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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 $L^2$?


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Probably $S$ contains a dense subspace of $L^2$ like, say, the $C^\infty$ functions with compact support, but easy examples show that it is not all of $L^2$, so ... –  Nik Weaver May 8 '13 at 4:27

1 Answer 1

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.

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