A challenging problem related to characterisation on BV-functions.

Question

I was trying to find the answer to the following question:

Let $f$ be a BV-function on $[a,b]$,with total variation $V_a^b(f)$.Then does it imply that for any sequence $\{P_n\}$ of partitions of $[a,b]$,with $||P_n||\to 0$,we have $V(f,P_n)\to V_a^b(f)$.Where $||P_n||$ is the diameter of the largest subinterval in $P_n$.I am not sure how to proceed.First I thought that I can do it using concept of net and subnets.Notice that the partition of $[a,b]$ along with the inclusion $\subset$ relation is a directed set and $V$ defined by $V(P):=V(f,P)$ is a net from $\mathcal P[a,b]$ to $\mathbb R$.

We can index elements of this net by $P$ i.e. $V_P$ denotes $V(f,P)$.Now,$(V_P)$ converges to $V_a^b(f)$.

What next,how do I proceed now?

If $P$ is $a = t_0 < t_1 < \ldots < t_n = b$, then $$V(f,P) = \sum_{k = 1}^n \lvert f(t_k) - f(t_{k-1})\rvert,,$$ right? I'd suggest looking for a counterexample. — Daniel Fischer, May 14 '20 at 14:56

score 1 · Accepted Answer · answered May 14 '20 at 17:17

1

No. If $f(x) = 1$ for $\frac{b+a}{2}$ and $f(x) = 0$ for all other $x$, then if P is a partition with $\frac{b+a}{2} \notin P$, then $$V(f,P) = 0$$ On the other hand $V_a^b(f) = 2$.

answered May 14 '20 at 17:17

cha21

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A challenging problem related to characterisation on BV-functions.

1 Answers1