Timeline for Does weak convergence implies weak convergence of the positive part?
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4 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2016 at 12:42 | comment | added | Denis Serre | Of course. I just wrote too fast by saying "$f$ continuous". I meant "$f$ locally Lipschitzian". | |
| Apr 23, 2016 at 6:36 | comment | added | gerw | @DenisSerre: I do not get that point, $f(u)$ might even fail to belong to $W^{1,p}(\Omega)$: $\Omega = (0,1)$, $p = 2$, $u(x) = x$ and $f(t) = \sqrt{t}$. Then, $\nabla(f(u))(x) = x^{-1/2}/2$, which is not square-integrable. It might work if $f$ is Lipschitz. | |
| Apr 22, 2016 at 20:34 | comment | added | Denis Serre | I see. Actually, if $p>\frac1n$ (so that $W^{1,p}(\Omega)$ is an algebra), your argument works even if we replace the positive part by every continuous function of $u$. Namely $u_r\rightharpoonup u$ implies $f(u_r)\rightharpoonup f(u)$. My error was that I wanted to use the fact, in a Hilbert space, that if in addition $\|u_r\|\rightarrow\|u\|$, then the convergence is strong. In $L^2$ this amounts to $u_r^2\rightharpoonup u^2$, but not in $H^1$ !! | |
| Apr 22, 2016 at 19:26 | history | answered | gerw | CC BY-SA 3.0 |