Additivity of the Total Variation of a Function

# Additivity of the Total Variation of a Function

Recall from the Total Variation of a Function page that if $f$ is of bounded variation on the interval $[a, b]$ then the total variation $V_f (a, b)$ of $f$ on $[a, b]$ is the supremum of the variations of $f$ associated with each partition $P \in \mathscr{P}[a, b]$, that is:

(1)\begin{align} \quad V_f(a, b) = \sup \left \{ V_f(P) : P \in \mathscr{P}[a, b] \right \} \end{align}

Now consider the interval $[a, b]$ and some point $c \in (a, b)$. Then $[a, b] = [a, c] \cup [c, b]$ and $[a, c] \cap [c, b] = \{ c \}$ and we would expect that the total variation of $f$ on $[a, b]$ would equal the total variation of $f$ on $[a, c]$ plus the total variation of $f$ on $[c, b]$. Fortunately, this is indeed the case as we prove in the following theorem.

Lemma 1: If $f$ is of bounded variation on the interval $[a, b]$, $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ and for some $c \in (a, b)$ we have that $\hat{P} = \{ a = x_0, x_1, ..., x_{k-1}, c, x_k, ..., x_n = b \} \in \mathscr{P}[a, b]$ then $V_f(P) \leq V_f(\hat{P})$. |

**Proof:**Let $f$ be a function of bounded variation on the interval $[a, b]$ and let $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ and let $c \in (a, b)$ be such that $\hat{P} = \{ a = x_0, x_1, ..., x_{k-1}, c, x_k, ..., x_n = b \} \in \mathscr{P}[a, b]$ is a refinement of $P$. Then:

\begin{align} \quad V_f(\hat{P}) = \mid f(x_1) - f(x_0) \mid + \mid f(x_2) - f(x_1) \mid + ... + \mid f(c) - f(x{k-1}) \mid + \mid f(x_k) - f(c) \mid + ... + \mid f(x_n) - f(x_{n-1}) \mid \\ \end{align}

- By the triangle inequality we have that $\mid f(c) - f(x_{k-1}) \mid + \mid f(x_k) - f(c) \mid \geq \mid f(x_k) - f(x_{k-1}) \mid$ and so:

\begin{align} \quad V_f(\hat{P}) \geq \mid f(x_1) - f(x_0) \mid + \mid f(x_2) - f(x_1) \mid + ... + \mid f(x_k) - f(x_{k-1}) \mid + ... + \mid f(x_n) - f(x_{n-1}) \mid = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid = V_f(P) \end{align}

- Therefore $V_f(P) \leq V_f(\hat{P})$ as desired. $\blacksquare$

Theorem 1: If $f$ is of bounded variation on the interval $[a, b]$ and $c \in (a, b)$ then $V_f(a,b) = V_f (a, c) + V_f (c, b)$. |

**Proof:**We will prove that $V_f(a, b) \leq V_f(a, c) + V_f(c, b)$ and then $V_f(a, b) \geq V_f(a, c) + V_f(c, b)$ to conclude that $V_f(a, b) = V_f(a, c) + V_f(c, b)$.

- Let $P_1 = \{ a = x_0, x_1, ..., x_n = c \} \in \mathscr{P}[a, c]$ and $P_2 = \{ c = y_0, y_1, ..., y_m = b \} \in \mathscr{P}[c, b]$. Then the union $P = \{ a = x_0, x_1, ..., x_n = c = y_0, y_1, ..., y_m = b \} = P_1 \cup P_2 \in \mathscr{P}[a, b]$. So:

\begin{align} \quad V_f(a, b) \geq V_f(P) = \sum_{k=1}^n \mid f(x_k) - f(x_{k-1}) \mid + \sum_{k=1}^{m}\mid f(y_k) - f(y_{k-1}) \mid = V_f(P_1) + V_f(P_2) \end{align}

- So $V_f(a, b) \geq V_f(P_1) + V_f(P_2)$ for ALL $P_1 \in \mathscr{P}[a, c]$ and for ALL $P_2 \in \mathscr{P}[c, b]$ so:

\begin{align} \quad V_f(a,b) \geq V_f(a, c) + V_f(c, b) \quad (*) \end{align}

- Now let $P = \{ a = x_0, x_1, ..., x_n \} \in \mathscr{P}[a, b]$. Then since $P$ is a partition of the interval $[a, b]$ and $c \in (a, b)$ we have that there exists a $k$ such that $x_{k-1} \leq c \leq x_k$. Define the new partition $\hat{P} = \{ a = x_0, x_1, ..., x_{k-1}, c, x_k, ..., x_n \} \in \mathscr{P}[a, b]$. Then for $P_1 = \{ a = x_0, x_1, ..., x_{k-1}, c \} \in \mathscr{P}[a, c]$ and $P_2 = \{ c, x_k, x_{k+1}, ..., x_n \} \in \mathscr{P}[c, b]$ we have that $\hat{P} = P_1 \cup P_2$ and so by applying Lemma 1:

\begin{align} \quad V_f(P) \leq V_f(\hat{P}) = V_f(P_1) + V_f(P_2) \end{align}

- So $V_f(P) \leq V_f(P_1) + V_f(P_2)$ for all $P_1 \in \mathscr{P}[a, c]$ and for all $P_2 \in \mathscr{P}[c, b]$ so:

\begin{align} \quad V_f(a, b) \leq V_f(a, c) + V_f(c, b) \quad (**) \end{align}

- From $(*)$ and $(**)$ we conclude that $V_f(a, b) = V_f(a, c) + V_f(c, b)$. $\blacksquare$