From Handwiki 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 [math]\displaystyle{ X }[/math] is a map [math]\displaystyle{ d : X \times X \rarr \R }[/math] satisfying the following properties:
A pseudometric is called a metric if it satisfies:
Ultrapseudometric
A pseudometric [math]\displaystyle{ d }[/math] on [math]\displaystyle{ X }[/math] is called a ultrapseudometric or a strong pseudometric if it satisfies:
Pseudometric space
A pseudometric space is a pair [math]\displaystyle{ (X, d) }[/math] consisting of a set [math]\displaystyle{ X }[/math] and a pseudometric [math]\displaystyle{ d }[/math] on [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ X }[/math]'s topology is identical to the topology on [math]\displaystyle{ X }[/math] induced by [math]\displaystyle{ d. }[/math] We call a pseudometric space [math]\displaystyle{ (X, d) }[/math] a metric space (resp. ultrapseudometric space) when [math]\displaystyle{ d }[/math] is a metric (resp. ultrapseudometric).
If [math]\displaystyle{ d }[/math] is a pseudometric on a set [math]\displaystyle{ X }[/math] then collection of open balls: [math]\displaystyle{ B_r(z) := \{ x \in X : d(x, z) \lt r \} }[/math] as [math]\displaystyle{ z }[/math] ranges over [math]\displaystyle{ X }[/math] and [math]\displaystyle{ r \gt 0 }[/math] ranges over the positive real numbers, forms a basis for a topology on [math]\displaystyle{ X }[/math] that is called the [math]\displaystyle{ d }[/math]-topology or the pseudometric topology on [math]\displaystyle{ X }[/math] induced by [math]\displaystyle{ d. }[/math]
Pseudometrizable space
A topological space [math]\displaystyle{ (X, \tau) }[/math] is called pseudometrizable (resp. metrizable, ultrapseudometrizable) if there exists a pseudometric (resp. metric, ultrapseudometric) [math]\displaystyle{ d }[/math] on [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ \tau }[/math] is equal to the topology induced by [math]\displaystyle{ d. }[/math][1]
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 [math]\displaystyle{ \tau }[/math] on a real or complex vector space [math]\displaystyle{ X }[/math] 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 [math]\displaystyle{ X }[/math] into a topological vector space).
Every topological vector space (TVS) [math]\displaystyle{ X }[/math] is an additive commutative topological group but not all group topologies on [math]\displaystyle{ X }[/math] are vector topologies. This is because despite it making addition and negation continuous, a group topology on a vector space [math]\displaystyle{ X }[/math] 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.
If [math]\displaystyle{ X }[/math] is an additive group then we say that a pseudometric [math]\displaystyle{ d }[/math] on [math]\displaystyle{ X }[/math] is translation invariant or just invariant if it satisfies any of the following equivalent conditions:
If [math]\displaystyle{ X }[/math] is a topological group the a value or G-seminorm on [math]\displaystyle{ X }[/math] (the G stands for Group) is a real-valued map [math]\displaystyle{ p : X \rarr \R }[/math] with the following properties:[2]
where we call a G-seminorm a G-norm if it satisfies the additional condition:
If [math]\displaystyle{ p }[/math] is a value on a vector space [math]\displaystyle{ X }[/math] then:
Theorem[2] — Suppose that [math]\displaystyle{ X }[/math] is an additive commutative group. If [math]\displaystyle{ d }[/math] is a translation invariant pseudometric on [math]\displaystyle{ X }[/math] then the map [math]\displaystyle{ p(x) := d(x, 0) }[/math] is a value on [math]\displaystyle{ X }[/math] called the value associated with [math]\displaystyle{ d }[/math], and moreover, [math]\displaystyle{ d }[/math] generates a group topology on [math]\displaystyle{ X }[/math] (i.e. the [math]\displaystyle{ d }[/math]-topology on [math]\displaystyle{ X }[/math] makes [math]\displaystyle{ X }[/math] into a topological group). Conversely, if [math]\displaystyle{ p }[/math] is a value on [math]\displaystyle{ X }[/math] then the map [math]\displaystyle{ d(x, y) := p(x - y) }[/math] is a translation-invariant pseudometric on [math]\displaystyle{ X }[/math] and the value associated with [math]\displaystyle{ d }[/math] is just [math]\displaystyle{ p. }[/math]
Theorem[2] — If [math]\displaystyle{ (X, \tau) }[/math] is an additive commutative topological group then the following are equivalent:
If [math]\displaystyle{ (X, \tau) }[/math] is Hausdorff then the word "pseudometric" in the above statement may be replaced by the word "metric." A commutative topological group is metrizable if and only if it is Hausdorff and pseudometrizable.
Let [math]\displaystyle{ X }[/math] be a non-trivial (i.e. [math]\displaystyle{ X \neq \{ 0 \} }[/math]) real or complex vector space and let [math]\displaystyle{ d }[/math] be the translation-invariant trivial metric on [math]\displaystyle{ X }[/math] defined by [math]\displaystyle{ d(x, x) = 0 }[/math] and [math]\displaystyle{ d(x, y) = 1 \text{ for all } x, y \in X }[/math] such that [math]\displaystyle{ x \neq y. }[/math] The topology [math]\displaystyle{ \tau }[/math] that [math]\displaystyle{ d }[/math] induces on [math]\displaystyle{ X }[/math] is the discrete topology, which makes [math]\displaystyle{ (X, \tau) }[/math] into a commutative topological group under addition but does not form a vector topology on [math]\displaystyle{ X }[/math] because [math]\displaystyle{ (X, \tau) }[/math] is disconnected but every vector topology is connected. What fails is that scalar multiplication isn't continuous on [math]\displaystyle{ (X, \tau). }[/math]
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 [math]\displaystyle{ \mathcal{N} }[/math] of subsets of a vector space is called additive[5] if for every [math]\displaystyle{ N \in \mathcal{N}, }[/math] there exists some [math]\displaystyle{ U \in \mathcal{N} }[/math] such that [math]\displaystyle{ U + U \subseteq N. }[/math]
Continuity of addition at 0 — If [math]\displaystyle{ (X, +) }[/math] is a group (as all vector spaces are), [math]\displaystyle{ \tau }[/math] is a topology on [math]\displaystyle{ X, }[/math] and [math]\displaystyle{ X \times X }[/math] is endowed with the product topology, then the addition map [math]\displaystyle{ X \times X \to X }[/math] (i.e. the map [math]\displaystyle{ (x, y) \mapsto x + y }[/math]) is continuous at the origin of [math]\displaystyle{ X \times X }[/math] if and only if the set of neighborhoods of the origin in [math]\displaystyle{ (X, \tau) }[/math] is additive. This statement remains true if the word "neighborhood" is replaced by "open neighborhood."[5]
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 [math]\displaystyle{ U_{\bull} = \left(U_i\right)_{i=0}^{\infty} }[/math] be a collection of subsets of a vector space such that [math]\displaystyle{ 0 \in U_i }[/math] and [math]\displaystyle{ U_{i+1} + U_{i+1} \subseteq U_i }[/math] for all [math]\displaystyle{ i \geq 0. }[/math] For all [math]\displaystyle{ u \in U_0, }[/math] let [math]\displaystyle{ \mathbb{S}(u) := \left\{ n_{\bull} = \left(n_1, \ldots, n_k\right) ~:~ k \geq 1, n_i \geq 0 \text{ for all } i, \text{ and } u \in U_{n_1} + \cdots + U_{n_k}\right\}. }[/math]
Define [math]\displaystyle{ f : X \to [0, 1] }[/math] by [math]\displaystyle{ f(x) = 1 }[/math] if [math]\displaystyle{ x \not\in U_0 }[/math] and otherwise let [math]\displaystyle{ f(x) := \inf_{} \left\{ 2^{- n_1} + \cdots 2^{- n_k} ~:~ n_{\bull} = \left(n_1, \ldots, n_k\right) \in \mathbb{S}(x)\right\}. }[/math]
Then [math]\displaystyle{ f }[/math] is subadditive (meaning [math]\displaystyle{ f(x + y) \leq f(x) + f(y) \text{ for all } x, y \in X }[/math]) and [math]\displaystyle{ f = 0 }[/math] on [math]\displaystyle{ \bigcap_{i \geq 0} U_i, }[/math] so in particular [math]\displaystyle{ f(0) = 0. }[/math] If all [math]\displaystyle{ U_i }[/math] are symmetric sets then [math]\displaystyle{ f(-x) = f(x) }[/math] and if all [math]\displaystyle{ U_i }[/math] are balanced then [math]\displaystyle{ f(s x) \leq f(x) }[/math] for all scalars [math]\displaystyle{ s }[/math] such that [math]\displaystyle{ |s| \leq 1 }[/math] and all [math]\displaystyle{ x \in X. }[/math] If [math]\displaystyle{ X }[/math] is a topological vector space and if all [math]\displaystyle{ U_i }[/math] are neighborhoods of the origin then [math]\displaystyle{ f }[/math] is continuous, where if in addition [math]\displaystyle{ X }[/math] is Hausdorff and [math]\displaystyle{ U_{\bull} }[/math] forms a basis of balanced neighborhoods of the origin in [math]\displaystyle{ X }[/math] then [math]\displaystyle{ d(x, y) := f(x - y) }[/math] is a metric defining the vector topology on [math]\displaystyle{ X. }[/math]
Proof
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Assume that [math]\displaystyle{ n_{\bull} = \left(n_1, \ldots, n_k\right) }[/math] always denotes a finite sequence of non-negative integers and use the notation: [math]\displaystyle{ \sum 2^{- n_{\bull}} := 2^{- n_1} + \cdots + 2^{- n_k} \quad \text{ and } \quad \sum U_{n_{\bull}} := U_{n_1} + \cdots + U_{n_k}. }[/math] For any integers [math]\displaystyle{ n \geq 0 }[/math] and [math]\displaystyle{ d \gt 2, }[/math] [math]\displaystyle{ U_n \supseteq U_{n+1} + U_{n+1} \supseteq U_{n+1} + U_{n+2} + U_{n+2} \supseteq U_{n+1} + U_{n+2} + \cdots + U_{n+d} + U_{n+d+1} + U_{n+d+1}. }[/math] From this it follows that if [math]\displaystyle{ n_{\bull} = \left(n_1, \ldots, n_k\right) }[/math] consists of distinct positive integers then [math]\displaystyle{ \sum U_{n_{\bull}} \subseteq U_{-1 + \min \left(n_{\bull}\right)}. }[/math] It will now be shown by induction on [math]\displaystyle{ k }[/math] that if [math]\displaystyle{ n_{\bull} = \left(n_1, \ldots, n_k\right) }[/math] consists of non-negative integers such that [math]\displaystyle{ \sum 2^{- n_{\bull}} \leq 2^{- M} }[/math] for some integer [math]\displaystyle{ M \geq 0 }[/math] then [math]\displaystyle{ \sum U_{n_{\bull}} \subseteq U_M. }[/math] This is clearly true for [math]\displaystyle{ k = 1 }[/math] and [math]\displaystyle{ k = 2 }[/math] so assume that [math]\displaystyle{ k \gt 2, }[/math] which implies that all [math]\displaystyle{ n_i }[/math] are positive. If all [math]\displaystyle{ n_i }[/math] are distinct then this step is done, and otherwise pick distinct indices [math]\displaystyle{ i \lt j }[/math] such that [math]\displaystyle{ n_i = n_j }[/math] and construct [math]\displaystyle{ m_{\bull} = \left(m_1, \ldots, m_{k-1}\right) }[/math] from [math]\displaystyle{ n_{\bull} }[/math] by replacing each [math]\displaystyle{ n_i }[/math] with [math]\displaystyle{ n_i - 1 }[/math] and deleting the [math]\displaystyle{ j^{\text{th}} }[/math] element of [math]\displaystyle{ n_{\bull} }[/math] (all other elements of [math]\displaystyle{ n_{\bull} }[/math] are transferred to [math]\displaystyle{ m_{\bull} }[/math] unchanged). Observe that [math]\displaystyle{ \sum 2^{- n_{\bull}} = \sum 2^{- m_{\bull}} }[/math] and [math]\displaystyle{ \sum U_{n_{\bull}} \subseteq \sum U_{m_{\bull}} }[/math] (because [math]\displaystyle{ U_{n_i} + U_{n_j} \subseteq U_{n_i - 1} }[/math]) so by appealing to the inductive hypothesis we conclude that [math]\displaystyle{ \sum U_{n_{\bull}} \subseteq \sum U_{m_{\bull}} \subseteq U_M, }[/math] as desired. It is clear that [math]\displaystyle{ f(0) = 0 }[/math] and that [math]\displaystyle{ 0 \leq f \leq 1 }[/math] so to prove that [math]\displaystyle{ f }[/math] is subadditive, it suffices to prove that [math]\displaystyle{ f(x + y) \leq f(x) + f(y) }[/math] when [math]\displaystyle{ x, y \in X }[/math] are such that [math]\displaystyle{ f(x) + f(y) \lt 1, }[/math] which implies that [math]\displaystyle{ x, y \in U_0. }[/math] This is an exercise. If all [math]\displaystyle{ U_i }[/math] are symmetric then [math]\displaystyle{ x \in \sum U_{n_{\bull}} }[/math] if and only if [math]\displaystyle{ - x \in \sum U_{n_{\bull}} }[/math] from which it follows that [math]\displaystyle{ f(-x) \leq f(x) }[/math] and [math]\displaystyle{ f(-x) \geq f(x). }[/math] If all [math]\displaystyle{ U_i }[/math] are balanced then the inequality [math]\displaystyle{ f(s x) \leq f(x) }[/math] for all unit scalars [math]\displaystyle{ s }[/math] such that [math]\displaystyle{ |s| \leq 1 }[/math] is proved similarly. Because [math]\displaystyle{ f }[/math] is a nonnegative subadditive function satisfying [math]\displaystyle{ f(0) = 0, }[/math] as described in the article on sublinear functionals, [math]\displaystyle{ f }[/math] is uniformly continuous on [math]\displaystyle{ X }[/math] if and only if [math]\displaystyle{ f }[/math] is continuous at the origin. If all [math]\displaystyle{ U_i }[/math] are neighborhoods of the origin then for any real [math]\displaystyle{ r \gt 0, }[/math] pick an integer [math]\displaystyle{ M \gt 1 }[/math] such that [math]\displaystyle{ 2^{-M} \lt r }[/math] so that [math]\displaystyle{ x \in U_M }[/math] implies [math]\displaystyle{ f(x) \leq 2^{-M} \lt r. }[/math] If the set of all [math]\displaystyle{ U_i }[/math] form basis of balanced neighborhoods of the origin then it may be shown that for any [math]\displaystyle{ n \gt 1, }[/math] there exists some [math]\displaystyle{ 0 \lt r \leq 2^{-n} }[/math] such that [math]\displaystyle{ f(x) \lt r }[/math] implies [math]\displaystyle{ x \in U_n. }[/math] [math]\displaystyle{ \blacksquare }[/math] |
If [math]\displaystyle{ X }[/math] is a vector space over the real or complex numbers then a paranorm on [math]\displaystyle{ X }[/math] is a G-seminorm (defined above) [math]\displaystyle{ p : X \rarr \R }[/math] on [math]\displaystyle{ X }[/math] that satisfies any of the following additional conditions, each of which begins with "for all sequences [math]\displaystyle{ x_{\bull} = \left(x_i\right)_{i=1}^{\infty} }[/math] in [math]\displaystyle{ X }[/math] and all convergent sequences of scalars [math]\displaystyle{ s_{\bull} = \left(s_i\right)_{i=1}^{\infty} }[/math]":[6]
A paranorm is called total if in addition it satisfies:
If [math]\displaystyle{ p }[/math] is a paranorm on a vector space [math]\displaystyle{ X }[/math] then the map [math]\displaystyle{ d : X \times X \rarr \R }[/math] defined by [math]\displaystyle{ d(x, y) := p(x - y) }[/math] is a translation-invariant pseudometric on [math]\displaystyle{ X }[/math] that defines a vector topology on [math]\displaystyle{ X. }[/math][8]
If [math]\displaystyle{ p }[/math] is a paranorm on a vector space [math]\displaystyle{ X }[/math] then:
If [math]\displaystyle{ X }[/math] is a vector space over the real or complex numbers then an F-seminorm on [math]\displaystyle{ X }[/math] (the [math]\displaystyle{ F }[/math] stands for Fréchet) is a real-valued map [math]\displaystyle{ p : X \to \Reals }[/math] with the following four properties: [11]
An F-seminorm is called an F-norm if in addition it satisfies:
An F-seminorm is called monotone if it satisfies:
An F-seminormed space (resp. F-normed space)[12] is a pair [math]\displaystyle{ (X, p) }[/math] consisting of a vector space [math]\displaystyle{ X }[/math] and an F-seminorm (resp. F-norm) [math]\displaystyle{ p }[/math] on [math]\displaystyle{ X. }[/math]
If [math]\displaystyle{ (X, p) }[/math] and [math]\displaystyle{ (Z, q) }[/math] are F-seminormed spaces then a map [math]\displaystyle{ f : X \to Z }[/math] is called an isometric embedding[12] if [math]\displaystyle{ q(f(x) - f(y)) = p(x, y) \text{ for all } x, y \in X. }[/math]
Every isometric embedding of one F-seminormed space into another is a topological embedding, but the converse is not true in general.[12]
Every F-seminorm is a paranorm and every paranorm is equivalent to some F-seminorm.[7] Every F-seminorm on a vector space [math]\displaystyle{ X }[/math] is a value on [math]\displaystyle{ X. }[/math] In particular, [math]\displaystyle{ p(x) = 0, }[/math] and [math]\displaystyle{ p(x) = p(-x) }[/math] for all [math]\displaystyle{ x \in X. }[/math]
Theorem[11] — Let [math]\displaystyle{ p }[/math] be an F-seminorm on a vector space [math]\displaystyle{ X. }[/math] Then the map [math]\displaystyle{ d : X \times X \to \Reals }[/math] defined by [math]\displaystyle{ d(x, y) := p(x - y) }[/math] is a translation invariant pseudometric on [math]\displaystyle{ X }[/math] that defines a vector topology [math]\displaystyle{ \tau }[/math] on [math]\displaystyle{ X. }[/math] If [math]\displaystyle{ p }[/math] is an F-norm then [math]\displaystyle{ d }[/math] is a metric. When [math]\displaystyle{ X }[/math] is endowed with this topology then [math]\displaystyle{ p }[/math] is a continuous map on [math]\displaystyle{ X. }[/math]
The balanced sets [math]\displaystyle{ \{x \in X ~:~ p(x) \leq r\}, }[/math] as [math]\displaystyle{ r }[/math] ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of closed set. Similarly, the balanced sets [math]\displaystyle{ \{x \in X ~:~ p(x) \lt r\}, }[/math] as [math]\displaystyle{ r }[/math] ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of open sets.
Suppose that [math]\displaystyle{ \mathcal{L} }[/math] is a non-empty collection of F-seminorms on a vector space [math]\displaystyle{ X }[/math] and for any finite subset [math]\displaystyle{ \mathcal{F} \subseteq \mathcal{L} }[/math] and any [math]\displaystyle{ r \gt 0, }[/math] let [math]\displaystyle{ U_{\mathcal{F}, r} := \bigcap_{p \in \mathcal{F}} \{x \in X : p(x) \lt r\}. }[/math]
The set [math]\displaystyle{ \left\{U_{\mathcal{F}, r} ~:~ r \gt 0, \mathcal{F} \subseteq \mathcal{L}, \mathcal{F} \text{ finite }\right\} }[/math] forms a filter base on [math]\displaystyle{ X }[/math] that also forms a neighborhood basis at the origin for a vector topology on [math]\displaystyle{ X }[/math] denoted by [math]\displaystyle{ \tau_{\mathcal{L}}. }[/math][12] Each [math]\displaystyle{ U_{\mathcal{F}, r} }[/math] is a balanced and absorbing subset of [math]\displaystyle{ X. }[/math][12] These sets satisfy[12] [math]\displaystyle{ U_{\mathcal{F}, r/2} + U_{\mathcal{F}, r/2} \subseteq U_{\mathcal{F}, r}. }[/math]
Suppose that [math]\displaystyle{ p_{\bull} = \left(p_i\right)_{i=1}^{\infty} }[/math] is a family of non-negative subadditive functions on a vector space [math]\displaystyle{ X. }[/math]
The Fréchet combination[8] of [math]\displaystyle{ p_{\bull} }[/math] is defined to be the real-valued map [math]\displaystyle{ p(x) := \sum_{i=1}^{\infty} \frac{p_i(x)}{2^{i} \left[ 1 + p_i(x)\right]}. }[/math]
Assume that [math]\displaystyle{ p_{\bull} = \left(p_i\right)_{i=1}^{\infty} }[/math] is an increasing sequence of seminorms on [math]\displaystyle{ X }[/math] and let [math]\displaystyle{ p }[/math] be the Fréchet combination of [math]\displaystyle{ p_{\bull}. }[/math] Then [math]\displaystyle{ p }[/math] is an F-seminorm on [math]\displaystyle{ X }[/math] that induces the same locally convex topology as the family [math]\displaystyle{ p_{\bull} }[/math] of seminorms.[13]
Since [math]\displaystyle{ p_{\bull} = \left(p_i\right)_{i=1}^{\infty} }[/math] is increasing, a basis of open neighborhoods of the origin consists of all sets of the form [math]\displaystyle{ \left\{ x \in X ~:~ p_i(x) \lt r\right\} }[/math] as [math]\displaystyle{ i }[/math] ranges over all positive integers and [math]\displaystyle{ r \gt 0 }[/math] ranges over all positive real numbers.
The translation invariant pseudometric on [math]\displaystyle{ X }[/math] induced by this F-seminorm [math]\displaystyle{ p }[/math] is [math]\displaystyle{ d(x, y) = \sum^{\infty}_{i=1} \frac{1}{2^i} \frac{p_i( x - y )}{1 + p_i( x - y )}. }[/math]
This metric was discovered by Fréchet in his 1906 thesis for the spaces of real and complex sequences with pointwise operations.[14]
If each [math]\displaystyle{ p_i }[/math] is a paranorm then so is [math]\displaystyle{ p }[/math] and moreover, [math]\displaystyle{ p }[/math] induces the same topology on [math]\displaystyle{ X }[/math] as the family [math]\displaystyle{ p_{\bull} }[/math] of paranorms.[8] This is also true of the following paranorms on [math]\displaystyle{ X }[/math]:
The Fréchet combination can be generalized by use of a bounded remetrization function.
A bounded remetrization function[15] is a continuous non-negative non-decreasing map [math]\displaystyle{ R : [0, \infty) \to [0, \infty) }[/math] that has a bounded range, is subadditive (meaning that [math]\displaystyle{ R(s + t) \leq R(s) + R(t) }[/math] for all [math]\displaystyle{ s, t \geq 0 }[/math]), and satisfies [math]\displaystyle{ R(s) = 0 }[/math] if and only if [math]\displaystyle{ s = 0. }[/math]
Examples of bounded remetrization functions include [math]\displaystyle{ \arctan t, }[/math] [math]\displaystyle{ \tanh t, }[/math] [math]\displaystyle{ t \mapsto \min \{t, 1\}, }[/math] and [math]\displaystyle{ t \mapsto \frac{t}{1 + t}. }[/math][15] If [math]\displaystyle{ d }[/math] is a pseudometric (respectively, metric) on [math]\displaystyle{ X }[/math] and [math]\displaystyle{ R }[/math] is a bounded remetrization function then [math]\displaystyle{ R \circ d }[/math] is a bounded pseudometric (respectively, bounded metric) on [math]\displaystyle{ X }[/math] that is uniformly equivalent to [math]\displaystyle{ d. }[/math][15]
Suppose that [math]\displaystyle{ p_\bull = \left(p_i\right)_{i=1}^\infty }[/math] is a family of non-negative F-seminorm on a vector space [math]\displaystyle{ X, }[/math] [math]\displaystyle{ R }[/math] is a bounded remetrization function, and [math]\displaystyle{ r_\bull = \left(r_i\right)_{i=1}^\infty }[/math] is a sequence of positive real numbers whose sum is finite. Then [math]\displaystyle{ p(x) := \sum_{i=1}^\infty r_i R\left(p_i(x)\right) }[/math] defines a bounded F-seminorm that is uniformly equivalent to the [math]\displaystyle{ p_\bull. }[/math][16] It has the property that for any net [math]\displaystyle{ x_\bull = \left(x_a\right)_{a \in A} }[/math] in [math]\displaystyle{ X, }[/math] [math]\displaystyle{ p\left(x_\bull\right) \to 0 }[/math] if and only if [math]\displaystyle{ p_i\left(x_\bull\right) \to 0 }[/math] for all [math]\displaystyle{ i. }[/math][16] [math]\displaystyle{ p }[/math] is an F-norm if and only if the [math]\displaystyle{ p_\bull }[/math] separate points on [math]\displaystyle{ X. }[/math][16]
A pseudometric (resp. metric) [math]\displaystyle{ d }[/math] is induced by a seminorm (resp. norm) on a vector space [math]\displaystyle{ X }[/math] if and only if [math]\displaystyle{ d }[/math] is translation invariant and absolutely homogeneous, which means that for all scalars [math]\displaystyle{ s }[/math] and all [math]\displaystyle{ x, y \in X, }[/math] in which case the function defined by [math]\displaystyle{ p(x) := d(x, 0) }[/math] is a seminorm (resp. norm) and the pseudometric (resp. metric) induced by [math]\displaystyle{ p }[/math] is equal to [math]\displaystyle{ d. }[/math]
If [math]\displaystyle{ (X, \tau) }[/math] is a topological vector space (TVS) (where note in particular that [math]\displaystyle{ \tau }[/math] is assumed to be a vector topology) then the following are equivalent:[11]
If [math]\displaystyle{ (X, \tau) }[/math] is a TVS then the following are equivalent:
Birkhoff–Kakutani theorem — If [math]\displaystyle{ (X, \tau) }[/math] is a topological vector space then the following three conditions are equivalent:[17][note 1]
By the Birkhoff–Kakutani theorem, it follows that there is an equivalent metric that is translation-invariant.
If [math]\displaystyle{ (X, \tau) }[/math] is TVS then the following are equivalent:[13]
Let [math]\displaystyle{ M }[/math] be a vector subspace of a topological vector space [math]\displaystyle{ (X, \tau). }[/math]
If [math]\displaystyle{ X }[/math] is Hausdorff locally convex TVS then [math]\displaystyle{ X }[/math] with the strong topology, [math]\displaystyle{ \left(X, b\left(X, X^{\prime}\right)\right), }[/math] is metrizable if and only if there exists a countable set [math]\displaystyle{ \mathcal{B} }[/math] of bounded subsets of [math]\displaystyle{ X }[/math] such that every bounded subset of [math]\displaystyle{ X }[/math] is contained in some element of [math]\displaystyle{ \mathcal{B}. }[/math][22]
The strong dual space [math]\displaystyle{ X_b^{\prime} }[/math] of a metrizable locally convex space (such as a Fréchet space[23]) [math]\displaystyle{ X }[/math] is a DF-space.[24] The strong dual of a DF-space is a Fréchet space.[25] The strong dual of a reflexive Fréchet space is a bornological space.[24] The strong bidual (that is, the strong dual space of the strong dual space) of a metrizable locally convex space is a Fréchet space.[26] If [math]\displaystyle{ X }[/math] is a metrizable locally convex space then its strong dual [math]\displaystyle{ X_b^{\prime} }[/math] has one of the following properties, if and only if it has all of these properties: (1) bornological, (2) infrabarreled, (3) barreled.[26]
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.[14] 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 [math]\displaystyle{ M }[/math] is a metrizable locally convex TVS that possess a countable fundamental system of bounded sets, then [math]\displaystyle{ M }[/math] is normable.[27]
If [math]\displaystyle{ X }[/math] is a Hausdorff locally convex space then the following are equivalent:
and if this locally convex space [math]\displaystyle{ X }[/math] is also metrizable, then the following may be appended to this list:
In particular, if a metrizable locally convex space [math]\displaystyle{ X }[/math] (such as a Fréchet space) is not normable then its strong dual space [math]\displaystyle{ X^{\prime}_b }[/math] is not a Fréchet–Urysohn space and consequently, this complete Hausdorff locally convex space [math]\displaystyle{ X^{\prime}_b }[/math] is also neither metrizable nor normable.
Another consequence of this is that if [math]\displaystyle{ X }[/math] is a reflexive locally convex TVS whose strong dual [math]\displaystyle{ X^{\prime}_b }[/math] is metrizable then [math]\displaystyle{ X^{\prime}_b }[/math] is necessarily a reflexive Fréchet space, [math]\displaystyle{ X }[/math] is a DF-space, both [math]\displaystyle{ X }[/math] and [math]\displaystyle{ X^{\prime}_b }[/math] are necessarily complete Hausdorff ultrabornological distinguished webbed spaces, and moreover, [math]\displaystyle{ X^{\prime}_b }[/math] is normable if and only if [math]\displaystyle{ X }[/math] is normable if and only if [math]\displaystyle{ X }[/math] is Fréchet–Urysohn if and only if [math]\displaystyle{ X }[/math] is metrizable. In particular, such a space [math]\displaystyle{ X }[/math] is either a Banach space or else it is not even a Fréchet–Urysohn space.
Suppose that [math]\displaystyle{ (X, d) }[/math] is a pseudometric space and [math]\displaystyle{ B \subseteq X. }[/math] The set [math]\displaystyle{ B }[/math] is metrically bounded or [math]\displaystyle{ d }[/math]-bounded if there exists a real number [math]\displaystyle{ R \gt 0 }[/math] such that [math]\displaystyle{ d(x, y) \leq R }[/math] for all [math]\displaystyle{ x, y \in B }[/math]; the smallest such [math]\displaystyle{ R }[/math] is then called the diameter or [math]\displaystyle{ d }[/math]-diameter of [math]\displaystyle{ B. }[/math][14] If [math]\displaystyle{ B }[/math] is bounded in a pseudometrizable TVS [math]\displaystyle{ X }[/math] then it is metrically bounded; the converse is in general false but it is true for locally convex metrizable TVSs.[14]
Theorem[29] — All infinite-dimensional separable complete metrizable TVS are homeomorphic.
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 [math]\displaystyle{ X }[/math] is a metrizable TVS and [math]\displaystyle{ d }[/math] is a metric that defines [math]\displaystyle{ X }[/math]'s topology, then its possible that [math]\displaystyle{ X }[/math] is complete as a TVS (i.e. relative to its uniformity) but the metric [math]\displaystyle{ d }[/math] is not a complete metric (such metrics exist even for [math]\displaystyle{ X = \R }[/math]). Thus, if [math]\displaystyle{ X }[/math] is a TVS whose topology is induced by a pseudometric [math]\displaystyle{ d, }[/math] then the notion of completeness of [math]\displaystyle{ X }[/math] (as a TVS) and the notion of completeness of the pseudometric space [math]\displaystyle{ (X, d) }[/math] are not always equivalent. The next theorem gives a condition for when they are equivalent:
Theorem — If [math]\displaystyle{ X }[/math] is a pseudometrizable TVS whose topology is induced by a translation invariant pseudometric [math]\displaystyle{ d, }[/math] then [math]\displaystyle{ d }[/math] is a complete pseudometric on [math]\displaystyle{ X }[/math] if and only if [math]\displaystyle{ X }[/math] is complete as a TVS.[36]
Theorem[37][38] (Klee) — Let [math]\displaystyle{ d }[/math] be any[note 2] metric on a vector space [math]\displaystyle{ X }[/math] such that the topology [math]\displaystyle{ \tau }[/math] induced by [math]\displaystyle{ d }[/math] on [math]\displaystyle{ X }[/math] makes [math]\displaystyle{ (X, \tau) }[/math] into a topological vector space. If [math]\displaystyle{ (X, d) }[/math] is a complete metric space then [math]\displaystyle{ (X, \tau) }[/math] is a complete-TVS.
Theorem — If [math]\displaystyle{ X }[/math] is a TVS whose topology is induced by a paranorm [math]\displaystyle{ p, }[/math] then [math]\displaystyle{ X }[/math] is complete if and only if for every sequence [math]\displaystyle{ \left(x_i\right)_{i=1}^{\infty} }[/math] in [math]\displaystyle{ X, }[/math] if [math]\displaystyle{ \sum_{i=1}^{\infty} p\left(x_i\right) \lt \infty }[/math] then [math]\displaystyle{ \sum_{i=1}^{\infty} x_i }[/math] converges in [math]\displaystyle{ X. }[/math][39]
If [math]\displaystyle{ M }[/math] is a closed vector subspace of a complete pseudometrizable TVS [math]\displaystyle{ X, }[/math] then the quotient space [math]\displaystyle{ X / M }[/math] is complete.[40] If [math]\displaystyle{ M }[/math] is a complete vector subspace of a metrizable TVS [math]\displaystyle{ X }[/math] and if the quotient space [math]\displaystyle{ X / M }[/math] is complete then so is [math]\displaystyle{ X. }[/math][40] If [math]\displaystyle{ X }[/math] is not complete then [math]\displaystyle{ M := X, }[/math] but not complete, vector subspace of [math]\displaystyle{ X. }[/math]
A Baire separable topological group is metrizable if and only if it is cosmic.[23]
Banach-Saks theorem[45] — If [math]\displaystyle{ \left(x_n\right)_{n=1}^{\infty} }[/math] is a sequence in a locally convex metrizable TVS [math]\displaystyle{ (X, \tau) }[/math] that converges weakly to some [math]\displaystyle{ x \in X, }[/math] then there exists a sequence [math]\displaystyle{ y_{\bull} = \left(y_i\right)_{i=1}^{\infty} }[/math] in [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ y_{\bull} \to x }[/math] in [math]\displaystyle{ (X, \tau) }[/math] and each [math]\displaystyle{ y_i }[/math] is a convex combination of finitely many [math]\displaystyle{ x_n. }[/math]
Mackey's countability condition[14] — Suppose that [math]\displaystyle{ X }[/math] is a locally convex metrizable TVS and that [math]\displaystyle{ \left(B_i\right)_{i=1}^{\infty} }[/math] is a countable sequence of bounded subsets of [math]\displaystyle{ X. }[/math] Then there exists a bounded subset [math]\displaystyle{ B }[/math] of [math]\displaystyle{ X }[/math] and a sequence [math]\displaystyle{ \left(r_i\right)_{i=1}^{\infty} }[/math] of positive real numbers such that [math]\displaystyle{ B_i \subseteq r_i B }[/math] for all [math]\displaystyle{ i. }[/math]
Generalized series
As described in this article's section on generalized series, for any [math]\displaystyle{ I }[/math]-indexed family family [math]\displaystyle{ \left(r_i\right)_{i \in I} }[/math] of vectors from a TVS [math]\displaystyle{ X, }[/math] it is possible to define their sum [math]\displaystyle{ \textstyle\sum\limits_{i \in I} r_i }[/math] as the limit of the net of finite partial sums [math]\displaystyle{ F \in \operatorname{FiniteSubsets}(I) \mapsto \textstyle\sum\limits_{i \in F} r_i }[/math] where the domain [math]\displaystyle{ \operatorname{FiniteSubsets}(I) }[/math] is directed by [math]\displaystyle{ \,\subseteq.\, }[/math] If [math]\displaystyle{ I = \N }[/math] and [math]\displaystyle{ X = \Reals, }[/math] for instance, then the generalized series [math]\displaystyle{ \textstyle\sum\limits_{i \in \N} r_i }[/math] converges if and only if [math]\displaystyle{ \textstyle\sum\limits_{i=1}^\infty r_i }[/math] converges unconditionally in the usual sense (which for real numbers, is equivalent to absolute convergence). If a generalized series [math]\displaystyle{ \textstyle\sum\limits_{i \in I} r_i }[/math] converges in a metrizable TVS, then the set [math]\displaystyle{ \left\{i \in I : r_i \neq 0\right\} }[/math] is necessarily countable (that is, either finite or countably infinite);[proof 1] in other words, all but at most countably many [math]\displaystyle{ r_i }[/math] will be zero and so this generalized series [math]\displaystyle{ \textstyle\sum\limits_{i \in I} r_i ~=~ \textstyle\sum\limits_{\stackrel{i \in I}{r_i \neq 0}} r_i }[/math] is actually a sum of at most countably many non-zero terms.
If [math]\displaystyle{ X }[/math] is a pseudometrizable TVS and [math]\displaystyle{ A }[/math] maps bounded subsets of [math]\displaystyle{ X }[/math] to bounded subsets of [math]\displaystyle{ Y, }[/math] then [math]\displaystyle{ A }[/math] is continuous.[14] Discontinuous linear functionals exist on any infinite-dimensional pseudometrizable TVS.[46] Thus, a pseudometrizable TVS is finite-dimensional if and only if its continuous dual space is equal to its algebraic dual space.[46]
If [math]\displaystyle{ F : X \to Y }[/math] is a linear map between TVSs and [math]\displaystyle{ X }[/math] is metrizable then the following are equivalent:
Open and almost open maps
A vector subspace [math]\displaystyle{ M }[/math] of a TVS [math]\displaystyle{ X }[/math] has the extension property if any continuous linear functional on [math]\displaystyle{ M }[/math] can be extended to a continuous linear functional on [math]\displaystyle{ X. }[/math][22] Say that a TVS [math]\displaystyle{ X }[/math] has the Hahn-Banach extension property (HBEP) if every vector subspace of [math]\displaystyle{ X }[/math] has the extension property.[22]
The Hahn-Banach theorem guarantees that every Hausdorff locally convex space has the HBEP. For complete metrizable TVSs there is a converse:
Theorem (Kalton) — Every complete metrizable TVS with the Hahn-Banach extension property is locally convex.[22]
If a vector space [math]\displaystyle{ X }[/math] has uncountable dimension and if we endow it with the finest vector topology then this is a TVS with the HBEP that is neither locally convex or metrizable.[22]
Proofs
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Categories: [Topological vector spaces]