\section{Pseudonorms}
\label{section:tvs-pseudonorm}

\begin{definition}[Pseudonorm]
\label{definition:pseudonorm}
    Let $E$ be a vector space over $K \in \RC$, then a \textbf{pseudonorm} is a function $\rho: E \to [0, \infty)$ such that
    \begin{enumerate}
        \item[(PN1)] $\rho(0) = 0$.
        \item[(PN2)] For any $x \in X$ and $\lambda \in K$ with $\abs{\lambda} \le 1$, $\rho(\lambda x) \le \rho(x)$.
        \item[(PN3)] For any $x, y \in X$, $\rho(x + y) \le \rho(x) + \rho(y)$.
        \item[(PN4)] For any $x \in X$, $\lim_{\lambda \to 0}\rho(\lambda x) = 0$.
        \item[(PN5)] For any $\lambda \in K$, and $\seq{x_n} \subset X$ with $\rho(x_n) \to 0$ as $n \to \infty$, $\limv{n}\rho(\lambda x_n) = 0$.
    \end{enumerate}
\end{definition}

\begin{definition}[Topology Induced by Pseudonorm]
\label{definition:tvs-pseudonorm-topology}
    Let $E$ be a vector space over $K \in \RC$ and $\seqi{\rho}$ be pseudonorms on $E$. For each $i \in I$, let
    \[
    d_i: E \times E \quad (x, y) \mapsto \rho_i(x - y)
    \]

    then $d_i$ is a pseudometric. The uniform topology on $E$ induced by $\seqi{d}$ is the \textbf{topology induced by $\seqi{\rho}$}, and
    \begin{enumerate}
        \item The topology induced by $\seqi{\rho}$ is a vector space topology.
        \item The uniformity induced by $\seqi{\rho}$ is the translation-invariant uniformity for its topology.
        \item 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 $0$.
    \end{enumerate}
\end{definition}
\begin{proof}
    (3): By \autoref{definition:pseudometric-uniformity}.

    (2): Each $d_i$ is translation-invariant.

    (1):
    \begin{enumerate}
        \item[(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).
        \item[(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 $\rho_i(\lambda x - \lambda' x') < 3\eps$.
    \end{enumerate}
\end{proof}


\begin{proposition}
\label{proposition:tvs-pseudonorm-compatible}
    Let $E$ be a TVS over $K \in \RC$, and $\rho: E \to [0, \infty)$ be a pseudonorm, then the following are equivalent:
    \begin{enumerate}
        \item $\rho \in UC(E; [0, \infty))$.
        \item $\rho \in C(E; [0, \infty))$.
        \item $\rho$ is continuous at $0$.
        \item The topology on $E$ contains the topology induced by $\rho$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    $(4) \Rightarrow (1)$: By \autoref{definition:tvs-pseudonorm-topology}, 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 $\rho \in UC(E; [0, \infty))$.
\end{proof}

\begin{lemma}[{{\cite[Theorem I.6.1]{SchaeferWolff}}}]
\label{lemma:tvs-sequence-pseudonorm}
    Let $E$ be a vector space over $K \in \RC$, $\seq{U_n} \subset 2^E$ such that
    \begin{enumerate}
        \item[(a)] For each $n \in \natp$, $U_n$ is circled, radial, and contains $0$.
        \item[(b)] For each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$.
    \end{enumerate}
    then there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $n \in \natp$,
    \[
    U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}
    \]

\end{lemma}
\begin{proof}
    For each $H \subset \natp$ finite, let
    \[
    U_H = \sum_{n \in H}V_n \quad \rho_H = \sum_{n \in H}2^{-n}
    \]

    Define
    \[
    \rho: E \to [0, 1] \quad x \mapsto \inf\bracs{\rho_H|x \in U_H}
    \]

    then
    \begin{enumerate}
        \item[(PN1)] Since $0 \in \bigcap_{H \subset  \natp \text{ finite}}U_H$, $\rho(0) = 0$.
        \item[(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 $\rho(\lambda x) \le \rho(x)$.
        \item[(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 $\rho(x + y) \le \rho(x) + \rho(y)$.
    \end{enumerate}
    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 \autoref{proposition:dyadic-semigroup-order}. 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 $\bracs{U_n|n \in \natp}$.
    \begin{enumerate}
        \item[(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 $\rho(x) \le 2^{-n}$.
        \item[(PN5)] Let $\lambda \in K$ and $n \in \natp$. By assumption (b), there exists $m \in \nat$ such that $\lambda U_{n-m} \subset \sum_{j = 1}^m U_{n-m}^j \subset U_n$.
    \end{enumerate}
\end{proof}

\begin{remark}
	\label{remark:tvs-sequence-pseudonorm}

	As discussed in \autoref{remark:uniform-sequence-pseudometric} on the proof of \autoref{lemma:uniform-sequence-pseudometric}, 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.
\end{remark}

\begin{theorem}[Metrisability of Topological Vector Spaces]
\label{theorem:tvs-metrisable}
    Let $E$ be a TVS over $K \in \RC$, then the following are equivalent:
    \begin{enumerate}
        \item There exists a pseudonorm that induces the topology on $E$.
        \item There exists a translation-invariant pseudometric that induces the topology on $E$.
        \item $E$ admits a countable fundamental system of entourages.
        \item There exists a pseudometric that induces the topology on $E$.
        \item $E$ admits a countable fundamental system of neighbourhoods at $0$.
    \end{enumerate}
\end{theorem}
\begin{proof}
    (3) $\Rightarrow$ (4): By \autoref{theorem:uniform-metrisable}.

    (4) $\Rightarrow$ (1): By \autoref{proposition:tvs-good-neighbourhood-base}, 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 \autoref{lemma:tvs-sequence-pseudonorm}, 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 $E$.
\end{proof}

\begin{remark}
    \label{remark:tvs-metrisable}

    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 \hyperref[Minkowski functionals]{definition:gauge}, $\seqi{\rho}$ 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.
\end{remark}

\begin{definition}[Locally Bounded]
\label{definition:locally-bounded}
    Let $E$ be a TVS over $K \in \RC$, then $E$ is \textbf{locally bounded} if there exists $U \in \cn^o(0)$ bounded.
\end{definition}

\begin{proposition}[{{\cite[I.6.2]{SchaeferWolff}}}]
\label{proposition:locally-bounded-metrisable}
    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 $E$.
\end{proposition}
\begin{proof}
    Let $U \in \cn^o(0)$ be bounded. Using \autoref{proposition:tvs-good-neighbourhood-base}, 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 \autoref{theorem:tvs-metrisable}, the topology on $E$ is induced by a pseudonorm.
\end{proof}
