# The Unapologetic Mathematician

## Functions of Bounded Variation II

Let’s consider the collection of functions of bounded variation on $\left[a,b\right]$ a little more deeply. It turns out that they form a subring of the ring of all real-valued functions on $\left[a,b\right]$. Just to be clear, the collection of all real-valued functions on an interval becomes a ring by defining addition and multiplication pointwise.

Okay, so to check that we’ve got a subring we just have to check that the sum, difference, and product of two functions of bounded variation is again of bounded variation. Let’s take $f$ and $g$ to be two functions of bounded variation on $\left[a,b\right]$, and let $(x_0,...,x_n)$ be a partition of $\left[a,b\right]$. Then we calculate

$|f(x_i)g(x_i)-f(x_{i-1})g(x_{i-1})|=$
$|f(x_i)g(x_i)-f(x_{i-1})g(x_i)+f(x_{i-1})g(x_i)-f(x_{i-1})g(x_{i-1})|\leq$
$|f(x_i)g(x_i)-f(x_{i-1})g(x_i)|+|f(x_{i-1})g(x_i)-f(x_{i-1})g(x_{i-1})|=$
$|g(x_i)||f(x_i)-f(x_{i-1})|+|f(x_{i-1})||g(x_i)-g(x_{i-1})|\leq$
$A|f(x_i)-f(x_{i-1})|+B|g(x_i)-g(x_{i-1})|$

where $A$ is the least upper bound of $|g(x)|$ on $\left[a,b\right]$, and $B$ is the least upper bound of $|f(x)|$. Then we find $AV_f+BV_g$ is an upper bound for the sum over the partition. In fact, this not only shows that the product $fg$ is of bounded variation, it establishes the inequality $V_{fg}\leq AV_f+BV_g$.

The proofs for the sum and difference are similar. You should be able to work them out, and to establish the inequality $V_{f\pm g}\leq V_f+V_g$.

We can’t manage to get quotients of functions because we can’t generally divide functions. The denominator might be ${0}$ at some point, after all. But if $f(x)$ is bounded away from ${0}$ — if there is an $m$ with $0 — then $g(x)=\frac{1}{f(x)}$ is of bounded variation, and $V_g\leq\frac{V_f}{m^2}$. Indeed, we can check that

$\displaystyle\left|g(x_i)-g(x_{i-1})\right|=\left|\frac{1}{f(x_i)}-\frac{1}{f(x_{i-1})}\right|=$
$\displaystyle\left|\frac{f(x_i)-f(x_{i-1})}{f(x_i)f(x_{i-1})}\right|\leq\frac{\left|f(x_i)-f(x_{i-1})\right|}{m^2}$