# The Unapologetic Mathematician

## 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<... 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 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 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