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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?

Davide Giraudo
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nullUser
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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
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    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
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    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 00:03
  • @nullUser Why don't you write an answer to your question? – Norbert Jun 23 '12 at 15:23
  • but, how did you prove that $|f-f_n|\to 0$ ? –  Oct 07 '12 at 04:05

1 Answers1

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Let $\{f_n\}_{n=1}^{ +\infty} $ be 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 $$\lVert f_k-f_j\rVert\geqslant\operatorname{Var}(f_k-f_j)\geqslant \left|f_k(1)-f_j(1)\right|-\left|f_k(0)-f_j(0)\right|,$$ proving that $\left\{f_k(1)\right\}$ 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 $t_0=0<t_1<\dots<t_n=1$ be a partition of $[0,1]$. Then $$\sum_{j=0}^{n-1}\left|f(t_{j+1})-f(t_j)\right|\leqslant \sum_{j=0}^{n-1}\left|f(t_{j+1})-f_N(t_{j+1})\right|+\sum_{j=0}^{n-1}\left|f(t_j)-f_N(t_j)\right|+\operatorname{Var}(f_N).$$ Let $n_0$ be an integer such that $\operatorname{Var}(f_j-f_k)\leqslant 1$ if $j,k\geqslant n_0$. Then for each $N$, $\operatorname{Var}\left(f_N\right)\leqslant \max\left\{1+\operatorname{Var}\left(f_{n_0}\right),\operatorname{Var}(f_1),\dots,\operatorname{Var}\left(f_{n_0-1}\right)\right\}=:M$. Consequently, $$\sum_{j=0}^{n-1}\left|f(t_{j+1})-f(t_j)\right|\leqslant \sum_{j=0}^{n-1}\left|f(t_{j+1})-f_N(t_{j+1})\right|+\sum_{j=0}^{n-1}\left|f(t_j)-f_N(t_j)\right|+M.$$ Taking in last displayed equation the $\limsup_{N\to +\infty}$, we get that $f$ is of bounded variation. Indeed, for each $j\in\{0,\dots,n\}$, $f_N(t_j)\to f(t_j)$. Therefore, $$\sum_{j=0}^{n-1}\left|f(t_{j+1})-f(t_j)\right|\leqslant M,$$ and $M$ is independent on the choice of the partition.

  • $\lVert f-f_N\rVert\to 0$. 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 $m,n\geqslant N$ and $0=t_0<t_1<\dots<t_l=1$ is a partition of $[0,1]$ then $$\sum_{j=0}^{l-1}|(f_m-f_n)(x_{j+1})-(f_m-f_n)(x_j)|\leqslant\varepsilon.$$ Let us take $\limsup_{m\to +\infty}$ in the previous inequality. Since $f_m(x_j)\to f(x_j)$ for each $j\in \{0,\dots,l\}$, we get for each $n\geqslant N(\varepsilon)$: $$\sum_{j=0}^{l-1}|(f-f_n)(x_{j+1})-(f-f_n)(x_j)|\leqslant\varepsilon.$$ Since this inequality is true for any partition, we get $\operatorname{Var}(f-f_n)\leqslant \varepsilon$ for $n \geqslant N(\varepsilon)$.

Davide Giraudo
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  • Just a quick question regarding your first point: If we knew that $\lim\limits_{m\to\infty}f_m = f$ uniformly, then we could abbreviate: for all $m,n>n_0$ we have $\Vert f_m\Vert-\Vert f_n\Vert\leq\Vert f_m-f_n\Vert<\epsilon\implies \Vert f_m\Vert<\epsilon +\Vert f_n \Vert$. Now letting $m\to\infty$ we get $\Vert f\Vert\leq\epsilon +\Vert f_n \Vert$. Is this correct? – Philipp Oct 01 '21 at 17:58