11.3 Pseudonorms
Definition 11.3.1 (Pseudonorm).label Let $E$ be a vector space over $K \in \RC$, then a pseudonorm is a function $\rho: E \to [0, \infty)$ such that
- (PN1)
$\rho(0) = 0$.
- (PN2)
For any $x \in X$ and $\lambda \in K$ with $\abs{\lambda}\le 1$, $\rho(\lambda x) \le \rho(x)$.
- (PN3)
For any $x, y \in X$, $\rho(x + y) \le \rho(x) + \rho(y)$.
- (PN4)
For any $x \in X$, $\lim_{\lambda \to 0}\rho(\lambda x) = 0$.
- (PN5)
For any $\lambda \in K$, and $\seq{x_n}\subset X$ with $\rho(x_{n}) \to 0$ as $n \to \infty$, $\limv{n}\rho(\lambda x_{n}) = 0$.
Definition 11.3.2 (Topology Induced by Pseudonorm).label Let $E$ be a vector space over $K \in \RC$ and $\seqi{\rho}$ be pseudonorms on $E$. For each $i \in I$, let
then $d_{i}$ is a pseudometric. The uniform topology on $E$ induced by $\seqi{d}$ is the topology induced by $\seqi{\rho}$, and
- (1)
The topology induced by $\seqi{\rho}$ is a vector space topology.
- (2)
The uniformity induced by $\seqi{\rho}$ is the translation-invariant uniformity for its topology.
- (3)
For each $i \in I$, $x \in E$, and $r > 0$, let $B_{i}(x, r) = \bracs{y \in E|d_i(x, y) < r}$, then
\[\bracs{\bigcap_{j \in J}B_j(0, r)|J \subset I \text{ finite}, r > 0}\]is a fundamental system of neighbourhoods at $0$.
Proof. (3): By Definition 6.3.3.
(2): Each $d_{i}$ is translation-invariant.
(1):
- (TVS1)
Let $x, x', y, y' \in E$, $J \subset I$ be finite, and $r > 0$. If for each $j \in J$, $d_{j}(x, x'), d_{j}(y, y') < r/2$, then $d_{j}(x + x', y + y') < r$ by (PN3).
- (TVS2)
Let $x, x' \in E$ and $\lambda, \lambda' \in K$, then
\[\lambda x - \lambda' x' = \lambda(x - x') + (\lambda - \lambda')x'\]Let $i \in I$ and $\eps > 0$. By (PN5) then there exists $\delta > 0$ such that if $\rho_{i}(x - x') < \delta$, then $\rho_{i}(\lambda (x - x')) < \eps$.
On the other hand,
\[(\lambda - \lambda')x' = (\lambda - \lambda')x + (\lambda - \lambda')(x' - x)\]By (PN4), there exists $\delta' \in (0, 1]$ such that if $\abs{\lambda - \lambda'}< \delta'$, then $\rho_{i}((\lambda - \lambda')x) < \eps$. In which case, since $\delta' \le 1$, (PN2) implies that
\[\rho_{i}((\lambda - \lambda')x') < \eps + \rho_{i}(x' - x) < 2\eps\]Therefore $\rho_{i}(\lambda x - \lambda' x') < 3\eps$.
$\square$
Proposition 11.3.3.label Let $E$ be a TVS over $K \in \RC$, and $\rho: E \to [0, \infty)$ be a pseudonorm, then the following are equivalent:
- (1)
$\rho \in UC(E; [0, \infty))$.
- (2)
$\rho \in C(E; [0, \infty))$.
- (3)
$\rho$ is continuous at $0$.
- (4)
The topology on $E$ contains the topology induced by $\rho$.
Proof. $(4) \Rightarrow (1)$: By Definition 11.3.2, for each $r > 0$, $\rho^{-1}([0, r)) \in \cn_{E}(0)$. Thus for any $x, y \in E$, if $x - y \in \rho^{-1}([0, r))$, then $\abs{\rho(x) - \rho(y)}\le r$. Therefore $\rho \in UC(E; [0, \infty))$.$\square$
Lemma 11.3.4.label Let $E$ be a vector space over $K \in \RC$, $\seq{U_n}\subset 2^{E}$ such that
- (a)
For each $n \in \natp$, $U_{n}$ is circled, radial, and contains $0$.
- (b)
For each $n \in \natp$, $U_{n+1}+ U_{n+1}\subset U_{n}$.
then there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $n \in \natp$,
Proof [Theorem I.6.1, SW99].For each $H \subset \natp$ finite, let
Define
then
- (PN1)
Since $0 \in \bigcap_{H \subset \natp \text{ finite}}U_{H}$, $\rho(0) = 0$.
- (PN2)
Let $x \in X$ and $\lambda \in K$ with $\abs{\lambda}\le 1$. By assumption (a), $\lambda U_{n} \subset U_{n}$ for each $n \in \natp$. Thus for any $H \subset \natp$ finite with $x \in U_{H}$,
\[\lambda x \in \sum_{n \in H}\lambda U_{n} \subset \sum_{n \in H}U_{n}\]so $\rho(\lambda x) \le \rho(x)$.
- (PN3)
Let $x, y \in X$ and $M, N \subset \natp$ finite such that $x \in U_{M}$ and $y \in U_{N}$. Assume without loss of generality that $\rho_{M} + \rho_{N} < 1$, then there exists a unique $P \subset \nat$ finite such that $\rho_{P} = \rho_{M} + \rho_{N}$. In which case, $U_{P} \supset U_{M} + U_{N}$ by assumption (b). Therefore $\rho(x + y) \le \rho(x) + \rho(y)$.
For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1}< 2^{n}$, so $U_{n+1}\subset \rho^{-1}([0, 2^{-n}))$ by Proposition 3.1.5. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}}= U_{n}$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
- (PN4)
Let $x \in X$ and $n \in \natp$. By assumption (a), there exists $\alpha > 0$ such that for any $\lambda \in K$ with $\abs{\lambda}\ge \alpha$, $x \in \lambda U_{n}$. Therefore for any $\lambda \in K$ with $\abs{\lambda}\le \alpha^{-1}$, $\lambda x \in U_{n}$, and $\rho(x) \le 2^{-n}$.
- (PN5)
Let $\lambda \in K$ and $n \in \natp$. By assumption (b), there exists $m \in \nat$ such that $\lambda U_{n-m}\subset \sum_{j = 1}^{m} U_{n-m}^{j} \subset U_{n}$.
$\square$
Remark 11.3.1.label As discussed in Remark 6.3.2 on the proof of Lemma 6.3.4, constructing a pseudometric from entourages by building its level sets is difficult because composing symmetric entourages does not necessarily lead to symmetric entourages. The topological vector space does not have this shortcoming, and as such allows this construction.
Theorem 11.3.5 (Metrisability of Topological Vector Spaces).label Let $E$ be a TVS over $K \in \RC$, then the following are equivalent:
- (1)
There exists a pseudonorm that induces the topology on $E$.
- (2)
There exists a translation-invariant pseudometric that induces the topology on $E$.
- (3)
$E$ admits a countable fundamental system of entourages.
- (4)
There exists a pseudometric that induces the topology on $E$.
- (5)
$E$ admits a countable fundamental system of neighbourhoods at $0$.
Proof. (3) $\Rightarrow$ (4): By Theorem 6.3.10.
(4) $\Rightarrow$ (1): By Proposition 11.1.6, there exists $\seq{U_n}\subset \cn_{E}(0)$ circled and radial such that for each $n \in \natp$, $U_{n+1}+ U_{n+1}\subset U_{n}$. By Lemma 11.3.4, there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $N \in \natp$, $U_{n+1}\subset \rho^{-1}([0, 2^{-n})) \subset U_{n}$. In which case, $\rho$ induces the topology on $E$.$\square$
Remark 11.3.2.label Let $E$ be a TVS over $K \in \RC$. Similar to the case of uniform spaces, there exists a family of pseudonorms $\seqi{\rho}$ that induces the topology on $E$. Using Minkowski functionals, $\seqi{\rho}$ can be taken such that $\rho_{i}(\lambda x) = \abs{\lambda}\rho_{i}(x)$ for all $x \in E$ and $\lambda \in K$. However, a single pseudonorm with this property cannot always induce the topology even if the space is metrisable, hence the difference between pseudonorms and seminorms.
Definition 11.3.6 (Locally Bounded).label Let $E$ be a TVS over $K \in \RC$, then $E$ is locally bounded if there exists $U \in \cn^{o}(0)$ bounded.
Proposition 11.3.7 ([I.6.2, SW99]).label Let $E$ be a locally bounded TVS over $K \in \RC$, then there exists a pseudonorm $\rho: E \to [0, \infty)$ that induces the topology on $E$.
Proof. Let $U \in \cn^{o}(0)$ be bounded. Using Proposition 11.1.6, assume without loss of generality that $U$ is circled. For each $n \in \natp$, let $U_{n} = n^{-1}U$. Let $V \in \cn^{o}(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$, $\abs{\lambda}^{-1}U \subset V$. For any $n \in \natp$ with $n^{-1}< \abs{\lambda}^{-1}$, $U_{n} \subset V$. Thus $E$ admits a countable fundamental system of neighbourhoods at $0$. By Theorem 11.3.5, the topology on $E$ is induced by a pseudonorm.$\square$
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