# 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

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

March 6, 2008 Posted by | Analysis | 2 Comments

## LaTeX screw-ups

Somehow this morning the renderer on WordPress decided to forget all of my backslashes in certain posts, with predictable results. I’m fixing them all back through the post on indefinite integration, but if you see any others earlier, send up a flag.

Incidentally, if you run into any LaTeX errors, particularly failures to render, please let me know in a comment on the offending page so I can fix them. It seems that every so often WordPress decides to play with those of us who use LaTeX, and I have to go back and fix all the mistakes they have introduced.

March 6, 2008 Posted by | Uncategorized | 7 Comments