# 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

1. […] Variation addenda I left a few things out of last Saturday’s post. Since I’ve spent all morning finishing off that paper (I’ll post the arXiv link […]

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2. […] back when we originally discussed integrators of bounded variation, we can write our integrator as the difference of two increasing functions. It’s no loss of generality, then, to assume that is increasing. We also remember that the […]

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3. […] is sort of like how we found that functions of bounded variation can be written as the difference between two strictly increasing func…. In fact, if we’re loose about what we mean by “function”, and […]

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4. hi , i am aya from egypt .it is good but i want to know how to construct a partition
to prove that a function is of bounded variation or not
and i have some problems that i cannot solve or at least have some idea about the solution via f(x)=(sin x)^2 on the interval [0,3.14].
thanks a lot

Comment by aya mohamed hussein | October 24, 2010 | Reply

• That function is pretty clearly continuous on the given interval. That should at least tell you the direction to try to prove.

Comment by John Armstrong | October 24, 2010 | Reply