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Let $\Vert f \Vert = |f(0)| + \mathrm{Var}f$ for all $f \in BV([0,1])$; we are given that it is a norm. Show that $BV([0,1])$ is a complete normed space with this norm.

I have shown that any Cauchy sequence in $BV([0,1])$ must converge to some function pointwise, but I am stuck at proving that the function must have bounded variation.

Could someone help me?

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I think you can show that if $u_n$ is a sequence of $BV~$ functions that converge pointwise to $u$, then $\mathrm{Var}(u)\leq\mathrm{liminf} ~ \mathrm{Var}(u_n)\in\mathbb{R}_+\cup\lbrace\infty\rbrace$. Since you have a Cauchy sequence, the right hand side is finite, and this shows that the limit function has bounded variation. – Olivier Bégassat Oct 17 '11 at 23:24
I did prove that $\mathrm{Var}(u) \leq \liminf \mathrm{Var}(u_n)$ a few days ago, but I don't see how the sequence being Cauchy implies the right hand side is finite. – nullUser Oct 17 '11 at 23:29
Cauchy sequences are bounded, thus for some $C>0$ and for all $n$, $$\mathrm{Var}(u_n)\leq ||u_n||\leq C.$$ – Olivier Bégassat Oct 17 '11 at 23:32
Beware, your definition of $||\cdot||$ is missing vertical bars around $f$ : It should read $$||f||=|f(0)|+\mathrm{Var}(f).$$ – Olivier Bégassat Oct 17 '11 at 23:34
Of course! I have been staring at this for so long that I neglected the simplest of tests! Thanks so much. – nullUser Oct 18 '11 at 0:03
up vote 3 down vote accepted

Let $\{f_n\}$ a Cauchy sequence for $\lVert\cdot\rVert$. In particular, the sequence of real numbers $\{f_n(0)\}$ is Cauchy, hence converges to a real number we call $f(0)$. Now, considering the partition $t_0=0<1=t_1$, we have $$\lVert f_k-f_j\rVert\geq\operatorname{Var}(f_k-f_j)\geq |f_k(1)-f_j(1)|-|f_k(0)-f_j(0)|,$$ proving that $\{f_k(1)$ is Cauchy, hence converges to a real number called $f(1)$. Now for $t\in(0,1)$, we consider the partition $t_0=0<t=t_1<t_2=1$ to get that $\{f_k(t)\}$ is Cauchy, hence it converges to a number called $f(t)$. Now, to conclude, we need two things:

  • $f$ is of bounded variation. Indeed, let $t_0=0<t_1<\dots<t_n=1$ a partition of $[0,1]$. Then $$\sum_{j=0}^{n-1}|f(t_{j+1})-f(t_j)|\leq \sum_{j=0}^{n-1}|f(t_{j+1})-f_N(t_{j+1})|+\sum_{j=0}^{n-1}|f(t_j)-f_N(t_j)|+\operatorname{Var}(f_N).$$ Let $n_0$ such that $\operatorname{Var}(f_j-f_k)\leq 1$ if $j,k\geq n_0$. Then for each $N$, $\operatorname{Var}(f_N)\leq \max\{1+\operatorname{Var}(f_{n_0}),\operatorname{Var}(f_1),\dots,\operatorname{Var}(f_{n_0-1})\}=:M$. Taking in last displayed equation the $\limsup_{N\to +\infty}$, we get that $f$ is of bounded variation.

  • $\lVert f-f_N\rVert\to 0$. We have by definition $f_n(0)\to f(0)$ so we have to show that $\operatorname{Var}(f_n-f)\to 0$. Let $\varepsilon>0$. We can find $N=N(\varepsilon)$ such that if $m,n\geq N$ and $0=t_0<t_1<\dots<t_l=1$ is a partition of $[0,1]$ then $$\sum_{j=0}^{l-1}|(f_m-f_n)(x_{j+1})-(f_m-f_n)(x_j)|\leq\varepsilon.$$ We get the result taking the $\limsup_{m\to +\infty}$.

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