# 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

1. The strict inequality |f|<1 later turns out to be not so strict in your post.

Comment by Johan Richter | March 5, 2008 | Reply

2. Thanks for catching that.

Comment by John Armstrong | March 5, 2008 | Reply

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8. Why is the upper bound M on alpha, and not on the set of all possible “distances” (i.e. absolute values of differences) between successive alpha values?

Comment by Kizi | November 14, 2012 | Reply

9. It may not be completely clear; if there’s an upper bound on the $\alpha$ values then there’s an upper bound on their differences. That’s in the motivation, but the actual definition that follows is written in terms of an upper bound on the sums of absolute differences.

Comment by John Armstrong | November 14, 2012 | Reply

• Thank you for the prompt response and clarification. In the motivation, do the alpha values need to be bounded below as well (or is there some property of the alpha values that I am missing, like non-negativity)? I do apologise if I am missing something obvious. Thanks.

Comment by Kizi | November 14, 2012 | Reply

10. Yeah, they’d probably need to be bounded above and below; the motivation is meant to give the idea, but the definition is precise.

Comment by John Armstrong | November 14, 2012 | Reply

• Thank you

Comment by Kizi | November 14, 2012 | Reply