We define the BV2 space by
$S = \lbrace f\in L^2:\textrm{TV}(f)<\infty\rbrace$, where $TV(f)=\sup_{g\in C_c^1,\g\_\infty\leq 1}\int f\cdot \textrm{div}g$.
My question is: is $S$ closed in $L^2$?
Thanks.
We define the BV2 space by
$S = \lbrace f\in L^2:\textrm{TV}(f)<\infty\rbrace$, where $TV(f)=\sup_{g\in C_c^1,\g\_\infty\leq 1}\int f\cdot \textrm{div}g$.
Thanks. 


The answer is no  take the sequence $f_n(x): n^\frac{1}{3} \chi_{[0,\frac{1}{n}}(x)$. Then $f_n \to 0$ in $L^2$ strongly but $TV(f_n) = 2n^\frac{1}{3} \to +\infty$. In general, an arbitrary $f \in L^2$ can be approximated by smooth functions with compact support, for which $TV$ is finite, which shows that in general this is false. Actually, if you look at a subset of $S$, say $f \in S$ with $TV(f)\leq C$, this is a closed subspace, since lowersemicontinuity of the $TV$ with respect to strong convergence in $L^2$ implies a sequence in the set converging to a limit $f$ necessarily has $TV(f)\leq C$. Hope this helps. Cheers. 

