i'm reading Leon Simon's lectures on geometric measure theory and I'm feeling a bit lost in the part on function of bounded variation:
Simon defines a function $u \in L_{\text{loc}}^1( U )$ ( where $U \subset R^N$ is an open set) to be in $BV_{\text{loc}}$ if for every $W \subset \subset U$ there exists a costant $c(W)$ such that for every $g:U \to R^n$, $g \in C_{\text{c}}^{\infty}(U, R^N)$ and $\text{spt}g \subset W$ one has:
$\int_{W} u\ \text{div}g\ dL^N \le c(W) \Vert g \Vert_{\infty}$
Then he says that this definition gives rise to a linear functional on $C_{\text{c}}(U, R^N)$ which is bounded on compact sets contained in $U$. I didn't see how he could say it so easily so i tried to give some details, please check if they are ok:
Given a $g \in C_{\text{c}}(U, R^N)$, one can have $g \in \overline{C_{\text{c}}^{\infty}(U, R^N)}$ or in the complement. in the latter case we can define $L(g) = 0$ Else observe that there exists a sequence $ g_n \in C_{\text{c}}^{\infty}(U, R^N)$ such that $ \Vert g_n - g \Vert_{\infty} \to 0$. Now observe that $\text{spt}\ g \subset U$ is compact so it exists $W \subset \subset U$ with $ \text{spt}\ g \subset W$. Take a function $f \in C_{\text{c}}^{\infty}(U, R)$ with $ \text{spt} f \in W$ such that $f = 1$ on $ \text{spt} g$ and with $ 0 \le f \le $. We have that $fg_n \to g$. And $ \text{spt} fg_n \subset W$. We can thus define $L(g) = lim_{n} \int_{W} u\ \text{div}(fg_n)\ dL^N$ which obviously exists. Then one checks that this is well defined and this should work, is it right?
Second question: Using the previous construction Simon deduce from the riesz representation theorem that for a $BV_{loc}$ function there exists a radon measure $|Du|$ which represent the functional over constructed, characterized by, for every open $V \subset U$:
$|Du| (V) = sup\{ \int_{U} u\ \text{div}g\ dL^N, g \in C_{\text{c}}^{\infty}(U, R^N), \text{spt} g \subset V, |g| \le 1\}$
Now he takes $ u \in BV_{loc}(U)$ and sets $U_n = \{ x \in U: d(x, U^c) > 2/n\}$. He set $u_n = u$ on $U_n$ and zero outside and takes a sequence of mollifiers $ \phi_n$ such that $ \phi_n \star u_n \to u$ in $L_{loc}^1(U)$. He claims that given $f \in C_{\text{c}}(U, R)$ non negative one has:
$\int_{U} f |D(\phi_n \star u_n)| = sup\{ \int_{U} g * \nabla(\phi_n \star u_n)\ dL^N, g \in C_{\text{c}}^{\infty}(U, R^N), |g |\le f\}$. I don't understand how to easily see this, can you help me? thank you
I think the easiest way to look at this is by first considering the case where $u\in C^\infty_c(U)$. In that case, we have $$\int_Uu\ \text{div}\, g\,dx=-\int_U g\cdot Du\,dx$$ for all $g\in C^\infty_c(U)$, and we can define a measure $\nu_u$ on $U$ by $$\nu_u(V):=\int_V|Du|\,dx.$$ By Cauchy-Schwarz we have $g\cdot Du\le|g||Du|$ and we know that this inequality is optimal (indeed, equality occurs if $g=\alpha Du$ for $\alpha>0$). We hence have that $$\nu_u(V)=\sup\left\{\int_Ug\cdot Du\,dx\,:\,g\in C^\infty_c(U),\ \text{spt}\,g\subset V, |g|\le 1\right\}\\ =\sup\left\{\int_Uu\ \text{div}\,g\,dx\,:\,g\in C^\infty_c(U),\ \text{spt}\,g\subset V,|g|\le1\right\}$$ for every open $V$, and using a similar calculation $$\int_Uf\,d\nu_u=\int_Uf|Du|\,dx=\sup\left\{\int_Uu\ \text{div}\,g\,dx\,:\,g\in C^\infty_c(U),|g|\le f\right\}$$ for every nonnegative $f\in C^\infty_c(U$). (We do not need the minus signs in front of the integrals because we can simply replace $g$ by $-g$.)
If we only assume $u\in L^1(U)$, we fail at the first step. So we define bounded variation functions to be exactly those which work, namely those for which the linear functional $g\mapsto\int_Uu\ \text{div}\,g\,dx$ is bounded. Under this assumption, $Du$ does not necessarily exist as a function, but by the Riesz representation theorem the measure $\nu_u$ does exist. In the smooth case we often abuse notation and write $|Du|$ for the measure $\nu_u$, and we continue to do so in this case. The characterizations of $|Du|(V)$ and $\int_Uf|Du|$ are also simple consequences of the Riesz representation theorem.
In your case, you are actually dealing with functions of local bounded variation, but it is not hard to see that $u\in BV_\text{loc}(U)$ if and only $u\big|_W\in BV(W)$ for every $W\subset\subset U$, and all the consequences are immediate. The only thing to verify is that if $\nu_u^W$ and $\nu_u^{W'}$ are the respective measure for the subsets $W$ and $W'$ then they agree on $W\cap W'$, but this is not difficult.