# Show that $V[a,b]$ is a Banach space

Let $V[a,b]$ be the space of all complex-valued functions of bounded variation on $[a,b]$, which we treat naturally as a $\Bbb C$-vector space. In addition, we equip it with the norm: $$\|x\|:=|x(a)|+Vx,$$ in which $Vx$ denotes the total variation of $x(t)$ on $[a,b]$. I have to show that $(V[a,b],\|\cdot\|)$ is a Banach space.

I think I'm almost there but stuck in one critical step, here are my thoughts:

Suppose $\{x_n\}$ is a Cauchy sequence in $V[a,b]$, then it is clear that both $|x_m(a)-x_n(a)|$ and $V(x_m-x_n)$ turn small when $m,n$ turn large. In particular, $\lim_{n\to\infty}x_n(a)$ exists; and given any $\epsilon>0$, we can find:
1). $N_1$ s.t. $\forall m,n>N_1$, $|x_m(a)-x_n(a)|<\epsilon/2$ and $V(x_m-x_n)<\epsilon/2$.
2). $N_2>N_1+1$ s.t. $\forall m,n>N_2$, $|x_m(a)-x_n(a)|<\epsilon/2^2$ and $V(x_m-x_n)<\epsilon/2^2$.
2). $N_3>N_1+1$ s.t. $\forall m,n>N_3$, $|x_m(a)-x_n(a)|<\epsilon/2^3$ and $V(x_m-x_n)<\epsilon/2^3$.
$$\cdots$$ This progress may continue indefinitely and we can pick $n_k\in (N_k,N_{k+1})$ so that $$|x_{n_{k+1}}(a)-x_{n_k}(a)|<\epsilon/2^k,\, V(x_{n_{k+1}}-x_{n_k})<\epsilon/2^k.$$ Define $g_N=x_{n_1}+\sum_{k=1}^N (x_{n_{k+1}}-x_{n_k})$. Note that for all $t\in [a,b]$ we have \begin{align} V(x_{n_{k+1}}-x_{n_k})&\ge |(x_{n_{k+1}}-x_{n_k})(b)-(x_{n_{k+1}}-x_{n_k})(t)|+|(x_{n_{k+1}}-x_{n_k})(t)-(x_{n_{k+1}}-x_{n_k})(a)|\\ &\ge |(x_{n_{k+1}}-x_{n_k})(t)-(x_{n_{k+1}}-x_{n_k})(a)| \\ &\ge |(x_{n_{k+1}}-x_{n_k})(t)|-|(x_{n_{k+1}}-x_{n_k})(b)|. \end{align} Hence we have $$|(x_{n_{k+1}}-x_{n_k})(t)|\le |(x_{n_{k+1}}-x_{n_k})(a)|+V(x_{n_{k+1}}-x_{n_k})<\epsilon/2^{k-1}.$$ Hence by Weierstrass M-test $g_N$ converges absolutely uniformly to some function $g$ on $[a,b]$. The hardest part is: how to show $g_N\to g$ by $\|\cdot\|$?
Naively I guess the following should be right: $$V(g-g_N)=V(\lim_{m\to\infty}g_m-g_N)=\lim_{m\to\infty}V(g_m-g_N).$$ However, even the uniform convergence isn't enough to justify the interchange of limits! So I really don't know any method else to get around it. Any help? Thanks in advance!

• Uniform convergence is more than sufficient. Take any partition, with say $M$ points, and look $g_m$ with $m$ large enough so that $|g_m(x_j)-g(x_j)|<\varepsilon/M$ for each of the points $x_j$ in the chosen partition. – SamM Sep 17 '16 at 14:36
• @SamM This isn't right. One can't vary $m$ while fixing $M$. In fact there is a quick counterexample: $$x_n(t):[0,1]\in t\mapsto \frac1{2^n}\sum_{i=0}^{2^n-1}\mathbb{1}_{[\frac{2i}{2^{n+1}},\frac{2i+1}{2^{n+1}}]}(t)\in \Bbb R,$$ then $x_n$ converges uniformly, yet $V(x_n)\equiv 1$. – Vim Sep 17 '16 at 14:44
• The choice of partition of $[a,b]$ is completely independent of the sequence $g_m$ and its limit. – SamM Sep 17 '16 at 14:47
• @SamM the total variation is defined for each $g_m$. So we may well for each $g_m$ choose a different partition. Anyway my example has furnished a counterexample where $x_m$ uniformly approaching $x$ doesn't mean $(x_m-x)$ has vanishing total variation. – Vim Sep 17 '16 at 14:54
• I believe I expressed myself quite poorly in my above; what I meant is, if you have BV-Cauchy and uniform convergence, then we can conclude that we pass the limit. – SamM Sep 17 '16 at 17:11

Fix a BV-norm Cauchy sequence $(g_n)$, $\varepsilon>0$, and $j,k\in\Bbb N$ such that $||g_j-g_k||<\varepsilon$. Let $R>0$ such that $||g_n||\leq R$ for all $n$.
You have shown that $(g_n)$ converges uniformly to a bounded function $g$ on $[a,b]$. (Here is another proof of this fact which is a bit easier: Let $t\in [a,b]$. We have $$|g_j(t)-g_k(t)|\leq |g_j(t)-g_j(a)|+|g_j(a)-g_k(a)|+|g_k(a)-g_k(t)|\\ \leq V(g_j)+|g_j(a)-g_k(a)|+V(g_k)<3\varepsilon$$ and so $(g_n)$ is uniformly Cauchy.)
Let $a=x_1<x_2<\dots<x_n=b$ be a partition of $[a,b]$. Let $N\in\Bbb N$ large enough so that $|g_k(x_i)-g(x_i)|<\varepsilon/n$ for all $i$ whenever $k\geq N$. (Pointwise convergence is enough for this.) Assuming $k\geq N$, we have $$\sum_{i=1}^{n-1}|g(x_{i+1})-g(x_i)|\leq \sum_{i=1}^{n-1}|g(x_{i+1})-g_k(x_{i+1})|+\sum_{i=1}^{n-1}|g_k(x_{i+1})-g_k(x_i)|+{\sum_{i=1}^{n-1}|g_k(x_{i})-g(x_i)|}\\ \leq 2\varepsilon + \sum_{i=1}^{n-1}|g_k(x_{i+1})-g_k(x_i)|.$$ Now we have $$\sum_{i=1}^{n-1}|g(x_{i+1})-g(x_i)|\leq R+2\varepsilon$$ and we conclude that $g\in V[a,b]$.
Now, by similar calculations assuming both $j\geq N$, we obtain $$\sum_{i=1}^{n-1}|(g_k-g)(x_{i+1})-(g_k-g)(x_i)|\leq \sum_{i=1}^{n-1}|(g_k-g_j)(x_{i+1})-(g_k-g_j)(x_i)|+2\varepsilon\\\leq V(g_j-g_k)+2\varepsilon<3\varepsilon.$$ This is independent of the choice of partition, so $V(g_k-g)\leq 3\varepsilon$.