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Jul 26 at 14:24 answer added Pietro Majer timeline score: 6
Jul 26 at 14:24 comment added Pietro Majer OK, thank you, I wasn't sure I had understood the question.
Jul 26 at 14:17 comment added Aleksei Kulikov @PietroMajer huh, this is way simpler and much more elegant than the monstrosity that I have created. Why don't you post this as an answer?
Jul 26 at 9:09 comment added Pietro Majer Yes, I asked just for being sure the final goal of the OP was not just approximating by $f_n$ below $f.$ Another possibility is an approximation of the identity via convolution; by linearity it smoothes the $1_{\Delta_i}$ as well, and of course $\|\phi_\epsilon*f\|_\infty \le \|f\|_\infty$
Jul 26 at 6:59 comment added Aleksei Kulikov @PietroMajer because this will do some unspeakable horrors to the functions $f_k^m$. I interpreted the goal (and the difficulty) of the question being that we want to preserve the existence of the functions $f_k^m$ while adding the extra norm condition.
Jul 26 at 6:51 comment added Pietro Majer Why don't you truncate $f_n$ at $M:=\|f\|_\infty$, and consider $(f_n\wedge M)\vee(-M)$ that has the same limit $f$ of $f_n$?
Jul 26 at 6:50 answer added Aleksei Kulikov timeline score: 3
Jul 26 at 6:32 history edited Daniele Tampieri CC BY-SA 4.0
Minor Math Jaxing
Jul 26 at 6:08 comment added mathlover yes, want almost everywhere
Jul 26 at 6:05 comment added Aleksei Kulikov I would also assume that you want convergence almost everywhere, not everywhere, right?
Jul 26 at 6:03 comment added mathlover $\Delta_i$ are measurable sets with $|\Delta_i|<\infty$
Jul 26 at 6:00 comment added Aleksei Kulikov Is $\Delta_i$ an open set, measurable set or something else?
Jul 26 at 5:49 history asked mathlover CC BY-SA 4.0