13.2 Functions of Bounded Variation

Definition 13.2.1 (Total Variation).label 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 variation of $f$ with respect to $\rho$ and $P$. The supremum over all such partitions

\[[f]_{\text{var}, \rho}= \sup_{P \in \scp([a, b])}V_{\rho, P}(f)\]

is the total variation of $f$ on $[a, b]$ with respect to $\rho$.

If $E$ is a normed vector space, then the variation and total variation of $f$ is taken with respect to its norm.

Definition 13.2.2 (Variation Function).label Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, $f: [a, b] \to E$, then the function

\[T_{f, \rho}(x) = \sup_{P \in \scp([a, x])}V_{\rho, P}(f) = [f|_{[a, x]}]_{\text{var}, \rho}\]

is the variation function of $f$ with respect to $\rho$, and:

(1)
$T_{f, \rho}: [a, b] \to [0, \infty]$ is a non-negative, non-decreasing function.
(2)
If $f \in BV([a, b]; E)$, then for any $[c, d] \subset [a, b]$, $[f]_{\text{var}, \rho}= T_{f, \rho}(d) - T_{f, \rho}(a)$.

Proof. (2): Let $P \in \scp([a, c])$ and $Q = \seqf{x_j}\in \scp([a, d])$ be partitions containing $P$, then

\[V_{\rho, Q}(f) - V_{\rho, P}(f) = \sum_{x_j > c}\rho(f(x_{j}) - f(x_{j - 1})) \le [f]_{\text{var}, \rho}\]

As this holds for all $Q \in \scp([a, d])$ containing $P$,

\[T_{f, \rho}(d) - T_{f, \rho}(c) \le T_{f, \rho}(d) - V_{\rho, P}(f) \le [f|_{[c, d]}]_{\text{var}, \rho}\]

On the other hand, for any $R \in \scp([c, d])$, $P \cup R \in \scp([a, d])$ and contains $P$. Therefore

\[T_{f, \rho}(d) - V_{\rho, P}(f) \ge V_{\rho, R \cup P}(f) - V_{\rho, P}(f) = V_{\rho, R}(f)\]

Since this holds for all $P \in \scp([a, c])$,

\[T_{f, \rho}(d) - T_{f, \rho}(c) \ge V_{\rho, R}(f)\]

and as the above holds for all $R \in \scp([c, d])$, $T_{f, \rho}(d) - T_{f, \rho}(c) \ge [f|_{[c, d]}]_{\text{var}, \rho}$.$\square$

Definition 13.2.3 (Bounded Variation).label 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

(1)
$BV([a, b]; E)$ is a vector space.
(2)
For each continuous seminorm $\rho$ on $E$, $[\cdot]_{\text{var}, \rho}$ is a seminorm on $BV([a, b]; E)$.
(3)
Let $\fF$ be a filter on $BV([a, b]; E)$ and $f: [a, b] \to E$. If
1. (1)
  $\pi_{x}(\fF) \to f(x)$ for all $x \in [a, b]$.
2. (2)
  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 $[f]_{\text{var}, \rho}\le M_{\rho}$.
(4)
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 vector space, then

(5)
$f$ has at most countably many discontinuities.

Proof [Proposition X.1.1, Lan93]. (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

(a)
$|E_{k}| \ge N - k$.
(b)
$E_{k} \subset I_{k}^{o}$.

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 $[f]_{\text{var}}= \infty$.$\square$

Proposition 13.2.4.label Let $E$ be a complete locally convex space and $f \in BV([a, b]; E)$, then for each $x \in [a, b]$, the limits $\lim_{y \downto x}f(y)$ and $\lim_{y \upto x}f(y)$ exist.

Proof. By flipping $f$, it is sufficient to consider the right-side limit $\lim_{y \downto x}f(y)$.

Let $\rho: E \to [0, \infty)$ be a continuous seminorm on $E$, and $T_{\rho, f}: [a, b] \to [0, \infty)$ be the variation function of $f$ with respect to $\rho$. For any $\eps > 0$, there exists $\delta > 0$ such that $T_{\rho, f}(z) - \lim_{y \downto x}T_{\rho, f}(y) < \eps$ for all $z \in (x, x + \delta)$. In which case, for any $x < y < z < x + \delta$,

\[\rho(f(z) - f(y)) \le [f|_{y, z}]_{\text{var}, \rho}\le T_{\rho, f}(z) - T_{\rho, f}(y) \le T_{\rho, f}(z) - \lim_{u \downto x}T_{\rho, f}(u) < \eps\]

By completeness of $E$, the limit $\lim_{y \downto x}f(y)$ exists.$\square$

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13.2 Functions of Bounded Variation

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