\section{Support Functions}
\label{section:support-function}

\begin{definition}[Support Function]
\label{definition:support-function}
    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $A \subset E$ be non-empty, then the mapping
    \[
    H_A: F \to (-\infty, \infty] \quad y \mapsto \sup_{x \in A}\dpn{x, y}{\lambda}
    \]
    
    the \textbf{support function} of $A$ with respect to $\dpn{E, F}{\lambda}$. 
\end{definition}

\begin{definition}[Indicator Function]
\label{definition:infinity-characteristic-function}
    Let $E$ be a vector space over $\real$ and $A \subset E$, then the mapping
    \[
    I_A: E \to (-\infty, \infty] \quad x \mapsto \begin{cases}
        \infty &x \not\in A \\
        0 & x \in A
    \end{cases}
    \]

    is the \textbf{infinity characteristic function/indicator function} of $A$. 
\end{definition}



\begin{lemma}
\label{lemma:support-function-gymnastics}
    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, then:
    \begin{enumerate}
        \item For any non-empty $A \subset E$, let $\ol{\conv}(A)$ be the $\sigma(E, F)$-closed convex hull of $A$, then $H_A = H_{\ol{\conv}(A)}$. 
        \item For any non-empty $A, B \subset E$, $H_A \le H_B$ if and only if $A$ is contained in the $\sigma(E, F)$-closed convex hull of $B$.
        \item For any non-empty $A \subset E$, $I_A^* = H_A$. 
        \item For any $A \subset E$ non-empty, $\sigma(E, F)$-closed, and convex, $H_A^* = I_A$. 
    \end{enumerate}
\end{lemma}
\begin{proof}
    (1): Since $\ol{\conv}(A) \supset A$, $H_A \le H_{\ol{\conv}(A)}$. On the other hand, for each $\phi \in F$, $A \subset \bracs{\phi \le H_A(\phi)}$. Since $\bracs{\phi \le H_A(\phi)}$ is a $\sigma(E, F)$-closed convex set, it contains $\ol{\conv}(A)$. Thus $\ol{\conv}(A) \subset \bracs{\phi \le H_A(\phi)}$ and $H_{\ol{\conv}(A)}(\phi) \le H_A(\phi)$. 

    (2): Using (1), assume without loss of generality that $B$ is $\sigma(E, F)$-closed and convex. Suppose that $H_A \le H_B$, then $A \subset \bigcap_{\phi \in F}\bracs{\phi \le H_B(\phi)}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, $B = \bigcap_{\phi \in F}\bracs{\phi \le H_B(\phi)}$, so $A \subset B$. 

    (3): Let $\phi \in F$, then since $I_A|_{A^c} = \infty$, 
    \begin{align*}
    I_A^*(\phi) &= \sup_{x \in E}\dpn{x, \phi}{\lambda} - I_A(x) = \sup_{x \in A}\dpn{x, \phi}{\lambda} - I_A(x) \\
    &= \sup_{x \in A}\dpn{x, \phi}{\lambda} = H_A(\phi)
    \end{align*}
    
    (4): Given that $A$ is $\sigma(E, F)$-closed and convex, $I_S$ is convex and lower semicontinuous. By the \hyperref[Fenchel-Moreau Theorem]{theorem:fenchel-moreau} and (3), $I_A = I_A^{**} = H_A^*$. 
\end{proof}

\begin{theorem}
\label{theorem:support-function-seminorm}
    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, $f: X \to (-\infty, \infty]$ be a $\sigma(E, F)$-lower semicontinuous, subadditive, and positively homogeneous function with $f(0) = 0$, and
    \[
    \Sigma = \bracs{\phi \in F| (\phi, 0) \le f}
    \]

    then:
    \begin{enumerate}
        \item $f^* = I_\Sigma$. 
        \item $\Sigma$ is the unique non-empty $\sigma(F, E)$-closed convex subset of $F$ such that $f = H_\Sigma$. 
        \item $\Sigma$ is equicontinuous if and only if there exists $U \in \cn_E(0)$ such that $\sup_{x \in U}f(x) < \infty$. 
    \end{enumerate}

    Conversely, 
    \begin{enumerate}[start=2]
        \item For any non-empty $\Sigma \subset A$, $H_\Sigma$ is a lower semicontinuous, subadditive, and positively homogeneous function with $H_\Sigma(0) = 0$. 
    \end{enumerate}
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 4.25]{Clarke}}}. ]
    (1): Let $\phi \in \Sigma$, then since $\dpn{x, \phi}{\lambda} \le f(x)$ for all $x \in E$, 
    \[
    0 \ge f^*(\phi) = \sup_{x \in E}\dpn{x, \phi}{\lambda} - f(x) = \dpn{0, \phi}{\lambda} - f(0) = 0
    \] 

    where the supremum is achieved at $0$. On the other hand, if $\phi \in F \setminus \Sigma$, then there exists $x \in E$ such that $\dpn{x, \phi}{\lambda} - f(x) > 0$. In which case, by positive homogeneity of $f$, 
    \[
    f^*(\phi) \ge \sup_{\mu > 0}\dpn{\mu x, \phi}{\lambda} - f(\mu x) = \sup_{\mu > 0}\mu\braks{\dpn{ x, \phi} - f(x)} = \infty
    \]

    Thus $f^{*} = I_\Sigma$. 
    
    (2): By \autoref{lemma:lsc-affine-minorant}, $I_\Sigma \ne \infty$, so $\Sigma \ne \emptyset$. Since
    \[
    \Sigma = \bigcap_{x \in E}\bracs{\phi \in F| \dpn{x, \phi}{\lambda} \le f(x)}
    \]

    is an intersection of $\sigma(F, E)$-closed and convex sets, it is also $\sigma(F, E)$-closed. 
    
    Now, let $x, y \in E$ and $t \in [0, 1]$, then since $f$ is subadditive and positively homogeneous, 
    \[
        f((1 - t)x + ty) \le f((1 - t)x) + f(ty) = (1 - t)f(x) + tf(y)
    \]

    so $f$ is convex. Given that $f$ is also $\sigma(E, F)$-lower semicontinuous, the \hyperref[Fenchel-Moreau Theorem]{theorem:fenchel-moreau} and (4) of \autoref{lemma:support-function-gymnastics} imply that
    \[
    f = f^{**} = I_\Sigma^* = H_\Sigma
    \]

    By (2) of \autoref{lemma:support-function-gymnastics}, $\Sigma$ is the unique closed $\sigma(F, E)$-convex set such that $f = H_\Sigma$. 

    (3): Let $U \in \cn_E(0)$ be circled such that $M = \sup_{x \in U}f(x) < \infty$, then
    \[
    \sup_{\substack{y \in \Sigma \\ x \in U}}\dpn{x, y}{\lambda} \le \sup_{x \in U}f(x) = M < \infty
    \]

    Since $U$ is circled, $\bigcup_{y \in \Sigma}\dpn{M^{-1}U, y}{\lambda} \subset \ol{B_\real(0, 1)}$, so $\Sigma$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. 

    Conversely, if $\Sigma$ is equicontinuous, then there exists $U \in \cn_E(0)$ such that $M = \sup_{y \in \Sigma, x \in U}\dpn{x, y}{\lambda} < \infty$. In which case, (2) implies that
    \[
    \sup_{x \in U}f(x) = \sup_{x \in U}H_\Sigma(x) = \sup_{\substack{y \in \Sigma \\ x \in U}} \dpn{x, y}{\lambda} = M < \infty
    \]

    (4): By \autoref{proposition:semicontinuous-properties}, $H_\Sigma$ is lower semicontinuous.

    Let $x, y \in E$ and $\mu > 0$, then
    \begin{align*}
        H_\Sigma(x + y) &= \sup_{z \in \Sigma}\dpn{x + y, z}{\lambda} \\
        &\le \sup_{z \in \Sigma}\dpn{x, z}{\lambda} + \sup_{z \in \Sigma}\dpn{y, z}{\lambda} = H_\Sigma(x) + H_\Sigma(y)
    \end{align*}

    and
    \[
    H_\Sigma(\mu x) = \sup_{z \in \Sigma}\dpn{\mu x, z}{\lambda} = \mu\sup_{z \in \Sigma}\dpn{x, z}{\lambda} = \mu H_\Sigma(x)
    \]

    so $H_\Sigma$ is subadditive and positively homogeneous. 
    
    
\end{proof}