Space Topologies and Their Dual Space Topologies for Conventional Functional Space Topologies ()
1. Introduction
On some diagrams, only a specific collection of nodes or edges (arcs) represent a structure. A topology on a collection of nodes and edges of a non-directed graph and a topology on a set of nodes and arcs of a directed graph are defined in [1] . In addition, Dr. Alqahtani. M highlighted some of the topological characteristics of these spaces, and we examine some of the connections between them and the graphs. We further demonstrate that this topology meets the requirement of being Alexandrov.
In the group of pretopological spaces with p-continuous maps, we present the T0-reflection as arrows. Following that, we’ll look at a few of this category’s separation axioms, [2] [3] . On the other hand, [4] if A is open (respectively closed) and (respectively
is a finite array, then the subset A of the topological space X is known as an F-open (respectively F-closed) set. He examined the key aspects of these definitions in the study and illustrated and F-closed groups relate to other kinds, including regularly open, regularly closed, closed, and open groups in topological spaces. Moreover, utilizing the ideas of F-open and F-closed groups, [5] I will explain several topological features for the group, including internal F, closure F, derivative F, etc. Eventually, I’ll discuss F-Continuous, F-compact, and other concepts and ideas that are relevant.
We now provide our lesson on some classical function spaces’ topology. There are standard spaces among them, as well as areas that can be measured and some that cannot. Yet, given that they are all topological vector spaces. We have decided to display this work in this setting. Both the topology of its dual spaces and the topology of those spaces are of interest to us. Fundamental topological features of topological spaces comprised our first section. Part 2, He investigated functions of class
over
in Frechite spaces, particularly the
space. In addition to all of its quickly deteriorating partial derivatives, it exists. Also, we looked at the diluted distributions’ Double
area. The final section seeks to specify the topology. The inductive limit of Frechet spaces is based on an increasing union of Frechet spaces. We focus on the area
of the class
functions on the plus
space distributions over the open group. In addition to all of its quickly deteriorating partial derivatives, it exists. Also, we looked at the diluted distributions’ Double
area.
2. Continuous Linear Forms and Dual Space
2.1. Standard Vector Spaces
Proposition 2.1. E is a vector space on
or
.
1) It is
. We call the end segment a and b the subset [ab] of E defined by
2) Let, Ω is a subset of E. They say that,
a) Ω is convex if for all
, we have
.
b) Ω is equilibrium, if for all
and all
, then
.
c) Ω is absorbent, if for all
, it exists
such that,
.
Remark 2.2.
A balanced or absorbing subset necessarily contains 0.
Vector sub-spaces are convex and balanced.
Only E is an absorbing vector subspace.
Definition 2.3.
1) A norm on E is a map
verifying,
a)
,
,
b)
and
,
,
c)
,
.
2) We say that the pair
is a norm vector space if
is a norm on E.
is a norm vector space on
. We call an open ball with center
and radius
the set,
Any ball of E is convex and the balls with center 0 are balanced and absorbent.
A norm vector space admits a metric space structure. For the topology associated with this metric the balls with center 0 form a fundamental system of neighborhood of 0.
Two norms
and
over E are said to be equivalent, if there exist two strictly positive real numbers
and
such that,
Two equivalent norms
on E define the same topology.
Proposition 2.4. Let
be a norm vector space.
1) The map
is continuous.
2) The map
is continuous.
3) Let
, then
a) The translation by the vector a defined by
is a homeomorphism on E,
b) The multiplication by a scalar defined by
is continued on
.
Example 2.5.
,
,
and
.
and
are two norms on
. They are both equivalent.
For
,
is a a vector space on norm by
On
, we define the following norms,
More generally
- Let
be a measure space and
. The spaces
are norm vector spaces.
- For
and Ω an open set of
, we note,
²
the space of functions
of class
on Ω.
²
the space of functions
shach that f and all its partial derivatives up to order k are bounded on Ω.
²
the space of functions
has compact support included in Ω.
²
the space of functions
shach that f and all its partial derivatives up to order k tend to 0 at infinity.
The map
is a norm on the spaces
and
.
Definition 2.6. Let
be a norm vector space and
. We say that K is a compact if for any covering of K by a family of open sets of E, we can extract a finite subcovering. More precisely, let I be any set
a family of open sets of E such that
, then there is
such that
.
Remark 2.7. It is easy to verify that
The finished sets are compacts.
If
converge to
. Then
is a compact.
If K is a compact, then K is a closed bound of E.
If K is a compact of E and
, then
is compact.
Definition 2.8. Let
be a norm vector space. We say that K locally compact if 0 admits a compact neighborhood.
Proposition 2.9.
is locally compact if and only if the closed unit ball
is compact.
Definition 2.10. Let
be a norm vector space.
1) Let
be a sequence in E. We say that
is a cauchy sequence if,
2) We say that
is complete if every cauchy sequence in E is convergent. In this case, we say that E is a Bannach space.
2.2. Continuous Linear Forms and Dual Space
denotes a norm vector space.
We call linear form on E any linear application defined on E and takes its values in the body
.
Proposition 2.11. Let l be a linear form on E. Then the following properties are equivalent
1) l is continuous on E.
2) l is uniformly continuous on.
3) l is continuous in 0.
4) Le noyau de l est un sous-espace vectoriel ferme de E.
5) There is a constant
, such that
Let l be a continuous linear form on E. Then l is bounded on the closed unit ball of E and we set:
Remark 2.12.
1) There are linear shapes that are not continuous. For example, consider the vector space
be defined by
. The map l is a linear form and we have: for all
and
.
So l is not continuous.
2) On the other hand, if E is not reduced to {0}, then there exist nonzero continuous linear forms on E. The existence of such linear forms is ensured by the following Hahn-Banach theorem.
Theorem 2.13. Hahn-Banach theorem
Let E be a norm vector space and F a vector subspace of E not reduced to {0}. If l is a continuous linear form on F, then there exists
a continuous linear form on E such that
1)
,
2)
,
Let
be nonzero. Then there exists l a continuous linear form on E such that
et
.
Corollary 2.14. Let E be a norm vector space and
a nonzero vector. Then there exists l a continuous linear form on E which satisfies
.
Proof. Let
be a nonzero vector. Let l be the linear form defined on
by
. So we have:
So l is continuous on F and
. The Hahn-Banach theorem ensures the existence of a continuous linear form
on E which verifies
Proposition 2.15. Let E be a norm vector space. So E’ is a space vectoril on
, the map
is a norm on E’ etr
is a Banach space.
We note E" the dual of E' also calls the bidual space of E. The space E" endowed with norm
is a Banach space.
Let
. The map
defined by
, is a linear form on E’ which verifies
We deduce that the map
is an isometric bijection of E in E".
In general, the map J is not surjective and particularly it is not surjective if E is not a Banach space. But the map J makes it possible to prove that any norm vector space E can be injected isometrically into a Banach space. Therefore it is dense in the Banach space
.
Definition 2.16. We say that E is a reflexive space if the isometric injection J from E into E" is surjective. We are going to define a topology on E weak topology site. Let
Definition 2.17. We call weak topology on E and we denote by
the topology on E generated by
.
Remark 2.18.
1) The topology
is less fine than the topology of the standard space E.
2) Let
. We obtain a fundamental system of neighborhoods of a for the topology
by considering the open sets of the form
where
,
,
.
3) We assume that E is not of finite dimension. Let
an open neighborhood of 0 for the weak topology. There exists a nonzero vector in E, such that
So the line
is included in
. This proves that the balls
are not open for the weak topology.
Remark 2.19. Let E be a norm vector space. therefore
1) The topology
is separated.
2) The map
of
in E and the map
of
in E are continuous when E is endowed with the weak topology.
Proof. Let a and b be two distinct vectors in E. By the corollary (2.14) there exists
such that
. So the disks
are disjoint.
We deduce that
and
are indeed two disjoint open sets and which contain respectively a and b.
Definition 2.20. Let
be a sequence in E and
.
1) We say that
converges (strongly) to x, if the sequence
converges to 0.
2) We say that
converges weakly to x, if the sequence
converges to x for the topology
.
Since the weak topology is less fine than the topology defined by the norm, then we have the following result:
Proposition 2.21. Let
be a sequence in E and
.
1) The sequence
converges weakly to x, if and only if for all
the sequence
converges to
.
2) If
converges strongly to x, then
converges weakly to x.
2.3. Finite Dimensional Vector Spaces
The characteristic properties of finite-dimensional vector spaces follow from the following properties of
,
endowed with the absolute value is a complete metric space.
The compacts of
are the closed bounded subsets.
Theorem 2.22. A finite dimensional norm vector space,
. Therefore,
E is compact.
All the norms on E are equivalent.
Corollary 2.23. E a finite dimensional norm vector space.
1) If E has finite dimension n, then it is homeomorphic to
.
2) Any linear map defined on a finite-dimensional norm vector space with values in a norm vector space is continuous.
3) Let F be a vector subspace of E. If F is of finite dimension then it is closed in E.
Proof.
1) Let
be a finite dimensional vector space and let
be a basis of E. We consider the map
defined by
The map
is an isomorphism of E on
. We endow
with the norm N defined by
Thus the application
verifies, for any vector
,
We deduce that
is a surjective isometry between the spaces
and
and therefore it is a homeomorphism.
2) Let
,
be two norm vector spacesa, and let
be a linear map. We assume that E is finite dimensional and let
be a basis of E. We consider on E the norm
defined by
Let
therefore
3) If F is a finite dimensional vector subspace, then F endowed with the norm of E is complete. So it is firm in E.
Proposition 2.24. Let E be a finite dimensional vector space n, then
1) E’ be a finite dimensional vector space n.
2) The
topology coincides with the usual topology.
Proof.
1) Let
be a basis of E. We consider the linear forms
defined on E by
It is easy to check that if l is a linear form on E, then
So
is a basis of E’, called dual basis of
, and hence E’ has dimension n.
2) To show this result, it suffices to prove that for all
, the ball
is a neighborhood of 0 for the topology
. Let
be a basis of E, satisfying
. We denote
its dual basis. For all
, we have
If
then
which implies
, so
2.4. Hilbert Space
Let H be a vector space over
.
Definition 2.25.
1) We call sesquilinear form on H any map
satisfying for all
and
,
and
.
2) A Hermitian form b on H is a ferifying sesquilinear form
3) A Hermitian form b on H is said to be positive if for all
we have,
4) A Hermitian form b on H is said to be positive define, if it is positive and if for all
we have,
In this case, we say that b is a scalar product (or hermitian product) on H. Such a product is generally denoted by
instead of
. When
, a sesquilinear form is a bilinear form and a Hermitian form is a symmetric bilinear form.
Theorem 2.26. Let b be a positive Hermitian form on H Then for all
we have,
1)
(Schwarz’s inequality). If moreover b is positive definite, then the equality holds if and only if x and y are collinear.
2) Minkowski’s Inequality,
3) If moreover b is positive definite,
is a norm on H.
Definition 2.27. A pre-Hilbertian space is a vector space H over
endowed with a scalar product
. It is implied standard by
.
A Hilbert space is a complete pre-Hilbert space.
Definition 2.28. Let
and
be two Hilbert spaces. They are said to be isomorphic if there exists an isomorphism of vector spaces
which satisfies
Example 2.29
The space
is a Hilbert space for the Hermitian product
defined for all
and
in
.
Every finite-dimensional pre-Hilbert space is a Hilbert space.
The set
is a Hilbert space for the Hermitian product
defined for all
and
in
.
Let
be a measure space. The vector space
is a Hilbert space for the scalar product
defined for all
by
1) The space
is a special case for
,
and
the measure of the cardinal.
2) Let Ω be an open set of
and
a positive measurable function on Ω for the Lebesgue measure dx. Then
into a Hilbert space.
Let H be a Hilbert space. For all fixed a in H, the map from H to
is a continuous linear form because
and we thus obtain all the continuous linear forms on H.
Theorem 2.30. (Riesz representation theorem)
1) Let l be a continuous linear form on a Hilbert space H. Then there exists a unique element
such that
for all
and we have
2) The map
of H in H' which has each a in H associates
is bijective antilinear. In particular
is a Hilbert space and
is an ant-isopmorphism of Hilbert space. So the canonical injection J of H into its bidual H’’ given by
is surjective and verifies
So H" is a Hilbert space and the injection J is an isomorphism of Hilbert spaces, we deduce the following result.
Corollary 2.31 Hilbert spaces are reflexive spaces.
2.5. Bannach Spaces
Banach spaces are complete vector spaces.
Example 2.32.
Finite dimensional spaces and Hilbert spaces are Bannach spaces.
For
, the spaces
are Bannach spaces.
More generally, if
a measure space and
, then the spaces
are Bannach spaces.
Let
, Ω be an open set of
, and
. We set
where
is a multi-index of length
and
is the partial derivative of f of order
. Then the map
is a norm on
and we have
-
is Bannach space.
-
is a norm vector subspace of
. We denote
its adherence which is a Bannach space.
- The space
corresponds to the spaces of functions f of class
on
which tend to 0 a l ‘infinity as well as their derivatives of order
such as
.
If E is a norm vector space, then E’ and E’’ are Bannach spaces. If E is dense without a Bannach space F, then
(we restrict the elements of F’ to E).
Corollary 2.33. Let E be a Bannach space. If
is a sequence in E’ such that, for all
, the sequence
converges to
, then
and
In general, a Bannach space is not necessarily reductive as shown by the following proposition.
Proposition 2.34. We consider the Bannach spaces
,
and
. So we have
Proof. We will start by showing that
.
We consider, for all
, the sequence
where
denotes the symbol by Kroneker. So
is the sequence of which all the terms are zero except the term of order
which is worth 1. Let
. We denote by
the linear form on
defined by
Then
is a continuous linear form on
and it is easy to show that
Conversely, let
. Then the sequence
is bounded and we have
on
the space of sequences which are zero from a certain rank. But
is dense in
which implies that
and
. So the map
is a one-to-one isometry from
to
.
Let’s show that
.
Let
and
be the linear form on
defined by
So
and hence
is a continuous linear form on
and
.
To show that
, we consider the sequence
defined for
by
or
Then the sequence
belongs to
and verifies
and
which implies by passing to the limit that
. So the map
is an isommetry of
in
. For surjectivity, we consider, for all
, the sequence
. Let
. We consider the sequence
defined by
and
the sequence in
defined by
or
therefore
Let
. We consider the sequence
in
defined by
So we have
therefore the sequence
converges to x in
. On the other hand, we have
. So
So the series
is convergent and equals
. To conclude, it suffices to show that the sequence
. We consider the sequence
defined for
for
where
The sequence
is in
and
. So, if
, then we have
or
We deduce that
and
.
Let us show that
.
We consider E the space of convergent sequences in
. So E is a vector subspace of
which contains
. Let l be the linear form on E defined by
Then l is continuous and
.
According to the Hahn-Banach theorem (2.13), there exists a linear form
is zero. There is no
such that
.
Theorem 2.35. Let E be a Banach space. So we have,
E is reflexive if and only if the closed unit ball is compact for the weak topology
.
E is reflexive if and only if E’ is reflexive.
We deduce from this theorem that the closed subspaces of a reflexive Banach space are reflexive.
2.6. Dual of Lp Spaces
The triple
denotes a measurable space and
two conjugate exponents.
Definition 2.36.
is said to be a finite measure if
.
is said to be a
-finite measure if there exists a sequence
in
such that
and
We say that
is a semi-finite measure if for all
with
, then there exists
included in A and such that
. For
, we denote
the linear form on
defined by,
Lemma 2.37. Suppose that p and q are two conjugate exponents and
.
If
, then
Moreover, if
is semi-finite then the previous equality is true for
.
We deduce that the linear map
is an isometry of
in
. More precisely, we have the following theorem,
Theorem 2.38. Suppose p and q are two conjugate exponents if
. then the linear map
is an isometric isomorphism from
to
. the result remains true if
and
is
-finite.
Corollary 2.39. If
then
is a reflexive Banach space.
For the case
, the map
may not be surjective.
In this case the spaces
and
are not reductive.
2.7. Dual of Space
Let
a space measures a signed measure on
is a mapping
or
such that
1)
2)
the series
is convergent and
If
is a signed measure, then there is a unique pair
of positive measures such that
. We call total variation of
the positive measure
and we define the spaces,
If
is a complex measure, then its real part
and its imaginary part
are two signed measures and
and we define the spaces,
where
is a positive measure, called the total change of
and it is defined by
In the following, we move to the case
where Ω is an open set of
and
is the Borelian tribe.
A (borelian) measure signed
on
is said to be of Radon if it is finite on compact sets. Let
be a complex (borelian) measure on B, then
and
are Radon measures. Similarly, its total variation
is a positive Radon measure.
We denote by
the set of complex measures on
, and we define for
the map
Proposition 2.40.
is a vector space over
and the map
is a norm over
.
designates the complete for the norm
of the space of continuous functions with compact support in Ω. The space
is the set of continuous functions which tend to 0 at infinity.
Theorem 2.41. (Riesz representation theorem) Let Ω be an open set of
. For
and
, let
Then the application
is an isometric isomorphism from
to
Corollary 2.42. If K is a compact of
, then
is isometrically isomorphic to
.
3. Introduction to Topological Vector Spaces
Definition 3.1. A topological vector space (EVT) E is a vector space over
endowed with a topology
such as the maps
and
are continues.
Proposition 3.2. Let
be a topological vector space,
and
nonzero, then
1) The translation by the vector a defined by
is a homeomorphism from E to E.
2) The
ratio scaling defined by
is a continuous isomorphism on E as well as its inverse.
Proof. The maps
and
are invertible and their inverses are given by
and
. The continuity of the map and the multiplication by a scalar imply respectively the continuity of
and
and likewise the continuity of their inverses.
From this proposition, we deduce that the topology
is inveriant under translation. More precisely, if
is an open set, then its translates
, with
are open sets of E. Consequently, the topology
is completely determined by the given basis of neighborhoods of any point of E, in particular at 0. Thus in a topological vector space
the term neighborhood base always means a neighborhood base of 0.
Proposition 3.3. Let E be a topological vector space and U an open neighborhood of 0 in E
1)The open set U is an absorbing subset of E.
2) There exists
an equilibrium open neighborhood of 0. Moreover, for all
we have
.
Proof. Let U be an open neighborhood of 0 in E. Let
be a real number. We denote
1) Let
. The map
defined by
is continuous. So the reciprocal image of U by
is an open neighborhood of 0 in
. So there exists
such that
so
which proves that
therefore U is absorbent.
2) Let
be the map defined, for all
, by
By definition of the topological vector space, the map F is continuous so
is an open neighborhood of
in
. Therefore, there is
and W an open neighborhood of 0 in E such that
Therefore
is an open neighborhood of 0 included in U and it is equilibrium. Let
and
be a real. Since V is balanced then
and since
then
. or
.
Definition 3.4. A topological space is said to be separate if for all distinct points x and y of E there exists a neighborhood
of x and a neighborhood
of y such that
Proposition 3.5. A topological vector space is separate if and only if the singleton {0} is closed.
Proof. Let E be a topological vector space.
If E is separated then it is easy to see that
is an open and therefore {0} is closed. Conversely, suppose that {0} is closed.
Let
and
defined by
The map f is continuous and we have
. So Ω is an open set of E. Let
be distinct, hence
. Since Ω is an open, then there exists
an open neighborhood of a and
an open neighborhood of b such that
. This implies that
and that E is a separate topological vector space.
In the following, all topological vector spaces will be assumed to be separate
Definition 3.6. Let E be a topological vector space.
1) Let
. We say that A is bounded, if for every neighborhood V of 0, there exists
such that
2) Let
be a sequence in E. We say that
is a Cauchy sequence if for every Ω a neighborhood of 0 in E, there exists an integer
such that
.
3) We say that E is complete if any cauchy sequence in E converges in E.
Example 3.7.
1) Any finite union of bounded sets of topological vector spaces TVS is bounded.
2) Any finite subset of a topological vector space is bounded.
3) Any Cauchy sequence is bounded.
Proof.
1) Let E be a TVS and
bound subsets. Let U be an open neighborhood of 0. Then, for all
, there exists
such that
implies that
. Let
. SO
this proves that
is bounded.
2) According to the proposition 3.3, any neighborhood of 0 is absorbing. so every singleton is bounded. We conclude with item 1. of 3.3.
3) Let
be a Cauchy sequence in E. Let U be a neighborhood of 0 in E. By proposition 3.3, there exists
an equilibrium open neighborhood of 0 which satisfies
for all
. There exists an integer
such that for any integer
, we have
. Since the singleton
is bounded, then there exists
such that
, for all
. This implies that the whole
We deduce that
is bounded because it is a union finish of ensembleornes.
Definition 3.8. Let E be a TVS and
. We say that K is said to be a compact if from any cover of K by open sets we can extract a finite subcover.
Proposition 3.9. Let E be a TVS and
. We say that K is said to be a compact if from any cover of K by open sets we can extract a finite subcover.
Proof. Let K be a compact of a TVS E. To begin, we will show that K is firm.
Let
. Then for all
, since E is separated, there exists
, open neighborhood of x in E and
an open neighborhood of y in E such that
. So
, since K is compact, then there exists
such that
The set
is an open neighborhood of y which verifies,
The open set U is included in Ω and contains y so Ω is an open set and therefore K is a closed set. Show that K is bounded, let U be an open neighborhood of 0 in E. By proposition 3.3, there exists
an equilibrium open neighborhood of 0 which satisfies
for all
. Therefore, for any integer
is an open and balanced neighborhood of 0. Moreover the sequence
is increasing and it satisfies
So there exists an integer
such that
and consequently, for all
we have
which shows the result.
Definition 3.10. Let E be a TVC. We say that,
1) E is locally convex if E admits a basis of convex neighborhoods.
2) E is locally bounded if 0 admits a bounded neighborhood.
3) E is locally compact if 0 admits a compact neighborhood.
4) E is metrisable if there is a distance on CE which defines the topology of E.
Example 3.11.
1) Norm vector spaces are topological vector spaces. Moreover, they are E is metrisable if there is a distance on E which defines the topology of E.
a) locally convex if E admits a basis of convex neighborhoods.
b) E is locally convex, because the balls are convex.
c) locally bounded, because
is a bounded neighborhood of 0.
d) E is locally compact only when they are of finite dimension.
e) metrizable .
2) The norm vector spaces endowed with the weak topology
are topological vector spaces.
Topology Defined by a Family of Semi-Norms
Definition 3.12. Let E be a vector space. A semi-norm on E is a map
verifying
1)
,
.
2)
,
and
.
In the following E denotes a vector space over
. Let p be a semi-norm on E,
and
.
We call p-ball with center a and radius r, the set,
More generally, let
be a family of semi-norms on E.
It is said to be separant if it verifies,
Let
. We call P-ball with center a any set of the form,
where
and
a real number. We denote by
the topology on E generated by the P-balls. The topology
is inveriant under translation and the P-balls with center 0 form a fundamental system of neighborhood of 0.
Proposition 3.13. Let
be a family of separating semi-norms on E. Then E endowed with the topology
is a locally convex and separate topological vector species. also for a sequence
converges to x in E, if and only if for all
, we have,
Proof. Let
defined over
and has values in E and Ω a neighborhood of 0 in E. Then there is
. inclusion,
Let
is a neighborhood of
IN
and hence
is continuous.
Let
defined over
and has values in E and Ω a neighborhood of 0 in E. Then there is
. inclusion,
implies that
is a neighborhood of
in
and therefore
is continuous. The continuity of the maps
and
implies that
is a topological vector space.
To show that this topology is separated, it suffices to show that {0} is a closed one. Let
be nonzero, then there exists
a semi-norm which satisfies
. The P- ball
is a neighborhood of a which does not contain 0, so {0} is firm and the topological vector space
is separate. Finally,
is locally convex because the P-balls are convex.
Proposition 3.14. Let
be a topological vector space defined by a family of semi-norms.
1) Balls with center 0 are balanced sets.
2) A set
is bounded if and only if
E denotes a topological vector space.
Proposition 3.15. Let l be a nonzero linear form on E. Then we have the following equivalences,
l is continued on E.
l is bounded on a neighborhood of 0.
l is continuous at 0.
The core of l is firm.
The kernel of l is not dense in E.
Definition 3.16. E' is the set of continuous linear forms on E, E' is a vector space on
, called dual (topological) space of E.
Example 3.17. Let
be a measure space. Let
and
be the space of classes of functions measurable on X such that,
The application
verifies
which implies that
is a vector space over
and the map
defines a translation invariant distance on
. Then
endowed with this distance is a locally bounded topological vector space and does not contain any nontrivial convex open set. Moreover, on
the only continuous linear form is the zero linear form, so
.
The following Hahn-Banach theorem ensures that
, if E is locally convex not reduced to {0}.
Theorem 3.18. (Hahn-Banach) Let E be a locally convex topological vector space. Then, for all nonzero
, there exists
satisfying
.
We are provided with two topologies defined by semi-norms,
Strong topology on E’,
Let
be the set of bounded subsets of E. For all
, we consider the semi-norm
defined by
We denote by
the topology on E’ defined by the family of semi-norms
. The space E’ endowed with the topology
is a topological vector space.
Weak topology on E’,
For all
, we consider the semi-norm
defined by,
We denote by
the topology on E’ defined by the family of semi-norms
. The space E’ endowed with the topology
is a topological vector space.
It is clear that the topology
is less fine than the topology
.
Proposition 3.19. The topological vector spaces
and
are locally convex and separate.
Proof. It suffices to show that the families of semi-norms
and
are separating.
4. Frechet Space
4.1. Definition and Properties of Frechet Spaces
Proposition 4.1. Let E be a vector space and
a sequence of separating semi-norms on E. So we have,
The application
is a distance on E.
For all
, the map
is continuous.
The topology on E defined by the family of semi-norms
coincides with the metric topology
.
Proof.
The continuity of a semi-norm
follows from the inequality
valid for all
such that
.
Let
and Ω be a neighborhood of
in
. Then there exists
such that
.
let
and m a natural integer satisfying
therefore
Then if
we have
so
. As a result.
. This proves that Ω is a neighborhood of
for the topology defined by the semi-norms. The reciprocal is deduced from the continuity of the
.
Definition 4.2. See [6] .
A Frechet space is a complete topological vector space whose topology is defined by a sequence of separate semi-norms.
Example 4.3. (Examples of Frechet spaces)
Finite dimensional vector spaces, Hilbert spaces and Bannach spaces are Frechet spaces.
he space of functions of class
on an open set of
. Let Ω be an open set of
and
. We consider
an increasing sequence of compact sets such that
,
and
.
We denote by
the family of semi-norms in
defined by,
if k is finite, else
The space
equipped with the topology
defined by the sequence of semi-norms
is a Frechet space. Moreover, this topology does not depend on the choice of the sequence of compacts
.
A sequence
convergent to f in
if and only if, for any multi-index
such that
, the sequences
converges uniformly on any compact from Ω to
.
the subspace of
of functions with support included in a compact K.
Let Ω be an open set of
and
a compact. We consider the space
The space
endowed with the induced topology is closed in
and it is therefore a Frechet space. Indeed, let
and
be a linear form on
. There exists
such that
. So
this proves that
is continuous, so its kernel is a farm and we have,
If
, then
is a Banach space or
When
, we also denote
the space
.
the space of functions locally
.
Let Ω be an open set of
and
the measure space where
is the tribe of Lebesgue and dx is the Lebesgue measure.
Let
. A function f measurable on Ω is said locally
if for any compact
the function
. We denote by
the set of these functions. We consider
an increasing sequence of compacts satisfying the properties (1). For all
, let
be the semi-norm in
defined by
or
designates the characteristic function of
. By considering this sequence of semi-norms we endow
with a Frechet space structure.
is the Schwartz space. A function
is said to have rapid decrease if for all
the function
is bounded on
. We denote by
the space of functions f of class
such that for
, the function
is fast decreasing over
. This space is also called Schwartz space.
We endow
with a sequence of semi-norms
defined by,
Proposition 4.4. The space
endowed with the sequence of semi-norms
is a Frechet space.
Proof. Let
be a cauchy sequence in
. So, for all
,
converges uniformly on
to a function
with g is of class
. It remains to show that the functions
are rapidly decreasing and that the sequence
converges to g for the topology of
. Let
and
. There exists
such that
then,
,
we have
We fix x and let p tend to infinity, then we get,
Therefore
So
is rapidly decreasing and verified,
This then proves that the Cauchy sequence
is convergent in
and therefore
is a Frechet space.
Proposition 4.5.
1) The space
is a subspace of the following topological spaces
,
and
. Moreover, the injection of
into each of the preceding spaces is continued.
2) Let K be a compact of
. The canonical injection of
into
is continuous.
4.2. Continuous Linear Forms and Dual Space (See [7] )
Proposition 4.6. Let E be a Frechet space. Then its dual E' is complete for the strong topology
.
Definition 4.7. Let E be a Frechet space. The strong bidual E" of E is the strong dual of the strong dual E' of E. We say that E is reflexive if the canonical injection of E into E" is an isomorphism of E onto E".
Theorem 4.8. Let E be a Frechet space. For E" to be reflexive, it suffices that every weakly closed and bounded set in E be weakly compact.
In particular, if the bounded firm sets of a Frechet space E are compact, then E is reflexive. In this case, we say that E is a Montel space. So the only norm vector spaces that are Montel spaces are the finite dimensional spaces.
Theorem 4.9. Let K be a compact of
. In
, any closed and bounded subset is a compact set. So
is a Montel space.
The most notable example of a Frechet space is the space
.
Proposition 4.10. The space
is a reflexive Frechet space.
Definition 4.11. We call tempered distribution the linear forms belonging to
the dual of
.
Let T be a linear form on
. Then T is a tempered distribution if there exists
and a pair of positive integers
such that for all
, we have
Since
is metrisable, a linear form T is a tempered distribution if it is sequentially continuous. So T is continuous if for any sequence
in
, the sequence
Let g be a measurable function on
such that for all
, the function
belongs to
. We denote by
the linear form on
, defined by
We then have,
Proposition 4.12. If
, then the map
is a continuous injection of
in
endowed with its weak topology.
5. Inductive Limit Spaces of Frechet Spaces (See [7] )
Definition 5.1. Let E be a vector space over
. We suppose that there exists a strictly increasing sequence
of vector subspaces of E satisfying
For all
,
is a Frechet space.
For all
, the restriction of the topology of
to its subspace
coincides with the initial topology of
.
.
Under these conditions, we endow E with a topology
as follows
Let
. We say that Ω is an open set of E, if and only if, for all
,
is an open set of
. We thus define a topology on E, called inductive limit of Frechet spaces and we denote the space
Proposition 5.2. Let
be an inductive limiting space of Frechet spaces. So we have,
1) E is a separate topological vector space.
2) For a set
to be bounded, it is necessary and sufficient that there exists an integer n such that
and that A is bounded in
.
3) A sequence
is convergent in E, if and only if there exists an integer n such that
is a convergent sequence in
.
4) E is complete but it is not metrisable.
Example 5.3. Let Ω be an open set of
and
.
The space
is the set of functions
which are of class
and have compact support.
We consider
an increasing sequence of compact sets such that
,
and
.
The spaces
are Frechet spaces and we have,
We thus provide
with an inductive limit space structure of Frechet spaces which does not depend on the choice of the sequence of compact sets
. When
, we denote
The space
.
5.1. Continuous Linear Forms and Dual Space
Theorem 5.4. Let
be a space
.
1) Let T be a linear form on E. Then T is continuous, if and only if the restriction of T to
is continuous for all
.
2) Let T be a linear form on E. Then T is continuous, if and only if it is sequentially continuous.
Proposition 5.5. Let E be a space
. Then its dual E’ is complete for the strong topology
.
Definition 5.6. Let E be a space
. The strong bidual E" of E is the strong dual of the strong dual E' of E. We say that E is reflexive if the canonical injection of E into E" is an isomorphism of E onto E".
Theorem 5.7. If E is a space
, for E to be reflexive, it is necessary and sufficient that every weakly closed and bounded set in E be weakly compact.
We deduce that, if
is a space
such that, for all
,
is reflexive, then E is reflexive.
5.2. The Space of Distributions
See [8]
Definition 5.8. Let Ω be an open set of
. A distribution T on Ω is a continuous linear form on
. We note
the dual of
.
Let T be a linear form on
. Then T is a distribution if and only if, for any compact K the sequence of compacts satisfying the property (1)
Therefore, for any compact
, there exists
and an integer
such that, for any
, we have,
Let g be a measurable function on Ω such that for all
, the function
belongs to
. We denote by
the linear form on
defined by,
Then we have,
Proposition 5.9.
1) If
, then the map
is a continuous injection of
daNS
endowed with its weak topology.
2)
and this injection is continuous.
Definition 5.10. Let
.
1) Let F be a farm of Ω). We say that the support of T is included in F if for all
the condition
implies
.
2) We say that T has compact support if its support is included in a compact. We denote by
the space of distributions
which have compact support.
3) We say that T is of finite order if there exists an integer
such that for every compact
, there exists a constant
such that, for every
, we have,
If T is of finite order, we call order of T the smallest integer k which satisfies the preceding inequality. We denote
the subspace of distributions of order less than or equal to k.
If T is not of finite order, we say that it is of infinite order.
Example 5.11. Let Ω be an open set of
and
.
Let
, then the map
is a distribution with support
. Moreover,
is of finite order equal to
.
Let
be defined by
Then, T is a distribution of infinite order and inclusive support
.
6. Conclusions
This work examines the topology of classical functional spaces, such as standard spaces, metrizable spaces, and those that cannot be metrizable. It presents the fundamental topological properties of topological vector spaces, as well as the space
of functions of class
on
and its rapidly decreasing partial derivatives. It also defines a topological structure on an increasing union of Frechet spaces, called the inductive limit of Frechet spaces, and studies the space
of functions of class
with compact supports on Ω.
Let
, let
be a distribution with support included in a compact K. Let
be a compact satisfying
and
such that
for all
. Then, for all
, we have,
this is what gives that any distribution with compact support is of finite order. Let
and Ω be an open set of
. Then the dual of
is
the space of distributions of order less than or equal to k.
Let
and an open set Ω an open set of
. Then the dual of
is
, the space of distributions of order less than or equal to k and has compact support. Finally, for E the space of continuous and rapidly decreasing functions
. The space E endowed with the family of semi-norms (see [9] [10] )
is a Frechet space. Its dual E’ is equal to
, the space of tempered distributions of order 0. We say that these distributions are temperate measures.