A function $f:[a,b]\rightarrow\mathbb{R}$ is said to be of bounded variation if $$V(f):=\text{sup }\lbrace \sum_{j=1}^{r}|f(t_j)-f(t_{j-1})|:r\in\mathbb{Z}^+\text{ and } a=t_0<t_1<\dots<t_r=b\rbrace$$ is finite. The set of all functions of bounded variation on the interval $[a,b]$ is denoted by $BV[a,b]$.
Prove that
- $|f|:=|f(a)|+V(f)$ defines a norm on $BV[a,b]$
- $BV[a,b]$ is compete with respect to the norm defined above
I have problem with the second bit. Any idea ?
Consider a Cauchy sequence $(f_n)$. If $|f_n-f_m|<\epsilon$ then they have to be $\epsilon$ uniformly close. So the Cauchy sequence converges pointwise (everywhere) to some $f$. Now, $V(f_n-f_m)<\epsilon$ for $n,m\geq N_\epsilon$, so for any fixed partition $P=(a=t_0<t_1<\cdots< t_r=b)$ we have $$V(f_n-f_m,P)=\sum_i |(f_n-f_m)(t_j)-(f_n-f_m)(t_{j-1})| <\epsilon$$ for every $n,m\geq N_\epsilon$. Taking the limit $m\rightarrow +\infty$ and using pointwise convergence: $$V(f_n-f,P)=\sum_i |(f_n-f)(t_j)-(f_n-f)(t_{j-1})| \leq \epsilon$$ which implies $V(f,P) \leq V(f_n,P)+\epsilon \leq V(f_n)+\epsilon< +\infty$. Take now the sup over $P$ on the LHS to get: $V(f) \leq V(f_n)+\epsilon< +\infty$.