\section{Functions of Bounded Variation}
\label{section:bv}


\begin{definition}[Total Variation]
\label{definition:total-variation}
    Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, $f: [a, b] \to E$, and $P \in \scp([a, b])$ be a partition, then
    \[
    V_{\rho, p}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
    \]
    is the \textbf{variation} of $f$ with respect to $\rho$ and $P$. The supremum over all such partitions
    \[
    [f]_{\var, \rho} = \sup_{P \in \scp([a, b])}V_{\rho, P}(f)
    \]
    is the \textbf{total variation} of $f$ on $[a, b]$ with respect to $\rho$.

    If $E$ is a normed space, then the variation and total variation of $f$ is taken with respect to its norm.
\end{definition}

\begin{definition}[Bounded Variation, {{\cite[Proposition X.1.1]{Lang}}}]
\label{definition:bounded-variation}
    Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, and $f: [a, b] \to E$. If $[f]_{\var, \rho} < \infty$, then $f$ is of \textbf{bounded variation} with respect to $\rho$.

    The space $BV([a, b]; E)$ is the set of functions $[a, b] \to E$ of bounded variation with respect to every continuous seminorm on $E$, and
    \begin{enumerate}
        \item $BV([a, b]; E)$ is a vector space.
        \item For each continuous seminorm $\rho$ on $E$, $[\cdot]_{\var, \rho}$ is a seminorm on $BV([a, b]; E)$.
        \item Let $\fF$ be a filter on $BV([a, b]; E)$ and $f: [a, b] \to E$. If
        \begin{enumerate}
            \item $\pi_x(\fF) \to f(x)$ for all $x \in [a, b]$.
            \item For every continuous seminorm $\rho$ on $E$, there exists $U \in \fF$ such that $\sup_{g \in U}[g]_{\var, \rho} = M_\rho < \infty$.
        \end{enumerate}
        then $f \in BV([a, b]; E)$ with $[f]_{\var, \rho} \le M_\rho$.
        \item For any $f \in BV([a, b]; E)$ and continuous seminorm $\rho$ on $E$, $\sup_{x \in [a, b]}\rho(f(x)) \le \rho(f(a)) + [f]_{\var, \rho}$.
    \end{enumerate}
    If $(E, \norm{\cdot}_E)$ is a normed space, then
    \begin{enumerate}
        \item[(5)] $f$ has at most countably many discontinuities.
    \end{enumerate}
\end{definition}
\begin{proof}
    (3): Let $\rho$ be a continuous seminorm on $E$ and $P \in \scp([a, b])$, then by assumption (a),
    \[
    V_{\rho, P}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
    = \lim_{g, \fF}\sum_{j = 1}^n \rho(g(x_j) - g(x_{j - 1}))
    = \lim_{g \in \fF}V_{\rho, P}(g)
    \]
    By assumption (b), $[0, M_\rho]$ is in the filter generated by $V_{\rho, P}(\fF)$. Thus $V_{\rho, P}(f) \le M_\rho$. As this holds for all $P \in \scp([a, b])$, $V_{\rho, P}(f) \le M_\rho$, and $f \in BV([a, b]; E)$.

    (5): For each $n \in \nat^+$, let
    \[
    D_n = \bracs{x \in [a, b]|\forall \eps > 0, \exists y \in (x - \eps, x + \eps): \norm{f(x) - f(y)}_E \ge 1/n}
    \]
    then $D = \bigcup_{n \in \nat^+}D_n$ is the set of discontinuity points of $f$. If $D$ is uncountable, then there exists $N \in \nat^+$ such that $D_n$ is infinite.

    Fix $N \in \nat^+$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then
    \begin{enumerate}
        \item[(a)] $|E_k| \ge N - k$.
        \item[(b)] $E_k \subset I_k^o$.
    \end{enumerate}
    for $k = 1$.

    Let $k \le N$ and suppose inductively that $E_k, I_k$ have been constructed. Let $x_k \in E_k$, then by (b), there exists $\eps > 0$ such that $[x_k - \eps, x_k + \eps] \subset I_k$ and $|E_k \setminus [x_k - \eps, x_k + \eps]| \ge N - k$. Let $y_k \in [x_k - \eps, x_k + \eps]$ such that $\norm{f(x_k) - f(y_k)} \ge 1/n$, $I_{k + 1} = I_k \setminus [x_k - \eps, x_k + \eps]$, and $E_{k+1} = E_k \setminus [x_k - \eps, x_k + \eps]$, then $I_k$ and $E_k$ satisfies (a) and (b).

    Therefore there exists pairs $\bracs{(x_k, y_k)|1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_E \ge 1/n$ for all $n$, and the smallest interval containing each $(x_k, y_k)$ are pairwise disjoint. Thus $[f]_{\var} \ge N/n$ for all $N \in \nat^+$, so $[f]_{\var} = \infty$.
\end{proof}