Multiples and Products of Functions of Bounded Variation

# Multiples and Products of Functions of Bounded Variation

Recall from the Functions of Bounded Variation page that $f$ is of bounded variation on the interval $[a, b]$ if there exists a positive real number $M > 0$ such that for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P} [a, b]$ we have that:

(1)\begin{align} \quad V_f (P) = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid \leq M \end{align}

We will now look at some nice theorems which tell us that if $f$ and $g$ are both of bounded variation on the interval $[a, b]$ then so is scalar multiple $kf$ for $k \in \mathbb{R}$ and the product $fg$.

Theorem 1: If $f$ is of bounded variation on the interval $[a, b]$ and $t \in \mathbb{R}$ then $tf$ is of bounded variation on $[a, b]$. |

**Proof:**Let $f$ be of bounded variation on the interval $[a, b]$. Then there exists a positive real number $M_0 > 0$ such that for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P} [a, b]$ we have that:

\begin{align} \quad V_f (P) = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid < M_0 \end{align}

- Let $h = tf$ for any $t \in \mathbb{R}$ and consider the variation of $h$ associated with $P \in \mathscr{P}[a, b]$:

\begin{align} \quad V_h (P) = \sum_{k=1}^{n} \mid h(x_k) - h(x_{k-1}) \mid = \sum_{k=1}^{n} \mid (tf)(x_k) - (tf)(x_{k-1}) \mid = \sum_{k=1}^{n} \mid t f(x_k) - tf(x_{k-1}) \mid \end{align}

- We note that $\mid t f(x_k) - tf(x_{k-1})\mid = \mid t \mid \mid f(x_k) - f(x_{k-1}) \mid$ and so:

\begin{align} \quad V_h(P) = \sum_{k=1}^{n} \mid t \mid \mid f(x_k) - f(x_{k-1}) \mid = \mid t \mid \sum_{k=1}^{n}\mid f(x_k) - f(x_{k-1}) \mid \leq \mid t \mid M_0 \end{align}

- Let $M = \mid t \mid M_0 > 0$. Then for all $P \in \mathscr{P}[a, b]$ there exists a positive real number $M > 0$ such that $V_h(P) = V_{tf}(P) \leq M$, so $h = tf$ is of bounded variation on $[a, b]$. $\blacksquare$

Theorem 2: If $f$ and $g$ are of bounded variation on the interval $[a, b]$ then $fg$ is of bounded variation on $[a, b]$. |

**Proof:**Let $f$ and $g$ be of bounded variation on the interval $[a, b]$. Then there exists positive real numbers $M_1, M_2 > 0$ such that for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P} [a, b]$ we have that:

\begin{align} \quad V_f (P) = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid < M_1 \quad \mathrm{and} \quad V_g (P) = \sum_{k=1}^{n} \mid g(x_k) - g(x_{k-1}) \mid < M_2 \end{align}

- Let $h = fg$ and consider the variation of $h$ associated with $P \in \mathscr{P}[a,b]$:

\begin{align} \quad V_h = \sum_{k=1}^{n} \mid h(x_k) - h(x_{k-1}) \mid = \sum_{k=1}^{n} \mid (fg)(x_k) - (fg)(x_{k-1}) \mid = \sum_{k=1}^{n} \mid f(x_k)g(x_k) - f(x_{k-1})g(x_{k-1}) \mid \end{align}

- We have that:

\begin{align} \quad \mid f(x_k)g(x_k) - f(x_{k-1})g(x_{k-1}) \mid = \mid f(x_k)g(x_k) - f(x_k)g(x_{k-1}) + f(x_k)g(x_{k-1}) - f(x_{k-1})g(x_{k-1}) \mid \\ \quad \mid f(x_k)g(x_k) - f(x_{k-1})g(x_{k-1}) \mid \leq \mid f(x_k)g(x_k) - f(x_k)g(x_{k-1}) \mid + \mid f(x_k)g(x_{k-1}) - f(x_{k-1})g(x_{k-1}) \mid \\ \quad \mid f(x_k)g(x_k) - f(x_{k-1})g(x_{k-1}) \mid \leq \mid f(x_k) \mid \mid g(x_k) - g(x_{k-1}) \mid + \mid g(x_{k-1}) \mid \mid f(x_k) - f(x_{k-1}) \mid \end{align}

- Since $f$ and $g$ are of bounded variation on the interval $[a, b]$ we have that $f$ and $g$ are bounded on $[a, b]$ and so there exists positive real numbers $A, B > 0$ such that for all $x \in [a, b]$ we have that $\mid f(x) \mid \leq A$ and $\mid g(x) \mid \leq B$. Since $x_k, x_{k-1} \in [a, b]$ for all $k \in \{ 1, 2, ..., n \}$ we have that $\mid f(x_k) \mid \leq A$ and $\mid g(x_{k-1}) \mid \leq B$ and so:

\begin{align} \quad \mid f(x_k)g(x_k) - f(x_{k-1})g(x_{k-1}) \mid \leq A \mid g(x_k) - g(x_{k-1}) \mid + B \mid f(x_k) - f(x_{k-1}) \mid + B \mid \end{align}

- Therefore we have that:

\begin{align} \quad V_h (P) \leq \sum_{k=1}^{n} [A \mid g(x_k) - g(x_{k-1}) \mid + B \mid f(x_k) - f(x_{k-1}) \mid + B \mid] \\ \quad V_h (P) \leq A \sum_{k=1}^{n} \mid g(x_k) - g(x_{k-1}) \mid + B \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid \\ \quad V_h (P) \leq AM_2 + BM_1 \end{align}

- Let $M = AM_2 + BM_1 > 0$. Then for all $P \in \mathscr{P}[a, b]$ there exists a positive real number $M > 0$ such that $V_h(P) = V_{fg}(P) \leq M$, so $h = fg$ is of bounded variation on $[a, b]$. $\blacksquare$