ANALELE STIINTIFICE ALE UNIVERSITATII “AL.I. CUZA” DIN IASI (S.N.)MATEMATICA, Tomul LVIII, 2012, f.2
KOTHE SPACES OF VECTOR FIELDS
BY
ION CHITESCU and LILIANA SIRETCHI
Abstract. The standard theory of Kothe spaces of measurable functions is extendedfor vector fields on a separated locally compact space. Our theory extends the theoryof Orlicz spaces of vector fields. Basic results, concerning completeness and relationsbetween different types of convergence are obtained. The paper ends with two examples,illustrating the theoretical facts previously presented.
Mathematics Subject Classification 2010: 46E30, 28A20, 46B25, 28C15.
Key words: function norm, Kothe space, vector field, continuous vector field, mea-surable vector field.
1. Introduction
Function spaces have been a very important object of study in FunctionalAnalysis. Successive generalizations appeared (e.g. from Lebesgue spacesLp to Orlicz spaces and continuing with Kothe spaces). At the same time,the generalizations involved the range of the functions in the function spaces,considering first scalar valued functions and then vector valued functions. Inthis respect, the idea of considering vector fields instead of vector functionsappeared in connection with differential geometry and physics.
A major step in the development of the theory of vector fields was theseries of papers [2]-[5] by Dinculeanu, where a theory of Orlicz spaces ofvector fields was introduced. The theory in [2]-[5] is sistematically presentedin the monograph [6] by the same author.
The aim of the present paper is to continue the ideas in [2]-[5], initiatinga more general theory: the theory of Kothe spaces of vector fields. To thisend, we have intensively used the monographs [6] ,[8] and [1]. In order
304 ION CHITESCU and LILIANA SIRETCHI 2
to have a suitable theory of measurability, compatible with the theory ofvector fields, the framework of (locally compact) topological spaces as basicspaces seems natural. It is our intention to continue the development ofthis theory.
2. Preliminary facts
Throughout the paper, K will be the scalar field (either K = R = the real
numbers, or K = C = the complex numbers). We shall write R+def= [0,∞)
and R+def= [0,∞]. As usual, N = {0, 1, 2, . . .} = the set of natural numbers
and N∗ = N \ {0}.
2.1. Assume (S,Σ,µ) is a measure space, i.e. S is a non empty set, Σis a σ - algebra of subsets of S and µ : Σ → R+ is a complete non nullmeasure. Write M+(µ) = {u : S → R+| u is µ−measurable}.
A function norm is a function ρ : M+(µ) → R+ having the followingproperties (here u,v are in M+(µ) and a is in R+):
(i) ρ(u) = 0 iff u(t) = 0µ - a.e.;
(ii) u ≤ v ⇒ ρ(u) ≤ ρ(v);
(iii) ρ(u+ v) ≤ ρ(u) + ρ(v);
(iv) ρ(au) = aρ(u),
with the convention 0 · ∞ = 0.
One knows that ρ(u) < ∞ ⇒ u(t) < ∞ µ - a.e. and u(t) ≤ v(t) µ - a.e⇒ ρ(u) ≤ ρ(v).
We say that ρ has the Riesz - Fischer property (and write ρ R − F ) incase
ρ(
∞∑n=0
un) ≤∞∑n=0
ρ(un),
for any sequence (un)n in M+(µ).
We say that ρ has the Fatou property (and write ρ F ) in case
ρ(supn
un) = supn
ρ(un),
for any increasing sequence (un)n in M+(µ).
3 KOTHE SPACES OF VECTOR FIELDS 305
It is known that ρ F ⇒ ρ R − F (the converse implication is not true).For any A ∈ Σ, we denote by φA the characteristic (indicator) function ofA. We shall write
ρ(A)def= ρ(φA)
M(µ) = {f : S → K| f is µ - measurable}
Lρ(µ) = Lρ = {f ∈ M(µ)| ρ|f | < ∞}
(where ρ|f | def= ρ(|f |)). Then Lρ is a vector subspace of M(µ), seminormedwith the seminorm f 7→ ρ|f |. The null space of this seminorm is
N(µ) = {f ∈ Lρ | ρ|f | = 0} = {f ∈ M(µ) | f(t) = 0 µ - a.e.}
(actually, N(µ) = {f : S → K| f(t) = 0µ - a.e.}).The associate normed space is Lρ(µ) = Lρ = Lρ(µ)/N(µ) (the equiva-
lence relation being given via f ∼ g ⇔ ρ|f − g| = 0 ⇔ f(t) = g(t) µ - a.e.)normed with the norm f 7→ ∥f∥ = ρ|f |, for any f ∈ f .
The spaces Lρ are called Kothe spaces. One knows that Lρ is Banachiff ρ R − F . The spaces Lρ generalize the Lebesgue spaces Lp (for thesespaces the generating function norm ∥ ∥p has the Fatou property) or theOrlicz spaces.
2.2. Now let us consider a separated locally compact topological spaceT . Let us consider also a family E = (Et)t∈T of Banach spaces. The normin each Et will be denoted by ∥z∥, for z ∈ Et. A more precise notationwould have been ∥z∥t, but we think there is no danger of confusion.
A vector field (with respect to E) is a function x : T →∪
t∈T Et suchthat x(t) ∈ Et for any t ∈ T . The set of all vector fields with respect to Ewill be denoted by C(E). Clearly, C(E) is a vector space with respect to thepointwise defined operations
(x, y) 7→ x+ y where (x+ y)(t)def= x(t) + y(t)
(α, x) 7→ αx where (αx)(t)def= αx(t),
for any x, y ∈ C(E) and any α ∈ K. Actually, one has
C(E) =∏t∈T
Et.
306 ION CHITESCU and LILIANA SIRETCHI 4
For any x ∈ C(E), one can define the function |x| : T → R+, via |x|(t) =∥x(t)∥.
A fundamental family of continuous vector fields is a vector subspace Aof C(E), satisfying the following axioms:
(A1) For any x ∈ A, the function |x| is continuous.(A2) For any t ∈ T , the set {x(t) | x ∈ A} is dense in Et.
Particular case. The unicity case C(E).Assume that E is such that Et = E (where E is a Banach space) for any
t ∈ T . Then we say that we are in the unicity case C(E). In this case:- A vector field is a function x : T → E.- A fundamental family of continuous vector fields isA = all the constant
functions x : T → E. We shall write A = E (identifying any function x ∈ Awith its constant value x(t) for all t ∈ T ).
2.3. Let us consider a fundamental family of continuous vector fieldsA. We shall say that x ∈ C(E) is continuous at t0 ∈ T with respect to A if,for any ϵ > 0, there exist V ∈ V(t0) = the set of all neighbourhoods of t0and y ∈ A such that ∥x(t)− y(t)∥ < ϵ for any t ∈ V .
If ∅ = A ⊂ T we say that x is continuous on A if x is continuous at anyt ∈ A. In case A = T we say that x is continuous (with respect to A).
Concerning the above definition, we remark that:- Any x ∈ A is continuous with respect to A (take y = x in definition).- The vector field x is continuous at t0 (with respect to A) iff for any
y ∈ A and any α ∈ K, the vector field x + αy is continuous at t0 (withrespect to A).
- The vector field x is continuous at t0 with respect to A iff |x − y| iscontinuous at t0 for any y ∈ A. Hence, if x is continuous at t0 with respectto A, it follows that |x| is continuous at t0.
Write: CA(E) = {x ∈ C(E) | x is continuous with respect to A}. ThenA ⊂ CA(E) and CA(E) is a fundamental family of continuous vector fields.
- In the unicity case C(E): a vector field (function) x : T → E iscontinuous at t0 with respect to E iff x is continuous at t0 in the usualsense.
2.4. From now on, we shall work within the following
Framework. Assume T is a separated locally compact topologicalspace with Borel sets B. Let T be a σ - algebra of subsets of T such
5 KOTHE SPACES OF VECTOR FIELDS 307
that B ⊂ T and let µ : T → R+ be a complete non null measure withµ(A) < ∞ for any A ∈ K = the compact sets of T . We also assume that µis regular. So, we have the measure space (T ,T ,µ). We consider also a fam-ily E = (Et)t∈T of Banach spaces, the vector space C(E) and a fundamentalfamily of continuous vector fields A.
A vector field x ∈ C(E) is called µ - measurable with respect to A (shortly,x is (A, µ) - measurable) if, for any A ∈ K and any ϵ > 0, there existsK ∋ Aϵ ⊂ A such that µ(A \Aϵ) < ϵ and x is continuous with respect to Aon Aϵ.
Write M(A, µ) = {x ∈ C(E) | x is (A, µ) - measurable} hence CA(E) ⊂M(A, µ). Notice that M(A, µ) is a vector subspace of C(E) having thefollowing properties:
- For any x, y ∈ C(E) such that x ∈ M(A, µ) and y = x µ - a.e., one hasy ∈ M(A, µ).
- For any x ∈ M(A, µ) the function |x| is µ - measurable.- If (xn)n is a sequence in M(A, µ) and x is in C(E) such that xn −→
nx µ
- a.e., then x ∈ M(A, µ). Moreover (Egorov’s theorem), it follows that, forany A ∈ K and any ϵ > 0, there exists K ∋ Aϵ ⊂ A such that µ(A \Aϵ) < ϵand (xn)n converges to x uniformly on Aϵ.
- An element x ∈ C(E) is in M(A, µ) iff φAx is in M(A, µ) for anyA ∈ K.
Particular case. Assume we are in the unicity case C(E), with A = E.Then, one can prove (Lusin’s theorem) that a vector field (i.e. a function)x : T → E is (E, µ) - measurable iff x is µ - measurable. In this case we
write M(E,µ)def= ME(µ).
3. Results
3.1. In this section we introduce the object of study of the presentpaper, namely the Kothe spaces of vector fields. We shall work within theframework of 4., section 1, and we shall assume, supplementarily, that ρ isa function norm on (T ,T ,µ).
The seminormed Kothe space Lρ(E ,A) is defined as follows:
Lρ(E ,A) = {x ∈ M(A, µ) | ρ|x| < ∞}.
The definition is meaningful, because, as we saw, for any x ∈ M(A, µ),one has |x| ∈ M+(µ). Clearly, Lρ(E ,A) is a vector subspace of M(A, µ),seminormed with the seminorm given via x 7→ ρ|x|.
308 ION CHITESCU and LILIANA SIRETCHI 6
The null space of this seminorm is
Nρ(E ,A) = {x ∈ M(A, µ)| ρ|x| = 0} = {x ∈ M(A, µ) | x(t) = 0 µ− a.e.}.
We get the quotient space Lρ(E ,A) = Lρ(E ,A)/Nρ(E ,A) (the equivalencerelation is given via x ∼ y ⇔ x(t) = y(t) µ - a.e.) normed with the normgiven via x 7→ ∥x∥ = ρ|x| for any x ∈ x.
Definition 1. The vector space Lρ(E ,A), normed with the norm fromabove, is called Kothe space of vector fields.
Particular cases. 1. Assume we are in the unicity case C(E). We shall
write Lρ(E , E)def= Lρ(E, µ). It is seen that Lρ(E, µ) = {x ∈ ME(µ) | ρ|x| <
∞} seminormed with the seminorm x 7→ ρ|x|. We shall write Lρ(E , E)def=
Lρ(E, µ) normed with x 7→ ∥x∥ = ρ|x| for any x ∈ x.
Of course, Lρ(K,µ) = Lρ and Lρ(K,µ) = Lρ.
2. Assume we have, for any t ∈ T , a measure space (St,Σt,µt) anda function norm ρt on (St,Σt,µt). We shall consider that, for any t ∈ T ,Et = Lρt .(hence all ρt R − F , because all Et must be Banach spaces). Inthis case we shall write:
Lρ(E ,A)def= Lρ((ρt)t,A) and Lρ(E ,A)
def= Lρ((ρt)t,A).
Concerning this particular case, we add the fact that, the most ”normal”situation is that one when all the measure spaces are equal, i.e. (St,Σt, µt) =(S,Σ, µ) for any t ∈ T . In this case, the variability is furnished by thefunction norms ρt, t ∈ T .
Our aim in this paper is to study some properties of the spaces Lρ(E ,A),(e.g. completeness), thus initiating a theory which extends the theory ofthe spaces Lρ, in particular the theory of the Lebesgue spaces Lp and of theOrlicz spaces.
3.2.
Lemma 2. Assume ρ R−F . Let (xn)n be a sequence in Lρ(E ,A) suchthat
∑∞n=0 ρ|xn| < ∞. Then:
a) The series∑∞
n=0 xn converges µ - a.e. , i.e. there exists A ∈ T withµ(A) = 0 such that, for any t ∈ T \ A, the series
∑∞n=0 xn(t) converges in
Et.
7 KOTHE SPACES OF VECTOR FIELDS 309
b) Define x ∈ C(E) via x(t) =
{∑∞n=0 xn(t), if t ∈ T \A
arbitrary in Et, if t ∈ A.. Then
x ∈ Lρ(E ,A) and, for any natural n
ρ|x−n∑
k=0
xk| ≤∞∑
k=n+1
ρ|xk| −→n
0.
Proof. a) Let us define f : T → R+ via f =∑∞
n=0 |xn|. Then f ∈M+(µ) and ρ(f) ≤
∑∞n=0 ρ|xn| < ∞ hence f(t) < ∞ µ - a.e. and the series∑∞
n=0 xn(t) converges (absolutely) µ - a.e. (i.e. on T \ A, for some A ∈ Twith µ(A) = 0).
b) Defining x as in the enunciation, one can see that x ∈ M(A, µ) and,for any t ∈ T \A:
∥x(t)∥ = limn
∥n∑
k=0
xk(t)∥ ≤ limn
n∑k=0
∥xk(t)∥ =
∞∑k=0
∥xk(t)∥ = f(t)
hence ρ|x| ≤ ρ(f) < ∞, i.e. x ∈ Lρ(E ,A). For any t ∈ T \ A, using thesame device:
∥x(t)−n∑
k=0
xk(t)∥ = ∥∞∑
k=n+1
xk(t)∥ ≤∞∑
k=n+1
∥xk(t)∥
hence
ρ|x−n∑
k=0
xk| ≤ ρ(∞∑
k=n+1
|xk|) ≤∞∑
k=n+1
ρ|xk|.
�With the aid of the preceding Lemma, we can prove
Theorem 3 (Completeness of Kothe spaces of vector fields). Assumeρ R− F . Then Lρ(E ,A) is Banach.
Proof. Let (xn)n≥0 be a Cauchy sequence in Lρ(E ,A) and take xn ∈ xnfor any n. We shall find x ∈ Lρ(E ,A) such that xn −→
nx, or, which is the
same, we shall find x ∈ Lρ(E ,A) such that xn −→n
x in Lρ(E ,A).
Let ϵ > 0. There exists n(ϵ) such that, for any n ≥ n(ϵ),m ≥ n(ϵ), onehas
(1) ∥xn − xm∥ = ρ|xn − xm| < ϵ
2.
310 ION CHITESCU and LILIANA SIRETCHI 8
We can find a strictly increasing sequence n1 < n2 < . . . < nk < . . . suchthat, for any k
(2) ρ|xnk+1− xnk
| < 1
2k
as follows. First, we find n1 such that, for any n ≥ n1, one has ρ|xn−xn1 | <12 . Then, we find n2 > n1 such that, for any n ≥ n2 ρ|xn − xn2 | < 1
22and
(of course) ρ|xn2 −xn1 | < 12 . Continuing, we find n3 > n2 such that, for any
n ≥ n3 ρ|xn − xn3 | < 123
and (of course) ρ|xn3 − xn2 | < 122.
The procedure continues and we find inductively (nk)k satisfying (2).Write, for all k, yk = xnk+1
− xnkand we have (see (2))
ρ|yk| <1
2k⇒
∞∑k=1
ρ|yk| < ∞.
Lemma 2 says that the series∑∞
k=1 yk(t) converges µ - a.e. and thefunction (defined µ - a.e.) y =
∑∞k=1 yk belongs to Lρ(E ,A). For any k ≥ 2,
one has
(3) y1 + y2 + . . .+ yk−1 = xnk− xn1 .
Let us define x = y + xn1 hence, we have pointwise µ - a.e. (see (3))
x = limk(y1 + y2 + . . .+ yk−1 + xn1) = lim
kxnk
.
It is seen that, for any k ≥ 2, one has
(4) ρ|xn − xnk| < 1
2k−1.
Indeed, using (3) and Lemma 2:
ρ|x− xnk| = ρ|(y + xn1)− (y1 + y2 + . . .+ yk−1 + xn1)|
= ρ|y − (y1 + y2 + . . .+ yk−1)| ≤∞∑p=k
ρ|yp| =1
2k−1.
We can find k such that:
(5) nk ≥ n(ϵ) and1
2k−1<
ϵ
2.
9 KOTHE SPACES OF VECTOR FIELDS 311
Finally, let n ≥ n(ϵ). One has, for this k:
(6) ρ|x− xn| ≤ ρ|x− xnk|+ ρ|xnk
− xn|.
But, using (5) we have
ρ|x− xnk| < 1
2k−1<
ϵ
2(with (4)), ρ|xnk
− xn| <ϵ
2(with (1))
and (6) gives: ρ|x− xn| < ϵ which shows that xn −→n
x, in Lρ(E ,A). �
3.3. We have introduced the space Lρ(E ,A) and the correspondingconvergence of sequences in this space. In this section we shall introduceanother type of convergence (on all of M(A, µ)) which is weaker than theconvergence of Lρ(E ,A) on this space.
Notice first that in case A,A1 are in T such that A1⊂A and µ(A \ A1)= 0, it follows that ρ(A) = ρ(A1) (because φA = φA1 +φA\A1
hence ρ(A) ≤ρ(A1) + ρ(A \ A1) = ρ(A)). Using φA∪B ≤ φA + φB, we get ρ(A ∪ B) ≤ρ(A) + ρ(B) for any A, B in T .
Definition 4. Let (fn)n be a sequence in M(A, µ).1. For f ∈ M(A, µ), we say that (fn)n converges to f in ρ - measure
(and write fnρ
−→n
f), if for any a > 0, one has
limn
ρ(|fn − f | > a) = 0, where ρ(|fn − f | > a)def= ρ(Aa
n)
and Aan = {t ∈ T | ∥fn(t)− f(t)∥ > a} ∈ T .
2. We say that (fn)n is Cauchy in ρ - measure if, for any a > 0, onehas: for any ϵ > 0, there exists a natural n(ϵ) such that, for any m ≥ n(ϵ),n ≥ n(ϵ),
ρ(|fm − fn| > a) < ϵ, where ρ(|fm − fn| > a)def= ρ(Aa
m,n)
and Aam,n = {t ∈ T | ∥fm(t)− fn(t)∥ > a} ∈ T .
Remark. We have Aan ∈ T and Aa
m,n ∈ T , because f ∈ M(A, µ) andfn ∈ M(A, µ).
Proposition 5. Assume (fn)n is convergent in ρ - measure (i.e. there
exists f ∈ M(A, µ), such that fnρ
−→n
f). Then (fn)n is Cauchy in ρ -measure.
312 ION CHITESCU and LILIANA SIRETCHI 10
Proof. For any ϵ and any a > 0, if m,n are large enough
Aam,n ⊂ A
a2m ∪A
a2n ⇒ ρ(Aa
m,n) ≤ ρ(Aa2m) + ρ(A
a2n ) <
ϵ
2+
ϵ
2< ϵ.
�Proposition 6. 1. If fn
ρ
−→n
f and f = g µ - a.e., then fnρ
−→n
g.
2. If fnρ
−→n
f and fnρ
−→n
g, then f = g µ - a.e..
Proof. 1. Notice first that g = f µ - a.e. and f ∈ M(A, µ), impliesg ∈ M(A, µ) and the enunciation is meaningful. Let us denote M = {t ∈T | f(t) = g(t)} hence M ∈ T and µ(M) = 0. It follows that, for anynatural n and any a > 0
(7) ρ(|fn − f | > a) = ρ(|fn − g| > a).
Equality (7) is obtained as follows: if we write:
Aan = {t ∈ T | ∥fn(t)− f(t)∥ > a} and Ba
n = {t ∈ T | ∥fn(t)− g(t)∥ > a}
we have Aan \M = Ba
n \M and, of course Aan \ (Aa
n \M) = Aan∩M and Ba
n \(Ba
n \M) = Ban∩M , which implies (see the Remark at the beginning of the
section) ρ(Aan) = ρ(Aa
n \M) = ρ(Ban \M) = ρ(Ba
n).
2. Fix a > 0 and write Aa = {t ∈ T | ∥f(t)− g(t)∥ > a}. We get
Aa ⊂ Aa2n ∪B
a2n ⇒ ρ(Aa) ≤ ρ(A
a2n ) + ρ(B
a2n ) −→
n0.
It follows that ρ(Aa) = 0, hence µ(Aa) = 0. The number a beingarbitrary, we obtain ρ(A) = 0 ⇒ µ(A) = 0, where
A =
∞∩n=1
A 1n= {t ∈ T | f(t) = g(t)}.
�The following theorem shows that convergence in Lρ(E ,A) is stronger
than convergence in ρ - measure.
Theorem 7. If fn −→n
f in Lρ(E ,A), then fnρ
−→n
f .
11 KOTHE SPACES OF VECTOR FIELDS 313
Proof. Take an arbitrary a > 0. Then, with previously introducednotations |fn − f | = |fn − f |φAa
n+ |fn − f |φT\Aa
n≥ |fn − f |φAa
nwhich
implies ρ|fn − f | ≥ ρ(|fn − f |φAan) ≥ ρ(aφAa
n) = aρ(Aa
n). Hence
ρ(Aan) ≤
1
aρ|fn − f | −→
n0
and a is arbitrary, which shows that fnρ
−→n
f . �Using Theorem 7 and Proposition 6, we get
Corollary 8. Assume fn −→n
f and fn −→n
g in Lρ(E ,A). Then f = g µ- a.e.
3.4. In this section we generalize the almost uniform (asymptotic) con-vergence and introduce relationships between all types of convergence.
Definition 9. Let (fn)n be a sequence in M(A, µ).1. For f ∈ M(A, µ), we say that (fn)n converges ρ - almost uniformly
(ρ - asymptotically) to f (and write fnρu
−→n
f) if, for any ϵ > 0, there exists
Aϵ ∈ T such that ρ(Aϵ) < ϵ and (fn)n converges uniformly to f on T \Aϵ.2. We say that (fn)n is ρ - almost uniformly (ρ-asymptotically) Cauchy
if, for any ϵ > 0, there exists Aϵ ∈ T such that ρ(Aϵ) < ϵ and (fn)n isuniformly Cauchy on T \Aϵ (this means that, for any δ > 0 , there exists anatural n(δ) having the property that, for m ≥ n(δ) and n ≥ n(δ) one has∥fm(t)− fn(t)∥ < δ, for any t ∈ T \Aϵ).
Proposition 10. Assume fnρu
−→n
f . Then (fn)n is almost uniformly
Cauchy.
Proof. Let ϵ > 0 and take Aϵ ∈ T like in the definition of the fact that
fnρu
−→n
f . Pick δ > 0 and find a natural n(δ) such that, for any n ≥ n(δ) and
any t ∈ T \ Aϵ, one has ∥fn(t) − f(t)∥ < δ2 . It follows that, for n ≥ n(δ),
m ≥ n(δ) and t ∈ T \Aϵ, one has ∥fn(t)− fm(t)∥ < δ. �Almost uniform convergence is stronger than almost everywhere conver-
gence and convergence in measure, as the following theorem shows.
Theorem 11. Let (fn)n be a sequence in M(A, µ), and f ∈ M(A, µ).
314 ION CHITESCU and LILIANA SIRETCHI 12
1. If fnρu
−→n
f ,then fn −→n
f µ - a.e.
2. If fnρu
−→n
f , then fnρ
−→n
f .
Proof. 1. For any natural n > 0, pick An ∈ T with µ(An) <1n , such
that (fm)m converges to f uniformly on T \An. Writing A =∩∞
n=1An, wehave ρ(A) = 0, hence µ(A) = 0. For t ∈ T \ A =
∪∞n=1(T \ An) we find n
such that t ∈ T \An. It follows that fm(t) −→m
f(t) a.s.o.
2. Let a > 0 and ϵ > 0. One can find Aϵ ∈ T with ρ(Aϵ) < ϵ, such thatthere exists n(a) with the property that for any n ≥ n(a) and any t ∈ T \Aϵ
one has ∥fn(t)− f(t)∥ ≤ a. Hence, we have the inclusion
{t ∈ T | ∥fn(t)− f(t)∥ > a} ⊂ Aϵ ⇒ ρ(|fn − f | > a) ≤ ρ(Aϵ) < ϵ.
�Using the preceding facts, we obtain (see also the similar Proposition 6)
Proposition 12. 1. If fnρu
−→n
f and f = g µ - a.e., then fnρu
−→n
g.
2. If fnρu
−→n
f and fnρu
−→n
g, then f = g µ - a.e..
Proof. 1. Write M = {t ∈ T | f(t) = g(t)} hence µ(M) = 0. Take ϵ > 0arbitrarily and find Aϵ ∈ T with ρ(Aϵ) < ϵ, such that (fn)n converges to funiformly on T \Aϵ, hence (fn)n converges to f uniformly on T \Bϵ, whereBϵ = Aϵ ∪M . But ρ(Bϵ) = ρ(Aϵ) < ϵ and, for t ∈ T \ Bϵ, f(t) = g(t). So,(fn)n converges to g uniformly on T \Bϵ, a.s.o.
2. Assume fnρu
−→n
f and fnρu
−→n
g . Using Theorem 11.1., one has fn −→n
f µ
- a.e. and fn −→n
g µ - a.e. and this implies f = g µ - a.e. �The next result shows that M(A, µ), is ”complete” with respect to ρ -
almost uniform convergence.
Theorem 13. Let (fn)n be a sequence in M(A, µ). The followingassertions are equivalent:
1. (fn)n is ρ - almost uniformly Cauchy.
2. (fn)n is ρ - almost uniformly convergent.
(i.e. there exists f ∈ M(A, µ) such that fnρu
−→n
f) .
13 KOTHE SPACES OF VECTOR FIELDS 315
Proof. The implication 2.⇒1. has been proved in Proposition 10. Theimplication 1.⇒2. will be proved in two steps.
Step 1. We shall find f ∈ M(A, µ) such that fn −→n
f µ - a.e.. The proof
is similar to the proof of Theorem 11.
For any natural k > 0, we find Ak ∈ T with ρ(Ak) <1k , such that, for
any δ > 0, there exists n(δ) having the property that, if m ≥ n(δ),n ≥ n(δ)and t ∈ T \ Ak, one has
(8) ∥fm(t)− fn(t)∥ < δ.
Writing A =∩∞
k=1Ak, we have µ(A) = 0. For any t ∈ T \A =∪∞
k=1(T \Ak), we find k such that t ∈ T \ Ak. It follows that (8) is true, hencethe sequence (fn(t))n is Cauchy in Et, hence convergent in Et. Writingf(t) = limn fn(t) we have defined f(t) for any t ∈ T \A. Defining f(t) = 0for t ∈ A, we got f : T →
∪t∈T Et, with f(t) ∈ Et for any t ∈ T , hence we
defined f ∈ C(E) and one can see that f ∈ M(A, µ) and fn −→n
f µ - a.e.
Step 2. We show that fnρu
−→n
f .
Let ϵ > 0. There exists Bϵ ∈ T with ρ(Bϵ) < ϵ and such that, for anyδ > 0, one can find n(δ) with the property that, if m ≥ n(δ), n ≥ n(δ) andt ∈ T \Bϵ,
(8′) ∥fm(t)− fn(t)∥ < δ.
Write Aϵ = A ∪ Bϵ, hence ρ(Aϵ) = ρ(Bϵ) < ϵ. We have seen that, forany t ∈ T \ Aϵ, one has fn(t) −→
nf(t). At the same time, for a given δ > 0,
we can find n(δ) as previously and, for m ≥ n(δ), n ≥ n(δ) and t ∈ T \ Aϵ,we have again (8′).
Keeping m ≥ n(δ) fixed in (8′) and letting n tend to ∞, we get from(8′):
∥fm(t)− f(t)∥ ≤ δ
and this shows that (fn)n converges to f uniformly on T \Aϵ. �
Corollary 14. Let (fn)n be a sequence in M(A, µ) and f ∈ M(A, µ)such that (fn)n is ρ - almost uniformly Cauchy and fn −→
nf µ - a.e. Then
fnρu
−→n
f .
316 ION CHITESCU and LILIANA SIRETCHI 14
Proof. Using Theorem 13, we find g ∈ M(A, µ) such that fnρu
−→n
g. So
fn −→n
g µ - a.e (Proposition 1.). At the same time fn −→n
f µ - a.e. It follows
that f = g µ - a.e. .
We have fnρu
−→n
g and g = f µ - a.e.. We apply Proposition 12. and
obtain that fnρu
−→n
f . �Before proceeding further, we make the following
Remark. Assume ρ R−F . Let (An)n≥1 be a sequence of sets An ∈ T .Then, one has
ρ(
∞∪n=1
An) ≤∞∑n=1
ρ(An).
Indeed: φA =∑∞
n=1 φBn (pointwise), where:
A =
∞∪n=1
An; B1 = A1; B2 = A2 \A1; B3 = A3 \ (A1 ∪A2); . . .
hence
ρ(A) ≤∞∑n=1
ρ(Bn) ≤∞∑n=1
ρ(An).
Theorem 15. Assume ρ R − F . Let (fn)n be a sequence in M(A, µ)which is Cauchy in ρ - measure. Then, one can find a subsequence (fnk
)k ⊂(fn)n and a function f ∈ M(A, µ) such that fnk
ρu
−→k
f (consequently fnk−→k
f µ - a.e. and fnk
ρ
−→k
f .)
Proof. The remarks in the brackets follow from Theorem 11.A. We shall show that there exists a sequence (fnk
)k ⊂ (fn)n which is ρ- almost uniformly Cauchy. This subsequence is constructed as follows.
Take k = 1. There exists n1 ∈ N∗ with the property that, for anym,n ≥ n1, one has ρ(|fm−fn| > 1
21) < 1
21. Take k = 2. There exists n2 > n1
such that, for any m,n ≥ n2, one has ρ(|fm − fn| > 122) < 1
22. Continuing
in the same manner, we obtain the sequence (nk)k, n1 < n2 < . . . < nk <nk+1 < . . . such that, for any m,n ≥ nk one has ρ(|fm − fn| > 1
2k) < 1
2k.
Because nk+1 > nk, one has for any k:
ρ(|fnk+1− fnk
| > 1
2k) <
1
2k.
15 KOTHE SPACES OF VECTOR FIELDS 317
Our further goal is to show that the subsequence (fnk)k ⊂ (fn)n fulfills the
conditions in the enunciation (i.e. is ρ - almost uniformly Cauchy). For anyk ∈ N∗, let
(9) Ek = {t ∈ T | ∥fnk+1(t)− fnk
(t)∥ >1
2k}
hence ρ(Ek) <12k.
Let ϵ > 0. There exists n(ϵ) ∈ N such that
∞∑k=n(ϵ)
1
2k< ϵ.
Put Aϵ =∪∞
k=n(ϵ)Ek. It follows that (see the preceding Remark)
ρ(Aϵ) ≤∞∑
k=n(ϵ)
ρ(Ek) <∞∑
k=n(ϵ)
1
2k< ϵ.
We close part A showing that, for any δ > 0, there exists m(δ) ∈ N suchthat, for any i, j ≥ m(δ) and any t ∈ T \ Aϵ one has ∥fni(t) − fnj (t)∥ < δ.Indeed, let δ > 0. We can find j0 ∈ N∗ such that 1
2j0−1 < δ. Take m(δ) =max(n(ϵ), j0). Let i > j ≥ m(δ) and let t ∈ T \Aϵ.
We shall show that ∥fni(t)− fnj (t)∥ < δ.Indeed: because t ∈ T \ (
∪∞k=n(ϵ)Ek) =
∩∞k=n(ϵ)(T \Ek), it follows that,
for any k ≥ n(ϵ), one has t /∈ Ek, which implies (see (9)):
∥fnk+1(t)− fnk
(t)∥ ≤ 1
2k.
So, we have i > j ≥ n(ϵ), i > j ≥ j0:
∥fnj (t)− fni(t)∥ ≤ ∥fnj (t)− fnj+1(t)∥+ ∥fnj+1(t)− fnj+2(t)∥
+ . . .+ ∥fni−1(t)− fni(t)∥ ≤ 1
2j+
1
2j+1+ . . .+
1
2i−1<
∞∑k=jo
1
2k=
1
2j0−1< δ.
B. Because (fnk)k is ρ - almost uniformly Cauchy, we apply Theorem
13 and find f ∈ M(A, µ) such that fnk
ρu
−→k
f . �Theorem 15 is very important. Here are some of its consequences. We
lay stress upon the fact that ρ R − F . The first consequence is Theorem
318 ION CHITESCU and LILIANA SIRETCHI 16
16, stating that M(A, µ), equipped with the convergence in ρ - measure,is ”complete” (analogue of Slutsky’s theorem). The proof follows (directly)from Theorem 15 and Proposition 5.
Theorem 16. Assume ρ R − F . Let (fn)n be a sequence in M(A, µ).The following assertions are equivalent:
1. The sequence (fn)n is convergent in ρ - measure (i.e. there exists
f ∈ M(A, µ) such that fnρ
−→n
f).
2. The sequence (fn)n is Cauchy in ρ - measure.
Proof. 1. ⇒ 2. Follows from Proposition 5.2. ⇒ 1. Theorem 15 implies the existence of a measurable vector field
f ∈ M(A, µ) and of a subsequence (fnk)k ⊂ (fn)n, such that fnk
ρ
−→k
f .
Using a standard device, we shall prove that fnρ
−→n
f .
To this end, let ϵ > 0 and a > 0. There exists n(ϵ) such that, for anym ≥ n(ϵ),n ≥ n(ϵ), one has
(10) ρ(|fm − fn| >a
2) <
ϵ
2.
There exists k(ϵ) such that nk(ϵ) ≥ n(ϵ) and for any k ≥ k(ϵ)
(11) ρ(|f − fnk| > a
2) <
ϵ
2.
Take n ≥ n(ϵ). Then we have, using (10) and (11) and previous and devices
ρ(|f − fn| > a) ≤ ρ(|f − fnk(ϵ)| > a
2) + ρ(|fnk(ϵ) − fn| >
a
2) < ϵ.
�The second consequence of Theorem 15 is Theorem 17, which relates
convergence in the seminormed Kothe space L(E ,A) to the other types ofconvergence.
Theorem 17. Assume ρ R − F . Let (fn)n be a sequence in Lρ(E ,A)and f ∈ Lρ(E ,A) such that fn −→
nf in Lρ(E ,A).
Then, there exists a subsequence (fnk)k ⊂ (fn)n such that fnk
ρu
−→k
f
(hence fnk−→k
f µ - a.e. and fnk
ρ
−→k
f).
17 KOTHE SPACES OF VECTOR FIELDS 319
Proof. We have seen (Theorem 11) that the remarks in the brackets are
true. Using Theorem 7, we see that fnρ
−→n
f . It follows (Proposition 5) that
(fn)n is Cauchy in ρ - measure. Using Theorem 15 we find F ∈ M(A, µ) and
a subsequence (fnk)k ⊂ (fn)n such that fnk
ρu
−→k
F . Consequently, fnk
ρ
−→k
F
(again Theorem 11). We have also fnk
ρ
−→k
f (Theorem 7). Then f = F µ -
a.e. (Proposition 6).
Finally, we have fnk
ρu
−→k
F and F = f µ - a.e. and we use Proposition
12 to conclude that fnk
ρu−→k
f . �
3.5. This last section or our paper is dedicated to some examples.
Example 18 (Trivial vector fields). We consider a non empty set T andthe locally compact (separated) space will be T equipped with the discretetopology. The σ - algebra T will be P(T ) = the set of all subsets of T andthe measure µ : P(T ) → R+ will be the counting measure (µ(A) = thenumber of elements in A, in case A is finite and µ(A) = ∞, in case A isinfinite). Because the only negligible set is ∅, it is clear that f(t) = g(t) µ -a.e. means f = g.
Let also E = (Et)t∈T be a family of Banach spaces. We shall take asfundamental family of continuous vector fields A = C(E). This is possible,because any function defined on T is continuous and, for any t ∈ T , wehave the equality {x(t)| x ∈ C(E)} = Et. Any x ∈ C(E) is continuous(with respect to A = C(E) - take y = x in the definition of continuity),hence we have the equalities C(E) = CA(E) = M(A, µ). Because M+(µ) ={u : T → R+}, we have for any function norm ρ : M+(µ) → R+ theequality Lρ(E ,A) = {x ∈ C(E) | ρ|x| < ∞} and also, identifying classes
with functions Lρ(E ,A) ≡ Lρ(E ,A)def= Lρ normed with the norm x 7→
∥x∥ = ρ|x|.We shall work in the concrete case T = N∗. To this end, recall first the
definition of the Banach spaces of sequences lp, for 1 ≤ p ≤ ∞:
- for 1 ≤ p < ∞:
lp = {x = (xn)n≥1 | xn ∈ K,
∞∑n=1
|xn|p < ∞}
320 ION CHITESCU and LILIANA SIRETCHI 18
equipped with the norm
∥x∥p = (
∞∑n=1
|xn|p)1p ;
- for p = ∞:
l∞ = {x = (xn)n≥1 | xn ∈ K, (xn)n is bounded }
equipped with norm∥x∥∞ = sup
n|xn|.
So, let us take T = N∗ and let, for any n ∈ N∗,En = ln. For any functionnorm ρ : M+(µ) → R+, an element x ∈ Lρ is identified with an infinitematrix (xmn)m≥1,n≥1 such that, for any n ∈ N∗,
x(n) = (xmn)m≥1 ∈ ln, |x(n)| = ∥x(n)∥n = (
∞∑m=1
|xmn|n)1n
and |x| = (|x(n)|)n≥1, ∥x∥ = ρ|x|. In case ρ = ∥ ∥p, for 1 ≤ p ≤ ∞, we shallhave:
- for 1 ≤ p < ∞:
∥x∥ = (
∞∑n=1
|x(n)|p)1p = (
∞∑n=1
(
∞∑m=1
|xmn|n)pn )
1p
- for p = ∞:
∥x∥ = supn
|x(n)| = supn(
∞∑m=1
|xmn|n)1n .
We shall exhibit an element x ∈ Lρ for any ρ = ∥ ∥p, 1 ≤ p ≤ ∞. Takeα > 1, β > 1 and let x ≡ (xmn)m≥1,n≥1 where xmn = 1
mαnβ . First, it is seenthat x ∈ Lρ, for ρ = ∥ ∥1. Namely, in this case
∥x∥ =
∞∑n=1
(
∞∑m=1
(1
mαnβ)n)
1n =
∞∑n=1
1
nβ(
∞∑m=1
1
mαn)1n =
∞∑n=1
1
nβA
1nn ,
where An =∑∞
m=11
mαn . It is seen that the sequence (An)n is decreasing,
with 1 < An ≤ A1 < ∞. Hence 1 < A1nn ≤ A
1n1 ≤ M , where M is a constant
(depending upon α), hence ∥x∥ < ∞.
19 KOTHE SPACES OF VECTOR FIELDS 321
We proved the fact that x ∈ Lρ, for ρ = ∥ ∥1, which means that(|x(n)|)n≤1 ∈ l1. But l1 ⊂ lp, for any 1 ≤ p ≤ ∞, hence (|x(n)|)n≥1 ∈ lp,1 ≤ p ≤ ∞, which proves the fact that x ∈ Lρ for ρ = ∥ ∥p, 1 ≤ p ≤ ∞.
Let us notice also that, in case ρ = ∥ ∥p, 1 ≤ p ≤ ∞, one has ρ F , henceρ R−F (see [1]). So, in this case, Lρ is Banach (Theorem 3) and convergencein Lρ implies pointwise convergence for a subsequence (Theorem 17).
Example 19. The (locally) compact separated space will be T = [0, 1],with natural topology. The σ - algebra T will be the set of Lebesguemeasurable sets of T and µ : T → R+ will be the Lebesgue measure. Weshall consider the family of Banach spaces E = (Et)t∈T , where, for any
t ∈ T = [0, 1], we take Et = L1/t (with E0 = L∞), writing Lp def= Lp(µ), for
1 ≤ p < ∞.So, for any t ∈ T = [0, 1], we have the measure space (St,Σt, µt),where
St = T,Σt = T , µt = µ. We also have, for any t ∈ T , the function normρt = ∥ ∥1/t, giving the Kothe space Et = Lρt(= L1/t). From now on,
1/0def= ∞.In order to complete our schema, we shall construct a fundamental fam-
ily A of continuous vector fields for C(E). To this end, write
E(T ) = {f : T → K | f is T - simple } and E(T ) = {f | f ∈ E(T )}
(here f = {g : T → K | g(t) = f(t) µ - a.e. }).One knows that E(T ) is dense in all Lp, 1 ≤ p ≤ ∞. This makes possible
the construction, for any f ∈ E(T ), of the element x(f) ∈ C(E) acting viax(f)(t) = f ∈ L1/t, for any t ∈ T .
Write A = {x(f) | f ∈ E(T )} and let us prove that A is a fundamentalfamily of continuous vector fields for C(E). Indeed:
a) A is a vector subspace of C(E) (we have x(f) + x(g) = x(f + g) andαx(f) = x(αf) with obvious notations).
b) Axiom (A1) is fulfilled: for any f ∈ E(T ), the function φ : [0, 1] →R+, φ(t) = |x(f)|(t) = ∥f∥1/t (with 1/0
def= ∞) is continuous. This is
mathematical folklore. (see, e.g., [7], pp. 2).c) Axiom (A2) is fulfilled: for any t ∈ T , the set {x(t) | x ∈ A} is dense
in Et = L1t .
Indeed, the last set is precisely {x(f)(t) | f ∈ E(T )} = E(T ).We completed our schema, so it is possible to speak about the Kothe
space of vector fields Lρ((ρt)t,A).
322 ION CHITESCU and LILIANA SIRETCHI 20
In the sequel, we shall present an example of measurable vector field withrespect to A. To this end, we shall consider a bounded Lebesgue measurable
function f : [0, 1] → K. For any t ∈ [0, 1], let us write x(t) = fφ[0,t] ∈ L1/t.We defined x ∈ C(E) and we shall prove that x is continuous with respect
to A, at any t0 ∈ (0, 1]. Before proceeding to the proof, let us remark thatfor any bounded measurable function g : [0, 1] → K and for any 0 < t ≤ 1,one has, if A ∈ T :
(12) ∥gφA∥1/t ≤ ∥g∥∞µ(A)t.
To prove the continuity of x at t0 ∈ T \ {0} means to prove the followingfact: for any ϵ > 0, there exists a (basic) neighbourhood V of t0 in [0, 1]
and an element h1 ∈ E(T ) such that, for any t ∈ V one has
(13) d(t) = ∥x(t)− x(h1)(t)∥1/t = ∥fφ[0,t] − h1∥1/t < ϵ.
Let 1 > ϵ > 0. We shall find h1 ∈ E(T ) and V ∈ V(t0) which will make(13) to be true. Notice first the existence of h ∈ E(T ) such that
(14) ∥h− f∥∞ <ϵ
2.
We shall see that one can take
h1 = hφ[0,t0] ∈ E(T ).(15a)
It remains to find V . Namely, we shall take V of the form
V = (t0 − δ, t0 + δ) ∩ [0, 1],(15b)
where t0 − δ > 0 (so 0 /∈ V ) and δ must satisfy some supplementary condi-tions which will be exhibited in the sequel. Because limδ→0 δ
t0−δ = 0 thereexists δ1 > 0 such that, if 0 < δ < δ1 , one has
(16) (∥f∥∞)δt0−δ <ϵ
2.
Because limδ→0 δt0 = 0 there exists δ2 > 0 such that, for any 0 < δ < δ2,
one has
(17) ∥f∥∞ · δt0 <ϵ
2.
21 KOTHE SPACES OF VECTOR FIELDS 323
The neighbouhood V will be constructed using δ > 0 such that (supple-mentary condition) δ < min(δ1, δ2) and this implies that (16) and (17) aretrue.
We finish the proof of the continuity at t0 by showing that h1 chosenaccording to (15a) and V chosen according to (15b) satisfy (13). So lett ∈ V .
In case 0 < t ≤ t0, we can write
d(t) = ∥fφ[0,t] − hφ[0,t] − hφ[t,t0] ∥1/t ≤ ∥(f − h)φ[0,t] ∥1/t + ∥hφ[t,t0] ∥1/t.
From (14) we get ∥h∥∞ < ∥f∥∞ + 1 and, using (12) and (14):
d(t) ≤ ϵ
2(µ([0, t]))t + (∥f∥∞ + 1)(t− t0)
t <ϵ
2tt + (∥f∥∞ + 1)δt
<ϵ
2+ (∥f∥∞ + 1)δt0−δ < ϵ
(according to (16)).In case t0 ≤ t ≤ 1, we can write
d(t) = ∥fφ[0,t0] + fφ[t0,t] − hφ[0,t0]∥1/t ≤ ∥(f − h)φ[0,t0]∥1/t + ∥fφ[t0,t]∥1/t.
Again (12) and (14) lead us to:
d(t) ≤ ϵ
2(µ([0, t0]))
t + ∥f∥∞(t− t0)t =
ϵ
2tt0 + ∥f∥∞(t− t0)
t
≤ ϵ
2+ ∥f∥∞(t− t0)
t0 ≤ ϵ
2+ ∥f∥∞δt0 < ϵ
(according to (17)).In all cases (13) is proved. So x is continuous on T \ {0} = (0, 1] with
respect to A. The reader might ask what happens for t0 = 0. In thiscase it is possible for x to be discontinuous at t0. For instance, let us takef ≡ 1. We shall see that is this case the vector field x is discontinuous at 0(with respect to A). Indeed, the continuity of x at t0 = 0 would imply thecontinuity of |x| at t0 = 0.
We have |x|(0) = 0. On the other hand, for any 0 < t ≤ 1, one hasx(t) = φ[0,t] hence |x(t)| = ∥φ[0,t]∥1/t = tt and limt→0 |x(t)| = 1 = |x|(0).On can prove (taking y = 0) that, in case limt→0 f(t) = 0, the vector fieldx is continuous at t0 = 0. So, generally speaking, x is continuous on (0, 1]and (possibly) discontinuous at t = 0. It follows immediately that x is(A, µ) - measurable. If ρ is a function norm on (T, T , µ) it is possible to
324 ION CHITESCU and LILIANA SIRETCHI 22
compute ρ|x|. Let us take, e.g., f ≡ 1 and ρ = ∥ ∥p ,1 ≤ p ≤ ∞. We shallsee that ρ|x| < ∞ for all such ρ, consequently we have x ∈ Lp((ρt)t,A)(and x ∈ Lρ((ρt)t,A)). As we have seen, x(t) = φ[0,t] for any t ∈ [0, 1].Consequently: |x(t)| = tt, if t > 0 and |x(0)| = 0. For ρ = ∥ ∥p,1 ≤ p < ∞,we have
ρ|x| = (
∫|x|pdµ)1/p = (
∫ 1
0+0tptdt)1/p < ∞
(because tpt ≤ 1) For ρ = ∥ ∥∞ , we have ρ|x| = supt∈[0,1] tt = 1 (with the
convention 00 = 1).
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Received: 30.VIII.2011 Faculty of Mathematics and Computer Science,
University of Bucharest,
Bucharest,
ROMANIA