Metrizable topological vector space

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 $X$ is a map $d : X \times X \rarr \R$ satisfying the following properties:

1. $d(x, x) = 0 \text x \in X$;


2. Symmetry: $d(x, y) = d(y, x) \text x, y \in X$;


3. Subadditivity: $d(x, z) \leq d(x, y) + d(y, z) \text x, y, z \in X.$

A pseudometric is called a metric if it satisfies:

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

Ultrapseudometric

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

5. Strong/Ultrametric triangle inequality: $d(x, z) \leq \max \ \text x, y, z \in X.$

Pseudometric space

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

Topology induced by a pseudometric

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

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

Pseudometrizable space

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

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

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

1. Translation invariance: $d(x + z, y + z) = d(x, y) \text x, y, z \in X$;


2. $d(x, y) = d(x - y, 0) \text x, y \in X.$

Value/G-seminorm

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

1. Non-negative: $p \geq 0.$


2. Subadditive: $p(x + y) \leq p(x) + p(y) \text x, y \in X$;


3. $p(0) = 0..$


4. Symmetric: $p(-x) = p(x) \text x \in X.$

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

5. Total/Positive definite: If $p(x) = 0$ then $x = 0.$

Properties of values

If $p$ is a value on a vector space $X$ then:

• $, p(x) - p(y), \leq p(x - y) \text x, y \in X.$


• $p(n x) \leq n p(x)$ and $\frac p(x) \leq p(x / n)$ for all $x \in X$ and positive integers $n.$


• The set $\$ is an additive subgroup of $X.$

Equivalence on topological groups

Pseudometrizable topological groups

An invariant pseudometric that doesn't induce a vector topology

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

This example shows that a translation-invariant (pseudo)metric is enough to guarantee a vector topology, which leads us to define paranorms and ''F''-seminorms.

Additive sequences

A collection $\mathcal$ of subsets of a vector space is called additive if for every $N \in \mathcal,$ there exists some $U \in \mathcal$ such that $U + U \subseteq N.$

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.

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

For any integers $n \geq 0$ and $d > 2,$
$U_n \supseteq U_ + U_ \supseteq U_ + U_ + U_ \supseteq U_ + U_ + \cdots + U_ + U_ + U_.$

From this it follows that if $n_ = \left(n_1, \ldots, n_k\right)$ consists of distinct positive integers then $\sum U_ \subseteq U_.$

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

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

Paranorms

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

1. Continuity of multiplication: if $s$ is a scalar and $x \in X$ are such that $p\left(x_i - x\right) \to 0$ and $s_ \to s,$ then $p\left(s_i x_i - s x\right) \to 0.$


2. Both of the conditions:
   * if $s_ \to 0$ and if $x \in X$ is such that $p\left(x_i - x\right) \to 0$ then $p\left(s_i x_i\right) \to 0$;
   * if $p\left(x_\right) \to 0$ then $p\left(s x_i\right) \to 0$ for every scalar $s.$


3. Both of the conditions:
   * if $p\left(x_\right) \to 0$ and $s_ \to s$ for some scalar $s$ then $p\left(s_i x_i\right) \to 0$;
   * if $s_ \to 0$ then $p\left(s_i x\right) \to 0 \text x \in X.$


4. Separate continuity:
   * if $s_ \to s$ for some scalar $s$ then $p\left(s x_i - s x\right) \to 0$ for every $x \in X$;
   * if $s$ is a scalar, $x \in X,$ and $p\left(x_i - x\right) \to 0$ then $p\left(s x_i - s x\right) \to 0$ .

A paranorm is called total if in addition it satisfies:

• Total/Positive definite: $p(x) = 0$ implies $x = 0.$

Properties of paranorms

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

If $p$ is a paranorm on a vector space $X$ then:

• the set $\$ is a vector subspace of $X.$


• $p(x + n) = p(x) \text x, n \in X$ with $p(n) = 0.$


• If a paranorm $p$ satisfies $p(s x) \leq , s, p(x) \text x \in X$ and scalars $s,$ then $p$ is absolutely homogeneity (i.e. equality holds) and thus $p$ is a seminorm.

Examples of paranorms

• If $d$ is a translation-invariant pseudometric on a vector space $X$ that induces a vector topology $\tau$ on $X$ (i.e. $(X, \tau)$ is a TVS) then the map $p(x) := d(x - y, 0)$ defines a continuous paranorm on $(X, \tau)$; moreover, the topology that this paranorm $p$ defines in $X$ is $\tau.$


• If $p$ is a paranorm on $X$ then so is the map q(x) := p(x) / + p(x)


• Every positive scalar multiple of a paranorm (resp. total paranorm) is again such a paranorm (resp. total paranorm).


• Every seminorm is a paranorm.


• The restriction of an paranorm (resp. total paranorm) to a vector subspace is an paranorm (resp. total paranorm).


• The sum of two paranorms is a paranorm.


• If $p$ and $q$ are paranorms on $X$ then so is $(p \wedge q)(x) := \inf_ \.$ Moreover, $(p \wedge q) \leq p$ and $(p \wedge q) \leq q.$ This makes the set of paranorms on $X$ into a conditionally complete lattice.


• Each of the following real-valued maps are paranorms on $X := \R^2$:
  * $(x, y) \mapsto , x,$
  * $(x, y) \mapsto , x, + , y,$


• The real-valued maps $(x, y) \mapsto \sqrt$ and $(x, y) \mapsto \left, x^2 - y^2\^$ are paranorms on $X := \R^2.$


• If $x_ = \left(x_i\right)_$ is a Hamel basis on a vector space $X$ then the real-valued map that sends $x = \sum_ s_i x_i \in X$ (where all but finitely many of the scalars $s_i$ are 0) to $\sum_ \sqrt$ is a paranorm on $X,$ which satisfies $p(sx) = \sqrt p(x)$ for all $x \in X$ and scalars $s.$


• The function $p(x) := , \sin (\pi x), + \min \$ is a paranorm on $\R$ that is balanced but nevertheless equivalent to the usual norm on $R.$ Note that the function $x \mapsto , \sin (\pi x),$ is subadditive.


• Let $X_$ be a complex vector space and let $X_$ denote $X_$ considered as a vector space over $\R.$ Any paranorm on $X_$ is also a paranorm on $X_.$


• 
  

''F''-seminorms

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

1. Non-negative: $p \geq 0.$


2. Subadditive: $p(x + y) \leq p(x) + p(y)$ for all $x, y \in X$


3. Balanced: $p(a x) \leq p(x)$ for $x \in X$ all scalars $a$ satisfying $, a, \leq 1;$
   * This condition guarantees that each set of the form $\$ or $\$ for some $r \geq 0$ is a balanced set.


4. For every $x \in X,$ $p\left(\tfrac x\right) \to 0$ as $n \to \infty$
   * The sequence $\left(\tfrac\right)_^\infty$ can be replaced by any positive sequence converging to the zero.

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

5. Total/Positive definite: $p(x) = 0$ implies $x = 0.$

An ''F''-seminorm is called monotone if it satisfies:

6. Monotone: $p(r x) < p(s x)$ for all non-zero $x \in X$ and all real $s$ and $t$ such that $s < t.$

''F''-seminormed spaces

An ''F''-seminormed space (resp. ''F''-normed space) is a pair $(X, p)$ consisting of a vector space $X$ and an ''F''-seminorm (resp. ''F''-norm) $p$ on $X.$

If $(X, p)$ and $(Z, q)$ are ''F''-seminormed spaces then a map $f : X \to Z$ is called an isometric embedding if $q(f(x) - f(y)) = p(x, y) \text x, y \in X.$

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

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 $p$ and $q$ are ''F''-seminorms on $X$ then so is their pointwise supremum $x \mapsto \sup \.$ The same is true of the supremum of any non-empty finite family of ''F''-seminorms on $X.$


• The restriction of an ''F''-seminorm (resp. ''F''-norm) to a vector subspace is an ''F''-seminorm (resp. ''F''-norm).


• A non-negative real-valued function on $X$ 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. In particular, every seminorm is an ''F''-seminorm.


• For any $0 < p < 1,$ the map $f$ on $\Reals^n$ defined by
  \left(x_1, \ldots, x_n\right)p = \left, x_1\^p + \cdots \left, x_n\^p
  is an ''F''-norm that is not a norm.


• If $L : X \to Y$ is a linear map and if $q$ is an ''F''-seminorm on $Y,$ then $q \circ L$ is an ''F''-seminorm on $X.$


• Let $X_\Complex$ be a complex vector space and let $X_\Reals$ denote $X_\Complex$ considered as a vector space over $\Reals.$ Any ''F''-seminorm on $X_\Complex$ is also an ''F''-seminorm on $X_\Reals.$

Properties of ''F''-seminorms

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

Topology induced by a single ''F''-seminorm

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

Suppose that $\mathcal$ is a non-empty collection of ''F''-seminorms on a vector space $X$ and for any finite subset $\mathcal \subseteq \mathcal$ and any $r > 0,$ let
$U_ := \bigcap_ \.$

The set $\left\$ forms a filter base on $X$ that also forms a neighborhood basis at the origin for a vector topology on $X$ denoted by $\tau_.$ Each $U_$ is a balanced and absorbing subset of $X.$ These sets satisfy
$U_ + U_ \subseteq U_.$

• $\tau_$ is the coarsest vector topology on $X$ making each $p \in \mathcal$ continuous.


• $\tau_$ is Hausdorff if and only if for every non-zero $x \in X,$ there exists some $p \in \mathcal$ such that $p(x) > 0.$


• If $\mathcal$ is the set of all continuous ''F''-seminorms on $\left(X, \tau_\right)$ then $\tau_ = \tau_.$


• If $\mathcal$ is the set of all pointwise suprema of non-empty finite subsets of $\mathcal$ of $\mathcal$ then $\mathcal$ is a directed family of ''F''-seminorms and $\tau_ = \tau_.$

Fréchet combination

Suppose that $p_ = \left(p_i\right)_^$ is a family of non-negative subadditive functions on a vector space $X.$

The Fréchet combination of $p_$ is defined to be the real-valued map
$p(x) := \sum_^ \frac.$

As an ''F''-seminorm

Assume that $p_ = \left(p_i\right)_^$ is an increasing sequence of seminorms on $X$ and let $p$ be the Fréchet combination of $p_.$
Then $p$ is an ''F''-seminorm on $X$ that induces the same locally convex topology as the family $p_$ of seminorms.

Since $p_ = \left(p_i\right)_^$ is increasing, a basis of open neighborhoods of the origin consists of all sets of the form $\left\$ as $i$ ranges over all positive integers and $r > 0$ ranges over all positive real numbers.

The translation invariant pseudometric on $X$ induced by this ''F''-seminorm $p$ is
$d(x, y) = \sum^_ \frac \frac.$

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

As a paranorm

If each $p_i$ is a paranorm then so is $p$ and moreover, $p$ induces the same topology on $X$ as the family $p_$ of paranorms.
This is also true of the following paranorms on $X$:

• $q(x) := \inf_ \left\.$


• $r(x) := \sum_^ \min \left\.$

Generalization

The Fréchet combination can be generalized by use of a bounded remetrization function.

A is a continuous non-negative non-decreasing map R : subadditive (meaning that $R(s + t) \leq R(s) + R(t)$ for all $s, t \geq 0$), and satisfies $R(s) = 0$ if and only if $s = 0.$

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

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

Characterizations

Of (pseudo)metrics induced by (semi)norms

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

Of pseudometrizable TVS

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

1. $X$ is pseudometrizable (i.e. the vector topology $\tau$ is induced by a pseudometric on $X$).


2. $X$ has a countable neighborhood base at the origin.


3. The topology on $X$ is induced by a translation-invariant pseudometric on $X.$


4. The topology on $X$ is induced by an ''F''-seminorm.


5. The topology on $X$ is induced by a paranorm.

Of metrizable TVS

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

1. $X$ is metrizable.


2. $X$ is Hausdorff space">Hausdorff and pseudometrizable.


3. $X$ is Hausdorff and has a countable neighborhood base at the origin.


4. The topology on $X$ is induced by a translation-invariant metric on $X.$


5. The topology on $X$ is induced by an ''F''-norm.


6. The topology on $X$ is induced by a monotone ''F''-norm.


7. The topology on $X$ is induced by a total paranorm.

Of locally convex pseudometrizable TVS

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

1. $X$ is locally convex and pseudometrizable.


2. $X$ has a countable neighborhood base at the origin consisting of convex sets.


3. The topology of $X$ is induced by a countable family of (continuous) seminorms.


4. The topology of $X$ is induced by a countable increasing sequence of (continuous) seminorms $\left(p_i\right)_^$ (increasing means that for all $i,$ $p_i \geq p_.$


5. The topology of $X$ is induced by an ''F''-seminorm of the form:
   $p(x) = \sum_^ 2^ \operatorname p_n(x)$
   where $\left(p_i\right)_^$ are (continuous) seminorms on $X.$

Quotients

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

• If $X$ is a pseudometrizable TVS then so is $X / M.$


• If $X$ is a complete pseudometrizable TVS and $M$ is a closed vector subspace of $X$ then $X / M$ is complete.


• If $X$ is metrizable TVS and $M$ is a closed vector subspace of $X$ then $X / M$ is metrizable.


• If $p$ is an ''F''-seminorm on $X,$ then the map $P : X / M \to \R$ defined by
  $P(x + M) := \inf_ \$
  is an ''F''-seminorm on $X / M$ that induces the usual quotient topology on $X / M.$ If in addition $p$ is an ''F''-norm on $X$ and if $M$ is a closed vector subspace of $X$ then $P$ is an ''F''-norm on $X.$

Examples and sufficient conditions

• Every seminormed space $(X, p)$ is pseudometrizable with a canonical pseudometric given by $d(x, y) := p(x - y)$ for all $x, y \in X.$.


• If $(X, d)$ is pseudometric ''TVS'' with a translation invariant pseudometric $d,$ then $p(x) := d(x, 0)$ defines a paranorm. However, if $d$ is a translation invariant pseudometric on the vector space $X$ (without the addition condition that $(X, d)$ is ), then $d$ need not be either an ''F''-seminorm nor a paranorm.


• If a TVS has a bounded neighborhood of the origin then it is pseudometrizable; the converse is in general false.


• If a Hausdorff TVS has a bounded neighborhood of the origin then it is metrizable.


• Suppose $X$ is either a DF-space or an LM-space. If $X$ is a sequential space then it is either metrizable or else a Montel DF-space.

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

The strong dual space $X_b^$ of a metrizable locally convex space (such as a Fréchet spaceGabriyelyan, S.S
"On topological spaces and topological groups with certain local countable networks
(2014)) $X$ is a DF-space.
The strong dual of a DF-space is a Fréchet space.
The strong dual of a reflexive Fréchet space is a bornological space.
The strong bidual (that is, the strong dual space of the strong dual space) of a metrizable locally convex space is a Fréchet space.
If $X$ is a metrizable locally convex space then its strong dual $X_b^$ has one of the following properties, if and only if it has all of these properties: (1) bornological, (2) infrabarreled, (3) barreled.

Normability

A topological vector space is seminormable if and only if it has a convex bounded neighborhood of the origin.
Moreover, a TVS is normable if and only if it is Hausdorff and seminormable.
Every metrizable TVS on a finite- dimensional vector space is a normable locally convex complete TVS, being TVS-isomorphic to Euclidean space. Consequently, any metrizable TVS that is normable must be infinite dimensional.

If $M$ is a metrizable locally convex TVS that possess a countable fundamental system of bounded sets, then $M$ is normable.

If $X$ is a Hausdorff locally convex space then the following are equivalent:

1. $X$ is normable.


2. $X$ has a (von Neumann) bounded neighborhood of the origin.


3. the strong dual space $X^_b$ of $X$ is normable.

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

4. the strong dual space of $X$ is metrizable.


5. the strong dual space of $X$ is a Fréchet–Urysohn locally convex space.

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

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

Metrically bounded sets and bounded sets

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

Properties of pseudometrizable TVS

• Every metrizable locally convex TVS is a quasibarrelled space, bornological space, and a Mackey space.


• Every complete metrizable TVS is a barrelled space and a Baire space (and hence non-meager). However, there exist metrizable Baire spaces that are not complete.


• If $X$ is a metrizable locally convex space, then the strong dual of $X$ is bornological if and only if it is barreled, if and only if it is infrabarreled.


• If $X$ is a complete pseudometrizable TVS and $M$ is a closed vector subspace of $X,$ then $X / M$ is complete.


• The strong dual of a locally convex metrizable TVS is a webbed space.


• If $(X, \tau)$ and $(X, \nu)$ are complete metrizable TVSs (i.e. F-spaces) and if $\nu$ is coarser than $\tau$ then $\tau = \nu$; this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete. Said differently, if $(X, \tau)$ and $(X, \nu)$ are both F-spaces but with different topologies, then neither one of $\tau$ and $\nu$ contains the other as a subset. One particular consequence of this is, for example, that if $(X, p)$ is a Banach space and $(X, q)$ is some other normed space whose norm-induced topology is finer than (or alternatively, is coarser than) that of $(X, p)$ (i.e. if $p \leq C q$ or if $q \leq C p$ for some constant $C > 0$), then the only way that $(X, q)$ can be a Banach space (i.e. also be complete) is if these two norms $p$ and $q$ are equivalent; if they are not equivalent, then $(X, q)$ can not be a Banach space.
  As another consequence, if $(X, p)$ is a Banach space and $(X, \nu)$ is a Fréchet space, then the map $p : (X, \nu) \to \R$ is continuous if and only if the Fréchet space $(X, \nu)$ the TVS $(X, p)$ (here, the Banach space $(X, p)$ 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.


• Any product of complete metrizable TVSs is a Baire space.


• A product of metrizable TVSs is metrizable if and only if it all but at most countably many of these TVSs have dimension $0.$


• 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 metrizable TVS is a barrelled space and a Baire space (and thus non-meager).


• The dimension of a complete metrizable TVS is either finite or uncountable.

Completeness

Every topological vector space (and more generally, a topological group) has a canonical uniform structure, induced by its topology, which allows the notions of completeness and uniform continuity to be applied to it.
If $X$ is a metrizable TVS and $d$ is a metric that defines $X$'s topology, then its possible that $X$ is complete as a TVS (i.e. relative to its uniformity) but the metric $d$ is a complete metric (such metrics exist even for $X = \R$).
Thus, if $X$ is a TVS whose topology is induced by a pseudometric $d,$ then the notion of completeness of $X$ (as a TVS) and the notion of completeness of the pseudometric space $(X, d)$ are not always equivalent.
The next theorem gives a condition for when they are equivalent:

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

A Baire separable topological group is metrizable if and only if it is cosmic.Gabriyelyan, S.S
"On topological spaces and topological groups with certain local countable networks
(2014)

Subsets and subsequences

• Let $M$ be a separable locally convex metrizable topological vector space and let $C$ be its completion. If $S$ is a bounded subset of $C$ then there exists a bounded subset $R$ of $X$ such that $S \subseteq \operatorname_C R.$


• Every totally bounded subset of a locally convex metrizable TVS $X$ is contained in the closed convex balanced hull of some sequence in $X$ that converges to $0.$


• In a pseudometrizable TVS, every bornivore is a neighborhood of the origin.


• If $d$ is a translation invariant metric on a vector space $X,$ then $d(n x, 0) \leq n d(x, 0)$ for all $x \in X$ and every positive integer $n.$


• If $\left(x_i\right)_^$ is a null sequence (that is, it converges to the origin) in a metrizable TVS then there exists a sequence $\left(r_i\right)_^$ of positive real numbers diverging to $\infty$ such that $\left(r_i x_i\right)_^ \to 0.$


• A subset of a complete metric space is closed if and only if it is complete. If a space $X$ is not complete, then $X$ is a closed subset of $X$ that is not complete.


• If $X$ is a metrizable locally convex TVS then for every bounded subset $B$ of $X,$ there exists a bounded disk $D$ in $X$ such that $B \subseteq X_D,$ and both $X$ and the auxiliary normed space $X_D$ induce the same subspace topology on $B.$

Generalized series

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

Linear maps

If $X$ is a pseudometrizable TVS and $A$ maps bounded subsets of $X$ to bounded subsets of $Y,$ then $A$ is continuous.
Discontinuous linear functionals exist on any infinite-dimensional pseudometrizable TVS. Thus, a pseudometrizable TVS is finite-dimensional if and only if its continuous dual space is equal to its algebraic dual space.

If $F : X \to Y$ is a linear map between TVSs and $X$ is metrizable then the following are equivalent:

1. $F$ is continuous;


2. $F$ is a (locally) bounded map (that is, $F$ maps (von Neumann) bounded subsets of $X$ to bounded subsets of $Y$);


3. $F$ is sequentially continuous;


4. the image under $F$ of every null sequence in $X$ is a bounded set where by definition, a is a sequence that converges to the origin.


5. $F$ maps null sequences to null sequences;

Open and almost open maps

:Theorem: If $X$ is a complete pseudometrizable TVS, $Y$ is a Hausdorff TVS, and $T : X \to Y$ is a closed and almost open linear surjection, then $T$ is an open map.

:Theorem: If $T : X \to Y$ is a surjective linear operator from a locally convex space $X$ onto a barrelled space $Y$ (e.g. every complete pseudometrizable space is barrelled) then $T$ is almost open.

:Theorem: If $T : X \to Y$ is a surjective linear operator from a TVS $X$ onto a Baire space $Y$ then $T$ is almost open.

:Theorem: Suppose $T : X \to Y$ is a continuous linear operator from a complete pseudometrizable TVS $X$ into a Hausdorff TVS $Y.$ If the image of $T$ is non- meager in $Y$ then $T : X \to Y$ is a surjective open map and $Y$ is a complete metrizable space.

Hahn-Banach extension property

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

The Hahn-Banach theorem guarantees that every Hausdorff locally convex space has the HBEP.
For complete metrizable TVSs there is a converse:

If a vector space $X$ 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.

See also

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Notes

Proofs

References

Bibliography

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{{Metric spaces

Metric spaces
Topological vector spaces