The Unapologetic Mathematician

Let’s say we take an integrator of bounded variation on an interval and a function that’s Riemann-Stieltjes integrable with respect to over that interval. Then we know that is also integrable with respect to over the subinterval . Let’s use this to define a function on by

We can immediately say some interesting things about this function. First of all, is, like , of bounded variation. Next, wherever is continuous, so is . Finally, if is increasing, then is differentiable wherever is differentiable and is continuous. At such points, we have . Notice that, as usual, the first two results will hold if we show that they hold for increasing integrators.

These results are similar to those we get from the Fundamental Theorem of Calculus, and we can use some of the same techniques to prove them. In particular, we call on the Integral Mean-Value Theorem for Riemann-Stieltjes integrals. If we take points and in , this tells us that

where is some point between and .

Now we let be the supremum of on . For any partition of we calculate the variation

thus giving an upper bound to the variation of .

Similarly, let be a point where is continuous. Then given an we can find a so that implies that . Thus , and we see that is continuous at as well.

Finally, we can divide by to find

Then as gets closer to , gets squeezed in towards as well. If is continuous at and is differentiable there, then the limit of this difference quotient exists, and has the value stated.

Over at the new, improved Mathematics Under the Microscope (off of Blogger, onto WordPress), Alexandre makes an interesting point. word problems take place in a graded ring.

Sting is right: he used to be kind of cool once.

From their 1980 album Zenyattà Mondatta, The Police with “Don’t Stand So Close to Me”.
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We’ve got two different analogues of the integral mean value theorem for the Riemann-Stieltjes integral.

The first one says that if is increasing on and is integrable with respect to , with supremum and infimum in the interval, then there is some “average value” between and . This satisfies

In particular, we should note that if is continuous then the intermediate value theorem tells us that there is some with . That is, there is some such that

When this gives us the old integral mean value theorem back again.

So why does this work? Well, if then both sides are zero and the theorem is trivially true. Now, the lowest lower sum is , while the highest upper sum is . The integral itself, which we’re assuming to exist, lies between these bounds:

So we can divide through by to get the result we seek.

We can get a similar result which focuses on the integrator by using integration by parts. Let’s assume is continuous and is increasing on . Our sufficient conditions tell us that the integral of with respect to exists, and the integration by parts formula says

But the first integral mean value theorem tells us that the integral on the right is equal to for some . Then we can rearrange the above formula to read

So there is some point so that the integral of is the same as the integral of the step function taking the value until and the value after it.

Someone got ahold of one of my students’ homeworks from last semester.

Isabel at God Plays Dice cites (uncharacteristically credulously) this story of a 13th-century monk who supposedly discovered the Mandelbrot set. How surprising, since we didn’t see it until we had computer graphics. But there’s a lot more anachronism just beneath the surface. There are so many topics that feed into the standard representation of the Mandelbrot set that we have so completely internalized we hardly know how to think without them.

First, this story asks us to believe that Brother Udo worked with complex numbers 300 years before Tartaglia and Cardano hesitantly advanced them. They were motivated by problems that could not be solved without them, while here they’re just being used for multiplications that can be perfectly well defined over the reals. Why would he have even thought to use complex numbers? Sure, maybe he was using a geometric description that happened to exactly correspond, but he wouldn’t have thought of it as complex multiplication. Still, more on that possibility later.

Next the story asks us to believe that Brother Udo graphed algebraic relations 400 years before Descartes introduced analytic geometry. So he somehow had the idea that algebra and geometry were interlinked so far in advance of everybody else and didn’t tell anyone? Again, maybe the construction was wholly geometrical.

Next the story asks us to believe that Brother Udo was not only graphing any old algebraic relations, but using the graphical representation of complex numbers 500 years before Argand described that interpretation. Again, this is something we internalize in high school (or earlier) and so we forget how late in the game it actually came along.

Okay, so could there have been a totally geometrical construction that these historians are simplifying into the function ? I doubt it. Mostly because you run it 70 times and then.. what? In rendering of the Mandelbrot set you know whether to keep or toss a point by whether you eventually go outside the circle of radius 2. So now we need to posit an unspoken proof that once a point leaves that circle it can never return. No reference is made to that result either.

Of course, the page cites sources. But the only source directly on point for the main content of the story is this one:

[5] Schipke, R.J. and Eberhardt, A. “The forgotten genius of Udo von Aachen”, Harvard Journal of Historical Mathematics, 32, 3 (March 1999), pp 34-77.

but this “Harvard Journal of Historical Mathematics” doesn’t seem to exist either. In fact, a Google search on that journal’s name gives… a bunch of references back to this very story! Let me be clear about this:

Just because someone includes a bibliography doesn’t mean they’re not forging the bibliography too!

And, of course, the page itself says it was published April 1, 1999. It’s not even a new hoax. But as I show above, you don’t need to see that cue to realize it’s a joke. You just have to stop, think, and really understand what someone is trying to sell you. You have to be skeptical about any assertion of fact. And it seems a lot of people aren’t.

This story got included in course materials for a course on fractal geometry taught at Yale by my teaching mentor Michael Frame, with help by Benoît Mandelbrot himself. John Allen Paulos wrote it up for ABC news. Sure, maybe they knew it was a hoax and passed it along as such. Paulos wrote his article on April 1, 2001, after all. But then it gets swept up by woo like “Life Technology”, “The All-Oneness Hadron Materia”, and badly-researched psychology texts. And nobody tags it as the hoax it is.

So here I am. This story is false. Unquestionably false, like faking the moon landing or a face on Mars. False like psychic powers and divine cameos on corn chips. False like that girl who got so high on LSD she microwaved the baby she was taking care of. Those of you who love linking to big-name skepti-bloggers might want to point them here so we can squelch this thing once and for all.

As Chris Hillman just pointed out in a comment, the 2008 Abel Prize went to John Griggs Thompson and Jacques Tits “for their profound achievements in algebra and in particular for shaping modern group theory”. The comment went on my recent throwaway post about the 7×7x7 Rubik’s Cube, but a more appropriate one might have been this one from over a year ago, in which I discuss the Feit-Thompson theorem in passing.

Incidentally, I think I’ve met both of the winners. Tits I’m sure of, since I tried and failed horribly to take a short course he gave at Yale on “buildings”. Thompson I believe showed up for Walter Feit’s memorial, but I could be wrong about that. I wish I could say I was particularly close to one or the other, but I suppose that will have to wait until Adams and Vogan win the prize.

Now that we know how a Riemann-Stieltjes integral behaves where the integrand has a jump, we can put jumps together into more complicated functions. The ones we’re interested in are called “step functions” because their graphs look like steps: flat stretches between jumps up and down.

More specifically, let’s say we have a sequence of points

and define a function to be constant in each open interval . We can have any constant values on these intervals, and any values at the jump points. The difference we call the “jump” of at . We have to be careful here about the endpoints, though: if then the jump at is , and if then the jump at is . We’ll designate the jump of at by .

So, as before, the function may or may not be continuous from one side or the other at a jump point . And if we have a function discontinuous on the same side of the same point, then the integral can’t exist. So let’s consider any function so that at each , at least one of or is continuous from the left, and at least one is continuous from the right. We can chop up the interval into chunks so that each one contains only one jump, and then the result from last time (along with the “linearity” of the integral in the interval) tells us that

That is, each jump gives us the function at the jump point times the jump at that point, and we just add them all together. So finite weighted sums of function evaluations are just special cases of Riemann-Stieltjes integrals.

Here’s a particularly nice family of examples. Let’s start with any interval and some natural number . Define a step function by starting with and jumping up by at , , , and so on. Then the integral of any continuous function on gives

But notice that this is just a Riemann sum for the function . Since is continuous, we know that it’s Riemann integrable, and so as gets larger and larger, these Riemann sums must converge to the Riemann integral. That is

But at the same time we see that converges to . Clearly there is some connection between convergence and integration to be explored here.

In the discussion of necessary conditions for Riemann-Stieltjes integrability we saw that when the integrand and integrator are discontinuous from the same side of the same point, the integral can’t exist. But how close can we come to that situation? It turns out that as long as one of the two functions is continuous from each side, then things generally work out.

Specifically, let’s 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 jump there. We let it be constant on either side, with for and for . We’ll let be anything at all. Generally it will be discontinuous from both sides at , but if then we’ll have continuity from the left, and similarly on the right. Of course, we could have , with being the only point with a different value for .

Now let’s let be any other function on . We know that we can’t let it be discontinuous from the left at if is, or from the right either, so let’s assume it’s continuous from the left or the right at to satisfy these conditions, but put no other assumptions on it. I assert that is then integrable with respect to on , and has the value

Where is the limit of as we approach from the right. In our situation this will be , but we’ll telegraph a bit by writing it like this. Similarly, is the limit of as we approach from the left, which here is .

To see that this is the case, take a tagged partition , and if it doesn’t already contain just refine it by throwing the new point in. Now every term in the Riemann-Stieltjes sum

is zero except for the one on either side of . We can then write the difference as

Since either or is continuous from the left, the first term above must go to zero. Similarly, because at least one is continuous from the right, the second term must also go to zero. Thus the sums converge to the value asserted.

We’ve talked about sufficient conditions for integrability, which will tell us that a given integral does exist. Now we’ll consider the situation where we know that a given integral exists, and see what we can tell about its integrand or integrator.

First, let’s consider an increasing integrator over an interval , and an interior point . Let’s assume that both and the integrand are discontinuous from the right at . Then the integral cannot exist.

Consider what the assumption of discontinuity means: there is some so that for every , no matter how small, there are points so that and . That is, as we approach from the right we can always find a sample point that differs from the value we want by at least , both in the integrand and in the integrator. So let’s see what happens when we throw in as a partition point.

Suddenly we’re stuck! Riemann’s condition tells us to look at the sum

where each term is positive. So to make this sum go down to zero in Riemann’s condition we need to make each term go down to zero. But for the subinterval whose left endpoint is , we can’t make this happen! The change in will always be at least for some smaller subinterval, and the difference in sample values of must always exceed as well. Thus the difference between the upper and lower sum will always be at least , violating Riemann’s condition, and thus integrability.

Of course, we could run through the same argument approaching from the left. We could also use an integrator of bounded variation, as usual.

What this tells us is that while we may allow integrands or integrators to be discontinuous, they can’t be discontinuous from the same side at the same point. For example, any discontinuity in an integrator of bounded variation is fine as long as we’re pairing it off with a continuous integrand, as we did last time.