Reciprocal of Function of Bounded Variation Bounded away from Zero is of Bounded Variation

Theorem
Let $a, b$ be real numbers with $a < b$.

Let $f: \closedint a b \to \R$ be functions of bounded variation.

Let $f$ be bounded away from zero.

That is, there exists $M \in \R$ such that:


 * $\size {\map f x} \ge M > 0$

for all $x \in \closedint a b$.

Let $\map {V_f} {\closedint a b}$ be the total variation of $f$ on $\closedint a b$.

Then:


 * $\dfrac 1 f$ is of bounded variation

with:


 * $\map {V_{1 / f} } {\closedint a b} \le \dfrac {\map {V_f} {\closedint a b} } {M^2}$

where $\map {V_{1 / f} } {\closedint a b}$ denotes the total variation of $\dfrac 1 f$ on $\closedint a b$.

Proof
For each finite subdivision $P$ of $\closedint a b$, write:


 * $P = \set {x_0, x_1, \ldots, x_n }$

with:


 * $a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$

Note that:


 * $\dfrac 1 {\map f x} \le \dfrac 1 M$

for all $x \in \closedint a b$.

We then have:

Since $f$ is of bounded variation, there exists $K \in \R$ such that:


 * $\map {V_f} {P ; \closedint a b} \le K$

for every finite subdivision $P$.

Therefore:


 * $\map {V_{1 / f} } {P ; \closedint a b} \le \dfrac K {M^2}$

So $\dfrac 1 f$ is of bounded variation.

We also have: