Timeline for Is the set of the absolutely continuous functions a Borel set of the space of the continuous functions?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 20, 2013 at 9:29 | vote | accept | Theluze | ||
| Mar 19, 2013 at 17:30 | vote | accept | Theluze | ||
| Mar 19, 2013 at 18:21 | |||||
| Mar 19, 2013 at 17:17 | comment | added | Theluze | Thank you for your elegant proof. However, i don't know how to prove that $\phi$ is lower semi-continuous. Indeed if, for $F(t)= a +\int_0^t f(t) dt$ i write $$|F|_{1,1} = G(F) + H(F) $$ where $$G(F) =\int_0^1 |F(t)|dt $$ and $$H(F) = \int_0^1 |f(t)|dt $$ then $F\to G(F)$ clearly continuous for the norm of the uniform convergence, but i can't prove that $H$ is lower semi-continuous. Moreover $W^{1,1}$ does not seem to be closed in $C$. Maybe i don't take the proof in the good way. Could someone enlighten me on the good way to prove the lower semicontinuity of $\phi$ ? | |
| Mar 19, 2013 at 12:45 | history | answered | marcoromito | CC BY-SA 3.0 |