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Jul 23, 2021 at 21:29 comment added Bogdan Now I see. Thanks a lot for your answer!
Jul 23, 2021 at 21:28 vote accept Bogdan
Jul 23, 2021 at 20:33 comment added Leo Moos I adapted the argument according to your comments - it suggests that the counterexample I gave earlier would work for any $\omega \subset \Omega$ with bounded perimeter.
Jul 23, 2021 at 20:30 history edited Leo Moos CC BY-SA 4.0
adapted answer according to comments.
Jul 23, 2021 at 16:11 comment added Bogdan Yes, with non-empty $\omega$. I try to find an example with $\omega$ being an interval (in the 1 dimensional case -which can be viewed as a two dimensional case if $y$ is not involved) by considering as you said a molliefied version of the characteristic function + something like your $f_n$. But the variantion goes to 2 also. I do not see why adding $f_n$ to a mollified characteristic function will increase the perimeter...
Jul 23, 2021 at 15:43 comment added Leo Moos When you say other examples, do you mean non-empty sets $\omega$? That you can certainly get, for example by taking the sum of a mollified indicator $\mathbf{1}_\omega$ with the $f_n$ above. Could you explain what you mean when you say that you 'tried with intervals'?
S Jul 23, 2021 at 15:41 history suggested Bogdan CC BY-SA 4.0
I modify $+\infty$ to $0$.
Jul 23, 2021 at 15:40 review Suggested edits
S Jul 23, 2021 at 15:41
Jul 23, 2021 at 15:38 comment added Bogdan I wonder if there are other examples. I tried with intervals but seems ok. Thanks for your example!
Jul 23, 2021 at 15:25 history answered Leo Moos CC BY-SA 4.0