2. DISEÑO CENTRADO EN LAS PERSONAS
2.1. ESCUCHAR
2.1.6. Desarrollar tu paradigma
In this section, we endow the space Lν with a distance eδ and we show that the space
(Lν, eδ) is complete and separable. Let us recall that, given an admissible profile ν, one
has
Lν= Lν ,+∩ Lν ,−.
Since the spaces Lν ,+ and Lν ,− have been endowed with a topology, we directly get a topology on Lν. Let us also recall that α := (αn)n∈N denotes a dense sequence in
[0, αs], α0:= (α0p)p∈Ndenotes a dense sequence in [αs, +∞) and ε := (εm)m∈N denotes
a sequence of (0, +∞) which converges to 0.
Proposition 6.6.1. For every sequences α, α0 and ε chosen as above, e
δα,α0,ε:= eδα,ε+ eδα0,ε
is a distance on Lν. All these distances define the same topology.
Proof. Using Propositions 6.4.15 and 6.5.3, it is clear that eδα,α0,ε is a distance on Lν.
The independence on the choice of the sequences follows from the same propositions. This result allows us to write this distance eδ independently of these sequences α, α0 and ε. We directly get the following result.
Proposition 6.6.2.
1. The topology defined by eδ on Lν is the weakest topology such that the inclusions
Lν → Lν ,+ and Lν→ Lν ,− are continuous.
2. A sequence of Lν is a Cauchy sequence in (Lν, eδ ) if and only if it is a Cauchy
sequence in (Lν ,+, eδ+) and in (Lν ,−, eδ−).
3. A sequence of Lν converges in (Lν, eδ ) if and only if it converges in (Lν ,+, eδ+) and
in (Lν ,−, eδ−).
Thanks to Propositions 6.4.16 and 6.5.12, we have the next result.
Proposition 6.6.3.
1. The topology defined by eδ on Lν is the weakest topology such that, for every m, n, p ∈ N, the inclusions Lν→ eA+(m, n) and Lν → eA−(m, p) are continuous. 2. A sequence of Lν is a Cauchy sequence in (Lν, eδ ) if and only if it is a Cauchy
sequence in ( eA+(m, n), eδ+
m,n) and in ( eA−(m, p), eδm,p− ) for every m, n, p ∈ N.
3. A sequence of Lν converges in (Lν, eδ ) if and only if it converges in ( eA+(m, n), eδ+
m,n)
Proposition 6.6.4. The space (Lν, eδ ) is a complete metric space and thus a Baire
space.
Proof. Let (~c(l))
l∈N be a Cauchy sequence in (Lν, eδ ). Then, it is a Cauchy sequence
in (Lν ,+, eδ+) and from Proposition 6.4.17, it converges to ~c in (Lν ,+, eδ+). Moreover, it
is also a Cauchy sequence in (Lν ,−, eδ−) hence, from Corollary 6.5.13, −−→1
e(m)
m∈N
is a Cauchy sequence in (Sν−, δ). Since the space (Sν−, δ) is complete, there exists ~x ∈ Sν−
such that −→1
e(l)
l∈N
converges to ~x in (Sν−, δ). Since the topology of Sν− is stronger
than the pointwise topology, we know that xλ= lim
l→+∞
1 e(l)λ
, ∀λ ∈ Λ.
Moreover, since the convergence in Lν ,+ is stronger than the uniform convergence, we
have sup λ0⊆λ |cλ0| = lim l→+∞e (l) λ , ∀λ ∈ Λ.
It follows that that
xλ=
1 supλ0⊆λ|cλ0|
, ∀λ ∈ Λ.
In particular, supλ0⊆λ|cλ0| 6= 0 for every λ ∈ Λ and ~c ∈ Lν ,−. Proposition 6.6.2 gives
the conclusion.
Lemma 6.6.5. Assume that αmin > 0. For every m, n, p ∈ N, let C(m, n), C0(m, n),
D(m, p) and D0(m, p) be positive constants. Let us define e
K = eK+∩ eK− where eK+ (resp.
e
K−) is defined as in Lemma 6.4.22 (resp. Lemma 6.5.14). Every sequence of eK which converges pointwise converges also in (Lν, eδ) to an element of eK.
Proof. Since αmin > 0, there exists n ∈ N such that 0 < αn < αmin. By construction,
e
K ⊆ eK+
m,n which is bounded in (Cαn, || · ||Cαn). Using Lemma 6.4.21 (item 1), we
get that if a sequence of eK converges pointwise, it converge also uniformly. The result follows then from Corollary 6.4.24 and Lemma 6.5.14.
The next characterization of the compact sets of (Lν, eδ ) is immediate.
Proposition 6.6.6. Assume that αmin> 0. A subset of Lν is compact in (Lν, eδ ) if and
only if it is closed and included in some eK = eK+∩ eK−.
Proof. Since any compact set of a metric space is closed and bounded, the condition is necessary using Propositions 6.4.7 and 6.5.6. It suffices to show that eK is compact in (Lν, eδ ). Let us assume (~c(l))
l∈N is a sequence of eK. By Lemma 6.4.22, since eK ⊆ eK+,
we can extract a subsequence which converges pointwise. The conclusion follows from Lemma 6.6.5.
Let us now study the separability of the space (Lν, eδ ). We will see that it holds only
if αmin> 0. Let us recall that, while working with Lνas a function space, this condition
Lemma 6.6.7. Assume that αmin > 0 and let ~z ∈ Lν. For every ~c ∈ Lν and every N ∈ N, we set c(N )j,k := cj,k if j < N, ej,k if j = N, zj,k if j > N.
Then the sequence (~c(N ))
N ∈N converges to ~c in (Lν, eδ ).
Proof. It is clear that ~c(N ) ∈ Lν
for every N ∈ N since c(N )j,k = zj,k if j > N . Since
~c, ~z ∈ Lν ,+, there are C(m, n), C0(m, n) > 0 such that ~c, and ~z belong to
~ x ∈ Lν: #{λ ∈ Λj: sup λ0⊆λ |xλ0| > C(m, n)2−αnj} ≤ C0(m, n)2(ν(αn)+εm)j, ∀j ∈ N0
for every m, n ∈ N. Let us remark that if j ≤ N , then for every λ ∈ Λj, we have either
e(N )λ = eλ or e
(N )
λ = supλ0⊆λ
0|zλ0| ≤ supλ0⊆λ|zλ0| where λ0 ⊆ λ. It follows that for
m, n ∈ N and j ≤ N , #{λ ∈ Λj : e (N ) λ > C(m, n)2 −αnj} ≤ #{λ ∈ Λj : eλ> C(m, n)2−αnj} + #{λ ∈ Λj : sup λ0⊆λ |zλ0| > C(m, n)2−αnj} ≤ 2C0(m, n)2(ν(αn)+εm)j.
Since c(N )j,k = zj,k if j > N , we get that ~c(N ) ∈ eK+ where eK+ is the intersection over
m, n ∈ N of the sets ~x ∈ Lν: #{λ ∈ Λj: sup λ0⊆λ |xλ0| > C(m, n)2−αnj} ≤ 2C0(m, n)2(ν(αn)+εm)j, ∀j ∈ N0 .
For the decreasing part, since ~c, ~z ∈ Lν ,−, there are D(m, p), D0(m, p) > 0 such that
~c and ~z belong to the set eK− given by the intersection over m, p ∈ N of the sets ~ x ∈ Lν : #{λ ∈ Λj : sup λ0⊆λ |xλ0| < D(m, p)2−α 0 pj} ≤ D0(m, p)2(ν(α 0 p)+εm)j, ∀j ∈ N 0 .
Let us show that ~c(N ) ∈ eK−
for every N . Let us fix m, p ∈ N and N ∈ N. If j ≤ N , then e(N )λ ≥ eλ for every λ ∈ Λj so that
#λ ∈ Λj: e (N ) λ < D(m, p)2 −α0 pj ≤ D0(m, p)2(ν(α 0 p)+εm)j.
Moreover, if j > N , then c(N )j,k = zj,kand if follows that ~c(N )∈ eK−.
Since (~c(N ))
N ∈N converges pointwise to ~c, Lemma 6.6.5 gives that (~c(N ))N ∈N con-
verges to ~c in (Lν, eδ).
Lemma 6.6.8. Assume that αmin > 0. Let B be a subset of Lν and ~z ∈ Lν. If B is
pointwise bounded and if there exists N ∈ N such that cλ = zλ for every λ ∈ Λj with
Proof. Since ~z ∈ Lν, there are C(m, n), C0(m, n) > 0 such that # λ ∈ Λj: sup λ0⊆λ |zλ0| > C(m, n)2−αnj ≤ C0(m, n)2(ν(αn)+εm)j, ∀j ∈ N 0.
There are also D(m, p), D0(m, p) > 0 such that # λ ∈ Λj : sup λ0⊆λ |zλ0| < D(m, p)2−α 0 pj ≤ D0(m, p)2(ν(α0p)+εm)j, ∀j ∈ N 0.
If j ≤ N and ~c ∈ B, one has
2αnj sup λ0⊆λ |cλ0| = 2αnjmax ( sup λ0⊆λ,j0>N |zλ0|, sup λ0⊆λ,j0≤N |cλ0| ) ≤ 2αnNmax ( sup λ0⊆λ,j0>N |zλ0|, sup λ0⊆λ,j0≤N |cλ0| )
which is bounded by a constant independent of ~c ∈ B since B is pointwise bounded. Therefore, there is C(n) > 0 such that
#{λ ∈ Λj : sup λ0⊆λ
|cλ0| > C(n)2−αnj} = 0, ∀j ≤ N, ∀~c ∈ B.
Similarly, for every ~c ∈ B, if j ≤ N , one has sup λ0⊆λ |cλ0| ≥ sup λ0⊆λ,j0>N |zλ0| > 0, ∀λ ∈ Λj since ~z ∈ Lν
. Consequently, for every p ∈ N, there is a constant D(p) > 0 such that #λ ∈ Λj: sup
λ0⊆λ
|cλ0| < D(p)2−α 0
pj = 0, ∀j ≤ N, ∀~c ∈ B.
It suffices then to take the constants max{C(n), C(m, n)}, C0(m, n), min{D(p), D(m, p)} and D0(m, p) and to use Proposition 6.6.6.
We can now prove that, if αmin> 0, the space (Lν, eδ ) is separable.
Proposition 6.6.9. Assume that αmin > 0. Let us fix ~z ∈ Lν and let us consider the
set U of sequences ~c ∈ Ω for which there exists N ∈ N such that |cλ| = |zλ| if λ ∈ Λj
with j > N , and cλ ∈ Q +i Q if λ ∈ Λj with j ≤ J . Then U ⊆ Lν and U is dense
in (Lν, eδ ). In particular, the space (Lν, eδ ) is separable.
Proof. It is clear that U ⊆ Lν since ~z ∈ Lν. Let us fix ~c ∈ Lν and let us consider
the sequence (~c(N ))
N ∈N given by Lemma 6.6.7. For every N ∈ N, it suffices to find a
sequence of U which converges to ~c(N )in (Lν, e
δ). Using the density of Q +i Q in C, there is a sequence (~q(N,l))
l∈Nof U which converges pointwise to ~c(N ). From Lemma 6.6.8, this
sequence can be included in a compact set eK of Lν and since it is pointwise convergent, it converges also to ~c(N ) in (Lν, eδ ) using Lemma 6.6.5.
Let us now consider the case where the admissible profile ν is such that αmin = 0.
The previous result is no longer valid. Indeed, with the admissible profile considered in Remark 6.4.25, the space Lν is C0 which is not separable. More generally, we have the
Proposition 6.6.10. If αmin= 0, the metric space (Lν, eδ ) is not separable.
Proof. This result uses classical considerations concerning sup-norms. Indeed, let us consider the uncountable set A of sequences ~c of C0
such that for each scale j ∈ N0,
cj,0 ∈ {0, 1} and the other coefficients are equal to 2−αsj. Using the assumption on
αmin, we easily prove that A is a subset of Lν. Moreover, k~c − ~c0kC0 = 1 for all distinct
elements ~c and ~c0 of A.
Let D be a dense subset of (Lν, eδ ). For every ~c ∈ A, there exists a sequence (~c(m))m∈Nof elements of D which converges in (Lν, eδ ) to ~c ∈ Lν. From Remark 6.4.19, the convergence also holds in C0. Consequently, there exists M ∈ N such that
k~c − ~c(m)k
C0<
1
2, ∀m ≥ M. In particular, there exists ~a ∈ D such that
k~c − ~akC0<
1 2.
Since the C0norm between two distinct elements of A is equal to 1, D must contain at
least as many elements as A and cannot be countable.