8.2 Pseudonorms
Definition 8.2.1 (Pseudonorm). Let $E$ be a vector space over $K \in \RC$, then a pseudonorm is a function $\rho: E \to [0, \infty)$ such that
$\rho(0) = 0$.
For any $x \in X$ and $\lambda \in K$ with $\abs{\lambda}\le 1$, $\rho(\lambda x) \le \rho(x)$.
For any $x, y \in X$, $\rho(x + y) \le \rho(x) + \rho(y)$.
For any $x \in X$, $\lim_{\lambda \to 0}\rho(\lambda x) = 0$.
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 8.2.2 (Topology Induced by Pseudonorm). 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
The topology induced by $\seqi{\rho}$ is a vector space topology.
The uniformity induced by $\seqi{\rho}$ is the translation-invariant uniformity for its topology.
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 5.3.3.
(2): Each $d_{i}$ is translation-invariant.
(1):
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).
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 8.2.3. Let $E$ be a TVS over $K \in \RC$, and $\rho: E \to [0, \infty)$ be a pseudonorm, then the following are equivalent:
$\rho \in UC(E; [0, \infty))$.
$\rho \in C(E; [0, \infty))$.
$\rho$ is continuous at $0$.
The topology on $E$ contains the topology induced by $\rho$.
Proof. $(4) \Rightarrow (1)$: By Definition 8.2.2, 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 8.2.4. Let $E$ be a vector space over $K \in \RC$, $\seq{U_n}\subset 2^{E}$ such that
For each $n \in \natp$, $U_{n}$ is circled, radial, and contains $0$.
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. For each $H \subset \natp$ finite, let
Define
then
Since $0 \in \bigcap_{H \subset \natp \text{ finite}}U_{H}$, $\rho(0) = 0$.
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 $\rho(\lambda x) \le \rho(x)$.
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 $\bracs{U_n|n \in \natp}$.
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 $\rho(x) \le 2^{-n}$.
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 8.2.5. As discussed in Remark 5.3.6 on the proof of Lemma 5.3.4, 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 8.2.6 (Metrisability of Topological Vector Spaces). Let $E$ be a TVS over $K \in \RC$, then the following are equivalent:
There exists a pseudonorm that induces the topology on $E$.
There exists a translation-invariant pseudometric that induces the topology on $E$.
$E$ admits a countable fundamental system of entourages.
There exists a pseudometric that induces the topology on $E$.
$E$ admits a countable fundamental system of neighbourhoods at $0$.
Proof. (3) $\Rightarrow$ (4): By Theorem 5.3.12.
(4) $\Rightarrow$ (1): By Proposition 8.1.11, 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 8.2.4, 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 8.2.7. 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 8.2.8 (Locally Bounded). Let $E$ be a TVS over $K \in \RC$, then $E$ is locally bounded if there exists $U \in \cn^{o}(0)$ bounded.
Proposition 8.2.9. 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 8.1.11, 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 8.2.6, the topology on $E$ is induced by a pseudonorm.$\square$