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 $d:X\times X\to ℝ$ satisfying the following properties:

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

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: $d(x,z)\le \max \{d(x,y),d(y,z)\}\text{ for all }x,y,z\in X.$

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 $d.$

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

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 $p:X\to ℝ$ with the following properties:

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

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:

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

Equivalence on topological groups

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

1. $\tau$ is induced by a pseudometric; (i.e. $(X,\tau )$ is pseudometrizable);
2. $\tau$ is induced by a translation-invariant pseudometric;
3. 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. $X\ne \{0\}$) 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 $x\ne y.$ 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 $𝒩$ of subsets of a vector space is called additive if for every $N\in 𝒩,$ there exists some $U\in 𝒩$ 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."

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 $U_{\bullet }={\left(U_i\right)}_{i=0}^{\mathrm{\infty }}$ 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 $i\ge 0.$ For all $u\in U_0,$ let $$𝕊(u):=\{n_{\bullet }=(n_1,\dots ,n_k):k\ge 1,n_i\ge 0\text{ for all }i,\text{ and }u\in U_{n_1}+\cdots +U_{n_k}\}.$$

Define $f:X\to [0,1]$ by $f(x)=1$ if $x\in ̸U_0$ and otherwise let $$f(x):=\underset{}{\inf }\{2^{-n_1}+\cdots 2^{-n_k}:n_{\bullet }=(n_1,\dots ,n_k)\in 𝕊(x)\}.$$

Then $f$ is subadditive (meaning $f(x+y)\le f(x)+f(y)\text{ for all }x,y\in X$) and $f=0$ on $\bigcap _{i\ge 0}{U}_i,$ 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 $f(sx)\le f(x)$ for all scalars $s$ such that $|s|\le 1$ 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 $n_{\bullet }=(n_1,\dots ,n_k)$ always denotes a finite sequence of non-negative integers and use the notation: $$\sum 2^{-n_{\bullet }}:=2^{-n_1}+\cdots +2^{-n_k}\text{ and }\sum U_{n_{\bullet }}:=U_{n_1}+\cdots +U_{n_k}.$$

For any integers $n\ge 0$ 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 $n_{\bullet }=(n_1,\dots ,n_k)$ 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 $n_{\bullet }=(n_1,\dots ,n_k)$ consists of non-negative integers such that $\sum 2^{-n_{\bullet }}\le 2^{-M}$ for some integer $M\ge 0$ 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 $m_{\bullet }=(m_1,\dots ,m_{k-1})$ 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 $0\le f\le 1$ so to prove that $f$ is subadditive, it suffices to prove that $f(x+y)\le f(x)+f(y)$ 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 $f(-x)\le f(x)$ and $f(-x)\ge f(x).$ If all $U_i$ are balanced then the inequality $f(sx)\le f(x)$ for all unit scalars $s$ such that $|s|\le 1$ 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 $f(x)\le 2^{-M}<r.$ 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 $0<r\le 2^{-n}$ such that $f(x)<r$ implies $x\in U_n.$ $◼$

Paranorms

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

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

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 $d:X\times X\to ℝ$ defined by $d(x,y):=p(x-y)$ is a translation-invariant pseudometric on $X$ that defines a vector topology on $X.$

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

• the set $\{x\in X:p(x)=0\}$ is a vector subspace of $X.$
• $p(x+n)=p(x)\text{ for all }x,n\in X$ with $p(n)=0.$
• If a paranorm $p$ satisfies $p(sx)\le |s|p(x)\text{ for all }x\in X$ 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 $q(x):=p(x)/[1+p(x)].$
• 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 $(p\wedge q)(x):=\underset{}{\inf }\{p(y)+q(z):x=y+z\text{ with }y,z\in X\}.$ Moreover, $(p\wedge q)\le p$ and $(p\wedge q)\le q.$ 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 $(x,y)\mapsto \sqrt{|x^2-y^2|}$ and $(x,y)\mapsto {|x^2-y^2|}^{3/2}$ are not paranorms on $X:={ℝ}^2.$
• 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 $s.$
• The function $p(x):=|\sin (\pi x)|+\min \{2,|x|\}$ is a paranorm on $ℝ$ that is not balanced but nevertheless equivalent to the usual norm on $R.$ Note that the function $x\mapsto |\sin (\pi x)|$ is subadditive.
• Let $X_{ℂ}$ be a complex vector space and let $X_{ℝ}$ denote $X_{ℂ}$ considered as a vector space over $ℝ.$ Any paranorm on $X_{ℂ}$ is also a paranorm on $X_{ℝ}.$

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 $p:X\to ℝ$ with the following four properties:

1. Non-negative: $p\ge 0.$
2. Subadditive: $p(x+y)\le p(x)+p(y)$ for all $x,y\in X$
3. Balanced: $p(ax)\le p(x)$ for $x\in X$ all scalars $a$ satisfying $|a|\le 1;$
   • This condition guarantees that each set of the form $\{z\in X:p(z)\le r\}$ or $\{z\in X:p(z)<r\}$ for some $r\ge 0$ is a balanced set.
4. For every $x\in X,$ $p\left(\frac{1}{n}x\right)\to 0$ as $n\to \mathrm{\infty }$
   • The sequence ${\left(\frac{1}{n}\right)}_{n=1}^{\mathrm{\infty }}$ 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: $p(rx)<p(sx)$ 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 $q(f(x)-f(y))=p(x,y)\text{ for all }x,y\in X.$

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

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 $X.$
• 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 ${ℝ}^n$ defined by $$[f(x_1,\dots ,x_n){]}^p={\left|x_1\right|}^p+\cdots {\left|x_n\right|}^p$$ 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 $X.$
• Let $X_{ℂ}$ be a complex vector space and let $X_{ℝ}$ denote $X_{ℂ}$ considered as a vector space over $ℝ.$ Any F-seminorm on $X_{ℂ}$ is also an F-seminorm on $X_{ℝ}.$

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

Theorem—Let $p$ be an F-seminorm on a vector space $X.$ Then the map $d:X\times X\to ℝ$ 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 $\{x\in X:p(x)\le r\},$ 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 $ℒ$ is a non-empty collection of F-seminorms on a vector space $X$ and for any finite subset $ℱ\subseteq ℒ$ and any $r>0,$ let $$U_{ℱ,r}:=\bigcap _{p\in ℱ}\{x\in X:p(x)<r\}.$$

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

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

Fréchet combination

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

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

As an F-seminorm

Assume that $p_{\bullet }={\left(p_i\right)}_{i=1}^{\mathrm{\infty }}$ 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.

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

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

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. This is also true of the following paranorms on $X$:

• $q(x):=\underset{}{\inf }\{\sum _{i=1}^n{p}_i(x)+\frac{1}{n}:n>0\text{ is an integer }\}.$
• $r(x):=\sum _{n=1}^{\mathrm{\infty }}\min \{\frac{1}{2^n},p_n(x)\}.$

Generalization

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

A bounded remetrization function is a continuous non-negative non-decreasing map $R:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ that has a bounded range, is subadditive (meaning that $R(s+t)\le R(s)+R(t)$ for all $s,t\ge 0$), 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 $t\mapsto \frac{t}{1+t}.$ 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 $p_{\bullet }={\left(p_i\right)}_{i=1}^{\mathrm{\infty }}$ is a family of non-negative F-seminorm on a vector space $X,$ $R$ is a bounded remetrization function, and $r_{\bullet }={\left(r_i\right)}_{i=1}^{\mathrm{\infty }}$ is a sequence of positive real numbers whose sum is finite. Then $$p(x):=\sum _{i=1}^{\mathrm{\infty }}{r}_{i}R(p_i(x))$$ defines a bounded F-seminorm that is uniformly equivalent to the $p_{\bullet }.$ 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 $i.$ $p$ is an F-norm if and only if the $p_{\bullet }$ 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 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.

Birkhoff–Kakutani theorem—If $(X,\tau )$ is a topological vector space then the following three conditions are equivalent:

1. The origin $\{0\}$ is closed in $X,$ and there is a countable basis of neighborhoods for $0$ in $X.$
2. $(X,\tau )$ is metrizable (as a topological space).
3. 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:

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 ${\left(p_i\right)}_{i=1}^{\mathrm{\infty }}$ (increasing means that for all $i,$ $p_i\ge p_{i+1}.$
5. The topology of $X$ is induced by an F-seminorm of the form: $$p(x)=\sum _{n=1}^{\mathrm{\infty }}{2}^{-n}\arctan p_n(x)$$ where ${\left(p_i\right)}_{i=1}^{\mathrm{\infty }}$ 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 $$P(x+M):=\underset{}{\inf }\{p(x+m):m\in M\}$$ 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 pseudometric TVS), 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, $(X,b(X,X^{\prime })),$ is metrizable if and only if there exists a countable set $ℬ$ of bounded subsets of $X$ such that every bounded subset of $X$ is contained in some element of $ℬ.$

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

Normability

A topological vector space is seminormable if and only if it has a convex bounded neighborhood of the origin. Moreover, a TVS is normable if and only if it is Hausdorff and seminormable. Every metrizable TVS on a finite-dimensional vector space is a normable locally convex complete TVS, being TVS-isomorphic to Euclidean space. Consequently, any metrizable TVS that is not normable must be infinite dimensional.

If $M$ 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 $X_b^{\prime }$ 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 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 $d(x,y)\le R$ 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

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

• Every metrizable locally convex TVS is a quasibarrelled space, bornological space, and a Mackey space.
• Every complete pseudometrizable TVS is a barrelled space and a Baire space (and hence non-meager). However, there exist metrizable Baire spaces that are not complete.
• If $X$ 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 $p\le Cq$ or if $q\le Cp$ 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 $p:(X,\nu )\to ℝ$ 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.
• 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 pseudometrizable TVS is a barrelled space and a Baire space (and thus non-meager).
• The dimension of a complete metrizable TVS is either finite or uncountable.

Completeness

Main article: 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 $X$ 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 $X=ℝ$). 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.

Theorem (Klee)—Let $d$ be any 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 ${\left(x_i\right)}_{i=1}^{\mathrm{\infty }}$ in $X,$ if $\sum _{i=1}^{\mathrm{\infty }}p\left(x_i\right)<\mathrm{\infty }$ then $\sum _{i=1}^{\mathrm{\infty }}{x}_i$ converges in $X.$

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 complete 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.

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 $S\subseteq {\mathrm{cl}}_{C}R.$
• 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 $d(nx,0)\le nd(x,0)$ for all $x\in X$ and every positive integer $n.$
• If ${\left(x_i\right)}_{i=1}^{\mathrm{\infty }}$ is a null sequence (that is, it converges to the origin) in a metrizable TVS then there exists a sequence ${\left(r_i\right)}_{i=1}^{\mathrm{\infty }}$ of positive real numbers diverging to $\mathrm{\infty }$ such that ${\left(r_i{x}_i\right)}_{i=1}^{\mathrm{\infty }}\to 0.$
• 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 $B.$

Banach-Saks theorem—If ${\left(x_n\right)}_{n=1}^{\mathrm{\infty }}$ is a sequence in a locally convex metrizable TVS $(X,\tau )$ that converges weakly to some $x\in X,$ then there exists a sequence $y_{\bullet }={\left(y_i\right)}_{i=1}^{\mathrm{\infty }}$ 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—Suppose that $X$ is a locally convex metrizable TVS and that ${\left(B_i\right)}_{i=1}^{\mathrm{\infty }}$ is a countable sequence of bounded subsets of $X.$ Then there exists a bounded subset $B$ of $X$ and a sequence ${\left(r_i\right)}_{i=1}^{\mathrm{\infty }}$ 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 $\sum _{i\in I}{r}_i$ as the limit of the net of finite partial sums $F\in \mathrm{FiniteSubsets}(I)\mapsto \sum _{i\in F}{r}_i$ where the domain $\mathrm{FiniteSubsets}(I)$ is directed by $\subseteq .$ If $I=\mathbb{N}$ and $X=ℝ,$ for instance, then the generalized series $\sum _{i\in \mathbb{N}}{r}_i$ converges if and only if $\sum _{i=1}^{\mathrm{\infty }}{r}_i$ converges unconditionally in the usual sense (which for real numbers, is equivalent to absolute convergence). If a generalized series $\sum _{i\in I}{r}_i$ converges in a metrizable TVS, then the set $\{i\in I:r_i\ne 0\}$ 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 $\sum _{i\in I}{r}_i=\sum _{\stackrel{i\in I}{r_i\ne 0}}{r}_i$ 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. 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 null sequence 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

Main article: Hahn-Banach theorem

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:

Theorem (Kalton)—Every complete metrizable TVS with the Hahn-Banach extension property is locally convex.

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

• Asymmetric norm – Generalization of the concept of a norm
• Complete metric space – Metric geometry
• Complete topological vector space – Structure in functional analysis
• Equivalence of metrics – Mathematical notion
• F-space – Topological vector space with a complete translation-invariant metric
• Fréchet space – Locally convex topological vector space that is also a complete metric space
• Generalised metric – Metric geometry
• Locally convex topological vector space – Vector space with a topology defined by convex open sets
• Metric space – Mathematical space with a notion of distance
• Pseudometric space – Generalization of metric spaces in mathematics
• Relation of norms and metrics – Mathematical space with a notion of distancePages displaying short descriptions of redirect targets
• Seminorm – Mathematical function
• Sublinear function – Type of function in linear algebra
• Uniform space – Topological space with a notion of uniform properties
• Ursescu theorem – Generalization of closed graph, open mapping, and uniform boundedness theorem

Notes

Proofs