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Consider the space $(BV[0,1];||.||)$ with the norm

$||f||=|f(0)|+V_{f}[0,1]$ Where $V_{f}[0,1]$ is the variation of $f$. My questions

what is the closure of $C^1[0,1]$ with respect to this norm?

Another question is how to prove that this norm is Banach?

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    Yeah, but elsewhere I looking for the answer of the first question, rather than the second!2012-03-24

2 Answers 2

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sIt is the space $W^{1,1}[0, 1]$. Note initially that $W^{1,1}[0, 1]$ is a subspace of $BV$ moreover in there the norms are equivalent. In fact if $f\in W^{1,1}[0, 1]$, for its continuous representative

f(x)=f(0)+\int_{0}^{x}f'(t)dt\tag{1} Then |f(x)|\leq|f(0)|+|\int_{0}^{x}f'(t)dt|\leq |f(0)|+\int_{0}^{1}|f'(t)|dt.

But V_f[0,1]=\int_{0}^{1}|f'(t)|dt. So $||f||_L^1\leq||f||_{\infty}\leq||f||_{BV}$ since ||f'||_L^1\leq||f||_{BV} then

$||f||_{W^{1,1}}\leq 2||f||_{BV}.$ Using $(1)$ again we get f(0)=-f(x)+\int_{0}^{x}f'(t)dt

then |f(0)|\leq |f(x)|+\int_{0}^{1}|f'(t)|dt,

integrating from $0$ to $1$ $|f(0)|\leq ||f||_{W^{1,1}}$ and analogously $||f||_{BV}\leq 2||f||_{W^{1,1}}$ ergo both norms are equivalent.

If $f_n \to f$ in $BV$ with $(f_n)\subset C^{1}$ then $(f_n)$ is a Cawchy sequence in $W^{1,1}$, since it is a complete space $f_n\to g$ in $W^{1,1}$.

Then $f_n\to g$ in $BV$ because the norms are equivalents and by the unity of the limit $g=f\in BV$.

This proves that the closure of ${C^1}$ in the $BV$ norm is a subspace of $W^{1,1}$.

The other inclusion is given because any function $f\in W^{1,1}$ can be approximated by a sequence $f_n\to f $ in $W^{1,1}$ what is the same $f_n\to f $ in $BV$ $\blacksquare$

PS: That is my answer it may contains some English problems but it is mathematically correct! And all details are included! It is of course inspired on @Jonas answer! And he deserved to gain the bounty, and I thanks to him! But how he was not so sure ( he wrote I think it is W1,1) I had to put a complete answer. My first thought were to edit his answer but I realized it was a complete rewriting. I decided to put a new answer.

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    Hey downvoters, I gave up 50 points to this question do you think I'll mind for some improper downvotes?2012-05-13
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I think the space is $W^{1,1}[0, 1]$. We clearly have that the closure (say $B$) is in $W^{1, 1}$. Furthermore, $W^{1, 1}$ is a proper subset of $\text{BV}$.

So, take a function $f$ in $W^{1, 1}$ and take an approximating sequence $f_n$ consisting of $C^\infty$ functions in the $W^{1, 1}$ norm.

So, we have $\|f_n - f\|_{\text{BV}} \lesssim \|f_n - f\|_{W^{1, 1}} \to 0$.

As $f_n$ are all in $C^1$ we also have that $f$ is in $B$.

Now we have

f(x) = f(0) + \int_0^x f'(t) \, \textrm{d}t.

So, \|f\|_{W^{1, 1}} = \|f\|_{L^1} + \|f'\|_{L^1}.

And \|f\|_{L^1} \leqslant |f(0)| + \int_0^1 \left | \int_0^x f'(t) \, \textrm{d}t \right | \leqslant |f(0)| + \int_0^1 |f'(t)| \, \textrm{d}t.

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    Interesting. Why the downvote?2012-03-26