## Functions of Bounded Variation II

Let’s consider the collection of functions of bounded variation on a little more deeply. It turns out that they form a subring of the ring of *all* real-valued functions on . 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 and to be two functions of bounded variation on , and let be a partition of . Then we calculate

where is the least upper bound of on , and is the least upper bound of . Then we find is an upper bound for the sum over the partition. In fact, this not only shows that the product is of bounded variation, it establishes the inequality .

The proofs for the sum and difference are similar. You should be able to work them out, and to establish the inequality .

We can’t manage to get quotients of functions because we can’t generally divide functions. The denominator might be at some point, after all. But if is bounded away from — if there is an with — then is of bounded variation, and . Indeed, we can check that

## 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.

## Functions of Bounded Variation I

In our coverage of the Riemann-Stieltjes integral, we have to talk about Riemann-Stieltjes sums, which are of the form

How can we choose a function to make sums like this get really big? Well, we could make the function get big, but that’s sort of a cheap trick. Let’s put a cap that . Now how can we make a Riemann-Stieltjes sum big?

Well, sometimes and sometimes . In the first case, we can try to make , and in the second we can try to make . In either case, we’re adding as much as we possibly can, subject to our restriction.

We’re saying a lot of words here, but what it boils down to is this: given a partition we can get up to in our sum. As we choose finer and finer partitions, it’s not hard to see that this number can only go up. So, if there’s an upper bound for our , then we’re never going to get a Riemann-Stieltjes sum bigger than . That will certainly make it easier to get nets of sums to converge.

So let’s make this definition: a function is said to be of “bounded variation” on the interval if there is some so that for any partition we have

That is, is an upper bound for the set of variations as we look at different partitions of . Then, by the Dedekind completeness of the real numbers, we will have a *least* upper bound , which we call the “total variation” of on

Let’s make sure that we’ve got some interesting examples of these things. If is monotonic increasing — if implies that — then we can just drop the absolute values here. The sum collapses, and we’re just left with for every partition. Similarly, if is monotonic decreasing then we always get . Either way, is clearly of bounded variation.

An easy consequence of this condition is that functions of bounded variation are bounded! In fact, if has variation on , then for all . Indeed, we can just use the partition and find that

from which the bound on easily follows.

## The Riemann-Stieltjes Integral IV

Let’s do one more easy application of the Riemann-Stieltjes integral. We know from last Friday that when our integrator is continuously differentiable, we can reduce to a Riemann integral:

So where else have we seen derivatives as factors in integrands? Right! integration by parts! Here our formula says that

We can rewrite this using Riemann-Stieltjes integrals as

So if and are both continuously differentiable, this formula gives back our rule for integration by parts. But we can prove this without making those assumptions. In fact, we just need to assume that one of the two integrals exists, and the existence of the other one (and the formula) will follow.

Let’s assume that exists. That is, for every there is some tagged partition so that for every finer partition we have

Now let’s take any partition finer than and use it to set up the Rieman-Stieltjes sum

We can also use this partition to rewrite as

So subtracting the one from the other we find

But this is a Riemann-Stieltjes sum for the partition we get by throwing together all the and as partition points, and using as tags. This is a finer partition than , and so we see that

whenever is a partition finer than . This shows that the Riemann-Stieltjes integral of with respect to exists, and has the value we want.

## The Riemann-Stieltjes Integral III

Last Friday we explained the change of variables formula for Riemann integrals by using Riemann-Stieltjes integrals. Today let’s push it a little further and prove a change of variables formula for Riemann-Stieltjes integrals.

We start with a function which we assume to be Riemann-Stieltjes integrable by the function . Now, instead of the full generality we used before, let’s just let be a strictly increasing continuous function with and . Define and to be the composite functions and . Then is Riemann-Stieltjes integrable by on , and we have the equality

For decreasing functions we get almost the exact same thing, so you should figure out the parallel statement and proof yourself.

Since is strictly increasing, it must be one-to-one, and it’s onto by assumption. In fact, is an explicit homeomorphism of the intervals and , and its inverse is also a strictly increasing continuous function. We can now use and its inverse to set up a bijection between partitions of and : if is a partition, then is a partition, and vice versa. Further, refinements of partitions of one side correspond to refinements of partitions on the other side.

So if we’re given an then there’s some partition of so that for any finer partition we have . Let be the corresponding partition of , and let be a partition of finer than it. Then it’s easily verified that the Riemann-Stieltjes sum is equal to the Riemann-Stieltjes sum . Everything else follows quickly from here.