# The Unapologetic Mathematician

## Mathematics for the interested outsider

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

1. John,

Thank you for your very interesting stuff on bdd- variation.

My question is : Say f and g are 2 monotonically increasing

functions. Is f-g mono increasing or decreasing. In my

case it looks as if f-g is decreasing. Is there any

proof of what f-g has to be ?

How do you determine the number that is the total variation

of f-g ?

Eugene

Comment by eugene lalonde | April 5, 2008 | Reply

2. The best you can say for that difference is that it’s of bounded variation. It might be increasing in some areas and decreasing in others. As for the variation of the difference, the best you can say is that it’s less than the sum of the variations of $f$ and $g$ separately. Luckily, those variations are easy, since for monotonic functions it’s just the difference in values at the two endpoints.

Comment by John Armstrong | April 5, 2008 | Reply

3. […] consider jump discontinuities. These are especially useful to understand, since they’re the only sort a function of bounded variation can have. So let’s say we take an interior point and define as simple a function as we can with a […]

Pingback by Integrating Across a Jump « The Unapologetic Mathematician | March 14, 2009 | Reply

4. Hi! Thanks for posting such interesting stuff. I would like to be able to say that all discontinuity points of a bounded variable function are isolated. Your observation “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” would be sufficient. But… are you sure it is true? (I am not sure). Any reference? Thanks!

Comment by Luis Mendo | August 26, 2009 | Reply

5. … Sorry, my previous comment is nonsense. My real problem is: I would like to order the set of discontinuities of a bounded-variation function , i.e. to put them in a monotone sequence. Do you think that can be done?

Comment by Luis Mendo | August 26, 2009 | Reply

6. I’m not sure what you mean. They’re already ordered because they lie on the real line, which is totally ordered.

Comment by John Armstrong | August 26, 2009 | Reply

• Sorry, I didn’t explain well. Given a bounded variation function $f$, I would like to say “let $(x_i)$ be the sequence of discontinuity points of $f$, with $x_i < x_{i+1}$ for all $n \in \mathbb N$." But I don't see if that can be done. For example, the set of discontinuity points can be $\{1/n | n \in \mathbb N}$. Or worse, $\{1/n+1/m | m,n \in \mathbb N}$ (which is also countable), in which every point is an accumulation point. Can you order those points in an increasing sequence? I do not see how there is a first (i.e. lowest) element, or a second, or… So it seems it can´t be done in general. What do you think?

Comment by Luis Mendo | August 28, 2009 | Reply

7. You’re asking, in effect, if the order induced by the order on the real line (the one I was talking about) is a well-order. And your examples show that, no, in general it’s not.

Comment by John Armstrong | August 28, 2009 | Reply