Let [a,b] be an interval in R,and denote by E the Vector space of functions f:[a,b]->R such that f is of bounded variation over [a,b] and f(a)=0.Prove that by setting $||f||=Var|_{a}^{b}(f)$ for each f $\in$ V,we define a norm on V,and that V is complete in this norm.
Given a cauthy sequence{ $f_{n}$},I tried to firstly prove a weak case where $f_{n}$ is monotone case for any integer n, and I want to check whether I can prove this case,but I am stuck. I feel when $f_{n}$ is monotone, I can prove there is a f such that $f_{n}(x)->f(x)$ in the sense of real sysytem,but I coudn't prove $f_{n}(x)->f(x)$ in this norm vector system,any help? Also,if $f_{n}$ is not necessarily monotone, how to prove it? Any help?
I assume you've proved $\|\,\|$ is a norm on $V.$ Suppose $f_n$ is Cauchy in this norm. Fix $x$. Then
$$|f_n(x) - f_m(x)| = |f_n(x) - f_m(x)- (f_n(a) - f_m(a))| \le \text {Var} (f_n-f_m).$$
Hence $f_n(x)$ is Cauchy in $\mathbb {R}$ and thus converges. Thus $f_n \to $ some $f$ on $[a,b]$ pointwise everywhere. Now $(f_n)$ Cauchy in this norm implies $\sup_n \|f_n\| = M < \infty.$ So now fix a partition $\{x_0, \dots , x_p\}$ of $[a,b].$ Then
$$\sum_{k=1}^{p}|f(x_k)-f(x_{k-1})| = \lim \sum_{k=1}^{p}|f_n(x_k)-f_n(x_{k-1})|\le M.$$
Thus $\text {Var} (f) < \infty,$ hence $f\in V.$ Now you still have something left to prove, which is $\|f_n-f\| \to 0.$
Added later: to show $\|f_n-f\| \to 0,$ let $\epsilon >0.$ Choose $N$ such that $m,n \ge N \implies \|f_m-f_n\| < \epsilon.$ Let $\{x_0, \dots ,x_p\}$ be a partition of the interval. Let $n>N.$ Then
$$ \sum_{k=1}^{p}|(f-f_n)(x_k)-(f-f_n)(x_{k-1})| = \lim_{m\to \infty} \sum_{k=1}^{p}|(f_m-f_n)(x_k)-(f_m-f_n)(x_{k-1})| \le \limsup_{m\to \infty}\|f_m-f_n\|\le \epsilon.$$
This shows $\|f_n-f\| <\epsilon,n>N$ as desired.