# The Unapologetic Mathematician

## Functions of Bounded Variation III

I’ve been busy the last couple of days, so this post got delayed a bit.

We continue our study of functions of bounded variation by showing that total variation is “additive” over its interval. That is, if $f$ is of bounded variation on $\left[a,b\right]$ and $c\in\left[a,b\right]$, then $f$ is of bounded variation on $\left[a,c\right]$ and on $\left[c,b\right]$. Further, we have $V_f(a,b)=V_f(a,c)+V_f(c,b)$.

First, let’s say we’ve got a partition $(x_0,...,x_m)$ of $\left[a,c\right]$ and a partition $(x_m,...,x_n)$ of $\left[c,b\right]$. Then together they form a partition of $\left[a,b\right]$. The sum for both partitions together must be bounded by $V_f(a,b)$, and so the sum of each partition is also bounded by this total variation. Thus $f$ is of bounded variation on each subinterval. This also establishes the inequality $V_f(a,c)+V_f(c,b)\leq V_f(a,b)$.

On the other hand, given any partition at all of $\left[a,b\right]$ we can add the point $c$ to it. This may split one of the parts of the partition, and thus increase the sum for that partition. Then we can break this new partition into a partition for $\left[a,c\right]$ and a partition for $\left[c,b\right]$. The first will have a sum bounded by $V_f(a,c)$, and the second a sum bounded by $V_f(c,b)$. Thus we find that $V_f(a,b)\leq V_f(a,c)+V_f(c,b)$.

So, with both of these inequalities we have established the equality we wanted. Now we can define the “variation function” $V$ on the interval $\left[a,b\right]$. Just set $V(x)=V_f(a,x)$ (and $V(a)=0$). It turns out that both $V$ and $D=V-f$ are increasing functions on $\left[a,b\right]$.

Indeed, given points $x in $\left[a,b\right]$ we can see that $V_f(a,y)=V_f(a,x)+V_f(x,y)$, and so $V(x)\leq V(y)$. On the other hand, $D(y)-D(x)=V(y)-V(x)-(f(y)-f(x))=V_f(x,y)-(f(y)-f(x))$. But by definition we must have $f(y)-f(x)\leq V_f(x,y)$! And so $D(x)\leq D(y)$.

Given a function $f$ of bounded variation, we have constructed two increasing functions $V$ and $D$. It is easily seen that $f=V-D$, so any function of bounded variation is the difference between two increasing functions. On the other hand, we know that increasing functions are of bounded variation. And we also know that the difference of two functions of bounded variation is also of bounded variation. And so the difference between two increasing functions is a function of bounded variation. Thus this condition is both necessary and sufficient!

Even better, since many situations behave nicely with respect to differences of functions, it’s often enough to understand how increasing functions behave. Then we can understand the behavior of functions of bounded variation just by taking differences. For example, we started talking about functions of bounded variation to discuss integrators $\alpha$ in Riemann-Stieltjes integrals. If we study these integrals when $\alpha$ is increasing, then we can use the linearity of the integral with respect to the integrator to understand what happens when $\alpha$ is of bounded variation!

March 9, 2008 Posted by | Analysis | 5 Comments