\section{Seminorms}
\label{section:seminorms}


\begin{definition}[Convex]
\label{definition:convex}
    Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.
\end{definition}

\begin{definition}[Convex Hull]
\label{definition:convex-hull}
    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then the set
    \[
    \text{Conv}(A) = \bracs{\sum_{j = 1}^n t_j x_j \bigg | \seqf{t_j} \subset [0, 1], \seqf{x_j} \subset E, \sum_{j = 1}^n t_j = 1 }
    \]

    is the \textbf{convex hull} of $A$. 
\end{definition}

\begin{definition}[Convex Circled Hull]
\label{definition:convex-circled-hull}
    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then the set
    \[
    \Gamma(A) = \bracs{\sum_{j = 1}^n t_j x_j \bigg | \seqf{t_j} \subset K, \seqf{x_j} \subset E, \sum_{j = 1}^n |t_j| \le 1 }
    \]

    is the \textbf{convex circled hull} of $A$. 
\end{definition}



\begin{lemma}[{{\cite[II.1.1]{SchaeferWolff}}}]
\label{lemma:convex-interior}
    Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^o$, and $y \in \ol{A}$, then
    \[
    \bracs{tx + (1 - t)y|t \in (0, 1)} \subset A^o
    \]

\end{lemma}
\begin{proof}
    Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^o \in \cn^o(y)$. Since $y \in \ol{A}$, $\alpha A^o \cap A \ne \emptyset$, so there exists $z \in A^o$ such that $\alpha z \in A$.

    Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
    \[
    \mu z + (1 - \mu)\alpha z = \frac{\alpha z}{\alpha - 1} + \frac{(\alpha - 1 - \alpha)\alpha z}{\alpha - 1} = 0
    \]

    By (TVS1) and (TVS2),
    \[
    U = \underbrace{\bracs{\mu w + (1 - \mu)\alpha z|w \in A^o}}_{\subset A} \in \cn^o(0)
    \]

    so $0 \in A^o$.
\end{proof}

\begin{lemma}
\label{lemma:convex-gymnastics}
    Let $E$ be a TVS over $K \in \RC$, $A, B \subset E$ be convex, then the following sets are convex:
    \begin{enumerate}
        \item $A^o$.
        \item $\ol{A}$.
        \item $A + B$.
        \item For any $\lambda \in K$, $\lambda A$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    (1): By \autoref{lemma:convex-interior}.

    (2): Let $x, y \in \ol{A}$. By \autoref{definition:closure}, there exists filters $\fF, \mathfrak{G} \subset 2^A$ such that $\fF$ converges to $x$ and $\mathfrak{G}$ converges to $y$. In which case,
    \[
    \fU = \bracs{tE + (1 - t)F|t \in [0, 1], E \in \fF, F \in \mathfrak{G}} \subset 2^A
    \]

    converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
\end{proof}

\begin{proposition}[{{\cite[II.1.3]{SchaeferWolff}}}]
\label{proposition:convex-interior-closure}
    Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^o \ne \emptyset$, then $\ol{A} = \ol{A^o}$.
\end{proposition}
\begin{proof}
    Since $A^o \subset A$, $\ol{A^o} \subset \ol{A}$. Let $x \in A^o$, then for any $y \in \ol{A}$,
    \[
    y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}
    \]

    by \autoref{lemma:convex-interior}.
\end{proof}



\begin{definition}[Sublinear Functional]
\label{definition:sublinear-functional}
    Let $E$ be a vector space over $K \in \RC$, then a \textbf{sublinear functional} is a mapping $\rho: E \to \real$ such that:
    \begin{enumerate}
        \item $\rho(0) = 0$.
        \item For any $x \in E$ and $\lambda \ge 0$, $\rho(\lambda x) = \lambda \rho(x)$.
        \item For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
    \end{enumerate}
\end{definition}


\begin{definition}[Seminorm]
\label{definition:seminorm}
    Let $E$ be a vector space over $K \in \RC$, then a \textbf{seminorm} on $E$ is a mapping $\rho: E \to [0, \infty)$ such that:
    \begin{enumerate}
        \item[(SN1)] $\rho(0) = 0$.
        \item[(SN2)] For any $x \in E$ and $\lambda \in K$, $\rho(\lambda x) = \abs{\lambda} \rho(x)$.
        \item[(SN3)] For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
    \end{enumerate}
\end{definition}

\begin{lemma}
\label{lemma:continuous-seminorm}
    Let $E$ be a TVS over $K \in \RC$ and $[\cdot]: E \to [0, \infty)$ be a seminorm on $E$, then the following are equivalent:
    \begin{enumerate}
        \item $[\cdot]$ is uniformly continuous.
        \item $[\cdot]$ is continuous.
        \item $[\cdot]$ is continuous at $0$.
        \item $\bracs{x \in E| [x] < 1} \in \cn_E(0)$.
        \item $\bracs{x \in E| [x] \le 1} \in \cn_E(0)$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    $(5) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
    \[
    x - y \in \bracs{x \in E|[x] \le r} = r\bracs{x \in E|[x] \le 1} \in \cn_E(0)
    \]
    
    then $[x - y] \le r$.
\end{proof}


\begin{definition}[Topology Induced by Seminorm]
\label{definition:seminorm-topology}
    Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then:
    \begin{enumerate}
        \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudo-metric.
        \item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
        \item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
    \end{enumerate}
    The topology induced by $\seqi{d}$ is the \textbf{vector space topology induced by} $\seqi{[\cdot]}$. In addition,
    \begin{enumerate}
        \item[(U)] For any family $\bracsn{[\cdot]_j}_{j \in J}$ of continuous seminorms on $E$, the vector space topology induced by $\bracsn{[\cdot]_j}_{j \in J}$ is contained in the vector space topology induced by $\seqi{[\cdot]}$.
    \end{enumerate}
\end{definition}


\begin{definition}[Gauge/Minkowski Functional]
\label{definition:gauge}
    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be radial, then the mapping
    \[
    [\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
    \]

    is the \textbf{gauge/Minkowski functional} of $A$, and
    \begin{enumerate}
        \item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
        \item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
        \item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
        \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$.
    \end{enumerate}
    In particular,
    \begin{enumerate}[start=4]
        \item If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
        \item If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
    \end{enumerate}
\end{definition}
\begin{proof}
    (2): Let $\lambda, \mu > 0$ such that $\lambda^{-1}x, \mu^{-1}y \in A$. By convexity, $t\lambda^{-1} + (1 - t)\mu^{-1}y \in A$ for all $t \in [0, 1]$. Let $t \in [0, 1]$ such that
    \[
    (\lambda + \mu)^{-1}(x + y) = t\lambda^{-1}x + (1 - t)\mu^{-1}y
    \]

    then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.

    (4): Let $x \in \bracs{\rho \le 1}$, then $\lambda x \in A$ for all $\lambda \in (0, 1)$. Therefore
    \[
    x \in \overline{\bracs{\lambda x|\lambda \in (0, 1)}} \subset A
    \]

    so $x \in \overline{A}$. 
\end{proof}

\begin{definition}[Locally Convex Space]
\label{definition:locally-convex}
    Let $E$ be a TVS over $\RC$, then the following are equivalent:
    \begin{enumerate}
        \item There exists a fundamental system of neighborhoods at $0$ consisting of convex sets.
        \item There exists a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets.
        \item There exists a family of seminorms $\seqi{[\cdot]}$ that induces the topology on $E$.
    \end{enumerate}
    If the above holds, then $E$ is a \textbf{locally convex} space.
\end{definition}
\begin{proof}
    $(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \text{Conv}(V)$ be the convex hull of $V$, then $W \subset U$ is convex and circled. 

    $(2) \Rightarrow (3)$: For each $V \in \cn(0)$ convex, circled, and radial, let $[\cdot]_V: E \to [0, \infty)$ be its gauge, then $[\cdot]_V$ is a seminorm. For each $x, y \in X$ and $r > 0$, $[x - y]_V < r$ if and only if $x - y \in rV$. Thus the uniformity induced by $[\cdot]_V$ corresponds to the uniformity generated by $\bracs{U_{rV}| r > 0}$, where $U_V = \bracs{(x, y) \in E|x - y \in V}$. Since this holds for all $U \in \cn(0)$, the topology of $E$ and the topology induced by $\bracs{[\cdot]_V| V \in \cn(0), \text{ convex, circled, radial}}$ coincide.

    $(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.
\end{proof}
