# 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:

 $\ds \map {V_{1 / f} } {P ; \closedint a b}$ $=$ $\ds \sum_{i \mathop = 1}^n \size {\frac 1 {\map f {x_i} } - \frac 1 {\map f {x_{i - 1} } } }$ using the notation from the definition of bounded variation $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \size {\frac {\map f {x_{i - 1} } - \map f {x_i} } {\map f {x_i} \map f {x_{i - 1} } } }$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \frac {\size {\map f {x_i} - \map f {x_{i - 1} } } } {\size {\map f {x_i} \map f {x_{i - 1} } } }$ $\ds$ $\le$ $\ds \frac 1 {M^2} \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }$ since $\size {\map f {x_i} \map f {x_{i - 1} } } \le M^2$ $\ds$ $=$ $\ds \frac {\map {V_f} {P ; \closedint a b} } {M^2}$

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:

 $\ds \map {V_{1 / f} } {\closedint a b}$ $=$ $\ds \sup_P \paren {\map {V_{1 / f} } {P ; \closedint a b} }$ Definition of Total Variation of Real Function on Closed Bounded Interval $\ds$ $\le$ $\ds \sup_P \paren {\frac {\map {V_f} {P ; \closedint a b} } {M^2} }$ $\ds$ $=$ $\ds \frac 1 {M^2} \sup_P \paren {\map {V_f} {P ; \closedint a b} }$ Multiple of Supremum $\ds$ $=$ $\ds \frac {\map {V_f} {\closedint a b} } {M^2}$ Definition of Total Variation of Real Function on Closed Bounded Interval

$\blacksquare$