In functional analysis and related areas of mathematics, a metrizable (resp. pseudometrizable) topological vector space (TVS) is a TVS whose topology is induced by a metric (resp. pseudometric). An LM-space is an inductive limit of a sequence of locally convex metrizable TVS.
A pseudometric on a set is a map satisfying the following properties:
- ;
- Symmetry: ;
- Subadditivity:
A pseudometric is called a metric if it satisfies:
- Identity of indiscernibles: for all if then
Ultrapseudometric
A pseudometric on is called a ultrapseudometric or a strong pseudometric if it satisfies:
- Strong/Ultrametric triangle inequality:
Pseudometric space
A pseudometric space is a pair consisting of a set and a pseudometric on such that 's topology is identical to the topology on induced by We call a pseudometric space a metric space (resp. ultrapseudometric space) when is a metric (resp. ultrapseudometric).
Topology induced by a pseudometric
If is a pseudometric on a set then collection of open balls:
as ranges over and ranges over the positive real numbers,
forms a basis for a topology on that is called the -topology or the pseudometric topology on induced by
- Convention: If is a pseudometric space and is treated as a topological space, then unless indicated otherwise, it should be assumed that is endowed with the topology induced by
Pseudometrizable space
A topological space is called pseudometrizable (resp. metrizable, ultrapseudometrizable) if there exists a pseudometric (resp. metric, ultrapseudometric) on such that is equal to the topology induced by
An additive topological group is an additive group endowed with a topology, called a group topology, under which addition and negation become continuous operators.
A topology on a real or complex vector space is called a vector topology or a TVS topology if it makes the operations of vector addition and scalar multiplication continuous (that is, if it makes into a topological vector space).
Every topological vector space (TVS) is an additive commutative topological group but not all group topologies on are vector topologies.
This is because despite it making addition and negation continuous, a group topology on a vector space may fail to make scalar multiplication continuous.
For instance, the discrete topology on any non-trivial vector space makes addition and negation continuous but do not make scalar multiplication continuous.
Translation invariant pseudometrics
If is an additive group then we say that a pseudometric on is translation invariant or just invariant if it satisfies any of the following equivalent conditions:
- Translation invariance: ;
Equivalence on topological groups
Pseudometrizable topological groups
An invariant pseudometric that doesn't induce a vector topology
Let be a non-trivial (i.e. ) real or complex vector space and let be the translation-invariant trivial metric on defined by and such that
The topology that induces on is the discrete topology, which makes into a commutative topological group under addition but does not form a vector topology on because is disconnected but every vector topology is connected.
What fails is that scalar multiplication isn't continuous on
This example shows that a translation-invariant (pseudo)metric is not enough to guarantee a vector topology, which leads us to define paranorms and F-seminorms.
A collection of subsets of a vector space is called additive if for every there exists some such that
Continuity of addition at 0 — If is a group (as all vector spaces are), is a topology on and is endowed with the product topology, then the addition map (i.e. the map ) is continuous at the origin of if and only if the set of neighborhoods of the origin in is additive. This statement remains true if the word "neighborhood" is replaced by "open neighborhood."
All of the above conditions are consequently a necessary for a topology to form a vector topology.
Additive sequences of sets have the particularly nice property that they define non-negative continuous real-valued subadditive functions.
These functions can then be used to prove many of the basic properties of topological vector spaces and also show that a Hausdorff TVS with a countable basis of neighborhoods is metrizable. The following theorem is true more generally for commutative additive topological groups.
Theorem — Let be a collection of subsets of a vector space such that and for all
For all let
Define by if and otherwise let
Then is subadditive (meaning ) and on so in particular
If all are symmetric sets then and if all are balanced then for all scalars such that and all
If is a topological vector space and if all are neighborhoods of the origin then is continuous, where if in addition is Hausdorff and forms a basis of balanced neighborhoods of the origin in then is a metric defining the vector topology on
More information Assume that ...
Proof |
Assume that always denotes a finite sequence of non-negative integers and use the notation:
For any integers and
From this it follows that if consists of distinct positive integers then
It will now be shown by induction on that if consists of non-negative integers such that for some integer then
This is clearly true for and so assume that which implies that all are positive.
If all are distinct then this step is done, and otherwise pick distinct indices such that and construct from by replacing each with and deleting the element of (all other elements of are transferred to unchanged).
Observe that and (because ) so by appealing to the inductive hypothesis we conclude that as desired.
It is clear that and that so to prove that is subadditive, it suffices to prove that when are such that which implies that
This is an exercise.
If all are symmetric then if and only if from which it follows that and
If all are balanced then the inequality for all unit scalars such that is proved similarly.
Because is a nonnegative subadditive function satisfying as described in the article on sublinear functionals, is uniformly continuous on if and only if is continuous at the origin.
If all are neighborhoods of the origin then for any real pick an integer such that so that implies
If the set of all form basis of balanced neighborhoods of the origin then it may be shown that for any there exists some such that implies |
Close
If is a vector space over the real or complex numbers then a paranorm on is a G-seminorm (defined above) on that satisfies any of the following additional conditions, each of which begins with "for all sequences in and all convergent sequences of scalars ":
- Continuity of multiplication: if is a scalar and are such that and then
- Both of the conditions:
- if and if is such that then ;
- if then for every scalar
- Both of the conditions:
- if and for some scalar then ;
- if then
- Separate continuity:
- if for some scalar then for every ;
- if is a scalar, and then .
A paranorm is called total if in addition it satisfies:
- Total/Positive definite: implies
Examples of paranorms
- If is a translation-invariant pseudometric on a vector space that induces a vector topology on (i.e. is a TVS) then the map defines a continuous paranorm on ; moreover, the topology that this paranorm defines in is
- If is a paranorm on then so is the map
- Every positive scalar multiple of a paranorm (resp. total paranorm) is again such a paranorm (resp. total paranorm).
- Every seminorm is a paranorm.
- The restriction of an paranorm (resp. total paranorm) to a vector subspace is an paranorm (resp. total paranorm).
- The sum of two paranorms is a paranorm.
- If and are paranorms on then so is Moreover, and This makes the set of paranorms on into a conditionally complete lattice.
- Each of the following real-valued maps are paranorms on :
- The real-valued maps and are not paranorms on
- If is a Hamel basis on a vector space then the real-valued map that sends (where all but finitely many of the scalars are 0) to is a paranorm on which satisfies for all and scalars
- The function is a paranorm on that is not balanced but nevertheless equivalent to the usual norm on Note that the function is subadditive.
- Let be a complex vector space and let denote considered as a vector space over Any paranorm on is also a paranorm on
If is a vector space over the real or complex numbers then an F-seminorm on (the stands for Fréchet) is a real-valued map with the following four properties:
- Non-negative:
- Subadditive: for all
- Balanced: for all scalars satisfying
- This condition guarantees that each set of the form or for some is a balanced set.
- For every as
- The sequence can be replaced by any positive sequence converging to the zero.
An F-seminorm is called an F-norm if in addition it satisfies:
- Total/Positive definite: implies
An F-seminorm is called monotone if it satisfies:
- Monotone: for all non-zero and all real and such that
F-seminormed spaces
An F-seminormed space (resp. F-normed space) is a pair consisting of a vector space and an F-seminorm (resp. F-norm) on
If and are F-seminormed spaces then a map is called an isometric embedding if
Every isometric embedding of one F-seminormed space into another is a topological embedding, but the converse is not true in general.
Examples of F-seminorms
- Every positive scalar multiple of an F-seminorm (resp. F-norm, seminorm) is again an F-seminorm (resp. F-norm, seminorm).
- The sum of finitely many F-seminorms (resp. F-norms) is an F-seminorm (resp. F-norm).
- If and are F-seminorms on then so is their pointwise supremum The same is true of the supremum of any non-empty finite family of F-seminorms on
- The restriction of an F-seminorm (resp. F-norm) to a vector subspace is an F-seminorm (resp. F-norm).
- A non-negative real-valued function on is a seminorm if and only if it is a convex F-seminorm, or equivalently, if and only if it is a convex balanced G-seminorm. In particular, every seminorm is an F-seminorm.
- For any the map on defined by
is an F-norm that is not a norm.
- If is a linear map and if is an F-seminorm on then is an F-seminorm on
- Let be a complex vector space and let denote considered as a vector space over Any F-seminorm on is also an F-seminorm on
Topology induced by a single F-seminorm
Topology induced by a family of F-seminorms
Suppose that is a non-empty collection of F-seminorms on a vector space and for any finite subset and any let
The set forms a filter base on that also forms a neighborhood basis at the origin for a vector topology on denoted by Each is a balanced and absorbing subset of These sets satisfy
- is the coarsest vector topology on making each continuous.
- is Hausdorff if and only if for every non-zero there exists some such that
- If is the set of all continuous F-seminorms on then
- If is the set of all pointwise suprema of non-empty finite subsets of of then is a directed family of F-seminorms and
Suppose that is a family of non-negative subadditive functions on a vector space
The Fréchet combination of is defined to be the real-valued map
Generalization
The Fréchet combination can be generalized by use of a bounded remetrization function.
A bounded remetrization function is a continuous non-negative non-decreasing map that has a bounded range, is subadditive (meaning that for all ), and satisfies if and only if
Examples of bounded remetrization functions include and
If is a pseudometric (respectively, metric) on and is a bounded remetrization function then is a bounded pseudometric (respectively, bounded metric) on that is uniformly equivalent to
Suppose that is a family of non-negative F-seminorm on a vector space is a bounded remetrization function, and is a sequence of positive real numbers whose sum is finite.
Then
defines a bounded F-seminorm that is uniformly equivalent to the
It has the property that for any net in if and only if for all
is an F-norm if and only if the separate points on
- Every seminormed space is pseudometrizable with a canonical pseudometric given by for all .
- If is pseudometric TVS with a translation invariant pseudometric then defines a paranorm. However, if is a translation invariant pseudometric on the vector space (without the addition condition that is pseudometric TVS), then need not be either an F-seminorm nor a paranorm.
- If a TVS has a bounded neighborhood of the origin then it is pseudometrizable; the converse is in general false.
- If a Hausdorff TVS has a bounded neighborhood of the origin then it is metrizable.
- Suppose is either a DF-space or an LM-space. If is a sequential space then it is either metrizable or else a Montel DF-space.
If is Hausdorff locally convex TVS then with the strong topology, is metrizable if and only if there exists a countable set of bounded subsets of such that every bounded subset of is contained in some element of
The strong dual space of a metrizable locally convex space (such as a Fréchet space[23]) is a DF-space.
The strong dual of a DF-space is a Fréchet space.
The strong dual of a reflexive Fréchet space is a bornological space.
The strong bidual (that is, the strong dual space of the strong dual space) of a metrizable locally convex space is a Fréchet space.
If is a metrizable locally convex space then its strong dual has one of the following properties, if and only if it has all of these properties: (1) bornological, (2) infrabarreled, (3) barreled.
Normability
A topological vector space is seminormable if and only if it has a convex bounded neighborhood of the origin.
Moreover, a TVS is normable if and only if it is Hausdorff and seminormable.
Every metrizable TVS on a finite-dimensional vector space is a normable locally convex complete TVS, being TVS-isomorphic to Euclidean space. Consequently, any metrizable TVS that is not normable must be infinite dimensional.
If is a metrizable locally convex TVS that possess a countable fundamental system of bounded sets, then is normable.
If is a Hausdorff locally convex space then the following are equivalent:
- is normable.
- has a (von Neumann) bounded neighborhood of the origin.
- the strong dual space of is normable.
and if this locally convex space is also metrizable, then the following may be appended to this list:
- the strong dual space of is metrizable.
- the strong dual space of is a Fréchet–Urysohn locally convex space.[23]
In particular, if a metrizable locally convex space (such as a Fréchet space) is not normable then its strong dual space is not a Fréchet–Urysohn space and consequently, this complete Hausdorff locally convex space is also neither metrizable nor normable.
Another consequence of this is that if is a reflexive locally convex TVS whose strong dual is metrizable then is necessarily a reflexive Fréchet space, is a DF-space, both and are necessarily complete Hausdorff ultrabornological distinguished webbed spaces, and moreover, is normable if and only if is normable if and only if is Fréchet–Urysohn if and only if is metrizable. In particular, such a space is either a Banach space or else it is not even a Fréchet–Urysohn space.
- Every metrizable locally convex TVS is a quasibarrelled space, bornological space, and a Mackey space.
- Every complete pseudometrizable TVS is a barrelled space and a Baire space (and hence non-meager). However, there exist metrizable Baire spaces that are not complete.
- If is a metrizable locally convex space, then the strong dual of is bornological if and only if it is barreled, if and only if it is infrabarreled.
- If is a complete pseudometrizable TVS and is a closed vector subspace of then is complete.
- The strong dual of a locally convex metrizable TVS is a webbed space.
- If and are complete metrizable TVSs (i.e. F-spaces) and if is coarser than then ; this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete. Said differently, if and are both F-spaces but with different topologies, then neither one of and contains the other as a subset. One particular consequence of this is, for example, that if is a Banach space and is some other normed space whose norm-induced topology is finer than (or alternatively, is coarser than) that of (i.e. if or if for some constant ), then the only way that can be a Banach space (i.e. also be complete) is if these two norms and are equivalent; if they are not equivalent, then can not be a Banach space.
As another consequence, if is a Banach space and is a Fréchet space, then the map is continuous if and only if the Fréchet space is the TVS (here, the Banach space is being considered as a TVS, which means that its norm is "forgetten" but its topology is remembered).
- A metrizable locally convex space is normable if and only if its strong dual space is a Fréchet–Urysohn locally convex space.[23]
- Any product of complete metrizable TVSs is a Baire space.
- A product of metrizable TVSs is metrizable if and only if it all but at most countably many of these TVSs have dimension
- A product of pseudometrizable TVSs is pseudometrizable if and only if it all but at most countably many of these TVSs have the trivial topology.
- Every complete pseudometrizable TVS is a barrelled space and a Baire space (and thus non-meager).
- The dimension of a complete metrizable TVS is either finite or uncountable.
Completeness
Every topological vector space (and more generally, a topological group) has a canonical uniform structure, induced by its topology, which allows the notions of completeness and uniform continuity to be applied to it.
If is a metrizable TVS and is a metric that defines 's topology, then its possible that is complete as a TVS (i.e. relative to its uniformity) but the metric is not a complete metric (such metrics exist even for ).
Thus, if is a TVS whose topology is induced by a pseudometric then the notion of completeness of (as a TVS) and the notion of completeness of the pseudometric space are not always equivalent.
The next theorem gives a condition for when they are equivalent:
If is a closed vector subspace of a complete pseudometrizable TVS then the quotient space is complete.
If is a complete vector subspace of a metrizable TVS and if the quotient space is complete then so is If is not complete then but not complete, vector subspace of
A Baire separable topological group is metrizable if and only if it is cosmic.[23]
Subsets and subsequences
- Let be a separable locally convex metrizable topological vector space and let be its completion. If is a bounded subset of then there exists a bounded subset of such that
- Every totally bounded subset of a locally convex metrizable TVS is contained in the closed convex balanced hull of some sequence in that converges to
- In a pseudometrizable TVS, every bornivore is a neighborhood of the origin.
- If is a translation invariant metric on a vector space then for all and every positive integer
- If is a null sequence (that is, it converges to the origin) in a metrizable TVS then there exists a sequence of positive real numbers diverging to such that
- A subset of a complete metric space is closed if and only if it is complete. If a space is not complete, then is a closed subset of that is not complete.
- If is a metrizable locally convex TVS then for every bounded subset of there exists a bounded disk in such that and both and the auxiliary normed space induce the same subspace topology on
Generalized series
As described in this article's section on generalized series, for any -indexed family family of vectors from a TVS it is possible to define their sum as the limit of the net of finite partial sums where the domain is directed by
If and for instance, then the generalized series converges if and only if converges unconditionally in the usual sense (which for real numbers, is equivalent to absolute convergence).
If a generalized series converges in a metrizable TVS, then the set is necessarily countable (that is, either finite or countably infinite);[proof 1]
in other words, all but at most countably many will be zero and so this generalized series is actually a sum of at most countably many non-zero terms.
In fact, this is true for topological group, for the proof doesn't use the scalar multiplications.
Not assumed to be translation-invariant.
Proofs
Suppose the net converges to some point in a metrizable TVS where recall that this net's domain is the directed set
Like every convergent net, this convergent net of partial sums is a Cauchy net, which for this particular net means (by definition) that for every neighborhood of the origin in there exists a finite subset of such that
for all finite supersets
this implies that for every (by taking and ).
Since is metrizable, it has a countable neighborhood basis at the origin, whose intersection is necessarily (since is a Hausdorff TVS).
For every positive integer pick a finite subset such that for every
If belongs to then belongs to
Thus for every index that does not belong to the countable set
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