5.2 Functions of Bounded Variation

Definition 5.2.1 (Total Variation).label Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, $f:[a,b]\to E$ and $P\in\scr{P}([a,b])$ be a partition, then

\begin{align*}V_{\rho,P}(f)=\sums{n}{j=1}\rho(f(x_{j})-f(x_{j-1}))\end{align*}

is the variation of $f$ with respect to $\rho$ and $P$. The supremum over all such parititions

\begin{align*}V_{\rho}(f)=\SUP{P\in\scr{P}([a,b])}V_{\rho,P}(f)\end{align*}

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 and denoted $V_{P}(f),V(f)$ respectively.

Definition 5.2.2 (Bounded Variation).label Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, and $f:[a,b]\to E$. If $V_{\rho}(f)<\infty$, then $f$ is of bounded variation with respect to $\rho$. The space of functions of bounded variation can be defined as

\begin{align*}BV([a,b];E)\define\curl{f:[a,b]\to E|\fall \text{continuous seminorm }\rho\text{ on } E,V_\rho(f)<\infty}\end{align*}

Observe

(1)
$BV([a,b];E)$ is a vector space with pointwise operations induced from $E$
(2)
For each continuous seminorm $\rho$ on $E$, $V_{\rho}(\cd)$ is a seminorm on $BV([a,b];E)$
(3)
For each continuous seminorm $\rho$ and each $M_{\rho}>0$, the sublevel set $S_{\rho}\define \curl{f\in E^{[a,b]}:V_\rho(f)\leq M_\rho}$ is closed with respect to the product topology on $E^{[a,b]}$. In particular if a filter $\fraks{F}$ is eventually contained in $S_{\rho}$ then any pointwise limit $f$ of $\fraks{F}$ also lies in it, and so $f\in BV([a,b];E)$ with $V_{\rho}(f)\leq M_{\rho}$.
(4)
For any $f\in BV([a,b];E)$ and continuous seminorm $\rho$ on $E$
\begin{align*}\SUP{x\in [a,b]}\rho(f(x))\leq \rho(f(a))+V_{\rho}(f)\end{align*}
(5)
If $(E,\norm{\cd}{E})$ is a normed vector space, then each $f\in BV([a,b];E)$ has at most countably many discontinuities.

Proof.

(1)
Follows from every continuous seminorm $\rho$ on $E$ is subadditive, absolutely homogeneous and definition of $BV([a,b];E)$ using such $\rho$.
(2)
Follows from seminorm properties of $\rho$ and definition of total variation.
(3)
Let $\rho$ be a continuous seminorm on $E$ and $P\in\scr{P}([a,b])$. In product topology on $E^{[a,b]}$, the map $V_{\rho,P}:E^{[a,b]}\to \R$ is continuous since $\rho$ and the evaluation maps $f\mto f(x)$ are continuous for each $x\in [a,b]$. Let $M_{\rho}>0$ then
\begin{align*}f\in S_{\rho}\iff V_{\rho}(f)\leq M_{\rho}\iff \fall P\in\scr{P}([a,b]),V_{\rho,P}(f)\leq M_{\rho}\end{align*}

thus $S_{\rho}=\caps{}{P\in\scr{P}([a,b])}\curl{f\in E^{[a,b]}:V_{\rho,P}(f)\leq M_\rho}$ is closed. Let $\fraks{F}$ be a filter on $E^{[a,b]}$ such that $S_{\rho}\in\fraks{F}$ and $\fraks{F}$ converges pointwise to some $f\in E^{[a,b]}$. Since $S_{\rho}$ is closed and belongs to $F$, its limit point must lie in $S_{\rho}$ hence $f\in S_{\rho}$ and $V_{\rho}(f)\leq M_{\rho}$.
(4)
Follows from triangle inequality for seminorms
(5)
For each $n\in\N^{+}$ define
\begin{align*}D_{n}\define \curl{x\in [a,b]:\fall \e>0,\exists y\in (x-\e,x+\e),\norm{f(x)-f(y)}{E}\geq \frac{1}{n}}\end{align*}

Then $D\define \cups{}{n\in \N^+}D_{n}$ is the set of discontinuity points of $f$. If $D$ is uncountable then there exists $N\in\N^{+}$ such that $D_{N}$ is uncountable. Let $E_{1}=D_{N}\cap (a,b)$ and $I_{1}=(a,b)$ and pick $x_{1}\in E_{1}\suf D_{N}$. Since $x_{1}\in D_{N}$ pick $y_{1}$ such that $\norm{f(x_1)-f(y_1)}{E}\geq \frac{1}{n}$ and $\norm{x_1-y_1}{}=\e_{1}< \text{diam}(I_{1})$, so the closed ball $B_{1}\define \cl{B}(x_{1},\e_{1})$ is contained in $I_{1}$. Inductively we define $I_{k+1}$ (resp. $E_{k+1}$) by removing $B_{k}$ from $I_{k}$ (resp. $E_{k}$). By taking smaller $\e_{k}$ if necessary we can guarantee $\abs{E_k}\geq N-k$ and $E_{k}\suf I^{o}_{k}$. We have constructed $N$ pairs $(x_{k},y_{k})$ such that $\norm{f(x_k)-f(y_k)}{E}\geq \frac{1}{n}$ the intervals containing each pair are disjoint. Thus
\begin{align*}V(f)\geq \sums{N}{k=1}\norm{f(x_k)-f(y_k)}{E}\geq \frac{N}{n}\end{align*}

Since $N$ was arbitrary, $V(f)=\infty$.

$\square$

Definition 5.2.3 (Variation Function).label Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$ and $f:[a,b]\to E$, then the function

\begin{align*}T_{f,\rho}(x)=\SUP{P\in\scr{P}([a,x])}V_{\rho,P}(f)=V_{\rho}(f|_{[a,x]})\end{align*}

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 $[c,d]\suf [a,b]$, $V_{\rho}(f|_{[c,d]})=T_{f,\rho}(d)-T_{f,\rho}(c)$

Proof. Let $P\in\scr{P}([a,c])$ and $Q=\curl{x_j}^{n}_{j=0}\in \scr{P}([a,d])$ be partitions containing $P$ then

\begin{align*}V_{\rho,Q}(f)-V_{\rho,P}=\sums{}{x_j>c}\rho(f(x_{j})-f(x_{j-1}))\leq V_{\rho}(f)\end{align*}

Taking supremum over $Q$ and $P$ respectively we have

\begin{align*}T_{f,\rho}(d)-T_{f,\rho}(c)\leq T_{f,\rho}(d)-V_{\rho,P}(f)\leq V_{\rho}(f|_{[c,d]})\end{align*}

OTOH for any $R\in\scr{P}([c,d])$, $P\cup R\in\scr{P}([a,d])$ and contains $P$ hence

\begin{align*}T_{f,\rho}(d)-V_{\rho,P}(f)\geq V_{\rho,P\cup R}(f)-V_{\rho,P}(f)=V_{\rho,R}(f)\end{align*}

Taking supremum over $P$ and $R$ respectively we obtain

\begin{align*}T_{f,\rho}(d)-T_{f,\rho}(c)\geq T_{f,\rho}(d)-V_{\rho,P}(f)\geq V_{\rho}(f|_{[c,d]})\end{align*}

$\square$

Proposition 5.2.4 (One Sided Limits).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 $\limit{y\searrow x}f(y)$ and $\limit{y\nearrow x}f(y)$ exists.

Proof. I’ll prove this whenever I reference it.$\square$

The Eigenspace

5.2 Functions of Bounded Variation

The Eigenspace