The space $BPV(0,1)$ is a separable metric space under certain metric.

Question

This is exercise 2.42 from Leoni's book A First Course in Sobolev Spaces.

The BPV is defined as the space contain the function $u$ such that $$ \text{Var}[u]:=\sup\left\{ \sum_{i=1}^n|u(x_i)-u(x_{i-1})|\right\} <\infty$$ where the supremum is taken over all partitions.

We give a metric in $BPV([0,1])$ by the following:

$$ d(u,v):=\int_0^1|u(x)-v(x)|dx+|\text{Var}[u]-\text{Var}[v]| $$

The exercise claims that endowed with this metric, $(BPV([0,1]),\,d)$ is separable, and it breaks down this claim into some sub questions.

However, I do NOT think the first sub-question is correct. The first question asks that if given a sequence $(u_n)\subset BPV([0,1])$ and $\sup_n d(u_n,0)<\infty$, then there exists a subsequence $(u_{n_k})$ such that $d(u_{n_k},u)\to 0$ for some $u\in BPV([0,1])$. In another word, the Ball in $BPV([0,1])$ under metric $d$ is pre-compact.

I tried with the hint that using Helly's selection theorem. I was sucsscifruly proved that, up to a subsequence, $u_n\to u$ in $L^1$ but I was stacked on proving $\text{Var}[{u_n}]\to \text{Var}[u]$.

After I tried a while, I start to think this may not be right. After all, in higher dimension, by Compactness, we will only have $Du_n$ weakly start to $Du$ as a Radon measure and it do happens that $\liminf\|Du_n\|>\|Du\|$.

And now I come up with a counterexample.

Take $u_n$ defined as follows: for each $n$ we partition $[0,1]$ into $n$ small pieces $(I_n)$, each has length $1/n$. i.e., $I_1=[0,1/n]$, $I_2=[1/n,2/n]$

$$ u_n(x)= \begin{cases} 1/n & \text{on }I_i\text{ if $i$ is even}\\ -1/n & \text{on }I_i\text{ if $i$ is odd} \end{cases} $$ Then clearly $u_n\to 0$ in $L^1$. However, we have $\text{Var}[u_n]\equiv 2$ and hence $\liminf\text{Var}[u_n]>\text{Var}[u]=0.$

Please help me check whether my counterexample makes sense! Thx!

Answer 1

Your counterexample is indeed a valid counterexample to 2.42 (i).

On the other hand, 2.42 (iii) is correct: the space is separable. A quick way to show this is to embed it into the following separable Banach space $X$: the direct sum $L^1([0,1])\oplus \mathbb R$ with the norm $$ \|(f,c) \| = |c|+\int_0^1 |f(x)| \,dx $$ Since both $L^1([0,1])$ and $\mathbb R$ are separable, so is $X$. The map $$ F(u) = (u, \operatorname{Var}u) $$ is an isometric embedding of $(BPV([0,1]),d)$ into $X$. A subspace of a separable metric space is separable; hence, $(BPV([0,1]),d)$ is separable.

Answer 2

The counterexample does not work. If $u_n\to 0$ in $L^1$, then $u_n\to u$ in $L^1$, where $u$ is any function which is zero $\mathcal{L}^1$ a.e. in $(0,1)$ (Lebesgue integration does not see sets of measure zero). Hence if you take $u(x)=0$ for $x\ne\frac12$ and $u(\frac12)=1$ you have that $u_n\to u$ in $L^1$ and Var$u$=2.

The exercise is correct. After you use Helly's theorem, you can always change the function $u$ in a finite number of points to make its pointwise variation the number that you want.