Let $-\infty < a \le b < \infty$, $(E, \norm{\cdot}_E)$ be a [[Normed Vector Space|normed vector space]] over $F$, and $f: [a, b] \to E$ be a [[Function|function]]. > [!definition] > > Let $P = \seqfz{x_j}$ be a [[Partition of Interval|partition]] of $[a, b]$, then > $ > V_P(f) = \sum_{j = 1}^n \norm{f(x_j) - f(x_{j- 1})}_E > $ > is the **variation** of $f$ with respect to $P$. If $Q \in \scp([a, b])$ with $Q \ge P$, then $V_Q(f) \ge V_P(f)$. The [[Supremum and Infimum|supremum]] over all partitions > $ > \norm{f}_{\text{var}} = \norm{f}_{\text{var}, [a, b]} = \sup_{P \in \scp([a, b])}V_P(f) = \lim_{P \in \scp([a, b])}V_P(f) > $ > is the **total variation** of $f$ on $[a, b]$. > > If $\norm{f}_{\text{var}, [a, b]} < \infty$, then $f$ is of **bounded variation**. The set $BV([a, b]; E)$ is the space of all functions from $[a, b]$ to $E$ with bounded variation, where > > 1. For every $f \in BV([a,b]; E)$, $f$ has at most countably many discontinuities. > 2. $BV([a, b]; E)$ is a [[Vector Space|vector space]] and $\norm{\cdot}_{\text{var}, [a, b]}$ is a [[Seminorm|seminorm]] on $BV([a, b]; E)$. > 3. Let $\seq{f_n} \subset BV([a, b]; E)$ and $f: [a, b] \to E$ such that $f_n \to f$ pointwise and $M = \sup_{n \in \nat}\norm{f_n}_{\text{var}} < \infty$, then $f \in BV([a, b]; E)$ and $\norm{f}_{\text{var}} \le M$. > 4. For any $f \in BV([a,b]; E)$, $\norm{f}_{u} \le \norm{f(a)}_{E} + \norm{f}_{\text{var}}$. > > *Proof of Equality of Limits*. First note that if $\seqfz[m]{y_j} = Q \ge P$, then for each $0 \le j \le n$, there exists $0 \le k_j \le m$ such that $x_j = y_{k_j}$. In which case, > $ > \begin{align*} > V_P(f) &= \sum_{j = 1}^n \norm{f(x_j) - f(x_{j- 1})}_E \le \sum_{j = 1}^n\sum_{k = k_{j - 1} + 1}^{k_{j}}\norm{f(y_{k}) - f(y_{k - 1})}_E \\ > &= \sum_{k = 1}^m\norm{f(y_{k}) - f(y_{k - 1})}_E = V_Q(f) > \end{align*} > $ > Since $V_P(f) \le \sup_{Q \in \scp([a, b])}V_Q(f)$ for all $P \in \scp([a, b])$, and > $ > \sup_{Q \in \scp([a, b])}V_Q(f) \ge \lim_{Q \in \scp([a,b])}V_Q(f) > $ > On the other hand, for every $P \in \scp([a, b])$, $V_Q(f) \ge V_P(f)$ whenever $Q \ge P$. Therefore > $ > \sup_{Q \in \scp([a, b])}V_Q(f) \le \lim_{Q \in \scp([a,b])}V_Q(f) > $ > and the limits are equal. > > *Proof of Properties*. $(1)$: 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. > > Let $N \in \nat$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then $|E_1| \ge N$ and for every $x \in E_1$, there exists $\delta > 0$ such that $(x - \delta, x + \delta) \subset I_k$. > > For each $k \le N$, let $x_k \in E_k$. By inductive assumptions, there exists $\eps > 0$ such that $[x_k - \eps, x_k + \eps] \subset I_k$ and $E_k \setminus [x_k - \eps, x_k + \eps]$ has at least $N - k$ elements. Let $y_k \in I_k \cap [x_k - \eps, x_k + \eps]$ such that $\norm{f(x_k) - f(y_k)} \ge 1/n$. Lastly, let $I_{k + 1} = I_{k} \setminus [x_k - \eps, x_k + \eps]$ and $E_{k + 1} = E_k \setminus [x_k - \eps, x_k + \eps]$, then for every $x \in E_{k + 1}$, there exists $\delta > 0$ such that $(x - \delta, x + \delta) \subset I_{k + 1}$. > > Therefore there exists pairs $\seqf[N]{(x_k, y_k): 1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_{E} \ge 1/n$ for all $n \in \nat$, and the smallest interval containing each pair $(x_k, y_k)$ are pairwise disjoint. Let $P = \seqfz[m]{z_j} \in \scp([a, b])$ such that > $ > \bracs{z_j: 0 \le j \le m} \supset \bracs{x_k: 1 \le k \le N} \cup \bracs{y_k: 1 \le k \le N} > $ > then $V_P(f) \ge N/n$. As $N$ is arbitrary, $\norm{f}_{\text{var}} = \infty$. > > $(2)$: For any $f, g \in BV([a, b]; E)$ and $\lambda \in F$, > $ > V_P(\lambda f) = \sum_{j = 1}^n \abs{\lambda}\norm{f(x_j) - f(x_{j- 1})}_E = \abs{\lambda} V_P(f) > $ > and > $ > V_P(f + g) \le \sum_{j = 1}^n \braks{\norm{f(x_j) - f(x_{j - 1})}_E + \norm{g(x_j) - g(x_{j - 1})}_E} = V_P(f) + V_P(g) > $ > By taking the limit, $\norm{\lambda f}_{\text{var}} = \abs{\lambda}\norm{f}_{\text{var}}$ and $\norm{f + g}_{\text{var}} \le \norm{f}_{\text{var}} + \norm{g}_{\text{var}}$. Thus $\lambda f + g \in BV([a,b]; E)$. > > $(3)$: Let $P = \seqfz[m]{x_j} \in \scp([a, b])$, then > $ > \begin{align*} > V_P(f) &= \sum_{j = 1}^m \norm{f(x_j) - f(x_{j - 1})}_E = \limv{n}\sum_{j = 1}^m \norm{f_n(x_j) - f_n(x_{j - 1})}_E \\ > &= \lim_{n \to \infty}V_P(f_n) \le \sup_{n \in \nat}\norm{f_n}_{\text{var}} = M > \end{align*} > $ > As the above holds for all partitions, $\norm{f}_{\text{var}} \le M$. > > $(4)$: For any $t \in [a, b]$, $\norm{f(t) - f(a)}_E \le \norm{f}_{\text{var}}$. > [!definition] > > For each $t \in [a, b]$, let $T_f(t) = \norm{f}_{\text{var}, [a, t]}$, then $T_f$ is a non-decreasing function, known as the **variation function** of $f$. > > If $f \in BV([a, b]; E)$, then for any $t \in [a, b]$, $T_f$ is left/right-continuous at $t$ if and only if $f$ is left/right-continuous at $t$. > > *Proof*. For any $s \in [a, b]$, > $ > \abs{f(t) - f(s)} \le \begin{cases} > T_f(t) - T_f(s) &t \ge s \\ > T_f(s) - T_f(t) &t \le s > \end{cases} > $ > so if $T_f$ is left/right-continuous at $t$, then so is $F$. > > On the other hand, suppose for contradiction that$f$ is right-continuous at $t$ but $T_f$ is not, then there exists $\eps > 0$ such that $T_f(s) - T_f(t) = \norm{f}_{\text{var}, [t, s]} > \eps$ for all $s \ge t$. > > Let $s_1 \in (t, b]$. For each $n \in \nat$, there exists $P_n = \seqfz[m_n]{x_{n, j}} \in \scp([t, s_n])$ such that > $ > V_n = \sum_{j = 1}^{m_n}\abs{f(x_{n, j}) - f(x_{n, j - 1})} > \eps > $ > assume without loss of generality that $x_{n, j} > x_{n, j - 1}$ for all $1 \le j \le m_n$. By right-continuity of $f$, there exists $t < s_{n + 1} < x_{n, 1} < s_n$ such that $\norm{f(s_{n + 1}) - f(t)}_E < V_n - \eps$, then > $ > \abs{f(s_{n+1}) - f(x_{n, 1})} + \sum_{j = 2}^{m_n}\abs{f(x_{n, j}) - f(x_{n, j - 1})} > \eps > $ > so $\norm{f}_{\text{var}, [s_{n+1}, s_n]} > \eps$, and > $ > \norm{f}_{\text{var}, [t, s_1]} \ge \sum_{j = 1}^N\norm{f}_{\text{var}, [s_{j+1}, s_j]} \ge N\eps > $ > which contradicts the fact that $\norm{f}_{\text{var}, [t, s_1]} < \infty$. The same argument works in reverse for left-continuity.