# The Unapologetic Mathematician

## Functions of Bounded Variation II

Let’s consider the collection of functions of bounded variation on $\left[a,b\right]$ a little more deeply. It turns out that they form a subring of the ring of all real-valued functions on $\left[a,b\right]$. Just to be clear, the collection of all real-valued functions on an interval becomes a ring by defining addition and multiplication pointwise.

Okay, so to check that we’ve got a subring we just have to check that the sum, difference, and product of two functions of bounded variation is again of bounded variation. Let’s take $f$ and $g$ to be two functions of bounded variation on $\left[a,b\right]$, and let $(x_0,...,x_n)$ be a partition of $\left[a,b\right]$. Then we calculate

$|f(x_i)g(x_i)-f(x_{i-1})g(x_{i-1})|=$
$|f(x_i)g(x_i)-f(x_{i-1})g(x_i)+f(x_{i-1})g(x_i)-f(x_{i-1})g(x_{i-1})|\leq$
$|f(x_i)g(x_i)-f(x_{i-1})g(x_i)|+|f(x_{i-1})g(x_i)-f(x_{i-1})g(x_{i-1})|=$
$|g(x_i)||f(x_i)-f(x_{i-1})|+|f(x_{i-1})||g(x_i)-g(x_{i-1})|\leq$
$A|f(x_i)-f(x_{i-1})|+B|g(x_i)-g(x_{i-1})|$

where $A$ is the least upper bound of $|g(x)|$ on $\left[a,b\right]$, and $B$ is the least upper bound of $|f(x)|$. Then we find $AV_f+BV_g$ is an upper bound for the sum over the partition. In fact, this not only shows that the product $fg$ is of bounded variation, it establishes the inequality $V_{fg}\leq AV_f+BV_g$.

The proofs for the sum and difference are similar. You should be able to work them out, and to establish the inequality $V_{f\pm g}\leq V_f+V_g$.

We can’t manage to get quotients of functions because we can’t generally divide functions. The denominator might be ${0}$ at some point, after all. But if $f(x)$ is bounded away from ${0}$ — if there is an $m$ with $0 — then $g(x)=\frac{1}{f(x)}$ is of bounded variation, and $V_g\leq\frac{V_f}{m^2}$. Indeed, we can check that

$\displaystyle\left|g(x_i)-g(x_{i-1})\right|=\left|\frac{1}{f(x_i)}-\frac{1}{f(x_{i-1})}\right|=$
$\displaystyle\left|\frac{f(x_i)-f(x_{i-1})}{f(x_i)f(x_{i-1})}\right|\leq\frac{\left|f(x_i)-f(x_{i-1})\right|}{m^2}$

March 6, 2008 Posted by | Analysis | 2 Comments

## LaTeX screw-ups

Somehow this morning the renderer on WordPress decided to forget all of my backslashes in certain posts, with predictable results. I’m fixing them all back through the post on indefinite integration, but if you see any others earlier, send up a flag.

Incidentally, if you run into any LaTeX errors, particularly failures to render, please let me know in a comment on the offending page so I can fix them. It seems that every so often WordPress decides to play with those of us who use LaTeX, and I have to go back and fix all the mistakes they have introduced.

March 6, 2008 Posted by | Uncategorized | 7 Comments

## 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

## The Riemann-Stieltjes Integral IV

Let’s do one more easy application of the Riemann-Stieltjes integral. We know from last Friday that when our integrator is continuously differentiable, we can reduce to a Riemann integral:

$\displaystyle\int\limits_{\left[a,b\right]}f(x)d\alpha(x)=\int\limits_{\left[a,b\right]}f(x)\alpha'(x)dx$

So where else have we seen derivatives as factors in integrands? Right! integration by parts! Here our formula says that

$\displaystyle\int\limits_a^bf(x)g'(x)dx+\int\limits_a^bf'(x)g(x)dx=f(b)g(b)-f(a)g(a)$

We can rewrite this using Riemann-Stieltjes integrals as

$\displaystyle\int\limits_{\left[a,b\right]}f(x)dg(x)+\int\limits_{\left[a,b\right]}g(x)df(x)=f(b)g(b)-f(a)g(a)$

So if $f$ and $g$ are both continuously differentiable, this formula gives back our rule for integration by parts. But we can prove this without making those assumptions. In fact, we just need to assume that one of the two integrals exists, and the existence of the other one (and the formula) will follow.

Let’s assume that $\int_{\left[a,b\right]}f,dg$ exists. That is, for every $\epsilon>0$ there is some tagged partition $x_\epsilon$ so that for every finer partition $x$ we have

$\displaystyle\left|\int\limits_{\left[a,b\right]}f(x)dg(x)-f_{g,x}\right|<\epsilon$

Now let’s take any partition $x$ finer than $x_\epsilon$ and use it to set up the Rieman-Stieltjes sum

$\displaystyle g_{f,x}=\sum\limits_{i=1}^ng(t_i)f(x_i)-\sum\limits_{i=1}^ng(t_i)f(x_{i-1})$

We can also use this partition to rewrite $A=f(b)g(b)-f(a)g(a)$ as

$\displaystyle A=\sum\limits_{i=1}^nf(x_i)g(x_i)-\sum\limits_{i=1}^nf(x_{i-1})g(x_{i-1})$

So subtracting the one from the other we find

$\displaystyle A-g_{f,x}=\sum\limits_{i=1}^nf(x_i)(g(x_i)-g(t_i))+\sum\limits_{i=1}^nf(x_{i-1})(g(t_i)-g(x_{i-1}))$

But this is a Riemann-Stieltjes sum $f_{g,\bar{x}}$ for the partition $\bar{x}$ we get by throwing together all the $x_i$ and $t_i$ as partition points, and using $x_i$ as tags. This is a finer partition than $x_\epsilon$, and so we see that

$\displaystyle\left|\left(A-\int\limits_{\left[a,b\right]}f(x)dg(x)\right)-g_{f,x}\right|=\left|\int\limits_{\left[a,b\right]}f(x)dg(x)-\left(A-g_{f,x}\right)\right|<\epsilon$

whenever $x$ is a partition finer than $x_\epsilon$. This shows that the Riemann-Stieltjes integral of $g$ with respect to $f$ exists, and has the value we want.

March 4, 2008 Posted by | Analysis, Calculus | 1 Comment

## The Riemann-Stieltjes Integral III

Last Friday we explained the change of variables formula for Riemann integrals by using Riemann-Stieltjes integrals. Today let’s push it a little further and prove a change of variables formula for Riemann-Stieltjes integrals.

We start with a function $f:\left[a,b\right]$ which we assume to be Riemann-Stieltjes integrable by the function $\alpha$. Now, instead of the full generality we used before, let’s just let $g:\left[c,d\right]\rightarrow\left[a,b\right]$ be a strictly increasing continuous function with $g(c)=a$ and $g(d)=b$. Define $h$ and $\beta$ to be the composite functions $h(x)=f(g(x))$ and $\beta(x)=\alpha(g(x))$. Then $h$ is Riemann-Stieltjes integrable by $\beta$ on $\left[c,d\right]$, and we have the equality

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\int\limits_{\left[c,d\right]}hd\beta$

For decreasing functions we get almost the exact same thing, so you should figure out the parallel statement and proof yourself.

Since $g$ is strictly increasing, it must be one-to-one, and it’s onto by assumption. In fact, $g$ is an explicit homeomorphism of the intervals $\left[a,b\right]$ and $\left[c,d\right]$, and its inverse $g^{-1}:\left[a,b\right]\rightarrow\left[c,d\right]$ is also a strictly increasing continuous function. We can now use $g$ and its inverse to set up a bijection between partitions of $\left[a,b\right]$ and $\left[c,d\right]$: if $a=x_0 is a partition, then $c=g^{-1}(x_0) is a partition, and vice versa. Further, refinements of partitions of one side correspond to refinements of partitions on the other side.

So if we’re given an $\epsilon>0$ then there’s some partition $y_\epsilon$ of $\left[a,b\right]$ so that for any finer partition $y$ we have $|f_{\alpha,y}-\int_{\left[a,b\right]}f,d\alpha|<\epsilon$. Let $x_\epsilon=g^{-1}(y_\epsilon)$ be the corresponding partition of $\left[c,d\right]$, and let $x$ be a partition of $\left[c,d\right]$ finer than it. Then it’s easily verified that the Riemann-Stieltjes sum $h_{x,\beta}$ is equal to the Riemann-Stieltjes sum $f_{g(x),\alpha}$. Everything else follows quickly from here.

March 3, 2008 Posted by | Analysis, Calculus | 12 Comments