Metrizable TVS

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 $d:X\times X\rightarrow \mathbb {R}$ satisfying the following properties:

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

A pseudometric is called a metric if it satisfies:

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:

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

Pseudometric space

A pseudometric space is a pair $(X,d)$ 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: $B_{r}(z):=\{x\in X:d(x,z)<r\}$ 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 $d.$

Convention: 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 $d.$ ^[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 $\tau$ 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:

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

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 $p:X\rightarrow \mathbb {R}$ with the following properties:^[2]

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

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

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

Properties of values

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

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

Equivalence on topological groups

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

Pseudometrizable topological groups

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

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

If $(X,\tau )$ 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 $X$ be a non-trivial (i.e. $X\neq \{0\}$ ) real or complex vector space and let $d$ be the translation-invariant trivial metric on $X$ defined by $d(x,x)=0$ and $d(x,y)=1{\text{ for all }}x,y\in X$ 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 not 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 not enough to guarantee a vector topology, which leads us to define paranorms and F-seminorms.

Additive sequences

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

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

Define $f:X\to [0,1]$ by $f(x)=1$ if $x\not \in U_{0}$ and otherwise let $f(x):=\inf _{}\left\{2^{-n_{1}}+\cdots 2^{-n_{k}}~:~n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)\in \mathbb {S} (x)\right\}.$

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

Proof
Assume that $n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)$ always denotes a finite sequence of non-negative integers and use the notation: $\sum 2^{-n_{\bullet }}:=2^{-n_{1}}+\cdots +2^{-n_{k}}\quad {\text{ and }}\quad \sum U_{n_{\bullet }}:=U_{n_{1}}+\cdots +U_{n_{k}}.$ For any integers $n\geq 0$ and $d>2,$ $U_{n}\supseteq U_{n+1}+U_{n+1}\supseteq U_{n+1}+U_{n+2}+U_{n+2}\supseteq U_{n+1}+U_{n+2}+\cdots +U_{n+d}+U_{n+d+1}+U_{n+d+1}.$ From this it follows that if $n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)$ consists of distinct positive integers then $\sum U_{n_{\bullet }}\subseteq U_{-1+\min \left(n_{\bullet }\right)}.$ It will now be shown by induction on $k$ that if $n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)$ consists of non-negative integers such that $\sum 2^{-n_{\bullet }}\leq 2^{-M}$ for some integer $M\geq 0$ then $\sum U_{n_{\bullet }}\subseteq U_{M}.$ 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 $m_{\bullet }=\left(m_{1},\ldots ,m_{k-1}\right)$ from $n_{\bullet }$ by replacing each $n_{i}$ with $n_{i}-1$ and deleting the $j^{\text{th}}$ element of $n_{\bullet }$ (all other elements of $n_{\bullet }$ are transferred to $m_{\bullet }$ unchanged). Observe that $\sum 2^{-n_{\bullet }}=\sum 2^{-m_{\bullet }}$ and $\sum U_{n_{\bullet }}\subseteq \sum U_{m_{\bullet }}$ (because $U_{n_{i}}+U_{n_{j}}\subseteq U_{n_{i}-1}$ ) so by appealing to the inductive hypothesis we conclude that $\sum U_{n_{\bullet }}\subseteq \sum U_{m_{\bullet }}\subseteq U_{M},$ 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 $x\in \sum U_{n_{\bullet }}$ if and only if $-x\in \sum U_{n_{\bullet }}$ 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 $f(sx)\leq f(x)$ 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 $2^{-M}<r$ so that $x\in U_{M}$ implies $f(x)\leq 2^{-M}<r.$ 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 $0<r\leq 2^{-n}$ such that $f(x)<r$ implies $x\in U_{n}.$ $\blacksquare$

Proof

Assume that $n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)$ always denotes a finite sequence of non-negative integers and use the notation: $\sum 2^{-n_{\bullet }}:=2^{-n_{1}}+\cdots +2^{-n_{k}}\quad {\text{ and }}\quad \sum U_{n_{\bullet }}:=U_{n_{1}}+\cdots +U_{n_{k}}.$

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

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

It will now be shown by induction on $k$ that if $n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)$ consists of non-negative integers such that $\sum 2^{-n_{\bullet }}\leq 2^{-M}$ for some integer $M\geq 0$ then $\sum U_{n_{\bullet }}\subseteq U_{M}.$ 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 $m_{\bullet }=\left(m_{1},\ldots ,m_{k-1}\right)$ from $n_{\bullet }$ by replacing each $n_{i}$ with $n_{i}-1$ and deleting the $j^{\text{th}}$ element of $n_{\bullet }$ (all other elements of $n_{\bullet }$ are transferred to $m_{\bullet }$ unchanged). Observe that $\sum 2^{-n_{\bullet }}=\sum 2^{-m_{\bullet }}$ and $\sum U_{n_{\bullet }}\subseteq \sum U_{m_{\bullet }}$ (because $U_{n_{i}}+U_{n_{j}}\subseteq U_{n_{i}-1}$ ) so by appealing to the inductive hypothesis we conclude that $\sum U_{n_{\bullet }}\subseteq \sum U_{m_{\bullet }}\subseteq U_{M},$ 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 $x\in \sum U_{n_{\bullet }}$ if and only if $-x\in \sum U_{n_{\bullet }}$ 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 $f(sx)\leq f(x)$ 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 $2^{-M}<r$ so that $x\in U_{M}$ implies $f(x)\leq 2^{-M}<r.$ 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 $0<r\leq 2^{-n}$ 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) $p:X\rightarrow \mathbb {R}$ on $X$ that satisfies any of the following additional conditions, each of which begins with "for all sequences $x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }$ in $X$ and all convergent sequences of scalars $s_{\bullet }=\left(s_{i}\right)_{i=1}^{\infty }$ ":^[6]

Continuity of multiplication: if $s$ is a scalar and $x\in X$ are such that $p\left(x_{i}-x\right)\to 0$ and $s_{\bullet }\to s,$ then $p\left(s_{i}x_{i}-sx\right)\to 0.$
Both of the conditions:
- if $s_{\bullet }\to 0$ 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 $p\left(x_{\bullet }\right)\to 0$ then $p\left(sx_{i}\right)\to 0$ for every scalar $s.$
Both of the conditions:
- if $p\left(x_{\bullet }\right)\to 0$ and $s_{\bullet }\to s$ for some scalar $s$ then $p\left(s_{i}x_{i}\right)\to 0$ ;
- if $s_{\bullet }\to 0$ then $p\left(s_{i}x\right)\to 0{\text{ for all }}x\in X.$
Separate continuity:^[7]
- if $s_{\bullet }\to s$ for some scalar $s$ then $p\left(sx_{i}-sx\right)\to 0$ for every $x\in X$ ;
- if $s$ is a scalar, $x\in X,$ and $p\left(x_{i}-x\right)\to 0$ then $p\left(sx_{i}-sx\right)\to 0$ .

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 $d:X\times X\rightarrow \mathbb {R}$ defined by $d(x,y):=p(x-y)$ is a translation-invariant pseudometric on $X$ that defines a vector topology on $X.$ ^[8]

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

the set $\{x\in X:p(x)=0\}$ is a vector subspace of $X.$ ^[8]
$p(x+n)=p(x){\text{ for all }}x,n\in X$ with $p(n)=0.$ ^[8]
If a paranorm $p$ satisfies $p(sx)\leq |s|p(x){\text{ for all }}x\in X$ and scalars $s,$ then $p$ is absolutely homogeneity (i.e. equality holds)^[8] 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 $\tau .$ ^[8]
If $p$ is a paranorm on $X$ then so is the map $q(x):=p(x)/[1+p(x)].$ ^[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 $p$ and $q$ are paranorms on $X$ then so is $(p\wedge q)(x):=\inf _{}\{p(y)+q(z):x=y+z{\text{ with }}y,z\in X\}.$ 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.^[8]
Each of the following real-valued maps are paranorms on X := R 2 {\displaystyle X:=\mathbb {R} ^{2}} $X:=\mathbb {R} ^{2}$ :
- $(x,y)\mapsto |x|$
- $(x,y)\mapsto |x|+|y|$
The real-valued maps $(x,y)\mapsto {\sqrt {\left|x^{2}-y^{2}\right|}}$ and $(x,y)\mapsto \left|x^{2}-y^{2}\right|^{3/2}$ are not paranorms on $X:=\mathbb {R} ^{2}.$ ^[8]
If $x_{\bullet }=\left(x_{i}\right)_{i\in I}$ is a Hamel basis on a vector space $X$ then the real-valued map that sends $x=\sum _{i\in I}s_{i}x_{i}\in X$ (where all but finitely many of the scalars $s_{i}$ are 0) to $\sum _{i\in I}{\sqrt {\left|s_{i}\right|}}$ is a paranorm on $X,$ which satisfies $p(sx)={\sqrt {|s|}}p(x)$ for all $x\in X$ and scalars $s.$ ^[8]
The function $p(x):=|\sin(\pi x)|+\min\{2,|x|\}$ is a paranorm on $\mathbb {R}$ that is not balanced but nevertheless equivalent to the usual norm on $R.$ Note that the function $x\mapsto |\sin(\pi x)|$ is subadditive.^[10]
Let $X_{\mathbb {C} }$ be a complex vector space and let $X_{\mathbb {R} }$ denote $X_{\mathbb {C} }$ considered as a vector space over $\mathbb {R} .$ Any paranorm on $X_{\mathbb {C} }$ is also a paranorm on $X_{\mathbb {R} }.$ ^[9]

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 $p:X\to \mathbb {R}$ with the following four properties: ^[11]

Non-negative: $p\geq 0.$
Subadditive: $p(x+y)\leq p(x)+p(y)$ for all $x,y\in X$
Balanced: p ( a x ) ≤ p ( x ) {\displaystyle p(ax)\leq p(x)} $p(ax)\leq p(x)$ for x ∈ X {\displaystyle x\in X} $x\in X$ all scalars a {\displaystyle a} $a$ satisfying | a | ≤ 1 ; {\displaystyle |a|\leq 1;} $|a|\leq 1;$
- This condition guarantees that each set of the form $\{z\in X:p(z)\leq r\}$ or $\{z\in X:p(z)<r\}$ for some $r\geq 0$ is a balanced set.
For every x ∈ X , {\displaystyle x\in X,} $x\in X,$ p ( 1 n x ) → 0 {\displaystyle p\left({\tfrac {1}{n}}x\right)\to 0} $p\left({\tfrac {1}{n}}x\right)\to 0$ as n → ∞ {\displaystyle n\to \infty } $n\to \infty$
- The sequence $\left({\tfrac {1}{n}}\right)_{n=1}^{\infty }$ 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:

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

An F-seminorm is called monotone if it satisfies:

Monotone: $p(rx)<p(sx)$ for all non-zero $x\in X$ and all real $s$ and $t$ such that $s<t.$ ^[12]

F-seminormed spaces

An F-seminormed space (resp. F-normed space)^[12] 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^[12] if $q(f(x)-f(y))=p(x,y){\text{ for all }}x,y\in X.$

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 $p$ and $q$ are F-seminorms on $X$ then so is their pointwise supremum $x\mapsto \sup\{p(x),q(x)\}.$ The same is true of the supremum of any non-empty finite family of F-seminorms on $X.$ ^[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 $X$ 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 $0<p<1,$ the map $f$ on $\mathbb {R} ^{n}$ defined by $[f\left(x_{1},\ldots ,x_{n}\right)]^{p}=\left|x_{1}\right|^{p}+\cdots \left|x_{n}\right|^{p}$ is an F-norm that is not a norm.
If $L:X\to Y$ is a linear map and if $q$ is an F-seminorm on $Y,$ then $q\circ L$ is an F-seminorm on $X.$ ^[12]
Let $X_{\mathbb {C} }$ be a complex vector space and let $X_{\mathbb {R} }$ denote $X_{\mathbb {C} }$ considered as a vector space over $\mathbb {R} .$ Any F-seminorm on $X_{\mathbb {C} }$ is also an F-seminorm on $X_{\mathbb {R} }.$ ^[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 $X$ 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

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

The balanced sets $\{x\in X~:~p(x)\leq r\},$ as $r$ ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of closed set. Similarly, the balanced sets $\{x\in X~:~p(x)<r\},$ as $r$ 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 ${\mathcal {L}}$ is a non-empty collection of F-seminorms on a vector space $X$ and for any finite subset ${\mathcal {F}}\subseteq {\mathcal {L}}$ and any $r>0,$ let $U_{{\mathcal {F}},r}:=\bigcap _{p\in {\mathcal {F}}}\{x\in X:p(x)<r\}.$

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

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

Fréchet combination

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

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

As an F-seminorm

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

Since $p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty }$ is increasing, a basis of open neighborhoods of the origin consists of all sets of the form $\left\{x\in X~:~p_{i}(x)<r\right\}$ 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 $d(x,y)=\sum _{i=1}^{\infty }{\frac {1}{2^{i}}}{\frac {p_{i}(x-y)}{1+p_{i}(x-y)}}.$

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

$q(x):=\inf _{}\left\{\sum _{i=1}^{n}p_{i}(x)+{\frac {1}{n}}~:~n>0{\text{ is an integer }}\right\}.$ ^[8]
$r(x):=\sum _{n=1}^{\infty }\min \left\{{\frac {1}{2^{n}}},p_{n}(x)\right\}.$ ^[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 $R:[0,\infty )\to [0,\infty )$ that has a bounded range, is 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 $\arctan t,$ $\tanh t,$ $t\mapsto \min\{t,1\},$ and $t\mapsto {\frac {t}{1+t}}.$ ^[15] 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 $d.$ ^[15]

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

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:^[11]

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

Of metrizable TVS

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

$X$ is metrizable.
$X$ is Hausdorff and pseudometrizable.
$X$ is Hausdorff and has a countable neighborhood base at the origin.^[11]^[12]
The topology on $X$ is induced by a translation-invariant metric on $X.$ ^[11]
The topology on $X$ is induced by an F-norm.^[11]^[12]
The topology on $X$ is induced by a monotone F-norm.^[12]
The topology on $X$ is induced by a total paranorm.

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

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

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

Of locally convex pseudometrizable TVS

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

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

Quotients

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

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

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 $x,y\in X.$ ^[19].
If $(X,d)$ is pseudometric TVS with a translation invariant pseudometric $d,$ then $p(x):=d(x,0)$ defines a paranorm.^[20] However, if $d$ is a translation invariant pseudometric on the vector space $X$ (without the addition condition that $(X,d)$ is pseudometric TVS), then $d$ 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 $X$ 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, $\left(X,b\left(X,X^{\prime }\right)\right),$ is metrizable if and only if there exists a countable set ${\mathcal {B}}$ of bounded subsets of $X$ such that every bounded subset of $X$ is contained in some element of ${\mathcal {B}}.$ ^[22]

The strong dual space $X_{b}^{\prime }$ of a metrizable locally convex space (such as a Fréchet space^[23]) $X$ 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 $X$ is a metrizable locally convex space then its strong dual $X_{b}^{\prime }$ 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 $M$ is a metrizable locally convex TVS that possess a countable fundamental system of bounded sets, then $M$ is normable.^[27]

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

$X$ is normable.
$X$ has a (von Neumann) bounded neighborhood of the origin.
the strong dual space $X_{b}^{\prime }$ of $X$ is normable.^[28]

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

the strong dual space of $X$ is metrizable.^[28]
the strong dual space of $X$ is a Fréchet–Urysohn locally convex space.^[23]

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

Another consequence of this is that if $X$ is a reflexive locally convex TVS whose strong dual $X_{b}^{\prime }$ is metrizable then $X_{b}^{\prime }$ is necessarily a reflexive Fréchet space, $X$ is a DF-space, both $X$ and $X_{b}^{\prime }$ are necessarily complete Hausdorff ultrabornological distinguished webbed spaces, and moreover, $X_{b}^{\prime }$ 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 $B.$ ^[14] 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.^[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 $X$ 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.^[26]
If $X$ is a complete pseudometrizable TVS and $M$ is a closed vector subspace of $X,$ then $X/M$ is complete.^[11]
The strong dual of a locally convex metrizable TVS is a webbed space.^[32]
If $(X,\tau )$ and $(X,\nu )$ are complete metrizable TVSs (i.e. F-spaces) and if $\nu$ is coarser than $\tau$ then $\tau =\nu$ ;^[33] this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete.^[34] Said differently, if $(X,\tau )$ 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 $p\leq Cq$ or if $q\leq Cp$ 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 $p:(X,\nu )\to \mathbb {R}$ is continuous if and only if the Fréchet space $(X,\nu )$ is 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.^[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 $0.$ ^[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

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 not a complete metric (such metrics exist even for $X=\mathbb {R}$ ). 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:

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

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

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

If $M$ is a closed vector subspace of a complete pseudometrizable TVS $X,$ then the quotient space $X/M$ is complete.^[40] If $M$ is a complete vector subspace of a metrizable TVS $X$ and if the quotient space $X/M$ is complete then so is $X.$ ^[40] 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.^[23]

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 $S\subseteq \operatorname {cl} _{C}R.$ ^[41]
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.^[42]
If $d$ is a translation invariant metric on a vector space $X,$ then $d(nx,0)\leq nd(x,0)$ for all $x\in X$ and every positive integer $n.$ ^[43]
If $\left(x_{i}\right)_{i=1}^{\infty }$ is a null sequence (that is, it converges to the origin) in a metrizable TVS then there exists a sequence $\left(r_{i}\right)_{i=1}^{\infty }$ of positive real numbers diverging to $\infty$ such that $\left(r_{i}x_{i}\right)_{i=1}^{\infty }\to 0.$ ^[43]
A subset of a complete metric space is closed if and only if it is complete. If a space $X$ 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 $B.$ ^[44]

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

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

Generalized series

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

Linear maps

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

$F$ is continuous;
$F$ is a (locally) bounded map (that is, $F$ maps (von Neumann) bounded subsets of $X$ to bounded subsets of $Y$ );^[12]
$F$ is sequentially continuous;^[12]
the image under $F$ of every null sequence in $X$ is a bounded set^[12] where by definition, a null sequence is a sequence that converges to the origin.
$F$ 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.^[47]

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.^[47]

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.^[47]

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.^[47]

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 $X.$ ^[22] Say that a TVS $X$ has the Hahn-Banach extension property (HBEP) if every vector subspace of $X$ 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 $X$ 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

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

Proofs

Suppose the net ${\textstyle \textstyle \sum \limits _{i\in I}r_{i}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\textstyle \lim \limits _{A\in \operatorname {FiniteSubsets} (I)}}\ \textstyle \sum \limits _{i\in A}r_{i}=\lim \left\{\textstyle \sum \limits _{i\in A}r_{i}\,:A\subseteq I,A{\text{ finite }}\right\}}$ converges to some point in a metrizable TVS $X,$ where recall that this net's domain is the directed set $(\operatorname {FiniteSubsets} (I),\subseteq ).$ Like every convergent net, this convergent net of partial sums $A\mapsto \textstyle \sum \limits _{i\in A}r_{i}$ is a Cauchy net, which for this particular net means (by definition) that for every neighborhood $W$ of the origin in $X,$ there exists a finite subset $A_{0}$ of $I$ such that ${\textstyle \textstyle \sum \limits _{i\in B}r_{i}-\textstyle \sum \limits _{i\in C}r_{i}\in W}$ for all finite supersets $B,C\supseteq A_{0};$ this implies that $r_{i}\in W$ for every $i\in I\setminus A_{0}$ (by taking $B:=A_{0}\cup \{i\}$ and $C:=A_{0}$ ). Since $X$ is metrizable, it has a countable neighborhood basis $U_{1},U_{2},\ldots$ at the origin, whose intersection is necessarily $U_{1}\cap U_{2}\cap \cdots =\{0\}$ (since $X$ is a Hausdorff TVS). For every positive integer $n\in \mathbb {N} ,$ pick a finite subset $A_{n}\subseteq I$ such that $r_{i}\in U_{n}$ for every $i\in I\setminus A_{n}.$ If $i$ belongs to $(I\setminus A_{1})\cap (I\setminus A_{2})\cap \cdots =I\setminus \left(A_{1}\cup A_{2}\cup \cdots \right)$ then $r_{i}$ belongs to $U_{1}\cap U_{2}\cap \cdots =\{0\}.$ Thus $r_{i}=0$ for every index $i\in I$ that does not belong to the countable set $A_{1}\cup A_{2}\cup \cdots .$ $\blacksquare$

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Pseudometrics and metrics

Topology induced by a pseudometric

Pseudometrics and values on topological groups

Translation invariant pseudometrics

Value/G-seminorm

Properties of values

Equivalence on topological groups

Pseudometrizable topological groups

An invariant pseudometric that doesn't induce a vector topology

Additive sequences

Paranorms

Properties of paranorms

Examples of paranorms

F-seminorms

F-seminormed spaces

Examples of F-seminorms

Properties of F-seminorms

Topology induced by a single F-seminorm

Topology induced by a family of F-seminorms

Fréchet combination

As an F-seminorm

As a paranorm

Generalization

Characterizations

Of (pseudo)metrics induced by (semi)norms

Of pseudometrizable TVS

Of metrizable TVS

Of locally convex pseudometrizable TVS

Quotients

Examples and sufficient conditions

Normability

Metrically bounded sets and bounded sets

Properties of pseudometrizable TVS

Completeness

Subsets and subsequences

Linear maps

Hahn-Banach extension property

See also

Notes

References

Bibliography