# Functions of Bounded Variation

## Homework Statement

Given a sequence of scalars (cn) and a sequence of distinct points (xn) in (a, b), define f(x) = cn if x = xn for some n, and f(x) = 0 otherwise. Under what condition(s) is f of bounded variation on [a,b]?

## Homework Equations

Vbaf = supp($$\Sigma$$lf(ti) - f(ti-1)l< +inf, then f is of Bounded Variation.

## The Attempt at a Solution

My understanding of the question is that we have points at x in a vertical line and points along y=0 at every other value of x. So the only way that this function is of bounded variation is if there is a supremum of the constants so that they can't go on vertically to infinitum. If we merely say that they are bounded above we won't have bounded variation, so we must say that the supremum is a member of the set of cn's.

No, you're on the wrong track. The total variation $$V_a^b f = \sup_P \sum_j |f(x_j+1) - f(x_j)|$$ is the supremum of all possible sums of jumps in the value of $$f$$ where the function is measured at finitely many points. So you should answer two questions:

1. What can a jump of the value of $$f$$ from one point to another look like?
2. What analytic idea should you use to think about the "supremum of all finite sums of something"?

You can't answer this question by examining the values $$c_n$$ one by one; there is a sense in which you must consider them collectively.

So, we need to use norms? If we add the assumption of completeness, we will satisfy bounded variation.

Since we have Thm: BV[a,b] is complete under llfllBV = lf(a)l + vbaf
and
Lemma: llfllinf $$\leq$$ llfllBV
then we can prove that llfllBV is complete and therefore the supremum of f is convergent?

So, we need to use norms? If we add the assumption of completeness, we will satisfy bounded variation.

Since we have Thm: BV[a,b] is complete under llfllBV = lf(a)l + vbaf
and
Lemma: llfllinf $$\leq$$ llfllBV
then we can prove that llfllBV is complete and therefore the supremum of f is convergent?

The sentences "If we add the assumption of completeness, we will satisfy bounded variation" and "we can prove that $$\|f\|_{BV}$$ is complete and therefore the supremum of $$f$$ is convergent" don't make any sense to me.

You do not need to think about the BV-norm or completeness to solve this problem -- you can apply the definition of bounded variation rather directly.