Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
7  
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
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.