The Unapologetic Mathematician

Mathematics for the interested outsider

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

\displaystyle\sum\limits_{i=1}^nf(t_i)(\alpha(x_i)-\alpha(x_{i-1}))

How can we choose a function f to make sums like this get really big? Well, we could make the function f get big, but that’s sort of a cheap trick. Let’s put a cap that |f(x)|\leq1. Now how can we make a Riemann-Stieltjes sum big?

Well, sometimes \alpha(x_i)>\alpha(x_{i-1}) and sometimes \alpha(x_i)<\alpha(x_{i-1}). In the first case, we can try to make f(t_i)=1, and in the second we can try to make f(t_i)=-1. 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 \sum\limits_{i=1}^n|\alpha(x_i)-\alpha(x_{i-1})| 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 M for our \alpha, then we’re never going to get a Riemann-Stieltjes sum bigger than \max\limits_{x\in\left[a,b\right]}(f(x))M. That will certainly make it easier to get nets of sums to converge.

So let’s make this definition: a function \alpha is said to be of “bounded variation” on the interval \left[a,b\right] if there is some M so that for any partition a=x_0<...<x_n=b we have

\displaystyle\sum\limits_{i=1}^n|\alpha(x_i)-\alpha(x_{i-1})|\leq M

That is, M is an upper bound for the set of variations as we look at different partitions of \left[a,b\right]. Then, by the Dedekind completeness of the real numbers, we will have a least upper bound V_\alpha(a,b), which we call the “total variation” of \alpha on \left[a,b\right]

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

An easy consequence of this condition is that functions of bounded variation are bounded! In fact, if \alpha has variation V_\alpha(a,b) on \left[a,b\right], then |\alpha(x)|\leq|\alpha(a)|+V_\alpha(a,b) for all x\in\left[a,b\right]. Indeed, we can just use the partition a<x<b and find that

\displaystyle|\alpha(x)-\alpha(a)|\leq|\alpha(x)-\alpha(a)|+|\alpha(b)-\alpha(x)|\leq V_\alpha(a,b)

from which the bound on |\alpha(x)| easily follows.

March 5, 2008 Posted by | Analysis | 12 Comments

   

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