Definition 11.2.2 (Bounded Variation, [Proposition X.1.1, Lan93]). Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, and $f: [a, b] \to E$. If $[f]_{\text{var}, \rho}< \infty$, then $f$ is of 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
$BV([a, b]; E)$ is a vector space.
For each continuous seminorm $\rho$ on $E$, $[\cdot]_{\text{var}, \rho}$ is a seminorm on $BV([a, b]; E)$.
Let $\fF$ be a filter on $BV([a, b]; E)$ and $f: [a, b] \to E$. If
$\pi_{x}(\fF) \to f(x)$ for all $x \in [a, b]$.
For every continuous seminorm $\rho$ on $E$, there exists $U \in \fF$ such that $\sup_{g \in U}[g]_{\text{var}, \rho}= M_{\rho} < \infty$.
then $f \in BV([a, b]; E)$ with $[f]_{\text{var}, \rho}\le M_{\rho}$.
For any $f \in BV([a, b]; E)$ and continuous seminorm $\rho$ on $E$, $\sup_{x \in [a, b]}\rho(f(x)) \le \rho(f(a)) + [f]_{\text{var}, \rho}$.
If $(E, \norm{\cdot}_{E})$ is a normed space, then
$f$ has at most countably many discontinuities.
Proof. (3): Let $\rho$ be a continuous seminorm on $E$ and $P \in \scp([a, b])$, then by assumption (a),
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
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
$|E_{k}| \ge N - k$.
$E_{k} \subset I_{k}^{o}$.
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 $[f]_{\text{var}}\ge N/n$ for all $N \in \nat^{+}$, so $[f]_{\text{var}}= \infty$.$\square$