# Metrizable topological vector space

Short description: A topological vector space whose topology can be defined by a metric

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.

## Pseudometrics and metrics

A pseudometric on a set $\displaystyle{ X }$ is a map $\displaystyle{ d : X \times X \rarr \R }$ satisfying the following properties:

1. $\displaystyle{ d(x, x) = 0 \text{ for all } x \in X }$;
2. Symmetry: $\displaystyle{ d(x, y) = d(y, x) \text{ for all } x, y \in X }$;
3. Subadditivity: $\displaystyle{ d(x, z) \leq d(x, y) + d(y, z) \text{ for all } x, y, z \in X. }$

A pseudometric is called a metric if it satisfies:

1. Identity of indiscernibles: for all $\displaystyle{ x, y \in X, }$ if $\displaystyle{ d(x, y) = 0 }$ then $\displaystyle{ x = y. }$

Ultrapseudometric

A pseudometric $\displaystyle{ d }$ on $\displaystyle{ X }$ is called a ultrapseudometric or a strong pseudometric if it satisfies:

1. Strong/Ultrametric triangle inequality: $\displaystyle{ d(x, z) \leq \max \{ d(x, y), d(y, z) \} \text{ for all } x, y, z \in X. }$

Pseudometric space

A pseudometric space is a pair $\displaystyle{ (X, d) }$ consisting of a set $\displaystyle{ X }$ and a pseudometric $\displaystyle{ d }$ on $\displaystyle{ X }$ such that $\displaystyle{ X }$'s topology is identical to the topology on $\displaystyle{ X }$ induced by $\displaystyle{ d. }$ We call a pseudometric space $\displaystyle{ (X, d) }$ a metric space (resp. ultrapseudometric space) when $\displaystyle{ d }$ is a metric (resp. ultrapseudometric).

### Topology induced by a pseudometric

If $\displaystyle{ d }$ is a pseudometric on a set $\displaystyle{ X }$ then collection of open balls: $\displaystyle{ B_r(z) := \{ x \in X : d(x, z) \lt r \} }$ as $\displaystyle{ z }$ ranges over $\displaystyle{ X }$ and $\displaystyle{ r \gt 0 }$ ranges over the positive real numbers, forms a basis for a topology on $\displaystyle{ X }$ that is called the $\displaystyle{ d }$-topology or the pseudometric topology on $\displaystyle{ X }$ induced by $\displaystyle{ d. }$

Convention: If $\displaystyle{ (X, d) }$ is a pseudometric space and $\displaystyle{ X }$ is treated as a topological space, then unless indicated otherwise, it should be assumed that $\displaystyle{ X }$ is endowed with the topology induced by $\displaystyle{ d. }$

Pseudometrizable space

A topological space $\displaystyle{ (X, \tau) }$ is called pseudometrizable (resp. metrizable, ultrapseudometrizable) if there exists a pseudometric (resp. metric, ultrapseudometric) $\displaystyle{ d }$ on $\displaystyle{ X }$ such that $\displaystyle{ \tau }$ is equal to the topology induced by $\displaystyle{ d. }$[1]

## Pseudometrics and values on topological groups

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 $\displaystyle{ \tau }$ on a real or complex vector space $\displaystyle{ X }$ 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 $\displaystyle{ X }$ into a topological vector space).

Every topological vector space (TVS) $\displaystyle{ X }$ is an additive commutative topological group but not all group topologies on $\displaystyle{ X }$ are vector topologies. This is because despite it making addition and negation continuous, a group topology on a vector space $\displaystyle{ X }$ 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 $\displaystyle{ X }$ is an additive group then we say that a pseudometric $\displaystyle{ d }$ on $\displaystyle{ X }$ is translation invariant or just invariant if it satisfies any of the following equivalent conditions:

1. Translation invariance: $\displaystyle{ d(x + z, y + z) = d(x, y) \text{ for all } x, y, z \in X }$;
2. $\displaystyle{ d(x, y) = d(x - y, 0) \text{ for all } x, y \in X. }$

### Value/G-seminorm

If $\displaystyle{ X }$ is a topological group the a value or G-seminorm on $\displaystyle{ X }$ (the G stands for Group) is a real-valued map $\displaystyle{ p : X \rarr \R }$ with the following properties:[2]

1. Non-negative: $\displaystyle{ p \geq 0. }$
2. Subadditive: $\displaystyle{ p(x + y) \leq p(x) + p(y) \text{ for all } x, y \in X }$;
3. $\displaystyle{ p(0) = 0.. }$
4. Symmetric: $\displaystyle{ p(-x) = p(x) \text{ for all } x \in X. }$

where we call a G-seminorm a G-norm if it satisfies the additional condition:

1. Total/Positive definite: If $\displaystyle{ p(x) = 0 }$ then $\displaystyle{ x = 0. }$

#### Properties of values

If $\displaystyle{ p }$ is a value on a vector space $\displaystyle{ X }$ then:

• $\displaystyle{ |p(x) - p(y)| \leq p(x - y) \text{ for all } x, y \in X. }$[3]
• $\displaystyle{ p(n x) \leq n p(x) }$ and $\displaystyle{ \frac{1}{n} p(x) \leq p(x / n) }$ for all $\displaystyle{ x \in X }$ and positive integers $\displaystyle{ n. }$[4]
• The set $\displaystyle{ \{ x \in X : p(x) = 0 \} }$ is an additive subgroup of $\displaystyle{ X. }$[3]

### Equivalence on topological groups

Theorem[2] — Suppose that $\displaystyle{ X }$ is an additive commutative group. If $\displaystyle{ d }$ is a translation invariant pseudometric on $\displaystyle{ X }$ then the map $\displaystyle{ p(x) := d(x, 0) }$ is a value on $\displaystyle{ X }$ called the value associated with $\displaystyle{ d }$, and moreover, $\displaystyle{ d }$ generates a group topology on $\displaystyle{ X }$ (i.e. the $\displaystyle{ d }$-topology on $\displaystyle{ X }$ makes $\displaystyle{ X }$ into a topological group). Conversely, if $\displaystyle{ p }$ is a value on $\displaystyle{ X }$ then the map $\displaystyle{ d(x, y) := p(x - y) }$ is a translation-invariant pseudometric on $\displaystyle{ X }$ and the value associated with $\displaystyle{ d }$ is just $\displaystyle{ p. }$

### Pseudometrizable topological groups

Theorem[2] — If $\displaystyle{ (X, \tau) }$ is an additive commutative topological group then the following are equivalent:

1. $\displaystyle{ \tau }$ is induced by a pseudometric; (i.e. $\displaystyle{ (X, \tau) }$ is pseudometrizable);
2. $\displaystyle{ \tau }$ is induced by a translation-invariant pseudometric;
3. the identity element in $\displaystyle{ (X, \tau) }$ has a countable neighborhood basis.

If $\displaystyle{ (X, \tau) }$ 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.

### An invariant pseudometric that doesn't induce a vector topology

Let $\displaystyle{ X }$ be a non-trivial (i.e. $\displaystyle{ X \neq \{ 0 \} }$) real or complex vector space and let $\displaystyle{ d }$ be the translation-invariant trivial metric on $\displaystyle{ X }$ defined by $\displaystyle{ d(x, x) = 0 }$ and $\displaystyle{ d(x, y) = 1 \text{ for all } x, y \in X }$ such that $\displaystyle{ x \neq y. }$ The topology $\displaystyle{ \tau }$ that $\displaystyle{ d }$ induces on $\displaystyle{ X }$ is the discrete topology, which makes $\displaystyle{ (X, \tau) }$ into a commutative topological group under addition but does not form a vector topology on $\displaystyle{ X }$ because $\displaystyle{ (X, \tau) }$ is disconnected but every vector topology is connected. What fails is that scalar multiplication isn't continuous on $\displaystyle{ (X, \tau). }$

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 $\displaystyle{ \mathcal{N} }$ of subsets of a vector space is called additive[5] if for every $\displaystyle{ N \in \mathcal{N}, }$ there exists some $\displaystyle{ U \in \mathcal{N} }$ such that $\displaystyle{ U + U \subseteq N. }$

Continuity of addition at 0 — If $\displaystyle{ (X, +) }$ is a group (as all vector spaces are), $\displaystyle{ \tau }$ is a topology on $\displaystyle{ X, }$ and $\displaystyle{ X \times X }$ is endowed with the product topology, then the addition map $\displaystyle{ X \times X \to X }$ (i.e. the map $\displaystyle{ (x, y) \mapsto x + y }$) is continuous at the origin of $\displaystyle{ X \times X }$ if and only if the set of neighborhoods of the origin in $\displaystyle{ (X, \tau) }$ 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 $\displaystyle{ U_{\bull} = \left(U_i\right)_{i=0}^{\infty} }$ be a collection of subsets of a vector space such that $\displaystyle{ 0 \in U_i }$ and $\displaystyle{ U_{i+1} + U_{i+1} \subseteq U_i }$ for all $\displaystyle{ i \geq 0. }$ For all $\displaystyle{ u \in U_0, }$ let $\displaystyle{ \mathbb{S}(u) := \left\{ n_{\bull} = \left(n_1, \ldots, n_k\right) ~:~ k \geq 1, n_i \geq 0 \text{ for all } i, \text{ and } u \in U_{n_1} + \cdots + U_{n_k}\right\}. }$

Define $\displaystyle{ f : X \to [0, 1] }$ by $\displaystyle{ f(x) = 1 }$ if $\displaystyle{ x \not\in U_0 }$ and otherwise let $\displaystyle{ f(x) := \inf_{} \left\{ 2^{- n_1} + \cdots 2^{- n_k} ~:~ n_{\bull} = \left(n_1, \ldots, n_k\right) \in \mathbb{S}(x)\right\}. }$

Then $\displaystyle{ f }$ is subadditive (meaning $\displaystyle{ f(x + y) \leq f(x) + f(y) \text{ for all } x, y \in X }$) and $\displaystyle{ f = 0 }$ on $\displaystyle{ \bigcap_{i \geq 0} U_i, }$ so in particular $\displaystyle{ f(0) = 0. }$ If all $\displaystyle{ U_i }$ are symmetric sets then $\displaystyle{ f(-x) = f(x) }$ and if all $\displaystyle{ U_i }$ are balanced then $\displaystyle{ f(s x) \leq f(x) }$ for all scalars $\displaystyle{ s }$ such that $\displaystyle{ |s| \leq 1 }$ and all $\displaystyle{ x \in X. }$ If $\displaystyle{ X }$ is a topological vector space and if all $\displaystyle{ U_i }$ are neighborhoods of the origin then $\displaystyle{ f }$ is continuous, where if in addition $\displaystyle{ X }$ is Hausdorff and $\displaystyle{ U_{\bull} }$ forms a basis of balanced neighborhoods of the origin in $\displaystyle{ X }$ then $\displaystyle{ d(x, y) := f(x - y) }$ is a metric defining the vector topology on $\displaystyle{ X. }$

Proof

Assume that $\displaystyle{ n_{\bull} = \left(n_1, \ldots, n_k\right) }$ always denotes a finite sequence of non-negative integers and use the notation: $\displaystyle{ \sum 2^{- n_{\bull}} := 2^{- n_1} + \cdots + 2^{- n_k} \quad \text{ and } \quad \sum U_{n_{\bull}} := U_{n_1} + \cdots + U_{n_k}. }$

For any integers $\displaystyle{ n \geq 0 }$ and $\displaystyle{ d \gt 2, }$ $\displaystyle{ U_n \supseteq U_{n+1} + U_{n+1} \supseteq U_{n+1} + U_{n+2} + U_{n+2} \supseteq U_{n+1} + U_{n+2} + \cdots + U_{n+d} + U_{n+d+1} + U_{n+d+1}. }$

From this it follows that if $\displaystyle{ n_{\bull} = \left(n_1, \ldots, n_k\right) }$ consists of distinct positive integers then $\displaystyle{ \sum U_{n_{\bull}} \subseteq U_{-1 + \min \left(n_{\bull}\right)}. }$

It will now be shown by induction on $\displaystyle{ k }$ that if $\displaystyle{ n_{\bull} = \left(n_1, \ldots, n_k\right) }$ consists of non-negative integers such that $\displaystyle{ \sum 2^{- n_{\bull}} \leq 2^{- M} }$ for some integer $\displaystyle{ M \geq 0 }$ then $\displaystyle{ \sum U_{n_{\bull}} \subseteq U_M. }$ This is clearly true for $\displaystyle{ k = 1 }$ and $\displaystyle{ k = 2 }$ so assume that $\displaystyle{ k \gt 2, }$ which implies that all $\displaystyle{ n_i }$ are positive. If all $\displaystyle{ n_i }$ are distinct then this step is done, and otherwise pick distinct indices $\displaystyle{ i \lt j }$ such that $\displaystyle{ n_i = n_j }$ and construct $\displaystyle{ m_{\bull} = \left(m_1, \ldots, m_{k-1}\right) }$ from $\displaystyle{ n_{\bull} }$ by replacing each $\displaystyle{ n_i }$ with $\displaystyle{ n_i - 1 }$ and deleting the $\displaystyle{ j^{\text{th}} }$ element of $\displaystyle{ n_{\bull} }$ (all other elements of $\displaystyle{ n_{\bull} }$ are transferred to $\displaystyle{ m_{\bull} }$ unchanged). Observe that $\displaystyle{ \sum 2^{- n_{\bull}} = \sum 2^{- m_{\bull}} }$ and $\displaystyle{ \sum U_{n_{\bull}} \subseteq \sum U_{m_{\bull}} }$ (because $\displaystyle{ U_{n_i} + U_{n_j} \subseteq U_{n_i - 1} }$) so by appealing to the inductive hypothesis we conclude that $\displaystyle{ \sum U_{n_{\bull}} \subseteq \sum U_{m_{\bull}} \subseteq U_M, }$ as desired.

It is clear that $\displaystyle{ f(0) = 0 }$ and that $\displaystyle{ 0 \leq f \leq 1 }$ so to prove that $\displaystyle{ f }$ is subadditive, it suffices to prove that $\displaystyle{ f(x + y) \leq f(x) + f(y) }$ when $\displaystyle{ x, y \in X }$ are such that $\displaystyle{ f(x) + f(y) \lt 1, }$ which implies that $\displaystyle{ x, y \in U_0. }$ This is an exercise. If all $\displaystyle{ U_i }$ are symmetric then $\displaystyle{ x \in \sum U_{n_{\bull}} }$ if and only if $\displaystyle{ - x \in \sum U_{n_{\bull}} }$ from which it follows that $\displaystyle{ f(-x) \leq f(x) }$ and $\displaystyle{ f(-x) \geq f(x). }$ If all $\displaystyle{ U_i }$ are balanced then the inequality $\displaystyle{ f(s x) \leq f(x) }$ for all unit scalars $\displaystyle{ s }$ such that $\displaystyle{ |s| \leq 1 }$ is proved similarly. Because $\displaystyle{ f }$ is a nonnegative subadditive function satisfying $\displaystyle{ f(0) = 0, }$ as described in the article on sublinear functionals, $\displaystyle{ f }$ is uniformly continuous on $\displaystyle{ X }$ if and only if $\displaystyle{ f }$ is continuous at the origin. If all $\displaystyle{ U_i }$ are neighborhoods of the origin then for any real $\displaystyle{ r \gt 0, }$ pick an integer $\displaystyle{ M \gt 1 }$ such that $\displaystyle{ 2^{-M} \lt r }$ so that $\displaystyle{ x \in U_M }$ implies $\displaystyle{ f(x) \leq 2^{-M} \lt r. }$ If the set of all $\displaystyle{ U_i }$ form basis of balanced neighborhoods of the origin then it may be shown that for any $\displaystyle{ n \gt 1, }$ there exists some $\displaystyle{ 0 \lt r \leq 2^{-n} }$ such that $\displaystyle{ f(x) \lt r }$ implies $\displaystyle{ x \in U_n. }$ $\displaystyle{ \blacksquare }$

## Paranorms

If $\displaystyle{ X }$ is a vector space over the real or complex numbers then a paranorm on $\displaystyle{ X }$ is a G-seminorm (defined above) $\displaystyle{ p : X \rarr \R }$ on $\displaystyle{ X }$ that satisfies any of the following additional conditions, each of which begins with "for all sequences $\displaystyle{ x_{\bull} = \left(x_i\right)_{i=1}^{\infty} }$ in $\displaystyle{ X }$ and all convergent sequences of scalars $\displaystyle{ s_{\bull} = \left(s_i\right)_{i=1}^{\infty} }$":[6]

1. Continuity of multiplication: if $\displaystyle{ s }$ is a scalar and $\displaystyle{ x \in X }$ are such that $\displaystyle{ p\left(x_i - x\right) \to 0 }$ and $\displaystyle{ s_{\bull} \to s, }$ then $\displaystyle{ p\left(s_i x_i - s x\right) \to 0. }$
2. Both of the conditions:
• if $\displaystyle{ s_{\bull} \to 0 }$ and if $\displaystyle{ x \in X }$ is such that $\displaystyle{ p\left(x_i - x\right) \to 0 }$ then $\displaystyle{ p\left(s_i x_i\right) \to 0 }$;
• if $\displaystyle{ p\left(x_{\bull}\right) \to 0 }$ then $\displaystyle{ p\left(s x_i\right) \to 0 }$ for every scalar $\displaystyle{ s. }$
3. Both of the conditions:
• if $\displaystyle{ p\left(x_{\bull}\right) \to 0 }$ and $\displaystyle{ s_{\bull} \to s }$ for some scalar $\displaystyle{ s }$ then $\displaystyle{ p\left(s_i x_i\right) \to 0 }$;
• if $\displaystyle{ s_{\bull} \to 0 }$ then $\displaystyle{ p\left(s_i x\right) \to 0 \text{ for all } x \in X. }$
4. Separate continuity:[7]
• if $\displaystyle{ s_{\bull} \to s }$ for some scalar $\displaystyle{ s }$ then $\displaystyle{ p\left(s x_i - s x\right) \to 0 }$ for every $\displaystyle{ x \in X }$;
• if $\displaystyle{ s }$ is a scalar, $\displaystyle{ x \in X, }$ and $\displaystyle{ p\left(x_i - x\right) \to 0 }$ then $\displaystyle{ p\left(s x_i - s x\right) \to 0 }$ .

A paranorm is called total if in addition it satisfies:

• Total/Positive definite: $\displaystyle{ p(x) = 0 }$ implies $\displaystyle{ x = 0. }$

### Properties of paranorms

If $\displaystyle{ p }$ is a paranorm on a vector space $\displaystyle{ X }$ then the map $\displaystyle{ d : X \times X \rarr \R }$ defined by $\displaystyle{ d(x, y) := p(x - y) }$ is a translation-invariant pseudometric on $\displaystyle{ X }$ that defines a vector topology on $\displaystyle{ X. }$[8]

If $\displaystyle{ p }$ is a paranorm on a vector space $\displaystyle{ X }$ then:

• the set $\displaystyle{ \{ x \in X : p(x) = 0 \} }$ is a vector subspace of $\displaystyle{ X. }$[8]
• $\displaystyle{ p(x + n) = p(x) \text{ for all } x, n \in X }$ with $\displaystyle{ p(n) = 0. }$[8]
• If a paranorm $\displaystyle{ p }$ satisfies $\displaystyle{ p(s x) \leq |s| p(x) \text{ for all } x \in X }$ and scalars $\displaystyle{ s, }$ then $\displaystyle{ p }$ is absolutely homogeneity (i.e. equality holds)[8] and thus $\displaystyle{ p }$ is a seminorm.

### Examples of paranorms

• If $\displaystyle{ d }$ is a translation-invariant pseudometric on a vector space $\displaystyle{ X }$ that induces a vector topology $\displaystyle{ \tau }$ on $\displaystyle{ X }$ (i.e. $\displaystyle{ (X, \tau) }$ is a TVS) then the map $\displaystyle{ p(x) := d(x - y, 0) }$ defines a continuous paranorm on $\displaystyle{ (X, \tau) }$; moreover, the topology that this paranorm $\displaystyle{ p }$ defines in $\displaystyle{ X }$ is $\displaystyle{ \tau. }$[8]
• If $\displaystyle{ p }$ is a paranorm on $\displaystyle{ X }$ then so is the map $\displaystyle{ q(x) := p(x) / [1 + p(x)]. }$[8]
• Every positive scalar multiple of a paranorm (resp. total paranorm) is again such a paranorm (resp. total paranorm).
• Every seminorm is a paranorm.[8]
• The restriction of an paranorm (resp. total paranorm) to a vector subspace is an paranorm (resp. total paranorm).[9]
• The sum of two paranorms is a paranorm.[8]
• If $\displaystyle{ p }$ and $\displaystyle{ q }$ are paranorms on $\displaystyle{ X }$ then so is $\displaystyle{ (p \wedge q)(x) := \inf_{} \{ p(y) + q(z) : x = y + z \text{ with } y, z \in X \}. }$ Moreover, $\displaystyle{ (p \wedge q) \leq p }$ and $\displaystyle{ (p \wedge q) \leq q. }$ This makes the set of paranorms on $\displaystyle{ X }$ into a conditionally complete lattice.[8]
• Each of the following real-valued maps are paranorms on $\displaystyle{ X := \R^2 }$:
• $\displaystyle{ (x, y) \mapsto |x| }$
• $\displaystyle{ (x, y) \mapsto |x| + |y| }$
• The real-valued maps $\displaystyle{ (x, y) \mapsto \sqrt{\left|x^2 - y^2\right|} }$ and $\displaystyle{ (x, y) \mapsto \left|x^2 - y^2\right|^{3/2} }$ are not a paranorms on $\displaystyle{ X := \R^2. }$[8]
• If $\displaystyle{ x_{\bull} = \left(x_i\right)_{i \in I} }$ is a Hamel basis on a vector space $\displaystyle{ X }$ then the real-valued map that sends $\displaystyle{ x = \sum_{i \in I} s_i x_i \in X }$ (where all but finitely many of the scalars $\displaystyle{ s_i }$ are 0) to $\displaystyle{ \sum_{i \in I} \sqrt{\left|s_i\right|} }$ is a paranorm on $\displaystyle{ X, }$ which satisfies $\displaystyle{ p(sx) = \sqrt{|s|} p(x) }$ for all $\displaystyle{ x \in X }$ and scalars $\displaystyle{ s. }$[8]
• The function $\displaystyle{ p(x) := |\sin (\pi x)| + \min \{ 2, |x| \} }$is a paranorm on $\displaystyle{ \R }$ that is not balanced but nevertheless equivalent to the usual norm on $\displaystyle{ R. }$ Note that the function $\displaystyle{ x \mapsto |\sin (\pi x)| }$ is subadditive.[10]
• Let $\displaystyle{ X_{\Complex} }$ be a complex vector space and let $\displaystyle{ X_{\R} }$ denote $\displaystyle{ X_{\Complex} }$ considered as a vector space over $\displaystyle{ \R. }$ Any paranorm on $\displaystyle{ X_{\Complex} }$ is also a paranorm on $\displaystyle{ X_{\R}. }$[9]

## F-seminorms

If $\displaystyle{ X }$ is a vector space over the real or complex numbers then an F-seminorm on $\displaystyle{ X }$ (the $\displaystyle{ F }$ stands for Fréchet) is a real-valued map $\displaystyle{ p : X \to \Reals }$ with the following four properties: [11]

1. Non-negative: $\displaystyle{ p \geq 0. }$
2. Subadditive: $\displaystyle{ p(x + y) \leq p(x) + p(y) }$ for all $\displaystyle{ x, y \in X }$
3. Balanced: $\displaystyle{ p(a x) \leq p(x) }$ for $\displaystyle{ x \in X }$ all scalars $\displaystyle{ a }$ satisfying $\displaystyle{ |a| \leq 1; }$
• This condition guarantees that each set of the form $\displaystyle{ \{z \in X : p(z) \leq r\} }$ or $\displaystyle{ \{z \in X : p(z) \lt r\} }$ for some $\displaystyle{ r \geq 0 }$ is a balanced set.
4. For every $\displaystyle{ x \in X, }$ $\displaystyle{ p\left(\tfrac{1}{n} x\right) \to 0 }$ as $\displaystyle{ n \to \infty }$
• The sequence $\displaystyle{ \left(\tfrac{1}{n}\right)_{n=1}^\infty }$ can be replaced by any positive sequence converging to the zero.[12]

An F-seminorm is called an F-norm if in addition it satisfies:

1. Total/Positive definite: $\displaystyle{ p(x) = 0 }$ implies $\displaystyle{ x = 0. }$

An F-seminorm is called monotone if it satisfies:

1. Monotone: $\displaystyle{ p(r x) \lt p(s x) }$ for all non-zero $\displaystyle{ x \in X }$ and all real $\displaystyle{ s }$ and $\displaystyle{ t }$ such that $\displaystyle{ s \lt t. }$[12]

### F-seminormed spaces

An F-seminormed space (resp. F-normed space)[12] is a pair $\displaystyle{ (X, p) }$ consisting of a vector space $\displaystyle{ X }$ and an F-seminorm (resp. F-norm) $\displaystyle{ p }$ on $\displaystyle{ X. }$

If $\displaystyle{ (X, p) }$ and $\displaystyle{ (Z, q) }$ are F-seminormed spaces then a map $\displaystyle{ f : X \to Z }$ is called an isometric embedding[12] if $\displaystyle{ q(f(x) - f(y)) = p(x, y) \text{ for all } x, y \in X. }$

Every isometric embedding of one F-seminormed space into another is a topological embedding, but the converse is not true in general.[12]

### 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 $\displaystyle{ p }$ and $\displaystyle{ q }$ are F-seminorms on $\displaystyle{ X }$ then so is their pointwise supremum $\displaystyle{ x \mapsto \sup \{p(x), q(x)\}. }$ The same is true of the supremum of any non-empty finite family of F-seminorms on $\displaystyle{ X. }$[12]
• The restriction of an F-seminorm (resp. F-norm) to a vector subspace is an F-seminorm (resp. F-norm).[9]
• A non-negative real-valued function on $\displaystyle{ X }$ 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.[10] In particular, every seminorm is an F-seminorm.
• For any $\displaystyle{ 0 \lt p \lt 1, }$ the map $\displaystyle{ f }$ on $\displaystyle{ \Reals^n }$ defined by $\displaystyle{ [f\left(x_1, \ldots, x_n\right)]^p = \left|x_1\right|^p + \cdots \left|x_n\right|^p }$ is an F-norm that is not a norm.
• If $\displaystyle{ L : X \to Y }$ is a linear map and if $\displaystyle{ q }$ is an F-seminorm on $\displaystyle{ Y, }$ then $\displaystyle{ q \circ L }$ is an F-seminorm on $\displaystyle{ X. }$[12]
• Let $\displaystyle{ X_\Complex }$ be a complex vector space and let $\displaystyle{ X_\Reals }$ denote $\displaystyle{ X_\Complex }$ considered as a vector space over $\displaystyle{ \Reals. }$ Any F-seminorm on $\displaystyle{ X_\Complex }$ is also an F-seminorm on $\displaystyle{ X_\Reals. }$[9]

### Properties of F-seminorms

Every F-seminorm is a paranorm and every paranorm is equivalent to some F-seminorm.[7] Every F-seminorm on a vector space $\displaystyle{ X }$ is a value on $\displaystyle{ X. }$ In particular, $\displaystyle{ p(x) = 0, }$ and $\displaystyle{ p(x) = p(-x) }$ for all $\displaystyle{ x \in X. }$

### Topology induced by a single F-seminorm

Theorem[11] — Let $\displaystyle{ p }$ be an F-seminorm on a vector space $\displaystyle{ X. }$ Then the map $\displaystyle{ d : X \times X \to \Reals }$ defined by $\displaystyle{ d(x, y) := p(x - y) }$ is a translation invariant pseudometric on $\displaystyle{ X }$ that defines a vector topology $\displaystyle{ \tau }$ on $\displaystyle{ X. }$ If $\displaystyle{ p }$ is an F-norm then $\displaystyle{ d }$ is a metric. When $\displaystyle{ X }$ is endowed with this topology then $\displaystyle{ p }$ is a continuous map on $\displaystyle{ X. }$

The balanced sets $\displaystyle{ \{x \in X ~:~ p(x) \leq r\}, }$ as $\displaystyle{ r }$ ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of closed set. Similarly, the balanced sets $\displaystyle{ \{x \in X ~:~ p(x) \lt r\}, }$ as $\displaystyle{ r }$ ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of open sets.

### Topology induced by a family of F-seminorms

Suppose that $\displaystyle{ \mathcal{L} }$ is a non-empty collection of F-seminorms on a vector space $\displaystyle{ X }$ and for any finite subset $\displaystyle{ \mathcal{F} \subseteq \mathcal{L} }$ and any $\displaystyle{ r \gt 0, }$ let $\displaystyle{ U_{\mathcal{F}, r} := \bigcap_{p \in \mathcal{F}} \{x \in X : p(x) \lt r\}. }$

The set $\displaystyle{ \left\{U_{\mathcal{F}, r} ~:~ r \gt 0, \mathcal{F} \subseteq \mathcal{L}, \mathcal{F} \text{ finite }\right\} }$ forms a filter base on $\displaystyle{ X }$ that also forms a neighborhood basis at the origin for a vector topology on $\displaystyle{ X }$ denoted by $\displaystyle{ \tau_{\mathcal{L}}. }$[12] Each $\displaystyle{ U_{\mathcal{F}, r} }$ is a balanced and absorbing subset of $\displaystyle{ X. }$[12] These sets satisfy[12] $\displaystyle{ U_{\mathcal{F}, r/2} + U_{\mathcal{F}, r/2} \subseteq U_{\mathcal{F}, r}. }$

• $\displaystyle{ \tau_{\mathcal{L}} }$ is the coarsest vector topology on $\displaystyle{ X }$ making each $\displaystyle{ p \in \mathcal{L} }$ continuous.[12]
• $\displaystyle{ \tau_{\mathcal{L}} }$ is Hausdorff if and only if for every non-zero $\displaystyle{ x \in X, }$ there exists some $\displaystyle{ p \in \mathcal{L} }$ such that $\displaystyle{ p(x) \gt 0. }$[12]
• If $\displaystyle{ \mathcal{F} }$ is the set of all continuous F-seminorms on $\displaystyle{ \left(X, \tau_{\mathcal{L}}\right) }$ then $\displaystyle{ \tau_{\mathcal{L}} = \tau_{\mathcal{F}}. }$[12]
• If $\displaystyle{ \mathcal{F} }$ is the set of all pointwise suprema of non-empty finite subsets of $\displaystyle{ \mathcal{F} }$ of $\displaystyle{ \mathcal{L} }$ then $\displaystyle{ \mathcal{F} }$ is a directed family of F-seminorms and $\displaystyle{ \tau_{\mathcal{L}} = \tau_{\mathcal{F}}. }$[12]

## Fréchet combination

Suppose that $\displaystyle{ p_{\bull} = \left(p_i\right)_{i=1}^{\infty} }$ is a family of non-negative subadditive functions on a vector space $\displaystyle{ X. }$

The Fréchet combination[8] of $\displaystyle{ p_{\bull} }$ is defined to be the real-valued map $\displaystyle{ p(x) := \sum_{i=1}^{\infty} \frac{p_i(x)}{2^{i} \left[ 1 + p_i(x)\right]}. }$

### As an F-seminorm

Assume that $\displaystyle{ p_{\bull} = \left(p_i\right)_{i=1}^{\infty} }$ is an increasing sequence of seminorms on $\displaystyle{ X }$ and let $\displaystyle{ p }$ be the Fréchet combination of $\displaystyle{ p_{\bull}. }$ Then $\displaystyle{ p }$ is an F-seminorm on $\displaystyle{ X }$ that induces the same locally convex topology as the family $\displaystyle{ p_{\bull} }$ of seminorms.[13]

Since $\displaystyle{ p_{\bull} = \left(p_i\right)_{i=1}^{\infty} }$ is increasing, a basis of open neighborhoods of the origin consists of all sets of the form $\displaystyle{ \left\{ x \in X ~:~ p_i(x) \lt r\right\} }$ as $\displaystyle{ i }$ ranges over all positive integers and $\displaystyle{ r \gt 0 }$ ranges over all positive real numbers.

The translation invariant pseudometric on $\displaystyle{ X }$ induced by this F-seminorm $\displaystyle{ p }$ is $\displaystyle{ d(x, y) = \sum^{\infty}_{i=1} \frac{1}{2^i} \frac{p_i( x - y )}{1 + p_i( x - y )}. }$

This metric was discovered by Fréchet in his 1906 thesis for the spaces of real and complex sequences with pointwise operations.[14]

### As a paranorm

If each $\displaystyle{ p_i }$ is a paranorm then so is $\displaystyle{ p }$ and moreover, $\displaystyle{ p }$ induces the same topology on $\displaystyle{ X }$ as the family $\displaystyle{ p_{\bull} }$ of paranorms.[8] This is also true of the following paranorms on $\displaystyle{ X }$:

• $\displaystyle{ q(x) := \inf_{} \left\{ \sum_{i=1}^n p_i(x) + \frac{1}{n} ~:~ n \gt 0 \text{ is an integer }\right\}. }$[8]
• $\displaystyle{ r(x) := \sum_{n=1}^{\infty} \min \left\{ \frac{1}{2^n}, p_n(x)\right\}. }$[8]

### Generalization

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 $\displaystyle{ R : [0, \infty) \to [0, \infty) }$ that has a bounded range, is subadditive (meaning that $\displaystyle{ R(s + t) \leq R(s) + R(t) }$ for all $\displaystyle{ s, t \geq 0 }$), and satisfies $\displaystyle{ R(s) = 0 }$ if and only if $\displaystyle{ s = 0. }$

Examples of bounded remetrization functions include $\displaystyle{ \arctan t, }$ $\displaystyle{ \tanh t, }$ $\displaystyle{ t \mapsto \min \{t, 1\}, }$ and $\displaystyle{ t \mapsto \frac{t}{1 + t}. }$[15] If $\displaystyle{ d }$ is a pseudometric (respectively, metric) on $\displaystyle{ X }$ and $\displaystyle{ R }$ is a bounded remetrization function then $\displaystyle{ R \circ d }$ is a bounded pseudometric (respectively, bounded metric) on $\displaystyle{ X }$ that is uniformly equivalent to $\displaystyle{ d. }$[15]

Suppose that $\displaystyle{ p_\bull = \left(p_i\right)_{i=1}^\infty }$ is a family of non-negative F-seminorm on a vector space $\displaystyle{ X, }$ $\displaystyle{ R }$ is a bounded remetrization function, and $\displaystyle{ r_\bull = \left(r_i\right)_{i=1}^\infty }$ is a sequence of positive real numbers whose sum is finite. Then $\displaystyle{ p(x) := \sum_{i=1}^\infty r_i R\left(p_i(x)\right) }$ defines a bounded F-seminorm that is uniformly equivalent to the $\displaystyle{ p_\bull. }$[16] It has the property that for any net $\displaystyle{ x_\bull = \left(x_a\right)_{a \in A} }$ in $\displaystyle{ X, }$ $\displaystyle{ p\left(x_\bull\right) \to 0 }$ if and only if $\displaystyle{ p_i\left(x_\bull\right) \to 0 }$ for all $\displaystyle{ i. }$[16] $\displaystyle{ p }$ is an F-norm if and only if the $\displaystyle{ p_\bull }$ separate points on $\displaystyle{ X. }$[16]

## Characterizations

### Of (pseudo)metrics induced by (semi)norms

A pseudometric (resp. metric) $\displaystyle{ d }$ is induced by a seminorm (resp. norm) on a vector space $\displaystyle{ X }$ if and only if $\displaystyle{ d }$ is translation invariant and absolutely homogeneous, which means that for all scalars $\displaystyle{ s }$ and all $\displaystyle{ x, y \in X, }$ in which case the function defined by $\displaystyle{ p(x) := d(x, 0) }$ is a seminorm (resp. norm) and the pseudometric (resp. metric) induced by $\displaystyle{ p }$ is equal to $\displaystyle{ d. }$

### Of pseudometrizable TVS

If $\displaystyle{ (X, \tau) }$ is a topological vector space (TVS) (where note in particular that $\displaystyle{ \tau }$ is assumed to be a vector topology) then the following are equivalent:[11]

1. $\displaystyle{ X }$ is pseudometrizable (i.e. the vector topology $\displaystyle{ \tau }$ is induced by a pseudometric on $\displaystyle{ X }$).
2. $\displaystyle{ X }$ has a countable neighborhood base at the origin.
3. The topology on $\displaystyle{ X }$ is induced by a translation-invariant pseudometric on $\displaystyle{ X. }$
4. The topology on $\displaystyle{ X }$ is induced by an F-seminorm.
5. The topology on $\displaystyle{ X }$ is induced by a paranorm.

### Of metrizable TVS

If $\displaystyle{ (X, \tau) }$ is a TVS then the following are equivalent:

1. $\displaystyle{ X }$ is metrizable.
2. $\displaystyle{ X }$ is Hausdorff and pseudometrizable.
3. $\displaystyle{ X }$ is Hausdorff and has a countable neighborhood base at the origin.[11][12]
4. The topology on $\displaystyle{ X }$ is induced by a translation-invariant metric on $\displaystyle{ X. }$[11]
5. The topology on $\displaystyle{ X }$ is induced by an F-norm.[11][12]
6. The topology on $\displaystyle{ X }$ is induced by a monotone F-norm.[12]
7. The topology on $\displaystyle{ X }$ is induced by a total paranorm.

Birkhoff–Kakutani theorem — If $\displaystyle{ (X, \tau) }$ is a topological vector space then the following three conditions are equivalent:[17][note 1]

1. The origin $\displaystyle{ \{ 0 \} }$ is closed in $\displaystyle{ X, }$ and there is a countable basis of neighborhoods for $\displaystyle{ 0 }$ in $\displaystyle{ X. }$
2. $\displaystyle{ (X, \tau) }$ is metrizable (as a topological space).
3. There is a translation-invariant metric on $\displaystyle{ X }$ that induces on $\displaystyle{ X }$ the topology $\displaystyle{ \tau, }$ which is the given topology on $\displaystyle{ X. }$

By the Birkhoff–Kakutani theorem, it follows that there is an equivalent metric that is translation-invariant.

### Of locally convex pseudometrizable TVS

If $\displaystyle{ (X, \tau) }$ is TVS then the following are equivalent:[13]

1. $\displaystyle{ X }$ is locally convex and pseudometrizable.
2. $\displaystyle{ X }$ has a countable neighborhood base at the origin consisting of convex sets.
3. The topology of $\displaystyle{ X }$ is induced by a countable family of (continuous) seminorms.
4. The topology of $\displaystyle{ X }$ is induced by a countable increasing sequence of (continuous) seminorms $\displaystyle{ \left(p_i\right)_{i=1}^{\infty} }$ (increasing means that for all $\displaystyle{ i, }$ $\displaystyle{ p_i \geq p_{i+1}. }$
5. The topology of $\displaystyle{ X }$ is induced by an F-seminorm of the form: $\displaystyle{ p(x) = \sum_{n=1}^{\infty} 2^{-n} \operatorname{arctan} p_n(x) }$ where $\displaystyle{ \left(p_i\right)_{i=1}^{\infty} }$ are (continuous) seminorms on $\displaystyle{ X. }$[18]

## Quotients

Let $\displaystyle{ M }$ be a vector subspace of a topological vector space $\displaystyle{ (X, \tau). }$

• If $\displaystyle{ X }$ is a pseudometrizable TVS then so is $\displaystyle{ X / M. }$[11]
• If $\displaystyle{ X }$ is a complete pseudometrizable TVS and $\displaystyle{ M }$ is a closed vector subspace of $\displaystyle{ X }$ then $\displaystyle{ X / M }$ is complete.[11]
• If $\displaystyle{ X }$ is metrizable TVS and $\displaystyle{ M }$ is a closed vector subspace of $\displaystyle{ X }$ then $\displaystyle{ X / M }$ is metrizable.[11]
• If $\displaystyle{ p }$ is an F-seminorm on $\displaystyle{ X, }$ then the map $\displaystyle{ P : X / M \to \R }$ defined by $\displaystyle{ P(x + M) := \inf_{} \{ p(x + m) : m \in M \} }$ is an F-seminorm on $\displaystyle{ X / M }$ that induces the usual quotient topology on $\displaystyle{ X / M. }$[11] If in addition $\displaystyle{ p }$ is an F-norm on $\displaystyle{ X }$ and if $\displaystyle{ M }$ is a closed vector subspace of $\displaystyle{ X }$ then $\displaystyle{ P }$ is an F-norm on $\displaystyle{ X. }$[11]

## Examples and sufficient conditions

• Every seminormed space $\displaystyle{ (X, p) }$ is pseudometrizable with a canonical pseudometric given by $\displaystyle{ d(x, y) := p(x - y) }$ for all $\displaystyle{ x, y \in X. }$[19].
• If $\displaystyle{ (X, d) }$ is pseudometric TVS with a translation invariant pseudometric $\displaystyle{ d, }$ then $\displaystyle{ p(x) := d(x, 0) }$ defines a paranorm.[20] However, if $\displaystyle{ d }$ is a translation invariant pseudometric on the vector space $\displaystyle{ X }$ (without the addition condition that $\displaystyle{ (X, d) }$ is pseudometric TVS), then $\displaystyle{ d }$ need not be either an F-seminorm[21] nor a paranorm.
• If a TVS has a bounded neighborhood of the origin then it is pseudometrizable; the converse is in general false.[14]
• If a Hausdorff TVS has a bounded neighborhood of the origin then it is metrizable.[14]
• Suppose $\displaystyle{ X }$ is either a DF-space or an LM-space. If $\displaystyle{ X }$ is a sequential space then it is either metrizable or else a Montel DF-space.

If $\displaystyle{ X }$ is Hausdorff locally convex TVS then $\displaystyle{ X }$ with the strong topology, $\displaystyle{ \left(X, b\left(X, X^{\prime}\right)\right), }$ is metrizable if and only if there exists a countable set $\displaystyle{ \mathcal{B} }$ of bounded subsets of $\displaystyle{ X }$ such that every bounded subset of $\displaystyle{ X }$ is contained in some element of $\displaystyle{ \mathcal{B}. }$[22]

The strong dual space $\displaystyle{ X_b^{\prime} }$ of a metrizable locally convex space (such as a Fréchet space[23]) $\displaystyle{ X }$ 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 $\displaystyle{ X }$ is a metrizable locally convex space then its strong dual $\displaystyle{ X_b^{\prime} }$ has one of the following properties, if and only if it has all of these properties: (1) bornological, (2) infrabarreled, (3) barreled.[26]

### 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.[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 $\displaystyle{ M }$ is a metrizable locally convex TVS that possess a countable fundamental system of bounded sets, then $\displaystyle{ M }$ is normable.[27]

If $\displaystyle{ X }$ is a Hausdorff locally convex space then the following are equivalent:

1. $\displaystyle{ X }$ is normable.
2. $\displaystyle{ X }$ has a (von Neumann) bounded neighborhood of the origin.
3. the strong dual space $\displaystyle{ X^{\prime}_b }$ of $\displaystyle{ X }$ is normable.[28]

and if this locally convex space $\displaystyle{ X }$ is also metrizable, then the following may be appended to this list:

1. the strong dual space of $\displaystyle{ X }$ is metrizable.[28]
2. the strong dual space of $\displaystyle{ X }$ is a Fréchet–Urysohn locally convex space.[23]

In particular, if a metrizable locally convex space $\displaystyle{ X }$ (such as a Fréchet space) is not normable then its strong dual space $\displaystyle{ X^{\prime}_b }$ is not a Fréchet–Urysohn space and consequently, this complete Hausdorff locally convex space $\displaystyle{ X^{\prime}_b }$ is also neither metrizable nor normable.

Another consequence of this is that if $\displaystyle{ X }$ is a reflexive locally convex TVS whose strong dual $\displaystyle{ X^{\prime}_b }$ is metrizable then $\displaystyle{ X^{\prime}_b }$ is necessarily a reflexive Fréchet space, $\displaystyle{ X }$ is a DF-space, both $\displaystyle{ X }$ and $\displaystyle{ X^{\prime}_b }$ are necessarily complete Hausdorff ultrabornological distinguished webbed spaces, and moreover, $\displaystyle{ X^{\prime}_b }$ is normable if and only if $\displaystyle{ X }$ is normable if and only if $\displaystyle{ X }$ is Fréchet–Urysohn if and only if $\displaystyle{ X }$ is metrizable. In particular, such a space $\displaystyle{ X }$ is either a Banach space or else it is not even a Fréchet–Urysohn space.

## Metrically bounded sets and bounded sets

Suppose that $\displaystyle{ (X, d) }$ is a pseudometric space and $\displaystyle{ B \subseteq X. }$ The set $\displaystyle{ B }$ is metrically bounded or $\displaystyle{ d }$-bounded if there exists a real number $\displaystyle{ R \gt 0 }$ such that $\displaystyle{ d(x, y) \leq R }$ for all $\displaystyle{ x, y \in B }$; the smallest such $\displaystyle{ R }$ is then called the diameter or $\displaystyle{ d }$-diameter of $\displaystyle{ B. }$[14] If $\displaystyle{ B }$ is bounded in a pseudometrizable TVS $\displaystyle{ X }$ then it is metrically bounded; the converse is in general false but it is true for locally convex metrizable TVSs.[14]

## Properties of pseudometrizable TVS

Theorem[29] — All infinite-dimensional separable complete metrizable TVS are homeomorphic.

• Every metrizable locally convex TVS is a quasibarrelled space,[30] bornological space, and a Mackey space.
• Every complete pseudometrizable TVS is a barrelled space and a Baire space (and hence non-meager).[31] However, there exist metrizable Baire spaces that are not complete.[31]
• If $\displaystyle{ X }$ is a metrizable locally convex space, then the strong dual of $\displaystyle{ X }$ is bornological if and only if it is barreled, if and only if it is infrabarreled.[26]
• If $\displaystyle{ X }$ is a complete pseudometrizable TVS and $\displaystyle{ M }$ is a closed vector subspace of $\displaystyle{ X, }$ then $\displaystyle{ X / M }$ is complete.[11]
• The strong dual of a locally convex metrizable TVS is a webbed space.[32]
• If $\displaystyle{ (X, \tau) }$ and $\displaystyle{ (X, \nu) }$ are complete metrizable TVSs (i.e. F-spaces) and if $\displaystyle{ \nu }$ is coarser than $\displaystyle{ \tau }$ then $\displaystyle{ \tau = \nu }$;[33] this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete.[34] Said differently, if $\displaystyle{ (X, \tau) }$ and $\displaystyle{ (X, \nu) }$ are both F-spaces but with different topologies, then neither one of $\displaystyle{ \tau }$ and $\displaystyle{ \nu }$ contains the other as a subset. One particular consequence of this is, for example, that if $\displaystyle{ (X, p) }$ is a Banach space and $\displaystyle{ (X, q) }$ is some other normed space whose norm-induced topology is finer than (or alternatively, is coarser than) that of $\displaystyle{ (X, p) }$ (i.e. if $\displaystyle{ p \leq C q }$ or if $\displaystyle{ q \leq C p }$ for some constant $\displaystyle{ C \gt 0 }$), then the only way that $\displaystyle{ (X, q) }$ can be a Banach space (i.e. also be complete) is if these two norms $\displaystyle{ p }$ and $\displaystyle{ q }$ are equivalent; if they are not equivalent, then $\displaystyle{ (X, q) }$ can not be a Banach space. As another consequence, if $\displaystyle{ (X, p) }$ is a Banach space and $\displaystyle{ (X, \nu) }$ is a Fréchet space, then the map $\displaystyle{ p : (X, \nu) \to \R }$ is continuous if and only if the Fréchet space $\displaystyle{ (X, \nu) }$ is the TVS $\displaystyle{ (X, p) }$ (here, the Banach space $\displaystyle{ (X, p) }$ 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.[31]
• A product of metrizable TVSs is metrizable if and only if it all but at most countably many of these TVSs have dimension $\displaystyle{ 0. }$[35]
• 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).[31]
• The dimension of a complete metrizable TVS is either finite or uncountable.[35]

### Completeness

Main page: Complete topological vector space

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 $\displaystyle{ X }$ is a metrizable TVS and $\displaystyle{ d }$ is a metric that defines $\displaystyle{ X }$'s topology, then its possible that $\displaystyle{ X }$ is complete as a TVS (i.e. relative to its uniformity) but the metric $\displaystyle{ d }$ is not a complete metric (such metrics exist even for $\displaystyle{ X = \R }$). Thus, if $\displaystyle{ X }$ is a TVS whose topology is induced by a pseudometric $\displaystyle{ d, }$ then the notion of completeness of $\displaystyle{ X }$ (as a TVS) and the notion of completeness of the pseudometric space $\displaystyle{ (X, d) }$ are not always equivalent. The next theorem gives a condition for when they are equivalent:

Theorem — If $\displaystyle{ X }$ is a pseudometrizable TVS whose topology is induced by a translation invariant pseudometric $\displaystyle{ d, }$ then $\displaystyle{ d }$ is a complete pseudometric on $\displaystyle{ X }$ if and only if $\displaystyle{ X }$ is complete as a TVS.[36]

Theorem[37][38] (Klee) — Let $\displaystyle{ d }$ be any[note 2] metric on a vector space $\displaystyle{ X }$ such that the topology $\displaystyle{ \tau }$ induced by $\displaystyle{ d }$ on $\displaystyle{ X }$ makes $\displaystyle{ (X, \tau) }$ into a topological vector space. If $\displaystyle{ (X, d) }$ is a complete metric space then $\displaystyle{ (X, \tau) }$ is a complete-TVS.

Theorem — If $\displaystyle{ X }$ is a TVS whose topology is induced by a paranorm $\displaystyle{ p, }$ then $\displaystyle{ X }$ is complete if and only if for every sequence $\displaystyle{ \left(x_i\right)_{i=1}^{\infty} }$ in $\displaystyle{ X, }$ if $\displaystyle{ \sum_{i=1}^{\infty} p\left(x_i\right) \lt \infty }$ then $\displaystyle{ \sum_{i=1}^{\infty} x_i }$ converges in $\displaystyle{ X. }$[39]

If $\displaystyle{ M }$ is a closed vector subspace of a complete pseudometrizable TVS $\displaystyle{ X, }$ then the quotient space $\displaystyle{ X / M }$ is complete.[40] If $\displaystyle{ M }$ is a complete vector subspace of a metrizable TVS $\displaystyle{ X }$ and if the quotient space $\displaystyle{ X / M }$ is complete then so is $\displaystyle{ X. }$[40] If $\displaystyle{ X }$ is not complete then $\displaystyle{ M := X, }$ but not complete, vector subspace of $\displaystyle{ X. }$

A Baire separable topological group is metrizable if and only if it is cosmic.[23]

### Subsets and subsequences

• Let $\displaystyle{ M }$ be a separable locally convex metrizable topological vector space and let $\displaystyle{ C }$ be its completion. If $\displaystyle{ S }$ is a bounded subset of $\displaystyle{ C }$ then there exists a bounded subset $\displaystyle{ R }$ of $\displaystyle{ X }$ such that $\displaystyle{ S \subseteq \operatorname{cl}_C R. }$[41]
• Every totally bounded subset of a locally convex metrizable TVS $\displaystyle{ X }$ is contained in the closed convex balanced hull of some sequence in $\displaystyle{ X }$ that converges to $\displaystyle{ 0. }$
• In a pseudometrizable TVS, every bornivore is a neighborhood of the origin.[42]
• If $\displaystyle{ d }$ is a translation invariant metric on a vector space $\displaystyle{ X, }$ then $\displaystyle{ d(n x, 0) \leq n d(x, 0) }$ for all $\displaystyle{ x \in X }$ and every positive integer $\displaystyle{ n. }$[43]
• If $\displaystyle{ \left(x_i\right)_{i=1}^{\infty} }$ is a null sequence (that is, it converges to the origin) in a metrizable TVS then there exists a sequence $\displaystyle{ \left(r_i\right)_{i=1}^{\infty} }$ of positive real numbers diverging to $\displaystyle{ \infty }$ such that $\displaystyle{ \left(r_i x_i\right)_{i=1}^{\infty} \to 0. }$[43]
• A subset of a complete metric space is closed if and only if it is complete. If a space $\displaystyle{ X }$ is not complete, then $\displaystyle{ X }$ is a closed subset of $\displaystyle{ X }$ that is not complete.
• If $\displaystyle{ X }$ is a metrizable locally convex TVS then for every bounded subset $\displaystyle{ B }$ of $\displaystyle{ X, }$ there exists a bounded disk $\displaystyle{ D }$ in $\displaystyle{ X }$ such that $\displaystyle{ B \subseteq X_D, }$ and both $\displaystyle{ X }$ and the auxiliary normed space $\displaystyle{ X_D }$ induce the same subspace topology on $\displaystyle{ B. }$[44]

Banach-Saks theorem[45] — If $\displaystyle{ \left(x_n\right)_{n=1}^{\infty} }$ is a sequence in a locally convex metrizable TVS $\displaystyle{ (X, \tau) }$ that converges weakly to some $\displaystyle{ x \in X, }$ then there exists a sequence $\displaystyle{ y_{\bull} = \left(y_i\right)_{i=1}^{\infty} }$ in $\displaystyle{ X }$ such that $\displaystyle{ y_{\bull} \to x }$ in $\displaystyle{ (X, \tau) }$ and each $\displaystyle{ y_i }$ is a convex combination of finitely many $\displaystyle{ x_n. }$

Mackey's countability condition[14] — Suppose that $\displaystyle{ X }$ is a locally convex metrizable TVS and that $\displaystyle{ \left(B_i\right)_{i=1}^{\infty} }$ is a countable sequence of bounded subsets of $\displaystyle{ X. }$ Then there exists a bounded subset $\displaystyle{ B }$ of $\displaystyle{ X }$ and a sequence $\displaystyle{ \left(r_i\right)_{i=1}^{\infty} }$ of positive real numbers such that $\displaystyle{ B_i \subseteq r_i B }$ for all $\displaystyle{ i. }$

Generalized series

As described in this article's section on generalized series, for any $\displaystyle{ I }$-indexed family family $\displaystyle{ \left(r_i\right)_{i \in I} }$ of vectors from a TVS $\displaystyle{ X, }$ it is possible to define their sum $\displaystyle{ \textstyle\sum\limits_{i \in I} r_i }$ as the limit of the net of finite partial sums $\displaystyle{ F \in \operatorname{FiniteSubsets}(I) \mapsto \textstyle\sum\limits_{i \in F} r_i }$ where the domain $\displaystyle{ \operatorname{FiniteSubsets}(I) }$ is directed by $\displaystyle{ \,\subseteq.\, }$ If $\displaystyle{ I = \N }$ and $\displaystyle{ X = \Reals, }$ for instance, then the generalized series $\displaystyle{ \textstyle\sum\limits_{i \in \N} r_i }$ converges if and only if $\displaystyle{ \textstyle\sum\limits_{i=1}^\infty r_i }$ converges unconditionally in the usual sense (which for real numbers, is equivalent to absolute convergence). If a generalized series $\displaystyle{ \textstyle\sum\limits_{i \in I} r_i }$ converges in a metrizable TVS, then the set $\displaystyle{ \left\{i \in I : r_i \neq 0\right\} }$ is necessarily countable (that is, either finite or countably infinite);[proof 1] in other words, all but at most countably many $\displaystyle{ r_i }$ will be zero and so this generalized series $\displaystyle{ \textstyle\sum\limits_{i \in I} r_i ~=~ \textstyle\sum\limits_{\stackrel{i \in I}{r_i \neq 0}} r_i }$ is actually a sum of at most countably many non-zero terms.

### Linear maps

If $\displaystyle{ X }$ is a pseudometrizable TVS and $\displaystyle{ A }$ maps bounded subsets of $\displaystyle{ X }$ to bounded subsets of $\displaystyle{ Y, }$ then $\displaystyle{ A }$ 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 $\displaystyle{ F : X \to Y }$ is a linear map between TVSs and $\displaystyle{ X }$ is metrizable then the following are equivalent:

1. $\displaystyle{ F }$ is continuous;
2. $\displaystyle{ F }$ is a (locally) bounded map (that is, $\displaystyle{ F }$ maps (von Neumann) bounded subsets of $\displaystyle{ X }$ to bounded subsets of $\displaystyle{ Y }$);[12]
3. $\displaystyle{ F }$ is sequentially continuous;[12]
4. the image under $\displaystyle{ F }$ of every null sequence in $\displaystyle{ X }$ is a bounded set[12] where by definition, a null sequence is a sequence that converges to the origin.
5. $\displaystyle{ F }$ maps null sequences to null sequences;

Open and almost open maps

Theorem: If $\displaystyle{ X }$ is a complete pseudometrizable TVS, $\displaystyle{ Y }$ is a Hausdorff TVS, and $\displaystyle{ T : X \to Y }$ is a closed and almost open linear surjection, then $\displaystyle{ T }$ is an open map.[47]
Theorem: If $\displaystyle{ T : X \to Y }$ is a surjective linear operator from a locally convex space $\displaystyle{ X }$ onto a barrelled space $\displaystyle{ Y }$ (e.g. every complete pseudometrizable space is barrelled) then $\displaystyle{ T }$ is almost open.[47]
Theorem: If $\displaystyle{ T : X \to Y }$ is a surjective linear operator from a TVS $\displaystyle{ X }$ onto a Baire space $\displaystyle{ Y }$ then $\displaystyle{ T }$ is almost open.[47]
Theorem: Suppose $\displaystyle{ T : X \to Y }$ is a continuous linear operator from a complete pseudometrizable TVS $\displaystyle{ X }$ into a Hausdorff TVS $\displaystyle{ Y. }$ If the image of $\displaystyle{ T }$ is non-meager in $\displaystyle{ Y }$ then $\displaystyle{ T : X \to Y }$ is a surjective open map and $\displaystyle{ Y }$ is a complete metrizable space.[47]

### Hahn-Banach extension property

A vector subspace $\displaystyle{ M }$ of a TVS $\displaystyle{ X }$ has the extension property if any continuous linear functional on $\displaystyle{ M }$ can be extended to a continuous linear functional on $\displaystyle{ X. }$[22] Say that a TVS $\displaystyle{ X }$ has the Hahn-Banach extension property (HBEP) if every vector subspace of $\displaystyle{ X }$ 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 $\displaystyle{ X }$ 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]

## Notes

1. In fact, this is true for topological group, for the proof doesn't use the scalar multiplications.
2. Not assumed to be translation-invariant.

Proofs

1. Suppose the net $\displaystyle{ \textstyle\sum\limits_{i \in I} r_i ~\stackrel{\scriptscriptstyle\text{def}}{=}~ {\textstyle\lim\limits_{A \in \operatorname{FiniteSubsets}(I)}} \ \textstyle\sum\limits_{i \in A} r_i = \lim \left\{\textstyle\sum\limits_{i\in A} r_i \,: A \subseteq I, A \text{ finite }\right\} }$ converges to some point in a metrizable TVS $\displaystyle{ X, }$ where recall that this net's domain is the directed set $\displaystyle{ (\operatorname{FiniteSubsets}(I), \subseteq). }$ Like every convergent net, this convergent net of partial sums $\displaystyle{ A \mapsto \textstyle\sum\limits_{i \in A} r_i }$ is a Cauchy net, which for this particular net means (by definition) that for every neighborhood $\displaystyle{ W }$ of the origin in $\displaystyle{ X, }$ there exists a finite subset $\displaystyle{ A_0 }$ of $\displaystyle{ I }$ such that $\displaystyle{ \textstyle\sum\limits_{i \in B} r_i - \textstyle\sum\limits_{i \in C} r_i \in W }$ for all finite supersets $\displaystyle{ B, C \supseteq A_0; }$ this implies that $\displaystyle{ r_i \in W }$ for every $\displaystyle{ i \in I \setminus A_0 }$ (by taking $\displaystyle{ B := A_0 \cup \{i\} }$ and $\displaystyle{ C := A_0 }$). Since $\displaystyle{ X }$ is metrizable, it has a countable neighborhood basis $\displaystyle{ U_1, U_2, \ldots }$ at the origin, whose intersection is necessarily $\displaystyle{ U_1 \cap U_2 \cap \cdots = \{0\} }$ (since $\displaystyle{ X }$ is a Hausdorff TVS). For every positive integer $\displaystyle{ n \in \N, }$ pick a finite subset $\displaystyle{ A_n \subseteq I }$ such that $\displaystyle{ r_i \in U_n }$ for every $\displaystyle{ i \in I \setminus A_n. }$ If $\displaystyle{ i }$ belongs to $\displaystyle{ (I \setminus A_1) \cap (I \setminus A_2) \cap \cdots = I \setminus \left(A_1 \cup A_2 \cup \cdots\right) }$ then $\displaystyle{ r_i }$ belongs to $\displaystyle{ U_1 \cap U_2 \cap \cdots = \{0\}. }$ Thus $\displaystyle{ r_i = 0 }$ for every index $\displaystyle{ i \in I }$ that does not belong to the countable set $\displaystyle{ A_1 \cup A_2 \cup \cdots. }$ $\displaystyle{ \blacksquare }$

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