Eberhard Malkowsky
and
Vladimir Rakočević
AN INTRODUCTION INTO
THE THEORY OF SEQUENCE SPACES
AND MEASURES OF NONCOMPACTNESS
CONTENTS
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1. FK Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1. Linear metric and paranormed spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2. Introduction into the theory of FK spaces . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3. Matrix transformations into l∞ , c and c0 . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4. The α– and β–duals of sets of sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5. The continuous duals of the classical sequence spaces . . . . . . . . . . . . . . .
1.6. Matrix transformations between some classical sequence spaces . . . . .
145
147
147
151
154
156
159
160
2. Measures of noncompactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2. The Kuratowski measure of noncompactness . . . . . . . . . . . . . . . . . . . . . . .
2.3. The Hausdorff measure of noncompactness . . . . . . . . . . . . . . . . . . . . . . . . .
2.4. Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
161
161
164
168
174
3. Matrix domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1. Ordinary and strong matrix domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2. Matrix transformations into matrix domains . . . . . . . . . . . . . . . . . . . . . . . .
3.3. Bounded and convergent difference sequences of order m . . . . . . . . . . .
3.4. Matrix transformations in the spaces c0 (∆(m) ), c(∆(m) ), l∞ (∆(m) )
and their measures of noncompactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5. Sequences of weighted means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6. Matrix transformations in the spaces (N, q)0 , (N, q) and (N, q)∞
and their measures of noncompactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.7. Spaces of strongly summable and convergent sequences . . . . . . . . . . . . .
3.8. Further results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
177
178
180
182
190
197
199
207
217
4. Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
4.1. Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
4.2. The closed graph theorem and the Banach–Steinhaus theorem . . . . . . 228
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
Preface
This paper gives a self–contained, comprehensive treatment of the theories of
sequence spaces and measures of noncompactness, as well as a survey of some of
the authors’ recent research results in these fields. It contains subjects of lectures
at the universities of Niš, Novi Sad and Belgrade, Giessen (Germany) and Irbid
(Jordan), and talks given by the authors at various international conferences in the
Czech Republic, Germany, Hungary, India, Italy, Jordan, Poland and Yugoslavia.
For the first time, methods from the fields of summability, in particular of
sequence spaces and matrix transformations on one hand, and of measures of noncompactness on the other are successfully linked on a large scale to obtain necessary
and sufficient conditions for matrix maps between certain sequence spaces of a general class to be compact operators. The original idea for research in this field dates
back to the classical paper of L. W. Cohen and N. Dunford [10]. In this paper they
gave necessary and sufficient conditions for matrix transformations from l1 to lp , lp
to c0 and lp to l1 , and found the norm of these transformations. Furthermore they
established necessary and sufficient conditions for these operators to be compact.
Although the concept of measure of noncompactness is not explicitly mentioned in
their paper, their studies and techniques are very closely related to our research.
These notes are addressed to both experts and nonexperts with an interest in
getting acquainted with sequence spaces and measures of noncompactness. They
could also be used as a guideline for research and teaching at graduate and post
graduate levels.
Sections 1 and 2 deal with the necessary basic concepts and results of the
theory of FK spaces, their duals, matrix transformations and measures of noncompactness. Although most of the results presented are well known and can be found,
for instance, in [105, 108, 107, 91] concerning Section 1, and in [1, 7, 86] concerning
Section 2, proofs are given in almost all cases to make the paper self–contained.
In Section 3, the authors give their own research results and apply the methods and results of the first two chapters to characterize matrix transformations
between sequence spaces closely related to various concepts of summability, such as
ordinary and strong summability, spaces of difference sequences of higher order and
of strongly convergent and bounded sequences. Finally they apply the Hausdorff
measure of noncompactness to give necessary and sufficient conditions for a matrix
map between these spaces to be a compact operator.
Although there is a very wide range of problems for further research related to the presented topics, only the most closely related and possibly interesting
will be mentioned here. We hope that the results presented here will be a useful
introduction to further studies in the following fields. Concerning measures of noncompactness it seems most interesting to study when operators between sequence
spaces are strictly singular [45, 104], Fredholm or semi–Fredholm [16, 17, 19, 21].
Results in this direction could also be applied in the perturbation theory of Fredholm and semi–Fredholm operators. The authors’ research is also connected with
A. Wilansky’s results [106], and will be in this direction. Concerning the theory
146
Malkowsky and Rakočević
of sequence spaces and matrix transformations it would be interesting to extend
the author’s results to mixed normed spaces [22, 23, 35, 36, 38]. Furthermore, a
generalization should be considered of the spaces lp , w0p to l(p), w0 (p) by replacing
the constant p in the exponent by a sequence p = (pk )∞
k=0 of positive reals [95, 43,
44, 49, 50, 52, 53, 55, 56, 48, 90, 61, 57, 58, 60, 62].
Finally, matrix domains of general infinite matrices should be considered instead of matrix domains of triangles.
Acknowledgements
The authors express their sincere gratitude to Academician Bogoljub Stanković
and Professor Stevan Pilipović for their interest in our work and their encouragement to present our results, to Professor Ljiljana Gajić for her thorough proofreading of our manuscript and to Professor Ivan Jovanović for his help in preparing
the final version of the manuscript. They also made many valuable comments and
suggestions which the authors are grateful for.
The work of the first author was completed at the University of Niš under the
DAAD (Deutscher Akademischer Austauschdienst) grant 9011031028.
The work of the second author was supported by Science Fund of Serbia, grant
04M03, through Matematički institut.
Theory of sequence spaces
147
1. FK spaces
In this section, we shall give a short introduction into the general theory of FK
spaces and apply the results to characterize matrix transformations between the
classical sequence spaces. For further studies the reader is referred to [108, 91, 37,
114, 54, 111, 112, 113]. Results closely related to our studies can be found in [33].
For a comprehensive survey on results on matrix transformations we recommend
[97].
1.1. Linear metric and paranormed spaces. In this subsection, we shall
introduce the concepts of linear metric and paranormed spaces. They will play an
important role in our studies of sequence spaces.
The concept of a linear space involves an algebraic structure given by the
definition of two operations, namely the sum of any two of its vectors and the
product of any scalar with any vector. On the other hand a topological structure
of a set may be given by a metric. If a set is both a linear and metric space, then
it will be natural to require the algebraic operations to be continuous with respect
to the metric.
Let X be a linear space and d a metric on X. Then (X, d), or X for short,
is said to be a linear metric space, if the algebraic operations on X are continuous
functions. A complete linear metric space is said to be a Fréchet space (cf. [105,
Definition 5.3.2, p. 78]). (Unfortunately this terminology is not universally agreed
on. Some authors call a complete linear metric space an F space and a locally
convex F space a Fréchet space (see e.g. [92, p. 8], [40, p. 208] and [96, p. 8]),
which Wilansky calls an F –space. We shall follow Wilansky’s terminology in [105,
107, 108]. The continuity of the algebraic operations of a linear metric space (X, d)
means the following: If (xn ) and (yn ) are two sequences in X and (λn ) is a sequence
of scalars with xn → x, yn → y and λn → λ (n → ∞), then xn + yn → x + y
and λn xn → λx (n → ∞). This means that d(xn , x), d(yn , y) → 0 and λn → λ
(n → ∞) together imply d(xn + yn , x + y) → 0 and d(λn xn , λx) → 0 (n → ∞). By
the completeness of a metric space we mean that every Cauchy sequence converges.
The concept of paranorm is closely related to linear metric spaces. It is a
generalization of that of absolute value. The paranorm of a vector x may be thought
of as the distance from x to the origin 0.
Definition 1.1. Let X be a linear space. A function p : X 7→ R is called
paranorm, if
(P.1) p(0) = 0
(P.2) p(x) ≥ 0 for all x ∈ X
(P.3) p(−x) = p(x) for all x ∈ X
(P.4) p(x + y) ≤ p(x) + p(y) for all x, y ∈ X (triangle inequality)
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Malkowsky and Rakočević
(P.5) if (λn ) is a sequence of scalars with λn → λ (n → ∞) and (xn ) is a
sequence of vectors with p(xn −x) → 0 (n → ∞), then p(λn xn −λx) → 0 (n → ∞)
(continuity of multiplication by scalars).
If p is a paranorm on X, then (X, p), or X for short, is called a paranormed
space. A paranorm p for which p(x) = 0 implies x = 0 is called total. For any
two paranorms p and q, p is called stronger than q if, whenever (xn ) is a sequence
such that p(xn ) → 0 (n → ∞), then also q(xn ) → 0 (n → ∞). If p is stronger
than q, then q is said to be weaker than p. If p is stronger than q and q is stronger
than p, then p and q are called equivalent. If p is stronger than q, but p and q are
not equivalent, then p is said to be strictly stronger than q, and q is called strictly
weaker than p.
It is easy to see that every totally paranormed space is a linear metric space.
The converse is also true. The metric of any linear metric space is given by some
total paranorm (cf. [105, Theorem 10.4.2, p. 183]). A sequence of paranorms may
be used to define a paranorm.
Theorem 1.2. Let (pk )∞
k=1 be a sequence of paranorms on a linear space X.
We define the so-called Fréchet combination of (pk ) by
(1.1)
p(x) =
∞
X
1 pk (x)
2k 1 + pk (x)
n=0
for all x ∈ X.
Then:
(a) p is a paranorm on X and satisfies
(1.2)
p(xn ) → 0 (n → ∞) if and only if pk (xn ) → 0 (n → ∞)
for each k;
(b) p is the weakest paranorm which is stronger than every pk ;
(c) p is total if and only if every pk is total.
Proof. (a) Conditions (P.1), (P.2) and (P.3) in Definition 1.1 are obvious,
since every pk is a paranorm. To prove condition (P.4), we observe that, for all
reals a and b with 0 ≤ a ≤ b, we have a(1 + b) = a + ab ≤ b + ab = b(1 + a) and so
a/(1 + a) ≤ b/(1 + b). Applying this with 0 ≤ a = pk (x + y) ≤ pk (a) + pk (y) = b,
we conclude
pk (x + y)
pk (x) + pk (y)
pk (x)
pk (y)
≤
≤
+
1 + pk (x + y)
1 + pk (x) + pk (y)
1 + pk (x) 1 + pk (y)
for all k,
and from this p(x + y) ≤ p(x) + p(y). To prove the statement in (1.2), we first
assume pk (xn ) → 0 (n → ∞) for each k. Since
0≤
pk (xn )
≤1
1 + pk (xn )
for all n, k
Theory of sequence spaces
and
P∞
k=1
149
1/2k converges, the series
∞
X
1 pk (xn )
2k 1 + pk (xn )
k=1
converges uniformly in n. Thus limn→∞ p(xn ) = 0.
Conversely, we assume p(xn ) → 0, (n → ∞) and fix k. Then
1 pk (xn )
≤ p(xn )
2k 1 + pk (xn )
implies pk (xn ) ≤ 2k p(xn ) + 2k pk (xn )p(xn ). Since p(xn ) → 0 (n → ∞), it follows
that 2k p(xn ) < 1 for all sufficiently large n, hence pk (xn )(1 − 2k p(xn )) ≤ 2k p(xn )
for all sufficiently large n, and consequently
pk (xn ) ≤
2k p(xn )
1 − 2k p(xn )
for all sufficiently large n.
This implies pk (xn ) → 0 (n → ∞).
To prove condition (P.5), let λn → λ and p(xn − x) → 0 (n → ∞). By the
statement in (1.2), pk (xn − x) → 0 (n → ∞) for all k, and, since every pk is a
paranorm, this implies pk (λn xn − λx) → 0 (n → ∞) for all k. Now it follows from
the statement in (1.2) that p(λn xn − λx) → 0 (n → ∞).
(b) Let q be a paranorm which is stronger than every pk . Then q(xn ) → 0
(n → ∞) implies pk (xn ) → 0 (n → ∞) for all k, and, by the statement in (1.2),
p(xn ) → 0 (n → ∞). Thus q is stronger than p.
(c) Part (c) is trivial.
¤
Let us recall that a subset S of a linear space X is said to be absorbing if for
each x ∈ X there is ε > 0 such that λx ∈ S for all scalars λ with |λ| ≤ ε.
Remark 1.3. Let (X, p) be a paranormed space. Then the open neighbourhoods of 0, Nr (0) = {x ∈ X : p(x) < r}, are absorbing for all r > 0.
Proof. We assume that is Nr (0) is not absorbing for some r > 0. Then
there are x ∈ X and a sequence (λn )∞
n=0 of scalars with λn → 0 (n → ∞) and
λn x ∈
/ Nr (0) for all n = 0, 1, . . . . But this means p(λn x) ≥ r for all n contradicting
condition (P.5) in Definition 1.1.
¤
Example 1.4. The set C of complex numbers with the usual algebraic operations and p = | · |, the modulus, is a totally paranormed space. If we put
d(z, w) = |z − w| for all z, w ∈ C, then (C, d) is a Fréchet space.
By ω, we denote the set of all complex sequences x = (xk )∞
k=0 which becomes
∞
a linear space with x + y = (xk + yk )∞
k=0 and λx = (λxk )k=0 or all x, y ∈ ω and
λ ∈ C. As an immediate consequence of Theorem 1.2 and Example 1.4, we obtain
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Malkowsky and Rakočević
Theorem 1.5. The set ω is a Fréchet space with respect to the metric d
defined by
(1.3)
d(x, y) =
∞
X
1 |xk − yk |
2k 1 + |xk − yk |
for all x, y ∈ ω.
k=0
Furthermore convergence in (ω, d) and coordinatewise convergence are equivalent,
(n)
that is x(n) → x (n → ∞) in (ω, d) if and only if xk → xk (n → ∞) for every k.
Now we introduce the concept of a Schauder basis. For further studies on bases
we refer the reader to [46, 74].
Definition 1.6. A Schauder basis of a linear metric space X is a sequence
(bn ) of vectors
such that for each vector x ∈ P
is a unique sequence (λn ) of
P∞
Pthere
m
scalars with n=1 λn bn = x, that is limm→∞ n=1 λn bn = x.
For finite dimensional spaces, the concepts of Schauder and algebraic bases
coincide. In most cases of interest, however, the concepts differ. Every linear space
has an algebraic basis. But there are linear metric spaces without a Schauder basis,
as we shall see later in this subsection.
(n)
Example 1.7. For each n = 0, 1, . . . , let e(n) be the sequence with en = 1
(n)
More precisely,
and ek = 0 for k 6= n. Then (e(n) )∞
n=0 is a Schauder basis of ω.
P∞
∞
every sequence x = (xk )k=0 ∈ ω has a unique representation x = k=0 xk e(k) that
Pm
is limm→∞ x[m] = x for x[m] = k=0 xk e(k) , the m–section of x.
A metric space (X, d) is called separable if it has a countable dense set. That
means there is a countable set A ⊂ X such that for all ε > 0 and for all x ∈ X
there is an element a ∈ A with d(x, a) < ε.
Theorem 1.8. Every complex linear metric space X with Schauder basis is
separable.
Proof. Let (bn ) be a Schauder basis of X. For each m ∈ N, we put
Am =
½X
m
n=1
¾
∞
[
ρn bn : ρn ∈ Q + iQ (n = 1, 2, . . . , m) and A =
Am .
m=1
Then A is a countable set in X and it is easy to see that A is dense in X.
¤
Example 1.9. The set l∞ = {x ∈ ω : supk |xk | < ∞} of all bounded sequences
is a Banach space with kxk∞ = supk |xk | (x ∈ l∞ ) which has no Schauder basis.
Proof. The proof that (l∞ , k·k∞ ) is a Banach space is standard and left to the
reader. To show that l∞ has no Schauder basis, we show that l∞ is not separable
and apply Theorem 1.8. We assume that l∞ is separable. Then there is a countable
dense set A = {an : n = 0, 1, . . . } ⊂ l∞ . For every n, let Un = N1/3 (an ) = {x ∈
S∞
l∞ : kx − an k∞ < 1/3}. Since A ⊂ l∞ is dense, l∞ ⊂ n=0 Un . The set
©
ª
B = {0, 1}N0 = x ∈ ω : xk ∈ {0, 1} for all k = 0, 1, . . . ⊂ l∞
Theory of sequence spaces
151
is uncountable. Therefore there must be a set Um which contains at least two
distinct sequences x and x0 of B. Then
kx − x0 k∞ ≥ 1 and
kx − x0 k∞ ≤ kx − am k∞ + kam − x0 k∞ < 2/3,
a contradiction. Therefore l∞ cannot be separable.
¤
At the end of this subsection we study the so–called classical sequence spaces
©
ª
l∞ = x ∈ ω : sup |xk | < ∞ ,
k
©
ª
c = x ∈ ω : lim xk = l for some l ∈ C ,
k→∞
©
ª
c0 = x ∈ ω : lim xk = 0
k→∞
of all bounded, convergent and null sequences, and
½
lp =
x∈ω:
∞
X
¾
p
|xk | < ∞
for 1 ≤ p < ∞.
k=0
The following result gives the algebraic and topological properties of the sets l∞ ,
c, c0 and lp .
Theorem 1.10. (a) Each of the sets l∞ , c0 and c is a Banach space with
k · k∞ defined by kxk∞ = supk |xk |. Moreover |xk | ≤ kxk∞ for all k = 0, 1, . . . .
(b) The sets lp are Banach spaces for 1 ≤ p < ∞ with k · kp defined by
¡P∞
¢
p 1/p
kxkp =
. Moreover |xk | ≤ kxkp for all k = 0, 1, . . . .
k=0 |xk |
(c) The sequence (e(n) )∞
n=0 is a Schauder basis for each of the spaces c0 and lp
for 1 ≤ p < ∞. More precisely, every
x = (xn )∞
n=0 in any of these spaces
P∞ sequence
has a unique representation x = n=0 xn e(n) .
(d) Let e be the sequence with ek = 1 for all k = 0, 1, . . . . We put b(0) = e
and b(n) = e(n−1) for n = 1, 2, . . . . Then the sequence (b(n) )∞
n=0 is a Schauder basis
∞
∈
c
has
a
unique representation
for c. More
precisely,
every
sequence
x
=
(x
)
n
n=0
P∞
x = le + n=0 (xn − l)e(n) where l = l(x) = limn→∞ xn .
(e) The space l∞ has no Schauder basis.
Proof. Part (e) is Example 1.9. Parts (a) to (d) are standard and therefore left
to the reader. (The triangle inequality for k · kp follows by Minkowski’s inequality
(see appendix A.4.2).)
¤
1.2. Introduction into the theory of FK spaces. In this subsection, we shall
give an introduction into the general theory of FK spaces. It is the most powerful
tool for the solution of problems of various kinds in summability, in particular in
the characterization of matrix transformations between sequence spaces. Most of
the results of this subsection can be found in [108].
We saw in Theorem 1.5 that the set ω is a Fréchet space with the metric
d defined in (1.3) and that convergence in ω and coordinatewise convergence are
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Malkowsky and Rakočević
equivalent. Furthermore, by Theorem 1.10, the spaces l∞ , c0 , c and lp (1 ≤ p < ∞)
are Banach spaces with the norms k · k∞ and k · kp , and convergence in any one
of these spaces implies coordinatewise convergence by the inequalities in Theorem
1.10 parts (a) and (b). Thus the metric generated by these norms is stronger than
the metric of ω on them.
Definition 1.11. A Fréchet sequence space (X, dX ) is said to be an FK space
if its metric dX is stronger than the metric d|X of ω on X. A BK space is an FK
space which is a Banach space.
Remark 1.12. (a) Some authors include local convexity in the definition of
FK spaces. But much of the theory can be developed without local convexity.
(b) By definition, an FK space X is continuously embedded in ω, that is the
inclusion map ι : (X, dX ) 7→ (ω, d) defined by ι(x) = x (x ∈ X) is continuous. An
FK space X is a Fréchet sequence space with continuous coordinates Pk : X 7→ C
defined by Pk (x) = xk (k = 0, 1, . . . ) for all x ∈ X.
Example 1.13. The space ω is an FK space with its natural metric d. The
spaces l∞ , c, c0 and lp (1 ≤ p < ∞) are BK spaces with their natural norms.
Theorem 1.14. Let (X, dX ) be a Fréchet space, (Y, dY ) an FK space and
f : X 7→ Y a linear map. Then f : (X, dX ) 7→ (Y, d|Y ) is continuous if and only if
f : (X, dX ) 7→ (Y, dY ) is continuous.
Proof. First we assume that f : (X, dX ) 7→ (Y, dY ) is continuous. Since
Y is an FK space its metric dY is stronger than the metric d|Y of ω on Y . So
f : (X, dX ) 7→ (Y, d|Y ) is continuous.
Conversely we assume that f : (X, dX ) 7→ (Y, d|Y ) is continuous. Since (Y, d|Y )
is a Hausdorff space and f is continuous, the graph of f , graph(f ) = {(x, f (x)) :
x ∈ X}, is a closed set in (X, dX ) × (Y, d|Y ) by the closed graph lemma (see
appendix A.4.4), hence a closed set in (X, dX ) × (Y, dY ), since the FK metric dY
is stronger than d|Y . By the closed graph theorem (see appendix A.4.5), the map
f : (X, dX ) 7→ (Y, dY ) is continuous.
¤
Corollary 1.15. Let X be a Fréchet space, Y an FK space, f : X 7→ Y
a linear map and Pn the n-th co-ordinate, that is Pn (y) = yn (y ∈ Y ) for all
n = 0, 1, . . . . If each map Pn ◦ f : X 7→ C is continuous, so is f : X 7→ Y .
Proof. Since Pn ◦ f : X 7→ C is continuous for each n, the map f : X 7→ ω
is continuous by the equivalence of coordinatewise convergence and convergence in
ω. By Theorem 1.14, f : X 7→ Y is continuous.
¤
By φ we denote the set of all finite sequences that is of sequences that terminate
in zeros.
We shall frequently make use of the following result.
P∞
Remark 1.16. Let X ⊃ φ be an FK space and a ∈ ω. If the series k=0 ak xk
converges for each x ∈ X, then the linear functional fa : X 7→ C defined by
fa (x) =
∞
X
k=0
ak xk
for all x ∈ X
Theory of sequence spaces
153
is continuous.
Proof.PFor each n ∈ N0 , we define the linear functional fa,n : X 7→ C by
n
fa,n (x) =
k=0 ak xk for all x ∈ X. Since X is an FK space, the coordinates
Pk : X P
7→ C are continuous on X for all k = 0, 1, . . . , and so are the functionals
n
fa,n = k=0 ak Pk (n = 0, 1, . . . ). For each x ∈ X, fa (x) = limn→∞ fa,n (x) exists,
and so fa : X 7→ C is continuous by the Banach–Steinhaus theorem (see appendix
A.4.6).
¤
For the next result we shall need some notations.
Given any two subsets X and Y of ω and any infinite matrix A = (ank )∞
n,k=0
of complex numbers, we shall write An = (an,k )∞
k=0 for the sequence in the n-th
row of A,
∞
X
An (x) =
ank xk (x ∈ X) for all n = 0, 1, . . .
k=0
(provided the series converge) and
A(x) = (An (x))∞
n=0 .
Furthermore let (X, Y ) be the class of all matrices A that map X into Y , that is
for which the series An (x) converge for all x ∈ X and for all n, and A(x) ∈ Y for
all x ∈ X.
Theorem 1.17. Any matrix map between FK spaces is continuous.
Proof. Let X and Y be FK spaces, A ∈ (X, Y ) and the map fA : X 7→ Y
be defined by fA (x) = A(x) for all x ∈ X. Since the maps Pn ◦ fA : X 7→ C
are continuous for all n ∈ N0 by Remark 1.16, the linear map fA is continuous by
Corollary 1.15.
¤
Definition 1.18. An FK space X ⊃ φ has AK if, for every sequence x =
(xk )∞
k=0 ∈ X,
x=
∞
X
k=0
xk e(k) ,
that is
x[m] =
m
X
xk e(k) → x (m → ∞),
k=0
and X has AD if φ is dense in X. If an FK space has AK or AD we also say that
it is an AK or AD space.
Remark 1.19. Every AK space has AD. The converse is not true in general.
Proof. The first part is trivial, and the second part can be found in [108,
Example 5.2.5, p. 78].
¤
Example 1.20. The spaces ω, c0 and lp (1 ≤ p < ∞) all have AK by Example
1.7 and Theorem 1.10.
The FK metric of an FK space will turn out to be unique.
154
Malkowsky and Rakočević
Theorem 1.21. Let X and Y be FK spaces and X ⊂ Y . Then the metric
dX on X is stronger than the metric dY |X of Y on X. The metrics are equivalent
if and only if X is a closed subspace of Y . In particular, the metric of an FK space
is unique, this means there is at most one way to make a linear subspace of ω into
an FK space.
Proof. Let ι : (X, dX ) 7→ (Y, dY ) be the inclusion map. Since X is an FK
space, ι : (X, dX ) 7→ (Y, d|Y ) is continuous, and so is ι : (X, dX ) 7→ (Y, dY ) by
Theorem 1.14. Thus dX is stronger than dY |X . The uniqueness of an FK space
is shown in exactly the same way. Let X be closed in Y , then X becomes an FK
space with dY |X , and the uniqueness of an FK metric implies that dX and dY |X
are equivalent.
Conversely, if dX and dY |X are equivalent, then X is a complete subspace of
Y , hence a closed subspace of Y .
¤
Example 1.22. The BK spaces c0 and c are closed subspaces of l∞ . Thus
the BK norms on c0 , c and l∞ must be the same. The BK space l1 is a subspace
of l∞ which is not closed in l∞ . Thus its BK norm k · k1 is strictly stronger than
the BK norm k · k∞ on l∞ .
1.3. Matrix transformations into l∞ , c and c0 . In this subsection we shall
apply the results of Subsection 1.2 to characterize classes (X, Y ) where X is any FK
space and Y is any of the spaces l∞ , c and c0 . We shall need some more notations.
If X ⊂ ω is a linear metric space with respect to dX and a, x0 ∈ X, then we
shall write
Sδ [x0 ] = SX,δ [x0 ] = {x ∈ X : dX (x, x0 ) ≤ δ} (δ > 0)
¯
½¯X
¾
¯∞
¯
∗
∗
¯
¯
kakD = kakX,D = sup ¯
ak xk ¯ : x ∈ S1/D [0] (D > 0)
k=0
provided the expression on the right exists and isPfinite. By Remark 1.6, this is the
∞
case whenever X is an FK space and the series k=0 ak xk converge for all x ∈ X.
If X is a BK space we write
∗
kak =
kak∗X
¯
½¯X
¾
¯∞
¯
¯
= sup ¯
ak xk ¯¯ : kxk = 1 .
k=0
Let X and Y be two Fréchet spaces. By B(X, Y ) we denote the set of all continuous
linear operators L : X 7→ Y , and we write X 0 = B(X, C) for the set of all continuous
linear functionals on X, the set X 0 is called the continuous dual of X. If X and Y
are normed spaces and L ∈ B(X, Y ), then we write
(1.4)
kLk = sup{kL(x)k : kxk = 1}
for all L ∈ B(X, Y ).
for the operator norm of L; furthermore we write X ∗ for X 0 with the norm in (1.4),
that is kf k = sup{|f (x)| : kxk = 1} for all f ∈ X 0 .
155
Theory of sequence spaces
Let A be an infinite matrix, D a positive real and X an FK space. Then we
put
∗
MA,D
(X, l∞ ) = sup kAn k∗D
n
and, if X is a BK space, then we write
MA∗ (X, l∞ ) = sup kAn k∗ .
n
Theorem 1.23. Let X and Y be FK spaces.
(a) Then (X, Y ) ⊂ B(X, Y ), that is, every A ∈ (X, Y ) defines a linear operator
LA ∈ B(X, Y ) where LA (x) = A(x) for all x ∈ X.
(b) Then A ∈ (X, l∞ ) if and only if
∗
kAk∗D = MA,D
(X, l∞ ) < ∞
(1.5)
for some D > 0.
If X is a BK space and A ∈ (X, l∞ ), then kAk∗ = MA∗ (X, l∞ ) = kLA k < ∞.
(c) If (bk )∞
k=0 is a Schauder basis for X, and Y1 a closed FK space in Y , then
A ∈ (X, Y1 ) if and only if A ∈ (X, Y ) and A(b(k) ) ∈ Y1 for all k = 0, 1, . . . .
Proof. Part (a) is Theorem 1.17. (b) First we assume that condition (1.5)
holds. Then, for all x ∈ S1/D [0], the series An (x) (n = 0, 1, . . . ) converge and
A(x) ∈ l∞ . Since the set S1/D [0] is absorbing by Remark 1.3, we conclude that
An (x) converges for each x ∈ X and A(x) ∈ l∞ for all x ∈ X, hence A ∈ (X, l∞ ).
Conversely, we assume A ∈ (X, l∞ ). Then LA is continuous by part (a). Hence
there exist a neighbourhood N of 0 in X and a real D > 0 such that S1/D [0] ⊂ N
and kLA (x)k < 1 for all x ∈ N . This implies condition (1.5). If X is a BK space,
then LA ∈ B(X, Y ) implies
kA(x)k∞ = sup |An (x)| = kLA (x)k∞ ≤ kLA k
n
for all x ∈ X with kxk = 1.
Thus |An (x)| ≤ kLA for all n and for all x ∈ X with kxk = 1, and, by the definition
of the norm k · k∗ ,
(1.6)
kAk∗ = sup kAn k∗ ≤ kLA k.
n
Further, given ε > 0, there is x ∈ X with kxk = 1 such that kA(x)k∞ ≥ kLA k−ε/2,
and there is n(x) ∈ N0 with |An(x) (x)| > kA(x)k∞ − ε/2, consequently |An(x) (x)| ≥
kLA k − ε. Therefore kAk∗ = supn kAn k∗ ≥ kLA k − ε. Since ε > 0 was arbitrary,
kAk∗ ≥ kLA k, and, with (1.6), we have kAk∗ = kLA k.
(c) The necessity of the conditions for A ∈ (X, Y1 ) is trivial.
Conversely, if A ∈ (A, Y ), then LA ∈ B(X, Y ). Since Y1 is a closed subspace
of Y , the FK metrics of Y1 and Y are the same by Theorem 1.21. Consequently, if
S is any subset in Y1 , then, for its closures closY1 (S) and closY |Y1 (S) with respect
to the metrics dY1 and dY |Y1 , we have
(1.7)
closY1 (S) = closY |Y1 (S).
156
Malkowsky and Rakočević
©P m
ª
(k)
Let x ∈ X and SB =
: m ∈ N0 , λk ∈ C (k = 0, 1, . . . ) denote the
k=0 λk b
span of {b(k ) : k = 0, 1, . . . }. Since LA (b(k)¯) ∈ Y1 for all k = 0, 1, .. and the metrics
dY1 and dY |Y1 are equivalent, the map LA ¯SB : (X, dX ) 7→ (Y1 , dY1 ) is continuous.
Further, since (bk )∞
k=0
¯ is a basis of X, we have SB = X. Therefore, by (1.7) and
the continuity of LA ¯SB , we have
¡ ¯
¢
¡ ¯
¢
LA (X) = LA (SB) = closY1 LA ¯SB (SB) = closY |Y1 LA ¯SB (SB)
⊂ closY |Y1 (Y1 ) = Y1
Thus A(x) ∈ Y for all x ∈ X.
¤
1.4. The α– and β–duals of sets of sequences. In this subsection we shall
study the so–called α–, β– and continuous dual spaces of sets of sequences. The
first two kinds of dual spaces naturally arise in the study of absolute and ordinary
convergence of sequences from a subset of ω.
Furthermore the conditions given in Subsection 1.3 for an infinite matrix A
to be in the classes (X, l∞ ), (X, c) and (X, c0 ) for arbitrary FK spaces X involved
the norm of the operator
LA defined by LA (x) = A(x). Since A ∈ (X, Y ) can only
P∞
hold if An (x) = k=0 ank xk converges for all x ∈ X and P
for all n = 0, 1, . . . , it is
∞
essential to know the set of all sequences a ∈ ω for which k=0 ak xk converges for
all x ∈ X, the so–called β–dual of X. Finally, if X and Y are given FK spaces,
then we intend to replace the operator norm in the conditions for A ∈ (X, Y ) by
conditions for the entries of the matrix A. In many cases this can be achieved by
replacing the operator norm by the natural norm on the β–dual of X.
The α– and β–duals are special cases of the so–called multiplier spaces.
Definition 1.24. Let X and Y be subsets
© of ω.
ª
(a) For all z ∈ ω, we write z −1 ∗©Y = x ∈ ω : xz = (xk zk )∞
∈ Y . The
k=0
ª
T
set Z = M (X, Y ) = x∈X x−1 ∗ Y = a ∈ ω : ax ∈ Y for all x ∈ X is called the
multiplier space of X and Y .
(b) By cs and bs, we denote
respecP∞ the set of all convergent and boundedPseries,
n
tively, that is cs = {x ∈ ω : k=0 xk converges} and bs = {x ∈ ω : ( P
x
)∞
k
n=0 ∈
k=0
n
l∞ }, and we define the norm k · kbs on cs and bs by kxkbs = supn | k=0 xk |. In
the special case where Y = l1 or Y = cs, the multiplier spaces X α = M (X, l1 )
and X β = M (X, cs) are called the α– or Köthe–Toeplitz and β–duals of X. If
† denotes either of the symbols α or β, then X ⊂ ω is said to be †–perfect if
X †† = (X † )† = X.
Lemma 1.25. Let X, Y, Z ⊂ ω and {Xδ : δ ∈ A} be any collection of subsets
of ω. Then:
(i) X ⊂ M (M (X, Y ), Y )
(ii) X ⊂ Z implies M (Z, Y ) ⊂ M (X, Y )
(iii) M (X,
¡S Y ) = M (M
¢ (M
T(X, Y ), Y ), Y )
(iv) M
X
,
Y
=
δ∈A δ
δ∈A M (Xδ , Y ).
Proof. (i) If x ∈ X, then ax ∈ Y for all a ∈ M (X, Y ), and consequently x ∈
M (M (X, Y ), Y ).
157
Theory of sequence spaces
(ii) Let X ⊂ Z. If a ∈ M (Z, Y ), then ax ∈ Y for all x ∈ Z, hence ax ∈ Y for
all x ∈ X, since X ⊂ Z. Thus a ∈ M (X, Y ).
(iii) We apply (i) with X replaced by M (X, Y ) to obtain
M (X, Y ) ⊂ M (M (M (X, Y ), Y ), Y ).
Conversely, by (i), X ⊂ M (M (X, Y ), Y ), and so (ii) with Z = M (M (X, Y ), Y )
yields M (M (M (X, Y ),SY ), Y ) ⊂ M (X, Y ).
³ S
´
T
(iv) First Xδ ⊂
Xδ for all δ ∈ A implies M
Xδ , Y ⊂
M (Xδ , Y )
δ∈A
δ∈A
δ∈A
by part (i).
T
Conversely, if a ∈ δ M (Xδ , Y ), then a ∈ M (Xδ , Y ) for all δ ∈ A, and so we
have
ax ∈ Y for all δ ∈ S
A and for all x ∈ XT
δ . This implies ax ∈ Y
S
S for all x ∈
X
,
hence
a
∈
M
(
X
,
Y
).
Thus
M
(X
,
Y
)
⊂
M
(
δ
δ
δ
δ∈A
δ∈A
δ∈A
δ∈A Xδ , Y ).
¤
As an immediate consequence of Lemma 1.25 we obtain
Corollary 1.26. Let X, Y ⊂ ω and {Xδ : δ ∈ A} be a collection of subsets of
ω. If † denotes either of the symbols α or β, then
(i) X ⊂ X ††
(ii) X ⊂ Y implies Y † ⊂ X †
¢† T
¡S
= δ∈A Xδ† .
(iii) X † = X †††
(iv)
δ∈A Xδ
A subset X of ω is said to be normal if x ∈ X and |x̃k | ≤ |xk | (k = 0, 1, . . . )
together imply x̃ ∈ X.
Remark 1.27. Obviously X α ⊂ X β for arbitrary X ⊂ ω. If X is a normal
subset of ω, then X α = X β .
Proof. The first part is obvious. For the second part, we have to show X β ⊂
X α . Let a ∈ X β and x ∈ X be given. We define the sequence y by yk = sgn(xk )|xk |
for k = 0, 1, . . . . Then obviously |yk | ≤ |xk | for all k, and consequently y ∈ X,
since X is normal, and so ax ∈ cs. Further, by the definition of the sequence y,
α
ay = (|ak ||xk |)∞
k=0 = |ax| ∈ cs, hence ax ∈ l1 . Since x ∈ X was arbitrary, a ∈ X .
This shows X β ⊂ X α .
¤
Example 1.28. We have:
(i) M (c0 , c) = l∞ , (ii) M (c, c) = c, (iii) M (l∞ , c) = c0 .
Proof. (i) If a ∈ l∞ , then ax ∈ c for all x ∈ c0 , and so l∞ ⊂ M (c0 , c).
Conversely we assume a ∈
/ l∞ . Then there is a subsequence (akj )∞
j=0 of the
sequence a such that |akj | > j + 1 for all j = 0, 1, . . . . We define the sequence x by
½
(1.8)
xk =
(−1)j /akj
0
for k = kj
for k =
6 kj
(j = 0, 1, . . . ).
Then x ∈ c0 and akj xkj = (−1)j for all j = 0, 1, . . . , hence ax 6∈ c. This shows
M (c0 , c) ⊂ l∞ .
(ii) If a ∈ c, then ax ∈ c for all x ∈ c, and so c ⊂ M (c, c).
158
Malkowsky and Rakočević
Conversely we assume a ∈
/ c. Since e ∈ c and ae = a ∈
/ c, we have a ∈
/ M (c, c).
This shows M (c, c) ⊂ c.
(iii) If a ∈ c0 , then ax ∈ c for all x ∈ l∞ , and so c0 ⊂ M (l∞ , c).
Conversely we assume a ∈
/ c0 . Then there are a real b > 0 and a subsequence
(akj )∞
j=0 of the sequence a such that |akj | > b for all j = 0, 1, . . . . We define the
sequence x as in (1.8). Then x ∈ l∞ and akj xkj = (−1)j for j = 0, 1, . . . , hence
a∈
/ M (l∞ , c). This shows M (l∞ , c) ⊂ c0 .
¤
Now we shall give the α– and β–duals of the classical sequence spaces.
Theorem 1.29. Let † denote either of the symbols α or β. Then
(a) ω † = φ and φ† = ω.
(b) l1† = l∞ , lp† = lq for 1 < p < ∞ and q = p/(p − 1), and for all a ∈ lpβ ,
∗
kakl1 = kak∞ and kak∗lp = kakq for 1 < p < ∞.
β
†
.
(c) c†0 = c† = l∞
= l1 and kak∗c0 = kak∗c = kak∗l∞ = kak1 for all a ∈ l∞
The multiplier space of two BK spaces will turn out to be a BK space.
Theorem 1.30. Let (X, k · kX ) and (Y, k · kY ) be BK spaces with X ⊃ φ and
Z = M (X, Y ). Then Z is a BK space with k · k defined by
kzk = kzk∗X = sup{kxzkY : kxkX = 1}
for all z ∈ Z.
Proof. It is well known that B = B(X, Y ) is a Banach space. Each z ∈ Z
defines a diagonal matrix map ẑ : X 7→ Y where ẑ(x) = xz for all x ∈ X which
is continuous by Theorem 1.17. This embeds Z in B, for if ẑ = 0, then ẑ(e(n) ) =
(zn )∞
n=0 = 0 = z. To see that the coordinates are continuous, we fix n ∈ N0 and
put u = 1/ke(n) kX and v = ke(n) kY . Then kue(n) kX = 1 and
uv|zn | = ukzn e(n) kY = uke(n) zkY = k(ue(n) )zkY ≤ kzk∗X = kzk
for all n.
It remains to show that Z is a closed subspace of B. Let (ẑ (m) )∞
m=0 be a sequence
in B with ẑ (m) → T ∈ B. For each fixed X ∈ X, we obtain ẑ (m) (x) → T (x) ∈ Y
(m → ∞) and since Y is a BK space, this implies (ẑ (m) (x))k → (T (x))k , that is
(m)
(m)
zk xk → (T (x))k (m → ∞) for each fixed k. We put x = e(k) . Then zk →
(m)
(m)
(T (e(k) ))k = tk (m → ∞), and so xk zk = (T (x))k (m → ∞) and xk zk →
(T (x))k (m → ∞). Therefore T (x) = xt for all x ∈ X, and so T = t̂.
¤
Corollary 1.31. The α– and β– duals P
of a BK space X are BK spaces with
∞
respect to kakα = kakX,α = P
sup{kaxk1 = k=0 |ak xk | : kxk ≤ 1} and kakβ =
n
kakX,β = sup{kaxkbs = supn | k=0 ak xk | : kxk ≤ 1}.
Example 1.32. Let X be any of the spaces l∞ , c, c0 and lp for 1 ≤ p < ∞.
Then the norms k · kX β , k · k∗X , k · kX,α and k · kX,β are equivalent on X β .
Proof. The norm k · k∗X and the natural norm k · kX β are equal on X β by
Theorem 1.29. Since each set X β is a BK space with its natural norm, k · kX β and
Theory of sequence spaces
159
k ·kX,β are equivalent by Corollary 1.31 and Theorem 1.21. Finally, since X α = X β
for each set X, the norms k · kX,α and k · kX,β are equivalent by Corollary 1.31 and
Theorem 1.21.
¤
The analogues of Theorem 1.30 and Corollary 1.31 do not hold for FK spaces
in general.
Remark 1.33. The space ω is an FK space and ω α = ω β = φ and φ has no
Fréchet metric (cf. [108, 4.0.5, p. 51]).
1.5. The continuous duals of the classical sequence spaces. In this subsection we shall give the continuous duals of the spaces lp for 1 ≤ p < ∞, c and
c0 .
There is a close relation between the β–dual and the continuous dual of an
FK space which is very useful in the determination of the continuous duals of the
spaces lp , c and c0 .
Theorem 1.34. Let X be a BK space and X ⊃ φ. Then there is a linear
one-to-one map T : X β 7→ X 0 ; we denote this by X β ⊂ X 0 . If X has AK, then T
is onto.
Proof. We define the map P
T on X β as follows. For every a ∈ X β , let T a :
∞
X 7→ C be defined by (T a)(x) = k=0 ak xk for all x ∈ X. Since a ∈ X β , the series
P
∞
k=0 ak xk converge for all x ∈ X, and obviously T a is linear. Further, since X is
an FK space, T a ∈ X 0 for ach a ∈ X β . Therefore T : X β 7→ X 0 . Further it is easy
to see that T is linear.
To show that T is one-to-one, we assume a, b ∈ X β with T a = T b. This means
(T a)(x) = (T b)(x) for all x ∈ X. Since φ ⊂ X, we may choose x = e(k) for each
k ∈ N0 and obtain (T a)(e(k) ) = ak = bk = (T b)(e(k) ) for k = 0, 1, . . . , and so a = b.
Now we assume that X has AK and
f ∈ X 0 . We put an = f (e(n) ) for n =
P∞
0, 1, . . . . Let x ∈ X be given. Then x = k=0 xk e(k) , since X has AK, and f ∈ X 0
P∞
P∞
implies f (x) = k=0 xk f (e(k) ) = k=0 ak xk = (T a)(x). As x ∈ X was arbitrary
and the series converge, a ∈ X β and f = T a. This shows that T is onto X 0 .
¤
Now we shall give the continuous duals of the classical sequence spaces.
Two linear spaces (X, k · kX ) and (Y, k · kY ) are called norm isomorphic if
there is an isomorphism T : X 7→ Y such that kT (x)kY = kxkX for all x ∈ X; we
shall write X ' Y .
Theorem 1.35. We have:
(a) lp∗ ' l∞ for 0 < p ≤ 1 and lp∗ ' lq for 1 < p < ∞ where q = p/(p − 1);
(b) c∗0 ' l1 ;
P∞
(c) f ∈ c∗ if and only P
if f (x) = lχf + k=0 ak xk with a ∈ l1 where l =
∞
limk→∞ xk and χf = f (e) − k=0 ak . Furthermore kf k∗ = |χf | + kak1 .
It is worth mentioning that the continuous dual of l∞ is not isomorphic to a
sequence space (cf. [40, 31.1, pp. 427, 428] or [105, Example 6.4.8, pp. 93, 94]).
For further studies concerning multiplier spaces, some important special cases,
f – and continuous duals, we refer the reader to [108, 91, 22, 23].
160
Malkowsky and Rakočević
1.6. Matrix transformations between some classical sequence spaces. We
now apply the results of the previous subsections to characterize certain classes of
matrix transformations between some classical sequence spaces by giving necessary
and sufficient conditions on the entries of a matrix to belong to the respective class.
Let A be an infinite matrix. We write q = p/(p − 1) for 1 < p < ∞, q = ∞ for
p = 1 and q = 1 for p = ∞, put
(
(p = 1)
kAk∞ = supn,k |ank |
´
³P
MA (lp , l∞ ) =
∞
q
kAkq = supn
(1 < p ≤ ∞)
k=0 |ank |
and consider the conditions
(1.9)
(1.10)
(1.11)
(1.12)
lim ank = 0 (k = 0, 1, . . . ),
µX
¶
∞
lim
ank = 0,
n→∞
n→∞
k=0
lim ank = lk for some lk ∈ C (k = 0, 1, . . . )
n→∞
µX
¶
∞
lim
ank = l for some l ∈ C.
n→∞
k=0
Theorem 1.36. We have
(a) (c0 , l∞ ) = (c, l∞ ) = (l∞ , l∞ ) and A ∈ (l∞ , l∞ ) if and only if
(1.13)
MA (l∞ , l∞ ) = sup
n
µX
∞
¶
|ank |
< ∞,
k=0
(b) A ∈ (c0 , c0 ) if and only if conditions (1.13) and (1.9) hold;
(c) A ∈ (c, c0 ) if and only if conditions (1.13), (1.9) and (1.10) hold;
(d) A ∈ (c0 , c) if and only if conditions (1.13) and (1.11) hold;
(e) A ∈ (c, c) if and only if conditions (1.13), (1.11) and (1.12) hold.
Proof. (a) We have A ∈ (l∞ , l∞ ) if and only if condition (1.13) holds by
Theorems 1.23 and 1.29.
Further, if condition (1.13) holds, then A ∈ (l∞ , l∞ ) ⊂ (c0 , c), since c0 ⊂ l∞ .
Conversely, let A ∈ (c0 , l∞ ). Then supn kAn k∗c0 < ∞ by Theorem 1.23 (b).
Since the series A
(x) converge for all x and n, we have fAn ∈ c∗0 for all n
Pn∞
∗
where fAn (x) =
k=0 ank xk for all x ∈ c0 , hence |fAn (x)| ≤ kfAn k = kAn kc0 .
[m,n]
We fix nP∈ N0 . Let m ∈ N0 be arbitrary. We define the sequence x
by
m
(k)
[m,n]
[m,n]
[m,n]
[m,n]
sgn(a
k)e
.
Then
x
∈
c
,
kx
k
≤
1
and
|f
(x
)|
=
x
=
n
0
∞
An
k=0
Pm
P∞
∗
∗
|a
|
≤
kA
k
|a
|
≤
kA
k
.
Since
m
∈
N
was
arbitrary,
kA
k
=
n c0
n c0
0
n 1
k=0 nk
k=0 nk
for all n = 0, 1, . . . . Therefore condition (1.13) must hold. Finally c0 ⊂ c ⊂ l∞ and
(c0 , l∞ ) = (l∞ , l∞ ) together imply (c, l∞ ) = (l∞ , l∞ ).
Parts (b) to (e) follow from part (a), Theorem 1.23 (c) and Theorem 1.10. ¤
Similarly, we obtain
Theory of sequence spaces
161
Theorem 1.37. Let 1 < p < ∞. Then:
(a) A ∈ (lp , l∞ ) if and only if
(1.14)
M (lp , l∞ ) < ∞;
(b) A ∈ (lp , c0 ) if and only if conditions (1.14) and (1.9) hold;
(c) A ∈ (lp , c) if and only if conditions (1.14) and (1.11) hold.
2. Measures of concompactness
In Section 1 we developed and applied parts of the FK space theory to give
necessary and sufficient conditions for A ∈ (X, Y ) for given sequence spaces. The
most important result was that matrix transformations between FK spaces are continuous. It is quite natural to find conditions for a matrix map between FK spaces
to define a compact operator. This can be achieved by applying the Hausdorff
measure of noncompactness. The first measure of noncompactness, the function α,
was defined and studied by Kuratowski [41] in 1930. It is surprising that later in
1955 Darbo [12] was the first who continued to use the function α. Darbo proved
that if T is a continuous self-mapping of a nonempty, bounded, closed and convex
subset C of a Banach space X such that α(T (Q)) ≤ kα(Q) for all Q ⊂ C, where
k ∈ (0, 1) is a constant, then T has at least one fixed point in the set C. Darbo’s
fixed point theorem is a very important generalization of Schauder’s fixed point
theorem and it includes the existence part of Banach’s fixed point theorem.
Other measures were introduced by Goldenstein, Gohberg and Markus (the ball
measures of noncompactness, Hausdorff measure of noncompactness) [19] in 1957
(later studied by Goldenstein and Markus [20] in 1968), Istrǎţesku [30] in 1972
and others. Apparently Goldenstein, Gohberg and Markus were not aware of the
work of Kuratowski and Darbo. It is surprising that Darbo’s theorem was almost
never noticed and applied, not till in the seventies mathematicians working in operator theory, functional analysis and differential equations begun to apply Darbo’s
theorem and developed the theory connected with measures of noncompactness.
The use of these measures is discussed for example in the monographs [1, 6, 7,
24, 25, 28, 31, 42, 86, 99, 100], Ph. D. theses [2, 4, 75, 77, 83, 102] and expository
papers [47, 93, 104]. We refer the reader to these works with references given there.
2.1. Introduction. Let us recall some definitions and results which are probably well known. If M and S are subsets of a metric space (X, d) and ² > 0, then the
set S is called ²-net of M if for any x ∈ M there exists s ∈ S, such that d(x, s) < ².
If the set S is finite, then the ²-net S of M is called finite ²-net. The set M is
said to be totally bounded if it has a finite ²-net for every ² > 0. It is well known,
that a subset M of a metric space X is compact if every sequence (xn ) in M has
a convergent subsequence, and in this case the limit of that subsequence is in M .
The set M is said to be relatively compact if the closure M of M is a compact set.
If the set M is relatively compact, then M is totally bounded. If the metric space
(X, d) is complete, then the set M is relatively compact if and only if it is totally
bounded. It is easy to prove that a subset M of a metric space X is relatively
162
Malkowsky and Rakočević
compact if and only if every sequence (xn ) in M has a convergent subsequence; in
that case the limit of that subsequence need not be in M .
If x ∈ X and r > 0, then the open ball with centre at x and radius r is denoted
by B(x, r), B(x, r) = {y ∈ X : d(x, y) < r}. If X is a normed space, then we denote
by BX the closed unit ball in X and by SX the unit sphere in X. Let MX (or
simply M) be the set of all nonempty and bounded subsets of a metric space (X, d),
and let McX (or simply Mc ) be the subfamily of MX consisting of all closed sets.
Further, let NX (or simply N ) be the set of all nonempty and relatively compact
subsets of (X, d). Let dH : MX × MX 7→ R be the function defined by
(2.1)
dH (S, Q) = max{sup d(x, Q), sup d(y, S)}
x∈S
y∈Q
(S, Q ∈ MX ).
The function dH is called Hausdorff distance, and dH (S, Q) (S, Q ∈ MX ) is the
Hausdorff distance of sets S and Q.
Let us remark that if ∅ 6= F ⊂ X, r > 0 and
[
B(F, r) =
B(x, r) = {y ∈ X : d(y, F ) < r}
x∈F
is the open ball with centre in F and radius r, then (2.1) is equivalent to
dH (S, Q) = inf{² > 0 : S ⊂ B(Q, ²)
and
Q ⊂ K(S, ²)},
(S, Q ∈ MX ).
It is well known that (MX , dH ) is a pseudometric space and that (McX , dH ) is a
metric space.
Let X and Y be infinite-dimensional complex Banach spaces and denote the set
of bounded linear operators from X into Y by B(X, Y ). We put B(X) = B(X, X).
For T in B(X, Y ), N (T ) and R(T ) will denote, respectively, the null space and the
range space of T . A linear operator L from X to Y is called compact (or completely
continuous) if D(L) = X for the domain of L, and for every sequence {xn } ⊂ X
such that kxn k ≤ C, the sequence {L(xn )} has a subsequence which converges in
Y . A compact operator is bounded. An operator L in B(X, Y ) is of finite rank
if dim R(L) < ∞. An operator of finite rank is clearly compact. Let F (X, Y ),
K(X, Y ) denote the set of all finite rank and compact operators from X to Y ,
respectively. Set F (X) = F (X, X) and K(X) = K(X, X).
Let X be a vector space over the field F. A subset E of X is said to be convex
if λx + (1 − λ)y ∈ E for all x, y ∈ E and for all λ ∈ (0, 1).
Clearly the intersection of any family of convex sets is a convex set. If F is a
subset of X, then the intersection of all convex sets that contain F is called convex
cover or convex hull of F denoted by co(F ).
The vector subspace linF is the set of all linear combinations of elements in
F . We shall prove that there is an analogous representation of the set co(F ). Let
us mention that a convex combination of elements of the set F is an element of the
form
µ
¶
n
X
λ1 x1 + · · · + λn xn xi ∈ F, λi ≥ 0 (i = 1, . . . , n),
λi = 1 (n ∈ N) .
i=1
Let us write cvx(F ) for the set of all convex combinations of elements of the set F .
163
Theory of sequence spaces
Theorem 2.1. If X is a vector space over the field F and E, E1 , . . . En are
convex subsets of X and F ⊂ X, then
(2.2)
cvx(E) ⊂ E,
(2.3)
(2.4)
co
µ[
n
¶
Ei
=
½X
n
i=1
co(F ) = cvx(F ),
n
X
λi Ei : λi ≥ 0,
i=1
¾
λi = 1, i = 1, . . . , n .
i=1
Proof. To prove (2.2), it suffices to show that for any n ≥ 2
(2.5)
xi ∈ E, λi ≥ 0 (i = 1, . . . n)
n
X
and
λi = 1
together
i=1
imply
λ1 x1 + · · · + λn xn ∈ E.
We shall use the method of mathematical induction. For n = 2 the statement
clearly is true. Suppose that the statement in (2.5) is true for a natural number
n > 2, and let us prove the statement for n + 1. If xi ∈ E, λi ≥ 0 (i = 1, . . . , n + 1)
Pn+1
Pn
and i=1 λi = 1, then there are two cases: first, if i=1 λi = 0, P
then λi = 0
n
(i = 1, . . . , n) and λ1 x1 + · · · + λn+1 xn+1 = xn+1 ∈ E; second, if λ ≡ i=1 λi 6= 0,
−1
−1
then λ1 x1 + · · · + λn+1 xn+1 = λ(λ1 λ x1 + · · · + λn λ xn ) + λn+1 xn+1 ∈ E. Thus
we have shown inclusion (2.2).
It follows from (2.2) that cvx(F ) ⊂ co(F ). Hence, since co(F ) is a convex
subset of X, it suffices to show that cvx(F ) is convex. Suppose that λ ∈P
(0, 1), and
n
x, y ∈ cvx(F ). Then there exist
n,
m
∈
N,
α
,
x
(i
=
1,
.
.
.
,
n)
with
i
i
i=1 αi =
Pn
Pm
α
x
and
y =
β
=
1
such
that
x
=
1, βj , yj (j = 1, . . . , m) with
i
i
j
i=1
Pm j=1
Pn
Pm
(1−λ)β
=
λ+(1−λ)
=
1
implies
λx+(1−λ)y
∈
λα
+
β
y
.
Now
j
i
j=1
i=1
j=1 j j
cvx(F ). Hence we have
Pn
Pn proved (2.3).
= 1, . . . , n) i=1 λi = 1}. By (2.2)
We put S = { i=1
Snλi Ei : λi ≥ 0,S(i
n
it follows that S ⊂ co( i=1 Ei ). Since i=1 Ei ⊂ S, to prove (2.4), it suffices to
show that S is convex. Suppose
that λ ∈ (0, 1) and x, y ∈ S. P
Now there exist
Pn
n
αi , xi (i = P
1, . . . , n) with Pi=1 αi = 1, βi , yi (i = 1, . . . , n, ) with i=1 βi = 1 such
n
n
that x = i=1 αi xi , y = i=1 βi yi . We put γi = λαi + (1 − λ)βi (i = 1, . . . n).
Since E1 , . . . En , are convex, there exist zi ∈ Ei (i = 1, . . . n) such that
(2.6)
λαi xi + (1 − λ)βi yi = γi zi
for i = 1, . . . , n.
Let us remark
(2.7)
n
X
i=1
γi = λ
n
X
i=1
αi + (1 − λ)
n
X
βi = λ + (1 − λ) = 1.
i=1
By (2.6) and (2.7) we have λx + (1 − λ)y =
Pn
i=1
γi zi ∈ S.
We continue with the study of convex sets in normed spaces.
¤
164
Malkowsky and Rakočević
Lemma 2.2. Let Q be a bounded subset of a normed space X. Then for any
x∈X
(2.8)
sup kx − yk = sup kx − zk.
y∈co(Q)
z∈Q
Proof. To prove (2.8), it suffices to show the inequality
Pn“≤”. If y ∈ co (Q),
then
there
exist
x
∈
Q,
λ
≥
0
(i
=
1,
.
.
.
n)
such
that
i
i=1 λi = 1 and y =
Pn
Pin
Pn
Pn
λ
x
.
From
x
−
y
=
λ
x
−
λ
x
=
λ
(x
− xi ), it follows that
i=1 i i P
i=1 i
i=1 i i
i=1 i
n
kx − yk ≤ i=1 λi kx − xi k ≤ supz∈Q kx − zk.
¤
Corollary 2.3. Let Q be a bounded subset of a normed space X. Then the
sets Q and co(Q) have equal diameter, that is diam(Q) = diam(co(Q)).
Proof. This follows by Lemma 2.2.
¤
Let Q be a nonempty and bounded subset of a normed space X. Then the
convex closure of Q, is denoted by Conv(Q) , and Conv(Q) is the smallest convex
and closed subset of X that contains Q. It is easy to prove that Conv(Q) = co(Q).
Corollary 2.4. Let Q be a bounded subset of a normed space X. Then the
sets Q and Conv(Q) have equal diameters, that is diam(Q) = diam(Conv(Q)).
Proof. This follows from Corollary 2.3.
¤
2.2. The Kuratowski measure of noncompactness. The notation of measure of noncompactness (α– measure or set-measure), introduced by Kuratowski
[41], and the associated notion of an α– contraction, have proved useful in several areas of functional analysis, operator theory and differential equations (see for
example, [1, 6, 7]). We start with some results from Kuratowski [41, 42].
Definition 2.5. Let (X, d) be a metric space and Q a bounded subset of X.
Then the Kuratowski measure of noncompactness of Q , denoted by α(Q), is the
infimum of the set of all numbers ² > 0 such that Q can be covered by a finite
number of sets with diameters < ², that is
(2.9)
½
¾
n
[
α(Q) = inf ² > 0 : Q ⊂
Si , Si ⊂ X, diam(Si ) < ² (i = 1, . . . , n; n ∈ N) .
i=1
The function α is called Kuratowski’s measure of noncompactness. Clearly
(2.10)
α(Q) ≤ diam(Q) for each bounded subset Q of X.
As an immediate consequence of Definition 2.5, we obtain.
Theory of sequence spaces
165
Lemma 2.6. Let Q, Q1 and Q2 be bounded subsets of a complete metric
space (X, d). Then:
(2.11)
α(Q) = 0
if and only if
(2.12)
α(Q) = α(Q),
(2.13)
Q1 ⊂ Q2
(2.14)
α(Q1 ∪ Q2 ) = max{α(Q1 ), α(Q2 )},
(2.15)
α(Q1 ∩ Q2 ) ≤ min{α(Q1 ), α(Q2 )}.
implies
Q is compact,
α(Q1 ) ≤ α(Q2 ),
Proof. The statements in (2.11) and (2.13) follow from Definition 2.5.
Clearly α(Q) ≤ α(Q). Let ² > 0, Si be a bounded subset of X with diam(Si ) <
Sn
Sn
Sn
² for i = 1, . . . , n, and Q ⊂ i=1 Si . Then Q ⊂ i=1 Si = i=1 Si . Since
diam(Si ) = diam(Si ), we conclude α(Q) ≤ α(Q). This proves equality (2.12).
From (2.13), we have α(Q1 ) ≤ α(Q1 ∪ Q2 ) and α(Q2 ) ≤ α(Q1 ∪ Q2 ), and so
(2.16)
max{α(Q1 ), α(Q2 )} ≤ α(Q1 ∪ Q2 ).
Let max{α(Q1 ), α(Q2 )} = s and ² > 0. By Definition 2.5 we know that Q1 and
Q2 can be covered by a finite number of subsets of diameter smaller than s + ².
Obviously, the union of these covers is a finite cover of Q1 ∪ Q2 . Hence, we have
α(Q1 ∪ Q2 ) ≤ s + ², and now we obtain (2.14) from (2.16). From Q1 ∩ Q2 ⊂ Q1
and Q1 ∩ Q2 ⊂ Q2 we obtain α(Q1 ∩ Q2 ) ≤ α(Q1 ) and α(Q1 ∩ Q2 ) ≤ α(Q2 ). Hence
α(Q1 ∩ Q2 ) ≤ min{α(Q1 ), α(Q2 )}. This proves inequality (2.15).
¤
The next theorem is a generalization of the well–known Cantor intersection
theorem.
Theorem 2.7. (Kuratowski [41]) Let (X, d) be a complete metric space. If
(Fn ) is a decreasing sequence of nonempty, closed andTbounded subsets of X such
∞
that limn→∞ α(Fn ) = 0, then the intersection F∞ = n=1 Fn is a nonempty and
compact subset of X.
Proof. The set F∞ is a closed subset of X. Since F∞ ⊂ Fn for all n = 1, 2, . . . ,
we obtain from (2.11) and (2.1.3) that F∞ is a compact set. Now we show F∞ 6= ∅.
Let xn ∈ Fn (n = 1, 2, . . . ) and Xn = {xi : i ≥ n} for n = 1, 2, . . . . Since Xn ⊂ Fn ,
we obtain from (2.11), (2.13) and (2.14)
(2.17)
α(X1 ) = α(Xn ) ≤ α(Fn ) for each n.
The assumption of the theorem and (2.17) together imply α(X1 ) = 0, hence X1
is a relatively compact set. Thus the sequence (xn ) has a convergent subsequence
with limit x ∈ X, say. Since Fn is closed in X, we get x ∈ Fn for all n = 1, 2, . . . ,
that is x ∈ F∞ .
¤
If X is a normed space, then the function α has some additional properties
connected with the vector (linear) structures of a normed space [12].
166
Malkowsky and Rakočević
Theorem 2.8. (Darbo [12]) Let Q, Q1 and Q2 be bounded subsets of a
normed space X. Then:
(2.18)
α(Q1 + Q2 ) ≤ α(Q1 ) + α(Q2 ),
(2.19)
α(Q + x) = α(Q)
for each x ∈ X,
(2.20)
α(λQ) = |λ|α(Q)
for each λ ∈ F,
(2.21)
α(Q) = α(co(Q)).
Proof. Let Si S
be a bounded subset of X with diam(Si ) < d for each i =
n
1, . . . , n and Q1 ⊂ i=1 Si . Furthermore, let G
Sjmbe a bounded subset of X with
diam(Gj ) < p for each j = 1, . . . , m and Q2 ⊂ j=1 Gj . Then
(2.22)
Q1 + Q2 ⊂
n [
m
[
(Si + Gj )
and diam(Si + Gj ) < d + p.
i=1 j=1
It follows from (2.22) that α(Q1 + Q2 ) < d + p. This shows inequality (2.18). Let
x ∈ X. By (2.18) it follows that
(2.23)
α(Q + x) ≤ α(Q) + α({x}) = α(Q),
and by the same argument we have
(2.24)
α(Q) = α((Q + x) + (−x)) ≤ α(Q + x) + α({−x}) = α(Q + x).
Now we obtain (2.19) from (2.23) and (2.24).
For λ = 0, equality (2.20) is obvious. Let SSi be a bounded subset of X
n
with diam(S
Sn i ) < d for i = 1, . . . , n and Q1 ⊂ i=1 Si . Then for any λ ∈ F,
λQ ⊂ i=1 λSi and diam(λSi ) = |λ| diam Si . Hence it follows that α(λQ) ≤
|λ| α(Q). If λ 6= 0, analogously we have α(Q) = α(λ−1 (λQ)) ≤ |λ−1 | α(λQ), that
is |λ|α(Q) ≤ α(λQ). This proves (2.20).
Now we prove (2.21). Clearly α(Q) ≤ α(co Q), and it suffices to show α(co
Q) ≤ α(Q). Let S
Si be a bounded subset of X with diam(Si ) < d for each i =
n
1, . . . , n and Q = i=1 Si . By Theorem 2.1 it follows that
(2.25)
co(Q) =
½X
n
λi xi : λi ≥ 0,
i=1
n
X
¾
λi = 1, xi ∈ co(Si ) (i = 1, . . . , n) .
i=1
Pn
Let ² > 0 and S={(λ1 , . . . , λn ) :
i=1 λi = 1, λi ≥ 0 (i = 1, . . . , n)}. Then S is
a compact subset of (Rn , k · k∞ ), where k(λ1 , . . . , λn )k∞ = sup1≤i≤n |λi |. We put
Sn
M = sup{kxk : x ∈ i=1 co(Si )}. Let T = {(tj,1 , . . . , tj,n ) : j = 1, . . . , m}
Pn⊂ S be
a finite ²/(M n) -net for S, with respect to the k · k∞ -norm. Hence, if i=1 λi xi
is a convex combination of elements of Q, where we suppose that xi ∈ co(Si ) for
i = 1, . . . , n, then there exists (tj,1 , . . . , tj,n ) ∈ T such that
(2.26)
k(λ1 , . . . , λn ) − (tj,1 , . . . , tj,n )k∞ <
²
n.
M
167
Theory of sequence spaces
Since
(2.27)
n
X
λi xi =
i=1
n
X
tj,i xi +
i=1
n
X
(λi − tj,i )xi ,
i=1
it follows from (2.25), (2.26) and (2.27) that
(2.28)
co(Q) ⊂
m ½X
n
[
¾
n
² X
tj,i co(Si ) +
Bi ,
M n i=1
i=1
j=1
where Bi = {x ∈ X : kxk ≤ M } for i = 1, 2, . . . , n. Now, by (2.4), (2.5), (2.18),
(2.20), Corollary 2.3 and (2.28), we have
µ[
m ½X
n
¾¶
µ
¶
n
² X
α(co(Q)) ≤ α
tj,i co(Si )
+α
Bi
M n i=1
j=1 i=1
¶
µX
n
n
² X
≤ max α
tj,i co(Si ) +
α(Bi )
1≤j≤m
M n i=1
i=1
< max
1≤j≤m
n
X
tj,i α(co(Si )) +
i=1
n
X
< d max
1≤j≤m
²
2nM
Mn
tj,i + 2² < d + 2².
i=1
¤
Let us remark that Darbo [12] proved (2.21) and, then applied it in the proof
of his famous fixed point theorem [12, 1, 7, 86, 100]. His fixed point theorem is a
very important generalization of the Schauder fixed point theorem, and is the first
important result with applications of Kuratowski’s measure of noncompactness.
Let X be an infinite–dimensional normed space and BX the closed unit ball
in X. Then, clearly α(BX ) ≤ 2, but Furi and Vignoli [18] and Nussbaum [79] have
shown more precisely:
Theorem 2.9. (Furi-Vignoli [18], Nussbaum [79]) Let X be an infinite-dimensional normed space. Then α(BX ) = 2.
Proof. Clearly α(BX ) ≤ 2. If α(BX ) < 2, then there exist bounded
Sn and
closed subsets Qi of X with diam(Qi ) < 2 for i = 1, . . . , n such that BX ⊂ i=1 Qi .
Let {x1 , . . . xn } be a linearly independent subset of X and En be the set of all linear
combinations of elements of the set {x1 , . . . , xn } with real coefficients. Clearly,
En is a real n-dimensional normed space (the norm on En , of course, being the
restriction of the norm on X). By Sn = {xS∈ En : kxk = 1}, we denote the unit
n
sphere of En . Let us mention that Sn ⊂ i=1 Sn ∩ Qi , diam(Sn ∩ Qi ) < 2 and
Sn ∩ Qi is a closed subset of En for each i = 1, . . . n. This is a contradiction to the
168
Malkowsky and Rakočević
well–known Ljusternik-Šnirelman-Borsuk theorem (see the proof in [86] or in [15,
pp. 303–307]: If Sn is the unit sphere of an n-dimensional
Snreal normed space En ,
Fi a closed subset of En for each i = 1, . . . , n and Sn ⊂ i=1 Fi , then there exists
i0 ∈ {1, . . . , n} such that the set Sn ∩ Fi0 contains a pair of antipodial points, that
is, there exists x0 ∈ Sn ∩ Fi0 , such that {x0 , −x0 } ⊂ Sn ∩ Fi0 .
¤
2.3. The Hausdorff measure of noncompactness. Usually it is complicated to find the exact value of α(Q). Another measure of noncompactness, which
is more applicable in many cases, were introduced and studied by Goldenstein,
Gohberg and Markus (the ball measures of noncompactness, Hausdorff measure
of noncompactness) [19] in 1957 (later studied by Goldenstein and Markus [20] in
1968), is given in the next definition.
Definition 2.10. Let (X, d) be a metric space and Q a bounded subset of X.
Then the Hausdorff measure of noncompactness of the set Q, denoted by χ(Q) is
defined to be the infimum of the set of all reals ² > 0 such that Q can be covered
by a finite number of balls of radii < ², that is
(2.29) χ(Q) = inf{² > 0 : Q ⊂
n
[
B(xi , ri ), xi ∈ X, ri < ² (i = 1, . . . n) n ∈ N}.
i=1
The function χ is called Hausdorff measure of noncompactness.
Let us remark that in the definition of the Hausdorff measure of noncompactness of the set Q it is not supposed that centres of the balls which cover Q belong
to Q. Hence, (2.29) can equivalently be stated as follows:
(2.30)
χ(Q) = inf{² > 0 : Q has a finite ²–net in X}.
The Hausdorff measure of noncompactness is often called ball measure of noncompactness. The next lemma and theorem could be proved analogously as in the
case of the Kuratowski measure of noncompactness.
Lemma 2.11. Let Q, Q1 and Q2 be bounded subsets of the metric space
(X, d). Then
χ(Q) = 0
if and only if
Q is totally bounded,
χ(Q) = χ(Q),
Q1 ⊂ Q2
implies χ(Q1 ) ≤ χ(Q2 ),
χ(Q1 ∪ Q2 ) = max{χ(Q1 ), χ(Q2 )},
χ(Q1 ∩ Q2 ) ≤ min{χ(Q1 ), χ(Q2 )}.
Proof. The proof is left as an exercise for the reader.
¤
Theory of sequence spaces
169
Theorem 2.12. Let Q, Q1 and Q2 be bounded subsets of the normed space
X. Then
χ(Q1 + Q2 ) ≤ χ(Q1 ) + χ(Q2 ),
(2.31)
χ(Q + x) = χ(Q)
for each x ∈ X,
χ(λQ) = |λ|χ(Q)
for each λ ∈ F
χ(Q) = χ(co(Q)).
Proof. The proof is left as an exercise for the reader.
¤
The next theorem shows that the functions α and χ are in some sense equivalent.
Theorem 2.13. Let (X, d) be a metric space and Q be a bounded subset of
X. Then
(2.32)
χ(Q) ≤ α(Q) ≤ 2χ(Q).
Proof. Let ² > 0. If {x1 , . . . , xn } is an ²-net of Q, then {Q ∩ B(xi , ²)}ni=1 is
a cover of Q with sets of diameter < 2². This shows α(Q) ≤ 2χ(Q). To prove the
left side inequality in (2.32), let us suppose that {Si }ki=1 is a cover of Q with sets
of diameter < ² and yi ∈ Si for i = 1, . . . k. Now {y1 , . . . , yk } is an ²-net of Q.
This proves χ(Q) ≤ α(Q).
¤
Let us remark that the inequalities (2.32) are best possible in general, as an
example shows. These measures are closely related to geometric properties of the
space and it is possible to improve the inequality χ(Q) ≤ α(Q) in certain spaces
(see e.g. Dominguez Benavides and Ayerbe
√ [14], Webb and Weiyu Zhao [103]). For
example (see√[1], [7]) in Hilbert space, 2χ(Q) ≤ α(Q) ≤ 2χ(Q), and in lp for
1 ≤ p < ∞, p 2χ(Q) ≤ α(Q) ≤ 2χ(Q).
Theorem 2.14. Let X be an infinite-dimensional normed space and BX be
the closed unit ball of X. Then χ(BX ) = 1.
Proof. Obviously χ(BX ) ≤ 1. If χ(BX ) = q < 1, then we choose ² > 0 such
that q + ² < 1. Now there exists a (q + ²)-net of BX , say {x1 , . . . , xk }. Hence
(2.33)
BX ⊂
k
[
{xi + (q + ²)BX }.
i=1
Now it follows from Lemmas 2.11 and 2.12 that
(2.34)
q = χ(BX ) ≤ max χ({xi + (q + ²)BX }) = (q + ²)q.
1≤i≤k
Since q + ² < 1, by (2.33) we have q = 0, that is BX is a totally bounded set. But
this is impossible since X is an infinite-dimensional space. Hence χ(BX ) = 1. ¤
Let us remark that Theorem 2.14 follows from Theorems 2.9 and 2.13. (But
we offer another proof.)
Now we shall show how to compute the Hausdorff measure of noncompactness
in the spaces `p for 1 ≤ p < ∞ and c0 .
170
Malkowsky and Rakočević
Theorem 2.15. Let Q be a bounded subset of the normed space X, where
X is `p for 1 ≤ p < ∞ or c0 . If Pn : X 7→ X is the operator defined by
Pn (x1 , x2 , . . . ) = (x1 , x2 , . . . , xn , 0, 0, . . . ) for (x1 , x2 , . . . ) ∈ X; then
(2.35)
χ(Q) = lim sup k(I − Pn )xk.
n→∞ x∈Q
Proof. Clearly
(2.36)
Q ⊂ Pn Q + (I − Pn )Q.
It follows from Lemma 2.11, Theorem 2.12 and (2.36) that
(2.37)
χ(Q) ≤ χ(Pn Q) + χ((I − Pn )Q) = χ((I − Pn )Q) ≤ sup k(I − Pn )xk.
x∈Q
Since the limit in (2.35) clearly exists, we have by (2.37)
(2.38)
χ(Q) ≤ lim sup k(I − Pn )xk.
n→∞ x∈Q
Now we prove the converse inequality in (2.38). Let ² > 0 and {z1 , . . . , zk } be a
[χ(Q) + ²]-net of Q. It is easy to prove that
(2.39)
Q ⊂ {z1 , . . . , zk } + [χ(Q) + ²]BX .
It follows from (2.39) that for any x ∈ Q there exist z ∈ {z1 , . . . , zk } and s ∈ BX
such that x = z + [χ(Q) + ²]s. Hence
(2.40)
sup k(I − Pn )xk ≤ sup k(I − Pn )zi k + [χ(Q) + ²].
x∈Q
1≤i≤k
Finally, (2.40) implies limn→∞ supx∈Q k(I − Pn )xk ≤ χ(Q) + ².
¤
Concerning the space `∞ (R), to the best of our knowledge is the following
theorem [13, Proposition 3.5].
Theorem 2.16. (Dominguez Benavides [13]) Let `∞ be the real normed space
of bounded sequences with sup-norm and Q be a bounded subset of `∞ . Then
α(Q) = 2χ(Q).
Proof. We know thatSα(Q) ≤ 2χ(Q). Let ² > 0 and Q1 , . . . , Qn be subsets
n
of `∞ (R) such that Q ⊂ i=1 Qi and diam Qi < α(Q) + ². For any k ∈ N we
put αk,i = inf{xk : (xj ) ∈ Qi }, βk,i = sup{xk : (xj ) ∈ Qi }, ck,i = (αk,i + βk,i )/2,
Bi = B((ck,i )∞
k=1 , (α(Q) + ²)/2) for i = 1, . . . , n. It is easy to prove that Qi ⊂ Bi .
Hence χ(Q) ≤ (α(Q) + ²)/2, that is 2χ(Q) ≤ α(Q).
¤
We shall prove that the Hausdorff measure of noncompactness is connected
with the Hausdorff distance.
Theory of sequence spaces
171
Theorem 2.17. Let (X, d) be a metric space. Then (McX , dH ) is a metric
space.
Proof. Clearly dH (S, Q) = 0 if and only if S = Q, and dH (S, Q) = dH (Q, S)
for all S, Q ∈ McX .
To show the triangle inequality, suppose S, Q, F ∈ McX , x ∈ S, y ∈ Q and
z ∈ F . It is easy to prove d(x, F ) ≤ d(x, y) + d(y, F ) ≤ d(x, y) + dH (Q, F ), and
this implies
d(x, F ) ≤ inf d(x, y) + dH (Q, F ) = d(x, Q) + dH (Q, F )
y∈Q
(2.41)
≤ dH (S, Q) + dH (Q, F ).
Replacing x and F by z and S in (2.41), respectively, we obtain
(2.42)
d(z, S) ≤ dH (F, Q) + dH (Q, S)
Finally, (2.41) and (2.42) together imply dH (S, F ) ≤ dH (S, Q) + dH (Q, F ).
¤
c
be
Theorem 2.18. Let (X, d) be a metric space, Q, Q1 , Q2 ∈ MX , and NX
the set of all nonempty and compact subsets of (X, d). Then
(2.43)
|χ(Q1 ) − χ(Q2 )| ≤ dH (Q1 , Q2 ),
(2.44)
c
).
χ(Q) = dH (Q, NX
Proof. Let ² > 0 and d = dH (Q1 , Q2 ). Then it follows from (2.29) and (2.1)
that there exists a finite set S ⊂ X, such that
(2.45)
Q1 ⊂ B(Q2 , d + ²) and
Q2 ⊂ B(S, χ(Q2 ) + ²).
Furthermore, (2.45) implies
(2.46)
Q1 ⊂ B(S, d + χ(Q2 ) + 2²),
and so we conclude
(2.47)
χ(Q1 ) ≤ χ(Q2 ) + d + 2².
Now (2.43) clearly follows from (2.47).
To prove (2.44), let us remark that the inequality ≤ in (2.44) follows from
(2.43). Therefore it suffices to show the inequality ≥. If ² > 0, then there exists a
finite set F ⊂ X, such that
(2.48)
Q ⊂ B(F, χ(Q) + ²) and
F ⊂ B(Q, χ(Q) + ²).
c
Now (2.48) and (2.1) together imply dH (Q, NX
) ≤ dH (Q, F ) ≤ χ(Q) + ².
¤
172
Malkowsky and Rakočević
c
Corollary 2.19. Let NX
be the set of all nonempty and compact subsets of
c
a complete metric space (X, d). Then NX
is a closed subset of (McX , dH ).
Proof. This is an immediate consequence of (2.44).
¤
If the centres of the balls in Definition 2.10 are in Q we have
Definition 2.20. Let (X, d) be a metric space and Q a bounded subset of X.
Then the inner Hausdorff measure of noncompactness of the set Q, denoted by
χi (Q) is defined to be the infimum of the set of all reals ² > 0 such that Q can be
covered by a finite number of balls of radii < ² and centers in Q, that is
½
χi (Q) = inf ² > 0 : Q ⊂
¾
B(xi , ri ), xi ∈ Q, ri < ² (i = 1, . . . n) n ∈ N .
n
[
i=1
The function χi is called inner Hausdorff measure of noncompactness. Hence
the formula in Definition 2.20 can equivalently be stated as follows:
χi (Q) = inf{² > 0 : Q has a finite ²–net in Q}.
If Q, Q1 and Q2 are bounded subsets of the metric space (X, d), then
χi (Q) = 0
if and only if Q
is totally bounded,
χi (Q) = χi (Q),
but in general
Q1 ⊂ Q2
does not imply
χi (Q1 ) ≤ χi (Q2 ),
and
χi (Q1 ∪ Q2 ) 6= max{χi (Q1 ), χi (Q2 )}.
Let Q, Q1 and Q2 be bounded subset of the normed space X. Then
χi (Q1 + Q2 ) ≤ χi (Q1 ) + χi (Q2 ),
χi (Q + x) = χi (Q)
for each x ∈ X,
χi (λQ) = |λ|χi (Q)
for each λ ∈ F,
but in general
χi (Q) 6= χi (co(Q)).
In the fixed point theory in normed space (or more generally in locally convex
spaces) the relation α(Q) = α(co(Q)) is of great importance. Let us remark that
O. Hadžić [26], among other things, studied the inner Hausdorff measure of noncompactness in paranormed spaces. She proved under some additional conditions
the inequality χi (co(Q)) ≤ ϕ[χi (Q)], where ϕ : [0, ∞) 7→ [0, ∞), and, then got some
fixed point theorems for multivalued mappings in general topological vector spaces.
173
Theory of sequence spaces
Istrǎţescu’s measure of noncompactness is closely related to the Hausdorff and
Kuratowski measures of noncompactness. Before we give its definition, we need to
recall that a bounded subset Q of a complete metric space (X, d) is to be said ²–
discrete if d(x, y) ≥ ² for all x, y ∈ Q with x 6= y. Obviously, the set Q is relatively
compact if and only if every ²–discrete set is finite for all ² > 0.
Definition 2.21. (Istrǎţescu, [30]) Let (X, d) be a complete metric space
and Q a bounded subset of X. Then the Istrǎţescu measure of noncompactness
(β–measure, I –measure) of Q, is denoted by β(Q), and defined by
β(Q) = inf{² > 0 : Q has no infinite ²–discrete subsets}.
The function β is called Istrǎţesku’s measure of noncompactness. Let us remark
[11] that β can be defined also by
β(Q) = sup{² > 0 : Q contains an infinite ²–discrete set},
and the above mentioned properties of α are also valid for β (see e.g. [1, 7, 11]).
Theorem 2.22. (Daneš, [11]) Let (X, d) be a metric space and Q be a bounded subset of X. Then
χ(Q) ≤ χi (Q) ≤ β(Q) ≤ α(Q) ≤ 2χ(Q).
1
α(Q) ≤ β(Q) ≤ α(Q) and χ(Q) ≤ β(Q) ≤ 2χ(Q).
2
Now we shall point out the well–known result of Goldenštein, Gohberg and
Markus [19, Theorem 1] (see also [7, Theorem 6.1.1] or [1, 1.8.1]) concerning the
Hausdorff measure of noncompactness in Banach spaces with Schauder basis. Let X
be a Banach space with a Schauder
P∞ basis {e1 , e2 , . . . }. Then each element x ∈ X has
a unique representation x = i=1 φi (x)ei where the functions φi are the basis functionals. Let Pn :P
X 7→ X be the projector onto the linear span of {e1 , e2 , . . . , en },
n
that is Pn (x) = i=1 φi (x)ei . Then, in view of the Banach-Steinhaus theorem, all
operators Pn and I − Pn are equibounded. Now we shall prove
Hence, in particular,
Theorem 2.23. (Goldenštein, Gohberg and Markus [19]) Let X be a Banach
space with a Schauder basis {e1 , e2 , . . . }, Q be a bounded subset of X, and Pn :
X 7→ X the projector onto the linear span of {e1 , e2 , . . . , en }. Then
(2.49)
³
´
1
lim sup sup k(I − Pn )(x)k ≤ χ(Q) ≤
a n→∞ x∈Q
³
´
≤ inf sup k(I − Pn )(x)k ≤ lim sup sup k(I − Pn )(x)k ,
n x∈Q
n→∞
x∈Q
where a = lim supn→∞ kI − Pn k.
Proof. Clearly, for any natural number n we have
(2.50)
Q ⊂ Pn Q + (I − Pn )Q.
174
Malkowsky and Rakočević
It follows from Lemma 2.11, Theorem 2.12 and (2.50) that
(2.51)
χ(Q) ≤ χ(Pn Q) + χ((I − Pn )Q) = χ((I − Pn )Q) ≤ sup k(I − Pn )(x)k.
x∈Q
Now we obtain
µ
¶
χ(Q) ≤ inf sup k(I − Pn )(x)k ≤ lim sup sup k(I − Pn )(x)k ,
(2.52)
n x∈Q
n→∞
x∈Q
Hence it suffices to show the first inequality in (2.49). Let ² > 0 and {z1 , . . . , zk }
be a [χ(Q) + ²]-net of Q. It is easy to show that Q ⊂ {z1 , . . . , zk } + [χ(Q) + ²]BX .
This implies that for any x ∈ Q there exist z ∈ {z1 , . . . , zk } and s ∈ BX such that
x = z + [χ(Q) + ²]s, and so
sup k(I − Pn )(x)k ≤ sup k(I − Pn )(zi )k + [χ(Q) + ²] k(I − Pn )k.
x∈Q
1≤i≤k
This implies
´
³
lim sup sup k(I − Pn )(x)k ≤ (χ(Q) + ²) lim sup kI − Pn k.
n→∞
n→∞
x∈Q
¤
Let us mention that concerning the number a in Theorem 2.23, if X = c0 , then
a = 1, but if X = c, then a = 2 (see e.g. [7, p. 22]).
2.4. Operators. So far we “measured” the noncompactness of a bounded
subset of a metric space. Now we “measure” the noncompactness of an operator.
Definition 2.24. Let κ1 and κ2 any of the measures of noncompactness
defined above on the Banach spaces X and Y , respectively. An operator L : X 7→ Y
is said to be (κ1 , κ2 )–bounded if
(2.53)
L(Q) ∈ MY
for each Q ∈ MX
and there exists a real k with 0 ≤ k < ∞ such that
(2.54)
κ2 (L(Q)) ≤ kκ1 (Q)
for each Q ∈ MX .
If an operator L is (κ1 , κ2 )–bounded then the number kKkκ1 ,κ2 defined by
(2.55)
kLkκ1 ,κ2 = inf{k ≥ 0 : κ2 (L(Q)) ≤ kκ1 (Q)
for each Q ∈ MX }
is called (κ1 , κ2 )–operator norm of L, or (κ1 , κ2 )–measure of noncompactness of
L, or simply measures of noncompactness of L.
If κ1 = κ2 = κ, then we write kLkκ instead of kLkκ,κ .
The next theorem is related to the Hausdorff measure of noncompactness.
175
Theory of sequence spaces
Theorem 2.25. Let X and Y be Banach spaces and L ∈ B(X, Y ). Then
kLkχ = χ(L(SX )) = χ(L(BX )).
Proof. We write B = BX and S = SX . Since co(S) = BX and L(co(S)) =
co(L(S)), it follows from (2.31) that
(2.56)
χ(L(B)) = χ(L(co(S)) = χ(coL(S)) = χ(L(S)),
hence we have by (2.55) and Theorem 2.14 χ(L(B)) ≤ kLkχ . Now we S
show kLkχ ≤
n
χ(L(B)). Let Q ∈ M and {xi }ni=1 be a finite r-net of Q. Then Q ⊂ i=1 B(xi , r)
and obviously
(2.57)
n
[
L(Q) ⊂
L(B(xi , r)).
i=1
It follows from (2.57), Lemma 2.11 and Theorem 2.12 that
χ(L(Q)) ≤ χ
µ[
n
¶
L(B(xi , r))) = χ(L(B(0, r)
= rχ(L(B)),
i=1
and we have χ(L(Q)) ≤ χ(Q)χ(L(B))
¤
Corollary 2.26. Let X, Y and Z be Banach spaces, L ∈ B(X, Y ), L̃ ∈
B(Y, Z) and k · kK the quotient norm on the Banach space B(X, Y )/K(X, Y ).
Then k · kχ is a seminorm on B(X, Y ) and
(2.58)
kLkχ = 0
(2.59)
kLkχ ≤ kLk,
(2.60)
if and only if
L ∈ K(X, Y ),
kL + Kkχ = kLkχ , for each K ∈ K(X, Y ),
(2.61)
kL̃ ◦ Lkχ ≤ kL̃kχ kLkχ .
(2.62)
kLkχ ≤ kLkK .
Proof. The proof is left as an exercise to the reader.
¤
The following results will give a technique for the evaluation of the Hausdorff
measure of noncompactness of an operator on the space l1 .
Theorem 2.27. We have L ∈ B(l1 , l1 ) if and only if there exists an infinite
matrix A = (ank )∞
n,k=0 of complex numbers such that
(2.63)
kAk = sup
k
(2.64)
∞
X
|ank | < ∞
n=0
L(x) = A(x)
for all x ∈ l1 .
176
Malkowsky and Rakočević
In this case
(2.65)
kLk = kAk,
and the operator L uniquely determines the matrix A = (ank )∞
n,k . The operator L
is said to be given (defined) by the matrix A.
Proof. First we assume L ∈ B(X, Y ). We write Ln = Pn ◦ L for all n where
Pn denotes the n–th coordinate, and put ank = Ln (e(k) ) for all n, k = 0, 1, . . . .
Since l1 is a BK space, we have Ln ∈ l1∗ for each n and so Ln (x) = An (x) for each
n by Theorem 1.35. This yields the representation in (2.64). If we choose x = e(k) ,
then
kL(e(k) )k1 =
∞
X
|Ln (e(k) )| =
n=0
∞
X
|ank | ≤ kLkke(k) k1 = kLk
for all ,
n=0
that is
(2.66)
kAk = sup
k
∞
X
|ank | ≤ kLk < ∞
n=0
and (2.63) holds. Further
(2.67)
kL(x)k1 =
∞
X
k=0
|An (x)| ≤
∞
X
|xk |
∞
X
|ank | ≤ kAkkxk1
for all x ∈ l1 ,
n=0
k=0
and so kLk ≤ kAk. This and (2.66) together yield (2.65).
Conversely let condition (2.63) hold. Then obviously supk |ank | < ∞ for all n,
that is An ∈ X β for all n. Let x ∈ l1 . As in (2.67), we obtain A(x) ∈ l1 by (2.63),
whence A ∈ (l1 , l1 ). We define the linear operator L : l1 7→ l1 by (2.64). Then
L ∈ B(l1 , l1 ) by Theorem 1.23 (a).
¤
Theorem 2.28. (Goldenštein, Gohberg and Markus [19]) Let L ∈ B(l1 , l1 )
be given by an infinite matrix A = (ank )∞
n,k=0 . Then
(2.68)
kLkχ = lim sup
m→∞
k
∞
X
|ank |.
n=m
Proof. We write S = S l1 . It follows from Theorems 2.15 and 2.27 that
(2.69)
kLkχ = χ(L(S)) = lim sup
¯
∞ ¯X
X
¯∞
¯
¯
¯
a
x
nk k ¯.
¯
m→∞ x∈S
n=m k=0
177
Theory of sequence spaces
The limit in (2.68) obviously exists. From
¯
∞ ¯X
∞ X
∞ X
∞
∞
X
X
X
¯∞
¯
¯
¯ ≤ sup
sup
a
x
|a
x
|
=
sup
|ank ||xk |
nk
k
nk
k
¯
¯
x∈S n=m
k=0
x∈S n=m
≤ sup
k
∞
X
x∈S
k=0
k=0 n=m
|ank |
n=m
and (2.69) we obtain
(2.70)
∞
X
kLkχ ≤ lim sup
m→∞
k
|ank |.
n=m
To prove the converse inequality, we choose x = e(k) ∈ l1 . Since L(e(k) ) =
A = (ank )∞
n=0 , Theorem 2.15 implies
k
χ({L(e(k) ) : k = 0, 1, . . . }) = lim sup
m→∞
k
∞
X
|ank | ≤ χ(L(S)).
n=m
This and inequality (2.70) together yield (2.68).
¤
As an immediate consequence of Theorem 2.28, we have
Corollary 2.29. Let L ∈ B(l1 , l1 ) be given by the infinite matrix A =
(ank )∞
n,k=0 . Then L is compact if and only if
lim sup
m→∞
k
∞
X
|ank | = 0.
n=m
Let us mention that measures of noncompactness are of special interest in
spectral theory, the theory of Fredholm and semi–Fredholm operators (see e.g. [8,
9, 17, 19, 21, 45, 78, 87, 88, 89, 93, 94, 101, 102, 110, 115, 116].
3. Matrix domains
In this section, we shall deal with sequence spaces related to the concepts of
ordinary and strong summability, spaces of sequences of differences and sequences
that are strongly convergent and bounded. We shall characterize matrix transformations between these spaces and apply the Hausdorff measure of noncompactness
to give necessary and sufficient conditions for these matrix maps to be compact
operators. This section contains some of our recent research results which can be
found in [34, 64, 65, 68, 69, 70, 71, 72, 73] and in the survey articles [32, 66, 67].
Let A be an infinite matrix and x = (xk )∞
k=0 be a sequence. The sequence x is
said to be A–summable to l ∈ C if
An (x) =
∞
X
k=0
ank xk → l
(n → ∞);
we shall write x → l(A).
178
Malkowsky and Rakočević
This means An ∈ xβ = x−1 ∗ cs for all n and A(x) ∈ c.
The sequence x is said to be strongly summable A to l ∈ C if
∞
X
ank |xk − l| → 0 (n → ∞);
we shall write x → l[A].
k=0
The sequence x is said to be absolutely summable A if
∞
X
|An (x)| < ∞.
n=0
We shall mainly be interested in the first two concepts.
3.1. Ordinary and strong matrix domains. In this subsection, we define
ordinary and strong matrix domains and study their topological properties.
Definition 3.1. Let X be a set of sequences and A an infinite matrix. The
sets
©
ª
XA = x ∈ ω : A(x) ∈ X
and
∞
©
¡X
¢∞
ª
X[A] = x ∈ ω : A(|x|) =
ank |xk | n=0 ∈ X
k=0
are called the (ordinary) matrix domain and strong matrix domain of A. In the
special case where X = c, the sets cA and c[A] are called convergence domain and
strong convergence domain of A.
The sets cA and c[A] are closely related to the concepts of ordinary and strong
summability. Obviously x → l(A) if and only if x ∈ cA and x → l[A] if and only if
x − le ∈ (c0 )[A] .
It is known that the ordinary matrix domain of an FK space again is an FK
space [108, Theorem 4.3.12, p. 63] or [91, Proposition 4.2.1, p. 101]. Since we shall
here confine our studies to BK spaces and matrix domains of triangles, we shall
only prove the special result. We need the following
Lemma 3.2. Let X be a linear space, (Y, k·k) a normed space and T : X 7→ Y
a linear one-to-one map. Then X becomes a normed space with kxkX = kT (x)k.
If, in addition, Y is a Banach space and T is onto Y , then (X, k · kX ) is a Banach
space.
Proof. The proof is elementary and left to the reader.
¤
Theorem 3.3. Let T be a triangle and (X, k · k) be a BK space. Then XT is
a BK space with kxkT = kT (x)k.
Proof. We define the map LT : XT 7→ X by LT (x) = T (x) for all x ∈
XT . Then LT is linear, one–to–one, since T is a triangle, and onto X, since
XT = L−1
T (X) and LT is one–to–one. By Lemma 3.2, XT is a Banach space.
Theory of sequence spaces
179
To show that the coordinates are continuous in XT , let x(n) → x in XT . Then
(n)
yk = Tk (x(n) ) → yk = Tk (x), since X is a BK space. Let S be the inverse of T ,
Pk
Pk
(n)
(n)
also a triangle. Then xk = j=0 skj yj → j=0 skj yj = xk . This shows that
the coordinates are continuous on XT .
¤
As a special case of Theorem 3.3, we obtain
Corollary 3.4. [108, Theorem 4.3.13, p. 64] Let T be a triangle. Then cT is
a BK space with kxkT,∞ = kT (x)k∞ .
Theorem 3.5. [108, Theorem 4.3.14, p. 64] If X is a closed subspace of Y ,
then XA is a closed subspace of YA .
Proof. Define the map f : YA 7→ Y by f (y) = A(y), a continuous map. Then
fA is continuous by Theorem 1.17, and so XA = f −1 (X) is closed.
¤
A result similar to Theorem 3.3 holds for the strong matrix domains of triangles. We call a norm k · k a sequence space X monotone, if |x̃k | ≤ |xk | (k = 0, 1, . . . )
implies kx̃k ≤ kxk.
Theorem 3.6. [34, Theorem 1] Let X be a normal BK space with monotone
norm k · k, T a triangle and B a positive triangle. Then X[B] is a BK space with
kxkX[B] = kB(|x|)k for all x ∈ X[B] .
Proof. We write k · k0 = k · kX[B] for short. Obviously, k · k0 is a norm on X[B] .
Further, since X is a BK space,
kx(m) − xk0 = kB(|x(m) − x|)k → 0 (m → ∞)
Pn
(m)
implies Bn (|x(m) − x|) = k=0 bnk |xk − xk | → 0 (m → ∞) for all n. Thus
|x(m)
− xn ≤
n
1
Bn (|x(m) − x|) → 0 (m → ∞) for all n.
bnn
Hence the norm k · k0 is stronger than the metric of ω on X[B] . Let (x(m) )∞
m=0 be
a Cauchy sequence in X[B] , hence in ω by what we have just shown. Then there is
y ∈ ω such that
x(m) → y
(3.1)
in ω.
Further, by the completeness of X, there is z ∈ X such that
B(|x(m) |) → z
(3.2)
in X.
(m)
From (3.1), we conclude xk → yk (m → ∞) for each fixed k, hence Bn (|x(m) |) →
Bn (|y|) (m → ∞) for all n, and consequently
(3.3)
B(|x(m) |) → B(|y|) in ω.
Finally (3.2) and (3.3) together imply z = B(|y|) ∈ X, that is y ∈ X[B] .
¤
There is no general method to find the Schauder basis of a matrix domain XA
or X[A] from that of X not even when A is a positive triangle. We give a special
result which will be applied later.
180
Malkowsky and Rakočević
Theorem 3.7. [34, Theorem 2] (a) Let X be a BK space with basis (bk )∞
k=0 ,
U = {u ∈ ω : uk 6= 0 for all k} u ∈ U and c(k) = (1/u) · b(k) (k = 0, 1, . . . ) where
(k) ∞
1/u = (1/uk )∞
)k=0 is a basis for Y = u−1 ∗ X.
k=0 . Then (c
(b) Let u ∈ U be a sequence such that |u0 | ≤ |u1 | ≤ . . . and |un | → ∞ for
n → ∞, and T a triangle with tnk = 1/un (0 ≤ k ≤ n) and tnk = 0 (k > n) for all
n = 0, 1, . . . . Then (c0 )T has AK.
Proof. (a) Let k · k be the BK norm on X. Then Y is a BK space with
kyku = ku · yk (y ∈ Y ) by Theorem 3.3. Further u · c(k) = b(k) ∈ X (k = 0, 1, . . . )
implies c(k) ∈
(k = 0, 1, . . . ). Finally let y ∈ Y be given. Then u · y = x ∈ X
PY
m
and xhmi = k=0 λk b(k) → x (m → ∞) in X. We put y hmi = (1/u) · xhmi . Then
P∞
u · y hmi = xhmi → x = u · y inX, hence y hmi → y in Y , that is y = k=0 λk c(k) .
Obviously, this representation is unique.
¯ 1 P
¯
n
¯
¯
(b) (c0 )T is a BK space with respect to kxk(c0 )T = sup¯
xk ¯, by Theorem
n un k=0
3.3. Further |un | → ∞ (n → ∞) implies φ ⊂ (c0 )T . Let ε > 0 and x ∈ (c0 )T be
given. Then there is a nonnegative integer n0 such that |Tn (x)| < ε/2 for all n ≥ n0 .
Let m > n0 . Then
[m]
kx − x
k(c0 )T
¯
¯ 1
= sup ¯¯
n≥m+1 un
n
X
k=m+1
¯
¯
xk ¯¯ ≤ sup |Tn (x) < +|Tm (x)| < ε.
n≥m+1
Obviously, the representation is unique.
¤
3.2. Matrix transformations into matrix domains. In this subsection, we
shall show that, for triangles T , the characterizations of the classes (X, Y ) and
(X, Y[T ] ) can be reduced to that of (X, Y ).
Theorem 3.8. [65, Theorem 1], [71, Proposition 3.4] Let T be a triangle.
(a) Then, for arbitrary subsets X and Y of ω, A ∈ (X, YT ) if and only if
B = T A ∈ (X, Y ).
(b) Further, if X and Y are BK spaces and A ∈ (X, YT ), then
(3.4)
kLA k = kLB k.
Proof. (a) The proof of part (a) is straightforward and can be found in [65,
Theorem 1].
(b) Let A ∈ (X, YT ). Since Y is a BK space and T a triangle, YT is a BK space
with
(3.5)
kykYT = kT (y)kY
(y ∈ YT )
by Theorem 3.3. Thus A is continuous by Theorem 1.17 and consequently
(3.6)
kLA k = sup{kLA (x)kYT : kxk = 1} = sup{kA(x)kYT : kxk = 1} < ∞.
181
Theory of sequence spaces
Further, since B is continuous,
(3.7)
kLB k = sup{kLB (x)kY : kxk = 1} = sup{kB(x)kY : kxk = 1} < ∞.
Let x ∈ X. Since An ∈ X β for all n = 0, 1, . . . , we have x ∈ ωA . Further Tn ∈ φ
(n = 0, 1, . . . ), since T is a triangle. Thus B(x) = (T A)(x) = T (A(x)) (cf. [108,
Theorem 1.4.4, p. 8]), and (3.4) follows from (3.5), (3.6) and (3.7).
¤
For the characterization of the class (X, Y[T ] ), we need the following lemma.
Lemma 3.9. [81] Let a0 , a1 , . . . , an ∈ C. Then
n
X
|ak | ≤ 4 ·
k=0
max
N ∈{0,...,n}
¯X ¯
¯
¯
¯
ak ¯¯.
¯
k∈N
Proof. First we consider the case where a0 , a1 , . . . , an ∈ R. We put N + =
{k ∈ {0, . . . , n} : ak ≥ 0} and N − = {k ∈ {0, . . . , n} : ak < 0}. Then
n
X
k=0
¯
¯ ¯
¯
¯ X
¯ ¯ X
¯
|ak | = ¯¯
ak ¯¯ + ¯¯
ak ¯¯ ≤ 2 · max
N ∈{0,...,n}
k∈N +
k∈N −
¯
¯
¯X ¯
¯
¯
a
k ¯.
¯
k∈N
Now let a0 , a1 , . . . , an ∈ C. We write ak = αk + iβk (k = 0, 1, . . . , n). For any
subset N of {0, . . . , n}, we write
xN =
X
αk , yN =
k∈N
X
βk
and
zN = xN + iyN =
k∈N
X
ak .
k∈N
Now we choose subsets Nr , Ni and N∗ of {0, . . . , n} such that
|xNr | =
max
N ⊂{0,...,n}
|xN |, |yNi | =
max
N ⊂{0,...,n}
|yN |
and
|zN∗ | =
max
N ⊂{0,...,n}
|zN |.
Then, for all N ⊂ {0, . . . , n}, we have |xN |, |yN | ≤ |zN∗ | and |xNr |+|yNi | ≤ 2·|zN∗ |.
Thus, by the first part of the proof,
n
X
k=0
|ak | ≤
n
X
k=0
|αk | +
n
X
|βk | ≤ 2(|xNr | + |yNi |) ≤
k=0
¯X ¯
¯
¯
≤ 4|zN∗ | = 4 · max ¯¯
zk ¯¯.
N ⊂{0,...,n}
k∈N
¤
182
Malkowsky and Rakočević
Theorem 3.10. [70, Theorem 2] Let A be an infinite matrix, B a positive
triangle. For each m ∈ N0 , let Nm be a subset of the set {0, 1, . . . , m}, N =
(Nm )∞
m=0 the sequence of the subsets Nm and N the set of all such sequences N .
Furthermore, for each N ∈ N , we define the matrix S N = S N (A) by
X
sN
bmn ank (m, k = 0, 1, . . . ).
mk =
n∈Nm
Then, for arbitrary subsets X of ω and any normal set Y of sequences, A ∈ (X, Y[B] )
if and only if S N (A) ∈ (X, Y ) for all sequences N in N .
Proof. First we assume A ∈ (X, Y[B] ). Then An ∈ X β (n = 0, 1, . . . ) implies
∈ X β for all m and all N ∈ N . For each x ∈ X, we put y = B(|(A(x)|). Then
A(x) ∈ Y[B] , that is y ∈ Y , and
¯
¯X
¯ ¯ X
∞
X
¯∞ N
¯ ¯
¯
N
¯
¯
¯
bmn
|Sm (x)| = ¯
smk xk ¯ = ¯
ank xk ¯¯ ≤ |ym | (m = 0, 1, . . . )
N
Sm
n∈Nm
k=0
k=0
N
for all N ∈ N together imply S (x) ∈ Y for all N ∈ N , since Y is normal. Thus
S N ∈ (X, Y ) for all N ∈ N .
N
∈ X β for all m
Conversely we assume S N ∈ (X, Y ) for all N ∈ N . Then Sm
N
∞
and for all N ∈ N , in particular, for N = ({m})m=0 , Sm = bmm Am ∈ X β , hence
Am ∈ X β , since bmm 6= 0. Further, let x ∈ X be given. For every m = 0, 1, . . . , we
(0)
choose the set Nm ⊂ {0, . . . , } such that
¯
¯ ¯
¯
¯ X
¯ ¯
¯
¯
¯
¯
¯
b
A
(x)
=
max
b
A
(x)
mn n
¯
¯ ¯Nm ⊂{0,...,m} mn n ¯.
(0)
n∈Nm
Then, by Lemma 3.9,
¯
¯
¯ X
¯
¯ (0) ¯
|ym | ≤ 4 · ¯¯
bmn An (x)¯¯ = 4 · ¯S N (x)¯.
(0)
n∈Nm
(0)
By hypothesis, S N (x) ∈ Y , and the normality of Y implies y = B(|A(x)|) ∈ Y ,
that is A ∈ (X, Y[B] ).
¤
3.3. Bounded and convergent difference sequences of order m. Now we
apply the results of the previous subsections to sets of of bounded and convergent
sequences of order m which may be considered as ordinary matrix domains of a
certain triangle. We shall give their Schauder bases and their α– and β–duals. The
results may be found in [69] and [39, 63] in the special case m = 1.
P(m)
Let m denote a positive integer throughout and the operators ∆(m) ,
:
ω 7→ ω be defined by
(∆(1) x)k = ∆(1) xk = xk − xk−1 ,
∆(m) = ∆(1) ◦ ∆(m−1) ,
³P
(1) ´
k
P
x =
xj
k
(m)
P
=
(k = 0, 1, . . . ),
j=0
(1)
P
◦
m−1
P
(m ≥ 2).
183
Theory of sequence spaces
We shall write ∆ = ∆(1) for short and use the convention that any term with a
negative subscript is equal to naught. For any subset X of ω let
©
ª
X(∆(m) ) = x ∈ ω : ∆(m) x ∈ X .
We shall be interested in the cases where X = c0 , X = c or X = l∞ . The following
results are well known and can be found in [27]:
(3.8) (∆(m) x)k =
m
X
(−1)j
j=0
µ ¶
m
xk−j =
j
µ
m
X
(−1)k−j
j=max{0,k−m}
¶
m
xj
k−j
(k = 0, 1, . . . ),
(3.9)
(3.10)
(3.11)
µX
¶
(m) ¶
k µ
X
m+k−j−1
=
x
xj
k−j
k
j=0
(m)
X
◦∆
=∆
(m)
◦
¶
k µ
X
m+j−1
j=0
(3.12)
(m)
j
(m)
X
=
(k = 0, 1, . . . ),
= id, the identity on ω,
µ
¶
m+k
(k = 0, 1, . . . )
k

positive constants M1 , M2 such that
 there are µ
¶
m+k
m
 M1 k ≤
≤ M2 k m for all k = 1, 2 . . .
k
As an immediate consequence of Example 1.13 and Theorems 3.3 and 3.5, we
obtain
Corollary 3.11. [69, Proposition 1] Let m be a positive integer. Then the
sets l∞ (∆(m) ), c(∆(m) ) and c0 (∆(m) ) are BK spaces with k · k defined by
¯m
¯
µ ¶
¯¡ (m) ¢ ¯
¯X
¯
j m
¯
¯
¯
kxk = sup ∆ x k = sup¯ (−1)
xk−j ¯¯
j
k
k j=0
and c0 (∆(m) ) and c(∆(m) ) are closed subspaces of l∞ (∆(m) ).
Now we shall give Schauder bases for the spaces c0 (∆(m) ) and c(∆(m) ).
184
Malkowsky and Rakočević
Theorem 3.12. [69, Theorem 1] Let m be a positive integer. We define the
sequences bk (m) by
µ
¶
m+n
(−1)
bn (m) =
(n = 0, 1, . . . ),
n
½
0
(n ≤ k − 1)
¡m+n−k−1¢
b(k)
(k = 0, 1, . . . ).
n (m) =
(n ≥ k)
n−k
(m)
(a) Then (b(k) (m))∞
). More precisely, every sequence
k=0 is a basis of c0 (∆
∞
(m)
x = (xk )k=0 ∈ c0 (∆ ) has a unique representation
(3.13)
x=
∞
X
λk (m)b(k) (m) where λk (m) = (∆(m) x)k
(k = 0, 1, . . . ).
k=0
(m)
(b) Then (b(k) (m))∞
). More precisely, every sequence
k=−1 is a basis of c(∆
∞
(m)
x = (xk )k=0 ∈ c(∆ ) has a unique representation
(3.14)
x = lb(−1) (m) +
∞
X
(λk (m) − l)b(k) (m) where l = lim (∆(m) x)k .
k→∞
k=0
Proof. (a) For k = 0, 1, . . . , we put
b(k) = e −
k−1
X
½
(k)
e(p) , that is bj
=
p=0
0
(j ≤ k − 1)
1
(j ≥ k).
Then by (3.9) and (3.11),
µ(m−1)
X
¶
b(k)
=
n
=
¶
n µ
X
m − 1 + n − j − 1 (k)
bj
n−j
j=0
( Pn ¡m−1+n−j−1¢
(n ≥ k)
j=k
n−j
0
(n ≤ k − 1),
µ
¶
µ
¶
µ
¶
n
n−k
X
X m−1+l−1
m−1+n−j−1
m−1+n−k
=
=
(n ≥ k),
n−j
l
n−k
j=k
hence b(k) (m) =
(3.15)
l=0
P(m−1)
b(k)
(k = 0, 1, . . . ), and by (3.10),
∆(m) b(k) (m) = ∆b(k) = e(k) ∈ c0
(k = 0, 1, . . . )
Thus
(3.16)
b(k) (m) ∈ c0 (∆(m) )
(k = 0, 1, . . . ).
185
Theory of sequence spaces
Let x = (xk )∞
∈ c0 (∆(m) ) be given. For every nonnegative integer p, we put
Pp k=0
hpi
x = k=0 λk (m)b(k) (m). Then by the linearity of ∆(m) and by (3.15)
∆(m) xhpi =
p
X
λk (m)∆(m) b(k) (m) =
½
∆
(∆(m) x)k e(k)
k=0
k=0
(m)
p
X
hpi
(x − x
)n =
0
(n ≤ p)
(m)
(∆
x)n
(n ≥ p + 1).
Given ε > 0 there is an integer p0 such that |(∆(m) x)p | < ε/2 for all p ≥ p0 , hence
kx − xhpi k = sup |(∆(m) x)n | ≤ sup |(∆(m) )n | ≤ ε/2 < ε
n≥p
n≥p0
for all p ≥ p0 . This proves the representation
in (3.13). To show the uniqueness
P∞
of this representation we assume x = k=0 µk b(k) . Since ∆(m) : c0 (∆(m) ) → c0
obviously is a continuous linear operator, we have by (3.15)
(∆
(m)
x)n =
∞
X
k=0
(m) (k)
µk (∆
b
)n =
∞
X
µk e(k)
n = µn
(n = 0, 1, . . . ).
k=0
P(m)
e implies ∆(m) b(−1) (m) = e ∈ c, that is b(−1) (m) ∈
(b) First b(−1) (m) =
(m)
c(∆ ). In view of (3.16) and the fact that c0 (∆(m) ) ⊂ c(∆(m) ), we have b(k) (m) ∈
(m)
c(∆(m) ) for all k = −1, 0, 1, . . . . Let x = (xk )∞
) be given. Then there
k=0 ∈ c(∆
is a unique number l such that (3.14) holds. We put y = x − l · b(−1) (m). Then
∆(m) y = ∆(m) (x − lb(−1) (m)) = ∆(m) x − le, that is y ∈ c0 (∆(m) ),
and it follows from part (a) that x has a unique representation (3.14).
¤
Now we shall give the α–duals of the sets c0 (∆(m) ), c(∆(m) ) and l∞ (∆(m) ). If
u ∈ U , then obviously
(3.17)
(u−1 ∗ X)† = (1/u)−1 ∗ X † († ∈ {α, β}) for every subset X of ω.
Theorem 3.13. [69, TheoremP
2] Let m be a positive integer.
∞
(a) We put M α (m) = {a ∈ ω : k=0 |ak |k m < ∞}. Then
(3.18)
¡
¢α ¡
¢α ¡
¢α
c0 (∆(m) ) = c(∆(m) ) = l∞ (∆(m) ) = M α (m).
(b) We put M αα (m) = {a ∈ ω : supk≥1 |ak |k −m < ∞}. Then
(3.19)
¡
¢αα ¡
¢αα ¡
¢αα
c0 (∆(m) )
= c(∆(m) )
= l∞ (∆(m) )
= M αα (m).
186
Malkowsky and Rakočević
Proof. (a) First we assume a ∈ M α (m). Then
∞
X
(3.20)
|ak |k m < ∞.
k=0
Let x ∈ l∞ (∆(m) ). Then there is a positive constant M such that |(∆(m) x)k | ≤ M
(k = 0, 1, . . . ), and by (3.10), (3.9), (3.11), (3.12) and (3.20)
∞
X
∞
X
|ak xk | =
k=0
¯µX
µ
¶¶
¯ (m)
∆(m) x
|ak |¯¯
k
k=0
¯
¯
¯
¯
¶ ¯
¶¯µ
¯
m + k − j − 1 ¯¯
(m)
≤
|ak |
∆ x ¯¯
¯
k−j
j
j=0
k=0
µ
¶
∞
∞
X
X
m+k
≤M
|ak |
≤ M · M2
|ak |k m < ∞.
k
k µ
X
∞
X
k=0
k=0
Thus we have shown
¡
¢α
M α (m) ⊂ l∞ (∆(m) ) .
(3.21)
Conversely let a ∈
/ M α (m). By (3.12), there is a sequence (k(s))∞
s=0 of integers
0 = k(0) < k(1) < . . . such that
k(s+1)−1
X
(3.22)
k=k(s)
µ
¶
m+k
|ak |
≥s+1
k
(s = 0, 1, . . . ).
We define the sequence x by
xk =
s−1
X
l=0
1
l+1
k(l+1)−1 µ
¶
µ
¶
k
X
m+k−j−1
1
m+k−j−1
+
k−j
k−j
s+1
X
j=k(l)
j=k(s)
(k(s) ≤ k ≤ k(s + 1) − 1; s = 0, 1, . . . ).
If we define the sequence y ∈ c0 by yk = 1/(s + 1) for k(s) ≤ k ≤ k(s + 1) − 1 (s =
P(m)
y. Thus ∆(m) x = y ∈ c0
0, 1, . . . ), then it easily follows from (3.9) that x =
(m)
and x ∈ c0 (∆ ). On the other hand by (3.11) and (3.22)
k(s+1)−1
X
k=k(s)
k(s+1)−1
|ak xk | ≥
X
|ak |
k=k(s)
1
=
s+1
¶
k µ
1 X m+j−1
s + 1 j=0
j
k(s+1)−1
X
k=k(s)
µ
¶
m+k
|ak |
≥ 1 (s = 0, 1, . . . ).
k
187
Theory of sequence spaces
Thus a ∈
/ c0 (∆(m) ), and we have shown
¡
¢α
c0 (∆(m) ) ⊂ M α (m).
(3.23)
Since c0 (∆(m) ) ⊂ c(∆(m) ) ⊂ l∞ (∆(m) ), (3.18) follows from (3.21) and (3.23).
α
m ∞
−1
(b) Since y = (xk+1 )∞
∗ l1 , and
k=0 ∈ M (m) if and only if y ∈ ((k + 1) )k=0 )
α
since l1 = l∞ , identity (3.19) follows from (3.17) and part (a).
¤
To determine the β–duals of the sets c0 (∆(m) ), c(∆(m) ) and l∞ (∆(m) ), we need
a few results.
Lemma 3.14. [69, Lemma 1] Let m be a positive integer. Then for arbitrary
sequences a
µ
µ
¶¶∞
m+k
ak
∈ cs
k
k=1
¡
if and only if
ak k m
¢∞
k=1
∈ cs.
Proof. We define the sequences b and c by
¡m+k¢
bk =
k
km
(k = 1, 2, . . . )
and
c = 1/b.
Since csβ = bv [108, Theorem 7.3.5(v), p. 110], it suffices to show that b, c ∈ bv. It
is well known that limk→∞ bk = 1/m! [27, p. 97]. Therefore we have to show that b
is monotone. We define the function f on [0, 1/2] by f (x) = (1+mx)(1−x)m . Then
f 0 (x) = −m(1 − x)m−1 (m + 1)x ≤ 0 for all x ∈ [0, 1/2], whence f (x) ≤ f (0) = 1
for all x ∈ [0, 1/2]. Thus
bk+1
k + 1 + m km
=
=
bk
k + 1 (k + 1)m
µ
1+
m
k+1
¶µ
1−
1
k+1
¶m
µ
=f
1
k+1
¶
for all k ≥ 1.
≤1
¤
Lemma 3.15. [63, Lemma 1] Let (Pn ) be a sequence of non decreasing positive reals. Then y ∈ cs implies
µ X
¶
∞
yn+k−1
lim Pn
= 0.
n→∞
Pn+k
k=1
Proof. The proof can be found in [39, Lemma 3] and [63, Lemma 1].
¤
(Pn )∞
n=1
Corollary 3.16. [63, Corollary 1] Let
be a sequence of nondecreasing
P∞
positive reals. Then a ∈ (Pn )−1 ∗cs implies R ∈ (Pn )−1 ∗co where Rn = k=n+1 ak
(n = 1, 2, . . . ).
Proof. Put yk = Pk+1 ak+1 (k = 1, 2, . . . ) in Lemma 3.15.
¤
188
Malkowsky and Rakočević
We shall frequently apply the following two versions of Abel’s summation by
parts: Let b, c ∈ ω. We put
s = s(c) =
(1)
X
c and, if c ∈ cs, Rk = Rk (c) =
∞
X
ck (k = 0, 1, . . . ).
j=k
Then
n
X
(3.24)
k=0
n
X
(3.25)
bk ck = −
bk ck =
k=0
n
X
sk ∆bk+1 + sn bn+1
k=0
n
X
Rk ∆bk − bn Rn+1
(n = 0, 1, . . . )
(n = 0, 1, . . . ).
k=0
Theorem 3.17. [69, Theorem 3] Let m be a positive integer.
P∞
P∞
(m)
(m−1)
(a) We put Rk1 = Rk =
=
(k = 0, 1, . . . ) for
j=k aj , Rk
j=k Rj
m ≥ 2 and
½
β
(m) =
M∞
a∈ω:
∞
X
ak k m converges and
k=0
∞
X
¾
(m)
|Rk | < ∞ .
k=0
Then
¡
(3.26)
¢β ¡
¢β
β
c(∆(m) ) = l∞ (∆(m) ) = M∞
(m).
(b) Further, let c+
0 denote the set of all positive sequences in c0 . We put
½
M0β (m) =
¶
¾
k µ
X
m+k−j−1
vj converges for all v ∈ c+
0
k−j
j=0
k=0
¾
∞
\½
X
(m)
a∈ω:
|Rk | < ∞ .
a∈ω:
∞
X
ak
k=0
Then
(3.27)
¡
¢β
c0 (∆(m) ) = M0β (m).
P(p)
Proof. (a) For all positive integers p let s(p) =
e; we write s = s(1) . First
β
β
β
we assume m = 1 and write M∞ = M∞ (1). Let a ∈ M∞ . Then
(3.28)
R ∈ l1
(3.29)
as ∈ cs
189
Theory of sequence spaces
Now condition (3.29) and Corollary 3.16 together imply
¡
(3.30)
Rn+1 sn
¢∞
n=0
∈ c0 .
Let x ∈ l∞ (∆). From (3.25) with b = x and c = a, we have
n
X
(3.31)
ak xk =
k=0
n
X
Rk (∆x)k − Rn+1 xn (n = 0, 1, . . . ).
k=0
Since ∆x ∈ l∞ , condition (3.28) implies
(3.32)
R∆x ∈ cs.
Further there is a constant M > 0 such that |(∆x)k | ≤ M for all k and so |xn | ≤
P(1) (1)
˙ + 1) = M ṡn for all n. Now condition (3.30) implies
|(
(∆ x))k | ≤ M (n
¡
(3.33)
Rn+1 xn
¢∞
n=0
∈ c0 .
Finally (3.31), (3.32) and (3.33) together imply ax ∈ cs for all x ∈ l∞ , that is
a ∈ (l∞ (∆))β .
Conversely, let a ∈ (c(∆))β . Then ax ∈ cs for all x ∈ c(∆). First e ∈ c(∆)
implies a = ae ∈ cs, hence the sequence R is defined. Further, for x = s, we have
∆x = e ∈ c, that is x ∈ c(∆), and condition (3.29) holds. By Corollary 3.16, we
have (3.30), and again this yields (3.33) for all x ∈ c(∆). From (3.31) we conclude
R∆x ∈ cs for all x ∈ c(∆), and so R ∈ cβ = l1 . Now we assume that identity (3.26)
β
holds for some integer m ≥ 1. Let a ∈ M∞
(m + 1). Then by
R(m+1) = Rm (R) ∈ l1 .
(3.34)
and, by Lemma 3.14,
as(m+1) ∈ cs
(3.35)
Applying identity (3.24) with b = R and c = s(m) we obtain
(3.36)
n
X
k=0
(m)
sk Rk =
n
X
(m+1)
ak sk
+ Rn+1 s(m+1)
n
k=0
By Corollary 3.16, condition (3.35) implies
(3.37)
¡
Rn+1 s(m+1)
n
¢∞
n=0
and consequently by (3.36)
(3.38)
s(m) R ∈ cs
∈ c0
(n = 0, 1, . . . ).
190
Malkowsky and Rakočević
Now, by assumption, (3.34) and (3.38) together imply
(3.39)
¡
¢β
R ∈ l∞ (∆(m) ) .
Let x ∈ l∞ (∆(m+1) ) be given. Since x ∈ l∞ (∆(m+1) ) if and only if y = ∆x ∈
l∞ (∆(m) ), condition (3.39) implies
(3.40)
R∆x ∈ cs
for all x ∈ l∞ (∆(m+1) ).
Further there is a positive constant M such that |(∆(m+1) x)j | ≤ M (j = 0, 1, . . . )
and thus
¯µ(m+1)
¶
¯ X
(m+1)
¯
|xk | = ¯
(∆
(x))
k
¯ X
¶
¶¯µ
k µ
¯
m + k − j ¯¯
(m+1)
¯≤
x
¯ ∆
¯
k−j
j=0
j
¯
¯
¯
¯
¶
µ(m+1)
k µ
X
X ¶
m+k−j
(m+1)
≤M·
=M
e
= M · sk
k
−
j
k
j=0
for k = 0, 1, . . . , and condition (3.37) implies
(3.41)
¡
Rn+1 xn
¢∞
n=0
∈ c0 .
Finally (3.31), (3.40) and (3.41) together imply ax ∈ cs for all x ∈ l∞ (∆(m+1) ),
¡
¢β
consequently a ∈ l∞ (∆(m+1) ) .
¡
¢β
Conversely let a ∈ c(∆(m+1) ) . Then ax ∈ cs for all x ∈ c(∆(m+1) ). First,
e ∈ c(∆(m+1) ) implies a = ae ∈ cs, hence the sequence R is defined. Further for
P(m+1)
e) = e ∈ c, that is x ∈ c(∆(m+1) ),
x = s(m+1) we have ∆(m+1) x = ∆(m+1) (
and condition (3.35) is satisfied. By Corollary 3.16, we have (3.37) and again this
yields (3.41) for all x ∈ c(∆(m+1) ). From (3.31) we conclude R∆x ∈ cs for all x ∈
c(∆(m+1) ) and consequently R ∈ (c(∆(m) ))β . This implies R(m+1) = R(m) (R) ∈ l1
by assumption.
P(p)
(v);
(b) For all positive integers p and all sequences v ∈ c0 , let t(p) (v) =
we write t(v) = t(1) (v). The proof of part (b) is exactly the same as that of part
(a) with s, s(m) and s(m+1) replaced by t(v), t(m) (v) and t(m+1) (v).
¤
Remark 3.18. By [63, Theorem 2 (c)] it is obvious that (c0 (∆(m) ))β 6=
(l∞ (∆(m) ))β .
3.4. Matrix transformations in the spaces c0 (∆(m) ), c(∆(m) ) and l∞ (∆(m) )
and their measures of noncompactness. In this subsection we shall characterize
matrix transformations between the spaces of bounded and convergent m–th order
difference sequences and apply the Hausdorff measure of noncompactness to give
necessary and sufficient conditions for these matrix maps to be compact operators.
191
Theory of sequence spaces
Lemma 3.19. [69, Lemma 4] Let m be a positive integer Then
kak∗ = kR(m) k1 =
(3.42)
∞
X
(m)
|Rk |
k=0
on any of the spaces (c0 (∆(m) ))β , (c(∆(m) ))β and (l∞ (∆(m) ))β .
Proof. Let X be any of the sequences c0 , c or l∞ . If m = 1 and a ∈
(X(∆(1) ))β , then
∞
X
(3.43)
ak xk =
k=0
∞
X
(k)
Rk ∆(1) xk
for all x ∈ X(∆(1) )
k=0
by the proof of Theorem 3.17. Since x ∈ X(∆(1) ) if and only if ∆(1) x ∈ X, this
implies R(1) ∈ X β = l1 . It is well known that k · k∗ = k · k1 on X β , and (3.42)
follows from the definition of the norm on X(∆(1) ).
Now we assume that (3.42) holds for some integer m ≥ 1. Let a ∈ X(∆(m+1) ).
Again, by the proof of Theorem 3.17, (3.43) holds for all x ∈ X(∆(m+1) ). Since
x ∈ X(∆(m+1) ) if and only if ∆(1) ∈ X(∆(m) ), this implies R(1) ∈ (X(∆(m) ))β ,
and by assumption kak∗ = kR(m) (R(1) )k1 = kR(m+1) k1 .
¤
Theorem 3.20. [69, Theorem 4] Let m be a positive integer and A be an infiP∞
P∞
(m−1)
(1)
(m)
nite matrix. For each n, we put Rnk = Rnk = j=k anj and Rn = j=k Rnj
for m ≥ 2.
(a) Then A ∈ (l∞ (∆(m) ), l∞ ) if and only if
∞
X
(3.44)
k m ank
converges for all n = 0, 1, . . .
k=0
and
(3.45)
sup
n
∞
X
(m)
|Rnk | < ∞.
k=0
Further (l∞ (∆(m) ), l∞ ) = (c(∆(m) ), l∞ ).
(b) Then A ∈ (c0 (∆(m) ), l∞ ) if and only if condition (3.45) holds and
(3.46)
∞
X
k=0
¶
k µ
X
m+k−j−1
ank
vj
k−j
j=0
converges for all v ∈ c+
0
and for all n = 0, 1, . . . .
(c) Then A ∈ (c0 (∆(m) ), c0 ) if and only if conditions (3.45) and (3.46) hold
and
(3.47)
µX
¶ ¶
∞ µ
m−1+j−k
lim
anj = 0
n→∞
j−k
j=k
(k = 0, 1, . . . ).
192
Malkowsky and Rakočević
(d) Then A ∈ (c0 (∆(m) ), c) if and only if conditions (3.45) and (3.46) hold and
µX
¶ ¶
∞ µ
m−1+j−k
(3.48)
lim
anj = lk (k = 0, 1, . . . ).
n→∞
j−k
j=k
(e) Then A ∈ (c(∆(m) ), c0 ) if and only if conditions (3.45), (3.46), (3.47) hold
and
(3.48)
µX
¶ ¶
∞ µ
m+j
anj = 0.
n→∞
j
j=0
lim
(f) Then A ∈ (c(∆(m) ), c) if and only if conditions (3.45), (3.46), (3.48) hold
and
(3.50)
µX
¶ ¶
∞ µ
m+j
anj = l−1 .
n→∞
j
j=0
lim
Proof. (a) Let A ∈ (l∞ (∆(m) ), l∞ ). Then An ∈ (l∞ (∆(m) ))β for n = 0, 1, . . . ,
P∞
(m)
and, by Theorem 3.17 (a), condition (3.44) holds for all n and n=0 |Rnk | < ∞
P
(m)
∞
(n = 0, 1, . . . ). Further kAk∗ = supn ( k=0 |Rnk |) < ∞ by Theorem 1.23 (b)
and Lemma 3.19. Conversely let conditions (3.44) and (3.45) hold. By Theorem
3.17 (a), this implies An ∈ (l∞ (∆(m) ))β for all n, and again Theorem 1.23 (b)
and Lemma 3.19 together imply A ∈ (l∞ (∆(m) ), l∞ ). Clearly (l∞ (∆(m) ), l∞ ) ⊂
(c(∆(m) ), l∞ ). If A ∈ (c(∆(m) ), l∞ ), then condition (3.45) follows from Theorems
1.23 (b) and 3.17 (a) and Lemma 3.19. Further An ∈ (c(∆(m) ))β = (l∞ (∆(m) ))β
and condition (3.44) holds (see Theorem 3.17 (a)). Therefore A ∈ (l∞ (∆(m) ), l∞ )
by what we have shown above.
(b) The proof of part (b) is exactly the same as that of the first part of part
(a) with condition (4.3) and Theorem 3.17 (a) replaced by condition (3.46) and
Theorem 3.17 (b).
Parts (c) to (f) follow from Theorem 1.23 (c) and parts (a) or (b), since
c0 (∆(m) ) and c(∆(m) ) are closed subspaces of l∞ (∆(m) ) by Corollary 3.11.
¤
As a corollary of Theorems 1.23 and 3.8, we have
Corollary 3.21. ([71, Corollary 3.5] Let X be a BK space.
(a) Then A ∈ (X, l∞ (∆(m) )) if and only if
°
µ
¶ °∗
n
X
°
°
m
(m)
n−l
°
(3.51)
M (X, l∞ (∆ )) = sup°
(−1)
Al °
° < ∞.
n−l
n
l=max{0,n−m}
(m)
(b) Further, if (bk )∞
)) if and only if
k=0 is a basis of X, then A ∈ (X, c0 (∆
condition (3.51) holds and
µ
µ
¶
¶
n
X
m
n−l
(k)
(3.52)
lim
(−1)
Al (b ) = 0 for each k;
n→∞
n−l
l=max{0,n−m}
193
Theory of sequence spaces
A ∈ (X, c(∆(m) )) if and only if condition (3.51) holds and
µ
µ
¶
¶
n
X
m
n−l
(k)
(3.53) lim
(−1)
Al (b ) = αk for each k = 0, 1, . . . ;
n→∞
n−l
l=max{0,n−m}
Remark 3.22. (a) If X = lp (1 ≤ p < ∞) and Y is any of the spaces l∞ (∆(m) ),
c(∆ ) and c0 (∆(m) ), then the conditions for A ∈ (X, Y ) follow from the respective
ones in Corollary 3.21 by replacing the norm k·k∗ in condition (3.51) by the natural
norm on the β–dual of lp , that is on lq (q = p/(p − 1), 1 < p < ∞; q = ∞, p = 1)
which is norm isomorphic to lp∗ . Hence we have
¯ ¶
µ∞¯

n
¡ m ¢ ¯q
P ¯¯
P

n−l
¯

(1 < p < ∞)
(−1)

n−l alk ¯
¯
 sup
(m)
M (lp , l∞ (∆(m) )) =
n
¯
¯



 sup¯¯
k=0 l=max{0,n−m}
n
P
n−l
(−1)
n,k l=max{0,n−m}
¡
m
n−l
¯
¢ ¯
alk ¯¯
(p = 1).
(b) Let s be a nonnegative integer. If X is any of the spaces l∞ (∆(s) ), c(∆(s) )
and c0 (∆(s) ), and Y is any of the spaces l∞ (∆(m) ), c(∆(m) ) and c0 (∆(m) ), then
the conditions for A ∈ (X, Y ) are obtained from the respective ones in Theorem
3.20 by replacing the entries of the matrix A by those of the matrix B = T A, for
instance
sup kBn k∗ = sup kR(s) (Bn )k1 < ∞
n
where
Bn =
n
µ
X
(−1)n−l
l=max{0,n−m}
¶
m
Al .
n−l
Theorem 3.23. [71, Theorem 4] Let A be as in Theorem 3.20, and for any
integers m, n, r, n > r, set
(3.54)
kAk(r) = sup kR(m) (An )k1
n>r
Let X be either c0 (∆(m) ) or X = c(∆(m) ), and let A ∈ (X, c0 ). Then we have
(3.55)
kLA kχ = lim kAk(r) .
r→∞
Let X be either c0 (∆(m) ) or X = c(∆(m) ), and let A ∈ (X, c). Then we have
(3.56)
1
· lim kAk(r) ≤ kLA kχ ≤ lim kAk(r) .
r→∞
2 r→∞
Let X be either l∞ (∆(m) ), c0 (∆(m) ) or X = c(∆(m) ), and let A ∈ (X, l∞ ). Then
we have
(3.57)
0 ≤ kLA kχ ≤ lim kAk(r) .
r→∞
194
Malkowsky and Rakočević
Proof. Let us remark that the limits in (3.55), (3.56) and (3.57) exist. We
put B = {x ∈ X : kxk ≤ 1}. In the case A ∈ (X, c0 ) for X = c0 (∆((m) ) or
X = c(∆((m) ), we have by Theorem 2.23
·
(3.58)
kLA kχ = χ(A(B)) = lim
¸
sup k(I − Pr )(A(x))k ,
r→∞ x∈B
where Pr : c0 7→ c0 for r = 0, 1, . . . is the projector on the first r + 1 coordinates,
that is Pr (x) = (x0 , x1 , . . . , xr , 0, 0, . . . ) for x = (xk ) ∈ c0 ; (let us remark that
kI − Pr k = 1 for r = 0, 1, . . . ). Further we have by Theorem 3.20
kAk(r) = sup k(I − Pr )(A(x))k,
(3.59)
x∈B
and by (3.58) we get (3.55).
To prove (3.56) let us remark that every sequence x = (xk )∞
k=0 ∈ c has a unique
P∞
representation x = le + k=0 (xk − l)e(k) where l ∈ C is such that x − le ∈ c0 . Let
Pr
us define Pr : c 7→ c by Pr (x) = le + k=0 (xk − l)e(k) for r = 0, 1, . . . . It is easy to
prove that kI − Pr k = 2 for r = 0, 1, . . . . Now the proof of (3.56) is similar as in
the case (3.55), and we omit it.
To prove (3.57), we define Pr : l∞ 7→ l∞ by Pr (x) = (x0 , x1 , . . . , xr , 0, 0, . . . ) for
x = (xk ) ∈ l∞ and r = 0, 1, . . . . It is clear that A(B) ⊂ Pr (A(B))+(I −Pr )(A(B)).
Now, by the elementary properties of function the χ we have
χ(A(B)) ≤ χ(Pr (A(B))) + χ((I − Pr )(A(B))) = χ((I − Pr )(A(B))
(3.60)
≤ sup k(I − Pr )(A(x))k.
x∈B
Finally we get (3.57) by Theorem 3.20.
¤
As a corollary of the theorem above, we have
Corollary 3.24. [71, Corollary 4.3] Let A be as in Theorem 3.23. Then if
A ∈ (X, c0 ) for X = c0 (∆(m) ) or X = c(∆(m) ), or if A ∈ (X, c) for X = c0 (∆(m) )
or X = c(∆(m) ), then in all cases we have
(3.61)
LA
is compact if and only if
lim kAk(r) = 0.
r→∞
Further, if A ∈ (X, l∞ ) for X = l∞ (∆(m) ), X = c0 (∆(m) ) or X = c(∆(m) ), then we
have
(3.62)
LA
is compact if
lim kAk(r) = 0.
r→∞
The following example will show that it is possible for LA in (3.62) to be
compact in the case limr→∞ kAk(r) > 0, and hence in general we have just “if” in
(3.62).
195
Theory of sequence spaces
Example 3.25. Let the matrix A be defined by An = e(0) (n = 0, 1, . . . ).
Then obviously R(m) (An ) = e(0) for all n, and A ∈ (l∞ (∆(m) ), l∞ ). Further,
kAk(r) = sup kR(m) (An )k1 = sup ke(0) k1 = 1 > 0 for all r,
n>r
n>r
whence limr→∞ kAk(r) > 0. Since A(x) = x0 e for all x ∈ l∞ (∆(m) ), A is a compact
operator.
Concerning Corollary 3.2.1 and the measures of noncompactness we have
Theorem 3.26. [71, Theorem 4.5] Let X be a BK space and let A be as in
Corollary 3.21 Then for all integers m, n, r, n > r, we put
°
µ
¶ °∗
n
X
°
°
m
(r)
n−j
°
(−1)
Aj °
(3.63)
kAk∆ = sup°
° .
n−j
n>r
j=max{0,n−m}
Further, if X has a Schauder basis, and A ∈ (X, c0 (∆(m) )), then we have
(r)
(3.64)
kLA kχ = lim kAk∆ .
r→∞
If X has a Schauder basis, and A ∈ (X, c(∆(m) )), then we have
1
(r)
(r)
· lim kAk∆ ≤ kLA kχ ≤ lim kAk∆ .
r→∞
2 r→∞
(3.65)
Finally, if A ∈ (X, l∞ (∆(m) )), then we have
(r)
(3.66)
0 ≤ kLA kχ ≤ lim kAk∆ .
r→∞
Proof. Let us remark that the limits in (3.64), (3.65) and (3.66) exist. We
put B = {x ∈ X : kxk ≤ 1}. To prove (3.64), we have by Theorems 3.12 and 2.23
·
¸
(3.67)
kLA kχ = χ(A(B)) = lim sup k(I − Pr )(A(x))k ,
r→∞ x∈B
where Pr : c0 (∆(m) ) 7→ c0 (∆(m) ) (r = 0, 1, . . . ) is the projector defined by
(3.68)
Pr (x) =
r
X
λk (m)b(k) (m),
k=0
P∞
(k)
(m) ∈ c0 (∆(m) ) and the Schauder basis (b(k) (m))∞
for x =
k=0 of
k=0 λk (m)b
(m)
c0 (∆ ) (see Theorem 3.12). Let us remark that kI − Pr k = 1 for (r = 0, 1, . . . ).
Further we have by Theorem 3.8
(3.69)
(r)
kAk∆ = sup k(I − Pr )(A(x))k,
x∈B
196
Malkowsky and Rakočević
To prove (3.65), let us remark (see Theorem 3.12) that c(∆(m) ) has the Schauder basis b(k) (m) k = −1, 0, 1, . . . , and every x ∈ c(∆(m) ) has a unique representation
∞
X
(3.70)
x = lb(−1) (m) +
(λk (m) − l)b(k) (m) where l = lim (∆(m) x)k .
k→∞
k=0
Now let us define Pr : c(∆(m) ) 7→ c(∆(m) ) (r = 0, 1, . . . ) by
(3.71)
Pr (x) = lb
(−1)
(m) +
r
X
(λk (m) − l)b(k) (m).
k=0
It is easy to show that kI − Pr k = 2 for r = 0, 1, . . . . Now the proof of (3.65) is
similar as in the case (3.64), and we omit it.
Finally in order to prove (3.66), we define Pr : l∞ (∆(m) ) 7→ l∞ (∆(m) ), by
Pr (x) = (x0 , x1 , . . . , xr , 0, 0, . . . ) for x = (xk ) ∈ l∞ (∆(m) ) and r = 0, 1, . . . . It is
clear that A(B) ⊂ Pr (A(B))+(I −Pr )(A(B)). Now, by the elementary properties of
the function χ, we again have (3.60) and, then (3.66) by Theorem 3.8 and Corollary
3.21.
¤
As a corollary of the theorem above, we have
(r)
Corollary 3.27. [71, Corollary 4.6] Let X be a BK space and let A and kAk∆
be as in Theorem 3.26. If X has a Schauder basis, and either A ∈ (X, c0 (∆(m) )) or
(r)
A ∈ (X, c(∆(m) )), then LA is compact if and only if limr→∞ kAk∆ = 0. Further,
(r)
if A ∈ (X, l∞ (∆(m) )), then LA is compact if limr→∞ kAk∆ = 0.
Finally we obtain several corollaries concerning Remark 3.22.
Corollary 3.28. [71, Corollary 4.7] If either A ∈ (lp , c0 (∆(m) )) or A ∈
(l , c(∆(m) )) (1 < p < ∞), then
p
LA is compact if and only if
µX
µ
¶ ¯q ¶
∞ ¯
n
X
¯
¯
m
n−j
¯
lim sup
(−1)
ajk ¯¯ = 0,
¯
r→∞ n>r
n−j
q = p/(p − 1).
k=0 j=max{0,n−m}
Further, if either A ∈ (l∞ , c0 (∆(m) )) or A ∈ (l∞ , c(∆(m) )), then
LA is compact if and only if
¯
µ
¶ ¯
n
X
¯
¯
m
lim sup ¯¯
(−1)n−j
ajk ¯¯ = 0.
r→∞ n>r,k
n−j
j=max{0,n−m}
If A ∈ (lp , l∞ (∆(m) )) for 1 < p < ∞, then
LA is compact if
µX
∞ ¯
¯
¯
lim sup
¯
r→∞ n>r
n
X
k=0 j=max{0,n−m}
¶ ¯q ¶
¯
m
ajk ¯¯ = 0,
n−j
µ
n−j
(−1)
q = p/(p − 1).
197
Theory of sequence spaces
Finally, if A ∈ (l∞ , l∞ (∆(m) )), then
LA is compact if
¯
¯
lim sup ¯¯
r→∞ n>r,k
n
X
µ
(−1)n−j
j=max{0,n−m}
¶ ¯
¯
m
ajk ¯¯ = 0.
n−j
From Corollary 3.24, Theorem 3.20 and Remark 3.22, we have
Corollary 3.29. [71, Corollary 4.8] Let s and m be non negative integers.
If A ∈ (X, c0 (∆(m) ) for X = c0 (∆(s) ) or X = c(∆(s) ), or if A ∈ (X, c(∆(m) ) for
X = c0 (∆(s) ) or X = c(∆(s) ), then in all cases we have
LA is compact if and only if
°
µ
° (s)
lim sup °
R
r→∞ n>r °
µ
n
X
¶ ¶°
°
m
Aj °
° = 0.
n−j
1
(−1)n−j
j=max{0,n−m}
Further, if A ∈ (X, l∞ (∆(m) ) for X = l∞ (∆(s) ), X = c0 (∆(s) ) or X = c(∆(s) ), then
we have
LA is compact if
°
µ
° (s)
°
lim sup °R
r→∞ n>r
n
X
j=max{0,n−m}
µ
(−1)
n−j
¶ ¶°
°
m
Aj °
° = 0.
n−j
1
3.5. Sequences of weighted means. In this subsection, we shall study sets of
weighted means sequences , give their bases and determine their β– and continuous
duals. The results can be found in [34].
Pn
Let (qk )∞
k=0 be a positive sequence, Q be the sequence with Qn =
k=0 qk
(n = 0, 1, . . . ) and the matrix N̄q be defined by
½
(N̄q )n,k =
qk /Qn
0
(0 ≤ k ≤ n)
(k > n)
(n = 0, 1, . . . ).
Then we define the sets (N̄ , q)0 = (c0 )N̄q , (N̄ , q) = cN̄q and (N̄ , q)∞ = (l∞ )N̄q of
sequences that are (N̄ , q) summable to naught, summable and bounded, respectively.
For any x ∈ X, we write τ = τ (x) for the sequence defined by
n
1 X
τn = (N̄q )n (x) =
qk xk (n = 0, 1, . . . ),
Qn
k=0
and τ is called the sequence of the N̄q or weighted means of x. As an immediate
consequence of Example 1.13 and Theorems 3.3 and 3.5, we obtain
198
Malkowsky and Rakočević
Proposition 3.30. (cf. [34, Corollary 1]) Each of the sets (N̄ , q)0 , (N̄ , q) and
(N̄ , q)∞ is a BK space with
¯
¯
n
¯ 1 X
¯
¯
qk xk ¯¯.
kxkN̄q = sup¯
n Qn
k=0
Further, if Qn → ∞ (n → ∞), then (N̄ , q)0 has AK, and
x =
P∞every sequence
(k)
(xk )∞
∈
(
N̄
,
q)
has
a
unique
representation
x
=
le
+
(x
−
l)e
where
k
k=0
k=0
l ∈ C is such that x − le ∈ (N̄ , q)0 .
∞
We define the operator ∆+ : ω 7→ ω by ∆+ x = ((∆+ x)k )∞
k=0 = (xk −xk+1 )k=0 .
Theorem 3.31. [34, Theorem
6] Let q = (qk )∞
k=0 be a positive sequence and
Pn
Q the sequence with Qn = k=0 qk (n = 0, 1, . . . ). We write 1/q for the sequence
+
−1
(1/qn )∞
∗(M1 ∩(Q−1 ∗l∞ )),
n=0 , and put M1 = {a ∈ ω : Q(∆ a) ∈ l1 }, N0 = (1/q)
−1
−1
−1
N = (1/q) ∗ (M1 ∩ (Q ∗ c)) and N∞ = (1/q) ∗ (M1 ∩ (Q−1 ∗ c0 )). Then
(N̄ , q)β0 = N0 , (N̄ , q)β = N and (N̄ , q)β∞ = N∞ .
Proof. We put X1 = (1/q)−1 ∗ M1 and observe that
1
(3.72)
xk = (Qk τk − Qk−1 τk−1 ) (k = 0, 1, . . . ) for all x ∈ ω
qk
and for all n = 0, 1, . . .
µ ¶¶
n
n
n−1
X
X
Xµ
ak
an Qn
+ ak
(3.73)
ak xk =
∆(Qk τk ) =
+
τn .
Qk τk ∆
qk
qk
qn
k=0
k=0
+
k=0
β
l∞
.
Let a ∈ X1 , that is Q∆ (a/q) ∈ l1 =
Thus τ · (Q∆+ (a/q)) ∈ cs for all τ ∈ l∞ ,
hence for all τ ∈ c and τ ∈ c0 . Further a ∈ (Q/q)−1 ∗ c0 implies (aQ/q)τ ∈ c0
for all τ ∈ l∞ . Since τ = τ (x) ∈ l∞ if and only if x ∈ (N̄ , q)∞ , ax ∈ cs for
all x ∈ (N̄ , q)∞ by (3.73), that is a ∈ (N̄ , q)β∞ . Similarly a ∈ (Q/q)−1 ∗ c or
a ∈ (Q/q) ∗ l∞ imply a ∈ (N̄ , q)β or a ∈ (N̄ , q)β0 . Thus we have proved N0 ⊂
(N̄ , q)β0 , N ⊂ (N̄ , q) and N∞ ⊂ (N̄ , q)β∞ .
To prove the converse inclusions we first assume ax ∈ cs for all x ∈ (N̄ , q)0 .
Then ax ∈ c0 for all x ∈ (N̄ , q)0 , hence (a/q)∆(Qτ ) ∈ c0 for all τ = τ (x) ∈ c0 ,
whence
ak
ak
∆(Qk (−1)k |τk |) = (−1)k (Qk |τk | + Qk−1 |τk−1 |) → 0 for all τ ∈ c0 .
qk
qk
This implies (aQ/q)τ ∈ c0 for all τ ∈ c0 , and thus aQ/q ∈ l∞ by Example 1.28.
From (3.73), we conclude Q∆+ (a/q)τ ∈ cs for all τ ∈ c0 , that is Q∆+ (a/q) ∈ cβ0 =
l1 . Thus a ∈ X1 , and we have proved (N̄ , q)β0 ⊂ N0 .
Now let ax ∈ cs for all x ∈ (N̄ , q). Then ax ∈ cs for all x ∈ (N̄ , q)0 , and
consequently a ∈ (N̄ , q)β0 ⊂ X1 . Thus by (3.73), (aQ/q)τ ∈ c for all τ ∈ c, hence
aQ/q ∈ c by Example 1.28. This proves (N̄ , q)∞ ⊂ N .
Finally let ax ∈ cs for all x ∈ (N̄ , q)∞ . Then again a ∈ X1 , and by (3.73),
(aQ/q)τ ∈ c for all τ ∈ l∞ , hence aQ/q ∈ c0 by Example 1.28. This proves
(N̄ , q)β ⊂ N∞ .
¤
Theory of sequence spaces
199
3.6. Matrix transformations in the spaces (N̄ , q)0 , (N̄ , q) and (N̄ , q)∞
and their measures of noncompactness. We need the following proposition
Proposition 3.32. [72, Proposition 3.1] On any of the spaces (N̄ , q)β0 , (N̄ , q)β
and (N̄ , q)β∞ , we have
¯ ¯
¯¶
µn−1
X ¯¯ ak
ak+1 ¯¯ ¯¯ an Qn ¯¯
kak∗ = sup
Qk ¯¯ −
+
.
qk
qk+1 ¯ ¯ qn ¯
n
k=0
Proof. Given any sequence x we shall write τ [n] = τ (x[n] ) (n = 0, 1, . . . )
where x[n] is the n–section of x. Let a ∈ N0 and n be a nonnegative integer. We
define the sequence b[n] by
[n]
bk

+

 Qk ∆ (a/q)k
=
an Qn /qn


0
(0 ≤ k ≤ n)
(k = n)
(k > n)
P∞
[n]
and put kakN = supn kb[n] k1 = supn ( k=0 |bk |). Then
¯∞
¯ n−1¯
¯
¯ ¯ n
¯ ¯
¯X
¯ X¯
¯X ak
¯ ¯ an Qn ¯ [n]
[n] ¯¯
[n] +
[n] ¯
¯
¯
¯
¯
¯
¯
∆(Qτ )k ¯ ≤
ak xk ¯ = ¯
¯
¯Qk τk ∆ (a/q)k ¯ + ¯ qn ¯|τn |
qk
k=0
k=0
k=0
¯
¯ ¯
¯¶
µn−1
X¯
¯ ¯
¯
[n]
¯Qk ∆+ (a/q)k ¯ + ¯ an Qn ¯
≤ sup |τk | ·
¯
¯
¯
qn ¯
k
[n]
= kx
k=0
[n]
kN̄q kb k1 =
kakN kx[n] kN̄q .
Thus
kak∗ ≤ kakN .
(3.74)
To prove the converse inequality let n be an arbitrary integer. We define the
[n]
sequence x(n) by τk (x(n) ) = sign(bk ) for k = 0, 1, . . . . Then τk (x(n) ) = for k > n,
that is x(n) ∈ (N̄ , q)0 , kx(n) kN̄n = kτ (x(n) )k∞ ≤ 1 and
¯X
¯ ¯ n
¯
n
¯∞
¯X [n] (n) ¯ X
(n) ¯¯
[n]
¯
¯
¯=
a
x
=
b
x
|bk | ≤ kak∗ .
k
k ¯
k
k ¯
¯
¯
k=0
k=0
k=0
Since n was arbitrary, we have kakN ≤ kak∗ . This and (3.74) together yield the
conclusion.
¤
As a corollary of Theorems 1.23 and 3.31 and Proposition 3.32 we obtain
200
Malkowsky and Rakočević
∞
Corollary
Pn 3.33. [72, Corollary 3.4] Let q = (qk )k=0 be a positive sequence
and Qn = k=0 qk → ∞ (n → ∞).
(a) Then A ∈ ((N̄ , q)∞ , l∞ ) if and only if
(3.75)
M ((N̄ , q)∞ , l∞ ) = sup
n,m
µm−1
X
k=0
¯
¯
¶
¯ ank
an,k+1 ¯¯
|anm |
Qk ¯¯
−
<∞
+
Q
m
qk
qk+1 ¯
qm
and
(3.76)
An Q/q ∈ c0
for all n = 0, 1, . . . .
(b) Then A ∈ ((N̄ , q), l∞ ) if and only if condition (3.75) holds and
(3.77)
An Q/q ∈ c for all n = 0, 1, . . . .
(c) Then A ∈ ((N̄ , q)0 , l∞ ) if and only condition (3.75) holds.
(d) Then A ∈ ((N̄ , q)0 , c0 ) if and only condition (3.75) holds and
(3.78)
lim ank = 0
n→∞
for all k = 0, 1, . . . .
(e) Then A ∈ ((N̄ , q)0 , c) if and only if condition (3.75) holds and
(3.79)
lim ank = lk
n→∞
for all k = 0, 1, . . . .
(f) Then A ∈ ((N̄ , q), c0 ) if and only if conditions (3.75), (3.77) and (3.78) hold
and
(3.80)
lim
n→∞
∞
X
ank = 0.
k=0
(g) Then A ∈ ((N̄ , q), c) if and only if conditions (3.75), (3.77) and (3.79) hold
and
(3.81)
lim
n→∞
∞
X
ank = l.
k=0
As a corollary of Theorem 3.8 and Corollary 3.33, we obtain
∞
Corollary 3.34.
Pn[72, Corollary 3.5] Let X be a BK space, (pk )k=0 a positive
sequence and Pn = k=0 pk (n = 0, 1, . . . ). Then A ∈ (X, (N̄ , p)∞ ) if and only if
(3.82)
°
°∗
m
° 1 X
°
°
M (X, (N̄ , p)∞ ) = sup°
p
A
n n ° < ∞.
°
Pm n=0
m
201
Theory of sequence spaces
Further, if (bk )∞
k=0 is a basis of X, then A ∈ (X, (N̄ , p)0 ) if and only if condition
(3.82) holds and
µ
¶
m
1 X
(k)
(3.83)
lim
pn An (b ) = 0 for all k = 0, 1, . . . ,
m→∞ Pm
n=0
and A ∈ (X, (N̄ , p)) if and only if condition (3.83) holds and
µ
¶
m
1 X
(k)
(3.84)
lim
pn An (b ) = lk for all k = 0, 1, . . . .
m→∞ Pm
n=0
Remark 3.35. (a) If X = lr (1 ≤ r < ∞) and Y is any of the spaces (N̄ , p)∞ ,
(N̄ , p) and (N̄ , p)0 , then the conditions for A ∈ (X, Y ) follow from the respective
ones in Corollary 3.34 by replacing the norm k·k∗ in condition (3.82) by the natural
norm on ls where s = ∞ for r = 1 and s = r/(r − 1) for 1 < r < ∞, that is
¯
¯

¯ 1 Pm
¯

¯
¯
 supm,k ¯
(r = 1)

n=0 pn ank ¯
Pm
¯
¯s
M (lr , (N̄ , p)∞ ) =
¯ 1 Pm
¯

P

¯
 supm ∞
pn ank ¯¯
(1 < r < ∞),
k=0 ¯
Pm n=0
and by replacing the terms An (b(k) ) in conditions (3.83) and (3.84) by the terms
ank .
(b) We consider the conditions
(3.85) M ((N̄ , q)∞ , (N̄ , p)∞ )
¯¶
¯
µn−1
µ
¶ ¯
m
m
X ¯¯ 1 X
¯
¯
¯ Qn X
+
¯
¯
¯
= sup
Qk ¯
pl ∆ Al /q ¯ + ¯
pl aln ¯¯ < ∞,
P
q
P
m,n
m
n m
k
k=0
µ
(3.86)
µ
(3.87)
(3.88)
(3.89)
(3.90)
(3.91)
ank Qk
qk
l=0
l=0
¶∞
∈ c0
(n = 0, 1, . . . ),
¶k=0
∞
ank Qk
∈c
qk
k=0
¶
µ
m
1 X
lim
pn ank = 0
m→∞ Pm
n=0
µ
¶
m
1 X
lim
pn ank = lk
m→∞ Pm
n=0
µX
¶¶
µ
m
∞
1 X
lim
pn
ank
= 0,
m→∞ Pm
n=0
k=0
µ
µX
¶¶
m
∞
1 X
lim
pn
ank
= l.
m→∞ Pm
n=0
k=0
(n = 0, 1, . . . ),
(k = 0, 1, . . . ),
(k = 0, 1, . . . ),
202
Malkowsky and Rakočević
Then
A ∈ ((N̄ , q)∞ , (N̄ , p)∞ )
and only if
(3.85) and (3.86);
A ∈ ((N̄ , q), (N̄ , p)∞ )
and only if
(3.85) and (3.87);
A ∈ ((N̄ , q)0 , (N̄ , q)∞ )
and only if
(3.85);
A ∈ ((N̄ , q)0 , (N̄ , p)0 )
and only if
(3.85) and (3.88);
A ∈ ((N̄ , q)0 , (N̄ , p))
and only if
(3.85) and (3.89);
A ∈ ((N̄ , q), (N̄ , p)0 )
and only if
(3.85), (3.87), (3.88) and (3.90);
A ∈ ((N̄ , q), (N̄ , p))
and only if
(3.85), (3.87), (3.89) and (3.91).
Theorem 3.36. [72, Theorem 4.2] Let A be as in Corollary 3.33, and for all
integers n, r, n > r, set
¯
¯
¶
µm−1
X
¯ ank
an,k+1 ¯¯
|anm |
+
Q
(3.92)
kAk(r) = sup
−
Qk ¯¯
m
m
qk
qk+1 ¯
qm
n>r
k=0
Let X be either (N̄ , q)0 or X = (N̄ , q), and let A ∈ (X, c0 ). Then we have
kLA kχ = lim kAk(r) .
r→∞
Let X be either (N̄ , q)0 or X = (N̄ , q), and let A ∈ (X, c). Then we have
1
· lim kAk(r) ≤ kLA kχ ≤ lim kAk(r) .
r→∞
2 r→∞
Let X be either (N̄ , q)0 , (N̄ , q) or X = (N̄ , q)∞ , and let A ∈ (X, l∞ ). Then we
have 0 ≤ kLA kχ ≤ limr→∞ kAk(r) .
Proof. The proof follows exactly the same lines as that of Theorem 3.26.
¤
As a corollary of the theorem above, we have
Corollary 3.37. [72, Corollary 4.3] Let A be as in Theorem 3.36. Then if
A ∈ (X, c0 ) for X = (N̄ , q)0 or X = (N̄ , q), or if A ∈ (X, c) for X = (N̄ , q)0 or X =
(N̄ , q), then in all cases we have LA is compact if and only if limr→∞ kAk(r) = 0.
Further, if A ∈ (X, l∞ ) for X = (N̄ , q)0 , X = (N̄ , q) or X = (N̄ , q)∞ , then we have
(3.93)
LA
is compact if
lim kAk(r) = 0.
r→∞
The following example will show that it is possible for LA in (3.93) to be
compact in the case limr→∞ kAk(r) > 0, and hence in general we have just ” if” in
(3.93).
Example 3.38. Let the matrix A be defined by An = e(0) (n = 0, 1, . . . ) and
qn = 2n for n = 0, 1, 2, . . . . Then M ((N̄ , q)∞ , l∞ ) = supn [1 + (2 − 2−n )] < 3, and
by Corollary 3.33 we know that A ∈ ((N̄ , q)∞ , l∞ ). Further
1
1
kAk(r) = sup [1 + (2 − n )] = 3 − r+1 for all r,
2
2
n>r
whence limr→∞ kAk(r) = 3 > 0. Since A(x) = x0 e0 for all x ∈ (N̄ , q)∞ , LA is a
compact operator.
Now we continue with the following auxiliary result.
203
Theory of sequence spaces
Pn
Lemma 3.39. [72, Lemma 4.5] Let qk > 0 (k = 0, 1, . . . ) and Qn = k=0 qk
(r)
→ ∞ (n → ∞). Let r ≥ 0 and the operators B0 : (N̄ , q)0 7→ (N̄ , q)0 and
B (r) : (N̄ , q) 7→ (N̄ , q) be defined by
∞
X
(r)
B0 x =
B (r) x =
xk e(k) ,
k=r+1
∞
X
(x ∈ (N̄ , q)0 )
(xk − l)e(k) ,
(x ∈ (N̄ , q))
k=r+1
where l = limn→∞ τn (x). Then
(r)
(3.94)
kB0 k = 1 +
(3.95)
kB (r) k = 2.
Qr
Qr+1
(r)
Proof. First we show identity (3.94). Let x ∈ (N̄ , q)0 . Since τn (B0 (x)) = 0
for 0 ≤ n ≤ r, and, for n ≥ r + 1,
¯
¯ ¯
¯
n
¯ 1 X
¯ ¯
¯
Qr
(r)
¯
¯
¯
|τn (B0 (x))| = ¯
qk xk ¯ = ¯τn (x) −
τr (x)¯¯
Qn
Qn
k=r+1
µ
¶
Qr
≤ 1+
kxk(N̄ ,q)∞ ,
Qr+1
it follows that
° (r) °
°B (x)°
0
(N̄ ,q)∞
¶
µ
Qr
kxk(N̄ ,q)∞ ,
≤ 1+
Qr+1
and consequently
° (r) °
°B ° ≤ 1 + Qr .
0
Qr+1
(3.96)
Defining the sequence x by

−1





Qr + Qr+1



qr+1
xk =
Q

r + Qr+1


−


qr+2



0
(0 ≤ k ≤ r)
(k = r + 1)
(k = r + 2)
(k ≥ r + 3),
we conclude τn (x) = −1 for 0 ≤ n ≤ r
Qr
Qr
+
+1=1
Qr+1
Qr+1
µ
¶
1
Qr
τn (x) =
−Qr + Qr + Qr+1 − (Qr + Qr+1 ) = −
Qn
Qn
τr+1 (x) = −
(n ≥ r + 2).
204
Malkowsky and Rakočević
Since Qn → ∞ (n → ∞), we have x ∈ (N̄ , q)0 and kxk(N̄ ,q)∞ = 1. Further
(r)
τr+1 (B0 (x)) =
¢
1 ¡
Qr
Qr + Qr+1 = 1 +
Qr+1
Qr+1
(r)
and τn (B0 (x)) = 0 for n 6= r + 1. Therefore
° (r) °
Qr
°B (x)°
=1+
=
0
(N̄ ,q)∞
Qr+1
µ
¶
Qr
1+
kxk(N̄ ,q)∞
Qr+1
(r)
and kB0 k ≥ 1 + Qr /Qr+1 . Now this and (3.96) together yield identity (3.94).
Now we prove identity (3.95). Let x ∈ (N̄ , q). Since τn (B (r) (x)) = 0 for
0 ≤ n ≤ r and, for n ≥ r + 1,
¯
¯
¯ ¯
n
X
¯
¯ ¯
¯ ¯
Qr
Qr ¯¯
¯
¯τn (B (r) (x))¯ = ¯ 1
¯
τ
(x)
−
=
τ
(x)
−
l
+
l
q
(x
−
l)
r
k k
¯ Qn
¯ ¯ n
Qn
Qn ¯
k=r+1
¯
¯
¯
¯
¯
¯
Qr ¯¯
Qr ¯¯
¯
≤ ¯¯1 +
kxk
+
1
−
|l|,
(N̄ ,q)∞
¯
Qn ¯
Qn ¯
since |l| = limn→∞ |τn (x)| ≤ kxk(N̄ ,q)∞ , it follows that |τn (B (r) (x))| ≤ 2kxk(N̄ ,q)∞
for n ≥ r + 1 and consequently
kB (r) k ≤ 2.
(3.97)
Defining the sequence x by


 −1
xk =
2Qr+1 /qr+1 − 1


−1
(0 ≤ k ≤ r)
(k = r + 1)
(k ≥ r + 2),
we conclude τn (x) = −1 for 0 ≤ n ≤ r,
¢
1 ¡
−Qr + 2Qr − qr+1 = 1
τr+1 (x) =
Qr+1
µ
¶
n
X
¢
1
1 ¡
τn (x) =
−Qr + 2Qr+1 −
qk =
−Qn + 2Qr+1
Qn
Qn
k=r+1
= −1 + 2
Qr+1
≤1
Qn
(n ≥ r + 2).
Hence kxk(N̄ ,q)∞ = 1 and limn→∞ τn (x) = −1, that is x ∈ (N̄ , q).
τn (B (r) (x)) = 0 (0 ≤ n ≤ r),
qr+1
τr+1 (B (r) (x)) =
(xr+1 + 1) = 2
Qr+1
Qr+1
≤ 2 (n ≥ r + 2).
τn (B (r) (x)) = 2
Qn
Finally
This implies kB (r) k ≥ 2, and together with (3.97) we obtain (3.95).
Concerning Corollary 3.34 and the measures of noncompactness we have
¤
Theory of sequence spaces
205
Theorem 3.40. [72, Theorem 4.6] Let X be a BK space, let A be as in
Corollary 3.34, and let Pm → ∞ (m → ∞). Then for all integers m, r, m > r, we
put
°
°∗
m
° 1 X
°
(r)
°
pn An °
kAk(N̄ ,p)∞ = sup °
° .
m>r Pm
n=0
Further, if X has a Schauder basis, and A ∈ (X, (N̄ , p)0 ), then we have
1
(r)
(r)
(3.98)
· lim kAk(N̄ ,p)∞ ≤ kLA kχ ≤ lim kAk(N̄ ,p)∞ ,
r→∞
b r→∞
where b = lim supn→∞ (2 − pn /Pn ).
If X has a Schauder basis, and A ∈ (X, (N̄ , p)), then we have
1
(r)
(r)
(3.99)
· lim kAk(N̄ ,p)∞ ≤ kLA kχ ≤ lim kAk(N̄ ,p)∞ .
r→∞
2 r→∞
Finally, if A ∈ (X, (N̄ , p)∞ ), then we have
(r)
(3.100)
0 ≤ kLA kχ ≤ lim kAk(N̄ ,p)∞ .
r→∞
Proof. Let us remark that the limits in (3.98), (3.99) and (3.100) exist. We
(r)
put B = {x ∈ X : kxk ≤ 1}. Suppose that A ∈ (X, (N̄ , p)0 ). Let B0 : (N̄ , p)0 7→
(N̄ , p)0 be the projector defined in Lemma 3.39. Then by (3.94) we have that
(r)
kB0 k = 2−pr /Pr . Now, to prove (3.98), we have by Theorem 2.23 and Proposition
3.30
µ
¶
µ
¶
1
(r)
(r)
lim sup sup kB0 (A(x))k ≤ χ(A(B)) ≤ lim sup sup kB0 (A(x))k ,
b r→∞ x∈B
r→∞
x∈B
(r)
where b = lim supr→∞ kB0 k. This proves (3.98), since
(r)
(r)
sup kB0 (A(x))k = kAk(N̄ ,p)∞ .
x∈B
To prove (3.99) let us remark (see Proposition 3.30) that (N̄ , p) has the Schauder
basis e, e(k) , k = 0, 1, . . . , and every (xk )∞
k=0 ∈ (N̄ , q) has a unique representaP∞
(k)
tion x = le + k=0 (xk − l)e , where l ∈ C is such that x − le ∈ (N̄ , p)0 . Let
(r)
B
(N̄ , p)0 be the projector defined by (see Lemma 3.39) B (r) (x) =
P∞ : (N̄ , p)0 7→ (k)
. Then we have kB (r) k = 2 by (3.95). Now the proof of (3.99)
k=r+1 (xk − l)e
is similar as in the case (3.98), and we omit it. Let us prove (3.100). Now define
Pr : (N̄ , p)∞ 7→ (N̄ , p)∞ , by Pr (x) = (x0 , x1 , . . . , xr , 0, 0, . . . ) for all x = (xk ) ∈
(N̄ , p)∞ and r = 1, 2, . . . . It is clear that A(B) ⊂ Pr (A(B)) + (I − Pr )(A(B)). By
Remark 3.22 (b) it follows that Pr is a bounded operator, and since it is obviously a
finite-rank, it is a compact operator. Now, by the elementary properties of function
χ we have
χ(A(B)) ≤ χ(Pr (A(B))) + χ((I − Pr )(A(B))) = χ((I − Pr )(A(B))
(r)
≤ sup k(I − Pr )(A(x))k = kAk(N̄ ,p)∞ .
x∈B
¤
As a corollary of the theorem above we have
206
Malkowsky and Rakočević
Corollary 3.41. [72, Corollary 4.7] Let X be a BK space and let A and
(r)
kAk(N̄ ,p) be as in Theorem 4.6. If X has a Schauder basis, and either A ∈
(r)
(X, (N̄ , p)0 ) or A ∈ (X, (N̄ , p)), then LA is compact if and only if lim kAk(N̄ ,p) = 0.
(r)
r→∞
Further, if A ∈ (X, (N̄ , p)∞ ), then LA is compact if lim kAk(N̄ ,p) = 0.
r→∞
Now we get several corollaries concerning Remark 3.22.
Corollary 3.42. [72, Corollary 4.8] If either A ∈ (lu , (N̄ , p)0 ) or A ∈ (lu , (N̄ , p))
for 1 < u < ∞, then
LA
is compact if and only if
¯v ¶1/v ¸
µX
·
∞ ¯
m
¯ 1 X
¯
¯
lim sup
=0
pn ank ¯¯
¯ Pm
r→∞ m>r
n=0
where v = u/(u − 1).
k=0
Further, if either A ∈ (l1 , (N̄ , p)0 )) or A ∈ (l1 , (N̄ , p))), then
¯
¯¶
µ
m
¯ 1 X
¯
¯
LA is compact if and only if
lim
sup
pn ank ¯¯ = 0.
r→∞ n>r,k¯ Pm
n=0
If A ∈ (lu , (N̄ , p)) for 1 < u < ∞, then
LA
is compact if
¯v ¶1/v ¸
·
µX
m
∞ ¯
¯
¯ 1 X
¯
¯
lim sup
p
a
=0
n nk ¯
¯
r→∞ m>r
Pm n=0
where v = u/(u − 1).
k=0
Finally, if A ∈ (l1 , (N̄ , p)), then
¯¶
¯
m
¯
¯ 1 X
¯
pn ank ¯¯ = 0.
sup ¯
P
m
n>r,k
µ
LA
is compact if lim
r→∞
n=0
From Corollary 3.41, Theorem 3.8 and Remark 3.22 (b), we have
Corollary 3.43. [72, Corollary 4.9] If A ∈ (X, (N̄ , p)0 ) for X = (N̄ , q)0 or
X = (N̄ , q), or if A ∈ (X, (N̄ , p)) for X = (N̄ , q)0 or X = (N̄ , q), then we have in
all cases
LA
is compact if and only if
¯ ¯
¯¶¸
·
µn−1
m
m
X ¯¯ 1 X
¯ ¯ Qn X
¯
pl (∆+ Al /q)k ¯¯ + ¯¯
pl aln ¯¯
lim sup
Qk ¯¯
= 0.
r→∞ m>r,n
Pm
qn Pm
k=0
l=0
l=0
Further, if A ∈ (X, (N̄ , p)∞ ) for X = (N̄ , q)∞ , X = (N̄ , q)0 or X = (N̄ , q), then
we have
LA
is compact if
¯ ¯
¯¶¸
µn−1
·
m
m
X ¯¯ 1 X
¯ ¯ Qn X
¯
+
¯
¯
¯
Qk ¯
lim sup
pl (∆ Al /q)k ¯ + ¯
pl aln ¯¯
= 0.
r→∞ m>r,n
Pm
qn Pm
k=0
l=0
l=0
207
Theory of sequence spaces
3.7. Spaces of strongly summable and convergent sequences. In this
subsection, we shall study spaces of strongly summable and strongly convergent
sequences and give their dual spaces. The results of this subsection can be found
in [59, 64, 65, 70].
For X ⊂ ω and any real p > 0, we write
©
ª
X[A]p = x ∈ ω : A(|x|p ) ∈ X .
If p = 1, then we omit the index p and write X[B] = X[B]1 for short.
Let C1 be the Cesáro matrix of order 1, that is (C1 )nk = 1/n for 1 ≤ k ≤ n
and 0 for k > n (n = 0, 1, . . . ). For 0 < p < ∞, Maddox [51, 54] defined the sets
w0p
= (c0 )[C1 ]p
½
µ X
¶
¾
n
1
p
= x ∈ ω : lim
|xk | = 0 ,
n→∞ n
k=1
©
ª
wp = x ∈ ω : x − le ∈ w0p for some complex number l ,
p
= (l∞ )[C1 ]p .
w∞
These sets are special cases of the so-called mixed normed spaces (see e.g. [38, 35,
36, 59, 22, 23]). Here we shall only deal with the cases 1 ≤ p < ∞. The following
result is well known.
p
Proposition 3.44. [51, 59] Each of the set w0 , wp and w∞
is a BK space for
1 ≤ p < ∞ with
µ
1
kxk = sup ν
ν≥0 2
(3.101)
2ν+1
X−1
¶1/p
|xk |
p
;
k=2ν
p
w0p has AK; every sequence x = (xk )∞
k=1 ∈ w has a unique representation x =
P∞
(k)
le + k=1 (xk − l)e where l ∈ C is such that x − le ∈ w0p .
Let µ = (µn )∞
n=0 be a nondecreasing sequence of positive reals tending to
infinity. If (n(ν))∞
ν=0 is a sequence such that 0 = n(0) < n(1) < n(2) < . . . , then
we denote
K hνi , and we
P the set of all integers k with n(ν) ≤ k ≤ n(ν + 1) − 1 by hνi
write ν and maxν for the sum and maximum taken over all k in K . We define
∞
the matrices B = (bnk )∞
n,k=0 and B̃ = (b̃νk )ν,k=0 by
½
bnk =
1/λn
0
(0 ≤ k ≤ n)
(k < n)
½
and b̃νk =
Further, let ∆(µ) be the matrix with

(k = n − 1)

 −µn−1
∆nk (µ) =
µn
(k = n)


0
(otherwise)
1/λn(ν+1)
(k ∈ K <ν> )
0
(k ∈
/ K <ν> ).
(n = 0, 1, . . . )
where µ−1 = 0.
208
Malkowsky and Rakočević
We define the sets [76, 64]
c0 (µ) = ((c0 )[B] )∆(µ) ,
©
ª
c(µ) = x ∈ ω : x − le ∈ c0 (µ) ,
c∞ (µ) = ((l∞ )[B] )∆(µ) ,
c˜0 (µ) = ((c0 )[B̃] )∆(µ) ,
©
ª
c̃(µ) = x ∈ ω : x − le ∈ c̃0 (µ) ,
c˜∞ (µ) = ((l∞ )[B̃] )∆(µ) .
The following result is well known.
Proposition 3.45. [64, Theorem 2 (c)] Let µ = (µn )∞
n=0 be a nondecreasing
sequence of positive reals tending to infinity. Then each of the spaces c0 (µ), c(µ)
and c∞ (µ) is a BK space
µ
¶
n
1 X
0
kxk = kB(|∆(µ)x|))k∞ = sup
|µk xk − µk−1 xk−1 | ;
n≥0 µn
k=0
c0 (µ) has AK; every sequence x = (xk )∞
k=0 ∈ cµ has a unique representation x =
P∞
le + k=1 (xk − l)e(k) where l ∈ C is such that x − le ∈ c0 (µ).
A sequence Λ = (λn )∞
n=0 of positive reals is called exponentially bounded if
there is an integer m ≥ 2 such that for all integers ν there is at least one λn in the
interval [mν , mν+1 ). It is known (cf. [64, Lemma 1]) that a nondecreasing sequence
Λ = (λn )∞
n=0 of positive reals is exponentially bounded if and only if there are reals
s ≤ t such that 0 < s ≤ λn(ν) /λn(ν+1) ≤ t < 1 for some subsequence (λn(ν+1) )∞
ν=0
for all ν = 0, 1, . . . ; such a subsequence is called an associated subsequence.
The following result is well known.
Proposition 3.46. [64, Theorem 2] Let Λ = (λn )∞
n=0 be a nondecreasing
exponentially bounded sequence of positive reals and (λn(ν+1) )∞
ν=0 an associated
subsequence. Then c0 (Λ) = c̃0 (Λ), c(Λ) = c̃(Λ) and c∞ (Λ) = c̃∞ (Λ). The norms
kxk0 and
µ
¶
1 X
kxk = lkB̃(|∆(Λ)x|)k∞ = sup
|λk xk − λk−1 xk−1 |
ν≥0 λn(ν+1) ν
are equivalent on c0 (Λ), c(Λ) and c∞ (Λ). Thus each of the spaces c0 (Λ), c(Λ) and
c∞ (Λ) is a BK space with k · k.
Proposition 3.47. [51] and [59, Theorems 4 and 6] Let K hνi = [2ν , 2ν+1 − 1]
(ν = 0, 1, . . . ). We put
n
o
∞
P
ν/p

a
∈
ω
:
2
max
|a
|
<
∞
(p = 1)

ν
k

ν=0
p
M = n
´1/q
o ³
³P
∞
P

p ´

 a∈ω:
2ν/p
|ak |q
<∞
1 < p < ∞; q =
p−1
ν
ν=0
 P
∞

2ν/p maxν |ak |
(p = 1)


ν=0
kakMp =
for all a ∈ Mp .
´1/q
³P
∞
 P


|ak |q
(1 < p < ∞)
2ν/p
ν=0
ν
p β
Then (w0p )β = (wp )β = (w∞
) = Mp and kak∗ = kakMp on Mp .
209
Theory of sequence spaces
Proposition 3.48. [65, Lemma 2] Let Λ = (λn )∞
n=0 be a nondecreasing exponentially bounded sequence of positive reals and (λn(ν+1) )∞
ν=0 an associated subsequence. We put
¯∞
¯
¾
¯ X ak ¯
¯
¯
C(Λ) = a ∈ ω :
λn(ν+1) max¯
<∞
ν
λk ¯
ν=0
k=n
¯X
¯
∞
X
¯ ∞ ak ¯
¯
¯ for all a ∈ C(Λ).
kakC(Λ) =
λn(ν+1) max¯
¯
ν
λ
k
ν=0
½
∞
X
k=n
Then (c0 (Λ))β = (c(Λ))β = (c∞ (Λ))β = C(Λ) and kak∗ = kakC(Λ) on C(Λ).
As an immediate consequence of Theorems 3.8 and 3.10 we obtain
Corollary 3.49. [70, Corollary 1] Let X be an arbitrary FK space. Further,
let µ = (µn )∞
n=0 be a nondecreasing sequence of positive reals tending to infinity.
We write w0 = w01 etc., for short and put
°∗ ¶
°
°1 X
°
°
°
A
n°
°
Nm ⊂{1,...,m} m
µ
M (X, w∞ ) = sup
m≥1
max
D
n∈Nm
°
°∗ ¶
° 1 X
°
°
M (X, c∞ (µ)) = sup
max
(µn An − µn−1 An−1 °
°
° .
µ
N
⊂{0,...,m}
m
m
m≥0
µ
D
n∈Nm
(a) Then A ∈ (X, w∞ ) if and only if
(3.102)
M (X, w∞ ) < ∞
for some D > 0.
Furthermore, if (bk )∞
k=0 is a basis of X, then A ∈ (X, w0 ) if and only if condition
(3.102) holds and
µ
(3.103)
lim
m→∞
¶
m
1 X
(k)
|An (b )| = 0
m n=1
for all k = 0, 1, . . . ;
A ∈ (X, w) if and only if condition (3.102) holds and there are complex numbers
lk (k = 0, 1, . . . ) such that
µ
(3.104)
lim
m→∞
¶
m
1 X
(k)
|An (b ) − lk )| = 0
m n=1
for all k = 0, 1, . . . .
Finally, if X is a normed space and A ∈ (X, Y ) for Y = w0 , w or w∞ , then, for
°
°∗ ¶
°1 X
°
°
= sup
max
An °
°
° ,
m≥1 Nm ⊂{1,...,m} m
µ
kAk∗w∞
n∈Nm
210
Malkowsky and Rakočević
we have
kAk∗w∞ ≤ kLA k ≤ 4 · kAk∗w∞ .
(3.105)
(b) Then A ∈ (X, c∞ (µ)) if and only if
(3.106)
M (X, c∞ (µ)) < ∞
for some D > 0.
Further, if (bk )∞
k=0 is a basis of X, then A ∈ (X, c0 (µ)) if and only if condition
(3.106) holds and
µ
(3.107)
lim
m→∞
¶
m
1 X
|µn An (b(k) ) − µn−1 An−1 (b(k) )| = 0
µm n=0
for all k = 0, 1, . . . ; A ∈ (X, c(µ)) if and only if condition (3.107) holds and there
are complex numbers lk such that
µ
(3.108)
lim
m→∞
¶
m
1 X
(k)
(k)
|µn (An (b ) − lk ) − µn−1 (An−1 (b ) − lk )| = 0
µm n=0
for all k = 0, 1, . . . .
Finally, if X is normed and A ∈ (X, Y ) for Y = c0 (µ), c(µ) or c∞ (µ), then, for
°
°∗ ¶
° 1 X
°
°
= sup
max
(µn An − µn−1 An−1 °
°
° ,
m≥0 Nm ⊂{0,...,m} µm
µ
kAk∗c∞
n∈Nm
we have
(3.109)
kAk∗c∞ (µ) ≤ kLA k ≤ 4 · kAk∗c∞ (µ) .
Proof. All we have to show are inequalities (3.105) and (3.109). Let A ∈
(X, Y ) where Y = w0 , w or w∞ . Then
¯
¯
m
X
¯
¯1 X
¯≤ 1
¯
A
(x)
|An (x)| ≤ kLA k
n
¯ m
¯m
n=1
n∈Nm
for all m = 1, 2, . . . , all N ⊂ Nm and all kxk = 1. This implies
kAk∗w∞ ≤ kLA k.
(3.110)
Further, given ε > 0 there is x ∈ X with kxk = 1 such that
µ
kA(x)k = sup
m≥1
¶
m
1 X
|An (x)| ≥ kLA k − ε/2,
m n=1
211
Theory of sequence spaces
and there is an integer m(x) such that
m(x)
1 X
|An (x)| ≥ kA(x)k − ε/2,
m(x) n=1
consequently
m(x)
1 X
|An (x)| ≥ kLA k − ε.
m(x) n=1
By Lemma 3.9
¯
1 ¯¯ X
m(x) ¯
µ
4·
max
Nm(x) ⊂{1,...,m(x)}
¯¶
¯
An (x)¯¯ ≥
n∈Nm(x)
m(x)
1 X
|An (x)| ≥ kLA k − ε
m(x) n=1
and so 4 · kAk∗w∞ ≥ kLA k − ε. Since ε > 0 was arbitrary, 4 · kAk∗w∞ ≥ kLA k.
Together with inequality (3.110) this yields (3.105). The inequalities in (3.109) are
proved similarly.
¤
Remark 3.50. If X is a given BK space, and Y is any of the spaces w0 ,
w, w∞ , c0 (µ), c(µ) or c∞ (µ), then the conditions for A ∈ (X, Y ) follow from the
respective ones in Corollary 3.49 by replacing the norms k · k∗D in conditions (3.102)
and (3.106) by the natural norms on the β–duals of X. We shall write


max
Nm
for

max
if Y = w0 , w, w∞
max
if Y = c0 (µ), c(µ), c∞ (µ),
Nm ⊂{0,...,m}
Nm ⊂{1,...,m}
q = p/(p − 1) for 1 < p < ∞, ∆n (µn ank ) = µn ank − µn−1 an−1,k
max
ν
for
max
2ν ≤k≤2ν+1 −1
,
X
ν
for
X
.
2ν ≤k≤2ν+1 −1
(a) For X = lp , we have
¯¶¶
¯
µ
µ

¯
¯1 P

¯

ank ¯¯
(p = 1)
max sup¯

 sup
Nm
m
k m n∈Nm
¯q ¶¶
M (lp , w∞ ) =
µ
µ∞¯
¯

P ¯¯ 1 P


ank ¯¯
(1 < p ≤ ∞),
 sup max
¯
Nm
m
m
¯¶¶
µ
µk=1 ¯ n∈Nm

¯
¯ 1 P

¯
¯

∆
(µ
a
)
(p = 1)
sup
max
sup

n
n
nk
¯
¯
 m Nm
k µm n∈Nm
¯q ¶¶
M (lp , c∞ (µ)) =
µ∞¯
µ
¯

P¯ 1 P

¯

∆n (µn ank )¯¯
(1 < p < ∞).
 sup max
¯
Nm
µm
m
k=1
n∈Nm
212
Malkowsky and Rakočević
p
(b) For X = w∞
, we have
¯¶¶
¯
µ
µ∞

¯
¯1 P
P ν/p

¯
¯

sup
a
for p = 1
max
2
max
nk ¯

¯
 m Nm ν=0
ν
m n∈Nm
p
M (w∞
, w∞ ) =
¯q ¶1/q ¶¶
µ ¯
µ
µ∞

¯
¯

 sup max P 2ν/p P¯ 1 P ank ¯
for 1 < p < ∞,

¯
¯
Nm
m
ν m n∈Nm
ν=0
and
p
, c∞ (µ)) =
M (w∞
¯¶¶
¯
µ
µ∞

¯
¯ 1 P
P ν/p

¯
¯

sup
max
∆
(µ
a
)
for p = 1
2
max
n n nk ¯

¯
 m Nm ν=0
ν
µm n∈Nm
¯q ¶1/q ¶¶
µ∞
µ
µ ¯

¯
P ν/p P¯¯ 1 P


¯
for 1 < p < ∞.
2
∆
(µ
a
)
 sup max
n n nk ¯
¯
Nm
m
ν µm n∈Nm
ν=0
(c) Let Λ = (λk )∞
of positive reals and
k=0 be an exponentially bounded sequence P
(λk(ν) )∞
be
an
associated
subsequence.
We
write
max
and
ν
ν=0
ν for the maximum
and the sum taken over all integers k such that k(ν) ≤ k ≤ k(ν + 1) − 1. Then for
X = c∞ (Λ) we have
¯∞
µ
µX
µ
¶¯¶¶
∞
¯
¯X 1 1 X
¯
M (c∞ (Λ), w∞ ) = sup max
anj ¯¯ ,
λk(ν+1) max¯
ν
N
λj m
m
m
ν=0
n∈Nm
j=k
and
M (c∞ (Λ), c∞ (µ)) =
¯∞
µ
¶¯¶¶
µ
µX
∞
¯
¯X 1
1 X
∆n (µn ank ) ¯¯ .
sup max
λk(ν+1) max¯¯
ν
Nm
λj µ m
m
ν=0
j=k
n∈Nm
The main result of this subsection is the following theorem (see Corollary 3.49).
Theorem 3.51. [70, Theorem 3] Let A, X and Y be as in Corollary 3.49
(a) If X is a normed space and A ∈ (X, Y ) for Y = w0 , w and w∞ , then, for
°∗ ¶
°
µ
°
°1 X
(m)
°
Ai °
kAkw∞ = sup
max
°
°
k>m Nm,k ⊂{m+1,...,k} k
i∈Nm,k
we have
(3.111)
(3.112)
(m)
lim kAk(m)
w∞ ≤ kLA kχ ≤ 4 · lim kAkw∞
m→∞
m→∞
1
(m)
· lim kAk(m)
w∞ ≤ kLA kχ ≤ 4 · lim kAkw∞
m→∞
2 m→∞
if Y = w0 ,
if Y = w,
213
Theory of sequence spaces
0 ≤ kLA kχ ≤ 4 · lim kAk(m)
w∞
(3.113)
m→∞
if Y = w∞ .
(b) If X is a normed space and A ∈ (X, Y ) for Y = c0 (µ), c(µ) and c∞ (µ),
then, for
°
°∗ ¶
µ
° 1 X
°
(m)
°
(3.114)
kAkc∞ = sup
max
µi Ai − µi−1 Ai−1 °
°
°
k>m Nm,k ⊂{m+1,...,k} µk
i∈Nm,k
we have
(m)
lim kAk(m)
c∞ ≤ kLA kχ ≤ 4· lim kAkc∞
(3.115)
m→∞
m→∞
if Y = c0 (µ),
(3.116)
1
(m)
· lim kAk(m)
c∞ ≤ kLA kχ ≤ 4· lim kAkc∞
m→∞
2 m→∞
0 ≤ kLA kχ ≤ 4· lim kAk(m)
c∞
(3.117)
m→∞
if Y = c(µ),
if Y = c∞ (µ).
Proof. Let us remark that the limits in (3.111) and (3.115) exist. We put
B = {x ∈ X : kxk ≤ 1}. In the case Y = w0 we have by Theorem 2.23
·
¸
(3.118)
kLA kχ = χ(A(B)) = lim sup k(I − Pm )(A(x))k ,
m→∞ x∈B
where Pm : w0 7→ w0 , m = 1, 2, . . . , is the projector on the first m coordinates,
that is Pm (x) = (x1 , x2 , . . . , xm , 0, 0, . . . ) for x = (xi ) ∈ w0 ; (let us remark that
kI − Pm k = 1 for m = 1, 2, . . . ). For given ² > 0 there is x ∈ B such that
(3.119)
k(I − Pm )(A(x))k > k(I − Pm )(A)k − ²/2.
Now there is an integer k(x) > m such that
k(x)
X
²
1
|Ai (x)| > k(I − Pm )(A(x))k − .
k(x) i=m+1
2
(3.120)
Further by Lemma 3.9
µ
4·
max
Nm,k(x) ⊂{m+1,...,k(x)}
¯
1 ¯¯
k(x) ¯
¯¶
¯
Ai (x)¯¯ ≥
X
i∈Nm,k(x)
Now, by (3.119) and (3.120) we get
¯
µ
1 ¯¯
(3.121) 4 ·
max
Nm,k(x) ⊂{m+1,...,k(x)} k(x) ¯
X
i∈Nm,k(x)
k(x)
X
1
|Ai (x)|.
k(x) i=m+1
¯¶
¯
Ai (x)¯¯ ≥ k(I − Pm )(A)k − ².
214
Malkowsky and Rakočević
Since ² > 0 was arbitrary and x ∈ B, from (3.121) we have for each m
(3.122)
°
°¶¸
°1 X
°
°
° .
A
i°
°
Nm,k ⊂{m+1,...,k} k
·
µ
k(I − Pm )(A)k ≤ 4 · sup
k>m
max
i∈Nm,k
Hence, by (3.118) and (3.122) we get the right inequality in (3.111). To prove the
left inequality in (3.111), suppose that m is an integer, k > m, Nm,k ⊂ {m + 1,
. . . , k} and x ∈ B. Then
¯
¯
k
¯1 X
¯ 1 X
1 X
¯
¯
≤
A
(x)
|A
(x)|
≤
|Ai (x)| ≤ k(I − Pm )(A(x))k.
i
i
¯k
¯ k
k i=m+1
i∈Nm,k
i∈Nm,k
Thus for all m and k > m we have
°
°
°1 X
°
°
°
A
i ° ≤ k(I − Pm )(LA )k,
°k
i∈Nm,k
and by (3.118) we get the left inequality in (3.111).
To prove (3.112) we recall that every sequence x = (xk )∞
k=1 ∈ w has a unique
P∞
(k)
representation x = le + k=1 (xk − l)e where l ∈ C is such that x − le ∈ w. Let
Pm
us define Pm : w 7→ w by Pm (x) = le + k=1 (xk − l)e(k) for m = 1, 2, . . . . It is easy
to prove that kI − Pm k = 2 for m = 1, 2, . . . . Now the proof of(3.112) is similar as
in the case (3.111), and we omit it.
Let us prove (3.113). Now define Pm : w∞ 7→ w∞ by Pm (x) = (x1 , x2 , . . . , xm ,
0, . . . ) for all x = (xi ) ∈ w∞ and m = 1, 2, . . . . It is clear that A(B) ⊂ Pm (A(B)) +
(I − Pm )(A(B)). Now, by the elementary properties of the function χ we have
χ(A(B)) ≤ χ(Pm (A(B))) + χ((I − Pm )(A(B))) = χ((I − Pm )(A(B))
(3.123)
≤ sup k(I − Pm )(A(x))k.
x∈B
Since the limit in (3.113) obviously exists, by (3.123) and from the proof of the
right inequality in (3.111) we get (3.113).
Let us mention that inequalities (3.115), (3.116) and (3.117) are proved similarly as the inequalities (3.111), (3.112) and (3.113).
¤
Now as a corollary of the theorem above we have
Corollary 3.52. [70, Corollary 2] Let A, X and Y be as in Theorem 3.51.
Then for A ∈ (X, Y ) we have
A is compact if and only if kAkw∞ < ∞ and
lim kAk(m)
w∞ = 0,
m→∞
if Y = w0 and w,
215
Theory of sequence spaces
A is compact if kAkw∞ < ∞ and
lim kAk(m)
w∞ = 0,
if Y = w∞ ,
m→∞
A is compact if and only if kAkc∞ (µ) < ∞ and
(m)
lim kAkc∞ (µ) = 0,
if Y = c0 (µ) and c(µ),
m→∞
A is compact if kAkc∞ (µ) < ∞ and
(m)
lim kAkc∞ (µ) = 0,
if Y = c∞ (µ).
m→∞
Now, concerning Remark 3.50, we get several corollaries.
Corollary 3.53. [70, Corollary 3] Let A, X and Y be as in Theorem 3.51
and in Remark 3.50 (a). We shall write maxNm,k for maxNm,k ⊂{m+1,...,k} . For
A ∈ (X, Y ) and X = lp , we set for each m
¯
¯¶¶
µ
µ

¯
¯1 P

¯
sup max sup¯
aij ¯¯
for p = 1


 k>m Nm,k j k i∈Nm,k
(m)
M (lp , w∞ )
=
¯q ¶1/q ¶
µ
µ∞¯

¯
P¯1 P


¯
aij ¯¯
for 1 < p < ∞,
 sup max
¯
N
k
m,k
k>m
j=1
i∈Nm,k
and
(m)
M (lp , c∞ (µ))
¯
µ
µ

¯ 1

sup max sup¯¯



Nm,k
µk
j
P
k>m
=
µ
µ∞¯

P¯ 1


¯
 sup max
¯
N
m,k
k>m
j=1 µk
i∈Nm,k
P
i∈Nm,k
¯¶¶
¯
∆i (µi aij )¯¯
for p = 1
¯q ¶1/q ¶
¯
∆i (µi aij )¯¯
for 1 < p < ∞.
Now we have
A is compact if and only if M (lp , w∞ ) < ∞ and
lim M (lp , w∞ )(m) = 0,
m→∞
if Y = w0 and w,
A is compact if M (lp , w∞ ) < ∞ and
lim M (lp , w∞ )(m) = 0,
m→∞
if Y = w∞ ,
A is compact if and only if M (lp , c∞ (µ)) < ∞ and
lim M (lp , c∞ (µ))(m) = 0,
m→∞
if Y = c0 (µ) and c(µ),
A is compact if M (lp , c∞ (µ)) < ∞ and
lim M (lp , c∞ (µ))(m) = 0,
m→∞
if Y = c∞ (µ).
216
Malkowsky and Rakočević
Corollary 3.54. [70, Corollary 4] Let A, X and Y be as in Theorem 3.51
and in Remark 3.50 (b). We shall write maxNm,k for maxNm,k ⊂{m+1,...,k} . For
p
A ∈ (X, Y ) and X = w∞
, we set for each m
p
M (w∞
, w∞ )(m)
¯
¯¶¶
µ
µ∞

¯1 P
¯
P ν/p

¯
¯

ν
ν+1
max
2
max
a
sup
ij ¯
2
≤j≤2
−1

¯

k i∈Nm,k

k>m Nm,k ν=0



 for p = 1
=
¯
¯q ¶1/q ¶¶
µ
µ∞
µ
¯1 P
¯

P ν/p
P

¯

sup max
aij ¯¯
2

¯

N

m,k
k>m
ν=0
2ν ≤j≤2ν+1 −1 k i∈Nm,k



for 1 < p < ∞,
and
p
M (w∞
, c∞ (µ))(m)
¯¶¶
¯
µ
µ∞

¯
¯ 1 P
P ν/p

¯
¯

∆
(µ
a
)
sup
max
2
max
i i ij ¯

¯
ν
ν+1

2 ≤j≤2
−1 µk i∈Nm,k

k>m Nm,k ν=1



 for p = 1
=
¯q ¶1/q ¶¶
¯
µ
µ∞
µ
¯
¯ 1 P

P ν/p
P

¯

∆i (µi aij )¯¯
sup max
2

¯

N
µ

m,k
k i∈Nm,k
k>m
ν=1
2ν ≤j≤2ν+1 −1



for 1 < p < ∞.
Now we have
p
A is compact if and only if M (w∞
, w∞ ) < ∞ and
p
lim M (w∞
, w∞ )(m) = 0,
m→∞
if Y = w0 and w,
p
, w∞ ) < ∞ and
A is compact if M (w∞
p
lim M (w∞
, w∞ )(m) ,
m→∞
if Y = w∞ ,
p
A is compact if and only if M (w∞
, c∞ (µ)) < ∞ and
p
lim M (w∞
, c∞ (µ))(m) = 0,
m→∞
if Y = c0 (µ) and c(µ),
p
A is compact if M (w∞
, c∞ (µ)) < ∞ and
p
lim M (w∞
, c∞ (µ))(m) = 0,
m→∞
if Y = c∞ (µ).
217
Theory of sequence spaces
Corollary 3.55. [70, Corollary 5] Let A, X and Y be as in Theorem 3.51
and in Remark 3.50 (c). We shall write maxNm,k for maxNm,k ⊂{m+1,...,k} . For
A ∈ (X, Y ), if X = c0 (Λ), c(Λ) or c∞ (Λ), we set for each m
M (c∞ (Λ), w∞ )(m)
µ
µX
∞
= sup max
λr(ν+1)
Nm,k
k>m
ν=0
¯X
µ
¶¯¶¶
¯
¯∞ 1 1 X
¯
¯
a
ij ¯
¯
λj k
r(ν)≤r≤r(ν+1)−1
max
j=r
i∈Nm,k
and
M (c∞ (Λ), c∞ (µ))(m)
µ
µX
∞
= sup max
λr(ν+1)
k>m
Nm,k
ν=0
¯X
µ
¶¯¶¶
¯∞ 1
¯
1 X
¯
∆i (µi aij ) ¯¯
max
¯
λj µk
r(ν)≤r≤r(ν+1)−1
j=r
i∈Nm,k
Now we have
A is compact if and only if M (c∞ (Λ), w∞ ) < ∞ and
lim M (c∞ (Λ), w∞ )(m) = 0,
m→∞
if Y = w0 and w,
A is compact if M (c∞ (Λ), w∞ ) < ∞ and
lim M (c∞ (Λ), w∞ )(m) = 0,
m→∞
if Y = w∞ ,
A is compact if and only if M (c∞ (Λ), c∞ (µ)) < ∞ and
lim M (c∞ (Λ), c∞ (µ))(m) = 0,
m→∞
if Y = c0 (µ) and c(µ),
A is compact if M (c∞ (Λ), c∞ (µ)) < ∞ and
lim M (c∞ (Λ), c∞ (µ))(m) = 0,
m→∞
if Y = c∞ (µ).
3.8. Further results. In this subsection, we shall give the characterizations
p
of the classes (X, Y ) where X = l1 and Y = w∞
, wp , w0p (1 ≤ p < ∞), or X = w0 ,
p
w, w∞ and Y = lp (1 ≤ p < ∞), or X = w0 , w, w∞ and Y = w0p , wp and w∞
(1 ≤ p < ∞). Furthermore we shall apply the Hausdorff measure of compactness to
give necessary and sufficient conditions for a linear operator between these spaces
to be compact. The results can be found in [73].
Let a ∈ ω. Then we write
¯
½¯X
¾
¯∞
¯
∗∗
∗
¯
¯
kak = kakMp = sup ¯
ak xk ¯ : kxkMp = 1 .
k=1
218
Malkowsky and Rakočević
Lemma 3.56. [73, Lemma 1] Let 1 ≤ p < ∞.
p β
(a) Then (w0p )β = (wp )β = (w∞
) = Mp and kak∗ = kakMp on Mp (cf. [49]
or [65, Lemma 2]). Further the sets Mp are BK spaces with the norms k · kMp (cf.
[59, Theorem 2 (a)]), and it is easy to see that the spaces Mp have AK.
p
p ββ
p
p
(b) Then w∞
is β–perfect, that is (w∞
) = w∞
and (w0p )ββ = (wp )ββ = w∞
∗∗
p
β
p
p
on (M ) = w∞ (cf. [63,
(cf. [59, Theorem 4 (b) and (c)]) and kak = kakw∞
Theorem 6 (b)]).
If A is an infinite matrix, then we write AT for its transpose.
Theorem 3.57. [73, Theorem 1] Let 1 ≤ p < ∞. Then
p
(a) A ∈ (l1 , w∞
) if and only if
(3.124)
p
M (l1 , w∞
)
µ X
¶
m
1
p
= sup
|ank | < ∞;
m,k m n=1
(b) A ∈ (l1 , w0p ) if and only if condition (3.124) holds and
µ
(3.125)
lim
m→∞
¶
m
1 X
|ank |p = 0
m n=1
for all k;
(c) A ∈ (l1 , wp ) if and only if condition (3.124) holds and there is a sequence
∈ ω such that
(λk )∞
k=1
µ
(3.126)
lim
m→∞
¶
m
1 X
p
|ank − λk | = 0
m n=1
for all k.
Proof. (a) condition (3.124) follows from [108, Example 8.4.1, p. 126] with
p
Y = w∞
.
(b) Parts (b) and (c) follow from part (a) and [70, Theorem 1 (c)].
¤
By T we denote the set of all strictly increasing sequences (tν )∞
ν=0 of integers
such that for each ν there is one and only one tν with 2ν ≤ tν ≤ 2ν+1 − 1. We put
¯
¯¶
µ∞

¯P
¯
P ν

¯

sup
2 max¯
ank ¯¯


ν
N ⊂N

ν=0
n∈N

N finite


¯p ¶¶
µ
µ∞¯

¯
P¯ P ν
¯
¯
M (w0 , lp ) =
sup
sup
2
a
n,t
ν
¯
¯


N ⊂N0 t∈T n=1 ν∈N


¯
¯
µ
µ
¶¶


¯P ν
¯


¯
¯
 sup sup sup¯
2 an,tν ¯
N ⊂N0
t∈T
n
ν∈N
(p = 1)
(1 < p < ∞)
(p = ∞).
Theory of sequence spaces
219
Theorem 3.58. [73, Theorem 2] Let 1 ≤ p ≤ ∞. Then
(3.127)
(w0 , lp ) = (w, lp ) = (w∞ , lp );
further A ∈ (w0 , lp ) if and only if
(3.128)
M (w0 , lp ) < ∞.
Proof. The case p = 1 follows from Lemma 3.56 (a) and from [63, Theorem
1] with X = w0 , w, w∞ .
For 1 < p ≤ ∞, we apply [108, Theorem 8.3.9, p.124] with X = w0 and Z = lq
where q = 1 for p = ∞ and q = p/(p − 1) for 1 < p < ∞. Since X and Z are BK
spaces with AK, we obtain (w0 , lp ) = (w0ββ , lp ) = (w∞ , lp ). (The second equality
holds in view of Lemma 3.56 (b).) Since w0 ⊂ w ⊂ w∞ , we have established the
identities in (3.127). Further, by [108, Theorem 8.3.9], A ∈ (w0 , lp ) if and only
if AT ∈ (lq , w0β ) = (lq , M1 ), by Lemma 3.56 (a). Finally, by [59, Theorem 7],
AT ∈ (lq , M1 ) if and only if M (w0 , lp ) < ∞.
¤
Let us remark that an application of [70,
P∞Theorem 1 (b)] and Lemma 3.56
(a) yields A ∈ (w0 , l∞ ) if and only if supn ν=0 2ν maxν |ank | < ∞, a condition
equivalent to condition (3.128) in Theorem 3.58 for p = ∞.
We write N hµi for the set of all integers n with 2µ ≤ n ≤ 2µ+1 − 1, and we put
¯
¯¶¶
µ
µ∞

¯1 P
¯
P ν

¯
 sup
2 max¯ µ
ank ¯¯
(p = 1)

 µ∈N0 Nµmax
hµi
ν
2 n∈Nµ
⊂N
ν=0
p
M (w0 , w∞ ) =
¯
¯p ¶¶¶
µ
µ
µ
m ¯ P 1
¯

1 P


¯
¯
a
(1 < p < ∞).
 sup sup sup
n,t
ν
¯
¯
m n=1 ν∈N 2ν
m
N ⊂N0 t∈T
Theorem 3.59. [73, Theorem 3] Let 1 ≤ p < ∞.
(a) Then
(3.129)
p
p
p
(w0 , w∞
) = (w, w∞
) = (w∞ , w∞
);
p
further A ∈ (w0 , w∞
) if and only if
p
M (w0 , w∞
) < ∞;
(3.130)
(b) A ∈ (w0 , w0p ) if and only if conditions (3.130) and (3.125) hold; A ∈
(w0 , wp ) if and only if conditions (3.130) and (3.126) hold; A ∈ (w, w0p ) if and only
if conditions (3.130) and (3.125) hold and
µ
(3.131)
lim
m→∞
¯p ¶
m ¯ ∞
¯
1 X ¯¯X
¯ = 0;
a
nk ¯
¯
m n=1
k=1
220
Malkowsky and Rakočević
A ∈ (w, wp ) if and only if conditions (3.130) and (3.126) hold and
µ
(3.132)
lim
m→∞
¯p ¶
m ¯ ∞
¯
1 X ¯¯X
¯ =0
a
−
λ
nk
¯
¯
m n=1
for some complex number λ.
k=1
Proof. (a) For p = 1, part (a) is an immediate consequence of [63, Korollar
2] and Lemma 3.56 (a).
For 1 < p < ∞, the identities in (3.129) follow by an argument similar to that
in the proof of Theorem 3.58. We apply [108, Theorem 8.3.9] with X = w0 and
p
Z = Mp to conclude A ∈ (w0 , w∞
) if and only if AT ∈ (Mp , w0p ) = (Mp , M1 ).
p
Finally, by [59, Theorem 7], AT ∈ (Mp , M1 ) if and only if M (w0 , w∞
) < ∞.
(b) Part (b) follows from [70, Theorem 1 (c)], the fact that w0 has AK and
the representation for sequences in w given in Proposition 3.45.
¤
Now we shall give estimates for the operator norm kLA k. We put
¶1/p
m
1 X
p
= sup
|ank |
m,k m n=1
µ
p
MA∗ (l1 , w̃∞
)
(1 ≤ p < ∞),
and for any BK space X
MA∗ (X, l1 )
MA∗ (X, l∞ )
°
°∗
°X
°
°
= sup °
An °
° ,
N ⊂N
=
N finite n∈N
sup kAn k∗ ,
n
°∗ ¶
°
°
°1 X
°
An °
= sup
max ° µ
°
hµi 2
N
⊂N
µ
µ
µ
MA∗ (X, w∞ )
n∈Nµ
° ¶
µ° X
°
°∗
µ
° .
MA∗ (X, M1 ) = sup °
2
A
tµ °
°
N ⊂N
0
µ∈N
p the norm on w p
Theorem 3.60. [73, Theorem 4] (a) Let 1 ≤ p < ∞, k · kw̃∞
∞
defined by
µ
(3.133)
p
kxkw̃∞
n
1X
|xk |p
= sup
n
n
¶1/p
k=1
p
and A ∈ (l1 , w∞
). Then
(3.134)
p
kLA k = MA∗ (l1 , w̃∞
).
(b) Let X be an arbitrary BK space. If A ∈ (X, l1 ), then
(3.135)
MA∗ (X, l1 ) ≤ kLA k ≤ 4 · MA∗ (X, l1 ).
221
Theory of sequence spaces
If A ∈ (X, l∞ ), then
kLA k = MA∗ (X, l∞ ).
(3.136)
If k · kw∞ is the norm on w∞ defined in (3.101) and A ∈ (X, w∞ ), then
MA∗ (X, w∞ ) ≤ kLA k ≤ 4 · MA∗ (X, w∞ ).
(3.137)
(c) Let X be an arbitrary BK space and A ∈ (X, M1 ), then
MA∗ (X, M1 ) ≤ kLA k ≤ 4 · MA∗ (X, M1 ).
(3.138)
p
Proof. (a) Let A ∈ (l1 , w∞
), x ∈ l1 with kxk1 =
be given. Then we have by Minkowski’s inequality
µ
P∞
k=1
|xk | = 1 and m ∈ N
¯p ¶1/p X
¶1/p µ X
µ X
¶1/p
m
m ¯X
∞
m
¯∞
¯
1
1
1 X
p
p
¯
¯
=
≤
|An (x)|
ank xk ¯
|xk |
|ank |
m n=1
m n=1¯
m n=1
k=1
k=1
p
≤ MA∗ (l1 , w̃∞
),
p ≤ M ∗ (l , w̃ p ) and consequently
hence, since m was arbitrary, kA(x)kw̃∞
∞
A 1
p
p : kxk = 1} ≤ M (l , w̃
kLA k = sup{kA(x)kw̃∞
1
A 1
∞ ).
(3.139)
Now let x = e(k) (k = 1, 2, . . . ). Then x ∈ l1 , kxk1 = 1 and
µ X
¶1/p
m
1
p = sup
kA(x)kw∞
|ank |p
≤ kLA k
m n=1
together imply
p
) ≤ kLA k.
MA∗ (l1 , w̃∞
(3.140)
Finally, from (3.139) and (3.140), we conclude (3.134).
(b) First we show (3.135). Let A ∈ (X, l1 ), x ∈ X with kxk = 1 and m ∈ N be
given. Then
¯∞µ
¶ ¯
m
X
¯X X
¯
|An (x)| ≤ 4 · max ¯¯
ank xk ¯¯ ≤ 4 · MA∗ (X, l1 ).
n=1
N ⊂{1,...,m}
k=1
n∈N
Since m was arbitrary, we conclude kA(x)k1 ≤ 4 · MA∗ (X, l1 ) and consequently
(3.141)
kLA k ≤ 4 · MA∗ (X, l1 ).
Conversely, let N ⊂ N be an arbitrary finite °set. Then° given
¯ ε > 0 there
¯ is a
° P
°∗ ¯ P
¯
sequence x = x(N, ε) ∈ X such that kxk = 1 and °
An ° ≤ ¯
An (x)¯ + ε.
n∈N
n∈N
222
Malkowsky and Rakočević
Therefore
°∗ X
°X
∞
°
°
°
° ≤
A
|An (x)| + ε ≤ kA(x)k1 + ε ≤ kLA k + ε.
n
°
°
n=1
n∈N
Since N ⊂ N and ε > 0 were arbitrary,
MA∗ (X, l1 ) ≤ kLA k.
(3.142)
Finally, from (3.141) and (3.142), we conclude (3.135) Equality (3.136) is Theorem
1.23 (b). The inequalities in (3.137) are shown in exactly the same way as those
in [70, Corollary 1 (a), (2.8)] with m, Nm ⊂ {1, . . . , m} and 1/m replaced by µ,
Nµ ⊂ N hµi and 1/2µ .
(c) Let A ∈ (X, M1 ), x ∈ X with kxk = 1 and µ0 ∈ N0 be given. We choose
nµ ∈ N hµi (µ = 0, 1, . . . ) such that |Anµ (x)| = maxn∈N hµi |An (x)|. Then we have
µ0
X
µ=0
¯X
¯
¯
¯
µ
¯
2 Anµ (x)¯¯
2 |Anµ (x)| ≤ 4 ·
max ¯
N ⊂{0,...,µ0 }
µ
µ∈N
¯∞µ
¶ ¯
¯X X µ
¯
¯
=4·
max ¯
2 anµ ,k xk ¯¯
N ⊂{0,...,µ0 }
k=1
µ∈N
°
°∗ ¶
µ
°X µ
°
° = 4 · MA∗ (X, M1 ).
≤ 4 · sup sup°
2
A
tµ °
°
t∈T
N ∈N
0
µ∈N
Since this holds for all µ0 ∈ N0 , we conclude kA(x)kM1 ≤ 4 · MA∗ (X, M1 ) and
consequently
kLA k ≤ 4 · MA∗ (X, M1 )
(3.143)
Conversely, let N ∈ N0 , t ∈ T and ε >
P0 be given. ThenPthere is a sequence
x = x(N, t, ε) ∈ X such that kxk = 1 and k µ∈N 2µ Atµ k∗ ≤ | µ∈N 2µ Atµ (x)|+ε.
Therefore
°
°∗ X
∞
°X µ
°
°
°
2
A
≤
2µ max |An (x)| + ε = kA(x)kM1 + ε ≤ kLA k + ε.
t
µ°
°
n∈N hµi
µ∈N
µ=0
Since N ∈ N0 , t ∈ T and ε > 0 were arbitrary, we have MA∗ (X, M1 ) ≤ kLA k.
Finally, from this and (3.143), we conclude (3.138).
¤
Now we apply the previous results to estimate the operator norms of the matrix
transformations characterized in Theorems 3.57, 3.58 and 3.59.
223
Theory of sequence spaces
Let X be any of the spaces w0 , w and w∞ . We put
¯X
¯¶
µX
∞
¯
¯
MA∗ (X, l1 ) = sup
2ν max¯¯
ank ¯¯ ,
ν
N ⊂N
N finite ν=0
µX
∞
ν
¶
2 max|ank | ,
MA∗ (X, l∞ ) = sup
n
ν
ν=0
µ
MA∗ (X, w∞ )
= sup
µ
n∈N
max
Nµ ⊂N hµi
µX
∞
ν=0
¯
¯¶¶
¯1 X
¯
¯
2 max¯ µ
ank ¯¯ ,
ν
2
ν
n∈Nµ
and, for 1 < p < ∞ and q = p/(p − 1),
¯p ¶1/p ¶
µ
µX
∞ ¯X
¯
¯
ν
¯
¯
MA∗ T (lq , M1 ) = sup sup
2
a
,
n,tν ¯
¯
N ⊂N0
t∈T
n=1 ν∈N
µ
µ
µ
1
MA∗ T (Mp , M1 ) = sup sup sup µ
2
µ
N ⊂N0 t∈T
¯p ¶1/p ¶¶
X ¯¯ X
¯
ν
¯
2 an,tν ¯¯
.
¯
n∈N hµi ν∈N
Corollary 3.61. [73, Corollary 1] Let X be any of the spaces w0 , w and w∞
p the norm defined in (3.101).
and k · kw∞
If A ∈ (X, l1 ), then MA∗ (X, l1 ) ≤ kLA k ≤ 4 · MA∗ (X, l1 ).
If A ∈ (X, l∞ ), then kLA k = MA∗ (X, l∞ ).
p
If A ∈ (X, lp ) (1 < p < ∞, q = p−1
), then MA∗ T (lq , M1 ) ≤ kLA k ≤ 4·MA∗ T (lq , M1 ).
∗
If A ∈ (X, w∞ ), then MA (X, w∞ ) ≤ kLA k ≤ 4 · MA∗ (X, w∞ ).
p
) (1 < p < ∞), then MA∗ T (Mp , M1 ) ≤ kLA k ≤ 4 · MA∗ T (Mp , M1 ).
If A ∈ (X, w∞
We need with the following auxiliary lemma.
Lemma 3.62. [73, Lemma 2] (a) Let Pm : w0p 7→ w0p for 1 ≤ p < ∞ and m =
1, 2, . . . be the projector on the first m coordinates, that is Pm (x) = (x1 , x2 , . . . , xm ,
0, 0, . . . ) for x = (xi ) ∈ w0p . Then kI − Pm k = 1, m = 1, 2, . . . .
(b) For x ∈ wp , we useP
the representation in Proposition 3.44 and define Pm :
m
wp 7→ wp by Pm (x) = le + k=1 (xk − l)e(k) for m = 1, 2, . . . . Then kI − Pm k = 2
for m = 1, 2, . . . .
Proof. (a) It is clear that kI − Pm k ≤ 1. Since I − Pm 6= O is a bounded
linear operator and projector, we have kI − Pm k ≥ 1. This proves (a).
p
(b) Let x = (xk )∞
k=1 ∈ w . Then x has the representation in Proposition 3.44,
and we obtain
k(I − Pm )(x)k = k (0, . . . , 0, xm+1 − l, xm+2 − l, . . . k ≤ kxk + |l| ≤ 2kxk.
| {z }
m
Hence kI − Pm k ≤ 2, m = 1, 2, . . . . To prove that kI − Pm k ≥ 2, let ² > 0. Then,
since
¶1/p
µ
k
→ 2 (k → ∞),
2
m+k
224
Malkowsky and Rakočević
there exists k0 ∈ N such that
µ
2
k0
m + k0
¶1/p
> 2 − ².
Let u0 ∈ wp be defined by
u0 = (1, . . . , 1, −1, . . . , −1, 1, 1, 1, . . . ).
| {z } | {z }
k0
m
Then ku0 k = 1, l = 1 and
µ
k(I − Pm )(u0 )k ≥
1
· 2p k0
m + k0
¶1/p
µ
=2
k0
m + k0
¶1/p
> 2 − ².
Hence kI − Pm k > 2 − ², that is kI − Pm k ≥ 2.
p the norm on
Theorem 3.63. [73, Theorem 5] Let 1 ≤ p < ∞, k · kw̃∞
p
and w∞
defined in (3.133). We put
µ
p
MA∗ (l1 , w̃∞
)(m) = sup
k
u>m
¤
w0p ,
wp
¶1/p
u
1 X
|ank |p
.
u n=m+1
(a) If A ∈ (l1 , w0p ), then
(3.144)
p
kLA kχ = lim MA∗ (l1 , w̃∞
)(m) .
m→∞
(b) If A ∈ (l1 , wp ), then
(3.145)
1
p
p
· lim M ∗ (l1 , w̃∞
)(m) ≤ kLA kχ ≤ lim MA∗ (l1 , w̃∞
)(m) .
m→∞
2 m→∞ A
p
(c) If A ∈ (l1 , w∞
), then
(3.146)
p
)(m) .
0 ≤ kLA kχ ≤ lim MA∗ (l1 , w̃∞
m→∞
Proof. Let us remark that the limits in (3.144), (3.145) and (3.146) exist. We
put B = {x ∈ l1 : kxk ≤ 1}. In the case (a) we have by the inequality in Theorem
2.23
·
¸
(3.147)
kLA kχ = χ(A(B)) = lim sup k(I − Pm )(A(x))k ,
m→∞ x∈B
where Pm : w0p 7→ w0p for m = 1, 2, . . . is the projector on the first m coordinates,
that is Pm (x) = (x1 , x2 , . . . , xm , 0, 0, . . . ) for x = (xk ) ∈ w0p . Let us recall that by
Theory of sequence spaces
225
Lemma 3.62 (c) we have kI − Pm k = 1, m = 1, 2, . . . . Let A(m) = (ãnk ) be the
infinite matrix defined by ãnk = 0 if 1 ≤ n ≤ m and ãnk = ank if m < n. Now, by
(3.134) we have
p
p
sup k(I − Pm )(A(x))k = kLA(m) k = MA∗ (m) (l1 , w̃∞
)(m) = MA∗ (l1 , w̃∞
)(m)
x∈B
Part (a) now follows from this and (3.147).
p
(b) Let x = (xk )∞
representation in Proposition 3.44,
k=1 ∈ w0 . Then x has the P
m
p
p
and we define Pm : w 7→ w by Pm (x) = le + k=1 (xk − l)e(k) for m = 1, 2, . . . .
By Lemma 3.62 (b) we know that kI − Pm k = 2 for m = 1, 2, . . . . Now the proof
of (b) is similar as in the case (a), and we omit it.
p
p
Let us prove (3.146). Now define Pm : w∞
7→ w∞
by Pm (x) = (x1 , x2 , . . . , xm ,
p
0, . . . ) for x = (xi ) ∈ w∞ and m = 1, 2, . . . . It is clear that A(B) ⊂ Pm (A(B)) +
(I − Pm )(A(B)). Now, by the elementary properties of the function χ we have
χ(A(B)) ≤ χ(Pm (A(B))) + χ((I − Pm )(A(B))) = χ((I − Pm )(A(B))
(3.148)
≤ sup k(I − Pm )(A(x))k = kLA(m) k.
x∈B
Since the limit in (3.146) obviously exists, by (3.148) and (3.135) we get (3.146).
¤
Now as a corollary of the above theorem we have
Corollary 3.64. [73, Corollary 2] If either A ∈ (l1 , w0p ) or A ∈ (l1 , w0p ), then
LA
p
is compact if and only if lim MA∗ (l1 , w̃∞
)(m) = 0.
m→∞
p
If A ∈ (l1 , w∞
), then
(3.149)
LA
p
is compact if lim MA∗ (l1 , w̃∞
)(m) = 0.
m→∞
The following example will show that it is possible for LA in (3.149) to be
p
compact in the case limm→∞ MA∗ (l1 , w̃∞
)(m) > 0, and hence in general we have
just “if” in (3.149).
Example 3.65. [73, Example 1] Let the matrix A be defined by ank = 1 if
p
p
n = 1 and ank = 0 if n 6= 1. Then MA∗ (l1 , w̃∞
) = 1 and A ∈ (l1 , w̃∞
). Further
µ
p
MA∗ (l1 , w̃∞
)(m)
=
sup
k≥1,u>m
¶1/p
¶1/p
µ
u
1 X
u−m
p
= 1.
|ank |
= sup
u n=m+1
u
k≥1,u>m
p
Whence limm→∞ MA∗ (l1 , w̃∞
)(m) = 1 > 0. Since A(x) = x1 e for all x ∈ l1 , LA is a
compact operator.
226
Malkowsky and Rakočević
Now, concerning Corollary 3.61 we continue to study the measures of noncomp
pactness of operators when the final spaces are the spaces lp and w∞
. Let X be
any of the spaces w0 , w and w∞ . For m ∈ N we put
¯X
¯¶
µX
∞
¯
¯
∗
ν
¯
MA (X, l1 )(m) =
sup
2 max¯
ank ¯¯ ,
ν
N ⊂Nr{1,2,...,m}
N finite
µX
∞
MA∗ (X, l∞ )(m) = sup
n>m
µ
MA∗ (X, w∞ )(m) = sup
µ>m
ν=0
n∈N
¶
ν
2 max|ank | ,
ν=0
max
ν
Nµ ⊂N hµi
¯¶¶
¯
¯
¯1 X
2ν max¯¯ µ
ank ¯¯ ,
ν
2
ν=0
µX
∞
n∈Nµ
and, for 1 < p < ∞ and q = p/(p − 1),
¯p ¶1/p ¶
µ
µ X
∞ ¯X
¯
¯
∗
1
ν
¯
MAT (lq , M )(m) = sup sup
2 an,tν ¯¯
,
¯
t∈T
N ⊂N
0
MA∗ T (Mp , M1 )(m)
n=m+1 ν∈N
µ
µ
µ
1
=
sup
sup sup µ
2
µ
N ⊂N0 r{1,2,...,m} t∈T
¯p ¶1/p ¶¶
X ¯¯ X
¯
ν
¯
2 an,tν ¯¯
.
¯
n∈N hµi ν∈N
Theorem 3.66. [73, Theorem 6] Let X be any of the spaces w0 , w and w∞
p the norm defined in (3.101). If A ∈ (X, l ), then
and k · kw∞
1
(3.150)
lim MA∗ (X, l1 )(m) ≤ kLA kχ ≤ 4 · lim MA∗ (X, l1 )(m) .
m→∞
m→∞
If A ∈ (X, l∞ ), then
kLA kχ ≤ lim MA∗ (X, l∞ )(m) .
(3.151)
m→∞
If A ∈ (X, lp ) (1 < p < ∞, q = p/(p − 1)), then
(3.152)
lim MA∗ T (lq , M1 )(m) ≤ kLA kχ ≤ 4 · lim MA∗ T (lq , M1 )(m) .
m→∞
m→∞
If A ∈ (X, w∞ ), then
(3.153)
If A ∈
p
(X, w∞
)
(3.154)
kLA kχ ≤ 4 · lim MA∗ (X, w∞ )(m) .
m→∞
(1 < p < ∞), then
kLA kχ ≤ 4 · lim MA∗ T (Mp , M1 )(m) .
m→∞
Proof. Let us remark that the limits in (3.150) to (3.154) exist. Let Pm : lp 7→
lp for m = 1, 2, . . . and 1 ≤ p < ∞ be the projector on the first m coordinates,
that is Pm (x) = (x1 , x2 , . . . , xm , 0, 0, . . . ) for x = (xi ) ∈ lp . It is easy to check
that kI − Pm k = 1, m = 1, 2, . . . . Now the proof of (3.150) and (3.152) (when final
spaces have a basis) can be given by the method of proof of Theorem 3.63 (a), while
in the proof of (3.151), (3.153), and (3.154) (when final spaces have no basis) we
can use the method of the proof of Theorem 3.63 (c).
¤
Now as a corollary of the theorem above we have
227
Theory of sequence spaces
Corollary 3.67. [73, Corollary 3] Let X be any of the spaces w0 , w and w∞
p the norm defined in (3.101). If A ∈ (X, l ), then
and k · kw∞
1
(3.155)
LA
lim MA∗ (X, l1 )(m) = 0.
is compact if and only if
m→∞
If A ∈ (X, l∞ ), then
(3.156)
LA
lim MA∗ (X, l∞ )(m) = 0.
is compact if
m→∞
If A ∈ (X, lp ) (1 < p < ∞, q = p/(p − 1)), then
(3.157)
LA
lim MA∗ T (lq , M1 )(m) = 0.
is compact if and only if
m→∞
If A ∈ (X, w∞ ), then
(3.158)
LA
lim MA∗ (X, w∞ )(m) = 0.
is compact if
m→∞
p
If A ∈ (X, w∞
) (1 < p < ∞), then
(3.159)
LA
lim MA∗ T (Mp , M1 )(m) = 0.
is compact if
m→∞
Let us remark that it is possible for LA in (3.156), (3.158) and (3.159) to be
compact in the cases limm→∞ MA∗ (X, l∞ )(m) > 0, limm→∞ MA∗ (X, w∞ )(m) > 0 and
limm→∞ MA∗ T (Mp , M1 )(m) > 0, respectively. This can be proved by Example 3.65.
4
Appendix
In this appendix, we collect the results from Functional Analysis needed in the
previous sections.
4.1. Inequalities.
Theorem A.4.1. (Hölder’s inequality) Let 1 < p < ∞, q = p/(p − 1) and
x0 , x1 , . . . , xn , y0 , y1 , . . . , yn ∈ C. Then
n
X
|xk yk | ≤
µX
∞
k=0
|xk |p
¶1/p µX
∞
k=0
¶1/q
|yk |q
.
k=0
If x ∈ lp and y ∈ lq , then xy = (xk yk )∞
k=0 ∈ l1 and kxyk1 ≤ kxkp kykq .
Theorem A.4.2. (Minkowski’s inequality) Let 1 ≤ p < ∞ and x0 , x1 , . . . , xn ,
y0 , y1 , . . . , yn ∈ C. Then
µX
n
k=0
¶1/p
|xk + yk |
p
≤
µX
n
k=0
¶1/p
p
|xk |
+
µX
n
k=0
If x, y ∈ lp , then x + y ∈ lp and kx + ykp ≤ kxkp + kykp .
¶1/p
|yk |
p
228
Malkowsky and Rakočević
Theorem A.4.3. (Jensen’s inequality) Let p > 0 and x0 , x1 , . . . , xn ∈ C.
Then
µX
¶1/p
n
|xk |p
is a decreasing function in p,
k=0
that is, if r > s > 0, then
µX
n
¶1/r
|xk |r
≤
µX
n
k=0
¶1/s
|xk |s
.
k=0
If p > p0 , then lp0 ⊂ lp .
4.2. The closed graph theorem and the Banach–Steinhaus theorem.
Theorem A.4.4. (Closed graph lemma) Any continuous map into a Hausdorff
space has closed graph [105, Theorem 11.1.1, p. 195].
Theorem A.4.5. (Closed graph theorem) If X and Y are Fréchet spaces and
f : X 7→ Y is a linear map with closed graph, then f is continuous [105, Theorem
11.2.2, p. 200].
Theorem A.4.6. (Banach–Steinhaus theorem) Let (fn )∞
n=0 be a pointwise
convergent sequence of continuous linear functionals on a Fréchet space X. Then
f defined by
f (x) = lim fn (x) for all x ∈ X,
n→∞
is continuous [105, Corollary 11.2.4, p. 200].
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