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.
March 27, 2008
Posted by
John Armstrong |
Uncategorized |
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19 Comments
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
![\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\sum\limits_{i=1}^nf(c_k)\alpha_k \displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\sum\limits_{i=1}^nf(c_k)\alpha_k](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha%3D%5Csum%5Climits_%7Bi%3D1%7D%5Enf%28c_k%29%5Calpha_k&bg=e6e6e6&fg=000000&s=0)
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
![\displaystyle\int\limits_{\left[a,b\right]}fd\alpha_n=\sum\limits_{i=1}^nf\left(a+(i-\frac{1}{2})\frac{b-a}{n}\right)\frac{b-a}{n} \displaystyle\int\limits_{\left[a,b\right]}fd\alpha_n=\sum\limits_{i=1}^nf\left(a+(i-\frac{1}{2})\frac{b-a}{n}\right)\frac{b-a}{n}](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha_n%3D%5Csum%5Climits_%7Bi%3D1%7D%5Enf%5Cleft%28a%2B%28i-%5Cfrac%7B1%7D%7B2%7D%29%5Cfrac%7Bb-a%7D%7Bn%7D%5Cright%29%5Cfrac%7Bb-a%7D%7Bn%7D&bg=e6e6e6&fg=000000&s=0)
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
![\displaystyle\lim\limits_{n\to\infty}\int\limits_{\left[a,b\right]}fd\alpha_n=\int\limits_{\left[a,b\right]}fdx \displaystyle\lim\limits_{n\to\infty}\int\limits_{\left[a,b\right]}fd\alpha_n=\int\limits_{\left[a,b\right]}fdx](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha_n%3D%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfdx&bg=e6e6e6&fg=000000&s=0)
But at the same time we see that
converges to
. Clearly there is some connection between convergence and integration to be explored here.
March 27, 2008
Posted by
John Armstrong |
Analysis, Calculus |
|
1 Comment