# The Unapologetic Mathematician

## Bounded Variation addenda

I left a few things out of last Saturday’s post. Since I’ve spent all morning finishing off that paper (I’ll post the arXiv link tomorrow when it shows up) I’m sort of mathed out for the moment. I’ll just tack up these addenda and take a nap or something to brace for tomorrow’s Clifford Lectures (which is the only day of them I’ll be able to see, due to airline pricing).

Okay, so we said that we can represent any function $\alpha$ of bounded variation as the difference $\alpha_+-\alpha_-$ of two increasing functions. But we should notice here that this decomposition is by no means unique. In fact, if $\beta$ is an increasing function, then we can also write $\alpha=(\alpha_++\beta)-(\alpha_-+\beta)$, and get a different representation of $\alpha$.

Usually nonuniqueness gets messy, but there’s a way this can come in handy. If we pick $\beta$ to be strictly increasing, then so will $\alpha_++\beta$ and $\alpha_-+\beta$ be. So any function of bounded variation can be written as the difference between two strictly increasing functions. This restriction may be useful in some situations.

It also turns out that wherever the function $\alpha$ is continuous, then so will its variation $V_\alpha$ be. Thus any continuous function of bounded variation can be written as the difference of two continuous, (strictly) increasing functions. I’ll leave this fact to you, though it’s not really hard.

Now let’s look at bounded increasing functions for a moment. Such a function $f$ might jump up at certain points in its domain, like the Heaviside function $H(x)$ that sends all negative numbers to ${0}$ and all nonnegative numbers to $1$. However, a monotone function can’t have any other kinds of discontinuities. Further, since each jump must increase by a finite amount, we can only have a countable number of them in a finite interval, or else the function would have to take arbitrarily large values and would no longer be bounded!

So increasing functions can only have a countable number of jump discontinuities. But any function of bounded variation is the difference of two increasing functions. Thus any function of bounded variation can only have a countable number of discontinuities, where at worst the function jumps up or down by a finite amount at each one. The only other sort of discontinuity is a point where the function has a limit, but takes a different value. For instance, $H(x)-(-H(-x))$ takes the value $1$ at every positive or negative number, but $H(0)=2$.

March 11, 2008 Posted by | Analysis | 8 Comments