Quotients of Functions of Bounded Variation

# Quotients 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 $P \in \mathscr{P} [a, b]$ 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 have already seen that if $f$ and $g$ are functions of bounded variation on $[a, b]$ and $k \in \mathbb{R}$ then the following functions are also of bounded variation on $[a, b]$:

• The sum $f + g$.
• The difference $f - g$.
• The multiple $kf$.
• The product $fg$.

Of course, the case regarding the quotient of functions is always a bother since $\displaystyle{\frac{f}{g}}$ may be undefined if $g$ equals zero, or may not be of bounded variation of $g$ approaches $0$.

To look at these cases more carefully, we will first prove a lemma telling us under what conditions the function $\displaystyle{\frac{1}{g}}$ is of bounded variation provided that $g$ is of bounded variation.

 Lemma 1: Let $g$ be a function of bounded variation on the interval $[a, b]$. If there exists an $m \in \mathbb{R}$, $m > 0$ such that for all $x \in [a, b]$ we have that $m \leq \mid g(x) \mid$ then $\frac{1}{g}$ is of bounded variation on $[a, b]$ and $\displaystyle{V_{1/g} (P) \leq \frac{V_g (P)}{m^2}}$.

If such an $m$ exists then we say that $f$ is bounded away from $0$.

• Proof: Let $g$ be a function of bounded variation on $[a, b]$. Then there exists a positive real number $M > 0$ such that for all partitions $P \in \mathscr{P}[a, b]$ we have that:
(2)
\begin{align} \quad V_g (P) = \sum_{k=1}^{n} \mid g(x_k) - g(x_{k-1}) \mid \leq M \end{align}
• Now suppose that there exists an $m \in \mathbb{R}$, $m > 0$ such that for all $x \in [a, b]$ we have that $m \leq \mid g(x) \mid$, and consider the variation of $\frac{1}{g}$ with respect to $P$:
(3)
\begin{align} \quad V_{1/g} (P) = \sum_{k=1}^{n} \biggr \lvert \frac{1}{g(x_k)} - \frac{1}{g(x_{k-1})} \biggr \rvert \end{align}
• When we take common denominators we get that:
(4)
\begin{align} \quad V_{1/g} (P) = \sum_{k=1}^{n} \biggr \lvert \frac{g(x_{k-1})}{g(x_k)g(x_{k-1})} - \frac{g(x_k)}{g(x_k)g(x_{k-1})} \biggr \rvert = \sum_{k=1}^n \frac{\mid g(x_{k-1}) - g(x_k) \mid}{\mid g(x_k)g(x_{k-1}) \mid} = \frac{\sum_{k=1}^{n} \mid g(x_{k-1}) - g(x_k) \mid}{\mid g(x_k)g(x_{k-1}) \mid} = \frac{V_g(P)}{\sum_{k=1}^{n} \mid g(x_k)g(x_{k-1}) \mid} \end{align}
• Now since $m \leq g(x)$ for all $x \in [a, b]$ we have that $m^2 \leq \mid g(x_k)g(x_{k-1}) \mid$ so $\frac{1}{\mid g(x_k) g(x_{k-1}) \mid} \leq \frac{1}{m^2}$ and so:
(5)
 Theorem 1: Let $f$ and $g$ be of bounded variation on the interval $[a, b]$ and let $g$ be such that there exists an $m \in \mathbb{R}$, $m > 0$ such that for all $x \in [a, b]$ we have that $m \leq \mid g(x) \mid$. Then $\displaystyle{\frac{f}{g}}$ is a function of bounded variation on $[a, b]$.
• Proof: By Lemma 1 we have that $\frac{1}{g}$ is a function of bounded variation, and we've already proven that products of functions of bounded variation are of bounded variation, so $\displaystyle{f \cdot \frac{1}{g} = \frac{f}{g}}$ is of bounded variation. $\blacksquare$