# The closure of $C^1$ in the  functions of bounded variation

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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– t.b. Mar 9 '12 at 0:26
I guess this space is of absolute continuous functions! – checkmath Mar 9 '12 at 20:16
As @t.b. points out, your 2nd question has already been asked elsewhere, and there have been some hints. – user16299 Mar 24 '12 at 2:08
Yeah, but elsewhere I looking for the answer of the first question, rather than the second! – checkmath Mar 24 '12 at 2:18

## 2 Answers

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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Is the space of Lipschitzian functions closed in this odd norm? – checkmath Mar 24 '12 at 20:41
I hope so 8-). Let me try. – Jonas Teuwen Mar 24 '12 at 20:43
@chessmath What do you think now? – Jonas Teuwen Mar 25 '12 at 16:50
I'll check! But if you explain the first " we clearly have" will not hurt! – checkmath Mar 25 '12 at 17:49

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Interesting. Why the downvote? – Jonas Teuwen Mar 26 '12 at 13:48
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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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That is my answer it may contain some English trouble 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! But how he was not so sure ( he wrote I think it is $W^{1,1}$). My first thought were to edit his answer but I realized it was a complete rewriting. I decided to put a new answer. – checkmath Mar 26 '12 at 11:36
Well, I say that I think the space is $W^{1, 1}$ and then give a reasonably complete argument 8-). Maybe it would have been better if you would have added this to your post instead of giving an answer? – Jonas Teuwen Mar 26 '12 at 12:03
Come on your answer was lacking some pieces, No offense but you answered wrong first! I had to be sure! – checkmath Mar 26 '12 at 12:08

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Yes, I have answered it wrong first, and then I have fixed it. I understand that you want to be sure if it is correct. By the way, I'm not the one that downvoted you. – Jonas Teuwen Mar 26 '12 at 12:29
Hey downvoters, I gave up 50 points to this question do you think I'll mind for some improper downvotes? – checkmath May 13 '12 at 5:19
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