(|a|α), l_{p}(α)).

Proof. i), ii) and iii) are direct consequences of Corollary 9. iv) We get

∥(I − D_{1/a}M(γ, a, η))x∥_{l_{p}(α)} = ( ∑_{n=1}^{∞} \frac{1}{α_{n}^{p}} |\frac{γ_{n}}{a_{n}} x_{n−1} + \frac{η_{n}}{a_{n}} x_{n+1}|^{p})^{1/p}

= ( ∑_{n=1}^{∞}|\frac{γ_{n}}{a_{n}} \frac{α_{n−1}}{α_{n}} \frac{x_{n−1}}{α_{n−1}} + \frac{η_{n}}{a_{n}} \frac{α_{n+1}}{α_{n}} \frac{x_{n+1}}{α_{n+1}} |^{p})^{1/p}

⩽ ( ∑_{n=1}^{∞}(K_{1}|\frac{x_{n−1}}{α_{n−1}} | + K_{2}|\frac{x_{n+1}}{α_{n+1}} |)^{p})^{1/p};

Applying Minkowski’s inequality we get

( ∑_{n=1}^{∞}(K_{1}|\frac{x_{n−1}}{α_{n−1}} | + K_{2}|\frac{x_{n+1}}{α_{n+1}} |)^{p})^{1/p} ⩽ K_{1}( ∑_{n=1}^{∞}|\frac{x_{n−1}}{α_{n−1}} |^{p})^{1/p} + K_{2}( ∑_{n=1}^{∞}|\frac{x_{n+1}}{α_{n+1}} |^{p})^{1/p}
⩽ (K_{1} + K_{2})∥x∥_{l_{p}(α)}.

we conclude that

∥I − D_{1/a}M(γ, a, η)∥_{B(l_{p}(α))}^{∗} ⩽ (K_{1} + K_{2}) < 1.

So D_{1/a}M(γ, a, η) is invertible in B(l_{p}(α)) and A = D_{a}(D_{1/a}M(γ, a, η)) is bijective from l_{p}(α) into l_{p}(|a|α). Since B(l_{p}(α)) = (l_{p}(α), l_{p}(α)), we conclude

[D_{1/a}M(γ, a, η)]^{−1} ∈ (l_{p}(α), l_{p}(α))

and M(γ, a, η)^{−1} ∈ (l_{p}(|a|α), l_{p}(α)).

We deduce the next corollary.

Corollary 18. If K̃_{1,α} < 1, then M(γ, a, η) is bijective from l_{1}(α) to l_{1}(|a|α) and bijective from s_{α} to s_{|a|α}.

Proof. First taking p = 1 in Proposition 17 iv), we deduce that A is bijective from l_{1}(α) to l_{1}(|a|α). Then from

∥I − D_{1/a}M(γ, a, η)∥_{S_{α}} = K̃′_{α} ⩽ K̃_{1,α} < 1,

we conclude that M(γ, a, η) is bijective from s_{α} to s_{|a|α}.

Remark 3. Note that in the case when p = 1, the condition

∥I − [D_{1/a}M(γ, a, η)]^{t}∥_{S_{α}} = sup_{n⩾1} (|\frac{γ_{n+1}}{a_{n+1}} | \frac{α_{n+1}}{α_{n}} + |\frac{η_{n−1}}{a_{n−1}} | \frac{α_{n−1}}{α_{n}} ) < 1,

also implies M(γ, a, η) is bijective from l_{1}(α) to l_{1}(|a|α).

5.2. The inverse of an infinite tridiagonal matrix.

In this subsection, among other things, we are interested in the calculation of the inverse of M(γ, η) defined in Example 14.

For this we need to recall the next results. We can associate with any power series f(z) = ∑_{k=0}^{∞} a_{k}z^{k} defined in the open disk |z| < R the upper triangular infinite matrix A = φ(f) ∈ ⋃_{0<r<R} S_{r} defined by

φ(f) = ( \begin{matrix}a_{0} & a_{1} & a_{2} & . \\ & a_{0} & a_{1} & . \\ O & & a_{0} & . \\ & & & . \end{matrix} ) ,

(see [6]). Practically we shall write φ[f(z)] instead of φ(f). We have

Lemma 19. i) The map φ: f ↦ A is an isomorphism from the algebra of the power series defined in |z| < R, into the algebra of the corresponding matrices Ā.

ii) Let f(z) = ∑_{k=0}^{∞} a_{k}z^{k}, with a_{0} ≠ 0, and assume that 1/f(z) = ∑_{k=0}^{∞} a′_{k}z^{k} admits R′ > 0 as radius of convergence. We then have

φ (\frac{1}{f} ) = [φ(f)]^{−1} ∈ ⋃_{0<r<R′} S_{r}.

From the previous results we get

Proposition 20. Let γ, η be reals with 0 < γ + η < 1. Then

i) M(γ, η) : x ⟼ M(γ, η)x is bijective from χ into itself, for χ ∈ {s_{1}, c_{0}, c}. ii) a) Let χ be either one of the sets s_{1}, or c_{0}, or c and put

u = (1 − \sqrt{1 − 4γη})/2γ and v = (1 − \sqrt{1 − 4γη})/2η.

Then for any given b ∈ χ the equation M(γ, η)x = b admits a unique solution x^{∘} = (xn^{∘})_{n⩾1} in χ given by

xn^{∘} = (\frac{uv + 1}{uv − 1}) (−1)^{n}v^{n} ∑_{m=1}^{∞}[1 − (uv)^{−l}](−1)^{m}u^{m}b_{m} for all n.  (23)

with l = min(n, m).

b) The inverse [M(γ, η)]^{−1} = (a′_{nm})_{n,m⩾1} is given by

a′_{nm} = (\frac{uv + 1}{uv − 1}) (−1)^{n+m}v^{n−m}[(uv)^{l} − 1] for all n, m ⩾ 1 and l = min(n, m).

(24)

Proof. i) We have ∥I − M(γ, η)∥_{S_{1}} = γ + η < 1, so M(γ, η) ∈ Γ_{1} and we conclude by Corollary 9 i) a) in which α = e. ii) Let u, v be reals with 0 < u < 1, 0 < v < 1 and consider the matrices

Δ_{u}^{+} = ( \begin{matrix}1 & u & & O \\ & 1 & u \\ O & & . & . \end{matrix} ) and Δ_{v} = ( \begin{matrix}1 & & & O \\ v & 1 \\ O & v & 1 \\ & & . & . \end{matrix} ) .

By a short calculation we get

\frac{1}{1 + uv} Δ_{u}^{+}Δ_{v} = ( \begin{matrix}1 & \frac{u}{1+uv} & & O \\ \frac{v}{1+uv} & 1 & \frac{u}{1+uv} \\ O & \frac{v}{1+uv} & 1 & \frac{u}{1+uv} \\ & & . & . \end{matrix} ) .

Thus the identity M(γ, η) = \frac{1}{1+uv} Δ_{u}^{+}Δ_{v} is equivalent to η = u/(1 + uv) and γ = v/(1 + uv). Putting ξ = 1 + uv, we get η = u/ξ and γ = v/ξ, so ξ = 1 + γηξ^{2}. Since 0 < γ + η < 1, we have 4γη < 1 and ξ = (1 ± \sqrt{1 − 4γη})/2γη and the condition |u|, |v| < 1 implies u = (1 − \sqrt{1 − 4γη})/2γ and v = (1 − \sqrt{1 − 4γη})/2η. Then

∥I − \frac{1}{1 + uv} Δ_{u}^{+}Δ_{v}∥_{S_{1}} = γ + η = \frac{u + v}{1 + uv} < 1.

Furthermore we have ∥I − Δ_{u}^{+}∥_{S_{1}} = u < 1 and ∥I − Δ_{v}∥_{S_{1}} = v < 1. So Δ_{u}^{+} and Δ_{v} are invertible in S_{1} and by Lemma 19, we have

(Δ_{u}^{+})^{−1} = φ(1/(1 + uz)) = φ( ∑_{k=0}^{∞}(−1)^{k}u^{k}z^{k}),
(Δ_{v})^{−1} = [φ( ∑_{k=0}^{∞}(−1)^{k}v^{k}z^{k})]^{t},

with |uz|, |vz| < 1. This means that

(Δ_{u}^{+})^{−1} = Σ_{u}^{+} = ( \begin{matrix}1 & −u & u^{2} & . \\ & 1 & −u & u^{2} \\ O & & . & . \\ & & & . \end{matrix} ) and (Δ_{v})^{−1} = Σ_{v} = ( \begin{matrix}1 & & O \\ −v & 1 \\ v^{2} & −v & 1 \\ . & . & . & . \end{matrix} ) .

Then (\frac{1}{1+uv} Δ_{u}^{+}Δ_{v})^{−1} = (1 + uv)Σ_{v}Σ_{u}^{+}. For any given b ∈ s_{1}, we successively get

Σ_{u}^{+}b = ( ∑_{k=0}^{∞}(−1)^{k}u^{k}b_{n+k})_{n⩾1},
Σ_{v}(Σ_{u}^{+}b) = ( ∑_{s=1}^{n}(−1)^{n−s}v^{n−s}( ∑_{k=0}^{∞}(−1)^{k}u^{k}b_{s+k}))_{n⩾1}

and the unique solution is given by

xn^{∘} = (1 + uv) ∑_{s=1}^{n} ∑_{k=0}^{∞}(−1)^{n−s+k}v^{n−s}u^{k}b_{s+k}.  (25)

Hence writing

xn^{∘} = (1 + uv)(−v)^{n}σ_{n} with σ_{n} = ∑_{s=1}^{n}(uv)^{−s} ∑_{k=0}^{∞}(−u)^{k+s}b_{s+k},

putting l = min(n, m), we get

σ_{n} = ∑_{s=1}^{n}(uv)^{−s} ∑_{m=0}^{∞}(−u)^{m}b_{m}

= ∑_{s=1}^{n}(uv)^{−s} ∑_{m=s}^{n}(−u)^{m}b_{m} + ∑_{s=1}^{n}(uv)^{−s} ∑_{m=n+1}^{∞}(−u)^{m}b_{m}

= ∑_{m=1}^{∞}(−u)^{m}b_{m} ∑_{s=1}^{m}(uv)^{−s} + ∑_{m=n+1}^{∞}(−u)^{m}b_{m} ∑_{s=1}^{n}(uv)^{−s}

= ∑_{m=1}^{∞}(−u)^{m}b_{m} ∑_{s=1}^{l}(uv)^{−s} = ∑_{m=1}^{∞}(−u)^{m}b_{m} \frac{(uv)^{−1}−(uv)^{−l−1}}{1−(uv)^{−1}}

= \frac{1}{uv − 1} ∑_{m=1}^{∞}(−u)^{m}b_{m}(1 − (uv)^{−l}),

that is (23) and ii) a) holds. ii) b) is a direct consequence of the identity xn^{∘} = ∑_{m=1}^{∞} a′_{nm}b_{m} for all n, where M^{−1} = (a′_{nm})_{n,m≥1}.

Acknowledgements. The author thanks the referee for his contribution to improve some results and the presentation of the paper.

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60 B. de Malafosse (received 02.07.2004, in revised form 03.12.2004) LMAH Université du Havre, BP 4006 I.U.T Le Havre, 76610 Le Havre, France

E-mail: bdemalaf@wanadoo.fr